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I am trying to solve this PDE using Method of characteristics:

$$(u+e^x)u_x+(u+e^y)u_y=u^2-e^{x+y}$$

I don't know how the next equation is called in English, but it is used to solve the PDE:

$$\frac{dx}{u+e^x}=\frac{dy}{u+e^y}=\frac{du}{u^2-e^{x+y}}$$

I attempted to find $f(x)$, $g(y)$ and $h(u)$ such that

$$f(x)(u+e^x)+g(y)(u+e^y)+h(u)(u^2-e^{x+y})=0$$

other attempt I tried was, given that $d(e^{-x})=-e^{-x}dx$, then I get a fourth equcation

$$\frac{e^{-x}dx-e^{-y}dy}{ue^{-x}+1-ue^{-y}-1}=\frac{d(e^{-y}-e^{-x})}{u(e^{-x}-e^{-y})}$$

I am not 100% sure about the last one.

PonchoALV98
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    Where did you encounter this problem? Do you have any reason to believe that it can be solved explicitly? – Hans Lundmark Sep 15 '21 at 10:25
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    It was shown by our teacher as a Challenge when we were studying how to solve that equations. He said yes, it can be solved – PonchoALV98 Sep 15 '21 at 18:18
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    And was the challenge only to find the characteristic curves (which can be done), or to actually find an explicit expression for the general solution of the PDE (which seems very hard)? – Hans Lundmark Sep 16 '21 at 15:28
  • Yes, only find the two characteristic curves. Is the implicit function of the two characteristic curves not considerated a general solution? – PonchoALV98 Sep 16 '21 at 18:50
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    Well, perhaps in principle, but it can be quite a big step to actually go from that to an explicit solution $u(x,t)$ satisfying some given condition such as $u(x,0)=f(x)$, for example. – Hans Lundmark Sep 16 '21 at 19:11

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I usually find it easier to write the ODEs for the parametrized characteristic curves $(x(t),y(t),u(t))$: $$ \dot x = u+e^x ,\qquad \dot y = u+e^y ,\qquad \dot u = u^2 - e^{x+y} . $$ Now after a bit of playing around with this, you might (if you're lucky) notice that $$ \tfrac{d}{dt} (e^{-x}) = - e^{-x} \dot x = - e^{-x} (u+e^x) = -u e^{-x} - 1 $$ and then that $$ \tfrac{d}{dt} (u e^{-x}) = \dot u \, e^{-x} + u (-u e^{-x} - 1) = (u^2 - e^{x+y}) \, e^{-x} - u^2 e^{-x} - u = - (u + e^y), $$ where suddenly a very familiar expression appeared on the right-hand side!

Do you think you can take it from there? (I don't want to spoil your teacher's challenge entirely...)

Hans Lundmark
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  • We didn't saw them from that side, not a justify, I'll review some things. Do you know any good bibliography? – PonchoALV98 Sep 16 '21 at 22:46
  • Also, thanks for taking your time to respond. – PonchoALV98 Sep 16 '21 at 22:49
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    It's explained in many PDE textbooks, like McOwen, F. John, Evans... But writing the equations for the characteristics the way I did is nothing deep; it's just what you get if you set your three expressions $dx/(u+e^x)$ etc. all equal to $dt$ and solve for $dx/dt$, $dy/dt$ and $du/dt$. – Hans Lundmark Sep 17 '21 at 05:27