Converting an integral equation to a differential equation I was recently working on a problem and ended up with an integral equation that I was hoping can be solved or at least be converted to a differential equation. I have no experience in integral equation so I have no Idea how to proceed. Consider the following integral equation:
$$\phi(x,y) = \sigma(\int \phi(x',y')w(x,y,x',y') dx'dy')$$
Where the integration bounds and the functions  are defined over a rectangle. The function $\sigma$ in general a continuous non-linear function. Assuming the function $w(x,y,x',y')$ is known, is there any general solution or differential equation form of this equation?
Is it possible to convert this to a differential equation in case $\sigma$ is linear?
Edit: As far as I can tell, it seems that there is no general method to approaching this problem except through numerical methods. In that case, are there any general theorems for equations of this form on the existence of solutions given boundary conditions? For instance, given a set of boundary conditions for the relevant function, does there exist a function that satisfies the above relation (unique -or not-solution for $\phi(x,y)$ given $w(x,y,x',y')$ and boundary conditions for $\phi$ and unique -or not- solution for $w(x,y,x',y')$ given $\phi(x,y)$ and the boundary conditions for $w$)?
 A: In full generality I would say that the answer is no, but in some cases you can associate a differential equation with that expression. However, the integral equation can be just as nice (or even nicer) to work with.
I will use the convention that $x,y\in \mathbb{R}^2$ and call the integration domain $D$, instead of looking at the coordinates separately, which, in my opinion, complicates the notation.
For the Laplacian (and many other second order differential operators) in a (nice) domain $D$, it is possible to show that there exists a Green function $G$ (also called the fundamental solution), which yields an operator inverse to $\Delta$ with null boundary conditions on $D$. Namely, if $u\in C^2_0(D)$ and $\Delta u = f$, then for $x\in D$,
$$u(x) = \int_D G(x,y)  f(y)\, dy.$$
In other words, $u$ solves the following Dirichlet problem
$$\begin{cases} \Delta u = f,\quad &{\rm in}\ D,\\ u = 0,\quad &{\rm on}\ \partial D.\end{cases} $$
Similarly, the problem
$$\begin{cases} \Delta u = u,\quad &{\rm in}\ D,\\ u = 0,\quad &{\rm on}\ \partial D,\end{cases} $$
corresponds to the integral equation
$$u(x) = \int_D G(x,y)u(y)\, dy.$$
Note that it is exactly what you are asking for in case that $\sigma$ is a linear function. The integral form is quite convenient for getting existence, uniqueness, and regularity result, because it gives a good ground for fixed point arguments. You can get properties like boundedness, or differentiability for $u$ from similar properties of $G$. For example, we can formally get
$$\nabla u(x) = \int_D \nabla_x G(x,y) u(y)\, dy.$$
A very simple example of a fixed point argument can go as follows: we want to prove the existence of the solution of the equation
$$u(x) = \int_D w(x,y) u(y)\, dy,$$
where $w$ is a continuous non-negative function which satisfies $\int_D w(x,y)\, dy \leq 1/2$ for all $x\in D$. We define a linear map $T$ by saying that $T(v) = u$, if
$$u(x) = \int_D w(x,y) v(y)\, dy.$$
Note that being the solution to our equation is equivalent to $T(u) = u$, so it suffices to show that $T$ has a fixed point. We will use Banach's fixed point theorem in the space $L^\infty(D)$. Assume that $v\in L^\infty(D)$. Then for all $x\in D$,
$$|u(x)| \leq \int_D w(x,y) |v(y)|\, dy \leq \|v\|_{\infty} \int_D w(x,y)\, dy \leq \frac 12\|v\|_{\infty}.$$
Therefore $\|u\|_{\infty}= \|T(v)\|_{\infty} \leq \frac 12 \|v\|_{\infty}$. This proves that $T$ maps $L^\infty(D)$ into itself and that it is a contraction (because it's linear). Therefore we can use Banach's fixed point theorem to see that there exists $u\in L^{\infty}(D)$ such that $T(u) = u$, and so we get a solution to our equation (it is easy to see that $u=0$, which perhaps makes the example a bit trivial :).
Other fixed point theorems often used for integral equations include Schauder or Leray--Schauder. You may find some more advanced examples, including differentiability properties, e.g., in chapter 10 of Gilbarg--Trudinger.
Fixed point arguments may work just as well even if $\sigma$ is a nonlinear function, provided that we know something about its regularity.
