Overlapping Schwarz

From Polymath1Wiki

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 $Ω 0$ the interior of the incircle. This is tangent to the segments $A B, B C$ and $C A$ at $F, D, E$ respectively. Denote by $Γ 10$ the segment of the circle connecting $E, F$ . Denote by $Ω 1$ the sector $A E F$ , and by $Γ 01$ the arc connecting $E, F$ . Note that $Ω 1$ and $Ω 2$ 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 $(u_{i}^n$ , $λ i)$ 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 $Γ 0 i$ , then one can use local Fourier-Bessel expansions to get $u_i^n$ . Then one uses their traces onto $Γ 10$ , 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 $Ω i$

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 $Γ 01$ 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 $Ω 1$ 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 $A E, A F < m a t h > , a s w e l l a s s a t i s f y t h e e q u a t i o n < m a t h > - Δ w k = λ w k$ .

So, we suppose $u_1^n = \sum_{k=1}^M c_k w_k (r,\theta)$ . We want to find the coefficients $c k$ so as to satisfy the boundary condition on $Γ 01$ . Now, $Γ 01$ is an arc of radius $ρ 1 = A E$ . Let $(ρ 1,θ j)$ be $2 M$ 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 $Ω 0$

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 $Γ i 0$ .

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 $z m (r,θ) = J m (λ r) e i m θ$ , 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 $Γ i 0$ .

{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.

Overlapping Schwarz

From Polymath1Wiki

Solving on the wedge $Ω i$

Solving on the disk $Ω 0$

{Convergence?}

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Overlapping Schwarz

From Polymath1Wiki

Solving on the wedge Ωi

Solving on the disk Ω0

{Convergence?}

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Navigation

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Solving on the wedge $Ω i$

Solving on the disk $Ω 0$