Once again I have to solve a PDE:
$e^{2y} u_{xx} + u_y = u_{yy}$
I have found this is hyperbolic, with canonical form:
$u_{\phi\psi}=\frac{1}{\phi-\psi} u_\psi$
I think this is how to do it: let $z=u_\psi$
Then $z_\phi = \frac{1}{\phi-\psi} z$. This is separable if we treat $\psi$ as a constantso we get $u_\psi=A(\phi-\psi)$ where A is some constant. Then we get $u=\frac{-A\psi^2}{2}+A\psi\phi+B$ for some constants A and B.
My question is, is this allowed? Can I treat $\phi$ as a constant in the first part of this calculation?
Thanks in advance.
Best Answer
$$u_{\phi\psi}=\frac{1}{\phi-\psi} u_\psi$$ Yes substitute $z=u_{\psi}$: $$(\phi-\psi)z_{\phi}=z$$ $$(\phi-\psi)z_{\phi}-z=0$$ $$\left ( \dfrac {z}{\phi-\psi} \right)'=0$$ $$ \dfrac {z}{\phi-\psi}=C(\psi)$$ Note that $C$ is not a constant but a function of $\psi$: $$ {u}_{\psi}=C(\psi)(\phi-\psi)$$ Integrating by part gives: $$ {u}{(\psi,\phi)}=g(\psi)(\phi-\psi)+\int g(\psi)d\psi+f(\phi)$$