Multiple solutions for fractional differential equations with nonlinear boundary conditions

Computers and Mathematics with Applications 59 (2010) 2880–2886

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Multiple solutions for fractional differential equations with nonlinear boundary conditionsI Xiping Liu ∗ , Mei Jia College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China

article

info

Article history: Received 10 June 2009 Received in revised form 3 February 2010 Accepted 4 February 2010

abstract In this paper, we study certain fractional differential equations with nonlinear boundary conditions. By means of the Amann theorem and the method of upper and lower solutions, some new results on the multiple solutions are obtained. © 2010 Elsevier Ltd. All rights reserved.

Keywords: Caputo derivative Fractional differential equations Nonlinear boundary conditions Amann theorem Method of upper and lower solutions Multiple solutions

1. Introduction In this paper, we are concerned with the existence of multiple solutions for the fractional differential equations with nonlinear boundary conditions

C q D x(t ) = f (t , x(t ), x0 (t )), t ∈ (0, 1),   g0 (x(0), x0 (0)) = 0, 0  g001 (x(1), x 000(1)) = 0, x (0) = x (0) = · · · = x(n−1) (0) = 0,

(1.1)

where C Dq is the standard Caputo derivative, n > 2 is an integer, q ∈ (n − 1, n]. f ∈ C ([0, 1] × R × R, R), and g0 , g1 are given nonlinear functions. Fractional differential equations have been of great interest recently. This is because of both the intensive development of the theory of fractional calculus itself and the applications of such constructions in various scientific fields such as physics, mechanics, chemistry, engineering, etc. For details, see [1–3] and the references therein. Recently, there have been some papers which deal with the existence of the solutions of initial value problems or linear boundary value problems for fractional differential equations. In [4], the basic theories for the fractional calculus and the fractional differential equations are discussed. In [5–7], the basic theory for the initial value problems for fractional differential equations or fractional functional differential equations involving Riemann–Liouville differential operators are discussed. The results on general existence and uniqueness are proved by means of the monotone iterative technique and the method of upper and lower solutions.

I Supported by the Innovation Program of Shanghai Municipal Education Commission (No. 10ZZ93).

∗

Corresponding author. Tel.: +86 021 65683623-107. E-mail address: [email protected] (X. Liu).

0898-1221/$ – see front matter © 2010 Elsevier Ltd. All rights reserved. doi:10.1016/j.camwa.2010.02.005

X. Liu, M. Jia / Computers and Mathematics with Applications 59 (2010) 2880–2886

2881

In [8], by using some fixed-point theorems on a cone, Bai investigates the existence and multiplicity of positive solutions for nonlinear fractional differential equations with linear boundary conditions



Dα u(t ) + f (t , u(t )) = 0, u(0) = u(1) = 0,

t ∈ (0, 1),

where 1 < α ≤ 2 is a real number, Dα is the standard Riemann–Liouville derivative. In [9], the authors study the nonlinear fractional differential equation with linear boundary conditions



Dα u(t ) = f (t , u(t )), t ∈ (0, 1), u(0) = u(1) = u0 (0) = u0 (1) = 0,

where 3 < α ≤ 4 is a real number, Dα is the standard Riemann–Liouville derivative. Some multiple positive solutions for singular and nonsingular boundary value problems are given. However, research on the multiple solutions of the fractional differential equations with nonlinear boundary conditions is proceeding very slowly. In this paper, we focus on the multiple solutions for fractional differential equations with nonlinear boundary conditions. By means of the famous Amann theorem and the lower and upper solutions method, we obtain a new result on the existence of at least three distinct solutions under certain conditions. 2. Preliminaries In this section, we introduce preliminary facts which are used throughout this paper. The definitions concerning the fractional integral and the Caputo fractional derivative can be found in the literature [4]. Lemma 2.1. Suppose that y ∈ C [0, 1] and mi , ni ∈ R, i = 1, 2, with m1 (n1 + n2 ) − m2 n1 6= 0. Then the following linear boundary value problem:

C q D x(t ) = y(t ), t ∈ (0, 1),   m1 x(0) + m2 x0 (0) = λ0 , 0 n1 x(1) + n2 x (1) = λ1 ,  00 000 x (0) = x (0) = · · · = x(n−1) (0) = 0

(2.1)

is equivalent to the following fractional integral:



m2 − m1 t

x(t ) = ρ λ0 n2 + n1 (1 − t ) − λ1 (m2 − m1 t ) +

+




Γ (q)

n1 (1 − s) + n2 (q − 1) (1 − s)q−2 y(s)ds






0

t

Z

1

Γ (q)

1

Z

(t − s)q−1 y(s)ds,

(2.2)

0

where ρ = m (n +n1 )−m n . 1 1 2 2 1 That is, every solution of (2.1) is also a solution of (2.2) and vice versa. Proof. By C Dq x(t ) = y(t ), t ∈ (0, 1), and the boundary conditions x00 (0) = x000 (0) = · · · = x(n−1) (0) = 0, we have x(t ) = I q y(t ) + x(0) + x0 (0)t +

=

Γ (q)

2!

t2 + · · · +

x(n−1) (0) n−1 t (n − 1)!

t

Z

1

x00 (0)

(t − s)q−1 y(s)ds + x(0) + x0 (0)t , 0

where I q y is the fractional integral of order q of the function y. According to the proposition of the Caputo derivative, we get x0 (t ) =

t

Z

1

Γ (q − 1)

(t − s)q−2 y(s)ds + x0 (0).

0

Then x(1) =

1

Z

1

Γ (q)

(1 − s)q−1 y(s)ds + x(0) + x0 (0), 0

and x0 (1) =

1

Γ (q − 1)

1

Z

(1 − s)q−2 y(s)ds + x0 (0). 0

(2.3)

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X. Liu, M. Jia / Computers and Mathematics with Applications 59 (2010) 2880–2886

By the boundary conditions m1 x(0) + m2 x0 (0) = λ0 and n1 x(1) + n2 x0 (1) = λ1 , and noting that Γ (q) = (q − 1)Γ (q − 1), we have

 0 m1 x(0) + m2 x (0) = λ0 ,

1

Z

1

n1 x(0) + (n1 + n2 )x0 (0) = λ1 −

Γ (q)

n1 (1 − s) + n2 (q − 1) (1 − s)q−2 y(s)ds.




0

Hence, x(0) = ρ λ0 (n1 + n2 ) − λ1 m2 +

m2

1

Z

Γ (q)

!
n1 (1 − s) + n2 (q − 1) (1 − s)q−2 y(s)ds ,

0

and x (0) = ρ λ1 m1 − λ0 n1 − 0

m1

Γ (q)

1

Z

!
q −2 n1 (1 − s) + n2 (q − 1) (1 − s) y(s)ds .

0

We can easily get that


m2 − m1 t x(t ) = ρ λ0 n2 + n1 (1 − t ) − λ1 (m2 − m1 t ) + Γ (q) Z t 1 + (t − s)q−1 y(s)ds.  Γ (q) 0

Z

!

1


n1 (1 − s) + n2 (q − 1) (1 − s)q−2 y(s)ds

0

Lemma 2.2. Suppose that m1 , m2 , n1 , n2 ≥ 0 are constants with 0 < m1 < m2 < (1 + function with

n2 n1

)m1 , and x ∈ C n−1 ([0, 1]) is a

C q D x(t ) ≥ 0, t ∈ (0, 1),   m1 x(0) + m2 x0 (0) ≥ 0, 0 n1 x(1) + n2 x (1) ≤ 0,  00 000 x (0) = x (0) = · · · = x(n−1) (0) = 0.

(2.4)

Then x(t ) ≥ 0 and x0 (t ) ≤ 0, t ∈ [0, 1]. Proof. For any y ∈ C ([0, 1]) and y(t ) ≥ 0, t ∈ [0, 1], let λ0 ≥ 0, λ1 ≤ 0 be constants. Consider the following boundary value problem:

C q D x(t ) = y(t ), t ∈ (0, 1),   m1 x(0) + m2 x0 (0) = λ0 , 0  n001 x(1) + n0002 x (1) = λ1 , (n−1) x (0) = x (0) = · · · = x (0) = 0.

(2.5)

In view of Lemma 2.1, we have


m2 − m1 t x(t ) = ρ λ0 n2 + n1 (1 − t ) − λ1 (m2 − m1 t ) + Γ (q) Z t 1 + (t − s)q−1 y(s)ds. Γ (q) 0 Notice that 0 < m1 < m2 < (1 +

ρ=

1 m1 (n1 + n2 ) − m2 n1

n2 n1

Z

!

1


n1 (1 − s) + n2 (q − 1) (1 − s)q−2 y(s)ds

0

)m1 ; we have

> 0.

And then x(t ) ≥ 0 for t ∈ [0, 1]. Since x ( t ) = ρ λ 1 m 1 − λ 0 n1 − 0

≤ ρ λ 1 m 1 − λ 0 n1 −

m1

Γ (q) m1

Γ (q)

1

Z

!
q −2 n1 (1 − s) + n2 (q − 1) (1 − s) y(s)ds +

0 1

Z 0

!
q −2 n1 (1 − s) + n2 (q − 1) (1 − s) y(s)ds +

1

Γ (q − 1) 1

Γ (q − 1)

Z

t

(t − s)q−2 y(s)ds

0

Z 0

1

(1 − s)q−2 y(s)ds

X. Liu, M. Jia / Computers and Mathematics with Applications 59 (2010) 2880–2886

= ρ λ1 m1 − λ0 n1 −

m1 n1

1

Z

Γ (q − 1)

! 1−s


q−1

y(s)ds

+

0

1 − ρ m1 n2

Γ (q − 1)

2883

1

Z

(1 − s)q−2 y(s)ds, 0

and 1 − ρ m1 n2 = ρ n1 (m1 − m2 ) < 0, it is easy to see that x0 (t ) ≤ 0 for t ∈ [0, 1].



For convenience, throughout the paper, we suppose that the following hypotheses are satisfied: (H1 ) f (t , x1 , y1 ) ≤ f (t , x2 , y2 ),

for t ∈ [0, 1], x1 ≤ x2 and y2 ≤ y1 ,

and furthermore, f (t , x1 , y1 ) < f (t , x2 , y2 ), when at least one of x1 ≤ x2 and y2 ≤ y1 holds strictly. (H2 ) There exist constants mi , ni > 0 (i = 1, 2), with 0 < m1 < m2 < (1 + y2 ≤ y1 such that

n2 n1

)m1 , and x1 , x2 , y1 , y2 ∈ R with x1 ≤ x2 ,

g0 (x2 , y2 ) − g0 (x1 , y1 ) ≥ −m1 (x2 − x1 ) − m2 (y2 − y1 ) and g1 (x2 , y2 ) − g1 (x1 , y1 ) ≤ −n1 (x2 − x1 ) − n2 (y2 − y1 ). ˚ Every cone P in E Let E be a Banach space, P ⊂ E be a cone. A cone P is called solid if it contains interior points, i.e., P. defines a partial ordering in E given by x  y iff y − x ∈ P. If x  y and x 6= y, we write x ≺ y; if a cone P is solid and ˚ we write x ≺≺ y. A cone P is said to be normal if there exists a constant N > 0 such that 0  x  y implies y − x ∈ P, kxk ≤ N kyk. If P is normal, then every ordered interval [x, y] = {z ∈ E |x  z  y} is bounded. In this paper, the partial ordering ‘‘’’ is always given by P. The following Lemma 2.3 is the famous Amann three-solution theorem (see [10,11]), which will be used in the later proof of our main results about the multiple solutions of the boundary value problem. Lemma 2.3. Let E be a Banach space, and P ⊂ E be a normal solid cone. Suppose that there exist α1 , β1 , α2 , β2 ∈ E with

α1 ≺ β1 ≺ α2 ≺ β2 , and A : [α1 , β2 ] → E is a completely continuous strongly increasing operator such that

α1  Aα1 ,

Aβ1 ≺ β1 ,

α2 ≺ Aα2 ,

Aβ2  β2 .

Then the operator A has at least three fixed points x1 , x2 , x3 such that

α1  x1 ≺≺ β1 ,

α2 ≺≺ x2  β2 ,

α2 6 x3 6 β1 .

3. Multiple solutions of the boundary value problem Let the Banach space E = C 1 [0, 1] be endowed with the norm

kxk := max{ max |x(t )|, max |x0 (t )|}, t ∈[0,1]

t ∈[0,1]

and define the cone P ⊂ E by P := {x ∈ E | x(t ) ≥ 0, x0 (t ) ≤ 0, t ∈ [0, 1]}. Obviously, P is a normal solid cone in E, and x  y ∈ E iff x(t ) ≤ y(t ) and x0 (t ) ≥ y0 (t ), for t ∈ [0, 1]. Definition 3.1. Let α, β ∈ E. α is called a lower solution of boundary value problem (1.1) if it satisfies

C q D α(t ) ≤ f (t , α(t ), α 0 (t )), t ∈ (0, 1),   g0 (α(0), α 0 (0)) ≥ 0, 0  g100(α(1), α 000(1)) ≤ 0, α (0) = α (0) = · · · = α (n−1) (0) = 0.

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X. Liu, M. Jia / Computers and Mathematics with Applications 59 (2010) 2880–2886

And β is called an upper solution of boundary value problem (1.1) if it satisfies

C q D β(t ) ≥ f (t , β(t ), β 0 (t )), t ∈ (0, 1),   g0 (β(0), β 0 (0)) ≤ 0, 0  g100(β(1), β 000(1)) ≥ 0, β (0) = β (0) = · · · = β (n−1) (0) = 0. Theorem 3.1. Suppose that (H1 ) and (H2 ) hold, and there exist two lower solutions α1 , α2 and two upper solutions β1 , β2 of boundary value problem (1.1) such that α2 , β1 are not the solutions of the boundary value problem (1.1) with

α1 ≺ β1 ≺ α2 ≺ β2 . Then the boundary value problem (1.1) has at least three distinct solutions x1 , x2 , x3 , and satisfies

α1  x1 ≺≺ β1 ,

α2 ≺≺ x2  β2 ,

α2 6 x3 6 β1 .

Proof. For any u ∈ C 1 [0, 1], we consider the boundary value problem

C q D x(t ) = f (t , u(t ), u0 (t )), t ∈ (0, 1),   m1 x(0) + m2 x0 (0) = g0 (u(0), u0 (0)) + m1 u(0) + m2 u0 (0) := λ0 (u), 0 u0 (1)) + n1 u(1) + n2 u0 (1) := λ1 (u),  n001 x(1) + n0002 x (1) = g1 (u((1n), x (0) = x (0) = · · · = x −1) (0) = 0.

(3.1)

By Lemma 2.1, the boundary value problem (3.1) has a unique solution x ∈ C [0, 1] and



x(t ) = ρ λ0 (u) n2 + n1 (1 − t ) − λ1 (u)(m2 − m1 t )

+




m2 − m1 t

Z

Γ (q)

1


q−2 0 n1 (1 − s) + n2 (q − 1) (1 − s) f (s, u(s), u (s))ds +

0

1

Γ (q)

t

Z

(t − s)q−1 f (s, u(s), u0 (s))ds 0

:= Tu(t ).

(3.2)

Then T : E → E is a continuous, compact and bounded operator, and it is easy to see that x is the solution of the boundary value problem (1.1) if and only if x is the fixed point of T . Firstly, we prove that T is a strongly increasing operator. For any u1 , u2 ∈ E with u1 ≺ u2 , i.e., u1 (t ) ≤ u2 (t ), u01 (t ) ≥ u02 (t ) and u1 (t ) 6≡ u2 (t ). By (H1 ), we have f (t , u2 (t ), u02 (t )) − f (t , u1 (t ), u01 (t )) ≥ 0,

t ∈ [0, 1].

Since u1 (t ) 6≡ u2 (t ), there exists an interval [a, b] ⊂ [0, 1] such that u1 (t ) < u2 (t ), for t ∈ [a, b]. And we have f (t , u2 (t ), u02 (t )) − f (t , u1 (t ), u01 (t )) > 0, t ∈ [a, b].

(3.3)

By (H2 ), we have


λ0 (u2 ) − λ0 (u1 ) = g0 (u2 (0), u02 (0)) − g0 (u1 (0), u01 (0)) + (m1 u2 (0) + m2 u02 (0)) − (m1 u1 (0) + m2 u01 (0)) ≥ 0.
λ1 (u2 ) − λ1 (u1 ) = g1 (u2 (1), u02 (1)) − g1 (u1 (1), u01 (1)) + (n1 u2 (1) + n2 u02 (1)) − (n1 u1 (1) + n2 u01 (1)) ≤ 0. From (3.2) and (3.3), for any t ∈ [0, 1],

(Tu2 )(t ) − (Tu1 )(t ) = ρ + + >

m2 − m1 t

Γ (q) Z

m2 − m1

> 0.

Γ (q)




λ0 (u2 ) − λ0 (u1 ) n2 + n1 (1 − t ) − λ1 (u2 ) − λ1 (u1 ) (m2 − m1 t )

1

n1 (1 − s) + n2 (q − 1) (1 − s)q−2 f (s, u2 (s), u02 (s)) − f (s, u1 (s), u01 (s)) ds







0 t

1

Γ (q)

Z




(t − s)q−1 f (s, u2 (s), u02 (s)) − f (s, u1 (s), u01 (s)) ds

0 b

Z

n1 (1 − s)q−1 f (s, u2 (s), u02 (s)) − f (s, u1 (s), u01 (s)) ds




a



X. Liu, M. Jia / Computers and Mathematics with Applications 59 (2010) 2880–2886

2885

We have



m1

Z

1


n1 (1 − s) + n2 (q − 1) (1 − s)q−2 f (s, u(s), u0 (s))ds

(Tu) (t ) = ρ λ1 (u)m1 − λ0 (u)n1 − Γ (q) 0 Z t 1 (t − s)q−2 f (s, u(s), u0 (s))ds + Γ (q − 1) 0  Z t Z 1 1 q −2 0 q −2 0 (1 − s) f (s, u(s), u (s))ds (t − s) f (s, u(s), u (s))ds − ρ m1 n2 = Γ (q − 1) 0 0   Z 1 m 1 n1 q−1 0 (1 − s) f (s, u(s), u (s))ds . + ρ λ1 (u)m1 − λ0 (u)n1 − Γ (q) 0 0



Then noting that 1 − ρ m1 n2 < 0, λ0 (u2 ) − λ0 (u1 ) ≥ 0, λ1 (u2 ) − λ1 (u1 ) ≤ 0 and (3.3), we can get that for t ∈ [0, 1]

(Tu2 ) (t ) − (Tu1 ) (t ) = 0

− ρ m 1 n2

t

Z

1

0

Γ (q − 1)


(t − s)q−2 f (s, u2 (s), u02 (s)) − f (s, u1 (s), u01 (s)) ds

0

1

Z

(1 − s)

q −2

f (s, u2 (s), u2 (s)) − f (s, u1 (s), u1 (s)) ds 0

0








+ ρ (λ1 (u2 ) − λ1 (u1 ))m1

0

− (λ0 (u2 ) − λ0 (u1 ))n1 − ≤

1 − ρ m1 n2

Γ (q − 1)

< 0.

m1 n1

Γ (q)

Γ (q)

1

Z

(1 − s)

q −2

1

(1 − s)

q −1

f (s, u2 (s), u2 (s)) − f (s, u1 (s), u1 (s)) ds 0

0






0

 f (s, u2 (s), u2 (s)) − f (s, u1 (s), u1 (s)) ds + ρ (λ1 (u2 ) − λ1 (u1 ))m1 0

0




0

− (λ0 (u2 ) − λ0 (u1 ))n1 − ≤−

m 1 n1

Z

m 1 n1

Γ (q)

1

Z

(1 − s)

q −1

f (s, u2 (s), u2 (s)) − f (s, u1 (s), u1 (s)) ds 0

0






0

b

Z


(1 − s)q−1 f (s, u2 (s), u02 (s)) − f (s, u1 (s), u01 (s)) ds a

Therefore, (Tu1 )(t ) < (Tu2 )(t ) and (Tu2 )0 (t ) < (Tu1 )0 (t ) for any t ∈ [0, 1] and we can show that T is a strongly increasing operator. Secondly, we prove α1  T α1 . From the definition of T , we have

C q D (T α1 )(t ) = f (t , α1 (t ), α10 (t )), t ∈ (0, 1),   m1 (T α1 )(0) + m2 (T α1 )0 (0) = g0 (α1 (0), α10 (0)) + m1 α1 (0) + m2 α10 (0), 0 0 0  1 (1)) + n1 α1 (1) + n2 α1 (1), n1 (T α001 )(1) + n2 (T α0001 ) (1) = g1 (α1 (1), (αn− 1) (T α1 ) (0) = (T α1 ) (0) = · · · = (T α1 ) (0) = 0. Define α(t ) = (T α1 )(t ) − α1 (t ). Since α1 is a lower solution of the boundary value problem (1.1), then C

Dq α(t ) = C Dq (T α1 )(t ) − C Dq α1 (t ) = f (t , α1 (t ), α10 (t )) − C Dq α1 (t ) ≥ 0,

t ∈ (0, 1),

and m1 α(0) + m2 α 0 (0) = m1 ((T α1 )(0) − α1 (0)) + m2 ((T α1 )0 (0) − α10 (0)) = (m1 (T α1 )(0) + m2 (T α1 )0 (0)) − (m1 α1 (0) + m2 α10 (0)) = g0 (α1 (0), α10 (0)) + m1 α1 (0) + m2 α10 (0) − (m1 α1 (0) + m2 α10 (0)) = g0 (α1 (0), α10 (0)) ≥ 0. n1 α(1) + n2 α 0 (1) = n1 ((T α1 )(1) − α1 (1)) + n2 ((T α1 )0 (1) − α10 (1)) = (n1 (T α1 )(1) + n2 (T α1 )0 (1)) − (n1 α1 (1) + n2 α10 (1)) = g2 (α1 (1), α10 (1)) + n1 α1 (1) + n2 α10 (1) − (n1 α1 (1) + n2 α10 (1)) = g1 (α1 (1), α10 (1)) ≤ 0. It is obvious that α 00 (0) = α 000 (0) = · · · = α n−1 (0) = 0. By Lemma 2.2, we have

α(t ) = (T α1 )(t ) − α1 (t ) ≥ 0,

α 0 (t ) = (T α1 )0 (t ) − α10 (t ) ≤ 0.

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X. Liu, M. Jia / Computers and Mathematics with Applications 59 (2010) 2880–2886

Therefore

α1  T α1 . Similarly, we can get that α2  T α2 . Because α2 is a lower solution of (1.1), and not a solution of (1.1), we have (T α2 ) 6= α2 . Thus

α2 ≺ T α2 . Using the same method, we can also get that T β1 ≺ β1 ,

T β2  β2 .

It follows that T has at least three fixed points x1 , x2 , x3 ∈ [α1 , β2 ] from Lemma 2.3, and moreover,

α1  x1 ≺≺ β1 ,

α2 ≺≺ x2  β2 ,

α2 6 x3 6 β1 .

Hence, the boundary value problem (1.1) has at least three distinct solutions.



4. Illustration To illustrate our main results, we present an example. Example 4.1. Consider the following nonlinear boundary value problem:

 2t 3  C 72  x ( t ) = arctan(x(t ) − x0 (t )), t ∈ (0, 1), D    − cos(π x(0))π+ (2.22 − 3.2x0 (0)) = 0,   cos(14.06x0 (1) − 5.41) + 0.93 − x(1) = 0,    4   00 x (0) = x000 (0) = 0,

(4.1)

3

where q = 72 , f (t , x, y) = 2tπ arctan(x − y), g0 (x, y) = − cos(π x)+(2.22 − 3.2y), g1 (x, y) = cos(14.06y − 5.41)+(0.93 − 4x ). Obviously, f ∈ C ([0, 1] × R × R, R). It is easy to see that f satisfies (H1), and g0 , g1 satisfy (H2) with m1 = π , m2 = 3.2, n1 = 14 and n2 = 14.06.

1 4 1 5 Set α1 (t ) = 1 + t, α2 (t ) = 3 + 21 t, β1 (t ) = 2 + 32 t + 20 t and β2 (t ) = 4 + 25 t + 60 t for t ∈ [0, 1]. We can easily verify that α1 (t ) and α2 (t ) are lower solutions, β1 (t ) and β2 (t ) are upper solutions of the boundary value problem (4.1), and α1 ≺ β1 ≺ α2 ≺ β2 . The conditions of Theorem 3.1 are all satisfied. So by Theorem 3.1, the boundary value problem (4.1) has at least three distinct solutions x1 , x2 , x3 ∈ [α1 , β2 ], and moreover,

α1  x1 ≺≺ β1 ,

α2 ≺≺ x2  β2 ,

α2 6 x3 6 β1 . 

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

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