Numer. Math. (1999) 84: 275–303 Digital Object Identifier (DOI) 10.1007/s002119900103

Numerische Mathematik

c Springer-Verlag 1999

Transport equation with boundary conditions for free surface localization E. Maitre? , P. Witomski Laboratoire de Mod´elisation et Calcul, Universit´e de Grenoble I, F-38041 Grenoble, France (e-mail: [email protected], [email protected]) Received November 1, 1997 / Revised version received December 9, 1998 / Published online September 24, 1999

Summary. During the filling stage of an injection moulding process, which consists in casting a melt polymer in order to manufacture plastic pieces, the free interface between polymer and air has to be precisely described. We set this interface as a zero level set of an unknown function. This function satisfies a transport equation with boundary conditions, where the velocity field has few regularity properties. In a first part, we obtain existence and uniqueness result for these equations, under weaker regularity assumptions than C. Bardos [Bar70], and C. Bardos, Y. Leroux and J.C. Nedelec [BLN79] in previous articles, but stronger assumptions than R.J. DiPerna and P.L. Lions [DL89b] who studied the case without boundary condition. We also study some regularity properties of the interface. A second part is devoted to an application to injection molding of melt polymer. We give a numerical experiment which shows that our method leads to an accurate localization of interface, which is robust, since it easily handles changes of topology of the free interface, as bubble formation or fusion of two fronts of melt polymer.

Mathematics Subject Classification (1991): 65M60, 35R35

?

Present address: Laboratoire de Math´ematiques et Applications, Universit´e de Mulhouse, 4 rue des Fr`eres Lumi`ere, F-68100 Mulhouse, France Correspondence to: E. Maitre

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1. Introduction Our problem has been drawn out of an industrial process which aims to manufacture plastic parts of cars. In order to explain quickly the process, let us assume we get melt polymer under high pressure. This fluid is injected into a mold by channels, and the air, which initially filled the mold, gets out through small holes. Our study ends when the polymer fills completely the mold, and all these holes are obstructed. One of the main difficulty of this problem is to provide a good localization of the interface between polymer and air, because physical properties of these two fluids are extremely different. Indeed, the air is almost as compressible as an ideal fluid, and his viscosity is very small, whereas polymers are almost incompressible but of high viscosity. These physical coefficients are thus discontinuous through the interface, and a good numerical calculation depends on its accurate localization. We study the flow of a non-newtonian viscous compressible fluid into a mold, under Hele Shaw assumption [Mai93], which is valid thanks to geometrical assumptions on the mold thickness and physical properties of the polymers (one could also refer to [Tuc89] or [Ken95]). In this framework the velocity is proportional to the pressure gradient, i.e. (1)

v = −S(x, t, p)∇p

where S depends on the fluid (air or polymer), and on the pressure through the viscosity (it is discontinuous across the interface, whereas the velocity is continuous). Our model consists in a pressure equation, coupled with energy equation, and with front propagation at the velocity v above. The energy equation is crucial since in the filling stage the control of temperature is essential to avoid formation of solid polymer obstacles. However the driving force for the interface displacement is definitively the pressure; in our study of interface displacement we kept only the pressure and front displacement equations, with the thermal effects appearing in a given source term. We already studied the pressure equation, (2)

∂p − χ(x, t, p)S(x, t, p) k∇pk2 − div(S(x, t, p)∇p) = Φth ∂t which can be set as a doubly nonlinear equation [MW96] (in the above equation, χ(x, t, p) accounts for the local compressibility of the air or polymer, depending where (x, t) is). Now we assume that the pressure is known, thus the velocity field is given, and we study the front propagation. There is several ways to localize free interfaces (cf. J.M. Floryan [Flo90] and R.H. Nochetto [Noc90] for a review). The one we used is the level sets method, introduced by J. Sethian [Set90, Set96], in the context of motion by χ(x, t, p)

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mean curvature. We describe the interface as the zero level set of an unknown function ψ. Thus in place of an initial fraction of the mold Ω occupied by the polymer, say Ωp0 , with inner boundary Γ0 , we choose an initial function ψ 0 such that Ωp0 = {x ∈ Ω : ψ 0 (x) > 0}. Then we seek a function ψ which satisfy the following tranport equation (cf [Mai97] for details) (3)

ψt + v · ∇ψ = 0 ψ=ψ

(4)

on Ω×]0, T [ 0

on Ω × {0},

and we defined the interface Γ (t) by def

Γ (t) = {x ∈ Ω : ψ(x, t) = 0}. Remark 1. – Computing the normal and the normal velocity of the curves ψ(x, t) = k, one may easily show that equation (3) means that each level set of ψ moves with velocity v (for a level set only the normal component of the velocity is relevant). – In [Set90], the equation verified by ψ was a parabolic equation. Only the numerical point of view is considered. A theoretical work on this equation through viscosity solutions may be found in an article of L.C. Evans and J. Spruck [ES91]. Coming back to the transport equation, we know from C. Bardos [Bar70] that some boundary condition should be prescribed on the part Γi (t) of the boundary where the velocity field is inward, i.e. has negative scalar product with the outer normal. Let this boundary condition be g; the solution of our problem must satisfy: [ ψ=g on (5) Γi (t) × {t}. t∈]0,T [

We are going to study four questions which arise in this setting. 1. Does a (at least distribution) solution to problem (3)-(5) exists under weak assumption on the velocity field? 2. What about unicity of this solution? 3. Is the zero level set of ψ independent of the choice of ψ 0 representing a given initial curve Γ0 ? (ψ globally depends on ψ 0 , of course). 4. Can we prove for Γ (t), the zero level set of ψ, some regularity properties (is it a real curve)? After this theoretical part of our work, we present the numerical simulation of the thermo-injection process. Other questions arisising in this simulation are:

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1. Which choice of ψ0 and g is convenient? 2. What kind of numerical method is relevant to compute ψ? 3. How to get a precise localization of interface from discrete values of ψ on a mesh?

2. Existence of a solution

First we recall some existence results related to such problems. In [Bar70], the case of a smooth and time independent velocity field is solved by semigroup theory. We showed in [Mai97] that this result could be extended to the time dependent case, but we had to make a non-realistic hypothesis on this time dependence. We already mentioned the paper [BLN79] in which smoothness of the velocity is assumed, and a solution of entropy kind is obtain in the spirit of Kruzkov’s results. We did not want to go further in that direction, because we expected to obtain existence of a renormalized solution to derive regularity results more easily. Renormalized solutions were introduced by [DL89b,DL89a], where the case without boundary condition, i.e. Ω = RN or when the velocity field is tangential to the boundary, was studied. Quasi-minimal assumptions were made over velocity regularity. We choose this approach and extend it to the case of non-tangent velocity field. To apply results which hold in RN , we extend our velocity field and our initial condition (using the boundary condition) to RN . The point is to construct an extension which give sufficient regularity on this initial condition and on the velocity. We encountered a technical difficulty, arising from our extension. Indeed, without an additional assumption on the velocity field (namely (12)), we could not ensure that extended characteristics out of Ω do not make a loop from linking the inward and outward boundaries. As will be mentioned below, this could make difficult the definition of our extended initial condition. This provides us with a distribution solution. Then we follow an approach of Nouri, Poupaud and Demay[NPD93] who studied the incompressible case, to prove existence of trace for this distribution solution on the inward and outward boundaries. We then derive existence and uniqueness of renormalized solution. Note that this definition of solution has strong similarities with finite volume calculation of the solution. Properties of solutions in term of maximum principle and level sets dependence with respect to data are new as far as we know. At last we stretch out that we allow the different parts of the boundary (with an inward, outward or tangential velocity) to be time dependent.

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It has a physical motivation: this corresponds to close an injection boundary (in multiple injection points) or to the end of injection, when there is an obstruction of holes in the bottom of the mold. In the following, we consider an open bounded set Ω of RN , whose boundary is piecewise C 1 . We know from [MS76,Neˇc67]) that the normal to such boundary exists almost everywhere (in fact all our results hold for a Lipschitz boundary). We denote by k.k a norm on RN . 2.1. Extension of the velocity field to RN Our method relies on an appropriate extension of the velocity field in order to apply results of R.J. DiPerna and P.L. Lions in RN . We assume that v verifies in Ω (6)

v ∈ L1 (0, T ; L∞ (Ω)) ∩ L1 (0, T ; W 1,1 (Ω))

(7)

div v ∈ L1 (0, T ; L∞ (Ω))

Remark 2. These assumptions are verified by a Lipschitz velocity field, i.e. such that v ∈ L1 (0, T ; Lip(Ω)). The gap between Lipschitz regularity and assumptions (6) and (7) could seem very thin, but it is far more easy to show that a velocity field included in a partial differential equation verifies these two conditions than to show it is Lipschitz. Moreover we assume further regularity on the boundary of Ω, i.e. v ∈ Lip(∂Ω×]0, T [),

(8)

which will be useful to extend our initial condition on RN . Let us then introduce the inward, outward and tangential boundaries1 . We denote by χ the part of ∂Ω where the outward normal n does not exist. We already stated that in case of a piecewise C 1 boundary, χ is of null measure in ∂Ω. (9) (10) (11)

Γi (t) = {x ∈ ∂Ω \ χ : v (x, t) · n(x) < 0} Γo (t) = {x ∈ ∂Ω \ χ : v (x, t) · n(x) > 0} Γlat (t) = {x ∈ ∂Ω \ χ : v (x, t) · n(x) = 0}

We also introduce the time-space counterparts of these boundaries, i.e. [ Σi = Γi (t) × {t}, t∈]0,T [ 1

we do not assume that these boundaries are time-independent.

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with the same notation for Σo and Σlat . Our method is first to extend the velocity field to RN ×]0, T [, such that it is completely Lipschitz on (RN \ Ω)×]0, T [ (we use (8) for that). The following lemma, close to Kirszbraun’s extension theorem for Lipschitz functions ([Fed69], p. 201) provides us with this extension: Lemma 1. A velocity field v verifying (6)(7)(8) admits an extension outside Ω, with compact support in RN , belonging to Lip((RN \ Ω)×]0, T [), and verifying conditions (*) and (**) of [DL89b] on RN ×]0, T [, i.e. (still denoting this extension by v ), (*) (**)

v ∈ L1 (0, T ; W 1,1 (RN )), div v ∈ L1 (0, T ; L∞ (RN )), kv (x, t)k ∈ L1 (0, T ; L1 (RN )) + L1 (0, T ; L∞ (RN )). 1 + kxk

Proof. We postpone the problem of compact support to the end of the construction, see below. For each component vi of v we define its extension by wi (y, t) =

inf

(x,s)∈∂Ω×]0,T [

{vi (x, s) + λi (kx − yk + |s − t|)}

where λi denotes the Lipschitz constant of vi on the boundary. First we verify that wi coincide with vi on ∂Ω×]0, T [. Indeed let (y, t) ∈ ∂Ω×]0, T [, we have by definition of Lipschitz function vi (x, s) + λi (kx − yk + |s − t|) ≥ vi (y, t) for all (x, s) ∈ ∂Ω×]0, T [. This means exactly vi (y, t) = wi (y, t). Then we show that wi ∈ Lip((RN \ Ω)×]0, T [). Let (y, t), (y 0 , t0 ) ∈ (RN \ Ω)×]0, T [, we have wi (x, t) ≤ vi (x, s) + λi (kx − yk + |s − t|), ∀(x, s) ∈ ∂Ω×]0, T [,



≤ vi (x, s) + λi ( x − y 0 + y 0 − y + s − t0 + t0 − t ). As this inequality stands for every (x, s), we have

wi (y, t) ≤ wi (y 0 , t0 ) + λi ( y 0 − y + t0 − t ). The opposite inequality is obtained similarly, and we thus show the first part of our lemma. Now we may multiply this velocity field by a bounded smooth function which vanishes outside a bounded open set O containing Ω and equal 1 on Ω. For sake of simplicity, we still denote by v this extension with compact support. Assumption (*) of [DL89b] is trivially satisfied: Inside Ω Sobolev regularity was assumed in (6), outside v is Lipschitz, and at the boundary it is continuous. A simple distribution calculus show that the first part of (*)

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holds. Outside Ω the velocity has bounded divergence since it is Lipschitz, and inside (7) holds, thus the second part of (*) holds too. Let us introduce λ = maxi λi . We show that a stronger condition than (**) holds, i.e. kv (x, t)k ∈ L1 (0, T ; L∞ (RN )). 1 + kxk Indeed, this is true outside Ω since from the construction of the extension, kv (x, t)k ≤ kv(t)kL∞ (Ω) + λ(T + max kx − yk)

a.e on ]0, T [,

y∈∂Ω

≤ kv(t)kL∞ (Ω) + λ(T + kxk + max kyk) y∈∂Ω

a.e on ]0, T [

with maxy∈∂Ω kyk bounded by a constant C > 0, since Ω is bounded.

t u

The extension we constructed may generate characteristic lines issued from Σi which intersect Σo , and then enter into Ω where characteristics do not exist everywhere. To overcome this difficulty, we modify the previous extension by truncature (only outside Ω). A structure condition on v on the boundary was necessary for technical reason (see the proof below and those of Proposition 1). We assume (12) There exists an open set U in ∂Ω containing

[

Γi (t), such that

t∈]0,T [

∃α > 0, ∀t ∈ [0, T ], dist(∂U, ∂Γi (t)) ≥ α,

v = 0 on U \ Γi (t).

Remark 3. This is a physical relevant assumption, when Γi does not depend on time, because near the injection point, the melt polymer which is viscous does not slip along the boundary (it may slip elsewhere on the boundary during the filling). When Γi (t) does depend of time, then it means for example that when an injection zone is closed, the fluid does not slip along the part of boundary where was this injection zone. Lemma 2. If (12) holds, then for δ > 0 small enough, one may construct an extension vanishing for points in (RN \ Ω)×]0, T [ between two open sets Ω1δ ×]0, T [, Ω2δ ×]0, T [ and whose boundaries are at distance from Σi less than δ. Proof. We take for a δ > 0 such that δ ≤ α2 , a smooth function ψδ : [0, +∞[−→ [0, 1], with bounded derivatives, such that  4  1 on [0, δ ], ψδ = 0 on [δ 3 , δ 2 ],   1 on [δ, +∞[.

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Fig. 1. Construction of ψδ

θδ =1

Ω

θδ=0

Γ (t) in

U

U

θδ =1

{t}

Fig. 2. Construction of θδ

Then we introduce θδ : RN ×]0, T [−→ [0, 1] defined by θδ (x, t) = ψδ (dist((x, t), Σi )). We claim that this function is bounded and belongs to Lip(RN ×]0, T [). To show that, since ψδ has bounded derivatives, it suffices to notice that d : (x, t) −→ dist((x, t), Σi ) is Lipschitz. Then the velocity field obtained from v by multiplying it by θδ outside Ω still verifies (*), (**) of [DL89b]. Indeed first (*) is verified: v ∈ L1 (0, T ; W 1,1 (Ω)) and θδ v belongs to L1 (0, T ; W 1,1 (RN \ Ω)) since θδ ∈ Lip(RN )) and has bounded support. At the boundary θδ v and v coincide thanks to (12). Thus θδ v ∈ L1 (0, T ; W 1,1 (RN )). Finally, the second part of

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(*) is satisfied since the divergence of θδ v is still bounded. Condition (**) is clearly satisfied since θδ v still has a bounded support. It is still Lipschitz outside Ω×]0, T [. Indeed v and θδ are bounded and Lipschitz on that domain, thus their product is still Lipschitz thanks to the following inequality: |ab − cd| ≤ max(a, b, c, d)(|a − c| + |b − d|). Now θδ v vanishes at distance less than δ from Σi . For sake of simplicity, we still denote by v this extension. Note that is is relevant since it coincides with v on Ω×]0, T [. u t Thanks to that extension of v to RN which is completely Lipschitz on \ Ω)×]0, T [, characteristics exist as long as they remain outside Ω. Let us see what happens in Ω. (RN

2.2. Existence of characteristics on RN We study over RN ×]0, T [ the following nonlinear ordinary differential system: (13)

∂X = v (X(s), s), ∂s

X(t) = x,

with a velocity field verifying (*) and (**). Then, according to R.J. DiPerna and P.L. Lions the differential system (13) admits an unique (distribution) solution s → X(s; x, t) such that 1. X ∈ C([0, T ] × [0, T ]; L(RN )) where L denote measurables functions finite a.e. 2. X(t3 ; x, t1 ) = X(t3 ; X(t2 ; x, t1 ), t2 ) a.e. on RN , for all (t1 , t2 , t3 ) ∈ [0, T ]3 which will be refered later as the group property. 3. For all t ≥ 0, for almost all x ∈ RN , s → X(s) ∈ W 1,1 (0, T ) and ∂X = v (X(s), s) on ]0, T [. ∂s N 4. X ∈ C([0 ≤ t ≤ T ]; L∞ loc (R ; C([0 ≤ s ≤ T ]))) ∩ C([0 ≤ s ≤ N T ]; L∞ loc (R ; C([0 ≤ t ≤ T ]))). From the property 3, this solution coincides with the classical one on as long as it remains outside Ω. Remark 4. The group property implies that x → X(s; x, t) is one-to-one on RN .

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2.3. Regularity properties of characteristics outside Ω Proposition 1. Provided that (8) and (12) hold, (x, t) −→ X(0; x, t) is Lipschitz from Σi to RN , and (x, t) −→ X(T ; x, t) is Lipschitz from Σo to RN . Proof. We know that v has been extended to a Lipschitz vector field outside Ω×]0, T [, thus if we can show that for (x, t) ∈ Σi , X(s; x, t) remains outside Ω for s ∈ [0, t], our claim will derive from the classical Lipschitz dependance of solution to the ordinary differential equation from the data. So let (x, t) ∈ Σi , i.e. x ∈ Γi (t). As v is Lipschitz in time and space on (RN \ Ω)×]0, T [, and v (x, t) · n(x) < 0, there exists a neighborhood V of (x, t) in (RN \ Ω)×]0, T [ such that ∀(y, s) ∈ V, v (y, s) · n(x) < 0. Thus characteristics are entering into Ω at the point (x, t). This means that for s < t close enough of t, X(s; x, t) is outside Ω. Therefore it suffices to show that until s = 0, it remains outside Ω. But thanks to lemma 1, v vanishes as close to Σi as we wish (we can choose the constant δ as small as we want). Thus no characteristic line could get out of Σi to re-enter Ω, because it should re-enter Ω by Σo , and that the velocity field vanishes before reaching it. The same holds for Σo . As the image of a measurable set by a Lipschitz operator is measurable t ([EG92], p. 92), X(0, Σi ) and X(T, Σo ) are measurable. u 2.4. Definition of a solution We introduce an initial condition, (14)

ψ0 ∈ L∞ (Ω),

and an inner boundary condition (15)

g ∈ L∞ (Σi ).

Our method is to extend the initial condition to RN using the boundary condition conveniently, thanks to Lipschitz regularity of v outside Ω×]0, T [, to invoke the result in RN of DiPerna and Lions, and then to come back to Ω by restriction. c0 , is constructed as follows. Our extended initial condition, ψ  c  for x ∈ Ω, ψ0 (x) = ψ0 (x) c0 (X(0; y, t)) = g(y, t) for t ∈]0, T [, y ∈ Γi (t), (16) ψ  c elsewhere. ψ0 (x) = 0

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Remark 5. This definition is not ambiguous: indeed on Σi , if (y, t) 6= (y 0 , t0 ) then X(0; y, t) 6= X(0; y 0 , t0 ) since v is Lipschitz ouside Ω. Indeed let us suppose that (y, t) 6= (y 0 , t0 ) with X(0; y, t) = X(0; y 0 , t0 ), with t < t0 . Then thanks to the group property one has X(t; y 0 , t0 ) = y. But as y 0 ∈ Σi and t < t0 , y should be outside Ω. c0 inherits assumptions on ψ0 and g. We first prove that ψ Lemma 3. Assume (8)(12) and (14)(15). Then c0 ∈ Lp (RN ), ψ

∀p ∈ [1; +∞].

c0 has a bounded support, because Ω is bounded, and by Proof. Clearly ψ c0 beconstruction X(0; Σi ) is bounded. Thus it is sufficient to show that ψ ∞ N longs to L (R ). c0 is measurable, since its bounds are clearly We only have to show that ψ the maximum of those of g and ψ0 . It suffices to show that the set n o c0 (x) > α ω α = x ∈ RN : ψ is measurable. The measurability of ωα ∩ Ω which is {x ∈ Ω : ψ0 (x) > α} follows from assumption (14). On the complementary domain of Ω we can c0 is zero. So we look also restrict ourselves to X(0, Σi ), since elsewhere ψ for the measurability of o n c0 (x) > α x ∈ X(0, Σi ) : ψ which is also X(0, {(y, t) ∈ Σi : g(y, t) > α}). As g is a measurable function we just want to know whether X(0, .) maps measurable sets to measurable sets. This follows from Proposition 1, which ensures that X(0, .) is Lipschitz (in both x and t variables), and to a theorem of [EG92], p. 92. u t 2.5. Distribution solution The results of [DL89b] ensure existence and uniqueness of a solution ψ ∈ L∞ (0, T ; L∞ (RN )) ∩ C(0, T ; Lploc (RN )) for p < ∞ to the transport equation in RN , i.e. such that Z TZ Z ∂µ c0 (x)µ(x, 0) dx, − + div(v µ)) dx dt = ψ (17) ψ( ∂t 0 RN RN

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for all test function µ ∈ D([0, T [×RN ). Moreover, they proved that this solution is defined on ]0, T [×RN by c0 (X(0; x, t)), ψ(x, t) = ψ

(18)

a.e. in RN

∀t ∈ [0, T ],

Using results of [Rav87] p. 37, ψ defined by (18) is solution, on (RN \ Ω)×]0, T [ where v is lipschitz in time and space, of the following backward problem (with unprescribed final condition):  ∂ψ N   ∂t + v · ∇ψ = 0 on (R \ Ω)×]0, T [, ψ(T ) = ψ(T ) on (RN \ Ω) × {t = T },   ψ=g on Σi . These equations are to be understood in the distribution sense, i.e. (a)

−

Z TZ 0

∂η + div(v η)) dx dt ∂t RN \Ω Z Z TZ =− ψ(T )η(T ) dx + ψ(

RN \Ω

0

Γi (t)

v · n gη dσ dt,

for all η ∈ D(]0, T ] × RN ) vanishing on Σo . Note that η vanishes for t = 0 whereas the test function µ of (17) vanished for t = T . But for ν ∈ D(]0, T [×RN ) vanishing on Σo , (17) and (a) give Z TZ Z TZ ∂ν + div(v ν)) dx dt = − − (b) ψ( v · n gν dσ dt. ∂t 0 Ω 0 Γi (t) To recover the initial condition, we consider for a µ ∈ D([0, T [×RN ), a sequence of test functions for (b) given by µε = αε µ where ε  1, αε = 1 on [ε, T ], and αε = 0 on [0, ε2 ]. This gives −

Z TZ ε

Ω

∂µ + div(v µ)) dx dt ∂t   Z εZ ∂µ + αε div(v µ) dx dt − ψ αε0 µ + αε ∂t ε2 Ω Z Z T =− αε v · n gµ dσ dt.

ψ(

0

When ε goes to zero, the only non trivial term is Z εZ ψµαε0 dx dt. ε2 Ω

Γi (t)

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As ψµ ∈ C(0, T ; Lp (RN )), we can find hδ ∈ C 1 (0, T ; Lp (RN )) uniformly convergent towards ψµ on [0, T ], integrate by parts, using αε (ε2 ) = 0, and letting ε → 0 to obtain Z Z hδ (0) dx which converges towards ψ(0)µ(0) dx Ω

Ω

as δ → 0. As ψ(0) is ψ0 on Ω by construction, we prove the following existence result. Proposition 2. Assume (6)(7)(8)(12), and (14)(15), then there exists a solution ψ belonging to L∞ (0, T ; Lp (Ω)) for all p ∈ [1; +∞], to the following formulation: −

Z TZ 0

Ω

ψ(

∂µ + div(v µ)) dx dt ∂t Z TZ Z ψ0 (x)µ(x, 0) dx − = Ω

0

Γi (t)

v · n gµ dσ dt,

for every test function µ ∈ C ∞ (Ω × [0, T ]) which vanishes on Σo and for t = T . Moreover, for all p < ∞, ψ ∈ C(0, T ; Lp (Ω)). For the previous solutions we did not get an uniqueness result. Following the work of DiPerna and Lions, and strongly influenced by a preprint of Nouri and Poupaud, we introduce a notion of renormalized solution and prove uniqueness for it. 2.6. Renormalized solution The approximation Theorem II.1 of [DL89b] says that the regularized funcdef tions ψε = ψ ∗ ρε satisfies to ∂ψε + v · ∇ψε = rε ∂t with rε → 0 in L1 (]0, T [×RN ) strong. Now let β be a C 1 function on R, ψε is regular in time and space since ∂ψε = −(v · ∇ψ) ∗ ρε , ∂t thus β(ψε ) verifies (19)

∂β(ψε ) + v · ∇β(ψε ) = rε β 0 (ψε ). ∂t

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Taking two differents approximations of ψ, corresponding to ε, δ > 0, we subtract the transport equations verified by both and then apply our function β, to get ∂β(ψε − ψδ ) + v · ∇β(ψε − ψδ ) = (rε − rδ )β 0 (ψε − ψδ ). ∂t Let µ ∈ D([0, T [×Ω) we have, integrating by parts, Z

T

0

 ∂µ + div(µv ) dx dt −β(ψε − ψδ ) ∂t Ω Z TZ − v · nµβ(ψε − ψδ ) dσ dt 0 ∂Ω Z TZ (rε − rδ )β 0 (ψε − ψδ )µ dx dt. = 

Z

0

Ω

For a test function µ which vanishes on Σo , and for β(r) = |r|α , α > 1, we get Z 0

T

Z Γi (t)

|v · n| |ψε − ψδ |α µ dσ dt Z =

0

T

Z Ω

α



|ψε − ψδ | Z +

0

T

Z Ω

 ∂µ + div(µv ) dx dt ∂t α(rε − rδ ) |ψε − ψδ |α−1 µ dx dt.

We know from the properties of convolution that ψε → ψ in L∞ (0, T ; Lp (RN )) strong

∀p < ∞.

Thus if we can take µ > 0 on Σi (as µ vanishes on Σo , we can find such a function provided that Σi ∩ Σo = ∅, which is ensured by (12)), the previous estimate proves that the trace of ψε on Σi is a Cauchy sequence in Lα (0, T ; Lα (Γi (t), |v · n| dσ)). Thus this trace converges towards a function hi which belongs to Lα (0, T ; Lα (Γi (t), |v · n| dσ)) for all 1 < α < ∞. The same procedure holds for the trace of ψε on Σo , and we get a corresponding ho . Furthermore, hi and ho are bounded functions, since they are Lα limits of uniformly bounded sequences (ψε is uniformly bounded as ψ is bounded, from the definition of convolution by a cut-off function). Remember that we wanted to study the traces of the function ψ. To do that, we pass to the limit in (19). Let µ ∈ D([0, T [×Ω) we get, multiplying

Transport equation with boundary conditions for free surface localization

(19) by µ,   Z TZ ∂µ + div(µv ) dx dt −βM (ψε ) ∂t 0 Ω Z − βM (ψε0 )µ(0, x) dx Ω Z TZ Z + v · nβM (ψε )µ dσ dt = 0

T

Z

0

∂Ω

Ω

289

0 rε βM (ψε )µ dx dt,

where, since the derivative of β could be unbounded, we introduced a smooth troncature of β to a step M > 0, denoted by βM . Now we can pass to the limit, when ε goes to zero, in the previous formulation, to get   Z TZ Z ∂µ + div(µv ) dx dt − −βM (ψ) βM (ψ 0 )µ(0, x) dx ∂t 0 Ω Ω Z TZ Z TZ + v · nβM (hi )µ dσ dt + v · nβM (ho )µ dσ dt = 0. 0

0

Γi (t)

Γo (t)

Now, as ψ, hi , ho are bounded functions, for M large enough we get the same equality for β in place of βM . Furthermore, as there is no more derivative of β in this formulation, it remains valid for functions β, which are only continuous. Taking the identity for β, and identifying with the former distribution solution we got, we deduce that hi = g. Moreover ψ has a bounded trace on Σo . Therefore, the first part of the following result has already been proved. Proposition 3. Assume (6)(7)(8)(12), and (14)(15). Then a distribution solution ψ defined in Proposition 2 admits a bounded trace on Σo and verifies Z TZ Z ∂µ β(ψ)( β(ψ0 )(x)µ(x, 0) dx − + div(v µ)) dx dt = ∂t 0 Ω Ω Z TZ Z TZ − v · n β(g)µ dσ dt − β(ψ)µv · n dσ dt, 0

Γi (t)

0

Γo (t)

for every test function µ ∈ C ∞ (Ω × [0, T ]) which vanishes for t = T , and for all function β ∈ C(R). Moreover, for such β the following integral conservation holds for all t ∈ [0, T ], Z Z tZ Z (20) β(ψ)(t) dx − β(ψ) div v dx ds = β(ψ 0 ) dx 0 Ω Ω Ω Z tZ Z tZ + |v · n| β(g) dσ ds − |v · n| β(ψ) dσ ds. 0

Γi (s)

0

Γo (s)

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Proof. The first part of the proof was shown just above. For the second part, we fix a t ∈]0, T ], and a valid test function ν(x, t), and we introduce for any integer n > 1t ,   s) on [0, t − n1 ], ν(., 
 1 µn (., s) = 1 − n s − t + n ν(., s) on [t − n1 , t],   0 on [t, T [. Remark 6. This function should be regularized in time to be a real test function for our distribution formulation, but as only first order time derivative occurs in this formulation, one could show by density that this function can be chosed. Thus for this function, −

Z TZ 0

Z tZ ∂µn ∂µn dx ds = − dx ds β(ψ) β(ψ) ∂s ∂s 0 Ω Ω Z t Z =− β(ψ)(−n)ν dx ds Z −

−

t

   ∂ν 1 β(ψ) 1 − n s − t + dx ds n ∂s Ω Z t Z β(ψ)ν dx ds =n

1 t− n

Z

t

1 t− n Ω

Z

t− 1 Ω

  n  ∂ν 1 dx ds. β(ψ) 1 − n s − t + n ∂s Ω

Z

1 t− n

Passing to the limit gives, from the continuity in time of ψ, Z β(ψ)(x, t)ν(x, t) dx. Ω

The same holds for other terms, and we thus show that a solution ψ verifies everywhere on ]0, T ] the following integral identity: Z Z tZ β(ψ)(t)ν(x, t) dx − β(ψ) div(v ν) dx ds 0 Ω Ω Z tZ Z 0 β(ψ )ν(x, 0) dx + |v · n| β(g)ν(σ, t) dσ ds = Ω

−

Z tZ 0

Γo (s)

0

Γi (s)

|v · n| β(ψ)ν(σ, t) dσ ds.

Transport equation with boundary conditions for free surface localization

291

However the previous integral identity is not really a conservation law because of the remaining test function inside. But since this test function is not constrained to vanish on the boundary, we can pass to the limit to a constant function on Ω, to get (20). u t Definition 1. A function ψ ∈ L∞ (0, T ; Lp (Ω)) for all p ∈ [1; +∞] solution of (3)–(5), which verifies (20) is called a renormalized solution. Remark 7. Let us point out that (20) has a physical meaning in term of inward flux, outward flux and inside compression or dilatation (these terms being interpreted differently depending of which quantity β(ψ) represents). Moreover one could also extend this formulation to open subsets of Ω instead of the whole domain, and this leads to a finite-volume formulation. See below the link with numerical calculations. This notion of solution provides us with a maximum principle, and thus an uniqueness result in case of bounded data. Proposition 4. Assume (6)(7)(8),(14) and (15). Then a renormalized solution ψ to (20) verifies for all t ∈]0, T [, (21) min(ess inf ψ0 , ess inf g) ≤ ψ(x, t) ≤ max(ess sup ψ0 , ess sup g), Ω

Σi

Ω

Σi

a.e. on Ω. Thus the renormalized solution of (20) for a given (ψ0 , g) is unique. Proof. For sake of simplicity we denote αmin = min(ess inf ψ0 , ess inf g), Ω

Σi

αmax = max(ess sup ψ0 , ess sup g). Ω

Σi

We construct a sequence of continuous functions βn which will be used in (20) in place of β, verifying β(r) ∈ [0, 1] on R and ( 0 on ]αmin − n1 , αmax + n1 [, βn (r) = 1 on ] − ∞, αmin − n2 ] ∪ [αmax + n2 , +∞[. Using this function in (20) makes the initial and boundary terms cancel, so that leads to Z Z tZ βn (ψ)(t) dx − βn (ψ) div v dx ds 0 Ω Ω Z tZ =− |v · n| βn (ψ) dσ ds. 0

Γo (s)

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As βn is positive, we can estimate Z Z tZ βn (ψ)(t) dx ≤ βn (ψ) div v dx ds. 0

Ω

Ω

When n goes to infinity, βn (ψ) converges towards the indicator function of the measurable set ω(t) = {x ∈ Ω : ψ(x, t) 6∈ [αmin , αmax ]} , thus we get meas {ω(t)} ≤

Z tZ 0

ω(s)

div v dx ds

and using assumptions (7), Z t meas {ω(t)} ≤ meas {ω(s)} kdiv v kL∞ (Ω) (s) ds. 0

The Gronwall lemma thus leads to meas {ω(s)} = 0, which means exactly (21).

on]0, T [,

t u

2.7. Properties of the renormalized solution (case v · n = 0) We now turn to the properties we are expecting from our solution. Given two continuous initial conditions ψ10 and ψ20 verifying the assumptions of existence of associated renormalized solutions, we assume that their initial zero level set is identical:  def def  Γ10 = x ∈ Ω : ψ10 (x) = 0 = x ∈ Ω : ψ20 (x) = 0 = Γ20 . (22) We are asking whether the solutions ψ1 and ψ2 these initial conditions generate possess the same zero level set, i.e. whether (23) def

def

Γ1 (t) = {x ∈ Ω : ψ1 (x, t) = 0} = {x ∈ Ω : ψ2 (x, t) = 0} = Γ2 (t). We first assume that ψ1 and ψ2 are continuous. The technic of [ES91] is to construct an even continuous function β such that β(0) = 0, β is increasing on R+ , such that β(ψ1 ) ≥ |ψ2 |

Transport equation with boundary conditions for free surface localization

293

on Ω×]0, T [. As β(ψ1 ) has the same zero level set as ψ1 , it will provide us an inclusion Γ1 (t) ⊂ Γ2 (t). To construct function β, we look for the same inequality on initial conditions, i.e. β(ψ10 ) ≥ ψ20 on Ω, and use the maximum principle we just proved. Now let start this construction, 0 1 by introducing the following sets: for k ≥ 1, Ek = {x ∈ Ω : ψ (x) ≥ }. We have 1 k E1 ⊂ E2 · · · ⊂ Ek ⊂ Ek+1 ⊂ . . . ,

Ω\

Γ10

=

∞ [

Ek .

k=1

We then introduce (E0 = ∅)

ak = max ψ20 Ω\Ek−1

which is a decreasing sequence, from the preceding inclusions, and verifying limk→∞ ak = 0, because of (22). Let the following β function, defined on R with values in [0; +∞[, which is even and such that β(0) = 0,

1 β( ) = ak , k

β linear on [

1 1 , ], k+1 k

constant on [1, +∞[.

f0 = β(ψ 0 ) f1 = β(ψ1 ) solves the transport equation with ψ We notice that ψ 1 1 as initial condition. Now on the set Ek \ Ek−1 we have f0 (x) = β(ψ 0 (x)) ≥ β( 1 ) = a ≥ ψ 0 (x) ψ k 1 2 1 k thus from the maximum principle theorem, and as the ψi are continuous, we have f1 ≥ |ψ2 | on Ω × [0, T [. ψ f1 (x, t) = 0, Thus, if some x belong to Γ1 (t), then ψ1 (x, t) = 0, so ψ which implies ψ2 (x, t) = 0, so x belongs to Γ2 (t). The opposite inclusion is symetric, thus the two zero level sets are equal. Let us state this result precisely: Proposition 5. Let ψ1 and ψ2 be two continuous solutions associated to the same boundary condition g verifying (15) and two continuous initial conditions ψ10 and ψ20 verifying (14), and such that   x ∈ Ω : ψ10 (x) = 0 = x ∈ Ω : ψ20 (x) = 0 . Then for each t ∈ [0, T ] we have {x ∈ Ω : ψ1 (x, t) = 0} = {x ∈ Ω : ψ2 (x, t) = 0} .

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Now we could study the dependence of the zero level set of the solution with respect to the choice of the boundary condition. What we want to show is that given an initial condition ψ 0 and two boundary conditions g1 and g2 , the zero level set of the solutions associated with g1 and g2 should coincide. Due to the maximum principle theorem we must assume that both gi are positive on the inner boundary. However we did not succeed in proving what is therefore a conjecture Conjecture 1. Assume that g1 and g2 are strictly positives on Σi and verify (15). Let ψ0 verifying (14), and ψ1 and ψ2 the two renormalized solutions associated to g1 and g2 . Then the zero level sets of these solutions are identical. At last we give a result of non-spreading of the interface. The integral formulation (20) allows us to show that the zero level set of the solution ψ is of null measure provided that those of ψ 0 had the same property. First let us show a boundness result. Corollary 1. The renormalized solution ψ verifies for all t ∈ [0, T ],  kh(ψ(t))kL1 (Ω) ≤ kh(ψ0 )kL1 (Ω)  + kv · nkL1 (Σi ) kh(g)kL∞ (Σi ) exp kdiv v kL1 (0,T ;L∞ (Ω)) for all bounded positive and continuous function h on R. Proof. It suffices to apply Gronwall lemma ([CH90], p. 55) to the following estimate, obtained from (20) as h is positive: kh(ψ(t))kL1 (Ω) ≤ kh(ψ0 )kL1 (Ω) + kv · nkL1 (Σi ) kh(g)kL∞ (Σi ) Z t + kdiv v (s)kL∞ (Ω)) kh(ψ(s))kL1 (Ω) ds. 0

t u We derive the following Proposition 6. Let ψ the renormalized solution of (3)-(5) with an initial condition ψ0 verifying (24)

meas {x ∈ Ω : ψ0 (x) = 0} = 0,

and with a boundary condition g of absolute value strictly greater than a constant α > 0 on Σi . Then for all t ∈ [0, T ] we have (25)

meas {x ∈ Ω : ψ(x, t) = 0} = 0.

Transport equation with boundary conditions for free surface localization

295

Fig. 3. Function hε

Proof. We use Corollary 1 with the following sequence of h functions, for ε > 0:

hε (r) =

  1

if |r| ≤ ε2 , if ε2 ≤ |r| ≤ ε, if |r| ≥ ε.

ε−|r|

ε−ε2   0

We then have khε (ψ0 )kL1 (Ω) ≤ meas {x ∈ Ω : |ψ0 (x)| ≤ ε} , and  khε (ψ(t))kL1 (Ω) ≥ meas x ∈ Ω : |ψ(x, t)| ≤ ε2 . Moreover, since |g| is bounded from below by a strictly positive number, hε (g) is the null function for ε small enough. Thus using (24) leads to lim khε (ψ0 )kL1 (Ω) = 0,

ε→0

and Corollary 1 implies lim khε (ψ(t))kL1 (Ω) = 0,

ε→0

and a fortiori  lim meas x ∈ Ω : |ψ(x, t)| ≤ ε2 = 0.

ε→0

This last convergence means (25).

t u

296

E. Maitre, P. Witomski

3. Link with numerical simulation We have already introduced the model we are studying. Remember it consists in a pressure equation and the transport equation of ψ. Let us state this pressure equation. (26)

χ(ψ, p)

∂p − χ(ψ, p)S(ψ, p) k∇pk2 − div(S(ψ, p)∇p) = Φth . ∂t

In the above equation, χ(ψ, p) accounts for the local compressibility of the fluid, which is the air if ψ < 0, or the polymer, if ψ > 0. The same holds for the coefficients S(ψ, p) which link the pressure gradient with the velocity by (1). It must be highlighted that these coefficients are discontinuous on the interface (i.e. ψ = 0), due to the dissemblant physical properties of polymer and air. Numerical calculation is thus crucial at this place. Equation (26) has associated initial-boundary conditions, which are p = p0 on Ω × {0}, on Σo , p = ps ∂p = Ge S(ψ, p) on Σi , ∂n ∂p S(ψ, p) =0 on Σlat . ∂n For sake of completeness, we recall the transport equation verified by ψ, (27) (28) (29)

∂ψ − S(ψ, p)∇p · ∇ψ = 0, ∂t ψ=g ψ = ψ0

on Ω×]0, T [, on Σi , on Ω × {0}.

Note that we have to assume that the inner boundary is alimented by polymer fluid at all time. This is expressed by (30)

g>0

◦

on Σ i .

Even if we did not made a numerical analysis of our numerical method, one could see that the all our theoretical results about renormalized solution take a very natural meaning in terms of finite volume method for the transport equation. Indeed the uniqueness result, as well as the non dependence with respect to the initial condition in the transport equation of the zero level set was a necessary consistence condition before turning to calculations. The non-spreading of interface result gives also some garanty about the behaviour of the numerical interface. But the essential point is that our definition of solution in term of renormalization is very close to the real calculation if the finite volume method

Transport equation with boundary conditions for free surface localization

297

i

j

Fig. 4. Triangulation and dual mesh

Fig. 5. Approximation of interface

(especially is the donnor cell algorithm). In this method one consider for each cell what is entering the cell from upstream, what is getting outside downstream, and how the data remaining in that cell change. And that is exactly what says the formulation (20), see Remark 7. So even if we did not carry on a real numerical analysis of the method, our Theoretical result let us expect a well behaved simulation.

3.1. Numerical calculations The pressure is computed by a finite-element method P1, the pressure being linear on a triangular element [Pir89]. We used a public domain mesh generator to triangulate the mold cavity. The function ψ is computed on a dual finite-volume mesh obtained from the triangulation by linking the gravity centers of two adjacent elements: see Fig. 4.

298

E. Maitre, P. Witomski

Interface

i

Approximated interface j

Fig. 6. Approximated zero level set of ψ

The choice of a finite volume method for the computation of ψ is relevant as it is solution of a hyperbolic equation. Let us recall the transport equation ∂ψ − S(ψ, p)∇p · ∇ψ = 0, ∂t

on Ω×]0, T [.

This equation is not in divergence form so we rewrite it as follows, ∂ψ − div(ψS(ψ, p)∇p) = −ψ div(S(ψ, p)∇p), ∂t and then we integrate it on a cell Ci , applying the divergence theorem, which leads to the following semi-discrete scheme ψ n+1 − ψin |Ci | i − ∆t

Z ∂Ci

n ψup S(ψin , pni )∇pn · n dσ Z S(ψin , pni )∇pn · n dσ. = ψin ∂Ci

Here ψup stands for the upwind value of ψ, depending on which vertex of the cell is considered and on the value of ψ on the two cells adjacent to that vertex. Our algorithm is to solve successively the pressure and ψ. Assume p and ψ are known at time tn , on the mesh nodes. We get the interface position by the trace of the zero level set of ψ on the triangles. This may be obtained by a linear approximation between two mesh points where ψ has different signs (see Fig. 5). Between two mesh points, the fraction of polymer filling the vertice is computed by αij =

[ψi ]+ + [ψj ]+ , |ψi | + |ψj |

where [r]+ = max(r, 0). Of course another (i.e. nonlinear) approximation could be implemanted easily, if more relevant to physical considerations.

Transport equation with boundary conditions for free surface localization

299

Isovalues of psi -1.960e-01 -1.779e-01 -1.597e-01 -1.415e-01 -1.233e-01 -1.051e-01 -8.693e-02 -6.874e-02 -5.056e-02

Isovalues of psi -7.968e-01

-7.226e-01

-3.237e-02 4.000e-03

-6.484e-01

Time t = 0.0000 s.

-5.741e-01

-4.999e-01

-4.257e-01

Fig. 7. Minimum distance

-3.515e-01

Isovalues of psi -2.865e-01

-2.773e-01

-2.583e-01

-2.031e-01

-2.301e-01

-1.289e-01

-2.019e-01

1.957e-02 -1.738e-01 -1.456e-01 -1.174e-01

Time t = 0.0000 s.

-8.922e-02 -6.105e-02 -3.287e-02 2.348e-02

Time t = 0.0000 s.

Fig. 9. Minimum of euclidian distances to connected components Fig. 8. Euclidian distance Isovalues of psi -6.622e-01

Isovalues of psi -5.831e-01

-6.003e-01

-5.283e-01

-5.383e-01

-4.735e-01

-4.763e-01

-4.187e-01

-4.143e-01

-3.639e-01

-3.523e-01

-3.091e-01

-2.903e-01

-2.544e-01

-2.284e-01

-1.996e-01

-1.664e-01

-1.448e-01

-1.044e-01

-9.000e-02

1.957e-02

1.957e-02

Time t = 2.5216 s.

Time t = 4.0076 s.

Isovalues of psi

Isovalues of psi

-5.017e-01

-4.257e-01

-4.543e-01

-3.852e-01

-4.069e-01

-3.447e-01

-3.595e-01

-3.042e-01

-3.122e-01

-2.638e-01

-2.648e-01

-2.233e-01

-2.174e-01

-1.828e-01

-1.700e-01

-1.423e-01

-1.226e-01

-1.019e-01

-7.521e-02

-6.138e-02

1.957e-02

1.957e-02

Time t = 6.0114 s.

Fig. 10. Isovalues of ψ from t = 0 to t = 8 s

Time t = 8.0153 s.

300

E. Maitre, P. Witomski Isovalues of psi -3.433e-01

-3.103e-01

-2.773e-01

Pressure (Pa) 1.000e+05

1.855e+06

3.609e+06

-2.443e-01 5.364e+06

-2.113e-01 7.118e+06

-1.784e-01 8.873e+06

-1.454e-01 1.063e+07

-1.124e-01 1.238e+07

-7.939e-02 1.414e+07

-4.640e-02 1.589e+07

1.957e-02

Time t = 10.0190 s.

Time t = 10.0190 s.

Velocity scale :

3.777e-01 m/s

Fig. 11. Interface position and velocity field at t = 10 s Isovalues of psi

Isovalues of psi

-2.739e-01

-2.160e-01

-2.472e-01

-1.946e-01

-2.206e-01

-1.732e-01

-1.939e-01

-1.518e-01

-1.672e-01

-1.303e-01

-1.405e-01

-1.089e-01

-1.138e-01

-8.751e-02

-8.716e-02

-6.609e-02

-6.048e-02

-4.468e-02

-3.379e-02

-2.326e-02

1.957e-02

1.957e-02

Time t = 12.0003 s.

Time t = 14.0041 s.

Fig. 12. Isovalues of ψ at t = 12 and t = 14 s

Doing that computation for each vertice leads to a piecewise affine approximation of the free boundary (Fig. 6). Once the interface has been numerically localized, we compute the coefficients S and χ in either sides, using an average law in the case of a cell or a triangle which is intersected by the front. For example, in order to ensure the velocity continuity at interface, the S coefficient must be computed on an element as the harmonic average of the Sij for each vertice ij of the triangle. This coefficient Sij are obtained by the following law Sij =

S1 S2 , αij S2 + (1 − αij )S1

where S1 and S2 correspond respectively to the polymer and air.

Transport equation with boundary conditions for free surface localization Isovalues of psi

301 Isovalues of psi

-1.151e-01

-6.453e-02

-1.029e-01

-5.688e-02

-9.062e-02

-4.924e-02

-7.838e-02

-4.159e-02

-6.614e-02

-3.395e-02

-5.389e-02

-2.630e-02

-4.165e-02

-1.866e-02

-2.941e-02

-1.101e-02

-1.716e-02

-3.366e-03

-4.918e-03

4.279e-03

1.957e-02

1.957e-02

Time t = 16.0079 s.

Time t = 18.0116 s.

Fig. 13. Isovalues of ψ at t = 16 and t = 18 s Pressure (Pa) 1.000e+05

3.338e+06

6.576e+06

Interface position / time 0.000e+00

1.818e+00

3.636e+00

5.455e+00 9.814e+06

7.273e+00 1.305e+07

9.091e+00 1.629e+07

1.091e+01 1.953e+07

1.273e+01 2.277e+07

1.455e+01 2.600e+07

1.636e+01 2.924e+07

2.000e+01

Time t = 20.0154 s.

Velocity scale :

3.779e-01 m/s

Fig. 14. Last velocity field and successive positions of interface

3.2. Choice of initial and boundary conditions The function ψ must be zero on the interface. Given an initial curve we choose for ψ0 a signed distance from this curve. If this initial curve is not given, we may construct ψ0 as follows, ψ0 (x) = α − d(x, Γi (0)), where α > 0 is computed from the Γi (0) size. The choice of the distance (i.e. euclidian, or minimum distance) depends on the mold geometry. Indeed we sometimes want to simulate a flow with an initial flat front (so in this case we use the minimum distance function), but in general we prefer the more regular euclidian distance (see Fig. 7). However,

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in case of multiple injection, i.e. when Γi (0) is not connected, we choose the minimum of the euclidian distances to each connected component. Another point is the numerical choice of g. We choose a fixed in time Dirichlet condition which equals the trace of ψ0 on the inner boundary. 3.3. Results We present the numerical results corresponding to the mold of Fig. 9. The holes into the mold correspond to obstacles during the polymer flow, that means the final plastic thing we are injecting will have these holes. In this case, the melt polymer is injected by two parts of the left boundary, which are represented with thick lines. The air initially filling the mold flow out of it through the right boundary. Note that this is done thanks to small holes, which are obstructed by the polymer as soon as it reaches them (to this correspond a change of the boundary condition on Σo to an homogeneous Neumann condition). In this model case, the physical properties of the two fluids, are summarized below, for polymers used to make plastic parts involved in cars. S χ Polymer 2.10−9 10−8 Air 10−3 10−5 Concerning boundary conditions, we set Ge = 2.109 and ps = 105 . The following Fig. 10 represent the interface position between initial time and t = 8 s. At this time (t = 10 s), we have represented the pressure and velocity fields (Fig. 11). Following the flow, we now represent on Figs. 12 and 13 the front propagation up to t = 18 s. At this moment, we can see that some of the holes on the lower right corner have been obstructed by the melt polymer, as the velocity vanishes in this area. At the end of the calculation, we represented successive positions of the interface every 0.2 s. (Fig. 14). Note that this representation is useful to get an idea of the acceleration or decelaration of the fluid around obstacles. We may also see the final acceleration of the flow when the holes on the bottom of the mold are partially obstructed. References ´ Bardos, C. (1970): Probl`emes aux Limites pour les Equations aux D´eriv´ees Partielles du Premier Ordre a` Coefficients R´eels; Th´eor`emes d’Approximation; Ap´ ´ plication a` l’Equation de Transport. Ann. Sci. Ecole Norm. Sup. 3, 185–233 [BLN79] Bardos, C., Leroux, A., Nedelec, J.C. (1979): First Order Quasilinear Equations with Boundary Conditions. Comm. Partial Differential Equations 4(9), 1017– 1034

[Bar70]

Transport equation with boundary conditions for free surface localization

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