Open Access

Existence and multiplicity of positive bound states for Schrödinger equations

Boundary Value Problems20132013:271

DOI: 10.1186/1687-2770-2013-271

Received: 5 September 2013

Accepted: 19 November 2013

Published: 12 December 2013

The Erratum to this article has been published in Boundary Value Problems 2016 2016:199

Abstract

In this paper, we study the existence and multiplicity of positive bound states of non-autonomous systems of nonlinear Schrödinger equations. The proof is based on the fixed point theorems in a cone.

MSC:34B15, 35J20.

Keywords

nonlinear Schrödinger systems positive solutions fixed point theorems

1 Introduction

Because of the important background in nonlinear optics and other fields, many authors pay more attention to the study of different types of vector nonlinear Schrödinger equations, we refer the readers to [15]. Most of these results have been proven using critical point theory, variational approaches or a fixed point theorem. More recently, Chu has applied another topological approach, a nonlinear alternative principle of Leray-Schauder, to establish some new existence results for the following Schrödinger equations
u ¨ ( x ) + a ( x ) u ( x ) = b ( x ) f ( u ( x ) ) + e ( x ) ,
where a L ( R , R N ) is nonnegative almost everywhere, e L 1 ( R , R N ) . The author considered two different cases. One is the singular case, that is, f C ( R N { 0 } , R N ) and
lim u 0 f i ( u ) = + , i = 1 , 2 , , n .

The other is the regular case, that is, f C ( R N , R N ) . However, the references [46] are not concerned with multiplicity of positive solutions for the scalar Schrödinger equation or system.

Motivated by the study of solitary wave solutions, in this paper we mainly aim to study the existence and multiplicity of positive bound states of the more general system of non-autonomous Schrödinger equations
{ u ¨ ( x ) + A ( x ) u ( x ) = C ( x ) F ( u ( x ) ) , lim | x | u ( x ) = lim | x | u ˙ ( x ) = 0 , + u 2 ( x ) d x + + u ˙ 2 ( x ) d x < + ,
(1)
where
u ( x ) = ( u 1 ( x ) u 2 ( x ) u n ( x ) ) , A ( x ) = ( a 11 ( x ) a 12 ( x ) a 1 n ( x ) a 21 ( x ) a 22 ( x ) a 2 n ( x ) a n 1 ( x ) a n 2 ( x ) a n n ( x ) ) , C ( x ) = ( c 1 ( x ) 0 0 0 c 2 ( x ) 0 0 0 c n ( x ) ) , F ( u ( x ) ) = ( f 1 ( u 1 ( x ) , u 2 ( x ) , , u n ( x ) ) f 2 ( u 1 ( x ) , u 2 ( x ) , , u n ( x ) ) f n ( u 1 ( x ) , u 2 ( x ) , , u n ( x ) ) ) .

The methods used here are the Krasnoselskii fixed point theorem and the Leggett-Williams fixed point theorem together with a compactness criterion due to Zima.

We organize the paper as follows. In Section 2, we give some preliminaries; in Section 3, we discuss the existence and multiplicity of positive solutions for (1).

2 Preliminaries

For convenience, we assume that the following conditions hold throughout this paper.

(H1) a i j L ( R , R ) , and a i j ( x ) satisfies the following property
a i j ( x ) = { a i i ( x ) 0 , and  inf a i i ( x ) > 0  if  i = j , a i j ( x ) 0 , and  Supp ( a i j )  is a nonempty compact set if  i j .
(H2) The support of c i ( x ) > 0 denoted by Supp ( c i ) is a nonempty compact set, and
0 < M G i ( s , s ) c i ( s ) d s < + .
(H3) f i ( u ( x ) ) C ( R N , R + ) is continuous, and the following notations are introduced:
f i , 0 = lim u 0 f i ( u ) i = 1 n u i , f i , = lim u f i ( u ) i = 1 n u i .
Since (H1) holds, then for the homogeneous problem
{ ϕ ¨ ( x ) + a i i ( x ) ϕ ( x ) = 0 , ϕ ( ) = 0 , ϕ ( + ) = 0 ,
the associated Green’s function is expressed by
G i ( x , s ) = { ϕ i 1 ( x ) ϕ i 2 ( s ) , < x s < + , ϕ i 1 ( s ) ϕ i 2 ( x ) , < s x < + ,
where ϕ i 1 , ϕ i 2 are solutions such ϕ i 1 ( ) = 0 , ϕ i 2 ( + ) = 0 . Moreover, ϕ i 1 , ϕ i 2 can be chosen as positive increasing and positive decreasing functions, respectively. Note that ϕ i 1 , ϕ i 2 intersect at a unique point x 0 . Therefore, we can define a function p i ( x ) BC ( R ) by
p i ( x ) = { 1 ϕ i 2 ( x ) , x x 0 , 1 ϕ i 1 ( x ) , x > x 0 ,

where BC ( R ) denotes the space of bounded continuous functions.

Lemma 2.1 [4]

For each i = 1 , 2 , , n , Green’s function G i ( x , s ) satisfies the following properties:
  1. (i)

    G i ( x , s ) > 0 for every ( x , s ) R × R ;

     
  2. (ii)

    G i ( x , s ) G i ( s , s ) for every ( x , s ) R × R ;

     
  3. (iii)
    Given a nonempty compact subset P R , we have
    G i ( x , s ) m i ( P ) p i ( s ) G i ( s , s ) for all  ( x , s ) P × R ,
     

where m i ( P ) = min { ϕ i 1 ( inf P ) , ϕ i 2 ( inf P ) } .

Lemma 2.2 [4]

Assume that (H1) holds and e ( x ) L 1 ( R ) . Then the unique solution of
{ u i ¨ ( x ) + a i i ( x ) u i ( x ) = e ( x ) , u i ( ) = 0 , u i ( + ) = 0
belongs to H 1 ( R ) , and the solution can be expressed as
u i ( x ) = R G i ( x , s ) e ( s ) d s .

The proof of our main results is based on the following fixed points, which can be found in [7].

Lemma 2.3 Let E be a Banach space, and let K E be a cone in E. Assume that Ω 1 , Ω 2 are open subsets of E with 0 Ω 1 , Ω ¯ 1 Ω 2 , and let T : K ( Ω ¯ 2 Ω 1 ) K be a completely continuous operator such that either
  1. (i)

    T u u , u K Ω 1 and T u u , u K Ω 2 ; or

     
  2. (ii)

    T u u , u K Ω 1 and T u u , u K Ω 2 .

     

Then T has a fixed point in K ( Ω ¯ 2 Ω 1 ) .

Let E be a real Banach space and P be a cone in E. A map α is said to be a nonnegative continuous concave functional on P if
α : P [ 0 , + )
is continuous and
α ( t x + ( 1 t ) y ) t α ( x ) + ( 1 t ) α ( y )

for all x , y P and t [ 0 , 1 ] .

For numbers a, b such that 0 < a < b , letting α be a nonnegative continuous concave functional on P, we define the following convex sets:
P a = { x P : x < a }
and
P ( α , a , b ) = { x P : a α ( x ) , x b } .
Lemma 2.4 Let T : P ¯ c P ¯ c be completely continuous and α be a nonnegative continuous concave functional on P such that α ( x ) x for all x P ¯ c . Suppose that there exist 0 < d < a < b c such that
  1. (i)

    { x P ( α , a , b ) : α ( x ) > a } and α ( T x ) > a for x P ( α , a , b ) ;

     
  2. (ii)

    T x < d for x d ;

     
  3. (iii)

    α ( T x ) > a for x P ( α , a , c ) with T x > b .

     
Then T has at least three fixed points x 1 , x 2 , x 3 satisfying
x 1 < d , a < α ( x 2 ) , x 3 > d and α ( x 3 ) < a .

In addition, the following compactness criterion proved by Zima in [8] is also used in our proof.

Lemma 2.5 Let Ω BC ( R ) . Let us assume that the functions u Ω are equicontinuous in each compact interval of R and that for all u Ω , we have
| u ( x ) | ξ ( x ) , x R ,
where ξ BC ( R ) verifies
lim | x | + ξ ( x ) = 0 .

Then Ω is relatively compact.

3 Main results

From now on, we assume that M = i j Supp ( a i j ) Supp ( c i ) is a nonempty compact set. Let E denote the Banach space https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-271/MediaObjects/13661_2013_Article_497_IEqa_HTML.gif with the norm u = i = 1 n | u i | , | u i | = max x M | u i ( x ) | for u = ( u 1 , u 2 , , u n ) E . Define a cone K E as
K = { u = ( u 1 , u 2 , , u n ) E : u i ( x ) 0  and  min x M i = 1 n u i ( x ) δ u } ,

where δ = min i = 1 , 2 , , n { m i p 0 i } ( 0 , 1 ) , p 0 i = inf M p i ( x ) and the constants m i m i ( M ) , i = 1 , 2 , , n , are defined by property (iii) of Lemma 2.1. Since M is compact, then p 0 i > 0 , i = 1 , 2 , , n . Moreover, from (iii) of Lemma 2.1 it follows that m i p 0 i < 1 for i = 1 , 2 , , n .

Let T : K E be a map with components ( T 1 , , T n ) defined by
T i ( u ) ( x ) = M G i ( x , s ) [ j i a i j u j + c i ( s ) f i ( u ( s ) ) ] d s .

A fixed point of T is a solution of (1) which belongs to https://static-content.springer.com/image/art%3A10.1186%2F1687-2770-2013-271/MediaObjects/13661_2013_Article_497_IEqb_HTML.gif .

Lemma 3.1 Assume that (H1)-(H3) hold. Then T ( K ) K , and T : K K is completely continuous.

Proof The continuity is trivial. Since M is compact, there exists a point x m where min x M T i ( u ) ( x ) is attained. Then, for any x R , we have
T i ( u ) ( x m ) = M G i ( x , s ) [ j i a i j u j + c i ( s ) f i ( u ( s ) ) ] d s m i M p i ( s ) G i ( s , s ) [ j i a i j u j + c i ( s ) f i ( u ( s ) ) ] d s m i p 0 i M G i ( s , s ) [ j i a i j u j + c i ( s ) f i ( u ( s ) ) ] d s m i p 0 i M G i ( x , s ) [ j i a i j u j + c i ( s ) f i ( u ( s ) ) ] d s m i p 0 i T i ( u ) ( x ) ,
namely,
min x M T i ( u ) ( x ) m i p 0 i T i ( u ) ( x ) δ T i ( u ) ( x ) .

Therefore, it is clear that T ( K ) K .

Finally, we prove that each component of T is compact. Let Ω K be a bounded set, then there exists a constant C > 0 which is uniformly bounded for its element. Since the derivative is bounded in compacts, the functions of T i ( Ω ) are equicontinuous on each compact interval. On the other hand, for any u Ω ,
| T i ( u ) ( x ) | C R G i ( x , s ) ( j i a i j ( s ) ) d s + max u C f i ( u ) R G i ( x , s ) c i ( s ) d s = ξ ( x ) .

 □

Theorem 3.2 Assume that (H1)-(H3) hold. In addition, a i j ( x ) ( i j ) satisfies
M G i ( s , s ) ( j i a i j ) d s < 1 2 n .
  1. (a)

    If f i , 0 = 0 , f i 0 , = for some i 0 { 1 , 2 , , n } , then (1) has at least one positive solution.

     
  2. (b)

    If f i , = 0 , f i 0 , 0 = for some i 0 { 1 , 2 , , n } , then (1) has at least one positive solution.

     
Proof (a) On the one hand, since f i , 0 = 0 , then there exists r > 0 such that
f i ( u ) ϵ i = 1 n u i for  0 < u 1 + + u n r ,
where ϵ > 0 is sufficiently small such that
ϵ M G i ( s , s ) c i ( s ) d s < 1 2 n .
Set Ω r = { u E : u < r } . Then, for any u Ω r K , we have
T i ( u ) ( x ) = M G i ( x , s ) [ j i a i j u j + c i ( s ) f i ( u ( s ) ) ] d s M G i ( s , s ) [ j i a i j ( s ) u j ( s ) + c i ( s ) ϵ i = 1 n u i ( s ) ] d s [ M G i ( s , s ) ( j i a i j ( s ) ) d s + ϵ M G i ( s , s ) c i ( s ) d s ] u 1 n u .
Furthermore, for any u Ω r K , we have
T ( u ) = i = 1 n | T i ( u ) | u .
On the other hand, since f i 0 , = for some i 0 { 1 , 2 , , n } , then there exists R ¯ such that
f i 0 ( u ) η i = 1 n u i for  i = 1 n u i R ¯ ,
where η > 0 is sufficiently large such that
η δ 2 M G i 0 ( s , s ) c i 0 ( s ) d s > 1 .
Let R = max { 2 r , R ¯ δ } and set Ω R = { u E : u < R } . Then, for any u K Ω 2 , min x M i = 1 n u i ( x ) δ u R ¯ , and we have
T i 0 ( u ) ( x ) = M G i 0 ( x , s ) [ j i 0 a i 0 j u j + c i 0 ( s ) f i 0 ( u ( s ) ) ] d s m i 0 p 0 i 0 M G i 0 ( s , s ) c i 0 ( s ) η i = 1 n u i ( s ) d s η δ 2 M G i 0 ( s , s ) c i 0 ( s ) d s u > u .
Furthermore, we have
T ( u ) > u for  u K Ω R .

Now by Lemma 2.3, T has a fixed point u = ( u 1 , u 2 , , u n ) K ( Ω ¯ R Ω r ) , namely, (1) has a positive solution.

(b) On the one hand, since f i 0 , 0 = for some i 0 { 1 , 2 , , n } , there exists r > 0 such that
f i 0 ( u ) η i = 1 n u i for  0 i = 0 n u i r ,
where η > 0 is sufficiently large such that
η δ 2 M G i 0 ( s , s ) c i 0 ( s ) d s > 1 .
Set Ω r = { u E : u < r } . Then, for any u K Ω r , we have
T i 0 ( u ) ( x ) = M G i 0 ( x , s ) [ j i 0 a i 0 j u j + c i 0 ( s ) f i 0 ( u ( s ) ) ] d s m i 0 p 0 i 0 M G i 0 ( s , s ) c i 0 ( s ) η i = 1 n u i ( s ) d s η δ 2 M G i 0 ( s , s ) c i 0 ( s ) d s u > u .
Furthermore, we have
T ( u ) > u for  u K Ω r .
On the other hand, since f i , = 0 , then there exists R ¯ such that
f i ( u ) ϵ i = 1 n u i for  u 1 + + u n R ¯ ,
where ϵ > 0 is sufficiently small such that
ϵ M G i ( s , s ) c i ( s ) d s < 1 2 n .
Let R = max { 2 r , R ¯ δ } and set Ω R = { u E : u < R } . Then, for any u K Ω R , min x M i = 1 n u i ( x ) δ u R ¯ , and we have
T i ( u ) ( x ) = M G i ( x , s ) [ j i a i j u j + c i ( s ) f i ( u ( s ) ) ] d s M G i ( s , s ) [ j i a i j ( s ) u j ( s ) + c i ( s ) ϵ i = 1 n u i ( s ) ] d s [ M G i ( s , s ) ( j i a i j ( s ) ) d s + ϵ M G i ( s , s ) c i ( s ) d s ] u 1 n u .
Furthermore, for any u Ω R K , we have
T ( u ) = i = 1 n | T i ( u ) | u .

Now, by Lemma 2.3, T has a fixed point u = ( u 1 , u 2 , , u n ) K ( Ω ¯ R Ω r ) , namely, (1) has a positive solution. □

Corollary 3.3 Assume that (H1)-(H3) hold. a i j ( x ) ( i j ) satisfies
M G i ( s , s ) ( j i a i j ) d s < 1 2 n .

In addition, the following conditions hold.

(H4) If there exist constants R ˆ , ρ > 0 such that for some i 0 { 1 , 2 , , n } ,
f i 0 ( u ) ρ R ˆ for  δ R ˆ i = 1 n u i R ˆ ,
where ρ satisfies
ρ δ M G i 0 ( s , s ) c i 0 ( s ) d s > 1 ;

(H5) f i , 0 = 0 , f i , = 0 .

Then (1) has at least two positive solutions.

Corollary 3.4 Assume that (H1)-(H3) hold. a i j ( x ) ( i j ) satisfies
M G i ( s , s ) ( j i a i j ) d s < 1 2 n .

In addition, the following conditions hold.

(H6) If there exist constants R ˜ , ϑ > 0 such that
f i ( u ) ϑ R ˜ for  δ R ˜ i = 1 n u i R ˜ ,
where ϑ satisfies
ϑ M G i ( s , s ) c i ( s ) d s < 1 2 n ;

(H7) f i 0 , 0 = , f j 0 , = for some i 0 , j 0 { 1 , 2 , , n } .

Then (1) has at least two positive solutions.

Theorem 3.5 Assume that (H1)-(H3) hold and a i j ( x ) ( i j ) satisfies
M G i ( s , s ) ( j i a i j ) d s < 1 2 n .

In addition, there exist numbers a, c and d with 0 < d < a < c 4 such that the following conditions are satisfied:

(H8) f i ( u ) < 1 2 n M G i ( s , s ) c i ( s ) d s d for u i 0 and 0 i = 1 n u i < d ;

(H9) there exists i 0 { 1 , 2 , , n } such that
f i 0 ( u ( x ) ) > a δ 2 Γ for  x M , u i 0  and  i = 1 n u i [ a , a δ ] ,

where Γ = min { M G i ( s , s ) c i ( s ) d s } ;

(H10) f i ( u ) 1 2 n M G i ( s , s ) c i ( s ) d s c for u i 0 and i = 1 n u i < c .

Then (1) has at least three positive solutions.

Proof For u = ( u 1 , u 2 , , u n ) K , define
α ( u ) = min x M ( u 1 ( x ) + u 2 ( x ) + + u n ( x ) ) ,

then it is easy to know that α is a nonnegative continuous concave functional on K with α ( u ) u for u K .

Set b = a δ . First, we show that T : K ¯ c K ¯ c with c > b . For any u K ¯ c , we have
T i ( u ) ( x ) = R G i ( x , s ) [ j i a i j u j + c i ( s ) f i ( u ( s ) ) ] d s = M G i ( x , s ) [ j i a i j u j + c i ( s ) f i ( u ( s ) ) ] d s M G i ( s , s ) [ j i a i j u j + c i ( s ) f i ( u ( s ) ) ] d s M G i ( s , s ) [ j i a i j u j + c i ( s ) c 2 n M G i ( s , s ) c i ( s ) d s ] d s c 2 n + c 2 n = c n .

So T ( u ) = i = 1 n | T i ( u ) | c .

In a similar way, we also can prove that T : K ¯ d K d . Then (ii) of Lemma 2.4 holds.

Next, we shall show that (i) of Lemma 2.4 is satisfied. It is clearly seen that u = ( a + b 2 n , , a + b 2 n ) { u = ( u 1 , u 2 , , u n ) K ( α , a , b ) : α ( u ) > a } . Then, for any u K ( α , a , b ) and x M , it is easy to obtain that
b i = 1 n | u i | i = 1 n u i ( x ) min x M ( i = 1 n u i ( x ) ) = α ( u ) > a .
Then, by (H9), we can have
α ( T ( u ) ( x ) ) = min x M ( i = 1 n T i ( u ) ( x ) ) min x M T i 0 ( u ) ( x ) δ max x M T i 0 ( u ) ( x ) δ M G i 0 ( x , s ) [ j i 0 a i 0 j u j + c i 0 ( s ) f i 0 ( u ( s ) ) ] d s δ M m i 0 p i 0 ( s ) G i 0 ( s , s ) c i 0 ( s ) f i 0 ( u ( s ) ) d s δ 2 M G i 0 ( s , s ) c i 0 ( s ) a δ 2 Γ d s = a .
Finally, we verify that (iii) of Lemma 2.4 is satisfied. Suppose that u K ( α , a , c ) with T ( u ) > b , then we can have
α ( T ( u ) ) = min M ( i = 1 n T i ( u ) ( x ) ) δ T ( u ) > b δ = a .
From the above, the hypotheses of Leggett-Williams theorem are satisfied. Hence (1) has at least three positive solutions u 1 , u 2 and u 3 such that u 1 < d , a < min x M ( i = 1 n u i 2 ( x ) ) , and u 3 > d with
min x M ( i = 1 n u i 3 ( x ) ) < a .

 □

Notes

Declarations

Authors’ Affiliations

(1)
College of Science, Hohai University

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Copyright

© Sheng et al.; licensee Springer. 2013

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