# reducing a pde to a canonical form

I'm really struggling with this one and I can't seem to find what's wrong with my approach. I am given a PDE in the form $$U_{xx} + x y U_{yy} = 0,$$ and I am supposed to bring it to its canonical form. I've set $$ξ = y^{1/2} + (-x)^{1/2}$$ and $$η = y^{1/2} - (-x)^{1/2}.$$

I'm solving it for the domain where it's hyperbolic. But some terms don't cancel out. What am I doing wrong?

Can anyone tell me how $U_{xx}$ and $U_{yy}$ would look after we perform the change of variables?

Edit: There may be also the possibility that I've chosen the wrong change of variables. If someone could point out my mistake, I think I can continue from there.

The characteristic curves satisfy:

$$\frac{dy}{dx}=\frac{b\pm\sqrt{b^2-ac}}{a}$$

where $a,b,c$ are coefficients of $U_{xx}, U_{xy}, U_{yy}$, respectively.

With your example, you should get

$$\frac{dy}{dx}=\pm\sqrt{-xy}$$

Solving these two ODE's you should see the following change of variables:

$$\xi=y^{1/2}+\frac{1}{3}(-x)^{3/2}\\ \eta=y^{1/2}-\frac{1}{3}(-x)^{3/2}$$

In the integration, the right hand side with respect to $x$ is not in the denominator. It is $\sqrt{-x}$ instead.

• thank you, but can you tell me how you reached this transformation? Because I don't think I can obtain this with the method in my textbook.. – gspddcv May 26 '15 at 18:26
• @gspddcv: I thought you got your change of variables the same way.... I will edit the answer. – KittyL May 26 '15 at 18:27
• oh god, I see it now. It was an algebraic mistake all along. Thanks for that edit. – gspddcv May 26 '15 at 18:29
• that square root of negative x, somehow ended up in the denominator when I performed the integration for the ODE's, need to pay more attention to these things apparently. – gspddcv May 26 '15 at 18:35
• @gspddcv: Yes, I guessed so. :) – KittyL May 26 '15 at 18:36