Infinitely many solutions for Schrödinger equations with Hardy potential and Berestycki-Lions conditions

Theorem 1.1

Assume that N ≥ 3 , 0 < μ < ( N − 2 ) 2 4 , and ( g 1 )–( g 4 ) hold. Then problem ( P ) possesses an infinite sequence of distinct solutions.

Remark 1.2

We point out, as far as we are concerned, that this article seems to be the first attempt to obtain the existence of infinitely many solutions to problem ( P ). About Schrödinger equations with the Hardy potential, most of the authors consider critical nonlinearities, see [19–23]. Our aim here is to illustrate the existence of solutions for the nonlinearity under Berestycki-Lions conditions, which is different from those results in the literature involving a potential having a different shape. For subcritical nonlinearities, there are many similar research studies, please refer to the literature [24], where the assumptions on the nonlinear term are stronger than Berestycki-Lions conditions.

Compared with [4,6], we want to obtain the existence of infinitely many solutions when V ( x ) is a Hardy potential, which clearly does not satisfy ( V 2 ) or ( V 5 ) . Of course, the methods dealing with our problem are different from ones in [4,6]. Our work extend and complete some results in [4,6].

Lemma 2.1

Assume that N ≥ 3 and ( g 1 ) − ( g 4 ) hold. Then

1. M t + g ( t ) ≤ h ( t ) ≤ f ( t ) for all t ≥ 0 ;

2. h ( t ) ≥ 0 , f ( t ) ≥ 0 for all t ≥ 0 ;

3. There exists δ 0 > 0 such that h ( t ) = f ( t ) = 0 for t ∈ [ 0 , δ 0 ] ;

4. There exists t 0 > 0 such that 0 < h ( t 0 ) ≤ f ( t 0 ) ;

5. t ↦ f ( t ) ∕ t p 0 is non-decreasing on ( 0 , ∞ ) ;

6. h ( t ) , f ( t ) satisfies ( g 3 ′ ) .

Proof

The proof is similar to the proof of [3, Lemma 2.1]. We omit the details.□

Lemma 2.2

Assume that N ≥ 3 and ( g 1 )–( g 3 ) hold. Then ℐ possesses the mountain pass geometry, namely

1. there exists ρ > 0 such that inf ‖ u ‖ = ρ ℐ ( u ) ≥ inf ‖ u ‖ = ρ J ( u ) > 0 ,

2. For any n ∈ N , there exists an odd continuous mapping

   γ 0 , n : S n − 1 = { σ = ( σ 1 , … , σ n ) ∈ R n ; ∣ σ ∣ = 1 } → H r 1 ( R N )

   such that

   I ( γ 0 n ( σ ) ) < 0 , J ( γ 0 n ( σ ) ) < 0 for a l l σ ∈ S n − 1 .

Proof

(i): Obviously, ℐ ( u ) ≥ J ( u ) for all u ∈ H r 1 ( R N ) . It follows from (e)–(f) in Lemma 2.1 that there exist C such that for any s ∈ R

Thus, one has

Since p 0 ∈ ( 1 , ( N + 2 ) ∕ ( N − 2 ) ) , 2 * > 2 , there exists ρ > 0 such that ℐ ( u ) ≥ J ( u ) ≥ 1 4 min m , 1 − μ 2 N − 2 2 ρ 2 > 0 for ‖ u ‖ = ρ > 0 .

(ii): We can argue as in [2, Theorem 10] and find for any n ∈ N an odd continuous mapping π n : S n − 1 → H r 1 ( R N ) such that

For t ≥ 1 , set

Then

Thus for sufficiently large t = t n ≥ 1 , γ 0 , n ( σ ) has the desired property for ℐ . Moreover, γ 0 , n ( σ ) also has the desired property for J . This completes the proof.□

Lemma 2.3

Assume that N ≥ 3 and ( g 1 )–( g 4 ) hold. Then there exists { u m , n } satisfying

where

Proof

Following Jeanjean [5,11] (see also [3,25]), we define the map Φ : R × H r 1 ( R N ) → H r 1 ( R N ) for t ∈ R , v ∈ H r 1 ( R N ) , and x ∈ R N by

For every t ∈ R and v ∈ H r 1 ( R N ) , the functional ℐ ∘ Φ is computed as

In view of ( g 1 ) –( g 4 ), ℐ ∘ Φ is continuously Fréchet-differentiable on R × H r 1 ( R N ) . We equip a standard product norm ‖ ( θ , u ) ‖ R × H r 1 = ( ∣ θ ∣ 2 + ‖ u ‖ 2 ) 1 ∕ 2 to R × H r 1 ( R N ) . We define minimax values b ˜ n for ℐ ∘ Φ by

where Γ ˜ n = { γ ˜ ( σ ) ∈ C ( B n , R × H r 1 ( R N ) ) : γ ˜ ( σ ) = ( θ ( σ ) , η ( σ ) ) satisfies ( θ ( − σ ) , η ( − σ ) ) = ( θ ( σ ) , − η ( σ ) ) for all σ ∈ B n and ( θ ( σ ) , η ( σ ) ) = ( 0 , γ 0 , n ( σ ) ) for all σ ∈ ∂ B n } .

Then we have b ˜ n = b n for all n ∈ N . Indeed, for any γ ∈ Γ n we can see that ( 0 , γ ( σ ) ) ∈ Γ ˜ n and we may regard Γ n ⊂ Γ ˜ n . Thus by the definitions of b n and b ˜ n , we have b ˜ n ≤ b n . On the other hand, for any given γ ˜ ( σ ) = ( θ ( σ ) , η ( σ ) ) ∈ Γ ˜ n , we set γ ( σ ) = η ( σ ) ( e − θ ( σ ) x ) . We can verify that γ ( σ ) ∈ Γ n , ℐ ( γ ( σ ) ) = ℐ ∘ Φ ( γ ˜ ( σ ) ) for all σ ∈ B n . Thus, we also have b ˜ n ≥ b n .

As in [3, Proposition 4.2], there exists a sequence { ( t m , n , v m , n ) } ⊂ R × H r 1 ( R N ) such that as m → ∞

Since for every ( h , w ) ∈ R × H r 1 ( R N )

Then, by taking u m , n = Φ ( t m , n , v m , n ) we have

This completes the proof.□

Proof

Step 1. We show that for all n there exists a nontrivial solution u n such that

In Lemma 2.3, we recall that { u m , n } satisfies as m → ∞ ,

Indeed, we know

This implies that { u m , n } is bounded in D 1 , 2 ( R N ) . On the other hand, we have

Then { u m , n } is bounded in L 2 ( R N ) from the Sobolev embedding. So that { u m , n } is bounded in H r 1 ( R N ) . Furthermore, there exists u n ∈ H r 1 ( R N ) such that u m , n ⇀ u n in H r 1 ( R N ) , u m , n → u in L loc 2 ( R N ) , u m , n ( x ) → u n ( x ) a.e. in R N . For any φ ∈ C 0 ∞ ( R N ) , one sees

Namely, u n is a solution of problem ( P ). We claim that u m , n → u n in H r 1 ( R N ) . Indeed,

We observed that

and

Therefore, by Fatou’s lemma, as n large enough we have

Combining with the Hardy inequality, from (8) we have

This implies u m , n → u n strongly in H r 1 ( R N ) as m → ∞ .

Step 2. Finally, we shall prove that b n → ∞ as n → ∞ . This implies that ℐ has infinitely many critical points.

We can also see that c n ( n ∈ N ) are critical values of J (see [3,26]). We apply an argument in [26, Chapter 9]. We set

Here Y m is the family of closed sets A ⊂ R m \ { 0 } such that − A = A and genus ( A ) is the Krasnoselskii’s genus of A . We define another sequence of minimax values by

Then we have c n ≥ d n for all n ∈ N , d 1 ≤ d 2 ≤ … ≤ d n ≤ d n + 1 ≤ … . Note that J satisfies the Palais-Smale condition. By modifying the argument in [26, Chapter 9] slightly, we have d n → ∞ as n → ∞ . Thus, c n → ∞ as n → ∞ . By (6), the minimax values b n satisfy b n → ∞ as n → ∞ .