Partial Differential Equations, 2nd Edition, L.C.Evans Chapte Chapterr 6 Second Second-Or -Order der Ellipt Elliptic ic Equati Equations ons Shih-Hsin Chen∗, Yung-Hsiang Huang † 2017/03/27
√
1. (a) Direct computation. (b) c = ∆ a. 2. Proof. Define
n
B [u, v ] =
aij uxi vxj + cuv dx, u, v
U i,j =1
1 0
U ). ∈ H (U )
Clearly, ij
|B[u, v]| ≤ a Du
L2 (U )
∞
Dv
L2 (U )
for some positive constant α. For c(x)
+ c
u ∞
L2 (U )
v
≤ αu
L2 (U )
H 01 (U )
v
H 01 (U )
,
≥ −µ, where µ is a fixed constant to be assigned later.
Using the uniform ellipticity, we have θ
U
n
2
|Du|
dx
≤
aij uxi uxj
U i,j =1
= B[ B [u, u] − cu dx ≤ B[u, u] + µ u dx 2
U
2
U
2
≤ B[u, u] + C µ |Du|
dx,
U
θ where the Poincar´ Poincar´e inequality inequality is applied applied in the last inequality inequality. Now we set µ = and see 2C θ that if c(x) , then 2C θ Du 2 dx B [u, u]. 2 U
≥
| |
≤
∗
Department of Math., National Taiwan Taiwan University. University. Email:
[email protected] [email protected] w
†
Department of Math., National Taiwan Taiwan University. University. Email:
[email protected] [email protected] w
1
Therefore,
u
2
2
= u
+ Du
2
≤ (1 + C )Du ≤ θ2 (1 + C )B[u, u].
H 01 (U )
L2 (U )
L2 (U )
2
L2 (U )
3. Proof. Define
n
B[u, v] =
uxi xj vxi xj dx, u,v
U i,j =1
2 0
∈ H (U ).
Use the integration by part (divergence theorem),
U
for u, v
n
2
2
|D u|
dx =
uxixj uxixj dx =
U i,j =1
uxi xi uxj xj dx =
U i,j =1
2 0
∈ H (U ). (Fix i, consider F = u 2
Du
L2 (U )
U
xi Du xi and then use
div U
F dx =
|∆u|
∂U
2
dx
F
· ν dS .)
− u∆u dx ≤ εu + 4ε1 ∆u ≤ εC Du + 4ε1 ∆u =
U
2
2
L2 (U )
L2 (U )
2
2
L2 (U )
L2 (U )
the Poincar’e inequality is used in the last inequality. We choose ε = 2
Du
2
≤ C ∆u
L2 (U )
L2 (U )
1 so that 2C
.
Therefore, 2
u
H 02 (U )
= u
2
L2 (U )
+ Du
2
2
+ D u ≤ (1 + C )Du + ∆u ≤ (1 + C + C )∆u
L2 (U )
2
L2 (U )
2
L2 (U )
2
2
L2 (U )
2
L2 (U )
= (1 + C + C 2 )B[u, u]. Thus, 1 u 1 + C + C 2
2
Remark
2 0
≤ B[u, u], u ∈ H (U ).
H 02 (U )
0.1. The validity of the definition of weak solution to Biharmonic equation with zero
Dirichlet and zero Neumann boundary is based on the zero trace theorem for W 02,p function, which is not proved in Evans’ book. 2
4. Proof. ” ”
⇒ ” take v = 1.
⇐ ” Define B[u, v] =
Du Dv dx, for u, v
·
U
1
∈ H (U ).
1
Consider a subspace of H (U ) 1
{ ∈ H (U )(u)
A = u Let l(u) =
=0
U
} 1
u dx, then l is a continuous linear functional on H (U ). Then A = l
−1
U
closed in H 1 (U ), so A is a Hilbert space with induced inner product. Note
|B[u, v]| ≤ u v
.
A
A
( 0 ) is
{}
and by Poincar´e inequality, 2
u
L2 (U )
=
U
2
|u − (u) | U
dx
2
≤ C |Du|
dx.
U
Therefore 2
u
A
2
= u
L2 (U )
+ Du
2
2
≤ (1 + C )Du
L2 (U )
L2 (U )
= (1 + C )B[u, u].
2
∈ L (U ) (with f dx = 0), there exists a unique u ∈ A such that B[u , v] = (f, v) , ∀v ∈ A. Let v ∈ H (U ), then v − (v) ∈ A. Thus, we have (f, v) = (f, (v) ) + (f, v − (v) ) = 0 + B[u , v − (v) ], (since f dx = 0) By Lax-Milgram, for all f
f
U
L2 (U )
f
1
U
L2 (U )
U L2 (U )
U L2 (U )
f
U
U
= B[uf , v] =
Duf Dv dx,
·
U
1
Therefore, uf
1
∀v ∈ H (U ).
∈ A ⊂ H (U ) is a weak solution. In general, u
f
+ C is always a weak solution
for any constant C . 5. Motivation. If u
U
∞
∈ C
(U ) solves the given boundary value problem, then for each v
fv =
( ∆u)v =
U
−
∂U
∂u v + ∂n
−
U
∇u∇v =
vu +
∂U
U
∞
∈ C
(U ),
∇u∇v.
By density theorem and trace theorem, we define B on (H 1 (U ))2 by B[u, v] =
(T v)(T u) +
∇u∇v, where T is the trace operator with finite operator norm T . We call u ∈ H (U ) is a weak solution of the given Robin boundary value problem if B[u, v] = f v for all v ∈ H (U ). ∂U
U
1
1
U
3
Proof. We prove uniqueness and existence by Lax-Milgram. Given u, v 2
|B[u, v]| ≤ max{1, T }u v H 1
H 1
1
∈ H (U ), it’s easy to see
. Next, we prove B is strictly coercive by a contradiction
argument similar to the one for Poincar´e inequality. Assume for each k
1
2
∈ N, there exists u ∈ H such that u > kB[u , u ]. Let v = u /u , then 1 = v > kB[v , v ], so > B[v , v ] ≥ Dv , > T v and hence v < 1 + ≤ 2. Rellich’s compactness theorem asserts there exists a subsequence, still denoted by {v }, converging to a function v in L sense. Note Dv → 0, so for every φ ∈ C (U ), k
2
k H 1
k
1
k
k H 1
2
k
1
k
k
k
k
k
2
k
k L2
1
2
k L2 (∂U )
k
k H 1
k H 1
k
2
k
∞
k L2
c
Hence v
1
∈ H ,
Dv
vDφ = lim
k→∞
vk Dφ = lim
k→∞
Now we know vk
≡ 0.
1
→ v in H
−(Dv )φ = 0 k
sense, so v
H 1
= 1. Moreover, v is
constant in each component of U . But
≤ T (v − v )
T v
L2 (∂U )
k
L2 (∂U )
+ T vk
≤ T v − v
L2 (∂U )
k H 1
+
1
which means v
k
→ 0, as k → ∞,
≡ 0 on ∂U and then v ≡ 0 on U . This contradicts to v
H 1
= 1. Therefore B
is strictly coercive. 6. (This boundary value problem is called Zaremba problem sometimes) Definition. Similar to Exercise 5, we define B[u, v] =
1
∇u∇v on H := H ∩ {u : T u
Γ = 0 . 1
}
The linearity and continuity of T ensures that H is a closed subspace of H 1 , so H is a Hilbert space with induced inner product. u
∈ H is a weak solution of the given Robin boundary value problem if B[u, v] = f v for all v ∈ H . (The zero Newmann boundary condition is U
involved.)
Uniqueness and Existence. They are established by Lax-Milgram again, the coercivity of B is
established by almost the same argument as Exercise 5.
Take v =
−D
−h
i
1
n
∈ H (R ), ∇u∇v + c(u)v = f v. D u, 0 < |h| 1. Then v ∈ H and has compact support. Hence
7. Proof. By definition, we know for each v
1
h i
h i
|D Du|
2
h i
h i
+ D c(u)D u =
−
f (Di−h Dih u).
By mean value theorem, Dih c(u)(x)Dih u(x) =
c(u(x + hei )) h
− c(u(x)) D u(x) = c (··· )|D u(x)| ≥ 0. h i
4
h i
2
Hence
1 |D Du| ≤ | f (D D u)| ≤ |f | + 2 |D D u| 1 ≤ |f | + 2 |DD u| The last inequality follows easily from mean value theorem. Hence |D Du | ≤ |f | , for all small enough h. This implies |D u| ≤ |f | . (Evans, Page 292)
2
h i
−h
1 2 1 2
h i
i
−h
2
h i
i
2
2
h i
h i
2
2
2
2
2
2
8. Proof. Differentiate the equation Lu = 0, we get (Daij )uxi xj + aij (Du)xi xj = 0. Multiply this equation by 2Du, we have 2Du (Daij )uxi xj =
−2a
·
ij
Du (Du)xi xj . Adding
·
ij
−2a
(Du)xi (Du)xj
to both sides, we find 2Du (Daij )uxi xj
·
ij
− 2a
(Du)xi (Du)xj =
2
ij
−a (|Du| )
(1)
xi xj
On the other hand, multiply the origin equation by 2u and add 2aij uxi uxj to both sides, we have 2aij uxi uxj = 2aij uxi uxj + 2aij uuxi xj = a ij (u2 )xi xj (1)
− λ(2) implies −2λa u u ij
xi
xj
+ 2Du (Daij )uxi xj
− 2a
·
ij
(Du)xi (Du)xj =
ij
−a
(2)
(λu2 + Du 2 )xi xj
| |
By uniform ellipticity, boundedness of Daij and Young’s inequality, the left hand side is nonpositive for large λ. The second assertion is a consequence of weak maximum principle to L( Du 2 + λu2 )
| |
≤0
9. Proof. By weak maximum principle, min w = min w = w(x0 ). Let v1 = u + f v2 = u
− f
∞ w,
then Lv1
and
≥ 0, Lv ≤ 0. 2
By weak maximum principle, min v1 = min v1 = f ∂U
U
max v2 = max v2 = U
∞w
∂U
U
−f
∞ min w =
∂U
∂U
−f
∞ w(x
0
∞ min w ∂U
(x ) + f ≥ ∂v∂ν (x ) = ∂u ∂ν
0
≤ ∂v∂ν (x ) = ∂u (x ) − ∂ν 2
0
0
0
0
∂w 0 (x ), ∂ν ∂w f ∞ (x0 ). ∂ν ∞
Since u = 0 on ∂U ,
∇u//ν . So ∂w |∇u(x )| = | ∂u (x )| ≤ f | (x )|. ∂ν ∂ν 0
0
5
∞
0
0 ∞ w(x )
) = v 2 (x0 ). Therefore,
0
1
= f
= v1 (x0 ) and
10. Proof. We omit (a) since is standard. For (b), if u attains an interior maximum, then the conclusion follows from strong maximum principle. If not, then for some x0
0
∈ ∂U, u(x ) > u(x) ∀x ∈ U . Then Hopf’s lemma implies
∂u ∂ν
(x0 ) > 0,
which is a contradiction. 0.2. A generalization of this problem to mixed boundary conditions is recorded in
Remark
Gilbarg-Trudinger, Elliptic PDEs of second order, Problem 3.1.
11. Proof. Define B[u, v] =
1
aij uxi vxj dx for u
U i,j
By Exercise 5.17, φ(u)
∞
1
1 0
∈ H (U ), v ∈ H (U ).
∈ H (U ). Then, for all v ∈ C
c
(U ), v
≥ 0,
B[φ(u), v] = a (φ(u)) v dx = a φ (u)u v dx, (φ (u) is bounded since u is bounded) = a u (φ (u)v) − a φ (u)u u v dx ij
xi xj
U i,j
ij
xi xj
U i,j
ij
xi
xj
ij
U i,j
xi
xj
i,j
φ (u)v Du 2 dx
≤0−
| |
U
≤ 0, by convexity of φ.
(We don’t know whether the product of two H 1 functions is weakly differentiable. This is why 1 0
we do not take v
∈ H .) Now we complete the proof with the standard density argument. ¯ with Lu ≤ 0 in U and u ≤ 0 on ∂U . Since U is ¯ compact and 12. Proof. Given u ∈ C (U ) ∩ C (U ) ¯ v ≥ c > 0. So w := ∈ C (U ) ∩ C (U ). ¯ Brutal computation gives us v ∈ C (U ), a v u a v u − a u v 2 v u − vu + −a w = −a u v + − a v v v v v (Lu − b u − cu)v + ( −Lv + b v + cv)u 2 = + 0 + a v w , since a = a . 2
u v
ij
ij
ij
xi xj
xi xj
2
ij
xi xj
xi
2
i
=
ij
xj
xi xj
ij
xj
2
i
xi
Lu v
− uLv −bw v
ij
wxi xj + wxi bi
i
2
xi
ij
v2
xi
v
2 + aij vxj wxi v
xj
xi
xi
2
xi
ij
ji
Therefore, Mw :=
−a
ij
−a
2 Lu vxj = v v
− uLv ≤0 v 2
{ ∈ U ¯ : u > 0} ⊆ U
on x
{ ∈ U ¯ : u > 0} is not empty, Weak maximum principle to the operator M with bounded coefficeints (since v ∈ C (U )) will lead a contradiction that If x
1
0 < max w = max w = {u>0}
Hence u
∂ {u>0}
≤ 0 in U . 6
0 =0 v
13. (Courant-Fischer’s principle) Proof. First we need to note the same proof of Theorem 2 on Page 356 leads the following
theorem: Theorem 0.3. For
the symmetric operator L =
−(a
ij
uxi )xj on a bounded smooth domain
U where (aij ) is symmetric and satisfies the uniform ellipticity. Denote k-th zero Dirichlet eigenvalue and eigenfunction by λk and ωk respectively, then 1 0
λk = min B[u, u] u
| ∈ H , u
{
L2
= 1, (u, ωi )L = 0, 1 2
≤ i ≤ k − 1}
Through this theorem, we know the desired identity is true if we replace equality by
≤. Now
we prove the converse: Given k
− 1 dimensional subspace S . By spectral theory for compact self-adjoint operators, there exists y = Σ c ω ∈ span{ω · ·· ω } with y = 1, such that y ∈ S , so k
1
1 i i
⊥
min B[u, u] u
| ∈ S , u
{
L2
⊥
L2
k
=1
k
2
k
2
} ≤ B[y, y] = Σ |c | λ ≤ Σ |c | λ ≤ λ . i
1
i
1
i
k
k
Since S is arbitrary chosen, we are done.
∞
14. Proof. Let Y := u
{ ∈ C
(U )
| u > 0 in U, u = 0 on ∂U }.
Let u1 > 0 be the zero Dirichlet
eigenfunction with respect to the eigenvalue λ1 which exists by Krein-Rutman theorem (Page Lu u u∈Y x∈U
361, Theorme 3). By classical Schauder’s theory, we know u1
∈ Y and therefore sup inf ≥ λ .
Given v
1
∈ Y , then w = v − u ∈ Y . Note that 1
Lv L(u1 + w) L(w) λ1w = = λ 1 + v u1 + w u1 + w
−
Consider the adjoint operator L∗ w = ∂ ij (aij w)
i
− ∂ (b w) + cw = a i
ij
∂ ij w + [∂ j (aij + a ji )
i
ij
i
− ∂ b ]∂ w + (c + ∂ a − ∂ b )w, i
i
ij
i
the Krein-Rutman theorem tells us that, for m > sup U c+∂ ij aij ∂ i bi , there exists λ∗1 , such that
|
− |
λ∗1 +m is the simple principle zero Dirichlet eigenvalue of L ∗ +m with positive eigenfunction u∗1 . Since λ∗1(u∗1 , u1 )L = (L∗ u∗1, u1)L = (u∗1, Lu1 )L = λ 1(u∗1 , u1)L , the fact that u∗1 , u1 > 0 implies 2
2
2
2
λ1 = λ ∗1 . Consider (L(w)
∗ 1
∗ 1
∗ 1
∗ 1
− λ w, u ) = λ (w, u ) − λ (w, u ) = 0 1
1
Due to u∗1 > 0, there are only two possibilities: (1) If L(w)
− λ w ≡ 0. Since λ is simple, w = u . Then 1
1
1
7
Lv v
=
L(2u1 ) 2u1
= λ 1.
(2) If L(w) and hence
− λ w changes sign in U . Then there exist x ∈ U such that (L(w) − λ w)(x) < 0 ≤λ . = λ + < λ . Since v is arbitrary, sup inf 1
L(v )(x) v (x)
1
(L(w)−λ1 w)(x) (u1 +w)(x)
1
Lu u u∈Y x∈U
1
1
Hence sup inf Lu = λ 1. u u∈Y x∈U
15. Proof. (We use Reynolds transport theorem without mentions. ) λ(τ ) = λ(τ )
|ω(x, τ )| = 2
U (τ )
U (τ )
−(∆ ω)(x, τ )¯ω(x, τ )
U (τ )
|∇ ω(x, τ )| .
x
=
Differentiate the original equation and w
= 1 with respect to τ , we get ˙ −∆ ω = λω + λω (ω ) = 0 + (ω ) , ω v · ν + x
0=
L2 (U τ )
τ
τ
2
∂U (τ )
2
x
2
2
τ
U (τ )
τ
U (τ )
Hence,
˙ λ(τ ) = = = =
2
· ν + · ν + · ν +
∂U (τ )
|∇ ω| v
∂U (τ )
| ∂ω |v ∂ν
∂U (τ )
| ∂ω |v ∂ν
∂U (τ )
| ∂ω | v · ν + ∂ν
∂U (τ )
| ∂ω | v · ν + λ · 0 + 2λ˙ ∂ν
x
2
2
2
U (τ )
2
U (τ )
2
|∇ ω| )
(
x
τ
∇ ω∇ ω , Since u = 0 on ∂U. x
x
τ
2ω( ∆x ωτ )
−
U (τ )
˙ 2ω(λωτ + λω)
U (τ )
2
=
16. Proof. Direct computations tell us e−iσt v := e iσ(ω ·x−t) satisfies the wave equation utt σ 2 ∆u = 0
−
and r( ∂v ∂r r( ∂ ∂rΦ
− iσv) = iσ(x · ω − |x|)v which did not tend to 0 as |x| → 0. On the other hand, − iσΦ) = −Φ which tends to 0 as |x| → ∞. The proof to show Φ is a fundamental solution
is almost the same as usual Laplacian case. Try to mimic the proof Evans gives on page 24. 17. I think we need to assume w = O(R−1 ). and the second expression Evans gives in this problem is wrong, the integrand on the left-hand side should be wr 2 + σ 2 w Proof. Given x0
Φ(x
3
∈ R , let B
R
2
¯ ). | | − 2σIm(ww
| |
r
denote the ball centered at x0 with radius R and Φx (x) := 0
− x ), where Φ is defined in Exercise 16 (b). Hence w(x ) = (∆ + σ )Φ w − (∆ + σ )w Φ 0
2
0
2
x0
x0
=
BR
x0
∂ Φ = ∂B R
∂r
x0
− iσΦ
∂B R
w − ∂w − iσwΦ
x0
∂r
8
∂ Φx w ∂r 0
Φ − ∂w ∂r
x0
(3) (4)
the magnitude of the integrand is o(R−1 ) O(R−1), so its integral tends to 0 as R
∗
Remark
→ ∞.
0.4. Further results for Radiation condition on general dimension N
≥ 2 can be found
in G. Eskin, Lectures on Linear PDEs, Section 19. (Need some knowledges on stationary phase formula.)
9
