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Solutions to Exercises

of x) in the limit t → ∞. Thus, lim u(x, t) = A 0 . It is instructive to compute
t→∞ 2
A by an alternative method. Notice that
0
Z L Z L Z L
d
u(x, t)dx ≡ u (x, t)dx ≡ k u (x, t)dx =
t
xx
dt 0 0 0
= k[u (L, t) − u (0, t)] = 0,
x
x
where the last equality follows from the Neumann boundary condition. Hence,

L L L
Z Z Z
u(x, t)dx = u(x, 0)dx = f(x)dx
0 0 0
holds for all t > 0. Since the uniform convergence of the series implies the

convergence of the integral series, you can infer A 0 = 1 R L f(x)dx.
2 L 0
L
R
A physical interpretation The quantity u(x, t)dx was shown to be con-
0
served in a one-dimensional insulated rod. The quantity ku (x, t) measures
x
the heat flux at a point x and time t. The homogeneous Neumann condition
amounts to stating that there is zero flux at the rod’s ends. Since there are
no heat sources either (the equation is homogeneous), the temperature tends
to equalize its gradient, and therefore it converges to a constant temperature,

such that the total stored energy is the same as the initial energy.

Exercise 192. To obtain a homogeneous equation write u = v+w, where w = w(t)
satisfies w − kw xx = A cos αt, w(x, 0) ≡ 0. Therefore, w(t) = A sin αt. Solving
t
α
2
1
for v, the complete solution is u(x, t) = 3 + 2 cos 2πxe −4kπ t + A sin αt.
2 / α

Exercise 194. a
∞
n
2
P (−n +h)t 2 R π 4((−1) −1)
(a) u(x, t)= B e sin nx, where Bn= π 0 x(π−x) sin nxdx= πn 3 .
n
n=1
(b) The limit lim u(x, t) exists if and only if h ≤ 1. When h < 1 the series
t→∞
converges uniformly to 0. If h = 1, the series converges to B sin x.
1

∞ 2 2 R 1
P
Exercise 197. u(x, t)=e −t B e −(2n+1) tπ /4 sin 2n+1 πx , where B = 2 0 x(2−
n
n
2
n=0

x) sin 2n+1 πx = 32 .
3 3
2 (2n+1) π

∞
a 0 P r n

Exercise 207. u(r, θ) = + (a cos nθ + b sin nθ) ,
n
n
2 n=1 a
π
1 R
a = f(θ) cos nθ dθ = 0 f(θ) is odd
n
π −π
108