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 thenon-existence of smooth solutions) of the following system of Eikonal-like equations:

Find u H10(Ω) such that

(EIK-L)

|∂u/∂xi| = 1 a.e. on Ω, i = 1, …, d,

with Ω a bounded domain of Rd 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 ≥ ∫Ωvdx, v E,

with

E = {v|v H10(Ω), |∂v/∂xi| = 1 a.e. on Ω, i = 1, …, d }.

Some exact solutions (useful for numerical validation):

Suppose that Ω = {x| x = {x1,x2}, |x1 x2| < 1}; the uniquesolution of (EIK-L-MAX) is given on Ω by:

u(x) = 1 – |x1| – |x2|.

In general, (EIK-L-MAX) has no solution. Assuming howeer that such u exists, we can easily show that

u ≥ 0 on ΩΓ.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|2dx – C ∫Ωvdx,

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/∂xi| = 1 by exterior

penalty. Moreover, in order to control mesh-related

oscillations, we are going to bound ||2v||L2(Ω) (without thisadditional constraint our method did not work; PAG did something like that, too). This leads to approximate the functional J in (UP) by the

following Jε :

Jε (v) = ½ε1 ∫Ω|Δv|2dx + J(v) +

¼ ε2– 1Σd

i = 1 ∫Ω(|∂v/∂xi|2 –1)2dx;

above ε = {ε1, ε2 } with ε1, ε2 both positive and small. Then,we approximate (UP) by (UP)ε defined as follows:

The problem (UP)ε

uε K+ H2(Ω), (UP)ε

Jε(uε) Jε (v), K+H2(Ω),

with K+ = {v| v H10(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 (L2(Ω))d by Λ and uε by pε; problem (UP)ε

is clearly equivalent to:

pε Λ ,

(UP-E)ε

jε(pε) jε(q), q Λ ,

with q = {qi}d

I = 1 and, q Λ ,

jε(q) = ½ ∫Ω|q|2dx – C ∫Ω1.qdx +

¼ ε2– 1Σd

i = 1∫Ω(|qi|2 –1)2dx + I+(q);here:

1 is the unique solution in H10(Ω) of

– Δ1 = 1 in Ω, 1 = 0 on Γ.

I+(.) is defined by

½ ε1 ∫Ω |.q|2dx if q (K+H2(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 + ε2– 1Σd

i = 1∫Ω(|pi|2 –1)pi qi dx + <∂I+(p),q> =

(E-L)

C ∫Ω1.qdx, q Λ ; p Λ .

To (E-L) we associate the following flow to capture asymptotically solutions of (EL)

p(0) = p0,

∫Ω(∂p/∂t).q dx + ∫Ω p.q dx + ε2– 1Σd

i = 1∫Ω(|pi|2 –1)pi qi dx

(E-L-F)

+ <∂I+(p),q> = C ∫Ω1.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) p0 = p0 ( = 0, or 1, for exemple);

then, for n ≥ 0, pn → pn + ½ → pn + 1 as follows:

(1/Δt)∫Ω(pn + ½ – pn).q dx + ∫Ω pn + ½ .q dx +

(1) ε2– 1Σd

i = 1∫Ω(|pin + ½ |2 –1)pi

n + ½ qi

dx = C ∫Ω1.qdx,

q Λ; pn + ½ Λ ,

Operator-splitting solution of (E-L-F)(II)

(1/Δt)∫Ω(pn +1 – pn + ½).q dx + + <∂I+(pn +1 ),q> = 0,

(2)

q Λ , pn +1.

What about the solution of (1) and (2)?

(i) Problem (1) has a unique solution “as soon” as:

Δt ε2

Problem (1) can be solved pointwise; indeed,

pin+ ½(x) is then the unique solution of a single variable

cubic equation of the following type:

(1 – Δtε2– 1 + Δt)z + Δtε2

– 1z3 = RHS

whose Newton’s solution is trivial.

(ii) The solution of (ii) is given by

pn+1 = un+1

where pn+1 is the solution of the following obstacle type

variational inequality:

(EVI) un +1 K+H2(Ω),

∫Ωun +1.(v – un +1)dx + Δtε1 ∫Ω2un +12(v – un +1)dx ≥

∫Ωpn+ ½.(v – un +1)dx , v K+H2(Ω).

In order to facilitate the numerical solution of (EVI) we

perform a ‘variational crime’ by ‘approximately’

factoring (EVI) as follows:

ωn +1 K+,

(P1) ∫Ωωn +1.(v – ωn +1)dx ≥ ∫Ωpn+ ½.(v – ωn +1)dx ,

v K+

followed by

(P2) un +1 – Δtε12un +1 = ωn +1 in Ω, un +1 = 0 on Γ.

The maximum principle implies the ≥ 0 of un +1( H2(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/∂x1| = |∂u/∂x2| = 1 in Ω, u = g on Γ,

with

g(x1,0) = g(x1,1) = min(x1,1 – x1), 0 x1 1,

g(0, x2) = g(1, x2) = min(x2,1 – x2), 0 x 2 1.

We have then: umax(x) = min(x1,1 – x1) + min(x2,1 – x2),

umin(x) = |x1 – x2| or |1 – (x1 – x2)| depending where x is in Ω.

C = 10 or – 10, h = 1/128, Δtε1= h2/36, ε2 = 10– 3, Δt = 10– 4

||uh– u||2≈ 10–3, ||uh– u||∞ ≈ 10–2

Test problem 3: Ω = (0,1) × (0,1)

|∂u/∂x1| = |∂u/∂x2| = 1 in Ω, u = 0 on Γ,

with C = 10, h = 1/512 and 1/1024, Δtε1= h2/9, ε2 = 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, Δtε1= h2/49, ε2 = 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(x1,0) = 0, g(x1,1) = min(x1,1 – x1), 0 x1 1,

g(0, x2) = g(1, x2) = 0, 0 x 2 1.

Parameters: C = 10, h = 1/256, Δtε1= h2/49, ε2 = 10– 3,

Δt = 10– 4