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Multiplicity of periodic solutions for a delayed ratio-dependent predator–prey model with monotonic functional response and harvesting terms
Applied Mathematics and Computation 244 (2014) 878–894
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Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Multiplicity of periodic solutions for a delayed ratio-dependent predator–prey model with monotonic functional response and harvesting terms q Yong-Hong Fan ⇑, Lin-Lin Wang School of Mathematics and Statistics Sciences, Ludong University, Yantai, Shandong 264025, People’s Republic of China
a r t i c l e
i n f o
Keywords: Predator–prey model Functional response Positive periodic solution Generalized continuation theorem
a b s t r a c t The existence of four positive periodic solutions for a delayed ratio-dependent predator– prey model with monotonic functional response
8 xðtÞ > < x0 ðtÞ ¼ xðtÞ½aðtÞ bðtÞxðtÞ cðtÞg yðtÞ yðtÞ h1 ðtÞ; h i > : y0 ðtÞ ¼ yðtÞ f ðtÞg xðtsðtÞÞ dðtÞ h2 ðtÞ; yðtsðtÞÞ is established by using the generalized continuation theorem, where aðtÞ; bðtÞ; cðtÞ; dðtÞ; f ðtÞ; sðtÞ; h1 ðtÞ and h2 ðtÞ are all nonnegative periodic continuous functions with period x > 0. When h1 ðtÞ ¼ h2 ðtÞ 0, the conditions that guarantee the existence of four positive periodic solutions reduces exactly to that in Fan et al. (2004) [1]. And our main result also improves the corresponding results in Zhang and Hou (2010) [2] and Fan and Wang (2001) [3]. Ó 2014 Elsevier Inc. All rights reserved.
1. Introduction During the last few decades, much work has been devoted to dynamical characteristics of population models, among these models, predator–prey systems play an important role in population theory. To describe a predator–prey relationship, it is necessary to specify the rate of prey consumption by an average predator. This rate is known as the functional response. Usually, the functional response can be classified as: (i) prey dependent, when prey density alone determines the response; (ii) predator dependent, when both predator and prey populations affect the response. For a long time, predation theory was dominated by prey-dependent models and by Holling’s three type classification of these responses. Arditi and Ginzburg [4] stimulated recent interest in alternative forms for functional responses, with their suggestion that a ratio-dependent functional response was a better starting point for modeling predation. The ratio-dependence is a particular type of predator dependence in which the response only depends on the ratio of prey population size to predator population size. Recently, there is a growing explicit biological and physiological evidences [5,6] that in many situations, especially when predators have to search for food (and therefore have to share or compete for food), a more suitable general predator–prey theory should be based on the so-called ratio-dependent theory. q Supported by NSF of China (11201213), NSF of Shandong Province (ZR2010AM022), and the outstanding young and middle-aged scientists research award fund of Shandong Province (BS2011SF004). ⇑ Corresponding author. E-mail addresses: [email protected] (Y.-H. Fan), [email protected] (L.-L. Wang).
http://dx.doi.org/10.1016/j.amc.2014.07.046 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
879
Many previous investigations (such as [7,8]) focus on the following ratio-dependent predator–prey system
8 > < x0 ¼ rx 1 Kx g yx y; > : y0 ¼ y lg yx D ; where gðxÞ is the functional response function, which reflects the capture ability of the predator to prey. For more biological meaning, the reader may consult Freedman [9], May [10] and Murry [11]. On the other hand, when we consider the effect of human activity, usually, a harvesting term should be added. Xiao et al. [12] investigated the following Michaelis–Menten type ratio-dependent predator–prey system
8 axy h1 ; < x0 ¼ rx 1 Kx myþx fx 0 : y ¼ y d þ h2 ; myþx
ð1:1Þ
when h1 ¼ 0. They gave a nice systematic work on the global qualitative analysis of this system. In addition, most efforts on ratio-dependent predator–prey systems with harvesting terms concentrate on the models with Holling II functional response, such as [13,14] etc. In what follows we shall use the notation
f ¼
Z
1
x
x
f ðtÞdt; 0
f M ¼ max f ðtÞ; t2½0;x
f L ¼ min f ðtÞ; t2½0;x
where f is a periodic continuous function with period x. More recently, for a similar ratio-dependent model with Holling II functional response, in [2], by using the Mawhin’s continuation theorem combined with the homotopy invariance of the Brouwer degree, the authors obtained that. Theorem 1.1. Assume that the following conditions hold: qffiffiffiffiffiffiffiffiffiffiffiffi M M M (A1) aL > mc þ 2 b h1 , and qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L M L h M dM mM hM 2 a (A2) ðf dÞ aM1 > h2 mM þ 2 . bL
Then the equations
8 cðtÞxðtÞyðtÞ < x0 ðtÞ ¼ xðtÞ½aðtÞ bðtÞxðtÞ mðtÞyðtÞþxðtÞ h1 ðtÞ; h i f ðtÞxðt s ðtÞÞ : y0 ðtÞ ¼ yðtÞ dðtÞ h2 ðtÞ; mðtÞyðtsðtÞÞþxðtsðtÞÞ
ð1:2Þ
has at least four positive x-periodic solutions when sðtÞ 0. Remark. When sðtÞ is not always equal to zero, the methods they used do not work. When the harvesting terms vanish, from Theorem 1.1, we can obtain. Corollary 1.1. Assume that sðtÞ 0; h1 ðtÞ ¼ h2 ðtÞ 0 and the following conditions hold: (B1) aL >
c M m L
, and
(B2) ðf dÞ > 0.
Then the Eqs. (1.2) has at least one positive x-periodic solution. Remark. Comparing the above result with that in [3], we find that there is a gap between them. That is, when h1 ðtÞ ¼ h2 ðtÞ 0, the conditions for the existence here are the maximal–minimal type conditions, while the conditions used in [3] are of the average type. And this will be one of the main problems we want to resolve in the present paper. In this paper, we consider the following delayed ratio-dependent predator–prey system with monotonic functional response
8 xðtÞ > < x0 ðtÞ ¼ xðtÞ½aðtÞ bðtÞxðtÞ cðtÞg yðtÞ yðtÞ h1 ðtÞ; h i xðtsðtÞÞ > : y0 ðtÞ ¼ yðtÞ f ðtÞg yðt sðtÞÞ dðtÞ h2 ðtÞ;
ð1:3Þ
880
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
where xðtÞ and yðtÞ represent the densities of the prey population and predator population respectively; aðtÞ; bðtÞ; cðtÞ; dðtÞ; f ðtÞ; sðtÞ; h1 ðtÞ and h2 ðtÞ are all nonnegative periodic continuous functions with period x > 0. Here aðtÞ stands for prey intrinsic growth rate, dðtÞ stands for the death rate of the predator, cðtÞ and f ðtÞ stand for the conversion xðtÞ rates; the function xðtÞ½aðtÞ bðtÞxðtÞ represents the specific growth rate of the prey in the absence of predator; and g yðtÞ denotes the predator response function, which reflects the capture ability of the predator and satisfies the following monotonic condition (for short, we call them M): (i) g 2 C 1 ½0; þ1Þ; gð0Þ ¼ 0; (ii) g 0 ðuÞ > 0 for u 2 ð0; þ1Þ; (iii) limu!þ1 gðuÞ ¼ 1 (if limu!þ1 gðuÞ ¼ k – 1, obviously we can use c ðtÞ ¼ kcðtÞ; f ðtÞ ¼ kf ðtÞ, replace cðtÞ and f ðtÞ in (1.3)). When h1 ðtÞ ¼ h2 ðtÞ 0, system (1.3) becomes
8 xðtÞ > < x0 ðtÞ ¼ xðtÞ½aðtÞ bðtÞxðtÞ cðtÞg yðtÞ yðtÞ; h i xðtsðtÞÞ > : y0 ðtÞ ¼ yðtÞ f ðtÞg yðt sðtÞÞ dðtÞ :
ð1:4Þ
In [1], Fan etc. obtained a sufficient condition for the existence of at least one positive x-periodic solution, for convenience, we list it below. Theorem 1.2. Assume that the following conditions hold: (C1) a > mc, and (C2) f > d, where
m ¼ sup u>0
gðuÞ : u
ð1:5Þ
Then (1.4) has at least one positive x-periodic solution. In the present paper, we want to solve the following problems: (1) to gain the sufficient conditions for the existence of four positive periodic solutions of system (1.3), here sðtÞ may not always equal to zero; (2) to recuperate the gap as mentioned above when h1 ðtÞ ¼ h2 ðtÞ 0. The main method usually used to gain the existence of periodic solutions for periodic systems is the coincidence degree theory developed by Gaines and Mawhin [15], we refer to [16–25] and references cited therein. While in this paper, we will use the generalized continuation theorem. Although they are the same in nature, the latter is more convenient. And our main results generalize the corresponding results in [2,1]. The tree of this paper is: in the next section, we establish the main existence results; and in the third section, some suitable examples are given to illustrate the main theorem, also some remarks are given. 2. Multiplicity of positive periodic solutions In order to obtain the existence of positive periodic solutions for the system (1.3), first we make the following preparations. Let X Rn be an open bounded set with closure X and f 2 C 1 ðX; Rn Þ \ CðX; Rn Þ. For x 2 X, let J f ðxÞ denote the Jacobian determinant of f at x and Sf be the set of all critical points of f, i.e., Sf ¼ fx 2 X : J f ðxÞ ¼ 0g. For y 2 Rn n f ð@ X [ Sf Þ, i.e., y is a regular point of f, the degree of f at y is defined as
degff ; X; yg ¼
X
sgn J f ðxÞ:
x2f 1 ðyÞ
Let X and Z be two Banach spaces, L: DomL X ! Z be a Fredholm mapping of index zero and
N : X ½0; 1 ! Z; ðx; kÞ # Nðx; kÞ be a L-compact mapping in X ½0; 1. Notice that if L is a Fredholm mapping of index zero, then there exist continuous projections P : X ! X and Q : Z ! Z such that
ImP ¼ KerL and ImL ¼ KerQ ¼ ImðI Q Þ:
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
881
Theorem A [15, p. 26-30], generalized continuation theorem. Let X, Z, L and Q be mentioned above and J: ImQ ! KerL is an isomorphism, furthermore, assume that (a) for each k 2 ð0; 1Þ, x 2 @ X\DomL,
Lx – kNðx; kÞ; (b) for each x 2 @ X\KerL,
QNðx; 0Þ – 0; (c) degfJQNð; 0ÞjKerL ; X\KerL; 0g – 0. Then, for each k 2 ½0; 1Þ, the equation Lx ¼ kNðx; kÞ has at least one solution in X and equation Lx ¼ Nðx; 1Þ has at least one solution in X. For simplicity, we set
expfk1 g d u h2 expff xg; f1 ðuÞ ¼ f g u
expfk1 g d u h2 ; f2 ðuÞ ¼ f g u
u > 0;
u > 0;
where
k1 ¼ ln
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b
:
If there exist a positive number u such that f1 ðu Þ > 0 (a necessary condition that guarantee this conclusion is f > d), notice that f1 ðuÞ 2 Cð0; þ1Þ; f 1 ðþ1Þ ¼ 1 and f1 ð0þ Þ ¼ h2 expff xg, then the set fujf1 ðuÞ ¼ 0; u > 0g is bounded, nonempty and closed. Now we set
u1 ¼ minfujf1 ðuÞ ¼ 0; u > 0g and
u2 ¼ maxfujf1 ðuÞ ¼ 0; u > 0g: It is obvious that
0 < u1 < u < u2 < þ1: Since
expfk1 g d u h2 expff xg f1 ðuÞ ¼ f g u
expfk1 g d u h2 ¼ f2 ðuÞ; < fg u and
f2 ðuÞ 2 Cð0; þ1Þ;
f 2 ðþ1Þ ¼ 1 and f 2 ð0þ Þ ¼ h2 ;
f2 ðuÞ ¼ 0 has at least one root in ð0; u1 Þ, also it has at least one root in ðu2 ; þ1Þ. Now we are in a position to state one of our main results. Theorem 2.1. Assume that the following conditions hold: qffiffiffiffiffiffiffiffi (H1) ða mcÞ P ð1 þ expfaxgÞ b h1 ; (H2) 9u > 0 such that f1 ðu Þ > 0 and u2 P u1 expff xg, furthermore assume that f2 ðuÞ ¼ 0 has only one root in ð0; u1 Þ and one root in ðu2 ; þ1Þ. Then (1.3) has at least four positive x-periodic solutions. Proof. For biological meaning, we only need to study the positive solution of (1.3), so make change of variables
xðtÞ ¼ exp fx1 ðtÞg;
yðtÞ ¼ exp fx2 ðtÞg;
ð2:1Þ
882
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
then (1.3) becomes
x01 ðtÞ ¼ aðtÞ bðtÞ expfx1 ðtÞg cðtÞhðexp fx1 ðtÞ x2 ðtÞgÞ h1 ðtÞ exp fx1 ðtÞg;
ð2:2Þ
x02 ðtÞ ¼ f ðtÞg ðexp fx1 ðt sðtÞÞ x2 ðt sðtÞÞgÞ dðtÞ h2 ðtÞ exp fx2 ðtÞg: Here hðuÞ ¼ gðuÞ=u. In order to apply Theorem A to system (2.2), we take
n o X ¼ Z ¼ xðtÞ ¼ ðx1 ðtÞ; x2 ðtÞÞT 2 CðR; R2 Þ : xðt þ xÞ ¼ xðtÞ and denote
jjxjj ¼ jjðx1 ðtÞ; x2 ðtÞÞT jj ¼ max max jx1 ðtÞj; max jx2 ðtÞj : t2½0;x
t2½0;x
Then X and Z are Banach spaces when they are endowed with the norm k k. Set
2
aðtÞ mcðtÞ bðtÞ exp fx1 ðtÞg h1 ðtÞ exp fx1 ðtÞg
3
7 6 þkcðtÞðm hðexp fx1 ðtÞ x2 ðtÞgÞÞ 7 6 Nðx; kÞ ¼ 6 7; 4 kf ðtÞ½g ðexp fx1 ðt sðtÞÞ x2 ðt sðtÞÞgÞ gðexpfk1 x2 ðt sðtÞÞgÞ 5 þf ðtÞgðexpfk1 x2 ðt sðtÞÞgÞ dðtÞ h2 ðtÞ exp fx2 ðtÞg where m is defined as in (1.5), and
Lx ¼ x0 ;
Px ¼
1
x
Z
x
xðtÞdt;
x 2 X;
0
Qz ¼
1
x
Z
x
zðtÞdt;
z 2 Z:
0
Rx Evidently, KerL ¼ R2 ; ImL ¼ zjz 2 Z; 0 zðtÞdt ¼ 0 is closed in Z and dimKerL ¼ codim ImL ¼ 2. Hence, L is a Fredholm mapping of index zero. Furthermore, the generalized inverse (to L) K p : ImL ! KerP\DomL has the form
K p ðzÞ ¼
Z
t
zðsÞds
0
Z
1
x
x
0
Z
t
zðsÞdsdt:
0
Thus
2 1 Rx x
0
½aðtÞ mcðtÞ bðtÞ expfx1 ðtÞg h1 ðtÞ exp fx1 ðtÞg
3
6 7 þkcðtÞðm hðexp fx1 ðtÞ x2 ðtÞgÞÞdt 6 7 QNðx; kÞ ¼ 6 1 R x 7; 4 5 ½ kf ðtÞ ½ g ð exp x ðt s ðtÞÞ x ðt s ðtÞÞ Þ gðexpfk x ðt s ðtÞÞgÞ f g 1 2 1 2 x 0 þf ðtÞgðexpfk1 x2 ðt sðtÞÞgÞ dðtÞ h2 ðtÞ exp fx2 ðtÞgdt and
2Rt
3 ½aðsÞ mcðsÞ bðsÞ expfx1 ðsÞg h1 ðsÞ exp fx1 ðsÞg 0 7 6 7 6 þkcðsÞðm hðexp fx1 ðsÞ x2 ðsÞgÞÞds 7 6 K p ðI Q ÞNðx; kÞ ¼ 6 R 7 6 t ½kf ðsÞðg ðexp fx ðs sðsÞÞ x ðs sðsÞÞgÞ gðexpfk x ðs sðsÞÞgÞÞ 7 5 4 0 1 2 1 2 þf ðsÞgðexpfk1 x2 ðs sðsÞÞgÞ dðsÞ h2 ðsÞ exp fx2 ðsÞgds 3 2 1 Rx Rt ½aðsÞ mcðtÞ bðsÞ expfx1 ðsÞg h1 ðsÞ exp fx1 ðsÞg 0 x 0 7 6 7 6 þkcðsÞðm hðexp fx1 ðsÞ x2 ðsÞgÞÞdsdt 7 6 6 R R 7 x t 7 61 4 x 0 0 ½kf ðsÞðg ðexp fx1 ðs sðsÞÞ x2 ðs sðsÞÞgÞ gðexpfk1 x2 ðs sðsÞÞgÞÞ 5 þf ðsÞgðexpfk1 x2 ðs sðsÞÞgÞ dðsÞ h2 ðsÞ exp fx2 ðsÞgdsdt R x 1 x 2 0 ½aðsÞ mcðsÞ bðsÞ expfx1 ðsÞg h1 ðsÞ exp fx1 ðsÞg
2t
3
7 6 7 6 þkcðsÞðm hðexp fx1 ðsÞ x2 ðsÞgÞÞds 7 6 6 7: R x 7 6 t 1 ½ kf ðsÞ ð g ð exp x ðs s ðsÞÞ x ðs s ðsÞÞ Þ gðk x ðs s ðsÞÞÞ Þ f g 1 2 1 2 5 4 x 2 0 þf ðsÞgðk1 x2 ðs sðsÞÞÞ dðsÞ h2 ðsÞ exp fx2 ðsÞgds
Clearly, QN and K p ðI Q ÞN are continuous, moreover, QN X ½0; 1 ; K p ðI Q ÞNðX ½0; 1Þ are relatively compact for any open bounded set X X. Hence, N is L-compact on X ½0; 1. Now we reach the position to search for an appropriate open bounded subset X for the application of Theorem A. Corresponding to equation Lx ¼ kNðx; kÞ; k 2 ð0; 1Þ, we have
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
8 0 x1 ðtÞ ¼ k½aðtÞ mcðtÞ bðtÞ exp fx1 ðtÞg h1 ðtÞ exp fx1 ðtÞg > > > < þkcðtÞðm hðexp fx1 ðtÞ x2 ðtÞgÞÞ; > x02 ðtÞ ¼ k½kf ðtÞðg ðexp fx1 ðt sðtÞÞ x2 ðt sðtÞÞgÞ gðexpfk1 x2 ðt sðtÞÞgÞÞ > > : þf ðtÞgðexpfk1 x2 ðt sðtÞÞgÞ dðtÞ h2 ðtÞ exp fx2 ðtÞg:
883
ð2:3Þ
Suppose that xðtÞ ¼ ðx1 ; x2 ÞT 2 X is an x-periodic solution of system (2.3) for a certain k 2 ð0; 1Þ. Then there exist ni ; gi 2 ½0; x such that
xi ðgi Þ ¼ max xi ðtÞ for i ¼ 1; 2:
xi ðni Þ ¼ min xi ðtÞ; t2½0;x
ð2:4Þ
t2½0;x
And
Z
x
½aðtÞ mcðtÞ bðtÞ exp fx1 ðtÞg h1 ðtÞ exp fx1 ðtÞg þ kcðtÞðm hðexp fx1 ðtÞ x2 ðtÞgÞÞdt ¼ 0;
ð2:5Þ
0
Z
x
½kf ðtÞðg ðexp fx1 ðt sðtÞÞ x2 ðt sðtÞÞgÞ gðexpfk1 x2 ðt sðtÞÞgÞÞ þ f ðtÞgðexpfk1 x2 ðt sðtÞÞgÞ
0
dðtÞ h2 ðtÞ exp fx2 ðtÞgdt ¼ 0:
ð2:6Þ
From (2.5), we have
Z
x
½h1 ðtÞ exp fx1 ðtÞg þ bðtÞ exp fx1 ðtÞg ðaðtÞ mcðtÞÞdt ¼
0
Z
x
½kcðtÞðm hðexp fx1 ðtÞ x2 ðtÞgÞÞdt > 0;
0
which implies that
expfx1 ðn1 Þgh1 þ expfx1 ðg1 Þgb > a mc:
ð2:7Þ
The first equation of (2.3) implies that for any t 2 ½n1 ; x,
x1 ðtÞ ¼ x1 ðn1 Þ þ
Z
t
x01 ðtÞdt < x1 ðn1 Þ þ ax
n1
and for any t 2 ½0; n1 ,
x1 ðtÞ ¼ x1 ðn1 Þ þ
Z
t
x01 ðtÞdt ¼ x1 ðn1 Þ
n1
< x1 ðn1 Þ þ
Z
Z t
n1
x01 ðtÞdt
n1
½ð1 kÞmcðtÞ þ bðtÞ expfx1 ðtÞg þ h1 ðtÞ exp fx1 ðtÞg þ kcðtÞhðexp fx1 ðtÞ x2 ðtÞgÞdt
t
6 x1 ðn1 Þ þ
Z
x
½ð1 kÞmcðtÞ þ bðtÞ expfx1 ðtÞg þ h1 ðtÞ exp fx1 ðtÞg þ kcðtÞhðexp fx1 ðtÞ x2 ðtÞgÞdt 0
¼ x1 ðn1 Þ þ ax: Thus, for any t 2 ½0; x, we have
x1 ðtÞ < x1 ðn1 Þ þ ax; especially,
x1 ðg1 Þ < x1 ðn1 Þ þ ax;
ð2:8Þ
thus (2.7) and (2.8) show that
b expfaxg expf2x1 ðn1 Þg ða mcÞ expfx1 ðn1 Þg þ h1 > 0; then (H1) leads to
x1 ðn1 Þ P ln
ða mcÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1 expfaxg 2b expfaxg
or
x1 ðn1 Þ 6 ln
ða mcÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1 expfaxg 2b expfaxg
:¼ ln l1
:
Using (2.7) and (2.8), we have
b expf2x1 ðg1 Þg ða mcÞ expfx1 ðg1 Þg þ h1 expfaxg > 0;
ð2:9Þ
884
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
by (H1), we know
x1 ðg1 Þ > ln
ða mcÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1 expfaxg
ð2:10Þ
2b
or
x1 ðg1 Þ < ln
ða mcÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1 expfaxg 2b
þ
:¼ ln k1 :
ð2:11Þ
On the other hand, from (2.5), we also have
Z
Z
x
½h1 ðtÞ exp fx1 ðtÞg þ bðtÞ exp fx1 ðtÞg aðtÞdt ¼
0
x
½mcðtÞ þ kcðtÞðm hðexp fx1 ðtÞ x2 ðtÞgÞÞdt
0
Z
¼
x
½ð1 kÞmcðtÞ kcðtÞhðexp fx1 ðtÞ x2 ðtÞgÞdt < 0;
0
which implies that
h1 exp fx1 ðg1 Þg þ b exp fx1 ðn1 Þg a < 0; then (2.8) gives
b exp f2x1 ðn1 Þg a exp fx1 ðn1 Þg þ h1 expfaxg < 0; by (H1), we have
ln
k1
¼: ln
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b
< x1 ðn1 Þ < ln
aþ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg
:
2b
ð2:12Þ
Similarly, we can obtain
ln
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b expfaxg
þ
< x1 ðg1 Þ < ln
þ
þ
aþ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b expfaxg
þ
:¼ ln l1 :
ð2:13Þ
We claim that k1 < k1 ; l1 < l1 and k1 6 l1 . Constructing a function
pffiffiffiffiffiffiffiffiffiffiffiffiffi qðxÞ ¼ x x2 a;
a > 0 and x >
pffiffiffi a;
then
qðxÞ q0 ðxÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi < 0; x2 a which implies that
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b
<
ða mcÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1 expfaxg 2b
ð2:14Þ
;
also in view of
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b
>
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b
ð2:15Þ
:
By (2.11), (2.12), (2.14) and (2.15), we know k1
¼
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4bh1 expfaxg 2b
<
ða mcÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1 expfaxg 2b
In addition, constructing two functions
hðxÞ ¼
aþ
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 bx ; x
0
a2 ; b
and
sðxÞ ¼ x þ
pffiffiffiffiffiffiffiffiffiffiffiffiffi x2 a;
it is easy to verify
a > 0; x >
pffiffiffi a;
a > 0; b > 0
þ
¼ k1 :
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
885
0
h ðxÞ < 0 and s0 ðxÞ > 0; thus l1
¼
<
ða mcÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1 expfaxg
2b expfaxg qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ þ ða mcÞ2 4b h1 expfaxg
2b expfaxg qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a þ a2 4b h1 expfaxg þ < ¼ l1 : 2b expfaxg þ
þ
Now we show that k1 6 l1 when (H1) holds true. In fact, k1 6 l1 is equivalent to
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ ða mcÞ2 4b h1 expfaxg
6
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ þ ða mcÞ2 4b h1 expfaxg 2b expfaxg
2b
;
that is,
ða mcÞðexpfaxg 1Þ 6
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1 expfaxgðexpfaxg þ 1Þ;
obviously, the above inequality is satisfied when (H1) hold true. From (2.9) and (2.11)–(2.13), we have for any t 2 ½0; x,
þ
ln k1 < x1 ðtÞ < ln k1
or
þ
ln l1 < x1 ðtÞ < ln l1 :
By (2.6) and the monotonicity of the function g, we have
Z 0
x
½f ðtÞgðexp fk1 x2 ðt sðtÞÞgÞ dðtÞ h2 ðtÞ exp fx2 ðtÞgdt Z x kf ðtÞðg ðexp fx1 ðt sðtÞÞ x2 ðt sðtÞÞgÞ gðexpfk1 x2 ðt sðtÞÞgÞÞdt < 0: ¼
ð2:16Þ
0
Then
f gðexpfk1 x2 ðg2 ÞgÞ < d þ h2 expfx2 ðn2 Þg: Notice that
x2 ðtÞ ¼ x2 ðn2 Þ þ
Z
t
x02 ðsÞds;
n2
thus by the second equation of (2.3) and notice that for any t 2 ½0; x; x1 ðtÞ P k1 , we know for any t 2 ½n2 ; x,
x2 ðtÞ ¼ x2 ðn2 Þ þ < x2 ðn2 Þ þ þ k2
x02 ðsÞds
n2 Z t
kf ðsÞgðexpfk1 x2 ðs sðsÞÞgÞds
t
f ðtÞ½gðexpfx1 ðs sðsÞÞ x2 ðs sðsÞÞgÞ gðexpfk1 x2 ðs sðsÞÞgÞds
n2
6 x2 ðn2 Þ þ þk
t
n2
Z
Z
Z
Z
t
kf ðsÞgðexpfk1 x2 ðs sðsÞÞgÞds n2
t
f ðsÞ½gðexpfx1 ðs sðsÞÞ x2 ðs sðsÞÞgÞ gðexpfk1 x2 ðs sðsÞÞgÞds
n2
¼ x2 ðn2 Þ þ k
Z
t
f ðsÞgðexpfx1 ðs sðsÞÞ x2 ðs sðsÞÞgÞds
n2
6 x2 ðn2 Þ þ f x: Here we use the following properties of the function g,
gðuÞ 6 1 and
@gðexpfx1 x2 gÞ > 0: @x1
ð2:17Þ
886
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
On the other hand, for any t 2 ½0; n2 , we have
x2 ðtÞ ¼ x2 ðn2 Þ þ
Z
t
x02 ðsÞds
n2
¼ x2 ðn2 Þ þ
Z
n2
t
< x2 ðn2 Þ þ
Z
x02 ðsÞ ds
n2
½dðtÞ þ h2 ðtÞ exp fx2 ðtÞgds
t
6 x2 ðn2 Þ þ
Z
x
½dðtÞ þ h2 ðtÞ exp fx2 ðtÞgds
0
¼ x2 ðn2 Þ þ f x; which implies that for any t 2 ½0; x,
x2 ðtÞ < x2 ðn2 Þ þ f x: Therefore
x2 ðg2 Þ < x2 ðn2 Þ þ f x: So
f gðexpfk1 x2 ðg2 ÞgÞ < d þ h2 expff xg expfx2 ðg2 Þg
ð2:18Þ
f gðexpfk1 f x x2 ðn2 ÞgÞ < d þ h2 expfx2 ðn2 Þg:
ð2:19Þ
and
By (2.18), we have
f gðexpfk1 x2 ðg2 ÞgÞ expfx2 ðg2 Þg d expfx2 ðg2 Þg h2 expff xg < 0: The above inequality and (H2) implies that
x2 ðg2 Þ > ln u2
or x2 ðg2 Þ < ln u1 :
ð2:20Þ
Set
" f3 ðuÞ ¼ f g
! # expfk1 f xg d u h2 ; u
u > 0;
then
f3 ðuÞ ¼ f1 ðu expff xgÞ; thus f3 ðuÞ < 0 implies
x2 ðn2 Þ > ln u2 f x or x2 ðn2 Þ < ln u1 f x:
ð2:21Þ
By (2.20) and (2.21), we have for any t 2 ½0; x, þ
x2 ðtÞ < ln u1 :¼ ln k2
or x2 ðtÞ > ln u2 f x :¼ ln l2 :
ð2:22Þ
And the second equation of (2.6) implies
Z
x
½dðtÞ þ h2 ðtÞ exp fx2 ðtÞgdt ¼
0
Z 0
þ
x
kf ðtÞðg ðexp fx1 ðt sðtÞÞ x2 ðt sðtÞÞgÞ gðexpfk1 x2 ðt sðtÞÞgÞÞdt Z
x
f ðtÞgðexpfk1 x2 ðt sðtÞÞgÞdt
0
< f x;
ð2:23Þ
which leads to
x2 ðg2 Þ > ln
h2 f d
ð2:24Þ
:
Thus for any t 2 ½0; x,
x2 ðtÞ > ln
h2 f d
f x :¼ ln k2 :
ð2:25Þ
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
887
On the other hand, from (2.23), we also have
Z
x
f ðtÞg ðexp fx1 ðt sðtÞÞ x2 ðt sðtÞÞgÞdt þ l1 < fg x; exp fx2 ðn2 Þg
dx <
0
which shows that þ
x2 ðn2 Þ < ln
l1 : d f
g 1
Thus þ
x2 ðg2 Þ < ln
l1 þ f x :¼ ln lþ2 :
g 1
ð2:26Þ
d f
We claim that
þ
k2 < k2 Since
þ f1 ðk2 Þ
þ
and l2 < l2 :
ð2:27Þ
¼ f1 ðu1 Þ ¼ 0,
"
! # expfk1 g þ k2 d þ k2 þ < f d k2 ;
h2 expff xg ¼ f g
thus
h2 expff xg
þ
k2 >
>
f d
h2 expff xg f d
¼ k2 :
On the other hand, f1 ðu2 Þ ¼ f1 ðl2 expff xgÞ ¼ 0, therefore
"
expfk1 g
!
#
d l2 l2 expff xg " ! # þ l1 d l2 ; < fg l2 expff xg
0 < h2 ¼ f g
which shows þ
l1
l2 expff xg
>g
1
! d f
;
thus þ
l2 <
g 1
d f
þ
l1
expff xg
<
l1 expff xg ¼ lþ2 : g 1 d f
This complete the claim (2.27). Note that the condition (H2) implies that
u1 6 u2 expff xg; that is þ
k2 6 l2 : From (2.22) and (2.25)–(2.27), we know for any t 2 ½0; x,
þ
ln k2 < x2 ðtÞ < ln k2
or
þ
ln l2 < x2 ðtÞ < ln l2 :
Now we construct four open sets:
þ
ln k2 < x2 ðtÞ < ln k2 g;
þ
ln l2 < x2 ðtÞ < ln l2 g;
X1 ¼ fðx1 ðtÞ; x2 ðtÞÞjðx1 ðtÞ; x2 ðtÞÞ 2 X;
ln k1 < x1 ðtÞ < ln k1 ;
X2 ¼ fðx1 ðtÞ; x2 ðtÞÞjðx1 ðtÞ; x2 ðtÞÞ 2 X;
ln k1 < x1 ðtÞ < ln k1 ;
þ
þ
888
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
þ
ln k2 < x2 ðtÞ < ln k2 g;
þ
ln l2 < x2 ðtÞ < ln l2 g:
X3 ¼ fðx1 ðtÞ; x2 ðtÞÞjðx1 ðtÞ; x2 ðtÞÞ 2 X;
ln l1 < x1 ðtÞ < ln l1 ;
X4 ¼ fðx1 ðtÞ; x2 ðtÞÞjðx1 ðtÞ; x2 ðtÞÞ 2 X;
ln l1 < x1 ðtÞ < ln l1 ;
þ
þ
þ
þ
þ
þ
Clearly, li ; li and ki ; ki (i ¼ 1; 2) are independent of k, from k1 6 l1 and k2 6 l2 , we know that Xi \ Xj ¼ /; i – j ði; j ¼ 1; 2; 3; 4Þ. Now we show under the assumptions of Theorem 2.1, the system of algebraic equations
8 < ða mcÞu bu2 h1 ¼ 0; h i : f g expfk1 g d v h2 ¼ 0 v
ð2:28Þ
has a unique root in every Xi ði ¼ 1; 2; 3; 4Þ. In fact, from the first equation of (2.28) and (H1), we know the equation
ða mcÞu bu2 h1 ¼ 0 þ þ e 1 and u e2 (u e1 < u e 2 ). We claim u e 1 2 ðk e has only two positive solutions, namely, u 1 ; k1 Þ and u 2 2 ðl1 ; l1 Þ. The reason is,
e1 ¼ u >
>
e1 ¼ u <
e2 ¼ u
ða mcÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1
2bffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 a a 4b h1 2b qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi a a2 4b h1 expfaxg 2b ða mcÞ
¼ k1 ;
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1
2b qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ða mcÞ ða mcÞ 4b h1 expfaxg 2b ða mcÞ þ
þ
¼ k1 :
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ða mcÞ 4b h1
2bffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a þ a2 4b h1 a þ a2 4b h1 expfaxg þ < ¼ l1 ; < 2b 2b
e2 ¼ u >
ða mcÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi c 2 a 2m 4b h1
2b qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi c 2 ða mcÞ þ a 2m 4b h1 expfaxg 2b expf2axg
¼ l1 :
On the other hand, from the second equation of (2.28) and (H2), we know that the equation
expfk1 g d v h2 ¼ 0 fg
v
e 1 and v e 2 (v e1 < v e 2 ), obviously, v e 1 < u1 ¼ kþ e has only two roots in ð0; u1 Þ [ ðu2 ; þ1Þ, namely, v 2 and v 2 > u2 ¼ l2 expff xg. We claim that
ve 1 > k2
and
ve 2 < lþ2 :
Notice that
expfk1 g d vei ; 0 < h2 ¼ f g vei thus
h2 < f d vei ; and
i ¼ 1; 2
i ¼ 1; 2;
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
d < fg
þ expfk1 g l < fg 1 ; vei vei
889
i ¼ 1; 2;
which implies that
vei >
h2 f d
>
h2 expff xg f d
¼ k2 ;
i ¼ 1; 2
and þ
vei <
þ
l1 l expff xg < 1 ¼ lþ2 ; g 1 d g 1 d f
i ¼ 1; 2:
f
So far, we reach the conclusion of the claim. According to the above estimation of priori boundary, it is clear that Xi satisfies the condition (a) of Theorem A. When
x ¼ ðx1 ; x2 ÞT 2 @ Xi \ KerL ¼ @ Xi \ R2 ; x is a constant vector in R2 , thus
" QNðx; 0Þ ¼
ða mcÞ b exp fx1 g h1 exp fx1 g f g ðexpfk1 x2 gÞ d h2 expfx2 g
# – 0:
Furthermore, let J: ImQ !KerL; x ! x, in view of the assumptions in Theorem 2.1, it is easy to see that
deg fJQN; Xi \ KerL; 0g – 0: By now we know that Xi verifies all the requirements of Theorem A. thus system (2.2) has at least four x-periodic solutions. By the medium of (2.1), we derive that (1.3) has at least four positive x-periodic solutions. The proof is complete. h As a degenerate case, we can obtain from Theorem 2.1, when a > mc and f > d, system (1.3) has at least one positive xperiodic solution provided that h1 ðtÞ ¼ h2 ðtÞ 0. Obviously, this condition is the same as that in Theorem 1.2, so our results generalize this conclusion in [1]. From the proof of Theorem 2.1, we can easily obtain that. Corollary 2.1. Assume that the following conditions hold: qffiffiffiffiffiffiffiffi (1) ða mcÞ P ð1 þ expfaxgÞ b h1 ; (2) f1 ðuÞ ¼ 0 has only two positive roots, namely, u1 ; u2 and u2 P u1 expff xg, furthermore assume that f2 ðuÞ ¼ 0 has only two positive roots. Then (1.3) has at least four positive x-periodic solutions.
Theorem 2.2. Assume that sðtÞ 0 and the following conditions hold: qffiffiffiffiffiffiffiffiffiffiffiffi M M (H10 ) ða mcÞL > 2 b h1 ; 0 (H2 ) f2 ðuÞ ¼ 0 has only two positive roots, furthermore assume that the equation f4 ðuÞ ¼ 0 has at least two positive roots, where
expfk2 g M M d u h2 ; f4 ðuÞ ¼ f L g u qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L L 2 aM ðaM Þ 4b h1 : k2 ¼ ln L 2b Then (1.3) has at least four positive x-periodic solutions. Proof. In (2.3) and (2.4), replace k1 by k2 , then it is obvious that x0i ðni Þ ¼ 0 and x0i ðgi Þ ¼ 0; i ¼ 1; 2. Therefore
aðn1 Þ mcðn1 Þ bðn1 Þ exp fx1 ðn1 Þg h1 ðn1 Þ exp fx1 ðn1 Þg þ kcðn1 Þðm hðexp fx1 ðn1 Þ x2 ðn1 ÞgÞÞ ¼ 0;
ð2:29Þ
kf ðn2 Þðgðexpfx1 ðn2 Þ x2 ðn2 ÞgÞ gðexpfk2 x2 ðn2 ÞgÞÞ þ gðexpfk2 x2 ðn2 ÞgÞ dðn2 Þ h2 ðn2 Þ exp fx2 ðn2 Þg ¼ 0 ð2:30Þ
890
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
and
aðg1 Þ mcðg1 Þ bðg1 Þ exp fx1 ðg1 Þg h1 ðg1 Þ exp fx1 ðg1 Þg þ kcðg1 Þðm hðexp fx1 ðg1 Þ x2 ðg1 ÞgÞÞ ¼ 0;
ð2:31Þ
kf ðg2 Þðgðexpfx1 ðg2 Þ x2 ðg2 ÞgÞ gðexpfk2 x2 ðg2 ÞgÞÞ þ gðexpfk2 x2 ðg2 ÞgÞ dðg2 Þ h2 ðg2 Þ exp fx2 ðg2 Þg ¼ 0: ð2:32Þ By (2.29), we have
aðn1 Þ mcðn1 Þ bðn1 Þ exp fx1 ðn1 Þg h1 ðn1 Þ exp fx1 ðn1 Þg ¼ kcðn1 Þðm hðexp fx1 ðn1 Þ x2 ðn1 ÞgÞÞ < 0; which leads to M
M
ða mcÞL b exp fx1 ðn1 Þg h1 exp fx1 ðn1 Þg < 0; thus (H10 ) gives
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi h i2 M M L ða mcÞ þ ða mcÞ 4b h1 L
x1 ðn1 Þ > ln
M
2b
:¼ ln l3
ð2:33Þ
or
ða mcÞL x1 ðn1 Þ < ln
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi h i2 M M ða mcÞL 4b h1 M
2b
þ
:¼ ln k3 :
ð2:34Þ
From (2.29), we obtain
aðn1 Þ bðn1 Þ exp fx1 ðn1 Þg h1 ðn1 Þ exp fx1 ðn1 Þg ¼ mcðn1 Þ kcðn1 Þðm hðexp fx1 ðn1 Þ x2 ðn1 ÞgÞÞ ¼ mð1 kÞcðn1 Þ þ kcðn1 Þhðexp fx1 ðn1 Þ x2 ðn1 ÞgÞ > 0; therefore L
L
b expf2x1 ðn1 Þg aM exp fx1 ðn1 Þg þ h1 < 0; by using (H10 ) once more, we know
ln
k3
¼: ln
aM
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L L 2 ½aM 4b h1 2b
L
< x1 ðn1 Þ < ln
aM þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L L 2 ½aM 4b h1 L
2b
þ
:¼ ln l3 :
ð2:35Þ
It is easy to see that
þ
þ
k3 < k3 < l3 < l3 ; thus by (2.33)–(2.35), we have
þ
ln k3 < x1 ðn1 Þ < ln k3
or
þ
ln l3 < x1 ðn1 Þ < ln l3 :
Similarly, by (2.31) and the above analysis, we can obtain
þ
ln k3 < x1 ðg1 Þ < ln k3
or
þ
ln l3 < x1 ðg1 Þ < ln l3 :
Which implies that for any t 2 ½0; x,
þ
ln k3 < x1 ðtÞ < ln k3
or
þ
ln l3 < x1 ðtÞ < ln l3 :
ð2:36Þ
From (2.30) and (2.36), we have
f ðn2 Þgðexpfk2 x2 ðn2 ÞgÞ dðn2 Þ h2 ðn2 Þ exp fx2 ðn2 Þg ¼ kf ðn2 Þðgðexpfx1 ðn2 Þ x2 ðn2 ÞgÞ gðexpfk2 x2 ðn2 ÞgÞÞ < 0; thus M
M
f L gðexpfk2 x2 ðn2 ÞgÞ d h2 exp fx2 ðn2 Þg < 0: þ
Considering (H20 ), we denote the two positive solutions of f4 ðuÞ ¼ 0 as k4 (the minimal one) and l4 (the maximal one), then
x2 ðn2 Þ > ln l4
þ
or x2 ðn2 Þ < ln k4 :
ð2:37Þ
891
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
Notice that (2.30) also implies
dðn2 Þ < dðn2 Þ þ h2 ðn2 Þ exp fx2 ðn2 Þg ¼ kf ðn2 Þðgðexpfx1 ðn2 Þ x2 ðn2 ÞgÞ gðexpfk2 x2 ðn2 ÞgÞÞ þ gðexpfk2 x2 ðn2 ÞgÞ þ expfx1 ðn2 Þg l3 < f ðn2 Þg < f ðn2 Þ; < f ðn2 Þg expfx2 ðn2 Þg expfx2 ðn2 Þg which shows that L
þ
h2
ln k4 ¼: ln
< x2 ðn2 Þ < ln
M
ðf dÞ
g 1
l3 þ L :¼ ln l4 :
ð2:38Þ
d f
By similar analysis as above, also in view of (2.32), we have
x2 ðg2 Þ > ln l4
þ
or x2 ðg2 Þ < ln k4
and
þ
ln k4 < x2 ðg2 Þ < ln l4 : Obviously,
þ
þ
k4 < k4 < l4 < l4 ; thus for any t 2 ½0; x, we have
þ
ln k4 < x2 ðtÞ < ln k4
or
þ
ln l4 < x2 ðtÞ < ln l4 :
ð2:39Þ
The rest of the proof is similar to that of Theorem 2.1, we omit it here. h
3. Examples and remarks In this section, we give some examples to illustrate our main result. First, we set
f5 ðuÞ ¼ mM du2 where
k3 ¼
a
h i f d k3 mM h2 expff xg u þ h2 k3 expff xg;
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b
:
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi If ðf dÞk3 > h2 mM expff xg þ 2 mM dh2 k3 expff xg, then the equation f5 ðuÞ ¼ 0 has only two roots, denote the two positive b2; u b1 ðu b1 < u b 2 Þ. Obviously, solution of equation f5 ðuÞ ¼ 0 as u
h b1 ¼ u
;
2mM d h
b2 ¼ u
ffi i rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi h i2 M M M f d k3 m h2 expff xg f d k3 m h2 expff xg 4m dh2 k3 expff xg ffi i rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi h i2 f d k3 mM h2 expff xg þ f d k3 mM h2 expff xg 4mM dh2 k3 expff xg :
2mM d
For system (1.2), we have the following conclusion. Theorem 3.1. Assume that the following conditions hold: qffiffiffiffiffiffiffiffi ; ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (D1) a P mc þ ð1 þ expfaxgÞ b h1q (D2) ðf dÞk3 > h2 mM expff xg þ 2 mM dh2 k3 expff xg: Then the Eqs. (1.2) has at least four positive x-periodic solutions. Remark. When h1 ðtÞ ¼ h2 ðtÞ 0, the conditions in Theorem 3.1 reduce to a > in [3].
c m
and f d > 0, it is exactly the same as that
892
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
Example 3.1. Consider the following model
8 h i ð4pt Þ yðtÞ > < x0 ðtÞ ¼ xðtÞ 2 cos ð4pt Þ 1sin10ð4ptÞ xðtÞ 1þcos ð1 þ cos ð4ptÞÞ; 100 xðtÞþ0:01yðtÞ h i xðtÞ > : y0 ðtÞ ¼ yðtÞ 1þsin10ð4ptÞ þ ð1 þ sin ð4ptÞÞ xðtÞþ0:01yðtÞ 1sin10ð4ptÞ : It is easy to verify that all the conditions in Theorem 2.1 hold true. Therefore, this system has at least four positive 1=2periodic solutions. Remark. The above example does not satisfy the conditions in Theorem 1.1. On the other hand, the Theorem 2.1 in [2] can also be generalized to the following theorem. Theorem 3.2. Assume that the following conditions hold: qffiffiffiffiffiffiffiffiffiffiffiffi M M M (E1) aL P mc þ 2 b h1 , and qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L M M M (E2) ðf dÞ k4 > h2 mM þ 2 mM d h2 k4 , where
k4 ¼
aL
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 M M ðaL Þ 4b h1 M
2b
:
Then the equations (1.2) (when sðtÞ 0) has at least four positive x-periodic solutions. Remark. When (A2) holds, (E2) also holds true. Example 3.2. Consider the following system
8 2 ðtÞyðtÞ < x0 ðtÞ ¼ xðtÞ½aðtÞ bðtÞxðtÞ m2cðtÞx h1 ðtÞ; ðtÞy2 ðtÞþx2 ðtÞ h i eðtÞx2 ðtsðtÞÞ : y0 ðtÞ ¼ yðtÞ dðtÞ h2 ðtÞ: m2 ðtÞy2 ðtsðtÞÞþx2 ðtsðtÞÞ
ð3:1Þ
In order to give an existence result, we set
2 2 2 2 f6 ðuÞ ¼ mM du3 þ mM h2 expfexgu2 k5 e d u þ k5 h2 expfexg and
2 mM h2 expfexg þ e¼ u
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 2 4 2 ðmM Þ h2 2 expf2exg þ 3ðmM Þ dk5 e d 2
3ðmM Þ d
;
where
k5 ¼
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b
:
e Þ < 0; f 6 ðuÞ ¼ 0 has only two positive roots, namely, u1 ; u2 ðu1 < u2 Þ, from Corollary 2.1, we have When f6 ð u Theorem 3.3. Assume that the following conditions hold: qffiffiffiffiffiffiffiffi (F1) a P 12 mc þ ð1 þ expfaxgÞ b h1 ; e Þ < 0 and u2 P u1 expfexg. (F2) f6 ð u Then (3.1) has at least four positive x-periodic solutions. c Remark. Notice that when h1 ðtÞ ¼ h2 ðtÞ 0, the conditions in Theorem 3.3 now reduce to a 2m > 0 and e d > 0, in this þ þ case, k1 ¼ k1 ¼ 0, k2 ¼ k2 ¼ 0, thus the bounded open sets X1 ; X2 ; X3 now disappear, only X4 is left, and the corresponding algebraic equations has only one root, therefore, we can only obtain the existence of at least one positive solution. This improves the corresponding conclusion in [26]. Similarly, from Theorem 2.2, we can deduce.
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
893
Theorem 3.4. Assume thatsðtÞ 0 and the following conditions hold: qffiffiffiffiffiffiffiffiffiffiffiffi M M c L (G3) a 2m > 2 b h1 ; (G4) f7 ðu Þ < 0, where
2 M 2 M 2 L 2 M f7 ðuÞ ¼ mM d u3 þ mM h2 u2 k6 ðe dÞ u þ k6 h2 and
k6 ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L L 2 aM ðaM Þ 4b h1 L
2b
;u ¼
2 M mM h2
þ
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ðmM Þ
4
M
h2
2
2
2 M 2
þ 3ðmM Þ d k5 ðe dÞ
3ðmM Þ d
L
:
Then (3.1) has at least four positive x-periodic solutions. Compare the condition (H1) with (H10 ) and (H2) with (H20 ), also from some numerical examples, we find that the periodic
x has no influence on the topological structure for the solution of system (1.3), thus we may conclude. Conjecture. Assume that the following conditions hold: qffiffiffiffiffiffiffiffi (I1) ða mcÞ > 2 b h1 ; (I2) the equation f5 ðuÞ ¼ 0 has at least two positive roots, where
expfk7 g d u h2 ; f5 ðuÞ ¼ f g u
k7 ¼ ln
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 2b
:
Then (1.3) has at least four positive x-periodic solutions.
Acknowledgments The authors are grateful to the reviewers for their careful reading and valuable suggestions, which helps us to improve this paper. References [1] Y.H. Fan, W.T. Li, L.L. Wang, Periodic solutions of delayed ratio-dependent predator–prey models with monotonic or nonmonotonic functional responses, Nonlinear Anal.: Real World Appl. 5 (2) (2004) 247–263. [2] Z.Q. Zhang, Z.T. Hou, Existence of four positive periodic solutions for a ratio-dependent predator prey system with multiple exploited (or harvesting) terms, Nonlinear Anal.: Real World Appl. 11 (3) (2010) 1560–1571. [3] M. Fan, K. Wang, Periodicity in a delayed ratio-dependent predator–prey system, J. Math. Anal. Appl. 262 (2001) 179–190. [4] R. Arditi, L.R. Ginzburg, Coupling in predator–prey dynamics: ratio-dependence, J. Theor. Biol. 139 (1989) 311–326. [5] R. Arditi, H. Saiah, Empirical evidence of the role of heterogeneity in ratio-dependent consumption, Ecology 73 (1992) 1544–1551. [6] A.P. Gutierrez, The physiological basis of ratio-dependent predator–prey theory: a metabolic pool model of Nicholson’s blowflies as an example, Ecology 73 (1992) 1552–1563. [7] C. Jost, O. Arino, R. Arditi, About deterministic extinction in ratio-dependent predator–prey models, Bull. Math. Biol. 61 (1999) 19–32. [8] Y. Kuang, E. Beretta, Global qualitative analysis of a ratio-dependent predator–prey system, J. Math. Biol. 36 (1998) 389–406. [9] H.I. Freedman, Deterministic Mathematical Models in Population Ecology, Marcel Dekker, New York, 1980. [10] R.M. May, Complexity and Stability in Model Ecosystems, Princeton University Press, Princeton, NJ, 1973. [11] J.D. Murry, Mathematical Biology, Springer-Verlag, New York, 1989. [12] D. Xiao, W. Li, M. Han, Dynamics in a ratio-dependent predator–prey model with predator harvesting, J. Math. Anal. Appl. 324 (1) (2006) 14–29. [13] D. Xiao, S. Ruan, Bogdanov–Takens bifurcations in predator–prey systems with constant rate harvesting, Fields Inst. Commun. 21 (1999) 493–506. [14] D. Xiao, L. Jennings, Bifurcations of a ratio-dependent predator–prey system with constant rate harvesting, SIAM Appl. Math. 65 (2005) 737–753. [15] R.E. Gaines, J.L. Mawhin, Coincidence Degree and Nonlinear Differential Equations, Lecture Note in Mathematics, Springer, Berlin, 1977. [16] J.K. Hale, J. Mawhin, Coincidence degree and periodic solutions of neutral equations, J. Differ. Equ. 15 (2) (1974) 295–307. [17] Y.M. Chen, Multiple periodic solutions of delayed predator–prey systems with type IV functional responses, Nonlinear Anal.: Real World Appl. 5 (1) (2004) 45–53. [18] W.T. Li, Y.H. Fan, Periodic solutions in a delayed predator–prey model with nonmonotonic functional response, Discrete Contin. Dyn. Syst. – Ser. B 8 (1) (2007) 175–185. [19] Y.K. Li, Periodic solutions of a periodic delay predator–prey system, Proc. Am. Math. Soc. 127 (1999) 1331–1335. [20] Y.K. Li, Y. Kuang, Periodic solutions of periodic delay Lotka–Volterra equations and systems, J. Math. Anal. Appl. 255 (2001) 260–280. [21] B.R. Tang, Y. Kuang, Existence, uniqueness and asymptotic stability of periodic solutions of periodic functional differential systems, Tohoku Math. J. 49 (1997) 217–239. [22] L.L. Wang, W.T. Li, Existence and global stability of positive periodic solutions of a predator–prey system with delays, Appl. Math. Comput. 146 (1) (2003) 67–185. [23] T. Zhao, Y. Kuang, H.L. Smith, Global existence of periodic solutions in a class of delayed Gause-type predator–prey systems, Nonlinear Anal. TMA 28 (1997) 1373–1394.
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[24] S.J. Yang, B. Shi, Periodic solution for a three-stage-structured predator–prey system with time delay, J. Math. Anal. Appl. 341 (1) (2008) 287–294. [25] X.Q. Ding, J.F. Jiang, Periodicity in a generalized semi-ratio-dependent predator–prey system with time delays and impulses, J. Math. Anal. Appl. 360 (1) (2009) 223–234. [26] L.L. Wang, W.T. Li, Periodic solutions and permanence for a delayed nonautonomous ratio-dependent predator–prey model with Holling type functional response, J. Comput. Appl. Math. 162 (2) (2004) 341–357.
Contents lists available at ScienceDirect
Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc
Multiplicity of periodic solutions for a delayed ratio-dependent predator–prey model with monotonic functional response and harvesting terms q Yong-Hong Fan ⇑, Lin-Lin Wang School of Mathematics and Statistics Sciences, Ludong University, Yantai, Shandong 264025, People’s Republic of China
a r t i c l e
i n f o
Keywords: Predator–prey model Functional response Positive periodic solution Generalized continuation theorem
a b s t r a c t The existence of four positive periodic solutions for a delayed ratio-dependent predator– prey model with monotonic functional response
8 xðtÞ > < x0 ðtÞ ¼ xðtÞ½aðtÞ bðtÞxðtÞ cðtÞg yðtÞ yðtÞ h1 ðtÞ; h i > : y0 ðtÞ ¼ yðtÞ f ðtÞg xðtsðtÞÞ dðtÞ h2 ðtÞ; yðtsðtÞÞ is established by using the generalized continuation theorem, where aðtÞ; bðtÞ; cðtÞ; dðtÞ; f ðtÞ; sðtÞ; h1 ðtÞ and h2 ðtÞ are all nonnegative periodic continuous functions with period x > 0. When h1 ðtÞ ¼ h2 ðtÞ 0, the conditions that guarantee the existence of four positive periodic solutions reduces exactly to that in Fan et al. (2004) [1]. And our main result also improves the corresponding results in Zhang and Hou (2010) [2] and Fan and Wang (2001) [3]. Ó 2014 Elsevier Inc. All rights reserved.
1. Introduction During the last few decades, much work has been devoted to dynamical characteristics of population models, among these models, predator–prey systems play an important role in population theory. To describe a predator–prey relationship, it is necessary to specify the rate of prey consumption by an average predator. This rate is known as the functional response. Usually, the functional response can be classified as: (i) prey dependent, when prey density alone determines the response; (ii) predator dependent, when both predator and prey populations affect the response. For a long time, predation theory was dominated by prey-dependent models and by Holling’s three type classification of these responses. Arditi and Ginzburg [4] stimulated recent interest in alternative forms for functional responses, with their suggestion that a ratio-dependent functional response was a better starting point for modeling predation. The ratio-dependence is a particular type of predator dependence in which the response only depends on the ratio of prey population size to predator population size. Recently, there is a growing explicit biological and physiological evidences [5,6] that in many situations, especially when predators have to search for food (and therefore have to share or compete for food), a more suitable general predator–prey theory should be based on the so-called ratio-dependent theory. q Supported by NSF of China (11201213), NSF of Shandong Province (ZR2010AM022), and the outstanding young and middle-aged scientists research award fund of Shandong Province (BS2011SF004). ⇑ Corresponding author. E-mail addresses: [email protected] (Y.-H. Fan), [email protected] (L.-L. Wang).
http://dx.doi.org/10.1016/j.amc.2014.07.046 0096-3003/Ó 2014 Elsevier Inc. All rights reserved.
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
879
Many previous investigations (such as [7,8]) focus on the following ratio-dependent predator–prey system
8 > < x0 ¼ rx 1 Kx g yx y; > : y0 ¼ y lg yx D ; where gðxÞ is the functional response function, which reflects the capture ability of the predator to prey. For more biological meaning, the reader may consult Freedman [9], May [10] and Murry [11]. On the other hand, when we consider the effect of human activity, usually, a harvesting term should be added. Xiao et al. [12] investigated the following Michaelis–Menten type ratio-dependent predator–prey system
8 axy h1 ; < x0 ¼ rx 1 Kx myþx fx 0 : y ¼ y d þ h2 ; myþx
ð1:1Þ
when h1 ¼ 0. They gave a nice systematic work on the global qualitative analysis of this system. In addition, most efforts on ratio-dependent predator–prey systems with harvesting terms concentrate on the models with Holling II functional response, such as [13,14] etc. In what follows we shall use the notation
f ¼
Z
1
x
x
f ðtÞdt; 0
f M ¼ max f ðtÞ; t2½0;x
f L ¼ min f ðtÞ; t2½0;x
where f is a periodic continuous function with period x. More recently, for a similar ratio-dependent model with Holling II functional response, in [2], by using the Mawhin’s continuation theorem combined with the homotopy invariance of the Brouwer degree, the authors obtained that. Theorem 1.1. Assume that the following conditions hold: qffiffiffiffiffiffiffiffiffiffiffiffi M M M (A1) aL > mc þ 2 b h1 , and qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L M L h M dM mM hM 2 a (A2) ðf dÞ aM1 > h2 mM þ 2 . bL
Then the equations
8 cðtÞxðtÞyðtÞ < x0 ðtÞ ¼ xðtÞ½aðtÞ bðtÞxðtÞ mðtÞyðtÞþxðtÞ h1 ðtÞ; h i f ðtÞxðt s ðtÞÞ : y0 ðtÞ ¼ yðtÞ dðtÞ h2 ðtÞ; mðtÞyðtsðtÞÞþxðtsðtÞÞ
ð1:2Þ
has at least four positive x-periodic solutions when sðtÞ 0. Remark. When sðtÞ is not always equal to zero, the methods they used do not work. When the harvesting terms vanish, from Theorem 1.1, we can obtain. Corollary 1.1. Assume that sðtÞ 0; h1 ðtÞ ¼ h2 ðtÞ 0 and the following conditions hold: (B1) aL >
c M m L
, and
(B2) ðf dÞ > 0.
Then the Eqs. (1.2) has at least one positive x-periodic solution. Remark. Comparing the above result with that in [3], we find that there is a gap between them. That is, when h1 ðtÞ ¼ h2 ðtÞ 0, the conditions for the existence here are the maximal–minimal type conditions, while the conditions used in [3] are of the average type. And this will be one of the main problems we want to resolve in the present paper. In this paper, we consider the following delayed ratio-dependent predator–prey system with monotonic functional response
8 xðtÞ > < x0 ðtÞ ¼ xðtÞ½aðtÞ bðtÞxðtÞ cðtÞg yðtÞ yðtÞ h1 ðtÞ; h i xðtsðtÞÞ > : y0 ðtÞ ¼ yðtÞ f ðtÞg yðt sðtÞÞ dðtÞ h2 ðtÞ;
ð1:3Þ
880
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
where xðtÞ and yðtÞ represent the densities of the prey population and predator population respectively; aðtÞ; bðtÞ; cðtÞ; dðtÞ; f ðtÞ; sðtÞ; h1 ðtÞ and h2 ðtÞ are all nonnegative periodic continuous functions with period x > 0. Here aðtÞ stands for prey intrinsic growth rate, dðtÞ stands for the death rate of the predator, cðtÞ and f ðtÞ stand for the conversion xðtÞ rates; the function xðtÞ½aðtÞ bðtÞxðtÞ represents the specific growth rate of the prey in the absence of predator; and g yðtÞ denotes the predator response function, which reflects the capture ability of the predator and satisfies the following monotonic condition (for short, we call them M): (i) g 2 C 1 ½0; þ1Þ; gð0Þ ¼ 0; (ii) g 0 ðuÞ > 0 for u 2 ð0; þ1Þ; (iii) limu!þ1 gðuÞ ¼ 1 (if limu!þ1 gðuÞ ¼ k – 1, obviously we can use c ðtÞ ¼ kcðtÞ; f ðtÞ ¼ kf ðtÞ, replace cðtÞ and f ðtÞ in (1.3)). When h1 ðtÞ ¼ h2 ðtÞ 0, system (1.3) becomes
8 xðtÞ > < x0 ðtÞ ¼ xðtÞ½aðtÞ bðtÞxðtÞ cðtÞg yðtÞ yðtÞ; h i xðtsðtÞÞ > : y0 ðtÞ ¼ yðtÞ f ðtÞg yðt sðtÞÞ dðtÞ :
ð1:4Þ
In [1], Fan etc. obtained a sufficient condition for the existence of at least one positive x-periodic solution, for convenience, we list it below. Theorem 1.2. Assume that the following conditions hold: (C1) a > mc, and (C2) f > d, where
m ¼ sup u>0
gðuÞ : u
ð1:5Þ
Then (1.4) has at least one positive x-periodic solution. In the present paper, we want to solve the following problems: (1) to gain the sufficient conditions for the existence of four positive periodic solutions of system (1.3), here sðtÞ may not always equal to zero; (2) to recuperate the gap as mentioned above when h1 ðtÞ ¼ h2 ðtÞ 0. The main method usually used to gain the existence of periodic solutions for periodic systems is the coincidence degree theory developed by Gaines and Mawhin [15], we refer to [16–25] and references cited therein. While in this paper, we will use the generalized continuation theorem. Although they are the same in nature, the latter is more convenient. And our main results generalize the corresponding results in [2,1]. The tree of this paper is: in the next section, we establish the main existence results; and in the third section, some suitable examples are given to illustrate the main theorem, also some remarks are given. 2. Multiplicity of positive periodic solutions In order to obtain the existence of positive periodic solutions for the system (1.3), first we make the following preparations. Let X Rn be an open bounded set with closure X and f 2 C 1 ðX; Rn Þ \ CðX; Rn Þ. For x 2 X, let J f ðxÞ denote the Jacobian determinant of f at x and Sf be the set of all critical points of f, i.e., Sf ¼ fx 2 X : J f ðxÞ ¼ 0g. For y 2 Rn n f ð@ X [ Sf Þ, i.e., y is a regular point of f, the degree of f at y is defined as
degff ; X; yg ¼
X
sgn J f ðxÞ:
x2f 1 ðyÞ
Let X and Z be two Banach spaces, L: DomL X ! Z be a Fredholm mapping of index zero and
N : X ½0; 1 ! Z; ðx; kÞ # Nðx; kÞ be a L-compact mapping in X ½0; 1. Notice that if L is a Fredholm mapping of index zero, then there exist continuous projections P : X ! X and Q : Z ! Z such that
ImP ¼ KerL and ImL ¼ KerQ ¼ ImðI Q Þ:
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
881
Theorem A [15, p. 26-30], generalized continuation theorem. Let X, Z, L and Q be mentioned above and J: ImQ ! KerL is an isomorphism, furthermore, assume that (a) for each k 2 ð0; 1Þ, x 2 @ X\DomL,
Lx – kNðx; kÞ; (b) for each x 2 @ X\KerL,
QNðx; 0Þ – 0; (c) degfJQNð; 0ÞjKerL ; X\KerL; 0g – 0. Then, for each k 2 ½0; 1Þ, the equation Lx ¼ kNðx; kÞ has at least one solution in X and equation Lx ¼ Nðx; 1Þ has at least one solution in X. For simplicity, we set
expfk1 g d u h2 expff xg; f1 ðuÞ ¼ f g u
expfk1 g d u h2 ; f2 ðuÞ ¼ f g u
u > 0;
u > 0;
where
k1 ¼ ln
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b
:
If there exist a positive number u such that f1 ðu Þ > 0 (a necessary condition that guarantee this conclusion is f > d), notice that f1 ðuÞ 2 Cð0; þ1Þ; f 1 ðþ1Þ ¼ 1 and f1 ð0þ Þ ¼ h2 expff xg, then the set fujf1 ðuÞ ¼ 0; u > 0g is bounded, nonempty and closed. Now we set
u1 ¼ minfujf1 ðuÞ ¼ 0; u > 0g and
u2 ¼ maxfujf1 ðuÞ ¼ 0; u > 0g: It is obvious that
0 < u1 < u < u2 < þ1: Since
expfk1 g d u h2 expff xg f1 ðuÞ ¼ f g u
expfk1 g d u h2 ¼ f2 ðuÞ; < fg u and
f2 ðuÞ 2 Cð0; þ1Þ;
f 2 ðþ1Þ ¼ 1 and f 2 ð0þ Þ ¼ h2 ;
f2 ðuÞ ¼ 0 has at least one root in ð0; u1 Þ, also it has at least one root in ðu2 ; þ1Þ. Now we are in a position to state one of our main results. Theorem 2.1. Assume that the following conditions hold: qffiffiffiffiffiffiffiffi (H1) ða mcÞ P ð1 þ expfaxgÞ b h1 ; (H2) 9u > 0 such that f1 ðu Þ > 0 and u2 P u1 expff xg, furthermore assume that f2 ðuÞ ¼ 0 has only one root in ð0; u1 Þ and one root in ðu2 ; þ1Þ. Then (1.3) has at least four positive x-periodic solutions. Proof. For biological meaning, we only need to study the positive solution of (1.3), so make change of variables
xðtÞ ¼ exp fx1 ðtÞg;
yðtÞ ¼ exp fx2 ðtÞg;
ð2:1Þ
882
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
then (1.3) becomes
x01 ðtÞ ¼ aðtÞ bðtÞ expfx1 ðtÞg cðtÞhðexp fx1 ðtÞ x2 ðtÞgÞ h1 ðtÞ exp fx1 ðtÞg;
ð2:2Þ
x02 ðtÞ ¼ f ðtÞg ðexp fx1 ðt sðtÞÞ x2 ðt sðtÞÞgÞ dðtÞ h2 ðtÞ exp fx2 ðtÞg: Here hðuÞ ¼ gðuÞ=u. In order to apply Theorem A to system (2.2), we take
n o X ¼ Z ¼ xðtÞ ¼ ðx1 ðtÞ; x2 ðtÞÞT 2 CðR; R2 Þ : xðt þ xÞ ¼ xðtÞ and denote
jjxjj ¼ jjðx1 ðtÞ; x2 ðtÞÞT jj ¼ max max jx1 ðtÞj; max jx2 ðtÞj : t2½0;x
t2½0;x
Then X and Z are Banach spaces when they are endowed with the norm k k. Set
2
aðtÞ mcðtÞ bðtÞ exp fx1 ðtÞg h1 ðtÞ exp fx1 ðtÞg
3
7 6 þkcðtÞðm hðexp fx1 ðtÞ x2 ðtÞgÞÞ 7 6 Nðx; kÞ ¼ 6 7; 4 kf ðtÞ½g ðexp fx1 ðt sðtÞÞ x2 ðt sðtÞÞgÞ gðexpfk1 x2 ðt sðtÞÞgÞ 5 þf ðtÞgðexpfk1 x2 ðt sðtÞÞgÞ dðtÞ h2 ðtÞ exp fx2 ðtÞg where m is defined as in (1.5), and
Lx ¼ x0 ;
Px ¼
1
x
Z
x
xðtÞdt;
x 2 X;
0
Qz ¼
1
x
Z
x
zðtÞdt;
z 2 Z:
0
Rx Evidently, KerL ¼ R2 ; ImL ¼ zjz 2 Z; 0 zðtÞdt ¼ 0 is closed in Z and dimKerL ¼ codim ImL ¼ 2. Hence, L is a Fredholm mapping of index zero. Furthermore, the generalized inverse (to L) K p : ImL ! KerP\DomL has the form
K p ðzÞ ¼
Z
t
zðsÞds
0
Z
1
x
x
0
Z
t
zðsÞdsdt:
0
Thus
2 1 Rx x
0
½aðtÞ mcðtÞ bðtÞ expfx1 ðtÞg h1 ðtÞ exp fx1 ðtÞg
3
6 7 þkcðtÞðm hðexp fx1 ðtÞ x2 ðtÞgÞÞdt 6 7 QNðx; kÞ ¼ 6 1 R x 7; 4 5 ½ kf ðtÞ ½ g ð exp x ðt s ðtÞÞ x ðt s ðtÞÞ Þ gðexpfk x ðt s ðtÞÞgÞ f g 1 2 1 2 x 0 þf ðtÞgðexpfk1 x2 ðt sðtÞÞgÞ dðtÞ h2 ðtÞ exp fx2 ðtÞgdt and
2Rt
3 ½aðsÞ mcðsÞ bðsÞ expfx1 ðsÞg h1 ðsÞ exp fx1 ðsÞg 0 7 6 7 6 þkcðsÞðm hðexp fx1 ðsÞ x2 ðsÞgÞÞds 7 6 K p ðI Q ÞNðx; kÞ ¼ 6 R 7 6 t ½kf ðsÞðg ðexp fx ðs sðsÞÞ x ðs sðsÞÞgÞ gðexpfk x ðs sðsÞÞgÞÞ 7 5 4 0 1 2 1 2 þf ðsÞgðexpfk1 x2 ðs sðsÞÞgÞ dðsÞ h2 ðsÞ exp fx2 ðsÞgds 3 2 1 Rx Rt ½aðsÞ mcðtÞ bðsÞ expfx1 ðsÞg h1 ðsÞ exp fx1 ðsÞg 0 x 0 7 6 7 6 þkcðsÞðm hðexp fx1 ðsÞ x2 ðsÞgÞÞdsdt 7 6 6 R R 7 x t 7 61 4 x 0 0 ½kf ðsÞðg ðexp fx1 ðs sðsÞÞ x2 ðs sðsÞÞgÞ gðexpfk1 x2 ðs sðsÞÞgÞÞ 5 þf ðsÞgðexpfk1 x2 ðs sðsÞÞgÞ dðsÞ h2 ðsÞ exp fx2 ðsÞgdsdt R x 1 x 2 0 ½aðsÞ mcðsÞ bðsÞ expfx1 ðsÞg h1 ðsÞ exp fx1 ðsÞg
2t
3
7 6 7 6 þkcðsÞðm hðexp fx1 ðsÞ x2 ðsÞgÞÞds 7 6 6 7: R x 7 6 t 1 ½ kf ðsÞ ð g ð exp x ðs s ðsÞÞ x ðs s ðsÞÞ Þ gðk x ðs s ðsÞÞÞ Þ f g 1 2 1 2 5 4 x 2 0 þf ðsÞgðk1 x2 ðs sðsÞÞÞ dðsÞ h2 ðsÞ exp fx2 ðsÞgds
Clearly, QN and K p ðI Q ÞN are continuous, moreover, QN X ½0; 1 ; K p ðI Q ÞNðX ½0; 1Þ are relatively compact for any open bounded set X X. Hence, N is L-compact on X ½0; 1. Now we reach the position to search for an appropriate open bounded subset X for the application of Theorem A. Corresponding to equation Lx ¼ kNðx; kÞ; k 2 ð0; 1Þ, we have
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
8 0 x1 ðtÞ ¼ k½aðtÞ mcðtÞ bðtÞ exp fx1 ðtÞg h1 ðtÞ exp fx1 ðtÞg > > > < þkcðtÞðm hðexp fx1 ðtÞ x2 ðtÞgÞÞ; > x02 ðtÞ ¼ k½kf ðtÞðg ðexp fx1 ðt sðtÞÞ x2 ðt sðtÞÞgÞ gðexpfk1 x2 ðt sðtÞÞgÞÞ > > : þf ðtÞgðexpfk1 x2 ðt sðtÞÞgÞ dðtÞ h2 ðtÞ exp fx2 ðtÞg:
883
ð2:3Þ
Suppose that xðtÞ ¼ ðx1 ; x2 ÞT 2 X is an x-periodic solution of system (2.3) for a certain k 2 ð0; 1Þ. Then there exist ni ; gi 2 ½0; x such that
xi ðgi Þ ¼ max xi ðtÞ for i ¼ 1; 2:
xi ðni Þ ¼ min xi ðtÞ; t2½0;x
ð2:4Þ
t2½0;x
And
Z
x
½aðtÞ mcðtÞ bðtÞ exp fx1 ðtÞg h1 ðtÞ exp fx1 ðtÞg þ kcðtÞðm hðexp fx1 ðtÞ x2 ðtÞgÞÞdt ¼ 0;
ð2:5Þ
0
Z
x
½kf ðtÞðg ðexp fx1 ðt sðtÞÞ x2 ðt sðtÞÞgÞ gðexpfk1 x2 ðt sðtÞÞgÞÞ þ f ðtÞgðexpfk1 x2 ðt sðtÞÞgÞ
0
dðtÞ h2 ðtÞ exp fx2 ðtÞgdt ¼ 0:
ð2:6Þ
From (2.5), we have
Z
x
½h1 ðtÞ exp fx1 ðtÞg þ bðtÞ exp fx1 ðtÞg ðaðtÞ mcðtÞÞdt ¼
0
Z
x
½kcðtÞðm hðexp fx1 ðtÞ x2 ðtÞgÞÞdt > 0;
0
which implies that
expfx1 ðn1 Þgh1 þ expfx1 ðg1 Þgb > a mc:
ð2:7Þ
The first equation of (2.3) implies that for any t 2 ½n1 ; x,
x1 ðtÞ ¼ x1 ðn1 Þ þ
Z
t
x01 ðtÞdt < x1 ðn1 Þ þ ax
n1
and for any t 2 ½0; n1 ,
x1 ðtÞ ¼ x1 ðn1 Þ þ
Z
t
x01 ðtÞdt ¼ x1 ðn1 Þ
n1
< x1 ðn1 Þ þ
Z
Z t
n1
x01 ðtÞdt
n1
½ð1 kÞmcðtÞ þ bðtÞ expfx1 ðtÞg þ h1 ðtÞ exp fx1 ðtÞg þ kcðtÞhðexp fx1 ðtÞ x2 ðtÞgÞdt
t
6 x1 ðn1 Þ þ
Z
x
½ð1 kÞmcðtÞ þ bðtÞ expfx1 ðtÞg þ h1 ðtÞ exp fx1 ðtÞg þ kcðtÞhðexp fx1 ðtÞ x2 ðtÞgÞdt 0
¼ x1 ðn1 Þ þ ax: Thus, for any t 2 ½0; x, we have
x1 ðtÞ < x1 ðn1 Þ þ ax; especially,
x1 ðg1 Þ < x1 ðn1 Þ þ ax;
ð2:8Þ
thus (2.7) and (2.8) show that
b expfaxg expf2x1 ðn1 Þg ða mcÞ expfx1 ðn1 Þg þ h1 > 0; then (H1) leads to
x1 ðn1 Þ P ln
ða mcÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1 expfaxg 2b expfaxg
or
x1 ðn1 Þ 6 ln
ða mcÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1 expfaxg 2b expfaxg
:¼ ln l1
:
Using (2.7) and (2.8), we have
b expf2x1 ðg1 Þg ða mcÞ expfx1 ðg1 Þg þ h1 expfaxg > 0;
ð2:9Þ
884
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
by (H1), we know
x1 ðg1 Þ > ln
ða mcÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1 expfaxg
ð2:10Þ
2b
or
x1 ðg1 Þ < ln
ða mcÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1 expfaxg 2b
þ
:¼ ln k1 :
ð2:11Þ
On the other hand, from (2.5), we also have
Z
Z
x
½h1 ðtÞ exp fx1 ðtÞg þ bðtÞ exp fx1 ðtÞg aðtÞdt ¼
0
x
½mcðtÞ þ kcðtÞðm hðexp fx1 ðtÞ x2 ðtÞgÞÞdt
0
Z
¼
x
½ð1 kÞmcðtÞ kcðtÞhðexp fx1 ðtÞ x2 ðtÞgÞdt < 0;
0
which implies that
h1 exp fx1 ðg1 Þg þ b exp fx1 ðn1 Þg a < 0; then (2.8) gives
b exp f2x1 ðn1 Þg a exp fx1 ðn1 Þg þ h1 expfaxg < 0; by (H1), we have
ln
k1
¼: ln
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b
< x1 ðn1 Þ < ln
aþ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg
:
2b
ð2:12Þ
Similarly, we can obtain
ln
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b expfaxg
þ
< x1 ðg1 Þ < ln
þ
þ
aþ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b expfaxg
þ
:¼ ln l1 :
ð2:13Þ
We claim that k1 < k1 ; l1 < l1 and k1 6 l1 . Constructing a function
pffiffiffiffiffiffiffiffiffiffiffiffiffi qðxÞ ¼ x x2 a;
a > 0 and x >
pffiffiffi a;
then
qðxÞ q0 ðxÞ ¼ pffiffiffiffiffiffiffiffiffiffiffiffiffi < 0; x2 a which implies that
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b
<
ða mcÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1 expfaxg 2b
ð2:14Þ
;
also in view of
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b
>
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b
ð2:15Þ
:
By (2.11), (2.12), (2.14) and (2.15), we know k1
¼
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4bh1 expfaxg 2b
<
ða mcÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1 expfaxg 2b
In addition, constructing two functions
hðxÞ ¼
aþ
pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 bx ; x
0
a2 ; b
and
sðxÞ ¼ x þ
pffiffiffiffiffiffiffiffiffiffiffiffiffi x2 a;
it is easy to verify
a > 0; x >
pffiffiffi a;
a > 0; b > 0
þ
¼ k1 :
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
885
0
h ðxÞ < 0 and s0 ðxÞ > 0; thus l1
¼
<
ða mcÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1 expfaxg
2b expfaxg qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ þ ða mcÞ2 4b h1 expfaxg
2b expfaxg qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a þ a2 4b h1 expfaxg þ < ¼ l1 : 2b expfaxg þ
þ
Now we show that k1 6 l1 when (H1) holds true. In fact, k1 6 l1 is equivalent to
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ ða mcÞ2 4b h1 expfaxg
6
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ þ ða mcÞ2 4b h1 expfaxg 2b expfaxg
2b
;
that is,
ða mcÞðexpfaxg 1Þ 6
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1 expfaxgðexpfaxg þ 1Þ;
obviously, the above inequality is satisfied when (H1) hold true. From (2.9) and (2.11)–(2.13), we have for any t 2 ½0; x,
þ
ln k1 < x1 ðtÞ < ln k1
or
þ
ln l1 < x1 ðtÞ < ln l1 :
By (2.6) and the monotonicity of the function g, we have
Z 0
x
½f ðtÞgðexp fk1 x2 ðt sðtÞÞgÞ dðtÞ h2 ðtÞ exp fx2 ðtÞgdt Z x kf ðtÞðg ðexp fx1 ðt sðtÞÞ x2 ðt sðtÞÞgÞ gðexpfk1 x2 ðt sðtÞÞgÞÞdt < 0: ¼
ð2:16Þ
0
Then
f gðexpfk1 x2 ðg2 ÞgÞ < d þ h2 expfx2 ðn2 Þg: Notice that
x2 ðtÞ ¼ x2 ðn2 Þ þ
Z
t
x02 ðsÞds;
n2
thus by the second equation of (2.3) and notice that for any t 2 ½0; x; x1 ðtÞ P k1 , we know for any t 2 ½n2 ; x,
x2 ðtÞ ¼ x2 ðn2 Þ þ < x2 ðn2 Þ þ þ k2
x02 ðsÞds
n2 Z t
kf ðsÞgðexpfk1 x2 ðs sðsÞÞgÞds
t
f ðtÞ½gðexpfx1 ðs sðsÞÞ x2 ðs sðsÞÞgÞ gðexpfk1 x2 ðs sðsÞÞgÞds
n2
6 x2 ðn2 Þ þ þk
t
n2
Z
Z
Z
Z
t
kf ðsÞgðexpfk1 x2 ðs sðsÞÞgÞds n2
t
f ðsÞ½gðexpfx1 ðs sðsÞÞ x2 ðs sðsÞÞgÞ gðexpfk1 x2 ðs sðsÞÞgÞds
n2
¼ x2 ðn2 Þ þ k
Z
t
f ðsÞgðexpfx1 ðs sðsÞÞ x2 ðs sðsÞÞgÞds
n2
6 x2 ðn2 Þ þ f x: Here we use the following properties of the function g,
gðuÞ 6 1 and
@gðexpfx1 x2 gÞ > 0: @x1
ð2:17Þ
886
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
On the other hand, for any t 2 ½0; n2 , we have
x2 ðtÞ ¼ x2 ðn2 Þ þ
Z
t
x02 ðsÞds
n2
¼ x2 ðn2 Þ þ
Z
n2
t
< x2 ðn2 Þ þ
Z
x02 ðsÞ ds
n2
½dðtÞ þ h2 ðtÞ exp fx2 ðtÞgds
t
6 x2 ðn2 Þ þ
Z
x
½dðtÞ þ h2 ðtÞ exp fx2 ðtÞgds
0
¼ x2 ðn2 Þ þ f x; which implies that for any t 2 ½0; x,
x2 ðtÞ < x2 ðn2 Þ þ f x: Therefore
x2 ðg2 Þ < x2 ðn2 Þ þ f x: So
f gðexpfk1 x2 ðg2 ÞgÞ < d þ h2 expff xg expfx2 ðg2 Þg
ð2:18Þ
f gðexpfk1 f x x2 ðn2 ÞgÞ < d þ h2 expfx2 ðn2 Þg:
ð2:19Þ
and
By (2.18), we have
f gðexpfk1 x2 ðg2 ÞgÞ expfx2 ðg2 Þg d expfx2 ðg2 Þg h2 expff xg < 0: The above inequality and (H2) implies that
x2 ðg2 Þ > ln u2
or x2 ðg2 Þ < ln u1 :
ð2:20Þ
Set
" f3 ðuÞ ¼ f g
! # expfk1 f xg d u h2 ; u
u > 0;
then
f3 ðuÞ ¼ f1 ðu expff xgÞ; thus f3 ðuÞ < 0 implies
x2 ðn2 Þ > ln u2 f x or x2 ðn2 Þ < ln u1 f x:
ð2:21Þ
By (2.20) and (2.21), we have for any t 2 ½0; x, þ
x2 ðtÞ < ln u1 :¼ ln k2
or x2 ðtÞ > ln u2 f x :¼ ln l2 :
ð2:22Þ
And the second equation of (2.6) implies
Z
x
½dðtÞ þ h2 ðtÞ exp fx2 ðtÞgdt ¼
0
Z 0
þ
x
kf ðtÞðg ðexp fx1 ðt sðtÞÞ x2 ðt sðtÞÞgÞ gðexpfk1 x2 ðt sðtÞÞgÞÞdt Z
x
f ðtÞgðexpfk1 x2 ðt sðtÞÞgÞdt
0
< f x;
ð2:23Þ
which leads to
x2 ðg2 Þ > ln
h2 f d
ð2:24Þ
:
Thus for any t 2 ½0; x,
x2 ðtÞ > ln
h2 f d
f x :¼ ln k2 :
ð2:25Þ
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
887
On the other hand, from (2.23), we also have
Z
x
f ðtÞg ðexp fx1 ðt sðtÞÞ x2 ðt sðtÞÞgÞdt þ l1 < fg x; exp fx2 ðn2 Þg
dx <
0
which shows that þ
x2 ðn2 Þ < ln
l1 : d f
g 1
Thus þ
x2 ðg2 Þ < ln
l1 þ f x :¼ ln lþ2 :
g 1
ð2:26Þ
d f
We claim that
þ
k2 < k2 Since
þ f1 ðk2 Þ
þ
and l2 < l2 :
ð2:27Þ
¼ f1 ðu1 Þ ¼ 0,
"
! # expfk1 g þ k2 d þ k2 þ < f d k2 ;
h2 expff xg ¼ f g
thus
h2 expff xg
þ
k2 >
>
f d
h2 expff xg f d
¼ k2 :
On the other hand, f1 ðu2 Þ ¼ f1 ðl2 expff xgÞ ¼ 0, therefore
"
expfk1 g
!
#
d l2 l2 expff xg " ! # þ l1 d l2 ; < fg l2 expff xg
0 < h2 ¼ f g
which shows þ
l1
l2 expff xg
>g
1
! d f
;
thus þ
l2 <
g 1
d f
þ
l1
expff xg
<
l1 expff xg ¼ lþ2 : g 1 d f
This complete the claim (2.27). Note that the condition (H2) implies that
u1 6 u2 expff xg; that is þ
k2 6 l2 : From (2.22) and (2.25)–(2.27), we know for any t 2 ½0; x,
þ
ln k2 < x2 ðtÞ < ln k2
or
þ
ln l2 < x2 ðtÞ < ln l2 :
Now we construct four open sets:
þ
ln k2 < x2 ðtÞ < ln k2 g;
þ
ln l2 < x2 ðtÞ < ln l2 g;
X1 ¼ fðx1 ðtÞ; x2 ðtÞÞjðx1 ðtÞ; x2 ðtÞÞ 2 X;
ln k1 < x1 ðtÞ < ln k1 ;
X2 ¼ fðx1 ðtÞ; x2 ðtÞÞjðx1 ðtÞ; x2 ðtÞÞ 2 X;
ln k1 < x1 ðtÞ < ln k1 ;
þ
þ
888
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
þ
ln k2 < x2 ðtÞ < ln k2 g;
þ
ln l2 < x2 ðtÞ < ln l2 g:
X3 ¼ fðx1 ðtÞ; x2 ðtÞÞjðx1 ðtÞ; x2 ðtÞÞ 2 X;
ln l1 < x1 ðtÞ < ln l1 ;
X4 ¼ fðx1 ðtÞ; x2 ðtÞÞjðx1 ðtÞ; x2 ðtÞÞ 2 X;
ln l1 < x1 ðtÞ < ln l1 ;
þ
þ
þ
þ
þ
þ
Clearly, li ; li and ki ; ki (i ¼ 1; 2) are independent of k, from k1 6 l1 and k2 6 l2 , we know that Xi \ Xj ¼ /; i – j ði; j ¼ 1; 2; 3; 4Þ. Now we show under the assumptions of Theorem 2.1, the system of algebraic equations
8 < ða mcÞu bu2 h1 ¼ 0; h i : f g expfk1 g d v h2 ¼ 0 v
ð2:28Þ
has a unique root in every Xi ði ¼ 1; 2; 3; 4Þ. In fact, from the first equation of (2.28) and (H1), we know the equation
ða mcÞu bu2 h1 ¼ 0 þ þ e 1 and u e2 (u e1 < u e 2 ). We claim u e 1 2 ðk e has only two positive solutions, namely, u 1 ; k1 Þ and u 2 2 ðl1 ; l1 Þ. The reason is,
e1 ¼ u >
>
e1 ¼ u <
e2 ¼ u
ða mcÞ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1
2bffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 a a 4b h1 2b qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi a a2 4b h1 expfaxg 2b ða mcÞ
¼ k1 ;
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ða mcÞ2 4b h1
2b qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ða mcÞ ða mcÞ 4b h1 expfaxg 2b ða mcÞ þ
þ
¼ k1 :
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 ða mcÞ 4b h1
2bffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a þ a2 4b h1 a þ a2 4b h1 expfaxg þ < ¼ l1 ; < 2b 2b
e2 ¼ u >
ða mcÞ þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi c 2 a 2m 4b h1
2b qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi c 2 ða mcÞ þ a 2m 4b h1 expfaxg 2b expf2axg
¼ l1 :
On the other hand, from the second equation of (2.28) and (H2), we know that the equation
expfk1 g d v h2 ¼ 0 fg
v
e 1 and v e 2 (v e1 < v e 2 ), obviously, v e 1 < u1 ¼ kþ e has only two roots in ð0; u1 Þ [ ðu2 ; þ1Þ, namely, v 2 and v 2 > u2 ¼ l2 expff xg. We claim that
ve 1 > k2
and
ve 2 < lþ2 :
Notice that
expfk1 g d vei ; 0 < h2 ¼ f g vei thus
h2 < f d vei ; and
i ¼ 1; 2
i ¼ 1; 2;
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
d < fg
þ expfk1 g l < fg 1 ; vei vei
889
i ¼ 1; 2;
which implies that
vei >
h2 f d
>
h2 expff xg f d
¼ k2 ;
i ¼ 1; 2
and þ
vei <
þ
l1 l expff xg < 1 ¼ lþ2 ; g 1 d g 1 d f
i ¼ 1; 2:
f
So far, we reach the conclusion of the claim. According to the above estimation of priori boundary, it is clear that Xi satisfies the condition (a) of Theorem A. When
x ¼ ðx1 ; x2 ÞT 2 @ Xi \ KerL ¼ @ Xi \ R2 ; x is a constant vector in R2 , thus
" QNðx; 0Þ ¼
ða mcÞ b exp fx1 g h1 exp fx1 g f g ðexpfk1 x2 gÞ d h2 expfx2 g
# – 0:
Furthermore, let J: ImQ !KerL; x ! x, in view of the assumptions in Theorem 2.1, it is easy to see that
deg fJQN; Xi \ KerL; 0g – 0: By now we know that Xi verifies all the requirements of Theorem A. thus system (2.2) has at least four x-periodic solutions. By the medium of (2.1), we derive that (1.3) has at least four positive x-periodic solutions. The proof is complete. h As a degenerate case, we can obtain from Theorem 2.1, when a > mc and f > d, system (1.3) has at least one positive xperiodic solution provided that h1 ðtÞ ¼ h2 ðtÞ 0. Obviously, this condition is the same as that in Theorem 1.2, so our results generalize this conclusion in [1]. From the proof of Theorem 2.1, we can easily obtain that. Corollary 2.1. Assume that the following conditions hold: qffiffiffiffiffiffiffiffi (1) ða mcÞ P ð1 þ expfaxgÞ b h1 ; (2) f1 ðuÞ ¼ 0 has only two positive roots, namely, u1 ; u2 and u2 P u1 expff xg, furthermore assume that f2 ðuÞ ¼ 0 has only two positive roots. Then (1.3) has at least four positive x-periodic solutions.
Theorem 2.2. Assume that sðtÞ 0 and the following conditions hold: qffiffiffiffiffiffiffiffiffiffiffiffi M M (H10 ) ða mcÞL > 2 b h1 ; 0 (H2 ) f2 ðuÞ ¼ 0 has only two positive roots, furthermore assume that the equation f4 ðuÞ ¼ 0 has at least two positive roots, where
expfk2 g M M d u h2 ; f4 ðuÞ ¼ f L g u qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L L 2 aM ðaM Þ 4b h1 : k2 ¼ ln L 2b Then (1.3) has at least four positive x-periodic solutions. Proof. In (2.3) and (2.4), replace k1 by k2 , then it is obvious that x0i ðni Þ ¼ 0 and x0i ðgi Þ ¼ 0; i ¼ 1; 2. Therefore
aðn1 Þ mcðn1 Þ bðn1 Þ exp fx1 ðn1 Þg h1 ðn1 Þ exp fx1 ðn1 Þg þ kcðn1 Þðm hðexp fx1 ðn1 Þ x2 ðn1 ÞgÞÞ ¼ 0;
ð2:29Þ
kf ðn2 Þðgðexpfx1 ðn2 Þ x2 ðn2 ÞgÞ gðexpfk2 x2 ðn2 ÞgÞÞ þ gðexpfk2 x2 ðn2 ÞgÞ dðn2 Þ h2 ðn2 Þ exp fx2 ðn2 Þg ¼ 0 ð2:30Þ
890
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
and
aðg1 Þ mcðg1 Þ bðg1 Þ exp fx1 ðg1 Þg h1 ðg1 Þ exp fx1 ðg1 Þg þ kcðg1 Þðm hðexp fx1 ðg1 Þ x2 ðg1 ÞgÞÞ ¼ 0;
ð2:31Þ
kf ðg2 Þðgðexpfx1 ðg2 Þ x2 ðg2 ÞgÞ gðexpfk2 x2 ðg2 ÞgÞÞ þ gðexpfk2 x2 ðg2 ÞgÞ dðg2 Þ h2 ðg2 Þ exp fx2 ðg2 Þg ¼ 0: ð2:32Þ By (2.29), we have
aðn1 Þ mcðn1 Þ bðn1 Þ exp fx1 ðn1 Þg h1 ðn1 Þ exp fx1 ðn1 Þg ¼ kcðn1 Þðm hðexp fx1 ðn1 Þ x2 ðn1 ÞgÞÞ < 0; which leads to M
M
ða mcÞL b exp fx1 ðn1 Þg h1 exp fx1 ðn1 Þg < 0; thus (H10 ) gives
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi h i2 M M L ða mcÞ þ ða mcÞ 4b h1 L
x1 ðn1 Þ > ln
M
2b
:¼ ln l3
ð2:33Þ
or
ða mcÞL x1 ðn1 Þ < ln
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi h i2 M M ða mcÞL 4b h1 M
2b
þ
:¼ ln k3 :
ð2:34Þ
From (2.29), we obtain
aðn1 Þ bðn1 Þ exp fx1 ðn1 Þg h1 ðn1 Þ exp fx1 ðn1 Þg ¼ mcðn1 Þ kcðn1 Þðm hðexp fx1 ðn1 Þ x2 ðn1 ÞgÞÞ ¼ mð1 kÞcðn1 Þ þ kcðn1 Þhðexp fx1 ðn1 Þ x2 ðn1 ÞgÞ > 0; therefore L
L
b expf2x1 ðn1 Þg aM exp fx1 ðn1 Þg þ h1 < 0; by using (H10 ) once more, we know
ln
k3
¼: ln
aM
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L L 2 ½aM 4b h1 2b
L
< x1 ðn1 Þ < ln
aM þ
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L L 2 ½aM 4b h1 L
2b
þ
:¼ ln l3 :
ð2:35Þ
It is easy to see that
þ
þ
k3 < k3 < l3 < l3 ; thus by (2.33)–(2.35), we have
þ
ln k3 < x1 ðn1 Þ < ln k3
or
þ
ln l3 < x1 ðn1 Þ < ln l3 :
Similarly, by (2.31) and the above analysis, we can obtain
þ
ln k3 < x1 ðg1 Þ < ln k3
or
þ
ln l3 < x1 ðg1 Þ < ln l3 :
Which implies that for any t 2 ½0; x,
þ
ln k3 < x1 ðtÞ < ln k3
or
þ
ln l3 < x1 ðtÞ < ln l3 :
ð2:36Þ
From (2.30) and (2.36), we have
f ðn2 Þgðexpfk2 x2 ðn2 ÞgÞ dðn2 Þ h2 ðn2 Þ exp fx2 ðn2 Þg ¼ kf ðn2 Þðgðexpfx1 ðn2 Þ x2 ðn2 ÞgÞ gðexpfk2 x2 ðn2 ÞgÞÞ < 0; thus M
M
f L gðexpfk2 x2 ðn2 ÞgÞ d h2 exp fx2 ðn2 Þg < 0: þ
Considering (H20 ), we denote the two positive solutions of f4 ðuÞ ¼ 0 as k4 (the minimal one) and l4 (the maximal one), then
x2 ðn2 Þ > ln l4
þ
or x2 ðn2 Þ < ln k4 :
ð2:37Þ
891
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
Notice that (2.30) also implies
dðn2 Þ < dðn2 Þ þ h2 ðn2 Þ exp fx2 ðn2 Þg ¼ kf ðn2 Þðgðexpfx1 ðn2 Þ x2 ðn2 ÞgÞ gðexpfk2 x2 ðn2 ÞgÞÞ þ gðexpfk2 x2 ðn2 ÞgÞ þ expfx1 ðn2 Þg l3 < f ðn2 Þg < f ðn2 Þ; < f ðn2 Þg expfx2 ðn2 Þg expfx2 ðn2 Þg which shows that L
þ
h2
ln k4 ¼: ln
< x2 ðn2 Þ < ln
M
ðf dÞ
g 1
l3 þ L :¼ ln l4 :
ð2:38Þ
d f
By similar analysis as above, also in view of (2.32), we have
x2 ðg2 Þ > ln l4
þ
or x2 ðg2 Þ < ln k4
and
þ
ln k4 < x2 ðg2 Þ < ln l4 : Obviously,
þ
þ
k4 < k4 < l4 < l4 ; thus for any t 2 ½0; x, we have
þ
ln k4 < x2 ðtÞ < ln k4
or
þ
ln l4 < x2 ðtÞ < ln l4 :
ð2:39Þ
The rest of the proof is similar to that of Theorem 2.1, we omit it here. h
3. Examples and remarks In this section, we give some examples to illustrate our main result. First, we set
f5 ðuÞ ¼ mM du2 where
k3 ¼
a
h i f d k3 mM h2 expff xg u þ h2 k3 expff xg;
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b
:
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi If ðf dÞk3 > h2 mM expff xg þ 2 mM dh2 k3 expff xg, then the equation f5 ðuÞ ¼ 0 has only two roots, denote the two positive b2; u b1 ðu b1 < u b 2 Þ. Obviously, solution of equation f5 ðuÞ ¼ 0 as u
h b1 ¼ u
;
2mM d h
b2 ¼ u
ffi i rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi h i2 M M M f d k3 m h2 expff xg f d k3 m h2 expff xg 4m dh2 k3 expff xg ffi i rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi h i2 f d k3 mM h2 expff xg þ f d k3 mM h2 expff xg 4mM dh2 k3 expff xg :
2mM d
For system (1.2), we have the following conclusion. Theorem 3.1. Assume that the following conditions hold: qffiffiffiffiffiffiffiffi ; ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi (D1) a P mc þ ð1 þ expfaxgÞ b h1q (D2) ðf dÞk3 > h2 mM expff xg þ 2 mM dh2 k3 expff xg: Then the Eqs. (1.2) has at least four positive x-periodic solutions. Remark. When h1 ðtÞ ¼ h2 ðtÞ 0, the conditions in Theorem 3.1 reduce to a > in [3].
c m
and f d > 0, it is exactly the same as that
892
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
Example 3.1. Consider the following model
8 h i ð4pt Þ yðtÞ > < x0 ðtÞ ¼ xðtÞ 2 cos ð4pt Þ 1sin10ð4ptÞ xðtÞ 1þcos ð1 þ cos ð4ptÞÞ; 100 xðtÞþ0:01yðtÞ h i xðtÞ > : y0 ðtÞ ¼ yðtÞ 1þsin10ð4ptÞ þ ð1 þ sin ð4ptÞÞ xðtÞþ0:01yðtÞ 1sin10ð4ptÞ : It is easy to verify that all the conditions in Theorem 2.1 hold true. Therefore, this system has at least four positive 1=2periodic solutions. Remark. The above example does not satisfy the conditions in Theorem 1.1. On the other hand, the Theorem 2.1 in [2] can also be generalized to the following theorem. Theorem 3.2. Assume that the following conditions hold: qffiffiffiffiffiffiffiffiffiffiffiffi M M M (E1) aL P mc þ 2 b h1 , and qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L M M M (E2) ðf dÞ k4 > h2 mM þ 2 mM d h2 k4 , where
k4 ¼
aL
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 2 M M ðaL Þ 4b h1 M
2b
:
Then the equations (1.2) (when sðtÞ 0) has at least four positive x-periodic solutions. Remark. When (A2) holds, (E2) also holds true. Example 3.2. Consider the following system
8 2 ðtÞyðtÞ < x0 ðtÞ ¼ xðtÞ½aðtÞ bðtÞxðtÞ m2cðtÞx h1 ðtÞ; ðtÞy2 ðtÞþx2 ðtÞ h i eðtÞx2 ðtsðtÞÞ : y0 ðtÞ ¼ yðtÞ dðtÞ h2 ðtÞ: m2 ðtÞy2 ðtsðtÞÞþx2 ðtsðtÞÞ
ð3:1Þ
In order to give an existence result, we set
2 2 2 2 f6 ðuÞ ¼ mM du3 þ mM h2 expfexgu2 k5 e d u þ k5 h2 expfexg and
2 mM h2 expfexg þ e¼ u
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi 2 4 2 ðmM Þ h2 2 expf2exg þ 3ðmM Þ dk5 e d 2
3ðmM Þ d
;
where
k5 ¼
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 expfaxg 2b
:
e Þ < 0; f 6 ðuÞ ¼ 0 has only two positive roots, namely, u1 ; u2 ðu1 < u2 Þ, from Corollary 2.1, we have When f6 ð u Theorem 3.3. Assume that the following conditions hold: qffiffiffiffiffiffiffiffi (F1) a P 12 mc þ ð1 þ expfaxgÞ b h1 ; e Þ < 0 and u2 P u1 expfexg. (F2) f6 ð u Then (3.1) has at least four positive x-periodic solutions. c Remark. Notice that when h1 ðtÞ ¼ h2 ðtÞ 0, the conditions in Theorem 3.3 now reduce to a 2m > 0 and e d > 0, in this þ þ case, k1 ¼ k1 ¼ 0, k2 ¼ k2 ¼ 0, thus the bounded open sets X1 ; X2 ; X3 now disappear, only X4 is left, and the corresponding algebraic equations has only one root, therefore, we can only obtain the existence of at least one positive solution. This improves the corresponding conclusion in [26]. Similarly, from Theorem 2.2, we can deduce.
Y.-H. Fan, L.-L. Wang / Applied Mathematics and Computation 244 (2014) 878–894
893
Theorem 3.4. Assume thatsðtÞ 0 and the following conditions hold: qffiffiffiffiffiffiffiffiffiffiffiffi M M c L (G3) a 2m > 2 b h1 ; (G4) f7 ðu Þ < 0, where
2 M 2 M 2 L 2 M f7 ðuÞ ¼ mM d u3 þ mM h2 u2 k6 ðe dÞ u þ k6 h2 and
k6 ¼
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi L L 2 aM ðaM Þ 4b h1 L
2b
;u ¼
2 M mM h2
þ
rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi ffi ðmM Þ
4
M
h2
2
2
2 M 2
þ 3ðmM Þ d k5 ðe dÞ
3ðmM Þ d
L
:
Then (3.1) has at least four positive x-periodic solutions. Compare the condition (H1) with (H10 ) and (H2) with (H20 ), also from some numerical examples, we find that the periodic
x has no influence on the topological structure for the solution of system (1.3), thus we may conclude. Conjecture. Assume that the following conditions hold: qffiffiffiffiffiffiffiffi (I1) ða mcÞ > 2 b h1 ; (I2) the equation f5 ðuÞ ¼ 0 has at least two positive roots, where
expfk7 g d u h2 ; f5 ðuÞ ¼ f g u
k7 ¼ ln
a
qffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi a2 4b h1 2b
:
Then (1.3) has at least four positive x-periodic solutions.
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