2
$\begingroup$

I am dealing with Volterra-Fredholm integral equations of the following form: \begin{equation} \tag{1} \phi(x) = f(x) + n \int_a^x K(x,t) \phi(t) dt + \int_a^b K(x,t)\phi(t)dt, \end{equation} where $a,b \in \mathbb{R}, n \in \mathbb{N}$, $f$ is continuous, and $K \in L^2([a,b];\mathbb{R})$ is symmetric and positive semidefinite. $\phi$ is the unknown function to be solved for. I have already proved existence and uniqueness of a solution and would like to solve the equation numerically.

One could (1) as a pure Fredholm equation with kernel $\tilde{K}(x,t) = K(x,t)(1 + n \, 1_{\{x \ge t\}}),$ but $\tilde{K}$ is neither symmetric nor continuous when $n \neq 0$. This appears overly complicated to me.

Most research papers on Volterra-Fredholm equations seem to deal with nonlinear or mixed equations. There is some nice theory about Fredholm equations with symmetric, positive semidefinite kernels, but I could not find similar results for Volterra-Fredholm equations.

Can you recommend literature that helps me solve (1) numerically?

$\endgroup$
  • $\begingroup$ I didn't even know "Volterra-Fredholm" equations were a thing; most of the time it's either-or. Not much of the literature develops this theory, I guess. Thanks for posting. $\endgroup$ – SZN Nov 17 '16 at 5:21
0
$\begingroup$

It seems that viewing (1) as a pure Fredholm equation with asymmetric and discontinuous kernel is the method of choice.

This paper by K.E. Atkinson (1967) uses a quadrature method (also known as Nystrom method) to approximate Fredholm equations numerically. The only assumptions on the kernel $K$ are:

  • $\sup_{x \in [a,b]} \int_a^b K(x,t) dt < \infty$ and
  • $\int_a^b |K(x_1,t) - K(x_2,t)| dt \to 0$ uniformly for $x_1, x_2 \in [a,b]$ as $|x_1 - x_2| \to 0$.

If the kernel $K$ in (1) satisfies these assumptions, so does $\tilde{K}(x,t) = K(x,t)(1+n1_{\{x\ge t\}})$.

I am still grateful for anything you might know about Volterra-Fredholm equations. Judging from the age of Atkinson's paper, I guess there are more powerful methods available.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.