Page 90 - 4811

Laplace Transform Method for Partial differential equations
.

Example 5.16, Solve the integral equation

t
∫
κ(t) = e −t + sin(t − τ)κ(τ)dτ.
0
.
. . . . .
Solution. We apply the Laplace transform to both sides of the equation so
that
−t
L(κ(t)) = L(e ) + L(sin t)L(κ(t))
and
−t
2
L(e ) s + 1 2 1 1
L(κ(t)) = = = + − .
2
1 − L(sin t) s (s + 1) s + 1 s 2 s
Hence
κ(t) = 2e −t + t − 1.

. . . .

Laplace Transform Method for Partial differential

equations

Partial differential equations, like their one-variable counterpart, ordinary
differential equations, are ubiquitous throughout the scientific spectrum.

However, they are, in general, more difficult to solve. Yet here again, we may
apply the Laplace transform method to solve PDE’s by reducing the initial
problem to a simpler ODE.
Partial differential equations come in three types. For a function of two
variables u = u(x, y), the general second-order linear PDE has the form

2
2
2
∂ u ∂ u ∂ u ∂u ∂u
a + 2b + c + d + e + fu = g, (5.8)
∂x 2 ∂x∂y ∂y 2 ∂x ∂y
where a, b, c, d, e, f, g may depend on x and y only. We call (5.8)
2
• elliptic if b − ac < 0,
2
• hyperbolic if b − ac > 0,
2
. • parabolic if b − ac = 0.

Example 5.17,
1. The heat equation

2
∂u ∂ u
= c
∂t ∂x 2
is parabolic.
. . . . .

89