Characteristics of a PDE As I continue working through lecture notes for my DE course, I encounter the following as an exercise:
Looking at the PDE
$$e^{2y}u_{xx}+u_y=u_{yy}$$
how can we find the differential equation satisfied by its characteristic curves and show that $$\lambda =x+e^y \text{ and } \mu =x-e^y $$are canonical variables for the PDE?
Any help would be very appreciate. Best regards, MM
 A: I will follow a Hamilton-Jacobi argument. Let us consider a solution in the form
$$u=e^{S(x,y)}.$$
The equation will take the form
$$S_{xx}+(S_x)^2=e^{-2y}(S_{yy}+(S_y)^2-S_y)$$
but now we are in a situation to operate a variable separation as
$$S=S_1(x)+S_2(y)$$
that will yield the two equations
$$S_{1xx}+(S_{1x})^2=k^2$$
and
$$e^{-2y}(S_{2yy}+(S_{2y})^2-S_{2y})=k^2$$
and it is not difficult to show that $S_1=kx$ and $S_2=ke^y$ are solutions.
A: Let's write your equation in the standard form :
$$e^{2y}u_{xx}-u_{yy}=-u_y$$
Then the 'characteristic equation' is :
$\displaystyle e^{2y}(dy)^2-(dx)^2=0$
which splits into the equations
$$e^y dy=dx$$
$$e^y dy=-dx$$
After integration we get your canonical variables :
$$e^y =C+x$$
$$e^y =D-x$$
More generally for the equation :
$$a(x,y)u_{xx}+2b(x,y)u_{xy}+c(x,y)u_{yy}=F(x,y,u,u_x,u_y)$$
the characteristic equation would be :
$\displaystyle a(dy)^2-2b(dx)(dy)+c(dx)^2=0$
(sources Polyanin "Handbook of Mathematics for engineers and scientists" ch.14 p585-)
