# Overlapping Schwarz

(Difference between revisions)
 Revision as of 01:36, 19 July 2012 (view source) (New page: Our goal is to either analytically solve, or numerically approximate, the 2nd Neumann eigenfunction of the Laplacian on a generic acute triangle $\Omega \equiv$ABC (see figure)....)← Older edit Current revision (01:39, 19 July 2012) (view source)m Line 26: Line 26: - + ==Solving on the wedge $\Omega_i$== - + - + - \begin{figure}[htbp] %  figure placement: here, top, bottom, or page + - \centering + - \includegraphics[width=2in]{ddmethod.eps} + - \caption{Domains} + - \label{fig:example} + - \end{figure} + - ===Solving on the wedge $\Omega_i$=== + We are at step $n$ of the iteration. We are at step $n$ of the iteration. Suppose $i=1$ is fixed for concreteness, and the traces of $u_0^{n-1}$ and $\frac{\partial u_0^{n-1}}{\partial \nu}$ on $\Gamma_{01}$ are known. Suppose $i=1$ is fixed for concreteness, and the traces of $u_0^{n-1}$ and $\frac{\partial u_0^{n-1}}{\partial \nu}$ on $\Gamma_{01}$ are known.

## Current revision

Our goal is to either analytically solve, or numerically approximate, the 2nd Neumann eigenfunction of the Laplacian on a generic acute triangle $\Omega \equiv$ABC (see figure). Since we don't have an analytic solution on this domain, but can solve problems on sectors or the circle, we propose a domain decomposition strategy in the spirit of the overlapping Schwarz iteration.

Let 0 be the incenter of the triangle, and denote by Ω0 the interior of the incircle. This is tangent to the segments AB,BC and CA at F,D,E respectively. Denote by Γ10 the segment of the circle connecting E,F. Denote by Ω1 the sector AEF, and by Γ01 the arc connecting E,F. Note that Ω1 and Ω2 overlap. Repeat this process for the other three vertices. Let us denote the exact second Neumann eigenvalue as λ.

Now we'll proceed by iteration. At step n, suppose $u_i^n$ for i = 1,2,3 satisfies

$-\Delta u_i^n = \lambda^n_i u_i, x\in \Omega_i, \frac{\partial u_i^n}{\partial \nu}=0, x\in \partial \Omega_i \setminus{\Gamma_{0i}}$ and the Robin condition $u_0^{n-1} \frac{\partial u_i^{n}}{\partial \nu} - \frac{\partial u_0^{n-1}}{\partial \nu} u_i^n =0, x\in \Gamma_{0i}$

The function $u_0^n$ solves $-\Delta u_0^n = \lambda^n_0 u_0^n, x\in \Omega_0, u_i^{n} \frac{\partial u_0^{n}}{\partial \nu} - \frac{\partial u_i^{n}}{\partial \nu} u_0^n =0, x\in \Gamma_{i0},$

Now $(u_{i}^n$, λi) solve generalized eigenfunction problems on sectors of circles. If one knows $u_0^{n-1}, \frac{\partial u_0^{n-1}}{\partial \nu}$ on the curves Γ0i, then one can use local Fourier-Bessel expansions to get $u_i^n$. Then one uses their traces onto Γ10, and has the correct data to solve the eigenvalue problem for $u_0^n$ on the disk. One can use Fourier-Bessel expansions to do this as well.

The claim is that as $n\rightarrow \infty$, the sequences $u_i^n$ converge to the restriction of the actual eigenfunction u on the sub-domains. Clearly we have to prescribe a starting guess for the iteration.

## Solving on the wedge Ωi

We are at step n of the iteration. Suppose i = 1 is fixed for concreteness, and the traces of $u_0^{n-1}$ and $\frac{\partial u_0^{n-1}}{\partial \nu}$ on Γ01 are known. $-\Delta u_1^n = \lambda^n_1 u_1, x\in \Omega_1, \frac{\partial u_1^n}{\partial \nu}=0 x\in \partial \Omega_1\setminus{\Gamma_{01}}, u_0^{n-1} \frac{\partial u_1^{n}}{\partial \nu} - \frac{\partial u_0^{n-1}}{\partial \nu} u_1^n =0, x\in \Gamma_{01}$

We will try the method of particular solutions of Fox and Henrici, adapted to this problem. We know that since the opening angle is α, in Ω1 the functions $w_k(r,\theta):= J_{\frac{\pi k}{\alpha}} \sqrt{\lambda} r) \cos(\frac{\pi k}{\alpha} \theta)$ will satisfy the Neumann conditions on the line segments AE,AF < math > ,aswellassatisfytheequation < math > − Δwk = λwk.

So, we suppose $u_1^n = \sum_{k=1}^M c_k w_k (r,\theta)$. We want to find the coefficients ck so as to satisfy the boundary condition on Γ01. Now, Γ01 is an arc of radius ρ1 = AE. Let 1j) be 2M collocation points along this curve. At each point, we want to enforce $0= u_0^{n-1}(\rho_1,\theta_j) \frac{\partial u_1^{n}}{\partial \nu} (\rho_1,\theta_j)- \frac{\partial u_0^{n-1}}{\partial \nu}(\rho_1,\theta_j) u_1^n(\rho_1,\theta_j)$ $= u_0^{n-1}(\rho_1,\theta_j) \sum_{k=1}^M c_k \frac{\partial}{\partial r} J_{\frac{\pi k}{\alpha}} (\sqrt{\lambda} \rho_1) \cos(\frac{\pi k}{\alpha} \theta_j) - \frac{\partial u_0^{n-1}}{\partial \nu}(\rho_1,\theta_j) \sum_{k=1}^M c_k J_{\frac{\pi k}{\alpha}} (\sqrt{\lambda} \rho_1) \cos(\frac{\pi k}{\alpha} \theta_j) .$

This is equivalent to solving the rectangular nonlinear system $A(\lambda) \vec{c}=0$ where $a_{jk}(\lambda) = u_0^{n-1}(\rho_1,\theta_j) \frac{\partial}{\partial r} J_{\frac{\pi k}{\alpha}} (\sqrt{\lambda} \rho_1) \cos(\frac{\pi k}{\alpha} \theta_j) - \frac{\partial u_0^{n-1}}{\partial \nu}(\rho_1,\theta_j)J_{\frac{\pi k}{\alpha}} (\sqrt{\lambda} \rho_1) \cos(\frac{\pi k}{\alpha} \theta_j)$

We find the solutions by looking for values of λ so that the smallest singular value of A(λ) approaches 0. This is the Moler approach to the original Fox-Henrici-Moler paper.

Once we locate the solution $\vec{c}$, we have the iterate $u_1^n$. We do this same process for the other wedges as well.

### Solving on the disk Ω0

We are at step n of the iteration, and have solved for the functions $u_i^n$. We therefore have their traces on the arcs Γi0.

The function $u_0^n$ solves $-\Delta u_0^n = \lambda^n_0 u_0^n, x\in \Omega_0, u_i^{n} \frac{\partial u_0^{n}}{\partial \nu} - \frac{\partial u_i^{n}}{\partial \nu} u_0^n =0, x\in \Gamma_{i0}$ We shall again use a Fourier-Bessel ansatz: let zm(r,θ) = Jmr)eimθ, and assume $u_0^n(r,\theta) = \sum_{k=0}^M d_m z_m(r,\theta)$. Repeat the process above of enforcing the (non-standard) boundary conditions at collocation points along Γi0.

### {Convergence?}

At the end of the nth step, we have 4 functions: $u_{i}^n$, i=0,1,2,3 and 4 eigenvalues. We repeat the process until the eigenvalues are all the same number.