The multi-layer free boundary problem for the p-laplacian in convex domains

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1 The multi-layer free boundary problem for the p-laplacian in convex domains Andrew ACKER Dept. of Mathematics and Statistics Wichita State University Wichita, KS USA Mikayel POGHOSYAN Department of Mathematics Yerevan State University Alex Manoogian, , Yerevan, ARMENIA Antoine HENROT Ecole des Mines and Institut Elie Cartan, UMR CNRS 7502 and INRIA BP 239, Vandoeuvre-les-Nancy Cedex, FRANCE Henrik SHAHGHOLIAN Department of Mathematics Royal Institute of Technology Stockholm SWEDEN August 2, 2004 Abstract The main result of this paper concerns existence of classical solutions to the multi-layer Bernoulli free boundary problem with nonlinear joining conditions and the p-laplacian as governing operator. The present treatment of the 2-layer case involves technical refinements of the one-layer case, studied earlier by two of the authors. The existence treatment of the multi-layer case is largely based on a reduction to the two-layer case, in which uniform separation of the free boundaries plays a key role. Introduction and statement of the problem. The mathematical setting In this paper we continue the study of the free boundary problem arising in connection with potential flow with power-law nonlinearity (see [HS 4] for backgrounds). Mathematically, our starting point is an annular region, bounded by two convex surfaces in R N (N 2): K = K m+2 \ K with K, K m+2 convex and K K m+2. The aim is to show that for given positive integer m and data λ i (, ), F i (x, p, q) : K R + R + R, (i =,, m) with λ i > λ i+, one can find convex domains K K 2 K m+ K m+2 The second author thanks Göran Gustafsson Foundation for several visiting appointments, to RIT in Stockholm. The third author thanks Swedish Royal Academy of Sciences for visiting appointment to RIT. Supported in part by Swedish Research Council.

2 such that the p-capacitary potential u i for each annular convex region K i+ \ K i satisfies a nonlinear joining Bernoulli condition (see the main theorems, Theorem 4., and Theorem 6.) F i (x, u i (x), u i+ (x) ) = 0 on K i+, (i =,, m). () The p-capacitary potential refers to the solution of the following Dirichlet problem p u = 0 in K i+ \ K i u = λ i on K i u = λ i+ on K i+, where p, < p < is the p-laplace operator defined by.2 Applications p u := div( u p 2 u). The above described problem appears in several physical situations and can be appropriately interpreted in many industrial applications. A general way of interpreting the above problem is to consider u as the potential function of several adjacent flows in convex rings with prescribed pressure on the free streamlines. A more interesting application, however, is related to the so-called Stefan problem, for large time. In this connection, the two phase model describes crystallization (freezing) or melting of some physical substance. Multi-phase Stefan problem refers to materials capable of assuming any of three or more different states (solid, liquid, gaseous, in particular). We expand this in more details for the two phase case. Let us consider a cylindrical container with the horizontal cut as the domain K = K 3 \ K (this is the two dimensional case). The exterior wall K 3 is kept at temperature u =, and the interior wall K at temperature u =. The container is also filled with liquid, and the temperature of the liquid is assumed to be known initially. Suppose the material (liquid) solidifies at temperature < λ <. For simplicity we take λ = 0. By continuity of the temperature for positive times, we know that there must be a curve Γ(x, t) (for each time t) on which u(x, t) = 0. Hence on the subregion {u > 0} the material is in liquid form and on the subregion {u < 0} the material is in solid form. Let us also assume that the temperature u (depending on the material) also satisfies the (nonlinear) heat equation p u D t u = 0, in K \ {u = 0}. (2) On the transition phase Γ(x, t) the Stefan condition (Bernoulli condition), which follows from the energy conservation law, gives the dynamic equation of the moving curve u = g(x, u 2, V ), (3) where u and u 2 represent the function u on {u > 0} and on {u < 0} respectively. Here V is the normal velocity of the curve Γ(x, t), and the nonlinear joining condition (3) may depend on the density of the heat source over the inter-phase boundary (due for instance to an extra super-heating). For large time, the heat flux tends to stabilize and becomes stationary. Hence u t, and V both become approximately zero. Therefore the realistic model for the stationary problem is the one given by p u = 0, in K \ {u = 0}, u = g(x, u 2, 0) on {u = 0}. (4) It is noteworthy that the p-laplace operator constitutes a subclass of a larger class of operators, appearing in many modeling problems in industrial applications, due to non-newtonian behavior of fluids. For further applications, and backgrounds in the case p = 2, we refer the reader to [A,2], and the references therein. 2

3 .3 Main result We prove existence and C regularity of the free boundary in the two-phase case. More precisely, the main result of this paper is the following: Theorem. (two phases) Let K, K 3 be two convex domains, such that K K 3 and g : (K 3 \K ) R + R + a continuous positive function, non-decreasing with respect to its second argument and satisfying some concavity property (see Definition 2.3 for a precise statement). Then there exists a convex C domain ω, K ω K 3, which is a classical solution of the two-layer free boundary problem, the latter means that the p-capacitary potentials u and u 2 of the sets ω \ K and K 3 \ ω respectively, i.e. solutions of p u = 0 in K 2 \ K u = on K u = 0 on ω with p, < p < + the p-laplace operator, satisfy, p u 2 = 0 in K 3 \ ω u 2 = on K 3 u 2 = 0 on ω lim u z x (z) = lim z ω\k y K 3\ω g(y, u 2 (y) ) x ω. Section 2 is devoted to describe the possible nonlinear joining conditions we are able to handle. In section 3, we give some useful auxiliary results. Section 4 is devoted to the proof of the main theorem, section 5 is the separation result and section 6 describes extension to the multi-phase case. 2 The nonlinear joining condition In this section, we discuss what could be the nonlinear joining condition involving u i and u i+ at the interface γ i = K i+ between the two phases. We recall that this condition is written in the general form F i (x, u i (x), u i+ (x) ) = 0, (i =,, m) (5) with F i : K R + R + R. We will always assume that F i (x, p, q) is a continuous function on K R + R +, and that F i (x, p, q) is strictly increasing regarding as a function of variable p for all x, q. This assumption and the implicit function theorem allows us to write the joining condition (5) in the following equivalent form u i (x) = g i (x, u i+ (x) ), (6) where g i : K R + R + are given functions. An important tool, used in the proof of our main theorem and due to [LS] and [A4], is the following property: if γ i contains a line segment I then x / u i (x) is a convex function while x / u i+ (x) is a concave function on I (see Lemma 3.2). This in conjunction with concavity assumption on the function x /g i (x, u i+ (x) ) underlies one of the main techniques in the proof of our main result. Therefore, the property that g i must satisfy is the following: x /g i (x, q(x)) is a concave function as soon as /q(x) is a concave function. (7) For general functions g we cannot expect to have convexity of the level sets of the solution. In fact the first author (see [A3]) obtained an example of the convex two-layer problem in the plane for which no convex solution exists corresponding to the joining condition in the form: u 2 (x) 2 u (x) 2 = λ 2 (compare with eq. (8)). Laurence and Stredulinsky ([LS]) gave an example of the convex two-layer problem with the same joining condition such that the natural variational minimizer is not convex. 3

4 It is also an open question whether there exist classical solutions at all, for general (regular) functions g. In two space dimensions this was settled by H.W. Alt and L.A. Caffarelli [AC]. So the question to be raised is what are the necessary and sufficient conditions to have existence of convex classical solutions. Our condition (7) may seem somewhat artificial, but it is the only working condition at this moment. Let us remark that a similar convexity condition was assumed in [A2] and [A5]. Example : The classical nonlinear joining condition, see e.g. [A], [LS], is given by u i (x) α u i+ (x) α = a i (x) α, (8) where α and a i (x) > 0. This joining condition satisfies the convexity condition (7) provided that the function /a i is concave, as will follow from Lemma 2. below. (Regarding applicability of other convexity conditions, we refer to [A2], Example 2.9 and [A5], Example 4.7.) Lemma 2. Let a and q be positive functions defined on R N and such that /a and /q are concave. Then, for α, the function x is concave. (a(x) α + q(x) α ) /α Proof : It is sufficient to do the proof for C functions a and q, since the result will follow for less regular functions by a simple density argument just using pointwise convergence. Let us set f(x) := (a(x) α + q(x) α ), /α which is a C function. We want to prove the following inequality: Now x, y R N ( f(x), y x) f(y) f(x). (9) [ ( f(x), y x) = a(x) α ( a(x), y x) + q(x) α ( q(x), y x) ]. (a(x) α + q(x) α ) +/α By concavity of /a we have ( ) ( (x), y x) = a a 2 ( a(x), y x) (x) a(y) a(x) A similar inequality holds for /q. Putting these in (0) yields [ ( a(x) α+ a(y) a(x) ( f(x), y x) (a(x) α + q(x) α ) +/α that is ( a(x) α+ ( f(x), y x) (a(x) α + q(x) α ) +/α a(y) (0) ) ( + q(x) α+ q(y) )] q(x) + q(x)α+ q(y) ) f(x). So, to prove (9), it remains to prove the following inequality ( ) a(x) α+ + q(x)α+. () (a(x) α + q(x) α ) +/α a(y) q(y) (a(y) α + q(y) α /α ) Let us set x = a(x) a(y), x 2 = q(x) q(y), t = a(x) α, t 2 = q(x) α. Inequality () can be rewritten x α x α 2 t + t 2 x α t + xα 2 t2 ( t x + t 2 x 2 t + t 2 ) α 4

5 or t + t 2 t x α 2 + t 2x α ( t ) α x 2 + t2 x. (2) t + t 2 Now, the mean-value inequality between harmonic and arithmetic means yields ( ) α ( ) α t + t 2 t x 2 + t 2 x t + t 2 t x 2 + t2 x and the inequality (2) follows immediately using the convexity property of the function x x α. We present now another class of functions g i which satisfy the above-mentioned property. Lemma 2.2 Assume that the function g : R 2 + R + satisfies the following set of hypothesis: (H): g is concave, (H2): g satisfies the following inequality (ξ, ξ 2, η, η 2 ) R 4 + g(ξ, η )g(ξ 2, η 2 ) g 2 ( ξ ξ 2, η η 2 ). (3) Now if a : R N R +, q : R N R + are two given functions such that /a and /q are concave, then x is concave. g(a(x), q(x)) Proof : It suffices to prove this for C -functions g, a and q, since the result will follow for less regular functions by a simple density argument just using pointwise convergence. Let us set f(x) := g(a(x), q(x)), which is a C function. We prove the following inequality (cf. (9)) Now ( f(x), y x) = g 2 (a(x), q(x)) x, y R N ( f(x), y x) f(y) f(x). (4) [ ] g g (a(x), q(x)) ( a(x), y x) + (a(x), q(x)) ( q(x), y x). ξ η (5) and by concavity assumption on /a and /q (see the proof of Lemma 2.) we ll have [ ( ) ( f(x), y x) g g 2 (a(x),q(x)) ξ (a(x), q(x)) a2 (x) a(y) a(x) + ( )] + g η (a(x), q(x)) q2 (x) q(y) q(x) Next, using concavity of g we arrive at g ξ (ξ, η )(ξ 2 ξ ) + g η (ξ, η )(η 2 η ) g(ξ 2, η 2 ) g(ξ, η ), where (ξ, η ) = (a(x), q(x)) and (ξ 2, η 2 ) = ( a2 (x) a(y), q2 (x) q(y) ). Inequality (4) now follows immediately from this and inequality (3). Examples 5

6 . The function g(ξ, η) := (ξ α + η α ) /α which corresponds to the classical nonlinear joining condition already mentioned does not fall into the above framework when α (because g is convex and not concave), but for 0 α <!. Indeed, in this case assumption (H) is easily verified by proving that the Hessian of g is negative on R 2 +. As for (H2), it follows immediately from the inequality which gives (ξ η 2 ) α + (ξ 2 η ) α 2(ξ ξ 2 η η 2 ) α/2 [(ξ ξ 2 ) α + (η η 2 ) α + (ξ η 2 ) α + (ξ 2 η ) α ] /α [(ξ ξ 2 ) α + (η η 2 ) α + 2(ξ ξ 2 η η 2 ) α/2] /α which is inequality (3). 2. The function g(ξ, η) := ξ α η β with α 0, β 0 and α + β can also be considered. Conditions (H), (H2) are readily verified in this case. 3. More generally, we can consider a function like g(ξ, η) := i I a iξ α i η β i (finite or infinite sum) with a i 0, α i 0, β i 0 and α i + β i. Assumption (H) is elementary (g is a combination of concave functions with positive coefficients) while inequality (3) is obtained from the expansion of i I i 2 I ( ) a i a i2 ξ αi /2 ξ αi 2 /2 2 η βi /2 η βi 2 /2 2 ξ αi 2 /2 ξ αi /2 2 η βi 2 /2 η βi / Definition 2.3 Define G to be the family of all functions g : K R + R +, satisfying the following conditions (A): g is continuous and c 0 > 0 such that g(x, 0) c 0 for all x K, (A2): g is non-decreasing with respect to second argument, (A3): g satisfies the following concavity property: x g(x, q(x)) given function such that /q is concave, and is concave whenever q is a (A4): for any given value y 0 > 0, there exist constants 0 < C < C 2 such that C (g(x, y)/y) C 2, uniformly for all x K and all y y 0. Hence forward we will always consider the following nonlinear joining condition with g i G. 3 Preliminary results u i (x) = g i (x, u i+ (x) ), (6) In this section we will sum up some of the auxiliary results used in this paper. We remark that the usual comparison and maximum principle for elliptic partial differential equations, is one of the basic tools here; see [T]. Lemma 3. (Exterior Barrier) Let D be a convex domain in R N and suppose u is a continuous non-negative function on B(x 0, r), p-harmonic in B(x 0, r) D, with x 0 D. Let also u = 0 on D B(x 0, r). If D is not C at x 0, i.e. D has (at least) two supporting planes at x 0, then lim u(x) = 0, x D. x x0 6

7 Lemma 3.2 (Interior Barrier) Let D be a convex domain in R N and suppose u is a continuous non-negative function on B(x 0, r), p-harmonic in B(x 0, r) \ D, with x 0 D. Let also u = 0 on D B(x 0, r). If u C 0 in B(x 0, r) \ D, then D B(x 0, r/2) is C with a uniform C norm, i.e. there exists a constant C = C (C 0, N) such that ψ(x) ψ(y) C, where ψ is a map that represents D near x 0 D. The proofs of these lemmas follow from standard theory using barriers at conical boundary points. The existence of such barriers are proven in [Do], see also [K]. Remark 3.3 By Lemmas 3., and 3.2, if C 0 u C 0 in D, D must be C with C -norm depending on C 0. Definition 3.4 (Blow-up) For the functions u j : B(x j, ) R and for a sequence of nonnegative numbers {r j } (r j 0) we define the scaled functions on B(0, /r j ) by ũ j (x) = uj (r j x + x j ) u j (x j ) r j. Obviously, if all functions u j are Lipschitz-continuous in B(x j, ) with the same constant, then ũ j are uniformly Lipschitz in B(0, R) (R < /r j ). Thus, there exists a subsequence n k, such that ũ n k converges locally in C α (R N ) to a function u 0. Moreover, if u j are p-harmonic, then so is u 0 in {u 0 > 0} and u 0 (0) = 0. Lemma 3.5 Let S(C 0 ) be the set of all C domains D R + R N, such that B(0, ) D is convex, 0 D and D C (B(0,/2)) C 0. Then any blow-up of a sequence D j S(C 0 ) converge to a half space, i.e., if r j 0 and D j S(C 0 ), then for D j := r j D j = {x : r j x D j } we have lim sup D j = T, where T = {x > 0}, and lim sup means the set of all limit points of sequences {x j } with x j D j. Lemma 3.6 Let u j be the p-capacitary potential of an annular domain D j = D 2 j \ D j with convex uniform C boundaries. Suppose moreover the gradient of u j satisfy u j (x) Λ 0 <, uniformly both in j and x D j. Then any convergence blow-up ũ rj at any boundary point gives a linear function u 0 = αx +, after suitable rotation and translation. In particular, for any boundary point x j D j u j (y + x j ) = u j (x j ) + αy + + o(r j) in B(0, r j ), in some rotated system. The proof of this lemma is just the same as the proof of Lemma 2.4 in [HS2]. The uniformity in norms are crucial. Using these lemmas, we can prove the following (cf. Theorem.3 [HS4]). Lemma 3.7 Let D and D 2 be two nested open convex domains (D D 2 ), and u denote the p-capacitary potential of D = D 2 \ D. Then for x D lim u(y) exists non-tangentially (with values in [0, ]). In particular u can be defined (with values in [0, ]) up to the boundary D as non-tangential limit. Moreover, u is upper semi-continuous up to D 2 and lower semi-continuous up to D. 7

8 Proof: Since the problem is local, depending on whether we are close to D or D 2, we may start with point x 0 D. In case u is bounded in a neighborhood of x 0 the proof was given in Theorem.3 in [HS4]. So suppose there exists a sequence x j D with lim j x j = x 0, u(x j ) c 0 x j x 0 dist(x j, D ), for some c 0 > 0, where the last condition means non-tangential approach of x j to x 0. Obviously it suffices to show that for any such sequence {x j } we have u(y j ) y j B(x j, d j ), where 8d j = dist(x j, D ). To show this, we scale the function u by u j (x) := u(d jx + x j ) u(x j ) in B 8 (0). Since we have, by Harnack s inequality, p u j = 0, u j > 0 in B 8 (0), sup u j C inf u j Cu j (0) = C. B 4 B 4 In particular, u j is a bounded sequence in B 4. Hence by standard elliptic theory, a subsequence of u j converges to a solution u 0 in B 4, satisfying p u 0 = 0, u 0 (0) =, u 0 > 0 in B 4. Moreover, the level sets of u 0 are convex, since they are convex for all u j. Now suppose u(x j ) > j. Then by uniform C,α estimates Hence Now if for some C 0 u j (0) = d j u(x j ) u(x j ) y j = d j ỹ j + x j B(x j, d j ), we have u(y j ) C for some C > 0, then u j (ỹ j ) = d j u(d j ỹ j + x j ) u(x j ) jd j u(x j ). u(x j ) jd j C 0. (7) = d j u(y j ) u(x j ) (ỹj B ), d jc u(x j ) C C 0, j where in the last inequality we have used (7). Hence it follows that u 0 (ỹ) = 0, where ỹ = lim ỹ j B, for an appropriate subsequence. To summarize, we have a positive p-harmonic function u 0 in B 4, with convex level sets, and with the further property that for some ỹ B there holds u 0 (ỹ) = 0. This contradicts the Hopf s boundary point lemma (see [T]). And the proof is completed in this case. The second case x 0 D 2 is treated similarly, with reversed argument. We sketch some details. So we may start as we did in the previous case, and assuming now u(x j ) < /j, and u(y j ) C 0 > 0, with y j as before y j = d j ỹ j + x j B(x j, d j ), (ỹj B ). 8

9 Again all the above arguments are in order and we have the limit function u 0 and the limit point ỹ in B. Let us see what more information we can deduce. Indeed, on the one side, by elliptic estimates, and on the other side C u j (ỹ j ) = d j u(y j ) u(x j ) u j (0) = d j u(x j ) u(x j ) Upon combining these estimates, we arrive at u j (0) C 0C. j d jc 0 u(x j ), d j ju(x j ). As j tends to infinity we ll have u 0 (0) = 0. And again the Hopf s principle is violated. The lower and upper semi-continuity properties follow in the same way as in the proof of Theorem.3 in [HS4]. Lemma 3.8 Let u be a solution to p u = 0 in a domain Ω, and introduce the linear elliptic operator L u defined everywhere, except at critical points of u, by Then L u ( u p ) 0 in Ω. L u := u p 2 + (p 2) u p 4 N k,l= u u 2. x k x l x k x l This lemma is essentially proved, though stated differently, in the papers of Payne and Philippin, [PP] and [PP2], see also the discussion in [HS2]. For two nested convex sets D D 2, and for x D we denote by T x,a the supporting hyperplane at x with the normal a pointing away from D. Obviously, T x,a is not necessarily unique, depending on the geometry of D. Now for each x D there corresponds a point y x (not necessarily unique) on D 2 {z : a (z x) > 0} and such that a (y x x) = max a (z x), where the maximum has been taken over all z D 2 {z : a (z x) > 0}. Lemma 3.9 Let D and D 2 be two nested convex domains (D D 2 ) and denote by u the p-capacitary potential of D 2 \ D, i.e. the solution of where c and c 2 are two given constants c > c 2 0. Then p u = 0 in D 2 \D u = c on D (8) u = c 2 on D 2 lim sup z x z D 2 \D u(z) lim sup z y x u(z) x D, (9) z D 2\D where y x is the point indicated in the discussion preceding this lemma. For a proof of this lemma see [HS2], [HS3]. Definition 3.0 (Extremal points) For a domain D R N we say a point x D is an extremal point if there exists a supporting plane to D touching D at x only. We denote the set of all extremal points of D by E D. Lemma 3. Retain the hypothesis in Lemma 3.9 and suppose also that D and D 2 are C.Then u(x) inf u(y), for all x D 2 \ D. y E D2 9

10 This lemma is a consequence of Lemma 3.9 and geometric considerations. The next lemma was an important tool in the variational existence treatment of the multilayer problem by P. Laurence and E Stredulinsky [LS]: Lemma 3.2 (See [LS], Lemma 4. and [A4], Thm..) Retain the hypothesis in Lemma 3.9. Suppose moreover D i (i =, 2) contains a line segment l i, and that u c 0 > 0. Then u is convex on l 2 and it is concave on l. 4 The two-layer problem 4. Main result Let us consider two bounded convex domains K and K 3 in R N, such that K 3 strictly contains K (i.e. K K 3 ). We look for a convex domain K 2, such that K K 2 K 3 and the p-capacitary potentials u and u 2 of the sets K 2 \ K and K 3 \ K 2 respectively, i.e. solutions of p u = 0 in K 2 \ K u = on K, u = 0 on K 2 satisfy a nonlinear joining condition like We have the following result. p u 2 = 0 in K 3 \ K 2 u 2 = on K 3 (20) u 2 = 0 on K 2 u (x) = g(x, u 2 (x) ) on K 2. (2) Theorem 4. (two phases) Let K, K 3 be two convex domains, such that K 3 strictly contains K, and g G. Then there exists a convex C domain ω, K ω K 3, which is a classical solution of the two-layer free boundary problem. The latter means that the p-capacitary potentials u and u 2 of the sets ω \ K and K 3 \ ω respectively (i.e. solutions of (20) with K 2 = ω) satisfy lim u z x (z) = lim z ω\k y K 3 \ω g(y, u 2 (y) ) x ω. (22) 4.2 Notations, definitions 4.2. p-capacitary potentials For every subdomain ω such that K ω K 3, we set ω = ω \ K and ω 2 = K 3 \ ω. We introduce the p-capacitary potentials u ω (respectively u ω 2 ) or more simply u (respectively u 2 ) when there is no possible confusion, the solutions of the boundary value problems p u = 0 in ω u = on K u = 0 on ω p u 2 = 0 in ω 2 u 2 = on K 3 u 2 = 0 on ω In the sequel, we will refer to u as the inner potential and to u 2 as the outer potential of the set ω. We want to find a domain Ω satisfying a joining condition written u (x) = g(x, u 2 (x) ) as explained in the previous subsection. For that purpose, we introduce the following classes of domains: (23) 0

11 4.2.2 Subsolutions, supersolutions An open set ω (such that K ω K 3 ) is called a subsolution (of the problem) if its p-capacitary potentials u and u 2 satisfy: lim inf z x z ω u (z) lim sup y ω 2 g(y, u 2 (y) ) x ω. (24) An open set ω (such that K ω K 3 ) is called a supersolution (of the problem) if its p-capacitary potentials u and u 2 satisfy: lim sup z x z ω u (z) lim inf y ω 2 g(y, u 2 (y) ) x ω. (25) The classes A, B, C We are going to work only with convex domains, so let us denote by C = {ω convex bounded open subset of R N, K ω K 3 }. Then, we will denote by A the class of convex subsolutions and B the class of convex supersolutions: A = {ω C : lim inf z x z ω u (z) lim sup y ω 2 g(y, u 2 (y) ) x ω }. B = {ω C : lim sup z x z ω u (z) lim inf y ω 2 g(y, u 2 (y) ) x ω }. A classical solution of the two-phase free-boundary problem is obviously a domain Ω A B. 4.3 Stability results for the class B First we show that the class B is closed under intersection. Lemma 4.2 Let ω, ω 2 be in B. Then ω ω 2 B. Proof: As the intersection of two convex domains is convex, we need to prove the condition on the gradients for u := u ω ω 2 and u 2 := u ω ω 2 2 at the boundary of ω ω 2. By comparison principle, 0 u min(u ω, u ω2 ), which implies that for x (ω ω 2 ) ω ω 2 we have (for example, we choose the case where x ω ): u (x) = u ω (x) = 0 and lim sup y (ω ω 2 ) while, since u 2 min(u ω 2, u ω2 2 ) 0, we have u 2 (x) = u ω 2 (x) = 0 and lim inf y (ω ω 2 ) 2 u (y) lim sup y ω u 2 (y) lim inf y ω 2 u ω (y) u ω 2 (y). Now, by monotonicity of g with respect to its second argument, and the fact that ω belongs to B: lim sup u (y) lim sup u ω (y) y (ω ω 2 ) y ω lim inf y ω 2 g(y, u 2 (y) ) lim inf g(y, u 2 (y) ) y (ω ω 2 ) 2

12 Now, the technical and more difficult point is to prove that B is stable, in some sense, for decreasing sequences of convex domains. Indeed, our aim is to construct a solution to the free boundary problem, by taking a minimal element (for inclusion) in the class B. So, we need some stability of B by the constructing process that we are going to use. Theorem 4.3 Let ω ω 2, be a decreasing sequence of convex domains in B, and suppose ω = ω n (the interior of the closure) belongs to C. Then ω B. Proof: Since the domains involved are convex and they all contain K, they are uniformly Lipschitz. In particular, by standard up to boundary regularity (see [K]), the p-capacitary potentials u n, u n 2 are C α (α depending on the uniform cone property of ω n ) in the entire space R N (after appropriate extension). Since also u n, u n 2 are decreasing sequences we have a limit functions u, u 2 which are the p-capacitary potentials of ω = ω \ K, and ω 2 = K 3 \ ω, respectively. Moreover, by local C,α regularity (see [Le]), convergence takes place also for the gradients on every compact subset of ω and ω 2 respectively. We need to show ω B. Let ɛ > 0 be small enough and fix x 0 K. Now for each y ω let us denote by R(x 0, y) the ray emanating from x 0 and traveling through y. Then, by the choice of x 0, and the convexity of the sets ω, ω n we can choose unique points x = x(y, n) ω n R(x 0, y) and x ɛ {u n 2 = δ ɛ } R(x 0, y), where δ ɛ > 0 is to be chosen later. It follows that lim x(y, n) = y, lim n ɛ 0 xɛ = x(y, n) non-tangentially. where Next denote by v n the solution of the following boundary-value problem: L u n (v n ) = 0 in {0 < u n < 2 } v n (x) = u n (x) p on {u n = 2 }, v n (x) = G n,ɛ (x) on ω n G n,ɛ (x) = min (2M p, g p (x, u n 2 (x ɛ ) + ɛ)), with M = sup {0<u n < 2 } u n, and L u n is defined in Lemma 3.8. Observe that the boundedness of M follows by simple (linear) barrier argument. Fix a point y ω. Then two possibilities may arise (see Lemma 3.7) Case ) u 2 (y) = non-tangentially, Case 2) u 2 (y) = M non-tangentially. In both cases we ll have u n 2 (x ɛ ) u n 2 (x). In Case we obtain g p (x, u n 2 (x ɛ ) + ɛ) > 2M p u n (x) p, i.e., G n,ɛ (x) u n (x) p. In Case 2 we have (by non-tangential continuity of u n 2 ) u n 2 (x) u n 2 (x ɛ ) + ɛ, provided δ ɛ is small enough. And by non-decreasing property of g we have g p (x, u n 2 (x ɛ ) + ɛ) g p (x, u n 2 (x) ) u n (x) p, Hence G n,ɛ (x) u n (x) p. 2

13 Therefore upon applying the comparison principle (for the operator L u n ; see Lemma 3.8) we can obtain v n (x) u n (x) p in {0 < u n < /2}. Now as n, v(x) := lim n v n (x) u (x) p in {0 < u < /2}. Since x ɛ is compactly inside ω 2 and u n 2 (x ɛ ) u 2 (x ɛ ) in C α -norm (see [Le]) we have a uniform convergence for to v n ω n = G n,ɛ (x) = min (2M p, g p (x, u n 2 (x ɛ ) + ɛ) ), Therefore for z B(x, r ɛ ) ω and x ω G ɛ (x) = min (2M p, g p (x, u 2 (x ɛ ) + ɛ) ). u p (z) v(z) G ɛ (x) + ɛ g p (x, u 2 (x ɛ ) + ɛ) + ɛ, provided r ɛ is small enough. By Lemma 3.7, and continuity of g (as ɛ 0) we get lim sup z x z ω u (z) lim inf g(y, u 2 (y) ). y ω 2 Hence ω B. 4.4 Proof of Theorem 4. st step Existence of subsolutions and supersolutions. Let us consider the solution u of the boundary value problem (p-capacitary potential) p u = 0 in K 3 \ K u = on K u = on R N \ K 3 (26) For any < α <, let ω α = {u(x) > α}. Also define u,α (x) = ((u(x) α)/( α)) in the closure of ω,α = {α < u(x) < } and u 2,α (x) = ((u(x) α)/( + α)) in the closure of ω 2,α = { < u(x) < α}. Then ω α is a supersolution (resp. subsolution) if u(x) α = u,α(x) < (>)g(x, u 2,α (x) ) = g(x, u(x) + α ) for all x ω α. But a comparison argument involving the p-capacitary potential in any slab between parallel boundary planes tangent to ω α and K 3 shows that u(x) ((α+)/m) (( + α)/r) for all x ω α, where M = sup{dist(x, ω α ) : x K 3 }. Therefore, ω α is a supersolution (resp. subsolution) provided that + α y) < (>)g(x, α y for all x ω α and all y y 0 = (/R). Applying Assumption (A4), we see that ω α is a supersolution if (( + α)/( α)) C (true for α sufficiently close to ), and that ω α is a subsolution if (( + α)/( α)) C 2 (true for α sufficiently close to ). We remark that K and K 3 are regular, so that u(x) is both uniformly bounded and uniformly positive in K 3 \K, then the above argument obtains supersolutions and subsolutions without involking Assumption (A4) (one can replace it by the much weaker assumption that g(x, y) as y uniformly over x K). In the sequel Ω 0 will denote a given subsolution and Ω a given supersolution. 3

14 2nd step Construction of a minimal element in the class B. We introduce the class S := {ω B, with Ω 0 ω Ω }. Let I be the intersection of all domains in the class S and set Ω = I (the interior of the closure, which is still convex). To prove Ω B, we select a sequence of domains {ω n } n= in S such that ω n = I and we n consider the sequence of domains {Ω n } n= defined by Ω = ω and Ω n+ = Ω n ω n+ (n ). By Lemma 4.2 each Ω n is convex and belongs to B. Hence, since Ω i+ Ω i, Theorem 4.3 gives the desired result. 3rd step On E Ω, extremal points of Ω, we have lim sup u (z) = lim inf g(y, u 2 (y) ). This property can be proved in the same way as in [HS2], but since it is slightly more complicated and for sake of completeness, we give here the complete proof. Suppose the property fails. Then, there exists X 0 E Ω such that lim sup z X 0 z Ω u (z) = lim inf y X 0 g(y, u 2 (y) )( 4α), with α > 0. (27) y Ω 2 and We denote by l = lim sup z X 0 z Ω l 2 = lim inf y X 0 y Ω 2 u (z) u 2 (y). Note that since g is continuous and non decreasing, lim inf g(y, u 2 (y) ) = g(x 0, l 2 ). Therefore y X 0 assumption (27) can be written as l = g(x 0, l 2 )( 4α). (28) We assume first l 2 < +. Hence for some small neighborhood V of X 0 there holds and u (z) l ( + α) z V Ω (29) u 2 (y) l 2 ( α) y V Ω 2. (30) Let us fix a hyperplane T d, parallel to a supporting plane at X 0, with dist(x 0, T d ) = d and such that T d Ω V. This is possible due to the extremal property of X 0. By rotation and translation, we assume X 0 is the origin and T d = {x = d}. Let now T δ = {x = δ} and set Ω δ = Ω \ {x δ}. Then by comparison principle the (inner) p- capacitary potential u δ of Ω δ satisfies while the (outer) potential satisfies which implies that on points x belonging to Ω Ω δ : lim sup y Ω δ Now by (29) and (3) u δ (y) lim sup y Ω 0 u δ u in Ω δ, (3) u δ 2 u 2 0 in Ω 2, (32) u (y) lim inf y Ω 2 g(y, u 2 (y) ) lim inf g(y, u δ 2(y) ). y Ω δ 2 max u δ max u d sup u l ( + α)d. (33) T d T d {0 x d} 4

15 Define v := u δ + l ( + α)d d δ (d x ). Since the second derivatives of v and u δ coincide, we have L u δ v = L u δ u δ = 0 in Ω δ {x < d}. Therefore in Ω δ { x < d}, v takes its maximum on the boundary. By inspection and (33), it is easy to see that on (Ω δ { x < d}) T d T δ ( Ω {δ < x < d}), v l ( + α)d, with equality on T δ. Hence v x 0 on T δ, i.e., u δ l ( + α)d d δ on T δ. (34) Now, it remains to estimate u δ 2 on T δ. For that purpose, let us introduce a part of a level set of u δ 2 contained in the neighborhood V Ω 2 and consider, on that level set, one point, say x δ where the supporting hyperplane is parallel to T δ. By lemma 3.9, we have y T δ u δ 2(y) u δ 2(x δ ). (35) Now, by continuity of g, we can choose ε and δ small enough such that y T δ g(x 0, l 2 ) g(y, l 2 ( ε))( + α). (36) Now, by uniform convergence of u δ 2 to u 2 when δ 0 on the level set, we can choose δ small enough such that u δ 2(y) u δ 2(x δ ) l 2 ( ε). Replacing in (36) and using(28), (34) and the monotonicity of g yields y T δ u δ (y) g(y, uδ 2(y) )( + α) 2 ( 4α)d d δ Now, it suffices to choose δ even smaller so that ( + α)2 ( 4α)d, which in turn implies d δ Ω δ B. Since Ω δ Ω we have reached a contradiction. Now, if l 2 = +, we can choose the neighborhood V in such a way that u 2 (y) 2M y V Ω 2 where M = sup x Ω u (x). Then, we reach a contradiction exactly in the same way, by choosing δ small enough such that Ω δ will be in the class B. 4th step The boundary of Ω is C. It suffices to show that at each boundary point there exists a unique tangent plane. Suppose the latter fails. Let x 0 Ω, with two supporting planes Π, Π 2 at x 0. Then by barrier arguments (Lemma ) lim u (y) = 0 and lim u 2(z) = +. Ω y Ω Π Π 2. Ω 2 z Ω Π Π 2. Let Π 3 be a third plane supporting Ω at x 0 and such that Π 3 Ω Π Π 2, i.e., Π 3 does not touch any other boundary points of Ω than those on the intersection of the planes Π and Π 2. Now, move Π 3 towards the interior of Ω such that it cuts off Ω a small cap; it may well be a tub-like region. Then a similar argument as in the previous step will prove that this new domain is still in the class B. This contradicts the minimal property of Ω.. 5

16 5th step The nonlinear joining condition holds on E Ω. Let x E Ω be fixed. On one hand, we have the following chain of (in)-equalities (here n.t. means non-tangentially, see Lemma 3.7 for details): u (x) := lim z x n.t. z Ω lim inf y Ω 2 u (z) lim sup z x u z Ω (z) g(y, u 2 (y) ) lim n.t. y Ω 2 g(y, u 2 (y) ) := g(x, u 2 (x) ) (37) where the first and last equalities are due to Lemma 3.7, the second and fourth inequalities come from the definition of a liminf and limsup (we also use the continuity of g) and the third inequality comes from the fact that Ω belongs to the class B. On the other hand, we have the following chain of (in)-equalities: u (x) lim sup u (z) = lim inf g(y, u z x 2 (y) ) g(x, u 2 (x) ) (38) z Ω y Ω 2 where the first inequality is the upper semi-continuity of u at x, the equality is step 3 and the second inequality is the lower semi-continuity of u 2 at x. Now, (37) and (38) together give the desired result. 6th step The nonlinear joining condition holds at every boundary point. According to step 5, it remains to prove the equality u (x) = g(x, u 2 (x) ) on maximal line segments in I = [a, b] Ω. For any such line segment one readily verifies that a, b E Ω. Also at the points a, b we have equation (38) verified. In view of assumption (A3) for the function g in conjunction with Lemma 3.2 we claim that the function x u (x) g(x, u 2 (x) ) is convex, non-negative. The latter depends on the fact that Ω belongs to the class B and it vanishes at the extremities of any segment (by step 5 and n.t.-continuity). Therefore, this function vanishes identically. This completes the proof. 5 Uniform separation estimate Theorem 5. (compare to [A2], Lemma 4.4) Let H denote the set of all configurations (K, ω, K 3 ) such that K, ω, K 3 are convex, B ρ (0) K ω K 3 B R (0), and ω is a supersolution relative to K and K 3. Then there exists a value η > 0 such that uniformly for all (K, ω, K 3 ) H. dist( K, ω) ηdist( K, K 3 ) (39) This result follows directly from lemmas 5.2 and 5.3, which follow. Lemma 5.2 For any (K, ω, K 3 ) H, let α = max{u(x) : x ω} (, ), where u solves the Dirichlet problem (26). Then there exists a value α 0 (, ) such that α α 0 uniformly over all (K, ω, K 3 ) H. Proof: It suffices to consider only configurations in H such that α (0, ). Given such a configuration (and the corresponding value α),let u, u 2, ω, ω 2 be as defined in (23). Define the p-harmonic functions u,α (x) = ((u(x) α)/( α)) and u 2,α (x) = ((u(x) α)/( + α)), 6

17 both in the closure of the set Ω := K 3 \ K. Then u = u,α = on K and u,α 0 = u on ω. It follows by the comparison principle for p-harmonic functions that u,α u in ω. Similarly, we have u 2 = u 2,α = on K 3 and u 2,α 0 = u 2 on ω, from which it follows by the comparison principle that u 2,α u 2 in ω 2. We choose a point x 0 ω such that u(x 0 ) = α. Clearly the function u is regular near x 0 Ω (and therefore so are u,α and u 2,α ). For small δ > 0, let x δ = x 0 + δν 0 ω, where ν 0 denotes the unit vector with direction opposite u(x 0 ). Also let γ δ ω denote the directed line-segment of length δ joining x 0 to x δ. Clearly and ( u (x)/ ν 0 ) u (x) sup x γ δ u (x) ( u,α (x)/ ν 0 ) u,α (x 0 ) z(δ), both on γ δ, where z(δ) 0 as δ 0. Therefore 0 u (x δ ) u,α (x δ ) = ( / ν 0 )(u (x) u,α (x))ds γ δ from which it follows that ( sup x γ δ u (x) u,α (x 0 ) + z(δ))δ, lim sup u (x) u,α (x 0 ) = u(x 0) ω x x 0 α For small δ > 0, Let γ δ denote a directed arc of steepest ascent of u 2 of length δ, joining a point x δ ω 2 to the point x 0. Since u 2 (x)/ ν = u 2 (x) on γ δ, where ν denotes the forward unit tangent vector to the arc, we have 0 u 2,α (x δ ) u 2 (x δ ) = ( / ν)(u 2 (x) u 2,α (x))ds γ δ ( u 2 (x) u 2,α (x) )ds, γ δ from which it follows that inf u 2 (x) u 2,α (x 0 ) + z(δ), x γ δ and therefore that lim inf u 2 (x) u 2,α (x 0 ) = u(x 0) ω 2 x x 0 + α. (4) In view of the definition of an outer solution (see ), it follows from (40) and (4) that (40) ( u(x 0 ) /( α)) g(x 0, ( u(x 0 ) /( + α))). (42) A simple comparison argument involving the p-capacitary potential in a slab bounded by parallel planes, one tangent to the surface {u(x) = α} at x 0, the other tangent to K 3, shows that u(x 0 ) ((α + )/M) ((α + )/R), where M = sup x K3 dist(x, {u(x) = α}). It follows from (42) and Assumption (A4) that (/( α)) (( + α)/( α)) (g(x 0, y)/y) C 2, (43) where we set y = ( u(x 0 ) /(+α)) y 0 = (/R), and where C 2 depends only on R, y 0, and the function g. The assertion follows, since (43) cannot be satisfied unless α α 0 = ( (/C 2 )). Lemma 5.3 In the context of Lemma 5.2, there is a constant η > 0 such that dist( K, {u(x) = α 0 }) ηdist( K, K 3 ) for any convex sets K, K 3 such that B ρ (0) K K 3 B R (0). 7

18 Proof: For any r (0, ] and unit vector ν, let E(r, ν) = {x R N : dist(x, D(r, ν)) < r}, where λ = (ρ 2 /R) and D(r, ν) denotes the closure of the convex hull of the set {0} B λr ( rρν). Let u r,ν (x) denote the p-harmonic function in the annular domain Ω(r, ν) = E(r, ν)\ D(r, ν) whose continuous extension to the closure satisfies u r,ν ( D(r, ν)) =, u r,ν ( E(r, ν)) =. Then dist(0, {u r,ν (x) = α 0 }) = rη, where η = dist(0, {u,ν (x) = α 0 }) > 0, since u r,ν (x) = u,ν (x/r). For r = min{, dist( K, K 3 )} and any point x 0 K, we have x 0 + D(r, (x 0 / x 0 )) K and x 0 + E(r, (x 0 / x 0 )) K 3. By the comparison principle, we have u(x) u r,ν (x x 0 ) in Ω (x 0 + Ω r,ν ), where ν = (x 0 / x 0 ). It follows that B rη (x 0 ) K {u(x) < α 0 } for all x 0 K, from which the assertion follows. 6 The multi-layer case Let us recall the problem. We are given two strictly nested convex domains K K m+2, real numbers λ i, (i =, 2,, m + ) with λ i > λ i+, and continuous functions g i : (K m+2 \ K ) R + R + (i = 2,, m + ). We are looking for a sequence of nested convex domains K K 2 K m+ K m+2 such that the p-capacitary potentials u i (x) of the sets K i+ \ K i, i.e. solutions of p u i = 0 in K i+ \ K i u i = λ i on K i u i = λ i+ on K i+, satisfy the following joining conditions: u i (x) = g i (x, u i+ (x) ) on K i+ (i =,, m). For simplicity we set λ = and λ m+ =. The following is our main result in this paper. Theorem 6. (multi-layer) Let K, K m+2 be two bounded convex domains, such that K m+2 strictly contains K, λ i (, ), i = 2,, m+ are arbitrary real numbers with λ i > λ i+, and g i G, i =,, m. Then there exists a sequence of convex C domains {K i : < i < m + 2}, such that K K 2 K m+ K m+2, and which is a classical solution of the multilayer free boundary problem. The latter means that the p-capacitary potentials u i of the sets K i+ \ K i, i =,, m, i.e. solutions of (44) satisfy lim u z x i (z) = lim g(y, u i+ (y) ) x K i+ i =,, m. (45) z K i+\k i y K i+2 \K i+ (44) Definitions: We let B denote the family of all ordered m -tuples ω := (ω 2, ω 3,, ω m+ ), such that K ω 2 ω m+ K m+2, every ω i is a convex domain and each domain ω i is a supersolution of the two-layer problem relative to ω i and ω i+. The latter means that for every i =,, m, we have limsup u i (z) liminf g z x i (y, u i+ (y) ) x ω i+, (46) z ω i+ \ω i y ω i+2 \ω i+ where we take ω := K, ω m+2 := K m+2 and define u i to be the solution or (44) with K i replaced by ω i for each i = 2,, m +. st step: modified 2-layer existence result 8

19 In view of Theorem 5., the proof of Theorem 4. actually proves the following result for the 2-layer problem: (a)there exists a unique absolute minimizer among all outer solutions. (b) This absolute minimizer solves the 2-layer problem (in the same sense as in Theorem 4.). 2nd step: B is not empty. Under our assumptions, an outer (m )-surface outer solution can be easily obtained in the form: ω i = {u(x) > α i }, i = 2,, m +, where u denotes the solution of (26) with K 3 replaced by K m+2, and where the values α i are appropriately chosen so that each ω i is a supersolution relative to its neighbors ω i and ω i+ (same argument as in the first step in the proof of Theorem 4.). 3rd step Uniform separation in B Let ω := (ω 2,, ω m+ ) B. Then for each i = 2,, m +, ω i is a supersolution of the 2-layer problem relative to ω i, ω i+, and the function g i. Thus, by Theorem 5., we have dist( ω i, ω i ) ηdist( ω i, ω i+ ) ηdist( ω i, ω i+ ) for all i = 2,, m +. It follows that dist( ω i, ω i ) η m+2 i dist( ω m+, K m+2 ) η m dist( ω m+, K m+2 ) for all i = 2,, m +. Thus, if ω n, n =, 2,, is a weakly decreasing sequence of elements of B (so that the corresponding sequence of (m )-st components is also weakly decreasing and thus uniformly bounded away from K m+2 ), then there exists a value δ > 0 such that for all n =, 2,, the surface components of ω n are separated from each other (and from K and K m+2 by a distance of at least δ. Therefore the componentwise intersection has the same property. 4th step pairwise intersection; minimal sequence in B. B is closed under the operation of componentwise intersection. In fact, given ω, ω 2 B, let ω = ω ω 2 be the componentwise intersection. Then ω ω ω 2, and it is easy to see (using the standard comparison principle) that u ω u ωj, j =, 2, componentwise in the common domains of the component p-capacitary potentials. By repeated application of componentwise intersections, one defines a (componentwise) weakly decreasing minimal sequence of supersolutions ω n = (ω n 2,, ω n m+), n =, 2,, where the latter means that for any i = 2,, m and any x R N such that x ω i for some supersolution ω B, we have x ω n i for all sufficiently large n. 5th step minimal element in B For each fixed i = 2,, m +, the sequence of domains ωi n, n =, 2, 3,, is weakly decreasing under set inclusion and therefore convergent to a domain Ω i K (Ω i := the interior of the infinite intersection of the domains ωi n, n =, 2, 3, ). Clearly the domains Ω i are strictly ordered by inclusion, and in fact by step 3, we have dist( Ω i, Ω i+ ) δ for all i =,, m + (where we set Ω := K and Ω m+2 := K m+2 ). Since Ω i ωi n for all i, each ωi n, i = 2,, m +, is actually a supersolution of the 2-layer problem relative to Ω i, Ω i+, and g i. Therefore, Ω i (the interior of the infinite intersection of the ωi n ) is also a supersolution of the same 2-layer problem, due to Theorem 4.3. Therefore Ω B. In fact Ω is, by construction, the minimal supersolution in B. 6th step Ω solves the multi-layer problem Since Ω is a minimal element in B, each component Ω i of Ω must be the minimal supersolution of the 2-layer problem relative Ω i, Ω i+, and g i. Therefore, by step, Ω i is a solution of this 2-layer problem in the sense of Theorem 4.. Thus Theorem 6. is proved. Acknowledgement : The authors want to thank the referees whose suggestions improved significantly the writing of this paper. 9

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