Another second order PDE in canonical form 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.
 A: $$u_{\phi\psi}=\frac{1}{\phi-\psi} u_\psi$$
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:
$$  {u}{(\psi,\phi)}=g(\psi)(\phi-\psi)+\int g(\psi)d\psi+f(\phi)$$
A: Letting $V=\partial_{\eta} U$ we get
$$\partial_{\xi}V=\frac{1}{\xi-\eta}V$$
This is a straightforward application of characteristics to this semilinear PDE.
$$\begin{bmatrix}
\partial_{t}[ \xi ](t,s)\\
\partial_{t}[ \eta ](t,s)\\
\partial_{t}[ \zeta ](t,s)
\end{bmatrix} =\begin{bmatrix}
1\\
0\\
\zeta\cdot( \xi -\eta )^{-1}
\end{bmatrix}$$
So $\xi(t,s)=t+f_1(s)$, $\eta(t,s)=f_2(s)$ and
$$\zeta(t,s)=V(\xi(t,s),\eta(t,s))=f_3(s)(f_1(s)-f_2(s)+t)=f_3(s)(\xi(t,s)-\eta(t,s))$$
From $\eta(t,s)=f_2(s)$ we can write $s=f_2^{-1}(\eta)$
Hence $V(\xi,\eta)=f_4(\eta)\cdot(\xi-\eta)$, where $f_4=f_3\circ f_2^{-1}$. This can be easily verified. Then,
$$V=\partial_{\eta}U=f_4(\eta)\cdot(\xi-\eta)$$
We can now directly integrate.
$$u(x,y)=U(\xi,\eta)=\int f(\eta)\cdot (\xi-\eta)\mathrm{d}\eta+g(\eta)$$
Which can be integrated by parts from @Aryadeva 's answer. Substitute whatever $\xi,\eta$ are in terms of $x,y$ and you're done.
