Existence of positive solutions to third order differential equations with advanced arguments and nonlocal boundary conditions

Nonlinear Analysis 75 (2012) 913–923

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Existence of positive solutions to third order differential equations with advanced arguments and nonlocal boundary conditions Tadeusz Jankowski ∗ Gdansk University of Technology, Department of Differential Equations and Applied Mathematics, 11/12 G.Narutowicz Str., 80–233 Gdańsk, Poland

article

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Article history: Received 18 June 2011 Accepted 16 September 2011 Communicated by Enzo Mitidieri Keywords: Third order differential equations Nonlocal boundary value problems Advanced arguments Fixed point theorem The existence of multiple positive solutions

abstract We use a fixed point theorem due to Avery and Peterson to establish the existence of at least three non-negative solutions of some nonlocal boundary value problems to third order differential equations with advanced arguments. An example is given to illustrate the main results. © 2011 Elsevier Ltd. All rights reserved.

1. Introduction Put J = [0, 1]. In this paper, we are interested in the existence of multiple positive solutions to boundary value problems of type: x′′′ (t ) + h(t )f (t , x(α(t ))) = 0, t ∈ (0, 1), x(0) = x′′ (0) = 0, x(1) = β x(η) + λ[x], β > 0, η ∈ (0, 1),



(1)

where λ denotes a linear functional on C (J ) given by

λ[x] =

∫

1

x(t ) dΛ(t )

0

involving a Stieltjes integral with a suitable function Λ of bounded variation. It is important to indicate that it is not assumed that λ[x] is positive to all positive x. The measure dΛ can be a signed measure (see Remark 3 and Example). Let us introduce the following assumptions:

(H1 ) f ∈ C (J × R+ , R+ ), R+ = [0, ∞), α ∈ C (J , J ) and t ≤ α(t ) on J , (H2 ) h is a nonnegative continuous function defined on J; h is not identically zero on any subinterval on J , (H3 ) 0 < βη + λ[p] < 1 for p(t ) = t . There are a lot of papers concerning the problem of existence of positive solutions to boundary value problems. Yet, only in a few papers has the problem of third order differential equations been considered, see for example [1–8]. Third order differential equations have been discussed under the following Boundary Conditions (BCs):

∗

Fax: +48 58 3472821. E-mail address: [email protected].

0362-546X/$ – see front matter © 2011 Elsevier Ltd. All rights reserved. doi:10.1016/j.na.2011.09.025

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T. Jankowski / Nonlinear Analysis 75 (2012) 913–923

x(0) = x′ (0) = 0,

ax′ (1) + bx′′ (1),

a, b ≥ 0, a + b > 0,

x(0) = x (0) = x (1) = 0, ′

′′

x(0) = x(1) = x′′ (1) = 0, x(0) = x′ (0) = x′ (1) = 0, 1 ≤ η < 1, 2 see papers [1,4,5,8,6], respectively. The last BC has also been considered in paper [7]. In paper [2], differential equations have been discussed under BCs: x(0) = x′ (η) = x′′ (1) = 0,

x(0) = x′ (p) =

1

∫

w(s)x′′ (s) ds = 0, q

1 2

< p < q,

where w ∈ C [q, 1] → [0, ∞), nondecreasing and w(t ) > 0, t > q, while in paper [3], under more general BCs of the form: x(0) = λ1 [x],

x′ (p) = 0,

x′′ (1) + λ2 [x] = λ3 [x′′ ]

all of λ1 , λ2 , λ3 are signed measures, with some conditions. The BC in paper [2] does not require the sign restriction on w . In all the above mentioned papers, the authors used Krasnoselskii’s fixed point theorem of cone or fixed point index theory involving eigenvalues for problems with no advanced arguments. Note that the BC with functional λ in problem (1) covers some nonlocal BCs, for example:

λ[x] =

r −

βi x(γi ),

0 < γ1 < γ2 < · · · < γr < 1,

i =1

λ[x] =

1

∫

g (t )x(t ) dt . 0

In this paper, the assumption that the measure dΛ, in the definition of λ, is positive is not needed. More precisely, one needs to choose constants βi and function g in such a way that Assumption (H4 ) holds. It means that g can change the sign on J (see Remark 3 and Example). The situation with a signed measure dΛ has been first discussed in papers [9,10] for second order differential equations; it has also been discussed in papers [11,12] for second order impulsive differential equations. In paper [13], the authors have studied second order differential equations with the BC of the form x(1) =

 b¯ a¯

x(t ) dΛ(t ),

where 0 < a¯ < b¯ < 1 and Λ : [¯a, b¯ ] → R is increasing. Moreover, by using the fixed point index, in paper [14], the authors have obtained the results for existence of nonzero solutions for second order impulsive differential equations with the nonlocal BCs: x′ (0) + λ[x] = 0, ax′ (1) + x(η) = 0, for corresponding η. Motivated by [9,10,15], in this paper, we apply the Avery–Peterson fixed point theorem to obtain sufficient conditions of the existence of multiple (at least three) solutions to problems of type (1). Note that, in problem (1), an unknown x depends on an advanced argument α . The approach considered in [9,10,15] to study positive solutions of problems such as (1) is to find a corresponding Green’s function k to replace problem (1) by an integral equation with the integral operator Ax(t ) =

t

λ[u] +

βt ∆

∫

1

k(η, s)h(s)f (s, x(α(s))) ds +

∫

1

k(t , s)h(s)f (s, x(α(s))) ds ∆ 0 0 with ∆ ≡ 1 − βη > 0, and then to seek fixed points of operator A in a corresponding cone. Usually, we need to find a nonnegative function κ and a constant c ∈ (0, 1] such that k(t , s) ≤ κ(s) for t , s ∈ J ; and κ(t , s) ≥ c κ(t ) for t ∈ [η, η] ¯ ⊂ [0, 1] and s ∈ J (see for example [9,10,15]). Such an approach cannot be applied to problems with advanced arguments, so we do a little modification of this approach to apply the fixed point theorem to problems of type (1). Note that this paper is the first one when a fixed point theorem is applied to third order differential equations with advanced arguments. An example is given to illustrate the main results. In this example, we discuss assumptions of Theorem 2 for three fixed cases of the linear functional λ. 2. Some lemmas Consider the following boundary value problem u′′′ (t ) + y(t ) = 0, t ∈ (0, 1), u(0) = u′′ (0) = 0, u(1) = β u(η) + λ[u], β > 0, η ∈ (0, 1).



(2)

We require the following Lemma 1. Assume that βη ̸= 1 and y ∈ C (J , R). Then, problem (2) has the unique solution given by the following formula u(t ) =

t 1 − βη

λ[u] +

βt 1 − βη

1

∫

k(η, s)y(s) ds + 0

1

∫

k(t , s)y(s) ds, 0

t ∈ J,

(3)

T. Jankowski / Nonlinear Analysis 75 (2012) 913–923

915

where

(1 − t )(t − s2 ), if s ≤ t , 2 2 t (1 − s) , if t ≤ s. 

1

k(t , s) =

Proof. Integrating the differential equation in (2) three times from 0 to t and using the boundary conditions u(0) = u′′ (0) = 0, we have u(t ) = tu′ (0) −

1 2

t

∫

(t − s)2 y(s) ds.

(4)

0

Using formula (4) and condition u(1) = β u(η) + λ[u], we see that u′ (0) −

1

1

∫

2

(1 − s)2 y(s) ds = β u(η) + λ[u]. 0

Finding from this u′ (0) and substituting it into formula (4) we see that u(t ) = t (β u(η) + λ[u]) +

1

∫

k(t , s)y(s) ds.

(5)

0

Now, putting t = η, we see that u(η) =

1 1 − βη



ηλ[u] +

1

∫



k(η, s)y(s) ds . 0

Substituting it into formula (5) we finally get the assertion of this lemma.



Remark 1. It is easy to show that k(t , s) ≤

1 2

(1 + s)(1 − s)2 ,

t , s ∈ [0, 1].

Put

 K =



u ∈ C (J , R+ ) : min u(t ) ≥ Γ ‖u‖, λ[u] ≥ 0 , [η,1]

where ‖u‖ means the maximum norm on J and

 β(1 − η) , βη, η . Γ = min 1 − βη 

Note that such a cone was introduced in paper [9], see also [10,15]. Indeed, K is a cone of nonnegative continuous functions on J . For u ∈ C (J , R+ ), we define two operators T and S by t Tu(t ) = λ[u] + Fu(t )

∆

and Su(t ) =

t

λ[Fu] + Fu(t ), ∆−ρ where ∆ = 1 − βη, ρ = λ[p] for p(t ) = t, and ∫ ∫ 1 βt 1 Fu(t ) = k(η, s)h(s)f (s, u(α(s))) ds + k(t , s)h(s)f (s, u(α(s))) ds. ∆ 0 0 Lemma 2. Let Assumptions (H1 )–(H3 ) hold. Moreover, we assume that Assumption (H4 ) holds with (H4 ) Λ is of bounded variation and 1

∫

dΛ(t ) ≥ 0, 0

∫

1

t dΛ(t ) ≥ 0,

K (s) =

0

1

∫

k(t , s) dΛ(t ) ≥ 0. 0

Then T : K → K , S : K → K . Proof. Indeed, T , S : C (J , R) → C (J , R). Problem (1) has a solution u if and only if u solves the operator equation u = Tu. Then

β λ[u] = λ[p]λ[u] + λ[p] 1 − βη 1 − βη 1

1

∫

k(η, s)h(s)f (s, u(α(s))) ds + 0

∫

1

K (s)h(s)f (s, u(α(s))) ds.

0

This, Assumptions (H1 )–(H4 ) and the positivity of the Green’s function k proves that λ[u] ≥ 0.

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T. Jankowski / Nonlinear Analysis 75 (2012) 913–923

Let u ∈ K . Note that (Tu)′′ ≤ 0. It shows that Tu is concave down. Moreover, we observe that Fu(0) = 0, Fu(1) = Fu(η) =

β ∆

∫

1

∫

1

k(η, s)h(s)f (s, u(α(s))) ds ≥ 0, 0

∆

1

k(η, s)h(s)f (s, u(α(s))) ds ≥ 0, 0

so

(Tu)(0) = Fu(0) = 0, 1

λ[u] + Fu(1) ≥ 0, ∆ η (Tu)(η) = λ[u] + Fu(η) ≥ 0. ∆ It proves that Tu(t ) ≥ 0, t ∈ J. (Tu)(1) =

Next, we have to prove: min[η,1] Tu(t ) ≥ Γ ‖Tu‖. To do it we consider two steps. Step 1. Let Tu(η) ≤ Tu(1). Then ‖Tu‖ = Tu(t¯), t¯ ∈ (η, 1) and min[η,1] Tu(t ) = Tu(η). Moreover, t¯ − 0 ‖Tu‖ − Tu(0) ≤ . Tu(η) − Tu(0) η−0 This yields min Tu(t ) ≥ η‖Tu‖. [η,1]

Step 2. Let Tu(η) > Tu(1) and ‖Tu‖ = Tu(t¯). If t¯ ∈ (0, η), then

‖Tu‖ − Tu(1)

1 − t¯

≤

Tu(η) − Tu(1)

1−η

,

so

] 1 − βη ‖Tu‖ ≤ Tu(1) + (Tu(1) − λ[u]) − Tu(1) ≤ Tu(1) . 1−η β β(1 − η) [

1

1

This shows min Tu(t ) ≥ [η,1]

β(1 − η) ‖Tu‖. 1 − βη

If t¯ ∈ (η, 1), then t¯ − η ‖Tu‖ − Tu(η) ≤ , Tu(η) − Tu(0) η−0 so

‖Tu‖ ≤

1

η

Tu(η) =

1

βη

(Tu(1) − λ[u]) ≤

1

βη

Tu(1).

It gives min Tu(t ) ≥ βη‖Tu‖. [η,1]

Combining it, we finally obtain min Tu(t ) ≥ Γ ‖Tu‖. [η,1]

Now, we need to show that λ[Tu] ≥ 0. Note that

λ[Fu] =

1

∫



0

βρ ∆ ≥ 0.

βt ∆

1

∫

k(η, s)h(s)f (s, u(α(s))) ds 0

1

∫

k(η, s)h(s)f (s, u(α(s))) ds +

=

0



dΛ(t ) +

1

∫ 0

∫ 0

1

∫

k(t , s)h(s)f (s, u(α(s))) ds 0

1

K (s)h(s)f (s, u(α(s))) ds



dΛ(t )

T. Jankowski / Nonlinear Analysis 75 (2012) 913–923

917

So

λ[Tu] =

1

∆

λ[u]

1

∫

t dΛ(t ) + λ[Fu] ≥ 0. 0

This shows that T : K → K . Now we consider the operator S. Note that (Su)′′ ≤ 0 and Su(0) = 0, 1

Su(1) =

λ[Fu] + Fu(1) ≥ 0, ∆−ρ η Su(η) = λ[Fu] + Fu(η) ≥ 0. ∆−ρ Moreover,

λ[Su] =

∆ ρ λ[Fu] + λ[Fu] = λ[Fu] ≥ 0. ∆−ρ ∆−ρ

Similarly as for operator T , this shows S : K → K . The proof is complete.



Remark 2. Note that the inequalities in Assumption (H4 ) are trivially satisfied when dΛ is a positive measure. Remark 3. Take dΛ(t ) = (at − 1) dt , a ≥ 2. Note that the measure changes sign. Then Assumption (H4 ) holds. Indeed, we have 1

∫

dΛ(t ) =

a−2 2

0 1

∫

t dΛ(t ) =

2a − 3 6

0

K (s) =

1 24

≥ 0, > 0,

(1 − s)2 [as2 + (a − 2)(2s + 1)] > 0.

If a = 2, then 1

∫

1

∫

dΛ(t ) = 0, 0

t dΛ(t ) = 0

1 6

,

K (s) =

1 12

s2 (1 − s)2 .

The next lemma is similar to Lemma 2.5 [10]. Lemma 3. Let Assumptions (H1 )–(H4 ) hold. Then operators T and S have the same fixed points in K . Proof. Let u = Su. Then

λ[u] =

∆ ρ λ[Fu] + λ[Fu] = λ[Fu], ∆−ρ ∆−ρ

so u(t ) = Su(t ) =

=

t

∆

t

∆−ρ

λ[Fu] + Fu(t )

λ[u] + Fu(t ) = Tu(t ).

Let u = Tu. Then

λ[u] =

ρ λ[u] + λ[Fu], ∆

λ[u] =

∆ λ[Fu] ∆−ρ

so

and hence u(t ) = Tu(t ) =

=

t

∆−ρ

This ends the proof.

t

∆

λ[u] + Fu(t )

λ[Fu] + Fu(t ) = Su(t ). 

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T. Jankowski / Nonlinear Analysis 75 (2012) 913–923

3. Existence of positive solutions of problem (1) 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) ku ∈ 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 (tx + (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(tx + (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: P (ϕ, d) = {x ∈ P : ϕ(x) < d}, P (ϕ, Φ , b, d) = {x ∈ P : b ≤ Φ (x), ϕ(x) ≤ d}, P (ϕ, Θ , Φ , b, c , d) = {x ∈ P : b ≤ Φ (x), Θ (x) ≤ c , ϕ(x) ≤ d}, R(ϕ, Ψ , a, d) = {x ∈ P : a ≤ Ψ (x), ϕ(x) ≤ d}. We will use the following fixed point theorem of Avery and Peterson to establish multiple positive solutions to problem (1). Theorem 1 (See [16]). 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 (kx) ≤ kΨ (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 (Tx) > b for x ∈ P (ϕ, Θ , Φ , b, c , d); (S2 ) Φ (Tx) > b for x ∈ P (ϕ, Φ , b, d) with Θ (Tx) > c , (S3 ) 0 ̸∈ R(ϕ, Ψ , a, d) and Ψ (Tx) < a for x ∈ R(ϕ, Ψ , a, d) with Ψ (x) = a. Then, T has at least three fixed points x1 , x2 , x3 ∈ P (ϕ, d), such that

ϕ(xi ) ≤ d, for i = 1, 2, 3, b < Φ (x1 ), a < Ψ (x2 ),

with Φ (x2 ) < b

and

Ψ (x3 ) < a. We apply Theorem 1 with the cone K instead of P and let P¯ r = {x ∈ K : ‖x‖ ≤ r }. Now, we define the non-negative continuous concave functional Φ on K by

Φ (x) = min |x(t )|. [η,1]

Note that Φ (x) ≤ ‖x‖. Put Ψ (x) = Θ (x) = ϕ(x) = ‖x‖.

T. Jankowski / Nonlinear Analysis 75 (2012) 913–923

919

Theorem 2. Let Assumptions (H1 )–(H4 ) hold. In addition, we assume that there exist positive constants a, b, c , d, a < b and such that

µ>

1

D2 + D1 ,

∆−ρ

η βη + ∆ 0 < L < min(1, β) ∆−ρ ∆ 

1

∫

k(η, s)h(s) ds + D2



0

with

∫ 1 ∫ β 1 1 l(s)h(s) ds with l(s) = (1 + s)(1 − s)2 , k(η, s)h(s) ds + ∆ 0 2 0 ∫ 1 ∫ βρ 1 K (s)h(s) ds D2 = k(η, s)h(s) ds + ∆ 0 0 D1 =

and

(A1 ) f (t , u) ≤ (A2 ) f (t , u) ≥ (A3 ) f (t , u) ≤

d

µ

for (t , u) ∈ J × [0, d],

for (t , u) ∈ [η, 1] × b, for (t , u) ∈ J × [0, a]. µ

b L a



b

Γ



,

Then, problem (1) has at least three non-negative solutions x1 , x2 , x3 satisfying ‖xi ‖ ≤ d, i = 1, 2, 3, b ≤ Λ(x1 ),

a < ‖x2 ‖ with Λ(x2 ) < b

and ‖x3 ‖ < a. Proof. Based on the definitions of T and S, we see that T P¯ and S P¯ are equicontinuous on J, so T and S are completely continuous. Because operators T and S have the same fixed points (see Lemma 3), to prove this theorem we use operator S instead of T . Let x ∈ P (ϕ, d), so 0 ≤ x(t ) ≤ d, t ∈ J, and ‖x‖ ≤ d. By Assumption (A1 ), we see that

‖Fx‖ = max |Fx(t )| t ∈J

] ∫ ∫ 1 βt 1 = max k(η, s)h(s)f (s, x(α(s))) ds + k(t , s)h(s)f (s, x(α(s))) ds t ∈J ∆ 0 0 ∫ ∫ 1 β 1 k(η, s)h(s)f (s, x(α(s))) ds + l(s)h(s)f (s, x(α(s))) ds ≤ ∆ 0 0 [ ∫ 1 ] ∫ 1 β d ≤ k(η, s)h(s) ds + l(s)h(s) ds ∆ 0 µ 0 [

≤ λ[Fx] =

D1

d,

µ ∫ 1

Fx(t ) dΛ(t ) =

0 1

∫

1

∫

t dΛ(t ) 0

1

∫

k(η, s)h(s)f (s, x(α(s))) ds 0

1

∫

k(t , s)h(s)f (s, x(α(s))) ds

+ 0

β ∆



dΛ(t )

0

∫ ∫ 1 βρ 1 k(η, s)h(s)f (s, x(α(s))) ds + K (s)h(s)f (s, x(α(s))) ds ∆ 0 0 [ ] ∫ ∫ 1 βρ 1 d ≤ k(η, s)h(s) ds + K (s)h(s) ds ∆ 0 µ 0 =

≤

D2

µ

d.

Combining it we have

ϕ(Sx) = ‖Sx‖ = max |(Sx)(t )| = max(Sx)(t ) t ∈J t ∈J   t = max λ[Fx] + Fx(t ) t ∈J ∆−ρ

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T. Jankowski / Nonlinear Analysis 75 (2012) 913–923

1

λ[Fx] + ‖Fx‖ ] d 1 ≤ D2 + D1 < d. µ ∆−ρ ≤

∆−ρ [

This proves that S : P (ϕ, d) → P (ϕ, d). Now we need to show that condition (S1 ) is satisfied. Take 1

x(t ) =

 b+

2

b



Γ

,

t ∈ J.

Then b(Γ + 1)

‖x ‖ =

2Γ

b

<

Γ

= c so Φ (x) = min x(t ) =

b(Γ + 1) 2Γ

[η,1]

>b=

b

Γ

Γ > Γ ‖x‖.

Moreover λ[x] > 0. This proves that



 x∈P

ϕ, Θ , Φ , b,

Let b ≤ x(t ) ≤

b

Γ



 , d : b < Φ (x) ̸= ∅.

for t ∈ [η, 1]. Then, η ≤ t ≤ α(t ) ≤ 1 on [η, 1]. It yields b ≤ x(α(t )) ≤

b

Γ

b

on [η, 1]. Note that

b

= c. Then

Γ

min(Sx)(t ) = min((Sx)(η), (Sx)(1)) = min [(Sx)(η), β(Sx)(η) + λ[x]] [η,1]

≥ min(1, β)(Sx)(η). Moreover,

βρ λ[Fx] = ∆[

1

∫

k(η, s)h(s)f (s, x(α(s))) ds + 0

βρ ≥ L ∆ b

1

∫

k(η, s)h(s) ds + 0

K (s)h(s)f (s, x(α(s))) ds 0

1

∫

1

∫

]

K (s)h(s) ds = 0

b L

D2 ,

∫ βη + ∆ 1 k(η, s)h(s)f (s, x(α(s))) ds ∆ 0 ∫ 1 b βη + ∆ ≥ k(η, s)h(s) ds. L ∆ 0

Fx(η) =

It yields

 η Φ (Sx) = min(Sx)(t ) ≥ min(1, β) λ[Fx] + Fx(η) [η,1] ∆−ρ   ∫ 1 η βη + ∆ b ≥ min(1, β) k(η, s)h(s) ds + D2 ∆−ρ ∆ L 0 > 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 ‖Sx‖ >

Φ (Sx) = min(Sx)(t ) ≥ Γ ‖Sx‖ > Γ [η,1]

b

Γ

Γ

= 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‖ = a. Then

Ψ (Sx) = ‖Sx‖ = max |(Sx)(t )| = max(Sx)(t ) t ∈J t ∈J   t = max λ[Fx] + Fx(t ) t ∈J ∆−ρ 1

λ[Fx] + ‖Fx‖ ] a 1 ≤ D2 + D1 < a. µ ∆−ρ ≤

∆−ρ [

This shows that condition (S3 ) is satisfied.

T. Jankowski / Nonlinear Analysis 75 (2012) 913–923

921

Since all the conditions of Theorem 1 are satisfied, problem (1) has at least three non-negative solutions x1 , x2 , x3 such that ‖xi ‖ ≤ d for i = 1, 2, 3, and b ≤ min x1 (t ),

a < ‖x 2 ‖

[η,1]

This ends the proof.

with min x2 (t ) < b,

‖ x 3 ‖ < a.

[η,1]



Example. We consider the following example

 ′′′ x (t ) + Bf (x(α(t ))) = 0,     x(0) = x′′ (0) = 0,     1 1  x(1) = x + λ[x], 2

t ∈ (0, 1), (6)

2

where B is a positive constant, α ∈ C (J , J ), t ≤ α(t ) on J, and

f ( u) =

 u2   ,  10      1     (78u − 38),

0≤u≤ 1

40

2

u,      1   (u + 380),     96 5,

1

,

2

≤ u ≤ 1,

1 ≤ u ≤ 4, 4 ≤ u ≤ 100, u ≥ 100.

As a function α , we can take, for example. α(t ) = Moreover 1

∫

k(η, s)h(s) ds = 0

B 16

1

∫

,

√

l(s)h(s) ds = 0

t. Note that f ∈ C (R+ , R+ ), h(t ) = B, β = η = 5B 24

,

so D1 =

B 4

1 , 2

so ∆ =

3 4

, Γ = 41 .

.

1. Let λ[x] = 0. In this case, we have:

ρ = 0,

K (s) = 0,

D2 = 0,

B

µ>

4

,

L<

B 36

.

For example, if we take a = 12 , b = 1, c = 4, d = 100 and µ = 20, B = 72, L = 1, then all assumptions of Theorem 2 hold, so problem (6) has at least two positive solutions and zero solution too. 2. Now, let λ[x] =

ρ=

1 6

,

1 0

x(t )(2t − 1) dt. Note that the function 2t − 1 changes sign on the interval [0, 1]. Then,

K (s) =

1 12

s2 (1 − s)2 ,

D2 =

7 720

B,

µ>

4 15

B,

L<

409 10080

B.

Similarly as in point 1, if we take a = 21 , b = 1, c = 4, d = 100 and µ = 20, B = 72, L = 1, then all assumptions of Theorem 2 hold, so problem (6) has at least two positive solutions and a zero solution too. 3. Now, we discuss problem (6) for λ[x] = γ x(δ), γ ̸= β, δ ̸= η, δ ∈ (0, 1), 0 ≤ γ δ < 43 . Then,

ρ = γ δ, µ>

K (s) = γ k(δ, s), B

12(3 − 4γ δ)

[9 − 2γ δ − 8γ δ 3 ],

Assume that γ = 2, δ =

γδ =

1 2

,

D2 =

D2 =

1 . 4

19B 192

,

Bγ δ 24

(5 − 4δ 2 ),

L<

B 48(3 − 4γ δ)

[4 + γ δ(3 − 4γ δ)(5 − 4δ 2 )].

Then

µ>

31B 48

,

L<

51B 384

.

If we take B = 10, then, as in points 1 and 2, we can choose a = 12 , b = 1, c = 4, d = 100, µ = 20, L = 1 which guarantee that all assumptions of Theorem 2 hold, so problem (6) has at least two positive solutions and a zero solution too.

922

T. Jankowski / Nonlinear Analysis 75 (2012) 913–923

4. Some comments Assume that α(t ) ≡ t, so problem (1) has no advanced argument. Remark 1 gives the upper bound for the Green function k. It is also true that 1 k(t , s) ≤ Φ0 (s) ≡ (1 − s2 )2 , t , s ∈ [0, 1]. 8 Now, using the function Φ0 , we can find the lower bounds for k when t ∈ [η, ξ ], η, ξ ∈ (0, 1), η < ξ , s ∈ [0, 1]. Indeed, 1

k(t , s) =

2

(1 − t )(t − s2 ) ≥

1 2

t (1 − t )2 ≥

1 2

η(1 − ξ )2 ≥ η(1 − ξ )2 Φ0 (s)

if s ≤ t, and 1 1 t (1 − s)2 ≥ η(1 − s)2 ≥ ηΦ0 (s) 2 2 if t ≤ s. It shows that k(t , s) =

k(t , s) ≥ min[η, η(1 − ξ )2 ]Φ0 (s) = η(1 − ξ )2 Φ0 (s),

t ∈ [η, ξ ], s ∈ [0, 1].

In this case, we can discuss problem (1) with more general boundary conditions of the form:



x′′′ (t ) + h(t )f (t , x(t )) = 0, t ∈ (0, 1), x(0) = x′′ (0) = 0, x(1) = λ[x].

(7)

The cone K now has the form

 K =



u ∈ C (J , R+ ) : min u(t ) ≥ η(1 − ξ ) ‖u‖, λ[u] ≥ 0 . 2

[η,ξ ]

We see that for problem (7), the operators T and S have the form: Tu(t ) = t λ[u] + Fu(t ), t Su(t ) = λ[Fu] + Fu(t ), 1 − λ[p]

0 ≤ λ[p] < 1 for p(t ) = t ,

with Fu(t ) =

1

∫

k(t , s)h(s)f (s, u(s)) ds. 0

Now, replacing [η, 1] and Γ by [η, ξ ] and η(1 − ξ )2 , respectively, the assertion of Theorem 2 holds true if

µ>

1 1 − λ[p]

D2 + D1 ,

0 < L < D2

with 1

∫

Φ0 (s)h(s) ds with Φ0 (s) =

D1 = 0

1 8

(1 − s2 )2 ,

1

∫

K (s)h(s) ds.

D2 = 0

References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12]

M. El-Shahed, Positive solutions for nonlinear singular third order boundary value problem, Commun. Nonlinear Sci. Numer. Simul. 14 (2009) 424–429. J.R. Graef, L. Kong, Positive solutions for third order semipositone boundary value problems, Appl. Math. Lett. 22 (2009) 1154–1160. J.R. Graef, J.R.L. Webb, Third order boundary value problems with nonlocal boundary conditions, Nonlinear Anal. 71 (2009) 1542–1551. S. Li, Positive solutions of nonlinear singular third-order two-point boundary value problem, J. Math. Anal. Appl. 323 (2006) 413–425. Z. Liu, J.S. Ume, S.M. Kang, Positive solutions of a singular nonlinear third order two-point boundary value problems, J. Math. Anal. Appl. 326 (2007) 589–601. Y. Sun, Positive solutions of singular third-order three-point boundary value problem, J. Math. Anal. Appl. 306 (2005) 589–603. Q. Yao, The existence and multiplicity of positive solutions for a third-order three-point boundary value problem, Acta Math. Appl. Sin. 19 (2003) 117–122. Q. Yao, Y. Feng, The existence of solution for a third-order two-point boundary value problem, Appl. Math. Lett. 15 (2002) 227–232. J.R.L. Webb, G. Infante, Positive solutions of nonlocal boundary value problems: a unified approach, J. Lond. Math. Soc. 74 (2006) 673–693. J.R.L. Webb, G. Infante, Positive solutions of nonlocal boundary value problems involving integral conditions, NoDEA Nonlinear Differential Equations Appl. 15 (2008) 45–67. G. Infante, P. Pietramala, M. Zima, Positive solutions for a class of nonlocal impulsive BVPs via fixed point index, Topol. Methods Nonlinear Anal. 36 (2010) 263–284. T. Jankowski, Positive solutions for second order impulsive differential equations involving Stieltjes integral conditions, Nonlinear Anal. 74 (2011) 3775–3785.

T. Jankowski / Nonlinear Analysis 75 (2012) 913–923

923

[13] G.L. Karakostos, P.Ch. Tsamatos, Existence of multipoint positive solutions for a nonlocal boundary value problem, Topol. Methods Nonlinear Anal. 19 (2002) 109–121. [14] G. Infante, P. Pietramala, Nonlocal impulsive boundary value problems with solutions that change sign, in: A. Cabada, E. Liz, J.J. Nieto (Eds.), Mathematical Models in Engineering, Biology and Medicine, Proceedings of the International Conference on Boundary Value Problems, 2009, pp. 205–213. CP1124. [15] J.R.L. Webb, G. Infante, Non-local boundary value problems of arbitrary order, J. Lond. Math. Soc. 79 (2009) 238–259. [16] R.I. Avery, A.C. Peterson, Three positive fixed points of nonlinear operators on ordered Banach spaces, Comput. Math. Appl. 42 (2001) 313–322.