Positive Solutions Of Nonlinear Fourth-Order Two-Point Boundary

ROMANIAN JOURNAL OF MATHEMATICS AND COMPUTER SCIENCE, 2018, VOLUME 8, ISSUE 1, p.17-30

POSITIVE SOLUTIONS OF NONLINEAR FOURTH-ORDER TWO-POINT BOUNDARY VALUE PROBLEM WITH A PARAMETER NOUREDDINE BOUTERAA, SLIMANE BENAICHA, HABIB DJOURDEM AND MOHAMED ELARBI BENATTIA Abstract. In this paper, we study the existence and nonexistence of positive solutions of elastic beam equations with a parameter λ for fourth-order two-point boundary value problem u(4) (t) = λf (t, u (t)) , t ∈ (0, 1) , u (0) = u0 (0) = u0 (1) = u000 (1) + ψ (u (1)) = 0, where λ is a positive parameter. By using Krasnoselskii’s fixed point theorem of cone expansion-compression type we show that there exist λ∗ ≥ λ∗ > 0 such that the beam equation has at least two, one and no positive solutions for 0 < λ ≤ λ∗ , λ∗ < λ ≤ λ∗ and λ > λ∗ respectively. Mathematics Subject Classification (2010): 34B18; 34B15. Key words: Elastic beam equation; Nonexistence; Positive solutions; Green’s function; Fixed-point theorem of cone expansion and compression type

Article history: Received 08 January 2017 Received in revised form 02 November 2017 Accepted 21 November 2017 1. Introduction In this paper, we are interested in the existence of positive solutions to nonlinear fourth-order two-point boundary value problem (BVP) with a parameter: (1.1) (1.2)

u(4) (t) = λf (t, u (t)) ,

t ∈ (0, 1) ,

u (0) = u0 (0) = u0 (1) = u000 (1) + ψ (u (1)) = 0,

where λ ≥ 0 is a parameter, f ∈ C ([0, 1] × [0, ∞) , [0, ∞)) and ψ ∈ C ([0, ∞) , [0, ∞)). Fourth-order boundary value problems modeling bending equilibria of elastic beams have been considered in several papers, because they have important applications in mechanics and engineering; [1, 10, 11, 16]. Many authors have studied the beam equation under various boundary conditions and by different approaches. For example, Benaicha and Haddouchi [7] studied the existence of solutions for the following fourth-order boundary value problem ( u(4) (t) + f (u (t)) = 0, t ∈ (0, 1) , R1 u0 (0) = u0 (1) = u00 (0) = 0, u (0) = 0 a (s) u (s) ds. Some nonlinear elastic beam equations have been studied extensively. For a small sample of such work, we refer the reader to the work of Bai [5], Bai and Wang [6], Bonnano and Bella [8], Infante and Pietramala [17], Ma and Thompson [22], Ma and Xu [23], on elastic beams whose ends are simply supported, the works of Alves et al. [3] and Yang [27] on elastic beam of where one end is embedded and another end

17


is fastened with sliding clamp, the work of Graef et al. [14, 15] on multi-point boundary value problems and the works of Li [21], Sun and Zhu [25] and Wang et al. [26] with parameters. When the elastic beam equation does not contain parameter λ, the existence of multiple positive solutions and unique positive solution was presented in [9, 18, 20, 22] by variational methods, for some other results on boundary value problems for the beam equation, we refer the reader to [2, 4, 12, 13, 28]. The aim of this paper is to show that the existence and number of positive solutions of BVP (1.1)-(1.2) are affected by the parameter λ. Inspired and motivated by the works mentioned above, we deal with existence and nonexistence of positive solutions to the BVP (1.1)-(1.2) by using the fixed point theorem together with the properties of the Green’s function. The paper is organized as follows. In Section 2, we present that a nontrivial and nonnegative solution of BVP (1.1)-(1.2) is a positive solution. In Section 3, we obtain that some results about existence, multiplicity and nonexistence of positive solutions for BVP(1.1)-(1.2) depend on the parameter λ, and we give an example to illustrate our results. 2. Preliminaries We shall consider the Banach space C [0, 1] equipped with sup norm kuk = max |u (t)| and C + [0, 1] 0≤t≤1

is the cone of nonnegative functions in C [0, 1]. Definition 2.1. A nonempty closed and convex set P ⊂ E is called a cone of the Banach space E if it satisfies (i) u ∈ P , r > 0 implies ru ∈ P ; (ii) u ∈ P, −u ∈ P implies u = θ, where θ denote the zero element of E. Definition 2.2. A cone P is said to be normal if there exists a positive number N called the normal constant of P , such that θ ≤ u ≤ v implies kuk ≤ N kvk. In arriving our results, we need the following seven preliminary lemmas. The first one is well known. Lemma 2.3. Let y ∈ C [0, 1]. If u ∈ C 4 [0, 1], then the BVP ( u(4) (t) = y (t) , 0 ≤ t ≤ 1, (2.1) 0 0 000 u (0) = u (0) = u (1) = u (1) + ψ (u (1)) = 0, has a unique solution Z1 u (t) =

G (t, s) y (s) ds + ψ (u (1)) φ (t) , 0

where ( (2.2)

G (t, s) =

1 2 12 t 1 2 12 s

6s − 3s2 − 2t , 6t − 3t2 − 2s ,

0 ≤ t ≤ s ≤ 1, 0 ≤ s ≤ t ≤ 1,

and (2.3)

φ (t) =

1 2 1 3 t − t . 4 6

Lemma 2.4. For any (t, s) ∈ [0, 1] × [0, 1], we have 1 2 2 1 ∂G 1 (2.4) t s ≤ G (t, s) ≤ s2 2t − t2 and 0 ≤ (t, s) ≤ (1 − t) s2 , 12 4 ∂t 2 1 1 0 ≤ φ (t) ≤ 2t − t2 and 0 ≤ φ0 (t) ≤ (1 − t) . 4 2

18


Proof. Suppose 0 ≤ t ≤ s ≤ 1. Then 1 2 1 2 t 6s − 3s2 − 2t ≥ t 6s − 3s2 − 2s 12 12 1 2 t 4s − 3s2 ≥ 12 1 2 ≥ t 4s2 − 3s2 12 1 2 2 t s . ≥ 12 Analogously if 0 ≤ s ≤ t ≤ 1. Now for showing the upper bound about G we suppose 0 ≤ s ≤ t ≤ 1 ( it similar in the other case). Then 1 1 2 1 2 s 6t − 3t2 − 2s ≤ s 6t − 3t2 = s2 2t − t2 . 12 12 4 It is easy to prove that ∂G 1 0≤ (t, s) ≤ (1 − t) s2 , t, s ∈ [0, 1] × [0, 1] . ∂t 2 Now, it is obvious that φ (t) ≥ 0 for t ∈ [0, 1]. Let us check the inequality φ (t) ≤ 41 2t − t2 , t ∈ [0, 1]. For t ∈ [0, 1], we have t 2t2 t t2 2t2 1 φ (t) = t− ≤ t+ t− − = 2t − t2 . 4 3 4 3 3 4 Lemma 2.5. For y ∈ C + [0, 1]. Then the unique solution u (t) of BVP ( u(4) (t) = y (t) , 0 < t < 1, u (0) = u0 (0) = u0 (1) = u000 (1) = 0. is nonnegative and satisfies u (t) ≥

t3 kuk . 6

Proof. Let y (t) ∈ C + [0, 1], then from G (t, s) ≥ 0. We know u ∈ C + [0, 1]. Set u (t0 ) = kuk, t0 ∈ (0, 1] . We first prove that G (t, s) 1 ≥ t3 , t, t0 , s ∈ (0, 1] . G (t0 , s) 6 In fact, we can consider four cases: (1) if 0 < t, t0 ≤ s ≤ 1, then t2 6s − 3s2 − 2s t2 t (4 − 3s) t3 G (t, s) ≥ ≥ ≥ . 2 G (t0 , s) 6s − 3s − 2t0 6 6 (2) if 0 < t ≤ s ≤ t0 ≤ 1, then t2 6s − 3s2 − 2t t2 6s − 3s2 − 2s G (t, s) t2 s (4 − 3s) t2 (t) t3 ≥ ≥ ≥ ≥ ≥ . 2 G (t0 , s) 6t0 − 3t0 − 2s 6t0 6 6 6 (3) if 0 < s ≤ t, t0 ≤ 1, then 6t − 3t2 − 2s 4t − 3t2 G (t, s) t (4 − 3t) t t3 ≥ ≥ ≥ ≥ ≥ . G (t0 , s) 6t0 6 6 6 6

19

(4) if 0 < t0 ≤ s ≤ t ≤ 1, then


t20 6t − 3t2 − 2s 6t − 3t2 − 2t G (t, s) t (4 − 3t) t t3 ≥ 2 ≥ ≥ ≥ ≥ . G (t0 , s) t0 (6s − 3s2 − 2t0 ) 6 6 6 6 Therefore, for t ∈ [t0 , 1] we have Z 1 Z 1 Z 1 3 G (t, s) t t3 t3 u (t) = G (t, s) y (s) ds = G (t0 , s) y (s) ds ≥ G (t0 , s) y (s) ds = u (t0 ) = kuk . 6 6 0 0 G (t0 , s) 0 6 The proof is complete.

If we let P =

u ∈ C + [0, 1] : u (t) ≥

t3 kuk , 6

then it is easy to see that P is a cone in C [0, 1]. From [3] and [20], it is evident that BVP (1.1)-(1.2) has an integral formulation given by Z1 (2.5)

u (t) = λ

G (t, s) f (s, u (s)) ds + ψ (u (1)) φ (t) , 0

where G and φ defined in (2.2) and (2.3). Now, we define the integral operators A, B, Tλ : P → C [0, 1] by Z1 (λAu) (t) = λ

G (t, s) f (u (s)) ds; (Bu) (t) = ψ (u (1))

1 2 1 3 t − t , Tλ = λA + B. 4 6

0

Lemma 2.6. Let y ∈ C + [0, 1]. If u ∈ C 4 [0, 1] satisfies ( u(4) (t) = y (t) , 0 ≤ t ≤ 1, (2.6) 0 0 (3) u (0) = u (0) = u (1) = 0, u (1) + ψ (u (1)) = 0, then (i) u (t) ≥ 0 for t ∈ [0, 1]; (ii) u0 (t) ≥ 0 for t ∈ [0, 1]. Proof. From Lemma 2.4, we obtain u (t) ≥ 0 and u0 (t) ≥ 0 for t ∈ [0, 1].

It is clear from (2.5) that solution of the BVP (1.1)-(1.2) are fixed points of Tλ : P → P . In particular u is fixed point of B if and only if u is solution of the following BVP ( u(4) (t) = 0, 0 < t < 1, (2.7) u (0) = u0 (0) = u0 (1) = u000 (1) + ψ (u (1)) = 0, and u is fixed point of λA if and only if u is solution of the following BVP ( u(4) (t) = λf (t, u (t)) , 0 < t < 1, (2.8) u (0) = u0 (0) = u0 (1) = u000 (1) = 0. Lemma 2.7. A (P ) ⊂ P, B (P ) ⊂ P and Tλ (P ) ⊂ P .

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Proof. u ∈ P implies u (t) ≥ 0, so f (t, u (t)) ≥ 0 and ψ (u (1)) ≥ 0. Moreover, for u ∈ P , (4)

(Tλ u)

(t) = λf (t, u (t)) ≥ 0, t ∈ [0, 1] , 0

0


(Tλ u) (0) = (Tλ u) (0) = (Tλ u) (1) = 0, and

(3)

(Tλ u)

(1) = −ψ (u (1)) ≤ 0.

To prove Tλ (P ) ⊂ P , we show that (Tλ u) (t) − Let u (t) ∈ P, t ∈ [0, 1], we have λ 3 t3 t (Tλ u) (t) − (Tλ u) (1) ≥ 6 36

Z1

t3 6

(Tλ u) (1) ≥ 0, f or t ∈ [0, 1].

3

t2 t3 − 4 6

t2 13t3 − 4 6 × 12

s f (s, u (s)) ds + ψ (u (1))

t3 − 6

1 ψ (u (1)) 12

0

λ 3 ≥ t 36

Z1

3

s f (s, u (s)) ds + ψ (u (1))

0

≥

λ 3 t 36

Z1

s3 f (s, u (s)) ds ≥ 0.

0 3

Therefore (Tλ u) (t) ≥ t6 (Tλ u) (1) , f or t ∈ [0, 1]. Thus, we obtain Tλ (P ) ⊂ P . From the above proof, we can show that A (P ) ⊂ P, B (P ) ⊂ P . If we let K = {u ∈ P/u (t) is nondecreasing} , then, it is easy to show that K ⊂ P is also a cone in E, and if u ∈ K, then kuk = u (1). Lemma 2.8. Tλ (P ) ⊂ K, A (P ) ⊂ K and B (P ) ⊂ K. Proof. It follows from Lemma 2.6 (ii) and Lemma 2.7.

Lemma 2.9. (i) A : K → K is completely continuous. (ii) If ψ (u) is nondecreasing, then B : K → K is completely continuous. Proof. We know that A (P ) ⊂ K, B (P ) ⊂ K. Since the proof is similar we only prove that A is completely continuous. To do this let D be a bounded subset of K. Then there exists a positive constant M1 such that kuk ≤ M1 , ∀u ∈ D. Now, we shall prove that A (D) is relatively compact in K. Suppose that (yk )k∈N? ⊂ A (D). Then there exist (xk )k∈N? ⊂ D, such that yk = Axk .

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Let M2 = sup |f (t, u (t))| for all (t, u) ∈ [0, 1] × [0, M1 ]. For any k ∈ N∗ , by Lemma 2.4, we have 0≤t≤1

1 Z |yk (t)| = |(Axk ) (t)| = λ G (t, s) f (s, xk (s)) ds 0

Z1 ROMANIAN JOURNAL OF MATHEMATICS AND COMPUTER SCIENCE, 2018, VOLUME 8, ISSUE 1, p.17-30

≤ λM2

G (t, s) ds 0

1 ≤ λM2 2t − t2 4

Z1

s2 ds

0

λ ≤ M2 , 6 which implies that (yk (t))k∈N? is uniformly bounded. Now, we show that A (D) is equicontinuous. For any u ∈ K, n ≥ k, and t1 , t2 ∈ [0, 1] with |t1 − t2 | < δ, we have |yk (t1 ) − yk (t2 )| = |Au (t1 ) − Au (t2 )| 1 Z ≤ λ (G (t1 , s) − G (t2 , s)) f (s, xk (s)) ds 0

Z1 ≤ λM2

|G (t1 , s) − G (t2 , s)| ds. 0

It follows from the uniform continuity of Green’s function G on [0, 1] × [0, 1], that for any ε > 0, we have ε |G (t1 , s) − G (t2 , s)| ≤ , f or t1 , t2 , s ∈ [0, 1] , |t1 − t2 | < δ. λM2 Then |yk (t1 ) − yk (t2 )| = |Au (t1 ) − Au (t2 )| Z1 ≤ λM2

|G (t1 , s) − G (t2 , s)| ds 0

λM2 ε = ε. ≤ λM2 Therefore, A (D) is equicontinuous. By the Ascoli-Arzela Theorem, we know that A is completely continuous. By Lemmas 2.8 and 2.9, we know that if u ∈ P \ {θ} is solution for BVP (1.1)-(1.2), then u is positive solution for BVP (1.1)-(1.2) and it is obvious from Lemma 2.8 that if u ∈ P \ {θ} is a solution for BVP (1.1)-(1.2) then u ∈ K \ {θ}. 3. Existence and nonexistence results In this section we will apply a theorem due to Krasnoselskii to study the existence, multiplicity and nonexistence of solutions for BVP (1.1)-(1.2) in K \ {θ}.

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Theorem 3.1. (see [19]) Let E be a Banach space and K ⊂ E be a cone in E. Assume Ω1 , Ω2 are open subset of E with 0 ∈ Ω1 , Ω1 ⊂ Ω2 , and let


¯ 2 \ Ω1 → K T :K∩ Ω be a completely continuous operator such that (A) kT uk ≤ kuk, ∀u ∈ K ∩ ∂Ω1 and kT uk ≥ kuk, ∀u ∈ K ∩ ∂Ω2 ; or (B) kT uk ≥ kuk, ∀u ∈ K ∩ ∂Ω1 and kT uk ≤ kuk, ∀u ∈ K ∩ ∂Ω2 . ¯ 2 \ Ω1 . Then T has a fixed point in K ∩ Ω We adopt the following assumptions: (H1 ) f (t, u) is nondecreasing in u ∈ [0, ∞) for fixed t ∈ [0, 1]. (H2 ) ψ (u) is nondecreasing in u ∈ [0, ∞). R1 (H3 ) F0 = 0 s2 f (s, 0) ds > 0. (H4 ) ψ (1) < 4. = +∞. (H5 ) f∞ = lim min f (t,u) u→∞t∈[ 1 ,1] u 2 (H6 ) ψ∞ = limu→∞ inf ψ(u) u > 12. Set Λ = {λ > 0/there exists uλ ∈ K \ {θ} such that Tλ uλ = uλ } ,

(3.1) and

λ∗ = supΛ. 0

Lemma 3.2. Suppose that (H1 ), (H2 ) and (H3 ) hold. If λ ∈ Λ, then (0, λ0 ] ⊂ Λ. 0

Proof. λ ∈ Λ means that there exists uλ0 ∈ K \ {θ} such that Tλ0 uλ0 = uλ0 . Therefore, for any λ ∈ (0, λ0 ] we have Tλ uλ0 ≤ Tλ0 uλ0 = uλ0 Set w0 = uλ0 , wn = Tλ wn−1 , n = 1, 2, ... From (H1 ) and (H2 ), we obtain w0 (t) ≥ w1 (t) ≥ ... ≥ wn (t) ≥ ... ≥

F0 λ 2 t , 12

by Lemma 2.9 and (H3 ), {wn } converges to fixed point of Tλ in K \ {θ}. Thus (0, λ0 ] ⊂ Λ. The proof is complete. Now, we let λ∗ =

4−ψ(1) F1 ,

F1 =

R1 0

2s2 f (s, 1) ds, u0 (t) =

F0 λ 2 12 t ,

v0 (t) = t, F∞ = lim sup max u→∞

0≤t≤1

f (t,u) u

and Ψ∞ = lim sup ψ(u) u . u→∞

Theorem 3.3. Suppose that (H1 ), (H2 ) and (H3 ) hold. If (H4 ) holds, then Tλ has minimal and maximal fixed point in [u0 , v0 ] for λ ∈ (0, λ∗ ]. Moreover, there exists λ∗ ≥ λ∗ > 0 such that Tλ has at least one and has no fixed points in K \ {θ} for 0 < λ < λ∗ and λ > λ∗ , respectively.

23

Proof. From (H1 ) , (H2 ) , (H3 ) , (H4 ) and (2.4), we have λ∗ > 0. For any λ ∈ (0, λ∗ ], we obtain Z1 (Tλ u0 ) (t) = λ

t2 t3 − 4 6

t2 t3 − 4 6

G (t, s) f (s, u0 (s)) ds + ψ (u0 (1)) 0

Z1 ROMANIAN JOURNAL OF MATHEMATICS AND COMPUTER SCIENCE, 2018, VOLUME 8, ISSUE 1, p.17-30

≥λ

G (t, s) f (s, u0 (0)) ds + ψ (u0 (1)) 0

λ 2 t ≥ 12

Z1

2

s f (s, 0) ds + ψ (u0 (1))

t2 t3 − 4 6

0

≥

λ 2 t F0 = u0 (t) , 12

and Z1 (Tλ v0 ) (t) = λ

G (t, s) f (s, v0 (s)) ds + ψ (v0 (1))

t3 t2 − 4 6

0

λ∗ 2t − t2 ≤ 4

Z1

2

s f (s, v0 (1)) ds + ψ (v0 (1))

t t2 − 4 6

t t2 − 4 6

t

0

≤

λ∗ 2t − t2 4

Z1

s2 f (s, v0 (1)) ds + ψ (v0 (1))

0

Z1

t λ∗ (2t) s2 f (s, v0 (1)) ds + ψ (v0 (1)) 4 4 0   Z1 t ≤ λ∗ 2s2 f (s, v0 (1)) ds + ψ (1) , v0 (1) = 1 4 ≤

0

t ≤ [λ∗ F1 + ψ (1)] ≤ v0 (t) . 4 Set un = Tλ un−1 , vn = Tλ vn−1 , n = 1, 2, ..., then from (H1 ) , (H2 ), we have u0 (t) ≤ u1 (t) ≤ ... ≤ un (t) ≤ ... ≤ vn (t) ... ≤ v1 (t) ≤ v0 (t) .

(3.2)

Lemma 2.9 implies that {un } and {vn } converge to fixed points uλ and vλ of Tλ , respectively. From (3.2) it is evident that uλ , vλ ∈ K \ {θ} are the minimal fixed point and maximal fixed point of Tλ in [u0 , v0 ], respectively. +∞ By the definition of λ∗ , there exists a nondecreasing sequence {λn }1 such that lim λn = λ∗ . Let n→+∞

+∞

{uλn }1

is bounded subset in K. Then there exists a constant M > 0 such that kuλn k ≤ M, f or n ∈ N∗ , +∞

which implies that {uλn }1 is uniformly bounded. +∞ Now, we show that {uλn }1 is equicontinuous. For any uλn ∈ K, n ∈ N∗ and t1 , t2 ∈ [0, 1], with

24

|t1 − t2 | < δ, we have |uλn (t1 ) − uλn (t2 )| ≤ λ

∗

Z1 |G (t1 , s) − G (t2 , s)| f (s, M ) ds


0

1 1 1 2 2 + ψ (M ) |t1 − t2 | |t1 + t2 | + t1 + t1 t2 + t2 2 2 3 1 Z 5 ≤ λ∗ |G (t1 , s) − G (t2 , s)| f (s, M ) ds + ψ (M ) |t1 − t2 | , 6 0 +∞ {uλn }1

which implies that is equicontinuous subset in K. Consequently, by an application of the +∞ Arzela-AscoliTheorem we conclude that {uλn }1 is a relatively compact set in K. So, there exists a subsequence uλni ⊂ {uλn } converging to u∗ ∈ K. Note that Z1

uλni (t) = λni

G (t, s) f s, uλni (s) ds + ψ uλni

(1)

t3 t2 − 4 6

.

0 ∗

By taking the limit we have u (t) = (Tλ∗ u∗ ) (t). Therefore Tλ has at least one fixed point for 0 < λ < λ∗ . Finally, Tλ has no fixed point for λ > λ∗ . The proof is complete. Theorem 3.4. Suppose that (H1 ), (H2 ), (H3 ), (H4 ) and (2.4) hold. If (F+∞ < +∞, Ψ∞ < 4), then, ∞) when F∞ > 0, there exists λ∗ ≥ 3(4−Ψ > 0 such that Tλ has at least one and has no fixed points in 2F∞ ∗ ∗ K \ {θ} for 0 < λ < λ and λ > λ , respectively. When F∞ = 0, Tλ has at least one fixed points in K \ {θ} for λ > 0. Proof. Since F∞ < ∞, Ψ∞ < 4, for any 0 < ε < 4 − Ψ∞ , there exists N0 > 0 such that f (t, u) ≤ ∞ −ε) (F∞ + ε) u and ψ (u) ≤ (Ψ∞ + ε) u for u > N0 , t ∈ [0, 1]. Let w0 (t) = 4N0 t and λ0 = 3(4−Ψ 2(F∞ +ε) , then λ0 > 0 and Z1 (Tλ0 w0 ) (t) = λ0

G (t, s) f (s, w0 (s)) ds + ψ (w0 (1))

t2 t3 − 4 6

0

λ0 (2t) ≤ 4

Z1

2

s f (s, w0 (s)) ds + ψ (w0 (1))

t2 4

0

λ0 ≤ (2t) 4

Z1

2

s (F∞ + ε) w0 (t) ds + w0 (1) (Ψ∞ + ε) 0

Z1 λ0 t ≤ w0 (t) (F∞ + ε) (2t) s2 ds + (Ψ∞ + ε) 4 4 0 1 2λ0 ≤ tw0 (t) (F∞ + ε) + (Ψ∞ + ε) 4 3 1 2λ0 ≤ w0 (t) (F∞ + ε) + (Ψ∞ + ε) ≤ w0 (t) . 4 3 Now, set w0 (t) = 4N0 t, wn = Tλn−1 wn−1 , n = 1, 2, ...

25

t2 4

From (H1 ) and (H2 ), we obtain w0 (t) ≥ w1 (t) ≥ ... ≥ wn (t) ≥ ... ≥

F0 λ 2 t . 12


Therefore, the sequence {wn } is bounded in K \ {θ}. By Lemma 2.9 and the definition of λ∗ , the operator Tλn completely continuous. Hence the sequence {wn } is compact in K \ {θ}, it is also monotone. Then it is uniformly convergent to fixed points u∗ of Tλn in K \ {θ}. When we pass to the limit we get u∗ = Tλ∗ u∗ . +∞

For λ > λ∗ , there exists {λn }1 , with lim λn = λ, we prove that the problem has no positive solution. n→∞ Suppose the contrary that the problem has a positive solution uλn , then we get 1 kuλn k = (Tλn uλn ) (1) ≤ λn 4

Z1

1 2s2 f (s, uλn (s)) ds + ψ (uλn (1)) 4

0

Z1

λn ≤ 2s2 (F∞ + ε) uλn (1) ds + uλn (1) (Ψ∞ + ε) 4 0 1 2λn (F∞ + ε) + (Ψ∞ + ε) kuλn k < kuλn k . ≤ 4 3

1 4

Thus kuλ k < kuλ k , which is a contradiction. The proof is complete.

Lemma 3.5. Assume that (H1 ), (H2 ), (H3 ) and (H4 ) hold and one of (H5 ) and (H6 ) holds. If Λ is nonempty, then (i) Λ is bounded from above, that is λ∗ < +∞. (ii) λ∗ ∈ Λ. ∞

Proof. (i) Suppose to the contrary that there exists an increasing sequence {λn }1 lim λn = +∞. Set uλn ∈ K \ {θ} is a fixed point of Tλn that is,

⊂ Λ such that

n→+∞

Tλn uλn = uλn . There are two cases to be considered. +∞ Case 1. {uλn }1 is bounded, that is there exists a constant M > 0 such that kuλn k ≤ M, f or n = 1, 2, ... Hence, from (H1 ) , (H2 ) , (H3 ), and (2.4), we have M ≥ kuλn k ≥ (Tλn uλn ) (1) λn ≥ 12

Z1

s2 f (s, uλn (s)) ds

0

≥

λn 12

Z1

s2 f (s, 0) ds =

λn F0 → +∞, 12

0

which is a contradiction. +∞ +∞ +∞ Case 2. {uλn }1 is unbounded, that is there exists subsequence of {uλn }1 still denoted by {uλn }1

26

such that lim kuλn k = +∞. When (H5 ) holds, take L > 288 λ1 there exists N1 > 0 such that f (t, u) ≥ n→+∞

1 Lu, f or u ≥ N1 , t ∈ 2 , 1 . Choose n1 such that uλn1 > 24N1 . Thus

1 1 1

f t, uλn1 ≥ L uλn1 , t ∈ ,1 24 24 2


Moreover, from (H1 ) and the definition of K, we have

uλn ≥ Tλn uλn (1) 1

1

≥

λ n1 12

Z1

1

s2 f s, uλn1 (s) ds

1 2

λn ≥ 1 12

Z1

1

uλn1 (s) ds s f s, 6 2

1 2

>

λn1 L uλn1 > uλn1 , 36 × 8

which is contraduction. 1 When (H6 ) holds, choose ε > 0 such that 12

(ψ∞ −

ε) > 1. There exists N2 > 1 such that ψ (u) ≥ (ψ∞ − ε) u, f or u ≥ N2 . Choose n2 such that uλn2 > N2 . Then

ψ (uλn2 (1)) = ψ uλn2 ≥ (ψ∞ − ε) uλn2 . Moreover

uλn = Tλn uλn (1) ≥ 1 ψ uλn (1) ≥ 1 (ψ∞ − ε) uλn > uλn , 2 2 2 2 2 2 12 12 which is contraduction. Consequently, we find that Λ is bounded from above. ∞ (ii) From the definition of λ∗ , there exists a nondecreasing sequence {λn }1 such that lim λn = λ∗ . Let n→+∞

+∞

uλn ∈ K \ {θ} be a fixed point of Tλn . Arguing similarly as above in case 2, we can show that {uλn }1 is a bounded subset in K, that is there exists a constant M > 0. From (H1 ), (H3 ) and (H4 ), we have kuλn k = (Tλn uλn ) (1) λn ≤ 12

Z1

s2 f (s, uλn (s)) ds +

0

 1  ≤ λn 12

Z1

1 ψ (1) 12 

s2 f (s, 0) ds + 4

0

1 1 (λn F0 + 4) → (λ∗ F0 + 4) = M, as n → ∞. = 12 12 Therefore kuλn k ≤ M, n = 1, 2, ... +∞

From the proof of Theorem 3.3 we know that {uλn }1 is equicontinuous subset in K and by an application ∞ of the Arzela-Ascoli Theorem we conclude that {uλn }1 is a relatively compact set in K. So, there exists ∗ a subsequence uλni ⊂ {uλn } converging to u ∈ K. Note that

Z1

uλni (t) = λni

G (t, s) f s, uλni (s) ds + ψ uλni

(1)

t3 t2 − 4 6

.

0

By taking the limit we have u∗ (t) = (Tλ∗ u∗ ) (t) ≥

27

λ1 2 12 F0 t ,

that is λ∗ ∈ Λ. The proof is complete.

Theorem 3.6. Suppose that (H1 ), (H2 ), (H3 ) and (H4 ) hold and that one of (H5 ) and (H6 ) holds. Then there exists λ∗ ≥ λ∗ > 0 such that BVP (1.1)-(1.2) has at least two, one and no positive solutions for 0 < λ ≤ λ∗ , λ∗ < λ ≤ λ∗ and λ > λ∗ respectively.


Proof. From (H1 ), (H2 ), (H3 ) and (H4 ) we have (0, λ∗ ] ⊂ Λ. So λ∗ ≥ λ∗ > 0. From Lemmas 3.2 and 3.5, we have (0, λ∗ ] = Λ. Therefore, from the definition of λ∗ we only to prove that Tλ has at least two fixed points in K \ {θ} for λ ∈ (0, λ∗ ]. Now, given λ ∈ (0, λ∗ ]. Theorem 3.3 means that Tλ has at least one fixed point uλ,1 ∈ K \ {θ} which satisfies kuλ,1 k ≤ 1. Let K1 = {u ∈ K : kuk < 1} . 3 Note that 2 − t ≤ 1, f or t ∈ [0, 1], so for u ∈ K with kuk = 1, i.e u ∈ ∂K1 , we have Z1 kuk = kTλ uk = (Tλ u) (1) ≤ λ

G (t, s) f (s, u (s)) ds +

1 ψ (u (1)) 12

0

λ ≤ 12

Z1

s2 (3 − 2s) f (s, u (s)) ds +

1 ψ (u (1)) 12

0

(3.3)

≤

λ 12

Z1

s2 (3 − 2s) f (s, 1) ds +

1 ψ (u (1)) 12

0



 Z1 1  3 ≤ λ∗ 2s2 − s f (s, 1) ds + ψ (u (1)) 12 2 0 1 4 − ψ (1) 1 ≤ F1 + ψ (u (1)) = < kuk , 12 F1 3 When (H5 ) holds, take L > 288 λ there exists N1 > 0 such that f (t, u) ≥ Lu, f or u ≥ N1 , t ∈ K2 = {u ∈ K : kuk < 24N1 }. Then K 1 ⊂ K2 . If u ∈ ∂K2 , we have

1

. Set

2, 1

kTλ uk = (Tλ u) (1) ≥

1 λ ψ (u (1)) ≥ 12 12

Z

1

s2 f 1 2

s,

1 kuk ds 24

λL kuk > kuk . 288 Consequently, applying Theorem 3.1 that Tλ has a fixed point uλ,2 ∈ K 2 \ K1 . 1 When (H6 ) holds, choose ε > 0 such that 12 (ψ∞ − ε) > 1. There exists N2 > 1 such that ψ (u) ≥ (ψ∞ − ε) u, f or u ≥ N2 . 0 0 0 Set K2 = {u ∈ K | kuk < N2 }. Then K 1 ⊂ K2 . If u ∈ ∂K2 , we have 1 kTλ uk = (Tλ u) (1) ≥ ψ (u (1)) 12 1 ≥ (ψ∞ − ε) u (1) 12 > kuk . ≥

Consequently, applying Theorem 3.1 that Tλ has a fixed point uλ,2 ∈ K 0 2 \ K1 . Equation (3.3) implies that Tλ has no fixed points in ∂K1 . In conclusion, for λ ∈ (0, λ∗ ], Tλ has at least two fixed points uλ,1 and uλ,2 in K with 0 < kuλ,1 k < 1 < kuλ,2 k. The proof is complete.

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We present an example to illustrate the applicability of the results shown before. Example 3.7. Consider the following boundary value problem: (3.4)

u(4) (t) = λ (t + u + ln (1 + u)) ,

(3.5)

u (0) = u0 (0) = u0 (1) = u000 (1) + ψ (u (1)) = 0,


where

   2 arcsin u, π ψ (u) = 4 + arcsin u,   0,

0 < t < 1,

√

0 ≤ u ≤ 22 , 2 2 ≤ u ≤ 1, u > 1.

√

It is easy to compute that F∞ = 2, ψ∞ = 0, F0 =

1 1 2 3π 48 − 9π , F1 = + (1 + ln 2)) , ψ (1) = and λ∗ = . 4 2 3 4 6 + 8(1 + ln 2)

So, the assumptions (H1 ), (H2 ), (H3 ) and (H4 ) are satisfied, it follows from Theorem 3.4 there exists λ∗ = 3 ≥ λ∗ such that BVP (3.4)-(3.5) has at least one positive solution for 0 < λ ≤ 3 and has no positive solution for λ > λ∗ . Acknowledgments. The authors want to thank the anonymous referee for the throughout reading of the manuscript and several suggestions that help us improve the presentation of the paper. References [1] A. R. Aftabizadeh, Existence and uniqueness theorems for fourth-order boundary value problems, J. Math. Anal. Appl. 116 (1986), 415-426. [2] R. P. Agarwal, D. O’Regan and P. J. Wang, Positive solutions of differential, Differance, and integral equations, Kluwer Academic, Boston, Ma, 1999. [3] E. Alves, T. F. Ma. and M. L. Pelicer, Monotone positive solutions for a fourth order equation with nonlinear boundary conditions, Nonlinear anal. 71 (2009), 3834-3841. [4] D. R. Anderson and R. I. Avery, A fourth-order four-point right focal boundary value problem, Rocky Mountain of Mathematics. 36 (2006), 367-380. [5] Z. Bai, The upper and lower solution method for some fourth order boundary value problem, Nonlinear Anal. 67 (2007), 1704- 1709. [6] Z. Bai and H. Wang, Positive solutions of some nonlinear fourth order beam equations, J. Math. Anal. Appl. 270 (2002), 357-368. [7] S. Benaicha and F. Hadouchi, Positive solutions of a nonlinear fourth-order integral boundary value problem, Annals of West University of Timisoara - Mathematics and Computer Science, 54 (2016), 73- 86. [8] G. Bonanno and BD. Bella, A boundary value problem for fourth-order elastic beam equations, J. Math. Anal. Appl. 343 (2008), 1166-1176. [9] N. Bouteraa and S. Benaicha, Triple positive solutions of higher-order nonlinear boundary value problem, J. Comput. Sci . Comp. Math. 7 (2017), 25-30. [10] A. Cabada and S. Heikkila, Uniqueness comparison and existence results for third order functional initial-boundary value problems, Comput. Math. Appl. 41 (2001), 607-618. [11] A. Cabada and S. Tersian, Existence and multiplicity of solutions to boundary value problems for fourth-order impulsive differential equations. Bound. Value Probl. 105 (2014). [12] J. Davis and J. Henderson, Uniqueness implies existence for fourth-order Lidstone boundary value problems, Panamer. Math.J. 8 (1998), 23-35. [13] L. H. Erbe and H. Wang, On he existence of positive solutions of ordinary differential equations, Proc. Amer. Math. Soc. 120 (1994), 743-748.

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[14] J. R. Graef, C. Qian and B.Yang, A three point boundary value problem for nonlinear fourth order differential equations, J. Math. Anal. Appl. 287 (2003), 217-233. [15] J. R. Graef and B. Yang, Existence and nonexistence of positive solutions of fourth order nonlinear boundary value problems, Appl. Anal, 74 (2000), 201-214. [16] C. P. Gupta, Existence and uniqueness theorem for the bending of an elastics beam equation, Appl. Anal. 26 (1988), 289-304. [17] G. Infante and P. Pietramala, A cantilever equation with nonlinear boundary conditions, Electron. J. Qual. Theory Differe. Equ. Spec. Ed. 15 (2009), 1-14. [18] N. Kosmatov, Countably meany solutions of a fourth order boundary-value problems, Electron. J. Qual. Theory Differ. Equ. 12 (2004), 1-15. [19] M. A. Krasnosel’skii, Positive solutions of operator equations, Noordhoff, Groningen, 1964. [20] S. Li and Q. Zhang, Existence and uniqueness of monotone positive solutions for an elastic beam equation with nonlinear boundary conditions Comput. Math. Appl. 63 (2012), 1355-1360. [21] Y. Li, Positive solutions of fourth-order boundary value problems with two parameters, J. Math. Anal. Appl. 281 (2003), 477-484. [22] R. Ma and B. Thompson, Nodal solutions for a nonlinear fourth-order eigenvalue problem, Acta Math. sin. Engl. Ser. 24 (2008), 27-34. [23] R. Ma and L. Xu, Existence of positive solutions for a nonlinear fourth-order boundary value problem. Appl. Math. Lett. 23 (2010), 537-543. [24] R. Ma and H. Wang, On the existence of positive solutions of fourth order ordinary differential equations, Appl. Anal. 59 (1995), 225-231. [25] Y. Sun and C. Zhu, Existence of positive solutions for singular fourth-order three-point boundary value problems. Adv. Differ. Equ. 51(2013), 1-13. [26] W. Wang, Y. Zheng, H. Yang and J. Wang, Positive solutions for elastic beam equations with nonlinear boundary conditions and a parameter, Bound. Value Prob. 80 (2014), 17 pages. [27] B.Yang, Positive solutions for a fourth order boundary-value problem, Electron. J. Qual. Theory Differ. Equ. 3 (2005), 1-17 [28] X. Zhang, Existence and iteration of monotone positive solutions for an elastic beam equation with a corner, Nonlinear Anal. Real World appl. 10 (2009), 2097-2103. Laboratory of Fundamental and Applied Mathematics of Oran (LMFAO), University of Oran1, Ahmed Benbella, Oran, Algeria E-mail address: [email protected] Laboratory of Fundamental and Applied Mathematics of Oran (LMFAO), University of Oran1, Ahmed Benbella, Oran, Algeria E-mail address: [email protected] Laboratory of Fundamental and Applied Mathematics of Oran (LMFAO), University of Oran1, Ahmed Benbella, Oran, Algeria E-mail address: [email protected] Laboratory of Mathematics and its Applications (LAMAP), University of Oran1, Ahmed Benbella, Oran, Algeria E-mail address: [email protected]

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