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Laplace Transform Method for Partial differential equations

Note that in the present context the derivative theorem 2 reads

( )
∂u
+
+
L = sL(u(x, t)) − u(x, 0 ) = sU(x, s) − u(x, 0 ).
∂t
The Laplace transform method applied to the solution of PDEs consists of first
applying the Laplace transform to both sides of the equation as we have done
before. This will result in an ODE involving U as a function of the single variable

x.
For example, if
∂u ∂u
= , (5.11)
∂x ∂t

then
( ) ( )
∂u ∂u
L = L
∂x ∂t
implying
d
+
U(x, s) = sU(x, s) − u(x, 0 ), (5.12)
dx
The ODE obtained is then solved by whatever means avail themselves. If, say,
+
u(x, 0 ) = x for equation (5.11), we find that the general solution is given by

x 1
sx
U(x, s) = ce + + . (5.13)
s s 2
PDEproblemsinphysicalsettingscomewithoneormoreboundaryconditions,
say for (5.11) that

u(0, t) = t. (5.14)

Since the boundary conditions also express u as a function of t, we take the rather
unusual step of taking the Laplace transform of the boundary conditions as well.
So for (5.14)
1
U(0, s) = L(u(0, t)) = .
s 2
Feeding this into (5.13) gives c = 0 so that

x 1
U(x, s) = + .
s s 2

Since this is the transform of the desired function u(x, t), inverting gives the
+
solution to (5.11) and (5.14) [(and u(x, 0 ) = x]:

u(x, t) = x + t.

This simple example illustrates the basic techniques involved in solving partial
differential equations.

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