A ”SUP + C INF” INEQUALITY FOR LIOUVILLE-TYPE EQUATIONS WITH SINGULAR POTENTIALS DANIELE BARTOLUCCI(1) Abstract. Let Ω ⊂ R2 be open and bounded. We prove that for any α ∈ (−1, 1), for any C1 > max{1,
1+α } 1−α
and for any compact subset K ⊂⊂ Ω, there exist C2 > 0 such that sup u + C1 inf u ≤ C2 , Ω
K
for any solution of, −∆u = |x|2α V eu
in
Ω,
with V any continuous function which satisfies 0 < a ≤ V ≤ b < +∞.
1. Introduction We are interested in the analysis of the qualitative properties of solutions for, −∆u = |x|2α V eu
(1.1)
in
Ω,
with Ω ⊂ R2 open and bounded, α ∈ (−1, 1). We always assume that 0 ∈ Ω. Let us suppose for the moment that V satisfies, (1.2)
V ∈ L∞ , 0 < a ≤ V (x) ≤ b < +∞, a.e. in Ω,
for some given 0 < a ≤ b < ∞. We will say that u is a solution for (1.1), if u ∈ L1 (Ω), |x|2α V eu ∈ L1 (Ω) and u is a distribution solution for (1.1). By the results in [10] and standard elliptic regularity theory, any such solution 1,γ 2,p 1,γ for (1.1) satisfies u ∈ Cloc (Ω), for any γ ∈ (0, 1) if α > 0, while u ∈ Wloc (Ω \ {0}) ∩ Cloc (Ω \ 2,q 0 (Ω), for any p ∈ [1, +∞), γ ∈ (0, 1), q ∈ [1, 1 ) if α ∈ (−1, 0). {0}) ∩ Wloc (Ω) ∩ Cloc α
2000 Mathematics Subject classification: 35B45, 35J60, 35J99. (1)
Daniele Bartolucci, Department of Mathematics, University of Rome ”La Sapienza”,
Piazzale A. Moro n.5, 00185 Roma, Italy. 1
2
The analysis of (1.1) is motivated by many problems in pure and applied mathematics. We mention, among others, the conformal geometry of surfaces with conical singularities [15], [17], [33], the statistical mechanics of point vortices in 2d turbulence [11], [12], [13] and of self-gravitating clouds [34], [35] interacting with singular sources [5], [6] and the electroweak theory of GlashowSalam-Weinberg [8], [28]. We also refer to the monographs [32], [36] for many other applications in gauge field theories. Our concern is to establish a ”sup +C inf” inequality [27], for solutions of (1.1)-(1.2) in the case where α ∈ (−1, 1). In particular, by assuming V to be any continuous function satisfying (1.2), we will obtain, in case α ∈ (−1, 1), a result similar to the one provided by I. Shafrir [27]. The analysis of such kind of inequalities has also attracted a lot of attention, see [9], [18], [27], [31], due to their consequences on the subtle quantization properties [24], [29], [31] and pointwise estimates [4], [23] for sequence of solutions for Liouville type elliptic problems, and their deep consequences on the existence theory for equations from mathematical physics [8], [19], [20], [21].
For a better understanding of our motivations, let us briefly recall some known facts concerning the ”regular” case, i.e. α = 0. In their seminal paper [10], among many other things, H. Brezis and F. Merle conjectured that for any compact subset K ⊂ Ω, there exist C1 ≥ 1 and C2 > 0 such that sup u + C1 inf u ≤ C2 for any solution of (1.1)-(1.2) with α = 0. Then, I. Shafrir [27], K
Ω
was able to prove that their conjecture was correct, by showing that for any compact subset K ⊂ Ω, there exist C1 ≥ 1 depending indeed only by a, b, and C2 > 0, depending also by dist(K, ∂Ω), such that, (1.3)
sup u + C1 inf u ≤ C2 , K
Ω
for any solution of (1.1)-(1.2) with α = 0. His result has been further improved by H. Brezis, Y.Y. Li and I. Shafrir in [9]. Indeed, they proved the so-called ”sup + inf” inequality, i.e., by assuming (1.4)
k∇V k∞ ≤ A,
3
they obtained that (1.3) holds true with C1 = 1 and, of course, C2 depending only by a, b, A and dist(K, ∂Ω). Finally C.C. Chen and C.S. Lin [18] proved that the ”sup + inf” inequality holds true by assuming an even weaker condition on V . Indeed, they answered an open problem in [9] and proved, among other things, that it is enough to assume V to be locally uniformly bounded in any H¨older continuous norm; (actually their sufficient conditions are much weaker, see theorem 1.3 in [18]). We remark that they also obtain the optimal constant in (1.3) for solutions of q (1.1)-(1.2) and α = 0, i.e. C1 = ab . In literature, author’s also refers to the inequality (1.3) as the Harnack type inequality, see [23].
We observe at this point that, in case α 6= 0, the translation invariance of the problem is lost, and the blow up analysis [10], [24] for problem (1.1) is more delicate. We refer to [8], [4], [31] for a complete discussion concerning this problem in case α > 0. On the other side, it seems that the case α ∈ (−1, 0) is easier. Indeed, we may observe for example that, as in the case α = 0, solutions for (1.1) with V ≡ 1 and α ∈ (−1, 0) on either Ω = Br (0) with u = 0 on ∂Br (0) or on Ω = R2 are necessarily radial, [6], [26]. This is not the case if α > 0, see [5], [26]. Moreover, the Brezis-Merle concentration compactness theory for solutions of (1.1)-(1.2) can be generalized to the case α ∈ (−1, 0) more or less by the same arguments provided in [10], see [6]. Once more, the case α > 0 seems to be more delicate. Indeed, the corresponding results have been obtained only in case V satisfies to the much stronger assumption (1.4), see [8]. A partial extension of the Brezis-Merle theory, limited to the case α ∈ [0, 1], has been obtained by assuming V to be continuous in [1]. Then, it is not too surprising that the Shafrir’s inequality (1.3) has been recently generalized to cover the case α ∈ (−1, 0), for solutions of (1.1)-(1.2), see [3]. So far, the only known result in this direction concerning the case α > 0, has been obtained in [30], by G. Tarantello. She obtained a ”milder” version of the ”sup + inf” inequality, that is, she proved that for any α > 0,
4
there exist C2 > 0 such that, u(0) + inf u ≤ C2 ,
(1.5)
Ω
for any solution of (1.1)-(1.2)-(1.4). Another partial result concerning the Harnack type inequality in case α > 0 has been obtained in [4], where the ”sup + inf” inequality [9] (i.e. (1.3) R with C1 = 1) has been obtained by virtue of the further assumptions Ω |x|2α eu ≤ c and max u − min u ≤ c0 , for some given uniform constants c > 0 and c0 ≥ 0. ∂Ω
∂Ω
So far, at least to our knowledge, there is no proof of the fact that the inequality (1.3) holds true for solutions of (1.1)-(1.2)-(1.4) with α > 0. Our main aim is to show, at least in case α ∈ (0, 1), that indeed (1.3) holds true for any C1 >
1+α 1−α ,
for solutions of (1.1)-(1.2) whenever
V is further assumed to be continuous. In particular, C2 depends by a, b dist(K, Ω) and the modulus of continuity of V on K. More exactly, our main result is the following, Theorem 1.1. Fix α ∈ (−1, 1), and let 0 (Ω). V ∈ Cloc
(1.6)
1+α }, there exist C2 > 0 depending by For any compact subset K ⊂ Ω and for any C1 > max{1, 1−α
a, b, α, dist(K, ∂Ω) and the uniform modulus of continuity of V on K, such that, (1.7)
sup u + C1 inf u ≤ C2 , K
Ω
for any solution u of (1.1)-(1.2)-(1.6). Remark 1.2. It is clear from the statement that (1.7) holds true uniformly in V ∈ F as far 0 (Ω). In other words, if we assume that V belongs to a family F which satisfies as F ⊂⊂ Cloc
(1.2) and admits a (locally uniform) modulus of continuity in Ω, then (1.7) holds true for 1+α any C1 > max{1, 1−α } and C2 depending by a, b, α, dist(K, ∂Ω) and the uniform modulus of
continuity of F on K. Remark 1.3. Notice that in case α = 0, theorem 1.1 asserts that indeed (1.3) holds true for any C1 > 1. Of course, we may obtain this improved version of the result in [27] only because of the
5
much stronger (then (1.2) of course) assumption (1.6). By means of a counterexample, it was proven in [9] that (1.3) with C1 = 1 does not holds true for solutions of (1.1)-(1.2) with α = 0 0 (Ω). We refer to [9] pg. 355-357 for further details. Inspired by [9], we will and V ∈ F ⊂⊂ Cloc
provide an example which yields a similar result in our situation as well, see section 3 for more details. Another open problem posed in [9] was that to establish whether or not (1.3) holds true for C1 = 1 and V a fixed continuous function. To our knowledge this problem is still open. It turns out that the result of theorem 1.1 is not sharp in case α > 0. Indeed, the analysis of radial solutions yields a version of (1.7) which holds true for any C1 > 1 . In particular we obtain the following, Theorem 1.4. For any r0 > 0 such that Br0 (0) ⊂⊂ Ω, for any compact subset K ⊂ Br0 (0) and for any C1 > 1, there exist C2 > 0 depending by a, b and the uniform modulus of continuity of V on K, such that, sup u + C1 inf u ≤ u(0) + C1 inf u ≤ C2 , Ω
K
Ω
for any radial solution u of (1.1)-(1.2)-(1.6). As a consequence of (1.7), we have a generalization of the Li and Shafrir’s [24] quantization theorem to the case where α is assumed to satisfy α ∈ (−1, 1). Theorem 1.5. Assume that {un } is a sequence of solutions for −∆un = |x|2α Vn eun in
B1 (0),
α ∈ (−1, 1),
0 ≤ Vn , in Ω, {Vn } ⊂ C 0 (B 1 (0)), Vn → V in C 0 (B 1 (0)), and Z B1 (0)
|x|2α eun ≤ C, ∀ n ∈ N.
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Suppose in addiction that max un → +∞, as n → +∞, B 1 (0)
and ∀ r ∈ (0, 1),
max
un → −∞, as n → +∞,
B 1 (0)\Br (0)
|x|2α Vn eun * βδp=0 , as n → ∞, weakly in the sense of measures in B1 (0). Then, β ∈ 8πN ∪ {8π(1 + α) + 8πZ+ }. Here Z+ = N ∪ {0}. Let us recall that in the ”regular” case α = 0, theorem 1.5 has been proved in [24], then providing an affirmative answer to the corresponding ”quantization” conjecture in [10]. One of the crucial tools in the subtle Li and Shafrir’s proof [24], was indeed the Harnack type inequality (1.3). It was then proved in [14] that in fact one can construct sequences where β = 8πm, where m ≥ 1 can be chosen to be any given integer number satisfying of course 8πm ≤ kV k∞ C. Motivated by the analysis of multivortex configurations in the Glashow-Salam-Weinberg electroweak theory, and by further assuming, k∇Vn k∞ ≤ A, ∀ n ∈ N,
(1.8)
and max un − min un ≤ c, for some uniform constants A > 0, c > 0, it was proved in [8] that, ∂Ω
∂Ω
for any given α > 0, then β = 8π(1 + α). The same argument works in case α ∈ (−1, 0), and we may conclude that β = 8π(1 + α) as well, see [6]. The quantization problem has been completely solved in case α > 0 by G. Tarantello in [29], where the author was able to drop the further boundary condition assumption. Indeed, by assuming (1.8), and by using (1.5), she proved that, for any α > 0, then either β = 8πn, N 3 n ≥ 1 or β = 8π(1 + α) + 8πn, Z 3 n ≥ 0. It has also been shown in [29] that this result is sharp, i.e. there exist sequences such that any one of the allowed values for β is achieved. Theorem 1.5 provides ( in case α ∈ (−1, 1) ) a full generalization of the quantization result in [24] and also a slight improvement of the result in [29], since we do not assume (1.8), but only the weaker condition Vn → V in C 0 (B 1 (0)).
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Since, for α ∈ (0, 1), (1.7) implies (1.5), then, in this particular case, the proof of theorem 1.5 can be worked out by a step by step adaptation of the argument in [29]. It seems moreover that this argument can be easily extended to cover the case α ∈ (−1, 0). Otherwise, by using (1.7), and by arguing directly as in [24], one may also work out another proof which covers the full range α ∈ (−1, 1). Hence, we will not discuss the proof of theorem 1.5, and refer the reader to [24] and [29] for further details.
This paper is organized as follows. In section 2 we prove theorems 1.1 and 1.4. In section 3 we provide a counterexample relevant for the analysis of (1.7).
2. The ”sup +C inf” inequality In this section we will prove theorems 1.1 and 1.4. We will first state a result concerning blow up sequences of solutions for (1.1)-(1.2)-(1.6), suitable to be applied to our problem. Lemma 2.1. Let {un } be a sequence of solutions for, −∆un = |x|2αn Vn eun in
Ω,
{αn } ⊂ (−1, 1], αn → α ∈ (−1, 1], as n → +∞, 0 ≤ Vn , in Ω, {Vn } ⊂ C 0 (Ω), Vn → V in C 0 (Ω), and Z
|x|2αn eun ≤ C, ∀ n ∈ N.
Ω
Suppose in addiction that there exist x0 ∈ Ω, and r0 > 0 such that there exist a sequence of points {xn } ⊂ B2r0 (x0 ) ⊂⊂ Ω such that xn → x0 , as n → +∞, and un (xn ) → +∞, as n → +∞.
8
Then, either α ∈ [0, 1], and we have, Z lim inf
n→+∞
|x|2αn Vn eun ≥ 8π, ∀ 0 < r ≤ r0 ,
Br (x0 )
or α ∈ (−1, 0) and x0 = 0, and we have, Z
|x|2αn Vn eun ≥ 8π(1 + α), ∀ 0 < r ≤ r0 .
lim inf
n→+∞ Br (0)
Proof of lemma 2.1 If x0 6= 0, the conclusion follows by arguing as in lemma 1 in [24], while, in case x0 = 0 and α ≥ 0, then we may argue as in lemma 1 in [1]. If x0 = 0 and α ∈ (−1, 0), we simply observe that, by using the classification of solutions in [26], the blow up argument in [1] yields the desired conclusion as well. In what follows, we will often extract subsequences. To avoid cumbersome notations, and in case there is no matter of confusion, we will not relabel the corresponding indices. Proof of theorem 1.1 Clearly, we may assume without loss of generality that 0 ∈ K. Notice that, for any t > 0, and any sequence of solutions {un } for (1.1)-(1.2)-(1.6) in Ω, then (t)
(t)
un (y) = un (ty)+2(1+α) log t is a sequence of solutions for (1.1) with Vn (y) = Vn (ty) satisfying (1.2) and (1.6) in Ωt := {y ∈ R2 | ty ∈ Ω} as well. Then, it is not difficult to verify that we may pass to a smaller K if necessary, and assume that K ⊂ B1 (0) ⊂ B3 (0) ⊂⊂ Ω. 1+α We argue by contradiction and assume that, for any given C1 > max 1, and K ⊂⊂ 1−α B1 (0), there exist a sequence {Vn } satisfying, 0 < a ≤ Vn (x) ≤ b < +∞, ∀ x ∈ Ω, and {Vn } ⊂ F ⊂⊂ C 0 (B 3 (0)) such that there exist a sequence of solutions {un } of −∆un = |x|2αn Vn eun
in
Ω,
9
such that sup un + C1 inf un → +∞, as n → +∞. Ω
K
Clearly, passing to a subsequence, we may assume that Vn → V, in C 0 (B 3 (0)).
(2.1)
Extracting a further subsequence if necessary, we may find {yn } ⊂ K such that yn → y0 ∈ K and un (yn ) + C1 inf un = max un + C1 inf un → +∞, as n → +∞.
(2.2)
Ω
K
Ω
Step 1 In step 1 we will consider the case y0 = 0. We first use a slightly modified version of the arguments provided in [27] and [9] at the very beginning of their proofs. For any r ∈ (0, 1] and n ∈ N, we define, Z C1 (2.3) ψn (r; α) := un (yn ) + un dσ + 2(C1 + 1)(1 + α) log r. 2πr ∂Br (yn )
Integrating by parts, we conclude that, Z
d C1 ψn (r; α) = dr 2πr
∂un C1 + 1 dσ + 2(1 + α) = ∂ν r
∂Br (yn )
C1 2πr
Z ∆un + 2(1 + α)
C1 + 1 = r
Br (yn )
−
C1 2πr
Z
C1 + 1 |x|2αn Vn eun dx + 2(1 + α) , ∀ r ∈ (0, 1), ∀ n ∈ N, r
Br (yn )
so that d ψn (r; α) ≥ 0, if and only if dt
Z
C1 + 1 |x|2α Vn eun dx ≤ 4π(1 + α) . C1
Br (yn )
Notice that, since C1 > max 1, (2.4)
1+α , then, 1−α
4π(1 + α)
C1 + 1 < min{8π, 8π(1 + α)}. C1
10
R At this point, for any n ∈ N, either |x|2α Vn eun dx ≤ 4π(1 + α) CC1 +1 , and then we de1 B1 (yn ) R fine sn := 1, or |x|2α Vn eun dx > 4π(1 + α) CC1 +1 and we may choose sn < 1, such that 1 B1 (yn ) R . In any case, ∀ n ∈ N, ψn has a unique maximum point |x|2α Vn eun dx = 4π(1 + α) CC1 +1 1
Bsn (yn )
at sn , ψn (1; α) ≤ ψn (sn ; α), ∀ n ∈ N,
(2.5)
and, by using also (2.4), Z C1 + 1 |x|2α Vn eun dx ≤ 4π(1 + α) (2.6) < min{8π, 8π(1 + α)}, ∀ n ∈ N. C1 Bsn (yn )
Clearly we have un (yn ) + C1 inf un ≤ un (yn ) + C1 inf un ≤ ψn (1; α), Ω
B1 (yn )
while, since un is superharmonic, ψn (sn ; α) ≤ (1 + C1 )un (yn ) + 2(1 + C1 )(1 + α) log sn . By using (2.2) and (2.5), we conclude that (2.7)
un (yn ) + 2(1 + α) log sn → +∞, as n → +∞.
Clearly, un satisfies, (2.8)
−∆un = |x|2α Vn eun in Bsn (yn ).
At this point, we have to analyze the following two situations: either: CASE (A) there exist a subsequence {snk } such that ∃ s > 0 such that snk ≥ 2s > 0, ∀k ∈ N, or, CASE (B) sn → 0+ , as n → +∞.
CASE (A) If there exist a subsequence {snk } such that snk ≥ 2s > 0, ∀ k ∈ N, then, since yn → 0, by using
11
R
(2.6), we can pass to a further subsequence such that
Bs (0)
< |x|2αnk Vnk eunk ≤ 4π(1 + α) CC1 +1 1
min{8π, 8π(1 + α)}, for any k ∈ N. The assumption inf Vn ≥ a > 0, ∀ n ∈ N, reveals here its Ω
crucial role. Indeed, it implies that Z
C1 + 1 |x|2α eunk ≤ 4π(1 + α) , ∀ k ∈ N. aC1
Bs (0)
As a consequence, by using (2.1), (2.7) and (2.8), we see that {unk } satisfies to all the assumptions of lemma 2.1 in Bs (0) with x0 = 0. We then conclude that, along a subsequence, Z lim inf k→+∞
|x|2αnk Vnk eunk ≥ min{8π, 8π(1 + α)},
Bρ (0)
for any ρ < s small enough, which is of course a contradiction.
CASE (B) We have to distinguish two further cases: either: CASE (B.1) there exist C > 0 such that
|yn | ≤ C, ∀ n ∈ N, sn
or, CASE (B.2) there exist subsequences {yn } and {sn }, such that
|yn | → +∞, as n → +∞. sn
CASE (B.1) We may extract a further subsequence if necessary, so that ξn :=
yn sn
→ ξ0 ∈ R2 . Next, let us
define vn (z) = un (yn + sn (z − ξn )) + 2(1 + α) log sn , for any z ∈ B1 (ξn ). Clearly, for any n large enough, B 1 (ξ0 ) ⊂ B1 (ξn ). Hence, by using (2.8) and (2.6), it is not difficult to verify that vn 2
satisfies −∆vn = |z|2αn V˜n evn in B 1 (ξ0 ) 2
Z B 1 (ξ0 ) 2
C1 + 1 |z|2αn V˜n evn ≤ 4π(1 + α) < min{8π, 8π(1 + α)}, ∀ n ∈ N, C1
12
where, V˜n (z) = Vn (yn + sn (z − ξn )), V˜n → V (0) uniformly in B 1 (ξ0 ). 2
Note that, by using (2.7), we have, vn (ξn ) = un (yn ) + 2(1 + αn ) log sn → +∞, as n → +∞. Since inf Vn ≥ a > 0, ∀ n ∈ N, we also have, Ω
Z
C1 + 1 |z|2α evn ≤ 4π(1 + α) , ∀ n ∈ N. aC1
B 1 (ξ0 ) 2
Then, we conclude that {vn } satisfies to all the assumptions of lemma 2.1 in B 1 (ξ0 ) with x0 = ξ0 . 2
Hence, along a subsequence, Z
|z|2α V˜n evn ≥ min{8π, 8π(1 + α)},
lim inf
n→+∞ Bρ (ξ0 )
for any ρ > 0 small enough, which is the desired contradiction.
CASE (B.2) We discuss the cases where α ∈ (0, 1) and α ∈ (−1, 0) separately. CASE (B.2) and α ∈ (0, 1). |yn | Since → +∞, as n → +∞, we may extract a further subsequence if necessary, so that sn |yn | ≥ 4, ∀ n ∈ N. Then, by using (2.7), we see that, sn (2.9)
un (yn ) + 2 log sn + 2α log |yn | ≥ un (yn ) + 2(1 + α) log sn → +∞, as n → +∞.
For any z ∈ B1 (0), we define, yn sn vn (z) = un (yn + sn z) + 2 log sn + 2α log |yn | + 2α log + z. |yn | |yn | By using (2.8) and (2.6), it is not difficult to verify that vn satisfies, −∆vn = V˜n evn in B1 (0)
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Z
C1 + 1 V˜n evn ≤ 4π(1 + α) < 8π, ∀ n ∈ N, C1
B1 (0)
where, V˜n (z) = Vn (yn + sn z), V˜n → V (0) uniformly in B1 (0). Note that by (2.9) we have, vn (0) = un (yn ) + 2 log sn + 2α log |yn | → +∞, as n → +∞. Since inf Vn ≥ a > 0, ∀ n ∈ N, we also have, Ω
Z
C1 + 1 evn ≤ 4π(1 + α) , ∀ n ∈ N. aC1
B1 (0)
We conclude that {vn } satisfies to all the assumptions of lemma 2.1 in B1 (0) with {αn } ≡ {0} and x0 = 0. Hence, along a subsequence, Z lim inf
n→+∞
V˜n evn ≥ 8π,
Bρ (0)
for any ρ > 0 small enough, which is the desired contradiction.
CASE (B.2) and α ∈ (−1, 0). We observe that the argument provided in case α ∈ (0, 1) does not work in this case, since the reversed inequality holds true in (2.9). Let ψn (r; α) be defined by (2.3). By arguing as in the lines following (2.3), ∀ n ∈ N we define tn ∈ (0, 1], to be the unique maximum point of ψn (r; 0), which then satisfies, ψn (1; 0) ≤ ψn (tn ; 0), ∀ n ∈ N,
(2.10) and, Z (2.11) Btn (yn )
C1 + 1 |x|2α Vn eun dx ≤ 4π < 8π, ∀ n ∈ N. C1
14
As above, since un is superharmonic, and by using (2.2) and (2.10), we conclude that, un (yn ) + 2 log tn → +∞, as n → +∞.
(2.12)
Of course, (2.11) and (2.12) does not coincide with the corresponding expressions (2.6) and (2.7) since tn is the maximum of ψn (r; α)|α=0 . Put, δn un (yn ) , and τn = . δn = exp − 2 |yn |α Since |yn |/sn → +∞, and using (2.2), we conclude that, (2.13)
τn |yn |
2
=
δn |yn |1+α
2 = exp [−(un (yn ) + 2(1 + α) log |yn |)] ≤
exp [−(un (yn ) + 2(1 + α) log sn )] → 0, as n → +∞. Moreover, by setting dn = (2.14)
(dn )2 =
tn τn
2
tn τn ,
=
and using (2.12) we have,
tn |yn |α δn
2
= |yn |2α exp [(un (yn ) + 2 log tn )] → +∞, as n → +∞.
At this point, let us define vn (z) = un (yn + τn z) − un (yn ) for any z ∈ Bdn (0). Clearly, vn satisfies −∆vn = V˜n evn in Bdn (0) Z
C1 + 1 < 8π, ∀ n ∈ N, V˜n evn ≤ 4π C1
(2.15) Bdn (0)
where, yn τn 2α ˜ Vn (z) = + z Vn (yn + τn z). |yn | |yn | Moreover, vn (z) ≤ vn (0) = 0, ∀ z ∈ B tn (0), τn
and, by using (2.13) and (2.14), we see that for any R ≥ 1, and for any n large enough, V˜n → V (0) uniformly on BR (0).
15
It follows in particular that, for any R ≥ 1, and for any n large enough, V˜n (z) ≥ a ˜ > 0, ∀ z ∈ BR (0) and then, Z
C1 + 1 evn ≤ 4π , a ˜C1
BR (0)
for some uniform a ˜ > 0. Thus, by using the Brezis-Merle concentration compactness theorem [10], we may find a diagonal subsequence {vnk } which converges uniformly on BR (0), for any R > 0, to the well known, see [16], and unique solution of −∆v = V (0)ev on R2 , v(z) ≤ v(0) = 0, R v e < +∞. It follows in particular that,
∀ z ∈ R2 ,
R2
Z V (0)
ev = 8π,
R2
and then, Z lim inf lim
R→+∞ k→+∞ Bdn (0)
V˜nk evnk ≥ lim
k
Z lim
R→+∞ k→+∞ BR (0)
V˜nk evnk = 8π,
which is a contradiction to (2.14)-(2.15). Step 2 In step 2 we will consider the case y0 6= 0. The proof of step 2 is easier. Indeed, since y0 6= 0, we may define Kn (x) = |x|2α Vn (x) and use (2.1) to conclude that {un } is a sequence of solutions for, −∆un = Kn eun in B |y0 | (y0 ), 2
with Kn satisfying 0 < a ≤ Kn ≤ b < +∞, ∀ n ∈ N, Kn → K, uniformly in B |y0 | (y0 ), 2
for suitable 0 < a ≤ b < +∞. Then, the proof of step 2 reduces to a particular case of the proof of step 1 where one sets α = 0. In particular, it is easy to verify that we may fix C1 to be any positive number strictly greater then 1. We skip the details to avoid repetitions. Proof of theorem 1.4 As in the proof of theorem 1.1, without loss of generality we may assume that r0 ≥ 1, B1 (0) ⊂ 1,γ Br0 (0) ⊂⊂ Ω and d0 = dist(0, ∂Ω) > r0 . Since u ∈ Cloc (Ω) for any γ ∈ (0, 1), for any K ⊂⊂ Ω
16
we have max |∆u| ≤ CK . Hence, using the radial assumption and integrating (1.1) on (0, ρ), K
ρ ∈ (0, d0 ), it is not difficult to verify that, d ρ u=− dρ
Zρ
r2α+1 V (r)eu dr.
0
It follows that u is strictly decreasing in [0, d0 ) and in particular that for any K ⊂⊂ Ω, max u ≤ u(0). K
At this point, we may argue by contradiction as in the proof of theorem 1.1, and assume that, for any given C1 > 1, there exist a subsequence {Vn } which satisfies, 0 < a ≤ Vn (x) ≤ b < +∞, ∀ x ∈ Ω, and Vn → V, in C 0 (B 1 (0)) such that there exist a sequence of solutions {un } for, −∆un = |x|2α Vn eun
in
Ω,
such that un (0) + C1 inf un → +∞, as n → +∞. Ω
By arguing as in the proof of theorem 1.1, we conclude that there exist a sequence of positive numbers {sn } ⊂ (0, 1] satisfying un (0) + 2(1 + α) log sn → +∞, and Z
C1 + 1 |x|2α Vn eun ≤ 4π(1 + α) < 8π(1 + α), ∀ n ∈ N, C1
B1 (0)
where we have used C1 > 1. Define vn (z) = un (sn z) + 2(1 + α) log sn , which satisfies, −∆vn = |z|2α Vn (sn z)evn
in
B1 (0),
Vn (sn z) → V (0), unfiormly in B1 (0), Z (2.16) B1 (0)
C1 + 1 |z|2α Vn (sn z)evn ≤ 4π(1 + α) < 8π(1 + α), ∀ n ∈ N, C1
17
and vn (0) = un (0) + 2(1 + α) log sn → +∞, as n → +∞. Since inf Vn ≥ a > 0, ∀ n ∈ N, we also have Ω
Z
C1 + 1 |z|2α evn ≤ 4π(1 + α) , ∀ n ∈ N. aC1
B1 (0)
Then, by using a standard blow up argument, see for example lemma 1 in [1], and the result of Prajapat-Tarantello [26], we conclude that there exist a subsequence {vnk } such that Z lim
k→+∞ B1 (0)
|z|2α Vnk (snk z)evnk ≥ 8π(1 + α).
This fact clearly contradicts (2.16).
3. A counterexamples to (1.7) In this section, we provide an example relevant for the analysis of (1.7). Example 1 Let {αn } ⊂ (0, 1) satisfy αn → 0+ . Inspired by [9], we are going to construct a sequence of solutions {un } for (1.1)-(1.2)-(1.6), which violates (1.7) with C1 = 1. For any given α ∈ (0, +∞), 0 < b < +∞ and x ∈ R2 , let r = |x|. We define, ! 8(1 + α)2 b−1 r ∈ [0, 1], log 2 , 1 + r2(1+α) u(r; α) = 8(1 + α)2 b−1 r−2α , r ∈ (1, +∞). log (1 + r2 )2 It is easy to verify that u ∈ C 1 (R2 ), u(1; α) = log (2(1 + α)2 b−1 ), u(1; ˙ α) = −2(1 + α). In particular u is a weak solution for, −∆u = r2α V eu
in
R2 ,
18
where,
V (r) =
b,
b , (1 + α)2
r ∈ [0, 1],
r ∈ (1, +∞).
For any given n ≥ 2, we define, 1 δn = exp exp and un (r) = u(nr; αn ) + 2(1 + αn ) log δn . αn Clearly δn → +∞, and αn log δn → +∞, as n → +∞. Moreover, we have −∆un = r2αn Vn eun with
Vn (r) =
in
b,
b , (1 + αn )2 Since un is radial and super harmonic, we have,
B1 (0),
r ∈ [0, δ1n ], r ∈ ( δ1n , 1),
un (0) + inf un = un (0) + un (1) = 2 log (8(1 + αn )2 b−1 ) + log B1 (0)
2(1+αn ) 2 δn 2 (1 + δn )2
δn
! ≥
2 log (8(1 + α)2 b−1 )+2((1+αn )−1) log δn = 2 log (8(1 + α)2 b−1 )+αn log δn → +∞, as n → +∞. Since, Vn → b uniformly, the desired sequence can be constructed by a smoothing argument, see for example [9].
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[email protected]
