Volume 2010, Article ID 195916,14pages doi:10.1155/2010/195916
Research Article
Fixed Points and Stability in Nonlinear
Equations with Variable Delays
Liming Ding,
1, 2Xiang Li,
1and Zhixiang Li
11Department of Mathematics and System Science, College of Science,
National University of Defence Technology, Changsha 410073, China
2Air Force Radar Academy, Wuhan 430010, China
Correspondence should be addressed to Liming Ding,[email protected]
Received 9 July 2010; Accepted 18 October 2010
Academic Editor: Hichem Ben-El-Mechaiekh
Copyrightq2010 Liming Ding et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.
We consider two nonlinear scalar delay differential equations with variable delays and give some new conditions for the boundedness and stability by means of the contraction mapping principle. We obtain the differences of the two equations about the stability of the zero solution. Previous results are improved and generalized. An example is given to illustrate our theory.
1. Introduction
Fixed point theory has been used to deal with stability problems for several years. It has conquered many difficulties which Liapunov method cannot. While Liapunov’s direct method usually requires pointwise conditions, fixed point theory needs average conditions.
In this paper, we consider the nonlinear delay differential equations
xt −atxt−r1t btgxt−r2t, 1.1
xt −atfxt−r1t btgxt−r2t, 1.2
wherer1t, r2t :0,∞ → 0,∞,r max{r10, r20},a, b :0,∞ → R,f, g : R → R are continuous functions. We assume the following:
A1r1tis differentiable,
Many authors have investigated the special cases of 1.1 and 1.2. Since Burton
1 used fixed point theory to investigate the stability of the zero solution of the equation
xt −atxt−r, many scholars continued his idea. For example, Zhang2 has studied the equation
xt −atxt−rt, 1.3
Becker and Burton3 have studied the equation
xt −atfxt−rt, 1.4
Jin and Luo4 have studied the equation
xt −atxt−r1t btx1/3t−r2t. 1.5
Burton 5 and Zhang 6 have also studied similar problems. Their main results are the following.
Theorem 1.1Burton1 . Suppose thatrt r, a constant, and there exists a constantα <1such that
t
t−r
|asr|ds t
0
|asr|e−staurdu
s
s−r
|aur|du ds≤α, 1.6
for allt≥ 0and0∞asds ∞. Then, for every continuous initial functionψ :−r,0 → R, the solutionxt xt,0, ψof1.3is bounded and tends to zero ast → ∞.
Theorem 1.2Zhang2 . Suppose thatris differentiable, the inverse functionhtoft−rtexists, and there exists a constantα∈0,1such that fort≥0
i
lim inf t→ ∞
t
0
ahs>−∞, 1.7
ii
t
t−rt|ahs|ds
t
0
e−stahudu|ahs|
s
s−rs|ahv|dv dsθs, 1.8
where θt 0te−stahudu|as||rs|ds. Then, the zero solution of 1.3 is asymptotically stable if and only if
iii
t
0
Theorem 1.3Burton7 . Suppose thatrt r, a constant. Letf be odd, increasing on0, L , and satisfies a Lipschitz condition, and letx−fxbe nondecreasing on0, L. Suppose also that for eachL1∈0, L, one has
L1−fL1sup t≥0
t
0
e−staurdu|asr|dsfL1sup t≥0
t
t−r
|aur|du
fL1sup t≥0
t
0
e−staurdu|asr|
s
s−r
|aur|du ds < L1,
1.10
and there existsJ >0such that
−
t
0
asrds≤J fort≥0. 1.11
Then, the zero solution of 1.4is stable.
Theorem 1.4 Becker and Burton 3 . Suppose f is odd, strictly increasing, and satisfies a Lipschitz condition on an interval−l, l and thatx−fxis nondecreasing on0, l . If
sup t≥t1
t
t−rtaudu < 1
2, 1.12
wheret1is the unique solution oft−rt 0, and if a continuous functiona :0,∞ → Rexists such that
at at1−rt, 1.13
on 0,∞, then the zero solution of 1.5 is stable at t 0. Furthermore, if f is continuously differentiable on−l, l withf0/0and
t
0
audu−→ ∞ as t−→ ∞, 1.14
then the zero solution of 1.4is asymptotically stable.
In the present paper, we adopt the contraction mapping principle to study the boundedness and stability of 1.1and 1.2. That means we investigate how the stability
property will be when 1.3 and 1.4 are added to the perturbed termbtgxt−r2t. We obtain their differences about the stability of the zero solution, and we also improve and generalize the special caser1t r1. Finally, we give an example to illustrate our theory.
2. Main Results
xt,0, ψ ψton−r,0 . LetCS1, S2denote the set of all continuous functionsφ:S1 → S2 andψmax{|ψt|:t∈−r,0 }. Stability definitions can be found in8 .
Theorem 2.1. Suppose that the following conditions are satisfied:
ig0 0, and there exists a constantL >0so that if|x|,|y| ≤L, then
gx−g
y≤x−y, 2.1
iithere exists a constantα∈0,1and a continuous functionh:−r,∞ → Rsuch that
t
t−r1t
|hs|ds t
0
e−tshudu|hs|
s
s−r1s
|hu|du ds
t
0
e−tshuduhs−r1s1−r 1s
−as|bs| ds≤α,
2.2
iii
lim inf t→ ∞
t
0
hsds >−∞. 2.3
Then, the zero solution of 1.1is asymptotically stable if and only if
iv
t
0
hsds−→ ∞, ast−→ ∞. 2.4
Proof. First, suppose thativholds. We set
Jsup
t≥0
−
t
0
hsds
. 2.5
LetS{φ|φ∈C−r,∞, R,φsupt≥−r|φt|<∞}, thenSis a Banach space. Multiply both sides of1.1bye0thsds, and then integrate from 0 totto obtain
xt x0e−
t 0hsds
t
0
e−sthuduhsxsds
−
t
0
e−sthuduasxs−r1sds
t
0
e−sthudubsgxs−r2sds.
By performing an integration by parts, we have
xt x0e−
t 0hsds
t
0
e−sthudu
s
s−r1s
huxudu
ds
t
0
e−sthuduhs−r1s1−r 1s
−as xs−r1sds
t
0
e−sthudubsgxs−r2sds,
2.7
or
xt x0e−
t
0hsds−e− t
0hsds
0
−r10
hsxsds t
t−r1t
hsxsds
−
t
0
e−sthuduhs
s
s−r1s
huxudu ds
t
0
e−sthuduhs−r1s1−r 1s
−as xs−r1sds
t
0
e−sthudubsgxs−r2sds.
2.8
Let
M
φ|φ∈S,sup t≥−r
φt≤L, φt ψtfort∈−r,0 , φt−→0 ast−→ ∞. 2.9
Then,Mis a complete metric space with metric supt≥0|φt−ηt|forφ, η∈M. For allφ∈M, define the mappingP
P φt ψt, t∈−r,0 ,
P φt ψ0e−0thsds−e− t
0hsds
0
−r10
hsψsds t
t−r1t
hsφsds
−
t
0
e−sthuduhs
s
s−r1s
huφudu ds
t
0
e−sthuduhs−r1s1−r 1s
−as φs−r1sds
t
0
e−sthudubsgφs−r2sds, t≥0.
Byiandg0 0,
P φt≤ψ
1
0
−r10
|hs|ds
e−0thsds
L
t
t−r1t
|hs|ds t
0
e−sthudu|hs|
s
s−r1s
|hu|du ds
t
0
e−sthuduhs−r1s1−r 1s
−as|bs| ds
≤ψ
1
0
−r10
|hs|ds
eJαL.
2.11
Thus, whenψ ≤δ 1−αL/1−0r
10|hs|ds e
J,|P φt| ≤L.
We now show thatP φt → 0 as t → ∞. Sinceφt → 0 andt−r1t → ∞ as
t → ∞, for eachε >0, there exists aT1>0 such thatt > T1implies|φt−r1t|< ε. Thus, for
t≥T1,
|I1|
t
t−r1t
hsφsds≤ε t
t−r1t
|hs|ds≤αε. 2.12
Hence,I1 → 0 ast → ∞. And
|I2|
t
0
e−sthuduhs
s
s−r1s
huφudu ds
≤
T1
0
e−sthudu|hs|
s
s−r1s
|hu|φudu ds
t
T1
e−sthudu|hs|
s
s−r1s
|hu|φudu ds
≤L T1
0
e−sthudu|hs|
s
s−r1s
|hu|du dsε t
T1
e−sthudu|hs|
s
s−r1s
|hu|du ds,
2.13
Byiiandiv, there existsT2> T1such thatt≥T2implies
L T1
0
e−sthudu|hs|
s
s−r1s
|hu|du ds < ε. 2.14
Applyiito obtain|I2|< ε2ε <2ε. Thus,I2 → 0 ast → ∞. Similarly, we can show that the rest term in2.10approaches zero ast → ∞. This yieldsP φt → 0 ast → ∞, and hence
Also, byii,Pis a contraction mapping with contraction constantα. By the contraction mapping principle,Phas a unique fixed pointxinMwhich is a solution of1.1withxs
ψson−r,0 andxt → 0 ast → ∞.
In order to prove stability att 0, let ε > 0 be given. Then, choose m > 0 so that
m <min{ε, L}. ReplacingLwithminM, we see there is aδ >0 such that ψ< δimplies that the unique continuous solutionxagreeing withψon−r,0 satisfies|xt| ≤m < εfor all
t≥ −r. This shows that the zero solution of1.1is asymptotically stable ifivholds.
Conversely, supposeivfails. Then, byiii, there exists a sequence{tn},tn → ∞as
n → ∞such that limn→ ∞0tnhsdslfor somel∈R. We may choose a positive constantN satisfying
−N≤ tn
0
hsds≤N, 2.15
for alln≥1. To simplify the expression, we define
ωs hs−r1s
1−r1s−as|bs|hs s
s−r1s
|hu|du, 2.16
for alls≥0. Byii, we have
tn
0
e−stnhuduωsds≤α. 2.17
This yields
tn
0
e0shuduωsds≤αe tn
0 hudu ≤αeN. 2.18
The sequence{tn 0 e
s
0huduωsds}is bounded, so there exists a convergent subsequence. For
brevity of notation, we may assume that
lim n→ ∞
tn
0
e0shuduωsdsγ, 2.19
for someγ ∈Rand choose a positive integerkso large that
tn
tk
e0shuduωsds < δ0
4N, 2.20
Byiii,J in2.5is well defined. We now consider the solutionxt xt, tk, ψ of 1.1withψtk δ0and|ψs| ≤δ0 fors≤tk. We may chooseψ so that|xt| ≤Lfort≥tk and
ψtk− t
k
tk−r1tk
hsψsds≥ 1
2δ0. 2.21
It follows from2.10withxt P xtthat forn≥tk,
xtn−
tn
tn−r1tn
hsxsds≥ 1
2δ0e
−tn t
khudu−
tn
tk
e−stnhuduωsds
1 2δ0e
−tn
tkhudu−e−0tnhudu
tn
tk
e0shuduωsds
e− tn
tkhudu
1 2δ0−e
−tk 0 hudu
tn
tk
e0shuduωsds
≥e− tn
tkhudu
1 2δ0−N
tn
tk
e0shuduωsds
≥ 1
4δ0e
−tn
tkhudu≥ 1
4δ0e
−2N>0.
2.22
On the other hand, if the solution of1.1xt xt, tk, ψ → 0 ast → ∞, sincetn−r1tn →
∞asn → ∞andiiholds, we have
xtn−
tn
tn−r1tn
hsxsds−→0 asn−→ ∞, 2.23
which contradicts2.22. Hence, conditionivis necessary for the asymptotically stability of the zero solution of1.1. The proof is complete.
Whenr1t r1, a constant,ht atr1, we can get the following.
Corollary 2.2. Suppose that the following conditions are satisfied:
ig0 0, and there exists a constantL >0so that if|x|,|y| ≤L, then
gx−g
iithere exists a constantα∈0,1such that for allt≥0, one has
t
t−r1
|asr1|ds
t
0
e−staur1du|asr
1|
s
s−r1
|aur1|du ds
t
0
e−staur1du|bs|ds≤α,
2.25
iii
lim inf t→ ∞
t
0
asr1ds >−∞. 2.26
Then, the zero solution of 1.1is asymptotically stable if and only if
iv
t
0
asr1ds−→ ∞, ast−→ ∞. 2.27
Remark 2.3. We can also obtain the result thatxtis bounded byLon−r,∞. Our results
generalize Theorems1.1and1.2.
Theorem 2.4. Suppose that a continuous functiona:0,∞ → Rexists such thatat at1− r1tand that the inverse functionhtoft−r1texists. Suppose also that the following conditions are satisfied:
ithere exists a constantJ >0such thatsupt≥0{−0tahsds}< J,
iithere exists a constantL >0such thatfx, x−fx, gxsatisfy a Lipschitz condition with constantK >0on an interval−L, L ,
iiifandgare odd, increasing on0, L .x−fxis nondecreasing on0, L,
ivfor eachL1∈0, L , one has
L1−fL1sup
t≥0
t
0
e−stahudu|ahs|dsfL1sup t≥0
t
t−r1t
|ahs|ds
gL1sup t≥0
t
0
e−tsahudu|bs|ds < L1.
2.28
Proof. Byiv, there existsα∈0,1such that
L−fLsup t≥0
t
0
e−tsahudu|ahs|dsfLsup t≥0
t
t−r1t
|ahs|ds
gLsup
t≥0
t
0
e−tsahudu|bs|ds≤αL.
2.29
LetSbe the space of all continuous functionsφ:−r,∞ → Rsuch that
φ
K:sup
e−dK20t|ahs||bs|dsφt : t∈−r,∞
<∞, 2.30
where d > 3 is a constant. Then,S,| · |K is a Banach space, which can be verified with Cauchy’s criterion for uniform convergence.
The equation1.2can be transformed as
xt −ahtfxt d
dt
t
t−r1t
ahsfxsdsbtgxt−r2t
−ahtxtahtxt−fxt
d
dt t
t−r1t
ahsfxsdsbtgxt−r2t.
2.31
By the variation of parameters formula, we have
xt x0e−
t
0ahsds−e− t
0ahsds
0
−r10
ahsfxsds t
0
e−stahuduxs−fxs ds
t
t−r1t
ahsfxsds t
0
e−stahudubsgxs−r2sds.
2.32
Let
M
φ|φ∈S, sup
t≥−r
φt≤L, φt ψt, t∈−r,0
thenMis a complete metric space with metric|φ−η|K forφ, η ∈ M. For allφ ∈M, define the mappingP
P φt ψt, t∈−r,0 ,
P φt ψ0e−0tahsds−e− t
0ahsds
0
−r10
ahsfψsds
t
0
e−stahuduφs−fφs ds
t
t−r1t
ahsfφsds t
0
e−stahudubsgφs−r2sds.
2.34
Byi,iii, and2.29, we have
P φ
t≤ψeJeJfψ 0
−r10
|ahs|dsL−fLsup t≥0
t
0
e− t
s−r1sahuduahsds
fLsup
t≥0
t
t−r1t
|ahs|dsgLsup
t≥0
t
0
estahudu|bs|ds
≤ψeJeJfψ 0
−r10
|ahs|dsαL.
2.35
Thus, there existsδ∈0, Lsuch thateJ1Kδ0
−r10|ahs|ds <1−αLand|P φt| ≤L.
Hence,P φ∈M.
We now show thatP is a contraction mapping inM. For allφ, η∈M,
P φt−P ηt≤ t
0
e−stahudu|ahs|φs−fφs−ηs fηsds
t
t−r1t
|ahs|fφs−fηsds
t
0
e−stahudu|bs|gφs−r2s−gηs−r2sds.
Since
e−dK20t|ahs||bs|ds
t
0
e−stahudu|ahs|φs−fφs−ηs fηsds
≤
t
0
e−tsahudu|ahs|Kφs−ηse−dK2 s
0|ahu||bu|du
×e−dK2st|ahu||bu|duds
≤
t
0
e−tsahudu|ahs|Ke−dK2 t
s|ahu||bu|dudsφ−η K
≤ 1
dφ−ηK,
e−dK20t|ahs||bs|ds
t
t−r1t
|ahs|fφs−fηsds
≤
t
t−r1t
|ahs|Kφs−ηse−dK20s|ahu||bu|due−dK2 t
s|ahu||bu|duds
≤
t
t−r1t
|ahs|Ke−dK2st|ahu||bu|dudsφ−η K
≤ 1
dφ−ηK,
e−dK20t|ahs||bs|ds
t
0
e−stahudu|bs|gφs−r2s− gηs−r2sds
≤
t
0
e−tsahudu|bs|Kφs−r2s−ηs−r2se−dK2 s−r2s
0 |ahu||bu|du
e−dK2 s
s−r2s|ahu||bu|due−dK2st|ahu||bu|duds
≤
t
0
e−tsahudu|bs|Ke−dK2 t
s|ahu||bu|dudsφ−η K
≤ 1
dφ−ηK,
2.37
we havee−dK20t|ahs||bs|ds|P φt−P ηt| ≤ 3/d|φ−η|
K. That means|P φ−P η| ≤ 3/d|φ−η|K. Hence,Pis a contraction mapping inMwith constant 3/d. By the contraction mapping principle, P has a unique fixed point x in M, which is a solution of 1.2 with
xs ψson−r,0 and supt≥−r|xt| ≤L.
In order to prove stability att 0, let ε > 0 be given. Then, choose m > 0 so that
m < min{ε, L}. ReplacingLwithminM, we see there is aδ > 0 such thatψ< δimplies that the unique continuous solutionxagreeing withψon−r,0 satisfies|xt| ≤m < εfor all
Whenr1t r1, a constant, we have the following.
Corollary 2.5. Suppose that the following conditions are satisfied:
ithere exists a constantJ >0such thatsupt≥0{−0tasr1ds}< J,
iithere exists a constantL >0such thatfx, x−fx, gxsatisfy a Lipschitz condition with constantK >0on an interval−L, L ,
iiifandgare odd, increasing on0, L .x−fxis nondecreasing on0, L,
ivfor eachL1∈0, L , one has
L1−fL1sup
t≥0
t
0
e−staur1du|asr
1|dsfL1sup t≥0
t
t−r1
|aur1|du
gL1sup t≥0
t
0
e−staur1du|bs|ds < L
1.
2.38
Then, the zero solution of the equation
xt −atfxt−r1 btgxt−r2t 2.39
is stable.
Corollary 2.6. Suppose that the following conditions are satisfied:
ithere exists a constantJ >0such thatsupt≥0{−0tasds}< J,
iithere exists a constantL > 0such thatfx,x−fx,gxsatisfy a Lipschitz condition with constantK >0on an interval−L, L ,
iiifandgare odd, increasing on0, L .x−fxis nondecreasing on0, L,
ivfor eachL1∈0, L , one has
L1−fL1sup
t≥0
t
0
e−staudu|as|dsgL1sup t≥0
t
0
e−staudu|bs|ds < L1. 2.40
Then, the zero solution of
xt −atfxt btgxt−rt 2.41
is stable.
Remark 2.7. The zero solution of1.2is not as asymptotically stable as that of1.1. The key
is thatMis not complete under the weighted metric when added the condition toMthat
φt → 0 ast → ∞.
3. An Example
We use an example to illustrate our theory. Consider the following differential equation:
xt −atxt−r1t btgxt−r2t. 3.1
wherer1t 0.281t,r2∈CR, R,gx x3,at 1/0.719t1, andbt μsint/t1,
μ >0. This equation comes from4 .
Choosinght 1.2/t1, we have
t
t−r1t
|hs|ds t
0.719t 1.2
s1ds1.2 ln
t1
0.719t1 <0.396,
t
0
e−sthuduhs−r1s1−r 1s
−asds
t 0e−
t
shudu|hs|s
s−r1s|hu|du ds <0.396,
t 0e−
t
s1.2/u1du1−1.2×0.719
0.719s1 ds
< 1−1.2× −0.719
0.719s1
t 0e−
t
s1.2/u1du 1.2
s1ds <0.1592,
t
0
e−sthudu|bs|ds≤ μ
1.2.
3.2
Letα:0.3960.3960.1592μ/1.2, whenμis sufficiently small,α <1. Then, the condition iiofTheorem 2.1is satisfied.
LetL√3/3, then the conditioniofTheorem 2.1is satisfied.
And0thsds 0t1.2/s1ds 1.2 lnt1, then the conditioniiiandivof
Theorem 2.1are satisfied.
According toTheorem 2.1, the zero solution of3.1is asymptotically stable.
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