Hindawi Publishing Corporation International Journal of Diο¬erential Equations Volume 2016, Article ID 6906049, 10 pages http://dx.doi.org/10.1155/2016/6906049
Research Article Multiplicity of Positive Solutions for Fractional Differential Equation with π-Laplacian Boundary Value Problems Sabbavarapu Nageswara Rao Department of Mathematics, Jazan University, P.O. Box 114, Jazan, Saudi Arabia Correspondence should be addressed to Sabbavarapu Nageswara Rao; sabbavarapu
[email protected] Received 15 October 2015; Accepted 6 April 2016 Academic Editor: Kanishka Perera Copyright Β© 2016 Sabbavarapu Nageswara Rao. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited. We investigate the existence of multiple positive solutions of fractional differential equations with π-Laplacian operator π½ π·π+ (ππ (π·ππΌ+ π’(π‘))) = π(π‘, π’(π‘)), π < π‘ < π, π’(π) (π) = 0, π = 0, 1, 2, . . . , π β 2, π’(πΌ1 ) (π) = ππ’(πΌ1 ) (π), ππ (π·ππΌ+ π’(π)) = 0 = π½
π·π+1 (ππ (π·ππΌ+ π’(π))), where π½ β (1, 2], πΌ β (π β 1, π], π β₯ 3, π β (0, β), π β (π, π), π½1 β (0, 1], πΌ1 β {1, 2, . . . , πΌ β 2} is a fixed integer, and ππ (π ) = |π |πβ2 π , π > 1, ππβ1 = ππ , (1/π) + (1/π) = 1, by applying LeggettβWilliams fixed point theorems and fixed point index theory.
1. Introduction
with the boundary conditions
The goal of differential equations is to understand the phenomena of nature by developing mathematical models. Fractional calculus is the field of mathematical analysis, which deals with investigation and applications of derivatives and integrals of an arbitrary order. Among all, a class of differential equations governed by nonlinear differential operators appears frequently and generated a great deal of interest in studying such problems. In this theory, the most applicable operator is the classical π-Laplacian, given by ππ (π’) = |π’|πβ2 , π > 1. The positive solutions of boundary value problems associated with ordinary differential equations were studied by many authors [1β4] and extended to π-Laplacian boundary value problems [5β8]. Later, these results are further extended to fractional order boundary value problems [9β12] by applying various fixed point theorems on cones. Recently, researchers are concentrating on the theory of fractional order boundary value problems associated with π-Laplacian operator [13β19]. The above few papers motivated this work. In this paper, we are concerned with the existence of multiple positive solutions for the fractional differential equation with π-Laplacian operator π½
π·π+ (ππ (π·ππΌ+ π’ (π‘))) = π (π‘, π’ (π‘)) , π < π‘ < π,
(1)
π’(π) (π) = 0,
π = 0, 1, 2, . . . , π β 2,
π’(πΌ1 ) (π) = ππ’(πΌ1 ) (π) ,
(2)
π½
ππ (π·ππΌ+ π’ (π)) = π·π+1 (ππ (π·ππΌ+ π’ (π))) = 0, where ππ (π ) = |π |πβ2 π , π > 1, ππβ1 = ππ , (1/π) + (1/π) = 1, π½ β (1, 2], πΌ β (π β 1, π], π β₯ 3, π½1 β (0, 1], πΌ1 β [1, πΌ β 2] is a fixed integer, and π β (0, β), π β (π, π) are constants. The function π : [π, π] Γ π
+ β π
+ is continuous and π·ππΌ+ , π½ π½ π·π+ , π·π+1 are the standard Riemann-Liouville fractional order derivatives. The rest of this paper is organized as follows. In Section 2, the Green functions for the homogeneous BVPs corresponding to (1)-(2) are constructed and the bounds for the Green functions are estimated. In Section 3, sufficient conditions for the existence of at least two or at least three positive solutions are established, by using fixed point index theory and LeggettWilliams fixed point theorems. In Section 4, as an application, an example is presented to illustrate our main result.
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2. Greenβs Function and Bounds In this section, we construct Greenβs function for the homogeneous boundary value problem and estimate bounds for Greenβs function that will be used to prove our main theorems. Let πΊ(π‘, π ) be Greenβs function for the homogeneous BVP
From π’(π) (π) = 0, 0 β€ π β€ π β 2, we have ππ = ππβ1 = ππβ2 = β
β
β
= π2 = 0. Then π’ (π‘) =
β1 π‘ β« (π‘ β π )πΌβ1 π¦ (π ) ππ + π1 (π‘ β π)πΌβ1 , Ξ (πΌ) π πΌ1
π’(πΌ1 ) (π‘) = π1 β (πΌ β π) (π‘ β π)πΌβπΌ1 β1
(9)
π=1
βπ·ππΌ+ π’ (π‘) = 0, π < π‘ < π,
πΌ1
π’(π) (π) = 0, 0 β€ π β€ π β 2,
(3)
πΌβπΌ1 β1
Lemma 1. Let π = (π β π) β π(π β π) πΆ[π, π], then the fractional order BVP
> 0. If π¦ β
π’
πΌ1
πΌ1
π=1
π=1
π1 β (πΌ β π) (π β π)πΌβπΌ1 β1 β β (πΌ β π)
1 Ξ (πΌ)
π
β
β« (π β π )πΌβπΌ1 β1 π¦ (π ) ππ
π·ππΌ+ π’ (π‘) + π¦ (π‘) = 0, π < π‘ < π, (π)
π=1
π‘ 1 β« (π‘ β π )πΌβπΌ1 β1 π¦ (π ) ππ . Ξ (πΌ) π
From π’(πΌ1 ) (π) = ππ’(πΌ1 ) (π), we have
π’(πΌ1 ) (π) = ππ’(πΌ1 ) (π) . πΌβπΌ1 β1
β β (πΌ β π)
π
(4)
(π) = 0, 0 β€ π β€ π β 2,
π’(πΌ1 ) (π) = ππ’(πΌ1 ) (π)
(10)
πΌ1
= π [π1 β (πΌ β π) (π β π) π=1
(5)
πΌ1
β β (πΌ β π) π=1
π
has a unique solution, π’(π‘) = β«π πΊ(π‘, π )π¦(π )ππ , where π (π‘ β π)πΌβ1 πΊ (π‘, π ) = πΊ1 (π‘, π ) + πΊ2 (π, π ) ; π
πΌβπΌ1 β1
π 1 πΌβπΌ β1 β« (π β π ) 1 π¦ (π ) ππ ] . Ξ (πΌ) π
Therefore (6)
π1 =
1 πΌβπΌ1 β1
Ξ (πΌ) [(π β π)πΌβπΌ1 β1 β π (π β π)
]
π
Γ [β« (π β π )πΌβπΌ1 β1 π¦ (π ) ππ
here
π
π
π
(π‘ β π)πΌβ1 (π β π )πΌβπΌ1 β1 { , π β€ π‘ β€ π β€ π, { { { (π β π)πΌβπΌ1 β1 β
{ πΌβπΌ β1 πΌβ1 { { (π‘ β π) (π β π ) 1 { β (π‘ β π )πΌβ1 , π β€ π β€ π‘ β€ π, { (π β π)πΌβπΌ1 β1 1 πΊ2 (π, π ) = Ξ (πΌ) πΌβπΌ1 β1
πΌβπΌ1 β1
β π β« (π β π )
1 πΊ1 (π‘, π ) = Ξ (πΌ)
π¦ (π ) ππ ] =
1 Ξ (πΌ) [(π β π)πΌβπΌ1 β1 ]
π
β
β« (π β π )πΌβπΌ1 β1 π¦ (π ) ππ π
πΌβπΌ1 β1
(7)
+
π (π β π)
Ξ (πΌ) (π β π)πΌβπΌ1 β1 [(π β π)πΌβπΌ1 β1 β π (π β π) π
(π β π) (π β π ) { { , π β€ π β€ π, { { πΌβπΌ1 β1 { (π β π) β
{ πΌβπΌ β1 { { (π β π) 1 (π β π )πΌβπΌ1 β1 πΌβπΌ β1 { { β (π β π ) 1 , π β€ π β€ π. (π β π)πΌβπΌ1 β1 {
π
β
π
Proof. Assume that π’ β πΆ[πΌ]+1 [π, π] is a solution of fractional order BVP (4)-(5) and is uniquely expressed as πΌππΌ+ π·ππΌ+ π’(π‘) = βπΌππΌ+ π¦(π‘), so that β1 π‘ π’ (π‘) = β« (π‘ β π )πΌβ1 π¦ (π ) ππ + π1 (π‘ β π)πΌβ1 Ξ (πΌ) π + π2 (π‘ β π)
πΌβ2
+ β
β
β
+ ππ (π‘ β π)
πΌβπ
.
π Ξ (πΌ) [(π β π)πΌβπΌ1 β1 β π (π β π) πΌβπΌ1 β1
β
β« (π β π ) π
(8)
πΌβπΌ1 β1
π
π
π [(π β π) π
πΌβπΌ1 β1
π
πΌβπΌ1 β1
β π (π β π)
β
β« πΊ2 (π, π ) π¦ (π ) ππ .
]
1 Ξ (πΌ) (π β π)πΌβπΌ1 β1
π¦ (π ) ππ =
β
β« (π β π )πΌβπΌ1 β1 π¦ (π ) ππ +
]
(11)
β
β« (π β π )πΌβπΌ1 β1 π¦ (π ) ππ
πΌβπΌ1 β1
πΌβπΌ1 β1
]
International Journal of Differential Equations
3 Here π»(π‘, π ) is Greenβs function for
Thus, the unique solution of (4)-(5) is
π½
βπ·π+ (ππ (π₯ (π‘))) = 0, π < π‘ < π,
β1 π‘ π’ (π‘) = β« (π‘ β π )πΌβ1 π¦ (π ) ππ Ξ (πΌ) π +
+
ππ (π₯ (π)) = 0, π½
π (π‘ β π)πΌβ1 β« (π β π )πΌβπΌ1 β1 π¦ (π ) ππ πΌβπΌ1 β1 π Ξ (πΌ) (π β π)
π·π+1 (ππ (π₯ (π))) = 0. Proof. An equivalent integral equation for (13) is given by
π (π‘ β π)πΌβ1 [(π β π)πΌβπΌ1 β1 β π (π β π)
πΌβπΌ1 β1
] (12)
π
π
π
π
ππ (π·ππΌ+ π’ (π‘)) =
π‘ 1 β« (π‘ β π)π½β1 β (π) ππ Ξ (π½) π
+ π1 (π‘ β π)
β
β« πΊ2 (π, π ) π¦ (π ) ππ = β« πΊ1 (π‘, π ) π¦ (π ) ππ +
[(π β π)
π½β1
+ π2 (π‘ β π)
π½β2
(18) .
By (14), one can determine π2 = 0 and π1 = (β1/(π β
π (π‘ β π)πΌβ1 πΌβπΌ1 β1
(17)
π
β π (π β π)
π
π
π
π
πΌβπΌ1 β1
π)π½βπ½1 β1 )(1/Ξ(π½)) β«π (π β π)π½βπ½1 β1 β(π)ππ. Thus, the unique solution of (13), (2) is
]
ππ (π·ππΌ+ π’ (π‘))
β
β« πΊ2 (π, π ) π¦ (π ) ππ = β« πΊ (π‘, π ) π¦ (π ) ππ ,
= where πΊ(π‘, π ) is given in (6).
π‘ 1 β« (π‘ β π)π½β1 β (π) ππ Ξ (π½) π
β
Lemma 2. If β β πΆ[π, π], then the fractional order differential equation
(π‘ β π)π½β1
π 1 β« (π β π)π½βπ½1 β1 β (π) ππ π½βπ½1 β1 Ξ (π½) π (π β π)
(19)
π
= β β« π» (π‘, π) β (π) ππ.
π½
π·π+ (ππ (π·ππΌ+ π’ (π‘))) = β (π‘) ,
π < π‘ < π,
π
(13)
π
Therefore, ππβ1 (ππ (π·ππΌ+ π’(π‘))) = βππβ1 (β«π π»(π‘, π)β(π)ππ). Conπ
sequently, π·ππΌ+ π’(π‘) + ππ (β«π π»(π‘, π)β(π)ππ) = 0. Hence,
satisfying (5) and ππ (π·ππΌ+ π’ (π)) = 0, (14)
π½
π·π+1 (ππ (π·ππΌ+ π’ (π))) = 0
π
π’ (π‘) = β« πΊ (π‘, π ) ππ (β« π» (π , π) β (π) ππ) ππ , π
π
(15)
where π» (π‘, π ) =
π
π
π
(20)
Lemma 3. Greenβs function πΊ(π‘, π ) satisfies the following inequalities:
has a unique solution, π
π
π’ (π‘) = β« πΊ (π‘, π ) ππ (β« π» (π , π) β (π) ππ) ππ .
(ii) πΊ(π‘, π ) β₯ ((π β π)/(π β π))πΌβ1 πΊ(π, π ), πππ πππ (π‘, π ) β [π, π] Γ [π, π]. Proof. Consider Greenβs function given by (6). Let π β€ π‘ β€ π β€ π. Then, we have
1 Ξ (π½)
(π‘ β π)π½β1 (π β π )π½βπ½1 β1 { { , { { { (π β π)π½βπ½1 β1 β
{ π½βπ½ β1 { { (π‘ β π)π½β1 (π β π ) 1 { { β (π‘ β π )π½β1 , π½βπ½1 β1 β π) (π {
(i) πΊ(π‘, π ) β€ πΊ(π, π ), πππ πππ (π‘, π ) β [π, π] Γ [π, π],
1 π (π‘ β π)πΌβ2 (π β π )πΌβπΌ1 β1 [ ] πΊ1 (π‘, π ) = ππ‘ Ξ (πΌ β 1) (π β π)πΌβπΌ1 β1
π β€ π‘ β€ π β€ π, (16) π β€ π β€ π‘ β€ π.
β₯
(1 β π) π‘πΌβ2 [ Ξ (πΌ β 1)
β
(1 β π )πΌβ2 β₯ 0.
πΌβ2 πΌβπΌ1 +1
π
(1 β (1 β πΌ1 ) π + π (π 2 ))
(π β π)πΌβπΌ1 β1
]
(21)
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Let π β€ π β€ π‘ β€ π. Then, we have π 1 (π‘ β π)πΌβ2 (π β π )πΌβπΌ1 β1 β (π‘ β π )πΌβ2 ] πΊ1 (π‘, π ) = [ ππ‘ Ξ (πΌ β 1) (π β π)πΌβπΌ1 β1 β₯
πΌβ2
π‘ [ Ξ (πΌ β 1)
(1 β π)πΌβ2 ππΌβπΌ1 +1 (1 β (1 β πΌ1 ) π + π (π 2 )) β (π β π)πΌβπΌ1 β1 (π β π)πΌβπΌ1 β1
Therefore πΊ1 (π‘, π ) is increasing in π‘, which implies πΊ1 (π‘, π ) β€ πΊ1 (π, π ). Now we prove πΊ2 (π, π ) β₯ 0, π β [π, π] .
β₯
(π β π) 1 (π β π ) 1 [ Ξ (πΌ) (π β π)πΌβπΌ1 β1 (π β π) 1 [ Ξ (πΌ)
] (1 β π )πΌβ2 β₯ 0.
In fact, if π β₯ π, obviously, (23) holds. If π β€ π, one has
(23)
πΌβπΌ β1
πΊ2 (π, π ) =
(22)
πΌβπΌ1 β1
(π β π )
πΌβπΌ1 β1
πΌβπΌ1 β1
β (π β π )
πΌβπΌ1 β1
β (π β ππ )
]
πΌβπΌ1 β1
(24) πΌβπΌ1 β1
(π β π)
πΌβπΌ1 β1
(π β π)
That implies that (23) is also true. Therefore, by (6), (21), and (23), we find
If π β€ π‘ β€ π β€ π, we have πΊ1 (π‘, π ) =
π π (πΌ β 1) π (π‘ β π)πΌβ2 πΊ (π‘, π ) = πΊ1 (π‘, π ) + πΊ2 (π, π ) ππ‘ ππ‘ π (25) β₯ 0. Therefore πΊ(π‘, π ) is increasing with respect to π‘ β [π, π]. Hence the inequality (i) is proved. Now, we establish the inequality (ii). On the other hand, if π β€ π β€ π‘ β€ π, we have
] β₯ 0.
=
1 (π‘ β π)πΌβ1 (π β π )πΌβπΌ1 β1 [ ] Ξ (πΌ) (π β π)πΌβπΌ1 β1
π‘ β π πΌβ1 (π β π)πΌβ1 (π β π )πΌβπΌ1 β1 1 ( ] ) [ Ξ (πΌ) π β π (π β π)πΌβπΌ1 β1
β₯(
(27)
π β π πΌβ1 ) πΊ1 (π, π ) . πβπ
Therefore πΊ1 (π‘, π ) β₯ (
π β π πΌβ1 ) πΊ1 (π, π ) . πβπ
(28)
From (6) and (28) we have πΊ1 (π‘, π )
πΊ (π‘, π ) = πΊ1 (π‘, π ) +
=
1 (π‘ β π)πΌβ1 (π β π )πΌβπΌ1 β1 [ β (π‘ β π )πΌβ1 ] Ξ (πΌ) (π β π)πΌβπΌ1 β1
β₯
1 π‘ β π πΌβ1 ( ) Ξ (πΌ) π β π
(π β π)πΌβ1 (π β π )πΌβπΌ1 β1 β
[ β (π β π )πΌβ1 ] (π β π)πΌβπΌ1 β1 β₯(
πβπ ) πβπ
πΌβ1
πΊ1 (π, π ) .
(26)
π (π‘ β π)πΌβ1 πΊ2 (π, π ) π
β₯(
π β π πΌβ1 π (π‘ β π)πΌβ1 ) πΊ1 (π‘, π ) + πΊ2 (π, π ) (29) πβπ π
β₯(
π β π πΌβ1 ) πΊ (π, π ) . πβπ
Therefore πΊ (π‘, π ) β₯ (
π β π πΌβ1 ) πΊ (π, π ) πβπ
(30) β (π‘, π ) β [π, π] Γ [π, π] .
Hence the inequality (ii) is proved.
International Journal of Differential Equations Lemma 4. Greenβs function π»(π‘, π ) satisfies the following inequalities: (i) π»(π‘, π ) β€ π»(π , π ), πππ πππ (π‘, π ) β [π, π] Γ [π, π], (ii) π»(π‘, π ) β₯ πΎπ»(π , π ), πππ πππ (π‘, π ) β πΌ Γ [π, π],
5 is a compact map such that π₯ =ΜΈ ππ₯ for π₯ β πΞ©π . Thus, one has the following conclusions: (π·1) If βπ₯β < βππ₯β for π₯ β πΞ©π , then π(π, Ξ©π , π) = 0. (π·2) If βπ₯β β₯ βππ₯β for π₯ β πΞ©π , then π(π, Ξ©π , π) = 1.
where πΌ = [(3π + π)/4, (π + 3π)/4] and πΎ = (1/4)π½β1 .
3. Main Results
The method of proof is similar to that [20], and we omit it here.
In this section, the existence of at least two or at least three positive solutions for fractional differential equation with π-Laplacian operator BVP (1)-(2) is established by using fixed point index theory and Leggett-Williams fixed point theorems. Let πΈ = {π’ : π’ β πΆ[π, π]} be the real Banach space equipped with the norm βπ’β = maxπ‘β[π,π] |π’(π‘)|. Define the cone π β πΈ by
Theorem 5 (Leggett-Williams [3]). Let π : ππ β ππ be completely continuous and let π be a nonnegative continuous concave functional on π such that π(π¦) β€ βπ¦β for all π¦ β ππ . Suppose that there exist π, π, π, and π with 0 < π < π < π β€ π such that (π΄1) {π¦ β π(π, π, π) : π(π¦) > π} =ΜΈ 0 and π(ππ¦) > π for π¦ β π(π, π, π),
π = {π’ β πΈ : π’ (π‘) β₯ 0, βπ‘ β [π, π] , min π’ (π‘) π‘β[π,π]
(π΄3) π(ππ¦) > π for π¦ β π(π, π, π) with βππ¦β > π. Then π has at least three fixed points π¦1 , π¦2 , and π¦3 in ππ satisfying βπ¦1 β < π, π < π(π¦2 ), βπ¦3 β > π, and π(π¦3 ) < π. Theorem 6 (see [3]). Let π : ππ β π be a completely continuous operator and let π be a nonnegative continuous concave functional on π such that π(π¦) β€ βπ¦β for all π¦ β ππ . Suppose that there exist π, π, and π with 0 < π < π < π such that (π΅1) {π¦ β π(π, π, π) : π(π¦) > π} =ΜΈ 0 and π(ππ¦) > π for π¦ β π(π, π, π), (π΅2) βππ¦β < π for βπ¦β β€ π, (π΅3) π(ππ¦) > (π/π)βππ¦β for π¦ β ππ with βππ¦β > π. Then π has at least two fixed points π¦1 and π¦2 in ππ satisfying βπ¦1 β < π, βπ¦2 β > π and π(π¦2 ) < π. Theorem 7 (see [21]). Let π be a closed convex set in a Banach space πΈ and let Ξ© be a bounded open set such that Ξ©π fl Ξ©β©π =ΜΈ 0. Let π : Ξ©π β π be a compact map. Suppose that π₯ =ΜΈ ππ₯ for all π₯ β ππ :
(31)
π β π πΌβ1 β₯( ) βπ’β} . πβπ
(π΄2) βππ¦β < π for βπ¦β β€ π,
Let π : π β πΈ be the operator defined by ππ’ (π‘) π
π
π
π
= β« πΊ (π‘, π ) ππ (β« π» (π , π) π (π, π’ (π)) ππ) ππ .
If π’ β π is a fixed point of π, then π’ satisfies (32) and hence π’ is a positive solution of π-Laplacian fractional order BVP (1)-(2). Lemma 9. The operator π defined by (32) is a self-map on π. Proof. Let π’ β π. Clearly, ππ’(π‘) β₯ 0, for all π‘ β [π, π] and π
π
ππ’ (π‘) = β« πΊ (π‘, π ) ππ (β« π» (π , π) π (π, π’ (π)) ππ) ππ π
π
π
π
π
π
(33)
β€ β« πΊ (π, π ) ππ (β« π» (π , π) π (π, π’ (π)) ππ) ππ so that π
π
π
π
βππ’β β€ β« πΊ (π, π ) ππ (β« π» (π , π) π (π, π’ (π)) ππ) ππ . (34)
(πΆ1) Existence: if π(π, Ξ©π , π) =ΜΈ 0, then π has a fixed point in Ξ©π .
On the other hand, by Lemma 3, we have
(πΆ2) Normalization: if π’ β Ξ©π , then π(Μ π’, Ξ©π , π) = 1, where Μ (π₯) = π’ for π₯ β Ξ©π . π’
min ππ’ (π‘) = min β« πΊ (π‘, π )
(πΆ3) Homotopy: let V : [0, 1] Γ Ξ©π β π be a compact map such that π₯ =ΜΈ V(π‘, π₯) for π₯ β πΞ©π and π‘ β [0, 1]. Then π(V(0, β
), Ξ©π , π) = π(V(1, β
), Ξ©π , π). (πΆ4) Additivity: if π1, π2 are disjoint relatively open subsets of Ξ©π such that π₯ =ΜΈ ππ₯ for π₯ β Ξ©π \ (π1 βͺ π2), then π(π, Ξ©π , π) = π(π, π1, π) + π(π, π2, π), where π(π, ππ, π) = π(π | π’π, ππ, π) (π = 1, 2). Theorem 8 (see [22]). Let π be a cone in a Banach space πΈ. For π > 0, define Ξ©π = {π₯ β π : βπ₯β < π}. Assume that π : Ξ©π β π
(32)
π
π‘β[π,π]
π‘β[π,π] π
π
β
ππ (β« π» (π , π) π (π, π’ (π)) ππ) ππ π
β₯(
π β π πΌβ1 π ) β« πΊ (π, π ) πβπ π π
β
ππ (β« π» (π , π) π (π, π’ (π)) ππ) ππ π
β₯(
π β π πΌβ1 ) βππ’β . πβπ
(35)
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International Journal of Differential Equations
Hence ππ’ β π and so π : π β π. Standard argument involving the Arzela-Ascoli theorem shows that π is completely continuous.
π
β₯ π΄πσΈ β« πΊ (π, π ) ππ (β« π
= 2πσΈ (
For convenience of the reader, we denote
πβπΌ
πΎπ» (π, π) ππ) ππ
π β π πΌβ1 π β π πΌβ1 ) ) . > πσΈ ( πβπ πβπ (38)
πΌβ1
π΄=
π΅=
2 ((π β π) / (π β π)) π
β«π πΊ (π, π ) ππ (β«πβπΌ πΎπ» (π, π) ππ) ππ 1 π
π
β«π πΊ (π, π ) ππ (β«π π» (π, π) ππ) ππ
π0 = lim+ min
π’β0 π‘β[π,π]
,
Consequently, min ππ’ (π‘) β₯ (
π‘β[π,π]
,
π (ππ’) > πσΈ ,
Theorem 10. Let π(π‘, π’) be nonnegative continuous on [π, π] Γ [0, β). Assume that there exist constants πσΈ , πσΈ with πσΈ > πσΈ > 0 such that the following conditions are satisfied: σΈ
σΈ
σΈ
(π»1) π(π‘, π’(π‘)) β₯ ππ (π΄π ) for all (π‘, π’) β [π, π] Γ [π , π ((π β π)/(π β π))2(πΌβ1) ]. (π»2) π(π‘, π’(π‘)) < ππ (π΅πσΈ ) for all (π‘, π’) β [π, π] Γ [0, πσΈ ].
Proof. Let π : π β [0, β) be the nonnegative continuous concave functional defined by π(π’) = minπ‘β[π,π] π’(π‘), π’ β π. Evidently, for each π’ β π, we have π(π’) β€ βπ’β. It is easy to see that π : ππσΈ ((πβπ)/(πβπ))2(πΌβ1) β π is completely continuous and πσΈ ((π β π)/(π β π))2(πΌβ1) > πσΈ > πσΈ > 0. We choose π’(π‘) = πσΈ ((π β π)/(π β π))2(πΌβ1) ; then
πβπ π (π’) = π ( ) πβπ σΈ
π
π‘β[π,π]
π‘β[π,π]
π
π
π
π
π
π
π
π
π
< π΅πσΈ β« πΊ (π, π ) ππ (β« π» (π, π) ππ) ππ = πσΈ , which shows that π : ππσΈ β ππσΈ , that is, βππ’β < πσΈ for π’ β ππσΈ . This shows that condition (π΅2) of Theorem 6 is satisfied. Finally, we show that (π΅3) of Theorem 6 also holds. Assume that π’ β ππσΈ ((πβπ)/(πβπ))2(πΌβ1) with βππ’β > πσΈ ((πβπ)/(πβ π))2(πΌβ1) ; then by the definition of cone π, we have
>( =
) | π(π’) > π } =ΜΈ 0. So {π’ β π(π, π , π ((π β π)/(π β π)) 2(πΌβ1) σΈ σΈ Hence, if π’ β π(π, π , π ((πβπ)/(πβπ)) ), then πσΈ β€ π’(π‘) β€ 2(πΌβ1) σΈ π ((π β π)/(π β π)) for π‘ β [π, π]. Thus for π‘ β [π, π], from assumption (π»1), we have
(41)
β
ππ (β« π» (π, π) π (π, π’ (π)) ππ) ππ
σΈ
σΈ
π
β
(β« π» (π , π) π (π, π’ (π)) ππ) ππ ] β€ β« πΊ (π, π )
π‘β[π,π]
(37)
(40)
βππ’β = max |ππ’ (π‘)| = max [β« πΊ (π‘, π ) ππ
>π.
2(πΌβ1)
σΈ
π β π 2(πΌβ1) ). ) πβπ
Therefore, condition (π΅1) of Theorem 6 is satisfied. Now if π’ β ππσΈ , then βπ’β β€ πσΈ . By assumption (π»2), we have
π (ππ’) = min ππ’ (π‘) β₯ (
π β π 2(πΌβ1) ), ) πβπ
2(πΌβ1)
βπ’ β π (π, πσΈ , πσΈ (
π
Then fractional order BVP (1)-(2) has at least two positive solutions π’1 and π’2 satisfying βπ’1 β < πσΈ , minπ‘β[π,π] π’2 (π‘) < πσΈ , and βπ’2 β > πσΈ .
σΈ
(39)
for π β€ π‘ β€ π, πσΈ β€ π’(π‘) β€ πσΈ ((π β π)/(π β π))2(πΌβ1) . That is,
π (π‘, π’) πβ = lim max . π’ββ π‘β[π,π] π (π’) π
π’ β π (π, πσΈ , πσΈ (
π β π πΌβ1 π β π πΌβ1 = πσΈ >( ) Γ πσΈ ( ) πβπ πβπ
(36)
π (π‘, π’) , ππ (π’)
π β π πΌβ1 ) βππ’β πβπ
π β π (πΌβ1) ) βππ’β πβπ
π β π 2(πΌβ1) ) βππ’β πβπ
(42)
πσΈ 2(πΌβ1)
πσΈ ((π β π) / (π β π))
βππ’β .
So condition (π΅3) of Theorem 6 is satisfied. Thus using Theorem 6, π has at least two fixed points. Consequently, boundary value problem (1)-(2) has at least two positive solutions π’1 and π’2 in ππσΈ ((πβπ)/(πβπ))2(πΌβ1) satisfying σ΅©σ΅© σ΅©σ΅© σΈ σ΅©σ΅©π’1 σ΅©σ΅© < π ,
ππ’ (π) π
π
π
π
π
π
π
π
= β« πΊ (π, π ) ππ (β« π» (π , π) π (π, π’ (π)) ππ) ππ β₯ β« πΊ (π, π ) ππ (β« π» (π , π) π (π, π’ (π)) ππ) ππ
min π’2 (π‘) < πσΈ ,
π‘β[π,π]
σ΅©σ΅© σ΅©σ΅© σΈ σ΅©σ΅©π’2 σ΅©σ΅© > π .
(43)
International Journal of Differential Equations
7
Theorem 11. Let π(π‘, π’) be nonnegative continuous on [π, π] Γ [0, β). Assume that there exist constants πσΈ , πσΈ , πσΈ with ((π β π)/(π β π))2(πΌβ1) πσΈ > πσΈ > πσΈ > 0 such that σΈ
σΈ
(π»3) π(π‘, π’(π‘)) < ππ (π΅π ) for all (π‘, π’) β [π, π] Γ [0, π ], (π»4) π(π‘, π’(π‘)) β₯ ππ (π΄πσΈ ) for all (π‘, π’) β [π, π] Γ [πσΈ , πσΈ ((π β π)/(π β π))2(πΌβ1) ], (π»5) π(π‘, π’(π‘)) β€ ππ (π΅πσΈ ) for all (π‘, π’) β [π, π] Γ [0, πσΈ ]. Then fractional order BVP (1)-(2) has at least three positive solutions π’1 , π’2 , and π’3 with βπ’1 β < πσΈ , minπ‘β[π,π] π’2 (π‘) > πσΈ , βπ’3 β > πσΈ , and minπ‘β[π,π] π’3 (π‘) < πσΈ . σΈ
Proof. If π’ β ππσΈ , then βπ’β β€ π . By assumption (π»5), we have
π‘β[π,π]
(π»6) π0 = πβ = β; (π»7) there exists a constant π1 > 0 such that π(π‘, π’) < ππ (π΅π1 ), for (π‘, π’) β [π, π] Γ [0, π1 ], then boundary value problem (1)-(2) has at least two positive solutions π’1 and π’2 such that 0 < βπ’1 β < π1 < βπ’2 β. Proof. From Lemma 1, we obtain π : π β π being completely continuous. In view of π0 = β, there exists π1 β (0, π1 ) such that π(π‘, π’) β₯ ππ (π1 π’), for π β€ π‘ β€ π, 0 < π’ β€ π1 , where π1 β (π΄/2, β). Let Ξ©π1 = {π’ β π | βπ’β < π1 }. Then, for any π’ β πΞ©π1 , we have π
π
π
π
ππ’ (π) = β« πΊ (π, π ) ππ (β« π» (π , π) π
π
β
π (π, π’ (π)) ππ) ππ β₯ β« πΊ (π, π )
βππ’β = max |ππ’ (π‘)| = max [β« πΊ (π‘, π ) ππ π‘β[π,π]
Theorem 12. Let π(π‘, π’) be nonnegative continuous on [π, π] Γ [0, β). If the following assumptions are satisfied:
π
π
π
π
π
π
π
π
π
π
β
ππ (β« π» (π , π) ππ (π1 π’) ππ) ππ β₯ β« πΊ (π,
β
(β« π» (π , π) π (π, π’ (π)) ππ) ππ ] β€ β« πΊ (π, π ) (44)
π
π ) ππ (β«
β
ππ (β« π» (π, π) π (π, π’ (π)) ππ) ππ
πβπΌ
π
σΈ
π
π
π
π
β
ππ (π1 (
σΈ
β€ π΅π β« πΊ (π, π ) ππ (β« π» (π, π) ππ) ππ = π . This shows that π : ππσΈ β ππσΈ . Using the same arguments as in the proof of Theorem 10, we can show that π : ππσΈ β ππσΈ is a completely continuous operator. It follows from conditions (π»3) and (π»4) in Theorem 11 that πσΈ > πσΈ ((π β π)/(π β π))2(πΌβ1) > πσΈ > πσΈ . Similarly to the proof of Theorem 10, we have π : ππσΈ β ππσΈ and π β π 2(πΌβ1) ) | π (π’) > πσΈ } =ΜΈ 0, {π’ β π (π, π , π ( ) πβπ σΈ
(45)
π (ππ’) > πσΈ , for all π’ β π(π, πσΈ , πσΈ ((π β π)/(π β π))2(πΌβ1) ). Moreover, for π’ β π(π, πσΈ , πσΈ ) and βππ’β > πσΈ ((π β π)/(π β π))2(πΌβ1) , we have
π β π πΌβ1 > πσΈ . >π ( ) πβπ
π β π πΌβ1 ) βπ’β) ππ) ππ πβπ
π π β π πΌβ1 ) βπ’β β« πΊ (π, π ) πβπ π
β
ππ (β«
πβπΌ
πΎπ» (π, π) ππ) ππ =
2π1 βπ’β > βπ’β , π΄
which implies βππ’β > βπ’β for π’ β πΞ©π1 . Hence, Theorem 8 implies π (π, Ξ©π1 , π) = 0.
σΈ
π β π πΌβ1 π (ππ’) = min ππ’ (π‘) β₯ ( ) βππ’β π‘β[π,π] πβπ
= π1 (
(47)
πΎπ» (π, π)
On the other hand, since πβ = β, there exists π3 > π1 such that π(π‘, π’) β₯ ππ (π2 π’), for π’ β₯ π3 , where π2 β (π΄/2, β). Let π2 > max{π3 ((π β π)/(π β π))πΌβ1 , π1 } and Ξ©π2 = {π’ β π | βπ’β < π2 }. Then minπ‘β[π,π] π’(π‘) β₯ ((π β π)/(π β π))πΌβ1 βπ’β > π3 , for any π’ β πΞ©π2 . By using the method to get (48), we obtain ππ’(π) > (2π1 /π΄)βπ’β > βπ’β, which implies βππ’β > βπ’β for π’ β πΞ©2 . Thus, from Theorem 8, we have π (π, Ξ©π2 , π) = 0.
(46)
σΈ
So all the conditions of Theorem 5 are satisfied. Thus using Theorem 5, π has at least three fixed points. So, th boundary value problem (1)-(2) has at least three positive solutions π’1 , π’2 , and π’3 with βπ’1 β < πσΈ , minπ‘β[π,π] π’2 (π‘) > πσΈ , βπ’3 β(π‘) > πσΈ , and minπ‘β[π,π] βπ’3 β(π‘) < πσΈ .
(48)
(49)
Finally, let Ξ©π1 = {π’ β π | βπ’β < π1 }. Then, for any π’ β πΞ©π1 , by (π»7), we then get π
π
π
π
ππ’ (π‘) = β« πΊ (π‘, π ) ππ (β« π» (π , π) π (π, π’ (π)) ππ) ππ π
π
π
π
π
π
π
π
β€ β« πΊ (π, π ) ππ (β« π» (π, π) π (π, π’ (π)) ππ) ππ < β« πΊ (π, π ) ππ (β« π» (π, π) ππ (π΅π1 ) ππ) ππ
8
International Journal of Differential Equations π
π
π
π
Next, since πβ = 0, there exists πΏ3 > π2 such that π(π‘, π’) β€ ππ (π3 π’), for π’ β₯ πΏ3 , where π3 β (0, π΅). We consider two cases.
= π΅π1 β« πΊ (π, π ) ππ (β« π» (π, π) ππ) ππ = π1 = βπ’β , (50) which implies βππ’β < βπ’β for π’ β πΞ©π1 . Using Theorem 8 again, we get π (π, Ξ©π1 , π) = 1.
(51)
Note that π1 < π1 < π2 , by the additivity of fixed point index and (48)β(51); we obtain π (π, Ξ©π1 \ Ξ©π1 , π) = π (π, Ξ©π1 , π) β π (π, Ξ©π1 , π) = 1, π (π, Ξ©π2 \ Ξ©π1 , π) = π (π, Ξ©π2 , π) β π (π, Ξ©π1 , π)
(52)
= β1.
ππ’ (π‘) π
π
π
π
π
π
π
π
β€ β« πΊ (π, π ) ππ (β« π» (π, π) π (π, π’ (π)) ππ) ππ β€ β« πΊ (π, π ) ππ (β« π» (π, π) ππ (π) ππ) ππ π
π
π
π
= π β« πΊ (π, π ) ππ (β« π» (π, π) ππ) ππ =
(55)
π < πΏ4 π΅
= βπ’β .
Hence, π has a fixed point π’1 in Ξ©π1 \ Ξ©π1 , and it has a fixed point π’2 in Ξ©π2 \ Ξ©π1 . Clearly, π’1 and π’2 are positive solutions of boundary value problem (1)-(2) and 0 < βπ’1 β < π1 < βπ’2 β. Theorem 13. Let π(π‘, π’) be nonnegative continuous on [π, π] Γ [0, β). If the following assumptions are satisfied: (π»8) π0 = πβ = 0; (π»9) there exists a constant π2 > 0 such that π(π‘, π’) > ππ (π΄π2 ), for (π‘, π’) β [π, π]Γ[((πβπ)/(πβπ))πΌβ1 π2 , π2 ], then boundary value problem (1)-(2) has at least two positive solutions π’1 and π’2 such that 0 < βπ’1 β < π2 < βπ’2 β. Proof. From Lemma 9, we obtain π : π β π being completely continuous. In view of π0 = 0, there exists πΏ1 β (0, π2 ) such that π(π‘, π’) β€ ππ (π2 π’), for π β€ π‘ β€ π, 0 < π’ β€ πΏ1 , where π2 β (0, π΅). Let Ξ©πΏ1 = {π’ β π | βπ’β < πΏ1 }. Then, for any π’ β πΞ©πΏ1 , we have
Case (ii). Suppose that π is unbounded. In view of π : [π, π] Γ [0, β) β [0, β) being continuous, there exist π‘β β [π, π] and πΏ5 > max{((π β π)/(π β π))πΌβ1 πΏ3 , π2 } such that π(π‘, π’) β€ π(π‘β , πΏ5 ), for π β€ π‘ β€ π, 0 β€ π’ β€ πΏ5 . Then, for π’ β π with βπ’β = πΏ5 , we obtain ππ’ (π‘) π
π
π
π
π
π
π
π
π
π
π
π
β€ β« πΊ (π, π ) ππ (β« π» (π, π) π (π, π’ (π)) ππ) ππ β€ β« πΊ (π, π ) ππ (β« π» (π, π) π (π‘β , πΏ5 ) ππ) ππ (56)
β€ β« πΊ (π, π ) ππ (β« π» (π, π) ππ (π3 βπ’β) ππ) ππ π
π
π
π
β€ π3 πΏ5 β« πΊ (π, π ) ππ (β« π» (π, π) ππ) ππ =
π3 πΏ5 π΅
< πΏ5 = βπ’β .
ππ’ (π‘) π
π
π
π
π
π
π
π
So, in either case, if we always choose Ξ©πΏ2 = {π’ β πΈ | βπ’β < πΏ2 = max{πΏ4 , πΏ5 }}, then we have βππ’β < βπ’β, for π’ β πΞ©2 . Thus, from Theorem 8, we have
β€ β« πΊ (π, π ) ππ (β« π» (π, π) π (π, π’ (π)) ππ) ππ β€ β« πΊ (π, π ) ππ (β« π» (π, π) ππ (π2 π’) ππ) ππ π
π
π
π
(57)
Finally, let Ξ©π2 = {π’ β π | βπ’β < π2 }. Then, for any π’ β πΞ©π2 , minπ‘β[π,π] π’(π‘) β₯ ((πβπ)/(πβπ))πΌβ1 βπ’β = ((πβπ)/(πβπ))πΌβ1 π2 , by (π»9), and we then obtain
π2 βπ’β < βπ’β , π΅
ππ’ (π)
which implies βππ’β < βπ’β for π’ β πΞ©πΏ1 . Hence, Theorem 8 implies π (π, Ξ©πΏ1 , π) = 0.
π (π, Ξ©πΏ2 , π) = 1.
(53)
β€ π2 βπ’β β« πΊ (π, π ) ππ (β« π» (π, π) ππ) ππ =
Case (i). Suppose that π is bounded, which implies that there exists π > 0 such that π(π‘, π’) β€ ππ (π) for all π‘ β [π, π] and π’ β [0, β). Take πΏ4 > max{π/π΅, πΏ3 }. Then, for π’ β π with βπ’β = πΏ4 , we get
(54)
π
π
π
π
π
π
π
π
= β« πΊ (π, π ) ππ (β« π» (π , π) π (π, π’ (π)) ππ) ππ β₯ β« πΊ (π, π ) ππ (β« π» (π , π) ππ (π΄π2 ) ππ) ππ
International Journal of Differential Equations π
β₯ π΄π2 β« πΊ (π, π ) ππ (β« π
πβπΌ
9 (ii) π(π‘, π’(π‘)) β₯ 68.364 = ππ (π΄πσΈ ) for all (π‘, π’) β [0.5, 2] Γ [2, 16.0049],
πΎπ» (π, π) ππ) ππ
(iii) π(π‘, π’(π‘)) β€ 104.52 = ππ (π΅πσΈ ) for all (π‘, π’) β [0, 1] Γ [0, 100].
π β π πΌβ1 ) π2 > π2 = βπ’β , = 2( πβπ (58) which implies βππ’β > βπ’β for π’ β πΞ©π2 . An application of Theorem 8 again shows that π (π, Ξ©π2 , π) = 0.
Consequently, all of the conditions of Theorem 11 are satisfied. With the use of Theorem 11, boundary value problem (61) has at least three positive solutions π’1 , π’2 , and π’3 with σ΅©σ΅© σ΅©σ΅© σ΅©σ΅©π’1 σ΅©σ΅© < 1,
(59)
min π’2 (π‘) > 2,
π‘β[1/2,1]
Note that πΏ1 < π2 < πΏ2 by the additivity of fixed point index and (54)β(59); we obtain
σ΅©σ΅© σ΅©σ΅© σ΅©σ΅©π’3 σ΅©σ΅© > 1,
π (π, Ξ©π2 \ Ξ©πΏ1 , π) = π (π, Ξ©π2 , π) β π (π, Ξ©πΏ1 , π) = β1,
min π’3 (π‘) < 2.
π‘β[1/2,1]
(60)
Competing Interests
π (π, Ξ©πΏ2 \ Ξ©π2 , π) = π (π, Ξ©πΏ2 , π) β π (π, Ξ©π2 , π) = 1. Hence, π has a fixed point π’1 in Ξ©π2 \ Ξ©πΏ1 , and it has a fixed point π’2 in Ξ©πΏ2 \ Ξ©π2 . Consequently, π’1 and π’2 are positive solutions of boundary value problem (1)-(2) and 0 < βπ’1 β < π2 < βπ’2 β.
4. Example In this section, we consider boundary value problem of the fractional differential equation 2.5 π·01.5 + (ππ (π·0+ π’ (π‘))) = π (π‘, π’ (π‘)) ,
σΈ
π’ (0) = 0, 1 1 π’σΈ (1) = π’σΈ ( ) , 3 2
(61)
The author expresses his gratitude to his guide Professor K. Rajendra Prsasd.
References
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= 0,
where π‘ { 0 β€ π’ β€ 2, + 12π’5 , { { 100 π (π‘, π’) = { { { π‘ + π’ + 58, π’ β₯ 2. { 100
Acknowledgments
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ππ (π·02.5 + π’ (0)) = 0, (ππ (π·02.5 + π’ (1)))
The author declares that he has no competing interests regarding the publication of this paper.
[1] R. P. Agarwal, D. OβRegan, and P. J. Wong, Positive Solutions of Differential, Difference and Integral Equations, Kluwer Academic, Dordrecht, The Netherlands, 1999.
0 < π‘ < 1,
π’ (0) = 0,
π·00.5 +
(63)
(62)
Let π = 1/2. We note that π = 0, π = 1, π = 1/2, π = 1/3, π½ = 3/2, π½1 = 1/2, π = 3, πΌ = 5/2, and πΌ1 = 1. By a simple calculation, we obtain ((π β π)/(π β π))πΌβ1 = 0.3535, πΎ = 0.5, π΅ = 1.0452, and π΄ = 34.1482. Choosing πσΈ = 1, πσΈ = 2, and πσΈ = 100, then 0 < πσΈ < πσΈ < ((π β π)/(π β π))2(πΌβ1) πσΈ and π satisfies (i) π(π‘, π’(π‘)) < 1.0452 = ππ (π΅πσΈ ) for all (π‘, π’) β [0, 1] Γ [0, 1],
[5] S. Nageswararao, βSolvability of second order delta-nabla pLaplacian m-point eigenvalue problem on time scales,β TWMS Journal of Applied and Engineering Mathematics, vol. 5, no. 1, pp. 98β109, 2015. [6] M. Ramaswamy and R. Shivaji, βMultiple positive solutions for classes of p-Laplacian equations,β Differential and Integral Equations, vol. 17, no. 11-12, pp. 1255β1261, 2004. [7] C. Yang and J. Yan, βPositive solutions for third-order SturmLiouville boundary value problems with p-Laplacian,β Computers & Mathematics with Applications, vol. 59, no. 6, pp. 2059β 2066, 2010. [8] C. Zhai and C. Guo, βPositive solutions for third-order SturmLiouville boundary-value problems with p-Laplacian,β Electronic Journal of Differential Equations, vol. 2009, no. 154, pp. 1β9, 2009.
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