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So I am attempting to solve this problem from Partial Differential Equations 2nd Edition, by Evans, from Chapter 4. The issue is that I am not sure how to get past a certain point:

Use separation of variables to find a nontrivial solution of $u$ of the PDE

$$u_{x}^{2}u_{xx}+2u_{x}u_{y}u_{xy}+u_{y}^{2}u_{yy}=0$$

I started by assuming the solution $u$ is given by $u(x,y)=w(x)v(y)$. Then we have

$$u_{x} = w'(x)v(y)$$ $$u_{y} = w(x)v'(y)$$ $$u_{xx} = w''(x)v(y)$$ $$u_{yy} = w(x)v''(y)$$ $$u_{xy} = w'(x)v'(y)$$

So the equation may be rewritten (after some algebraic manipulation) as

$$\frac{(w')^{2}w''}{w^{3}}+\frac{2(w')^{2}(v')^{2}}{w^{2}v^{2}}+\frac{(v')^{2}v''}{v^{3}}=0$$

I've tried multiple things from here but each thing gives me a dead-end. Any hints as to what to do here? I imagine it may have to do with rewriting some of these terms as differentials and then using a factoring, but I am not seeing it. Perhaps I am not even in the right ballpark. I'd love some help with this.

EDIT: The solution is obtained upon assuming $u(x,y)=w(x)+v(y)$ rather than $u(x,y)=w(x)\cdot v(y)$. Whereupon the middle term of the PDE vanishes and we obtain the equation

$$(w')^{2}w'' + (v')^{2}v''=0$$

This is solved by noting that $(w')^{2}w''=-(v')^{2}v''=constant$ and using integration to find the solution.

MrStormy83
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  • The fact that these are sort of homogenous in $u$ makes me want to write $u(x,y)=e^{f(x,y)}.$ Not sure that helps, though. Maybe a mix of your approach, with $u(x,y)=e^{w(x)+v(y)}.$ But that approach, assuming $u(x,y)=w(x)v(y),$ seems best when the equation is linear. – Thomas Andrews Nov 30 '22 at 18:33
  • Where did the $2$ go in your manipulation? Seems like you should still get a $2$ coefficient for the middle term. – Thomas Andrews Nov 30 '22 at 18:34
  • @ThomasAndrews a minor typo, good catch. I'll give your suggestion a try and see if it yields anything meaningful. – MrStormy83 Nov 30 '22 at 18:43
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    @ThomasAndrews I think I have something by using the additive separation $u(x,y)=w(x)+v(y)$ instead. I will post my results as soon as I come up with a possible solution. – MrStormy83 Nov 30 '22 at 19:04
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    Yeah, $u(x,y)=w(x)+v(y)$ gives $u_{xy}=0,$ so the middle term vanishes, and you get $(w')^2w''+(v')^2v'',$ but this just gives constants. So $(v')^2v''=c$ must be a constant, and $(v')^3=3cy +d$ or $v'=\sqrt[3]{3cy+d}$ or $v=\frac{1}{4c}(3cx+d)^{4/3}+b.$ – Thomas Andrews Nov 30 '22 at 19:10
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    Similarly, $w=\frac{-1}{4c}(-3cx+d')^{4/3}+b'.$ – Thomas Andrews Nov 30 '22 at 19:15
  • @ThomasAndrews those are the same results I obtained. That was fun. – MrStormy83 Nov 30 '22 at 19:19
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    These won't be all results, I suspect. And you can't linearly combine different $u$ to get a new solution, so this only finds a subset of solutions. – Thomas Andrews Nov 30 '22 at 19:20
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    Look to theorem 9 in the paper by Gunnar Aronsson in Arkiv For Mathematik Band 7 nr 28 from 12 April 1967 titled "On the partial differential equation $u_x^2 u_{xx} + 2 u_x u_y u_{xy} + u_y^2 u_{yy}$". The only global solutions to this equation are linear. – discretephenom Nov 30 '22 at 21:29
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    One obvious solution is exp(x+iy). – Ivan Neretin Nov 30 '22 at 22:10

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