# Existence and Monotone Iteration of Positive Pseudosymmetric Solutions for a Third-Order Four-Point BVP withp-Laplacian

Authors: Dan Li; Libo Wang; Minghe Pei
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We study the existence and monotone iteration of solutions for a third-order four-point boundary value problem withp-Laplacian. An existence result of positive, concave, and pseudosymmetric solutions and its monotone iterative scheme are established by using the monotone iterative technique. Meanwhile, as an application of our result, an example is given.


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## Body

## 1. Introduction

The third-order equations arise in many areas of applied mathematics and physics [1] and thus have been discussed by many authors and many excellent results were obtained; see [1–31] and the references therein. Recently, wide attention has been paid to the third-order boundary value problems with the p-Laplace operator. In fact, the third-order equations involving the p-Laplace operator can be seen as a generalized model for various physical, natural or physiological phenomena such as the flow of a thin film of viscous fluid over a solid surface, the solitary wave solution of the Korteweg-de Vries equation or a thyroid-pituitary interaction [17].In 2005, Cabada et al. [7] studied the one-dimensional nonlinear third-order ϕ-Laplacian equation
(1)-(ϕ(u′′(t)))′=f(t,u(t)),t∈[a,b]
with the boundary conditions
(2)u(a)=A,u′′(a)=B,u′′(b)=C,
where ϕ:ℝ→ℝ is an increasing homeomorphism with ϕ(0)=0. By applying the monotone iterative technique based on suitable antimaximum principles, they obtained the existence of extremal solutions for the problem.In 2006, using the monotone iterative technique, Zhou and Ma [30] obtained the existence of positive solutions and established a corresponding iterative scheme for the following third-order p-Laplacian problem of the form:
(3)(ϕp(u′′(t)))′=q(t)f(t,u(t)),t∈[0,1],u(0)=∑i=1m‍αiu(ξi),u′(η)=0,u′′(1)=∑i=1n‍βiu′′(θi).
In 2007, Wang and Ge [26] considered third-order differential equation
(4)(ϕ(u′′(t)))′+f(t,u(t),u′(t),u′′(t))=0,t∈(0,1)
subject to the following integral boundary conditions:
(5)u(0)=0,u′(0)-k1u′′(0)=∫01‍h1(u(s))ds,u′(1)+k2u′′(1)=∫01‍h2(u(s))ds.
The existence result to the problem is obtained by applying the method of upper and lower solutions and Leray-Schauder degree theory.In 2009, Sun et al. [24] studied the existence of positive solutions for the following third-order p-Laplacian problem:
(6)(ϕp(u′′(t)))′=q(t)f(t,u(t),u′(t),u′′(t)),t∈[0,1],u(0)=∑i=1m‍αiu(ξi),u′(η)=0,u′′(1)=∑i=1n‍βiu′′(θi).
By applying a monotone iterative method, the authors obtained the existence of positive solutions for the problem and established iterative schemes for approximating the solutions.In 2010, Jin and Lu [17] considered the following third-order p-Laplacian resonant problem of the form:
(7)(ϕp(x′′(t)))′=f(t,x(t),x′(x),x′′(t)),t∈(0,1),x(0)=0,x′(1)=∑i=1m-2‍aix′(ξi),x′′(0)=0.
The authors obtained the existence of solutions for the problem by using Mawhin's continuation theorem.In 2010, by using the fixed point index method, Yang and Yan [31] established the existence of at least one or at least two positive solutions for the following third-order p-Laplacian problem:
(8)(ϕp(u′′(t)))′+f(t,u(t))=0,t∈(0,1),αu(0)-βu′(0)=0,γu(1)+δu′(1)=0,u′′(0)=0.
Motivated by the above works and [32, 33], in this paper, we consider the existence and monotone iteration of positive, pseudosymmetric solutions of the following third-order four-point p-Laplacian boundary value problem:
(9)(ϕp(u′′(t)))′+q(t)f(t,u(t),u′(t))=0,t∈(0,1)
subject to boundary conditions
(10)u(0)=0,u(1)=u(η),u′′(1+η2)=0,
where ϕp(s)=|s|p-2s,p>1, and η∈(0,1) be constant. Here u*(t)  is said to be a positive solution of BVP (9), (10) if and only if u*(t) is the solution of BVP (9), (10) and satisfies u*(t)>0 for t∈(0,1]. BVP (9), (10) can model the static deflection of an elastic beam with linear supports at both endpoints.To the best of our knowledge, the existence results of the pseudosymmetric solutions for the third-order boundary value problem has not been considered.This work is organized as follows. In Section2, some notations and preliminaries are introduced. The main results are discussed in Section 3. As applications of our results, an example is given in the last section.


## 2. Preliminary

In this section, we give some definitions and lemmas which help to simplify the presentation of our main result.Definition 1 (see [32]).
Letu∈C[0,1], η∈(0,1). One says that u is pseudosymmetric about η on [0,1], if u is symmetric on [η,1], that is,
(11)u(t)=u(1+η-t),∀t∈[η,1].Definition 2.
Letu∈C[0,1], η∈(0,1). One says that u is pseudo-antisymmetric about η on [0,1], if u is antisymmetric on [η,1], that is,
(12)u(t)+u(1+η-t)=0,∀t∈[η,1].
Let the Banach space E=C1[0,1] be endowed with the norm
(13)∥u∥=max0≤t≤1(u2(t)+u′2(t))1/2,
and define the cone P⊂E by
(14)P={u∈E:uisnonnegative,concaveandpseudosymmetricaboutηon[0,1]},
and let
(15)P¯a={u∈P:∥u∥≤a}.
For convenience, we consider the following.  (H0)
q(t) is a nonnegative continuous function defined on (0,1), q(t)≢0 on any subinterval of (0,1). In addition, ∫01‍q(t)dt<+∞ and q(t) is pseudosymmetric about η on [0,1].(H1)
f(t,u,v):[0,1]×[0,+∞)×ℝ→ℝ is continuous and
(16)f(t,u,v)≥0,∀(t,u,v)∈[0,1+η2]×[0,+∞)×ℝ.(H2)
f(t,u,v)+f(1+η-t,u,-v)=0 for all (t,u,v)∈[η,1]×[0,+∞)×ℝ.(H3)
f(t,0,0)≢0 on [0,1].
Now, we define an operatorT:C1[0,1]→C1[0,1] as follows: for u∈C1[0,1],
(17)(Tu)(t)={∫0t‍∫τ(1+η)/2‍ϕp-1(∫s(1+η)/2‍q(r)×f(r,u(r),u′(r))dr∫s(1+η)/2‍)dsdτ,0≤t≤1+η2,∫0η‍∫τ(1+η)/2‍ϕp-1(∫s(1+η)/2‍q(r)×f(r,u(r),u′(r))dr∫s(1+η)/2‍)dsdτ-∫t1‍∫(1+η)/2τ‍ϕp-1(∫(1+η)/2s‍q(r)×f(r,u(r),u′(r))dr∫s(1+η)/2‍)dsdτ,1+η2≤t≤1.
Obviously under assumptions(H0) and (H1), the operator T is well defined and it is easy to verify that BVP (9), (10) has a solution if and only if T:C1[0,1]→C1[0,1] has a fixed point.
The next lemmas are some properties of the operatorT.Lemma 3.
Assume that(H0), (H1), and (H2) hold. Then TP⊂P.Proof.
From the definition ofT, it is easy to check that Tu is nonnegative on [0,1] and satisfies (10) for all u∈P. Furthermore, since
(18)(Tu)′′(t)={-ϕp-1(∫t(1+η)/2‍q(r)f(r,u(r),u′(r))dr)≤0,0≤t≤1+η2,ϕp-1(∫(1+η)/2t‍q(r)f(r,u(r),u′(r))dr)≤0,1+η2≤t≤1,
it follows that Tu is concave on [0,1].
Next we prove thatTu is pseudosymmetric about η on [0,1]. In fact, if t∈[η,(1+η)/2], then 1+η-t∈[(1+η)/2,1], and it follows that
(19)(Tu)(1+η-t)=∫0η‍∫τ(1+η)/2‍ϕp-1×(∫s(1+η)/2‍q(r)f(r,u(r),u′(r))dr)dsdτ-∫1+η-t1‍∫(1+η)/2τ‍ϕp-1×(∫(1+η)/2s‍q(r)f(r,u(r),u′(r))dr)dsdτ=∫0η‍∫τ(1+η)/2‍ϕp-1×(∫s(1+η)/2‍q(r)f(r,u(r),u′(r))dr)dsdτ+∫1η‍∫(1+η)/2τ‍ϕp-1×(∫(1+η)/2s‍q(r)f(r,u(r),u′(r))dr)dsdτ+∫ηt‍∫(1+η)/2τ‍ϕp-1×(∫(1+η)/2s‍q(r)f(r,u(r),u′(r))dr)dsdτ+∫t1+η-t‍∫(1+η)/2τ‍ϕp-1×(∫(1+η)/2s‍q(r)f(r,u(r),u′(r))dr)dsdτ.
Also since u is pseudosymmetric about η on [0,1], that is, u(t)=u(1+η-t) for t∈[η,1], then
(20)u′(t)=-u′(1+η-t),t∈[η,1].
Thus, for all t∈[η,1], from (H2), we have
(21)q(r)f(r,u(r),u′(r))=-q(1+η-r)f(1+η-r,u(r),-u′(r))=-q(1+η-r)×f(1+η-r,u(1+η-r),u′(1+η-r)).
Hence q(r)f(r,u(r),u′(r)) is pseudo-antisymmetric about η on [0,1], and thus ∫(1+η)/2s‍q(r)f(r,u(r),u′(r))dr is pseudosymmetric about η on [0,1]. Furthermore ϕp-1(∫(1+η)/2s‍q(r)f(r,u(r),u′(r))dr) is pseudosymmetric about η on [0,1]. Thus the function ∫(1+η)/2τ‍ϕp-1(∫(1+η)/2s‍q(r)f(r,u(r),u′(r))dr)ds is pseudo-antisymmetric about η on [0,1], and hence
(22)∫1η‍∫(1+η)/2τ‍ϕp-1(∫(1+η)/2s‍q(r)f(r,u(r),u′(r))dr)dsdτ=0.
Using the similar technique, we can get
(23)∫t1+η-t‍∫(1+η)/2τ‍ϕp-1(∫(1+η)/2s‍q(r)f(r,u(r),u′(r))dr)dsdτ=0.
From (19), (22), and (23), it follows that
(24)(Tu)(1+η-t)=∫0η‍∫τ(1+η)/2‍ϕp-1×(∫s(1+η)/2‍q(r)f(r,u(r),u′(r))dr)dsdτ+∫ηt‍∫(1+η)/2τ‍ϕp-1×(∫(1+η)/2s‍q(r)f(r,u(r),u′(r))dr)dsdτ=∫0t‍∫(1+η)/2τ‍ϕp-1×(∫(1+η)/2s‍q(r)f(r,u(r),u′(r))dr)dsdτ=(Tu)(t),t∈[η,1+η2].
If t∈[(1+η)/2,1], then 1+η-t∈[η,(1+η)/2]. From (24), it follows that
(25)(Tu)(1+η-t)=(Tu)(1+η-(1+η-t))=(Tu)(t),t∈[1+η2,1].
This together with (24) implies that
(26)(Tu)(t)=(Tu)(1+η-t),t∈[η,1].
In summary, Tu∈P, and then TP⊂P.The following lemma can be easily verified by a standard argument.Lemma 4.
Assume that(H0), (H1), and (H2) hold. Then T:P→P is completely continuous.Lemma 5.
Assume that(H0), (H1), and (H2) hold. Suppose also that there exists a>0 such that for 0≤t≤(1+η)/2, 0≤u1≤u2≤a, 0≤|v1|≤|v2|≤a,
(27)f(t,u1,v1)≤f(t,u2,v2).
Then for u1, u2∈P¯a with
(28)u1(t)≤u2(t),|u1′(t)|≤|u2′(t)|,t∈[0,1],
we have
(29)(Tu1)(t)≤(Tu2)(t),|(Tu1)′(t)|≤|(Tu2)′(t)|,(Tu1)(t)≤(Tu2)(t),|(Tu1)′(t)|t∈[0,1].Proof.
First we prove that, for allt∈[0,(1+η)/2],
(30)(Tu1)(t)≤(Tu2)(t),|(Tu1)′(t)|≤|(Tu2)′(t)|.
From assumptions, we have
(31)f(r,u1(r),u1′(r))≤f(r,u2(r),u2′(r)),r∈[0,1+η2],
and hence
(32)∫s(1+η)/2‍f(r,u1(r),u1′(r))dr≤∫s(1+η)/2‍f(r,u2(r),u2′(r))dr,s∈[0,1+η2].
Since ϕp-1 is strictly increasing on ℝ, then for all s∈[0,(1+η)/2], we have
(33)ϕp-1(∫s(1+η)/2‍f(r,u1(r),u1′(r))dr)≤ϕp-1(∫s(1+η)/2‍f(r,u2(r),u2′(r))dr).
Thus for t∈[0,(1+η)/2],
(34)(Tu1)(t)-(Tu2)(t)=∫0t‍∫τ(1+η)/2‍[ϕp-1(∫s(1+η)/2‍q(r)f(r,u1(r),u1′(r))dr)-ϕp-1(∫s(1+η)/2‍q(r)f(r,u2(r),u2′(r))dr)]dsdτ≤0,|(Tu1)′(t)|-|(Tu2)′(t)|=∫t(1+η)/2‍[ϕp-1(∫s(1+η)/2‍q(r)f(r,u1(r),u1′(r))dr)-ϕp-1(∫s(1+η)/2‍q(r)f(r,u2(r),u2′(r))dr)]ds≤0.
Therefore, (30) holds for t∈[0,(1+η)/2].
Next we prove that (30) holds for t∈[(1+η)/2,1]. In fact, if t∈[(1+η)/2,1], then 1+η-t∈[0,(1+η)/2], and hence from the fact that  Tu1 and Tu2 are pseudosymmetric about η on [0,1], it follows that, for t∈[(1+η)/2,1],
(35)(Tu1)(t)-(Tu2)(t)=(Tu1)(1+η-t)-(Tu2)(1+η-t)≤0,|(Tu1)′(t)|-|(Tu2)′(t)|=|(Tu1)′(1+η-t)|-|(Tu2)′(1+η-t)|≤0.
In summary,
(36)(Tu1)(t)≤(Tu2)(t),|(Tu1)′(t)|≤|(Tu2)′(t)|,(Tu1)(t)≤(Tu2)(t),,|(Tu1)′(t)|t∈[0,1].Now, we introduce some notations as follows:(37)A1=∫0(1+η)/2‍ϕp-1(∫s(1+η)/2‍q(r)dr)ds,A2=ϕp-1(∫0(1+η)/2‍q(r)dr),A=max{2A1,2A2}=2A2.Lemma 6.
Assume that(H0), (H1), and (H2) hold. Suppose also that there exists a>0 such that for 0≤t≤(1+η)/2, 0≤u1≤u2≤a, 0≤|v1|≤|v2|≤a,
(38)f(t,u1,v1)≤f(t,u2,v2),max0≤t≤(1+η)/2f(t,a,a)≤ϕp(aA).
Then T:P¯a→P¯a.Proof.
Define two functionals onE as follows:
(39)α(u):=max0≤t≤1|u(t)|,β(u):=max0≤t≤1|u′(t)|.
Then
(40)∥u∥≤2max{α(u),β(u)}.
If u∈P¯a, then
(41)0≤u(t)≤max0≤t≤1|u(t)|≤∥u∥≤a,t∈[0,1],0≤|u′(t)|≤max0≤t≤1|u′(t)|≤∥u∥≤a,t∈[0,1].
From the assumptions, for all t∈[0,(1+η)/2],
(42)0≤f(t,u(t),u′(t))≤f(t,a,a)≤max0≤t≤(1+η)/2f(t,a,a)≤ϕp(aA).
Then,
(43)α(Tu)=max0≤t≤1|(Tu)(t)|=(Tu)(1+η2)=∫0(1+η)/2‍∫τ(1+η)/2‍ϕp-1×(∫s(1+η)/2‍q(r)f(r,u(r),u′(r))dr)dsdτ≤∫0(1+η)/2‍∫τ(1+η)/2‍ϕp-1×(∫s(1+η)/2‍q(r)ϕp(aA)dr)dsdτ≤A1·aA≤22a,β(Tu)=max0≤t≤1|(Tu)′(t)|=(Tu)′(0)=∫0(1+η)/2‍ϕp-1(∫s(1+η)/2‍q(r)f(r,u(r),u′(r))dr)ds≤∫0(1+η)/2‍ϕp-1(∫s(1+η)/2‍q(r)ϕp(aA)dr)ds≤A2·aA≤22a.
So we have
(44)∥Tu∥≤2max{α(Tu),β(Tu)}≤2·22a=a.
Thus T:P¯a→P¯a.


## 3. Main Result

Now we establish existence result of positive, concave, and pseudosymmetric solutions and its monotone iterative scheme for BVP (9), (10).Theorem 7.
Assume that(H0), (H1), (H2), and (H3) hold. Suppose also that there exists a>0 such that for 0≤t≤(1+η)/2, 0≤u1≤u2≤a, 0≤|v1|≤|v2|≤a,
(45)f(t,u1,v1)≤f(t,u2,v2),max0≤t≤(1+η)/2f(t,a,a)≤ϕp(aA).
Then BVP (9), (10) has two positive, concave, and pseudosymmetric solutions w* and v* with
(46)0<w*(t)≤22a,t∈(0,1],limn→∞wn=limn→∞Tnw0=w*(inC1norm),0<v*(t)≤a,t∈(0,1],limn→∞vn=limn→∞Tnv0=v*(inC1norm),
where
(47)w0(t)={22at,0≤t≤1+η2,22a(1+η-t),1+η2≤t≤1;v0(t)≡0,0≤t≤1,(Tu)(t)={∫0t‍∫τ(1+η)/2‍ϕp-1(∫s(1+η)/2‍q(r)×f(r,u(r),u′(r))dr∫s(1+η)/2‍)dsdτ,0≤t≤1+η2,∫0η‍∫τ(1+η)/2‍ϕp-1(∫s(1+η)/2‍q(r)×f(r,u(r),u′(r))dr∫s(1+η)/2‍)dsdτ-∫t1‍∫(1+η)/2τ‍ϕp-1(∫(1+η)/2s‍q(r)×f(r,u(r),u′(r))dr∫s(1+η)/2‍)dsdτ,1+η2≤t≤1.Proof.
Let(48)w0(t)={22at,0≤t≤1+η2,22a(1+η-t),1+η2≤t≤1,w1(t)={∫0t‍∫τ(1+η)/2‍ϕp-1(∫s(1+η)/2‍q(r)×f(r,w0(r),w0′(r))dr∫s(1+η)/2‍)dsdτ,0≤t≤1+η2,∫0η‍∫τ(1+η)/2‍ϕp-1(∫s(1+η)/2‍q(r)×f(r,w0(r),w0′(r))dr∫s(1+η)/2‍)dsdτ,-∫t1‍∫(1+η)/2τ‍ϕp-1(∫(1+η)/2s‍q(r)×f(r,w0(r),w0′(r))dr∫s(1+η)/2‍)dsdτ,1+η2≤t≤1.
Then w1(t)∈C1[0,(1+η)/2]∩C1[(1+η)/2,1].
Next we prove that(49)limt→((1+η)/2)-w1(t)=limt→((1+η)/2)+w1(t),(50)limt→((1+η)/2)-w1′(t)=limt→((1+η)/2)+w1′(t).
In fact, from (H2) it follows that
(51)limt→((1+η)/2)+w1(t)=∫0η‍∫τ(1+η)/2‍ϕp-1×(∫s(1+η)/2‍q(r)f(r,22ar,22a)dr)dsdτ-∫(1+η)/21∫(1+η)/2τ‍ϕp-1‍×(∫(1+η)/2s‍q(r)f(r,22a(1+η-r),-22a)dr)dsdτ=(∫0(1+η)/2‍+∫(1+η)/2η‍)×∫τ(1+η)/2‍ϕp-1(∫s(1+η)/2‍q(r)f(r,22ar,22a)dr)dsdτ-∫(1+η)/21‍∫(1+η)/2τ‍ϕp-1×(∫(1+η)/2s‍q(r)×f(r,22a(1+η-r),-22a)dr)dsdτ=∫0(1+η)/2‍∫τ(1+η)/2‍ϕp-1×(∫s(1+η)/2‍q(r)f(r,22ar,22a)dr)dsdτ+∫(1+η)/2η‍∫τ(1+η)/2‍ϕp-1×(∫s(1+η)/2‍q(r)f(r,22ar,22a)dr)dsdτ-∫(1+η)/21‍∫(1+η)/2τ‍ϕp-1×(∫(1+η)/2s‍q(r)×f(r,22a(1+η-r),-22a)dr)dsdτ=∫0(1+η)/2‍∫τ(1+η)/2‍ϕp-1×(∫s(1+η)/2‍q(r)f(r,22ar,22a)dr)dsdτ=limt→((1+η)/2)-w1(t).
Then (49) holds. Equation (50) can be obtained in a similar way. Thus from (49) and (50), it follows that
(52)w1(t)∈C1[0,1].
We note that for t∈[0,(1+η)/2],
(53)0≤w1(t)=∫0t‍∫τ(1+η)/2‍ϕp-1×(∫s(1+η)/2‍q(r)f(r,22ar,22a)dr)dsdτ≤∫0t‍∫τ(1+η)/2‍ϕp-1(∫s(1+η)/2‍q(r)ϕp(aA)dr)dsdτ≤aA∫0t‍∫τ(1+η)/2‍ϕp-1(∫0(1+η)/2‍q(r)dr)dsdτ≤aAA2t=w0(t),
and for t∈[(1+η)/2,1],
(54)w1(t)=∫0η‍∫τ(1+η)/2‍ϕp-1×(∫s(1+η)/2‍q(r)f(r,22ar,22a)dr)dsdτ-∫t1‍∫(1+η)/2τ‍ϕp-1×(∫(1+η)/2s‍q(r)×f(r,22a(1+η-r),-22a)dr)dsdτ≤∫0η‍∫τ(1+η)/2‍ϕp-1(∫s(1+η)/2‍q(r)ϕp(aA)dr)dsdτ+∫t1‍∫(1+η)/2τ‍ϕp-1(∫(1+η)/2s‍q(r)ϕp(aA)dr)dsdτ≤aA∫0η‍∫τ(1+η)/2‍ϕp-1(∫0(1+η)/2‍q(r)dr)dsdτ+aA∫t1‍∫(1+η)/2τ‍ϕp-1(∫(1+η)/21‍q(r)dr)dsdτ≤aA∫0η‍∫τ(1+η)/2‍A2dsdτ+aA∫t1‍∫(1+η)/2τ‍ϕp-1(∫0(1+η)/2‍q(r)dr)dsdτ≤aA∫0η‍A2dτ+aA∫t1‍∫(1+η)/2τ‍A2dsdτ≤aAA2η+aAA2(1-t)=w0(t).
So,
(55)w1(t)≤w0(t)≤22a,t∈[0,1].
Thus,
(56)α(w1):=max0≤t≤1|w1(t)|≤22a.
From assumptions, for t∈[0,(1+η)/2],
(57)|w1′(t)|=∫t(1+η)/2‍ϕp-1×(∫s(1+η)/2‍q(r)f(r,22ar,22a)dr)ds≤∫t(1+η)/2‍ϕp-1(∫s(1+η)/2‍q(r)ϕp(aA)dr)ds≤aA∫t(1+η)/2‍ϕp-1(∫0(1+η)/2‍q(r)dr)ds≤aAA2≤22a=|w0′(t)|,
and for t∈[(1+η)/2,1],
(58)|w1′(t)|=|∫(1+η)/2t‍ϕp-1(∫(1+η)/2s‍q(r)×f(r,22a(1+η-r),-22a)dr)ds|≤∫(1+η)/2t‍ϕp-1(∫(1+η)/21‍q(r)ϕp(aA)dr)ds≤aA∫(1+η)/2t‍ϕp-1(∫0(1+η)/2‍q(r)dr)ds≤aAA2≤22a=|w0′(t)|.
Hence from (57) and (58), we have
(59)β(w1):=max0≤t≤1|w1′(t)|≤22a.
Consequently, from (56) and (59), it follows that
(60)∥w1∥≤2max{α(w1),β(w1)}≤a.
From the proof of Lemma 3, we see that w1 is nonnegative, concave, and pseudosymmetric about η on [0,1], and hence
(61)w1∈P¯a.
Define {wn} as follows:
(62)wn+1=Twn=Tnw1=Tn+1w0,n=0,1,….
Then {wn} is well defined and for n=1, 2, …,
(63)wn+1(t)≤wn(t),|wn+1′(t)|≤|wn′(t)|,t∈[0,1].
In fact, for t∈[0,(1+η)/2],
(64)w2(t)=Tw1(t)=∫0t‍∫τ(1+η)/2‍ϕp-1×(∫s(1+η)/2‍q(r)×f(r,w1(r),w1′(r))dr∫s(1+η)/2‍)dsdτ≤∫0t‍∫τ(1+η)/2‍ϕp-1×(∫s(1+η)/2‍q(r)×f(r,22ar,22a)dr∫s(1+η)/2‍)dsdτ=w1(t),(65)|w2′(t)|=|Tw1′(t)|=∫t(1+η)/2‍ϕp-1×(∫s(1+η)/2‍q(r)×f(r,w1(r),w1′(r))dr∫s(1+η)/2‍)ds≤∫t(1+η)/2‍ϕp-1×(∫s(1+η)/2‍q(r)×f(r,22ar,22a)dr∫s(1+η)/2‍)ds=w1′(t).
For t∈[(1+η)/2,1], since w1, w2∈P¯a, it follows from (64) and (65) that
(66)w2(t)=w2(1+η-t)≤w1(1+η-t)=w1(t),(67)|w2′(t)|=|w2′(1+η-t)|≤|w1′(1+η-t)|=|w1′(t)|.
So from (64)–(67), we have
(68)w2(t)≤w1(t),|w2′(t)|≤|w1′(t)|,t∈[0,1],
that is, (63) holds when n=1. Assume that (63) holds when n=k, that is,
(69)wk+1(t)≤wk(t),|wk+1′(t)|≤|wk′(t)|,t∈[0,1].
Then from Lemma 5, we obtain
(70)wk+2(t)=(Twk+1)(t)≤(Twk)(t)=wk+1(t),t∈[0,1],|wk+2′(t)|=|(Twk+1)′(t)|≤|(Twk)′(t)|=|wk+1′(t)|,t∈[0,1].
So by induction (63) holds.
SinceT:P¯a→P¯a is completely continuous, it follows that {wn}n=1∞ is relative compact. Then {wn} has a convergent subsequence {wnk} and w*∈P¯a such that
(71)wnk⟶w*(k⟶∞),
that is,
(72)wnk(t)⇉w*(t)(k⟶∞),wnk′(t)⇉w*′(t)(k⟶∞)on[0,1].
While from (63) and the fact for each n=1,2,…, wn′((1+η)/2)=0 and  wn′′(t)≤0 on [0,1], it follows that
(73)w1(t)≥w2(t)≥⋯≥wn(t)≥wn+1(t)≥⋯,n=1,2,…,on[0,1],w1′(t)≥w2′(t)≥⋯≥wn′(t)≥wn+1′(t)≥⋯,n=1,2,…,on[0,1+η2],w1′(t)≤w2′(t)≤⋯≤wn′(t)≤wn+1′(t)≤⋯,n=1,2,…,on[1+η2,1].
Hence,
(74)wn(t)⇉w*(t)(n⟶∞),wn′(t)⇉w*′(t)(n⟶∞)on[0,1],
that is,
(75)wn⟶w*(n⟶∞).
This together with the continuity of T and wn+1=Twn, implies that
(76)Tw*=w*.
Also, since
(77)0≤wn(t)≤w0(t)={22at,0≤t≤1+η2,22a(1+η-t),1+η2≤t≤1,0≤|wn′(t)|≤|w1′(t)|≤22a,t∈[0,1],
we have
(78)0≤w*(t)≤22a,0≤|w*′(t)|≤22a,t∈[0,1].
Furthermore, we have
(79)w*(t)>0,t∈(0,1].
In fact, from (H3) and w*(t)≢0 on [0,1], we have w*((1+η)/2)>0. Since w*(t) is concave on [0,1], then
(80)w*(t)≥w*((1+η)/2)-0((1+η)/2)-0t=21+ηw*(1+η2)t>0,t∈(0,1+η2].
Consequently from the fact w* is pseudosymmetric on [0,1], we have
(81)w*(t)>0,t∈(0,1].
Let v0(t)≡0 on [0,1], then v0∈P¯a. Set vn+1=Tvn, n=0, 1, 2, …. Then from Lemma 6, it follows that
(82)vn∈P¯a,n=1,2,….
From Lemma 4, we see that {vn}n=1∞ is relative compact, and hence there exists a convergent subsequence {vnk}⊂{vn} and v*∈P¯a such that
(83)vnk⟶v*(k⟶∞),
that is,
(84)vnk(t)⇉v*(t)(k⟶∞)on[0,1],(85)vnk′(t)⇉v*′(t)(k⟶∞)on[0,1].
Since v1=Tv0=T0∈P¯a, then
(86)v1(t)=Tv0(t)=(T0)(t)≥0,t∈[0,1],|v1′(t)|=|(Tv0)′(t)|=|(T0)′(t)|≥0,t∈[0,1].
Thus from Lemma 5,
(87)v2(t)=Tv1(t)≥Tv0(t)=v1(t),t∈[0,1],|v2′(t)|=|(Tv1)′(t)|≥|(Tv0)′(t)|=|v1′(t)|,t∈[0,1].
By induction, it is easy to see that for n=1, 2, …,
(88)vn+1(t)≥vn(t),t∈[0,1],(89)|vn+1′(t)|≥|vn′(t)|,t∈[0,1].
From (84)–(89), we see that
(90)vn(t)⇉v*(t)(n⟶∞),vn′(t)⇉v*′(t)(n⟶∞)on[0,1].
Therefore, vn→v*(n→∞),  v*∈P¯a. By the continuity of T and vn+1=Tvn, we have
(91)Tv*=v*.
Again from (H3), we have v*(t)>0 on (0,1].
Since every fixed point ofT in P is the solution of BVP (9), (10), then w* and v* are two positive, concave and pseudosymmetric solutions of BVP (9), (10). This completes the proof of the theorem.


## 4. An Example

Consider the following third-order four-point boundary value problem:(92)u′′′(t)+f(t,u(t),u′(t))=0,t∈(0,1),u(0)=0,u(1)=u(12),u′′(34)=0,
where
(93)f(t,u,v)=284(34-t)(uv2+162),(t,u,v)∈[0,1]×[0,+∞)×ℝ.
It is easy to see that BVP (92) corresponds to BVP (9), (10) when p=2, q(t)≡1, and η=1/2. Take a=62, and then A=32/4.Next we verify that all conditions of Theorem7 are satisfied. In fact, obviously the conditions (H0), (H1), (H2), and (H3) hold. In addition, for 0≤t≤3/4, 0≤u1≤u2≤62, 0≤|v1|≤|v2|≤62,
(94)f(t,u1,v1)≤f(t,u2,v2),max0≤t≤3/4f(t,a,a)=f(0,62,62)=8=ϕ2(aA).
Hence, from Theorem 7, BVP (92) has two positive, concave, and pseudosymmetric solutions w* and v* such that
(95)0<w*(t)≤6,0<|w*′(t)|≤6,t∈[0,1],limn→∞wn=limn→∞Tnw0=w*,limn→∞wn′=limn→∞(Tnw0)′=w*′,
where
(96)w0(t)={6t,0≤t≤34,9-6t,34≤t≤1,
for n=0,1,2,…,
(97)wn+1(t)={∫0t‍∫τ3/4‍∫s3/4‍284(34-r)×(wn(r)wn′2(r)+162)drdsdτ,(wn(r)wn′2(r)+162)dr0≤t≤34,∫01/2‍∫τ3/4‍∫s3/4‍284(34-r)×(wn(r)wn′2(r)+162)drdsdτ-∫t1‍∫3/4τ‍∫3/4s‍284(34-r)×(wn(r)wn′2(r)+162)drdsdτ,(wn(r)wn′2wn′2(r)+62)dr34≤t≤1,0<v*(t)≤62,0<|v*′(t)|≤62,t∈[0,1],limn→∞vn=limn→∞Tnv0=v*,limn→∞vn′=limn→∞(Tnv0)′=v*′,
where v0(t)≡0 on [0,1] and for n=0,1,2,…,
(98)vn+1(t)={∫0t‍∫τ3/4‍∫s3/4‍284(34-r)×(vn(r)vn′2(r)+162)drdsdτ,(wn(r)wn′2(r)+162)dr0≤t≤34,∫01/2‍∫τ3/4‍∫s3/4‍284(34-r)×(vn(r)vn′2(r)+162)drdsdτ-∫t1‍∫3/4τ‍∫3/4s‍284(34-r)×(vn(r)vn′2(r)+162)drdsdτ,(wn(r)wn′2(r)+162)dr34≤t≤1.
The first two terms of {wn(t)} and three terms of {vn(t)}, respectively, are as follows:
(99)w0(t)={6t,0≤t≤34,9-6t,34≤t≤1;(100)w1(t)={-3270t5+(92112-163)t4+121t3-(812896+356)t2+(24323584+3112)t,0≤t≤34,3270t5-(272112+163)t4+(27256+121)t3-(4052896+356)t2+(72923584+3112)t-729235840,34≤t≤1;v0(t)≡0,t∈[0,1],v1(t)=-163t4+121t3-356t2+3112t,t∈[0,1],v2(t)=22867038902t14-2273051324t13+256010528t12-218670176t11+472426746880t10-652398297088t9+25619712t8-592413048832t7+132157351936t6-5121573519360t5+(921258815488-163)t4+121t3-(812161128382464+356)t2+(812598476849152+3112)t,t∈[0,1].



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*Source: 192509-2013-06-03.xml*