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One-Dimensional Heat Equation

The transformed equation is

2
d U
− sU = 0,
dx 2
whose solution is given by
√
U(x, s) = c e − sx
2
in view of condition 3). By 2),

U(0, s) = L(f(t)) = F(s),

so that c = F(s) and √
2
U(x, s) = F(s)e − sx .

Invoking Theorem 5 and the convolution theorem 2, we have

∫
t x
2
u(x, t) = √ e −x /4τ f(t − τ)dτ.
0 2 πτ 3
2
2
Making the substitution σ = x /4τ, we find that
∫ ( )
2 ∞ 2 x 2
u(x, t) = √ e −σ f t − dσ,
√
π x/2 t 4σ 2
.
which is the desired solution.

Example 5.22, Solve

2
∂ u ∂u
= , 0 < x < l, t > 0,
∂x 2 ∂t

for
+
1) u(x, 0 ) = u ,
0
2) ∂ u(0, t) = 0 (i.e., left end insulated),
∂x
3) u(l, t) = u .
1
. . . . .
.
Solution. Taking the Laplace transform gives

2
d U
= sU − u .
0
dx 2
Then
√ √ u 0
U(x, s) = c cosh sx + c sinh sx ,
1
2
s
and by 2), c = 0, so that
2
√ u 0
U(x, s) = c cosh sx + .
1
. . . . s

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