Electronic Journal of Differential Equations, Vol. 2014 (2014), No. 62, pp. 1–13.
ISSN: 1072-6691. URL: http://ejde.math.txstate.edu or http://ejde.math.unt.edu
ftp ejde.math.txstate.edu
AN APPLICATION OF VARIATIONAL METHODS TO
SECOND-ORDER IMPULSIVE DIFFERENTIAL EQUATION
WITH DERIVATIVE DEPENDENCE
JIAN LIU, ZENGQIN ZHAO
Abstract. In this article, we study the existence of solutions for nonlinear
impulsive problems. We show the existence of classical solutions by using
variational methods.
1. Introduction
Variational methods are used in the modeling of certain nonlinear problems from
biological neural networks, elastic mechanics, anisotropic problems, and so forth.
During the previous decade, variational methods have been applied to boundary
value problems for differential equations. Recently, impulsive differential equation
has been studied in many classical works, for example, [1, 2, 6, 8, 15, 17]. The
study of impulsive differential equation via variational methods was initiated by
Nieto and O’Regan [5], Tian and Ge [11]. The study of second order impulsive
differential equation with derivative dependence ordinary differential equations via
variational methods was initiated by Nieto [4]. Since then there is a trend to
study differential equation via variational methods which leads to many meaningful
results, see [9, 10, 13, 12, 14, 18, 19, 20, 21] and the references therein.
Recently, Nieto [4] studied the following damped linear Dirichlet boundary value
problem with impulses:
−u00 (t) + g(t)u0 (t) + λu(t) = σ(t),
0
∆u (tj ) = dj ,
a.e. t ∈ [0, T ],
j = 1, 2, . . . , p,
(1.1)
u(0) = u(T ) = 0,
where λ, dj ∈ R, σ ∈ C[0, 1], the author introduced a variational formulation for the
damped linear Dirichlet problem with impulses and the concept of a weak solution
for such a problem.
We would also like to mention that Xiao and Nieto [13], considered the following nonlinear boundary value problems for second order impulsive differential
2000 Mathematics Subject Classification. 34B37, 35B38.
Key words and phrases. Variational method; derivative dependence;
impulsive differential equation.
c
2014
Texas State University - San Marcos.
Submitted August 4, 2013. Published March 5, 2014.
1
2
J. LIU, Z. ZHAO
EJDE-2014/62
equations:
−u00 (t) + g(t)u0 (t) + λu(t) = f (t, u(t)),
0
−∆u (tj ) = Ij (u(tj )),
a.e. t ∈ [0, T ],
j = 1, 2, . . . , p,
(1.2)
u(0) = u(T ) = 0,
where T > 0, 0 = t0 < t1 < · · · < tp < tp+1 = T , f : [0, T ] × R → R is
continuous, g ∈ C[0, T ], and Ij : R → R, j = 1, 2, . . . , p are continuous, and
0 −
0 ±
∆u0 (tj ) = u0 (t+
u0 (t). Authors used critical point
j ) − u (tj ), for u (tj ) = limt→t±
j
theory and variational methods to obtain the above second order impulsive differential equations has at least one positive solution.
Motivated by the above mentioned work, in this paper we consider the impulsive
boundary value problem
−u00 (t) + λu(t) + g(t)u0 (t) = f (t, u),
0
−∆u (ti ) = Ii (u(ti )),
u(0) = 0,
a.e. t ∈ [0, T ],
i = 1, 2, . . . , p,
(1.3)
0
αu(T ) + βu (T ) = 0,
where λ is a parameter, T > 0, g ∈ C[0, T ], f ∈ C([0, T ] × R, R) and Iij : R → R,
i = 1, 2, . . . , p are continuous, 0 = t0 < t1 < · · · < tp < tp+1 = T , ∆u0 (ti ) =
0 −
u0 (t+
u0 (t) − limt→t− u0 (t), α ≥ 0, β > 0 (or β = 0).
i ) − u (ti ) = limt→t+
i
i
We consider the existence of classical solutions for the nonlinear impulsive problems and obtain some new existence theorems of solutions by using variational
methods. We obtain the equation (1.3) has at least one classical solution, at least
two classical solutions and infinitely many classical solutions under different conditions, and the conditions in our paper is easy to verify compared with the papers
in the literature.
The rest of the article is organized as follows: In Section 2, we give some preliminaries, lemmas and variational structure. The main theorems are formulated
and proved in Section 3. In Section 4, some examples are presented to illustrate
our results.
2. Preliminaries, and variational structure
We first recall some basic results in eigenvalue problems. For linear problem
−u00 (t) = λu(t),
u(0) = 0,
0
a.e. t ∈ [0, T ],
αu(T ) + βu (T ) = 0,
the eigenvalue λ satisfies
√
α sin
α ≥ 0, β > 0,
√
√
λT + β λ cos λT = 0.
(2.1)
(2.2)
2
Solving (2.2), we obtain λ = ( kπ−α
T ) , k = 0, 1, . . . , where α satisfies cos α =
√ 2α 2 .
α +λβ
Let λ1 be the first eigenvalue of the above linear problem (2.1). For linear
problem
−u00 (t) = λu(t), a.e. t ∈ [0, T ],
(2.3)
u(0) = 0, αu(T ) + βu0 (T ) = 0, α ≥ 0, β = 0,
the eigenvalue λ satisfies
√
α sin( λT ) = 0.
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AN APPLICATION OF VARIATIONAL METHODS
3
2
Solving this equation, we have λ = ( kπ
T ) , k = 1, 2, . . . , thus the first eigenvalue of
2
2
the linear problem is Tπ 2 ; i.e. λ1 = Tπ 2 . In the remaining part of this paper, we
assume that λ > −mλ1 /M , where m = mint∈[0,T ] eG(t) , M = maxt∈[0,T ] eG(t) , and
Rt
G(t) = − 0 g(s)ds.
We denote the Sobolev space H := H01 (0, T ) = u : [0, T ] → R|u is absolutely
continuous, u0 ∈ L2 (0, T ) and u(0) = 0 with the inner product and the corresponding norm
Z T
(u, v) =
eG(t) u0 (t)v 0 (t)dt,
kuk =
Z
0
T
1/2
eG(t) (u0 (t))2 dt
.
0
0
Let H (0, T ) = {u : [0, T ] → R|u, u are absolutely continuous, u00 ∈ L2 (0, T )}. For
u ∈ H 2 (0, T ), we have that u, u0 are both absolutely continuous, and u00 ∈ L2 (0, T ).
Hence ∆u0 (t) = u0 (t+ ) − u0 (t− ) = 0 for any t ∈ (0, T ). If u ∈ H 1 (0, T ), we have
that u is absolutely continuous, and u0 ∈ L2 (0, T ), thus the one side derivatives
u0 (t+ ), u0 (t− ) may not exist, which leads to the impulsive effects.
So by a classical solution to (1.3) we mean a function u ∈ C[0, T ] satisfying the differential equation in (1.3) such that ui = u|(ti ,ti+1 ) ∈ H 2 (ti , ti+1 ) and
0 +
u0 (t−
i ), u (ti ) exist for every i = 1, 2, . . . , p and verify the impulsive and the boundary conditions. The weak solution to (1.3) is given below and it is inspired by the
weak solution defined in [4].
Multiply the first equation of (1.3) by eG(t) , we obtain
2
0
−(eG(t) u0 (t)) + λeG(t) u(t) = eG(t) f (t, u(t)).
Now multiply by v ∈ H at both sides,
0
− (eG(t) u0 (t)) v(t) + λeG(t) u(t)v(t) = eG(t) f (t, u(t))v(t).
(2.4)
Integrate (2.4) on the interval [0, T ] and use the boundary condition u(0) = 0,
αu(T ) + βu0 (T ) = 0 to obtain
Z T
Z T
eG(t) u0 (t)v 0 (t)dt + λ
eG(t) u(t)v(t)dt
0
−
0
p
X
i=1
Z T
=
eG(ti ) Ii (u(ti ))v(ti ) +
α G(T )
e
u(T )v(T )
β
(2.5)
eG(t) f (t, u(t))v(t)dt.
0
Thus, a weak solution of the impulsive boundary value problem (1.3) is a function
u ∈ H such that (2.5) holds for any v ∈ H.
Define
Z T
Z T
α
eG(t) u(t)v(t)dt + eG(T ) u(T )v(T ),
A(u, v) =
eG(t) u0 (t)v 0 (t)dt + λ
β
0
Z0 u
F (t, u) =
f (t, s)ds.
0
4
J. LIU, Z. ZHAO
EJDE-2014/62
Consider ϕ : H → R defined by
Z T
Z u(ti )
p
X
1
ϕ(u) = A(u, u) −
eG(t) F (t, u(t))dt.
Ii (t)dt −
eG(ti )
2
0
0
i=1
(2.6)
Using the continuity of f and Ii , i = 1, 2, . . . p, we obtain the continuity and differentiability of ϕ and ϕ ∈ C 1 (H, R). For any v ∈ H, one has
Z T
Z T
α
eG(t) u(t)v(t)dt + eG(T ) u(T )v(T )
eG(t) u0 (t)v 0 (t)dt + λ
ϕ0 (u)v =
β
0
0
(2.7)
Z T
p
X
G(ti )
G(t)
−
e
Ii (u(ti ))v(ti ) −
e
f (t, u(t))v(t)dt.
0
i=1
Hence, a critical point of ϕ defined by (2.6), gives us a weak solution of (1.3).
Definition 2.1. Let E be a Banach space and ϕ : E → R, is said to be sequentially
weakly lower semi-continuous if limk→+∞ inf ϕ(xk ) ≥ ϕ(x) as xk * x in E.
Definition 2.2 ([16, p. 81]). Let E be a real reflexive Banach space. For any
sequence {uk } ⊂ E, if ϕ(uk ) is bounded and ϕ0 (uk ) → 0, as k → +∞ possesses a
convergent subsequence, then we say ϕ satisfies the Palais-Smale condition.
Lemma 2.3. If u ∈ H is a weak solution of (1.3), then u is a classical solution of
(1.3).
Proof. By the definition of weak solution, one has
Z T
Z T
α
eG(t) u0 (t)v 0 (t)dt + λ
eG(t) u(t)v(t)dt + eG(T ) u(T )v(T )
β
0
0
Z T
p
X
−
eG(ti ) Ii (u(ti ))v(ti ) −
eG(t) f (t, u(t))v(t)dt = 0.
(2.8)
0
i=1
For i ∈ {0, 1, 2, . . . p}, we choose v ∈ H with v(t) = 0 for every t ∈ [0, ti ] ∪ [ti+1 , T ].
Then we have
Z ti+1
Z ti+1
Z ti+1
eG(t) u0 (t)v 0 (t)dt + λ
eG(t) u(t)v(t)dt =
eG(t) f (t, u(t))v(t)dt.
ti
ti
ti
By the definition of weak derivative, the above equality implies
0
− (eG(t) u0 (t)) + λeG(t) u(t) = eG(t) f (t, u(t)),
a.e. t ∈ (ti , ti+1 ).
(2.9)
i.e.
−u00 (t) + λu(t) + g(t)u0 (t) = f (t, u), a.e. t ∈ (ti , ti+1 ).
Hence, ui = u|(ti ,ti+1 ) ∈ H 2 (ti , ti+1 ) and u satisfies the first equation in (1.3) a.e.
on [0, T ]. Now, multiplying by v ∈ H, v(T ) = 0 and integrating between 0 and T ,
we obtain
p
p
X
X
−
∆(eG(ti ) u0 (ti ))v 0 (ti ) =
(eG(ti ) I(ti ))v 0 (ti ).
i=1
i=1
Hence
−∆u0 (ti ) = Ii (u(ti )), i = 1, 2, . . . , p,
thus, u satisfies the impulsive conditions. It is easy to verify u satisfies the boundary
conditions u(0) = 0, αu(T ) + βu0 (T ) = 0. Therefore, u is a classical solution to
(1.3).
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AN APPLICATION OF VARIATIONAL METHODS
5
Lemma 2.4 ([16, 7, Theorem 38]). For the functional F : M ⊆ X → [−∞, +∞]
with M 6= ∅, minu∈M F (u) = α has a solution when the following conditions hold:
(i) X is a real reflexive Banach space;
(ii) M is bounded and weak sequentially closed; i.e., by definition, for each
sequence un in M such that un * u as n → ∞, we always have u ∈ M ;
(iii) F is weak sequentially lower semi-continuous on M .
Next we sate the mountain pass theorem [3, Theorem 4.10].
Lemma 2.5. Let E be a Banach space and ϕ ∈ C 1 (E, R) satisfy Palais-Smale
condition. Assume there exist x0 , x1 ∈ E, and a bounded open neighborhood Ω of
x0 such that x1 6∈ Ω and
max{ϕ(x0 ), ϕ(x1 )} < inf ϕ(x).
x∈∂Ω
Then there exists a critical value of ϕ; that is, there exists u ∈ E such that ϕ0 (u) = 0
and ϕ(u) > max{ϕ(x0 ), ϕ(x1 )}.
Now we have the symmetric mountain pass theorem [7, Theorem 9.12].
Lemma 2.6. Let E be an infinite dimensional real Banach space. Let ϕ ∈ C 1 (E, R)
be an even functional which satisfies the Palais-Smale condition, and ϕ(0) = 0.
Suppose that E = V ⊕ X, where V is infinite dimensional, and ϕ satisfies that
(i) there exist γ > 0 and ρ > 0 such that ϕ(u) ≥ γ for all u ∈ X with kuk = ρ,
(ii) for any finite dimensional subspace W ⊂ E there is R = R(W ) such that
ϕ(u) ≤ 0 on W \ BR (W ).
Then ϕ possesses an unbounded sequence of critical values.
Lemma 2.7. There exists δ > 0 such that if u ∈ H, then kuk∞ ≤ δkuk, where
kuk∞ = maxt∈[0,T ] |u(t)|.
Proof. It follows from Hölder’s inequality that
Z t
|u(t)| = |
u0 (s)ds|
Z
0
t
|u0 (s)|ds ≤
≤
0
≤
T
0
thus, we can choose δ =
T
|u0 (t)|dt
0
Z
1/2 Z
ds
G(t)
1
e
r
T
kuk,
m
q
T
m
≤
Z
T
1/2
eG(t) |u0 (t)|2 ds
0
such that Lemma 2.7 holds.
Lemma 2.8. There exist two constants θ2 > θ1 > 0 such that if u ∈ H, then
θ1 kuk2 ≤ A(u, u) ≤ θ2 kuk2 .
Proof. Firstly, when λ ≥ 0, we obtain the following results by Poincaré’s inequality,
Z T
Z T
α
eG(t) (u(t))2 dt + eG(T ) u2 (T )
A(u, u) =
eG(t) (u0 (t))2 dt + λ
β
0
0
Z T
Z T
α
≤
eG(t) (u0 (t))2 dt + λM
(u(t))2 dt + eG(T ) u2 (T )
β
0
0
6
J. LIU, Z. ZHAO
T
Z
eG(t) (u0 (t))2 dt +
≤
0
λM
λ1
EJDE-2014/62
T
Z
eG(t) 0
α
(u (t))2 dt + eG(T ) u2 (T )
m
β
0
Z T
λM
Mα
≤ (1 +
eG(t) (u0 (t))2 dt +
)
ku(t)k2∞
λ1 m 0
β
λM
M αδ 2
≤ (1 +
+
)kuk2 .
λ1 m
β
and
T
Z
eG(t) (u0 (t))2 dt + λ
A(u, u) =
Z
T
eG(t) (u(t))2 dt +
0
0
α G(T ) 2
e
u (T )
β
T
Z
eG(t) (u0 (t))2 dt
≥
0
= kuk2 .
2
thus, θ1 = 1, θ2 = 1 + λλM
+ M βαδ .
1m
Secondly, when 0 > λ > −mλ1 /M , by using Poincaré’s inequality, one has
Z T
Z T
α
A(u, u) =
eG(t) (u0 (t))2 dt + λ
eG(t) (u(t))2 dt + eG(T ) u2 (T )
β
0
0
Z T
Z T
≥
eG(t) (u0 (t))2 dt + λ
eG(t) (u(t))2 dt
0
0
T
Z
eG(t) (u0 (t))2 dt + λM
≥
0
eG(t) (u0 (t))2 dt +
≥
0
T
Z
(u(t))2 dt
0
T
Z
T
Z
eG(t) (u0 (t))2 dt +
≥
0
T
λM
λ1
Z
λM
λ1 m
Z
(u0 (t))2 dt
0
T
eG(t) (u0 (t))2 dt
0
λM
)kuk2 ,
= (1 +
λ1 m
and
T
Z
A(u, u) =
e
G(t)
0
2
(u (t)) dt + λ
0
Z
Z
T
eG(t) (u(t))2 dt +
0
α G(T ) 2
e
u (T )
β
T
α
≤
eG(t) (u0 (t))2 dt + eG(T ) u2 (T )
β
0
M αδ 2
≤ (1 +
)kuk2 ,
β
thus, θ1 = 1 +
λM
λ1 m ,
θ2 = 1 +
M αδ 2
β .
Lemma 2.9. The functional ϕ is continuous, continuously differentiable and weakly
lower semi-continuous.
Proof. By the continuity of f and Ii (i = 1, 2, . . . , p), it is easy to check that
functional ϕ is continuous, continuously differentiable. To show that ϕ is weakly
lower semi-continuous, let {un } be a weakly convergent sequence to u in H, then
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AN APPLICATION OF VARIATIONAL METHODS
7
kuk ≤ lim inf n→∞ kun k, and {un } converges uniformly to u in [0, T ], so when
n → ∞, we have
Z
α G(T ) 2
1
λ T G(t)
e
(un (t))2 dt +
lim inf ϕ(un ) = lim inf ( kun k2 +
e
un (T )
n→∞
n→∞ 2
2 0
2β
Z un (ti )
Z T
p
X
G(t)
G(ti )
Ii (t)dt)
e
F (t, un )dt −
e
−
0
≥
0
i=1
Z
T
1
λ
α G(T ) 2
eG(t) (u(t))2 dt +
kuk2 +
e
u (T )
2
2 0
2β
Z T
Z u(ti )
p
X
−
eG(t) F (t, u)dt −
eG(ti )
Ii (t)dt
0
i=1
0
= ϕ(u).
Thus, by Definition 2.1, ϕ is weakly lower semi-continuous.
Now, we introduce the well-known Ambrosetti-Rabinowitz condition: There exist
µ > 2 and r > 0 such that
0 < µF (t, u) ≤ f (t, u)u,
∀u ∈ R \ {0}, t ∈ [0, T ].
It is well known that the Ambrosetti-Rabinowitz condition is quite natural and
convenient not only to ensure the Palais-Smale sequence of the functional ϕ is
bounded but also to guarantee the functional ϕ has a mountain pass geometry.
Lemma 2.10. Suppose that Ambrosetti-Rabinowitz condition holds. Furthermore,
we assume
Z u
Ii (u)u ≥ µ
Ii (t)dt, u ∈ R \ {0}.
0
then the functional ϕ satisfies Palais-Smale condition.
Proof. Let {uk } be a sequence in H such that {ϕ(uk )} is bounded and ϕ0 (uk ) → 0,
as k → +∞, then we will prove {uk } possesses a convergent subsequence.
First we prove
R uthat {uk } is bounded. By the Ambrosetti-Rabinowitz condition
and Ii (u)u ≥ µ 0 Ii (t)dt, we have
µϕ(uk ) − ϕ0 (uk )uk
Z uk (ti )
p
p
X
X
µ
− 1)A(uk , uk ) − µ
eG(ti )
Ii (t)dt +
eG(ti ) Ii (uk (ti ))uk (ti )
2
0
i=1
i=1
Z T
Z T
−µ
eG(t) F (t, uk )dt +
eG(t) f (t, uk )uk dt
=(
0
0
µ
− 1)θ1 kuk k2 ,
2
which implies that {uk } is bounded. Hence there exists a subsequence of {uk } (for
simplicity denoted again by {uk }) such that {uk } weakly converges to some u in
H, then the sequence {uk } converges uniformly to u in [0, T ]. Hence
≥(
(ϕ0 (uk ) − ϕ0 (u))(uk − u) → 0,
Z
0
T
eG(t) (F (t, u) − f (t, u))(uk − u)dt → 0,
8
J. LIU, Z. ZHAO
EJDE-2014/62
[Ii (uk (ti )) − Ii (u(ti ))](uk (ti ) − u(ti )) → 0,
as k → +∞. Thus, we have
(ϕ0 (uk ) − ϕ0 (u))(uk − u)
= ϕ0 (uk )(uk − u) − ϕ0 (u)(uk − u)
Z T
Z
G(t) 0
0
2
=
e
(uk (t) − u (t)) dt + λ
0
−
T
eG(t) (uk (t) − u(t))2 dt +
0
p
X
α
(uk (T ) − u(T ))2
β
eG(ti ) [Ii (uk (ti )) − Ii (u(ti ))](uk (ti ) − u(ti ))
i=1
T
Z
−
eG(t) (f (t, uk ) − f (t, u))(uk (t) − u(t))dt
0
= A(uk (t) − u(t), uk (t) − u(t)) −
p
X
Z
− Ii (u(ti ))](uk (ti ) − u(ti )) −
eG(ti ) [Ii (uk (ti ))
i=1
T
G(t)
e
(f (t, uk ) − f (t, u))(uk (t) − u(t))dt
0
≥ θ1 kuk − uk2 −
p
X
eG(ti ) [Ii (uk )(ti )
i=1
Z
T
− Ii (u(ti ))](uk (ti ) − u(ti )) −
eG(t) (f (t, uk ) − f (t, u))(uk (t) − u(t))dt,
0
which means kuk − uk → 0, as k → +∞. That is, {uk } converges strongly to u in
H.
The following Lemma was proved in [20].
Lemma 2.11. Denote M̄ = maxt∈[0,T ],|u|=1 F (t, u), m̄ = mint∈[0,T ],|u|=1 F (t, u).
Suppose that Ambrosetti-Rabinowitz condition holds. Then, for every t ∈ [0, T ], the
following inequalities hold.
(i) F (t, u) ≤ M̄ |u|µ , if |u| < 1,
(ii) For any finite dimensional subspace W ∈ H and any u ∈ W , there exist
RT
constants A, B > 0, such that 0 F (t, u)dt ≥ m̄B µ kukµ − AT .
3. Main results
Our main results are the following theorems.
Theorem 3.1. Suppose that λ > −mλ1 /M , f and Ii (i = 1, 2, . . . , p) are bounded,
and furthermore f (t, 0) 6≡ 0, then (1.3) has at least one classical solution.
Proof. Take C > 0, Ci > 0, i = 1, 2, . . . , p, such that
|f (t, u)| ≤ C,
|Ii (u)| ≤ Ci ,
∀(t, u) ∈ [0, T ] × R,
∀(t, u) ∈ [0, T ] × R, i = 1, 2, . . . , p.
For any u ∈ H, one has
Z u(ti )
Z T
p
X
1
G(ti )
e
Ii (t)dt −
eG(t) F (t, u(t))dt
ϕ(u) = A(u, u) −
2
0
0
i=1
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AN APPLICATION OF VARIATIONAL METHODS
9
Z u(ti )
Z T
p
X
1
2
G(ti )
Ii (t)dt −
eG(t) F (t, u(t))dt
≥ θ1 kuk −
e
2
0
0
i=1
Z T
p
X
1
≥ θ1 kuk2 −
eG(ti ) Ci |u(ti )| −
eG(t) F (t, u(t))dt
2
0
i=1
Z T
p
X
1
2
G(ti )
|u(t)|dt
e
Ci |u(ti )| − CM
≥ θ1 kuk −
2
0
i=1
≥
p
X
1
θ1 kuk2 −
eG(ti ) Ci kuk∞ − CM T kuk∞
2
i=1
≥
X
1
θ1 kuk2 − M
Ci kuk∞ − CM T kuk∞
2
i=1
≥
X
1
θ1 kuk2 − M
Ci δkuk − CM T δkuk,
2
i=1
p
p
which implies that lim inf kuk→∞ ϕ(u) = +∞, thus, ϕ is coercive. Hence, by [3,
Lemma 2.7 and Theorem 1.1], ϕ has a minimum, which is a critical point of ϕ,
then (1.3) has at least one solution.
Analogously we have the following result.
Theorem 3.2. Suppose that λ > −mλ1 /M , f and Ii (i = 1, 2, . . . , p) have sublinear
growth, and furthermore f (t, 0) 6≡ 0, then (1.3) has at least one classical solution.
Proof. Let a, b, ai , bi > 0, and γ, γi ∈ [0, 1), i = 1, 2, . . . , p, such that
|f (t, u)| ≤ a + b|u|γ ,
γi
|Ii (u)| ≤ ai + bi |u| ,
∀(t, u) ∈ [0, T ] × R,
∀u ∈ R, i = 1, 2, . . . , p.
By using the same methods as in the above proof, there exists η > 0, such that
ϕ(u) ≥
1
θ1 kuk2 − ηkukγ+1 ,
2
which implies that lim inf kuk→∞ ϕ(u) = +∞, thus, ϕ is coercive. Hence, by [3,
Lemma 2.7 and Theorem 1.1], ϕ has a minimum, which is a critical point of ϕ,
then (1.3) has at least one solution.
Theorem 3.3. Suppose the Ambrosetti-Rabinowitz condition
holds, λ > −mλ1 /M ,
Ru
and there exist δi > 0, µ > 2, i = 1, 2, . . . p such that 0 Ii (t)dt ≤ δi |u|µ , Ii (u)u ≥
Ru
µ 0 Ii (t)dt > 0, u ∈ R \ {0}. Then the impulsive problem (1.3) has at least two
classical solutions.
Proof. Firstly, We will show that there exists ρ > 0 such that the functional ϕ has
a local minimum u0 ∈ Bρ = {u ∈ H : kuk < ρ}. By the same methods used in
[20] show that B ρ is a bounded and weak sequentially closed. Noting that ϕ is
weak sequentially lower semi-continuous on B ρ and H is a reflexive Banach space.
Then by Lemma 2.4 we can know that ϕ has a local minimum u0 ∈ Bρ ; that is,
ϕ(u0 ) = minu∈B ρ ϕ(u).
10
J. LIU, Z. ZHAO
EJDE-2014/62
In the following, we will show that ϕ(u0 ) < inf u∈∂Bρ ϕ(u). Choose ρ small
enough such that
p
X
θ1 2
ρ −M
δi ρµ − M M̄ ρµ δ µ T > 0.
2
i=1
For all u = ρω, ω ∈ H with kωk = 1, we have kuk = kρωk = ρkωk = ρ, thus
u ∈ ∂Bρ . By Lemma 2.8 and (i) of Lemma 2.11, one has
ϕ(u) = ϕ(ρω)
Z ρω(ti )
Z T
p
X
1
Ii (t)dt −
eG(t) F (t, ρω(t))dt
A(ρω, ρω) −
eG(ti )
2
0
0
i=1
Z ρω(ti )
Z T
p
X
θ1
Ii (t)dt − M
M̄ |ρω|µ dt
≥ ρ2 −
eG(ti )
2
0
0
i=1
Z T
p
X
θ1
≥ ρ2 − M
δi |ρω|µ − M M̄ ρµ
|ω|µ dt
2
0
i=1
=
p
≥
X
θ1 2
ρ −M
δi ρµ − M M̄ ρµ δ µ T,
2
i=1
thus we obtain ϕ(u) > 0 = ϕ(0) ≥ ϕ(u0 ) for u ∈ ∂Bρ , which implies ϕ(u0 ) <
inf u∈∂Bρ ϕ(u).
Secondly, we will show that there exists u1 with kuk > ρ, such that ϕ(u1 ) <
inf u∈∂Bρ ϕ(u). By Lemma 2.8, the sublinear growth of Ii , i = 1, 2 . . . , p and (ii) of
Lemma 2.11, one has
Z u(ti )
Z T
p
X
1
G(ti )
ϕ(u) = A(u, u) −
e
Ii (t)dt −
eG(t) F (t, u(t))dt
2
0
0
i=1
≤
1
θ2 kuk2 − m(m̄B µ kukµ − AT ).
2
Therefore, we can choose u1 with ku1 k sufficiently large such that ϕ(u1 ) < 0. Then
we have
max{ϕ(u0 ), ϕ(u1 )} < inf ϕ(u).
u∈∂Bρ
Lemma 2.10 shows that u satisfies Palais-smale condition. Hence, by Lemma 2.5
there exists a critical point û. Therefore, u0 and û are two critical points of ϕ, and
they are also classical solutions of (1.3).
Theorem 3.4. Suppose the Ambrosetti-Rabinowitz condition
holds, λ > −mλ1 /M ,
Ru
and there exist δi > 0, µ > 2, i = 1, 2, . . . p such that 0 Ii (t)dt ≤ δi |u|µ , Ii (u)u ≥
Ru
µ 0 Ii (t)dt > 0, u ∈ R \ {0}. Moreover, f (t, u) and Ii are odd about u, then the
impulsive problem (1.3) has infinitely many classical solutions.
Proof. For any u ∈ H, we know that kuk ≤ 1δ implies kuk∞ ≤ 1 by Lemma 2.7,
thus when kuk ≤ 1δ , one has the following inequality by (i) of Lemma 2.11,
Z u(ti )
Z T
p
X
1
G(ti )
ϕ(u) = A(u, u) −
e
Ii (t)dt −
eG(t) F (t, u(t))dt
2
0
0
i=1
EJDE-2014/62
AN APPLICATION OF VARIATIONAL METHODS
p
X
1
≥ θ1 kuk2 − M
δi |u|µ −
2
i=1
p
≥
T
Z
X
1
θ1 kuk2 − M
δi kukµ∞ −
2
i=1
11
eG(t) F (t, u(t))dt
0
T
Z
eG(t) F (t, u(t))dt
0
p
X
1
δi δkukµ − M M̄ T δ µ kukµ .
≥ θ1 kuk2 − M
2
i=1
Thus we can choose u with kuk sufficiently small such that ϕ(u) ≥ γ > 0. Thus ϕ
satisfies condition (i) of Lemma 2.6.
In the following, it is turn to verify condition (ii) of Lemma 2.6. In fact, we can
get the following inequality by (ii) of Lemma 2.11,
Z u(ti )
Z T
p
X
1
A(u, u) −
eG(ti )
Ii (t)dt −
eG(t) F (t, u(t))dt
2
0
0
i=1
Z T
1
≤ θ2 kuk2 − m
F (t, u(t))dt
2
0
1
≤ θ2 kuk2 − m(m̄B µ kukµ − AT ).
2
ϕ(u) =
Noting that µ > 2, the above inequality implies that ϕ(u) → −∞ as kuk → ∞
with u ∈ W . Therefore, there exists R = R(W ) such that ϕ(u) ≤ 0 on W \ BR .
According to Lemma 2.6, the functional ϕ(u) possesses infinitely many critical
points; i.e., the impulsive problem (1.3) has infinitely many classical solutions. Remark 3.5. Equation (1.3) when β = 0, i.e. (1.2), which has been studied in
[13]. By defining a new functional
Z
Z
1 T G(t) 0
λ T G(t)
0
ϕ̃(u) =
e
u (t)v (t)dt +
e
u(t)v(t)dt
2 0
2 0
Z u(ti )
Z T
p
X
−
eG(ti )
Ii (t)dt −
eG(t) F (t, u(t))dt,
0
i=1
0
and using the same methods, we can obtain the same results as the above-proved
mπ 2
four theorems when λ > − M
T2 .
4. Examples
Example 4.1. Take T > 0, t1 ∈ (0, T ), g(t) = t, a(t), b(t) ∈ C([0, T ], R), c ∈ R,
α ≥ 0, β > 0. Consider the equation
−u00 (t) + λu(t) + g(t)u0 (t) = a(t) sin u(t) + b(t),
a.e. t ∈ [0, T ],
0
−∆u (t1 ) = c cos u(t1 ),
u(0) = 0,
when λ > −eT
2
/2
0
αu(T ) + βu (T ) = 0,
λ1 , equation (4.1) is solvable according to Theorem 3.1.
(4.1)
12
J. LIU, Z. ZHAO
EJDE-2014/62
Example 4.2. Take T > 0, t1 ∈ (0, T ), g(t) = t, a(t), b(t) ∈ C([0, T ], R), c, d ∈ R,
α ≥ 0, β > 0. Consider the equation
p
−u00 (t) + λu(t) + g(t)u0 (t) = a(t) 3 u(t) + b(t) sin u(t), a.e. t ∈ [0, T ],
p
(4.2)
−∆u0 (t1 ) = c 5 u(t1 ) + d cos u(t1 ),
u(0) = 0,
when λ > −eT
2
/2
αu(T ) + βu0 (T ) = 0,
λ1 , equation (4.2) is solvable according to Theorem 3.2.
Example 4.3. Take T > 0, t1 ∈ (0, T ), g(t) = t, a(t) ∈ C([0, T ], (0, +∞)), c > 0,
µ = 3, δ1 = 1/6, α ≥ 0, β > 0. Consider the equation
2
2
2
2
−u00 (t) + λu(t) + g(t)u0 (t) = 2a(t)(eu − e−u )u5 + 4a(t)(eu − e−u )u3 ,
a.e. t ∈ [0, T ],
(4.3)
−∆u0 (t1 ) = cu5 (t1 ),
u(0) = 0,
when λ > −eT
Theorem 3.3.
2
/2
αu(T ) + βu0 (T ) = 0,
λ1 , equation (4.3) has at least two classical solutions according to
Example 4.4. Take T > 0, t1 ∈ (0, T ), g(t) = t, a(t), b(t) ∈ C([0, T ], (0, +∞)),
c > 0, µ = 3, δ1 = 1/4, α ≥ 0, β > 0. Consider the equation
−u00 (t) + λu(t) + g(t)u0 (t) = a(t)u5 (t) + b(t)u7 (t),
0
3
−∆u (t1 ) = cu (t1 ),
u(0) = 0,
a.e. t ∈ [0, T ],
(4.4)
0
αu(T ) + βu (T ) = 0,
2
when λ > −eT /2 λ1 , equation (4.4) has infinitely many classical solutions according
to Theorem 3.4.
Acknowledgments. Jian Liu was supported by the Shandong Provincial Natural
Science Foundation, China (ZR2012AQ024). Zengqin Zhao was supported by the
Doctoral Program Foundation of Education Ministry of China (20133705110003)
and Program for Scientific research innovation team in Colleges and universities of
Shandong Province.
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Jian Liu
School of Mathematics and Quantitative Economics, Shandong University of Finance
and Economics, Jinan, Shandong 250014, China
E-mail address: liujianmath@163.com
Zengqin Zhao
School of Mathematical Sciences, Qufu Normal University, Qufu, Shandong 273165,
China
E-mail address: zqzhao@mail.qfnu.edu.cn