23.4 CLOSED-FORM SOLUTIONS
so we can write
y(x)=f(x)+√
2 πλ ̃y(x). (23.25)If we now take the Fourier transform of (23.25) but continue to denote the
independent variable byx(i.e. rather thank, for example), we obtain
̃y(x)= ̃f(x)+√
2 πλy(−x). (23.26)Substituting (23.26) into (23.25) we find
y(x)=f(x)+√
2 πλ[
̃f(x)+√
2 πλy(−x)]
,but on making the changex→−xand substituting back in fory(−x), this gives
y(x)=f(x)+√
2 πλ ̃f(x)+2πλ^2[
f(−x)+√
2 πλf ̃(−x)+2πλ^2 y(x)]
.Thus the solution to (23.24) is given by
y(x)=1
1 −(2π)^2 λ^4[
f(x)+(2π)^1 /^2 λ ̃f(x)+2πλ^2 f(−x)+(2π)^3 /^2 λ^3 ̃f(−x)]
.(23.27)Clearly, (23.24) possesses a unique solution providedλ=± 1 /
√
2 πor±i/√
2 π;these are easily shown to be the eigenvalues of the corresponding homogeneous
equation (for whichf(x)≡0).
Solve the integral equationy(x)=exp(
−
x^2
2)
+λ∫∞
−∞exp(−ixz)y(z)dz, (23.28)whereλis a real constant. Show that the solution is unique unlessλhas one of two particular
values. Does a solution exist for either of these two values ofλ?Following the argument given above, the solution to (23.28) is given by (23.27) with
f(x)=exp(−x^2 /2). In order to write the solution explicitly, however, we must calculate
the Fourier transform off(x). Using equation (13.7), we find ̃f(k)=exp(−k^2 /2), from
which we note thatf(x) has the special property that its functional form is identical to
that of its Fourier transform. Thus, the solution to (23.28) is given by
y(x)=1
1 −(2π)^2 λ^4[
1+(2π)^1 /^2 λ+2πλ^2 +(2π)^3 /^2 λ^3]
exp(
−
x^2
2)
.
(23.29)
Sinceλis restricted to be real, the solution to (23.28) will be unique unlessλ=± 1 /√
2 π,
at which points (23.29) becomes infinite. In order to find whether solutions exist for either
of these values ofλwe must return to equations (23.25) and (23.26).
Let us first consider the caseλ=+1/
√
2 π. Putting this value into (23.25) and (23.26),
we obtain
y(x)=f(x)+ ̃y(x), (23.30)
̃y(x)=f ̃(x)+y(−x). (23.31)