Overlapping Schwarz

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 $\Omega_0$ the interior of the incircle. This is tangent to the segments $AB, BC$ and $CA$ at $F,D, E$ respectively. Denote by $\Gamma_{10}$ the segment of the circle connecting $E,F$. Denote by $\Omega_1$ the sector $AEF$, and by $\Gamma_{01}$ the arc connecting $E,F$. Note that $\Omega_1$ and $\Omega_2$ overlap. Repeat this process for the other three vertices. Let us denote the exact second Neumann eigenvalue as $\lambda$.

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 $(u_{i}^n$, $\lambda_i)$ 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 $\Gamma_{0i}$, then one can use local Fourier-Bessel expansions to get $u_i^n$. Then one uses their traces onto $\Gamma_{10}$, 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 $\Omega_i$

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 $\Gamma_{01}$ 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 $\alpha$, in $\Omega_1$ 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 $AE, AF\ltmath\gt, as well as satisfy the equation \ltmath\gt-\Delta w_k =\lambda w_k$.

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 $\Gamma_{01}$. Now, $\Gamma_{01}$ is an arc of radius $\rho_1=AE$. Let $(\rho_1,\theta_j)$ 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 $\lambda$ so that the smallest singular value of $A(\lambda)$ 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 $\Omega_0$

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 $\Gamma_{i0}$.

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 $z_m(r,\theta) = J_m(\lambda r) e^{im\theta}$, 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 $\Gamma_{i0}$.

{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.