I have a 2D linear non-homogeneous Volterra integral equation of the second kind that I am painfully trying to solve. Here is my equation:
$L_o(x,y) = L_e(x,y) + \frac{1}{\pi} \int_0^{2\pi} \int_0^{\theta_\text{max}(x,y,\phi)} L_o(x+h \tan \theta \cos \phi, y + h \tan \theta \sin \phi) \cos \theta \sin \theta \ \text{d}\theta \ \text{d}\phi$
The unknown function, $L_o(x,y)$ is defined over $(x,y) \in \ ]0 \ ; h[ \ \times \ ]0 \ ; h/2[ $. $h$ and $d$ are known-parameters in $\mathbb{R}^{+*}$ and $L_e(x,y)$ is perfectly known, and can be chosen as a constant if needed. $\theta_{\text{max}}$ does not simplify the thing, and is equal to
$\theta_\text{max}(x,y,\phi) = \begin{cases} \arctan\left(\frac{(h-x)\sqrt{1+\tan^2 \phi}}{d}\right) &\text{if } 0 < \phi < \arctan\left( \frac{h-y}{h-x} \right)\\ \arctan\left(\frac{(h-y)\sqrt{1+\cot^2 \phi}}{d}\right) &\text{if } \arctan\left( \frac{h-y}{h-x} \right) < \phi < \frac{\pi}{2}+\arctan \left( \frac{x}{h-y}\right)\\ \arctan\left(\frac{x\sqrt{1+\tan^2 \phi}}{d}\right) &\text{if } \frac{\pi}{2}+\arctan \left( \frac{x}{h-y}\right) < \phi < \pi + \arctan \left( \frac{y}{x}\right)\\ \arctan\left(\frac{y\sqrt{1+\cot^2 \phi}}{d}\right) &\text{if } \pi + \arctan \left( \frac{y}{x}\right) < \phi < 3\pi/2 + \arctan \left( \frac{h-y}{x}\right)\\ \arctan\left(\frac{(h-x)\sqrt{1+\tan^2 \phi}}{d}\right) &\text{if } 3\pi/2 + \arctan \left( \frac{h-y}{x}\right) < \phi < 2\pi \end{cases}$
I haven't found anything about 2D Volterra integral equations on Mathematica. Is there a simple way to solve it, at least numerically ? From what I read in the scientific papers, it appers to be a really tough problem...
I tried for example a simple one:
eqn = L[x,y] == ∫_0^x ∫_0^y L[t,p] dt dp;
DSolveValue[eqn, L[x,y], {x,y}]
but it gave me as an output $\int_0^x \int_0^y L[t,p] \, \text{d}t \, \text{d}p$