Local Well-Posedness of a Quasi-Incompressible Two-Phase Flow

We show well-posedness of a diffuse interface model for a two-phase flow of two viscous incompressible fluids with different densities locally in time. The model leads to an inhomogeneous Navier-Stokes/Cahn-Hilliard system with a solenoidal velocity field for the mixture, but a variable density of the fluid mixture in the Navier-Stokes type equation. We prove existence of strong solutions locally in time with the aid of a suitable linearization and a contraction mapping argument. To this end we show maximal $L^2$-regularity for the Stokes part of the linearized system and use maximal $L^p$-regularity for the linearized Cahn-Hilliard system.


Introduction and Main Result
In this contribution we study a thermodynamically consistent, diffuse interface model for two-phase flows of two viscous incompressible system with different densities in a bounded domain in two or three space dimensions. The model was derived by A., Garcke and Grün in [6] and leads to the following inhomogeneous Navier-Stokes/Cahn-Hilliard system: in Q T := Ω × (0, T ) together with the initial and boundary values v |∂Ω = ∂ n ϕ |∂Ω = ∂ n µ |∂Ω = 0 on (0, T ) × ∂Ω, in Ω. (6) Here Ω ⊆ R d , d = 2, 3, is a bounded domain with C 4 -boundary. In this model the fluids are assumed to be partly miscible and ϕ : Ω × (0, T ) → R denotes the volume fraction difference of the fluids. v, q, and ρ denote the mean velocity, the pressure and the density of the fluid mixture. It is assumed that the density is a given function of ϕ, more precisely whereρ 1 ,ρ 2 are the specific densities of the (non-mixed) fluids. Moreover, µ is a chemical potential and W (ϕ) is a homogeneous free energy density associated to the fluid mixture, ε > 0 is a constant related to "thickness" of the diffuse interface, which is described by {x ∈ Ω : |ϕ(x, t)| < 1−δ} for some (small) δ > 0, and m(ϕ) is a mobility coefficient, which controls the strength of the diffusion in the system. Finally η(ϕ) is a viscosity coefficient and Dv = 1 2 (∇v + ∇v T ). Existence of weak solution for this system globally in time was shown by A., Depner, and Garcke in [4] and [5] for non-degenerate and degenerate mobility in the case of a singular free energy density W . Moreover, Grün showed in [13] convergence (of suitable subsequences) of a fully discrete finite-element scheme for this system to a weak solution in the case of a smooth W : R → R with suitable polynomial growth. In the case of dynamic boundary conditions, which model moving contact lines, existence of weak solutions for this system was shown by Gal, Grasselli, and Wu in [11]. In the case of non-Newtonian fluids of suitable p-growth existence of weak solutions was proved by A. and Breit [3]. For the case of a non-local Cahn-Hilliard equation and Newtonian fluids the corresponding results was derived by Frigeri in [10] and for a model with surfactants by Garcke and the authors in [7]. Recently, Giorgini [12] showed in the two-dimensional case well-posedness locally in time for general bounded domains and globally in time under periodic boundary conditions. Remark 1.1 In [4] it is shown that the first equation is equivalent to ρ∂ t v + ρv +ρ 1−ρ2 2 m(ϕ)∇( 1 ε W ′ (ϕ) − ε∆ϕ) · ∇v + ∇p − div(2η(ϕ)Dv) = −ε∆ϕ∇ϕ.
This reformulation will be useful in our analysis.
For the following we assume: Let Ω ⊆ R d be a bounded domain with C 4 -boundary and d = 2, 3.
With these assumptions we will show our main existence result on short time existence of strong solutions for (1)-(6):
The structure of this contribution is as follows: In Section 2 we introduce some basic notation and recall some results used in the following. The main result is proved in Section 3. For its proof we use suitable estimates of the non-linear terms, which are shown in Section 4, and a result on maximal L 2regularity of a Stokes-type system, which is shown in Section 5. Acknowledgements: The authors acknowledge support by the SPP 1506 "Transport Processes at Fluidic Interfaces" of the German Science Foundation (DFG) through grant GA695/6-1 and GA695/6-2. The results are part of the second authors PhD-thesis [16].

Preliminaries
For an open set U ⊆ R d , m ∈ N 0 and 1 ≤ p ≤ ∞ we denote by W m p (U ) the L p -Sobolev space of order m and W m p (U ; X) its X-valued variant, where X is a Banach space. In particular, L p (U ) = W 0 p (U ) and L p (U ; X) = W 0 p (U ; X). Moreover, B s pq (Ω) denotes the standard Besov space, where s ∈ R, 1 ≤ p, q ≤ ∞, and L 2 σ (Ω) is the closure of C ∞ 0,σ (Ω) = {u ∈ C ∞ 0 (Ω) d : div u = 0} in L 2 (Ω) d and P σ : L 2 (Ω) d → L 2 σ (Ω) the orthogonal projection onto it, i.e., the Helmholtz projection.
We will frequently use:

Theorem 2.1 (Composition with Sobolev functions)
Let Ω ⊆ R d be a bounded domain with C 1 -boundary, m, n ∈ N and let 1 ≤ p < ∞ such that m − dp > 0. Then for every f ∈ C m (R N ) and every R > 0 there exists Proof: The first part follows from [14, Chapter 5, Theorem 1 and Lemma]. The second part can be easily reduced to the first part.
In particular we have uv ∈ W m p (Ω) for all u, v ∈ W m p (Ω) under the assumptions of the theorem.
Let X 0 , X 1 be Banach spaces such that X 1 ֒→ X 0 densely. It is well known that continuously for I = [0, T ], 0 < T < ∞, and I = [0, ∞), cf. Amann [8,Chapter III,Theorem 4.10.2]. Here (X 0 , X 1 ) θ,p denotes the real interpolation space of (X 0 , X 1 ) with exponent θ and summation index p. Moreover, BU C(I; X) is the space of all bounded and uniformly continuous f : I → X equipped with the supremum norm, where X is a Banach space. Moreover, we will use: for every x ∈ X 0 and a constant C > 0, where θ ∈ (0, 1). Then continuously.
The result is well-known and can be proved in a straight forward manner.

Proof of the Main Result
We prove the existence of a unique strong solution (v, ϕ) ∈ X T for small T > 0, where the space X T will be specified later. The idea for the proof is to linearize the highest order terms in the equations above at the initial data and then to split the equations in a linear and a nonlinear part such that where we still have to specify in which sense this equation has to hold. To linearize it formally at the initial data we replace v, p and ϕ by v 0 + εv, p 0 + εp and ϕ 0 + εϕ and then differentiate with respect to ε at ε = 0. In (1) and the equivalent equation (7), the highest order terms with respect to t and x are ρ∂ t v, div(2η(ϕ)Dv) and ∇p. Hence the linearizations are given by where ρ 0 := ρ(ϕ 0 ) and ρ ′ 0 := ρ ′ (ϕ 0 ). Moreover, we omit the term div(2η ′ (ϕ 0 )ϕDv 0 ) in the second linearization since it is of lower order. For the last equation we get the linearization We can omit the first term since it is of lower order. The second term can formally be reformulated as Here the first summand is of lower order again. Hence, the linearization is given by −εm(ϕ 0 )∆ 2 ϕ upto terms of lower order. Due to these linearizations we define the linear operator L : where L consists of the principal part of the lionization's, i.e., of the terms of the highest order. Furthermore, we define the nonlinear operator F : It still remains to define the spaces X T and Y T . To this end, we set We equip Z 1 T and Z 2 T with the norms · ′ We use these norms since they guarantee that for all embeddings we will study later the embedding constant C does not depend on T , cf. Lemma 3.1. To this end we use: Lemma 3.1 Let 0 < T 0 < ∞ be given and X 0 , X 1 be some Banach spaces such that X 1 ֒→ X 0 densely. For every 0 < T < T0 2 we define where 1 ≤ p < ∞, equipped with the norm Then there exists an extension operator E : X T → X T0 and some constant C > 0 independent of T such that Eu |(0,T ) = u in X T and for every u ∈ X T and every 0 < T < T0 2 . Moreover, there exists a constant for every u ∈ X T and every 0 < T < T0 2 . Proof: The result is well-known. In the case u(0) = 0, one can prove the result with the aid of the extension operator defined by The case u(0) = 0 can be easily reduced to the case u(0) = 0 by substracting a suitable extension of u 0 to [0, ∞). We refer to [16,Lemma 5.2] for the details.
The last preparation before we can start with the existence proof is the definition of the function spaces X T := X 1 T × X 2 T and Y T by and are the initial values from (6). Note that in the space X 2 T we have to ensure that ϕ |t=0 = ϕ 0 ∈ [−1, 1] since we will use this property to show the Lipschitz continuity of F : X T → Y T in Proposition 3.2. Moreover, we note that X T is not a vector space due to the condition ϕ |t=0 = ϕ 0 . It is only an affine linear subspace of Proposition 3.2 Let the Assumptions 1.2 hold and ϕ 0 be given as in Theorem 1.3. Then there is a constant C(T, R) > 0 such that The proposition is proved in Section 4 below.
Theorem 3.3 Let T > 0 and L, X T and Y T be defined as before. Then This theorem is proved in Section 5 below.
Proof of Theorem 1.3: First of all we note that (1)-(4) is equivalent to The fact that L is invertible will be proven later. Equation (10) implies that we have rewritten the system to a fixed-point equation which we want to solve by using the Banach fixed-point theorem.
To this end, we consider some (ṽ,φ) ∈ X T and define Now let R > 0 be given such that (ṽ,φ) ∈ B XT R (0) and R > 2M . Then it follows from Proposition 3.2 that there exists a constant C = C(T, R) > 0 such that Here we have to ensure that L −1 L(YT ,XT ) does not converge to +∞ as T → 0. But since Lemma 5.6 and Lemma 5.8 below yield L −1 L(YT ,XT ) < C(T 0 ) for every 0 < T < T 0 and for a constant that does not depend on T , this is not the case and we can choose T > 0 in such a way that the previous estimate holds. Note that T depends on R and in general T has to become smaller the larger we choose R. Since we want to apply the Banach fixed-point theorem on . From the considerations above we know that there exists (ṽ,φ) ∈ B XT R (0) such that Then a direct calculation shows . For applying the Banach fixed-point theorem it remains to show that the map- which shows the statement. Hence, the Banach fixed-point theorem can be applied and yields some (v, ϕ) ∈ B XT R (0) ⊆ X T such that the fixed-point equation (10) holds, which implies that (v, ϕ) is a strong solution for the equations (1)- (4).
Finally, in order to show uniqueness in X T , let (v,φ) ∈ X T be another solution. ChooseR ≥ R such that (v,φ) ∈ B XT R (0). Then by the previous arguments we can find someT ∈ (0, T ] such that (10)

Lipschitz Continuity of F
Before we continue we study in which Banach spaces v, ϕ, ∇ϕ, m(ϕ) and so on are bounded.
Note that in the definition of X 2 T , p has to be larger than 4 because we will need to estimate terms like ∇∆ϕ · ∇v, where p = 2 is not sufficient for the analysis and therefore we need to choose p > 2. But for most terms in the analysis p = 2 would be sufficient and 4 < p < 6 would not be necessary. Nevertheless, for consistency all calculations are done for the case 4 < p < 6.
Let ϕ ∈ X 2 T be given. From it (8) follows (Ω)). (15) This implies since p > 4. Note that when we write "ϕ is bounded in Z" for some function space Z, we mean that the set of all functions {ϕ ∈ X 2 T : ϕ X 2 T ≤ R} is bounded in Z in such a way that the upper bound only depends on R and not on T , i.e., there exists C(R) > 0 such that together with the estimate for every t ∈ [0, T ]. Hence, Lemma 2.2 implies Because of W 3 p (Ω) ֒→ C 2 (Ω) for d = 2, 3 due to 4 < p < 6, we obtain that In the nonlinear operator F : X T → Y T the terms η(ϕ), η(ϕ 0 ), m(ϕ), m(ϕ 0 ) and W ′ (ϕ) appear. Hence, we need to know in which spaces these terms are bounded in the sense that there is a constant C(R) > 0, which does not depend on T , such that the norms of these terms in a certain Banach space are bounded by Due to (17) and because the embedding constant only depends on R, it holds Moreover, Theorem 2.1 yields the existence of L > 0 such that for every t ∈ [0, T ], ϕ 1 , ϕ 2 ∈ X 2 T and f ∈ {η, m, W ′ }. In the next step, we want to show that f (ϕ) is bounded in X 2 T and therefore the same embeddings hold as for ϕ, where f ∈ {η, m, W ′ }. Note that from now on until the end of the proof of the interpolation result for f (ϕ), we always use some general f ∈ C 4 b (R). But all these embeddings are valid for f ∈ {η, m, W ′ }. We want to prove that if it holds ϕ ∈ X 2 T with ϕ X 2 T ≤ R, then there exists a constant C(R) > 0 such that f (ϕ) X 2 T ≤ C(R). To this end, let ϕ ∈ X 2 T be given with ϕ X 2 T ≤ R. Since we already know ϕ ∈ C([0, T ]; C 2 (Ω)), cf. (18), we can conclude for a.e. t ∈ (0, T ) and every ϕ ∈ X 2 T with ϕ X 2 T ≤ R. Since f ′ (ϕ) is bounded in L ∞ (0, T ; W 3 p (Ω)) and ∇ϕ is bounded in L p (0, T ; W 3 p (Ω)), the estimate above implies the boundedness of ∇f (ϕ) in L p (0, T ; W 3 p (Ω)), i.e., there exists C(R) > 0 such that Altogether this implies that f (ϕ) is bounded in L p (0, T ; W 4 p (Ω)). Analogously we can conclude from the boundedness of ϕ in W 1 Thus the same interpolation result holds as in (17), i.e., where θ := For the sake of clarity we study both summands in (22) separately and begin with the first one. Recall that the operator F 1 is defined by and that it holds P σ L(L 2 (Ω) d ,L 2 σ (Ω)) ≤ 1 for the Helmholtz projection P σ . We For the first two terms we can calculate Since it holds ∂ t v i ∈ L 2 (0, T ; L 2 σ (Ω)), i = 1, 2, we need to estimate every ρ-term in the L ∞ -norm. To this end, we use that ρ is affine linear and for i = 1, 2 and θ = Analogously the second term can be estimated by Here we used the fact that ϕ In the next step we apply the divergence on the η(ϕ i )-and Dv i -terms and for the sake of clarity we study both terms in the previous inequality separately.
For the first one we use η(ϕ) Analogously as before we can estimate the second summand by For the sake of clarity we study these three terms separately again. Firstly, where we used in the first step that η ′ (ϕ) is bounded in C([0, T ]; C 2 (Ω)). Furthermore, (20) together with Analogously to the second summand we can estimate the third one by T , which shows the statement for the second term.
Finally, we study the last term of P σ (F 1 (v 1 , ϕ 1 ) − F 1 (v 2 , ϕ 2 )) L 2 (QT ) . It holds Using ∆ϕ i ∈ C([0, T ]; C 0 (Ω)) and ∇ϕ i ∈ C 0,(1− 1 p )θ ([0, T ]; W 2 p (Ω)), i = 1, 2, cf. (17), the first term can be estimated by Analogously the second term can be estimated by Remember that we study the nonlinear operator F : X T → Y T given by and we want to show its Lipschitz continuity such that (9) holds. We already showed its Lipschitz continuity for the first part. Now we continue to study the second one. This part has to be estimated in L p (0, T ; L p (Ω)) for 4 < p < 6. For the analysis we use the boundedness of ∇ϕ in C([0, T ]; C 1 (Ω)) and of v in L ∞ (0, T ; L 6 (Ω)). Then it holds Next we study the term div(m(ϕ)∇W ′ (ϕ)). We use the boundedness of f (ϕ) in where we used (20) again in the penultimate step. Finally, we study the last term in (24). Here we get Since ∇m(ϕ 1 ) is bounded in C([0, T ]; C 1 (Ω)) and ∇∆ϕ i is bounded in C([0, T ]; L p (Ω)) for i = 1, 2, we can estimate the second summand by Thus it remains to estimate the first term of (25). Here we get .
Hence, (25) is Lipschitz continuous and therefore also the second part of F is Lipschitz continuous. Together with the Lipschitz continuity of the first part of F we have shown and a constant C(T, R) > 0 such that C(T, R) → 0 as T → 0.

Existence and Continuity of L −1
In the following we need: Theorem 5.1 Let the linear, symmetric and monotone operator B be given from the real vector space E to its algebraic dual E ′ , and let E ′ b be the Hilbert space which is the dual of E with the seminorm Let A be the subdifferential, ∂ϕ, of a convex lower-semi-continuous function ϕ : and Bu(0) = Bu 0 . If in addition u 0 ∈ dom(ϕ), then The proof of this theorem can be found in [15, Chapter IV, Theorem 6.1]. To prove Theorem 3.3 we need to show the existence of (ṽ,φ) ∈ X T such that (11) holds and to prove that L : X T → Y T is invertible with uniformly bounded inverse, i.e., there exists a constant C > 0 which does not depend on T such that L −1 L(YT ,XT ) ≤ C. Recall that the linear operator L : We note that the first part only depends on v while the second part only depends on ϕ. Thus both equations can be solved separately.
To show the existence of a unique solution v for every right-hand side f in the first equation we use Theorem 5.1.
So we have to specify what E, E ′ b , ϕ and so on are in the problem we study and show that the conditions of Theorem 5.1 are fulfilled. Then Theorem 5.1 yields the existence of a solution. More precisely, we obtain the following lemma. such that Proof: Since we want to solve (26)-(29) with Theorem 5.1, we define for u ∈ E, where we still need to specify the real vector space E. But as we want to have d dt Bu ∈ L 2 (0, T ; L 2 σ (Ω)), the dual space E ′ b has to coincide with L 2 σ (Ω). But this can be realized by choosing E = L 2 σ (Ω). Then E ′ b ∼ = L 2 σ (Ω) and with the notation in Theorem 5.1 we get the Hilbert space E ′ b equipped with the seminorm for every v ∈ L 2 σ (Ω) and We note ψ(0) = 0 and v 0 is in the L 2 -closure of dom(ψ), i.e., in L 2 σ (Ω). Hence, it remains to show that ψ is convex and lower-semi-continuous and that A is the subdifferential of ψ. Then we can apply Theorem 5.1. But the first two properties are obvious. Thus it remains to show the subdifferential property, which is satisfied by Lemma 5.3 below. Hence, we are able to apply Theorem 5.1 which yields the existence. Moreover, the initial condition is also fulfilled as Theorem 5.1 yields In particular we can conclude for every ψ ∈ C ∞ 0,σ (Ω). By approximation this identity also holds for ψ := v(0) − v 0 ∈ L 2 σ (Ω) and we get This implies v(0) = v 0 in L 2 σ (Ω). For the uniqueness we consider v 1 , v 2 ∈ W 1 2 (0, T ; L 2 σ (Ω))∩L ∞ (0, T ; H 1 0 (Ω)∩ L 2 σ (Ω)) such that (26) holds for a.e. (t, x) ∈ (0, T ) × Ω. Then v := v 1 − v 2 solves the homogeneous equation a.e. in (0, T ) × Ω. Testing this homogeneous equation with v we get Hence, it follows v ≡ 0 and therefore v 1 = v 2 , which yields the uniqueness.
For the regularity of the Stokes system with variable viscosity we used the following lemma.
We still need to ensure that L −1 L(YT ,XT ) remains bounded. This is shown in the next lemma.
Due to Lemma 5.2 together with Lemma 5.5 there exists a unique solutioñ v ∈ X 1 T0 of So let v ∈ X 1 T be the solution of the previous equations with T 0 replaced by T . Thenṽ and v solve these equations on (0, T ) × Ω. Since the solution is unique, we can deduceṽ |(0,T ) = v. Hence, which shows the statement since it holds L −1 ) < ∞ by the bounded inverse theorem.
Finally, we have to show invertibility of the second part of L.