# Numerical Solution of a Non-Smooth Eigenvalue Problem

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Numerical Solution of a Non-Smooth Eigenvalue ProblemAn Operator-Splitting Approach

A. Caboussat & R. Glowinski

1. Formulation. MotivationOur main objective is the numerical solution of the following problem from Calculus of Variations

Compute

= inf v |v|dx (NSEVP)

where: is a bounded domain of R2 and

= {v| v H01(), |v|2dx = 1}.

Actually, = 2 , independently of the shape and sizeof (holds even for non-simply connected and in fact for unbounded ) (G. Talenti).

A natural question is then:

Why solve numerically a problem whose exact solution is known ?

(i) If I claim that it is a new method to compute nobody will believe me.(ii) (NSEVP) is a fun problem to test solution methods for non-smooth & non-convex optimization problems.

(iii) |v|dx arises in a variety of problems from Image Processing and Plasticity.

Actually, our motivation for investigating (NSEVP) arises from the following problem from visco-plasticity :

u L2(0,T; H01()) C0([0,T ]; L2()); u(0) = u0, (BFP) (u/t)(t)(v u(t))dx + u(t).(v u(t))dx + g[ |v|dx |u(t)|dx ] C(t)(v u(t))dx, v H01(), a.e. t (0, T), with > 0, > 0, g > 0, a bounded domain of R2 and u0 L2().

(BFP) models the flow of a Bingham visco-plastic fluid in an infinitely long cylinder of cross section , C being the pressure drop per unit length. Suppose that C = 0 and that T = +; we can show that

(C-O.PR) u(t) = 0, t Tc,

with Tc = (/0)ln[1 + (0/g)||u0||L2()],

0 being the smallest eigenvalue of 2 in H01().

A similar cut-off property holds if after space discretization we usethe backward Euler scheme for the time discretization of (BFP),with 0 and replaced by their discrete analogues 0h and h.

Suppose that the space discretization is achieved via C0-piecewiselinear finite element approximations, we have then

|0h 0| = O(h2).

But what can we say about |h | ?

The main goal of this lecture is to look for answers to the above question !

2. Some regularization procedures There are several ways to approximate (NSEVP) at the continuous level by a better posed and/or smoother variational problem. The most obvious candidate is clearly

= inf v (|v|2 + 2)dx, (NSEVP.1)

a regularization quite popular in Image Processing. Assuming that the above problem has a minimizer u, thisminimizer verifies the following Euler-Lagrange equation (reminiscent of the mean curvature equation):

First regularized problem:

(RP.1) is clearly a nonlinear eigenvalue problem for a close variant of the mean curvature operator, the eigenvalue being .

Another regularization, more sophisticated in some sense, since this time the regularized problem has minimizers, is provided (with > 0) by = min v [ |v|2dx + |v|dx ]. (NSEVP.2)

An associated Euler-Lagrange (multivalued) equation reads as follows, also of the nonlinear (in fact, non-smooth) eigenvalue type (as above the eigenvalue is ):

2u + j(u) u in ,(RP.2) u = 0 on , |u|2dx = 1;

in (RP.2), j(u) is the subgradient at u of the functional j : H01() R defined by j(v) = |v|dx. The solution of problems such as (RP.2) is discussed inGKM (2007); the method used in the above referenceis of the operator-splitting/inverse power method type.

In order to avoid handling simultaneously two small parameters, namely and h, we will address the solutionof = inf v |v|dx

without using any regularization (unless we consider the space approximation as a kind of regularization, that it is indeed).

3. Finite Element Approximation(i) First, we introduce a family {h}h of polygonal approxi- mations of , such that

limh0 h = .

(ii) With each h we associate a triangulation Th verifying the usual assumptions of: (a) compatibility between triangles, and (b) regularity.

(iii) With each Th we associate the finite dimensional space V0h defined (classically) as follows:

V0h = {v| v C0(hh), v|T P1, T Th, v = 0 on h}.

(iv) We approximate = inf v |v|dx (NSEVP)by h = min v h h |v|dx (NSEVP)h

with h = {v| v V0h, ||v||L2(h) = 1}. It is easy to prove that:

(i) Problem (NSEVP)h has a solution, that is there exists uh h such that

h |uh|dx = h.

(ii) limh0 h = ( = 2).

We would like to investigate (computationally) the orderof the convergence of h to . From the non-smoothness of the problem, we do not expect O(h2).

4. An iterative method for the solutionof (NSEVP)h We are going to look for robustness, modularity and simplicity of programming instead of performance measured in number of elementary operations (this is not image processing and/or real time). At ADI 50 ( December 2005, at Rice University), we showed that the inverse power method for eigenvalue computations has an operator-splitting interpretation; we also showed the equivalence between some augmented Lagrangian algorithms and ADI methods such as Douglas-Rachfords and Peaceman-Rachfords. For problem (NSEVP)h we think that it is simpler to take the AL approa-ch, keeping in mind that it will lead to a disguised ADI method.

For formalism simplicity, we will use the continuous problem notation. We observe that there is equivalencebetween = inf v |v|dxand = inf {v, q, z} E |q|dx,where

E = {{v, q, z}| v H01(), q (L2())2, z L2(), v q = 0, v z = 0, ||z||L2() = 1}.

The above equivalence suggests introducing the followingaugmented Lagrangian functional

Lr : (H01()QL2())(QL2()) R

defined as follows, with Q = (L2())2 and r = {r1, r2}, ri > 0,

Lr(v, q, z; 1, 2) = |q|dx + r1 |v q|2dx + r2 |v z|2dx + (v q).1dx + (v z)2dx

We consider then, the following saddle-point problem

Find {{u, p, y}, {1, 2}} (H01()QS)(QL2())such that

Lr(u, p, y; 1, 2) Lr(u, p, y; 1, 2) Lr(v, q, z; 1, 2), (SDP-P) {{v, q, z}, {1, 2}} (H01()QS)(QL2()),

with S = {z| z L2(), ||z||L2() = 1}.

Suppose that the above saddle-point problem has a solution. We have then p = u, y = u, u being a minimizer for the original mimimization problem (the primal one).

To solve the above saddle-point problem, we advocate the algorithm below (called ALG 2 by some practitioners (BB)):

(1) {u 1, {10, 20}} is given in H01()(QL2());

for n 0, assuming that {un 1, {1n, 20}} is known, solve:

(2) {pn, yn} = arg min{q, z} QS Lr(un 1, q, z; 1n, 2n),then(3) un = arg minv Lr(v, pn, yn; 1n, 2n), v H01(),

(4) 1n+1 = 1n + r1(un pn), 2n+1 = 2n + r2(un yn).

The above algorithm is easy to implement since:(i) Problem (3) is equivalent to the following linear variational problem in H01()

un H01(), r1un.v dx + r2 unv dx = (r1pn 1n ).v dx + (r2yn 2n )v dx, v H01().

The solution of the discrete analogue of the above problem is a simple task nowadays.

(ii) Problem (2) decouples as

(a) pn = arg min q Q [ r1 |q|2dx + |q|dx (r1un + 1n).qdx ].(b) yn = arg min z S [ r2 |z|2dx (r2un + 2n)zdx ].

Both problems have closed form solutions; indeed, since ||z||L2() = 1, z S, one has yn = (r2un + 2n) / ||r2un + 2n ||L2().

Similarly, the minimization problem in (a) can be solved point-wise (one such elementary problem for each triangle of Th, in practice). We obtain then, a.e. on ,

pn(x) = (1/r1) (1 1/|Xn(x)|)+ Xn(x),where Xn(x) = r1un(x) + 1n(x).

5. Numerical experimentsFirst Test Problem: is the unit disk

Unit Disk Test ProblemVariation of h versus h

Unit Disk Test ProblemVariation of h versus h

Unit Disk Test ProblemVisualisation of the coarse mesh solution

Unit Disk Test ProblemVisualisation of the fine mesh solution

Unit Disk Test ProblemCoarse mesh solution contours

Unit Disk Test ProblemFine mesh solution contours

Unit Disk Test ProblemFine mesh solution contours (details)

Second Test Problem: is the unit squareCoarse mesh

Unit Square Test ProblemFine mesh

Unit Square Test ProblemVariation of h versus h

Unit Square Test ProblemVariation of h versus h

Unit Square Test Problem

Visualisation of the coarse mesh solution

Unit Square Test ProblemVisualisation of the fine mesh solution

Unit Square Test ProblemContours of the coarse mesh solution

Unit Square Test ProblemContours of the fine mesh solution

Unit Square Test Problem

Contours of the fine mesh solution (details)

Circular Ring Test Problem (coarse mesh)

Circular Ring Test Problem (fine mesh)

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