# Lecture notes on the Stefan problem

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1 Lecture notes on the Stefan problem Daniele Andreucci Dipartimento di Metodi e Modelli Matematici Università di Roma La Sapienza via Antonio Scarpa Roma, Italy

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3 Introduction These notes cover a portion of a course of the Doctorate Modelli e metodi matematici per la tecnologia e la società, given in 21 in Rome. The title of the course was Evolution equations and free boundary problems and its topics included, essentially, an introduction to Stefan and Hele-Shaw problems. Here only the material concerning the Stefan problem is partially reproduced. The present notes assume the reader has some knowledge of the elementary theory of L p and Sobolev spaces, as well as of the basic results of existence and regularity of solutions to smooth parabolic equations. The bibliography is minimal; only books and articles quoted in the text are referenced. See [12], [18] and [2] for further references. I thank the audience of the course for many stimulating comments and questions, and prof. R. Ricci for interesting discussions on the subject of these notes. Rome, January 22 Revised version: Rome, January 24 iii

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5 Notation The notation employed here is essentially standard. Appendix D contains a list of the main symbols used in the text. Constants denoted by γ, C, C δ,...may change fom line to line. v

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7 Contents Introduction Notation iii v Chapter 1. The classical formulation 1 1. The Stefan condition 1 2. The free boundary problem 4 3. Explicit examples of solutions 5 4. Basic estimates 7 5. Existence and uniqueness of the solution 8 6. Qualitative behaviour of the solution Regularity of the free boundary 15 Chapter 2. Weak formulation of the Stefan problem From the energy balance to the weak formulation Comparing the weak and the classical formulations 2 3. Definition of weak solution Uniqueness of the weak solution Existence of weak solutions A comparison result 37 Appendix A. Maximum principles for parabolic equations The weak maximum principle The strong maximum principle Hopf s lemma (the boundary point principle) Maximum principle for weak solutions 48 Appendix B. A theorem by Kruzhkov Mollifying kernels The main estimate 53 m+α m+α, Appendix C. The spaces H 2 (Q T ) Comments 55 Appendix D. Symbols used in text 57 Bibliography 59 vii

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9 CHAPTER 1 The classical formulation In this chapter we consider the formulation of the Stefan problem as a classical initial boundary value problem for a parabolic partial differential equation. A portion of the boundary of the domain is a priori unknown (the free boundary), and therefore two boundary conditions must be prescribed on it, instead than only one, to obtain a well posed problem. 1. The Stefan condition The Stefan problem ([17]) is probably the simplest mathematical model of a phenomenon of change of phase. When a change of phase takes place, a latent heat is either absorbed or released, while the temperature of the material changing its phase remains constant. In the following we denote by L > the latent heat per unit of volume (p.u.v.), and neglect for the sake of simplicity any volume change in the material undergoing the change of phase. We also assume the critical temperature of change of phase to be a constant, θ. To be specific, consider at time t = t a domain A divided by the plane x 1 = s into two subdomains. At time t = t the sub-domain A 1 = A {x 1 < s } is filled by water, while A 2 = A {x 1 > s } is filled by ice. In the terminology of problems of change of phase, A 1 is the liquid phase and A 2 is the solid phase. The surface separating the two phases is referred to as the interface. Assume also the setting is plane symmetric, that is the temperature θ is a function of x 1 only, besides the time t, and the interface is a plane at all times. Denote by x 1 = s(t) the position of the interface at time t. Note that, due to the natural assumption that temperature is continuous, θ(s(t)+, t) = θ(s(t), t) = θ, for all t. (1.1) Assume ice is changing its phase, that is the interface is advancing into the solid phase. Due to the symmetry assumption we stipulate, we may confine ourselves to consider any portion D, say a disk, of the interface at time t. At a later time t 1 > t the interface occupies a position s(t 1 ) > s(t ) = s. The cylinder D (s(t 1 ), s(t )) has been melted over the time interval (t, t 1 ) (see Figure 1). The change of phase has therefore absorbed a quantity of heat volume of the melted cylinder latent heat p.u.v. = area(d)(s(t 1 ) s(t ))L. (1.2) The heat must be provided by diffusion, as we assume that no heat source or sink is present. We adopt for heat diffusion Fourier s law heat flux = k i Dθ, (1.3) where k 1 > is the diffusivity coefficient in water, and k 2 > is the diffusivity coefficient in ice (in principle k 1 k 2 ). Thus, the quantity of heat in (1.2) must 1

10 2 DANIELE ANDREUCCI water ice s(t ) s(t 1 ) e 1 Figure 1. Melting ice equal t 1 [ k 1 Dθ(s(t), t) e 1 k 2 Dθ(s(t)+, t) ( e 1 )] dx 2 dx 3 dt = t D(t) area(d) t 1 t [ k 1 θ x1 (s(t), t) + k 2 θ x1 (s(t)+, t)] dt. (1.4) Equating the two quantities, dividing the equation by t 1 t and letting t 1 t, we finally find k 1 θ x1 (s(t), t) + k 2 θ x1 (s(t)+, t) = Lṡ(t), (1.5) where we have substituted t with the general time t, as the same procedure can be obviously carried out at any time. This is called the Stefan condition on the free boundary. We stress the fact that the Stefan condition is merely a law of energetical balance. Several remarks are in order. Remark 1.1. Note that, although we did not assume anything on the values of θ(x 1, t) inside each one of the two phases, on physical grounds we should expect θ θ, in water, i.e., in A 1 ; θ θ, in ice, i.e., in A 2. (1.6) The equality θ θ in either one of the two phases (or in both) can not be ruled out in the model. Rather, it corresponds to the case when a whole phase is at critical temperature. Diffusion of heat, according to (1.3), can not take place in that phase, as θ x1 there. Thus (1.5) reduces to, e.g., if the solid phase is at constant temperature, k 1 θ x1 (s(t), t) = Lṡ(t). (1.7) Note that if θ > θ in water, (1.7) predicts that ṡ(t) >. In other words, melting of ice is predicted by the model, instead of solidification of water. This is consistent with obvious physical considerations. Problems where one of the two phases is everywhere at the critical temperature

11 1. THE STEFAN CONDITION 3 are usually referred to (somehow misleadingly) as one phase problems, while the general case where (1.5) is prescribed is the two phases problem. In the latter case, the sign of ṡ(t), and therefore the physical behaviour of the system water/ice predicted by the mathematical model, depends on the relative magnitude of the two heat fluxes at the interface. Remark 1.2. On the interface, which is also known as the free boundary two conditions are therefore prescribed: (1.1) and (1.5). In the case of a one phase problem, this fact has the following meaningful interpretation in terms of the general theory of parabolic PDE: The boundary of the domain where the heat equation (a parabolic equation based on (1.3)) is posed, contains an a priori unknown portion, corresponding to the interface separating the liquid phase from the solid one. Clearly, if only the Dirichlet boundary condition (1.1) was imposed on it, we could choose arbitrarily this part of the boundary, and solve the corresponding initial value boundary problem. Apparently the solution would not, in general, satisfy (1.7). A similar remark applies to solutions found imposing just (1.7) (where now the functional form of the arbitrarily given boundary x 1 = s(t) is explicitly taken into account). It is therefore evident that on the free boundary both conditions (1.1) and (1.7) should be prescribed in order to have a well posed problem. (Or, anyway, in more general free boundary problems, two different boundary conditions are required.) Incidentally, this circle of ideas provides the basic ingredient of a possible proof of the existence of solutions: we assign arbitrarily a candidate free boundary s and consider the solution θ to the problem, say, corresponding to the data (1.1). Then we define a transformed boundary s exploiting (1.7), i.e., k 1 θ x1 (s (t), t) = Lṡ (t). A fixed point of this transform corresponds to a solution of the complete problem. Remark 1.3. The meaning of conditions (1.1) and (1.5) in the context of two phases problems is probably better understood in terms of the weak formulation of the Stefan problem, which is discussed below in Chapter 2. We remark here that, actually, one phase problems are just two phases problems with one phase at constant temperature, so that the discussion in Chapter 2 applies to them too. Remark 1.4. Problems where the temperature restriction (1.6) is not fulfilled, are sometimes called undercooled Stefan problems. We do not treat them here, albeit their mathematical and physical interest (see however Subsection 2.4 of Chapter 2); let us only recall that they are, in some sense, ill posed Exercises Write the analogs of Stefan condition (1.5) in the cases of cylindrical and spherical symmetry in R Note that if we assume, in (1.5), θ < θ in A 1 and θ θ in A 2, we find ṡ(t) <. This appears to be inconsistent with physical intuition: a phase of ice at sub critical temperature should grow into a phase of water at identically critical temperature. Indeed, (1.5) is not a suitable model for the physical setting considered here. In other words, the Stefan condition in the form given above keeps memory of which side of the interface is occupied by which phase. Find out where we implicitly took into account this piece of information and write the Stefan condition when ice and water switch places.

12 4 DANIELE ANDREUCCI 1.3. Prove that Stefan condition (1.5) does not change its form if bounded volumetric heat sources are present (i.e., if the heat equation is not homogeneous). 2. The free boundary problem Keeping the plane symmetry setting considered above, we may of course assume the problem is one dimensional. Denoting by x the space variable, the complete two phases problem can be written as c 1 θ t k 1 θ xx =, in Q 1, (2.1) c 2 θ t k 2 θ xx =, in Q 2, (2.2) k 1 θ x (, t) = h 1 (t), < t < T, (2.3) k 2 θ x (d, t) = h 2 (t), < t < T, (2.4) θ(x, ) = Θ(x), < x < d, (2.5) k 1 θ x (s(t), t) + k 2 θ x (s(t)+, t) = Lṡ(t), < t < T, (2.6) θ(s(t), t) = θ(s(t)+, t) = θ, < t < T, (2.7) s() = b. (2.8) Here < b < d, T and c 1, c 2, k 1, k 2 are given positive numbers. The c i represent the thermal capacities in the two phases. The liquid phase occupies at the initial time t = the interval (, b), while the solid phase occupies (b, d). The problem is posed in the time interval (, T). Moreover we have set Q 1 = {(x, t) < x < s(t), < t < T }, Q 2 = {(x, t) s(t) < x < d, < t < T }. We are assuming that < s(t) < d for all < t < T. If the free boundary hits one of the two fixed boundaries x = and x = d, say at time t, of course the formulation above should be changed. In practice, one of the two phases disappears at t = t. We leave to the reader the simple task of writing the mathematical model for t > t. One could impose other types of boundary data, instead of (2.3), (2.4), e.g., Dirichlet data. If we are to attach the physical meaning of a change of phase model to the problem above, the data must satisfy suitable conditions. At any rate Θ(x) θ, < x < b; Θ(x) θ, b < x < d. Essentially, we need θ θ in Q 1 and θ θ in Q 2. Actually, we will deal mainly with the one phase version of (2.1) (2.8) where the solid phase is at constant temperature. Namely, after adimensionalization, we look at u t u xx =, in Q s,t, (2.9) u x (, t) = h(t) >, < t < T, (2.1) u(x, ) = u (x), < x < b, (2.11) u x (s(t), t) = ṡ(t), < t < T, (2.12) u(s(t), t) =, < t < T, (2.13) s() = b. (2.14)

13 3. EXPLICIT EXAMPLES OF SOLUTIONS 5 (We have kept the old names for all variables excepting the unknown u.) Here we denote for each positive function s C([, T]), such s() = b, Q s,t = {(x, t) < x < s(t), < t < T }. We regard the rescaled temperature u as a function defined in Q s,t. The solid phase therefore does not appear explicitly in the problem. As a matter of fact we assume it to be unbounded in the positive x direction (i.e., d = + ), so that no upper limit has to be imposed on the growth of the free boundary s. The sign restrictions in (2.1) and in (2.11) are imposed so that u > in Q s,t, see Proposition 4.1 below. Definition 2.1. A solution to problem (2.9) (2.14) is a pair (u, s) with s C 1 ((, T]) C([, T]), s() = b, s(t) >, t T ; u C(Q s,t ) C 2,1 (Q s,t ), u x C(Q s,t {t = }), and such that (2.9) (2.14) are satisfied in a classical pointwise sense. We prove a theorem of existence and uniqueness of solutions to (2.9) (2.14), under the assumptions h C([, T]), h(t) >, t T ; (2.15) u C([, b]), u (x) H(b x), x b. (2.16) We also study some qualitative behaviour of the solution. The adimensionalization of the problem does not play a substantial role in the mathematical theory we develop here. For further reading on the one-phase Stefan problem, we refer the reader to [5], [3]; we employ in this chapter the techniques found there, with some changes. Remark 2.1. The free boundary problem (2.9) (2.14) is strongly non linear, in spite of the linearity of the PDE and of the boundary conditions there. Indeed, recall that the free boundary s itself is an unknown of the problem; its dependence on the data is not linear (as, e.g., the explicit examples of Section 3 show) Exercises Prove that the change of variables θ αu + θ, x βξ, t γτ, allows one to write the one phase problem in the adimensionalized form (2.9) (2.14), for a suitable choice of the constants α, β, γ. Also note that adimensionalizing the complete two phases problem similarly is in general impossible. 3. Explicit examples of solutions Example 1. An explicit solution of the heat equation (2.9) is given by ( ) x v(x, t) = erf 2 = 2 x/2 t π t v(x, ) = 1, x >. e z2 dz, x,t >,

14 6 DANIELE ANDREUCCI Here erf denotes the well known error function. Fix C > arbitrarily, and set, for a α > to be chosen presently, { ( )} x u(x, t) = C erf α erf 2. t Define also s(t) = 2α t; note that s() =. Thus u > in Q s,t, and (2.9) as well as (2.13) are satisfied. By direct calculation u x (x, t) = C πt e x2 4t. Hence, u x (s(t), t) = u x (2α t, t) = C πt e α2 = ṡ(t) = α t, if and only if C = παe α2. Note that on the fixed boundary x = we may select either one of the conditions u x (, t) = C πt, or u(, t) = C erf α. We have proven that, when α is chosen as above, u(x, t) = 2αe α2 α x/2 t e z2 dz (3.1) solves the problem sketched in Figure 2. Note however that u is not continuous at (, ); the notion of solution in this connection should be suitably redefined. t s(t) = 2α t u x = C/ πt or u = C erf α u t u xx = u = u x = ṡ(t) = α/ t x Figure 2. The Stefan problem solved by u in (3.1) Example 2. It is obvious by direct inspection that the function u(x, t) = e t x 1, (3.2) solves the Stefan problem in Figure 3, corresponding to the free boundary s(t) = t. Note that we are forced to prescribe an exponentially increasing flux on x = in order to obtain a linear growth for s(t).

15 4. BASIC ESTIMATES 7 t s(t) = t u x = e t or u = e t 1 u t u xx = u = u x = ṡ(t) = 1 x Figure 3. The Stefan problem solved by u in (3.2) 3.1. Exercises Convince yourself that the solutions corresponding to u x (, t) = 2C/ πt, in the case of Example 1, and to u x (, t) = 2e t in the case of Example 2, can not be obtained by linearity from the ones given above. 4. Basic estimates Proposition 4.1. If (u, s) is a solution to (2.9) (2.14), then u(x, t) >, in Q s,t ; (4.1) ṡ(t) >, for all t >. (4.2) Proof. By virtue of the weak maximum principle, u must attain its maximum on the parabolic boundary of Q s,t, i.e., on Q s,t {(x, t) t = T, < x < s(t)}. The data h being positive, the maximum is attained on t = or on x = s(t). Therefore u (remember that u ). If we had u( x, t) = in some ( x, t) Q s,t, invoking the strong maximum principle we would obtain u in Q s,t {t < t}. This is again inconsistent with h >. Thus (4.1) is proven. Then, the value u = attained on the free boundary is a minimum for u. Recalling the parabolic version of Hopf s lemma, we infer ṡ(t) = u x (s(t), t) >, for all t >. Remark 4.1. The proof of estimate (4.1) does not make use of the Stefan condition (2.12). In the same spirit, we consider in the following results solutions to the initial value boundary problem obtained removing Stefan condition from the formulation. The rationale for this approach is that we want to apply those results to approximating solutions constructed according to the ideas of Remark 1.2. Lemma 4.1. Let u be a solution of (2.9), (2.1), (2.11), (2.13), where s is assumed to be a positive non decreasing Lipschitz continuous function in [, T], such that s() = b. Let (2.15) and (2.16) be in force. Then u x is continuous up to all the points of the boundary of Q s,t of the form (, t), (s(t), t), with T t >.

16 8 DANIELE ANDREUCCI The regularity of u x up to x = is classical; the proof of this result will be completed in Section 7, see Lemma 7.1. Proposition 4.2. Let u, s be as in Lemma 4.1. Then where M = max( h, H). < u(x, t) M(s(t) x), in Q s,t, (4.3) Proof. Define v(x, t) = M(s(t) x). It follows immediately v t v xx = Mṡ(t), in Q s,t ; v(s(t), t) = u(s(t), t) =, t T, v x (, t) = M h = u x (, t), < t < T, v(x, ) = M(b x) u (x) = u(x, ), x b. Therefore, taking into account the results of Appendix A, v(x, t) u(x, t), in Q s,t. Corollary 4.1. If u, s are as in Proposition 4.2, then > u x (s(t), t) M, < t < T, (4.4) where M is the constant defined in Proposition 4.2. Proof. A trivial consequence of Proposition 4.1 and of Proposition 4.2, as well as of Lemma 4.1. We also keep in mind Remark 4.1, and of course make use of Hopf s lemma for the strict inequality in (4.4). Remark 4.2. We stress the fact that the barrier construction of Proposition 4.2 is made possible by the fact that ṡ. In turn, this is for solutions of the Stefan problem a consequence of the positivity of u. Thus, such a barrier function, and in general any similar barrier function, does not exist in the case of the undercooled Stefan problem Exercises Note that the function v defined in the proof of Proposition 4.2 is just Lipschitz continuous in t. In spite of this fact, one may apply to v u the weak maximum principle in the form given in Section 4 of Appendix A. Carry out the proof in detail. We prove here 5. Existence and uniqueness of the solution Theorem 5.1. Assume (2.15), (2.16). Then there exists a unique solution to (2.9) (2.14). Let u, s be as in Proposition 4.2. Moreover assume that s Σ := {σ Lip([, T]) σ M, σ() = b}. Here M is the constant defined in Proposition 4.2. The set Σ is a convex compact subset of the Banach space C([, T]), equipped with the max norm. A useful property of all s Σ is b + Mt s(t) b, t T.

17 Define the transform T (s) by 5. EXISTENCE AND UNIQUENESS OF THE SOLUTION 9 t T (s)(t) = b u x (s(τ), τ) dτ, T t. Note that, as a consequence of Lemma 4.1 and of Corollary 4.1, and T (s) Lip([, T]) C 1 ((, T]), M d dt T (s)(t) = u x(s(t), t), t >. Then T : Σ Σ. Also note that a fixed point of T corresponds to a solution of our Stefan problem. We use the divergence theorem to transform the boundary flux integral defining T (s). Namely we write = t s(τ) (u τ u xx ) dxdτ = b t u (x) dx u(s(τ), τ)ṡ(τ) dτ + s(t) u(x, t) dx t t u x (s(τ), τ) dτ h(τ) dτ. Therefore t T (s)(t) = b b b + u x (s(τ), τ) dτ = t u (x) dx + h(τ) dτ s(t) u(x, t) dx =: F(t) s(t) u(x, t) dx. (5.1) Note that this equality allows us to express T (s) in terms of more regular functions than the flux u x (s(t), t), which appeared in its original definition. We are now in a position to prove that T is continuous in the max norm. Let s 1, s 2 Σ. Let us define α(t) = min(s 1 (t), s 2 (t)), β(t) = max(s 1 (t), s 2 (t)), i = 1, if β(t) = s 1 (t), i = 2, otherwise. (The number i is a function of time; this will not have any specific relevance.) Let us also define v(x, t) = u 1 (x, t) u 2 (x, t). Then v satisfies v t v xx =, in Q α,t, (5.2) v x (, t) =, < t < T, (5.3) v(x, ) =, < x < b, (5.4) v(α(t), t) = u i (α(t), t) M(β(t) α(t)), < t < T. (5.5)

18 1 DANIELE ANDREUCCI Therefore, we may invoke the maximum principle to obtain where we also denote On the other hand, we have v,t := max Q α,t v M s 1 s 2,t, (5.6) s 1 s 2,t = max τ t s 1(τ) s 2 (τ). T (s 1 )(t) T (s 2 )(t) = Therefore s 2 (t) s 1 (t) u 2 (x, t) dx u 1 (x, t) dx = α(t) v(x, t) dx + ( 1) i β(t) α(t) u i (x, t) dx. T (s 1 )(t) T (s 2 )(t) α(t) v,t + M(β(t) α(t)) 2 (b + MT)M s 1 s 2,t + M s 1 s 2 2,t, (5.7) and the continuity of T : Σ Σ is proven. By Schauder s theorem, it follows that a fixed point of T exists, and thus existence of a solution. Uniqueness might be proven invoking the monotone dependence result given below (see Theorem 6.1). However, mainly with the purpose of elucidating the role played in the theory of free boundary problems by local integral estimates, we proceed to give a direct proof of uniqueness of solutions. More explicitly, we prove the contractive character (for small t) of T, thereby obtaining existence and uniqueness of a fixed point Local estimates vs the maximum principle. Estimates like (5.6), obtained through the maximum principle, have the advantage of providing an immediate sup estimate of the solution in the whole domain of definition. However, the bound they give may be too rough, at least in some regions of the domain. Consider for example, in the setting above, a point (b/2, ε), with ε 1. The maximum principle predicts for v(b/2, ε) a bound of order M s 1 s 2,ε, that is, obviously, the same bound satisfied by the boundary data for v. On the other hand, taking into account (5.3), (5.4) one might expect v(b/2, ε) to be much smaller than v(α(ε), ε). This is indeed the case, as we show below. In order to do so, we exploit local integral estimates of the solution, that is, estimates involving only values of v in the region of interest, in this case, away from the boundary. We aim at proving that T is a contraction, that is, T (s 1 ) T (s 2 ),t d s 1 s 2,t, (5.8) with d < 1 for small enough t. A quick glance at (5.7) shows that (5.8) does not, indeed, follow from there (unless bm < 1). This failure is due only to the term

19 5. EXISTENCE AND UNIQUENESS OF THE SOLUTION 11 originating from the estimate of α(t) v(x, t) dx. Thus, a better estimate of this integral is needed. A key step in any local estimation is a good choice of cut off functions. These are, typically, non negative smooth functions equal to 1 in the region we want to single out, and identically vanishing away from it. Let b/2 > δ >, and define the cut off function ζ(x), such that ζ(x) = 1, x b 2δ, ζ(x) =, b δ x, 2 δ ζ x(x). Multiply (5.2) by vζ 2 and integrate by parts in Q b,t = [, b] [, t]. We get = (v τ vζ 2 v xx vζ 2 ) dxdτ Q b,t = 1 2 b v(x, t) 2 ζ(x) 2 dx + vxζ 2 2 dxdτ + 2 Q b,t whence (using the inequality 2ab εa 2 + b 2 /ε, ε > ) Q b,t vv x ζζ x dxdτ, (5.9) 1 b v(x, t) 2 ζ(x) 2 dx + v 2 xζ 2 2 dxdτ 2 v x v ζ x ζ dxdτ Q b,t Q b,t 2 v 2 ζ 2 1 x dxdτ + v 2 xζ 2 2 dxdτ. Q b,t Q b,t By absorbing the last integral into the left hand side, we find b v(x, t) 2 ζ(x) 2 dx + vxζ 2 2 dxdτ 4 Q b,t Let us now go back to t = 4 T (s 1 )(t) T (s 2 )(t) = = α(t) b 2δ b δ b 2δ Q b,t v 2 ζ 2 x dxdτ v 2 ζ 2 x dxdτ 16 δ 2 tδ v 2,t = 16 δ t v 2,t. (5.1) v(x, t) dx + ( 1) i v(x, t) dx + α(t) b 2δ β(t) α(t) u i (x, t) dx v(x, t) dx + ( 1) i β(t) α(t) u i (x, t) dx.

20 12 DANIELE ANDREUCCI Use Hölder s inequality and (5.1) to bound the integral over (, b 2δ) (recall that ζ 1 there), and find T (s 1 )(t) T (s 2 )(t) ( b 2δ v(x, t) 2 dx) 1/2 b + (α(t) b + 2δ) v,t + M(β(t) α(t)) 2 b 4 δ t v,t + (Mt + 2δ) v,t + M(β(t) α(t)) 2. Take now δ = t (t is fixed in this argument), and apply again the sup estimate (5.6), finally obtaining T (s 1 ) T (s 2 ),t { 4M b 4 t + M(Mt + 2 t) + M 2 t } s 1 s 2,t. Clearly, for t = t (M, b), T is a contractive mapping Exercises In Subsection 5.1 we proved existence and uniqueness of a fixed point of T in a small time interval (locally in time). Show how the argument can be completed to give existence and uniqueness of a fixed point in any time interval (global existence) Why uniqueness of solutions (u, s) with s Σ is equivalent to uniqueness of solutions in the class of Definition 2.1, without further restrictions? 5.3. Give an interpretation of (5.1) as an energetical balance Prove that T has the property (see also Figure 4) s 1 (t) < s 2 (t), < t < T = T (s 1 )(t) > T (s 2 )(t), < t < T Extend the proof of existence and uniqueness of solutions to the case when h. What happens if h, u? 5.6. To carry out rigorously the calculations in (5.9) actually we need an approximation procedure: i.e., we need first perform integration in a smaller 2- dimensional domain, bounded away from the boundaries x =, t =. Recognize the need of this approach, and go over the (easy) details. 6. Qualitative behaviour of the solution Theorem 6.1. (Monotone dependence) Let (u i, s i ) be solutions of (2.9) (2.14), i = 1, 2, respectively corresponding to data h = h i, b = b i, u = u i. Assume both sets of data satisfy (2.15), (2.16). If h 1 (t) h 2 (t), < t < T ; b 1 b 2 ; u 1 (x) u 2 (x), < x < b 1 ; (6.1) then s 1 (t) s 2 (t), < t < T. (6.2) Proof. 1) Let us assume first b 1 < b 2. Reasoning by contradiction, assume t = inf{t s 1 (t) = s 2 (t)} (, T). Then the function v = u 2 u 1 is strictly positive in { < t < t, < x < s 1 (t)},

21 6. QUALITATIVE BEHAVIOUR OF THE SOLUTION 13 t T (s) = s x = σ(t) x = T (σ)(t) b x Figure 4. Behaviour of the transform T by virtue of the strong maximum principle and Hopf s lemma. Indeed, v(s 1 (t), t) >, < t < t. Then v attains a minimum at (s 1 ( t), t) = (s 2 ( t), t), where Thus, due to Hopf s lemma, But we compute v(s 1 ( t), t) =. v x (s 1 ( t), t) <. v x (s 1 ( t), t) = u 2x (s 2 ( t), t) u 1x (s 1 ( t), t) = ṡ 2 ( t) + ṡ 1 ( t). Hence ṡ 2 ( t) > ṡ 1 ( t), which is not consistent with the definition of t. 2) Assume now b 1 = b 2. Let us extend the data u 2 to zero over (b, b + δ), where < δ < 1 is arbitrary. Let (u δ, s δ ) be the solution of problem (2.9) (2.14) corresponding to the data h = h 2, b = b 2 + δ, u = u 2. Then, by the first part of the proof, s 2 < s δ, s 1 < s δ, and for all < t < T s δ (t) s 2 (t) = δ s 2 (t) [u δ u 2 ](x, t) dx s δ (t) s 2 (t) u δ (x, t) dx δ. Therefore s δ s 2 + δ, so that s 1 < s δ s 2 + δ. On letting δ we recover s 1 s 2. Let us investigate the behaviour of the solution of the Stefan problem for large times. In doing so, we of course assume that T =. Note that the result of existence and uniqueness applies over each finite time interval; a standard extension technique allows us to prove existence and uniqueness of a solution defined for all positive times. Owing to Proposition 4.2, we have s(t) S, < t < = u(x, t) MS, in Q s,. (6.3)

22 14 DANIELE ANDREUCCI Moreover, s being monotonic, certainly there exists s = lim t s(t). (6.4) Theorem 6.2. Let (u, s) be the solution of Theorem 5.1. Then Proof. 1) Assume first b s = lim s(t) = b + t u (x) dx + h(t) dt. (6.5) h(t) dt = +. (6.6) We only need show s is unbounded. Let us recall that, for all t >, b s(t) = b + t u (x) dx + h(τ) dτ s(t) u(x, t) dx. (6.7) From (6.3), it follows that, if s is bounded over (, ), then u is also bounded over (, ). This is clearly inconsistent with (6.7), when we keep in mind (6.6). 2) Assume h(t) dt < +. (6.8) The balance law (6.7), together with u >, immediately yields b s(t) < b + u (x) dx + h(t) dt < +. (6.9) Then we have s(t) s < ; it is only left to identify s as the quantity indicated above. Owing to (6.7) again, we only need show lim t s(t) u(x, t) dx =. (6.1) On multiplying (2.9) by u and integrating by parts in Q s,t, we get 1 2 s(t) u(x, t) 2 dx + u 2 x dxdτ = Q s,t 1 b t u (x) 2 dx + u(, τ)h(τ) dτ. (6.11) 2 Recalling (6.3) and (6.9), the last integral in (6.11) is majorised by Thus a consequence of (6.11) is Ms h(τ) dτ <. Q s, u x (x, t) 2 dxdt < +. (6.12)

23 Elementary calculus then shows that u(x, t) 2 dxdt = Q s, 7. REGULARITY OF THE FREE BOUNDARY 15 Q s, s Q s, [ s(t) x s(t) x u ξ (ξ, t) dξ] 2 dxdt u ξ (ξ, t) 2 dξ dxdt s 2 Then there exists a sequence {t n }, t n, such that But the function s(t n ) t u(x, t n ) 2 dx. s(t) u(x, t) 2 dx, Q s, u ξ (ξ, t) 2 dξ dt <. when we take into account (6.11), is easily seen to have limit as t, so that this limit is. By Hölder s inequality, s(t) u(x, t) dx [ s(t) 1/2 s u(x, t) dx] 2, t, completing the proof of (6.1) Exercises Find conditions ensuring that the inequality in (6.2) is strict Discuss the necessity of assumptions (2.15), (2.16) in Theorem Regularity of the free boundary The approach in this Section is taken from [16], and provides an example of bootstrap argument, i.e., of an inductive proof where any given smoothness of the solution allows us to prove even more regularity for it. Our first result will become the first step in the induction procedure, and is however required to prove Lemma 4.1. Lemma 7.1. Let u C 2,1 (Q s,t ) C(Q s,t ), where s Lip([, T]), and s(t) > for t T. Assume u fulfils u t u xx =, in Q s,t, (7.1) u(s(t), t) =, t T. (7.2) Then for each small enough ε >, u x is continuous in P ε, where P ε = {(x, t) ε < x < s(t), ε < t < T }.

24 16 DANIELE ANDREUCCI Proof. The proof is based on standard local regularity estimates for solutions of parabolic equations. Introduce the following change of variables { y = x s(t), v(y, τ) = u(ys(τ), τ). τ = t ; The set Q s,t is mapped onto R = (, 1) (, T), where v solves v τ 1 s(τ) 2v yy ṡ(τ) s(τ) yv y =, in R, (7.3) v(1, τ) =, τ T. (7.4) More explicitly, (7.3) is solved a.e. in R, as v is locally a Sobolev function in R. Classical results, see [11] Chapter IV, Section 1, imply that for any fixed ε >, v τ, v y, v yy L q ((ε, 1) (ε, T)), for all q > 1. Then we use the embedding Lemma 3.3 of [11] Chapter II, to infer that, for q > 3, v y H α, α 2 ([ε, 1] [ε, T]), (7.5) where α = 1 3/q (see also Remark 11.2 of [11], p. 218; the space H α, α 2 is defined in Appendix C). Since u x (x, t) = 1 s(t) v ( x y s(t), t), the result follows. Our next result implies that the free boundary in the Stefan problem (2.9) (2.14) is of class C (, T). Theorem 7.1. Assume u and s are as in Lemma 7.1, and moreover u x (s(t), t) = cṡ(t), < t < T, (7.6) where c is a given constant. Then s C (, T). Proof. For v defined as in the proof of Lemma 7.1, we rewrite (7.6) as ṡ(τ) = 1 cs(τ) v y(1, τ), < τ < T. (7.7) Choose α (, 1). Then (7.5) and (7.7) yield at once ṡ H α, α 2 ([ε, 1] [ε, T]), for each fixed ε >. (7.8) Next we make use of the following classical result: If the coefficients in (7.3) (i.e., ṡ), are of class m+α m+α, H 2 ([ε, 1] [ε, T]), then v is of class H 2+m+α,2+m+α 2 ([2ε, 1] [2ε, T]). Here m is any integer. We prove by induction that for all m (7.9) m+α m+α, ṡ H 2 ([ε, 1] [ε, T]), for each fixed ε >. (7.1) We already know this is the case when m =, from (7.8). Assume then (7.1) is in force for a given m. Then, owing to (7.9), v H 2+m+α,2+m+α 2 ([ε, 1] [ε, T]), for each fixed ε >. (7.11)

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