Positive solutions for boundary value problems of a class of second-order differential equation system

Lemma 2.1

[22] Let h ∈ L 2 ( I ) . Then, the solution u = S h ∈ H 2 ( I ) of LBVP (2.1) satisfies ‖ u ‖ 2 ≤ 1 π ‖ u ′ ‖ 2 , ‖ u ′ ‖ 2 ≤ 1 π ‖ u ″ ‖ 2 .

Lemma 2.2

[13] Let h ∈ C + ( I ) . Then, the solution u = S h of LBVP (2.1) has the following properties:

1. u ( t ) ≥ t ( 1 − t ) ‖ u ‖ C for all t ∈ I ;

2. ‖ u ‖ C ≤ π 3 4 ∫ 0 1 u ( t ) sin π t d t ;

3. ‖ u ‖ C ≤ ‖ u ′ ‖ C , and ‖ u ‖ C 1 = ‖ u ′ ‖ C ;

4. there exists a constant ξ ∈ ( 0 , 1 ) such that u ′ ( ξ ) = 0 , u ′ ( t ) ≥ 0 for all t ∈ [ 0 , ξ ) , and u ′ ( t ) ≤ 0 for all t ∈ ( ξ , 1 ] . Moreover, ‖ u ′ ‖ C = max { u ′ ( 0 ) , − u ′ ( 1 ) } .

Lemma 2.3

Let Ω be a bounded open subset of E × E with ( θ , θ ) ∈ Ω (namely, Ω contains zero elements ( θ , θ ) ), and mapping A : Ω ¯ ∩ K → K is completely continuous. If A satisfies

then, the fixed point index i ( A , Ω ∩ K , K ) = 1 on cone K .

Lemma 2.4

Let Ω be a bounded open subset of E × E , and mapping A : Ω ¯ ∩ K → K are completely continuous. If there exist ( x 0 , y 0 ) ∈ K \ { ( θ , θ ) } , such that

then, i ( A , Ω ∩ K , K ) = 0 .

Lemma 2.5

Let Ω be a bounded open subset of E × E , Ω ¯ ∩ K ≠ ∅ , and mappings A , A 1 : Ω ¯ ∩ K → K are completely continuous. If A and A 1 satisfy

then i ( A , Ω ∩ K , K ) = i ( A 1 , Ω ∩ K , K ) .

Theorem 3.1

Let f , g : [ 0 , 1 ] × R + × R + × R → R + be continuous. If f and g satisfy the following conditions:

( F 1 ) there exist positive constants a, b, c, and d satisfying a + b π 2 + c + d π < 1 and δ > 0 , such that

where ( t , x , y ) ∈ I × R + × R + , p , q ∈ R , ∣ ( x , p ) ∣ = ∣ x ∣ 2 + ∣ p ∣ 2 < δ , ∣ ( y , q ) ∣ < δ ;

( F 2 ) there exist positive constants e > π 2 and H > 0 , such that

where ( t , x , y ) ∈ I × R + × R + , p , q ∈ R , ∣ ( x , p ) ∣ > H , ∣ ( y , q ) ∣ > H ;

( F 3 ) given any M > 0 , there exists a positive continuous function G M : R + → R + satisfies

such that f and g satisfy

where ( t , x , y ) ∈ I × [ 0 , M ] × [ 0 , M ] , p , q ∈ R ;

then BVP (1.1) has at least one positive solution.

Proof

Choose E = C 1 ( I ) , K ⊂ E × E is a closed convex cone defined by equation (2.3). Let 0 < r < R < + ∞ and set

We prove that A has a fixed point in ( Ω 2 \ Ω ¯ 1 ) ∩ K when r is small enough or R is large enough in two steps.

Step I. We prove that i ( A , Ω 1 ∩ K , K ) = 1 when r is sufficiently small. Let r ∈ 0 , δ 2 , where δ is the positive constant in condition ( F 1 ) . Applying Lemma 2.3 to A on ∂ Ω 1 ∩ K , namely,

In fact, if (3.5) does not hold, there exists μ 0 ∈ ( 0 , 1 ] and ( u 0 , v 0 ) ∈ ∂ Ω 1 ∩ K , such that ( u 0 , v 0 ) = μ 0 A ( u 0 , v 0 ) . Since ( u 0 , v 0 ) = μ 0 A ( u 0 , v 0 ) = μ 0 ( S ∗ F ( u 0 , v 0 ) ) = S ∗ ( μ 0 F ( u 0 , v 0 ) ) , then by definition of S ∗ , ( u 0 , v 0 ) is a solution of LBVP (2.2) for ( h 1 , h 2 ) = μ 0 F ( u 0 , v 0 ) ∈ C + ( I ) × C + ( I ) . Hence, ( u 0 , v 0 ) ∈ C 2 ( I ) × C 2 ( I ) satisfies the differential equation

Since ( u 0 , v 0 ) ∈ ∂ Ω 1 , for all t ∈ I , we have

Therefore,

Since

therefore,

Hence,

By condition ( F 1 ) , for all t ∈ I , we have

Add the first formula of equation (3.6) to the second formula of equation (3.6), from the aforementioned inequality, we can obtain

Taking ‖ ⋅ ‖ 2 on both sides, we obtain that

Owing to a + b π 2 + c + d π < 1 , therefore ‖ u 0 ″ ‖ 2 + ‖ v 0 ″ ‖ 2 = 0 , namely, u 0 = 0 , v 0 = 0 , this contradicts with ( u 0 , v 0 ) ∈ ∂ Ω 1 . So (3.5) holds, namely, A satisfies the condition of Lemma 2.3 in ∂ Ω 1 ∩ K . By Lemma 2.3, we can see

Step II. We prove that i ( A , Ω 2 ∩ K , K ) = 0 when R is sufficiently large. Let

By condition ( F 2 ) , we have

where ( t , x , y ) ∈ I × R + × R + , p , q ∈ R . Define a nonlinear mapping F 1 : K → C + ( I ) × C + ( I ) as follows:

and set A 1 = S ∗ ∘ F 1 , then A 1 : K → K is completely continuous.

First, we prove that i ( A 1 , Ω 2 ∩ K , K ) = 0 . Choose ( x 0 ( t ) , y 0 ( t ) ) = ( sin π t , sin π t ) . Since ( − x 0 ″ ( t ) , − y 0 ″ ( t ) ) = ( π 2 sin π t , π 2 sin π t ) = π 2 ( x 0 ( t ) , y 0 ( t ) ) , then by the definition of S ∗ , there are ( x 0 , y 0 ) = S ∗ ( π 2 ( x 0 , y 0 ) ) ∈ K . Now, we show that A 1 satisfies Lemma 2.4, namely,

In fact, if (3.9) does not hold, there exist τ 0 ≥ 0 and ( u 1 , v 1 ) ∈ ∂ Ω 2 ∩ K , such that ( u 1 , v 1 ) − A 1 ( u 1 , v 1 ) = τ 0 ( x 0 , y 0 ) . Since ( u 1 , v 1 ) = A 1 ( u 1 , v 1 ) + τ 0 ( x 0 , y 0 ) = S ∗ ( F 1 ( u 1 , v 1 ) + τ 0 π 2 ( x 0 , y 0 ) ) , then by definition of S ∗ , ( u 1 , v 1 ) is the solution of LBVP (2.2) for ( h 1 , h 2 ) = F 1 ( u 1 , v 1 ) + τ 0 π 2 ( x 0 , y 0 ) ∈ C + ( I ) × C + ( I ) . Hence, ( u 1 , v 1 ) ∈ C 2 ( I ) × C 2 ( I ) satisfies the differential equation

Add the first formula of equation (3.10) to the second formula of equation (3.10), from (3.8), we can obtain

Multiply both sides of the aforementioned inequality by sin π t and integrate on I , we have

Since ∫ 0 1 u 1 ( t ) sin π t d t ≥ 4 π 3 ‖ u 1 ‖ C > 0 , ∫ 0 1 v 1 ( t ) sin π t d t ≥ 4 π 3 ‖ v 1 ‖ C > 0 by Lemma 2.2 (2), it follows that π 2 ≥ e , this contradicts with e > π 2 . Hence (3.9) holds, namely, A 1 satisfies the condition of Lemma 2.4 in ∂ Ω 2 ∩ K . By Lemma 2.4, we can see

Second, we show that i ( A , Ω 2 ∩ K , K ) = i ( A 1 , Ω 2 ∩ K , K ) . When R is sufficiently large, A and A 1 satisfy Lemma 2.5, namely,

In fact, if (3.12) does not hold, there exist s 0 ∈ [ 0 , 1 ] and ( u 2 , v 2 ) ∈ ∂ Ω 2 ∩ K , such that ( 1 − s 0 ) A ( u 2 , v 2 ) + s 0 A 1 ( u 2 , v 2 ) = ( u 2 , v 2 ) . Since ( u 2 , v 2 ) = ( 1 − s 0 ) S ∗ ( F ( u 2 , v 2 ) ) + s 0 S ∗ ( F 1 ( u 2 , v 2 ) ) = S ∗ ( ( 1 − s 0 ) F ( u 2 , v 2 ) + s 0 F 1 ( u 2 , v 2 ) ) , then by definition of S ∗ , ( u 2 , v 2 ) is the solution of LBVP (2.2) for ( h 1 , h 2 ) = ( 1 − s 0 ) F ( u 2 , v 2 ) + s 0 F 1 ( u 2 , v 2 ) = F ( u 2 , v 2 ) + s 0 ( e 1 , e 2 ) ∈ C + ( I ) × C + ( I ) . Hence, ( u 2 , v 2 ) ∈ C 2 ( I ) × C 2 ( I ) satisfies the differential equation