Existence and multiplicity of positive solutions for multiparameter periodic systems

Lemma 2.1

Let E be a Banach space and P a cone in E. For r > 0 , define P r = { x ∈ P : ∣ ∣ x ∣ ∣ < r } . Assume that T : P r ¯ → P r is completely continuous such that T x ≠ x for x ∈ ∂ P r = { x ∈ P : ∣ ∣ x ∣ ∣ = r } .

1. If ∣ ∣ T x ∣ ∣ ≥ ∣ ∣ x ∣ ∣ for x ∈ ∂ P r , then i ( T , P r , P ) = 0 .

2. If ∣ ∣ T x ∣ ∣ ≤ ∣ ∣ x ∣ ∣ for x ∈ ∂ P r , then i ( T , P r , P ) = 1 .

Lemma 3.1

Suppose that (3.1) has an upper solution ( β u , β v ) and a lower solution ( α u , α v ) . Let f 1 ( x , u , v ) (resp. f 2 ( x , u , v ) ) be quasi-monotone nondecreasing with respect to v (resp. u ) and define

Then,

1. there exists at least one solution of problem (3.1) in A α , β ;

2. if ( u 0 , v 0 ) ∈ A α , β is the unique solution of (3.1) and there exists ρ 0 > 0 such that B ( ( u 0 , v 0 ) , ρ 0 ) = { ( u , v ) ∈ K : ∣ ∣ ( u − u 0 , v − v 0 ) ∣ ∣ ≤ ρ 0 } ⊂ A α , β , then

   i ( F , B ( ( u 0 , v 0 ) , ρ ) , K ) = 1 , for a l l 0 ≤ ρ ≤ ρ 0 ,

   where F ( u , v ) = ( F 1 ( u , v ) , F 2 ( u , v ) ) and F i : K → K defined by

   F i ( u , v ) = ∫ 0 T G ( x , r ) f i ( r , u , v ) d r .

Proof

(i) We define the continuous functions Γ 1 , Γ 2 : [ 0 , T ] × [ 0 , ∞ ) 2 → [ 0 , ∞ ) ,

with γ i given by

And we consider the modified problem

Next we write (3.3) as a system of integral equations

The operator F ¯ i : K → K defined by

is completely continuous and bounded. By Schauder’s theorem, F ¯ ( u , v ) = ( F ¯ 1 ( u , v ) , F ¯ 2 ( u , v ) ) has a fixed point, which is a solution of (3.3). We prove that any solution ( u , v ) of (3.3) satisfies ( u , v ) ∈ A α , β . Here we only establish the inequality α u ≤ u on [ 0 , T ] (a similar argument can be made for α v ≤ v ).

Suppose by contradiction that there exists x 0 ∈ [ 0 , T ] such that

If x 0 ∈ ( 0 , T ) , then there exists a sequence { x k } ⊂ ( 0 , x 0 ) converging to x 0 such that α u ′ ( x 0 ) = u ′ ( x 0 ) and α u ′ ( x k ) − u ′ ( x k ) ≥ 0 . This implies

which yields

Since ( α u , α v ) is a lower solution of (3.1) and f 1 is quasi-monotone nondecreasing with respect to v , we have

which is a contradiction. If max 0 ≤ x ≤ T ( α u − u ) = α u ( 0 ) − u ( 0 ) = α u ( T ) − u ( T ) , then α u ′ ( 0 ) − u ′ ( 0 ) ≤ 0 , α u ′ ( T ) − u ′ ( T ) ≥ 0 . Using that α u ′ ( 0 ) ≥ α u ′ ( T ) , we deduce that α u ′ ( 0 ) − u ′ ( 0 ) = 0 = α u ′ ( T ) − u ′ ( T ) . Applying similar reasoning as for x 0 = 0 , we have

Then, using the fact of α u ′ ( 0 ) = u ′ ( 0 ) and following a similar approach as in the case of x 0 ∈ ( 0 , T ) , we can once again get a contradiction. Therefore, α u ( x ) ≤ u ( x ) for all x ∈ [ 0 , T ] and similarly, we can show that β u ( x ) ≥ u ( x ) for all x ∈ [ 0 , T ] .

(ii) Observe that the operator F ¯ = F on A α , β and, by the result of (i), any fixed point ( u , v ) of F ¯ satisfies ( u , v ) ∈ A α , β . In particular, it is also a fixed point of F . Therefore, ( u 0 , v 0 ) is the unique fixed point of F ¯ . Since

for sufficiently large d , and combining this fact with the excision property and [19], we obtain

Since F ¯ = F on A α , β and B ¯ ( ( u 0 , v 0 ) , ρ 0 ) ⊂ A α , β , the conclusion is immediate.□

Lemma 4.1

Assume that (H1) and (H2) hold. Then, the following are true:

1. there exist Λ 1 , Λ 2 > 0 such that Σ ⊂ [ Λ 1 , + ∞ ) × [ Λ 2 , + ∞ ) and for all ( λ 1 , λ 2 ) ∈ ( 0 , + ∞ ) 2 \ ( [ Λ 1 , + ∞ ) × [ Λ 2 , + ∞ ) ) , problem (1.1) has no positive solution;

2. if ( λ 1 ̲ , λ 2 ̲ ) ∈ Σ , then [ λ 1 ̲ , + ∞ ) × [ λ 2 ̲ , + ∞ ) ⊂ Σ ;

3. if ( λ 1 ̲ , λ 2 ̲ ) ∈ Σ , then for all ( λ 1 , λ 2 ) ∈ ( λ 1 ̲ , + ∞ ) × ( λ 2 ̲ , + ∞ ) , there exist at least two positive solutions of problem (1.1).

Proof

(i) For ( u , v ) ∈ K and ∣ ∣ ( u , v ) ∣ ∣ = p , let

Choose a number r 1 > 0 , let λ 0 = r 1 2 m ( r 1 ) and set

Then, for λ 1 , λ 2 ≥ λ 0 and ( u , v ) ∈ K ∩ ∂ Ω r 1 , we have

which implies

for ( u , v ) ∈ K ∩ ∂ Ω r 1 . Hence, Lemma 2.1 implies

Since g i , 0 = 0 , we may choose r 2 ∈ ( 0 , r 1 ) so that g i ( u , v ) ≤ η ( u + v ) for 0 < u , v < r 2 , where the constant η > 0 satisfies

Set Ω r 2 = { ( u , v ) : ( u , v ) ∈ K , ∣ ∣ ( u , v ) ∣ ∣ < r 2 } . If ( u , v ) ∈ K ∩ ∂ Ω r 2 , we have

Hence, ∣ ∣ F λ ( u , v ) ∣ ∣ = ∣ ∣ F 1 , λ ( u , v ) ∣ ∣ + ∣ ∣ F 2 , λ ( u , v ) ∣ ∣ ≤ ∣ ∣ ( u , v ) ∣ ∣ for ( u , v ) ∈ K ∩ ∂ Ω r 2 . Using Lemma 2.1 once again, we have

Now, it follows from (4.1), (4.2), and the additivity of the fixed point index that for λ i > λ 0 ,

Consider now the nonempty sets

and let

It follows that Σ ⊂ [ Λ 1 , + ∞ ) × [ Λ 2 , + ∞ ) and for all ( λ 1 , λ 2 ) ∈ ( 0 , + ∞ ) 2 \ ( [ Λ 1 , + ∞ ) × [ Λ 2 , + ∞ ) ) , system (1.1) has no positive solution.

(ii) Let ( λ 1 0 , λ 2 0 ) ∈ [ λ 1 ̲ , + ∞ ) × [ λ 2 ̲ , + ∞ ) be arbitrarily chosen and suppose that ( α u , α v ) is a positive solution of (1.1) when λ 1 = λ 1 ̲ , and λ 2 = λ 2 ̲ . Then, for fixed λ 1 = λ 1 0 and λ 2 = λ 2 0 , ( α u , α v ) is a lower solution of (1.1). Similarly, let ( λ ¯ 1 , λ ¯ 2 ) ∈ [ λ 1 0 , + ∞ ) × [ λ 2 0 , + ∞ ) be arbitrarily chosen and suppose that ( β u , β v ) is a positive solution for (1.1) when λ 1 = λ ¯ 1 , and λ 2 = λ ¯ 2 . Then, for fixed λ 1 = λ 1 0 and λ 2 = λ 2 0 , ( β u , β v ) is an upper solution of (1.1). According to Lemma 3.1 (i) and the positivity of ( α u , α v ) , we conclude that ( λ 1 0 , λ 2 0 ) ∈ Σ .

(iii) From (ii) we obtain that ( λ 1 ̲ , + ∞ ) × ( λ 2 ̲ , + ∞ ) ⊂ Σ and let

It remains to show that system (1.1) with λ 1 = λ 1 0 and λ 2 = λ 2 0 has a second positive solution. For this, we define ( α u , α v ) as the lower solution and ( β u , β v ) as the upper solution, both constructed as above. We fix ( u 0 , v 0 ) a positive solution of (1.1) with λ 1 = λ 1 0 and λ 2 = λ 2 0 such that ( u 0 , v 0 ) ∈ A α , β .

Now, we claim that there exists ε > 0 such that B ¯ ( ( u 0 , v 0 ) , ε ) ⊂ A α , β . For all x ∈ [ 0 , T ] , we have