The method of lower and upper solutions for mixed fractional four-point boundary value problem with p -Laplacian operator

Applied Mathematics Letters 65 (2017) 56–62

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Applied Mathematics Letters www.elsevier.com/locate/aml

The method of lower and upper solutions for mixed fractional four-point boundary value problem with p-Laplacian operator✩ Xiping Liua, *, Mei Jiaa , Weigao Ge b a b

College of Science, University of Shanghai for Science and Technology, Shanghai 200093, China Department of Mathematics, Beijing Institute of Technology, Beijing 100081, China

article

info

Article history: Received 26 August 2016 Accepted 3 October 2016 Available online 14 October 2016 Keywords: Fractional differential equation Mixed fractional derivatives Four-point boundary value problem Positive solution p-Laplacian operator Lower and upper solutions

abstract Based on the monotone iterative technique, a new method of lower and upper solutions which is used to study the multi-point boundary value problem of nonlinear fractional differential equations with mixed fractional derivatives and p-Laplacian operator is proposed. Some new results on the existence of positive solutions for the four-point boundary value problem are established. Finally, an example is presented to illustrate the wide range of potential applications of our main results. © 2016 Elsevier Ltd. All rights reserved.

1. Introduction In the past decades, the fractional differential equation theory has gained considerable popularity and importance due to its demonstrated applications in numerous widespread fields of science and engineering, see [1–3]. Driven by the wide range of the applications, the boundary value problems for fractional order differential equations have been studied by more and more researchers, see [4–14]. On the other hand, for studying the turbulent flow in a porous medium, the model of differential equation with the p-Laplacian operator is introduced, see [15]. Many important results related to the boundary value problem of fractional differential equations with p-Laplacian operator have also been obtained, see [16–22] and references therein. In this paper, we consider the following four-point boundary value problem of fractional differential equation with mixed fractional derivatives and p-Laplacian operator ( ( ( ) ⎧ )) c β α c β ⎪ ⎨D0+ φp D0+ u(t) = f t, u(t), D0+ u(t) , t ∈ (0, 1), c β (1.1) D u(0) = u′ (0) = 0, ⎪ ⎩ 0+ c β c β u(1) = r1 u(η), D0+ u(1) = r2 D0+ u(ξ), ✩

This work is supported by the National Natural Science Foundation of China (No. 11171220).

* Corresponding author. E-mail address: [email protected] (X. Liu). http://dx.doi.org/10.1016/j.aml.2016.10.001 0893-9659/© 2016 Elsevier Ltd. All rights reserved.

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where 1 < α, β ≤ 2, r1 , r2 ≥ 0, D0α+ is the Riemann–Liouville fractional derivative operator, and D0β+ is the Caputo fractional derivative operator, p > 1, φp is the p-Laplacian operator and f ∈ C([0, 1]×[0, +∞)×(−∞, 0], [0, +∞)). The purpose of this paper is to establish a method of lower and upper solutions which is used to study the existence of positive solutions of boundary value problem (1.1). c

2. Preliminary results We say a function u = u(t) is a positive solution of boundary value problem (1.1) if it satisfies boundary value problem (1.1) and u(t) ≥ 0, for t ∈ [0, 1]. Lemma 2.1. For any given function h ∈ C[0, 1] and real numbers a, b ∈ R, the following boundary value problem ( ( ⎧ )) α c β ⎪ D φ D u(t) = h(t), t ∈ (0, 1), + p ⎨ 0 0+ c β ′ (2.1) D u(0) = u (0) = 0, ⎪ ⎩ 0+ c β u(1) = a, D0+ u(1) = b, has a unique solution, which is given by ( ∫ 1 ∫ α−1 u(t) = a − G(t, s)φq φp (b)s − 0

1

) H(s, τ )h(τ )dτ

ds,

(2.2)

0

and ( D0β+ u(t) = φq φp (b)tα−1 −

c

∫

1

) H(t, s)h(s)ds ,

(2.3)

0

where 1 G(t, s) = Γ (β)

{

(1 − s)β−1 − (t − s)β−1 , (1 − s)β−1 ,

0 ≤ s ≤ t ≤ 1, 0 ≤ t < s ≤ 1,

(2.4)

and H(s, τ ) =

1 Γ (α)

{

sα−1 (1 − τ )α−1 − (s − τ )α−1 , sα−1 (1 − τ )α−1 ,

0 ≤ τ ≤ s ≤ 1, 0 ≤ s < τ ≤ 1.

(2.5)

) ( Proof . Let φp cD0β+ u(t) = v(t), we can easily show that boundary value problem (2.1) can be decomposed into the following coupled boundary value problems: { α D0+ v(t) = h(t), t ∈ (0, 1), (2.6) v(0) = 0, v(1) = φp (b), and {c β D0+ u(t) = φq (v(t)), t ∈ (0, 1), u′ (0) = 0, u(1) = a.

(2.7)

By the standard way, we can get that boundary value problem (2.6) has a unique solution, which is given by t

∫ 1 ) (t − s)α−1 h(s)ds − tα−1 (1 − s)α−1 h(s)ds + φp (b)tα−1 0 0 ∫ 1 ( ) α−1 = φp (b)t − H(t, s)h(s)ds = φp cD0β+ u(t) .

v(t) =

1 ( Γ (α)

∫

0

(2.8)

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And boundary value problem (2.7) has a unique solution, which is given by ∫ t ∫ 1 1 1 u(t) = a + (t − s)β−1 φq (v(s))ds − (1 − s)β−1 φq (v(s))ds Γ (β) 0 Γ (β) 0 ∫ 1 G(t, s)φq (v(s))ds. = a−

(2.9)

0

Therefore, boundary value problem (2.1) has a unique solution which is given by (2.2) and we can easily get that cD0β+ u(t) is given by (2.3). ■ From Lemma 2.1, it is obvious that boundary value problem (1.1) is equivalent to the following integral equation ( ) ∫ 1 ∫ 1 ( ) c β α−1 c β u(t) = r1 u(η) − G(t, s)φq φp (r2 D0+ u(ξ))s H(s, τ )f τ, u(τ ), D0+ u(τ ) dτ ds. (2.10) − 0

0

From (2.4) and (2.5), we can easily prove that G(t, s) and H(t, s) satisfy the following lemma. Lemma 2.2. Suppose the functions G(t, s) and H(t, s) are defined by (2.4) and (2.5), then G(t, s) and H(t, s) are continuous and G(t, s) ≥ 0, H(t, s) ≥ 0, for (t, s) ∈ [0, 1]×[0, 1].

3. The method of lower and upper solutions In this section, we present the method of lower and upper solutions and existence theorems of positive solutions for boundary value problem (1.1) based on the monotone iterative technique. Definition 3.1. Let x ∈ AC 2 [0, 1], we say that x is a lower solution of boundary value problem (1.1), if ( ( ( ) ⎧ )) α c β c β ⎪ D φ D x(t) ≤ f t, x(t), D x(t) , t ∈ (0, 1), ⎪ + p 0+ 0+ ⎨ 0 D0β+ x(0) = x′ (0) = 0,

c

⎪ ⎪ ⎩

x(1) ≤ r1 x(η),

c β D0+ x(1)

(3.1)

≥

r2 cD0β+ x(ξ).

Let y ∈ AC 2 [0, 1], we say that y is an upper solution of boundary value problem (1.1) , if ( ( ( ) ⎧ )) c β α c β ⎪ y(t) ≥ f t, y(t), D y(t) , t ∈ (0, 1), D φ D ⎪ + p + + 0 0 0 ⎨ c β ′ D0+ y(0) = y (0) = 0, ⎪ ⎪ ⎩ y(1) ≥ r1 y(η), cD0β+ y(1) ≤ r2 cD0β+ y(ξ).

(3.2)

Denote that E = {u : u ∈ C[0, 1], cD0β+ u ∈ C[0, 1], u′ (0) = 0} and endowed with the norm ∥u∥β = ∥u∥∞ + ∥cD0β+ u∥∞ , where ∥u∥∞ = max0≤t≤1 |u(t)| and ∥cD0β+ u∥∞ = max0≤t≤1 |cDβ u(t)| . Then (E, ∥ · ∥β ) is a Banach space. We denote that P = {u : u ∈ E, u(t) ≥ 0, cD0β+ u(t) ≤ 0, t ∈ [0, 1]}. Then P is a normal cone on E. We denote x ⪯ y if and only if y − x ∈ P , for x, y ∈ E. In this section, we assume the following condition holds: (H1) f ∈ C([0, 1]×[0, +∞)×(−∞, 0], [0, +∞)), f (t, w1 , z1 ) ≤ f (t, w2 , z2 ), for 0 ≤ w1 < w2 , z1 > z2 ≥ 0 and any t ∈ [0, 1]. Theorem 3.1. Assume that (H1) holds, boundary value problem (1.1) has a nonnegative lower solution x0 ∈ P and an upper solution y0 ∈ P such that x0 ⪯ y0 . Then boundary value problem (1.1) has the

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maximal lower solution x∗ and the minimal upper solution y ∗ on [x0 , y0 ] ⊂ P , both x∗ and y ∗ are positive solutions of boundary value problems (1.1). Furthermore, 0 ≤ x0 (t) ≤ x∗ (t) ≤ y ∗ (t) ≤ y0 (t) and D0β+ y0 (t) ≤ cD0β+ y ∗ (t) ≤ cD0β+ x∗ (t) ≤ cD0β+ x0 (t) ≤ 0.

c

Proof . The proof is divided into the following three steps. Step 1. We will obtain the lower solution sequence {xk } and the upper solution sequence {yk }. According to Lemma 2.1, for the given x0 ∈ P , the following boundary value problem ( ( ( ) ⎧ )) α c β c β ⎪ ⎨D0+ φp D0+ x1 (t) = f t, x0 (t), D0+ x0 (t) , t ∈ (0, 1), c β ′ ⎪ D0+ x1 (0) = x1 (0) =β 0, ⎩ x1 (1) = r1 x0 (η), cD0+ x1 (1) = r2 cD0β+ x0 (ξ) has a unique solution x1 . Since x0 is a lower solution of boundary value problems (1.1), then ( ( ( ) ⎧ )) α c β c β ⎪ ⎨D0+ φp D0+ x0 (t) ≤ f t, x0 (t), D0+ x0 (t) , c β D + x0 (0) = x′0 (0) = 0, ⎪ ⎩ 0 x0 (1) ≤ r1 x0 (η), cD0β+ x0 (1) ≥ r2 cD0β+ x0 (ξ).

(3.3)

t ∈ (0, 1), (3.4)

(3.3) minus (3.4), and we can get that ( ( ⎧ ) (c β )) α c β ⎪ D φ D x (t) − φ D x (t) ≥ 0, t ∈ (0, 1), + p 1 p 0 ⎨ 0 0+ 0+ c β c β ′ ′ (3.5) D x1 (0) − D0+ x0 (0) = x1 (0) − x0 (0) = 0, ⎪ ⎩ 0+ c β c β x1 (1) − x0 (1) ≥ 0, D0+ x1 (1) − D0+ x0 (1) ≤ 0. ) ) ) ( ( ( Let φp cD0β+ x1 (t) − φp cD0β+ x0 (t) := v(t). Since cD0β+ x1 (0) − cD0β+ x0 (0) = 0, then φp cD0β+ x1 (0) − ) ) ( ( φp cD0β+ x0 (0) = φp cD0β+ x1 (t) − cD0β+ x0 (t) |t=0 = 0 which is v(0) = 0. And since cD0β+ x1 (1) − ) ) ( ( c β D0+ x0 (1) ≤ 0, we get that v(1) = φp cD0β+ x1 (1) − φp cD0β+ x0 (1) ≤ 0. ( ( ) ( )) Let h(t) = D0α+ φp cD0β+ x1 (t) − φp cD0β+ x0 (t) and φp (b) = v(1), then we obtain the following boundary value problem { α D0+ v(t) = h(t) ≥ 0, t ∈ (0, 1), (3.6) v(0) = 0, v(1) = φp (b) ≤ 0. ( ) ( ) ∫1 By (2.8) and Lemma 2.2, v(t) = φp cD0β+ x1 (t) − φp cD0β+ x0 (t) = φp (b)tα−1 − 0 H(t, s)h(s)ds ≤ 0, for t ∈ [0, 1]. From the monotonicity of p-Laplacian operator φp , we have D0β+ x1 (t) − cD0β+ x0 (t) = cD0β+ (x1 (t) − x0 (t)) := δ(t) ≤ 0.

c

Then we obtain the following boundary value problem {c β D0+ (x1 (t) − x0 (t)) := δ(t) ≤ 0, t ∈ (0, 1), x′1 (0) − x′0 (0) = 0, x1 (1) − x0 (1) := a ≥ 0. Similar to (2.9), and by Lemma 2.2, x1 (t) − x0 (t) = a −

∫1 0

(3.7)

G(t, s)δ(s)ds ≥ 0. So, we can get that x0 ⪯ x1 .

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From the condition (H1), we get ( ( ( ) ( ) ⎧ )) α c β c β c β ⎪ ⎨D0+ φp D0+ x1 (t) = f t, x0 (t), D0+ x0 (t) ≤ f t, x1 (t), D0+ x1 (t) , c β D + x1 (0) = x′1 (0) = 0, ⎪ ⎩ 0 x1 (1) = r1 x0 (η) ≤ r1 x1 (η), cD0β+ x1 (1) = r2 cD0β+ x0 (ξ) ≥ r2 cD0β+ x1 (ξ). Then x = x1 (t) is a lower solution of boundary value problem (1.1). Starting from the initial function x0 , by the following iterative scheme ( ( ) ( ⎧ )) α c β c β ⎪ D φ x (t) , t ∈ (0, 1), D x (t) = f t, x (t), D + p k−1 ⎨ 0 0+ k 0+ k−1 ′ c β D + xk (0) = xk (0) = 0, ⎪ ⎩ 0 xk (1) = r1 xk−1 (η), cD0β+ xk (1) = r2 cD0β+ xk−1 (ξ),

t ∈ (0, 1),

k = 1, 2, . . . ,

(3.8)

(3.9)

we can obtain the sequence {xk }, where x = xk (t) are lower solutions of boundary value problem (1.1), and xk−1 ⪯ xk , so that {xk } is monotonically increasing. Starting from the initial function y0 , by the following iterative scheme ( ( ( ) ⎧ )) α c β c β ⎪ ⎨D0+ φp D0+ yk (t) = f t, yk−1 (t), D0+ yk−1 (t) , t ∈ (0, 1), c β k = 1, 2, . . . , (3.10) D yk (0) = yk′ (0) = 0, ⎪ ⎩ 0+ c β c β yk (1) = r1 yk−1 (η), D0+ yk (1) = r2 D0+ yk−1 (ξ), we can get a sequence {yk }, where y = yk (t) are upper solutions of boundary value problem (1.1), and {yk } is monotonically decreasing. Step 2. We prove that xk ⪯ yk , if xk−1 ⪯ yk−1 , k = 1, 2, . . .. Since xk−1 ⪯ yk−1 , then xk−1 (t) ≤ yk−1 (t) and cD0β+ xk−1 (t) ≥ cD0β+ yk−1 (t), and from (H1), ( ) ( ) f t, xk−1 (t), cD0β+ xk−1 (t) ≤ f t, yk−1 (t), cD0β+ yk−1 (t) . By (3.9) and (3.10), we can get ( ( ⎧ ) (c β )) α c β ⎪ D φ D y (t) − φ D x (t) = f (t, yk−1 (t), cD0β+ yk−1 (t)) − f (t, xk−1 (t), cD0β+ xk−1 (t)) ≥ 0, p p ⎨ 0+ 0+ k 0+ k c β D yk (0) − cD0β+ xk (0) = yk′ (0) − x′k (0) = 0, ⎪ ⎩ 0+ yk (1) − xk (1) ≥ 0, cD0β+ yk (1) − cD0β+ xk (1) ≤ 0. Similarly, we can show that xk ⪯ yk , in the same way as the above. Therefore, x0 ⪯ x1 ⪯ · · · ⪯ xk ⪯ · · · ⪯ · · · ⪯ yk ⪯ · · · ⪯ y1 ⪯ y0 . Since P is a normal cone on E, the {xk } is uniformly bounded. Because H, G, φp , φq and f are continuous, we can easily get that {xk } is equicontinuous. Hence, the {xk } is relatively compact. Then there exist x∗ and y ∗ such that lim xk = x∗ ,

k→∞

lim cD0β+ xk = cD0β+ x∗ ,

(3.11)

lim cD0β+ yk = cD0β+ y ∗ ,

(3.12)

k→∞

and lim yk = y ∗ ,

k→∞

k→∞

which imply that x∗ is the maximum lower solution, y ∗ is the minimal upper solution of boundary value problem (1.1) in [x0 , y0 ] ⊂ P , and x∗ ⪯ y ∗ .

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Step 3. We prove that both x∗ and y ∗ are the solutions of boundary value problem (1.1). According to Lemma 2.1 and (3.9), we can get that ∫ 1 ∫ 1 ( ( ) ) ( ) α−1 c β G(t, s)φq φp r2 D0+ xk−1 (ξ) s H(s, τ )f τ, xk−1 (τ ), cD0β+ xk−1 (τ ) dτ ds. xk (t) = r1 xk−1 (η) − − 0

0

From (3.11), and by the continuity of φp , f , G, H and Lebesgue dominated convergence theorem, we have ) ( ∫ 1 ∫ 1 ) ( ( ) G(t, s)φq φp r2 cD0β+ x∗ (ξ) sα−1 − H(s, τ )f τ, x∗ (τ ), cD0β+ x∗ (τ ) dτ ds, (3.13) x∗ (t) = r1 x∗ (η) − 0

0

∗

which implies that x is a solution of boundary value problem (1.1). In the same way, we can show that y ∗ is a solution of boundary value problem (1.1), too. Furthermore, 0 ≤ x0 (t) ≤ x∗ (t) ≤ y ∗ (t) ≤ y0 (t), cD0β+ y0 (t) ≤ cD0β+ y ∗ (t) ≤ cD0β+ x∗ (t) ≤ cD0β+ x0 (t) ≤ 0. The proof is completed. ■ 4. Illustration We consider the following four-point boundary value problem of nonlinear fractional differential equation with mixed fractional derivatives and p-Laplacian operator ) ( ) ⎧ 3 ( ( 3 3 ) c 2 c 2 2 ⎪ ⎪ D D φ u(t) = f t, u(t), D u(t) , t ∈ (0, 1), 3 + ⎪ 0+ 0+ 2 ⎪ ⎨ 03 c 2 (4.1) D0+ u(0) =(u′ (0) ( ) ) = 0, ⎪ ⎪ 3 3 ⎪ ⎪ ⎩u(1) = 5 u 1 , cD 2+ u(1) = 2 cD 2+ u 2 , 0 4 3 3 0 3 3

c 2 ( 3 ) D u(t) where φ 3 cD02+ u(t) = √ 0+3 2

Assume that f (t, w, z)

|cD 2+ u(t)| 0 1 = 2(1+t) 2

3 3 ( 3 ) if cD02+ u(t) ̸= 0, and φ 3 cD02+ u(t) = 0 if cD02+ u(t) = 0. 2

(

)

1 + tetw−2 − z . We can easily check that x0 = x0 (t) ≡ 0 is a lower

5 4t 2 (84−48t+8t2 )

√ solution and y0 (t) = 1 + t − is an upper solution of boundary value problem (4.1). All 315 π conditions in Theorem 3.1 hold. Then boundary value problem (4.1) has the maximal lower solution x∗ and the minimal upper solution y ∗ , both x∗ and y ∗ are solutions of boundary value problems (4.1). Furthermore,

0 ≤ x∗ (t) ≤ y ∗ (t) ≤ 1 + t − 4t

5 2 (84−48t+8t2 )

√ 315 π

and − 12 ≤ − 12 (−2 + t)2 t = cD0β+ y0 (t) ≤ cD0β+ y ∗ (t) ≤ cD0β+ x∗ (t) ≤

D0β+ x0 (t) = 0.

c

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