# PDE: Solving using the Method of characteristics

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.

• Where did you encounter this problem? Do you have any reason to believe that it can be solved explicitly? Sep 15, 2021 at 10:25
• 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 Sep 15, 2021 at 18:18
• 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)? Sep 16, 2021 at 15:28
• Yes, only find the two characteristic curves. Is the implicit function of the two characteristic curves not considerated a general solution? Sep 16, 2021 at 18:50
• 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. Sep 16, 2021 at 19:11

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!
• 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 $du/dt$. Sep 17, 2021 at 5:27