Page 634 - Advanced_Engineering_Mathematics o'neil

614 CHAPTER 17 The Heat Equation

we can omit the even values of n in the summation to write the solution
∞
4A 1 (2n − 1)πx 2 2 2
u(x,t) = sin e −(2n−1) π kt/L .
π 2n − 1 L
n=1
17.2.2 Insulated Ends

Suppose the bar has insulated ends, hence no energy is lost across the ends. The temperature
distribution is modeled by the initial-boundary value problem
2
∂u ∂ u
= k for 0 < x < L,t > 0,
∂t ∂x 2
∂u ∂u
(0,t) = (L,t) = 0for t ≥ 0,
∂x ∂x
and
u(x,0) = f (x) for 0 ≤ x ≤ L.
As before, separation of variables yields

X + λX = 0 and T + λkT = 0.

The insulation conditions give us
∂u ∂u

(0,t) = X (0)T (t) = (L,t) = X (L)T (t) = 0,
∂x ∂x
so X (0) = X (L) = 0, and the problem for X is

X + λX = 0; X (0) = X (L) = 0.

2
2
2
Previously, we solved this problem for the eigenvalues λ n = n π /L and eigenfunctions
cos(nπx/L) for n = 0,1,2,···.
The equation for T is
2
2
n π k

T + T = 0.
L 2
For n = 0, we get T 0 (t) = constant. For n = 1,2,···,
2 2
T n (t) = e −n π kt/L 2 ,
or constant multiples of this function. We now have a function
nπx 2 2 2

u n (x,t) = c n cos e −n π kt/L ,
L
which satisfies the heat equation and the insulation boundary conditions for n = 0,1,2,···.To
satisfy the initial condition, use the superposition
∞
1 nπx −n π kt/L 2
2 2
u(x,t) = c 0 + c n cos e .
2 L
n=1
Then
∞
1 nπx
u(x,0) = f (x) = c 0 + c n cos ,
2 L
n=1
so choose the c n ’s to be the Fourier cosine coefficients of f on [0, L]:
2 L nπξ
c n = f (ξ)cos dξ.
L 0 L
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October 14, 2010 15:25 THM/NEIL Page-614 27410_17_ch17_p611-640