R E S E A R C H
Open Access
Multiple solutions of three-point boundary
value problems for second-order impulsive
differential equation at resonance
Yulin Zhao
1*, Haibo Chen
2and Qiming Zhang
1*Correspondence:
zhaoylch@sina.com
1School of Science, Hunan
University of Technology, Zhuzhou, Hunan 412007, PR China Full list of author information is available at the end of the article
Abstract
In this paper, by using the coincidence degree theory and the upper and lower solutions method, we deal with the existence of multiple solutions to three-point boundary value problems for second-order differential equation with impulses at resonance. An example is given to show the validity of our results.
Keywords: resonance; coincidence degree; upper and lower solutions; impulsive; three-point boundary value problems
1 Introduction
The purpose of the present paper is to investigate the following second-order impulsive differential equations:
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
(ρ(t)u(t))=f(t,u(t),u(t)), t∈J,t=tk,
u(tk) =Ik(tk,u(tk)), k= , , . . . ,m,
u(tk) =Jk(tk,u(tk)),
(.)
together with the boundary conditions:
u() = , u() =u(η), (.)
whereJ= [, ],ρ:J→(, +∞) is a continuous differentiable function,f :J×R→Ris
continuous, <η< ,Ik,Jk∈C(J,R) for ≤k≤m,mis a fixed positive integer, =t<
t<t<· · ·<tm<tm+= ,η=tk,u(tk) =u(t+k) –u(tk–) denotes the jump ofu(t) att=tk,
u(tk) =u(tk+) –u(t–k).u(tk+),u(t+k) (u(tk–),u(t–k)) represent the right limit (left limit) of
u(t) andu(t) att=tk, respectively.
Impulsive differential equations describe processes which experience a sudden change of their state at certain moments. The theory of impulse differential equations has been a significant development in recent years and played a very important role in modern applied mathematical models of real processes rising in phenomena studied in physics, population dynamics, chemical technology, biotechnology, and economics; see [–] and the references therein.
Recently, several authors (see [, –] and the references therein) have studied the existence of nontrivial or positive solutions for second-order three-point boundary value problem of the type
⎧ ⎨ ⎩
u=f(t,u,u),
u() = , u() =au(η). (.)
Note that the nonlinear termfdepends onuand its derivativeu, then the relative prob-lem becomes more complicated. A general method to deal with this difficulty is to add some conditions to restrict the growth of theuterm. One condition is the Caratheodory nonlinearity, the other usual condition is Nagumo condition or Nagumo-Winter condi-tion (see [, , , –]). Whena= , the linear operatorLu=uis invertible, this is the so-called non-resonance case. Guptaet al.made use of the Leray-Schauder continuation theorem to get the results on the existence of the solution for the problems (.) when
a= in []. By using the Leray-Schauder continuation theorem and in the presence of
two pairs of upper and lower solutions, Khan and Webb [] established the existence of at least three solutions for the problem (.) whena= . The linear operatorLu=uis non-invertible whena= , this is the so-called resonance case, and the Leray-Schauder continuation theorem cannot be applied. In [], by using the coincidence degree theory of Mawhin [] and some linear or non-linear growth assumptions onf, Feng and Webb obtained the existence of the solution of the problem (.) whena= . By applying the nonlinear alternative of Leray-Schauder, Ma [] have showed the existence of at least one solution for the problem (.) whena= .
Recently, using the coincidence degree theory and the concept of autonomous curvature bound set, Liu and Yu [] have studied the existence of at least one solution for the problem (.)-(.) whenu(tk) =Ik(u(tk),u(tk)),u(tk) =Jk(u(tk),u(tk)).
In the present paper, we assume that there existn(n∈Nandn≥) pairs of upper and lower solutions for problem (.)-(.) and the nonlinearf satisfies a Nagumo-like growth condition with respect tou. By considering a suitably modified nonlinearity and applying the coincidence degree method of Mawhin [], the existence of multiple solutions for the problem (.)-(.) is given.
2 Preliminaries
Let
X=PC(J)∩u() = ,u() =u(η), Z=PC(J)×Rm, where
PC(J) =u∈CJ* ,ut– andut+ exist, andut–k =u(tk)
.
PC(J) =u:J→R:u(t) is continuously differentiable fort= , ,tk;u
t–
andut+ exist, andutk– =u(tk)
, J*=J\{t,t, . . . ,tm}.
Obviously,Xis a Banach space with the following norm:
uX=max
sup
t∈J
u(t),sup
t∈J
In the following, we recall the concept of strict upper and lower solutions for problem (.)-(.).
Definition . A functionα(t)∈PC(J)∩C(J*) is said to be a strict lower solution of the
problem (.)-(.) if
ρ(t)α(t) >ft,α(t),α(t), t∈J*, (.)
α(tk) =Ik
tk,α(tk) , α(tk)≥Jk
tk,α(tk), k= , , . . . ,m, (.)
α()≥, α() –α(η)≤. (.)
Similarly, a functionβ(t)∈PC(J)∩C(J*) is said to be a strict upper solution of the
problem (.)-(.) if
ρ(t)β(t)<ft,β(t),β(t), t∈J*, (.)
β(tk) =Ik
tk,β(tk) , β(tk)≤Jk
tk,β(tk) , k= , , . . . ,m, (.)
β()≤, β() –β(η)≥. (.)
Remark . Letf :J×R→Rbe continuous,Ik,Jk∈C(J,R), andu∈PC(J)∩C(J*) is a
solution of the problem (.)-(.), ifα(β) is a strict lower solution (strict upper solution) for the problem (.)-(.) withα≤u(u≤β), thenα<u(u<β) on (, ).
Definition . Letα be a strict lower solution andβ be a strict upper solution for the problem (.)-(.) satisfyingα(t) <β(t) onJ. We say thatf :J×R→Rhas property (H)
relative toαandβ, if there exists a functionψ∈C([, +∞), (, +∞)) such that
f(t,u,p)<ψ|p| , (.)
for allu(t)∈(–β(t), –α(t))∪(α(t),β(t)),t∈J, and
+∞
s
θs+ψ(s)ds= +∞, (.)
where ≤θ< +∞with|ρ(t)| ≤θ,t∈J.
3 The key lemmas
LetdomL=C(J*)∩X, and
L:domL→Z, u→ρ(t)u(t) ,u(t), . . . ,u(tm),u(t), . . . ,u(tm) ,
N:u→z, u→ft,u,u ,It,u(t) , . . . ,Im
tm,u(tm) ,
J
t,u(t) , . . . ,Jm
tm,u(tm)
.
Then the problem (.)-(.) can be written as
Lemma . Suppose that L be defined in the above.Then L is a Fredholm operator of index zero.Furthermore
Ker(L) ={u∈X:u=c,c∈R}, (.)
and
Im(L) =(y,a, . . . ,am,b, . . . ,bm) :
ρ(t)u(t) =y(t),u(tk) =ak,u(tk) =bk,
k= , . . . ,m, for some y∈domL
=
(y,a, . . . ,am,b, . . . ,bm) :
η
ρ(s)
s
y(τ)dτds+
η
ρ(s)
tk<s
ρ(tk)bkds
+
η<tk< ak=
. (.)
Proof Firstly, it is clear that (.) holds. Next, we shall prove that (.) holds. The following problem:
⎧ ⎨ ⎩
(ρ(t)u(t))=y(t),
u(tk) =ak, u(tk) =bk, k= , . . . ,m
(.)
has a solutionu(t) satisfyingu() = andu() =u(η) if and only if
η
ρ(s)
s
y(τ)dτds+
η
ρ(s)
tk<s
ρ(tk)bkds+
η<tk<
ak= . (.)
In fact, if (.) has a solutionu(t) satisfyingu() = ,u() =u(η), then from (.) we have
u(t) =u() +
t
ρ(s)
s
y(τ)dτds+
t
ρ(s)
tk<s
ρ(tk)bkds+
tk<t ak.
According tou() = ,u() =u(η), we get (.). On the other hand, if (.) holds, setting
u(t) =c+
t
ρ(s)
s
y(τ)dτds+
t
ρ(s)
tk<s
ρ(tk)bkds+
tk<t ak,
wherecis an arbitrary constant, thenu(t) is a solution of (.) withu() = ,u() =u(η).
Hence (.) holds.
Take the projectorQ:Z→Zas follows:
Q(y,a, . . . ,am,b, . . . ,bm)
=
φ() –φ(η)
η
ρ(s)
s
y(τ)dτds+
η
ρ(s)
tk<s
ρ(tk)bkds+
whereφ(t) =tρs(s)ds,t∈(, ). For every (y,a, . . . ,am,b, . . . ,bm)∈Z, set
z= (y,a, . . . ,am,b, . . . ,bm) = (y,a, . . . ,am,b, . . . ,bm) –Q(y,a, . . . ,am,b, . . . ,bm).
Thus, we obtain
η
ρ(s)
s
y(τ)dτds+
η
ρ(s)
tk<s
ρ(tk)bkds+
η<tk< ak
=
η
ρ(s)
s
y(τ)dτds+
η
ρ(s)
tk<s
ρ(tk)bkds+
η<tk< ak
×
–
φ() –φ(η)
η
ds ρ(s)
= .
Then z∈ImL. HenceZ=ImL+R. SinceImL∩R={}, we haveZ=ImL⊕R, which implies dim Ker(L) =domR=codim ImL= . HenceLis a Fredholm operator of index zero.
TakeP:Z→Z,Pu=u(). So the generalized inverseKP:ImL→domL∩KerPofL
can be written as
KPz(t) =KP(y,a, . . . ,am,b, . . . ,bm)
=
t
ρ(s)
s
y(τ)dτds+
t
ρ(s)
tk<s
ρ(tk)bkds+
tk<t
ak. (.)
Setδ:=mint∈Jρ(t) > ,d. Then for the functionψdefined by (.), leth(u) be the
solution of the following initial value problem:
δyy+θy+ψ(y) = , y() =d, (.)
andh(u) be the solution of the following initial value problem:
δyy–θy–ψ(y) = , y() =d. (.)
Lemma . Suppose that there exists a constant M> ,then h(u)is well defined in[,M]
and positive on this interval,h(u)is also well defined and positive in[–M, ].Moreover,if d,then h(u)for any u∈[,M];h(u)for any u∈[–M, ].
Proof We only consider the casey=h(u) (in the casey=h(u), the proof is similar). Assume that there exists au∈[,M] such thaty(u) =h(u) = . Letu=inf{u:h(u) = ,u∈[,M]}. It follows from (.) that
– yy
θy+ψ(y)=δ
–, y> .
Integrating the above equation over [,u] we get (letτ=y(s))
–
u
y(s)y(s)ds θy(s) +ψ(y(s))=
d
τdτ θ τ+ψ(τ)=δ
–u
However, the left side of the above equation equals∞by (.). We reach a contradiction. Henceh(u) > for everyu≥. Fromd, by the continuity of solution of differential equations on the initial values, we obtainh(u) foru∈[,M]. The proof is complete.
Define the following sets:
G=(t,u,p) :t∈J,|u|<M,|p|<h(u) foru∈[,M],
and|p|<h(u) foru∈[–M, ],
=u∈PC(J) :t,u(t),u(t) ∈G,t∈J;tk+,ut+k ,ut+k ∈G,k= , . . . ,m. Define the functionh(u) as
h(u) =
⎧ ⎨ ⎩
h(u), u∈[,M],
h(u), u∈[–M, ]. (.)
Lemma . LetDegdenote the coincidence degree.Let the following conditions hold: (i) f(t, –M, ) < <f(t,M, ),∀t∈J(Mis given in Lemma.);
(ii) |f(t,u,p)|<ψ(|p|),∀t∈J,|u| ≤M,p∈R;
(iii) Ik(tk,±M) = ,andJk(tk, –M) < <Jk(tk,M),k= , . . . ,m.
Then
Deg(L,N),= –.
Proof Consider the following family of equations:
Lu=λNu, λ∈(, ]. (.)
We will show
Lu=λNu, ∀u∈∂,λ∈(, ]. (.)
If not, then there exist some λ∈(, ] andu∈∂ such that (.) holds. Note that
u∈∂if and only if (t,u(t),u(t))∈ ¯Gand either (¯t,u(¯t),u(¯t))∈∂Gfor some¯t∈J, or (t+k,u(tk+),u(t+k))∈∂for somek∈ {, , . . . ,m}. There are two possibilities.
Case (I). If (¯t,u(¯t),u(¯t))∈∂G,¯t∈[, ],t¯=tk+. In this case,|u(¯t)|=M, or|u(¯t)|=h(u(¯t)). Subcase (). Suppose |u(¯t)|=M. Letg(t) = (u(t))–
M. Then g(t)≤, t∈J and g(¯t) = . Whent≤ ¯t∈(, ),g(t)≤, we get
≤g(¯t– ) =u(¯t)u(¯t).
Ifg(¯t– ) = , thenu(¯t) = . From condition (i), we have
≥g(¯t– ) =u(¯t) +u(¯t)
ρ(¯t)·
ρ(¯t)u(¯t) =± M
ρ(¯t)·λf(¯t,±M, ) > ,
Ifg(¯t– ) > , then¯t=tkfor somek∈ {, . . . ,m}andg(tk– ) > . Thus from (iii), we
have
g(tk+ ) =
u(tk) +λIk
tk,u(tk) ·
u(tk) +λJk
tk,u(tk)
=u(tk)u
(t
k) +u(tk)·λJk
tk,u(tk) +u
(t
k)·λIk
tk,u(tk)
+λIk
tk,u(tk) ·Jk
tk,u(tk)
=g(tk– ) +
±M·λJk(tk,±M)
> .
On the other hand,g(t)≤,t∈Jandg(tk+ ) = , thus
≥g(tk+ ) =
u(tk) +λIk
tk,u(tk) ·
u(tk) +λJk
tk,u(tk) > ,
which is a contradiction.
Ift¯= , it is easy to see thatg() = . Sinceu() = , we haveg() = , thus we can obtaing()≤. However, from condition (i) we know
≥ρ()g() =u()·λf,u(), > ,
which is a contradiction.
If¯t= , then|u()|=M. This means thatu()∈∂G. Sinceu(η) =u(), we haveu(η)∈∂G. However, according to the above arguments, we knowu(η) /∈∂G, which is a contradiction. Subcase (). Suppose|u(¯t)|=h(u(¯t)) fort¯∈(tk,tk+],k∈ {, , . . . ,m}. Obviously,¯t= .
Since|(ρ(t)u(t))|=|f(t,u(t),u(t))|<ψ(|u(t)|). Thus we get
ρ(t)u(t)–ρ(t)u(t)<ψu(t) ,
u(t)<
δρ
(t)u(t)+ψu(t) ≤
δ
θu(t)+ψu(t) . (.)
Let
p(t) =
u(¯t) –
hu(t) .
Without loss of generality, we suppose thatu(¯t) =h(u(¯t)) > for¯t∈(tk,tk+]. Then we
have three possibilities.
(i) Ifu(¯t) = , then there exists a sufficiently small neighborhood of¯tsuch thatu(t) > . Thusp(t) = (u(t))–
(h(u(t))), andp(t) has a local maximum value on¯t, which implies p(¯t+ )≤. But in this case, from (.), (.), and (ii), we have
p(¯t+ ) =u(¯t)·u(¯t) –hu(¯t)hu(¯t)
>u(¯t)·
–
δ
θu(¯t) +ψu(¯t) +θh(u(t)) +ψ(h(u(t)))
δ
= , (.)
which is a contradiction.
(ii) Ifu(¯t) < , then there exists a sufficiently small neighborhood of¯tsuch thatu(t) < . Thusp(t) =(u(t))–
p(¯t)≥. But in this case, from (.), (.), and (ii), we have
p(¯t) =u(¯t)·u(¯t) –h
u(¯t) hu(¯t)
<u(¯t)·
δ
θu(¯t) +ψu(¯t) –θh(u(t)) +ψ(h(u(t)))
δ
= ,
which is a contradiction.
(iii) Ifu(¯t) > , then there exists a sufficiently small neighborhood of¯tsuch thatu(t) > . Thusp(t) =(u(t))–
(h(u(t))). By the same argument as in (.), we reach a
contra-diction. Case (II). (t+
k,u(t
+
k),u
(t+
k))∈∂for somek∈ {, , . . . ,m}. In this case,|u(t
+
k)|=M, or
|u(t+k)|=h(u(tk+)).
Subcase (). Suppose|u(t+k)|=Mfor somek∈ {, , . . . ,m}. Letg(t) be defined in the above. Obviously, we haveg(t)≤ fort∈Jandg(tk+ ) = . It follows from (iii) that
≥g(tk+ ) =
u(tk) +λIk
tk,u(tk) ·
u(tk) +λJk
tk,u(tk)
=u(tk)u(tk) +u(tk)·λJk
tk,u(tk) .
Ifg(tk+ ) = , thenu(t
+
k) = . Form (.) and (i), we have
≥g(tk+ ) =
ut+k +u(t
+
k)
ρ(tk+)·
ρ(tk)u
tk+ =± M
ρ(tk+)·λf(tk,±M, ) > , which is a contradiction.
Ifg(tk+ ) < , together with (iii), we getg(tk– ) =u(tk)u(tk) < . Butg(tk) = and
g(t)≤ fort∈J, which yieldsg(tk– ) > , and we reach a contradiction.
Similar to subcase (), we can show that|u(tk+)|=h(u(tk+)) is also impossible. Combin-ing the results of case (I) and case (II) we obtain (.).
On the other hand, for λ∈(, ], it follows from [] that (.) is equivalent to the following family of operator equations:
u=Pu+QNu+λKP(E–Q)Nu, (.)
whereEis the identity mapping. Note thatKerL=R, and
∩KerL=c∈R: (t,c,p)∈G,t∈J;tk+,c,p ∈G, for somek= , . . . ,m.
From (.) and (.), we have
QNu=
φ() –φ(η)
η
ρ(s)
s
fτ,u(τ),u(τ) dτds
+
η
ρ(s)
tk<s ρ(tk)Jk
tk,u(tk) ds+
η<tk< Ik
tk,u(tk) , , . . . ,
and
KP(E–Q)Nu=
φ() –φ(η)
t
ρ(s)
s
fτ,u(τ),u(τ) dτds
+
t
ρ(s)
tk<s ρ(tk)Jk
tk,u(tk) ds+
tk<t Ik
tk,u(tk)
– φ(t)
φ() –φ(η)
η
ρ(s)
s
fτ,u(τ),u(τ) dτds
+
η
ρ(s)
tk<s ρ(tk)Jk
tk,u(tk) ds+
η<tk< Ik
tk,u(tk) .
By the Ascoli-Arzela theorem, it is easy to show thatQN(¯) is bounded andKP(E–Q)N:
¯
→Xis compact. ThusNisL-compact on¯. Define a mappingH:¯ ×[, ]→X,
H(u,λ) =Pu+QNu+λKP(E–Q)Nu, ∀(u,λ)∈ ¯×[, ].
Then it is easy to prove thatH(u,λ) is completely continuous and we claim that
u=H(u,λ), ∀u∈∂, ≤λ≤. (.)
In fact, for <λ≤, it follows from (.) and (.) that (.) holds. Forλ= , if there exists au¯∈∂such thatu¯=H(u¯, ), that isu¯=Pu¯+QNu¯, in this case,QNu¯= ,u¯∈KerL, henceu¯=Moru¯= –M. However, it follows from (iii) that
QN(c) =
φ() –φ(η)
η
ρ(s)
s
f(τ,c, )dτds
+
η
ρ(s)
tk<s
ρ(tk)Jk(tk,c)ds+
η<tk< Ik(tk,c)
=
φ() –φ(η)
η
ρ(s)
s
f(τ,c, )dτds+
η
ρ(s)
tk<s
ρ(tk)Jk(tk,c)ds, c∈R.
Thus we getQN(M) > andQN(–M) < , which contradictsu=Pu+QNuforu∈KerL. Therefore (.) holds. As follows from [] and by using the invariance of Leray-Schauder degree under homotopy, we obtain
DegE–H(·, ),, =DegE–H(·, ),,
=DegB(E–P–QN)|KerL∩ ¯,KerL∩, =DegB(–QN)|KerL∩ ¯,KerL∩, . SinceKerLis one dimensional andQN(M) > ,QN(–M) < , we get
DegB(–QN)|KerL∩ ¯,KerL∩, = –.
Lemma . Assume that
(c) there exist lower and upper solutionsα(t),β(t)of the problem(.)-(.), respectively,withα(t) <β(t);
(c) f :J×R→Ris continuous and has property(H)relative toα,β; (c) Ik,Jkare continuous for eachk= , , . . . ,m,and satisfy
Ik
tk,α(tk) =Ik
tk,β(tk) = , and Jk
tk,α(tk) < <Jk
tk,β(tk).
ThenDeg[(L,N),αβ] = –.
Proof ChooseM> large enough foru,p∈R,u∈(–β(t), –α(t))∪(α(t),β(t)),t∈J, such that
ft,β(t), +M–β(t) > , M>β(t),∀t∈J,
ft,α(t), –M–α(t) < , –M<α(t),∀t∈J,
Jk
tk,α(tk) –M–α(tk) < <Jk
tk,β(tk) +M–β(tk), k= , . . . ,m,
and lethi(u) (i= , ) be defined in (.) and (.), then it follows from Lemma . that we
can choosed> large enough such that
min
[,M]h(u) >max
max
t∈J
β(t),max
t∈J
α(t),
min
[–M,]h(u) >max
max
t∈J
β(t),max
t∈J
α(t).
Define a setαβas
αβ=
u∈PC(J)|(t,u,p) :α(t) <u<β(t),|p|<h(u),∀t∈J,
whereh(u) is given in (.).
Define the auxiliary functionsFand¯Ik,¯Jkas follows:
F(t,u,p) =ft,n(t,u),q(t,p) +u(t) –n(t,u), t∈J*,
¯
Ik
tk,u(tk) =Ik
tk,n
tk,u(tk)
, k∈ {, . . . ,m},
¯
Jk
tk,u(tk) =Jk
tk,n
tk,u(tk)
+u(tk) –n
tk,u(tk) , k∈ {, . . . ,m},
where
nt,u(t) =
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
β(t), β(t) <u≤M,
u(t), α(t)≤u≤β(t),
α(t), –M≤u<α(t),
q(t,p) =
⎧ ⎪ ⎪ ⎨ ⎪ ⎪ ⎩
h(u), p>h(u),|u| ≤M,
p, |p| ≤h(u),|u| ≤M,
We then generalizeFtoJ×Rand¯I
k,J¯k:J×R→R. It is easy to see thatF,I¯k,J¯kare
continuous and satisfy
F(t, –M, ) < <F(t,M, ), ¯
Ik(tk,±M) = , and J¯k(tk, –M) < <J¯k(tk, +M), k∈ {, . . . ,m}.
Moreover, when|u| ≤M,F,¯Ik,¯Jkare bounded. It follows from Lemma . that
Deg(L,N¯),= –,
where
¯
Nu=Ft,u,u ,I¯
t,u(t) , . . . ,¯Im
tm,u(tm) ,¯J
t,u(t), . . . ,J¯m
tm,u(tm)
.
Next we show
Deg(L,N¯),αβ
= –. (.)
It suffices to show that
Lu=Nu¯ , ∀u∈ ¯\αβ. (.)
In fact, letu∈ ¯such thatLu=Nu¯ and assume that
max
t∈J
u(t) –β(t)=u(τ) –β(τ)≥.
Case (). If there existsτ∈(, ),τ=tk such that the functiony(t) =u(t) –β(t) attains
its maximum valuey(τ)≥, which impliesy(τ) = andy(τ)≤. But on the other hand,
ρ(τ)y(τ) =ρ(τ)u(τ) –ρ(τ)β(τ)
=Fτ,β(τ),u(τ) +u(τ) –β(τ) –ρ(τ)β(τ)
=fτ,β(τ),β(τ) +u(τ) –β(τ) –ρ(τ)β(τ) > , (.)
which impliesy(τ) > . We reach a contradiction.
Ifτ= , theny()≥,y()≤. On the other hand, sinceu() = andβ(t) is a strict upper solution of problem (.)-(.), we see thaty() =u() –β()≥. Thusy() = . Note thaty(t) assumes the maximum value att= , there existsτ∈(, ),τ=tk such
thaty(τ)≥ andy(τ)≤. By the same argument as in (.) whereτ=τ, we reach a
contradiction.
Ifτ= , theny() =u() –β()≥. According to (.), we get
y() =u() –β()≤u() –β(η) =u(η) –β(η) =y(η),
Hence the functiony(t) cannot have any nonnegative maximum value on the interval (, ),t=tkfork∈ {, , . . . ,m}.
Case (). If there exists aτ∈Jsuch thaty(τ) =u(τ) –β(τ) =ε≥, from case (), we get
τ =tkfor somek= , , . . . ,m. Hencen(tk,u(tk)) =β(tk),β(tk)≤u(tk), and consequently
we havey(t+
k) =y(tk) =ε, which impliesy(t+k)≥, because
ut+k –u(tk) =¯Jk
tk,u(tk) =Jk
tk,β(tk) +u(tk) –β(tk)
≥Jk
tk,β(tk) ≥β
tk+ –β(tk).
Consequently, we getu(t+
k) =β(t+k) ory(tk+) = andD+y(t+k)≤.
By the continuity offandβ(t) is a strict upper solution of problem (.)-(.), there exists a sequenceζj∈Rwithζj> ,ζj→, asj→ ∞such that
D+ρ(tk)β
tk+ = lim
j→∞sup
ρ(tk+ζj)β(tk+ζj) –ρ(tk)β(tk)
ζj
= lim
j→∞
ρ(tk+ζj)β(tk+ζj)
< lim
j→∞
ftk+ζj,β(tk+ζj),β(tk+ζj) =f
tk,β
t+k ,βt+k , (.)
whereζj∈(tk,tk+ζj) are from the mean value theorem. As before, we also get
D+ρ(tk)u
tk+ =Ftk,u
tk+ ,utk+ ,
whereD+ρ(tk)u(tk+) =limj→∞infρ(tk+ζj)u (t
k+ζj)–ρ(tk)u(tk)
ζj . As a result, we can obtain
D+ρ(tk)β
tk+ ≥D+ρ(tk)u
t+k =ftk,β
tk+ ,βtk+ +ytk+ ≥ftk,β
tk+ ,βtk+
which is a contradiction to (.).
Case (). Ify(t) =u(t) –β(t) <εfor allt∈J, then there must be ak(≤k≤m) such thaty(t+
k) =u(t
+
k) –β(t
+
k) =ε, andy(tk) <ε, which impliesβ(t
+
k) –β(tk) <u(t
+
k) –u(tk).
Namely,β(tk) <u(tk). However, this is impossible because
u(tk) =I¯k
tk,n
tk,u(tk)
=Ik
tk,β(tk) =β(tk).
Thus we have proved thatu(t) <β(t) onJ. Similarly we can show thatα(t) <u(t) onJ. It then follows thatα(t) <u(t) <β(t) onJ.
We now shall prove that
u(t)<hu(t) fort∈[tk,tk+],k= , , . . . ,m. (.)
Assume that (.) cannot hold. Thus there are two possibilities:
(a) There existsτ∈(tk,tk+]for somek∈ {, , . . . ,m}such that|u(τ)| ≥h(u(τ)).
(b) There exists somek∈ {, , . . . ,m}such that|u(t+
k)| ≥h(u(t
+
k)).
Since|(ρ(t)u(t))|=|F(t,u(t),u(t))|=|f(t,u(t),h(u(t)))|<ψ(h(u(t))). Thus we have
ρ(t)u(t)–ρ(t)u(t)<ψhu(t) ,
u(t)<
δρ
(t)u(t)+ψhu(t) ≤
δ
θu(t)+ψhu(t) . (.)
Ifu(τ) = , then there exists a sufficiently small neighborhood ofτsuch thatu(t) > . Thus
r(t) =u(t) –h(u(t)) andr(τ+ )≤. However, it follows from (.) and (.) that
≥r(τ+ ) =u(τ) –hu(τ) u(τ)
> –
δ
θu(τ) +ψhu(τ) +ψ(h(u(τ))) +θh(u(τ))
δh(u(τ)) ·u (τ)
= ψ(h(u(τ)))
δh(u(τ))
u(τ) –h
u(τ) ≥, (.)
which is a contradiction.
Ifu(τ) < , then there exists a sufficiently small neighborhood ofτ such thatu(t) < . Thusr(t) =u(t) –h(u(t)) andr(τ) = . However, it follows from (.) and (.) that
=r(τ) =u(τ) –hu(τ)u(τ)
<
δ
θu(τ) +ψhu(τ) –ψ(h(u(τ))) +θh(u(τ))
δh(u(τ))
·u(τ)
= –ψ(h(u(τ)))
δh(u(τ))
u(τ) –h
u(τ) ≤,
a contradiction.
Ifu(τ) > , then there exists a sufficiently small neighborhood ofτ such thatu(t) > . Thusr(t) =u(t) –h(u(t)) andr(τ) = . By the similarly argument as in (.), we reach a contradiction.
Henceu(t) <h(u(t)),t∈(tk,tk+],k= , , . . . ,m. Similarly, we can prove –h(u(t)) <u(t), t∈(tk,tk+],k= , , . . . ,m.
Likewise, we can show that case (b) is also impossible, and thus (.) holds. Combining the above results, we see that ifu∈,Lu=Nu¯ , thenu∈αβ. Hence (.) is proved. From the property of coincidence, we know that (.) holds. Since inαβ,F=f, we haveN=N¯.
The proof is complete.
4 Main results
We are now in a position to prove our main result on the existence of at least n– solu-tions of boundary value problem (.)-(.).
Theorem . Assume that
(H) there existn(n∈N andn≥)pairs of strict lower and upper solutions{αi(t)}ni=,
{βi(t)}ni=of the problem(.)-(.)such that
(H) f :J×R→R,f(t,u,p)is continuous onJ*×Rand has property(H)relative toα(t), βn(t);
(H) Ik,Jkare continuous for eachk= , , . . . ,m,and satisfy
Ik
tk,αi(tk) =Ik
tk,βi(tk) = , and
Jk
tk,αi(tk) < <Jk
tk,βi(tk) , i= , , . . . ,n.
Then BVP(.)-(.)has at leastn– solutions u(t),u(t), . . . ,un–(t)such thatα(t) < u(t) <β(t),α(t) <u(t) <β(t), . . . ,αn(t) <un(t) <βn(t),and
min
t∈J un+(t) <α(t), maxt∈J un+(t) >β(t), . . . , mint∈J un–(t) <αn(t),
max
t∈J un–(t) >βn–(t).
Proof ChooseM> large enough such that, fori= , , . . . ,n,
ft,βi(t), +M–βi(t) > , M>βi(t),∀t∈J,
ft,αi(t), –M–αi(t) < , –M<αi(t),∀t∈J,
Jk
tk,αi(tk) –M–αi(tk) < <Jk
tk,βi(tk) +M–βi(tk), k= , . . . ,m,
and leth(u) be defined in (.), then it follows from Lemma . that we can taked> large enough such that, for allu∈[–M,M],
min
[,M]h(u) >max
max
t∈J
βi(t),max
t∈J
αi(t),i= , , . . . ,n
,
min
[–M,]h(u) >max
max
t∈J
β
i(t),max t∈J
α
i(t),i= , , . . . ,n
.
Define the setGi(i= , , . . . ,n) as
Gi=
u∈PC(J)|(t,u,p) :αi(t) <u<βi(t),|p|<h(u),∀t∈J
.
Also
Gn+=
u∈PC(J)|(t,u,p) :α(t) <u<βn+(t),|p|<h(u),∀t∈J
.
Fori= , , . . . ,n+ , we define the sets
i=
u∈PC(J) :t,u(t),u(t) ∈Gi,t∈J;
t+k,utk+ ,utk+ ∈Gi,k= , . . . ,m
.
Then by Lemma ., we have
Deg(L,N),i
= –, i= , , . . . ,n+ .
From the additive property of coincidence degree, we obtain
Deg(L,N),n+\∪∪ · · · ∪n
Sincen≥ is arbitrary, we first deal with the casen= , the above discussion implies that the equationLu=Nu, that is, the problem (.)-(.) has at least one solution in the set
,and\∪respectively. That is, there exist at least three different solutions u(t),u(t) andu(t) such that
α(t) <u(t) <β(t), α(t) <u(t) <β(t), onJ,
andmint∈Ju(t) <α(t),maxt∈Ju(t) >β(t).
Forn= , replacingα,α,β,βbyα,α,β,βrespectively, then we can obtain another
two different solutionsu(t),u(t) of the problem (.)-(.) such that
α(t) <u(t) <β(t), and min
t∈J u(t) <α(t), maxt∈J u(t) >β(t).
Along this way, we can complete the proof by the induction method.
We now present an example to illustrate that the assumptions of our theorem can be verified.
Example Consider the following boundary value problem:
⎧ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎨ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎪ ⎩
(ρ(t)u)=a– asinπu(t)cosπu(t) +bt
+(u)( +sinu), t∈[, ],t=t
k,
u(tk) = (tk–u(tk))cosπu(tk), k= , . . . ,m,
u(tk) =tktgπu(tk), k= , . . . ,m,
u() = , u() =u(η),
(.)
whereρ(t) =c+sinπt,c> , <b<a≤
, <η< , <t<t<· · ·<tm< ,η=tk.
First, we haveρ(t)≥c– > and|ρ(t)|=|cost| ≤, thusθ= ,δ=c– . Next, it is clear that{αi}ni=={–,
, . . . ,n–
}and{βi}
n i=={,
, . . . ,n–
}aren(n∈N
andn≥) pairs of strict lower and upper solutions of the problem (.), respectively. It can be seen thatIk(tk,αi(tk)) =Ik(tk,βi(tk)) = ,Jk(tk,αi(tk)) < ,Jk(tk,βi(tk)) > ,i= , . . . ,n.
Thus conditions (H) and (H) of Theorem (.) hold.
Sincef(t,u,p) =a– asinπucosπu+bt+
p( +sin
p), it follows that
f
t,i– ,
=a– asin
π
i–
cos
π
i–
+bt
=a– a+bt≤b–a< , i= , . . . ,n.
f
t,i– ,
=a– asin
π
i–
cos
π
i–
+bt
=a+ a+bt> , i= , . . . ,n.
For allu∈(–(n–), –)∪(, (n–)), we have
f(t,u,p)≤a+bt+p≤a+b+p< +p
Takeψ(p) =+p, then +∞
s
θs+ψ(s)ds=
+∞
s
s++sds= +∞.
Hencef has property (H). It follows from Theorem . that the problem (.) has at least n– different solutionsu(t), . . . ,un(t), . . . ,un–(t) such that
–
<u(t) < ,
<u(t) <
, . . . ,
n–
<un(t) <
n–
, and
min
t∈J un+(t) <
, maxt∈J un+(t) >
, . . . , mint∈J un–(t) <n–
,
max
t∈J un–(t) >n–
.
Competing interests
The authors declare that they have no competing interests.
Authors’ contributions
The authors declare that the study was realized in collaboration with the same responsibility. All authors read and approved the final manuscript.
Author details
1School of Science, Hunan University of Technology, Zhuzhou, Hunan 412007, PR China.2Department of Mathematics,
Central South University, Changsha, Hunan 410075, PR China.
Acknowledgements
The authors are highly grateful to the referees for careful reading and comments on this paper. The research is supported by Hunan Provincial Natural Science Foundation of China (No. 13JJ3106, and 12JJ2004); it is also supported by the National Natural Science Foundation of China (No. 61074067, 11271372, and 11201138).
Received: 20 December 2013 Accepted: 8 April 2014 Published:07 May 2014
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