Mathematical Methods for Physics and Engineering : A Comprehensive Guide

23.8 Exercises

thus Hermitian. In order to solve this inhomogeneous equation using SH theory, however,
we must first find the eigenvalues and eigenfunctions of the corresponding homogeneous
equation.
In fact, we have considered the solution of the corresponding homogeneous equation
(23.13) already, in subsection 23.4.1, where we found that it has two eigenvaluesλ 1 =2/π
andλ 2 =− 2 /π, with eigenfunctions given by (23.16). The normalised eigenfunctions are

y 1 (x)=

1

√

π

(sinx+cosx)andy 2 (x)=

1

√

π

(sinx−cosx) (23.58)

and are easily shown to obey the orthonormality condition (23.49).
Using (23.54), the solution to the inhomogeneous equation (23.57) has the form
y(x)=a 1 y 1 (x)+a 2 y 2 (x), (23.59)

where the coefficientsa 1 anda 2 are given by (23.53) withf(x)=sin(x+α). Therefore,
using (23.58),

a 1 =

1

1 −πλ/ 2

∫π

0

1

√

π

(sinz+cosz) sin(z+α)dz=

√

π

2 −πλ

(cosα+sinα),

a 2 =

1

1+πλ/ 2

∫π

0

1

√

π

(sinz−cosz) sin(z+α)dz=

√

π

2+πλ

(cosα−sinα).

Substituting these expressions fora 1 anda 2 into (23.59) and simplifying, we find that
the solution to (23.57) is given by

y(x)=

1

1 −(πλ/2)^2

[

sin(x+α)+(πλ/2) cos(x−α)

]

.

23.8 Exercises

23.1 Solve the integral equation
∫∞

0

cos(xv)y(v)dv=exp(−x^2 /2)

for the functiony=y(x)forx>0. Note that forx<0,y(x) can be chosen as
is most convenient.
23.2 Solve
∫∞

0

f(t)exp(−st)dt=

a

a^2 +s^2

.

23.3 Convert

f(x)=expx+

∫x

0

(x−y)f(y)dy

into a differential equation, and hence show that its solution is

(α+βx)expx+γexp(−x),
whereα,βandγare constants that should be determined.
23.4 Use the fact that its kernel is separable, to solve fory(x) the integral equation

y(x)=Acos(x+a)+λ

∫π

0

sin(x+z)y(z)dz.

[ This equation is an inhomogeneous extension of the homogeneous Fredholm

equation (23.13), and is similar to equation (23.57). ]