A pide

**shawsend** · October 18th 2009, 06:46 AM

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?

**Opalg** · October 19th 2009, 12:58 PM

Spoiler:

**shawsend** · October 19th 2009, 01:45 PM

Originally Posted by Opalg

Spoiler:

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:

**Opalg** · October 20th 2009, 03:01 AM

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:

**shawsend** · October 20th 2009, 05:35 AM

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?

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