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Application of Laplace Transform
.

for
+
1) u(x, 0 ) = 1, x > 0,
2) u(0, t) = 0, t > 0,
3) lim u(x, t) = 1.
x→∞
. . . . .
.

Solution. Taking the Laplace transform of (5.17) yields

2
d U
+
= sU − u(x, 0 ) = sU − 1. (5.18)
dx 2
Transforming the boundary conditions 2) and 3) gives

U(0, s) = L(u(0, t)) = 0,

1
lim U(x, s) = lim L(u(x, t)) = L( lim u(x, t)) = .
x→∞ x→∞ x→∞ s
Now (5.18) is an ODE whose solution is given by

√ √ 1
U(x, s) = c e sx + c e − sx + .
2
1
s
The boundary condition lim U(x, s) = 1/s implies c = 0, and U(0, s) = 0
1
x→∞
implies
√
1 e sx
U(x, s) = − .
s s
By (4.27),
√
( ) ∫ x/2 t
x 2 2
u(x, t) = erf √ = √ e −u du.
2 t π 0
Direct calculation shows that u(x, t) indeed satisfies (5.17) and that the

. . . . initial and boundary conditions are satisfied.
.

Example 5.21, Solve

2
∂ u ∂u
= , x > 0, t > 0,
∂x 2 ∂t

for
+
1) u(x, 0 ) = 0,
2) u(0, t) = f(t), t > 0,

3) lim u(x, t) = 0.
x→∞
. . . . .

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