)')
(p +
+ xP+\342\200\234\"\342\200\231<9\342\200\231(x). l)xP\342\200\234\"\342\200\235
that
F (0)Z'(0)
= =
F(x)Z'(x)) %
ligio
[(19
+
= 0-)) + x\342\200\235*\"\342\200\2312\342\200\231(x)
230
FUNCTIONS
BBSSEL
facts, we
all these
Using
have
X\342\200\231, I;
noted,
already
F(x) =
= 0.
near x
This
1=\"(x))
3:1\342\200\231,
from
follows
it
that
< 1))
of (19.3) is
the integrand
that
implies
(19.3))
formula Tay1or\342\200\231s
< 0
(0
\302\247x2F\"(0x)
2 dx.
bounded.
then)
But
1 \342\200\224
Ll, (F(x)
Schwarz
1
2
|z|
U0 also
2
< L
dx)
dx)
1
1
< Jo 23
=
dx
dx
xJ\302\247()\\x)
L
<
Ch.
M
T
2])
(M =
const))
(14.1)]. and therefore
equation
If we use (19.4),it
from
follows
|I| the inequality
proves
(L = const).)
dx
(4.1) of
equation
[see
inequality
dx f; |z|
which
Io |z|
x\357\254\202z F\"(x))
by the
Thus,
[see
8)
\357\254\201nd that)
(F(x)
we
CHAP.
-
1 = As
seams
FOURIBR-BESSEL
AND
<
(19.4)
\\/M/X.
(19.3)
that)
Lx/It)
s
-372-.)
(19.1).)
Let f (x) be a bounded and twice di\357\254\202erentiable function = 0, interval [0, 1], such that f (0) = f \342\200\231(0) f (1) = 0, and such that f \"(x) is bounded derivative may not exist at certain (the second points). Then the Fourier-Bessel f (x) satisfy coe\357\254\201icientsof the function 1.
THEOREM
on de\357\254\201ned
the
the
.) .
inequality C |c,,|
(C =
$75
(19.5))
the conditions of the theorem, so doesthe (x) satis\357\254\201es = 1/} f (x). Therefore, the lemma, we obtain) applying
F(x)
ff,
xf(x)J,0.x)
\\/3: f (1)
ax)
R \342\202\254
W2 II)
By
const).
If f
Proof.
function
<
equation
F(x)J,,(7.,,x)
(R =
const).
(14.1))
dx If, xJ:
2 .~?'|><))
(K 95
0).)
dx)
SEC. 19
so
AND FOURIBR-BESSBL
FUNCTIONS
BESSEL
that)
xfJ. dx
I;
=)
lcnl
1)
<) Nlbu)1:12\342\200\231)
dx) I; xI:
which
the required
proves
we
on the
theorem
The
2. Let
THEOREM
[8,
2
>
If p
1 of
interval
[0,
f
absolutely
> -1,
Order
The
a Function
-1,
from the
follows
assertion
the
1] follows
and uniformly on the whole
If p 2 0,
uniform
the
preceding theorem on the
convergence
from the preceding theorem and
Theorem
f (x)
is
be a function
f (x)
di\357\254\202erentiable
2s times
\302\260 \302\260 = \302\260
1)
f(0)
2)
is bounded fl?-*)(x)\302\260
3) f(1)
=f\342\200\231(0)
on de\357\254\201ned
uniform
conditions convergence
=
inequality
[0, 1] such
0;)
(this derivative may
= - - -
the interval
1) and such that)
(s >
=f\"*\342\200\234\"(0)
=f\342\200\231(l)
Then the following
these
with
Coe\357\254\201icients of Magnitude of the Fourier-Bessel Which is Differentiable Several Times)
1. Let
THEOREM that
and
18.)
Sec.
of
twice
=
Remark. If we use Theorem 2 of Sec. 18, then on f (x) and with p 2 \342\200\224 we obtain absolute and \302\247, of the series (18.5) on the whole interval [0, 1].)
\"20.
and
0, f (1) = 0, and \342\200\231(0) not exist at certain points).
(0) = f
0.)
and Theorem3 of Sec.18. whole
Sec. 16is a
is continuous
which
second derivative may series of f (x) converges 1] where 0 < 8 if p
Fourier-Bessel
on every subinterval interval [0, 1] If p Proof.
be a function [0, 1], let
f (x)
be bounded (the
the
Then,
2 of
Theorem
1:)
the interval
on
dzjferentiable let f \"(x)
to F(x).) supplementing
Theorem
of
true if we impose the requirements of) = \\/:6 f (x) instead of on f (x), since
F(x)
lemma
proposition
following
consequence
function
the
applied
actually
(19.5).)
inequality
1 remains
Theorem
Remark. the
SERIES)
not
exist
at certain
points);
=
0=f\"-\342\200\234\342\200\235(1) is
by satis\357\254\201ed
the Fourier-Bessel
coe\357\254\202icients
of f (x):) C)
lC,,|
S
-Kim
= const).
(20.1)))
BESSBL FUNCTIONS
to see that the function F(x) = 1/} f (x) also of the theorem. In particular, F(x) satis\357\254\201es the the lemma of Sec.19and hence satis\357\254\201es i.e.,) (19.3),
conditions
satis\357\254\201es the
of
conditions
dx = J; x:r
I =
I
= m =
where
CHAP.
SERIES
is easy
It
Proof.
AND FOURIBR-BESSBL
parentheses
[oF(x)zdx \342\200\224 z
p3
1,
=
F(x)J,(A.x>
1)
o(\302\247F\342\200\224 F\342\200\235)zdx,) =5\342\200\230-3
\\/3cJ,()\\,,x . then we
last integral,
the
in
1
dx)
\302\253/3:
I;
denotes
the
function
in
have)
1
1
I =
If F,
Flzdx.)
\342\200\230[0 \342\200\230E
the function
Since
F,
of the lemma, this
the conditions
satis\357\254\201es all
time
gives)
(19.3)
1 I =
1
T: Io
we have
where
F22
dx,
written
n=\302\247n\342\200\224m)
> 2, then F3 again satis\357\254\201es the conditions of the lemma, and the s repeat argument times, \357\254\201nally obtaining)
If s
we can
1
1
I=:'2-;J\342\200\230oF_,Zdx,)
where)
m)
is a
bounded
function.
follows
that)
It
];F,z
dx|
< L]; |z| dx
(L
=
const).
By (19.4)
dx < x/\342\200\224M/A, I; |z| so
that
we
have)
\"'
Lx/Tl? \342\200\230))
(M
= const),)
in
fact
8)
20
sec.
\302\253FUNCTIONS AND
BESSEL
FOURIER-BESSEL
233
SERIES
But
I 0\" l
| f;xf(x)Jp(1..x) _\342\200\224 I
L:
dx| ,)
xJ\302\247(A,,x)
dxl
since)
and
1
dx [0 xJ,2,().,,x) (14.1) we
to equation
according
2
K
(K > 0),)
E
obtain) \357\254\201nally
Lx/I11) < |\342\200\230\302\273| \342\200\230T $2.-\342\200\224<1/53') which
the inequality
proves
The next
2.
THEOREM
1) For p
consequence of
is a
result
(20.1).)
hypotheses of Theorem 1 are
If the
2)
(0 < x p 2
For
met
for
s 2
1,
then)
2 0
|c,,J,,(7.,,x)I< foranyx
Theorem1:)
= const)
(H
(20.2)
)-\\2';,\342\200\2241_LI(\342\200\2241,-2-)
<1);
\342\200\224 \302\253}
|c,,J,(7\302\273,,x)I
S
= const)
(L
(20.3))
Va?\342\200\230-3
for all x
(0 < x <
3) For if n
p >
> n(x). Proof.
-1,
(In
uniformly;)
1)
this
the
(20.3)
inequality
is no
there
case,
In Sec. 18 (seethe J,(x) is bounded for p
proof
holds for
the
inequality
for
p >
-1,
(18.6), we it
follows
need
x<
1)
Hence,
<
the
-
(20.1)
apply
(L
x<
1)
inequality L
since
to get
the
that
(20.2) is an satis\357\254\201es
J,(7.,,x)
(20.3).
Finally,
formula (9.16)that)
the asymptotic
from
|J,(A,,x)I
for every x (0 < (20.3).))
only
1), we saw
of Theorem
2 0. immediate consequenceof (20.1). Forp 2
function
each x (0 <
in x.))
uniformity
= const),
(20.4))
-\342\200\230%C
and
n >
n(x).
Again
using
(20.1),
we obtain
234
FUNCTIONS
BESSEL
*2|. Term by Given a
AND
Term
of
Differentiation
=
for the
conditions \357\254\201nd su\357\254\202icient
from
formula
2
(7.8)
=
> -1,
that p
equality
fl c.1.J;(2.x>.
- 71..xJ,.+1(k..x)|
|pJ,(>~.x)
< assume
of the
(21.2)
that)
|1..xJ;.(1..x)|
We
(211)
validity
(c..J.'=
=
f\342\200\231(x)
follows
Series)
c.J.<2.x>.
\"Z1
It
Fourier-Bessel
8
Fourier-Besselseries) \357\254\201x)
we now
cmp.
seams
rounmn-nnssm.
+
In-7p(l..x)|
so that
+
p
0. Therefore,the
l >
(21.3))
|>~.x\342\200\224\342\200\231p+1(>~..x)|-
quantity
'1/7?:-7\342\200\230 -7p+1(7\\nx)|
by the
is bounded,
where
asymptotic formula
(9.16),
+ I pJ,(7.,,x)| |).,,xJ;(7\\,,x)|\342\202\254 consider only values of x such
we
hence)
and
(H = const),
H
\\/7 that
1. If
x <
0 <
(21.4))
C |C,,|
where s
> 0 and
C are
constants, .
\342\202\254
9
then for x 012
<
>
0)
J 0~..x)
+
CH
|cn)\342\200\230nJp()\342\200\230nx)| 2)+\302\242 \342\200\230Ly.\342\200\231 x']'+\302\242x\342\200\231) 19
or
by (20.4))
+)
<
Ic.2.J;(x.x)I
(3535
CH
A
7\\,',+\302\242x')
[see (18.3)]that the series (21.2) converges for 0 < x < l on every subinterval [8, 1] where 0 < 8 < l. The uniformly converges last fact implies the validity of (21.2) for 0 < x < 1. As for the validity of (21.2) for x = 0, if p < 1, p a\303\251 0, it is easy to see for x = 0, since J,,(x) = x\342\200\231.,,x) where
\342\200\230This implies
at once
and
=
H-7p-1(7\302\273.x) -\342\200\231L(7~nx)
\342\200\230 JP\342\200\230!-l()\342\200\230nx)]s))
21
sec.
the functions
where
hencearebounded.
the
in
indices and
side have nonnegative
right-hand
235)
SERIES
Then)
H |c..?\302\273.|
<
|c..?\302\273J,',(7~.x)|
so that
AND rounmn-nessan
FUNCTIONS
BESSBL
= const).
(H
if)
C
Ion] <
(21-6)) \342\200\230A?\342\200\231
is
series (21.2)
the holds
on
everywhere
the interval
[0, 1].
the
on [0, 1], i.e., (21.2)
if p
Finally,
=
O
we
have
= |1..J1(A..x)|
|1..J6(h.x)|
insteadof (21.3). Since
(18.3)]
[cf.
convergent
uniformly
function
J, is
bounded, CH)
<
lcnxn J()(x)|
then
')F\303\251')
holds, and the series in (21.2) is again uniformly convergent, is valid for all x in the interval [0, 1]. (21.2) theorem:) we have proved the following Thus, \357\254\201nally,
if (21.6)
so that
the condition 1 and ifc,, satis\357\254\201es (21.5), then the If p > \342\200\224 series can be dz\357\254\201erentiated term by term for 0 < x < 1. (21.1) = 0 or the condition (21.6), then the series (21.1) p 2 1 and c,,satisg\357\254\201es can be differentiated term by term everywhere on [0, 1].3 the series (21.1) for differentiating We now \357\254\201nd a su\357\254\202icientcondition of the relation) twice for the term i.e., validity by term,
THEOREM 1.
If
p
f\"(x) = Since J,,(x) is a
:1 (c.J.(x..x))'\"Z =
of
solution
(21.7)
c,.x:J;o.x>.
we have
Besse1\342\200\231s equation,
-
=
+ + ?~?.x\342\200\231J.'.f(R.x) 1..xJ,\342\200\231.(R..x) 03.x\342\200\231p\342\200\231)Jp(l..x)
0-
Thus
-
=
I-?~..xJ;(l..x) |7~.\342\200\231.x\342\200\231J;(A..x)| <
so that
by
3
IpJ,.(A.x)I+
that the convergenceof theoremsof Sec.18.)
Note
+ IA:x=J..(A..x)I+
Ip2Jp(x.x)I.
(21.4)
IA:x2J,';(A,.x)I < and the
Ix..xJ,;(x.x)I
+ p\342\200\231J,,(7~..x)| ?~?.x\342\200\231Jp(7~..x)
the
H + W\342\200\230. series
|x3x*J.(A.x)I
+ Ip2J,(x.x>I-)
(21.1) itself follows from
(21.5) and
(21.6)
236
AND FOURIER-BESSEL
wucrrous
BESSEL
CHAP.
seams
8)
we have)
Therefore,
,,
< no.1
Ic.x:J.o.x>|
H
A
3
+
\342\200\235
+
(\"\"5
x3)
Ic.J.I.
If
C
|c,,| \342\202\254
o)
> 0 and
where e
C are
then
constants,
CH
2 II
x;+=x2
we obtain)
(20.4),
+ CL(|pI
\342\200\230
'\302\260\"\"'J'0\"\'")
by
+p2)
CL)
).\302\247,+\302\260x3\\/.7:
x,I,+=x/3:)
together with (18.3) implies the convergenceof the series in (21.7) on the condition every interval [8, 1], where 0 < 8 < 1. Since (21.8) implies the convergence of the series (21.2) for 0 < x < 1,the uniform convergence of the series in (21.7) on every interval [8, 1] implies that (21.7) is valid for This
0<
x s 1.
we observe that for -1 < p < 2,p 96 0, p a\303\251 1, all the functions become in\357\254\201nite at x = 0, since J,,(x) = xP
we note that an argument Finally, with Theorem 1 can be used to show
meaningless.
connection
p = 1and
similar to for p
that
that
in
given
2 2, p
= 0 or
for)
C
|C,,| <
53?\342\200\230:)
0 and C Thus we have)
e >
where
[0, 1].
THEOREM
then
(21.8),
0
1.4 Ifp by
corollaryof
THEOREM
s =
2,
then
3. the
=
0,p term
Theorem
18.))
(21.7) holds
=
I orp
If the hypotheses of Theorem 1 Fourier-Bessel series of the function
twice
4 The convergence of Sec.
relation
2 2, andifthe c,,satisfy is possible everywhere differentiation 2, we obtain)
term by term [0,1]ifp=0,p=lorp>2.)
tiated
the
of the
jbr 0
< x < 1 ifp > -1
series (21.1)itself
follows
the on
and
for
condition [0, 1].
of Sec.20are met f (x)
on
everywhere
2. If p > -1 and if the coe\357\254\201icientsc,, satisfy the inequality the series (21.1) can be differentiated twice term by term
(21.9), then term
As a
are constants),
for
can be di\357\254\201\342\200\231e everywhere
on
from (21.8) or (21.9)and the theorems
22
sec.
In fact,
this
in
237)
scares
to (20.1))
according
case,
AND rooms:-nassm.
FUNCTIONS
BESS!-IL
C
= const), (C |c,,| < 5\342\200\230.,\342\200\224,2 0)
only apply Theorem
we need
and
let M, 7.3,. . .
now
We
,
in
= 0, this
existence
particular
to Sec.
-1 the
>
p
an
given
below
the roots of we We
call such
c,, =
which
function
[0 f
the
in
according
Moreover,
(22.3))
x. The considerations the roots of (22.1); all the
to
(22.3) corresponds
system
(1)
THEOREM. function
Let
xJ\302\247(x,,x)
here the
\342\200\224
p2>J3
the system
+ ---
Then
(22.3). (22.4)
constants
general), theorem,
which
be a
\"\" \342\200\230225\ \"\"\"\342\200\235\"\342\200\230\"\"\342\200\231 Ll\302\273
coe\357\254\202icients
can be
f (x)
to
dx
2).:) + (A:
on de\357\254\201ned
c2J,(A2x)
type;
integrable on [0, 1].
respect
dx
Fourier-Bessel
Fourier coef\357\254\201cientsin We have the following
+
xf(x)J,()\\,,x)
xzJ,;2(A..)
ordinary
is absolutely
c,J,()\\,x)
second
'\342\200\235'
tinuous)
weight 7., are
Fourier series of f (x) with a series)
a Fourier-Bessel seriesof the 1
like
10.
(22.2).
f (x) ~
just
of equation (22.1)[and
J,(x,,x), . ..
with
case where the valid for the special case where to the
pertain
Let f (x) be any can form the shall
. . .,
J,,(A2x),
system on [0, l],
orthogonal
are
results
(22.2))
functions)
J,(x,x), form
becomes
equation
of an in\357\254\201nite set of positive roots of equation (22.2)] was proved in Sec.
12, for
(22.1)
= 0.
J;,(x) The
of the equation (H = const),
= 0
For H
order.
increasing
Type)
the roots
. . . be
7.,,,
HJ,(x) xJ;(x) \342\200\224 arranged
Second
of the
Series
Fourier-Bessel
22.
2.)
(or
obtained
by
usual
closely resembles
piecewisesmooth [0, 1].
that
for the
Then
the
matter, just like formal argument.
Theorem I of Sec.16:)
(continuous Fourier-Bessel
or
disconseries
of))
nnssnr.
238
of the second type its sum equals f over,
f(x)
\342\200\224> p \302\253k,
2
(p
(x) at
More-
of f (x).)
of discontinuity
point
every
< x < 1. (x) and)
8)
- 0)]
+f(x
+ 0)
%lf(x at
H) converges for0 of continuity of f
point
every
can-.
seams
AND FOURIER-BESS!-IL
FUNCTIONS
\342\200\224if For x = 1, the series convergesto the value f(l 0); f(x) is con= tinuous at x the series con1, the series convergesto f of the system (22.3) vanish verges to zero, sincein this case all the functions at zero. Sinceall the functions (22. 3) become in\357\254\201nite at x = 0, it is meaningless to talk about the convergenceof the series at x = 0.
(1). If p > 0,
theorem, noting only that the condition proof of this no analog in the theorems of Sec. because required of complications that arise when p < H. It turns out that if p < H, then the theorem is true only if we add a new function to the system (22.3). For is xv, and instead of (22.3)we have to example,if p = H the new function consider the new orthogonal system)
the
omit
We
p > H (which
18)is
has
xv, J,(}.1x), In
this
case,
of the
the orthogonality consequence of
is a
functions
= 0 isa root
1) x
of the
the
equation
2)
(7.8)];
x9 (with
function
weight
1) to the other
facts:
two
following
equation)
xJ;,(x) [see
J,(A2x),. . ..)
=
\342\200\224
pJ,,(x)
0)
' with
x!\342\200\231 satis\357\254\201es Bessel\342\200\231s equation
7. =
parameter
0, i.e., the
equation)
+ xy\342\200\231 xzy\342\200\235 pzy = 0.
-
as can
be veri\357\254\201ed directly.)
Therefore, of Sec.
tions
of
form
we can apply 12 (where the
Bessel\342\200\231s equation
is
function
much
to
the functions
was
Sec. 16,
with
in
with
connection
Example. with
the of the
elimination
respect
xv and J,().,,x) all the considerathe solutions of the parametric
For p
discussed).
more complicated.
ourselves to the case p > H. A remark 1 of Sec. 16 also applies to Theorem
there are also two
of
orthogonality
Thus,
the
simplicity,
\342\200\234additional\342\200\2
we
restrict
made after the statement of the present theorem, and moreover, theorems completely analogous to Theorems 2 and 3 of condition p > H in both cases and the supplementary condition f (1) = 0 in the second case. (The remark made Theorems 2 and 3 of Sec. 16remains in force.))
Make a seriesexpansion system)
like that
of the
function f
(x) =
x!\342\200\231 (0 <
x <
1)
to the
J,,(7.,x), J,(7.2x),. . ., J,().,,x), where
< H, the
for
1,, are the
roots of the
equation
. . .,
= 0 J ,\342\200\231,(x) (p
(22.6)
>
0).))
23
sec.
BESSEL
By formula
SERIES
239)
(22.5)) 1
21,2,
and
AND FOURIER-BESSEL
FUNCTIONS
by (16.1))
f (1)xP+1J,(A,,x) by the
Therefore,
\302\243\342\200\224nJ,+,(A,,).
theorem given above, we can write)
ask whether since for p = 0)
now
We
negative,
dx =
this
holds
expansion
=
Jp-I-l()\342\200\230n J10\302\273:
=
for p
= 0.
The
answer
is
=
0) -J(\342\200\231)0\342\200\230n)
used equation (7.8)]. Thus, the side of (22.7) right-hand side is f(x) = x0 = 1. The reason why left-hand the = = = H fails for 0 In our case that H. is the so 0, expansion p following: p must the be Then, as remarked above, the system (22.6) supplemented by = x0 = 1. If we denote the function to this x\342\200\235 coef\357\254\201cientcorresponding we have
[where
is zero, while
function
by
the
co, we
obtain)
= co
sincef(x) =
1.
Moreover,
c,,
dx) J: xf(x)-l I x-13 dx I0
= 0 if n
_.
= 1, 2, . .
..
Therefore,
instead
of
(22.7) we obtain the expansion)
l=l+0+0+---,) which
is obviously
*23.
Extension Series
valid!)
of the Results of Secs. |7\342\200\2242|to of the Second Type)
Fourier-Bessel
of Secs. 17and 18, we essentially made no use whatsoever that the numbers 1,,are the roots of the equation J,,(x) = 0. assumption series of the these are equally true for Fourier-Bessel theorems Therefore, second type. However,in the lemma of Sec. 19, we did use the condition = 0 to prove the formula we now need a new lemma, Thus, (19.3). J,,(A) which we proceed to prove.)) In the
of the
theorems
240
that
=
F(0)
di\357\254\202erentiable function \342\200\224 + \302\247)F(l) F\342\200\231(1)(H
(the second
a root
if A is
of the
derivative
p >
-1,
it
arrive 1
1:57
at the
=
HJ,(x
<
0,)
(R =
1%
dx|
(23.1))
const).
1) \342\200\224 zdx o)
(F2; x)
[F '(1)Z(1) -
+
\342\200\224 [F\342\200\231zFz\342\200\231]\"\342\200\234 x==0)
F\342\200\235)
19,
that
that
equality)
as in the lemma of Sec. the last term vanishes.
just
[0, 1],
and such
at certain points).
not exist
may
\342\200\224
F(x)J,().x)
\302\2431/.7
We
0,
that)
follows
I=)
Proof.
on de\357\254\201ned =
8)
equation)
xJ;(x) where
CHAP.
SERIES
twice
=
F\342\200\231(0)0,
is bounded F\342\200\235(x) Then,
a
F(x) be
Let
LEMMA.
such
AND FOURIER-BBSSEL
FUNCTIONS
BESSBL
whole
the
and
F(1)z'(l)l-
term We begin by calculating the \357\254\201rst
reduces to showing
problem
This last term
difference)
is the
-
[F'(0)Z(0)
(23-2))
F(0)z'(0)l.
Since 2
in brackets.
=
,
\\/:_cJ,().x
we have)
2(1) =
-\342\200\231p(7\\).
=
+
w.=<~>1..,)
+ >J;,m =
= we
where
\357\254\201rst term
used the condition in brackets is just) have
-
term in
second
follows
proof The
by the
vanishes
which
lemma
brackets
by just
the
just proved
HJ,0)
the
But
then
leads to the following
The fact that indeed the rest of
lemma.
and vanishes, as in the method
(23.2)
same
= 0.
\342\200\224
the
%)F(l)l-7p(?\\).)
of
hypothesis in
)J;,(7t)
(H +
[F\342\200\231(1)
(H + 91.0),)
the the
lemma of Sec.19.)
result:)
on the Let f (x) be a twice differentiable function de\357\254\201ned = = = such) and such that O, f \342\200\231(1) 0,5 f \342\200\231(0) Hf(l) [0, 1], f(0)
THEOREM 1.
interval 5
The
naturally
\342\200\224 f\342\200\231(l) Hf (1) the applications.))
condition in
= 0 appears
arti\357\254\201cial at this
point, but
it
arises
quite
sec. 24 that f
is bounded \342\200\235(x)
Then
the
(the second
derivative
at certain points).
not exist
may
24!)
saunas
inequality)
IC,,| S is
AND rounmn-nessan
FUNCTIONS
BESSBL
Fourier
by the
obeyed
C
=
W
const),)
f (x)
coe\357\254\201icients of
with
respect
to the
system
=
and
(22.3).
Proof. If F(x) = 1/;f (x), bounded.
F\"(x) is
then
F(0) =
obviously
F\342\200\231(0)0
Moreover)
Fee) =
+
~/:'cf'(x>.
{}\342\200\230-\342\200\224f;
so that -
(H +
F\342\200\231(1)
~i)F(1)
= lf(1) Thus, the
lemma
dx
formula
(H +
be applied to the
can
L1)xf(x)J(A,,x)
Since by
-
+f\342\200\231(1)
I;
1 and
THEOREM
[0, 1],
such
f \"(x) is Then
the
lutely
and
and
that
const).)
2
(K
>
0),)
T\"
= f
twice
=
0, \342\200\231(0)
dt\357\254\202erentiable function \342\200\224 = f\342\200\231(1)Hf(l)
secondderivative may series of f (x) of the
not
[0, 1]if p 2 0.5) 1 and 2 of Sec.21apply
exist at certain
second
type
points). abso-
converges
1,ifp > -1
< 8<
where0
[8, 1],
on the interval 0 and such that
whole interval
Theorems second
equally
well to
Fourier-Bessel
type.)
Fourier-Bessel
Expansions
of Functions
De\357\254\201ned
on
the
[0,1])
Let f (x) be an [0, I], and set x = It need
dx
on every subinterval
uniformly
Finally
Interval
(R =
%
of Sec. 18imply)
f (x) be a
f(0)
Fourier-Bessel
series of the 24.
<
is)
account of (22.5).
the results
2. Let
bounded (the
on the
dx
= 0.
Hf(1)
the result
F(x);
K xJ\302\247(A,,x)
(23.3), taking
Theorem
function
We F(x)J(A,,x)
1
obtain
-
=f\342\200\231(1)
(14.1))
Jo we
l)f(1)
integrable
absolutely or
t =
x/I.
5 If the conditionp > H appearing not have 1\342\200\230 (x) as its sum.))
function
Then the function in the
on de\357\254\201ned
= f(l
theorem of Sec.22isnot
the
interval
on) t) is de\357\254\201ned
met,
then the
series
massm.
242
the
interval
AND FOURIER-BESSEL
FUNCTIONS
[0, 1]
of the t-axis,and
~
we
CI-IAP. 8)
write
can '
4'
\342\200\230P(t) C1Jp(7\\1t)
seams
4' C2Jp(7~2\342\200\230)
\302\260 \302\260 \302\260 + cnJp0\342\200\230ut) 4' \302\260 '9
(24-1)
where
c,, = case of
in the
cu)
a Fourier-Besselseriesof
= + \357\254\201_\"Jmn)
$3
_
the
case of a Fourier-Besselseriesof variable x, we obtain) ~
1,2,...) or)
\357\254\201rst type,
1,2,...))
Returning to the
type.
+---,
+---+ c,,J,
c2J,
c1J,,+
second
the
=
(n
@130\") j;up(z)J,(1,z)dz
in the
f(x)
(n =
dt
:\302\242(:)J,(21,,z) f\342\200\231 0) J,\342\200\2242\342\200\224 p+1(7\\n)
(24.2))
where)
c,, =
xf(x)J,
I;
x)
dx
=
(n
1, 2,.
. .),
(24.3)
?,\342\200\224J\342\200\224\357\254\201m\342\200\224) Or)
1
27%
=
\302\260\"
+
Iztxzlm.)
(1%
I o
p7>J.%(x..)1
1,,
d\
\"f\"\"\342\200\231P (\342\200\230T \")
(n =
1, 2,...).
(24.4))
series (24.1) converges, then the series (24.2) converges, and conversely. It should be noted that we can also obtain the expansion (24.2)directly,
If the
the auxiliary
avoiding
function
if we
observe
xJ,,
[L where
with
x)
we have
x on the
weight
J,
x)
dx =
interval
12
changed the variable
we have establishedthe
orthogonality
I
(1)
[0,1].
tJ().,,,t)J(7.,,t)
of integration of the
system)
(24.5))
-1)
(p>
J,(5llx),...,J,(%'x),...
is orthogonal
that the
we
In fact, dt =
o
have)
(m ;e n),)
x = It. Once we calculatethe system (24.5), by
setting
usual way, obtaining or (24.4), depending (24.3) = 0 or of the 1,, are the roots of the equation J,(x) \342\200\224 = 0. Just as in the case of the interval equation [0, l], the HJ,,(x) xJ;,(x) series (24.2) with the coe\357\254\202icients(24.3) is called a Fourier-Besselseriesof the and the series (24.2) with the coefficients \357\254\201rst type, (24.4) is called a FourierBessel series of the second type. we can go from the series(24.2) to the)) Since Fourier
on
whether
coe\357\254\202icients in
the
numbers
the
PROBLEMS
series (24.1) interval
the substitution
making
by
convergence criteria are valid, interval
[0,
x =
caseof the
hold for the
1] also
[0,
for the
AND 1-\342\200\230uncnous
BESSEL
all
It,
FOURIER-BESSBL
243)
results for the In particular, the
our previous
interval
[0, I].
that
of course
provided
seams
are
they
formulated
of for [0, 1].)
1] instead
PROBLEMS
1. Write
the
solution of
general
differential
the
5 , .. + ,1\342\200\231 Ey
+ y =
0.)
Calculate)
2.
a positive
where n is 3. Find
an
4. Find
5. Show Hint.
6. Show
J,,(x) (p
functions
Use formulas
1/}
preceding
J,(x)
(p 2
problem.) \342\200\224 are
1})
bounded
(4.3) and
(9.16).
that)
By multiplying
tJo(t) 4:
=
xJ,(x).)
and e-1\342\200\230/2', the power seriesfor ex\342\200\230/3 show
\342\200\224 =
exp
and
> 0) and
the
co.)
J:
8.
J,\342\200\231,(x).
(9.16).)
for J ;(x). (1.1), formula (9.16)and
equation the
that
(7.9) and formula
asymptotic
Use
for
formula
asymptotic
an
Hint.
for 0
integer.)
Use formulas
Hint.
7.
equation
By substituting t imaginary parts,
that)
J,,(x)t\".)
(t
\"-3200
= e\342\200\230\302\260 in the formula of the preceding problem, taking real) and using the formula for trigonometric Fourier coemcients,
that)
show
J2,,(x)
=
cos (x
;\342\200\230-r ['5
sin 0) cosn0
d0,
m=mLz\342\200\235o
J,,,.,,(x) =
9.
Show
a)
sin (x
;\342\200\230-r I:
sin
0) sin
allx
(k,
n6 d0.)
that)
J,,(x) < 1 for
n =
0,1,2,
. . .);))
runcnons
nassu
244
b)
J,,(x) = 0
lim
c)
(n =
= 0
J2,,+,(0)
seams
FOURIER-BBSSEL
AND
8)
CI-IAP.
0,1, 2,. . .);
for all x.
ll->0
10.
the function
Expand
x <
x\342\200\230? (0 <
1) in
series
Fourier-Bessel
of the
\357\254\201rst
kind.
the function
11. Expand the
x!\342\200\231 (0 <
x <
1) in
series with
Fourier-Bessel
respect to
system \302\260 - \302\260 9 Jp()\342\200\230lx)9 Jp()\342\200\2302x)9
are the roots Use the example
Hint.
12. Expand
= 3.
Use
Hint.
13. Let
function
the
with p
kind,
of the
the 1,,
where
1,,
A2,
. . . be
x3 (0
(24.3) and
formula
= O.
equation xJ,',(x)\342\200\224 HJ,(x) 22.
in Sec.
< 2) in
the
results
Fourier-Bessel
series
of the
in
of the
the positive roots
example
equation
Jo(x)
Sec.
= 0.
of the
\357\254\201rst
16.) that
Show
@
i(1-
x=)=
(0
1).
equation
J,(x)
gl\357\254\201
Hint.
Use formulas
14. Let 1,,1.2,
. . . be
(7.2), and
(7.1),
the positive roots
(15.3).)
of the
= 0
(p >
\342\200\224
-}).
Show that)
;
+ 1)
a)
x\302\273+1=
2=(p
b)
xP+3 =
23(p + l)(p
+
(1))
2)
\"El
;) %\342\200\234)%
\302\260\302\260 L-:\302\253_z\302\243L\302\273_x.> = \302\260') \302\260\342\200\231
x.J,+.
,,,,, (In
each case, Hint.
to x,
0 <
x <1.)
a) Multiply using
x\342\200\230(P+1) and
the last
b) Multiply differentiate, using (7.1);
formula
of Sec.
(I) by xP+3 (7.2).))
16 by xp\342\200\234, and integrate integrate; c) Multiply
and
from 0 (I)
by
9)
EIGENFUNCTION
THE
AND ITS APPLICATIONS
METHOD
TO
PHYSICS)
MATHEMATICAL
Part
THEORY)
I.
I. The Gistof the Method) Many
of mathematical physics lead to linear Examples of such equations are)
problems
equations.
Ba
83::
6314
P-3-;_5+R-5+ P3314
-5?+
differential
partial
(l.l)
Qu\342\200\224-E5:
_au
Ran
12
'5-I-Qu\342\200\224-8-in)
of the variable x, and u = u(x, t) is P, R, Q are continuous functions \357\254\201rst arises in and an unknown x t. The of the variables function equation second arises in while the of and vibrations rods, strings problems involving to our attention shall con\357\254\201ne of heat \357\254\202ow. We equation (1.1), problems since this will be quite suf\357\254\201cientto explain the gist of the method to be where
discussed
here.
In every concrete a solution a requires
suppose and
that
all t 2
we
have
0, which
problem leadingto of (1.1) which to
an
of the equation conditions.
satis\357\254\201es certain
a = u(x, t), conditions) boundary
\357\254\201nd a solution
satis\357\254\201es the
atu(a, t)
+
yu(b,t)
+
[3
8u(a, t) _) _ Bx
6
For example, b a S x \342\202\254
for de\357\254\201ned
0,)
(1.3))) b,
0
3-\342\200\224u-a(-x%)=
245)
form (1.1),one
246
METHOD
\342\200\230rm: EIGENFUNCTION
any t 2 0, where initial conditions)
for
a, (3,
APPLICATIONS
CI-IAP.
constants, and
8 are
and
7,
rrs
AND
the satis\357\254\201es
which
= go)
u(x,o> =f(x).
9)
(1.4))
a < x < b, wheref (x) and g(x) are given continuous functions. Usually, 2:stands for length and t for time, which explains the terms boundary conditions and initial conditions. We shall assume that the pair at, (3 nor neither the pair 7, 8 can vanish simultaneously [for otherwise, instead of the relations 0 = 0]. This assumption can be (1.3), we would have the vacuousidentities
for
as)
written
a2 +
To solve the equation
:32 as o,
+
82
look (1.1), we \357\254\201rst
for
y2
75
o.
(1.5))
solutions of the
particular
form1)
u =
(1.6))
(x)T(t),
only the boundary conditions (1.3). (We satisfy in solutions which do not vanish identically.) With this entiate (1.6) and substitute in (1. 1), obtaining) the result
are
which
+
P\"T
R\342\200\231T + QT
=
only in mind,
interested
we differ-
T\342\200\235,
whence
Pd)\" +
Rd)\342\200\231 + Q _ W\342\200\231 \" T'\342\200\231_)
?
(D
Sincethe left-hand sideof side
right-hand
is a
is a function only is possible only equality
last equation
this
function of
t, the
of x, while the if both sides
equal a constant:) Pd>\342\200\235+R\302\242'+Q\302\243_1\342\200\235__
<11
leads
This second
to the
two
following
A
\342\200\234T\342\200\235)
(A =
const).)
linear differential
ordinary
Pd)\"
+
Obviously, a function tions (1.3) if and only
= = 0.
MD\342\200\231 + Q4) T\342\200\235 + AT
\342\200\224).,
if the
condithe form (1.6) satis\357\254\201es the boundary function (x) satis\357\254\201es conditions the boundary
w@+mm=m
is known
as the
(1.7)
(1.8)
u \302\242 0 of
y(b) 1 This
equations of the
order:)
+
method of separation
8'(b)
=
') 0\342\200\235
O.
of variables.
(Translator)))
SEC. I
The
of
problem
tions (1.9)
will
METHOD
AND ITS APPLICATIONS)
of the equation the boundary value problem
a solution \357\254\201nding
called
be
conditions
the
with
EIGENFUNCTION
THE
(1.7)satisfying
the
condi-
the equation
for
(1.7)
(1.9).
In general, the boundary value problemfor a linear differential equation second for every value of A; in particular, order does not have a solution this is true of the boundary value problem for the equation (1.7)with the conditions Nevertheless, it can be shown that there exists an in\357\254\201nite (1.9). set of values 7.0, 711, . . ., 7.,,, . . . for which the boundary value has a problem solution, provided only that P a6 0. Every value A for which the boundary (D $ 0 is called an eigenvalue, and value problem has a solution the solution (1)corresponding to 7. is called an eigenfunction. It will be shown below that in our to each eigenvalue (to within case only one eigenfunction corresponds is an in\357\254\201nite set of eigena constant for there our Thus, problem, factor). values 7.0,71,,. . ., 7.,,, . . ., with corresponding eigenfunctions)
of the
,,(x),. . ..
1(x), . . .,
o(x),
(1.10))
an orthogonal will be shown in Sec. 4, the functions (1.10) form system on [a, b], with a certain weight. with Having \357\254\201nished equation (1.7), we next solve equation (1.8) for each on two 7. = 1,, and function T,,(t), which \357\254\201nd the corresponding depends arbitrary constants A,, and B,,. Thus, if 1,, > 0 for n = 0, 1, 2, . . . (and we obviously have) this case occurs quite often in concrete problems),
As
A,, and
where
B,, are arbitrary un(xs
will
be a
constants.
Then,
(1.11)
each function
('3 = 0: ls 2: -
= (Dn(x)Tn(t) t)
solution of equation
Because of the
+ 3,, sin 1/7,,t,
= A,,cosx/1\",:
T,,(t)
-)
1.1)
conditions (1.3). the boundary ( satisfying and homogeneity (with respect to u and its derivatives) of solutions of (1.1) is also a solution. sum every \357\254\201nite
linearity
of equation (1.1), The same is also true of the
series) in\357\254\201nite
Q)
a =
2
=
u,.(x. t)
n=0
if it
3
n=
converges and if it can be differentiated t. If this is the case, we have)
(1.12)
T.(:>,.(x>. 0)
term
by
twice
term
with respect
to x and
8214
83a
Bu
Qu--5t\342\200\2242) P-'\342\200\224+R5;+ ax? \302\260\302\260 \342\200\224
8711,,
+
Page}?
\302\260\302\260
3a,, Rngo\357\254\201
''\302\260
''\302\260 \342\200\2248211,,
+
Qngoun
go?)
\302\260\302\260
=
0214,,
8n,
6 8x) \"575)) 07ru,,)_ Z(P\342\200\224x\342\200\2242+R\342\200\224\342\200\224+Qu,,\342\200\224) \"=0
me
248
METHOD BIGBN1-\342\200\230UNCTION
can-.
APPLICATIONS
rrs
AND
9)
term in parentheses in the last sum'vanishes [because u,, is a of (l.1)], the entire sum vanishes, which means that the function is a solution of (1.1). Moreover, each term of the series (1.12) since (1.12) conditions (1.3), the sum of the series, i.e., the function satis\357\254\201es the boundary each
Since
solution
u,
also
by
conditions.
satis\357\254\201es these
We must
now
expressions
t).
u,,(x,
This can be achieved
conditions (1.4).
values of the
choosing the for the functions
suitably
initial
the
satisfy
constants
A,, and
With this
in
mind,
B,, appearing in the we require that the
relations)
0) =
u(x.
f(x) =
i
..(x>T.(o).
n=0) (1.13) =
=
so)
3\342\200\224\"<,2;\302\243>
that the functions f (x) and g(x) can to the eigenfunctions (1.10). The possimaking such expansionscan be proved under rather broad conditions
which is
hold,
be expanded of
bility
on the
\302\247o..(x>T:.(o))
in
to
equivalent
requiring
with respect
series
coe\357\254\202icients in
(1.1) and on the
equation
which
functions
are
to be
expanded. Thus let) =
f(x)
c...(x>.)
3,0
g(x) = we need
Then
c...(x>.
Z0
only set
: in
order
to
(1.14)
\357\254\201nd A,, and
B,,.
we.
(n
=
Hence, if (1.11) holds, we
0) =
f(x) =
in
\302\260\342\200\231
=
Mg;
=
we
$0
(1.15)
..),
0, 1,2,.
have
A.,..
B..~/T...(x>.
so that
A,, =
C,,, B,,=
c,, (n
_
\342\200\224
0,1,
(1.16))
...).
2,
VA.\"
Our results
are
based
and can be differentiated
fore, the
coe\357\254\202icientsA,,
on the supposition by term twice
term
and
B,, just
found
that with
must
the
series
respect
be such
(1.12) converges
and t. Thereas to guarantee that)) to x
me
sec.
1
this
is the
do
often converge
make
case. However,in this
have
not
that
know
We
order to avoid
words
functions.
boundary
concerning
By the conditions
in order.
are
the coefficients A,, and B, problems, Whether or not the series (1.12)will in Sec. 7. In the meantime, discussed we
de\357\254\201ne discontinuous
confusion,a few
conditions
initial
be
often
series
249)
actual
property.
case will remarks:
in such a the following
AND 113 APPLICATIONS
METHOD
EIGENFUNCTION
(1.3) and
Hence,
in
conditions and (1.4),
we mean
more precisely) at
lim .
Y 1
\"(x, 0
3},
of (1.3),
instead
t) +
u(x,
(3
o,)
.
8u(x, t) = 0 + 8 l \342\200\224\342\200\224-\342\200\224 .13}, 3x)
t)
=
lim
f(x),
=
g(x)
3\342\200\230i\302\247-\"*\342\200\224\342\200\230) 1 :\342\200\224>o
:\342\200\224>0
In other words, in as etc., are to be interpreted
of (1.4).
0u(a, t)/ax,
3\342\200\230$\"*\342\200\224\342\200\2303 =
and lim u(x,
instead
lim
a) X\342\200\224>d
(1.3)
and
the limits
the values
(1.4),
to
which
u(x,
of
u(a,
t),
t), au(x, t)/Bx, b, t > 0 con-
etc., convergeas the point (x, t) lying in the region a < x < It is quite clear that such verges to the correspondingboundary point. only an interpretation of the boundary and initial conditions can correspond to the physical of the problem. In the same way , when we say content that in the region the function a < J: < b, t 2 0, we u(x, t) is continuous mean that u(x, t) is continuous in the region a < x < b, t > 0 and that the limit
lim
u(x,
t)
(a <
x < b,
t >
0))
x\342\200\230Vxo (\342\200\230\"510)
a
has
the
for every point (xo, to) on the boundary of the region} to show that the boundary conditions vary continuously as to) is moved along the boundary.
\357\254\201nite value
Then, it point
is easy (xo,
similar by the solution of equation (1.1),or of some equasolution which is mean a continuous in the sense just always It is easy to see that if such a solution is to exist, then the bound-
Subsequently,
tion, we shall
indicated. ary conditions and the
and (b, 0) in continuity.
t) = 0,
such Thus,
initial
conditions
a way that the boundary for example, if the
must
at the \342\200\234agree\342\200\235 values
do
points not undergo
conditions (1.3)
have
the
(a, 0) a disform
u(0, u(l, t) 0, and the conditions (1.4) have the form u(x, 0) = x + 3u(x, 0)/at = x2, then it is clear that the boundary values a undergo discontinuity at the points (0, 0) and (1, 0), so that the problem certainly cannot have a continuous solution.) =
1,
2
a jump
The analytic expression on the boundary.))
for u(x,
t) may
not be a continuous
function,
i.e., may
undergo
THE mcmruncnou
250
Usual
The
2.
AND rrs APPLICATIONS
METHOD
of the
Statement
9) emu\302\273.
BoundaryValue
Problem)
P in equation (1.1) does not vanish. that the function nor gains eigenvalues and eigenloses equation (1.7) neither function. We now if we multiply all its terms by a nonvanishing we can transform out such a multiplication, (1.7) into by carrying
shall assume
We
Obviously, functions
show that the form)
=
+ q<1\342\200\231 -M1\342\200\231. (p\342\200\2301\'")
where tion
functions p, q, and r are continuous with a continuous derivative, and r
of x
is a
(2-1))
on [a, b],p isa positive
positive
(This can be done without all the terms of since we need only multiply generality, otherwise \342\200\224 \342\200\224l and replace A by A.) Then we solve the system) First we
Proof.
assume that
p
P
> 0.
= rP,
=
rR, p\342\200\231
some point of the interval be zero.) Obviously, p need only consider
x0 is
where
to
integration
we
Now,
rP'
[a,
b].
func-
function.) loss
(1.7) by (2.2))
(We take the constant and r >
> 0, p\342\200\231 is continuous
+ rQIl> =
of
of
0.
\342\200\224).r
set)
and
q = according
Then,
(2.3)
rQ.
to (2.2), =
\342\200\224w9) p\302\242\342\200\235+p\342\200\231d)\342\200\231
is just
which The
with
(2.1),
are still
3. The We
basic
(real
given
by
posed
the requirements same formulas
coe\357\254\202icientssatisfying
conditions
boundary
the desired equation (2.1). value problem is usually
boundary
the
equation of the form just stated. The boundary (1.9).) for an
Existenceof Eigenvalues)
shall
not give
value
a completeproof of the
problem
under
existence
consideration;
idea of such a proof. Thus, or complex) and \357\254\201nd a solution =
3.
in equation
satisfying
of eigenvalues
for the
instead we just describethe value (2.1) we give 1.a \357\254\201xed the = -a\
conditions)
sec. 3
rm:
i.e.,
(3.1))
boundary conditions (1.9). (The prime respect to x.) As A changes, d>(x, A) also continues to satisfy the condition (3.1). Thus, a
with
differentiation
changes, but
= 0,
A) [3\342\200\231(a,
\357\254\201rst of the
the A) satis\357\254\201es
(x,
denotes
25|)
Obviously
+
oc
AND 113 APPLICATIONS
METHOD
this solution by d>(x, A).
denote
We
monnruncrron
nevertheless
of x and the parameterA satis\357\254\201es the \357\254\201rst of the conditions it is shown that (In the theory of differential equations, of a power series in A, and hence is d>(x, A) can be represented in the form an analytic function of A for all values of A.) We now form the function) function
known
(1.9) for
A.
any
A) +
y(x,
(3.2)
8\342\200\231(x, A)
and set
D(A) = a
D(A) is
D(A) =
is
an
obviously
and (3.3)
met.
Thus,
variable
of one
function
known
y(b,
of our
eigenvalue
are satis\357\254\201edsimultaneously, the
problem
of the roots of D(A).
By
has an in\357\254\201nite set of sequenceof the form)
A) +
y(b,
A) +
8\342\200\231(b, A). A,
and
A
= 0
existence of eigenvalues
using
this fact,
it
can
which
of A, (3.1)
values
such
both of the
i.e.,
of the
for
(3.3))
8\342\200\231(b, A)
problem, since for
real eigenvalues
value of
every
conditions (1.9) are to a study that the problem can be written as a reduces
be shown
(see Sec.4), which
)\\o<)\\l<...<)\\n<...,
where
limA,,= +00.)
4. Eigenfunctionsand Let
A
eigenfunction
be an
eigenvalue
Their
Orthogonality)
of our
boundary
to A.
corresponding
value
Then, it is
form Cd>(x),where C is an
problem,
and let
easy to see that
every
an
function
nonzero constant, is also an eigenarbitrary shall We not consider such linearly corresponding dependent to be different, and in fact any of them can be regarded as a eigenfunctions of the family of functions of the form C(x),where C 96 0. \342\200\234representative\342\200\235 We now ask whether two linearly independent eigenfunctions d>(x)and can belong to the same eigenvalue.3 With our hypotheses, the answer) \342\200\230I\342\200\231(x)
of the
to A.
function
3
a 0 implies If a(x) + b\342\200\230P'(x)
independent.
Cf. Ch.
2, Prob. 9.
a = b = 0, then (Translator)))
are said (x) and \342\200\230I\"(x)
to be linearly
monuruncrrou
me
252
is negative.
value.
Then,
differential
AND rrs APPLICATIONS
METHOD
cmur.
In fact, suppose that (x) and \342\200\230P'(x) to the same belong solutions independent by a familiar property of linearly would we have) equation,
'(x)
(x)
9)
eigenof a
0 a\303\251
(4.1))
\342\200\230I\342\200\231(x) \342\200\230F'(x))
on [a,
everywhere
conditions
of the
b], and in particular (1.9), we have)
= a.4
at x
But
according
to the
\357\254\201rst
w@+mm=m
w@+wm=a) which
would
(4.1)
by
Thus,
(1.5).
hypothesis
to each
corresponds
that
imply to
=
a function
is
variables, e.g., on the
Then,
=
0, [3 = 0, thereby constant factor, only
depending on x. the partial
we write
t,
(If
(4.2))
an\302\273.
(D also
depends on other to x.) respect
with
-derivative
identity)
,%
twice
holds for any The
+
5 (p
\342\200\224 = L<\342\200\230I*> \342\200\230PL<>
their
as
2.
LEMMA
d1 and
functions
dijferentiable
and
(4.3))
\342\200\230F.)
L() and
replacing
L(\342\200\230I\342\200\ by
(4.2).
by
given (D
If
\342\200\224 [p(\342\200\230I\"\"I\342\200\231)1
immediate consequenceof
is an
proof
expressions
\342\200\230I\342\200\231 the satisfy
boundary
conditions (1.9),
\342\200\224 \342\200\224 = = [d>\342\200\230P\" [\342\200\230I\" \"I\342\200\231],,_.,, d>\"I\342\200\231],,.,,, 0.
The
Proof.
on and do not vanish [3, which satisfy the homogeneous system)
numbers
to (1.5),
according
the
contradicting one eigenfunction
eigenvalue.)
Ltd\302\273)
<13
a
1. Let)
LEMMA
where
at
within
oc(a) +
=
[i'(a)
then (4.4))
simultaneously
0,
=
0. + $\342\200\230I\342\200\235(a) ac\342\200\230I\342\200\231(a)
This is
4 The
possible only
a
symbol
(4.1) is usually
O))
called
(a)
\342\200\231(a)) [<1>~1~
WI)
ma)
denotes 3
determinant
if the
=
the
a Wronskian.
determinant
of the system vanishes, i.e., -\342\200\224
<1>'\\1r],._.,.,
=
The ad \342\200\224 bc.
(\357\254\202anslator))
o.)
functional
determinant)
4
sec.
equation (4.4)is proved
part of
second
The
now show that any two to different eigenvalues
We
ing
[a, b],with
p.,
and
(x)
253)
way.)
correspond\342\200\230I\342\200\231(x)
on
are orthogonal
respectively,
r.)
weight
and
Let Q
Proof.
and
same
the
in
eigenfunctions 7.
AND 113 APPLICATIONS
mrmon
\342\200\230ran BIGBNFUNCTION
\342\200\230I\342\200\231 the satisfy
equations
= -Artps =
L01\342\200\231) \342\200\230WW9)
by
boundary conditions second by (D and then
the same
with
\342\200\230P\342\200\231 and the
By Lemma 1, we
second.
[P(<1\"1\"
(1.9).
We
\357\254\201rst equation
from
\357\254\201rst equation
the
obtain)
-
-
=
\302\242\"1\342\200\231)l\342\200\231 0\302\273 t\302\242)'\302\242\342\200\230P'-)
we have
Therefore
= (A
\342\200\224 [p((D\342\200\230I\342\200\235 \"P') 3::
By Lemma
\342\200\224
p.)
that
r\342\200\230P' dx.
(4.5)
\"
2 \342\200\224 [p(<1>\342\200\230P\" \"I*1::::
so
the
multiply
the
subtract
that)
implies
(4.5)
= 0,
o\342\200\224mf@Ta=o
Since A
a\303\251 p., it
follows
that =
r
as
to be
was
shown.)
In Sec.
Remark.
3, we said
the
that
eigenvalues
are real,
but
we
did
eigenvalues can be deducedfrom reality the of the eigenfunctions just proved. In fact, if A = orthogonality is an eigenvalue and if d>(x) =
a proof.
of the
The
ww+ww+\302\253vH\302\256=\342\200\224m+Mw+w>)
But
then
we also
have the equation)
\342\200\230 4' q(
which
means
that
X
=
p.
-
iv
\"'
=
-(l-'ill\342\200\231)
is also
\"
\342\200\234 l\302\260V)\"(
an eigenvalue
and
that
the
function))
=
5(x)
METHOD
EIGENFUNCTION
THE
-
implies
5.
to
corresponding
this
But
X.
X.) a\303\251
Sign
of the
=
#115 dx
J:
for as just
is impossible,
sinceA
r(
dx 96 0,) 4:\342\200\231)
+
proved, d)
be orthogonal
5 must
and
Eigenvalues)
The following
theorem
values for the
case
encountered very
often If r
THEOREM.
more precise
gives
for a the applications.) q < 0
where
in
< 0 and
> 0, q
if the
a
b,
eigenwhich is
situation
conditions
boundary
the
about
infonnation
that)
imply
< 0. [p\342\200\230D\342\200\2301\"\302\247Zf\342\200\231.
then all
the
boundary value problem for
of the
eigenvalues
(5-1)) (2.1)
equation
are nonnegative.)
Proof. Let eigenfunction.
an
be
A
[p(IXI>\342\200\231 3:3
It follows
the
be
(x)
and
integrating,
corresponding we obtain)
=
fqozdx
\342\200\224
Therefore
-=- 0, i.e., <1)\342\200\231
+
f\"p<1>'2dx
from (5.1) and
(5.2) is < 0.
by
(2.1)
,\342\200\224ALbr3dx,
by parts:
integrating
whence,
and let
eigenvalue
Multiplying
dx + j:(p')'<1>
q 5 0,
f\"q<1>2dx q <
condition
the
0
=
-1 the
that
A ,>. 0, and
=
moreover, only if the equation (2.1) has the A
j\342\200\235r<1>2dx.
left-hand
0 is
possible
(5.2))
side of only
if
form
= -M1\342\200\231 (p\342\200\2301\342\200\235)\342\200\231 and
9)
that)
J: which
the eigenfunction
i\302\242(x)is
CHAP.
ITS APPLICATIONS
AND
the function
(D
=
const
is an eigenfunction.
Remark. The condition it is not. In fact, it is actually most often encounteredin the =
(5.1)
seems
satis\357\254\201ed for
applications,
to
be quite arti\357\254\201cial,but boundary conditions
the very
namely
=
1)
0; 4\342\200\231(a) 4\342\200\231(b)
2)
0; d>\342\200\231(b) \342\200\231(a)
3)
\342\200\231(a) h(a)
=
=
-
=
O, '(b)
+ H(b)=
0,))
5
sec.
EIGENFUNCTION
THE
where h and H are nonnegative and second In the third cases.
=
< hp(a)\342\200\231(a)
A0, A1, .
Let
. ., An,
arranged
problem,
be
all the
increasing
order,
eigenfunctions,
corresponding
Eigenfunctions)
eigenvalues of and
for
which
our
value
boundary
let)
\302\260 \302\260 '9 (pn(x)9
(pl(x)9
(p0(x)9 be the
the
Respect to
. . .
in
0.
\342\200\231(b) \342\200\224H(b).)
with
Series
\357\254\201rst
=
\342\200\231(a) h(a),
Fourier
6.
255)
n\342\200\230s APPLICATIONS
is obvious for the
This
constants.
[p\342\200\231]\302\247:\302\243\342\200\231.
since
AND
case,)
Hp(b)2(b)-
-
=
METHOD
\302\260 \302\260 \302\260)
we regard
simplicity
as having
normalization)
the
for
Then,
form
can
(n =
= 1
j\"r:(x)dx every function f (x) the Fourier series)
is
which
o, 1,2,...).
(6.2))
on [a, b],
integrable
absolutely
we
+ \302\242\342\200\2301\342\200\230p1(x) 4' ' \302\260 \302\260.) f(x) \"' \302\242'o\342\200\230po(-7\342\200\230) where
c,, = the following
and
THEOREM smooth
(but
boundary
propositions, f (x)
1. If
is continuous
of our
the Fourier
and
= 0.
THEOREM continuous
2. If or
proof, are
valid:)
respect
i.e.,)
3f'(b) = 0.
to the
(6-4))
eigenfunctions converges
uniformly.)
the
function
discontinuous),
if we recall how our and (l.l4)] and regardf(x) as for writing f(x) = u(x, 0), then
arti\357\254\201cial,but
in (1.4)
value of the function u(x, the conditions 0, (1.3)becomejust
either
problem,
Yf(b) +
initial
t =
(6.3))
[a, b],
value
The conditions (6.4) may appear formula problem arose [seethe \357\254\201rst the
. .),
if f (x) has a piecewise and derivative, if f (x) satis\357\254\201esthe
boundary
series of f (x) with
0, 1,2,.
cite without
on
discontinuous)
possibly
conditions
to f (x) absolutely
we
which
+ l3f'(a) \302\260\302\242f(a) then
(n =
dx
rf(x),,(x)
f\342\200\231
f (x)
t), the
conditions
(6.4).)
is piecewise smooth then the
on
[a, b]
(but
Fourier series of f (x) with))
of
point
+ 0)
~Hf(x at
Instead of the consider the system)
in the
is orthogonal
<1>.,(x), \\/F
be
a square
integrable
agree
and
the
with
(x) at
with
r, we
weight
can
(6.5))
which)
=
(n
o. 1.
with
Then,
2....).)
respect to the
new
the form)
ax.)
rf(x)<1>..(x)
If
Fourier
for
=1
function.
coe\357\254\202icientshave
c. = i.e., they
orthogonal
sense,
ordinary
,/ ffr?.(x)dx
its Fourier
system,
value f
,(x),. . .,
=
|IW..(x)|I
Let 1/;f (x)
to the
- 0)]
+f(x
which is
(6.1),
system
x/F which
a
9)
of discontinuity.)
point
every
can.
APPLICATIONS
us
AND
eigenfunctions convergesfor continuity and to the value)
to the
respect every
mmroo
aronuwucrron
me
256
coe\357\254\202icientsof
respect to the
f (x) with
system
(6.1).
to the function \\/r f (x), the (6.5) becomes [seeequation (7.1)of
completeness condition for
Applied system
b
L If this
Ch.
\302\260\302\260
Q)
..
dx = rf\342\200\231(x)
= Zoc\357\254\202lx/r.(x>II= n=
equation holds for any the problem to the
square
the
2])
integrable
(6.6) Zoe: n=-
function we simply
f (x),
instead
then
say system prove that this is actually the case. It is clear from what has just been said that it is su\357\254\202icientto prove that the ordinary (6.5) is complete. Now, any continuous orthogonal system function
(6.1) is
complete with
weight
(6.5),
the
We now
r.
derivatives) satisfying the boundary value we can choose boundary problem. [For example, = = = such that g(a) = g\342\200\231(a) We shall g(b) g\342\200\231(b)0.]
a function
g(x) (with
conditions
of our
for g(x)
a
not
a detailed
give
system
that
function
continuous
two
proof of
this
fact,
which
is quite
clear
from
geometric
considerations.
Thus,
let)
ff
where
s >
seriesof
g(x)
0 is
arbitrarily
with
respect
r(x>
\342\200\224
g(x)?
dx
<
(6.7))
3\302\273
to Theorem 1, the small. According to the system (6.1) converges uniformly
Fourier to
g(x).))
me
6
sec.
there existsa linear
means that
This
=
0n(x)
such
This
combination)
\302\260 \302\260 4' Y1\342\200\230D1(x) 4' \302\260 4' Yn(Dn(x) Yo\342\200\230Do(x)
-
c..(x)| <
(5-3)
\342\200\224
elementary
(6.7) and
from
\342\200\224
M)
If
+
2
This
=
[<1>(x)
-
degree
Now let F(x) be any is also continuous. If
g(x>12 dx
by
what
has just
\342\200\224 a.(x\302\2731= dx)
a.(x>12
be approximated form
<
s.)
in the
mean
F(x)/V
r(x)
(6.8).
Then the
function.
dx
function
been proved, there
exists
a
combination
a linear
a,,(x)
which)
Fla)
_
o',,(x)
W
dx s
]\342\200\231
2. h
we have)
the
function
system (6.5), and
this systems
complete, as wasto be shown. THEOREM
(6.6)
3.
dx =
W o,,(x) is
hence
orthogonal
relation
can
r(x),
\342\200\224 1/r o,,(x)]2 I: [F(x)
ordinary
If [goo
max
L
then
+ 2
function
continuous
\"
Therefore,
+ (see)
we write)
(6.1) for
functions
g(x))
expression of the
h =
of the
\342\200\224
[((x>
If
that any continuous of accuracy by an
proves
to any
If
+ B2),
< 2(A3 B)\342\200\231
(6.9) that)
dx
a..(x>1=
s
(6.9))
\302\247-
inequality
(A follows
dx <
a..(x>12
{got}
By the
(a < x < b).
./4-(T333
I\342\200\235
But
257)
that
implies
then
AND rrs APPLICATIONS
that
|g(x)
it
METHOD
EIGENFUNCTION
The
holds for
system
a
E
r
2 \342\200\224
dx
of functions
combination
linear
the completeness system satis\357\254\201es the (see Ch. 2, Sec. 9), we have proved) Thus,
i.e.,
(6.1) is
every square
< e.)
o',,(x)]
completewith
integrable
function
weight
f
(x).))
of the for
criterion
(6.5)
system
r,
i.e.,
the
is
rm:
258
The
METHOD
EIGENFUNCTION
result is an
following
lim
r
Fd
It-bw
with
Ch.2 to the
THEOREM
5.
weight r to all
the
Any
series
Fourier
continuous
The eigenfunctions
as
well
proved
7.
of f
need
function
coef\357\254\201cients are
to
zero.
Then)
0, f(x)
mean to
(7.3) of
(6.1)
(6.5).) is
which must
be
dx tf\342\200\231(x)
by
with zero.)
orthogonal
identically
functions of the
But then
respect
in the
converges
f (x)
all the
with
only apply equation
system
function
9)
3:)
the function
(x) always
we
of the system
If
whence f (x) 5
this,
(x) and to the
In fact, if f (x)is orthogonal all its
dx =
2 kao c,,,,(x)]2
coe\357\254\201icientsof
Fourier
\\/7 f
function
CI-IAP.
function.
\342\200\224
To prove
r.
weight
APPLICATIONS
integrable
square
[f(x)
where the c,, are the Fourier to the system (6.1).)
In other words,the
rrs
easy consequenceof Theorem
Let f (x) be any
THEOREM 4.
f(x),
AND
system
(6.1),
then
(6.6))
= o.
0.)
in series with respect to the expansion of a function value problem (under quite broad hypotheses),
on the
theorem
of a
boundary
theorem as the related completeness by the prominent mathematician
and V.
A.
Does the Eigenfunction Method Always Solution of the Problem?)
its consequences,
were
first
Steklov.5)
Lead
to
a
of the problem will certainly lead to a solution The method eigenfunction of all, the functions f (x) and de\357\254\201ned by (1.4) have g(x) posed in Sec. 1 if \357\254\201rst which to the eigenfunctions ,,(x), series expansions (1.14),with respect the coefficients A,, and B,, de\357\254\201ned to f (x) and g(x), and if secondly, converge are such as to guarantee the convergenceof the series (1.12) and by (1.16) or not these justify differentiating it twice term by term. However, whether can be the solution conditions are met, every time the problem has a solution, in the form of a series(1.12) indicated in Sec. 1. This the method by found be deduced by an can often implies that the solution is unique, a fact which based on the physical content of the problem. This physical argument the problem has a solution in the) content also allows us to decide whether 5 The theory given ticians J. C. F. Sturm
in this and
chapter is usually associated (Translator)))
J. Liouville.
with the
namesof the
mathema-
ma
sec. 7 In view
of
AND rrs APPLICATIONS
METHOD
arcauruncrron
259)
been said, this explains why the physicist method and manipulating the series using eigenfunction as if term by term differentiation and other operations were justi\357\254\201ed,even when in fact this is not the case, nevertheless arrives at correct results. A more formulation of these remarks goes as follows:) precise
\357\254\201rst place.
or engineer,by
has
what
just
the
Let the function u(x, t) be a continuous solution of in the region a < x < b, t 2 0, satisfying the boundary (1.1) conditions Then) (1.3) and the initial (1.4).
THEOREM. equation conditions
=
u(x, t)
the
(7.1)
T,,(t),,(x), \"$0
where the ,,(x) value problem.\342\200\230
to
subject
associated with eigenfunctions T,,(t) can be found from the
the
are The
T;
+ A,,T,,=
the initial
conditions
0
T,,(0) = C,,, T,',(0)
< I <
0
and
continuous
are
(It
,,(x).
functions
conditions
initial
=
to the
derivatives
of the
region
every
f (x) and system of eigenBu/at and 8214/ at\342\200\231
coe\357\254\202icients of
respect the
that
(7.2)
0, l, 2, . . .),)
Fourier
the with
boundary
equation
2, . . .),
0, l,
(n =
c,,
(l.4)]
is assumed bounded in
=
(n
C,, and c,, are
the quantities
where
g(x) [cf. the
the
functions
form a
< x < b,
to.))
Proof
We
equation
multiply
(1.1) by the \" R
1 _ 1' \342\200\224
function)
_\342\200\224 \302\243 P
'I\342\200\224_.,CXp{J\342\200\230xo-}3dX}
(see Sec.
2).
Then accordingto (2.2)and 8214 ,au
we obtain)
(2.3),
63::
P\342\200\230a372+P3;e+\342\200\2301\342\200\234='7a7)
or
Eu
3 53: By
(4.2),
this can
be
(P
written
5)
+
-_
instead
of (2.1) we
can
simplicity, we assume that
azu (7.3))
72-.
write
L() 6 For
Bzu \342\200\230in\342\200\231
as
L(u) = r and
'
4\342\200\234
=
\342\200\224).r.)
the eigenfunctions
are normalized as in
(6.2).))
260
METHOD
\342\200\230rm: monnruncuou
CHAP.)
the relation
Therefore,
L(,,)=
for the eigenfunctions By the hypothesis of the b and a< every t > 0, series of the form (7.1),where)
of the boundary theorem and
x<
from (7.4)
follows
value
function
the
. .)
1, 2,. Theorem
by
(7.4))
problem. 2 of
t) can
u(x,
Sec. 6, for be expandedin a
0, 1,2,. . .),
(n =
ru(x, t),,(x)dx
=
T,,(t)
= 0,
(n
\342\200\2247.,,r,,
holds
It
APPLICATIONS
rrs
AND
(7.5)
that
=
r,.
\342\200\224 ;\342\200\230;L<.>.
so
that
no
=
-
If
z>L<..> ax,
u(x,
or by (4.3)) x=-b)
=
T, (z)
The
last term
Thus we
\342\200\224
vanishes becauseof
other hand,
where the differentiation of our hypotheses obtain
a <
region
x <
..(x>L(u)
-,1-I\342\200\235
b
1
\342\200\224
X30)
Sec. 4.
(7.6)
dx.
(7.7)
the
behind
2
<1>,,(x)
twice
93-;,,(x)
fr
with respect to
t
dx,
gives
(7.3))
integral sign is legitimate Comparing 83:4/atz.
because
(7.7) and since u(x, t) is continuous in the Furthermore, 0 and since lim u(x, t) = f(x), it follows from
concerning 314/
(7.2).
:r,,(z) =
r\342\200\224>0
dx.
8214
3;
(7.5)
differentiating
b, t
r
in
at
and
\342\200\234\342\200\231\302\260)
(7.5)that lim
2 of
Lemma
-
T,';(z) =
(7.8),we
(<1>,,g\342\200\224\302\247 <1>;.u)])
we obtain
(7.3),
using
7',,(z) = On the
[p
have)
T..(z) =
whence,
-\342\200\224
dx +
,,(x)L(u)
lim
ru(x,
f\"a) r\342\200\224>0
rf(x),,(x)
t),,(x) dx dx =
c,,
(n =
o, 1,2,. . .),
(7.9)))
ms eromnmcnou
7
sec.
the C,
where
are the Fourier coe\357\254\202icientsof
T,,(t) is
function
continuous,
we
show
=
c,, are the
the
completesthe
proof
Fourier
On the
be found
(n =
0, 1,2,. ..).
(n =
0, l, 2, . . .),)
f (x).
Sincethe
relations)
in
coe\357\254\202icients
of
the
function
This
g(x).
theorem.)
of the
Thus, if the problemin solution can
to the
26!)
that c,, T',\342\200\231.(0)
where
AND rrs APPLICATIONS
the function
is equivalent
this
= C,,
T,,(0) Similarly,
METHOD
we are
which
of a
the form
interested has
series(1.12)by
a solution
the
often
all, the of Sec. 1. u(x, t) which as be regarded at
method
this method leads to a function everywhere. Such a u(x,t) cannot of the problem in the exact senseof the word, since u(x, t) certainly must the differential equation! However, becauseof the theorem satisfy to look for an exact solution, since if just proved, in this case it is useless such it would have to coincidewith u(x, t). This compels us to be existed, satis\357\254\201ed with the function u(x, t) that we have already found; we shall call this function solution of the problem. that a generalized It can be shown with our hypotheses, the series (l.12) obtained of Sec. 1 by the method or de\357\254\201nes a function to which it converges either in the ordinary sense always in the mean, and thereforethe eigenfunction a method always gives generalized if it fails to give an exact solution.) solution,
other
hand, quite a derivative
does not a solution
have
8. The
Generalized Solution)
is the practical value of the generalizedsolution described above? it represent anything or engineer, of use to the physicist or is it of mathematical is in fact valuable, interest? The generalizedsolution purely as will emerge from the following theorem:) What
Does
Let)
THEOREM.
u(x,
0 ~
3
T.(:>.
n==0)
be
either
the exact
the conditions um
or
the
generalized
(1.3) and (1.4).
f\"rLr(x>
solution
of equation
(1. 1), subject
to
= o.
(8.1)))
If)
\342\200\224f...(x)]2dx
=
um
f\"rtg(x>
\342\200\224
g...(x>1=dx
262
METHOD
\342\200\230rm: EIGENFUNCTION
and g,,,(x) converge in
if f,,,(x)
i.e.,
g(x), respectively,
m
as
is either
exact
the
the
to
boundary
t) =
We recall
Proof.
in
T...(z>.
= fm(x).
convergesto u(x, t)
=
mean as
in the
g..(x>.)
m \342\200\224> oo.)
Tn(0) =
0: )\342\200\230nTn
=
Cm
cn
= 0, (n +
Tfnn
( 1.1), subject
that)
=
T7: +
f(x) and
if)
or the generalized solution of the equation conditions (1.3) and the initial conditions) u..
then u,,,(x, t)
Cl-IAP.
(with weight r) to
mean
the
\342\200\224> co,\" and
u..(x.
APPLICATIONS
rrs
AND
)\\nTmn = 0:
=
Tmn(o)
T
Cmns
(n
l,2,...),
= = 0,
(8.2))
cmn
l,2,...),)
where C,,,c,,,C,,,,,, c,,,,, are the Fourier coeflicients of g(x), f,,,(x), g,,,(x),respectively. Since)
the
functions
f (x),
Ao<)l1<7.,<---<7.,,<---
and
+oo
lim A,,=
n-no)
(see
Sec.
1,,
s 0 for
a
3), only n
N and
<
T,, =
1,, >
0 for
n
+
%(C,, x/0:1)
> N.
1,, can be Then by (8.2),
we have3)
e-V\342\200\231-T.\302\273 (n
< N),
x/fl\") cn
T,,=C,,cos\\/T,,t+
sinx/7T,,.t
V 7\302\273. 7
In particular,
3 It
may
happen
this is the that
KN
case = 0,
if f,,.(x)
in
(n>N),))
\342\200\224> f(x) and g,,.(x)\342\200\224+ g(x)
which
TN =
Let
negative.
e\342\200\230/3-Tn\342\200\230)
\342\200\224
+
the
of
\357\254\201nite number
case we
have
Cu + cut.)
uniformly.)
(8.3))
9)
sec. 8 and
METHOD
EIGENFUNCTION
THE
263)
APPLICATIONS
ns
AND
similarly)
_\342\200\224 \302\243
Tm,
+
cmn
+
2(C,,,,,
\342\200\230/T An\342\200\230)
V-jj-\342\200\230)8
-
' 8\342\200\234/:7;
% (Cm
(n <
(8.4))
N),
x/c:\"_\\342\200\230n)")
(n>N). T,,,,,=C,,,,,cosx/X,-,t+\342\200\230c/'\"%sin\\/7\302\273:,t I!)
consider
Now
r[f(x)
If
-
dx =
\342\200\224
r[g(x)
3 of
Theorem
-
=
dx fm(x)]\342\200\231
c,,...>=.)
,$:\342\200\230o(c,.
If (see
-
g...1=
in.
Sec. 6). In view
of
c,,.,.>=)
these formulas
(8.1),
imply
that)
Q)
=
\342\200\224
2 (0,, m\342\200\224>ao lim
c,,,,,)2
o,)
\"=0)
(8.5)
00 \"\"
2 (cu
=
0:
cm\302\273):
m\342\200\224>ao \"=0
whence
lim C,,,,, =
C,,,
\"\"\342\200\231\302\260\302\260
c,,,,, =
lim
Then,
(8.4)
(8.3),
by
(8.6)
c,,.
and (8.6) \"
Tmn] = 09
\"1'i_I\342\200\231n\302\260\302\260[Tn
where =
:r,,,,,]2 [11,\342\200\224
\342\200\224
cos
c,,,,,)
[(c,,
sin
\302\253/11: +
\302\253/17,
t]2
\"\"\"/'f?\"'\" \"
(8.8)
g
for
n >
N.
r[u(x.
J\342\200\235 \"
All
0
2 that
_ [(53,
4.
cm\:")
remains
- u.,.1= dx
VAT.)
is to
consider the
= E (21.n=0)
relation)
T.....>2 Q)
=
i
n=0
(Tn
\"'
Tmn)2 +
2
n==N+l))
(Tu
_
Tmn)2)
1113 momruucrron
264
Sec. 6).
3 of
Theorem
(see
AND rrs APPLICATIONS
METHOD
(8.5),
By
-
If r[u(x.2) that the
which
means
as m
\342\200\224> oo.
function
9)
and (8.8)
u..(x. 012 dx\342\200\224>.o.)
u,,,(x,
in the
t) converges
mean to u(x, t)
theorem.)
proves the
This
(8.7),
cruur.
as follows: we have just proved can be summarized If f,,,(x) is close or even closeness close to the sense is and g(x) of uniform g,,,(x) (in f (x) closeness in the mean), then the function u,,,(x, t) is close to u(x, t) in the mean. and engineering, the that in actual \302\273problems of physics We now observe not exact, but rather represent approxifunctions f (x) and g(x) are in general to mations certain exact functions. Nevertheless, by the theorem just if of to the conditions even the solution (1.1), equation subject proved, is and not solution but a an exact generalized solution, it will only (1.3) (1.4), of uniform closeness or closeness in the mean) di\357\254\201'er the sense only slightly (in value of from the true solution of the problem. Thisconstitutes the practical What
to
generalized solutions. we
Finally,
consequence of the
one further
note
theorem
proved
above:)
and g,,,(x) are chosenin such a way that the exact solutions of the problems corresponding be chosen !9), then the exact or general(such f,,,(x) and g,,,(x) can always ized solution of equation (1.1), subject to the conditions (1.3) and (1.4), is the limit of the exact solutions u,,,(x, t) as f,,,(x) \342\200\224> g(x) f (x) and g,,,(x)-\342\200\224>
If
the
either
u,,,(x,
uniformly
It follows
9.
The
f,,,(x)
functions
functions
t) are
or
in
mean.)
the
immediately that
is
solution
generalized
unique.)
Problem)
Inhomogeneous
Instead of equation
the
(1.1),
consider
the more Bzu
Bu
0714
general equation)
P5?+REc+Qu=-aT5+F(x,t)
subject to vibration
tions,
the
same
problems,
while
multiplied
and initial
boundary equation
equation (1.1) by the function)
corresponds to the \342\200\230R
__1. 9 For
example, we can choose the series of f (x) and g(x).))
Fourier
the
(9.-1) \342\200\230corresponds to
r\342\200\224Pexp{Lo
of the
conditions
functions
(1.3) and (1.4). In forced vibra-
case of
of free
case
vibrations.
When
-3
}gdx}\342\200\224P)
f.,.(x) and
g,..(x)to be the
mth partial
sums
sec. 9
EIGBNFUNCTION
THE
(see Sec.
2), (9.1) takes the
Suppose now has
that
=
of the
+
value problem correspondingto equation of a series expansionin terms for the equation) value problem
that
t) has
F(x,
boundary
L()
Then, for t >
Sec. 2).
(see
(9.2))
t).
rF(x,
rg\303\251\342\200\230
the boundary
a solution and
eigenfunctions
265
form
L(u)
(9.1)
AND rrs APPLICATIONS
METHOD
0, we
u(x, :)
=
the
\342\200\224m1>)
write
u(x,
=3 n=0
series)
t) as the
(9.3)
T..(t>,..
where
r,,(z) = which
given
the
proof
=
mo [see (7.6)]
=
ff
\342\200\224 .1. 3.3.22
3:2),,(x)
1,, .,
where we have used(9.2). di\357\254\201erentiating
if we F(x,
the argument
(9.4)
dx
\342\200\224
\" Bu/at
b
rF(x,
t),,(x)
dx,
and 8114/8:3 are
to t
(9.5))
bounded,we
that)
azu
\342\200\234P-a7(D,,(x)dX.)
set)
t) = =
1\302\253*,,(:)
(9.5)
Repeating
obtain)
with respect
twice ,,
then
6.
(9.4))
,.(x>L
that
Assuming
T,,= Finally,
\342\200\224
o, 1,2, . . .),
or)
T,,(t)
\357\254\201nd after
(n =
because of Theorem2 of Sec. of the theorem of Sec.7, we
is possible in
t),,(x) dx
ru(x, L\342\200\235
\":0
F,,(t),,(x)dx,
rF(x,
t),,(x)
dx
(9.6) (n = o,
1, 2,
.),
gives
1:, =
\342\200\224)
or
T$',+).,,T,,+F,,=0
(n=0,l,2,...).
(9.7)))
266 me moanruncnou
as in the
proceeding
Then,
seq.], we
mrmoo
APPLICATIONS
rrs
proof of
the
of
theorem
9)
CI-IAP.
Sec. 7
[see (7.9) et
\357\254\201nd)
=
T,,(0)
C,, and
where
AND
series (9.3),where conditions
= T ,\342\200\231,(0) c,,
0, 1, 2, . . .),
(n =
(9.8))
Fourier coe\357\254\202icients of the functions f (x) and g(x), solution of the problem exists, it is given by the to the from the equation (9.7), subject determined
c,, are the if a
Thus,
respectively.
C,,,
T,, is
(9.8).
It is remarkable that just as in the case of equation we can arrive at (1.1), the series (9.3)by carrying out formal operations with without series, regard to whether or not these operations are legitimate. In fact, if in equation the series (9.3) and the series (9.6)for F(x, (9.2) we substitute t), differentiate term of and then the coef\357\254\201cients term, (9.3) ,,(x) to zero, we by equate obtain (9.7). Moreover,if we set t = 0 in (9.3) and require that) 0) =
u(x,
we
the
obtain
in
= r.) r...(x)
\357\254\201rst of the
term and again set
t
=
equations (9.8). that) 0, we \357\254\201nd =
If we differentiate
(9.3)
term
by
= gcx).) T:.(o>..(x>
3\342\200\230%\342\200\2313)
:0
second of the equations(9.8). case of equation (1.1)in Sec. 7, we can introduce the concept of a generalized solution for equation (9.1). Then it turns out that for it continuous a function any u(x, t) to which F(x, t), the series (9.3) de\357\254\201nes in or so that either in the the the sense mean, converges problem ordinary this solution solution. Moreover, always has either an exact or a generalized is unique, if U and V are two solutions of the boundary since value problem for equation (9.1), the function u = U \342\200\224 Vis also a solution of the boundary value for equation (1.1) with But as obinitial conditions. zero problem in Secs. 7 and 8, the boundary served for equation (1.1) has a value problem which is identically solution Sincethe function unique (exact or generalized). zero is a solution of the boundary value problem for equation(1.1)with zero initial conditions, the As concerns it follows that u =.-_- 0, i.e., that U E V. we can repeat the considerations of) practical value of generalizedsolutions, is the
which
Just
as
in the
Sec.
8.
I0.
Supplementary We
our
considered
considerations
Remarks) equation
are also
(1.1) only for
applicable to
the
sake
equation
of being (1.2),
explicit. with
the
All
same))
nu: EIGENFUNCTION
sec. 10
In
and
conditions
boundary
this
the
case,
u(x, 0) = f(x). method of Sec.1 gives)
(10.1)
0 =
nor.
267)
condition)
initial
the
with
AND rrs APPLICATIONS
METHOD
(10.2)) \302\247o1:.(:>.(x).
T,, are
where
the quantities
where
O, T}, + 7.,,T,, = 0, 1, 2, . . .) are the C,, (n theorem of Sec. 7 needs only
The
the
from
found
differential
order \357\254\201rst
equation
(n = 0, 1,2,. . .),
T,,(0) = C,,
=
of f
coefficients
Fourier
(10.3))
(x).
in its statement, and require the continuity and the new that the reader suggest prove a corresponding
i.e., one has to consider(10.3)instead
change
of (7.2)
of only Bu/at. We version of the theorem as an exercise. can also be introduced in the case The conceptof a generalized solution of equation (1.2). But this case is essentially different from the case of out to be the exact equation (1.1), in that the generalized solution turns of the number solution as well. This results from the fact that all but a \357\254\201nite such are for and 1,, 1,, (10.3) gives) positive,
boundedness
T,,
out
it turns
Consequently,
term
differentiated
= the
that
C,,e\342\200\2307\342\200\230n'.)
(10.2) is convergent and can be of times. i.e., (10.2)gives the exact
series
any number
by term
solution.
The considerations equation
of Sec.9 are also
02:; with the
is given
Qll\342\200\2245?
same boundary and initial the
by
conditions
T,, is
now defined by
the
vergence
The solution
and (10.1).
series
= 3 r.<:>.(x). n=-0)
in
the
(10.2))
(n=0,l,2,...).)
T,,(0)=C,,
results of Secs.7-10,we
shall
not
worry
solved in problems obtained as a result of solving
now know that the series a solution (exact always de\357\254\201nes some of the problems in speaking, that but it can be shown studied,
as well.))
(1.3)
equations
T,\342\200\231.+7.,,T+F,,=0,
In view of the of
+F(x,t),
series)
u
inhomogeneous
\357\254\202u
Bu
P-3-x'3+R-5;+
to the
applicable
to be
Part
about the conII, since we
such
a problem
or generalized). It is true that strictly Part [I do not fall into the category already the previous considerations apply to them
me
268
a Vibrating
of
Equation\342\200\230
Consider a
supposethat this
straight
stretched
equilibrium
line
to be the
= 0 and
the points x
x
9)
String) fastened
string,
homogeneous
the
CHAP.
APPLICATIONS)
II.
Part
ll.
AND rrs APPLICATIONS
METHOD
BIGBNFUNCTION
of the
position
is a
string
x-axis, and let the ends of the is the
= I, where I
the
of
length
at both ends, and line. Take be located at string If the string string. straight
velocities are imparted it begins to vibrate. We shall consider vibrations; then, the length of the only the case of small can be regarded as unchanged. Moreover, we shall regard the string vibrations that as taking place in one plane, in such a way each point of the in to the x-axis. moves the direction string perpendicular with Let u(x, t) be the displacementat time t of the point of the string of t, the graph of the function value x. Then, for every \357\254\201xed abscissa at the time t. The u(x, t) obviously represents the form of the string element AB of the string (see Fig. 47) is acted upon by the tension forces T1) from its
is displaced
points of the
to the
and being,
the string
and
is
then
T2, which are directed along the tangent we assume that no other forces act on
position,
the tension
T is the
assumethat
that we can remark),
position
equilibrium
string),
(or if certain
can
we
the
sameat that
the
to the the
in the
Therefore, T1 and T, have the same magnitude different directions, and because of the curvature direction is not quite the negative of the other. the element AB in the direction of the u-axis is) T [sin ((9
+
Acp)
z T[tan
In the To
string.
not
does
string
tension
For the
string.
string.
of the
points
of the
length
also assume
all
released,
change does
string
as T,
although
the extent
(see above not
change. have
they
element AB, one the force acting on
of the Hence,
\342\200\224 sin
<9]
(<9
+
Acp)
\342\200\224 tan cp]
= T
+ Ax, t)
3u(x
=
81u(x + ex,
0Ax,)
_ 8u(x, t) 8x)
0x
T[) 0
Ax
time
equilibrium
(0 <
0<
1),))
me
11
sec.
AND rrs APPLICATIONS
METHOD
EIGBNFUNCFION
z denotes approximateequality. Regarding and using Newton's second law of motion,
where
32
Ax as being
the element
write
can
we
small,
very
269)
8114
=
pAx5t\342\200\224;f
where
p
the
is
a2 =
Setting
density of the
linear
T/p and dividing
8214
is the
which
for free
equation
Next, suppose of amount F(x, by a force equation (11.1), we obtain
per
t)
to the
32a
=
unit
length).
8314
W9
of the
vibrations
addition
in
that
unit
obtain)
= 02
'87
mass per
the
(i.e.,
string
Ax, we
by
string.
tension T, the of the
length
82
+
Ax
is acted
string
Then,
string.
upon
instead
of
t)Ax,
F(X,
or
2 8311+
Bzu _
F(x, t),)
\342\200\224 0 -3-\342\200\230?
W
equation for forced
is the
which
We now study the velocity of its time
arbitrary
t?
following at the initial points
and
vibrations,
and
x
t
=
reducesto (11.3)
equation
of the string and is its form at the
form
0, what in
solving
=
u(l,
t)
(11.4)
= goo.
= f(x).
(11.5))
x=0
equations
in Part I.)
Vibrations of a String) of starting
once more go through solution
vanish for
casesof the
I2. Free Instead
equation
of forced
= 0
continuous which functions, f (x) and g(x) are given = 1. Equations (11.2) and (11.3) are special
studied
case
the
conditions
u(x, o) where
string.
conditions)
boundary
u(0, t) initial
time
this problem
Mathematically,
case of free vibrations, subject to the
the
the
problem: Given the
the
(11.2) in the
and
P)
of
vibrations
(different
from the formulas the
from
derivation
u 5
already found
given in Sec.
1.
in We
Part are
I, we shall for a
looking
0) of the form) u(x,
t) =
(12.1)))
which
METHOD
EIGBNFUNCTION
TI-IE
satis\357\254\201es the
conditions.
boundary
CHAP.
ITS APPLICATIONS
AND
=
T\"
Substituting (12.1) in
(11.2)
9)
gives)
a2\"T,
whence
= -1
=
%-
\342\200\224
const,
3%-.
so that =
<1)\"
(12.2)
\342\200\2247.,
= \342\200\224a17\\T. T\342\200\235
If a function of the form the conditions (11.4),then
is not
which
(12.1),
(12.3)) zero,
identically
is to
satisfy
the condition)
obviously
(0) =
=
(l)
0
(12.4))
Thus,we obtain
value problem for equation (12.2), a boundary (12.4). In view of the theorem of Sec. 5 (seealso the are positive.1\302\260 remark made there), all the eigenvalues of our problem )3 instead 7.. of it is to write Then, equations(12.2) Therefore, permissible and take the form) (12.3)
must be met. to the
subject
condition
23(1) =
+
<1)\"
(12.5)
0,
T\342\200\235 + a2A2T = 0.
The solution =
(D
x =
for
where
is
of (12.5)
xx +
C1 cos
0 and x
C2sin Ax
= [we must
that
C;
a\303\251 0, since
that
AI =
and
the corresponding
1m, where
n
is an
10
This
0.
can also By
doing
satis\357\254\201ed.))
be
CD
would
Setting C,
be identically
= 1 gives)
zero, we
\357\254\201nd
eigenfunctions are) d>,,(x) =
1<
const),
= 0. M = 0.)
sin
C2
otherwise
integer.
C2 =
have =
(D0) =
= const,
(C1
C1 \342\200\230D(0)
Assuming
(12.6)
veri\357\254\201ed directly
this, the
(n
=
1, 2,
. . .).)
sin\342\200\231-5;\342\200\231-\342\200\230
reader can
(12.2) by examining the solution of equation himself that the condition (12.4)cannot
convince
for be
sec.
12
EIGENFUNCTION
THE
AND rrs APPLICATIONS
METHOD
271)
We do not consider negative of n, since they values give the same eigenfunctions (up to a constant as the corresponding factor) positive values of n. in the sense indicated in Sec. 4, only one eigenfunction to Thus, corresponds each value of 73. For A = 1,, equation (12.6) gives) T,, =
A,, cos
B,, sin
+
a7l,,t
aA,,t)
(n=1,2,...),
=A,,cosa\342\200\224\357\254\202;\342\200\231-\342\200\224t+B,,sin%\"\342\200\224t
so that \302\260
u,,(x, z) =
+
(A,
3,,
sin
cos\342\200\234\"\342\200\224l'\"
\342\200\231-\342\200\230-73 (12.7)
=1,2,...).
(n
Thus, to
our
solve
ace. 0 =
,,fl
=
Z
It)= I)
and
require
we set
problem,
o
u.(x.
(12.8)) (A,
B,, sin)
+
.
cosa\342\200\224nI\302\243\302\243-7\"
aunt)
-ttnx T\342\200\231
that\342\200\234)
0) =
u(x,
A,,
:1
sin
= f(x),)
5'15
\302\260\302\260
3u(x,0)
at
=
[ n=l)
=
=
E
n=-1)
B \"
T
(_Ana1tnsing_1cln__t+
B,,\302\243nI\342\200\224\" sing;-J5
we have to expand f (x) and the system {sin (nnx/1)}. The formulas I
=
.
sin
\342\200\224_ \302\260s
I a1cnt)si
I
-nznx
\342\200\224)
1
:=o
g(x).
Therefore,
2 A,, = 7 Iof(x)
a_1tLzc
series
in Fourier
g(x) for
the
rcnx
-1- dx
(n
Fourier
with
respect
to
coe\357\254\202icients give)
=
1, 2,.
. .),
(12.9)
=
1, 2,.
. .),
(12.10))
%I:,so\302\273
or
,,
=
I
2
. g(x)
\357\254\201lo
i.e.,
the
B,, are
solution
of our
determined from
11As in
the method
-rcnx
(n
s1n\342\200\2247\342\200\224dx
problem is formulas
of Sec.
1, we
given
(12.9) differentiate
by
and
the
series (12.8), where
(12.10).)
the series term
by
term.))
A,,
and
272
THE
we see
Thus,
METHOD AND ITS APPLICATIONS
BIGBNFUNCTION
of separate
motion that the vibrational vibrations of the form
of the
harmonic
u,,
= H,,
(12.7),
+
sin
CHAP.
sin an\")
string is a superposition the equivalent
or of
9)
form)
#.)
where) H,, =
The amplitude
of the
1/11}+
B3,
Sinai,
=
\302\2435
H,,
of the point
vibration
=)
cosat.,,
H,,
coordinate
with
x is
9) H, sin 71;\342\200\231-c
x = 0,1/n, 21/n, . . ., and as nodes. \357\254\201xed are known motion, (n during is a whose vibrations are formula described Hence, string by (12.7) divided into n segments, the end points of which do not vibrate. Moreover, in the of the have adjacent signs, and displacements string opposite segments, the midpoints of the segments,the so-called vibrate with the largest antinodes, is known as a standing wave. amplitudes. This whole phenomenon 48 shows consecutive positionsof a string are vibrations whose Figure described by formula case, where) (12.7), where n = 1, 2, 3, 4. In the general
is
and
independent I, remain 1)l/n,
The points for
of t.
\342\200\224
which
the
\302\273=3)
n=4) Froune
the
vibrations
(mode)
of the
corresponds
string
to the
are
described
component ul are
48)
by formula with
1:
(12.8), the fundamental
frequency)
7
\342\200\230\302\260\342\200\230=T=7\302\253/E)
and
period))
_2_'-=-
&/ET) \342\200\230l'1\342\200\224(o1\342\200\2242I
me
12
sec.
vibrational modes of the
The other
,,
and
METHOD
AND rrs APPLICATIONS
i.e., the
overtones,
EIGENFUNCTION
\"
_-
string,
52;! _ _
with
273)
frequencies
1\"
/.T'
I
I
p)
periods)
of the sound. If the string is held \357\254\201xed the timbre or \342\200\234color\342\200\235 of the string, for which then clearly the even overtones the is of a are the the fundamental node, midpoint string preserved. However, and the odd overtones are immediately since by holding the extinguished, \357\254\201xed we essentially of length I to a midpoint of the string go from a string of length Ito 1/2 in (12.8) leads to a seriescontaining string I/2, and changing even components. Then the overtone with period 1-, = 21!/(02 = 1-1/2 only of the the role fundamental.) plays
characterize
at
its midpoint,
Forced
I3.
Next
of a
Vibrations
String)
the case of a periodicperturbing
we consider
=
M
A
force,
i.e.,
we write)
cot.
sin
P
Then
=
A
cot =
sin
51:49
2
sin
F,,(t)
(13.1)
n=l
where
F,,(t) =
A sin %-
= 133-
If we
cot
dx sin\342\200\231-ti?
L,
[1
(n =
\342\200\224 sincot (\342\200\2241)'-1
write)
u(x,
t) =
T,,(t)
sin
(13.2) and (13.1) in we obtain) differentiation,
substitute
Q)
,,
\"Z! (T,
+
a741:2n3
(11.3),
2A
and
(13.2))
15$\342\200\230.
\"$1
term
1, 2,...).
carry
\342\200\224 \342\200\224 [1 (-1) -17- T, n\342\200\224n
out the required
,, . ]s1ncot)
.
term
rmx _ \342\200\224 0,)) s1n\342\200\224I
by
274
METHOD AND rrs APPLICATIONS
\342\200\230ml; EIGBNFUNCFION
can.
9)
whence
-
+ T,\342\200\235. T,, \302\24321]?-'3
\342\200\224 23 [1 (-l)\"]
= 0.
sin at
(13.3)
Writing
o),,=
for simplicity (these will characteristic vibrations
this
T,,
if (0,, ye
on.
A, cos
w,,t + B,, sin
satisfy the
To
3u(x, 0)
(13.4))
=
=
oo,,B,, +
2
=
z
7 I0
f(x).)
= g(x).) and g(x)
f (x)
sin
f(x)
%
that)
\342\200\231i\342\200\235,-\342\200\231-\342\200\230-
=
= 4,,
:r.'.(o_)
=
(13.5))
cot,
we require
(11.5),
T,'.(0) sin 1:?
S
(-1)\"])sin
_
\342\200\234no\342\200\235, 0),) ll)
Fourier coe\357\254\202icientsof T,,(0)
2/111 -
+
w,,t
= 3 mo) sin
at)
of the
at.
(\342\200\224 1)\") sin
conditions (11.4)and
u(x. 0)
A calculation
-
$11
or
free
the
equation (13.3) as)
we obtain
equation, =
frequencies of rewrite
can
we
string).
=
1,2,...))
as the
recognized
of the
+ mar.
T:
Solving
be
=
(n
$
gives)
dx,)
\"-ll\342\200\2313\342\200\230\342\200\224
2Aco[l
(-1)\"])
(13.6)
_
0),) \357\254\201n\342\200\235 .
g(x) sin 5'13.
01' 13,,
where
we have
substituting
the
=
g(x)
used (13.5).
-
dx
(13.7))
\"it\"
2:n\342\200\230:fn'(o;,'
and
(13.6)
Substituting
expression for T,, in
resulting
u(x, t) =
sin
Z (.4,cosco,,t
+
(13.2),
(o,,t) sin
3,, sin
n=l
4
\302\260\302\260
.
\"\" S\342\200\234 \342\200\231'
?
.2.
sin
[(2k
(21: +
7\342\200\230
k=o
(276 +
then
\342\200\2303;\342\200\224\342\200\231-\342\200
+ l)11:x/I]
1)
\342\200\224 \302\253>33)
(13.8)))
sin wz,,,,_,t sin
_
(13.7) in (13.5), and we obtain
[(2k +
1)1I:x/I] ')
\342\200\230 002) 1)\302\260\302\2602k+1(\302\242\302\2603k+1
sec. 13 we have
where
METHOD AND rrs
\342\200\230nu: EIGENFUNCFION
275)
APPLICATIONS
written
B\"
2
=
E
I
.
-ttnx S111 dx.) g(x) -\342\200\224-1\342\200\230[0
for co,,, the reader will easily recognize that the side of (13.8)is the function giving the free of the string, vibrations to the conditions and subject (11.4) (11.5). [See the second and third sums (12.8),(12.9),and (12.10).] Therefore, give the \342\200\234correction\342\200\235 caused force. The second by the presence of the perturbing \342\200\234 term represents what is sometimes forced referred to as the pure\342\200\235 vibrations, the expression the right-hand
Recalling
in
\357\254\201rst sum
since they occur with of the perturbing the frequency force. 0). We now examine the situation when Equation (13.5) holds if 0),, a\303\251 = the as when of the force is the same one co, i.e., co,, frequency perturbing of the characteristic of the Then, string. equation (13.4) gives) frequencies
T,, = This
shows
term
T,,(t)
that sin
out +
cos
A,,
n is
when
of the
(nnx/I)
B,, sin
cot
-
7% [1
\342\200\224 cos (\342\200\2241)\"]
amplitude of sum (13.2)is)
odd, the
.
the
cot.)
in the
vibration
nth
rtnx
S11\342\200\230!
-7-)
1371(1)) H=j(A,,\342\200\2242\302\243-t-)2+B,f)
unbounded as
becomes
which
t
In this
increases.
case, we say
that
resonance
occurs.)
I4.
of the
Equation
LongitudinalVibrations of a Rod) rod of
a homogeneous
Consider
compressed along
length
axis, and
I.
If the
rod is
stretched or
it executes
longithe x-axis along the axis of the rod, and we assume that in the equilibrium position, the ends of the rod are located at the of the rod points x = 0 and x = -I. Let x be the abscissa of a crosssection at the section of this cross at rest, and let u(x, t) be the displacement at the AB of the rod, whoseends are located time t. Consideran element points x and x + Ax when the rod is at rest, and let the ends of this element x + u(x, t) and A\342\200\231 and B\342\200\231 at the time t, with abscissas have the new positions x + Ax + u(x + Ax, t), respectively(seeFig.49). Thus, at the time t, the element AB has length Ax + u(x + Ax, t) \342\200\224 u(x, t), i.e., its absolute increase tudinal
in
length
its
vibrations.
longitudinal We choose
then
released,
is)
u(x +
Ax,
t)
\342\200\224
u(x,
8u(x t) =
+ 0Ax,)
ax
t)Ax
(0 < 0 <
1),))
THE mcnuruncnon
276
its relative
while
increase in
9 cum\302\273.
AND ITS APPLICATIONS)
METHOD
is)
length
+ 0Ax,
8u(x
t)_)
Bx
as
In the limit
\342\200\224> 0, the
Ax
increase
relative
in
of the
length
cross section of
at x is)
rod
the
3u(x, t). Bx)
to
According
given by the
cross sectionat
force T acting on the
the
Hooke\342\200\231s law,
x
is
formula)
2!!
T=Es)
ex\342\200\231
where
E is
which
the
the modulus of elasticity is made, and s is the
rod
material from
of the
(Y oung\342\200\231s modulus)
cross-sectionalarea of
rod.)
the
1?
4
x-Ij-Ax
fr
iii
4|
2:+1/'(x,f)
x+Ax+z/t(x+Ax,r))
49
Home
at time t, to the element AB which has the new position A\342\200\231B\342\200\231 is acted upon by forces T1 and T2, applied at the points A\342\200\231 and B\342\200\231 and directed along the x-axis.) (For the time being, we consider no other forces.) The resultant of these forces is)
Returning
we note
that
AB
6u(x,
8x
0x
T,-T,=E,(@\302\243x_+_A\302\247\302\273_t_>_)
= ES 37*u(x and is also directed very small, we can
along
the
x-axis.
E/p
p
is the
= a3
and
density dividing
Ax,) the
Regarding
81\302\242: -\"'= ES '5?\342\200\231
being)
(14.1)
Ax,)
of the material from which by Ax and s, we obtain 3:2
is the equation for free for as the equation
sameform
AB as
azu
Ex-5
the
\342\200\224 .32.\342\200\234 _ a2 _a.2_.u
which
element
write)
pS Ax
where
t) 5|.-x:Ax,
t))
vibrations
free vibrations
the of
rod. a
made.
Setting
(14.2)
Bx?\342\200\231)
of
rod is
string.))
This equation has the
SEC. 14
rod is acted upon
If the
volume,
is the
This
We now rod
the
boundary
'5?
The
rod
t)s Ax,)
F(x,
Case 2) One end of the
on the freeend of the 3) Both endsare free.)
(the force
We
(14.4)
=
u(l, t)
= 0; and the other
= -0,
t) =) 0
8u(I,
(14.4))
ex)
hence Bu/Bx
is zero and
rod
end is free:)
=
shall devote our attention to Case 2, where the boundary are met. The initial conditions have the familiar form) u(x.
i.e.,the
initial
are
0) =
= sec).
f(x).
and initial velocities
displacements
0);
conditions
(14.5))
of the crosssectionsof the
speci\357\254\201ed.)
Free
Vibrations
in the
As
t)_)
P)
is fastened,
rod
u(0, t)
Case
+ F(x,
at both ends:
is fastened
u(0, t)
solutions
of a
case of the form)
Rod)
(see Sec.
string
vibrating
and arrive at the
=
for particular
(x)T(t),
equations d)\"
+
12(1) =
T\342\200\235 + a3A3T
the
12), we look
of the
u(x, t)
with
t) per unit
equation for forced oscillationsof the rod [cf. (1 1.3)]. the problem of \357\254\201nding the displacement of the cross sections pose at any time t, with speci\357\254\201ed and boundary initial conditions. The conditions can take various forms:)
Case 1)
I5.
F(x,
that)
3314 _ Bzu \342\200\224 G2
rod
+
-a7:5Ax
-372'
of
APPLICATIONS)
have)
= Es 3214
811:
psAx -57
so
we
ITS
force of amount
an additional
by
of (14.1),
instead
then
METHOD AND
EIGENFUNCTION
THE
boundary
0,
= 0,
(15.1) (15.2)
conditions (0) =
'(I)=
0.
(15.3)))
THE EIGENFUNCTION
has the
(15.1)
Equation
= (x)
C, cos
Assuming \357\254\201nd that
=0
C1 =
=
0,
C2
=
const).
must have
= Iwe
x
and
const,
C21 \342\200\231(I)
=
cos M
0.)
=
(n
o,1,2,...),
=Q\"\342\200\2243;7l)\342\200\224\342\200\231\342\200\224\342\200\230
sin 7t,,x =
=
,,(x)
values of
negative
9)
be identically zero, we C, 96 0, since otherwise (x) would n is an integer. where We write) (2n + 1)1c/2,
that M =
A,,
(The
(C1 =
C2 sin Ax
+
Ax
=
(0)
CHAP.
solution
to (15.3), for x
According
ITS APPLICATIONS
METHOD AND
sin
Q-n-'\342\200\224\"\342\200\224l)1-x= 21)
:1give
no
For
new eigenfunctions.)
(15.4)
2, . . .).)
0,1,
(n
A
=
1,, equation
(15.2) leads to) = A,,
T,,(t) and
cos
+
aA,,t
B,, sin
(n =
aA,,t
0, 1,2, . . .),
therefore
u,,(x, t)
To
= our
solve
a).,,t)sin A,,x form the series) B,, sin
a7\\,,t +
cos
(A,,
we
problem,
(n
= 0,
l, 2, . . .).
Q)
= 14%\342\200\230)
and
Z
11-:0)
=
u,,(x, t)
a7t,,t +
B,, sin
sin
aA,,t)
7t,,x,
(15.5))
that)
require
u(x, 0)
M
(A,, cos \"go
= Z [ \":0
at
=
=
(-A,,aA,,
B,,ax,, sin
2
go
sin
A,,
)\\,,x
a)x,,t +
sin
= f
(x),)
B,,a)x,,cos aA,,t)
sin
Ax]
(=0)
= g(x).)
x,,x
n=-0)
A calculation
of the
system {sin 7\\,,x}
Fourier coefficientsof
f (x)
and g(x)
with
gives)
_
L,
f (x) sin
= J:
dx
(n=0,1,2,...),)
Ll
B,,a7t,,
)t,,x
sin?
A,,x
g(x) sin [1
dx)
)\\,,x
sin? ).,,x dx)
dx)
(n =
0,1, 2,.
. .).))
respect
to the
SEC. 15
AND ITS APPLICATIONS)
EIGENI-\342\200\230UNCTION METHOD
THE
But) . sin?
7.,,x dx
=
L\342\200\231
and
1
I
dx =
\342\200\224 cos 2A,,x) 51; (1
5.)
therefore)
B,, =
1,2,...),
(n=0,
.4,,=%fo'f(x)sinx,,xdx
.
2
257L:
g(x)s1n
(2n +
7.,,x dx
(15.6) (n=0l2...) 9
0
=-\342\200\2244j\342\200\224J\342\200\230lg(x)sin7\\\"xdx
9
9
(1 5.4)].
[see
Thus, the solution of our problem is given by and B,, are determined from (15.6),i.e.,the vibrational of separate harmonic vibrations) superposition (A,, cos
=
u,,
aA,,t + B,,sin a7.,,t)
formula motion
sin
(15.5), where A,,) of the rod is a
(15.7)
Aux,
or u,,
=
H,, sin
(aA,,t +
a,,)sin Aux,
.
A
where
H,, =
\342\200\224\342\200\224
+
x/A3
B},
sin a:,, =
cos
a.,,
=
B
F:\302\273
The amplitude
which
depends
As for
t.
time
of the
vibrational
motion
H,,|sinA,,x|
=
described
H,, sin
on the position x of the frequency, it is given
is
|.) the
only
by (15.7)
cross
section,
and not on the
by)
_(2n+l)a1r_(2n+l)1t
E)
a),,=a7\\,,\342\200\224-p\342\200\224\342\200\224\342\200\224\342\200\224\302\247I\342\200\224\342\200\224 P9)
and
hence
the period is
T\" =
2;: = a),,
In n =
case of a 0; it has amplitude) the
vibrating
41 (2n + rod,
l)a the
A0 sin
=
41
2n + l J1
fundamental
TEX) 21:))
E)
mode is
obtained for
280
\342\200\230rm: BIGENFUNCTION
METHOD
AND
APPLICATIONS
rrs
9) cum\302\273.
frequency)
and
period)
\342\200\230l'o=4lJ%\302\260)
Therefore,
to a node at
(x =
as
mode corresponds the fundamental and an at the free end (x = I), antinode 0),
II]]]]I|l|lll||Jl||||||lIl||ll
I
II
I I
IJ
x=O
the
end \357\254\201xed
in Fig. 50.)
shown
ll x=l)
l\357\254\201oUne50)
Next,
x =
0 and
of a
Vibrations
Forced
I6.
the case where the rod is suspended force is the force of gravity, i.e.,)
we consider the
Rod)
perturbing
F (x.
where F(x, t) is the
is the
force per
due to
acceleration
unit
gravity.
end
t) = 93.) this
the density
p is
volume,
In
the
from
the equation
case,
of
the
rod,
and
g
for the vibrations
takes the form azu
33a
=
0257 -372\342\200\231
[see (l4.3)], again conditions
to the
subject
(14.5).
+ g
boundary conditions
(14.4)and
t) =
u(x,
:0 \"3
T,,(t)
t) = = F(x, \342\200\224\342\200\224
sin
initial
7t,,x,
\302\260\302\260
.
F,, sin ).,,x,)
g
\"20
P)
where) 1
gsinxuxdx
F,,
the
We set)
-I\302\260\342\200\224\342\200\224\342\200\224\342\200\224-\342\200\224\342\200\224=3g-
J;
sin? 7t,,x
dx))
A\
(n=0,l,2,...).)
(16.2)
16
sec.
the series
Substituting
the
(16.2) in
and
(16.1)
AND rrs APPLICATIONS
terms to
all the
transposing
28|)
obtain)
we
left,
METHOD
EIGENFUNCTION
THE
\342\200\224
+
[V18) It) o) (T;
a2)\\,2,T,,
sin
= 0,
lax
11)
whence)
(n=0,l,2,...).)
T;+a27\\,2,T,,\342\200\224l2\342\200\224f-=0
The solution T,,
of this
= A,,
has
the form)
B,, sin
aA,,t +
equation
cos
+
aA,,t
2g
=
(n
0,1,
2,. .
.).)
la2A,3,)
To
satisfy
(14.5), we require that)
the conditions
u(x, 0) =
3
x..x =
sin
12(0)
f(x).)
n=-0)
0) =
0u(x, at)
of the
A calculation
system {sin Aux}
Fourier
coef\357\254\201cients of
= A.
+
=
T,\342\200\231,(0) B,,aA,,
so
to the
respect
sin x..x ax.)
=
% ,\342\200\224f,g7,_
=
fo'f
g(x)
I
B,,
we
with
7\\,,x dx
sin
that)
2 . A,, = 7 f(x) sin [0
Therefore,
and g(x)
f(x)
gives)
mo)
[cf. (15.6)],
= g(x).)
sin A,,x T,\342\200\231,(O) S\342\200\230 n=-0)
g)
.
2
=
2
-
)\\,,x dx
g(x) s1nA,,xdx
ml;
(n
=
0,1,2,...).)
have)
0
u(x,
1) =
Z (A,, cos a).,,t +
B,,
sin
aA,,t) sin
Aux
n=0)
0)
_ ;g_ la? The giving
reader
will recognize at
the solution
2
cosa7\\,,t
once that
sin
\302\260\302\260
A,,x
1;\342\200\231;
the
of the problem of the
\357\254\201rst sum
free
sin
E on the
vibrations
7\\,,x. A},)
Ia3n=0
right is the function
of a
rod with
the
same))
ml; nromruncnon
282 conditions
the second and
Hence,
(14.5).
due to the
AND 113 APPLICATIONS
METHOD
terms
third
CHAP.
9)
the corrections
give
of gravity.)
force
of a Rectangular Membrane)
I7. Vibrations
we mean an elastic\357\254\201lm by a closed plane curve. supported is at rest, all its points lie in one plane, which we take to be the xy-plane. If the membrane is displaced from its equilibrium it begins to vibrate. We consider released, position and then only small vibrations of the membrane, assuming that the area of the membrane does not to and that each of its points vibrates in a direction change perpendicular the xy-plane. Let u(x,y, t) denote the displacement at time t of the point from its equilibrium position. Then, by a derivation (x, y) of the membrane similar to that made in the case of the string, one \357\254\201nds that the equation for the free vibrations of the membrane has the form) By a
membrane,
membrane
the
When
87-1:
3314
=
8324
+
C2
-875
while the equation for
forced vibrations
8114=
and
F(x,
solution
the
of
the
in
area
time
any
of the u(x.y.
initial
surface density, membrane. 9 is its
membrane,
on the
membrane can be posedas follows: the (17.1)or (17.2),i.e.,\357\254\201nd displacement
boundary (\357\254\201xed)
the
y, t)
To)
acting
Find
of the
condition)
0
(17.3)
membrane, and the 0) =f(x.
initial
conditions
(17-4))
y)
of the membrane) and
displacement Bu
to the
t, subject 14 =
(specifying
is
vibrating
of equation
points of the membrane at met on the
membrane F(x,
+
+
is the forceper unit
The problem the
3311
0311
the tension
T is
T/p,
y, t)
of the
C2
W where c2 =
W):
,
, 0
=
(17.5))
g(x.y)
-\342\200\224\342\200\230\342\200\224\342\200\231g-31-\342\200\231
(specifying the We
of a
now rectangle
initial
study
of the
velocities
the case
R:
considered
0
s x
points of the
of the free vibrations
< a, 0 < y
<
b.
of
membrane). a membrane
The problem
differs
in the from
shape the
on three rather that the function u depends problem than two variables. method Nevertheless, we again apply the eigenfunction and begin by looking for particular solutions of the form) in
Part
I in
14
=
4\342\200\231(x. J\342\200\231)T(t).
(17-6)))
SEC. 17 which
are
boundary
result
EIGENFIJNCTION
THE
in
AND ITS APPLICATIONS)
METHOD
not identically zero and which of the rectangle R. Differentiating we obtain) (17.1),
=
c3
satisfy
+
the
condition
(17.6)
and
(17.3) on the the
substituting
T,
so that\342\200\235 +
gig
83;?
T\342\200\235
=
= -7.3
E5,
\342\200\224-3-L
= const.
Hence
no
+
+
-_-
T\342\200\235 + c27.2T
where, as is easily
=
(I)
on the boundary Thus, we
(17.9).
We
the
of
= 0,
(17.8))
(I) satis\357\254\201es the
the function
seen,
(17.7)
o,
%:\342\200\224\342\200\230f -21-\342\200\230?
condition
0
(17.9))
rectangle.
equation (17.7), with the boundary look for particular solutions of (17.7)of the
consider
now
A and \357\254\201x
=
satisfy the
(17.7)
gives)
condition (17.9)on
the
Substituting
boundary.
form) (17~l0))
<1\342\200\231(x. J\342\200\231)
which
condition
(17.10)
in
?\342\200\235\342\200\230l\342\200\231+q\"l\342\200\231\"+7\342\200\2302\342\200\230Pq\342\200\231=
O1-13)
'
'
+ M!\342\200\231 \342\200\230P\342\200\231, \342\200\230V - _
=
'6
_T\"'
k3 = const,)
whence + kzcp <9\342\200\235
where we have
= 0,
tla\342\200\235 +134:
= 0,
(17.11)
written
I3
= A2
\342\200\224 k3.
(17.12))
12 The fact that this constant should not be chosen to be positive is clear just from the the fact that otherwise the coe\357\254\202icient of Tin equation (17.8) would be negative, so that solution of (17.8) would not be periodic, not have i.e., contrary to experience,we would
vibrations. 13If the constant were
positive,
we could
not satisfy
the boundary
condition.))
ms
284
METHOD BIGENI-\342\200\230UNCTION
of (17.11),we
For the solutions = C1
+
cos kx
\357\254\201nd
sin
C2
kx,
v.Iz(x)
C3, and C4 are constants.
C1, C2,
CHAP. 9
APPLICATIONS
rrs
AND
=
C3 cos
The
C4 sin
+
ly
ly,)
condition
boundary
implies
that)
9(0) =
=
9(0)
=
0.
W?)
\302\242(0)
= 0.
and hence)
C2sinka= C4SiI'llb=0,)
C1-'= C3=0,
so
ka
that
and
we
= mt, where m and
= m11:,lb
are
n
Setting C2
integers.
= C4 =
1
writing)
\357\254\201nd)
.
=
sin k,,,x
.
=
-ttmx
sin -\342\200\224a\342\200\224\302\273)
do not
(We functions
(17.12),
=
sin l,,x
sin)
if
a constant
to within \302\253[2,, -)
A =
A,,,,,
we obtain
the
boundary
condition:)
=
Vk,3,,
1,3,
=
=
Therefore, the
1: 1/(m3/a2)
=
sin
4\302\273.
A,,,,, cos
+
(17.13))
(n3/b\357\254\201,
(17.7) satisfying
the
sin
\342\200\231-?\342\200\224';3\342\200\230#-)
1 =
every c)\\,,,,,t
+
of equation
solutions
solve equation (17.8)for T,,,,,(t)
u,,,,,(x,
+
particular
following
...
values of m and n, since they give the same factor.) According to (17.10)and
consider-negative and
n= 1,2,...)
l,2,...;
(m= =
v.|\302\273,,(y)
).,,,,,; the result is
B,,,,, sin
cA,,,,,t.
functions
y, t)
=
(A,,,,,
cos
cA,,,,,t
+ B,,,,, sin
cA,,,,,t)sin
sin
13-)\342\200\2307%!) 07\302\26014))
(m=l,2,...;n=l,2,...) are
particular
solutions
of (17.1) satisfying
the
boundary
condition
(17.3).))
SEC.17
AND ITS APPLICATIONS)
of equation(17.1)satisfying
the solution
obtain
To
we
METHOD
EIGENFUNCTION
THE
initial
the
conditions,
set)
\342\200\234(xi ya 0
=
=
lumn(xs
Z m.
and
ya t)
2
n=)
'
c).,,,,,t +
cos
(A,,,,,
B,,,,,sin
sin
c)\\,,,,,t)
sin
(17.15)
121. E-2-\342\200\231-x
that)
require
=
u(x.y.o)
A...sin\342\200\231i\342\200\231;\342\200\231,1sin\342\200\231%\342\200\231 =f(x.y).)
2
m. n= 1
y. 0)
0u(x.
at \302\260\302\260
=
[ m, 2n=l
=
Z
Assuming with
.
sin
B,,,,,c)\\,,,,,
that f
respect
Ch. 7, Sec.4,
sin c7.,,,,,t +
(\342\200\224 A,,,,,cA,,,,,
. sin
2-cmx
. sin
-itny
-\342\200\224a\342\200\224 \342\200\224b\342\200\ t=0
sin) 7i\342\200\231?
(x, y) and g(x,y)
to the we
cos c7.,,,,,t) B,,,,,c7i,,,,,
in double
be expanded
can
system {sin (-itmx/a) sin
and
(nny/b)}
using
Fourier series the results of
obtain)
A,,,,,
4
=
35 LJ
f(x, y)
.
. sin
7%
sin
'3?
dx dy
(17.16)
and
B,,,,,c).,,,,,
4
=
35 LJg(x,
. . dx dy, y) sin 71? sin \342\200\234-22
01-14)
B,,,,, =
4
W
Thus, the solution of our and B,,,,, are calculated by
.
.
sin sin LJg(x, y) 73'? problem
using
dx
dy.
7-5%)-\342\200\231
is given by the series (17.15),where formulas (17.16) and (17.17).)
(17.17))
A,,,,,
series in a rectangle of the Sec. 4, we expanded functions in double Fourier < x < I, \342\200\224h < y < h. If instead, we want to expand a function f (x, y) in a recof the form 0 < x < a, 0 < y < b, with respect to the system {sin (nmx/a) sin (itny/b)}, tangle in the variable x and then in this can be done by making the odd extensionof f (x, y), \357\254\201rst take the form (17.16).)) the variable Then, the formulas for the Fourier coe\357\254\201icients y. 14
form
In Ch. 7,
-I
ml: moannmcnon
286
The
=
a),,,,,
c)\\,,,,,
=
of
vibrations
the
9)
rectangular
form)
the corresponding
with
(17.14)],
21:
(m =1, 2, . . .;
+ (n3/bl)
1tc\\/(m27a:2)
[see(17.13)and
CHAP.
APPLICATIONS
characteristic
of the
frequencies
membrane have the
AND us
METHOD
n
=1,
2,. .
.),
periods)
2 (m\342\200\224l,2,...,n\342\200\224l,2,...).
1m\342\200\224mmn\342\200\224cv
is an important difference betweenthe case of the vibrating membrane characteristic of the vibrating string: In the case of a string, every to its own whereas in the case frequency corresponds con\357\254\201guration of nodes, of a membrane, the same characteristic frequency can correspondto several of nodal lines (i.e., lines or curveswhich remain \357\254\201xed con\357\254\201gurations during the course of the motion). We illustrate this situation, using the particularly case of a square simple where of vibration we set a = b = 1. Then, the membrane, frequencies There
and
that
are)
=
co,,,,,
can
write)
H,,,,, sin
(o),,,,,t
of (17.14), we
instead
and
u,,,,, =
c}.,,,,,
=
1:c\\/ m2
+
+
a,,,,,) sin
rtmx sin
n3
-n:ny,)
where
,
H,,,,, = sin
the
For
fundamental
from m
=
which 1, n
= H11
all)
1, we
have
sin rcx sin
-try,)
modes
two
= H12 sin
an = H2, the
(\302\253out+
-
w=\302\242o12=m,1=1tc\\/3
to the
a1,
of which \357\254\201rst
x = }.
sin
=
it is clear that there are no nodal lines at all. Next, we set = 2 or m = 2, n = 1. Then, the same frequency) ~
corresponds
i.e., for m = n an = -:tc\\/2,)
mode,
u\"
cos a.,,,,,=
=
o:,,,,,
Moreover,
has the the
sin
(wt (wt
+ an) +
-nx
sin
an) sin 21tx sin
nodal line y =
frequency
sin
(0 also
and \302\247
the
21ty, -try,)
second
corresponds to the
the nodal line
\342\200\23 \342\200\234compoun
me sroanmmcnou
SEC. 17
mode an
um + = :12, =
an,
an +
=
H12 sin
form a sin
+
21tx
for
sin
equation
21cxsin -try
= H2, we
H1,
=
0
(17.18)
obtain)
2 sin
nx sin
(cos -rcx \342\200\230my
+ cos
-ny)
=
0,)
nodal line
gives the
x+y Similarly,
writing
-try
=
which
+ H2, sin
sin 2'.-ry \342\200\230ttx
sin
the
satisfying
nodal line. In particular, rcx sin 21ty
In fact,
line.
nodal
new
sin 21:y + H3, sin 21cxsin -try), sin \342\200\230ltx
sin o)t(H12
that the points
it follows
whence
a
287
APPLICATIONS
we obtain
simplicity,
uzl
leads to
in general
which
0 for
AND ns
METHOD
for H1,
=
=
\357\254\201nd the
we \342\200\224H,,,
1.
line
nodal
\342\200\224 = 0.) y
x
tone if we vary the coe\357\254\202icients H1, and H21 in the \342\200\234compound\342\200\235 number in\357\254\201nite an new leads to nodal + lines, i.e., u21, equation (17.18) an In Fig. 5l(a), we show) of nodal lines can correspondto a given frequency. co = new/3-.) the simple nodal lines just found for the frequency Thus,
I
\342\200\231 \\\342\200\230
E
(o)
/I
\\\\
lb)
1!
/
\"'**\"
//\\
ll
(c)
\342\200\230sa\342\200\231 \\
Frounuzsl)
If we
set m
= 2, n
=
2, we
obtain =
(on corresponding
In
this
case, the
frequency
1cc\\/'8-,
unique mode
to the 142;
the
=
H22 sin
(cont + 0:22)sin 21cxsin 21:y.)
nodal line is the
locus
of the
points
in
the
plane
for
which))
288
AND 11'sAPPLICATIONS
mrmon
\342\200\230run moauwncrrou
= 1-[seeFig.5l(b)]. to simplest nodallinescorresponding x =
either
} or y
CRAP.
51(c), we show the
in Fig.
Finally,
9)
the frequency)
(!)=C013=-\342\200\231(031=1CC\342\200\231\\/.17)..)
I8.
Next,
as
membrane of radius
coordinatesin
the
equation
the
convenience, we the center of the
For
1.
choosing
xy-plane,
The change
origin.
x= transforms
Circular Membrane)
a circular
consider
we
introduce polar membrane
of a
Vibrations
Radial
of variables) y=
rcos0,
rsin\357\254\202)
(17.1) into)
0114=
1 Eu
0114
+
+
1
8324 03-\342\200\230)
757 5:\342\200\230: am)\342\200\231 \302\260\342\200\231(\342\200\230a7z and
equation
(17.2) into
8114_
1
1 Ba
8314
8114
2972\342\200\234 \302\260\342\200\231(7a7+7'\303\251;+7z'a'o3)
0, t)
F(r,
(\342\200\2303-2)) -7-\342\200\230
of a circular membrane, discuss only the case of free vibrations \357\254\201rst that the initial described by equation (18.1), assuming velocities of the points of the membrane are displacements and initial of the angle 0. Then, it is clear that at any time t, the displaceindependent ment is also independent In this case, the vibraof 0, so that u = u(r, t). We
i.e., the
shall
case
tions are saidto be radial,
and
(18.1) reduces
equation
to)
(18.3)
rar\342\200\231) Dr\342\200\231 8t2\342\200\224 \302\247\357\254\202_cz(?_\342\200\231_\342\200\234 12'!)
The
boundary
condition has the
form
u(I,
and the initial
conditions
have
t) =
(18.4)
0,
the form =
f(r)9)
=
so).
\342\200\234(rs
(18.5))
-\"\342\200\234\342\200\230T';\"-\342\200\231
Following our previous method, we look for
of the
form)
14(r. t)
= R(r)T(t).))
solutions
of equation
(18.3)
ml: aromruucnou
sec. 18
satisfy the condition substituting the result in (18.3),
(18.4).
which
AND rrs APPLICATIONS
mrmon
this
Differentiating
expression
289)
and
we obtain)
+
= 82 RT\342\200\235
(R\":r
%R'T).)
-
whence
R\342\200\235 + (1/r)R\342\200\231 T\342\200\235 = = \342\200\224)\\2 = const, -\342\200\224\342\200\224-R\342\200\224-\342\200\224 \342\200\230Eff,
so that R\342\200\235 +
%R'
Equation (18.6)is the
the general
has =
Sec. 6).
Ch. 8, 96 0.
Then,
u.
=
is unbounded at r = 0, we must set C; = 0. different from R -=- 0, we have to assume that from the boundary condition (18.4)that)
follows
it
be a
A1 must
index
a solution
= 0,
Jo(7J)
i.e.,
with
Bessel\342\200\231s equation,
solution)
Since Yo()\\r)
Moreover,to obtain C1
(18.7))
4' C2 Yo(7\\\
R0\342\200\230)C1Jo(7\\'')
(see
of
form
parametric
0 (seeCh.8, Sec.11),and
(18.6)
= 0.
T\342\200\235 + c3x2T
p =
= o,
+ me
zero of the
function
C1 =
Setting
Jo(p.).
1, we obtain
_ A, _ b,I
R,,(r) = J.,(x,,r)= where
31,,
For A
= 1,1 =
is the
1,, equation = A,,
T,,(t) we
Thus,
u,,(r, t) satisfying
have
= the
cos
cos
cA,,t
cA,,t +
boundary
(n
+
B,, sin
condition
t) =
E
n=-1))
(A,,
cos
. . .),
(18.8))
1,2, . . .). of (18.3) of the form (n =
c7.,,t
solutions
(n
cA,,t)Jo(A,,r)
=
1, 2,.
. .),
(18.9))
(18.4).
To \357\254\201nd a solution of (18.3) satisfying and the initial conditi_o_ns (18.5), we write) u(r,
1, 2,
solution)
B,, sin
particular
=
function Jo(p.).
(18.7) has the
found \357\254\201nally
(A,,
of the
zero
nth
Jo
both
c7.,,t +
the boundary
B,, sin
cA,,t)Jo(A,,r),
condition (18.4)
(18.10))
nroanwncrron
290
um
and
require
AND ms APPLICATIONS
METHOD
9) emu\302\273.
that
ucr. 0)
=
=
f(r).)
S144\302\273:-7o(7~n\")
=
sin (\342\200\224A,,cA,,
[ S:
3-14%\302\273
+ c7\302\273,,t
B,,c71,, cos
c)\\,,t)Jo()\\,,r)L=o)
=
=
so).)
31311071..-7o(7\\n\")
A calculation
of the Fourier coe\357\254\201cients of
f (r)
system {Jo(71,,r)}gives (see Ch. 8, Sec.24)) =
An
with
to the
respect
41'.) '.'f('\342\200\230)Jo(7~n'') J: ,\342\200\224;J\342\200\224%2(-\302\2435
(18.11)
,
2
=
31.07\302\273.
and g(r)
W)
Io
2
I
4?. '\342\200\2318('')Jo(7~n')
O1\342\200\230
3,,
.\342\200\224..-
Thus, the solution of our coe\357\254\202icientsA,,
The separate vibrational
B, are
and
vibrations
of the
(18.12))
the the series (18.10),where (18.11) and (18.12). (18.9) which make up the complicated
formulas
the
membrane, can
u,,(r, t)
dr.
is given by
problem
determined from
harmonic
motion
rg(r).I.,(x,.r)
-0-[\357\254\201g-(E)[0
= H, sin (cA,,t
be representedin the +
form)
a1,,)Jo(7\\,,r),
where)
H, = and
the
=-ff.
amplitude
U
frequencies of the membrane
characteristic
= cl,
0),,
The
cosa1,, =
+ 3,3\342\200\230, sinoc,, 1/A\342\200\230\342\200\234\342\200\234g
of vibration
of the nth
-32. U)
are)
= cg\342\200\231?-)
mode
is
H..|Jo(>~..r)|.
and depends only
on
r.
The nodal lines
Jo(71,,r)=J(&;-r)
[see(18.8)].If
n
=
1, there
are obtainedfrom
= o
are no nodal
the
equation)
(0 < r <1)) lines. If n
=
2, a
nodal line
is))
19
sec.
for r
obtained
r =
METHOD EIGENI-\342\200\230UNCTION
THE
=
p.21/[J.3,
3, nodal lines are obtained 52, we show the nodal lines
if
p.11/uz,
In
etc.
Fig.
n =
/\\
a membrane
sections
of
=
2, 3.)
I9. Vibrations of a In
the
general of radius
membrane
beginning
(see
whose
with
the initial
29!)
r =
p.11/p.3,
corresponding)
\342\200\224
--
52)
are
vibrations
described
by (18.9),
for
Circular Membrane (GeneralCase))
18), with
the
vibrations
of a
circular
condition
boundary
= 0,
u(l, 0, t) and
-.
case, the problem of the free I reduces to solving the equation)
of Sec.
for and
\342\200\230 1;
X13\"
Fromm
l,
an
\342\200\224\302\247
\\::\302\2421J/
n
AND rrs APPLICATIONS
(19.2)
conditions) u(r, 0. 0) = f (r.0).) =
g(r, 9).
\302\2471\342\200\230\342\200\224(%\302\260\342\200\224)
We look for
particular
solutions u(r,
which (19.2).
are not
identically
Substituting
zero
0, t) and
in the
=
which
this expression in 324)
= \342\200\230W\342\200\231
\"(7>7i'+
1
form of a product 0)T(t),)
satisfy the
(19.1),
we
8(1)
(19.3))
boundary
obtain
1 T)
+7279\342\200\230:)) 78\342\200\224r 82\302\242)
condition
ms
292
so
METHOD
EIGENFUNCTION
AND
CHAP.
APPLICATIONS
rrs
9)
that15) 182(1)
18(1)
82(1)
\"8r_=+7'87+r'i'\303\251'62
T\302\273)
= _
__
=
7. 2
=
const.)
16
CZT
(D
Therefore
18(1)
82(1)
182(1))=
W + 7 3;
:3
307
= 0,
T\342\200\235 + c2)t2T
where to satisfy
the
Thus, we
at a
arrive
are
not (19.7)
stituting
identically zero and in (19.4) gives) R\"F +
for equation of (19.4)
solutions
particular
which
satisfy
form)
(19.7)) the
+ 7.2RF
%_R\342\200\231F+ $RF\"
To
(19.4).
of the
= R(r)F(0),
(I)(r, 0) which
(19.6))
boundary value problem
problem, we look for
that
= 0.
(1)(I, 0)
solve this
(19.5)
(19.2), we must require
condition
boundary
(19.4))
\342\200\224xz<1>,
condition
(19.6).
Sub-
= 0,
\342\200\224R\342\200\235+(lmR\342\200\231+A2R\342\200\224F\342\200\235\342\200\224 _ v2 _ \342\200\224 \342\200\224 const.\342\200\234 . -I-5
_
(1/r2)R
we have
Therefore,
r2R\342\200\235 + rR\342\200\231 + (7.2r2
\342\200\224
v2)R
= 0,
F\342\200\235 + v2F = 0.
Equation
change function v
must
has solutions of the fonn cos v0 and sin v0. Since if we (19.9) 0 by 21:, we come back to the same point of the membrane, the have u, and hence (I)and F, must period 21:. Therefore, the number be an integer, i.e.,)
v=n, and
(19.8)
(19.9))
the
solutions
of (19.9) cos
n6,
(n=O,l,2,...),
are\342\200\235)
sin n6,
(n =
0, 1,2,. . .).
(19.10))
take the constant to be negative, since otherwisethe function T would not turn periodic, and we would not have vibrations, contrary to what actually happens. 16 The constant is taken to be negative, since in a problem like this, the function F(0) must be periodic [see (19.9)et seq.]. 17 For v = 0, equation (19.9) has another solution of the form C0, which can be dissince it is not periodic.)) carded, 15 We
out to be
19
sec.
EIGENFUNCTION
1-115
values of
negative factor.)
(The
u
Thus, equation (19.8) takes the r3R\" + This
to
functions,
293)
constant
a
within
- n1)R = of
form
0.)
Bessel\342\200\231s equation,
with
integral
has
and
n,
AND rrs APPLICATIONS
form)
rR\342\200\231 + (Pr?
is the parametric the general solution)
equation
index
same
the
give
METHOD
= c,.I,,(xr)
no)
+ C, Y,,()\\r).)
must be bounded,we are compelled to set C, = 0, because r->0 (see Ch. 8, Sec.6). If we put C1 = 1, then by the condition (19.6))
the solution
Since
co as
Y,,(Ar)->
boundary
R0) = J..(7J)'=0.) M =
i.e.,
p. must be
We now
a zero
of
the
function
J,,(p.).
write)
= J.) R... = J.(.x....r> where
mth positive zero (in order of increasing size) of the function the boundary value problemfor equation (19.4), subject to the condition has the eigenvalues 7.,,,,, [see (19.11)], and the (19.6),
is the
y.,,,,,
Then,
J,,(u.). boundary
(19.7),
[sec
eigenfunctions
<1>,,,,,(r,
0)
=
J,,(71,,,,,r)
= J..(7~.....r) 9) \342\200\230D3\302\247..(r. For
A =
7.,,,,,,
we
Therefore,
boundary u,,,,,(r,
(19.10), (19.11)]) cos
n0,
Sin \"9)
(19.5)
equation
T,,,,, =
the
n=0,1,2,...),)
l,2,...;
(m=
(19.11) I)
A,,,,,
gives cos
have the following conditions (19.2):) 0, t)
0, t) u,\342\200\230,\",,,(r,
c7.,,,,,t +
=
cA,,,,,t.)
solutions
particular
= (A,,,,,cos c7.,,,,,t (m =1,2,...;n
B,,,,,sin
of (19.1),
B,,,,, sin c7.,,,,,t)J,,(7.,,,,,r)
+
n8
=0,1,2,...),) +
sin c7.,,,,,t)J,,(7.,,,,,r) cos c7.,,,,,t (A,\342\200\230}\342\200\230,,, B,\342\200\230,\",,,
(m = l,2,...;
cos
satisfying
n
=
l,2,...).))
sin n0
294
r
AND rm APPLICATIONS)
\342\200\230rm: METHOD EIGENI-\342\200\230UNCTION
To obtain a solution of equation series conditions (19.3), we form
which
(19.1)
also
A]
9
1.. .1n.1ary
satis\357\254\201r; the
the
u(r, 0, t)
=
Z
c).,,,,,t +
cos
[(A,,,,,
B,,,,,sin c7.,,,,,t)cos
n=0m==l
+ and
require 0,
u(r, =
(19.12))
sin
sin c71,,,,,t) cos cA,,,,,t + B,\342\200\230,'\342\200\230,,, (A,\342\200\230}\342\200\230,,,
J_
J,,{'rt,,,,,r)
that)
0))
S:
S: (A,,,,,cos n0
= f
sin n6)J,,(71,,,,,r) A:\342\200\230,,,
+
(r, 6),)
n=0m=l)
0,
6u(r,
O))
at \302\260\342\200\230\342\200\231 \342\200\230'\302\260
=
=
sin [(\342\200\224A,,,,,cA,,,,,
Z Z { n=0m=l)
S: S:
+
c7.,,,,,t
+
sin (-A,\342\200\230f,,,c71,,,,,
+
cos c7.,,,,,t) B,\342\200\231,\",,,c71,,,,,
cos
(B,,,,,c7.,,,,,
cos
B,,,,,c71,,,,,
cA,,,,,t) cos
n0
( 19.13 )
c}.,,,,,t)
sin
n0]J,,(71,,,,,r)}:=o)
n0
n=0m=l) = g(r,
sin n0)J,,(}.,,,,,r) +B,\342\200\231,\",,,c}.,,,,,
To \357\254\201nd the coe\357\254\202icientsi11 these
0).)
we now argue
expansions,
as follows:
Let)
0) =
}(r,
30
[f,,(r) cos n0
sin n0], + f ,\342\200\230f(r)
(19.14))
where
fo(r) *= f,,(r) =11:
_\"\"
6) de.) 51,; I_\342\200\231;f
e) cos
f(r,
n0 d0,
(19.15) (n =
=
f(r,
f,\342\200\231f(r) 11:
i.e., respect
f
we expand to the
the function variable 0.
0) sin n0
f (r, We
0)
then
series with respect
in Fourier ,\342\200\230\302\247\342\200\230(r)
f..(r)
=
\"2
C.
in
1,2,...)
d0)
a
expand
to the
trigonometric
Fourier
each of the system
.. 1:10) =
functions
{J,,(A,.,,,r)}.
c,:\"..J.o...r>.))
\"21
The
series f,(r)
result
with
and is)
' us. 19
me
AND rrs APPLICATIONS
METHOD
EIGENFUNCTION
295)
where)
c;m:=:\357\254\2015Zj%;;;yL:Ua0%LmhmW)dh
(19.16)
9) =
f (r,
Comparing (19.16), we
this
gives)
S S: (C,,,,,
n0 +
cos
sin n0)J,,(}.,,,,,r).) C,\342\200\231,\",,,
n=0m=1)
(19.13), and
formulas
of the \357\254\201rst
the
with
dr.) rf.*:\342\200\230J,.(x.....r>
Jo
(19.16) in (19.14)
Substituting
I
2
c:::.. =
(19.15)
using
and
obtain)
=
n
I
1
=
co...
40\302\273:
dr
rm.
9)-7o(7~om\") [0 ,7,\342\200\224,,\342\200\224Q;M_\342\200\224,
L
do.)
\342\200\231tm:=\342\200\230C%m
=
dr
1
If (r, 0) cosn0J,,(A,,,,,r)
J:
Io,
d0
(19.17))
(n=1,2,...;m=l,2,...),
=
2
same way,
1,2,...;
L
7
\342\200\224
=1,2,...).)
n
I
I
-
m
d0
\357\254\201nd that
we
Bom\342\202\2547~om
B,,,,, clam
.
rf(r, e) sin neJ,,(x,,,,,r) [0 L 1-d\342\200\224,.,,\342\200\224H(Tn;) dr
(n =
In just the
u
z
d7
L\"
'80\342\200\230. 9)-7o(7~om\")
I
2
dr
7tI2J%+1(1\"nm)\342\200\230[0
rg(r,
1'\342\200\230
9) cos
d9.)
n0J,,(A,,,,,r)d6) (19.18))
(n=l,2,...;m=1,2,...),
1
2
tr
.
l\357\254\201zdkm,==;gZEE:GZ;5J;\302\242bw[mrg030)su1n0[4XMg)aM
(n=1,2,...;m=1,2,...).)
Thus, the solution of
the is given by the series (19.12),where from formulas (19.17) and (19.18). to the simple are a great variety of nodal lines There corresponding vibration and Which make the harmonic vibrations up complex u,,,,, ufm. the nodal lines shown in For example, for um, uoz, and um, one \357\254\201nds (19. 12). lines shown in Fig. 53.)) the nodal Fig. 52, while for an, un, and an, one\357\254\201nds
coe\357\254\202icientsare
our
determined
problem
296 As
11-13
in the
case of
\357\254\201gurations of
on the
METHOD AND rrs
EIGBNFUNCTION
a rectangular
coe\357\254\202icientsAm,
an
membrane,
B,,,,,,
of different
in\357\254\201nite set
correspond to the
lines can
nodal
CHAP.
APPLICATIONS
same
frequency
9)
con-
(depending
B,\342\200\230,\",,,).) A,\342\200\231:',,,,
FIGURE 53)
20.
Flow in
of Heat
Equation
Rod)
a
a homogeneous cylindrical rod, whose lateral surface is insulated medium. Choose the x-axis along the axis of the rod, at time t of the cross sectionof the rod and let u(x, t) denotethe temperature x and between with abscissa x. Let AB be the element of the rod lying is so small that we can x + Ax (see Fig. 49). Let At be a time interval which assume that of the cross sectionsat x and x + Ax does not the temperature that the amount It been has established experimentally change (in time). a rod at two of heat q \357\254\202owing whose ends areheld constant temperathrough the temperatures, to the crosstures is proportionalto the difference between sectional of the rod and to the time interval area At, while q is inversely of heat proportional to the length of the rod. Therefore, the amount the element AB is) \357\254\202owing through Consider the
from
surrounding
q =
K[u(x
t)
+ Ax,
\342\200\224
u(x,
(0 <
= K6u(x+0Ax,t)
s
ax
where
K is
a constant of proportionality called which the rod is made,and s is
In the limit
through
the cross
Now,
it
can
easily
AQ
be seen
time At
that
< 1),
the thermal conductivity the cross-sectional area
as Ax -> 0, we obtain the amount of section at x in the time At:) _Q(x)
AB receives in the
0
At
from
material
rod.
t)]s At
Ax
=
heat Q(x)\357\254\202owing
.9 At.
(20.1))
Kg\342\200\230
the
of heat
amount
of the
of the)
AQ which
the
element
is)
=
Q06 + Ax)
=
K
-
Q(x)) \342\200\230
-\342\200\230A\342\200\231
=KsAtAx
(20.2)
03u(x
+
GIAX,
axz))
1)
< 01
< 1).)
SEC. 20
be kept
should
(It
being the same. AQ
AND ITS APPLICATIONS)
METHOD
opposite to the AQ can also be is so small that
direction
in the \357\254\202ows
temperature increases.) The quantity method: Suppose that the element
AB
of all itsfcrosssectionscan
the temperature
time
given
any
heat
that
mind
in
direction in which the calculated by another at
EIGENFUNCTION
THE
be
as
regarded
Then)
=
cps Ax
[u(x, t
=
cps Ax
At au(x,
+
- u(x,
At)
0, At))
ta?
c is the heat capacity and p material from which the rod is made.
the
is
where
t)])
< 02
(0
(per
density
(Thus, ps
is the
Ax
< 1), unit
(20.3)) of the
length)
mass of the
element
AB.) A
C9
6 u (x,
02 At)
t +
= K
at
pass to the
if we
and
and (20.3) shows that)
of (20.2)
comparison
Bu
0, then
6714)
a2
Ex-29
5;\342\200\231
where a2
= K/cp. In this in a rod.
we obtain
way,
Ax, t)\342\200\231)
Bx?!
as At\342\200\224> 0, Ax->
limit
+ 0
02u(x
the equation
for heat
\357\254\202ow (or heat
conduction)
now
We
pose
imposed on the
a variety of problems,corresponding of the rod.)
2|. Heat Flow This
boundary
problem
in
a Rod
consists
With
in
Ends Held at
(the endsof the
f (x) is of Part
rod
are
= 0 and
at x
a given function. I, and hence all
we
look
u(I,
x
t)
(10.4), with
= 0
= I),
the
for particular u(x,
with
the
initial
= f (x).
Equation (20.4) considerations
solutions of the t) =
(21.1)
and
the
present problem. Thus,
conditions
Zero Temperature)
of equation
solution
the \357\254\201nding
u(x, 0)
(1.2)
different
conditions)
u(0, t) =
where
to
ends
condition (21.2))
is a
specialcaseof
equation
given there apply to the form)
nromruucrron
me
298
do not
which
vanish
satisfy the
which
and
identically
CHAP.
APPLICATIONS
this expression in
Substituting
(21.1).
AND us
METHOD
conditions
boundary
gives
(20.4)
= a3 QT\342\200\231
9)
\"T,
whence\342\200\234
=
=
%:
= const.)
-7.3
a\342\200\22472'\342\200\224;.
Therefore, we have = 0, T\342\200\231 + a2A3T = 0.
(21.3) (21.4)
<1)\342\200\235 + 1.30
The solution
of
is
(21.3)
C1 cos Ax
(x) =
and because of the
condition
=
d>(I)
assuming C; = 1, then
Therefore,
If we
set
A,
=
=
To
sinx,,x
AI
=
t) =
0.)
we obtain M = ')
=
where
1m,
(n
=
1,2,.
sin\342\200\231-\342\200\230-\342\200\231%
_a=n\342\200\230-w,' I9 ,
.
A,, sin
n is
an integer.
. .).)
A,, =
(n =
const
1, 2,. ..).)
rrnx
_\342\200\230\302\260\"\"'\342\200\235t 19)
e
T
(n=1,2,...))
the boundary satisfy particular solutions of (20.4)which the initial condition, we form the series)
conditions.
satisfy
0\302\260
u(x,
13
=
gives)
A,,e-091?.\342\200\230A,,e
u,,(x,
represent
C2 sin
that
functions)
the
Hence,
= 0,
require
7.
(21.4)
1,, equation
T,,(t)
96 0,
C;
= C1
M,
\342\200\230Nil
_
=
,,(x)
For A
that
we must
(21.1), (0)
C2 sin
+
We
leave it
(see Sec.12).))
to
the
reader
t) =
Z
A,,
to decide
a'3n9n9
'
.2 3\342\200\230 sin\342\200\231-LE\342\200\231-\342\200\230
why
the constant
is chosen to
(215))
be negative
here
me
sec.
21
and
require
Fourier
of the
presence
are
of the
to the
respect
system {sin (nnx/1)}.
A
coe\357\254\202icients gives)
=
(n
% f(x) of
solution
coe\357\254\202icients A,,
f(x).)
1\342\200\231;-\342\200\231f
=
,,
i.e., the
=
sin
2.1,,
expandf (x)with
we have to
calculation
299)
that
u(x, o) = Thus,
AND 113 APPLICATIONS
METHOD
EIGBNFUNCTION
1, 2,.
sin\342\200\231i'l1\342\200\231-\342\200\230ax
our problem is given from the determined
series
the
by
(21.6))
..),
(21.5), where the
formulas (21.6).
Becauseof the
factors)
e it is easy to see that for any to > 0. The
_a_\342\200\234\342\200\230.~.\342\200\230*1-3. I9)
(21.5) is
for t 2 to > 0, uniformly convergent for the series obtained by term by term of times). number differentiation of (21.5) with to x and t (any respect term and the term sum of is the the series Therefore, continuous, by (21.5) is legitimate (cf. Sec. 10).) differentiation
Heat
22.
This
boundary
Flow problem
with
is true
same
in a
ConstantTemperatures)
Rod with Ends Held at in
consists
of equation
solution
the \357\254\201nding
the
initial
t) =
A =
const,
u(l,
= B
t)
= const,
the
(22.1)
condition
u(x, 0) = f (x). We look
(20.4), with
conditions\342\200\235)
u(0, and
series
the
for a solution
in
the
form
u(x, t)
=
of a
Z
(22.2)
series
T,,(t) sin
n==l
#7\342\200\230.
(22.3)
where
:r,(z) = 19
The
like
u(x,
z) sin
dx.
these.))
(22.4))
\342\200\234\342\200\224\342\200\231I\342\200\231\342\200\224\342\2
boundary conditions in this problem, and from those considered previously.
a di\357\254\202erent form cases
\302\247
also in We
the problem of Sec. 23, have show below how to deal with
THE EIGENFUNCTION
twice, we
by parts
Integrating
METHOD
obtain)
x='
I
T\"
2
=
I3 6u(x, t)
_I_l_I\302\2437_C_,_L) C0821}
1m
[_
I)
CHAP. 9)
ITS APPLICATIONS
AND
0x
[ nzznz
x=0
sin
-71tnx]x=\342\200\231 x=0)
.. .11.. 12.2-.\342\200\234 sin 1% 1r3n2 0 8x3
Since
u(x, t)
satis\357\254\201es equation
and the
(20.4)
2
I
Q.\342\200\230 sin TE
a21t2n3 o at
n=%u\342\200\224pwm-
have)
m)
I)
to t, we obtain)
with respect
(22.4)
Differentiating
conditions (22.1),we
.__g.._
3.
d
I)
2 tau . nnx , _ \342\200\224 7L a\342\200\224ts1n\342\200\224l\342\200\224dx,
Tn so
that
l
l ;'n[/1
-
57'\302\273
- (-1) ,,B]
I2
-
,
\302\2475;;5n\342\200\224zT...
or
2\"2 \342\200\234z\342\200\235 2\342\200\2342\"\342\200\235 \342\200\224 = [A 1\",, (\342\200\224 l)\"B].
+ 1\",\342\200\231,
12
This equation has the
solution)
T,,
To
satisfy
the
(22.5))
12)
= Ane\"
:52.
condition
initial
u(x,
o) =
A
+ 2
:9
\342\200\224 (\342\200\224l)\"B_)
(22.6))
\357\254\201n
we require
(22.2),
that
= f(x). 2 T,,(O)sin \342\200\231-\342\200\230ll\342\200\231-\342\200\231f
am!)
A calculation
{sin
(-nznx/1)}
of the
T,,(0) and
= A,,
coe\357\254\202icients of
respect to
f (x) with
the
system
+
= 2\302\243\342\200\230_1f:';'._1)\"\302\243
f(x)
sin
dx,
\342\200\231-5%-\342\200\2315
hence ,,
Thus,
Fourier
gives)
the solution
are determined from
I
.._2
7 [0
f(x)
. sin
-rcnx
-7-
dx
of our problemis given the
formulas
- 2A
\342\200\224(\342\200\224l)\"B
(22.7))
\342\200\224\342\200\224n\342\200\224,;-\342\200\224\342\200\22
by
(22.6) and
the series
(22.7).))
(22.3), where the T,,
sec.23
THE EIGENFUNCTION
Flow
Heat
23.
Variable
problem,
to
is required
it
\357\254\201nd the
u(0. t) = <90). and
<9
sp
are
T; +
by
same as
is the
q: and
=
(20.4), with
0) =
(23.1)
\302\253#0)
condition
initial
(23.2))
f(x).
look for a solution in the form of a series (22.3). of Sec. 22, we \357\254\201nd for T,,(t) the equation)
again
argument
which
of equation
solution
u(I. t)
and with the
given),
u(x, We
Speci\357\254\201ed
conditions)
boundary
(where
30l)
Temperatures)
In this the
at
Ends are
Whose
Rod
a
in
AND rrs APPLICATIONS
METHOD
=
T.
\342\200\224 (\342\200\2241>\"\302\242(:>1) I<9(t) -2-\342\200\230\302\247\342\200\234,i\342\200\235
\342\200\230#533
formula
we
B have been replaced
that A and
except
(22.5),
this equation,
Solving cl\302\273.
the
Repeating
obtain)
_a3u9n9,
T,, =
12)
A,,e
(23.3))
+ To
L;
the condition
satisfy
of the Fourier
A calculation
sin {(-nznx/1)}
the
Therefore,
24. Heat
from
the
in
a Rod
Flow
the
coe\357\254\202icients of
then
by the
= .4,
=
=
f(x).)
f (x) with
1,\"
problem is given
respect to the system
by
dx. the
(23.4)) series (22.3), where
(23.3) and (23.4).)
Whose
Ends Exchange
Heat Freely
Medium)
of a body exchanges
the amount
sin
f(x)
%
formulas
Surrounding
If the surface given
T,,(0)sin $
2
of our
solution
T,, is determined
medium,
(-1>\"4\302\273<:>1dz.)
gives)
:r,,(o)
with
312\342\200\230mo
(23.2), we requirethat)
o) =
u(x,
-
of heat
with
heat
through \357\254\202owing
a surrounding an area
gaseous
sin the time At
formula)
Q
= H(u
\342\200\224
uo)s
At,
(24.1)))
is
um
302
nrommncnou
u is
where
rounding
AND rrs APPLICATIONS
METHOD
the temperature of the and H is a constant
body,
medium
called
CHAP.
uo is the temperature the emissivity.
In the case of heat \357\254\202ow in a rod whose lateral ends exchange heat freely with the surrounding of (20.1)and (24.1) leads to the boundary conditions)
whose
H(u for
x =
0,
x =
1.
Setting
h =
medium,
but
a comparison
8
\342\200\224=
uo)
K-3-2)
\342\200\224=
uo)
(h >
H/K
-
-33 x
\342\200\224K%-\342\200\231:)
0), these
+
assume
that
=
(24.2) 0.)
uo)L-1
= 0.
uo
::=0
\342\200\224
h(u
become)
= 0,
\342\200\224
h(u
3-3\342\200\230
let us
conditions
110)]
an First,
surfaceis insulated,
sur-
and)
H(u
for
of the
9)
The
conditions
boundary
then
take the
form)
3
= 0,
\342\200\224 hu
[\303\251lc
*=\302\260
(24.3)
an
= 0,
-I-a\342\200\224 3\342\200\230 hit]
while
x=l
is
the initial condition
14(x.0) = f (x). as before.
Following our
equation (20.4)of
the
usual
method,
the
conditions
for particular solutions of
form)
=
u(x. t) satisfying
(24-4))
we look
(24.3).
Substituting
this
expression
in (20.4)
gives)
= a3\"T, IDT\342\200\231
whence \"
3
=
T\342\200\231
=
-7.
(\342\200\230T1,
= const,
(24.5)
so that <13\"
=
\342\200\224)x,
= -azu\". T\342\200\231
<2\342\200\234)))
me
24
sec.
To
the
satisfy
METHOD
BIGBNFUNCTION
conditions
boundary
AND rrs APPLICATIONS
303)
we must obviously
(24.3),
require
that)
= o, = 0.
\342\200\224
<1>'(o)
h(0)
+ h(l) \342\200\231(I)
the
this
of A;
instead
problem
thus, instead of the
= 13\302\242
+
T\342\200\231 + a3)3T
The solution
of
(24.8)
0,
(24.8)
= 0.
(24.9)
C1 cos Ax
+
C2 sin
= C37.\342\200\224 hC, sin 7.(\342\200\224C1
Ax,
have
we must
(24.7),
by
eigenwrite
can
is (x) =
and
(24.6),
the
we obtain)
(24.6),
equations
CD\"
for equation
problem
According to Sec.5, all are positive. Therefore,we
(24.7).
value
boundary
value
boundary
conditions
boundary
values of 2.2
arrived at a
we have
Thus, with
<2\342\200\234))
+ h(C1
C2 cos N)
M +
0,
C2 sin M)
Al +
cos
=
0.)
Therefore)
=
(24.10)
7-.
.7
so that tan
The positive is
worth
pointing
section (in :1
the
-
= (A2
we can
(24.10),
be the set C;
,,(x)
For 7.
=
A,,,
leads
u,,(x,
to the t)
=
root
positive
= 1,, C, = h.
\342\200\224-. 1,, cos
solution
the
nth
7.,,x +
=
our
problem.
It
points of interM and the hyperbola) of the
abscissas
2/tan
=
of equation (24.11). the eigenfunctions Then,
hsin 7.,,x
of equation
T,,(t) This
eigenvaluesof
that these roots are the of the function pt. pt).-plane) out
h\342\200\230)/Ah.
let 1,,
Now,
21h
K-7?)
equation give the
of this
roots
M =
(n =
cos
A,,e-4%\342\200\230?!
A,,x + h
become
l,2,...).
(24.9) is (n
=
l, 2,.
following particular solutions of A,,(7.,,
Accordingto
sin
. .).
(20.4):)
).,,x)e-02?\342\200\230?-I
(n
=
l, 2,
. . .).))
304
1'm.=.
To
satisfy
=
S:
CHAP.
(24.4), we form
the initial condition u(x, t)
)\\,,x +
cos
A,,()\\,,
the
sin
in
9)
series)
7.,,x)e-\342\200\230Wu\342\200\231,
(24.12))
require that)
and
= S: A,,()\\,,
u(x, 0) A
AND 113 APPLICATIONS
mmron
BIGBNFUNCTION
sin ).,,x) = f(x).)
Fourier coeflicients of the
of the
calculation
the
)\\,,x + h
cos
function
f (x) with
respect to
system\342\200\235)
= {A,, cos
{d>,,(x_)}
A,,x +
1:sin ).,,x}
gives
_ ,4 ,, _
dx
f(x)..(x)
fa\342\200\231 .j____)
,
-_-
(n
1, 2,.
. .).
(24.13)
[0 d>:(x)
dx)
the
Thus,
of our problem
solution
coeflicientsare determined The integral The equation)
the
in
from
is given
the
by
series (24.12), where the
(24.13).)
can be calculated
of (24.13)
denominator
(I); +
as
follows:
=
0 )\342\200\230?n(Dn
implies that =
A\357\254\201dlf, -\342\200\224,,;.)
we have)
Therefore,
=
\342\200\224
,'. 32:6
>.,=,fo\342\200\231<1>:dx
+
;,2dx.
Io\342\200\231
(24.14)
But
(D, =
).,,cos).,,x+ hsinxnx,
=
sin ).,,x 111,\342\200\231, -1,\342\200\231, which
means
+ In,
cos )\\,,x,
that
Ago: +
of = 13+ hzxg,
(24.15)
hence)
and
xi 20
According
dx <1\302\273:
to Sec. 4, this
+
system
;,= dx fa\342\200\231
is
orthogonal.))
=
(A2
+
h\342\200\231).,\342\200\231,)I.)
me
24
sec.
with (24.14),
together
This,
213
=
(A: +
191:): \342\200\224 [d>,,;,];:.',.
(24.16)
conditions and (24.15)
the boundary
from
follows
it
305
APPLICATIONS
implies
dx <1>\302\247
other hand,
On the
AND us
METHOD
EIGENFUNCTION
that
=
2.: +
+ h2\302\247 ).f,\302\247
hzx\357\254\201,
or
0%
at both x
= 0 and
= I.
x
= xi,
(24.17))
writing the
Therefore,
in the
conditions
boundary
form) \"
=
0. [\342\200\230Dn\342\200\230p:'a h\342\200\230D:2Jx=-o =
\"' hq\342\200\231;2.]x-=1 0. [\342\200\230pn\342\200\230p:'a
and using
that (24.17), we \357\254\201nd =
1,.:. \342\200\230:1. -21:13.
this expression in
Substituting
gives
(24.16)
of (24.13), we
instead
Thus,
2h_)
11:):
can write) cos A,,x +
2
f(x)(A,,
=
A\"
+
(13 +
= Lz\357\254\201dx
(x3, +
h
}.,,x) dx
sin
M): + 2h (n
we
Next, heat
consider
a medium
with
heat
exchanges
reduced
with
solvedby u =
function
1, 2,
. . .).
12
=
v
+
substitution)
w,
only on
v(x) depends
the
making
x and
satis\357\254\201es the
= 0, 12'\342\200\231
with the
boundary
-
w
satis\357\254\201es the
equation
(24.19)
conditions
\342\200\224 [v\342\200\231 h(v
while
(24.13))
the case where the end of the rod at x = 0 exchanges at the temperature uo, while the end of the rod at x = I a medium at the temperature ul. This problem can be
to the problem just
where the
=
uo)],,.o =
0,
+ h(v [v\342\200\231
- u1)],,., =
equation)
8w Bt
.3\302\273 =
32w
a 2 1--, Bx\342\200\231))
0,
(24.20)
306
METHOD AND 113 APPLICATIONS
\342\200\230ml; monuruncnou
the boundary
with
9) cum\302\273.
conditions = 0,
\342\200\224
hw] [-8527\342\200\230
x-=-0
= 0
+
x
:\342\200\224w hw] x-=-I
and
condition)
the initial
W(x. 0) to (24.19),
According
the
= f(x) v =
function
v
=
110:)-
v(x) has the
form
+ B,)
Ax
where the constants A and B are determined from the This leads to the system of equations)
conditions(24.20).
A\342\200\224h(B-uo)=0,
A+h(AI+B\342\200\224ll1)--0,)
which can be solvedvery easily. is of the type discussed above.)
25. Heat Flow
in
an
case of an in\357\254\201nite there rod, to reduces a solution problem \357\254\201nding and
t
>
0, and
are no boundary conditions, of equation (20.4) which is
we look for
(-00 < x < 00).
= f(x) solutions
particular
u(x, t) = Substituting
this
w
expression
and
the
for de\357\254\201ned
condition)
satis\357\254\201es the
u(x,0) As usual,
value problemfor
In\357\254\201niteRod)
In the
all x
the boundary
Then,
in (20.4)
of the
(25.1)
form
(x)T(t).
gives)
= a3 \"T, QT\342\200\231
whence <1)\"
-6
=
T\342\200\231
=
-13
-5-2-7-.
= const,
(25.2)
so that <1)\" +
= 73\302\242
0,
T\342\200\231 + a3A2T = 0.
(25.3))
(25.4)))
sec. 25
mozuruncnou
ms
These
307)
APPLICATIONS
have the solutions
equations
=
=
T(t)
C3sin Ax,
).x +
C1 cos
\302\242(x)
C3e-am\342\200\230.
particular solutions can be written
the required
Therefore,
AND ns
METHOD
u(x,
B sin
cos xx +
t; A) = (.4
the
in
form\342\200\235
).x)e-am\342\200\230.)
as values of regarded that for discrete values of A of \357\254\201nite we formed an in\357\254\201nite series rod), and in a we then of chose the coe\357\254\202icients the series such particular solutions, as to obtain a initial In solution the condition. the way satisfying present and we set) case,A varies continuously, A and
constants
The
B, being
= A0.) and B = (as in the case of a
t) =
u(x,
t; I: u(x, A)
=
If we to
t
can
di\357\254\202erentiate
this
equation
Ba
8314
_
__
\"273
0_t
co
co
Jo
the
To satisfy
initial
u(x, o) be the
will
This
integral
piecewise these
Sec. 5), a
and
smooth
With
8u(x,
=
B()\\) = [see equations 21 If
(5.5) to
we chose the increases
oil\342\200\230 as t
limit as t
(25.5)) with respect
willbe a solution
of
32u(x, t;
t; 7.) _
\"2\"\"'a'x2
xx +
t; A)
d)\342\200\230 Bx\342\200\231
d\"
1.))
_
\342\200\235 0')
that
3(1) sin 2.x):1).= f(x).)
f (x)
can be
integrable
representedas a Fourier
is that f (x) be a function on the whole x-axis. for which
assumptions)
A0)
falls
81u(x,
Jo
su\357\254\202icient condition
absolutely
co
t; 7.) _ a2 d)\342\200\230
we require cos
(Am
u(x, t)
d).. (once
sign
integral
function
the
case if we require that
(see Ch. 7,
sin xx)e-amt
'8: (\"\342\200\230
condition,
=
3(1)
have)
at
_ \342\200\230
+
Ax
then
8u(x,
o
be
behind the
function
with respect to x), In fact, we then (20.4).
twice
and
recalled
will
d}.
cos
(Am
be
can
arbitrary,
B(7.). It
A
functions
increases,
contrary
1
cosM
dv,)
(25.6)
00
sin Av 1-: _w f (:2)
dv)
(5.7)of Ch.7].)
constant would
f(v)
to be positive in (25.2), then the exponential factor which without by an exponential factor which increases to the physical of the problem.)) meaning be replaced
METHOD AND
Tl-IE EIGENFUNCTION
For
integral
of the the
ITS APPLICATIONS
CHAP.
9)
the can be differentiated behind (25.5) A0) and B0), the integral to x and t. In fact, because with respect sign any number of times in the integrand of (25.5) and becauseof of the factor e-027*\342\200\235! presence
such
inequalities)
|A(A)l <
dv ,1If; |f(v)|
<
|3(x)|
|f(v)|
1-:
dv
=
c.
=
c
(c =
const),)
obtained from it by differentiating the (25.5) and the integrals with of to x and t are uniformly number times integrand any respect > > 0 is for t 0 This follows from the 2 to convergent (where to arbitrary). the
integral
inequalities
|(A()t) cos
Ax
[(A(7t) cos Ax
B(7t) sin
+
+
sin Bo\302\273)
)tx)|e-am\342\200\230<
2Ce'0\342\200\235*\342\200\235' < 2Ce'\302\260\342\200\231*\342\200\235*o,)
).x)e\342\200\2300\342\200\235*\342\200\235'])
Hg.
g
K.
Ax +
cos
31\"\342\200\231)
sin
B(7\\)
2C)\\ne-(2212!
<
2C)\\ne-azlzto,
)tx)e\342\200\2342\342\200\2313']
<
2Ca2\"*A3\"'
e-am\342\200\230 < 2Ca2m).3me-\302\260\"7\\2'o,)
fact that the majorizing functions on the right are integrable in A 0 to co. Thus, we need only apply Theorems 4 and 3 of Ch. 7, Sec. 6. It should that although our argument be noted shows that u(x, t) is a solution of equation (20.4), it does not show that) and
the
from
\"Or. 0
133 However, Using
this
relation is true and can easilybe proved. we can rewrite the solution of our problem
(25.6),
u(x,
To
further
to change 3> 0
z) =
transform the
order
1
co
1-:f0
7t(x
su\357\254\202icientlylarge
dv
dx
co
f_wf(v)
this expression, of integration.
f(v)
we begin
To show this,
<
L00 v)e-\302\242\342\200\231*2'dA|
I (where
L\342\200\230-t L\342\200\235
cos we \342\200\224 o)e-am:
\342\200\224
Utoocos
for
= f (x)-)
t >
cos x(x
proving
by
we
note
as)
dv. that
(25.7) it is
\357\254\201rst that
e-0\342\200\235\342\200\231~\342\200\230\302\260\"dA < e
0 is \357\254\201xed).Therefore) \342\200\224
<
o)e-aw:
ail
possible
for every
|f(v)|
1-\342\200\230:
dv,))
SEC. 25
which it follows But then)
from
as
the
that
METHOD AND
EIGENFUNCTION
THE
in the
integral
APPLICATIONS)
ITS
side
left-hand
to zero
converges
I\342\200\224> oo.
dv
cos
11:
-
A(x
J:\302\260f(v)
0\302\260
lim 1
=
dx) v)e-am\342\200\230
dv
7: -co) l\342\200\224\302\273oo
dl 11:
Here the
(25.7)].
[sec the
f(v) cos
do v)e-am\342\200\230
= u(x,
is legitimate,
of integration
order
in
dl) }.(x \342\200\224 v)e-02*\342\200\235!
cos ).(x -
f(v)
J:
change
fol
t))
because
integral)
f (12)
I: is uniformly the
convergent in is dominated
integrand
u(x, z) = of (25.7). It
instead
0 <
A for
A
\342\200\224 dv) v)e-021\342\200\235!
from the fact that Ch. 7, Sec. 6).
This follows
I.
<
Theorems 4 and
(v)| (see
|f
by
)(x
2 of
write)
can
we
Thus,
cos
11:
the inner
that
out
turns
cos x(x
dv
f(v)
\342\200\224 dz, v)e\342\200\224am=
In
be evaluated.
can
integral
(25.8)
fact, set
am/t~=z, so
A(x\342\200\224v)=y.z
that)
dz J)\342\200\230-=
-
\342\200\230L
E
.
ax/?)
we have)
Then
no
L
cos 7x(x -
1
v)e
with
Differentiation
\342\200\234*1\342\200\235! dx =
co
e-=2 cos p.z dz
L a\342\200\224\342\200\230/.\342\200\224t
respect to p. behind the = \342\200\224e-=2 z I\342\200\231(p.) L\342\200\235
this
x\342\200\224v
=
,
is legitimate
differentiation
convergent We
now
in
integral
sign
=
1
Z7; I(p.). gives
sin y.z dz;
because the
resulting
integral
is uniformly
(1..)
integrate
I , (u)
=
1
by parts, . sm
5 Ir\342\200\234
M
u.z1,-..
obtaining)
\342\200\224
% If
(25.9))
e-=2 cos
uzdz =
gm\302\273).))
me
3 I0
It follows
that
=
I(p.) To
\357\254\201nd C, we
set p.
= 0.
Ce-it\342\200\235/4.
This
gives)
c =
1(0) =
which, as we know,
an integral
e-3\342\200\235 dz,
Ch.
(see \302\247\\/-T
equals
1(9)= W5 and
CHAP. 9
APPLICATIONS
rrs
METHOD AND
BIGBNFUNCTION
8, Sec.
8). Therefore)
e-'**\342\200\230\342\200\231'/\342\200\230\342\200\230.
by (25.9))
cos 7.(x \342\200\224 d). v)e-am\342\200\231
J?
this expression
Substituting
u(x, t)
in (25.8), we
=)
= 515
e-0:3?\342\200\230) \\/1-:/\342\200\224t
\357\254\201nd) \357\254\201nally
1
2a
f(v)e'(L4;'3-9
dv
\302\260) (2510)
\302\260
f\302\260\302\260 -00 Wt\342\200\230:
the one hand, formula (25.10) shows that as t increases, 0, u(x, t) \342\200\224> the heat the rod. On the other hand, (25.10) i.e., \342\200\234spreadsout\342\200\235 along \342\200\235 shows that the heat is \342\200\234transmitted In fact, instantaneously along the rod. let the initial this be for v and zero outside < < temperature positive xo x, interval. Then the subsequent distribution of temperature is given by the On
formula)
u(x, t) which it is
from
large
trarily
that
2171/; u(x,
t) >
f
0'0.) (\3-sag\342\200\230;")
\302\2433
0 for
small
arbitrarily
t >
0 and arbi-
x.)
Heat Flow
26.
clear
=
in
a
Circular
Cylinder
Whose
Surface is
Insulated)
of a circularcylinder of radius lbe directed along the z-axis, be insulated let the length of the cylinder be in\357\254\201nite). (or that the initial temperature distribution and the boundary conditions Suppose are independent of 2. Then, it can be shown that the equation for heat \357\254\202ow is) Let
and
the axis
its ends
let
au.
where
9)
,(a2u
K/cp, K is the conductivity c is its heat capacity, and
a3 =
is made,
a
azu)
of the material from which the rod 9 its density. Thus, the temperature is))
SEC. 26
EIGENFUNCTION
THE
METHOD
AND ITS APPLICATIONS
I)
of z (a fact which is of course a consequence of the assumptions and we are with a problem in the plane. If just made), essentially dealing we to polar coordinates by setting x = r cos 0, y = r sin 6, then, go over instead of equation (26.1),we obtain) independent
\342\200\224
We
now
heat
0.
\357\254\202ow equation
that the
further
assume
independent of
Then,
obviously
takes
the form Bu
2
T9?=\"
We also assume rounding
and
initial
u is
a
'-)
+
only
are and t, and the
conditions
boundary
function
of r
1
(26.2))
3\342\200\231: 7 7' (6324 8a).)
of the cylinder is insulated
surface
the
that
r3 802
r Br
Br?
at
1E .
191
E
2
_
Q
from
the
aua. o
=)
0
(26.3))
Br and (absence of heat \357\254\202ow), condition) the by
the initial
that
0) =
u(r, We
look
for particular
solutions of the u(r,
Substituting
sur-
medium, i.e.,)
this
t) =
in (16.2)
expression
= a2 RT\342\200\231
temperature
distribution
is given
(26.4)
f (r). form
R(r)T(t). gives)
+
(RT
$127).)
whence R\342\200\235 + (1/r)R\342\200\231 =
T
T\342\200\231
-12
'aTT=
_\342\200\224 COnSt,
so that +
R\"
121: =
+
\302\24312\342\200\231 T\342\200\231 + a27.3T
Equation
(26.5)
is the
= 0.
parametric form
p = 0 (seeCh.8, Sec.11). Itsgeneral =
(26.6))
of Bessel\342\200\231s equation,
solution
'|' R0\342\200\230)C1100\342\200\231)
(26.5)
o,
is)
C2 Yo(7\\\\302\253))
with
index
3 I2
\342\200\230run BIGENFUNCTION
nth positive
the
is
=
C,
Taking
l, we
= 0-
1604) M
= 0.
we r\342\200\224>0,
the boundary
Therefore, 1:.=
Cl-IAP. 9)
have to set C, condition (26.3) that)
as
co Yo().r)\342\200\224>
Since
\357\254\201nd from
AND 113 APPLICATIONS
METHOD
root of the
J
equation
=
{,(y.)
We
0.
write)
_Pm
*~-7')
R..(r> = where )1 =
p.,,
1,, in
=
the
).,,I is
nth
for
particular
of the
zero
positive
=
equation
(26.2), subject of the t) =
function
Setting
J(,(p.).
to
the
. . .).
1, 2,
condition
(26.7))
(26.3),
we have
found
form)
=
l, 2,.
(n A,,Jo(7.,,r)e-0\342\200\235?\342\200\230-\342\200\234at
. .).
(26.8)
the series)
form
now
=
(n
A,,e-02*?-I
solutions u,,(r,
We
1. 2. . - .).
(n =
-70
equation (26.6), we \357\254\201nd T,,(t)
Thus,
=
Jo(7~n\")
u(r, t) =
S 1 A,,.Io(7.,,r)e-0\342\200\235~3',
(26.9)
n= and
the
to satisfy
condition
initial
we require
(26.4),
0) =
u(r,
=
that
(26.10))
f(r).
21/1n\302\253\342\200\231o(7~n')
A calculation
of the
Fourier
coe\357\254\202icients of
respect to
f (r) with
the
system
{Jo(7\302\273.r)}gives) 2)
A,, =
(secCh.8, Sec.24).
Therefore,
series (26.9), where the
(n =
rf(r)Jo(7\\,.r)dr
Fm
the solution
coef\357\254\201cients A,,
are
of our
(26.11))
1,2,...)
problemis given
by
the
from the formula
determined
(26.11).) 27.
Heat
Flow
Heat
with
This problem
Circular Cylinder Whose Medium) Surrounding
in a the
reduces
to
solving
Surface
(26.2)
equation
Exchanges
with
the
boundary
condition) +
hu(l,
z) =
@g;\342\200\224\342\200\230)
o
(27.1)))
sec.27 the previous
with
and
Repeatingthe
of Sec.
argument
we again
and
3 I 3)
condition)
initial
u(r, o)
(26.6),
AND rrs APPLICATIONS
METHOD
\342\200\234rm; EIGENFUNCTION
= f(r).
(27.2))
26, we again obtain equations
(26.5) and
\357\254\201nd that)
= Jo(N)-
R(r)
The condition(27.1)gives = 0
+ hJo()J)
).J(,()J)
or = 0.)
+ hlJo()J) )JJ(\342\200\231,(7J)
Therefore, the
number
p.
= 11
must be
a root
of
it-16(9) + hUo(:t) = We now
the
equation)
0-
(27-3)
write 2.,
R,.
=
=
Jo(7~n\")
__
\357\254\201n, I)
(n =
1..
1. 2. . . .),)
is the For nth positive root of equation A = 1,, (27.3). p.,, = and (26.8) of equation 1,2, . . .), the solution (n by (26.7), (26.6) is given the condition We again givesparticular solutions of (26.2)satisfying (27.1). form the series(26.9), and that the relation (26.10) be satis\357\254\201ed. A require calculation of the Fourier coe\357\254\202icients of f (r) with respect to the system
where
{Jo(A,,r)}
leads
to the
A \"
formula) 2
=
+ l2[J6\342\200\231(u..)
(see Ch.
L:tf(r)Jo(A,,r)
dr
(27.4))
the of equation (26.2), subject to the solution Thus, and (27.2), is given by the series (26.9), where the coefficients from (27.4), and the numbers p.,, are the roots of equation
8, Sec. 24).
conditions
(27.1)
are detennined (27.3).) 28.
J%(w.)l)
Steady-State
We now assume of the cylinder, and
Heat Flow
in
that
a constant
that
the
a Circular
Cylinder)
temperature is maintained
distribution
of temperature
on
the surface
is independentof z.))
nroauruncnou
ma
3|-1
a
after
Then
AND rrs APPLICATIONS
METHOD
at every point of the cylinder, on t. Thus, instead of equation
or
in
alu
Byz-)
the
1 Ba 7 7,
+
372
on the
temperature
look
be
0) =
(23.1))
by speci\357\254\201ed
condition
the
=f(9).
(28-2)
gives)
= o,
+ imp\"
+ imp
MD
o.
R(r)(0).
in (28.1)
expression
=
of the form
u(r, this
8114
:5 5.72\342\200\231
boundary
for particular solutions
Substituting
1
+
140.0) and
to
ceases
coordinates
polar
Let
u
0.
@411\342\200\230-
3x3
is
de\357\254\201nite temperature
i.e., the function (26.1), we have)
established depend
a
of time,
interval
su\357\254\202iciently long
9) emu\302\273.
r
r3)
whence \342\200\224
=
%-
\342\200\224g-)5
= -13
= const,
(28.3)
so that r3R\"
lb\"
The solution of
follows
It
must
=
=
0.
+
Bsin
= A
(28.5)
cos
A0
A0.)
the physical meaning of the problem that the function 21:, and hence A must be an integer. (Incidentally, would not have been periodicif we had taken the constant
(0)
be positive.)
n, equation
we write)
Thus, A,,
cos
n0 +
B,
sinn\357\254\201
(28.4) takes the r3R\"
+ rR'
(n
= O, 1,
\342\200\224 n3R
..).
= 0,
is
direct
R,
2,.
(28.6)
form
a second order linear differential equation. r\" and r-\" satisfy substitution that the functions fore, for n > 0, the general solution of (28.7)is) which
(28.4)
period
d>,,(0)=
For A
12(1)
from
have
we note that in (28.3) to
+
(28.5)is) (9.)
113(0)
- HR = 0,
+ rR'
= C,.r\" +
D,,r'\".
(28.7)) It can this
be
equation.
veri\357\254\201ed by
There-
(28.8)))
sec. 28
run
Since r-n->
EIGBNFUNCTION
00 as r -> 0, we
D,, =
to set
have
AND rrs APPLICATIONS
METHOD
0. For n
=
0, we
3 I5
easily
\357\254\201nd
that
R0 =
C0 +
and hence we must again take D0 = 0. conditions D,, = 0 (n = 0, 1, 2, . . .), we (28.1)as)
0) =
u,,(r,
now
form the
the
satisfy
+
n0
(3,,
sin
n0)r\",)
21
(28.2), we
condition
+
g3
3
Fourier
+
(ot,,cosn0
coe\357\254\202icientsof
f(0)cosn6d0
=1}-cf\342\200\230
=
9,1\"
l, 2, . . .),
(n =
n0)r\"
(1,, cos
+ 9%\342\200\231
=
of the a,,z~
sin
series)
boundary
u(I,0) A calculation
(3,,
write
7\302\260
u(r, 0) = and to
+
(28.8), (28.9), and the the particular solutions of
(28.6),
Using
can
\342\200\230*0
=
110 We
(on,cos n0
I\342\200\230,) D0111
require that) =f(0
[3,,sinn6)I\"
f (0)
.)
gives)
= a,,
(n =
o,1,2,...),)
=
(n =
1, 2,.
b,,
:\342\200\224cf_\342\200\231;f(e)sinnede
. .),)
so that an =
u
bn
=
is\302\273 \342\200\230'77\342\200\231 F\302\260
Therefore
u(r, e)
For r < I, this with
respect
0
r0,
solution
the
This integral
(a,, cos
+ 12\302\260
n0 +
sin
b,,
(28.10))
n0)
2]
can be differentiated term by term any number of times and 0, sinceeach resulting is uniformly convergent for series < I is It follows that (28.10) actually r0 arbitrary. gives
series
to r where
of equation Ch.
(28.1).
a more compact 7). Then we have)
be given
can
solution (see
=
6, Sec.
_\342\200\230 _1_ \342\200\234(\" \302\260) 21::
\"
1
Lf
-
if we
(V/1)\342\200\231 \342\200\224
-
2(r/I) cos (:
I2
\342\200\224 2Ir cos
(\342\200\230)
1
form
0)
use the Poisson
\"\342\200\231)
+ (r/I)?
01'
=
u(r\342\200\231 6)
I
1:
2-1:
.L.f(t)
[2 _
'02)
(t -75-)
+ r3
dt\302\260))
3 I6
METHOD AND rrs APPLICATIONS
erosuruucnou
\"rue
Cl-IAP. 9)
Moreover
0) =
u(r, \357\254\201n}
is continuous, (28.2).)
wherever
f (6)
boundary
condition
i.e.,
f (0))
solution
the
just
written
satis\357\254\201es the
PROBLEMS)
1. Find
the
equation
+
=
>.(x)
and normalized 0 on the interval
of
eigenfunctions
[0, 1],
the
subject to
differential
the
following
conditions:)
boundary
a)
eigenvalues
\"(x)
=
b)
c)
em) =
a
2. Consider
=0
points x
=
(1)
=
0;
0; \342\200\231(1)= =
0.) h\342\200\231(1)
+
<1>(1)
at the I, tension T and linear density p, fastened Suppose that at time t = 0, the point x = c by an amount h and then released (the \342\200\234plucked displaced the subsequent the initial conditions and \357\254\201nd motion of the of length
string
x = 1.
and
(0 < c < I) is string\342\200\235).Write
string.)
T and linear density p, fastened at the of the string be a parabola which is initial diswith respect to the center of the string, with maximum symmetric placement equal to h, and let the initial velocity of the string be zero. Write the subsequent the initial conditions and \357\254\201nd motion of the string.) 3. Consider
points x
a
21, tension initial position
of length
string
= 1-I.
the
Let
of length 21, tension Tand linear at the 4. Consider a string p, fastened density at time t = 0, the string receives an impulse of x = 1 1. Supposethat points Find the subsequent motion of the string. P at its midpoint. magnitude Hint. Solve the problem with the initial conditions)
u(x, 0) = at
pass to the
then
5. Consider a string x = 0 and x =
I.
0 {\302\2433
6. Let point
a
x = Hint.
for e
e,
< |x|
< I,
at the points tension T and density p, fastened be initially at rest in its equilibrium position. between x, and x2 (0 < x, < I) is acted string force F(x, t) = 9/! sin cot (see Sec. 13). Find the Verify that when x, = 0, x2 = I, the answer string.
of length I, the string
Let
x; <
(13.8).)
concentrated
c of the Pass to
|x| <
e\342\200\224-> 0.)
limit
Suppose the section of the upon by a periodic perturbing motion of the subsequent
reducesto formula
for
o) =
3u(x,
and
0,)
perturbing
periodic string the
limit
x.
force F
5. Find the -> c, x2-> c in
of Prob.
=
subsequent Prob.
5.))
A
cot act upon the motion of the string.
sin
mcrmnmcnou
me
pnonuams
AND 113 APPLICATIONS
METHOD
3 I7)
7. Consider a rod of length I, Young\342\200\231s modulus E, cross section 3 and density fastened at the end x = 0. Suppose that the rod is stretched by a force F acting on the end x = I, and then is suddenly released at time t = 0. Write the initial conditions and \357\254\201nd the subsequent longitudinal vibrations of the rod. Hint. The amount by which the rod is initially stretched is Fl/Es.)
8. Let the free end of the rod of Prob. 7 receive a sudden impulse t = 0. Find the subsequent longitudinal vibrations of the rod. Hint. Solve the problem with the initial conditions)
P at
p,
time
u(x.0)=0. 0
_
for0
au(x\342\200\2310)_
at
and
then
to the
pass
force
assuming
that
10. Find the membrane
equation
A. sin cot.
the rod
0.)
linear
of the
this curve actually
of Prob.
rod
Find the
is initially
7 is acted upon
by
a periodic
vibrations
longitudinal
subsequent
of the
per-
rod,
at rest.)
of the
combinations
to the
corresponding
for!-esxsl,
cps
the free end of the
9. Suppose turbing
e->
limit
\342\200\224
modes
lines
nodal
\342\200\234nodal indicated ring\342\200\235
by
and un of a rectangular 5l(c) of Sec.17. Write the fourth sketch in Fig. 5l(c). um
of Fig.
the
Is
a circle?)
11. Consider a circular membrane Find the radial vibrations P at time t = 0 distributed over at rest in its equilibrium originally Hint. Solve the problemwith area.
u(r9
=
of radius
T and density 9 per unit 1, tension membrane if it receives a \342\200\230sudden impulse circle r < e, assuming that the membrane is
of the the
position. the initial
conditions)
0:)
for0
e,
at
O fore
12. Find the radial vibrations of the membrane of Prob.11if it is acted upon by distributed perturbing force F = A sin cot per unit area, uniformly over the entire membrane,assuming that at time t = 0 the membrane is at rest in its equilibrium position. a periodic
13. Consideran in\357\254\201nite slab of width 21 bounded by the planes x = i I, made with thermal conductivity heat K, speci\357\254\201c c and density p. The slab) is \357\254\201rst heated to temperature To and then at time t = 0, its faces are held at zero in the slab.) Find the subsequent temperature distribution temperature. of material
slab of Prob.13have an initial temperature distribution u(x, 0) = if starting from time Find the subsequent distribution f(x). temperature t = 0, the slab exchanges heat with the surrounding medium which is at freely zero temperature. (Let the emissivity of the slab be H (seeSec.24.)))) 14.
Let the
THE
8 15.
EIGENFUNCTION
the steady-state
Find
prism bounded the planes x = by
0 <
the
METHOD AND
plane
x <
a.
by
y = b has (Draw
9)
distribution in an in\357\254\201nite rectangular x = 0, x = a,y = 0, y = b, if the faces formed by = 0 are at zero temperature, while the face formed
temperature
the planes
0, x = a, y
CHAP.
ITS APPLICATIONS
the
temperature
a \357\254\201gure.) Specialize
u(x, b)
distribution the answer
to
the
casef(x)
= f(x), =
where
To.))
ANSWERS
PROBLEMS)
TO
CHAPTER])
1.
\342\200\230CR) ed\342\200\234...
a))
.,.x...
E
\342\200\224
[$4,)
21%?-1-)%(acosnx
nsinnx)])
(-1: b))
cosax
=
s1nax=
\"Z1 \302\260\302\260
2 .
.
c))
(-1)\
+
\303\251sinau
(10:35,
an
,
=
5\"
coshax=
2
(-1r
a2_n2
1
=
(_])n-1;,
'
smx\342\200\224
sm22x-
\302\260\302\260
1
.
9-21-\342\200\230cos3x
(_\357\254\202
sin33x_si1:\342\200\2304x
a))
GCOSILX)
nsinnx
.
2
f(x) =\302\247\342\200\224;cosx+
3.
a
+
1-:sInhan[-2-\342\200\234
\"Z!
(\342\200\2241)\"'-15-4-_\342\200\224a;cosnx)
(-1:
b))
sinnx]
sinhax=13:sinhan[\"Zl(-l)\"*1,1T_'|'_\342\200\224;'-5
4.
a))
.
l+cosau
1:
(-1r<
<
xs
cosznx)
Q)
Z
cos(2n+1)x - (2n + 1)?)
\":0a3
3|?)
1:),)
x< 1:).)
\302\260\302\260
l
l\342\200\224cosa1c
+2a
1:);)
x < 1:);) (\342\200\2241:< (\342\200\2241)\"a2\342\200\224_\342\200\224\342\200\224n2)
E81110?! Z -1
d))
(0
TO pnosmms
ANSWERS
320
(What
b)
f(x)=2;\342\200\235
except
\357\254\201\357\254\202ri-'\342\200\224hcosnx]
for the value
+ 2-1?
1
5. 3)
x =
h,
equals 1/2.
the sum
where
nx]
(si\357\254\201hnhr co
x<
(0 <
cos
E
4
1I:x
.
integer?)) (0
%+
=
f(x)
c)
a is an
when
happens
n
21mx
.
(Why?))
1:).)
(0
f(x)\342\200\224:s1n\342\200\224I-;-:ngl(-1)\"4-\342\200\231-\302\2432-Tl-S11\342\200\230!-T
except for
-\302\273
the
value
x =
1/2, where
the
1/2;)
sum equals
r=)
4.41tx
-
6.61tx
+
(osxsl),)
\342\200\224\302\260\302\260\302\260)
-1-38111-7\342\200\231 \302\2475-Sll'l\342\200\224I- ) \302\247Sll1\342\200\224l\342\200\224 :(2.21tx
except
a,, =
7.
for the value
x = 1/2,where
=
cosnxdx
Tl:
71: \357\254\201nf(x)
=
sum equals
the
-%
+ z_1:;/:3/\f(1;:")
frfc-:
+
t)
0.)
cos nt
cosntdt. t)
Hence an =
so
b)
rc-:
x>}cos-x.
that
|a,,|
8. a)
\342\200\224 + IZ\"{f
<
f(x)
fl\342\200\231-J?
+ \342\200\224f(-3
x)|
|cosnx|
dx <)
cosx=1\302\247:i\342\200\224:\342\200\224-)
x3 =
2:-.-2S
sin nx
(-1):-+1)n)
+12 E
n=l))
sin nx (\342\200\224l)\" ,,3)
dt
TO PROBLEMS
ANSWERS
2)
CHAPTER
3.
By
the
inequality
Cauchy
---+
|ao+ a.x+
<
a,,x\"|
x2 +---+
a:)|/2(1+
(a3:+---+
1 _ x2n+2 1/2 (1\342\200\224x=)
Now
32l
The sharper estimate is
integrate.
1
x2\302\273)!/2)
if0
\342\200\230x/\"1\342\200\224\342\200\234_ xz
obtained
by
use of
direct
the
Schwarz
inequality.)
4. Use and
Schwarz
the
5. Usethe
Schwarz
+ a,,q>,,(x)
such
be
a) No
+ ||P,,||3<
orthogonal
which
function
orthogonal exists. In
function
fact,
exists,
consider
0 = a0 = a;
C0 +
=
-%+
unless
\302\273\302\273
a-
[for
A polynomial
2
\"Bl
\302\253-
all
(4;2:,,gl(),E2:
the
= C1=
An
c,-
are
orthogonal
Cz=-'\302\260\302\260\3
zero; b) No function g
then
the
+ ---+
|an|\342\200\231(
of de\357\254\201nition
of degree n
by parts.)
;::;,::
if)
gq\302\273..(x).
identically.)
(-1)\"\"
and
1 <-1)\302\273
Z
=
\"21
a function
i.C., C3 =\302\260\302\260\302\260, '-co
(Sec. 6)
11. Use repeatedintegration 12. a)
is zero,
= -- - = a,, = 0
Ch. 2, Sec.10];b) it vanishes
---
K g(x)q>..(x)dx.)
n=0 + - - - + a,,
aocpo
= a,,
,7-
were such
Q
i.e.,
P,,(x)
by a similar proof; c) the continuous function) go)
9. a) If
let
-+ e\342\200\230\"u
0, i.e.,g(x)E 0,sinceg(x)iscontinuous.)
since if g
exists,
0 = C0 + C1= Co + C; = violates Bessel\342\200\231s inequality
g =
+
1
<
dx
P..
||g||3 =
so that
1/n,
-
[g(x)
6.. = thCl'l
emu\342\200\230 + --
to l]
the q>;(x), and
all
to
orthogonal
b
8.
equal
that)
fa
Then ||g||3
function
inequality.)
g(x) is
7. Suppose
with one
inequality
equal to 1.)
the other
)!P2n-l(x)3
: 23;.
.)) . \342\200\230.>,<.2;:.
has
Prob. 10 and real zeros, unless
see (<9, 4\302\273),
at most n
322
TO rnonwms)
ANSWERS
CHAPTER 2.
a) Neither
limit
0) f1~(0) =
= f\342\200\231(0+) 0-
3. This
2::
1
4.
1
|:.(x)|ax<\302\247;
0
2-\"
as
(6.3) just
from
follows
1 (0) =
b) f
exists;
=
21'!
ff,\"
2-1,,
If,\"
0, f '(0+)(1068
the 21':
1
_
|f(x
|e(t)|
exist;
r)g(r)Idtdx) lf(x
|e(t)l
=
not
of Sec. 6.)
proof
55],,
0
-21;
in
3)
|f(x)l
ff,\"
- 0!
dx)
dt)
4:. dx)
Moreover, we have) 1
..
2\"
0
C\"-5;
=
8\342\200\224;nx;,( x))dx g(t)
K\"
517-\342\200\230 K\"
1
=5;
2\342\200\234
0 g(t)
1
e-'\"xf(x
-
dt)
t)
dx)
2\342\200\234 \342\200\231 \342\200\224tnu -0\"
e
-27Jo
f(u)du e
dt=a,,b,,.
If
Q
Q
2
< co. Ian!\342\200\231
2
< co, |b,.|\342\200\231
n=l
n=l then Q
2 |a..b..|< 0\302\260
n=l
by the
6. We
Cauchy
inequality
(see
Ch. 2,
Prob. 2).
have)
Q
%2vial) 9,, f:(l It follows from
2 N,
< M(b
<
sin 2nx sin 26,.)dx (cos 2nx cos 20,,\342\200\224
> N.
a).
b E\342\200\234
i.e., \342\200\224 f:(1+ cos2nxcos20,,sinnxsin20,,)dx
if n
\342\200\224
(2.9)that I:
if n
\342\200\224 + cos2nxcos20,, sin2nxsin20,,)dx
>
\342\200\234)
52\"
Therefore)
Q
Q)
sothat \302\247Zp,,\342\200\2352\342\200\234'
2
n=-N))
p,,<4M.
ANSWERS
PROBLEMS
323
formula
7. a) The
g
+ s,,(x) =
(3.2) and
from equations
follows
b)
T0
I:
dt)
p,,(:)
some manipulation;
after
(4.1)
If)
.
1
-
= D,,(x)
D\342\200\230,\", (x)
cos
=)
nx
-2-
then
(D,,(t)
J: co (why
as n->
I:
%
cos nt dt -+ 0)
?). Moreover)
-
dt =
I:
sin
ntdt\342\200\224>0)
(S\342\200\234t\342\200\230\"\342\200\231\342\200\224 p:(:)) \303\251cot\303\251)
co.
as n->
\342\200\224 dt = D',\342\200\230.'(t))
(Why?)
Thus)
where
Si\342\200\230:
8. Usepreceding
b) when
from
follows 35\342\200\230
sin
a) x
7%
x
2. For
3. a) x 4. a),
+
(2k
b) forx
x; c)
x
75
+ l)r.-; c)
x) cosh
(sin x); b)
7. a)
cos (cosx) cosh (l +
x 75
4)
Zkn.
(sin
cosx)In
(2
c) for
Zkn;
(2k
75
sin
8. a)
b)
1):.-;
6. a)
(cos
inspecting
No.
Zkn.
at
a6
b) for all
2k1:;
by
t/t.)
CHAPTER l.
n-> co.)
can be derived
The inequality
problem.
y =
curve
the
=
dt
I:
under
dt +
I:
as n\342\200\224>oo.) \302\242o,,(x)\342\200\224>0 \"\342\200\231
c)
=
o,,(z) dt
[:
for x
75
all
x;
Zkn/3;
d) for
all
d) for
x
cos(cosx) sinh
(sin
x).)
x); b) sin (cos x) sinh
(sin
x).
cos
+
gsin
x. 75
(2k
x;)
b)
(\342\200\2241r
3-C(l+cosx)\342\200\224sinxln(2cos\302\247)
+ 1)::/2.
the area
324
TO PROBLEMS)
ANSWERS
1
9. 3)
-2-)
-
sinxlog
b)
10.21) b)
(-1: < x < 1:).
%sinx
(zcosg)
cosxln(2cos;)-%cosx\342\200\224%;)
J-Ecosx 2
(-1: < x < 1:).)
+ -lsinx 4
' cosnx .x . n ]+snnpx[ cospx[ln(2snn\302\247)+mg\342\200\230
1l.a)
_ b)
cos px
[R 2
p x \342\200\224
Use
13.
Imitate
Abel\342\200\231s lemma
the
14. Assume
proof 0 <
that
2
'sinnx_
\342\200\224nZl
.
smnnx] \357\254\201g\342\200\230
x<
px
[In (2 sin
+
\"Z1 coilnx]
21:).)
(Sec. 1).
of Abel's lemma.)
xo <
1:.
cos?
Since
0 <
\342\200\230loos 0|, we
have
Q nxo < |a,,|cos\342\200\231
2
n=l But 2 cos?nxo
=
1 +
cos Znxo, and
1 of
Theorem
by
co. Sec. 3,
Q
2 |a,,|cos
< co.
Znxo
n=-I)
the series)
Therefore,
\302\256 Ianl
2
<
\302\260\302\260-)
CHAPTER
on
co
T
on
I
(_l)n_H_
5)
1
\302\260\342\200\231
3
'..\302\247<\"\342\200\224\342\200\230\342\200\2342\302\273+r>~'..Z.<7\302\273T4)
l
n
p
+ sin
(0 < 12.
-n:\342\200\224x
1
-\"\342\200\230(\302\247\342\200\234s-M)-Tzrr))
],
ANSWERS
-ro pnonums
325)
1 l 1c\342\200\230_ 9) \342\200\230\342\200\2359\342\200\230B=\302\260\342\200\231m \"\"\342\200\230(s\357\254\201\342\200\230m)\302\260 1:5
3
3
3
2-a)-I-\342\200\230\302\247;b)135;c)3\302\260;)
the odd
of f(x) onto the interval extension [\342\200\224 1:, 0], we obtain an with equal (zero) values at the end points of the interval [- 1:, 1:]. is obviously an even differentiable is an odd funcMoreover, F\342\200\231(x) function, F\342\200\231(x) tion which at the end points of the interval again has equal (zero) values [- 1:, 1:], and F\"\342\200\231 three x) is a continuous even function. Therefore, F(x) and its \357\254\201rst 3.
By making
odd function
F(x)
can be derivatives smooth function.
function
those
and
extended continuously over the whole x-axis, the Fourier coe\357\254\202icientsof But we know that of its derivative are related by the formulas)
-
a,, =
Therefore, in
[see (8.3)].
smooth
for a
Moreover,
the
function,
5 = _
21.
n
n3)
the squares of the Sec. 10). Since)
sum of
(Ch. 3,
coe\357\254\202icients converges
Fourier
=%:'
Ibil = ii it follows
term
by
=
obtained
-
f\342\200\231(x)
5.
r'(x)=
series
last
series
the
4.
values
of
nzlbnl.)
-;-
converges. by one
or
two
cosnx_sinnx + S: n3) [(-1)~+l) n+1
Q)
Z
,
_
(-1)\302\273
1
\"x0 5;\342\200\230) +1)
\302\260\302\260
cosnx
\342\200\230-3\342\200\231 \342\200\234*\342\200\231-'\302\247')
b)
3
absolute
This in turn implies the uniform converterm by term differentiations of the the that of these series guarantees series, and the uniform convergence term differentiation is legitimate.)
gence of original
that the
a F\342\200\231(x)
case
our
,,\" =
the
b},
%-s
with
a continuous
f\342\200\231(x)= _T16...li\302\243L\342\200\231\302\243.))
[x 16
(2k
+ 01:].)
T0 pnonnms)
326
ANSWERS
7.
F(x)=c;cosx+czsinx+i
(0
where
Q
1
1
1
1
1
1
smcen3\342\200\224l=\302\247(tT:-1-\342\200\224n+1)\342\200\231
c\342\200\230=n=2n3-l+\302\247=2\342\200\231 \342\200\230N
c;=
8. a)
\"'5')
(In eachcasef(x) denotes
f(x) =
+
n 2 (1
n\302\243
sum of
the
the series.))
n3+1 \342\200\224l-\342\200\224-)sinnx)
\302\260\302\260 sin 1:\342\200\224x =\342\200\224\342\200\224+ (0
[see
b)
formula
(2) of
Sec. 12.])
f(x
(0
c)
f(x) =
-
1n(2sin\342\200\231-2?)
61tx-l-1%-(3x3\342\200\224
21:2))
Q
+032
(0
I-\342\200\230,\342\200\224(\302\260\342\200\230:!\"\342\200\231\342\200\224\342\200\224J\342\200\231 on
x
1
(_I)\"
d) f(x)=;\342\200\224aLln(2cos\302\247)dx+a3
.
\342\200\231;\342\200\230)
(-1:
e)
f(x) = ln
<1c);)
+ a2
+
\342\200\234'13\342\200\235:
\"2
(';)\",(;\342\200\224\342\200\2241\302\247\302\260\"%\
(\342\200\2241:
0
f(x) =
g
+ (2a\342\200\224 1)[:1ntan\302\247dx sin(2n+
+(2a-1)2
2)
f(x)=
\342\200\224-
mm;
l)x)
Z(2n+n,(n+a) +
(0
\342\200\230z\342\200\230'\342\200\230 \"\342\200\235) \"ff\"\"\342\200\230
+ (2a \342\200\224 1):
\"Z!
(o
< x
2\342\200\235\342\200\234) (2:\302\260:(21\342\200\231)',(\342\200\230;
<
1:).))
1'0
ANSWERS
Theorem 1 of Sec.3.
9. Use
10.
Use
Theorem
11. To
derive
formula
(I).)
formula
2\"
with
terms b)
Use the
c)
Sum
Sec. 8 and
1 of
Set N =
12. a)
(3.1). Then,
and apply the last part index n 2 N/2. (For
Cauchy
the
formula
use
(I),
(3.1).)
equation
b) with
inequalities
respect to
b)
for
1:/2N
preceding problem, keeping 2 ~}.)) sin3(1m/2N)
= 7}; b) o\342\200\230
(1
k.)
6)
_1__
= o\342\200\231
< x<
. the
o.) 1:,x?\342\200\230-'
series converges.)
(1 + p)\342\200\231)
- Zpcosx +123)?)
b) 5\302\273; c) i; d) Not summable by arithmetic to 0; e) Not summable by arithmetic means, but relation
the
a) Use and
the the
s,, =
relation
theorem
(n +
l)c,,+, -
t,, = (n + of Sec. 2.)
means,
but Abel-summable
Abel-summable
to i.)
nc,,.)
l)s,, -
(n
+
l)o,,+1;
b) This
follows from a)
CHAPTER 7 .
3
4. a)
c)
7.
only
2);
a) i;
8. Use
6.
in
p3cosx+pcosx\342\200\2242p3
5.
9.
-n
f(x)=%cot\342\200\231\302\247\342\200\230
4. a)
h =
for-1:
f(x)=0
l.a)
substitute
terms,
2, Prob.
(Ch.
inequality
of the such
CHAPTER
7.
327)
PROBLEMS
=
\302\253pm
=
\302\253Mn x/2F: ,,,\342\200\224\342\200\230+\342\200\2244;
1:)
=
f\"j\"f,; =
\302\253M\302\273 a) \302\253mu \302\253/7? ~/27.-\342\200\235\342\200\224\342\200\2345,l,\342\200\224\342\200\230\302\260'\342\200\224i\342\200\230;
a)
f(x) =
c)
f(x) =)
Imitate
~/W:
proofs for
b)
the
discrete
f(x)=%1 +1x,;)
case (Ch.
3, Prob. 4).))
328
TO rnonuzms)
ANSWERS
f
0
0,
0 forx<0,
_
d)
for0
%x3
\"-(x)-
for1
\342\200\224-}(x3-6x+4)
> 2.
for x
0 9. a)
f(x) =
b)
f(x) =
0 {m.,,
< O,
for
x
for
x 2
0;
e\"\"\".)
CHAPTER
1.
Setting = xzy
p =
2, we reducethe
[see (2.l)] reducesthe
equation
to the
equation
to the
u\"+1u\342\200\231+
x
y = 2\302\273+1
2
P(
Sec.
3.
where and
6.
+
C2
-
Y2(x))-
2
2
22P(2)) \342\200\224 3)---5-3-1\342\200\230/1-t 2\302\273)
l)(2n
=
+) \342\200\231
= const,
A , J,,(x) =
4.
1
,7 (c.Jz(x)
3).)
J;,(x)
where B
x \342\200\224-\302\2435)u=0.
(1
-2\302\273-1-2\302\273-3...2.2 )\342\200\2302
= (2n (see
form (2.2). Then, the substitution equation:) following Bessel\342\200\231s
once that)
It follows at
2'
8)
[3
and p(x) is bounded
= const
as x->
co.)
\302\260
+
\342\200\230\302\260)
+
smx\342\200\230/\342\200\231\302\247.c
A and co are the same constants is bounded as x-> co.)
as in
the
asymptotic
expression for
J,(x),
f(x)
Use formula
9. Usethe
10.
We
(7.1).) problem
preceding
(2.9) of Ch. 3.
and formula
have) =
\342\200\230~
1
2 I.
1
Jo
x\"P+lJp(;\342\200\230nx)dx
=
\342\200\234\342\200\231\342\200\234*) \"\"\342\200\231\342\200\234\"\302\273\342\200\234~\"\342\200\231 1)
W)
t-P+1J,(t)
L?\342\200\235
dt.))
TO rnonuams
ANSWERS
p by
Replacing
\342\200\224 1 in
p
formula (7.2),
51-
obtain)
we
= -t-r+1J,(t).)
[t-P+1J,-.(t)]
dt
329)
Therefore 7*
(4.3)],
[see
-
dt =
rI:+lJ,(t)
you
J
Jp\342\200\224lO\342\200\230n) _
A
13+ .0\302\273
c \" =
'
12.
x3
=
0 <
x <
-
+
to
E
2
x!\342\200\231 if p
- i. p
> H.
7\342\200\230nJ4()\342\200\230n)
2.
2 to 6, a2 =
In Problems
9
T/p.
Q
u(x. I) =
o
sm
(Sm/I)
;2;\342\200\231; C) ea\342\200\231:
\"2!
\302\260\302\260
3,
=
-1
32h
T
(\342\200\230(35, 1)
Z n=-0 Q
u(x, I) =
4.
(zit
5.
u(x, t) =
where
co,,
6.
u(x,
2A
In
7.
-\"iii!
11\"\342\200\231-
2n
1
1
1t.x\302\253.cos \357\254\202at. \342\200\224-2-IL
21
(2;n++
cos
sin
ll)\342\200\234,
\342\200\224 cos
1 2\";
war.
0
.
sin
'
t
1) =
f) =
2A
0\302\260
.
E '61- \"=1 7 to 8F
mtc
.
SID.
0\302\260 t .) 83:2\342\200\235,_\302\260\"('o\302\247;n
mtx as sin o:,,t \342\200\224 0),, sin wt \302\260 _ 02-) II)
T\342\200\230 m II(02 Sll\342\200\231l\342\200\224I\342\200\224
9, a2 =
E/p.)
\302\260\302\260 (-1)\302\273
;TE.\342\200\230sn\302\245o
.2n+l
S111
T
2n+l
1l.'x\302\260COS
T
'
\342\200\224
nu.-a/I.)
Problems
u(x,
\"-75
2n
cos
\302\260\302\260
(cos -;:\342\200\224nZI
sin
+)'l')\342\200\231
cos
ilf\357\254\201nzo
=
.
p\342\200\231)J3(>\302\273.)
CHAPTER
2.
1
x.:\302\273+=2\302\273-'I\342\200\230(p>
\"\342\200\231(A\"x/2)
15
\"=1 for
:\342\200\2242p_lr(p)
27\342\200\230nJp+l(7*n) f
(1% 7~?:J\342\200\2313(1..)
series converges
The
\342\200\224
and hence
c\" = 11
1
=
[t\342\200\230p+l] _1(t)]:_:\342\200\230n).;p+l],,_1(A,,)
nat.))
330
9.
u(x, t)
where
to\" = M13,
PROBLEMS) \"\302\260 -1 41\302\260
=
u(x, t)
8.
10.
to
ANSWERS
sin
2(n _|_)\"1
-\342\200\234E n=-0)
2n
1
nx-sin
-
((13 \342\200\230I\342\200\231 1131. The l'lOd&l
I131,
2p _\342\200\224 :5 \"Z!
is the
.r.(.1
ring
is the
IKX,
0,
s1n
I0
of the
or sin
marlin,
41-\302\273
Io(p.) =
equation
sin cot
nth positive root
of the
c3 = 779.)
and
u\342\200\224\".J.
equation
(\342\200\230W _ -15 I) \342\200\224 1\342\200\230 2-':-f CXp{ \"go
0
_ Jo(p.,.r/I),
-
ma) -\302\273..(\302\253\302\273*
Jo(p.) =
\302\260\302\260
13.
equation
. 1; ct
r
e/I)
nth positive root
_ 2_4_ \" \342\200\230 \"\342\200\230\"
p.,,
has the
+15
\342\200\224--p'3J\303\251'(p_\") (11,, -I
9 E, where
xx,)
21
circular.)
quite
\302\260\302\260
\"'
sm
_ m2sL-
21
no; t) 11,,
.
0)2n+]SinC0t\342\200\234(0Sin(I)z+1t
3EnZo(-1)\"\342\200\224rm2\"+l(wz%+1
\302\260\302\260
where
\302\253at. 2\";
cosznx + coszny
11.
1
2n
Q
-
_2A
((31, ((13
and is not
.
2\";
21_+_1
(
2\342\200\231 1:)
0
co, = p.,,c/I.)
and
2
2\"\"'1 CO8
21
1'}
xx.)
where -r = 09)
14.
u(x, :1
=
j:,r(a>cos1;'5dz)
ap1-(he/12>} \302\260\302\260
4' 1
'
sin (1-1..x/I1
fag\342\200\230
where
1- =
3,, is the
(Kt/cp), a:,,is
nth
positive
the
nth
of the
root
positive equation
. an:
Sm \302\242XP{-G331\342\200\230/1\302\260)} I_\342\200\231f(E)
-1-
tan x = root of the equation tan x = \342\200\224x/IH.)
(15.
IH/x,
15. sm%\342\200\235 f(E)sin\"\"7Ed\342\202\254.)
\302\253x,y>=\302\247\"\302\247l
where
sinh x
is the
hyperbolic
not * y) =
15 1:
sine (cf. Prob.
3 of Ch. 1).
Inthe
special
+ 1)
sinh sin [(211 -my/a] [(211 + ux/a]_ 2\302\273 + 1)) [(211 + '_o\"\"\342\200\230\342\200\224\342\200\230\342\200\224\342\200\234'sinh
S
1) 1) nb/a]
case)
and
BIBLIOGRAPHY)
translated by M. Tenenbaum and S., Lectures on Fourier Integrals, H. Pollard, Princeton Press, Princeton (1959). University Fourier Princeton University Bochner, S. and K. Chandrasekharan, Transforms, Press, Princeton (1949). Treatise on Fourier\342\200\231s Series and Spherical, Cylindrical, Byerly, W. A., An Elementary and Ellipsoidal Ginn and Co., Boston (1893).) Harmonics, H. S., Introduction to the Theory Series and Integral, Macmillan Carslaw, of Fourier\342\200\231s York.) & Co.,Ltd., London (1930), by Dover Publications, Inc., New reprinted H. S. and J. C. Jaeger, Conduction of Heat in Solids, second edition, Carslaw, Oxford University New York (1959).) Press, R. Value Problems, McGraw-Hill Book Churchill, V., Fourier Series and Boundary Bochner,
Co.,Inc.,New
York
Franklin, P., Fourier reprinted by Dover
(1941).)
Publications,
Goldberg, R. R., Fourier D., Fourier
Transforms,
Jackson,
No. 6,
Monograph
R. L.,
Jeffrey,
Press,
edition,
Series,
W., Bessel Functions New York (1955).
Oxford
Transforms,
E. C., Introduction University
Press,
of Toronto
For Engineers,
second
H. Cohn
by
York
(1949), (1961).)
Mathematical
Press, Toronto (1956). edition,
and F.
Oxford Univer-
Steinhardt,
second
(1959).
McGraw-Hill
New York
York
New
(1941).
University
to the
Co., Inc.,
University Press, New Polynomials, Carus
Cambridge
Orthogonal
Math. Assoc. America
Trigonometric
Sneddon, I. A., Fourier Titchmarsh,
ential
and
Fourier Series, translated W.W., Chelsea Publishing Co., New York
Rogosinski,
Titchmarsh,
Series
N.
McLachlan, sity
Book McGraw-Hill New York.) Inc.,
Methods,
Theory of
Book Co., Fourier
Inc., New
Integrals,
York
second
(1951).)
edition,
(1948).)
C., Eigenfunction Expansions Associated with Second-Order Equations, Oxford University Press, New York, Part I (1946), Part E.
on the Theory of BesselFunctions, second Watson, G. N., A Treatise Cambridge University Press, New York (1945). and Certain of Its Applications, Wiener, N., The Fourier Integral Cambridge New York (1933), reprinted Press, sity by Dover Publications, Inc., New in two volumes, CambridgeUniversity A., Trigonometric Series, Zygmund, New York (1959).)
33I))
Di\357\254\202'er-
II (1958).
edition,
UniverYork.) Press,
INDEX)
A)
C)
97 of summation, 162 functions, 8 Absolutely integrable Fourier coe\357\254\202icients of, 71
Cauchy inequality, 63 value, 189 Cauchy principal
Abel\342\200\231s lemma,
Abel\342\200\231s method
54 Completenesscondition, Complete systems, 54-60
Antinode, 272
criterion
of functions
Approximation
by polynomials,
120-122 Arithmetic
method
means: of,
58-60
for,
properties of, 57-58 in the mean, 55 Convergence Convolution of two functions, 196
156-162)
Corners,
18
Cosine series,
23
Cosinetransform,
B)
Basic trigonometric system, 10,41,175-177 of, 117-118 completeness of, 119-120 important consequences, in two variables, 175-177 154 Bernstein\342\200\231s theorem, 197 Bessel\342\200\231s equation,
general solution of,
D)
Derivative,
203-204
207-208 of negative order,
of the first
Discontinuity
of the
E)
Eigenfunction theory
method:
of, 245-267
de\357\254\201nition of,
Eigenvalues:
213-215
de\357\254\201nition of,
247
existence of, 250-251 of, 251, 253 reality sign of, 254-255
54, 174
245 ff. value problems, 247 ff. solutions of, 261-264 generalized conditions,
Euler\342\200\231s constant,
Boundary
204
Euler's formula, 33 of a Even extension Even function, 21)
inhomogeneous,264-266 inequality,
247
orthogonality of, 251-254
consequencesof, 66, 223
Buniakovski
kind, 17 second kind, 17)
Discontinuity
Eigenfunctions:
relations between,205-207
Boundary
73
applicationsof, 268-318
202-203 of nonnegative order, 198-201 of the second kind, 203-205 of, 216-218 orthogonality
Bessel\342\200\231s inequality,
73
left-hand,
right-hand,
parametric form of, 215 Bessel functions, 197-220 formulas for, 208-213 asymptotic of integrals involving, 218-220 evaluation of half-integral order, of the \357\254\201rst kind,
zerosof,
192)
51) 333))
function,
23
mnex)
334
G)
F)
221 coef\357\254\201cients,
Fourier-Bessel Fourier-Bessel
Gamma
of, 228-234
order of magnitude
220-244
series,
criteria for convergence
of, 221-223
differentiation
of, 234-237
of functions of the second
on [0,l], de\357\254\201ned
65)
H)
241-243 type, 237-241 of, 225-228 convergence
uniform
Fourier
calculation
of functions Fourier
integral,
series and
addition
(trigonometric), subtraction of,
means of 157-158
arithmetic
partial
complexform conditions for convergence
of, 19,
75-89,
178
with
different of,
periods, 180 by using functions
105-112 complex variable, improving the convergenceof,
of
a
of,
on,
operations
partial sums
products of, Fourier
of, 73
inverse,
9
differentiation of, 10 integration
of, 80-89
Fourier transform, 190-193
Fourier
convergent,
for, 73
of, 10
sum of, 9
_
series with respectto eigenfunctions, 255
Functions
uniformly
convergent,
21,35-38,94
coe\357\254\201icients of,
Fourier series of, 35 Fundamental mode, 272)
35
9
Initial conditions, 246 11\342\200\230. by parts, 8) Integration
190
of period
para-
uniformly convergent,182-185 In\357\254\201nite series of functions, 9 fl\342\200\230.
123-125
convergence
on a
of, 184
differentiation
of, 155-171
summation
integrals depending 182-185
continuity of, 183 number, 123
115-154
formula
integral
uniform
by a
40
276)
meter,
integration of, 125-129 list of, 147-150 multiplication
306-310
I)
Improper
144
tem-
variable speci\357\254\201ed
\357\254\202ow in an in\357\254\201nite rod,
Hooke\342\200\231s law,
double, 173-180 evaluation
with Heat
at
peratures, 301 ends at zero temperature,297-299
Holder (Lipschitz)condition,
of, 129-143
differentiation
ends
with
for, 158 of, 32-34
formula
integral
310-316
whosesurface exchanges heat, 312-313 whose surface is insulated, 310-312 in a \357\254\201nite Heat \357\254\202ow rod, 296-306 1-40, 66-196 with ends at constant 299temperature, 122-123 300 sums of, with ends exchanging heat, 301-306
proof of, 188 Fourier
Heat capacity, 297 \357\254\202ow Heat in a circular cylinder, steady-state, 313-316
of, 189-190
forms
different
3
of, 7
superposition
174
182
theorem, 182
of, 3
angular frequency initial phase of,
150-152
of,
of two variables, 180-196
de\357\254\201nition of,
of, 3
amplitude
34
complex,
2-6
Harmonics,
13 ff. coe\357\254\201icients,
approximate
96
Gram-Schmidt orthogonalization process,
221
de\357\254\201nition of,
199, 201-202
function,
Gibb\342\200\231s phenomenon,
J)
Jump
discontinuity,
Jump of a function,
17 17))
nmnx L
335
R)
polynomials, 65)
Legendre
points,
Regular
108
Restoringforce, 3 17
left-hand.
Linearly
independent
Liouville,
1., 258
functions,
64, 251
Localization principle, 90
S)
Scalar
A. M., 119)
Lyapunov,
70)
lemma,
Riemann-Lebesgue
17
right-hand,
product,
Schwarz
61, 65 51, 63
inequality,
246 Separation of variables, Simple harmonic motion, 3
M)
Sine
error, 51-53
Mean square
of, 52-53
minimum
of elasticity
modulus), (Young\342\200\231s
276) N)
Nodal
lines, 286-288,
Nodes,
295-296
272
Norm of a
42,
function,
105
Singularity,
Membrane, 282 Modulus
24
series,
Sine transform, 192 Smooth
functions, 18
Spectral
function,
of sines,
173)
by
0)
by
Operational
24
of Fourier
complete with weight r, 256 linearly independent,64
41-65, 173-174
systems,
complete,54, 174 examples
54-60, 64
complete,
Orthogonal functions, 12, 41 Orthogonal
series, 155-171
method, 164-170 method of arithmetic means (Cesaro\342\200\231s 156-162 method), Abel's
System of functions:
192
calculus,
vector
for,
analogy
60-63)
of, 44-50
Fourier coellicients with respect to, 43 Fourier series with respect to, 43
T)
normalized,42, 173 in two variables, 173-174
Thermal
conductivity, 296
Timbre,
273
Triangle
inequality, 64
Orthonormal
system.
42
Overtones, 273)
limits of, 67-71
integrals,
Trigonometric
Trigonometricpolynomials, approximation
P)
Trigonometric Parseval\342\200\231s theorem,
119, 177
Period, 1 standard,
50
98
Summability
Odd extensionof a function, Odd function, 21
194
Square integrable functions, Standing wave, 272 Steklov, V. A., 258 J. C. F., 258 Sturm, Sum of consines, 71
6
of functions by, 115-117 series with decreasing co-
97-114 e\357\254\202icients,
convergence
of,
100-105)
8
Periodic extensions, 15 fl\342\200\230. Periodic functions, de\357\254\201nition of, 1 properties of, 1-2 Piecewise smooth functions, 18 Poisson's 164) kernel,
U)
Unbounded functions, Fourier expansions
of, 91-94
Uniform
convergence,
9))
mnnx)
336
of
Vibrations
V)
a rectangular
282-288
Vibrating
string, 269-275 of, 273-275
forcedvibrations
free vibrations
of, 269-273
of a rod forced, 280-282
Vibrations
(longitudinal),
membrane, 277-282
free, 277-280)
Vibrations:
characteristic,
274
forced, 264
W)
free, 264 of
Vibrations
a circular
296 general
case,
membrane, 288-
Weight
(function),
217
Weierstrass\342\200\231 approximation
291-296
radial, 288-291)
Weierstrass\342\200\231 M-test,
Wronskian,
252))
10
theorem,
120