Now, let us state several lemmas which will be useful to prove our main conclusion.
Proof.
We divided the proof into four steps.
Step
1. We show
(10)limsupk→+∞x(k)≤M1.
From the first equation of (1), we have
(11)x(k+1)≤x(k)exp{a(k)-b(k)x(k)}.
By Lemma 2, we have
(12)limsupk→+∞x(k)≤1blexp(au-1)=M1.
Previous inequality shows that for any ε>0, there exists a k1>0, such that
(13)x(k)≤M1+ε, ∀k≥k1.
Step
2. We prove limsupk→+∞y(k)≤M2 by distinguishing two cases.
Case
1. There exists a l0≥k1, such that y(l0+1)≥y(l0).
By the second equation of system (1), we have
(14)-d(l0)+B(l0)x2(l0)r(l0)x2(l0)+y2R(l0)≥0,
which implies
(15)-d(l0)+B(l0)x2(l0)y2R(l0)≥0.
The previous inequality combined with (13) leads to y(l0)≤{Bu(M1+ε)2/dl}1/2R. Thus, from the second equation of system (1), again we have
(16)y(l0+1) =y(l0)exp{-d(l0)+B(l0)x2(l0)r(l0)x2(l0)+y2R(l0)} ≤{Bu(M1+ε)2dl}1/2Rexp{-dl+Burl} =defM2ε.
We claim that
(17)y(k)≤M2ε ∀k≥l0.
By a way of contradiction, assume that there exists a p0≥l0 such that y(p0)>M2ε. Then p0≥l0+2. Let p0~≥l0+2 be the smallest integer such that y(p0~)>M2ε. Then y(p0~)>y(p0~-1). The previous argument produces that y(p0~)≤M2ε, a contradiction. This proves the claim. Therefore, limsupk→+∞y(k)≤M2ε. Setting ε→0 in it leads to limsupk→+∞y(k)≤M2.
Case
2. Suppose y(k+1)<y(k) for all k≥k1. Since y(k) is nonincreasing and has a lower bound 0, we know that limk→+∞y(k) exists, denoted by y-, we claim that
(18)y-≤{BuM12dl}1/2R.
By a way of contradiction, assume that y->{BuM12/dl}1/2R.
Taking limit in the second equation in system (1) gives
(19)limk→+∞{-d(k)+B(k)x2(k)r(k)x2(k)+y2R(k)}=0,
however,
(20)limk→+∞{-d(k)+B(k)x2(k)r(k)x2(k)+y2R(k)} ≤limsupk→+∞{-d(k)+B(k)x2(k)r(k)x2(k)+y2R(k)} ≤-dl+BuM12y-2R <0,
which is a contradiction. It implies that y-≤{BuM12/dl}1/2R. By the fact Bu>dlrl, we obtain that
(21)y-≤{BuM12dl}1/2R≤{BuM12dl}1/2Rexp{-dl+Burl}=M2.
Therefore, we have
(22)limk→+∞y(k)=y-≤M2.
Then,
(23)limsupk→+∞y(k)≤M2.
Step
3. We verify
(24)liminfk→+∞x(k)≥m1.
Conditions (H1) imply that for enough small positive constant ε, we have
(25)al-Au(M2+ε)1-R2rl>0.
For the previous ε, it follows from Steps 1 and 2 that there exists a k2 such that for all k≥k2(26)x(k)≤M1+ε, y(k)≤M2+ε.
Then, for k≥k2, it follows from (26) and the first equation of system (1) that
(27)x(k+1)≥x(k)exp{al-Au(M2+ε)1-R2rl-bux(k)}.
According to Lemma 3, one has
(28)liminfk→+∞x(k) ≥min{m1*,al-Au(M2+ε)1-R/2rlbu} =m1*,
where
(29)m1*=al-Au(M2+ε)1-R/2rlbu ×exp{al-Au(M2+ε)1-R2rl-bu(M1+ε)}.
Setting ε→0 in (28) leads to
(30)liminfk→+∞x(k)≥al-AuM21-R/2rlbu ×exp{al-AuM21-R2rl-buM1}=m1.
By the fact that minx∈R+{[exp(x-1)]/x}=1, we see that M1=exp(au-1)/bl≥au/bl≥al/bu≥(al-AuM21-R/2rl)/bl≥m1.
This ends the proof of Step 3.
Step
4. We present two cases to prove that
(31)liminfk→+∞y(k)≥m2.
For any small positive constant ε<m1/2, from Step 1 to Step 3, it follows that there exists a k3≥k2 such that for all k≥k3(32)x(k)≥m1-ε, x(k)≤M1+ε, y(k)≤M2+ε.
Case
1. There exists a n0≥k3 such that y(n0+1)≤y(n0), then
(33)-d(n0)+B(n0)x2(n0)r(n0)x2(n0)+y2R(n0)≤0.
Hence,
(34)y(n0)≥{(Bl-rudu)(m1-ε)2du}1/2R=defc1ε,
and so,
(35)y(n0+1) ≥{(Bl-rudu)(m1-ε)2du}1/2R ×exp{-du+Bl(m1-ε)2ru(m1-ε)2+(M2+ε)2R} =defc2ε.
Set
(36)m2ε=min{c1ε,c2ε}.
We claim that
(37)y(k)≥m2ε ∀k≥n0.
By a way of contradiction, assume that there exists a q0≥n0, such that y(q0)<m2ε. Then q0≥n0+2. Let q0~≥n0+2 be the smallest integer such that y(q0~)<m2ε. Then y(q0~)<y(q0~-1), which implies that y(q0)≥m2ε, a contradiction, this proves the claim. Therefore, liminfk→+∞y(k)≥m2ε, setting ε→0 in it leads to liminfk→+∞y(k)≥m2.
Case
2. Assume that y(k+1)>y(k) for all k≥k3, then, limk→+∞y(k) exists, denoted by y_, then limk→+∞y(k)=y_. We claim that
(38)y_≥m2.
By a way of contradiction, assume that y_<m2. Taking limit in the second equation in system (1) gives
(39)limk→+∞{-d(k)+B(k)x2(k)r(k)x2(k)+y2R(k)}=0,
which is a contradiction since
(40)limk→+∞{-d(k)+B(k)x2(k)r(k)x2(k)+y2R(k)} ≥liminfk→+∞{-d(k)+B(k)x2(k)r(k)x2(k)+y2R(k)} ≥-du+Blm12rum12+y_2R >0.
This proves the claim, then we have
(41)limk→+∞y(k)=y_≥m2.
So,
(42)liminfk→+∞y(k)≥m2.
Obviously, M2={BuM12/dl}1/2Rexp{-dl+Bu/rl}≥{(Bl-rudu)m12/du}1/2R≥m2. This completes the proof of the theorem.