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:


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

With your example, you should get


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.

  • $\begingroup$ 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.. $\endgroup$ – gspddcv May 26 '15 at 18:26
  • 1
    $\begingroup$ @gspddcv: I thought you got your change of variables the same way.... I will edit the answer. $\endgroup$ – KittyL May 26 '15 at 18:27
  • $\begingroup$ oh god, I see it now. It was an algebraic mistake all along. Thanks for that edit. $\endgroup$ – gspddcv May 26 '15 at 18:29
  • $\begingroup$ 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. $\endgroup$ – gspddcv May 26 '15 at 18:35
  • $\begingroup$ @gspddcv: Yes, I guessed so. :) $\endgroup$ – KittyL May 26 '15 at 18:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.