Solve the following PDE using a change of variable:

$$ \alpha^2 \dfrac{\partial^2 z}{\partial x^2} - \beta^2 \dfrac{\partial^2 z}{\partial y^2} = 0$$

This is my attemp:

Let the following change of variables:

$$ \left\{ \begin{array}{ll} \phi = a x + by \\ \eta = cx + dy \end{array} \right.$$

Now I compute the derivatives according to that change:

$$\left\{ \begin{array}{ll} z_x = a z_\phi + c z_\eta \\ z_y = b z_\phi + d z_\eta \\ z_{xx} = a^2 z_{\phi \phi} + 2 a c z_{\phi \eta} + c^2 z_{\eta \eta} \\ z_{yy} = b^2 z_{\phi \phi} + 2 b d z_{\phi \eta} + d^2 z_{\eta \eta} \end{array} \right.$$

Thus the previous PDE has now been turned into

$$\alpha^2 (a^2 z_{\phi \phi} + 2 c a z_{\eta \phi}+ c^2 z_{\eta \eta}) - \beta^2 (b^2 z_{\phi \phi} + 2 b d z_{\eta \phi} + d^2 z_{\eta \eta}) = 0$$

or equivalently

$$ (\alpha^2 a^2 - \beta^2 b^2) z_{\phi \phi} + (2 c a \alpha^2 - 2 b d \beta^2)z_{\eta \phi} + (\alpha^2 c^2 - \beta^2 d^2) z_{\eta \eta} = 0$$

Now usually, in other example I am able to solve this by chosing $a,b,c,d$ such that the coefficients of $z_{\phi \phi}$ and $z_{\eta \eta}$ are both $0$. But this time that implies that my equation becomes just $0=0$

Have I done something wrong in my calculations? Should I have used another change of variables? Is there any other way (as simple as possible, as I have just started with PDEs) to solve this PDE?

  • 1
    $\begingroup$ You want to variables $\phi = \alpha x + \beta y$ and $\eta = \beta x - \alpha y$, and you'll get a nice canonical form solvable by integration. $\endgroup$
    – FireGarden
    Jun 14, 2014 at 12:55

3 Answers 3


I'm guessing that you get $0$ because you do the following:

$\alpha^2 a^2 - \beta^2 b^2 = 0 \Rightarrow \text{let}\ a = \beta, b = \alpha\\ \alpha^2 c^2 - \beta^2 d^2 = 0 \Rightarrow \text{let}\ c = \beta, d = \alpha $

Then $ca\alpha^2 - bd\beta^2 = 0$ as well.

So instead, observe that you just need, for example, $c^2 = \beta^2$. Instead of choosing $c = \beta$, therefore, you can just as well choose $c = -\beta$. Then $ca\alpha^2 - bd\beta^2 = -2\alpha^2 \beta^2 \ne 0$.

Then $\boxed{\phi = \alpha x + \beta y\\\ \eta = \alpha y - \beta x}$

Note: There are many other possible choices too.

  • $\begingroup$ It was very stupid from me to just ignore the fact that the variables were squared. Thanks! $\endgroup$
    – José D.
    Jun 14, 2014 at 13:08

$\xi=x/\alpha+y/\beta$; $\eta=x/\alpha-y/\beta$. You get $z_{\xi\eta}=0$.

  • $\begingroup$ Well, and you get just that, only the mixed term $z_{\xi\eta}$ remains, right? $\endgroup$
    – Vladimir
    Jun 14, 2014 at 13:25
  • $\begingroup$ Oh, I see what you mean. Nevermind. Sorry about that. $\endgroup$
    – M. Vinay
    Jun 14, 2014 at 13:27
  • $\alpha^2 D_x^2-\beta^2 D_y^2=(\alpha D_x-\beta D_y)(\alpha D_x+\beta D_y)$
  • solution of $\;\alpha z_x-\beta z_y=0\;$ is $\;z_1=f(\alpha y+\beta x)$
  • solution of $\;\alpha z_x+\beta z_y=0\;$ is $\;z_2=g(\alpha y-\beta x)$
  • Then solution of $\;\alpha^2 z_{xx}-\beta^2 z_{yy}=0\;$ is $$z=z_1+z_2=f(\alpha y+\beta x)+g(\alpha y-\beta x)$$

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