ENERGY ESTIMATES FOR A CLASS OF SEMILINEAR ELLIPTIC EQUATIONS ON HALF EUCLIDEAN BALLS. = h(u), B + 3

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1 ENERGY ESTIMTES FOR CLSS OF SEMILINER ELLIPTIC EQUTIONS ON HLF EUCLIDEN BLLS YING GUO ND LEI ZHNG BSTRCT. For a class of semi-linear elliptic equations with critical Sobolev exponents and boundary conditions, we prove point-wise estimates for blowup solutions and energy estimates. special case of this class of equations is a locally defined prescribing scalar curvature and mean curvature type equation. 1. INTRODUCTION In this article we consider u = g(u), B + 3, (1.1) u = h(u), B + 3 x Rn +, n where u > 0 is a positive continuous solution, B + 3 is the upper half ball centered at the origin with radius 3, g is a continuous function on (0, ) and h is locally Hölder continuous on (0, ). If g(s) = s n n+ n and h(s) = cs n, the equation (1.1) is a typical prescribing curvature equation. If we use δ to represent the Euclidean metric, then u n 4 δ is conformal to δ. Equation (1.1) in this special case means the scalar curvature under the new metric is 4(n 1)/(n ) and the boundary mean curvature under the new metric is n c. Equation (1.1) is very closely related to the well nown Yamabe problem and the boundary Yamabe problem. For g and h we assume GH 0 : g is a continuous function on (0, ), h is Hölder continuous on (0, ). and Let GH 1 : { lim s g(s)s n n+ is non-increasing, lims g(s)s n+ n (0, ). s n n h(s) is non-decreasing and lims s n n h(s) <. (1.) c h := lim s n n h(s). s Date: November 4, Mathematics Subject Classification. 35J15, 35J60. Key words and phrases. semi-linear, Harnac inequality, blowup solutions, mean curvature, scalar curvature, Yamabe type equations, energy estimate. 1

2 YING GUO ND LEI ZHNG Then if c h > 0 we assume GH : If c h 0 our assumption on g,h is GH 3 : sup g(s)s 1 <, and sup s 1 h(s) <. 0<s 1 0<s 1 sup g(s) <, and sup h(s) <. 0<s 1 0<s 1 The main result of this article is concerned with the case c h > 0: Theorem 1.1. Let u > 0 be a solution of (1.1) where g and h satisfy GH 0 and GH 1. Suppose c h > 0 and GH also holds, then (1.3) u + u n n C, for some C > 0 that depends only on g, h and n. Obviously if (1.4) g(s) = c 1 s n+ n, c1 > 0 B + 1 and h(s) = c h s n n, ch > 0, g and h satisfy the assumptions in Theorem 1.1. The energy estimate (1.3) for this special case has been proved by Li-Zhang [9]. It is easy to see that the assumptions of g and h in Theorem 1.1 include a much large class of functions. For example, for any non-increasing function c 1 (s) satisfying lim s c 1 (s) > 0 and lim s 0+ c 1 (s)s n 4 <, g(s) = c1 (s)s n n+ satisfies the assumptions of g. Similarly h(s) = c (s)s n n for a nondecreasing function c (s) with lim s c (s) = c h and lim s 0+ c (s) s n <, satisfies the requirement of h in Theorem 1.1. For the case c h 0 we have Theorem 1.. Let u > 0 be a solution of (1.1) where g and h satisfy GH 0 and GH 1. Suppose c h 0 and g,h satisfy GH 3, then the energy estimate (1.3) holds for C depending only on g, h and n. If we allow lim s s n n+ g(s) = 0, then the energy estimate (1.3) may not hold. For example, let g(s) = 1 4 (s + 1) 3, then g satisfies the assumption in Theorem 1. except that lim s s n n+ g(s) = 0. Let u j (x) = x 1 + j 1, it is easy to verify that u j satisfies u j = g(u j ) in B + 3, u j x n = 0, on B + 3 Rn +. Note that h = 0 in this case. Then clearly (1.3) does not hold for u j. The energy estimate (1.3) is closely related to the following Harnac type inequality: (1.5) (min B + 1 u)(maxu) C, B + which was proved by Li-Zhang [9] for the special case (1.4). Li-Zhang [9] also proved the (1.3) for (1.4) using (1.5) in their argument in a nontrivial way.

3 ENERGY ESTIMTE 3 In the past two decades Harnac type inequalities similar to (1.5) have played an important role in blowup analysis for semilinear elliptic equations with critical Sobolev exponents. Pioneer wors in this respect can be found in Schoen [13], Schoen-Zhang [14], Chen-Lin [], and further extensive results can be found in [3, 5, 7, 9, 10, 1, 14, 16] and the references therein. Usually for a semi-linear equation without boundary condition, for example the conformal scalar curvature equation u + K(x)u n+ n = 0, B3, a Harnac inequality of the type (max B 1 u )( minu ) C B immediately leads to the energy estimate u + u n n C B 1 by the Green s representation theorem and integration by parts. However, when the boundary condition as in (1.1) appears, using the Harnac inequality (1.5) to derive (1.3) is much more involved. In order to derive energy estimate (1.3) and pointwise estimates for blow up solutions, Li and Zhang prove the following results in [9]: Theorem (Li-Zhang). Let u > 0 be a solution of (1.1) where g and h satisfy GH 0, GH 1 and GH 3. Suppose in addition max B + 1, then 1 (max B + 1 u)(minu) C. B + Here we note that in Theorem no sign of c h is specified. One would expect the energy estimate (1.3) to follow directly from Li-Zhang s theorem. This is indeed the case if c h 0. However for c h > 0 substantially more estimates are needed in order to establish a precise point-wise estimate for blowup solutions. s a matter of fact we need to assume (GH ) instead of (GH 3 ) in order to obtain (1.3). The organization of this article is as follows. In section two we prove Theorem 1.1. The idea of the proof is as follows. First we use a selection process to locate regions in which the bubbling solutions loo lie global solutions. Then we consider the interaction of the bubbling regions. Using delicate blowup analysis and Pohozaev identity we prove that bubbling regions must be a positive distance apart. Then the energy estimate (1.3) follows. Even thought the main idea we use to prove Theorem 1.1 is similar to Li-Zhang s proof of the special case g(s) = s n n+,h(s) = ch s n n, there are a lot of technical difficulties for the more general case. For example Li-Zhang s proof relies heavily on the fact that the equation is invariant under scaling, thus they don t need any classification theorem in their moving sphere argument. However in the more general case the equation is not scaling invariant any more and we have to use the classification theorem of Caffarelli-Gidas-Spruc or Li-Zhu. In section three we prove Theorem 1. using Theorem and integration by parts.

4 4 YING GUO ND LEI ZHNG. PROOF OF THEOREM 1.1 The proof of Theorem 1.1 is by way of contradiction. Suppose there is no energy bound, then there exists a sequence u such that (.1) u B1 n + u n. We claim that max B + 3/ u. Indeed, if this is not the case, which means there is a uniform bound for u on B + 3/, we just tae a cut-off function η C such that η 1 on B + 1 and η 0 on B+ 3/ \ B+ 1 and η C. Multiplying the equation (1.1) by u η, using integration by parts and simple Cauchy s inequality we obtain a uniform bound of B 1 u, a contradiction to (.1). Definition.1. Let {u } be a sequence of solutions of (1.1). ssume that x x B + and lim u (x ) =. If there exist C > 0 (independent of ) and r > 0 (independent of ) such that u (x) x x n C for x x r, then we say that x is an isolated blow-up point of {u }. Proposition.1. Let {u } be a sequence of solutions of (1.1) and maxu (x) x n as, x B + 1 then there exist a sequence of local maximum points x such that along a subsequence (still denoted as {u }) v (y) := u ( x ) 1 u (u ( x ) n y + x ) either converges uniformly over all compact subsets of R n to V that satisfies (.) V + V n+ n = 0, (where = lim s g(s)s n+ n ), or converges to V1 defined on {y R n ; y n > T } ( T := lim u ( x ) n xn ) that satisfies { V (.3) 1 + V n+ n 1 = 0, R n {y n > T }, n V 1 / y n = c h V n 1, y n = T. where c h = lim s s n n h(s). Remar.1. ll solutions of (.) are described by the classification theorem of Caffarelli-Gidas-Spruc [1]. ll solutions of (.3) are described by Li-Zhu [11]. Proof of Proposition.1: Let x B + 1 be a sequence such that R n u (x ) x n as. Let d = x and S (y) = u (y)(d y x ) n, y B + 1. Set S ( x ) = max B(x,d ) {t> x n } S.

5 Then ENERGY ESTIMTE 5 (.4) S ( x ) S (0) = u (x ) x n. Let σ = 1 (d x x ), then clearly (.4) can be written as (.5) u ( x ) n For all x B + σ ( x ), since we have Using x ˆx σ, and we obtain n σ u (x )d n as. u (x)(d x x ) n u ( x )(d x x ) n, u (x) u ( x ) ( ) n d x x. d x x d x x d x ˆx x ˆx σ, (.6) u (x) n u ( x ), f or all x B + σ ( x ). Let M = u ( x ) and v (y) = M 1 u (M n y + x ), M n y + ˆx B + 3. Direct computation shows (.7) v (y) + (M v (y)) n+ n g(m v (y)) v (y) n+ n = 0 By (.6) we have (.8) 0 v (y) n y B(0,M n σ ) {y n M n xn } We consider the following two cases. Case one: lim M n xn =. n Since M n σ and M ˆx n both tend to infinity, (.7) is defined on y l for some l. By (.8) we assume that v is bounded above in B l. We claim that v V uniformly over all compact subsets of R n and V satisfies (.) with = lim s s n n+ g(s). Indeed, we claim that for any R > 1, (.9) v (y) C(R) > 0, y R. Once (.9) is established, we see clearly that M V over all B R, thus M n+ n g(m v ) = (M v ) n n+ n g(m v )v n+ V n+ n over all compact subsets of R n. Then it is easy to see that V solves (.). Therefore we only need to establish (.9) for fixed R > 1. Let and Ω R, := {y B R ; v (y) 3M 1 } a (y) = M n+ n g(m v )/v.

6 6 YING GUO ND LEI ZHNG It follows from (GH 1 ) that in B R \ Ω R, n a (y) g(3)v 4 4g(3). For y Ω R, we use (GH ) to obtain a (y) CM 4 n, y Ω R,. In either case a (y) is a bounded function. From v (y) + a (y)v (y) = 0 and standard Harnac inequality we have in B R 1 = v (0) max B R/ v C(R)min B R/ v. Thus (.9) is established. Consequently V, as the limit of v indeed solves (.). By the classification theorem of Caffarelli-Gidas-Spruc [1] ( ) n n(n ) V (y) = ( ) n µ µ y Obviously V has a maximum point x R n. Correspondingly there exists a sequence of local maximum points of u, denoted x, that tends to x after scaling. Thus v can be defined as in the statement of Proposition.1. Case two: lim M n ˆx n <. In this case we let T = lim n M x n. It is easy to verify that v satisfies v (y) + (M v (y)) n+ n n g(m v (y))v n+ (y) = 0, in M n y + ˆx B + 3, v y n = (M v (y)) n n h(m v (y))v (y) n n v (y), on y n = M ˆx n. We claim that for any R > 1, there exists C(R) > 0 such that (.10) v (y) C(R) in B R {y n M n ˆx n }. n The proof of (.10) is similar to the interior case. Let T = M x n and p = (0, T ). On B(p,R) {y n T } we write the equation for v as v + a v = 0, in B(p,R) {y n > T }, (.11) n v + b v = 0, on B(p,R) {y n = T }. where it is easy to use GH to prove that a + b C for some C independent of and R. By a classical Harnac inequality with boundary terms (for example, see Lemma 6. of [15] or Han-Li [6]), we have 1 = v (0) max v C(R) min v. B(p,R/) {y n T } B(p,R/) {y n T }

7 ENERGY ESTIMTE 7 Therefore v is bounded below by positive constants over all compact subsets. Thus the limit function V 1 solves (.3). By Li-Zhu s classification theorem [11], V 1 (y) = ( ( n(n ) ) n 4 λ 1 + λ ( y ȳ + y n ȳ n ) ) n, where ȳ n = c h (n )n//((n )λ), ȳ R n 1, λ is determined by V 1 (0) = 1. Thus the local maximum of V 1 can be used to defined v as in the statement of the proposition. Proposition.1 is established. Proposition.1 determines the first point in the blowup set Σ. The other points in Σ can be determined as follows: Consider the maximum of S (x) = u (x) n dist(x,σ ). If S (x) is uniformly bounded we stop. Otherwise the same selection process we get another blowup profile by either the classification theorem of Caffarelli-Gidas- Spruc or Li-Zhu. Eventually we have {q i } Σ (i = 1,,..,) that satisfy B + (q ri i ) B+ (q r j ) = /0, f or i j, j q i q j n u (q j), f or j > i, u (x) n dist(x,σ ) C. and ri are chosen so that in B + (q ri i ), the profile of u is either lie an entire standard bubble described in (.) or a part of the bubble described in (.3). Tae any q Σ, let σ = dist(q,σ \ {q }) and we let where Ω := {y; ũ (y) = σ n u (q + σ y), in Ω q + σ y B + 3 }. By the selection process we have (.1) ũ (y) C y n, y 3/4,y Ω and (.13) ũ (0). We further prove in the following proposition that ũ decays lie a harmonic function: Proposition.. (.14) ũ (0)ũ (y) y n C, for y B /3 Ω. Remar.. The meaning of Proposition. is each isolated blowup point is also isolated simple.

8 8 YING GUO ND LEI ZHNG Proof. Direct computation shows that ũ satisfies (.15) ũ (y) + σ n+ g(σ n ũ ) = 0, in Ω, n ũ (y) = σ n h(σ n ũ ), on Ω {y n = σ 1 q n }, Let M = ũ (0). By (.13) M. Set v (z) = M 1 ũ ( M n z), for z Ω, where n Ω := {z; z M, M n z Ω } Note that v is defined on a bigger set, but for the proof of Proposition. we only need to consider the part in Ω. Direct computation gives (.16) v (z) + l n+ n g(l v ) = 0, z Ω, v z n = l n n h(l v ), {z n = T } Ω. where l = σ n n M and T = l q n. We consider two cases. Case one: T. By the same argument as in the proof of Proposition.1, we now v V in C 1,α loc (Rn ) where V solves (.). Thus there exist R such that v V C 1,α (B R ) CR 1. Clearly (.14) holds for z M n R, we just need to prove (.14) for z > M n R. Since V (0) = 1 is the maximum point of V, V (z) = (1 + n(n ) z ) n, z R n. Lemma.1. There exists 0 > 1 such that for all 0 and r (R, M n ), n(n ) (.17) min v ( ) n r n. B r Ω Proof of Lemma.1: Suppose (.17) does not hold, then there exist r such that (.18) v (z) ( Clearly r R. Let n(n ) ) n r n, z = r,z Ω. v λ (z) = ( λ z )n v (z λ ), z λ = λ z z.

9 The equation of v λ, by direct computation, is ENERGY ESTIMTE 9 (.19) v λ (z) + ( λ z )n+ l n+ n g(l ( z λ )n v λ (z)) = 0, in where Σ λ := {z Ω ; λ < z < r }. Σ λ Clearly v λ V λ in C 1,α loc (Rn ) for fixed λ > 0. By direct computation V (z) > V λ n(n ) (z), for λ (0,( ) 1/ ), z > λ V (z) < V λ n(n ) (z), for λ > ( ) 1/, z > λ. We shall apply the method of moving spheres for λ ( 1 ( n(n ) ) 1/,( n(n ) ) 1/ ). First we prove that for λ 0 = 1 ( n(n ) ) 1/, (.0) v (z) > v λ 0 (z), z Σ λ. To prove (.0) we first observe that v > v λ 0 in B R \ B λ for any fixed R large. Indeed, v = v λ 0 on B λ0. On B λ0 we have ν V > ν V λ 0. Thus the C 1,α convergence of v to V gives that v > v λ 0 near B λ0. Then by the uniform convergence we further now that (.0) holds on B R \ B λ0. On B R, we have (.1) v (z) (( and n(n ) ) n ε) z n, z = R (.) v λ ) 0 (z) ((n(n ) n ε) z n, z R for some ε > 0 independent of. Next we shall use maximum principle to prove that n(n ) (.3) v (z) > (( ) n ε) z n > v λ 0 (z), z Σ λ 0 \ B R. The proof of (.3) is by contradiction. We shall compare v and n(n ) f := (( ) n ε) z n. Clearly v f is super harmonic in Σ λ0 B R and, by (.1),(.) and (.18), v f > 0 on B R and Σ λ0 (R n + \ B R ). If there exists z 0 Σ λ0 {z n = T } and we would have 0 > v (z 0 ) f (z 0 ) = min Σ λ0 \B R v f (.4) 0 < n (v f )(z 0 ) = l n n h(l v (z 0 )) n f (z 0 ).

10 10 YING GUO ND LEI ZHNG Easy to verify n f (z 0 ) > N f (z 0 ) n n for some N. However by GH 1 l n n h(l v (z 0 )) Cv (z 0 ) n n. Easy to see it is impossible to have v (z 0 ) < f (z 0 ) and (.4). (.3) is established. Before we employ the method of moving spheres we set ( O λ := {z Σ λ ; v (z) < min ( z ) λ )n, v λ (z) }. This is the region where maximum principle needs to be applied. By GH 1 we have Therefore in O λ we have ( λ z )n+ l n n+ ( = ( z λ )n l v λ n+ ) g(l ( z λ )n v λ ) z n g(( λ )n l v λ )(vλ ) n n+ (l v ) n+ n g(l v )(v λ ) n+ n, in Oλ. (.5) v λ + (l v ) n+ n g(l v )(v λ ) n+ n 0, in Oλ. The equation for v can certainly be written as (.6) v + (l v ) n+ n n g(l v )v n+ = 0. Let w λ, = v v λ, we have, from (.5) and (.6) (.7) w λ, + n(n )(l v ) n+ n n g(l v )ξ 4 w λ, 0, in O λ, where ξ is obtained from the mean value theorem. Now we apply the method of moving spheres to w λ,. Let λ = inf{λ [( n(n ) ) 1 n(n ) ε0,( ) 1 + ε0 ]; v > v µ in Σ µ, µ > λ } where ε 0 > 0 is chosen to be independent of and (.8) v (z) > v λ (z), z = r, z Ω, λ [λ 0,λ 1 ]. From (.18) we see that ε 0 can be chosen easily. By (.0), λ > ( n(n ) ) 1 ε 0. We claim that λ = ( n(n ) ) 1 + ε 0. Suppose this is not the case we have λ < ( n(n ) ) 1 + ε 0. By continuity w λ, 0 and by (.18) w λ, > 0 on the outside boundary: {z Σ λ \( B λ n = T }). By (.7), if min Σ λ w λ, = 0, the minimum will have to appear on. From (.8) we see that the minimum does not appear Σ λ on \ {z Σ λ ( B λ n = T }). If there exists x 0 {z Σ λ n = T }, we have 0 < zn w λ, (x 0)

11 ENERGY ESTIMTE 11 Note that we have strict inequality because of Hopf Lemma. On the other hand, using T and elementary estimates we have v λ z n For v, GH 1 implies > N (v λ ) n n, in O λ {z n = T }, for some N. zn v c h v n n, in O λ {z n = T } where c h = lim s s n n h(s). Then it is easy to see that w λ > 0 on {z n = T }. Then Hopf Lemma and the continuity lead to a contradiction of the definition of λ. Thus we have proved λ = ( n(n ) ) 1 + ε 0. However V < V λ for y > λ if λ > ( n(n ) ) 1. So it is impossible to have lim λ > ( n(n ) ) 1. This contradiction proves (.17) under Case one. Lemma.1 is established. From Lemma.1 we further prove the spherical Harnac inequality for v. For fixed, consider R r 1 M n and let ṽ (z) = r n v (rz). By (.1), ṽ (z) C. Direct computation yields ṽ (z) + r n+ l n+ n g(l r n 1 ṽ ) = 0, < z <,rz Ω, Let n ṽ = r n l n n h(r n l ṽ ), Ω. a = r n+ l n n+ b = r n l n n g(l r n ṽ )/ṽ h(r n l ṽ )/ṽ n By the definition of of l and r we see that r = o(1)l. Using the assumptions of g we have n g(1)ṽ 4 C, if l r n ṽ (z) 1, a (z) Cr l n 4 = o(1), if l r n ṽ (z) 1, and b (z) c h ṽ n C, if l r n ṽ (z) 1 Crl n = o(1), if l r n ṽ (z) 1, Hence a and b are both bounded functions. Consequently the equation for ṽ can be written as ṽ (z) + a ṽ = 0, 1 < z <,rz Ω, n ṽ = b ṽ, Ω {z n = T /r}.

12 1 YING GUO ND LEI ZHNG Clearly we apply classical Harnac inequality for two cases: either T /r > 1 or T /r 1. In the first case we have In the second case we have z =3/4ṽ(z) max C z =3/4ṽ. min max ṽ(z) C min ṽ. z =1,z n T /r z =1,z n T /r Clearly (.14) is implied. Proposition. is established for Case one. Case two: lim T = T Recall that v satisfies (.16). By the same argument as in the proof of Proposition.1, we now v V in C 1,α loc (Rn {y n T }) where V solves (.3). Thus there exist R such that v V C 1,α (B T R ) CR 1. Clearly (.14) holds for y M n R {y n T }, we just need to prove (.14) for { y > M n R } {y n T }. Since V (0) = 1 is the maximum point of V, V (z) = (1 + n(n ) z ) n, z R n. n Lemma.. There exists 0 > 1 such that for all 0 and r (R, M ), n(n ) (.9) min v ( ) n r n. B r Ω Proof of Lemma.: Just lie the interior case suppose there exist r R such that n(n ) (.30) min v > ( ) n r n. B r Ω Let ṽ (z) = v (z T e n ), ṽ λ (z) = ( λ z )n ṽ ( λ z z ) and D := {z; M n (z T e n ) Ω B r } be the domain of ṽ. Then D R n +. Set Σ λ := {z D ; Let Ṽ be the limit of ṽ in C loc (Rn +): Ṽ (z) = (1 + z > λ}. n(n ) z Te n ) n, then there exists λ 0 < λ 1 depending only on n,,t such that Ṽ > Ṽ λ 0 in R n + \ B λ0

13 and ENERGY ESTIMTE 13 Ṽ < Ṽ λ 1 in R n + \ B λ1. We shall employ the method of moving spheres to compare ṽ and ṽ λ on Σ λ for λ [λ 0,λ 1 ]. First we use the uniform convergence of ṽ to Ṽ to assert that, for any fixed R > 1 (.31) ṽ (y) > ṽ λ 0 (y), y Σ λ 0 B R. and For R large we have (let a 1 = ( n(n ) ) n ) To prove ṽ > ṽ λ 0 ṽ (y) (a 1 ε/5) y n on B R R n + ṽ λ 0 (y) (a 1 ε/5) y n, y > λ 0. in Σ λ0 \ B R we compare ṽ with w = (a 1 3ε/10) y 1 e n n where 1 = 1/(n )c h a n 1. For R chosen sufficiently large we have w ṽ λ 0 in Σ λ0 \ B R and ṽ > w on B R Σ λ0. To compare ṽ and w over Σ λ0 \B R, it is easy to see that ṽ > w on B R Σ λ0 and Σ λ0 \ (B R {z n > 0}). Since ṽ w is super-harmonic, the only thing we need to prove is on R n + \ B λ0 (.3) n (ṽ w) < ξ (ṽ w), on z n = 0. for some positive function ξ. Then standard maximum principle can be used to conclude that ṽ > w on Σ λ0 \ B R. To obtain (.3) first for ṽ we use GH to have n ṽ c h ṽ n n, z n = 0. On the other hand by the choice of 1 we verify easily that n w > c h w n n, zn = 0. Thus (.3) holds from mean value theorem. We have proved that the moving sphere process can start at λ = λ 0 : ṽ > ṽ λ 0 in Σ λ0. Let λ be the critical moving sphere position: λ := min{λ [λ 0,λ 1 ]; ṽ µ > ṽµ in Σ µ, µ > λ.}. s in Case one we shall prove that λ = λ 1, thus getting a contradiction from Ṽ < Ṽ λ 1 for z > λ 1. For this purpose we let w λ, = ṽ ṽ λ.

14 14 YING GUO ND LEI ZHNG To derive the equation for w λ, we first recall from (.16) and the definition of ṽ that ṽ (z) + l n+ n g(l ṽ ) = 0, z Ω, (.33) ṽ z n = l n n h(l ṽ ), {z n = 0} Ω. where l = σ n (.34) M. Correspondingly ṽ λ satisfies ṽ λ + ( λ z )n+ l n n+ ṽ λ z n = ( λ z )n l n n g(l ( z λ )n ṽ λ (z)) = 0, in Σ λ, h(l ( z λ )n ṽ λ (z)) on Σ λ {z n = 0}. Let O λ be defined as before. Then in O λ we have, by GH 1, ( λ z )n l n n ( λ z )n+ l n n+ g(l ( z λ )n ṽ λ (z)) (v l ) n+ n g(l v )(v λ ) n+ n, in Oλ h(l ( z λ )n ṽ λ (z)) (l v ) n n h(l v )(v λ ) n n, on Oλ {z n = 0}. The inequalities above yield w λ, + ξ 1, w λ, 0, in O λ n w λ, ξ, w λ,, on O λ {z n = 0}. where ξ 1, > 0 and ξ, are continuous functions obtained from mean value theorem. It is easy to see that the moving sphere argument can be employed to prove that λ = λ 1, which leads to a contradiction from the limiting function Ṽ. Thus Lemma. is established. Lemma. guarantees that on each radius R r 1 M the minimum of v is always comparable to z n. Re-scaling v as r n v (rz) we see the spherical Harnac inequality holds by the GH and GH 3. Thus Proposition. is established in Case Two as well. Lemma.3. Let {u } be a sequence of solutions of (1.1) and q q B + 1 be a sequence of points in Σ. Then there exist C > 0, r > 0 independent of such that u (q )u (x) C x q n in x q r, x B + 3. Proof. Without loss of generality we assume that q s are local maximum points of u. We consider two cases Case one: u (q ) n qn. Let M = u (q ) and (.35) v (y) = M 1 u (M n y + q ), y Ω := {y; M n y + q B + 3 }.

15 Then in this case v converges uniformly to n V (y) = (1 + y ) n(n ) over all compact subsets of R n. For ε > 0 small we let ENERGY ESTIMTE 15 n(n ) φ = ( ε) n ( y n M n ) y M on y R where R > 1 is chosen so that v > φ on B R. By direct computation we have φ > N φ n n, on yn = q n M y n for some N. It is easy to see that v φ on Ω \ {y n = M {y n = M n q n } we have yn (v φ) c h (v φ). n q n }. On Thus standard maximum principle implies v φ on Ω. Lemma.3 is established in this case. Now we consider n Case two: M q n C. Let v be defined as in (.35). In this case the boundary condition is written as yn v = (M n v ) n n h(m n n v )v n n, y n = M q n. v converges to V 1 over all compact subsets of R n {y n T } where T = lim n M q n. V 1 satisfies (.3). For R large and ε > 0 small, both independent of, we have n(n ) v (y) ( ε) n y n, y = R. In B R R n + we have the uniform convergence of v to V 1. Our goal is to prove that v is bounded below by O(1) y n outside B R. To this end let where Then it is easy to chec that n(n ) w(y) = ( ε) n y 1 e n n 1 = c h ( n(n ) ) T. w > c h w(y) n n, on n yn = M q n. y n By choosing R larger if needed we have n(n ) v (y) > ( ε) n y n > w(y), y = R, y R n +.

16 16 YING GUO ND LEI ZHNG Then it is easy to apply maximum principle to prove v > w in Ω \ B R. Lemma.3 is established. Let q 1 Σ and q be its nearest or almost nearest sequence in Σ : We claim that q q 1 = (1 + o(1))d(q 1,Σ \ {q 1}). Lemma.4. There exists C > 0 independent of such that Proof. Let σ = d(q 1,Σ \ {q 1 }) and 1 C u (q 1) u (q ) Cu (q 1). ũ (y) = σ n u (q 1 + σ y). We use e to denote the image of q after scaling (so e 1). Then in B 1, ũ (x) ũ (0) 1 x n for x 1/. On one hand, for x = 1 we have, by Lemma.3 applied to e, ũ (0) 1 ( 1 ) n Cũ (e ) 1 which is just u (q 1 ) Cu (q ). On the other hand, the same moving sphere argument can be applied to u near q with no difference. The Harnac type inequality gives min u C. Using and we have max B(q,1/4) B+ 3 u B(q,1/) B+ 3 max u u (q ) B(q,1/4) B+ 3 min B(q,1/) B+ 3 u min B(q 1,σ ) B + 3 (.36) ũ (e )ũ (0) 1 C. Thus (.36) gives u (q ) Cu (q 1 ). Lemma.4 is established. Remar.3. Proposition. is not needed in the proof of Lemma.4. The following lemma is concerned with Pohozaev identity that can be verified by direct computation. Lemma.5. Let u solve u + g(u) = 0, in B + σ, n u = h(u) on B + σ. u

17 Then (.37) = B + σ B + σ n 1 h(u)( where G(s) = s 0 g(t)dt. i=1 ENERGY ESTIMTE 17 x i i u + n B u) + (ng(u) n + σ ) ( σ(g(u) 1 u + ( ν u) ) + n u νu Proposition.3. There exists d > 0 independent of such that lim q 1 q d. g(u)u) Proof. Recall that σ = (1 + o(1)) q 1 q. We prove by way of contradiction. Suppose σ 0, et M = ũ (0). We claim that (.38) M ũ (y) a y n + b(y) in C loc(b 3/4 Ω \ {0}), with a > 0, b(0) > 0 where Ω = {y; σ y + q 1 B+ 3 }. Proof of (.38): s usual we consider the following two cases: Case one: lim q M n 1n, and Case two: lim q M 1n First we consider Case one. Let T = M n q 1n. n T < Recall the equation for ũ is (.15). Multiplying M on both sides and letting it is easy to see from the assumptions of g and h that M ũ h in C loc (B 1 \ {0}) where h is a harmonic function defined in B 1 \ {0}. Thus h(y) = a y n + b(y) for some harmonic function b(y) in B 1. From the pointwise estimate in Lemma.3 we see that a > 0. Given any ε > 0, we compare ũ and w := (a ε)( y n R n ) on y R. Here R is less than T. Observe that ũ > w on B R and y = ε 1 for ε 1 sufficiently small. Thus ũ > w by the maximum principle. Let we have, in B 1 a y n + b(y) (a ε) y n, B 1 \ B ε1. Then let ε 0, which implies ε 1 0 we have b(y) 0 in B 1. Next we claim that b(0) > 0 because by Lemma.3 and Lemma.4 we have a y n + b(y) a 1 y e n in B 1 for some a 1 > 0, where e = lim e. Thus b(y) > 0 when y is close to e, which leads to b(0) > 0. (.38) is established in Case one. Case two. gain we first have M ũ h in C loc (B T 1 \ {0}) and h is of the form h(y) = a y n + b(y), y n T.

18 18 YING GUO ND LEI ZHNG To prove b(y) 0 we compare, for fixed ε > 0, M ũ with w (y) = (a ε)( y b e n n (R 1) n ) where b 0 and R are chosen to satisfy and (n )b R > c 0 σ, c 0 = sup s h(s) 0<s 1 (n )b > c h M n a n. It is easy to see that such b and R can be found easily. Let h = M ũ, and Ω = Ω {y n = T }. We divide Ω into two parts: E 1 = {z Ω ; Then by the assumptions on h ũ (z)σ n 1 }, E = Ω \ E 1. { c0 σ h, x E, n h c h M n n h n, x E 1, With the choice of b and R it is easy to verify that n w max{c 0 σ w,c h M n w n n } on Ω B R. Thus standard maximum principle can be applied to prove h w on Ω B R. Letting first and ε 0 next we have b(y) 0 in B 1 {y n T }. Then by Proposition.4 we see that b(y) > 0 when y is close to e, the limit of e. The fact n b = 0 at 0 implies b(0) > 0. (.38) is proved in both cases. Finally to finish the proof of Proposition.3 we derive a contradiction from each of the following two cases: n Case one: lim M q 1n > 0. n In this case we use the following Pohozaev identity on B σ for σ < lim M q 1n : ( ng (ũ ) n ) (.39) ũ g (ũ ) B σ ( = σ(g (ũ ) 1 ũ + ν ũ ) + n ) ũ ν ũ where B σ g (s) = σ n+ First we claim that for s > 0, g(σ n t s), G(t) = g(s)ds, 0 (.40) G (s) n n sg (s). G (s) = σ n n G(σ s).

19 ENERGY ESTIMTE 19 Indeed, writing g(t) = c(t)t n+ n, we see from GH1 that c(t) is a non-increasing function, thus G (s) = σ n σ n n c(σ s) n G(σ s) = σ n n σ 0 s n σ 0 t n+ n dt = n s c(t)t n+ n dt n c(σ n s)s n n = n n sg (s). Replacing s by ũ we see that the left hand side of (.39) is non-negative. Next we prove that ( (.41) lim M σ(g (ũ ) 1 ũ + ν ũ ) + n ) ũ ν ũ < 0 B σ for σ > 0 small. Clearly after (.41) is established we obtain a contradiction to (.39). To this end first we prove that (.4) M G (ũ ) = o(1). Indeed, by GH 1 and GH 3 G (ũ ) = σ n n σ 0 n σ σ n σ n( 1 ũ g(t)dt ũ 0 ctdt, if σ n 0 ctdt + σ n ũ 1 ct n n+ dt), M 1 1, if σ n M 1 > 1. Therefore { Cσ G (ũ ) M, if σ n M 1 1, Cσ n +C M n n, if σ n M 1 > 1. Clearly (.4) holds in either case. Consequently we write the left hand side of (.41) as ( 1 B σ σ h + σ ν h + n h νh) + o(1) where h(y) = a y n + b(y), b(0) > 0,a > 0. By direct computation we have ( 1 B σ σ h + σ ν h + n h νh) ( ) (n ) = b(0) σ 1 n + O(σ n ) ds. a B σ Thus (.41) is verified when σ > 0 is small. The second case we consider is Case two: lim M n q 1n = 0.

20 0 YING GUO ND LEI ZHNG In this case we use the following Pohozaev identity on B + σ : Let then we have (.43) = B + σ R n + B + σ R n + ( n 1 h (ũ )( i=1 h (s) = σ n h(σ n s), x i i ũ + n ũ ) + (ng (ũ ) n B + σ g (ũ )ũ ) σ(g (ũ ) 1 ũ + ( ν ũ ) ) + n ) ũ ν ũ Multiplying M on both sides and letting we see by the same estimate as in Case one that the second term on the left hand side is non-negative, the right hand side is strictly negative. The only term we need to consider is lim M B + σ R n + n 1 h (ũ )( x i i ũ + n ũ ). Let H(s) = s 0 h(t)dt, then from integration by parts we have (.44) = B + σ R n + B σ R n + + B + σ R n + n 1 h (ũ )( σ n 1 i=1 i=1 x i i ũ + n ũ ) H(σ n ũ )σ ( (n 1)σ n 1 H(σ n ũ ) + n ũ h (ũ ))dx. For the first term on the right hand side of (.44) we claim (.45) M σ n 1 Indeed, by GH 1 and GH H(σ n ũ ) H(σ n ũ ) = o(1) on B σ. n σ ũ 0 ctdt, if σ n 1 0 ctdt + σ n ũ 1 ct n Using ũ = O(1/ M ) on B σ we then have { M σ n 1 H(σ n O(σ ũ ) ), ũ 1, n dt, if σ n ũ 1, O(σ ) + O( M n if σ n ũ > 1, ), if σ n ũ > 1. Thus (.45) is verified and the first term on the right hand side of (.44) is o(1). Therefore we only need to estimate the last term of (.44), which we claim is non-negative. Indeed, for t > 0, we write h(t) = b(t)t n n for some non-decreasing

21 function b. Then we have σ n 1 =σ n 1 H(σ n n σ 0 s ENERGY ESTIMTE 1 s) = σ n 1 n σ 0 s h(t)dt b(t)t n n dt σ n 1 b(σ n s) n σ 0 s t n n dt = n n b(σ n s)s n n n = n h (s)s. Replacing s by ũ in the above we see that the last term of (.44) is non-negative. Thus there is a contradiction in (.43) in Case two as well. Proposition.3 is established. We are in the position to finish the proof of Theorem 1.1. By Proposition.3 there is a positive distance between any two members of Σ. The uniform bound of follows readily from the pointwise estimates for blowup solutions near +1 B un/n an isolated blowup point. The uniform bound for B + 1 u can be obtained by scaling and standard elliptic estimates for linear equations. Thus we have obtained a contradiction to (.1). Theorem 1.1 is established. 3. PROOF OF THEOREM 1. If h is non-positive, the energy estimate follows from the Harnac inequality in a straight forward way. Indeed, let G(x,y) be a Green s function on B + 3 such that G(x,y) = 0 if x B + 3,y B+ 3 Rn + and yn G(x,y) = 0 for x B + 3,y B+ 3 Rn +. It is easy to see that G can be constructed by adding the standard Green s function on B 3 its reflection over R n +. It is also immediate to observe that G(x,y) C n x y n, x B + 3, y B+. Multiplying G on both sides of (1.1) and integrating by parts, we have u(x) + h(u(y))g(x,y)ds y + u(y) G(x,y) ds y B + 3 Rn + B + ν 3 Rn + = g(u(y))g(x, y)dy. Here ν represents the outer normal vector of the domain. Using h 0 and ν G 0, we have u(x) g(u(y))g(x,y)dy, x B + B In particular tae let u(x 0 ) = min B + u, then x 0 =, thus C maxu minu g(u(y))u(y)g(x 0,y)dy C g(u)udy. B + 1 B + B + 3/ Therefore we have obtained the bound on B + 3/ g(u)udy. To obtain the bound on B + 1 u, we use a cut-off function η which is 1 on B + 1 and is 0 on B+ \ B+ 3/ and B + 3 B + 3/

22 YING GUO ND LEI ZHNG η C. Multiplying uη to both sides of (1.1) and using integration by parts and Cauchy inequality we obtain the desired bound on B + 1 u. Theorem 1. is established. REFERENCES [1] Caffarelli, Luis.; Gidas, Basilis; Spruc, Joel symptotic symmetry and local behavior of semilinear elliptic equations with critical Sobolev growth. Comm. Pure ppl. Math. 4 (1989), no. 3, [] Chen, Chiun Chuan; Lin, Chang Shou, Estimates of the conformal scalar curvature equation via the method of moving planes. Comm. Pure ppl. Math. 50 (1997), no. 10, [3] Chen, Chiun Chuan; Lin, Chang Shou, Local behavior of singular positive solutions of semilinear elliptic equations with Sobolev exponent. Due Math. J. 78 (1995), no., [4] Djadli, Zindine; Malchiodi, ndrea; Ould hmedou, Mohameden, Prescribing scalar and boundary mean curvature on the three dimensional half sphere. J. Geom. nal. 13 (003), no., [5] Gui, Changfeng; Lin, Chang-Shou, Estimates for boundary-bubbling solutions to an elliptic Neumann problem. J. Reine ngew. Math. 546 (00), [6] Han, Zheng-Chao; Li, Yanyan The Yamabe problem on manifolds with boundary: existence and compactness results. Due Math. J. 99 (1999), no. 3, [7] Li, obing; Li, Yanyan On some conformally invariant fully nonlinear equations. Comm. Pure ppl. Math. 56 (003), no. 10, [8] Li, YanYan, Prescribing Scalar Curvature on S n and Related Problems, Part I, J. Differential Equations 10 (1995), no., [9] Li, YanYan; Zhang, Lei Liouville-type theorems and Harnac-type inequalities for semilinear elliptic equations. J. nal. Math. 90 (003), [10] Li, Yan Yan; Zhang, Lei Harnac type inequality for the Yamabe equation in low dimensions. Calc. Var. Partial Differential Equations 0 (004), no., [11] Li, YanYan; Zhu, Meijun Uniqueness theorems through the method of moving spheres. Due Math. J. 80 (1995), no., [1] Lin, Chang-Shou; Prajapat, Jyotshana V. Harnac type inequality and a priori estimates for solutions of a class of semilinear elliptic equations. J. Differential Equations 44 (008), no. 3, [13] Schoen, Richard; Lecture notes of Stanford University. [14] Schoen, Richard; Zhang, Dong; Prescribed scalar curvature on the n-sphere. Calc. Var. Partial Differential Equations 4 (1996), no. 1, 1-5. [15] Zhang, Lei, Classification of conformal metrics on R + with constant Gauss curvature and geodesic curvature on the boundary under various integral finiteness assumptions. Calc. Var. 16, (003). [16] Zhang, Lei Harnac type inequalities for conformal scalar curvature equation. Math. nn. 339 (007), no. 1, DEPRTMENT OF MTHEMTICS, UNIVERSITY OF FLORID, 358 LITTLE HLL P.O.BOX , GINESVILLE FL address: yguo@ufl.edu DEPRTMENT OF MTHEMTICS, UNIVERSITY OF FLORID, 358 LITTLE HLL P.O.BOX , GINESVILLE FL address: leizhang@ufl.edu