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I want to analytical Solve or numerical solve by Mathematica a biharmonic equation with homogeneous boundary conditions homogeneous?

I considered its solution using Fourier analysis, which is not responsive and its convergence speed is low and can not be used.

$$\nabla^4 U=f(x,y);\,\,U=\frac{\partial^2U}{\partial x^2}=0,\,\,\,at\,\, x=0,1;U=\frac{\partial^2U}{\partial y^2}=0,\,\,at\,\, y=0,1$$


marked as duplicate by Kuba Aug 16 '18 at 6:17

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  • 1
    $\begingroup$ Welcome to Mma.SE. Start by taking the tour now and learning about asking and what's on-topic. Always edit if improvable, show due diligence, give brief context, include minimal working example of code and data in formatted form. By doing all this you help us to help you and likely you will inspire great answers. The site depends on participation, as you receive give back: vote and answer questions, keep the site useful, be kind, correct mistakes and share what you have learned. $\endgroup$ – rhermans Aug 13 '18 at 11:06
  • $\begingroup$ It seems you could expand both $U$ and $f$ as $\sin(\pi p x ) \sin(\pi q y)$. then the $\nabla^4 U_{pq}$ yields $\pi ^4 \left(p^2+q^2\right)^2 U_{pq}$ $\endgroup$ – chris Aug 13 '18 at 12:27
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    $\begingroup$ 1. You should write more details about the domain in which you solve this equation. Is it a square? 2. Independently of this you should make a trivial substitution: introduce a new variable V(x,y) equal to the Laplacian of U(x,y). For this variable your equation will be harmonic, non-homogeneous with zero boundary conditions. 3. Then you solve the next harmonic, non-homogeneous equation. 4. If the domain is a square, this is easily done without Mathematica. $\endgroup$ – Alexei Boulbitch Aug 13 '18 at 12:31
  • $\begingroup$ I trasformation rectangular domain to normalize domail [0 1]*[0 1]. if I want solve by mathematica, how to solve this problem $\endgroup$ – Mohammad Bagher Bagheri Aug 13 '18 at 14:06
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    $\begingroup$ Please respond why the linked topic does not help with your case. And please do not delete the question once it is marked a duplicate. If you provide information that makes it distinct, it will be reopened. $\endgroup$ – Kuba Aug 16 '18 at 6:16