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The existence of positive solution for an elliptic problem with critical growth and logarithmic perturbation

  • Yinbin Deng , Qihan He EMAIL logo , Yiqing Pan and Xuexiu Zhong
Published/Copyright: February 21, 2023

Abstract

We consider the existence and nonexistence of the positive solution for the following Brézis-Nirenberg problem with logarithmic perturbation:

Δ u = u 2 2 u + λ u + μ u log u 2 x Ω , u = 0 x Ω ,

where Ω R N is a bounded open domain, λ , μ R , N 3 and 2 2 N N 2 is the critical Sobolev exponent for the embedding H 0 1 ( Ω ) L 2 ( Ω ) . The uncertainty of the sign of s log s 2 in ( 0 , + ) has some interest in itself. We will show the existence of positive ground state solution, which is of mountain pass type provided λ R , μ > 0 and N 4 . While the case of μ < 0 is thornier. However, for N = 3 , 4 , λ ( , λ 1 ( Ω ) ) , we can also establish the existence of positive solution under some further suitable assumptions. A nonexistence result is also obtained for μ < 0 and ( N 2 ) μ 2 + ( N 2 ) μ 2 log ( N 2 ) μ 2 + λ λ 1 ( Ω ) 0 if N 3 . Comparing with the results in the study by Brézis and Nirenberg (Positive solutions of nonlinear elliptic equations involving critical Sobolev exponents, Comm. Pure Appl. Math. 36 (1983), 437–477), some new interesting phenomenon occurs when the parameter μ on logarithmic perturbation is not zero.

MSC 2010: 35A01; 35A15; 35B09; 35B33; 35G30

1 Introduction and main results

In this article, we investigate the existence and nonexistence of positive solution for the following Brézis-Nirenberg problem with a logarithmic term:

(1.1) Δ u = u 2 2 u + λ u + μ u log u 2 x Ω , u = 0 x Ω ,

where Ω R N is a bounded open domain, λ , μ R , N 3 , and 2 = 2 N N 2 is the critical Sobolev exponent for the embedding H 0 1 ( Ω ) L 2 ( Ω ) . Here, H 0 1 ( Ω ) denotes the closure of C 0 ( Ω ) equipped with the norm u ( Ω u 2 d x ) 1 2 .

Our motivation to consider (1.1) is that it resembles some variational problems in geometry and physics, which is lack of compactness. The most notorious example is Yamabe’s problem: finding a function u satisfying

4 N 1 N 2 Δ u = R u 2 2 u R ( x ) u on , u > 0 on ,

where R is a constant, is an N -dimensional Riemannian manifold, and Δ denotes the Laplacian and R ( x ) represents the scalar curvature. For some other examples, we refer to [2,8,10, 1315] and the references therein.

When λ = μ = 0 , (1.1) is reduced to

(1.2) Δ u = u 2 2 u x Ω , u = 0 x Ω .

Pohoǎev [11] asserts that (1.2) has no nontrivial solutions when Ω is starshaped. But, as Brézis and Nirenberg have shown in [3], a lower-order terms can reverse this circumstance. Indeed, they considered the following classical problem:

(1.3) Δ u = u 2 2 u + λ u x Ω , u = 0 x Ω ,

with λ R , N 3 and Ω R N is a bounded domain. They found out that the existence of a solution depends heavily on the values of λ and N . Precisely, they showed that:

  1. when N 4 and λ ( 0 , λ 1 ( Ω ) ) , there exists a positive solution for problem (1.3);

  2. when N = 3 and Ω is a ball, problem (1.3) has a positive solution if and only if λ 1 4 λ 1 ( Ω ) , λ 1 ( Ω ) ;

  3. problem (1.3) has no solutions when λ 0 and Ω is starshaped;

where λ 1 ( Ω ) denotes the first eigenvalue of Δ with zero Dirichlet boundary value. Furthermore, Brézis and Nirenberg [3] also considered the following general case:

(1.4) Δ u = u 2 2 u + f ( x , u ) x Ω , u = 0 x Ω ,

where f ( x , u ) satisfies some of the following assumptions:

  1. f ( x , u ) = a ( x ) u + g ( x , u ) , a ( x ) L ( Ω ) ;

  2. lim u 0 + g ( x , u ) u = 0 , uniformly in x Ω ;

  3. lim u + g ( x , u ) u 2 1 = 0 , uniformly in x Ω ;

  4. α > 0 such that ( v 2 a ( x ) v 2 ) d x α v 2 d x for all v H 0 1 ( Ω ) ;

  5. f ( x , u ) 0 for a.e x ω 0 and for all u 0 , where ω 0 is some nonempty open subset of Ω ;

  6. f ( x , u ) δ 0 > 0 for a.e x ω 0 and for all u I , where ω 0 is given in ( f 5 ) , I ( 0 , + ) is some nonempty open interval and δ 0 > 0 is some constant;

  7. f ( x , u ) δ 1 u for a.e x ω 1 and for all u [ 0 , A ] , or, f ( x , u ) δ 1 u for a.e x ω 1 and for all u [ A , + ] , where ω 1 is some nonempty open subset of Ω and δ 1 , A are two positive constants;

  8. lim u + f ( x , u ) u 3 = + uniformly in x ω 2 , where ω 2 is some nonempty open subset of Ω .

They showed that if the assumptions ( f 1 ) ( f 4 ) hold and there exists some 0 u 0 H 0 1 ( Ω ) { 0 } such that sup t 0 I ( t u 0 ) < 1 N S N 2 , then (1.4) has a positive solution. More precisely, they proved that:
  1. If N 5 , (1.4) has a positive solution provided ( f 1 ) ( f 6 ) ;

  2. If N = 4 , (1.4) has a positive solution provided ( f 1 ) ( f 5 ) , and ( f 7 ) ;

  3. If N = 3 , (1.4) has a positive solution provided ( f 1 ) ( f 5 ) , and ( f 8 ) .

Some similar results can be seen in [1,5,7]. Barrios et al. [1] proved the existence of positive solution for a fractional critical problem with a lower-order term, and Gao and Yang [5] and Li and Ma [7] considered the existence of positive solution to a Choquard equation with critical exponent and lower-order term in a bounded domain Ω and in R N , respectively.

Remark 1.1

Compared with u 2 2 u , u log u 2 is a lower-order term at infinity. However, we note that the situation we considered in present article is not covered earlier. Indeed, in equation (1.1), f ( x , u ) = λ u + μ u log u 2 . So ( f 1 ) fails due to the fact of lim u 0 + u log u 2 u = . That is, λ u = o ( μ u log u 2 ) for u close to 0. So it is natural to believe that μ u log u 2 has much more influence than λ u on the existence of positive solutions to equation (1.1). Hence, our main goal in the present article is to make clear this guess.

To find a positive solution to equation (1.1), we define a modified functional:

(1.5) I ( u ) = 1 2 Ω u 2 d x 1 2 u + 2 d x λ 2 u + 2 d x μ 2 u + 2 ( log u + 2 1 ) d x , u H 0 1 ( Ω ) ,

which can be rewritten by

(1.6) I ( u ) = 1 2 Ω u 2 d x 1 2 u + 2 d x μ 2 u + 2 log u + 2 + λ μ 1 d x , u H 0 1 ( Ω ) ,

where u + = max { u , 0 } , u = max { u , 0 } . It is easy to see that I is well defined in H 0 1 ( Ω ) , and any nonnegative critical point of I corresponds to a solution of equation (1.1).

Before stating our results, we introduce some notations. Hereafter, we use to denote Ω d x , unless specifically stated, and let S and λ 1 ( Ω ) be the best Sobolev constant of the embedding H 1 ( R N ) L 2 ( R N ) and the first eigenvalue of Δ with zero Dirichlet boundary value, respectively, i.e.,

S inf u H 1 ( R N ) { 0 } R N u 2 d x R N u 2 d x 2 2

and

λ 1 ( Ω ) inf u H 0 1 ( Ω ) { 0 } Ω u 2 d x Ω u 2 d x .

We also set

v 2 v 2 , v H 0 1 ( Ω ) , N { u H 0 1 ( Ω ) { 0 } g ( u ) = 0 } ,

and

(1.7) c g inf u N I ( u ) , c M inf γ Γ max t [ 0 , 1 ] I ( γ ( t ) ) ,

where

g ( u ) u 2 u + 2 λ u + 2 μ u + 2 log u + 2

and

Γ { γ C ( [ 0 , 1 ] , H 0 1 ( Ω ) ) γ ( 0 ) = 0 , I ( γ ( 1 ) ) < 0 } .

Let

A 0 { ( λ , μ ) λ R , μ > 0 } , B 0 ( λ , μ ) λ [ 0 , λ 1 ( Ω ) ) , μ < 0 , 1 N λ 1 ( Ω ) λ λ 1 ( Ω ) N 2 S N 2 + μ 2 Ω > 0 , C 0 ( λ , μ ) λ R , μ < 0 , 1 N S N 2 + μ 2 e λ μ Ω > 0 .

Here, comes our main results.

Theorem 1.2

If ( λ , μ ) A 0 and N 4 , then problem (1.1) has a positive Mountain pass solution, which is also a ground state solution.

Denote f ( s ) s 2 2 s + λ s + μ s log s 2 , which is of odd. It is easy to see that N and c M c g if problem (1.1) has a positive mountain pass solution. On the other hand, when λ R and μ > 0 , f ( s ) s is strictly increasing in ( 0 , + ) and strictly decreasing in ( , 0 ) , which enable one to show that c M c g (see [18, Theorem 4.2]). Therefore, the ground state energy c g equals to the Mountain pass level energy c M , which implies that the mountain pass solution must be a ground state solution. So, in Theorem 1.2, we only need to show that problem (1.1) has a positive mountain pass solution.

The case of μ < 0 is thorny. Indeed for ( λ , μ ) B 0 C 0 , I ( u ) still has the mountain pass geometry (see Lemma 2.1). However, in such a case, it holds that c g < c M . Since we can not check the ( PS ) c M condition for I ( u ) , we apply the mountain pass theorem without the ( PS ) c M condition to gain a positive solution for equation (1.1) when ( λ , μ ) B 0 C 0 . However, we do not know whether this solution is of mountain pass type.

Theorem 1.3

Problem (1.1) possesses a positive solution provided one of the following condition holds:

  1. N = 3 , ( λ , μ ) B 0 C 0 ;

  2. N = 4 , ( λ , μ ) B 0 C 0 with 32 e λ μ ρ max 2 < 1 , where ρ max sup { r > 0 : x Ω s.t. B ( x , r ) Ω } .

For the nonexistence of positive solutions for problem (1.1), we have the following partial result.

Theorem 1.4

Assume that N 3 . If μ < 0 and ( N 2 ) μ 2 + ( N 2 ) μ 2 log ( N 2 ) μ 2 + λ λ 1 ( Ω ) 0 , then problem (1.1) has no positive solutions.

The existence and nonexistence results given by Theorems 1.21.4 can be described on the ( λ , μ ) plane in Figure 1. The pink regions stand for the existence of positive solution, while the blue regions correspond the nonexistence of the positive solution. Here, τ 1 , η 1 , η 2 , and η 3 are curves given by

τ 1 : ( N 2 ) μ 2 + ( N 2 ) μ 2 log ( N 2 ) μ 2 + λ λ 1 ( Ω ) = 0 , η 1 : 1 N λ 1 ( Ω ) λ λ 1 ( Ω ) N 2 S N 2 + μ 2 Ω = 0 , η 2 : 1 N S N 2 + μ 2 e λ μ Ω = 0 , η 3 : 32 e λ μ = ρ max 2 , ρ max sup { r ( 0 , + ) : x Ω s.t. B ( x , r ) Ω } .

Figure 1 
               Existence and nonexistence. (a) 
                     
                        
                        
                           N
                           =
                           3
                        
                        N=3
                     
                  . (b) 
                     
                        
                        
                           N
                           =
                           4
                        
                        N=4
                     
                  . (c) 
                     
                        
                        
                           N
                           ≥
                           5
                        
                        N\ge 5
                     
                  .
Figure 1

Existence and nonexistence. (a) N = 3 . (b) N = 4 . (c) N 5 .

Remark 1.5

Comparing the results of [3] and Figure 1, we find that for the case of N 4 , equation (1.1) possesses a positive solution only for λ ( 0 , λ 1 ( Ω ) ) if μ = 0 , while it has a positive solution for all λ R if μ > 0 . So we see that μ u log u 2 ( μ > 0 ) really plays a leading role (compared with λ u ) in the effect on the existence of positive solution to equation (1.1). A similar phenomenon occurs for N = 3 : equation (1.1) has a positive solution only for λ ( λ , λ 1 ( Ω ) ) ( 0 , λ 1 ( Ω ) ) if μ = 0 , while it has a positive solution for all λ , μ log 2 S N 2 N μ Ω 0 , 1 μ N Ω 2 S N 2 2 N λ 1 ( Ω ) if μ < 0 .

Before closing the introduction, we give the outline of this article. In Section 2, we will check the mountain pass geometry structure for I ( u ) , under different specific situations. We also give some other preliminaries. In Section 3, we are devoted to estimate the mountain pass level c M for different parameters λ , μ , and N . The proofs of our main Theorems 1.2, 1.3, and 1.4 are given in Section 4.

2 Preliminaries

In this section, first we verify the mountain pass geometry structure for I ( u ) when ( λ , μ ) A 0 B 0 C 0 . Second, we show that I ( u ) satisfies ( PS ) d condition provided d < 1 N S N 2 and μ > 0 . Finally, we deduce a existence result for equation (1.1) when c M 0 , 1 N S N 2 and ( λ , μ ) B 0 C 0 .

Lemma 2.1

Assume that N 3 and ( λ , μ ) A 0 B 0 C 0 . Then the functional I ( u ) satisfies the mountain pass geometry structure:

  1. there exist α , ρ > 0 such that I ( v ) α for all v = ρ ;

  2. there exists ω H 0 1 ( Ω ) such that ω ρ and I ( ω ) < 0 .

Proof

We divide the proof into three cases.

Case 1: ( λ , μ ) A 0 .

Since μ > 0 , it follows from the fact s 2 log s 2 C s 2 for all s [ 1 , + ) that

μ u + 2 log u + 2 + λ μ 1 μ u + 2 log u + 2 + λ μ = μ u + 2 log e λ μ u + 2 = μ e λ μ u + 2 1 u + 2 log e λ μ u + 2 + μ e λ μ u + 2 1 u + 2 log e λ μ u + 2

μ e λ μ u + 2 1 u + 2 log e λ μ u + 2 C μ e λ μ u + 2 1 e ( 2 2 ) λ 2 μ u + 2 C e ( 2 2 ) λ 2 μ μ u 2 C e ( 2 2 ) λ 2 μ μ u 2 .

So

I ( u ) 1 2 u 2 C 1 u 2 C 2 u 2 for some  C 1 , C 2 > 0 ,

which implies that there exist α > 0 and ρ > 0 such that I ( v ) α > 0 for all v = ρ .

Let 0 φ H 0 1 ( Ω ) { 0 } be a fixed function, then

I ( t φ ) = t 2 2 φ 2 t 2 2 φ 2 μ 2 t 2 φ 2 log ( t 2 φ 2 ) + λ μ 1 = t 2 2 φ 2 t 2 2 φ 2 μ 2 t 2 φ 2 ( log t 2 + log φ 2 + λ μ 1 ) = t 2 2 φ 2 t 2 2 φ 2 μ 2 t 2 log t 2 φ 2 μ 2 t 2 φ 2 ( log φ 2 + λ μ 1 ) as t + ,

since lim t + t 2 t 2 log t 2 = + . Therefore, we can choose t 0 R + large enough such that

I ( t 0 φ ) < 0 and t 0 φ > ρ .

So there is a function ω H 0 1 ( Ω ) such that ω ρ and I ( ω ) < 0 .

Case 2: ( λ , μ ) B 0 .

Since μ < 0 , we have

μ 2 u + 2 ( log u + 2 1 ) = μ 2 u + 2 log ( e 1 u + 2 ) = μ 2 { e 1 u + 2 1 } u + 2 log ( e 1 u + 2 ) μ 2 { e 1 u + 2 1 } u + 2 log ( e 1 u + 2 ) μ 2 { e 1 u + 2 1 } u + 2 log ( e 1 u + 2 ) μ 2 e { e 1 u + 2 1 } e 1 d x μ 2 Ω .

It follows that

(2.1) I ( u ) 1 2 λ 1 ( Ω ) λ λ 1 ( Ω ) u 2 1 2 S 2 2 u 2 2 2 + μ 2 Ω .

Put α 1 N λ 1 ( Ω ) λ λ 1 ( Ω ) N 2 S N 2 + μ 2 Ω and ρ λ 1 ( Ω ) λ λ 1 ( Ω ) N 2 4 S N 4 , then α > 0 and ρ > 0 due to the fact ( λ , μ ) B 0 . By (2.1),

I ( v ) 1 2 λ 1 ( Ω ) λ λ 1 ( Ω ) ρ 2 1 2 S 2 2 ρ 2 + μ 2 Ω = 1 N λ 1 ( Ω ) λ λ 1 ( Ω ) N 2 S N 2 + μ 2 Ω = α > 0

for any v = ρ . Applying a similar argument as Case 1, we can find a function ω H 0 1 ( Ω ) such that ω ρ and I ( ω ) < 0 .

Case 3: ( λ , μ ) C 0 .

Since μ < 0 , we have

λ 2 u + 2 μ 2 u + 2 ( log u + 2 1 ) = μ 2 u + 2 log u + 2 + λ μ 1 = μ 2 u + 2 log e λ μ 1 u + 2 = μ 2 e λ μ 1 u + 2 1 u + 2 log e λ μ 1 u + 2 μ 2 e λ μ 1 u + 2 1 u + 2 log e λ μ 1 u + 2 μ 2 { e λ μ 1 u + 2 1 } u + 2 log e λ μ 1 u + 2 μ 2 e 1 λ μ { e λ μ 1 u + 2 1 } e 1 d x μ 2 e λ μ Ω .

It follows that

(2.2) I ( u ) 1 2 u 2 1 2 S 2 2 u 2 2 2 + μ 2 e λ μ Ω .

Put α 1 N S N 2 + μ 2 e λ μ Ω and ρ S N 4 , then α > 0 due to that ( λ , μ ) C 0 . By (2.2),

I ( v ) 1 2 ρ 2 1 2 S 2 2 ρ 2 + μ 2 e λ μ Ω = 1 N S N 2 + μ 2 e λ μ Ω = α > 0

for any v = ρ . Similarly, it is not hard to find a function ω H 0 1 ( Ω ) such that ω ρ and I ( ω ) < 0 .□

Lemma 2.2

Assume that N 3 , λ R and μ R { 0 } . Then any ( PS ) d sequence { u n } of I must be bounded in H 0 1 ( Ω ) for all d R .

Proof

By the definition of the ( PS ) d sequence, we have that, as n + ,

I ( u n ) d and I ( u n ) 0 in H 1 ( Ω ) .

That is,

(2.3) 1 2 u n 2 1 2 ( u n ) + 2 μ 2 ( u n ) + 2 log ( u n ) + 2 + μ λ 2 ( u n ) + 2 = d + o n ( 1 ) ,

and

(2.4) u n 2 ( u n ) + 2 λ ( u n ) + 2 μ ( u n ) + 2 log ( u n ) + 2 = o n ( 1 ) u n

as n + . Now we divide the proof into two cases.

Case 1: μ > 0 .

It follows from (2.3) and (2.4) that

d + o n ( 1 ) + o n ( 1 ) u n = I ( u n ) 1 2 I ( u n ) , u n = 1 N ( u n ) + 2 + μ 2 ( u n ) + 2 μ 2 ( u n ) + 2 ,

and thus, ( u n ) + 2 2 C + C u n . By using (2.3) and (2.4) again, we have, for n large enough,

2 d + u n I ( u n ) 1 2 I ( u n ) , u n = 1 N u n 2 λ N ( u n ) + 2 + μ 2 ( u n ) + 2 1 N μ ( u n ) + 2 log ( u n ) + 2 .

Recalling the following inequality (see [12] or see [9, Theorem 8.14])

u 2 log u 2 a π u 2 + ( log u 2 2 N ( 1 + log a ) ) u 2 2 for u H 0 1 ( Ω ) and a > 0 ,

we have that

1 N u n 2 2 d + u n + C ( u n ) + 2 + 1 N μ ( u n ) + 2 log ( u n ) + 2 C + C u n + 1 N μ a π u n 2 + ( log ( u n ) + 2 2 N ( 1 + log a ) ) ( u n ) + 2 2 C + C u n + 1 2 N u n 2 + ( u n ) + 2 2 log ( u n ) + 2 2 + C ( u n ) + 2 2 C + C u n + 1 2 N u n 2 + C ( u n ) + 2 2 δ + C ( u n ) + 2 2 + δ + C ( u n ) + 2 2 C + C u n + 1 2 N u n 2 + C ( C + C u n ) 2 δ 2 + C ( C + C u n ) 2 + δ 2 ,

where a > 0 with a π μ < 1 2 and δ ( 0 , 1 ) . So there exists C > 0 such that u n < C .

Case 2: μ < 0 .

For n large enough, we have

(2.5) 2 d + u n 1 N u n 2 λ N ( u n ) + 2 + μ 2 ( u n ) + 2 1 N μ ( u n ) + 2 log ( u n ) + 2 = 1 N u n 2 1 N μ ( u n ) + 2 log ( e λ μ N 2 ( u n ) + 2 ) 1 N u n 2 1 N μ { e λ μ N 2 ( u n ) + 2 1 } ( u n ) + 2 log ( e λ μ N 2 ( u n ) + 2 ) 1 N u n 2 1 N μ { e λ μ N 2 ( u n ) + 2 1 } e N 2 λ μ 1 d x 1 N u n 2 + μ N e N 2 λ μ 1 Ω ,

which implies that { u n } is bounded in H 0 1 ( Ω ) .

Lemma 2.3

Let { u n } be a bounded sequence in H 0 1 ( Ω ) such that u n u a.e in Ω as n , then

(2.6) lim n Ω u n 2 log u n 2 d x = Ω u 2 log u 2 d x ,

and

(2.7) lim n Ω ( u n ) + 2 log ( u n ) + 2 d x = Ω u + 2 log u + 2 d x .

Proof

We only prove (2.6). And (2.7) can be proved similarly.

Under the conditions, there exists some C > 0 such that

Ω u n 2 log u n 2 C and Ω u 2 log u 2 C .

By [12, Lemma 3.1], we have

lim n Ω u n 2 log u n 2 u n u 2 log u n u 2 = Ω u 2 log u 2 .

Since s 2 log s 2 C s 2 δ + C s 2 + δ and the embedding of H 0 1 ( Ω ) L p ( 1 p < 2 ) is compact, we obtain that

Ω u n u 2 log u n u 2 d x C Ω u n u 2 δ + C Ω u n u 2 + δ 0 as n .

Hence,

lim n Ω u n 2 log u n 2 = Ω u 2 log u 2 .

Lemma 2.4

If N 3 , λ R , μ > 0 , and d < 1 N S N 2 , then I ( u ) satisfies the ( PS ) d condition.

Proof

Let { u n } be a ( PS ) d sequence of I . By Lemma 2.2, we know that { u n } is bounded in H 0 1 ( Ω ) . So there exists u H 0 1 ( Ω ) such that, up to a subsequence,

u n u in H 0 1 ( Ω ) , u n u in L q ( Ω ) , 1 q < 2 , u n u a.e. in Ω .

Since I ( u n ) , φ 0 as n for any φ C 0 ( Ω ) , u is a weak solution to

Δ u = u + 2 2 u + + λ u + + μ u + log u + 2 ,

which implies that

u 2 = u + 2 + λ u + 2 + μ u + 2 log u + 2

and

(2.8) I ( u ) = 1 2 u 2 1 2 u + 2 λ 2 u + 2 μ 2 u + 2 ( log u + 2 1 ) = 1 N u + 2 + μ 2 u + 2 0 .

Following from the definition of ( PS ) d sequence, we have

u n 2 ( u n ) + 2 λ ( u n ) + 2 μ ( u n ) + 2 log ( u n ) + 2 = o n ( 1 )

and

1 2 u n 2 1 2 ( u n ) + 2 λ 2 ( u n ) + 2 μ 2 ( u n ) + 2 ( log ( u n ) + 2 1 ) = d + o n ( 1 ) .

Set v n = u n u . Then

v n 2 ( v n ) + 2 = o n ( 1 )

and

I ( u ) + 1 2 v n 2 1 2 ( v n ) + 2 = d + o n ( 1 ) .

Let

v n 2 k , as n .

So

( v n ) + 2 k , as n .

By the definition of S , we have

u 2 2 S u 2 2 , u H 0 1 ( Ω )

and

k + o n ( 1 ) = v n 2 S ( v n ) + 2 2 2 = S k N 2 N + o n ( 1 ) .

If k > 0 , then k S N 2 . By (2.8), we have

0 I ( u ) d 1 2 1 2 k d 1 N S N 2 < 0 ,

which is impossible. So k = 0 , and thus,

u n u , in H 0 1 ( Ω ) .

Lemma 2.5

Assume that N 3 , λ R , μ < 0 , and c ( , 0 ) 0 , 1 N S N 2 . If { u n } is a ( PS ) c sequence of I , then there exists a u H 0 1 ( Ω ) { 0 } such that u n u weakly in H 0 1 ( Ω ) and u is a nonnegative weak solution of (1.1).

Proof

Let { u n } be a ( PS ) c sequence of I . By Lemma 2.2, we know that { u n } is bounded in H 0 1 ( Ω ) . So there exists u H 0 1 ( Ω ) such that, up to a subsequence,

u n u in H 0 1 ( Ω ) , u n u in L q ( Ω ) , 1 q < 2 , u n u a.e in Ω .

Since I ( u n ) , φ 0 as n for any φ C 0 ( Ω ) , u is a weak solution to

(2.9) Δ u = u + 2 2 u + + λ u + + μ u + log u + 2 .

Assume that u = 0 and set v n u n u . Following from the definition of ( PS ) c sequence and Brezis-Lieb lemma, we have

v n 2 ( v n ) + 2 = o n ( 1 )

and

(2.10) 1 2 v n 2 1 2 ( v n ) + 2 = c + o n ( 1 ) .

Let

v n 2 k , as n .

Then

( v n ) + 2 k , as n .

It is easy to see that k > 0 . In fact, if k = 0 , then u n 2 = v n 2 0 , which implies that I ( u n ) 0 , contradicting to c 0 . From Lemma 2.4, we can obtain that k S N 2 . So, by (2.10), we have

1 N S N 2 1 N k = 1 2 1 2 k = c < 1 N S N 2 ,

a contradiction. Hence, u 0 .

By the density of C 0 in H 0 1 ( Ω ) and (2.9), we have that

u 2 = 0 ,

which implies that u 0 . Therefore, we can see that u H 0 1 ( Ω ) { 0 } and u is a nonnegative weak solution of (1.1).□

3 Estimations on c M

In this section, we are going to give an estimation that c M < 1 N S N 2 , under different assumptions on parameters λ and μ and dimension N . Inspired by Brézis and Nirengberg [3], it is sufficient to find some suitable U ε H 0 1 ( Ω ) such that sup t 0 I ( t U ε ) < 1 N S N 2 . Without loss of generality, we may assume that 0 Ω , in particular, we suppose that 0 is the geometric center of Ω , i.e., ρ max = dist ( 0 , Ω ) .

It is well known (see [4,6,16]) that the following problem

Δ u = u 2 2 u , x R N , u > 0 , u ( 0 ) = max x R N u ( x ) ,

has a unique solution u ˜ ( x )

u ˜ ( x ) = [ N ( N 2 ) ] N 2 4 1 ( 1 + x 2 ) N 2 2 .

Correspondingly, up to a dilations,

u ε ( x ) = [ N ( N 2 ) ] N 2 4 ε ε 2 + x 2 N 2 2

is a minimizer for S .

We let φ ( x ) C 0 ( Ω ) be such that φ ( x ) 1 for x in some neighborhood B ρ ( 0 ) of 0, and define

(3.1) U ε ( x ) = φ ( x ) u ε ( x ) .

Lemma 3.1

If N 4 , then we have, as ε 0 + ,

(3.2) Ω U ε 2 = S N 2 + O ( ε N 2 ) ,

(3.3) Ω U ε 2 = S N 2 + O ( ε N ) ,

and

Ω U ε 2 = d ε 2 ln ε + O ( ε 2 ) , i f N = 4 , d ε 2 + O ( ε N 2 ) , i f N 5 ,

where d is a positive constant.

Proof

The proof can be found in [18].□

Lemma 3.2

If N 5 , then we have, as ε 0 + ,

Ω U ε 2 log U ε 2 = C 0 ε 2 log 1 ε + O ( ε 2 ) ,

where C 0 is a positive constant.

Proof

It follows from (3.1) that

Ω U ε 2 log U ε 2 = Ω φ 2 u ε 2 log φ 2 + Ω φ 2 u ε 2 log u ε 2 = I + II .

Since s 2 log s 2 C for 0 s 1 , we have

I C Ω u ε 2 = O ( ε 2 ) . II = B ρ ( 0 ) u ε 2 log u ε 2 + Ω B ρ ( 0 ) φ 2 u ε 2 log u ε 2 = II 1 + II 2 .

Since s log s C 1 s 1 δ + C 2 s 1 + δ for all s > 0 , where 0 < C 1 < C 2 and 0 < δ < 1 3 such that ( N 2 ) ( 1 δ ) 2 , we have that

II 2 Ω B ρ ( 0 ) u ε 2 log u ε 2 C Ω B ρ ( 0 ) ( u ε 2 ( 1 δ ) + u ε 2 ( 1 + δ ) ) C Ω ( ε ( N 2 ) ( 1 δ ) + ε ( N 2 ) ( 1 + δ ) ) = O ( ε 2 ) ,

and

II 1 = B ( 0 , ρ ) u ε 2 log u ε 2 d x = C ε 2 B ρ ε ( 0 ) 1 ( 1 + y 2 ) N 2 log C ε ( N 2 ) 1 ( 1 + y 2 ) N 2 d y = C ε 2 log 1 ε B ρ ε ( 0 ) 1 ( 1 + y 2 ) N 2 + C ε 2 B ρ ε ( 0 ) 1 ( 1 + y 2 ) N 2 log C ( 1 + y 2 ) N 2 = C ε 2 log 1 ε R N 1 ( 1 + y 2 ) N 2 d y + O ( ε 2 ) + ε 2 O R N 1 ( 1 + y 2 ) N 2 1 4 d y = C ε 2 log 1 ε + O ( ε 2 ) ,

where we have used the fact that

B ρ ε c ( 0 ) 1 ( 1 + y 2 ) N 2 d y = O ( ε N 4 ) .

Thus,

Ω U ε 2 log U ε 2 = C 0 ε 2 log 1 ε + O ( ε 2 ) .

Lemma 3.3

If N 5 , λ R and μ > 0 , then c M < 1 N S N 2 .

Proof

Let g ( t ) = I ( t U ε ) . By Lemma 2.1, g ( 0 ) = 0 and lim t + g ( t ) = , we can find t ε ( 0 , + ) such that

sup t 0 I ( t U ε ) = sup t 0 g ( t ) = g ( t ε ) = I ( t ε U ε ) .

So

U ε 2 t ε 2 2 U ε 2 λ U ε 2 μ U ε 2 log U ε 2 μ log t ε 2 U ε 2 = 0 ,

which implies that, as ε 0 + ,

2 S N 2 U ε 2 λ U ε 2 μ U ε 2 log U ε 2 = t ε 2 2 U ε 2 + μ log t ε 2 U ε 2 t ε 2 2 1 2 S N 2 c log t ε 2 .

So there exists c 1 > 0 such that t ε < c 1 .

On the other hand, as ε 0 + ,

1 2 S N 2 U ε 2 λ U ε 2 μ U ε 2 log U ε 2 = t ε 2 2 U ε ε 2 + μ log t ε 2 U ε 2 2 S N 2 t ε 2 2 + C t ε 2 2 ,

which implies that there exists c 2 > 0 such that t ε > c 2 .

Therefore, combining with the definition of c M , we have that, as ε 0 + ,

c M sup t 0 I ( t u ε ) = t ε 2 2 U ε 2 t ε 2 2 U ε 2 λ 2 t ε 2 U ε 2 μ 2 t ε 2 U ε 2 ( log ( t ε 2 U ε 2 ) 1 ) t ε 2 2 t ε 2 2 S N 2 + O ( ε 2 ) + μ 2 t ε 2 ( 1 log t ε 2 ) U ε 2 μ 2 t ε 2 U ε 2 log U ε 2 1 N S N 2 c μ ε 2 log 1 ε + O ( ε 2 ) < 1 N S N 2 .

The case for N = 4 : Let φ ( x ) C 0 ( Ω ) be a radial function satisfying that φ ( x ) = 1 for 0 x ρ , 0 φ ( x ) 1 for ρ x 2 ρ , φ ( x ) = 0 for x Ω B 2 ρ ( 0 ) , where 0 < ρ 1 .

Set

U ε = φ ( x ) u ε ( x ) .

Lemma 3.4

If N = 4 , then, as ε 0 + ,

Ω U ε 2 log U ε 2 8 log 8 ( ε 2 + ρ 2 ) e ( ε 2 + 4 ρ 2 ) 2 ω 4 ε 2 log 1 ε + O ( ε 2 )

and

Ω U ε 2 log U ε 2 8 log 8 e ( ε 2 + 4 ρ 2 ) ( ε 2 + ρ 2 ) 2 ω 4 ε 2 log 1 ε + O ( ε 2 ) ,

where ω 4 denotes the area of the unit sphere surface in R 4 .

Proof

Following from the definition of U ε , we have, as ε 0 + ,

(3.4) Ω U ε 2 log ( U ε 2 ) = 8 Ω φ 2 ε ε 2 + x 2 2 log 8 φ 2 ε ε 2 + x 2 2 d x = 8 Ω φ 2 ε ε 2 + x 2 2 log ε ε 2 + x 2 2 d x + 8 log ( 8 ) Ω φ 2 ε ε 2 + x 2 2 d x + 8 Ω B ρ ( 0 ) φ 2 ε ε 2 + x 2 2 log φ 2 d x = I 1 + I 2 + O ( ε 2 ) .

By direct computation, we obtain

(3.5) I 2 = 8 log 8 B ρ ( 0 ) ε ε 2 + x 2 2 d x + O ( ε 2 ) = 8 log 8 ω 4 ε 2 0 ρ ε 1 ( 1 + r 2 ) 2 r 3 d r + O ( ε 2 ) = 4 log 8 ω 4 ε 2 log ( r 2 + 1 ) + 1 1 + r 2 0 ρ ε + O ( ε 2 ) = 4 log 8 ω 4 ε 2 log ρ 2 + ε 2 ε 2 + 1 1 + ρ 2 ε 2 1 + O ( ε 2 ) = 8 log 8 ω 4 ε 2 log 1 ε + O ( ε 2 ) ,

(3.6) I 1 = 8 Ω φ 2 ε ε 2 + x 2 2 log ε ε 2 + x 2 2 d r = 8 B 2 ρ ( 0 ) φ 2 ε ε 2 + x 2 2 log ε ε 2 + x 2 2 d x = 8 ε 2 B 2 ρ ε ( 0 ) φ 2 ( ε x ) 1 ( 1 + x 2 ) 2 log 1 ε 2 1 ( 1 + x 2 ) 2 d x = 16 ε 2 log 1 ε B 2 ρ ε ( 0 ) φ 2 ( ε x ) 1 ( 1 + x 2 ) 2 d x + 8 ε 2 B 2 ρ ε ( 0 ) φ 2 ( ε x ) 1 ( 1 + x 2 ) 2 log 1 ( 1 + x 2 ) 2 d x = I 11 + I 12 ,

where

(3.7) I 11 16 ω 4 ε 2 log 1 ε 0 ρ ε 1 ( 1 + r 2 ) 2 r 3 d r = 8 ω 4 ε 2 log 1 ε log 1 ε 2 + log ( ρ 2 + ε 2 ) + ε 2 ρ 2 + ε 2 1 = 16 ω 4 ε 2 log 1 ε 2 + 8 ω 4 log ρ 2 + ε 2 e ε 2 log 1 ε + O ε 4 log 1 ε ,

(3.8) I 11 16 ω 4 ε 2 log 1 ε 0 2 ρ ε 1 ( 1 + r 2 ) 2 r 3 d r = 8 ω 4 ε 2 log 1 ε log 1 ε 2 + log ( 4 ρ 2 + ε 2 ) + ε 2 4 ρ 2 + ε 2 1 = 16 ω 4 ε 2 log 1 ε 2 + 8 ω 4 log 4 ρ 2 + ε 2 e ε 2 log 1 ε + O ε 4 log 1 ε ,

(3.9) I 12 8 ε 2 B 2 ρ ε ( 0 ) 1 ( 1 + x 2 ) 2 log ( 1 + x 2 ) 2 d x = 16 ω 4 ε 2 0 2 ρ ε 1 ( 1 + r 2 ) 2 log ( 1 + r 2 ) r 3 d r = 8 ω 4 ε 2 0 2 ρ ε r 2 + 1 1 ( 1 + r 2 ) 2 log ( 1 + r 2 ) d ( 1 + r 2 ) = 8 ω 4 ε 2 0 2 ρ ε 1 1 + r 2 log ( 1 + r 2 ) d ( 1 + r 2 ) + 8 ω 4 ε 2 0 2 ρ ε 1 ( 1 + r 2 ) 2 log ( 1 + r 2 ) d ( 1 + r 2 ) 4 ω 4 ε 2 ( log ( 1 + r 2 ) ) 2 0 2 ρ ε = 4 ω 4 ε 2 log ( 1 + 4 ρ 2 ε 2 ) 2 = 4 ω 4 ε 2 log ( ε 2 + 4 ρ 2 ) + 2 log 1 ε 2 = 16 ω 4 ε 2 log 1 ε 2 16 ω 4 log ( ε 2 + 4 ρ 2 ) ε 2 log 1 ε + O ( ε 2 ) .

and

(3.10) I 12 8 ε 2 B ρ ε ( 0 ) 1 ( 1 + x 2 ) 2 log ( 1 + x 2 ) 2 d x = 16 ω 4 ε 2 0 ρ ε 1 ( 1 + r 2 ) 2 log ( 1 + r 2 ) r 3 d r = 8 ω 4 ε 2 0 ρ ε r 2 + 1 1 ( 1 + r 2 ) 2 log ( 1 + r 2 ) d ( 1 + r 2 ) = 8 ω 4 ε 2 0 ρ ε 1 1 + r 2 log ( 1 + r 2 ) d ( 1 + r 2 ) + 8 ω 4 ε 2 0 ρ ε 1 ( 1 + r 2 ) 2 log ( 1 + r 2 ) d ( 1 + r 2 ) 4 ω 4 ε 2 ( log ( 1 + r 2 ) ) 2 0 ρ ε + 8 ω 4 ε 2 0 ρ ε 1 ( 1 + r 2 ) d ( 1 + r 2 ) = 4 ω 4 ε 2 log ( ε 2 + ρ 2 ) + 2 log 1 ε 2 + 8 ω 4 ε 2 log ( ε 2 + ρ 2 ) + 2 log 1 ε = 16 ω 4 ε 2 log 1 ε 2 16 ω 4 log ε 2 + ρ 2 e ε 2 log 1 ε + O ( ε 2 ) .

So, by (3.4)–(3.12), we have that

Ω U ε 2 log U ε 2 8 log 8 ω 4 ε 2 ln 1 ε + 16 ω 4 ε 2 log 1 ε 2 + 8 ω 4 log ε 2 + ρ 2 e ε 2 log 1 ε 16 ω 4 ε 2 log 1 ε 2 16 ω 4 log ( ε 2 + 4 ρ 2 ) ε 2 log 1 ε + O ( ε 2 ) = 8 log 8 ( ε 2 + ρ 2 ) e ( ε 2 + 4 ρ 2 ) 2 ω 4 ε 2 log 1 ε + O ( ε 2 )

and

Ω U ε 2 log U ε 2 8 log 8 ω 4 ε 2 ln 1 ε + 16 ω 4 ε 2 log 1 ε 2 + 8 ω 4 log ε 2 + 4 ρ 2 e ε 2 log 1 ε 16 ω 4 ε 2 log 1 ε 2 16 ω 4 log ε 2 + ρ 2 e ε 2 log 1 ε + O ( ε 2 ) = 8 log 8 e ( ε 2 + 4 ρ 2 ) ( ε 2 + ρ 2 ) 2 ω 4 ε 2 log 1 ε + O ( ε 2 ) .

Lemma 3.5

Assume that N = 4 . If ( λ , μ ) B 0 C 0 and 32 e λ μ ρ max 2 < 1 , or λ R and μ > 0 , then c M < 1 N S N 2 .

Proof

Let g ( t ) = I ( t U ε ) . Similar to the case of N 5 , we can find a t ε ( 0 , + ) such that

sup t 0 I ( t U ε ) = I ( t ε U ε )

and

U ε 2 t ε 2 2 U ε 2 λ U ε 2 μ U ε 2 log U ε 2 μ log t ε 2 U ε 2 = 0 .

Similar to the case of N 5 again, we can see that there exists 0 < C 2 such that t ε < C 2 for any μ R { 0 } and there exists C 1 > 0 such that t ε > C 1 for μ > 0 .

Case 1: μ > 0 .

From Lemmas 3.1 and 3.2, we have that

μ log t ε 2 U ε 2 = O ( ε 2 ln ε ) ,

and

t ε 2 2 = U ε 2 λ U ε 2 μ U ε 2 log U ε 2 μ log t ε 2 U ε 2 U ε 2 = S N 2 + O ( ε 2 ( log 1 ε ) 2 ) S N 2 + O ( ε N ) 1 as ε 0 + ,

which implies that

μ log t ε 2 U ε 2 = o ( ε 2 ln ε ) .

According to (3.5), we obtain

Ω U ε 2 = 8 ω 4 ε 2 log 1 ε + O ( ε 2 ) .

Therefore, we have that

c M I ( t ε U ε ) = t ε 2 2 U ε 2 t ε 2 2 U ε 2 + μ λ 2 t ε 2 U ε 2 μ 2 t ε 2 U ε 2 log U ε 2 + o ( ε 2 ln ε ) t ε 2 2 t ε 2 2 S N 2 + O ( ε 2 ) + μ λ 2 t ε 2 U ε 2 μ 2 t ε 2 U ε 2 log U ε 2 + o ( ε 2 ln ε ) 1 N S N 2 t ε 2 2 [ μ U ε 2 log U ε 2 + ( λ μ ) U ε 2 ] + o ( ε 2 ln ε ) 1 N S N 2 t ε 2 2 8 μ log 8 ( ε 2 + ρ 2 ) e ( ε 2 + 4 ρ 2 ) 2 ω 4 ε 2 log 1 ε + ( λ μ ) 8 ω 4 ε 2 log 1 ε + o ( ε 2 ln ε ) 1 N S N 2 4 t ε 2 log 8 μ ( ε 2 + ρ 2 ) μ e 2 μ λ ( ε 2 + 4 ρ 2 ) 2 μ ω 4 ε 2 log 1 ε + o ( ε 2 ln ε ) 1 N S N 2 4 t ε 2 log 8 μ 2 5 μ e 2 μ λ ρ 2 μ ω 4 ε 2 log 1 ε + o ( ε 2 ln ε ) 1 N S N 2 4 C 1 2 C ω 4 ε 2 log 1 ε + o ( ε 2 ln ε ) < 1 N S N 2 ,

for small ε > 0 , where we choose ρ > 0 small enough such that 8 μ 2 5 μ e 2 μ λ ρ 2 μ > 1 .

Case 2: μ < 0 .

When ( λ , μ ) B 0 C 0 , we choose ρ = ρ max . Similar to Case 1, we can see that t ε 1 and t ε 2 log t ε 2 = o ( 1 ) . Therefore,

(3.11) μ 2 t ε 2 log t ε 2 U ε 2 = o ( ε 2 ln ε )

as ε 0 + . Thus, for small ε > 0 , we have that

c M I ( t ε U ε ) = t ε 2 2 U ε 2 t ε 2 2 U ε 2 + μ 2 t ε 2 ( 1 log t ε 2 ) U ε 2 μ 2 t ε 2 U ε 2 log U ε 2 + λ μ t ε 2 2 t ε 2 2 S N 2 + μ 2 t ε 2 U ε 2 μ 2 t ε 2 U ε 2 log U ε 2 λ 2 t ε 2 U ε 2 + o ( ε 2 ln ε ) 1 N S N 2 μ 2 t ε 2 λ μ 1 8 ω 4 ε 2 log 1 ε + 8 log 8 e ( ε 2 + 4 ρ 2 ) ( ε 2 + ρ 2 ) 2 ω 4 ε 2 log 1 ε + o ( ε 2 ln ε ) 1 N S N 2 4 μ ω 4 t ε 2 λ μ 1 + log 8 e ( ε 2 + 4 ρ 2 ) ( ε 2 + ρ 2 ) 2 ε 2 log 1 ε + o ( ε 2 ln ε ) 1 N S N 2 4 μ ω 4 t ε 2 log 8 e λ μ ( ε 2 + 4 ρ 2 ) ( ε 2 + ρ 2 ) 2 ε 2 log 1 ε + o ( ε 2 ln ε ) 1 N S N 2 4 μ ω 4 log 32 e λ μ ρ 2 ε 2 log 1 ε + o ( ε 2 ln ε ) < 1 N S N 2

since the fact that 32 e λ μ ρ max 2 < 1 .□

The case for N = 3 :

Let φ ( x ) C 0 1 ( Ω ) be a radial function satisfying that φ ( x ) = 1 for 0 x ρ , 0 φ ( x ) 1 for ρ x 2 ρ , φ ( x ) = 0 for x Ω B 2 ρ ( 0 ) , where 0 < ρ is any fixed constant such that B 2 ρ ( 0 ) Ω .

Set

U ε = φ ( x ) u ε ( x ) .

Lemma 3.6

For N = 3 , we have that

(3.12) Ω U ε 2 d x = S 3 2 + 3 ω 3 ρ 2 ρ φ ( r ) 2 d r ε + O ( ε 3 ) ,

(3.13) Ω U ε 2 d x = S 3 2 + O ( ε 3 ) ,

(3.14) Ω U ε 2 d x = 3 ω 3 0 2 ρ φ 2 d r ε + O ( ε 2 ) ,

and

(3.15) U ε 2 log U ε 2 d x = 3 ω 3 0 2 ρ φ 2 d r ε log ε + O ( ε ) ,

as ε 0 + , where ω 3 denotes the area of the unit sphere surface in R 3 .

Proof

Following from the definition of U ε , direct computations implies that

Ω U ε 2 d x = B 2 ρ φ 2 3 ε ε 2 + x 2 2 φ φ ( r ) 3 ε x ( ε 2 + x 2 ) 2 + 3 ε φ 2 ( r ) x 2 ( ε 2 + x 2 ) 3 d x = 3 ω 3 ε ρ 2 ρ φ ( r ) 2 r 2 ε 2 + r 2 d r 2 3 ω 3 ε ρ 2 ρ φ ( r ) r φ ( r ) r 2 ( ε 2 + r 2 ) 2 d r + 3 ω 3 ε 0 ρ r 4 ( ε 2 + r 2 ) 3 d r + 3 ω 3 ε ρ 2 ρ φ 2 ( r ) r 4 ( ε 2 + r 2 ) 3 d r = 3 ω 3 ε ρ 2 ρ φ ( r ) 2 d r + O ( ε 3 ) 2 3 ω 3 ε ρ 2 ρ φ ( r ) r φ ( r ) 1 ε 2 + r 2 d r + O ( ε 3 ) + 3 ω 3 ε 0 + r 4 ( ε 2 + r 2 ) 3 d r 3 ω 3 ε ρ + r 4 ( ε 2 + r 2 ) 3 d r + 3 ω 3 ε ρ 2 ρ φ 2 ( r ) r 4 ( ε 2 + r 2 ) 3 d r = 3 ω 3 ε ρ 2 ρ φ ( r ) 2 d r 2 3 ω 3 ε ρ 2 ρ φ ( r ) φ ( r ) 1 r d r + O ( ε 3 ) + R N u ε 2 3 ω 3 ε ρ + 1 ε 2 + r 2 d r + 3 ω 3 ε ρ 2 ρ φ 2 ( r ) 1 ε 2 + r 2 d r + O ( ε 3 ) = S 3 2 + O ( ε 3 ) + 3 ω 3 ε ρ 2 ρ φ ( r ) 2 d r ρ 2 ρ 2 φ ( r ) φ ( r ) 1 r d r ρ + 1 r 2 d r + ρ 2 ρ φ 2 ( r ) 1 r 2 d r = S 3 2 + 3 ω 3 ε ρ 2 ρ φ ( r ) 2 d r + O ( ε 3 )

Ω U ε 2 d x = B ρ ( 0 ) u ε 2 d x + O ( ε 3 ) = R 3 u ε 2 d x + O ( ε 3 ) = S 3 2 + O ( ε 3 ) ,

and

Ω U ε 2 d x = 3 B 2 ρ ( 0 ) φ 2 ε ε 2 + x 2 d x = 3 ω 3 ε 0 2 ρ φ 2 1 ε 2 + r 2 r 2 d r = 3 ω 3 ε 0 2 ρ φ 2 d r 3 ω 3 ε 3 0 2 ρ φ 2 1 ε 2 + r 2 d r = 3 ω 3 ε 0 2 ρ φ 2 d r + O ( ε 2 ) .

On the other hand, we have

(3.16) Ω U ε 2 log U ε 2 = 3 Ω φ 2 ε ε 2 + x 2 log 3 φ 2 ε ε 2 + x 2 d x = 3 B 2 ρ ( 0 ) φ 2 ε ε 2 + x 2 log 3 φ 2 ε ε 2 + x 2 d x = 3 ω 3 ε 0 2 ρ φ 2 ( r ) r 2 ε 2 + r 2 log 3 + log ε + log φ 2 + log 1 ε 2 + r 2 d r = 3 log 3 ω 3 ε 0 2 ρ φ 2 ( r ) r 2 ε 2 + r 2 d r + 3 ω 3 ε log ε 0 2 ρ φ 2 r 2 ε 2 + r 2 d r + 3 ω 3 ε 0 2 ρ φ 2 log φ 2 r 2 ε 2 + r 2 d r + 3 ω 3 ε 0 2 ρ φ 2 ( r ) r 2 ε 2 + r 2 log 1 ε 2 + r 2 d r = I 1 + I 2 + I 3 + I 4 .

By direct computation, we obtain that

(3.17) I 1 = O ( ε ) , I 3 = O ( ε ) ,

(3.18) I 2 = 3 ω 3 ε log ε 0 2 ρ φ 2 d r 3 ω 3 ε log ε 0 2 ρ φ 2 ε 2 ε 2 + r 2 d r = 3 ω 3 ε log ε 0 2 ρ φ 2 d r + O ( ε 2 log ε ) ,

and

(3.19) I 4 3 ω 3 ε 0 ρ log ( ε 2 + r 2 ) d r + O ( ε ) = 3 ω 3 ε r log ( ε 2 + r 2 ) 0 ρ + 3 ω 3 ε 0 ρ r 1 ε 2 + r 2 2 r d r + O ( ε ) = O ( ε ) .

It follows from (3.16)–(3.19) that

U ε 2 log U ε 2 d x = 3 ω 3 0 2 ρ φ 2 d r ε log ε + O ( ε ) .

Lemma 3.7

If N = 3 and ( λ , μ ) B 0 C 0 , then c M < 1 N S N 2 .

Proof

Assume that g ( t ) I ( t U ε ) . Since g ( 0 ) = 0 , lim t + g ( t ) = and, Lemma 2.1, we can obtain a t ε ( 0 , + ) such that

sup t 0 I ( t U ε ) = I ( t ε U ε ) .

Similar to the case of N = 4 , we can see that there exists 0 < C 1 < C 2 such that t ε ( C 1 , C 2 ) . Therefore, for ε small enough,

c M I ( t ε U ε ) = t ε 2 2 U ε 2 t ε 2 2 U ε 2 μ 2 t ε 2 U ε 2 log U ε 2 + O ( ε ) = t ε 2 2 t ε 2 2 S 3 2 μ 2 t ε 2 U ε 2 log U ε 2 + O ( ε ) 1 N S 3 2 3 μ 2 C 1 2 ω 3 0 2 ρ φ 2 d r ε log ( ε ) + O ( ε ) < 1 N S 3 2 .

4 The proof of main theorems

The proof of Theorem 1.2

Assume that N 4 and ( λ , μ ) A 0 . By Lemma 2.1 and the mountain-pass theorem, there exists a sequence { u n } H 0 1 ( Ω ) such that, as n ,

I ( u n ) c M , I ( u n ) 0 , in ( H 0 1 ( Ω ) ) 1 ,

which, together with Lemma 2.2, implies that { u n } is bounded in H 0 1 ( Ω ) . By Lemmas 2.4, 3.3, and 3.5, we can see that there exists u H 0 1 ( Ω ) such that u n u in H 0 1 ( Ω ) , which implies that

I ( u ) = c M and I ( u ) = 0 .

Thus,

0 = I ( u ) , u = u 2 ,

which implies that u 0 .

Therefore, u is a nonnegative nontrivial weak solution of ( 1.1 ) . By Moser’s iteration, it is standard to prove that u L ( Ω ) , then the Hölder estimate implies that u C 0 , γ ( Ω ) ( 0 < γ < 1 ) . Let β : [ 0 , + ) R be defined by

β ( s ) 3 μ 2 s log s 2 , s > 0 , 0 , s = 0 ,

then for a > 0 small enough, one can see that

Δ ( u ) = u 2 1 λ u μ u log u 2 β ( u ) in { x Ω : 0 < u ( x ) < a } .

We may also assume that a < 1 e , then β ( s ) = 3 μ 2 ( log s 2 2 ) > μ ( log a 1 ) > 0 for s ( 0 , a ) . So we have that β ( 0 ) = 0 and β ( s ) is nondecreasing in ( 0 , a ) . Furthermore,

0 a 2 ( β ( s ) s ) 1 2 d s = 2 3 μ ( 2 log s ) 1 2 0 a 2 = + .

Hence, by [17, Theorem 1], we have that u ( x ) > 0 in Ω . In particular, for any compact K Ω , there exists c = c ( K ) > 0 such that u ( x ) c , x K . Take K K 1 Ω and put f ( x ) u ( x ) 2 1 λ u ( x ) μ u ( x ) log u ( x ) 2 , then Δ u = f ( x ) in K 1 and f is of C 0 , γ in K 1 . So by the standard Schauder estimate, we see that u C 2 , γ ( K ) . By the arbitrariness of K , we obtain that u C 2 ( Ω ) and u > 0 in Ω . The proof is completed.

The proof of Theorem 1.3

By Lemma 2.1 and the mountain-pass theorem, there exists a sequence { u n } H 0 1 ( Ω ) such that, as n ,

I ( u n ) c M , I ( u n ) 0 , in ( H 0 1 ( Ω ) ) 1 ,

combining with Lemmas 2.5, 3.5, and 3.7, problem (1.1) has a nonnegative nontrivial weak solution u . Applying a similar argument as the proof of Theorem 1.2, we obtain that u C 2 ( Ω ) and u ( x ) > 0 in Ω .

The proof of the Theorem 1.4

Assume that problem (1.1) has a positive solution u 0 , and let φ 1 ( x ) > 0 be the first eigenfunction corresponding to λ 1 ( Ω ) . Then

Ω ( u 0 2 1 + λ u 0 + μ u 0 log u 0 2 ) φ 1 ( x ) = Ω Δ u 0 φ 1 ( x ) = Ω Δ φ 1 ( x ) u 0 = Ω λ 1 ( Ω ) φ 1 ( x ) u 0 ,

which implies that

(4.1) Ω ( u 0 2 2 + λ λ 1 ( Ω ) + μ log u 0 2 ) u 0 φ 1 ( x ) = 0 .

Define

f ( s ) s 2 2 + μ log s 2 + λ λ 1 ( Ω ) , s > 0 ,

then

f ( s ) = ( 2 2 ) s 2 3 + 2 μ 1 s .

By a direct computation, f ( s ) = 0 has a unique root, s 0 = ( N 2 ) μ 2 N 2 4 . Furthermore, f ( s ) < 0 in 0 , ( N 2 ) μ 2 N 2 4 and f ( s ) > 0 in ( N 2 ) μ 2 N 2 4 , + . Hence,

(4.2) f ( s ) f ( N 2 ) μ 2 N 2 4 = ( N 2 ) μ 2 + ( N 2 ) μ 2 log ( N 2 ) μ 2 + λ λ 1 ( Ω ) 0 .

Since u 0 H 0 1 ( Ω ) and u 0 , φ 1 > 0 , we have Ω f ( u 0 ( x ) ) u 0 φ 1 ( x ) > 0 . Otherwise, f ( u 0 ( x ) ) = 0 a.e in Ω . That is, u 0 ( x ) = ( N 2 ) μ 2 N 2 4 a.e in Ω , which contradicts to u 0 H 0 1 ( Ω ) . By (4.1) and (4.2), we have that

0 = Ω ( u 0 2 2 + λ λ 1 ( Ω ) + μ log u 0 2 ) u 0 φ 1 ( x ) = Ω f ( u 0 ( x ) ) u 0 φ 1 ( x ) > 0 ,

a contradiction. Hence, problem (1.1) has no positive solutions.

Acknowledgement

This work was supported by the Natural Science Foundation of China (Nos. 12271196, 12061012, 11931012, and 12271184), the special foundation for Guangxi Ba Gui Scholars and Guangdong Basic and Applied Basic Research Foundation (2021A1515010034), and Guangzhou Basic and Applied Basic Research Foundation (202102020225).

  1. Conflict of interest: The authors state no conflict of interest.

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Received: 2022-06-30
Revised: 2022-10-08
Accepted: 2022-12-25
Published Online: 2023-02-21

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  13. Non-degeneracy of multi-peak solutions for the Schrödinger-Poisson problem
  14. Gagliardo-Nirenberg-type inequalities using fractional Sobolev spaces and Besov spaces
  15. Ground states of Schrödinger systems with the Chern-Simons gauge fields
  16. Quasilinear problems with nonlinear boundary conditions in higher-dimensional thin domains with corrugated boundaries
  17. A system of equations involving the fractional p-Laplacian and doubly critical nonlinearities
  18. A modified Picone-type identity and the uniqueness of positive symmetric solutions for a prescribed mean curvature problem
  19. On a version of hybrid existence result for a system of nonlinear equations
  20. Special Issue: Geometric PDEs and applications
  21. Preface for the special issue on “Geometric Partial Differential Equations and Applications”
  22. Convex hypersurfaces with prescribed Musielak-Orlicz-Gauss image measure
  23. Total mean curvatures of Riemannian hypersurfaces
  24. On degenerate case of prescribed curvature measure problems
  25. A curvature flow to the Lp Minkowski-type problem of q-capacity
  26. Aleksandrov reflection for extrinsic geometric flows of Euclidean hypersurfaces
  27. A note on second derivative estimates for Monge-Ampère-type equations
  28. The Lp chord Minkowski problem
  29. Widths of balls and free boundary minimal submanifolds
  30. Smooth approximation of twisted Kähler-Einstein metrics
  31. The exterior Dirichlet problem for the homogeneous complex k-Hessian equation
  32. A Carleman inequality on product manifolds and applications to rigidity problems
  33. Asymptotic behavior of solutions to the Monge-Ampère equations with slow convergence rate at infinity
  34. Pinched hypersurfaces are compact
  35. The spinorial energy for asymptotically Euclidean Ricci flow
  36. Geometry of CMC surfaces of finite index
  37. Capillary Schwarz symmetrization in the half-space
  38. Regularity of optimal mapping between hypercubes
  39. Special Issue: In honor of David Jerison
  40. Preface for the special issue in honor of David Jerison
  41. Homogenization of oblique boundary value problems
  42. A proof of a trace formula by Richard Melrose
  43. Compactness estimates for minimizers of the Alt-Phillips functional of negative exponents
  44. Regularity properties of monotone measure-preserving maps
  45. Examples of non-Dini domains with large singular sets
  46. Sharp inequalities for coherent states and their optimizers
  47. Gradient estimates and the fundamental solution for higher-order elliptic systems with lower-order terms
  48. Propagation of symmetries for Ricci shrinkers
  49. Linear extension operators for Sobolev spaces on radially symmetric binary trees
  50. The Neumann problem on the domain in 𝕊3 bounded by the Clifford torus
  51. On an effective equation of the reduced Hartree-Fock theory
  52. Polynomial sequences in discrete nilpotent groups of step 2
  53. Integral inequalities with an extended Poisson kernel and the existence of the extremals
  54. On singular solutions of Lane-Emden equation on the Heisenberg group
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