Mathematical Methods for Physics and Engineering : A Comprehensive Guide

23.4 CLOSED-FORM SOLUTIONS

23.4.2 Integral transform methods

If the kernel of an integral equation can be written as a function of the difference

x−zof its two arguments, then it is called adisplacementkernel. An integral

equation having such a kernel, and which also has the integration limits−∞to

∞, may be solved by the use of Fourier transforms (chapter 13).

If we consider the following integral equation with a displacement kernel,

y(x)=f(x)+λ

∫∞

−∞

K(x−z)y(z)dz, (23.17)

the integral overzclearly takes the form of a convolution (see chapter 13).

Therefore, Fourier-transforming (23.17) and using the convolution theorem, we

obtain

̃y(k)= ̃f(k)+

√

2 πλK ̃(k)y ̃(k),

which may be rearranged to give

̃y(k)=

f ̃(k)

1 −

√

2 πλK ̃(k)

. (23.18)

Taking the inverse Fourier transform, the solution to (23.17) is given by

y(x)=

1

√

2 π

∫∞

−∞

̃f(k) exp(ikx)

1 −

√

2 πλK ̃(k)

dk.

If we can perform this inverse Fourier transformation then the solution can be

found explicitly; otherwise it must be left in the form of an integral.

Find the Fourier transform of the function

g(x)=

{

1 if|x|≤a,

0 if|x|>a.

Hence find an explicit expression for the solution of the integral equation

y(x)=f(x)+λ

∫∞

−∞

sin(x−z)

x−z

y(z)dz. (23.19)

Find the solution for the special casef(x)=(sinx)/x.

The Fourier transform ofg(x) is given directly by

̃g(k)=

1

√

2 π

∫a

−a

exp(−ikx)dx=

[

1

√

2 π

exp(−ikx)

(−ik)

]a

−a

=

√

2

π

sinka

k

.

(23.20)

The kernel of the integral equation (23.19) isK(x−z) = [sin(x−z)]/(x−z). Using
(23.20), it is straightforward to show that the Fourier transform of the kernel is

K ̃(k)=

{√

π/2if|k|≤1,

0if|k|>1.

(23.21)