Chapter 5 483
The first three nonzero terms are, fora=b, those with(m,n)=
( 1 , 1 ), ( 1 , 3 )=( 3 , 1 ), ( 3 , 3 ). All terms with an even index are 0.u(a/ 2 ,a/ 2 ,t)∼=16 T
π^2(
e−^2 τ−2
3 e− 10 τ+^1
9 e− 18 τ)
,
whereτ=ktπ^2 /a^2.5.u(r)=(
a^2 −r^2)
/2andu(r)=∑∞
1CnJ 0 (λnr), withCn=^2 a2
αn^3 J 1 (αn).
7.w(x,t)=a 0 +∑∞
n= 1ancos(λnx)exp(
−λ^2 nkt)
,
v(y,t)=∑
bmsin(μmy)exp(
−μ^2 mkt)
,
whereμm=mπ/b,λn=nπ/a, and initial conditions are
v(y, 0 )= 1 , 0 <y<b; w(x, 0 )=Tx/a, 0 <x<a.9.J 0 (λr)exp(−λ^2 kt).11.Bk=bk/k(k+ 1 )fork= 1 , 2 ,...;b 0 must be 0, andB 0 is arbitrary.
((
1 −x^2)
y′)′
− m2
1 −x^2 y+μ(^2) y=0.
15.u(r,z)=
∑∞
n= 1ansinhsinh(λ(λnz)
nb)J 0 (λnr),whereλn=αn
aandan=^2 U^0
αnJ 1 (αn).
17.u(r,z,t)=sin(μz)J 0 (λr)sin(νct)is a product solution ifμ=mπ/b,λ=
αn/a,andν=
√
μ^2 +λ^2. The frequencies of vibration are thereforeνc
orc√(
mπ
b) 2
+
(
αn
a) 2
.
- Each of the two terms satisfies∇^2 φ=−( 5 π^2 )φ.Ony=0andx=1,
both terms are 0; ony=xthey are obviously equal in value, opposite in
sign. - Each term satisfies∇^2 φ=−( 16 π^2 / 3 )φ.
Ony=0,φ=sin( 2 nπx)−sin( 2 nπx);
ony=√
3 x,φ=sin( 4 nπx)+ 0 −sin( 2 nπ· 2 x);