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Application of Laplace Transform

In what follows we will demonstrate the utility of the Laplace transform

method when applied to a variety of PDEs. However, be- fore proceeding further,
we require two more inverses based upon (4.28):

( √ ) ( )
e −a s a
−1
L = erfc √ , a > 0.
s 2 t

Theorem 5.3È

The following equalities are true:
√ 2
−1
1. L (e −a s ) = √ a e −a /4t (a > 0).
2 πt 3
( √ )
2
−1 e −a s 1 −a /4t
2. L √ = √ e (a > 0).
s πt
. . . . . . .
PROOF. 1. Applying the derivative theorem to (4.28) and noting that
√
+
erfc(a/2 t) → 0 as t → 0 , we have
( ( ))
d a √
L erfc √ = e −a s ,
dt 2 t

that is,
( )
a 2 √
L √ e −a /4t = e −a s , (5.15)
2 πt 3
as desired.
2. For 2, we differentiate (5.15) with respect to s,

√
( )
d a −a /4t ae −a s
2
L √ e = − √ ,
ds 2 πt 3 2 s
and by Theorem 2,

√
( ) −a s
at 2 ae
L − √ e −a /4t = − √ ,
2 πt 3 2 s

which after cancellation gives 2. 2
. . . .
.

Example 5.19, Solve the boundary-value problem

∂y ∂y
2
x + + ay = bx , (5.16)
∂x ∂t
+
where x > 0, t > 0, a, b be constants, y(0, t) = 0, y(x, 0 ) = 0.
. . . . .

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