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Solvability of fractional differential equations with p-Laplacian at resonance
Applied Mathematics and Computation 260 (2015) 48–56
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Solvability of fractional differential equations with p-Laplacian at resonance Weihua Jiang∗ College of Sciences, Hebei University of Science and Technology Shijiazhuang, Hebei 050018, PR China
a r t i c l e
i n f o
MSC: 34A08 34B15 Keywords: Continuous theorem Fractional differential equation Resonance p-Laplacian Boundary value problem
a b s t r a c t In order to study boundary value problems with p-Laplacian, the extension for the continuous theorem of Ge and Ren is generalized. By using the new results and constructing suitable operators, we investigate the existence of solutions for p-Laplacian fractional differential equations at resonance. Examples are given to illustrate our results. © 2015 Elsevier Inc. All rights reserved.
1. Introduction The fractional differential equations have become an important area of investigation because of the intensive development of the theory of fractional calculus itself and its wide applications in various sciences such as physics, mechanics, chemistry, engineering, etc. (see [1–8]). On the other hand, p-Laplacian boundary value problems have been widely studied owing to its importance in theory and application of mathematics and physics (see [9–14] and the results cited therein). Mawhin’s continuous theorem [15] is an effective tool in studying the existence of solutions for the abstract equation Lx = Nx (see [16–25]), where L is a noninvertible linear operator. Ge and Ren [26] extended Mawhin’s results in [15] and obtained the existence of solutions when L is a noninvertible nonlinear one. This is a wonderful instrument to solve the p-Laplacian boundary value problems at resonance. Using this theorem, the authors need to construct two projectors P and Q (see [26,27]). But sometimes it is difficult to give the projector Q in many boundary value problems with p-Laplacian. We will generalize the extension of the continuous theorem in [26] and show that the equation Lx = Nx is solvable when Q is not a projector but satisfies some conditions. And, using our new results, we will prove the existence of solutions for the boundary value problem with p-Laplacian at resonance:
β D0+ ϕp Dα0+ u (t) + f t, u(t), Dα0+−1 u(t), Dα0+ u(t) = 0, 1 u(0) = Dα0+ u(0) = 0, u(1) = 0 h(t)u(t)dt,
(1.1)
1 where 0 < β ≤ 1, 1 < α ≤ 2, 0 h(t)tα −1 dt = 1, ϕp (s) = |s|p−2 s, p > 1. To the best of our knowledge, this is the first paper to study fractional differential equations with p-Laplacian and integral boundary value conditions at resonance when Q is not a projector.
∗
Tel.: +8613785419695. E-mail address: [email protected], [email protected], [email protected]
http://dx.doi.org/10.1016/j.amc.2015.03.036 0096-3003/© 2015 Elsevier Inc. All rights reserved.
W. Jiang / Applied Mathematics and Computation 260 (2015) 48–56
49
2. Preliminaries and a generalized theorem Definition 2.1 ([26]). Let X and Z be two Banach spaces with norms · X , · Z , respectively. A continuous operator M : X ∩ domM → Z is said to be quasi-linear if (i) ImM := M(X ∩ domM) is a closed subset of Z, (ii) KerM := {x ∈ X ∩ domM : Mx = 0} is linearly homeomorphic to Rn , n < ∞, where domM denotes the domain of the operator M. Let X1 = KerM and X2 be the complement space of X1 in X. Then X = X1 ⊕ X2 . Let P : X → X1 be projector and ⊂ X be an open and bounded set with the origin θ ∈ . Definition 2.2. Suppose that Nλ : → Z, λ ∈ [0, 1] is a continuous and bounded operator. Denote N1 by N. Let λ = {x ∈ : Mx = Nλ x}. Nλ is said to be M-quasi-compact in if there exists a vector subspace Z1 of Z satisfying dimZ1 = dimX1 and two operators Q and R such that for λ ∈ [0, 1], (a) (b) (c) (d)
KerQ = ImM, QNλ x = θ , λ ∈ (0, 1) ⇔ QNx = θ , R(·, 0) is the zero operator and R(·, λ)| λ = (I − P )| λ , M[P + R(·, λ)] = (I − Q )Nλ ,
where Q : Z → Z1 , QZ = Z1 is continuous, bounded and satisfies Q (I − Q ) = 0 and R : × [0, 1] → X2 is continuous and compact. Theorem 2.1. Let X and Z be two Banach spaces with the norms · X , · Z , respectively, and ⊂ X be an open and bounded nonempty set. Suppose
M : X ∩ domM → Z is a quasi-linear operator and that Nλ : → Z, λ ∈ [0, 1] is M-quasi-compact. In addition, if the following conditions hold:
(C1 ) Mx = Nλ x, ∀x ∈ ∂ ∩ domM, λ ∈ (0, 1), (C2 ) deg{JQN, ∩ KerM, 0} = 0, then the abstract equation Mx = Nx has at least one solution in domM ∩ , where N = N1 , J : ImQ → KerM is a homeomorphism with J(θ ) = θ . The proof used in this paper is similar to the one in [26]. Proof. Define Sλ : ∩ domM −→ X, λ ∈ [0, 1] by
Sλ = P + R(·, λ) + JQN.
(2.1)
Since Q and N are continuous and bounded, P, R, J are continuous and compact, Sλ is a continuous compact mapping for
(x, λ) ∈ × [0, 1].
Firstly, we prove that x ∈ is a solution of
Mx = Nλ x, λ ∈ (0, 1] if and only if x is a fixed point of Sλ . If x ∈ is a solution of (2.2), by Definition 2.2(a), we get
Nλ x = Mx ∈ ImM = KerQ. Hence,
QNλ x = θ , λ ∈ (0, 1]. This, together with (2.2) and Definition 2.2(c), means
(I − P)x = R(x, λ), JQNx = θ . Therefore,
x = Px + R(x, λ) = Px + R(x, λ) + JQNx = Sλ x. So, x is a fixed point of Sλ . On the other hand, if x ∈ is a fixed point of Sλ , λ ∈ (0, 1], considering R(x, λ) ∈ X2 , we get
Px = PSλ x = Px + P (JQNx), which implies JQNx = θ ∈ X1 and QNx = θ ∈ Z1 . Hence
QNλ x = θ ∈ Z1 , x = Px + R(x, λ), λ ∈ (0, 1].
(2.2)
50
W. Jiang / Applied Mathematics and Computation 260 (2015) 48–56
These, together with Definition 2.2(d), mean that
Mx = M(Px + R(x, λ)) = (I − Q )Nλ x = Nλ x − QNλ x = Nλ x. Now, we will prove that I − Sλ is a homotopic mapping. By the above results and condition (C1 ), we can get that
λ ∈ (0, 1), x ∈ ∂ .
x = Sλ x,
(2.3)
Condition (C2 ) means that
x ∈ ∂ .
x = S0 x,
(2.4)
In fact, if x = S0 x, x ∈ ∂ , then x = Px + JQNx ∈ X1 . So, we have x = x + JQNx and then JQNx = θ which contradicts (C2 ). For λ = 1, if there is x0 ∈ ∂ such that x0 = S1 x0 then we have Mx0 = Nx0 , x0 ∈ ∂ ⊂ . Theorem 2.1 is proved. Without loss generality, suppose
x ∈ ∂ .
x = S1 x,
(2.5)
Combining (2.3)–(2.5), we have
λ ∈ [0, 1], x ∈ ∂ .
x = Sλ x,
This means that I − Sλ is a homotopic mapping. Thus,
deg{I − S1 , ∩ domM, 0} = deg{I − S0 , ∩ domM, 0} = deg{I − P − JQN, ∩ domM, 0} = deg{I − P − JQN, ∩ kerM, 0} = deg{−JQN, ∩ kerM, 0} = 0. So, S1 has a fixed point x0 ∈ , i.e. Mx0 = Nx0 . The proof is completed. The following definitions can be found in [1–3]. Definition 2.3. The fractional integral of order α > 0 of a function y : (0, ∞) → R is given by
1
I0α+ y(t) =
(α)
0
t
(t − s)α−1 y(s)ds,
(2.6)
provided the right-hand side is pointwise defined on (0, ∞). Definition 2.4. The fractional derivative of order α > 0 of a function y : (0, ∞) → R is given by
Dα0+ y(t) =
1 dn
(n − α) dtn
t 0
(t − s)n−α−1 y(s)ds,
provided the right-hand side is pointwise defined on (0, ∞), where n = [α ] + 1. In order to obtain our main results, we need the following lemmas. Lemma 2.1 ([2,8]). Assume f ∈ L[0, 1], q ≥ p ≥ 0, q > 1,then
D0+ I0+ f (t) = I0+ f (t). p
q
q−p
Lemma 2.2 ([8]). Assume α > 0, λ > −1, then
Dα0+ tλ =
(λ + 1) dn n+λ−α (t ),
(n + λ − α + 1) dtn
where n = [α ] + 1. Example. Dα tα −i = 0, i = 1, 2, . . . , n. 0+ Lemma 2.3 ([8]). Dα u(t) = 0 if and only if 0+
u(t) = c1 tα −1 + c2 tα −2 + · · · + cn tα −n , where n is the smallest integer greater than or equal to α , ci ∈ R, i = 1, 2, . . . , n. +m Lemma 2.4 ([8]). If the fractional derivatives Dα y(t) and Dα y(t) exist, then 0+ 0+
(Dα0+ y(t))(m) = Dα0++m y(t), where α > 0, m is a positive integer.
(2.7)
W. Jiang / Applied Mathematics and Computation 260 (2015) 48–56
51
Lemma 2.5 ([28]). For any u, v ≥ 0, then (1) ϕp (u + v) ≤ ϕp (u) + ϕp (v), 1 < p ≤ 2; (2) ϕp (u + v) ≤ 2p−2 (ϕp (u) + ϕp (v)), p ≥ 2, where ϕp (s) = |s|p−2 s = sp−1 , s ≥ 0. 3. The solutions for the problem (1.1) 1 In the following, we will always suppose that f : [0, 1] × R3 → R is continuous, h(t) ≥ 0, t ∈ [0, 1], 0 h(t)tα −1 dt = 1 and that q satisfies 1/p + 1/q = 1. −1 Let X = {u|Dα u ∈ C[0, 1], u(0) = Dα u(0) = 0} with norm u = max{u∞ ,Dα u∞ ,Dα u∞ }, Y = C[0, 1] with norm 0+ 0+ 0+ 0+ y∞ = maxt∈[0,1] |y(t)|. We know that (X, · ) and (Y, · ∞ ) are Banach spaces. Define operators M : X ∩ domM → Y, Nλ : X → Y as follows: β
Mu(t) = D0+
ϕp (Dα0+ u) (t), Nλ u(t) = −λf t, u(t), Dα0+−1 u(t), Dα0+ u(t) ,
t,
λ ∈ [0, 1],
1 β where domM = {u ∈ X |D0+ ϕp Dα u ∈ C[0, 1], u(1) = 0 h(t)u(t)dt}. 0+ Lemma 3.1. M is a quasi-linear operator. Proof. It is easy to get that KerM = {ktα −1 |k ∈ R}. For u ∈ X ∩ domM, if Mu = y, then y satisfies
1 0
h(t)
+
0
1
0
t
[(t − ts)α −1 − (t − s)α −1 ]ϕq
h(t)tα −1
1
t
(1 − s)α−1 ϕq
s
0
s 0
(s − r)β −1 y(r)dr dsdt
(s − r)β −1 y(r)dr dsdt = 0.
On the other hand, if y ∈ Y satisfies (3.1), take
u(t) =
1
(α)
t 0
(t − s)α−1 ϕq
1
(β)
s
0
(3.1)
(s − r)β −1 y(r)dr ds + ctα−1 ,
where c is a constant. By simple calculations, we get u ∈ X ∩ domM and Mu = y. Thus
ImM = {y ∈ Y |y satisfies (3.1)}. Obviously, ImM ⊂ Y is closed. So, M is quasi-linear. The proof is completed. Take a projector P : X → KerM and an operator Q : Y → R as follows:
Dα0+−1 u(0) α −1 t ,
(Pu)(t) =
(Qy)(t) = c,
(α)
where c satisfies
1 0
h(t)
+
1 0
t 0
[(t − ts)α −1 − (t − s)α −1 ]ϕq
h(t)tα −1
1
(1 − s)α−1 ϕ
s
q
t
0
s 0
(s − r)β −1 [y(r) − c]dr dsdt
(s − r)β −1 [y(r) − c]dr dsdt = 0.
We will prove that c is the unique constant satisfying (3.2). For y ∈ Y, let
F (c) =
1 0
+
h(t)
1 0
t 0
[(t − ts)α −1 − (t − s)α −1 ]ϕq
h(t)tα −1
1 t
(1 − s)α−1 ϕq
s 0
s 0
(3.2)
(s − r)β −1 [y(r) − c]dr dsdt
(s − r)β −1 [y(r) − c]dr dsdt.
Obviously F (c) is continuous and strictly decreasing in R. Take c1 = mint∈[0,1] y(t), c2 = maxt∈[0,1] y(t). It is easy to see that F (c1 ) ≥ 0, F (c2 ) ≤ 0. So, there exists a unique constant c ∈ [c1 , c2 ] such that F (c) = 0. Furthermore, Q () is bounded if ⊂ Y is bounded, i.e. Q is bounded. Remark. By the definition of Q, we can easily obtain that Q is not a projector but satisfies Q (I − Q )y = Q (y − Qy) = 0, y ∈ Y. Lemma 3.2. Q is continuous in Y.
52
W. Jiang / Applied Mathematics and Computation 260 (2015) 48–56
Proof. For y1 , y2 ∈ Y, assume Qy1 = c1 , Qy2 = c2 . Noticing h(t) ≥ 0 and that ϕq is strictly increasing, we obtain that if c2 − c1 > maxt∈[0,1] (y2 (t) − y1 (t)), then
0=
0
1
+ =
0
<
0
1 0
1
1 0
1
+
1 0
t 0
[(t − ts)α −1 − (t − s)α −1 ]ϕq
t 0
t 0
1 t
(1 − s)α−1 ϕq
s 0
1
(1 − s)α−1 ϕ
s
q
t
0
1
(1 − s)α−1 ϕ
s
q
t
0
s 0
(s − r)β −1 [y2 (r) − c2 ]dr dsdt
s 0
(s − r)β −1 [y1 (r) − c1 + y2 (r) − y1 (r) − (c2 − c1 )]dr dsdt
(s − r)β −1 [y1 (r) − c1 + y2 (r) − y1 (r) − (c2 − c1 )]dr dsdt
[(t − ts)α −1 − (t − s)α −1 ]ϕq
h(t)tα −1
(s − r)β −1 [y2 (r) − c2 ]dr dsdt
[(t − ts)α −1 − (t − s)α −1 ]ϕq
h(t)tα −1
h(t)
h(t)tα −1
h(t)
+
h(t)
s 0
(s − r)β −1 [y1 (r) − c1 ]dr dsdt
(s − r)β −1 [y1 (r) − c1 ]dr dsdt = 0.
A contradiction. On the other hand, if c2 − c1 < mint∈[0,1] (y2 (t) − y1 (t)), then
0=
0
1
+ =
0
+
1 0
1
1 0
t 0
[(t − ts)α −1 − (t − s)α −1 ]ϕq
t 0
t 0
1 t
(1 − s)α−1 ϕq
s 0
1
(1 − s)α−1 ϕ
s
q
t
0
h(t)tα −1
1 t
(1 − s)α−1 ϕq
s 0
s 0
(s − r)β −1 [y2 (r) − c2 ]dr dsdt
s 0
(s − r)β −1 [y1 (r) − c1 + y2 (r) − y1 (r) − (c2 − c1 )]dr dsdt
(s − r)β −1 [y1 (r) − c1 + y2 (r) − y1 (r) − (c2 − c1 )]dr dsdt
[(t − ts)α −1 − (t − s)α −1 ]ϕq
(s − r)β −1 [y2 (r) − c2 ]dr dsdt
[(t − ts)α −1 − (t − s)α −1 ]ϕq
h(t)tα −1
h(t)
h(t)tα −1
h(t)
> 0
1 0
1
+
h(t)
s 0
(s − r)β −1 [y1 (r) − c1 ]dr dsdt
(s − r)β −1 [y1 (r) − c1 ]dr dsdt = 0.
A contradiction, too. So, we have mint∈[0,1] (y2 (t) − y1 (t)) ≤ c2 − c1 ≤ maxt∈[0,1] (y2 (t) − y1 (t)), i.e. |c2 − c1 | ≤ y2 − y1 ∞ . The proof is completed. Lemma 3.3. Define an operator R : X × [0, 1] → X2 as
R(u, λ)(t) =
1
(α)
0
t
(t − s)α−1 ϕq
1
(β)
s 0
(s − r)β −1 (Nλ u(r) − QNλ u(r))dr ds,
where KerM ⊕ X2 = X. Then R : × [0, 1] → X2 is continuous and compact,where ⊂ X is an open bounded set. Proof. Obviously, R is continuous. By the continuity of f and the boundedness of Q, we can get that there exist constants u(t))| ≤ k1 , |Qf | ≤ k2 for u ∈ . Thus, k1 > 0, k2 > 0 such that |f (t, u(t), D0α+−1 u(t), Dα 0+
|R(u, λ)| ≤
1 ϕ
(α + 1) q
k1 + k2 ,
(β + 1)
|Dα0+−1 R(u, λ)| ≤ ϕq
k1 + k2 ,
(β + 1)
[|Dα0+ R(u, λ)| ≤ ϕq
i.e. R is bounded in × [0, 1]. For (u, λ) ∈ × [0, 1], t1 , t2 ∈ [0, 1], t1 < t2 , we have
|R(u, λ)(t2 ) − R(u, λ)(t1 )| =
(α) −
=
t2
1
t1 0
0
(t1 − s)α−1 ϕq
(α)
t1
1
+
t2 t1
(t2 − s)α−1 ϕq
0
1
(β) 1
(β)
s 0
s 0
(t2 − s)α−1 ϕq
1
(β)
s 0
k1 + k2 ,
(β + 1)
(s − r)β −1 (Nλ u(r) − QNλ u(r))dr ds
(s − r)β −1 (Nλ u(r) − QNλ u(r))dr ds
[(t2 − s)α −1 − (t1 − s)α −1 ]ϕq
1
(β)
s 0
(s − r)β −1 (Nλ u(r) − QNλ u(r))dr ds
(s − r)β −1 (Nλ u(r) − QNλ u(r))dr ds
W. Jiang / Applied Mathematics and Computation 260 (2015) 48–56
≤
ϕq
k1 + k2
1
(β + 1) (α + 1)
α −1 D0+ R(u, λ)(t2 ) − Dα0+−1 R(u, λ)(t1 ) =
t2 0
ϕq
− = ≤
t1
t1
ϕq
ϕq
α t2 − t1α ,
1
(β)
ϕq
0 t2
s 0
(β)
k1 + k2
(β + 1)
s 0
s 0
1
(s − r)β −1 (Nλ u(r) − QNλ u(r))dr ds
1
(β)
53
− ϕq Since
1 (β)
t2 0
(β)
0
t1
t2 0
(t2 − t1 ).
(β)
(t2 − s)β −1 (Nλ u(s) − QNλ u(s))ds
1
(t2 − s)β −1 (Nλ u(s) − QNλ u(s))ds −
1 =
(β) ≤
1
(s − r)β −1 (Nλ u(r) − QNλ u(r))dr ds
−1 So, {R(u, λ)|(u, λ) ∈ × [0, 1]} and {Dα R(u, λ)|(u, λ) ∈ × [0, 1]} are equicontinuous. 0+
|Dα0+ R(u, λ)(t2 ) − Dα0+ R(u, λ)(t1 )| = ϕq
(s − r)β −1 (Nλ u(r) − QNλ u(r))dr ds
t1
0
1
(t1 − s)β −1 (Nλ u(s) − QNλ u(s))ds .
(β)
t1 0
(t1 − s)β −1 (Nλ u(s) − QNλ u(s))ds
[(t2 − s)β −1 − (t1 − s)β −1 ](Nλ u(s) − QNλ u(s))ds +
t2
t1
(t2 − s)β −1 (Nλ u(s) − QNλ u(s))ds
k1 + k2 β β t2 − t1 ,
(β + 1)
t
1 2 1 2 − s)β −1 (Nλ u(s) − QNλ u(s))ds| ≤ (β , u ∈ , λ ∈ [0, 1], and ϕq is uniformly continuous in [− (β , 1 2 ], we +1) +1) (β +1) α know that {D0+ R(u, λ)|(u, λ) ∈ × [0, 1]} is equicontinuous, too. By Arzela-Ascoli Theorem, we get that R : × [0, 1] → X2 is compact.
1 | (β)
k +k
0 (t
k +k
k +k
Lemma 3.4. Assume that ⊂ X is an open and bounded set. Then Nλ is M-quasi-compact in . Proof. It is clear that ImP = KerM, dimKerM = dimImQ, Q (I − Q ) = 0, KerQ = ImM, R(·, 0) = 0 and that Definition 2.2(b) holds. For β u ∈ λ = {u ∈ : Mu = Nλ u}, we have QNλ u = 0 and Nλ u = D0+ (ϕp (Dα u)). It follows from Dα u(0) = u(0) = Dα R(u, λ)(0) = 0+ 0+ 0+ R(u, λ)(0) = 0 that
β β β β R(u, λ) = I0α+ ϕq I0+ (Nλ u − QNλ u) = I0α+ ϕq I0+ Nλ u = I0α+ ϕq I0+ D0+ ϕp Dα0+ u Dα+−1 u(0) α −1 t = (I − P )u, = I0α+ Dα0+ u = u − 0
(α)
i.e. Definition 2.2(c) holds. For u ∈ , we have
M[Pu + R(u, λ)] = Nλ u − QNλ u = (I − Q )Nλ u. So, Definition 2.2(d) holds. Therefore, Nλ is M-compact in . The proof is completed. Theorem 3.1. Assume that the following conditions hold:
(H1 ) There exists a constant K > 0 such that one of the following inequalities holds: (1) B f (t, A, B, C ) > 0, t ∈ [0, 1], |B| > K, A, C ∈ R; (2) B f (t, A, B, C ) < 0, t ∈ [0, 1], |B| > K, A, C ∈ R. (H2 ) There exist nonnegative functions a(t), b(t), c(t), d(t) ∈ C[0, 1] such that
|f (t, x, y, z)| ≤ a(t)ϕp (|x|) + b(t)ϕp (|y|) + c(t)ϕp (|z|) + d(t), t ∈ [0, 1], x, y, z ∈ R
54
W. Jiang / Applied Mathematics and Computation 260 (2015) 48–56
with
2q−2 ϕ ϕq ( (β + 1)) q
A1 :=
1
A2 :=
ϕq ( (β + 1))
ϕq
a∞ + b∞ + c∞ < 1, if 1 < p ≤ 2; ϕp ( (α)) 2p−2 a∞
ϕp ( (α))
+ 2p−2 b∞ + c∞ < 1, if p ≥ 2;
Then boundary value problem (1.1) has at least one solution. In order to prove Theorem 3.1, we show two lemmas. Lemma 3.5. Suppose (H1 ) and (H2 ) hold. Then the set
1 = {u ∈ domM|Mu = Nλ u, λ ∈ (0, 1)} is bounded in X. −1 u(t0 )| ≤ K. Since D0α+−1 u(t) = Proof. For u ∈ 1 , we have QNλ u = 0. It follows (H1 ) that there exists t0 ∈ [0, 1] such that |Dα 0+ t α α −1 D0+ u(t0 ) + t D0+ u(s)ds, then 0
α −1
|D0+ u(t)| ≤ K + Dα0+ u∞ , t ∈ [0.1]. It follows from u(0) = 0 that
1
(α − 1)
u(t) = I0α+−1 Dα0+−1 u(t) = So, we have
|u(t)| ≤
1
(α − 1)
t 0
t 0
(3.3)
(t − s)α−2 Dα0+−1 u(s)ds.
(t − s)α−2 ds K + Dα0+ u∞ ≤
K + Dα0+ u∞
(α)
.
u(0) = 0, we get that By Mu = Nλ u, Dα 0+
Dα0+ u(t) = ϕq
1
(β)
0
t
(3.4)
(t − s)β −1 (−λf (s, u(s), Dα0+−1 u(s), Dα0+ u(s))ds) .
This, together with (H2 ), (3.2), (3.3), means
|Dα0+ u(t)| ≤
1
ϕq ( (β + 1))
If 1 < p ≤ 2, then
|Dα0+ u(t)| ≤
2q−2 ϕ ϕq ( (β + 1)) q
If p ≥ 2, then
|Dα0+ u(t)| ≤
ϕq a∞ ϕp u∞ + b∞ ϕp Dα0+−1 u∞ + c∞ ϕp Dα0+ u∞ + d∞ .
1
ϕq ( (β + 1))
ϕq
a∞ ϕp (K ) + b∞ ϕp (K ) + d∞ + A1 Dα0+ u∞ . ϕp ( (α)) 2p−2 a∞ ϕp (K )
ϕp ( (α))
+ 2p−2 b∞ ϕp (K ) + d∞ + A2 Dα0+ u∞ .
These, together with (3.2) and (3.3), mean that 1 is bounded. The proof is completed. Lemma 3.6. Assume (H1 ) holds. Then
2 = {u ∈ KerM|QNu = 0} is bounded in X, where N = N1 . −1 Proof. For u ∈ 2 , we have u = ktα −1 and Q (Nu) = 0. By (H1 ), we get |Dα u| = |k (α)| ≤ K. So, 2 is bounded. The proof is 0+ completed.
The proof of Theorem 3.1. Let ⊃ 1 ∪ 2 ∪ {x|x ∈ X, x ≤ max{K/ (α), K } + 1} be an open and bounded set of X. By Lemmas 3.5 and 3.6, we know Mu = Nλ u, u ∈ domM ∩ ∂ and QNu = 0, u ∈ KerM ∩ ∂ . α −1 , Let H(u, δ) = ρδ u + (1 − δ)JQNu, δ ∈ [0, 1], u ∈ KerM ∩ , where J : R = ImQ → KerM is a homeomorphism with Jk = kt −1, if (H1 )(1) holds, ρ= 1, if (H1 )(2) holds. For u ∈ KerM ∩ ∂ , we have u = k0 tα −1 and |k0 (α)| > K. Thus
H(u, δ) = ρδ k0 tα −1 + (1 − δ)[Q (−f )]tα −1 .
W. Jiang / Applied Mathematics and Computation 260 (2015) 48–56
55
δ If δ = 1, H(u, 1) = ρ k0 tα −1 = 0. If δ = 0, by (H1 ), we get H(u, 0) = −(Qf )tα −1 = 0. For 0 < δ < 1, if H(u, δ) = 0, then k0 = 1− δ ρ(Qf ). By (H1 ), we get
(α)k20 =
1−δ
δ
ρ (α)k0 (Qf ) < 0.
A contradiction. So, H(u, δ) = 0, u ∈ KerM ∩ ∂ , δ ∈ [0, 1]. By the homotopy of degree, we get that
KerM, 0 = deg H(·, 0), KerM, 0 deg JQN, = deg H(·, 1), KerM, 0 = deg ρ I, KerM, 0 = 0.
By Theorem 2.1, we can get that Mu = Nu has at least one solution in . The proof is completed. 4. Example Example 4.1. Let’s consider the following boundary value problem at resonance
⎧ 1 3 1 3 ⎨ D 2+ ϕ D 2+ u (t) + f (t, u(t), D 2+ u(t), D 2+ u(t)) = 0, p 0 0 0 0 3 1 ⎩ u(0) = D02+ u(0) = 0, u(1) = 0 h(t)u(t)dt,
(4.1)
1
1 2 1 2 3 where p = 4, h(t) = 2t 2 , f (t, x, y, z) = 24 t sin x3 + 16 t y + 14 t2 sin(tz)3 + t3 . Corresponding to the problem (1.1), we have β = 12 , α = 32 , q = 43 . 1 2 1 2 t , b(t) = 16 t , c(t) = 14 t5 , d(t) = t3 , K = 3. By simple calculations, we can get that the conditions (H1 ), (H2 ) hold. Take a(t) = 24 By Theorem 3.1, we obtain that the problem (4.1) has at least one solution.
Example 4.2. Considering the boundary value problem
⎧ 1 5 1 5 ⎨ D 4+ ϕ D 4+ u (t) + f (t, u(t), D 4+ u(t), D 4+ u(t)) = 0, p 0 0 0 0 5 1 ⎩ u(0) = D04+ u(0) = 0, u(1) = 0 h(t)u(t)dt,
where p =
5 3 , h(t) =
f (t, x, y, z) =
(4.2)
3
2t 4 ,
a(t) min{1, x2 } + b(t)y3 + c(t) sin z 3 + d(t), t ∈ [0, 1], x, z ∈ R, |y| ≤ 1, 2
a(t) min{1, x2 } + b(t)y + c(t) sin z 3 + d(t), t ∈ [0, 1], x, z ∈ R, |y| > 1. 1 3
2
If a(t), b(t), c(t), d(t) ∈ C[0, 1] and there exists a constant l > 0 such that the following conditions hold:
( (1.25)) 3 5
(h1 ) a∞ + (b∞ + c∞ )( (1.25)) 3 < 2
(h2 )
√ 3 2
.
|a(t)| + |c(t)| + |d(t)| < l, t ∈ [0, 1]. |b(t)|
Then the problem (4.2) has at least one solution. Proof. Corresponding to the problem (1.1), we have β =
1 4,α
=
5 4,q
=
5 2.
Let K = max{l3 , 1}. For |y| > K, by (h2 ), we have
|a(t)| + |c(t)| + |d(t)| < |b(t)|l ≤ |b(t)|K 3 < |b(t)||y| 3 . 1
1
(4.3)
By b(t) ∈ C[0, 1] and (h2 ), we get that b(t) > 0, t ∈ [0, 1] or b(t) < 0, t ∈ [0, 1]. It follows from (4.3) that if b(t) < 0, then yf (t, x, y, z) < 0 for |y| > K, and yf (t, x, y, z) > 0 for |y| > K, if b(t) > 0, i.e. the condition (H1 ) holds. 2
2
2
Obviously, |f (t, x, y, z)| ≤ |a(t)||x| 3 + |b(t)||y| 3 + |c(t)||z| 3 + |d(t)|, and
A1 =
2q−2 ϕ ϕq ( (β + 1)) q
22 a∞ + b ∞ + c ∞ = 3 ϕp ( (α)) ( (1.25)) 2 1
a∞ + b ∞ + c ∞ 2 ( (1.25)) 3
32 .
It follows from (h1 ) that A1 < 1, i.e. the condition (H2 ) holds. By Theorem 3.1, we obtain that the problem (4.2) has at least one solution. Acknowledgments The author is grateful to anonymous referees for their constructive comments and suggestions which led to improvement of the original manuscript. This work is supported by the Natural Science Foundation of China (11171088) and the Natural Science Foundation of Hebei Province (A2013208108).
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References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28]
I. Podlubny, Fractional Differential Equations, Academic Press, New York, London, 1999. K.S. Miller, B. Ross, An Introduction to the Fractional Calculus and Fractional Differential Equations, Wiley, New York, 1993. A.A. Kilbas, H.M. Srivastava, J.J. Trujillo, Theory and Applications of Fractional Differential Equations, Elsevier B.V., 2006. R.P. Agarwal, J.P. dos Santos, C. Cuevas, Analytic resolvent operator and existence results for fractional integro-differential equations, J. Abstr. Differ. Equ. Appl. 02 (2) (2012) 26–47. R.P. Agarwal, B. de Andrade, C. Cuevas, Type of periodicity and ergodicity to a class of fractional order differential equations, Adv. Differ. Equ. 2010 (2010) 1–25. R.P. Agarwal, M. Belmekki, M. Benchohra, Survey on semi-linear differential equations and inclusions involving Riemann–Liouville fractional derivative, Adv. Differ. Equ. 2009 (2009) 1–47. R.P. Agarwal, M. Benchohra, S. Hamani, A survey on existence result for boundary value problems of nonlinear fractional differential equations and inclusions, Acta Appl. Math. 109 (2010) 973–1033, doi:10.1007/s10440-008-9356-6. Z. Bai, H. Lu, Positive solutions for boundary value problem of nonlinear fractional differential equation, J. Math. Anal. Appl. 311 (2005) 495–505, doi:10.1016/j.jmaa.2005.02.052. S. Marano, D. Motreanu, Infinitely many critical points of non-differentiable functions and applications to a Neumann type problem involving the p-Laplacian, J. Differ. Equ. 182 (2002) 108–120, doi:10.1006/jdeq.2001.4092. D. Pina, M. Elgueta, R. Manasevich, A homotopic deformation along p of a Leray–Schauder degree result and existence for (|u |p−2 u ) + f (t, u) = 0, u(0) = u(t) = 0, p > 1, J. Differ. Equ. 80 (1) (1997) 3227–3238. M. Garcia-Huidobro, R. Manasevich, F. Zanolin, A Fredholm-like result for strongly nonlinear second order odes, J. Differ. Equ. 114 (1994) 132–167. M. Garcia-Huidobro, P. Ubilla, Multiplicity of solutions for a class of nonlinear second order equations, Nonlinear Anal. 28 (9) (1997) 1509–1520. D.R. Anderson, R.I. Avery, J. Henderson, Existence of solutions for a one-dimensional p-Laplacian on time scales, J. Differ. Equ. Appl. 10 (2004) 889–896, doi:10.1080/10236190410001731416. C.S. Goodrich, The existence of a positive solution to a second-order delta-nabla p-Laplacian bvp on a time scale, Appl. Math. Lett. 25 (2012) 157–162, doi:10.1016/j.aml.2011.08.005. J. Mawhin, Topological degree methods in nonlinear boundary value problems, in: NSFCBMS Regional Conference Series in Mathematics, American Mathematical Society, Providence, RI, 1979. W. Feng, J.R.L. Webb, Solvability of m-point boundary value problems with nonlinear growth, J. Math. Anal. Appl. 212 (1997) 467–480. Y. Liu, W. Ge, Solvability of nonlocal boundary value problems for ordinary differential equations of higher order, Nonlinear Anal. 57 (2004) 435–458. Z. Du, X. Lin, W. Ge, Some higher-order multi-point boundary value problem at resonance, J. Comput. Appl. Math. 177 (2005) 55–65. B. Liu, Solvability of multi-point boundary value problem at resonance (II), Appl. Math. Comput. 136 (2003) 353–377. X. Zhang, M. Feng, W. Ge, Existence result of second-order differential equations with integral boundary conditions at resonance, J. Math. Anal. Appl. 353 (2009) 311–319. N. Kosmatov, Multi-point boundary value problems on an unbounded domain at resonance, Nonlinear Anal. 68 (2008) 2158–2171. N. Kosmatov, A boundary value problem of fractional order at resonance, Electron. J. Differ. Equ. 2010 (2010) 135 1–10. W. Jiang, The existence of solutions to boundary value problems of fractional differential equations at resonance, Nonlinear Anal. (TMA). 74 (2011) 1987–1994. W. Jiang, Solvability for a coupled system of fractional differential equations at resonance, Nonlinear Anal. (RWA). 13 (2012) 2285–2292. W. Jiang, Solvability of (k,n-k) conjugate boundary-value problems at resonance, Electron. J. Differ. Equ. 2012 (2012) 114 1–10. W. Ge, J. Ren, An extension of Mawhin’s continuation theorem and its application to boundary value problems with a p-Laplacian, Nonlinear Anal. (TMA). 58 (2004) 477–488. W. Jiang, Solvability for p-Laplacian boundary value problem at resonance on the half-line, Bound. Value Probl. 2013 (2013) 207. J. Kuang, Applied Inequalities, Shandong Science and Technology Press, Jinan, 2014, (P132).
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Solvability of fractional differential equations with p-Laplacian at resonance Weihua Jiang∗ College of Sciences, Hebei University of Science and Technology Shijiazhuang, Hebei 050018, PR China
a r t i c l e
i n f o
MSC: 34A08 34B15 Keywords: Continuous theorem Fractional differential equation Resonance p-Laplacian Boundary value problem
a b s t r a c t In order to study boundary value problems with p-Laplacian, the extension for the continuous theorem of Ge and Ren is generalized. By using the new results and constructing suitable operators, we investigate the existence of solutions for p-Laplacian fractional differential equations at resonance. Examples are given to illustrate our results. © 2015 Elsevier Inc. All rights reserved.
1. Introduction The fractional differential equations have become an important area of investigation because of the intensive development of the theory of fractional calculus itself and its wide applications in various sciences such as physics, mechanics, chemistry, engineering, etc. (see [1–8]). On the other hand, p-Laplacian boundary value problems have been widely studied owing to its importance in theory and application of mathematics and physics (see [9–14] and the results cited therein). Mawhin’s continuous theorem [15] is an effective tool in studying the existence of solutions for the abstract equation Lx = Nx (see [16–25]), where L is a noninvertible linear operator. Ge and Ren [26] extended Mawhin’s results in [15] and obtained the existence of solutions when L is a noninvertible nonlinear one. This is a wonderful instrument to solve the p-Laplacian boundary value problems at resonance. Using this theorem, the authors need to construct two projectors P and Q (see [26,27]). But sometimes it is difficult to give the projector Q in many boundary value problems with p-Laplacian. We will generalize the extension of the continuous theorem in [26] and show that the equation Lx = Nx is solvable when Q is not a projector but satisfies some conditions. And, using our new results, we will prove the existence of solutions for the boundary value problem with p-Laplacian at resonance:
β D0+ ϕp Dα0+ u (t) + f t, u(t), Dα0+−1 u(t), Dα0+ u(t) = 0, 1 u(0) = Dα0+ u(0) = 0, u(1) = 0 h(t)u(t)dt,
(1.1)
1 where 0 < β ≤ 1, 1 < α ≤ 2, 0 h(t)tα −1 dt = 1, ϕp (s) = |s|p−2 s, p > 1. To the best of our knowledge, this is the first paper to study fractional differential equations with p-Laplacian and integral boundary value conditions at resonance when Q is not a projector.
∗
Tel.: +8613785419695. E-mail address: [email protected], [email protected], [email protected]
http://dx.doi.org/10.1016/j.amc.2015.03.036 0096-3003/© 2015 Elsevier Inc. All rights reserved.
W. Jiang / Applied Mathematics and Computation 260 (2015) 48–56
49
2. Preliminaries and a generalized theorem Definition 2.1 ([26]). Let X and Z be two Banach spaces with norms · X , · Z , respectively. A continuous operator M : X ∩ domM → Z is said to be quasi-linear if (i) ImM := M(X ∩ domM) is a closed subset of Z, (ii) KerM := {x ∈ X ∩ domM : Mx = 0} is linearly homeomorphic to Rn , n < ∞, where domM denotes the domain of the operator M. Let X1 = KerM and X2 be the complement space of X1 in X. Then X = X1 ⊕ X2 . Let P : X → X1 be projector and ⊂ X be an open and bounded set with the origin θ ∈ . Definition 2.2. Suppose that Nλ : → Z, λ ∈ [0, 1] is a continuous and bounded operator. Denote N1 by N. Let λ = {x ∈ : Mx = Nλ x}. Nλ is said to be M-quasi-compact in if there exists a vector subspace Z1 of Z satisfying dimZ1 = dimX1 and two operators Q and R such that for λ ∈ [0, 1], (a) (b) (c) (d)
KerQ = ImM, QNλ x = θ , λ ∈ (0, 1) ⇔ QNx = θ , R(·, 0) is the zero operator and R(·, λ)| λ = (I − P )| λ , M[P + R(·, λ)] = (I − Q )Nλ ,
where Q : Z → Z1 , QZ = Z1 is continuous, bounded and satisfies Q (I − Q ) = 0 and R : × [0, 1] → X2 is continuous and compact. Theorem 2.1. Let X and Z be two Banach spaces with the norms · X , · Z , respectively, and ⊂ X be an open and bounded nonempty set. Suppose
M : X ∩ domM → Z is a quasi-linear operator and that Nλ : → Z, λ ∈ [0, 1] is M-quasi-compact. In addition, if the following conditions hold:
(C1 ) Mx = Nλ x, ∀x ∈ ∂ ∩ domM, λ ∈ (0, 1), (C2 ) deg{JQN, ∩ KerM, 0} = 0, then the abstract equation Mx = Nx has at least one solution in domM ∩ , where N = N1 , J : ImQ → KerM is a homeomorphism with J(θ ) = θ . The proof used in this paper is similar to the one in [26]. Proof. Define Sλ : ∩ domM −→ X, λ ∈ [0, 1] by
Sλ = P + R(·, λ) + JQN.
(2.1)
Since Q and N are continuous and bounded, P, R, J are continuous and compact, Sλ is a continuous compact mapping for
(x, λ) ∈ × [0, 1].
Firstly, we prove that x ∈ is a solution of
Mx = Nλ x, λ ∈ (0, 1] if and only if x is a fixed point of Sλ . If x ∈ is a solution of (2.2), by Definition 2.2(a), we get
Nλ x = Mx ∈ ImM = KerQ. Hence,
QNλ x = θ , λ ∈ (0, 1]. This, together with (2.2) and Definition 2.2(c), means
(I − P)x = R(x, λ), JQNx = θ . Therefore,
x = Px + R(x, λ) = Px + R(x, λ) + JQNx = Sλ x. So, x is a fixed point of Sλ . On the other hand, if x ∈ is a fixed point of Sλ , λ ∈ (0, 1], considering R(x, λ) ∈ X2 , we get
Px = PSλ x = Px + P (JQNx), which implies JQNx = θ ∈ X1 and QNx = θ ∈ Z1 . Hence
QNλ x = θ ∈ Z1 , x = Px + R(x, λ), λ ∈ (0, 1].
(2.2)
50
W. Jiang / Applied Mathematics and Computation 260 (2015) 48–56
These, together with Definition 2.2(d), mean that
Mx = M(Px + R(x, λ)) = (I − Q )Nλ x = Nλ x − QNλ x = Nλ x. Now, we will prove that I − Sλ is a homotopic mapping. By the above results and condition (C1 ), we can get that
λ ∈ (0, 1), x ∈ ∂ .
x = Sλ x,
(2.3)
Condition (C2 ) means that
x ∈ ∂ .
x = S0 x,
(2.4)
In fact, if x = S0 x, x ∈ ∂ , then x = Px + JQNx ∈ X1 . So, we have x = x + JQNx and then JQNx = θ which contradicts (C2 ). For λ = 1, if there is x0 ∈ ∂ such that x0 = S1 x0 then we have Mx0 = Nx0 , x0 ∈ ∂ ⊂ . Theorem 2.1 is proved. Without loss generality, suppose
x ∈ ∂ .
x = S1 x,
(2.5)
Combining (2.3)–(2.5), we have
λ ∈ [0, 1], x ∈ ∂ .
x = Sλ x,
This means that I − Sλ is a homotopic mapping. Thus,
deg{I − S1 , ∩ domM, 0} = deg{I − S0 , ∩ domM, 0} = deg{I − P − JQN, ∩ domM, 0} = deg{I − P − JQN, ∩ kerM, 0} = deg{−JQN, ∩ kerM, 0} = 0. So, S1 has a fixed point x0 ∈ , i.e. Mx0 = Nx0 . The proof is completed. The following definitions can be found in [1–3]. Definition 2.3. The fractional integral of order α > 0 of a function y : (0, ∞) → R is given by
1
I0α+ y(t) =
(α)
0
t
(t − s)α−1 y(s)ds,
(2.6)
provided the right-hand side is pointwise defined on (0, ∞). Definition 2.4. The fractional derivative of order α > 0 of a function y : (0, ∞) → R is given by
Dα0+ y(t) =
1 dn
(n − α) dtn
t 0
(t − s)n−α−1 y(s)ds,
provided the right-hand side is pointwise defined on (0, ∞), where n = [α ] + 1. In order to obtain our main results, we need the following lemmas. Lemma 2.1 ([2,8]). Assume f ∈ L[0, 1], q ≥ p ≥ 0, q > 1,then
D0+ I0+ f (t) = I0+ f (t). p
q
q−p
Lemma 2.2 ([8]). Assume α > 0, λ > −1, then
Dα0+ tλ =
(λ + 1) dn n+λ−α (t ),
(n + λ − α + 1) dtn
where n = [α ] + 1. Example. Dα tα −i = 0, i = 1, 2, . . . , n. 0+ Lemma 2.3 ([8]). Dα u(t) = 0 if and only if 0+
u(t) = c1 tα −1 + c2 tα −2 + · · · + cn tα −n , where n is the smallest integer greater than or equal to α , ci ∈ R, i = 1, 2, . . . , n. +m Lemma 2.4 ([8]). If the fractional derivatives Dα y(t) and Dα y(t) exist, then 0+ 0+
(Dα0+ y(t))(m) = Dα0++m y(t), where α > 0, m is a positive integer.
(2.7)
W. Jiang / Applied Mathematics and Computation 260 (2015) 48–56
51
Lemma 2.5 ([28]). For any u, v ≥ 0, then (1) ϕp (u + v) ≤ ϕp (u) + ϕp (v), 1 < p ≤ 2; (2) ϕp (u + v) ≤ 2p−2 (ϕp (u) + ϕp (v)), p ≥ 2, where ϕp (s) = |s|p−2 s = sp−1 , s ≥ 0. 3. The solutions for the problem (1.1) 1 In the following, we will always suppose that f : [0, 1] × R3 → R is continuous, h(t) ≥ 0, t ∈ [0, 1], 0 h(t)tα −1 dt = 1 and that q satisfies 1/p + 1/q = 1. −1 Let X = {u|Dα u ∈ C[0, 1], u(0) = Dα u(0) = 0} with norm u = max{u∞ ,Dα u∞ ,Dα u∞ }, Y = C[0, 1] with norm 0+ 0+ 0+ 0+ y∞ = maxt∈[0,1] |y(t)|. We know that (X, · ) and (Y, · ∞ ) are Banach spaces. Define operators M : X ∩ domM → Y, Nλ : X → Y as follows: β
Mu(t) = D0+
ϕp (Dα0+ u) (t), Nλ u(t) = −λf t, u(t), Dα0+−1 u(t), Dα0+ u(t) ,
t,
λ ∈ [0, 1],
1 β where domM = {u ∈ X |D0+ ϕp Dα u ∈ C[0, 1], u(1) = 0 h(t)u(t)dt}. 0+ Lemma 3.1. M is a quasi-linear operator. Proof. It is easy to get that KerM = {ktα −1 |k ∈ R}. For u ∈ X ∩ domM, if Mu = y, then y satisfies
1 0
h(t)
+
0
1
0
t
[(t − ts)α −1 − (t − s)α −1 ]ϕq
h(t)tα −1
1
t
(1 − s)α−1 ϕq
s
0
s 0
(s − r)β −1 y(r)dr dsdt
(s − r)β −1 y(r)dr dsdt = 0.
On the other hand, if y ∈ Y satisfies (3.1), take
u(t) =
1
(α)
t 0
(t − s)α−1 ϕq
1
(β)
s
0
(3.1)
(s − r)β −1 y(r)dr ds + ctα−1 ,
where c is a constant. By simple calculations, we get u ∈ X ∩ domM and Mu = y. Thus
ImM = {y ∈ Y |y satisfies (3.1)}. Obviously, ImM ⊂ Y is closed. So, M is quasi-linear. The proof is completed. Take a projector P : X → KerM and an operator Q : Y → R as follows:
Dα0+−1 u(0) α −1 t ,
(Pu)(t) =
(Qy)(t) = c,
(α)
where c satisfies
1 0
h(t)
+
1 0
t 0
[(t − ts)α −1 − (t − s)α −1 ]ϕq
h(t)tα −1
1
(1 − s)α−1 ϕ
s
q
t
0
s 0
(s − r)β −1 [y(r) − c]dr dsdt
(s − r)β −1 [y(r) − c]dr dsdt = 0.
We will prove that c is the unique constant satisfying (3.2). For y ∈ Y, let
F (c) =
1 0
+
h(t)
1 0
t 0
[(t − ts)α −1 − (t − s)α −1 ]ϕq
h(t)tα −1
1 t
(1 − s)α−1 ϕq
s 0
s 0
(3.2)
(s − r)β −1 [y(r) − c]dr dsdt
(s − r)β −1 [y(r) − c]dr dsdt.
Obviously F (c) is continuous and strictly decreasing in R. Take c1 = mint∈[0,1] y(t), c2 = maxt∈[0,1] y(t). It is easy to see that F (c1 ) ≥ 0, F (c2 ) ≤ 0. So, there exists a unique constant c ∈ [c1 , c2 ] such that F (c) = 0. Furthermore, Q () is bounded if ⊂ Y is bounded, i.e. Q is bounded. Remark. By the definition of Q, we can easily obtain that Q is not a projector but satisfies Q (I − Q )y = Q (y − Qy) = 0, y ∈ Y. Lemma 3.2. Q is continuous in Y.
52
W. Jiang / Applied Mathematics and Computation 260 (2015) 48–56
Proof. For y1 , y2 ∈ Y, assume Qy1 = c1 , Qy2 = c2 . Noticing h(t) ≥ 0 and that ϕq is strictly increasing, we obtain that if c2 − c1 > maxt∈[0,1] (y2 (t) − y1 (t)), then
0=
0
1
+ =
0
<
0
1 0
1
1 0
1
+
1 0
t 0
[(t − ts)α −1 − (t − s)α −1 ]ϕq
t 0
t 0
1 t
(1 − s)α−1 ϕq
s 0
1
(1 − s)α−1 ϕ
s
q
t
0
1
(1 − s)α−1 ϕ
s
q
t
0
s 0
(s − r)β −1 [y2 (r) − c2 ]dr dsdt
s 0
(s − r)β −1 [y1 (r) − c1 + y2 (r) − y1 (r) − (c2 − c1 )]dr dsdt
(s − r)β −1 [y1 (r) − c1 + y2 (r) − y1 (r) − (c2 − c1 )]dr dsdt
[(t − ts)α −1 − (t − s)α −1 ]ϕq
h(t)tα −1
(s − r)β −1 [y2 (r) − c2 ]dr dsdt
[(t − ts)α −1 − (t − s)α −1 ]ϕq
h(t)tα −1
h(t)
h(t)tα −1
h(t)
+
h(t)
s 0
(s − r)β −1 [y1 (r) − c1 ]dr dsdt
(s − r)β −1 [y1 (r) − c1 ]dr dsdt = 0.
A contradiction. On the other hand, if c2 − c1 < mint∈[0,1] (y2 (t) − y1 (t)), then
0=
0
1
+ =
0
+
1 0
1
1 0
t 0
[(t − ts)α −1 − (t − s)α −1 ]ϕq
t 0
t 0
1 t
(1 − s)α−1 ϕq
s 0
1
(1 − s)α−1 ϕ
s
q
t
0
h(t)tα −1
1 t
(1 − s)α−1 ϕq
s 0
s 0
(s − r)β −1 [y2 (r) − c2 ]dr dsdt
s 0
(s − r)β −1 [y1 (r) − c1 + y2 (r) − y1 (r) − (c2 − c1 )]dr dsdt
(s − r)β −1 [y1 (r) − c1 + y2 (r) − y1 (r) − (c2 − c1 )]dr dsdt
[(t − ts)α −1 − (t − s)α −1 ]ϕq
(s − r)β −1 [y2 (r) − c2 ]dr dsdt
[(t − ts)α −1 − (t − s)α −1 ]ϕq
h(t)tα −1
h(t)
h(t)tα −1
h(t)
> 0
1 0
1
+
h(t)
s 0
(s − r)β −1 [y1 (r) − c1 ]dr dsdt
(s − r)β −1 [y1 (r) − c1 ]dr dsdt = 0.
A contradiction, too. So, we have mint∈[0,1] (y2 (t) − y1 (t)) ≤ c2 − c1 ≤ maxt∈[0,1] (y2 (t) − y1 (t)), i.e. |c2 − c1 | ≤ y2 − y1 ∞ . The proof is completed. Lemma 3.3. Define an operator R : X × [0, 1] → X2 as
R(u, λ)(t) =
1
(α)
0
t
(t − s)α−1 ϕq
1
(β)
s 0
(s − r)β −1 (Nλ u(r) − QNλ u(r))dr ds,
where KerM ⊕ X2 = X. Then R : × [0, 1] → X2 is continuous and compact,where ⊂ X is an open bounded set. Proof. Obviously, R is continuous. By the continuity of f and the boundedness of Q, we can get that there exist constants u(t))| ≤ k1 , |Qf | ≤ k2 for u ∈ . Thus, k1 > 0, k2 > 0 such that |f (t, u(t), D0α+−1 u(t), Dα 0+
|R(u, λ)| ≤
1 ϕ
(α + 1) q
k1 + k2 ,
(β + 1)
|Dα0+−1 R(u, λ)| ≤ ϕq
k1 + k2 ,
(β + 1)
[|Dα0+ R(u, λ)| ≤ ϕq
i.e. R is bounded in × [0, 1]. For (u, λ) ∈ × [0, 1], t1 , t2 ∈ [0, 1], t1 < t2 , we have
|R(u, λ)(t2 ) − R(u, λ)(t1 )| =
(α) −
=
t2
1
t1 0
0
(t1 − s)α−1 ϕq
(α)
t1
1
+
t2 t1
(t2 − s)α−1 ϕq
0
1
(β) 1
(β)
s 0
s 0
(t2 − s)α−1 ϕq
1
(β)
s 0
k1 + k2 ,
(β + 1)
(s − r)β −1 (Nλ u(r) − QNλ u(r))dr ds
(s − r)β −1 (Nλ u(r) − QNλ u(r))dr ds
[(t2 − s)α −1 − (t1 − s)α −1 ]ϕq
1
(β)
s 0
(s − r)β −1 (Nλ u(r) − QNλ u(r))dr ds
(s − r)β −1 (Nλ u(r) − QNλ u(r))dr ds
W. Jiang / Applied Mathematics and Computation 260 (2015) 48–56
≤
ϕq
k1 + k2
1
(β + 1) (α + 1)
α −1 D0+ R(u, λ)(t2 ) − Dα0+−1 R(u, λ)(t1 ) =
t2 0
ϕq
− = ≤
t1
t1
ϕq
ϕq
α t2 − t1α ,
1
(β)
ϕq
0 t2
s 0
(β)
k1 + k2
(β + 1)
s 0
s 0
1
(s − r)β −1 (Nλ u(r) − QNλ u(r))dr ds
1
(β)
53
− ϕq Since
1 (β)
t2 0
(β)
0
t1
t2 0
(t2 − t1 ).
(β)
(t2 − s)β −1 (Nλ u(s) − QNλ u(s))ds
1
(t2 − s)β −1 (Nλ u(s) − QNλ u(s))ds −
1 =
(β) ≤
1
(s − r)β −1 (Nλ u(r) − QNλ u(r))dr ds
−1 So, {R(u, λ)|(u, λ) ∈ × [0, 1]} and {Dα R(u, λ)|(u, λ) ∈ × [0, 1]} are equicontinuous. 0+
|Dα0+ R(u, λ)(t2 ) − Dα0+ R(u, λ)(t1 )| = ϕq
(s − r)β −1 (Nλ u(r) − QNλ u(r))dr ds
t1
0
1
(t1 − s)β −1 (Nλ u(s) − QNλ u(s))ds .
(β)
t1 0
(t1 − s)β −1 (Nλ u(s) − QNλ u(s))ds
[(t2 − s)β −1 − (t1 − s)β −1 ](Nλ u(s) − QNλ u(s))ds +
t2
t1
(t2 − s)β −1 (Nλ u(s) − QNλ u(s))ds
k1 + k2 β β t2 − t1 ,
(β + 1)
t
1 2 1 2 − s)β −1 (Nλ u(s) − QNλ u(s))ds| ≤ (β , u ∈ , λ ∈ [0, 1], and ϕq is uniformly continuous in [− (β , 1 2 ], we +1) +1) (β +1) α know that {D0+ R(u, λ)|(u, λ) ∈ × [0, 1]} is equicontinuous, too. By Arzela-Ascoli Theorem, we get that R : × [0, 1] → X2 is compact.
1 | (β)
k +k
0 (t
k +k
k +k
Lemma 3.4. Assume that ⊂ X is an open and bounded set. Then Nλ is M-quasi-compact in . Proof. It is clear that ImP = KerM, dimKerM = dimImQ, Q (I − Q ) = 0, KerQ = ImM, R(·, 0) = 0 and that Definition 2.2(b) holds. For β u ∈ λ = {u ∈ : Mu = Nλ u}, we have QNλ u = 0 and Nλ u = D0+ (ϕp (Dα u)). It follows from Dα u(0) = u(0) = Dα R(u, λ)(0) = 0+ 0+ 0+ R(u, λ)(0) = 0 that
β β β β R(u, λ) = I0α+ ϕq I0+ (Nλ u − QNλ u) = I0α+ ϕq I0+ Nλ u = I0α+ ϕq I0+ D0+ ϕp Dα0+ u Dα+−1 u(0) α −1 t = (I − P )u, = I0α+ Dα0+ u = u − 0
(α)
i.e. Definition 2.2(c) holds. For u ∈ , we have
M[Pu + R(u, λ)] = Nλ u − QNλ u = (I − Q )Nλ u. So, Definition 2.2(d) holds. Therefore, Nλ is M-compact in . The proof is completed. Theorem 3.1. Assume that the following conditions hold:
(H1 ) There exists a constant K > 0 such that one of the following inequalities holds: (1) B f (t, A, B, C ) > 0, t ∈ [0, 1], |B| > K, A, C ∈ R; (2) B f (t, A, B, C ) < 0, t ∈ [0, 1], |B| > K, A, C ∈ R. (H2 ) There exist nonnegative functions a(t), b(t), c(t), d(t) ∈ C[0, 1] such that
|f (t, x, y, z)| ≤ a(t)ϕp (|x|) + b(t)ϕp (|y|) + c(t)ϕp (|z|) + d(t), t ∈ [0, 1], x, y, z ∈ R
54
W. Jiang / Applied Mathematics and Computation 260 (2015) 48–56
with
2q−2 ϕ ϕq ( (β + 1)) q
A1 :=
1
A2 :=
ϕq ( (β + 1))
ϕq
a∞ + b∞ + c∞ < 1, if 1 < p ≤ 2; ϕp ( (α)) 2p−2 a∞
ϕp ( (α))
+ 2p−2 b∞ + c∞ < 1, if p ≥ 2;
Then boundary value problem (1.1) has at least one solution. In order to prove Theorem 3.1, we show two lemmas. Lemma 3.5. Suppose (H1 ) and (H2 ) hold. Then the set
1 = {u ∈ domM|Mu = Nλ u, λ ∈ (0, 1)} is bounded in X. −1 u(t0 )| ≤ K. Since D0α+−1 u(t) = Proof. For u ∈ 1 , we have QNλ u = 0. It follows (H1 ) that there exists t0 ∈ [0, 1] such that |Dα 0+ t α α −1 D0+ u(t0 ) + t D0+ u(s)ds, then 0
α −1
|D0+ u(t)| ≤ K + Dα0+ u∞ , t ∈ [0.1]. It follows from u(0) = 0 that
1
(α − 1)
u(t) = I0α+−1 Dα0+−1 u(t) = So, we have
|u(t)| ≤
1
(α − 1)
t 0
t 0
(3.3)
(t − s)α−2 Dα0+−1 u(s)ds.
(t − s)α−2 ds K + Dα0+ u∞ ≤
K + Dα0+ u∞
(α)
.
u(0) = 0, we get that By Mu = Nλ u, Dα 0+
Dα0+ u(t) = ϕq
1
(β)
0
t
(3.4)
(t − s)β −1 (−λf (s, u(s), Dα0+−1 u(s), Dα0+ u(s))ds) .
This, together with (H2 ), (3.2), (3.3), means
|Dα0+ u(t)| ≤
1
ϕq ( (β + 1))
If 1 < p ≤ 2, then
|Dα0+ u(t)| ≤
2q−2 ϕ ϕq ( (β + 1)) q
If p ≥ 2, then
|Dα0+ u(t)| ≤
ϕq a∞ ϕp u∞ + b∞ ϕp Dα0+−1 u∞ + c∞ ϕp Dα0+ u∞ + d∞ .
1
ϕq ( (β + 1))
ϕq
a∞ ϕp (K ) + b∞ ϕp (K ) + d∞ + A1 Dα0+ u∞ . ϕp ( (α)) 2p−2 a∞ ϕp (K )
ϕp ( (α))
+ 2p−2 b∞ ϕp (K ) + d∞ + A2 Dα0+ u∞ .
These, together with (3.2) and (3.3), mean that 1 is bounded. The proof is completed. Lemma 3.6. Assume (H1 ) holds. Then
2 = {u ∈ KerM|QNu = 0} is bounded in X, where N = N1 . −1 Proof. For u ∈ 2 , we have u = ktα −1 and Q (Nu) = 0. By (H1 ), we get |Dα u| = |k (α)| ≤ K. So, 2 is bounded. The proof is 0+ completed.
The proof of Theorem 3.1. Let ⊃ 1 ∪ 2 ∪ {x|x ∈ X, x ≤ max{K/ (α), K } + 1} be an open and bounded set of X. By Lemmas 3.5 and 3.6, we know Mu = Nλ u, u ∈ domM ∩ ∂ and QNu = 0, u ∈ KerM ∩ ∂ . α −1 , Let H(u, δ) = ρδ u + (1 − δ)JQNu, δ ∈ [0, 1], u ∈ KerM ∩ , where J : R = ImQ → KerM is a homeomorphism with Jk = kt −1, if (H1 )(1) holds, ρ= 1, if (H1 )(2) holds. For u ∈ KerM ∩ ∂ , we have u = k0 tα −1 and |k0 (α)| > K. Thus
H(u, δ) = ρδ k0 tα −1 + (1 − δ)[Q (−f )]tα −1 .
W. Jiang / Applied Mathematics and Computation 260 (2015) 48–56
55
δ If δ = 1, H(u, 1) = ρ k0 tα −1 = 0. If δ = 0, by (H1 ), we get H(u, 0) = −(Qf )tα −1 = 0. For 0 < δ < 1, if H(u, δ) = 0, then k0 = 1− δ ρ(Qf ). By (H1 ), we get
(α)k20 =
1−δ
δ
ρ (α)k0 (Qf ) < 0.
A contradiction. So, H(u, δ) = 0, u ∈ KerM ∩ ∂ , δ ∈ [0, 1]. By the homotopy of degree, we get that
KerM, 0 = deg H(·, 0), KerM, 0 deg JQN, = deg H(·, 1), KerM, 0 = deg ρ I, KerM, 0 = 0.
By Theorem 2.1, we can get that Mu = Nu has at least one solution in . The proof is completed. 4. Example Example 4.1. Let’s consider the following boundary value problem at resonance
⎧ 1 3 1 3 ⎨ D 2+ ϕ D 2+ u (t) + f (t, u(t), D 2+ u(t), D 2+ u(t)) = 0, p 0 0 0 0 3 1 ⎩ u(0) = D02+ u(0) = 0, u(1) = 0 h(t)u(t)dt,
(4.1)
1
1 2 1 2 3 where p = 4, h(t) = 2t 2 , f (t, x, y, z) = 24 t sin x3 + 16 t y + 14 t2 sin(tz)3 + t3 . Corresponding to the problem (1.1), we have β = 12 , α = 32 , q = 43 . 1 2 1 2 t , b(t) = 16 t , c(t) = 14 t5 , d(t) = t3 , K = 3. By simple calculations, we can get that the conditions (H1 ), (H2 ) hold. Take a(t) = 24 By Theorem 3.1, we obtain that the problem (4.1) has at least one solution.
Example 4.2. Considering the boundary value problem
⎧ 1 5 1 5 ⎨ D 4+ ϕ D 4+ u (t) + f (t, u(t), D 4+ u(t), D 4+ u(t)) = 0, p 0 0 0 0 5 1 ⎩ u(0) = D04+ u(0) = 0, u(1) = 0 h(t)u(t)dt,
where p =
5 3 , h(t) =
f (t, x, y, z) =
(4.2)
3
2t 4 ,
a(t) min{1, x2 } + b(t)y3 + c(t) sin z 3 + d(t), t ∈ [0, 1], x, z ∈ R, |y| ≤ 1, 2
a(t) min{1, x2 } + b(t)y + c(t) sin z 3 + d(t), t ∈ [0, 1], x, z ∈ R, |y| > 1. 1 3
2
If a(t), b(t), c(t), d(t) ∈ C[0, 1] and there exists a constant l > 0 such that the following conditions hold:
( (1.25)) 3 5
(h1 ) a∞ + (b∞ + c∞ )( (1.25)) 3 < 2
(h2 )
√ 3 2
.
|a(t)| + |c(t)| + |d(t)| < l, t ∈ [0, 1]. |b(t)|
Then the problem (4.2) has at least one solution. Proof. Corresponding to the problem (1.1), we have β =
1 4,α
=
5 4,q
=
5 2.
Let K = max{l3 , 1}. For |y| > K, by (h2 ), we have
|a(t)| + |c(t)| + |d(t)| < |b(t)|l ≤ |b(t)|K 3 < |b(t)||y| 3 . 1
1
(4.3)
By b(t) ∈ C[0, 1] and (h2 ), we get that b(t) > 0, t ∈ [0, 1] or b(t) < 0, t ∈ [0, 1]. It follows from (4.3) that if b(t) < 0, then yf (t, x, y, z) < 0 for |y| > K, and yf (t, x, y, z) > 0 for |y| > K, if b(t) > 0, i.e. the condition (H1 ) holds. 2
2
2
Obviously, |f (t, x, y, z)| ≤ |a(t)||x| 3 + |b(t)||y| 3 + |c(t)||z| 3 + |d(t)|, and
A1 =
2q−2 ϕ ϕq ( (β + 1)) q
22 a∞ + b ∞ + c ∞ = 3 ϕp ( (α)) ( (1.25)) 2 1
a∞ + b ∞ + c ∞ 2 ( (1.25)) 3
32 .
It follows from (h1 ) that A1 < 1, i.e. the condition (H2 ) holds. By Theorem 3.1, we obtain that the problem (4.2) has at least one solution. Acknowledgments The author is grateful to anonymous referees for their constructive comments and suggestions which led to improvement of the original manuscript. This work is supported by the Natural Science Foundation of China (11171088) and the Natural Science Foundation of Hebei Province (A2013208108).
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W. Jiang / Applied Mathematics and Computation 260 (2015) 48–56
References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28]
I. Podlubny, Fractional Differential Equations, Academic Press, New York, London, 1999. K.S. Miller, B. Ross, An Introduction to the Fractional Calculus and Fractional Differential Equations, Wiley, New York, 1993. A.A. Kilbas, H.M. Srivastava, J.J. Trujillo, Theory and Applications of Fractional Differential Equations, Elsevier B.V., 2006. R.P. Agarwal, J.P. dos Santos, C. Cuevas, Analytic resolvent operator and existence results for fractional integro-differential equations, J. Abstr. Differ. Equ. Appl. 02 (2) (2012) 26–47. R.P. Agarwal, B. de Andrade, C. Cuevas, Type of periodicity and ergodicity to a class of fractional order differential equations, Adv. Differ. Equ. 2010 (2010) 1–25. R.P. Agarwal, M. Belmekki, M. Benchohra, Survey on semi-linear differential equations and inclusions involving Riemann–Liouville fractional derivative, Adv. Differ. Equ. 2009 (2009) 1–47. R.P. Agarwal, M. Benchohra, S. Hamani, A survey on existence result for boundary value problems of nonlinear fractional differential equations and inclusions, Acta Appl. Math. 109 (2010) 973–1033, doi:10.1007/s10440-008-9356-6. Z. Bai, H. Lu, Positive solutions for boundary value problem of nonlinear fractional differential equation, J. Math. Anal. Appl. 311 (2005) 495–505, doi:10.1016/j.jmaa.2005.02.052. S. Marano, D. Motreanu, Infinitely many critical points of non-differentiable functions and applications to a Neumann type problem involving the p-Laplacian, J. Differ. Equ. 182 (2002) 108–120, doi:10.1006/jdeq.2001.4092. D. Pina, M. Elgueta, R. Manasevich, A homotopic deformation along p of a Leray–Schauder degree result and existence for (|u |p−2 u ) + f (t, u) = 0, u(0) = u(t) = 0, p > 1, J. Differ. Equ. 80 (1) (1997) 3227–3238. M. Garcia-Huidobro, R. Manasevich, F. Zanolin, A Fredholm-like result for strongly nonlinear second order odes, J. Differ. Equ. 114 (1994) 132–167. M. Garcia-Huidobro, P. Ubilla, Multiplicity of solutions for a class of nonlinear second order equations, Nonlinear Anal. 28 (9) (1997) 1509–1520. D.R. Anderson, R.I. Avery, J. Henderson, Existence of solutions for a one-dimensional p-Laplacian on time scales, J. Differ. Equ. Appl. 10 (2004) 889–896, doi:10.1080/10236190410001731416. C.S. Goodrich, The existence of a positive solution to a second-order delta-nabla p-Laplacian bvp on a time scale, Appl. Math. Lett. 25 (2012) 157–162, doi:10.1016/j.aml.2011.08.005. J. Mawhin, Topological degree methods in nonlinear boundary value problems, in: NSFCBMS Regional Conference Series in Mathematics, American Mathematical Society, Providence, RI, 1979. W. Feng, J.R.L. Webb, Solvability of m-point boundary value problems with nonlinear growth, J. Math. Anal. Appl. 212 (1997) 467–480. Y. Liu, W. Ge, Solvability of nonlocal boundary value problems for ordinary differential equations of higher order, Nonlinear Anal. 57 (2004) 435–458. Z. Du, X. Lin, W. Ge, Some higher-order multi-point boundary value problem at resonance, J. Comput. Appl. Math. 177 (2005) 55–65. B. Liu, Solvability of multi-point boundary value problem at resonance (II), Appl. Math. Comput. 136 (2003) 353–377. X. Zhang, M. Feng, W. Ge, Existence result of second-order differential equations with integral boundary conditions at resonance, J. Math. Anal. Appl. 353 (2009) 311–319. N. Kosmatov, Multi-point boundary value problems on an unbounded domain at resonance, Nonlinear Anal. 68 (2008) 2158–2171. N. Kosmatov, A boundary value problem of fractional order at resonance, Electron. J. Differ. Equ. 2010 (2010) 135 1–10. W. Jiang, The existence of solutions to boundary value problems of fractional differential equations at resonance, Nonlinear Anal. (TMA). 74 (2011) 1987–1994. W. Jiang, Solvability for a coupled system of fractional differential equations at resonance, Nonlinear Anal. (RWA). 13 (2012) 2285–2292. W. Jiang, Solvability of (k,n-k) conjugate boundary-value problems at resonance, Electron. J. Differ. Equ. 2012 (2012) 114 1–10. W. Ge, J. Ren, An extension of Mawhin’s continuation theorem and its application to boundary value problems with a p-Laplacian, Nonlinear Anal. (TMA). 58 (2004) 477–488. W. Jiang, Solvability for p-Laplacian boundary value problem at resonance on the half-line, Bound. Value Probl. 2013 (2013) 207. J. Kuang, Applied Inequalities, Shandong Science and Technology Press, Jinan, 2014, (P132).