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Multiple solutions to multi-critical Schrödinger equations

  • Ziyi Xu and Jianfu Yang EMAIL logo
Published/Copyright: July 11, 2022

Abstract

In this article, we investigate the existence of multiple positive solutions to the following multi-critical Schrödinger equation:

(0.1) Δ u + λ V ( x ) u = μ u p 2 u + i = 1 k ( x ( N α i ) u 2 i ) u 2 i 2 u in R N , u H 1 ( R N ) ,

where λ , μ R + , N 4 , and 2 i = N + α i N 2 with N 4 < α i < N , i = 1 , 2 , , k are critical exponents and 2 < p < 2 min = min { 2 i : i = 1 , 2 , , k } . Suppose that Ω = int V 1 ( 0 ) R N is a bounded domain, we show that for λ large, problem (0.1) possesses at least cat Ω ( Ω ) positive solutions.

MSC 2010: 35J10; 35J20; 35J61

1 Introduction

In this article, we investigate the existence of multiple positive solutions to the following multi-critical Schrödinger equation:

(1.1) Δ u + λ V ( x ) u = μ u p 2 u + i = 1 k ( x ( N α i ) u 2 i ) u 2 i 2 u in R N , u H 1 ( R N ) ,

where λ , μ R + , N 4 , and 2 i = N + α i N 2 with N 4 < α i < N , i = 1 , 2 , , k are critical Hardy-Littlewood-Sobolev exponents and 2 < p < 2 min = min { 2 i : i = 1 , 2 , , k } .

In the case k = 1 , problem (1.1) is the so-called Choquard equation. It has been extensively studied for subcritical and critical cases in [11,13,14,18,19,20] and references therein. Particularly, multiple solutions for the nonlinear Choquard problem

(1.2) Δ u + λ u = ( x ( N α ) u q ) u q 2 u in Ω , u H 0 1 ( Ω )

in the bounded Ω are obtained in [13] by the Lusternik-Schnirelman theory. It was proved that if q is closed to the critical exponent 2 α = N + α N 2 , problem (1.2) possesses at least cat Ω ( Ω ) positive solutions; similar results were obtained in [14] for the critical case q = 2 α . This sort of result extends early works for elliptic problem with local nonlinearities by [1,3,5,6,10,15] etc.

A counterpart for the Schrödinger equation

(1.3) ε 2 Δ u + V ( x ) u = u p 2 u in R N

was considered in [7] and [8]. A graph of the potential function V ( x ) affects the existence of number of solutions. Precisely, assume, among other things, that Ω V 1 ( 0 ) = { x R N : V ( x ) = 0 } is bounded, it is proved in [8] for the subcritical case and in [7] for the critical case that problem (1.3) possesses at least cat Ω ( Ω ) positive solutions if ε > 0 is small.

Analogous problem for the Schrödinger equation

(1.4) Δ u + ( λ V ( x ) + 1 ) u = u p 1 u H 1 ( R N ) ,

was investigated in [4] and [9] for subcritical and critical cases, respectively. Under certain conditions on V ( x ) , it is obtained in [4] and [9]  cat Ω ( Ω ) positive solutions for problem (1.4) if λ > 0 large. Various extensions of the results in [4] and [9] can be found in [1,2,18].

In this article, we focus on problem (1.1) in the case k > 2 , multi-critical terms then involved in. In the bounded domain Ω , the existence of positive solution of the multi-critical Sobolev-Hardy problem

(1.5) Δ u = λ x s u p 1 + i = 1 l λ i x s i u 2 ( s i ) 1 + u 2 1 in Ω , u > 0 in Ω , u = 0 on Ω

was studied in [12], while the number of positive solutions of the Choquard-type problem

(1.6) Δ u = μ u p 2 u + i = 1 k ( x ( N α i ) u 2 i ) u 2 i 2 u in Ω , u H 0 1 ( Ω )

is described in [17] by the Lusternik-Schnirelman category cat Ω ( Ω ) of the domain Ω . Such a result seems to be difficult to establish for problem (1.5).

Inspired by [4] and [9], we will show that problem (1.1) has at least cat Ω ( Ω ) positive solutions based on the result in [17].

Our hypotheses are as follows.

( V 1 ) V ( x ) C ( R N , R ) , V ( x ) 0 , and Ω int V 1 ( 0 ) is a nonempty bounded set with smooth boundary, and Ω ¯ = V 1 ( 0 ) .

( V 2 ) There exists M 0 > 0 such that

mes { x R N : V ( x ) M 0 } < .

Let μ 1 be the first eigenvalue of Δ on Ω with zero Dirichlet condition. We present our first result as follows.

Theorem 1.1

Suppose ( V 1 ) and ( V 2 ) hold, then for every 0 < μ < μ 1 , there exists λ ( μ ) > 0 such that problem (1.1) has at least one ground state positive solution for each λ λ ( μ ) .

Positive solutions of problem (1.1) are found as critical points of the associated functional

(1.7) I λ , μ ( u ) = 1 2 R N ( u 2 + λ V ( x ) u 2 ) d x μ p R N ( u + ) p d x i = 1 k 1 2 2 i R N R N u + ( x ) 2 i u + ( y ) 2 i x y N α i d x d y ,

where u + max { u , 0 } , in the Hilbert space

(1.8) E u H 1 ( R N ) : R N V ( x ) u 2 d x <

endowed with the norm

(1.9) u 2 = u H 1 2 + R N V ( x ) u 2 d x .

It is standard to show that a nontrivial critical point of I λ , μ is nonnegative the maximum principle then implies that it is positive.

Theorem 1.1 is proved by the mountain pass theorem. A crucial ingredient is to verify the ( PS ) c condition for the functional I λ , μ ( u ) . By a ( PS ) c condition we mean a sequence { u n } in E satisfying I λ , μ ( u n ) c , I λ , μ ( u n ) 0 as n contains a convergent subsequence.

We will show that I λ , μ ( u ) satisfies ( PS ) c condition if c < m ( R N ) with

(1.10) m ( R N ) inf { J ( u ) : u N R N } ,

where

(1.11) J ( u ) = 1 2 R N u 2 d x i = 1 k 1 2 2 i R N R N u + ( x ) 2 i u + ( y ) 2 i x y N α i d x d y

and

(1.12) N R N { u D 1 , 2 ( R N ) \ { 0 } : J ( u ) , u = 0 } .

It is shown in [17] that m ( R N ) is uniquely achieved up to translations and dilations by the function

U ( x ) = C ε ε 2 + x 2 N 2 2 ,

which is a solution for the limit problem

(1.13) Δ u = i = 1 k ( x ( N α i ) u 2 i ) u 2 i 2 u in R N .

Next, we consider the limit behavior of solutions (1.1). Let λ n and { u n } be the corresponding solutions of (1.1). Then, we will show that { u n } concentrates at a solution of the limit problem

(1.14) Δ u = μ u p 2 u + i = 1 k ( x ( N α i ) u 2 i ) u 2 i 2 u in Ω , u H 0 1 ( Ω ) .

Theorem 1.2

Suppose (V1) and (V2) hold. Let { u n } be a sequence of solutions of (1.1) such that 0 < μ < μ 1 , λ n , and I λ n , μ ( u n ) c < m ( R N ) as n . Then, { u n } concentrates at a solution u of (1.14), that is, u n u in E as n .

Finally, we derive from the Lusternik-Schnirelman theory a multiplicity result for problem (1.1).

Theorem 1.3

Suppose (V1) and (V2) hold and N 4 . Then there exists 0 < μ < μ 1 , for each 0 < μ μ , there is Λ ( μ ) > 0 such that problem (1.1) has at least cat Ω ( Ω ) positive solutions whenever λ Λ ( μ ) .

The article is organized as follows. In Section 2, we present preliminary results. Then we prove Theorems 1.1 and 1.2 in Section 3. The multiple result in Theorem 1.3 is established in Section 4.

2 Preliminaries

In the sequel, we denote by u q the norm of u in L q ( R N ) and

u λ 2 = u H 1 2 + λ R N V ( x ) u 2 d x

for λ > 0 , which is equivalent to that defined in (1.9).

Now, by the Hardy-Littlewood-Sobolev inequality in [16], we see that the functionals I λ , μ ( u ) and J ( u ) are well-defined and differentiable in E and D 1 , 2 ( R N ) , respectively.

Lemma 2.1

Let λ n 1 and u n E be such that u n λ n 2 K for λ n . Then there is a u H 0 1 ( Ω ) such that, up to a subsequence, u n u weakly in E and u n u in L 2 ( R N ) as n . Moreover, if u = 0 , we have u n 0 in L p ( R N ) for 2 p < 2 N N 2 as n .

Proof

Since for λ n 1 , u n 2 u n λ n 2 K , we may assume u n u weakly in E and u n u in L loc 2 ( R N ) . Set C m x : V ( x ) 1 m for m N . Then

R N \ Ω ¯ = m = 1 + C m ,

and for all m N ,

C m u n 2 d x m C m V u n 2 d x = m λ n C m λ n V u n 2 d x m λ n u n λ n 2 m λ n K .

Hence, u n 0 in L 2 ( C m ) as n . By a diagonal process, we may assume that u n 0 in L 2 ( R N \ Ω ¯ ) as n .

In order to show that u n u in L 2 ( R N ) , we set for R > 0 that

A ( R ) { x R N : x > R , V ( x ) M 0 }

and

B ( R ) { x R N : x > R , V ( x ) < M 0 } ,

where M 0 is given in ( V 2 ) . Since A ( R ) C m if m large, we have

(2.1) A ( R ) u n 2 d x C m u n 2 d x 0 , as n .

By the Hölder inequality, for 1 < p < N N 2 , 1 p + 1 p = 1 there holds

B ( R ) u n u 2 d x R N u n u 2 p d x 1 p B ( R ) 1 p C u n u λ n 2 B ( R ) 1 p = o R ( 1 ) ,

where o R ( 1 ) 0 as R . Then, for ε > 0 , there exists N > 0 such that for n > N ,

R N u n u 2 d x B R u n u 2 d x + A ( R ) u n 2 d x + A ( R ) u 2 d x < ε .

It yields that u n u in L 2 ( R N ) .

Now, we turn to the last assertion. If u n 0 , for 2 < p < 2 N N 2 and any subset Ω ˜ R N , the interpolation inequality yields

(2.2) Ω ˜ u n p d x C Ω ˜ u n 2 d x θ p 2 Ω ˜ u n 2 d x ( 1 θ ) p 2 C u n λ n θ p A ( R ) u n 2 d x + B ( R ) u n 2 d x ( 1 θ ) p 2 < ε ,

for n large, where θ = N ( p 2 ) 2 p . Choosing Ω ˜ to be A ( R ) and B ( R ) , respectively, we obtain u n 0 in L p ( R N ) as n .□

Let L λ Δ + λ V be the self-adjoint operator acting on L 2 ( R ) with the domain E for all λ 0 . We denote by ( , ) the L 2 -inner product and write

( L λ u , v ) = R N ( u v + λ V ( x ) u v ) d x

for u , v E . It implies that

(2.3) ( L λ u , u ) = R N ( u 2 + λ V ( x ) u 2 ) d x u λ 2 .

Denote by a λ inf σ ( L λ ) the infimum of the spectrum of L λ . It follows that

0 a λ = inf { ( L λ u , u ) : u E , u 2 = 1 } ,

which is nondecreasing about the parameter λ . It is proved in [9] the following result.

Lemma 2.2

For every 0 < μ < μ 1 , there exists λ ( μ ) > 0 such that

a λ μ + μ 1 2

if λ λ ( μ ) . Moreover, for all u E and λ λ ( μ ) there exists a constant C > 0 such that

C u λ 2 ( L λ u , u ) ,

where C ( μ + μ 1 ) / ( 2 + μ + μ 1 ) .

3 Positive solutions and concentration

In this section, we will prove the existence of positive solution for problem (1.1) and investigate the limit behavior of the solution as the parameter λ .

Lemma 3.1

If μ ( 0 , μ 1 ) and λ λ ( μ ) , then every ( PS ) c sequence { u n } E for I λ , μ is bounded in E .

Proof

Since 2 < p < 2 min , by Lemma 2.2 we have

c + 1 + o ( 1 ) u n λ I λ , μ ( u n ) 1 p I λ , μ ( u n ) , u n 1 2 1 p ( L λ u n , u n ) + i = 1 k 1 p 1 2 2 i R N R N u n + ( x ) 2 i u n + ( y ) 2 i x y N α i d x d y 1 2 1 p C u n λ 2 .

This implies that { u n } is bounded in E .□

Now we verify that I λ , μ ( u ) has the mountain pass geometry.

Lemma 3.2

For all μ ( 0 , μ 1 ) and λ λ ( μ ) , the functional I λ , μ satisfies that

  1. I λ , μ ( 0 ) = 0 and there exist ρ , δ > 0 , such that inf u λ = ρ I λ , μ ( u ) δ > 0 ;

  2. There exists e E such that e λ > ρ and I λ , μ ( e ) 0 .

Proof

By the Hardy-Littlewood-Sobolev inequality, we have

I λ , μ ( u ) C u λ 2 u λ p i = 1 k u λ 2 2 i .

Then, ( i ) follows by choosing u λ = ρ sufficiently small. ( i i ) is valid since for any fixed positive u E \ { 0 } , I λ , μ ( t u ) as t + .□

By Lemma 3.2 and the mountain pass theorem, we know that there exists a ( PS ) c sequence of the functional I λ , μ ( u ) at the mountain pass level

c λ , μ = inf γ Γ λ max t [ 0 , 1 ] I λ , μ ( γ ( t ) ) ,

where

Γ λ = { γ C ( [ 0 , 1 ] , H 1 ( R N ) ) : γ ( 0 ) = 0 , I λ , μ ( γ ( 1 ) ) 0 , γ ( 1 ) 0 } .

Define

m λ , μ inf u N λ , μ I λ , μ ( u )

and

c λ , μ s = inf u H 1 ( R N ) \ { 0 } sup t > 0 I λ , μ ( t u ) .

We may show as [21] that

(3.1) m λ , μ = c λ , μ = c λ , μ s .

In the same spirit, we may define for the functional

(3.2) I μ , Ω ( u ) = 1 2 Ω u 2 d x 1 p Ω μ ( u + ) p d x i = 1 k 1 2 2 i Ω Ω u + ( x ) 2 i u + ( y ) 2 i x y N α i d x d y

associated with problem (1.14) the mountain pass level

c μ , Ω = inf γ Γ 0 max t [ 0 , 1 ] I μ , Ω ( γ ( t ) ) ,

where

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

and

c μ , Ω s = inf u H 0 1 ( Ω ) \ { 0 } sup t > 0 I μ , Ω ( t u )

as well as

(3.3) m μ , Ω inf { I μ , Ω ( u ) : u N μ , Ω } ,

where

(3.4) N μ , Ω { u H 0 1 ( Ω ) \ { 0 } : I μ , Ω ( u ) , u = 0 }

is the corresponding Nehari manifold. Similarly,

m μ , Ω = c μ , Ω = c μ , Ω s .

Lemma 3.3

There exist τ = τ ( μ ) > 0 and a constant σ ( μ ) > 0 such that the mountain pass level c λ , μ of I λ , μ satisfies that

σ ( μ ) c λ , μ < m ( R N ) τ

for all λ > 0 .

Proof

It is proved in [17] that c μ , Ω < m ( R N ) and that there exists a critical point u H 0 1 ( Ω ) of I μ , Ω such that I μ , Ω ( u ) = c μ , Ω . Extend the function u to R N such that u = 0 outside Ω , then u N λ , μ , which implies

c λ , μ I λ , μ ( u ) = I μ , Ω ( u ) = c μ , Ω .

So there exists τ = τ ( μ ) > 0 such that

c λ , μ c μ , Ω < m ( R N ) τ

for all λ > 0 .

Next, for each u N λ , μ ,

0 = I λ , μ ( u ) , u = ( L λ u , u ) μ u + p p i = 1 k R N R N u + ( x ) 2 i u + ( y ) 2 i x y N α i d x d y u λ 2 C u λ p C i = 1 k u λ 2 2 i ,

where C depending only on μ , yields that there exists σ > 0 independent of λ such that u λ σ , and then

C σ C u λ 2 ( L λ u , u ) = μ u + p p + i = 1 k R N R N u + ( x ) 2 i u + ( y ) 2 i x y N α i d x d y .

Hence,

c λ , μ = m λ , μ = inf u N λ , μ I λ , μ ( u ) inf u N λ , μ 1 2 1 p μ u + p p + i = 1 k R N R N u + ( x ) 2 i u + ( y ) 2 i x y N α i d x d y 1 2 1 p C σ ,

the conclusion follows.□

Now we show that I λ , μ satisfies the ( PS ) c condition for certain c .

Proposition 3.4

There exist μ ( 0 , μ 1 ) and λ ( μ ) > 0 such that I λ , μ satisfies the ( PS ) c condition for

c < m ( R N ) τ ,

whenever λ λ ( μ ) .

Proof

It is known from Lemma 3.1 that the ( PS ) c sequence { u n } of I λ , μ is bounded in E . We may assume that

u n u 0 in E , u n u 0 in L loc 2 ( R N ) , u n ( x ) u 0 ( x ) a.e. on R N

as n . By the Hardy-Littlewood-Sobolev inequality and Hölder inequality,

(3.5) ( x ( N α i ) u n 2 i ) u n 2 i 1 ( x ( N α i ) u 0 2 i ) u 0 2 i 1 in L 2 N N + α i ( R N ) ,

for i = 1 , 2 , , k . Thus, for every φ H 1 ( R N ) ,

0 = lim n I λ , μ ( u n ) , φ = I λ , μ ( u 0 ) , φ ,

that is, u 0 is a weak solution of problem (1.1).

Let w n = u n u 0 . The Brézis-Lieb-type lemma in [11]

(3.6) lim n R N R N u n + ( x ) 2 i u n + ( y ) 2 i x y N α i d x d y R N R N w n + ( x ) 2 i w n + ( y ) 2 i x y N α i d x d y = R N R N u 0 + ( x ) 2 i u 0 + ( y ) 2 i x y N α i d x d y

enables us to deduce

o ( 1 ) = I λ , μ ( u n ) , u n = I λ , μ ( u 0 ) , u 0 + I λ , μ ( w n ) , w n + o ( 1 ) ,

namely,

(3.7) I λ , μ ( w n ) , w n = o ( 1 ) .

Similarly,

(3.8) c + o ( 1 ) = I λ , μ ( u n ) = I λ , μ ( u 0 ) + I λ , μ ( w n ) I λ , μ ( w n ) ,

since I λ , μ ( u 0 ) 0 .

In order to prove that u n converges strongly to u 0 in E , we claim that for each nontrivial nonnegative u E the function f u ( t ) = I λ , μ ( t u ) has a unique critical point t u such that f u ( t u ) = max t 0 f ( t ) and t u u N λ , μ . Indeed,

f u ( t ) = t ( L λ u , u ) t p 1 μ u + p p i = 1 k t 2 2 i 1 R N R N u + ( x ) 2 i u + ( y ) 2 i x y N α i d x d y = t [ ( L λ u , u ) m u ( t ) ] ,

where the function

m u ( t ) t p 2 μ u + p p + i = 1 k t 2 2 i 2 R N R N u + ( x ) 2 i u + ( y ) 2 i x y N α i d x d y > 0

satisfies lim t + m u ( t ) = + and m u ( 0 ) = 0 . The claim follows readily.

Now we show u n converge strongly to u 0 in E . Suppose, by contradiction, that u n does not converge strongly to u 0 , then w n 0 in E as n . So there exists a unique t n > 0 such that t n w n N λ , μ and f w n ( t ) achieves the maximum at t = t n . By (3.7) and

(3.9) R N ( w n 2 + λ V ( x ) w n 2 ) d x = t n p 2 μ w n + p p + i = 1 k t n 2 2 i 2 R N R N w n + ( x ) 2 i w n + ( y ) 2 i x y N α i d x d y ,

we find

( t n p 2 1 ) μ w n + p p + i = 1 k ( t n 2 2 i 2 1 ) R N R N w n + ( x ) 2 i w n + ( y ) 2 i x y N α i d x d y = o ( 1 ) ,

which implies t n 1 since w n 0 in E . Therefore,

(3.10) I λ , μ ( w n ) = I λ , μ ( t n w n ) + o ( 1 ) .

For any Ω ˜ R N , we define

(3.11) J Ω ˜ ( u ) = 1 2 Ω ˜ u 2 d x i = 1 k 1 2 2 i Ω ˜ Ω ˜ u + ( x ) 2 i u + ( y ) 2 i x y N α i d x d y

and denote J ( u ) = J R N ( u ) . Set v n = t n w n . We may show similarly that the function g u ( s ) = J ( s u ) for each u E { 0 } has unique critical point s u > 0 which is the maximum point of g u ( s ) . Particularly, there exists a unique s n R + such that g v n ( s n ) = 0 and s n v n N R N . Since v n N λ , μ , we obtain

I λ , μ ( v n ) = sup s 0 I λ , μ ( s v n ) .

By Lemma 2.1, (3.8), and (3.10),

m ( R N ) τ I λ , μ ( u n ) I λ , μ ( w n ) = I λ , μ ( t n w n ) + o ( 1 ) I λ , μ ( s n v n ) + o n ( 1 ) J ( s n v n ) + o n ( 1 ) + o λ ( 1 ) m ( R N ) + o n ( 1 ) + o λ ( 1 ) .

This yields a contradiction by choosing λ large enough. Hence, I λ , μ satisfies the ( PS ) c condition.□

Now, we are ready to prove Theorem 1.1.

Proof of Theorem 1.1

Let { u n } be a minimizing sequence of I λ , μ restricting on N λ , μ . By the Ekeland variational principle, we may assume that { u n } is a ( PS ) c λ , μ sequence, that is,

I λ , μ ( u n ) c λ , μ and I λ , μ ( u n ) 0 .

It follows from Proposition 3.4 and Lemma 3.3 that { u n } has a subsequence converging to a least energy solution u λ of (1.1).

By Theorem 1.1, for all λ n λ ( μ ) , the problem

(3.12) Δ u + λ n V ( x ) u = μ u p 2 u + i = 1 k ( x ( N α i ) u 2 i ) u 2 i 2 u in R N

has at least one positive solution u λ n .□

Proof of Theorem 1.2

Let u λ n be the positive solution to (3.12) for every λ n λ ( μ ) . Then we have I λ n , μ ( u λ n ) = c λ n , μ and I λ n , μ ( u λ n ) = 0 . By Lemma 3.3, c λ n , μ is uniformly bounded. We may show as Lemma 3.1 that there is C > 0 such that u λ n λ n 2 C . By Lemma 2.1, there exists u H 0 1 ( Ω ) such that, up to a subsequence, u λ n u weakly in E , u λ n u in L 2 ( R N ) and u λ n u in L p ( R N ) .

For each φ H 0 1 ( Ω ) , we extend φ to R N by setting φ = 0 outside Ω . Then we have

(3.13) 0 = lim n I λ n , μ ( u λ n ) , φ = I μ , Ω ( u ) , φ .

This means that u is a weak solution of problem (1.14).

Now we show u λ n converges to u in E . Suppose on the contrary that v n u λ n u 0 in E as n . By (3.6) and (3.13),

lim n I λ n , μ ( v n ) , v n = lim n I λ n , μ ( u λ n ) , u λ n I μ , Ω ( u ) , u = 0 .

Arguing as the proof of Proposition 3.4, there is unique t n > 0 such that t n v n N λ n , μ and t n 1 as n . Hence,

I λ n , μ ( v n ) + o ( 1 ) = I λ n , μ ( t n v n ) = sup t 0 I λ n , μ ( t v n ) .

On the other hand, there exists s n > 0 such that s n t n v n N R N . We derive from Proposition 3.4 that

m ( R N ) τ > c λ n , μ + o ( 1 ) = I λ n , μ ( u λ n ) I λ n , μ ( v n ) = I λ n , μ ( t n v n ) + o ( 1 ) I λ n , μ ( s n t n v n ) + o ( 1 ) J ( s n t n v n ) + o ( 1 ) m ( R N ) + o ( 1 ) ,

a contradiction. The assertion follows.□

Corollary 3.5

If μ ( 0 , μ 1 ) , then lim λ n c λ n , μ = c μ , Ω .

Proof

Obviously, c λ n , μ c μ , Ω . On the other hand, by Theorems 1.1 and 1.2, there exists { u n } such that c λ n , μ = I λ n , μ ( u n ) and lim λ n I λ n , μ ( u n ) = I μ , Ω ( u ) m μ , Ω = c μ , Ω .□

4 Multiple solutions

In this section, we show that problem (1.1) has multiple solutions in connection with the domain V 1 ( 0 ) = Ω by the Lusternik-Schnirelman theory. Such a problem is actually related to the limit problem (1.14). We know from [17] that (1.14) possesses at least cat Ω ( Ω ) positive solutions, and eventually so does problem (1.1) as we will show.

Let us fix r > 0 small enough so that

Ω 2 r + { x R N : dist ( x , Ω ) < 2 r }

and

Ω r { x Ω : dist ( x , Ω ) > r }

are homotopically equivalent to Ω . Denote B r ( 0 ) { x R N : x < r } Ω and define c μ , r c μ , B r ( 0 ) . Therefore, we may verify as Lemma 3.3 that for μ ( 0 , μ 1 ) ,

c μ , Ω < c μ , r < m ( R N ) τ .

Let η C c ( R N ) be such that η ( x ) = x for all x Ω . We introduce the barycenter of a function 0 u H 1 ( R N ) as

β ( u ) R N η ( x ) u 2 d x R N u 2 d x .

Denote I μ , Ω c = { u N μ , Ω : I μ , Ω c } the level set. It is proved in [17] the following result.

Lemma 4.1

There exists μ ( 0 , μ 1 ) such that if 0 < μ < μ , then

β ( u ) Ω r + for every u I μ , Ω c μ , r .

For any domain Ω ˜ R N , we define

m ( Ω ˜ ) inf u N Ω ˜ J Ω ˜ ( u ) ,

where

N Ω ˜ { u D 0 1 , 2 ( Ω ˜ ) \ { 0 } : J Ω ˜ ( u ) , u = 0 } .

Lemma 4.2

There holds

m ( Ω ˜ ) = m ( R N ) ,

and m ( Ω ˜ ) is never achieved unless Ω ˜ = R N .

Proof

Obviously, m ( Ω ˜ ) m ( R N ) . To prove the inverse inequality, let { u n } C 0 ( R N ) be a minimizing sequence of m ( R N ) . Choosing y n R N and λ n > 0 so that v n ( x ) = λ n N 2 2 u n ( λ n x + y n ) C 0 ( Ω ˜ ) , hence we obtain m ( Ω ˜ ) m ( R N ) .□

Lemma 4.3

For μ ( 0 , μ 1 ) , we have lim μ 0 c μ , r = lim μ 0 c μ , Ω .

Proof

For any μ n ( 0 , μ 1 ) and μ n 0 as n , by [17] there exists a solution u n of (1.14) such that I μ n , Ω ( u n ) = c μ n , Ω . We may show as Lemma 3.1 that { u n } is uniformly bounded in H 0 1 ( Ω ) . Moreover, there is a unique t n R + such that t n u n N Ω , that is,

Ω u n 2 d x = i = 1 k t n 2 2 i 2 Ω Ω u n + ( x ) 2 i u n + ( y ) 2 i x y N α i d x d y .

Because u n is bounded, so does t n . Since u n N μ n , Ω , we deduce

i = 1 k t n 2 2 i 2 Ω Ω u n + ( x ) 2 i u n + ( y ) 2 i x y N α i d x d y = μ n Ω ( u n + ) p d x + i = 1 k Ω Ω u n + ( x ) 2 i u n + ( y ) 2 i x y N α i d x d y i = 1 k Ω Ω u n + ( x ) 2 i u n + ( y ) 2 i x y N α i d x d y

implying t n 1 .

Note that for q > 2 , the function h ( t ) = 1 2 t 2 1 q t q is decreasing if t 1 and increasing if t 1 . In particular, h ( t ) is decreasing for both q = p and q = 2 2 i , i = 1 , 2 , k whenever t 1 . Therefore,

I μ n , Ω ( u n ) I μ n , Ω ( t n u n ) = 1 2 1 p t n 2 2 t n p p μ n Ω ( u n + ) p d x + i = 1 k 1 2 1 2 2 i t n 2 2 t n 2 2 i 2 2 i Ω Ω u n + ( x ) 2 i u n + ( y ) 2 i x y N α i d x d y

yields

m ( Ω ) I μ n , Ω ( t n u n ) I μ n , Ω ( u n ) c μ n , Ω + μ n p Ω ( t n u n + ) p d x ,

and then

m ( Ω ) liminf n c μ n , Ω .

On the other hand, for each u N Ω , there exists s n > 0 such that s n u N μ n , Ω , alternatively,

i = 1 k Ω Ω u + ( x ) 2 i u + ( y ) 2 i x y N α i d x d y = μ n s n p 2 Ω ( u + ) p d x + i = 1 k s n 2 2 i 2 Ω Ω u + ( x ) 2 i u + ( y ) 2 i x y N α i d x d y i = 1 k s n 2 2 i 2 Ω Ω u + ( x ) 2 i u + ( y ) 2 i x y N α i d x d y

yielding s n 1 . As a result,

c μ n , Ω = m μ n , Ω I μ n , Ω ( s n u ) I μ n , Ω ( u ) J ( u ) ,

which yields

limsup n c μ n , Ω m ( Ω ) .

Consequently, m ( Ω ) = lim n c μ n , Ω . Similarly, m ( B r ) = lim n c μ n , B r . Lemma 4.2 then implies

lim μ 0 c μ , r = lim μ 0 c μ , Ω .

Similarly, denote I λ , μ c = { u N λ , μ : I λ , μ c } the level set.

Proposition 4.4

There exists μ ( 0 , μ 1 ) such that for each 0 < μ < μ there is Λ ( μ ) > 0 so that if u I λ , μ c μ , r there holds β ( u ) Ω 2 r + , provided that λ Λ ( μ ) .

Proof

We argue indirectly. Suppose on the contrary that for μ arbitrarily small, there is a sequence { u n } such that u n N λ n , μ , λ n , I λ n , μ ( u n ) c c μ , r as n and β ( u n ) Ω 2 r + . We may show as Lemma 3.1 that { u n } is uniformly bounded in E . By Lemma 2.1, there is u μ H 0 1 ( Ω ) such that u n u μ weakly in E and u n u μ in L 2 ( R N ) as n . The interpolation inequality yields u n u μ in L p ( R N ) . Now, we distinguish two cases to discuss.

Case 1: i = 1 k Ω Ω u μ + ( x ) 2 i u μ + ( y ) 2 i x y N α i d x d y Ω u μ 2 d x μ Ω ( u μ + ) p d x .

Let w n = u n u μ . First, we show w n 0 in E as n . Indeed, were it not the case, we would have w n 0 in E as n . By the Brézis-Lieb-type lemma,

(4.1) c + o ( 1 ) = I λ n , μ ( u n ) = I λ n , μ ( w n ) + I λ n , μ ( u μ ) + o ( 1 ) I λ n , μ ( w n ) + o ( 1 )

and by the assumption,

(4.2) o ( 1 ) = I λ n , μ ( u n ) , u n = I λ n , μ ( w n ) , w n + Ω u μ 2 d x μ Ω ( u μ + ) p d x i = 1 k Ω Ω u μ + ( x ) 2 i u μ + ( y ) 2 i x y N α i d x d y + o ( 1 ) I λ n , μ ( w n ) , w n + o ( 1 ) .

That is,

(4.3) R N ( w n 2 + λ n V ( x ) w n 2 ) d x μ R N ( w n + ) p d x + i = 1 k R N R N w n + ( x ) 2 i w n + ( y ) 2 i x y N α i d x d y + o ( 1 ) .

As before, we can find a unique constant t n > 0 such that t n w n N λ n , μ , which and (4.3) yield

(4.4) t n p 2 μ w n + p p + i = 1 k t n 2 2 i 2 R N R N w n + ( x ) 2 i w n + ( y ) 2 i x y N α i d x d y μ w n + p p + i = 1 k R N R N w n + ( x ) 2 i w n + ( y ) 2 i x y N α i d x d y + o ( 1 ) ,

then t n 1 . By (4.1) and (4.4),

c + o ( 1 ) = I λ n , μ ( u n ) 1 2 1 p μ R N ( w n + ) p d x + i = 1 k 1 2 1 2 2 i R N R N w n + ( x ) 2 i w n + ( y ) 2 i x y N α i d x d y + o ( 1 ) 1 2 1 p μ R N ( t n w n + ) p d x + i = 1 k 1 2 1 2 2 i t n 2 2 i R N R N w n + ( x ) 2 i w n + ( y ) 2 i x y N α i d x d y + o ( 1 ) = I λ n , μ ( t n w n ) + o ( 1 ) .

Since v n t n w n N λ n , μ , we have

(4.5) I λ n , μ ( v n ) = sup s 0 I λ n , μ ( s v n ) .

Similarly, there exists a unique s n R + such that s n v n N R N . It follows that

(4.6) i = 1 k s n 2 2 i 2 R N R N v n + ( x ) 2 i v n + ( y ) 2 i x y N α i d x d y = R N v n 2 d x R N ( v n 2 + λ n V v n 2 ) d x = μ R N ( v n + ) p d x + i = 1 k R N R N v n + ( x ) 2 i v n + ( y ) 2 i x y N α i d x d y ,

which implies { s n } is bounded. Therefore, by (4.5) and the fact that u n u μ in L p ( R N ) as n we deduce

m ( R N ) τ > I λ n , μ ( v n ) + o n ( 1 ) I λ n , μ ( s n v n ) + o n ( 1 ) J ( s n v n ) + o n ( 1 ) m ( R N ) + o n ( 1 ) ,

a contradiction. Hence, u n u μ in E as n . As a result, β ( u n ) β ( u μ ) as n . Moreover, by Lemma 2.1,

I μ , Ω ( u μ ) lim n I λ n , μ ( u n ) c μ , r .

Whence by Lemma 4.1, β ( u μ ) Ω r + , which is a contradiction to β ( u n ) Ω 2 r + .

Case 2: i = 1 k Ω Ω u μ + ( x ) 2 i u μ + ( y ) 2 i x y N α i d x d y > Ω u μ 2 d x μ Ω ( u μ + ) p d x .

It is known that there exists t μ > 0 such that t μ u μ N μ , Ω . By the assumption,

t μ p 2 μ Ω ( u μ + ) p d x + i = 1 k t μ 2 2 i 2 Ω Ω u μ + ( x ) 2 i u μ + ( y ) 2 i x y N α i d x d y = Ω u μ 2 d x < μ Ω ( u μ + ) p d x + i = 1 k Ω Ω u μ + ( x ) 2 i u μ + ( y ) 2 i x y N α i d x d y ,

which implies t μ ( 0 , 1 ) . Since p < 2 min , we have

c μ , Ω I μ , Ω ( t μ u μ ) t μ p 1 2 1 p μ Ω ( u μ + ) p d x + i = 1 k 1 2 1 2 2 i i = 1 k Ω Ω u μ + ( x ) 2 i u μ + ( y ) 2 i x y N α i d x d y t μ p lim n 1 2 1 p μ R N ( u n + ) p d x + i = 1 k 1 2 1 2 2 i i = 1 k R N R N u n + ( x ) 2 i u n + ( y ) 2 i x y N α i d x d y lim n I λ n , μ ( u n ) c μ , r .

Thus, by Lemma 4.3, for n large we obtain

(4.7) I λ n , μ ( u n ) I μ , Ω ( t μ u μ ) c μ , r c μ , Ω 0

as μ 0 . Therefore,

I λ n , μ ( u n ) I μ , Ω ( t μ u μ ) = 1 2 1 p μ R N ( u n + ) p d x t μ p Ω ( u μ + ) p d x + i = 1 k 1 2 1 2 2 i R N R N u n + ( x ) 2 i u n + ( y ) 2 i x y N α i d x d y t μ 2 2 i Ω Ω u n + ( x ) 2 i u n + ( y ) 2 i x y N α i d x d y

and (4.7) yield that

(4.8) R N R N u n + ( x ) 2 i u n + ( y ) 2 i x y N α i d x d y t μ 2 2 i Ω Ω u n + ( x ) 2 i u n + ( y ) 2 i x y N α i d x d y 0

for n large enough and as μ 0 , i = 1 , 2 , , k .

Extending u μ to R N by setting u μ = 0 outside Ω , we deduce from I λ n , μ ( u n ) , u n I μ , Ω ( t μ u μ ) , t μ u μ = 0 and (4.8) that

R N ( u n 2 t μ u μ 2 ) d x = o μ ( 1 ) .

Hence,

R N u μ 2 d x lim n R N u n 2 d x = R N t μ u μ 2 d x + o μ ( 1 ) .

That is, t μ 1 as μ 0 and

lim n R N u n 2 d x = R N u μ 2 d x + o μ ( 1 ) .

Since u n u μ in E as n , we have

R N ( u n u μ ) 2 d x = R N ( u n 2 2 u n u μ + u μ 2 ) d x = o n ( 1 ) + o μ ( 1 ) .

This readily yields for μ > 0 small enough and n large that

β ( u n ) β ( t μ u μ ) < r .

It is known from Lemma 4.1 that β ( t μ u μ ) Ω r + , whereas β ( u n ) Ω 2 r + . The contradiction completes the proof.□

We know from [17] and [19] that there exists a radially symmetric minimize v > 0 of I μ , B r on N μ , B r for μ ( 0 , μ 1 ) . It allows us to estimate the category of the level set of I λ , μ .

Lemma 4.5

If N 4 and μ ( 0 , μ ) , for λ Λ ( μ ) , then

cat I λ , μ c μ , r ( I λ , μ c μ , r ) cat Ω ( Ω ) .

Proof

Define γ : Ω r I λ , μ c μ , r by

γ ( y ) ( x ) = v ( x y ) , x B r ( y ) , 0 , x B r ( y ) .

We may verify that γ ( y ) ( x ) N λ , μ , I λ , μ ( γ ( y ) ( x ) ) c μ , r , and β γ = i d : Ω r Ω r .

Assume that cat I λ , μ c μ , r ( I λ , μ c μ , r ) = n , and

I λ , μ c μ , r = j = 1 n A j ,

where A j , j = 1 , 2 , , n , is closed and contractible in I λ , μ c μ , r , i.e., there exists h j C ( [ 0 , 1 ] × A j , I λ , μ c μ , r ) such that, for every u , v A j ,

h j ( 0 , u ) = u , h j ( 1 , u ) = h j ( 1 , v ) .

Let B j γ 1 ( A j ) , j = 1 , 2 , , n . For each x Ω r ,

γ ( x ) I λ , μ c μ , r j = 1 n A j .

So there exists j 0 such that γ ( x ) A j 0 , that is, x γ 1 ( A j 0 ) = B j 0 . Therefore,

Ω r j = 1 n B j .

For x , y B j , γ ( x ) , γ ( y ) A j , the deformation

g j ( t , x ) = β 0 ( h j ( t , γ ( x ) ) ) , j = 1 , 2 , , n ,

fulfills

g j ( 0 , x ) = β 0 ( h j ( 0 , γ ( x ) ) ) = β 0 ( γ ( x ) ) = x

and

g j ( 1 , x ) = β 0 ( h j ( 1 , γ ( x ) ) ) = β 0 ( h j ( 1 , γ ( y ) ) ) = g j ( 1 , y ) .

Hence, B j is contractible in Ω 2 r + . It follows that

cat Ω ( Ω ) = cat Ω 2 r + ( Ω r ) k = 1 n cat Ω 2 r + ( B k ) = n .

Lemma 4.6

If I λ , μ constraint to N λ , μ denoted by I λ , μ N λ , μ is bounded from below and satisfies the ( PS ) c condition for any c [ c λ , μ , c μ , r ] , then I λ , μ N λ , μ has a minimum and I λ , μ c μ , r contains at least cat I λ , μ c μ , r ( I λ , μ c μ , r ) critical points of I λ , μ N λ , μ .

Proof of Theorem 1.3

For 0 < μ μ and λ Λ ( μ ) , we defined two maps

Ω r γ I λ , μ c μ , r β Ω 2 r + .

The conclusion follows from Proposition 3.4, Proposition 4.4, Lemma 4.5, and Lemma 4.6.□

  1. Funding information: This work was supported by the National Natural Science Foundation of China (No: 12171212).

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

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Received: 2022-01-11
Revised: 2022-05-31
Accepted: 2022-06-01
Published Online: 2022-07-11

© 2022 Ziyi Xu and Jianfu Yang, published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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