Fast-reaction asymptotics for a two-scale reaction-diffusion system

Fast-reaction asymptotics for a two-scale reaction-diffusion system Sebastian Meier∗

Adrian Muntean†‡

October 8, 2008

Abstract We investigate a reaction–diffusion process in a two-phase medium with microscopic length scale ε. The diffusion coefficients in the two phases are highly different (d1 /D = ε2 ) and the reaction constant k is large. First, the homogenisation limit ε → 0 is taken, which leads to a two-scale model. Afterwards, we pass to the fast-reaction limit k → ∞ and obtain a two-scale reaction-diffusion system with a moving boundary traveling within the microstructure.

Keywords: two-scale model, reaction–diffusion system, homogenization, fast-reaction limit

1

Introduction

In this paper we consider a semilinear parabolic system which arises in the modeling of gas-solid reactions taking place in wet porous media. The relevant physical problem is the following: A gaseous species A penetrates a non-saturated porous medium via the air phase of its pore space, dissolves in the pore water where A reacts very fast with a species B. For simplicity, in our model the species B is already present in the pore water from the beginning. In general, B will become available by a dissolution mechanism or any ∗

Centre for Industrial Mathematics, FB 3, University of Bremen, Postfach 330 440, 28334 Bremen, Germany, email [email protected] † Corresponding author. ‡ CASA -Centre for Analysis, Scientific computing and Applications, Department of Mathematics and Computer Science, Eindhoven University of Technology, PO Box 513, 5600 MB, Eindhoven, The Netherlands, email [email protected]

1

1 INTRODUCTION

2

other depletion mechanism that can bring B from the solid matrix into the pore solution. It is known that the homogenization limit of the corresponding pore-scale model is of two-scale structure, provided that diffusion of A in the pore water is sufficiently slow [1, 2]. On the other hand, regarding the fast reaction, we are also interested in the singular-limit analysis k → ∞, where k is the reaction constant. In [3], we have shown by formal asymptotics that it does not matter in which order we take limits, with other words, if we first pass to the fast-reaction limit k → ∞ and then take the homogenization limit ε → 0 or vice versa. If no diffusion of B is allowed, the (formal) result is a two-scale problem with a family of one-phase Stefan problems on the micro scale. Here we target at proving rigorously the singular limit analyses ε → 0 and then k → ∞. In order to simplify the analysis, we allow for diffusion of B in the pore water and replace the instantaneous (Dirichlet) exchange condition between the two phases by a regularized (Robin) condition. We perform the analysis in two stages: • Firstly, we use a standard homogenization-based approach (very much inspired by the works [1, 2, 4]) to prove the convergence as ε → 0 of the starting micro PDE system towards a two-scale reaction-diffusion system; • Secondly, we adapt to the setting of two-scale parabolic equations some of the fast-reaction asymptotics techniques developed in [5, 6, 7, 8], and [9], and then prove the limiting behavior k → ∞. We refer the reader to [10, 11, 12] for a collection of fast-reaction–(slow)diffusion settings playing an important role in pattern formation and corrosion of porous materials, and [13, 14] for further conceptually related scenarios arising in the modeling of catalytic reactors and deformation in hydrophilic swelling porous media. The organization of this paper is as follows: In section 2, we show the existence and uniqueness of weak solutions to the reaction-diffusion system defined at the microscale. In section 3 we prove the homogenization limit ε → 0, while in section 4 we investigate the fast-reaction asymptotics for the two-scale reaction-diffusion problem. Section 5 contains a strong formulation of the resulting two-scale moving-boundary problem. Finally, in section 6 we illustrate numerically the fact that as k → ∞ the production by reaction within the microstructure spatially shrinks to a Radon measure localized at the position of the moving boundary.

2 THE MICRO PROBLEM

2

3

The micro problem

Let L > 0 and R ∈ (0, 1) be given lengths and let ε > 0 be a small number. We consider a layered medium G := (0, L)2 in two space dimensions as divided into periodic cells εZ, which are ε-scaled versions of the unit cell Z := (0, 12 ) × (−1, 1) as depicted in Figure 1. Let Z2 := (0, 21 ) × (−R, R) be the material where diffusion is slow and Z1 := Z \ Z 2 the remaining part. Then G is separated into subdomains [ ε(k + Zi ), i = 1, 2. Gεi := G ∩ k2 (2k1 , 2 )∈Z2 The interface between the two phases is denoted by Γε := ∂Gε1 ∩ ∂Gε2 . Its unit normal pointing towards Gε1 is ν := (0, ±1)T . Moreover, we specify two different parts of the exterior boundary ∂G via Γext := {0} × (0, L) and ΓN := ∂G \ Γext .

Figure 1: Geometry of the model (Pkε ). The right figure is scaled by ε−1 . Let D, d1 , d2 , k, b and α be positive constants. We consider the following reaction–diffusion problem on the micro scale ∂t U ε − D∆U ε = 0, ∂t uε − ε2 d1 ∆uε = −kuε v ε , ∂t v ε − ε2 d2 ∆v ε = −αkuε v ε ,

x ∈ Gε1 , t > 0 x ∈ Gε2 , t > 0 x ∈ Gε2 , t > 0

(1) (2) (3)

x ∈ Γε , t ≥ 0

(4) (5) (6)

with interior boundary conditions −D∇U ε · ν = −ε2 d1 ∇uε · ν, = εb(uε − U ε ), d2 ∇v ε = 0,

2 THE MICRO PROBLEM

4

exterior boundary conditions U ε = U ext (t),

x ∈ ∂Gε1 ∩ Γext , t ≥ 0

(7)

D∇U ε = 0, d1 ∇uε = d2 ∇v ε = 0,

x ∈ ∂Gε1 ∩ ΓN , t ≥ 0 x ∈ ∂Gε2 ∩ ∂G, t ≥ 0

(8) (9)

x = (x1 , x2 ) ∈ G.

(10) (11) (12)

and initial conditions U ε (0, x) = U0 (x1 ), uε (0, x) = u0 (x1 , x2 /ε), v ε (0, x) = v0 (x1 , x2 /ε),

The system (10)–(12) is referred to as (Pkε ). For typical fast-reaction–slowtransport scenarios, the parameter k is large and defines a high Thiele modulus. The case b → ∞ formally corresponds to the Dirichlet case uε |y=R = U ε . Let S := [0, T ] be a bounded time interval. We also denote Ω := (0, L) and Y := (0, R). We assume for the data U ext ∈ H 1 (S),

U0 ∈ H 1 (Ω),

u0 , v0 ∈ H 1 (Ω × Y ),

and there is a constant CM > 0 such that 0 ≤ U ext , U0 , U0 , v0 ≤ CM

a.e.

(13)

and the following compatibility conditions: U ext (0) = U0 (0), u0 (x1 , R) = U0 (x1 ), ∂x2 u0 (x1 , 0) = ∂x2 v0 (x1 , 0) = ∂x2 v0 (x1 , R) = 0. We can therefore extend u0 (x1 , x2 ) and v0 (x1 , x2 ) first by symmetry to x2 ∈ (−R, R), afterwards by U0 for x2 ∈ (−1, 1), and finally periodically to x2 ∈ .

R

Remark 2.1. Note that, due to the symmetry of the layered geometry and of the data, it is actually sufficient to consider only one single strip of height ε. However, in order to apply known homogenization results later on, it is more convenient to work with an ε-independent domain G. The symmetry will be taken into account after the homogenization step. In order to formulate the problem in a weak setting, we introduce the function space VDε := {u ∈ H 1 (Gε1 ) : u = 0 at ∂Gε1 ∩ Γext }.

2 THE MICRO PROBLEM

5

The weak formulation of Problem (Pkε ) is: Find functions uε ∈ U ext + L2 (S; VDε ) and U ε , V ε ∈ L2 (S; H 1 (Gε2 )) such that (10)–(12) are satisfied, and it holds Z Z Z d ε ε u ϕ+ D∇u ∇ϕ + b(U ε − uε )ϕ = 0 (14) ε dt Gε1 ε G1 Γ Z Z Z Z d ε 2 ε ε ε u Φ+ ε d1 ∇u ∇Φ + b(u − U )Φ + kuε v ε Φ = 0 (15) dt Gε2 Gε2 Γε Gε2 Z Z Z d ε 2 ε v Ψ+ ε d2 ∇v ∇Ψ + αkuε v ε Ψ = 0 (16) ε ε dt Gε2 G2 G2 for all ϕ ∈ VDε and Φ, Ψ ∈ H 1 (Gε2 ). The following boundedness result is obvious. Proposition 2.2. Any weak solution is nonnegative and essentially bounded via 0 ≤ uε , U ε , v ε ≤ CM a.e., where CM is given from (13). Proposition 2.3. The following energy estimates hold independently of ε: Z t Z t ε 2 ε 2 ε 2 kU (t)kGε1 + ku (t)kGε2 + k∇U kGε1 + kε∇uε k2Gε2 0 0 (17) Z t 2 2 ext 2 ≤ C(kU0 kΩ + ku0 kΩ×Y + (U ) ), 0 Z t kv ε (t)k2Gε2 + kε∇v ε k2Gε2 ≤ Ckv0 k2Ω×Y , (18) 0

Proof. Test (14) with ϕ := uε − U ext and (15) with Φ := uε and adding the result yields (17). The second estimate (18) is obtained by testing (15) with Ψ := v ε . Proposition 2.4. The following improved estimate holds independently of ε: k∂t U ε (t)k2Gε1 + k∂t uε (t)k2Gε2 + k∂t v ε (t)k2Gε2 Z t Z t Z t ε 2 ε 2 + k∇∂t U kGε1 + kε∇∂t u kGε2 + kε∇∂t v ε k2Gε2 ≤ C. 0

0

0

Proof. Differentiate the system with respect to time and test with the time derivatives; see also Proposition 5 in [2].

2 THE MICRO PROBLEM

6

Remark 2.5. Most likely, we can perform the homogenization step without that much regularity. It is probably sufficient to ensure ∂t U ε ∈ L2 (S × Gε1 ), which gives compactness for U ε . Note that we cannot get compactness for uε and v ε due the ε2 -scaling in the elliptic terms. Theorem 2.6. There exists a unique weak solution of (Pkε ). Proof. This is obtained by standard techniques, for example by cutting off the nonlinearities and using Schauder’s fixed point theorem. We refer the reader to [2], Thm. 1, for the precise proof for the case of a quite similar PDE system. Remark 2.7. Strong solutions of (Pkε ), or even classical solutions can in general not be expected, due to the mixed boundary conditions for U ε . Also, the domains are not smooth (rectangles). However, these regularity restrictions will be removed naturally in the limit system! Proposition 2.8. There exist extensions U ε , uε and v ε defined on all of G such that for a subsequence w

Uε * U Uε → U Uε → U ∇U ε → ∇x U (x) + ∇y U1 (x, y)

weakly in L2 (S; H 1 (G)), strongly in L2 (S × G), two-scale, two-scale,

w

weakly in L2 (S; H 1 (G)), two-scale, two-scale,

w

weakly in L2 (S; H 1 (G)), two-scale, two-scale.

uε * u uε → u ε∇uε → ∇y u(x, y) vε * v vε → v ε∇v ε → ∇y v(x, y)

Proof. All convergence result are basically standard; for more details, see Proposition 8 and Proposition 9 in [2]. In order to prove them, first an extension theorem needs to be applied in the same way as reported in [2] or [1]. Then, as next and final step, the two-scale convergence results stated above follow from well-known convergence results; see, e.g., Proposition 1.14 in [15].

3 HOMOGENIZATION LIMIT

3

7

Homogenization limit

Let θ := (1 − R)/2. Recall that Ω = (0, L) and Y = (0, R). We denote by Problem (Pk ) the following two-scale system in one space dimension: θ∂t U − θD∂xx U = −b(U − u|y=R ), ∂t u − d1 ∂yy u = −kuv, ∂t v − d2 ∂yy v = −αkuv,

x ∈ Ω, t ∈ S x ∈ Ω, y ∈ Y, t ∈ S x ∈ Ω, y ∈ Y, t ∈ S

(19) (20) (21)

with boundary conditions on the microscale d1 ∂y u(t, x, R) = −b(u(t, x, R) − U (t, x)), ∂y u(t, x, 0) = ∂y v(t, x, 0) = ∂y v(t, x, R) = 0,

x ∈ Ω, t ∈ S

(22) (23)

t∈S

(24) (25)

exterior boundary conditions U (t, 0) = U ext (t), D∂x U (t, R) = 0, and initial conditions U (0, x) = U0 (x), u(0, x, y) = u0 (x, y), v(0, x, y) = v0 (x, y),

x ∈ Ω, y ∈ Y.

(26) (27) (28)

We formulate (Pk ) in a weak setting as follows: Let HL1 (Ω) denote the functions in H 1 (Ω) vanishing at x = 0. Define V1 := HL1 (Ω), V2 := L2 (Ω; H 1 (Y )), V := V1 × [V2 ]2 , H := L2 (Ω) × [L2 (Ω × Y )]2 , Vi := L2 (S; Vi ), i = 1, 2,

V := L2 (S; V ).

The weak formulation of Problem (Pk ) is derived by multiplying the equations (19)–(21) with test functions (ϕ, Φ, ψ) ∈ V , respectively, and integrating (19) over Ω and (20),(21) over Ω × Y . The result is: Find essentially bounded functions U ∈ U ext + V1 and u, v ∈ V2 that satisfy (26)–(28) and Z Z Z d θU ϕ + θD∂x U ∂x ϕ + b(U − u|y=R )(ϕ − Φ|y=R ) dt Ω Ω Ω Z Z Z Z Z Z d + uΦ + d1 ∂y u∂y Φ + kuvΦ = 0 ∀(ϕ, Φ) ∈ V1 × V2 , dt Ω Y Ω Y Ω Y (29)

3 HOMOGENIZATION LIMIT d dt

Z Z

Z Z vΨ +

Ω

Y

8 Z Z

d2 ∂y v∂y Ψ + Ω Y

∀Ψ ∈ V2 .

αkuvΨ = 0

(30)

Ω Y

Remark 3.1. Note that in this formulation, the boundedness of the involved functions is only needed in order to give a meaning to the nonlinear reaction terms. Theorem 3.2. The limit functions U , u and v satisfy the two-scale problem (Pk ) in the weak sense. Proof. Step 1: The limit system in space dimension 2 Using the a priori estimates from Propositions 2.2–2.4, we can apply the same monotonicity trick as in [2] (Prop. 12) in order to pass to the limit ε → 0 via two-scale convergence. In the strong formulation, the limit equations are given by Z ∗ |Z1 |∂t U − ∇ · (D ∇U ) = − b(U − u) x ∈ G, Γ

∂t u − d1 ∆u = −kuv ∂t v − d2 ∆v = −αkuv

x ∈ G, y ∈ Z2 , x ∈ G, y ∈ Z2 .

The macro diffusion tensor is given by Z 1 + ∂y1 ζ1 ∂y1 ζ2 ∗ D := D dy, ∂y2 ζ1 1 + ∂y2 ζ2 Z1 where ζ1 , ζ2 are Z-periodic solutions of ∆y ζj = 0 − ∇y ζ · ν = ej · ν

y ∈ Z1 y∈Γ

(31) (32)

Moreover, the limit functions are subject to the same L∞ -bounds as that from Proposition 2.2. Step 2: Dimension reduction in the micro problem Due to the symmetry of the geometry and data, it follows immediately that u and v are independent of y1 , i.e., u = u(t, x, y2 ) and v = v(t, x, y2 ). Moreover, it is sufficient to consider only y2 ∈ (0, R) = Y . From that we obtain (20) and (21). The macro problem is still two-dimensional, but it simplifies to (1 − R)∂t U − ∇ · (D∗ ∇U ) = −2b(U − u|y=R ), Step 3: Dimension reduction in the macro problem

x ∈ G.


9

The solutions of (31)–(32) are given by ζ2 (y) = −y2 + c1

and

ζ1 (y) = c1 .

This gives ∂y1 ζ2 = 0 = ∂y1 ζ1 = ∂y2 ζ1

∂y2 ζ2 = −1.

and

Hence, we get Z

∗

D =D Z1

1 0 0 0

dy =

D(1 − R) 0 0 0

.

This points out that diffusion is restricted to x1 -direction. It follows by symmetry that u = u(t, x1 ) is actually independent of x2 and satisfies (20). Remark 3.3. The positivity and L∞ -bounds of the solutions follow from the homogenization procedure. On the other hand, they can be also proved directly for (Pk ). Lemma 3.4. It holds kkuvkL1 (S×Ω×Y ) ≤

1 kv0 kL1 (Ω×Y ) . α

(33)

Proof. Integrating (21) along S × Ω × Y we directly obtain Z TZ Z Z TZ Z α kuv = − ∂t v − d2 ∂yy v 0 Ω Y 0 Ω Y Z Z Z Z Z tZ Z v(T, x, y) + = − v0 (x, y) + d2 ∂y v Ω Y Ω Y 0 Ω ∂Y | {z } | {z } =0 ≥0 Z Z ≤ v0 (x, y). (34) Ω

Y

Lemma 3.5. (Energy estimates independent of k) There exists a constant c > 0, which is independent of k, such that following estimates hold: Z Z Z TZ Z 2 max v + |∂y v|2 ≤ c (35) t∈S Ω Y 0 Ω Y Z Z TZ Z Z Z TZ Z 2 2 2 max U + u + |∂x U | + |∂y u|2 ≤ c. (36) t∈S

Ω

Ω

Y

0

Ω

0

Ω

Y


10

Proof. Testing (21) by v, we obtain Z 0

T

1d 2 dt

Z Z

Z

2

T

Z Z

Y

uv 2 = 0 Ω Y {z }

|∂y v| + αk

v + d2 Ω

Z tZ Z

2

0

Ω

Y

0

|

(37)

≥0

from which we deduce Z tZ Z Z Z Z Z 1 1 2 2 v(t) + d2 |∂y v| ≤ v02 2 Ω Y 2 0 Ω Y Ω Y which gives (35). Furthermore, testing (19) by U − U ext gives Z Z Z Z θ d 2 ext 2 U − θ U ∂t U + θD (∂x U ) = −b (U − u|y=R )(U − U ext ) (38) 2 dt Ω Ω Ω Ω and testing (20) by u − U ext gives 1d 2 dt

Z Z

2

Z Z

u − Ω Y

u∂t U Ω YZ

−b

ext

Z Z + d1

(∂y u)2

Ω Y

(U − u|y=R )(u − U

ext

Z Z )=−

Ω

kuv(u − U ext ) (39)

Ω Y

Adding (38) and (39) and integration from 0 to t yields with standard estimates Z Z tZ Z Z Z tZ Z 2 2 2 θ U (t) + 2θD (∂x U ) + u(t) + 2d1 (∂y u)2 Ω 0 Ω Ω Y Z Z Z Z t 0 Ω YZ t ≤ θ U02 + u20 + k∂t U ext k2S + θ2 kU k2Ω + kuk2Ω×Y Ω Ω Y 0 0 Z tZ Z ext + kU k∞ kuv. 0

Ω Y

Gronwall’s inequality and Lemma 3.4 yield (35). Denoting by w the expression u − α1 v, we see that ∂t w = d1 ∆y u −

d2 ∆y v. α

(40)

Proposition 3.6. There exists a constant c1 , which is independent of k, such that k∂t U kV10 + k∂t wkV20 ≤ c1 . (41)


11

Proof. Take (ϕ, ξ) ∈ V1 × V2 arbitrarily. Testing (40) with ξ gives Z 0

T

Z

T

Z Z

d2 ∂y u∂y ξ + α

Z

T

Z Z

h∂t w, ξi = −d1 ∂y v∂y ξ 0 Ω Y 0 Ω Y Z TZ (u(t, x, R) − U (t, x)ξ(t, x, R) − b 0

(42)

Ω

and testing (19) with ϕ yields Z

T

T

Z

Z

h∂t U, ϕi = −θ

θ 0

Z

T

Z

∂x U ∂x ϕ − b 0

Ω

(U (x, t) − u(t, x, R))ϕ(t, x). 0

Ω

(43) Using Cauchy-Schwarz and Hölder inequalities as well as the (k-independent!) energy estimates stated in Lemma 3.5, we get the estimate Z T h∂t w, ξi| ≤ c0 ||∂y u||L2 (S×Ω×Y ) + ||∂y v||L2 (S×Ω×Y ) ||∂y ξ||L2 (S×Ω×Y ) | 0 Z TZ + bCM ||ξ||H 1 (Y ) 0

Ω

≤ c||ξ||L2 (S×Ω×H 1 (Y )) .

(44)

where the constant c is independent of k. Therefore k∂t wkL2 (S×Ω;(H 1 (Y ))0 ) ≤ c < ∞. Proceeding in a similar way, we also obtain that k∂t U kL2 (S;HL1 (Ω)0 ) ≤ c < ∞. Finally, adding the latter two estimates we obtain (41). The estimate (41) encourages us to use the compactness criterion provided via the Lions-Aubin’s Lemma [as stated in the following Lemma 3.7]. Lemma 3.7. (Lions-Aubin) Let B0 ,→ B ,→ B1 be Banach spaces such that B0 and B1 are reflexive and the embedding B0 ,→ B is compact. Fix p, q > 0 and let dz p q W = z ∈ L (S; B0 ) : ∈ L (S; B1 ) dt with ||z||W := ||z||Lp (S;B0 ) + ||∂t z||Lq (S;B1 ) . Then W ,→,→ Lp (S; B).

4 FAST-REACTION LIMIT

12

Since the embedding V2 ,→ L2 (Ω × Y ) is not compact, Lemma 3.7 can not be applied for the whole two-scale system. However, it is applicable for the choice (B0 , B, B1 ) := (H 1 (Y ), L2 (Y ), H 1 (Y )0 ). This implies that W := {w ∈ L2 (S, H 1 (Y )) : ∂t w ∈ L2 (S; H 1 (Y )0 )} ,→,→ L2 (S × Y ). Remark 3.8. Depending on the choice of the micro-macro coupling in the twoscale problem, additional regularity of the micro-variables with respect to the macro-variable x is sometimes needed in order to get the desired compactness for the problem stated at the micro-level. This can be done by differentiating the micro system with respect to x ∈ Ω and using a characterization of the spaces H 1 (Ω, B) by translation estimates (cf. Lemma 7.23 and Lemma 7.24 in [16] for the case B = R).

4

Fast-reaction limit

We deduce from Lemma 3.5 that the family {Uk } is bounded in L2 (S; H 1 (Ω)) and in L∞ (S; L2 (Ω)) and that the families {uk } and {vk } are bounded in V2 and in L∞ (S; L2 (Ω × Y )). Thus, there exist subsequences of {Uk }, {uk } and {vk }, which we denote the same, and functions (U , u) ∈ V1 and v ∈ V2 such that 0 ≤ U , u, v ≤ CM a.e. and w

Uk * U w

Uk (T ) * κ1 w

w

uk * u, vk * v w

w

uk (T ) * κ2 , vk (T ) * κ3

weakly in L2 (S; H 1 (Ω)),

(45)

weakly in L2 (Ω),

(46)

weakly in V2 ,

(47)

2

weakly in L (Ω × Y )

(48)

Results from the previous section provide, for a.e. x ∈ Ω, the needed relative compactness of wk (·, x, ·) in L2 (S × Y ) which ensures that vk →w in L2 (S × Y ) as k → ∞. (49) wk = uk − α It follows for a subsequence that wk → w for almost every (t, x, y) ∈ S×Ω×Y , and therefore, by the uniform boundedness of uk and vk , wk → w

in Lp (S × Ω × Y )

as k → ∞

for any p ≥ 1. Moreover, from the estimate in Lemma 3.4 we deduce that uk v k → 0

in L1 (S × Ω × Y ) as k → ∞.

(50)


13

Lemma 4.1. (a) The subsequences uk and vk are such that uk → w+ , and vk → αw− as k → ∞ in L1 (S × Ω × Y ) and a.e. in S × Ω × Y . (b) It holds: u¯ = w+ and v¯ = αw− . Proof. Since (b) is a straightforward consequence of (a), we only show a proof for (a). To this end, we follow the lines of the proof of Lemma 3.1 in [6]. Recall vk that for a.e. (t, x, y) ∈ S ×Ω×Y we have wk = uk − α (t, x, y) → w(t, x, y) and (uk vk )(t, x, y) → 0 as k → ∞. We distinguish between three distinct cases: (1) w(t, x, y) > 0: In this case, for all η > 0 there exists a rank kη ∈ N such that for all k ≥ kη we have |uk − vαk − w| ≤ η. In particular, we obtain uk − vαk ≥ w − η. Choosing now η := w2 , we use the positivity of vk to get that uk ≥ w2 . (2) w(t, x, y) < 0: We proceed analogously as in case (1). (3) w(t, x, y) = 0: We argue now via reductio ad absurdum. Assume that there exists a subsequence uk → ρ > 0. Then the separation in space of the reactants forces vk → 0 as k → ∞, and hence, uk − vαk → ρ. On the other hand, we recall that uk − vαk → 0 as k → ∞, which is contradicting our sign assumption on ρ. Since all mentioned subsequences are essentially bounded, we obtain by Lebesgue’s theorem of dominated convergence that these subsequences are also Lp convergent for all p ≥ 1. In what follows we want to derive both weak and strong formulations for the limit problem satisfied by w and U . Proposition 4.2. The pair of functions (U , w) defined via (45) and (49) satisfies the Problem (P), which is Z TZ

Z

Z TZ

θD∂x U ∂x ϕ θU ∂t ϕ − θU0 ϕ(0) + Ω 0 Ω Z TZ Z Z Z Z TZ Z v0 d2 − w∂t Φ − (u0 − )Φ(0) + (d1 ∂y u − ∂y v)∂y Φ α α 0 Ω Y Ω Y 0 Ω Y Z T = −b (U¯ − u¯|y=R )(ϕ − Φ|y=R ) (51)

−

0

Ω

0

for all test functions ϕ ∈ C 1 (S; HL1 (Ω)) and Φ ∈ C 1 (S; L2 (Ω; H 1 (Y ))) such that ϕ(T ) = Φ(T ) = 0.


14

Proof. We multiply the strong wk -problem by test functions ϕ and Φ as stated above. Integration by parts gives Z TZ

Z

Z TZ

θUk ∂t ϕ −

− Z

θU0 ϕ(0) + θD∂x Uk ∂x ϕ 0 Ω Z Z Z TZ Z v0 d2 wk ∂t Φ − (u0 − )Φ(0) + (d1 ∂y uk − ∂y vk )∂y Φ α α Y Ω Y 0 Ω Y Z TZ (Uk − uk |y=R (ϕ − Φ|y=R ). = −b

0 Ω TZ Z

− 0

Ω

Ω

0

Ω

Relying on the weak convergence (45)–(48), we pass in each term on the left-hand side to its corresponding limit for k → ∞. Thus, we obtain (51) in the limit. Theorem 4.3. Assume that d1 = d2 /α. Then there exists at most a couple (U¯ , w) satisfying the limit problem (P ) in a weak sense. Proof. Since problem (P) is linear, we choose w0 = 0, U ext = 0, and U¯0 = 0 and discuss the uniqueness problem for the case b ∈]0, 1]. Testing in the weak formulation of problem (P) by U¯ and w we obtain Z Z Z θ d 2 2 ¯ ¯ (52) |U | + θD |∂x U | = −b (U¯ − w+ |y=R )U¯ 2 dt Z Ω Z Z Ω Z ZΩ 1d (53) |w|2 + d1 |∂y w|2 = b (U¯ − w+ |y=R )w. 2 dt Ω Y Ω Ω Y Summing up (52) and (53), we obtain: Z Z Z Z Z Z θ d 1d 2 2 2 ¯ ¯ |U | + |w| + θD |∂x U | + d1 |∂y w|2 2 dt Ω 2 dt Ω Y Ω Ω Y Z = −b (U¯ − w+ )(U¯ − w) ZΩ Z + 2 = −b (U¯ − w ) −b (U¯ − w+ )w− Ω

Ω

| ≤0 Integrating now (54) along (0, t) we have Z Z Z 1 1 2 ¯ θU (t) + w(t)2 ≤ 0, 2 Ω 2 Ω Y which concludes the proof of the theorem.

{z

≤0

}

(54)


15

Let us now recover the strong formulation of problem (P ). Note however that in general the second derivative of w may not exist! Choosing ϕ ≡ 0 and Φ ∈ C0∞ (S × Ω × Y ) and partial integration back yields d2 in Y. ∂t w − d1 ∂yy u − ∂yy v = 0 α Testing afterwards with some Φ being nonzero at t = 0 and y = 0, respectively, yields the initial condition w(0) = u0 − α1 v0 and the boundary condition d2 d1 ∂y u|y=0 − ∂y v|y=0 = 0. α Furthermore, testing with a function which is not vanishing at y = R, we obtain d2 u(t, x, R) − U¯ (t, x)). −d1 ∂y u¯ + ∂y v¯ = −b(¯ α Let’s assume that we have w(t, x, R) > 0

for all t and x.

(55)

then it follows the continuity of w in y = R that ∂y v¯|y=R = 0.

(56)

Note that (55) can be removed if instead more regularity of the initial data u0 , v0 and U0 is provided; For related remarks, see Remark 1.1. and Theorem 1.3 in [17], p. 43, Lemma 3.5 in [6], or Theorem 1.1 in [13]. Letting now ϕ ∈ C0∞ (S × Ω) and Φ ≡ 0, the integration by parts yields θ∂t U − θD∂xx U = −b(U¯ − u¯|y=R ) in Ω. Testing afterwards with nonzero ϕ at x = L we recover the boundary condition ∂y U |x=L = 0. Finally, we can recover the initial condition U (0) = U 0 by the standard technique (see e.g. pp. 356-357 in [18] an example for the derivation of U (0) = U 0 ). Finally, we arrive at the following strong formulation for (P ). Define d(w) := d1 if w ≥ 0 and d(w) := dα2 otherwise. Note that d(w)∂yy w = ∂y (d(w)∂y w). θ∂t U − θD∂xx U = −b(U¯ − w+ |y=R ), x ∈ Ω, t ∈ S ∂t w − ∂y (d(w)∂y w) = 0, −d(w)∂y w(t, x, R) = b(w+ (t, x, R) − U (t, x)), ∂y w− (t, x, R) = 0, ∂y w(t, x, 0) = 0, U (t, 0) = U ext (t), ∂x U (t, L) = 0,

x ∈ Ω, y ∈ Y, t ∈ S x ∈ Ω, t ∈ S x ∈ Ω, t ∈ S x ∈ Ω, t ∈ S t∈S t∈S

5 RECOVERING THE TWO-SCALE FREE-BOUNDARY PROBLEM16 with initial conditions U (0) = U0 and w(0) = u0 − α1 v0 .

5

Recovering the two-scale free-boundary problem

Under sufficient regularity assumptions on the involved functions and geometry, the limit problem (P) can be written as a two-scale free boundary problem with the free boundary concentrated in the microscale as we conjectured for the case of a related problem with matched micro-macro boundary conditions; see [3] (section 3.2). We denote by Σ the non-empty support of v and assume Σ = {{t} × {x} × (0, s(t, x)) : t ∈ S, x ∈ Ω}

(G)

where s(t, x) ∈ (0, R) denotes the point separating the support of u¯(·, x, ·) from that one of v¯(·, x, ·) for a.e. x ∈ Ω. We refer to s(t, x) as the position of the a priori unknown free boundary at time t ∈ S corresponding to the point x ∈ Ω. Moreover, we assume that u > 0 outside Σ, i.e., the interface is in fact sharp and the reaction in each local cell concentrates on the point y = s(t, x). Remark 5.1. The initial position s0 of the free boundary is immediately recovered since the initial values of the micro-concentrations, solutions of the (Pk ) problem, have initially disjoint supports, i.e. there is a function s0 : Ω → (0, R) such that w < 0 in (0, s0 (x)) and w > 0 in (s0 (x), R). It is worth mentioning at this point that in Remark 1 from [9] the choice of s0 is related to an upper estimation of the size of the initial transient layer for a reaction-diffusion system with fast reaction. U¯ , u¯, v¯ and s are sufficiently smooth.

(R)

If assumption (R) is satisfied, then problem (P) can be reformulated as follows: Find (U¯ , u¯, v¯, s) satisfying following model equations θ∂t U¯ − θD∂xx U¯ = −d1 ∂y u¯(t, x, R) ∂t u¯ − d1 ∂yy u¯ = 0 ∂t v¯ − d2 ∂yy v¯ = 0

in Ω, in (s(t, x), R), in (0, s(t, x)),

(57) (58) (59)

5 RECOVERING THE TWO-SCALE FREE-BOUNDARY PROBLEM17 boundary conditions d1 ∂y u¯(t, x, R) = −b(¯ u(t, x, R) − U¯ (t, x)), −d2 ∂y v¯(t, x, 0) = 0 d2 −d1 ∂y u¯(t, x, s(t, x)) = ∂y v¯(t, x, s(t, x)) α u¯(t, x, s(t, x)) = 0 = v¯(t, x, s(t, x))

(60) (61) (62) (63)

as well as initial conditions U¯ (0, x) = U0 (x), u¯(0, x, y) = u0 (x, y), v¯(0, x, y) = v0 (x, y), s(0, x) = s0 (x), (64) for all (x, y) ∈ Ω × Y and t ∈ S. We refer to (57)–(64) as (FBP). Remark 5.2. The presence of two spatial scales makes the free-boundary problem (FBP) nonstandard. If we would forget for a while the macroequation (57), then the remaining equations would constitute the classical two-phase Stefan problem with zero latent heat and instantaneous reaction. We refer the reader to [19] for the existence, uniqueness and regularity study of such problem in one-space dimension and to [20] for the analysis of its n-dimensional counterpart. Remark 5.3. By assuming that s(t, x) > 0, we exclude the case where the moving boundary reaches the left end of the interval. Note that in this case, the PDE (59) and the boundary conditions (61) and (62) lose their meaning. In the next result, we show that the production by reaction will concentrate on a (Radon) measure localized within the microstructure. In this sense, we adapt some ideas from [8]. Denote by µk the expression −kuk vk . Theorem 5.4. (Convergence of the reaction term to a measure) Assume that (R) and (G) hold. Then µk converges to the Radon measure µ given by 1 (65) µ(t, x, y) = (−d1 ∂y u(t, x, y) + d2 ∂y v(t, x, y)) δ(y − s(t, x)), 2 where (t, x, y) ∈ S × Ω × Y . Proof. Let take ψ ∈ C0∞ (S × Y ). Integrating by parts we get Z Z Z TZ 1 T 2 vkt ψ − d2 ∂yy vk ψ µk ψdydt = α 0 Y 0 Y Z Z Z Z 1 T d2 T = − vk ∂t ψ + ∂y vk ∂y ψ α 0 Y α 0 Y Z TZ = ukt ψ − d1 ∂yy uk ψ 0 Y Z TZ Z TZ = −uk ∂t ψ + d1 ∂y uk ∂y ψ. 0

Y

0

Y

6 NUMERICAL EXAMPLE ILLUSTRATING THE CASE K → ∞

18

Passing to the limit k → ∞ in the last expression yields Z 0

T

Z Z Z Z 1 T s(t) d2 T s(t) µψdydt = − v¯∂t ψ + ∂y v¯∂y ψ α 0 0 α 0 0 Z TZ R Z TZ R −¯ u∂t ψ + d1 ∂y u¯∂y ψ. =

Z Y

0

s(t)

0

(66) (67)

s(t)

Putting now in (66) and (67) the derivatives on u¯ and v¯ and using the limit equations (58) and (59) for u¯ and v¯, we obtain Z

T

Z

Z

T

−d1 ∂y u¯(t, x, s(t, x))ψ(t, s(t, x))

µψ = 0

Y

0

Z = 0

T

d2 ∂y v¯(t, x, s(t, x))ψ(t, s(t, x)). α

This proves (65).

6

Numerical example illustrating the case k → ∞

For illustration, typical solution profiles of the two-scale model with finite reaction constant (Pk ) are plotted in Figure 2 for k = 103 and in Figure 3 for k = 104 . The remaining parameters are chosen as L = 1, R = 0.6, D = 1, d1 = d2 = 0.5, α = 0.5, b = 100. The initial values for u and v have disjoint supports. In the left column, the macroscopic concentration U is plotted at different times. In the central and right column, the local concentrations u and v at two different fixed points x = 0.1 and x = 0.5 are shown. It can be seen that the overlapping zone is small, such the reaction in each local cell concentrates on a narrow region. This reaction zone shrinks for larger values of k.


19

Figure 2: Solution profiles of the two-scale model (Pk ) at different times (k = 103 ). Left columns: Profiles of U . Central and right column: Local cell profiles of u and v at x = 0.1 and x = 0.5.


20

Figure 3: Solution profiles of the two-scale model (Pk ) at different times (k = 104 ). Left columns: Profiles of U . Central and right column: Local cell profiles of u and v at x = 0.1 and x = 0.5.

REFERENCES

21

Acknowledgments A.M. acknowledges fruitful discussions with Dr. Maria-Neuss Radu (Heidelberg) and Dr. Sébastian Martin (Paris, Orsay) on two-scale reaction-diffusion models and singular limit techniques.

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