How does one solve a fourth-order PDE of the form $\frac{\partial^4y}{\partial x^4}=c^2\frac{\partial^2y}{\partial t^2}$? It looks like a one dimensional wave equation, but I'm unfortunately very bad at PDEs.

  • $\begingroup$ Well, you can use the good old method of se $\endgroup$ – RougeSegwayUser Sep 11 '12 at 4:10
  • 1
    $\begingroup$ As Chris was saying, just suppose $y(x,t)=X(x)T(t)$ and subsitute that into the problem to obtain $X''''T=c^2XT''$. Divide by $XT$ etc... $\endgroup$ – James S. Cook Sep 11 '12 at 4:17

You do almost the same thing as people explained in your other question. Unfortunately, you can only factor the operator into

$$ \left(\frac{\partial^2}{\partial x^2} - c\frac{\partial}{\partial t}\right) \left(\frac{\partial^2}{\partial x^2} + c\frac{\partial}{\partial t}\right)y = 0. $$

Then you have to solve a heat-equation like equation.

If your domain is finite, you should try separation of variables.

  • $\begingroup$ Why does the finitiness of the domain matter for the separation of variables? $\endgroup$ – Lurco Jul 21 '15 at 13:35
  • 1
    $\begingroup$ When the domain is finite, it may be possible to find a countable complete orthogonal basis that consists of functions that satisfy given homogeneous boundary conditions. $\endgroup$ – Tunococ Aug 10 '15 at 7:07
  • $\begingroup$ I see, so if the domain is not bounded then the separation of variables still works, but the solutions obtained by it do not generate the whole space of solutions? $\endgroup$ – Lurco Aug 10 '15 at 10:29
  • $\begingroup$ @Lurco There are many things that won't work. The most obvious reason is probably that when the number of solutions is uncountable, summation does not work. People have tried to use integration instead, and that yields two common alternatives to Fourier series: Laplace transform (half real line) and Fourier transform (whole real line). Of course there are many other alternatives (even for the finite case), but I think you get the picture. $\endgroup$ – Tunococ Aug 11 '15 at 3:24

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.