Extremal functions for singular Trudinger–Moser inequalities in the entire Euclidean space

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ScienceDirect J. Differential Equations 264 (2018) 4901–4943 www.elsevier.com/locate/jde

Extremal functions for singular Trudinger–Moser inequalities in the entire Euclidean space Xiaomeng Li a,b , Yunyan Yang a,∗ a Department of Mathematics, Renmin University of China, Beijing 100872, PR China b School of Information, Huaibei Normal University, Huaibei, 235000, PR China

Received 8 October 2016 Available online 2 January 2018

Abstract In a previous work (Adimurthi and Yang, 2010 [2]), Adimurthi–Yang proved a singular Trudinger–Moser inequality in the entire Euclidean space RN (N ≥ 2). Precisely, if 0 ≤ β < 1 and 0 < γ ≤ 1 − β, then there holds for any τ > 0,  u∈W 1,N (RN ),

 sup N N RN (|∇u| +τ |u| )dx≤1

RN

⎛ ⎞ kN N−2 N  α k γ k |u| N−1 1 ⎝ αN γ |u| N−1 N ⎠ dx < ∞, − e k! |x|Nβ k=0

1/(N−1)

where αN = N ωN−1 and ωN−1 is the area of the unit sphere in RN . The above inequality is sharp in the sense that if γ > 1 − β, all integrals are still finite but the supremum is infinity. In this paper, we concern extremal functions for these singular inequalities. The regular case β = 0 has been considered by Li and Ruf (2008) [12] and Ishiwata (2011) [11]. We shall investigate the singular case 0 < β < 1 and prove that for all τ > 0, 0 < β < 1 and 0 < γ ≤ 1 − β, extremal functions for the above inequalities exist. The proof is based on blow-up analysis. © 2017 Elsevier Inc. All rights reserved. MSC: 46E35 Keywords: Singular Trudinger–Moser inequality; Extremal function; Blow-up analysis

* Corresponding author.

E-mail addresses: [email protected] (X. Li), [email protected] (Y. Yang). https://doi.org/10.1016/j.jde.2017.12.028 0022-0396/© 2017 Elsevier Inc. All rights reserved.

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X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

1. Introduction and main results Let  ⊂ RN (N ≥ 2) be a bounded smooth domain, W01,N () be the usual Sobolev space. De1/(N−1) note αN = N ωN −1 , where ωN−1 is the area of the unit sphere in RN . The famous Trudinger– Moser inequality [29,17,16,23,14] reads  eα|u|

sup u∈W01,N (),





|∇u|N dx≤1

N N−1

dx < ∞,

∀α ≤ αN .

(1)



This inequality is sharp in the sense that all integrals are still finite when α > αN , but the supremum is infinity. It was extended by Cao [3], J.M. do Ó [8], Panda [15], Ruf [18], and Li–Ruf [12] to the entire Euclidean space RN (N ≥ 2). Namely  u∈W 1,N (RN ),



sup

N

e

N N RN (|∇u| +|u| )dx≤1

α|u| N−1

−

N−2  k=0

RN

Nk

α k |u| N−1 k!

dx < ∞,

∀α ≤ αN .

(2)

Recently, several interesting developments of (2) has been obtained by J.M. do Ó and M. de Souza [6,9]. Using a rearrangement argument and a change of variables, Adimurthi–Sandeep [1] generalized the Trudinger–Moser inequality (1) to a singular version as follows:  sup u∈W01,N (),



N  |∇u| dx≤1



N N−1

eαN γ |u| |x|Nβ

dx < ∞,

0 ≤ β < 1, 0 < γ ≤ 1 − β.

(3)

This inequality is also sharp in the sense that all integrals are still finite when γ > 1 − β, but the supremum is infinity. Obviously, if β = 0, then (3) reduces to (1). Later, (3) was extended to the entire RN by Adimurthi–Yang [2]. Precisely there holds for constants τ > 0, 0 ≤ β < 1 and 0 < γ ≤ 1 − β,  

sup N N RN (|∇u| +τ |u| )dx≤1 N R

1 |x|Nβ

e

αN γ |u|N/(N−1)

−

N−2  k=0

(αN γ )k |u| kN/(N−1) k!

dx < ∞.

(4)

Clearly, (2) is a special case of (4). It should be remarked that in [2], the proof of (4) is essentially based on the Young inequality; while in [12], (2) is proved via the method of blow-up analysis. Such kind of singular Trudinger–Moser inequalities are very important in analysis of partial differential equations, see for examples [24–26]. An interesting problem on Trudinger–Moser inequalities is whether or not extremal functions exist. Existence of extremal functions for the Trudinger–Moser inequality (1) was first obtained by Carleson–Chang [4] when  is the unit ball, by M. Struwe [20] when  is close to the ball in the sense of measure, then by M. Flucher and K. Lin [10,13] when  is a general bounded smooth domain. For recent developments, we refer the reader to Yang [27]. On extremal functions for (2), it was proved by Ruf [18] and Ishiwata [11] that if N = 2, then there exists some 0 > 0 such that for all 0 < α ≤ 2π , the supremum

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

 u∈W 1,2 (R2 ),



2

(eαu − 1)dx

sup 2 2 R2 (|∇u| +u )dx≤1

4903

R2

can be attained by some function u ∈ W 1,2 (R2 ) satisfying u W 1,2 (R2 ) ≤ 1. While for sufficiently small α > 0, the above supremum can not be attained. If N ≥ 3, then for any 0 ≤ α < αN , the supremum in (2) can be achieved. While Li–Ruf [12] proved that when α = αN , extremal function exists for the above supremum. Our aim is to find extremal functions for the singular Trudinger–Moser inequality (4) in the case 0 < β < 1. Note that the case β = 0 has been studied by Ruf [18], Ishiwata [11] and Li–Ruf [12]. While these two situations are quite different in analysis. Throughout this paper, we write for all τ ∈ (0, ∞), ⎛ ⎜

u 1,τ = ⎝



⎞1/N



⎟ |u|N dx ⎠

|∇u|N dx + τ

RN

(5)

.

RN

Obviously · 1,τ is equivalent to the standard Sobolev norm on W 1,N (RN ). Define a function ζ : N × R → R by ζ (N, s) = es −

N−2  sk k=0

k!

=

∞  sk . k!

(6)

k=N−1

Our main results are the existence of extremal functions for subcritical or critical singular Trudinger–Moser inequality, which can be stated as the following two theorems respectively. Theorem 1 (Subcritical case). Let N ≥ 2, τ > 0, · 1,τ and ζ : N × R → R be defined as in (5) and (6) respectively. Then for any 0 < β < 1 and 0 <  < 1 − β, the supremum  N,β,τ, =

N

sup u∈W 1,N (RN ), u 1,τ ≤1

RN

ζ (N, αN (1 − β − )|u| N−1 ) dx |x|Nβ

(7)

can be attained by some nonnegative decreasing radially symmetric function u ∈ C 1 (RN \{0}) ∩ C 0 (RN ) ∩ W 1,N (RN ) with u 1,τ = 1. Theorem 2 (Critical case). Let N ≥ 2, τ > 0, · 1,τ and ζ : N × R → R be defined as in (5) and (6) respectively. Then for any 0 < β < 1, the supremum  N,β,τ =

sup u∈W 1,N (RN ), u 1,τ ≤1

RN

N

ζ (N, αN (1 − β)|u| N−1 ) dx |x|Nβ

(8)

can be attained by some nonnegative decreasing radially symmetric function u∗ ∈ C 1 (RN \{0}) ∩ C 0 (RN ) ∩ W 1,N (RN ) with u∗ 1,τ = 1.

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X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

Trudinger–Moser inequalities involved the norm · 1,τ was first introduced by Adimurthi– Yang [2]. This type of inequalities are easy to use in analysis of partial differential equations with exponential growth. It should be remarked that both the above inequalities and existence of extremal functions are independent of τ . Let us give the outline of proving Theorems 1 and 2. The proof of Theorem 1 is based on a direct method of variation. By a rearrangement argument, we can take a maximizing sequence uj satisfying uj ≥ 0 and decreasing radially symmetric. Clearly uj u weakly in W 1,N (RN ) for some u . Since 0 <  < 1 − β and 0 < β < 1, for any ν > 0, there exists sufficiently large R > 0 such that  |x|>R

N

ζ (N, αN (1 − β − )|uj | N−1 ) dx < ν. |x|Nβ

Since αN (1 − β − ) < αN (1 − β), we have by the singular Trudinger–Moser inequality (4) that  lim

N

j →∞ |x|≤R

ζ (N, αN (1 − β − )|uj | N−1 ) dx = |x|Nβ



|x|≤R

N

ζ (N, αN (1 − β − )|u | N−1 ) dx. |x|Nβ

Then the conclusion of Theorem 1 follows from the above two estimates. Following Li–Ruf [12] and thereby closely following Carleson–Chang [4] and Ding–Jost–Li– Wang [7], we prove Theorem 2 via the method of blow-up analysis. Particularly we divide the proof into several steps: Step 1. For any 0 <  < 1 − β, the supremum N,β,τ, can be attained by some function u (this is the content of Theorem 1 exactly). The Euler–Lagrange equation of u is semi-linear elliptic when N = 2, or quasi-linear elliptic when N ≥ 3; Step 2. Denote c = u (0) = maxRN u . If c is a bounded sequence, then applying elliptic estimates to the equation of u , we conclude that u converges to a desired extremal function in 1 (RN \ {0}) ∩ C 0 (RN ). If c → +∞, then by a delicate analysis on u , we derive Cloc   loc  N,β,τ = lim

→0 RN

N

ζ (N, αN (1 − β − )uN−1 ) 1 ωN−1 N−1 dx ≤ e k=1 Nβ |x| 1−β N

1 k +αN (1−β)A0

.

Here A0 = limx→0 (G(x) + (N/αN ) log |x|), G is a Green function satisfying −div(|∇G|N−2 ∇G) + τ GN−1 = δ0

in RN ,

where δ0 is a Dirac measure centered at 0. Step 3. We construct a sequence of functions φ ∈ W 1,N (RN ) satisfying φ 1,τ = 1 and if  is sufficiently small, then  RN

N

ζ (N, αN (1 − β)φN−1 ) 1 ωN−1 N−1 dx > e k=1 Nβ |x| 1−β N

1 k +αN (1−β)A0

.

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

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Comparing Steps 2 and 3, we conclude that c must be bounded and thus the existence of extremal function follows from elliptic estimates. It should be remarked that in Step 2, we shall use an estimate of Carleson–Chang [4]: Lemma 3. Let B1 be the unit ball in RN , v ∈ W01,N (B1 ) satisfy

 B1

weakly in W01,N (B1 ). Then  lim sup →0

(eαN |v |

N/(N−1)

− 1)dx ≤

|∇v |N dx ≤ 1, and v 0

ωN−1 N−1 1 e k=1 k . N

B1

Before ending this introduction, we mention Csato–Roy [5] and Yang–Zhu [28] who studied the same topic in bounded planar domain. Throughout this paper, we do not distinguish sequence and subsequence, the reader can easily see it from the context. We denote a ball centered at 0 with radius r by Br , o (1) → 0 as  → 0, or (1) → 0 as r → 0, and oR (1) → 0 as R → ∞. The remaining part of this paper is devoted to the proof of Theorems 1 and 2 and organized as follows: Since the proof is transparent in R2 , we show it in Section 2. In Section 3, we prove Theorems 1 and 2 in N (≥ 3) dimensions. 2. Two dimensional case When N = 2, extremal functions for subcritical singular Trudinger–Moser inequalities are distributional solutions of elliptic partial differential equations of second order. Compared with N ≥ 3, analysis in two dimensions becomes much easier and transparent, so we deal with this case first. 2.1. Proof of Theorem 1 We rephrase Theorem 1 as below: Theorem 4. Let τ > 0 and 0 < β < 1 be fixed. Then for any 0 <  < 1 − β, there exists some nonnegative decreasing radially symmetric function u ∈ C 1 (R2 \ {0}) ∩ C 0 (R2 ) ∩ W 1,2 (R2 ) satisfying u 1,τ = 1 and  R2

2

e4π(1−β−)u − 1 dx = 2,β,τ, = sup |x|2β u∈W 1,2 (R2 ), u 1,τ ≤1

 R2

2

e4π(1−β−)u − 1 dx. |x|2β

(9)

Proof. Let τ > 0, 0 < β < 1 and 0 < u is the decreasing   <2 1 − β be fixed. Suppose2 that  rearrangement of |u|. It is known that R2  u dx = R2 u2 dx, R2 |∇ u| dx ≤ R2 |∇u|2 dx and  R2

2

u −1 e4π(1−β−) dx ≥ |x|2β

 R2

2

e4π(1−β−)u − 1 dx. |x|2β

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X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

Here we used the Hardy–Littlewood inequality in the last estimate. Therefore we have  2,β,τ, = sup

u∈S

R2

2

e4π(1−β−)u − 1 dx, |x|2β

where S is a set consisting of all nonnegative decreasing radially symmetric functions u ∈  2 W 1,2 (R2 ) with u 1,τ ≤ 1. Take uj ∈ S such that R2 (e4π(1−β−)uj − 1)/|x|2β dx → 2,β,τ, as j → ∞. Without loss of generality, we can assume that there exists some function u ∈ W 1,2 (R2 ) such that up to a subsequence, as j → ∞, there holds uj u weakly in W 1,2 (R2 ), uj → u p in Lloc (R2 ) for any p > 0 and uj → u a.e. in R2 . Hence up to a set of zero measure, u is nonnegative decreasing radially symmetric on R2 . Moreover, we have that u 1,τ ≤ lim supj →∞ uj 1,τ ≤ 1. Note that 0 < β < 1 and 0 <  < 1 − β. Given any ν > 0, in view of the Trudinger–Moser inequality (2), there exists a sufficiently large r > 0 such that for all u ∈ W 1,2 (R2 ) with u 1,τ ≤ 1, 

1 r 2β

|x|>r



1

2

(e4π(1−β−)u − 1)dx ≤

2

(e4π(1−β−)u − 1)dx < ν.

r 2β

(10)

R2

p

Since uj → u in Lloc (R2 ) for any p > 0, we have by using the mean value theorem, 

2

|x|≤r

e4π(1−β−)u − 1 dx = lim j →∞ |x|2β



2

|x|≤r

e4π(1−β−)uj − 1 dx. |x|2β

(11)

Combining (10) and (11), we obtain  R2

2

e4π(1−β−)u − 1 dx − ν ≤ lim sup |x|2β j →∞



2

R2

e4π(1−β−)uj − 1 dx ≤ |x|2β



2

R2

e4π(1−β−)u − 1 dx + ν. |x|2β

Since ν is arbitrary, there holds  lim

j →∞ R2

2

e4π(1−β−)uj − 1 dx = |x|2β

 R2

2

e4π(1−β−)u − 1 dx. |x|2β

This leads to (9). Noting that  R2

2 e4π(1−β−)u

|x|2β

−1

 dx ≤ R2

4π(1−β−)

e

u2 

u 2 1,τ

|x|2β

−1

dx,

we get the extremal function u , which is nonnegative and decreasing radially symmetric, and satisfies u 1,τ = 1. A straightforward calculation shows that u satisfies the following Euler– Lagrange equation

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

4907

⎧ 2 1 u e4π(1−β−)u ⎪ + τ u = in R2 , −u ⎪   2β ⎪ λ |x| ⎪ ⎪ ⎨ u > 0 in R2 ,  ⎪ ⎪

u 1,τ = 1, ⎪ ⎪ ⎪  ⎩ 2 λ = R2 |x|−2β u2 e4π(1−β−)u dx.

(12)

Applying elliptic estimates to (12), we have u ∈ C 1 (R2 \ {0}) ∩ C 0 (R2 ). Here u > 0 follows from the classical maximum principle and the fact that u (0) = maxR2 u . This completes the proof of the theorem. 2 From now on, we prove Theorem 2 by using the method of blow-up analysis. 2.2. Elementary properties of u In view of the equation (12), it is important to know whether λ has a positive lower bound or not. For this purpose, we have the following: Lemma 5. Let λ be as in (12). Then there holds lim inf→0 λ > 0. Proof. For any u ∈ W 1,2 (R2 ) with u 1,τ ≤ 1, we calculate by employing Theorem 4,  R2

2

e4π(1−β)u − 1 dx = lim →0 |x|2β



R2

2

e4π(1−β−)u − 1 dx ≤ lim →0 |x|2β



R2

2

e4π(1−β−)u − 1 dx. |x|2β

This leads to  2,β,τ =

sup u∈W 1,2 (R2 ), u 1,τ ≤1

R2

2

e4π(1−β)u − 1 dx ≤ lim →0 |x|2β



R2

2

e4π(1−β−)u − 1 dx. |x|2β

(13)

But one can easily see that  R2

2

e4π(1−β−)u − 1 dx ≤ 2,β,τ . |x|2β

Moreover, using the inequality et ≤ 1 + tet for t ≥ 0, we get λ ≥

1 4π(1 − β − )

 R2

2

e4π(1−β−)u − 1 dx. |x|2β

This together with (13) and (14) leads to 1 →0 4π(1 − β − )



lim inf λ ≥ lim →0

This ends the proof of the lemma. 2

R2

2

2,β,τ e4π(1−β−)u − 1 dx = > 0. 2β |x| 4π(1 − β)

(14)

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X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

Denote c = u (0) = maxR2 u . If c is bounded, then applying elliptic estimates to (12), we 1 (R2 \ {0}) ∩ C 0 (R2 ). Clearly u∗ is the can find some u∗ ∈ W 1,2 (R2 ) such that u → u∗ in Cloc loc desired extremal function satisfying  R2

∗2

e4π(1−β)u − 1 dx = sup |x|2β u∈W 1,2 (R2 ), u 1,τ ≤1



2

e4π(1−β)u − 1 dx. |x|2β

R2

Hence the proof of Theorem 2 terminates. In the following, we assume c → +∞. Since u is bounded in W 1,2 (R2 ), we can assume without loss of generality, u converges to u0 weakly in q W 1,2 (R2 ), strongly in Lloc (R2 ) for any q > 0, and a.e. in R2 . Then we have the following: Lemma 6. u0 ≡ 0 and |∇u |2 dx δ0 weakly in the sense of measure, where δ0 denotes the Dirac measure centered at 0 ∈ R2 . Moreover, u → 0 strongly in Lp (R2 ) for all p ≥ 2. Proof. For any a ≥ 0, b ≥ 0 and p ≥ 1, there holds ([24], Lemma 2.1) (ea − 1)p ≤ epa − 1.

(15)

Using ea+b − 1 = (ea − 1)(eb − 1) + (ea − 1) + (eb − 1), the Hölder inequality and (15), we estimate 

2

R2

eβ pu − 1 dx ≤ |x|2βp



eβ p

  (1+ν)(u −u0 )2 +(1+ν −1 )u20

−1

|x|2βp

R2



(eβ p(1+ν)(u −u0 ) − 1)(eβ p(1+ν |x|2βp 2

= R2



+ ⎛ ⎜ ≤⎝

R2



R2

 + R2

eβ p(1+ν)(u −u0 ) − 1 dx + |x|2βp 2

 R2

dx

−1 )u2 0

|x|2βp

−1

⎟ dx ⎠

dx

−1

2

eβ p(1+ν )u0 − 1 dx |x|2βp

⎞1/p1 ⎛ 2 eβ pp1 (1+ν)(u −u0 )

− 1)

⎜ ⎝



e

β pp2 (1+ν −1 )u20

|x|2βp

R2

eβ p(1+ν)(u −u0 − 1 dx + |x|2βp )2

 R2

−1

⎞1/p2 −1

⎟ dx ⎠

2

eβ p(1+ν )u0 − 1 dx, |x|2βp

(16)

where β = 4π(1 − β − ), p > 1, ν > 0, p1 > 1 and 1/p1 + 1/p2 = 1. We first prove that u0 ≡ 0. Suppose not. Since 

u − u0 21,τ = u 21,τ + u0 21,τ − 2 R2

(∇u ∇u0 + τ u u0 )dx = 1 − u0 21,τ + o (1),

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

4909

one can choose p, p1 sufficiently close to 1 and ν sufficiently close to 0 such that β pp1 (1 + ν) u − u0 21,τ 4π

+

2βp < 1. 2

In view of the singular Trudinger–Moser inequality (4), we conclude that all integrals on the right hand side of (16) are bounded. Therefore  R2

2

e4π(1−β−)pu − 1 dx ≤ C |x|2βp 2

for some constant C depending only on β and p. It follows that e4π(1−β−)u /|x|2β is bounded in Lp (B1 ). This together with Lemma 5 and u is bounded in Lq (B1 ) for all q > 0 implies that  u is bounded in Lp (B1 ) for some p  > 1. Applying elliptic estimate to (12), we conclude that u is uniformly bounded in B1/2 . This contradicts c → +∞. Therefore u0 ≡ 0. We next prove that |∇u |2 dx δ0 . For otherwise, we can choose sufficiently small r¯ > 0 such that  (|∇u |2 + τ u2 )dx ≤ η < 1 Br¯ 

for sufficiently small  > 0. Hence u is bounded in Lp (Br¯ ) for some p  > 1, and thus elliptic estimate leads to u is uniformly bounded in Br¯/2 contradicting c → +∞. For the last assertion, noting that u 1,τ = 1 and |∇u |2 dx δ0 , we obtain u L2 (R2 ) = o (1). Taking M > 0 such that if |x| > M, then u < 1, one has for any p > 2, 





up dx = R2

up dx +

|x|>M



up dx

|x|≤M

u2 dx + o (1)

≤ |x|>M



≤

u2 dx + o (1) = o (1). R2 q

Here we have used the fact that u → 0 in Lloc (R2 ) for any q > 0. This completes the proof of the lemma. 2 2.3. Blow-up analysis √ 2 1/(1−β) 1/(1−β) Set r = λ c−1 e−2π(1−β−)c , ψ (x) = c−1 u (r x) and ϕ (x) = c (u (r x) − c ). Then we have the following:

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X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943 2

1 (R2 \ Lemma 7. (i) For any γ < 2π(1 −β), there holds r eγ c → 0 as  → 0; (ii) ψ → 1 in Cloc 1 0 0 1 2 2 2 {0}) ∩ Cloc (R ); (iii) ϕ → ϕ in Cloc (R \ {0}) ∩ Cloc (R ), where ϕ(x) = − 4π(1−β) log(1 +  π 2(1−β) ) and −2β e8π(1−β)ϕ dx = 1. R2 |x| 1−β |x|

Proof. (i) By definition of r , we have γ

r2 e2γ c = c−2 e−4π(1−β−− 2π )c 2

≤ c−2

 R2

≤ c−2



R2

2

 R2

2

u2 e4π(1−β−)u dx |x|2β

2γ u2

u2 e dx |x|2β 

2

u2 (e2γ u − 1) dx + c−2 |x|2β

R2

u2 dx. |x|2β

(17)

By Lemma 6, we know that u Lp (R2 ) = o (1) for any p ≥ 2. As an easy consequence, there holds for any p ≥ 2  lim

→0 R2

p

u dx = 0. |x|2β

(18)

Noting that γ < 2π(1 − β), we can choose p1 > 1 such that γp1 < 2π(1 − β). In view of (4), (15) and (18), we have by the Hölder inequality,  R2

⎛ 2 u2 (e2γ u − 1) dx |x|2β

⎜ ≤⎝



R2

⎞1/p1 ⎛ 2 e2γp1 u

−1 ⎟ dx ⎠ |x|2β

⎜ ⎝



R2

⎞1/p2 2p2 u ⎟ dx ⎠ |x|2β

= o (1),

(19)

2

where 1/p2 + 1/p1 = 1. Inserting (18) and (19) into (17), we obtain r eγ c → 0 as  → 0. (ii) It can be easily checked that ψ satisfies the equation 1/(1−β)

−ψ (x) = −τ r2/(1−β) ψ (x) + c−2 |x|−2β ψ (x)e4π(1−β−)(u (r 2

x)−c2 )

.

(20)

Since |ψ | ≤ 1, u2 ≤ c2 and r → 0 as  → 0, we have by applying elliptic estimates to (20), 1 (R2 \ {0}) ∩ C 0 (R2 ), where ψ is a bounded harmonic function on R2 . Then the ψ → ψ in Cloc loc Liouville theorem leads to ψ ≡ ψ(0) = 1. (iii) A straightforward calculation shows −ϕ (x) = −τ c2 r2/(1−β) ψ (x) + |x|−2β ψ (x)e4π(1−β−)(1+ψ (x))ϕ (x) .

(21)

Note that ϕ (x) ≤ 0 = maxR2 ϕ . Applying elliptic estimates to (21), we conclude that ϕ → ϕ 1 (R2 \ {0}) ∩ C 0 (R2 ), where ϕ is a distributional solution to in Cloc loc

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943



−ϕ(x) = |x|−2β e8π(1−β)ϕ(x)

4911

R2 ,

in

(22)

ϕ(0) = 0.

Since u is decreasing symmetric and u (0) = maxR2 u = c , ϕ must be decreasing symmetric and ϕ(0) = maxR2 ϕ. If we set ϕ(r) ¯ = ϕ(x) for any x ∈ R2 and r = |x|, then (22) reduces to 

−(r ϕ¯  ) = r 1−2β e8π(1−β)ϕ¯ ,

(23)

ϕ(0) ¯ = 0. Clearly, this equation has a special solution ϕ(r) ¯ =−

1 π 2(1−β) ). log(1 + r 4π(1 − β) 1−β

By the standard uniqueness result of the ordinary differential equation (23), we have ϕ(x) = −

1 π log(1 + |x|2(1−β) ), 4π(1 − β) 1−β

x ∈ R2 .

It follows that  |x|

−2β 8π(1−β)ϕ

e

+∞ dx =

R2

0

2πr 1−2β π 2(1−β) 2 dr = (1 + 1−β r )

+∞ 0

1 dt = 1. (1 + t)2

(24)

This completes the proof of the lemma. 2 Lemma 7 gives convergence behavior of u near 0. To reveal the convergence behavior of u away from 0, following [12], we define u,γ = min{u , γ c } for any 0 < γ < 1. Then we have the following: Lemma 8. For any 0 < γ < 1, there holds lim→0

 R2

|∇u,γ |2 dx = γ .

Proof. Testing the equation (12) by u,γ , we have for any fixed R > 0, 

 |∇u,γ |2 dx = −τ

R2

u u,γ dx + R2

1 ≥ λ B



1 λ



2

u u,γ R2

e4π(1−β−)u dx |x|2β

2

e4π(1−β−)u γ c u dx + o (1) |x|2β

1/(1−β) Rr



= (1 + o (1))γ BR

e8π(1−β)ϕ(x) dx + o (1). |x|2β

4912

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

Hence 

 |∇u,γ |2 dx ≥ γ

lim inf →0

R2

BR

e8π(1−β)ϕ(x) dx. |x|2β

In view of (24), passing to the limit R → +∞, we obtain  |∇u,γ |2 dx ≥ γ .

lim inf →0

(25)

R2

Testing the equation (12) by (u − γ c )+ , we obtain for any fixed R > 0, 



+ 2

|∇(u − γ c ) | dx = −τ R2

u (u − γ c ) dx + R2

R2



1 λ

≥



+

2

e4π(1−β−)u (u − γ c ) u dx λ |x|2β +

2

u (u − γ c )+

B

1/(1−β) Rr



= (1 + o (1))(1 − γ ) BR

e4π(1−β−)u dx + o (1) |x|2β

e8π(1−β)ϕ(x) dx + o (1). |x|2β

Similarly as above, we have  lim inf →0

|∇(u − γ c )+ |2 dx ≥ 1 − γ .

(26)

R2

Note that 

 |∇u,γ | dx +

R2

+ 2



|∇(u − γ c ) | dx =

2

R2

|∇u |2 dx = 1 + o (1).

(27)

R2

Combining (25), (26) and (27), we conclude the lemma. 2 Lemma 9. There holds  lim

→0 R2

2

e4π(1−β−)u − 1 λ dx = lim 2 . 2β →0 c |x|

(28)

Proof. Let 0 < γ < 1 be fixed. Using the inequality et − 1 ≤ tet (t ≥ 0) and the definition of u,γ , we obtain

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943



2

u ≤γ c

e4π(1−β−)u − 1 dx ≤ |x|2β



4913

2

e4π(1−β−)u,γ − 1 dx |x|2β

R2



2

u2,γ e4π(1−β−)u,γ

≤ 4π(1 − β)

|x|2β

R2

= 4π(1 − β)

⎧ ⎪ ⎨

u2,γ

⎪ ⎩

e

dx

4π(1−β−)u2,γ

−1

|x|2β

R2

 dx + R2

⎫

⎪ ⎬ u2,γ dx . |x|2β ⎪ ⎭

(29)

It follows from (18) that 



u2,γ |x|

R2

u2 dx = o (1). |x|2β

dx ≤ 2β R2

(30)

Moreover, combining Lemma 6 and Lemma 8, we have lim→0 u,γ 21,τ = γ < 1. Let 1 < p < 1/γ be fixed and 1/p + 1/p  = 1. Using the Hölder inequality and the singular Trudinger–Moser inequality (4), we have ⎛

 u2,γ

e

4π(1−β−)u2,γ

|x|2β

R2

−1

⎜ dx ≤ ⎝

⎞1/p ⎛



e

4π(1−β−)pu2,γ

−1

|x|2β

R2

⎛

⎜ ≤C⎝



⎜ ⎝

⎟ dx ⎠

R2

⎞1/p

2p 



u,γ ⎟ dx ⎠ |x|2β

R2

⎞1/p

2p  u,γ ⎟ dx ⎠ |x|2β

(31)

for some constant C depending only on β, p and γ . Inserting (30) and (31) into (29), one has  u ≤γ c

2

e4π(1−β−)u − 1 dx = o (1). |x|2β

(32)

Moreover, we estimate  u >γ c

2

e4π(1−β−)u − 1 dx = |x|2β



2

e4π(1−β−)u dx + o (1) |x|2β

u >γ c

1 ≤ 2 γ

 u >γ c

≤

2

u2 e4π(1−β−)u dx + o (1) c2 |x|2β

1 λ + o (1). γ 2 c2

(33)

4914

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

Combining (32) and (33), we have 

2

e4π(1−β−)u − 1 λ 1 dx ≤ 2 lim inf 2 . |x|2β γ →0 c

lim

→0 R2

Letting γ → 1, we conclude 

2

lim

→0 R2

e4π(1−β−)u − 1 λ dx ≤ lim inf 2 . →0 c |x|2β

(34)

On the other hand, λ = c2

 R2



= R2



≤

2

u2 e4π(1−β−)u dx c2 |x|2β 

2

u2 e4π(1−β−)u − 1 1 dx + 2 c2 |x|2β c e

4π(1−β−)u2

−1

|x|2β

R2

R2

u2 dx |x|2β

dx + o (1).

Thus lim sup →0

λ ≤ lim c2 →0



R2

2

e4π(1−β−)u − 1 dx. |x|2β

(35)

Combining (34) and (35), we get the desired result. 2 Corollary 10. If θ < 2, then λ /cθ → +∞ as  → 0. Proof. An obvious consequence of Lemma 9.

2

Lemma 11. For any function φ ∈ C00 (R2 ), there holds  lim

→0 R2

2

c u e4π(1−β−)u φdx = φ(0). λ |x|2β

−2β c u e4π(1−β−)u . Given Proof. Let φ ∈ C00 (R2 ) be fixed. Write for simplicity h = λ−1    |x| 0 < γ < 1. Firstly we calculate 2

 h φdx = u ≤γ c

c λ



2

u φ u ≤γ c

e4π(1−β−)u − 1 c dx + 2β λ |x|

 u ≤γ c

u φ dx. |x|2β

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

4915

In view of an obvious analog of (31), there holds       

u ≤γ c



  2 2 e4π(1−β−)u,γ − 1 e4π(1−β−)u − 1  u φ dx  ≤ sup |φ| u,γ dx = o (1). |x|2β |x|2β  R2 2 R

q

Note that u → 0 in Lloc (R2 ) for any q > 0. We derive       

u ≤γ c



   u φ  dx ≤ sup |φ| |x|2β  R2

supp φ

u dx = o (1). |x|2β

By Corollary 10, c /λ = o (1). Therefore  h φdx = o (1).

(36)

u ≤γ c

It follows from Lemma 7 that BRr 1/(1−β) ⊂ {u > γ c } for sufficiently small  > 0, that 



⎛ ⎜ h φdx = φ(0)(1 + o (1)) ⎝

B



⎞ e8π(1−β)ϕ |x|2β

BR

1/(1−β) Rr

⎟ dx + o (1)⎠

= φ(0)(1 + o (1) + oR (1)), and that    

    2  1  u2 e4π(1−β−)u   h φdx  ≤ dx sup |φ|  λ |x|2β  γ R2  {u >γ c }\B 1/(1−β)  {u >γ c }\BRr 1/(1−β) Rr  ⎛ ⎞

 8π(1−β)ϕ e 1 ⎜ ⎟ dx + o (1)⎠ sup |φ| ⎝1 − ≤ γ R2 |x|2β BR

= o (1) + oR (1). It then follows that  lim

→0 u >γ c

h φdx = φ(0).

Combining (36) and (37), we complete the proof of the lemma. 2

(37)

4916

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

1 (R2 \ {0}) and weakly in W 1,q (R2 ) for all 1 < q < 2, where G is Lemma 12. c u → G in Cloc a distributional solution to

−G + τ G = δ0

in R2 .

(38)

Moreover, G ∈ W 1,2 (R2 \ Br ) for any r > 0 and G takes the form G(x) = −

1 log |x| + A0 + w(x), 2π

(39)

where A0 is a constant, w ∈ C 1 (R2 ) and w(0) = 0. Proof. Multiplying both sides of the equation (12) by c , one has 2

c u e4π(1−β−)u −(c u ) + τ (c u ) = λ |x|2β

in R2 .

(40)

−2β c u e4π(1−β−)u is bounded in L1 (R2 ). Using an In view of Lemma 11, h = λ−1    |x| loc argument of Li–Ruf ([12], Proposition 3.7), which is adapted from that of Struwe ([21], The1,q orem 2.2), one concludes that c u is bounded in Wloc (R2 ) for all 1 < q < 2. Hence c u G 1,q weakly in Wloc (R2 ) for any 1 < q < 2 and G is a distributional solution to (38). Since p 1 log |x|) ∈ Lloc (R2 ) for any p > 2, (39) follows from elliptic estimates immedi(G(x) + 2π 1 (R2 \ {0}). ately. Applying elliptic estimates to the equation (40), we obtain c u → G in Cloc Note that c u ∈ W 1,2 (R2 ). Multiplying both sides of (40) by c u and integrating by parts on the domain R2 \ Br for some r > 0, we get    ∂(c u ) (|∇(c u )|2 + τ (c u )2 )dx = − c u h c u dx dσ + ∂ν 2

R2 \Br

R2 \Br

∂Br

 ≤−

2

c u

∂(c u ) c2 e4πu (r) dσ +  ∂ν λ

∂Br

 R2 \Br

u2 dx |x|2β

≤ Cr 1 (R2 \ {0}). This also leads to for some constant Cr depending only on r, since c u → G in Cloc

 (|∇G|2 + τ G2 )dx ≤ Cr ,

∀R > r.

r≤|x|≤R

Passing to the limit R → ∞, we have  (|∇G|2 + τ G2 )dx ≤ Cr . R2 \Br

This gives the desired result.

2

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

4917

2.4. Upper bound estimate We need a singular version of Carleson–Chang’s upper bound estimate, namely Lemma 3. Lemma 13. Let w ∈ W01,2 (Br ) satisfies w is radially symmetric. Then 

Br

|∇w |2 dx ≤ 1, w 0 weakly in W01,2 (Br ), and 2

e4π(1−β)w − 1 eπ 2(1−β) dx ≤ . r |x|2β 1−β

lim sup →0



Br

(41)

Proof. We first prove (41) for r = 1. √ Denote w (|x|) = w (x). Let v (x) = 1 − βw (|x|1/(1−β) ). Then   |∇v |2 dx = |∇w |2 dx. B1

B1

Clearly we can assume up to a subsequence, v v0 weakly in W01,2 (B1 ), v → v0 strongly in L2 (B1 ), and v → v0 a.e. in B1 . Also, we can assume w → 0 a.e. in B1 . Hence we conclude v0 = 0 a.e. in B1 . By a change of variable t = s 1/(1−β) , there holds  B1

2

e4π(1−β)w − 1 dx = |x|2β

1

2

e4π(1−β)w (t) − 1 2πtdt t 2β

0

1

2π = 1−β =

1/(1−β) )

− 1)s β/(1−β) ds

0

1

2π 1−β

1 = 1−β

2

s (1−2β)/(1−β) (e4π(1−β)w (s

2

s(e4πv (s) − 1)ds 0



2

(e4πv − 1)dx. B1

It follows from Lemma 3 that  lim sup →0

B1

2

e4π(1−β)w − 1 eπ dx ≤ . |x|2β 1−β

(42)

We next prove (41) for the case of general r. Set w˜  (x) = w (rx) for x ∈ B1 . One can check that   |∇ w˜  |2 dx = |∇w |2 dx B1

Br

4918

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

and that 



2

e4π(1−β)w − 1 dx = r 2(1−β) |x|2β

Br

B1

e4π(1−β)w˜  − 1 dx. |x|2β 2

2

This together with (42) gives the desired result.

By the equation (12) and u 1,τ = 1, we have 





|∇u |2 dx = 1 −

(|∇u |2 + τ u2 )dx − τ

R2 \Br

Br



Br 2

=1− R2 \Br

u2 dx

u2 e4π(1−β−)u dx + λ |x|2β

 u

∂u dσ − τ ∂r

∂Br

 u2 dx.

(43)

Br

Since 



2

R2 \Br

u2 e4π(1−β−)u 1 c2 dx = λ |x|2β c2 λ = ⎛

 u ∂Br

∂u 1 ⎜ dσ = 2 ⎝ ∂r c

2

u2 R2 \Br

e4π(1−β−)u dx |x|2β

o (1) , c2  G

⎞

∂G ⎟ dσ + o (1)⎠ , ∂r

∂Br

and ⎛

 u2 dx = Br

1 ⎜ ⎝ c2



⎞ ⎟ G2 dx + o (1)⎠ .

Br

1 Inserting these equations into (43) and noting that G(x) = − 2π log |x| + A0 + w(x), we conclude

 |∇u |2 dx = 1 − Br

1 c2



 1 1 log + A0 + o (1) + or (1) . 2π r

(44)

Denote s,r = sup∂Br u = u (r) and u,r = (u − s,r )+ , the positive part of u − s,r . Clearly we have u,r ∈ W01,2 (Br ). In view of Lemma 13,  lim sup →0

Br

2

e4π(1−β)u,r /τ,r − 1 eπ 2(1−β) dx ≤ , r 2β |x| 1−β

(45)

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

4919

 where τ,r = Br |∇u |2 dx. Moreover, we know from Lemma 7 that u = c + o (1) on BRr 1/(1−β) . Hence, in view of (44), there holds on BRr 1/(1−β) ⊂ Br , 



4π(1 − β − )u2 ≤ 4π(1 − β)(u,r + s,r )2 = 4π(1 − β)u2,r + 8π(1 − β)s,r u,r + o(1) = 4π(1 − β)u2,r − 4(1 − β) log r + 8π(1 − β)A0 + o(1) = 4π(1 − β)u2,r /τ,r − 2(1 − β) log r + 4π(1 − β)A0 + o(1), where o(1) → 0 as  → 0 first and next r → 0. Therefore 



2

e4π(1−β−)u − 1 dx ≤ r −2(1−β) e4π(1−β)A0 +o(1) |x|2β

B

B

1/(1−β) Rr

2

e4π(1−β)u,r /τ,r dx |x|2β

1/(1−β) Rr

=r



−2(1−β) 4π(1−β)A0 +o(1)

e

B

2

e4π(1−β)u,r /τ,r − 1 dx |x|2β

1/(1−β) Rr

+ o(1) ≤ r −2(1−β) e4π(1−β)A0 +o(1)



2

e4π(1−β)u,r /τ,r − 1 dx |x|2β

Br

+ o(1).

(46)

Combining (45) with (46), one concludes for any fixed R > 0, 

2

e4π(1−β−)u − 1 π 1+4π(1−β)A0 dx ≤ . e 2β |x| 1−β

lim sup →0

B

1/(1−β) Rr

In view of Lemma 7, we calculate  B

2

e4π(1−β−)u − 1 dx = r2 |x|2β



2

BR

1/(1−β) Rr

⎛ =

λ ⎜ ⎝ c2



BR

=

1/(1−β)

e4π(1−β−)u (r |y|2β

y)

dy + o (1) ⎞

e8π(1−β)ϕ(y) |y|2β

⎟ dy + o (1)⎠ + o (1)

λ (1 + oR (1) + o (1)) + o (1). c2

(47)

4920

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

As a consequence, 

2

e4π(1−β−)u − 1 λ dx = lim 2 . →0 c |x|2β

lim lim

R→∞ →0 B

(48)

1/(1−β) Rr

Combining (47), (48) and (28), in view of (13) and (14), we arrive at  sup u∈W 1,2 (R2 ), u 1,τ ≤1

R2

2

e4π(1−β)u − 1 dx = lim →0 |x|2β



2

R2

e4π(1−β−)u − 1 π 1+4π(1−β)A0 dx ≤ . e |x|2β 1−β (49)

2.5. Test function computation We now construct test functions such that (49) does not hold. Precisely we construct a sequence of functions φ ∈ W 1,2 (R2 ) satisfying φ 1,τ = 1 and  R2

2

e4π(1−β)φ − 1 π 1+4π(1−β)A0 dx > e 1−β |x|2β

for sufficiently small  > 0. For this purpose we set ⎧   π |x|2(1−β) ⎨c+ 1 − 1 log(1 + ) + b , c 4π(1−β) 1−β  2(1−β) φ (x) = ⎩ G, c

(50)

x ∈ B R x ∈ R2 \ BR ,

where G is given as in Lemma 12, R = (− log )1/(1−β) , b and c are constants depending only on  to be determined later. To ensure φ ∈ W 1,2 (R2 ), we let       1 1 π 1 1 2(1−β) c+ − log 1 + R − log(R) + A0 + w(R) . +b = c 4π(1 − β) 1−β c 2π This leads to c2 =

1 1 π 1 log + A0 − b − log  + O( 2−2β ) + O(R). 4π(1 − β) 1−β 2π R

Now we calculate  (|∇φ |

2

+ τ φ2 )dx



1 = 2 c

R2 \BR

(51)

(|∇G|2 + τ G2 )dx R2 \BR

=−



1 c2

G

∂G dσ ∂r

∂BR

 1 1 + O(R log(R)) log(R) + A − 0 2π c2 

=

(52)

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

4921

and 

1 |∇φ | dx = 4πc2

R

2r 3−4β

2

(r 2(1−β) +

0

BR

1 = 4π(1 − β)c2

1−β 2(1−β) 2 ) π 

dr



π R 2−2β ) + log(1 + 1−β 1+



1 π 2−2β 1−β R

−1

  1 1 π 2−2β = − 1 + O( 2−2β ) . + log R log 4π(1 − β)c2 1−β R

(53)

Moreover, we require b to be bounded with respect to . It then follows that  φ2 dx

1 = 2 c

BR



1 π |x|2(1−β) )+b log(1 + c − 4π(1 − β) 1 − β  2(1−β)

2

2

BR

dx = O(R).

(54)

Combining (52), (53) and (54), we obtain

φ 21,τ

  1 1 1 π 1 1 = 2 − log  + A0 − + log + O( 2−2β ) . c 2π 4π(1 − β) 4π(1 − β) 1−β R

Setting φ 1,τ = 1, we have c2 = −

1 1 1 π 1 log  + A0 − + log + O( 2−2β ), 2π 4π(1 − β) 4π(1 − β) 1−β R

(55)

which together with (51) leads to b=

1 1 + O( 2−2β ). 4π(1 − β) R

(56)

For all x ∈ BR , it follows from (55) and (56) that 4π(1 − β)φ2 (x) ≥ 4π(1 − β)c2

 + 8π(1 − β)b − 2 log 1 +



π |x|2−2β = −2 log 1 + 1 − β  2−2β + 4π(1 − β)A0 + log



π |x|2−2β 1 − β  2−2β

− 2(1 − β) log 

π 1 + 1 + O( 2−2β ). 1−β R

Hence  BR

2

1 e4π(1−β)φ − 1 π 1+4π(1−β)A0 +O( 2−2β ) R dx ≥  −2(1−β) e 2β |x| 1−β



4922

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

 × BR

1 (1 +

π |x|2(1−β) 2 2β 1−β  2(1−β) ) |x|

dx + O((R)2−2β )

1 π 1+4π(1−β)A0 +O( 2−2β ) R e = 1−β  1 × dy + O((R)2−2β ) π (1 + 1−β |y|2(1−β) )2 |y|2β

BR

=

π 1+4π(1−β)A0 1 + O( 2−2β ). e 1−β R

(57)

Also we calculate 



2

R2 \BR

e4π(1−β)φ − 1 4π(1 − β) dx ≥ 2β |x| c2

R2 \BR

⎛

=

4π(1 − β) ⎜ ⎝ c2



R2

G2 dx |x|2β ⎞

G2 ⎟ dx + o (1)⎠ . |x|2β

(58)

Combining (57) and (58) and noting that c2 /R 2−2β = o (1), we have  R2

⎛ 2 e4π(1−β)φ

−1

|x|2β

dx ≥

π 1+4π(1−β)A0 4π(1 − β) ⎜ + e ⎝ 1−β c2



R2

⎞ G2 |x|2β

⎟ dx + o (1)⎠ .

Therefore we conclude (50) for sufficiently small  > 0. 2.6. Completion of the proof of Theorem 2 Comparing (50) with (49), we conclude that c must be bounded. Then applying elliptic estimates to (12), we get the desired extremal function. This ends the proof of Theorem 2. 2 3. N -dimensional case In this section, we prove Theorems 1 and 2 in the case that N ≥ 3. We put emphasis on the essential difference between 2 dimensions and N dimensions. In the sequel, we denote N u = div(|∇u|N−2 ∇u) for any u ∈ W 1,N (RN ) (N ≥ 3). Let ζ : N × R → R be defined as in (6). Obviously, one has d ζ (N, t) = ζ (N − 1, t). dt

(59)

In view of ([24], Lemma 2.1), there holds for all p ≥ 1 and t ≥ 0, (ζ (N, t))p ≤ ζ (N, pt).

(60)

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

4923

3.1. A priori estimates We need elliptic estimates for quasi-linear equations as below. Theorem 14. Let R > 0 be fixed. Suppose that u ∈ W 1,N (BR ) is a weak solution of −N u = f

in BR ⊂ RN .

Then the following a priori estimates hold: • (Harnack inequality) If u ≥ 0 and f ∈ Lp (BR ) for some p > 1, then there exists some constant C depending only on N , R and p such that supBR/2 u ≤ C(infBR/2 u + f Lp (BR ) ); • (C α -estimate) If u L∞ (BR ) ≤ L and f Lp (BR ) ≤ M for some p > 1, then there exists two constants 0 < α ≤ 1 and C depending only on N , R, p, L and M such that u ∈ C α (B R/2 ) and u C α (B R/2 ) ≤ C; • (C 1,α -estimate) If u L∞ (BR ) ≤ L and f L∞ (BR ) ≤ M, then there exists two constants 0 < α ≤ 1 and C depending only on N , R, L and M such that u ∈ C 1,α (B R/2 ) and

u C 1,α (B R/2 ) ≤ C. In the above theorem, the first two estimates were obtained by J. Serrin ([19], Theorems 6 and 8), while the third estimate was proved by Tolksdorf ([22], Theorem 1). 3.2. Extremal functions for subcritical Trudinger–Moser inequalities In this subsection, we prove Theorem 1 in the case N ≥ 3. The proof is based on a direct method of variation. Throughout this section, we denote for simplicity βN, = αN (1 − β − ).

(61)

Proof of Theorem 1. Let SN be a subset of W 1,N (RN ) consisting of all functions, which are nonnegative decreasing radially symmetric almost everywhere. By a rearrangement argument, we have  N,β,τ, =

sup

u∈SN , u 1,τ ≤1

RN

N

ζ (N, βN, u N−1 ) dx, |x|Nβ

where N,β,τ, is defined as in (7) and βN, is defined as in (61). Take uj ∈ SN with uj 1,τ ≤ 1 such that  ζ (N, β u N−1 ) N, j N

lim

j →∞ RN

|x|Nβ

dx = N,β,τ, .

Up to a subsequence, we can find some function u such that uj converges to u weakly in q W 1,N (RN ), strongly in Lloc (RN ) for any q > 0, and a.e. in RN . Obviously u ∈ SN . It follows from the weak convergence of uj in W 1,N (RN ) that

4924

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943



 |∇u | dx = lim

|∇u |N−2 ∇uj ∇u dx,

N

j →∞ RN

RN

which together with the Hölder inequality leads to 

 |∇u |N dx ≤ lim sup j →∞

RN

|∇uj |N dx.

(62)

RN

q

While it follows from uj → u in Lloc (RN ) for any q > 0 that for any fixed R > 0, 

 uN  dx = lim

j →∞ BR

BR

uN j dx.

(63)

Combining (62) and (63), one can easily see that u 1,τ ≤ lim supj →∞ uj 1,τ ≤ 1. Given any ν > 0, there hold  |x|>ν

N/(N −1)

ζ (N, βN, uj

|x|Nβ

1 − Nβ

 |x|>ν

)

N/(N −1)

)dx ≤ νN,0,τ ,

(64)

−1) ζ (N, βN, uN/(N )dx ≤ νN,0,τ , 

(65)

ζ (N, βN, uj

RN

N/(N −1)

ζ (N, βN, u |x|Nβ

1 − Nβ

 dx ≤ ν

)

 dx ≤ ν RN

where N,0,τ is defined as in (8). In view of (59), we have by the mean value theorem, N

N

N

N

ζ (N, βN, ujN−1 ) − ζ (N, βN, uN−1 ) = ζ (N − 1, ϑ)βN, (ujN−1 − uN−1 ) N

N

≤ max{ζ (N − 1, βN, ujN−1 ), ζ (N − 1, βN, uN−1 )} N

N

× βN, (ujN−1 − uN−1 ), N/(N −1)

where ϑ lies between βN, uj 1 for some p, 1 < p < min{ 1− , β1 },  |x|≤ν

1 − Nβ

(ζ (N − 1, ϑ))p dx ≤ |x|Nβp



|x|≤ν

1 − Nβ



= |x|≤ν

N/(N −1)

and βN, u

1 − Nβ

(66)

. Employing (60) and (4), we calculate

ζ (N − 1, pϑ) dx |x|Nβp ζ (N, pϑ) dx + |x|Nβp



|x|≤ν

1 − Nβ

1 (pϑ)N−2 dx (N − 2)! |x|Nβp

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

 ζ (N, β pu N−1 ) N, j



N

≤

|x|Nβp

RN

4925 N

dx + RN

ζ (N, βN, puN−1 ) dx + C1 ≤ C, |x|Nβp

where C1 is a constant depending only on N, β, p, while C is a constant depending on N, β,  q and p. This together with (64)–(66), the Hölder inequality and the fact that uj → u in Lloc (RN ) for any q > 0 implies that  ζ (N, β u N−1 ) N, j



N

N,β,τ, = lim

j →∞ RN

|x|Nβ

N

ζ (N, βN, uN−1 ) dx. |x|Nβ

dx = RN

Clearly we must have u 1,τ = 1. Moreover, by a straightforward calculation, we derive the Euler–Lagrange equation of u as follows: ⎧ N/(N −1) ⎨ − u + τ uN−1 = 1 u1/(N−1)  ) N   λ |x|Nβ ζ (N − 1, βN, u  ⎩ N/(N −1) N/(N −1) λ = RN |x|−Nβ u ζ (N − 1, βN, u )dx.

in RN ,

(67)

Applying Theorem 14 to (67), we have u ∈ C 1 (RN \ {0}) ∩ C 0 (RN ). This completes the proof of the theorem. 2 The remaining part of this section is devoted to the proof of Theorem 2. 3.3. Elementary properties of u Similar to Lemma 5, we have the following: Lemma 15. Let λ be defined as in (67). Then there holds lim inf→0 λ > 0. Proof. Employing the Lebesgue dominated convergence theorem and noting that u is a maximizer for subcritical Trudinger–Moser inequalities, we have for all u ∈ W 1,N (RN ) with

u 1,τ ≤ 1,  RN

N

ζ (N, αN (1 − β)|u| N−1 ) dx = lim →0 |x|Nβ



N

ζ (N, βN, |u| N−1 ) dx |x|Nβ

RN

 ≤ lim sup →0

RN

N

ζ (N, βN, uN−1 ) dx. |x|Nβ

One easily concludes  N,β,τ = lim

→0 RN

N

ζ (N, βN, uN−1 ) dx. |x|Nβ

(68)

4926

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

Since for any t ≥ 0, tζ (N − 1, t) =

∞ ∞ ∞    t k+1 tk tk = ≥ = ζ (N, t), k! (k − 1)! k!

k=N −2

k=N −1

k=N−1

one has 

1

λ ≥

βN,

RN

N

ζ (N, βN, uN−1 ) 1 dx = N,β,τ + o (1). Nβ |x| αN (1 − β)

Thus we get the desired result since N,β,τ > 0.

2

Since u 1,τ = 1, one can find some function u0 such that u converges to u0 weakly in q W 1,N (RN ), strongly in Lloc (RN ) for any q > 0, and a.e. in RN . Denote c = u (0). If c is a bounded sequence, then applying a priori estimates in Theorem 14 to (67), we conclude that 0 (RN ) ∩ C 1 (RN \ {0}). It is not difficult to see that u → u0 in Cloc loc  RN

N

ζ (N, αN (1 − β)u0N−1 ) dx = lim →0 |x|Nβ

 RN

N

ζ (N, αN (1 − β)uN−1 ) dx = N,β,τ . |x|Nβ

This also implies that u0 1,τ = 1 and thus u0 is the desired maximizer for the critical Trudinger– Moser functional. In the following, without loss of generality, we assume c → +∞ as  → 0. Lemma 16. u0 ≡ 0 and up to a subsequence, |∇u |N dx δ0 weakly in the sense of measure. Proof. We first prove that |∇u |N dx δ0 . Suppose not. There exists r0 > 0 such that  lim sup →0

|∇u |N dx ≤ η < 1.

Br0

Note that u is decreasing radially symmetric. Let  u (x) = u (x) − u (r0 ) for x ∈ Br0 . Then 1,N u LN (Br ) ≤ η < 1. Denote  u ∈ W0 (Br0 ) satisfies ∇ 0

1/(N −1)

f (x) =

N (x) 1 u ζ (N − 1, βN, uN−1 (x)). Nβ λ |x|

There holds for any p > 1, p1 > 1 and 1/p1 + 1/p2 = 1, 

 fp (x)dx

Br0

≤ Br0

p/(N −1)

N 1 u N−1 ζ (N − 1, β pu (x))dx N,  p λ |x|Nβp

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

⎛ ≤

1 ⎜ p ⎝ λ



Br0

⎛ ≤

1 ⎜ p ⎝ λ



Br0

⎞1/p1 pp1 /(N −1) u ⎟ dx ⎠ Nβp |x|

⎛



⎜ ⎝

Br0

⎞1/p1 ⎛

pp /(N −1) u 1 ⎟ dx ⎠ |x|Nβp



⎜ ⎝

N N−1

ζ (N − 1, βN, pp2 u |x|Nβp

|x|Nβp

Br0

u (r0 ) ≤

N ωN−1

1/N u N  L (Br

0)

r0

 ≤

 Br0

)

⎟ dx ⎠

⎟ dx ⎠

.

(69)

ωN−1 N N uN  dx ≥ u (r0 ) N r0 . It

1/N

N

⎞1/p2

⎞1/p2

N

N−1 eαN (1−β)pp2 u

Since u is nonnegative decreasing radially symmetric, one has follows that 

4927

ωN−1 τ

1 . r0

(70)

Here we have used u 1,τ = 1. For any ν > 0, there exists some constant C0 depending only on N and ν such that for all x ∈ Br0 , N

N

N

uN−1 (x) ≤ (1 + ν) uN−1 (x) + C0 uN−1 (r0 ).

(71)

Choosing p > 1, p2 > 1 sufficiently close to 1 and ν > 0 sufficiently small such that (1 − β)pp2 (1 + ν) + βp < 1, inserting (70) and (71) into (69), and noting that u is bounded in Lq (Br0 ) for any fixed q > 0, one can see from (3) and Lemma 15 that f is bounded in Lp (Br0 ). By the elliptic estimate (Theorem 14), u is uniformly bounded in Br0 /2 contradicting c → +∞. This confirms that |∇u |N dx δ0 in the sense of measure. Next we prove u0 ≡ 0. It follows from u 1,τ = 1 and |∇u |N dx δ0 that u LN (RN ) = o (1), which leads to 

 uN 0 dx ≤ lim sup

RN

→0

uN  dx = 0. RN

Therefore u0 ≡ 0 and the proof of the lemma is completed. 2 3.4. Blow-up analysis Let N/(N−1)

−1/(N−1) −βN, c r = λ1/N e  c

/N

.

(72)

Define ψN, (x) = c−1 u (r1/(1−β) x)

(73)

ϕN, (x) = c1/(N −1) (u (r1/(1−β) x) − c ).

(74)

and

4928

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

Analogous to Lemma 7, we have the following: Lemma 17. Let r , ψN, and ϕN, be defined as in (72)–(74). Then (i) for any γ < αN (1 −β)/N , N/(N−1)

1 (RN \{0}) ∩C 0 (RN ); (iii) ϕ → 0 as  → 0; (ii) ψN, → 1 in Cloc there holds r eγ c N, → loc 0 1 N N ϕN in Cloc (R \ {0}) ∩ Cloc (R ), where

ϕN (x) = −

  N N −1 αN N−1 (1−β) . |x| log 1 + N/(N −1) αN (1 − β) N (1 − β)1/(N−1)

Moreover  RN

N

e N−1 αN (1−β)ϕN dx = 1. |x|Nβ

(75)

Proof. (i) In view of (72), one has N

N−1 rN eN γ c

N

− N −α (1−β−− Nγ )cN−1 αN = c N−1 e N



N

RN N

− N −α (1−β−− Nγ )cN−1 αN ≤ c N−1 e N



 RN

N N−1

N

RN − N ≤ c N−1

N

uN−1 ζ (N − 1, βN, uN−1 ) dx |x|Nβ uN−1 eβN, u |x|Nβ

dx

N N−1

N

uN−1 eN γ u |x|Nβ

(76)

dx

Since Nγ < αN (1 − β), one can see from (3) that  RN

N N−1

u

N

N−1 e N γ u

|x|Nβ

dx ≤ C

N/(N−1)

for some constant C, which together with (76) implies that r eγ c (ii) Clearly ψN, is a distributional solution to N

1

= o (1).

N/(N−1)

N−1 N−1 −N ψN, (x) = −τ r1−β ψN, (x) + c−N |x|−Nβ ψN, (x)e−βN, c N

1

× ζ (N − 1, βN, uN−1 (r1−β x)).

(77)

1 (RN \ {0}) ∩ C 0 (RN ), where ψ is Applying Theorem 14 to (77), we have ψN, → ψN in Cloc N loc N a distributional solution to N ψN = 0 in R . Clearly ψN ≡ 1 on RN .

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

4929

(iii) In view of (67), we derive the equation of ϕN, as follows. −N ϕN, (x) = gN, (x) 1

N/(N−1)

N−1 (x)e−βN, c = |x|−Nβ ψN,

1

N

ζ (N − 1, βN, uN−1 (r1−β x))

N

N−1 − τ r1−β cN ψN, (x).

(78)

Let R, r be any two positive numbers such that R > 4r. Clearly gN, is bounded in Lp (BR ) for some p > 1. Moreover, −ϕN, ≥ 0. Theorem 14 implies that ϕN, is uniformly bounded in BR/2 . While gN, is bounded in L∞ (BR \ Br ). Hence we have by applying Theorem 14 to (78), ϕN, is bounded in C 1,α (BR/2 \ B2r ) for some 0 < α < 1. Therefore up to a subsequence, there exists 1 (RN \ {0}) ∩ C 0 (RN ). To derive the equation some function ϕN such that ϕN, → ϕN in Cloc loc of ϕN , we estimate

0≤e

N/(N−1) −βN, c

N−3 

1

Nk

k u N−1 (r 1−β x) βN,  

k!

k=0

≤e

N/(N−1) −βN, c

N−3 

Nk

k c N−1 βN, 

k=0

k!

= o (1)

uniformly on BR for any R > 0. Moreover, by the mean value theorem, we have 1

N

1 1 N ξN−1 (u (r1−β x) − c ) N −1 1 N (ξ /c ) N−1 ϕN, (x) = N −1 N ϕN (x) + o (1), = N −1

N

uN−1 (r1−β x) − cN−1 =

1/(1−β)

where ξ lies between u (r Hence N/(N−1)

e−βN, c

(79)

x) and c , and o (1) → 0 uniformly on BR for any fixed R > 0. N

1

N

ζ (N − 1, βN, uN−1 (r1−β x)) = eαN (1−β) N−1 ϕN (x) + o (1).

Furthermore, we obtain the equation of ϕN as follows: ⎧ ⎨ ⎩

−N ϕN (x) =

e

N ϕ (x) αN (1−β) N−1 N

|x|Nβ

in

RN ,

(80)

ϕN (0) = maxRN ϕN = 0.

Since ϕN, is decreasingly symmetric on RN , ϕN is also decreasingly symmetric. Denote ϕN (r) = ϕN (x), where r = |x| and x ∈ RN . Then (80) can be reduced to an ordinary differential equation, namely 

 N ϕN (r)  (r))N−1  = r N−1−Nβ eαN (1−β) N−1 (−rϕN

ϕN (0) = 0.

(81)

4930

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

By a standard uniqueness result of ordinary differential equations (see for example [12]), we can solve (81) as ϕN (r) = −

  N N −1 log 1 + cN r N−1 (1−β) , αN (1 − β)

where cN = αN N −N/(N −1) (1 − β)−1/(N −1) . It then follows that 

N

RN

eαN (1−β) N−1 ϕN (x) dx = ωN−1 |x|Nβ

∞ 0

r N−1−Nβ dr (1 + cN r N (1−β)/(N−1) )N

N −1 = ωN−1 N (1 − β)

∞ 0

t N−2 dt. (1 + cN t)N

(82)

Integration by parts gives ∞ IN = (N − 1) 0

=−

1 cN

t N−2 dt (1 + cN t)N

∞ t N−2 d(1 + cN t)1−N 0

∞ ∞  1 N−2 N −2 1−N  =− t (1 + cN t) t N−3 (1 + cN t)1−N dt  + c cN N 0 0

1 = IN−1 . cN Iteration leads to IN =

1

I = N−2 2

cN

1

∞

N−2 cN 0

1 1 dt = N−1 . (1 + cN t)2 cN

Inserting (83) into (82), we obtain  RN

N

e N−1 αN (1−β)ϕN (x) ωN−1 1 dx = = 1. N−1 |x|Nβ N (1 − β) cN

This completes the proof of the lemma. 2 For any 0 < γ < 1, we set u,γ = min{u , γ c }. Then we have the following:  Lemma 18. For any 0 < γ < 1, there holds lim→0 RN |∇u,γ |N dx = γ .

(83)

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

4931

Proof. Testing the equation (67) by u,γ , we have for any fixed R > 0, 

 |∇u,γ |N dx = −τ

RN

uN−1 u,γ dx +  RN

≥

1 λ



1 λ



1/(N−1)

u,γ RN

u −1) ζ (N − 1, βN, uN/(N )dx  |x|Nβ

1/(N −1)

γ c B

N/(N−1) u eβN, u dx + o (1) Nβ |x|

1/(1−β) Rr

 = (1 + o (1))γ BR

N

eαN (1−β) N−1 ϕN dx + o (1). |x|Nβ

Hence 



N

|∇u,γ | dx ≥ γ N

lim inf →0

RN

BR

eαN (1−β) N−1 ϕN dx. |x|Nβ

In view of (75), passing to the limit R → +∞, we obtain  |∇u,γ |N dx ≥ γ .

(84)

|∇(u − γ c )+ |N dx ≥ 1 − γ .

(85)

lim inf →0

RN

Similarly we have  lim inf →0

RN

Noting that u LN (RN ) = o (1), we have 

 |∇u,γ |N dx +

RN

|∇(u − γ c )+ |N dx =

RN

 |∇u |N dx = 1 + o (1).

(86)

RN

Combining (84)–(86), we conclude the lemma. 2 Lemma 19. We have  lim

→0 RN

N/(N −1)

ζ (N, βN, u |x|Nβ

)

λ . N/(N −1) →0 c 

dx = lim

As a consequence, for any θ < N/(N − 1), there holds λ /cθ → +∞ as  → 0.

(87)

4932

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

Proof. Let 0 < γ < 1 be fixed and u,γ be defined as before. Applying the mean value theorem to the function ζ (N, t) and recalling (59), we have −1) −1) −1) −1) ζ (N, βN, uN/(N ) = ζ (N − 1, ξ )βN, uN/(N ≤ ζ (N − 1, βN, uN/(N )βN, uN/(N , ,γ ,γ ,γ ,γ N/(N −1)

where ξ lies between βN, u,γ and 0. Since ζ (N − 1, t) = ζ (N, t) + t N−2 /(N − 2)! for all t ≥ 0, it follows from the above inequality that N−1 N −1) −1) −1) ζ (N, βN, uN/(N ) ≤ ζ (N, βN, uN/(N )βN, uN/(N + βN, u /(N − 2)!. ,γ ,γ ,γ

(88)

It is easy to see that 

q

u dx = o (1), |x|Nβ

RN

∀q ≥ N.

(89)

In view of Lemma 18, one can find some p > 1 such that 

N/(N −1)

ζ (N, pβN, u,γ |x|Nβ

lim sup →0

RN

)

dx < ∞.

(90)

By the Hölder inequality and (60), one has  RN

N

N

N−1 N−1 ζ (N, βN, u,γ )βN, u,γ dx Nβ |x|

⎛ ⎜ ≤⎝



RN

⎞1/p ⎛

N N−1

ζ (N, pβN, u,γ ) ⎟ dx ⎠ |x|Nβ

⎜ ⎝



N N−1

(βN, u,γ |x|Nβ

RN

 )p

⎞1/p ⎟ dx ⎠

,

(91)

where 1/p + 1/p  = 1. Combining (88)–(91), one concludes  lim

→0 RN

N N−1 ζ (N, βN, u,γ ) dx = 0. Nβ |x|

(92)

q

Since u → 0 in Lloc (RN ) for any q > 0, we obtain  u >γ c

N

ζ (N, βN, uN−1 ) dx = |x|Nβ



u >γ c

≤

1

N N−1

eβN, u |x|Nβ 

N

γ N−1 u

 >γ c

dx + o (1) N

N N−1

uN−1 eβN, u N |x|Nβ cN−1

dx + o (1)

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

1

≤ γ

N N−1

λ N

4933

+ o (1).

(93)

cN−1

Combining (92) and (93), we have  lim

→0 RN

N

ζ (N, βN, uN−1 ) λ 1 dx ≤ N/(N −1) lim inf N/(N −1) . →0 c |x|Nβ γ 

Letting γ → 1, we conclude 

N

ζ (N, βN, uN−1 ) λ dx ≤ lim inf N/(N −1) . Nβ →0 c |x| 

lim

→0 RN

(94)

An obvious analog of (35) is

lim sup →0



λ N/(N −1)

c

≤ lim

→0 RN

N

ζ (N, βN, uN−1 ) dx. |x|Nβ

(95)

N/(N −1)

Combining (94) and (95), we obtain (87), which together with (68) implies that λ /c a positive lower bound. Then for any θ < N/(N − 1), there holds

has

λ /cθ = cN/(N −1)−θ λ /cN/(N −1) → ∞. This proves the second assertion of the lemma. 2 1

1 (RN \ {0}) and weakly in W 1,q (RN ) for any 1 < q < N , where Lemma 20. cN−1 u → G in Cloc G is a distributional solution to

−N G + τ GN−1 = δ0

in RN .

Moreover, G ∈ W 1,N (RN \ Br ) for any r > 0 and G takes the form G(x) = −

N log |x| + A0 + w(x), αN

where A0 is a constant, and w ∈ C 0 (RN ) ∩ C 1 (RN \ {0}) satisfies w(x) = O(|x|N logN−1 |x|) as |x| → 0. Proof. Multiplying both sides of (67) by c , we have 1

1 N−1

−N (c

u ) + τ c uN−1 

N

c uN−1 ζ (N − 1, βN, uN−1 ) = λ |x|Nβ

in RN .

4934

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

Replacing Lemma 7 and Corollary 10 with Lemma 17 and Lemma 19 respectively in the proof of Lemma 11, we obtain for any φ ∈ C01 (RN ),  lim

→0 RN

1

N

c uN−1 ζ (N − 1, βN, uN−1 ) φdx = φ(0). λ |x|Nβ

Since the remaining part of the proof is completely analogous to that of ([12], Proposition 3.7 and Lemma 3.8), we omit the details but refer the reader to [12]. 2 To estimate the supremum N,β,τ , we need the following:  Lemma 21. Let w ∈ W01,N (Br ) satisfy Br |∇w |N dx ≤ 1, w 0 weakly in W01,N (Br ), and w is nonnegative and radially symmetric. Then  lim sup →0

Br

N/(N−1)

eαN (1−β)w |x|Nβ

−1

1 ωN−1 N (1−β) N−1 1 e k=1 k . r 1−β N

dx ≤

(96)

Proof. We first prove (96) for r = 1. Denote w (|x|) = w (x). Let v (x) = (1 − β)(N−1)/N w (|x|1/(1−β) ). Then 

 |∇v | dx =

|∇w |N dx.

N

B1

B1

Clearly we can assume up to a subsequence, v v0 weakly in W01,N (B1 ), v → v0 strongly in LN (B1 ), and v → v0 a.e. in B1 . Also, we can assume w → 0 a.e. in B1 . Hence we conclude v0 = 0 a.e. in B1 . By a change of variable t = s 1/(1−β) , there holds  B1

N N−1

eαN (1−β)w |x|Nβ

−1

1 dx =

N N−1

eαN (1−β)w t Nβ

(t)

−1

ωN−1 t N−1 dt

0

1 = 1−β 1 = 1−β =

1 1−β

1

N N−1

(eαN (1−β)w

(s 1/(1−β) )

− 1)ωN−1 s N−1 ds

0

1

N N−1

(eαN v

(s)

− 1)ωN−1 s N−1 ds

0



N N−1

(eαN v B1

− 1)dx.

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

4935

This together with Lemma 3 implies that  lim sup →0

B1

N N−1

eαN (1−β)w |x|Nβ

−1

1 ωN−1 N−1 1 e k=1 k . 1−β N

dx ≤

(97)

We next prove (96) for the case of general r. Set w˜  (x) = w (rx) for x ∈ B1 . One can check that   |∇ w˜  |N dx = |∇w |N dx B1

Br

and that  Br

N N−1

eαN (1−β)w |x|Nβ

−1

N N−1

 dx = r

N (1−β) B1

eαN (1−β)w˜  |x|Nβ

−1

dx.

2

This together with (97) gives the desired result.

By the equation (67) and u 1,τ = 1, we have    |∇u |N dx = 1 − (|∇u |N + τ uN )dx − τ uN   dx RN \Br

Br

 =1− RN \Br

Br N

 +

N

uN−1 ζ (N − 1, βN, uN−1 ) dx λ |x|Nβ N −2 ∂u

u |∇u |

∂r

 dσ − τ

∂Br

uN  dx.

(98)

Br

We estimate the right three terms on the above equation respectively. The first term can be calculated by  RN \Br

N

N

N/(N −1)

uN−1 ζ (N − 1, βN, uN−1 ) c 1 dx = N/(N −1) λ |x|Nβ λ c =



N

N N−1

u RN \Br

ζ (N − 1, βN, uN−1 ) dx |x|Nβ

o (1) . N/(N −1) c

(99)

A straightforward calculation on the second term reads ⎛

 u |∇u |N−2 ∂Br

∂u 1 ⎜ dσ = N/(N −1) ⎝ ∂r c

⎞



∂Br

G|∇G|N−2

∂G ⎟ dσ + o (1)⎠ ∂r

4936

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

⎛ =

⎜ ⎝G(r)

1 N/(N −1) c

⎞ ⎟ N Gdx + o (1)⎠

Br

⎛ =



⎜ ⎝−G(r) + τ G(r)

1 N/(N −1) c



⎞ ⎟ GN−1 dx + o (1)⎠ ,

(100)

Br

since G is a distributional solution of −N G + τ GN−1 = δ0 . Concerning the third term, one has ⎛

 uN  dx = Br

1 N/(N −1) c

⎜ ⎝



⎞ ⎟ GN dx + o (1)⎠ .

(101)

Br

Inserting (99)–(101) into (98) and noting that G(x) = − αNN log |x| + A0 + w(x), we conclude  |∇u | dx = 1 − N

Br



1 N/(N −1)

c

 1 N log + A0 + o (1) + or (1) . αN r

(102)

Define u,r = (u − u (r))+ , the positive part of u − u (r). Obviously u,r ∈ W01,N (Br ). It follows from Lemma 21 that  lim sup →0

Br

N/(N−1)

eαN (1−β)u,r |x|Nβ

/τ,r

−1

dx ≤

1 ωN−1 N (1−β) N−1 1 e k=1 k , r 1−β N

(103)

N/(N −1)

where τ,r = ∇u LN (B ) . One can see from Lemma 17 that u = c + o (1) on BRr 1/(1−β) . r  This together with Lemma 20 and (102) leads to that on BRr 1/(1−β) ⊂ Br , 

N

N

βN, uN−1 ≤ αN (1 − β)(u,r + u (r)) N−1 N

N−1 = αN (1 − β)u,r + N

1 N N−1 u (r) + o (1) αN (1 − β)u,r N −1

N αN (1 − β)G(r) + o (1) N −1   N N 1 + log + A0 + or (1) + o (1) αN (1 − β) N −1 αN r

N−1 = αN (1 − β)u,r + N

N−1 = αN (1 − β)u,r N

N−1 /τ,r + N (1 − β) log = αN (1 − β)u,r

1 + αN (1 − β)A0 + or (1) + o (1). r

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

4937

This together with (103) leads to N N−1

 B

N N−1



eβN, u −1 dx ≤ r −N (1−β) eαN (1−β)A0 +o(1) Nβ |x|

eαN (1−β)u,r |x|Nβ

B

1/(1−β) Rr

/τ,r

dx

1/(1−β) Rr N N−1



= r −N (1−β) eαN (1−β)A0 +o(1)

eαN (1−β)u,r /τ,r − 1 dx + o(1) |x|Nβ

B

1/(1−β) Rr

≤

1 ωN−1 N−1 e k=1 1−β N

1 k +αN (1−β)A0

+ o(1).

(104)

In view of (79), we obtain  B



N

ζ (N, βN, uN−1 ) dx = rN |x|Nβ

1/(1−β) Rr

BR

N N−1

⎛ =

=

1 1−β

eβN, u (r |y|Nβ

λ N/(N −1) c

⎜ ⎝



y)

dy + o (1) ⎞

e

N αN (1−β) N−1 ϕN (y)

|y|Nβ

BR

⎟ dy + o (1)⎠ + o (1)

λ (1 + oR (1) + o (1)) + o (1). N/(N −1) c

Therefore  lim lim

R→∞ →0 B

1/(1−β) Rr

N

ζ (N, βN, uN−1 ) λ dx = lim N/(N −1) . →0 c |x|Nβ 

(105)

Combining (104), (105) and (87), we conclude  N,β,τ = lim

→0 RN

N

ζ (N, βN, uN−1 ) 1 ωN−1 N−1 dx ≤ e k=1 |x|Nβ 1−β N

1 k +αN (1−β)A0

.

(106)

3.5. Test function computation We now construct test functions such that (106) does not hold. Precisely we construct a sequence of functions φ ∈ W 1,N (RN ) satisfying φ 1,τ = 1 and  RN

N

ζ (N, αN (1 − β)φN−1 ) 1 ωN−1 N−1 dx > e k=1 Nβ |x| 1−β N

1 k +αN (1−β)A0

(107)

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X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

for sufficiently small  > 0. For this purpose we set φ (x) =

⎧ ⎨c+ ⎩

1 c1/(N−1)

  N N−1 (1−β) ) + b , − αNN−1 (1−β) log(1 + cN (|x|/)

x ∈ B R x ∈ RN \ BR ,

G , c1/(N−1)

where cN = αN /(N N/(N −1) (1 − β)1/(N −1) ), G is given as in Lemma 20, R = (− log )1/(1−β) , b and c are constants depending only on  and β to be determined later. Note that G ∈ W 1,N (RN \ Br ) for any r > 0. To ensure φ ∈ W 1,N (RN ), we let c+

1 c1/(N−1)

 −

 N G(R) N −1 log(1 + cN R N−1 (1−β) ) + b = 1/(N−1) . αN (1 − β) c

By Lemma 20, we have G(x) = −(N/αN ) log |x| + A0 + w(x), where w(x) = O(|x|N × logN−1 |x|) as |x| → 0. Then the above equality leads to N

c N−1 =

N 1 N ωN−1 log  + O(R − N−1 (1−β) ). log + A0 − b − αN (1 − β) N (1 − β) αN

(108)

Now we calculate by the equation of G,  (|∇φ |

N

+ τ φN )dx



1

=

(|∇G|N + τ GN )dx

N

c N−1

RN \BR

c

∂BR

N N−1

⎛

⎜ G(R) ⎝1 − τ

−

N

c N−1



∂G dσ ∂ν ⎞ ⎟ GN−1 dx ⎠

BR



1

=

G|∇G|N−2

N N−1

1 c



1

=−

=

RN \BR

 N log(R) + A0 + O((R)N logN (R)) . (109) αN

Note that for any T > 0, there holds T IN (T ) ≡ 0

t N−1 dt (1 + t)N

1 = 1−N 1 = 1−N

T t N−1 d(1 + t)1−N 0



T 1+T

N−1

T + 0

t N−2 dt (1 + t)N−1

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

1 = 1−N



T 1+T

4939

N−1 + IN−1 (T ).

Since I1 (T ) = log(1 + T ), we have by iteration IN (T ) = log(1 + T ) −

N−1  k=1

1 k



T 1+T

k .

Hence, by a change of variables t = cN (r/)N (1−β)/(N−1) , we obtain 

R

1

|∇φ | dx = N

1

BR

N

N

−1 N−1 (1−β) N (r N−1 (1−β) + cN  )

N

N−1 N−1 ωN−1 c

N2

r N−1 (1−β)−1

0

dr

N (1−β)

1

=

1

N

N−1 N−1 ωN−1 c

= =

N −1 N (1 − β)

N −1 N

αN (1 − β)c N−1 1 αN (1 − β)c

cN R N−1 

0

t N−1 dt (1 + t)N

⎧

k ⎫ N ⎨  ⎬  N−1 1 N−1 (1−β) N R c N log 1 + cN R N−1 (1−β) − N (1−β) ⎩ ⎭ k 1 + c R N−1 k=1



N N−1

log

N

ωN−1 + N (1 − β) log R N (1 − β)

− (N − 1)

N−1  k=1

 1 1 ) . + O( N k R N−1 (1−β)

Moreover, we require b to be bounded with respect to . It then follows from (108) that  φN dx = O((R)N (log )N−1 ).

(110)

(111)

BR

Combining (109)–(111), we obtain

φ N 1,τ

=

1 N

c N−1



N−1 N −1  1 1 ωN−1 N log  + A0 − + log αN αN (1 − β) k αN (1 − β) N (1 − β) k=1  1 + O( N ) + O((R)N (log )N ) . R N−1 (1−β)

−

Setting φ 1,τ = 1, we have N

c N−1 = −

N−1 N N −1  1 1 ωN−1 1 log  + A0 − ), + log + O( N αN αN (1 − β) k αN (1 − β) N (1 − β) R N−1 (1−β) k=1

(112)

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X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

which together with (108) leads to N−1 N −1  1 1 ). + O( N αN (1 − β) k R N−1 (1−β)

b=

(113)

k=1

Denote b (x) = −

N N −1 log(1 + cN (|x|/) N−1 (1−β) ) + b. αN (1 − β)

(114)

Then c−N/(N −1) b (x) = O((log log  −1 )/ log ) uniformly in x ∈ BR , where R = (log  −1 )1/(1−β) . We have by the Taylor formula of (1 + t)N/(N −1) near t = 0,  N  N N N N−1 φN−1 (x) = c N−1 1 + c− N−1 b (x)   N N 2−N N 1 N N − N−1 − N−1 2 N−1 N−1 =c b (x) + (1 + ξ ) (c b (x)) c 1+ N −1 2 (N − 1)2 N

≥ c N−1 +

N b (x), N −1

(115)

N

where ξ lies between c− N−1 b (x) and 0. Inserting (112)–(114) into (115), we obtain for all x ∈ BR , N

αN (1 − β)φN−1 (x) ≥ −N (1 − β) log  + αN (1 − β)A0 +

N−1  k=1

− N log(1 + cN (|x|/)

N N−1 (1−β)

1 ωN−1 + log k N (1 − β) 1

) + O( R

N N−1 (1−β)

).

(116)

N

Also we have by a change of variables t = cN r N−1 (1−β) and integration by parts, 

1 N

BR

(1 + cN |y| N−1 (1−β) )N |y|Nβ

R dy =

ωN−1 r N−1−Nβ N

0

(1 + cN r N−1 (1−β) )N

dr

cN R N−1 (1−β)   =− +  N−1 (1 + t) N

t N−2

0

N (1−β) cN R N−1



= 0

R

0

(N − 2)t N−3 dt (1 + t)N−1

1 1 dt + O( N ) (1 + t)2 R N−1 (1−β) 1

= 1 + O(

N (1−β)

cN R N−1 

N N−1 (1−β)

).

(117)

X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

4941

Combining (116) and (117), we obtain 

N

BR

ζ (N, αN (1 − β)φN−1 ) dx = |x|Nβ



BR

≥

N N−1

eαN (1−β)φ |x|Nβ

dx + O(c

N(N−2) N−1

(R)N (1−β) )

N (1−β)

N−1 1 − N−1 ωN−1 ) k=1 k +αN (1−β)A0 +O(R e N (1 − β) N (1−β)  1 × dx N N−1 (1−β) )N |x|Nβ (1 + c (|x|/) N B R

+ O(c =

N(N−2) N−1

(R)N (1−β) )

N (1−β) ωN−1 N−1 1 +αN (1−β)A0 +O(R − N−1 ) e k=1 k N (1 − β)  1 × dy N (1 + cN |y| N−1 (1−β) )N |y|Nβ B R

+ O(c =

N(N−2) N−1

(R)N (1−β) )

ωN−1 N−1 e k=1 N (1 − β)

1 k +αN (1−β)A0

1

+ O( R

N N−1 (1−β)

).

(118)

Moreover, 



N

RN \BR

N−1 αN (1 − β)N−1 ζ (N, αN (1 − β)φN−1 ) dx ≥ N |x|Nβ (N − 1)!c N−1

RN \BR

⎛

=

N−1 (1 − β)N−1 αN

(N − 1)!c N

N N−1

⎜ ⎝



RN

GN dx |x|Nβ ⎞

GN ⎟ dx + o (1)⎠ . (119) |x|Nβ

N

Combining (118), (119) and noting that R − N−1 (1−β) c N−1 = o (1), we have  RN

N−1

N

1

ζ (N, αN (1 − β)φN−1 ) ωN−1 e k=1 k +αN (1−β)A0 dx ≥ |x|Nβ N (1 − β) ⎛ +

(αN

(1 − β))N−1

(N − 1)!c

N N−1

Therefore we conclude (107) for sufficiently small  > 0.

⎜ ⎝



RN

⎞ GN |x|Nβ

⎟ dx + o (1)⎠ .

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X. Li, Y. Yang / J. Differential Equations 264 (2018) 4901–4943

3.6. Completion of the proof of Theorem 2 Under the assumption that c → +∞, there holds (106). While it follows from (107) that N,β,τ >

1 ωN−1 N−1 e k=1 1−β N

1 k +αN (1−β)A0

.

This contradicts (106) and implies that c must be bounded. Then applying Theorem 14 to the equation (67), we get the desired extremal function. 2 Acknowledgments X. Li is supported by Natural Science Foundation of the Education Department of Anhui Province (No. KJ2016A641); Y. Yang is supported by the National Science Foundation of China (Grant Nos. 11171347 and 11471014). References [1] Adimurthi, K. Sandeep, A singular Moser–Trudinger embedding and its applications, Nonlinear Differential Equations Appl. 13 (2007) 585–603. [2] Adimurthi, Y. Yang, An interpolation of Hardy inequality and Trudinger–Moser inequality in RN and its applications, Int. Math. Res. Not. 13 (2010) 2394–2426. [3] D. Cao, Nontrivial solution of semilinear elliptic equations with critical exponent in R2 , Comm. Partial Differential Equations 17 (1992) 407–435. [4] L. Carleson, A. Chang, On the existence of an extremal function for an inequality of J. Moser, Bull. Sci. Math. 110 (1986) 113–127. [5] G. Csato, P. Roy, Extremal functions for the singular Moser–Trudinger inequality in 2 dimensions, Calc. Var. 54 (2015) 2341–2366. [6] M. de Souza, J.M. do Ó, A sharp Trudinger–Moser type inequality in R2 , Trans. Amer. Math. Soc. 366 (2014) 4513–4549. [7] W. Ding, J. Jost, J. Li, G. Wang, The differential equation u = 8π − 8π heu on a compact Riemann surface, Asian J. Math. 1 (1997) 230–248. [8] J.M. do Ó, N -Laplacian equations in RN with critical growth, Abstr. Appl. Anal. 2 (1997) 301–315. [9] J.M. do Ó, M. de Souza, A sharp inequality of Trudinger–Moser type and extremal functions in H 1,n (Rn ), J. Differential Equations 258 (2015) 4062–4101. [10] M. Flucher, Extremal functions for Trudinger–Moser inequality in 2 dimensions, Comment. Math. Helv. 67 (1992) 471–497. [11] M. Ishiwata, Existence and nonexistence of maximizers for variational problems associated with Trudinger–Moser type inequalities in RN , Math. Ann. 351 (2011) 781–804. [12] Y. Li, B. Ruf, A sharp Trudinger–Moser type inequality for unbounded domains in RN , Indiana Univ. Math. J. 57 (2008) 451–480. [13] K. Lin, Extremal functions for Moser’s inequality, Trans. Amer. Math. Soc. 348 (1996) 2663–2671. [14] J. Moser, A sharp form of an inequality by N. Trudinger, Indiana Univ. Math. J. 20 (1971) 1077–1091. [15] R. Panda, Nontrivial solution of a quasilinear elliptic equation with critical growth in Rn , Proc. Indian Acad. Sci. Math. Sci. 105 (1995) 425–444. [16] J. Peetre, Espaces d’interpolation et theoreme de Soboleff, Ann. Inst. Fourier (Grenoble) 16 (1966) 279–317. [17] S. Pohozaev, The Sobolev embedding in the special case pl = n, in: Proceedings of the Technical Scientific Conference on Advances of Scientific Research 1964–1965, Mathematics Sections, Moscov. Energet. Inst., Moscow, 1965, pp. 158–170. [18] B. Ruf, A sharp Trudinger–Moser type inequality for unbounded domains in R2 , J. Funct. Anal. 219 (2005) 340–367. [19] J. Serrin, Local behavior of solutions of quasi-linear equations, Acta Math. 111 (1964) 247–302.

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