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I solved a simple test example by overlapping domain decomposition. The problem domain is a rectangular that is decomposed to two domains.

So the value on the intersection boundary is guessed at the first time and updated at each step.

Now I want to solve this problem by non-overlapping method. But when I use this procedure, the answer in all steps are the same. Am I wrong? or I should solve this, differently?

Please hint me.

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  • $\begingroup$ Could you share your non-overlapping algorithm? $\endgroup$ – Paul Jun 17 '14 at 15:31
  • $\begingroup$ @paul I did not use any special algorithm for this, the domain is rectangle '[-1,1]*[0 1]'. the subdomains are '[-1 0.2]*[0 1]' and '[-0.2 1]*[0 1]'. at the first I guessed the unknown boundary condition (as Dirichlet)for the left domain, then solved on this domain, for the unknown boundary in the right domain, I updated my initial guess by approximate it with nodal value of the left domain's solution. $\endgroup$ – rosa Jun 18 '14 at 9:15
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Short answer: yes, you have to use something different, e.g. a Neumann-Neumann method. A good reference is Widlund's book. Non-overlapping methods are based on the principle that, if $u$ solves the Poisson equation in a domain $\Omega$ which has been partitioned into two domains $\Omega_1$, $\Omega_2$, at the shared boundary $\Gamma$ of both domains, $u$ and its gradient have to match up across $\Gamma$.

By contrast, for an overlapping DD method, the intersection of the domains $\Omega_1$ and $\Omega_2$ has a non-empty interior. You can use the interior values of $u$ in $\Omega_1$ as Dirichlet boundary conditions to solve the Poisson equation in $\Omega_2$, then use the interior values in $\Omega_2$ to update the solution in $\Omega_1$, and so forth. If you try and apply an overlapping method to domains that don't actually overlap, there's no room for the solutions on each sub-domain to disagree, and so you'll update neither.

More to the point, the convergence rate of an overlapping DD method goes like $H/h$, where $H$ is the width of the overlap and $h$ is the mesh width*. There's a trade-off: you can have a really small overlap and cheap subdomain solves, but the method won't converge fast; or you can have a large overlap with more expensive subdomain solves, but the whole method converges faster.

*Unless you use a global coarse solve, but now we're getting off topic.

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