Consider the telegraph or Klein-Gordon equation on a rectangle*,

$$ \begin{align} \left(\frac{\partial^2}{\partial x^2}-\frac{\partial^2}{\partial y^2}\right)\psi(x,y)=\gamma^2\psi(x,y), \end{align} $$

with $\gamma$ some arbitrary (maybe complex) constant . Suppose that the necessary initial conditions are given ($\psi$ or its $x$ derivative at $x_1$ and $x_2$ if $x$ varies between $[x_1,x_2]$, and $\psi(x,t_0)$ and $\partial\psi(x,t_0)/\partial t$ at some arbitrary $t_0$), then it is convenient to separate variables in the form $\psi=A(x)B(y)$, which yields the ODEs

$$ \begin{align} \frac{\mathrm{d}^2A}{\mathrm{d}x^2}=-\alpha^2A\;\;\;,\;\;\frac{\mathrm{d}^2B}{\mathrm{d}y^2}=-(\gamma^2+\alpha^2)B, \end{align} $$

for $\alpha$ some (maybe complex) constant. These are second order equations and then the problem is essentially solved thanks to Sturm-Liouville: along $x$ and $y$ lines we have complete bases that can span generic initial/boundary data.

Now we rotate the coordinates:

$$ \begin{align} u=x+y\;\;\;,\;\;v=x-y \end{align} $$ and observe that the equation becomes

$$ \begin{align} \frac{\partial}{\partial v}\frac{\partial}{\partial u}\psi(u,v)=\frac{\gamma^2}{2}\psi(u,v), \end{align} $$

and after separating variables as $\psi=U(u)V(v)$ we have that

$$ \begin{align} \frac{\mathrm{d}\log U}{\mathrm{d} u}\frac{\mathrm{d}\log V}{\mathrm{d} v}=\frac{\gamma^2}{2}. \end{align} $$

There are several ways to split the constants (e.g. $\frac{\mathrm{d}\log U}{\mathrm{d} u}=\beta,\;\frac{\mathrm{d}\log V}{\mathrm{d} v}=\frac{\gamma^2}{2\beta}$, with $\beta$ an arbitrary constant), but the point is that

Now we have 1st rather than 2nd order equations

Then, is it true that the eigenfunctions form a basis that can span arbitrary initial/boundary data (specified on the $(u,v)$ plane)? I suspect the answer is no because the family of real exponentials are not complete. Moreover, is there any deep reason for the separated ODEs to become 1st order equations after the rotation ?

*The rectangle is just to illustrate the problem, but something similar may happen on several different domains.

  • $\begingroup$ Are you the separable ansatz actually solves the problem after changing variables to $u, v$? Because it might be that it doesn't work and something else like $\psi(u, v) = U(u) + V(v)$ is now the correct ansatz (though I don't actually know which one works to be honest). $\endgroup$
    – mattos
    Jul 23 at 15:16

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