I have been working on a problem in queuing theory, and in order to understand the steady state behaviour, I have obtained a PDE. The equation is
$(\mu + 2\lambda)f(x, y) + \displaystyle\frac{\partial f}{\partial x} + \displaystyle\frac{\partial f}{\partial y} = \lambda \bigg(f(x, w_{c}) + f(w_{c}, y)\bigg) + \mu\lambda^{2}\int_{x}^{w_{c}}\int_{y}^{w_{c}}f(u, v)dv du$
Here, $\mu$, $\lambda$ and $w_{c}$ are all positive constants, and $0\leq x$, $y\leq w_{c}$.
$f(x, y) = 0$ $x > w_{c}$ or $y > w_{c}$. Also, only solutions where $f(x, y) \geq 0$ make sense.
I have attempted a power series expansion of $f(x, y)$ to see if I can obtain the coefficients of the series, but I just get stuck with a double-indexed linear recurrence.
Using a 2-D Laplace transform has also not made it any simpler.
Any advice would be greatly appreciated! If you have come across a problem like this, please let me know.
EDIT
In case anyone can solve this, or suggest an approach:
The recurrence relation that I obtained from the PDE was as follows -
$$(\mu + 2\lambda)c(i+1, j+1) + c(i+2, j+1) + c(i+1, j+2) = \mu\lambda^{2}c(i, j) \hspace{4mm} \forall i, j \geq 0$$
Also,
$$(\mu + \lambda)c(i, 0) + c(i+1, 0) + c(i, 1) = 0 \hspace{4mm} \forall i\geq 1$$
and $$\mu c(0, 0) + c(1, 0) + c(0, 1) = 0$$
$c(i, j)$ is symmetric in $i$ and $j$.
I have tried out a geometric solution to the recurrence, but it does not seem to work. A 2D Z-Transform gives 2 polynomial equations in 2 variables, and leads nowhere.
I have run out of options to try, and cannot resort to numerical solutions. Please advise!