(ProQuest: ... denotes non-US-ASCII text omitted.)
Academic Editor:Bashir Ahmad
School of Mathematics Science, University of Electronic Science and Technology of China, Chengdu 611731, China
Received 28 February 2014; Accepted 26 April 2014; 15 May 2014
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.
1. Introduction
In this paper, we consider boundary value problem for a fractional differential equation with p -Laplacian operator [figure omitted; refer to PDF] where 2 < α ...4; 3 , 0 < η < 1 is a constant and λ > 0 is a parameter. p > 1 , [varphi] p ( s ) is the p -Laplacian operator; that is, [varphi] p ( s ) = | s | p - 2 s , [varphi] p - 1 ( s ) = [varphi] q ( s ) , 1 / p + 1 / q = 1 . D denotes the Caputo fractional derivative.
Fractional differential equations have been of great interest recently. It is caused both by the intensive development of the theory of fractional calculus itself and by the applications of such constructions in various fields of sciences and engineering such as physics, chemistry, aerodynamics, electrodynamics of complex medium, electrical circuits, and biology (see [1-7] and their references). Differential equations with p -Laplacian arise naturally in non-Newtonian mechanics, nonlinear elasticity, glaciology, population biology, combustion theory, and nonlinear flow laws. Since the p -Laplacian operator and fractional calculus arise from so many applied fields, the fractional p -Laplacian differential equations are worth studying. Recently, there have appeared a very large number of papers which are devoted to the existence of solutions of boundary value problems and initial value problems for the fractional differential equations (see [8-12]), and the existence of solutions of boundary value problems for the fractional p -Laplacian differential equations has just begun in recent years (see [13-22]). On the other hand, there are few papers that consider the eigenvalue intervals of fractional boundary value problems (see [23, 24]). In [23], the author discussed the following system: [figure omitted; refer to PDF] By the use of appropriate conditions with respect to lim ... t [arrow right] 0 ( f ( t ) / t ) = f 0 and lim ... t [arrow right] ∞ ( f ( t ) / t ) = f ∞ , the author proved that the above problem has at least one or two positive solutions for some λ , where f 0 = 0 , l , ∞ , f ∞ = 0 , l , ∞ , 0 < l < ∞ . But almost all the results which the author obtained depend on both f 0 and f ∞ ; the case depends on one of f 0 and f ∞ ; the author only discussed f 0 = 0 , ∞ and f ∞ = 0 , ∞ ; the case f 0 = l or f ∞ = l has not been discussed. On the other hand, there exist several results on the existence of one solution to fractional p -Laplacian boundary value problems (BVPs); there are, to the best of our knowledge, relatively few results on the nonexistence and the uniqueness of positive solutions to fractional p -Laplacian differential equation with parameter.
Motivated by the above questions, in this paper, we will establish several sufficient conditions for the existence of positive solutions of (1) by using fixed point theorem and fixed point index theory.
The work is organized in the following fashion. In Section 2, we provide some necessary background. In particular, we will introduce some lemmas and definitions associated with fixed point index theory. The main results will be stated and proved in Section 3. Two examples are given in Section 4.
2. Preliminaries
In this section, we introduce definitions and preliminary facts which are used throughout this paper.
Definition 1 (see [1]).
The Riemann-Liouville fractional integral of order α > 0 of a function x : ( 0 , + ∞ ) [arrow right] R is given by [figure omitted; refer to PDF] provided that the right side is point-wise defined on ( 0 , + ∞ ) and Γ is the Gamma function.
Definition 2 (see [1]).
The Caputo fractional derivative of order α > 0 of a continuous function x : ( 0 , + ∞ ) [arrow right] R is given by [figure omitted; refer to PDF] where n = [ α ] + 1 , provided that the right side is point-wise defined on ( 0 , + ∞ ) .
Lemma 3 (see [1]).
Let α > 0 , and assume that x ∈ C [ 0,1 ] , and then the fractional differential equation D 0 α x ( t ) = 0 has unique solutions: [figure omitted; refer to PDF]
Lemma 4 (see [1]).
Let α > 0 . Then, [figure omitted; refer to PDF] where c i ∈ R , i = 0,1 , 2 , ... , n - 1 , n - 1 < α ...4; n .
Lemma 5 (see [25]).
Let P be a cone in a real Banach space E , and let Ω be a bounded open set of E . Assume that operator A : P ∩ Ω - [arrow right] P is completely continuous. If there exists a x 0 > 0 such that [figure omitted; refer to PDF] then i ( A , P ∩ Ω , P ) = 0 .
Lemma 6 (see [26]).
Let E be a Banach space and P ⊆ E a cone, and Ω 1 and Ω 2 are open set with 0 ∈ Ω 1 , Ω - 1 ⊆ Ω 2 , and let T : P ∩ ( Ω - 2 \ Ω 1 ) [arrow right] P be completely continuous operator such that either
(i) || T u || ...4; || u || , u ∈ P ∩ ∂ Ω 1 , and || T u || ...5; || u || , u ∈ P ∩ ∂ Ω 2
or
(ii) || T u || ...5; || u || , u ∈ P ∩ ∂ Ω 1 , and || T u || ...4; || u || , u ∈ P ∩ ∂ Ω 2
holds. Then T has a fixed point in P ∩ ( Ω - 2 \ Ω 1 ) .
3. Main Results
In this section, we present some new results on the existence, multiplicity, nonexistence, and the uniqueness of positive solution of problem (1) and dependence of the positive solution x λ ( t ) on the parameter λ .
Let E = C [ 0,1 ] ; then E is a real Banach space with the norm || · || defined by || x || = max ... 0 ...4; t ...4; 1 | x ( t ) | .
Lemma 7.
Let y ( t ) ∈ C ( 0,1 ) ∩ L 1 ( 0,1 ) and y ( t ) ...5; 0 ; then the solution of the problem [figure omitted; refer to PDF] is given by [figure omitted; refer to PDF] where 2 < α ...4; 3 , 0 < η < 1 , 1 - η ∫ 0 1 ... g ( s ) d s > 0 and σ = η ∫ 0 1 ... ( 1 - s ) g ( s ) d s / ( 1 - η ∫ 0 1 ... s g ( s ) d s ) .
Proof.
It is easy to see by integration of (8) that [figure omitted; refer to PDF] By the boundary condition x [variant prime] ( 0 ) = x [variant prime][variant prime] ( 0 ) = 0 , we can easily get that D 0 + α x ( 0 ) = 0 , and we obtain that [figure omitted; refer to PDF] that is [figure omitted; refer to PDF] By Lemmas 3 and 4, we get that [figure omitted; refer to PDF] Using the boundary condition x [variant prime] ( 0 ) = x [variant prime][variant prime] ( 0 ) = 0 , x ( 1 ) = η ∫ 0 1 ... g ( s ) x ( s ) d s , we get c 1 = c 2 = 0 and [figure omitted; refer to PDF] Thus, [figure omitted; refer to PDF] Direct differentiation of (9) implies [figure omitted; refer to PDF] By differentiation of (17), we get [figure omitted; refer to PDF]
Thus the solution of problem (8) is nonincreasing and concave on [ 0,1 ] , and [figure omitted; refer to PDF] On the other hand, as we assume that y ( t ) ...5; 0 , we see that [figure omitted; refer to PDF] Therefore, x ( t ) ...5; 0 and is concave for t ∈ [ 0,1 ] . So for every t ∈ [ 0,1 ] , [figure omitted; refer to PDF] Therefore, [figure omitted; refer to PDF] Since x ( 1 ) = η ∫ 0 1 ... x ( t ) g ( t ) d t , 1 - η ∫ 0 1 ... g ( t ) d t > 0 , we have [figure omitted; refer to PDF] that is, [figure omitted; refer to PDF] The Lemma is proved.
We construct a cone P in E by [figure omitted; refer to PDF] where σ is defined in Lemma 7. It is easy to see P is a closed convex cone of E .
Define T : P [arrow right] E by [figure omitted; refer to PDF]
It is clear that the fixed points of the operator [varphi] q ( λ ) T are the solutions of the boundary value problems (1).
We make the following hypotheses:
(H1) f : [ 0,1 ] × [ 0 , ∞ ) [arrow right] [ 0 , ∞ ) is continuous;
(H2) ω : ( 0,1 ) [arrow right] [ 0 , ∞ ) is continuous and not identical zero on any closed subinterval of ( 0,1 ) with 0 < ∫ 0 1 ... ω ( t ) d t < + ∞ ;
(H3) 1 - η ∫ 0 1 ... g ( s ) d s > 0 ;
(H4) f [varphi] 0 ∈ [ 0 , ∞ ] , where f [varphi] 0 : = lim ... x [arrow right] 0 ( f ( t , x ) / [varphi] p ( x ) ) uniformly for t ∈ [ 0,1 ] ;
(H5) f [varphi] ∞ ∈ [ 0 , ∞ ] , where f [varphi] ∞ : = lim ... x [arrow right] ∞ ( f ( t , x ) / [varphi] p ( x ) ) uniformly for t ∈ [ 0,1 ] ;
(H6) f ( t , x ) > 0 for any t ∈ [ 0,1 ] and x > 0 .
Lemma 8.
Assume that (H1)-(H3) hold. Then T : P [arrow right] E is completely continuous.
Proof.
First, we show that T is continuous. It is easy to see T ( P ) ⊂ P . For x n ( t ) ∈ E and x n ( t ) [arrow right] x ( t ) , as n [arrow right] ∞ , by the continuous of f ( t , x ( t ) ) , we get f ( t , x n ( t ) ) [arrow right] f ( t , x ( t ) ) , as n [arrow right] ∞ . This implies that [figure omitted; refer to PDF] So we have [figure omitted; refer to PDF] Denote Π = | ( T x n ) ( t ) - ( T x ) ( t ) | ; then [figure omitted; refer to PDF]
Hence, T is continuous.
Second, we show that T is compact.
For x ( t ) ∈ B r = { x ∈ P : || x || ...4; r } , by condition (H1), f ( t , x ( t ) ) < ∞ . Denote M = max ... 0 ...4; t ...4; 1 , || x || ...4; r f ( t , x ( t ) ) .
Then [figure omitted; refer to PDF] that is, T maps bounded sets into bounded sets in P .
For x ( t ) ∈ B r , 0 < t 1 < t 2 < 1 , [figure omitted; refer to PDF] By mean value theorem, we obtain that [figure omitted; refer to PDF] Thus [figure omitted; refer to PDF] This shows that | T x ( t 2 ) - T x ( t 1 ) | [arrow right] 0 , as t 2 - t 1 [arrow right] 0 , so { T x : x ∈ B r } is equicontinuous. Therefore, the operator T : P [arrow right] P is completely continuous by the Arzela-Ascoli theorem.
Now for convenience we introduce the following notations. Let [figure omitted; refer to PDF]
Theorem 9.
Assume that (H1)-(H5) hold.
(i) If 0 < f [varphi] 0 < + ∞ , then there exists ξ 0 > 0 such that, for every 0 < r < ξ 0 , problem (1) has a positive solution x r ( t ) satisfying || x r || = r associated with [figure omitted; refer to PDF] where λ - 0 and λ 0 are two positive finite numbers.
(ii) If 0 < f [varphi] ∞ < + ∞ , then there exists ξ 0 * > 0 such that, for every R > ξ 0 * , problem (1) has a positive solution x R ( t ) satisfying || x R || = R for any [figure omitted; refer to PDF] where λ - 0 * and λ 0 * are two positive finite numbers.
(iii): If f [varphi] 0 = + ∞ , then there exists ξ 1 > 0 such that, for any 0 < r * < ξ 1 , problem (1) has a positive solution x r * ( t ) satisfying || x r * || = r * for any [figure omitted; refer to PDF] where λ * is a positive finite number.
(iv) If f [varphi] ∞ = + ∞ , then there exists ξ - 1 > 0 such that, for every R * > ξ - 1 , problem (1) has a positive solution x R * ( t ) satisfying || x R * || = R * for any [figure omitted; refer to PDF] where λ * is a positive finite number.
(v) If there exists ξ > 0 such that m ( r ~ ) > ξ , then problem (1) has positive solution x r ~ ( t ) satisfying || x r ~ || = r ~ for any [figure omitted; refer to PDF] where λ ~ is a positive finite number.
Proof.
(i) It follows from 0 < f [varphi] 0 < + ∞ that there exists l 1 , l 2 , ρ > 0 , such that [figure omitted; refer to PDF]
Let ξ 0 = ρ / σ ; we show that ξ 0 is required. When x ∈ P ∩ ∂ Ω r , we have [figure omitted; refer to PDF] we need x ( t ) < ρ ; this implies that σ r < ρ , as r < ξ 0 , so σ r < σ ξ 0 = ρ .
Let λ 0 = ( 1 / l 1 ) ( Γ ( α ) / σ γ 1 ) p - 1 . Then we may assume that [figure omitted; refer to PDF] if not, then there exists x r ∈ P ∩ ∂ Ω r such that [varphi] q ( λ 0 ) T x r = x r , and then (35) already holds for λ r = λ 0 .
Define ψ ( t ) ...1; 1 , ∀ t ∈ [ 0,1 ] ; then ψ ( t ) ∈ P , and || ψ || ...1; 1 . We now show that [figure omitted; refer to PDF] In fact, if there exists x 1 ∈ P ∩ ∂ Ω r , κ 1 ...5; 0 such that x 1 - [varphi] q ( λ 0 ) T x 1 = κ 1 ψ , then (42) implies that κ 1 > 0 . On the other hand x 1 = [varphi] q ( λ 0 ) T x 1 + κ 1 ψ ...5; κ 1 ψ ; we may choose κ * = max ... { κ |" x 1 ...5; κ ψ } } ; then κ 1 ...4; κ * < + ∞ , x 1 ...5; κ * ψ . Therefore [figure omitted; refer to PDF] Consequently, for any t ∈ [ 0,1 ] , (26) and (44) imply that [figure omitted; refer to PDF] which implies that x 1 ( t ) ...5; ( κ * + κ ) ψ ( t ) , t ∈ [ 0,1 ] , which is a contradiction to the definition of κ * . Thus, (42) holds and, by Lemma 5, the fixed point index [figure omitted; refer to PDF] On the other hand, by the fact that the fixed point index of constants operator is 1, so [figure omitted; refer to PDF] where [vartheta] is the zero operator. It follows therefore from (46) and (47) and the homotopy invariance property that there exist x r ∈ P ∩ ∂ Ω r and 0 < ν 0 < 1 such that ν 0 [varphi] q ( λ 0 ) T x r = x r , but x r = [varphi] q ( λ r ) T x r , which implies that ν 0 [varphi] q ( λ 0 ) T x r = [varphi] q ( λ r ) T x r ; it follows that we get ν 0 [varphi] q ( λ 0 ) = [varphi] q ( λ r ) ) ; that is, [figure omitted; refer to PDF] and then [figure omitted; refer to PDF] From the proof above, for any r < ξ 0 , there exists a positive solution x r ∈ P ∩ ∂ Ω r associated with λ = λ r > 0 .
Thus [figure omitted; refer to PDF] with || x r || = r .
Next, we show that λ r ...5; λ - 0 . In fact, [figure omitted; refer to PDF] which implies that [figure omitted; refer to PDF] that is, [figure omitted; refer to PDF] it follows that we get [figure omitted; refer to PDF] In conclusion, λ r ∈ [ λ - 0 , λ 0 ] .
(ii) It follows from 0 < f [varphi] ∞ < ∞ that there exist l 1 * , l 2 * , ρ * > 0 such that [figure omitted; refer to PDF] Let ξ 0 * = ρ * / σ , the following proof are similar to (i).
(iii) It follows from f [varphi] 0 = + ∞ , there exists l * > 0 , ρ * > 0 , such that [figure omitted; refer to PDF] Let ξ 1 = ρ * / σ ; we show that ξ 1 is required. When x ∈ P ∩ ∂ Ω r * , we have x ( t ) ...5; σ || x || = σ r * ; since x ...4; ρ * , then we need σ r * < ρ * , as r * < ξ 1 , so σ r * < ρ * holds.
Let λ * = ( 1 / l * ) ( Γ ( α ) / σ γ 1 ) 1 / ( q - 1 ) ; we proceed in the same way as in the proof of (i): replacing (42) we may assume that x - [varphi] q ( λ * ) T x ...0; 0 , ( ∀ x ∈ P ∩ ∂ Ω r * ) , and replacing (43) we can prove x - [varphi] q ( λ * ) T x ...0; κ ψ , ( ∀ x ∈ P ∩ ∂ Ω r * , κ ...5; 0 ) . It follows from Lemma 5 that i ( [varphi] q ( λ * ) T , P ∩ Ω r * , P ) = 0 . Note that i ( θ , P ∩ Ω r * , P ) = 1 ; we can easily show that there exists x r * ∈ P ∩ ∂ Ω r * and 0 < ν r * < 1 such that ν r * [varphi] q ( λ * ) T x r * = x r * . Hence (37) holds for λ r * = λ * ν r * 1 - p < λ * .
The proof of Theorem 9(iv) follows by the method similar to Theorem 9(iii); we omit it here.
(v) It follows that m ( r ~ ) > ξ ; for any x ∈ P ∩ ∂ Ω r ~ , we have σ r ~ ...4; x ...4; r ~ and f ( t , x ) / [varphi] p ( x ) ...5; f ( t , x ) / [varphi] p ( r ~ ) ...5; m ( r ~ ) ...5; ξ ; then [figure omitted; refer to PDF] Let λ ~ = ( 1 / ξ ) ( Γ ( α ) / σ γ 1 ) p - 1 . The following proof is similar to that of (iv). This finished the proof of (v).
Remark 10.
In Theorem 9, all the criteria obtained depend on one of f [varphi] 0 and f [varphi] ∞ .
Let Ψ ( x ) = [varphi] q ( λ ) T x ; the following theorems give out the multiply, nonexistence, and the dependence of parameter.
Theorem 11.
Assume that (H1)-(H5) hold.
(i) If f [varphi] 0 = 0 and f [varphi] ∞ = 0 , lim ... _ x [arrow right] ∞ f ( t , x ) = f ~ ∞ ∈ ( 0 , + ∞ ] uniformly for t ∈ [ 0,1 ] , then there exists λ * > 0 such that the problem (1) has two solutions for any λ > λ * .
(ii) If f [varphi] 0 = 0 and f [varphi] ∞ = 0 , then there exists λ _ > 0 such that, for any λ < λ _ , problem (1) has no solution.
Proof.
(i) It follows from f ~ ∞ ∈ ( 0 , + ∞ ] that, for any ζ ~ > 0 , there exists ρ ~ > 0 , such that [figure omitted; refer to PDF]
Since f [varphi] 0 = 0 , there exists [straight epsilon] 1 > 0 , ρ - 1 > 0 such that [figure omitted; refer to PDF] where [straight epsilon] 1 satisfied [varphi] q ( λ [straight epsilon] 1 ) γ 2 / Γ ( α + 1 ) ( 1 - η ∫ 0 1 ... g ( s ) d s ) ...4; 1 .
For ∀ x ∈ P ∩ ∂ Ω ρ - 1 , we have [figure omitted; refer to PDF] This implied [figure omitted; refer to PDF] Let ρ - 3 > max ... { ρ ~ / σ , ρ - 1 } and λ * = ( 1 / ζ ~ ) ( ρ - 3 Γ ( α ) / σ γ 1 ) p - 1 ; then ∀ x ∈ P ∩ ∂ Ω ρ - 3 ; we have σ || x || ...4; x ...4; ρ - 3 and [figure omitted; refer to PDF] which implies [figure omitted; refer to PDF]
Next, for f [varphi] ∞ = 0 , there exist [straight epsilon] 2 > 0 , ρ - 2 > ρ - 3 > 0 such that [figure omitted; refer to PDF] where [straight epsilon] 2 satisfied [varphi] q ( λ [straight epsilon] 2 ) γ 2 / Γ ( α + 1 ) ( 1 - η ∫ 0 1 ... g ( s ) d s ) ...4; 1 .
Similar to the above proof, we get [figure omitted; refer to PDF]
Applying Lemma 7 to (61), (65), and (63) yields that Ψ has two fixed points x 1 , x 2 such that x 1 ∈ P ∩ ( Ω - ρ - 3 \ Ω ρ - 1 ) and x 2 ∈ P ∩ ( Ω - ρ - 2 \ Ω ρ - 3 ) .
(ii) It follows from f [varphi] 0 = 0 and f [varphi] ∞ = 0 that there exists [straight epsilon] 1 , [straight epsilon] 2 , ρ - 1 , ρ - 2 > 0 , such that [figure omitted; refer to PDF] then for x ∈ [ ρ - 1 , ρ - 2 ] , by the continuous of f ( t , x ( t ) ) , there exists x - ∈ [ ρ - 1 , ρ - 2 ] such that [figure omitted; refer to PDF] Let Υ _ = max ... { [straight epsilon] 1 , [straight epsilon] 2 , f ( t , x - ) / [varphi] p ( x - ) } > 0 ; then [figure omitted; refer to PDF] Assuming x ( t ) is a positive solution of problem (1), we will show that this leads to a contradiction for λ < λ _ , where λ _ = ( 1 / Υ _ ) ( Γ ( α + 1 ) ( 1 - η ∫ 0 1 ... g ( s ) d s ) / γ 2 ) p - 1 . It follows from (26) that [figure omitted; refer to PDF] This implies that 1 ...4; [varphi] q ( λ Υ _ ) γ 2 / Γ ( α + 1 ) ( 1 - η ∫ 0 1 ... g ( s ) d s ) = ( λ Υ _ ) q - 1 γ 2 / Γ ( α + 1 ) ( 1 - η ∫ 0 1 ... g ( s ) d s ) < ( λ _ Υ _ ) q - 1 γ 2 / Γ ( α + 1 ) ( 1 - η ∫ 0 1 ... g ( s ) d s ) = 1 , which is a contradiction. This finishes the proof of (ii).
Remark 12.
In (i), the condition lim ... _ x [arrow right] ∞ f ( t , x ) = f ~ ∞ ∈ ( 0 , + ∞ ] uniformly for t ∈ [ 0,1 ] can be replaced with condition (H6). The same methods can be used to prove the results.
Remark 13.
From the above proof, the condition f ~ ∞ ∈ ( 0 , + ∞ ] is important to keep the problem (1) having at least two positive solutions, if this condition is not satisfied, then there exists λ _ > 0 small enough such that the problem (1) has no positive solutions for all λ < λ _ .
Theorem 14.
Assume that (H1)-(H6) hold.
(i) If f [varphi] 0 = + ∞ and f [varphi] ∞ = + ∞ , then there exists λ * 0 > 0 such that problem (1) has at least two solutions for λ ∈ ( 0 , λ * 0 ) .
(ii) If f [varphi] 0 = + ∞ and f [varphi] ∞ = + ∞ , then there exists λ - > 0 such that problem (1) has no solution for all λ > λ - .
Proof.
(i) It follows from f [varphi] 0 = + ∞ that there exists r 1 > 0 , such that [figure omitted; refer to PDF] where L * 1 satisfies σ 2 [varphi] q ( λ L * 1 ) γ 1 / Γ ( α ) ...5; 1 .
Then for x ∈ P ∩ ∂ Ω r 1 , we have σ || x || ...4; x ...4; r 1 for t ∈ [ 0,1 ] and [figure omitted; refer to PDF] which implies that [figure omitted; refer to PDF] Next, from f [varphi] ∞ = + ∞ , there exists r - 2 > 0 , such that [figure omitted; refer to PDF] where L * 2 satisfies [figure omitted; refer to PDF] Let r 2 > max ... { r 1 , r - 2 / σ } ; then for x ∈ P ∩ ∂ Ω r 2 we have [figure omitted; refer to PDF] Similar to the above proof, we can get [figure omitted; refer to PDF] Let r 1 < r 3 < r 2 , M r 3 = max ... { f ( t , x ) |" t ∈ [ 0,1 ] , || x || ...4; r 3 } + 1 > 0 , and λ * 0 = ( 1 / M r 3 ) ( Γ ( α + 1 ) r 3 ( 1 - η ∫ 0 1 ... g ( s ) d s ) / γ 2 ) p - 1 . Then for x ∈ P ∩ ∂ Ω r 3 , we have [figure omitted; refer to PDF] Hence, [figure omitted; refer to PDF] This implies that [figure omitted; refer to PDF] Applying Lemma 7 to (72), (76), and (79) yields that Ψ has two fixed points x 1 , x 2 such that x 1 ∈ P ∩ ( Ω - r 3 \ Ω r 1 ) and x 2 ∈ P ∩ ( Ω - r 2 \ Ω r 3 ) .
(ii) By the proof of (i) and the continuity of f ( t , x ) , there exists x ~ ∈ [ r 1 , r - 2 ] such that min ... x ∈ [ r 1 , r - 2 ] ( f ( t , x ) / [varphi] p ( x ) ) = f ( t , x ~ ) / [varphi] p ( x ~ ) .
Let Υ - = min ... { L * 1 , L * 2 , f ( t , x ~ ) / [varphi] p ( x ~ ) } ; since (H6) holds, then Υ - > 0 and [figure omitted; refer to PDF] Assuming x ( t ) is a positive solution of problem (1), we will show that this leads to contradiction for all λ > λ - , where λ - = ( 1 / Υ - ) ( Γ ( α ) / γ 1 σ ) p - 1 . It follows form (26) that [figure omitted; refer to PDF] This implies that 1 ...5; [varphi] q ( λ Υ - ) γ 1 σ / Γ ( α ) = ( λ Υ - ) q - 1 γ 1 σ / Γ ( α ) > ( λ - Υ - ) q - 1 γ 1 σ / Γ ( α ) = 1 , which is a contradiction. This completes the proof of (ii).
Corollary 15.
Assume that (H1)-(H5) holds.
(i) If lim ... _ x [arrow right] ∞ f ( t , x ) = f ~ ∞ ∈ ( 0 , + ∞ ] uniformly for t ∈ [ 0,1 ] , and one of f [varphi] 0 = 0 and f [varphi] ∞ = 0 is satisfied, then there exists λ * > 0 such that the problem (1) has at least one positive solution for any λ > λ * .
(ii) If one of f [varphi] 0 = + ∞ and f [varphi] ∞ = + ∞ holds, then there exists λ * 0 > 0 such that problem (1) has at least one positive solutions for any λ ∈ ( 0 , λ * 0 ) .
Proof.
(i) The conclusion is a direct consequence of Theorem 11(i).
(ii) The conclusion is a direct consequence of Theorem 14(i).
Theorem 16.
Assume that (H1)-(H5) hold. Then the following conclusions hold.
(i) If f [varphi] 0 = 0 and f [varphi] ∞ = ∞ , then the problem (1) has a positive solution x λ ( t ) for all λ > 0 .
(ii) If f [varphi] 0 = ∞ and f [varphi] ∞ = 0 , then the problem (1) has a positive solution x λ ( t ) for all λ > 0 .
Proof.
We only prove (i); the proof of (ii) is similar, so we omit it here.
Let λ > 0 ; since f [varphi] 0 = 0 , there exists r > 0 , l > 0 such that [figure omitted; refer to PDF] where l satisfied [figure omitted; refer to PDF] Similar to the proof of Theorem 14, we obtain that [figure omitted; refer to PDF] From f [varphi] ∞ = ∞ , there exists R > r > 0 , L > 0 such that [figure omitted; refer to PDF] where L satisfied [figure omitted; refer to PDF] We get that [figure omitted; refer to PDF] Applying Lemma 7 to (84), (87) yields that Ψ has a fixed point x such that x ∈ P ∩ ( Ω - R \ Ω r ) .
Theorem 17.
Assume that (H1)-(H6) hold. Then the following conclusions hold.
If 0 < f [varphi] 0 < ∞ and 0 < f [varphi] ∞ < ∞ , then there exists λ ^ 1 , λ ^ 2 > 0 such that problem (1) has no positive solution for any λ > λ ^ 1 and 0 < λ < λ ^ 2 .
Proof.
It follows from 0 < f [varphi] 0 < ∞ and 0 < f [varphi] ∞ < ∞ that there exist L ^ 1 > 0 , L ^ 2 > 0 , l ^ 1 > 0 , l ^ 2 > 0 , and r ^ 2 > r ^ 1 > 0 such that [figure omitted; refer to PDF] Let M ^ = max ... { f ( t , x ) / [varphi] p ( x ) , r ^ 1 ...4; x ...4; r ^ 2 , 0 ...4; t ...4; 1 } , m ^ = min ... { f ( t , x ) / [varphi] p ( x ) , r ^ 1 ...4; x ...4; r ^ 2 , 0 ...4; t ...4; 1 } . By the condition (H6), m ^ > 0 . Then ∀ x ...5; 0 ; we have [figure omitted; refer to PDF] where L ^ = max ... { L ^ 1 , L ^ 2 , M ^ } > 0 and l ^ = min ... { l ^ 1 , l ^ 2 , m ^ } > 0 .
Let λ ^ 1 = ( 1 / l ^ ) ( Γ ( α ) / γ 1 σ ) p - 1 and λ ^ 2 = ( 1 / L ^ ) ( Γ ( α + 1 ) ( 1 - η ∫ 0 1 ... g ( s ) d s ) / γ 2 ) p - 1 . If problem (1) has positive solution x ( t ) in P , then [figure omitted; refer to PDF] which is a contradiction. This finishes the proof of Theorem 17.
4. Examples
In this section, we give out same examples to illustrate our main results can be used in practice.
Example 1.
Consider the following fractional differential equation: [figure omitted; refer to PDF] where α = 5 / 2 , p = 3 / 2 , q = 3 , ω ( t ) = t , f ( t , x ) = ( 1 + t 2 ) 1 / 2 x p - 1 arctan x , η = 1 / 2 , and g ( t ) = 1 / 2 .
By simple computation, l 1 * = π / 2 , l 2 * = 2 π / 2 , γ 1 = ∫ 0 1 ... ( 1 - s ) 3 / 2 [varphi] 3 ( ∫ 0 s ... τ d τ ) d s [approximate] 0.0043 , γ 2 = [varphi] 3 ( ∫ 0 1 ... τ d τ ) = 0.25 , σ [approximate] 0.1429 , λ 0 * = ( 1 / l 1 * ) ( Γ ( α ) / σ γ 1 ) p - 1 [approximate] 2.5149 , λ - 0 * = ( 1 / l 2 * ) ( Γ ( α + 1 ) ( 1 - η ∫ 0 1 g ( s ) d s ) ... / γ 2 ) p - 1 [approximate] 1.4222 . Then, it is easy to see all the conditions of Theorem 9(ii) are satisfied; thus ∀ λ ∈ [ 1.4222 , 2.5149 ] ] ; the problem (91) has at least one positive solution.
Example 2.
Consider the boundary value problem [figure omitted; refer to PDF] where α = 5 / 2 , p = 3 / 2 , q = 3 , ω ( t ) = t , f ( t , x ) = ( 1 + t 2 ) 1 / 2 x p e - x , η = 1 / 2 , and g ( t ) = 1 / 2 : [figure omitted; refer to PDF] Let ρ - 1 = 1 / 2 , ρ - 2 = 4 ; then [figure omitted; refer to PDF] so [straight epsilon] 1 = 1 / 2 e , [straight epsilon] 2 = 4 2 / e 4 , and ∀ x ∈ [ 1 / 2 , 4 ] , max ... 1 / 2 ...4; x ...4; 4 ( f ( t , x ) / [varphi] p ( x ) ) = 2 e - 1 . We have Υ _ = max ... { 1 / 2 e , 4 2 / e 4 , 2 e - 1 } = 2 e - 1 , γ 2 = 0.25 , and λ _ = ( 1 / Υ _ ) ( Γ ( α + 1 ) ( 1 - η ∫ 0 1 ... g ( s ) d s ) / γ 2 ) p - 1 = 3 5 π 1 / 4 e / 4 . By Theorem 11(ii), for any λ < λ _ = 3 5 π 1 / 4 e / 4 , the problem has no positive solution.
5. Conclusions
Recently, differential equations with p -Laplacian operator were widely discussed by several authors. In this paper, by using fixed point techniques combining with partially ordered structure of Banach space, we obtained the existence, multiply, and the dependence of parameter. These new results we presented can be used in numerical computation and analyze mathematical models of physical phenomena, mechanics, nonlinear dynamics, and many other related fields.
Acknowledgments
The authors are very grateful to anonymous referees for their valuable and detailed suggestions and comments to improve the original paper. This work is supported by the Program for New Century Excellent Talents in University (NCET-10-0097).
Conflict of Interests
The authors declare that there is no conflict of interests regarding the publication of this paper.
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Abstract
This paper investigates the existence, multiplicity, nonexistence, and uniqueness of positive solutions to a kind of two-point boundary value problem for nonlinear fractional differential equations with p -Laplacian operator. By using fixed point techniques combining with partially ordered structure of Banach space, we establish some criteria for existence and uniqueness of positive solution of fractional differential equations with p -Laplacian operator in terms of different value of parameter. In particular, the dependence of positive solution on the parameter was obtained. Finally, several illustrative examples are given to support the obtained new results. The study of illustrative examples shows that the obtained results are applicable.
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