# A pide

• Oct 18th 2009, 06:46 AM
shawsend
A pide
Find a non-trivial solution in the first quadrant for:

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

Hint: Can you think of a trial function of two parameters $\displaystyle f(x,y;a,b)$ that would make the right side converge?
• Oct 19th 2009, 12:58 PM
Opalg
Spoiler:
I tried $\displaystyle f(x,y) = \exp(-\alpha x-\beta y)$. This seems to work provided that $\displaystyle \alpha-\tfrac1\alpha + \beta-\tfrac1\beta = 1$. That has lots of solutions, for example $\displaystyle \alpha=\beta=\tfrac14(1+\sqrt{17})$.
• Oct 19th 2009, 01:45 PM
shawsend
Quote:

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

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:
• Oct 20th 2009, 03:01 AM
Opalg
Quote:

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 $\displaystyle \alpha-\tfrac1\alpha + \beta-\tfrac1\beta = 1$ should have been $\displaystyle 1-\alpha - \beta = \tfrac1{\alpha\beta}$. That has no real solutions for positive $\displaystyle \alpha$ and $\displaystyle \beta$. But it does have complex solutions with positive real parts (which are necessary for the integral to converge), for example $\displaystyle \alpha = \tfrac12,\ \beta = \tfrac14(1+i\sqrt{31})$. 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: $\displaystyle f(x,y) = \exp\bigl(-\tfrac12x-\tfrac14(1+i\sqrt{31})y\bigr)$.

Real solution (?): $\displaystyle f(x,y) = \exp\bigl(-\tfrac12x-\tfrac14y\bigr)\cos\bigl(\tfrac{\sqrt{31}}4y\bigr)$.
• Oct 20th 2009, 05:35 AM
shawsend
That's the right relationship between a and b. So there are only complex solutions. So if the operator is linear:

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

with $\displaystyle f=u+iv$ 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?