Iterative solutions for a coupled system of fractional differential–integral equations with two-point boundary conditions

Applied Mathematics and Computation 244 (2014) 903–911

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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc

Iterative solutions for a coupled system of fractional differential–integral equations with two-point boundary conditions q Na Xu, Wenbin Liu ⇑ School of Science, China University of Mining and Technology, Xuzhou, Jiangsu 221116, PR China

a r t i c l e

i n f o

Keywords: Fractional differential systems Boundary value problems Monotone iterative technique Upper and lower solutions

a b s t r a c t In this paper, we deal with a coupled system of fractional differential–integral equations with two-point boundary conditions. The maximal and minimal solutions are obtained by using the monotone iterative technique combined with the method of upper and lower solutions, meantime the error estimate of the solutions are given. In the end, some examples are given to illustrate the results. Ó 2014 Elsevier Inc. All rights reserved.

1. Introduction Fractional calculus has gained considerable popularity due to its frequent appearance in various fields such as physics, aerodynamics, electrodynamics of complex medium, polymer rheology, etc. For details, see [1–3] and the references therein. And there has been a significant development in fractional differential equations in recent years. On the other hand, the study of coupled systems involving fractional differential equations is also important as such systems occur in various problems of applied nature (see [4–7]). So considerable work has been done to study the existence result for them nowadays (see [8–12]). And the authors got the existence solutions by the method of fixed-point theorem or coincidence degree theorem. The monotone iterative technique, combined with the method of upper and lower solutions, is a powerful tool for proving the existence of solutions of nonlinear differential equations, for instance, see [13–15] and the references therein. In [16–18], the authors used the method of upper and lower solutions investigated the existence of solutions for initial value problems with fractional differential equations. By the same method some people got the solutions of boundary value problems for fractional differential equations, such as [19–21]. To the best of our knowledge, only few papers (such as [22,23]) considered the method of upper and lower solutions for boundary value problems with fractional coupled systems. Motivated by [22,23], in this paper, we use the monotone iterative technique, combined with the method of upper and lower solutions to study the coupled system of fractional differential–integral equations with two-point boundary conditions, which is given by

q

This work is sponsored by NNSF of China (11271364).

⇑ Corresponding author.

E-mail addresses: [email protected] (N. Xu), [email protected] (W. Liu). http://dx.doi.org/10.1016/j.amc.2014.07.043 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.

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8 a > D uðtÞ þ f ðt; v ðtÞ; Ib v ðtÞÞ ¼ 0; t 2 ½0; 1; > > > b < D v ðtÞ þ gðt; uðtÞ; Ia uðtÞÞ ¼ 0; t 2 ½0; 1; 3a a2 > > I uðtÞjt¼0 ¼ D uðtÞjt¼0 ¼ uð1Þ ¼ 0; > > : 3a I v ðtÞjt¼0 ¼ Da2 v ðtÞjt¼0 ¼ v ð1Þ ¼ 0;

ð1:1Þ

where 2 < a; b  3; uðtÞ; v ðtÞ 2 C½0; 1; f ; g : I  R  R # R satisfy Carathéodory conditions. Da ; Db and Ia ; Ib are the standard Riemam–Liouville fractional derivative and fractional integral respectively. As applying, we get the existence of solutions for a coupled system of fractional differential equations with two-point boundary problems

8 a D uðtÞ þ f ðt; v ðtÞÞ ¼ 0; t 2 ½0; 1; > > > b > < D v ðtÞ þ gðt; uðtÞÞ ¼ 0; t 2 ½0; 1; > I3a uðtÞjt¼0 ¼ Da2 uðtÞjt¼0 ¼ uð1Þ ¼ 0; > > > : 3a I v ðtÞjt¼0 ¼ Da2 v ðtÞjt¼0 ¼ v ð1Þ ¼ 0;

ð1:2Þ

where 2 < a; b  3; uðtÞ; v ðtÞ 2 C½0; 1; f ; g : I  R # R satisfy Carathéodory conditions. The rest of this paper is organized as follows. In Section 2, we give some necessary notations, definitions and lammas. In Section 3, we establish theorems of solutions for the problems (1.1) and (1.2). In Section 4, we give two examples to demonstrate our result. 2. Preliminaries In this section, we present some definitions, lemmas and assumptions that will be used in the whole paper, and the definitions about fractional calculus theory can be found in [1–3]. Definition 2.1. Let a > 0, the operator Ia , defined on L½a; b by a

a It

uðtÞ ¼

1 CðaÞ

Z

t

ðt  sÞa1 uðsÞ ds

a

for a 6 t 6 b, is called the Riemann–Liouville fractional integral operator of order a, where a 2 R and C is the Gamma function. Definition 2.2. Let a > 0 and n ¼ ½a þ 1, the operator Da , defined by a

a Dt

uðtÞ ¼

1 Cðn  aÞ



d dt

n Z

t

ðt  sÞna1 uðsÞ ds

a

is called the Riemann–Liouville fractional differential operator of order a. Definition 2.3 [24]. Let E and F be partially ordered Banach spaces, D  E; A : D # F be an operator. For all x; y 2 D with x 6 y, if we have Ax 6 Ay, then A is called an increasing operator. Definition 2.4 [24]. Let E be a Banach space, D  E; A : D # F be an operator. If x0 6 Ax0 (resp., x0 P Ax0 ) holds for all x0 2 D, then x0 is called the lower (resp., upper) solution of operator equation x ¼ Ax. Definition 2.5 [24]. We say the function f ðt; u; v Þ : ½0; 1  R  R # R satisfies Carathéodory conditions, if the following conditions are satisfied: (1) for each u; v 2 R; f ðt; u; v Þ is Lebesgue measurable for t; (2) for almost every t 2 ½0; 1; f ðt; u; v Þ is continuous for u; v .

Definition 2.6 [25]. Functions ðu0 ðtÞ; v 0 ðtÞÞ 2 C½0; 1  C½0; 1 are called a lower solution of (1.1) if they satisfy

8 a > D u ðtÞ þ f ðt; v 0 ðtÞ; Ib v 0 ðtÞÞ 6 0; t 2 ð0; 1Þ; > > b 0 > < D v ðtÞ þ gðt; u ðtÞ; Ia u ðtÞÞ 6 0; t 2 ð0; 1Þ; 0 0 0 > I3a u0 ðtÞjt¼0 6 0; Da2 u0 ðtÞjt¼0 6 0; u0 ð1Þ 6 0; > > > : 3a I v 0 ðtÞjt¼0 6 0; Da2 v 0 ðtÞjt¼0 6 0; v 0 ð1Þ 6 0:

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N. Xu, W. Liu / Applied Mathematics and Computation 244 (2014) 903–911

Analogously, functions ðu0 ðtÞ; v 0 ðtÞÞ are said to be a upper solution of problem (1.1), if they satisfy

8 a  D u0 ðtÞ þ f ðt; v 0 ðtÞ; Ib v 0 ðtÞÞ P 0; t 2 ð0; 1Þ; > > > > b  < D v 0 ðtÞ þ gðt; u0 ðtÞ; Ia u0 ðtÞÞ P 0; t 2 ð0; 1Þ; 3a a2 > > I u0 ðtÞjt¼0 P 0; D u0 ðtÞjt¼0 P 0; u0 ð1Þ P 0; > > : 3a  I v 0 ðtÞjt¼0 P 0; Da2 v 0 ðtÞjt¼0 P 0; v 0 ð1Þ P 0:

Lemma 2.7 [24]. Let E be a partially ordered Banach space, for each n 2 Zþ ; xn 6 yn , if xn ! x and yn ! y , we have x 6 y . Lemma 2.8 [10]. Assume that u 2 Cð0; 1Þ \ Lð0; 1Þ with a fractional derivative of order a > 0. Then Ia Da uðtÞ ¼ uðtÞ þ c1 t a1 þ c2 ta2 þ    þ cN t aN for some ci 2 R; i ¼ 1; 2; . . . N, where N ¼ ½a. Let us illustrate some symbols that used in our paper. Let I ¼ ½0; 1, E ¼ C½0; 1 with the norm kuk ¼ maxt2½0;1 juðtÞj, clearly it is a Banach space. Make P be a cone in E, we define a partial ordering in E by u 6 v iff v  u 2 P, then E is a partially ordered Banach space. For any u0 ; u0 2 E and they are lower 0 ; u  0 2 E and they are lower and upper and upper solutions of (1.1) respectively, with u0 6 u0 , let D ¼ ½u0 ; u0 . For any u 0 6 u  0 , let D1 ¼ ½u 0 ; u  0 . solutions of (1.2) respectively, with u The assumptions throughout the paper are as follows: (H1) f ; g : I  R  R # R satisfy Carathéodory conditions; (H2) ðu0 ðtÞ; v 0 ðtÞÞ 2 E  E and ðu0 ðtÞ; v 0 ðtÞÞ 2 E  E are lower and upper solutions of (1.1) respectively with u0 ðtÞ 6 u0 ðtÞ; v 0 ðtÞ 6 v 0 ðtÞ; (H3) For any ui ; v i 2 D with ui 6 v i ði ¼ 1; 2Þ, there exist four nonnegative constants Li and M i ði ¼ 1; 2Þ such that

0 6 f ðt; v 1 ; v 2 Þ  f ðt; u1 ; u2 Þ 6 L1 ðv 1  u1 Þ þ L2 ðv 2  u2 Þ; 0 6 gðt; v 1 ; v 2 Þ  gðt; u1 ; u2 Þ 6 M 1 ðv 1  u1 Þ þ M2 ðv 2  u2 Þ;

t 2 I; t 2 I;

 0 ðtÞ; v  0 ðtÞ; v  0 ðtÞÞ 2 E  E and ðu  0 ðtÞÞ 2 E  E are lower and upper solutions of (1.2) respectively, with (H4) ðu  0 ðtÞ 6 u  0 ðtÞ; v  0 ðtÞ 6 v  0 ðtÞ; u (H5) For any u; v 2 D1 with u 6 v , there exist two nonnegative constants L and M such that

0 6 f ðt; v Þ  f ðt; uÞ 6 Lðv  uÞ;

t 2 I;

0 6 gðt; v Þ  gðt; uÞ 6 Mðv  uÞ;

t 2 I:

In the paper, we let

K¼

L1 M1

Cða þ 1ÞCðb þ 1Þ

þ

L1 M 2

C2 ða þ 1ÞCðb þ 1Þ

þ

L2 M1

Cða þ 1ÞC2 ðb þ 1Þ

þ

L2 M2

C2 ða þ 1ÞC2 ðb þ 1Þ

:

3. Main results First, we construct the Green’s function of (1.1). Lemma 3.1. If f ; g : I  R  R # R satisfy Carathéodory conditions, then the solutions of (1.1) are exactly the solutions of coupled integral system of equations as follows

(

uðtÞ ¼

v ðtÞ ¼

R1 0

G1 ðt; sÞf ðs; v ðsÞ; Ib v ðsÞÞ ds;

0

G2 ðt; sÞgðs; uðsÞ; Ia uðsÞÞÞ ds;

R1

1 where Ib v ðtÞ ¼ CðbÞ

G1 ðt; sÞ ¼

Rt 0

t 2 I;

ðt  sÞb1 v ðsÞ ds; Ia uðtÞ ¼ Cð1aÞ

Rt 0

: ta1 ð1sÞa1 ;

CðbÞ

ð3:2Þ

0 6 t 6 s 6 1:

8 b1 b1 b1 < ðtsÞ þt ð1sÞ ; 0 6 s 6 t 6 1; CðbÞ : tb1 ð1sÞb1 ;

ð3:1Þ

ðt  sÞa1 uðsÞ ds and Gi ðt; sÞ ði ¼ 1; 2Þ are the Green’s functions defined by

8 a1 a1 a1 < ðtsÞ þt ð1sÞ ; 0 6 s 6 t 6 1; CðaÞ CðaÞ

G2 ðt; sÞ ¼

t 2 I;

0 6 t 6 s 6 1:

ð3:3Þ

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N. Xu, W. Liu / Applied Mathematics and Computation 244 (2014) 903–911

Proof. By Lemma 2.8, the solutions of (1.1) can be written as

uðtÞ ¼ c1 t a1  c2 ta2  c3 ta3  Ia f ðt; v ðtÞ; Ib v ðtÞÞ;

v ðtÞ ¼ d1 tb1  d2 tb2  d3 tb3  Ib gðt; uðtÞ; Ia uðtÞÞ: Using the boundary conditions, we find that c2 ; c3 ; d2 ; d3 ¼ 0 and

Z

1 CðaÞ

c1 ¼

d1 ¼

1

ð1  sÞa1 f ðs; v ðsÞ; Ib v ðsÞÞ ds;

0

Z

1 CðbÞ

1

ð1  sÞb1 gðs; uðsÞ; Ia uðsÞÞ ds:

0

Thus, the solutions of (1.1) are

Z 1 Z t t a1 1 ð1  sÞa1 f ðs; v ðsÞ; Ib v ðsÞÞ ds  ðt  sÞa1 f ðs; v ðsÞ; Ib v ðsÞÞ ds CðaÞ 0 CðaÞ 0 # Z t " a1 Z 1 a1 t ð1  sÞa1 ðt  sÞa1 t ð1  sÞa1 ¼ f ðs; v ðsÞ; Ib v ðsÞÞ ds f ðs; v ðsÞ; Ib v ðsÞÞ ds þ  CðaÞ CðaÞ CðaÞ 0 t Z 1 G1 ðt; sÞf ðs; v ðsÞ; Ib v ðsÞÞ ds: ¼

uðtÞ ¼

0

Similarly, v ðtÞ ¼

R1 0

G2 ðt; sÞgðs; uðsÞ; Ia uðsÞÞ ds.

h

The following properties of the Green’s functions play an important role in this paper. Lemma 3.2. The Green’s functions G1 ðt; sÞ and G2 ðt; sÞ defined by (3.2) and (3.3) satisfies: (1) Gi ðt; sÞ P 0; i ¼ 1; 2, for all t; s 2 ½0; 1; R1 R1 1 (2) 0 G1 ðt; sÞ ds 6 Cða1þ1Þ ; 0 G2 ðt; sÞ ds 6 Cðbþ1Þ for all t 2 ½0; 1.

Proof. For all t; s 2 ½0; 1, we have t  ts P t  s, so ½ðt  sÞa1 þ ta1 ð1  sÞa1 =CðaÞ P 0 as 0 < s 6 t < 1. Then G1 ðt; sÞ P 0 for all t; s 2 ½0; 1. For all t 2 ½0; 1, we get that

Z

1

G1 ðt; sÞ ds ¼

Z

0

ta1 ð1  sÞa1 ds  CðaÞ

1

0

Z 0

t

ðt  sÞa1 ta1 ð1  tÞ 1 ds ¼ 6 : CðaÞ CðaÞ Cða þ 1Þ

It is similar to G2 ðt; sÞ. h It is easy to see that (3.1) is equivalent to the following nonlinear integral equation

uðtÞ ¼

Z

1

 Z G1 ðt; sÞf s;

0

1

G2 ðs; zÞgðz; uðzÞ; Ia uðzÞÞ dz;Ib

0

Z

1

 G2 ðs; zÞgðz; uðzÞ; Ia uðzÞÞ dz ds:

ð3:4Þ

0

Theorem 3.3. Assume that the conditions ðH1Þ—ðH3Þ hold and K < 1, then there exist a unique minimal solution ðu ; v  Þ and maximal solution ðu ; v  Þ for (1.1). We use u0 ; u0 as initial iterations, and get the iterative sequences as follows for each n 2 Zþ ,

un ðtÞ ¼

Z

1

 Z G1 ðt; sÞf s;

0

un ðtÞ ¼

Z

1

G2 ðs; zÞgðz; un1 ðzÞ; Ia un1 ðzÞÞ dz; Ib

0 1

 Z G1 ðt; sÞf s;

0

0

1

G2 ðs; zÞgðz; un1 ðzÞ; Ia un1 ðzÞÞ dz; Ib

u0 6 u1 6 u2 6    6 un 6    6 un 6    6 u2 6 u1 6 u0 ;

n!1

1

 G2 ðs; zÞgðz; un1 ðzÞ; Ia un1 ðzÞÞ dz ds;

ð3:5Þ

 G2 ðs; zÞgðz; un1 ðzÞ; Ia un1 ðzÞÞ dz ds;

ð3:6Þ

0

and we have

u ðtÞ ¼ lim un ðtÞ;

Z

v  ðtÞ ¼

Z 0

1

G2 ðt; sÞgðs; u ðsÞ; Ia u ðsÞÞ ds;

Z 0

1

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N. Xu, W. Liu / Applied Mathematics and Computation 244 (2014) 903–911

v  ðtÞ ¼

u ðtÞ ¼ lim un ðtÞ; n!1

Z

1

G2 ðt; sÞgðs; u ðsÞ; Ia u ðsÞÞ ds:

0

Moreover, the error estimate of the minimal and maximal solutions are given by

ku ðtÞ  un ðtÞk 6

Kn ku1 ðtÞ  u0 ðtÞk; 1K

kun ðtÞ  u ðtÞk 6

Kn ku ðtÞ  u1 ðtÞk: 1K 0

Proof. We define an operator A : D # E by

Z

AuðtÞ ¼

1

 Z G1 ðt; sÞf s;

0

1

G2 ðs; zÞgðz; uðzÞ; Ia uðzÞÞ dz; Ib

Z

0

1

 G2 ðs; zÞgðz; uðzÞ; Ia uðzÞÞ dz ds:

0

It can be seen that the fixed point of A is the solution of (3.4), and A is continuous. Because (3.4) is equivalent to (3.1), we only need to prove the existence of the fixed point for A. If ðu0 ðtÞ; v 0 ðtÞÞ 2 E  E is the lower solution of (1.1), we get

Z

v 0 ðtÞ 6

1

G2 ðt; sÞgðs; u0 ; Ia u0 Þ ds;

u0 ðtÞ 6

0

Z

1

G1 ðt; sÞf ðs; v 0 ; Ib v 0 Þ ds:

0

Furthermore, Ib v 0 ðtÞ 6 Ib

R1 0

 Z f ðs; v 0 ; Ib v 0 Þ 6 f t;

G2 ðt; sÞgðs; u0 ; Ia u0 Þ ds. By (H3), we have 1

G2 ðt; sÞgðs; u0 ; Ia u0 Þ ds; Ib

0

Z

1

 G2 ðt; sÞgðs; u0 ; Ia u0 Þ ds :

0

Thus,

u0 ðtÞ 6

Z

Z

1

G1 ðt; sÞf ðs; v 0 ; Ib v 0 Þ ds 6

0

1

G1 ðt; sÞf ðt;

0

Z

1

G2 ðt; sÞgðs; u0 ; Ia u0 Þ ds; Ib

0

Z

1

G2 ðt; sÞgðs; u0 ; Ia u0 Þ dsÞ;

0

i.e., u0 is the lower solution of operator A. For any u; v 2 D with u 6 v , from (H3), we get Au 6 Av , i.e., A is an increasing operator. From the definition of A, (3.5) can be written as un ¼ Aun1 for each n 2 Zþ . Since A is increasing, in view of (H2), we get

u0 6 u1 6 u2 6    6 un :

ð3:7Þ a

For any u; v 2 D with u 6 v , obviously, I u 6 I

Z

1

G2 ðs; zÞgðt; u; Ia uÞ dz 6

0

Z

1

a

v . By (H3) we get gðt; u; I

G2 ðs; zÞgðt; v ; Ia v Þ dz; Ib

0

Z

1

a

uÞ 6 gðt; v ; I

a

v Þ. Since G2 ðt; sÞ P 0, we have

G2 ðs; zÞgðt; u; Ia uÞ dz 6 Ib

0

Z

1

G2 ðs; zÞgðt; v ; Ia v Þ dz:

0

It follows from (H3) that

 Z 0 6 f t;

1

G2 ðt; sÞgðs; v ; Ia v Þ ds; Ib

Z

0

G2 ðs; sÞgðs; v ; Ia v Þ ds



0

 Z  f t;

1

a

G2 ðt; sÞgðs; u; I uÞ ds; I 0

Z

1

b

Z

1

G2 ðs; sÞgðs; u; Ia uÞ ds



0

Z t Z 1   L2 G2 ðt; sÞ M 1 ðv  uÞ þ M2 ðIa v  Ia uÞ ds þ ðt  sÞb1 G2 ðs; sÞ½M 1 ðv  uÞ þ M2 ðIa v  Ia uÞ ds ds 6 L1 CðbÞ 0 0 0 Z 1 Z t Z 1 L1 L2 6 ½M 1 ðv  uÞ þ M2 ðIa v  Ia uÞ ds þ ðt  sÞb1 ½M 1 ðv  uÞ þ M2 ðIa v  Ia uÞ ds ds: Cðb þ 1Þ 0 CðbÞCðb þ 1Þ 0 0 1

Furthermore, we get

  Z  f t; 

1

G2 ðt; sÞgðs; v ; Ia v Þ ds; Ib

0

Z 0

1

  Z G2 ðs; sÞgðs; v ; Ia v Þ ds f t;

1 0

G2 ðt; sÞgðs; u; Ia uÞ ds; Ib

Z 0

1

  G2 ðs; sÞgðs; u; Ia uÞ ds  

L1 M 1 L1 M 2 L2 M1 L2 M2 6 kv  uk þ kv  uk þ 2 kv  uk þ kv  uk; Cðb þ 1Þ Cðb þ 1ÞCða þ 1Þ C ðb þ 1Þ Cða þ 1ÞC2 ðb þ 1Þ then

kAv  Auk 6 Kkv  uk: Noticing (3.7) and (3.8), we get

ð3:8Þ

908

N. Xu, W. Liu / Applied Mathematics and Computation 244 (2014) 903–911

ku2  u1 k ¼ kAu1  Au0 k 6 Kku1  u0 k; ku3  u2 k ¼ kAu2  Au0 k 6 K2 ku1  u0 k; ku4  u3 k ¼ kAu3  Au2 k 6 K3 ku1  u0 k;  Then we have kunþ1  un k 6 Kn ku1  u0 k. Therefore, for each n; p 2 Zþ , we have

kunþp  un k 6 kunþp  unþp1 k þ kunþp1  unþp2 k þ    þ kunþ1  un k 6 Kn

1  Kp ku1  u0 k: 1K

ð3:9Þ

For 0 < K < 1, we have kunþp  un k ! 0 as n ! 1. Thus fun g is a Cauchy sequence in D. Let u ðtÞ ¼ limn!1 un ðtÞ, thus Au ¼ u . R1 So (1.1) has a pair of solution ðu ðtÞ; v  ðtÞÞ, and v  ðtÞ ¼ 0 G2 ðt; sÞgðs; u ðsÞ; Ia u ðsÞÞ ds. n K ku  u k. Let p ! 1 in (3.9), we have ku  un k 6 1 1 0 K If we use the upper solution u0 as an initial iterative value, it is similar to get the solution ðu ; v  Þ of (1.1), and R1 u ðtÞ ¼ limn!1 un ðtÞ; v  ðtÞ ¼ 0 G2 ðt; sÞgðs; u ðsÞ; Ia u ðsÞÞ ds. It is clear to see that u0 6 u1 6 u2 6    6 un 6    6 un 6       6 u2 6 u1 6 u0 . Next, let us prove ðu ðtÞ; v  ðtÞÞ is the minimal solution and ðu ðtÞ; v  ðtÞÞ is the maximal solution of (1.1). For any xðtÞ 2 D with Ax ¼ x, we have un 6 x 6 un since A is an increasing operator. From Lemma 2.7, we get u ðtÞ 6 xðtÞ 6 u ðtÞ, i.e., u ðtÞ and u ðtÞ are the minimal and maximal fixed point of A respectively. So, ðu ðtÞ; v  ðtÞÞ and ðu ðtÞ; v  ðtÞÞ are the minimal and maximal solution of (1.1) respectively. In the end, let us prove the uniqueness of maximal and minimal solutions for (1.1). Suppose l0 ðtÞ; m0 ðtÞ 2 E are lower and upper solutions of operator equation u ¼ Au respectively, i.e., l0 6 Al0 ; m0 P Am0 ; t 2 I. Then we use l0 ; m0 as an initial iterations respectively, and we have ln ! l ; mn ! m by iteration as n ! 1. Furthermore, we have Al ¼ l ; Am ¼ m . Now let us prove l ¼ u . Since u0 6 l and A is increasing, we get un ¼ An u0 6 An l for each n 2 Zþ . So u0 6 u1 6 u2 6    6 un 6    6 l . From (3.8) and mathematical induction, we get kl  un k ¼ kAn l  An u0 k 6 Kn kl  u0 k ! 0; ðn ! 1Þ. Thus kl  u k ! 0; ðn ! 1Þ. It shows that u ¼ l . It is similar to get u ¼ m . So the minimal and maximal solutions of (1.1) are unique. Analogously, we have the error estimate of maximal solution as follows

kun ðtÞ  u ðtÞk 6

Kn ku ðtÞ  u1 ðtÞk: 1K 0



If f ; g : I  R # R satisfy Carathéodory conditions, by Lemma 3.1 we know that the solutions of (1.2) is equivalent to the solutions of the following coupled integral system of equations

(

R1

uðtÞ ¼

0

G1 ðt; sÞf ðs; v ðsÞÞ ds;

0

G2 ðt; sÞgðs; uðsÞÞ ds;

R1

v ðtÞ ¼

ð3:10Þ

where Gi ðt; sÞði ¼ 1; 2Þ are the Green’s functions defined by (3.2) and (3.3). It is easy to see that (3.10) is equivalent to the nonlinear integral equation given by

uðtÞ ¼

Z

 Z G1 ðt; sÞf s;

1

0

 G2 ðs; zÞgðz; uðzÞÞ dz ds:

1

ð3:11Þ

0

We define an operator A : D1 # E, by

AuðtÞ ¼

Z

1

 Z G1 ðt; sÞf s;

0

1

 G2 ðs; zÞgðz; uðzÞÞ dz ds:

0

It can be proved that the fixed point of A is the solution of (3.11). As a promotion of Theorem 3.3, we can get the following result and the proof is similar to Theorem 3.3. Theorem 3.4. Assume f ; g : I  R # R satisfy Carathéodory conditions, ðH4Þ; ðH5Þ hold, and ML < Cða þ 1ÞCðb þ 1Þ, then there  ; v   Þ and a unique maximal solution ðu   ; v   Þ for (1.2). We give the iterative sequences as exist a unique minimal solution ðu follows

 n ðtÞ ¼ u

Z

1

 Z G1 ðt; sÞf s;

0

 n ðtÞ ¼ u

Z 0

1

 n1 ðzÞÞ dz ds; G2 ðs; zÞgðz; u

n ¼ 1; 2; 3 . . .

0 1

G1 ðt; sÞf ðs;

Z 0

1

 n1 ðzÞÞ dzÞ ds; G2 ðs; zÞgðz; u

n ¼ 1; 2; 3 . . .

N. Xu, W. Liu / Applied Mathematics and Computation 244 (2014) 903–911

909

Fig. 1. A plot of the lower and upper solutions 1.

and we have

0 6 u 1 6 u 2 6    6 u n 6    6 u  n 6    6 u  2 6 u  1 6 u  0 ; u   ðtÞ ¼ lim u  n ðtÞ; u n!1

v  ðtÞ ¼

Z

1

  ðsÞÞ ds; G2 ðt; sÞgðs; u

0

  ðtÞ ¼ lim u  n ðtÞ; u n!1

v  ðtÞ ¼

Z

1

  ðsÞÞ ds: G2 ðt; sÞgðs; u

0

Moreover, we have the following error estimate

  u n k 6 ku

ðLMÞn Cða þ 1ÞCðb þ 1Þ 1  u  0 k; ku C ða þ 1ÞCn ðb þ 1Þ Cða þ 1ÞCðb þ 1Þ  LM

n  u   k 6 ku

n

ðLMÞn Cða þ 1ÞCðb þ 1Þ   u  1 k: ku C ða þ 1ÞCn ðb þ 1Þ Cða þ 1ÞCðb þ 1Þ  LM 0 n

4. Some examples Example 4.1. Consider the following problem

8 5 7 2 D2 uðtÞ þ v ðtÞ sin t þ t 2 I3 v ðtÞ ¼ 0; t 2 ½0; 1; > > > 7 > 5 < 3 D v ðtÞ þ uðtÞ cos2 t þ ð1  tÞ2 I2 uðtÞ ¼ 0; t 2 ½0; 1; > I3a uðtÞj ¼ Da2 uðtÞj ¼ uð1Þ ¼ 0; > > t¼0 t¼0 > : 3a I v ðtÞjt¼0 ¼ Da2 v ðtÞjt¼0 ¼ v ð1Þ ¼ 0:

ð4:1Þ

Let f ðt; v ðtÞ; Ib v ðtÞÞ ¼ v ðtÞ sin t þ t 2 Ib v ðtÞ; gðt; uðtÞ; Ia uðtÞÞ ¼ uðtÞ cos2 t þ ð1  tÞ2 Ia uðtÞ; a ¼ 5=2; b ¼ 7=3. It is easy to deduce that Li ¼ 1; Mi ¼ 1 (i ¼ 1; 2), i.e., for any uðtÞ 6 v ðtÞ, we have 2

0 6 f ðt; uðtÞ; Ib uðtÞÞ  f ðt; v ðtÞ; Ib v ðtÞÞ 6 ðv ðtÞ  uðtÞÞ þ ðIb ðv ðtÞ  uðtÞÞÞ; 0 6 gðt; uðtÞ; Ia uðtÞÞ  gðt; v ðtÞ; Ia v ðtÞÞ 6 ðv ðtÞ  uðtÞÞ þ ðIa ðv ðtÞ  uðtÞÞÞ; and

L1 M 1

Cðaþ1ÞCðbþ1Þ

L1 M 2 L2 M 1 L2 M 2 þ C2 ðaþ1Þ þ Cðaþ1Þ þ C2 ðaþ1Þ < 1. Cðbþ1Þ C2 ðbþ1Þ C2 ðbþ1Þ

910

N. Xu, W. Liu / Applied Mathematics and Computation 244 (2014) 903–911

Fig. 2. A plot of the lower and upper solutions 2.

We use the lower solution ðu0 ðtÞ; v 0 ðtÞÞ ¼ ð1; 1Þ and upper solution ðu0 ðtÞ; v 0 ðtÞÞ ¼ ð1; 1Þ as initial iterations respectively. From Theorem 3.3, we can obtain the iterative sequences un and un , then we can use them as approximate minimal and maximal solutions when n is large enough. From Fig. 1, we get that

u ðtÞ ¼ u5 ðtÞ; v  ðtÞ ¼

Z

1

5

G2 ðt; sÞ½u5 ðsÞ cos2 s þ ð1  sÞ2 I2 u5 ðsÞ ds;

0

u ðtÞ ¼ u5 ðtÞ; v  ðtÞ ¼

Z

1

0

5

G2 ðt; sÞ½u5 ðsÞ cos2 s þ ð1  sÞ2 I2 u5 ðsÞ ds:

The error estimates are as follows

kuðtÞ  u5 ðtÞk 6

K5 K5 ku1 ðtÞ  u0 k 6 maxju1 ðtÞ þ 1j ’ 3:2  104 ; 1K 1  K t2½0;1

kuðtÞ  u5 ðtÞk 6

K5 K5 ku1 ðtÞ  u0 k 6 maxju ðtÞ  1j ’ 3:2  104 : 1K 1  K t2½0;1 1

Example 4.2. Consider the following problem

8

4 > D9=4 uðtÞ þ t  12 v ðtÞ ¼ 0; t 2 ½0; 1; > > > >

< 14=5 D v ðtÞ þ t  12 2 uðtÞ ¼ 0; t 2 ½0; 1; > > I3a uðtÞjt¼0 ¼ Da2 uðtÞjt¼0 ¼ uð1Þ ¼ 0; > > > : 3a I v ðtÞjt¼0 ¼ Da2 v ðtÞjt¼0 ¼ v ð1Þ ¼ 0: 4

ð4:2Þ

2

Let f ðt; v ðtÞÞ ¼ ðt  12Þ v ðtÞ;gðt;uðtÞÞ ¼ ðt  12Þ uðtÞ; a ¼ 9=4; b ¼ 14=5. For any uðtÞ 6 v ðtÞ, we can get 0 6 f ðt; v ðtÞÞ  f ðt; uðtÞÞ 6 v ðtÞ  uðtÞÞ; 0 6 gðt; v ðtÞÞ  gðt; uðtÞÞ 6 14 ðv ðtÞ  uðtÞÞ. Let L ¼ 1=16; M ¼ 1=4, and they satisfy LM 6 Cða þ 1ÞCðb þ 1Þ. We  0 ðtÞ; v  0 ðtÞÞ ¼ ð1; 1Þ and upper solution ðu  0 ðtÞ; v  0 ðtÞÞ ¼ ð1; 1Þ as initial iterations respectively. use the lower solution ðu

1 ð 16

 n and u  n as approximate minimal and maximal solutions when n From Theorem 3.4, we can obtain the iterative sequence u is large enough. From Fig. 2, we get that

  ðtÞ ¼ u  5 ðtÞ; u

v  ðtÞ ¼

Z 0

1

G2 ðt; sÞðs  1=2Þ2 uðsÞ ds;

N. Xu, W. Liu / Applied Mathematics and Computation 244 (2014) 903–911

v  ðtÞ ¼

  ðtÞ ¼ u 5 ðtÞ; u

Z

1

911

G2 ðt; sÞðs  1=2Þ2 uðsÞ ds:

0

The error estimates are as follows

 5 ðtÞk 6 kuðtÞ  u 6

 5 ðtÞk 6 kuðtÞ  u 6

ðLMÞ5

Cða þ 1ÞCððb þ 1Þ  1 ðtÞ  u  0 ðtÞk ku C ða þ 1ÞC5 ðb þ 1Þ Cða þ 1ÞCððb þ 1Þ  LM 5

ðLMÞ5

Cða þ 1ÞCððb þ 1Þ  1 ðtÞ þ 1j ’ 3:8  1015 ; maxju C ð a þ 1ÞCððb þ 1Þ  LM t2½0;1 C ða þ 1ÞC ðb þ 1Þ 5

5

ðLMÞ5

Cða þ 1ÞCððb þ 1Þ  1 ðtÞ  u  0 ðtÞk ku C ða þ 1ÞC5 ðb þ 1Þ Cða þ 1ÞCððb þ 1Þ  LM 5

ðLMÞ5

Cða þ 1ÞCððb þ 1Þ  1 ðtÞ  1j ’ 3:8  1015 : maxju C ða þ 1ÞC5 ðb þ 1Þ Cða þ 1ÞCððb þ 1Þ  LM t2½0;1 5

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