1. ## A pide

Find a non-trivial solution in the first quadrant for:

$f+\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}=\int_{x}^{\infty}\int_{y}^{\infty} f(u,v)dvdu$

Hint: Can you think of a trial function of two parameters $f(x,y;a,b)$ that would make the right side converge?

2. Spoiler:
I tried $f(x,y) = \exp(-\alpha x-\beta y)$. This seems to work provided that $\alpha-\tfrac1\alpha + \beta-\tfrac1\beta = 1$. That has lots of solutions, for example $\alpha=\beta=\tfrac14(1+\sqrt{17})$.

3. Originally Posted by Opalg
Spoiler:
I tried $f(x,y) = \exp(-\alpha x-\beta y)$. This seems to work provided that $\alpha-\tfrac1\alpha + \beta-\tfrac1\beta = 1$. That has lots of solutions, for example $\alpha=\beta=\tfrac14(1+\sqrt{17})$.
You have the right idea but your solution is not correct: if I back-substitute your particular solution into the DE, I do not get equality. Also, if you guys are interested, explain what the plot below has to do with this PIDE:

4. Originally Posted by shawsend
You have the right idea but your solution is not correct: if I back-substitute your particular solution into the DE, I do not get equality.
That was me being careless, trying to do something in a hurry late at night (excuses, excuses...). The equation $\alpha-\tfrac1\alpha + \beta-\tfrac1\beta = 1$ should have been $1-\alpha - \beta = \tfrac1{\alpha\beta}$. That has no real solutions for positive $\alpha$ and $\beta$. But it does have complex solutions with positive real parts (which are necessary for the integral to converge), for example $\alpha = \tfrac12,\ \beta = \tfrac14(1+i\sqrt{31})$. Unless I'm still making silly mistakes, that should give a complex solution to the problem. I don't have time to check it now, but I suppose that the real part of that ought to give a real solution.

Spoiler:
Complex solution: $f(x,y) = \exp\bigl(-\tfrac12x-\tfrac14(1+i\sqrt{31})y\bigr)$.

Real solution (?): $f(x,y) = \exp\bigl(-\tfrac12x-\tfrac14y\bigr)\cos\bigl(\tfrac{\sqrt{31}}4y\bigr)$.

5. That's the right relationship between a and b. So there are only complex solutions. So if the operator is linear:

$\left(1+\frac{\partial}{\partial x}+\frac{\partial}{\partial y}-I_2\right)f=0$

with $f=u+iv$ a solution, then won't u and v be independent solutions as well?

Anyone care to explain the plot? Where are the solutions in complex a-b space?