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17.2 The Heat Equation on [0, L] 623

0.5
0.5
0.4
0.4
0.3
0.3

0.2
0.2

0.1
0.1

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 17.4 t = 1.2. FIGURE 17.5 t = 1.3.

EXAMPLE 17.5
2
A homogeneous bar of length π has initial temperature function f (x) = x cos(x/2) and ends
maintained at temperature zero. The temperature distribution function satisfies
2
∂u ∂ u
= k for 0 < x <π,t > 0,
∂t ∂x 2
u(0,t) = u(π,t) = 0for t > 0,
and
2
u(x,0) = x cos(x/2) for 0 ≤ x ≤ π.
The solution is
2 π 2
∞
2
u(x,t) = ξ cos(ξ/2)sin(nξ)dξ sin(nx)e −n kt
π
n=1 0
3
n
∞
n
4 16πn(−1) − 64πn (−1) − 48n − 64n 3 2
= sin(nx)e −n kt .
2
4
6
π 64n − 48n + 12n − 1
n=1
To gauge the effect of the diffusivity constant k on the solution, Figure 17.6 shows graphs of
y = u(x,t) for t = 0.2 and for k = 0.3,0.6, 1.1, and 2.7. Figure 17.7 has the graphs for the same
values of k,but t = 1.2. For each k, the temperature function decays with time, as we expect.
However, for each time the temperature function has a smaller maximum as k increases.
EXAMPLE 17.6
We will examine the effects on u(x,t), depending on whether the ends of the bar are kept at
2
temperature zero, or are insulated. Suppose the initial temperature function is f (x) = x (π − x)
and L = π and k = 1/4. For the ends at temperature zero, we find that

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October 14, 2010 15:25 THM/NEIL Page-623 27410_17_ch17_p611-640