# Existence of Positive Solution for a Third-Order BVP with Advanced Arguments and Stieltjes Integral Boundary Conditions

Authors: Jian-Ping Sun; Ping Yan; Ya-Hong Zhao; Fang-Di Kong
Journal: Mathematical Problems in Engineering
 (2013)

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## Abstract

A class of third-order boundary value problems with advanced arguments and Stieltjes integral
boundary conditions is discussed. Some existence criteria of at least one positive solution are
established. The main tool used is the Guo-Krasnoselskii fixed point theorem.


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## Body

## 1. Introduction

Third-order differential equations arise in a variety of different areas of applied mathematics and physics, for example, in the deflection of a curved beam having a constant or varying cross-section, a three-layer beam, electromagnetic waves or gravity-driven flows, and so on [1].Recently, third-order boundary value problems (BVPs for short) with integral boundary conditions, which cover third-order multipoint BVPs as special cases, have attracted much attention from many authors; see [2–6] and the references therein. In particular, in 2012, by using a fixed point theorem due to Avery and Peterson [7], Jankowski [4] established the existence of at least three nonnegative solutions to the following BVP:
(1)u′′′(t)+h(t)f(t,u(α(t)))=0,t∈(0,1),u(0)=u′′(0)=0,u(1)=βu(η)+λ[u],
where λ denoted a linear functional on C[0,1] given by
(2)λ[u]=∫01‍u(t)dΛ(t)
involving a Stieltjes integral with a suitable function Λ of bounded variation. The measure dΛ could be a signed one. The situation with a signed measure dΛ was first discussed in [8, 9] for second-order differential equations; it was also discussed in [10, 11] for second-order impulsive differential equations. For some other related results, one can refer to [12–14].Among the boundary conditions in (1), only u(1) is related to a Stieltjes integral. In this paper, we are concerned with the following third-order BVP with advanced arguments and Stieltjes integral boundary conditions:
(3)u′′′(t)+f(t,u(α(t)))=0,t∈(0,1),u(0)=γu(η)+λ[u],u′′(0)=0,u(1)=βu(η)+λ[u].
Throughout this paper, we always assume that α:[0,1]→[0,1] is continuous and α(t)≥t for t∈[0,1], 0<η<1, 0≤γ<β<1, Λ is a suitable function of bounded variation, and λ[u] is defined as in (2). It is important to indicate that it is not assumed that λ[u] is positive to all positive u. Some existence criteria of at least one positive solution to the BVP (3) are obtained by using the following well-known Guo-Krasnoselskii fixed point theorem [15, 16].Theorem 1.
LetE be a Banach space, and let K be a cone in E. Assume that Ω1 and Ω2 are bounded open subsets of E such that 0∈Ω1, Ω¯1⊂Ω2, and let T:K∩(Ω¯2∖Ω1)→K be a completely continuous operator such that either(1)
∥Tu∥≤∥u∥ for u∈K∩∂Ω1 and ∥Tu∥≥∥u∥ for u∈K∩∂Ω2 or(2)
∥Tu∥≥∥u∥ for u∈K∩∂Ω1 and ∥Tu∥≤∥u∥ for u∈K∩∂Ω2. Then, T has a fixed point in K∩(Ω¯2∖Ω1).


## 2. Preliminaries

LetΔ=1-γ-(β-γ)η. Then, Δ>0.Lemma 2.
For anyy∈C[0,1], the BVP
(4)u′′′(t)=-y(t),t∈(0,1),u(0)=γu(η)+λ[u],u′′(0)=0,u(1)=βu(η)+λ[u]
has the unique solution
(5)u(t)=1-(β-γ)η+(β-γ)tΔλ[u]+γ+(β-γ)tΔ∫01‍k(η,s)y(s)ds+∫01‍k(t,s)y(s)ds,t∈[0,1],
where
(6)k(t,s)=12{(1-t)(t-s2),0≤s≤t≤1,t(1-s)2,0≤t≤s≤1.Proof.
By integrating the differential equation in (4) three times from 0 to t and using the boundary condition u′′(0)=0, we get
(7)u(t)=u(0)+u′(0)t-12∫0t‍(t-s)2y(s)ds,t∈[0,1].
So,
(8)u′(0)=u(1)-u(0)+12∫01‍(1-s)2y(s)ds.
In view of (7), (8), and the boundary conditions u(0)=γu(η)+λ[u] and u(1)=βu(η)+λ[u], we have
(9)u(t)=[γ+(β-γ)t]u(η)+λ[u]+∫01‍k(t,s)y(s)ds,t∈[0,1].
So,
(10)u(η)=1Δλ[u]+1Δ∫01‍k(η,s)y(s)ds.
Substituting (10) into (9), we get
(11)u(t)=1-(β-γ)η+(β-γ)tΔλ[u]+γ+(β-γ)tΔ∫01‍k(η,s)y(s)ds+∫01‍k(t,s)y(s)ds,t∈[0,1].Lemma 3 (see [4]).
Consider that0≤k(t,s)≤(1/2)(1+s)(1-s)2, (t,s)∈[0,1]×[0,1].Throughout, we assume that the following conditions are fulfilled:(C1)f∈C([0,1]×[0,+∞),[0,+∞)),(C2)(12)∫01‍dΛ(t)≥0,∫01‍tdΛ(t)≥0,κ(s)=∫01‍k(t,s)dΛ(t)≥0,s∈[0,1].For convenience, we denote(13)ρ=[1-(β-γ)η]∫01‍dΛ(t)+(β-γ)∫01‍tdΛ(t),ρ′=γ∫01‍dΛ(t)+(β-γ)∫01‍tdΛ(t).
Obviously, ρ,ρ′≥0. In the remainder of this paper, we always assume that ρ<Δ.LetC[0,1] be equipped with the maximum norm. Then, C[0,1] is a Banach space. Define
(14)K={mint∈[η,1]u∈C[0,1]:u(t)≥0,t∈[0,1],mint∈[η,1η]u(t)≥Γ∥u∥,λ[u]≥0},
where
(15)Γ=min{β(1-η)1-βη,βη1-γ(1-η)}.
Then, K is a cone in C[0,1].Now, we define operatorsT and S on K by
(16)(Tu)(t)=1-(β-γ)η+(β-γ)tΔλ[u]+(Fu)(t),(Tu)(t)=1-(β-γ)η+(β-γ)tΔ5λ[u]t∈[0,1],(Su)(t)=1-(β-γ)η+(β-γ)tΔ-ρλ[Fu]+(Fu)(t),(Tu)(t)=1-(β-γ)η+(β-γ)tΔ55λ[u]t∈[0,1],
where
(17)(Fu)(t)=γ+(β-γ)tΔ∫01‍k(η,s)f(s,u(α(s)))ds+∫01‍k(t,s)f(s,u(α(s)))ds,t∈[0,1].Lemma 4.
Consider thatT,S:K→K.Proof.
Letu∈K. Then, it is easy to verify that
(18)(Tu)′′(t)=-∫0t‍f(s,u(α(s)))ds≤0,t∈[0,1],
which shows that Tu is concave down on [0,1]. In view of
(19)(Fu)(0)=γΔ∫01‍k(η,s)f(s,u(α(s)))ds≥0,(Fu)(1)=βΔ∫01‍k(η,s)f(s,u(α(s)))ds≥0,
we have
(20)(Tu)(0)=1-(β-γ)ηΔλ[u]+(Fu)(0)≥0,(Tu)(1)=1+(β-γ)(1-η)Δλ[u]+(Fu)(1)≥0.
So, (Tu)(t)≥0, t∈[0,1].
Now, we prove thatmint∈[η,1](Tu)(t)≥Γ∥Tu∥. To do it, we consider two cases.
Case  1. Let (Tu)(η)≤(Tu)(1). Then mint∈[η,1](Tu)(t)=(Tu)(η), and there exists t-∈[η,1] such that ∥Tu∥=(Tu)(t-). Moreover,
(21)(Tu)(t-)-(Tu)(0)t--0≤(Tu)(η)-(Tu)(0)η-0.
So,
(22)∥Tu∥≤1η(Tu)(η)-1-ηη(Tu)(0),
which together with
(23)(Tu)(0)=γ(Tu)(η)+λ[u]
implies that
(24)∥Tu∥≤1-γ(1-η)η(Tu)(η);
that is,
(25)mint∈[η,1](Tu)(t)≥η1-γ(1-η)∥Tu∥.
Case  2. Let (Tu)(η)>(Tu)(1) and ∥Tu∥=(Tu)(t-). Note that in this case mint∈[η,1](Tu)(t)=(Tu)(1).
Ift-∈[0,η], then
(26)(Tu)(1)-(Tu)(t-)1-t-≥(Tu)(1)-(Tu)(η)1-η.
So,
(27)∥Tu∥≤11-η(Tu)(η)-η1-η(Tu)(1),
which together with
(28)(Tu)(η)=1β((Tu)(1)-λ[u])
implies that
(29)∥Tu∥≤1-βηβ(1-η)(Tu)(1);
that is,
(30)mint∈[η,1](Tu)(t)≥β(1-η)1-βη∥Tu∥.
Ift-∈(η,1), then
(31)(Tu)(t-)-(Tu)(η)t--η≤(Tu)(η)-(Tu)(0)η-0.
So,
(32)∥Tu∥≤1η(Tu)(η)-1-ηη(Tu)(0),
which together with (23) and (28) implies that
(33)∥Tu∥≤1-γ(1-η)βη(Tu)(1);
that is,
(34)mint∈[η,1](Tu)(t)≥βη1-γ(1-η)∥Tu∥.
It follows from (25), (30), and (34) that
(35)mint∈[η,1](Tu)(t)≥Γ∥Tu∥.
Finally, we need to show thatλ[Tu]≥0. In view of
(36)λ[Fu]=∫01‍γ+(β-γ)tΔ∫01‍k(η,s)f(s,u(α(s)))dsdΛ(t)+∫01‍∫01‍k(t,s)f(s,u(α(s)))dsdΛ(t)=ρ′Δ∫01‍k(η,s)f(s,u(α(s)))ds+∫01‍κ(s)f(s,u(α(s)))ds≥0,
we have
(37)λ[Tu]=ρΔλ[u]+λ[Fu]≥0.
This shows thatT:K→K. Similarly, we can prove that S:K→K.Lemma 5.
The operatorsT and S have the same fixed points in K.Proof.
Suppose thatu∈K is a fixed point of S. Then,
(38)λ[u]=∫01‍(1-(β-γ)η+(β-γ)tΔ-ρλ[Fu]+(Fu)(t)1-(β-γ)η+(β-γ)tΔ-ρ)dΛ(t)=ΔΔ-ρλ[Fu],
which shows that
(39)λ[Fu]=Δ-ρΔλ[u].
So,
(40)u(t)=(Su)(t)=1-(β-γ)η+(β-γ)tΔ-ρλ[Fu]+(Fu)(t)=1-(β-γ)η+(β-γ)tΔλ[u]+(Fu)(t)=(Tu)(t),t∈[0,1],
which indicates that u is a fixed point of T.
Suppose thatu∈K is a fixed point of T. Then,
(41)λ[u]=∫01‍(1-(β-γ)η+(β-γ)tΔλ[u]+(Fu)(t)1-(β-γ)η+(β-γ)tΔ)dΛ(t)=ρΔλ[u]+λ[Fu],
which shows that
(42)λ[u]=ΔΔ-ρλ[Fu].
So,
(43)u(t)=(Tu)(t)=1-(β-γ)η+(β-γ)tΔλ[u]+(Fu)(t)=1-(β-γ)η+(β-γ)tΔ-ρλ[Fu]+(Fu)(t)=(Su)(t),t∈[0,1],
which indicates that u is a fixed point of S.Lemma 6.
T
,
S
:
K
→
K is completely continuous.Proof.
First, by Lemma4, we know that T(K)⊂K.
Next, we show thatT is compact. Let D⊂K be a bounded set. Then, there exists M1>0 such that ∥u∥≤M1 for any u∈D. Since Λ is a function of bounded variation, there exists M2>0 such that vΔ′=∑i=1n‍|Λ(ti)-Λ(ti-1)|≤M2 for any partition Δ′:0=t0<t1<⋯<tn-1<tn=1. Let
(44)M3=sup{f(t,u):(t,u)∈[0,1]×[0,M1]}.
Then, for any u∈D,
(45)∥Tu∥=maxt∈[0,1](Tu)(t)≤1+(β-γ)(1-η)Δλ[u]+βΔ∫01‍k(η,s)f(s,u(α(s)))ds+12∫01‍(1+s)(1-s)2f(s,u(α(s)))ds≤1+(β-γ)(1-η)ΔM1M2+βM3Δ∫01‍k(η,s)ds+524M3,
which shows that T(D) is uniformly bounded.
On the other hand, for anyε>0, since k(t,s) is uniformly continuous on [0,1]×[0,1], there exists δ1(ε)>0 such that for any t1,t2∈[0,1] with |t1-t2|<δ1(ε),
(46)|k(t1,s)-k(t2,s)|<ε3M3,s∈[0,1].
Let δ=min{δ1(ε),εΔ/3(β-γ)M1M2,εΔ/3(β-γ)M3∫01‍k(η,s)ds}. Then, for any u∈D, t1,t2∈[0,1] with |t1-t2|<δ, we have
(47)|(Tu)(t1)-(Tu)(t2)|=|(β-γ)(t1-t2)Δλ[u]+(β-γ)(t1-t2)Δ×∫01‍k(η,s)f(s,u(α(s)))ds+∫01‍(k(t1,s)-k(t2,s))×f(s,u(α(s)))ds(β-γ)(t1-t2)Δ|≤(β-γ)|t1-t2|Δλ[u]+(β-γ)|t1-t2|Δ×∫01‍k(η,s)f(s,u(α(s)))ds+∫01‍|k(t1,s)-k(t2,s)|f(s,u(α(s)))ds≤(β-γ)|t1-t2|M1M2Δ+(β-γ)|t1-t2|M3Δ∫01k(η,s)ds‍+M3∫01‍|k(t1,s)-k(t2,s)|ds<ε,
which shows that T(D) is equicontinuous. It follows from the Arzela-Ascoli theorem that T(D) is relatively compact. Thus, we have shown that T is a compact operator.
Finally, we prove thatT is continuous. Suppose that un, u∈K and limn→∞un=u. Then, there exists M4>0 such that ∥u∥≤M4 and ∥un∥≤M4(n=1,2,…). For any ε>0, since f(s,x) is uniformly continuous on [0,1]×[0,M4], there exists δ>0 such that for any x1,x2∈[0,M4] with |x1-x2|<δ,
(48)|f(s,x1)-f(s,x2)|<ε(2β/Δ)∫01‍k(η,s)ds+(5/12),<ε(2β/Δ)∫01‍k(η,s)ds+(5/12)f(s,x2)555s∈[0,1].
At the same time, since limn→∞un=u, there exists positive integer N such that for any n>N,
(49)∥un-u∥<min{δ,εΔ2[1+(β-γ)(1-η)]|Λ(1)-Λ(0)|}.
It follows from (48) and (49) that for any n>N,
(50)∥Tun-Tu∥=maxt∈[0,1]|(Tun)(t)-(Tu)(t)|≤1+(β-γ)(1-η)Δ|λ[un]-λ[u]|+βΔ∫01‍k(η,s)|f(s,un(α(s)))-f(s,u(α(s)))|ds+12∫01‍(1+s)(1-s)2|f(s,un(α(s)))-f(s,u(α(s)))|ds≤1+(β-γ)(1-η)Δ∥un-u∥|Λ(1)-Λ(0)|+∫01‍(βΔk(η,s)+12(1+s)(1-s)2)×|f(s,un(α(s)))-f(s,u(α(s)))|ds<ε,
which indicates that T is continuous. Therefore, T:K→K is completely continuous. Similarly, we can prove that S:K→K is also completely continuous.


## 3. Main Results

For convenience, we define(51)f0=limsupx→0+maxt∈[0,1]f(t,x)x,f0=liminfx→0+mint∈[η,1]f(t,x)x,f∞=limsupx→+∞maxt∈[0,1]f(t,x)x,f∞=liminfx→+∞mint∈[η,1]f(t,x)x,H1=1+(β-γ)(1-η)Δ-ρ[ρ′Δ∫01‍k(η,s)ds+∫01‍κ(s)ds]+βΔ∫01‍k(η,s)ds+524,H2=Γ{1Δ-ρ[ρ′Δ∫η1‍k(η,s)ds+∫η1‍κ(s)ds]+1Δ∫η1‍k(η,s)dsρ′Δ∫η1‍k(η,s)ds}.Theorem 7.
IfH1f0<1<H2f∞, then the BVP (3) has at least one positive solution.Proof.
SinceH1f0<1, there exists ε1>0 such that
(52)H1(f0+ε1)≤1.
By the definition of f0, we may choose ρ1>0 so that
(53)f(t,x)≤(f0+ε1)x,t∈[0,1],x∈[0,ρ1].
Let Ω1={u∈C[0,1]:∥u∥<ρ1}. Then, for any u∈K∩∂Ω1,
(54)0≤u(t)≤∥u∥=ρ1,t∈[0,1],
which together with (52) and (53) implies that
(55)(Su)(t)=1-(β-γ)η+(β-γ)tΔ-ρ×[ρ′Δ∫01‍k(η,s)f(s,u(α(s)))ds+∫01‍κ(s)f(s,u(α(s)))dsρ′Δ∫01‍k(η,s)f(s,u(α(s)))]+γ+(β-γ)tΔ∫01‍k(η,s)f(s,u(α(s)))ds+∫01‍k(t,s)f(s,u(α(s)))ds≤1+(β-γ)(1-η)Δ-ρ×[ρ′Δ∫01‍k(η,s)f(s,u(α(s)))ds+∫01‍κ(s)f(s,u(α(s)))dsρ′Δ∫01‍k(η,s)f(s,u(α(s)))]+βΔ∫01‍k(η,s)f(s,u(α(s)))ds+12∫01‍(1+s)(1-s)2f(s,u(α(s)))ds≤(f0+ε1){ρ′Δ1+(β-γ)(1-η)Δ-ρ×[ρ′Δ∫01‍k(η,s)u(α(s))ds+∫01‍κ(s)u(α(s))dsρ′Δ∫01‍k(η,s)f(s,u(α(s)))]+βΔ∫01‍k(η,s)u(α(s))ds+12∫01‍(1+s)(1-s)2u(α(s))dsρ′Δ1+(β-γ)(1-η)Δ-ρ}≤(f0+ε1)∥u∥{1+(β-γ)(1-η)Δ-ρ×[ρ′Δ∫01‍k(η,s)ds+∫01‍κ(s)ds]+βΔ∫01‍k(η,s)ds+5241+(β-γ)(1-η)Δ-ρ}=H1(f0+ε1)∥u∥≤∥u∥,t∈[0,1].
This shows that
(56)∥Su∥≤∥u∥,u∈K∩∂Ω1.
On the other hand, since1<H2f∞, there exists ε2>0 such that
(57)H2(f∞-ε2)≥1.
By the definition of f∞, we may choose ρ2¯>0 so that
(58)f(t,x)≥(f∞-ε2)x,t∈[η,1],x∈[ρ2¯,+∞).
Let ρ2=max{2ρ1,ρ2¯/Γ} and Ω2={u∈C[0,1]:∥u∥<ρ2}. Then, for any u∈K∩∂Ω2,
(59)u(t)≥Γ∥u∥=Γρ2≥ρ2¯,t∈[η,1],
which together with (57) and (58) implies that
(60)(Su)(η)=1Δ-ρ[ρ′Δ∫01‍k(η,s)f(s,u(α(s)))ds+∫01‍κ(s)f(s,u(α(s)))dsρ′Δ∫01‍k(η,s)f(s,u(α(s)))]+1Δ∫01‍k(η,s)f(s,u(α(s)))ds≥1Δ-ρ[ρ′Δ∫η1k(η,s)f(s,u(α(s)))ds‍+∫η1‍κ(s)f(s,u(α(s)))dsρ′Δ∫01‍k(η,s)f(s,u(α(s)))]+1Δ∫η1‍k(η,s)f(s,u(α(s)))ds≥(f∞-ε2){1Δ-ρ[ρ′Δ∫η1‍k(η,s)u(α(s))ds+∫η1‍κ(s)u(α(s))dsρ′Δ∫01‍k(η,s)f(s,u(α(s)))]+1Δ∫η1‍k(η,s)u(α(s))dsρ′Δ∫η1‍k(η,s)u}≥(f∞-ε2)Γ∥u∥{1Δ-ρ[ρ′Δ∫η1‍k(η,s)ds+∫η1‍κ(s)ds]+1Δ∫η1‍k(η,s)dsρ′Δ∫η1‍k(η,s)u}=H2(f∞-ε2)∥u∥≥∥u∥.
This indicates that
(61)∥Su∥≥∥u∥,u∈K∩∂Ω2.
Therefore, it follows from (56), (61), and Theorem 1 that the operator S has one fixed point u∈K∩(Ω2¯∖Ω1), which is a positive solution of the BVP (3).Theorem 8.
IfH1f∞<1<H2f0, then the BVP (3) has at least one positive solution.Proof.
SinceH2f0>1, there exists ε3>0 such that
(62)H2(f0-ε3)≥1.
By the definition of f0, we may choose ρ3>0 so that
(63)f(t,x)≥(f0-ε3)x,t∈[η,1],x∈[0,ρ3].
Let Ω3={u∈C[0,1]:∥u∥<ρ3}. Then, for any u∈K∩∂Ω3,
(64)Γ∥u∥≤u(t)≤∥u∥=ρ3,t∈[η,1],
which together with (62) and (63) implies that
(65)(Su)(η)=1Δ-ρ[ρ′Δ∫01‍k(η,s)f(s,u(α(s)))ds+∫01‍κ(s)f(s,u(α(s)))dsρ′Δ∫01‍k(η,s)f(s,u(α(s)))]+1Δ∫01‍k(η,s)f(s,u(α(s)))ds≥1Δ-ρ[ρ′Δ∫η1‍k(η,s)f(s,u(α(s)))ds+∫η1‍κ(s)f(s,u(α(s)))dsρ′Δ∫01‍k(η,s)f(s,u(α(s)))]+1Δ∫η1‍k(η,s)f(s,u(α(s)))ds≥(f0-ε3){1Δ-ρ[ρ′Δ∫η1‍k(η,s)u(α(s))ds+∫η1‍κ(s)u(α(s))dsρ′Δ∫01‍k(η,s)f(s,u(α(s)))]+1Δ∫η1‍k(η,s)u(α(s))ds}≥(f0-ε3)Γ∥u∥{1Δ-ρ[ρ′Δ∫η1‍k(η,s)ds+∫η1‍κ(s)dsρ′Δ∫01‍k(η,s)f(s,u(α(s)))]+1Δ∫η1‍k(η,s)ds}=H2(f0-ε3)∥u∥≥∥u∥.
This shows that
(66)∥Su∥≥∥u∥,u∈K∩∂Ω3.
On the other hand, sinceH1f∞<1, there exists ε4>0 so that
(67)H1(f∞+ε4)<1.
By the definition of f∞, we may choose ρ4¯>0 such that
(68)f(t,x)≤(f∞+ε4)x,t∈[0,1],x∈[ρ4¯,+∞),
which implies that
(69)f(t,x)≤M+(f∞+ε4)x,t∈[0,1],x∈[0,+∞),
where
(70)M=max{f(t,x):t∈[0,1],x∈[0,ρ4¯]}.
Now, we choose ρ4=max{2ρ3,MH1/(1-H1(f∞+ε4))}, and let Ω4={u∈C[0,1]:∥u∥<ρ4}. Then, for any u∈K∩∂Ω4,
(71)0≤u(t)≤∥u∥,t∈[0,1],
which together with (69) implies that
(72)(Su)(t)=1-(β-γ)η+(β-γ)tΔ-ρ×[ρ′Δ∫01‍k(η,s)f(s,u(α(s)))ds+∫01‍κ(s)f(s,u(α(s)))dsρ′Δ∫01‍k(η,s)f(s,u(α(s)))]+γ+(β-γ)tΔ∫01‍k(η,s)f(s,u(α(s)))ds+∫01‍k(t,s)f(s,u(α(s)))ds≤1+(β-γ)(1-η)Δ-ρ×[ρ′Δ∫01‍k(η,s)f(s,u(α(s)))ds+∫01‍κ(s)f(s,u(α(s)))dsρ′Δ∫01‍k(η,s)f(s,u(α(s)))]+βΔ∫01‍k(η,s)f(s,u(α(s)))ds+12∫01‍(1+s)(1-s)2f(s,u(α(s)))ds≤[M+(f∞+ε4)∥u∥]{1+(β-γ)(1-η)Δ-ρ×[ρ′Δ∫01‍k(η,s)ds+∫01‍κ(s)dsρ′Δ∫01‍k(η,s)f(s,u(α(s)))]+βΔ∫01‍k(η,s)ds+5241+(β-γ)(1-η)Δ-ρ}=[M+(f∞+ε4)∥u∥]H1≤∥u∥,t∈[0,1].
This indicates that
(73)∥Su∥≤∥u∥,u∈K∩∂Ω4.
Therefore, it follows from (66), (73), and Theorem 1 that the operator S has one fixed point u∈K∩(Ω4¯∖Ω3), which is a positive solution of the BVP (3).Example 9.
Consider the following BVP:(74)u′′′(t)+t[3359u(t)913eu(t)+5881(u(t))217(u(t)+1)]=0,+t[3359u(t)913eu(t)+81(u(t))217(u(t)+1)]t∈(0,1),u(0)=14u(12)+∫01‍u(t)(2t-1)dt,u′′(0)=0,u(1)=12u(12)+∫01‍u(t)(2t-1)dt.
In view ofdΛ(t)=(2t-1)dt, we have
(75)∫01‍dΛ(t)=0,∫01‍tdΛ(t)=16,κ(s)=112s2(1-s)2,∫01‍dΛ(t)=0,∫01‍tdΛ(t)=16,κ=11255s∈[0,1].
At the same time, since γ=1/4 and β=η=1/2, a simple calculation shows that
(76)Δ=58,ρ=ρ′=124,Γ=27,∫01‍k(η,s)ds=116,∫η1‍k(η,s)ds=196,∫01‍κ(s)ds=1360,∫η1‍κ(s)ds=1720.
So,
(77)H1=9133360,H2=172940.
If we let f(t,x)=t[(3359x/913ex)+(5881x2/17(x+1))], (t,x)∈[0,1]×[0,+∞), then it is easy to compute that
(78)f0=3359913,f∞=588134,
which together with (77) implies that
(79)H1f0=33593360<1<H2f∞=58815880.
Therefore, it follows from Theorem 7 that the BVP (74) has at least one positive solution.



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*Source: 157849-2013-09-02.xml*