mpccdist1 details:

Keywords: analytic solution

Global classification: nonlinear-quadratic

Functional: convex quadratic

Geometry: easy, fixed

Design: coupled via volume data

Differential operator:

Design constraints:

State constraints:

Mixed constraints:



Submitted on 2012-08-21 by Roland Herzog. Published on 2012-08-21


mpccdist1 description:


Introduction

The problem at hand is an optimal control problem in which the state is determined by variational inequality, viz. the elliptic obstacle problem, rather than by a partial differential equation. In fact, the variational inequality is formulated equivalently as an elliptic equation plus a complementarity system. Consequently, the optimal control problem is a function space MPCC (mathematical program with equilibrium constraints).

The problem and its solution are taken from [Meyer and Thoma2013, Example 7.1].

Variables & Notation

Unknowns

u L2(Ω) control variable y H01(Ω)state variable ξ L2(Ω) slack variable

Given Data

The given data is chosen in a way which admits an analytic solution.

Ω = (0,1)2 computational domain Γ its boundary Ω1 (see below) subdomain of Ω Ω2 = (0.0,0.5) × (0.0,0.8)subdomain of Ω Ω3 = (0.5,1.0) × (0.0,0.8)subdomain of Ω
yd(x) = 400q1(y1) + q2(y2)y=Qx,x Ω1 z1(x1)z2(x2), x Ω2 0 elsewhere desired state (discontinuous) ud = p1(Qx), x Ω1 z1(x1) z2(x2), x Ω2 z1(x1 0.5)z2(x2),x Ω3 0 elsewhere desired control (discontinuous)

The subdomain Ω1 is a square with midpoint x̂ = (0.8,0.9) and edge length 0.1, which has been rotated about its midpoint by 30 degrees in counter-clockwise direction. The four vertices of Ω1 can thus be obtained from

x̂x̂x̂x̂ +Q 0.05 0.050.050.05 0.05 0.05 0.05 0.05 0.78170.86830.81830.7317 0.8317 0.8817 0.9683 0.9183

with the rotation matrix

Q = cos π 6 sin π 6 sin π 6 cos π 6 .

Note that Ω1 does not intersect Ω2 nor Ω3. The remaining pieces of data are

z1(x1) = 4096x16 + 6144x 15 3072x 14 + 512x 13 z2(x2) = 244.140625x26 + 585.937500x 25 468.750x 24 + 125x 23 q1(y1) = 200(y1 0.8)2 + 0.5 q2(y2) = 200(y2 0.9)2 + 0.5 p1(y1,y2) = q1(y1)q2(y2).

Problem Description

Minimize1 2y ydL2(Ω)2 + 1 2u udL2(Ω)2 s.t. y = u + ξ in Ω y = 0 on Ω y 0,ξ 0,yξ = 0in Ω.

Optimality System

Besides the state y H01(Ω), control u L2(Ω) and slack variable ξ L2(Ω), the optimality system consists of the adjoint state p H01(Ω) and a Lagrange multiplier μ H1(Ω) pertaining to the constraint y 0. The adjoint state p serves a double role, since it also acts as Lagrange multiplier for the pointwise constraint ξ 0. As usual for MPCCs, no multiplier is introduced for the constraint yξ = 0.

It should be noted that for MPCCs, a canocical first-order optimality condition does not exist. The following system represents a particular set of first-order necessary conditions, viz. of strongly stationary type.

y = u + ξ in Ω y = 0 on Ω p = y yd + μin Ω p = 0 on Ω u ud p = 0 in Ω y 0,ξ 0,yξ = 0 in Ω μy = 0 in Ω a weak sense pξ = 0 in Ω p 0 in B μ 0 in B in a weak sense.

The set B = {x Ω : y(x) = ξ(x) = 0} is termed the bi-active set. It is the last two conditions on the signs of p and μ which are particular for the concept of strong stationarity.

Since μ belongs only to H1(Ω), two of the conditions above must be imposed in a weak sense. This can be done in the following way:

μ,vH1(Ω),H01(Ω) = 0for all v H01(Ω) satisfying v(x) = 0 where y(x) = 0 μ,vH1(Ω),H01(Ω) 0for all v H01(Ω) satisfying v(x) 0 where y(x) = 0 and v(x) = 0 where ξ(x) > 0.

Supplementary Material

The following functions given in [Meyer and Thoma2013, Example 7.1] satisfy the set of necessary optimality conditions of strongly stationary type above. An important feature of this selection is that there is a nontrivial bi-active set:

B = {x Ω : y(x) = ξ(x) = 0} = (0.0,1.0) × (0.8,1.0).

Moreover, second-order optimality conditions have been verified, and thus (y,ξ,u) is guaranteed to represent a local minimum.

y = z1(x1)z2(x2),x Ω2 0 elsewhere (of class C2(Ω¯)) u = z1(x1) z2(x2), x Ω2 z1(x1 0.5)z2(x2),x Ω3 0 elsewhere ξ = z1(x1 0.5)z2(x2),x Ω3 0 elsewhere (continuous) p = p1(Qx),x Ω1 0 elsewhere (continuous, but not C1(Ω)) μ,vH1(Ω),H01(Ω) =Ω1p|Ω1 n1vds,

where n1 is the unit outer normal to the rotated square subdomain Ω1. Note that μ is a line functional concentrated on Ω1. In more explicit terms, it can be expressed as

μ,vH1(Ω),H01(Ω) =0.750.85Q 0.5q1(x1) 20q1(x1) 0 1 v(x1,0.85)dx1 +0.750.85Q 0.5q1(x1) 20q1(x1) 0 1 v(x1,0.95)dx1 +0.850.95Q 20q2(x2) 0.5q2(x2) 1 0 v(0.75,x2)dx2 +0.850.95Q 20q2(x2) 0.5q2(x2) 1 0 v(0.85,x2)dx2.

The remaining data are

z1(x 1) = 122880x14 + 122880x 13 36864x 12 + 3072x 1 z2(x 2) = 7324.218750x24 + 11718.75x 23 5625x 22 + 750x 21 q1(x 1) = 400(x1 0.8) q2(x 2) = 400(x2 0.9).

References

   C. Meyer and O. Thoma. A priori finite element error analysis for optimal control of the obstacle problem. SIAM Journal on Numerical Analysis, 51(1):605–628, 2013. doi: 10.1137/110836092.