Electronic Journal of Differential Equations, Vol. 2009(2009), No. 140, pp. 1–12.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
PERIODIC SOLUTIONS FOR A LIÉNARD EQUATION WITH
TWO DEVIATING ARGUMENTS
YONG WANG, JUNKANG TIAN
Abstract. In this work, we prove the existence and uniqueness of periodic
solutions for a Liénard equation with two deviating arguments. Our main
tools are the Mawhin’s continuation theorem and the Schwarz inequality. We
obtain our results under weaker conditions than those in [14], as shown by an
example in the last section of this artticle.
1. Introduction
The Liénard equation can be derived from many fields, such as physics, mechanics
and engineering technique fields. An important question is whether this equation
can support periodic solutions. In the past several years, the existence of periodic
solutions to Liénard equation has been widely discussed, notably by Liénard [6]
and by Levinson and Smith [5]. Recently, Zhou and Long [14] studied the existence
and uniqueness of periodic solutions of the following Liénard equation with two
deviating arguments
x00 (t) + f (x(t))x0 (t) + g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) = e(t),
(1.1)
where f, τ1 , τ2 , e ∈ C(R, R), g1 , g2 ∈ C(R2 , R), τ1 (t), τ2 (t), g1 (t, x), g2 (t, x), e(t) are
periodic functions with period T , with respect to t, T -periodic for short.
In recent years, there have been many publications on the existence of periodic
solutions of the Liénard equation of the type (1.1); see for example [2, 7, 8, 1, 9, 10,
11, 12, 13]. However, as far as we know, there are fewer results on the existence and
uniqueness of periodic solutions to (1.1). Applying Mawhin’s continuation theorem
and some analysis techniques, Zhou and Long [14] provided a sufficient condition
for the existence and uniqueness of periodic solutions to (1.1), but their results can
be improved.
The main purpose of this paper is to provide a new sufficient condition for
guaranteeing the existence and uniqueness of T -periodic solutions to (1.1), by using
Mawhin’s continuation theorem and Schwarz inequality. Our results hold under
2000 Mathematics Subject Classification. 34C25, 34D40.
Key words and phrases. Periodic solution; Liénard equation; deviating argument.
c
2009
Texas State University - San Marcos.
Submitted July 7, 2009. Published October 30, 2009.
Supported by the SWPU Science and Technology Fund of China and by the Open
Fund of State Key Laboratory of Oil and Gas Researvoir Geology and Exploitation
(Southwest Petroleum University).
1
2
Y. WANG, J. TIAN
EJDE-2009/140
weaker conditions than those in [14], and that are verifiable as shown by an example
in the last section.
2. Preliminaries
For convenience, we define
|x0 |∞ = max |x0 (t)|,
|x|∞ = max |x(t)|,
t∈[0,T ]
|x|k =
Z
T
k
|x(t)| dt
t∈[0,T ]
1/k
,
0
1
ē =
T
Z
T
e(t)dt
0
Let
CT1 := {x ∈ C 1 (R, R) : x is T -periodic},
CT := {x ∈ C(R, R) : x is T -periodic},
which are Banach spaces with the norms
kxkCT1 = max{|x|∞ , |x0 |∞ },
kxkCT = |x|∞ .
The following conditions will be used in this paper:
(H0) There exist C1 ≥ 0, C2 ≥ 0, b1 ≥ 0 and b2 ≥ 0 such that |f (x1 ) −
f (x2 )| ≤ C1 |x1 − x2 |, |f (x)| ≤ C2 and |gi (t, u) − gi (t, v)| ≤ bi |u − v|,
for all x1 , x2 , x, t, u, v ∈ R, i = 1, 2.
The following Mawhin’s continuation theorem is useful in obtaining the existence
of T -periodic solutions of (1.1).
Lemma 2.1 ([3, p. 40]). Let X and Y be two Banach spaces. Suppose that
L : D(L) ⊂ X → Y is a Fredholm operator with index zero and N : X → Y is
L-compact on Ω, where Ω is an open bounded subset of X. Moreover, assume that
all the following conditions are satisfied:
(i) Lx 6= λN x, for all x ∈ ∂Ω ∩ D(L), λ ∈ (0, 1);
(ii) N x ∈
/ Im L, for all x ∈ ∂Ω ∩ ker L;
(iii) the Brouwer degree deg{JQN, Ω ∩ ker L, 0} =
6 0, where J : Im Q → ker L is
an isomorphism.
Then equation Lx = N x has at least one solution on Ω ∩ D(L).
Lemma 2.2. If x ∈ C 2 (R, R) with x(t + T ) = x(t), then
T 2 00 2
|x |2 .
|x0 |22 ≤
2π
The proof of the above lemma is a direct consequence of the Wirtinger inequality;
see for example [4]. Consider the homotopic equation of (1.1), for λ ∈ (0, 1),
x00 (t) + λf (x(t))x0 (t) + λg1 (t, x(t − τ1 (t))) + λg2 (t, x(t − τ2 (t))) = λe(t).
We have the following lemma.
Lemma 2.3. Suppose that the following conditions are satisfied:
(H1) one of the following conditions holds:
(1) (gi (t, u) − gi (t, v))(u − v) > 0 for all t, u, v ∈ R, u 6= v, i = 1, 2,
(2) (gi (t, u) − gi (t, v))(u − v) < 0 for all t, u, v ∈ R, u 6= v, i = 1, 2;
(H2) there exists d ≥ 0 such that one of the following conditions holds:
(1) x(g1 (t, x) + g2 (t, x) − ē) > 0, for all t ∈ R, |x| > d,
(2) x(g1 (t, x) + g2 (t, x) − ē) < 0, for all t ∈ R, |x| > d;
(2.1)
EJDE-2009/140
PERIODIC SOLUTIONS
If x(t) is a T -periodic solution of (2.1), then
√
T 0
|x|∞ ≤ d +
|x |2 .
2
3
(2.2)
Proof. Let x(t) be an arbitrary T -periodic solution of (2.1). Then, integrating (2.1)
from 0 to T , we have
Z T
[g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − e(t)]dt = 0,
(2.3)
0
which implies that there exists t1 ∈ R such that
g1 (t1 , x(t1 − τ1 (t1 ))) + g2 (t1 , x(t1 − τ2 (t1 ))) − ē = 0.
(2.4)
Now we show the following statement.
Claim. If x(t) is a T -periodic solution of (2.1), then there exists t2 ∈ R such that
|x(t2 )| ≤ d.
(2.5)
Assume, by way of contradiction, that (2.5) does not hold. Then
|x(t)| > d for all t ∈ R,
(2.6)
which, together with (H1), (H2) and (2.4), imply that one of the following four
relations holds:
x(t1 − τ1 (t1 )) > x(t1 − τ2 (t1 )) > d,
(2.7)
x(t1 − τ2 (t1 )) > x(t1 − τ1 (t1 )) > d,
(2.8)
x(t1 − τ1 (t1 )) < x(t1 − τ2 (t1 )) < −d,
(2.9)
x(t1 − τ2 (t1 )) < x(t1 − τ1 (t1 )) < −d.
(2.10)
Suppose that (2.7) holds, in view of (H1)(1), (H1)(2), (H2)(1) and (H2)(2), we
consider fours cases as follows:
Case (i): If (H1)(1) and (H2)(1) hold, according to (2.7), we have
0 < g1 (t1 , x(t1 − τ2 (t1 ))) + g2 (t1 , x(t1 − τ2 (t1 ))) − ē
< g1 (t1 , x(t1 − τ1 (t1 ))) + g2 (t1 , x(t1 − τ2 (t1 ))) − ē,
which contradicts (2.4). Thus (2.5) is true.
Case (ii): If (H1)(2) and (H2)(1) hold, according to (2.7), we have
0 < g1 (t1 , x(t1 − τ1 (t1 ))) + g2 (t1 , x(t1 − τ1 (t1 ))) − ē
< g1 (t1 , x(t1 − τ1 (t1 ))) + g2 (t1 , x(t1 − τ2 (t1 ))) − ē,
which contradicts (2.4). Thus (2.5) is true.
Case (iii): If (H1)(1) and (H2)(2) hold, according to (2.7), we have
0 > g1 (t1 , x(t1 − τ1 (t1 ))) + g2 (t1 , x(t1 − τ1 (t1 ))) − ē
> g1 (t1 , x(t1 − τ1 (t1 ))) + g2 (t1 , x(t1 − τ2 (t1 ))) − ē,
which contradicts (2.4). Thus (2.5) is true.
Case (iv): If (H1)(2) and (H2)(2) hold, according to (2.7), we have
0 > g1 (t1 , x(t1 − τ2 (t1 ))) + g2 (t1 , x(t1 − τ2 (t1 ))) − ē
> g1 (t1 , x(t1 − τ1 (t1 ))) + g2 (t1 , x(t1 − τ2 (t1 ))) − ē,
which contradicts (2.4). Thus (2.5) is true.
4
Y. WANG, J. TIAN
EJDE-2009/140
Suppose that (2.8)(or (2.9), or (2.10)) holds; using methods similar to those used
in Case (i)–(iv), we can show that (2.5) is true. This completes the proof of the
above claim.
Let t2 = kT + e
t2 , where e
t2 ∈ [0, T ] and k is an integer. Then noticing x(t) =
x(t + T ) and (2.5), for any t ∈ [e
t2 , e
t2 + T ], we obtain
Z t
Z t
t2 ) +
|x(t)| = x(e
x0 (s)ds ≤ d +
|x0 (s)|ds
e
t2
e
t2
and
Z
t2 + T ) +
|x(t)| = x(e
t
e
t2 +T
Z
x0 (s)ds ≤ d + −
t
e
t2 +T
Z
x0 (s)ds ≤ d +
e
t2 +T
|x0 (s)|ds.
t
Combining the two inequalities above, we obtain
Z
1 T 0
|x(t)| ≤ d +
|x (s)|ds.
2 0
Using Schwarz inequality yields
|x|∞ =
|x(t)| ≤ d +
max
t∈[e
t2 ,e
t2 +T ]
T
Z
1
2
0
This completes the proof.
1√
1
T |x0 |2 .
|x0 (s)|ds ≤ d + |1|2 |x0 |2 = d +
2
2
(2.11)
Lemma 2.4. Suppose (H0)–(H2) hold. Also suppose the following condition holds
2
T
(H3) C2 2π
+ (b1 + b2 ) T4π < 1.
If x(t) is a T -periodic solution of (1.1), then |x0 |∞ ≤ D, where
D=
[(b1 + b2 )d + max{|g1 (t, 0)| + |g2 (t, 0)| : 0 ≤ t ≤ T } + |e|∞ ]T
.
2
2
T
2 1 − C2 2π
− b1 T4π − b2 T4π
Proof. Let x(t) be a T -periodic solution of (1.1). From (H1) and (H2), we can
easily show that (2.2) also holds. Multiplying x00 (t) and (1.1) and then integrating
it from 0 to T , by Lemma 2, (H0), (2.2) and Schwarz inequality, we have
|x00 |22
T
Z
0
=−
T
Z
00
g1 (t, x(t − τ1 (t)))x00 (t)dt
f (x(t))x (t)x (t)dt −
0
Z T
−
0
Z
00
T
e(t)x00 (t)dt
g2 (t, x(t − τ2 (t)))x (t)dt +
0
T
Z
0
|f (x(t))kx0 (t)kx00 (t)|dt +
≤
0
Z
T
|g1 (t, x(t − τ1 (t)))kx00 (t)|dt
0
T
Z
|g2 (t, x(t − τ2 (t)))kx00 (t)|dt +
+
0
T
|e(t)kx00 (t)|dt
0
Z
T
≤ C2
0
00
Z
|x (t)kx (t)|dt +
Z
Z
0
T
T
[|g1 (t, x(t − τ1 (t))) − g1 (t, 0)| + |g1 (t, 0)|]|x00 (t)|dt
0
00
Z
[|g2 (t, x(t − τ2 (t))) − g2 (t, 0)| + |g2 (t, 0)|]|x (t)|dt +
+
0
0
T
|e(t)kx00 (t)|dt
EJDE-2009/140
PERIODIC SOLUTIONS
≤ C2 |x0 |2 |x00 |2 + b1
Z
T
|x(t − τ1 (t))kx00 (t)|dt +
T
Z
|g1 (t, 0)kx00 (t)|dt
00
T
Z
00
|g2 (t, 0)kx (t)|dt +
|x(t − τ2 (t))kx (t)|dt +
+ b2
T
Z
0
0
Z
5
|e(t)kx00 (t)|dt
0
0
0
T
√
T 00 2
|x |2 + (b1 + b2 ) T |x|∞ |x00 |2
2π
√
+ [max{|g1 (t, 0)| + |g2 (t, 0)| : 0 ≤ t ≤ T } + |e|∞ ] T |x00 |2
T
T 2 00 2
≤ C2
+ (b1 + b2 )
|x |2
2π
4π
√
+ (b1 + b2 )d + max{|g1 (t, 0)| + |g2 (t, 0)| : 0 ≤ t ≤ T } + |e|∞ T |x00 |2 ,
≤ C2
which, together with (H3), implies
√
[(b1 + b2 )d + max{|g1 (t, 0)| + |g2 (t, 0)| : 0 ≤ t ≤ T } + |e|∞ ] T
.
|x |2 ≤
2
2
T
1 − C2 2π
− b1 T4π − b2 T4π
00
(2.12)
Since x(0) = x(T ), there exists t0 ∈ [0, T ] such that x0 (t0 ) = 0, for any t ∈
[t0 , t0 + T ], we obtain
Z t
Z t
0
0
00
|x (t)| = x (t0 ) +
x (s)ds ≤
|x00 (s)|ds,
t0
Z
|x0 (t)| = x0 (t0 + T ) +
t
t0 +T
t0
Z
x00 (s)ds ≤ −
t0 +T
t
Z
x00 (s)ds ≤
t0 +T
|x00 (s)|ds.
t
Combining these two inequalities, we obtain
Z
1 T 00
0
|x (s)|ds.
|x (t)| ≤
2 0
Using Schwarz inequality yields
|x0 |∞ =
max
|x0 (t)| ≤
t∈[t0 ,t0 +T ]
1
2
Z
T
|x00 (s)|ds ≤
0
1√
1
|1|2 |x00 |2 =
T |x00 |2 .
2
2
(2.13)
By (2.12) and (2.13), we obtain
|x0 |∞ ≤
[(b1 + b2 )d + max{|g1 (t, 0)| + |g2 (t, 0)| : 0 ≤ t ≤ T } + |e|∞ ]T
+ : D.
2
2
T
2 1 − C2 2π
− b1 T4π − b2 T4π
This completes the proof.
Lemma 2.5. Suppose (H0)–(H3) hold. Also assume the condition
2
2
T
(H4) C1 D T4π + C2 2π
+ (b1 + b2 ) T4π < 1.
Then (1.1) has at most one T -periodic solution.
Proof. Suppose that x1 (t) and x2 (t) are two T -periodic solutions of (1.1). Set
Z(t) = x1 (t) − x2 (t). Then, we have
Z 00 (t) + [f (x1 (t))x01 (t) − f (x2 (t))x02 (t)] + [g1 (t, x1 (t − τ1 (t))) − g1 (t, x2 (t − τ1 (t)))]
+ [g2 (t, x1 (t − τ2 (t))) − g2 (t, x2 (t − τ2 (t)))] = 0.
(2.14)
6
Y. WANG, J. TIAN
EJDE-2009/140
Since x1 (t) and x2 (t) are two T -periodic solutions of (1.1), integrating (2.14) from
0 to T , we obtain
Z T
g1 (t, x1 (t − τ1 (t))) − g1 (t, x2 (t − τ1 (t)))
0
+ g2 (t, x1 (t − τ2 (t))) − g2 (t, x2 (t − τ2 (t))) dt = 0.
Thus, in view of Mean Value Theorem for integrals, it follows that there exists
t̃ ∈ R such that
g1 (t̃, x1 (t̃ − τ1 (t̃))) − g1 (t̃, x2 (t̃ − τ1 (t̃))) + g2 (t̃, x1 (t̃ − τ2 (t̃))) − g2 (t̃, x2 (t̃ − τ2 (t̃))) = 0.
(2.15)
By (H1), (2.15) implies
Z(t̃−τ1 (t̃))Z(t̃−τ2 (t̃)) = (x1 (t̃−τ1 (t̃))−x2 (t̃−τ1 (t̃)))(x1 (t̃−τ2 (t̃))−x2 (t̃−τ2 (t̃))) ≤ 0.
Since Z(t) = x1 (t) − x2 (t) is a continuous function in R, it follows that there exists
t̂ ∈ R such that
Z(t̂) = 0.
(2.16)
Set t̂ = nT + t̄, where t̄ ∈ [0, T ] and n is an integer. Noticing Z(t + T ) = Z(t), we
get
Z(t̄) = Z(nT + t̄) = Z(t̂) = 0.
(2.17)
Hence, for any t ∈ [t̄, t̄ + T ], we obtain
Z t
Z t
0
|Z(t)| = Z(t̄) +
Z (s)ds ≤
|Z 0 (s)|ds
t̄
t̄
and
Z
|Z(t)| = Z(t̄ + T ) +
t
t̄+T
0
Z (s)ds = −
Z
t̄+T
0
Z (s)ds ≤
t
Z
t̄+T
|Z 0 (s)|ds.
t
Combining these two inequalities, we obtain
Z
1 T 0
|Z(t)| ≤
|Z (s)|ds.
2 0
Using Schwarz inequality yields
Z
1 T 0
1√
1
|Z|∞ = max |Z(t)| ≤
|Z (s)|ds ≤ |1|2 |Z 0 |2 =
T |Z 0 |2 .
2 0
2
2
t∈[t̄,t̄+T ]
(2.18)
Multiplying Z 00 (t) and (2.14) and then integrating it from 0 to T , by Lemma 2,
Lemma 4, (H0), (2.18) and Schwarz inequality, we have
Z T
00 2
|Z |2 = −
[f (x1 (t))x01 (t) − f (x2 (t))x02 (t)]Z 00 (t)dt
Z
0
T
[g1 (t, x1 (t − τ1 (t))) − g1 (t, x2 (t − τ1 (t)))]Z 00 (t)dt
−
0
T
Z
[g2 (t, x1 (t − τ2 (t))) − g2 (t, x2 (t − τ2 (t)))]Z 00 (t)dt
−
0
T
Z
|f (x1 (t))kx01 (t) − x02 (t)kZ 00 (t)|dt
≤
0
Z
+
0
T
|f (x1 (t)) − f (x2 (t))kx02 (t)kZ 00 (t)|dt
EJDE-2009/140
PERIODIC SOLUTIONS
7
T
Z
|x1 (t − τ1 (t)) − x2 (t − τ1 (t))kZ 00 (t)|dt
+ b1
0
T
Z
|x1 (t − τ2 (t)) − x2 (t − τ2 (t))kZ 00 (t)|dt
+ b2
0
Z
T
≤
T
Z
C2 |Z 0 (t)kZ 00 (t)|dt +
0
C1 D|Z(t)kZ 00 (t)|dt
0
Z T
|Z(t − τ1 (t))kZ 00 (t)|dt + b2
|Z(t − τ2 (t))kZ 00 (t)|dt
0
0
√
√
≤ C2 |Z 0 |2 |Z 00 |2 + C1 D T |Z|∞ |Z 00 |2 + (b1 + b2 ) T |Z|∞ |Z 00 |2
T
T 2 00 2
T2
+ C2
+ (b1 + b2 )
|Z |2 .
≤ C1 D
4π
2π
4π
Since Z(t), Z 0 (t), Z 00 (t) are continuous T -periodic functions, by (H4), (2.18) and
the above inequality, we obtain
Z
T
+ b1
Z(t) = Z 0 (t) = Z 00 (t) = 0
for all t ∈ R.
Thus, x1 (t) ≡ x2 (t), for all t ∈ R. Hence, (1.1) has at most one T -periodic solution.
This completes the proof.
Lemma 2.6. Suppose (H0)–(H3) hold. Then the set of T -periodic solutions of
(2.1) are bounded in CT1 .
Proof. Let S ⊂ CT1 be the set of T -periodic solutions of (2.1). If S = ∅, the proof
is complete. Suppose S 6= ∅, and let x ∈ S. Multiplying x00 (t) and (2.1) and then
integrating it from 0 to T , by Lemma 2, (H0), (2.2) and Schwarz inequality, we
have
|x00 |22
T
Z
T
Z
f (x(t))x0 (t)x00 (t)dt − λ
= −λ
0
g1 (t, x(t − τ1 (t)))x00 (t)dt
0
T
Z
g2 (t, x(t − τ2 (t)))x00 (t)dt + λ
−λ
Z
0
e(t)x00 (t)dt
0
T
Z
|f (x(t))kx0 (t)kx00 (t)|dt +
≤
T
0
Z
T
|g1 (t, x(t − τ1 (t)))kx00 (t)|dt
0
T
Z
Z
00
|g2 (t, x(t − τ2 (t)))kx (t)|dt +
+
0
|e(t)kx00 (t)|dt
0
Z
T
|x0 (t)kx00 (t)|dt +
≤ C2
Z
T
0
T
Z
T
[|g1 (t, x(t − τ1 (t))) − g1 (t, 0)| + |g1 (t, 0)|]|x00 (t)|dt
0
Z
[|g2 (t, x(t − τ2 (t))) − g2 (t, 0)| + |g2 (t, 0)|]|x00 (t)|dt +
+
0
|e(t)kx00 (t)|dt
0
≤ C2 |x0 |2 |x00 |2 + b1
Z
T
|x(t − τ1 (t))kx00 (t)|dt +
0
Z
0
T
Z
|g1 (t, 0)kx00 (t)|dt
0
T
00
Z
T
|x(t − τ2 (t))kx (t)|dt +
+ b2
≤ C2
T
00
Z
|g2 (t, 0)kx (t)|dt +
0
√
T 00 2
|x |2 + (b1 + b2 ) T |x|∞ |x00 |2
2π
0
T
|e(t)kx00 (t)|dt
8
Y. WANG, J. TIAN
EJDE-2009/140
√
+ [max{|g1 (t, 0)| + |g2 (t, 0)| : 0 ≤ t ≤ T } + |e|∞ ] T |x00 |2
T
T 2 00 2
≤ C2
+ (b1 + b2 )
|x |2
2π
4π
√
+ (b1 + b2 )d + max{|g1 (t, 0)| + |g2 (t, 0)| : 0 ≤ t ≤ T } + |e|∞ T |x00 |2 ,
which, together with (H3), implies that there exists M0 > 0 such that
|x00 |2 < M0 .
(2.19)
This, together with Lemma 2 and Lemma 3, leads to
√
T3
|x|∞ < d +
M0 .
(2.20)
4π
On the other hand, since x(0) = x(T ), there exists t̄0 ∈ [0, T ] such that x0 (t̄0 ) = 0.
For any t ∈ [t̄0 , t̄0 + T ], from (2.19), we obtain
Z T
Z t
√
0
00
0
x (s)ds ≤
|x (t)| = x (t̄0 ) +
|x00 (s)|ds ≤ |1|2 |x00 |2 < T M0 ,
t̄0
0
which implies
|x0 |∞ =
max
|x0 (t)| <
√
t∈[t̄0 ,t̄0 +T ]
T M0 .
(2.21)
√
T3
Let M = max d + 4π
M0 , T M0 , by (2.20) and (2.21), we have kxk < M . This
completes the proof.
√
3. Main results
Now we are in the position to give our main results.
Theorem 3.1. Suppose (H0)–(H2), (H4) hold. Then (1.1) has a unique T -periodic
solution.
Proof. Lemma 5 states that (1.1) has at most one T -periodic solution. Thus, to
prove Theorem 1, it suffices to show that (1.1) has at least one T -periodic solution.
To do this, we apply Lemma 1.
By Lemma 6, there exists M > d such that, for any T -periodic solution x(t) of
(2.1)
kxk < M.
(3.1)
Set
Ω = {x : x ∈ CT1 , kxk < M }.
(3.2)
Define a linear operator L : D(L) ⊂ CT1 → CT by setting D(L) = {x : x ∈ CT1 , x00 ∈
C(R, R), for x ∈ D(L), and
Lx = x00 .
(3.3)
1
We also define a nonlinear operator N : CT → CT , by
N x = −f (x(t))x0 (t) − g1 (t, x(t − τ1 (t))) − g2 (t, x(t − τ2 (t))) + e(t).
(3.4)
Then (2.1) is equivalent to the operator equation
Lx = λN x,
λ ∈ (0, 1).
(3.5)
It is easy to see that
Z
ker L = R
, and
Im L = {x : x ∈ CT ,
T
x(s)ds = 0},
0
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PERIODIC SOLUTIONS
9
then, L is a Fredholm operator with index zero. Also let projectors P : CT1 → ker L
and Q : CT → CT / Im L defined by
P x = x(0) where x ∈ CT1 ,
Z
1 T
Qx =
x(s)ds where x ∈ CT ,
T 0
hence, Im P = Im Q = ker L = R and ker Q = Im L. Define the isomorphism as
follows
J : Im Q → ker L, J(x) = x.
(3.6)
Let
LP := LD(L)∩ker P : D(L) ∩ ker P → Im L,
then we know, by [3, p. 41–42], that LP has a continuous inverse L−1
P on Im L
defined by
Z T
−1
(LP y)(t) =
G(s, t)y(s)ds,
(3.7)
0
where
(
− Ts (T − t), 0 ≤ s ≤ t;
G(s, t) =
− Tt (T − s), t ≤ s ≤ T.
Using Ascoli-Arzela theorem we have, from (3.2) and (3.7), that L−1
P (I − Q)N (Ω)
is compact. On the other hand, QN (Ω) is bounded by the continuity of function
QN . Thus N is L-compact on Ω. By (3.2) and (3.5), condition (i) of Lemma 1 is
satisfied.
In view of (H2)(1) and (H2)(2), we will consider two cases:
Case(i): If (H2)(1) holds. Since
Z
1 T
QN x = −
[f (x(t))x0 (t) + g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − e(t)]dt
T 0
Z
1 T
[f (x(t))x0 (t) + g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − ē]dt;
=−
T 0
for any x ∈ ∂Ω ∩ ker L, x = M or x = −M , x0 = 0, we obtain
Z
1 T
QN (M ) = −
[g1 (t, M ) + g2 (t, M ) − ē]dt < 0,
T 0
Z
1 T
QN (−M ) = −
[g1 (t, −M ) + g2 (t, −M ) − ē]dt > 0
T 0
which implies the condition (ii) of Lemma 1 is satisfied. Define
H(x, µ) = −µx + (1 − µ)QN x
Z
1 T
= −µx − (1 − µ)
f (x(t))x0 (t) + g1 (t, x(t − τ1 (t)))
T 0
+ g2 (t, x(t − τ2 (t))) − e(t) dt
Z
1 T
= −µx − (1 − µ)
f (x(t))x0 (t) + g1 (t, x(t − τ1 (t)))
T 0
+ g2 (t, x(t − τ2 (t))) − ē dt
(3.8)
(3.9)
10
Y. WANG, J. TIAN
EJDE-2009/140
in view of (3.8) and (3.9), we get xH(x, µ) < 0, for all x ∈ ∂Ω ∩ ker L and µ ∈ [0, 1].
Hence, H(x, µ) is a homotopic transformation, together with (3.6) and by using
homotopic invariance theorem, we have
deg {JQN, Ω ∩ ker L, 0} = deg{QN, Ω ∩ ker L, 0} = deg{−x, Ω ∩ ker L, 0} =
6 0,
so condition (iii) of Lemma 1 is satisfied.
Case(ii): If (H2)(2) holds. Since
Z
1 T
QN x = −
[f (x(t))x0 (t) + g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − e(t)]dt
T 0
Z
1 T
[f (x(t))x0 (t) + g1 (t, x(t − τ1 (t))) + g2 (t, x(t − τ2 (t))) − ē]dt;
=−
T 0
for any x ∈ ∂Ω ∩ ker L, x = M or x = −M , x0 = 0, we obtain
Z
1 T
[g1 (t, M ) + g2 (t, M ) − ē]dt > 0,
QN (M ) = −
T 0
Z
1 T
QN (−M ) = −
[g1 (t, −M ) + g2 (t, −M ) − ē]dt < 0
T 0
(3.10)
(3.11)
which implies the condition (ii) of Lemma 1 is satisfied. Define
H(x, µ) = µx + (1 − µ)QN x
Z
1 T
f (x(t))x0 (t) + g1 (t, x(t − τ1 (t)))
= µx − (1 − µ)
T 0
+ g2 (t, x(t − τ2 (t))) − e(t) dt
Z
1 T
= µx − (1 − µ)
f (x(t))x0 (t) + g1 (t, x(t − τ1 (t)))
T 0
+ g2 (t, x(t − τ2 (t))) − ē dt,
in view of (3.10) and (3.11), we get xH(x, µ) > 0, for all x ∈ ∂Ω ∩ ker L and
µ ∈ [0, 1]. Hence, H(x, µ) is a homotopic transformation, together with (3.6) and
by using homotopic invariance theorem, we have
deg{JQN, Ω ∩ ker L, 0} = deg{QN, Ω ∩ ker L, 0} = deg{x, Ω ∩ ker L, 0} =
6 0,
so condition (iii) of Lemma 1 is satisfied. Therefore, it follows from Lemma 1 that
(1.1) has at least one T -periodic solution. This completes the proof.
In [14], Zhou and Long studied (1.1) and obtained the got the following results.
Theorem 3.2. Assume (H0), (H1), and that the following conditions hold:
(A2) there exists d ≥ 0 such that one of the following conditions holds:
(1) x(g1 (t, x) + g2 (t, x) − e(t)) > 0, for all t ∈ R, |x| > d,
(2) x(g1 (t, x) + g2 (t, x) − e(t)) < 0, for all t ∈ R, |x| > d;
2
2
T
+ (b1 + b2 ) T2π < 1, where
(A4) C1 D1 T2π + C2 2π
D1 =
[(b1 + b2 )d + max{|g1 (t, 0)| + |g2 (t, 0)| : 0 ≤ t ≤ T } + |e|∞ ]T
.
2
2
T
1 − C2 2π
− b1 T2π − b2 T2π
Then (1.1) has a unique T -periodic solution.
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PERIODIC SOLUTIONS
11
If e(t) 6= constant, it is easy to verify that the condition (H2) is weaker than the
condition (A2) since mint∈R e(t) < ē < maxt∈R e(t). On the other hand, noticing
1
1
1
4π < 2π and D < 2 D1 , we can see that the condition (H4 ) is also weaker than the
condition (A4 ). Therefore, our results improve those in [14].
4. Example and remark
In this section, we apply the main results obtained in previous sections to an
example.
Consider the existence and uniqueness of a 2π-periodic solution to the Liénard
equation
1
cos tx0 (t) + g1 (t, x(t − cos t)) + g2 (t, x(t − sin t)) = e(t),
(4.1)
x00 (t) +
10
1
where T = 2π, τ1 (t) = cos t, τ2 (t) = sin t, g1 (t, x) = 80π(1+cos
2 t) arctan x, g2 (t, x) =
2
1
1
(1
+
sin
t)
arctan
x
and
e(t)
=
sin
t.
60π
π
It is obvious that the conditions (A0) and (A1) in [14, Theorem 1] hold. However,
we can easily check that (A2) does not hold, which implies that (A4) does not hold.
Hence, [14, Theorem 1] can not be applied. Meanwhile, Theorem 1 in this paper
remains applicable, as we show now.
1
1
1
By (4.1), we can get b1 = 80π
, b2 = 30π
and C1 = C2 = 10
. Noticing ē =
R
R
2π 1
1 T
1
1
T 0 e(t)dt = 2π 0 π sin tdt = 0, we can get d = 10 (Actually, d can be an
arbitrarily small positive constant.) and check that (H0)–(H2) hold. On the other
hand, noticing that
D=
=
[(b1 + b2 )d + max{|g1 (t, 0)| + |g2 (t, 0)| : 0 ≤ t ≤ T } + |e|∞ ]T
2
2
T
2 1 − C2 2π
− b1 T4π − b2 T4π
1
1
1
[ 10
( 80π
+ 30π
) + π1 ]2π
≈ 1.176,
1
1
1
2(1 − 10 − 80 − 30
)
2
2
1
T
it is easy to verify that (H4) holds since C1 D T4π + C2 2π
+ (b1 + b2 ) T4π = 10
× 1.176 ×
1
1
1
π + 10 + 80 + 30 ≈ 0.515 < 1. Thus, Theorem 1 in this study shows that (1.1) has
a unique 2π-periodic solution. Hence our results improve those in [14].
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Yong Wang
School of Sciences, Southwest Petroleum University, Chengdu, Sichuan 610500, China
E-mail address: odinswang@yahoo.com.cn, wangyong@swpu.edu.cn
Junkang Tian
School of Sciences, Southwest Petroleum University, Chengdu, Sichuan 610500, China
E-mail address: tianjunkang1980@163.com