Page 639 - Advanced_Engineering_Mathematics o'neil

17.2 The Heat Equation on [0, L] 619

The problem for U is:
2
∂U ∂ U
= k for 0 < x < L,t > 0,
∂t ∂x 2
U(0,t) = U(L,t) = 0for t > 0,
and
1
U(x,0) = f (x) − (T 2 − T 1 )x − T 1 for 0 ≤ x ≤ L.
L
We know the solution of this problem for U:

∞
2 nπx 2 2 2
U(x,t) = c n sin e −n π kt/L
L L
n=1
where
2 L 1
nπξ
c n = f (ξ) − (T 2 − T 1 )ξ − T 1 sin dξ.
L 0 L L
The solution of the original problem is
1
u(x,t) = U(x,t) + (T 2 − T 1 )x + T 1 .
L
As a specific example, suppose T 1 = 1, T 2 = 2 and f (x) = 3/2. Now
1
ψ(x) = x + 1,
L
and
2 L 3 1
nπξ
c n = − x − 1 sin dξ
L 0 2 L L
2 L 1 1 nπξ
= − x sin dξ
L 0 2 L L
1 + (−1) n
= .
nπ
The solution is
∞ n
1 + (−1) nπx 1
2 2
−n π kt/L 2
u(x,t) = sin e + x + 1.
nπ L L
n=1
Sometimes an initial-boundary value problem involving the heat equation can be trans-
formed into a simpler problem by multiplying by an exponential function e αx+βt and making
appropriate choices for α and β. We will pursue this idea in the problems.
17.2.5 The Heat Equation with a Source Term

We will determine the solution of the initial-boundary value problem
2
∂u ∂ u
= k + F(x,t) for 0 < x < L,t > 0,
∂t ∂x 2
u(0,t) = u(L,t) = 0for t > 0,
and
u(x,0) = f (x) for 0 ≤ x ≤ L.

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October 14, 2010 15:25 THM/NEIL Page-619 27410_17_ch17_p611-640