Hi! I ran into this PDE working on a question in cake cutting. Here it is:
$x\partial_1f(x,y)(1x)\partial_2f(y,x)=0$
for all $(x,y) \in [0,1]\times[0,1]$.
Thanks!
Hi! I ran into this PDE working on a question in cake cutting. Here it is: $x\partial_1f(x,y)(1x)\partial_2f(y,x)=0$ for all $(x,y) \in [0,1]\times[0,1]$. Thanks! 


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 $g(x,y) = \frac{(1x)}{x}\partial_2 f(y,x)$. On that domain your PDE reduces to $$ \partial_1 f(x,y) = g(x,y) $$ $$ f(x,y) = f(y,y) + \int_y^x g(x',y) dx' $$ 

