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Appendix: All Solutions to the Problems
8.18 Let the vibration equation of a string be
(l + x)
∂
2 u
∂t 2 = a
2 ∂
2 u
∂ x 2
where, l is the length of the string, find the vibration frequency of the string when
both ends are fixed.
Solution Let u(x, t) = v(x) sin ωt, the vibration equation of a string is transformed
to
∂
2 v
∂ x 2 +
ω
2
a 2 (l + x)v = 0
Putting λ =
ω
2
a 2 , then above problem is converted into the characteristic problem
of the following differential equation
∂
2 v
∂ x 2 + λ(l + x)v = 0
v(0) = v(l) = 0
Let the approximate solution be
v 2 = a 1 sin
π
l
x + a 2 sin
2π
l
x
The second derivative and the residual are
v 2
= −a 1
π
2
l 2 sin
π
l
x − a 2
4π
2
l 2 sin
2π
l
x
R = v 2
+ λ(l + x)u 2 = −a 1
π
2
l 2 sin
π
l
x − a 2
4π
2
l 2 sin
2π
l
x
+ λ(l + x)
a 1 sin
π
l
x + a 2 sin
2π
l
x
The Galerkin equations are
1
0
R sin
nπ
l
xdx = 0 (n = 1, 2)
Substituting the concrete expression of into the above expression, the integral
results are the homogeneous linear equations about a 1 and a 2