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Is my solution to this exercise correct?

Consider the pde $e^xu_{xx}+e^yu_{yy}=u.$ Determine the region where the pde is elliptic, parabolic and hyperbolic.

My solution:

There's no region R where pde is parabolic since $e^{x+y}=0$ does not have solution.

Also there is no region R where pde is hyperbolic since $e^{x+y}>0,\forall (x,y)\in\mathbb R^2.$

There exists a region R when $x=0=y;x,y>0;x,y<0;x>0,y<0$ and $x<0,y>0,$ and in all this cases pde is elliptic.

(I'm going to write only one region since the rest of them are similar.)

If $x,y>0$,

$$R=\{(x,y)\in\mathbb R^2:x>0,y>0\}$$

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$$e^xu_{xx}+e^yu_{yy}=u.$$ The canonical form is written like this $$A\partial_{xx} \psi+B\partial_{xy} \psi +C\partial_{yy}\psi+...=0$$ so $$\Delta=B^2-4AC=0-4e^xe^y=-4e^{x+y}$$

Elliptic for $\Delta<0$

Since $\Delta <0$ and $\forall (x,y) \in \mathbb{R^2}$ the equation is elliptic. So you are correct. But it's the case for all $(x,y) \in \mathbb{R^2}$

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    $\begingroup$ Alright, just I think it's $\mathbb R^2$ instead of $\mathbb R$, I also made this typo in my question, I've edited it now, small typo. $\endgroup$
    – user486983
    Mar 25, 2018 at 8:35
  • $\begingroup$ Thanks @Isa corrected...but your answer is correct it's always elliptic...for all x y $\endgroup$ Mar 25, 2018 at 8:40
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    $\begingroup$ I see, thank you. $\endgroup$
    – user486983
    Mar 25, 2018 at 8:47

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