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854 Answers to Selected Problems

4
1
2
β k = ξ(ξ − 1)J 2 ( j 2k ξ) ,
[J 3 ( j 0k )] 2 0
p+1 1
4(−1)
δ pq = ξ J p ( j pq ξ)dξ .
pj pq [J p+1 ( j pq )] 2 0
Computing some of these terms, we have
z(r,θ,t) ≈
1.108022J 0 (1.202413r)cos(2.404826t) − 0.13977J 0 (2.760039r)cos(5.520078t)
+ 0.045476J 0 (4.32686r)cos(8.653728t) − 0.02099J 0 (5.895767r)cos(11.79153t)
+ 0.011636J 0 (7.465459t)cos(14.930918t) +···
+ cos(2θ)[−2.976777J 2 (2.567811r)cos(5.135622t)
− 1.434294J 2 (4.208622r)cos(8.417244t) +···]
+ sin(θ)[1.155175J 1 (1.915853r)sin(3.831706t)
− 0.14741J 1 (3.507794r)sin(7.015587t) + ···] + ··· .

Section 16.9 Vibrations in a Rectangular Membrane

n+1 2
8(−1) π 16 √
1 ∞ n 2
1. z(x, y,t) = + [(−1) − 1] sin(nx/2)sin(y)cos( n + 4t/2)
π n=1 n n 3
∞ ∞

3. z(x, y,t) = n=1 m=1 d nm cos((2n − 1)x/2)sin((2m − 1)y/2)sin(k nm t),
where
16
d nm = √
2
2
π (2n − 1)(2m − 1) (2n − 1) + (2m − 1) 2
and

2
2
k nm = (2n − 1) + (2m − 1) .
CHAPTER SEVENTEEN THE HEAT EQUATION
Section 17.1 Initial and Boundary Conditions
2
1. ∂u ∂ u
= k for t > 0,0 < x < L,
∂t ∂x 2
∂u
u(0,t)= (L,t) = 0for t > 0,
∂x
u(x,0)= f (x)
2
3. ∂u ∂ u
= k for t > 0,0 < x < L,
∂t ∂x 2
∂u
(0,t)= 0,u(L,t) = β(t) for t > 0,
∂x
u(x,0)= f (x)
Section 17.2 The Heat Equation on [0, L]
In these solutions, we sometimes use the notation exp(g(t)) = e g(t) .
8L 2
∞ 2 2 2

1. u(x,t) = sin((2n − 1)πx/L)exp(−(2n − 1) π kt/L )
n=1 3 3
(2n − 1) π
3.
∞

2
2
2
u(x,t) = d n sin((2n − 1)πx/L)exp(−3(2n − 1) π t/L )
n=1
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October 14, 2010 17:50 THM/NEIL Page-854 27410_25_Ans_p801-866