On No-Node Solutions of the Lazer-McKenna Suspension Bridge Models ()
1. Introduction
In [1], the Lazer-McKenna suspension bridge models are proposed as following
![](//html.scirp.org/file/57573x3.png)
If we look for no-node solutions of the form
and impose a forcing term of the form
, then via some computation, we can obtain the following system:
(1)
In this paper, by combining the analysis of the sign of Green's functions for the linear damped equation, together with a famous fixed point theorem, we will obtain some existence results for (1) if the nonlinearities satisfy the following semipositone condition
(H) The function
is bounded below, and maybe change sign, namely, there exists a sufficiently large constant M > 0 such that ![](//html.scirp.org/file/57573x9.png)
Such case is called as semipositone problems, see [2]. And one of the common techniques is the Krasnoselskii fixed point theorem on compression and expansion of cones.
Lemma 1.1 [3]. Let
be a Banach space,and
be a cone in
. Assume
,
are open subsets of
with
,
, Let
be a completely continuous operator such that either
(i)
; or
(ii)
;
Then,
has a fixed point in ![]()
2. Preliminaries
If the linear damped equation
(2)
is nonresonant, namely, its unique T-periodic solution is the trivial one, then as a consequence of Fredholm’s alternative in [4], the nonhomogeneous equation
admits a unique T-periodic solution
which can be written as
where G(t; s) is the Green’s function of (2). For convenience,
we will assume that the following standing hypothesis is satisfied throughout this paper:
(H1)
are T-periodic functions such that the Green’s function
, associated with the linear damped equation
![]()
is positive for all
, and ![]()
(H2)
are negative T-periodic functions, and satisfy:
![]()
Let E denote the Banach space
with the norm
for
. Define K to a cone in E by
where
. Also, for r > 0 a positive number, let
![]()
If (H), (H1) and (H2) hold, let
, (1) is transformed into
(3)
where
is chosen such that
![]()
Let
be a map, which defined by
, where
![]()
![]()
t is straightforward to verify that the solution of (1) is equivalent to the fixed point Equation ![]()
Lemma 2.1 Assume that (H), (H1) and (H2) hold. Then
is compact and continuous.
For convenience, define
, for any
.
Lemma 2.2 [2] Assume that (H), (H1) and (H2) hold. If
, then, for i = 1, 2, the functions
are continuous on
,
for
, and ![]()
Lemma 2.3 [2] Assume that (H), (H1) and (H2) hold. If
, then, for i = 1, 2, the functions
are continuous on
,
for
, and ![]()
3. Main Results
Theorem 3.1 Assume that (H), (H1) and (H2) hold.
(I) Then there exists a
such that (1) has a positive periodic solution for ![]()
(II) If
, then for an
, (1) has a positive periodic solution;
(III) If
, then (1) has two positive periodic solutions for all sufficiently small
.
Proof. (I) On one hand, take R > 0 such that
![]()
Set
Then, for each
, we have
![]()
Then from the above inequalities, it follows that there exists a
such that
![]()
Furthermore, for any
, we obtain ![]()
In the similar way, there exists a
, such that
and we also have
![]()
So let us choose
and we can obtain
![]()
On the other hand, from the condition
for all
, it follows that there is a sufficient small r > 0 such that
for
and
where
is chosen such that ![]()
Then, for any
, we obtain
![]()
![]()
So we have ![]()
Therefore, from Lemma 1.1, it follows that the operator B has at least one fixed point
in
, for ![]()
(II) Since
, then from Lemma 2.1, it follows that
Define a function
as
By Lemma 2.5 in [2], it is easy to see that
Thus by the definition, there is an
such that
where
satisfying ![]()
Then, for each
, we have
![]()
In the similar way, for any
, we also have
Furthermore, from The above inequalities, we get ![]()
Therefore, from Lemma 1.1, it follows that B has one fixed point
in
for any ![]()
(III) Since
, then from Lemma 2.2, it follows that
By the definition, there exists
such that
where
is chosen such that ![]()
Choosing
and for any
, we have
and
![]()
![]()
Thus from the above inequalities, we can get ![]()
Therefore, from Lemma 1.1, it follows that the operator B has at least two fixed points
in
and
in
. Namely, system (1) has two solutions for sufficiently small ![]()