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622 CHAPTER 17 The Heat Equation

Finally we need

2 π 40 π/4 40 1 − cos(nπ/4)
b n = f (ξ)sin(nξ)dξ = sin(nξ)dξ = .
π 0 π 0 π π
With all the components in place, the solution is

2
∞ 2 −4n t
1 −1 + 4n t + e
n+1
u(x,t) = (−1) sin(nx)
8 n 5
n=1
∞
2
40 1 − cos(nπ/4) −4n t
+ sin(nx)e .
π n
n=1
The second term on the right is the solution of the problem with the xt term omitted. Denote this
“no-source” solution as
∞
40 1 − cos(nπ/4)
2
u 0 (x,t) = sin(nx)e −4n t .
π n
n=1
Then the solution with the source term is

2
∞ 2 −4n t
1 −1 + 4n t + e

u(x,t) = u 0 (x,t) + (−1) n+1 sin(nx).
8 n 5
n=1
Writing the solution in this way enables us to gauge the effect of the xt term on the solution.
Figures 17.2 through 17.5 compare graphs of u(x,t) and u 0 (x,t) at times t = 0.3,0.8, 1.2 and
1.3. In each figure, u 0 (x,t) falls below u(x,t). Both solutions decay to zero as t increases, but
the effect of the xt term is to retard this decay.
17.2.6 Effects of Boundary Conditions and Constants
We will investigate how constants and terms appearing in diffusion problems influence the
behavior of solutions.

1.2
1
1
0.8
0.8

0.6
0.6
0.4
0.4
0.2
0.2

0
0
0 0.5 1 1.5 2 2.5 3
0 0.5 1 1.5 2 2.5 3
x
x
FIGURE 17.3 t = 0.8.
FIGURE 17.2 u(x,t) and u 0 (x,t) at t = 0.3.

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October 14, 2010 15:25 THM/NEIL Page-622 27410_17_ch17_p611-640