Page 633 - Advanced_Engineering_Mathematics o'neil

17.2 The Heat Equation on [0, L] 613

Put u(x,t) = X(x)T (t) into the heat equation to obtain
X T
= =−λ
X kT
in which λ is the separation constant. Now,
u(0,t) = X(0)T (t) = 0 = u(L,t) = X(L)T (t)
for all t ≥ 0, so X(0) = X(L) = 0, and the problem for X is
X + λX = 0; X(0) = X(L) = 0

2
with eigenvalues λ n = n π /L and eigenfunctions sin(nπx/L).
2
2
It is in the time dependence that the heat and wave equations differ. The equation for T is
2
2
n π k

T + T = 0,
L 2
2 2
which is a first-order differential equation with solutions that are constant multiples of e −n π kt/L 2 .
For n = 1,2,···, we have functions
nπx 2 2 2

u n (x,t) = c n sin e −n π kt/L ,
L
which satisfy the heat equation and the boundary conditions u(0,t) = u(L,t) = 0. To satisfy the
initial condition, we must (depending on f ) use a superposition
∞

nπx 2 2 2
u(x,t) = c n sin e −n π kt/L
L
n=1
and choose the coefficients so that
∞

nπx
u(x,0) = f (x) = c n sin .
L
n=1
This is the Fourier sine expansion of f on [0, L], so choose
2 L nπξ
c n = f (ξ)sin dξ.
L 0 L
With this choice of the c n ’s, the solution is
2 L −n π kt/L 2
∞
2 2
u(x,t) = f (ξ)sin(nπξ/L)dξ sin(nπx/L)e . (17.2)
L 0
n=1
EXAMPLE 17.1
Suppose the ends are kept at zero temperature and the initial temperature is f (x) = A, which is
constant. Compute
2 L 2A n
c n = A sin(nπξ/L)dξ = [1 − (−1) ].
L 0 nπ
The solution is
∞
2A 1 − (−1) n nπx 2 2 2
u(x,t) = sin e −n π kt/L .
π n L
n=1
Since

2if n is odd
n
1 − (−1) =
0if n is even,

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October 14, 2010 15:25 THM/NEIL Page-613 27410_17_ch17_p611-640