Mathematical Methods for Physics and Engineering : A Comprehensive Guide

21.4 INTEGRAL TRANSFORM METHODS

21.4 Integral transform methods

In the method of separation of variables our aim was to keep the independent

variables in a PDE as separate as possible. We now discuss the use of integral

transforms in solving PDEs, a method by which one of the independent variables

can be eliminated from the differential coefficients. It will be assumed that the

reader is familiar with Laplace and Fourier transforms and their properties, as

discussed in chapter 13.

The method consists simply of transforming the PDE into one containing

derivatives with respect to a smaller number of variables. Thus, if the original

equation has just two independent variables, it may be possible to reduce the

PDE into a soluble ODE. The solution obtained can then (where possible) be

transformed back to give the solution of the original PDE. As we shall see,

boundary conditions can usually be incorporated in a natural way.

Which sort of transform to use, and the choice of the variable(s) with respect

to which the transform is to be taken, is a matter of experience; we illustrate this

in the example below. In practice, transforms can be taken with respect to each

variable in turn, and the transformation that affords the greatest simplification

can be pursued further.

A semi-infinite tube of constant cross-section contains initially pure water. At timet=0,

one end of the tube is put into contact with a salt solution and maintained at a concentration

u 0. Find the total amount of salt that has diffused into the tube after timet, if the diffusion

constant isκ.

The concentrationu(x, t)attimetand distancexfrom the end of the tube satisfies the
diffusion equation

κ

∂^2 u

∂x^2

=

∂u

∂t

, (21.71)

which has to be solved subject to the boundary conditionsu(0,t)=u 0 for alltand
u(x,0) = 0 for allx>0.
Since we are interested only int>0, the use of the Laplace transform is suggested.
Furthermore, it will be recalled from chapter 13 that one of the major virtues of Laplace
transformations is the possibility they afford of replacing derivatives of functions by simple
multiplication by a scalar. If the derivative with respect to time were so removed, equation
(21.71) would contain only differentiation with respect to a single variable. Let us therefore
take the Laplace transform of (21.71) with respect tot:
∫∞

0

κ

∂^2 u

∂x^2

exp(−st)dt=

∫∞

0

∂u

∂t

exp(−st)dt.

On the LHS the (double) differentiation is with respect tox, whereas the integration is
with respect to the independent variablet. Therefore the derivative can be taken outside
the integral. Denoting the Laplace transform ofu(x, t)by ̄u(x, s) and using result (13.57)
to rewrite the transform of the derivative on the RHS (or by integrating directly by parts),
we obtain

κ

∂^2 ̄u

∂x^2

=s ̄u(x, s)−u(x,0).

But from the boundary conditionu(x,0) = 0 the last term on the RHS vanishes, and the