MADRID LECTURE # 5 Numerical solution of Eikonal equations

1. Historical Background. Motivated by the analysis of nonlinear models from micro-Mechanics and Material Science, B. Dacorogna (EPFL-Math) was lead to investigate the properties of the solutions (including the non-existence of smooth solutions) of the following system of Eikonal-like equations: Find u  H 1 0 (Ω) such that (EIK-L) |∂ u/ ∂ x i | = 1 a.e. on Ω,  i = 1, …, d, with Ω a bounded domain of R d of boundary Γ.

Since problem (EIK-L) has a infinity of solutions we are going to focus on those solutions which maximize (or “nearly” maximize) the functional v → ∫ Ω vdx. This leads us to consider the following problem from Calculus of Variations: u  E, (EIK-L-MAX) ∫ Ω udx ≥ ∫ Ω vdx,  v  E, with E = {v|v  H 1 0 (Ω), |∂ v/ ∂ x i | = 1 a.e. on Ω,  i = 1, …, d }.

Some exact solutions (useful for numerical validation): Suppose that Ω = {x| x = {x 1,x 2 }, |x 1  x 2 | < 1}; the unique solution of (EIK-L-MAX) is given on Ω by: u(x) = 1 – |x 1 | – |x 2 |. In general, (EIK-L-MAX) has no solution. Assuming howe er that such u exists, we can easily show that u ≥ 0 on Ω  Γ. Morever since |  v| 2 = d in E, problem (EIK-L-MAX) is equivalent to:

(UP) u  E + ; J(u)  J(v),  v  E +, with E + = {v| v  E, v ≥ 0 in Ω} and J(v) = ½ ∫ Ω |  v| 2 dx – C ∫ Ω vdx, C being an arbitrary positive constant. The iterative solution of (UP) will be discussed in the following sections. Our methodology will be elliptic in nature, unlike the one developed by PA Gremaud for the

solution of (EIK-L). Gremaud methology is hyperbolic in nature and relies on TVD schemes. PAG considers (EIK-L) as a kind of Hamilton-Jacobi equation. 2. Penalty/Regularization of (UP). Motivated by Ginzburg-Landau equation, we are going to treat,  i = 1, …, d, the relations |∂ v/ ∂ x i | = 1 by exterior penalty. Moreover, in order to control mesh-related oscillations, we are going to bound ||  2 v|| L 2 (Ω) (without this additional constraint our method did not work; PAG did something like that, too). This leads to approximate the functional J in (UP) by the following J ε :

J ε (v) = ½ε 1 ∫ Ω |Δv| 2 dx + J(v) + ¼ ε 2 – 1 Σ d i = 1 ∫ Ω (|∂v/∂x i | 2 –1) 2 dx; above ε = {ε 1, ε 2 } with ε 1, ε 2 both positive and small. Then, we approximate (UP) by (UP) ε defined as follows:

The problem (UP) ε u ε  K +  H 2 (Ω), (UP) ε J ε (u ε )  J ε (v),   K +  H 2 (Ω), with K + = {v| v  H 1 0 (Ω), v ≥ 0 in Ω}. (UP) ε is a ‘beautiful’ obstacle problem. Proving the existence of solutions by compactness methods is easy, the real difficulty being the actual computation of the solutions. r

3. An equivalent formulation of (UP) ε Let us denote (L 2 (Ω)) d by Λ and  u ε by p ε ; problem (UP) ε is clearly equivalent to: p ε  Λ, (UP-E) ε j ε (p ε )  j ε (q),  q  Λ, with q = {q i } d I = 1 and,  q  Λ,

j ε (q) = ½ ∫ Ω |q| 2 dx – C ∫ Ω  1.qdx + ¼ ε 2 – 1 Σ d i = 1 ∫ Ω (|q i | 2 –1) 2 dx + I + (q); here:   1 is the unique solution in H 1 0 (Ω) of – Δ  1 = 1 in Ω,  1 = 0 on Γ.  I + (.) is defined by ½ ε 1 ∫ Ω | .q| 2 dx if q   (K +  H 2 (Ω)), I + (q) = + ∞ elsewhere.

I + (.) is convex, proper, l.s.c. over the space Λ. The Euler-Lagrange equation of (UP-E) ε reads as follows (after dropping most εs): ∫ Ω p.q dx + ε 2 – 1 Σ d i = 1 ∫ Ω (|p i | 2 –1)p i q i dx + = (E-L) C ∫ Ω  1.qdx,  q  Λ ; p  Λ.

To (E-L) we associate the following flow to capture asymptotically solutions of (EL) p(0) = p 0, ∫ Ω (∂p/∂t).q dx + ∫ Ω p.q dx + ε 2 – 1 Σ d i = 1 ∫ Ω (|p i | 2 –1)p i q i dx (E-L-F) + = C ∫ Ω  1.qdx,  q  Λ ; p(t)  Λ, t  (0,+∞). The structure of (E-L-F) strongly calls for an Operator – Splitting solution; motivated by its simplicity*, we will apply the Marchuk-Yanenko scheme to the solution of (E-L-F).

Operator-splitting solution of (E-L-F)(I) (0) p 0 = p 0 ( = 0, or  1, for exemple); then, for n ≥ 0, p n → p n + ½ → p n + 1 as follows: (1/Δt) ∫ Ω (p n + ½ – p n ).q dx + ∫ Ω p n + ½.q dx + (1) ε 2 – 1 Σ d i = 1 ∫ Ω (|p i n + ½ | 2 –1)p i n + ½ q i dx = C ∫ Ω  1.qdx,  q  Λ; p n + ½  Λ,

Operator-splitting solution of (E-L-F)(II) (1/Δt) ∫ Ω (p n +1 – p n + ½ ).q dx + + = 0, (2)  q  Λ, p n +1. What about the solution of (1) and (2)? (i) Problem (1) has a unique solution “as soon” as: Δt  ε 2 Problem (1) can be solved pointwise; indeed,

p i n+ ½ (x) is then the unique solution of a single variable cubic equation of the following type: (1 – Δtε 2 – 1 + Δt)z + Δtε 2 – 1 z 3 = RHS whose Newton’s solution is trivial. (ii) The solution of (ii) is given by p n+1 =  u n+1 where p n+1 is the solution of the following obstacle type variational inequality:

(EVI) u n +1  K +  H 2 (Ω), ∫ Ω  u n +1.  (v – u n +1 )dx + Δtε 1 ∫ Ω  2 u n +1  2 (v – u n +1 )dx ≥ ∫ Ω p n+ ½.  (v – u n +1 )dx,  v  K +  H 2 (Ω). In order to facilitate the numerical solution of (EVI) we perform a ‘variational crime’ by ‘approximately’ factoring (EVI) as follows:

ω n +1  K +, (P1) ∫ Ω  ω n +1.  (v – ω n +1 )dx ≥ ∫ Ω p n+ ½.  (v – ω n +1 )dx,  v  K + followed by (P2) u n +1 – Δtε 1  2 u n +1 = ω n +1 in Ω, u n +1 = 0 on Γ. The maximum principle implies the ≥ 0 of u n +1 (  H 2 (Ω)).

The finite element implementation is easy, the main requirement being that the mesh preserves the maximum principle at the discrete level. Of course, everything we say applies to: (i) Non-homogeneous boundary conditions. (ii) The ‘true’ Eikonal equation |  u| = f (≥ 0).

4. Numerical experiments Test problems 1 & 2: Ω = (0,1) × (0,1) |∂u/∂x 1 | = |∂u/∂x 2 | = 1 in Ω, u = g on Γ, with g(x 1,0) = g(x 1,1) = min(x 1,1 – x 1 ), 0  x 1  1, g(0, x 2 ) = g(1, x 2 ) = min(x 2,1 – x 2 ), 0  x 2  1. We have then: u max (x) = min(x 1,1 – x 1 ) + min(x 2,1 – x 2 ), u min (x) = |x 1 – x 2 | or |1 – (x 1 – x 2 )| depending where x is in Ω. C = 10 or – 10, h = 1/128, Δtε 1 = h 2 /36, ε 2 = 10 – 3, Δt = 10 – 4

||u h – u|| 2 ≈ 10 –3, ||u h – u|| ∞ ≈ 10 –2 Test problem 3: Ω = (0,1) × (0,1) |∂u/∂x 1 | = |∂u/∂x 2 | = 1 in Ω, u = 0 on Γ, with C = 10, h = 1/512 and 1/1024, Δtε 1 = h 2 /9, ε 2 = 10 – 3, Δt = 10 – 4 This problem has no solution; the weak limit should be the distance function. The approximate solutions exhibit self- -similar multi-scale structures (fractals)?

Solution of the true Eikonal equation Test problem 4: Ω = (0,1) × (0,1) |  u| = 1 in Ω, u = 0 on Γ. The maximal solution is the distance to the boundary function Parameters: C = 10, h = 1/256, Δtε 1 = h 2 /49, ε 2 = 10 – 3, Δt = 10 – 4

Solution of the true Eikonal equation Test problem 5 : Ω = (0,1) × (0,1) |  u| = 1 in Ω, u = g on Γ, with g(x 1,0) = 0, g(x 1,1) = min(x 1,1 – x 1 ), 0  x 1  1, g(0, x 2 ) = g(1, x 2 ) = 0, 0  x 2  1. Parameters: C = 10, h = 1/256, Δtε 1 = h 2 /49, ε 2 = 10 – 3, Δt = 10 – 4