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Solved Problems

Solution:
a) The mathematical statement of the problem is

2
u = c u , 0 < x < L, t > 0,
tt
xx
u(0, t) = u(L, t) = 0,
(
v for |x − ξ| < d
u(x, 0) = 0, u (x, 0) =
t
0 for |x − ξ| > d.

Because we are interest in the harmonics of the motion, we will solve this problem
with an eigenfunction expansion in x. We substitute the separation of variables

u(x, t) = X(x)T(t) into the wave equation.

00
2
00
XT = c X T
T 00 X 00
= = −λ 2
2
c T X
The eigenvalue problem for X is,

2
00
X = −λ X, X(0) = X(L) = 0,
which has the solutions,
nπ nπx
λ = , X = sin , n ∈ N.
n
n
L L
The ordinary differential equation for the T are,
n
nπc
2
00
T = − T ,
n
n
L
which have the linearly independent solutions,

nπct nπct
cos , sin .
L L

The solution for u(x, t) is a linear combination of the eigen-solutions.

∞

X nπx nπct nπct
u(x, t) = sin a cos + b sin
n
n
L L L
n=1
Since the string initially has zero displacement, each of the a are zero.
n
∞

X nπx nπct
u(x, t) = b sin sin
n
L L
n=1
Now we use the initial velocity to determine the coefficients in the expansion. Be-
cause the position is a continuous function of x, and there is a jump discontinuity

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