# Integral equation involving convolution: is it possible to solve it?

I have to solve a finite difference equation, and I decided to attempt a solution through Fourier transform. What I obtain is an equation of the form (where I have to solve for $$\hat G$$ and $$\hat \Delta$$ is known)

$$\hat G(p,a)\mathrm e^{2\mathrm i p a}=\hat \Delta(p,a)*_p\hat G(p,a).$$ Here, $$\star_p$$ is the convolution product with respect to the variable $$p$$: $$\hat \Delta(p,a)*_p\hat G(p,a)=\int_{-\infty}^{+\infty}\hat \Delta(p-q,a)\hat G(q,a)\mathrm dq.$$ I have hints that a solution exists at the level of the finite difference equation (up to a multiplicative function of $$a$$ that I don't really need), but I have no idea about how to solve equation 1.

Is there some literature about this kind of equation? Does a solution (up to multiplicative factors) exist, and is it possible to write it in a closed form?

I am working at a purely formal level, meaning that I usually neglect all details about existence of the things I'm dealing with, convergence, existence of Fourier transforms et cetera (at least, until something breaks down!).

Thank you all!

## 1 Answer

This looks like a Fredholm equation of the second kind: https://en.wikipedia.org/wiki/Fredholm_integral_equation. Use the notation in the page, you could put $$k(p,q) = e^{-2ipa}\Delta(q-p,a)$$ and have that $$(KG) (p) = \int_{\mathbb{R}} k(p,q)G(q)dq$$

If $$||K||_2^2 \leq\int_{\mathbb{R}^2} | e^{-2ipa}\Delta(p-q,a)|^2 dpdq < 1$$ then there does exist a unique solution to the problem $$(I-K)G(p) = 0$$ But since $$I-K$$ is linear, then a unique solution to $$(I-K)G(p) = 0$$ must be $$G(p,a) =0$$. So for nontrivial solutions, you will need $$||K||\geq 1$$.

• Thank you for your insight. I am quite certain that non trivial solutions exist due to other very lengthy considerations (I am using a particular $\Delta$ here, but I don't think the form will help), and now I'm interested in the form of such non trivial solutions. This does not answer my question, but you gave me a reference to look for, and I thank you for that. Commented Sep 24, 2020 at 15:27