Linear Integral Equations

The two types

• **Fredholm$\ast \ast :y(x)=f(x)+\lambda {\int }_a$ᵇ K(x,t)y(t)dt — fixed limits.
• **Volterra$\ast \ast :y(x)=f(x)+\lambda {\int }_a$ˣ K(x,t)y(t)dt — variable upper limit.

Second kind has y outside the integral; first kind does not (and is ill-posed).

Volterra

The Neumann series always converges: a Volterra equation of the second kind has a unique solution for every $\lambda$, with no eigenvalues. Converting to an ODE is usually fastest: differentiate and use y(a) = f(a).

Example. y = eˣ $+{\int }_0$ˣ $((1+x^2)/(1+t^2))y$ dt. Put $I(x)={\int }_0$ˣ $y/(1+t^2)$; then I′ − I = eˣ$/(1+x^2)$, giving I = eˣ arctan x and y = eˣ$(1+(1+x^2)$arctan x).

Fredholm with a separable (degenerate) kernel

$K(x,t)=\sum a_i(x)b_i(t)$ reduces the equation to a finite linear system. Eigenvalues are where that system is singular.

Example. K = xt on $[0,1]:y=f+\lambda x{\int }_0^{1}ty$ dt. Setting $c=\int$ty gives $c(1-\lambda /3)=\int$tf, so $\ast \ast \lambda =3\ast \ast$ is the eigenvalue.

Fredholm alternative

For $y-\lambda \int$Ky = f with $\lambda$ an eigenvalue:

• a solution exists iff f is orthogonal to every solution of the homogeneous adjoint equation;
• then the solution is not unique (add any eigenfunction).

If $\lambda$ is not an eigenvalue, there is a unique solution for every f.

*Example.$\ast y-3{\int }_0^1$ tx y dt = f is solvable iff ${\int }_0^{1}tf(t)$ dt = 0 — true for $f=x^2-$ ½ and f = 2 − 3x, false for eˣ and x − 1.

Relation to ODE

A boundary value problem converts to a Fredholm equation with the Green's function as kernel; an initial value problem converts to a Volterra equation.

Key takeaways

• Volterra: always uniquely solvable. Fredholm: check for eigenvalues first.
• Separable kernels reduce everything to linear algebra.
• The alternative's solvability test is an orthogonality condition on f.