Fredholm and Volterra equations

The two types

• **Fredholm$**: y(x) = f(x) + \lambda\int_{a}$ᵇ K(x,t)y(t)dt — fixed limits.
• **Volterra$**: y(x) = f(x) + \lambda\int_{a}$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 $[0,1]: y = f + \lambda{}x\int_{0}^{1} t y$ dt. Setting $c = \int$ty gives $c(1 - \lambda/3) = \int$tf, so $**\lambda = 3**$ 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.$* y - 3\int_{0}^{1}$ tx y dt = f is solvable iff $\int_{0}^{1} t f(t)$ 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.