# Monge-Ampere PDE is hyperbolic ($f(x) < 0$) - General Solution

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!

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What you can do is to answer your own question, as you can see from here: blog.stackoverflow.com/2011/07/… –  Paul Feb 23 '12 at 5:11
Thanks! I had done it once, but without the knowledge that it was encouraged. I will do so. –  nate Feb 23 '12 at 5:15

For the answer to the second part:

Attempting to solve for the characteristics is unnecessary, an alternative approach can be used. It was noticed (by testing for separability with Maple's PDEtools) that the PDE is separable, by addition. Therefore, let the solution be represented by,

$u(x,y) = X(x)+Y(y),$

which has partial derivatives of,

$u_x = X'\\ u_y = Y'\\ u_{xx} = X''\\ u_{yy} = Y''\\ u_{xy} = 0\\$

The PDE can now be rewritten as,

$X''\,Y'' = -x^2,$

which is separable and results in the following two equations.

$\frac{X''}{x^2} = \frac{-1}{Y''} = k$

Solving each for their complementary and particular solutions yields,

$X'' = k\;x^2$

$\longrightarrow X_c(x) = C_1\;x+C_2$ (for $X'' = 0$),

and

$X_p(x) = C_3\;x^4$,

$X'_p(x) = 4\;C_3\;x^3$

$X''_p(x) = 12\;C_3\;x^2$

$\rightarrow C_3 = \frac{k}{12} \Rightarrow$

$X(x) = C_1\;x+ C_2 + \frac{k}{12}\;x^4.$

Similar results show that the solution for $Y(y)$ is,

$Y(y) = D_1\;y+D_2 - \frac{1}{2\;k}\;y^2$

Substituting these results into the PDE produces,

$u(x,y) = X(x) + Y(y) = C_1\;x+ C_2 + \frac{k}{12}\;x^4 + D_1\;y+D_2 - \frac{1}{2\;k}\;y^2$

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