Ann. I. H. Poincaré AN 19, 6 (2002) 889 901 2002 Éditions scientifiques et médicales Elsevier SAS. All rights reserved S0294-1449(02)00105-1/FLA LOCAL BEHAVIOR AND GLOBAL EXISTENCE OF POSITIVE SOLUTIONS OF au λ u u λ COMPORTEMENT LOCAL ET EXISTENCE GLOBALE DES SOLUTIONS POSITIVES DE au λ u u λ Steven D. TALIAFERRO Mathematics Department, Texas A&M University, College Station, TX 77843-3368, USA Received 5 December 2000, revised 22 October 2001 ABSTRACT. We study the behavior near the origin of C 2 positive solutions u(x) of au λ u u λ ( ) in a punctured neighborhood of the origin in R n (n > 2) where the constants λ and a satisfy n n+2 n 2 <λ< n 2 and 0 <a<1. We also study the existence of C2 positive solutions of ( )inr n. In both cases we show that changing a from one value in the open interval (0, 1) to another value in (0, 1) can have a dramatic effect. 2002 Éditions scientifiques et médicales Elsevier SAS RÉSUMÉ. On étudie le comportement prés de l origine des solutions positives de classe C 2 u(x) de au λ u u λ ( ) dans un voisinage épointé de l origine dans R n (n > 2) où les constantes λ et a satisfont n n+2 n 2 <λ< n 2 et 0 <a<1. On étudie aussi l existence de solutions positives de classe C2 de ( ) dansr n. Dans les deux cas nous montrons que changer la valeur de a dans l intervalle ouvert (0, 1) peut avoir un effet dramatique. 2002 Éditions scientifiques et médicales Elsevier SAS 1. Introduction Gidas and Spruck [3] prove the following two theorems: THEOREM A[3]. If u(x) is a C 2 positive solution of u = u λ (1.1) E-mail address: stalia@math.tamu.edu (S.D. Taliaferro).
890 S.D. TALIAFERRO / Ann. I. H. Poincaré AN 19 (2002) 889 901 in a punctured neighborhood of the origin in R n (n > 2) where the constant λ satisfies n n 2 <λ<n + 2 n 2 (1.2) then either u has a removable singularity at the origin or where l = l(n, λ) is a positive constant. x 2/(λ 1) u(x) l as x 0 + THEOREM B[3]. There does not exist a C 2 positive solution of (1.1) in R n (n > 2) when the constant λ satisfies 1 λ< n+2 n 2. In this paper we study the more general problem au λ u u λ (1.3) in various subsets of R n where a (0, 1). Henceforth we assume n>2. Our two results on the local behavior at the origin of C 2 positive solutions of (1.3) are THEOREM 1. Suppose λ>n/(n 2). Then there exists a = a(n,λ) (0, 1) such that for each continuous function ϕ : (0, 1) (0, ) there exists a C 2 positive solution u(x) of (1.3) in R n {0} satisfying u(x) O ( ϕ( x ) ) as x 0 +. THEOREM 2. Suppose λ satisfies (1.2). Then there exists a = a(n,λ) (0, 1) such that if u is a C 2 positive solution of (1.3) in a punctured neighborhood of the origin in R n then u(x) = O ( x 2/(λ 1)) as x 0 + (1.4) and u(x)/ū( x ) is bounded between positive constants for x small and positive where ū(r) is the average of u on the sphere x =r. Our two results on the local behavior at infinity of C 2 positive solutions of (1.3) are THEOREM 3. Suppose λ>n/(n 2). Then there exists a = a(n,λ) (0, 1) such that for each continuous function ϕ : (1, ) (0, ) there exists a C 2 positive solution u(x) of (1.3) in R n satisfying u(x) O ( ϕ( x ) ) as x. THEOREM 4. Suppose λ satisfies (1.2). Then there exists a = a(n,λ) (0, 1) such that if u is a C 2 positive solution of (1.3) in the complement of a compact subset of R n then u(x) = O ( x 2/(λ 1)) as x (1.5) and u(x)/ū( x ) is bounded between positive constants for x large where ū(r) is the average of u on the sphere x =r. Our two results on the global existence of C 2 positive solutions of (1.3) are
S.D. TALIAFERRO / Ann. I. H. Poincaré AN 19 (2002) 889 901 891 THEOREM 5. Suppose λ>n/(n 2). Then there exists a = a(n,λ) (0, 1) such that there exists a C 2 positive solution u(x) of (1.3) in R n. THEOREM 6. Suppose λ satisfies (1.2). Then there exists a = a(n,λ) (0, 1) such that there does not exist a C 2 positive solution u(x) of (1.3) in R n. When λ satisfies (1.2), these six theorems show that changing a from one value in the open interval (0, 1) to another value in (0, 1) can dramatically affect the local behavior and global existence of C 2 positive solutions of (1.3). Note that Theorem 3 implies Theorem 5. However we state Theorem 5 because it nicely complements Theorem 6 and because it facilitates our discussion below. Theorems 1 and 2 answer two of the seven open questions posed in [8] where we study in a general way C 2 positive solutions of af (u) u f(u) in a punctured neighborhood of the origin in R n where a [0, 1) is a constant and f : (0, ) (0, ) is a continuous function. Our proof of Theorem 1 in Section 3 is different and shorter than the proofs of two special cases of Theorem 1 given in [8]. Theorem 1 is not true when λ n/(n 2). In fact we prove in [9] that if u(x) is a C 2 positive solution of 0 u f(u) in a punctured neighborhood of the origin in R n where f : (0, ) (0, ) is a continuous function satisfying f(t)= O ( t n/(n 2)) as t then u(x) = O ( x 2 n) as x 0 + and u(x)/ū( x ) is bounded between positive constants for x small and positive where ū(r) is the average of u on the sphere x =r. When λ = (n + 2)/(n 2), which is of interest in geometry, we prove in [7] that the conclusion of Theorem 1 is true for each a (0, 2 4/(n 2) ) and we conjecture there that the conclusion of Theorem 2 is true for each a (2 4/(n 2), 1). Suppose λ satisfies (1.2) and a (0, 1). Then Gidas and Spruck [3, Theorem 3.3] prove that if u is a C 2 positive solution of (1.3) in a punctured neighborhood of the origin in R n such that h(x) := u(x) is C 1 (1.6) u(x) λ and ( ) 1 log h(x) =O as x 0 + (1.7) x then either u has a C 1 extension to the origin or x 2/(λ 1) u(x) is bounded between positive constants for x small and positive. (However we do not understand the third
892 S.D. TALIAFERRO / Ann. I. H. Poincaré AN 19 (2002) 889 901 sentence of their proof of Theorem 3.3 and we have contacted them for an explanation.) Conditions (1.6) and (1.7) in their result cannot both be omitted when a is near 0 by our Theorem 1. An open question is whether conditions (1.6) and (1.7) can be omitted when a is near 1. This open question reduces to an ODE problem by our Theorem 2 and the fact that every C 2 positive bounded solution of (1.3) in a puncture neighborhood of the origin has a C 1 extension to the origin. Theorem 3 is not true when λ n/(n 2). In fact Serrin and Zou [6, Theorem I] prove that if 0 λ n/(n 2) then there does not exist a C 2 positive solution of u u λ in R n B 1 (0) and this result can be easily extended to include all negative values of λ. Consider the Question. Forwhatvaluesofλ R and a (0, 1] does there exist a C 2 positive solution of (1.3) in R n? To answer this question we consider three mutually exclusive possibilities for the value of λ. Case I. Suppose <λ n/(n 2). Then for each a (0, 1], (1.3) does not have a C 2 positive solution in R n by Serrin and Zou s result mentioned above. Case II. Suppose λ satisfies (1.2). Then our Theorems 5 and 6 hold. Case III. Suppose (n + 2)/(n 2) λ<. Then for each a (0, 1], (1.3) has a C 2 positive solution in R n because Fowler [2] shows that (1.3) with a = 1hasaC 2 positive radial solution in R n. For λ satisfying (1.2) and for j = 1, 2,...,6, let I j = I j (n, λ) be the set of all a (0, 1) such that the conclusion of Theorem j is true. By Theorems 1 6, (i) I j is a nonempty subinterval of (0, 1) for j = 1, 2,...,6. (ii) The left endpoint of each of the intervals I 1, I 3,andI 5,is0. (iii) The right endpoint of each of the intervals I 2, I 4,andI 6,is1. (iv) I 1 I 2 = I 3 I 4 = I 5 I 6 =. (v) I 5 I 6 = (0, 1) and I 3 I 5. Some interesting open questions are: (i) Is I 1 I 2 = (0, 1) (respectively I 3 I 4 = (0, 1))? If not, what is the local behavior at the origin (respectively at infinity) of C 2 positive solutions of (1.3) when a (0, 1) (I 1 I 2 ) (respectively a (0, 1) (I 3 I 4 ))? (ii) Is I 1 = I 3 = I 5? (iii) Is I 2 = I 4 = I 6? (iv) Which of the intervals I j,j= 1, 2,...,6, are open? With regard to question (iii) we have THEOREM 7. Suppose λ satisfies (1.2). Then wherewedefinei λ ={a λ : a I}. I λ 2 I 6 and I λ 4 I 6
S.D. TALIAFERRO / Ann. I. H. Poincaré AN 19 (2002) 889 901 893 For later we need two lemmas. 2. Preliminary lemmas LEMMA 1. Suppose, for j = 1, 2,...,there exists a C 2 positive solution u j (x) of a j x σ u λ j u j x σ u λ j in Ɣ j (2.1a) u j (x) O ( x (2 σ)/(λ 1)) as x 0 + (2.1b) where Ɣ j is a punctured neighborhood of the origin in R n (n > 2), λ>n/(n 2) and σ [0, 2) are constants, and the sequence {a j } j=1 (0, 1] converges to a (0, 1].Then there exists a C 2 positive solution v of a λ v λ v v λ in R n. (2.2) When σ = 0anda j 1, Lemma 1 is well-known and easy to prove using a blow-up argument of Schoen [5]. See also Chen and Lin [1, p. 982]. However when a (0, 1), C 2 regularity is lost at a critical point in their proofs and new methods are needed. Proof of Lemma 1. Choose positive constants r j such that B 2rj (0) Ɣ j {0} and let j = B rj (0). Sinceu j is positive and superharmonic in Ɣ j, it is well-known (see for example [4]) that u j, u j L 1 ( j ) (2.3) and that there exist a nonnegative constant m j and a continuous function h j : j R which is harmonic in j such that u j (x) = m j x + α n 2 n j u j (y) x y n 2 dy + h j(x) for x j {0}. (2.4) 1 Here α n = n(n 2)ω n where ω n is the volume of the unit ball in R n. It follows from (2.3), (2.1a), and the nonnegativity of σ that u λ j L1 ( j ). Thus, since λ>n/(n 2) we have m j = 0. Let {ε j } j=1 (0, 1) be a sequence which converges to 0. Choose x j 1 2 j {0} such that lim j x j =0, lim ε j x j 2/(λ 1) x j σ/(λ 1) u j ( x j ) =, j max h j < x j (2 σ)/(λ 1). j and (2.5) Following Schoen [5], let B j = B εj x j ( x j ) and define ϕ j : B j [0, ) by ϕ j (x) = u j (x)d j (x) 2/(λ 1) x σ/(λ 1) where d j (x) = dist{x, B j }. Choose x j B j such that ϕ j (x j ) = max Bj ϕ j.then ϕ j (x j ) ϕ j ( x j ) as j (2.6)
894 S.D. TALIAFERRO / Ann. I. H. Poincaré AN 19 (2002) 889 901 by (2.5). Making the change of variables w j (ξ) = u j (x) u j (x j ), x = x j σ/2 u j (x j ) (λ 1)/2 ξ + x j (2.7) in (2.4) we obtain w j (ξ) = α n j w j (ζ ) ξ ζ n 2 dζ + H j(ξ) for ξ j {ξ j } (2.8) where j is the set of all ξ R n such that x, as given by (2.7), is in j and where ξ j is the inverse image of x = 0 in (2.7) and where By (2.5) and (2.6) we have H j (ξ) = h j(x) u j (x j ). max H j < (1 ε j) σ/(λ 1) 0 asj. (2.9) j ϕ j (x j ) Note that the ball given by x x j < 1 2 d j (x j ) maps under the transformation (2.7) onto the ball ξ <ρ j where ρ j := 1 2 ϕ j(x j ) (λ 1)/2 as j and that w j (ξ) = ϕ ( ) j(x) dj (x j ) 2/(λ 1) ( x ϕ j (x j ) d j (x) x j < 2 2/(λ 1) ( 1 + εj 1 ε j ) σ/(λ 1) ) σ/(λ 1) for ξ <ρ j because d j (x) > 1 2 d j(x j ) for x x j < 1 2 d j (x j ). Also by (2.1a) we have ( 1 εj 1 + ε j ) σ ( ) 1 + σ a j wj λ w εj j wj λ for ξ <ρ j. (2.10) 1 ε j It follows therefore from standard elliptic theory that there exists a continuously differentiable function w : R n [0, ) such that a subsequence of w j (which we again denote by w j ) converges uniformly on compact subsets of R n to w. (However unless a = 1 we cannot conclude that w is C 2.)
S.D. TALIAFERRO / Ann. I. H. Poincaré AN 19 (2002) 889 901 895 Letting j in (2.8) and using (2.9), (2.10), and Fatou s lemma we get aw(ζ) λ w(ξ) α n ζ ξ dζ for ξ n 2 Rn. (2.11) R n Thus lim inf ζ w(ζ) = 0 for otherwise the integral in (2.11) would be infinite for each ξ R n. We now show that (2.11) holds with the inequality reversed and the factor a omitted. Let ξ 1 R n and let ε>0. Choose ξ 2 R n such that w(ξ 2 )<ε/2 n 1 and then choose r>0suchthat ζ ξ 2 2 ζ ξ 1 for ζ >r. (2.12) For j large enough that B r (0) j {ξ j } we have by (2.8) that where w j (ξ 1 ) = α n I j := α n j B r (0) ζ <r w j (ζ ) dζ ζ ξ 1 n 2 + I j + H j (ξ 1 ) (2.13) w j (ζ ) dζ ζ ξ 1 n 2 2 n 2[ w j (ξ 2 ) H j (ξ 2 ) ], by (2.12) and (2.8) ε, for j large by (2.9). Thus letting j in (2.13) and using (2.10) we find w(ξ 1 ) α n ζ <r w(ζ) λ dζ ζ ξ 1 + ε α n 2 n Since ε>0 was arbitrary it can be omitted in (2.14). Thus letting w(ζ) λ dζ v(ξ) = α n ζ ξ n 2 R n R n w(ζ) λ dζ + ε. (2.14) ζ ξ 1 n 2 we have by (2.11) that v is a C 2 nonnegative solution of (2.2). By (2.7), w(0) = 1and thus v is positive on R n. LEMMA 2. Suppose, for j = 1, 2,...,there exists a C 2 positive solution v j (y) of a j v λ j v j v λ j in j v j (y) O( y 2/(λ 1) ) as y where j is the complement of a compact subset of R n (n > 2), λ is a constant satisfying (1.2), and the sequence {a j } j=1 (0, 1] converges to a (0, 1]. Then there exists a C 2 positive solution v of (2.2).
896 S.D. TALIAFERRO / Ann. I. H. Poincaré AN 19 (2002) 889 901 Proof. Let u j (x) = y n 2 v j (y) where y = x/ x 2. (This is the Kelvin transform.) Then u j (x) is a C 2 positive solution of (2.1a,b) where σ = n + 2 λ(n 2) (0, 2) and Ɣ j = { x R n : x/ x 2 } j is a punctured neighborhood of the origin. Thus Lemma 2 follows from Lemma 1. 3. Proofs In this section we prove Theorems 1 7. Proof of Theorem 1. Define W, w: R n R by 1, for y < 1 W(y)= 1, for y 1 y n 2 and w = N(W λ ) where N is the Newtonian potential operator over R n.thenw(y) equals [( ) ( 1 1 n 2 2 + 1 1 λ(n 2) 2 2 1 ] ) y 2 n or [( ) ( ) ] 1 1 1 n 2 y n 2 n + 1 1 λ(n 2) n λ(n 2) n 1 1 λ(n 2) 2 y λ(n 2) n depending on whether y < 1or y 1 respectively. Hence C 1 W(y) w(y) C 2 for y R n (3.1) where C 1 = C 1 (n, λ) and C 2 = C 2 (n, λ) are positive constants whose values will not change from line to line. For each h>0definev h, v h : R n R by V h (x) = KW(x/h), wherek = K(h) satisfies K λ 1 h 2 = 1, and v h = N(Vh λ).then ( ) x v h (x) = Kw for x R n. h Thus by (3.1) For later note that C 1 V h(x) v h (x) C 2 for x R n and h>0. (3.2) K, for x <h V h (x) = L x, for x h (3.3) n 2
S.D. TALIAFERRO / Ann. I. H. Poincaré AN 19 (2002) 889 901 897 where L = L(h) satisfies L λ 1 = h λ(n 2) n. (3.4) Let ε be a positive constant and let ϕ : (0, 1) (0, ) be a continuous function. Let U 0 = W and let {ε j } j=1 (0, 1) be a sequence such that Since, by (3.3) and (3.4), lim h 0 + ε j = ε. (3.5) j=1 V h (x sup xj ) = 0 for j = 1, 2,..., x x j r j U 0 (x) where x j = (2 2j, 0,...,0) R n and r j = 2 2j 1, there exists h j (0,r j ), j = 1, 2,...,such that L λ j < where L j = L(h j ) = h (λ(n 2) n)/(λ 1) j, (3.6) and j=1 lim j K j ϕ( x j ) = where K j = K(h j ) = h 2/(λ 1) j, (3.7) U j (x) C 1 ε j U 0 (x) for x x j r j and j = 1, 2,... (3.8) C 2 where U j (x) = V hj (x x j ) for j = 1, 2,... Letting u j = N(Uj λ ), j = 0, 1,...,and noting that (3.2) and (3.1) imply C 1 U j(x) u j (x) C 2 for x R n and j = 0, 1,... (3.9) it follows from (3.8) that u j 1 U j ε j U 0 ε j u 0 on R n ( B rj x j ) (3.10) C 1 C 2 for j = 1, 2,... Let r 0 > 0. Choose a positive integer j 0 such that x j 0 +r j0 <r 0. Then for j j 0 and x > 2r 0 we see by (3.3) that and U j (x) λ = V hj ( x x j ) λ = L λ j x x j (n 2)λ Lλ j r (n 2)λ 0 ( U j (x) λ) = (n 2)λLλ j x x j (n 2)λLλ j 1+(n 2)λ r (n 2)λ+1. 0 (3.11)
898 S.D. TALIAFERRO / Ann. I. H. Poincaré AN 19 (2002) 889 901 Thus by (3.6), j=j 0 U j (x) λ is C 1 and uniformly Lipschitz continuous on x > 2r 0. Hence j=0 U j(x) λ is uniformly Lipschitz continuous on x >ρfor each ρ>0. Also by the monotone convergence theorem, u := N ( Uj λ j=0 ) = N ( Uj λ ) = u j in R n {0} (3.12) j=0 which by (3.10) and (3.5) implies that u is finite at each point of R n {0}. Hence u is C 2 on R n {0} and u = j=0 Uj λ C2 λ uλ j Cλ 2 uλ in R n {0}. (3.13) j=0 j=0 Furthermore, it follows from (3.12), (3.9), and (3.7) that u(x j ) ϕ( x j ) u j(x j ) ϕ( x j ) 1 U j (x j ) C 2 ϕ x j ) = 1 K j C 2 ϕ( x j ) as j. To complete the proof of Theorem 1, we now show u C λ 1 (1 + ε) λ 2 λ 1 uλ in R n {0}. (3.14) First by (3.12), (3.10), (3.5), (3.9), and (3.13) we have in R n {0} j=1 B r j (x j ) that C λ 1 uλ C λ 1 (u 0 + εu 0 ) λ (1 + ε) λ U λ 0 (1 + ε)λ ( u). Next let j be a fixed positive integer. Then by (3.10), u i ε i u 0 in B rj (x j ) for i/ {0,j}. Thus by (3.12), (3.5), (3.9), and (3.13) we have in B rj (x j ) that C1 λ uλ C1 λ (u 0 + u j + εu 0 ) λ C1 λ (1 + ε)λ (u 0 + u j ) λ C1 λ (1 + ε)λ 2 λ 1( u λ 0 + ) uλ j (1 + ε) λ 2 λ 1( U0 λ + U j λ ) (1 + ε) λ 2 λ 1 ( u). Hence (3.14) holds. Remark. The proof of Theorem 1 gives an explicit expression for a(n,λ). Indeed, letting A 1 = 1 2 + 1 λ(n 2) 2, A 2 = 1 n + 1 λ(n 2) 2, and A 3 = 1 n + 1 λ(n 2) n,
S.D. TALIAFERRO / Ann. I. H. Poincaré AN 19 (2002) 889 901 899 the optimal choices for C 1 (n, λ) and C 2 (n, λ) in (3.1) are C 1 = n 2 max{a 1,A 3 } and C 2 = n 2 A 2 and the proof of Theorem 1 shows that a = a(n,λ) in the statement of Theorem 1 can be taken to be any positive number less than C λ 1 C λ 2 2 λ 1 = A λ 2 2 λ 1 max{a λ 1,A λ 3}. Proof of Theorem 2. Suppose for contradiction that for each constant a (0, 1) there exists a C 2 positive solution u(x) of (1.3) in a punctured neighborhood of the origin in R n such that (1.4) does not hold. Then letting {a j } j=1 (0, 1) be any sequence which converges to 1, we have by Lemma 1 with σ = 0 that there exists a C 2 positive solution of (1.1) in R n which contradicts Theorem B in the introduction. Let u(x) be a C 2 positive solution of (1.3) in a punctured neighborhood of the origin in R n which satisfies (1.4). To complete the proof of Theorem 2, it remains only to show that u(x)/ū( x ) is bounded between positive constants for x small and positive, and this is indeed the case by the asymptotic Harnack inequality (see Veron [10, Lemma 1.5]) and the fact that 0 < u(x) u(x) λ 1 = O ( x 2) as x 0 +. u(x) Proof of Theorem 3. A proof of Theorem 3 can be obtained by making the following changes to the proof of Theorem 1. The first change occurs above (3.6) where we now define x j = (2 2j, 0,...,0) and r j = 1. The next change is to replace the paragraph containing equation (3.11) with the following paragraph. Let r 0 > 1. Choose a positive integer j 0 such that x j 0 1 >r 0. Then for j j 0 and x <r 0 we see by (3.3) that U j (x) λ = V hj ( x x j ) λ = L λ j x x j (n 2)λ Lλ j and ( U j (x) λ) = (n 2)λLλ j x x j (n 1+(n 2)λ 2)λLλ j. Thus by (3.6), j=j 0 U j (x) λ is C 1 and uniformly Lipschitz continuous on x <r 0. Hence j=0 U j(x) λ is uniformly Lipschitz continuous on x <ρfor each ρ>0. Finally, in the remainder of the proof of Theorem 1 replace each occurrence of R n {0} with R n. Proof of Theorem 4. The proof of Theorem 4 is basically the same as the proof of Theorem 2 except we use Lemma 2 instead of Lemma 1. We omit the details.
900 S.D. TALIAFERRO / Ann. I. H. Poincaré AN 19 (2002) 889 901 Proof of Theorem 5. Theorem 5 has three short proofs. First, Theorem 3 implies Theorem 5. Second, by Theorem 1 there exists a constant a (0, 1) such that (1.3) has a C 2 positive solution u in R n {0} which does not satisfy (1.4). Applying Lemma 1 with σ = 0 and with a j = a and u j = u for j = 1, 2,..., we obtain Theorem 5. And third, a C 2 positive solution of (1.3) in R n with a = (C 1 /C 2 ) λ is u = C λ/(λ 1) 2 w, wherec 1, C 2, and w are as defined in the first paragraph of the proof of Theorem 1. Proof of Theorem 6. Suppose for contradiction there exists a sequence {a j } j=1 (0, 1) such that lim j a j = 1 and a sequence {u j : R n (0, )} j=1 of C2 functions satisfying a j u λ j u j u λ j in R n. (3.15) It follows then from Lemma 2 and Theorem B that (after omitting a finite number of terms of the sequence u j ) u j (x) = O ( x 2/(λ 1)) as x. (3.16) Thus u j assumes its maximum at some x j R n. By translating and scaling each u j we can assume x j = 0and u j (0) = 1 = max u j. (3.17) R n It follows from (3.15), (3.17), and standard elliptic theory that there exists a continuously differentiable function u : R n [0, ) such that a subsequence of u j (which we again denote by u j ) converges uniformly on compact subsets of R n to u. Thus multiplying (3.15) by a nonnegative test function ϕ C0 (Rn ), integrating over R n, integrating by parts, and letting j we obtain u λ u u λ in D ( R n). Hence u = u λ in D (R n ) and this together with the fact that u is C 1 implies u is C 2, which contradicts Theorem B. Proof of Theorem 7. Let {a j } j=1 (0, 1) I 2 be a sequence which converges to the left endpoint A 2 of I 2. Then by the last paragraph of the proof of Theorem 2 there exists a C 2 positive solution u j of a j u λ j u j u λ j in a punctured neighborhood Ɣ j of the origin such that u j (x) O ( x 2/(λ 1)) as x 0 +. Thus by Lemma 1 with σ = 0wehaveA λ 2 I 5. Hence I 6 ( A λ 2, 1) = (A 2, 1) λ I λ 2. Similarly, using Lemma 2 instead of Lemma 1, we obtain I 6 I λ 4.
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