Hi! I ran into this PDE working on a question in cake cutting. Here it is:
$x\partial_1f(x,y)-(1-x)\partial_2f(y,x)=0$
for all $(x,y) \in [0,1]\times[0,1]$.
Thanks!
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4$\begingroup$ You're right! That does look like some sort of differential equation! Uh, what is the question? Gerhard "Ask Me About System Design" Paseman, 2011.11.27 $\endgroup$– Gerhard PasemanCommented Nov 28, 2011 at 7:43
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1$\begingroup$ I'd call it a functional partial differential equation, since it has both $f(x,y)$ and $f(y,x)$. Anyway, among the solutions are $f(x,y) = F(x) + G(y)$ for arbitrary $F$ and $G$. $\endgroup$– Robert IsraelCommented Nov 28, 2011 at 7:57
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$\begingroup$ Write as $\partial_1\partial_2f(x,y)=\frac{(1-x)}{x}\partial_1\partial_2f(y,x)=0$ or vice versa. Give $f(x,y)$ any sufficiently regular value on $(x,y)$ in the domain with $y\ge x$. On the other half of the domain, you have a wave equation $\partial_1\partial_2f(x,y)=g(x,y)$ in characteristic coordinates with a source. The boundaries on this half of the domain are timelike ($x=y$) and characteristic (sides of the square). This could now be treated as a characteristic initial-boundary value problem. $\endgroup$– Igor KhavkineCommented Nov 28, 2011 at 8:26
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$\begingroup$ I made a slight change to the question. This should not have a single solution, but I'm looking for a characterization of possible solutions. I agree that a sum of functions of x and y is a solution - thanks! $\endgroup$– OmerCommented Nov 28, 2011 at 8:26
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$\begingroup$ Sorry for the edit, and thanks for the answers so far! $\endgroup$– OmerCommented Nov 28, 2011 at 8:28
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The updated PDE is simpler than the original one. The same argument I gave in the comments still applies, but now you don't need to solve a wave equation, just integrate along one of the coordinates.
Namely, give $f(x,y)$ any sufficiently regular value on $(x,y)\in[0,1]\times[0,1]$ with $y\ge x$. On the other half of the domain $(x,y)\in[0,1]\times[0,1]$ with $x\ge y$, define $g(x,y) = \frac{(1-x)}{x}\partial_2 f(y,x)$. On that domain your PDE reduces to $$ \partial_1 f(x,y) = g(x,y) $$ $$ f(x,y) = f(y,y) + \int_y^x g(x',y) dx' $$
answered Nov 28, 2011 at 8:36