# Successive Iteration of Positive Solutions for Fourth-Order Two-Point Boundary Value Problems

Authors: Yongping Sun; Xiaoping Zhang; Min Zhao
Journal: Abstract and Applied Analysis
 (2013)

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## Abstract

We are concerned with a fourth-order two-point boundary value problem. We prove the existence of positive solutions and establish iterative schemes for approximating the solutions. The interesting point of our method is that the nonlinear term is involved with all lower-order derivatives of unknown function, and the iterative scheme starts off with a known cubic function or the zero function. Finally we give two examples to verify the effectiveness of the main results.


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## Body

## 1. Introduction

The bending of an elastic beam can be described with some fourth-order boundary value problems. Owing to their importance in engineering, physics, and material mechanics, boundary value problems for elastic beam equations have attracted much attention of many authors. Many methods, such as upper and lower solutions method, shooting method, coincidence degree theory, or fixed point theorem and so on, have been developed to derive existence of solutions for fourth-order differential equations with two-point boundary conditions; see [1–20] and many references cited therein. However, almost all of the papers we mentioned focused their attention on the existence of solutions or positive solutions. In the existing literature, there are few papers that concern with the computational methods of solutions or positive solutions; see [21–24]. In this paper, by the successively iterative technique, we study the existence and iteration of monotone positive solutions for the following fourth-order differential equation:
(1)u′′′′(t)=q(t)f(t,u(t),u′(t),u′′(t),u′′′(t)),a.e.t∈[0,1],
together with two-point boundary conditions
(2)u(0)=u′(1)=u′′(0)=u′′′(1)=0.
Throughout this paper, the following conditions are satisfied: (A1)
f∈C([0,1]×[0,∞)×[0,∞)×(-∞,0]×(-∞,0],[0,∞));(A2)
q∈L1[0,1] is nonnegative and 0<∫01‍q(s)ds<∞.In material mechanics, (1) describes the deflection of a beam under a certain force. The boundary conditions of the forms (2) mean that the elastic beam is simply fixed at the end t=0 and fastened with a sliding clamp at the end t=1.Fourth-order differential equations with boundary conditions (2) have been studied. For example, Graef and Yang in [7] studied the nonlinear fourth-order differential equation:
(3)u′′′′(t)=λa(t)f(u(t)),0<t<1,
together with boundary conditions (2). The authors established some sufficient conditions for the existence and nonexistence of positive solutions. The main tool is the Kranosel’skii fixed point theorem. Using the Leggett-Williams fixed point theorem, Yang [18] established an existence criterion for triple positive solutions of the nonlinear fourth-order differential equation:
(4)u′′′′(t)=g(t)f(t,u(t),u′(t)),0<t<1,
with two-point boundary conditions (2). By applying iterative techniques, Sun and Wang [21] investigated the existence and iteration of monotone positive solution for the following elastic beam equation
(5)u′′′′(t)=f(t,u(t),u′(t)),0<t<1,
with boundary conditions (2). By using Leray-Schauder nonlinear alternate, Leray-Schauder fixed point theorem and a fixed point theorem due to Avery and Peterson, Sun [15] presented some results on the existence and multiplicity of positive solutions to problem (1)-(2). Bai [3] proposed a constructive method which yields two monotone sequences that converge uniformly to extremal solutions to the following differential equations
(6)u′′′′(t)-Mu′′(t)=f(t,u(t),u′(t),u′′(t),u′′′(t)),0<t<1,
or
(7)u′′′′(t)-Mu′′(t)=f(t,u(t),u′(t),u′′(t)),0<t<1,
with boundary conditions (2), if there exists an upper solution β and a lower solution α with β⩽α. By placing some restrictions on the nonlinear term f, Bai [4] obtained the existence results for the fourth-order two-point boundary value problem (1)-(2) via the lower and upper solution method. In particular, a new truncating technique and an appropriate Nagumo-type condition are introduced and employed. In addition, Yao studied (1) with nonhomogeneous boundary condition
(8)u(0)=a,u′(1)=b,u′′(0)=c,u′′′(1)=d.
In [20] the author obtained local existence theorem via Leray-Schauder fixed point theorem. In [22] Yao constructed an iterative sequence by the monotonic technique and proved that the sequence approximates successively to the solution of the equation under suitable assumptions.Motivated by the above-mentioned works, this paper investigates the existence and iteration of positive solutions for problem (1)-(2). It is noted that the successively iterative scheme starts off with the zero function or a simple cubic function, which is feasible for the computational purpose. At the end of the paper, two examples are included to illustrate the main results.


## 2. Preliminaries

In this section, we introduce some necessary definitions and properties of the Green function.Definition 1 (see [25, 26]).
LetE be a real Banach space. A nonempty closed convex set K⊂E is called a cone of E if it satisfies the following two conditions:(1)
x∈K,λ⩾0 imply λx∈K;(2)
x∈K,-x∈K imply x=0.Definition 2 (see [25, 26]).
An operator is called completely continuous if it is continuous and maps bounded sets into precompact sets.The Green’s functionG:[0,1]×[0,1]→[0,+∞) for problem (1)-(2) is given by
(9)G(t,s)=16{(6t-3t2-s2)s,0⩽s⩽t⩽1,(6s-3s2-t2)t,0⩽t⩽s⩽1,
so solving the boundary value problem (1)-(2) is equivalent to finding a solution to the integral equation
(10)u(t)=∫01‍G(t,s)q(s)f(s,u(s),u′(s),u′′(s),u′′′(s))ds.
Some properties of functions G(t,s) are summarized as follows.Lemma 3.
For allt,s∈[0,1], the following inequalities hold true:
(11)0⩽-∂3G(t,s)∂t3⩽1,0⩽-∂2G(t,s)∂t2⩽t,0⩽∂G(t,s)∂t⩽12(1-t2),0⩽G(t,s)⩽16(3t-t3).Proof.
For any fixeds∈[0,1], from (9) we obtain
(12)∂3G(t,s)∂t3={0,0⩽s<t⩽1,-1,0⩽t<s⩽1,∂2G(t,s)∂t2={-s,0⩽s⩽t⩽1,-t,0⩽t⩽s⩽1,
then,
(13)0⩽-∂3G(t,s)∂t3⩽1,0⩽-∂2G(t,s)∂t2⩽t,t,s∈[0,1].
For allt,s∈[0,1], it follows from (9) that
(14)0⩽∂G(t,s)∂t={(1-t)s,0⩽s⩽t⩽1,s-12s2-12t2,0⩽t⩽s⩽1,⩽12(1-t2).
For allt,s∈[0,1], if s⩽t, it yields
(15)G(t,s)=16(6t-3t2-s2)s=16[3-3(1-t)2-s2]s⩽16(3-s2)s⩽16(3t-t3).
While t⩽s, we have
(16)G(t,s)=16(6s-3s2-t2)t=16[3-3(1-s)2-t2]t⩽16(3t-t3).
The proof is completed.In what follows we will consider the Banach spaceE=C3[0,1] equipped with norm
(17)∥u∥=max{max0⩽t⩽1|u(t)|,max0⩽t⩽1|u′(t)|,max0⩽t⩽1|u′′(t)|,=maxmax0⩽t⩽1|u′′′(t)|}.
Define the cone K⊂C3[0,1] by
(18)K={u∈C3[0,1]:u(t)⩾0,u′(t)⩾0,u′′(t)⩽0,=u′′′(t)⩽0,t∈[0,1]}.
For u∈K, we give an operator T by
(19)(Tu)(t)=∫01‍G(t,s)q(s)f(s,u(s),u′(s),u′′(s),u′′′(s))ds.
Obviously, the fixed points of T are monotone and nonnegative solutions of the problem (1)-(2).Lemma 4.
Consider thatT:K→K is completely continuous.Proof.
Suppose thatu∈K. Using (A1), (A2), (19), and Lemma 3, we obtain (Tu)(t)⩾0, (Tu)′(t)⩾0, (Tu)′′(t)⩽0, (Tu)′′′(t)⩽0, for all t∈[0,1]. Then T(K)⊆K. Using arguments similar to those of [21, 22], one can prove that T is a completely continuous operator.


## 3. Main Results

In this section, we give the main results of this paper.Theorem 5.
Assume that there existsR>0 such that(H1)
f(t,x1,y1,z1,w1)⩽f(t,x2,y2,z2,w2) for any 0⩽t⩽1, 0⩽x1⩽x2⩽R, 0⩽y1⩽y2⩽R, 0⩽-z1⩽-z2⩽R, 0⩽-w1⩽-w2⩽R;(H2)
max0⩽t⩽1f(t,R,R,-R,-R)⩽ΛR, where Λ=(∫01‍q(s)ds)-1;(H3)
f(t,0,0,0,0)≢0.
Then problem (1)-(2) has two monotone positive solutions w* and v* satisfying ∥w*∥⩽R, ∥v*∥⩽R and limn→∞∥wn-w*∥=0, limn→∞∥vn-v*∥=0, where wn=Tnw0, vn=Tnv0, n=1,2,…, w0(t)=(R/6)(3t-t3),  and v0(t)=0, for t∈[0,1].Proof.
LetKR={u∈K:∥u∥⩽R}; it follows that T:KR→KR. In fact, if u∈KR, then ∥u∥⩽R; thus, for any t∈[0,1],
(20)0⩽u(t)⩽max0⩽t⩽1u(t)⩽∥u∥⩽R,0⩽u′(t)⩽max0⩽t⩽1u′(t)⩽∥u∥⩽R,0⩽-u′′(t)⩽max0⩽t⩽1(-u′′(t))⩽∥u∥⩽R,0⩽-u′′′(t)⩽max0⩽t⩽1(-u′′′(t))⩽∥u∥⩽R,
which together with the conditions (H1) and (H2) implies that
(21)0⩽(Tu)(t)=∫01‍G(t,s)q(s)f(s,u(s),u′(s),u′′(s),u′′′(s))ds⩽16(3t-t3)∫01‍q(s)f(s,R,R,-R,-R)ds⩽12ΛR∫01‍q(s)ds=12R,t∈[0,1],0⩽(Tu)′(t)=∫01‍∂G(t,s)∂tq(s)f(s,u(s),u′(s),u′′(s),u′′′(s))ds⩽12(1-t2)∫01‍q(s)f(s,R,R,-R,-R)ds⩽12ΛR∫01‍q(s)ds=12R,t∈[0,1],0⩽-(Tu)′′(t)=-∫01‍∂2G(t,s)∂t2q(s)f(s,u(s),u′(s),u′′(s),u′′′(s))ds⩽t∫01‍q(s)f(s,R,R,-R,-R)ds⩽ΛR∫01‍q(s)ds=R,t∈[0,1],0⩽-(Tu)′′′(t)=-∫01‍∂3G(t,s)∂t3q(s)f(s,u(s),u′(s),u′′(s),u′′′(s))ds⩽∫01‍q(s)f(s,R,R,-R,-R)ds⩽ΛR∫01‍q(s)ds=R,t∈[0,1].
Hence, we have shown that ∥Tu∥⩽R, that is, T:KR→KR.
(
1
) Let w0(t)=(R/6)(3t-t3),t∈[0,1]; then ∥w0∥=R and w0∈KR. Denote wn=Tnw0,n=1,2,3,…. Since T:KR→KR, we have wn∈T(KR)⊆KR,n=1,2,3,…. Since T is completely continuous, we assert that {wn}n=1∞ is a sequentially compact set.
Sincew1=Tw0∈KR, by Lemma 3, (H1), and (H2), we obtain
(22)w1(t)=(Tw0)(t)=∫01‍G(t,s)q(s)f(s,w0(s),w0′(s),w0′′(s),w0′′′(s))ds=∫01‍G(t,s)q(s)f(s,R6(3s-s3),=∫01‍G(t,s)q(s)fR2(1-s2),-Rs,-R)ds⩽16(3t-t3)∫01‍q(s)f(s,R,R,-R,-R)ds⩽R6(3t-t3)Λ∫01‍q(s)ds=R6(3t-t3)=w0(t),⩽R6(3t-t3)Λ∫01‍q(s)ds=R6(3t-t3)t∈[0,1],w1′(t)=(Tw0)′(t)=∫01‍∂G(t,s)∂tq(s)f(s,w0(s),w0′(s),w0′′(s),w0′′′(s))ds=∫01‍∂G(t,s)∂tq(s)f(s,R6(3s-s3),=∫01‍∂G(t,s)∂tq(s)fR2(1-s2),-Rs,-R)ds⩽12(1-t2)∫01‍q(s)f(s,R,R,-R,-R)ds⩽R2(1-t2)Λ∫01‍q(s)ds=R2(1-t2)=w0′(t),⩽R2(1-t2)Λ∫01‍q(s)ds=R2(1-t2)t∈[0,1],-w1′′(t)=-(Tw0)′′(t)=-∫01‍∂2G(t,s)∂t2q(s)f(s,w0(s),w0′(s),w0′′(s),w0′′′(s))ds=-∫01‍∂2G(t,s)∂t2q(s)f(s,R6(3s-s3),=-∫01‍∂2G(t,s)∂t2q(s)fR2(1-s2),-Rs,-R)ds⩽t∫01‍q(s)f(s,R,R,-R,-R)ds⩽tΛR∫01‍q(s)ds=Rt=-w0′′(t),t∈[0,1],-w1′′′(t)=-(Tw0)′′′(t)=-∫01‍∂3G(t,s)∂t3q(s)f(s,w0(s),w0′(s),w0′′(s),w0′′′(s))ds=-∫01‍∂3G(t,s)∂t3q(s)f(s,R6(3s-s3),=-∫01‍∂3G(t,s)∂t3q(s)fR2(1-s2),-Rs,-R)ds⩽∫01‍q(s)f(s,R,R,-R,-R)ds⩽ΛR∫01‍q(s)ds=R=-w0′′′(t),t∈[0,1].
Thus,
(23)w1(t)⩽w0(t),w1′(t)⩽w0′(t),-w1′′(t)⩽-w0′′(t),-w1′′′(t)⩽-w0′′′(t),t∈[0,1].
Furthermore, it follows from (H1) that
(24)w2(t)=(Tw1)(t)⩽(Tw0)(t)=w1(t),t∈[0,1],w2′(t)=(Tw1)′(t)⩽(Tw0)′(t)=w1′(t),t∈[0,1],-w2′′(t)=-(Tw1)′′(t)⩽-(Tw0)′′(t)=-w1′′(t),t∈[0,1],-w2′′′(t)=-(Tw1)′′′(t)⩽-(Tw0)′′′(t)=-w1′′′(t),t∈[0,1].
By the induction, we obtain
(25)wn+1(t)⩽wn(t),wn+1′(t)⩽wn′(t),-wn+1′′(t)⩽-wn′′(t),-wn+1′′′(t)⩽-wn′′′(t),t∈[0,1],n=0,1,2,….
Hence, there exists w*∈KR such that limn→∞∥wn-w*∥=0. Applying the continuity of T and using wn+1=Twn, we obtain Tw*=w*, which implies that w* is a nonnegative solution of problem (1)-(2). Furthermore, by (H3) we know that the zero function is not a solution of problem (1)-(2). Thus, max0⩽t⩽1w*(t)>0. From the definition of the cone K, it yields w*(t)⩾t∥w*∥>0,(w*)′(t)⩾0,t∈(0,1); that is, w*(t) is a monotone positive solution of problem (1)-(2).
(
2
) Let v0(t)=0,t∈[0,1]; then v0∈KR. Let vn=Tnv0,n=1,2,3,…. With respect to T:KR→KR, we obtain vn∈T(KR)⊆KR,n=1,2,3,…. Since T is completely continuous, we assert that {vn}n=1∞ is a sequentially compact set.
Sincev1=Tv0=T0∈KR, we obtain
(26)R⩾v1(t)=(Tv0)(t)=(T0)(t)⩾0=v0(t),t∈[0,1],R⩾v1′(t)=(Tv0)′(t)=(T0)′(t)⩾0=v0′(t),t∈[0,1],R⩾-v1′′(t)=-(Tv0)′′(t)=-(T0)′′(t)⩾0=-v0′′(t),t∈[0,1],R⩾-v1′′′(t)=-(Tv0)′′′(t)=-(T0)′′′(t)⩾0=-v0′′′(t),t∈[0,1].
So by (H1), we have
(27)v2(t)=(Tv1)(t)⩾(Tv0)(t)=v1(t),t∈[0,1],v2′(t)=(Tv1)′(t)⩾(Tv0)′(t)=v1′(t),t∈[0,1],-v2′′(t)=-(Tv1)′′(t)⩾-(Tv0)′′(t)=-v1′′(t),t∈[0,1],-v2′′′(t)=-(Tv1)′′′(t)⩾-(Tv0)′′′(t)=-v1′′′(t),t∈[0,1].
By the induction, we further obtain
(28)vn+1(t)⩾vn(t),vn+1′(t)⩾vn′(t),-vn+1′′(t)⩾-vn′′(t),-vn+1′′′(t)⩾-vn′′′(t),t∈[0,1],n=0,1,2,….
Hence, there exists v*∈KR such that limn→∞∥vn-v*∥=0, and v* is a monotone positive solution of problem (1)-(2). The proof is completed.Corollary 6.
Suppose that(H1) and (H3) hold. Assume(H4)limx→+∞max0⩽t⩽1f(t,x,x,-x,-x)/x<Λ.
Then problem (1)-(2) has at least two monotone positive solutions w* and v* satisfying the conclusion of Theorem 5.Remark 7.
w
* and v* may coincide with each other and then the problem (1)-(2) has only one monotone positive solution in KR.Remark 8.
By the proof of Theorem5, {wn}n=1∞ and {vn}n=1∞ converge in C3[0,1] to the monotone positive solution w* and v* of the problem (1)-(2), respectively. In addition, the sequences {wn}n=1∞,{vn}n=1∞,{wn′}n=1∞, and {vn′}n=1∞ are increasing, while the sequences {wn′′}n=1∞,{vn′′}n=1∞,{wn′′′}n=1∞, and {vn′′′}n=1∞ are decreasing.Remark 9.
The iterative schemes in Theorem5 start off with a known cubic function or the zero function, which is feasible for the computational purpose.


## 4. Examples

In this section we provide two examples to illustrate the application of our results.Example 1.
Consider the problem(29)u′′′′(t)=12t+118u2(t)+16u′(t)-16u′′(t)-16u′′′(t),t∈[0,1],u(0)=u′(1)=u′′(0)=u′′′(1)=0.
For this problem, f(t,x,y,z,w)=t+(1/9)x2+(1/3)y-(1/3)z-(1/3)w and q(t)=1/2. By the calculations, the conditions (A1),(A2), and (H1)–(H3) of Theorem 5 hold for R=6. From Theorem 5, we have that problem (29) has two monotone positive solutions w* and v* satisfying ∥w*∥⩽6,∥v*∥⩽6 and limn→∞∥wn-w*∥=0,limn→∞∥vn-v*∥=0.
Moreover, the two iterative schemes are(30)w0(t)=3t-t3,t∈[0,1],wn+1(t)=112∫0t‍(t-s)3=112∫0t‍×(s+19wn2(s)+13wn′(s)-13wn′′(s)-13wn′′′(s))ds+t12∫01‍(6s-3s2-t2)+t12∫01‍×(-13wn′′′(s)s+19wn2(s)+13wn′(s)-13wn′′(s)+t12∫01‍×-13wn′′′(s))ds,t∈[0,1],n=0,1,2,…,v0(t)=0,t∈[0,1],vn+1(t)=112∫0t‍(t-s)3=112∫0t‍×(s+19vn2(s)+13vn′(s)-13vn′′(s)-13vn′′′(s))ds+t12∫01‍(6s-3s2-t2)+t12∫01‍×(+t12∫01‍-13vn′′′(s)s+19vn2(s)+13vn′(s)-13vn′′(s)+t12∫01‍×-13vn′′′(s))ds,+t12∫01‍×t∈[0,1],n=0,1,2,….
Using Matlab 7.5, the second and third terms of the two schemes are as follows:(31)w1(t)=35594536t-27617560t3+116t4+180t5-15040t8+190720t10,t∈[0,1],w2(t)=548t-336279351111065984t3-13495477711110659840t5+288101058888527872t6+753646139258883200t7-228148577407106560t8-3175998124690355200t9+32921414115059200t10-177732659200t11-73501935554111488t12+4889349380710400t13-1227799888852787200t14+280289444426393600t15+10099148142131200t16-2235290790400t17+11567641600t18+1998761420800t19+110973491200t20+19444426393600t21-198761420800t22-1888852787200t23+13555411148800t24,t∈[0,1],v1(t)=548t-124t3+1240t5,t∈[0,1],v2(t)=548t-5159124416t3+48591244160t5+25995328t6+17776t7-5248832t8-19622080t9+1165888t10+113110400t11-11244160t12-16220800t13+124883200t14,t∈[0,1].Example 2 (see [22]).
Consider the problem(32)u′′′′(t)=min{t,1-t}48|1-2t|×{2+[u(t)+u′(t)]3-[u′′(t)+u′′′(t)]3},t∈[0,12)⋃‍(12,1],u(0)=u′(1)=u′′(0)=u′′′(1)=0.
In this problem, f(t,x,y,z,w)=(1/48)[2+(x+y)3-(z+w)3] and q(t)=(min{t,1-t})/|1-2t|. Obviously, q(t) and f(t,x,y,z,w) satisfy conditions (A1) and (A2). In addition, f(t,x,y,z,w) is increasing with regard to x and y and is decreasing with z and w, and f(t,0,0,0,0)≢0. Moreover,
(33)max0⩽t⩽1f(t,130,130,-130,-130)=148[2+23375]<32×130,∫01‍q(t)dt=∫01‍min{t,1-t}|1-2t|dt=23.
Therefore, assumptions (H1)–(H3) of Theorem 5 are satisfied with R=1/30. From Theorem 5, problem (32) has two monotone positive solutions w* and v* satisfying ∥w*∥⩽1/30,∥v*∥⩽1/30 and limn→∞∥wn-w*∥=0, limn→∞∥vn-v*∥=0.
Moreover, the two iterative schemes are(34)w0(t)=1180(3t-t3),t∈[0,1],wn+1(t)=1144∫0t‍(t-s)3min{s,1-s}|1-2s|=1144∫0t‍×{2+[wn(s)+wn′(s)]3=1144∫0t‍×-[wn′′(s)+wn′′′(s)]3}ds+t144∫01‍(6s-3s2-t2)min{s,1-s}|1-2s|+t144∫01‍×{[2+wn(s)+wn′(s)]3+t144∫01‍×-[wn′′(s)+wn′′′(s)]3}ds,t∈[0,1],n=0,1,2,…,v0(t)=0,t∈[0,1],vn+1(t)=1144∫0t‍(t-s)3min{s,1-s}|1-2s|=1144∫0t‍×{2+[vn(s)+vn′(s)]3=1144∫0t‍×-[vn′′(s)+vn′′′(s)]3}ds+t144∫01‍(6s-3s2-t2)min{s,1-s}|1-2s|+t144∫01‍×{2+[vn(s)+vn′(s)]3+t144∫01‍×-[vn′′(s)+vn′′′(s)]3}ds,t∈[0,1],n=0,1,2,….
Using Mathlab 7.5, the second terms of the two schemes are given as follows:(35)
w
1
(
t
)
=
-
8449074379661
23951656146600000
+
441468132942481
19516164267600000
t
-
10842948935033
1951616426760000
t
2
-
785738110303
169705776240000
t
3
+
(
8449074379661
23951656146600000
+
122149319510627
60221306882880000
t
+
+
2602305979889593
702581913633600000
t
2
-
265540119476783
150553267207200000
t
3
+
-
13518333667229
19161324917280000
t
4
-
1855160149
117096985605600000
t
5
+
-
74353511
17564547840840000
t
6
-
26984453
175645478408400000
t
7
+
+
1291391
117096985605600000
t
8
-
4131727
38322649834560000
t
9
+
-
31823
2105640100800000
t
10
+
3953
491316023520000
t
11
+
+
509
216757069200000
t
12
+
1
5704133400000
t
13
)
×
1
-
2
t
,
for
t
∈
[
0
,
1
2
]
,
w
1
(
t
)
=
-
8449074379661
23951656146600000
+
441468132942481
19516164267600000
t
-
10842948935033
1951616426760000
t
2
-
785738110303
169705776240000
t
3
+
(
-
929396364017203
2107745740900800000
+
62633084321047
22910279792400000
t
+
-
817863130365013
140516382726720000
t
2
+
2416385432114987
526936435225200000
t
3
+
-
10620690488597
15055326720720000
t
4
-
568554781
117096985605600000
t
5
+
-
686043443
175645478408400000
t
6
-
546781
3054703972320000
t
7
+
+
46120147
234193971211200000
t
8
-
1594511
19161324917280000
t
9
+
-
23767
867028276800000
t
10
+
11287
2456580117600000
t
11
+
+
13
6193059120000
t
12
+
1
5704133400000
t
13
)
×
2
t
-
1
,
for
t
∈
[
1
2
,
1
]
,
v
1
(
t
)
=
-
1
2835
+
19
840
t
-
1
180
t
2
-
1
216
t
3
+
(
1
2835
-
23
11340
t
+
1
270
t
2
-
1
576
t
3
+
-
2
2835
t
4
)
1
-
2
t
,
for
t
∈
[
0
,
1
2
]
,
v
1
(
t
)
=
-
1
2835
+
19
840
t
-
1
180
t
2
-
1
216
t
3
+
(
-
1
2268
+
31
11340
t
-
11
1890
t
2
+
13
2835
t
3
+
-
2
2835
t
4
)
2
t
-
1
,
for
t
∈
[
1
2
,
1
]
.Remark 10.
This example was carried out firstly in [22], in which the author showed that the solution u*∈C3[0,1] is nontrivial, ∥u*∥⩽1, and the iterative scheme starts off with
(36)u0(t)=-724∫01‍G(t,s)min{s,1-s}|1-2s|ds.In the present paper, we proved that the solutionu*∈C3[0,1] is monotone positive, ∥u*∥⩽1/30, and the iterative scheme starts off with u0(t)=(1/180)(3t-t3) or u0(t)=0, which is much more simple than that of [22].



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*Source: 621315-2013-07-09.xml*