Applied Mathematics Letters 97 (2019) 34–40
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Applied Mathematics Letters www.elsevier.com/locate/aml
Necessary conditions for the existence of positive solutions to fractional boundary value problems at resonance Yongqing Wang School of Mathematical Sciences, Qufu Normal University, Qufu 273165, Shandong, PR China
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Article history: Received 3 April 2019 Accepted 2 May 2019 Available online 13 May 2019 Keywords: Fractional differential equation Positive solution Resonance Necessary condition
abstract In this article, we obtain some properties of the positive solutions to a class of fractional differential equation multipoint boundary value problems at resonance. Necessary conditions for the existence of positive solutions are established by means of the spectral theory of linear operator. As application, necessary and sufficient conditions for the existence of positive solutions are given for the case that the nonlinearity has a fixed sign. Some examples are presented to illustrate our results. © 2019 Elsevier Ltd. All rights reserved.
1. Introduction In this paper, we consider the following fractional differential equation m-point boundary value problem (BVP for short): ⎧ α α−1 ⎪ D0+ u(t) + f (t, u(t), D0+ u(t)) = 0, 0 < t < 1, ⎨ m−2 ∑ (1.1) ⎪ u(0) = 0, u(1) = ηi u(ξi ), ⎩ i=1
α is where D0+ ∑m−2 α−1 η ξ i i i=1
the standard Riemann–Liouville derivative, 1 < α ≤ 2, ηi > 0, 0 < ξ1 < · · · < ξm−2 < 1, = 1, f : [0, 1] × [0, +∞) × R → R is continuous. BVP (1.1) happens to be at resonance since α λ = 0 is an eigenvalue of the linear problem −D0+ u = λu with the given boundary conditions. Due to the demonstrated applications in widespread fields of engineering and science, the fractional differential equation boundary value problems (FBVPs for short) have attracted a great deal of attention during the past decades (see [1–8]). While there are many papers considering the existence of solutions (or positive solutions) for fractional order non-resonant boundary value problems, the results dealing with resonant boundary value problems are relatively scarce. Problems at resonance are often discussed using the coincidence degree theory and Leggett–Williams norm-type theorem (see [9–13]). In [14], we considered BVP (1.1) for the case that f does not include the derivative term and m = 3. By using the technique of [15], we E-mail address:
[email protected].
https://doi.org/10.1016/j.aml.2019.05.007 0893-9659/© 2019 Elsevier Ltd. All rights reserved.
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rewrite the original resonant BVP as an equivalent non-resonant perturbed problems. Combining with the properties of the Green function, sufficient conditions for the existence of positive solutions are established by means of the fixed point index theory and iterative technique. For the case that α is an integer, Webb [16] used the concave (or convex) properties of the positive solution to establish necessary condition of the existence of positive solutions for second-order resonant BVP. It should be pointed out that the method of [16] does not apply to the fractional differential equations since α D0+ u ≥ 0 does not mean concavity or convexity. To the best of our knowledge, necessary conditions for the existence of positive solutions to resonant BVP (1.1) have not been considered before. The main purpose of this paper is to fill the gap. In this paper, by a positive solution to BVP (1.1), we mean a nonnegative function u ∈ C[0, 1] which does not vanish on [0, 1] α satisfies (1.1), and D0+ u ∈ L[0, 1]. 2. Preliminaries and several lemmas Lemma 2.1. Assume that b > 0, and y ∈ L[0, 1]. Then the unique solution of the problem ⎧ α − D0+ u(t) + bu(t) = y(t), 0 < t < 1, ⎪ ⎨ m−2 ∑ ⎪ u(0) = 0, u(1) = ηi u(ξi ), ⎩ i=1
can be expressed by ∫ u(t) =
1
K(t, s)y(s)ds, 0
where K(t, s) = K1 (t, s) + tα−1 Eα,α (btα )q(s), ∑m−2 i=1 ηi K1 (ξi , s) q(s) = , ∑ m−2 Eα,α (b) − i=1 ηi ξiα−1 Eα,α (bξiα ) ⎧ (t − ts)α−1 Eα,α (btα )Eα,α (b(1 − s)α ) ⎪ ⎪ , 0 ≤ t ≤ s ≤ 1, ⎪ ⎪ Eα,α (b) ⎪ ⎪ ⎪ ⎨ (t − ts)α−1 E (btα )E (b(1 − s)α ) α,α α,α K1 (t, s) = ⎪ E (b) α,α ⎪ ⎪ ⎪ ⎪ (t − s)α−1 Eα,α (b(t − s)α )Eα,α (b) ⎪ ⎪ ⎩− , 0 ≤ s ≤ t ≤ 1, Eα,α (b) in which Eα,α (x) =
+∞ ∑ k=0
xk . Γ ((k + 1)α)
∑m−2
ηi ξiα−1 = 1, we have [ ] ∑m−2 (k+1)α−1 k m−2 +∞ b 1 − i=1 ηi ξi ∑ ∑ Eα,α (b) − ηi ξiα−1 Eα,α (bξiα ) = Γ ((k + 1)α) i=1 k=0 ∑ +∞ m−2 α−1 ∑ bk (1 − ξikα ) i=1 ηi ξi = > 0. Γ ((k + 1)α)
Proof . By ηi > 0, 0 < ξi < 1 and
i=1
k=0
The rest of the proof is similar to Lemma 2.1 in [14], we omit it here. □
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From [14], we know that the function +∞
g(t) =:
∑ α−2 tk + Γ (α − 1) Γ ((k + 1)α − 2)
(2.1)
k=1
has a unique positive root b∗ > 0. Lemma 2.2. Assume that b ∈ (0, b∗ ]. Then K(t, s) satisfies: (1) K(t, s) > 0, ∀ t, s ∈ (0, 1); (2) h2 (s)tα−1 ≤ K(t, s) ≤ h1 (s)tα−1 , ∀ t, s ∈ [0, 1], where h1 (s) = (1 − s)α−1 Eα,α (b(1 − s)α ) + Eα,α (b)q(s), q(s) h2 (s) = . Γ (α)
Proof . The proof is similar to Lemma 2.2 in [14], so we omit it.
□
Let E = C[0, 1] be the Banach space with the maximum norm ∥u∥ = max0≤t≤1 |u(t)|, θ is the zero element. Define a cone P by P = {u ∈ E : u(t) ≥ 0, t ∈ [0, 1]} . Denote
∫ T u(t) =
1
K(t, s)u(s)ds. 0
Then T : P → P is a continuous linear operator. Similar to Lemma 2.3 in [14], we have the following lemma. Lemma 2.3. Assume that b ∈ (0, b∗ ], then the first eigenvalue of T is λ1 = b, and φ(t) = tα−1 is the positive eigenfunction corresponding to λ1 , that is φ = bT φ. Lemma 2.4. Assume that f has a fixed sign, and v is a positive solution of (1.1). Then there exist l1 > l2 > 0, such that l1 φ ≥ v ≥ l2 φ. α−1 Proof . Case (i): f ≥ 0. Select b ∈ (0, b∗ ]. It is clear that f (t, v(t), D0+ v(t)) ∈ L[0, 1] and (1.1) is equivalent to the following BVP: ⎧ α−1 α − D0+ u(t) + bu(t) = f (t, u(t), D0+ u(t)) + bu(t), 0 < t < 1, ⎪ ⎨ m−2 ∑ ⎪ ηi u(ξi ). ⎩ u(0) = 0, u(1) = i=1
Let
1
∫
[ ] α−1 K(t, s) f (s, u(s), D0+ u(s)) + bu(s) ds.
Au(t) =
(2.2)
0
It follows from Lemma 2.1 that v is a fixed point of A. From Lemma 2.2, we can get ∫ 1 v = Av ≥ φ h2 (s)bv(s)ds,
(2.3)
0
and
∫ v≤φ 0
1 α−1 h1 (s)[f (s, v(s), D0+ v(s)) + bv(s)]ds.
(2.4)
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Case (ii): f ≤ 0. From α−1 α D0+ v(t) = −f (t, v(t), D0+ v(t))
and u(0) = 0, we obtain v(t) = −
1 Γ (α)
t
∫
α−1 (t − s)α−1 f (s, v(s), D0+ v(s))ds + c1 tα−1 .
(2.5)
0
Noticing ∫ 0≥
t α−1 (t − s)α−1 f (s, v(s), D0+ v(s))ds ≥
∫
0
t α−1 (t − ts)α−1 f (s, v(s), D0+ v(s))ds,
0
we have v(t) = c1 , t→0+ tα−1 lim
which implies c1 ≥ 0. Next, we will show that c1 > 0. If c1 = 0. Then 1 Γ (α)
v(t) = − By v(1) =
∑m−2 i=1
t
∫
α−1 (t − s)α−1 f (s, v(s), D0+ v(s))ds.
(2.6)
0
ηi v(ξi ), we have ∫
=
1 α−1 (1 − s)α−1 f (s, v(s), D0+ v(s))ds
0 m−2 ∑
ηi ξiα−1
ξi
∫ 0
i=1
)α−1 ( s α−1 f (s, v(s), D0+ v(s))ds, 1− ξi
that is m−2 ∑
ηi ξiα−1
[∫
1 α−1 (1 − s)α−1 f (s, v(s), D0+ v(s))ds
0
i=1
ξi
∫ − 0
)α−1 ] ( s α−1 f (s, v(s), D0+ v(s))ds = 0. 1− ξi
(2.7)
Noticing that ∫
1 α−1 (1 − s)α−1 f (s, v(s), D0+ v(s))ds ≤ 0,
ξi
and ∫
ξi
[ α−1
(1 − s) 0
(
s − 1− ξi
)α−1 ]
α−1 f (s, v(s), D0+ v(s))ds ≤ 0.
(2.7) yields ∫
1 α−1 (1 − s)α−1 f (s, v(s), D0+ v(s))ds = 0,
ξi
and ∫ 0
ξi
[
( )α−1 ] s α−1 (1 − s)α−1 − 1 − f (s, v(s), D0+ v(s))ds = 0, ξi
α−1 which implies f (t, v(t), D0+ v(t)) ≡ 0, for t ∈ [0, 1]. Then it follows from (2.6) that v = θ, which contradicts with v is a positive solution. Therefore, we have proved c1 > 0.
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By (2.5), we have v(t) ≥ c1 tα−1 .
(2.8)
∫ t 1 α−1 (t − s)α−1 f (s, v(s), D0+ v(s))ds + c1 tα−1 v(t) = − Γ (α) 0 [ ] ∫ t 1 α−1 α−1 ≤ − (1 − s) f (s, v(s), D0+ v(s))ds + c1 tα−1 Γ (α) 0 [ ] ∫ 1 1 α−1 α−1 (1 − s) f (s, v(s), D0+ v(s))ds + c1 tα−1 . ≤ − Γ (α) 0
(2.9)
On the other hand, we have
It follows from (2.3), (2.4), (2.8) and (2.9) that the proof is complete. □ 3. Main result Theorem 3.1. Assume that ∀c > 0, f (t, ctα−1 , cΓ (α)) ̸≡ 0 on[0, 1]. Then the necessary condition for the existence of positive solutions to BVP (1.1) is f (t, x, y) cannot has a fixed sign. α−1 Proof . Suppose that f has a fixed sign and v is a positive solution of (1.1). Obviously, f (t, v(t), D0+ v(t)) ∈ L[0, 1]. Then v is a fixed point of A, where A is as defined in (2.2). It follows from Lemma 2.4 that there exist l1 > l2 > 0, such that
l1 φ ≥ v ≥ l2 φ. α−1 Firstly, we can see f (t, v(t), D0+ v(t)) ̸≡ 0 for t ∈ [0, 1]. If otherwise, v satisfies
⎧ α D v(t) = 0, ⎪ ⎨ 0+ ⎪ ⎩ v(0) = 0, v(1) =
m−2 ∑
ηi v(ξi ).
i=1
The solutions of the above equation can be expressed by v(t) = ctα−1 , which contradicts with the assumption. Case (i): f ≥ 0. By Lemma 2.2, we obtain ∫
1
α−1 K(t, s)f (s, v(s), D0+ v(s))ds + bT v ∫ 1 α−1 ≥ bT φ h2 (s)f (s, v(s), D0+ v(s))ds + bT v 0 ∫ v 1 α−1 h2 (s)f (s, v(s), D0+ v(s))ds + bT v ≥ bT l1 0 [ ∫1 ] α−1 b 0 h2 (s)f (s, v(s), D0+ v(s))ds = + b T v. l1
v = Av =
0
By induction, we can get [ ∫1 ]n α−1 b 0 h2 (s)f (s, v(s), D0+ v(s))ds v≥ + b T n v, l1
n = 1, 2, 3, . . .
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Then [ ∫1 ]n α−1 b 0 h2 (s)f (s, v(s), D0+ v(s))ds v≥ + b T n l2 φ l1 [∫ 1 ]n α−1 h (s)f (s, v(s), D0+ v(s))ds 0 2 = + 1 l2 φ. n = 1, 2, 3, . . . l1 Noticing
∫1 0
α−1 h2 (s)f (s, v(s), D0+ v(s))ds > 0, we have
[∫ 1
]n α−1 h2 (s)f (s, v(s), D0+ v(s))ds + 1 → +∞, (n → ∞), l1 which contradicts with v is a fixed positive solution. Case (ii): f ≤ 0. By Lemma 2.2, we have ∫ 1 α−1 v = Av = K(t, s)f (s, v(s), D0+ v(s))ds + bT v 0 ∫ 1 α−1 ≤ bT φ h2 (s)f (s, v(s), D0+ v(s))ds + bT v 0 ∫ 1 v α−1 ≤ bT h2 (s)f (s, v(s), D0+ v(s))ds + bT v l2 0 [ ∫1 ] α−1 b 0 h2 (s)f (s, v(s), D0+ v(s))ds = + b T v. l2 0
It is clear that 0<
b
∫1 0
α−1 h2 (s)f (s, v(s), D0+ v(s))ds + b < b. l2
By induction, we can get [ ∫1 ]n α−1 b 0 h2 (s)f (s, v(s), D0+ v(s))ds v≤ + b T n v, l2
(3.1)
n = 1, 2, 3, . . .
Then
]n [ ∫1 α−1 b 0 h2 (s)f (s, v(s), D0+ v(s))ds + b T n l1 φ l2 [∫ 1 ]n α−1 h (s)f (s, v(s), D0+ v(s))ds 0 2 = + 1 l1 φ. n = 1, 2, 3, . . . l1 Noticing (3.1), we have v = θ, which contradicts with v is a positive solution. □ v≤
Corollary 3.1. Assume that f has a fixed sign. Then BVP (1.1) has a positive solution if and only if there is a positive constant c such that f (t, ctα−1 , cΓ (α)) ≡ 0 for t ∈ [0, 1], and the solution is ctα−1 . 4. Examples Example 4.1.
Consider the following resonant BVP ⎧ 3 α−1 2 ⎪ u(t) + f (t, u(t), D0+ u(t)) = 0, 0 < t < 1, ⎨ D0+ ( ) √ ( ) 1 1 2 1 ⎪ ⎩ u(0) = 0, u(1) = u + u , 2 4 2 2
with f (t, x, y) = x2 y 2 −
√
tπxy +
π . 4
(4.1)
40
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1
√
It is easy to see that f : [0, 1] × [0, +∞) × R → [0, +∞) is continuous, and f (t, ct 2 , c 2 π ) ≡ 0 if and only if c = 1. Thus Corollary 3.1 ensures that tα−1 is the unique positive solution of the BVP (4.1). Example 4.2. Consider BVP (4.1) with f (t, x, y) = x2 y 2 −
√
tπxy + π.
It is easy to see that f : [0, 1] × [0, +∞) × R → [0, +∞) is continuous, and f ≥ ensures that the BVP (4.1) has no positive solution.
3π 4 .
Thus Theorem 3.1
Acknowledgments This work was supported by the Natural Science Foundation of Shandong Province of China (ZR2017MA036, ZR2014MA034), the National Natural Science Foundation of China (11571296, 11871302). References [1] A. Cabada, G. Wang, Positive solutions of nonlinear fractional differential equations with integral boundary value conditions, J. Math. Anal. Appl. 389 (2012) 403–411. [2] A. Cabada, Z. Hamdi, Nonlinear fractional differential equations with integral boundary value conditions, Appl. Math. Comput. 228 (2014) 251–257. [3] Y. Wang, L. Liu, Y. Wu, Positive solutions for a nonlocal fractional differential equation, Nonlinear Anal. 74 (2011) 3599–3605. [4] Y. Cui, Uniqueness of solution for boundary value problems for fractional differential equations, Appl. Math. Lett. 51 (2016) 48–54. [5] Y. Cui, W. Ma, Q. Sun, X. Su, New uniqueness results for boundary value problem of fractional differential equation, Nonlinear Anal. Model. Control. 23 (2018) 31–39. [6] B. Ahmada, R. Luca, Existence of solutions for sequential fractional integro-differential equations and inclusions with nonlocal boundary conditions, Appl. Math. Comput. 339 (2018) 516–534. [7] X. Zhang, L. Liu, Y. Wu, The uniqueness of positive solution for a singular fractional differential system involving derivatives, Commun. Nonlinear Sci. Numer. Simul. 18 (2013) 1400–1409. [8] X. Zhang, Q. Zhong, Uniqueness of solution for higher-order fractional differential equations with conjugate type integral conditions, Fract. Calc. Appl. Anal. 20 (2017) 1471–1484. [9] N. Kosmatov, A boundary value problem of fractional order at resonance, Electron. J. Differential Equations 135 (2010) 1–10. [10] N. Kosmatov, W. Jiang, Resonant functional problems of fractional order, Chaos Solitons Fractals 91 (2016) 573–579. [11] W. Jiang, The existence of solutions to boundary value problems of fractional differential equations at resonance, Nonlinear Anal. 74 (2011) 1987–1994. [12] G. Infantea, M. Zima, Positive solutions of multi-point boundary value problems at resonance, Nonlinear Anal. 69 (2008) 2458–2465. [13] L. Yang, C. Shen, On the existence of positive solution for a kind of multi-point boundary value problem at resonance, Nonlinear Anal. 72 (2010) 4211–4220. [14] Y. Wang, L. Liu, Positive solutions for a class of fractional 3-point boundary value problems at resonance, Adv. Differential Equations 2017 (2017) 13, http://dx.doi.org/10.1186/s13662-016-1062-5. [15] J. Webb, M. Zima, Multiple positive solutions of resonant and non-resonant nonlocal boundary value problems, Nonlinear Anal. 71 (2009) 1369–1378. [16] J. Webb, Remarks on nonlocal boundary value problems at resonance, Appl. Math. Comput. 216 (2010) 497–500.