Mathematical Methods for Physics and Engineering : A Comprehensive Guide

INTEGRAL EQUATIONS

Thus, using (23.18), we find the Fourier transform of the solution to be

̃y(k)=

{

̃f(k)/(1−πλ)if|k|≤1,

̃f(k)if|k|>1. (23.22)

Inverse Fourier-transforming, and writing the result in a slightly more convenient form,
the solution to (23.19) is given by

y(x)=f(x)+

(

1

1 −πλ

− 1

)

1

√

2 π

∫ 1

− 1

f ̃(k)exp(ikx)dk

=f(x)+

πλ

1 −πλ

1

√

2 π

∫ 1

− 1

̃f(k)exp(ikx)dk. (23.23)

It is clear from (23.22) that whenλ=1/π, which is the only eigenvalue of the
corresponding homogeneous equation to (23.19), the solution becomes infinite, as we
would expect.
For the special casef(x)=(sinx)/x, the Fourier transform ̃f(k) is identical to that in
(23.21), and the solution (23.23) becomes

y(x)=

sinx

x

+

(

πλ

1 −πλ

)

1

√

2 π

∫ 1

− 1

√

π

2

exp(ikx)dk

=

sinx

x

+

(

πλ

1 −πλ

)

1

2

[

exp(ikx)

ix

]k=1

k=− 1

=

sinx

x

+

(

πλ

1 −πλ

)

sinx

x

=

(

1

1 −πλ

)

sinx

x

.

If, instead, the integral equation (23.17) had integration limits 0 andx(so

making it a Volterra equation) then its solution could be found, in a similar way,

by using the convolution theorem for Laplace transforms (see chapter 13). We

would find

̄y(s)=

̄f(s)

1 −λK ̄(s)

,

wheresis the Laplace transform variable. Often one may use the dictionary of

Laplace transforms given in table 13.1 to invert this equation and find the solution

y(x). In general, however, the evaluation of inverse Laplace transform integrals

is difficult, since (in principle) it requires a contour integration; see chapter 24.

As a final example of the use of Fourier transforms in solving integral equations,

we mention equations that have integration limits−∞and∞and a kernel of the

form

K(x, z)=exp(−ixz).

Consider, for example, the inhomogeneous Fredholm equation

y(x)=f(x)+λ

∫∞

−∞

exp(−ixz)y(z)dz. (23.24)

The integral overzis clearly just (a multiple of) the Fourier transform ofy(z),