I would be very grateful if you could help me with seeing how things actually work in the elliptic case of second order semi-linear PDEs.
Such a PDE: $a(x,y)U_{xx}+2b(x,y)U_{xy} + c(x,y)U_{yy} = f(x,y,U, U_{x}, U_{y})$, where $U_{x}$ denotes partial differentiation of $U$ wrt $x$. If $ac > b^2$ then we get this elliptic type.
Now, I understand the change of coordinates we do, so that we simplify the equation to get $U_{z\overline{z}}$ for some function $z=z(x,y)$, though I am not exactly sure why do we need to solve $dy/dx = x_{1,2}$, where $x_{1,2}$ are the roots of the resulting characteristic polynomial.
Once we've done this we end up with $U_{z\overline{z}} = G(z,\overline{z}, U, U_{\overline{z}}, U_{z})$.
What my lecture notes suggest is to take $m=Re(z)$ and $n=Im(z)$ to turn the equation into $U_{mm}+U_{nn}=F(m,n,U,U_{m},U_{n})$. Could you please explain this step. If you also could help me with the example:
$yU_{xx} + U_{yy} = 0$, for $y > 0$
I'd be very grateful.
PS: I can see how you can go through the process by bashing out everything, but I am pretty sure I am missing something obvious.
Thank you in advance
Edit: Characteristic equation: $y(dy/dx)^2+1=0$
We get that $dy/dx = +-i/sqrt(y)$
So for the characteristic curves we can take $z=2y^{3/2} +- 3ix$
Now we should change the variables to $m=2y^{3/2}, n = 3ix$ and we should express our equation as $ U_{mm}+U_{nn}=F(...)$
I assume we should do the chain rule a few times but can't really see it.