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Transform the following PDE to the normalform $$ x^2u_{xx}-y^2u_{yy}=0~~~\text{ in }\Omega:=\left\{(x,y)\in\mathbb{R}^2\mid x>0,y>0\right\} $$

First of all, it is to say, that this is a hyperbolic PDE.

Furthermore, I already worked a lot and found the transformation $$ \xi:=\ln(y/x),~~~~~\eta:=-\ln(xy) $$ transforming the PDE above into the PDE $$ v_{\eta\xi}+u_{\xi}+v_{\xi\eta}=0, $$ where $v(\xi,\eta):=u(x,y)$.

Because of $v_{\eta\xi}=v_{\xi\eta}$ the "new" PDE is

$$ v_{\xi}+2v_{\xi\eta}=0. $$

Now the script says that it is easy to find another transformation that gives the normal form. But I do not see which transformation it is... do you see?

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    $\begingroup$ What is your definition of a normal form for a pde? $\endgroup$ – Kirill Oct 15 '13 at 18:37
  • $\begingroup$ For a hyperbolic PDE: $u_{\alpha \alpha}-u_{\beta \beta}=f(u,u_{\alpha},u_{\beta},\alpha,\beta)$ $\endgroup$ – math12 Oct 15 '13 at 18:41
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Take your calculations in a simplified form. Your solution is no-doubt true.

See here the characteristic equations are of the form

$$\frac{dy}{dx} = \frac{B -, + \sqrt{B^2 -4AC}}{2A}$$.

Putting the values of $A, B, C$ you shall get $\frac{dy}{dx} = \frac{y}{x}$ and $\frac{dy}{dx} = \frac{-y}{x}$ and solving you shall get $\frac{y}{x} = c_1$ and $xy = c_2$. I have just remove logarithms from the solutions.

Now take the transformations $\xi (x,y) = \frac{y}{x}$ and $\eta (x , y) = xy$.

Putting in the equation you shall get $u_{\xi \eta} = \frac{- u_\xi}{2xy}$. Hope my calculation is correct, but please check once.

Take another transformation $\alpha = \xi + \eta$ and $\beta = \xi - \eta$. Use it in your equation and get your standard form.

Or you could substitute $z = v_\xi$ in your equation given above, solve $z + 2z_\eta = 0$ first and then solve for $v$ by using $v_\xi = z$

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  • $\begingroup$ Thanks a lot, now I got it. $\endgroup$ – math12 Oct 16 '13 at 10:02

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