A SINGULAR PERTURBATION PROBLEM FOR THE p-laplace OPERATOR

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1 A SINGULAR PERTURBATION PROBLEM FOR THE -LAPLACE OPERATOR D. DANIELLI, A. PETROSYAN, AND H. SHAHGHOLIAN Abstract. In this aer we initiate the study of the nonlinear one hase singular erturbation roblem div( u ε 2 u ε ) = β ε (u ε ), (1 < < ) in a domain of R N. We rove uniform Lischitz regularity of uniformly bounded solutions. Once this is done we can ass to the limit to obtain a solution to the stationary case of a combustion roblem with a nonlinearity of ower tye. The case = 2 has been considered earlier by several authors. 1. Introduction Our objective in this aer is to study the singular erturbation roblem (P ε ) u ε = β ε (u ε ), u ε 0 in a domain of R N. Here, for 1 < <, denotes the -Lalace oerator, i.e. u = div( u 2 u). We recall that a solution to (P ε ) is a function u W 1, () L () such that (1.1) u ε 2 u ε ϕ dx = ϕβ ε (u ε ) dx for all ϕ C 0 (). We require β ε to be Li(R) and to satisfy (1.2) 0 β ε A ε χ (0,ε) and ε 0 β ε (s)ds = M for ositive constants A and M. In articular, these conditions are fulfilled when the functions β ε are constructed from a single nonnegative Lischitz function β suorted in [0, 1] setting (1.3) β ε (s) = 1 ( s ) ε β. ε Although our analysis alies to a general tye of oerators, as the ones considered in [To] of the form div(a(x, u, u)), for simlicity and clarity of the arguments we focus on the secific form of the -Lalacian. The motivation of the study in this aer comes from the alications to the one-hase case of the combustion roblem, aearing in the descrition of laminar flames as an asymtotic limit for high activation energy, that corresonds to the limit as ε 0 in (P ε ). For the case = 2 there is an extensive study of the roblem 2000 Mathematics Subject Classification. Primary 35R35, 35J60. H. Shahgholian was suorted by the Swedish Natural Science Research Council. This work started when all three authors were attending the secial PDE year at the Mittag-Leffler institute in The authors also thank Luis Caffarelli for valuable suggestions. 1

2 2 D. DANIELLI, A. PETROSYAN, AND H. SHAHGHOLIAN and more or less a comlete resolution of it; see [BCN], [LW] for the ellitic case and [CV], [Ca], [CLW1], [CLW2], [CK], [D] for the arabolic one. However, the nonlinear case, addressed here, has never been considered earlier. This might artly deend on the lack of an established theory for the -Lalace oerator, and artly on the fact that some of the earlier techniques fail in the absence of linearity. We show that, in a sense, the limit of (P ε ) as ε 0 is a free boundary roblem { u = 0 in {u > 0} (P ) u = c on {u > 0} with c = ( 1 M)1/. Namely, the main result of this aer (Theorem 4.3 in Section 4) asserts that the uniform limits u of u ε have the asymtotic develoment ( ) 1/ u(x) = 1 M x x 0, η + + o( x x 0 ), near x 0 {u > 0}, rovided {u > 0} admits a measure theoretic normal and u is not degenerate at x 0 (see Definitions ). The free boundary roblem (P ) for the -Lalacian was studied earlier under certain geometric (convexity) assumtions, by different techniques; see e.g. [AM] by Acker and Meyer and a series of aers [HS1] [HS3] by Henrot and Shahgholian. To rove the main theorem (Theorem 4.3) we need a uniform bound (Theorem 2.1) for the gradient of the solutions, in order to have some stability of the roblem as one asses to the limit. This tye of uniform bounds of the gradient usually constitutes the basics of the analysis to follow, and it is by no means an obvious generalization of earlier results. Indeed, it needs to be ointed out that one of the main difficulties in the consideration of oerators that do not admit linearization, as it is for the -Lalacian, aears in the deduction of the uniform gradient bound, which has its own indeendent interest. In this art of our analysis we aly techniques that have been recently develoed for related free boundary roblems, see [KKPS] and [CKS]. 2. The uniform gradient bound for solutions In this section we rove that solutions to u ε of the singular erturbation roblem (P ε ) are locally uniformly Lischitz. Our main theorem in this section is the following. Theorem 2.1. Let u ε be a nonnegative solution of (P ε ) in a domain of R N with β ε satisfying (1.2) and such that u ε L () L. Then for every comact K there is a constant C = C(N,, A, L, dist(k, )) indeendent of ε such that u ε L (K) C. It is also noteworthy that as far as the roof of Theorem 2.1 goes, one can relax the conditions on β ε. An imortant observation is that the same technique to follow shows that in the case of two-hase roblems (see [CLW1]-[CLW2]) one may deduce gradient bound for the non-negative art of the solution if one already knows that the negative art of the solution is Lischitz. In [Ca], L. Caffarelli alied this idea in combination with the monotonicity formulas to deduce gradient bound for the solution of the two-hase singular erturbation roblem for the Lalacian; see also

3 A SINGULAR PERTURBATION PROBLEM 3 [CK]. In the absence of the monotonicity formula we are not able to rove a similar result as that in [Ca]. It is aarent that some new technique is to be develoed to handle the sign change in the case of the -Lalacian or any other nonlinear case. This remains an oen and tantalizing roblem. The roof of Theorem 2.1 will be based on the following lemma. Lemma 2.2. Let v be a bounded nonnegative solution of 0 v Aχ {0<v<1} in the unit ball B 1 of R N with v(0) 1. Then there is a constant C = C(N,, A) such that v L (B 1/4 ) C. Remark 2.3. We exlicitely observe that v C 1,α () for some α > 0, thanks to the results in [To]. In the case when v = β(v) with β(s) c 0 s +c 1 s 1 for some constants c 0 and c 1, the conclusion of the Lemma 2.2 follows directly from Serrin s Harnack inequality for nonhomogeneous quasilinear oerators, see [Se]. Our roof, however, uses only Harnack inequality for homogeneous oerators and is based on comactness rather than energy methods, which allows to generalize it to a broad range of oerators. Proof. Indeed, assume the contrary. Then there exists a sequence of functions {v k }, k = 1, 2,..., satisfying the assumtions of the lemma and such that Consider the sets max v k (x) > 4 B 1/4 3 k. k = {x B 1 : v k (x) > 1} and Γ k = k B 1. Note that v k is -harmonic in k. Let now and define δ k (x) = dist(x, B 1 \ k ) O k = {x B 1 : δ k (x) 1 3 (1 x )} B 1 \ k. Observe that B 1/4 O k. In articular m k := su O k (1 x )v k (x) 3 4 max B 1/4 v k (x) > k. Since v k (x) is bounded (for fixed k), we will have (1 x )v k (x) 0 as x 1, and therefore m k will be attained at some oint x k O k : (2.1) (1 x k )v k (x k ) = max(1 x )v k (x). O k Clearly, v k (x k ) = m k 1 x k m k > k. Since x k O k, by the definition we will have (2.2) δ k := δ k (x k ) 1 3 (1 x k ).

4 4 D. DANIELLI, A. PETROSYAN, AND H. SHAHGHOLIAN Let now y k Γ k be a oint where δ k = dist(x k, Γ k ) is realized, so that (2.3) y k x k = δ k. Then we will have two inclusions both consequences of (2.2)-(2.3). inequality holds B 2δk (y k ) B 1 and B δk /2(y k ) O k, In articular, for z B δk /2(y k ) the following (1 z ) (1 x k ) x k z (1 x k ) 3 2 δ k 1 2 (1 x k ). This, in conjunction with (2.1), imlies that max v k 2v k (x k ). B δk /2(y k ) Next, since B δk (x k ) k, v k satisfies v k = 0 in B δk (x k ). By the Harnack inequality for -harmonic functions there is a constant c = c(n, ) > 0 such that In articular, Further, define min v k cv k (x k ). B 3δk /4(x k ) max v k cv k (x k ). B δk /4(y k ) w k (x) = v k(y k + δ k x) for x B 2. v k (x k ) Summarizing the roerties of v k above, we see that w k satisfies the following system 0 w k A(δ k ) /k 1 in B 2 max B1/2 w k 2, max B1/4 w k c > 0 w k 0, w k (0) 1/k. Therefore, from a riori estimates, we can conclude that a subsequence of {w k } will converge in C 1,α norm on every comact subset of B 1/2 to a function w 0 that satisfies w 0 = 0 in B 1/2 max B1/4 w 0 c > 0 w 0 0, w 0 (0) = 0. This, however, contradicts the strong maximum rincile for -harmonic functions. The lemma is roved. Proof of Theorem 2.1. We start with the observation that it is enough to rove the theorem in the case when = B 1, K = B 1/8, and under the assumtion Denote u ε (0) = ε and u ε L (B 1 ) = 1. ε = {x B 1 : u ε > ε} and Γ ε = ε B 1. Ste 1. Prove that there is a constant C = C(N,, A) such that (2.4) u ε (x) C for x B 1/2 \ ε.

5 A SINGULAR PERTURBATION PROBLEM 5 Indeed, take a oint x 0 B 1/2 with u ε (x 0 ) ε and consider a function v ε (x) = uε (x 0 + εx). ε Direct comutation shows that v ε satisfies 0 v ε Aχ {0<v ε <1}. All the assumtions of Lemma 2.2 are thus fulfilled for v = v ε and we can conclude (2.5) max B 1/4 v ε C(N,, A). From interior gradient estimates we obtain u ε (x 0 ) = v ε (0) C 1 (N,, A) max B 1/4 v ε C 2 (N,, A) for all x 0 B 1/2 \ ε. Hence (2.4) is roved. Ste 2. Prove that (2.6) u ε (x) ε + C(N,, A) dist(x, B 1 \ ε ) for x B 1/4 ε. Indeed, for x 0 B 1/4 ε denote m 0 = u ε (x 0 ) ε and δ 0 = dist(x 0, B 1 \ ε ) Notice that, since 0 Γ ε, δ 0 1/4. We want to rove that (2.7) m 0 C(N,, A)δ 0. Since B δ0 (x 0 ) is contained in ε, u ε ε will be nonnegative and -harmonic there. We can thus aly the Harnack inequality to conclude min B δ0 /2(x 0 ) (uε (x) ε) c 1 m 0 for c 1 = c 1 (N, ) > 0. Next, consider the -caacitary otential ϕ(x) of the ring B 1 \ B 1/2 which satisfies ϕ(x) = 0 in B 1 \ B 1/2, ϕ B1 = 0 and ϕ B1/2 = 1. The function ϕ will be sherically symmetric with ϕ = c 0 = c 0 (N, ) > 0 on B 1. Define ( ) x x0 ψ(x) = c 1 m 0 ϕ for x B δ0 (x 0 ) \ B δ0 /2(x 0 ). δ 0 From the comarison rincile for -harmonic functions we will have (2.8) ψ(x) u ε (x) ε for x B δ0 (x 0 ) \ B δ0 /2(x 0 ). Take y 0 B δ0 (x 0 ) Γ ε. Then y 0 B 1/2, and (2.9) ψ(y 0 ) = u ε (y 0 ) ε = 0. We infer from (2.8), (2.9), and (2.4) that ψ(y 0 ) u ε (y 0 ) c 2 (N,, A). Observe now that ψ(y 0 ) = c 1 m 0 c 0 /δ 0 and therefore we obtain m 0 c 2 c 0 c 1 δ 0. Thus, inequalities (2.7) and (2.6) are roved.

6 6 D. DANIELLI, A. PETROSYAN, AND H. SHAHGHOLIAN Ste 3. Prove that (2.10) u ε (x) C(n,, A) for x B 1/8 ε. Indeed, let x 0 B 1/8 ε, δ 0 = dist(x 0, B 1 \ ε ) and define w(x) = uε (x 0 + δ 0 x) ε δ 0 for x B 1. Then, from the inclusion B δ0 (x 0 ) ε and inequality (2.6) we will have 0 w C(N,, A) in B 1. Since also w is -harmonic in B 1, from the interior gradient estimates we obtain u ε (x 0 ) = w(0) C 1 (N,, A), which roves (2.10). Now the theorem follows from (2.4) and (2.10). 3. Passage to the limit: ε j 0 From now on we will assume that functions β ε in (P ε ) satisfy (1.2) (1.3). This section embodies the main technical tools that one needs to establish the main theorem (Theorem 4.3.) Lemma 3.1. Let {u ε } be an uniformly bounded family of solutions to (P ε ). Then for every sequence ε j 0 there exists a subsequence ε j 0 and u Li() such that: i) u ε j u uniformly on comact subsets of ; ii) u = 0 in \ {u > 0}; iii) u ε j u in L loc (). Proof. Part (i) follows by Theorem 2.1 and standard comactness argument. Let now E {u > 0} be oen. Then u c > 0 in E. By the uniform convergence, we will have u ε j > c/2 in E for small ε j. Hence, if also ε j < c/2, uε j will be -harmonic in E. This imlies that u is -harmonic in E and since E was arbitrary, ii) follows. Finally, we rove iii). Let ψ be a nonnegative C0 () function, and δ > 0. Take (u δ) + ψ as a test function. Since u = 0 in the ositivity set of u, integrating by arts, we obtain u ψ = u 2 u ψ u + δ u 2 u ψ. {u>δ} {u>δ} {u>δ} Letting δ 0 we find (3.1) {u>0} u ψ = u 2 u ψ u. {u>0} On the other hand, the observation β ε (u ε )u ε 0 yields (3.2) u ε ψ u ε 2 u ε ψ u ε.

7 A SINGULAR PERTURBATION PROBLEM 7 Using the uniform convergence of u ε to u and the weak convergence of u ε 2 u ε to u 2 u in L (3.3) 1 loc (), we infer from (3.1) and (3.2) that lim su j u ε j ψ u ψ. Since u ε j u in L loc (), we have (3.4) u ψ lim inf u ε j ψ. j It follows from (3.3), (3.4), and a simle comactness argument that u ε j u in L loc (). The conclusion of art iii) is roved, and so is the lemma. We now rove that limit solutions are solutions to the free boundary roblem in a very weak sense. Proosition 3.2. Let {u ε j } be a family of solutions to (P εj ). Assume that u ε j u uniformly on comact subsets of as ε j 0. Then there exists a locally finite measure µ suorted on the free boundary {u > 0} such that β εj (u ε j ) µ in. In articular, u = µ in, i.e., (3.5) u 2 u ϕ dx = ϕ dµ for all ϕ C 0 (). Proof. By definition of weak solutions to (P ε ), if ϕ C0 (), one has (3.6) u ε 2 u ε ϕ = β ε (u ε )ϕ. Since u ε j u uniformly on comact subsets of, by Lemma 3.1 we know u ε j u in L loc (), and so the left-hand side of (3.6) converges to the left-hand side of (3.5). Now let F be comact, and take ϕ C0 (), ϕ 0, ϕ 1 in F. The sequence { β ε j (u ε j )ϕdx } is convergent, and therefore it is bounded. Hence β εj (u ε j ) dx β εj (u ε j )ϕ dx C(ϕ). F This imlies that there exists a locally finite measure µ such that, assing to a subsequence (still denoted by ε j ) if necessary, β εj (u ε j ) µ as measures in. Passing to the limit in (3.6), we get (3.5). Moreover, since u = 0 in \ {u > 0} by Lemma 3.1, we conclude that µ is suorted in {u > 0}. The roof is thus comlete. Lemma 3.3. Let {u ε j } be a family of solutions to (P εj ) in such that u ε j u uniformly on comact subsets of and ε j 0 as j. Let x 0, x n {u > 0} be such that x n x 0 as n. Let λ n 0, u λn (x) = 1 λ n u(x n + λ n x), and (u ε j ) λn (x) = 1 λ n u ε j (x n + λ n x). Suose that u λn U as n uniformly on comact sets of R N. Then, there exists j(n) such that for every j n j(n) there holds that ε jn /λ n 0 and i) (u ε jn )λn U uniformly on comact sets of R N ; ii) (u ε jn )λn U in L loc (RN ); iii) u λn U in L loc (RN ).

8 8 D. DANIELLI, A. PETROSYAN, AND H. SHAHGHOLIAN Proof. The roof is along the lines of the one of Lemma 3.2 in [CLW1]. We discuss here only the relevant modifications. For simlicity we assume x n = x 0. Proceeding as in the cited reference, one can show that i) holds. The functions (u ε jn )λn are solutions to (u ε jn )λn = β εj n /λ n ((uε jn )λn ) in B k, where k is a fixed ositive number. By Lemma 3.1 there exists a subsequence, still denoted by j n, such that (u ε jn )λn U in L (B k ). Then also ii) holds. In order to rove iii), let δ > 0 and consider u λn U u λn (u ε j ) λn + (u ε j ) λn U = I + II, where all the norms are in L (B k ). By ii) we already know that II < δ if j j n and n is sufficiently large. Moreover, by virtue of Lemma 3.1 it holds I = u u ε j (x 0 + λ n x) dx B k = 1 λ N u u ε j (x) dx < δ n B λ nk(x 0 ) if j and n are sufficiently large. This roves iii). We now turn our attention to the secial case when the limit function u is onedimensional. Proosition 3.4. Let x 0, and let u ε k be solutions to u ε k = β εk (u ε k ) in. If u ε k converge to α(x x 0 ) + 1 uniformly on comact subsets of, with α R, and ε k 0 as k, then ( ) 1 0 α 1 M. Proof. Without loss of generality, we assume x 0 = 0. Since u ε k 0, we readily have α 0. Next, let ψ C0 (). Choosing u ε k x1 ψ as a test function in the weak formulation of u ε k = β εk (u ε k ) (see Remark 3.5 below) and integrating by arts we obtain 1 u ε k ψ x1 + u ε k 2 u ε k (3.7) x1 u εk ψ = B εk (u ε k )ψ x1. Here, B εk (s) = s 0 β ε k (τ) dτ. Since 0 B εk (s) M, there exists M(x) L (), 0 M(x) M, such that on a subsequence (still denoted by ε k ) B εk (u ε k ) M(x) -weakly in L (). If y {x 1 > 0}, then u ε k αy 1 /2 in a neighborhood of y for k sufficiently large. Hence, if u ε k (x) ε k we have, by (1.2), B εk (u ε k )(x) = u ε k (x)/εk 0 β(s) ds = M. Moreover, using Proosition 3.2 it is immediate to recognize that B εk (u ε k ) = β εk (u ε k ) u ε k 0

9 A SINGULAR PERTURBATION PROBLEM 9 in L 1 loc ( {x 1 < 0}). Hence M(x) = M [0, M] in {x 1 < 0}. Passing to the limit in (3.7) yields 1 α ψ x1 dx = M ψ x1 dx + M ψ x1 dx, {x 1 >0} and integrating by arts we find 1 α ψ dx = M {x 1 =0} {x 1 >0} {x 1 =0} ψ dx M {x 1 <0} {x 1 =0} ψ dx. The arbitrariness of ψ C0 () allows to conclude 1 α = M M M, because M 0. Hence α 1 M, and the roof is comlete. Remark 3.5. We recall that the weak solution u of the equation u = f in with bounded f has a reresentative in W 2,2 2, loc () if 2 < and in Wloc () for 1 < 2, see e.g. [To]. This, in conjunction with local L bounds on u ε k in, justifies the integration by arts in the roof of Proosition 3.4 above. Proosition 3.6. Let x 0, and let u ε k u ε k = β εk (u ε k ) be solutions to in. If u ε k converge to α(x x 0 ) γ(x x 0) 1 uniformly on comact subsets of, with α, γ > 0 and ε k 0 as k, then ( ) 1 α = γ 1 M. Proof. Without loss of generality we assume x 0 = 0. As in Proosition 3.4, u ε k satisfies (3.7), and it is immediate to recognize that B εk (u ε k ) M in L 1 loc (). Passing to the limit in (3.7), and integrating by arts in the resulting equation, we find that α = γ. ( ) 1 Now we assume that α > 1 M and show that this leads to a contradiction. Ste 1. Let R 2 := {x = (x 1, x ) R N : x 1 < 2, x < 2}. Without loss of generality, we may assume R 2. First of all, we construct a family {v ε j } of solutions to (P εj ) in R 2 with the roerty (3.8) v ε j (x 1, x ) = v ε j ( x 1, x ) in R 2 and such that v ε j u uniformly on comact subsets of R 2, where u(x) = α x 1. To this end, we let b εj = su R2 u ε j u and v ε j be the minimal solution (i.e. minimum of all suersolutions) to (P εj ) in R 2 with boundary values v ε j = u b εj on R 2. By virtue of Lemma 3.1, there exists v Li loc (R 2 ) such that, on a subsequence, v ε j v uniformly on comact subsets of R 2. From the minimality of v ε j it is immediate to recognize that u v. To rove the reverse inequality, we consider w C 2 (R), satisfying and let ( w 2 w ) = β(w) in R, w(0) = 1, w (0) = α ( w ε x1 j (x 1 ) = ε j w b ) ε j + s. ε j γε j

10 10 D. DANIELLI, A. PETROSYAN, AND H. SHAHGHOLIAN Here s < 0 is a constant, determined as in [CLW1, Proosition 5.3], such that { 1 + αs, s 0, w(s) = γ(s s), s s. Here α and γ are related by α γ = ( 1 M and therefore γ > 0 under the ) 1 assumtion α > 1 M. Moreover, we have w (x 1 ) γ for all x 1 R. Observe also that w ε j u b εj = v ε j on R 2. From Lemma 3.7 below, it follows that w ε j v ε j in R 2. Let us oint out that such an imlication is not immediate since, in general, there is no comarison rincile for β. As a consequence, u v in R 2 {x 1 > 0}. Using (3.8), finally, we see that u v in R 2, and the construction is comlete. Ste 2. Let R + = {x : 0 < x 1 < 1, x < 1}. Then using the weak formulation of (P εj ) in R + we have ( ) 1 E j = R x + 1 vε j dx = v ε j 2 v εj v ε j x 1 dx = R + = div( v ε j 2 v ε j v ε j x 1 )dx β εj (v ε j )v ε j x 1 dx = F j G j. R + R + Using the divergence theorem and that v ε j x 1 (0, x ) = 0 (from symmetry in the x 1 variable) we find that F j = v ε j 2 (v ε j x 1 ) 2 dx + v ε j 2 v ε j n v ε j x 1 ds, R + {x 1 =1} R + { x =1} where v ε j n is the exterior normal derivative of v ε j on R + { x = 1}. From the convergence v ε j u = αx αx 1 in R 2 and Lemma 3.1 it follows (at least for a subsequence) that v ε j x 1 αe 1 ointwise a.e. in R + 2 = R 2 {x 1 > 0}. Since v ε j are uniformly bounded, from the dominated convergence theorem we deduce that (3.9) lim F j = α dx. j On the other hand E j + G j = R + R + {x 1 =1} ( ) 1 x 1 vε j + B εj (v ε j ) dx ( ) 1 vε j + B εj (v ε j ) dx. R + {x 1 =1} Using again that v ε j u = αx αx 1 uniformly on comact subsets of R 2, we have v ε j α uniformly and B εj (v ε j ) = M on R + {x 1 = 1}, and therefore ( ) 1 (3.10) lim su(e j + G j ) j α + M dx. Combining (3.9) and (3.10) we obtain R + {x 1 =1} α 1 α + M

11 A SINGULAR PERTURBATION PROBLEM 11 or equivalently ( ) 1 α 1 M, which is a contradiction. The roof is thus comlete. Lemma 3.7. Let w ε (x 1 ) be strictly increasing solution of ( w ε 2 w ε ) = β ε (w ε ) on R and v ε (x) be solution of v ε = β ε (v ε ) in R = {x = (x 1, x ) : a < x 1 < b, x < r}, continuous u to R. Then the following comarison rincile holds: if v ε (x) w ε (x 1 ) for all x R then v ε (x) w ε (x 1 ) for all x R. Proof. Without loss of generality we assume that w ε (0) = 0. Since w ε is strictly increasing, we can find τ > max{ a, b } such that For η > 0 sufficiently small define w ε (x 1 τ) < v ε (x) on R. w ε,η (x 1 ) := w ε (ϕ η (x 1 c η )), where ϕ η (s) = s + ηs 2 and c η > 0 is the smallest constant such that ϕ η (s c η ) s on [ 2τ, 2τ]. By the construction we readily have w ε,η w ε on [ 2τ, 2τ] and the straightforward comutation shows that w ε,η = (ϕ η) ( w ε )(ϕ η ) + ((w ε ) (ϕ η )) 1 ϕ η > (ϕ η) β ε (w ε,η ) β ε (w ε,η ) on [ 2τ, 2τ]. The first inequality follows from the observation that ϕ η > 0 on [ 2τ, 2τ] (for small η > 0) and the second inequality follows from the fact that x 1 c η imlies β ε (w ε,η ) = 0 and that for x 1 c η we have ϕ η 1. Summarizing the construction above, we see that on the interval [ 2τ, 2τ] and for small η > 0 the function w ε,η is strictly increasing, satisfies w ε,η > β ε (w ε,η ), and w ε,η w ε. Moreover, as η 0, w ε,η converges uniformly to w ε. Let now τ 0 be the smallest constant with the roerty w ε,η (x 1 τ ) v ε (x) on R. Evidently, τ < τ, and in fact, we claim that τ = 0. Indeed, the minimality of τ imlies that there is a oint x R such that w ε,η (x 1 τ ) = v ε (x ). If τ > 0, we have w ε,η ( τ ) < w ε,η w ε v ε on R, and hence x is an interior oint of R. At this oint we observe that the gradient of w ε,η (x 1 τ ) is non-degenerate. We can thus aly the strong comarison rincile for the -Lalacian to obtain a contradiction, since at x we have w ε,η (x 1 τ ) > v ε (x ). This shows that τ = 0 and in articular that w ε,η v ε on R. Letting η 0, we conclude the roof of the lemma. 4. Asymtotic Behavior of Limit Solutions In this section we rove the asymtotic develoment of solutions to (P ε ). We begin with the relevant definitions. Definition 4.1. A unit vector η R N is said to be the inward unit normal in the measure theoretic sense to the free boundary {u > 0} at a oint x 0 {u > 0} if 1 (4.1) lim r 0 r N χ {u>0} χ {x x x0,η >0} dx = 0. B r (x 0 )

12 12 D. DANIELLI, A. PETROSYAN, AND H. SHAHGHOLIAN Definition 4.2. Let v be a continuous function in a domain R N. We say that v is non-degenerate at a oint x 0 {v = 0} if there exist c, r 0 > 0 such that 1 r N v dx cr for any r (0, r 0 ). B r (x 0 ) The main result in this section is the following. Theorem 4.3. Let u ε j be solutions to (P εj ) in a domain R N such that u ε j u uniformly on comact subsets of and ε j 0. Let x 0 {u > 0} be such that {u > 0} has an inward unit normal η in the measure theoretic sense at x 0, and suose that u is non-degenerate at x 0. Under these assumtions, we have ( ) 1/ u(x) = 1 M x x 0, η + + o( x x 0 ). The roof of Theorem 4.3 relies heavily on the following result. Theorem 4.4. Let u ε j be a solution to (P εj ) in a domain R N such that u ε j u uniformly on comact subsets of as ε j 0. Then ( ) 1/ lim su u(x) x x 0 1 M. Proof. Let α = lim su x x0 u(x). Since u Li(), clearly α <. If α = 0 there is nothing to rove, so we may assume α > 0. There exists a sequence x n x 0 such that u(x n ) α and u(x n ) > 0. Let z n {u > 0} be such that d n = z n x n = dist(x n, {u > 0}). Define u dn (x) = (1/d n )u(z n +d n x). Since u Li() and u dn (0) = 0 for every n, {u dn } is uniformly bounded on comact subsets of R N, and therefore for a subsequence (still denoted by d n ) u dn u 0 uniformly on comact subsets of R N, where u 0 Li(R N ). Now, set x n = (x n z n )/d n B 1. We may choose the subsequence d n so that x n x B 1. Then u 0 is -harmonic and nonnegative in B 1 (x). Consider now the sequence ν n = u d n (x n ) u dn (x n ) = u(x n) u(x n ). Passing to a subsequence, we assume, without loss of generality, that ν n e 1. At this oint we observe that B 2/3 (x) B 1 (x n ) for n sufficiently large, and therefore u dn is -harmonic in B 2/3 (x). By interior gradient estimates, u dn u 0 in C 1,σ (B 1/2 (x)) for some σ > 0. This suffices to show that u dn u 0 uniformly in B 1/3 (x) and thus, as a consequence, u(x n ) x1 u 0 (x). In articular, x1 u 0 (x) = α. Next, it is easy to show that u 0 α in R N. Indeed, let R > 1, δ > 0. Then there exists τ 0 > 0 such that u(x) α + δ for any x B τ0 R(x 0 ). Observe now that x n x 0 < τ 0 R/2 and d n < τ 0 /2 imly B dn R(x n ) B τ0 R(x 0 ) and therefore u dn (x) α + δ in B R for n large enough. In articular, u dn u 0 -weakly in L (B R ) and thus u 0 α+δ in B R. Since δ and R are arbitrary, we conclude u 0 α in R N. Let w = x1 u 0, which is a weak solution to the equation xj (θ ij ( u 0 ) xi w) = 0 in B 1 (x),

13 A SINGULAR PERTURBATION PROBLEM 13 with θ ij (ξ) = ( 2) ξ 4 ξ i ξ j + δ ij ξ 2 ξ ij. We also know that w α in B 1 (x), w(x) = α. By the maximum rincile (recall that x1 u 0 (x) = α > 0), w α in B 1 (x) and so u 0 (x) = α(x 1 y 1 ) in B 1 (x) for some y R N. It is not difficult to recognize that u 0 (x) = α(x 1 y 1 ) in {x 1 y 1 }. We now aly Lemma A.1 from the Aendix to u 0 in {x 1 y 1 < 0} and obtain u 0 (x) = γ(x y) 1 + o( x y ) in {x 1 y 1 < 0} for some γ 0. Define, for λ > 0, (u 0 ) λ (x) = (1/λ)u 0 (λx + y). There exist a sequence λ n 0 and a function u 00 Li(R N ) such that (u 0 ) λn u 00 uniformly on comact sets of R N. We have u 00 (x) = αx γx 1 in RN. By Lemma 3.3, there exists a sequence ε 0 j 0 such that uε0 j is a solution to (P ε 0 j ) and u ε0 j u 0 uniformly on comact subsets of B 1. Alying Lemma 3.3 again (the second blowu ) we find a sequence ε 00 j 0 and solutions u ε00 j to (P ε 00 ) converging j uniformly on comact subsets of B 1 to u 00 (x) = αx γx 1. Finally, we may aly Proosition 3.4 or Proosition 3.6, deending on whether γ = 0 or γ > 0, to ( ) conclude α 1/. 1 M We are now ready to rove Theorem 4.3. Proof of Theorem 4.3. Without loss of generality we may assume x 0 = 0 and η = e 1. Define, for λ > 0, u λ (x) = 1 λ u(λx), and let ρ > 0 be such that B ρ. Since u λ Li(B ρ/λ ) uniformly in λ, and u λ (0) = 0, there exist a subsequence λ j 0 and a function U Li(R N ) such that u λj U uniformly on comact subsets of R N. From Lemmas 3.1 and 3.3 it follows that u λ is -harmonic in its ositivity set {u λ > 0}. Next, rescaling (4.1) we see that, for every fixed k > 0 {u λ > 0} {x 1 < 0} B k 0 as λ 0. Hence, U is nonnegative in {x 1 > 0}, -harmonic in {U > 0}, and vanishes on {x 1 0}. By Lemma A.1, there exists α 0 such that U(x) = αx o( x ) in {x 1 > 0}. By virtue of Lemma 3.3, we can find a sequence ε j 0 and solutions uε j to (Pε j ) such that u ε j U uniformly on comact sets of R N as j. On the other hand, if we define U λ (x) = (1/λ)U(λx), U λ αx + 1 uniformly on comact subsets of R N as λ 0. Alying Lemma 3.3 the second time, we find now a sequence σ j 0 and solutions u σ j to (P σj ) such that u σ j αx + 1 uniformly on comact subsets of R N, and (4.2) u σ j αχ {x1 >0}e 1 in L loc (RN ) by virtue of Lemma 3.1. Further, we roceed as in the roof of Proosition 3.4. Let ψ C0 (R N ) and choose u σ j x 1 ψ as a test function in the weak formulation of u σ j = β σj (u σ j ). Then we find that B σj (u σ j ) Mχ {x1 >0}+Mχ {x1 <0} -weakly in L (), and ( 1 )α =

14 14 D. DANIELLI, A. PETROSYAN, AND H. SHAHGHOLIAN M M. We now claim that either M = 0 or M = M. To this end, let K {x 1 < 0}. Then for any ε > 0 there exists 0 < δ < 1 such that K {ε < B σj (u σ j ) < M ε} K {δ < u σ j /σ j < 1 δ} K {β σj (u σ j ) a/σ j } 0 as j, where a = inf [δ,1 δ] β > 0, since β σj (u σ j ) 0 in L 1 (K) by virtue of Proosition 3.2. At this oint we observe that, thanks to a simle comactness argument, the latter fact also imlies that the convergence of B σj (u σ j ) to M is actually in L 1 loc ({x 1 < 0}). We may thus conclude K {ε < M < M ε} = 0 for every ε > 0 and K {x 1 < 0}. Hence, the claim is roved. Let us see now that α > 0. By virtue of the non-degeneracy assumtion on u at 0, for every r > 0 and j sufficiently large, 1 r N and assing to the limit as j, 1 r N B r u λj cr, B r U cr. Clearly, this forces α > 0, and as a consequence M = 0 and α = have thus shown that ( ) 1/ (4.3) U(x) = 1 M x1 + o( x ), x 1 > 0; 0, x 1 0. Next, it follows from Theorem 4.4 that ( ) 1/ U 1 M in R N. ( 1 M ) 1/. We At this oint it suffices to observe that U 0 on {x 1 = 0} to conclude that ( ) 1/ U 1 M x1 in {x 1 > 0}. Alying Hof s boundary rincile we see that necessarily ( ) 1/ U(x) = 1 M x 1 in {x 1 > 0}. As an immediate consequence we finally obtain and the roof is comlete. u(x) = ( ) 1/ 1 M x o( x ), Aendix A Here we rove an analogue of Corollary A.1 in [CLW1] on asymtotic develoment of -harmonic functions near flat boundary oints by a slight modification of the roof of Lemma A.1 in [CLW1].

15 A SINGULAR PERTURBATION PROBLEM 15 Lemma A.1. Let U Li(B + 1 ) and assume that U is nonnegative in B 1 +, - harmonic in {U > 0} and vanishes on B 1 + {x 1 = 0}. Then in B 1 +, U has the asymtotic develoment U(x) = αx 1 + o( x ) with α 0. Proof. Let l k := inf{l : U(x) lx 1 in B + 2 k }. Since l k is a nonincreasing sequence of nonnegative finite numbers, there exists Then α = lim k l k. U(x) αx 1 + o( x ) in B + 1. If α = 0, the conclusion of the lemma will follow. Assume therefore that α > 0. Then there exists a sequence x k B 1 + with r k = x k 0 such that U(x k ) αx k 1 δ 0 r k, for some δ 0 > 0. Note that x k will belong to the cone C = { x (α/δ 0 )x 1 }, and thus we can assume that ν k := x k /r k converges to some ν 0 B 1 + C. Next, let U k (x) := U(r kx) r k. Since {U k } are uniformly Lischitz, we may assume that U k converges uniformly on B + 1 to a nonnegative function V. Then from the construction we will have and in addition V (x) αx 1 δ 0 2 V (x) αx 1 in B + 1 and U k (x) l k x 1 δ 0 2 on B + 1 B ε(ν 0 ) for a sufficiently small ε > 0 and large k. Let now w be a -harmonic function in B 1 + with smooth boundary values, such that w = x 1 on B 1 + \ B ε/2(ν 0 ), w = x 1 δ 0 4α on B+ 1 B ε/4(ν 0 ), x 1 δ 0 4α w x 1 on B + 1 B ε/2(ν 0 ). Then w vanishes on the flat boundary {x 1 = 0} B 1, is C 1,σ u to {x 1 = 0} B 1, and by the Hof boundary rincile w(x) (1 µ)x 1 in B + γ for some small µ, γ. Now, from the comarison rincile we will have U k l k w in B 1 + for large k, and consequently U k l k (1 µ)x 1 in B γ +. This imlies α (1 µ)α, which contradicts the assumtion that α > 0. The lemma is roved.

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