The half-planes problem for the level set equation

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The half-planes problem for the level set equation

Stéphane Clain, Malcom Djeno Ngomanda

To cite this version:

Stéphane Clain, Malcom Djeno Ngomanda. The half-planes problem for the level set equation. Inter- national Journal of Numerical Analysis and Modeling, Institute for Scientific Computing and Infor- mation, 2013, 10 (1), pp.99. �hal-00467651�

Mod´elisation Math´ematique et Analyse Num´erique

THE HALF-PLANES PROBLEM FOR THE LEVEL SET EQUATION

St´ephane Clain¹ and Malcom Djenno Ngomanda^2,3

Abstract. The paper is dedicated to the construction of an analytic solution for the level set equation in R² with an initial condition constituted by two half-planes. Such a problem can be seen as an equivalent Riemann problem in the Hamilton-Jacobi equation context. We first rewrite the level set equation as a non-strictly hyperbolic problem and obtain a Riemann problem where the line sharing the initial discontinuity corresponds to the half-planes junction. Three different solutions corresponding to a shock, a rarefaction and a contact discontinuity are given in function of the two half-planes configuration and we derive the solution for the level set equation. The study provides theoretical examples to test the numerical methods approaching the solution of viscosity of the level set equation.

We perform simulations to check the three situations using a classical numerical method on a structured grid.

1991 Mathematics Subject Classification. 65M10.

.

1. Introduction

The interface tracking problem takes place in various fields like front flame propagation, crystal growth in solidification process, fluid-structure interaction with moving solid boundary, computer vision, dynamics bubbles or drops for example. The level set method (see [Set96] for an overview) consists in representing the free boundary as the zero-level of a continuous functionφwhere the normal velocityF is a prescribed function.

Consider the Cauchy problem:

(∂_tφ(x, t) +F|∇φ(x, t)|= 0 inR²×[0, T],

φ(x, t= 0) =φ0(x) inR², (1)

where F(x, t) is a given Lipstick function onR²×[0, T] while φ0 is a Lipschitz function onR². Existence and uniqueness of the Lipschitz viscosity solution for problem (1) onR²×[0, T] is proved (see [Bar94, Lio82]).

Keywords and phrases: Riemann problem, Cauchy problem, level set equation, analytical solution, half-planes problem

1 Institut de math´ematiques, CNRS UMR 5219, Universit´e Paul Sabatier Toulouse 3, 118 route de Narbonne, F-31062 Toulouse cedex 4,France

e-mail: stephane.clain@math.univ-toulouse.fr

2 Laboratoire de math´ematiques, CNRS UMR 6620, Universit´e Blaise Pascal, BP 10448, F-63000 Clermont-Ferrand, France

e-mail: djenno@math.univ-bpclermont.fr

3 Universit des Sciences et Techniques de Masuku (USTM), B.P 941 Franceville, Gabon.

c EDP Sciences, SMAI 1999

Sinceφis a Lipschitz function, vectorU=∇φis a bounded vector-valued function and applying the gradient operator to equation (1), we derive the Cauchy problem for the conservation law system associated to the level set equation:

(∂_tU(x, t) +∇(F|U(x, t)|) = 0 inR²×[0, T],

U(x, t= 0) =∇φ0(x) inR². (2)

For the one dimensional situation, [Lio82] proves that the Lipschitz viscosity solution φ of equation (1) corresponds to the bounded entropic solutionU of equation (2) withU =∂xφ. Such a result is not established for higher dimension since we do not have the uniqueness of solution for the non-strictly hyperbolic system (2) [CFN95]. Nevertheless, if we regularize equations (1) and (2) adding a diffusion term ε∆φ and ε∆U respectively, the new solutions satisfyU_ε=∇φε. Passing to the limit assuming aL^∞convergence ofφεtoward φ and aL¹ convergence of U_ε toward U, we have that U = ∇φ where φ is the viscosity solution andU the entropy solution.

From a numerical point of view, the non-strictly hyperbolic system discretization using the finite volume method leads to solve Riemann problems for each cell interface. The problem is reduced to a one dimensional hyperbolic equation with a discontinuous flux function [CL83] making difficult a complete and explicit resolution.

We propose a different approach based on the following remark: a constant state for the non-strictly hyperbolic problem (2) corresponds to a plane for the level set equation (1). Hence we propose to study an equivalent Riemann problem in the level set equation context using two half-planes as an initial continuous condition [Cla02, Dje07].

2. The two half-planes problem

For the sake of simplicity, we assume in the following that functionF is reduced to a constant function. Such an assumption is not restrictive since we usually solve Riemann problems using the normal velocity evaluated at the interface midpoint.

Letπ1andπ2be two planes of the (x, z) = (x1, x2, z)∈R³space. We impose that the planes contain the origin point and the plane equations write

z=φ_i(x) =U_i.x, i= 1,2

where U_i are given vectors ofR². Of course, when the two planes are equal, one has φ0(x) =φ1(x) =φ2(x) as an initial condition and the solution is given by φ(x, t) = φ0(x)−F|U_L|t with U_L = U1 = U2 which corresponds to a simple translation of the initial plane with velocityF|U_L|(see figure (1)).

t = 0

|U_L|t

Figure 1: The trivial case when φ0 =φ1 = φ2. The solution representation in R³ corresponds to the initial plane translated following theOz axis with velocity F|U_L|=F|∇φ0|(we takeF = 1 in the figure).

Now, we consider the nontrivial case when the two planes are different. To construct an initial conditionφ0

for the Cauchy problem (1) we consider an arbitrary lineδ⊂R²passing to the origin, an arbitrary normal vector W∈R² and we define the left partPL and the right partPR ofR² such thatWgoes from left to right (see figure (2)). We then construct the initial condition byφ0(x) =φ1(x) ifx∈ PL andφ0(x) =φ2(x) ifx∈ PR. Such a definition gives rise to a function which is nota priori continuous onδand disqualify the construction since we would like to handle continuous solutions. It results that the interfaceδcan not be arbitrary but has to be defined such that we have a continuous connection between the two planes. To this end, we introduce the two following continuous Lipschitz functions

φ^m₀ = min(φ1, φ2), φ^M₀ = max(φ1, φ2)

which are the unique function linking π1 withπ2 continuously. Note that the particular caseU1 =U2 corre- sponds to a unique function φ^m₀ =φ^M₀ =φ1=φ2.

AssumingU₁6=U₂, we set

W= U₁−U₂

|U1−U2|

and define the two half-planes PL andPR ofR² by

PL={x∈R²; x.W<0}, PR={x∈R²; x.W>0}.

The lineδ_c=PL∩ PRis orthogonal toWand we choose the unit vectorVonδ_c such that the vectors{W,V} form a direct orthonormal basis ofR² (see figure (2)). Note that by constructionWgoes fromPL toPR.

δ _c P

P R L

x x

z

2

1

W V

Figure 2: Orientation and definition ofP^LandP^R.

InR³, the two planesπ1 and π2 have an intersection line ∆c =π1∩π2 ⊂R³ passing by the originO and δ_c⊂R²is the orthogonal projection of line ∆c⊂R³ on the plane (x1, x2) (see figure (3)).

z

x y π₂

π₁

∆c

δc

Figure 3: Representation of function φ^m0 in R³. The two half-planes intersection provides the line ∆^c which the orthogonal projection on (x¹, x2) corresponds to lineδc.

We now aim to study the Cauchy problem (1) using φ^m₀ or φ^M₀ as an initial condition. In the hyperbolic context, we have to consider Riemann problems (2) where the initial conditions are ∇φ^m₀ or ∇φ^M₀ with a discontinuity situated on the lineδc. Two cases arise whether we chooseφ^m0 orφ^M0 as an initial condition.

• If φ^m₀ is the initial condition, definition ofWyields that φ^m₀ =φ1 on the half-plane PL and φ^m₀ =φ2

on the half-plane PR. Indeed, we have W.W = 1 > 0 so (U1−U2).W > 0 then U1.W > U2.W which means that φ1(W)> φ2(W). Henceπ1 is aboveπ2 onPR whileπ1 is underπ2 onPL. In the

conservative law framework, we derive the following initial condition for the Riemann problemU_L=U1

onPL andU_R=U2 onPR.

• If we chooseφ^M₀ as the initial condition, we deduce thatU_L=U2 onPL andU_R=U1 onPR. In the sequel, for any arbitrary couple of vectorsU1andU2 we chooseφ^m₀ as the initial condition and we have U_L=U1 onPL andU_R=U2 onPR oriented from left to right by

W= U_L−U_R

|U_L−U_R|. (3)

2.1. The one dimensional problem reduction

The objective is to determine the theoretical solutionφof the level set equation withφ^m₀ as an initial condition using the solution U of the associated Riemann problem. To this end, we carry out the following change of variable where we introduce the coordinates in basis{W,V}setting

x=ζW+ηV.

The transformation corresponds to a rotation in the new basis where W=

1 0

, V=

0 1

.

By construction ofVandWwe haveU_L·V=U_R·V=ω0, thus, vectorsU_L andU_R are defined in the new base:

U_L=ω_LW+ω0V, U_R=ω_RW+ω0V, withωL=U_L·W,ωR=U_R·W.

Using the change of variables, the level set problem becomes







∂_tφ(ζ, η, t) +F|∇ζ,ηφ(ζ, η, t)|= 0, φ(ζ, η) =ωLζ+ω0η ifζ <0, η∈R, φ(ζ, η) =ω_Rζ+ω0η ifζ >0, η∈R,

(4)

and the associated Riemann problem:



























∂_tU(ζ, η, t) +∇ζ,η(F|U(ζ, η, t)|) = 0,

U(.,0) = ω_L ω0

!

ifζ <0, η∈R,

U(.,0) = ωR

ω0

!

ifζ >0, η∈R,

(5)

IfU(ζ, η, t) is a solution of (5) thenU(ζ, η+c, t) is also a solution of (5) for all c∈RthusU(ζ, η+c, t) = U(ζ, η, t) hence ∂_ηU= 0. It follows that

∂tU(ζ, t)· 0

1

= 0 henceU·

0 1

=ω0,∀η, ζ, t.

On the other hand, letω(ζ, t) =U· 1

0

, the component ofU following the directionW, functionωsatisfy the one-dimension scalar Riemann problem







∂_tω(ζ, t) +F ∂_ζp

w²(ζ, t) +w²₀= 0, ω(ζ,0) =ωL, ζ <0,

ω(ζ,0) =ω_R, ζ >0.

(6)

Note that the choice ofWyieldsU_L.W>U_R.Whenceω_L≥ω_R.

3. Solutions for the Riemann problem

We first give the solution for the situationω06= 0, caseω0= 0 will be studied as the limit situation whenω0

tends to 0. We denote byf(ω) =Fp

w²+w²₀, and a simple calculation provides

λ(ω) =f^′(ω) = F ω

pw²+w²₀, f^′′(ω) = F ω₀² (w²+w₀²)³².

We note thatf^′ is an increasing function for F ≥0 whilef^′ is a decreasing function ifF ≤0. SinceωL > ωR

it result thatλ(ωL)> λ(ωR) ifF >0 leading to an entropic shock configuration whileλ(ωL)< λ(ωR) ifF <0 which corresponds to a rarefaction [LeV92].

3.1. The shock case F ≥0

We first assumeω06= 0. The Rankine-Hugoniot condition yields σ= f(ωL)−f(ωR)

ω_L−ω_R =F|U_L| − |U_R|

|U_L−U_R| . (7)

ω=ω_L

ω=ω_L

ω =ω_R ξ =σt

Figure 4: Shock wave, caseω06= 0

We now consider the case where ω0 = 0 with ω_L 6=ω_R. Such a situation arises when vectorsU_L and U_R are colinear. Since ω_L 6= ω_R, then we can pass to the limit setting ω0 = 0 in relation (7) since f depends continuously onω0. The solution is still an entropic shock.

3.2. The rarefaction case F <0

We first assume thatω06= 0 and we seek for an autosimilar solution which satisfies

f^′(ω(ζ, t)) = ζ

t. (8)

Since f^′′(ω)< 0, we deduce that f^′(ω) is a one-to-one function mapping R onto ]F,−F[ and F < λ(ωL) <

λ(ωR)<−F sinceω_L> ω_R. From relation (8), we deduce

F²ω²= ζ

t 2

(ω²+ω²₀),

and we get an explicit expression ofωin the fan

ω(ζ, t) =−

ζ t|ω0| r

F²−

ζ t

², ζ

t ∈]λ(ωL), λ(ωR)[. (9)

Three cases arise in function of the ω_L andω_R signs that we display in figure (5).

(1) Ifω_L > ω_R ≥0 thenf^′(ωL)< f^′(ωR)≤0: the fan is contained in the left half-planePL and the flux on interface δ_c isF|U_R|.

(2) If 0≥ω_L> ω_R then 0≤f^′(ωL)< f^′(ωR): the fan is contained in the right half-planePR and the flux on interface δ_c isF|U_L|.

(3) At last, if ω_L > 0 > ω_R then f^′(ω_L) < 0 < f^′(ω_R), the fan crosses the line δ_c and the flux on the interface lineδ_c isF|ω0|.

ω =ωL

ω =ω(ξ, t)

ω=ωR

ξ=λ(ωL)t

ξ=λ(ωR)t

(a)

ω=ωR

ω=ω(ξ, t) ω=ωL

ξ=λ(ωL)t ξ=λ(ωR)t

(b)

ω=ωR

ω=ω(ξ, t)

ω=ωL

ξ=λ(ω_L)t ξ=λ(ωR)t

ω=ω(ξ, t)

(c)

Figure 5: Rarefaction situations withω06= 0. Three cases arise in function ofωLandωLsigns: ωR>0 (a), 0> ωL(b), ωL>0> ωR (c).

We now consider the situation whenω0vanishes. For all parametersω06= 0, functionf^′(ω;ω0) =λ(ω;ω0) is a one-to-one function ofωfromRonto ]F,−F[ and the rarefaction takes place betweenλ(ωL;ω0) andλ(ωR;ω0).

ForωL andωRfixed, we take the limitω0→0 and we have

ωlim0→0λ(ωi, ω0) =−F, ifω_i<0, lim

ω0→0λ(ωi;ω0) = +F, ifω_i>0 withi=L, R. In addition, relation (9) says

ωlim0→0w(ζ, t) = 0, ∀ζ∈]F,−F[, t >0.

Three cases again then arise.

(1) IfωL > ωR >0 then lim

ω0→0λ(ωL, ω0) = lim

ω0→0λ(ωR, ω0) = +F <0 and the rarefaction is reduced to a contact discontinuity moving with the velocityσ= +F (see figure (6-a)).

(2) If 0> ω_L < ω_R then lim

ω0→0λ(ωL, ω0) = lim

ω0→0λ(ωR, ω0) =−F >0 and the rarefaction is reduced to a contact discontinuity moving with the velocityσ=−F (see figure (6-b)).

(3) At last, if ω_L > 0 > ω_R then lim

ω0→0λ(ωL, ω0) = +F, lim

ω0→0λ(ωR, ω0) = −F and the rarefaction is reduced to the null solution in the coneF < ^ζ_t <−F (see figure (6-c)).

ω = ωR

ω = ωL

ξ = F t

(a)

ω=ωR

ξ =−F t ω =ωL

(b)

ω=ωR

ω= 0

ω =ω_L

ξ=F t ξ=−F t

ω = 0

(c)

Figure 6: The rarefaction degenerates into a contact discontinuity whenω0 = 0. Three situations arise in function of ωLandωR signs: ωL<0 (a), 0< ωR (b),ωR<0< ωL(c).

Remark 3.1. Two other cases arise whether ω_L = 0 case (i) or ω_R = 0 case (ii). This particular situation is similar to case (c) but the rarefaction cone is reduced to the left half-cone ]F,0] in case (i) and the right half-cone [0,−F[ in case (ii).

4. Solution for the level set equation

Based on the solution obtained with the Riemann problem, we build the solution for the level set equation.

We distinguish the cases whetherF is positive or negative.

4.1. The shock: case F >0

Fort≥0, we define the lineδ_c(t) parallel toδ_c(0) =δ_c moving with the velocity σ=F|U_g| − |U_d|

|U_g−U_d| .

i.e. δc(t) ={x∈R²; W.x=σt}. We then consider the two half-planes situated on the left and right side of δc(t) (see figure 7)

PL(t) ={x∈R²; x.W< σt}, PR(t) ={x∈R²; x.W> σt}.

From section 3.1, the solutionU(x, t) of the Riemann problem (2) is given by U(x, t) =U_L, x∈ PL(t),

U(x, t) =U_R, x∈ PR(t).

We then define

φ(x, t) =x.U_L−F|U_L|t, x∈ PL(t), φ(x, t) =x.UR−F|U_R|t, x∈ PR(t).

We easily check thatφis a continuous function thanks to the shock velocity definition and∇φ=U. Function φis the viscosity solution which satisfies the Cauchy problem (1) withφ0=φ^m₀ as an initial condition.

z

x y π_L

πR

P^L(t) δc(t)

P^R(t) δc

(a)

P^L(t)

δ_c(t)

P^R(t)

(b)

Figure 7: Solution for the half-planes problem: the shock case

4.2. The rarefaction: case F <0 with ω₀6= 0

Sinceω06= 0, solutionU of the Riemann problem (2) is a rarefaction situated in the coneλL< ^ζ_t < λR with λ_L=λ(ωL) =F ω_L

|U_L|, λ_R=λ(ωR) =F ω_R

|U_R|.

To describe the solutionφof the associated the Cauchy problem (1), we define the following lines and domains (see figure (8))

δ_L(t) ={x∈R²; x.W=λ_Lt}, δ_R(t) ={x∈R²; x.W=λ_Rt}, PL(t) ={x∈R²; x.W< λ_Lt}, PR(t) ={x∈R²; x.W> λ_Rt},

Pc(t) ={x∈R²; λ_Lt <x.W< λ_Rt}. The solutionU for the Riemann problem is a continuous function given by

U(x, t) =U_L=ω_LW+ω0V, x∈ PL(t), U(x, t) =U_R=ωRW+ω0V, x∈ PR(t), U(x, t) =ω(ζ, t)W+ω0V, ζ=x.W, x∈ Pc(t) where functionω(ζ, t) is given by relation (9).

On domainsPL(t) andPR(t), functionφis then given by

φ=φ_L(x, t) =x.U_L−F|U_L|t, x∈ PL(t), φ=φR(x, t) =x.UR−F|U_R|t, x∈ PR(t).

To give the analytical expression of the solution on the band Pc(t), we use the change of variables ζ, η and obtain a new functionφec(ζ, η, t) =φc(x, t) we have to explicit. On one hand, we have ∂ηφe=ω0 while on the other hand ∂_ζφe=ω(ζ, t) and∂_tφe=−Fp

ω²(ζ, t) +ω₀²with ω(ζ, t) =− ζ|ω0|

pt²F²−ζ², ζ

t ∈]λL, λR[.

We consider the functionγ(ζ, t) =|ω0|p

t²F²−ζ² and we claim thatφe_c(ζ, η, t) =γ(ζ, t) +ηω0 is the solution of the problem.

We first easily check that relations∂_ηφe=ω0and∂_ζφe_c=ω(ζ, t) are well-satisfied. We now check thatφe_csatisfies the level set equation

∂_tφe=−F|∇ζ,ηφe|=−F q

ω²(ζ, t) +ω²₀. Indeed, the time derivative provides

∂_tφe=∂_tγ(ζ, t) =F² t|ω0| pt²F²−ζ²

and, in addition, one has

F q

ω²(ζ, t) +ω₀²=F s

ζ²ω₀²

t²F²−ζ² +ω₀²=−F² t|ω0| pt²F²−ζ².

At last, we provide theφc expression in function of thexvariable:

φ_c(x, t) =|ω0| q

t²F²−(x·W)²+ (x·V)ω0, x∈ Pc(t).

To end the construction, we now prove that functionφ is continuous on the whole domain, in particularly on lineδ_L(t) andδ_R(t). Noting that lineδ_L(t) is characterized in theW,V basis by

ζ=ζ_L(t) :=F ω_L

|U_L|t and functionφeL(ζ, η, t) is given by

φ_L(ζ, η, t) =ω_Lζ+ω0ζ−F|U_L|t;

πL(t)

πR(t)

P^L(t)

P^R(t) δR(t) δL(t)

P^c(t)

(a)

P^L(t)

δ_L(t)

P^C(t)

δ_R(t)

P^R(t)

(b)

Figure 8: Solution for the half-planes problem: the rarefaction case withω06= 0

we then have on the lineδ_L(t):

φe_L(ζL(t), η, t)−φe_c(ζL(t), η, t) = ζ_L(t)ωL+ηω0−F|U_L|t−(−|ω0| q

t²F²−ζ_L²(t) +ηω0),

= F(ω_L)²

|U_L|t−F|U_L|t+|ω0| s

t²F²−

F t ω_L

|U_L| ²

,

= F t ω_L²

|U_L|− |U_L|+ |ω0|

|U_L| q

|U_L|²−ω²_L

,

= F t ω_L²

|U_L|− |U_L|+ ω₀²

|U_L|

= 0.

We deduce that functionφis continuous on lineδ_L(t) and the result also holds on lineδ_R(t)

4.3. The contact discontinuity: case F <0 withω₀= 0

The rarefaction degenerates into a contact discontinuity whenω0 tends to 0 and three situations have to be considered in functions of theω_L andω_Rsign.

First, assume thatωR< ωL<0, we denote byδc(t) the parallel line to δc given by δc(t) ={x∈R²; x.W=−F t}

and the two half-planes

PL(t) ={x∈R; x.W<−F t}, PR(t) ={x∈R; x.W>−F t}.

SolutionU(x, t) is given by

U(x, t) =U_L, x∈ PL(t), U(x, t) =U_R, x∈ PR(t).

and the viscosity solutionφis the continuous Lipschitz function defined by φ(x, t) =x.U_L−F|U_L|t, x∈ PL(t), φ(x, t) =x.U_R−F|U_R|t, x∈ PR(t).

The case where 0< ωR< ωL can be treated in the same way.

We now consider the situation where ω_R<0< ω_L, withF <0. We define the lines δL(t) ={x∈R²; x.W=F t}, δR(t) ={x∈R²; x.W=−F t} and the domain

PL(t) ={x∈R²; x.W< F t}, PR(t) ={x∈R²; x.W>−F t}, Pc(t) ={x∈R²; F t <x.W<−F t}.

πL(t)

πR(t)

P^L(t)

P^R(t) δR(t) δL(t)

P^c(t)

(a)

P^L(t)

δ_L(t)

P^C(t)

δ_R(t)

P^R(t)

(b)

Figure 9: Solution for the half-planes problem, the rarefaction case withω0= 0 andωR<0< ωL.

The solutionU for the Riemann problem is a piecewise constant function given by U(x, t) =U_L, x∈ PL(t),

U(x, t) =U_R, x∈ PR(t), U(x, t) = 0, x∈ Pc(t)