INTEGRAL EQUATIONS
By examining the special casesx=0andx= 1, show thatf(x)=2
(e+3)(e+1)[(e+2)ex−ee−x].23.13 The operatorMis defined by
Mf(x)≡∫∞
−∞K(x, y)f(y)dy,whereK(x, y) = 1 inside the square|x|<a,|y|<aandK(x, y) = 0 elsewhere.
Consider the possible eigenvalues ofMand the eigenfunctions that correspond
to them; show that the only possible eigenvalues are 0 and 2aand determine the
corresponding eigenfunctions. Hence find the general solution off(x)=g(x)+λ∫∞
−∞K(x, y)f(y)dy.23.14 For the integral equation
y(x)=x−^3 +λ∫bax^2 z^2 y(z)dz,show that the resolvent kernel is 5x^2 z^2 /[5−λ(b^5 −a^5 )] and hence solve the
equation. For what range ofλis the solution valid?
23.15 Use Fredholm theory to show that, for the kernel
K(x, z)=(x+z)exp(x−z)over the interval [0,1], the resolvent kernel isR(x, z;λ)=exp(x−z)[(x+z)−λ(^12 x+^12 z−xz−^13 )]
1 −λ− 121 λ^2,
and hence solvey(x)=x^2 +2∫ 1
0(x+z)exp(x−z)y(z)dz,expressing your answer in terms ofIn,whereIn=∫ 1
0 unexp(−u)du.23.16 This exercise shows that following formal theory is not necessarily the best way
to get practical results!
(a) Determine the eigenvaluesλ±of the kernelK(x, z)=(xz)^1 /^2 (x^1 /^2 +z^1 /^2 )and
show that the corresponding eigenfunctions have the formsy±(x)=A±(√
2 x^1 /^2 ±√
3 x),whereA^2 ±=5/(10± 4√
6).
(b) Use Schmidt–Hilbert theory to solvey(x)=1+^52∫ 1
0K(x, z)y(z)dz.(c) As will have been apparent, the algebra involved in the formal method used
in (b) is long and error-prone, and it is in fact much more straightforward
to use a trial function 1 +αx^1 /^2 +βx. Check your answer by doing so.