# Triple Positive Solutions for Third-Orderm-Point Boundary Value Problems on Time Scales

Authors: Jian Liu; Fuyi Xu
Journal: Discrete Dynamics in Nature and Society
 (2009)

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## Abstract

We study the following third-orderm-point boundary value problems on time scales (φ(uΔ∇))∇+a(t)f(u(t))=0,  t∈[0,T]T,  u(0)=∑i=1m−2biu(ξi),  uΔ(T)=0, φ(uΔ∇(0))=∑i=1m−2ciφ(uΔ∇(ξi)),  where φ:R→R is an increasing homeomorphism and homomorphism and φ(0)=0,  0<ξ1<⋯<ξm−2<ρ(T). We obtain the existence of three positive solutions by using fixed-point theorem in cones. The conclusions in this paper essentially extend and improve the known results.


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## Body

## 1. Introduction

The theory of time scales was initiated by Hilger [1] as a mean of unifying and extending theories from differential and difference equations. The study of time scales has led to several important applications in the study of insect population models, neural networks, heat transfer, and epidemic models; see, for example [2–6]. Recently, the boundary value problems with p-Laplacian operator have also been discussed extensively in the literature, for example, see [7–15].A time scaleT is a nonempty closed subset of R. We make the blanket assumption that 0,T are points in T. By an interval (0,T), we always mean the intersection of the real interval (0,T) with the given time scale; that is, (0,T)∩T.In [16], Anderson considered the following third-order nonlinear boundary value problem (BVP):
(1.1)x′′′(t)=f(t,x(t)),t1≤t≤t3,x(t1)=x′(t2)=0,γx(t3)+δx′′(t3)=0.
He used the Krasnoselskii and Leggett-Williams fixed-point theorems to prove the existence of solutions to the nonlinear boundary value problem.In [9, 10], He considered the existence of positive solutions of the p-Laplacian dynamic equations on time scales
(1.2)(ϕp(uΔ))∇+a(t)f(u(t))=0,t∈[0,T]T,
satisfying the boundary conditions
(1.3)u(0)-B0(uΔ(η))=0,uΔ(T)=0,
or
(1.4)uΔ(0)=0,u(T)-B1(uΔ(η))=0,
where η∈(0,ρ(T)). He obtained the existence of at least double and triple positive solutions of the problems by using a new double fixed point theorem and triple fixed point theorem, respectively.In [15], Zhou and Ma firstly studied the existence and iteration of positive solutions for the following third-order generalized right-focal boundary value problem with p-Laplacian operator
(1.5)(ϕp(u′′))′(t)=q(t)f(t,u(t)),0≤t≤1,u(0)=∑i=1mαiu(ξi),u′(η)=0,u′′(1)=∑i=1nβiu′′(θi).
They established a corresponding iterative scheme for the problem by using the monotone iterative technique.However, to the best of our knowledge, little work has been done on the existence of positive solutions for the increasing homeomorphism and positive homomorphism operator on time scales. So the goal of the present paper is to improve and generatep-Laplacian operator and establish some criteria for the existence of multiple positive solutions for the following third-order m-point boundary value problems on time scales
(1.6)(φ(uΔ∇))∇+a(t)f(u(t))=0,t∈[0,T]T,u(0)=∑i=1m-2biu(ξi),uΔ(T)=0,φ(uΔ∇(0))=∑i=1m-2ciφ(uΔ∇(ξi)),
where φ:R→R is an increasing homeomorphism and homomorphism and φ(0)=0, and bi,ci,a,f  satisfy(H1)
bi,ci∈[0,+∞), 0<ξ1<⋯<ξm-2<ρ(T), 0<∑i=1m-2bi<1,0<∑i=1m-2ci<1;(H2)
f:[0,+∞)→R+ is continuous, a∈Cld([0,T]T,R+) and there exits t0∈[0,T)T such that a(t0)>0, where R+=[0,+∞).A projectionφ:R→R is called an increasing homeomorphism and homomorphism, if the following conditions are satisfied:(i)
ifx≤y, then φ(x)≤φ(y),∀x,y∈R;(ii)
φ is continuous bijection and its inverse mapping is also continuous;(iii)
φ(xy)=φ(x)φ(y),∀x,y∈R.


## 2. Preliminaries and Lemmas

For convenience, we list the following definitions which can be found in [1–5].Definition 2.1.
A time scaleT is a nonempty closed subset of real numbers R. For t<supT and r>infT, define the forward jump operator σ and backward jump operator ρ, respectively, by
(2.1)σ(t)=inf{τ∈T∣τ>t}∈T,ρ(r)=sup{τ∈Tτ<r}∈T,
for all t,r∈T. If σ(t)>t, t is said to be right scattered, and if ρ(r)<r, r is said to be left scattered; if σ(t)=t, t is said to be right dense, and if ρ(r)=r, r is said to be left dense. If T has a right scattered minimum m, define Tk=T-{m}; otherwise set Tk=T. If T has a left scattered maximum M, define Tk=T-{M}; otherwise set Tk=T.Definition 2.2.
Forf:T→R and t∈Tk, the delta derivative of f at the point t is defined to be the number fΔ(t) (provided it exists) with the property that for each ϵ>0, there is a neighborhood U of t such that
(2.2)|f(σ(t))-f(s)-fΔ(t)(σ(t)-s)|≤ϵ|σ(t)-s|,
for all s∈U.
Forf:T→R and t∈Tk, the nabla derivative of f at t, denoted by f∇(t) (provided it exists) with the property that for each ϵ>0, there is a neighborhood U of t such that
(2.3)|f(ρ(t))-f(s)-f∇(t)(ρ(t)-s)|≤ϵ|ρ(t)-s|,
for all s∈U.Definition 2.3.
A functionf is left-dense continuous (i.e., ld-continuous), if f is continuous at each left-dense point in T and its right-sided limit exists at each right-dense point in T.Definition 2.4.
IfϕΔ(t)=f(t), then we define the delta integral by
(2.4)∫abf(t)Δt=ϕ(b)-ϕ(a).
If F∇(t)=f(t), then we define the nabla integral by
(2.5)∫abf(t)∇t=F(b)-F(a).Definition 2.5.
LetE be a real Banach space over R. A nonempty closed set P⊂E is said to be a cone provided that
(i)
u∈P, a≥0 implies au∈P;(ii)
u,-u∈P implies u=0.Definition 2.6.
Given a coneP in a real Banach space E, a functional ψ:P→P is said to be increasing on P, provided ψ(x)≤ψ(y), for all x,y∈P with x≤y.Definition 2.7.
Given a coneP in a real Banach space E, we define for each a>0 the set
(2.6)Pa={x∈P∣∥x∥<a}.Definition 2.8.
A mapα is called nonnegative continuous concave functional on a cone P of a real Banach space X if α:P→[0,+∞) is continuous and
(2.7)α(λx+(1-λ)y)≥λα(x)+(1-λ)α(y)
for all x,y∈P and λ∈[0,1]. Similarly we say that the map β is called nonnegative continuous concave functional on a cone P of a real Banach space X if β:P→[0,+∞) is continuous and
(2.8)β(λx+(1-λ)y)≤λβ(x)+(1-λ)β(y)
for all x,y∈P and λ∈[0,1].Letγ,θ be nonnegative continuous convex functionals on P, let α be a nonnegative continuous concave functional on P, and let  ψ be a nonnegative continuous functional on P. For nonnegative real numbers a,b,k, and c we define the following convex set:
(2.9)P(γ,c)={x∈P:γ(x)<c},P(α,b;γ,c)={x∈P:b≤α(x),γ(x)≤c},R(ψ,a;γ,c)={x∈P:a≤ψ(x),γ(x)≤c},P(α,b;θ,k;γ,c)={x∈P:b≤α(x),θ(x)≤k,γ(x)≤c}.Theorem 2.9 ([17]).
LetP be a cone in a real Banach space X. Let γ and θ be nonnegative continuous convex functionals on P, let  α be a nonnegative continuous concave functional on P, and let ψ be a nonnegative continuous functional on P satisfying ψ(λx)≤λψ(x) for 0≤λ≤1, such that for some positive numbers M and c, α(x)≤ψ(x) and ∥x∥≤Mγ(x) for all x∈P(γ,c)¯. Suppose that  Φ:P(γ,c)¯→P(γ,c)¯ is a completely continuous operator and there exist nonnegative numbers a,b, and k with 0<a<b such that
(i)
{x∈P(α,b;θ,k;γ,c):α(x)>b}≠∅ and α(Φ(x))>b for x∈P(α,b;θ,k;γ,c);(ii)
α(Φ(x))>b for x∈P(α,b;γ,c) with θ(Φ(x))>k;(iii)
0∈¯R(ψ,a;γ,c) and ψ(Φ(x))<a for x∈R(ψ,a;γ,c) with ψ(x)=a.Then Φ has at least three fixed points x1,x2,x3∈P(γ,c)¯ satisfying
(2.10)γ(xi)≤c,i=1,2,3,b<α(x1),α(x2)<b,a<ψ(x2),ψ(x3)<a.Theorem 2.10 ([18]).
LetA be a bounded closed convex subset of a Banach space E. Assume that A1,A2 are disjoint closed convex subsets of A and U1,U2 are nonempty open subsets of A with U1⊂A1 and U2⊂A2. Suppose that Φ:A→A is completely continuous and the following conditions hold:
(i)
Φ(A1)⊂A1,Φ(A2)⊂A2;(ii)
Φ has no fixed points in (A1∖U1)∪(A2∖U2).Then Φ has at least three points x1,x2,x3 such that x1∈U1,x2∈U2, and x3∈A∖(A1∪A2).Lemma 2.11.
If condition(H1) holds, then for h∈Cld[0,T]T, the boundary value problem (BVP)
(2.11)uΔ∇+h(t)=0,t∈(0,T),u(0)=∑i=1m-2biu(ξi),uΔ(T)=0
has the unique solution
(2.12)u(t)=∫0t(T-s)h(s)∇s+∑i=1m-2bi∫0ξi(T-s)h(s)∇s1-∑i=1m-2bi.Proof.
By caculating, we can easily get (2.12). So we omit it.Lemma 2.12.
If condition(H1) holds, then for h∈Cld[0,T]T, the boundary value problem (BVP)
(2.13)(φ(uΔ∇))∇+h(t)=0,t∈(0,T),u(0)=∑i=1m-2biu(ξi),uΔ(T)=0,φ(uΔ∇(0))=∑i=1m-2ciφ(uΔ∇(ξi))
has the unique solution
(2.14)u(t)=∫0t(T-s)φ-1(∫0sh(r)∇r+C)∇s+∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sh(r)∇r+C)∇s1-∑i=1m-2bi,
where C=∑i=1m-2ci∫0ξih(r)∇r/(1-∑i=1m-2ci), φ-1(s) is the inverse function to φ(s).Proof.
Integrating both sides of equation in (2.13) on [0,t], we have
(2.15)φ(uΔ∇(t))=φ(uΔ∇(0))-∫0th(r)∇r.
So,
(2.16)φ(uΔ∇(ξi))=φ(uΔ∇(0))-∫0ξih(r)∇r.
By boundary value condition φ(uΔ∇(0))=∑i=1m-2ciφ(uΔ∇(ξi)), we have
(2.17)φ(uΔ∇(0))=-∑i=1m-2ci∫0ξih(r)∇r1-∑i=1m-2ci.
By (2.15) and (2.17) we know
(2.18)uΔ∇(t)=-φ-1(∑i=1m-2ci∫0ξih(r)∇r1-∑i=1m-2ci+∫0th(r)∇r).
This together with Lemma 2.11 implies that
(2.19)u(t)=∫0t(T-s)φ-1(∫0sh(r)∇r+C)∇s+∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sh(r)∇r+C)∇s1-∑i=1m-2bi,
where C=∑i=1m-2ci∫0ξih(r)∇r/(1-∑i=1m-2ci). The proof is complete.Lemma 2.13.
Let condition(H1) hold. If h∈Cld[0,T]T and h(t)≥0, then the unique solution u(t) of (2.13) satisfies
(2.20)u(t)≥0,t∈[0,T]T.Proof.
ByuΔ∇(t)=-φ-1((∑i=1m-2ci∫0ξih(r)∇r/(1-∑i=1m-2ci))+∫0th(r)∇r)≤0, we can know that the graph of u(t) is concave down on (0,T)T and uΔ(t) is nonincreasing on [0,T]T. This together with the assumption that the boundary condition is uΔ(T)=0 implies that uΔ(t)≥0 for t∈[0,T]T. This implies that
(2.21)∥u∥=u(T),mint∈[0,T]Tu(t)=u(0).
So we only prove u(0)≥0. By condition (H1) we have
(2.22)u(0)=∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sh(r)∇r+C)∇s1-∑i=1m-2bi≥0.
The proof is completed.


## 3. Triple Positive Solutions

In this section, some existence results of positive solutions to BVP (1.6) are established by imposing some conditions on f and defining a suitable Banach space and a cone.LetE=Cld[0,T]T be endowed with the ordering x≤y if x(t)≤y(t) for all t∈[0,T]T, and ∥u∥=maxt∈[0,T]T|u(t)| is defined as usual by maximum norm. Clearly, it follows that (E,∥u∥) is a Banach space.We define a cone by(3.1)P={u:u∈E,u(t)isconcave,nondecreasing,andnonnegativeon[0,T]T,uΔ(T)=0}.
Let
(3.2)η=max{t∈T:t≥T2},
and fix l∈T such that
(3.3)0<η<l<T,
and define the nonnegative continuous convex functionals γ and θ, the nonnegative continuous concave functional α, and the nonnegative continuous functional ψ on the cone P by
(3.4)γ(u)=θ(u)=maxt∈[0,l]Tu(t)=u(l),α(u)=mint∈[η,T]Tu(t)=u(η),ψ(u)=maxt∈[0,η]T=u(η).
For notational convenience, denote
(3.5)Ĉ=∑i=1m-2ci∫0ξia(r)∇r1-∑i=1m-2ci,mη=∫0η(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s,m=∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s1-∑i=1m-2bi,Ml=∫0l(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s+∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s1-∑i=1m-2bi,Mη=∫0η(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s+∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s1-∑i=1m-2bi,M=∫0T(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s+∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s1-∑i=1m-2bi.Lemma 3.1 ([9]).
Ifu∈P, then
(1)
u(t)≥(t/T)||u|| for all t∈[0,T]T;(2)
u(s)/s≥u(t)/t for t,s∈[0,T]T with s≤t.
Define an operatorΦ:P→E by
(3.6)(Φu)(t)=∫0t(T-s)φ-1(∫0sa(r)f(u(r))∇r+C̃)∇s+∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sa(r)f(u(r))∇r+C̃)∇s1-∑i=1m-2bi,
where C̃=∑i=1m-2ci∫0ξia(r)f(u(r))∇r/(1-∑i=1m-2ci). Then, u is a solution of boundary value problem (1.6) if and only if u is a fixed point of operator Φ. Obviously, for u∈P one has (Φu)(t)≥0 for t∈[0,T]T. In addition, (Φu)Δ∇(t)≤0 for t∈[0,T]T and (Φu)Δ(T)=0, this implies ΦP⊂P. With standard argument one may show that Φ:P→P is completely continuous.Theorem 3.2.
Suppose conditions(H1) and (H2) hold, and there exist positive numbers a<(η/T)b<b<(l/T)c,Mlb<mc such that
(B1)
f(u)≤φ(c/Ml),u∈[0,Tc/l];(B2)
f(u)>φ(b/mη),u∈[b,T2b/l2];(B3)
f(u)<φ(a/Mη),u∈[0,Ta/η].Then, the BVP (1.6) has at least three positive solutions u1,u2,u3∈P(γ,c)¯ satisfying
(3.7)γ(ui)≤c,i=1,2,3,b<α(u1),α(u2)<b,a<ψ(u2),ψ(u3)<a.Proof.
Based on Lemma3.1, it is clear that for u∈P and λ∈[0,1], there are α(u)=ψ(u),ψ(λu)=λψ(u) and ∥u∥≤(T/l)u(l)=(T/l)γ(u). Furthermore, ψ(0)=0<a and therefore 0∈¯R(ψ,a;γ,c).
Takeu∈P(γ,c)¯, then 0≤u≤∥u∥≤(T/l)γ(u)≤(T/l)c. By means of (B1) one derives
(3.8)γ(Φu)=Φu(l)=∫0l(T-s)φ-1(∫0sa(r)f(u(r))∇r+C̃)∇s+∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sa(r)f(u(r))∇r+C̃)∇s1-∑i=1m-2bi≤cMl(∫0l(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s+∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s1-∑i=1m-2bi)=c.
Thus Φ:P(γ,c)¯→P(γ,c)¯.
Setu≡Tb/l and k=Tb/l, it follows that
(3.9)α(u)=u(η)=Tbl>b,θ(u)=u(l)=Tbl,γ(u)=Tbl<c,
which means {u∈P(α,b;θ,Tb/l;γ,c):α(u)>b}≠∅.
Foru∈P(α,b;θ,Tb/l;γ,c), we have b≤u(t)≤T2b/l2 for t∈[η,T]T. By condition (B2) we have
(3.10)α(Φu)=Φu(η)=∫0η(T-s)φ-1(∫0sa(r)f(u(r))∇r+C̃)∇s+∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sa(r)f(u(r))∇r+C̃)∇s1-∑i=1m-2bi≥bmη∫0η(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s=b.
So, (i) of Theorem 2.9 is fulfilled.
Ifu∈P(α,b;γ,c) and θ(Φu)>c, then due to (2) of Lemma 3.1(3.11)α(Φu)=Φu(η)≥ηl(Φu)(l)=ηlθ(Φu)>ηcl>Tηbl2>b.
Therefore, (ii) of Theorem 2.9 is fulfilled.
Takeu∈R(ψ,a;γ,c) and ψ(u)=a, then 0≤u≤∥u∥≤(T/η)u(η)=(T/η)ψ(u)=Ta/η, it then follows from (B3) that
(3.12)ψ(Φu)=Φu(η)=∫0η(T-s)φ-1(∫0sa(r)f(u(r))∇r+C̃)∇s+∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sa(r)f(u(r))∇r+C̃)∇s1-∑i=1m-2bi≤aMη(∫0η(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s+∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s1-∑i=1m-2bi)=a.
As a result, all the conditions of Theorem 2.9 are verified. This completes the proof.Theorem 3.3.
Suppose that conditions (H1) and (H2) hold. Let 0<a<b<c,Mb<mc and assume that the following conditions are satisfied:
(C1)
f(u)<φ(a/M), u∈[0,a];(C2)
there exists a numberd>c such that f(u)<φ(d/M), u∈[0,d];(C3)
φ(b/m)<f(u)<φ(c/M), u∈[b,c].Then, the BVP (1.6) has at least three positive solutions u1,u2, and u3 such that
(3.13)b<u1(t)<c,∥u2∥<a,∥u3∥>a.
Where for real number b, ϕb:[0,T]T→[0,+∞) is continuous, ϕb(t)=b, for t∈[0,T]T.Proof.
We first show thatA(P¯a)⊆Pa⊂P¯a  if condition (C1) holds. If u∈P¯a, then 0≤u≤∥u∥≤a, which implies f(u)<φ(a/M). We have
(3.14)∥Φu∥=(Φu)(T)≤∫0T(T-s)φ-1(∫0sa(r)f(u(r))∇r+C̃)∇s+∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sa(r)f(u(r))∇r+C̃)∇s1-∑i=1m-2bi<aM(∫0T(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s+∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s1-∑i=1m-2bi)=a.
This implies that Φ(P¯a)⊆Pa⊂P¯a.
Next, condition (C2) indicates that there exists d>c such that Φ(P¯d)⊂P¯d. Now we let
(3.15)A=P¯d,A1=[φb,φc],U1=int(A1),A2=P¯a,U2=Pa,
where int(A1) is the interior of A1. Then we have Φ(A)⊂A,Φ(A2)⊂A2. Moreover, Φ(P¯a)⊆Pa⊂P¯a means Φ(A2)⊆U2⊂A2. Thus Φ has no fixed point in (A2∖U2).
To showΦ(A1)⊂A1 and Φ has no fixed point in (A1∖U1), set u∈A1, following the definition of φb, we can know b≤u(t)≤c, for t∈[0,T]T. Condition (C3) then gives rise to φ(b/m)<f(u)<φ(c/M), which in turn produces
(3.17)(Φu)(t)≥(Φu)(0)=∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sa(r)f(u(r))∇r+C̃)∇s1-∑i=1m-2bi>bm∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s1-∑i=1m-2bi=b,(Φu)(t)≤(Φu)(T)2≤∫0T(T-s)φ-1(∫0sa(r)f(u(r))∇r+C̃)∇s+∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sa(r)f(u(r))∇r+C̃)∇s1-∑i=1m-2bi<cM(∫0T(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s+∑i=1m-2bi∫0ξi(T-s)φ-1(∫0sa(r)∇r+Ĉ)∇s1-∑i=1m-2bi)=c.
Combining the above two inequalities one achieves ϕb(t)=b<(Φu)(t)<c=ϕc(t), for t∈[0,T]T. That is, Φu∈U1. So Φ(A1)⊆U1⊂A1 and Φ has no fixed point in (A1∖U1). Therefore, all conditions of Theorem 2.10 are fulfilled, and the BVP (1.6) has at least three positive solutions u1,u2, and u3 such that
(3.18)b<u1(t)<c,∥u2∥<a,∥u3∥>a.


## 4. Some Examples

In the section, we present some simple examples to explain our results. We only study the caseT=R,(0,T)T=(0,1).Example 4.1.
Consider the following third-order three-point boundary value problem:(4.1)(φ(u′′))′+a(t)f(u)=0,0<t<1,u(0)=13u(12),u′(1)=0,φ(u′′(0))=14φ(u′′(12)),
where φ(x)=x,a(t)≡1,b1=1/3,c1=1/4,ξ1=1/2.We chooseη=1/2, by computing we can know mη=11/24,Ml=351/256,Mη=33/48. Let a=100,b=245,c=770,l=7/8, then a<ηb<b<lc. Obviously, Mlb<mc. We define a nonlinearity f as follows:
(4.2)f(u)={140,u∈[0,200],140+41045(u-200),u∈[200,245],550,u∈[245,320],550+5560(u-320),u∈[320,+∞).
Then, by the definition of f, we have(i)
f(u)≤φ(c/Ml)≈557.2,u∈[0,880];(ii)
f(u)>φ(b/mη)≈534.5,u∈[245,320];(iii)
f(u)<φ(a/Mη)≈145.4,u∈[0,200].By Theorem3.2, BVP (4.1) has at least three positive solutions.Example 4.2.
Consider the following third-order three-point boundary value problem:(4.3)(φ(u′′))′+a(t)f(u)=0,0<t<1,u(0)=13u(12),u′(1)=0,φ(u′′(0))=14φ(u′′(12)),
where φ(x)=x,a(t)≡1,b1=1/3,c1=1/4,ξ1=1/2,η=1/2.By computing, we can knowm=11/48,M=83/48. Let a=7,b=12,c=336,l=7/8, then a<b<c. Obviously, Mb<mc. We define a nonlinearity f as follows:
(4.4)f(u)={3,u∈[0,7],3+9725(u-7)2,u∈[7,12],100,u∈[12,336],100+11001764(u-336),u∈[336,+∞).
Then, by the definition of f, we have(i)
f(u)<φ(a/M)≈4.2,u∈[0,7];(ii)
and there existsd=2100>c such that f(u)≤φ(d/M)≈1214.4,u∈[0,2100];(iii)
φ(b/m)≈52.4<f(u)<φ(c/M)≈194.3,u∈[12,336].By Theorem3.3, BVP (4.3) has at least three positive solutions.Remark 4.3.
Consider following nonlinear m-point boundary value problem:(4.5)(φ(uΔ∇))∇+a(t)f(u(t))=0,0<t<T,u(0)=∑i=1m-2biu(ξi),uΔ(T)=0,φ(uΔ∇(0))=∑i=1m-2ciφ(uΔ∇(ξi)),
where
(4.6)φ(u)={u3,u≤0,u2,u>0,f and a satisfy the conditions (H1) and (H2). It is clear that φ:R→R is an increasing homeomorphism and homomorphism and φ(0)=0. Because p-Laplacian operators are odd, they do not apply to our example. Hence we generalize boundary value problem with p-Laplacian operator, and the results [8–11, 13–15] do not apply to the example.Remark 4.4.
In a similar way, we can get the corresponding results for the following boundary value problem:(4.7)(φ(uΔ∇))∇+a(t)f(u(t))=0,t∈(0,T)T,u(T)=∑i=1m-2aiu(ξi),uΔ(0)=0,φ(uΔ∇(0))=∑i=1m-2ciφ(uΔ∇(ξi)).



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*Source: 123283-2009-08-02.xml*