Hindawi Publishing Corporation
Abstract and Applied Analysis
Volume 2012, Article ID 307968, 14 pages
doi:10.1155/2012/307968
Research Article
Existence of Positive Solutions of
Neutral Differential Equations
B. Dorociaková, M. Kubjatková, and R. Olach
Department of Mathematics, University of Žilina, 010 26 Žilina, Slovakia
Correspondence should be addressed to B. Dorociaková, bozena.dorociakova@fstroj.uniza.sk
Received 10 November 2011; Accepted 14 December 2011
Academic Editor: Josef Diblı́k
Copyright q 2012 B. Dorociaková 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.
The paper contains some suffcient conditions for the existence of positive solutions which are
bounded below and above by positive functions for the nonlinear neutral differential equations
of higher order. These equations can also support the existence of positive solutions approaching
zero at infinity.
1. Introduction
This paper is concerned with the existence of a positive solution of the neutral differential
equations of the form:
dn
xt − atxt − τ −1n1 ptfxt − σ,
dtn
t ≥ t0 ,
1.1
where n > 0 is an integer, τ > 0, σ ≥ 0, a ∈ Ct0 , ∞, 0, ∞, p ∈ CR, 0, ∞, f ∈ CR, R, f
is a nondecreasing function and xfx > 0, x /
0.
By a solution of 1.1 we mean a function x ∈ Ct1 − τ, ∞, R for some t1 ≥ t0 , such
that xt − atxt − τ is n-times continuously differentiable on t1 , ∞ and such that 1.1 is
satisfied for t ≥ t1 .
The problem of the existence of solutions of neutral differential equations has been
studied and discussed by several authors in the recent years. For related results we refer
the reader to 1–17 and the references cited therein. However, there is no conception which
guarantees the existence of positive solutions which are bounded below and above by
positive functions. Maybe it is due to the technical difficulties arising in the analysis of
the problem. In this paper we presented some conception. The method also supports the
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existence of positive solutions which approaching zero at infinity. Some examples illustrating
the results.
The existence and asymptotic behavior of solutions of the nonlinear neutral differential
equations and systems have been also solved in 1–7, 12, 15.
As much as we know for 1.1 in the literature, there is no result for the existence of
solutions which are bounded by positive functions. Only the existence of solutions which
are bounded by constants is treated and discussed, for example, in 10, 15, 17. It seems that
conditions of theorems are rather complicate, but cannot be simpler due to Corollaries 2.4,
2.8, and 3.3.
The following fixed point theorem will be used to prove the main results in the next
section.
Lemma 1.1 see 7, 10, 12 Krasnoselskii’s fixed point theorem. Let X be a Banach space, let Ω
be a bounded closed convex subset of X, and let S1 , S2 be maps of Ω into X such that S1 x S2 y ∈ Ω
for every pair x, y ∈ Ω. If S1 is a contractive and S2 is completely continuous then the equation:
S1 x S2 x x
1.2
has a solution in Ω.
2. The Existence of Positive Solution
In this section, we will consider the existence of a positive solution for 1.1 which is bounded
by two positive functions. We will use the notation m max{τ, σ}.
Theorem 2.1. Suppose that there exist bounded functions u, v ∈ C1 t0 , ∞, 0, ∞, constant c > 0,
and t1 ≥ t0 m such that
ut ≤ vt,
t ≥ t0 ,
vt − vt1 − ut ut1 ≥ 0, t0 ≤ t ≤ t1 ,
∞
1
1
ut s − tn−1 psfvs − σds
ut − τ
n − 1! t
∞
1
1
≤ at ≤
vt s − tn−1 psfus − σds
vt − τ
n − 1! t
2.1
2.2
≤ c < 1,
2.3
t ≥ t1 .
Then 1.1 has a positive solution which is bounded by the functions u, v.
Proof. Let Ct0 , ∞, R be the set of all continuous bounded functions with the norm ||x|| supt≥t0 |xt|. Then Ct0 , ∞, R is a Banach space. We define a closed, bounded, and convex
subset Ω of Ct0 , ∞, R as follows:
Ω {x xt ∈ Ct0 , ∞, R : ut ≤ xt ≤ vt, t ≥ t0 }.
2.4
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3
We now define two maps S1 and S2 : Ω → Ct0 , ∞, R as follows:
S1 xt ⎧
⎨atxt − τ, t ≥ t1 ,
⎩S xt ,
t0 ≤ t ≤ t 1 ,
1
1
∞
⎧
1
⎪
⎨−
s − tn−1 psfxs − σds, t ≥ t1 ,
−
1!
n
t
S2 xt ⎪
⎩
t0 ≤ t ≤ t 1 .
S2 xt1 vt − vt1 ,
2.5
We will show that for any x, y ∈ Ω we have S1 x S2 y ∈ Ω. For every x, y ∈ Ω and t ≥ t1 we
obtain
S1 xt S2 y t ≤ atvt − τ −
1
n − 1!
∞
s − tn−1 psfus − σds ≤ vt.
2.6
t
For t ∈ t0 , t1 we have
S1 xt S2 y t S1 xt1 S2 y t1 vt − vt1 ≤ vt1 vt − vt1 vt.
2.7
Furthermore for t ≥ t1 we get
1
S1 xt S2 y t ≥ atut − τ −
n − 1!
∞
s − tn−1 psfvs − σds ≥ ut.
2.8
t
Finally let t ∈ t0 , t1 and with regard to 2.2 we get
vt − vt1 ut1 ≥ ut,
t0 ≤ t ≤ t 1 .
2.9
Then for t ∈ t0 , t1 and any x, y ∈ Ω we get
S1 xt S2 y t S1 xt1 S2 y t1 vt − vt1 ≥ ut1 vt − vt1 ≥ ut.
2.10
Thus, we have proved that S1 x S2 y ∈ Ω for any x, y ∈ Ω.
We will show that S1 is a contraction mapping on Ω. For x, y ∈ Ω and t ≥ t1 we have
S1 xt − S1 y t |at|xt − τ − yt − τ ≤ cx − y.
2.11
This implies that
S1 x − S1 y ≤ cx − y.
2.12
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Abstract and Applied Analysis
Also for t ∈ t0 , t1 the inequality above is valid. We conclude that S1 is a contraction mapping
on Ω.
We now show that S2 is completely continuous. First we will show that S2 is
continuous. Let xk xk t ∈ Ω be such that xk t → xt as k → ∞. Because Ω is closed,
x xt ∈ Ω. For t ≥ t1 we have
|S2 xk t − S2 xt|
1 ∞
n−1
ps
fx
≤
−
t
−
σ
−
fxs
−
σ
ds
s
s
k
n − 1! t
∞
1
≤
s − t1 n−1 psfxk s − σ − fxs − σds.
n − 1! t1
2.13
According to 2.8 we get
∞
s − t1 n−1 psfvs − σds < ∞.
2.14
t1
Since |fxk s − σ − fxs − σ| → 0 as k → ∞, by applying the Lebesgue dominated
convergence theorem we obtain that
lim S2 xk t − S2 xt 0.
2.15
k→∞
This means that S2 is continuous.
We now show that S2 Ω is relatively compact. It is sufficient to show by the ArzelaAscoli theorem that the family of functions {S2 x : x ∈ Ω} is uniformly bounded and
equicontinuous on t0 , ∞. The uniform boundedness follows from the definition of Ω. For
the equicontinuity we only need to show, according to Levitan result 8, that for any given
ε > 0 the interval t0 , ∞ can be decomposed into finite subintervals in such a way that on
each subinterval all functions of the family have change of amplitude less than ε. With regard
to the condition 2.14, for x ∈ Ω and any ε > 0 we take t∗ ≥ t1 large enough so that
1
n − 1!
∞
t∗
s − t1 n−1 psfxs − σds <
ε
.
2
2.16
Then for x ∈ Ω, T2 > T1 ≥ t∗ we have
|S2 xT2 − S2 xT1 |
∞
∞
1
1
≤
s − t1 n−1 psfxs − σds s − t1 n−1 psfxs − σds
n − 1! T2
n − 1! T1
<
ε ε
ε.
2 2
2.17
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5
For x ∈ Ω, t1 ≤ T1 < T2 ≤ t∗ and n ≥ 2 we get
1 ∞
s − T1 n−1 psfxs − σds
|S2 xT2 − S2 xT1 | n − 1! T1
−
∞
s − T2 n−1
T2
psfxs − σds
1 T2
s − T1 n−1 psfxs − σds
n − 1! T1
∞
s − T1 n−1 psfxs − σds
T2
−
∞
s − T2 T2
1
≤
n − 1!
≤ max∗
t1 ≤s≤t
psfxs − σds
sn−1 psfxs − σds
T1
1
n − 1!
T2
n−1
∞
s − T1 n−1 − s − T2 n−1 psfxs − σds
T2
1
n−1
s psfxs − σ T2 − T1 n − 1!
1
n − 1!
∞
s − T1 − s − T2 T2
× s − T1 n−2 s − T1 n−3 s − T2 · · · s − T1 s − T2 n−3
s − T2 n−2 psfxs − σds
≤ max∗
t1 ≤s≤t
1
sn−1 psfxs − σ T2 − T1 n − 1!
1
n − 2!
∞
T2 − T1 s − T1 n−2 psfxs − σds.
T2
2.18
With regard to the condition 2.14 we have that
1
n − 2!
∞
T2
s − T1 n−2 psfxs − σds < B,
B > 0.
2.19
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Abstract and Applied Analysis
Then we obtain
|S2 xT2 − S2 xT1 | <
max∗
t1 ≤s≤t
1
sn−1 psfxs − σ B T2 − T1 .
n − 1!
2.20
Thus there exists a δ1 ε/M B, where
M max∗
t1 ≤s≤t
1
sn−1 psfxs − σ ,
n − 1!
2.21
such that
|S2 xT2 − S2 xT1 | < ε
if 0 < T2 − T1 < δ1 .
2.22
For n 1 we proceed by the similar way as above. Finally for any x ∈ Ω, t0 ≤ T1 < T2 ≤ t1
there exists a δ2 > 0 such that
T2
v sds
|S2 xT2 − S2 xT1 | |vT1 − vT2 | T1
≤ max v s T2 − T1 < ε
t0 ≤s≤t1
2.23
if 0 < T2 − T1 < δ2 .
Then {S2 x : x ∈ Ω} is uniformly bounded and equicontinuous on t0 , ∞ and hence S2 Ω is
relatively compact subset of Ct0 , ∞, R. By Lemma 1.1 there is an x0 ∈ Ω such that S1 x0 S2 x0 x0 . We conclude that x0 t is a positive solution of 1.1. The proof is complete.
Corollary 2.2. Suppose that all conditions of Theorem 2.1 are satisfied and
lim vt 0.
t→∞
2.24
Then 1.1 has a positive solution which tends to zero.
Corollary 2.3. Suppose that there exist bounded functions u, v ∈ C1 t0 , ∞, 0, ∞, constant c > 0
and t1 ≥ t0 m such that 2.1, 2.3 hold and
v t − u t ≤ 0,
t0 ≤ t ≤ t 1 .
2.25
Then 1.1 has a positive solution which is bounded by the functions u, v.
Proof. We only need to prove that condition 2.25 implies 2.2. Let t ∈ t0 , t1 and set
Ht vt − vt1 − ut ut1 .
2.26
Abstract and Applied Analysis
7
Then with regard to 2.25, it follows that H t v t − u t ≤ 0, t0 ≤ t ≤ t1 . Since Ht1 0
and H t ≤ 0 for t ∈ t0 , t1 , this implies that
Ht vt − vt1 − ut ut1 ≥ 0,
t0 ≤ t ≤ t1 .
2.27
Thus all conditions of Theorem 2.1 are satisfied.
Corollary 2.4. Suppose that there exists a bounded function v ∈ C1 t0 , ∞, 0, ∞, constant c > 0
and t1 ≥ t0 m such that
∞
1
1
vt at s − tn−1 psfvs − σds ≤ c < 1,
vt − τ
n − 1! t
t ≥ t1 .
2.28
Then 1.1 has a solution xt vt, t ≥ t1 .
Proof. We put ut vt and apply Theorem 2.1.
Theorem 2.5. Suppose that p is bounded and there exist bounded functions u, v ∈ C1 t0 , ∞,
0, ∞, constant c > 0 and t1 ≥ t0 m such that 2.1, 2.2 hold and
t
1
n−1
ut −
t − s psfus − σds
n − 1! t1
t
1
1
n−1
vt −
≤ at ≤
t − s psfvs − σds
vt − τ
n − 1! t1
1
ut − τ
≤ c < 1,
2.29
t ≥ t1 ,
if n is odd,
t
1
n−1
ut t − s psfvs − σds
n − 1! t1
t
1
1
n−1
vt ≤ at ≤
t − s psfus − σds
vt − τ
n − 1! t1
1
ut − τ
≤ c < 1,
2.30
t ≥ t1 ,
if n is even, and
t
t − sn−2 psfvs − σds ≤ K,
t ≥ t1 , K > 0, n ≥ 2.
t1
Then 1.1 has a positive solution which is bounded by the functions u, v.
2.31
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Abstract and Applied Analysis
Proof. Let Ct0 , ∞, R be the set as in the proof of Theorem 2.1. We define a closed, bounded,
and convex subset Ω of Ct0 , ∞, R as in the proof of Theorem 2.1. We define two maps S1
and S2 : Ω → Ct0 , ∞, R as follows:
S1 xt ⎧
⎨atxt − τ,
t ≥ t1 ,
⎩
t0 ≤ t ≤ t 1 ,
S1 xt1 ,
⎧
n1 t
⎪
⎪
⎨ −1
t − sn−1 psfxs − σds, t ≥ t1 ,
−
1!
n
S2 xt t1
⎪
⎪
⎩S xt vt − vt ,
t 0 ≤ t ≤ t1 .
2
1
1
2.32
We shall show that for any x, y ∈ Ω we have S1 x S2 y ∈ Ω. For n odd, every x, y ∈ Ω and
t ≥ t1 we obtain
1
S1 xt S2 y t ≤ atvt − τ n − 1!
t
t − sn−1 psfvs − σds ≤ vt.
t1
2.33
For t ∈ t0 , t1 , we have
S1 xt S2 y t S1 xt1 S2 y t1 vt − vt1 ≤ vt1 vt − vt1 vt.
2.34
Furthermore for t ≥ t1 , we get
S1 xt S2 y t ≥ atut − τ 1
n − 1!
t
t − sn−1 psfus − σds ≥ ut.
t1
2.35
Let t ∈ t0 , t1 and according to 2.2 we have
vt − vt1 ut1 ≥ ut.
2.36
Then for t ∈ t0 , t1 and any x, y ∈ Ω we get
S1 xt S2 y t S1 xt1 S2 y t1 vt − vt1 ≥ ut1 vt − vt1 ≥ ut.
2.37
Thus we have proved that S1 x S2 y ∈ Ω for any x, y ∈ Ω.
For n even by the similar way as above we can prove that S1 x S2 y ∈ Ω for any
x, y ∈ Ω.
Abstract and Applied Analysis
9
As in the proof of Theorem 2.1, we can show that S1 is a contraction mapping on Ω.
We now show that S2 is completely continuous. First, we will show that S2 is
continuous. Let xk xk t ∈ Ω be such that xk t → xt as k → ∞. Because Ω is closed,
x xt ∈ Ω. For t ≥ t1 we have
|S2 xk t − S2 xt| ≤
1
n − 1!
t
t − sn−1 psfxk s − σ − fxs − σds.
2.38
t1
According to 2.33 there exists a positive constant M such that
t
t − sn−1 psfvs − σds ≤ M
for t ≥ t1 .
2.39
t1
The inequality above also holds for n even.
Since |fxk s − σ − fxs − σ| → 0 as k → ∞, by applying the Lebesgue dominated
convergence theorem we obtain that
lim S2 xk t − S2 xt 0.
k→∞
2.40
This means that S2 is continuous.
We now show that S2 Ω is relatively compact. It is sufficient to show by the ArzelaAscoli theorem that the family of functions {S2 x : x ∈ Ω} is uniformly bounded and
equicontinuous on t0 , ∞. The uniform boundedness follows from the definition of Ω. For
n ≥ 2 and with regard to 2.31 we have
t
d
1
S2 xt t − sn−2 psfxs − σds
dt
n − 2!
t1
t
1
≤
t − sn−2 psfvs − σ ≤ M1 ,
n − 2! t1
2.41
and for n 1 we obtain
d
S2 xt ptfvt − σ ≤ M2 ,
dt
2.42
for t ≥ t1 , M2 > 0 and |d/dtS2 xt| |v t| ≤ M3 for t0 ≤ t ≤ t1 , M3 > 0, which shows
the equicontinuity of the family S2 Ω, cf. 7, page 265. Hence S2 Ω is relatively compact and
therefore S2 is completely continuous. By Lemma 1.1, there is x0 ∈ Ω such that S1 x0 S2 x0 x0 . Thus x0 t is a positive solution of 1.1. The proof is complete.
Corollary 2.6. Suppose that all conditions of Theorem 2.5 are satisfied and
lim vt 0.
t→∞
Then 1.1 has a positive solution which tends to zero.
2.43
10
Abstract and Applied Analysis
Corollary 2.7. Suppose that p is bounded and there exist bounded functions u, v ∈ C1 t0 , ∞,
0, ∞, constant c > 0 and t1 ≥ t0 m such that 2.1, 2.29, 2.30, 2.31 hold and
v t − u t ≤ 0,
t0 ≤ t ≤ t 1 .
2.44
Then 1.1 has a positive solution which is bounded by the functions u, v.
Proof. The proof is similar to that of Corollary 2.3 and we omit it.
Corollary 2.8. Suppose that p is bounded and there exists a bounded function v ∈ C1 t0 , ∞,
0, ∞, constant c > 0 and t1 ≥ t0 m such that 2.31 holds and
1
at vt − τ
≤ c < 1,
−1n
vt n − 1!
t
n−1
t − s
psfvs − σds
2.45
t1
t ≥ t1 .
Then 1.1 has a solution xt vt, t ≥ t1 .
Proof. We put ut vt and apply Theorem 2.5.
3. Applications and Examples
In this section, we give some applications of the theorems above.
Theorem 3.1. Suppose that 0 < k1 ≤ k2 and there exist γ ≥ 0, c > 0, t1 ≥ t0 m such that
t0
k1
exp k2 − k1 ptdt ≥ 1,
k2
t0 −γ
exp −k2
t
psds
t−τ
×
∞
t−τ
1
exp k2
psds
n − 1!
t0 −γ
s − t
n−1
exp −k1
psf
t
≤ exp −k1
psds
×
t
s − t
n−1
psf
s−σ
t0 −γ
1
exp k1
n − 1!
exp −k2
t0 −γ
3.2
t−τ
psds
s−σ
t0 −γ
ds ≤ at
pξdξ
t
t−τ
∞
3.1
pξdξ
ds ≤ c < 1,
t ≥ t1 .
Then 1.1 has a positive solution which is bounded by two exponential functions.
Abstract and Applied Analysis
11
Proof. We set
ut exp −k2
t
t0 −γ
psds ,
vt exp −k1
t
t0 −γ
psds ,
t ≥ t0 .
3.3
We will show that the conditions of Corollary 2.3 are satisfied. With regard to 3.1 for t ∈
t0 , t1 we get
v t − u t −k1 ptvt k2 ptut
ptvt −k1 k2 ut exp k1
t0 −γ
ptvt −k1 k2 exp k1 − k2 t
≤ ptvt −k1 k2 exp k1 − k2 psds
t
t0 −γ
t0
t0 −γ
3.4
psds
psds
≤ 0.
Other conditions of Corollary 2.3 are also satisfied. The proof is complete.
Corollary 3.2. Suppose that all conditions of Theorem 3.1 are satisfied and
∞
ptdt ∞.
3.5
t0
Then 1.1 has a positive solution which tends to zero.
Corollary 3.3. Suppose that k > 0, c > 0, t1 ≥ t0 m and
at exp −k
t
psds
t−τ
×
∞
t−τ
1
exp k
psds
n − 1!
t0
s − tn−1 psf
exp −k
t
3.6
s−σ
pξdξ
ds ≤ c < 1,
t ≥ t1 .
t0
Then 1.1 has a solution:
xt exp −k
t
psds ,
t0
Proof. We put k1 k2 k, γ 0 and apply Theorem 3.1.
t ≥ t1 .
3.7
12
Abstract and Applied Analysis
Example 3.4. Consider the nonlinear neutral differential equation:
xt − atxt − 2 px3 t − 1,
3.8
t ≥ t0 ,
where p ∈ 0, ∞. We will show that the conditions of Theorem 3.1 are satisfied. The condition
3.1 has a form:
k1
exp k2 − k1 pγ ≥ 1,
k2
3.9
0 < k1 ≤ k2 , γ ≥ 0. For function at, we obtain
1
exp −2pk2 exp p k2 γ − t0 − 2 − 3k1 γ − t0 − 1 k2 − 3k1 t
3k1
≤ at ≤ exp −2pk1
3.10
1
exp p k1 γ − t0 − 2 − 3k2 γ − t0 − 1 k1 − 3k2 t ,
3k2
t ≥ t0 .
For p 1, k1 1, k2 2, γ 1, t0 1, the condition 3.9 is satisfied and
e−4 1 −t
e4
e ≤ at ≤ e−2 e−5t ,
3e
6
t ≥ t1 ≥ 3.
3.11
If the function at satisfies 3.11, then 3.8 has a solution which is bounded by the functions
ut exp−2t, vt exp−t, t ≥ 3.
Example 3.5. Consider the nonlinear differential equation:
xt − atxt − π ptfxt − π,
where fx √
t ≥ 0,
3.12
x, x > 0, pt 0.8 expπ − t 0.05 cos t, t ≥ 0, and
0.8 e0.1 cos t e0.1 cos t 0.8 eπ−t − 1 ≤ at ≤ e0.1 cos t e0.1 cos t √ eπ−t − 1
< 1,
b
3.13
for t ≥ π, b ∈ 1, 2. Set
ut e0.1 cos t ,
vt be0.1 cos t ,
t ≥ 0.
3.14
Then we have
v t − u t −0.1be0.1 cos t sin t 0.1e0.1 cos t sin t
−0.1b − 1e0.1 cos t sin t ≤ 0 for t ∈ 0, π.
3.15
Abstract and Applied Analysis
13
By Corollary 2.7, 3.12 has a solution which is bounded by the functions e0.1 cos t and
be0.1 cos t , t ≥ π. If
at e0.1 cos t e0.1 cos t 0.8 eπ−t − 1
for t ≥ π,
3.16
then 3.12 has the positive periodic solution xt ut e0.1 cos t , t ≥ π.
Acknowledgment
The research was supported by the Grants 1/0090/09 and 1/1260/12 of the Scientific Grant
Agency of the Ministry of Education of the Slovak Republic.
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