Legendre Transformation, Hodograph Method Please help me to solve this problem
Let $u(x,y)$ be a solution of a quasi-linear equation of the form:
$a(u_x,u_y)u_{xx} + 2b(u_x,u_y)u_{xy} + c(u_x,u_y)u_{yy}=0$
Introduce new independent variable $\xi, \eta$ and a new unknown function $\phi$ by:
$\xi=u_x(x,y), \eta=u_y(x,y) , \phi=xu_x + yu_y - u $     
Prove that $\phi$ as a function of $\xi, \eta$ satisfies $x=\phi_{\xi}, y=\phi_{\eta}$ and linear differential equation
$a(\xi,\eta)\phi_{\eta \eta} -  2b(\xi, \eta)\phi_{\xi \eta} + c(\xi, \eta)\phi_{\xi \xi}$
Here what I did : $\phi=xu_x + yu_y - u $ or $\phi=x\xi + y\eta - u $
Taking derivative $\phi$ w.r.t $\xi$ and $\eta$ we get: $\phi_{\xi}= x$ and $\phi_{\eta}= y$
But then I do not know how to get the linear differential equation 
$a(\xi,\eta)\phi_{\eta \eta} -  2b(\xi, \eta)\phi_{\xi \eta} + c(\xi, \eta)\phi_{\xi \xi}$
Any help I really appreciate.
 A: Ended up spending all day on this, but finally understand the solution! (Found this question based on a exercise problem in An Introduction to Nonlinear Partial Differential Equations by J. David Logan)
Anyways, here's the solution:
Really, the solution is not difficult, but can be difficult to see or easy to confuse if you're not used to it (as I was). 
We know: $u_x=\xi$ and $u_y=\eta$. (ignoring brackets for ease of reading)
From that, we can easily see that 
$$u_{xx}=\xi_x\\
 u_{xy}=\xi_y=\eta_x\\
u_{yy}=\eta_y$$
This is where I had trouble continuing.
This trick is to invert the partials acting on the new coordinates in order to get them instead acting on the old coordinates (Ex. $\xi_x \rightarrow x_\xi=\phi_{\xi\xi}$).
But note that you cannot simply inverse the derivative as you could if there was only one dependent variable($ \xi_x \neq \frac{1}{x_\xi}$), But instead have to complete the matrix inversion (this is really explained well here). (Also, this question expresses changing the partial derivatives in a detailed form) 
Therefore, the algebra becomes as follows:
$$
\begin{bmatrix}
u_{xx} & u_{xy} \\
u_{yx} & u_{yy} \\
\end{bmatrix}
=
\begin{bmatrix}
\xi_x & \xi_y \\
\eta_x & \eta_y \\
\end{bmatrix}
=
{
\begin{bmatrix}
x_\xi & x_\eta \\
y_\xi & y_\eta \\
\end{bmatrix}
}^{-1}
=
{
\begin{bmatrix}
\phi_{\xi\xi} & \phi_{\xi\eta} \\
\phi_{\eta\xi} & \phi_{\eta\eta} \\
\end{bmatrix}
}^{-1}
=
\frac{1}{\phi_{\eta\eta}\phi_{\xi\xi} -\phi_{\eta\xi}\phi_{\xi\eta}}
\begin{bmatrix}
\phi_{\eta\eta} & -\phi_{\eta\xi} \\
-\phi_{\xi\eta} & \phi_{\xi\xi} \\
\end{bmatrix}
$$
Assuming $u_{xy}=u_{yx}$ and $\phi_{\xi\eta}=\phi_{\eta\xi}$, and letting $\Delta=\frac{1}{\phi_{\eta\eta}\phi_{\xi\xi} -\phi_{\eta\xi}\phi_{\xi\eta}}$. We can see that:
$$
u_{xx} = \Delta\phi_{\eta\eta}\\
u_{xy} = -\Delta\phi_{\eta\xi}\\
u_{yy} = \Delta\phi_{\xi\xi}\\
$$
plugging this into the original equation:
$$
a(\xi,\eta)\Delta\phi_{\eta\eta} - 2b((\xi,\eta)\Delta\phi_{\eta\xi} + c((\xi,\eta)\Delta\phi_{\xi\xi} = 0
$$
and finally, dividing by $\Delta$, assuming $\Delta\neq 0$
$$
a(\xi,\eta)\phi_{\eta\eta} - 2b(\xi,\eta)\phi_{\eta\xi} + c(\xi,\eta)\phi_{\xi\xi} = 0
$$

For reference, another page I used as reference while trying to figure this out: 1
