Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 94, pp. 1–10.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
NECESSARY AND SUFFICIENT CONDITIONS FOR THE
EXISTENCE OF PERIODIC SOLUTION TO SINGULAR
PROBLEMS WITH IMPULSES
JUNTAO SUN, JIFENG CHU
Abstract. In this article we give a necessary sufficient conditions for the
existence of periodic solutions to impulsive periodic solution for a singular
differential equation. The proof is based on the variational method.
1. Introduction
In this article we discuss the T -periodic solution for the second-order singular
problem with impulsive effects
1
= e(t), a.e. t ∈ (0, T ),
uα (t)
∆u0 (tj ) = bj , j = 1, 2, . . . , p − 1,
u00 (t) −
(1.1)
0 −
0 ±
where α ≥ 1, e ∈ L1 ([0, T ], R) is T -periodic, ∆u0 (tj ) = u0 (t+
j )−u (tj ) with u (tj ) =
0
limt→t± u (t); tj , j = 1, 2, . . . , p − 1, are the instants where the impulses occur and
j
0 = t0 < t1 < t2 < · · · < tp−1 < tp = T , tj+p = tj + T ; and bj (j = 1, 2, . . . , p − 1)
are constants.
Impulsive effects occur widely in many evolution processes in which their states
are changed abruptly at certain moments of time. In the past few decades, impulsive differential equations have been extensively studied by many researchers
[6, 11, 13, 14, 15, 16, 17]. In particular, In 2008, Tian and Ge [17] studied the existence of solutions for impulsive differential equations by using a variational method.
Later, Nieto and O’Regan [11] further developed the variational framework for impulsive problems and established existence results for a class of impulsive differential equations with Dirichlet boundary conditions. From then on, the variational
method has been a powerful tool in the study of impulsive differential equations.
On the other hand, singular differential equations with different kinds of boundary
conditions have also been investigated widely in the literature by using either topological methods or variational methods; see [1, 2, 3, 4, 5, 7, 8, 9] and the references
therein.
2000 Mathematics Subject Classification. 34B15.
Key words and phrases. Positive periodic solution; singular differential equations;
impulses; variational methods.
c
2014
Texas State University - San Marcos.
Submitted October 30, 2013. Published April 10, 2014.
1
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J. SUN, J. CHU
EJDE-2014/94
In 1987, Lazer and Solimini [10] considered a the second order singular problem
1
u00 (t) − α
= e(t), t ∈ (0, T ).
(1.2)
u (t)
By using the method of upper and lower solutions, they obtained a famous sufficient
and necessary condition on positive T -periodic solution for Problem (1.2) as follows
Theorem 1.1 ([10]). Assume that e ∈ L1 ([0, T ], R) is T -periodic. Then Problem
RT
(1.2) has a positive T -periodic weak solution if and only if 0 e(t)dt < 0.
Motivated by the above fact, in the present paper we shall consider Problem
(1.2) with impulsive effects, i.e., Problem (1.1), and also obtain a sufficient and
necessary condition on T -periodic solution. It is worth emphasizing that the method
used by us is a variational method, which is different from that in Theorem 1.1.
Furthermore, we also point out the dynamical differences between singular problems
and singular problems with impulses.
Our results are presented as follows.
Theorem 1.2. Assume that e ∈ L1 ([0, T ], R) is T -periodic. Then Problem (1.1)
RT
Pp−1
has a positive T -periodic weak solution u ∈ HT1 if and only if j=1 bj + 0 e(t)dt <
0.
RT
Pp−1
Remark 1.3. From Theorem 1.2 we can see that if 0 e(t)dt ≥ 0, but j=1 bj +
RT
e(t)dt < 0, then Problem (1.1) still admits a positive T -periodic solution. This
0
shows that the existence of positive T -periodic solution for Problem (1.1) depends
on the forced term e and impulsive functions bj together, not single one.
2. Preliminaries
Set
HT1 = u : R → R|u is absolutely continuous, u0 ∈ L2 ((0, T ), R)
and u(t) = u(t + T ) for t ∈ R
with the inner product
Z
(u, v) =
T
T
Z
u0 (t)v 0 (t)dt,
u(t)v(t)dt +
0
∀u, v ∈ HT1 .
0
The corresponding norm is defined by
Z
Z T
kukHT1 =
|u(t)|2 dt +
0
T
1/2
|u0 (t)|2 dt
,
∀u ∈ HT1 .
0
Then HT1 is a Banach space (in fact it is a Hilbert space).
To study Problem (1.1), for any λ ∈ (0, 1) we consider the following modified
problem
u00 (t) + fλ (u(t)) = e(t), a.e. t ∈ (0, T ),
(2.1)
∆u0 (tj ) = bj , j = 1, 2, . . . , p − 1,
where fλ : R → R is defined by
(
− s1α , s ≥ λ,
fλ (s) =
− λ1α , s < λ.
Now we introduce the following concept of a weak solution for Problem (2.1).
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NECESSARY AND SUFFICIENT CONDITIONS
3
Definition 2.1. We say that a function u ∈ HT1 is a weak solution of Problem
(2.1) if
Z T
Z T
Z T
p−1
X
0
0
e(t)v(t)dt = 0
fλ (u(t))v(t)dt +
u (t)v (t)dt +
bj v(tj ) −
0
0
0
j=1
holds for any v ∈ HT1 .
Let Fλ ∈ C 1 (R, R) be defined by
Z
s
fλ (t)dt
Fλ (s) =
1
and consider the functional Φλ : HT1 → R defined by
Z T
Z T
Z
p−1
X
1 T 0 2
Φλ (u) :=
bj u(tj ) −
Fλ (u(t))dt +
e(t)u(t)dt.
|u (t)| dt +
2 0
0
0
j=1
(2.2)
Clearly, Φλ is well defined on HT1 , continuously Gâteaux differentiable functional
whose derivative is
Z T
Z T
Z T
p−1
X
Φ0λ (u)v =
u0 (t)v 0 (t)dt +
bj v(tj ) −
fλ (u(t))v(t)dt +
e(t)v(t)dt,
0
0
j=1
0
HT1 .
Moreover, it is easy to verify that Φλ is weakly lower semifor any v ∈
continuous. Furthermore, by the standard discussion, the critical points of Φλ are
the weak solutions of Problem (2.1).
3. Proof of Theorem 1.2
Proof. First we show that if u ∈ HT1 is a positive T -periodic weak solution of
RT
Pp−1
Problem (1.1), then j=1 bj + 0 e(t)dt < 0.
Integrating the first equation of Problem (1.1) from 0 to T , one has
Z T
Z T
Z T
1
00
u (t)dt −
dt =
e(t)dt.
(3.1)
α
0
0 u (t)
0
The first term one the left-hand side satisfies
Z T
p−1 Z tj+1
X
00
u (t)dt =
u00 (t)dt,
0
and
Z
tj+1
j=0
tj
0 +
u00 (t)dt = u0 (t−
j+1 ) − u (tj ).
tj
Thus,
Z
0
T
u00 (t)dt =
p−1
X
0 +
(u0 (t−
j+1 ) − u (tj ))
j=0
=−
p−1
X
∆u0 (tj ) + u0 (T ) − u0 (0)
j=1
=−
p−1
X
j=1
bj .
(3.2)
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J. SUN, J. CHU
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By (3.1) and (3.2) we have
Z
0>−
0
T
p−1
X
1
dt =
bj +
α
u (t)
j=1
T
Z
e(t)dt.
0
RT
Pp−1
Now we prove that if j=1 bj + 0 e(t)dt < 0, then Problem (1.1) has a positive
T -periodic weak solution u ∈ HT1 . The proof is based on the mountain pass theorem,
see [12]. We divide it into four steps.
Step 1. Let a sequence {un } in HT1 satisfy Φλ (un ) be bounded and Φ0λ (un ) → 0,
i.e., there exist a constant c1 > 0 and a sequence {n }n∈N ⊂ R+ with n → 0 as
n → +∞ such that for all n,
p−1
Z T 1
X
|u0n (t)|2 − Fλ (un (t)) + e(t)un (t) dt +
(3.3)
bj un (tj ) ≤ c1 ,
2
0
j=1
and for every v ∈ HT1 ,
p−1
Z T
X
[u0n (t)v 0 (t)) − fλ (un (t))v(t) + e(t)v(t)]dt +
bj v(tj ) ≤ n kvkHT1 .
0
(3.4)
j=1
Now we show that {un } is bounded in HT1 . Taking v(t) ≡ −1 in (3.4) one has
p−1 Z T
X
√
[fλ (un (t)) − e(t)]dt −
bj ≤ n T for all n,
0
j=1
which implies
T
Z
√
fλ (un (t))dt ≤ n T +
Z
0
T
e(t)dt +
0
p−1
X
|bj | := c2 .
j=1
Note that for any t ∈ [0, T ], fλ (un (t)) < 0. Thus
Z T
Z
T
|fλ (un (t))|dt = fλ (un (t))dt ≤ c2 .
0
0
On the other hand, take, in (3.4), v(t) ≡ wn (t) := un (t) − ūn , where ūn =
R
1 T
T 0 un (t)dt, by [12, Proposition 1.1] we have
Z
c3 kwn kHT1 ≥
T
[wn0 (t)2 − fλ (un (t))wn (t) + e(t)wn (t)]dt +
0
p−1
X
j=1
≥ kwn0 k2L2 − (c2 + kekL1 )kwn kL∞ −
p−1
X
|bj |kwn kL∞
j=1
= kwn0 k2L2 − (c2 + kekL1 +
p−1
X
|bj |)kwn kL∞
j=1
≥ kwn0 k2L2 − c4 kwn kHT1 ,
where c3 and c4 are two positive constants. Thus,
kwn0 k2L2 ≤ (c3 + c4 )kwn kHT1 .
bj wn (tj )
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NECESSARY AND SUFFICIENT CONDITIONS
5
Consequently, using the Wirtinger inequality, we obtain the existence of a positive
constant c5 such that
(3.5)
ku0n k2L2 ≤ c5 .
Now, suppose that kun kHT1 → +∞ as n → +∞. Since (3.5) holds, we have, passing
to subsequence if necessary, that either
Mn := max un → +∞ as n → +∞,
or
mn := min un → −∞ as n → +∞.
(i) Assume that the first possibility occurs. In view to the fact that fλ < 0, one
has
Z T
p−1
X
[Fλ (un (t)) − e(t)un (t)]dt −
bj un (tj )
0
j=1
T
Z
h Z
=
0
Z
Mn
fλ (s)ds −
1
p−1
X
−
Mn
p−1
i
X
fλ (s)ds − e(t)un (t) dt − Mn
bj
un (t)
j=1
bj (un (tj ) − Mn )
j=1
T
Z
T
Z
≥
Fλ (Mn )dt −
0
T
Z
Mn e(t)dt − max |Mn − un (t)|
t∈[0,T ]
0
− max |Mn − un (t)|
p−1
X
t∈[0,T ]
|e(t)|dt − Mn
0
p−1
X
bj
j=1
|bj |
j=1
≥ T Fλ (Mn ) − Mn
Z
T
e(t)dt +
0
= T Fλ (Mn ) − Mn
Z
T
e(t)dt +
0
≥ T Fλ (Mn ) − Mn
Z
T
e(t)dt +
0
p−1
X
p−1
X
bj − kekL1 +
|bj | |Mn − mn |
j=1
j=1
p−1
X
p−1
X
bj − kek
L1
+
j=1
j=1
p−1
X
p−1
X
bj − kekL1 +
j=1
j=1
t̂n
Z
|bj | t̄n
|bj |
Z
t̂n
u0n (t)dt
|u0n (t)|dt,
t̄n
where un (t̂n ) = Mn and un (t̄n ) = mn . Thus, using the Hölder inequality, one has
− Mn
Z
T
e(t)dt +
0
Z
≤
p−1
X
bj + T Fλ (Mn )
j=1
T
[Fλ (un (t)) − e(t)un (t)]dt −
0
p−1
X
p−1
X
√ bj un (tj ) + T kekL1 +
|bj | ku0n kL2 .
j=1
j=1
(3.6)
If α = 1, then Fλ (Mn ) = − ln Mn . By (3.6) one has
−Mn
Z
0
T
e(t)dt +
p−1
X
j=1
bj − T ln Mn → +∞ as n → +∞.
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J. SUN, J. CHU
EJDE-2014/94
1
1
If α > 1, then Fλ (Mn ) = − α−1
( M α−1
− 1). By (3.6) we obtain
n
−Mn
T
Z
e(t)dt +
0
p−1
X
bj −
j=1
1
1
(
− 1) → +∞ as n → +∞.
α − 1 Mnα−1
From (3.3) and (3.5), we see that the right hand side of (3.6) is bounded, which is
a contradiction.
(ii) Assume the second possibility occurs; i.e., mn → −∞ as n → +∞. We
replace Mn by −mn in the preceding arguments, and we also get a contradiction.
So {un } is bounded in HT1 .
Since HT1 is a reflexive Banach space, there exists a subsequence of {un }, denoted
again by {un } for simplicity, and u ∈ HT1 such that un * u in HT1 ; then, by the
Sobolev embedding theorem, we get un → u in C([0, T ]) and un → u in L2 ([0, T ]).
So
Z
T
(fλ (un (t)) − fλ (u(t)))(un (t) − u(t))dt → 0,
0
p−1
X
bj (un (tj ) − u(tj )) → 0,
j=1
Z
(3.7)
T
e(t)(un (t) − u(t))dt → 0,
0
(Φ0λ (un ) − Φ0λ (u))(un − u) → 0,
as n → ∞.
2
By (3.6), (3.7) and the fact that un → u in L ([0, T ]), we have kun − ukHT1 → 0
as n → ∞. That is, {un } strongly converges to u in HT1 , which means that the
Palais-Smale condition holds for Φλ .
Step 2. Let
Ω = u ∈ HT1 | min u(t) > 1 ,
t∈[0,T ]
and
∂Ω = {u ∈ HT1 |u(t) ≥ 1 for all t ∈ (0, T ), ∃tu ∈ (0, T ) : u(tu ) = 1}.
We show that there exists d > 0 such that inf u∈∂Ω Φλ (u) ≥ −d whenever λ ∈ (0, 1).
For any u ∈ ∂Ω, there exists some tu ∈ (0, T ) such that mint∈[0,T ] u(t) = u(tu ) =
1. By (2.2), and extending the functions by T -periodicity, we have
Z tu +T
p−1
X
1 0 2
|u (t)| − Fλ (u(t)) + e(t)u(t) dt +
bj u(tj )
Φλ (u) =
2
tu
j=1
Z
Z tu +T
Z tu +T
1 tu +T 0 2
≥
|u (t)| dt +
e(t)(u(t) − 1)dt +
e(t)dt
2 tu
tu
tu
+
p−1
X
j=1
bj (u(tj ) − 1) +
p−1
X
bj
j=1
≥
p−1
p−1
X
X
1 0 2
ku kL2 − kekL1 +
|bj | max (u(t) − 1) − kekL1 +
bj
2
t∈[0,T ]
j=1
j=1
=
p−1
p−1
Z ťu
X
X
1 0 2
ku kL2 − kekL1 +
|bj |
u0 (t)dt − kekL1 +
bj
2
tu
j=1
j=1
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NECESSARY AND SUFFICIENT CONDITIONS
≥
7
p−1
p−1
Z tu +T
X
X
1 0 2
ku kL2 − kekL1 +
|bj |
|u0 (t)|dt − kekL1 +
bj ,
2
tu
j=1
j=1
where ťu ∈ [0, T ] and maxt∈[0,T ] u(t) = u(ťu ). Applying the Hölder inequality, we
get
p−1
p−1
X
X
√ 1
|bj | ku0 kL2 − kekL1 +
bj .
Φλ (u) ≥ ku0 k2L2 − T kekL1 +
2
j=1
j=1
The above inequality shows that
Φλ (u) → +∞ asku0 kL2 → +∞.
For any u ∈ ∂Ω, it is easy to verify the fact that kukHT1 → +∞ is equivalent
to ku0 kL2 → +∞. Indeed, when ku0 kL2 → +∞, it is clear that kukHT1 → +∞.
When kukHT1 → +∞. Assume that ku0 kL2 is bounded, then kukL2 → +∞. Since
mint∈[0,T ] u(t) = 1, we have
Z t
Z T
√ Z T 0 2 1/2
|u (t)| dt
.
u(t) − 1 =
u0 (s)ds ≤
|u0 (s)|ds ≤ T
tu
0
0
2
Therefore, u is bounded in L (0, T ), which is a contradiction. Hence
Φλ (u) → +∞ as kukHT1 → +∞, ∀u ∈ ∂Ω,
which shows that Φλ is coercive. Thus it has a minimizing sequence. The weak
lower semi-continuity of Φλ yields
inf Φλ (u) > −∞.
u∈∂Ω
It follows that there exists d > 0 such that inf u∈∂Ω Φλ (u) > −d for all λ ∈ (0, 1).
Step 3. We prove that there exists λ0 ∈ (0, 1) with the property that for every
λ ∈ (0, λ0 ), any solution u of Problem (2.1) satisfying Φλ (u) > −d is such that
minu∈[0,T ] u(t) ≥ λ0 , and hence u is a solution of Problem (1.1).
Assume on the contrary that there are sequences {λn }n∈N and {un }n∈N such
that
(i) λn ≤ n1 ;
(ii) un is a solution of Problem (2.2) with λ = λn ;
(iii) Φλn (un ) ≥ −d;
(iv) mint∈[0,T ] un (t) < n1 .
RT
Since fλn < 0 and 0 [fλn (un (t)) − e(t)]dt = 0, one has
kfλn (un (·))kL1 ≤ c7 ,
for some constant c7 > 0.
Hence
ku0n kL∞ ≤ c8 ,
for some constant c8 > 0.
(3.8)
From Φλn (un ) ≥ −d it follows that there must exist two constants l1 and l2 , with
0 < l1 < l2 such that
max{un (t); t ∈ [0, T ]} ⊂ [l1 , l2 ].
If not, un would tend uniformly to 0 or +∞. In both cases, by (3.8), we have
Φλn (un ) → −∞ as n → +∞,
which contradicts Φλn (un ) ≥ −d.
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J. SUN, J. CHU
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Let τn1 , τn2 be such that, for n large enough
1
< l1 = un (τn2 ).
n
Multiplying the differential equation in (2.1) by u0n and integrating it on [τn1 , τn2 ],
or on [τn2 , τn1 ], we get
Z τn2
Z τn2
Z τn2
Ψ :=
u00n (t)u0n (t)dt +
fλn (un (t))u0n (t)dt =
e(t)u0n (t)dt.
(3.9)
un (τn1 ) =
1
τn
1
τn
1
τn
It is easy to verify that
1
1
(τ 2 ) − u02
Ψ = Ψ1 + [u02
n (τn )],
2 n n
where
2
τn
Z
Ψ1 =
1
τn
fλn (un (t))u0n (t)dt.
From (3.5) and (3.9) it follows that Ψ is bounded, and consequently Ψ1 is bounded.
On the other hand, it is easy to see that
d
[Fλn (un (t))].
dt
fλn (un (t))u0n (t) =
Thus, we have
1
.
n
From the fact that Fλn n1 → +∞ as n → +∞, we obtain Ψ1 → −∞, i.e., Ψ1 is
unbounded. This is a contradiction.
Step 4. Φ has a mountain-pass geometry for λ ≤ λ0 . Fix λ ∈ (0, λ0 ], one has
Z 0
Z 1
Fλ (0) =
fλ (s)ds = −
fλ (s)ds
Ψ1 = Fλn (l1 ) − Fλn
1
0
Z
λ
=−
Z
0
1
=
1
fλ (s)ds −
λα−1
fλ (s)ds
(3.10)
λ
Z
1
−
fλ (s)ds,
λ
which implies that
Z
1
Fλ (0) > −
Z
fλ (s)ds =
λ
λ
fλ (s)ds = Fλ (λ).
1
Thus we have
(
Φλ (0) = −T Fλ (0) < −T Fλ (λ) =
T ln λ,
T
− α−1
1
λα−1
if α = 1,
− 1 , if α > 1.
(3.11)
T
We choose λ ∈ (0, λ0 ] ∩ (0, e−d ) ∩ (0, [ T +d(α−1)
]1/(α−1) ), then it follows from (3.11)
that Φλ (0) < −d.
Also, we can choose a constant R > 1 enough large such that
Z T
p−1
X
T
1 −
bj +
e(t)dt R −
1 − α−1 > d,
α−1
R
0
j=1
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NECESSARY AND SUFFICIENT CONDITIONS
and
−
p−1
X
j=1
Thus, R ∈
HT1
Z
bj +
T
9
e(t)dt R − T ln R > d.
0
and
Φλ (R) = R
p−1
X
Z
bj − T Fλ (R) + R
T
e(t)dt
0
j=1
P
R
 p−1 bj R + T ln R + R T e(t)dt,
j=1
0
≤ Pp−1
RT
T
1

j=1 bj R + α−1 1 − Rα−1 + R 0 e(t)dt,
 P
RT
p−1

j=1 bj + 0 e(t)dt R + T ln R,
≤
1
 Pp−1 bj + R T e(t)dtR + T 1 − α−1
,
j=1
α−1
R
0
if α = 1,
if α > 1.
if α = 1,
if α > 1.
< −d.
Since Ω is a neighborhood of R, 0 6∈ Ω and
max{Φλ (0), Φλ (R)} < inf Φλ (u),
x∈∂Ω
Step 1 and Step 2 imply that Φλ has a critical point uλ such that
Φλ (uλ ) = inf max Φλ (h(s)) ≥ inf Φλ (u),
h∈Γ s∈[0,1]
x∈∂Ω
where
Γ = {h ∈ C([0, 1], HT1 ) : h(0) = 0, h(1) = R}.
Since inf u∈∂Ω Φλ (uλ ) ≥ −d, it follows from Step 3 that uλ is a positive solution of
Problem (1.1). The proof of the main result is complete.
Acknowledgments. Juntao Sun was supported by the National Natural Science
Foundation of China (Grant No. 11201270), Shandong Natural Science Foundation
(Grant No. ZR2012AQ010), and Young Teacher Support Program of Shandong
University of Technology. Jifeng Chu was supported by the National Natural Science Foundation of China (Grant Nos. 11171090, 11271078, and 11271333), the
Program for New Century Excellent Talents in University (Grant No. NCET10-0325), and China Postdoctoral Science Foundation funded project (Grant Nos.
20110491345 and 2012T50431).
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Juntao Sun
School of Science, Shandong University of Technology, Zibo, 255049 Shandong, China
E-mail address: sunjuntao2008@163.com
Jifeng Chu
College of Science, Hohai University, Nanjing, 210098 Jiangsu, China
E-mail address: jifengchu@126.com