MADRID LECTURE

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MADRID LECTURE 5

• Numerical solution of Eikonal equations

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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 ? H10(O) such that
• (EIK-L)
• ?u/?xi 1 a.e. on O, ? i
  1, , d,
• with O a bounded domain of Rd of boundary G.

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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 ? ?O vdx.
• This leads us to consider the following problem
  from
• Calculus of Variations
• u ? E,
• (EIK-L-MAX)
• ?O udx
  ?Ovdx, ? v ? E,
• with
• E vv? H10(O), ?v/?xi 1 a.e. on
  O, ? i 1, , d .

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Some exact solutions (useful for numerical
validation)

• Suppose that O x x x1,x2, x1 ? x2 lt 1
  the unique
• solution of (EIK-L-MAX) is given on O by
• u(x) 1 x1
  x2.
• In general, (EIK-L-MAX) has no solution. Assuming
  howe
• er that such u exists, we can easily show that
• u 0 on O?G.
• Morever since ?v2 d in E, problem (EIK-L-MAX)
  is
• equivalent to

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(UP) u ? E J(u) ? J(v), ? v ? E,

• with
• E v v ? E, v 0
  in O
• and
• J(v) ½ ?O?v2dx C
  ?Ovdx,
• 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

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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 ?2vL2(O)
  (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 Je

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Je (v) ½e1 ?O?v2dx J(v)

• ¼ e2 1Sdi 1 ?O(?v/?xi2 1)2dx
• above e e1, e2 with e1, e2 both positive and
  small. Then,
• we approximate (UP) by (UP)e defined as follows

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The problem (UP)e

• ue ? K ?H2(O),
• (UP)e
• Je(ue) ? Je (v), ? ? K?H2(O),
• with K v v ? H10(O), v 0 in O. (UP)e 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

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3. An equivalent formulation of (UP)e

• Let us denote (L2(O))d by ? and ?ue by pe
  problem (UP)e
• is clearly equivalent to
• pe ? ? ,
• (UP-E)e
• je(pe) ? je(q), ? q ? ? ,
• with q qidI 1 and, ? q ? ? ,

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je(q) ½ ?Oq2dx C ?O??1.qdx

• ¼ e2 1Sdi 1?O(qi2 1)2dx
  I(q)
• here
• ? ?1 is the unique solution in H10(O) of
• ??1 1 in O, ?1 0 on G.
• ? I(.) is defined by
• ½ e1 ?O ?.q2dx if q ? ?(K?H2(O)),
• I(q)
• 8 elsewhere.

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I(.) is convex, proper, l.s.c. over the space ? .

• The Euler-Lagrange equation of (UP-E)e reads as
  follows
• (after dropping most es)
• ?O p.q dx e2 1Sdi 1?O(pi2 1)pi
  qi dx lt?I(p),qgt
• (E-L)
• C ?O??1.qdx,? q ? ? p ? ? .

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

• p(0) p0,
• ?O(?p/?t).q dx ?O p.q dx e2 1Sdi
  1?O(pi2 1)pi qi dx
• (E-L-F)
• lt?I(p),qgt C ?O??1.qdx,? q ? ? p(t) ? ? ,
  t ? (0,8).
• 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).

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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/?t)?O(pn ½ pn).q dx ?O pn ½ .q
  dx
• (1) e2 1Sdi 1?O(pin ½ 2 1)pin ½ qi dx
  C ?O??1.qdx,
• ? q ? ? pn ½ ? ? ,

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

• (1/?t)?O(pn 1 pn ½).q dx lt?I(pn
  1 ),qgt 0,
• (2)
• ? q ? ? , pn 1.
• What about the solution of (1) and (2)?
• (i) Problem (1) has a unique solution as soon
  as
• ?t ? e2
• Problem (1) can be solved pointwise indeed,

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pin ½(x) is then the unique solution of a single
variable cubic equation of the following type

• (1 ?te2 1 ?t)z ?te2 1z3 RHS
• whose Newtons solution is trivial.
• (ii) The solution of (ii) is given by
• pn1 ?un1
• where pn1 is the solution of the following
  obstacle type
• variational inequality

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(EVI) un 1 ? K?H2(O),

• ?O?un 1.?(v un 1)dx ?te1 ?O?2un 1?2(v un
  1)dx
• ?Opn ½.?(v un 1)dx , ? v ? K?H2(O).
• In order to facilitate the numerical solution of
  (EVI) we
• perform a variational crime by approximately
• factoring (EVI) as follows

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?n 1 ? K,

• (P1) ?O??n 1.?(v ?n 1)dx ?Opn ½.?(v ?n
  1)dx ,
• ? v ? K
• followed by
• (P2) un 1 ?te1?2un 1 ?n 1 in O, un 1
  0 on G.
• The maximum principle implies the 0 of un 1(?
  H2(O)).

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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).

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4. Numerical experiments

• Test problems 1 2 O (0,1) (0,1)
• ?u/?x1 ?u/?x2 1 in O, u g on
  G,
• 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
  O.
• C 10 or 10, h 1/128, ?te1 h2/36,
  e2 10 3, ?t 10 4

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uh u2 103, uh u8 102

• Test problem 3 O (0,1) (0,1)
• ?u/?x1 ?u/?x2 1 in
  O, u 0 on G,
• with C 10, h 1/512 and 1/1024, ?te1 h2/9, e2
  10 3, ?t 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)?

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Solution of the true Eikonal equation

• Test problem 4 O (0,1) (0,1)
• ?u 1 in O, u 0 on G.
• The maximal solution is the distance to the
  boundary
• function
• Parameters C 10, h 1/256, ?te1 h2/49, e2
  10 3, ?t 10 4

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• 
  

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Solution of the true Eikonal equation

• Test problem 5 O (0,1) (0,1)
• ?u 1 in O, u g on G,
• 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, ?te1 h2/49, e2
  10 3,
• ?t 10 4

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