I have looked around for help and I have read and re-read my book but cannot get anywhere with this.
My question:
For $u_{xx}\,u_{yy} - u_{xy}^2 = f(x)$, show it is elliptic when $f(x) > 0$ and find a solution if $f(x)= -x^2$.
So, I found a method for fully nonlinear and some quasi-linear PDEs where you rewrite the equation as $F(x,y,u,u_x,u_y,u_{xx},u_{xy},u_{yy}) = 0$ and then assign $a:=\frac{\partial F}{\partial u_{xx}}$, $b:=\frac{\partial F}{\partial u_{xy}}$, and $c:=\frac{\partial F}{\partial u_{xx}}$. Using this method, which I am still reading about, I see that it is very simple to conclude that convex inhomogeneous terms lead to an elliptic PDE:
From the equation, $\frac{d\,y}{d\,x} = \frac{b \;\pm\;\sqrt[2]{b^2-4\,a\,c}}{2\,a},$
the discriminant is seen to be,
$b^2 - 4\,a\,c = 4\,u_{xy}^2 - 4\,u_{xx}\,u_{yy} = 4\,\left( u_{xy}^2 -u_{xx}\,u_{yy} \right) = 4\,\left(-f(x)\right) \overset{f(x)>0}{=} -4\,\left|f(x)\right| \Rightarrow u_{xx}\,u_{yy} > u_{xy}^2 \Rightarrow \textrm{ elliptic. }$
Now for $f(x)= -x^2$, the PDE is now hyperbolic and should give two characteristics. For the general solution, I now use, $\frac{dy}{dx} = \frac{-u_{xy} \;\pm\; x}{u_{yy}},$ to solve for the characteristics, but am again stuck because I am not sure how I can solve this. My plan is to find two solutions $y_1(x)=c_1$ and $y_2(x)=c_2$ and set these characteristics equal to the new variables from the change of variables, $\xi$ and $\eta$. Ultimately, I would like to be able to get this PDE into a canonical form. This is where I am stuck.
Thanks much!