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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

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2 Answers 2

up vote 3 down vote accepted

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.

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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-)

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Many thanks! Great help. –  Mathmo Jan 16 '12 at 17:56
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