R E S E A R C H
Open Access
New existence results for periodic
boundary value problems with impulsive
effects
Weibing Wang
*and Li Guo
*Correspondence:
Department of Mathematics, Hunan University of Science and Technology, Xiangtan, Hunan 411201, P.R. China
Abstract
New results as regards the existence of positive solutions for first order impulsive differential equations are provided. The method of proof relies on the fixed point theorem and degree theory. Some examples are presented to illustrate the main results.
MSC: 34B18; 34B37
Keywords: positive solutions; periodic boundary value problem; fixed point; degree
1 Introduction
In this paper, we study the existence of positive solutions for the following boundary value problem with impulsive effects:
⎧ ⎪ ⎨ ⎪ ⎩
x(t) +a(t)x(t) =f(t,x(t)), t=tk,t∈J,
x(tk) =Ik(x(tk)), k= , , . . . ,p,
x() =x(T),
(.)
whereJ= [,T], =t<t<t<· · ·<tp< =tp+,x(tk) =x(tk+) –x(tk–),x(t+k), andx(t–k)
represent the right limit and left limits ofu(t) attk, respectively.
Impulsive differential equations serve as basic models to study the dynamics of processes that are subject to sudden changes in their states and its theory has developed fast during the past few years. There has been increasing interest in the investigation for boundary value problems of nonlinear impulsive differential equations and much literature has been published about the existence of solutions for impulsive differential equations, see [–] and the references therein. There are some common techniques to approach those prob-lems: the fixed point theorems [–], the method of upper and lower solutions [–], the topological degree theory [, ], the variational method [–] and so on. Recently, using the fixed point theorem, Zhanget al.[] obtained the existence of a positive solu-tion of (.), where they required that the funcsolu-tionais of definite sign. In this paper, we continue to discuss (.). By using the fixed point theorem in a cone different from the one in [] and degree theory, we obtain some new conditions which guarantee the existence of single and multiple positive solutions for (.). Our results are different from the results in [] and are new even ifIk≡.
2 Main results
LetJ∗=J\{t,t, . . . ,tp},PC(J) ={u:J→R|u∈C(J∗),u(ti+),u(t–i) exist andu(t–i) =u(ti),i=
, , . . . ,p}.PC(J) is a Banach space with the normu=sup{|u(t)|:t∈J}.
Lemma .[] The function x∈PC(J)is a solution of(.)if and only if x is a solution of the following impulsive integral equation:
x(t) = T
G(t,s)fs,x(s)ds+
p k=
G(t,tk)Ik
x(tk)
,
where
G(t,s) = ⎧ ⎪ ⎨ ⎪ ⎩
e
s
t a(r)dr
e
T
a(s)ds–, if≤t<s≤T,
e
s t a(r)dre
T
a(r)dr
e
T
a(s)ds–
, if≤s≤t≤T.
Set
+=
a∈C(J,R) : T
a(r)dr>
, –=
a∈C(J,R) : T
a(r)dr<
,
M=maxG(t,s) :t,s∈J, m=minG(t,s) :t,s∈J.
The Green’s functionGis of definite sign ifa∈+∪–. Moreover,M>m> ifa∈+;
m<M< ifa∈–.
Define the operatorAand coneKonPC(J) by
(Ax)(t) = T
G(t,s)fs,x(s)ds+
p k=
G(t,tk)Ik
x(tk)
,
K=x∈PC(J) :x(t)≥δx,
whereδ=m
M ifa∈
+;δ=M
m ifa∈
–.
Lemma .[] Let X be a Banach space and K be a cone in X.Supposeandare
open subsets of X such that∈⊂ ¯⊂and suppose that
:K∩(¯\)→K
is a completely continuous operator such that:
(i) infu> ,u=μuforu∈K∩∂andμ≥,andu=μuforu∈K∩∂ and <μ≤,or
(ii) infu> ,u=μuforu∈K∩∂andμ≥,andu=μuforu∈K∩∂ and <μ≤.
Thenhas a fixed point in K∩(¯\).
Set
ϕ(s) =T sup (t,u)∈J×[δs,s]
f(t,u) u +sup
p
k= ln
+Ik(vk) vk
:vk∈[δs,s]
ψ(s) =T inf (t,u)∈J×[δs,s]
f(t,u) u +inf
p
k= ln
+Ik(vk) vk
:vk∈[δs,s]
,
ϕ= lim
s→+ϕ(s), ϕ∞=s→lim+∞ϕ(s), ψ=slim→+ψ(s), ψ∞=s→lim+∞ψ(s). The following theorems are the main results of this paper.
Theorem . Assume that a∈+and there exist two positive constants r<R such that
f ∈CJ×[δr,R], [, +∞), Ik∈C
[δr,R], [, +∞) (≤k≤p). (.)
Then(.)has at least one solution x with r≤ x ≤R if one of the following conditions is satisfied:
(H) ϕ(r) <
T
a(s)dsandψ(R) >
T
a(s)ds;
(H) ϕ(R) <
T
a(s)dsandψ(r) >
T
a(s)ds.
Proof Here, we only prove the case in which (H) is satisfied. LetR={x∈K:x<R},
r={x∈K:x<r}. At first, we show thatA:¯R\r→K. For anyx∈ ¯R\r,δr≤
x(t)≤R,t∈J. From <m≤G(t,s)≤Mand (.), we obtain, forx∈ ¯R\r,
≤(Ax)(t)≤M
T
fs,x(s)ds+
p k=
Ik
x(tk)
≤MTmaxf(t,u) :t∈J,δr≤u≤R
+pMmaxIk(u) :δr≤u≤R, ≤k≤p
< +∞,
(Ax)(t)≥m
T
fs,x(s)ds+
p k=
Ik
x(tk)
≥δAx.
Hence,A:¯R\r→K. In addition, one easily checks thatAis completely continuous.
Next, we show that:
(a) x=μAxforx∈K∩∂rand <μ≤,
(b) infAx> ,x=μAuforx∈K∩∂Randμ≥.
If (a) is not true, there existx∈K∩∂rand <μ≤ withx=μAx. Hence,
⎧ ⎪ ⎨ ⎪ ⎩
x(t) +a(t)x(t) =μf(t,x(t)), t=tk,t∈J,
x(t+
k) =x(tk) +μIk(x(tk)), k= , , . . . ,p,
x() =x(T).
(.)
Sincex(t)≥δr> , we rewrite (.) as
⎧ ⎪ ⎨ ⎪ ⎩
(lnx(t))+a(t) =μf(xt(,tx)(t)), t=tk,t∈J,
lnx(tk) =ln( +
μIk(x(tk))
x(tk) ), k= , , . . . ,p,
x() =x(T).
Integrating the first equality in (.) from toT, we obtain
which is a contradiction.
Suppose thatinfx∈K∩∂RAx= . There exists the sequencexn∈K∩∂R such that
which implies thatψ(R) = , a contradiction.
Suppose that there existu∈K∩∂Randμ≥ withu=μAu. Then
Integrating the first equality in (.) from toT, we obtain
T
which is a contradiction.
By Lemma ., there existsx∈ ¯R\rwithAx=x, which is the positive solution of (.).
The proof is complete.
Theorem . Assume that a∈+and there exist N+ positive constants p
<p<· · ·<
pN<pN+such that
f ∈CJ×[δp,pN+], [, +∞)
, Ik∈C
[δp,pN+], [, +∞)
(≤k≤p).
Further suppose that one of the following conditions is satisfied: () ϕ(pk–) <
T
a(s)ds,k= , , . . . , [(N+ )/],ψ(pk) >
T
a(s)ds,
k= , , . . . , [(N+ )/],or
() ψ(pk–) >
T
a(s)ds,k= , , . . . , [(N+ )/],ϕ(pk) <
T
a(s)ds,
k= , , . . . , [(N+ )/],
where[d]denotes the integer part of d.Then(.)has at least N positive solutions xk∈X,
k= , , . . . ,N with pk<xk<pk+.
Proof Assume that () holds. The case in which () holds is similar. Sinceϕ,ψare contin-uous functions, for any ≤j≤N, there existrj,Rjsuch thatpj<rj<Rj<pj+and
ϕ(rj) <
T
a(s)ds, ψ(Rj) >
T
a(s)ds, jis odd,
ϕ(Rj) <
T
a(s)ds, ψ(rj) >
T
a(s)ds, jis even.
By Theorem ., (.) has at least one positive solutionxjwithrj≤ xj ≤Rj. This ends the
proof.
Theorem . Assume that a∈–and there exist two positive constants r<R such that
f ∈CJ×[δr,R], (–∞, ], Ik∈C
[δr,R], (–∞, ] (≤k≤p). (.)
Further suppose that(H)or(H)is satisfied,wherelnα= –∞ifα≤.Then(.)has at
least one solution x with r≤ x ≤R.
The proof of Theorem . is similar to that of Theorem . and we omit it.
Theorem . Assume that a∈+and the following conditions are satisfied:
(D) There exist constants <α<βsuch thatf(t,u),Ik(u)are nondecreasing inu∈[α,β]
and
T
f(t,α)dt+
p k=
Ik(α) >
α
m,
T
f(t,β)dt+
p k=
Ik(β) <
β
M.
(D) There existsγ >βsuch that
f ∈CJ×[,γ], [, +∞), Ik∈C
[,γ], [, +∞),
T
sup ≤u≤γ
f(t,u)dt+
p k=
sup ≤u≤γ
Ik(u)≤
γ
(D)
lim
x→+ f(t,x)
x = , xlim→+ Ik(u)
u = .
Then(.)has at least two positive solutions in Uγ ={x∈PC(J) :x ≤γ}.
Proof Set
f =
f(t,|u|), |u| ≤γ,
f(t,γ) |u|>γ, Ik=
Ik(|u|), |u| ≤γ,
Ik(γ), |u|>γ,
(Tλu)(t) =λ
T
G(t,s)fs,u(s)ds+λ
p k=
G(t,tk)Ik
u(tk)
, λ∈[, ],u∈PC(J),
λ=I–Tλ, V=
x∈PC(J),α<x<β,
whereI denotes the identity map. It is easy to check thatTλ: [, ]×PC(J)→PC(J) is
compact. We show: () T(V)⊂V.
() λ(u) = implies thatu ≤γ.
() There existsc∈(,α)such thatλ(u) = admits only the trivial solution inUc.
The proof of (). From (D), we obtain, for∀u∈V,
f(t,α)≤f(t,u)≤f(t,β),
Ik(α)≤Ik(u)≤Ik(β).
Thus, for allu∈V,
T(u) =
T
G(t,s)fs,u(s)ds+
p k=
G(t,tk)Ik(u)≤M
T
f(s,β)ds+
p k=
Ik(β)
<β,
T(u)≥m
T
f(s,α)ds+
p k=
Ik(α)
>α.
That is,T(V)⊂V.
The proof of (). Ifϕλ(u) = , then
|u|=Tλ(u)=
T
G(t,s)fs,u(s)ds+
p k=
G(t,tk)Ik
u(tk)
≤M T
sup ≤u≤γ
f(s,u)ds+M
p k=
sup ≤u≤γ
Ik(u)≤γ.
The proof of (). From (D), there exists <c<αsuch that
f(t,u)≤ u
MT, Ik(u)≤ u
Ifu∈Ucwithu=Tλu, we have By the additivity property of the degree, one obtains
d,Ucγ,
positive solutions inUγ. The proof is complete.
Seth∗=min{h(t) :t∈J}andh∗=max{h(t) :t∈J}. We claim that (.) has at least one positive solution provided that the following conditions hold:
γ=
p k=
ln( +ck) <κT, ifh∗≥, (.)
γ=:
h∗–eλκTh∗eκT
–h∗ a
λ
<κ– T
p k=
ln( +ck), ifh∗< . (.)
In fact,f(t,u) =au–λ+h(t),ψ
= +∞. Moreover,ϕ∞=γifh∗≥. Ifh∗< , chooseb=
λ
√
a/(–h∗), ≤f(t,u)≤au–λ+h∗≤γufor (t,u)∈J×[δb,b] andϕ(b) =γT+γ<κT. By
Theorem ., (.) has one positive solution.
Remark . The conditions (.) and (.) are different from those of Corollary . in [].
Example . Consider the differential equation
⎧ ⎪ ⎨ ⎪ ⎩
x(t) + x(t) =x
(t)exp(–.x(t)), t∈[,t
)∪(t, ], x(t) = x
(t )
+x(t ), x() =x().
(.)
Letf(t) =te–.t,I
(t) = t/( +t),α= .,β= ,γ = . It is easy to check that the
conditions (D)-(D) hold. Hence, (.) has at least two positive solutions.
3 Application
In this section, we consider the differential equation
⎧ ⎪ ⎨ ⎪ ⎩
x(t) = –xaλ((tt))+f(x(t)), t=tk,t∈J,
x(tk) =ckx(tk), k= , , . . . ,p,
x() =x(T),
(.)
whereλ≥.
Sety=exp(xλ+(t)/(λ+ )) andF(u) =uλf(u), then
⎧ ⎪ ⎨ ⎪ ⎩
y(t) +a(t)y(t) =y(t)F(((λ+ )lny(t))λ+), t=tk,t∈J,
y(tk) =y(tk)(+ck)
λ+
–y(tk), k= , , . . . ,p,
y() =y(T).
(.)
If (.) has a solutiony(t) > ,t∈J, thenx(t) = ((λ+ )lny(t))λ+ is the positive solution of (.). By Theorem ., we have the following result.
Theorem . Assume that a∈+,f∈C((, +∞), [, +∞)),c
k> .If there exist constants
δ–<r<R such that
T sup δr≤u≤r
F(λ+ )lnu
λ++r
p k=
( +ck)λ+–
<
T
T inf δR≤u≤RF
(λ+ )lnu
λ++lnδ–R
p k=
( +ck)λ+–
>
T
a(t)dt
or
T sup δR≤u≤R
F(λ+ )lnu
λ++lnR
p k=
( +ck)λ+–
<
T
a(t)dt,
T inf δr≤u≤rF
(λ+ )lnu
λ++lnδ–r
p k=
( +ck)λ+–
>
T
a(t)dt,
then(.)has at least one positive solution.
Example . Consider the differential equation
⎧ ⎪ ⎨ ⎪ ⎩
x(t) = –xaλ((tt))+μf(x(t)), t=tk,t∈J,
x(tk) =μx(tk), k= , , . . . ,p,
x() =x(T),
(.)
whereλ≥,a∈+,f ∈C((, +∞), [, +∞)),μis a positive real parameter.
Corollary . There existsμ∗> such that(.)has at least a positive solution for any
μ∈(,μ∗).
Proof Choosingr=δ–+. Sinceuλf(u) is continuous in (, +∞) andlnu> foru∈[δr,r],
we obtain
≤ sup δr≤u≤r
(λ+ )lnu
λ
+λf(λ+ )lnuλ+≤C
for someC> . On the other hand, since [( +μ)λ+– ]/μ→ +λasμ→+, there is σ> such that
( +μ)λ+– ≤( +λ)μ, <μ≤σ.
Setμ∗=min{σ,Ta(s)ds/[CT+ p(λ+ )lnr]}, forμ<μ∗, we have
T sup δr≤u≤r
F(λ+ )lnu
λ++lnr
p k=
( +μ)λ+–
<μ∗CT+ p(λ+ )lnr<
T
a(t)dt.
Taking
R=max
exp
T
a(t)dt
p( +μ)+λ–p,δ
one easily checks that
T inf δR≤u≤RF
(λ+ )lnu
λ++lnδ–R
p k=
( +μ)λ+– >
T
a(t)dt.
By Theorem ., (.) has one positive solution.
Remark . Corollary . admits the case that the functionachanges sign. As far as we know, no paper discussed (.) whenachanges sign.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
Authors typed, read, and approved the final draft.
Acknowledgements
The authors wish to express their thanks to the referee for his/her very valuable suggestions and careful corrections. The work is supported by the NNSF of China (11171085), Hunan Provincial Natural Science Foundation of China (2015JJ2068).
Received: 1 March 2015 Accepted: 10 August 2015 References
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