Nontrivial homoclinic solutions for prescribed mean curvature

Zheng and Li Advances in Difference Equations (2015) 2015:77
DOI 10.1186/s13662-015-0417-7
RESEARCH
Open Access
Nontrivial homoclinic solutions for prescribed
mean curvature Rayleigh equations
Meilin Zheng1 and Jin Li2*
*
Correspondence:
[email protected]
2
School of Science, Jiujiang
University, Jiujiang, 332005, China
Full list of author information is
available at the end of the article
Abstract
In this paper, some sufficient conditions on the existence of 2kπ -periodic solutions
for a kind of prescribed mean curvature Rayleigh equations are given. Then the
existence of nontrivial homoclinic solutions for prescribed mean curvature Rayleigh
equations is obtained.
Keywords: prescribed mean curvature Rayleigh equation; homoclinic solution;
Mawhin’s continuation theorem
1 Introduction
In this paper, we are concerned with the existence of homoclinic solutions for the following
prescribed mean curvature Rayleigh equation:
x (t)
 + x (t)
+ f x (t) + g x(t) = e(t),
(.)
where f , g, e ∈ C(R, R) with f () =  and g() = .
A solution x(t) of (.) is named homoclinic (to ) if x(t) →  and x (t) →  as |t| → .
Furthermore, x(t) is called a nontrivial homoclinic solution (.) if x(t) is a homoclinic
solution of (.) and x = .
Various types of prescribed mean equations have been studied widely by some authors
in many papers (see [–]) because of their having appeared in some scientific fields, such
as differential geometry and physics. Recently, some results on the existence of solutions
for prescribed mean equations were obtained (see [–] and references cited therein).
Feng in [] discussed a delay prescribed mean curvature Liénard equation of the form
x (t)
 + x (t)
+ f x(t) x (t) + g t, x t – τ (t) = e(t),
(.)
estimated a priori bounds by eliminating the nonlinear term ( √ x (t) ) and established
+x (t)
sufficient conditions on the existence of periodic solutions for (.) by using Mawhin’s
continuation theorem. The main difficulty overcome by Feng in [] lies in the nonlinear
term ( √ x (t) ) , the existence of which obstructs the usual method of finding a priori
+x (t)
© 2015 Zheng and Li; licensee Springer. This is an Open Access article distributed under the terms of the Creative Commons
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Zheng and Li Advances in Difference Equations (2015) 2015:77
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bounds. He transformed (.) into the following equivalent system:
x (t) = ψ(x (t)) = √x (t) ,
–x (t)
(.)
x (t) = –f (x (t))ψ(x (t)) – g(t, x (t)) + e(t),
and used Mawhin’s continuation theorem to prove the existence result. From then on,
similar approaches were used by Li and Wang [] and Li et al. []. Clearly, |x | < c < 
(c is a constant) is necessarily satisfied when Mawhin’s continuation theorem is applied to
the system (.). But, as pointed out by Liang and Lu in [], the proof of |x | < c <  was
not done in [, , ]. To do it, Li and Wang in [] gave a complementary proof.
In [], Liang and Lu investigated the following prescribed mean curvature Duffing
equation:
x (t)
 + x (t)
+ cx (t) + f x(t) = p(t),
(.)
where f ∈ C  (R, R), c > . Assume
(A ) there exist constants m > , α >  such that xf (x) ≤ –m |x|α and f (x) < , ∀x ∈ R,
and
(A ) p ∈ C(R, R) is a bounded function with p(t) =  and
 β 
β
+ sup |p(t)| < +∞,
B := max
p(t) dt ,
p(t) dt
R
where

α
+

β
R
t∈R
= .
If (A ), (A ), and condition
α

B α– + TBmα– <

√
Tmα–
(.)
hold, they obtained the existence of homoclinic solutions for (.). For the method of obtaining homoclinic solution, we also see the papers [–]. It is not difficult to see that the
condition (.) is complex and strong. In this paper, we will consider more general equation (.) and obtain the existence of homoclinic solutions under the more simple and
reasonable conditions. To find a homoclinic solution for (.), we seek a limit of a certain
sequence of kT-periodic solutions xk (t) for the following equations:
x (t)
 + x (t)
+ f x (t) + g x(t) = ek (t),
(.)
where k ∈ N, ek : R → R is a kT-periodic function (T >  is a constant) defined by
e(t),
ek (t) =
e(kT – ε ) +
e(–kT)–e(kT–ε )
(t
ε
t ∈ [–kT, kT – ε ),
– kT + ε ), t ∈ [kT – ε , kT],
ε ∈ (, T) is a constant independent of k.
(.)
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2 Preliminaries
Let X and Y be real Banach spaces and L : X ⊃ Dom L → Y be a linear operator. L is said
to be a Fredholm operator with index zero provided that
(i) Im L is closed subset of Y ,
(ii) dim ker L = codim Im L < +∞.
Set X = ker L ⊕ X , Y = Im L ⊕ Y . Let P : X → ker L and Q : Y → Y be the nature projections. It is easy to see that ker L ∩ (Dom L ∩ X ) = . Thus the restriction LP := L|Dom L∩X
is invertible. We denote by K the inverse of LP .
Let be an open bounded subset of X with Dom L ∩ = φ. A map N : → Y is said to
be L-compact in if QN : → Y and K(I – Q)N : → X are compact.
The following lemma due to Mawhin (see []) is a fundamental tool to prove the existence of kT-periodic solutions for (.).
Lemma . Let L be a Fredholm operator of index zero and Let N be L-compact on . If
the following conditions hold.
(h ) Lx = λNx, ∀(x, λ) ∈ [(D(L) \ ker L) ∩ ∂] × (, );
(h ) Nx ∈/ Im L, ∀x ∈ ker L ∩ ∂;
(h ) deg(JQN|ker L , ∩ ker L, ) = , where J : Im Q → ker L is an isomorphism.
Then Lx = Nx has at least one solution in D(L) ∩ .
The following lemma is a special case of Lemma . in [].
Lemma . If x : R → R is continuously differentiable on R, a > , then the following inequality holds:
x(t)
≤ (a)– 
t+a x(s)
 ds

+ a(a)
t–a
– p
x (s)
 ds
t+a 
.
(.)
t–a

be kT-periodic function for each k ∈ N with
Lemma . [] Let xk ∈ CkT
x ≤ A ,
k ∞
xk ∞ ≤ A ,
x ≤ A ,
k ∞
where A , A , and A are constants independent of k ∈ N. Then there exists a function
x ∈ C  (R, R) such that for each interval [c, d] ⊂ R, there is a subsequence {xkj } of {xk }k∈N
with xkj (t) → x (t) uniformly on [c, d].
In order to apply Mawhin’s continuation theorem to study the existence of kT-periodic
solution of (.), we rewrite (.) as
⎧
⎨x (t) = √ y(t) ,
–y (t)
⎩y (t) = –f ( √ y(t)
–y (t)
) – g(x(t)) + ek (t).
(.)
Obviously, if z(t) = (x(t), y(t)) is a kT-periodic solution of (.), then x(t) must be a
kT-periodic solution of (.). Hence, the problem of finding a kT-periodic solution of
(.) reduces to finding one of (.).
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Now, we set
Xk = Yk = z : z(t) = x(t), y(t) ∈ C R , R , z(t) = z(t + kT) ,
with the norm z = max{x∞ , y∞ }, where
x∞ = max x(t)
,
t∈[,kT]
y∞ = max y(t)
.
t∈[,kT]
Clearly, Xk and Yk are Banach spaces. Meanwhile, let
L : Xk ⊃ Dom L → Yk ,
Lz = z = x (t), y (t) ,
where
Dom L = z : z = x(t), y(t) ∈ C  R, R , z(t) = z(t + kT) .
Define a nonlinear operator N : Xk → Yk by
⎛
Nz = ⎝
⎞
√ y(t)
–y (t)
–f ( √ y(t) ) – g(x(t)) + e(t)
⎠.
–y (t)
Then the system (.) can be written to Lz = Nz.
kT
It is easy to see that ker L = R and Im L = {u ∈ Yk : –kT u(s) ds = }. So L is a Fredholm
operator with index zero.
Let P : Xk → ker L and Q : Yk → Im Q be defined by

Pz =
kT
kT
z(s) ds,
–kT

Qu =
kT
kT
u(s) ds,
–kT
and denote by K the inverse of L|ker P∩Dom L . Then, ker L = Im Q = R and
kT
Ku(t) =
G(t, s)u(s) ds,
(.)
–kT
where
G(t, s) =
s
,
kT
s–kT
,
kT
–kT ≤ s < t ≤ kT,
–kT ≤ t ≤ s ≤ kT.
It follows from (.) that N is L-compact on , where is an open, bounded subset of Xk .
3 Main result
For the sake of convenience, we give the following fundamental assumptions.
(H ) There exist α and β with β > α >  such that ∀x ∈ R,
αx ≤ xf (x) ≤ βx .
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(H ) There exists γ >  such that
xg(x) ≤ –γ x .
(H ) e(t) is bounded on R and
 < d := max
+∞
e (t) dt

, sup
e(t)
< +∞.
t∈R
–∞
Obviously, the conditions (H ), (H ), and (H ) are simplistic and reasonable.
Theorem . Let (H ), (H ), and (H ) hold. Then for each k ∈ N, (.) has at least one
kT-periodic solution.
Proof We consider the auxiliary system of the system (.),
x (t) = λ √ y(t)
–y (t)
= λψ(y(t)),
(.)
y (t) = –λf (ψ(y(t))) – λg(x(t)) + λek (t),
where λ ∈ (, ] is a parameter. Firstly, we will prove that the set of all possible kT-periodic
solutions of the system (.) is bounded.
Obviously, the system (.) is equivalent to the following equation:
 x (t)
λ
+
 
x (t)
λ
 x (t) + λg x(t) = λek (t).
+ λf
λ
(.)
Multiplying (.) by x and integrating from –kT to kT, we have
kT
f
λ
–kT
kT
kT
 x (t) x (t) dt + λ
g x(t) x (t) dt = λ
ek (t)x (t) dt.
λ
–kT
–kT
(.)
By (H ), we obtain
λ
kT   x (t)

x (t) x (t) dt ≥ λ
x (t)
dt
f
f
λ
λ
λ
–kT
–kT
kT

x (t)
dt.
≥α
kT
–kT
By (.), we get
kT –kT
ek (t)
 dt =
kT–ε e(t)
 dt +
–kT
kT
kT–ε
e(kT – ε )

e(–kT) – e(kT – ε )
· (t – kT + ε )
dt
ε
kT
kT–ε

  e(t)
dt +
max e(kT – ε )
, e(–kT)
dt
≤
+
–kT
kT–ε
(.)
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+∞ e(t)
 dt +
≤
kT
d dt
–∞
kT–ε
≤ d ( + ε ).
Noticing that
kT
–kT
g(x(t))x (t) dt = , we have from (.), (.), and (.)
x (t)
 dt ≤ λ
 x (t) x (t) dt f
λ
–kT
kT
g x(t) x (t) dt + ≤λ kT α
–kT
(.)
kT
–kT
≤λ
kT
–kT
 kT
ek (t) dt
·
–kT
≤ λd  + ε ·
ek (t)x (t) dt kT x (t)
 dt

–kT
x (t)
 dt
kT 
,
–kT
i.e.,
kT x (t)
 dt

≤
–kT
λd  + ε .
α
(.)
Hence, there exists a positive constant D independent of k and λ such that
x ≤ D .

(.)
Multiplying (.) by x and integrating from –kT to kT, we obtain
kT –kT
+
 x (t)
λ
 
x (t)
λ
kT
 x (t) x(t) dt + λ
x(t) dt + λ
f
g x(t) x(t) dt
λ
–kT
–kT
kT
kT
ek (t)x(t) dt.
=λ
(.)
–kT
Since x (t) = λ √ y(t) , we get y(t) =
–y (t)
kT –kT
 λ
+
x (t)
 
x (t)
λ
 x (t)
λ
.
+  x (t)
Then
λ
kT
y (t)x(t) dt
x(t) dt =
–kT
kT
=–
y(t)x (t) dt
–kT
kT
= –λ
–kT
y (t)
dt.
 – y (t)
(.)
It follows from (.) and (.) that
kT
–
–kT
g x(t) x(t) dt ≤ –
kT
f
–kT
kT
 x (t) x(t) dt +
ek (t)x(t) dt.
λ
–kT
(.)
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By (H ), we have
kT
g x(t) x(t) dt ≤ –
–kT
kT
–kT
γ x (t) dt = –γ x .
(.)
By (H ) and (.), we obtain
kT   –
x (t) x(t) dt ≤ β
x (t)
x(t)
dt
f
λ
–kT
–kT λ
 kT

kT 
 
x(t)
 dt
x (t)
dt
·
≤β
–kT λ
–kT
β ≤ · x  · x
λ
βd ≤
 + ε x .
α
kT
(.)
By (.), we get
kT
ek (t)x(t) dt ≤
–kT
kT
–kT
ek (t) dt
 ·
≤ d  + ε x .
kT x(t)
 dt

–kT
(.)
Then we get from (.), (.), (.), and (.)
γ x ≤
βd  + ε x + d  + ε x ,
α
i.e.,
x ≤
α+β d  + ε .
αγ
Therefore, there exists a positive constant D independent of k and λ such that
x ≤ D .
(.)
Thus, by using Lemma ., we have
x(t)
≤ (T)– 
t+T x(s)


+ T(T)
– 
t–T
≤ (T)
– 

= (T)– 
t+kT x(s)

+ T(T)
kT ≤ (T)

– 
x(s)

D + T(T)
x (s)

t+kT t–kT


+ T(T)– 
–kT
– 
x (s)

t–T

t–kT
t+T kT –kT
– 
D .
x (s)



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Hence, there exists a positive constant M independent of k and λ such that
x∞ ≤ M .
In what follows, we prove that there exists a constant ε with  < ε  such that
y(t)
<  – ε ,
∀t ∈ R.
(.)
Since x∞ ≤ M and g is continuous, there exists M >  such that
–M < –g x(t) + ek (t) < M ,
∀t ∈ R.
By (H ), we get
f (x) ≥ αx,
∀x > .
Now we prove by contradiction that
y(t) ≤ M
M + α 
∀t ∈ R.
,
Assume that there exist t∗ > t∗ such that
M
,
y t∗ = M + α 
M
y t∗ > M + α 
and
y(t) > M
M
+ α
∀t ∈ t∗ , t∗ .
,
Noticing that λ ∈ (, ], we have ∀t ∈ (t∗ , t∗ ),
y (t) = λ –f ψ y(t) – g x(t) + ek (t) < ,
which is a contradiction. By (H ), we get
f (x) ≤ αx,
∀x < .
By using a similar argument, we can prove that
y(t) ≥ – M
M + α 
,
∀t ∈ R.
Thus
M
M
–
≤ y(t) ≤ ,


M + α
M + α 
Therefore, (.) holds.
∀t ∈ R.
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Putting
=
x(t), y(t) ∈ C R , R : x∞ < M + , y∞ <  – ε ,
we can give a standard argument on by using Lemma . (also see [] or []) and find
that the system (.) has at least one kT-periodic solution. Equivalently, (.) has at least
one kT-periodic solution.
Theorem . Let (H ), (H ), and (H ) hold. Then (.) has at least one nontrivial homoclinic solution.
Proof If (xk (t), yk (t)) is a solution of (.), then
x (t)
k  + (xk (t))
+ f xk (t) + g xk (t) = ek (t).
(.)
From Theorem ., we have
xk ∞ < M + ,
where yk (t) = √
yk ∞ <  – ε ,
xk (t)
+(xk (t))
(.)
. Equation (.) is equivalent to
⎧
⎪
⎨xk (t) =
y (t)
k
,
–yk (t)
yk (t)
⎪
⎩yk (t) = –f ( –y (t) ) – g(xk (t)) + ek (t).
k
Since f , g, ek are continuous, we have yk (t) is continuous differentiable on R. Moreover, it
follows that xk (t) is also continuous differentiable on R. Hence,
xk (t) =
yk (t)

( – yk (t)) 
.
(.)
Since f , g, ek are continuous, we find from (.) that there exists a positive constant M
independent of k such that
y < M .
k ∞
Hence, by (.), there exists a positive constant M independent of k such that
x < M .
k ∞
By using Lemma ., we find that there is a function x ∈ C  (R, R) such that, for each
interval [a, b] ⊂ R, there is a subsequence {xkj (t)} of xk (t) with xkj (t) → x (t) uniformly on
[a, b]. In what follows, we will show that x (t) is just a homoclinic solution of (.).
For all a, b ∈ R with a < b, there exist a positive integer j and a positive real number ε
such that, for j > j , [a – ε , b + ε ] ⊂ [–kj T, kj T – ε ]. Then for j > j ,
xkj (t)
+ f xkj (t) + g xkj (t) = e(t).
 + (xkj (t))
(.)
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Since xkj (t) → x (t) uniformly on [a, b], we have
–f xkj (t) – g xkj (t) + e(t) → –f x (t) – g x (t) + e(t),
Since
xk (t)
→√
j
+(xk (t))
j
x (t)
+(x (t))
uniformly on [a, b] and
xk (t)
j
+(xk (t))
uniformly on [a, b].
= ykj (t) is continuous dif-
j
ferentiable for t ∈ [a, b], we get
xkj (t)
x (t)
→  ,
 + (x (t))
 + (xkj (t))
uniformly on [a, b].
Hence, we obtain
x (t)
 + (x (t))
+ f x (t) + g x (t) = e(t),
t ∈ [a, b].
(.)
Noticing that a, b are two arbitrary constants with a < b, we find that x : R → R is a
solution of (.).
In the following, we prove that x (t) →  and x (t) →  as |t| → +∞. Firstly, we prove
x (t) →  as |t| → +∞. Let i be a positive integer with i < kj , then
x (t)
 + x (t)
 dt = lim

iT –iT
j→+∞ –iT
≤ lim
xk (t)
 + x (t)
 dt
kj
j
iT xk (t)
 + x (t)
 dt
kj
j
kj T j→+∞ –k T
j
≤ D + D .
Let i → +∞, then
x (t)
 + x (t)
 dt = lim

∞ –∞
x (t)
 + x (t)
 dt ≤ D + D ,



iT j→+∞ –iT
which implies
lim
r→+∞ |t|≥r
x (t)
 + x (t)
 dt = .

Using Lemma ., we have
x (t)
≤ (T)– 
t+T x (s)
 dt

+ T(T)
t–T


≤  (T)–  + T(T)– 
as |t| → +∞.
t+T t–T
x (s)
 dt

x (s)
 + x (s)
 dt
t+kT t–kT
→ ,
– 



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Finally, we prove that
x (t) →  as |t| → +∞.
(.)
From (.), we get
x (t)
≤ M + ,
x (t)
≤  – ε ,

ε – ε
for t ∈ R.
Then we have
x (t)
≤ f x (t) + g x (t) + |e(t)| ≤ M ,

 + (x (t))

where M is a positive constant.
If (.) does not hold, then there exists a sequence tk satisfying
|t | < |t | < |t | < · · ·
with |tk+ | – |tk | > , k = , , . . .
and ε ∈ (,  ) such that
x (tk )
≥ ε
,

 – (ε )
k = , , . . . .
Hence we have, for t ∈ [tk , tk +
ε
],
+M
x (t)
≥ x (t)

 + (x (t)) 
t
x (tk )
x (s)
 +
ds
=  

 + (x (s))
 + (x (tk ))
tk
t
x (tk )
– x (s)
ds
≥   + (x (tk ))
 + (x (s)) tk
≥ ε –
M
· ε
 + M
> ε .
Therefore,
∞
x (t)
 dt ≥

+∞ –∞
k=
ε

tk + +M
tk


x (t)
dt = ∞,

which is a contradiction. Then (.) holds.
By (H ), we have e = . Thus it follows from f () =  and g() =  that x is nontrivial.
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4 An example
In this section, we shall construct an example to show the applications of Theorem ..
Example . Let f (x) =
equation
x
√
 + x

√x +x ,
+x
 +x
.
+x
g(x) = – √x
The prescribed mean curvature Rayleigh
+ f x (t) + g x(t) = exp –t 
(.)
has at least one nontrivial homoclinic solution.
Proof Obviously, one has

√ · x ≤ xf (x) ≤ x

and

xg(x) ≤ – √ · x .

Then (H ) and (H ) hold. Since
d = max
+∞ 
exp –t  dt
–∞

π
, sup exp –t  =  ,

t∈R
(H ) holds. Therefore, (.) has at least one nontrivial homoclinic solution.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All authors have equally contributed in obtaining new results in this article and also read and approved the final
manuscript.
Author details
1
Department of Scientific Research Management, Jiujiang University, Jiujiang, 332005, China. 2 School of Science, Jiujiang
University, Jiujiang, 332005, China.
Received: 5 September 2014 Accepted: 17 February 2015
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