R E S E A R C H
Open Access
Existence and continuity of positive
solutions on a parameter for second-order
impulsive differential equations
Yafang Tian and XueMei Zhang
**Correspondence: [email protected]
Department of Mathematics and Physics, North China Electric Power University, Beijing, 102206, People’s Republic of China
Abstract
Applying the eigenvalue theory and theory of
α
-concave operator, we establish some new sufficient conditions to guarantee the existence and continuity of positive solutions on a parameter for a second-order impulsive differential equation.Furthermore, two nonexistence results of positive solutions are also given. In particular, we prove that the unique solutionuλ(t) of the problem is strongly increasing and depends continuously on the parameter
λ
.Keywords: continuity on a parameter; impulsive differential equations; transformation technique;Lp-integrable; eigenvalue
1 Introduction
We consider the second-order impulsive differential equation
⎧ ⎪ ⎨ ⎪ ⎩
u(t) +λω(t)f(u(t)) = , t∈(, ),t=tk,
u(t+
k) –u(tk) =cku(tk), k= , , . . . ,n,
u() = , au() +bu() =g(t)u(t)dt,
(.)
where λ> ,ω∈ Lp[, ] for some ≤ p≤+∞, f ∈ C(R+,R+),R+ = [, +∞],t
k (k = , , . . . ,n) are fixed points with <t<t<· · ·<tk<· · ·<tn< ,a,b> ,{ck} is a real sequence withck> –,k= , , . . . ,n,x(tk+) (k= , , . . . ,n) denotes the right-hand limit of
x(t) att=tk, andg∈C[, ] is a nonnegative function. In addition, we assume thatω,f,ck, andgsatisfy
(H) ω∈Lp[, ]for some≤p≤+∞, and there existsξ> such thatω(t)≥ξa.e. onJ;
(H) f ∈C([, +∞), [, +∞))withf() = andf(u) > foru> ,{ck}is a real sequence
withck> –,k= , , . . . ,n, andc(t) :=<tk<t( +ck);
(H) g∈C[, ]is nonnegative with
μ:=
g(t)c(t)dt∈,ac() . (.)
Remark . We always assume that the productc(t) :=<tk<t( +ck) equals unity if the number of factors is equal to zero, and let
cM=max
t∈J c(t), cm=mint∈J c(t), c
–(t) =
<tk<t( +ck)
–.
Remark . Combining (H) and the definition ofc(t), we know thatc(t) is a step function,
which is bounded onJ, and
c(t) > , ∀t∈J, c(t) = , ∀t∈[,t].
Such problems were first studied by Zhang and Feng []. By using transformation tech-nique to deal with impulse term of second-order impulsive differential equations, the au-thors obtained existence results of positive solutions by using fixed point theorems in a cone. However, they only considered the caseω(t)≡ ont∈[, ]. The other related re-sults can be found in [–]. However, there are almost no papers on second-order bound-ary value problems, especially second-order boundbound-ary value problems with impulsive ef-fects, using the eigenvalue theory. In this paper, we solve this problem.
The first goal of this paper is to establish several criteria for the optimal intervals of the parameterλso as to ensure the existence of positive solutions for problem (.). Our method is based on transformation technique, Hölder’s inequality, and the eigenvalue the-ory and is completely different from those used in [–].
Another contribution of this paper is to study the expression and properties of Green’s function associated with problem (.). It is interesting to point out that the Green’s func-tion associated with problem (.) is positive, which is different from that of [].
Moreover, we give two nonexistence results. The arguments that we present here are based on geometric properties of the super-sublinearity off at zero and infinity, which was first used by Sánchez in [] (see Properties .-.).
For convenience, we introduce the following notations:
f= lim
u→+ f(u)
u , f∞=u→lim+∞ f(u)
u .
The following geometric Properties .-. will be very important in our arguments.
Property . Iff= +∞andf∞= +∞, then there existsR> such that
f(R)
R =mint> f(t)
t . (.)
LetR¯be a point wheref attains its maximum on the interval (,R].
Property . Iff= andf∞= , then there existsR> such that
f(R)
R =maxu> f(u)
u . (.)
solutionuλ(t) of the above problem is strongly increasing and depends continuously on
the parameterλ.
The rest of this paper is organized as follows. In Section , we provide some necessary background. In particular, we introduce some lemmas and definitions associated with the eigenvalue theory and theory ofα-concave (or –α-convex) operators. Several technical lemmas are given in Section . In Section , we establish the existence and nonexistence of positive solutions for problem (.). In Section , we prove the uniqueness of a positive solution for problem (.) and its continuity on a parameter . In Section , we offer some remarks and comments on the associated problem (.). Finally, in Section , two examples are also included to illustrate the main results.
2 Preliminaries
In this section, we collect some known results, which can be found in the book by Guo and Lakshmikantham [].
Definition . LetEbe a real Banach space over R. A nonempty closed setP⊂Eis said to be a cone if
(i) au+bv∈Pfor allu,v∈Pand alla≥,b≥, and (ii) u, –u∈Pimpliesu= .
Definition . A conePof a real Banach spaceEis a solid cone ifP◦is not empty, where
P◦is the interior ofP.
Every coneP⊂E induces a semiorder inE given by “≤”. That is,x≤yif and only if
y–x∈P. If a conePis solid andy–x∈P◦, then we writexy.
Definition . A conePis said to be normal if there exists a positive constantδsuch that
x+y ≥δ, ∀x,y∈P,x= ,y= .
Geometrically, normality means that the angle between two positive unit vectors is bounded away fromπ. In other words, a normal cone cannot be too large.
Lemma . Let P be a cone in E.Then the following assertions are equivalent:
(i) Pis normal;
(ii) There exists a constantγ > such that
x+y ≥γmaxx,y, ∀x,y∈P;
(iii) There exists a constantη> such that≤x≤yimplies thatx ≤ηy,that is,
the norm · is semimonotone;
(iv) There exists an equivalent norm · onEsuch that≤x≤yimplies that x≤ y,that is,the norm · is semimonotone;
(v) xn≤zn≤yn(n= , , , . . .)andxn–x →,yn–x →imply that
zn–x →;
(vi) The set(B+P)∩(B–P)is bounded,where
(vii) Every order interval[x,y] ={z∈E:x≤z≤y}is bounded.
Remark . Some authors use assertion (iii) as the definition of normality of a conePand call the smallest numberηthe normal constant ofP.
Definition . LetPbe a solid cone of a real Banach spaceE. An operatorA:P◦→P◦is called anα-concave operator (–α-convex operator) if
A(tx)≥tαAxA(tx)≤t–αAx , ∀x∈P◦, <t< ,
where ≤α< . The operatorAis increasing (decreasing) ifx,x∈P◦andx≤ximply Ax≤Ax(Ax≥Ax), and further, the operatorAis strongly increasing (decreasing) if x,x∈P◦andx<ximplyAx–Ax∈P◦(Ax–Ax∈P◦). Letxλbe a proper element of
an eigenvalueλofA, that is,Axλ=λxλ. Thenxλis called strongly increasing (decreasing)
ifλ>λimplies thatxλ–xλ∈P◦(xλ–xλ∈P◦), which is denoted byxλxλ (xλ xλ).
Definition . An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.
Lemma .(Arzelà-Ascoli) A set M⊂C(J,R)is said to be a precompact set if the following two conditions are satisfied:
(i) All the functions in the setMare uniformly bounded,which means that there exists a constantr> such that|u(t)| ≤r,∀t∈J,u∈M;
(ii) All the functions in the setMare equicontinuous,which means that for everyε> ,
there isδ=δ(ε) > ,which is independent of the functionu∈M,such that u(t) –u(t)<ε
whenever|t–t|<δ,t,t∈J.
Lemma . Suppose D is an open subset of an infinite-dimensional real Banach space E, θ ∈D,and P is a cone of E.If the operator:P∩D→P is completely continuous with
θ=θ and satisfies
inf
x∈P∩∂Dx> ,
thenhas a proper element on P∩∂D associated with a positive eigenvalue.That is,there exist x∈P∩∂D andμsuch thatx=μx.
Lemma . Suppose that P is a normal cone of a real Banach space and A:P◦→P◦is anα-concave increasing(or–α-convex decreasing)operator.Then A has exactly one fixed point in P◦.
3 Some lemmas
LetJ= [, ]. A functionu(t) is said to be a solution of problem (.) onJif:
(i) u(t)is absolutely continuous on each interval(,t]and(tk,tk+],k= , , . . . ,n;
(ii) for anyk= , , . . . ,n,u(t+
k)andu(t–k)exist, andu(t–k) =u(tk);
We shall reduce problem (.) to a system without impulse. To this goal, firstly, by means of the transformation
u(t) =c(t)y(t) (.)
we convert problem (.) into
–y(t) =λω(t)c–(t)f(c(t)y(t)), t∈J,
y() = , ac()y() +bc()y() =g(s)c(s)y(s)ds. (.) The following lemmas will be used in the proof of our main results.
Lemma . Assume that(H)-(H)hold.Then
(i) Ify(t)is a solution of problem(.)onJ,thenu(t) =c(t)y(t)is a solution of problem
(.)onJ;
(ii) Ifu(t)is a solution of problem(.)onJ,theny(t) =c–(t)u(t)is a solution of problem
(.)onJ.
Proof (i) Lety(t) be a solution of (.) onJ. It is easy to see thatu(t) =c(t)y(t) is absolutely continuous on each interval (tk,tk+],k= , , . . . ,n. By the definition ofc(t) we havec(t) =
fort=tk. Then, fort=tk, we have
u(t) =c(t)y(t) +c(t)y(t) =c(t)y(t),
u(t) =c(t)y(t) +c(t)y(t) =c(t)y(t).
It follows that
–u(t) = –c(t)y(t) =λω(t)fc(t)y(t) =λω(t)fu(t) .
Fort=tk, we have
utk+ = lim
t→tk+c(t)y(t) =≤ti≤tk( +ci)y(tk),
utk– =lim
t→t–
k
c(t)y(t) =≤ti≤tk–( +ci)y(tk).
By (ii) of Definition .,u(t–
k) =u(tk), so we have
utk+ –u(tk) =u
tk+ –ut–k =≤ti≤tk–( +ci)cky(tk) =ckc(tk)y(tk) =cku(tk).
Thus,u(t+
k) –u(tk) =cku(tk).
It is obvious thatu(t) satisfies the boundary conditions. Thenu(t) is a solution of problem (.) onJ.
(ii) It is easy to see that, fort∈J,
Fort=tk,
ytk+ –ytk– =c–tk+ ut+k –c–tk– ut–k
=c–t+k u(tk) +cku(tk) –c–
tk– ut–k
=c–t–k ut–k –c–t–k utk–
= .
Theny(t) is continuous onJ. It is easy to prove thaty(t) is absolutely continuous onJand satisfies the boundary conditions.
Theny(t) is a solution of problem (.) onJ.
Lemma . If(H)-(H)hold,then problem(.)has a solution y,and y can be expressed in the form
y(t) =λ
H(t,s)ω(s)c–(s)fc(s)y(s) ds, (.)
where
H(t,s) =G(t,s) +
ac() –μ
G(τ,s)g(τ)c(τ)dτ+ bc()
ac() –μ, (.)
G(t,s) =
–t, ≤s≤t≤,
–s, ≤t≤s≤. (.)
Proof First, suppose thatyis a solution of problem (.). Integrating problem (.) from tot, by the boundary conditions we obtain that
y(t) = – t
z(s)ds, (.)
wherez(s) =λω(s)c–(s)f(c(s)y(s)). Integrating (.) from tot, we have
y(t) =y() – t
(t–s)z(s)ds. (.)
Lettingt= in (.) and (.), we find
y() =y() –
( –s)z(s)ds, y() = –
z(s)ds.
Combining these equalities with (.) and the boundary conditionsac()y() +bc()y() =
g(t)c(t)y(t)dt, we obtain
y(t) =
( –s)z(s)ds+
ac()
g(s)c(s)y(s)ds– t
(t–s)z(s)ds+b
a
z(s)ds
=
G(t,s)z(s)ds+
ac()
g(s)c(s)y(s)ds+b
a
and further
g(s)c(s)y(s)ds
=
g(s)c(s)
G(s,τ)z(τ)dτ+
ac()
g(τ)c(τ)y(τ)dτ+b
a
z(τ)dτ
ds = ac()
g(s)c(s)ds
g(τ)c(τ)y(τ)dτ+
g(s)c(s)
G(s,τ)z(τ)dτ ds +b a
g(s)c(s)ds
z(τ)dτ. (.)
Therefore, we have
g(s)c(s)y(s)ds= ac()
ac() –μ
g(s)c(s)
G(s,τ)z(τ)dτ ds +b a
g(s)c(s)ds
z(τ)dτ
. (.)
Substituting (.) into (.), we obtain
y(t) =
G(t,s)z(s)ds+
ac() –μ
g(s)c(s)
G(s,τ)z(τ)dτ ds +b a
g(s)c(s)ds
z(τ)dτ+b
a
z(s)ds
=
H(t,s)z(s)ds. (.)
Then
y(t) =λ
H(t,s)ω(s)c–(s)fc(s)y(s) ds, (.)
and the proof of sufficiency is complete. Conversely, from (.) it is easy to obtain
–y(t) =λω(t)c–(t)fc(t)y(t) ,
y() = , ac()y() +bc()y() =
g(t)c(t)y(t)dt.
Lemma . is proved.
Lemma . Letμ∈[,ac()),G,and H be given as in Lemma..Then we have the fol-lowing results:
H(t,s) > , G(t,s)≥, ∀t,s∈J, (.)
where≤e(t) = –t≤,and
<α∗≤H(t,s)≤βh(s)≤β∗, ∀t,s∈J, (.)
where
α∗= bc()
ac() –μ, β
= c()
ac() –μ, h(s) =ae(s) +b, β
∗=(a+b)c()
ac() –μ. (.)
Proof Relation (.) is simple to prove. For ≤s≤t≤, we have
e(t)e(s)≤e(t) =G(t,s) = –t≤ –s=e(s).
For ≤t≤s≤, we have
e(t)e(s)≤e(s) =G(t,s) = –s.
Then,
e(t)e(s)≤G(t,s)≤e(s).
This gives the proof of (.).
For anyt,s∈J, by (.), (.), and (.) we have
H(t,s) =G(t,s) +
ac() –μ
G(τ,s)g(τ)c(τ)dτ+ bc()
ac() –μ
≥ bc() ac() –μ
=α∗. On the other hand,
H(t,s) =G(t,s) +
ac() –μ
G(τ,s)g(τ)c(τ)dτ+ bc()
ac() –μ
≤e(s) +
ac() –μ
e(s)g(τ)c(τ)dτ+ bc()
ac() –μ
=e(s)
+
ac() –μ
g(τ)c(τ)dτ
+ bc()
ac() –μ
=e(s) ac()
ac() –μ+
bc()
ac() –μ
=βae(s) +b ≤βh(s)≤β∗.
Therefore, the proof of (.) is complete.
Lemma .(Hölder) Let e∈Lp[a,b]with p> ,h∈Lq[a,b]with q> ,and
p+
q= .Then
eh∈L[a,b]and
eh≤ ephq.
Let e∈L[a,b],h∈L∞[a,b].Then eh∈L[a,b],and eh≤ eh∞.
LetE=C[, ]. ThenEis a real Banach space with the norm · defined by
x=max
t∈J
x(t), x∈E.
Define two conesKandKinEby
K=y∈E:y(t)≥,y(t)≥δy,t∈J (.)
and
K=
y∈E:y(t)≥,t∈J,
whereδ=α∗ β∗=
b
a+b. It is easy to see thatKandKare two solid normal cones and
K=y∈E:y(t) > ,y(t)≥δy,t∈J,
K=y∈E:y(t) > ,t∈J.
Forr> , definerby
r=
y∈K:y<r,
∂r=
y∈K:y=r.
DefineT:K→Kby
(Ty)(t) =
H(t,s)ω(s)c–(s)fc(s)y(s) ds, t∈J. (.)
Lemma . Assume that(H)-(H)hold.Then T(K)⊂K,and T :K→K is completely continuous.
Proof Fory∈K, it follows from (.) and (.) that
(Ty)(t) =
H(t,s)ω(s)c–(s)fc(s)y(s) ds
≥α∗
≥α∗
β∗maxt∈J
H(t,s)ω(s)c–(s)fc(s)y(s) ds
=δTy.
Thus,T(K)⊂K.
Next, we prove that the operatorT:K→Kis completely continuous by standard meth-ods and the Arzelà-Ascoli theorem.
LetBr={y∈E| y ≤r}be a bounded set. Then, for ally∈Br, we have
Ty=max
t∈J
H(t,s)ω(s)c–(s)fc(s)y(s) ds≤c–mβhqωpL,
whereL=maxc(s)y(s)≤cMrf(c(s)y(s)). Therefore,T(Br) is uniformly bounded.
On the other hand, noticing thatH(t,s) is uniformly continuous onJ×J, we have that, for anyε> , there existsδ> such that if|t–t|<δ, then
H(t,s) –H(t,s)<
ε
ωc–mL .
Then, for anyy∈Br, taking|t–t|<δ, we get
(Ty)(t) – (Ty)(t)=
H(t,s)ω(s)c–(s)f
c(s)y(s) ds
–
H(t,s)ω(s)c–(s)f
c(s)y(s) ds
=
H(t,s) –H(t,s)ω(s)c–(s)f
c(s)y(s) ds
≤ ε
ωc–mL
ω(s)c–(s)fc(s)y(s) ds
≤ ωc–mL
ε
ωc–mL
=ε.
Thus, the set{T:y∈Br}is equicontinuous. The Arzelà-Ascoli theorem implies thatT
is completely continuous, and Lemma . is proved.
4 Existence and nonexistence of positive solutions on a parameter
In this section, we establish some sufficient conditions for the existence and nonexis-tence of positive solutions of problem (.). We consider the following three cases for ω∈Lp[, ] :p> ,p= , andp=∞. The casep> is treated in the following theorem.
Theorem . Assume that(H)-(H)hold.If <f∞< +∞,then there exists R> such that for any r>R,problem(.)has a positive solution ur(t)satisfyingur(t)=cMr for
any
λ=λr∈[λ,λ], (.)
Proof By (.) and (.) problem (.) has a positive solutionur(t) associated withλ> if and only if the operatorT has a proper elementyr associated with the eigenvalue
λ> .
Since <f∞< +∞, there existl>l> andη> such that
ly<f(y) <ly, ∀y≥η. (.)
Now, we prove thatR=cmηδ is required. Thus, for allr>R, ify∈K∩∂r, we have
y(t)≥δy=δr, t∈J.
Noticingr>R, we have
c(t)y(t)≥cmδy=cmδr>cmδR=η, t∈J.
Together with Lemma ., we have thatT:K∩ ¯r→Kis completely continuous with
Tθ=θ. In addition,
(Ty)(t) =
H(t,s)ω(s)c–(s)fc(s)y(s) ds
≥α∗ξc–M
lc(s)y(s)ds
≥α∗ξc–Mlcmδy =α∗ξc–Mlcmδr> .
Therefore, for anyr>Randy∈K∩∂r, we have
inf
y∈K∩∂rTy ≥
α∗ξc–Mlcmδr> .
By Lemma ., for anyr>R, the operatorT has a proper elementyr∈K associated with the eigenvalueγ> ; further,yrsatisfiesyr=r. Letλ=γ. Then problem (.) has a positive solutionyr(t) associated withλ.
Hence, it follows from Lemma . that problem (.) has a positive solutionur(t) asso-ciated withλand satisfyingur=cMr.
From the proof above, for anyr>R, there exists a positive solutionyr∈K∩∂r asso-ciated withλ> , that is,
yr(t) =λ
H(t,s)ω(s)c–(s)fc(s)y(s) ds
withyr=r. On the one hand,
and, further,
yr=r≤λc–mβhqωplcMr,
which means that
λ≥
lc–mcMβhqωp =λ.
On the other hand,
yr(t)≥λα∗ξc–Mlcmδr, (.)
and thus
yr=r≥λα∗ξc–Mlcmδr,
which leads to
λ≤
lδα∗ξc–Mcm =λ.
It is easy to see by calculating thatλ<λ.
In conclusion,λ∈[λ,λ]. The proof is complete.
The following Corollary . deals with the casep=∞.
Corollary . Assume that(H)-(H)hold.If <f∞< +∞,then there exists R> such that for any r>R,problem(.)has a positive solution ur(t)satisfyingur(t)=cMr for
any
λ∈λ,λ
,
where
λ=
lc–mcMβhω∞.
Proof Replacinghqωp byhω∞ and repeating the argument above, we get the
corollary.
Finally, we consider the case ofp= .
Corollary . Assume that(H)-(H)hold.If <f∞< +∞,then there exists R> such that for any r>R,problem(.)has a positive solution ur(t)satisfyingur(t)=cMr for
any
λ∈λ,λ
where
λ =
lc–mcMβ∗ω
.
Proof Replacingβhqωp by β∗ω and repeating the argument above, we get the
corollary.
In the following theorems, we only consider the case <p< +∞.
Theorem . Assume that(H)-(H)hold.If f∞= +∞,then there exists R> such that for any r>R,problem(.)has a positive solution ur(t)satisfyingur=cMr for any
λ=λr∈(,λ],
whereλis a positive finite number.
Proof Similarly to the proof of Theorem ., it is easy to see from (.) and (.) that
Theorem . is also true.
Theorem . Assume that(H)-(H)hold.If <f< +∞,then there exists r> such that for any <r<r,problem(.)has a positive solution ur(t)satisfyingur(t)=cMr for any
λ∈[λˆ,λˆ],
whereλˆandλˆare two positive finite numbers.
Proof By (.) and (.) problem (.) has a positive solutionur(t) associated withλ> if and only if the operatorThas a proper elementyrassociated with the eigenvalueλ> .
Since <f< +∞, there existη> and constantsc>c> such that
cu<f(u) <cu, ∀ <u<η.
Set
Ur=
y∈E:y<r,
where <r<r.
ThenUris a bounded open subset of the Banach spaceE, andθ∈Ur. Now, we prove thatr= η
cM is required.
Thus, fory∈K∩∂Ur, noticing <r<r, we have
y(t)≥δy=δr, t∈J,
and
Together with Lemma ., we note thatT:K∩ ¯Ur→Kis completely continuous with
Tθ=θand that
(Ty)(t) =
H(t,s)ω(s)c–(s)fc(s)y(s) ds
≥α∗ξc–M
cc(s)y(s)ds
≥α∗ξc–Mccmδy =α∗ξc–Mccmδr> .
So, for any <r<randy∈K∩∂Ur, we have
inf
y∈K∩∂UrTy ≥
α∗ξc–Mccmδr> .
By Lemma ., for any <r<r, the operatorThas a proper elementyr∈Kassociated with the eigenvalue γ > ; further,yr satisfiesyr=r. Lettingλ=γ and following the
proof of Theorem ., we complete the proof of Theorem ..
Theorem . Assume that(H)-(H)hold.If f= +∞,then there exists r> such that for any <r<r,problem(.)has a positive solution ur(t)satisfyingur=cMr for any
λ=λr∈(,λˆ],
whereλˆis a positive finite number.
Proof The proof is similar to that of Theorem ., so we omit it here.
Theorem . Assume that(H)-(H)hold.If f=f∞= +∞,then there existsλ¯> such that problem(.)has no positive solutions for allλ∈[λ¯, +∞).
Proof We argue by contradiction. Suppose that there exists a sequence{λn} withλn>
nsuch that for eachn, problem (.) has a positive solutionyn∈K. Letμn=λn. Since (Tyn)(t) =μnyn(t) fort∈Jandf(u)≥Nufor allu> , whereN=f(RR), we have
yn=max
t∈J λnTyn(t)
≥λn
H(t,s)ω(s)c–(s)fc(s)yn(s) ds
≥λnα∗ξc–M
Ncmyn(s)ds
≥λnα∗ξc–MNcmδyn >nα∗ξc–MNcmδyn, which implies that >nα∗ξc–MNcmδ.
Therefore, by Lemma . problem (.) has no positive solutions for allλ≥ ¯λ. This gives
the proof of Theorem ..
Theorem . Assume that(H)-(H)hold.If f=f∞= ,then there existsλ> such that problem(.)has no positive solutions forλ∈(,λ).
Proof It follows fromf=f∞= and (.) that there existsv¯> such that f(v¯)
¯ v
=max
v> f(v)
v .
Let
M=f(v¯)
¯ v
+ .
Then M > and
f(v)≤Mv, ∀v> . (.)
Lety(t) be a positive solution of problem (.). We will show that this leads to a con-tradiction forλ<λ, whereλ= (βhqωpc–mcMM)–. Letμ=λ. Since (Ty)(t) =μy(t) for
t∈J, it follows from (.) that
y(t) =λ
H(t,s)ω(s)c–(s)fc(s)y(s) ds
≤λβhqωpc–m
Mc(s)y(s)ds
≤λβhqωpc–mcMMy,
which shows that
y ≤λβhqωpc–mcMMy <λβhqωpc–mcMMy =y,
which is a contradiction. This finishes the proof.
Remark . The method to study the existence and nonexistence results of positive so-lutions is completely different from those of Zhang and Feng [].
5 Uniqueness and continuity of positive solution on a parameter
Theorem . Suppose that f(u) : [, +∞)→[, +∞)is a nondecreasing function with f(u) > for u> and satisfies f(ρu)≥ραf(u)for any <ρ< ,where≤α< .Then,
for anyλ∈(,∞),problem(.)has a unique positive solution uλ(t).Furthermore,such a solution uλ(t)satisfies the following properties:
(i) uλ(t)is strongly increasing inλ,that is,λ>λ> impliesuλ(t)uλ(t)fort∈J.
(ii) limλ→+uλ= ,limλ→+∞uλ= +∞.
(iii) uλ(t)is continuous with respect toλ,that is,λ→λ> impliesuλ–uλ →.
Proof Set=λT, whereTis the same as in (.). Similarly to Lemma ., the operator mapsKintoK. In view ofH(t,s) > ,ω(s) > ,c–(s) > , andf(u) > foru> , it is easy
to see that:K
→K. We assert that:K→Kis anα-concave increasing operator.
Indeed,
(ρy) =λ
H(t,s)ω(s)c–(s)fc(s)ρy(s) ds
≥ραλ
H(t,s)ω(s)c–(s)fc(s)y(s) ds
=ρα(y), ∀ <ρ< ,
where ≤α< . Sincef(u) is nondecreasing, we have
(y∗)(t) =λ
H(t,s)ω(s)c–(s)fc(s)y
∗(s) ds
≤λ
H(t,s)ω(s)c–(s)fc(s)y∗∗(s) ds
= (y∗∗)(t) fory∗≤y∗∗,y∗,y∗∗∈X.
In view of Lemma .,has a unique fixed pointyλ∈K. This shows that problem (.)
has a unique positive solutionyλ(t). It follows from Lemma . that problem (.) has a
unique positive solutionuλ(t).
Next, we give a proof for (i)-(iii). Letγ =λ and denoteλTyλ=yλbyTyγ =γyγ. Assume
that <γ<γ. Thenyγ≥yγ. Indeed, set
¯
η=sup{η:yγ≥ηyγ}. (.)
We assertη¯≥. If this is not true, then <η¯< , and further
γyγ=Tyγ≥T(η¯yγ)≥ ¯η αTy
γ=η¯ αγ
yγ,
which implies
yγ≥ ¯η αγ
γ yγ ¯η
α
yγ ¯ηyγ.
In view of the discussion above, we have
yγ=
γ
Tyγ≥
γ
Tyγ =
γ
γ
yγyγ. (.)
Hence, yγ(t) is strongly decreasing in γ. Namely, yλ(t) is strongly increasing in λ. By
Lemma ., (i) is proved.
Settingγ=γ and fixingγin (.), we haveyγ≥ γ
γyγ forγ >γ. Further,
yγ ≤
γN
γ yγ, (.)
whereN> is a normal constant. Noting thatγ =λ, we havelimλ→+yλ(t)= . Then
it follows from Lemma . thatlimλ→+uλ(t)= .
Similarly, lettingγ=γ and fixingγ, again by (.) and the normality of K we have
limλ→+∞yλ(t)= +∞. Then, it follows from Lemma . thatlimλ→+∞uλ(t)= +∞.
This gives the proof of (ii).
Next, we show the continuity ofuγ(t). For givenγ> , by (i),
yγyγ for anyγ >γ. (.)
Letlγ =sup{ν> |yγ≥νyγ,γ>γ}. Obviously, <lγ < andyγ ≥lγyγ. So, we have
γyγ =Tyγ≥T(lγyγ)≥l α γTyγ=l
α γγyγ,
and further
yγ≥
γ
γ l
α γyγ.
By the definition oflγ,
γ
γ l
α
γ≤lγ or lγ≥
γ
γ
–α .
Again by the definition oflγ, we have
yγ≥
γ
γ
–α
yγ for anyγ >γ. (.)
Noticing thatKis a normal cone, in view of (.) and (.), we obtain
yγ–yγ ≤N
–
γ
γ
–α
yγ →, γ→γ+ .
In the same way,
yγ–yγ →, γ →γ– ,
Therefore, by Lemma . we have
uγ–uγ ≤cMyγ–yγ ≤cMN
–
γ
γ
–α
yγ →, γ →γ+ .
uγ–uγ ≤cMyγ –yγ →, γ →γ– .
Consequently, (iii) holds. The proof is complete.
6 Remarks and comments
In this section, we offer some remarks and comments on the associated problem (.).
Remark . Some ideas of the proof of Theorem . come from Theorem .. in [] and Theorem in [], but there are almost no papers considering the uniqueness of positive solution for second impulsive differential equations, especially in the case whereω(t) is
Lp-integrable.
Remark . Generally, it is difficult to study the uniqueness of a positive solution for nonlinear second-order differential equations with or without impulsive effects (see, e.g., [, , ] and references therein).
For example, we consider the following problem:
⎧ ⎪ ⎨ ⎪ ⎩
u(t) +λω(t)f(t,u(t)) = , t∈J,t=tk,
u(t+k) –u(tk) =cku(tk), k= , , . . . ,n,
u() =u() =h(s)u(t)dt,
(.)
whereλ> is a positive parameter,J= [, ],ω∈Lp[, ] for some ≤p≤+∞,f ∈C(J× R+,R+),R+= [, +∞),t
k(k= , , . . . ,n) are fixed points with <t<t<· · ·<tk<· · ·<tn< ,{ck}is a real sequence withck> –,k= , , . . . ,n,x(tk+) (k= , , . . . ,n) is the right-hand limit ofx(t) attk, andh∈C[, ] is nonnegative.
By means of transformation (.) we can convert problem (.) into
–y(t) =λc–(t)ω(t)f(t,c(t)y(t)), t∈J,
y() =c()y() =h(s)c(s)y(s)ds. (.)
Using a proof similar to that of Lemma ., we can obtain the following results.
Lemma . If(H)-(H)hold,then problem(.)has a solution y,and y can be expressed in the form
y(t) =λ
H∗(t,s)ω(s)c–(s)fc(s)y(s) ds, (.)
where
H∗(t,s) =G∗(t,s) + –ν
G∗(t,s) =
t( –s), ≤t≤s≤,
s( –t), ≤s≤t≤. (.)
It is not difficult to prove that H∗(t,s)and G∗(t,s)have similar properties to those of H(t,s)
and G(t,s).However,we cannot guarantee that H∗(t,s) > for any t,s∈J.This implies that we cannot apply Lemma.to study the uniqueness of a positive solution for problem(.).
Remark . In Theorem ., even though we do not assume thatTis completely contin-uous or even contincontin-uous, we can assert thatuλdepends continuously onλ.
Remark . If we replaceK,KbyK,K, respectively, then Theorem . also holds.
7 Examples
To illustrate how our main results can be used in practice, we present two examples.
Example . Letn= ,t=,p= . It follows fromp= thatq=. Consider the following
boundary value problem: ⎧
⎪ ⎪ ⎨ ⎪ ⎪ ⎩
u(t) +λ(
|t–|
)(u+arctanu) = , t∈J,t=tk,
u(+) –u() =u(),
u() = , u() +u() =u(t)dt,
(.)
Conclusion Problem (.) has at least one positive solution for anyλ∈[., .].
Proof Problem (.) can be regarded as a problem of the form (.), where
ω(t) =
|t–|
∈L[, ], fu(t) = u+arctanu,
and
a=b= , g(t) = .
We convert problem (.) into ⎧
⎨ ⎩
–y(t) =λ(
|t–|
)c–(t)(c(t)y(t) +arctanc(t)y(t)), t∈J,
y() = , y() +y() =c(s)y(s)ds,
(.)
where
c(t) =
, ≤t≤,
,
<t≤.
Fromω(t) =
|t–|
,t∈J, choosingp= ,q=, it follows that
ωp=ω=
|t–|
dt
=( + ×
)
hq=h =
( –t) dt
=
×
–
≈..
Thus, it is easy to see by calculating thatω(t)≥ξ=
for a.e.t∈Jand that
μ=
g(t)c(t)dt=
, cM=
, cm= ,
and
α∗= bc()
ac() –μ= , β
= c()
ac() –μ= ,
β∗=(a+b)c()
ac() –μ = , δ= α∗ β∗ =
.
Therefore, it follows from the definitionsω(t),f, andgthat (H)-(H) hold and
f∞= lim
u→+∞ f(u)
u =u→lim+∞
+arctanu
u
= + lim
u→+∞
arctanu u = ,
so <f∞= < +∞. Thus, we have
<f∞< .
Setl= andl= . Then
λ=
lc–mcMβh ω
≈.,
λ=
lδα∗ξc–Mcm
≈..
Hence, by Theorem . the conclusion follows, and the proof is complete.
Example . Letn= ,t=,p= . It follows fromp= thatq=∞. Consider the
follow-ing boundary value problem: ⎧
⎪ ⎨ ⎪ ⎩
u(t) +λ(t+ )(u+arctanu) = , t∈J,t=t, u(+) –u() =u(),
u() = , u() +u() =u(t)dt.
(.)
Conclusion Problem (.) has at least one positive solution for anyλ∈[ ,].
Proof Problem (.) can be regarded as a problem of the form (.), where
ω(t) = t+ ∈L[, ], f(u) = u+arctanu,
and
We convert problem (.) into
–y(t) =λ(t+ )c–(t)(c(t)y(t) +arctanc(t)y(t)), t∈J,
y() = , y() +y() =c(s)y(s)ds, (.) where
c(t) =
, ≤t≤,
,
<t≤.
Thus, it is easy to see by calculating thatω(t)≥ξ= for a.e.t∈Jand that
μ=
g(t)c(t)dt=
, cM=
, cm= ,
and
α∗= bc()
ac() –μ= , β
∗=(a+b)c()
ac() –μ = , δ= α∗ β∗ =
.
Therefore, it follows from the definitionsω(t),f, andgthat (H)-(H) hold.
On the other hand, it follows fromω(t) = t+ that
ω=
(t+ )dt= .
Thus, we have
<f∞< .
Setl= andl= . Then
λ =
lc–mcMβ∗ω
=
,
λ=
lδα∗ξc–Mcm =
.
Hence, by Corollary . the conclusion follows, and the proof is complete.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
All results belong to YT and XZ. Both authors read and approved the final manuscript.
Acknowledgements
This work is sponsored by the National Natural Science Foundation of China (11301178, 11371117), the Beijing Natural Science Foundation of China (1163007), and the Scientific Research Project of Construction for Scientific and Technological Innovation Service Capacity (71E1610973). The authors are grateful to anonymous referees for their constructive comments and suggestions, which have greatly improved this paper.
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