Applied Mathematics and Computation 218 (2012) 8526–8536
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The eigenvalue problem for a singular higher order fractional differential equation involving fractional derivatives q Xinguang Zhang a,⇑, Lishan Liu b,c, Yonghong Wu c a
School of Mathematical and Informational Sciences, Yantai University, Yantai, 264005 Shandong, People’s Republic of China School of Mathematical Sciences, Qufu Normal University, Qufu, 273165 Shandong, People’s Republic of China c Department of Mathematics and Statistics, Curtin University of Technology, Perth, WA 6845, Australia b
a r t i c l e
i n f o
Keywords: Fractional differential equation Positive solution Green function Eigenvalue problem
a b s t r a c t In this paper, we study the following singular eigenvalue problem for a higher order fractional differential equation
8 a l1 l2 ln1 xðtÞÞ; 0 < t < 1; > < D xðtÞ ¼ kf ðxðtÞ; D xðtÞ; D xðtÞ; . . . ; D p2 P l l l i > aj D xðnj Þ; 1 6 i 6 n 1; : xð0Þ ¼ 0; D xð0Þ ¼ 0; D xð1Þ ¼ j¼1
where n 3; n 2 N, n 1 < a 6 n; n l 1 < a ll < n l, for l ¼ 1; 2; . . . ; n 2, and lP ln1 > 0; a ln1 2; a l > 1, aj 2 ½0; þ1Þ; 0 < n1 < n2 < < np2 < 1, al1 < 1, Da is the standard Riemann–Liouville derivative, and 0 < p2 j¼1 aj nj n f : ð0; þ1Þ ! ½0; þ1Þ is continuous. Firstly, we give the Green function and its properties. Then we established an eigenvalue interval for the existence of positive solutions from Schauder’s fixed point theorem and the upper and lower solutions method. The interesting point of this paper is that f may be singular at xi ¼ 0; for i ¼ 1; 2; . . . ; n. Ó 2012 Elsevier Inc. All rights reserved.
1. Introduction The purpose of this paper is to establish the existence of positive solutions to the following higher order fractional differential equation
8 a l1 l2 ln1 xðtÞÞ; 0 < t < 1; > < D xðtÞ ¼ kf ðxðtÞ; D xðtÞ; D xðtÞ; . . . ; D p2 P l l > aj Dl xðnj Þ; 1 6 i 6 n 1; : xð0Þ ¼ 0; D i xð0Þ ¼ 0; D xð1Þ ¼
ð1:1Þ
j¼1
where n 3; n 2 N, n 1 < a 6 n; n l 1 < a ll < n l, for l ¼ 1; 2; . . . ; n 2, and P al1 l ln1 > 0; a ln1 2; a l > 1, aj 2 ½0; þ1Þ; 0 < n1 < n2 < . . . < np2 < 1, 0 < p2 < 1, Da is the standard j¼1 aj nj Riemann–Liouville derivative, and f : ð0; þ1Þn ! ½0; þ1Þ is continuous. Recently, fractional differential equations have been of great interest. It is caused both by the theory of fractional calculus itself and by the applications in various sciences such as physics, mechanics, chemistry, engineering, for details, see [1–22]
q The authors were supported financially by the National Natural Science Foundation of China (11071141) and the Natural Science Foundation of Shandong Province of China (ZR2010AM017). ⇑ Corresponding author. E-mail address:
[email protected] (X. Zhang).
0096-3003/$ - see front matter Ó 2012 Elsevier Inc. All rights reserved. doi:10.1016/j.amc.2012.02.014
X. Zhang et al. / Applied Mathematics and Computation 218 (2012) 8526–8536
8527
and the references therein. In [1], by using the fixed point index theory, Bai considered the following fractional differential equation with three-point boundary condition
Da uðtÞ þ f ðt; uðtÞÞ ¼ 0; uð0Þ ¼ 0; uð1Þ ¼ buðgÞ;
0 < t < 1;
ð1:2Þ
where 1 < a 6 2, 0 < bga1 < 1, 0 < g < 1, Da is the standard Riemann–Liouville derivative, f : ½0; 1 ½0; þ1Þ ! ½0; þ1Þ is continuous. In [1], Bai got the property of the Green function and the existence of positive solutions for (1.2) when f is sublinear. In [2], Zhang considered the following problem whose nonlinear term and boundary condition contain integer order derivatives of the unknown function
(
Da xðtÞ þ qðtÞf ðx; x0 ; . . . ; xðn2Þ Þ ¼ 0; 0
xð0Þ ¼ x ð0Þ ¼ . . . ¼ x
ðn2Þ
ð0Þ ¼ x
ðn2Þ
0 < t < 1; n 1 < a 6 n; ð1Þ ¼ 0;
ð1:3Þ
where Da is the standard Riemann–Liouville fractional derivative of order a, q may be singular at t ¼ 0, and f may be singular at x ¼ 0; xprime ¼ 0; . . . ; xðn2Þ ¼ 0: And then, Goodrich [3] was concerned with a partial extension of the problem (1.3)
Da xðtÞ ¼ f ðt; xðtÞÞ;
0 < t < 1; n 1 < a 6 n;
xðiÞ ð0Þ ¼ 0; 0 6 i 6 n 2; Da xð1Þ ¼ 0;
1 6 a 6 n 2;
ð1:4Þ
and the author derived the Green function for the problem (1.4) and showed that it satisfies certain properties. Then by using the cone theoretic techniques the author deduce a general existence theorem for this problem provided that f ðt; xÞ satisfies some growth conditions. Next, a significative work is also developed by Goodrich [4] to study a system of nonlinear differential equations of fractional order having the form of (1.4). Another fractional problem of nonlocal-type similar to (1.4) is treated by Wang et al. [5] through the cone theoretic techniques, where f ðt; xÞ can be singular at x ¼ 0. In recent work [15], M. Rehman and R. Khan investigated the multi-point boundary value problems for fractional differential equations of the form
8 a > < Dt yðtÞ ¼ f ðt; yðtÞ; > : yð0Þ ¼ 0;
Dbt yðtÞÞ; t 2 ð0; 1Þ; m2 P Dbt yð1Þ fi Dbt yðni Þ ¼ y0 ;
ð1:5Þ
i¼1
P ab1 where 1 < a 6 2, 0 < b < 1, 0 < ni < 1, fi 2 ½0; þ1Þ with m2 < 1. By using the Schauder fixed point theorem and i¼1 fi ni the contraction mapping principle, the authors established the existence and uniqueness of nontrivial solutions for the BVP (1.5) provided that the nonlinear function f : ½0; 1 R R is continuous and satisfies certain growth conditions. However when the f and boundary condition involve fractional derivatives of the unknown function and the nonlinear term possesses singularity at xi ¼ 0; for i ¼ 1; 2; . . . ; n, fewer results are established on fractional differential equations. Inspired by the above works, the aim of this paper is to establish the existence of positive solutions for the higher nonlocal fractional differential equations (1.1). The present paper has the following features. Firstly, the nonlinear term f involves fractional derivatives of unknown functions; Secondly, the BVP (1.1) possesses singularity, that is, f ðx1 ; x2 ; . . . ; xn Þ may be singular at xi ¼ 0; for i ¼ 1; 2; . . . ; n. In the end, the boundary conditions involving fractional derivatives of the unknown function are more general case, which include two-point,three point, multi-point and some nonlocal problems as special cases. Our technique is also rather different from the ones presented previously such as [1–7,11,13,15,18,22]. The rest of the paper is organized as follows. In Section 2, we present some preliminaries and lemmas on fractional calculus theory, and then give the associated Green function and develop some properties of the Green function, finally, establish a maximal principle. In Section 3, by constructing the upper and lower solutions, we establish an eigenvalue interval for the existence of positive solutions for the BVP (1.1) from Schauder’s fixed point theorem, we also give an example to demonstrate our main results. 2. Basic definitions and preliminaries We first wish to collect some basic lemmas that will be important to us in what follows. These and other related results and their proofs can be found, for example, in [3,4,9,10,18]. Definition 2.1 (see [3,4]). Let a > 0 with a 2 R: Suppose that x : ½a; 1Þ ! R. Then the ath Riemann–Liouville fractional integral is defined to be
Ia xðtÞ ¼
1 CðaÞ
Z
t
ðt sÞa1 xðsÞ ds;
a
whenever the right-hand side is defined. Similarly, with a > 0 with a 2 R, we define the ath Riemann–Liouville fractional derivative to be
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ðnÞ Z t 1 d ðt sÞna1 xðsÞ ds; Cðn aÞ dt a
Da xðtÞ ¼
where n 2 N is the unique positive integer satisfying n 1 6 a < n and t > a. Remark 2.1. If x; y : ð0; þ1Þ ! R with order a > 0, then
Da ðxðtÞ þ yðtÞÞ ¼ Da xðtÞ þ Da yðtÞ: Lemma 2.1 (see [10]). (1) If x 2 L1 ð0; 1Þ; q > r > 0; then
Iq Ir xðtÞ ¼ Iqþr xðtÞ; Dr Iq xðtÞ ¼ Iqr xðtÞ; Dr Ir xðtÞ ¼ xðtÞ: (2) If q > 0; m > 0; then
Dq t m1 ¼
CðmÞ mq1 : t Cðm qÞ
Lemma 2.2 (see [8,18]). Assume that x 2 Cð0; 1Þ \ L1 ð0; 1Þ with a fractional derivative of order a > 0, then Ia Da xðtÞ ¼ xðtÞ þ c1 ta1 þ c2 t a2 þ þ cn tan ; where ci 2 R; i ¼ 1; 2; . . . ; n ðn ¼ ½a þ 1Þ. Here Ia stands for the standard Riemann–Liouville fractional integral of order a > 0 and Da denotes the Riemann–Liouville fractional derivative as Definition 2.1. Let 8
k1 ðt; sÞ ¼
k2 ðt; sÞ ¼
aln1 1 ð1sÞal1 ðtsÞaln1 1
Cðaln1 Þ
; 0 6 s 6 t 6 1;
: ; Cðaln1 Þ 8 al1 ðtsÞal1 ðtð1sÞÞ < ; t aln1 1 ð1sÞal1
0 6 t 6 s 6 1; ð2:1Þ 0 6 s 6 t 6 1;
Cðaln1 Þ
: ðtð1sÞÞal1 ;
0 6 t 6 s 6 1;
Cðaln1 Þ
obviously for t; s 2 ½0; 1, we have
ki ðt; sÞ 6
ð1 sÞal1 ; i ¼ 1; 2: Cða ln1 Þ
ð2:2Þ
Lemma 2.3.
Pp2 j¼1 aj k2 ðnj ; sÞ ; GðsÞ ¼ P al1 Cða ln1 Þ 1 p2 j¼1 aj nj
1þ M¼
1
Pp2 aj Pp2j¼1 al1 j¼1
aj nj
Cða ln1 Þ
:
If h 2 L1 ½0; 1, then the boundary value problem
8 aln1 wðtÞ ¼ hðtÞ; > < D ll > : wð0Þ ¼ 0; D n1 wð1Þ ¼
p2 P
ð2:3Þ
aj Dlln1 wðnj Þ;
j¼1
has the unique solution
wðtÞ ¼
Z
1
Kðt; sÞhðsÞ ds; 0
where
Kðt; sÞ ¼ k1 ðt; sÞ þ
p2 X taln1 1 aj k2 ðnj ; sÞ; Pp2 al1 1 j¼1 aj nj j¼1
ð2:4Þ
is the Green function of the boundary value problem (2.3). Proof. By applying Lemma 2.2, we may reduce (2.3) to an equivalent integral equation
wðtÞ ¼ Ialn1 hðtÞ þ c1 t aln1 1 þ c2 taln1 2 ;
c1 ; c2 2 R:
ð2:5Þ
X. Zhang et al. / Applied Mathematics and Computation 218 (2012) 8526–8536
8529
From wð0Þ ¼ 0 and (2.5), we have c2 ¼ 0. Consequently the general solution of (2.3) is
wðtÞ ¼ Ialn1 hðtÞ þ c1 t aln1 1 :
ð2:6Þ
By (2.6) and Lemma 2.1, we have
Dlln1 wðtÞ ¼ Dlln1 Ialn1 hðtÞ þ c1 Dlln1 t aln1 1 ¼ Ial hðtÞ þ c1 ¼
Z
Cða ln1 Þ al1 t Cða lÞ
ðt sÞal1 Cða ln1 Þ al1 : hðsÞds þ c1 t Cða lÞ Cða lÞ
t
0
ð2:7Þ
So, from (2.7), we have
Dlln1 wð1Þ ¼
Z
1
0
D
lln1
wðnj Þ ¼
Z
nj
0
By Dlln1 wð1Þ ¼
R1 c1 ¼
0
Pp2 j¼1
ð1 sÞal1 Cða ln1 Þ hðsÞds þ c1 ; Cð a l Þ Cða lÞ ðnj sÞal1 Cða ln1 Þ al1 ; hðsÞds þ c1 n Cða lÞ Cða lÞ j
ð2:8Þ for j ¼ 1; 2; . . . ; p 2:
aj Dlln1 wðnj Þ, combining with (2.8), we obtain
R nj P al1 ð1 sÞal1 hðsÞ ds p2 hðsÞ ds j¼1 aj 0 ðnj sÞ : Pp2 al1 Cða ln1 Þ 1 j¼1 aj nj
So, substituting c1 into (2.6), one has the unique solution of problem (2.3)
Z
ðt sÞaln1 1 t aln1 1 hðsÞds þ Pp2 al1 Cða ln1 Þ 1 j¼1 aj nj 0 (Z ) Z nj p2 al1 1 X ðnj sÞal1 ð1 sÞ aj hðsÞds hðsÞds Cða ln1 Þ 0 0 Cða ln1 Þ j¼1 Z 1 Z t ðt sÞaln1 1 ð1 sÞal1 t aln1 1 t aln1 1 hðsÞds þ hðsÞds þ ¼ Pp2 al1 Cða ln1 Þ Cða ln1 Þ 1 j¼1 aj n 0 0
wðtÞ ¼
t
j
p2 X
aj
¼
0
1
1
al1
ð1 sÞal1 nj
0
j¼1
Z
Z
aln1 1
hðsÞds
Cða ln1 Þ
1
t Pp2
p2 X
al1
j¼1 aj nj !
j¼1
Lemma 2.4. The function Kðt; sÞ has the following properties: (1) Kðt; sÞ > 0; for t; s 2 ð0; 1Þ; (2) t aln1 1 GðsÞ 6 Kðt; sÞ 6 Mð1 sÞal1 ; for t; s 2 ½0; 1, where
Pp2
aj k2 ðnj ; sÞ Pp2 al1 ; 1 j¼1 aj nj j¼1
1þ M¼
Pp2 aj Pp2j¼1 al1
1
j¼1
aj nj
Cða ln1 Þ
:
Proof. It is obvious that (1) holds. In the following, we will prove (2). Firstly, by (2.4),
Kðt; sÞ P
0
nj
p2 X t aln1 1 aj k2 ðnj ; sÞ ¼ t aln1 1 GðsÞ: Pp2 al1 1 j¼1 aj nj j¼1
al1
ðnj sÞ hðsÞds Cða ln1 Þ
Z 1 p2 X t aln1 1 k1 ðt; sÞ þ aj k2 ðnj ; sÞ hðsÞds ¼ Kðt; sÞhðsÞds: Pp2 al1 1 j¼1 aj nj 0 j¼1
The proof is completed. h
GðsÞ ¼
aj
Z
ð2:9Þ
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X. Zhang et al. / Applied Mathematics and Computation 218 (2012) 8526–8536
On the other hand, it follows from (2.2) that
Pp2 al1 p2 X taln1 1 ð1 sÞal1 j¼1 aj ð1 sÞ Kðt; sÞ ¼ k1 ðt; sÞ þ aj k2 ðnj ; sÞ 6 þ Pp2 al1 P Cða ln1 Þ Cða l Þ 1 p2 aj nal1 1 j¼1 aj nj j¼1 n1 j¼1 j ! Pp2 ð1 sÞal1 j¼1 aj : 6 1þ Pp2 al1 Cða ln1 Þ 1 j¼1 aj n j
This completes the proof. h For the convenience of expression in the rest of the paper, we let l0 ¼ 0. Now let us consider the following modified problem of the BVP (1.1)
8 aln1 v ðtÞ ¼ kf ðIln1l0 v ðtÞ; Iln1 l1 v ðtÞ; . . . ; Iln1 ln2 v ðtÞ; v ðtÞÞ; > < D p2 P ll > aj Dlln1 v ðnj Þ: : v ð0Þ ¼ 0; D n1 v ð1Þ ¼
ð2:10Þ
j¼1
l
Lemma 2.5. Let xðtÞ ¼ I n1l0 v ðtÞ; v ðtÞ 2 C½0; 1. Then we can transform (1.1) into (2.10). Moreover, if solution of problem (2.10), then the function xðtÞ ¼ Iln1l0 v ðtÞ is a positive solution of problem (1.1).
v 2 Cð½0; 1; ½0; þ1ÞÞ is a
Proof. Substituting xðtÞ ¼ Iln1l0 v ðtÞ into (1.1), by the definition of Riemann–Liouville fractional derivative and Lemmas 2.1 and 2.2, we can obtain that n
n
n
d na d d xðtÞ ¼ n Ina Iln1l0 v ðtÞ ¼ n Inaþln1 v ðtÞ ¼ Daln1 v ðtÞ; nI dt dt dt Dl1 xðtÞ ¼ Dl1 Iln1l0 v ðtÞ ¼ Iln1 l1 v ðtÞ; Dl2 xðtÞ ¼ Dl2 Iln1l0 v ðtÞ ¼ Iln1 l2 v ðtÞ; ...
Da xðtÞ ¼
ð2:11Þ
Dln2 xðtÞ ¼ Dln2 Iln1l0 v ðtÞ ¼ Iln1 ln2 v ðtÞ; Dln1 xðtÞ ¼ Dln1 Iln1l0 v ðtÞ ¼ v ðtÞ: we have Dln1 xð0Þ ¼ v ð0Þ ¼ 0, it follows from Dl xðtÞ ¼ Dl Iln1l0 v ðtÞ ¼ Dlln1 v ðtÞ that Dlln1 v ð1Þ ¼ lln1 a D v ðnj Þ. Hence, by xðtÞ ¼ Iln1l0 v ðtÞ; v 2 C½0; 1, we can transform (1.1) into (2.10). j¼1 j Now, let v 2 Cð½0; 1; ½0; þ1ÞÞ be a solution for problem (2.10). Then, from the definition of the Riemann–Liouville factional derivative and Lemmas 2.1 and (2.10) and (2.11), one has Also, Pp2
n
n
n
d na d d xðtÞ ¼ n Ina Iln1l0 v ðtÞ ¼ n Inaþln1 v ðtÞ ¼ Daln1 v ðtÞ nI dt dt dt ¼ kf ðIln1l0 v ðtÞ; Iln1 l1 v ðtÞ; . . . ; Iln1 ln2 v ðtÞ; v ðtÞÞ ¼ kf ðxðtÞ; Dl1 xðtÞ; Dl2 xðtÞ; . . . ; Dln1 xðtÞÞ;
Da xðtÞ ¼
Noting
Ia v ðtÞ ¼
1 CðaÞ
Z
t
ðt sÞa1 v ðsÞds;
0
which implies that Ia v ð0Þ ¼ 0; from (2.11), for i ¼ 1; 2; . . . ; n 1; we have
xð0Þ ¼ 0;
Dli xð0Þ ¼ 0;
Dl xð1Þ ¼
p2 X
aj Dl xðnj Þ:
j¼1
Moreover, it follows from the monotonicity and property of Iln1l0 that
Iln1l0 v 2 Cð½0; 1; ½0; þ1ÞÞ: Consequently, xðtÞ ¼ Iln1l0 v ðtÞ is a positive solution of problem (1.1). h Definition 2.2. A continuous function wðtÞ is called a lower solution of the BVP (2.10), if it satisfies
8 aln1 wðtÞ 6 kf ðIln1l0 wðtÞ; Iln1 l1 wðtÞ; . . . ; Iln1 ln2 wðtÞ; wðtÞÞ; > < D p2 P ll > aj Dlln1 wðnj Þ: : wð0Þ P 0; D n1 wð1Þ P j¼1
0 < t < 1:
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X. Zhang et al. / Applied Mathematics and Computation 218 (2012) 8526–8536
Definition 2.3. A continuous function /ðtÞ is called a upper solution of the BVP (2.10), if it satisfies
8 aln1 /ðtÞ P kf ðIln1l0 /ðtÞ; Iln1 l1 /ðtÞ; . . . ; Iln1 ln2 /ðtÞ; /ðtÞÞ; > < D p2 P ll > aj Dlln1 /ðnj Þ: : /ð0Þ 6 0; D n1 /ð1Þ 6 j¼1
Now we make the following assumptions throughout this paper: ðH1 Þ f : ð0; þ1Þn ! ½0; þ1Þ is continuous and is non-increasing in xi > 0 for i ¼ 1; 2; . . . ; n; ðH2 Þ For all r 2 ð0; 1Þ, there exists constant > 0 such that, for any ðx1 ; x2 ; . . . ; xn Þ 2 ð0; þ1Þn ,
f ðrx1 ; rx2 ; . . . ; rxn Þ 6 r f ðx1 ; x2 ; . . . ; xn Þ: Remark 2.2. If ðH2 Þ holds, then for r 1 we have the following result: There exists constant ðx1 ; x2 ; . . . ; xn Þ 2 ð0; þ1Þn ,
>0
such that, for any
f ðrx1 ; rx2 ; . . . ; rxn Þ P r f ðx1 ; x2 ; . . . ; xn Þ: In fact, if r 1, then for any ðx1 ; x2 ; . . . ; xn Þ 2 ð0; þ1Þn , we have
1 1 1 1 f r x1 ; r x2 ; . . . ; r xn 6 f ðrx1 ; rx2 ; . . . ; rxn Þ; r r r r that is
f ðrx1 ; rx2 ; . . . ; rxn Þ P r f ðx1 ; x2 ; . . . ; xn Þ:
Lemma 2.6 (Maximal principle). If
Dlln1 v ð1Þ ¼
v ð0Þ ¼ 0;
v 2 Cð½0; 1; RÞ satisfies
p2 X
aj Dlln1 v ðnj Þ
j¼1
and Daln1 v ðtÞ 0 for any t 2 ½0; 1, then
v ðtÞ 0;
t 2 ½0; 1:
Proof. By Lemma 2.4, the conclusion is obvious, we omit the proof. h
3. Main results Let en1 ðtÞ ¼ t aln1 1 ; and for i ¼ 0; 1; 2; . . . ; n 2, define
ei ðtÞ ¼
Z 0
t
ðt sÞln1 li 1 saln1 1 Cða ln1 Þ ali 1 : ds ¼ t Cðln1 li Þ Cða li Þ
Theorem 3.1. Suppose ðH1 Þ and ðH2 Þ hold, and the following condition is satisfied ðH3 Þ f ð1; 1; . . . ; 1Þ – 0, and
0<
Z
1
ð1 sÞal1 f ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds < þ1:
0
Then there is a constant k > 0 such that for any k 2 ðk ; þ1Þ, the BVP (1.1) has at least one positive solution xðtÞ, and xðtÞ satisfies
e0 ðtÞ 6 xðtÞ 6 Ntln1 ; where
N¼
k ln1 M Cðln1 Þ
Z
1
ð1 sÞal1 f ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds;
0
and M is defined by Lemma 2.4.
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X. Zhang et al. / Applied Mathematics and Computation 218 (2012) 8526–8536
Proof. Let E ¼ C½0; 1, and denote a set P:
P ¼ fv ðtÞ 2 E : there exists a positive number lv such that v ðtÞ P lv en1 ðtÞ; t 2 ½0; 1g:
ð3:1Þ
Clearly, P is a nonempty set since en1 ðtÞ 2 P. Now define the operator T k in E:
ðT k v ÞðtÞ ¼ k
Z
1
Kðt; sÞf ðIln1 v ðsÞ; Iln1 l1 v ðsÞ; . . . ; Iln1 ln2 v ðsÞ; v ðsÞÞds:
ð3:2Þ
0
Obviously T k is well defined and T k ðPÞ P: To see this, for any . 2 P, by the definition of P, there exists a positive number l. 0 0 such that .ðtÞ P l. en1 ðtÞ for any t 2 ½0; 1, take l. ¼ minf12 ; l. g, then .ðtÞ P l. en1 ðtÞ for any t 2 ½0; 1 and 0 < l. < 1. It follows from Lemma 2.4 and ðH2 Þ ðH3 Þ that
ðT k .ÞðtÞ ¼ k
Z
1
0
Kðt; sÞf ðIln1 .ðsÞ; Iln1 l1 .ðsÞ; . . . ; Iln1 ln2 .ðsÞ; .ðsÞÞds
Z
1
6 kM 0
ð1 sÞal1 f ðl. e0 ðsÞ; l. e1 ðsÞ; . . . ; l. en2 ðsÞ; l. en1 ðsÞÞds
Z
6 kMl.
1
ð1 sÞal1 f ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds < þ1:
ð3:3Þ
0
Let B ¼ maxf2; maxt2½0;1 .ðtÞg, for i ¼ 0; 1; 2; . . . ; n 2,
Z
Iln1 li .ðtÞ ¼
t
0
ðt sÞln1 li 1 .ðsÞ B ds 6 6 B; Cðln1 li Þ Cðln1 li Þ
then from ðH3 Þ, we have
ðT k .ÞðtÞ ¼ k
Z
1
Kðt; sÞf ðIln1 .ðsÞ; Iln1 l1 .ðsÞ; . . . ; Iln1 ln2 .ðsÞ; .ðsÞÞds
0
P ktaln1 1
Z
1
0
P kB taln1 1
GðsÞf ðB; B; . . . ; B; BÞds Z
1
GðsÞf ð1; 1; . . . ; 1; 1Þds
0
¼ kB f ð1; 1; . . . ; 1; 1Þ
Z
1
GðsÞdst
aln1 1
¼ kAB f ð1; 1; . . . ; 1; 1Þen1 ðtÞ;
ð3:4Þ
0
R1 where A ¼ 0 GðsÞds. (3.3) and (3.4) imply T k is well defined and T k ðPÞ P: Next we shall attempt to find the upper and lower solutions of the BVP (2.10). By ðH2 Þ and (3.2), we know the f is decreasing on xi for i ¼ 1; 2; . . . ; n. It follows
Z
1
Kðt; sÞf ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds P t aln1 1
Z
0
that
1
GðsÞf ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds;
8 t 2 ½0; 1;
0
Z
1
k1
Kðt; sÞf ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds P t aln1 1 ;
8t 2 ½0; 1;
0
where
k1 ¼ R 1 0
1 GðsÞf ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds
On the other hand, let bðtÞ ¼ for any k > k1 , by ðH3 Þ, we have
Z
1
R1 0
:
Kðt; sÞf ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds; since f ðx1 ; x2 ; . . . ; xn1 ; xn Þ is decreasing in xi > 0,
Kðt; sÞf ðkIln1 bðsÞ; kIln1 l1 bðsÞ; . . . ; kIln1 ln2 bðsÞ; kbðsÞÞds
0
Z
1
6 0
6M
Kðt; sÞf ðk1 Iln1 bðsÞ; k1 Iln1 l1 bðsÞ; . . . ; k1 Iln1 ln2 bðsÞ; k1 bðsÞÞds 6
Z
Z 0
1
ð1 sÞal1 f ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds < þ1:
0
Let C ¼ maxf2; maxt2½0;1 bðtÞg, for i ¼ 0; 1; 2; . . . ; n 2; we have
1
Kðt; sÞf ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds
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X. Zhang et al. / Applied Mathematics and Computation 218 (2012) 8526–8536
C
Iln1 li bðtÞ 6
Cðln1 li Þ
6 C:
Now take
( k > max 1; k1 ;
1þ1 ) C : Af ð1; 1; . . . ; 1; 1Þ
By ðH2 Þ, for any t 2 ½0; 1, we have
k f ðk Iln1 bðsÞ; k Iln1 l1 bðsÞ; . . . ; k Iln1 ln2 bðsÞ; k bðsÞÞ P ðk Þþ1 f ðC; C; . . . ; C; CÞ P ðk Þþ1 C f ð1; 1; . . . ; 1; 1Þ P A1 : Thus by Lemma 2.4, we obtain
k
Z
1
Kðt; sÞf ðk Iln1 bðsÞ; k Iln1 l1 bðsÞ; . . . ; k Iln1 ln2 bðsÞ; k bðsÞÞ P A1
0
Z
1
taln1 1 GðsÞds ¼ A1 t aln1 1
0
¼ t aln1 1 ; Let
/ðtÞ ¼ k
Z
Z
1
GðsÞds
0
8t 2 ½0; 1:
1
Kðt; sÞf ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞ; wðtÞ
0
¼ k
Z
1
Kðt; sÞf ðk Iln1 bðsÞ; k Iln1 l1 bðsÞ; . . . ; k Iln1 ln2 bðsÞ; k bðsÞÞds;
0
then by Lemma 2.3, for any t 2 ½0; 1, we have
8 R 1 aln1 1 ; > < /ðtÞ ¼ k 0 Kðt; sÞf ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds P t p2 P ll > aj Dlln1 /ðnj Þ; : /ð0Þ ¼ 0; D n1 /ð1Þ ¼
ð3:5Þ
j¼1
8 R 1 ln1 bðsÞ; k Iln1 l1 bðsÞ; . . . ; k Iln1 ln2 bðsÞ; k bðsÞÞds P t aln1 1 ; > < wðtÞ ¼ k 0 Kðt; sÞf ðk I p2 P ll > aj Dlln1 wðnj Þ: : wð0Þ ¼ 0; D n1 wð1Þ ¼
t 2 ½0; 1; ð3:6Þ
j¼1
Obviously, /ðtÞ; wðtÞ 2 P. By (3.5) and (3.6), we have
taln1 1 6 wðtÞ ¼ ðT k /ÞðtÞ; which implies
wðtÞ ¼ ðT k /ÞðtÞ ¼ k 6 k
Z
Z
1
taln1 1 6 /ðtÞ;
8t 2 ½0; 1;
ð3:7Þ
Kðt; sÞf ðIln1 /ðsÞ; Iln1 l1 /ðsÞ; . . . ; Iln1 ln2 /ðsÞ; /ðsÞÞds
0 1
Kðt; sÞf ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds ¼ /ðtÞ;
8t 2 ½0; 1:
ð3:8Þ
0
Thus, by (3.7) and (3.8), we have
Daln1 wðtÞ þ k f ðIln1l0 wðtÞ; Iln1 l1 wðtÞ; . . . ; Iln1 ln2 wðtÞ; wðtÞÞ P Daln1 ðT k /ÞðtÞ þ k f ðIln1l0 /ðtÞ; Iln1 l1 /ðtÞ; . . . ; Iln1 ln2 /ðtÞ; /ðtÞÞ ¼ k f ðIln1l0 /ðtÞ; Iln1 l1 /ðtÞ; . . . ; Iln1 ln2 /ðtÞ; /ðtÞÞ þ k f ðIln1l0 /ðtÞ; Iln1 l1 /ðtÞ; . . . ; Iln1 ln2 /ðtÞ; /ðtÞÞ ¼ 0;
ð3:9Þ
Daln1 /ðtÞ þ k f ðIln1l0 /ðtÞ; Iln1 l1 /ðtÞ; . . . ; Iln1 ln2 /ðtÞ; /ðtÞÞ ¼ k f ðe0 ðtÞ; e1 ðtÞ; . . . ; en2 ðtÞ; en1 ðtÞÞ þ k f ðIln1l0 /ðtÞ; Iln1 l1 /ðtÞ; . . . ; Iln1 ln2 /ðtÞ; /ðtÞÞ 6 k f ðe0 ðtÞ; e1 ðtÞ; . . . ; en2 ðtÞ; en1 ðtÞÞ þ k f ðe0 ðtÞ; e1 ðtÞ; . . . ; en2 ðtÞ; en1 ðtÞÞ ¼ 0:
ð3:10Þ
It is obvious that /; w satisfy the boundary conditions of the BVP (2.10). Thus it follows from (3.8)–(3.10) that wðtÞ; /ðtÞ are the upper and lower solutions of the BVP (2.10), and wðtÞ; /ðtÞ 2 P. Define the function F and the operator Ak in E by
8 l n1l0 wðtÞ; Iln1 l1 wðtÞ; . . . ; Iln1 ln2 wðtÞ; wðtÞÞ; u < wðtÞ; > < f ðI ln1l0 uðtÞ; Iln1 l1 uðtÞ; . . . ; Iln1 ln2 uðtÞ; uðtÞÞ; wðtÞ 6 u 6 /ðtÞ; FðuÞ ¼ f ðI > : n1l0 /ðtÞ; Iln1 l1 /ðtÞ; . . . ; Iln1 ln2 /ðtÞ; /ðtÞÞ; u > /ðtÞ f ðI
ð3:11Þ
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X. Zhang et al. / Applied Mathematics and Computation 218 (2012) 8526–8536
and
ðAk uÞðtÞ ¼ k
Z
1
8u 2 E:
Kðt; sÞFðuðsÞÞds; 0
Clearly, F : ½0; þ1Þ ! ½0; þ1Þ is continuous by (3.11). Consider the following boundary value problem
8 aln1 uðtÞ ¼ k FðuÞ; > < D
0 < t < 1; p2 P Dlln1 uð1Þ ¼ aj Dlln1 uðnj Þ:
> : uð0Þ ¼ 0;
ð3:12Þ
j¼1
Obviously, a fixed point of the operator Ak is a solution of the BVP (3.12). For all u 2 E, it follows from Lemma 2.4 and (3.11) and wðtÞ P t aln1 1 that
ðAk uÞðtÞ 6 k M
Z
1
sð1 sÞal1 FðuðsÞÞds 6 k M
Z
0
6 k M
Z
1
ð1 sÞal1 f ðIln1l0 wðsÞ; Iln1 l1 wðsÞ; . . . ; Iln1 ln2 wðsÞ; wðsÞÞds
0 1
ð1 sÞal1 f ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds < þ1:
0
So Ak is bounded. It is easy to see Ak : E ! E is continuous from the continuity of FðuÞ and Kðt; sÞ. On the other hand, let X E be bounded, by using the fact that the function Kðt; sÞ is uniformly continuous on [0, 1], it is easy to see Ak ðXÞ is equicontinuous. Thus by the means of the Arzela–Ascoli theorem, we have Ak : E ! E is completely continuous. Thus, by using the Schauder fixed point theorem, Ak has at least one fixed point w such that w ¼ Ak w. Now we prove
wðtÞ 6 wðtÞ 6 /ðtÞ;
t 2 ½0; 1:
Let zðtÞ ¼ /ðtÞ wðtÞ; t 2 ½0; 1. By /ðtÞ is the upper solution of the BVP (2.10) and w is a fixed point of Ak , we know
Dlln1 wð1Þ ¼
wð0Þ ¼ 0;
p2 X
aj Dlln1 wðnj Þ:
ð3:13Þ
j¼1
From the definition of F, (3.7) and (3.8), we obtain
f ðIn1l0 /ðtÞ; Iln1 l1 /ðtÞ; . . . ; Iln1 ln2 /ðtÞ; /ðtÞÞ 6 FðuðtÞÞ 6 f ðIln1l0 wðtÞ; Iln1 l1 wðtÞ; . . . ; Iln1 ln2 wðtÞ; wðtÞÞ 6 f ðe0 ðtÞ; e1 ðtÞ; . . . ; en2 ðtÞ; en1 ðtÞÞ; 8u 2 E; 8t 2 ½0; 1:
ð3:14Þ
Thus (3.5) and (3.14) imply
Daln1 zðtÞ ¼ Daln1 /ðtÞ Daln1 wðtÞ ¼ k f ðe0 ðtÞ; e1 ðtÞ; . . . ; en2 ðtÞ; en1 ðtÞÞ þ k FðwðtÞÞ 6 0:
ð3:15Þ
aln1
It follows from (3.15) that D zðtÞ 0, by Lemma 2.6, we have zðtÞ P 0 which implies wðtÞ 6 /ðtÞ on ½0; 1: By the same way, it is easy to prove wðtÞ P wðtÞ on ½0; 1: So we obtain
wðtÞ 6 wðtÞ 6 /ðtÞ;
t 2 ½0; 1: ln1 l0
Consequently, FðwðtÞÞ ¼ f ðI BVP (2.10). Finally, by (3.16), we know
wðtÞ; I
ð3:16Þ ln1 l1
wðtÞ; . . . ; I
ln1 ln2
wðtÞ; wðtÞÞ; t 2 ½0; 1: Then wðtÞ is a positive solution of the
wðtÞ P wðtÞ P taln1 1 ¼ en1 ðtÞ:
ð3:17Þ
On the other hand, it follows from (3.17) and Lemma 2.4 that
wðtÞ ¼ k
Z
1
Kðt; sÞf ðIln1 l0 wðsÞ; Iln1 l1 wðsÞ; . . . ; Iln1 ln2 wðsÞ; wðsÞÞds
0
6 k
Z
1
Kðt; sÞf ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds
0
6 k M
Z
1
ð1 sÞal1 f ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds ¼ N0 ;
0
where
N 0 ¼ k M
Z
1 0
ð1 sÞal1 f ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds:
ð3:18Þ
X. Zhang et al. / Applied Mathematics and Computation 218 (2012) 8526–8536
8535
(3.17) and (3.18) yield
taln1 1 6 wðtÞ 6 N0 : Consequently,
Z
e0 ðtÞ 6 xðtÞ ¼ Iln1 wðtÞ 6 N 0
ðt sÞln1 1 l N0 ln1 ¼ Ntln1 ; ds ¼ n1 t Cðln1 Þ Cðln1 Þ
t
0
where N ¼
ln1 N0 Cðln1 Þ :
h
If we adopt a stronger condition instead of ðH3 Þ, then we have the following nice result. Corollary 3.1. Suppose that conditions ðH1 ÞðH2 Þ hold. Furthermore, suppose f satisfies the following condition: ðH03 Þ f ð1; 1; . . . ; 1Þ – 0, and
0<
Z
1
f ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds < þ1:
0
Then there is a constant k > 0 such that for any k 2 ðk ; þ1Þ, the BVP (1.1) has at least one positive solution xðtÞ, and xðtÞ satisfies
e0 ðtÞ 6 xðtÞ 6 Ntln1 ; where
N¼
k ln1 M Cðln1 Þ
Z
1
f ðe0 ðsÞ; e1 ðsÞ; . . . ; en2 ðsÞ; en1 ðsÞÞds:
0
An example consider the following problem
8 1 1 > < D52 xðtÞ þ k x29 ðtÞ þ D98 xðtÞ 2 þ D54 xðtÞ 8 ¼ 0; > :
9
5
11
xð0Þ ¼ D8 xð0Þ ¼ D4 xð0Þ ¼ 0;
D 8 xð1Þ ¼
0 < t < 1;
ð3:19Þ
pffiffi 11 2 D 8 12 : 2
Proof. Let 2
1
1
f ðx1 ; x2 ; x3 Þ ¼ x1 9 þ x2 2 þ x3 8 ; then ðH1 Þ holds, and for all r 2 ð0; 1Þ and for any ðx1 ; x2 ; x3 Þ 2 ð0; þ1Þ3 , 2
2
1
1
1
1
1
f ðrx1 ; rx2 ; . . . ; rxn Þ ¼ r 9 x1 9 þ r 2 x2 2 þ r8 x3 8 6 r2 f ðx1 ; x2 ; x3 Þ; which implies that ðH2 Þ also holds. 1 On the other hand, by direct calculation, we have f ð1; 1; 1Þ ¼ 3 – 0, e2 ðtÞ ¼ t4 ; and then
e0 ðtÞ ¼
Z
t
0
e1 ðtÞ ¼
Z 0
t
1 1
ðt sÞ4 s4 1 ds P 5 Cð54Þ Cð4Þ 7 1
ðt sÞ8 s4 ds P Cð18Þ
Z
t
0
Z
t
1
ðt sÞ4 sds ¼
0
Z
1 Cð54Þ
7
ðt sÞ8 s 1 ds ¼ 1 Cð18Þ Cð8Þ
t
9
1
s4 ðt sÞds ¼
0
Z
t
16 9 t4 t4 P ; 45Cð54Þ 3Cð54Þ 9
7
s8 ðt sÞds ¼
0
9
64t 8 t8 P 1 ; 9Cð18Þ Cð8Þ
thus
0<
Z
1
1 8
ð1 sÞ f ðe0 ðsÞ; e1 ðsÞ; e2 ðsÞÞds 6
Z
0
Z 6 0
0 1
2 4
1 3Cð54Þ
!29 1
1
s2 ð1 sÞ8 þ
1
1
2
!12
Cð18Þ
9
s4 ð1 sÞ 4 3Cð54Þ 1 8
9
1
!29
1
3 !12 9 1 18 s8 þ þ s4 5ds Cð18Þ 3 1
s16 ð1 sÞ8 þ s32 ð1 sÞ8 5ds < þ1:
Hence ðH3 Þ holds. Then by Theorem 3.1 there is a constant k > 0 such that for any k 2 ðk ; þ1Þ the BVP (3.19) has at least one positive solution xðtÞ , and there exists a constant N > 0 such that 1
5
t4 6 xðtÞ 6 Nt4 :
8536
X. Zhang et al. / Applied Mathematics and Computation 218 (2012) 8526–8536
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