Fourier Series and Partial Differential Equations Lecture Notes

Chapter 3. The heat equation 29
From equation (3.34) we see that
T(x,0) =
∞
nπx
ansin . (3.36)
L
n=1
Comparing terms we see that a1 = 1, a2 = 1/2 and an = 0 (n ≥ 3) so that the solution is
πx
T(x,t) = sin e
L
−π2κt/L2 + 1
2 sin
2πx
e
L
−4π2κt/L2 . (3.37)
3.4.1 Application of Fourier series
To solve for more general initial conditions, we can use Fourier series to determine the
constants an:
∞
nπx
T(x,0) = ansin , 0 ≤ x ≤ L.
L
(3.38)
n=1
The question is now, given f(x), can it be expanded as a Fourier sine series
f(x) =
∞
nπx
ansin , 0 ≤ x ≤ L? (3.39)
L
n=1
From the lectures on Fourier series, we know that such an expansion exists if e.g. f is
piecewise continuously differentiable on [0,L]. The coefficients an are determined by the
orthogonality relations:
Thus
L
mπx
nπx
sin sin dx =
L L
0
0
0, m = n,
L, m = n.
1
2
(3.40)
an = 2
L
nπx
f(x)sin dx. (3.41)
L L
Example 3.2 Find the solution of the IBVP when
0 for 0 ≤ x ≤ L1 and L2 ≤ x ≤ L,
f(x) =
1 for L1 < x < L2.
Here f(x) has the Fourier sine expansion
2
π
∞
n=1
and the solution of IBVP is
T(x,t) = 2
π
∞
n=1
1
n
1
cos
n
cos
nπL1
−cos
L
nπL1
−cos
L
nπL2
sin
L
nπL2
sin
L
(3.42)
nπx
, (3.43)
L
nπx
e
L
−n2π2κt/L2 . (3.44)