I have the following PDE:

$$x^2u_{xx} - y^2u_{yy}-2yu_y = 0 .$$

after seperating variables, I obtain after separating variables, I obtain $$\frac{x^2}{\phi} \phi '' = - \lambda ,$$ and $$\frac{y^2}{g} g '' -\frac{2y}{g} g' = -\lambda,$$ where $u(x,y) = \phi(x)g(y)$ and $\lambda$ is the separation constant. How should I proceed?


It is more powerful to use the method of characteristics (MoC) than separation of variables (SoV) for OP's problem. In this answer we give a sketched derivation of the full solution using the MoC.

  1. Notice that OP's linear 2nd order PDE $$x^2 u_{xx}-y^2 u_{yy}-2yu_y~=~0 $$ is form-invariant under scalings $(x,y)\to (\lambda z, \mu y)$. This suggests that we should define new variables $$X~=~\ln|x|\quad\text{and}\quad Y~=~\ln|y|.$$ Then OP's linear 2nd order PDE becomes a linear 2nd order PDE $$u_{XX}-u_X-u_{YY}-u_Y~=~0 $$ with constant coefficient functions.

  2. If we define $$ X^{\pm}~=~X\pm Y, $$ the PDE factorizes $$ (4\partial_--2)\partial_+u~=~0, $$ or equivalently, $$ \partial_- e^{-X^-/2}\partial_+u~=~0, $$ with full solution $$u~=~ F(X^-)+ e^{X^-/2}G(X^+),$$ where $F,G$ are arbitrary functions.

  3. Going back to the original variables, the solution reads $$ u~=~ f(x/y) + \sqrt{\left| \frac{x}{y}\right|} g(xy), $$ where $f,g$ are arbitrary functions. This can in turn be beautified to$^1$ $$ u~=~ f(x/y) + x g(xy) $$ for arbitrary functions $f,g$. The above full solution would have been more difficult to derive using SoV.


$^1$ Hat tip: Robert Israel.

  • $\begingroup$ How did you know to choose $X= ln |x|$ and $Y=ln|y|$? $\endgroup$ – Dr. John A Zoidberg Feb 17 '16 at 21:52
  • $\begingroup$ From the form-invariance under scaling. I updated the answer. $\endgroup$ – Qmechanic Feb 17 '16 at 22:10
  • $\begingroup$ I don't understand. I can see why you chose to find new variables, but how did you find $X=ln|x|$? $\endgroup$ – Dr. John A Zoidberg Feb 17 '16 at 22:17
  • $\begingroup$ So that $x\partial_x=\partial_X$. $\endgroup$ – Qmechanic Feb 17 '16 at 22:23
  • 1
    $\begingroup$ A slightly nicer form, avoiding apparent problems at $x=0$, is $u = f(x/y) + x g(xy)$. $\endgroup$ – Robert Israel Feb 17 '16 at 22:25

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