Fredholm and Volterra equations

In one sentence

Fredholm equations integrate over a fixed interval and have eigenvalues; Volterra equations integrate up to the variable point and have none, so a Volterra equation of the second kind is always uniquely solvable.

Why the exam asks it

Because one difference — a fixed limit against a variable one — produces completely different solvability theory, and because the classification into kinds is a definition that can be tested directly.

The idea

The four types

Fredholm, fixed limits:

$$\text{first kind: } f(x) = \int_a^b K(x,t)y(t)\,dt, \qquad \text{second kind: } y(x) = f(x) + \lambda\int_a^b K(x,t)y(t)\,dt.$$

Volterra, variable upper limit: the same two with $\int_a^x$ in place of $\int_a^b$.

The kind is about where the unknown appears: in the first kind $y$ occurs only inside the integral; in the second it also appears outside. That distinction matters because first-kind equations are ill-posed — the integral operator smooths, so inverting it amplifies error — while second-kind equations are well behaved.

Why Volterra is easier

Solve by the Neumann series: substitute repeatedly to get

$$y = f + \lambda \hat K f + \lambda^2 \hat K^2 f + \cdots$$

For Fredholm this converges only when $|\lambda|$ is small enough — roughly $|\lambda| < 1/\|K\|$ — and outside that range the equation may have no solution or many.

For Volterra the $n$-th iterated kernel carries a factor $\frac{(x-a)^{n}}{n!}$, because the repeated integration is over a shrinking simplex. Factorials beat any geometric growth, so the series converges for every $\lambda$. Hence:

A Volterra equation of the second kind with continuous kernel has a unique continuous solution, for every $\lambda$ and every $f$.

Equivalently, a Volterra operator has no non-zero eigenvalues. That single sentence answers a large share of the questions in this topic.

Fredholm alternative

For the second kind, exactly one of these holds:

• $\lambda$ is not an eigenvalue: a unique solution for every $f$.
• $\lambda$ is an eigenvalue: the homogeneous equation has non-trivial solutions, and the inhomogeneous one is solvable only when $f$ is orthogonal to every solution of the transposed homogeneous equation — and then the solution is not unique.

This is the same alternative as for $Ax=b$ in linear algebra and for Green's functions in the ODE chapter. Recognising it as one theorem in three costumes is the useful move.

For a symmetric kernel $K(x,t)=K(t,x)$, the operator is self-adjoint: eigenvalues are real, the eigenfunctions are orthogonal and complete, and everything from the Sturm–Liouville chapter applies.

Connection to differential equations

An initial value problem integrates into a Volterra equation; a boundary value problem becomes a Fredholm equation with the Green's function as kernel.

That is the structural reason Volterra behaves better: an IVP has a unique solution under mild hypotheses, and a BVP need not — the same fact seen from the other side. $y''+\lambda y=0$ with $y(0)=y(\pi)=0$ has non-trivial solutions exactly at the eigenvalues, and the Fredholm alternative is that statement.

Where intuition breaks

"A Volterra equation may have no solution." The second kind always has exactly one, for every $\lambda$. The first kind can still be ill-posed.

"Volterra operators have eigenvalues like any other." None that are non-zero. $\int_0^x y\,dt = \lambda y$ forces $y\equiv0$.

"The Neumann series needs $|\lambda|$ small." For Fredholm, yes. For Volterra the factorial makes it converge always.

"First and second kind differ by which side the known function is on." By whether $y$ appears outside the integral. First-kind equations are ill-posed; that is the consequence that matters.

"The Fredholm alternative says a solution exists or does not." It says more: at an eigenvalue, solvability requires an orthogonality condition and the solution is then non-unique. Both halves are examined.

"A symmetric kernel is needed for the alternative." The alternative holds generally; symmetry gives the extra structure — real eigenvalues, orthogonal complete eigenfunctions.

"Converting an ODE to an integral equation loses the initial conditions." They are absorbed into $f$, which is the advantage: the integral equation carries them automatically rather than as side conditions.

The exam's angle

1. Classify first: fixed or variable limit, $y$ inside only or also outside. The type decides the theory.
2. If it is Volterra second kind, the answer to "is it solvable" is yes, uniquely, always.
3. For Fredholm, ask whether $\lambda$ is an eigenvalue. That is the whole question, and the alternative gives both branches.
4. Convert ODEs deliberately: IVP to Volterra, BVP to Fredholm with the Green's function.
5. Check symmetry of the kernel when the question mentions real eigenvalues or orthogonal eigenfunctions.

The night before

• Fredholm: $\int_a^b$. Volterra: $\int_a^x$. First kind: $y$ only inside — ill-posed. Second kind: $y$ also outside.
• Volterra second kind: unique solution for every $\lambda$; the iterated kernel carries $(x-a)^n/n!$, so the Neumann series always converges.
• Volterra operators have no non-zero eigenvalues.
• Fredholm second kind: unique unless $\lambda$ is an eigenvalue; then solvable only for $f$ orthogonal to the transposed homogeneous solutions, and non-uniquely.
• Symmetric kernel $\Rightarrow$ self-adjoint: real eigenvalues, orthogonal complete eigenfunctions.
• IVP $\to$ Volterra; BVP $\to$ Fredholm with Green's function kernel.