An Intro to Variational Inequalities

Leonid Prigozhin

Ben-Gurion University of the Negev, Blaustein Inst. for Desert Research, Israel

Lanzhou, September 2016

Outlook: Introduction: examples, extensions, comments, history. Superconductivity: cylinder in a parallel ﬁeld. The primal, dual, and mixed v.i. formulations. Electric ﬁeld calculation.

Lanzhou 2016 Intro to V.I. All these conditions can be written as a variational inequality: 0 Find x0 ∈ [a, b]: f (x0)(x − x0) ≥ 0 for all x ∈ [a, b]. If f is strictly convex, the solution is unique.

Part I. Example 1

Suppose f is a smooth function and we solve

f (x0) = min f (x). [a,b] Three cases are possible:

0 if a < x0 < b then f (x0) = 0; 0 if x0 = a then f (x0) ≥ 0; 0 if x0 = b then f (x0) ≤ 0;

Lanzhou 2016 Intro to V.I. Part I. Example 1

Suppose f is a smooth function and we solve

f (x0) = min f (x). [a,b] Three cases are possible:

0 if a < x0 < b then f (x0) = 0; 0 if x0 = a then f (x0) ≥ 0; 0 if x0 = b then f (x0) ≤ 0;

All these conditions can be written as a variational inequality: 0 Find x0 ∈ [a, b]: f (x0)(x − x0) ≥ 0 for all x ∈ [a, b]. If f is strictly convex, the solution is unique.

Lanzhou 2016 Intro to V.I. Two equivalent formulations: (1) Minimization of energy (2) Poisson equation:

min{E(u): u|∂Ω = 0} −λ∆u = f , u|∂Ω = 0.

Here (2) is the Euler-Lagrange equation for (1) and is the necessary and suﬃcient condition for minimum.

Example 2: a membrane

Elastic membrane y = u(x), u|∂Ω = 0. The elastic energy ∼ to the change of membrane area Z q Z 2 2 λ 2 U = λ 1 + ux1 + ux2 − 1 dΩ ≈ |∇u| dΩ, Ω 2 Ω where λ characterizes the elastic properties. R External force work A = Ω{fu}dΩ. The total potential energy is R λ 2 E(u) = U − A = Ω 2 |∇u| − fu dΩ.

Lanzhou 2016 Intro to V.I. Example 2: a membrane

min{E(u): u|∂Ω = 0} −λ∆u = f , u|∂Ω = 0.

Here (2) is the Euler-Lagrange equation for (1) and is the necessary and suﬃcient condition for minimum.

Lanzhou 2016 Intro to V.I. —————————————————————————— Optimality conditions: I. necessary Let u ∈ K be the (unique) solution of (1). If v ∈ K then v α = (1 − α)u + αv ∈ K for any 0 < α < 1 so E(u) < E(v α). α Then dE(v ) = R [λ∇u · ∇(v − u) − f (v − u)] dΩ ≥ 0. dα α=0+ Ω We arrived at the variational inequality: u ∈ K s.t. a(u, v − u) − (f , v − u) ≥ 0 for any v ∈ K, (2) R R where a(φ, ψ) = λ Ω ∇φ · ∇ψdΩ and (φ, ψ) := Ω φ ψdΩ.

Example 2: a membrane and an obstacle

An obstacle: u(x) ≥ g(x), where g|∂Ω < 0. Free boundary problem for Poisson equation? Better to solve a constr. min problem for energy:

min{E(u): u ∈ K}, (1)

where K = {v : v ≥ g, v|∂Ω = 0} is the admissible set.

Lanzhou 2016 Intro to V.I. Example 2: a membrane and an obstacle

An obstacle: u(x) ≥ g(x), where g|∂Ω < 0. Free boundary problem for Poisson equation? Better to solve a constr. min problem for energy:

min{E(u): u ∈ K}, (1)

where K = {v : v ≥ g, v|∂Ω = 0} is the admissible set. —————————————————————————— Optimality conditions: I. necessary Let u ∈ K be the (unique) solution of (1). If v ∈ K then v α = (1 − α)u + αv ∈ K for any 0 < α < 1 so E(u) < E(v α). α Then dE(v ) = R [λ∇u · ∇(v − u) − f (v − u)] dΩ ≥ 0. dα α=0+ Ω We arrived at the variational inequality: u ∈ K s.t. a(u, v − u) − (f , v − u) ≥ 0 for any v ∈ K, (2) R R where a(φ, ψ) = λ Ω ∇φ · ∇ψdΩ and (φ, ψ) := Ω φ ψdΩ.

Lanzhou 2016 Intro to V.I. Example 2: a membrane and an obstacle

An obstacle: u(x) ≥ g(x), where g|∂Ω < 0. Free boundary problem for Poisson equation? Better to solve a constr. min problem for energy:

min{E(u): u ∈ K}, (1)

where K = {v : v ≥ g, v|∂Ω = 0} is the admissible set. —————————————————————————— Optimality conditions: II. suﬃcient If u solves the v.i. u ∈ K s.t. a(u, v − u) − (f , v − f ) ≥ 0 for any v ∈ K (2) i.e. λ(∇u, ∇{v − u}) − (f , v − u) ≥ 0 for any v ∈ K then E(v) = E(u + {v − u}) = λ 2 E(u) + 2 |∇(v − u)| + λ(∇u, ∇{v − u}) − (f , v − u) ≥ E(u), Hence u solves (1) and v.i. is also a suﬃcient opt. condition.

Lanzhou 2016 Intro to V.I. Variational inequalities: comments...

1. If the solution is smooth, a(u, v − u) = (−λ∆u, v − u). Hence, the v.i. is a weak form of

Find u ∈ K s.t. (−λ∆u − f , v − u) ≥ 0 for any v ∈ K,

However, often solutions of a v.i. are less smooth than that of the b.v.p. without obstacle and weak formulations are employed. We can still write Find u ∈ K s.t. (Au − f , v − u) ≥ 0 for any v ∈ K, where Au is deﬁned by: (Au, v) = λ(∇u, ∇v) for any v such that v|∂Ω = 0.

Lanzhou 2016 Intro to V.I. Variational inequalities: comments...

2. Are the constrained optimization problems suﬃcient? Not every equation Au = 0 is the Euler-Lagrange equation for some functional. A simple example: du/dt − f (t) = 0. Another example: the diﬀusion equation. Hence, there can be no appropriate equivalent optimization problem. However, if a solution has to satisfy an a priory constraint, u ∈ K, the v.i. formulation

Find u ∈ K :(Au, φ − u) ≥ 0 for any φ ∈ K

still makes sense. Such are diﬀusion problems with a unilateral constraint on the boundary if there is a semi-penetrable membrane.

Lanzhou 2016 Intro to V.I. Variational inequalities: comments...

3. Models for growing sandpiles and type-II superconductors also give rise to evolutionary problems with unilateral constraints. After discretization in time the stationary v.i.-s on each time layer are equivalent to constrained optimization problems. These evolutionary v.i.-s are also limits of highly nonlinear equations. Both approaches can be employed for numerical solution. Nevertheless, the v.i. formulations are exact and convenient.

Lanzhou 2016 Intro to V.I. Variational inequalities: comments...

4. In the v.i. Find u ∈ K :(Au, v − u) ≥ 0 for any v ∈ K K should be a non-empty closed convex set. Its indicator function

0 v ∈ K, Φ(v) = ∞ v ∈/ K

is convex (and lower semi-continuous, i.e. limv→v0 Φ(v) ≥ Φ(v0)). The inequality can be written as an unconstrained problem

Find u :(Au, v − u) + Φ(v) − Φ(u) ≥ 0 for any v. (3)

This is an example of extended variational inequalities which appear also with other non-diﬀerentiable functionals Φ; we will use such formulations too.

Lanzhou 2016 Intro to V.I. Let sand be discharged onto a rigid support surface y = h0(x). The slope of a growing sandpile y = h(x, t) should not exceed the repose angle of sand, |∇h| ≤ tan αrep wherever h > h0. The uncovered by sand parts of the support, where h = h0, can be steeper. If the support has such steep parts, the inequality is quasi-variational: what part of h0 is uncovered depends on the solution, i.e. K = K(h).

Variational inequalities: comments...

5. More complicated formulations are quasi-variational inequalities,

Lanzhou 2016 Intro to V.I. Variational inequalities: comments...

5. More complicated formulations are quasi-variational inequalities,

Find u ∈ K(u):(Au, v − u) ≥ 0 for any v ∈ K(u), Here the set K depends on the unknown solution itself. Examples: The Kim model in superconductivity: the current density in a sc j ∈ K where K consists of functions satisfying the constraint |u| ≤ jc (|b|). Since the induction b depends on j, the formulation is a quasi-variational inequality: K = K(j). Let sand be discharged onto a rigid support surface y = h0(x). The slope of a growing sandpile y = h(x, t) should not exceed the repose angle of sand, |∇h| ≤ tan αrep wherever h > h0. The uncovered by sand parts of the support, where h = h0, can be steeper. If the support has such steep parts, the inequality is quasi-variational: what part of h0 is uncovered depends on the solution, i.e. K = K(h).

Lanzhou 2016 Intro to V.I. History

The study of v.i.-s has started in 60-s and continued in 70-s by Signorini, Fichera, Stampacchia, Lions, Baiocchi, Brezis ... Variational inequalities are a natural generalization of the boundary value problems and arise in various problems of Mechanics, Physics, Math. Economy, and Control Theory. First quasi-variational inequality was introduced by Bensoussan and Lions in 1973. Many problems leading to v.i.-s are considered here: G. Duvaut and J.-L. Lions, “Inequalities in Mechanics and Physics”, 1972. J.-F. Rodrigues, “Obstacle Problems in Mathematical Physics”, 1987. The classical book on numerical methods for v.i.-s is R. Glowinski, J.-L. Lions, R. Tremolieres, “Numerical Analysis of Variational Inequalities”, 1981.

Lanzhou 2016 Intro to V.I. Part II. SC cylinder in a parallel ﬁeld

2 Let the sc occupy {x ∈ Ω ⊂ R , −∞ < z < ∞} and he = he (t)iz . Then h = h(x, t)iz , e, j(x, t) ⊥ iz and h|∂Ω = he (t). In the cylinder cross-section Ω we have: Faraday law Amp`erelaw µ0∂t h + ∇ × e = 0, ∇ × h = j

and h|t=0 = h0. The Bean critical-state model:

|j| ≤ jc , ekj, |j| < jc ⇒ e = 0.

Set h = u + he (t).

Then u|∂Ω = 0 and j = ∇ × u = (∂x2 u, −∂x1 u). Hence

|j| ≤ jc ⇔ |∇u| ≤ jc . The admissible set: for any t

u(., t) ∈ K = {v : |∇v| ≤ jc in Ω, v|∂Ω = 0}. . Lanzhou 2016 Intro to V.I. SC cylinder: the primal v.i. formulation

Let u ∈ K, j = ∇ × u and e satisfy the Bean model, v ∈ K. From the Faraday law µ0(∂t (u + he ), v − u) = −(∇ × e, v − u) and

(∇ × e, v − u) = (e, ∇ × {v − u}) = (e, ∇ × v) − (e, j) ≤ (|e|, {|∇v| − jc }) ≤ 0.

We arrived at a variational inequality (the primal v.i. formulation):

Find u(x, t) s.t. u(., t) ∈ K for all t, u|t=0 = h0 − he (0), and (∂t {u + he }, v − u) ≥ 0 for any v ∈ K.

Solution for a simply-connected Ω (Barrett and P., 00):

non-decreasing he (t): u(x, t) = max{−dist(x, ∂Ω), h0(x) − he (t)}, non-increasing he (t): u(x, t) = min{dist(x, ∂Ω), h0(x) − he (t)}.

Extension of this solution for multiply-connected domains and the q.v.i. in the case of the Kim model: Barrett and P., 10.

Lanzhou 2016 Intro to V.I. Diﬀerent current-voltage relations

We assumed ekj and employed the multivalued e(j) law a. |E| |E| |E|

|J| |J| |J|

Jc Jc Jc a b c Are there similar formulations for other popular choices? Bossavit, 94: Let e(j) increase monotonically. Then R |j| R φ(j) := 0 e(s)ds is a convex function and Φ(j) := Ω φ(j(x))dx is a convex functional. For any w, j we obtain:

φ(w) − φ(j) ≥ e(|j|)(|w| − |j|) ⇒ R Φ(w) − Φ(j) ≥ Ω e(|j|)(|w| − |j|)dx. If we assume e and j have the same direction, then e(|j|)|j| = e · j while e(|j|)|w| ≥ e · w. Hence for any w Φ(w) − Φ(j) ≥ (e, w − j).

Lanzhou 2016 Intro to V.I. General current-voltage relation: e as a subgradient

Let Φ be a convex functional and for any w hold Φ(w) − Φ(j) ≥ (e, w − j). Then e is a subgradient of Φ at the point j, or an element of the subdiﬀerential: e ∈ ∂Φ(j). This seems to be the most general form of the current - voltage relation leading to a variational formulation. Since, from the Faraday law µ0(∂t (u + he ), v − u) + (e, ∇ × {v − u}) = 0 for any v s.t. v|∂Ω = 0, we obtain

µ0(∂t (u + he ), v − u) + Φ(∇ × v) − Φ(∇ × u) ≥ 0.

If Φ is diﬀerentiable, one gets an equation: e(j) = δΦ/δj. Otherwise a variational inequality should be solved.

Lanzhou 2016 Intro to V.I. Comments

1. Choosing an appropriate functional Φ and setting e ∈ ∂Φ(j) is the most natural way to account for, e.g., flux-cutting in the double critical-state model (Kashima 08) or material anisotropy. 2. For the critical-state models Φ is the non-differentiable indicator function of the set K. The e(j) relation is multivalued and, although h = u + he can be computed or even found analytically, the electric field e remains unknown. p−1 3. For the power law e = e |j| j one solves 0 jc jc

e0 p−1 µ0∂t h − p ∇ · (|∇h| ∇h) = 0. jc The solution h tends to that of the Bean model as p → ∞ (Barrett and P. 00). However, if p is large, accurate calculation of e is diﬃcult.

Lanzhou 2016 Intro to V.I. Dual V.I. in terms of the electric ﬁeld

Let e and j = ∇ × h obey the Bean model, |j| ≤ jc , ekj, |j| < jc ⇒ e = 0. For h = he (t) + u and any test ﬁeld v we obtain

(∇ × u) · (v − e) = j · (v − e) ≤ jc |v| − j · e = jc |v| − jc |e|.

Integrating the Faraday law µ0∂t h + ∇ × e = 0 we obtain h = u + h (t) = u + h (0) + 1 ∇ × R t edt. Hence e 0 e µ0 0 (∇ × u, v − e) = (u, ∇ × {v − e}) = 1 Z t f + ∇ × edt, ∇ × {v − e} , µ0 0

where f = u0 − he (t) + he (0). Finally, for any time t and any v 1 Z t f + ∇ × edt, ∇ × {v − e} ≤ (jc , |v| − |e|) µ0 0 Can be solved. But to use a mixed formulation (e and h) is easier.

Lanzhou 2016 Intro to V.I. Advantages: All variables can be calculated (as for the dual v.i. form.); Much simpler ﬁnite elements can be used for e (vectorial p-w constant instead of the Raviart-Thomas el).

Can be extended to the q.v.i. case jc (h) (as the primal v.i.);

Mixed V.I. formulation

We already showed that for any test ﬁeld v

(∇ × h, v − e) ≤ (jc , |v| − |e|). (1)

The Faraday law µ0∂t h + ∇ × e = 0 and the Green formula yield

µ0(∂t h, φ) + (e, ∇ × φ) = 0, (2)

where φ is any smooth enough scalar test function s.t. φ|∂Ω = 0. To complete the formulation: h|t=0 = h0, h|∂Ω = he (t).

Lanzhou 2016 Intro to V.I. Mixed V.I. formulation

We already showed that for any test ﬁeld v

(∇ × h, v − e) ≤ (jc , |v| − |e|). (1)

The Faraday law µ0∂t h + ∇ × e = 0 and the Green formula yield

µ0(∂t h, φ) + (e, ∇ × φ) = 0, (2)

where φ is any smooth enough scalar test function s.t. φ|∂Ω = 0. To complete the formulation: h|t=0 = h0, h|∂Ω = he (t).

Advantages: All variables can be calculated (as for the dual v.i. form.); Much simpler ﬁnite elements can be used for e (vectorial p-w constant instead of the Raviart-Thomas el).

Can be extended to the q.v.i. case jc (h) (as the primal v.i.);

Lanzhou 2016 Intro to V.I. Numerical solution of the mixed v.i.

1 Rotated variables: let E = [e2, −e1], V = [v2, −v1]. Then |E| = |e|, |V | = |v| and ∇ × φ · e = −∇φ · E. Hence (∇h, V − E) + (jc , |V | − |E|) ≥ 0, µ0(∂t h, φ) − (∇φ, E) = 0.

2 1 1+ε Smoothing: |E| ≈ 1+ε |E| for 0 < ε 1. ε−1 The inequality boils down to ∇h + jc |E| E = 0.

3 Time discretization: µ0(h − hˆ, φ) − ∆t(∇φ, E) = 0. 4 F.E. discretization: nonconforming linear el. for h, φ; vectorial p-w constant el. for E.

5 Iterations to resolve the nonlinearity: on the i-th iteration ε−1 i−1 ε−1 i−1 i−1 ε−1 i i−1 |E| E ≈ |E | E + (|E |δ) (E − E ) p 2 2 where |E|δ = |E| + δ with 0 < δ 1. Over-relaxation.

Lanzhou 2016 Intro to V.I. Scaling

In the numerical examples we assumed h(x, 0) = 0 and dhe /dt = const. We used the dimensionless variables: 0 0 0 0 0 x t j h e t0 x = , t = , j = , h = , e = 2 . L t0 jc Ljc L j0 Here L is the characteristic cross-section size, t0 is chosen to make dhe /dt = 1.

Lanzhou 2016 Intro to V.I. Simulation: homogeneous cylinders in a growing ﬁeld

Levels of |e|:

Levels of h and current contours:

Magnetization of homogeneous cylinders of diﬀerent cross-sections. Dimensionless variables, he (t) = t, shown for t = 0.3.

Lanzhou 2016 Intro to V.I. Simulation: inhomogeneous cylinder in a growing ﬁeld

he = 0.3

white: jc = 1/3, gray: jc = 1

h levels e and |e| levels

Lanzhou 2016 Intro to V.I. Simulation: hollow cylinder in a growing ﬁeld (he = 0.6)

Left – e and contours of j, right – |e| levels, bottom – h levels

The magnetic ﬁeld penetrates the hole through a thin channel; the electric ﬁeld is very strong along this path.

Lanzhou 2016 Intro to V.I. Simulation: hollow cylinder in a growing ﬁeld (he = 0.6) Left – e and contours of j, right – |e| levels, bottom – h levels

תודה רבה! !谢谢！ Thank you! Спасибо

Lanzhou 2016 Intro to V.I.