I'm trying to solve a fourth order pde similar to the biharmonic equation

$ 0=\frac{\partial ^4}{\partial x^4}u(x,y)+Q\frac{\partial ^4}{\partial x^2 \partial y^2}u(x,y)+\frac{\partial ^4}{\partial y^4}u(x,y) $

Where Q is a constant. The boundary condition are the following,

$ \frac{\partial^2}{\partial x^2}u(x,y)=0\quad \quad \frac{\partial^2}{\partial y^2}u(x,y)=0 \quad \textit{and} \quad \frac{\partial^2}{\partial x \partial y}u(x,y)=0 $

while the boundary is a rectangle.

So far neither Fourier series nor separation of variables seems to work on this one and I'm no expert to the topic.

  • $\begingroup$ Do you have boundary conditions? $\endgroup$ – EpicMochi Oct 21 '14 at 13:27
  • $\begingroup$ Yes, I just edited the question accordingly. Thanks for pointing out. $\endgroup$ – Cedric Oct 21 '14 at 13:52
  • $\begingroup$ Do you have some boundary conditions on $u$ itself? $\endgroup$ – spatially Oct 21 '14 at 13:57
  • $\begingroup$ no, only on the derivatives. $\endgroup$ – Cedric Oct 21 '14 at 14:00
  • $\begingroup$ $\dfrac{\partial^2}{\partial x^2}u(x,y)=0$ , $\dfrac{\partial^2}{\partial y^2}u(x,y)=0$ and $\dfrac{\partial^2}{\partial x\partial y}u(x,y)=0$ are not belonging to the boundary conditions since they are already restricted the form of $u(x,y)$ should be. $\endgroup$ – doraemonpaul Oct 22 '14 at 21:13

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