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r
T
If the string has density ρ and tension T , then we have seen above that c = .
ρ
So to increase the frequency of oscillation of a string you increase the tension and/or
decrease the density and/or shorten the string.
Example 5.1 Solve the following IBVP:
for all 0 < x < 1 and t > 0
u tt = u xx
u(0, t) = u(1, t) = 0 for all t > 0
u(x, 0) = x(1 − x) for all 0 < x < 1
u t (x, 0) = 0 for all 0 < x < 1.
This is a special case of equations (8.1)–(8.3) with l = 1, ϕ(x) = x(1−x) and ψ(x) = 0.
So, by (8.12),
∞
X
u(x, t) = (A n cos (πnt) + B n sin (πnt)) sin (πnx)
n=1
with
1
Z
A n = 2 x(1 − x) sin (πnx) dx, B n = 0.
0
Using the integration by parts formula

b b
Z Z
b

u dv = uv − v du
a
a a
we obtain:
1
Z

2
2

A n = 2 (x − x ) sin (πnx) dx = u = x − x ; dv = sin (πnx) dx; du = (1 − 2x)dx;
0
 
1
Z

1 1 2 1 1

v=− cos (πnx) =2  − (x − x ) cos (πnx) + (1 − 2x) cos (πnx) dx  =
πn πn 0 πn
0
1
Z
(2 − 4x) sin (πnx)

= cos (πnx)dx=u=1 − 2x; dv=cos (πnx)dx; du=−2dx; v= =
πn πn
0

1
Z
2 1 1 2 4 1

= (1 − 2x) sin (πnx) + sin (πnx) dx  = − cos (πnx) =
πn πn 0 πn π n 0
3 3
0

8
, n = 2k − 1

3
= π (2k − 1) 3
0, n = 2k

Thus,
∞
8 X cos (πkt) sin (πkx)
u(x, t) = .
π 3 (2k − 1) 3
k=1
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