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Successive iteration and positive solutions for nonlocal higher-order fractional differential equations with p-Laplacian

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Successive iteration and positive solutions for nonlocal higher-order fractional differential equations with p-Laplacian

QIUYAN ZHONG Jining Medical College

Department of Information Engineering Hehua Road No. 16, 272067, Jining

CHINA

[email protected]

XINGQIU ZHANG Jining Medical College

Department of Information Engineering Hehua Road No. 16, 272067, Jining

CHINA

[email protected]

Abstract: In this paper, we investigate the existence of positive solutions for higher-order fractional differential equations with p-Laplacian operator and nonlocal boundary conditions. By means of the properties of the corre- sponding Green function together with monotone iterative technique, we obtain not only the existence of positive solutions for the problems, but also establish iterative schemes for approximating the solutions. The nonlinearity permits singularities at t = 0 and/or t = 1.

Key–Words: Fractional differential equations; Integral boundary value problem; p-Laplacian; Positive solution;

Successive iteration

1 Introduction

The purpose of this paper is to consider the existence of positive solutions for the following singular frac- tional differential equations involving p-Laplacian op- erator (PFDE, for short) and integral conditions

D0+βp(D0+α u(t))) + a(t)f (t, u(t)) = 0, t ∈ J, u(0) = u0(0) = · · · = u(n−2)(0) = 0,

D0+α u(0) = 0, u(i)(1) = λ Z η

0

h(t)u(t)dt,

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where D0+α , Dβ0+ are standard Riemann-Liouville derivative, ϕp(s) = |s|p−2s, p > 1, a ∈ C((0, 1), R+), J = (0, 1), f ∈ C((0, 1) × R+, R+), f (t, u) may be singular at t = 0 and/or t = 1, R+ = [0, +∞), h ∈ L1[0, 1] is nonnegative and h(t) may be singular at t = 0 and t = 1, i ∈ [1, n − 2] is a fixed integer, n − 1 < α ≤ n, n ≥ 3, 0 < β ≤ 1 0 < η ≤ 1, 0 ≤ λR0ηh(t)tα−1dt < ∆, here

∆ = (α − 1)(α − 2) · · · (α − i).

Due to both by the intensive development of the theory of fractional calculus itself and by the wide ap- plications such as in control, porous media, aerody- namics, electrodynamics of complex medium, poly- mer rheology, electromagnetic, and so on, fractional differential equations have attracted more and more researchers’s much attention in recent years. We refer the readers to [1-3] for an extensive collection of such results.

In [4-12], by means of the fixed point index the- ory, fixed point theory together with the relevant re-

sults on u0bounded operator and lattice structure, the authors investigated the existence and multiplicity of positive and nontrivial solutions for fractional differ- ential equation

D0+α u(t) + a(t)f (t, u(t)) = 0, 0 < t < 1 (A) subject to different boundary conditions. A natural question is “How can we find the solutions when they are known to exist?”. There are few results on the computation of positive solutions for fractional differ- ential equations at present, see [8, 9, 12-15]. In [16], using the fixed point index theorem in cones, under some weak conditions concerning the first eigenvalue corresponding to the relevant linear operator, the au- thor obtained the existence and multiplicity of positive solutions for some singular higher-order fractional d- ifferential equations. Motivated by above papers, the aim of this paper is to give the iterative procedure of positive solutions for BVP (1).

This paper has the following three new features.

First, compared with [4-11], the boundary conditions contain the i-th order of the unknown function and a parameter λ. Particularly, a Lebesgue integrable func- tion h is involved in the boundary condition. Second, compared with [16], p-Laplacian operator is involved in differential operator. This means that the problems discussed in this paper have more general form. Fi- nally, we obtain not only the existence of positive so- lutions for the problems, but also establish iterative schemes for approximating the solutions. The itera- tive sequences begin with simple functions which is

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convenient in applications. Obviously, it is possible to replace the Riemann integrals in the boundary condi- tions by Riemann-Stieltjes integrals with minor mod- ifications.

The rest of this paper is organized as follows. In Section 2, we give some preliminaries and lemmas.

The main result is formulated in section 3 and an ex- ample is given in section 4 to illustrate how to use the main result.

2 Preliminaries and several lemmas

Let E = C[0, 1], kuk = max0≤t≤1|u(t)|, then (E, k · k) is a Banach space. For the reader’s convenience, we present some necessary definitions and lemmas from fractional calculus theory which can be found in the recent literature, see [1-3].

Definition 1 The Riemann-Liouville fractional inte- gral of orderα > 0 of a function y : (0, ∞) → R is given by

I0+α y(t) = 1 Γ(α)

Z t 0

(t − s)α−1y(s)ds provided the right-hand side is pointwise defined on (0, ∞).

Definition 2 The Riemann-Liouville fractional derivative of order α > 0 of a continuous function y : (0, ∞) → R is given by

Dα0+y(t) = 1 Γ(n − α)

d dt

nZ t 0

y(s) (t − s)α−n+1ds wheren = [α] + 1, [α] denotes the integer part of the numberα, provided that the right-hand side is point- wise defined on(0, ∞).

Now, we consider the following fractional differ- ential equation

Dα0+u(t) + y(t) = 0, 0 < t < 1, u(0) = u0(0) = · · · = u(n−2)(0) = 0, u(i)(1) = λ

Z η 0

h(t)u(t)dt.

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Lemma 3 [16] Assume that λR0ηh(t)tα−1dt 6= ∆.

Then for anyy ∈ L1[0, 1], the unique solution of the boundary value problems (2) can be expressed in the form

u(t) = Z 1

0

G(t, s)y(s)ds, t ∈ [0, 1], where

G(t, s) = G1(t, s) + G2(t, s), (3)

G1(t, s) =

tα−1(1 − s)α−1−i− (t − s)α−1

Γ(α) ,

0 ≤ s ≤ t ≤ 1, tα−1(1 − s)α−1−i

Γ(α) , 0 ≤ t ≤ s ≤ 1,

G2(t, s) = λtα−1

∆ − λR0ηh(t)tα−1dt Z η

0

h(t)G1(t, s)dt.

Here,G(t, s) is called the Green function of BVP (2).

Obviously,G(t, s) is continuous on [0, 1] × [0, 1].

Lemma 4 [16] If 0 ≤ λR0ηh(t)tα−1dt < ∆, then the functionG(t, s) defined by (3) satisfies

(a1) G(t, s) ≥ m1tα−1s(1 − s)α−1−i, ∀ t, s ∈ [0, 1];

(a2) G(t, s) ≤ M1tα−2s(1 − s)α−1−i, ∀ t, s ∈ [0, 1];

(a3) G(t, s) ≤ M1tα−1(1 − s)α−1−i, ∀ t, s ∈ [0, 1];

(a4) G(t, s) > 0, ∀ t, s ∈ (0, 1);

where m1 = Γ(α)1 1 + λ

∆−λRη

0 h(t)tα−1dt

Rη

0 h(t)tα−1dt, M1 = Γ(α)n 1 + λ

∆−λRη

0 h(t)tα−1dt

Rη

0 h(t)tα−2dt.

Let q > 1 satisfy 1p + 1q = 1. Then, ϕ−1p (s) = ϕq(s). To study the PFDE (1), we first consider the associated linear PFDE

Dβ0+p(Dα0+u(t))) + y(t) = 0, 0 < t < 1, u(0) = u0(0) = · · · = u(n−2)(0) = 0, Dα0+u(0) = 0, u(i)(1) = λ

Z η 0

h(t)u(t)dt, (4)

for y ∈ L1[0, 1] and h ≥ 0.

Lemma 5 The unique solution for the associated lin- ear PFDE (4) can be written by

u(t) =  1 Γ(β)

q−1Z 1 0

G(t, s) ϕ−1p 

Z s 0

(s − τ )β−1y(τ )dτds.

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Proof. Let w = Dα0+u, v = ϕp(w). Then, the initial value problem

( Dβ0+v(t) + y(t) = 0, t ∈ (0, 1),

v(0) = 0 (6)

has the solution v(t) = c1tβ−1− Iβy(t), t ∈ [0, 1].

Noticing that v(0) = 0, 0 < β ≤ 1, we have that c1 = 0. As a consequence,

v(t) = −Iβy(t), t ∈ [0, 1]. (7)

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Considering that D0+α u = w, w = ϕ−1p (v), we have from (7) that

Dα0+u(t) = ϕ−1p (−Iβ(y(t))), 0 < t < 1, u(0) = u0(0) = · · · = u(n−2)(0) = 0, u(i)(1) = λ

Z η 0

h(t)u(t)dt.

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By Lemma 3, the solution of (8) can be expressed by (5).

We list below assumptions used in this paper for convenience.

(H1) h ∈ L1[0, 1] is nonnegative;

(H2) f (t, 0) 6≡ 0 on [0, 1], f : (0, 1)×R+→ R+ is continuous and nondecreasing on x, and there exists constant r > 0 such that, for any t ∈ (0, 1), u ∈ R+,

f (t, cu) ≥ crf (t, u), ∀ 0 < c ≤ 1; (9) (H3) a : (0, 1) → R+ is continuous, a(t) 6≡ 0 with 0 < R01(1 − s)α−1−iϕ−1p R0s(s − τ )β−1f (τ, 1)dτa(s)ds < +∞.

Remark 6 If c ≥ 1, then it is not difficulty to see that (9) is equivalent to

f (t, cu) ≤ crf (t, u), ∀ c ≥ 1. (10) Define a subset P in E as follows

P = {u ∈ C([0, 1], R+) : there exist two positive numbers lu < 1 < Lusuch that lutα−1≤ u(t) ≤ Lutα−1, t ∈ [0, 1]}.

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Clearly, P is nonempty since tα−1 ∈ P.

Now define an operator A as follows:

(Au)(t) = 1 Γ(β)

q−1Z 1 0

G(t, s)ϕ−1p  Z s

0

(s − τ )β−1f (τ, u(τ ))dτds, t ∈ [0, 1].

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Lemma 7 Suppose that (H1) − (H3) hold. Then A : P → P is completely continuous and nondecreasing.

Proof. For any u ∈ P , there exist two positive num- bers 0 < lu < 1 < Lusuch that

lutα−1 ≤ u(t) ≤ Lutα−1, t ∈ [0, 1]. (13)

Thus, we have from Lemma 4, (9), (10), (H2), (H3) that

(Au)(t)

= 1 Γ(β)

q−1Z 1 0

G(t, s) ϕ−1p 

Z s 0

(s − τ )β−1f (τ, u(τ ))dτds

 1 Γ(β)

q−1

M1tα−1 Z 1

0

(1 − s)α−1−i ϕ−1p 

Z s 0

(s − τ )β−1f (τ, Luτα−1)dτds

 1 Γ(β)

q−1

M1tα−1 Z 1

0

(1 − s)α−1−i ϕ−1p 

Z s 0

(s − τ )β−1f (τ, Lu)dτds

 1 Γ(β)

q−1

M1tα−1 Z 1

0

(1 − s)α−1−i ϕ−1p 

Z s 0

(s − τ )β−1f (τ, 1)Lruds

= 1 Γ(β)

q−1

M1Lr(q−1)u Z 1

0

(1 − s)α−1−i ϕ−1p 

Z s 0

(s − τ )β−1f (τ, 1)dτds · tα−1

< +∞,

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(Au)(t)

= 1 Γ(β)

q−1Z 1 0

G(t, s) ϕ−1p 

Z s 0

(s − τ )β−1f (τ, u(τ ))dτds

 1 Γ(β)

q−1

m1tα−1 Z 1

0

s(1 − s)α−1−i ϕ−1p 

Z s 0

(s − τ )β−1f (τ, luτα−1)dτds

 1 Γ(β)

q−1

m1tα−1 Z 1

0

s(1 − s)α−1−i ϕ−1p 

Z s 0

(s − τ )β−1lruτr(α−1)f (τ, 1)dτds

= 1 Γ(β)

q−1

m1lr(q−1)u Z 1

0

s(1 − s)α−1−i ϕ−1p 

Z s 0

(s − τ )β−1τr(α−1)f (τ, 1)dτds · tα−1 (15)

which means that A is well defined, uniformly bound- ed and A(P ) ⊂ P . By standard argument, according to the Lebesgue dominated convergence theorem and the Arzela-Ascoli theorem, it is not difficult to see that A is completely continuous. Noticing the monotonic- ity of f on x, we know that A is nondecreasing.

3 Main result

For notational convenience, we denote Λ = 1

βΓ(β)

q−1

M1 Z 1

0

(1 − s)α−1−isβ(q−1)ds. (16)

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Theorem 8 Suppose that conditions (H1) − (H3) hold. In addition, if there existsa > 0, such that

f (t, u) ≤ ϕpa Λ

, (t, u) ∈ [0, 1] × [0, a]. (17) Then the BVP (1) has two positive solutions u and u; and there exist two positive numbersli < Li (i = 1, 2) such that

l1tα−1≤ u(t) ≤ L1tα−1,

l2tα−1≤ u(t) ≤ L2tα−1, t ∈ [0, 1]. (18) Moreover, for initial values u0 = 0, v0 = atα−1, monotone sequences {un}n=1 and {vn}n=1 satisfy limn→∞un = u, limn→∞vn = u, whereun+1 = Aun, vn+1 = Avn, n = 0, 1, 2, · · · .

Proof. Let Pa = {u ∈ P : kuk ≤ a}. For u ∈ Pa, we have 0 ≤ u(s) ≤ kuk ≤ a, s ∈ [0, 1]. Thus, for t ∈ [0, 1], it follows from Lemma 4, (16) and (17) that

(Au)(t)

= 1 Γ(β)

q−1Z 1 0

G(t, s) ϕ−1p 

Z s 0

(s − τ )β−1f (τ, u(τ ))dτds

 1 Γ(β)

q−1

M1tα−1 Z 1

0

(1 − s)α−1−i ϕ−1p 

Z s 0

(s − τ )β−1dτ · ϕpa Λ

ds

= 1 Γ(β)

q−1

M1 Z 1

0

(1 − s)α−1−i

·ϕ−1p 1

βsβds · a Λ

= 1 βΓ(β)

q−1

M1 Z 1

0

(1 − s)α−1−i sβ(q−1)ds · a

= a, Λ

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which implies that kAuk ≤ a, i.e., A(Pa) ⊂ Pa. Let u0(t) = 0, t ∈ [0, 1]. Similar to (19), by (17) we get that u1 = Au0∈ Pa. Denote

un+1= Aun= An+1u0, n = 1, 2, · · · . It follows from A(Pa) ⊂ Pa that un ∈ Pa (n = 1, 2, · · ·). By Lemma 7, we know that {un} is a se- quentially compact set.

Since u1 = Au0= A0 ∈ Pa, we have

u1(t) = (Au0)(t) = (A0)(t) ≥ 0 = u0(t), t ∈ [0, 1].

By induction, we get

un+1(t) ≥ un(t), n = 1, 2, · · · .

Consequently, there exists u ∈ Pa such that un → u. Let n → ∞, by the continuity of A and un+1 = Aun, we know that Au = u. This is to say that u ≥ 0 is a fixed point of A, i.e., u is a nonnegative solution for BVP (1). By (H2), it is not difficult to see that 0 is not the solution for BVP (1). Thus, u is a positive solution for BVP (1).

Let v0(t) = atα−1, t ∈ [0, 1], then v0 ∈ Pa. It follows from A(Pa) ⊂ Pathat v1 ∈ Pa. Denote

vn+1= Avn= An+1v0, n = 1, 2, · · · . Similarly, we get that

vn∈ Pa, n = 0, 1, 2, · · · . On the other hand, we have

v1(t) = (Av0)(t) = 1 Γ(β)

q−1Z 1 0

G(t, s) ϕ−1p 

Z s 0

(s − τ )β−1f (τ, v0(τ ))dτds

 1

Γ(β)

q−1

M1tα−1 Z 1

0

(1 − s)α−1−i ϕ−1p 

Z s 0

(s − τ )β−1dτ · ϕpa Λ

ds

 1

βΓ(β)

q−1

M1tα−1 Z 1

0

(1 − s)α−1−i

·sβ(q−1)ds · a

= atα−1= v0(t).Λ

By Lemma 7, we know that v2 = Av1 ≤ Av0 = v1. By induction, we get that

vn+1 ≤ vn, n = 0, 1, 2, · · · .

Consequently, there exists u ∈ Pa such that vn → u. Let n → ∞, by the continuity of A and vn+1 = Avn, we know that Au = u. This is to say that u ≥ 0 is a fixed point of A, i.e., u is a nonneg- ative solution for BVP (1). Moreover, considering that the zero function is not a solution of the prob- lem, we have that u is a positive solution for BVP (1). Since u, u ∈ Pa, there exist positive numbers li < Li(i = 1, 2) such that (18) holds.

Remark 9 The iterative sequences in Theorem 8 be- gin with simple functions which is significant for com- putational purpose.

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4 An example

Consider the following singular fractional differential equations

D

1 4

0+3(D

5 2

0+u(t))) + 1−t1 u32 = 0, 0 < t < 1, u(0) = u0(0) = 0, D

5 2

0+u(0) = 0, u0(1) = 2

3 Z 34

0

√1

tu(t)dt.

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Let α = 52, β = 14, p = 3, q = 32, λ = 23, η = 34, i = 1, n = 3, a(t) = 1

1−t, h(t) = 1

t, f (t, u) = u32, then ∆ = 32, 0 < λR0ηh(t)tα−1dt = 163 < ∆. Clear- ly, (H1), (H2) hold for h(t) = 1t, r = 32. On the other hand, we have

0 <

Z 1 0

(1 − s)α−1−iϕ−1p  Z s

0

(s − τ )β−1 f (τ, 1)dτa(s)ds

Z 1

0

ϕ−1p 1

βsβds < ϕ−1p 1 β

< +∞

which implies that (H3) also holds. By simple com- putation, we get

M1 = n Γ(α)

1 + λ

∆ − λR0ηh(t)tα−1dt Z η

0

h(t)tα−2dt

≈ 3.1166,

Λ = 4 Γ(14)

1

2×3.1166× Z 1

0

(1−s)12s18ds≈ 1.8752.

Take a = 103, then f (t, u) = u32 ≤ (103)32 = 3.1623 × 104 < 2.8438 × 105 = ϕ31.8752103 . Thus, (17) holds. It follows from Theorem 8 that BVP (20) has the positive minimal and maximal solution- s u and u; and there exist some positive numbers li < Li(i = 1, 2) such that (18) holds.

Acknowledgements: The project is supported fi- nancially by the Natural Science Foundation of Shandong Province of China (ZR2015AL002), a Project of Shandong Province Higher Education- al Science and Technology Program (J15LI16), the Foundations for Jining Medical College Natural Sci- ence (JYQ14KJ06, JY2015KJ019, JY2015BS07), and the National Natural Science Foundation of Chi- na (11571197, 11571296, 11371221, 11071141). The corresponding author of this paper is: Dr. Xingqiu Zhang.

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