Shen et al. Advances in Difference Equations 2013, 2013:295 http://www.advancesindifferenceequations.com/content/2013/1/295
RESEARCH
Open Access
Solvability of fractional boundary value problem with p-Laplacian operator at resonance Tengfei Shen, Wenbin Liu* and Xiaohui Shen * Correspondence:
[email protected] College of Sciences, China University of Mining and Technology, Xuzhou, 221116, P.R. China
Abstract In this paper, a class of multi-point boundary value problems for nonlinear fractional differential equations at resonance with p-Laplacian operator is considered. By using the extension of Mawhin’s continuation theorem due to Ge, the existence of solutions is obtained, which enriches previous results. MSC: 34A08; 34B15 Keywords: fractional differential equation; boundary value problem; p-Laplacian operator; coincidence degree theory; resonance
1 Introduction In the recent years, fractional differential equations played an important role in many fields such as physics, electrical circuits, biology, control theory, etc. (see [–]). Thus, many scholars have paid more attention to fractional differential equations and gained some achievements (see [–]). For example, Wang [] considered a class of fractional multipoint boundary value problems at resonance by Mawhin’s continuation theorem (see []): ⎧ α α– ⎪ t ∈ (, ), a.e. t ∈ (, ), ⎪ ⎨D+ u(t) = f (t, u(t), D+ u(t)), m α– u() = , D+ u() = i= ai Dα– + u(ξi ), ⎪ ⎪ ⎩ α– m α– D+ u() = i= bi D+ u(ηi ),
(.)
n where < α ≤ , < ξ < ξ < · · · < ξm < , < η < η < · · · < ηn < , m i= ai = , i= bi = , n α b η = , D is the standard fractional derivative, f : [, ] × R → R satisfies the + i= i i Carathéodory condition. But Mawhin’s continuation theorem is not suitable for quasi-linear operators. In [], Ge and Ren had extended Mawhin’s continuation theorem, which was used to deal with more general abstract operator equations. In [], Pang et al. considered a higher order nonlinear differential equation with a p-Laplacian operator at resonance: ⎧ (n–) ⎪ (t))) = f (t, u(t), . . . , u(n–) (t)) + e(t), ⎪ ⎨(ϕp (u (i)
u () = , i = , , . . . , n – , ⎪ ⎪ ⎩ u() = u(s) dg(s),
t ∈ (, ), (.)
©2013 Shen et al.; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons Attribution License (http://creativecommons.org/licenses/by/2.0), which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
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where ϕp (s) = |s|p– s, p > , f : [, ] × Rn → Rn and e : [, ] → R are continuous, n ≥ is an integer. g : [, ] → R is a nondecreasing function with dg(s) = , the integral in the second part of (.) is meant in the Riemann-Stieltjes sense. However, there are few articles which consider the fractional multi-point boundary value problem at resonance with p-Laplacian operator and dim Ker M = , because p-Laplacian operator is a nonlinear operator, and it is hard to construct suitable continuous projectors. In this paper, we will improve and generalize some known results. Motivated by the work above, our article is to investigate the multi-point boundary value problem at resonance for a class of Riemanne-Liouville fractional differential equations with p-Laplacian operator and dim Ker M = by constructing suitable continuous projectors and using the extension of Mawhin’s continuation theorem: ⎧ β α α– α ⎪ u(t), Dα– ⎪ + u(t), D+ u(t)), ⎨D+ ϕp (D+ u(t)) = f (t, u(t), D+ u() = Dα+ u() = , u() = m i= ai u(ξi ), ⎪ ⎪ ⎩ α– m α– D+ u() = i= bi D+ u(ηi ),
t ∈ (, ), (.)
where < α ≤ , < β ≤ , < α + β ≤ , < ξ < ξ < · · · < ξm < , < η < η < · · · < ηm < , α– α– p– = , m = , m s, ai ∈ R, bi ∈ R, < m, m ∈ N , m i= ai ξi i= ai ξi i= bi = , ϕp (s) = |s| α ϕp () = , < p, /p + /q = , ϕp is invertible and its inverse operator is ϕq , D+ is RiemannLiouville standard fractional derivative, f : [, ] × R → R is continuous. In order to investigate the problem, we need to suppose that the following conditions hold: = – = , where
m (α)q (αq + βq – q – α – β + ) αq+βq–q–β+ , – a i ξi = (α + β)q– (αq + βq – q – β + ) i=
m (α – )q– (α)(αq + βq – q – α – β + ) αq+βq–q–β+ , – = a i ξi (α + β – )q– (αq + βq – q – β + ) i=
m (α)q– αq+βq–q–α–β+ , – = bi ηi (α + β)q– (αq + βq – q – α – β + ) i=
m (α – )q– αq+βq–q–α–β+ . – = bi ηi (α + β – )q– (αq + βq – q – α – β + ) i= The rest of this article is organized as follows: In Section , we give some notations, definitions and lemmas. In Section , based on the extension of Mawhin’s continuation theorem due to Ge, we establish a theorem on existence of solutions for BVP (.).
2 Preliminaries For the convenience of the reader, we present here some necessary basic knowledge and definitions for fractional calculus theory that can be found in the recent literature (see [, , , , , ]).
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Let X and Y be two Banach spaces with norms · X and uY , respectively. A continuous operator M|dom M∩X : X ∩ dom M → Y is said to be quasi-linear if (i) Im M := M(X ∩ dom M) is a closed subset of Y , (ii) Ker M := {u ∈ X ∩ dom M : Mu = } is linearly homeomorphic to Rn , n < ∞. Let X = Ker M and X be the complement space of X in X, then X = X ⊕ X . On the other hand, suppose that Y is a subspace of Y , and Y is the complement space of Y in Y , so that Y = Y ⊕ Y . Let P : X → X be a projector and Q : Y → Y a semi-projector, and ⊂ X an open and bounded set with origin θ ∈ . θ is the origin of a linear space. Suppose that Nλ : → Y , λ ∈ [, ] is a continuous operator. Denote N by N . Let λ = {u ∈ : Mu = Nλ u}. Nλ is said to be M-compact in if there is an Y ⊂ Y with dim Y = dim X and an operator R : × [, ] → X continuous and compact such that for λ ∈ [, ], (I – Q)Nλ ( ) ⊂ Im M ⊂ (I – Q)Y ,
(.)
λ ∈ (, )
(.)
QNλ x = θ ,
⇔
QNx = θ ,
R(·, λ)| λ = (I – P)| λ
(.)
and R(·, ) is the zero operator, M P + R(·, λ) = (I – Q)Nλ .
(.)
Lemma . (Ge-Mawhin’s continuation theorem []) Let (X, · X ) and (Y , · Y ) be two Banach spaces, and ⊂ X an open and bounded nonempty set. Suppose that M : X ∩ dom M → Y is a quasi-linear operator Nλ : → Y , λ ∈ [, ] is M-compact in . In addition, if (i) Lu = Nλ u, ∀(u, λ) ∈ (dom M ∩ ∂ ) × (, ), (ii) deg(JQN, Ker M ∩ , ) = , where J : Im Q → Ker M is a homeomorphism with J(θ ) = θ and N = N , then the equation Mu = Nu has at least one solution in dom M ∩ . Definition . The Riemann-Liouville fractional integral of order α > of a function u is given by Iα+ u(t) =
(α)
t
(t – s)α– u(s) ds,
provided the right-hand side integral is pointwise almost everywhere defined on (, +∞). Definition . The Riemann-Liouville fractional derivative of order α > of a function u is given by Dα+ u(t) =
n t d u(s) ds, α–n+ (n – α) dt (t – s)
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provided the right-hand side integral is pointwise everywhere defined on (, +∞), where n = [α] + . Definition . Let X be a Banach space, and X ⊂ X is a subspace. A mapping Q : X → X is a semi-projector if Q satisfies (i) Q x = Qx, ∀x ∈ X, (ii) Q(μx) = μQx, ∀x ∈ X, μ ∈ R. Lemma . Assume that u ∈ C(, ) ∩ L (, ) with a fractional derivative of order α > that belongs to C(, ) ∩ L (, ). Then α Dα+ u(t) = u(t) + c t α– + c t α– + · · · + cN t α–N I+
for some ci ∈ R, i = , , . . . , N , where N = [α] + . Lemma . Assume that u(t) ∈ C[, ], ≤ p ≤ q, then q
p
p–q
D+ I+ u(t) = I+ u(t). Lemma . Assume that α ≥ , then: (i) If λ > –, λ = α – i, i = , , . . . , [α] + , we have that Dα+ t λ =
(λ + ) λ–α t . (λ – α + )
(ii) Dα+ t α–i = , i = , , . . . , [α] + . α– α In this paper, we take X = {u | u, Dα– + u, D+ u, D+ u ∈ C[, ]} with the norm uX = α– α max{u∞ , Dα– + u∞ D+ u∞ , D+ u∞ }, where u∞ = maxt∈[,] |u(t)|, and Y = C[, ] with the norm yY = y∞ . By means of the linear functional analysis theory, it is easy to prove that X and Y are Banach spaces, so we omit it. Define the operator M : dom M → Y by
β Mu = D+ ϕp Dα+ u(t) , β dom M = u ∈ X D+ ϕp Dα+ u ∈ Y , u() = Dα+ u() = , u() =
m i=
ai u(ξi ), Dα– + u() =
m
(.)
bi Dα– + u(ηi ) .
(.)
i=
Based on the definition of dom M, it is easy to find that dom M = ∅ such as u(t) = ct α– ∈ dom M, c ∈ R. Define the operator Nλ : X → Y , λ ∈ [, ], α– α Nλ u(t) = λf t, u(t), Dα– + u(t), D+ u(t), D+ u(t) ,
t ∈ [, ].
Then BVP (.) is equivalent to the operator equation Mu = Nu, where N = N .
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3 Main result In this section, a theorem on existence of solutions for BVP (.) will be given. Define operators Tj : Y → Y , j = , as follows:
T y =
( – s)
–
m
α–
ai
i=
ξi
ϕq
(β)
(s – τ )
(ξi – s)α– ϕq
s
(β)
β–
s
y(τ ) dτ ds (s – τ )β– y(τ ) dτ ds,
s β– ϕq (s – τ ) y(τ ) dτ ds T y = (β)
ηi
s m bi ϕq (s – τ )β– y(τ ) dτ ds. – (β) i=
Let us make some assumptions, which will be used in the sequel. (H) There exist nonnegative functions r, d, e, h, k ∈ Y such that for all t ∈ [, ], (u, v, w, z) ∈ R , f (t, u, v, w, z) ≤ r(t) + d(t)|u|p– + e(t)|v|p– + h(t)|w|p– + k(t)|z|p– . (H) There exists a constant A > such that for u ∈ dom M, if |Dα– + u(t)| > A for all t ∈ [, ], then T Nu(t) – T Nu(t) > sgn Dα– + u(t) or T Nu(t) – T Nu(t) < . sgn Dα– + u(t) (H) There exists a constant B > such that for u ∈ dom M, if |Dα– + u(t)| > B for all t ∈ [, ], then sgn Dα– – T Nu(t) + T Nu(t) > + u(t) or – T Nu(t) + T Nu(t) < . sgn Dα– + u(t) Theorem . Let f : [, ] × R → R be continuous and condition (H)-(H) hold, then BVP (.) has at least one solution, provided that A d∞ + e∞ + h∞ + k∞ < , (β + ) where A =
(α+)
+
(α)
+
. (α–)
In order to prove Theorem ., we need to prove some lemmas below.
(.)
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Lemma . The operator M : dom M ∩ X → Y is quasi-linear. Ker M = u ∈ X | u(t) = c t α– + c t α– , c , c ∈ R ,
(.)
Im M = {y ∈ Y | Tj y = , j = , }.
(.) β
Proof Suppose that u(t) ∈ dom M, by D+ ϕp (Dα+ u(t)) = , we have Dα+ u(t) = ϕq c t β– . Based on Dα+ u() = , one has u(t) = c t α– + c t α– + c t α– , which together with u() = yields that Ker M = u ∈ X | u(t) = c t α– + c t α– , c , c ∈ R . It is clear that dim Ker M = . So, Ker M is linearly homeomorphic to R . β If y ∈ Im M, then there exists a function u ∈ dom M such that y(t) = D+ ϕp (Dα+ u(t)). Based on Lemmas . and ., we have β u(t) = Iα+ ϕq I+ y(s) + c t α– + c t α– , β α– α Dα– + u(t) = D+ I+ ϕq I+ y(s) + c (α), α– α– which together with m = , m = and m i= ai ξi i= ai ξi i= bi = yields that Tj y(t) = , j = , . β On the other hand, suppose that y ∈ Y and satisfies (.), and let u(t) = Iα+ ϕq (I+ y(t)), β then u ∈ dom M and Mu(t) = D+ ϕp (Dα+ u(t)) = y, so y ∈ Im M and Im M := M(dom M) is a closed subset of Y . Thus, M is a quasi-linear operator. Lemma . Let ⊂ X be an open and bounded set, then Nλ is M-compact in . Proof Define the continuous projector P : X → X by Pu(t) =
α– α– Dα– Dα– + , + u()t + u()t (α) (α – )
Define the continuous projector Q : Y → Y , by Qy(t) = Q y(t) t α– + Q y(t) t α– ,
t ∈ [, ],
where T y(t) – T y(t) , Q y(t) = ϕp – T y(t) + T y(t) . Q y(t) = ϕp
t ∈ [, ].
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Obviously, X = Ker M = Im P and Y = ImQ. Thus, we have dim Y = dim X = . For any y ∈ Y , we have
Q Q y(t)t
α–
α– α– T Q y(t)t = ϕp – T Q y(t)t ( – ) = Q y(t). = Q y(t)ϕp
Similarly, we can get Q Q y(t)t α– = ,
Q Q y(t)t α– = ,
Q Q y(t)t α– = Q y(t).
Hence, the map Q is idempotent. Similarly, we can get Q(μy) = μQy, for all y ∈ Y , μ ∈ R. Thus, Q is a semi-projector. For any y ∈ Im M, we can get that Qy = and y ∈ Ker Q, conversely, if y ∈ Ker Q, we can obtain that Qy = , that is to say, y ∈ Im M. Thus, Ker Q = Im M. Let ⊂ X be an open and bounded set with θ ∈ , for each u ∈ , we can get Q[(I – Q)Nλ (u)] = . Thus, (I – Q)Nλ (u) ∈ Im M = Ker Q. Take any y ∈ Im M in the type y = (y – Qy) + Qy, since Qy = , we can get y ∈ (I – Q)Y . So, (.) holds. It is easy to verify (.). Define R : × [, ] → X by R(u, λ)(t) = (α)
t
(t – s)
α–
ϕq
(β)
s
(s – τ )
β–
(I – Q)Nλ u(τ ) dτ ds.
By the continuity of f , it is easy to get that R(u, λ) is continuous on × [, ]. Moreover, β for all u ∈ , there exists a constant T > such that |I+ (I – Q)Nλ u(τ )| ≤ T, so, we can easα– α ily obtain that R( , λ), Dα– + R( , λ), D+ R( , λ) and D+ R( , λ) are uniformly bounded. By Arzela-Ascoli theorem, we just need to prove that R : × [, ] → X is equicontinuous. For u ∈ , < t < t ≤ , < α ≤ , < β ≤ , < α + β ≤ , we have R(u, λ)(t ) – R(u, λ)(t ) β t = (t – s)α– ϕq I+ (I – Q)Nλ u(τ ) ds (α)
t β (t – s)α– ϕq I+ (I – Q)Nλ u(τ ) ds –
ϕq (L) ≤ (α) =
t
(t – s)
α–
– (t – s)
α–
(t – s)
ds +
α–
ds
t
ϕq (T) α α t –t , (α + )
α– D + R(u, λ)(t ) – Dα– + R(u, λ)(t ) t
β = (t – s)ϕq I+ (I – Q)Nλ u(τ ) ds –
≤ ϕq (T)
t
ϕq (T) t – t
t
β (t – s)ϕq I+ (I – Q)Nλ u(τ ) ds
t
(t – s) – (t – s) ds +
=
t
(t – s) ds t
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and α– D + R(u, λ)(t ) – Dα– + R(u, λ)(t ) t
β = ϕq I+ (I – Q)Nλ u(τ ) ds –
t
β ϕq I+ (I – Q)Nλ u(τ ) ds
≤ ϕq (T)(t – t ). α– Since t α is uniformly continuous on [, ], so, R( , λ), Dα– + R( , λ) and D+ R( , λ) are β equicontinuous. Similarly, we can get that I+ ((I – Q)Nλ u(τ )) ⊂ C[, ] is equicontinuous. Considering that ϕq (s) is uniformly continuous on [–T, T], we have that Dα+ R( , λ) = β I+ ((I – Q)Nλ ( )) is also equicontinuous. So, we can obtain that R : × [, ] → X is compact. β For each u ∈ λ , we have D+ ϕp (Dα+ u(t)) = Nλ (u(t)) ∈ Im M. Thus,
R(u, λ)(t) = =
(α) (α)
t
(t – s)α– ϕq
t
(t – s)α– ϕq
(β) (β)
s
(s – τ )β– (I – Q)Nλ u(τ ) dτ ds
s
β (s – τ )β– D+ ϕp Dα+ u(τ ) dτ ds,
which together with Dα+ u() = u() = yields that R(u, λ)(t) = u(t) –
α– α– Dα– Dα– – = (I – P)u(t). + u()t + u()t (α) (α – )
It is easy to verify that R(u, )(t) is the zero operator. So, (.) holds. Besides, for any u ∈ , M Pu + R(u, λ) (t)
t
s α– β– (t – s) ϕq (s – τ ) =M (I – Q)Nλ u(τ ) dτ ds (α) (β) α– α– α– Dα– D u()t + u()t + + + (α) (α – ) = (I – Q)Nλ u(t), which implies (.). So, Nλ is M-compact in . Lemma . Suppose that (H), (H) hold, then the set = u ∈ dom M \ Ker M | Mu = Nλ u, λ ∈ (, ) is bounded. Proof By Lemma ., for each u ∈ dom M, we have u(t) = Iα+ Dα+ u(t) + c t α– + c t α– + c t α– .
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Combined with u() = , we get c = . Thus, u(t) = Iα+ Dα+ u(t) + c t α– + c t α– , α Dα– + u(t) = I+ D+ u(t) + c (α), α Dα– + u(t) = I+ D+ u(t) + c (α)t + c (α – ).
By simple calculation, we get
t α Dα– u(t) – D u(s) ds , + + (α)
t
t α– α– α– α D+ u(t) – (t – s)D+ u(s) ds – D+ u(t) – D+ u(s) ds t . c = (α – ) c =
Take any u ∈ , then Nu ∈ Im M = Ker Q and QNu = . It follows from (H) and (H) α– that there exist ε , ε ∈ [, ] such that |Dα– + u(ε )| ≤ A, |D+ u(ε )| ≤ B. Thus,
α– Dα– + u(t) = D+ u(ε ) +
t
α– Dα– + u(t) = D+ u(ε ) +
Dα+ u(t) dt,
ε t
ε
Dα– + u(t) dt,
α– D + u ≤ A + Dα+ u , ∞ ∞ α– α D + u ≤ B + Dα– + u ∞ ≤ A + B + D+ u ∞ . ∞ So, we get α Dα– A + Dα+ u∞ , + u ∞ + D+ u ∞ ≤ (α) (α) Dα– + Dα– + Dα+ u u u |c | ≤ + + ∞ ∞ ∞ (α – ) A + B + Dα+ u∞ , ≤ (α – ) α u∞ ≤ A D+ u∞ + B , |c | ≤
+ (α) + (α–) , B = where A = (α+) α Based on D+ u() = , we have
A (α)
+
A (α–)
+
B . (α–)
β ϕp Dα+ u(t) = λI+ Nu(t). From (H) and λ ∈ (, ), we have α ϕp D + u(t)
t α– α ≤ (t – s)β– f s, u(s), Dα– + u(t), D+ u(s), D+ u(s) ds (β)
t p– p– ≤ (t – s)β– r(s) + d(s)u(s) + e(s)Dα– + u(s) (β)
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p– + k(s)Dα+ u(s)p– ds + h(s)Dα– + u(s) ≤
α– p– r∞ + d∞ up– ∞ + e∞ D+ u ∞ (β + ) p– + k∞ Dα+ up– , + h∞ Dα– + u
∞
∞
which together with |ϕp (Dα+ u(t))| = |Dα+ u(t)|p– , we can get α p– D + u ≤
∞
≤
α– p– r∞ + d∞ up– ∞ + e∞ D+ u ∞ (β + ) p– α p– + h∞ Dα– + u ∞ + k∞ D+ u ∞ p– r∞ + d∞ A Dα+ u∞ + B + e∞ A + Dα+ u∞ (β + ) p– p– + h∞ A + Dα+ u∞ + k∞ Dα+ u∞ .
In view of (.), we can obtain that there exists a constant M > such that α α– D + u ≤ M , D + u ≤ A + M := M , ∞ ∞ α– D + u ≤ A + M := M , u∞ ≤ A M + B := M . ∞ Thus, we have α– α uX = max u∞ , Dα– + u ∞ , D+ u ∞ , D+ u ∞ ≤ max{M , M , M , M } := M. So, is bounded.
Lemma . Suppose that (H) holds, then the set = {u | u ∈ Ker M, Nu ∈ Im M} is bounded. Proof For each u ∈ , we have that u(t) = c t α– + c t α– , c , c ∈ R and QNu = . It follows from (H) and (H) that there exists an ε , ε ∈ [, ] such that |Dα– + u(ε )| ≤ A, A A+B |Dα– u(ε )| ≤ A, which implies that |c | ≤ and |c | ≤ . So, is bounded. + (α) (α–) Define the isomorphism J – : Ker M → Im Q by J – (c t α– + c t α– ) = c t α– + c t α– , c , c ∈ R, t ∈ [, ]. In fact, for each c , c ∈ R, suppose that (Q y(t), Q y(t)) = (c , c ), we have ⎧ ⎨ T y(t) – T y(t) = ϕ (c ) := c, q ⎩– T y(t) + T y(t) = ϕq (c ) := c ,
(.)
where c , c ∈ R, by the condition – = , there exists a unique solution for (.), which is (T y(t), T y(t)) = (m , m ), m , m ∈ R. Now, we will prove that there exists β y ∈ Y such that (T y(t), T y(t)) = (m , m ). Based on y(t) ∈ C[, ], we choose y(t) = D+ y(t),
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where y(t) = ϕp (l t (α+β–)(q–) + l t (α+β–)(q–) ), l =
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(α)q– ,l (α+β)q–
=
(α–)q– , (α+β–)q–
=
m –m ,
t ∈ [, ], which together with < α ≤ , < β ≤ , < α + β ≤ , q > , = we have (α + β – )(q – ) > and (α + β – )(q – ) > . So, we have y() = . Thus, we can obtain that m –m ,
β
β
β
I+ y(t) = I+ D+ y(t) = y(t) + ct β– , where c ∈ R, which together with y() = , we have c = . So, we have ⎧ m ξi α– α– ⎪ ⎪ ⎨T y(t) = ( – s) ϕq (y(s)) ds – i= ai (ξi – s) ϕq (y(s)) ds = + = m , ⎪ ⎪ ηi ⎩ T y(t) = ϕq (y(s)) ds – m i= bi ϕq (y(s)) ds = + = m . Thus, there exists y ∈ Y such that (T y(t), T y(t)) = (m , m ). So J – : Ker L → Im Q is well defined. Lemma . Suppose that the first part of (H) holds, then the set = u ∈ Ker M | λJ – u + ( – λ)QNu = , λ ∈ [, ] is bounded. Proof For each u ∈ , we can get that u(t) = c t α– + c t α– , c , c ∈ R. By the definition of the set , we can obtain that λ c t α– + c t α– + ( – λ) Q N c t α– + c t α– t α– + Q N c t α– + c t α– t α– = . Thus,
α– α– α– α– T N c t + c t = , – T N c t + c t λc + ( – λ)ϕp – T N c t α– + c t α– + T N c t α– + c t α– = . λc + ( – λ)ϕp
(.) (.)
If λ = , then is bounded because of the first part of (H) and (H). If λ = , we get c = c = , obviously, is bounded. If λ ∈ (, ), by the first part of (H) and (.), we can A A+B , by the first part of (H) and (.), we have |c | ≤ (α–) . So, is obtain that |c | ≤ (α) bounded. Remark . If the second part of (H) holds, then the set = u ∈ Ker M | –λJ – u + ( – λ)QNu = , λ ∈ [, ] is bounded.
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Proof of Theorem . Assume that is a bounded open set of X with i= i ∪ ⊂ . By Lemmas . and ., we can obtain that M : dom M ∩ X → Y is quasi-linear, and Nλ is M-compact on . By the definition of , we have Lu = Nλ u,
∀(u, λ) ∈ (dom M ∩ ∂ ) × (, ),
H(u, λ) = ±λJ – (u) + ( – λ)QN(u) = ,
(∂ ∩ Ker M) × [, ].
Thus, by the homotopic property of degree, we can get deg(JQN, ∩ Ker M, ) = deg H(·, ), ∩ Ker M, = deg H(·, ), ∩ Ker M, = deg(±I, ∩ Ker M, ) = . So Lemma . is satisfied, and Mu = Nu has at least one solution in dom M ∩ . Namely, BVP (.) have at least one solution in the space X.
Competing interests The authors declare that they have no competing interests. Authors’ contributions The authors contributed equally in this article. All authors read and approved the final manuscript. Acknowledgements The authors are grateful to those who gave useful suggestions about the original manuscript. This research is supported by the National Natural Science Foundation of China (No. 11271364) and the Fundamental Research Funds for the Central Universities (2010LKSX09). Received: 1 January 2013 Accepted: 2 September 2013 Published: 07 Nov 2013 References 1. Kilbas, AA, Srivastava, HM, Trujillo, JJ: Theory and Applications of Fractional Differential Equations. Elsevier, Amsterdam (2006) 2. Sabatier, J, Agrawal, OP, Machado, JAT: Advances in Fractional Calculus: Theoretical Developments and Applications in Physics and Engineering. Springer, Dordrecht (2007) 3. Samko, SG, Kilbas, AA, Marichev, OI: Fractional Integrals and Derivatives: Theory and Applications. Gordon & Breach, New York (1993) 4. Magin, R: Fractional calculus models of complex dynamics in biological tissues. Comput. Math. Appl. 59, 1586-1593 (2010) 5. Metzler, R, Klafter, J: Boundary value problems for fractional diffusion equations. Physica A 278, 107-125 (2000) 6. Mainardi, F: Fractional diffusive waves in viscoelastic solids. In: Wegner, JL, Norwood, FR (eds.) Nonlinear Waves in Solids, pp. 93-97. ASME, Fairfield (1995) 7. Bai, J, Feng, X: Fractional-order anisotropic diffusion for image denoising. IEEE Trans. Image Process. 16, 2492-2502 (2007) 8. Agarwal, RP, O’Regan, D, Stanek, S: Positive solutions for Dirichlet problems of singular nonlinear fractional differential equations. J. Math. Anal. Appl. 371, 57-68 (2010) 9. Kosmatov, N: A boundary value problem of fractional order at resonance. Electron. J. Differ. Equ. 2010, 135 (2010) 10. Jafari, H, Gejji, VD: Positive solutions of nonlinear fractional boundary value problems using Adomian decomposition method. Appl. Math. Comput. 180, 700-706 (2006) 11. Bai, Z: Solvability for a class of fractional m-point boundary value problem at resonance. Comput. Math. Appl. 62, 1292-1302 (2011) 12. Bai, Z: On solutions of some fractional m-point boundary value problems at resonance. Electron. J. Qual. Theory Differ. Equ. 2010, 37 (2010) 13. Bai, Z, Zhang, Y: Solvability of fractional three-point boundary value problems with nonlinear growth. Appl. Math. Comput. 218, 1719-1725 (2011) 14. Zhang, Y, Bai, Z: Existence of positive solutions for nonlinear fractional three-point boundary value problem at resonance. Appl. Math. Comput. 36, 417-440 (2011) 15. Hu, Z, Liu, W: Solvability for fractional order boundary value problem at resonance. Bound. Value Probl. 2011, 20 (2011) 16. Rui, W: Existence of solutions of nonlinear fractional differential equations at resonance. Electron. J. Qual. Theory Differ. Equ. 2012, 66 (2012)
Shen et al. Advances in Difference Equations 2013, 2013:295 http://www.advancesindifferenceequations.com/content/2013/1/295
17. Wang, G, Liu, W, Yang, J, Zhu, S, Zheng, T: The existence of solutions for a fractional 2m-point boundary value problems. J. Appl. Math. (2012). doi:10.1155/2012/841349 18. Wang, W: Solvability for a coupled system of fractional differential equations at resonance. Nonlinear Anal. 13, 2285-2292 (2012) 19. Zhou, H, Kou, C, Xie, F: Existence of solutions for fractional differential equations with multi-point boundary conditions at resonance on a half-line. Electron. J. Qual. Theory Differ. Equ. 2011, 27 (2011) 20. Chen, Y, Tang, X: Solvability of sequential fractional order multi-point boundary value problems at resonance. Appl. Math. Comput. 218, 7638-7648 (2012) 21. Wang, J: The existence of solutions to boundary value problems of fractional differential equations at resonance. Nonlinear Anal. 74, 1987-1994 (2011) 22. Chen, T, Liu, W, Hu, Z: A boundary value problem for fractional differential equation with p-Laplacian operator at resonance. Nonlinear Anal. 75, 3210-3217 (2012) 23. Mawhin, J: Topological Degree and Boundary Value Problems for Nonlinear Differential Equations in Topological Methods for Ordinary Differential Equations. Lecture Notes in Math., vol. 1537, pp. 74-142 (1993) 24. Ge, W, Ren, J: An extension of Mawhin’s continuation theorem and its application to boundary value problems with a p-Laplacian. Nonlinear Anal. 58, 477-488 (2004) 25. Pang, H, Ge, W, Tian, M: Solvability of nonlocal boundary value problems for ordinary differential equation of higher order with a p-Laplacian. Comput. Math. Appl. 56, 127-142 (2008)
10.1186/1687-1847-2013-295 Cite this article as: Shen et al.: Solvability of fractional boundary value problem with p-Laplacian operator at resonance. Advances in Difference Equations 2013, 2013:295
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