Well-Posedness and qualitative behaviour of the Mullins-Sekerka problem with ninety-degree angle boundary contact

We show local well-posedness for the Mullins-Sekerka system with ninety degree angle boundary contact. We will describe the motion of the moving interface by a height function over a fixed reference surface. Using the theory of maximal regularity together with a linearization of the equations and a localization argument we will prove well-posedness of the full nonlinear problem via the contraction mapping principle. Here one difficulty lies in choosing the right space for the Neumann trace of the height function and showing maximal $L_p-L_q$-regularity for the linear problem. In the second part we show that solutions starting close to certain equilibria exist globally in time, are stable, and converge to an equilibrium solution at an exponential rate.


Introduction
In this article we study the Mullins-Sekerka problem inside a bounded, smooth domain Ω ⊂ R n , n = 2, 3, where the interface separating the two materials meets the boundary of Ω at a constant ninety degree angle. This leads to a free boundary problem involving a contact angle problem as well.
We assume that the domain Ω can be decomposed as Ω = Ω + (t)∪Γ(t)∪Ω − (t), whereΓ(t) denotes the interior of Γ(t), a (n − 1)-dimensional submanifold with boundary. We interpret Γ(t) to be the interface separating the two phases, Ω ± (t), which will be assumed to be connected. The boundary of Γ(t) will be denoted by ∂Γ(t). Furthermore we assume Γ(t) to be orientable, the unit vector field on Γ(t) pointing from Ω + (t) to Ω − (t) will be denoted by n Γ(t) .
(1.4) Furthermore, the volume of each of the two phases is conserved, Here, Ω ± (t) denote the two different phases separated by the sharp interface, Ω = Ω + (t) ∪Γ(t) ∪ Ω − (t). Then (1.5) stems from n Γ(t) · ∇µ dH n−1 (1.6) = Ω + (t) ∆µdx = 0. (1.7) However, the energy given by the surface area of the free interface Γ(t) satisfies d dt |Γ(t)| ≤ 0, t ∈ R + . (1.8) Indeed, an integration by parts readily gives µ| Γ(t) n Γ(t) · ∇µ dH n−1 (1.9) = − Ω |∇µ| 2 dx ≤ 0. (1.10) In this article we are concerned with existence of strong solutions of the Mullins-Sekerka problem (1.1). To this end we will later pick some reference surface Σ inside the domain Ω, also intersecting the boundary with a constant ninety degree angle, and write the moving interface as a graph over Σ by a height function h, depending on x ∈ Σ and time t ≥ 0. Pulling back the equations then to the time-independent domain Ω\Σ we reduce the problem to a nonlinear evolution equation for h. The corresponding linearization for the spatial differential operator for h then turns out to be a nonlocal pseudodifferential operator of order three, cf. [11]. We also refer to the introduction of Escher and Simonett [11] for further properties of the Mullins-Sekerka problem.
In the following, we will be interested in height functions h with regularity h ∈ W 1 p (0, T ; W 1−1/q q (Σ)) ∩ L p (0, T ; W 4−1/q q (Σ)), (1.11) where p and q are different in general. We will choose q < 2 and p finite but large, to ensure that the real interpolation space continuously embeds into C 2 (Σ), cf. Amann [4]. By an ansatz where p = q < 2, this is not achievable. We need however the restriction q < 2 to avoid additional compatibility conditions for the elliptic problem, cf. also Section 4.2. This however requires an L p − L q maximal regularity result of the underlying linearized problem, which we will also show in this article.
Outline of this paper. In Section 2 we will briefly introduce function spaces and techniques we work with and give references for further discussion. In section 3 we rewrite the free boundary problem of the moving interface as a nonlinear problem for the height function parametrizing the interface. Section 4 is devoted to the analysis of the underlying linear problem, where an extensive analysis is made on the half-space model problems. This is needed since these model problems at the contact line are not well-understood until now. The main result of this section is L p − L q maximal regularity for the linear problem. Section 5 contains that the full nonlinear problem is well-posed and Section 6 is concerned with the stability properties of solutions starting close to certain equilibria.

Preliminaries and Function Spaces
In this section we give a very brief introduction to the function spaces we use and techniques we employ in this thesis. For a more detailed approach we refer the reader to the books of Triebel [25] and Prüss and Simonett [22].

Bessel-Potential, Besov and Triebel-Lizorkin Spaces
As usual, we will denote the classical L p -Sobolev spaces on R n by W k p (R n ), where k is a natural number and 1 ≤ p ≤ ∞. The Bessel-potential spaces will be denoted by H s p (R n ) for s ∈ R and the Sobolev-Slobodeckij spaces by W s p (R n ). We will also denote the usual Besov spaces by B s pr (R n ), where s ∈ R, 1 ≤ p, r ≤ ∞. Lastly, as usual the Triebel-Lizorkin spaces are denoted by F s pr (R n ). These function spaces on a domain Ω ⊂ R n are defined in a usual way by restriction. The Banach space-valued versions of these spaces are denoted by L p (Ω; X), W k p (Ω; X), H s p (Ω; X), W s p (Ω; X), B s pr (Ω; X), F s pr (Ω; X), respectively. For precise definitions we refer to [20].
The following lemma is very well known and can easily be shown by using paraproduct estimates, see [6].

R-Boundedness, R-Sectoriality and H ∞ -Calculus
We first define the notion of sectorial operators as in Definition 3.1.1 in [22]. Definition 2.2. Let X be a complex Banach space and A be a closed linear operator on X. Then A is said to be sectorial, if both domain and range of A are dense in X, the resolvent set of A contains (−∞, 0), and there is some The concept of R-bounded families of operators is next. We refer to Definition 4.1.1 in [22]. Definition 2.3. Let X and Y be Banach spaces and T ⊆ L(X, Y ). We say that T is R-bounded, if there is some C > 0 and p ∈ [1, ∞), such that for each N ∈ N, {T j : j = 1, ..., N } ⊆ T , {x j : j = 1, ..., N } ⊆ X and for all independent, symmetric, ±1-valued random variables ε j on a probability space (Ω, A, µ) the inequality is valid. The smallest C > 0 such that (2.2) holds is called R-bound of T and denote it by R(T ).
We can now define R-sectoriality of an operator as is done in Definition 4.4.1 in [22]. Definition 2.4. Let X be a Banach space and A a sectorial operator on X. It is then said to be R-sectorial, if R A (0) := R{t(t + A) −1 : t > 0} is finite. We can then define the R-angle of A by means of ϕ R A := inf{θ ∈ (0, π) : We now define the important class of operators which admit a bounded H ∞calculus as in Definition 3.3.12 in [22]. For the well known Dunford functional calculus and an extension of which we refer to Sections 3.1.4 and 3.3.2 in [22]. Let 0 < ϕ ≤ π and Σ ϕ := {z ∈ C : | arg z| < ϕ} be the open sector with opening angle ϕ. Let H(Σ ϕ ) be the set of all holomorphic functions f : Σ ϕ → C and Definition 2.5. Let X be a Banach space and A a sectorial operator on X.
The class of operators admitting a bounded H ∞ -calculus on X will be denoted by H ∞ (X). The H ∞ -angle of A is defined by the infimum of all ϕ > ϕ A , such that (2.5) is valid, ϕ ∞ A := inf{ϕ > ϕ A : (2.5) holds}.

Maximal Regularity
Let us recall the property of an operator having maximal L p -regularity as is done in Definition 3.5.1 in [22].
Definition 2.6. Let X be a Banach space, J = (0, T ), 0 < T < ∞ or J = R + and A a closed, densely defined operator on X with domain D(A) ⊆ X. Then the operator A is said to have maximal L p -regularity on J, if and only if for in an almost-everywhere sense in L p (J; X).
There is a wide class of results on operators having maximal regularity, we refer to sections 3.5 and 4 in [22] for further discussion. For results on Rboundedness and interpolation we refer to [15].

Reduction to a Fixed Reference Surface
In this section we transform the problem (1.1a)-(1.1h) to a fixed reference configuration. To this end we construct a suitable Hanzawa transfrom, taking into account the possibly curved boundary of ∂Ω, by locally introducing curvilinear coordinates.
Let Σ ⊂ Ω be a smooth reference surface and ∂Ω be smooth at least in a neighbourhood of ∂Σ. Furthermore, let ∠(Σ, ∂Ω) = π/2 on ∂Σ. From Proposition 3.1 in [26] we get the existence of so called curvilinear coordinates at least in a small neighbourhood of Σ, that is, there is some possibly small a > 0 depending on the curvature of Σ and ∂Ω, such that is a smooth diffeomorphism onto its image and X(., .) is a curvilinear coordinate system. This means in particular that points on the boundary ∂Ω only get transported along the boundary, X(p, w) ∈ ∂Ω for all p ∈ ∂Σ, w ∈ (−a, a). We need to make use of these coordinates since the boundary ∂Ω may be curved. Therefore a transport only in normal direction of n Σ is not sufficient here. For details we refer to [26].
With the help of these coordinates we may parametrize the free interface as follows. We assume that at time t ≥ 0, the free interface is given as a graph over the reference surface Σ, that is, there is some h : for small T > 0, at least. With the help of this coordinate system we may construct a Hanzawa-type transform as follows.
Proof. The proof is straightforward. It is easy to check that for x ∈ Γ h we have that Θ h (x) = X(p, 0), where p ∈ Σ is determined by the identity x = X(p, h(p)).
Hence Θ h (Γ h ) = Σ. Furthermore it is easy to see that DF h and hence DΘ h is invertible in every point which concludes the proof since X γ ֒→ C 2 (Σ).
The following lemma gives a decomposition of the transformed curvature operator K(h) := H Γ h • Θ h for h ∈ U. The result and proof are an adpation of the work in Lemma 2.1 in [2] and Lemma 3.1 in [11]. Lemma 3.2. Let n = 2, 3, q ∈ (3/2, 2), p > 3/(2 − 3/q) and U ⊂ X γ be as before. Then there are functions (3.6) such that Moreover, where ∆ Σ denotes the Laplace-Beltrami operator with respect to the surface Σ.
Remark 3.3. Note that the orthogonality relations (3.2) in [11] do not hold if we take X to be curvilinear coordinates, since in X we not only have a variation in normal but also in tangential direction. Therefore we have to modify the proofs in [2], [11].
Proof. The curvilinear coordinates X are of form where the tangential correction τ T is as in [26]. More precisely, n Σ denotes the unit normal vector field of Σ with fixed orientation, T is a smooth vector field defined on the closure of Σ with the following properties: it is tangent to Σ, normal to ∂Σ, of unit length on ∂Σ and vanishing outside a neighbourhood of ∂Σ. In particular, T is bounded. Furthermore, τ = τ (s, r) is a smooth scalar function such that X(s, r) lies on ∂Ω whenever s ∈ ∂Σ. It satisfies τ (s, 0) = 0 for all s ∈ Σ. Moreover, since Σ and ∂Ω have a ninety degree contact angle, we have that ∂ r τ (s, 0) = 0, s ∈ ∂Σ. (3.10) Hence we may choose τ in [26] to satisfy (3.10) for all s ∈ Σ. We will now derive a formula for the transformed mean curvature K(h) in local coordinates. We follow the arguments of [11]. The surface Γ h (t) is the zero level set of the function We obtain that since X : Σ × (−a, a) → R n is a smooth diffeomorphism onto its image it induces a Riemannian metric g X on Σ × (−a, a). We denote the induced differential operators gradient, Laplace-Beltrami and the hessian with respect to (Σ × (−a, a), g X ) by ∇ X , ∆ X and hess X . As in equation (3.1) in [11] we find that for all s ∈ Σ, where ∇ X Φ h X := (g X (∇ X Φ h , ∇ X Φ h )) 1/2 . Note at this point that since X induces also a variation in tangential direction, the orthogonality relations (3.2) in [11] do not hold in general. However, we get in local coordinates that and (∂ n X|∂ n X) = 1 + (∂ r τ ) 2 ( T | T ). In particular we see that on the surface Σ the relations (3.2) in [11] still hold, but not away from Σ in general. By using well-known representation formulas for ∇ X , ∆ X , and hess X in local coordinates, one finds that as well as and in local coordinates. Mimicking the proof of Lemma 2.1 in [2], K(h) = P (h)h + Q(h) is the desired decompostion of K, since X γ ֒→ C 2 (Σ). The fact that P (0) = −∆ Σ follows from (3.22) and the formulas for a jk and a j .
We are now able now transform the problem (1.1a)-(1.1h) to a fixed reference domain Ω\Σ by means of the Hanzawa transform. This however yields a highly nonlinear problem for the height function. The transformed differential operators are given by 24) and the transformed normal by n h ∂Ω := n ∂Ω • Θ t h . This leads to the equivalent system where h 0 is a suitable description of the initial configuration such that Γ| t=0 = Γ 0 and β(h) := ∂ t hn Σ + ∂ r τ T , cf. (3.9). Note that by the initial condition (1.1h) we have that n h0 ∂Ω · n Γ h 0 = 0, which is a necessary compatibility condition for the system (3.25a)-(3.25f).
The following lemma states important differentiability properties of the transformed differential operators.
4 Maximal L p −L q Regularity for Linearized Problem

Reflection Operators
We denote the upper half space of R n by R n + := {x ∈ R n : x n > 0}. We will denote by R the even reflection of a function defined on R n + across the boundary ∂R n + in x n direction, that is, we define R as an extension operator via Ru(t, x 1 , ..., x n ) := u(t, x 1 , ..., −x n ) for all x n < 0. Note that R admits a bounded extension R : L q (R n + ) → L q (R n ). The following theorems state that even more is true.
Proof. It is straightforward to verify that for a given u ∈ W 1+α q (R n + ), and ∂ n Ru(x 1 , ..., x n ) = −∂ n u(x 1 , ..., −x n ). Hence also R : To show the claim for the fractional order space of order 1 + α, it remains to show that the odd reflection of Du ∈ W α q (R n + ), that is, say T Du, is again W α q (R n ) and that the corresponding bounds hold true. We first note that T Du( where e 0 denotes the extension by zero to the lower half plane. Note that by real interpolation method, since 0 < α < 1/q, cf. [25]. Now, both zero extension operators are bounded and linear. From Theorem 1.1.6 in [18] we obtain that e 0 is therefore also a bounded and linear operator between the corresponding interpolation spaces, hence the theorem is proven. Note that the above proof makes essential use of the fact that the derivative of u ∈ W 1+α q (R n + ) has no trace on ∂R n + since α < 1/q. If one has a trace it needs to be zero to reflect appropriately, which is the statement of the next theorem. The proof follows similar lines, we omit it here.
Theorem 4.2. Let q and R be as above. Then R induces a bounded linear operator W 1+β for all β ∈ (1/q, 1).
We also need a reflection argument for the initial data in X γ . The result reads as follows.
Proof. The second statement follows from the first one for The first claim is shown as in the proof of Theorem 4.1, using additionally that ∂ xn ∂ xn Ru = R∂ xn ∂ xn u.

The Shifted Model Problem on the Half Space
Let n = 2, 3. In this section we will be concerned with the linearized problem on the whole upper half space R n + with a flat interface Σ := {x ∈ R n + : x 1 = 0}.
More precisely, we will consider Here, x ′ = (x 2 , ..., x n ) and ω > 0 is a fixed shift parameter we need to introduce to get maximal regularity results on the unbounded time-space domain R + ×R n + . Let us discuss the optimal regularity classes for the data. We seek a solution h of this evolution equation in the space where p and q are specified below. In particular, µ ∈ L p (R + ; W 2 q (R n + \Σ)). Let (4.8) and the real interpolation space By simple trace theory, we may deduce the necessary conditions It is now a delicate matter to find the optimal regularity condition for g 5 , which turns out to be cf. Theorem B.1 in the Appendix. Note that g 5 has a time trace at t = 0, whenever 1 − 2/(3q) − 1/p > 0. Hence there is a compatibility condition inside the system whenever this inequality is satisfied, namely Note that there is no compatibility condition stemming from (4.6b) and (4.6d) on ∂Σ, whenever q < 2. The following theorem now states that these conditions are also sufficient. Note that the assumptions in Theorem 4.4 imply that q < 2 and 1 − 2/(3q) − 1/p > 0 hold.
and ω > 0. Then (4.6a)-(4.6f) has maximal L p − L q -regularity on R + , that is, for every (g 1 , g 2 , g 3 , g 4 , g 5 , h 0 ) satisfying the regularity conditions (4.10)-(4.12) and the compatibility condition (4.13), there is a unique solution (h, µ) ∈ (W 1 p (R + ; X 0 )∩ L p (R + ; X 1 ))×L p (R + ; W 2 q (R n + \Σ)) of the shifted half space problem (4.6a)-(4.6f). Furthermore, is bounded by Proof. We first reduce to a trivial initial value by extending h 0 toΣ = {0}×R n−1 using standard extension results of [25] and solving an L p −L q auxiliary problem on R n−1 using results of Section 4 in [23] by the compatibility condition (4.13). This allows us to use Theorem B.1 to By simple trace theory we may find µ 4 ∈ L p (R + ; W 2 q (R n + \Σ)) such that ∂ n µ 4 | ∂R n + = g 4 on ∂R n + . LetΣ := RΣ := {x ∈ R n : x 1 = 0}. We then solve the elliptic auxiliary problem by a uniqueμ ∈ L p (R + ; W 2 q (R n \Σ)), cf. [3]. Note at this point that we used that due to q < 2 and Theorem 4.1 we have that the data in (4.17b) is in L p (R + ; W 2−1/q q (Σ)). Note that by constructionμ is even in x n direction since both the data in (4.17) are.
We have reduced the problem to the case where (g 2 , g 3 , g 4 , g 5 , h 0 ) = 0, that is, we are left to solve for possibly modified g 1 not to be relabeled in an L p − L q -setting. We reflect the problem once more across the boundary ∂R n + using the even reflection in x n direction R and by doing so we obtain a full space problem with a flat interface and that the conditions (4.18d) and (4.18e) are fulfilled automatically. We obtain the problem where ). Let us denote by S(h) the unique solution of the elliptic problem (4.19b)-(4.19c). Then we can write the system as an abstract evolution equation as follows. Define Ah(x) := n Σ · ∇S(h) − ω 3 h and its realization in W Then we can modify the results of [23] to obtain that the operator A has the property of maximal L q -regularity on the whole half line R + , whence a general principle of maximal regularity going back to Dore [9] and the works of Bourgain and Benedek, Calderon and Panzone [7] now gives that A has also maximal L pregularity on R + , since 1 < p < ∞, cf. [22]. We give the full details below.
Having this at hand we can solve the initial value problem . By choosing we obtain a unique solution (h, S(h)) of the problem (4.19a)- (4.19d) in the proper L p − L q -regularity classes on R + × R n−1 . The estimate easily follows and the proof is complete. Let us give the details on how we obtain maximal L q -regularity for A on R + . We take Fourier transform with respect to (x 2 , ..., x n ) ∈ R n−1 to obtain a system whereπ =π(t, x 1 , ξ),ĥ =ĥ(t, ξ) andf =f (t, ξ) denote the Fourier transforms of π, h, and f with respect to the last n − 1 variables (x 2 , ..., x n ) ∈ R n−1 . We can now solve the second order differential equation forπ and together with boundary and decay conditions we finally obtain whence we obtain a modified version of the evolution equation in [23], namely = 0, cf. the proof of Proposition 8.3.1 in [22]. Let furthermore B 2 be the operator given by ( . Then by Example 4.5.16(i) in [22] we know that B 2 is invertible and admits a bounded H ∞ -calculus on L q (R n−1 ) and the H ∞ -angle is zero, ϕ ∞ B2 = 0. We now apply Corollary 4.5.12(iii) in [22] to get that P := 2B 1 B 2 is a closed, sectorial operator which itself admits a bounded H ∞ -calculus on L q (R n−1 ) as well and that the H ∞ -angle of P is zero. The fact that B 1 and B 2 commute stems from the fact that these are given as Fourier multiplication operators.
We now show that P admits a bounded H ∞ -calculus also on W s q (R n−1 ) for all 0 < s < 1, in particular for s = 1 − 1/q. To this end we show the claim for s = 1 and use real interpolation method. We will use the fact that ( . Taking Fourier transform just as in the proof of Theorem 6.1.8 in [22] gives where P(ξ) = 2|ξ| 2 ω 2 + |ξ| 2 is the corresponding symbol of P . Whence clearly we have the representation formula for all u ∈ C ∞ 0 (R n−1 ), in other words, the symbol of h(P ) is in fact h(P(ξ)). Since h(P ) and the shift operators (I − ∆) ±1/2 commute we easily see that P admits a bounded H ∞ -calculus on W 1 q (R n−1 ) and hence on all W s q (R n−1 ), where 0 < s < 1. The constant extension to L p (R + ; W s q (R n−1 )) which we will also denote by P then admits a bounded H ∞ -calculus on L p (R + ; W s q (R n−1 )) for all 0 < s < 1 with angle zero.
We now apply a version of Dore-Venni theorem, cf. [21]. To this end let B be the operator on L p (R + ; W Then B is sectorial and admits a bounded H ∞ -calculus on L p (R + ; W s q (R n−1 )) of angle π/2. Furthermore, B : D(B) → L p (R + ; W s q (R n−1 )) is invertible. Let as above P be the operator on L p (R + ; W 1−1/q q (R n−1 )) with domain D(P ) = L p (R + ; W 4−1/q q (R n−1 )) given by its symbol 2|ξ| 2 (ω 2 + |ξ| 2 ) 1/2 . Now, by the Dore-Venni theorem we get that the sum B+P with domain D(B+P ) = D(B)∩ D(P ) is closed, sectorial and invertible. In other words, the evolution equation Bu + P u = f posesses for every f ∈ L p (R + ; W 1−1/q q (R n−1 )) a unique solution u ∈ D(B) ∩ D(P ), hence the proof of maximal regularity is complete.
Dependence of the maximal regularity constant on the shift parameter. Note that at this point it is a priori not clear how the maximal regularity constant depends on the shift parameter ω > 0. However, we will need a good understanding of this dependence later on when we want to solve the bent halfspace problems.
We will now introduce suitable ω-dependent norms in both data and solution space and show that the maximal regularity constant is then independent of ω.
To this end we will proceed with a scaling argument. Fix ω > 0 and let (h, µ) be the solution on R + of the ω-shifted half space problem (4.6a)-(4.6f). Define new functions It is then easy to check that (h,μ) solves Since the operator on the left hand side is independent of ω, we get by the previous theorem that there is some constant M > 0 independent of ω, such that is bounded by Clearly, the ω-dependence is now hidden in the norms, whenceforth a careful calculation entails ω 4−1/q |h| Lp(R+;L q (Σ)) + ω 3−1/q |Dh| Lp(R+;Lq(Σ)) + ω 2−1/q |D 2 h| Lp(R+;Lq(Σ)) + for some K(ω) > 0 stemming from interpolating the estimates for X 0 and X 1 , the value of which does not matter. The calculations involving the Triebel-Lizorkin seminorm ofg 5 stem from the characterization via differences, see Proposition 2.3 in [19].

Bent Half Space Problems
In this section we consider the shifted model problem (4.6) on a bent half space R n γ := {x ∈ R n : x n > γ(x 1 , ..., x n−1 )}, where γ : R n−1 → R is a sufficiently smooth function with sufficiently small C 1 (R n−1 )-norm. Since also the reference surface may be curved, we consider a slightly bent interface Σ β := {x ∈ R n γ : x 1 = β(x 2 , ..., x n )}. Again, β : R n−1 → R is suitably smooth and the C 1 (R n−1 )norm is sufficiently small. The bent half space problem reads as where n γ denotes the normal at ∂R n γ . The smallness assumption on |β| C 1 +|γ| C 1 implies that the bent domain and interface are only a small perturbation of the half space and the flat interface. We will now solve this problem on the bent half space by transforming it back to the regular half space.
Lemma 4.5. Let k ∈ N and β, γ ∈ C k (R n−1 ). Then there is some F ∈ C k (R n ; R n ), such that F : R n → R n is a C k -diffeomorphism and such that additionally, F | R n γ : R n γ → R n + is a C k -diffeomorphism as well. Furthermore, F maps Σ β to the flat interface R n + ∩{x 1 = 0}. We also have that |I −DF | C l (R n ) |β| C l+1 (R n−1 ) + |γ| C l+1 (R n−1 ) for all l = 0, ..., k − 1.
Proof. To economize notation, let n = 3. We first transform in x 3 -direction via Φ 1 : R 3 → R 3 , x → (x 1 , x 2 , x 3 − γ(x 1 , x 2 )). It is then easy to see that the surface Φ 1 (Σ β ) is given by the set Note that this is equivalent to where Now note that whenever |(β, γ)| C 1 is sufficiently small, |H − id R 2 | C 1 is small. Then | det DH| ≥ 1/2 on R 2 and H : R 2 → R 2 is globally invertible. Hence the surface Φ 1 (Σ β ) can be parametrized by β • H −1 , Note that by the inverse function theorem, . We easily check that F := Φ 2 Φ 1 satisfies the desired properties.

Localization Procedure
Let us now be concerned with the shifted problem on a bounded smooth domain Ω ⊂ R n , where Σ is a perpendicular smooth surface inside. More precisely, the system reads as where ω ≥ ω 0 and ω 0 > 0 is as in Theorem 4.6. The main result reads as follows.
We will now show existence and uniqueness of the solution of this system via the localization method. To this end let (ϕ j ) j=0,...,N ⊆ C ∞ 0 (R n ) be a smooth partition of unity with respect to Ω and the open sets (U j ) j=0,...,N ⊆ R n , that is, the support of ϕ j is contained in U j for each j = 0, ..., N and Ω ⊆ j=0,...,N U j . Furthermore, let (ψ j ) j=0,...,N ⊆ C ∞ 0 (R n ) be smooth functions with compact support in U j such that ψ j ≡ 1 on supp ϕ j for every 0 ≤ j ≤ N. Now, by choosing N finite but sufficiently large and, corresponding to that, the open sets U j sufficiently small, we can assume that, up to a rotation, for each j = 0, ..., N there exist smooth curves γ j , β j such that Furthermore, again by a smallness argument, we can choose γ j and β j such that the C 1 -norm is as small as we like. We now assume for a moment that we have a solution (h, µ) of (4.57) to derive an explicit representation formula. We therefore multiply every equation with ϕ j and get corresponding equations for the localized functions (h j , µ j ) := ϕ j (h, µ). By doing so, we obtain where Σ j := Σ βj , Ω j := R n γj , (g lm ) is the first fundamental form of Σ j with respect to the surface Σ and (g lm ) its inverse. This way, we obtain a finite number of bent half space problems. Denote by L j the linear operator on the right hand side of the above system, as well as the data G j := (ϕ j f, 0, 0, 0, ϕ j b, 0) and the perturbation operator R j such that the right hand side equals G j + R j (h, µ), we can write the system of localized equations as (4.60) Since each L j is invertible, we may derive the representation formula Since now R := N j=0 ψ j (L j ) −1 R j is of lower order, a Neumann series argument now yields that I − R is invertible if T > 0 is small enough, hence we can rewrite (4.61) as (4.62) Letting L be the linear operator from the left hand side of (4.57), we obtain from (4.62) that L is injective, has closed range and a left inverse. It remains to show that L : E(T ) → F(T ) has a right inverse. To this end let z ∈ F(T ) be arbitrary. Define Applying I −R to both sides of (4.63) yields a formula for Sz, which then entails we can show, using again a Neumann series argument involving the fact that the commutator is lower order, that I+S R is invertible if T > 0 is small enough. The right inverse of L is therefore given by S(I + S R ) −1 . This then concludes the proof.

The non-shifted model problem on bounded domains
In this section we are concerned with problem (4.56) for ω = 0. The main result is the following.
Proof. As in the previous section we may reduce to the case (g 2 , g 3 , g 4 , h 0 ) = 0. It is also clear that the ω 3 -shift in equation (4.56a) can easily be resolved to the case ω = 0 by an exponential shift in solution and data. We are therefore left to solve where T 0 g is the unique solution of the two-phase elliptic problem This implies that problem (4.65) is equivalent to on ∂Σ, (4.67b) h| t=0 = 0, on Σ, (4.67c) provided η ≥ η 0 . Now choose large enough η to render the left hand side of (4.67) to be an invertible operator. We may estimate

Nonlinear Well-Posedness
In this section we will show local well-posedness for the full nonlinear (transformed) system (3.25) and therefore obtain that also the system (1.1) is wellposed. We will use the maximal L p − L q regularity result for the underlying linear problem and a contraction argument via the Banach's fixed point principle.
The main result reads as follows.
Theorem 5.1. Let 6 ≤ p < ∞, q ∈ (5/3, 2) ∩ (2p/(p + 1), 2p) and h 0 ∈ X γ sufficiently small. Then there is some possibly small τ > 0, such that (3.25) has a unique strong solution on (0, τ ), that is, there are solving (3.25) on (0, τ ), whenever h 0 satisfies the initial compatibility condition We now reduce to trivial initial data as follows, cf. [16]. Since h 0 satisfies the compatibility condition, we may solve where the nonlinear part is given by We may now define . By restricting to functions with vanishing trace at time zero, we get that the operator norm |L −1 | B(0F(T );0E(T )) stays bounded as T → 0.
Lemma 5.2. The mapping N : E(T ) → F(T ) is well-defined and bounded. Furthermore, N ∈ C 2 (E(T ); F(T )). Furthermore, N allows for contraction estimates in a neighbourhood of zero, that is, for all z 1 , z 2 ∈ B(r; 0) ⊂ 0 E(T ), if r > 0 and T = T (r) > 0 are sufficiently small. Here, B(r; 0) denotes the closed ball around 0 with radius r > 0.
Let now δ > 0, such that |h 0 | Xγ ≤ δ. By choosing r > 0, T = T (r) > 0 and δ = δ(T ) > 0 sufficiently small, we ensure K to be a 1/2-contraction on B(r, 0) ⊂ 0 E(T ). Note at this point that |z * | E(T ) ≤ C(T )δ. Let us note that Now we note thatÑ (0) is quadratic in z * = (h * , µ * ) except for the term Q(0) inÑ (0) 2 . Using By performing the above fixed point argument on the larger ball B(r ′ , 0) on a smaller time interval we see that T * > 0. Then solve the nonlinear problem with initial valuez(T * ) ∈ X γ . Note thatz(T * ) satisfies the compatibility condition ∂ ν [z(T * )] = 0 on ∂Σ, hence we may obtain a unique solution on a larger time interval (0, T * + ǫ * ) for some ǫ * > 0. This contradicts (5.10) and the fixed point has to be unique. This then in turn yields the uniqueness of the solution to (3.25).
Let us comment on how to prove Lemma 5.2. Using the differentiability properties from Lemma 3.4, the statement easily follows for the components N 1 , N 3 and N 4 . The decomposition K(h) = P (h)h + Q(h) from Lemma 3.2 renders a proof for N 3 . For N 5 we note that Depner in [8] calculated the linearization of the ninety-degree angle boundary condition (3.25e), which turns out to be n ∂Σ ·∇ Σ h| ∂Σ = 0, which in turn then allows for estimates for N 5 . Note that we also use the Banach algebra property for the trace space, cf. Theorem B.4 in the appendix. For sake of readability we omit the details here. Remark 5.3. We point out that the proof of Theorem 5.1 also gives wellposedness of (3.25) in the case where Ω = G × (L 1 , L 2 ) is a bounded container in R n , n = 2, 3. Hereby G ⊂ R n−1 is a smooth, bounded domain. In this case there is another model problem in the localization procedure for the linear problem stemming from when the top and bottom of the container G × {L 1 , L 2 } intersect the walls ∂G×(L 1 , L 2 ). This elliptic problem, although being a problem on a domain with corners, admits full regularity for the solution, cf. the appendix in Section A.2.

Convergence to equilibria
This section is devoted to the long-time behaviour of solutions to (1.1) starting close to equilibria. We will characterize the set of equilibria, study the spectrum of the linearization of the transformed Mullins-Sekerka equations around an equilibrium and apply the generalized principle of linearized stability to show that solutions starting sufficiently close to certain equilibria converge to an equilibirum at an exponential rate in X γ .
We note that the potential µ can always be reconstructed by Γ(t) by solving the elliptic two-phase problem on Γ(t), (6.1a) ∆µ = 0, in Ω\Γ(t), (6.1b) n ∂Ω · ∇µ| ∂Ω = 0, on ∂Ω. (6.1c) Whence we may concentrate on the set of equilibria for Γ(t). It now can easily be shown that for a stationary solution Γ of (1.1) with V Γ = 0 the corresponding chemical potential µ is constant, since then µ and ∇µ have no jump across the interface Γ and µ ∈ W 2 q (Ω) solves a homogeneous Neumann problem on Ω. By (6.1a), the mean curvature H Γ is constant. The set of equilibria for the flow Γ(t) is therefore given by Let us now consider the case where Ω ⊂ R n , n = 2, 3, is a bounded container, that is, Ω := Σ × (L 1 , L 2 ), where −∞ < L 1 < 0 < L 2 < ∞ and Σ ⊂ R n−1 × {0} is a bounded domain and ∂Σ is smooth. Note that flat interfaces are equilibria. Arcs of circles intersecting ∂Ω at a ninety degree angle also belong to E, since then (1.1g) is also satisfied.
If we now additionally assume that the contact points between Γ and ∂Ω are only on the walls of the cylinder and Γ is given as a graph over Σ, we may even deduce that H Γ = 0, that is, Γ is a flat interface described by a constant height function over the reference surface. This follows from the fact that we can describe Γ as graph of a height function h over Σ. Then using the well-known formula and the boundary condition (1.1g) on ∂Γ renders H Γ = 0. Indeed, assume that Γ = Γ h is a graph of h over Σ. We may assume that h has mean value zero and we already know H Γ is constant, but may be nonzero. Then an integration by parts entails The boundary integral vanishes due to (1.1g) and renders ∇h to be zero in Σ, hence h is constant. This implies H Γ = 0.
We will now study the problem for the height function in an L p -setting. We now rewrite the geometric problem (1.1g) as an abstract evolution equation for the height function h, cf. [2], [23], [11]. As seen before, by means of Hanzawa transform, the full system reads as in Ω\Σ, (6.5c) n ∂Ω · ∇ h µ| ∂Ω = 0, on ∂Ω, (6.5d) n ∂Σ · ∇ Σ h| ∂Σ = 0, on ∂Σ, (6.5e) Let us note that due to working in a container, the highly nonlinear angle condition (3.25e) reduces to a linear one, condition (6.5e). Define B(h)g := n h Σ · ∇g and S(h)g as the unique solution of the elliptic problem µ| Σ = g, on Σ, (6.6a) Recalling Lemma 3.2, we may rewrite (6.5) as an abstract evolution equation, We now interpret problem (6.7) as an evolution equation in L p (R + ; X 0 ), fitting in the setting of Prüss, Simonett and Zacher [23]. Regarding the linearization of (1.1), we have the following result.

Let h ∈ D(A) satisfy A(h)h = F (h). It then follows that B(h)S(h)H
Note that S(h)H Γ (h) is the unique solution of an h-perturbed elliptic problem with homogeneous Neumann boundary conditions. Therefore S(h)H Γ (h) has to be constant. Since S(h)H Γ (h) equals H Γ (h) on Σ, also H Γ (h) is constant. We then obtain that the mean curvature H Γ of the interface given as a graph of h over Σ is constant. Due to (6.5e) we may even deduce using formula (6.3) that H Γ = 0. Then h has to be constant. 5. This stems from Theorem 4.8. 6. Clearly, every constant is an element of N (A 0 ). For the converse, let h ∈ D(A 0 ), such that A 0 h = 0. Hence χ = T ∆ Σ h is constant and thererfore ∆ Σ h is constant. Since h ∈ D(A 0 ), an integration by parts shows ∆ Σ h = 0. Again since h ∈ D(A 0 ), h has to be constant. 7. We only need to show N ( Since every element in the range of A 0 has mean value zero, it follows that A 0 h = 0, whence h ∈ N (A 0 ). The proof is complete.
More precisely: 1. Near h * = 0 the set of equilibria E is a C 1 -manifold in X 1 of dimension one.
2. The tangent space of E at h * = 0 is given by the kernel of the linearization, Proof.
The following theorem is the main result on stability of solutions. It is an application of the generalized principle of linearized stability of Prüss, Simonett, and Zacher [23] to the evolution equation (6.7). Theorem 6.3. The trivial equilibrium h * = 0 is stable in X γ , and there is some δ > 0 such that the evolution equation 19) which converges at an exponential rate in X γ to some h ∞ ∈ E as t → +∞.
Theorem 6.4. (Geometrical version) Suppose that the initial surface Σ 0 is given as a graph, Σ 0 = {(x, h 0 (x)) : x ∈ Σ} for some function h 0 ∈ X γ . Then, for each ε > 0 there is some δ(ε) > 0, such that if the initial value h 0 ∈ X γ satisfies |h 0 − h * | Xγ ≤ δ(ε) for some constant function h * , there exists a global-in-time strong solution h on R + of the evolution equation, precisely h ∈ L p (R + ; D(A 0 )) ∩ W 1 p (R + ; X 0 )), and it satisfies |h(t)| Xγ ≤ ε for all t ≥ 0. Moreover, there is some constant h ∞ , such that Σ h(t) → Σ h∞ in the sense of h(t) → h ∞ in X γ and the convergence is at an exponential rate.
Note that by the following theorem we can characterize the limit. It is a priori not clear to which equilibrium the solution converges by the generalized principle of linearized stability.
Theorem 6.5. The limit h ∞ from above has the same mean value as h 0 , in other words, Σ h 0 dx/|Σ| = h ∞ .
Proof. The theorem is a consequence of the fact that the Mullins-Sekerka system conserves the measure of the domains separated by the interface in time. Hence the solution h from Theorem 6.3 satisfies d dt Σ h(t, x)dx = 0. (6.20) In particular, Let Ω ⊂ R n be a bounded domain with smooth boundary ∂Ω. Furthermore let Σ be a smooth submanifold of R n with boundary such that the interiorΣ is inside Ω and meets ∂Ω at a constant ninety degree angle.
In this chapter we are concerned with problems of elliptic type, namely, where η > 0 is a fixed shift parameter, as well as the non-shifted version, We will show optimal solvability of this problem via a localization method. To this end we consider first the model problem of (A.1) on R n + with flat interface {x n > 0, x 1 = 0}.
Theorem A.1. Let η > 0, q ∈ (3/2, 2) and Σ := {x n > 0, x 1 = 0}. Then, for every f ∈ L q (R n + ), g 1 ∈ W 2−1/q q (Σ) and g 2 ∈ W 1−1/q q (∂R + ) there exists a unique solution u ∈ W 2 q (R n + \Σ) of (A.1) with R n + replacing Ω. Furthermore, there is some C(η) > 0 and some K > 0 independent of η, such that Proof. We first solve an auxiliary upper half space problem to reduce the problem to for possibly modified g 1 not to be relabeled. Since ∂ n u = 0 on the boundary, we may reflect the problem via an even reflection to obtain an elliptic problem onṘ × R n−1 . By Theorem 4.1 using q < 2 we obtain that Rg 1 ∈ W 2−1/q q (Σ), whereΣ := {x 1 = 0}. Here, R denotes the aforementioned even reflection in x n -direction. The problem we are left to solve is now Let x ′ := (x 2 , ..., x n ). It is now well known that the operator (η − ∆ x ′ ) 1/2 with domain W 1 q (R n−1 ) has maximal regularity on the half line R + with respect to the base space L q (R n−1 ) and the induced semigroup is analytic. Note that by real interpolation method, We obtain |v| R n + | W 1 To obtain the dependence of the shift parameter one proceeds by a scaling argument as in Section 4.2. The proof is complete.
By a standard localization argument we can now show that the shifted problem is solvable in the case of a bounded, smooth domain.
We will now concern solvability of the non-shifted problem (A.2).
. Furthermore, there is some constant C > 0, such that Since D(A) compactly embeds into L q (Ω) by Sobolev embeddings, A has compact resolvent and the spectrum σ(A) only consists of eigenvalues of A with finite multiplicity. We will show that zero is not a possible eigenvalue, hence A is invertible. Suppose u = 0 is a nontrivial eigenfunction to the eigenvalue λ. Since by well-known results the spectrum is independent of q, we may let q = 2, cf. [5]. Testing the resolvent equation with u in L 2 (Ω) and invoking the boundary condition yields Whence if λ = 0, then u ∈ D(A) has to be a constant function, hence zero since u vanishes on Σ. This is a contradiction, hence λ = 0 is not a possible eigenvalue. Therefore we may uniquely solve (A.10) and the proof is complete.

A.2 Cylindrical domains.
In the case where n = 3 and Ω ⊂ R 3 is a bounded container, one needs a result for the elliptic model problem in the case where the top of the container meets the walls at a ninety degree angle. So let G := R + × R × R + .
(A. 17) This yields necessary conditions for the data. We see that on the set ∂S 1 ∩ ∂S 2 = {x 1 = x 3 = 0} where the two boundary conditions meet, there is no compatibility condition for the data g 1 and g 2 in the system. This is due to the fact that since q < 2 the functions ∇u| Sj do not have a trace on ∂S j . So let the given data satisfy By a simple reflection we can recude the problem to a upper half-space problem with one Neumann condition and obtain full W 2 q (G) regularity for the solution. Let us state this observation in the following theorem.

B The Neumann trace of the height function
In this section we characterize the optimal trace space for the Neumann trace of the height function h and show that it is a Banach algebra with respect to pointwise multiplication.
It remains to construct a continuous right inverse. This follows now by similar arguments using Corollary 2.7 in [13]. We omit the details here.
We again point out that the constant is only independent of T since we restrict ourselves to functions having vanishing trace at t = 0.
The boundedness of the trace operator can easily be generalized to the case of a curved interface by a standard argument involving a partition of unity and a localization argument.
The next result states that the Neumann trace space is a Banach algebra under pointwise multiplication.