Mathematical Methods for Physics and Engineering : A Comprehensive Guide

INTEGRAL EQUATIONS

Substituting (23.31) into (23.30) we find

y(x)=f(x)+f ̃(x)+y(−x),

but on changingxto−xand substituting back in fory(−x), this gives

y(x)=f(x)+ ̃f(x)+f(−x)+ ̃f(−x)+y(x).

Thus, in order for a solution to exist, we require that the functionf(x)obeys

f(x)+f ̃(x)+f(−x)+ ̃f(−x)=0.

This is satisfied iff(x)=− ̃f(x), i.e. if the functional form off(x) is minus the form of its
Fourier transform. We may repeat this analysis for the caseλ=− 1 /

√

2 π, and, in a similar
way, we find that this time we requiref(x)= ̃f(x).
In our casef(x)=exp(−x^2 /2), for which, as we mentioned above,f(x)= ̃f(x).
Therefore, (23.28) possesses no solution whenλ=+1/

√

2 πbut has many solutions when
λ=− 1 /

√

2 π.

A similar approach to the above may be taken to solve equations with kernels

of the formK(x, y)=cosxyor sinxy, either by considering the integral overyin

each case as the real or imaginary part of the corresponding Fourier transform

or by using Fourier cosine or sine transforms directly.

23.4.3 Differentiation

A closed-form solution to a Volterra equation may sometimes be obtained by

differentiating the equation to obtain the corresponding differential equation,

which may be easier to solve.

Solve the integral equation

y(x)=x−

∫x

0

xz^2 y(z)dz. (23.32)

Dividing through byx,weobtain

y(x)

x

=1−

∫x

0

z^2 y(z)dz,

which may be differentiated with respect toxto give

d

dx

[

y(x)

x

]

=−x^2 y(x)=−x^3

[

y(x)

x

]

.

This equation may be integrated straightforwardly, and we find

ln

[

y(x)

x

]

=−

x^4

4

+c,

wherecis a constant of integration. Thus the solution to (23.32) has the form

y(x)=Axexp

(

−

x^4

4

)

, (23.33)

whereAis an arbitrary constant.
Since the original integral equation(23.32) contains no arbitrary constants, neither
should its solution. We may calculate the value of the constant,A, by substituting the
solution (23.33) back into (23.32), from which we findA=1.