Fourier Series and Partial Differential Equations Lecture Notes

Chapter 4. The wave equation 38
Example 4.1 Solve the IBVP for the case
Since
f(x) = Asin
πx
L
f(x) = Asin
+Bsin
πx
cos
L
πx
+
L
1
2 Bsin
πx
, g(x) = 0. (4.44)
L
2πx
L
, (4.45)
the solution is obtained by taking a1 = A, a2 = B/2, an = 0 for n ≥ 3 and bn = 0 ∀n to
give
πx
πct
y(x,t) = Asin cos +
L L
1
2 Bsin
2πx 2πct
cos .
L L
(4.46)
Example 4.2 Solve the IBVP for the case
Since
we take an = 0 ∀n and identify
πc
b1 =
L
3
4 ,
2πc
b2 = 0,
L
to arrive at
y(x,t) = 3L
4πc sin
f(x) = 0, g(x) = sin 3 πx
L
g(x) = 3
4 sin
πx
−
L
1
4 sin
πx
sin
L
4.4.1 Application of Fourier series
3πx
L
. (4.47)
, (4.48)
3πc
b3 = −
L
1
4 , bn = 0 for n ≥ 4, (4.49)
πct
−
L
L
12πc sin
3πx
sin
L
3πct
L
. (4.50)
To solve for more general initial conditions, we again look for a solution as a superposition
of normal modes:
∞
nπx
y(x,t) = sin
L
nπct nπct
ancos +bnsin , (4.51)
L L
n=1
so that we arrive at the problem: given f(x) and g(x) can they be expanded as Fourier
sine series
f(x) =
g(x) =
∞
nπx
ansin , 0 ≤ x ≤ L,
L
n=1
(4.52)
∞
nπc
nπx
bnsin , 0 ≤ x ≤ L?
L L
(4.53)
n=1
From the lectures on Fourier Series we know that such an expansion as (4.52) exists if e.g.
f and g are piecewise continuously differentiable on [0,L]. The coefficients are determined
by the orthogonality relations:
L
mπx
nπx
sin sin dx =
L L
0
0 m = n,
L m = n.
1
2
(4.54)