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What is the best way to impose a "constant constraint" for a PDE? Specifically, I want to solve an eigenvalue problem $Au=\lambda u$ on the rectangle $(0,2\pi)\times(-\pi/2,\pi/2)$ with periodicity constraints horizontally and constant constraint on the lines $y=\pi/2, y=-\pi/2$. Here a constant constraint means that $u(x,\pi/2)$ is constant and the constant is not fixed. The same in $-\pi/2$ with possibly a different constant. I use a finite difference grid.

Is there any way to impose the constant constraint in the matrix $M$ which is the discretized of $A$?

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Let's say you have numbered your vertices along the line $y=-\pi/2$ as $u_1,u_2,\ldots,u_n$ (so that you'll have a total of $N=nm$ vertices, with $m$ rows of vertices covering your domain). Then your constraint can be written as $$ u_2=u_1, \\ u_3=u_1, \\ \vdots \\ u_n=u_1. $$ (And the same, obviously, for the vertices at the upper boundary.) This can be written as a homogenous constraint $CU=0$ where $C$ is a rectangular matrix where the constraints above will yield $m-1$ rows and $N$ columns.

Solving linear systems $AU=F$ where $CU=0$ is exactly the kind of operation we always do when we deal with hanging nodes, for example. Typically, we would eliminate $u_2, \ldots, u_n$ from the linear system.

You can find a number of descriptions of this in the literature. The earliest, I believe, is in Babuska and Reinboldt from the 1970s. I'm quite sure it's in Graham Carey "Computational Grids" book. I know it's in mine and Oliver Kayser-Herold's 2009 paper on hp, see # 27 at http://www.math.tamu.edu/~bangerth/publications.html#x-reviewed .

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$$ \left.\frac{\partial u}{\partial x}\right|_{y=\pi/2} = 0 $$ would be one strategy, though it's a little unusual to impose a tangential derivative boundary condition, and your system will likely be under-constrained if you don't fix the value of $u$ at at least one point.

If your $A$ is a decent operator, then your combination of periodic and fixed constant conditions will likely remove the $y$ dimension from the problem all together, allowing you to solve a 1-D problem instead. Can you say a little more about what $A$ is?

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  • $\begingroup$ $A$ is the Laplace-Beltrami operator on the sphere written in polar coordinates. My goal is to compute the Laplace-Beltrami eigenvalues of a piece of a sphere $\Omega$ by considering the auxiliary problem $-\Delta_{LB} u + \mu u = \lambda u$ where $\mu$ is $+\infty$ on the complementary of $\Omega$ (numerically, something large). This works reasonably well in 2D on a rectangle, so I thought if I could "flatten" the sphere to a rectangle I could get it to work like this. $\endgroup$
    – Beni Bogosel
    Commented Dec 4, 2013 at 11:27
  • $\begingroup$ You said $x$ and $y$ in your original question. Are you including the curvature terms in your operator, or are your variables really $\theta$ and $\phi$? I suspect that if you analyze your problem from a theoretical perspective, you'll find that it's 1D or quasi-1D and amenable to analytic solution. $\endgroup$
    – Bill Barth
    Commented Dec 4, 2013 at 22:38
  • $\begingroup$ Yes, I included the boundary terms in the operator, and the variables are the angles $\theta,\phi$. I will look into what you are saying. $\endgroup$
    – Beni Bogosel
    Commented Dec 4, 2013 at 23:06

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