. Nowwithx = r0, (141) =: '>€ H K,&n x H K, L^> ii) u satisfies the "outgoing wave condition", i.e., the propagation condition (179) towards x3 > 0 or < 0 out of the slab, has a unique solution u which is given by the Kirchhoffformula (222) u « * W • f ■ * ( p ) * ) dx, **
/s2e^^d«=/^eitopd«=2«/1eik™du=4n5i^=^*en(r).
Thus Uf = (4it)2 k~1 Uen, with a similar result in the Maxwell case.
®
LetG•*en be the space of traces: (142) G ^ - R U J , * ^;Pl r )» U cn = $ cn ,f, f=p% + div(pn«r)> (p, p')€ Y(see(59)}. Then G**1 is regular (C00 if T is C00, analytic if T is analytic) and contained in G|. Proposition 18. The operator (S out - S"1) is a regularizing (thus compact) operator with the essentialfollowing properties (in the Helmholtz and the Maxwell cases), with (Smt - S™)2 = 0 ker(S 0Ut -S in )«G i , I m ( S o u t - S ^ G * 1 -if^O^-Pf^Gj 1 )isdenseinG { . Moreover (Sout - Sm) is a one-to-one mappingfrom G? and G^ut onto G**.
3.1 WA VES DUE TO BOUNDED DOMAINS
97
We first prove that Gf flG°ut = {0}. Let ( v ^ ) e G^nG°Mt. Let u be the solution of the exterior Helmholtz problem with (u | _, g—| ) = (v0, Vj). PROOF (for Helmholtz).
Then the behavior of u at infinity is (up to o(l/r)) given by *(r)f(ka) and
<( J°i ) , ( $ )>y, Y =
where <,> denotes the duality H1/2(r), H"1/2(0. b) The Maxwell case. Now in V = H" 1/2(div,D we introduce the duality pairing (so that we can identify V with its dual space V = H ~ 1/2(curl,0) by: (144) = vv = y v ,=/ r u.nAvdT, Vu, vinV. Then we define the duality pairing in Ydiv (see (81)) (so that we can identify Ydiv with its dual space Y9 = Ycurf) by
(145)<[ Y ] , f *Jo ]>Y,Y =
Thus in the Maxwell and the Helmholtz cases we have: X (146) S = - S and thus: % = P{, % = Pe, with S = S* or Sout andtL = L,tR = R,tK = -J, t J = -K, and also: l T=-T, lR = -R. Now (146) implies that the polar set of Gen = Im (Sout - S"1) is ker (Sout - S"1), i.e. G., and the polar set of G. is again G.. This proves the density of Gen in G..
3 STATIONARY SCATTERING PROBLEMS
98
Remark 9. We can also use an antiduality pairing by (144)* =yy, = / u.nA ? dT, giving: (146)' S = (Sout) = - Sm, and thus T = (T°ut) - - 1", R - (R om ) = - R m . These formulas are again obtained using Green formulas in Q and Cl\ but with $ two waves, one incoming and the other outgoing. 2 . SCATTERING PROBLEMS WITH A COMPLEX WAVENUMBER k
Except in the free space case, physical media are dissipative (see chap. 1 section 3.1.6, with the H14) hypothesis). Of course dissipative media are bounded but in a first step we assume that such a medium occupies the whole space. This allows us to define the Calderon projectors and then to tackle physical problems in free space by the "limiting absorption principle". 2.1. Helmholtz equation in R3 with a complex wavenumber We again consider the Helmholtz equation (1) with a complex wavenumber k. Proposition 19. When Im k > 0, the Helmholtz equation (I), where f is a distribution with compact support, or a tempered distribution in R3. has a unique solution u in the space of tempered distributions, with u in H2(R ) when f is in L2(R3). PROOF.
(147)
Using a Fourier transform method we have: (-5 2 + k2)u = - f inR 3 .
Since(-5 2 + k 2 )*0foraUUR 3 ,with(-e 2 + k 2 )" l -^0when|£| — oo, wecan divide by (- S2 + k2), thus obtaining: (148)
fi
= - (- £2 + k 2 )'* \ € S *(R3)
with fi € L2(R3) when ? € L (R ). The inverse Fourier transform is the solution u of (l)inS'(R 3 ); forf inL2(R3), we write: (149)
u = RAf,
withRA = (-A-X)" 1 , X^k2;
Rx is the resolvent of the positive selfadjoint operator ( - A) since \ is not in the spectrum of ( - A). When f is in S '(R3), we also have by convolution (150)
u =
w i t h ^ ^ - Z - V - ^ + k2)"1, O k ( x ) = ^ , r = | x | ;
k is the tempered elementary solution of (1). With Im k > 0, it is exponentially decreasing with (high) r; then for f in E '(R3), u satisfies, with n = Im k:
3.2 COMPLEX WA VENUMBER
(151)
u(x) =
0(|SF),
99
(g-iku)(x) = o ( ^ F ) ,
whenr-oo.
There are no conditions at infinity to impose on u: Sommerfeld conditions appear when X tends to a real number with Im X positive or negative. This is easily seen on the elementary solution: Lemma 2. The elementary solution of the Helmholtz equation satisfies \^
=
k2^]^withlm\>09
O k ^ O ^ = *_^in^R 3 )or5XR 3 )wtenX==k 2 ^k^wir/fImX<0.
Then using the continuity properties of the convolution product (see Schwartz [1] p. 157 and 247) we obtain: Proposition 20. Limiting Absorption Principle in R3. When the complex number k tends to a real value k , with Im k2 positive (resp. negative), the solution u. of the Helmholtz equation (1) (with f a distribution with compact support) tends to the solution of the same equation for kQ with the outgoing (resp. incoming) Sommerfeld condition in S '(R3); thus (153)
ukd=fk*f—u.
= f O ko *fin5 , (R 3 )whenk-»k o withImk>0.
When fis in L2(R3) with compact support, (uk) converges in H2loc(R3). Obviously we cannot have Rxf-> RA f in L2(R3) for all f € L2(R3) when X0 = k2 is in o(- A), the spectrum of operator (-A). PROOF of Lemma 2. As an exercise, we prove (152) by a Fourier transform method, to specify the Fourier transform of the elementary solutions. For all * in £(R3), we have: (154) < ^ , < > > = / R 3 ? l - ?
with p = 11|, Im k * 0. Let 6(p) be defined on R by: (155) Then:
9(p) = p 2 M«(|p|), withM«(p)=/s24>(pa)da, p>0.
(156)
<<& k ,*>4/ R i^dp.
The limit for k —* k0 with Im k > 0 or Im k < 0 is given by the Plemelj formulas with respect to the real axis in C (see Dautray-Lions [1] chap. 11A.2; we could also use the variable t = p2, and the half-line R* instead of R). When k0 * 0, we have
100
(157)
3 STATIONARY SCATTERING PROBLEMS
k . v fco, t >o
^'*
A >=<
>= 1 pv
9(0)
+ in91 k
\-* i 4^*> i < o).
and 9(P) (158) k - n , - l ! % , . > o < ^ » # > - < * t ' * > - ^ ^ « p r^*-2tw
<\>*>
=
and )' VO-pv^r+it^-k^). (159)' r - ksolution ; The incoming elementary <&]£ is given by: K in
(160)
.
2
* j£«) = P v p - U " * *« " >£>•
PROOF of Proposition 20. Lemma 2 implies (153), and thanks to (158)', the Fourier transform of Uj. is:
(161)
\(i)=\(Qte>=Pvp-Trfto+fc**1- $ ««>• r-»5 r2 3.
n Note that this implies when f € L /(R ) with bounded support: Re ( u v f ) = (2n)"3 Re (fl^, f) = (2n)"3 P V / R 3 - J -1^ - J i^«|2 J f«)| dS, (162) I m ( u v 0 = (2n)-3Im(flko, f) = ( 2 « r 3 ^ / s 2 |t l (k o a)| 2 da>0.
Now we prove the convergence of (uk) in H20C(R3). Let Ba be a ball which contains the support off. We have: (163)
(u k -u ko Xx)=/ B
e
^
tlx')dx',
Fork = ko + in,withn>0,wehave: \-—^—| ikr
r=|x-x'|, s|
r"'|
^n.thus:
^o^
(164) |(u k - Uko Kx)| 2 ^C/ Ba Ie 4~f
| 2 |Rx')| 2 dx^Cn 2 / B a |«x')| 2 dx\
hence uk —»Uv in L (R ), and in Lj0C(R ). Thus: A (uk - u^) converges to 0 in L^R 3 ), hence in LJ^R3), since A (uk - u^) = - k 2 ^ - u^) + (k2, - k2) u^. Furthermore: (165)
grad(u k - U k o )=/ B a^ffcOdx',
3.2 COMPLEX WA VENUMBER
101
ikr
**o**
with: a = £ ^ £ , and
(166)
V - ^ - T ) -
But we have
Kr)|<[ 1 ~ e " V f J T ! r e T *'-g^i^—]« n 0(l),
thus grad (uk - u^ ) ~> 0 in L°°(R3), hence in Lfoc(R3). 2.2. Maxwell equations with complex coefficients Proposition 21. Let
2.3. The Calderon projectors for a complex wavenumber k We consider the "standard" problems (61) in the Helmholtz case, and (86)' in the Maxwell case with charges concentrated on a regular surface T as in 1.3.1. Proposition 22. Let w (or e, or y) be complex with Im k 2 * 0 . Then Proposition 3 in the Helmholtz case and Theorem 3 in the Maxwell case are also valid without condition at infinity if we replace the space X by the space: i) Helmholtz case: X = {u€D9(R\ (167)
U|
€H*(A,Q), ul
CH^O')}
ii) Maxwell case: X = X 0 x X 0 , with: X 0 = {u€D'(R 3 ) 3 , u| Q €H(curl,Q), u| a ,€H(curl,n'), divu| n €L 2 (Q), divu| Q I €L 2 (Q%
Then this allows us to define the Calderon projectors P. and P like in (65) and in (93) and thus the operators S (see (21) and (102)) with thee same properties. Furthermore we have for instance in the Helmholtz case as a consequence of the limiting absorption principle:
102
3 STA TIONARYSCA TTERING PROBLEMS
Corollary 3. When the complex number k tends to a real value kQ, with Im k2 positive {resp negative), then (168)
Sk( J . j - S j J £,] (wp.s£( £,])inH1/2(OxH-,/2(T),
and thus the corresponding Calderon projectors and the integral operators Kk, 1^, RR and J k strongly converge in their naturalfunctional spaces. PROOF. Let u = u k be the solution of (61) with k a complex number with Im k > 0. We use a lifting U of the jump conditions (63), with U in H^AjQ) and in H ^ A , ^ ) with compact support. Then vk = u - U satisfies: (169) Av k +k 2 v k =F,
withF€L 2 (R 3 )H£'(R 3 ).
Then v k =R^F converges to a limit denoted by R%F thanks to Proposition 20.
3 . VECTOR HELMHOLTZ EQUATION. KNAUFF-KRESS CONDITIONS
Using the (vector) Laplace operator A (as in chap. 2 section 11.2) instead of the curl curl operator in electromagnetism leads to some differences in the scattering theory by integral methods. We again assume that k is a real positive wavenumber. 3.1. Knauff-Kress conditions at infinity Let f be a given vector distribution with compact support in R3; then let u be the unique (vector) solution of the Helmholtz problem (13) with Sommerfeld conditions. Its behavior at infinity is given by (IS); then we have: (170)
curl u = 0 * curl f,
divu = **divf,
and thus their behaviors at infinity are given by: (171)
(curluXra) * *( r ). ika A ?(ka),
(divuXra) » 0(r) ika.f(ka);
then we obtain: (172) (adivu-aAcurlu-ikuXro) »
(adivu - aAcurlu - ikuXra) = o(y) whenr-*<».
These conditions are called Knauff-Kress conditions at infinity.
103
3.3 VECTOR HELMHOLTZEQUATION
3.2. Vector Helmholtz problems with jumps conditions Let u be a solution of the (homogeneous) vector Helmholtz equations in Q. (a bounded open set in R3) and its complement Q\ We suppose that u satisfies (174)
u | a € H\C1)\ u| ft , € H ^ Q ' ) \
with the Knauff-Kress conditions at infinity. Recalling the jumps formulas (53), (54), and then applying the curl and div operators to these formulas we obtain, with the same notations: (175)
curl curl u = (curl curl u) - [n A curl u] &J- - curl ([n A U] 6r), grad div u = (grad div u) - [div u] n5 r - grad ([n.u] 6j.).
With the usual relation: curl curl u = - Au + grad div u, we obtain that u satisfies inDXR3)3: (176)
Au + k2u = g, g = [n A curl u] f 6r - [div u] r n6r + curl ([n A u] r 6r) - grad ([n.u]r 8j-).
Then u is of the following form: (177)
u = -*g.
This formula is analogous to the KirchhofT formula (62) or the Stratton-Chu formula (88). We give some mathematical frameworks for these jumps: i) In the "regular" case (based on the H2 regularity): (178) [nAu] r €H t 3/2 (0, [n.u] r €H 3/2 (H, [nAcurlu] r €H t 1/2 (r), [divu] r €H 1/2 (r). ii) In the "variational" case (based on the H1 regularity): (179)[nAu] r €H t 1/2 (0, [n.u] r €H 1/2 (H, [nACurlu] r €H- 1/2 (0, [divu] r €H" 1/2 (r). Note that the space Ht1/2(0 and its dual space HJ" U2(T) are so that: rl/2,
r-1/2,
r-1/2,
1/z f-1/2/. Ht1,z(r) = H- 1,z (div,r)riH- 1/z (curl,r), Ht"1/2(n = H(div,r) + H- i/z (curl,D.
j-1/2,
l/2/i
Of course we can develop the integral method in these frameworks and define Calderon projectors. We can also consider complex wavenumbers k. Then we have the analogue of Proposition 19 in a domain Q.
104
3 STATIONARY SCATTERING PROBLEMS
4 . BOUNDARY PROBLEMS WITH REAL WAVENUMBER k
In chap. 2, section 12, we first studied some boundary problems which are "coercive" or "elliptic", that is, they may be solved thanks to a variational method and to Lax-Milgram lemma. This is possible for pure imaginary wavenumbers k and more generally for complex wavenumbers, but the case of real wavenumbers is more difficult. We can study it from various methods; here we develop the method based on the limiting absorption principle. 4.1. Limiting absorption principle We first consider "exterior" problems, i.e., in the complement Q* of a regular bounded set ft in R3. We assume that Q* is a connected set. Either in the (scalar) Helmholtz case with the Neumann (or also Dirichlet) condition or in the (vector) Helmholtz case, these problems can be written as: (180)
Au-Xu = f,
where A is a positive selfadjoint operator with a continuous spectrum only and X is in the spectrum of A. We specify in the scalar case, then in the vector case, with the definition (3) chap. 2: i)A = - A , withEK^^IucHVA^Xul^Oor^l^O}, ii) A = A" defined in (147) chap. 2. When f is a square integrable function with bounded support we have the following result, for which we refer to Wilcox [1], Bendali [2], Hormander [2]. (180)'
Lemma 3.
Let RQ , R be real positive numbers RQ < R with Q, contained in the
ball B Ro . Let 0 < a < b, a > 0, J a [a,b] x (0,o]. Then there exists M » M(J ,RQ, R), M > 0, so that: (181) IIR(z)fll ,
feL^ni^R3)
ii)u| r = 0 (resp.|j| r = 0 ) , iii) the outgoing Sommerfeld condition,
105
3.4 BOUNDARY PROBLEMS
has a unique solution u+, given thanks to the resolvent R(z) ofA= — A by: (183)
u* = R£f =
lim
R(z) f in H/0C(A,Q').
Obviously the theorem is true if we substitute incoming for outgoing and - 0 for 0 in (183). In the vector case we have: Theorem 4 \ Limiting absorption principle, vector case. Thefollowing problem: find u in Hj0C(A,Q')3, solution ofthe Helmholtz problem with k real and given f
(182)'
liJAu + k ^ - f uiQ', f€L2(Cl9?nE'(R3)3, u)nAu| r = 0, (divu)|r = 0, iii) u satisfies the KnauffiKress conditions (173),
has a unique solution u + which is given thanks to the resolvent R(z) ofA -A" (183)'
u+=
lim
by
R(z)f inH{0C(A,Qf.
z = lr + iaf o>0, a—»0
PROOF, i) Wefirstprove uniqueness in the scalar case. Using the Green formula on the domain Q'R = Q? n BR, we have (with f=0) (184)
(|gradu| 2 -k 2 |u|2)dx)
Re(-ikf
= R e ( - i k / r g u d T ) - R e ( - i k / s g u d S R ) = 0. Then using the Sommeifeld condition we have (185)
Re(-ik/ s HudS^^kVg |u| 2 dS R + o(l) = 0.
We conclude using the Rellich lemma. ii) In the vector Helmholtz case we can develop a similar proof, using the Green formula with: (186) Re(-ikf t (|curlu| 2 +|divu| 2 -k 2 |u|2)dx) = 0, Q
R
and then using the Knauff-Kress conditions and the Rellich Lemma. iii) The existence of the solution follows easily from lemma 3: uz « R(z)f tends to a limit u+ when z —♦ k2; u2 satisfies all the properties of the theorem but the conditions at infinity. These conditions are deduced from the KirchhofTfonnula (obtained from (62) and (177)) which gives the solution uz thanks to its traces on a sphere SR (successively in the scalar then in the vector case):
106
3 STATIONARY SCATTERING PROBLEMS
i) u(x)=/SR(u(y) §^(x - y) - |M>
ii) u(x) = - \fs (n A curl u + n div u) <&(x - y) dSR(y) + curl/ s nAiiy*(x-y)dS R (y)+grad/ s n.Uy*(x-y)dSR(y)}, R
R
with u = u and * « * 2 (in order to simplify). Passage to the limit in these formulas wnen z —» k2, gives the corresponding (Kirchhof!) formulas for k2 that is u + satisfies the required condition at infinity. 4.2. Exterior boundary problems Then we can solve some "usual" exterior boundary problems with inhomogeneous boundary conditions: Theorem 5. ExteriorscalarHelmholtzproblem. The following problem: find u in H/0C(A,Q') satisfying the Helmholtz problem with k real and given u0, or u{ i)Au + k2u = 0inQ', (188)
i0u|r-uo€HI/^(iwp.g|r-u1€irl/2(n)f iii) the outgoing Sommerfeld condition,
has a unique solution. PROOF. Uniqueness is proved in Theorem 4. In order to prove existence of the solution, we use a lifting to the boundary conditions obtained by the (variational) solution U in H^Q') of the problem (188) when k = i (for instance). Let C be in D(K3) with C(x) = 1 in a neighborhood of r. Let f be defined by: f = [2 grad C .grad U + U (At + (1 + k2) C)], then f is in L2(Q*) with compact support,
Thus v=u - CU is in HJ^.CA.Q'), and satisfy: |i)Av+k 2 v=-f, (189)
|ii)v|r-0 (resp.g|r = 0), iii) the outgoing Sommerfeld condition.
Then the limiting absorption principle gives the result. 0
Theorem 5 \ Exterior vector Helmholtz problem. The following problem:findu in r W satisfying the Helmholtz equation with k real and with given (m,g) Hl0C(A,Q')
107
3.4 BOUNDARY PROBLEMS lOAu + k ^ O inQ\ ii)nAu| r = m,m€H t 1/2 (0, (divu)|r = g,g€H- 1/2 (r),
(190)
iii) the Knauff-Kress conditions (173), has a unique solution. This is similar to the scalar case: uniqueness is proved in Theorem 4'. Existence is proved thanks to a lifting of the boundary conditions, which is the (variational) solution of (174) chap. 2 and then applying Theorem 4'. PROOF.
Then we can solve exterior Maxwell problem in the free space: Theorem 6. Exterior Maxwell problem. Let c, \nbe positive real numbers. The problem:find(E, H) satisfying, with a given m i) curl H + icoc E = 0, ii) - curl E + io>ji H = 0 in Q\ (191)
iii)nAE| r = m, m€H"1/2(div,0> iv) the Silver-Mutter conditions (43) at infinity I v) (E,H) € Hloc(curl,Q') x Hloc(curl,Q'),
has a unique solution. PROOF, i) We first prove uniqueness directly for zero boundary conditions. Using the Green formula on Q'R, like in (48), with the Maxwell equations, we have: Re(/ Q'
(192)
div(EAH)dx) = Re(/Sc n. E A H d S R - /l r n . E AHdT) R
[icop | H | 2 - i
= Re(/ a
R
Since we have /_ n. E A H dT = 0, thenS / n. E A H dSR = 0. Thus using the Silverr R Mu Her conditions (43) we obtain: (193)
|* r H| 2 dS R = / s
f$ K
|n r E| 2 dS R = 0. K
Since we also have (50), then: / s Rellich lemma and we have E
| H | 2 dSR = / s
| E | 2 dSR = 0. We can apply
A-o.
ii) The existence of the solution is proved thanks to a lifting of the boundary condition. Using the solution (u,v) of problem (151)' chap. 2 with K = 1, and a regular function £ with compact support equal to 1 in a neighborhood of T, we define:
108
3 STATIONARY SCATTERING PROBLEMS
(194) E = E-Cu, H = H-tv. Let(J,M)be: J = (-io)£ +l)Cu-(gradC)Av, Then (E, H) must satisfy:
(195)
M = ( - i
| i) - curl E + ioji H = M, ii) curl H + iwt E = J infl', Ui)nAE| r =0, iv) the SUver-Miiller conditions (43) at infinity, I v) E and H e Hl0C(curl,Q')-
We note that M and J are in H(curl,Q') with bounded supports. Furthermore, if T is smooth (C00): (196) divJanddivMeHs(Q')( VseR; E must satisfy the vector Helmholtz problem, with f given in L(£}') flF(R) :
(197)
|9AE + k 2 E - - f inQ\ ii)nAE| r = 0, divE| r = 0, iii) the Knauff-Kiess conditions (173) iv)E€Hloc(curl,Q').
1 Since divE=r^divJ €L2(Q^ with the boundary condition ii), then E € Hjoc(Q')3. Thus we can apply the limiting absorption principle (theorem 4'), which proves the existence of the solution of the Maxwell problem. & Remark 10. An application to scattering. Solving the boundary value problem (191) amounts to solve the following scattering problem: given an incident electromagnetic wave (EJ,HJ) on a (regular) bounded obstacle Q which we assume to be a perfect conductor,findthe reflected wave (E .H ). Since the total electromagneticfield(Et,Ht) (the sum of (EpHj) and (E^H^)) must satisfy, on the boundary T of Cl: n A Et | = 0, then the reflected wave (Er, HJ must be a solution of problem (191), with: n A E r | = m = - n A E j | given. We give below (see 4.8) another method (a surface integral method) to compute this electromagnetic field. # 4.3. Some consequences; the exterior Calderon operator We note that in the Maxwell problem (191), the magneticfieldH also satisfies (198)
nAH| r €H- |/4t1/2/. (div,r)
109
3.4 BOUNDARY PROBLEMS
Definition 4. The exterior (outgoing) Calderon operator. The mapping: (199)
C e (orC^): m = n A E | r ^ n A H | r
with (E,H) the solution of (191), is called the exterior (outgoing) Calderon operator. We also denote by C£ the mapping m = nAE|—»|g(nA curl E | ), thus C£=Z Cc does not depend on Z the impedance of the (exterior) medium, see (34), but only on the wavenumber k. There are many properties of these Calderon operators. First we have: Proposition 23.
The exterior Calderon operator is an isomorphism in the space
1/2
H" (div,r), with: (200)
(Cc)2 = - ^ I, (Ctf = - 1 ,
and (C^T 1 = - C£.
Furthermore the exterior Calderon operator is an isomorphism in Hs(div,0/br all real s i/T has a C00 regularity. PROOF. First we note that if we exchange c and ]i in the problem (191), then the solution (E,H) of (191) is changed into (H, - E); this easily gives (200). Then we prove that Cc is an isomorphism in H1/2(div,D, and the regularity result:
If m € H1/2(div,0, then the solution (E,H) of (191) is in H/OC(Q')3 X HJOC(Q')3. This is a consequence of Theorem 6 chap. 2,firstapplied to £E with C € D(R3), C = 1 in a neighborhood of r (since n A E | is in Ht1/2(0), then to H using (28), (30) chap. 2 io>P n.H | = n.curl E | f = curlr nTE = - divr (n A E | f ) € HV2(T). Then we have the regularity result for all half-integer s thanks to Proposition 7 chap. 2 and then for all real s by interpolation and duality. ®
Proposition 24. Positivity properties. The exterior Calderon operator satisfies: (201)
Re/ r C e m.(nAm)dT>0,
PROOF.
We apply the Green formula like in (192). Then we have:
(202)
Re/ r n. EAHdT = Re/ s n.EAHdS R >0, K.
Vm€H~l/2(div,r), m*0.
3 STATIONARY SCATTERING PROBLEMS
110
from the Silver-Muller conditions (and the Rellich lemma). This gives (201) (thanks to Propositions 3 and 4 chap. 2). We can give a more precise result: thanks to the formulas (30), (33), we have (202)'
Re/rn.EAHdT*-^/s2|j(ka)|2da,
with j(ka) = ik[aAJ(ka)+^n a M(ka)], J=mAH| r = C c (nAE| r ), M = - n A E | r . We can define, instead of (199), other Calderon operators such as: m —* icrH, which is an isomorphism from H~"1/2(div,D into H~ 1/2 (curl,D, and also an exterior incoming Calderon operator C c,in using incoming instead of outgoing conditions in Definition 4. Then using for instance the antiduality pairing (144)' we have the "reciprocity formula": (203)
nAE U €H" 1 / 2 (div,D,
thus C c = C c ' out = - (Ce'm)\ with C'tin = - C'_k = - C£. With the duality pairing (144), we obtain: ' C ^ - C * . We can also define an exterior Calderon operator C e k for complex k, with similar properties, and the limiting absorption principle gives: (204)
lim
c£m = C £ m , Vm€lT 1 / 2 (div,r).
The exterior Calderon operator above defined in (199) has numerous properties and is a basic tool in many applications to the scattering in electromagnetism (see below). 4.4. Interior boundary problems We first consider the inhomogeneous vector Helmholtz problem in a (regular) bounded domain Q. Since the selfadjoint operator A" defined in (147) chap. 2 has a compact resolvent (because the natural injection of the Sobolev space Hl(Q) into L2(Q) is compact), we know that the following interior Helmholtz or Maxwell problems depend on the Fredholm alternative: Theorem 7. Interior vector Helmholtz problem. The following problem: given a function f in L2(Q)3 and a complex number Y,find u in Hl(Q)3 satisfying (205)
OAu + l A i ^ - f inQ, ii)nAu| r = 0, (divu)| r = 0,
3.4 BOUNDARY PROBLEMS
111
depends on the Fredholm alternative: either k2 is not an eigenvalue of the selfadjoint operator A", then there exists a unique solution; or \? is an eigenvalue of the selfadjoint operator A", then there exists a solution u if and only iffis orthogonal to the corresponding eigenvector space, and then u is determined up to this eigenvector space. Theorem 7 \ Interior inhomogeneous vector Helmholtz problem. The following problem with given boundary conditions (m,g):find u in H*(Q)3 satisfying: (206)
i)Au + k2u = 0inQ, ii)n Au| r = m, (divu)|r = g, withm€Ht1/2(0, g€H- 1 / 2 (0,
depends on the Fredholm alternative: either k2 is not an eigenvalue of the selfadjoint operator A ' , then there exists a unique solution; or k2 is an eigenvalue of the selfadjoint operator A", then there exists a solution u if and only i/(m, g) satisfies'. (207)
J r (m.curl v + g n. v) dT = 0,
for all eigenvectors v relative to k2, and then u is determined up to this eigenvector space. The proof is classical using the Green formula: (208)
- / (Au + k2u) v - u (Av + k2v) dx=/r[(n A curl u). v - n A u. curl v - divu n. v + n.u div v] dT.
Theorem 8. Interior Maxwell problem. The problem: find the electromagnetic field (E,H) withfiniteenergy, i.e., E, H in H(curl,Q), satisfying, with given m i) curl H + io)£ E = 0, (209)
ii) - curl E + io>ji H = 0 in Q, 1/2
iii)nAE| r = m, m€H" (div,0,
depends on the Fredholm alternative: either k2 = co2€^ is not an eigenvalue of the selfadjoint operator A" or of A (see (134) chap. 2), then there exists a unique solution; or k2 is an eigenvalue of the selfadjoint operator A" (or A), then there exists a solution (E, H) if and only ifm satisfies: (210)
/ r (m.curl v)dT = 0,
for all eigenvectors v relative to k2, and then E is determined up to this eigenvector space.
112
3 STATIONARY SCATTERING PROBLEMS
PROOF. Thanks to a lifting of the boundary condition, using the solution (u,v) of (151)* chap. 2, we can reduce this problem to an inhomogeneous Maxwell problem which is directly relevant of the Fredholm alternative, like for the exterior Maxwell problem (see (191)). We can also reduce (209) to the vector Helmholtz problem (203). $ If c or fi is not a real number (i.e. if the medium is dissipative), the problem (209) has a unique solution for all m as in (209)iii). We can also solve the interior Maxwell problem with some heterogeneous media with € and p matrix valued and dependent on x, with the hypotheses H17) chap.l only. We can also solve problem (209) using a variational method on the unknown vectorfieldH. 4.5. Some consequences; the interior Calderon operator First we note that in the Maxwell problem (209), the magnetic field H also satisfies (198), then we can define in the "regular" case, (i.e., when k2 = ©2ey is not an eigenvalue of the selfadjoint operator A" or of A): Definition 5* The interior Calderon operator. The mapping: (211)
C^orCJp: m = n A E | r ^ n A H | r
with (E,H) the solution of(209), is called the interior Calderon operator. We also denote byC^ the mapping j m n A E| —»|g(n A curl E| ), thus Cj^ Z C1 does not depend on Z the impedance of the (interior) medium, see (34), but only on the wavenumber k, and satisfies (CJ)2 * - 1 . Note that it is possible to define interior Calderon operators when k2 is an eigenvalue of the selfadjoint operator A, we only have to eliminate the corresponding eigenspace but with these technical difficulties, it is less useful. There are also many properties of these Calderon operators. First, Proposition 23 is true for the interior Calderon operators in the regular case. Then, we also have positivity properties: Proposition 25. Positivity properties. The interior Calderon operator satisfies (212)
-Re/rCim.(nAm)dT>0,
Vm€H" 1/2 (div,0.
PROOF. This follows from the Green formula: (212)'
Re/ r n.EAHdT=/ ft [Re(-io)?)E.E + Re(ia>fi)H.H]dx<0.
If we have a conservative medium (the free space), then (212) is always zero. & From (201), we have the positivity property of the operator R (see (100)) for k > 0 -Re/ r RM.n A MdT=Tkf^^^ [M(x).M(y) - \ divrM(x) divrM (y)]dTxdry > 0.
3.4 BOUNDARY PROBLEMS
113
For a dissipative medium (e" = Im c> 0, p" > 0), then with (209) we have (213) -Re/ r C i m.(nAm)dTiC 0 IIEIlJ, (curlft) ,
VmeH~ ,/2 (div,n,
with C 0 a constant, and thus with the quotient norm: (214) -Kef
&m.(nAm)&
±C0knt_iniJt
_, Vm€H" 1/2 (div,r).
Therefore the sesquilinear form (215)
a(m,m') = - J r C1 m. (n A nV) dT 1/2
is coercive on VT (div,T)for a dissipative medium. Remark 11. Calderon operators for the Helmholtz equation. i) First in the scalar case, we can define the exterior Calderon operator Cc by:
(216)
c e (u 0 )=g;i r
u being the solution of the Helmholtz problem (188); n is the exterior normal to the bounded domain Q; Cc is an isomorphism from H1/2(r) onto H" 1/2 (D, and from H s (0 onto Hs ~ l(T) for all s if r is C°°; Cc has the positivity property (from Green formula, since Cc corresponds to outgoing waves), for all k > 0, (217)
Im(Ceu0, u0) = I m / r | ^ u dT>0,
Vu 0 €H 1/2 (r),
u o *0.
Then we can define the interior Calderon operator C1 like in (216), u being the solution of the interior Helmholtz problem corresponding to (188), generally with complex k. When k2 is not an eigenvalue of the interior Helmholtz problem, neither for the Dirichlet condition, nor for the Neumann condition, C1 is an isomorphism from H 1/2 (0 onto H" l / 2 (D, with the same regularity property. Furthermore it has the positivity property, from the Green formula (218)
Im (C 1 ^, u0) = - 2k,k" fQ \ u | 2 dx,
V u0 € H 1/2 (0, with k = k' + i k".
This implies the positivity properties for the operators L and R of (71) for k > 0 Im(Lp\p')>0, Vp'€H- 1/2 (r),P'*0,
Im(Rp,p)<0, Vp€H 1/2 (H, p*0.
We can verify easily using (71), (72), that the Calderon operators Cl and Cc are related to the integral operators L, J, K, R by: (219) Ce = - i L - 1 ( I + K) = - 2 ( I + Jr1R, Ci = - | L - I ( - I + K ) = - 2 ( - I + Jr 1 R.
114
3 STATIONARY SCATTERING PROBLEMS
ii) Then in the vector case, we can define Calderon operators. The exterior Calderon operator C c is given, with u the solution of the Helmholtz problem (190)) by: (220)
Cc(m,g) = (nAcuriu| r , n.ulp;
^isanisomorphfcmfromHj^ from H*/2(T) x Hl/2(T) onto H*/2(r) x Hm(T).
and also
4.6. Integral equations and boundary problems We can reduce the exterior or interior Maxwell problem (191) or (209) to an integral equation on the boundary in many ways. Here we give some of them. The general method is to choose an extension to the whole space, of the unknown (E,H) solution of the Maxwell equation. Then we only have to determine the jumps of E and H across the boundary T, or also the surface electric and magnetic currents J r and M r such that: (221)
Jr»-[nAH]r,
M r =[nAE] r
i) Solution with an electric layer only. The chosen extension (E9H) is such that there is no jump of the tangential electricfieldE across the boundary T: thus we have tofindthe surface electric current J p so that (using (101)) (222)
ZRJ r =m,
withM r =[nAE] r =0.
Then (E, H) is obtained from J r by (the restriction to Q' or to Q of) (223)
E^ZI^Jr,
H= -FmJr.
We note that this is valid in the exterior case as well as in the interior case. ii) Solution with a magnetic layer only. The chosen extension (E,H) is such that there is no jump of the tangential magnetic field H across the boundary T: thus we have to find the surface magnetic current M p so that a) for the exterior boundary problem (using (89), (93)): (224)
n^^1"]—m,
withJ r =-[nAH] r =0,
rij being the projection on thefirstcomponent, so that with (104) and (102): (224)'
|(I+T)Mr«-m.
b) for the interior boundary problem: (225)
|(I-T)Mr«m.
115
3.4 BOUNDARY PROBLEMS
Then in the two cases, (E, H) is given by (the restriction to Q or Cl' of) (226)
E = ZPmMr,
H^LmMr.
iii) Solution by extension by 0 a) for the exterior boundary problem, we have to find (227)
Jr=*nAH| ,
withMr = - m .
Then (Mr, Jr) must satisfy:
<» ;>('4* ")(*)■•• b) for the interior boundary problem, we have to find (227)'
Jr = - n A H | ,
withMr = m.
i
TTien (Mr, Jr) must satisfy
(229)
P
fMr-\
f I+T
-l'rJ-U*
- Z R y M
I*TJL
r
^
kh
and we obtain the electromagneticfieldin Q or Q' by formulas (96)'. Remark 12. There are other possibilities; for instance (see remark 10), when the exterior boundary problem is associated with a scattering problem of an incident wave in free space (which is naturally defined on the domain of the obstacle, like a plane wave) by a perfect conductor, then we can extend the reflected electromagnetic field by (-Ej,-H x ) in the obstacle. Then we are reduced tofindthe jump: Jr = -[nAH] r = nAH t | ,
whereas: Mr = [nAE]r = - nAE t | =0,
and then Jr must satisfy;
thus (230)'
ZRJr=-2nAE,|r,
(I-1)J r »2nAH I | . <8>
116
3 STATIONARY SCATTERING PROBLEMS
Remark 13. In fact for a given m, it is not obvious that the solution of the exterior problem is obtained only by an electric layer or a magnetic layer, and then it is possible that equations (222) and (224) have no solution, whereas method iii) (extension by 0) is always allowed. Therefore the system (228) is always solvable (but it is a system!). The difference is due to the eigenvalues of the selfadjoint operator A (see Theorem 8); these values are called irregular frequencies (for the exterior scattering problem). Thus to apply these methods, we have to avoid these irregular frequencies; for instance in a scattering problem, we use a linear combination of (230)' (see below), with a complex constant. 0 4.7. Some consequences for the integral operators i) From Theorem 6 (existence and uniqueness of the solution of the exterior boundary problem) we deduce (231)
Im(I-*D = ImR,
kerR H ker(I-T)*{0}.
ii) The "regular1* case. From Theorem 8 (existence and uniqueness of the solution of the interior boundary problem in the "regular" case) we deduce: (232) PROOF.
Im(I+T) = ImR,
kerR H ker(I+T) = {0}.
Let Ge be the set of the exterior boundary values for the
electromagnetic field: Gc ■ Im Pc » ker P.. Then for all u in X = H~ 1/2(div,D, there exists v in X so that (u,v) is in Gc, thus: (233)
Pi(;) + Pi[})-0.
Therefore the mappings: u € X - > P | ( J ) and v c X - ^ P j f J ] are injective and have the same range. Thanks to (102), (104) this gives (231). The proof of (232) is similar. Proposition 26. In the regular case (k2 is not an eigenvalue of A), R, (I + T) and (I -T) are isomorphisms in the space X = HP l/2(div,r)t and more generally (i/r has the C°° regularity) in Hs(div,D/or all real s. Thanks to relations (106) we have to prove the properties for the R operator only. That R is onto is a consequence of equation (222) and theorems 6 and 8 (or also of (231), (232), since X • Im (I + T) + Im (I - T)).
PROOF.
117
3.4 BOUNDARY PROBLEMS
Letj be such that Rj = 0; thenwehave: m = n A E | = 0 , thusE = 0 i n l a n d Q' (in the regular case), thus H = 0 and j = 0. ®
iii) The "singular" case (k2 is an eigenvalue of A) We define the following spaces: S k d # { n A H | r , H = icurlE, E C H W ^ E + I ^ E ^ , divE = 0, n A E | r = 0}, (234) 1 3 2 = {nAH| r ,H€H (Q) ,AH + k H = 0,divH = 0,nAcurlH| r = 0}, and (234)'
etf 5 ddef E k = (nAc,c€S k ),
E k H?{ u € X,/ r u.v dT = 0,
VveSJ.
Then instead of (232), we have (235)
kerRnker(I + T) = Sk, R(Sk) = (I+T)S k , ImR=E k |ImR + I m ( I - T ) 2 S k .
As a consequence we have: (236)
kerR = ker(I+T) = S k ,
ImR = Im(I-T) = Ek
(andalso: ker(I-T) = C e E k , Im(I + T) = C e S k ). Remark 14. From the properties of the S operator, we also have: (237)
kerR n kerT={0},
ImR + ImT=X.
<8> Remark 15, Let G and G. be the sets G„ = Im P = ker P., G. = Im P. = ker P . corresponding to the exterior and interior boundary values for the electromagnetic field, then their projections on their components are: (238)
Tlx Ge = fl2 Ge = X, with:
i) in the "regular" case: ITj Gx = n 2 Gj = X, ii) in the "singular" case: Il 1 G{ = n 2 Gj = S k .
These results are easily proved from the "reciprocity" formula: (239)
/jjiAE. H ' d r = / r n A H . E'dT
for all (E,H), (E\H') satisfying the Maxwell equations in Q; this is proved using the Green formula, with:
118
(240)
3 STATIONARY SCATTERING PROBLEMS
div (E A H*) * div (H A E').
Then taking E' an eigenvector of A, gives (238)ii). In the "regular" case, we can give the Calderon operators as function of the operators R, T, thanks to the formulas (228): (241)
cj-R"la-T)--a-,n"lR-(i-i-,i)R"1—Ra-i-T)"1, CIk=.R-1(I+T)=(I+T)"IR=-(I-T)R"1 = R(I-T)"1. ®
4.8. On the numerical solution of some scattering problems 4.8.1. The problem of irregular frequencies In order to solve the scattering problem of the remarks 10 and 12 thanks to an integral method, we have to solve equations (230)'. If k2 is (near to) an eigenvalue of the operator A, then we cannot solve only one of the two (vector) equations (230) since R and (I-T) are not invertible. This is the problem of "irregular" frequencies. We develop a method implemented by BonnemasonStupfel [1]. Let a be a (complex) number with Re a * 0. Then we replace (230) by the linear combination (242)
- ( I - T ) J r + aZnARJ r = 2f I H f 2[-nAH I | r + an r E I ].
First we know, thanks to Theorem 6 and Remark 10, that (242) has a solution. Then we prove that this problem always has at most one solution. i) First we consider the interior problem:find(E,H) with E in H!(Q)3 so that (243)
i)curlH + iwcE = 0, ii)-curlE + m\iH»0 inQ, iii) n A H | - a IIrE « f j, 1/2
with fj given in Ht (0 (for instance), with e and i* real positive numbers. Let a(E,E') be the sesquilinear form on Hl(Q)3 (244)
a(E,E')=/ n - [j^curlE .curll' + ioeE. E'Jdx- a / ^ E . E'dT.
Then we can write problem (243) in the variational form:findE in Hl(Q)3 with (245)
a(E,E')=/ r f f . E'dT,
VE'€HW.
Then taking the real part of (245) with E'» E, we obtain: (246)
Rea(E,E) = R e ( - a ) / r | n r E | 2 d r = Re(/ r f,.Edr).
This implies uniqueness: when fj« 0, (246) implies nrE = 0. Thus from (243)iii) n A H | = 0, and then, from section 13 chap. 2, E = H = 0 in Q.
119
3.4 BOUNDARY PROBLEMS
ii) Let Q 0 be defined by:
(247)
Qo£Yrr) —Jr+fltnAMr.
and thus
<*«>
C nAElr^ Qo[-nAH|jJ=nAHlr-aI1rEf nAE|.^
Fromtoi), we know that Q I _ n A j j i
I = 0 implies E = H = 0.
iii) Then applying Q to equation (230) gives:
< 249 >
QoPi[j0r]=-Qo(-nAH;![)=f..
and thus we obtain equation (242). Therefore uniqueness of the solution J r of (242) is deduced from that of (243) in the following way: (250)
Q o P i [ j ° r } - 0 implies Pi( £ ] = < > , thus ( ^ J € G e ,
and then J r = 0 from Theorem 6. We can show that problem (243) (and thus (242)) depends on the Fredholm alternative. This gives another proof of Theorem 6. 4.8.2. A saddle point method Here we briefly give a variational (and integral) method for solving a scattering problem, which is due to Bendali [1]. We consider the scattering problem of Remark 10, which is equivalent to (230). We only take the first equation of (230) (we assume that k2 is not near an eigenvalue). Let
(251)
c^-^IIrE,,
p^Jp
Then we have tofindp with given c in H" (252)
(curl, T) so that
- n A R p = 2ikc.
Thanks to the definition (100) with (95) (see also (144)), we have to solve (253)
Then we define: (254) and
±
\ = -^divrp, k2
V q € H" 1/2 (div,0.
120
3 STATIONARY SCATTERING PROBLEMS
(255) HT 1 / 2 (0 = {♦ e H" ,/2(T), / ♦ oT * 0 on all connected component Tj of rj. Thus X € KV2(T). Then we define, with* 0 (x)=^p, r= |x|: (256)
L0Mx)=/r*0(x-y)X(y)dry;
L0 is a continuous operator from H" 1/2 (0 into Hl/2(T). Then with these notations, problem (253) is:findp € H-,/2(div,r) and X e Km(T) so that: i)
a0(p, q) * KL^, q>,
b(X, q) =
b0(X, q ) .
c(X,*)=k2
Km(T),
i)a(p, q) + b(X, q)*
This system is solved by perturbation of the system:find(p,X) so that, for all q in H- 1/2 (div,0, and ♦ in H;1/2(r), (260)
i) a ^ , q) + b0(X, q) =
1/2
with given c in H" (curi,0 and x in the (anti) dual space of H.
(T) that is:
(HZ 1/2(T))' = Hi/2(D 4 e H1/2(D, / r L; V dT, V r4 component off).
3.5 DIELECTRIC OBSTACLE
121
Then we prove (see Bendali [1]) that the system (260) satisfies the "inf-sup condition" or "Brezzi condition11 (see for instance Girault-Raviart [1] p. 39) and thus (260) has a unique solution for all given (c,x) in H " 1/2(curl,H x H*1/2(H. Then we obtain the solution of the system (259) by perturbation of the system by a (regularizing) compact operator, and thus we can apply the Fredholm alternative. 5 . SCATTERING PROBLEMS BY A DIELECTRIC OBSTACLE
Let (Ej,H.) be an incident electromagnetic wave with an angular frequency o>, on a (regular) bounded obstacle Cl. Let the medium inside Q be a dielectric with permittivity and permeability t{ and i»1; then let kx be the wavenumber inside Cl; outside Cl is the free space (or some "conservative" medium) with permittivity and permeability t = eQ and p = u and wavenumber k. The problem is to find the electromagnetic wave (Ef,H ) reflected by this obstacle. Let (E,H) be the electromagnetic field inside Q; (Er,Hf) must satisfy the Maxwell equations in free space, with the Silver-Muller conditions at infinity and with locally finite energy; the problem is then reduced to find (E,H) satisfying (261)
I i) curl H + io)€i E = 0, iO-curlE + iwjij H = 0inO,
and the continuity relation on the boundary T of O (262)
nAEr + nAEpnAE,
nAHr + nAHj = nAHonr.
The fact that (Er,H ) is a reflected wave, gives us a relation between their boundary values on r as seen in section 4.2. We can obtain it in two ways. 5.1. A general variational formulation Here we could assume that inside Cl the medium is inhomogeneous and even anisotropic with usual dissipative conditions, see H14) chap. 1: (263)
Re (- io^) > 0,
Re (- mv{)> 0.
Using the exterior Calderon operator Cc, we have: (264)
nAH r | r -C e (nAE r | r ),
and then with (262), we obtain the boundary condition on T: (265)
nAH-C^nAE^f^nAHj-C^nAE!).
122
3 STATIONARY SCATTERING PROBLEMS
The problem: find (E,H) in Q satisfying (261) and (265), is solved by a variational method. Let a(E,E') be the sesquilinear form (266)
a(E,E*)=/Q - [ |jL curl E. curl E' + im E. I f ]dx - / r Cc(n A E). I* dT,
for E, E* in H(curl,Q). The problem (261), (265) is equivalent to:findE € H(curl,Q) so that (267)
a(E,E')=/r f,. E'dT, V E' € H(curl,Q).
The proof is "classical" with the Green formula. Thanks to (263) (if the medium is inhomogeneous we have to assume a slightly stronger hypothesis), and to the positivity property of the exterior Calderon operator Ce (see (201)), a(E,E') is coercive: there is a constant aQ so that (268) Rea(E,E) * fa(Re(-±)
|curlE| 2 + Re(-ia*)|E| 2 )dx fc a0IIEI^(curl$n).
Since a(E,E') and the form: E*-+/r fj.E'dT are continuous on H(curl,Q), we can apply the Lax-Milgram lemma: thus the scattering problem has a unique solution. Remark 16. We can solve other scattering problems, for instance when there is a perfect conductor with domain Qj in the obstacle. Then the electric field must satisfy the boundary condition on Ti9 the boundary of Qj: n A EL =0. We apply the variational method with V={E € KKcur^QVQj), n A E | = 0}. 5.2. Solution using an integral method Let (M, J) be defined by: (269)
M = nAE| r ,
J= -nAH|r
Let Pf be the interior Calderon projector associated to the exterior wavenumber k. Then we have (by definition):
(270
>
f n A Er| ^
^[-nAH^J*0
Then using (262) and (269) we have:
thus (with the integral operators T, R associated to k, and the impedance Z of free space):
3.5 DIELECTRIC OBSTACLE
123
|i)(I-T)M + ZRJ = 2nAE,, (271V i i ) - 2 R M + ( I - T ) J = -2nAH I onT. Now we assume that the medium is homogeneous with constant ej and v{. Let P* be the exterior Calderon projector associated to the interior wavenumber k{. Then we have (by definition): (272)
Pjf^^O,
that is also with Tl9 Rlf andZj relative to the interior medium: iHI + ' t y M - Z ^ J - O , (272)'
ii)j-R|M + (I + T|)J-0.
Thus the scattering problem amounts to find (M,J) solution of the integral equations (271) and (272), that is also (271)' and (272)'. We can take (270) into account in a weak form, (thanks to (100) with (95), see (253) for R): taking the scalar product of (271)'i) with n A J\ and (271)41) with n A M' and adding, give us for all NT, J' in H~ l/2(div,D iZk y (273)
r
[J(x).J'(y) - k-2 divrJ(x).divrJ'(y)] 4Kx -y) dTxdTy
- f/pfr [M(x).M'(y) - k~2 divrM(x).divrM'(y)]
We can also write equations (272)' in a weak form. The case of a dielectric layer against a perfect conductor would be solved in the same way. These problems depend on the Fredholm alternative, and uniqueness follows from the Rellich lemma. For some numerical developments on these methods, we refer to Bendali [1] and Heliot [1]. 5.3. Behavior at infinity. Radar cross section. Optical theorem The behavior at infinity of the reflected electromagnetic field (Ef,Hr) is given by the formulas (30), (31) with: (274)
M = -nAE r | r & r , J = nAH r | r 6 r .
Then we define the differential cross section for the direction a by ,2
(275)
2
o(a)V^ljm 4irp
^
|2'
IE,!'
,
withp = |x|.
124
3 STATIONARY SCATTERING PROBLEMS
Thus using (30) and (33), we have:
(2?6)
^^li^^li^'
since the incident wave is a plane wave, that is (EpHj) is given by: (277)
E,(x) = E,e ° , H,(x) = H,e ° ,
where a0 € S is the direction of the propagation of the plane wave, and Ev H{ arc constant vectors in C , Z being the free space impedance: (278)
a o ! , = 0, ao.H, = 0, H^J^oAE,,
1,^-Z^AH,.
The radar cross section, or back scattering of is defined by of » o( - aQ) where aQ is the incident direction of the plane wave. It is a surface and its unit is m2 (in the S.I.) but it is generally expressed in dbm2 (db decibels) by 10 log10 of. Then the total cross section or scattering coefficient o s is defined by: (279)
* ,a . o S = / s 2 ^ « ) d « =1 i _JL 7 ^ / s 2f l l I*?«/ I)~|U2 d
IE,I :
We also define the following powers: i) the scattered power Ps, as the real part of the flux of the complex Poynting vector of the reflected wave on the surface T: (280)
Fs = Re(f1 n . E r A H r d I ) = - ^ / 2 |5 j ( k a ) I 2 d a , (4it) Z
(see (202)')t and thus 05=4nZ | El1" Ps is proportional to the scattered power; ii) the absorbed power P t as the real part of the flux of the complex Poynting vector of the interior wave, on the surface T: (281)Pa = - R e / r n . E A H d T = / Q ^ (with H13, chap. 1); if Hj is real, Pa corresponds to the dissipation by Joule effect. iii) the toted power Pm = Ps + P . Using relations (262), we have: (282) Ptot = Re/ r [nAH r Ej - n AE r HJ dT=Re/ r [nAH. E, - nAE. H,J dr, since the flux of the Poynting vector of the incident wave plane on the obstacle is zero: Re (fr n.E, A H, dT) = 0. Then with (30), (31) and (269), we have:
3.5 DIELECTRIC OBSTACLE
125 -iko„.x
(283)
Ptot = - R e / r [ J . E , + M.HI]e" ° dT = - RfilJOw^^ + MOw^.H,]
=4rIm0(k«o).Ei)=4lm(A(kao).HI). This relation which gives the total power as a function of the forward scattering is generally called the optical theorem. Since the total power P is greater than P 9 we have the inequality s
1
(284)
l / s 2 lJ(M 2 da S £lm( j(ka0).Ii)-
(4*)'
Note that in the scalar Helmholtz case, we also define, with usual notations, a scattered power, an absorbed power, and a total power by /,
s = Hn/ r Sfn r dT,
Pa*-Im/r^udr=(Imk?)4|u|2dx,
|/»t = P s + i'a = - I m / r ( | u I . u ^ ) d r = I m / n ( k 2 - k g ) u u I d x , which we can also write in a form similar to (283) for an incident wave plane. Remark 17. Is it possible to have a differential cross section vanishing for all directions? First note that the reflected electromagnetic field would be OU/r2) at infinity, and thus would be zero from Rellich lemma. Then the electromagnetic field (E,H) in Q satisfies:
(285)
i)curlH + i
which is a Cauchy problem (see chap. 2). Let: Ex = E - Ej, Hj =* H - Hj in Q. Then (E^Hj) satisfy: I i) curl Hj + icoej Ej = iu>(E - zx) El, (286)
") - curl Ex + io^ Hx = iai(p - pj Hj, in Cl, |iii)nAE 1 | r = 0, n A H , | r - 0
Thus we can extend (E^Hj) by 0 to the whole space, and then apply formulas (30) and (31) with (M, J) given (thanks to the characteristic function of Q, lQ) by (287)
M d#i
Then j (ka) = 0, m(ka) = 0, V a € S% imply with (277):
126
(288)
3 STATIONARY SCATTERING PROBLEMS ^[^(e-c^Ej-kfti-^aAHji^k^aO,
V«€S 2 ,
when €., pj are constant; thus C| = e, p. = p, i.e., Q is occupied by free space! If the domain Q is occupied by a dissipative medium (with (213)), we can verify using the power balance (that is the variational method), that the problem (28S) cannot have a solution; (E,H) must satisfy (289)
/ r [Re(i^)H.H^Re(-ici)?)E.E]dx*Re(/p.E I AH I )dr*0,
that implies E ■ H = 0 (at least where the medium is dissipative), and then E = H = 0 anywhere, in contradiction to (285)iii)! Remark 18. It is possible to have a differential cross section vanishing for certain directions (that is for isolated directions) but it cannot be zero in the neighborhood of a direction (that is, in an open cone). This means that there is no perfect shadow domain. In other words: (290)
j(ka)*0 Va€S 2 sothata.a,>ii foraaj €S 2 , andn>0,
also implies Er = Hr = 0 ! We can prove this result, as an easy consequence of Muller's lemma 115, 117, pp. 341 and 342 (see Muller [1]). As a matter of fact, (290) implies: (291)
j^k^sOandn.dOc^AKka^srO, V«€S 2 suchthata.aj>n
(that is, the wave is circularly and linearly polarized respectively) and this implies thanks to analyticity properties (see Muller [1]) that (291) is satisfied for all a of the unit sphere. This immediately implies that (290) is also satisfied for all a of the unit sphere which proves the property. ® Remark 19. In radar problems, we are interested in having a zero radar cross section. This is possible for particular geometries (symmetry invariance with respect to an axis), as asserted by the well-known (among radarists at least) Weston Theorem; here we give a (small) generalization. Proposition 27. "Weston Theorem". Given an incident electromagnetic wave (EpHj) with an angular frequency ©, on a (regular) bounded obstacle Q which is invariant under a 90° rotation S about some axis. We assume that the direction of incident propagation is parallel to this axis. Furthermore we assume an "impedance" condition on the boundary T ofQ: (292)
UAH*!
-COtAE^
sothat 5" ! CS—2T 2 C" ! t
3.6 SEVERAL OBSTACLES
ill
with (Et,Ht) the total electromagneticfield,Z the impedance of free space\ C an operator on the boundary and S the rotation. Then the radar cross section is zero. We note that this contains the case where the medium inside Q is a dielectric with permittivity and permeability t{ and px with tx/tQ = P/PQ, since the operator C is then the interior Calderon operator; thus the impedance of the inside medium must be equal to that of free space. This also contains the case of the following usual impedance condition (which requires n = 1): (293)
7irE = nZnAH onT, n€C,
PROOF. First we note that the differential cross section is given by the following formula for all directions a, where (E,H) denotes the reflected field (294) o(a)=:—^-j r«mo ^ a A E f r a ^ H f r o ) ^ ^ ^ r !^4*^E(m).aAH(itt). l E ll r ~ > °° l E ll r~*°° This a simple consequence of (30), (31). Then we note that the problem relative to the reflected field is invariant under the 90° rotation about the incident axis so that (5E,5H) defined by (5E)(x) = SE(Sx), (SH)(x) = SH(Sx), and (ZH, - (1/Z)E) are solutions of the same problem. From uniqueness of the solution we deduce that (SE,SH) = (ZH, - (1/Z)E). Then this relation implies: o(- a 0 ) ^ - - ^ f Urn. 4m2 E(- ra0).SH(- ra0) (295) ^
rlim^
(- ^
E(- ra0).E(- ra0)) < 0 f
iE,r 6. SCATTERING AND INFLUENCE COEFFICIENTS FOR SEVERAL OBSTACLES
Let T be a "regular" bounded nonconnected surface with p components (rA), \ in I. We assume that T is the boundary of a bounded domain Q, with N components, Q j , . . , ^ . Let & be the (open) complement of Q with N' + 1 components Q'0, Q ^ , . . ^ ' ^ (with Q'0 unbounded). Note that Cl\ is not the complement of Q., nor is T. the boundary of Q.. We denote by 8Q. (resp. 3Q\) the boundary of Cl. (resp.Q*.). From Alexander's relation (see for instance Dautray-Lions [1] chap. 2.4.5) we have p = N + N \ We assume that there are electric and magnetic surface currents on T, that is, on each surface I\, (M., J.); we assume that the whole space outside T is occupied by free space. We study the electromagneticfielddue to these currents in R3 and its trace on the surfaces T., and the influence of currents on I\ onto another surface T .
3 STATIONARY SCATTERING PROBLEMS
128
6.1. Decomposition of the trace spaces The trace space X = H ~ 1/2(div,0 has the following decomposition: (296)
X=H _1/2 (div,I> $ XA» © H"l/2(dhr,r^.
Let (M, J) a (M v Jx), X € I, be magnetic and electric currents on T in this space. The electromagneticfield(E,H) induced by (M, J) is given by formulas (96): E=2EX, H = | H X , (297) |with E^PJA^ZI^^,
HK={LmMK-PmJk.
Then taking the traces on the boundary, we have (see (101) to (103)):
(29,)
1-
t\***»\ .4^ - v J
We see that S is decomposed into (S^); and S^ is a continuous operator from X x xX x intoX y xX y , and it is regularizing for X*p, t h a t i s I m S ^ i s i n H ^ r ^ x H ^ ) for all s if the surface Tp is C00. This implies that T^, R^ are also regularizing. Relation (105) implies: (299) 2SykS^6wi and on each surface Tk: (300)
VY,H€I,
CSJU^-I.
Remark 20. On the orientation of the normals to the surfaces I\. Let us denote by H. and Cl\ the domain of R3 inside and outside of Tk. We note that the natural onentation of the normal to the surface TK (from inside to outside) does not necessarily agree with that of the domain £1. Let S w be the matrix operator S given by (103) with (100) for a surface T with orientation t of the normal n. We have: S ( ~ € ) »-S ( i ) , and thus 5 = eS w is orientation independent. Choosing: e' = - e for i, e' = e for e, relations (93) and (104) are written: (301)
[-^Hir;J=-^<'-'«[J.
and thus the Calderon projectors are given by (302)
P€^(l-t>S(t))=%(l-s'*S),
w i t h e r - lfori, s'*=+lfore.
We denote by SA the operator S » S relative to the surface r x (for the natural orientation), and by P \ . the corresponding Calderon projectors. 0
3.6 SEVERAL OBSTACLES
129
Writting SA ^? S^, we have SA = t Sx according to the orientation of the normal to I\ with Cl. Thus the operators (303)
Pf=±(I + SA) = ±(I + eSA), P^=i(I-S A ) = i(I-eS A )
are the (exterior and interior for QA) Calderon projectors relatively to the surface Ty Then the (total) Calderon projectors relatively to the domain Cl are decomposed into operators (Pe .)A but only their diagonal terms are projectors. We denote by G c . (resp. G x c .) the graphs of the operators P c . (resp. PAc .), by riA the restriction to the surface TA. Then we have:
(304) GfeG^XxxX^Gfcn^G^cn^, or Gfcn^G^cnxGi, according to whether the natural orientation of the normal to the surface TA is the same as that for Cl, or not. Note that generally we don't have equality in (304)! We easily see that the graphs Ge . have the decompositions: (305) Gi =. ©Q ^ ( Q j ) , Ge = Gc(00)©.s ^
N
G ^ V ' with Cl0 = R\fi', o,
(with obvious notations), but GACL) is different from the sum
^
©
AGf.
rA£aa.
6.2. Screen effect. Extinction theorem In the general case, if (M,J) is in Ge (resp. G.) then the electromagnetic field (E,H) due to (M, J) is zero in the domain Cl (resp.Q'). This is generally known as the extinction theorem. Thus for each X, if (M.,J.) is in GA (resp. G\), the electromagnetic field (E.,H.) ._.
A
A
C
l
A A
satisfies: (306) EA(x) = 0, HA(x) = 0 forx€Q A (resp.a\). Thus currents in G\ (resp. Gxe) do not influence the exterior (resp.interior) of Clx. This implies: (307)
S^Pf = 0 if rM£Q'A,
S ^ O if rMcQA.
More generally, if (MA, JA) is in G^Cl^) for all X so that r x £9&j, then the field: (E,H) = 2 (EA,HA) is zero in R\£1J. Let Clyj0 be its unbounded component. Thus currents on its boundary prevent interior currents (on the boundary of Cl.) to act on points ofCl9. .
130
3 STATIONARY SCATTERING PROBLEMS
6.3. Coefficients of mutual influence of antennas If k2 is not an eigenvalue of the operator A in any bounded domain (Q, or Q'.), then we can define Calderon operators Ce>1 as above. We can write, using standard notation, for the exterior and interior domains respectively: (308)
C e - J ( (n AE| rx ) xeI ) = (nAH| r t e i ) A e I );
thus C e and C1 are matrices: C*'1 = (C*'') (309) d a .
©
C^Qj), C'-C^Q,,)©
- w - We have the decompositions: ©
CXCTj), but cVQj) *©€*(%).
Notice that the exterior Calderon operator for Q (the complementary set of the unbounded component of Q') always exists. This operator takes into account the mutual influence of the different bounded obstacles, i.e., the components of 7. MUTUAL INFLUENCE OF SHEETS
7.1. Introduction Let r be a (Lipschitz) connected surface which is the boundary of a bounded open set Cl in R3. Let I*A, X € I, be a partition of T. This partition can be made in order to apply a numerical method, or to decompose T into more regular pieces (r being a polyhedral surface), or into surfaces enlightened by a wave plane or in the shadow, in order to consider high frequencies problems. Then to a current J in H~1/2(div,D, corresponds the family (JA) X€l, with J. eX(div,r.) (see (87) chap. 2) with matching conditions in the case of two pieces see (89), in the more general case these matching conditions are more tedious and technical to write (obviously H""1/2(div,T) is not the sum of the spaces X(div,rx)). If we assume that each current on Tk is regular, that is in Hs(div,rA) for some positive number s, then these conditions are easy to write: Proposition 28. Let (TJ, X € I, be a partition ofY, with Tx regular. Let (JA) X € I, be regular currents on Tx (JA in Hs(div,rA) with positive s). Then (JA) X € I defines a current J in H " 1/2 (div,0 if the following matching conditions are satisfied (310)
vA. JA.vM.Jy = 0onAAM = rAnf>J, VX,n€l,
with vA the exterior unit normal in I\ to its boundary. PROOF. This results from the usual jump formula of the divergence on surfaces (the analogue to (54))
(311)
divrJ= 2 (divr JA)+ 2 exvA.JA,
131
3.7 MUTUAL INFLENCE OF SHEETS
with (divr JA) in the classical sense, EA = +1 or -1 according to the orientation of 8I\. Then we have J in L 2 t (0 and divf J in L 2 (0, thus J is in H°(div,D, and therefore inH" 1/2 (div,D. « Obviously we have an analogous proposition by changing divf into curl f , with the matching conditions: (312) VAAJA-VMAJ^0 onA^^^nf^, VX,n€l. 7.2. Electromagneticfielddue to currents on a sheet Here we study the electromagneticfieldproduced by a regular current J on a sheet r o which does not satisfy v.JaOon the boundary A of r o , and thus is not in the space H^1/2(div,ro ) (see (85) and (92) chap. 2), so that we know that (E,H) is not of locally finite energy in the neighborhood of the boundary A of r o . The electromagnetic field (E,H) produced by J is given by (96) (with M = 0). Its expression will be given thanks to the jump formula (which is a result of differential geometry, see (94), (273) and (278) in the Appendix and also for instance Schwartz [1] (IX, 3; 24)) (313)
curl J a (curl J) - v A J A 5 A , V the exterior unit normal to A in T0,
where ( ) cl is the "classical" part (on To), whereas the last term is concentrated on A. We can also prove (313) using the formula (see (310) in the Appendix) (314)
with C € D(R ) . Thus we can separate curl J into its transverse and tangential part. Applying the Stokes formula, we obtain the boundary term, and the expression of() cl isgivenby: (315)
<(curl J) c l , S>=/ r (curl^ J .Cn + J.| i (nAC) + 2RmJ. nAC)dT0 ,
for all C € D(R3)3. For the gradient we also have: (316) with
grad p = (grad p) d - v pA 5 A ,
(317) <(grad p) d , O = / r (grad^ p. ? - p ^
- 2Rmptn) dT0, V C € D(R3)3.
The expressions of (grad p)cl and (curl J)cl contain the transverse distributions G{p and G2J: (318)
VC€Z)(R3)3,
132
3 STATIONARY SCATTERING PROBLEMS
with the properties, for all regular p and J: + (* • Gjp) € H'CRVo), ♦ (* • G2J) € H ^ R V O ) 3 , V ♦ € D(R3). Proposition 29. The electromagneticfielddue to a "regular" electric current J on a sheet TQ which does not satisfy v. J = 0 on the boundary A ofTQ is decomposed into its regular part (E »H ) and its singular part (E sing »H si ) according to: I (E, H) = (Ereg, H^g) + (Esjng, H^g), (319)
Ereg = ikZ(ZJ+gL(gradp)d), H r e g ^ ^ E
sing s - * f
L
PWVPA*A
+ «wd (v. JA8A)], H^g = - L (v A J A S A ).
with (320)
♦ E^and^H^cHkRVo) 3 , V*€D(R 3 ), E^andH^€C~(R\A)3, withE^W-O*}), H ^ O f l o g r ) , r*d(x,A).
and thus ♦ EandtHeH^RVj3, (321) PROOF.
V*€D(R\A).
The electricfieldE is given by (96): |E*ikZ(U+4gradLdiv r J)
(322)
= ikZ(IJ+4gradL(div r J) cl )-i|grad£(vJ A 8 A ). We also have: (323) gradL(divrJ)d = i
H = curlU = curi**J=<&*curlJ,
and then we apply (313). Properties (320) on (E ,H ) follow from the usual regularity properties of the single layer potentials (see section 1.3.1). For ^sing'^sinP' tkis is a consequence of the properties of the Newtonian potentials relative to a charge density concentrated on a closed line (see below). We have a more precise result (325) E ^ x ) ^ . ^ ^ ^ ) ! + 0 ( l o g r ) , H stag (x) = :-^vAJ A logr + 0(l) > with a=j (x - nAx), r ■ d(x, A), nAx=the projection of x on A.
133
3.8 CURRENTS ON A LINE
7.3. Matrix elements; influence coefficients th sheet TQ9 we can decompose RJ and TJ into their regular and singular On the parts RJ = v(RJ)'reg +(RJ). , x 'sing'
TJ = (TJ)'reg +(TJ). 'sing',
(326) (RJ)rcg = 2iknA[U+gL(grad r p) c l ], (RJ)
2i siiuf " f
n A L [ ia,Vp
A*A +
(TJ)^ = - 2nAUcurl J) d>
g r 3 d (V J
" A8AM' Casing =
2 f l A UV
A J
A8A)-
When we have to calculate the matrix elements of the Calderon projectors, or of the S operator, we have to calculate matrix elements of R and T (see (102), (104) and (273)) between the sheets I\ and r (giving the mutual influence between Tk M andT) (327)
Ap
V J, J piecewise "regular" currents on T. We have four types of elements according to: i) I\nf M = 0 , ii) r A nr p has one point, iii) fA and fj, has a common line, iv)TA = T^ In cases ii) to iv) there are singular integrals to calculate for the matrix elements of T. We can use Stokes formula:
= / r x r *(x-y)divry(JA(y)AMll(x))dTydrx
(328)
+/8rxr ^(x-yJv^J^AM^dLydr,, A
p
and then with AA = drA, the last term can be calculated by: (329)
J
dLy(nyAJA(y))./r
'p
We refer for instance to Bendali [1], Heliot [1] for the explicit computation of such matrix elements with P{ functions. 8. FIELDS DUE TO CURRENTS ON A LINE
Let J a J $A be an electric current concentrated on a (closed, i.e. without boundary) bounded regular curve A, with JA regular, tangent to this curve. Then the electromagnetic field (E,H) due to this current in free space, which satisfies the Silver-Muller conditions at infinity is given by (44) with M = 0, that is: (330) E = io>p(A + k- 2 graddivA), H = curlA, with A = , J = ZJ = I(jeA5A),
134
3 STATIONARY SCATTERING PROBLEMS
with JA = j e A , eA unit vector tangent to A; A is called the Hertz potential. We have: (331)
divJ=§8A,
with s the usual curvilinear coordinate on A; we can write (330) in the form E = io>ji ( U + IT 2 grad L ( ^ 8A)),
(332)
H = curl (U).
We know that (E,H) cannot be of finite energy in the neighborhood of A, so we have to study the behavior of (E,H) in the neighborhood of A. This follows from the Lemma: Lemma 4. Let p dA be a distribution concentrated on the closed line A with regular density. The behavior of the Newtonian potential u 0/-p Q dA, in the neighborhood of A is given by (with x = (poc,s) in load coordinates): (333)
u ( x ) = J ? / A ] 7 - ^ p o ( y ( s ) ) d s = -p o ( J t A x)iLogp + 0(l),
nAx being the projection ofx onto A, p = d(x, A) = | x - wAx|, a.grad u(x) = - PQ(*Ax) TT~- + 0(log p), a unit vector orthogonal to A, (334)
a.grad u(x) = - PQ(*AX) ^ log f>, a = eA(s) along the tangent to A. PROOF.
(335)
4
First with the Lipschitz hypothesis on pQ we have: *u(x)-po(*Ax)/A]Iz^
Therefore the behavior of u(x) in the neighborhood of A is given by that of v(x): (336)
V(X) = P O (* A X)^W(X)
with
wfr^/^^jds;
with a < 0 < b and y(0) = 0, y*(0) » (0,0,1) = eA(0) = e 3 (choosing s = 0 for *Ax), we have:w(x) * J ^ + y J ) " 1 ' ^ * - 2 log p. The gradient of u satisfies: (337)
^(x)=^/A Ix-y^r^MsWds, y.-Xj
with p. =-^7—, r= |x - y|, and thus (with y=y(s)): (338) 4 * | | ( x ) - p o ( * A x ) / J x - y ^
135
3.8 CURRENTS ON A LINE Ifp i s C ' 1 on A, we have P„(y(s)) - Po(y(0)) = po(y) - Po(nAx) = p' (nAx) s + 0(s2), thus:
(339) 4«^(x)-p o (it A x)/ A P. |x-y|- 2 ds-p;(i. A x)/ A P. | x - y | - 2 s d s = 0(l). The gradient of w(x), given by ^ = /
A
| x - y | ~ 0. ds, i» 1 to 3, satisfies:
| ^ « - 2 ^ f o r i = i , 2, Xj orthogonal to A, 04 = (x-it A x)./p, (340)
| ~ « 0 along the tangent to A (03 is to first order an odd function of s). Then we have for i = 3: (341)
/ ^ | x - y | - 2 sds .
2
^ ^ d y
3
= ^ ^ d t . -21ogp,
and for i= 1, 2:
(34iy / A Mx-y|- 2 sds * ^i^f^s^-^i^^^ra^- T? from which we obtain (334). &
Proposition 30. The electromagneticfield(E,H) due to a line electric current J on a closed bounded curve, by (332) has the following behavior in the neighborhood of the line, with notations ofLemma 4 and with local coordinates x = (p
Htr(x) ~ ^ ( « A J e s )
(342)
Es(x) = i < ^ l o g p [ - j - k - ^ ^ ] ,
Hs(x) = (Xlogp),
with J = jes 5A, es = eA(s) a unit vector along the line A, and (Es, Hs) the components of (E, H) along es, (Etr, Htr) the transverse components. PROOF. Formula (332) for H gives: B ikr
(343)
H(x)=J A §f(|JAJe r )ds,
with f j . £
/
^
l
, r=|x-y|.
Then we obtain (342) for H, and also for E due to Lemma 4. ®
136
3 STATIONARY SCATTERING PROBLEMS
The electromagnetic field (E,H) due to a line current J has the following behavior at infinity: (344)
E(ro) * icoM [J(ko) - a (a. J(ka)] Q(r), a e S2, H(ra)sikaA?(ka)«(r)
with J(ka) * <&-*** J> =/ A c -
t e
forr= |x| -♦<», ^ j (s) es ds.
These formulas are an easy consequence of (30), and of: (345)
|(divJXka) =
Remark 21 • So far we have described a radiating antenna. The modelling by a line of a thin receiving antenna with a very conducting medium and the modelling of the induced current by a Dirac distribution concentrated on a line (according to the Faraday's effect) is outside the scope of this book.
9 . SCATTERING PROBLEMS BY A CHIRAL OBSTACLE
We consider the scattering problem of section 5, but with a linear chiral obstacle (see chap. 1.3.3) instead of a dielectric medium. Then we have to find an electromagnetic field described by (D,B), (E,H) in the obstacle Cl satisfying: (346)
i)curlH + io>D = 0, ii) - curlE + icoB = 0 inQ,
with the constitutive relations (see (96) chap. 1, where e, p, e, £ may be matrices): (347)
i)D = cE + eH, ii)B = jiE + nH,
with hypothesis HI8) chap. 1 of a dissipative medium, and with the boundary conditions (262), giving (265) on E. We emphasize that the boundary conditions are unchanged with the constitutive relations (at least if the medium is at rest), because they are due to the curl operator! We can easily prove uniqueness (at most) of the solution of this problem as follows.
3.9 CHIRAL OBSTACLE
137
Multiply (346)i) by E and ii) by H, integrate and add, we obtain: (348)
I R e ~ to KD'E> +
To prove uniqueness of the solution, we can take Ej = Hj = 0 and thus Re/rn.EAHdr = Re/rnAE.HdT=Re/rnAErHrdr>0 (see (202)). But from the dissipativity hypothesis H18) chap. 1 the l.h.s. of (348) is strictly positive except for E = H = 0, which proves uniqueness! Then proving that this problem depends on the Fredholm alternative (like in section (4.8.2)) implies existence and uniqueness of the solution. Here we develop a variational method with (slightly) different hypotheses. We write Maxwell equations (346) with (347) with E and H only: (349)
i) curl H + io)€ H + i
We can eliminate H applying -(curl + mt) (icon)"1 to (349)ii) adding to i) (350)
- (curl + iweXiw)*)" (- curl + ia>fi)E + icoe E = 0.
It is better to use a variational framework. We define the sesquilinear form (351)
a(E, F) = - f [(top)" ^curl E - ic*p E). (curl F + ia> xz F) + io>e E . F ] dx -/rCc(nAE).FdT,
with l€ the transposed matrix of?, and E, F in H(curl, Cl). The scattering problem is
(352) a(E,F)=/ r f,. FdT, VF€H(curi,n). We assume that (353) ? « ? • , and Re(- i
Re a(E, E) 2> a [llcurl E - ia>? EH2 + HEII2],
V E € H(curl,Q).
But we can obtain easily: there are positive numbers ot0, f>, so that (355)
llcurl E - icojl EH2 < a0 [llcurl Ell2 + IIEII2] < P [llcurl E - icojl EH2 + IIEII2]
for all E € H(curl,Q), and thus the norms:
138
3 STATIONARY SCATTERING PROBLEMS
E l , 1?[Ilcurl E - icoP EH2 ♦ 1EII2]1'2 and IIE«H(curla) d=*f[iicurl Ell2 + IE!2]1'2 are equivalent. Therefore the sesquilinear form a(E,F) is coercive on H(curl,Q). The Lax-Milgram lemma implies existence and uniqueness of the solution of the problem (352) like in section S. 1. Remark 22. The constitutive relations are often written in different forms, for instance: (356)
D = c c E+Y cm B,
H^B-rmcE,
with y _ - - y _ , which is equivalent to (347) with: (357)
P - ^ S - ^ y ^ «^c^cYme' * = W c = - 5 '
1 0 . CONCLUSION ON THE CALDERON OPERATORS FOR SCATTERING PROBLEMS
Let a be a bounded domain occupied by a homogeneous linear isotropic dielectric with permittivity e and permeability p. Then all the electromagnetic properties of this obstacle (with respect to an incident wave at frequency m) are contained in the interior Calderon operator C* (see (211)), if © is a "regular" frequency, that is k2 = o>2€|i is not an eigenvalue of the operator A defined by: Au»-Au,D(A)={u€L 2 (Q) 3 > curlu€L 2 (Q) 3 ,divu = 0,nAu| r =0}. Using a more physical minded terminology we can call it a surface admittance operator. The essential property of C1 is that its graph corresponds (up to the vector product with the normal) to the set of tangential boundary values of the electromagnetic field in Q. We recall also the positivity property (212). Less usual for physicists is the similar notion for the exterior domain, which is generally occupied by free space. Then all the electromagnetic properties of the exterior domain (at frequency CD) are contained in the exterior Calderon operator C e : this is a surface admittance operator for the free space domain; when a> = 0, this corresponds to the "usual" notion of a capacity operator. Like for C1 the essential property of C c is that its graph corresponds (up to the vector product with the normal) to the set of tangential boundary values of the electromagnetic field in the exterior domain Q'(taking into account the Silver-MQller condition and the local finite energy up to the boundary T of Qv). We recall also the positivity property (201). The main differences between C\ C c and the Calderon projectors P., P c are: i) P. and P keep informations on both sides of T, and exist in all frequencies, ii) P. and P c are integral operators (whereas C1 and Cc are pseudodifferential operators ofzeroth order, for this notion, see for instance Hormander [1], Gilkey [1], Taylor [1]),
3.10 CONCLUSIONONCALDERONOPERATORS
139
Hi) the range spaces G c = Im P e and G l * Im P l are (up to a minus sign) respectively the graph of C e and C1 for regular frequencies. Let r be a regular surface which is the boundary of a bounded domain Cl. The space Y of magnetic and electric currents (M,J) on T which produce an electromagnetic field (E,H) (at angular frequency
n A Ej = n A Et - n A Er, n A HJ = n A Ht - n A H r ,
that is to decompose space Y into the direct sum: (360) Y^G^eG^. Using the interior Calderon projector P? for the exterior wavenumber k0, we see from (271) section 3.5 that this problem is reduced to: the restriction off? to G\ is an isomorphism from G\ onto Gj,, i.e., there exists a > 0 such that IIP?(M, J)ll > a ||(M, J)H, V (M, J) € G/ . We can use the Calderon operators C l,e to solve numerically scattering problems by using special basis. We develop this in the Helmholtz case, for C c = Cj. Let ^ ( x ) as 4>(x - XQ) be the outgoing elementary solution of the Helmholtz equation centered at XQ. Let F be a subset of R3. We obtain the following, setting: (361)
MF-K«Sblr^l^«0€F|.
Proposition 31. We assume that F is a dense subset of the boundary TQ of an open subset Cl of Cl, strictly contained in Q, and such that - k2 is not an eigenvalue of the Dirichlet Laplacian in C1Q. Then Af- is a total family in G e = G(Ce), i.e. the vector space [Mp] generated by finite linear combinations of elements ofMF is dense in G e . Proposition 31 implies that given (u^u 1 ) in G , (thus u1 = Ceu°) and n > 0, there exist (x.) in To and complex numbers (a.) such that (362)
* llu -2aj*\»:Sn, 0
J
*i ta^ZcjWLysn.
140
3 STATIONARYSCATTERING PROBLEMS
of Proposition 31 (adapted from Petit-Cadilhac [2]). We show that the polar set [Mp]° of [MFJ for the pairing (143) is equal to the polar set of Gc, which is Ge. Let (vV 1 ) be in [MF]°; (v°\vl) satisfies: PROOF
(363) thus
<*%v^-<^lriS-0,
Vx 0 €F,
(363)'
/ r (v 1 (y)«Ky-x o ).v o (y)^(y.x o ))dr y S 0 > V x 0 € F .
Settingf=(vl Sr+div^nfij-)), andu=$«f, (363)*is:u0^)-0, Vx^,€F. Thus u satisfies the Helmholtz equation in Q, hence in Q0> with u | . Since - k2 is not an eigenvalue of the Dirichlet Laplacian in Q0 by hypothesis, this implies that u = 0 in Q0> thus in Q. Therefore (v°,v1) € Gc, i.e. the Proposition. ®
For the interior Calderon operator we have the analogue of Proposition 31: Let F be a dense set in the boundary Tx of an open set Qx with Q strictly contained in Qj. Then M^isa totalfamily in G. ■ 0(0*). Instead of elementary solutions we can use plane waves with wavenumber k for G.; we prove like in (363)' (but with Rellich lemma) (364) thesetM^{(eikaJi\T9 ika.ne****^), acS 2 } is a totalfamily in Gv Numerical methods based on expansions with elementary solutions such as (361) are called Fictitious Sources Methods (see Petit-Cadilhac [2]). 0 Calderon operators are useful for modelling many physical situations (see section 3.9 for example). We give another example. Excitation of a proper mode of a cavity by an incident wave. We assume that the cavity Cl is occupied by the free space (or a conservative medium) and is bounded by a perfectly conducting medium (occupying a bounded domain Qx) with a hole TQ so that an exterior incident wave (E{y Hj) at frequency
i)curlH + icncE = 0,
ii)-curlE + MiH«0 inQ.
Let T{ be the boundary of the conducting medium for Q. Let Cc be the Calderon operator exterior to the domain Q occupied by the cavity and the conducting medium, with boundary f. Then let C* be defined by c;
3.10 CONCLUSION ON CALDERON OPERA TORS
141
Recall that n A E is zero on I\r o . With the notations (SS), (86) of chap. 2 we have 1/2 H A E Ll €H; (div,r0), andC^(nAEL)€X(div,r 0 ). l o
o
Then (E,H) must satisfy the boundary conditions ji)nAE| r =0, '1
(366)
|ii)nAH|. - ^ ( B A E L ) - ^ ! . , with f ^ n A H , - C c (nAE,U).
We have proved (see section 5) that the scattering problem for a conducting obstacle has a unique solution and thus problem (365), (366) also has a unique solution. Now we assume that k2 = o>2ey is a simple eigenvalue for the operator A (see (147) chap.2). Let (EQ,H0) be a corresponding eigenmode with unit energy / a ( e | E 0 | 2 + n|H 0 | 2 )dx=l. Then (E,H) has an orthogonal decomposition: (E,H) = c 0 (E0,H0) + (E,H), with (E,H) orthogonal to (E0,H0). Then in many physical situations we are interested either to generate at best the eigenmode (E0,HQ), that is to maximize the modulus of c over all possible holes and incident waves with given energy or to minimize it in order to protect the interior from all incident beams. We note that the problem:find(E,H) satisfying Maxwell equations i) curl H + i(i>€ E = J,
ii) - curl E + io>y H = 0 in Q,
with the boundary condition on T = 8Q: nAE| =0 or nAE| =nAE 0 | and with given J and EQ, depends on the Fredholm alternative and does not generate eigenmodes! Remark 23. Interior Calderon operator and its Fourier transform. Let (E, H) be an electromagneticfieldin a domain Cl occupied by a medium with (complex) characteristics (c, p). Let M = n A E | , J = - n A H | . Generalizing (31) to complex k, we have (367)
- ika A M(ka) + ai>M riaJ(ka) = 0
that is also, with the (complex) impedance Z = cop/k, (367)'
naJ(ka) = Z"1 a A M(ka), with J = - C1 M.
142
3 STATIONARY SCATTERING PROBLEMS
This can be obtained from the behavior at infinity of the extension of (E, H) by 0 out of Q. In other words, the mapping: (M,-J) —• j(ka) (and &(ka)), given by (31) vanish on the space G1 (for the wavenumber k only). Of course we have similar properties in the scalar case for Heimholtz equation. 0 Remark 24. The Calderon operator for Heimholtz or for the Laplacian is also called the Dirichlet-Neumann operator or the Poincare-Stekloff operator. The Calderon projector for a domain Q is defined for instance in Chazarain-Piriou [1]. • Finally note that the Calderon operators and projectors have all the symmetries of the problem. The Calderon projectors well adapted to computation when we have a few number of homogeneous and linear media. 1 1 . MULTIPOLE EXPANSIONS. RAYLEIGH SERIES
Here we develop essentially a spectral method, based on expansions which are very familiar to physicists. But there are fine questions of convergence, well studied by Muller [1], which we follow. We begin by the simplest case, that is the scalar case with Heimholtz equation in two dimensions. 11.1. Multipole expansions for Heimholtz in R2 Let u be a solution of the (homogeneous) Heimholtz equation for the wavenumber k > 0 in a domain Q in R2. We assume that 0 is in Q, and thus there is a disc Ba of radius a contained in Q. Since u is regular on circles Cf of radius r ^ a, its restriction to Cf may be developed in Fourier series, and we have, in polar coordinates: (368)
u(r,9) = 2 cn(f)e™, n€Z
where cn(r) satisfies the differential equation: (369)
1^+1 ^
+
(k2-^)cn =0
for r
and thus c n is proportional to J n the Bessel function of order n: (370)
cn(r) = cnJn(kr), withc n €C.
Furthermore we have: (371) 2 Ic n (r)| 2 *2 |c n J n (kr)! 2 <+oo for r * a . When u is a solution of Heimholtz equation in the domain Q* (the complement of a bounded domain Q), and satisfies a Sommerfe i condition at infinity, cn(r) is proportional to the Hankel function of order n:
3.11 MULTIPOLE EXPANSIONS
(370)'
143
cn(r) = c n H f W with cn € C,
K = 1 for the outgoing condition, K * 2 for the incoming condition. We also have (371) (replacing the Bessel functions by the Hankel functions) for r > a. Let un be defined by: (372) un(r,0) = Jn(kr) e™ (resp. H J W ) e™). Thus un satisfies Helmholtz equation in R2 (resp. R2\{0}, and (A + k2)un = p is concentrated at the origin); u = I cnun is called the multipole expansion of u (or also a Rayleigh series). If u does not satisfy the homogeneous Helmholtz equation in the whole space R (or R2\{0}), then we see that the expansion u = I cnun cannot be convergent everywhere. Thus we have the question offindingthe domain of convergence of this series. The answer is similar to that for entire series I rV110: there is a radius of convergence p so that the series is convergent inside (resp. outside) the disc B for the expansion with Bessel functions (resp. Hankel functions), and is divergent outside (resp. inside) this disc. We recall that the Rayleigh hypothesis is that these series are convergent everywhere inside (resp. outside) Q, for any domain Q; we know that this Rayleigh hypothesis is wrong. We study this question of radius of convergence for three dimensions only. 11.2. Multipole expansions for Helmholtz in R First we recall some essential properties of spherical harmonics. For the study of such properties, we refer to Dautray-Lions [1] chap. 2.7.3 and 9.B 1.1, to Muller [1], and to Stein-Weiss [1]. These properties originate from the representation of the rotation group in R3. 2
We use a spherical coordinate system; let a = (aiya29a3) € S , with
O<0
The set of functions ( ^ ■), n€N, - n ^ j < n , defined with the Legendre functions (373) Knj(a)==(2nrl/2p|l(cos0)e1J0(also often denoted by Y/m(0,0), / = n, m=j) which are the traces on the unit sphere S2 of the homogeneous harmonic polynomials in R3, is an onhonormal basis o/L2(S2). Furthermore they are the eigenfunctions of the Laplace-Beitrami operator on S2 according to A
s2Kn,j(«)
= -n(n + l)K n
and thus the eigenspace Vn corresponding to the eigenvalue n(n + 1) has 2n + 1 dimensions. The elements of Vn are called spherical harmonics of order n and often denoted by Kfl. Using spherical harmonics allows us to find the solutions of the Helmholtz equation with separate variables (r,a) thanks to the usual expression of the Laplacian in (polar) spherical coordinates:
144
(374)
3 STATIONARY SCATTERING PROBLEMS
Au(nx)=|^+?f+^Aau,
w i t h A , , ^ , r= |x|, « - ? ,
we see that cn(r)K„(o) is a solution of the Heimholtz equation if y=cn(r) satisfies: d?
T
3F
(
-j—)y=0.
The solutions of this equation are well known: they are the "Bessel" function jn(kr) (for the regular solution up to 0) and the Hankel functions hn(kr), related to the previous J„(kr) and HQ(kr) by: (376)
j„(D=(£) 1/2 J ,(r),
hJt)(i)-(£),/2H<"),
wtthic=l,2.
Then let ^ be the function: (377)
uE(w)-hf)fla)Kll(«) (wsp.^kr^a)).
Expansions of the form Iu ft are called Rayleigh series (or also multipole expansions). We recall the main property of convergence of these series (see Muller [1], Theorems 11, 12) which will be a consequence of the following Lemmas. Theorem 9. Let (Kn) be a sequence ofspherical Harmonics. Let (378)
c i . / ^ l ^ a ) ! 2 * , c^^O.
We assume that there exists rQ so that the series: (379)
2 |h£W 0 )j 2 c* < + *
Then the series: (380) u(ia) - 2 hfW) ^(a) n "
"
(«*p. 2 U n 0» o )| 2 i^<+ » ) . (resp. u(ra) = 2 jJM ^(a)) n
n
"
converges absolutely and uniformly on every compact set K in R \Bt (resp. in Br ). This is equivalent to: if the series (380) is convergent in L2( Sr ), then it is also convergent on every compact set K as above. Furthermore it is also convergent in all "reasonable" topologies, for all Sobolev spaces for instance. This implies that the series u given by (380) satisfies the Heimholtz equation with wavenumberk. In the "exterior" case, that is for the expansion with Hankel functions, we will verify that such a expansion satisfies the Sommerfeld condition (outgoing or incoming for K s 1 or 2).
3.11 MULTIPOLE EXPANSIONS
145
Lemma 5. On any finite interval [a,b] with 0 < a < b < oo (resp. [0,DJ) we have (denoting by T the Eulerfunction): (381) h?(r) * < - l ) « ^ r ( n 4 uniformlyforn—oo, a ^ r
(?f
(resp. y
\
^ H S T ^ (§)\
(resp. 0 : £ r < D ) and then:
<*»• M^ -
$-c«'——
r
Corollary 4. IWift hypotheses (379) die coefficients c 0/ f/ie expansions (380) satisfy respectively fl
Lemma 6. ITie spherical harmonics satisfy the following inequality: (383)
2
2
iVa)! ^^!/^!^)! ^,
VneN.
PROOF of theorem 9. From the above lemmas 5, 6, there are an index N and a constant C such that for n > N we have: Q
Q
I h^wK^coi
s c 42inrr ( ^ )
(res . i ^ W K ^ C O I < c 42KTT [ £ ] ), n
P
which implies the convergence of the series in the conditions of the theorem.
Remark 25. It is possible to replace the topology of the L norm on the sphere of radius r by some other topology with the same convergence results; for instance we can assume that the expansion (380) converges in the sense of distributions on the sphere. In R this is equivalent to: there exists a number p such that Q
2
2
p
c |J (lcr )| ( l + n r - 0 w h e n | n | - o o . n
n
0
Then we define the radius ofconvergence of the series (with (378)) by: Ree^taffr, 2
(384)
IhSWc^+oo),
n
2
R c i ^ s u p f r , 2 |j (kr)| 2< oo}. n
C
+
146
3 STATIONARY SCATTERING PROBLEMS
Then if Q is a bounded obstacle which contains the origin and if u satisfies the Helmholtz equation outside O (resp. inside Cl) with the Sommerfeld condition at infinity, then u has an expansion given by (380) in the exterior (resp. interior) of all balls B which contain (resp. is contained in) Q, and thus: R ^ R ^ s u p |x| f
(resp.R i ^R m « infjx|).
When u is due to a distribution f with compact support (see (1)), we have a similar result with (supp f) replacing Q in the expression of RM. EXAMPLE
1. Elementary solution relative to a point xQ not at the origin.
Let <&• «
Au + k*u»-«^.
Then we can prove that $(x - XQ) has the expansions |^2(2n+l)Jn(kr0)h!ll)(kr)PIl(aa0) (386)
ifr>r 0 ,
*(x-x 0 ) = § | (2n+ 1) jn(kr) hjJWo) Pn(aa0) if r < r0,
withr= |x|, r0= ^ l , a=sx/r, a0 = x Pn the Legendre polynomial of degree n. We can find these formulas, using that $(x-x Q ) has a Rayleigh expansion with Bessel functions for r < rQ and with Hankel functions for r> r0, and then using that the jumps of $(x-x Q ) and its normal derivative across the sphere of radius r are (in the sense of distributions on the sphere): (387) W x - j g ] s
*m\W-Vm\M*09
[
g p ^ ] s =A*< a - a o>
with
^00-^0) | ^ P r f - ^ n
which is due to the relations: (388)
«^«-^ o (r)8(«-cg f
*a-a0)= | ^t^Pn(aa0).
This follows from the relation, for all regular functions f on the unit sphere: (389)
«« 0 ) . | / § 2 ^ J ± I Pft(oa0) f(a) da
(see for instance Muller [1] p. 53). Then we can see that thefirstseries in (386) is divergent for r < rQ, and the second series for r > rQ.
147
3.11 MULTIPOLE EXPANSIONS
Indeed the generic term Sn of thefirstseries is equivalent (thanks to lemma 5), for high n, to: S n « c [ ^ J Pn(t), t=a.cc0, Caconstant. Then the convergence is a consequence of the usual formula: (390)
(1 - 2zt + z2)- m = 2 PnW *> n
"
which defines a convergent series when | z| < 1, since the zeros of the polynomial ( l - 2 z t + z2)arez = eieande"iflfort = cos8. Thus the radius of convergence p of the entire series (390) is equal to 1 that is: P = limsup|P n (t)| 1/n =l, V t € [ - l , l ] . *-" * EXAMPLE 2. Entire solutions of the Helmholtz equation. Plane waves. Expansions of the form (380) (with Bessel functions) with an infinite radius of convergence define entire solutions of the Helmholtz equation. An example is given by plane waves, where the expansions are uniformly convergent on compact sets (391) e t o o X = e*"*0" = 2 in(2n + 1)Pn(a0.a)j (kr), witha0 eS2. n
"
Replacing in (391) the Bessel functions by Hankel functions, and using:
Jn(kr)-^(l4V) + h?(kr)), we might think that we obtain a decomposition of the plane wave into the sum of an outgoing and an incoming waves, but this is false since the expansions 2 in(2n+l)Pn(a0.a)hSc)(kr),K=l,2, ^ ^ / ^ ( P ^ a . a ^ d a ^ ^ J j are nowhere convergent! Then before studying the behavior at infinity of Rayleigh expansions, we will answer the question: what is the source of a wave un given by (377)? 11.3. The source of a wave u n
First note that un(m) = h j J W ^ a ) satisfies (A + k2) un * 0 in ZT(R\O). Furthermore the Hankel function of order n being equivalent in the neighborhood of 0 (up to a constant in r) to r~n, it can be extended (using finite parts, see Schwartz [1]) to a distribution on R3; thus (392)
- P n " ( A + k2)ull (in^XR3))
148
3 STATIONARY SCATTERING PROBLEMS
is a distribution with support {0}. Thus p is of the form p = Qn(D) 5, with Q a polynomial. Since u satisfies the outgoing condition at infinity, it is given by the convolution product (393)
u n «*»Q n (D)8 = Qn(D)*.
Thus we have to find O so that: (393)' ( ^ ( D ^ ^ ^ D J ^ ^ h i V ) K^a). We can see that this implies that Qn is a homogeneous polynomial of order n, so that its restriction to the unit sphere is proportional to Kfl:
094) Qn(it)=Qnam«)=vn m x w We can also find this result by Fourier transformation (see for instance SteinWeiss [1] p. 158, thanks to the commutation of the Fourier transformation with rotations !). Then vn is easily obtained from the behavior at infinity of u n and of the Hankel function (see Muller [1] p. 74, Petiau [1]); we have: u ^ O ^ V ^ a ) *^e-^^^a)^^)^),
r-oo,
with a e S , therefore: (395)
3n(ka) = Q^ika)=£ (-
if^K^a),
tbm v =
n lf [V] ' andthus:
(395)'
Q ^ l ^ f
( ^ ]
K^a).
We can also find another expression for pn using distributions in polar coordinates up to 0; for this we must specify the space of distributions. Space of distributions in polar coordinates in Rn\{0} and Rn. The mapping h: x —» (r,a) is a diffeomorphism from Rn\{0} onto R+ x S n ~ l (with R+ = (0, oo)) which allows us to identify distributions on Rn\{0} with distributions on R + x S n " l . But here we have precisely distributions on Rn and not only on Rn\{0}. Let u be a regular function on Rn. Then we denote: (396)
U(r,a) = u(w) = u(x),i.e., U - u o l T 1 .
Lemma 7. Let u be a regular (C°°(Rn)) function. Then U satisfies:
3.11 MULTIPOLE EXPANSIONS
(397)
|¥(0,a)=
2
149
rf« X D\i(0),
VX = (X1,..,Xn)eNn,
irftft |X| 8 ^ + ,,. + ^ , XlsXi!...^!. Thus —jr(0,a) & a sphericalharmonic oforder p anrf U(Ota) = u(0), & independent ofa, PROOF.
We have, with standard notation:
(398)
f g (r,a) =, 2 DjU(x).^ = 2 OjDjiKx),
with)^ = raj, Djssg^, hence i) . Then we easily verify that: (399)
(2«jD/u= j
J
J
2 a,
a: & ...R u,
ii-.Jp1
P
J|
P
hence (397).
®
Then we see that, if u is a C°°(Rn) function, U is also a Coo([0,oo)xSn""1) function. We define its extension to R x S n ~ l (also denoted by U) by (400)
U(-r,a) = U(r,-a),
Vr>0, Va€S n " ! .
Lemma 8. Let u be a regular (C°°(Rn)) function. Then its extension U to R x S n ~ l defined by (400) (and (396)) is such that: U€C 00 (RxS n " 1 ); furtheimore u € D(Rn) (resp. S(Rn)) implies U € D(R x Sn~l) (resp. S(R x Sn~*)). PROOF.
We only have to verify that lim 9?U(r,a) = lim 8{?U(r,a), which follows
from (397) and (400). Lemma 9. Lef u te iVi D(R3) or 5(R3). I7ie/i U(r,a) Aas an expansion in spherical harmonics (401)
U(r,a)« 2 Un ^ K ^ a ) ,
n € N, - n < j < n,
n,j
such that each U n j, satisfies: (401)'i) and also
Unj(0) = 0, . . . , - ^ ( 0 ) = 0, p = 0 t o n - l ,
apu •
(401)'ii)
- ^ ( 0 ) = 0 if(-l) n *(-l) p .
ISO
3 STATIONARY SCATTERING PROBLEMS
PROOF. (401)'i) is a simple consequence of Lemma 7, since we have: -57!l<0)-/a2S<0)K^a)d»-1?
ffDPu(0)/s2a*Knij(a)da = 0.
We also have K ^ - a M - l f K^a), with (400) which implies (401)'ii). Then (401)' with U n j € C^R*) implies that Unj(r) - i^gft 2 ), with Cnj € C~(R). Notation 1. Let DQ(R x Sn~ *) and SQ(R x S n ~') te *Ae subspaces of/unctions U /nD(RxSII"1)iwrfS(RxSn-l)wirA(400)aiirf: gpTj
(401)"
—jr (0,a) & a spherical harmonic oforder p.
We can identify the spaces of regular functions D(Rn) and 5(Rn) with the spaces D0(R x Sn~ *) and SQ(R x Sn~1), and then their dual spaces for the duality: <\J9&mf
jUfaaWw)^1
drda, resp. o=s/R+xSll.1U(r,a)#(r,a)dida
with subspaces of distributions on R x S n ~ l with support in [0, oo). Notation 2. DQ90(RxSn"l)
We denote by DQf(RxSR~l) f
tt
andSo 0(RxS "*))
and SQf(RxSn'1)
(respectively
these subspaces.
A distribution U on R x S n ~ l induces a distribution u on Rn by » , with $(i%a) = #(ra), but the mapping p: U -♦ u € Z>0(Rx Sn~l) or SQ(R x S n " l ), has a kernel. In the space D* (RxS n ~ ! ) of distributions with support in [0,oo) (with pairing <,>), this kernel is the space of distributions U so that: U(r,a) = 2 SW(r)Tv(a), with
2
^c^pS^xXwithc^^/^KptaJda,
in 2>0Q(R x S2) or D'(R3), with Kp a spherical harmonic of order p. Thus8(r)=8(0)(r)=4ii8(x), 5 ( 1 ) ( r ) c ^ J | ( x ) . Lemma 10. Let u^ro)« h ^ W ^ a ) in R3. Then un satisfies: (402)
(A + k ^ u ^ - p
151
3.11 MULT1POLEEXPANSIONS
with p ^ - v / ^ K ^ c O i n Z ^ R x S 2 ) , P B - - f / ^ W K ^ a ) inZ^
PROOF.
(403) with:
j (2n+Q! ^ 1 ^
„ '
„ i (2n+Q! „ " kn!(n+2)! ( - 2k) '
Let ♦ € D(R3). Then we can define using Lemma 9: L » tf <(A + k2) un> ♦>=
where * n (r)=/ 2 *(ra) K^a) da. Then integrate twice by part. We obtain: (405)
£(♦)— [ A i V ) ^ ! ) - i2 ( ^ W ^ r ) ] ^ = a-(n-1)[|(r»XV))^(r)-hi1)(kr)|
Using Lemma 9 and the recurrence formula for Hankel function:
(406)
| (i^hjfti)) = r^h^fr),
we get that the first term in (405) is equivalent to a2 when a —> 0. For the second term, using equivalence (see Abramovitz-Stegun [1] p. 437) (407)
h?>(W-
F(2n+I)
-
j
(2n)!
(407)
h, (ka) -
(ka)n+l2nr(n+1)
-
» (ka) n + l 2 » n , .
and also Lemma 9, we have (408)
^^—^--^-^-(O),
which gives us Lemma 10. At the first order we verify that h^ (kr)» ^ O(r). We note that for k = 0, we have: (409)
A(r-(n^1)Kn(a)) = . p n ,
with On - " T
8 ( n >
« ^(a) inD^toS 2 ), p„ = - ^ 8 ( n + 2 ) ( r ) !£„(«) inZ);(R*S2). ®
152
3 STATIONARY SCATTERING PROBLEMS
11.4. Multipole expansions and analytical functional From Lemma 10 we understand the expression "multipole expansion" for the Hankel functions expansion (376) of the solution u of the Helmholtz equation (1): each term u of the expansion is due to a distribution p of order n concentrated at the origin. The sum p m =Ip n cannot be a distribution, but we can define it as an analyticfunctional: Definition 6. Analytic functional. An analytic functional is a dual form of the space A(Cn) of entire analytic functions on Cn (or holomorphic functions on Cn) equipped with the topology of uniform convergence on compact sets. We denote by A'(Cn) the space of analyticfunctional. That pm is an analytic functional results from the Paley-Wiener Theorem for analytical functional, see below and for instance Hormander [2], [3], Treves [1] Ex. 22.7. We recall some notions Definition 7. An analytic functional y in Cn is said to be carried by a compact set K(orKis a carrier for p) if for every neighborhood
|p(0|:£C w sup|f|,
Vf€A(C n ).
Then we define the Fourier-Laplace (or Fourier-Borel) transform of an analytic functional ji in Cn by (411)
0(0 = FLVL(C) =
with C € C n ,
For all compact set K in Cn, we define the convex function on C n (412)
H K (t)=sup Re
WhenKistheballK = BainRn,thenHK(C) = a|C , |, wfthr-(C'i,...,C'n)-ReC. Definition 8. Let a a (a^.. .,an) with a. > 0. We denote by Exp (a) the space of the entire analytic functions of exponential type a, i.e., the space of the entire functions on Cn such that there is a constant A(f) so that: |f(z)|exp(-.a 1 |z 1 |....-a n |z n |)^A(f). We denote by Exp the vector space of all entire functions on Cn which are of exponential type, i. e., the union ofall the spaces Exp (a), with a. > 0. The "Paley-Wiener" theorem for analytical functional is:
3.11 MULTIPOLE EXPANSIONS
153
Theorem 10. If\i is an analyticfunctional in C n carried by the compact set K then its Fourier-Laplace transform is an entire analytic function of exponential type, and more precisely: for every 5 > 0, there is a constant C5 such that (413)
|£(t)|
VC€C n .
Conversely ifK is a convex compact set and (an entire function satisfying (413) for every 6 > 0, there exists an analytical functional \x carried by K whose FourierLaplace transform is f. Then we consider the expansion
(4i4)
r « - 2
where we assume that (cn), satisfies (379), with c n = /
2
| K^oc) | da.
Theorem 11 • The expansion (414) defines a harmonic function on R3 satisfying (415)
|p a n (y)|=o(re r V 2 )
forr= |y|-oo,
y€R
3
,
which can be extended to an entire analytic function (on C3 of exponential type. Thus fis the Fourier-Laplace transform of an analytical functional p*1 so that i) p M is the source of a wave u defined by (380) or also by (416)
u(x) =
withx€R 3 ,
(x^R^,
ii) u is an outgoing wave, and its behavior at infinity is given by (417)
u(ra) s O(r) p an(ka), forr-> «>.
Furthermore the radius of convergence of the series (380) is given by: (418)Rw==RMd=Ifmnsup[^^
.
Remark 26. Note that (416) is a way to write the Helmholtz equation (A + k2)u » pan, by the convolution product of 0 with pan. But as a function of z, 0(x - z) is analytical only if z * x. Thus (416) has to be feasible, for instance if pan can be identified to an analytical functional on an open set Q in C3 which does not contain x. Anyway we can define (416) by taking the limit of the series Z
154
3 STATIONARY SCATTERING PROBLEMS
Remaik 27. Let v be a hannonic function of exponential type (a), that is, for every 8 there is a constant C s such that: (419)
|v(y)| sC 8 exp «a+5)), r= |y|.
an Then (415) implies: p (y) is of exponential type a, with a = r 0 /2. 0 For the proof of Theorem 11, we use the lemma: Lemma 11. Lett be a harmonic function on Rn of exponential type. Then f is the restriction to Rn of a unique analyticfunction on Cn ofexponential type. PROOF for n at 3. We use the inequality (due to the Poisson formula), see Chazarain-Piriou [1] p. 23, for every harmonic function f:
(420)
|8 a ff0)|£(3n) n r n sup |f(x)|,
n=|a|,
x€S r
with S r the sphere of radius r. Thus for z € C3, |z|
(421)
12^3^(0)^1 < 2 ( 2 a4(3n) n f$)r- t t sup |f(x)| , a! * |«|«n ~ x€S r n X(3nf±(3t 0) r«sup ni
ft
x€S r
|f(x)|st2 n5^(^ 2 ) ,l ] sup |f(x)|. ft
-
x€S r
Then using the Stirling formula, we see that for large values of n: (422)
a/#g(^)n«^(9er0/r)n.
Thus for r=Cr 0 with C > 9e and with relation (419), we obtain: (423)
|«z)|
with |z|
Notice that we have, with (419) and (412), (413): (424)
a s inf H K (-ia)= inf H ImK (a). a€S 2
o€S 2
ITius when Im K={Im z, z € KJ is a convex set, the ball Ba satisfies B a £ Im K. ® of Theorem 11. The proof of the first part, that is, that the expansion (414) with (379) or (380) defines a harmonic function satisfying (41S) is fairly easy (see Muller [1] p . 89). Thanks to Lemma 11 this implies that f is of exponential type on C , with (424) and a = rQ/2. The Paley-Wiener theorem 10 implies that pm is an analytical functional. PROOF
3.11 MULTIPOLE EXPANSIONS
155
Then we can write (416) (see Remark 25). The behavior at infinity of u can be proved in a way similar to (15), Proposition 1. Thus u is an outgoing wave. There are many other ways to prove it (see for instance Muller [1] p. 90 using a Laplace transformation). Then we know (see Mffller [1] Chap. 3.5, Thms. 16 and 17 thenThm. 18)that for r > Rce the expansion of u defines an entire harmonic function so that R ^ Rce and conversely for r greater than RM (see (418)), the series (380) is convergent, thus R M > R • Furthermore we can characterize the radius of convergence Rce from the coefficients (cn) in (414), since we have: (425) thus: (426)
R^ = inf {r, so that c n = o [(kr/2)n+l T{n+\1/2)$ for n-^oo}, |kR a =^1i f limsup(r(n
+
i)c n ) 1 / n ;
P is the radius of convergence of the entire series 2 F(n + 1^2) Cn1^° <8> Note that it would be interesting also to use analytic functionals of the variable r in C , with values in L2(S2). Conclusion. Let X be the space of outgoing waves defined by: (427) X = {u€D'(R3), p-(A + k 2 )u€£'(R 3 ),/ s 2 |(^-ikuKra)| 2 da = o(l/r 2 )}. With B1a = R 3 /Bafl , we also define X = U Xa fl, where: a>0
2
(427)' X a = {u€Z>'(B;),(A + k )u = 0inB;, / g 2 | ( | j - ikuXra)|2da = o(l/r 2 )}. Thus for every u in X, there is an a > 0 such that u |
t €X . a Let Y be the space of traces on the sphere Sk = kS2 of harmonic functions on R 3 B
of exponential type: (428)
Y = {g€C°°(Sk), 3fonR 3 withAf=0and(419),fl
s
=g}. k
Y is also the space of traces on the sphere Sk = kS2 of the Fourier transforms of distributions with compact support on R3. Then we have: Theorem 12. Trace theorem at infinity. The mapping: y : u —► lim u(r
156
3 STATIONARY SCATTERING PROBLEMS
Definition 9. We denote by A'H(C ) the space ofanalytical functionate p a n so that the restriction ofp^toR
is harmonic ofexponential type.
Theorem 13. The mapping p811 - • p ** |
is (continuous) one-to-onefrom A'H(C3)
ontoY. We have to compare to the mapping on distributions with compact support: p €£'(R 3 ) — pi € Y , whichhas the kernel (A + k2)£*(R3)! Thenforeverype^XR^wedefmep^cA'H^withpL
■p a n L .
Let 2£(R ) be the space of distributions on R with support in the ball B a . The mapping p € £;(R3) -> u | , € Xa (with u » * * p), has the kernel (A + k^JS^R3), and we can identify the quotient mapping with the mapping p" 1 € A*H(C ) —• u, Note also that the radius of convergence of the multipole expansion of an outgoing wave u is given by the type of the Fourier-Laplace transform of the "source" of u (see Theorem 11). In the case of scattering of an incident wave by an obstacle, a conjecture of Bardos is that the radius of convergence corresponds to the distance of the origin to an envelop of the normals. 11.5. Multipole expansions for the electromagnetic field We consider Maxwell equations (22), with currents J and M having compact supports; the electromagnetic field (E,H) satisfies Helmholtz equations (28) with (m j ) given by (27). Then we can apply the results we obtained above on Helmholtz equation: let (j^.m^) be the analytical functional such that (429)
j M ( k a ) = j(ka), A ^(ka) =rfi(ka),
Va€S2,
with (j ***, ifi *") entire analytic vector functions on C3, of exponential type. Note that, thanks to (33), they also satisfy: (430) a. j ^(ka) = 0,
a. A ^(ka) = 0, V a € S 2 (and a A J ^(ka) = Z m ^(ka)).
Definition 10. We denote A' ^ t k ( C 3 ) ^ {j*11 € A'H(C3)3, a.j ^(ka) = 0, a € S2}. Thus (E, H) is obtained from (jan,man) by (431)
E(x) =
H(x) =
3.11 MULTIPOLE EXPANSIONS
157
Note thatrelations(430) imply (from Rellich Theorem) div E = 0, div H « 0 for r > R. Then from the expansion of j a n and m "" into spherical harmonics,
(432)
j "(pa)-5 2 (^) n K "(a), m an(p«) = f 1 f^fK
£(«),
we obtain the multipole expansion of (E, H): (433)
E(ro) = 2 h^'W) K °(
with the radii of convergence given by (see (418)): (434)R^ = l m s u p ^ l o g / s 2 |j" n (ia)| 2 da, RjJ = Itasup^log/ g 2 I n W ^ d a . E
H
Obviously, from Maxwell equations, we have R^ = R^ . Remark 28. First we have to note that the vector product or the scalar product by a of a (vector) spherical harmonic of order n is not a spherical harmonic. Furthermore a A Kn(a) and a A Km(a) (resp. a. Kn(a) and a. Km(a)) n * m, are not orthogonal. 0 Remark 29. Note also that we have the relations between j and m thanks to (27) (435)
i) icoy m + curl j «(A + k )M, ii) curl m + iwej s - (A + k2)J .
When M = 0 we have: icop m + curl j = 0, but we do not have: iwp m**1 + curl j 3 1 1 = 0 ! The differential operators (here the curl operator) do not operate in A' H t (C 3 )!
^
EXAMPLE 3. The (oscillating) dipole. We define the dipole oscillating at the angular frequency G> (and centred at the origin) as the charge density p(x,t) and the current J(x,t) given by: (436)
p(x,t) = e-iwtp(x), P(x) = p.grad6, J(x,t) = e"itt,t J(x), J(x) = io>p.6,
with p € R the dipole momentum; we have
158
(437)
3 STATIONARY SCATTERING PROBLEMS J(y) = iop,
j (y) = i^ [J - k"2 y (y. J)] = - co2 y [p - k~2 y (y.p)],
thus j (ka) = - a>2y [p - a(a.p)], which is also j ""(ka) from the definition. Then we easily find (438)
)m(y) =
-(*2l^){\p?-y(y.p)]-lptf-It*)},
and then: A(ka) = ikaAJ(ka) = -ka>aAp, giving: A " t y ^ - w y A p. We can write (438) also in the form: (438)'
j *%>«) = - (<*V k 2 ) {p2 K2(a) + K0(a)},
K2(a) and K0(a) spherical Harmonics of order 2 and 0 resp., lc 2 (a)=i p - a (a.p), K0(a) - 3 P k2. We have a. K2(a) = - | a.p = - a. K0(a); thus K2(a) is not tangent to the sphere S2, and a. K2(a) is not orthogonal to a. K0(a) in L2(S2). We also verify that: iwpA^yJ + i y A j ^ y ) * © and yJ"Ky)*0!
Conclusion. In the electromagnetic case we can conclude in a very similar way as in the Helmholtz case. Let X6"1 be the space of electromagnetic fields (E,H) on R3, satisfying: i) the Maxwell equations (22) with (J,M) distributions with compact support. ii) the Silver-Muller conditions (43) at infinity. ~em
~em
We also define X c m = U X a , where X a is the space of electromagnetic fields (E,H) on B; satisfying: i) the Maxwell equations: curlH + ia>£E = 0, -curlE + iam H = 0 inB a ii) the Silver-Muller conditions (43) at infinity. Thus for every (E, H) in X cm , there is an a > 0 such that (E, H) B| R , € X a ' a Then let Y*"1 be the trace space on S k Ycm = {g € C°°(Sk)3, a.g(ka) = 0, V a 6 S2, 3 f: R3 — C 3 harmonic of exponential type, i.e., with (419) and f | =* g}. 1
2
(The condition on f is equivalent to lim sup r- log/ 2 I f(ra)I da < oo.) r—» oo w
We have the following results:
s
159
3.12 SCATTERING BY A BALL
Theorem 14. Trace theorem at infinity. The mapping: y^: (E,H)-»rlim> E(iaV(T) is surjective from Xm (or Xm)
toY™.
But it is not injective ! Like in the Helmholtz case, the mapping (with (27), (29)) (J,M)€Fa(R3)3x£;(R3)3 -*(E,H)| B , €X
|g(a 0 )|
(Y being defined by (428) with k = 1 to simplify writing). The answer is negative. We can see it by taking the functions: gn(a) = Pn(a(0, with a given unit vector fl and Pfl the Legendre polynomial of degree n. We have (see Muller [1])
(440) /s2|gn(a)|2d« = 2nf J (Pn(t))2dt =
£j, Ml
butg (P) = P n (l)=l; for n— oo, we see that we cannot have 1
.
1 2 . SCATTERING BY A DIELECTRIC BALL
The scattering of an incident wave plane by a dielectric ball BR is often used as a reference to numerical implementation. Results are well known due to Rayleigh expansions. 12.1. Scattering with Helmholtz equation 12.1.1. Calderon operators and Calderon projectors on a sphere First we can obtain the Calderon operators C\ Ce and the Calderon projectors P\ Pe on the sphere SR (for a wavenumber k) thanks to the Rayleigh series (380) quite easily. The derivatives of these expressions with respect to r, for r = R, are for the (exterior) outgoing wave, and the interior wave: (441)
g (Rot) = 2 k hJfldOK^ resp. g (Ra) = 2 k j ^ k R ^ a ) ,
with h^ = h^'; the prime denotes the derivative. Thus ifj (kR) * 0, V n:
160
3 STATIONARY SCATTERING PROBLEMS
h' j' (442) (;%(<«) = 8^ K^a), d ^ a ) = 0*n K^a), 6^ = kt^(kR), ej, = k f . n " Jn Thus the Calderon operators are simple multiplication in the space of spherical Harmonics of order n. This is a consequence of the fact that the rotation group commutes with the Calderon operators. Now the exterior Calderon operator has the extra property (with k > 0): Re - (Ccu0,u0)>0, andeven
Re - (Ccu0,u0)> g- Hu0ll^2(s }-
Using (442), and the decomposition u0 = 2 u0n into spherical Harmonics, we have Re(Ceu0,u0) = 2 ReO£ | Uoil | 2 , Re9« =-L I £(|h n (kr)| 2 ), and | heeler)| is a decreasing function of r, see Abramowitz-Stegun [1] p.439. Then the Calderon projectors are obtained in the space of spherical Harmonics of order n also by solving the transmission problem: (443)
i) Jn(kR) K^a) - h^kR) K„(a)
= pn(a),
ii) kj;(kR) K^a) - k h;(kR) KJ.a) = P;(a), with given Pn(«)» P'n(a)- Thanks to the Wronskian (444) W(Jn,hnXkR) = [hAJn - h„ j^(kR) = i/(kR)2, we obtain,
I KnW J
Lkh; -hnJU;(«)J
with j for j (kR), hn for hn(kR). This gives the Calderon projectors Pj;, Pjj by: (446) P^ikR 2 and:
f u g ; - K>n\ kk 2 ig; -kjg n J
pi=ikR^f-khiJn L -k2h;j;
ps + pi-i. pi-^-s., withsn=ikR2
;n
Kin\ ug; J
^n
^ - 2k2ig; k(igny J
12.1.2. Scattering of an incident wave by a ball Let k be the wavenumber of the incident wave Up kj be the wavenumber of the scattered wave in the ball. We assume that u{ is an entire solution of Helmholtz equation. The usual transmission conditions on the sphere SR are (447) (447)
u* u e -u u - u -up
^ 3u!.!^l an-dn-an ■
161
3.12 SCATTERING BY A BALL
Using the Rayleigh expansion (380) and (442), we have (ifjn(kjR) * 0)
(448)
e1nIuj,-eX=e,nuI>n,
i4-nS-u,.n,
with (see 442):
9en = k^(kR),
(449)
^-kfcdcR), J
"n
ejf-kjJ^R). J
n
n
If k, is real, e" is also real, thus fn - e" * 0 since the Wronskian W(hn, h n ) * 0 . Then the solution of (448) is given by (450)
u« = R„u In , uj,=T n Ui n , Rn = - ^ - r r ,
T
«£-«
n
=^^,T e
n-°n
n
-R
n
= l.
Then for u,(ro) = I j (kr)Kj,(a), Uj n = Jn(kR)Kj,(a), the waves u1 and u e are ukw) = | f n j n (k,r)4(a),
with fn = T„Jn(kR)/Jn(k,R),
(451) ue(ra) = 2 R n h^flcr)^*), n
with Rn = R„ j (kRJ^kR), for r 2: R, "
n
with K^(a) = i (2n + l)P n (a.a 0 ) for a plane wave. The series for u e is convergent in L2(Sr) for all r> R. Its radius of convergence R ce is given by (418), with: ^ = /s2|Rnl^2n+l)2|Pn(«-«0)|2d« = 4n(2n+l)|Rn|2. Then we can verify that R ce = 0: the Rayleigh series converge for all r > 0 ! 12.2. Scattering with Maxwell equations 12.2.1. Debye potential of the electromagnetic field Let (E,H) be an electromagnetic field in a ball B a or its complementary, satisfying Maxwell equations (452)
curlH + io>eE = 0,
- curlE + io>y H = 0 in B a or B' a = R 3 \B a ,
with constant e,m with (locally) finite energy, and the Silver-Muller conditions at infinity. Then (see Schulenberger [1], Aydin-Hizal [1]), there exist two scalar (outgoing) solutions
ii) H = <^j curl curl (x<|> ) + curl (xJ.
We get <|) and <> | up to a function of r = | x| from E or H by solving the equations 2
- io)£x.E = L <>| , 2
-x.curlE = -
2 K«)JLIX.H = L
<> | ,
2
with L 0 = x.curl curl (x); L is the angular momentum operator (we can identify L2 with the Laplace-Beltrami operator - A 2 ); in L2(Q), O = B a or R 3 ^ , it is a
162
3 STATIONARY SCATTERING PROBLEMS
positive selfadjoint operator (see Dautray-Lions [1] chap. 9B.1.1). The functions c|>j and <|>2 are called the Debve potentials of the electromagnetic field (E,H). Furthermore the fields (E(1),H(1J) and (E (2) ,H (2) ) defined by: (454)
U) H(l) = 4 curl curl (x^),
E (1) = - curl (x^), E( } = (5eCurlcurl(x<|)A
H (2) = curl(x<|>2),
are respectively called TE (transverse electric) fields and TM (transverse magnetic) fields. Thanks to relation (28) chap. 2, we have: n.E (1) = 0, and n.H (2) * 0, when n is the normal to the sphere I = S f . Then taking the tangential trace of these fields on the sphere I , we obtain: i ) n A E | 2 = grad2(-r(|)1) + curl 2 (^(I + D)<|)2), (455) ii) n A H | 2 = curl2 ( ^ (I + D) ^) + grad2 (n|>2), with (456)
D
*=2xk^=r?F,
or also for the projections on the sphere £: i) * 2 E = curl2 (r^) + grad2 ( ^ (I + D) <|>2), (455)' I ii) * 2 H m grad2 (^r (I + D) y{) - curl2 (r<|>2). This is due to the relations (28), (30), (30)' chap. 2, with curl (xc|>) = - x A grad <|>. The Debye potentials <> | and c|>2 on £ are obtained from n A E | by: (457)
rA2 ^ = - div2 (n A E1 2 ),
A2 (I + D)<|>2 = - io>c curl2 (n A E1 2 ).
Formulas (455) or (455)' correspond to the Hodge decomposition of currents on the sphere I (see section 6.2, Appendix). Note that the sphere is simply connected, and thus the cohomoiogy space Hl(L) is reduced to 0. Let Il c and IT be the orthogonal projectors in the space L 2 t (l) of square integrable tangent fields on £, so that: (458) f = fc + fg = grad2 o 2 + curl2 *x with fg = IIgf = grad2 o 2 , fc = n / = curl2 *x, and *,,
with g(a,0) = - ^ l o g ^ ^ , a, f$ € 2 = S 2 .
3.12 SCATTERING BY A BALL
163
Note that the Debye potentials are related to the radial components Ef, Hf by (459)
A§2
A§2 <> | 2 = icoc rE r
This is easily seen from the radial component of Maxwell equations, with (28), (30) chap. 2. Furthermore rEf = x.E and rH = x.H satisfy the Helmholtz equation and converge to 0 at infinity (see (50)). Thus they can be used as potentials for the electromagnetic field, but it is not very easy to prove that they satisfy the outgoing condition, and the Debye potentials are more regular. For the converse of (459), we only have to eliminate constants functions on the sphere. 12.2.2. The Calderon operators on the sphere To obtain the Calderon operators on the sphere, we have to inverse the relations between (,,(|>2) and E 2 . Now D in (455) is also (up to factor r) the exterior or interior Calderon operator on the sphere S of the scalar case. In the exterior case, B er = (C* + y) is an isomorphism from HS(I) onto Hs~ (I). In the interior case, we have to eliminate irregular frequencies, see 4.7, 4.8). Now in the exterior case, we obtain from (455), (457), (460) n A H 2 = Ce(n A E2) = - -X-curl2A£lBercurl2E2 - icoe g r a d ^ 1 B~* div2 E 2 . We will have a similar formula for waveguides, see (68) chap.4.2. Now we give explicit formulas thanks to Rayleigh series. Using the orthonormal basis (Kn .(a) of spherical Harmonics (see (373)), then: gradjKnj,
cur^K^, n= 1, 2,..., - n < j < n ,
is an orthogonal family of L2t(I), giving an orthonormal basis by normalization. Now we use differential operators on the unit sphere S2, instead of I, thanks to (461)
grad2(r<|>(x)) = grad^ ¥(r,a),
c7rl2 (r<|>(x)) = clfrl^ ¥(r,a),
with <|>00 = ¥(r,a). Thanks to the exterior (resp. interior) Rayleigh series ofty{,ct>2 (462)
c|>(ra) = 2 h j J W ^ t o 1
n>0
(resp 2 jn(kr)4(«)), J= 1,2, n>0
n
in (455)', we obtain the exterior Rayleigh series: n E
l
= 2 K c"ris2 Ki + | cob hn grads2 KJ;,
(463) "2H = 2 4 r K « rad ,2 * i - 2 hn c"uri , K^,
164
3 STATIONARY SCATTERING PROBLEMS
with the notations:
(464)
1^ = hjJW
hn = (I + D)hjJW
We obtain the exterior Calderon operatorfor the TE and TM waves C™'e and c ™ ' e (465)
C™'c = - i § n « A n g ,
c^-hwCe^aAlIe,
with a 6 S2 the normal to the sphere, 9 £ * h n /l^, n g , Il c given by (458). The interior Rayleigh expansion is obtained by substituting j n for h n . The interior Calderon operators for TE and TM waves are given by: (466)
C™'1 — 4 r
5
n«Ang,
C^-iwrte^aAH.,
with the notations: (467)
j n =j n (kr),
j>(I
+
D)Jn
K = ln\.
12.2.3. Scattering of an incident wave by a dielectric ball Now let (EJ,HJ) be an incident wave on a ball Ba, with a dielectric medium of permittivity e{ and permeability ]i{. We assume that the incident wave is an entire solution of Maxwell equations in free space (for instance a plane wave). Then the incident wave is given thanks to the Debye potentials <|>n and (|>I2, with the Rayleigh expansions: (468)
(|)Xj(ra) = 2 Jn(kr)Kj;j(a),
j = 1,2.
Let (*,<|£) and (<|>*,<|>*) be the Debye potentials of the electromagnetic field inside the ball (E1,!!1), and of the reflectedfield(Ee,Hc). Let: (469)
^w)»2j n (kir)l4 j (a),
<|><(ra) = 2 h£V)Kj J <«). j - 1,2,
be their Rayleigh expansions with unknown harmonic functions KjjJ, K^' J . Then writing the transmission conditions on the sphere X, with r = a, forTE and TM waves we have: i> [*!], = ♦}-*?»♦[.]» 1.2 J
(470)
*»
J
j
J
| ii) [ i (I + D ) ^ =55^ (I + D ^ ; - 4r (I + D)4>« = ^ (I + D)*\ | iii) [ ^ (I + D>|)2]2=s^- (I + D ^ - A (I + D ^ = i (I + D ^ .
Now using the Rayleigh expansions of these Debye potentials, and the notations (471)
ji=J n (k,r), j i = (I + D)Jn(kir), e » = a i ) / j ; ,
3.12 SCATTERING BY A BALL
165
we obtain the system of equations: ;1 w-i,l
|- w-e,l
•
V I,1
(472) PjJnJSi
-jinn**
- P
j
^ '
n
We have the same system for the superscript 2, but with t instead of p. Their solutions are given by
(473)
ii^-1?^.
hX'^R^K^
j=1 2
''
and thus: JT ^ej „
Ij
" n V = ^ J n C j-1,2,
OV-'Ji Kl - *? V'J n 41), (.,r »Ji K2- liV'J, 4'2)(
with (1)
(475)
fj'Bn-P
M U _9in„(l) nRul M)_
flk
and also:
(475)
^ V e i ' - g - ' i s ' r--«»i7[r- '
with similar formulas for RJ, and T„ , butwithe, ej in place ofji, jij. Note that: (476)
li?-R®-l,
ij?-**-!,
j-1,2.
We finally obtain the reflected electromagnetic field (Ee,He), thanks to (463), by its components: ZE'TE,II = Rn
K
(477)
(
"2El)TE>n >
"ZH'TE.II = ^n
(nZHI*rE,n '
and thefield(ESH1) inside the ball by: (478)
* Z E TE.n = I n
(
*ZEl)TEn >
* 2 E T M , n = ^f<*Z E I>TM, n '
n
Z H T E , n = hi ( n Z H I>TE.n ' ^TM.n = ^ W V r M . n '
166
3 STATIONARY SCATTERING PROBLEMS
EXAMPLE. If the incident wave is a plane wave, given by: (479)
E^Ejoe 11 ™ 0 *", H,(x) = HI0e*""0", x = r
with given OQ € S2, E I0 and HI0 € C3, with aoEio = ^-Hio - °» ZH I0 ■ "o A EI0> a n d Z the impedance of free space. Taking aQ »(0,0,1), EJ0 ■ (1,0,0), we obtain the Debye potentials of this plane wave by the Rayleigh series (468), with the harmonic functions (373) (see Jones [1] p. 446):
(480)
K^-CJK^-K^],
l4 2 («)--Kt K n.l + K ».-^
with Cn = (2K) 1/2 2n(n+l) ' T W s g i v e s u s t h c s c a t t e r c d
field
*&&&& to (477).
3 . 1 3 . ADDENDUM. COMPACTNESS PROPERTIES IN SCATTERING PROBLEMS.
The scattering problems of scalar waves on bounded (regular) obstacles depend on the Fredholm alternative; this is a consequence of the compactness of the operators K, J and L (see (70)) in the framework H"" 1/2 (D (or Hl72(D, or C°(0), the range of these operators being in H1/2(r). Since we can often prove fairly easily uniqueness at most of the solution in scattering problems, then this also implies its existence! In electromagnetism, scattering problems also depend on the Fredholm alternative, and the sum: Cc + C1 = - R" lT (see (241)) is compact in H~ I/2(div,H, as a consequence of the proposition: Proposition 32. IfT is of class Cl'lt the operator T defined by (100)' is compact in H " i/2 (div,D, with Im T in Hl/2(div,D ifl is ofclass C2>l. PROOF, i) For all u f in H " 1/2(div,D, we verify that Tur is given by: (481) Tur = gradro + vr, 0€H 1/2 (O, v r €L 2 (0 3 ,withv r €H t I/2 (OifrisC 2,1 . By developing the double vector product in (100)' and using (70), we have: (482)
Tuj
'
which is of the form (481) with 0 = - 2^. Lur, thanks to the properties of K and J. ii) Now we verify that divr Tur € H 1/2 (0. Let (E, H) be defined by (223) with Jr = u r Using the notation {v} = v| + v| , we have like for (84) (see also (28), (30) chap.2) r
r
i
*e
(483) divrTur = - divr ({n A HJr) = - io>6 {n.E}r Then using (95)' and (70), we have: (484)
{n.E}r = ikZ{n.Lu^r + i|j(div r u r ) = 2ikZn.Lur + i|j(div r u r ).
Since Lur and J (divr ur) are in H 1/2 (0, this implies the proposition.
£>
CHAPTER 4
WAVEGUIDE PROBLEMS
1. WAVEGUIDES WITH HELMHOLTZ EQUATIONS
Here we study the propagation of scalar stationary waves in a "waveguide" which is a semi-infinite cylinder. The method that we develop, will allow us to deal with cases of more general geometry and later, with the electromagnetic case. Let Q+ = Cl x R+ be an infinite open set which is a semi-infinite cylinder in R3 with a regular bounded cross-section Q in R2, of boundary T. Let T+ = T x R+ be the lateral boundary of Q+. We first consider the Dirichlet (resp. Neumann) problem:findu such that
(1)
i)Au + k2u = 0inQ + , ii)u| = 0 ( r e s p . g | =0), iii) u(x,0) = u°(x), x € Q. c R2,
with a real wavenumber k and a given u° (for instance in L2(Q)). We denote by Aj, the Laplacian in the transverse variables xT = ( x ^ ) (in R2), by x3 the variable along the axis of the cylinder with origin at the end of the cylinder. Then we can write (l)i) in the form (l)i)' | T + V 1 + k2u = 0 in n + . 167
168
4 WAVEGUIDE PROBLEMS D
N
Let AT (resp. AT) be the Laplacian with Dirichlet (resp. Neumann) condition. Denoting by A either - Aj or - AT (both self adjoint operators), we can rewrite (1) in the form: :.3 2 U
(2)
i ) ^ - | - A u + k2u = 0, ii)u(.,0) = n° u(
Dx2 ADx Let (^.♦JJ) be a spectral decomposition of A (that is ((Xjj') ,$~) for A = - A^, and ((^) 2 ,^)) for A = - Aj, with on an orthonormal basis in L2(Q). We decompose u° and u(.,x3) on this basis:
(3)
u°=2XV
u(->x 3 )=!u n (x 3 )V
and then we have tofindthe unknown coefficients un(x3) so that: (4) ii)un(0) = uS.
Denoting by ocn and fl two arbitrary constants, the general solution of (4)i) is: a)ifX2>k2, un(x3) = a n exp(x 3 Jx^.k 2 ) + P n exp(.X3 i Jx2-k 2 ). If we require that the solution be tempered with respect to x3 , thus an = 0, then: (5)
un(x3) = exp ( - x 3 Jx 2 -k 2 ) ift
b)ifX2«k2, un(x3) = a n x 3 + pn. If we require that the solution be x3-bounded, then ocn = 0, un(x3) = u c)ifX2
4.1 WAVEGUIDES WITHHELMHOLTZ
169
For a time evolution u(x,t) = e,wtu(x), we make the converse choice. We define:
(7) end=efe;=(x2_k2),/2if^>k2, 9 n d #-ie-=-i(k 2 -^)^ifx^k 2 . Thus with the physical hypothesis of positive axis x3 wave propagation, we have to take a = 0 in case c), and then we obtain a (unique) solution of (1) in the form
(8)
uUx^Je"^^^
Theorem 1. Problem (1) with u° given in \}(Q) and with the condition: (9) u is an Xy bounded wave, propagating towards the positive axis xy has a unique solution u in C°([0,+oo[,L2(Q)), and it is obtained thanks to a holomorphic contraction semigroup of class C° in L2(Q), (G(x3)), x3 > 0, by (8). For these notions of semigroup, we refer to Dautray-Lions [1], Pazy [1]. We recall that a contraction semigroup in a Banach space X satisfies: IIG(t)u°ll
Vu°€X, t = x3.
We note G(x3) = GD(x3) for the Dirichlet boundary condition, G(xJ = GN(x3) for the Neumann boundary condition. Its infinitesimal generator C (C D or CN) is defined (on the basis of the cylinder) in L2(Q) by: (10)
J£-(.,0) = Cu° (withC = C D orC N ), D(C) = {u° € L2(a>, u(.,x3) = G(x3)u° € c\[0,+oo),L2(Q))l
Definition 1. The infinitesimal generator C of the semigroup (G(x3)) defined by (10) is called the Calderon (or also the capacity) operator of the guide at the wavenumber k. We can characterize this operator thanks to the spectral decomposition of the Laplacian by: (11)
O n = -e n * n ,
Vn, and D(CD) = Hj(Q)f D(CN) = H1(Q).
Thus C 2 = A - k2. Hence C is a square root of the self adjoint operator (A - k2I). Furthermore it has the following properties (for all v in D(C)): (12)Re(Cv,v)=
-0^|v n | 2
2 n.A^k
2
£ |v n | 2 > 0.
2 2
n,A^
170
4 WAVEGUIDE PROBLEMS
Thus L (Cl) has an orthogonal decomposition into: L2(C1) = H£®H£, with: H£ = {2cn0nwithX^
<>
aUlj i*r-cjV
Then using the continuity relations of the wave across the boundary I\: (15)
u = ur + u„
§H=aH"+a5-oneveiyrj,
we obtain from (14), (15) that the wave u must satisfy: (16)
S - q u - f 0 onl} with f y ^ - C j U y .
Thus we have tofindthe wave u in the junction, satisfying: (17)
i)Au + k2u = 0inQ, ii)u| =0and(16)onrj, j = 1 top.
This problem can be written in a variational form; we define the space:
4.1 WA VEGUIDES WITH HELMHOLTZ
(18)
H^PHueHH
171
u|_ =0}, iQ
O
and the sesquilinear form a(u, v) on this space by: (19)
a(u,v)=/_ (grad u . grad v - k2uv) dx - 2 / r Cu v dT, "
j
{
j
J
which is continuous on Hr (ft), and Hr (ft) coercive with respect to L (ft), thanks to the above properties of the capacity operators. The variational form of problem (17) is:findu in Hr (ft) satisfying l o (20) a(u,v) = 2/ r fijvdT, V V € H } ( Q ) . j
'j
°
For a regular domain ft, the natural injection Hr (ft) —* L (ft) is compact. Thus problem (17) depends on the Fredholm alternative: it has a unique solution except for wavenumbers k such that (17) has a nontrivial solution, that is, for the eigenmodes of the junction. Using the decomposition corresponding to (12) for the jth guide, we denote by: Cj = - Cj" + i Cj" the decomposition of Cj, by Vj" the space of propagating modes for the jth guide (above denoted by H£), and by P* the corresponding projection of uJ (the trace of u on Tj) onto the space Vj". Then u is an eigenmode of the junction if (20) is satisfied with fr = 0. Taking v = u in (20), and the real and imaginary parts, we easily obtain that u must satisfy: i)Au + k2u = 0inft, (21)
ii)u| r =0, g J . q U = 0onr j ,j=:ltop,and Pjuj = 0,(oruj €Vp.
Remark 1. The case of small junctions (or of weak wavenumber). If the domain of the junction is small, we can apply the Poincare inequality: (22)
Hul < C llgrad ull, V u € Hi (ft),
with C = \Q 1 , \ 0 being the smallest eigenvalue X of: i) Au + Xu = 0 in ft, (23)
Then taking the real part of (20) with v = u and f = 0, we obtain with (12): (24)
4(|gradu| 2 -k 2 |u|)dx<0;
and using (22) we have:
172
(25)
4 WAVEGUIDE PROBLEMS (l-k2C2)/aIgradu|2dxsO,
VU*H£Q;
thus for (1 - l^C2) > 0 (that is for X0 > k), we have u = 0. This implies that the wave problem (17) in a small junction always has a unique solution. Remark 2. Dissipative junctions. If the wavenumber k is complex with Im k2 > 0, we also have uniqueness of the solution of problem (20). This is obtained taking the imaginary part of (20) with v = u and f = 0 and using (12): (26)
Im (- k2) iluB2 = 2 Im (Cu,u) 2> 0, j
J
thus giving u = 0, and we have the same conclusion as in Remark 1. Remark 3. Cascades (see for instance Jones [1] p. 257). The case when the cross-section of the guide changes (with discontinuities or not) over a finite distance, is a special case ofjunction (with two different guides arriving to it). <8>
Remark 4. We can generalize to "inhomogeneous" waveguides that is when there are different media with different wavenumbers in the cross section of the guide (with the usual transmission conditions at the interfaces of the fibers). Remark 5. We can also generalize to "inhomogeneous" junctions with a wavenumber k dependent on x. The problem can be solved numerically by using finite elements inside the junction and spectral decompositions at the boundary for the capacity operators (for instance). 0
2 . WAVEGUIDES IN ELECTROMAGNETISM
We consider a semi-infinite cylinder Q+ = Q. x R+ in R3 with x3 axis, occupied by a conservative medium with permittivity and permeability t and ji (with positive real c, p), bounded by a perfect conductor. We assume that the cross-section Q is an open regular bounded and connected set (it can be simply connected or not). We first consider the following problem: find the electromagnetic field (E,H) in the waveguide Q+, at angular frequency co, satisfying (with the same notations as in section 1) |i)curlH + icD€E=:0, (27)
ii) - curl E + i
iii)nAE|
=0,
iv) E T (0) = E^ onQ,
where E T is a given transverse electricfield(for instance in H0(curlT,Q)).
4.2
WAVEGUIDESINELECTROMAGNETISM
173
We denote by (EpH-r) the transverse components of the electromagnetic field (E,H) and we use the index T for transverse derivations: let t be a function on Cl, v T = (Vj,v2) be a vector function on Q., we denote: (28)
gradT <> t = ( ^ - , ^ - ) = (3^, a2o),
curlT «> = (32o, - 3^)
curl T Vj= 3 ^ 2 - 3 2 Vj.
Let e 3 be the unit vector along x3, and let S be the operator: (29)
Sv T = S(v1,v2) = (-v 2 ,v,) = e 3 Av T .
With these notations, the Maxwell equations are: i) - 3 3 H T + gradT H3 + me SE T = 0, (30)
ii) 3 3 E T - gradT E 3 + ia>y SH T = 0,
with: (31)
i) curlT H T + ia>eE3 = 0, ii) - curlT E T + io)MH3 = 0,
and since n A E = (n2, - n j , njE2 - n2Ej) = - E 3 Sn + e 3 n T A E T , the boundary conditions are: |i)n T AE T | = 0andE 3 | = 0 , (32) ii)ET(0) = ET Like in the Helmholtz case we can see that this problem is ill-posed and it becomes a well-posed problem adding to it the physical hypothesis of propagation of the wave (E,H): (33)
(E, H) is an unbounded wave, propagating towards the positive axis x 3 .
We first formally replace E 3 and H 3 in (30) by their values in (31); thus: (34)
i) - 3 3 H T +A j E T = 0,
ii) 3 3 E T +A 2 H T = 0,
with: (35)
i) A j E T = | - gradT jj curlT E T + icoe SE T ii) A2Hj = jjj gradT j curlT H T + icop SHT .
We can write this system of equations (34), on E T or H T only: (36)
with:
3 | H T + ^ J ^ [ 2 H T = 0,
afEr+^jEraO,
174
(37)
4 WAVEGUIDE PROBLEMS i) A XA2 HT as (gradT p curlT pS + eS gradT j curly + o>2 epl) H T ,
1 1 ? ii) i4^! Ej = (gradj j curlT ES + pS gradj. p curlT + or epl) E j , or also using relations: (38) S gradT = - curlT, curlT S = divT, (39)
i) A XA2 HT =s (gradj p divT p - cS curlT j curlT + o>2 epl) H T , ii) A^A! ET = (gradT j divT E - pSZlt curl1T p curlT + or epl) ET
Now we make the following important hypothesis: The domain of the guide is occupied by a homogeneous and isotropic medium; thus e and p are constants in Q.Withk*= o2ep,(39)is: i) A ]i42HT = (gradj divT - curlT curlT + k I) H T , (40)
ii) A^iEj SB (gradT divT - curiT curlT + k2I) ET.
From now on we will not write subscripts T to differential operators grad, etc. The operators A1 A2 and A1 A1 are identical to the transverse Laplacian; we have verified only that ET and HT satisfy the Helmholtz equation! But these operators differ by their boundary conditions. We define them by the usual variational method. Wefirstdefine the following spaces: (41)
V = {v € L2(Q)2, div v € L2(Q), curl v € L2(Q)}, V- = {v€V,nAv| r =0},
V^fveV.n.vl^O}.
From the Appendix (see (121), (122)) we know that spaces V* and V~ are contained in H^Q). Then we define the sesquilinear form: (42)
a(u,v) a / (div u div v + curl u . curl v) dx.
Thus we can define the operators A+ and A " (see Appendix (156)") by: D(A+) (resp. D(A~)) * {u € V* (resp. V"), v-»a(u,v) is continuous on V* (resp. V"") equipped with the L topology} and thus D(A-)=s{u€H1(Q)2,Au€L2(Q)2, nAu| r = 0,divu| r = 0}, (44) D(A+) = {u € H W , AU € L2(Q)2, n.u | r = 0, curl u | r = 0}; (43)
A+ and A are positive selfadjoint operators, with compact resolvent, thus with discrete spectrum.
175
4.2 WAVEGUIDES IN ELECTROMAGNETISM
Then we verify (at least formally) that if (E, H) is a solution of the problem (27), (ET,HT)(x3) is in D(A ) x D(A~); condition (32) implies ^ - l r = 0, and from equation div E = 0, we have div E T | and (31)i), (32)i) imply curl H T |
= 0. Besides, (27)ii) implies n T .H T |
= 0,
= 0? *+
Determination of the transverse electric field ET We come back to problem (27) which, after (36), (32), and (44), amounts to determine the transverse electric field E T that satisfies: (45)
i)85E T = ( - A " - k 2 I ) E T , iOE^OJ^Ej.
But this is an ill-posed problem as we see it by spectral decomposition of the operator (A~ + k2I). It is transformed into a well-posed problem by adding the physical hypothesis (33) for E T . Using the decomposition of the self adjoint operator - ( A ~ +k 2 I) into its positive and negative parts corresponding to the space decomposition of L2(Q) (46)
L2(Q)2 = H+ ®H_ ,
- (A" + k2I) = (- A" - k2I)+ - (- A - - k2I)_ ,
we can define (from the symbolic calculus) two "square roots" of — (A~ +k I) denoted by A+ and A ~ , so that: (47)
on H+: A+ = A" = - (- A" - k2I)*/2 negative selfadjoint, on H_: A+ = - A" = - i(-A" - k 2 I) 1/2 with (-A" - k 2 I) 1/2 bounded, positive.
Thus A + and A" are adjoint normal operators with domain V" , with spectrum o(A+) c R" U -i[0, k] resp. o(A")c R" U i[0, k]. (The Hodge decomposition allows us a more explicit spectral decomposition using the spectral decomposition of the Dirichlet and the Neumann Laplacian.) Taking (33) into account, we then replace problem (45) by: (48)
i) 3 3 E T = A"ET in Q+, ii)ET(0) = ET.
The choice A+ would give the opposite direction of propagation. From now on we simply denote by A the operator A". With the above properties, the operator A is the infinitesimal generator of a contraction (and holomorphic) semigroup of class C° in L2(Q)2, denoted by (GA(x3)), x3 > 0; the solution of (48) is given by
176
4 WAVEGUIDE PROBLEMS
Ej(x3) = GA(x3) E j , x3 > 0. (49) For the transverse magnetic field HT, first assuming that it is given for x3 = 0, and denoting by (GB(x3», x3 > 0, the semigroup generated by the square root B" (like in (47)) of the operator - (A+ + k2I), we have: (49)' HT
-9 3 n T .H T +Qjp-io)£n T AE T = 0 onl+,
which implies (using (27)iii) and its consequence n.H | (51)
8HT°
onr
=0):
-
Furthermore the initial value of H3 is obtained from (31)ii): (52)
H3(0) = H5=jjJircuriET<0)=1lrcurlE°.
Thus we have to determine H3 satisfying: (53)
I i) AH3 + k2H3 = 0 in Cl+ (with A in R3 !), ii) the boundary condition (51) on T+, iii) the initial condition (52), iv) the propagation condition (33) for H3.
We have solved this Helmholtz problem in section 1; its solution is given using (a generalization of) Theorem 1 (note that for Ej in L2(Q)2, H3 is not in L2(C1)) (54)
H3(x3) = G N (x 3 )H5.
Determination ofE3. Wefirsthave from equation div E = 0 in R , for x3 = 0: (55)
^ ( 0 ) = -divE T (0) = -divE£. 0
7.
First assuming that E3(0) = E3 is given (in L (Q)), E3 must satisfy: (56)
2 u)E i) AE3 | r+ k=0 E(see in Q+, (with A in R3), 3 « 0(32)), iu)E3(0) = E 3 onQ, iv) the propagation condition (33) for E3.
4.2
177
WAVEGUIDESINELECTROMAGNETISM
Then, from Theorem 1, we know that this problem has a unique solution, which is given by: (57) E3(x3) = G D (x 3 )E°. We detennine the "initial" condition E 3 from (55) by: (58)
^ ( 0 ) = C D E° = -divE£.
If k is not an eigenvalue of{- A T ), C D is invertible and (58) has a unique solution in L2(Q) when E j is in L2(Q)2: (59)
E^-C^divE^.
Determination of the transverse magnetic field H T . We have to detennine the initial condition HT(0) = H° T . This is obtained from equation (30)ii) for x 3 = 0. Using (48) we have (60)
AE^ - grad E 3 + io>y SH^ = 0,
and thus, with (59) and using the impedance Z with kZ = u>\i (61)
S H ^ ^ I A E T + gradCDdivEj].
First properties. Calderon operator C of the guide. It will be useful to transform the basic relation (61) using the following Hodge decomposition (see chap.2 (114)") of u in L2(Q)2 in the form: (62)
u = grad A^ divu - curl Aj^ L^, + u,
with u € H2{C1\ where H2(Q) = {v € L2(Q)2, curl v=div v=0, n A V| = 0); A N is the isomorphism from H (Q)/R onto (H^Q))' associated with the Neumann i
2
2
Laplacian and Lu € (H (Q))' is defined for u in L (fir by: (63)
Lu(v) = (Su,gradv),
VveH^Q),
thus w=A£, L„ is the solution (up to constants) of: (63)'
(grad w.grad v) = (Su.grad v),
W e Hl(Q),
oralso: (curlw,curlv) = (u,curlv), V V C H ' ( Q ) . For u € H(curl,ft) , we have L„ = curl u - n A U | 5 r and for u € H0(curl, £2), we have A^ LjjsAj^curlu.
178
4 WAVEGUIDE PROBLEMS
From the spectral decomposition of the operators AD, AN, the space L2(Q)2 splits into H+ and H ~ (see (46)), with: (64) H. = grad(HpD©curl(Hi[)N©H2(Q), H+ = grad(V£)D®curl(V£)N, where (Hk) and (H£) arefinitedimensional spaces corresponding to the negative part of the operators (- AD - k2I) and (- AN - k2I) (see the decomposition in the Helmholtz case), and (V£) and (V£) are spaces corresponding to the negative part of these operators (in HQ(Q) or H^Q)). This decomposition gives us the relation (for u in V", with (62)): (65) with: (66)
Au = gradAp CDdivu-curlCN(AJ^ curlu) + iku, Re(Au,u)<0 and Im(Au,u) > 0, V u € V .
Yet we return to relation (61); we have: (67)
, and thus:
CD' + I ^ A D ' C ^ - A ^ C D .
Therefore (61) becomes (for ET in V" then by continuity, for ET in H0(curl, Q.)) (68) SH^=j^ [- k2 grad Ap CJ^div Ej - curl C N AN' curl E^+ik E §], with E T the projection of E^ on H2(&). A first consequence of (68) is: (69)
(SHT, E£) = (n A H£, E £ ) = J ^ [k2
-
This implies (for all Ej in V" and then in H0(curl,Q)): (69)*
Re (SH°, E%) = Re (n A H£, EJ) < 0.
Then using the term "mild" by reference to a solution of Maxwell equation in a "weak" sense (in the sense of semigroup theory, see Pazy [1]), we can prove: Theorem 2. Except when k2 is an eigenvalue of the Laplacian with Dirichlet condition, problem (27) with (33) and the initial condition (70) E!r€H0(curiT,fl) has a unique mild solution (E, H) which satisfies: 0 |ET€C ([0,+oo),Ho(curlT,Q)), 1
HTeC°([0,+oo),H(curlT,0)), E3andH3€C°([0,+oo),L2(a)).
4.2 WAVEGUIDES IN
ELECTROMAGNETISM
179
This solution is given by (49), (49)\ (68), (52), (54), (57), (59). Remark 6. When k2 is an eigenvalue of the "Dirichlet Laplacian", we verify easily that the foregoing problem has no unique solution, since the following electromagnetic field (E, H) is a solution for zero "initial" condition: E° = 0, H 3 = 0, but E3(x3) * E3(0) so that (A + k2)E3(0) = 0 with E3(0) in Kl0(Q) (an eigenvector of the Dirichlet Laplacian) and Hj(x3) = H^O) = j — curl T E 3 (0). Definition 2. The mapping C: SE° T — SH° T (or C: E°- -► H° T ) which is defined by (68) is called the Calderon (or also the admittance) operator of the guide at frequency
N
(x 3 ) A^1 d i v H j - curlG D (x 3 ) A^1 curl H j + e**3 H £(x3),
for E j in V", with E jinH2(Q)and (73)
H j » n H ^ Q ) . We note that the relation:
HT
=C SE^)),
which is true for x 3 = 0, is valid for all x3 > 0: the Calderon operator C commutes with the semigroups of (49), (49)', simply denoted by (GA(x3)) and (GB(x3)) (73)*
GB(x3)C = CGA(x3),
Vx3>0.
Remark 8. Note that when E T is in H 0 (curl,Q) and not only in D(A), we have 3H 3 , -g^- (x3) = C N H 3 (x 3 ) in (H (Cl)Y (from semigroup theory), whereas divT H T is in H" (Q), and this does not correspond to the usual sense of div H = 0 in D'(R3). Applying the curl and div operators to (68), we obtain: (74)
i) curl H° = | C^ 1 div E ° ,
ii) div H° = ^ A C N A^1 curl E ° ,
180
4 WAVEGUIDE PROBLEMS
that is also P93 H^ + div Hj = 0, with P = AAJ^1 which is the (orthogonal) projection from(H1(Q))' onto H'l(Q). Conversely, thanks to the Hodge decomposition ("conjugated" to (62)): (62)*
v = - grad A^1 L^ - curl A^1 curl v+v,
with v € Hjtfl), where Hj(Q) = {v € L2(Q)2, curl v=div v=0, n.v| f = 0}, applied to v = Hj, and then using (74), we easily obtain (68). We can also obtain the inverse relation to Ej with respect to HT either as above for SHT, or using the Hodge decomposition (62) with the inverse to (74) (this gives us the inverse to the Calderon operator C): SE£ = -curl A^1 div E^- grad AJ^1 curl E£ + S E $ (75)
= f [cTrl A ^ C D curl H^ - k2 grad C^1 A^1 L S H 0 - ik H $]
when Hj € H(curl,Q), with (76) H^Z^SE^; H j is the orthogonal projection of H j onto H^Cl) whereas E j is that of E T onto Hi(Cl). The inverse relation to (74)ii) is given by (77)
curl E£ = ikZ ACJi1 A^1 L
We recall that w » - A^1 L
0
0
.
satisfies Aw = div Hj since:
SHnp
(gradw, gradv) = (H^, gradv),
VveH^Q).
Using (75), we can see that conversely, given Hj € H(curl,Q), we have ET in H(curl, O) with the boundary condition nT A EJ S 0 on T; this is due to the relations nT A curl
4.2
WAVEGUIDESINELECTROMAGNETISM
181
and ~ yo _= L* *7""ici£ H OC oj .,
ii) the "Transverse-Electic field" (TE waves) ( E ^ , H TC ) with E 3 = 0, therefore: c u r l H j = 0, divE^ = 0, then H3(x3) = ^ £ G N ( x 3 ) c u r l E j , and: E ^ ) = - curl GN(x3) AN 1 curl E £ ,
H ^ ) = ^ grad G N (x 3 ) C N A^1 curl E j .
Thus the Calderon operatorfor TE-waves is given by SH
T " EZ ™tX C N A N
curl E
T5
iii) the "Transverse-Magnetic field" (TM waves) (E T M ,H T M ) with H 3 = 0, therefore: c u r l E j = 0, divHj = 0, then E3(x3) = G D (x 3 ) E 3 = - G D (x 3 ) C^ 1 div E j , and: ET(X 3 )
= grad GD(x3) A^1 div E j , H ^ ) = curl G D (x 3 ) A^1 curl
H£
with (74). Thus the Calderon operatorfor TM-waves is given by: S H j = =Y grad A^1 C^ 1 div E^. This decomposition of the electromagnetic field is due to the splitting of the domains V4" and V"" by the Hodge decomposition. Yet we give a more useful framework. For usual applications, we have to find an electromagnetic field with locally finite energy in the closure of Q + . This implies (see chap. 2.7.2) that in each cross-section of the cylinder we have, with notations (85), (86) chap. 2: (78)
E1(x3)\Q€
H; 1 / 2 (curl,a),
H^)^€X(curl,Q),
Vx3 > 0.
Thus we have to solve problem (27) with a given initial boundary condition E° T with (78). These spaces are not very easy to handle, so we extensively use the interpolation theory and the following Hodge decomposition (compare to the Hodge decomposition above): Lemma 1. We have equivalence between i) and ii): i)
u € H; 1/2 (div,Q) (see (85) chap. 2)
ii) u = grad o + curl <>| + u,
with <> t € D(A^/4), (J> € D(Ap 4 ), and u € H^Q)
and also, using the S operator, we have equivalence between: i)u€H;1/2(curl,Q), ii) u = grad 0 + curl <>| + u\
with 0 e
D(A{> /4 ),
<> | € DtA^ 4 ), and u' € H2(C1).
182
4 WAVEGUIDE PROBLEMS
Here A D and A N denote the generalizations of the Dirichlet Laplacian and the Neumann Laplacian to the interpolated spaces and the dual spaces. ForregularQ, we will use the domains D(A D ) - H 2 (Q)n H£(Q) , D ^ 2 ) - H*(Q) and by interpolation: LXA^4) = H^2(Q), D ^ ' 4 ) - H* /2 (Q)» H 3 / 2 (Q)n Hl0(O). We also have D(A N ) = {v e H2(Q), §J| - OJ, DCAjf2) = H ! (Q) and by interpolation: EKA},/4) = H 1/2 (Q), D(Ai!,/4)={u € H 3/2 (Q), n.grad u € H^ 2 (K r )}, VT being a neighborhood of T in £2. (SKETCH OF) PROOF of Lemma 1. i) implies ii). First * is determined (up to a constant) by solving: (79) A ^ = div u € H" 1/2(G) = D(A N 1/4 ), 3/4%
"■"*
which gives 0 in D(A N ). Then since the grad and curl operators are continuous from H 3/2 (Q) into H l/2 (Q) , by duality the div and curl operators arc continuous | is obtained by solving equation from H" 1/2 (Q) 2 into H" (Q). Thus <> (80) AD<|> = curl u € H"3/2(Q), 1/2
which has a unique solution in HQQ (Cl). ii) implies i) Thanks to the characterization of the interpolated domains, ♦ € D(A N *) implies v « grad 0 € H^ 1/2(div,Q) Lemma 1 implies for (locally) finite energy electromagnetic field in R3:
8
Theorem 3 . Except when k2 is an eigenvalue of the Laplacian with Dirichlet condition, the problem (27) with (33) and the initial condition: (81)
ET€H; 1/2 (curl T ,Q)
has a unique (mild) solution (E,H) which satisfies (with def.(S6) chap.2); (82)
I E T € C°([0,+oo), H;1/2(curlT,Q)),
H T € C°([0,+
E3€C°([0,+oo),(Hy 2 (a))') and H 3 €C°([0,+oo), H- 1/2 (Q)). Furthermore (E, H) has a locallyfiniteenergy up to the boundary in Q + : (83)
E and H € H loc (curl,5 + ), i.e., CE, CH € H(curl,n+), V C € Z)(R3).
Then we generalize Definition 2 and Proposition 1 on the Calderon operator to the present framework.
4,2 WAVEGUIDES IN
ELECTROMAGNETISM
183
Proposition 2. The Calderon operator C (resp. C) of the guide atfrequency a is a continuous mapping mapp from H^ 2(div,Q) onto X(div,Q) (resp. HjJ (curl,Q) onto continuous X(curl,n», with: (72)*
Re (CSEJ,ET) ^ 0 ,
V E^ € H; 1/2 (curl,Q).
Note that the electromagnetic field in Theorem 3 and the Calderon operator are obtained by the same formulas as above. Theorem 2 is a regularity result with respect to Theorem 3 for the electromagnetic field. Then we have another regularity result in the H 1 framework: Theorem 4 (Regularity), When k 2 is not an eigenvalue of the Laplacian with Dirichlet condition, the solution (E,H) of problem (27) with (33) and the "initial" condition: (84) E?.€Ho /2 (curl T ,n) is such that: (85)
H j € H 1/2 (curl T ,Q)riH; 1/2 (div T ,Q)), E° € H"02(Q) and H°3 € H 1 / 2 (n),
and thus (E,H) satisfies continuity properties from x- (in R+) into these spaces. Furthermore (E,H) has H 1 regularity up to the boundary, that is: (86)
E and H € H ^ Q J 3 , i.e. CE, CH € H 1 ^ ) 3 , V C € D(R3).
Proposition 3 . The Calderon operator Cisa continuous mapping from the space Ho/2(curlT,Q) onto H 1/2 (curl T ,Q)nH; l/2 (div T ,Q). Example 2. Junctions and Cascades in electromagnetism We consider the situation of Example 1 with a junction and p guides incoming to it. We assume that: i) the (side) boundaries of the guides and of the junction TQ are perfectly conducting; ii) a given regular incident electromagnetic wave (EJ,HJ) with angular frequency d> is coming from the guides to the junction; iii) the permittivity and the permeability of the jth guide and of the junction are respectively (e.,^) and (e,jj). Thus the reflected wave (E.,H .) in the jth guide must satisfy (from Propositions 2 and 3, using the Calderon operator C. of the jth guide) on top T. of the guide (87)
nAHr|=C(nAEr|). j
j
Then using the continuity relations of the wave across the boundary T.:
184
(88)
4 WAVEGUIDE PROBLEMS [n A E] =0,
[n A H] = 0, that is:
j
j
nAE = nAE r + nAEj, nAH = nAHr + nAHj, onlj, j = l top, we obtain that the electromagneticfield(E, H) has to satisfy on T-: (89)
nAHl -C j (nAEL) = fIj, with f„ «ii A H, | r - C O i A E , ! ) . j
*j
j
l
i
Thus we have to find the electromagnetic field (E,H) with finite energy in the junction satisfying: i) Maxwell equations in Cl: (90)
|i)curlH + i
ii) the boundary condition on ro, then on Ij (91)
nAE| r = 0, with(89)only
We write this problem in a variational form. We define: (92)
Hr (curl,Q) =f {u € H(curl,Q), n A u| _ = 0},
and the sesquilinear form on Hr (curl,Q): (93) a(E,E)=SQ(-XcuTlE
. curll - icoeE . E)dx- 2 / r C , ( n A E | r ) . E| r d^.
Then problem (90), (91) is equivalent tofindE in Hr (curl,Q) satisfying: (94)
a(E,E) = 2 / j
F„ . E dx,
V E € Hr (curl,Q).
lj
o
i) First if the medium of the junction is dissipative, that is if (95)
lme>0, and lmji>0,
problem (94) has a unique solution; this is easily proved as follows. Taking the real part of (94) with E = E, thanks to (72)', there is a constant C 0 so that: (9i)
Rea(E,E) > C 0 ! E £
(curla),
VE €H^cur!,^).
Thus the sesquilinear form a(E,E) is coercive on Hr (curl,Q). o Thanks to the Lax-Milgram lemma, problem (94) has a unique solution. ii) / / the medium of the junction is conservative (that is when c and \x are real positive) we will see that problem (94) depends on the Fredholm alternative.
4.2 WAVEGUIDES IN ELECTROMAGNETISM
185
Note that the natural injection Hr (curl,Q) —> L2(Q)3 is not compact. o
But for given regular incident electromagnetic field, fy is (for instance) in H (curl,rj)H H~ / (div,^); E must satisfy a variational problem in Hr (curl,Q) with div E = 0 and its boundary value must be in H1/2(curl,0; thus E is in Hl(Cl) . Since the natural mapping from H*(Q) into L2(Q) is compact (for regular Q), this problem (94) depends on the Fredholm alternative: (94) has a unique solution except for the eigenfrequencies of the junction (and also of the guides corresponding to the eigenvalues of the Dirichlet Laplacian). Remark 10. In these junction problems we are essentially interested in obtaining the electromagnetic field and above all the propagating modes which are outgoing from the junction through the guides. If the medium of the junction is homogeneous, we can write the problem on the boundary of the junction thanks to the Calderon projectors P. or Pe (see Def. 3 chap. 3) or the (interior) Calderon operator C1 (see Def. 5 and (211) chap. 3); thus we have to find the boundary value of the electromagneticfieldsatisfying: (97)
P e (nAE| r ,-nAH| r ) = 0, or nAH| r = C W E | r ) ,
with (89) and (91). Taking the value of the magnetic field on T, j = 1 to p, gives us a system of equations on the boundary value of the electric field. This is very simply written thanks to the Calderon operator (but less suitable for a numerical solution): (98) 2 (CJj,-Cj5jj.)Mj, = f Ij ,j = ltop,withM j = nAE| r €H; 1/2 (div,r j ). j ' = 1 to p
j
If the medium in the junction is inhomogeneous, then we have to connect (for instance) afiniteelement method in the domain Q with a spectral method on the boundary of Q. Such a method has been used and numerically implemented in order to evaluate the permittivity and permeability of a sample in a wave guide. Obviously there is a great variety of examples and of geometries for this study. <8>
Remark 11. We haven't studied the more difficult case of waveguides which are inhomogeneous in the fiber, that is, the cross-section of the guide up to infinity is inhomogeneous. It seems necessary to use microlocal analysis for such a case. 0 Remark 12. In Examples 1 (with Helmholtz) and 2 (with Maxwell), the Calderon operators (C.) depend on the angular frequency
A(tt>)u = FI(a,) with given F^,
186
4 WAVEGUIDE PROBLEMS
where (A^) is a holomorphic (in a>) family of closed operators in a Hilbert space H, with compact resolvents. Therefore we can apply theorem 1.10 (p.371) in Kato [1]. Since outside the real axis (for o>) the operator A(a)) has a kernel reduced to {0}, we deduce that the problem (99) depends also on the Fredholm alternative, with only a finite number of singular values to on everyfiniteinterval J (in the Maxwell case we have to assume that 0 is not in J).
CHAPTER 5
STATIONARY SCATTERING PROBLEMS ON UNBOUNDED OBSTACLES
1. PLANE GEOMETRY
The study of stationary waves scattered by infinite obstacles follows naturally from scattering by finite obstacles at high frequency. For a plane geometry, we have an ill-posed problem (as in the waveguides) if we don't take into account the sense of wave propagation. Here we extensively use the Fourier transform method, which replaces the discrete spectral decomposition in waveguides. First we study Helmholtz problems, then electromagnetism with Maxwell equations. 1.1. Plane geometry with Helmholtz equation First we consider a model Helmholtz problem with a Dirichlet condition (which corresponds to scattering by a soft wall in acoustics) as the starting point to understand the basic problem. 1.1.1. Helmholtz problem in a half-space with Dirichlet condition Let R* = ( X a ( x ^ ) € R n ~ l x R, xn>0} = Rn"1 xR + with (generally) n = 3, and let T = R n " l x {0} be its boundary. We consider the following Dirichlet problem:findu satisfying (1)
|i)Au + k2u = 0 inR? jii)u| r = u0 onT, 187
188
5 UNBOUNDED OBSTACLES
with a given wavenumber k > 0 and given IK on T, for instance in L2(D. We study this problem thanks to a (transverse) Fourier transformation. Let fv denote the Fourier transform of a tempered distribution v, with: (2)
v«) = .Fvtt)=/ n.ivMe-^dx,
forveSCR""1).
Then with the notation u(Xj) » uCx,^, we write (1) as: ,2A
:\d U
(3)
i ) ^ 3 + (k 2 -S 2 )fi = 0
forx n >0,
ii)fi(.,0) = fi0. Thus we are reduced to solving a differential equation in terms of the variable xn only. We get, with constant functions (with respect to xfl) a and 0 to specify i) for|t|>k, (4)
iKUn^attJe0* V K 9 e ~ 0 + \
with8+= tf - k2,
ii) for|t|
fitf^)
=
with 9. = ^k 2 -* 2 .
Wanting a solution which is tempered with respect to xn (and £) at infinity we have to choose (6)
ii&xj^e-****^)
for\i\>k.
When |£|
exp (i8jgvtf) (resp.expHe.x^vtf)),
with9_>0.
Then with the physical choice of waves propagating towards the positive axis xn, we get a unique solution for (3) (and then for (1» of the form: (10) fl(e,xn) = e"8(e)Xnfl0«), with:
189
5.1 PLANE GEOMETRY (11) 9(0 = e+ = (£ 2 -k 2 ) 1/2 when|£|>k,
8(£) = .ie. = -i(k 2 -£ 2 ) 1/2 when |t|
Theorem 1. The problem (1) with uQ given in L2(Rn~l) and with the condition (12) u is a tempered wave propagating towards the positive axis xn, has a unique solution u in C°([0,+oo),L2(Rn~l))y and it is obtained thanks to a contraction (and holomorphic) semigroup of class C° in L^R11""1), (G(xn)), xn > 0, by its Fourier transform (10), with (11) (13)
= G(xn)fi0«) = e" **** fi0«), £ € Rn"l.
fitf^)
PROOF.
The operator G(xn) is a contraction in L2(Rn"~l) since
(14)
sup logK-erosgi-i, Vx n >0.
Furthermore, we have: (15)
HuCxJ - U0l|2 = (2*)- n / R n-l Ifl&Xn) -U0«)|2d£ = (2«)* n / nD n.liao«)| 2 M-e- e X n | 2 dt 'R
11
1
Since UQCL^R "" ), for all 6i> 0, there exists a>0 such that: / . . and there exists n > 0 such that 0 < xn < n implies
|u 0 | dE
(16) |l-e" 0 X n | 2 <£ 1 , V£with|£|
W . ^ - U / ^ M O + IUOI2),
which implies that the semigroup is continuous. Proposition 1. The infinitesimal generator A of the semigroup (G(xfl))t xfl > 0, in L2(Rn~l) has the domain D(A) = H^R11"l) and is given by its Fourier transform: (18)
Au0(£) = - 9«)u0(S)
with 9 defined by (11).
PROOF .
The domain D(A) of the generator is given by its Fourier transform
(19)
I D(A) = {fl0 € LAR11-1), Afl0 6 L2(Rn-!)}, -|flo€LV-I),/,5,>k(l2-k2)|ao|2«<«>}f
from which we clearly have D(A) = H*(Rn~*). <8>
190
5 UNBOUNDED OBSTACLES
We also have:
(20)
A^O-tf-k2^©,
therefore: (21) A2 = -(A + k2),
withEKA^H^R11-1).
Thus A is a square root of the self adjoint (but not positive) operator - (A + k2). Like in the waveguide case, we can decompose the space L (R n ~ l ) into: (22) L^R^^H^eH^, so that A splits into: (23) A » - A+ + i A", A+, A" positive self adjoint operators resp. in H£ and H£, and A satisfies: (24)
Re (Au,u) <: 0,
Im (Au,u) > 0, V u € H^R11*1);
A is a normal operator: D(A) ■ D(A ) and A commutes with A , with spectrum (25)
o(A) = (-oo,0]U[0,ik].
Definition 1. The infinitesimal generator A of the semigroup (G(xn)), x n >0, defined by (IS) is called the Calderon (or the capacity) operator of the half-space, We recall that A is the mapping u0 —»g~ | , u the solution of (1) with (12). Remark 1. We have to verify that choosing waves propagating towards the positive axis xn as above, agrees with choosing outgoing waves defined by the Sommerfeld condition. It suffices to verify that the elementary outgoing wave ka centered at point X =(0,0, - a ) , "a" being positive, satisfies (12). Using a translation, we can take a = 0. Then using the Fourier transform of the elementary outgoing wave * k given by (159)' chap. 3, and taking the inverse Fourier transform with respect to variable xfl only, we obtain (for n = 3) (26) Thus: (27)
F^Fx^^^F^i^x^^^i^m^asp
(" «(© W>-
^S(£,x3) = Ok(£,x3 + a) = ^ e x p ( . 0 ( t ) | x 3 + a|),
which is of the form (13) forXn > 0 (but for a = 0, u0 = 1/(29) is not in L2). From formulas (13) and (26), we see that the semigroup is also given by:
5.1 PLANE GEOMETRY
(28)
191
G(x3)fl0tt) = - 2 g ^ ax 3 )a 0 «),
that is also with the convolution product * (a priori for u0 with compact support) (29)
G(x 3 )u 0 - - 2-gf (.,x 3 ).u 0 . ikR
The semigroup (G(t)), t = x3 > 0, is also given with <&fc(x,t) = | j g - , R = (x2 + t 2 ) i / 2 by (30)
G(t)u0(x) = - 2/ R 2^(x-x , ,t)u 0 (x , )dx'.
Thus the operator A is obtained by the convolution in R Au0 = ax*u0, with a(x) = - 2 ^ ( x , 0 ) = - 2 F P ( i | ( ^ ) ) , r = | x | , i . e . = lim/ e °°V5 , (r)dr, with 5Xr)«2~jJV(r,B)<», V*€Z)(R 2 ). Note that 5'(0) = 0 from (397) chap.3. When k = 0, the semigroup (G(t)) is given by (31)
G(t)u 0 (x)=^/ R 2 ( ( x _ x j
+ t 2 ) 3/ 2
«o(x*) dx',
that is, the Cauchy-Poisson semigroup which is holomorphic in ail LP(R2), p * oo (see Butzer-Berens f 1] p. 248). This seems not to be true for k * 0. Remark 2. We can find formula (29) using the "image" method: by symmetry we define in the whole space R3 (32) Then (33)
U(x,x3) = u(x,x3) when x3 > 0,
U(x,x3) = - u(x, - x3) when x3 < 0.
AU + k2U = div (2u0n6r) in R3, with n = e3 * (0,0,1).
Thus, for u0 with compact support, U is an outgoing wave given by (34)
1* 3 ) x *u 0 , U = .O k ^div(2u 0 n6 r ) = -gradO k ^(2u 0 n5 r ) = - 2 3^(.,x
that is (29); thus U is a double layer potential.
®
Remark 3. Some generalizations to other frameworks. First note that A is not a bijective mapping from H*(R2) onto L2(R2). This suggests that it is not the best framework to consider (from the point of view of applications also). We first define spaces with a special weight for tangent (or longitudinal) waves, i.e., for £ on the sphere Sk of radius k: (35)
H£/2(R2)={U€S>(R2),
fi€L/OC(R2),
fl€L2(R2,
|e| d«)},
Hk 1/2(R2) = {u €S\R 2 ), fi € L ^ R 2 ) , fi € L2(R2, |6| - 1 d£)}.
192
5 UNBOUNDED OBSTACLES
Note that if pk(() = d(£,Sk) is the distance from £ to the sphere Sk, then in the neighborhood of Sk: (36)
|e k (E)| 2 s |(|6| 2 -k 2 )| = | ( | £ | . k ) | ( | £ | + k ) « 2 k 6 ( t , S k ) = 2kpk(£).
An example is given by the trace on the plane T of a wave due to a point source not in the plane: thanks to (26), for a * 0,4>£| € H£ / 2 (R 2 ). Since 9 and 8"l are in L^R 2 ), then u € L2(R2, |9|d£) implies by Cauchy-Schwarz inequality
(37)
/K|fi|d^(/K|fi|2|9|d£)1/2(/K | 9 r W 2 ,
for aU compact sets KinR 2 , thus u € L^R 2 ). When u € L2(R2, | 9 | - 1 dt), we have the same result. The condition u € L/OC(R2) in (35) is only used to eliminate measures concentrated on the sphere Sk of radius k. Furthermore, using the inequality: |9|
2
1/2
Proposition 2, The mapping u -»Au is an isome tryfrom Hk (R ) onto Hk 1/2
2
2
(R ),
l/2
and is a continuous mapping from H (R ) into H" (R ). Moreoverfor u0 in Hk/2(R2), the solution u of(\\ (12), given by (13) is in H/ 0C (R I), that is,uis oflocallyfinite energy up to the boundary. The framework (35) does not allow us to directly tackle all useful examples such as plane waves. So we have to generalize. The framework of tempered distributions (in R2) is not adequate since exp(-0x 3 ) is not a multiplier in 5"(R2). But we can use other functional spaces, for instance Besov space B used by Hormander [1], T.l chap. XIV p. 227 and its dual space B*. For most applications in view, it seems interesting to work with the space Ek defined by: (40) Ek = (u € S*(R), JFxu is a Radon measure concentrated on the sphere Sk c R }.
5.7 PLANE GEOMETRY
193
From the Paley-Wiener Theorem, the elements of Ek are entire functions in R3, which satisfy the Helmholtz equation in the whole space. The set of elements of £ k propagating towards the positive axis x3 is (A\\ E£ = l u € Efr ^ x u *s a ^ a c , o n measure concentrated on the half sphere S£}, with S£={(5,£3) e Sk, with £3 > 0}. Of course, there are other functional spaces which may be useful, like: (42)
X = {u€L2(R2), fi€C0(R2)}, withC0(Rz) = {v€C(Rz), v(S)-> Owhen |E| — 00}
and its dual space X' also defined by (43)
X' = {f € ^(R 2 ), f € Ml(R2) + L2(R2)}, with Ml(R) the set of bounded Radon measures.
We can prove that X' is contained in the space: (44)
Cb(R2) + L2(R2), with Cb(R ) the space of continuous bounded functions on R
Furthermore, we can prove that the spaces X and X' are stable under the semigroup (G(t)), t > 0, which is a semigroup of class C° in these spaces; X' allows us to tackle the scattering of an incident plane waves by an infinite plane wall. But these spaces are not very easy to handle from the point of view of functional analysis. Plane waves are often used by the spectral theory as generalized eigenvectors (of the operator A). Remark 4. The Helmholtz problem (1) with Im k different from 0 (0 < arg k < n) is well-posed if we only require that u be a x3-tempered solution. It is also obtained by a semigroup of class C°, which is holomorphic, and given by its Fourier transform: (45) fl«,X3) = G(X3)u0a) = exp(-ex3)u0(£), e«) = tt2-k2)1/2 with Re 0(£) £ 0. Here 9*0, thus A is an isomorphism from H1/2(R2) onto H"1/2(R2). The limit of (45) when Im k tends to 0 with Re k positive (resp. negative) gives the wave propagating towards the positive (resp. negative) real axis x3, following the limiting absorption principle (see chap. 3). 0 1.1.2. Transmission Helmholtz problems in R3. Calderon projectors First we consider the following Helmholtz problem in the whole space: find u in R 3 satisfying (for a given real wavenumber k)
194
5 UNBOUNDED OBSTACLES i) Au + l^u = 0
(46)
forx3 * 0, i.e., in R* and R3. ,
ii)[u] r d ^u| r - u | r =p,
[§-] r =P* onr, 3
3
iii) the restrictions of u to each half-space R+ and R_ are "outgoing" waves, i.e., resp. propagating towards x3 > 0 and x3 < 0, with given p and p'onTsR x {0}, with hypotheses to be specified later. We know (for instance from chap. 3) that this problem has at most one solution. Using the above section, we are led to find the boundary values of the solution on r only. We can use different methods. Here we use an integral method, similar to chap. 3. Under usual (regularity) conditions on p and p' (see chap. 3), i) and ii) are equivalent to (47)
Au + k2u - - (p' 8r + div (pn 8r)) in Z)'(R3).
With the "outgoing" condition iii) (at least when p and p' have compact supports), problem (1) has a unique solution u given by the convolution product: (48)
u = * < (p , S r + div(pn6r)),
with O = * k the elementary outgoing solution (see (8) chap.3), or also (49)
u(.,x3) =
If we differentiate with respect to x3, we obtain: < 50 >
^<- x 3)=^^ x 3) x .P'^(-.x 3 );P.
Then, when x3 —»0+ and x3 -* 0_, using (29), we have:
m
['-£&)•'-[',)• (*&«)-'-(•)•
with P+ and P_ given by (52)
P+-±(I-S),
P . = i ( I + S),
Thus, we have (53)
P+ + P_=:I, and S2 = I.
We can verify that (54)
B^-AA-',
- ( : - ; >
5.1 PLANE GEOMETRY
195
is an integral operator given by the convolution product:
Bpw-«(..a>.p'-J^ 4r^T p ' ( x ' ) d x '-
(55)
Proposition 3. The operator Sis an isomorphism satisfying (53), in the space (56)
Y k d ^ H£/2(R2)x H£1/2(R2),
and thus P and P _ are continuous complementary projectors in this space, giving the boundary values of the problem (46) by (51) when (p, p') are given in Yfc. Definition 2. 77te operators P+ a/irf P_ are called Calderon projectors {for the Helmholtz problem) of the half space at wavenumber k. If we compare the half-space problem to that of a bounded domain (see chap. 3 Def. 2 and (71)), we see that we have, with notations of chap. 3: K = 0, J = 0, 2L = -A~ 1 ,
2R = - A .
1.1.3. Transmission problems with two different media We assume that the whole space is occupied by two different media, one on each side of a plane. Let k{ and k2 be (real for instance) wavenumbers in each medium. Then we consider the following transmission problem: find u in R3 satisfying i) Au + k2u = 0 forx 3 *0, withk = kj inR 3 andk = k2 inR 3 , (57) |ii)M r d = e f u| r _-u| r + = p,
[|y-] r = p >onr,
""
3
3
iii) the restrictions of u to each half-space R+ and R_ are "outgoing" waves, i.e., resp. propagating towards x3 > 0 and x3 < 0, with given p and p' on T = R x {0}, with hypotheses to be specified later. This problem is a scattering problem when we have a given incident wave Uj (in the first medium for instance) either produced by a source of compact support in R3+, or by a plane wave. Thus we have to find the reflected wave Uj in R + and the diffracted (or transmitted) wave u2 in R3 _, satisfying: 8u, 3uT 3u2 (58)
u 1 + u I = u2,
^-+5E-=^-
onr.
Thus taking u = ux in R +, u = u2 in R _, we see that u has to satisfy (57) with:
(59)
P
=Ul|r,
p'=gl| r .
196
5 UNBOUNDED OBSTACLES
Let A. and 0., j = 1, 2 resp. be the Calderon operator A and the function 0 defined by (18) and (11) for k.. Then problem (57) is reduced to finding the boundary values of u, u, and u2 on I", and these are obtained thanks to the Calderon operators of each half spaces from (57)ii) (with i) and Hi)) by: DOJOJ + M I - ? '
(60)
|ii)u 2 -fi, = p\
The solution of (60) is given by: A
61
< >
p'+e.p
A
*2-W'
p '--et2 _ p
^'T*2 *
When p and p' are given by (59) with an incident wave propagating towards the negative axis x3, we have (62)
p' = -A 1 (uil r ) = -A,(p).
We obtain the boundary values of the transmitted wave and the reflected wave by (63) with
u 2 =fu,,
u, = Ruj onT,
1 "^" 2 A
1 ^* 7.
A
R and T are called the reflection and transmission coefficients. They satisfy (65)
T-R=l.
Proposition 4. We assume that the incident wave u{ satisfies (66)
u,| r €H k /2 (R 2 ), t h u s ( U l , ^ ) | r € Y k i .
On the boundary T of the two media the reflected and the transmitted waves satisfy (67) u, | r € H k2(R2) and u 2 1 r e H k 2(R2)n H{/2(R2) (and even u21 f 6 H1/2(R2)) 3u, andthus ( U l , ^ i ) | r 6 Y k l ,
9u7 (u2,-g^eY^m^.
PROOF. We have (68)
/ R 2|fi2l 2 |e 2 l d 6 - / ^ | « , | 2 | t | 2 | e 2 | * - / ^ | « I | 2 | e 1 | t d t .
5.1 PLANE GEOMETRY
with T=
197
419,6? I 2 ' 2 . We easily see that T< 2, and thus:/ 2 |u 2 [ |S 2 |dt
Since | f | <2, we also have/ 2 |fi 2 | 2 |9il &< °°- Furthermore:
/^l^l^i.UlV^d^/^lii^iejad^ with g . 4 I W H ^ 1 / 2 . R
R
l e l+ 9 2l
But a is bounded on R2, thus we have u 2 | €H1/2(R ), and Proposition 3 follows. Generalizing to spaces such as that deflned by (41), or (43) allows us to deal with plane waves, giving the famous Snell-Descartes laws. The incident wave Uj is given by (69)
Uj(X)«iiI0e l*' ,
withuIO€C,a = (a T ,a 3 )€S 2 ,a 3 <0,
whose Fourier transform is the measure (69)' fi,(£,x3) = (2n)2 6({ - k ^ ) e*1"3*3 uI0, with |a T | 2 + a] = 1. Thus p and p' are given by their Fourier transforms: (70) £«) = {1,(5,0) = (2*)2 8« - klttT) uI0, $'«) = &T (£ ' 0) = (2n) m±i°^ ^ - kiotT) uioHere formulas (60) are also valid and imply that fl, and u2 are Dirac measures concentrated on { = kjaT. Then 8j and 02 are given by (71)
e ^ o ^ a i l ^ a j , e2(k!aT) =
(k24-^)1/2 |-i(k 2 -k 2 <4) 1 / 2
ftkl\^r\>k29 ifk!|a T |
Thus with the (relative) index n = k2/kj, we have (63) with: m\ (
*
2 g
l 3l I«3l+n9 12 '
A R=
I a 3l-n0 1 2 |«3|+n812'
' and (73) 912 = i02/k2 = i((4/ n 2 )- D 1/2 if K l >n,
e12 = (l -{al/T)2))1'2 if |a T | < n .
The inverse Fourier transforms of the reflected and the transmitted waves are, for the reflected wave A
(74)
ik,0.X
Uj(x,x3) = R uI0e
with 0 = (aT, - a3),
and the transmitted wave (74)' ^ u2(x,x3) = fu I0 e ik2 ^ X withg = (PT,P3), where P may be complex; we have
198
5 UNBOUNDED OBSTACLES
i)fork,
thus |P T | < K | , |P 3 | >|a 3 |,
or with respect to incident and refracted angles of wave propagations *j and *2, I 0 T | » sin *2 = I arl/n * (sin ♦,)/!!,
*2 < *p
ii)fork|>k2, i.e., n
1.1.4. Some examples of applications of the Calderon operator There are very many examples, where we have to use the Calderon operator A in order to obtain a well-posed problem, taking into account the sense of wave propagation, even for geometries which are not simple half-planes. Using the Calderon operator A often gives a boundary condition, which allows us to reduce the domain of the wave problem to a slab or to a bounded domain. Thus, we can treat multilayer scattering problems (that is, for slabs with different homogeneous media in the free space), screen scattering problems, where the screen can be thick or thin, infinite (for instance a half plane) or not, with apertures of arbitrary shape.... Here we give an example with a thin soft screen with bounded apertures. Let T0 be a "flat domain" that is an open subset of R2x{0}, modelling a thin screen; we assume that its complementary set Tl^R2x[0}\To is a "regular" bounded set (connected or not), modelling apertures in the screen. We assume that the free space is outside the screen and that a given incident stationary wave Uj with wavenumber k is propagating in R3+ towards the screen. Let \x{ and u2 resp. be the reflected wave in R3+, the wave in R3_ transmitted through the apertures. Let u be defined in R^R^O}) by u = Uj in R3+, u = u2 in R 3 _. Then u must satisfy: i)Au + k 2 u«0inR\r,withr=R 2 x{0}, 3o>.
(75)
3ui.
9uT
»I»lr1,,,r»«lrr,Iilrr,lilrI'ftrr2lrriSlr1-2lr1-0Bri Hi) U||
=-U||
>u2| =0 (soft screen conditions)
iv) u t (resp. u2) is propagating towards the positive (resp. negative) axis x3.
199
5.1 PLANE GEOMETRY
Using the Calderon operator A, this problem is reduced to finding u on both sides of r satisfying (76) Uj-Uj^UjOnTj, with (75)iii) on T0, and (Auj + Au^L ~--§5"l r onTj. Let w=u 2 + u1 + Uj on I\ Then, from (76), w = 2u2, and w must satisfy (77)
Aw|r=f,|r,
withw=0onr o ,
3Ur
withfjL ^-"gjrlr - A u i l r =-2AU|| (78)
u
llR2 J W € H i / 2 ( R 2 ) -
Now let V={v € H£ / 2 (R 2 ), v| find w in V such that (77)'
(Aw,v) = (f,,v),
given on Tj. We assume that Uj satisfies
thusf^HJ^R2). = 0}. Then we can write (77) in the variational form: W€V.
Lemma 1. The sesquilinear form a(u,v) defined by: (79)
u, v € H^/2(R2),
a(u,v) = (Au,v), 1/2
2
is "coercive" on H k (R ) and more precisely (80)
|a(u,u)|-lud l /a / .a 4 .
Vu€l4 /2
PROOF. From definition (79), we have (81)
a(u,u) = - f
2 0|G|
2
d*,
K.
and thus we obtain (80) from (81)' |a(u,u)|=/ R 2 |8||fi| 2 d£. Lemma 2. H£/2(R2)(resp. H^m(R2)) is contained in H/0/2(R2)(resp.Hf0lc/2(R2)). PROOF. Working on the Fourier transform of these spaces, we decompose the corresponding L2 spaces into the sum of two spaces, one WQ with functions equal to zero in a neighborhood of the sphere S. and the other Wj with functions equal to zero outside this neighborhood. Now we see that the inverse Fourier transform of functions in WQ are in the Sobolev space H1/2(R2) (resp. into H"~1/2(R2)), and that the functions in Wj having compact support are transformed into regular functions. <8>
200
5 UNBOUNDED OBSTACLES
Proposition 5. Let u} be an incident wave satisfying (78). Problem (77)' (thus (77)) has a unique solution w (thus u2 the transmitted wave) such that
(82)
wcH^OV,
giving the unique solution of (75) (with bounded apertures Tx in a screen in R3), that is the transmitted wave and the reflected wave thanks to the semigroups (G(x 3 ))am/(G(-x 3 ))M31). PROOF. First, using (39) and the lemma 2, we can identify H^O^) with the closed subspace V of H£ / 2 (R 2 ) by the mapping u —> 5, u being the extension by 0 of u 1/2
outside Tx. Furthermore, we have fj € V = (H0 0(r1)y. Then using Lemma 1, we can conclude thanks to the Lax-Milgram lemma (see for instance Lions [4]). 1.2. Plane geometry with Maxwell equations 1.2.1. A typical problem in a half-space We first consider the following problem: find the electromagnetic field (E,H) in free space, at angular frequency a>, satisfying i)curlH + io)£E = 0, (83)
ii)-curlE + i(i)jiH = 0
inR3 = R^xR£ ,
iii) n A E | = n A E j (orwrE = Ej), with given E T , with nT the orthogonal projection on the boundary r of R+. Like in the Helmholtz case, we can see by Fourier transformation with respect to x (under usual hypotheses on E°~) that this problem is an ill-posed problem, and it becomes well-posed with the following hypothesis (84) (E, H) is a wave propagating towards the positive axis x3 . Since the components of E and H satisfy the Helmholtz equation, they are obtained thanks to the semigroup (G(x3)) of section 1.1 due to (84), from their boundary values on I\ With usual notations (see also chap. 4) we write (85)ET
E3(x3) = G(x3)El HT
thus we only have to find E3, H3, Hj from the transverse component E^ of E. We can do it in many complementary ways. We first write Maxwell equations ID the form (30),(31) chap.4, decomposing the electromagnetic field into its transverse and longitudinal (along x3 axis) components. Using (31) and (38) in (30) gives us (34) with (35) chap. 4, then using the inverse of the infinitesimal generator A of the semigroup (G(x3)), we obtain (with
5.1 PLANE GEOMETRY
201
i) H T = 2 £ A " l (k2 + gmh d«vT) SE T ,
(86)
ii) Ej. = - Y A* l (k2 + gradf divT) SHT. Thus we get the following relations (87)
1)HT = C.ET,
or nAHT=C(nAE£)(i.e.SHT = C(SE£)),
ii) E j = - Z 2 C H £ , or n A E £ ■ - Z 2 C (n A H J ) , (i.e. SE^ = - Z 2 C ( S E £ ) ) ,
and thus: (88) C 2 = - Z ~ 2 I , C 2 = - Z - 2 I , and C = -SCS, C = -SCS, with C (and C) given by (86), that is, in Fourier transform, H j = C E T»
<89>
C=
zk9^
^.^J^zke^.^.^) . ^ J .
and C =-
— zke^(k2.$J)
.^J
Now we follow another method. Taking into account that E and H are divergence free (and satisfy Helmholtz equation), we can substitute to (30) chap.4: (90)
divE = 0,
divH = 0,
that is we only consider the equations: (91)
9E 3 »a^
+ div
9H3 a—+divTHT=o,
TET = 0 '
ii) curlT H T + io)6 E 3 = 0,
- curlT E T + io>y H 3 = 0.
In the following we drop subscript T to differential operators (when it is not confusing). From these equations, we easily get (at least formally) (92)
(93) divH^-AH^, and thus: (94) that is:
(94)'
E^-A^divEj,
H^-j^curlE^,
curlH^-^E^,
div H£ = j ^ A curl Ej,
curl H ^ = | A" l div E j ,
(diViil]4(° ^ curlH^ J
V. kA"'
^A1fdiV^l 0 J ^ curlE^ J
202
5 UNBOUNDED
OBSTACLES
Then we get H T using the Hodge decomposition: (95)
H £ = grad A"' div H^ - curl A" ! curl H^.
We define the following Riesz operators (this is useful to specify spaces) Rg* = grad (- A)- 1/2 ♦, i.e., Rg ♦«)=-j|r♦«), (96) R^cTrK-A)- 1 ' 2 *, i.e., Rc$tt) = - j | | $ « ) , and their adjoint operators R;u = -div(-A)- 1/2 u, i. e .,R>tt) = - « j H p , (96)' 1 ^ u = curl (-A)".1/2. " u,
i.e., R,.fl«)=
iUfl(t) —^-.
These operators satisfy: (97) R g = SR c , and also: (98)
Rc = - S R g )
| R g $ « ) | = !♦(«)!,
R^ = RgS,
Rg = - R ^ ,
|Rc$tt)M$tt)|,
and thus (with L norm for instance): IIRg0ll = 11*11, IIRc0ll = ll0ll, (99) i.e., R and R c are isometric operators in L 2 . Then the (orthogonal) projectors P a n d r c on the images of the grad and the (vector) curl operators associated to the Hodge decomposition are given by: (100)
P gu = R g R g u = grad A"l div u , Pcv=RCR* v = - curl A"" *curl v,
or with their Fourier transforms: (100)
pgU«)—np^-,
pcv(0—jjp^-,
which are associated with the decomposition: u({) = Pg u(£) + P c u(S). Using these operators, we can write (94) in the form (ioi)
R;H£=-J^AR^E£,
and also (since R* Pg = 0, Rg Pc = 0):
R^H£=-|A-1R;E£,
5.1 PLANE GEOMETRY
203
(102) PCH°=-§RCA-,RX=ZA-1PCS4 and thus the Calderon operator C is also given by: H ^ C E ^ - ^ R g A R ^ + kRcA^RjE^ (103) =
Z[-EAP« + kA " lp J S 4-
Now we define the following functional space: (104) H k (R 2 )^ {ue.SXR2 ) 2 ,fi€L^R2 )2 , R j u s H k 1/2(R2), R^ u eHk/2(R2)}. This is a Hilbert space with the norm: (105)
» ul W = Ml e l UAU«>I2+TOT U-a^)!2]—^ dt),/2 |0| vl/2
i«r
^"^"H^RV'V'H^'V) We also define the space Hk(R2) = SHk(R2): (104)'
H^RV^UCSXRV.U^L^R2)2,
Rju €Hk/2(R2), R^u eH k 1/2 (R 2 )j.
This is a Hilbert space with its natural norm, and S is an isomorphism from Hk(R ) onto Hk(R ). Now we assume that: (106)
Ej€Hk(R2).
Theorem 2. With hypothesis (106), problem (83) with (84) has a unique solution (E,H) with locally finite energy (up to the boundary, i.e., E and H are in Hloc(curl,R2 x [0, oo)) and the boundary value of the magneticfieldalso satisfies (107)
Hj€Hk(R2).
For the other components of the electromagneticfield,we define, like (35): (108)
X k ^ f {u€y(R 2 ),fi€l^ 0C (R 2 ), / R2 |fi| 2 (|0|/|£| 2 )d£
Note that if v satisfies/ 2 |v| 2 (|8|/|S| 2 )d£< oo, thus v € L^R 2 ), since from Cauchy-Schwarz inequality fK |v| d£<(fK |v| 2 (|0|/|«| 2 )d£) l/2 .(/ K (|5| 2 /|9|)d£) 1/2 , for all compact sets K in R2. Furthermore since |0|/|£| 2 ~ 1/|£| when |£| —> oo,
204
(108)'
5 UNBOUNDED OBSTACLES Xk c H f ^ R 2 ) .
Nowwith hypothesis (106), we have (-A)-1/2E3 and (-A)" 172 ^ € H£ / 2 (R 2 ), from (92), (93), that is: E^andH^cX,,. (109) E^ We also have: (109)' HkJR2)cHf0lf(dfcr>R2),
H^cHfJ^curlR2).
proof of (109)'. Let u € Hk(R2), then RgR^u € H£ / 2 (R 2 ) 2 , and R ^ u € Hk 1/2(R2)2, thus: u = RgRgU + R ^ u € Hf0lc/2(R2)2, thanks to lemma 2. Furthermore R^u€Hk/2(R2) implies divu€X k , andby(108)\ divucHfJ^R 2 ). This prove the inclusions in (109)', thanks to operator S. Then, like in Definition 4 chap. 3 (with (199), (200)), we define the "usual" surface operator (which depends on Z and k): Definition 3. The operator C (or C) defined by (87), or (89), or (103) is called the Calderon operator (or the surface admittance operator) for the half space (andfor Z, k, and the chosen sense of propagation). As a consequence of above properties, we have: Proposition 6. The Calderon operator C (resp. C) is an isomorphism in Hk(R2) (resp. SHk(R2)). Both commute with the semigroup (G(x3)), satisjy (8S), and (110)
Re (CSE£,E£) < 0, V E^ € Hk(R2).
This inequality is a consequence of the formula:
(111)
/R2nAH$J$dx=/R2£[-£|UE^§M^
which is easily deduced from (103), and the definition (11). Remark 5. Similar to waveguides the electromagnetic field is decomposed into transverse-electric waves and transverse-magnetic waves, thanks to the Hodge decomposition; using the decomposition Hk(R2) = Hk#g(R2)©Hkc(R2), with (112)
Hkg(R2) = {u€Hk(R2), curlu = 0}, H^R 2 ) = {u€Hk(R2), divu = 0},
we have, with hypothesis (106):
205
5.1 PLANE GEOMETRY E ^ . € H k c ( R 2 ) ^ E2 = 0 ~ H^.€H M (R 2 )forTEwaves
(113) E
T € Hk,g
H
T € H k,c( R2 ) for ™
waves
»
and thus the Calderon operator C is an isomorphism from H k C(R ) (resp. H^ (R )) onto H k g(R2) (resp. H^C(R2)) given by OforTEwaves: H £ = C E £ = J = ^ S A E J (114) ii)forTMwaves: HJ = CEJ^S
A
~* ET-
Note that there are no transverse electromagnetic (TEM) waves since the whole space is simply connected. Remark 6. Calderon projectors for the half-space We can easily verify that the Calderon operator for the inferior half space C. (or C_, for the same medium, the same angular frequency) is given by (115)
q*-C(orC|-~C).
Then we consider the usual transmission problem (or with given electric and magnetic currents, see (86) chap.3): find (E,H) with locally finite energy in each half space (up to the boundary) so that i) curl H + icoe E = 0, ii)-curlE + i
nAH|r =C(nAE|r),
n A H | r =C(nAE| r ).
Thus we get a linear system of equations which is very easy to solve explicitly, using (88) and (115). We can write the solution in the usual form
(,W with (119)
f n A E | > j
(M^
-1-"
P . = | ( I + S),
f n AE|
^
f MA
L—-'riJ-'-L ,J-
206
5 UNBOUNDED OBSTACLES
where S is given thanks to the Calderon operator C = Z" *C£ by:
«
s
-(c - ^ H z V i }
[Of course we can also use other forms, for instance:
L"AH|rJ *l J ' l"H'r.J "I J with: (119)'
P + = i ( I - S ) , P_=±(I + S), andS = -S.]
The operators P+ and P . are the Calderon projectors in Hk(R ) x Hk(R2) for the half space (with c, p, O>). Of course we also have S2 = I. 1.2.2. Scattering problems with two different media As in section 1.1.3, we assume that the whole space is occupied by two different media on each side of the plane T (x3 = 0), one with permittivity and permeability (Cjjjij) in the domain R3+ , the other with permittivity and permeability (E 2 ,IO in the domain R 3 _. We assume that there is an incident electromagnetic field (EpHj) in R3+, at angular frequency a>, for instance produced by charges and currents with compact support in R . We have to find the reflected and transmitted (or refracted) electromagnetic fields, resp. (E , H ) in R3 and (E t ,H t )inR 3 _. Let (E,H) be the electromagnetic field defined by: (E,H) » (Ef,Hr) in R3+ and (E t ,H t ) in R 3 _. Let (e,*i) be (t^)
in R3+, (e2,y2) in R 3 _. We first assume that
(£,,Pj), (*2»P2) are positive real numbers (the two media are conservative). Then (E, H) has to satisfy (116), with M and J given by: (121)
M = nAE,| r ,
J= -nAH,|r
Thanks to the Calderon operators in each half-space, Ct and C = - C2., we are led to find the boundary values of the field, which satisfy (116)iii), or also (122)
Hj2 - H j | » H j | ,
E*j<2 - EJJ = EJJ .
Applying the adjoint Riesz operators (see (96)') to (122), then using (101), we obtain two simple scalar systems. We define: (123) u f - R ^ E q J - l , 2, uf = RjE TI , u? = R*ETj, j = 1,2, uf-R^En,
207
5.1 PLANE GEOMETRY
and thus the Fourier transformation of (101) gives, keeping (e,|i) of the media: R
(124)
gHTl=Sirei^l»
R
g H T 2 = "®Jir82fi2'
R
g H f I = -Bpr6ifif
| R J ftT1 = iwEj 0]"1 flf, R % ft^ s - i(oe2 flj1 flf. R S H j , = - I**! S^1 fl f. We have to solve: (125)
k 0 2 lfi 2+ £ i°r lfl ? s£ i e r lfi ?
g
g Ti u2
A 8
g
ft i, - ui i i -=u
and Ifl
AC,
1 fl A C . 1 fl AC
p^«2 u 2 + f i J 0 I u l = p ^ ^ l u I
(125)
The solution of the system (125)c is given by: ft c - T ft c
c
ft
- P ii c
(126) ^P^+11,82*
c
"^ei+>1i02
Thus for an incident field such that: (127)
E I T and H IT € H k (R2), £
1/2
2
we have u, € H ^ (R ), and from Proposition 4, we get (128)
uf e H*k/2(R2) and u | e
H ^ R V H ' A R
Then we define (129)
Vj-af'flf.j-i.i
^-e^flf,
so that (125) becomes: g
(130)
|i)e 2 v 2 + €,v 1 = e1Vj |ii) 8 2 ^ - 0 , ^ = 8 ^ , .
The solution of the system (130) is given by v, = T v , , (131) T=
v.-Rv,,
2E,9
l°l
«1«2 + «2V
e,92 + e29,
2
)
(and even u£ e H 1/2 (R 2 )).
208
5 UNBOUNDED OBSTACLES
With the same hypothesis (127), we have Vj € H*k/2(R2) and from Proposition 4: (132) V! €H 1 k / 2 (R 2 )andv 2 €Hj c / 2 (R 2 )nHj/ 2 (R 2 )(andevenv 2 €H w Thus we have: (133) and
ufeH-^R2), u
2
s I
uf€H^ /2 (R 2 ),
guI»
u
l
s R
|
U
I»
(134)
$ ?2£ tefil £ £ £182 * c 2 e i +€ e g=S £ 0 +€ e ^Oj ei9 2 2 r l 2 2i ' (126) and (134) are Fresnel's formulas for electric field. Note that
(135)
Tg-Rg=l,
Tc-Rc=l.
Thus we get Theorem 3 . With hypothesis (127), the scattering problem (116) of an incident wave (E^Hj) by a (conservative) medium in a half-space (with boundary T) has a unique solution (E,H) of locally finite energy up to the boundary (on each side of T), which is obtained thanks to two semigroups (G(x 3 )) x 3 >0, and (G 2 (-x 3 )) x3 < 0 (see (85) and (29)) associated to the wavenumbers kIf k2 in each medium, and thanks to the transverse components of the field on T, so that: (136)
E ^ | and Hjj | r € H* (R2), j ■ 1,2 on each side of the plane T.
Moreover the transverse components of the solution on T split into: Ejj = E jj + E j j , Hjj = Hjj + Hjj (137) |E T j = R g R g E T j,
E^SR^EJJ,
H^sRgRgHjj, HTj = RcRcHjj,
with(R g E T j , R g H T j ) , (R;E T j , RCHV)given
by(123), (126)and(134).
Remark 7. The decomposition (137) of the transverse field also corresponds to the decomposition of the total field (E,H) into TE-waves and TM-waves. When E T » E£ on each side off, then div E ^ = 0, thus E 3 j = 0 on T and in R 3 . Thus H3j = | cop. J - curl E ^ satisfies, with u f given by (126) with respect to u f: <138>
ft
3nlr-Bfe:l«l«f
•
thusH3j|r€Xk..
5.1 PLANE GEOMETRY
209
When Ej = E | on each side of r, then curl ETj = 0, thus H3j = 0 on T and in R Pj satisfies, satisi Thus E3j = - Aj"1 div E^ with u * given by (134) with respect to u f: (139)
1
E 3j | r = m e rf f iAf *,
thusE^^cX,..
We call Tg, Tc, and R , Rc the transmission and the reflection coefficients of the electric field. We also define the transmission and reflection coefficients for the magneticfieldTp Tc, andR^, R; which satisfy (135) also: let (123)'
vf = R;HTj,j = l,2, v f ^ B n , V? = R;H TJ ,j= 1,2, vf-R^l,,.
Then we get Fresnel's formulas for magnetic field: ♦ S - * c H v? = R ; v f > v f = f ^ v f , O f s R ; v f ,
with:
(140) A
A £ ^ 8 I A
A P I " ?
T> _ -A—L x
T> — *—- T
A
A
D'__D
A
A
D ' — — 1?
Remark 8. We can extend these results and Theorem 3 to the case where the second medium is dissipative (with complex e, \i)9 and also where the two media are dissipative. The main point is to prove that the terms: e,92+ £29, and ^1^2+ ^2®i a r e n e v e r nu^- TWS *s t*le c a s e f° r instance when the first medium is conservative, and when Re E2, Re p2 (with Im E2, Im \i2) are all positive numbers. Furthermore for a dissipative medium, the wavenumber k has an imaginary part k2" * 0; This implies that the wave is exponentially decreasing from the boundary. We define the skin depth 6 as the distance in which the wave is attenuated by 1/e,. that is using the notations of (45): (141)
|ua,5)|/|fl«,0)|=exp(-(Ree 2 )6)
VUR2.
Since Re 92 > Im k2 = k2, we can take 6 = l/k2. For a conductor with y2 = ]xQ, we have e2 = c2 + (io/
5 UNBOUNDED OBSTACLES
210
The case of a "thin perfectly conducting screen11, that is a surface T (a part of a plane) on which we take n A E = 0 is reduced to the scalar case, since the electromagnetic field (E,H) also splits into TE and TM waves. 1.3. The slab 1.3.1. The scalar case with Helmholtz equation First we consider the Helmholtz problem in a slab of thickness T and with a real wavenumber k, for the Dirichlet condition: (141)'
OAu + l^usO inQ t £R 3 ,withO
U(.,T) = UT,
with given uQ, u f . It is natural to solve (141) using a Fourier transformation with respect to x = (x^x^ (if uQ, u t are tempered distributions). With usual notations we get i ) ~ ^ + (k 2 -e 2 )fl = 0,
0
(142) |ii)u(.,0) = u0,
U(.,T) = UT.
We note t = x3. Then (at least formally) the solution of (142) is given by
sh((x-t)4?^?) * „ s h ^ - k 2 )
i)u«,t) = u 0 «)(143) ii) u(E,t) = u 0 «)
sh(T^-k2)
sin((x-t)4 2 -£ 2 ) 2
2
sin(t4 -5 )
'
sh(t^2-k2)
when |E| > k ,
A/,xsin(tJ?^?) when | ( |
sin(t4l?-?)
or using 8 (£) defined by (11) t\Aw (143)
att) *\= Aflsh(x-t)6 fl(., 0__r- +
A
flsht8 Tii-g.
Let XQ and xt be functions defined by (144)
* " » - * & * •
< * » ■ & ■
When \i\ >k, x 0 «,t)< 1, x t (U)^ 1. But when |£|
k„ = Jk 2 - (nii/x)2- k J1 - (nA/2t)2, ^ ^ ^
ST= U Sk , 0
211
5.1 PLANE GEOMETRY
with Sj, the sphere of radius k^ the values a* with an = roi/x, n € N, n > 0 being the eigenvalues of - d /dt 2 on (0,x), with Dirichlet condition. Thus we have to eliminate the "irregular values" £ on these spheres, i.e., (€S T . Recall that the (generalized) eigenvectors (or eigenmodes of the slab) are: v M (x,t) = e Ux sin(tnn/x), with£2 + (nn/x)2 = k2, n € N, n>0. When k2 - £2 is equal to a 2 , the solution of (142) must satisfy the condition (146)
f0 (u"v -uv") dt = [u'v - uv']* = u(0)v'(0) - U(T)V'(T) = 0,
with v(t) = sin (nnt/x), and thus: (147) ( - l ) n u T - u 0 = 0. But since £ is a continuous parameter, we don't have to impose condition (147) in this form. We have to give a sense to (143) only. Thus when | £| < k we write: Hi 5 u 0 (E)cos(T^k 2 -£ 2 )-u T (£) r-y 5 2 2 (148) fi(£,t) = fl0«) cos (t^|k - £ ) - - ^ %-— — sin (t^k2 - t 2 ). sin (T ,|k 2 -£ 2 ) Thus we see that the hypotheses u 0 , uT € L (R ), and: (149)
r n
'(-1)nfio^>|2dt<
2
2
with Vn a neighbourhood of the sphere Sj, , will imply that u(., t) € L (R ) , 0 < t < x. We can also take space H £ / 2 ( R 2 ) instead of L2(R2) with (149). We note that: i) if the slab is thin enough, that is if x < X/2, problem (141) has a unique solution (for instance in L2(R2 x (0,x)) when uQ and UT are in L2(R2)); ii) if the wavenumber k is complex, with Im k > 0, we have the same conclusion. Then we can get the values of the normal derivative of u at the boundary of the slab (at least formally) thanks to (143)'. Their Fourier transforms are in matrix form:
d50) (150)
r fi,( -' T) i =c cr fl( - T) i U'(.,o)J- *U(.,o)J'
with
(15D (i5i)
£__flLfchTe <^-shTflL i A
We easily see that CT satisfies:
-n -cfnej-
212 (152)
5 UNBOUNDED OBSTACLES (CT)2 = 92I,
and thus C t has (at least formally) the inverse: (CT)-1 = 8"2 C t . The operator C t defined by (ISO) is the Calderon operator of the slab for k. For other developments on the slab, and for guided modes of the slab, see Petit [2]. It is interesting to consider the Cauchy problem (with x } = t), also (for instance in order to treat problems with many different slabs): (153)
» * O^f+AjU + lAisO inQ t cR 3 , withO
ii)u(.,0) = u0, §£(.,0) = u, with u0, U] given (at least temperate distributions). Using as above a Fourier transformation, we obtain the solution in the form:
For every given 8, (Ge(t)) (155) with
is a semigroup, which splits into:
G0(t) = e0tPt + e-0tP0_,
p»=|d+se), Pe_4(i-se), s e = [ o ^ J , that is, P+ and P_ are the Calderon projectors, since: s] = I, thus(P°)2 = P0, (PeJ2 = P9_, ?lP[ = ?lP{ = 0. By inverse Fourier transform, (GM) does not correspond to a semigroup in the usual L2 functional spaces, in other words: the Cauchy problem (153) is an illposed problem. But (153) is helpful because: i) we know that there is at most uniqueness of the solution of (153), ii) very often in physics, we work with given functions (or measures) uQ, u .whose Fourier transforms have bounded support, and then the inverse Fourier transform of (154) is well-defined. (Note that when the supports of the Fourier transforms of uQ and \x{ are contained in the ball of radius k, the semigroup (Ge(t)) is that of the wave evolution.) Thus if we have a heterogeneous slab composed of n homogeneous slabs with thickness T. and wavenumber k. , the values of the wave at the end of the heterogeneous slab is obtained in function of the values at the beginning by: (156>
( fiS ] =G'fO...G9fa) [ Jj ] , withx- 2xj.
213
5.1 PLANE GEOMETRY
Thus we easily obtain the solution of the scattering problem of a plane wave by a multilayer slab. 1.3.2. The slab with Maxwell equations Now we consider the Maxwell problem in a slab (with real £,p): I i)curlH + i
ii)-curlE + ia>jiH = 0 , inQ T ,
0
Eji.,1)» E\ with given E^, E j .
Since the components of (E,H) satisfy the Helmholtz equation, we obtain the transverse electric field Ej. from the above section. Thus we have to find the other components of (E,H) from E T . This is very easy from the usual decomposition into TE and TM waves, as follows: i) for TE waves, due to (30)ii) chap. 4, we have E^ = 0,
9 3 E T + i
and thus using the Calderon operator CT given by (151): (158)
H-p = — Qjp- C^SE^;
ii) for TM waves, from (30)ii) chap. 4, we have H 3 = 0, and thus: (159)
a 3 H T -ia>£SE T = 0,
H T = ia)ECT"lSET.
These two results are given by: (160)
HT = ^ h i c T P g + k C ; 1 P J S E T ,
which is relation (103) with CT in place of A, where E T and H T denote the couples: E T = ( E T ( . , 0 ) , ET(.,T)),
H T = (HT<.,0), HT<.,T)).
Furthermore the components E 3 , H 3 are obtained thanks to (91), by: (161)
H
3=15^ c u r l T E T»
E3 = - C ; 1 d i v T E T .
There are no new difficulties with respect to the scalar case: we only have to eliminate the "irregular values" of ( corresponding to the generalized eigenvectors of the slab with Dirichlet condition, thanks to a weight, like in (149). In order to treat multilayer problem, we consider the Cauchy problem:
214
5 UNBOUNDED OBSTACLES i)curl H + itoeE = 0,
ii) - curlE + icoy H = 0, inQ,, 0 < X 3 < T
(162) iii)ET<.,0) = ElJ., HT<.,0) = HT, with given E £ , H £ . This is also (at least formally) easily solved due to the decomposition into TE and TM waves, using the projectors P and P (see (100)) or the Riesz operators R and R . We first write Maxwell equations in the form (30), (31), then (34) cnap. 4, ue., (163)
8 3 H T = 5 |r( k 2 + 8 r a d d i v ) S E T»
93%—Ad^+graddWSHr.
Then applying Pg and Pc to (163), we have: (164) 3 3 P g H T =4 r (k 2 + A)PgSET, (165)
93PcHT = i»e P^Ep,
33PgET= - 4 (k2 + A)PgSHT,
SjPgEr = - iuji P£HT,
Then using the commutation relations: (166)
SPc = PgS,
P^-SPg,
and using relations (154) with u = PgHT, then PCET, with (164) and (165), gives us: (167)
f
Pc£T
=
ch (9T)
-'w*B~l sh (9T) S ^ f p c ET ^
^ -5ir8sh(et)S
ch(et
(i68) r P^T(T> 1 = r i
^ PCHT
f*F
W
l.
L *¥*) J
9V
< 0 , ,
ch«h) J (^ Pg ft° J
4esh(9t)s
I f p^ ].
ch (&r) J ^
P<;ftT
J
|.
\ H^O) J
Then we have the same conclusion as in the Helmholtz case. The formulas (167), (168) (or (169)) allow us to treat very easily the multilayer case like in (156). There are many applications of these formulas to scattering (Salisbury screen, Jaumann screen). 2 . PERIODIC GEOMETRY, 2-D GRATINGS
Periodic structures appear in numerous applications, especially in scattering by gratings (see for instance Petit [1]). As usual we first develop the scalar case, then electromagnetism.
215
5.2 PERIODIC GEOMETRY 2.1. Periodic geometry with Helmholtz equation We first define the notion of a quasiperiodic function: Definition 4. Let K = (Kj,K 2 )
wit
h K p K2 real numbers and L = (L p L 2 ) with Lx
and L 2 positive real numbers, be given. Let u be a function (or a distribution) in R 2 which satisfies "Bloch condition1': (170)
u(Xj + L p x 2 + l ^ ) » exp ifl^L, + K2L2) u(x p x 2 ), x = (xvx2) € R2,
(170)'
u(x + L) = exp (iK.L) u(x).
Then u is called a quasiperiodic function (or distribution) of type K-L, or a K-L (quasiperiodic) function (or distribution). When K = (0,0) or when K.L € 27iN, we obtain the L-periodic functions (distributions). Conversely given a quasiperiodic function u we obtain a periodic function v by: (171)
v(x) = exp(-iK.x)u(x).
This allows us to obtain the main properties of the quasiperiodic functions from those of the periodic functions. Let P denote the "elementary cell", that is the rectangle P = [0,Lj]x[0,L 2 ] with identified opposite faces; P is identified with the 2-D torus T 2 . Let IPI denote its area, IPI = Ljl^. We also define: Definition 5. The space of K-L "regular11 (C°°) quasiperiodic functions in R 2 (with the topology of uniform convergence of functions and all their derivatives) is denoted by DKL(R2)
(or DKL(?));
then its dual space, i.e.,
the space of
2
quasiperiodic distributions is denoted by D'K L(R ), and the space of quasiperiodic functions in R2 which are square integrable on the period is denoted by L2K L(R2) or simply L 2 .. We also denote by H* _ (R2) or H* . the Sobolev space of order s {real) of quasiperiodic functions (distributions), i.e., for s = man integer:
(172)
H™ L (R 2 ) = {u € 4 L ( R 2 ) , | J € 4 L ( R 2 ) , V J =
An orthonormal basis (up to a constant) of L2K is given by the functions: (173)
0J(x) = exp(iK.x)exp2ni( I i-x 1 +j^x 2 ),
withJ = ( J 1 , J 2 ) e Z 2 .
216
5 UNBOUNDED OBSTACLES
We also denote: (173)'
♦J(x)=expitJx,with£J = K+65 and t j ^ n ^ , A
J = (J,,J 2 )eZ 2 .
By expansion of quasiperiodic distributions on this basis, we have: (172)'
HsKL(R2) = {v€2?'KfL(R2), v = 2 v^j, 2 |K + ^| 2 s |v J | 2 < oo}, S €R.
We consider the standard problem in the half-space: find u satisfying (174)
OAu + k^OinR*, ii) u | _ = u0 given on the boundary r of R*,
with k a given wavenumber (k € R*) and u0 a K-L quasiperiodic function. The Helmholtz equation and the domain being invariant by (transverse) translations, we look for K-L quasiperiodic solutions. Thus we develop the unknown function u (and the given function uQ) on the above basis: (175)
u(x,x3) = 2 uj(x3) ♦jW,
u0 = 2 u0J Oj,
and problem (174) is reduced tofinding(Uj(x3)) satisfying (with Sj = K + £j): (176)
i ) ^ + (k 2 -£ 2 )u j3S 0, x 3 > 0 , *3
|u)uj(0) = uOJ.
When |£j| > k, there is only one bounded solution of (176): (177)
uj(x3) = u0J exp (- 9jx3),
with 9j =flj= «J - k2)1/2.
When | Ej | < k, the general solution of (176) is (with <*j + Pj = u0J): (178)
Uj(x3) = aj exp (- i9Jx3) + ^ exp (i9Jx3), with 9J - (k2 - l])U1.
Thus we cannot determine in a unique way the solution of (174), which is an illposed problem. In order to obtain a well-posed problem, we have to impose a sense to the wave propagation, like for waveguides, such as (179) u is a wave propagating towards the positive axis xy i.e., the propagating part ofu (178) is a superposition of waves Uj(x3»J(x) = e',j30J(x) withfij>0. Then taking:
5.2 PERIODIC GEOMETRY (180)
217
8j = « 2 - k2)172 when S2 > k2, 9j = - Kk2 - tym when tj ^ k2,
we obtain a unique solution of (174) with (178) by the Rayleigh series: (181)
u(x,x3) = 2 u0J e " 0 J \ ( x )
with J € Z2.
We have a situation quite similar to that of the waveguide, replacing the boundary condition on the waveguide by quasiperiodic conditions. Thus we have due to the Fourier expansion, or to the semigroup method Theorem 4. Problem (174) with a given quasiperiodic function uQ in L2K L(R2) and with condition (179) has a unique solution u in C°([0,oo),L2KL(R2)), given by (181) with a contraction holomorphicsemigroup (G(x3)), x3 > 0, in L2K L(R2). The infinitesimal generator A of the semigroup in L K L(R ) is defined by (182)
3u Au0 = |y-(.,0), EKA) = {u 0 €4 L (R 2 ), u€C\[09oo),L2KtL(R2))},
and using the spectral decomposition of the quasiperiodic Laplacian: (182)'
Au0 = - 2 6j uOJ0j, with u0 = 2 UQJOJ,
D(A) = KlK L(R2).
Definition 6. The infinitesimal generator A of the semigroup (G(x3)) is called the Calderon operator for the quasiperiodicfunctions and the wavenumber k. This operator has the same properties as that of the waveguides; A is a normal operator, a square root of the self adjoint operator - (A + lrl) in L2K L(R2). Let: (183)
U£ = {J€Z 2 sothat |£j|
and let: (183)' H£, H£, H£ the Hilbert spaces generated by *j , J € U£, U£, u£(resp.). Thus: ker A = H k , and Im A is its orthogonal space. Furthermore A satisfies the following inequalities for all v in D(A): (184)
I Re (Av,v) = - | P| 2 6j IVj|2 < 0, (the sum being on J in U£, 9j = 0j), Im (Av,v) = | P j 2 &j I Vj|2 > 0, (the sum being on J in U£, OJ = 18j| ).
218
5 UNBOUNDED OBSTACLES
According to (183), we have the decomposition of L^ L(R2) and of A into: (185)
L| L (R 2 ) = /4©H^,with^=H^©Hj, andA = - A + + iA~,
with A+ and A* positive selfadjoint operators in the orthogonal spaces H^ and H£, where H£ is of finite dimension and corresponds to the propagating modes. Remark 11. The Lg L(R ) framework is not "optimal" for all applications we have in view. More generally Theorem 4 is true in the quasiperiodic Sobolev spaces fig L(R2), with real s, and in this framework the Calderon operator A is continuous from Hf^R 2 ) into H^ L(R2). The case s m -1/2 corresponds to waves u with locally finite energy. Example 1. Scattering of a plane wave by aperiodic soft wall. Let r be a (regular) periodic surface in R (with period L = (Lj,L2)) which is a modelling of a periodic soft wall (or a 2D-grating) with free space on one side. We assume that T is contained in the half-space x3 < 0. Let Uj be an incident plane wave in the free space Q given by (186)
u,(X) = uI(x,x3) = u I 0 e t o X = u I0 e^ aTX ^ 3 * X3) , a€S 2 , a 3 <0.
Therefore u. satisfies the K-L quasiperiodic condition with K^koj, (thus | K | < k) and Uj(. ,0) = U^AQ (with (173)). We have to find the wave uf reflected by this structure. This is a K-L quasiperiodic function satisfying the following on the plane TQ (x3 » 0): (187)
9ur "gjf-Aur = 0 onT0.
The Calderon operator A reduces the problem in the infinite domain O to the bounded domain C1Q with disjoint boundaries TQ and T and L-periodic with respect to x. Let u = Uj + uf be the total wave. The transmission conditions on TQ are: »„ 3uP 3U| <187>' W3'W3+W3> «=«r+uionr 0 . Thus u has to satisfy on the boundary r0: (187)"
J|- - Au = fIf
with f, = ^ - Au,.
Therefore we have to find the total wave u, which is a K-L quasiperiodic function in QQ with finite energy (in O Q ) satisfying OAu+l^usOinQo, (188) ii) u | = 0, and u satisfies (187)" on T0.
5.2 PERIODIC GEOMETRY
219
We can write this problem in a variational framework. Let V be defined by: (188)* VsfueH^Qo), u l r = 0» u(.,x3) is a K-Lquasiperiodic function in QQ VX 3 ) Then let a(u, v) be the sesquilinear form on V: (189)
a(u,v)=/
(grad u.grad v - k2u v) dx - / r Au. v dT0.
The problem (188) amounts to finding u in V satisfying (for all v in V) (190)
a(u,v)=/ r f,vdT0.
Since the inclusion of V in L2(Q0) is compact, this problem depends on the Fredholm alternative. But we prove below (with fairly general hypotheses) that there is at most one solution, which implies its existence. ® Remark 12. We can treat more general scattering problems, for instance with a "hard" wall (with the Neumann condition) or a wall coated with some medium which may be heterogeneous (but periodic). If there are dissipative inclusions in this medium, that is domains where the wavenumber is not real, then these problems depend on the Fredholm alternative, and we can prove that there is at most one solution, thus we have a unique solution also. Example 2. Scattering by a slab with periodic inclusions, or by a grid We consider a system of periodic inclusions in a slab placed in free space, and with an incident plane wave arriving from one side. The inclusions can be hard, or soft, or made of a dissipative material. We can consider the case of a grid with thickness d, or the scattering by a dielectric medium also. We have to solve the standard problem in a (fictitious) slab: i) Au + k 2u = 0
(191)
in QT = R2 x (-T,0), or P X (-T,0),
i i ) | | - A u = fI onr0, f, given by (187)", t=x 3 iii)^«f AjU = 0 onTjsr^, iv) u is a K-L quasiperiodic function with finite energy in P x (- T, 0),
where T0, Tx are the boundaries of Qt; Aj is the Calderon operator for the domain with x3 < - T where the wavenumber is kj (here kj is a real number). In the simple dielectric grating case, we have (192) Q ^ ^ u f ^ , withiUkinty)* k^kjinQj. Like in Example 1, we can write this problem in a variational framework. We define (with usual notations) the space on the periodic domain Q :
220
5 UNBOUNDED OBSTACLES
(193)
V « HkL
and the sesquilinear form a(u, v) on V: (194) a(u,v)*/ (gradu.gradv-^u.^dx./j. Au.vdT 0 -/ r A^.vdTj. (If we have hard inclusions occupying a domain O in QQ, we replace V by the subspace of functions which are null in O). Then problem (191) is written: (195)
a(u,v)=f f^dTo,
VvcV.
Since the natural inclusion of V in L 2 (Q T ) is compact, this problem depends on the Fredholm alternative also. Under supplementary hypotheses, we can prove that there is at most one solution, which implies its existence. Proposition 7. We assume that one of the following hypotheses is satisfied: i) the grating is thin: the depth d(T is contained in a slab Q d of depth d) is small with respect to the wavelength X (d < A/4); ii) the grating is given by a Lipschitz periodic function x3 » g(x), except on vertical parts; Hi) the grating is coated by a (heterogeneous) medium with a dissipative part. Then the scattering problems (188) and (191) have a unique solution u with locally finite energy up to the boundary. Thanks to the variational method, we can prove that problems (190) and (195) with f» = 0 implies u = 0. Here we only give the proof in two cases. PROOF FOR (188), with hypothesis i). i) Taking first the real part of (190) with v s u gives, using (184), PROOF.
(196)
/ 0 (|gradu| 2 -k 2 |u| 2 )dxsRea(u,u) = 0.
Now assuming that T is above a plane T_d we can extend u by 0 up to r_d, and this extension u is in H^ L(Qd), Qd = Px (-d,0), thus: (197)
/
|gradu|2dx * ^ = inf/
|gradw|2dx,
the inf being taken on {w € HJ^ L(Qd), w=0 on r_d, llwil = l},and it is reached at the solution in this space, of |i)Aw+icfw=0 inQd, (198)
ii)w,
r_ d = °' £l ro -0-
Thus we have K{ £ n/2d. Since k = 2n/X, when d and X satisfy:
221
5.2 PERIODIC GEOMETRY (199)
K§-k 2 >(7t/2d) 2 -(27iA) 2 >0,
i.e., d
we have from (196): (200)
J
|u| 2 dx > J
|gradu|2dx > / ^ K 2 |u| 2 dx,
which implies u = 0 from (199). We note that this is a version of the Poincare inequality. PROOF for (188) with hypothesis ii). This is adapted from Cadilhac's proof (see Petit [1]). First taking the imaginary part of (190) with v = u gives (201)
u 0 1S f u| r €#£
(see (185), u is not propagating). Then applying the Green formula to u and v (another solution of the Helmholtz equation satisfying (201) also), we have (202)
/ no (Auv-uAv)dx = - / r | v d r + / r o ( ^ v - u f ) = 0.
But the integral on T0 is zero thanks to (201). Thus
(203)
/ r givdr=o.
Then taking v = ^ in (203), and since ^ = e3.grad u = n3 g~, with n3 > Owe obtain gj = 0 at last on a part of I\ and thus (with u = 0 on T) we have u = 0 on Q0. <$ Now given a regular periodic surface T in R3, we can define Sobolev spaces of quasiperiodic functions H5K L (0 for real s, like in (172) or (172)\ We take the situations of examples 1 and 2, but with inhomogeneous boundary conditions on obstacles. Let Q be a regular connected periodic domain with one of the following hypotheses, according to example 1 or 2: i) Q £ R x (-T,+OO), with boundary r Q R2 X (-T,0) (grating with one side) ii) Q' = K\h QR X (-T, 0); CT may be connected (grid) or not (inclusions). Proposition 8. We assume that one of the hypotheses of Proposition 7 is satisfied. Then the problem: find a K-L quasiperiodic function u with locallyfiniteenergy (up to the boundary) satisfying (204)
i) Au + k u = 0 in Q, with real k, il/2, ii)u| r = u0^ve/ii/iH^£(r) 3 Iiii) iii)uuisisaawave wavepropagating propagatingtox3>0 in R+, (also to x3 < 0 in R ),
has a unique solution. Furthermore we have
222
(205)
5 UNBOUNDED OBSTACLES
g|r€HiV&0.
This is a consequence of Proposition 7 and Theorem 4. Then Proposition 8 implies that the mapping A=A f defined by: (206)
UQ—1|| , with u the solution of (204),
is continuous from fl^ffF) into Hj£ ^(0Hie operator A of Definition 6 is a particular case of this mapping when Q is the half space. If T separate R3 into two connected components Qj and 0 2 (the grating case), we have to consider two operators Af+ and Af _. Definition 7. ffte operator Aj. defined by (206)frco/fetf dte Calderon operator of the 2-D periodic surface T. This operator has the following main property: (207) Im(A r v,v)£0, V v € H ^ ( r ) . The proof is an easy consequence of the property (184) for the half space, thanks to the Green formula. Furthermore (in the case of a grating for instance) if we define the sesquilinear form (208)
a^u0, v0) = - ± [(AfUQ, v0) - (u0, A^)]
on H^£(r),
which satisfies: a^UQ.UoJ-Im^^iio) > 0, we obtain due to the Green formula that (209)
aj
u. being the J component of the trace ofu on the plane I\ in the basis (173). Taking the conjugate of vQ in (209) leads to the reciprocity relations (see Cadilhac in Petit [1]). But note that if u is K-L quasiperiodic, its conjugate is not K-L quasiperiodic, but it is quasiperiodic for ( - K). Remark 13. In the case of 1-D gratings, that is if we assume that T is a periodic line in R2, as in Petit [1], all this is true under very little change: in particular we have to change double Fourier series with subscripts J in Z to simple Fourier series with subscripts n in Z. This corresponds to problems in R3 with cylindrical gratings (with axis x for instance) with given data which are independent of x2. But if the data depend on x2, without x periodicity, we have to make a Fourier transform with respect to x2 and we obtain a situation which is in-between the plane case and the periodic case.
223
5.2 PERIODIC GEOMETRY
We have to change the mathematical framework: we define spaces of K1-L1 quasiperiodic functions with respect to x(, using an orthonormal basis (up to a constant) ($n) of KJ-LJ quasiperiodic square integrable functions O^XjJsexpUjX!, with lx = K{ + In (n/Lj), n = Jj€Z, with a Fourier transformation with respect to x2 and then we have to substitute the following space for the Sobolev space (172)' with s = 1/2: (210)H$
L (R
2
) = {u = u(x1,x2)= 2 u „ ( x 2 ) V x l ) ' 2 / R | 8 n « ) | |fin(S)|2d£
where 9n «tf2 +t 2 - k2)1/2, £n = K{ + 2K (n/Lj), n € Z, and wherefln(£)is the Fourier transform of un(x2). We have a space in-between H^/2(R2) (see (37)) and H ^ L ( R 2 ) (see (172)'), which has the H ^ R 2 ) regularity. 2.2. An integral method for gratings Similar to bounded obstacle we can solve the scattering problems with a periodic obstacle from an integral method. Here we use the space D' (P x R) of distributions which are K-L quasiperiodic with respect to x. We win assume (with notations (173) and (183)) that K, L, k satisfy: (211)
|£j| * k , V J € Z 2 , i . e „ u £ = 0 , oralso: (Z^ + k2)u = 0 in P, (with u € L^ L(R2)) implies u = 0.
First we have to define an "elementary outgoing solution" of the Helmholtz equation in Z)' KL (PxR). Let 6^(x) be the Dirac (quasi) periodic distribution. We look forO*p)(x,x3) the "outgoing" quasiperiodic solution of: (212)
i) (A + k2) (P) (x,x3) = - 6(p) (x) «(x3), ii) <& p is a wave propagating from x3 = 0 towards x3 > 0 (resp. x3 < 0).
We develop
<*>(p)(x,x3) = | P | - 1 2 ^P)(x3>j(x),
J€Z 2 , | P | - L ^ ;
applying (212)i) to *., we see that * j (x3) must satisfy, with t = x3:
a2$(p) (214)
-g-f- - 02 * « = - 6(t),
with 0j2 = 52 - k2,
224
5 UNBOUNDED OBSTACLES
and thus it has jumps across t = x3 = 0: (P) »?* (215) [ * f ] 0 = 0, I-^-]0 = -l, with the outgoing condition. Since 6j * 0, Oj is given by: (216)
• ? ) ( x 3 ) - a j « P ( - e J lx3lX
with 8j given by (180). Thus * (p ' is given by its (double) Fourier series: (217)
*(p)(x,x3)= i P f 1 2 ^ e x p ^ O j |x 3 |)^(x).
We can split *
(218)
p
into two parts with J in U£ and J in U£:
• ( P ) . * ? ) + *J >) ,
the first term corresponding to the propagating modes, and the second to the diffusion (or evanescent modes). We easily see that: (219)
* (p) €L? oc (PxR), i.e.f • < p ) | p x h b J €L 2 (Px[a f b]) t
Va,b€R,
since 0(_p) is afinitesum of terms in Lj^P x R), thus *** is in Lj*0C(P x R), and
•? ) (x 3 )«$tt J ,x 3 ) f and thus *(p)(x,x3) = | P | - ! 2 *(£j,x3) ^(x),
A
with O the Fourier transform of O.
® oo
1
Remark 15. Let x be a smooth function (C ) with x(x,x3) = 1 in a ball Br of R , and x(x,x3) = 0 in the complementary of the ball B 2r , with 4r < min (LpL^). Then 4>x is a distribution with compact support in R3, and we can define its corresponding periodic distribution by translations (*x)^ = I Tj*X- We have defined a "parametrix" which is equal to exp(-iK.x)*^ up to a regular function. Thus 0 ( p ) has a 1/r singularity at the origin in PxR and it is C°° outside. #
5.2 PERIODIC GEOMETRY
225
Proposition 9. Let T be a regular (Lipschitz) 2-D periodic surface in R 3 which is contained in a (fictitious) slab of finite thickness d, R 2 x(-d,0). With hypothesis (211), the problem: find u a K-L quasiperiodic function with locallyfiniteenergy up to T, satisfying i)Au + k2u = - f inR 3 withf«p , 6 r + div(pn6r) (221) | w l t h « i v e n
[u]r = P ,
iglr-P'-
This proposition is similar to Proposition 3 chap. 3 for a bounded surface T (which is the boundary of a bounded domain). We have to note that the convolution product in (231) is that of D9 (P x R), that is with respect to x (for quasiperiodic distributions) and to x3; this makes sense since p and p' have compact supports with respect to x3. Then we can define single and double layer potentials like in (67) chap. 3 by (224)
L ^ V - * ® . ^ ) . i^P)P = <&(p)*div(pn6r),
with the "usual" properties (see (69) chap. 3). We also define the Calderon projectors P[ p) , P
(225)
Y(KP2 = < l(T) x Hj^ftO.
They will satisfy (72) chap. 3 with an operator S = S(p) so that S2 = I, given by the matrix (71) chap. 3 with four integral operators L(p), K(p), J (p) and R(p) which are obtained thanks to <&(p) by Fourier expansions. For their particular expressions in 1-D gratings, we refer to Maystre-Vincent and Petit in Petit [1]. This allows us to solve scattering problems with gratings like in chap. 3. Here we do not develop this topic. We only point out that the difficulties are quite similar to those of the bounded obstacles, in particular with respect to convergence of the Rayleigh series: for gratings we know that the solution of the usual scattering problem has a Rayleigh expansion (181), which is convergent in all (reasonable) senses, above 0 (and below - d also if we substitute + for- in (181)). For the questions of convergence of these expansions, see Cadihlac-Petit [1] (quite similar to those of chap. 3). #
226
5 UNBOUNDED OBSTACLES
Remark 16. Singular cases in periodic geometry. We often assume the wavenumber k, K and period L are such that (211) is satisfied, that is the kernel of the operator A is reduced to 0. Here we study what happens if it is not true. First we consider the Neumann problem in a half-space: find u with locally finite energy (up to the boundary) satisfying OAu + lAi^OinR*, (226)
iii) the propagation condition (178), with Uj given in ker A. It is an ill-posed problem, since the solution of (226) is a priori given by u(.,x3) = x3Uj + C, C constant, and is not bounded ! We have to eliminate this case, and thus to take: (227)
u 1 €(kerA) 1 =ImA, i.e.,/ p u 1 5 J dx=0,
VJ€lj£.
This condition written for the half-space allows us to obtain well-posed problems in singular cases. First in problems with given jumps (p,p') of u and its normal derivative (see (221)) on T = P x {0}, p' must also satisfy (227), and then the solution u of (221) is determined up to an element of ker A. As a consequence we have to change the definition of the Green function (212), since the Dirac (quasi)periodic distribution is not orthogonal to ker A. Instead of (212)i) we have: (228) (A + k2>D(P)(x,X3) = F, with F = - 5(p)(x) 8(x3) + \P\~l
2
♦J(0)*J(X)8(X 3 ).
J€U<>
Then O ^ must satisfy (224) for J out of u£, and: (229)
[^ p ) ] 0 = 0, [ - ^ ] 0 = 0,forJinu£,t = x3.
We can take *SP) = 0 for J in u£, so that * ( p ) , ^ - are orthogonal to ker A. Now if we go back to the jump problem (221) we have: i) in the simple case where r = P x {0}, we have a unique solution in the space(s) orthogonal to ker A, when the given data satisfy: (230)
(w>) € H ^ L X H K , L
wtth
/ p 5 J * = 0, /p*5j dx = 0, V J € u£.
ii) in the general case of a grating T, we can prove (using the Green formula successively in domains above and below the grating with u and +, then with x30j), that the orthogonality of p and p' to ker A must be replaced by the following conditions:
227
5.2 PERIODIC GEOMETRY
(231)
/r(p^J.p^i)dr=0,
/ r [ ( p ^ J . p ^ ) x 3 . p n 3 5 J ] d r = 0,
and then the boundary values of u and its normal derivative on each side of the grating will also satisfy these conditions (231). Thus the Calderon projectors P^ and P_ act in the subspace VK L defined by: VK L = {(p,P,)€Hjc^(r)xHK1/^(n, satisfying(231)}. 2.3. Periodic geometry with Maxwell equations First we consider like in section 1.2.1 (or in chap. 4 section 2) the following problem: find the electromagnetic field (E,H) in free space at angular frequency (o satisfying (232)
i) curl H + iwe E = 0 ii) - curl E + iwp H = 0 in the half-space R+, iiOE^.jOJsEj (ornAE = nAET)
with E° T a given K-L quasiperiodic transverse electric field. Similar to the Helmholtz case, using Fourier series, we see that (232) is an illposed problem (in K-L quasiperiodic spaces), and it becomes (generally) wellposed thanks to the usual physical assumption: (233)
(E, H) is a wave propagating towards the positive x3 axis.
Since each component of E and H satisfies the Helmholtz equation, they will be determined by the semigroup (G(x3)) of section 2.1 from their boundary values on the boundary T (x3 = 0) at least formally by: (234) ET
E^€HKL(P)2.
Since (G(x3)) is also a continuous semigroup in this space, this implies: Ej(x3) is in H^L(P)2, for all x3 > 0. Then (91) (for instance) and (92) imply: (236)
H° € L| L (P), H 3 (X 3 ) € l£ L (P), E°3 € H ^ P ) , and E3(x3) € H ^ P ) ,
thus (thanks to (30) chap.4): Hj(x3) and Hj € L£ L (P) 2 . We see that hypothesis (235) implies that the regularity of the electric field E(x3) is H1, whereas that of the magnetic field H(x3) is L2.
228
5 UNBOUNDED OBSTACLES
A less regular assumption is: (237) E^€HKL2(curl,P). This space corresponds to the "usual" regularity (see H.(R2) (104)). Using (double) Fourier series, it is defined by: (238)
H ^ c u i U P ) = {u = 2UJ0J , 2 j f j - (|uj| 2 + |5j A Uj|2) < oo},
whereas: HkL(P)2 = {u = 2 ^ j , 2 ( | $ J . U J | 2 + U J A U J | 2 > < °°lThen (237) implies for the other components: (239)
H^H^curl.P),
E^HJ^P),
H° € H J ^ P ) ,
and ET(x3), E3(x3), H1
E^€HKL(curl,P)
implies H T € H^L(curl,P); then E and H have the same regularity. 2.3.2. The Calderon operator We first assume that the operator A (see (182)) is invertible, i.e., (211) is satisfied. Then we have the usual relations between the transverse components of E and H given by (86) with A = A(p). Using Fourier series decomposition, we have (241)
H ^ ^ t f j t f j A E n ) + k 2 SB n ].
Definition 8. The mapping E°T — H°T (resp. SE°T — SH°T) defined by (86) (with invertible A given by (182)'), or in Fourier series by (241), is called the Calderon operator (or admittance operator) C or
(resp. C or C^)for the half space in
K-L quasiperiodic spaces (at angularfrequency o>). Proposition 10. The Calderon operator C (resp. C)isa continuous operator in: HK L(curl,P)(resp. H^/£(div,P)) and in H^ L(curl,P)(resp. H^ L(div,P)), s €R. Moreover it satisfies: (242) C 2 = - Z ~ 2 I , C 2 = - Z - 2 I , and C=-SCS, C = -SCS,
5.2 PERIODIC GEOMETRY
229
and also: (243) with:
Re(CSE^.,E^.) = Re(SCE!j.,ElJ.)<0,
(244) (CSE°,E°)=ZSH TJ .E TJ =X — j - ^ [ - e, ^ I ^ A E ^ 2 ^ ^.E^2] £iving only afinitesum (J € U£) in tfte real part (243). Furthermore from (241), we see that the Calderon operator C (or C) is naturally decomposed by Fourier series into C = ZCj (resp. C = I C}) with (see (89)):
and E^, H^ are given by (92), or with the Fourier series: (246)
E3j = i9j tj.ETJ,
^3J = £z^j A ^TJ,
and thus we can write formulas (94), (94)' between the div and curl:
Thus we obtain another expression of the Calderon operator (see (103)), corresponding to the Hodge decomposition of the transversal fields, or also of the total electromagnetic field into its TE (transverse electric, with E3 = 0) and TM (transverse magnetic, with H3 = 0) components, which is given with the Fourier series decomposition by: (248)HTJ = a j . H T J ) ^ +
ttJAHTJ)i=-^[^ajAETJ)fJ+^«J.ETJ)Sy
thanks to the relation Stj.u = tj A U , for all vectors u. This gives (244) directly. The Hodge decomposition of the space (237) into: (249)
HKL2(curl,P) = HKLtg(P)0HKLfC(P) (like in (112), %L,g( p )( res P%L,c( p )) = ( u€H KL 2 ( curl ' P )» curlu = 0, (resp. divu = 0)}
gives the TE and TM decomposition of (E,H) according to (113) and the decomposition of the Calderon operator C according to (114), given (from Fourier decomposition) by (250)
HT J = J ^ 0 J S E ^ J
for TE waves, H£J = - § OJ1 SE^forTM waves.
230
5 UNBOUNDED OBSTACLES
Now we assume that the operator A is not invertible. With notations (183), (185), we assume: i) E^j a grad ♦j, with ^ € H£ (thus A*j » 0, and H3 = 0). This implies Ejfa)« E^Jf and 3 3 E5 = - div ETJ » - Afy «lc2 * j , whose solution is the (unphysical) field E3(x3) = k20j x3 + constant. ii) ETJ = curl 4>j, withfy€ H k , thus div EJJ = 0, giving: E3 does not depend on x3 and E^ is not determined by E|J. (E^ is of the form: E^ » 2 Cj #j with +j € H£ and Cj constant). Furthermore we have ETJ(x3) = EJJ, and using (30) chap. 4, we have HT »jsp S grad E3 = - ^* curl E3 which is not determined by E£J. Thus in the two cases, we have an ill-posed problem. To eliminate these cases we assume that: (251)
ET is orthogonal to ker A x ker A.
With this assumption, we can develop the theory as in the regular case; for instance we can define the Calderon operator C in the space (and it is an isomorphism in this space): (252)
H£lL/2(curl,P)n (ker A ) 1 x(ker A ) 1 .
Remark 17. In these spaces of KL-quasiperiodic functions, we easily verify (for instance thanks to a Fourier expansion) that: (253) u € L^L(P), curl u » 0, div u « 0 implies u = 0, except in the trivial case where K.L/2n is an integer (then any constant vector u is a solution). This means that the cohomology space H (P) is reduced to {0} (thus there are no TEM waves), except for perioaic functions where dimiST (P)«2. 2.3.3. Some scattering problems Thanks to sections 2.3.1, 2.3.2, we can treat the scattering of an incident plane wave (EpHg) in free space by a P-periodic structure, like in Examples 1,2. From the geometrical point of view, we can tackle two types of scattering problems with a P-periodic structure: i) one-sided problems with a perfectly conducting grating, with a connected boundary T so that the domain of the scattering problem is on one side of this grating. The grating can be coated with some dielectric medium or not. ii) two-sided problems in one of the following cases: dielectric grating (in a domain which contains a half space), periodic dielectric medium contained in a slab, periodic dielectric (or perfectly conducting) inclusions or grid in a slab of finite thickness.
5.2 PERIODIC GEOMETRY
231
From the above theory, we reduce these scattering problems in an infinite domain to problems in afinitedomain with quasiperiodic conditions and with one or two Calderon operators at the boundary. Now from the mathematical point of view, for the solution of these problems we have the following cases: i) We are reduced to a problem in a bounded domain Cl occupied by a lossy medium, that is, with a permittivity c and a permeability ji, with B" = Im e > 0, ji" = Im ji > 0 (more generally the medium can be anisotropic, inhomogeneous). In this case we can apply a variational method in the usual space of electromagnetic fields withfiniteenergy in Cl, with a coercive sesquilinear form on a subspace V of the natural Hilbert space H (curl,&) of K-L quasiperiodic electricalfieldswithfiniteenergy. Thus from the Lax-Milgram lemma, there is a unique solution in Cl, and then for the scattering problem. ii) We are reduced to a problem in a bounded domain Cl occupied in part only by a lossy medium, so that we don't have a coercive sesquilinear form any more. But we can prove that the problem depends on the Fredholm alternative; this is generally due to the fact that the natural injection of V in L2KL(Q)3 is compact, but it is not true for H (curl,£2) and thus we have to substitute the Sobolev space of quasiperiodicfieldswith free divergence in H^Q) (see chap.2 section 11.2), to H (curl,&). Then we have one of the following situations: either we can prove that the problem has at most one solution: for instance this is the case where there is really a part with a lossy medium in Cl and a part with free space (or a conservative medium), and very likely in each of the cases of the Proposition 7). This implies (thanks to the Fredholm alternative) that there exists a unique solution in Cl and then in R3 to the scattering problem; or we cannot have uniqueness; the periodic structure has eigenmodes. From a numerical point of view, we often have to couple a spectral method (in order to take account of the Calderon operator) with afiniteelement method for instance to describe an inhomogeneous periodic medium, like in waveguides. But there are numerous other methods for gratings (see Petit [1]). We can also replace the boundary condition with the Calderon operator by "absorbing conditions", or we can use an integral method also (see Nedelec-Starling [1]) briefly described below. Example 3. "One sided problem". We assume that we have a lossy dielectric (with permittivity tx and permeability p.) in a P-periodic domain Cl contained in R2 x ( - T,0). This domain Cl is bounded on one side T by a perfectly conducting medium, on the other side I\ by free space. We will assume (in order to simplify) that T0 is the plane R2 x {0}. We have tofindthe electromagnetic field (E,H) in Cl due to the incidentfield(E„ FL) with angular frequency o>:
232
5 UNBOUNDED OBSTACLES ik(a T .x + a*x.*)
(254)
u,(x,x3) = uI0e
*
3
2
,u I 0 €C ,a = (aT,a3)€S , a 3 < 0 ,
u, * E, or Hj, E I0 s - jsz a A H IO , H,0 = ^ a A E IO , a.EI0 = 0, a.HI0 = 0 . Now we write the boundary conditions on rQ, like in the Helmholtz case, see (187) Example 1: the reflected field (Ef,Hf) satisfies thanks to C the Calderon operator of the half-space: (255)
nAH^CfciAEponlV
Then the electromagneticfield(E, H) in Cl satisfies the continuity relations on T0: (256)
nAHsnAHy + nAHp nAE = nAEr + nAEj.
Thus (E,H) must satisfy the boundary conditions on r0: (257)
nAH-C(nAE) = f,,
withf^nAfy-QnAE^onlV
Thus we have tofind(E,H) in Cl satisfying: I i) curl H + mtx E = 0, ii) - curl E + iaijij H = 0 in Cl, (258)
iii) n A E | r = 0, and (257) on T0. I iv) E and H € HKL(curl,0),
i.e., (E,H) is a K-L quasiperiodicfieldwithfiniteenergy; we save notation Clfor the elementary periodic domain. We can write this problem in a variational framework, with the sesquilinear form: (259) a(E,E)=/ n (- T±- curl E. curl E - icoe, E. E) dx - / r C (n A E|_ ). E dT0 in the Hilbert space (with the natural norm): (260)
V=VKL=:{E€HKL(curl,Q), withl| r = 0}.
Thus problem (258) is equivalent tofindingE in V so that: (261)
a(E,E)=/ r f,.EdT0,
VEeV.
Thanks to Proposition 10 on the Calderon operator C (that is (243) and that C is continuous in the natural trace space) and since the medium is lossy (thus with Im £j > 0, Im \i{ > 0), we obtain that the sesquilinear form a is a V-coercive form, i.e., there exists a constant CQ > 0 such that: (262)
Re a(E,E) > C 0 IIEIIy,
V E € V.
5.2 PERIODIC GEOMETRY
233
Thus we can apply the Lax-Milgram lemma: therefore problems (261), (258) and also the scattering problem, have unique solution (E,H). Now if we substitute free space for the lossy dielectric in Q, we obtain a problem which depends on the Fredholm alternative (in a subspace of the Sobolev space Hl(Q)3). Thus except for a countable set of singular values of o>, the scattering problem has one solution. This allows us to define a Calderon operator C f for this profile, with the properties of C for the half-space (Proposition 10). For numerical applications, an integral method is preferred. <8>
Example 4. "A two-sided problem". We consider the situation of Example 3 but the perfect conducting medium is replaced by free space (or by another dielectric medium). Now the scattering problem is in the whole space, with the electromagnetic field (E,H) propagating towards the negative axis x3 under Cl. Then we can reduce this problem to the domain £2 with the Calderon operator C ! r for the inferior domain (but generally we cannot use it for a numerical method), or to the domain Q = R 2 X ( - T , 0 ) with the Calderon operator C, relative to the half-space R 2 x ^ - < » , - T ) . In this case, we have to find (E,H) satisfying (258)i),ii),iv) and (257) and the condition (see (115)): (263)
nAH + CjfaAE^O onTx = R2x{-T}, withn = e3.
Now we define the sesquilinear form in V = HKL(curl,QT): a(E,E)- J ( - Xcurl E. curl E - me E. E) dx (264) ./roC(e3AE|r).Edr0-/riC1(e3AE|r).Edr1, with £ = 6j and \i = ]i{ in £t, t = c0, M = M0 in free space. This scattering problem is written in the variational form: find E in V such that (265)
a(E,E)=/ r ft.EdTo,
VEeV.
Since e and JJ in free space are positive real numbers, this new sesquilinear form is not V coercive. We will replace V by the following space (266)
WKL(QT) = {E€4L(QT)3, E^eH^Q) 3 ,
EI^HJU/Q')3,
[nAE] r = 0,
[n.E2E]r = 0},
where Q* is the complementary set of Q in QT. We note that the (transverse) traces of WKL(QT) on the boundaries T0 and T{ of Clx are in H^jfao)2 and H^CFi) 2 , with natural inclusions H^(r 0 ) 2 Q H j ^ d i v J o ) Q H ^ L V Q ) 2 and for Tx also.
234
J UNBOUNDED OBSTACLES
Thus the boundary terms in (264) which correspond to the Calderon operators C and Cj are continuous in this new space. Since the natural mapping of WKL(QT) in L2KL(ftx)3 is compact, the problem (265) depends on the Fredholm alternative in WRL(QJ. But uniqueness at most of the solution is an easy consequence of the dissipativity in the dielectric: M taking the real part of (265) with E = E when fj = 0 implies that E = 0 in Q, we have E a H * 0 in QT; the scattering problem (265) has a unique solution in WKL(QT).0 Remark 18. Integral method in periodic structures. We first consider, like in section 1.3.2. chap. 3, the "standard problem" with given electric and magnetic currents J p M- on a "regular11 periodic surface T (contained in R 2 X ( - T , 0 [ ) in free space:find(E,H) of locallyfiniteenergy up to T such that I i) curl H + ia>c E = Jr, Jr € HJ^2 (div,D, (267)
ii)-curiE + ia)|iH = M r inR 3 , Mr € HK L /2 (div,0; iii) E and H satisfy an "outgoing" wave condition above T (see (233)).
In order to simplify, we assume that the regularity condition (211) is satisfied (if not we have to assume that JL, ML satisfy supplementary conditions similar to (231) in the Helmholtz case, in order to satisfy the condition (251)). Then (267) has a solution (E,H) which has all the required properties and which is given by the convolution product (44) chap.3, but with the quasiperiodic elementary solution **p) given by (217) instead of * (and with the convolution product in D\ (P x R), which is allowed by thefinitethickness of the grating). Then (E,H) is obtained thanks to (quasiperiodic) electric and magnetic layers L*& and Pm*p) (like in (96) chap.3). By restriction to I\ on each side of T, we obtain the operators 1
Remark 19. ID-Gratings. Now we assume that T is a cylindrical periodic surface (with ^ axis for instance). If the incident plane wave (EJ,HJ) given by (254) is such that its propagation direction a is orthogonal to the x2 axis (i.e. a2 s 0), (EpH,) is independent of x^ then we can see: i) that the solution of the scattering problem is independent of x« also, ii) that this problem splits into two simple "scalar" problems (with Helmholtz equation) according to the polarization of the incident wave that is Ef parallel to x2 (P polarization) or orthogonal (S polarization), see for instance Petit [1]. We can easily apply the theory developed above for 2D-gratings (changing double Fourier series into the usual Fourier series).
5.3 CONICAL GEOMETRY
235
In more general cases, with an incident wave which depends on x~, it may be useful to work with a trace space on a plane (x3 = 0) which is in-between the plane situation (104) and the periodic situation (238) thanks to a Fourier expansion (with respect to xt) and a Fourier transformation. More precisely we can use definition (104), changing the Lebesgue measure dx » dfcjd^ into the sum of measures vn = 5 ^ - S^d^, and space H^/2(R2) into the space (210) in Remark 14. This (natural) trace space give the usual expected properties for an electromagnetic field with locally finite energy. Remark 20. On 2D-gratings with a small period. When K< | k | (which is the case for an incident plane wave), we always have the propagating mode J = (0,0). Since the propagating modes correspond to numbers J in Z so that | Sj| < k, there is no other such mode (with notations (173)') when |$J| > 2k, J*(0,0). But | $J| > 2tiL^!, with LM = max (L{,L2). Thus when 2nL^ > 2k, i.e., X = 2ir/k > 2LM the unique propagating mode is for J »(0,0). This simple remark, useful for applications with composite materials, could have been made before. 3. CONICAL GEOMETRY Here we consider stationary wave problems in domains Q in Rn such that Q =Q
r0,S = {x"(r>a)» r = M » a=x/r, r>r 0 , a € S }
with S a "regular" open subset of the sphere S n ~ l , and r0 € [0,00). Thus in polar coordinates these domains are of the form (rQ,oo)xS. We essentially work on the Helmholtz equation (this will reduce Maxwell problems to problems on the boundary). We use a method of separation of variables, which leads us to take into account Sommerfeld conditions, in a somewhat different way from the case of bounded obstacles. It will be interesting to compare the results of this powerful method with the usual "limiting absorption principle" (for which we must have sources of compact support) and to the other geometries. The fact that we have wellposed problems in the present framework implies the limiting absorption principle! First we consider the Helmholtz equation in a domain Cl of the form (0,00) x S: (268)
A v + k 2 v = - g inO,
with given g in Q, with a real wavenumber k, and with boundary conditions. Using polar coordinates, (268) is:
236 (269)
5 UNBOUNDED OBSTACLES r ^ ^ ^ r ^ - ^ + ^ A . V + ^ V ^ - G inR+xS,
with Aa the Laplace-Beltrami operator on S, and V(r,a)« v(x), G(r,a)» g(x). Now if we multiply (269) by r2, and we change variables and the unknown function (270)
p«kr, u(p,a) = r(n-3)/2V(r,a),
flpfa)-i^1,/20(rfa)f
we obtain the equation: (271)
^ u - B u = f inR+xS,
with A and B given by (keeping notation r for p from now on): (272)
i t u a - d 2 ! ! ? - ! 2 ! ! - ! ! / * and £ « Aa - ((n - 2V2)2I.
3.1. Properties of some unbounded operators associated to A First we study the main properties of the operator A. Obviously the main difficulties are at infinity and at 0 since A is a "degenerated" operator at 0 (that is the coefficient of the derivative of higher degree is zero at the boundary) with weight r2 at infinity. We choose the framework L2(R+). This choice is not at all obvious since, through the change (270), it does not correspond to the usual L2(Q) space but to the space L^C^dx/r2). But the properties obtained in such a space justify its use. We first define unbounded operators in L2(R+), with different domains associated to the differential operator/*. The first properties of these operators will be given with the usual notions on unbounded operators, see for instance Richtmyer [1], Dunford-Schwartz [1]. We define the "minimal" and "maximal" realizations of A, denoted by Am and AM by their different domains: (273)
EKiV) = D((0, oo)),
D(AM)«{u € L2(R+), An € L2(R+)}.
Proposition 11. The minimal operator A is symmetric with deficiency indices (1,1). Its adjoint operator is the maximal operator AM which has no boundary condition at 0, but two linearly independent boundary conditions at infinity B{ and B2, given by (274)
B,(u) = Dm e 2k (r u(r) e"*)', B2(u) = lim e' 2 * (r u(r) e*)',
In Weyl's terminology, we say that A is of limit point-type at 0, and A is of limitcircle type at oo, see Dunford-Schwartz [1] p. 1306. Definition 9. The boundary condition B} (resp. B2) is said to be the outgoing (resp. incoming) Sommerfeld condition (with respect to a time evolution e " lft*).
237
5.3 CONICAL GEOMETRY
PROOF of Proposition 11. The minimal operator Am is symmetric, i.e., (275) Wu,v) S5 /~[r 2 (u'v'-uv)-|uvJdr=(u,>4v),
Vu, v €l>((0,oo));
Am has (AJJJ)* « AM as adjoint; its deficiency indices are: n ± = dim ker (AM ± il). To determine n+, we note that there are two independent solutions of the equation (A + il) u = 0: u(r) = JA(r)/4?, X2 = i, thus X = ± exp (w/4), with Jx the Bessel function of order X. Its behavior at 0, then at «> is:
(276) |Jv(r)| «<«^> tev Jl^ToT W h e n r ^ 0 , and ,Jv(r>l
2rCr I/2whenr
"
^00-
Thus /!° |Jv(r)| dr/r< oo onlyifRev>0, therefore n+= 1. Similarlyn. = 1. Furthermore we obviously have D(AM) £ Hf0C(R+), and thus for all u, v in D(AM): (277) <-^u,v>£>R=/f{(r2u,), + r2u+|u}. v dr=t
R
+[rV.v - u.v ')£.
Then to obtain the limits of the last term in (277), we have to specify the behaviors of the elements of D(AM) at 0 and at infinity. We extensively use the Hardy inequality (see Dautray-Lions [1] chap. 8.2.7 lemma 3) for all Tfiniteor not (278) /J|w|2dr^4/J|(iw)'|2dr, Vw with (iw)' €L2(0,T). a) Behavior at 0 of u in D(AM). We have u and (rV)' in L2(0,T) for all T>0. From (278) r.u' is in L2(0,T), thus (r.u)' is in L2(0,T), i.e., w=r.u is in H ^ T ) ) . Thus w(0) exists and since u = w/r is in L2(0,T), this implies w(0) = 0. Furthermore thanks to Cauchy-Schwarz inequality, (279)
|w(T)| = |/Jw*(r)dr| S T 1 ' 2 ^ |w>(r)| W ^ T 1 ' 2 ^ .
Arguing similarly with w=r 2 .^, we obtain: (280)
u(r) = o(r I / 2 ),
u'(r) = o(r 3/2 ) whenr~>0,
and u € D(AM) implies: (281) u € L2(0,T), (r2.^)' € L2(0,T), i.e., u € L2(0,1), r.u € H^T), r*.u € H2(0,T). Now in (277), [r^u'. v - u. v *)](*) —> 0 with c, thus u has no boundary condition at 0. b) Behavior at infinity ofu in D(AM). Let: (282)
w*(r) = (e~k ru)' e2ir,
wj(r) = (e* ru)' e~2ir.
238
5 UNBOUNDED OBSTACLES
Then (for instance) w(r) ■ wjj(r) satisfies: rw"(r)=(- Axx - 1 u) e _ir € L2(0,oo). Therefore for all T > 0, W e L'CT, oo) and w
a ^ v ) * ^ - i [ru'.rv- - ru.rv"] =|[wjjwj- wjw|] ■OM-)^!
t(AMu,v)-(u,AMv)] = - i
Iim
r.mR [ r V v - uv*)]
(285) = 1™R^«, aR(u,v)=i[B2(u)B2(v)- BjOOB^v)]. This implies that the two mappings u —♦ BK(u), K = 1,2 are continuous linear forms on DCA^) with the graph norm. Furthermore let u be in DtA^; using (283) with B2 (and Bj) we have: (286) |(ru)' + iru - e ^ u ) ! *o(r 1 / 2 ),
|(ru)' - iru - e^B^u)! «o(r I / 2 ),
therefore giving when r —♦ oo: (286)' l u - ^ B ^ + ^ B ^ u ) ! =o(r"3/2), l u ' - ^ B ^ - ^ B ^ u ) ! =o(r 3/2 ). Example 5. We take u » Jv(r)/ Vr, with Re v > 0. From the usual properties of Bessel functions (see Abramowitz-Stegun [1] p. 364), u is in D(AM), with (287)
B1(u) = ^ e x p [ i | ( v - | ) ] ,
B 2 (u)=N|exp[-i§(v-|)]. 0
Definition 10. We define the operators A , tc«1,2 by restriction ofhu to (288)
D(AK)
Proposition 12. The operators A{ and A2 are adjoint to each other. Moreover the operators - iA} and iA2 are maximal dissipative operators and thus they are the infinitesimal generators of contraction semigroups of class C°. From relation (285), we easily see that (Aj)* = A^ (A^* « Ax. Furthermore, taking v = u in (285), we have
PROOF.
(289)
a(u,u) = i[(AMu,u) - (u,AMu)J =|[|B 2 (u)| 2 - |B^u)!2].
5.3 CONICAL GEOMETRY
239
Thus i)Re(iA,u,u)=± IBJOI)!2 > 0,
VuinlXA,),
(290) ii) Re(iA2u,u) = - \ jB^u)!2<,0,
Vu in D(A2).
Thus the operators iAf and ( - iA2) are accretive operators (and - iAt and iA2 are dissipative operators). Since (LA)* = -iA 2 , Proposition 12 is a consequence of Dautray-Lions [1] Thm. 8 chap. 17A.3. ® Then we define the operator Ag when 0 < 6 < 2 by: EKAg) = {u € D(AM), BL„u = 0}, AgU =Au, (291)
BI^u = exp [i | (| - 8)] B,(u) - exp [- i \ (\ - 8)] B2(u).
We can prove that it has the properties (see Dunford-Schwartz [1] 13, 9.15): Proposition 13. The operator \is a self adjoint operator whose spectrum o(Ag) has a continuous spectrum R+ and a set of discrete simple eigenvalues: (292)
o(Ag) = R+U{- (2n + 8)2, n e N},
with associated eigenvectors J^+gOMr. The resolvent R(- X, Aj) = (XI + Ag)~ of Ag when - X is not in ofAg) is given (with Re ^X > 0) by: (293) with:
R(- X,Ag)f(t)=/" Ke(t,s;X) f(s) ds,
(294)
Ke(t,s;X) =
f € L2(R+),
rt2(X)(J^(s)/^XJytVAlt), s
^ ( ^ ( t v ^ x ^ s ) / ^ ) . s>t, (295) J^sin5(X + 8)Jx + sin5(X-8)J_X( PROOF.
(296)
^
-
^
^
Jg
.
q
-
The fact that A^ is a selfadjoint operator is a simple consequence of B1(0 = B 2 (f),i^(f) s B l (f),
thus BL,(f) * - BL^f).
Furthermore thanks to (287), we have: (297)
( < [
Bl* (JA(r)ATr) = 2i ^ sin § (X - 6).
Thus when X is in R*f J^(r)/i/r is an eigenvector of A^ (for - X) if:
240
(298)
J UNBOUNDED OBSTACLES sin|(4X-9)*0, i.e., X = (9 + 2n)2, n€N.
This gives the discrete spectrum. The continuous spectrum is obtained by finding the generalized eigenfunctions so that (A + Xl)f =* 0, with BL0f« 0. We obtain: «r) » J*^r)A[T. The general theory of differential operators (see Dunford-Schwartz [1] p. 1326) yields the resolvent of the operator Ag as given in (293), (294). ® Proposition 14. The spectrum o(A|c) ofAK is continuous with o(A|c) = R+, K= 1,2. The resolvent R(- X, A^)«(XI+AIC)""l of A^ when X € C\R~ is an integral operator: R(-X,AK)f(t)=i^KK(t,s,X)f](s)ds,
Vf€L2(R+)
with kernel Kgfcs, A) given by (when Re (4X) > 0): (-l) K - ! |(J^(s)/^)(H ( ^(t)/# (299)
whens
K^A)* ( - i f 1 § (J^ty-TO (H ( $s)/^) when s > t,
where H* \ H* orerteuszia/ Hankel functions oforder v. PROOF.
(300)
The closure of D(R+) for the graph norm is the space D(Am) defined by EKAm)*{u€ D(AM), BK(u)-0, K . 1, 2}. "
Its range by (XI - A\ X € C\R+, is closed of codimension 1. Let: (301)
D ± c i f ker(A M ?iI), withdimD ± =l.
Then D(AM) is the orthogonal sum for the scalar product = (Au, Av) + (u, v): (302)
D(AM) = D(Am)®D+®D_,
and there are spaces D\ K= 1,2 of dimension 1 contained in D + §D - such that: (303)
D(Alc) = D(Am)©DK,
K=1,2.
In order to prove that (XI - AK)D(AK)» L2(R*), we only have to verify that: (304)
(XI - \)nK = 0, uK € D(\),
implies u = 0.
But the solutions of (304) are linear combinations of (J ~(s)A[5) and (J_ -(s)A[5») and the conditions u in L (R*), B^u = 0 imply u = 0. Thus we have proved that o(AK) £ R*. Furthermore the asymptotic behavior of the Hankel functions gives:
241
5.3 CONICAL GEOMETRY
BK((Hf(s)A[5) = 0. Thus for all X in C\R~, the equation: (A + XI) * = 0 has a unique (up to a multiplicative constant) generalized solution satisfying one of the conditions i) 0 is a square integrable function in a neighborhood of 0 only: $(r,X) = J -(r)/4F. ii) $ satisfies the condition at infinity Bt(0) « 0:
PA = R(-X,A^ - R(-X, A 2 ),
P* = R(-X,A,) - R(-X,AJ,
with kernel K(t,s,X) and K0(t,s,X):
I K(t,s,X) = i* (J^(s)/4s).(J^t)/4t), (306)
K*(t«K -exphi(n/2X4X-8)] K (t,s,X) = . pr-———K(t,s,X). sin[(n/2)(NX-8)] I7i^ norms of the operators Fk and PA are given with 4X = y0 + i jij by
(307) i p ^ . ^ - i p j } ! ,
llPx» = IiPAHexp(^,71/2) [ch 2 (| ji^ - cos 2 (|(n 0 - 0))]"1/2;
thus when X = X0 € R+, vQ = <|XjJ:
008)
IIPAOII=^,
,0 „
ir
1
K^w0wwm^m-
Proposition 15 is a simple consequence of (299). It implies the basic result: Proposition 16. The resolvents of the operators A satisfy: (309)
IIR(-X 0 ,AJ
K = (K+1)/2, VX 0 >0.
This result is the best in the sense that there cannot exist (K,a), K > 0, a > 1/2 so that (310) PROOF.
(311)
IR^A^KAjf. From (305), (308) we have for instance for Aj : IIR(-X0, A,)ll < IIR^II + IIP^II.
242
5 UNBOUNDED OBSTACLES
Since - A^ is a selfadjoint operator, the norm of its resolvent is given by: (312)
»R^|= l / d i s t f X ^ - A ^ 1/ inf |X 0 -(2n + 8) 2 |.
Let [PQ] be the integer part of i*0= ^ . When [JJ^ iseven, taking 9=|i 0 - [ ^ +1 gives |sin|(JI 0 - 8)| m 1, |X0 - (0 + 2n) 2 | > 2j*0. When [vj is odd, we obtain the same result choosing 8 = n0 -
[JIQ].
Finally we cannot have inequality (310) since by
difference, we would have IIP, II s 2K XjJ°, in contradiction with (308). 0 Remark 21. The so-called "Kantorovich-Lebedev" transformation (in the Hankel form) is used to diagonalize the operator ^. But we can expect that it is not so easy, because the operators A are not selfadjoint, nor normal, nor even spectral operators ! The definition of this transformation (with ji > 0) is
013)
$Koi) = iimA r mJW^w m ds, K. i, 2,
with H; the Hankel function. The mapping 0 —»(+|,+2)is a n isomorphism from L2(R+) onto H\ x H^ with (314)
/*>{g€Lk c (R + ), / R + Ig(p)| 2 e C T , 1 (thf )p * < < » } ,
«.<-lf,
with (unusual) properties of diagonalization of A . For developments on this transformation (which can be used to solve Helmholtz problems) see Cessenat [1]. [2], [3]. ® 3.2* Solution of Helmholtz problems Now from section 3.1 we can solve Helmholtz problems. First in the domain Q = R*xS, and in space L2(R+xS), (i.e., in L2(Q, dx/r2)) we define the following framework: (315)
DM(A - 5 ) = {u€L 2 (R*xS), (A - £ ) u €L 2 (R + xS)}.
This (maximal) space has traces on the boundary of Q: let u be in DM(A - B\ then: i) B\x € L2(R+, H"2(S)), thus^u € L2(R+, H'2(S)) also, and thus BKu exists in H~2(S). ii) An € L2(S, H,;2(R +)), (A maps H2C(R+) «{u e H^(R+), supp u is bounded} into L2(R*)). Thus B\x € L2(S, Hj~2(R +)), which implies that the traces yQn and YjU of U on the boundary R* x 8S make sense (at least in H^aS.Hj^R +)), for s = - 1/2 and s = - 3/2 respectively).
5.3 CONICAL GEOMETRY
243
Definition 11. We define the operators AK - B D , AK - B N , K = 1,2 by restriction of the differential operator A -Bto the following domains: I EKAc - BD) = {u € DM(A - B), BKu = 0, Y0u = u| R+ {316)
=0J,
a,, |EKAK-BN)={U€DMU-B),BKU=O,Y1U=^|R+X9S=OK
that is with the Dirichlet or the Neumann boundary conditions and with the outgoing or the incoming Sommerfeld conditions. Theorem 5. The Dirichlet (or Neumann) problem with the outgoing (or incoming) Sommerfeld condition at infinity: (317) ( ^ - B ^ u r r f ( r e s p . ^ - B ^ u ^ O , f€L2(R+xS), has a unique solution u in D(AK - BD) (resp. D(AK - BN)), except when n = 2, with the Neumann condition; in this case, the result is valid if we change (for u and f) L2(R+xS)i>if0 (318)
L2(R+xS) = {f€L2(R+xS), /sf(r,9)d9 = 0}.
We extensively use the following property: the Laplace-Beltrami operators Aap or AaN with the Dirichlet or Neumann condition are selfadjoint operators with compact resolvents, so that the operators BD and BN defined by (272) are strictly negative operators i.e. with pure point spectrum contained in ( - «>,0), except when n = 2 with the Neumann condition (in this case the supplementary condition (318) eliminates the eigenvalue 0, which correspond to constant eigenvectors). Thus the operators AK and B D or BN are commuting operators, with disjoint spectrum (thanks to Proposition 14). Under fairly general conditions (see Grisvard [3]), we can solve problems similar to (271), thanks to the Cauchy integral in the complex plane: (319)
u=~4/ Y (A K + zr1(B + z)-1fdz,
B=sBDorBN,
where y is a contour in the complex plane including the eigenvalues of - B (with horizontal asymptotes when Re z tends to +oo); a priori the use of (319) is purely formal, since the estimate on the resolvent of AK at infinity (see (309)) is not sufficient to apply this formula. We will justify it below. PROOF of Theorem 5. Let (\,*n) be a spectral resolution of - B, (with B = B D or BN) where the X are the eigenvalues (accounting their multiplicity) and ($n) a corresponding ortnonormal basis of eigenvectors, giving the decompositions
244
(320)
5 UNBOUNDED OBSTACLES
I«(«) = 1 ^(r)* (a), ««*)= 2 fn(r)*n(a), a € S, r> 0, " , , |with: br-2hlir
so that (271) is reduced tofind(un) satisfying (321)
(iWK-f,,.
Then since - Xn is not in the spectrum of Ay, (321) has a unique solution given by (322)
u n = (AK + V ) - 1 f n = R(-Xn,Alt)fn.
Moreover inequality (309) implies: (323)
hJ^H^I/^.
Thus the condition f € L2(R* x S) implies 2 Bunll2 < oo. The solution u of (317), is given by (324) u(.,a) = 2 R(-* n A> f n *n> and satisfies u € L2(R* x S). Thus u € DM(A - B) and also u € L2(R+, D(B1/2)) since (325)
2X n liu n !l 2
thanks to (323). Furthermore each u n is in D(AK) and thus we have BKU = 0 in a weak sense, but we don't have u in L2(S,D(AK)). Note also that (323) implies: (323)'
kl£K%;1/2m,
withK^^+D^^^infX^
The solution u of (317) is then given by the Cauchy integral: (326)
u = | u n = | (AK + X n ) - 4 n = ^
ton/^(AK+z)-1(B
+
z)- 1 fdz,
where y is a family of growing contours including the first v eigenvalues in the complex plane. This is a consequence of the usual formulae (with Q n the orthogonal projection on the eigenspace of - B for X and for large values of n) +z l (B n T
>~ =-2x^Q >
t\+\rt-A/
^(A.^zr^dz.
Remark 22. Rayleigh series. Using formula (299) of the resolvent, the solution u of (317) (see (324)) is written in the form (for K = 1): u(ra)=§ [ | ♦n(a)(J^_(r)/^^)J7, (H ( ^s)/,|5)f n (s)ds (327) + 2 ♦ n (a)(H ( ^-(r)/^)^(J^_(sy^)f n (s)ds],
5.3 CONICAL GEOMETRY
245
with f n (r)=/ s f(ra) «n(a) da. This gives the usual Rayleigh series, see (380) chap. 3: i) when supp f is contained in the ball Ba, then for r > a, (328)
u(rot)=11 cn *n(a) (H( .LWtf),
with c n = / ~ (J r^sy^s) fn(s) ds; ii) when supp f is contained in R"\ Ba, then for r < a, (328)'
u(ra)=f | c n ^(a) (J^r)A|f),
with c n = / ~ ( H ^ s y * ) fn(s) ds. In R , ^ = n + k, n € N, with multiplicity equal to 2n + 1, the eigenvectors of B are the spherical harmonics. Remark 23. In order to prove the following proposition, we note that the solution u in D(Aj) of (A{ + \I)u = f, with fin L2(R+), is obtained (with (305)) by: u as R( - X,Aj)f = R( - X,A^f + Pxf; thus its traces at infinity are BjU = 0, and (329) B2u = B2Pxf - in B2(J^(r)A[r) L^f,
with L^f=/~ (J^tWtWt) dt.
Hence from (287): (330)
B2u = i^exp[-f(
we have (thanks to the Cauchy-Schwaiz inequality): (331) using: (331)'
|B 2 u| 2 s2n|L 4X fi 2 < : |llfll 2 , J R+ |J v (t)| 2 dt/t=l/2v, v>0.
We note Y = D(A!-B) + D(A 2 -B),Y = YDwhenB = B D ,Y = YNwhen B = B N . We recall that D(B1/4) = H^2(S) when B = BD, D(B1/4) = H1/2(S) when B = B N . Proposition 17. Trace at infinity and Green formula. The mapping BK: U —► BKU is a continuous mapping from Y onto D(B1/4), K= 1,2. Furthermore all u, v in Y satisfying the Dirichlet or the Neumann condition, also satisfy the Green formula (332) aB(u,v) d=?f i {(A - 5)u,v) - (u, (A - B)v)=| / [B 2 u.I^ - BjU.I^] da.
246
5 UNBOUNDED OBSTACLES
PROOF. First we prove that u in D(Aj - B) implies B-u in D(B1/4). Using the decomposition (320) with (322), we have with (331) and f= (Aj - B)u: (333)
»B2ull2D(Bl/4)as 2 jfc | B 2 u n | 2 * * | lffn«2^w»fl2< °°-
Then given g in D(B1/4), we construct now a lifting of g into EKAj - B). Let g = 2 &n+n be an expansion of g with 2 ^ defined by: (334) with
| g | < °° • Then the function u
u-tAj-B)-1?,
F - ^ | ^* n «PP5<^-2>]
The first part of Proposition 17 follows easily. Now we prove the Green formula (332) using the spectral decomposition of B and (285), since we have: aB(u,v) = i 2 [((A + Xn)un,vn) - (u n ,U + Xn)vn)J (335) <8>
Proposition 18. The operators - i(Aj - B D ), KAj - BD) are maximal dissipative operators, adjoint to each other, and so that (336) EK(AK - BD)2)£L2(S,D(AK))nL2(RM)(BD))£D(AK - B D )£L 2 (R + ,EKB{) / 2 )). The same properties are valid if we replace the Dirichlet condition by the Neumann condition, but for n = 2, we have to substitute space L2(R*X2(S)0) for L2(R* x S), where L2(S)Q is the orthogonal space to constants in L2(S). PROOF. The Green formula (332) implies that: (337)
Re [ i((A! - BD)u,u)] > 0, V u € EKAj - BD).
Thus - i(A. - BD) is dissipative. Furthermore since this operator has a bounded inverse, it is a maximal dissipative operator (from Dautray-Lions [1] thm. 8 chap. 17A.4). We easily prove (336) and that i(A.-B D ), - i ( A 2 - B D ) are operators which are adjoint to each other from Green formula (332). 0
5.3 CONICAL GEOMETRY
247
We can prove (see Cessenat [1]) that if u is in D(AK~BD) (or in D(A K -B N )), then (338)
§£ € L?0C(R+ xS),
thus EKA, - B) £ H^tfO, oo)x S).
With this result (and (323)'), we can obtain an estimate on the norm of the solution u of (317) in Hjoc((0,oo) x S), like the estimate (181) of the limiting absorption Remark 24. Asymptotic behavior in DMC4 - B). Sommerfeld conditions. We can specify the behavior of functions u in DMU - B ) at infinity like in the proof of Proposition 12. Using the decomposition (320), we define (where ' is used for derivation with respect to r): (339)
w^(r,a)«(e* ru(r,a))'e-2ir, wj(r) - (e* run(r))'e-2ir,
thus w^(r,a) = 2 ^(r^Ja). Then we can prove that the formulas (286) are also valid, n " 3/2 the o(r"* ) refers to the norm H" *(S). This immediately implies: if u is in D(Aj - BD), i.e. B ^ = 0, then u(r,oc) = Q(t"l)t u'(r,a) - iu(r) = o(r~3/2), that is, u satisfies the outgoing Sommerfeld condition. We can also specify the behavior of functions u in DM(^4 - BD) at the origin like in (280), using the spectral decomposition of B = BD (then the o refers to the D(B"1/2)norm). Remark 25. Sources of compact support. Rellich lemma. i) If we have a source g (see (268)) with compact support and such that f = t^g (see (270)) is in L2(R* x s \ when n = 3, then we know (see (15) Prop. 1 chap. 3) that the asymptotic behavior of the outgoing solution u = v is given by: (340)
u(ra) = §^B2u(a) + o(r 3/2 ), with B 2 u(a)«i- g(a), a € S 2 , k = l .
ii) Proof of the Rellich Lemma 1 (chap. 3). Let u satisfy outside a ball Ba (341)
(A + k2)u = 0, i.e., (A - 5)u = 0, and/ 2 |u(rcc)|2da = o(r 2 ),
sothatu(ra)€L2(Bl). Furthermore (341) implies that B ^ B ^ O . Letu a =uL . Then using the spectral decomposition (320), we obtain that un satisfies: (342)
(A^ + y u ^ O ,
r>a, ic=l,2,
un(a) = u£.
Thus un is given by the two formulas (with K = 1 and 2):
248
5 UNBOUNDED OBSTACLES u„(r) = < (H ( ^-Wtf) (rf li(a)/4a)- l ,
V r 2: a,
which necessarily implies u*=un(r) = 0, for all n, thus u = 0 ! We emphasize that the Rellich lemma can be applied to any domain whose boundary is a conical surface when r is large enough. ® Then we solve completely inhomogeneous problems thanks to Theorem 5 and a duality method. First we define the following trace spaces with K » 1,2 |X5 = {u0=y0u = u | R + x 9 s , U E E K A . - B N ) } (343)
|XHul=*lu=lilR+x3S' ■•DCA.-BDH equipped with the quotient topology, and also with (315): (344)
X ^ l u o ^ u , u€D M W -B)}9
X ^ u ^ u , u€D M M - * ) } .
We can prove that XQ = XQ , x} = x\ (thus denoted by XQ, XJ), and
(345)
x 0 =(x 1 y,x 1 =(x 0 y, with x 0 cx 1 c(x 1 yc(x 0 )'.
Usual regularity results on elliptic problems in regular domains (see GilbargTrudinger [1] for instance) imply:
XocHf^xas^^cH^xas), JCocHf^Vxas), x^Hf^Vxas). Theorem 6. Let (f, g, uQ), resp. (f, g, Uj), be given such that: (346)
f€L2(R*xS), g€EKB1/4y, u 0 €Xo,
resp.u^Xp
Then there exists a unique solution u in T>M(A - B) of:
(347)
|i)04-5)u=:f inR + xS, ii) B2u = g (or BjU = g) I Hi) Y0u - u0
(resp. yp = Uj).
The only exception is the Neumann problem with n = 2. In this case the theorem is valid //(f,g,Uj) satisfy the conditions (with S identified with ( - 80,80)) g = 0,
i
jo"(ui+ + Uj-) + f € ImA! (or ImA2), with
(348) ^=^(.,+80), ^.=^(.,.90), f=^/^°f(.,e)d8,g=2^/^0g(e)de.
5.3 CONICAL GEOMETRY
249
The proof is based on the weak form of the problem (347): find u in L2(R+ x S) such that (349) with
(u, (4 - B)v) = LfgfU()(v),
V v € DfA, - B D ),
L
f,g,u0
Thanks to Theorem 6, we can define the Calderon operator C relative to Cl: when n * 2, this is the isomorphism defined, with u the solution of (347) for f = 0, g = 0,by: (350)
u 0 € Xo(resp. XQ) — YJU € Xx (resp. X^.
When n = 2, we have to replace spaces Xx and Xx by X \ and X | with (351)
XKresp.XlJ^fu^^^JeX^resp.X!),
u 1+ + u ^ e l m A ^ .
With the usual regularity properties of S (thus of Cl = R+ x S) we can extend the functions u in D^(AK - BD) or Dft(AK - BN) to Rn by symmetry and truncation. This implies that the spaces of traces (343) and (344) are the same for Cl and its complement Cl9. This allows us to solve the "transmission" problem (or with given sources on the boundary T of a cone Cl):findu in L2(R+ x S n ~ { ) such that i)04-B)u=rO inQandQ'^R+xS', S' = S n " \ s , (352)
ii) B ^ u l ^ O ,
B|(u| a > )s0 (i.e., u is outgoing),
with given jumps p and p' of u and its normal derivative across T. Theorem 7. Let p € X^ p* € XJ . Then (352) has unique solution u such that (353)
u|0€D&4-*),
u^eD^C*-2?)
(i/n ss 2 the condition onp'is: p'+ + p*_ € Im Ai with notations used in (351)). Furthermore we have the regularity result on the traces of u and its normal derivative on each side of the boundary T ofCl (354) (y0±u,yl±u)€X0xXl when (p, p')€XQXX X (when n = 2,we have to substitute Xj for X^. Let P ± be the mappings (pyf>9) —»(yQ ±u, Yj ± u), u the solution of (352), we verify that P * are continuous projectors in spaces XQ x Xj and in XQ x X, (or X 0 x X | w h e n n = 2).
250
5 UNBOUNDED OBSTACLES
These maps are the Calderon projectors in these spaces; they satisfy the usual relation: P + + P~ =1. These operators can be made explicit thanks to the spectral decomposition of the Laplace-Beltrami operator on Sn~*, and we can use them to solve scattering problems with an incident stationary field in free space on a cone occupied by a medium with wavenumber k.. We can generalize in many ways (see Cessenat [1]), for instance to domains Q = (a,oo)xS, a>0, to dissipative media (associated with the limiting absorption principle). Application to Maxwell equations. We can use the Calderon projectors above for Helmholtz equation to transform a boundary problem with Maxwell equations in a conical domain, into an (integrodifferential) problem on the boundary T of this domain. We can consider also the scattering of an incident wave by a dielectric cone. In order to do that, we have to transform relations between the electricfieldE and its normal derivative on T into relations between the tangential components of E and H on T, thanks to the differential formulas (310) in the Appendix. Using the same notations, the Maxwell equations at the boundary Tare: | i ) - ^ ( n A E ) + curl r E Il -2R m nAE + i(i)|iHr = 0, (355)
ii)^(nAH)-curl r H f l + 2RlllnAH + i£ En = 0,
with E r =t r E the tangential component of E. We recall the divergence formulas: (356)
d i v r E r + ^ + 2R m E n = 0, div r H r +^£ + 2R m H n = 0.
We emphasize that g- (n A E) and n A ^j E r (for instance) are different! Using an orthogonal system of coordinates (x^x 2 ^), the Riemannian metric is: g = ds2 = gjdxf + g2dx| + ds2 . If X = X ^ + X282 is a tangentfieldto T, then: (357) with:
^-(nAX) = n A ^ + f x - 2 R m n A X ,
« - ^ - x a 5 8 | + X | 5 % 1 , go=(gig2)"l/2 " , d 2 R --^4 , o 1 ' Of course, we would have to specify the functional framework for such Maxwell scattering problem in correspondence to the above scalar problems with Helmholtz equation. The functional spaces are based on weighted norms. We don't develop these (technical) spaces here, but results are especially interesting in the simplest case where the conical domain is the whole space! &
CHAPTER 6
EVOLUTION PROBLEMS Evolution problems in electromagnetism and more generally non-stationary wave problems are of many different types: First we consider Cauchy problems, i.e., evolution problems with given initial conditions (or also mixed Cauchy problems, i.e., with boundary conditions). These problems are fairly easy to solve and quite "standard" in free space, or if we assume an "ideal" linear isotropic media with constitutive relations D = eE, B = pH. In the general case the constitutive relations have to be written with a convolution in time, see (22) chap. 1. This implies that the usual Cauchy problem is ill-posed: we have to know all the past of the material to solve evolution problems, which are called Cauchy problems with delay or with thick initial conditions, see Dautray-Lions [1] chap. 18. Here also scattering problems lead us to define incoming and outgoing waves, associated with Cauchy problems under fewer initial conditions. Then we consider "causal" evolution problems, which are well-posed problems without any initial condition. Many evolution problems concerning waves and electromagnetism are of this type. Maxwell equations are an example of symmetric systems of Friedrichs. We do not develop this general theory, but only refer to Friedrichs [1], Lax-Phillips [3], Chazarain-Piriou [1], Colton-Kress [2]. We do not either use microlocal analysis (and the trace theorem of Hormander) in this book. 1. CAUCHY PROBLEMS
First we consider the following Cauchy problems. Let H be a Hilbert space; given UQ and F, an /f-valued function, find an H-valued function U such that: | i ) ^ + ^ U = F, with t>0 or with t€R, ii)U(0) = U0. 251
252
6 EVOLUTION PROBLEMS
Problem (1) for t > 0 only, is called a one-sided (or unilateral) problem. Problem (1) for real t is called a two-sided (or bilateral) Cauchyproblem. First we assume that -A is the infinitesimal generator of a semigroup (G(t)) of class C° in H. From usual theory (see Pazy [1], Dautray-Lions [1] chap. 17) we know that if F € C°([0, oo) ,H) (or if F € Lr([0,T) ,H) for all T) with UQ € H, the one sided problem (1) has a unique solution U € C°([0, oo) yH) (in a weak sense) given by: (2)
U(t) = G(t)U0+/0G(t - s)F(s) ds.
Then assuming that —A is the infinitesimal generator of a group (G(t)), and F is given for all real t, the two-sided Cauchy problem (1) has a unique solution with the same regularity properties. 1.1. Scalar wave Cauchy problems We recall some well-known results on Cauchy problems for waves. As a reference on this subject see Hormander [1], Chazarain-Piriou [1], DautrayLions [1]. The usual standard Cauchy problem for scalar waves is: given a (complex or real) Hilbert space H, find an H-valued function u so that: a2,, i ) ~ j + Au = f inR^orR^, (3)
ii)u(0) = u°, 3u, gW-u1, with given initial conditions u°, u1, and with a given source f. The operator A is a positive (or even bounded from below) selfadjoint operator defined using a variational framework (V, H, a(u,v)). The usual space with finite energy waves corresponds to u°€V, u*€H, and f€C°(Rt,H) (or more generally f€L l loc (R t ,H)).Thentaking# = VxH, and: <4, U
0
= ( ^ , U < , , . ( ; « ) . * . > = ( * > ) . A.[<
->).
- « . » .
problem (3) is written in the form (1), with A the infinitesimal generator of a group (G(t», and thus (3) has a unique solution U€C°(R,/0. giving u € C°(R, V), with u' e C°(R,H); u is given by (5) u(t) = cos(tAlV+A-y2sia(tAuW
+f0A~U2sin((t-s)Am)f(s)<1S'
We obtain very easily this formula of symbolic calculus, defining z and w by: (6)
z(t) - iA1/2 u(t) + u'(t), w(t) = -iA 1/2 u(t) + u'(t), Zo-iA^uO + u1,
w0 = -iA , / 2 u° + u1.
253
6.1 CAUCHY PROBLEMS
Then z and w satisfy: (7)
1/2 z* iA1/2 z(t) + f, z' == iA z(t)
v/ = - iA ,/2 w(t) + f,
and the solution of (7) is given as in (2) by: _/*%
(8)
iA1/2t 1
Izft^e * w(t) = e _lA
f*
Zo+io l
iA1/2(t-s)~ e
T won +J„e J0r
..
'f(s)ds, L1/2„
'f(s)ds.
We obtain (5) by taking the difference z(t) - w(t). We note that the operator: (9)
M(t)=A-1/2sin(tA1/2)
is a bounded operator, and that (5) is also given by: (5)'
u(t)=|j M(t)u° + M(t)u! + f0 M(t - s)f(s) ds.
Furthermore, multiplying (3)i) by u' (the time derivative of u) and integrating, we check that u satisfies the energy balance (for f=0 this is the energy conservation law) (10) \[H§jJ(t)ll2 + HA'^t)!2] =|IBu'U2 + IIA^VB2] + Re/ Q (f,g)ds. The first usual example is the Cauchy problem in free space occupying Rn associated with the wave equation or d'Alembert equation: 7
(ID
i ) D u d J f 4 | ^ . A u = f in^xRt, ii)u(0) = u°, fjW-u 1 ,
c being the velocity of light in free space, u°, u1 being given initial conditions, and f being a given source. The usual framework is the space offiniteenergy solutions. Using the Beppo Levi space: (12)
W1(Rn) = {v€Z)>(Rn), v€L?oc(Rn), /Rn(|gradv|2)dx
(see also Appendix, Def. 7 section 5.2), we assume that the data satisfy: (13)
u^W^R11), u!€L2(Rn), f€C°(Rt,L2(Rn)).
254
6 EVOLUTION PROBLEMS
Then taking V = W^R11), H = L2(Rn), H=W^R") x L2(Rn)f A » - c 2 A (but V is not contained in H), we see that problem (7) is of the form (3); thus it has a unique solution u obtained in the form (5), thanks to its Fourier transform (we take c = 1 in order to simplify the formulas): /nx */*** ^sin(|6|(t-s))j; / I H A A 0 m sin(|£|t) A i (14) u(t,t) = cos(|t|t)u (|) +—jt| u (*>+/o~—TH f&s)ds. We can write this formula using elementary solutions of the d'Alembert equation as will be seen below. We haw used the semigroup theory to solve this Cauchy problem, but there are many other ways to solve it (see Dautray-Lions [1] for instance). We can solve mixed Cauchy problems also, i.e., problems in a domain Cl with a Dirichlet or a Neumann condition. Let Q be the exterior of a bounded domain with boundary T. For a Dirichlet condition, we have to substitute for W*(Rn) the Beppo Levi space (see Def. 8 in the Appendix) which is also defined in the regular case by: (15)
wi(Q) = {u€D'(Q), u€l4 c (Q), / Q |gradu| 2 dx
u|r=0}.
We first give a general property of waves satisfying (3) when t goes to infinity. We define the kinetic energy Ec(t) and the potential energy E (t) of u by: Ec(t) = i l | j (t)ll2t
Ep(t) 4 IIAl/2u(t)ll2.
Then using the notion of spectral measure (see Kato [1]) and (5), we can prove the following property when there is no source (for instance when A = - A in the exterior of a bounded domain, see Goldstein [1], Lax-Phillips [1]) Proposition 1. Equipartition of energy. IfOis not an eigenvalue of A, then the solution u of'(3) with f = 0 satisfies equipartition in the mean: Moreover if the spectral measure dE(X) is absolutely continuous with respect to the Lebesgue measure dX, u satisfies: lim Ec(t)= lim EJt).
1.2. Cauchy problems In electromagnetism 1.2.1. Cauchy problems in free space R3 We consider the Cauchy problem: find the electromagnetic field (E,B) (or (E,H)) in free space Q = R3, satisfying the Maxwell equations (1) with (5) and (3) chap. 1, for positive t (one-sided problem), or for all real t (two-sided problem):
6.1 CAUCHY PROBLEMS
255
i ) - f ^ + curiH = J, §£ + curlE = 0, (16)
ii) divD = p, divB = 0, iii)D = £oE, B = »»0H, iv) E(0) = E0, B(0) = B0, (or H(0) = H0, D(0) = D0),
with given charge density p and current density J satisfying (3) chap. 1, and with given initial conditions for E, B (or D, H). It is more common to work with (E,B) in free space than with (E,H) for evolution problems, so we will use (E,B) only. As indicated in chap.l, these equations are partially redundant. We assume that: (17)
E0€L2(R3)3,
B0€L2(R3)3,
andJeL20c(Rt,L2(R3)3.
We define H = L2(R3)3 x L2(R3)3, and with e0ti0 = 1/c2,
Then the Cauchy problem (16), without taking ii) into account, is written in the form of (1). Moreover we can prove easily by Fourier transformation that -A is the infinitesimal generator of a unitary group (G(t)) in H, with domain D04) = H(curl,R3)xH(curl,R3). Then we know that the two-sided Cauchy problem (16) with (17) has a unique solution (E,B) in C°(Rt,/f) with finite energy (with conservation of energy when J = 0, see (7) chap. 1): (19)
W(t) = i/ R3 (D(t).E(t) + B(t).H(t))dx = W ( 0 ) - / 0 d s / R 3 E . J d x ,
but does not necessarily satisfy (16)ii). If the data satisfy, in addition to (3) chap. 1: (20)
div B0 = 0 and div E 0 = p/£0 with p € L ^ R ^ L2(R3)),
then we can prove that the solution (E,B) of the Cauchy problem also satisfies: div B(t) = 0, and div E(t) = p(t)/e0 for all t. 1.2.2. Cauchy problems in a domain fit Now let Q be a free space domain in R3 bounded by a perfect conductor, with boundary T. We have to solve a mixed Cauchy problem: find (E,H) satisfying the Maxwell problem (16) in O, with the boundary conditions (21)
n A E | r x R + = 0,
n . B | r x R + = 0.
We assume that the data satisfy (17) with Cl instead of R3. Then with H= L2(Q)3 x L2(Q)3, we can prove that operator 4, defined in (18), with domain D(A) = Ho(curl,Q) x H(curl,Q) is such that:
6 EVOLUTION PROBLEMS
256
A*=-A, with D(A*) m D(A) (U is a selfadjoint operator if we complexify the space). As a consequence of the Stone theorem, A is the infinitesimal generator of a unitary group (G(t)) in H> at least when fi satisfies one of the following conditions: i) Q is a bounded domain, or the complement of a bounded domain, ii) Q is a cylinder (see Duvaut-Lions [1], Dautray-Lions [1] chap. 9A and 17B.4). Of course we can take other frameworks which are subspaces of the above space H for these evolution problems, in order to take into account usual conditions of free divergence for B for instance, or regularity properties. This allows us to indicate in which sense the boundary conditions are satisfied (thanks to the notion of weak and strong solution). Thus we can use any variational framework ofchap.2.11.1. It is interesting to compare the Maxwell problem to the wave problem, for instance with respect to H (or B) only. We eliminate E by a usual trick to obtain: 2
9H i K1 ^ - A H = curlJ, c dtr
(22)
ii) H(0)=* H0,
divH=0,
fH(0) = - j^curl E 0 ,
iii)n.H| rxR+ =0,
nA curlH| rxR + =0.
The boundary conditions imply (see (147) chap. 2) that (-A) is the positive selfadjoint operator (A*-I). For E, we obtain (in the regular case) a similar result but with the operator (A~ -1). Of course we can generalize these results with respect to the Cauchy problem in many ways: for instance, thanks to a variational method described in DuvautLions [1], to the case of several linear (isotropic or not) media with different permittivities and permeabilities (but frequency independent in time Fourier transform). Now we recall some properties of the wave equation and above evolution problems. These properties are common to hyperbolic equations: the wave equation is a typical hyperbolic equation. 1.3. Some hyperbolic properties of wave evolution The main property of the d'Alembert equation is the existence of an elementary solution <&(x, t)(or 4> (x,t)) with support in the future cone orforward light cone: (23)
C+ m {(x,t) € R n x R, with |x| £ ct}.
This solution which is thus characterized (taking c = 1) by: (24)
D * d ^ ~ ? - A=6(x, t) m 6(x) 8(t),
with supp * £ C + ,
is unique: if
6A
(25)
CAUCHYPROBLEMS
257
□(<& ♦ 4>o)»(□*) ^ o = % = * * <□*<>) = • •
This existence is typical of hyperbolic equations (see Dautray-Lions [1] chap. 5). There are many ways to calculate this elementary solution, each method having its own interest (one of them is the Radon transform method). We can also obtain simply by its space Fourier transform (compare to (14)): (26)
A A A sin(Ult)
where Y(t) is the Heaviside function. Here we give the very simple result when n = 3: (27)
«(xft).i{«(r-t)-^Y(t)«(i2.t2),
r=|x|.
The operator M(t) given by (9) in this case is simply given by the convolution: (28)
M(t)v = R(.,t) ; v, with R(x,t)=^(signt)8(r 2 -1 2 ), t * 0 .
In the general case, results differ essentially with the evenness of space: when n is even the support of <& is the whole future cone, whereas when n is odd the support of O is the surface of this cone only. There is also a unique elementary solution 4> _ (., t) with support in the past cone or backward light cone, given by: <&_(. ,t) = - Y ( - t ) R ( . , t ) , and thus R(.,t) = 0 + (.,t) -_(., t) is a solution of the homogeneous wave equation. When Q is the complement of a bounded domain £1\ we obtain as a consequence of the properties of 4>: Proposition 2 (Finite velocity of propagation). Let R > 0 be such that Q' is contained in B R . If the initial conditions (u°, u 1 ) are zero outside B R , the solution uof: i ) ~ ^ - A u =0 (29)
inQxR^,
ii)u(0) = u°, IfCO)-!! 1 , iii)u
lrxR+ = 0
is zero outside the ball B
(or
(uV)€W£(0)xL2(Q),
SHlrxR+
= 0)
'
. t ,. Furthermore if the initial conditions (u°, u 1 ) are
zero inside the ball BR(xo) centered at xQ then the solution u of (29) is zero inside the ball BR _ ct(xQ) when t < R/c. Then we have the property of local propagation of energy: Let E(u,t,p) be the energy of the wave u at time t in a ball B (B being arbitrary):
258
(30)
6 EVOLUTION PROBLEMS E ( u , t , p ) = i / n n B (li|*
then we have (31)
E(u,T+ t,p) £ E(u,T,p + ct) for all T € R and t > 0.
1.4. Radon transforms The Radon transfonn method is a basic tool to obtain many properties in scattering and wave evolution. We briefly give the main definitions and properties, and refer to Gelfand-Graev-Vilenkin [1] and to Helgason [1] for more developments. First using regular functions we have: Definition 1. The Radon transform of a function f in 5(Rn) is the function on RxSn~l defined by; (32)
10l8 9 a).7(s 9 a)./ U a s f(i)dS 9
s€R, a € S n - \
dx being the surface element ofthe plane x. a = s, with ds dx = dx. Note that we can take a in Rn\{0} instead of S11"1. Then (32) would give a function satisfying the homogeneous condition: (33)
f(Xs,Xa)=|X|"1f(s,a),
V X * 0 , (s,a)€RxR?.
We first have the basic property: (34)
/ R e-^f(s,a)ds=/ R e-^ds^^^
where f is the Fourier transform off. Thus if we define (like in chap. 3, (396) (400)): (35)
F(r,a) = f(ro), r ;> 0, and F(r,a) = f((-rX-«))» r * 0 , V a € S n " \
F(r, a) is a function, defined on R x Sn"l, which is in space S0(R x S n " ) when f is in S(Rn) with the notation 1 of chap. 3.11.3. Then we define the following space: (36)
S~(R x Sn"l) is the space of inverse Fourier transforms of 50(R x S n " l ).
Proposition 3. The Radon transformation is an isomorphism from 5(Rn) onto S^RxS11"1); the Fourier transform of f and the Radon transform of f are related to each other through the Fourier transform (34) on s. The elements of S~(R x S n ~ l ) are characterized by |i)g(s,a) = g(-s, -a),
V(s,a)€RxS n - 1 ,
H) J R g(s,«) s* ds = Pk(a) a spherical harmonic of order k, k € N.
259
6.1 CAUCHY PROBLEMS
We note the properties of the Radon transformation with respect to translations, derivations and convolution: i) fa(s,a) = f(s + a.a,o) (38)
with fa(x) = f(x + a),
ii)(g|) ( s ^ c c ^ ^ a ) , andthus: (AfT(s,
The inverse Radon transformation is obtained using the inverse Fourier transformation: (39)
i W - ^ / ^ ^ i M - ^ / ^ ^ i ^ ^ ^ d n f c u
Then g(r,cc) = e1^" f(ra) has a natural extension to R x Sn~l by (36). Thus we have tw< cases according to the evenness of n: i) n is odd: (40)
W - 5 ^ / . ^ , e t a a F(r,«)r»-'drd«.
Then by the inverse Fourier transformation with respect to r, with m «(n - l)/2,
<4,)
» - ^ r V i $ S * - . I * " ^ ' - A > ° V- *"• •»""■
ii)niseve/i: (42)
fi(x) = 2(^jH/ RxS n.l e ^ ftr,«)(sign r)r^drda.
Then thanks to the Fourier transform of finite parts (see Schwartz [1] chap. 7.7): (42)' FK-lf^-^Pfs-^Msignr)^-1, we obtain
lA-n
«™ ( - l A n - l ) !
f
(43)
*).
/ R j t i B . 1 P f ^ — - ^ dsd«.
^
-JM..
We can define the Radon transformation on tempered distributions, either by using a Fourier transformation, or by duality. Let g be in 5(R x Sn~ *), and: (44)
g(x)=/ n -i g(x-a>«) da-
Then g —♦ g is a continuous mapping from 5(R x Sn~ ) onto S(Rn) so that (44)'
Vf€5'(R n ).
260
6 EVOLUTION PROBLEMS
That this mapping is onto is a consequence of inverse formulae (41) and (43). Then / R nf(x)g(x)dx=/ s n . 1 da/ R ds/ a X a s s f(x)g(s,a)dx=/ R x s I l - _ 1 f(s,a)g(s,a)dsda.
Thus (44)' defines the Radon transformation as a continuous mapping from SWimoSXRxS 1 1 - 1 ). Another basic property of the Radon transformation (and its advantage with respect to Fourier transformation) is that it preserves the boundness ofsupports. If suppf is bounded then it is obvious that supp jRf is also bounded. For the converse we refer to Helgason [1]. We also have the obvious property for odd n (45)
f(s,a) = 0for|s| s a , a > 0 , VacS"" 1 , impliesfi(x) = 0for |x|
Furthermore we have a "Plancherel" formula for Radon transformation; applying the Plancherel formula twice to Fourier transforms leads to: i) when n is odd (and m = (n - i)/2): (46)
/Rnf(x)g(x)dx=^T/RxSn.1p(s,«).^(s,a)dsd«.
ii)whenniseven; (-l) n/2 (n-l)! (47)fRllf(x)g(x)dx=sV" ' ( 2 | y '* J ^ ^ u i Pf «Pi.«>g
/R„J«x)|2dx=^^
6.1 CAUCHYPROBLEMS
261
Given two functions f0 and fj in L2(Rn), we define the function £ on R x S n ~' by: (50)
n(r,a)-?4(R) ifr>0,
h(r,a) = f ,((-rX-a)) ifr<0.
Let h = h(s, a) be its inverse Fourier transform (with respect to r). Then we note (51)
h=« 2 (f 0 ,f,).
Proposition 5.
3mh
The mapping: (f^fj) —»g-gr & * unitary (up to a constant factor)
mappingfrom L2(Rn)xL2(Rn) onto L2(R x Sn~'), with n odd, m - (n - 1 )/2: (52)
/ R n[|f 0 (x)| 2 + l f l( x )| 2 l d x =7T7rT/ R x S n-l I 0 ( s , « ) | 2 d s d « .
Definition 3. The unitary (up to a constant factor) mapping of Proposition 5 is denoted by JLd and is called a "two-sided differentiated" Radon transformation. We see that R^ and R^ are natural extensions of R? and R since /?2d(f,f) = and RJS,t)=Rf. Furthermore we have the following basic properties:
(53)
*2«o.fi>-|i*2Mi>.
*2((-A>1/2f' -<-A) l / 2 f)—it£.
/#(-A) 1/2 f 0 , - (-A)1/2)f,)= - i ^ ^ f , ) —
i^r«2(f0,f,)-
Definition 4. We denote by Rw the mapping (called "wave Radon transformation") definedfor (f,g) in Wl(Rn) x L2(Rn), with n odd, by (54)
*w(f,g) = / # - ^ ^ ^
Proposition 6. The mapping Rw is a unitary (up to a constant factor) transformation from W'(Rn) x L2(Rn) onto L2(R x S n l ):
(54)
' &&'***-* i*w««-«-
and satisfies, with A given by (4), and A = - A: (55)
Rw(A(f,g))^Rw((,g).
^^'KhK^Kw
262
6 EVOLUTION PROBLEMS
This Proposition is fairly easy to prove. Here we give its inverse only. Let h be in L2(Rx S n ~'); we have to find f and g so that h = rtw(f,g), that is with (54): <56>
" ^ 5 5 T + ^ w = h.
Using the different behaviors of each term with respect to symmetry, we have: ^ ( s , o ) = | ( h ( s , a) + (-l) m h(-s,-a)), ^ J | ( s , a ) = - ^(h(s,a) + (-l) m+1 h(-s,-a)). Then applying the inversion formula (41), we obtain: (57) «
p
/ s n.l^S(x.«,«)d«, «x) = i ^
F r
/ s n _ 1 ! s = T -(x.a,«)d«.
1.5. First applications of the Radon transform method Now the (usual) Radon transform of the wave problem (11) with c = 1, f = 0, and with (38), is the one dimensional wave problem: i)^|(s,a,t)-0(s,a,t) =O
inRsxSn-1xRt,
(58) ii) u(s,a,0) = u °(s,a), H (s,a,0) = u ! (s,a). at
Its solution is given by the d'Alembert formula: (59)
u(s,cc,t)=| (u °(s + t,a) + u ° ( s - t,a)) + ^f£
u V , « )
and we obtain the solution u of (11) with f=0 applying the inverse Radon transformation (41) or (44). Now let x,
a +a ,
be the characteristic function of the
interval [-a,+a] when a>0. When a<0, we define Xi_a+ai = "~Xr+a _ a i- Then the operator M(t) (see (9)) is easily obtained thanks to the convolution: (60)
M(t)v(x) = R ( . , t ) ) ? v = | ^ 1
thus (using (38)): (61)
R(.,t)=±R-l(X[_tM]).
Thus to obtain the elementary solution O with support in the future light cone which is related to R by 0(.,t) = Y(t)R(.,t) (see (26)), we have to calculate the inverse Radon transform of Xi_t+ti thanks to (41) or (44). We obtain the "Herglotz Petrovski" formulas for these elementary solutions:
263
6.1 CAUCHY PROBLEMS 0(x, t) = Y(t) (- l) m+1 — ^ / (2it) "
s
n -!
6(n ~2)(x.« - t) da, with odd n, m = (n - l)/2
(62) (x,t) = Y ( t ) ( - l ) n / 2 y ^ J J / s n - 1 Pf
L
^
withevenn.
We can obtain R(t) when n is odd (at least formally): we have 6 t W=/ s n.l «*•« - ^ d a - a ^ j / t 8(|x|s - t)(l - s2)"1"1 ds,
(63)
withx = inf(t/|x|, 1), o n - 1 = |S n ~ 2 | the area of the unit spheres11"2. Let: J(x)= x1 ( 1 - l ) " 1 - ^ | x | > l , I l |x| Then we have (64) hence: (65)
J(x) = 0 i f | x | < l .
*tW = o n . 1 | j ( f ) ; R(t) = - c n ( - A ) m - 1 1 | J(f), with c n = - on.1/2(27i)n-1.
We can also obtain usual formulas of the elementary solution (see Lax-Phillips [1] p. 125, Methee [1], Zieman [1]). Thus when n is odd, the support of this elementary solution is the surface of the future light cone. This implies: Proposition 7. Huyghens principle (with a velocity c). For odd dimension n of space (n > 1), if the supports of the initial data are contained in the ball B R , then the support of the solution u of( 11) with f=0w such that: supp u(t) = {x € Rn, 3 e € Sn~l, x + cte € supp u 0 U supp ux] Q KR t , (66) | K R t = {x€R n ,c|t| - R < | x | < c | t | + R } if |t| >R/c, K R t = {x€Rn, | x | < c | t | + R } i f | t | < R / c . PROOF.
Thanks to (5)' and (60) the solution u of (11) (with c = 1) is given by:
(67)
u(x,t) = | R(t)x* u° + R(t)x* u1,
and the property: supp (R(t) * v) £ supp R(t) + supp v, implies (66). We can verify (66) on the Radon transform of the solution using again (45).
®
Furthermore we can prove that (for an odd dimension of space) when f = 0 and when the initial conditions are regular, the solution u of (11) and all its derivatives converge to 0 at each point x when t goes to infinity (see for instance Goldstein [1] for generalization to hyperbolic operators). Although the energy of the wave is globally constant in time, it decreases locally with time.
264
6 EVOLUTION PROBLEMS
2 . SCATTERING PROBLEMS - INCOMING AND OUTGOING WAVES
We now study problems which are relevant to scattering, following Lax-Phillips theory. We assume that space dimension is odd. 2.1. Another application of the Radon transformation We consider the Cauchy problem: find u satisfying (68)
.•^u i)^-j±-Au = 0
inlTxRf,
ii) u(0) = u°, §f (0)=u 1 , with (u°, u1) given in H, and we apply the wave Radon transformation Rw. Then with (6) let: (69) k - * W ) ,
k(.,t)=/?w(u(t),v(t)) = ^(w(t),2(t)), withv(t)=^(t).
Using (54) we have: k s at
(70)
< - ' > - - f ^ T <«•«•«)+iim ft.0*1).
Thanks to (7) and (S3), (or directly with (55) on (1)), (68) is transformed into:
i)fjf(s,«,t) = -f k -(s,a,t), (71) ii)§£(s,a,0) = k(s,a), in L2(R x S n l ) . The solution is given by: (72)
k(s,a,t) = k(s-t,a).
Applying the inverse formulas (57), we obtain the solution of (68) by
(73) m
(-D | v ( x , t )9=uf, („ x. , , t ) = ^ j r r / rs n . 1 ^ ^( xk ., a - t , a ) d « ,
and thus for n = 3: (74) u(x,t) = ^ 2 / s 2 «*.« - t,a) da,
| f (x,t) = - ±f$2
g (x.a - t,a) da.
Furthermore the conservation of energy is given by (10) and (35):
6.2 SCA TTERING PROBLEMS
(75)
265
W(t)=A [ KOH2 + IK-A)1 /2u(t)B2 ] = —"THTT / ^ n - 1 |k(s,a)| 2 dsda. 4(2it)
Kx^
w
Thus R is a unitary transformation, which transforms the evolution group (G(t)) into translations; J?w(u0,u*) is called the translation representation of (u0,u*) by Lax-Phillips [1]. Definition 5. Outgoing waves, Incoming waves (with the velocity c = I) . A wave u is called outgoing (resp. incoming) if it satisfies (76)
u(x,t) = 0 when |x| < t, i.e., u is zero on the forward light cone C+ (resp. u(x,t) = 0 when |x| > t, i.e., u is zero on the backward light cone C_ ).
The corresponding initial conditions (n°f\xl) are called outgoing (resp. incoming), and we denote by D + (resp. D")the set of these initial conditions. Definition 6. p-outgoing waves, p-incoming waves. Let p be a real number. A wave u which satisfies (77) u(x,t) = 0, V(x,t)with|x|
(u^u1) € D + (resp.D~) ** k = /*w(u°,ul) satisfies k(s,a) = 0 if s < 0 (resp. s > 0), (u^u1) € D* (resp. D~) «■♦ ksatisfies k(s,a) = 0 if s < p (resp. s > - p).
Proposition 8. i) The spaces D*, D " are orthogonal supplementary spaces in H. ii) The outgoing waves (resp. incoming) satisfy equipartition of energy for all t > 0 (resp. t < 0). iii) The evolution of outgoing waves for t > 0 is given by a semigroup (g+(t)) in WJ(Rn), and if (u^u 1 ) is an outgoing initial condition, then we have (79)
/?du1(s,a) = -(signs)^i? d u 0 (s,a),
or u ! ( s , a ) - - (signs)^u°(s,a),
with a similar propertyfor incoming waves (change - into + in (79)).
266
6 EVOLUTION PROBLEMS
PROOF, i) is a straightforward consequence of the above equivalence (78). ii) When (u0,u*) is in D + , we have
(80)
(u°, - u1) € D* and for all t > 0 (u(t),u'(t)) € D* and (u(t), -u'(t)) € D".
Since D+ and D~ are orthogonal spaces we have: (81)
3/ R n|gradu(x,t)| 2 dx-A/ R n|g(x,t)| 2 dx = 0.
iii) From (70) (or from (56) with h = k(.,t), f = u(.,t), g = v(.,t)) we see that k(s,a,t) = 0 when t<0 implies: (82)
^%,a,t)=^(s,a,t) s -(signs)u ( m + 1 ) (s,a,t),
that is (79) when t s 0. Then if we consider L2(R+,L2(Sn~ *)) as a subspace of L2(R,L2(Sn~ *)), the translation group in L2(R,L2(Sn~*)) given by (72) induces on L2(R+,L2(Sn" *)) an isometric semigroup for t > 0 only, given by: (83)
k(s,oc,t) = k(s - t,
whose infinitesimal generator is (84)
Ao = - ^ ,
D(A0) = Hi(R+,L2(Sn-1)).
The translation semigoup gives, in Lm(R x Sn~ ) (see (48)), a semigroup (g(t)): |g(t)v(s,a) = v j s + t,a) + v + (s-t,a)if|s|>t, g(t)v(s,a)*0if | s |
C- = ( * V [ ( s i g n s ) ^ ] * d .
We can identify C with the infinitesimal generator A (thanks to R ).
6.2 SCA TTERING PROBLEMS
267
These Calderon operators are unitary operators from W^R") onto L2(Rn) (this follows from equipartition of energy) with: Re (Cu,u) = 0, u in W^R"), C = C* or C ~. Proposition 9. The problem:findan outgoing (resp. incoming) wave u with finite energy satisfying (87)
i)5-^-Au = 0 in R n xR^(resp.R n xR t -) ) ii) u(0) = u°, with the only data u° in w'(R n ),
has a unique solution, and thus is a well-posed Cauchy problem. Now going back to the p-outgoing or p-incoming waves we have the properties: (88)
DjcD;, and Dp-.cDp" ifp'
Thus (D*)
_ is a decreasing family of spaces, whereas (D~)
Pp€R
„ is increasing.
" p t K
Spaces D* and D~p are orthogonal supplementary spaces: H = D* ®D"p. Moreover p-outgoing (resp. p-incoming) waves satisfy equipartition of energy from time t = - p (resp. p). The evolution of p-outgoing waves is given from this time thanks to an isometric semigroup (g+ (t)), t > - p, by: g+p(t) = g+(t + p). We have similar results for p-incoming waves but with p > t. The main properties of these spaces for scattering (which are very easy to verify thanks to the wave Radon transformation) are: DGWDJCD;
(89)
ift>o, G(t)D;cD; ift
ii) OG(t)D!«0, p t>0
HG(t)D-=0, p
t<0
iii) u a t ) DpS U G(t)D;=if, p teR
t€R
that is for iii): the set of eventually outgoing (resp. eventually incoming) initial conditions is dense in H. 2.2. Incoming and outgoing waves in electromagnetism Now we consider the Maxwell Cauchy problem (16) in free space, without charges nor currents:find(E,B) satisfying i ) - ^ | ^ + curlB = 0, | ^ + curlE = 0 i n R ^ R ^ c (100) ii)divE = 0, divB = 0, iii) E(.,0) = E0, B(.,0) = B0, (with free divergence) and E0, B0 in L2(R3)3.
6 EVOLUTION PROBLEMS
268
We write the Radon transform of this problem. First we easily have: (101) *
ii)a.E = 0, iii)E(.,0) = E0,
^+^(«AE)=0
inR s xS 2 xR t ,
o.B=0, B(.,0) = B0.
Let: (102) n(.,t) = -E(.,t) + oAB(.,t). Thus n satisfies: (103)
i)gr=-^-. with
Note that for c * l , then n = -^E+aABsatisfies: gjf^-eg^Here it is better to use the transformation Rd instead of R; with the notation v = R\ I 0 and B0 are in L2(R x S2) (see (48)), thus n0 is in L2(R x S2) but not in Lf(R x S2): (104)
n0(-s, -a) m - E0(-s, -a) + (-a) A B 0 (-S, -a) = (E0(s,a) + a A B0(s,a)).
The evolution of n is given by a group of translations; if we define the outgoing electromagnetic waves as previously by n(s,a) = 0 when s < 0, we obtain that an electromagnetic wave is outgoing if it satisfies the relation: (105)
l(s,a,t) = - (sign s) a A B(s,a,t), for t > 0, (s,a) € Rx S2,
and thus the initial conditions must satisfy: (106)
E0(s,a) = -(signs)aAB0(s,a),
(s,a)€RxS 2 ,
which is equivalent (with (101)ii) to: (106)*
B0(s,a) = (sign s) a A E0(s,a).
Of course incoming electromagnetic waves will correspond to initial conditions with + instead of - in (106). Thus if (E,B) is an outgoing wave, then (E, - B) is an incoming wave for - 1 . Hence using orthogonality of these waves, we have: Proposition 10* Proposition 8 is valid in the electromagnetic case, by replacing only condition (79) by (106), concerning initial conditions of outgoing waves.
269
6.2 SCA TTERING PROBLEMS
We can also define outgoing and incoming Calderon projectors P+ and P~ in the space H of free divergence fields of L2(R3)3 x L2(R3)3 on the spaces D+ and D~" of outgoing and incoming initial conditions, then outgoing and incoming Calderon operators C+ and C " in space of free divergence fields of L2(R3)3) by: B0 = C+E0 given by (106)' (for C+). Furthermore C+ and C~ are unitary operators, with: C 2 = - I (C2= - c~2I whenc*l) and Re(CE0,E0) = 0, C = C + andC~. The Calderon projectors P+ and P~" are also given by:
(107)
P+ = I(I-S),
P-=1(I + S), with S=£ _° c+ ^ + J.
Like for scalar waves, we have: Proposition 11. The Cauchy problem: find an outgoing (resp. incoming) electromagnetic field (E,B) in R 3 xR + (resp. R 3 x R " ) with finite energy, satisfying (100)i),ii) and only the initial condition: (100)iii)'
E(.,0) = E0 with divE0 = 0,
E0 € L2(R3)3,
or with given initial condition on B only, has a unique solution (E,B) in C(R+, H) (resp.C(R~, H)), and thus it is a well posed problem. 2.3. Incoming and outgoing waves with a bounded obstacle; Lax-Phillips theory Let Q be the open complement of a bounded obstacle Cl\ and let B be a ball so that CT is contained in it. Let (G (t)) be the evolution group of waves in H = W!o(Q) x L2(Q) with a Dirichlet boundary condition. We first note that if u is a wave in Rn x R + which satisfies (77), that is u is zero in the truncated forward (backward) cone C + (resp.C ~), then u(t) satisfies the Dirichlet boundary condition when t > p (resp. t < - p). From this trivial remark we can consider D + (resp. D ~) as a p-outgoing (resp. p-incoming) space for the mixed problem. Then we apply the evolution group (Gfl(t)) to these spaces. We can prove (see Lax-Phillips[l]) that the essential properties (89) are also valid when substituting GQ(t) and HQ for G(t) and H. The only difficulty is (89)iii) which is proved to be equivalent to: (108) lim E(u,t,K) = 0, oronly liminf E(u,t,K) = 0, V K compact set, K c Q . Now let F = (f°,fl) € U G(t)D*. Then there is T> 0 so that F e G(-T)D£, and thus there is F + € HQ so that Gft(t)F+ = G(t)F, V t > T. This defines a mapping W+: F —* F+ which can be extended continuously from H into HQ. We define a mapping W _ in a similar way.
270
6 EVOLUTION PROBLEMS
Definition 8. Wave operators. The mappings W+, W_ from H into Ha defined by: (109) W ± F =
lim GQ(-t)G
wftJt J r/ie natural imbedding of HQ into H, are called the outgoing and the incoming wave operators. We can also define wave operators from L2(Rn) into L2(Q) using the groups generated by the operators iA1/2, A being the Laplacian in Rn or in Q, like in Wilcox [1]. Now these wave operators have the essential properties: Proposition 12. The wave operators are unitary operators from H onto Ha, with (110)
G a (t)W ± =W ± G(t),
Vt€R.
From this we can define the "scattering operator** SinH by: (111)
S = W;*W..
Then from Proposition 12, S is a unitary operator in /f, such that SG(t) = G(t)S, for all real t. These operators are a basic tool in scattering problems by a soft bounded obstacle. Obviously we can generalize to other boundary conditions, for example to hard obstacles with a Neumann condition. That the scattering operator S contains all the scattering information is seen from solving the inverse scattering problem (see Lax-Phillips [1] p. 173): Let Ox and 02 be two bounded soft obstacles; let Sf and S2 be the associated scattering operators in H. Then S% » S2 implies Ox = O . Furthermore we can easily deduce spectral properties from the wave operators on the selfadjoint operator A = - A with Dirichlet boundary condition: A has an absolutely continuous spectrum with no eigenvalue. Here we are essentially interested in outgoing and incoming waves. Thanks to the wave Radon transform Rw we define mappings R* and R"" (called outgoing and incoming representations) by:
(112) __
tf-rt^w;1,
R'^R^Z1* ^
i.
n i
They transform the evolution group in Ha into translations in L (R,L (S )), with: R+(Dp = L\p,+<*)X2(Sn-1)), ir(D;) = L2(( -oo, . p)9L2(Sn-{)). We can also define "outgoing and incoming initial conditions" DQ+ and DQ*" as the inverse images of L2((0,oo),L2(Sn"!)) and L2((-co,0)fLHsn~l)l respectively by R+ and R~, or DQ+ « W+D* and DQ~ = W^D", and the Calderon oroiectors P + . P ~ on these soaces bv:
271
6.2 SCATTERING PROBLEMS
,
/ J\ ~ 1 P ^ - W ^ C W ^ ) """1 ,1 oa«n^dDP Q- -_=WW _Dp~-/ («W but DQ+ and DQ " are not orthogonal, nor necessarily graphs of operators ! Then the problem: find an outgoing wave u with finite energy satisfying:
i ) ^ 4 - A u = 0 inQxR*, (113)
at2
ii)u| r x R + = 0, iii) u(.,0) m u° with u° given in wl(Q), and supp u° c B' = R M L
has a unique solution u in C(R£, W^Q)), with ^ in C(R^,L2(Q)). Thus it is a well-posed problem (as a trivial consequence of Proposition 9), but we cannot eliminate the condition on the support of u°! Application to stationary waves We can also define notions of outgoing or incoming waves in the stationary case in the following way. First we consider the stationary problem in a form similar to (1) in free space, in the whole space: (114)
4W + XW = F inR n ,
where A is given by (4) with A= - A, with given X in C and F = (f1,f2) given distributions in Rn with bounded supports. Using Definition 6, we have: Definition 9. A solution of (114) which is also an eventually outgoing (resp. incoming) initial condition for the evolution problem (68) is called a X-outgoing (resp. X-incoming) stationary wave. When F = (0,g), then W = (w,Xw), with w satisfying the Helmholtz equation: x\r - Aw = g, and w is called a X-outgoing (resp. X-incoming) stationary wave when W is X-outgoing (resp. X-incoming). Note that a X-outgoing stationary wave w is a ( - X)-incoming stationary wave also, and conversely. We can easily obtain results on stationary wave thanks to the wave Radon transformation Rw. We only give one example of application. Applying transformation Rw to (114) (this is possible up to now if F is in H, but we can generalize Rw to distributions, see Lax-Phillips) gives (115)
g + Xk = f inRjjeS11"1,
with k = /*wW and f=/*wF, f having a compact support. The solution with support bounded to the left (resp. to the right) which will give the outgoing (resp. incoming) solution of (114) by its inverse transform, is (116) k 0Ut (s,«)=/ S co e A(s, - s) fl[s',a)ds', kta(s,«) = - r e A ( s ' - s ) f ( s ' , « ) d s \
272
6 EVOLUTION PROBLEMS
WhenFisinff, thenfisinL^RxS 11 ' 1 ), and we see that if (Re X)>0, onlykout is in L2(R xS""1) whereas if (Re X) < 0, only k^ is in L2(R xS n " ! ), the other solution being not even tempered. (When X is real, both solutions are tempered, butnotinL^RxS 11 " 1 )) Then applying this to F = (0,-8) gives the elementary X-outgoing (or Xincoming) elementary solution of the Helmholtz equation * r When n = 3 we have the very simple result: (117)
*x(x)ss^S^withr=|x|.
When F = (0, -f)> f being a distribution with bounded support, the X-outgoing (resp. X-incoming) stationary wave w is obtained by convolution of f with $ x , and then we can verify that Definition 9 agrees with Definition 1 chap. 3 thanks to Theorem 1 chap. 3. Now we consider a domain Q exterior to a bounded soft obstacle. Then we can define X-outgoing (resp. X-incoming) stationary waves as in Definition 9. Here we have to use a more general framework taking waves into account which are not offiniteenergy, thus not in HQ = W!0(Q) x L2(Q). So we define: (118)
#loc(Q) = {U = (u°,ul), such that CU€ Ha,
VC€D(Rn)}.
Waves which are not of a finite energy may be of two very different types according to different situations: i) scattering states; they appear in usual stationary scattering problems as Xoutgoing (or X-incoming) stationary waves. They are generalized eigenvectors, w+(x,£) (resp. w_(x,t)) with £ in Rn, i.e., satisfying: i)(A + S2)w±(x,S)=0 inQ, (119)
ii)w ± (.,t)€H* loc (5), i.e., Cw±(.,t)€Hi(G), VC€2>(Rn), iii) wr+(x,£)=w+(x,S) - w0(x,£) is an i| 11 -outgoing wave, (resp. wr_(x,£) = wJx,{) - w0(x,S) is an i |S|-incoming wave),
with w0(x,6) as e a \ Recall that iii) means that wr ± (.,x) satisfies the outgoing (resp. incoming) Sommeifeld condition. This problem (119) has a unique solution thanks to the Rellich lemma (Lemma 1 chap. 3), wr being the usual reflected wave by the obstacle for the incident wave wQ. Then if we define for all fin L2(Q) with bounded support: (120) f+«)=/Qf(x)w+(x,?)dx,
resp.fJO=/Qf(x)wJx,«)dx,
the mapping f —> f+ (resp. f J has a continuous extension r (resp. F") from L (Q)
6.2 SCATTERING PROBLEMS
273
onto L2(Rn). Furthermore F* and F~ are unitary (up to a constant), and are natural generalizations of the Fourier transformation which diagonalize the Laplacian AD with Dirichlet conditions. They are called the incoming (resp. outgoing) spectral representation oftP. The operators F~ lF* and F~ lF~ with F the Fourier transformation, correspond to (inverse) wave operators W_ and W+ in L2, see Wilcox [1]. Note that w+(x,£), w_(x,£) are tempered distributions. ii) resonance states. Let v/bea non trivial solution with locallyfiniteenergy of: |i)-Aw + X2w=0 inO, withX€C, ii)w| r = 0, thusw€Hi loc (Q), (121)
iii) w is a \-outgoing stationary wave, i. e. w has the asymptotic behavior w(x)=^j- 9(a) + CHr-2")forr = | x | -♦ oo, 9(a) a regularfunction of a = x/r.
Definition 10. A wave w satisfying (121) is called a resonance state and \ a resonance. (We emphasize that this terminology is not universal: sometimes, see Sanchez Palencia [1], resonances are called scattering states.) From Rellich lemma, a resonance cannot be real. Furthermore it must satisfy Re \ < 0 (see Lax-Phillips [1] p. 161). Thus a resonance state cannot be tempered; its physical nature is quite different from that of a scattering state. There is at most a countable set of resonances, as a consequence of the Rellich compactness result: Proposition 13.
The unit ball of D(A) in Hft, {F € HQ, \\A Fll + IIFII
precompact set in Hloc(Q) equipped with the family ofsemi-norms for all compact setKinRn (122)
IIUIIK = ll(u 1 ,u 2 )ll K = s [|/ K n Q [|gradu 1 | 2 +|u 2 | 2 ]dx] 1 / 2 .
A complementary result on resonance states is the following (Lax-Phillips [1]) Proposition 14. The problem:findwwith locallyfiniteenergy satisfying i) X2 w - Aw = f, with f € L (Cl) with compact support, (123) ii)w| r = 0, thusw€H£ loc (Q), iii) w (or W = (w,Xw)) is a \-outgoing stationary wave, has a unique solution when \ is not a resonance. In order to study resonance we define the following spaces and projectors: (124) D P = D * © D ; , Kp the orthogonal of Dp in HQ9 thus# a = D
274
6 EVOLUTION PROBLEMS
Q* « I - P* (resp. Q" = I - PJ), the orthogonal projector in Ha on the orthogonal space to D* (resp. D~), and Q = Q+Q" = I - P p , the orthogonal projector on Kp. A basic tool to study resonance is the family: (125)
Zp(t) = QjG n (t)QJ,
t£0.
We can prove that (Zp(t)) has the properties: i)D p c kerZp(t), V t > 0 , Zp(t)KpcKp, V t * 0 , ii) (Zp(t)) is a contraction semigroup of class C° in K p , with: Zp(0) = Q , and Zp(t)U0-> 0 when t —+oo. Let Bp be its infinitesimal generator in Kp. Then the main point is: Proposition 15 (Lax-Phillips). The operator Zp(2p) (XI - Bp)~1 is compact in K p , forall\>0. This implies that the spectrum of Bp is a sequence of eigenvalues in half plane Re X < 0, with finite multiplicity without accumulation point at finite distance, and that the set of eigenvalues ofBP is the set ofresonances. Moreover if W*. is an eigenvector of Bp for the eigenvalue X. then we obtain a resonance state W. by: W.(x)»lim W* (x) when p -♦ oo, xfixedin Cl. Then resonances are useful to study evolution of waves when t goes to infinity: Proposition 16. Limiting Amplitude Principle, Let g be in L2(Q) with bounded support, and X a complex number which is not a resonance. Then the solution u with finite energy of the problem: i ) ^ | - A u = eAtg inQxRt, (126)
ii)u(.,t)| r = 0,
Vt€R,
iii)u(.,0) = u°,
^ ( . , 0 ) = u!,
with given (\x°tul) in W*(Q) x L2(Q), satisfies: (127) U(t) = (u(t), §J (t)) converge in Hloc(Q) to V(t) = (eAtv, XeAtv) when t -> +oo, where v is the \-outgoing stationary wave solution of: (128)
X ^ - A v s g inQ.
275
6.2 SCA TTERING PROBLEMS
Another result is the local decrease with time of wave with respect to energy. The first result is due to Morawetz [1]: Proposition 17. Let Q' be a bounded star-shaped domain. Then the solution u withfiniteenergy of: i ) ^ - A u = 0 inQxRj, (129) ii)u(.,t)| r = 0, V t € R ,
Cl^R^Q',
I iii) u(.,0) = u°, | * (.,0) = u1, with given U0 = (u^u1) in wi(Q)xL2(Q), satisfies (with C a constant and with t > 2p): (130) E(u I t,B p nQ) = i / B
2 na[|gradu(t)| +|l?
W h d x * jf
IIUQII^.
When the obstacle is star-shaped, and even when there is no captive ray (for this notion, see for instance Bardos [1]), the energy of the wave is exponentially decreasing on every compact set when t goes to inflnity (for an odd dimension of space). Moreover we have (Lax-Phillips [2]): Proposition 17*. Let Cly be a bounded star-shaped domain. Then for every LL given in HQ with bounded support, there are resonances (\X j = 1,...,N, with associated resonance states w., constant numbers C, j = i,...,N, so that the solution u of (129) satisfies for any compact set K: (131)
E((u(t)-
2 UltoN
QJ e J w.\uK)*C(K)e J
N+1
' .
Thus resonance states allow us to have an asymptotic behavior of the wave when t goes to inflnity. This formula would be generalized to more general domains. But we know (see Ralston [1]) that if there are arbitrarily long rays following optical geometry, contained in a ball of finite radius, energy may be arbitrarily slowly decreasing in compact sets. For a soft obstacle which has a cavity with a tiny hole (of diameter 6) resonances converge to the eigenvalues of the (closed) cavity when 8 goes to 0 (see Beale [1]). There are other similar examples with waves due to a drum in Sanchez Palencia [1]. Obviously all these results can be generalized, to hard obstacles with a Neumann condition, and to systems of equations, in elasticity and in electromagnetism. For Maxwell equations we can use potentials (scalar and vector potentials, and Debye potentials also) to obtain results thanks to those for the wave equation. We shall not develop these questions here, and we refer to Schmidt [1], Dassios [1], Bardos [1].
276
6 EVOLUTION PROBLEMS
3 . CAUSAL PROBLEMS
A great number of evolution problems in scalar waves and in electromagnetism have no initial condition but are of the type: Causal problem with a support condition. Given a distribution f in Rn x Rt with support bounded in the past: (132)
suppf c Q**{(x,t),x€R n ,t^a},
find a distribution u in Rn x Rt satisfying: (133)
|i)Ou=f inD,(RnxRt), ii)suppu c Q*,(orsuppu C supp f+C+with (23)), i.e., with support bounded in the past, with the same bound a.
Thanks to the elementary solution <& with support in the future light cone (see section 1.3) we have: Theorem 1. The causal problem (133) has a unique solution u, which is given by convolution: (134)
u=**,f. x,t
PROOF. We have to prove that the convolution product has a sense with the above hypotheses on the supports of f and $, that is their supports satisfy:
(135)
A n (K - B) is a bounded set, for all compact set K in Rn x Rt>
with A * supp f, B = C+ » supp <&. Let (x,t) be in A, (y,t) in C+, and (z,X) in K, with t = X-t,x = z-y.Wehavea^t, thustsX-a, and |y| ^ T S X - a , thus x and tare bounded, giving (135). Now if w is a solution of (133)i) with support bounded in the past only, then applying the convolution by $ to (133)i) gives: u=*»f=$,(Dw)=(D*)*w=i*w=w, which proves uniqueness of the solution with a weaker hypothesis on the support. ® Remark 1. Obviously we can transform a one-sided Cauchy problem (11) (for t > 0 only), into a causal problem, by usual extension by 0 for t < 0, so that it becomes: find U a distribution in R n xR t with support in Q* = RnxR+t satisfying: (136) DU - F + u18(t) + u° 8*(t),
277
6.3 CAUSAL PROBLEMS
F being the extension of f by 0 in the past. Equivalence between the two problems is realized if we assume usual regularity conditions on the data. Remark 2. For n = 3, the solution u of (133) is given at least formally by the "retarded potential": (137)
ufrO-^J^-jp^
We have solved the causal problem (133) by convolution. Under similar hypotheses we can apply a Laplace transformation method which gives also Theorem 1. For instance, let X be a Banach space of functions on Rn, let S'(RfX) be the space of tempered X-valued distributions. We define (138)
L+(R,,X) = {f € Z T ^ X ) , 3 E € R, e" * f € S>(Rt,X)}, D\(X) = {f € JD'(Rt>X) with suppt f bounded in the past}, L^(Rt,X) = L+(Rt,X)nDV(X)
Taking for instance fin L^CR^X) with X = L2(Rn), then we prove that (133) has a unique solution in the same space (see Dautray-Lions [1] chap. 16.1) which is obtained thanks to the resolvent of the operator A = - A. Furthermore the Laplace method allows us to define well-posed causal problems without condition on supports: we substitute for the condition of bounded support in the past a L2-condition with a weight in time, such as L2 (X) (also denoted by e^L^R, X)), with y > 0, defined by: (139)
L27(X) = {f: R ^ X measurable with Ilf1l2x=/Re"27t||fll^dt< oo}.
A typical example (which is relevant for wave equation and for Maxwell equations) is the following: let A be an operator in a Hilbert space X, which is the infinitesimal generator of a unitary group (G(t)) in X. Let f be a given function in L2 (X), with y > 0; find u in L (X) satisfying: (140)
| f + Au = f.
Proposition 19. The problem (140) has a unique solution u in L2 (X). PROOF.
The Laplace transform of a function 0 in L2 (X) being defined by:
(141)
$(p) = / R e-pt«t)dt,
p
the Laplace transform of (140) is: (142)
(pI + A)fi(p)=f(p).
= £ + in , t > Y,
278
6 EVOLUTION PROBLEMS
When U T > 0 , this equation has a unique solution, given by: (143)
u(p) = R(p)?(p) = (pI+A)-1?{p).
Thanks to the Stone theorem, we have: (144) IR(p)l*{, and thus: dn =/ = f_ DR(o)f(vriL dn** \4 HI** llf„ , , (145) /f_ Hufolll Ifi(p4dn R IR(p)f(p)lxdn which implies the Proposition. Furthermore if f has a bounded support in the past, we obtain that u also has this property, with the same bound (thanks to a Paley-Wiener theorem). 3.1. Some examples with the wave equation 3.1.1. A typical example of causal problem We first consider the simple problem which will be very useful for solving other problems with the d'Alembert equation:finda function v on R satisfying, with a given function f on R, and real X > 0 (146)i) vM+X2v=f, wherev H =|~, with either: (146)ii) supp v c supp f + R+, if supp f is bounded in the past, or (146)ii)' v€Lj(R) = eYtL2(R), iff€l^(R). In both cases, the solution for f = 0, v0(t) s A e ^ + B e"^1 must be zero. Thus (146) has the unique solution: (147)
v(t) s / o o S m M x t " S ) f(s)ds = (Gx.fXt),
with Gx(t) = Y(t) ^ ^ , Y(t) being the Heaviside function, and also: (147)'
e - v V t ) = / M e ^ - s ) . 5 ^ i > e - ^ s ) d s = G y , A< (e-%)),
with: GT>A(t) = Y(t)e- Yt 5l n : ^; thus GyX 6 L!(R), with / R |GyX(t)| d t < ^ . Therefore when f e L^R), we have v € LJ(R) and: (148)
My^Vly
withIMIy=/Re"2Yt |v| 2 dt.
6.3 CAUSAL PROBLEMS
279
3.1.2. An example with boundary conditions Now we consider the evolution problem: given a (regular) bounded domain Q in Rn, and given fin l^X), find u in Lj(X), X = L2(Q), so that (149)
i ) ^ "2 - A u = f inQxR., n 9t ii)u| 2 = 0, I = TxR.
Using the spectral decomposition (A2.,<>.) of the Laplacian with Dirichlet condition in L2(Q), we have: (150)
iKO-ZujCQftj, andRt) =2^(1)*. with I/ R e- 2Yt |fj(t)| 2 dt< oo,
and Uj has to satisfy: Uj e L^R), with: (151)
32u-^u.-f., j
€
N
.
From section 3.1.1, we know that this problem has the unique solution: (152)
, sin X.(t - s) UjCt)-/^ 1 fj(s)ds = (GAj,fXt),
and Uj satisfies (thanks to (148)): (153)
1 I I ^ S ^
This implies that u is in L;(X), X = L2(Q), with: (154)
1 llull^Cllfl^ C = ^ - .
Moreover since ((-AD)1/2u). = Xj Uj, we obtain with Y = HQ(Q), and C = y max (^-, 1) (155)
%
$ C
" , '
Furthermore differentiating (152) with respect to t, we have: (156)
u'j(t) = (Y(t)cos(Xjt)).fj
and thus: llu'jlly :£ | llf}^, giving: (157)
llu'll^llfl^.
280
6 EVOLUTION PROBLEMS
Thus we have obtained: (158) u € Lj(Y) with Y * H*(Q), U' € LJ(X) with X = L2(Q), thus e"Ytu € H ^ Q x R). Then we can prove also the trace result: e"Yt | ^ | € H"l/2 (I) on I ■ T x R. But this trace result is not optimal. Using the proof of Lions [3] p. 40, 41 in the framework of L2 (X) functions, we get:
<159>
| l z € e Y t L 2 ( Z ) , i.e., • ^ g l 2 « L a t e .
Remark 3, An a priori estimate. We multiply (149)i) by e~2ytu9 (the time derivative of u) and integrate by parts. We get: (160)
Y/ O x R e"^[|uM 2 +|giadurtd]cdt-/ O x R e- 2 y t fu»dxdt^0.
Then using the Cauchy-Schwaiz inequality, we obtain the estimate where IMI denotes theLj(L2(Q))normofv: (161)
Y[|Bu'll5 + IgradullJ]<^;llflJ.
This also implies (158), thanks to the Poincare inequality. ® Remark 4. We can solve in the same way the evolution problem (149) but with a (homogeneous) Neumann condition. The only new difficulty is with the eigenvalue 0. We obtainfinallythe same results (158) for the solution. Usual trace results give u | € eY H1/2(Z), but this is not an optimal result. If the unifonn Lopatinski condition (see Chazarain-Piriou [1] p. 362) were valid, we vt
1
could hope to have u | € e H (£). But this is not the case. From results of Lasiecka-Triggiani [1], we have u | € eYt H2/3(X), and when Q is a parallelepiped u | € eY H
(I), t > 0.
1
•
3.1.3. An example with transmission conditions, Integral methods Let Q be a regular bounded in Rn with boundary I\ We consider the problem: find u with given jumps (p, p') across I = T x R, so that a2,, i)^-Au =0 (162)
in(R n \r)xR t ,
» M 2 - P . [^}2 = P\ (P,P') e L ^ Y ) , Y=H I / 2 (OxH- 1 / 2 (n, in) supp u c supp (p9or) + C+,
281
6.3 CAUSAL PROBLEMS
where [v] is the jump of v across Z » T x R. Writing (162)i) in D'fR 1 ^), we see that u must satisfy: (163)
| ^ - A u = f, withf=-(p , 8 2 + div(pn62)),
and thus we obtain the solution of (162) in the form: (164)
u = $*f. x,t
But we have to specify the functional space, especially with respect to the boundary conditions. So it is better to use a Laplace transformation. Then problem (162) becomes:findu = u(p) (with Re p > $0), satisfying |i)(p 2 -A)iU0 inRV, (165)
r3fi
This problem has for each p a unique solution in: (166)
X = H1(A,RM^ = {u, u\a € Hl(AyCl), u|Q, € H!(A,Q%
Then we have a continuous lifting of the jump conditions into L^R^X) so that we are reduced to solving the problem:findU in L^(Rtf H2(Rn)) satisfying (167)
^
- AU = F,
F given in L^,L 2 (R n ».
This problem has a unique solution. Thus we obtain a unique solution u of (162), in L?(Rt,X) with X given by (166). Thus we can define causal Calderon projectors P? and P£ in L^R^ Y) by: (168)
Pffe>,p')-(u|ri, g y ,
Pec(p,P') = ( u | v g | F e ) .
with the usual property: P? + P£ = I. We get u in R3 x R from (164) and the retarded potentials by the KirchhofT formula (see Bamberger-Ha Duong [1]): u^LV + P^with:
282
6 EVOLUTION PROBLEMS
From these definitions, there is no jump of the trace of v = Lcp' across £, nor of the normal derivative of w=P°p across 2. From Lc and P°, we can define "usual retarded" integral operators on the surface £, and an operator Sc such that
(170)
Pf=^(I + Sc),
P°=4(I-SC),
withsCss
(2^c2jc].
and: (171)
LV«iV| 2 .
ICP-I^PL
+^PL,
3 /c»»
Of course it is interesting to specify these operators in a L2 (X) framework. We will do it, but optimal regularity results are missing. a) A regularframework. We define for real s and y > 0 (172)eYtH^(RW=f{f, e-ytf€S*(Rn+1), < y = / R n x R ( | p | 2 + £2)s|f«,p)|2cttdn< 00} with p a Y + in, ?(6,p) being the x-Fourier transform and t-Laplace transform off, Then we define for Cl a regular open set in Rn, by restriction, the space eY H^(Q x R), and (thanks to an atlas) eytH^(r x R), see Chazarain-Piriou [1] p. 412. Thus we have: (172)'
8u 2/T 2/ 2/i 2/ n l / x R)kdef eJyi*Hj(Q V [u € 1^(1/(0)), ^ ,du g € L^(I/(Q))}.
Now if we successively consider the wave problem (149) with Dirichlet boundary conditions, and Neumann conditions, we define the trace spaces: (173)
j o x ^ f u 1 ™ | 2 , u€e 7t Hj(axR) the solution of (149)}, lii) Yy = {u° = v | r v€eYtHj(QxR) the solution of (149)i) with §J | 2 = 0}.
We recall from the regularity results (Remark 4 and (159)) that: (174)
X^ceYtH2/3(2), Y^ceYtL2(I).
The elements of X x Y are traces at the boundary of waves u such that: u€D y (2?)^D/* D ) + Dy(5N), with: Dy(/?D) = {u€eYtHj(acR), D U € L J ( X ) , X = L2(Q), u| 2 = 0J, Dy(*NMu€eYtHj(QxR), DueLj(X), X = L2(Q), §f l 2 = 0}.
6.3 CAUSAL PROBLEMS
283
Let f be given in L^(X), X » L2(Q). Let u f (resp. vf) be the solution of the Dirichlet (Neumann) problem for f. Let U = U f « vf - uf. Then U satisfies: (175)
I i ) D U = 0,
!u)U|2 = v|2 = u<>, g , x . . g ,
I*t 9 (resp.9,) be the mapping f€ I^(X) ^ with: D*(B) = {U€Dy(B),niJ = 0}.
( ^
We can prove that ker 9 = ker 9I# Then taking the quotient space Ly(X)/ker 9, we obtain an isomorphism from L^XJ/ker 9 onto D°y(2?) and onto a subspace Gt of X^xY* Then we define an isomorphism C l = Clc from X^ onto Yly by: (176)
Ci(u°) = ^ l
2
(with(175)).
This operator C1 is the interior causal Calderon operator. Now we can operate in a similar way for the exterior domain. If the trace space X1 (resp., Y l ) is the same, for the interior domain as well as the exterior domain (which seems to be a very natural conjecture for regular domains), that is X1 = X e =X (resp. Y1 =Y* =Y ), we also define a subspace G e of X x Y and an operator C c for the exterior domain which is the exterior causal Calderon operator with G(Ce) = Gc; C e is also an isomorphism from X onto Y . Note that G1 and G c are closed subspaces in X xY with intersection reduced to {0}. A natural conjecture is that X x Y = G(C i )^G(C e ). b) A more general framework of "finite energy". Now we define: | w ^ = {u€e 7t H^(nxR), Du = 0 i n Q x R } , (177)
We can consider the space W as a natural space of waves with finite energy on both sides of I = T x R. The corresponding space of sources concentrated on 2 is (178)
Zy = {(p,p') = ([u] 2 ,[^] 2 ), u€W y }.
The mapping u —> (?,(>') is an isomorphism from Wy onto Zy, with inverse: (179)
u =
284
6 EVOLUTION PROBLEMS
Then Z is a Hilbert space, which is the direct sum of two closed subspaces G., Ge with intersection reduced to {0}, which are the set of boundary values of elements in W a and Wya : Z = G.eG c . The projectors on G. and Ge (along Gc and G.) are the causal Calderon projectors in Z . Furthermore we easily see that G. and Gc are the graphs of two operators, the causal Calderon operators, which are extensions of those defined in the preceding section. A reasonable conjecture is that Zy ■ Xy x Yy with the spaces of traces (180)
Xy = {v=u| r ,u€W^,Y Y = { v s s ^ | r , u € W f } ,
and that these spaces of traces for Q are equal to the spaces of traces for Q*. We refer to Bambeiger-Ha Duong [1] for numerical implementations, and for results with the space ^(R,X) = {u,/ R |p|2s»u(p)l^dfi< oo}, with s = 3/2, and X«H 1/2 (r). We note that proceeding as in Remark 3, we obtain the positivity property: (181) Y/ axR e- 2lft ([iu'| 2 +|gradui 2 l)dxdt=/ 2 e- 2Yt u'^drdt>0. 3.1,4. An example with a waveguide Let Q+ = QxR + z (with z=x3) be a semi-infinite cylinder (modelling a soft waveguide) with a regular bounded cross-section Q (with boundary O in R2. Then we consider the wave problem with a given value u° at the end of Q+: i
(182)
>^"
IH) u(xAt) = u ° & t ) o n °* R * u ° € L \ ( X ) > X = L 2 ( Q ) ' y > °* iii)u(x,z,t) = G onliXRjsTxR^xRt, iv)u€L2Y(L2(Q+)).
We can solve this problem due to a spectral decomposition of the Laplacian Ax with a Dirichlet condition. Let (X2.,^) be eigenvalues and eigenmodes so that we can decompose u and u° into: (183)
u(.,z, t) = I Uj(z,t) ^ ,
u° - X u? *,.
Then (182) is reduced tofindingu- satisfying: a 11
3 U
(184) ii)ui(0,t)-u,0(t), teR.
285
6.3. CAUSAL PROBLEMS
Now using the (time) Laplace transformation, we obtain:
(185)
i ) - ^ - < P 2 + X?)fij = 0
inR£,
ii)fij(0) = fij.
The solutions of (185)i) are given with two constants Aj, Bj by: (186)
uj(z) = A j e jZ + B j e" jZ , with 0? = p 2 + X?.
Now in order to have a solution in L y(X), we have to take Aj = 0, with Re 0j > y. Thus the solution Uj is given by: (187)
fi^-e^flj0.
The image of the half plane Re p £ y by the map p —»p2 is the set: 2
9
7
i.e., the exterior of a parabola/^, which contains the set {6j =p z + Xj, Rep > y}. We have the inequalities: (188)
|u j( z)M|uj>|,
/R+|fij(z)|2dz=I^|uJ0|2^^|flJ°|2.
This implies that (184) (and thus (182)) has a unique solution which is given thanks to a contraction semigroup (G(z)) (of class C°) in L2(X) by u(.,z,.) = G(z)u°. The infinitesimal generator A of (G(z)) is defined thanks to the spectral decomposition of the (transverse) Laplacian with Dirichlet condition and thanks to the time Laplace transformation by: (189)
Atyp) = -fl,flj(p)= (p2 + X?)1/2fij(p), with Re p > y, Refl,> y,
and its domain is: (189)'
D(A) = {v€L5(X),Z/ R |9 j | 2 |^ j (p)| 2 dn
Since |p| s 10j I , we have D(A) c eYtH^(QxRt). Thus the problem (182) is equivalent tofindingu in L^(X), X = L2(Q), satisfying: (190)
i ) | - A n s O , z>0, |ii)u(0) = u° onQxRt.
The operator A is the causal Calderon operator of the waveguide.
6 EVOLUTION PROBLEMS
286
3.2. Some examples with Maxwell equations 3.2.1. Causal Maxwell problems in the whole space We first consider the basic problem: let J and M be given electric and magnetic currents in free space, which are distributions in the whole space R3 x Rt, with supports bounded in the past. Like in the stationary case, we assume that there exists magnetic currents, which will be useful to consider in order to solve other problems in electromagnetism. We also assume the existence of related electric and magnetic charges p and C- Let cQ, M0 the permittivity and the permeability of the free space resp.. We have to find the electromagnetic field (E,B) satisfying i)||-c2curlB = - ^ J , (191)
i i ) | ^ + curlE = - M
inR^xR^
iii)divE=fjrp, anddivB = C, iv) supp (E, B) C supp (J, M) + C + ,
with: (192)
^ + divJ = 0,
§£ + divM = 0.
We first note that the equations (191) iii) are consequences of i), ii), iv) (and (192)). By the usual tricks, we eliminate B, then E, to obtain the d'Alembert equations (with (11)): (193)
DE=j,
DB = P0m,
with (j, m) given by: (194)
- j = i ( - y j + gradp) + cur!M,
-m=^(^^+grad?)-curlJ,
and thus we get the solution (E,B) of (191) thanks to the convolution product with the elementary solution O (see (24) and (135)) (with c = 1 to simplify): (195)
E = *j,
B-O.G^n).
We only note that taking the divergence of (195) implies (191) iii): (196)
divE = **divj =
divB =
Theorem 2. Let (J,M) be given currents with support bounded in the past. The causal problem (191) has a unique solution (E,B) in D'(R3 x R ) 3 x D'(R3 x R ) 3 given by (195) with (194). Remark 5. Causal vector and scalar potentials. When M = 0, we define the causal 1
1 o^
vector and scalar potentials: A = n0
6.3 CAUSAL PROBLEMS
287
3.2.2* Causal problems with currents on a surface. Transmission problems Now let (JL,Mr) be given electric and magnetic currents concentrated on a regular surface T in R3 which is the boundary of a bounded open set Q. Furthermore we assume that they satisfy, with notations (138): (197)
J r , M r € L°(Rt,X)
with X = H"1/2(div,r).
Then the electromagnetic field (E,B) due to these currents must satisfy (191). Theorem 2 implies its existence and uniqueness at least in the sense of distributions. Thanks to the hypothesis (197) we can prove (for instance thanks to a lifting) that thefield(E,B) has finite energy, that is: (198) (E,B)€L2(Rt,Y)xLj(Rt,Y), Y = L2(R3)3, thenE, B c L ^ H f c u r U R V ) ) . Thus E and B have traces (n A E | , n A B | ) on each side of the boundary I of Cl x R. Furthermore they satisfy (197). Using jumps formulas (25) chap. 2 we obtain that (E,B) satisfy (199) | w - c 2 c u r l B = c 2 [ n A B 1 2 S 2 ' S + curiE—[nAB] 2 6j f inR^xR,, div E * - [n.E] r div B = - [n.B]2, so that if we compare to (191), we obtain: (200)
J 2 = - ± [n A B ] r
M 2 = [n A EJ r
Moreover applying the divergence to (199), we obtain (201)
- 1 [n.E]2 * c 2 div ([n A B]2 6J) , | [n.B]2 = div ([n A EJ2 5^.
These basic relations correspond to those of the stationary case (see (84) chap.3). They are compatibility relations for traces, and they give the normal components of E and B thanks to the tangential components of B and E. Thus we have obtained with hypotheses (197), results similar to those of the stationary case (see Theorem 3 chap.3), but here (E,B) satisfies (198) and thus is of finite energy in the whole space. Furthermore (E,B) is obtained by the formulas (195) (with retarded potentials), with (j,m) given by (194), and with: (202)
p 2 = - e0 fa.E]2, Cj — [n.B]2.
These formulas (195) with surface currents are the "Stratton-Chu" formulas in evolution case (see (87) chap. 3 in the stationary case), which are simply written in a form similar to (96) chap. 3. We define like in (95) chap. 3: (203)
I ^ J = c ( 4 | f + gradp)*
/ £ J = - (curl J)♦
288
6 EVOLUTION PROBLEMS dp
with^ + divJ = 0, thusp = -Y.divJ, Y being the Heaviside function. Thus keeping the magneticfieldas unknown, we have (203)'
H= Z-1i4Mr-i^Jr.
E=^Mr+Zl4Jr,
Now we define the causal Calderon projectors for the electromagneticfield(E, H), P? and P* in L?(R,X) x L°(R,X) (see (197)) by: (204) P?(M,J)«(nAEL,-nAH|.),
P°(M,J)*-(nAE|_ , - D A H L ),
like in (93) chap. 3. These operators satisfy the usual relations (104), (105), (106) chap. 3 with an operator Sc given by: (205)
f-1* S°=|^R
C
ZRC^ _rj,
with 1*1 —
RcJss2nAl4j|r
We have to specify these operators in spaces with a L2 time behavior. Here also regularity results are missing. Like in the case of scalar waves we can define spaces with more or less regularity properties: here we define a general space with "finite energy" (but always with the time decreasing weight e " Yt) W?H(E,H)€l^(YQ), withYQ = L2(Q)3xL2(Q)3, (206)
£
o ! f - c u r i H = 0>
M0|{- + curlE = 0,inQxR},
| w y = {(E,H), (E,H)| ftxR €W?,
Then we define the space ofjumps across I = T x R: (207)
Zy-KMzJz). M2 = [nAE] r J2 = - [ n A H ] r with(E,H)€Wy}.
The mapping (E,H) —»(M2,J2) is an isomorphism from Wy onto Zy with inverse given by (195) with (194). Like in the scalar case Z is the direct sum of two closed subspaces G., Ge (with intersection reduced to {0}) corresponding to the set of boundary values of elements in W a and in W fl : Z ■ G.0G c . The projectors on G. and Ge (along Ge and G.) are the causal Calderon projectors Pc. and Pcc in Z (see (204)). Furthermore we easily see that G. and Gc are the graphs of two operators, - Cc. and - Cce, i.e., up to the sign, the causal Calderon operators, Cc. and Cce. Then let:
289
6.3 CAUSAL PROBLEMS (208)
X m Y tflnAE| 2 ,
Y m / # { n A H | 2 , (E,H)€W^}.
We easily see that Xm = Y1" . Here also a reasonable conjecture is that Z = X m x X m (we thus admit that the spaces of trace are the same for the interior as for the exterior). Then the Calderon operators Cc. and Cce are isomorphisms in X m . As usual, these operators are directly related to interior and exterior boundary problems, since Cc. and Cce are defined by: (209)
C?(n A E°) = n A H | , (resp. C\ for the exterior).
where (E,H) is the unique solution in L2 (YQ) (resp. Yft,) satisfying: i ) £ 0 ^ - c u r l H = 0, (210)
3H i i ) p 0 - ^ + curlE = 0, inQxR(resp. Q'xR), iii)nAE| =nAE° with given nAE inX m y . — 2vt
Furthermore multiplying!) by Ee
— 2vt
and ii) by He
then integrating, we get
(211) y/ a x R (e 0 E + M 0 H )e"^ dxdt = - / I n A E . H e-**"^d2>0 for the interior, and obviously a similar result for the exterior. This gives the usual positivity properties (see (201), (212) chap. 3): z
2
* *l.
.-2yt-
(212)
(C?m, nAm)=/ 2 (nAE).He^ 7 l dI^0 (C*m, nAm)=/ (nAE).He"
2Yt
(withm^nAE^),
dI > 0.
For a numerical implementation of these time integral methods, see for instance Pujols [1]. 3.2.3. Causal Maxwell problems with boundary conditions i) First we recall that the evolution problem in free space (for a regular domain Q bounded or not):find(E,H) in L2 (YQ), YQ = L2(Q)3 x L2(Q)3, such that (213)
i) e 0 ^ - c u r l H = J,
P 0 ^ + curlE = M inGxR,
ii) nAE| =0, with given electric and magnetic currents (J,M) in L 2 (Yft), has a unique solution thanks to Proposition 19. Furthemore it satisfies the energy relation:
290
(214)
6 EVOLUTION PROBLEMS
y fax
R
(c0 E2 + M0 H2) e'2Yt dx dt = - fQ% R (J.E + M.H) e'2yt dx dt.
ii) Now we solve a more general evolution problem: we assume that the boundary T is coated with a dielectric medium, in a domain Q p and the complementary domain in Q, Cl0 is occupied by free space. The dielectric medium is linear isotropic, its constitutive relations are given by (22), (or (23)) chap.l, with hypotheses H5) ((27),(28)) and H14) in chap.l. Thus with p «y + i n - - i w, <»>«ip = - n + iy, n € R, y > 0 fixed, we have: (215)
aw VRe (- ico?) > 0,
8 d i f Re <- A) > 0,
and there exists a constant C > 0 so that: (216)
aw > CE0Y,
PW * C y/(M0 | a> |
\
Note that (216) is also true in free space, with C= 1, and even with equalities. Now given an electric current J in L2 (Xft), XQ = L2(Q)3, we consider the evolution problem: And the electromagnetic field E, B, D, H in L2 (Xft), such that i ) ^ - c u r l H = J, (217)
ii)D = c*E,
^ + curlE = 0 inQxR,
B = p*H inQxR (witht*e06(t), y = M0S(t)inQQXR),
iii)nAE| =0. This is an evolution problem with delay. Using a time Fourier-Laplace transform method like in chap.l, with y = Rep = Ima>>0, problem (217) becomes: (218)
i)ia>6 E + curlH = J,
- ia>£H + curlE = 0 inQ,
ii)nAE| =0. We can solve this problem where o> is a complex parameter, with Im
♦, <> | € V = H0(curl,Q),
lop A
and then the problem (218) is equivalent tofindingE(o>) in V with: (220)
a(E(a>),(|>) = - fQ J(a>). $ dx,
V <> | € V.
The sesquilinear form a($,c|>) is coercive on V for all o> with Im a> * 0, thus (220) has a unique solution for all o> with Im o> * 0.
6.3 CAUSAL PROBLEMS
291
Now we integrate with respect to n, and take the real part of (220) with <> | = E; using (216) in ft, and the Cauchy-Schwarz inequality, we obtain with Cx = Ce0y: |E(c*)| 2 dxdn^ R BJ tt I c u r l E t ^ l ^ a ^ lE^l^dxdn^llJIIIIEII
(221) CJ^
with IIJII2=/ftxR I J(«)| 2 dx dn. Thus we have: (222) E(o>) € L2(C1 x R,,)3, and X curl E(o>) € L2(C1 x R / , with o>. that is: (223)
n
+ iy,
E € L^X), X = L2(Q)3, and E(t) = f / ^ E(s)ds satisfies: curl E € LJQC).
This implies: (224) thus:
H(o>) € L2(Q x R / , and X curl H(a>) € L2(Q x Fty3,
(225)
H € Lj(X), and H(t) = f / ^ H(s) ds satisfies: curlH € L^X).
Furthermore since 2s(ci>) = - X E(o>) and H(v>) * - ]jj H(w), (with |
IE € L 2 ^ , H0(curl,Q)), H € L 2 ^ , H(curl,Q)), I thus: E € Lj(H0(curl,a)), H € Lj(H(curl,Q)).
Proposition 20. The causal evolution problem (217) with delay, that is in a domain Cl occupied in part by free space and in part by a linear anisotropic medium with permittivity and permeability (E,JI) satisfying H5) and H14) chap. 1, has a unique solution (E, D, B, H) in L 2 (X Q ), with X = L2(C1)\ for a given electric current J in L2 (X), and this solution satisfies (226). y
O
3.2.4. A causal waveguide problem Like in section 3.1.4, let Q + = Qx Rz+ be a free space domain, which is a semiinfinite cylinder bounded on its lateral boundary by a perfect conductor (modelling a waveguide), with a regular bounded cross-section Cl (with boundary T) in R2. We assume that a tangential electric field is given at the end of the waveguide. Then we consider the evolution problem: find (E,H) in the waveguide, with (E,H) € L^(X3 x X 3 ), X = L2(Q+), satisfying: i ) - £ 0 ^ + curlH = 0, (227)
i * 0 ^ + curlE = 0 inQ + = Q + xR t ,
ii)nAE| =0 onl^TxR^xRt, iii)nAE| 2 =nAE° onZ0 = QxRt, Ex€l^(H0(curlT,Q)).
6 EVOLUTION PROBLEMS
292
Wefirstwrite Maxwell equations like in (30) chap. 4, using the same notations: € 9t SET 9SHT gradT H
I^
"
3" °
* °*
|ii)d 3 E T -grad T E3 + >i()atSHT«0 inQ+xR^ (where 8 t = ^ ) with: (229)
OcurljHj-co^Ea* 0 *
^curljEr+^a^sOinQ+xRt,
and with the boundary conditions: (230)
O^AEr! =0, E 3 | =0,
© B ^ = E£.
From now on we drop the subscript T for differential operators. The determination of the electromagneticfieldwill be made in several steps. Determination of the transverse electricfieldET In the same way as in chap. 4, wefirstobtain that E T satisfies: OafEj + A E T - c - ^ E T s O , (231)
inQ+xR^
ii)n T AE T | =0, (divEjH^O, iii) Ej(0) s E j o n I 0 , i^ErC^X^X^L2^);
the boundary condition on the divergence is a consequence, with (230)i), of: (232)
div
x,x 3 E = 8 3 E 3 +
div
TETs0-
Let AQ be the positive self adjoint operator in L (Q) defined by: AQU * - Au, with (233) ^ i V H u c H ^ Q ) 2 , curluand divueH 1 ^), nAu| r =0, (divu)|r=0} (see (147) chap. 2 with n = 3). Then using the spectral decomposition (X2.,*.) of AQ, with c = 1, we define the operator AT from its Laplace transform DA^pJ^-Gj^p), 9jSS(p2 + X?)1/2, withRee j>y , v^Iv^. inLj(XT), (234)
u)Dy(AT) = {v€Lj(XT),XT = L2(Q)2, Z/Rlflj^jVj^diK+oo, p = y + in}. We can prove that AT is the infinitesimal generator of a contraction semigroup (G^Xj)) in L2y(X,.), X^L^Q) 2 , of class C°. Thus like in section 3.1.4, the problem (231) has a unique solution Ej. which is given thanks to (G^x^) by:
6.3 CAUSAL PROBLEMS (235)
293
E^^G^Ef
2/ V 2x Then we can verify thanks to the Laplace transformation that E T € L~(X ).
Determination ofEy so that
First assuming that E 3 is given at x 3 = 0, we have to find E 3
i ) 8 3 E 3 + AE 3 -c*- 2 3 2 E 3 :=0 ii)E 3 | (236)
H
inQ + xR t ,
=0,
iii) E 3 1 2 = E 3 , with given E^ in L2y(L2(Q)), iv)E 3 €L?(X),X = L2(Q+). Then E 3 is obtained thanks to the semigroup (G(x3)) of section 3.1.4 (with generator A defined by (189), (189)') by: (237)
E 3 (X 3 ) = G ( X 3 ) E 5 .
Then we have to find E 3 . It is obtained (like in chap.4) from (232) by: (238)
33E3(0) = AE 3 = - div Ej.
We have div E j € L^(H~ l(Cl)) c (D(A))\ and thus, since A is an isomorphism from D(A)ontoLj(L 2 (Q)) (239)
E° * - A* W
Ej) € L}(L2(Q», thus E 3 € LJ(X).
Determination ofHy We get H 3 from <> | = curlj. E T by time integration of (229)ii). Applying curlj. to (231)i), then by taking the scalar product of (228)i) with the normal nT, we get
inQ+xR^
dip ,
I iii)
2,i 2, This gives H 3 e L~(I/(G)) and H3 e L^(X).
6 EVOLUTION PROBLEMS
294
Determination ofHT Now we can obtain H T by time integration of (228)ii); we have to verify that H- is in L 2(X2). We can prove this result like in the stationary case, using the Hodge decomposition for H p see (68) chap .4 (we can also use an inverse Laplace transformation on the formulas of chap. 4 taken with complex 03). Causal Calderon operator of the waveguide. Now we are especially interested in having the boundary value of H- at the end of the waveguide, i.e., for x3 = 0. This is also obtained thanks to (228)ii) with x3 = 0: (242)
8tSHT(.,0,t)=ATET(.,0,t) + gradA-ldivE!J<.,0,t).
We get a formula similar to (61) chap.4 (with (29) and (38) chap. 4): (243)
HrCAt) — £ ^ [SATE^.,0,s) + curlTA"1 divTE^<.,0,s)] ds.
We can also obtain a form similar to (68), chap.4, more useful for spaces. The operator C:E£—H^.,0,.)
(orCiSEj^SH^.A.))
is the causal Calderon operator ofthe waveguide. Of course we have to specify its domain. We can do so thanks to the operators Aj and A; we do not detail this question here. The interest of this operator is (like all previous Calderon operators) that it contains all informations at hand for the domain exterior to the waveguide. Furthermore we can prove that the causal Calderon operator satisfies the positivity properties: (244)
(CSEj, E£) = - (CEj, SE£) £ 0
which we get from t 245 * - ^ x R ^ o f
E +
>*of - W e - ^ d x d t ^ ^ d i v t H A ^ e - ^ d x d t ,
and integrating by parts: (246)
y / ^ x ^ ( £ 0 E 2 + M 0 H 2 )e- 2 y t dxdt=/ 2 o x R ^nAH.Ee- 2 Y t dI 0 dt.
3.2.5. Uniqueness at most In all these causal problems we have uniqueness (at most) results for the solution of the wave equation like the Maxwell equation (this is to compare to the stationary case, chap. 2.13; these results are given for a bounded domain Q, to simplify): Theorem 3. Uniqueness at most. Given a (regular bounded) domain Cl in Rn, with boundary T, the wave problem: find u so that
6.3 CAUSAL PROBLEMS
295
i) Du=f, f given in D'+(X), in Q x 1^, (247)
ii)u! 2 = u°, i J l ^ w i t h l ^ r x R , , withu°€D%(H-i/2(r)), u1 €D\(H'm(T)) liii)u€l)\(X),X = l/(Q),
has at most one solution. Theorem 4. Uniqueness at most. The Maxwell problem: find (E,H) so that i ) £ 0 ^ - c u r l H = J, (248)
j i 0 ^ + cur!E = M i n Q x R ,
with M and J given in D%(X), X = L2(Q)3 ii) n A E | 2 = M°, n A H | 2 = J° withM°andJ°€DV(H" 1/2 (div,r)) iii) (E, H) € Z)V(X), X = L2(Q)3xL2(Q)3,
has at most one solution. The essential difference with the stationary case is that these uniqueness results are not local in time: we can substitute a part Z 0 =sr o xR, with T0 a "regular" part of T, for I . However we cannot replace R by finite time intervals, except for bounded Q, in which case these time intervals have to be large enough. This is a consequence of the finite velocity of propagation. Finding the optimal time interval for which we have uniqueness is of great interest in control problems (see for instance Lions [3]). These uniqueness results imply that the causal Calderon operators and the causal Calderon projectors have the same importance as in the stationary case: these operators on the boundary T of a domain Q contains all the informations on the wave inside the domain Cl that we consider. Of course these operators are not very easy to handle especially for numerical implementations. So they are often replaced by approximate boundary conditions. However they are very useful in obtaining a priori estimates. CONCLUSION We have solved many evolution problems, but we would have to deal with many other linear interesting problems, which are not a simple application of the above theory, for example with a moving obstacle, and a moving antenna... Furthermore there are very often parameters giving different scales, so that an asymptotic analysis is necessary, with singular perturbations.
APPENDIX
DIFFERENTIAL GEOMETRY FOR ELECTROMAGNETISM
Using differential geometry in electromagnetism is quite natural, atfirstin the modelling of Maxwell equations by differential forms, according to the right transformation laws when changing a coordinate system into another one with a different orientation (this explains the notion of "polar vector" often used in electromagnetism). The notion of cohomology allows a deep understanding of many points, notably in the study of the differential operators grad, curl, div with their kernels and their images, in an open domain Q; and then in the trace spaces on the boundary T of Q for electromagneticfieldsoffiniteenergy. We have to use the manifold structure of T, and the differential operators gradp, curlp diVp on T, with the technical difficulty of having enough regularity in order to apply usual tools generally developed for very regular manifolds. Differential geometry is also useful in numerical methods in electromagnetism with Whitney forms (see Bossavit [1]). We suppose only elementary prerequisites onfinitedimensional manifolds. We review some simple notions in order to set notations, following Morrey [1], Arnold [1], Abraham-Marsden-Ratiu [1]. We also need some elementary notions on variational framework (the notion of a V-coercive bilinear form) and we often use the Peetre lemma (see chapter 2). We emphasize that this appendix is not a course on differential geometry; we refer to Malliavin [1] or to Kobayashi-Nomizu [1] for instance for that. 296
A.l INTRODUCTION
297
1. INTRODUCTION. MATHEMATICAL FRAMEWORK
1.1. M anifold with boundary The notion of manifold is supposed to be known, with the notions of charts, of mutually C^a-compatible charts, of atlases and of C*'a-compatible atlases (we refer to Dieudonne [1], Bourbaki [1]). We recall the standard definition of an n-dimensional manifold M with boundary T of class C**", k € N, 0 £ a <* 1 (C00, analytic). Each point of M is contained in some open set Fof M, which is the homeomorphic image either of the unit ball in Rn, or of the part of it for which ^ ^ 0, the points where ^ = 0 correspond to the boundary T of M; any two coordinate systems are related by a transformation of class C^a (C°°, analytic). Let Q be an open subset of Rn; Q. may be considered as an n-dimensional manifold with boundary T; following Grisvard [1], we then define: Definition 1. We say that Q. is cm n-dimensional continuous (resp Lipschitz, continuously differentiable, of class C**", k €N, 0 £ a < 1) submanifold with boundary in Rn, if for every x€T, there are a neighborhood V of x in Rn and an injective map yfrom Vinto Rn so that: i) Q n K= {y € Q, <>| (y) < 0}, where <>| (y) denotes the n-th component ofitfy); ii) <>| together (j)"1 defined on <|>(V) is continuous (resp. Lipschitz, continuously differentiable, ofclass C^a). o The boundary r of M = Cl is then locally defined by (> | (y) = 0. The interior M = Q, is a C°° (even analytic) manifold without boundary! The boundary T of an ndimensional continuously differentiable (resp.C , C '*) manifold M, has an induced structure of (n - l)-dimensional continuously differentiable (resp. ) manifold, according to the implicit function theorem. In the case of Lipschitz regularity, this is not true (see Grisvard [1]). Example 1. Polyhedrons, which are in common use for numerical applications, are only Lipschitz manifold, so we often have to make direct proofs of the main properties, like for the Stokes theorem! Let M be a regular (differentiable) manifold, we will use the following notations in this appendix.
A - DIFFERENTIAL GEOMETRY
298
TXM the tangent space at a point x of M (the space of tangent vectors at x), TM = U TxYM the tangent bundle of M; x€ M TXM the cotangent space at x € M (i.e., the dual space of TXM, or the space of covectors atx), T M = U TAYM the cotangent bundle of M (or the phase space); x€M r s * VSXM the tensor product &TXM # <8>TXM (the space of r-contravariant s-covariant tensors), T5M the tensor bundle oftype (r9s) over M,TfM = U TLM; x€M r , r ATXM and ATXM the spaces of v-covectors and of t-vectors, r * r * AT M = U ATAVM the r-exterior bundle over M (with A the exterior or wedge x€M product). A vectorfield(resp. a (r,s) tensorfield)is a section of TM (res p. of T^M), that is, a mapping v from M into TM (resp. T^M), such that « o v a l , with n the canonical projection from TM (resp. T^M) onto M; a differential form is a section of T*M, i.e., a mapping o>: M —> T*M such that it*ow = l, with n* the canonical projection from T M onto M. r r * A (differential) r-form » is a section of AT M, and an r-vector field is a r section of ATM, the regularity of these mappings being so far not specified. ft
i
ft
In the domain of any coordinate system (x 1 ,...^) we denote by Trff-fQZn (or simply by 3!,...,3 n ) the respectively associated vector fields, by dx 1 ,..., dx" ft
ft
i
the associated 1-forms, so that(-2r) ,...,(^i) is a basis of TXM, (dx1)v,...,(dxn) 0J£ X
OA
is the dual basis of TXM, with (dx*) . 3j-*y,
X
X
X
Vij.
An r-foim o> may be represented as follows: (1)
0)=
2
Wj ; dx1 A...Achr,
where (coj j) are the components of o> in that coordinate system and A denotes the exterior product. Let I be a sequence I = {ip.. ,,ir} with 1 < ij <.. .< i,. < n. We often abbreviate notations by: (iy
a) = 2 (Ojdxj. I
A A INTRODUCTION
299
We assume that M is an n-dimensional orientable manifold (i.e., there is a continuous n-form co on M such that
,~x
V1
,x
/
/ ~ x% <*vK » • • • » *
w,
)
^"■ ^ »^r?i'
dco^^dx'Adx1, I,a
1
or (3)'
9&>~
da> = 2 (ckO.*^, (*>),» 2 ( - i f " 1 — p , J
J
v=l
JV=J\UVK
g^v
dcox(v0,...,vr)= S^lrtD^x-ViXvo,...,^,...^,),
Vje^M,
where v{ denotes that Vj is missing, and Do>x is the derivative of co in charts. Note also d2 = 0, i.e.,d2a> = 0, V
gx(v,v)>0,
VveT x M,v*0.
In a local coordinate system x 1 ,..., x!\ g (which is often written as ds2) is given with fe«) = (& (9i > 9j)) a positive definite matrix, by: (5)f
g^gy^1®^-
300
A - DIFFERENTIAL GEOMETRY
A manifold M with a Riemannian stucture is called a Riemannian manifold. IfM is of class C^", g(x) is of class C*"1,01. For all x in M, e is an inner product onTxM; wenote: (6) gx (h,k) = (h,k) = 2 g ( x ^ , for all tangent vectors h, k of TXM, given in a local coordinate system x 1 ,..., x11 by h = Ih^i, k = Ikiai. Recall that (aj) is the dual basis of (dx i ),i=l,...,n. The length of the tangent vector h € TXM is defined by: |h|=(h,h) 1/2 . g x A tangent vector is said to be unitary if |h| = 1 (we say that h is a unit vector). For all x € M, the metric g assigns to each tangent vector h € TXM the covector a)h = Gx(h) € TxMby: (7)
(8)
coh(k)3SGx(hXk)«gx(h,k),
Vh,k € TXM,
and Gx is an isomorphism from TXM onto T*XM. The cotangent space T* M is therefore also equipped with a positive definite bilinear form (thus an inner product, corresponding to a tensorfieldg* on M): -1
(9)
-1
*
(cop
V»i, o)2 €TXM.
Thus using notations (8): (10)
K ,
V h, k€TxM
*x
(i.e., G x carries the metric of the tangent space onto the cotangent space), or in a local coordinate system: (11)
g* = I glj di®8j, (glj) inverse matrix of (g..).
r This can be extended more generally to exterior algebras: on the space ATXM of r tangent r-vectors at x, and on the space AT*XM of r-covectors at x, we define, for all hj, kj in TXM, and all coj, 3 j in TXM (hi A ... Ahr, kx A ... AkJ ^deta^hj) ), *x *x («! A ... A(or, S j A ... A 3 r ) adetftcoj, Sj) ); (12) ©X
J
©X
r r♦ Gx defines an isomorphism (again denoted by Gx) from ATXM onto ATXM by:
A.I
(13)
INTRODUCTION
301
G x (h,A...Ah r )1i f G x (h 1 )A...AG x (h r ) = 0 ) h i A . . . A a ) v
We will often denote by (,) x the inner products (6), or (9), or (12). For r = n, we can define an odd n-form v = v* which is positive (for the chosen orientation of the orientable manifold M): (14)
^ K o j A "-^n-l^A.I.Aa^r
±
"l A '"'
Aa)
nWetg*Ka)j)r1/2,
for a family C0ly...9(on of n independent 1-forms, where lo^ A ... Aa>n| is afunction ofx defined according to (12): Iwj A ... Ao)nlx=(a>1 A ... Aa>n, oj A ...Aw n ) x . This definition is independent of the chosen family. Thus (14) defines a Lebesgue measure on M. In a coordinate system x 1 ,.. .,x n , v is given by: (14)' v = v g = ± dx1 A ... Adx^det(g*(dx\dx*))r 1/2 = ± dx1 A ... Adx n (detg..) 1/2 . We often denote by dM the measure defined by Vs (and also by g the determinant of the matrix (g..)). But the notation dM may be misleading, since in general it is not the exterior derivative of any form. We then define a scalar product on the space of (smooth) r-forms (with compact support if M is only locally compact) by: (15)
(o),n) = (o),n)g d = f / M ^(G x n x ) dM,
with G x defined by (8) and (13). We also have, from the symmetry of g: -l
(16)
-l
a>(Gn) = n(Ga>),
(o>,n) = (n,
In a coordinate system, with r-forms a>, n, o) = I(o. dx1, I
l
n = 2 n v C^K» K
K
we have: (17) o>x (Gxtix) = IgIK(x)(uI(x)nK(x), with: ghh
,K gJK = det
...
g 'A|
dM = (det(g..))1/2dx1...dx11.
^ g ^ The scalar product (15) is independent of the system of local charts. Thus we can define a Hilbert space L r(M) as the closure of the space of smooth differential r-forms with compact support in M. It is also possible to define it as the space of r-forms with square integrable coefficients in every coordinate system. The elements of the space
302
(18)
A- DIFFERENTIAL GEOMETRY
LJ(M)»(
"generalizedr-formsM, (G>,G>) <+oo}
are called square integrable r-forms. We often do not specify whether o> is an even or odd r-form. But if necessary, we shall use the notations L2r e(M) and L2„ AM). Note that from the definition (14), we have (19) (v*,v«) x =l, Vx€M, and thus (19)' (v*,^) - J ^ - I M I (finite or not). For all n-forms © on M, we have (20)
a) = (co x ,v*) x v*,
/
M
o> = (a>, v*) g .
Then we define by duality to the exterior product the inner (or interior) product «-» of a p-covector u by an r-vector k, r s p, as the (p - r)-covector iku ■ k ^ u such that: (21)
k*«i u s G ^ k ' ^ u .
We then define the Hodge transform (or * transform) of an r-form n by: (24)
(•n)x=nx^v* = ( G - \ ) ^ v x \
Therefore the Hodge transform of an even (resp. odd) r-form is an odd (resp. even) (n - r)-form, with: (25)
(nAa,v*)f«(af.n)g,
V a € Lj_r(M), n € L*(M).
Note that the Hodge transformation has the following properties: (26)
.v*«l,
.l^v*.
One of the main properties of the Hodge transformation is: (27)
w A *n = (
303
A.l INTRODUCTION giving for all <■>, n in L2r(M), even or odd: (28)
/ M M A •n»(»,n).
This is expressed in a coordinate system, with (29) by: (30)
» 2 **i d*1' n = 2 n K d x K ,
card I s card K s r,
/ M » A . n = («,n)f = X / M 8IK «mK g 1/2 dx1.. .dx".
Taking the inner product of (27) with v*, and using (19) and the commutativity relation: (31)
(0A»ns(-l)
" *nAw,
Vr-formsa),n,
we obtain for all r-forms o>, n (32)
(a) ,n)x=(<■> A ♦ n, v \ = ( - l ) * - ^ ♦ n A a>, v*) x =(- l)*""^*, * * n)x,
hence for all r-form n: (33)
^
n
= (-D r ( n " r ) n.
Besides, we have: (34)
(oA (o A *n * n==( - 1)
(*w)An = n A *G>
(which may be verified by inner product with v?), thus (35)
(a>,n)g = (*n,*a>) g = (*a),*n) g ,
Va>,neL?(M).
The Hodge transformation is an isometry from Lr(M) onto Ln_r(M), changing even into odd differential forms, and conversely. Remark 1. Expression of the Hodge transformation in a coordinate system. For each sequence K C {1,..., n}, with card K = r, there are numbers (a KJ ), with card J = n - r, so that (36) *dxK = e 2 "KJ^* J
withe=*+l if(x1,...,x11) has the chosen orientation, E = - 1 if not. To specify (36), we define with the usual notations (see Bourbaki [1] A III 79-87) V = {1,.. .,n} \ I (37) p n , =5 ( - l)v, v ss number of couples (i,j) € I x V such that i > j , |pTV = 0 i f J n K * 0 , PIV = (-1) V i f J n K = 0 , (38)
JK
JK
|v s number of couples (\,\i) € J x K such that X > JJ, we have for all r-forms o>, n:
304
A - DIFFERENTIAL GEOMETRY
(39)
(40)
dx*A • dxK = ca K r dx I Adx r ,
d ^ A d ^ ^ p j p d x 1 A ... Adx".
Comparing with (30), we find: (41) (42)
«KI—«IKPir*l/2.
cardK = r, c a r t r = n - r ,
•dx K «£2ig I K P i r g 1 / 2 dx r ,
caidI = caidK=r.
Whenr=n, letN = {l,...,n}; we have P = 0 , d x 0 = 1, P N 0 = 1, a N 0 = g* 1/2 , thus *dxN = eg- 1 / 2 . A /\ Whenr=l, •dx k =c ^ g i k ( - D ^ ^ ^ d x 1 A . . . Adx i A... Adx11, wheredx* i
indicates that dx1 is missing; in a Cartesian system, we get • d x k * e ( - l ) k " W A ... Adb^A... Adx11.
®
1.3. Definition of the codifferential For any (regular) r-form
8cor = ( - l ) n ( r + 1 ) + 1 . d * o ) r .
We can easily show the following properties: (44)
58 = 0, 61=0, 5f=0 for all functions (O-forms)f, , 6 d = d6», *d6 = 6 d , , 6*d = d«5 = 0,
(45)
•(5o>r) = ( - l) r d( • cor),
r-l 8( ♦
From Stokes formula (see (81) below), 5 appears as the adjoint of the exterior derivative: (46)
(da,P) = (a,6(J),
o for all (smooth) (r - l)-fotms a and r-forms 0 with compact support in M, because: (46)'
(da,P)-(a,8P)=/ M daA*P-aA*6P=/ M daA*P + ( - l f - 1 a A d * P = / M d ( a A , 0 ) = O.
We will show the generalizations of (46) below.
305
A A INTRODUCTION
1.4. The gradient, divergence and Laplace-Beltrami operator Let f be a (smooth) function on M. Then its gradient, denoted by grad f, is the vector field defined by: (47)
gradf=G(df),
i.e., df=G(gradf) = Vadf •
Let X be a (smooth) vector field on M. Then its divergence denoted by div X is the function defined by: (48)
divX = -6GX = -6a>x.
Then we define the Laplace-Beltrami operator (also called the Laplace-de Rham operator) by: (49)
A = -(d6 + 6d).
Applying A to a (smooth) function f, we obtain: (49)'
Af=-Mf=-8GGdf=-SGgradf=divgradf;
the Laplace-Beltrami operator applied to the functions is identical to the Laplacian. o Note also that for all smooth r-forms $ with compact support in M (Aa,P) = -(6a,8P)-(da,dp) = (a,AW. or
Expression in a coordinate system (x{,.. .,xj. We have df = 2 Z I *^> thus: i ***
(50)
grad f = Gdf = X 5 Gdx1 = £ 3 gU 3j,
and for X = 2 X19}, we have: (51) »x = G(X)=2Xigdxi, (52)
.wx=±2(-iy"1Xjg1/2dx1A...Ad^A...Adxn.
We verify that (OXA ,(o x = ±(X,X) g 1/2 dx' A ... Adxn= ± ^ . x b c V 2 * 1 A ... Adxn, X
and we have Sa>x = -*d*u> x =-* ±2^r(X j g 1 / 2 )dx 1 A...Adx n , thus:
306
(53)
A- DIFFERENTIAL GEOMETRY
divX = g" 1/2 2 X (XV 2 ). j ftr"
Furthermore the Laplace-Beltrami operator is given by
(54)
Af-g-'^yfeV 2 ^.
1.5. Decomposition of the space of tangent p-vectors and tangent p-covectors into tangential and normal parts on the boundary Under the usual regularity hypotheses on the Riemannian manifold M, on its boundary T and on g, the tangent space TXM of M at a point x of T has an orthogonal decomposition with respect to gx into tangent and normal part to T (55)
T x M=T x TeR n ;
similarly the cotangent space TXM of M at x is orthogonally decomposed (with respect to g*) into: (55)*
T*M=T£©R;,
and we write the decomposition of elements with n a unit normal to TXM, a>£ = Gxn, (55)'
h = th + (nh)n, Vh€T x M, h* = th* +
There are corresponding decompositions of the spaces of tangent r-vectors and r-covectois: (56)
| ATxM = A(Txr©Rn) = ATxT© ^A T / I Rn) r^l
AT*M = A(Txr©R^) = ATxr©(rA T ^ I R^),
Vx€l\
r ♦
teforall
GtA=tGA, GnA=nGA,
VAcAT x M.
These decompositions involve decompositions of the restrictions to T of (smooth) r-fields and r-forms on an oriented manifold M (59)
0 - $ + i0A«ji-t0+iiP,
tjJ (resp. nfl) is called the tangential part (resp. normal part) of (5.
307
A A INTRODUCTION
Admissible boundary coordinate system; adapted chart. The decomposition (59) may be extended to a neighborhood of I\ using an "admissible boundary coordinate system" or a chart "adapted to the unit outgoing normal vector field". We recall the definition (Morrey [1]). An admissible boundary coordinate system on a manifold M with boundary T of class is a coordinate system (of class C; ,M) which maps its domain G^o c R^ (with o the part of 9G on xn = 0, o * 0 ) onto a boundary neighborhood N, in such a way that o is mapped onto NnT and the metric is given on o by: (60)
ds 2 = 2 g..(x,,0)dxi®dx? + (dxn)2, x ^ x 1 1 ) . lJ
i,j = i
[Also, if (W, <> | ) is a chart of r withy €W, there are a > 0 and a diffeomorphism: (t, y)€[0,
Xj(z) =
Then (^r I, )• , 9
Xj /j = l
*j(y),j = l,...,n- 1, xn(z) = t withz = Vy(t)€U. n
can be identified with the normal n.]J
If M is a manifold Q, Cl being an open set of Rn, we can use the Euclidean distance of z € Cl to the boundary T, d(z,0: t = s n = <>| (z) = d(z, 0 , and: w^ = dsn » d(|>n. This allows us to extend the decomposition (59) in a neighborhood V of I\ ® Let j be the canonical mapping I*—»M. IfjJisaC r-form on M, the pull-back j*P of 0 byj is the C k_1 r-form on T defined by (see Schwartz [1], Bourbaki [2]) (63)
0*P)x(v1,...,vr) = Px(v1,...,vr), Vx€r,v 1 ,...,v r €T x T.
This r-form j*(J can be identified with the restriction of the r-form t|J to T, and we write (64)
tr$=j*P = tPl r
Using notations (21), (23) we define an (r - l)-form n ^ on T for all r-form (J on M: (65) nrP = ( - l / -
1
^
» l r - ( - If-^W
A
Indeed for all (r - l)-forms a on M, we have: |(a, w l J (nfl A o>i)) = (o>l A a, nP A o>i) = ( - lf'l(a A «£, nfl A a>]>) (66)
= ( - l)r-'(taAo>i, npA • » ) - ( - l)T-\ta,a»)(mla, o^) = (-l) r - , (ta,np) = (-l) r - , («,nP).
308
A - DIFFERENTIAL GEOMETRY
g Let g denote the Riemannian metric induced by g on T, and v r =v r the (n - 1)form on r (corresponding to the Lebesgue measure dT on I\ see (14)); then (67) v r = a>£ J v g | r = ( - l f - V * (oralsov rss t r »o>i). It is thus possible to define the Hodge transformation on T as in (25). Lemma 1 • For all (smooth) r-forms f>onM, we have (68)
Ot^lM-ir'.nrP, |ii)nr
PROOF. Applying the Hodge transformation to (59), we obtain (69) *0 = *t0 + ,(n0A<4) s ,t0 + *n0. We have (70) U 0 - . S 0 , n*P = «#, t r «P*t r ,(n0A(i>i). Using definitions (25), (24), we have for all (n - r)-forms a on a neighborhood VofT (71)
(a, * (nP A a>£))g = (a, (np A a>£) ^ v) = (nfl A <4 A a, v)g «(-l) r " 1 (