We study the existence of positive solutions for the following nonlinear m-point boundary value
problem for an increasing homeomorphism and homomorphism with sign changing nonlinearity:
{(ϕ(u′(t)))′+a(t)f(t,u(t))=0, 0<t<1,
u′(0)=∑i=1m−2aiu′(ξi), u(1)=∑i=1kbiu(ξi)−∑i=k+1sbiu(ξi)−∑i=s+1m−2biu′(ξi), where ϕ:R→R is an increasing homeomorphism and homomorphism and ϕ(0)=0. The nonlinear term f may
change sign. As an application, an example to demonstrate our results is given. The conclusions in this paper essentially extend and improve the known results.

1. Introduction

In this
paper, we study the existence of positive solutions of the following nonlinear m-point boundary
value problem with sign changing nonlinearity:(ϕ(u′(t)))′+a(t)f(t,u(t))=0,0<t<1,u′(0)=∑i=1m−2aiu′(ξi),u(1)=∑i=1kbiu(ξi)−∑i=k+1sbiu(ξi)−∑i=s+1m−2biu′(ξi), where ϕ:R→R is an
increasing homeomorphism and homomorphism and ϕ(0)=0; ξi∈(0,1) with 0<ξ1<ξ2<⋯<ξm−2<1 and ai,bi,a,f satisfy

a(t):(0,1)→[0,+∞) does not vanish
identically on any subinterval of [0,1] and satisfies0<∫01a(t)dt<∞;

f∈C([0,1]×[0,+∞),(−∞,+∞)),f(t,0)≥0andf(t,0)≠0.

Definition 1.1.

A projection ϕ:R→R is called an
increasing homeomorphism and homomorphism, if the following conditions are
satisfied:

if x≤y, then ϕ(x)≤ϕ(y), for all x,y∈R;

ϕ is a continuous
bijection and its inverse mapping is also continuous;

ϕ(xy)=ϕ(x)ϕ(y), for all x,y∈R.

The study of multipoint boundary value problems for
linear second-order ordinary differential equations was initiated by Il'in and
Moiseev [1, 2]. Motivated by the study of [1, 2], Gupta [3] studied certain
three-point boundary value problems for nonlinear ordinary differential
equations. Since then, more general nonlinear multipoint boundary value
problems have been studied by several authors. We refer the reader to [4–12] for some references along this line. Multipoint boundary value
problems describe many phenomena in the applied mathematical sciences. For
example, the vibrations of a guy wire of a uniform cross-section and composed
of N parts of different densities can be set up as a multipoint boundary value
problems (see Moshinsky [13]); many problems in the theory of elastic stability
can be handled by the method of multipoint boundary value problems (see
Timoshenko [14]).

In 2001, Ma [6] studied m-point boundary
value problem (BVP):u′′(t)+h(t)f(u)=0,0≤t≤1,u(0)=0,u(1)=∑i=1m−2αiu′(ξi),where αi>0(i=1,2,…,m−2),∑i=1m−2αi<1,0<ξ1<ξ2<⋯<ξm−2<1, and f∈C([0,+∞),[0,+∞)), h∈C([0,1],[0,+∞)). Author established the existence of positive
solutions under the condition that f is either
superlinear or sublinear.

In [11], we considered the existence of positive
solutions for the following nonlinear four-point singular boundary value
problem with p-Laplacian:(ϕp(u′(t)))′+a(t)f(u(t))=0,0<t<1,αϕp(u(0))−βϕp(u′(ξ))=0,γϕp(u(1))+δϕp(u′(η))=0,where ϕp(s)=|s|p−2s,p>1,ϕq=(ϕp)−1,1/p+1/q=1,α>0,β≥0,γ>0,δ≥0,ξ,η∈(0,1),ξ<η,a:(0,1)→[0,∞). By using the fixed-point theorem of cone, the
existence of positive solution and many positive solutions for nonlinear
singular boundary value problem p-Laplacian is
obtained.

Recently, Ma et al. [5] used the monotone iterative
technique in cones to prove the existence of at least one positive solution
for m-point boundary value problem (BVP):(ϕp(u′(t)))′+a(t)f(t,u(t))=0,0<t<1,u′(0)=∑i=1m−2aiu′(ξi),u(1)=∑i=1m−2biu(ξi),where 0<∑i=1m−2bi<1,0<∑i=1m−2ai<1,0<ξ1<ξ2<⋯<ξm−2<1, a(t)∈L1[0,1], f∈C([0,1]×[0,+∞),[0,+∞)).

In [9], Wang and Hou investigated the following m-point BVP:(ϕp(u′(t)))′+f(t,u(t))=0,t∈(0,1),ϕp(u′(0))=∑i=1n−2aiϕp(u′(ξi)),u(1)=∑i=1n−2biu(ξi),where ϕp(u)=|u|p−2u, p>1, ξi∈(0,1) with 0<ξ1<ξ2<⋯<ξn−2<1 and ai,bi satisfy ai,bi∈[0,+∞), 0<∑i=1n−2ai<1,0<∑i=1n−2bi<1.

However, in all the above-mentioned paper, the authors
discuss the boundary value problem (BVP) under the key conditions that the
nonlinear term is positive continuous function. Motivated by the results
mentioned above, in this paper we study the existence of positive solutions of m-point boundary
value problem (1.1) for an increasing homeomorphism and homomorphism with sign
changing nonlinearity. We generalize the results in [4–12].

By a positive solution of BVP (1.1), we understand a
function u which is
positive on (0,1) and satisfies the differential equation as well as the
boundary conditions in BVP (1.1).

2. The Preliminary Lemmas

In this
section, we present some lemmas which are important to our main results.

Lemma 2.1.

Let (H1) and (H2) hold. Then for u≥0∈C[0,1], the problem(ϕ(u′(t)))′+a(t)f(t,u(t))=0,0<t<1,u′(0)=∑i=1m−2aiu′(ξi),u(1)=∑i=1kbiu(ξi)−∑i=k+1sbiu(ξi)−∑i=s+1m−2biu′(ξi)has a unique solution u(t) if and only if u(t) can be express
as the following equation:u(t)=−∫t1ωf(s)ds+B,where A,B satisfyϕ−1(A)=∑i=1m−2aiϕ−1(A−∫0ξia(s)f(s,u(s))ds),B=−11−∑i=1kbi+∑i=k+1sbi[∑i=1kbi∫ξi1ωf(s)ds−∑i=k+1sbi∫ξi1ωf(s)ds+∑i=s+1m−2biϕ−1(A−∫0ξia(s)f(s,u(s))ds)],whereωf(s)=ϕ−1(−∫0sa(r)f(r,u(r))dr+A).Define l=ϕ(∑i=1m−2ai)/(1−ϕ(∑i=1m−2ai))∈(0,1), then there exists a unique A∈[−l∫01a(s)f(s,u(s))ds,0] satisfying
(2.3).

Proof.

The method of the proof is similar to [5, Lemma 2.1],
we omit the details.

Lemma 2.2.

Let (H1) and (H2) hold. If u∈C+[0,1], the unique solution of the problem (2.1) satisfiesu(t)≥0,t∈[0,1].

Proof.

According to Lemma 2.1, we first have−A+∫0sa(r)f(t,u(r))dr≥0.Sou(1)=B=−11−∑i=1kbi+∑i=k+1sbi[∑i=1kbi∫ξi1ωf(s)ds−∑i=k+1sbi∫ξi1ωf(s)ds+∑i=s+1m−2biϕ−1(A−∫0ξia(s)f(t,u(s))ds)]=11−∑i=1kbi+∑i=k+1sbi[∑i=1kbi∫ξi1−ωf(s)ds−∑i=k+1sbi∫ξi1−ωf(s)ds+∑i=s+1m−2biϕ−1(−A+∫0ξia(s)f(t,u(s))ds)]≥11−∑i=1kbi+∑i=k+1sbi[∑i=1kbi∫ξk1−ωf(s)ds−∑i=k+1sbi∫ξk1−ωf(s)ds]=(∑i=1kbi−∑i=k+1sbi)∫ξk1−ωf(s)ds1−∑i=1kbi+∑i=k+1sbi≥0.If t∈[0,1), we haveu(t)=B−∫t1ϕ−1(A−∫0sa(r)f(r,u(r))dr)ds=u(1)+∫t1ϕ−1(−A+∫0sa(r)f(r,u(r))dr)ds≥u(1)≥0.So u(t)≥0,t∈[0,1]. The proof of Lemma 2.2 is completed.

Lemma 2.3.

Let (H1) and (H2) hold. If u∈C+[0,1], the unique solution of the problem (2.1) satisfiesinft∈[0,1]u(t)≥γ∥u∥,where γ=(∑i=1kbi−∑i=k+1sbi)(1−ξk)/(1−∑i=1kbiξk+∑i=k+1sbiξk)∈(0,1),∥u∥=maxt∈[0,1]|u(t)|.

Proof.

Clearlyu′(t)=ϕ−1(A−∫0ta(s)f(s,u(s))ds)=−ϕ−1(−A+∫0ta(s)f(s,u(s))ds)≤0.This implies that∥u∥=u(0),mint∈[0,1]u(t)=u(1).It is easy to see that u′(t2)≤u′(t1), for any t1,t2∈[0,1] with t1≤t2. Hence u′(t) is a decreasing
function on [0,1]. This means that the graph of u(t) is concave down
on (0,1). So we haveu(ξk)−u(1)ξk≥(1−ξk)u(0).Together with u(1)=∑i=1kbiu(ξi)−∑i=k+1sbiu(ξi)−∑i=s+1m−2biu′(ξi) and u′(t)≤0 on [0,1], we
getu(0)≤∑i=1kbiu(ξk)−u(1)∑i=1kbiξk−∑i=k+1sbiu(ξk)+u(1)∑i=k+1sbiξk(∑i=1kbi−∑i=k+1sbi)(1−ξk)≤∑i=1kbiu(ξi)−u(1)∑i=1kbiξk−∑i=k+1sbiu(ξi)+u(1)∑i=k+1sbiξk(∑i=1kbi−∑i=k+1sbi)(1−ξk)≤u(1)(1−∑i=1kbiξk+∑i=k+1sbiξk)(∑i=1kbi−∑i=k+1sbi)(1−ξk)=u(1)γ.The proof of Lemma 2.3 is
completed.

Lemma 2.4 (see [<xref ref-type="bibr" rid="B10">8</xref>]).

Let K be a cone in a Banach space X. Let D be an open
bounded subset of X with DK=D∩K≠ϕ and DK¯≠K. Assume that A:DK¯→K is a compact
map such that x≠AK for x∈∂DK. Then the following results hold.

If ∥Ax∥≤∥x∥, x∈∂DK, then i(A,DK,K)=1.

If there
exists x0∈K∖{θ} such that x≠Ax+λx0, for all x∈∂DK and all x>0, then i(A,DK,K)=0.

Let U be open in X such that U¯⊂DK. If i(A,DK,K)=1 and i(A,DK,K)=0, then A has a fixed
point in DK∖U¯K. The same results hold, if i(A,DK,K)=0 and i(A,DK,K)=1.

Let E=C[0,1], then E is Banach
space, with respect to the norm ∥u∥=supt∈[0,1]|u(t)|. DenoteK={u∣u∈C[0,1],u(t)≥0,inft∈[0,1]u(t)≥γ∥u∥},where γ is the same as
in Lemma 2.3. It is obvious that K is a cone in C[0,1].

We define φ(t)=min{t,1−t},t∈(0,1) andKρ={u(t)∈K:∥u∥<ρ},Kρ*={u(t)∈K:ρφ(t)<u(t)<ρ},Ωρ={u(t)∈K:minξm−2≤t≤1u(t)<γρ}={u(t)∈E:u≥0,γ∥u∥≤minξm−2≤t≤1u(t)<γρ}.

Lemma 2.5 (see [<xref ref-type="bibr" rid="B7">13</xref>]).

Ωρ defined above
has the following properties:

Kγρ⊂Ωρ⊂Kρ;

Ωρ is open
relative to K;

X∈∂Ωρ if and only if minξm−2≤t≤1cx(t)=γρ;

If x∈∂Ωρ, then γρ≤x(t)≤ρ for t∈[ξm−2,1].

Now, for the convenience, one introduces the following
notations:fγρρ=min{minξm−2≤t≤1f(t,u)ϕ(ρ):u∈[γρ,ρ]},f0ρ=max{max0≤t≤1f(t,u)ϕ(ρ):u∈[0,ρ]},fφ(t)ρρ=max{max0≤t≤1f(t,u)ϕ(ρ):u∈[φ(t)ρ,ρ]},fα=limu→αsupmax0≤t≤1f(t,u)ϕ(u),fα=limu→αinfminξm−2≤t≤1f(t,u)ϕ(u),(α:=∞or0+),m={(1+∑i=k+1sbi+∑i=s+1m−2bi)ϕ−1((l+1)∫01a(s)ds)1−∑i=1kbi+∑i=k+1sbi}−1,M={∑i=1kbi−∑i=k+1sbi1−∑i=1kbi+∑i=k+1sbi∫ξk1ϕ−1(∫0sa(r)dr)ds}−1.3. The Main Result

In the rest
of the section, we also assume the following conditions.

There exist ρ1,ρ2∈(0,+∞) with ρ1<γρ2 such that(1)f(t,u)>0,t∈[0,1],u∈[ρ1φ(t),+∞),(2)fφ(t)ρ1ρ1≤ϕ(m),fγρ2ρ2≥ϕ(Mγ).

There exist ρ1,ρ2∈(0,+∞) with ρ1<ρ2 such that(3)f(t,u)>0,t∈[0,1],u∈[min{γρ1,ρ2φ(t)},+∞),(4)fγρ1ρ1≥ϕ(Mγ),fφ(t)ρ2ρ2≤ϕ(m).

There exist ρ1,ρ2,ρ3∈(0,+∞) with ρ1<γρ2 and ρ2<ρ3 such that(1)f(t,u)>0,t∈[0,1],u∈[ρ1φ(t),+∞),(2)fφ(t)ρ1ρ1≤ϕ(m),fγρ2ρ2≥ϕ(Mγ),fφ(t)ρ3ρ3≤ϕ(m).

There exist ρ1,ρ2,ρ3∈(0,+∞) with ρ1<ρ2<γρ3 such that(3)f(t,u)>0,t∈[0,1],u∈[min{γρ1,ρ2φ(t)},+∞),(4)fγρ1ρ1≥ϕ(Mγ),fφ(t)ρ2ρ2≤ϕ(m),fγρ3ρ3≥ϕ(Mγ).

There exist ρ′,ρ∈(0,+∞) with ρ′<γρ such that(1)f(t,u)>0,t∈[0,1],u∈[ρ′φ(t),+∞),(2)fφ(t)ρ′ρ′≤ϕ(m),fγρρ≥ϕ(Mγ),0≤f∞<ϕ(m).

There exist ρ′,ρ∈(0,+∞) with ρ′<ρ such that(3)f(t,u)>0,t∈[0,1],u∈[min{γρ′,ρφ(t)},+∞),(4)fγρ′ρ′≥ϕ(Mγ),fφ(t)ρρ≤ϕ(m),ϕ(M)<f∞≤∞.

Our main results are the following theorems.

Theorem 3.1.

Assume that (H1),(H2),(H3),(A3) hold. Then BVP
(1.1) has at least three positive solutions.

Theorem 3.2.

Assume that (H1),(H2),(H3),(A4) hold. Then BVP
(1.1) has at least two positive solutions.

Theorem 3.3.

Assume that (H1),(H2),(H3) hold and also
assume that (A1) or (A2) hold. Then BVP
(1.1) has at least a positive solution.

Theorem 3.4.

Assume that (H1),(H2),(H3) hold and also
assume that (A5) or (A6) hold. Then BVP
(1.1) has at least two positive solutions.

Proof of Theorem <xref ref-type="statement" rid="thm3.1">3.1</xref>.

Without loss of
generality, we suppose that (A3) hold. Denotef*(t,u)={f(t,u),u≥ρ1φ(t),f(t,ρ1φ(t)),0≤u<ρ1φ(t),it is easy to check that f*(t,u)∈C([0,1]×[0,+∞),(0,+∞)).

Now define an operator T:K→C[0,1] by setting(Tu)(t)=−11−∑i=1kbi+∑i=k+1sbi[∑i=1kbi∫ξi1ω(s)ds−∑i=k+1sbi∫ξi1ω(s)ds+∑i=s+1m−2biϕ−1(A−∫0ξia(s)f*(s,u(s))ds)]−∫t1ω(s)ds,whereω(s)=ϕ−1(−∫0sa(r)f*(τ,u(r))dr+A).By Lemma 2.3, we have T(K)⊂K. So by applying Arzela-Ascoli's theorem, we can obtain
that T(K) is relatively
compact. In view of Lebesgue's dominated convergence theorem, it is easy to
prove that T is continuous.
Hence, T:K→K is completely
continuous.

Now, we consider the following modified BVP (1.1):(ϕ(u′))′+a(t)f*(t,u(t))=0,0<t<1,u′(0)=∑i=1m−2aiu′(ξi),u(1)=∑i=1kbiu(ξi)−∑i=k+1sbiu(ξi)−∑i=s+1m−2biu′(ξi).Obviously, BVP (3.10) has a
solution u(t) if and only if u is a fixed
point of the operator T. From the condition (A3)(2), we havefφ(t)ρ1*ρ1≤ϕ(m),fγρ2*ρ2≥ϕ(Mγ),fφ(t)ρ3*ρ3≤ϕ(m).

Next, we will show that i(T,Kρ1*,K)=1.

In fact, by fφ(t)ρ1*ρ1≤ϕ(m), for ∀u∈∂Kρ1*, we have(Tu)(t)=−11−∑i=1kbi+∑i=k+1sbi×(∑i=1kbi∫ξi1ω(s)ds−∑i=k+1sbi∫ξi1ω(s)ds+∑i=s+1m−2biϕ−1(A−∫0ξia(s)f*(s,u(s))ds))−∫t1ω(s)ds≤11−∑i=1kbi+∑i=k+1sbi(∑i=1kbi∫01ϕ−1((l+1)∫01a(r)f*(r,u(r))dr)ds+∑i=s+1m−2biϕ−1((l+1)∫01a(s)f*(s,u(s))ds))+∫01ϕ−1((l+1)∫01a(r)f*(r,u(r))dr)ds≤(1+∑i=k+1sbi+∑i=s+1m−2bi)ϕ−1((l+1)∫01a(s)ds)1−∑i=1kbi+∑i=k+1sbiϕ−1(ϕ(ρ1)ϕ(m))=ρ1=∥u∥.This implies that ∥Tu∥≤∥u∥ for u∈∂Kρ*. By Lemma 2.4(1), we havei(T,Kρ1*,K)=1.

Furthermore, we will show that i(T,Kρ2,K)=1.

Let e(t)≡1, for t∈[0,1], then e∈∂K1. We claim thatu≠Tu+λe,u∈∂Ωρ2,λ>0.

In fact, if not, there exist u0∈∂Ω2 and λ0>0 such that u0=Tu0+λ0e.

By (A3) and Lemma 2.1,
we have for t∈[0,1],−∫0sa(τ)f*(τ,u(τ))dτ+A≤−ϕ(ρ2)ϕ(Mγ)(∫0sa(τ)dτ),so that−ω(s)=ϕ−1(−∫0sa(τ)f*(τ,u(τ))dτ+A)≥ρ2Mγϕ−1[∫0sa(τ)dτ].Then, we have thatu0(t)=Tu0(t)+λ0e(t)≥11−∑i=1kbi+∑i=k+1sbi∑i=1k(bi∫ξk1(−ω(s))ds−∑i=k+1sbi∫ξk1(−ω(s))ds)+λ0≥∑i=1kbi−∑i=k+1sbi1−∑i=1kbi+∑i=k+1sbiρ2Mγ∫ξk1ϕ−1(∫0sa(r)dr)ds+λ0=γρ2+λ0.This implies that γρ2≥γρ2+λ0, this is a contradiction. Hence, by Lemma 2.4(2), it
follows thati(T,Ωρ2,K)=0.

Finally, similar to the proof of i(T,Kρ1*,K)=1, we can show that i(T,Kρ3*,K)=1.

By Lemma 2.5(a) and ρ1<γρ2 and ρ2<ρ3, we have K¯ρ1⊂Kγρ2⊂Ωρ2⊂Kρ2⊂Kρ3. It follows from Lemma 2.4(3) that T has three
positive fixed points u1,u2,u3 in Kρ1*,Ωρ2∖K*¯ρ1,Kρ3*, respectively.
Therefore, BVP (3.10) has three positive solutions u1,u2,u3 in Kρ1*,Ωρ2∖K*¯ρ1,Kρ3*, respectively.

Then, BVP (3.10) has three positive solutions u1,u2,u3∈[ρ1φ(t),∞), which means that u1,u2,u3 are also the
positive solutions of BVP (1.1).

Proof of Theorem <xref ref-type="statement" rid="thm3.2">3.2</xref>.

The proof of Theorem 3.2
is similar to that of Theorem 3.1, and so we omit it here. The proof of Theorem
3.2 is completed.

Proof of Theorem <xref ref-type="statement" rid="thm3.3">3.3</xref>.

Theorem 3.3 is corollary
of Theorem 3.1. The proof of Theorem 3.3 is completed.

Proof of Theorem <xref ref-type="statement" rid="thm3.4">3.4</xref>.

We show that condition (A5) implies
condition (A3). Let k∈(f∞,ϕ(m)), then there exists r>ρ such that maxt∈[0,1]f(t,u)≤kϕ(u),u∈[r,∞) since 0≤f∞<ϕ(m). Denoteβ=max{maxt∈[0,1]f(t,u):ρ′φ(t)≤u≤r},ρ3>max{ϕ−1(βϕ(m)−k),ρ}.Then we havemaxt∈[0,1]f(t,u)≤kϕ(u)+β≤kϕ(ρ3)+β≤ϕ(m)ϕ(ρ3),u∈[ρ′φ(t),∞).This implies that fφ(t)ρ3ρ3≤ϕ(m) and (A3) holds.
Similarly condition (A6) implies
condition (A4).

By an argument similar to that Theorem 3.1, we can
obtain the result of Theorem 3.4. The proof of Theorem 3.4 is completed.

4. ExamplesExample 4.1.

Consider the
following five-point boundary value problem with p-Laplacian:(ϕ(u′))′+f(t,u)=0,0<t<1,u′(0)=1128u′(14)+1256u′(12)+164u′(34),u(1)=18u(14)−164u(12),where a1=1/128,a2=1/256,a3=1/64,b1=1/8,b2=1/64,b3=0,ξ1=1/4,ξ2=1,/2,ξ3=3/4:ϕ(u)={−u2,u≤0,u2,u>0,f(t,u)={15(1+t)(u(t)−φ(t)2)30,(t,u)∈[0,1]×(0,2],15(1+t)(2−φ(t)2)30,(t,u)∈[0,1]×(2,+∞).

It is easy to check that f:[0,1]×[0,+∞)→[0,+∞) is continuous.
It follows from a direct calculation thatm={(1+∑i=k+1sbi+∑i=s+1m−2bi)ϕ−1((l+1)∫01a(s)ds)1−∑i=1kbi+∑i=k+1sbi}−1=0.96,γ=(∑i=1kbi−∑i=k+1sbi)(1−ξk)1−∑i=1kbiξk+∑i=k+1sbiξk=21250,M={∑i=1kbi−∑i=k+1sbi1−∑i=1kbi+∑i=k+1sbi∫ξk1ϕ−1(∫0sa(r)dr)ds}−1=0.76.

Choose ρ1=1,ρ2=250, it is easy to check that ρ1<γρ2 andf(t,u)>0,t∈[0,1],u∈[φ(t),+∞),fρ1φ(t)ρ1=max{max0≤t≤1(1/5)(1+t)(u(t)−φ(t)/2)3012}=(1/5)(1+1)13012=25<ϕ(m)=m2=0.92,t∈[0,1],u∈[φ(t)ρ1,ρ1],fγρ2ρ2=min{min3/4≤t≤1(1/5)(1+t)(2−φ(t)/2)302502}=(1/5)(1+3/4)(2−1/2)302502=1.0742>ϕ(Mγ)=(Mγ)2=0.004,t∈[34,1],u∈[γρ2,ρ2]. It follows that f satisfies the
condition (A1) of Theorem 3.3,
then problems (1.1) have at least two positive solutions.

Remark 4.2.

Let φ(u)=u, the problem is second-order m-point boundary
value problem.

Remark 4.3.

Let ϕp(s)=|s|p−2s,p>1, the problem is boundary value problem with p-Laplacian
operators.

Hence our results generalize boundary value problem
with p-Laplacian
operators.

Acknowledgment

This research is supported by the Doctor of Scientific Startup Foundation of Shandong University of Finance of
China (08BSJJ32).

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