Separable kernels and resolvent kernels

In one sentence

A separable kernel turns an integral equation into a finite linear system, so the whole apparatus of eigenvalues and the Fredholm alternative becomes ordinary linear algebra.

Why the exam asks it

Because it is the one case that can be solved completely by hand, so questions can ask for exact eigenvalues and explicit solutions — and because the reduction to a matrix problem is the moment the earlier linear algebra pays off.

The idea

A kernel is separable (or degenerate) if it is a finite sum of products:

$$K(x,t) = \sum_{i=1}^{n} a_i(x)\,b_i(t).$$

Then for the second-kind Fredholm equation $y(x) = f(x) + \lambda\int_a^b K(x,t)y(t)\,dt$,

$$y(x) = f(x) + \lambda\sum_{i=1}^n a_i(x)\,c_i, \qquad c_i = \int_a^b b_i(t)y(t)\,dt .$$

The unknown function is determined by the $n$ numbers $c_i$. Substituting back gives

$$c_i = \int b_i f + \lambda\sum_j \left(\int b_i a_j\right)c_j,$$

an $n\times n$ linear system $(I - \lambda A)\mathbf{c} = \mathbf{d}$ with $A_{ij} = \int_a^b b_i a_j$.

So:

• Eigenvalues of the integral equation are the $\lambda$ with $\det(I-\lambda A) = 0$ — at most $n$ of them.
• Unique solution exactly when $\det(I-\lambda A)\ne0$.
• At an eigenvalue, the Fredholm alternative in matrix form: solvable iff $\mathbf{d}$ is orthogonal to the null space of the transpose.

An infinite-dimensional problem has become an $n\times n$ determinant, which is why separable kernels are the ones that appear in examinations.

Worked

$$y(x) = f(x) + \lambda\int_0^1 x t\,y(t)\,dt .$$

Here $n=1$ with $a=x$, $b=t$. Write $c = \int_0^1 t\,y(t)\,dt$, so $y = f + \lambda x c$. Then

$$c = \int_0^1 t f + \lambda c\int_0^1 t^2 = \int_0^1 tf + \frac{\lambda c}{3},$$

giving $c\left(1-\frac\lambda3\right) = \int_0^1 tf$. So the single eigenvalue is $\lambda = 3$, and for every other $\lambda$ there is a unique solution

$$y = f + \frac{3\lambda x}{3-\lambda}\int_0^1 t f(t)\,dt .$$

At $\lambda=3$ the homogeneous equation has $y = x$ as a solution, and the inhomogeneous one is solvable only when $\int_0^1 tf = 0$.

The resolvent kernel

Writing the solution as

$$y(x) = f(x) + \lambda\int_a^b R(x,t;\lambda)f(t)\,dt$$

defines the resolvent $R$. For the example above, $R = \dfrac{3xt}{3-\lambda}$ — a rational function of $\lambda$ with a pole exactly at the eigenvalue. That is the general picture: the resolvent is meromorphic in $\lambda$ with poles at the eigenvalues, which is the same structure as $(A-\lambda I)^{-1}$ in finite dimensions.

For a Volterra equation the resolvent is entire in $\lambda$ — no poles, because there are no eigenvalues.

The resolvent also comes from the Neumann series, $R = \sum_{n\ge0}\lambda^n K_{n+1}$ with $K_n$ the iterated kernels, and summing that series is often the intended route when the kernel is simple.

Where intuition breaks

"A separable kernel can have infinitely many eigenvalues." At most $n$, the number of terms. Rank $n$ gives an $n\times n$ determinant.

"The eigenvalues are those of the matrix $A$." They satisfy $\det(I-\lambda A)=0$, so they are the reciprocals of the eigenvalues of $A$. Getting that inversion wrong is the standard slip.

"Every kernel is separable." Only finite sums of products. $K = e^{xt}$ is not, though it has a separable series approximation and that is a standard route.

"The resolvent exists for all $\lambda$." It has poles at the eigenvalues — for Fredholm. For Volterra it is entire.

"A rank-1 kernel has no eigenvalue." It has exactly one, $\lambda = 1/\int b\,a$, provided that integral is non-zero.

"At an eigenvalue there is no solution." There is none unless the orthogonality condition holds; when it does, there are infinitely many.

The exam's angle

1. Spot separability and set $c_i = \int b_i y$. The problem becomes finite immediately.
2. Compute $A_{ij} = \int b_i a_j$ and solve $\det(I-\lambda A)=0$ for the eigenvalues.
3. Remember the reciprocal: $\lambda$ eigenvalues of the equation, $1/\lambda$ of the matrix.
4. Write the resolvent and check its poles against the eigenvalues as a consistency check.
5. For Volterra, expect no poles — an entire resolvent is the signature.

The night before

• Separable $K = \sum_{i=1}^n a_i(x)b_i(t)$ reduces the equation to $(I-\lambda A)\mathbf{c} = \mathbf{d}$, $A_{ij} = \int b_i a_j$.
• At most $n$ eigenvalues, given by $\det(I-\lambda A)=0$ — reciprocals of the eigenvalues of $A$.
• $y = f + \lambda\int xt\,y$ on $[0,1]$: eigenvalue $\lambda=3$, resolvent $3xt/(3-\lambda)$, homogeneous solution $y=x$.
• Resolvent $R$: meromorphic in $\lambda$ with poles at the eigenvalues; entire for Volterra.
• $R = \sum \lambda^n K_{n+1}$ from the Neumann series.
• At an eigenvalue: solvable only under orthogonality, and then non-uniquely.