Singular limits for the two-phase Stefan problem

We prove strong convergence to singular limits for a linearized fully inhomogeneous Stefan problem subject to surface tension and kinetic undercooling effects. Different combinations of $\sigma \to \sigma_0$ and $\delta \to\delta_0$, where $\sigma,\sigma_0 \ge 0$ and $\delta,\delta_0 \ge 0$ denote surface tension and kinetic undercooling coefficients respectively, altogether lead to five different types of singular limits. Their strong convergence is based on uniform maximal regularity estimates.


Introduction
The aim of this note is to consider the fully inhomogeneous system which represents a linear model problem for the two-phase Stefan problem subject to surface tension and kinetic undercooling effects. Here v(t, x, y) = v + (t, x, y), y > 0, v − (t, x, y), y < 0, x ∈ R n , y ∈ R \ {0}, t ∈ J, denotes the temperature in the two bulk phases R n+1 ± = {(x, y); x ∈ R n , ±y > 0}, and we have setṘ n+1 = R n+1 + ∪ R n+1 − and J = (0, T ). The function ρ appearing in the boundary conditions describes the free interface, which is assumed to be given as the graph of ρ. We also admit the possibility of two different (but constant) diffusion coefficients c ± in the two bulk phases. The parameters σ and δ are related to surface tension and kinetic undercooling. The function ρ E is an extension of ρ chosen suitably for our purposes. Here it is always determined through (1.2) Using this notation, let [[c∂ y (v − ρ E )]] denote the jump of the normal derivatives across R n , that is, 1 where γ denotes the trace operator. The coefficient a is supposed to be a function of δ and σ, that is, a ± : [0, ∞) 2 → R, [(δ, σ) → a ± (δ, σ)]. It is further assumed to satisfy the conditions a ± ∈ C([0, ∞) 2 , R), a ± (0, 0) > 0. (1.3) Recall from [10] that the introduction of the additional term 'aρ E ' with a ± > 0 in the situation of the classical Stefan problem is motivated by the following two facts: for suitably chosen a (depending on the trace of the initial value and ∂ y ρ E ) it can be guaranteed that a certain nonlinear term remains small for small times.
On the other hand, the additional term 'aρ E ' is exactly the device that renders sufficient regularity for the linearized problem. Note that, concerning regularity, this additional term is not required if surface tension or kinetic undercooling is present. However, in order to obtain convergence in best possible regularity classes for the limit σ, δ → 0, we keep the term 'aρ E ' in all appearing systems. Since the data may (in general even must; see Remark 1.3) depend on σ and δ as well, a is a function of these two parameters. The natural and necessary convergence assumption (1.10) then implies that we can assume that a ± ∈ C([0, ∞) 2 , R). This continuity will be important in deriving maximal regularity estimates for related boundary operators; see the proof of Proposition 2.6. The results of this paper on system (1.1) represent an essential step in the treatment of singular limits for the nonlinear Stefan problem on general geometries. This will be the topic of a forthcoming paper.
(d) In the case δ 0 = σ 0 = 0 conditions (1.11) and (1.12) express that ρ µ 0 and respectively, but slower than σ and δ tend to zero. This seems to be natural in view of the fact that we do not have ρ from the regularity of solutions in the situation of the classical Stefan problem.
The Stefan problem is a model for phase transitions in liquid-solid systems that has attracted considerable attention over the last decades. We refer to the recent publications [5,10,11,12,13] by the authors, and the references contained therein, for more background information on the Stefan problem.
Previous results concerning singular limits for the Stefan problem with surface tension and kinetic undercooling are contained in [1,16]. Our work extends these results in several directions: we obtain sharp regularity results (for the linear model problems), we can handle all the possible combinations of singular limits, and we obtain convergence in the best possible regularity classes.
Our approach relies on the powerful theory of maximal L p -regularity, H ∞functional calculus, and R-boundedness, see for instance [2,8] for a systematic introduction.

Maximal regularity
First let us introduce suitable function spaces. Let Ω ⊆ R m be open and X be an arbitrary Banach space. By L p (Ω; X) and H s p (Ω; X), for 1 ≤ p ≤ ∞ and s ∈ R, we denote the X-valued Lebegue and the Bessel potential space of order s, respectively. We will also frequently make use of the fractional Sobolev-Slobodeckij spaces W s p (Ω; X), 1 ≤ p < ∞, s ∈ R \ Z, with norm where [s] denotes the largest integer smaller than s. Let T ∈ (0, ∞] and J = (0, T ). We set The spaces 0 H s p (J, X) are defined analogously. Here we remind that H k p = W k p for k ∈ Z and 1 < p < ∞, and that W s p = B s pp for s ∈ R \ Z. We refer to [14,15] for more information.
Before turning to the proofs of our main results, we add the following remarks on the linear two-phase Stefan problem (1.1) and the particularly chosen extension ρ E determined by equation (1.2).
Remarks 2.1. (a) (1.1)-(1.2) constitutes a coupled system of equations, with the functions (v, ρ, ρ E ) to be determined. We will in the sequel often just refer to a solution (v, ρ) of (1.1) with the understanding that the function ρ E also has to be determined.
). This follows, for instance, from [5, Proposition 5.1], thanks to (c) The solution ρ E (t, ·) of equation (1.2) provides an extension of ρ(t, ·) toṘ n+1 . We should remark that there are many possibilities to define such an extension. The chosen one is the most convenient for our purposes. We also remark that we have great freedom for the extension of ρ 0 .
Let T ∈ (0, ∞] and set J = (0, T ). By F T we always mean the space of given data (f, g, h, v 0 , ρ 0 ), i.e., F T is given by Analogously, we denote by E T the space of the solution (v, ρ, ρ E ). As was already pointed out in the introduction, we have, depending on the values of δ and σ, four different type of spaces. For this reason we set , and with E 2 T (δ, σ) as defined in (1.4) and equipped with the parameter dependent norm given in (1.5). Note that then the norm in E T (δ, σ) is given by . For fixed δ, σ > 0 by interpolation it can be shown that in the sense of isomorphisms. We remark that E 2 T (δ, σ) is the correct regularity class for the free surface if both, surface tension and kinetic undercooling are present.
is the proper class if just surface tension or just kinetic undercooling, respesctively, is present. Finally, E 2 T (0, 0) is the correct class if both of them are missing, i.e., E 2 T (0, 0) is the regularity class in the situation of the classical Stefan problem.
The corresponding spaces with zero time trace at the origin are denoted by 0 F 1 , and so on, that is,

Zero time traces.
We will first consider the special case that This allows us to derive an explicit representation for the solution of (1.1)-(1.2).
with C > 0 independent of the data, the parameters (δ, σ) ∈ [0, R] 2 , and T ∈ (0, T 0 ] for fixed T 0 > 0. Proof. (i) In order to be able to apply the Laplace transform in t, we consider the modified set of equations for the unknown functions (u, η, η E ) and for a fixed number κ ≥ 1 to be chosen later. We claim that system (2.3)-(2.4) admits for each ( satisfying inequality (2.2) in the corresponding norms for T = ∞.
(ii) In the following, the symbolˆdenotes the Laplace transform w.r.t. t combined with the Fourier transform w.r.t. the tangential space variable x. Applying the two transforms to equation (2.4) yields where we set with c(y) = c ± for (±y) > 0. Equation (2.5) can readily be solved to the result By employing the fundamental solution A simple computation shows that Inserting this and the fact that In order to show the claimed regularity for the Laplace Fourier inverse of the representation (û,η) we first show regularity properties of the symbols involved. To this end let us introduce the operators that correspond to the time derivative and the Laplacian in tangential direction. Let r, s ≥ 0 and Then by K s p we either mean the space H s p or the space W s p . On the space and i.e. both, G and D n admit an R-bounded H ∞ -calculus with RH ∞ -angle φ R,∞ G = π/2 and φ R,∞ Dn = 0, respectively. Recall that an operator A admits an R-bounded where R(T ) denotes the R-bound of an operator family T ⊂ L(X) for a Banach space X, see [2,8] for additional information.
The inverse transform of the occuring symbols can formally be regarded as functions of G and D n . We first consider the symbol ω ± . The corresponding operator is formally given by (2.14) Lemma 2.3. Let 1 < p < ∞ and r, s ≥ 0. Then we have that Proof. The assertion follows from [9, Proposition 2.9 and Lemma 3.1].
Next we show closedness and invertibility of the operator associated with the symbol m introduced in (2.10), in the space 0 F r p (R + , K s p (R n )). We will prove invertibility of L and derive uniform estimates with respect to the parameters (δ, σ) in various adapted norms. In view of (2.12), (2.13), and by the Theorem of Kalton and Weis [7,Theorem 4.4] it essentially remains to show the holomorphy and the boundedness of the symbols regarded as functions of λ and |ξ| 2 on certain complex sectors.
In order to obtain these estimates, the following simple lemma will be useful.
Then the following statements are equivalent: (ii) There exists a c 0 > 0 such that Proof. We set which is a well defined function. Observe that (ii) is equivalent to saying that 0 ∈ g(G). By contradiction arguments it is not difficult to show that this relation is equivalent to condition (i).
Remark 2.5. The assumption f 1 (z) = 0 for z ∈ G is just for technical reasons and can be removed. Now we prove closedness and invertibility of L.
In the following we let ϕ 0 ∈ (π/3, π/2) and ϕ ∈ (0, ϕ 0 − π/3). Note that by condition (1.3) on a there exist δ * , σ * > 0 and M, c 0 > 0 such that and Let m be as given in (2.10). We consider the function Note that by our choice of the angle ϕ for (λ, z, δ, σ, κ) By these two estimates we see that in any case we obtain Consequently, A similar argument holds for the case that (λ, z, δ, σ, κ) This implies that Lemma 2.4 now yields the existence of a c 1 > 0 such that . An iterative application of Lemma 2.4 on the summands of f 1 and f 2 and an application of inequality (2.16) then result in . This implies that the functions

Now consider the cases
The argumentation above shows that are still uniformly bounded functions and this even on The aim now is to show that the term a + (δ, σ) can be regarded as a perturbation of g, if κ is assumed to be large enough. Indeed, if δ ≥ δ * > 0, by using (2.18) we can estimate On the other hand, if σ ≥ σ * > 0, we deduce by virtue of (2.16) that Hence, for fixed κ chosen large enough we see that we can achieve . Thus, we may represent 1/f as and therefore the functions m 0 , . . . , m 6 are uniformly bounded for all (λ, z, δ, σ) ∈ The remaining argumentation is now analogous to the proof of Lemma 2.3. Employing (2.13) we obtain R m j (λ, D n , δ, σ) L (0F r p (R+,K s p (R n ))) : (λ, δ, σ) ∈ Σ π−ϕ0 × [0, R] 2 ≤ C, for j = 0, 1, . . . , 6. Consequently, m j (G, D n , δ, σ) L (0F r p (R+,K s p (R n ))) ≤ C ((δ, σ) ∈ [0, R] 2 ), by virtue of (2.12) and [7,Theorem 4.4]. The invertibility of the operators (G + 1) 1/2 : 0 F r+1/2 p (R + , K s p (R n )) → 0 F r p (R + , K s p (R n )), D 1/2 n + 1 : 0 F r p (R + , K s+1 p (R n )) → 0 F r p (R + , K s p (R n )), (see for instance Proposition 2.9 and Lemma 3.1 in [9]) then yields the assertion, since L −1 = m 0 (G, D n , σ, δ), and by employing the fact that h → h(G) is an algebra homomorphism from H ∞ (Σ π−ϕ0 , K G (X)) into L (X) for X = 0 F r p (R + , K s p (R n )) and where (iv) We turn to the proof of the corresponding regularity assertions in Theorem 2.2 for (u, η, η E ). According to the results in [5, pages 15-16], By the same arguments we also have Next, note that by Lemma 2.3 we have that Indeed, we obtain which is a consequence of the mixed derivative theorem. Thus all the terms inside the brackets on the right hand side of (2.9) belong to the space 0 F 3 ∞ . In the same way as we clarified the invertibility of F ± : 0 F 2 ∞ → 0 F 3 ∞ by applying Lemma 2.3, we can see that L : 0 E 2 ∞ (δ, σ) → 0 F 3 ∞ is invertible by an application of Proposition 2.6. For instance, if δ, σ > 0, this follows from the embedding , which is again a consequence of the mixed derivative theorem. Furthermore, Proposition 2.6 implies the estimate Altogether this gives us for (δ, σ) ∈ [0, R] 2 , which yields the desired regularity for η. Observe that u now can be regarded as the solution of the diffusion equation A trivial but important observation now is that this equation itself does not depend on δ and σ, but only the data. Therefore also the corresponding solution operator is independent of δ and σ. By well-known results (see e.g. [5, Proposition 5.1]) and in view of (2.22) we obtain Similarly we can proceed for η E . Since it satisfies equation (2.4), we deduce (v) Let T 0 > 0 be fixed, and let J : where E J is defined as It follows from [10, Proposition 6.1] and the fact g, h)), whose existence has been established in steps (i)-(iv) of the proof. We note that where R J denotes the restriction operator, defined by R J w := w| J for w : R + → X. Then it is easy to verify that and that there is a constant M = M (T 0 ) such that for 0 ≤ δ, σ ≤ R, and T ≤ T 0 . Finally, uniqueness follows by a direct calculation which is straight forward and therefore omitted here. This completes the proof.
We proceed with convergence results for the case of zero time traces. To indicate the dependence on the parameters δ and σ we label from now on the corresponding functions and operators by µ, as e.g. L µ , v µ , where µ = (δ, σ). that a is a function satisfying the conditions in (1.3), and let L µ be the operator defined in (2.15) corresponding to the parameter µ := (δ, σ). Then we have Proof. As pointed out in part (iv) of the proof of Theorem 2.2 the domain of the operator 1). Now pick f ∈ D(F 3 + ). From Proposition 2.6 we infer that This yields (2.26). In a very similar way (2.27) can be proved. In order to see (2.28) we write ) this representation shows that (2.28) is obtained as a consequence of (2.26)-(2.27), and (2.29) in conjunction with the continuity of a ± .
Based on this result we will now prove convergence of solutions of problem (1.1)-(1.2).
Suppose that a is a function satisfying the conditions in (1.3) and that Furthermore, denote by (v µ , ρ µ , ρ µ E ) the unique solution of (1.1)-(1.2) whose existence is established in Theorem 2.2 and that corresponds to the parameter µ = (δ, σ).

30)
we have that

31)
where µ 0 = (δ 0 , σ 0 ). In particular, if denotes the solution operator to system (1.1), we have that Proof. In view of the arguments in part (v) of the proof of Theorem 2.2 the solution (v µ , ρ µ , ρ µ E ) can be represented by (v µ , ρ µ , ρ µ E ) := (R J (e κt u µ ), R J (e κt η µ ), R J (e κt η µ E )), (2.33) where R J denotes the restriction operator and (u µ , η µ , η µ E ) is the solution of (2.3)-(2.4) with right hand side (R c J (f µ , g µ , h µ )) and R c J as defined in (2.23). Hence we see that it suffices to prove convergence for the vector (u µ , η µ , η µ E ). Clearly, (2.30) implies that Therefore, and for simplicity, we simlpy write (f µ , g µ , h µ ) for the data instead of (R c J (f µ , g µ , h µ )) in the remaining part of the proof. Next, recall from (2.9) that η µ is given by According to (2.21) we know that . This fact and relations (2.19) and (2.20) then imply, by virtue of assumption (2.30), that By the uniform boundedness of L −1 the operator that maps the solution to the data corresponding to system (2.3). From part (iv) of the proof of Theorem 2.2 we infer that Furthermore, observe that we have Relation (2.36) applied for µ = µ 0 then yields T is uniformly bounded in µ ∈ I 0 . Thus, by (1.3), (2.34), (2.35), and assumption (2.30) we conclude that The convergence of η µ E is easily obtained as a consequence of the convergence of η µ . Recall that η µ E is the solution of (1.2) with ρ replaced by η µ . Denote by T the solution operator of this diffusion equation which is obviously independent of µ. Then by [5, Proposition 5.1] we obtain Hence (2.32) readily follows from (2.31).

2.2.
Inhomogeneous time traces. Next we consider the fully inhomogeneous system (1.1)-(1.2) and we will prove Theorem 1.1. By introducing appropriate auxiliary functions, we will reduce this problem to the situation of Theorem 2.2.
Suppose we had a solution (v, ρ, ρ E ) of (1.1)-(1.2) as claimed in the statement of Theorem 1.1. Let v 1 be the solution of the two-phase diffusion equation Observe that by compatibility assumption (1.7) we have Next let ρ 1 be an extension function so that as constructed in Lemma 3.2, and let ρ 1,E be the solution of (1.2), with ρ replaced by ρ 1 . For the solvability of (2.38) and the existence of ρ 1 we have to check the required regularity and compatibility conditions for the data. By construction we have that g(0) + ζ = γv 0 and by the regularity assumptions on g and v 0 we deduce Then it follows from [5, Proposition 5.1] that there is a unique solution v 1 ∈ E 1 T of (2.38). Furthermore, if δ > 0, we may use compatibility condition (1.7) to obtain that If δ = 0, we may impose σ > 0 which gives Thus, in any case we can satisfy the assumptions of Lemma 3.2 which yields the existence of ρ 1 ∈ E 2 T (δ, σ) as claimed, and of ρ 1,E ∈ E 1 T by virtue of Remark 2.1(b). Now we set It is clear that ρ 2,E is the extension of ρ 2 given by (1.2) with ρ replaced by ρ 2 . Thus, (v 2 , ρ 2 , ρ 2,E ) satisfies (2.44) By construction, ρ 1 ∈ E 2 T (δ, σ), and by (2.42) one may readily check that as well as (i) We start with proving convergence of ρ µ 1 . This function is according to (2.41) an extension of the traces where we set (2.53) Since the extension operator in Lemma 3.2 is linear and independent of µ we can estimate for all µ ∈ I 0 , It is clear by (1.10) that the first term on the right hand side of (2.54) tends to zero. In order to see the convergence of the second term we distinguish the three cases δ 0 = σ 0 = 0, and δ 0 > 0, σ 0 ≥ 0, and δ 0 = 0, σ 0 > 0.
The case δ 0 = σ 0 = 0: Here we have F 6 T (µ 0 ) = W 1−3/p p (R n ) and we obtain by a direct estimate and (1.10) that The case δ 0 > 0, σ 0 ≥ 0: Then F 6 T (µ 0 ) = W 2−3/p p (R n ). In this case we can employ compatibility condition (1.7) in Theorem 1.1 which results in .(2.55) (1.10). This yields In the same way we see that the first and the second term on the right hand side of (2.55) vanish for µ → µ 0 .
The case δ 0 = 0, σ 0 > 0: Since δ → 0, here we cannot apply compatibility condition (1.7). This leads to condition (1.11) in the statement of the theorem. In fact, here we obtain . It is clear that for σ 0 > 0 condition (1.11) implies that the first term on the right hand side vanishes, whereas the second term tends to zero again by (1.10).
Also here the convergence of ρ µ 1,E follows by the convergence of ρ µ 1 in view of the fact that ρ µ 1,E is the solution of (1.2) with ρ replaced by ρ µ 1 . If T denotes again the solution operator of this diffusion equation, by [5, Proposition 5.1] we obtain by the just proved convergence of ρ µ 1 and (1.10). Observe that v µ 1 is, according to (2.38), the solution of the same diffusion equation with right hand side (f µ , g µ +e −(1−∆x)t (γv µ 0 −g µ (0)), v µ 0 ) for µ ∈ I 0 . Moreover, we have that by (1.10), and (i) is proved.
(ii) Note that (v µ 2 , ρ µ 2 , ρ µ 2,E ) is the solution of (2.43)-(2.44). According to Theorem 2.8 it therefore suffices to prove convergence for the corresponding data. To be precise, it remains to show that and we see that (2.58) follows from (i), (1.3), and (1.10). Forg µ we have g µ −g µ0 (2.59) By employing the convergence assumptions also here we will prove that each single term on the right hand side of (2.59) tends to zero for µ → µ 0 . In view of (2.39) and (1.10) the convergence of the third term in (2.59) is clear. The first two terms are more involved. In fact, this is the point where assumption (1.12) enters. In analogy to (i) we again distinguish the three cases δ 0 = σ 0 = 0, and δ 0 > 0, σ 0 ≥ 0, and δ 0 = 0, σ 0 > 0.
The case δ 0 = 0, σ 0 > 0: Here we have γv µ0 0 − σ∆ x ρ µ0 0 = g µ0 (0). (2.62) In a similar way as in the previous case we deduce, if δ > 0, that (2.63) Note that in the case σ 0 > 0 we also have that By this fact it is easy to see that the first two terms in (2.63) vanish for (µ → µ 0 ), whereas the convergence of the last two terms follows again by (1.10). That the second term in (2.59) tends to zero here follows easily from the inequality and the convergence of ρ µ 1 in E 2 T (µ 0 ) = E 2 T (0, σ 0 ) proved in (i). Observe that the last argument also implies convergence for the case δ = 0, since then the first term in (2.59) vanishes completely.