Put J = [ 0 , 1 ] , R + = [ 0 , ∞ ) . Let us consider the following boundary value problem:

{ x ″ ( t ) + h ( t ) f ( t , x ( α ( t ) ) , x ′ ( β ( t ) ) ) = 0 , t ∈ ( 0 , 1 ) , x ( 0 ) = γ x ( η ) + λ 1 [ x ] , x ( 1 ) = ξ x ( η ) + λ 2 [ x ] , η ∈ ( 0 , 1 ) ,

where λ 1 , λ 2 denote linear functionals on C ( J ) given by

λ 1 [ x ] = ∫ 0 1 x ( t ) d A ( t ) , λ 2 [ x ] = ∫ 0 1 x ( t ) d B ( t )

involving Stieltjes integrals with suitable functions A and B of bounded variation on J. It is not assumed that λ1, λ2 are positive to all positive x. As we see later, the measures dA, dB can be signed measures.

We introduce the following assumptions:

H₁: f ∈ C ( J × R + × R , R + ) , α , β ∈ C ( J , J ) , A and B are functions of bounded variation;

H₂: h ∈ C ( J , R + ) and h does not vanish identically on any subinterval;

H₃: 1 − γ − λ 1 [ p ] > 0 or 1 − ξ − λ 2 [ p ] > 0 for p ( t ) = 1 , t ∈ J , γ , ξ ≥ 0 .

Recently, the existence of multiple positive solutions for differential equations has been studied extensively; for details, see, for example, 12345678910111213141516171819202122232425262728293031. However, many works about positive solutions have been done under the assumption that the first-order derivative is not involved explicitly in nonlinear terms; see, for example, 36891011121314172025262730. From this list, only papers 9101112142030 concern positive solutions to problems with deviating arguments. On the other hand, there are some papers considering the multiplicity of positive solutions with dependence on the first-order derivative; see, for example, 24571516181921222324282931. Note that boundary conditions (BCs) in differential problems have important influence on the existence of the results obtained. In this paper, we consider problem (1) which is a problem with dependence on the first-order derivative with BCs involving Stieltjes integrals with signed measures of dA, dB appearing in functionals λ1, λ2; moreover, problem (1) depends on deviating arguments.

For example, in papers 2415182224, the existence of positive solutions to second-order differential equations with dependence on the first-order derivative (but without deviating arguments) has been studied with various BCs including the following:

by fixed point theorems in a cone (such as Avery-Peterson, an extension of Krasnoselskii’s fixed point theorem or monotone iterative method) with corresponding assumptions:

a i , b i ∈ ( 0 , 1 ) , i = 1 , 2 , … , n , ∑ i = 1 n a i , ∑ i = 1 n b i ∈ ( 0 , 1 ) ,

or 1 − α η > 0 , respectively.

For example, in papers 891011202230, the existence of positive solutions to second-order differential equations including impulsive problems, but without dependence on the first-order derivative, has been studied with various BCs including the following:

under corresponding assumptions by fixed point theorems in a cone (such as Avery-Peterson, Leggett-Williams, Krasnoselskii or fixed point index theorem). See also paper 13, where positive solutions have been discussed for second-order impulsive problems with boundary conditions

x ( 0 ) = 0 , x ( 1 ) = ∫ 0 1 x ( s ) d A ( s ) ;

here λ1 has the same form as in problem (1) with signed measure dA appearing in functional λ1.

Positive solutions to second-order differential equations with boundary conditions that involve Stieltjes integrals have been studied in the case of signed measures in papers 2526 with BCs including, for example, the following:

The main results of papers 2526 have been obtained by the fixed point index theory for problems without deviating arguments. The study of positive solutions to boundary value problems with Stieltjes integrals in the case of signed measures has also been done in papers 37131427 for second-order differential equations (also impulsive) or third-order differential equations by using the fixed point index theory, the Avery-Peterson fixed point theorem or fixed point index theory involving eigenvalues.

Note that BCs in problem (1) with functionals λ1, λ2 cover some nonlocal BCs, for example,

for some constants a i , b i and some functions g 1 , g 2 . In our paper, the assumption that the measures dA, dB in the definitions of λ1, λ2 are positive is not needed. More precisely, one needs to choose the above functions g1, g2 in such a way that the assumption H₄ holds. It means that g1, g2 can change sign on J.

A standard approach (see, for example, 252627) to studying positive solutions of boundary value problems such as (1) is to translate problem (1) to a Hammerstein integral equation

x ( t ) = Γ 1 ( t ) λ 1 [ x ] + Γ 2 ( t ) λ 2 [ x ] + Γ 3 ( t ) ∫ 0 1 G ( η , s ) h ( s ) f ( s , x ( α ( s ) ) , x ′ ( β ( s ) ) ) d s + ∫ 0 1 G ( t , s ) h ( s ) f ( s , x ( α ( s ) ) , x ′ ( β ( s ) ) ) d s ≡ W x ( t )

to find a solution as a fixed point of the operator W by using a fixed point theorem in a cone. Γ 1 , Γ 2 , Γ 3 are corresponding continuous functions while λ1 and λ2 have the same form as in problem (1). G denotes a Green function connected with our problem, so in our case it is given by

G ( t , s ) = { s ( 1 − t ) , 0 ≤ s ≤ t , t ( 1 − s ) , t ≤ s ≤ 1 .

In our paper, we eliminate λ1 and λ2 from problem (2) to obtain the equation x = W ¯ x with a corresponding operator W ¯ , and then we seek solutions as fixed points of this operator W¯.

Note that if we put γ = ξ = 0 in the BCs of problem (1), then this new problem is more general than the previous one because in this case someone, for example, can take λ 1 [ x ] = γ x ( η ) , λ 2 [ x ] = ξ x ( η ) . In this paper, we try to explain why for some cases we have to discuss problem (1) with constants γ > 0 or ξ > 0 .

To apply such a fixed point theorem in a cone to problem (1), we have to construct a suitable cone K. Usually, we need to find a nonnegative function κ and a constant ρ ¯ ∈ ( 0 , 1 ] such that G ( t , s ) ≤ κ ( s ) for t , s ∈ J and G ( t , s ) ≥ ρ ¯ κ ( t ) for t ∈ [ η , η ¯ ] ⊂ [ 0 , 1 ] and s ∈ J (see, for example, 252627) to work with the inequality

min [ η , η ¯ ] | x ( t ) | ≥ ρ ¯ max t ∈ J | x ( t ) | .

Indeed, for problems without deviating arguments, someone can use any interval [ η , η ¯ ] ⊂ [ 0 , 1 ] . It means that when α(t)=t on J, then we can take γ=ξ=0 in the boundary conditions of problem (1) to work with the inequality

min [ ζ , ϱ ] | x ( t ) | ≥ κ max t ∈ J | x ( t ) |

for ζ, ϱ such that ζ + ϱ < 1 , 0 < ζ < ϱ < 1 with κ = min ( ζ , 1 − ϱ ) ; see Section 5.

Note that for problems with delayed or advanced arguments, we have to use interval [ 0 , η ] ⊂ [ 0 , 1 ) or [ η , 1 ] ⊂ ( 0 , 1 ] , respectively. We see that if γ=ξ=0, then ρ ¯ = 0 for problem (1) with deviated arguments. It shows that the approach from papers 252627 needs a little modification to problems with delayed or advanced arguments. Consider the situation α ( t ) ≤ t on J. In this case, we can put ξ = 0 in the boundary conditions of problem (1) to find a constant ρ ∈ ( 0 , 1 ) to work with the inequality

min [ 0 , η ] | x ( t ) | ≥ ρ max t ∈ J | x ( t ) | ;

see Section 3. For the case α ( t ) ≥ t on J, we can put γ = 0 to work similarly as in Section 3; see Section 4. Note that in the above three cases for the argument β, we need only the assumption β ∈ C ( J , J ) , which means that β can change the character in J.

Note that in cited papers, positive solutions to differential equations with dependence on the first-order derivative have been investigated only for problems without deviating arguments, see 24571516181921222324282931. Moreover, BCs in problem (1) cover some nonlocal BCs discussed earlier.

Motivated by 252627, in this paper, we apply the fixed point theorem due to Avery-Peterson to obtain sufficient conditions for the existence of multiple positive solutions to problems of type (1). In problem (1), an unknown x depends on deviating arguments which can be both of advanced or delayed type. To the author’s knowledge, it is the first paper when positive solutions have been investigated for such general boundary value problems with functionals λ1, λ2 and with deviating arguments α, β in differential equations in which f depends also on the first-order derivative. It is important to indicate that problems of type (1) have been discussed with signed measures of dA, dB appearing in Stieltjes integrals of functionals λ1, λ2.

The organization of this paper is as follows. In Section 2, we present some necessary lemmas connected with our main results. In Section 3, we first present some definitions and a theorem of Avery and Peterson which is useful in our research. Also in Section 3, we discuss the existence of multiple positive solutions to problems with delayed argument α, by using the above mentioned Avery-Peterson theorem. At the end of this section, an example is added to verify theoretical results. In Section 4, we formulate sufficient conditions under which problems with advanced argument α have positive solutions. In the last section, we discuss problems of type (1) when α(t)=t on J.

Let us introduce the following notations:

∥ x ∥ = max ( ∥ x ∥ 1 , ∥ x ′ ∥ 1 ) with  ∥ z ∥ 1 = max t ∈ J | z ( t ) | .

Lemma 1 Let x ∈ C 1 ( J , R ) , p(t)=1, t∈J. Assume that A and B are functions of bounded variation and, moreover,

x ( 0 ) = γ x ( η ) + λ 1 [ x ] , x ( 1 ) = ξ x ( η ) + λ 2 [ x ] , γ , ξ ≥ 0 , η ∈ ( 0 , 1 )

with

(i) 1 − γ − λ 1 [ p ] ≠ 0 or

(ii) 1 − ξ − λ 2 [ p ] ≠ 0 .

Then

∥ x ∥ 1 ≤ M ∥ x ′ ∥ 1 , M = 1 + { Var A + γ | 1 − γ − λ 1 [ p ] | , in case  (i) , Var B + ξ | 1 − ξ − λ 2 [ p ] | , in case  (ii) .

Here, VarA denotes the variation of a function A on J.

Proof Note that in case (i), we have

x ( 0 ) = γ x ( η ) + λ 1 [ x ] = γ [ x ( η ) − x ( 0 ) ] + γ x ( 0 ) + ∫ 0 1 ( x ( t ) − x ( 0 ) ) d A ( t ) + λ 1 [ p ] x ( 0 ) = γ ∫ 0 η x ′ ( s ) d s + ∫ 0 1 ( ∫ 0 t x ′ ( s ) d s ) d A ( t ) + γ x ( 0 ) + λ 1 [ p ] x ( 0 ) ,

so

x ( 0 ) = 1 1 − γ − λ 1 [ p ] [ γ ∫ 0 η x ′ ( s ) d s + ∫ 0 1 ( ∫ 0 t x ′ ( s ) d s ) d A ( t ) ] .

Hence,

| x ( 0 ) | ≤ 1 | 1 − γ − λ 1 [ p ] | ( γ + Var A ) ∥ x ′ ∥ 1 .

Combining this with the relation

x ( t ) = x ( 0 ) + ∫ 0 t x ′ ( s ) d s ,

we obtain

∥ x ∥ 1 ≤ | x ( 0 ) | + ∥ x ′ ∥ 1 ≤ M ∥ x ′ ∥ 1 .

This proves case (i).

In case (ii), similarly,

x ( 1 ) = ξ x ( η ) + λ 2 [ x ] = ξ [ x ( η ) − x ( 1 ) ] + ξ x ( 1 ) − ∫ 0 1 ( x ( 1 ) − x ( t ) ) d B ( t ) + λ 2 [ p ] x ( 1 ) = − ξ ∫ η 1 x ′ ( s ) d s − ∫ 0 1 ( ∫ t 1 x ′ ( s ) d s ) d B ( t ) + ξ x ( 1 ) + λ 2 [ p ] x ( 1 ) ,

so

x ( 1 ) = − 1 1 − ξ − λ 2 [ p ] [ ξ ∫ η 1 x ′ ( s ) d s + ∫ 0 1 ( ∫ 0 1 x ′ ( s ) d s ) d B ( t ) ] .

Hence,

| x ( 1 ) | ≤ 1 | 1 − ξ − λ 2 [ p ] | ( ξ + Var B ) ∥ x ′ ∥ 1 .

Adding to this the relation

x ( t ) = x ( 1 ) − ∫ t 1 x ′ ( s ) d s ,

we get the result in case (ii). This ends the proof. □

Remark 1 If we assume that A and B are increasing functions, then there exists σ ∈ J such that

x ( 0 ) = 1 1 − γ − λ 1 [ p ] [ γ ∫ 0 η x ′ ( s ) d s + ∫ 0 1 ( ∫ 0 t x ′ ( s ) d s ) d A ( t ) ] = 1 1 − γ − λ 1 [ p ] [ γ ∫ 0 η x ′ ( s ) d s + ∫ 0 σ x ′ ( s ) d s ∫ 0 1 d A ( t ) ] .

Hence,

| x ( 0 ) | ≤ 1 | 1 − γ − λ 1 [ p ] | ( γ + | ∫ 0 1 d A ( t ) | ) ∥ x ′ ∥ 1 .

Similarly, we can show that

| x ( 1 ) | ≤ 1 | 1 − ξ − λ 2 [ p ] | ( ξ + | ∫ 0 1 d B ( t ) | ) ∥ x ′ ∥ 1 .

Now, the constant M from Lemma 1 has the form

M = 1 + { 1 | 1 − γ − λ 1 [ p ] | ( γ + | ∫ 0 1 d A ( t ) | ) , in case (i) , 1 | 1 − ξ − λ 2 [ p ] | ( ξ + | ∫ 0 1 d B ( t ) | ) , in case (ii) .

Consider the following problem:

{ u ″ ( t ) + y ( t ) = 0 , t ∈ ( 0 , 1 ) , u ( 0 ) = γ u ( η ) + λ 1 [ u ] , u ( 1 ) = ξ u ( η ) + λ 2 [ u ] , η ∈ ( 0 , 1 ) , γ , ξ ≥ 0 .

Let us introduce the assumption.

H₀: A and B are functions of bounded variation and

δ ≡ 1 − γ + η ( γ − ξ ) ≠ 0 , Δ ≡ A 1 ( B 2 − 1 + ξ η ) + A 2 ( 1 − ξ − B 1 ) + δ − η γ B 1 − ( 1 − γ ) B 2 ≠ 0

for

We require the following result.

Lemma 2 Let the assumption H₀ hold and let y ∈ L 1 ( J , R ) . Then problem (3) has a unique solution given by

u ( t ) = 1 Δ [ 1 − ξ η − B 2 − ( 1 − ξ − B 1 ) t ] λ 1 [ F ¯ y ] + 1 Δ [ η γ + A 2 + ( 1 − γ − A 1 ) t ] λ 2 [ F ¯ y ] + F ¯ y ( t )

with

Proof Integrating the differential equation in (3) two times, we have

u ( t ) = u ( 0 ) + t u ′ ( 0 ) − ∫ 0 t ( t − s ) y ( s ) d s .

Put t = 1 and use the boundary conditions from problem (3) to obtain

ξ u ( η ) + λ 2 [ u ] = γ u ( η ) + λ 1 [ u ] + u ′ ( 0 ) − ∫ 0 1 ( 1 − s ) y ( s ) d s .

Now, finding from this u ′ ( 0 ) and then substituting it to formula (4), we have

u ( t ) = [ γ + t ( ξ − γ ) ] u ( η ) + ( 1 − t ) λ 1 [ u ] + t λ 2 [ u ] + ∫ 0 1 G ( t , s ) y ( s ) d s .

Next, putting t = η , we can find u ( η ) , and then substitute it to formula (5) to obtain

u ( t ) = 1 δ ( [ 1 − ξ η − ( 1 − ξ ) t ] λ 1 [ u ] + [ η γ + ( 1 − γ ) t ] λ 2 [ u ] ) + F ¯ y ( t ) .

Now, we have to eliminate λ 1 [ u ] and λ 2 [ u ] from (6). If u is a solution of (6), then

{ λ 1 [ u ] = 1 δ [ ( 1 − ξ η ) A 1 − ( 1 − ξ ) A 2 ] λ 1 [ u ] + 1 δ [ η γ A 1 + ( 1 − γ ) A 2 ] λ 2 [ u ] + λ 1 [ F ¯ y ] , λ 2 [ u ] = 1 δ [ ( 1 − ξ η ) B 1 − ( 1 − ξ ) B 2 ] λ 1 [ u ] + 1 δ [ η γ B 1 + ( 1 − γ ) B 2 ] λ 2 [ u ] + λ 2 [ F ¯ y ] .

Solving this system with respect to λ1[u], λ2[u] and then substituting to (6), we have the assertion of this lemma. This ends the proof. □

Define the operator T by

T u ( t ) = 1 Δ [ 1 − ξ η − B 2 − ( 1 − ξ − B 1 ) t ] λ 1 [ F u ] + 1 Δ [ η γ + A 2 + ( 1 − γ − A 1 ) t ] λ 2 [ F u ] + F u ( t )

with

F u ( t ) = γ + t ( ξ − γ ) δ ∫ 0 1 G ( η , s ) h ( s ) f ( s , u ( α ( s ) ) , u ′ ( β ( s ) ) ) d s + ∫ 0 1 G ( t , s ) h ( s ) f ( s , u ( α ( s ) ) , u ′ ( β ( s ) ) ) d s .

We consider the Banach space E = ( C 1 ( J , R ) , ∥ ⋅ ∥ ) with the maximum norm ∥ x ∥ = max ( ∥ x ∥ 1 , ∥ x ′ ∥ 1 ) . Define the cone K ⊂ E by

K = { x ∈ E : x ( t ) ≥ 0 , t ∈ J , λ 1 [ x ] ≥ 0 , λ 2 [ x ] ≥ 0 , min [ 0 , η ] x ( t ) ≥ ρ ∥ x ∥ 1 }

with

ρ = min ( γ ( 1 − η ) , 1 − η , η γ 1 + γ ( η − 1 ) ) , γ > 0 .

Let us introduce the following assumption.

H₄: A and B are functions of bounded variation and

(i) δ > 0 , Δ > 0 , A j ≥ 0 , B j ≥ 0 , G j ( s ) ≥ 0 for j = 1 , 2 where A j , B j , G j , δ, Δ are defined as in the assumption H₀,

(ii) γ ( A 1 − A 2 ) + ξ A 2 ≥ 0 , γ ( B 1 − B 2 ) + ξ B 2 ≥ 0 , η γ B 1 + ( 1 − γ ) B 2 ≥ 0 , ( 1 − ξ η ) A 1 − ( 1 − ξ ) A 2 ≥ 0 , B 1 − B 2 ≥ 0 , δ − η γ B 1 − ( 1 − γ ) B 2 ≥ 0 , η γ A 1 + ( 1 − γ ) A 2 ≥ 0 , 1 − ξ η − B 2 ≥ 0 , δ − ( 1 − ξ η ) A 1 + ( 1 − ξ ) A 2 ≥ 0 , ( 1 − ξ η ) B 1 − ( 1 − ξ ) B 2 ≥ 0 .

Lemma 3 Let the assumptions H₁-H₄ hold. Then T : K → K .

Proof Clearly, u ∈ K is a positive solution of problem (1) if and only if u∈K solves the operator equation u = T u . Then

{ λ 1 [ F u ] = 1 δ [ γ ( A 1 − A 2 ) + ξ A 2 ] ∫ 0 1 G ( η , s ) h ( s ) f ( s , u ( α ( s ) ) , u ′ ( β ( s ) ) ) d s λ 1 [ F u ] = + ∫ 0 1 G 1 ( s ) h ( s ) f ( s , u ( α ( s ) ) , u ′ ( β ( s ) ) ) d s , λ 2 [ F u ] = 1 δ [ γ ( B 1 − B 2 ) + ξ B 2 ] ∫ 0 1 G ( η , s ) h ( s ) f ( s , u ( α ( s ) ) , u ′ ( β ( s ) ) ) d s λ 2 [ F u ] = + ∫ 0 1 G 2 ( s ) h ( s ) f ( s , u ( α ( s ) ) , u ′ ( β ( s ) ) ) d s .

Note that λ 1 [ F u ] ≥ 0 , λ 2 [ F u ] ≥ 0 in view of the assumptions H₁, H₂, H₄ and the positivity of Green’s function G.

Note that ( T u ) ″ ≤ 0 . Moreover,

Hence, Tu is concave and T u ( t ) ≥ 0 on J.

We next show that λ 1 [ T u ] ≥ 0 , λ 2 [ T u ] ≥ 0 . Indeed,

Finally, we show that

min [ 0 , η ] T u ( t ) ≥ ρ ∥ T u ∥ 1 .

To do it, we consider two steps. Let ∥ T u ∥ 1 = T u ( t ∗ ) .

Step 1. Let T u ( 0 ) < T u ( η ) . Then t ∗ ∈ ( 0 , η ) or t ∗ ∈ ( η , 1 ) and min [ 0 , η ] T u ( t ) = T u ( 0 ) .

Let t∗∈(0,η). Then

∥ T u ∥ 1 − T u ( 1 ) T u ( η ) − T u ( 1 ) ≤ 1 − t ∗ 1 − η ,

so

∥ T u ∥ 1 ≤ T u ( 1 ) + 1 1 − η [ T u ( η ) − T u ( 1 ) ] < 1 1 − η T u ( η ) = 1 γ ( 1 − η ) ( T u ( 0 ) − λ 1 [ u ] ) ≤ 1 γ ( 1 − η ) T u ( 0 ) .

It yields

min [ 0 , η ] T u ( t ) ≥ γ ( 1 − η ) ∥ T u ∥ 1 .

Let t ∗ ∈ ( η , 1 ) . Then

∥ T u ∥ 1 − T u ( 0 ) T u ( η ) − T u ( 0 ) ≤ t ∗ − 0 η − 0 ,

so

∥ T u ∥ 1 ≤ 1 η [ T u ( η ) + ( η − 1 ) T u ( 0 ) ] = 1 η [ 1 γ ( T u ( 0 ) − λ 1 [ u ] ) + ( η − 1 ) T u ( 0 ) ] .

It yields

min [ 0 , η ] T u ( t ) ≥ γ η 1 + γ ( η − 1 ) ∥ T u ∥ 1 .

Step 2. Let T u ( 0 ) ≥ T u ( η ) . Then t∗∈(0,η) and min [ 0 , η ] T u ( t ) = T u ( η ) . Then

∥ T u ∥ 1 − T u ( 1 ) T u ( η ) − T u ( 1 ) ≤ 1 − t ∗ 1 − η ,

so

∥ T u ∥ 1 ≤ T u ( 1 ) + 1 1 − η [ T u ( η ) − T u ( 1 ) ] < 1 1 − η T u ( η ) .

Hence,

min [ 0 , η ] T u ( t ) ≥ ( 1 − η ) ∥ T u ∥ 1 .

It shows T:K→K. This ends the proof. □

Remark 2 Take d B ( t ) = ( b t − 1 ) d t , b > 1 . Note that the measure changes the sign and is increasing. It is easy to show that

B 1 = 1 2 ( b − 2 ) , B 2 = 1 6 ( 2 b − 3 ) , G 2 ( s ) = s ( 1 − s ) 6 ( b s + b − 3 ) .

If we assume that b ≥ 3 , then B 1 > 0 , B 2 > 0 , G 2 ( s ) ≥ 0 , s∈J.

Remark 3 Take d A ( t ) = ( a t 2 − 1 ) d t , a > 1 . Note that the measure changes the sign and is increasing. It is easy to show that

A 1 = 1 3 ( a − 3 ) , A 2 = 1 4 ( a − 2 ) , G 1 ( s ) = s ( 1 − s ) 12 ( a s 2 + a s + a − 6 ) .

If we assume that a ≥ 6 , then A 1 > 0 , A 2 > 0 , G 1 ( s ) ≥ 0 , s∈J.

Remark 4 Let d A ( t ) = ( 3 t − 1 ) d t , d B ( t ) = ( 7 2 t − 1 ) d t , t∈J. Then the assumptions H₃, H₄ hold if one of the following conditions is satisfied:

(i) ξ=0, 0 < γ < 1 2 ,

(ii) γ=0, 0 < ξ < 1 4 ,

(iii) γ=ξ=0.

We consider only case (i). First of all, we see that dA, dB change the sign and are increasing. Indeed, for p = 1 , t∈J, we have

A 1 = A 2 = λ 1 [ p ] = 1 2 , B 1 = λ 2 [ p ] = 3 4 , B 2 = 2 3 .

It means that the assumption H₃ holds. Moreover,

It proves that the assumption H₄ holds.

By a similar way, we prove the assertion in case (ii) or (iii).

Now, we present the necessary definitions from the theory of cones in Banach spaces.

Definition 1 Let E be a real Banach space. A nonempty convex closed set P ⊂ E is said to be a cone provided that

(i) k u ∈ P for all u ∈ P and all k ≥ 0 , and

(ii) u , − u ∈ P implies u = 0 .

Note that every cone P⊂E induces an ordering in E given by x ≤ y if y − x ∈ P .

Definition 2 A map Φ is said to be a nonnegative continuous concave functional on a cone P of a real Banach space E if Φ : P → R + is continuous and

Φ ( t x + ( 1 − t ) y ) ≥ t Φ ( x ) + ( 1 − t ) Φ ( y )

for all x , y ∈ P and t∈[0,1].

Similarly, we say the map φ is a nonnegative continuous convex functional on a cone P of a real Banach space E if φ : P → R + is continuous and

φ ( t x + ( 1 − t ) y ) ≤ t φ ( x ) + ( 1 − t ) φ ( y )

for all x,y∈P and t∈[0,1].

Definition 3 An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.

Let φ and Θ be nonnegative continuous convex functionals on P, let Φ be a nonnegative continuous concave functional on P, and let Ψ be a nonnegative continuous functional on P. Then, for positive numbers a, b, c, d, we define the following sets:

We will use the following fixed point theorem of Avery and Peterson to establish multiple positive solutions to problem (1).

Theorem 1 (see 1)

Let P be a cone in a real Banach space E. Let φ and Θ be nonnegative continuous convex functionals on P, let Φ be a nonnegative continuous concave functional on P, and let Ψ be a nonnegative continuous functional on P satisfying Ψ ( k x ) ≤ k Ψ ( x ) for 0 ≤ k ≤ 1 such that for some positive numbers M ¯ and d,

Φ ( x ) ≤ Ψ ( x ) and ∥ x ∥ ≤ M ¯ φ ( x )

for all x ∈ P ( φ , d ) ¯ . Suppose

T : P ( φ , d ) ¯ → P ( φ , d ) ¯

is completely continuous and there exist positive numbers a, b, c with a < b such that

(S₁): { x ∈ P ( φ , Θ , Φ , b , c , d ) : Φ ( x ) > b } ≠ 0 and Φ ( T x ) > b for x ∈ P ( φ , Θ , Φ , b , c , d ) ,

(S₂): Φ(Tx)>b for x ∈ P ( φ , Φ , b , d ) with Θ ( T x ) > c ,

(S₃): 0 ∉ R ( φ , Ψ , a , d ) and Ψ ( T x ) < a for x ∈ R ( φ , Ψ , a , d ) with Ψ ( x ) = a .

Then T has at least three fixed points x 1 , x 2 , x 3 ∈ P ( φ , d ) ¯ such that

and

Ψ ( x 3 ) < a .

We apply Theorem 1 with the cone K instead of P and let P ¯ r = { x ∈ K : ∥ x ∥ ≤ r } . Now, we define the nonnegative continuous concave functional Φ on K by

Φ ( x ) = min [ 0 , η ] | x ( t ) | .

Note that Φ ( x ) ≤ ∥ x ∥ 1 . Put Ψ ( x ) = Θ ( x ) = ∥ x ∥ 1 , φ ( x ) = ∥ x ′ ∥ 1 .

Now, we can formulate the main result of this section.

Theorem 2 Let the assumptions H₁-H₄ hold with ξ=0, γ>0. Let α(t)≤t, t∈J. In addition, we assume that there exist positive constants a, b, c, d, M, a<b and such that

with

and

(A₁): f ( t , u , v ) ≤ d μ for ( t , u , v ) ∈ J × [ 0 , M d ] × [ − d , d ] ,

(A₂): f ( t , u , v ) ≥ b L for ( t , u , v ) ∈ [ 0 , η ] × [ b , b ρ ] × [ − d , d ] ,

(A₃): f ( t , u , v ) ≤ a μ for ( t , u , v ) ∈ J × [ 0 , a ] × [ − d , d ] .

Then problem (1) has at least three nonnegative solutions x 1 , x 2 , x 3 satisfying ∥ x i ′ ∥ 1 ≤ d , i = 1 , 2 , 3 ,

b ≤ Φ ( x 1 ) , a < ∥ x 2 ∥ 1 with   Φ ( x 2 ) < b

and ∥ x 3 ∥ 1 < a .

Proof Basing on the definitions of T, we see that T P ¯ is equicontinuous on J, so T is completely continuous.

Let x∈P(φ,d)¯, so φ ( x ) = ∥ x ′ ∥ 1 ≤ d . By Lemma 1, ∥ x ∥ 1 ≤ M d , so 0 ≤ x ( t ) ≤ M d , t∈J. Assumption (A₁) implies f ( t , x ( α ( t ) ) , x ′ ( β ( t ) ) ) ≤ d μ .

Moreover, in view of (7),

Combining it, we have

φ ( T x ) = max [ 0 , 1 ] | ( T x ) ′ ( t ) | ≤ 1 Δ ( 1 − B 1 ) λ 1 [ F x ] + 1 Δ ( 1 − γ − A 1 ) λ 2 [ F x ] + max [ 0 , 1 ] | ( F x ) ′ ( t ) | ≤ d μ ( 1 Δ ( 1 − B 1 ) D 1 + 1 Δ ( 1 − γ − A 1 ) D 2 + D 3 ) < d .

This proves that T : P ( φ , d ) ¯ → P ( φ , d ) ¯ .

Now, we need to show that condition (S₁) is satisfied. Take

x 0 ( t ) = 1 2 ( b + b ρ ) , t ∈ J .

Then x 0 ( t ) > 0 , t∈J, and

for p(t)=1, t∈J. Moreover,

This proves that

{ x 0 ∈ P ( φ , Θ , Φ , b , b ρ , d ) : b < Φ ( x 0 ) } ≠ ∅ .

Let b ≤ x ( t ) ≤ b ρ for t ∈ [ 0 , η ] . Then 0 ≤ α ( t ) ≤ t ≤ η for t∈[0,η], so b ≤ x ( α ( t ) ) ≤ b ρ , t∈[0,η]. Assumption (A₂) implies f ( t , x ( α ( t ) ) , x ′ ( β ( t ) ) ) ≥ b L . Hence,

Moreover,

It yields

Φ ( T x ) = min [ 0 , η ] ( T x ) ( t ) = min ( ( T x ) ( 0 ) , ( T x ) ( η ) ) ≥ γ ( T x ) ( η ) = γ Δ [ 1 − B 2 − ( 1 − B 1 ) η ] λ 1 [ F x ] + γ Δ [ η γ + A 2 + ( 1 − γ − A 1 ) η ] λ 2 [ F x ] + γ δ ∫ 0 1 G ( η , s ) h ( s ) f ( s , x ( α ( s ) ) , x ′ ( β ( s ) ) ) d s ≥ b γ L ( 1 Δ ( [ 1 − B 2 − ( 1 − B 1 ) η ] D 1 + [ η γ + A 2 + ( 1 − γ − A 1 ) η ] D 2 ) + D 4 δ ) > b .

This proves that condition (S₁) holds.

Now, we need to prove that condition (S₂) is satisfied. Take x∈P(φ,Φ,b,d) and ∥ T x ∥ 1 > b ρ = c . Then

Φ ( T x ) = min [ 0 , η ] ( T x ) ( t ) ≥ ρ ∥ T x ∥ 1 > ρ b ρ = b ,

so condition (S₂) holds.

Indeed, φ ( 0 ) = 0 < a , so 0∉R(φ,Ψ,a,d). Suppose that x∈R(φ,Ψ,a,d) with Ψ ( x ) = ∥ x ∥ 1 = a . Note that G ( t , s ) ≤ G ( s , s ) , t∈J. Then

∥ F x ∥ 1 ≤ γ δ ∫ 0 1 G ( η , s ) h ( s ) f ( s , x ( α ( s ) ) , x ′ ( β ( s ) ) ) d s + ∫ 0 1 G ( s , s ) h ( s ) f ( s , x ( α ( s ) ) , x ′ ( β ( s ) ) ) d s ≤ a μ [ γ δ ∫ 0 1 G ( η , s ) h ( s ) d s + ∫ 0 1 G ( s , s ) h ( s ) d s ] ≤ a μ D 5

and finally,

Ψ ( T x ) = max t ∈ J ( T x ) ( t ) ≤ 1 Δ [ 1 − B 2 ] λ 1 [ F x ] + 1 Δ [ η γ + A 2 + 1 − γ − A 1 ] λ 2 [ F x ] + ∥ F x ∥ 1 ≤ a μ ( 1 Δ ( [ 1 − B 2 ] D 1 + [ η γ + A 2 + 1 − γ − A 1 ] D 2 ) + D 5 ) < a .

This shows that condition (S₃) is satisfied.

Since all the conditions of Theorem 1 are satisfied, problem (1) has at least three nonnegative solutions x1, x2, x3 such that ∥ x i ′ ∥ ≤ d for i=1,2,3, and

b ≤ min [ 0 , η ] x 1 ( t ) , a < ∥ x 2 ∥ 1 with  min [ 0 , η ] x 2 ( t ) < b , ∥ x 3 ∥ 1 < a .

This ends the proof. □

Example Consider the following problem:

{ x ″ ( t ) + h f ( t , x ( α ( t ) ) , x ′ ( β ( t ) ) ) = 0 , t ∈ ( 0 , 1 ) , x ( 0 ) = 1 4 x ( 1 2 ) , x ( 1 ) = 1 2 ∫ 0 1 x ( t ) ( 7 t − 2 ) d t ,

where

f ( t , u , v ) = { 1 100 cos t + ( v 20 , 000 ) 2 , ( t , u , v ) ∈ [ 0 , 1 ] × [ 0 , 1 ] × [ − d , d ] , 1 100 cos t + 2 ( u − 1 ) + ( v 20 , 000 ) 2 , ( t , u , v ) ∈ [ 0 , 1 ] × [ 1 , 16 ] × [ − d , d ] , 1 100 cos t + 30 + ( v 20 , 000 ) 2 , ( t , v ) ∈ [ 0 , 1 ] × [ − d , d ] , u ≥ 16

with d = 2 , 000 . For example, we can take α ( t ) = ρ ¯ t , β ( t ) = t on J with fixed ρ ¯ ∈ ( 0 , 1 ) . Indeed, f ∈ C ( [ 0 , 1 ] × R + × [ − d , d ] , R + ) , γ = 1 4 , η = 1 2 , h ( t ) = h > 0 , ξ=0 and

λ 1 [ x ] = 0 , λ 2 [ x ] = 1 2 ∫ 0 1 x ( t ) ( 7 t − 2 ) d t , ρ = 1 8 .

Note that d B ( t ) = 1 2 ( 7 t − 2 ) d t , so the measure changes the sign on J. Moreover,

so the assumption H₄ holds; see Remark 4. Next,

Put a = 1 , b = 2 , h = 30 , then c = 16 , μ > 37.18 , L < 1.94 . Let μ = 40 , L = 1 . Then

and

f ( t , u , v ) ≤ 1 100 + 30 + ( 2 , 000 20 , 000 ) 2 = 30.02 < 50 = d μ