research article existence of positive solutions for an...

Research ArticleExistence of Positive Solutions for an Elastic Beam Equationwith Nonlinear Boundary Conditions

Ruikuan Liu and Ruyun Ma

Department of Mathematics Northwest Normal University Lanzhou 730070 China

Correspondence should be addressed to Ruikuan Liu liuruikuan2008163com

Received 9 December 2013 Accepted 18 February 2014 Published 20 March 2014

Academic Editor Ch Tsitouras

Copyright copy 2014 R Liu and R Ma This is an open access article distributed under the Creative Commons Attribution Licensewhich permits unrestricted use distribution and reproduction in any medium provided the original work is properly cited

We study the existence and nonexistence of positive solutions for the following fourth-order two-point boundary value problemsubject to nonlinear boundary conditions 1199061015840101584010158401015840(119905) = 120582119891(119905 119906(119905)) 119905 isin (0 1) 119906(0) = 0 119906

1015840(0) = 120583ℎ(119906(0)) 119906

10158401015840(1) = 0 119906

101584010158401015840(1) =

120583119892(119906(1)) where 120582 gt 0 120583 ge 0 are parameters and 119891 [0 1] times [0 +infin) rarr (0 +infin) ℎ [0 +infin) rarr [0 +infin) and 119892 [0 +infin) rarr

(minusinfin 0] are continuous By using the fixed-point index theory we prove that the problem has at least one positive solution for 120582 120583sufficiently small and has no positive solution for 120582 large enough

1 Introduction

In this paper we study the existence and nonexistence of pos-itive solutions for the following fourth-order boundary valueproblem with nonlinear boundary conditions

1199061015840101584010158401015840

(119905) = 120582119891 (119905 119906 (119905)) 119905 isin (0 1)

119906 (0) = 0 1199061015840

(0) = 120583ℎ (119906 (0))

11990610158401015840

(1) = 0 119906101584010158401015840

(1) = 120583119892 (119906 (1))

(1)

where 120582 gt 0 120583 ge 0 are parameters and 119891 isin 119862([0 1] times RR)

and ℎ 119892 isin 119862(RR) are real functions If 120582 = 1 120583 = 0 inmechanics problem (1) is called cantilever beam equation[1 2] The equation describes the deflection of an elasticbeam fixed at the left and freed at the right There are somepapers discussing the existence of solutions of the equation byusing various methods such as the lower and upper solutionmethod the Leray-Schauder continuation method fixed-point theory and the monotone iterative method see [3ndash11]If 120583 = 0 problem (1) has also been studied see [12ndash17]

In the case 120583 = 1 ℎ(119906(0)) = 0 Yang et al [12] obtainedsufficient conditions of the existence of two solutions of prob-lem (1) by using variational technique and a three-critical-point theorem Recently Li and Zhang [13] are concernedwith the existence and uniqueness of monotone positive

solution of problem (1) with 120582 = 120583 = 1 ℎ(119906(0)) = 0 by usinga new fixed-point theorem of generalized concave operatorsbut some monotone assumptions on 119891 and 119892 are needed

In 2013 by using a three-critical-point theorem Cabadaand Tersian [14] studied the existence and multiplicity ofsolutions of problem (1) with 120582 = 120583 ℎ(119906(0)) = 0

A natural question is what would happen if 120582 = 120583 andℎ(119906(0)) = 0

Motivated by the above papers wewill prove the existenceand nonexistence of positive solution for problem (1) byusing the fixed-point index theory with the nonlinearity 119891satisfying superlinear growth condition at infinity

We make the following assumptions

(A1) 119891 [0 1] times [0 +infin) rarr (0 +infin) is continuous(A2) ℎ [0 +infin) rarr [0 +infin) is continuous and 119892

[0 +infin) rarr (minusinfin 0] is continuous(A3) For any 119905 isin [0 1] 119891

infin= lim119906rarrinfin

(119891(119905 119906)119906) = +infin

The main results of the present paper are summarized asfollows

Theorem1 Assume that (A1)ndash(A3) holdThen problem (1) hasat least one positive solution for 120582 120583 sufficiently small

Theorem 2 Assume that (A1)ndash(A3) hold Then problem (1)has no positive solution for 120582 large enough

Hindawi Publishing CorporationJournal of Applied MathematicsVolume 2014 Article ID 972135 5 pageshttpdxdoiorg1011552014972135

2 Journal of Applied Mathematics

Remark 3 The results obtained in this paper are not a con-sequence of the previous theorem in the previous literatureClearly the boundary condition of (1) is more general thanthe above pieces of literature and problem (1) considers twodifferent parameters which is more extensive

Remark 4 It is pointed out thatwe donot need anymonotoneassumption on 119891 119892 and ℎ which is weaker than thecorresponding assumptions on119891119892 in [13] References [12 14]are only consideredwith the existence of solution for problem(1) with nonlinear boundary condition however we study theexistence and nonexistence of positive solution of (1)

The remainder of this paper is arranged as followsSection 2 presents some preliminaries The proofs of Theo-rems 1 and 2 are given in Section 3 Finally we give a simpleexample to illustrate our main results

2 Preliminaries

In this section we collect some preliminary results that willbe used in subsequent sections Let 119883 = 119862[0 1] then 119883 is aBanach space under the norm 119906 = max

119905isin[01]|119906(119905)| Define

119875 = 119906 isin 119883 | 119906(119905) ge 0 119905 isin [0 1] then119875 sub 119883 is a nonnegativecone

Lemma 5 (see [11 Lemma 22]) Let ℎ isin 119862[0 1] then thesolution 119906(119905) of the problem

1199061015840101584010158401015840

(119905) = ℎ (119905) 119905 isin (0 1)

119906 (0) = 1199061015840

(0) = 11990610158401015840

(1) = 119906101584010158401015840

(1) = 0

(2)

is

119906 (119905) = int

1

0

119866 (119905 119904) ℎ (119904) 119889119904 (3)

where

119866 (119905 119904) =

1

6

1199052(3119904 minus 119905) 0 le 119905 le 119904 le 1

1

6

1199042(3119905 minus 119904) 0 le 119904 le 119905 le 1

(4)

Now let us set

120601 (119905) =

1199052

2

minus

1199053

6

Φ (119905) = ℎ (119906 (0)) 119905 minus 119892 (119906 (1)) 120601 (119905)

forall119905 isin [0 1]

(5)

It follows from (3) and (5) with simple computation thatproblem (1) is equivalent to integral equation

119906 (119905) = 120582int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 + 120583Φ (119905) = 119879119906 (119905) (6)

Note that for any 119906 isin 119883 the function (119879119906)(119905) satisfiesthe boundary conditions in (1) by the definition of Greenrsquosfunction 119866(119905 119904) In view of Lemma 5 it is easy to see that119906 isin 119883 is a fixed point of the operator 119879 if and only if 119906 isin 119883 isa solution of problem (1)

Lemma 6 (see [6 Lemma 22] [13 Lemma 21]) 119866(119905 119904) 120601(119905)have the following properties

(i) 119866(119905 119904) gt 0 forall0 lt 119905 119904 lt 1(ii) max

119905isin[01]119866(119905 119904) le 119866(1 119904) = (16)119904

2(3minus119904) forall0 le 119904 le 1

(iii) 119866(119905 119904) ge 1199052119866(1 119904) ge (119905

23)119866(1 119904) forall0 le 119905 119904 le 1

(iv) 11990523 le 120601(119905) le 11990522 forall0 le 119905 le 1

Lemma 7 Φ(119905) has the following properties

(i) Φ(119905) le Φ(1) le ℎ(119906(0)) minus 119892(119906(1)) forall0 lt 119905 lt 1(ii) Φ(119905) ge Φ(119905

23) ge 0 forall0 lt 119905 lt 1

Proof First we will prove (i) For any 119905 isin [0 1] it follows that

Φ1015840

(119905) = ℎ (119906 (0)) minus 119892 (119906 (1)) (119905 minus

1

2

1199052) (7)

From (A2)Φ1015840(119905) ge 0 SoΦ(119905) is nondecreasing for 119905 isin [0 1]Moreover

Φ (119905) le Φ (1) = ℎ (119906 (0)) minus

1

3

119892 (119906 (1)) le ℎ (119906 (0)) minus 119892 (119906 (1))

(8)

so (i) holdsNowwe prove (ii) For any 119905 isin [0 1] obviously 119905 ge 119905

23 ge

0From (i)Φ(119905) is nondecreasing for 119905 isin [0 1] thenΦ(119905) ge

Φ(11990523) ge Φ(0) = 0

For any fixed 120591 isin (0 1) define the cone

119870 = 119906 isin 119875 | min119905isin[1205911]

119906 (119905) ge

1205912

3

119906 (9)

Then119870 sub 119883 is a positive cone letting119870119903= 119906 isin 119870 | 119906 lt 119903

120597119870119903= 119906 isin 119870 | 119906 = 119903 with 119903 gt 0 is a constant

Lemma 8 Assume that (A1) and (A2) hold then 119879 119870 rarr 119870

is completely continuous

Proof Obviously for any 119905 isin [0 1] 119906 isin 119870 it follows that119879119906(119905) ge 0

For any 119906 isin 119870 from (6) and Lemma 7(i) we have

max119905isin[01]

119879 (119906 (119905)) = max119905isin[01]

(120582int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 + 120583Φ (119905))

le 120582int

1

0

119866 (1 119904) 119891 (119904 119906 (119904)) 119889119904

+ 120583 (ℎ (119906 (0)) minus 119892 (119906 (1)))

(10)

Hence

119879119906 le 120582int

1

0

119866 (1 119904) 119891 (119904 119906 (119904)) 119889119904 + 120583 (ℎ (119906 (0)) minus 119892 (119906 (1)))

(11)

Journal of Applied Mathematics 3

For any fixed 120591 isin (0 1) (11) together with Lemma 7(ii)we get

min120591le119905le1

119879 (119906 (119905)) = min120591le119905le1

(120582int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904

+ 120583 (ℎ (119906 (0)) 119905 minus 119892 (119906 (1)) 120601 (119905)) )

ge

1205912

3

(120582int

1

0

119866 (1 119904) 119891 (119904 119906 (119904)) 119889119904

+120583 (ℎ (119906 (0)) minus 119892 (119906 (1))) )

ge

1205912

3

119879119906

(12)

So 119879(119870) sub 119870According to (A1) (A2) and Arzela-Ascoli theorem it is

not difficult to verify that 119879 119870 rarr 119870 is completely continu-ous

The following well-known fixed-point index theorem incones is crucial to our arguments

Lemma 9 (see [18]) Let 119864 be a Banach space and let119870 sub 119864 bea cone For 119903 gt 0 define 119870

119903= 119906 isin 119870 | 119906 lt 119903 Assume that

119879 119870119903rarr 119870 is compact map such that 119879119906 = 119906 for 119906 isin 120597119870

119903=

119906 isin 119870 | 119906 = 119903

(i) If 119879119906 ge 119906 for 119906 isin 120597119870119903 then 119894(119879 119870

119903 119870) = 0

(ii) If 119879119906 le 119906 for 119906 isin 120597119870119903 then 119894(119879 119870

119903 119870) = 1

3 Proof of Main Results

In this section we will prove our main results

Proof of Theorem 1 Let

119872 = max(119905119904)isin[01]times[01]

119866 (119905 119904) gt 0 (13)

Let 119902 gt 0 set

119868 (119902) = 119872 max119906isin119870119906=119902

(int

1

0

119891 (119904 119906 (119904)) 119889119904) gt 0 (14)

For any number 1199031gt 0 let 119870

1199031

= 119906 isin 119870 | 119906 le 1199031 Choose

120590 gt 0 so that

120590 le

1199031

2119868 (1199031)

forall119906 isin 1205971198701199031

120590 (ℎ (119906 (0)) minus 119892 (119906 (1))) le

1199031

2

forall119906 isin 1205971198701199031

(15)

Then for 120582 le 120590 120583 le 120590 and 119906 isin 1205971198701199031

combining (6) withLemma 7 we have

119879119906 (119905) le 120582119872int

1

0

119891 (119904 119906 (119904)) 119889119904

+ 120583 (ℎ (119906 (0)) minus 119892 (119906 (1)))

le 120590119872int

1

0

119891 (119904 119906 (119904)) 119889119904 + 120590 (ℎ (119906 (0)) minus 119892 (119906 (1)))

le 120590119868 (1199031) +

1199031

2

le 1199031

(16)

which implies that

119879119906 le 1199031= 119906 119906 isin 120597119870

1199031

(17)

Thus Lemma 9 implies that

119894 (1198791198701199031

119870) = 1 (18)

According to (A3) for any 119905 isin [0 1] there exists a 119901 gt 0

such that

119891 (119905 119906) ge 120578119906 forall119906 ge 119901 (19)

where 120578 is chosen so that

120582(

1205912

3

)

2

120578int

1

120591

119866 (1 119904) 119889119904 ge 1 forall120591 isin (0 1) (20)

For any 120591 isin (0 1) choose 1199032gt max(31205912)119901 2119903

1 and set

1198701199032

= 119906 isin 119870 | 119906 le 1199032 If 119906 isin 120597119870

1199032

then

min120591le119905le1

119906 (119905) ge

1205912

3

1199032gt 119901 (21)

It follows from (A2) and Lemma 6 that

min119905isin[1205911]

119879 (119906 (119905)) ge min119905isin[1205911]

(120582int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904)

ge 120582

1205912

3

int

1

0

119866 (1 119904) 119891 (119904 119906 (119904)) 119889119904

ge 120582

1205912

3

int

1

0

119866 (1 119904) 120578119906 (119904) 119889119904

ge 120582

1205912

3

120578int

1

120591

119866 (1 119904) 119906 (119904) 119889119904

ge 120582(

1205912

3

)

2

120578int

1

120591

119866 (1 119904) 119889119904 119906

ge 119906

(22)

which implies that

119879 (119906) ge 119906 119906 isin 1205971198701199032

(23)


Hence from Lemma 9 we get

119894 (1198791198701199032

119870) = 0 (24)

From (18) and (24) it follows that

119894 (1198791198701199032

1198701199031

119870) = 119894 (1198791198701199032

119870) minus 119894 (1198791198701199031

119870) = minus1 (25)

Therefore 119879 has a fixed point in1198701199032

1198701199031

that is problem (1)has at least one positive solution

Proof of Theorem 2 From (A1) and (A3) for any 119905 isin [0 1]there is a constant 119888 such that

119891 (119905 119906) ge 119888119906 forall119906 ge 0 (26)

Let 119906 isin 119883 be a positive solution of (1) then 119906 isin 119870 satisfies(6) Choose 120582 large enough such that

120582(

1205912

3

)

2

119888 int

1

120591

119866 (1 119904) 119889119904 gt 1 forall120591 isin (0 1) (27)

For any 120591 isin (0 1) we have

min119905isin[1205911]

119906 (119905) ge min119905isin[1205911]

(120582int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904)

ge 120582

1205912

3

int

1

120591

119866 (1 119904) 119888119906 (119904) 119889119904

ge 120582(

1205912

3

)

2

119888 int

1

120591

119866 (1 119904) 119889119904 119906

gt 119906

(28)

which is a contradiction Therefore problem (1) has nopositive solution for 120582 large enough

Finally we give an example to illustrate our main result

Example 10 Consider the following fourth-order two-pointproblem with nonlinear boundary conditions

1199061015840101584010158401015840

(119905) = 120582 (1199063

(119905) (3 + sin (119906 (119905))) + 1199052 + 1) 119905 isin (0 1)

119906 (0) = 0 1199061015840

(0) = 120583 (2[119906 (0)]2+ 1)

11990610158401015840

(1) = 0 119906101584010158401015840

(1) = 120583 ([minus (119906 (1))]12

minus 1)

(29)

Clearly the nonlinearity

119891 (119905 119906) = 1199063

(119905) (3 + sin (119906 (119905))) + 1199052 + 1

(119905 119906) isin [0 1] times [0infin)

ℎ (119904) = 21199042+ 1 119892 (119904) = minus119904

12minus 1 forall119904 isin [0 +infin)

(30)

It is easy to check that (A1)ndash(A3) are satisfied By simplecomputation we have119872 = 12 Set 119902 = 2 then 119868(119902) = 17

If 1199031= 4 (where 119903

1can be any real number greater than 0)

119906 isin 1205971198701199031

then 11990312119868(119902) = 217Φ(119905) le 36 and take 120590 = 120

then 119879(119906) le 119906 obviously holds for 120582 120583 lt 120Choosing 119903

2= 9 it is easy to verify that

119879 (119906) ge 119906 forall119906 isin 1205971198701199032

(31)

From Theorem 1 119879 has a fixed point in 1198701199032

1198701199031

that isproblem (29) has at least one positive solution

For any 119905 isin [0 1] there is a constant 119888 = 3 so that

119891 (119905 119906) ge 3119906 forall119906 ge 0 (32)

If 120591 = 12 then problem (29) has no positive solution for120582 gt 450

Conflict of Interests

The authors declare that there is no conflict of interestsregarding the publication of this paper

Acknowledgments

The authors are very grateful to the anonymous referees fortheir valuable suggestionsThis work was supported byNSFC(nos 11361054 and 11201378) SRFDP (no 2012 6203110004)and Gansu Provincial National Science Foundation of China(no 1208RJZA258)

References

[1] R P AgarwalBoundary Value Problems for Higher Order Differ-ential Equations World Scientific Publishing Singapore 1986

[2] A R Aftabizadeh ldquoExistence and uniqueness theorems forfourth-order boundary value problemsrdquo Journal of Mathemati-cal Analysis and Applications vol 116 no 2 pp 415ndash426 1986

[3] R P Agarwal ldquoOn fourth order boundary value problemsarising in beam analysisrdquo Differential and Integral Equationsvol 2 no 1 pp 91ndash110 1989

[4] J R Graef and B Yang ldquoExistence and nonexistence of positivesolutions of fourth order nonlinear boundary value problemsrdquoApplicable Analysis vol 74 no 1-2 pp 201ndash214 2000

[5] R Ma and H Wang ldquoOn the existence of positive solutionsof fourth-order ordinary differential equationsrdquo ApplicableAnalysis vol 59 no 1ndash4 pp 225ndash231 1995

[6] Q Yao ldquoPositive solutions of nonlinear beam equations withtime and space singularitiesrdquo Journal of Mathematical Analysisand Applications vol 374 no 2 pp 681ndash692 2011

[7] Q Yao ldquoMonotonically iterativemethod of nonlinear cantileverbeam equationsrdquo Applied Mathematics and Computation vol205 no 1 pp 432ndash437 2008

[8] A Cabada J Cid and L Sanchez ldquoPositivity and lower andupper solutions for fourth order boundary value problemsrdquoNonlinear Analysis Theory Methods amp Applications vol 67 no5 pp 1599ndash1612 2007

[9] Z Bai ldquoThe upper and lower solution method for some fourth-order boundary value problemsrdquo Nonlinear Analysis TheoryMethods amp Applications vol 67 no 6 pp 1704ndash1709 2007

[10] Z Bai B Huang and W Ge ldquoThe iterative solutions for somefourth-order p-Laplace equation boundary value problemsrdquoApplied Mathematics Letters vol 19 no 1 pp 8ndash14 2006


[11] Q Yao ldquoLocal existence of multiple positive solutions to asingular cantilever beam equationrdquo Journal of MathematicalAnalysis and Applications vol 363 no 1 pp 138ndash154 2010

[12] L Yang H Chen and X Yang ldquoThemultiplicity of solutions forfourth-order equations generated from a boundary conditionrdquoApplied Mathematics Letters vol 24 no 9 pp 1599ndash1603 2011

[13] S Li and X Zhang ldquoExistence and uniqueness of monotonepositive solutions for an elastic beam equation with nonlinearboundary conditionsrdquo Computers ampMathematics with Applica-tions vol 63 no 9 pp 1355ndash1360 2012

[14] A Cabada and S Tersian ldquoMultiplicity of solutions of a twopoint boundary value problem for a fourth-order equationrdquoAppliedMathematics andComputation vol 219 no 10 pp 5261ndash5267 2013

[15] E Alves T F Ma and M L Pelicer ldquoMonotone positive solu-tions for a fourth order equation with nonlinear boundary con-ditionsrdquo Nonlinear Analysis Theory Methods amp Applicationsvol 71 no 9 pp 3834ndash3841 2009

[16] T F Ma ldquoPositive solutions for a beam equation on a nonlinearelastic foundationrdquoMathematical and Computer Modelling vol39 no 11-12 pp 1195ndash1201 2004

[17] B Yang ldquoPositive solutions for the beam equation under certainboundary conditionsrdquo Electronic Journal of Differential Equa-tions vol 78 pp 1ndash8 2005

[18] D J Guo and V Lakshmikantham Nonlinear Problems inAbstract Cones Academic Press New York NY USA 1988

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Volume 2014 Hindawi Publishing Corporationhttpwwwhindawicom Volume 2014

Stochastic AnalysisInternational Journal of


Remark 3 The results obtained in this paper are not a con-sequence of the previous theorem in the previous literatureClearly the boundary condition of (1) is more general thanthe above pieces of literature and problem (1) considers twodifferent parameters which is more extensive

Remark 4 It is pointed out thatwe donot need anymonotoneassumption on 119891 119892 and ℎ which is weaker than thecorresponding assumptions on119891119892 in [13] References [12 14]are only consideredwith the existence of solution for problem(1) with nonlinear boundary condition however we study theexistence and nonexistence of positive solution of (1)

The remainder of this paper is arranged as followsSection 2 presents some preliminaries The proofs of Theo-rems 1 and 2 are given in Section 3 Finally we give a simpleexample to illustrate our main results

2 Preliminaries

In this section we collect some preliminary results that willbe used in subsequent sections Let 119883 = 119862[0 1] then 119883 is aBanach space under the norm 119906 = max

119905isin[01]|119906(119905)| Define

119875 = 119906 isin 119883 | 119906(119905) ge 0 119905 isin [0 1] then119875 sub 119883 is a nonnegativecone

Lemma 5 (see [11 Lemma 22]) Let ℎ isin 119862[0 1] then thesolution 119906(119905) of the problem

1199061015840101584010158401015840

(119905) = ℎ (119905) 119905 isin (0 1)

119906 (0) = 1199061015840

(0) = 11990610158401015840

(1) = 119906101584010158401015840

(1) = 0

(2)

is

119906 (119905) = int

1

0

119866 (119905 119904) ℎ (119904) 119889119904 (3)

where

119866 (119905 119904) =

1

6

1199052(3119904 minus 119905) 0 le 119905 le 119904 le 1

1

6

1199042(3119905 minus 119904) 0 le 119904 le 119905 le 1

(4)

Now let us set

120601 (119905) =

1199052

2

minus

1199053

6

Φ (119905) = ℎ (119906 (0)) 119905 minus 119892 (119906 (1)) 120601 (119905)

forall119905 isin [0 1]

(5)

It follows from (3) and (5) with simple computation thatproblem (1) is equivalent to integral equation

119906 (119905) = 120582int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 + 120583Φ (119905) = 119879119906 (119905) (6)

Note that for any 119906 isin 119883 the function (119879119906)(119905) satisfiesthe boundary conditions in (1) by the definition of Greenrsquosfunction 119866(119905 119904) In view of Lemma 5 it is easy to see that119906 isin 119883 is a fixed point of the operator 119879 if and only if 119906 isin 119883 isa solution of problem (1)

Lemma 6 (see [6 Lemma 22] [13 Lemma 21]) 119866(119905 119904) 120601(119905)have the following properties

(i) 119866(119905 119904) gt 0 forall0 lt 119905 119904 lt 1(ii) max

119905isin[01]119866(119905 119904) le 119866(1 119904) = (16)119904

2(3minus119904) forall0 le 119904 le 1

(iii) 119866(119905 119904) ge 1199052119866(1 119904) ge (119905

23)119866(1 119904) forall0 le 119905 119904 le 1

(iv) 11990523 le 120601(119905) le 11990522 forall0 le 119905 le 1

Lemma 7 Φ(119905) has the following properties

(i) Φ(119905) le Φ(1) le ℎ(119906(0)) minus 119892(119906(1)) forall0 lt 119905 lt 1(ii) Φ(119905) ge Φ(119905

23) ge 0 forall0 lt 119905 lt 1

Proof First we will prove (i) For any 119905 isin [0 1] it follows that

Φ1015840

(119905) = ℎ (119906 (0)) minus 119892 (119906 (1)) (119905 minus

1

2

1199052) (7)

From (A2)Φ1015840(119905) ge 0 SoΦ(119905) is nondecreasing for 119905 isin [0 1]Moreover

Φ (119905) le Φ (1) = ℎ (119906 (0)) minus

1

3

119892 (119906 (1)) le ℎ (119906 (0)) minus 119892 (119906 (1))

(8)

so (i) holdsNowwe prove (ii) For any 119905 isin [0 1] obviously 119905 ge 119905

23 ge

0From (i)Φ(119905) is nondecreasing for 119905 isin [0 1] thenΦ(119905) ge

Φ(11990523) ge Φ(0) = 0

For any fixed 120591 isin (0 1) define the cone

119870 = 119906 isin 119875 | min119905isin[1205911]

119906 (119905) ge

1205912

3

119906 (9)

Then119870 sub 119883 is a positive cone letting119870119903= 119906 isin 119870 | 119906 lt 119903

120597119870119903= 119906 isin 119870 | 119906 = 119903 with 119903 gt 0 is a constant

Lemma 8 Assume that (A1) and (A2) hold then 119879 119870 rarr 119870

is completely continuous

Proof Obviously for any 119905 isin [0 1] 119906 isin 119870 it follows that119879119906(119905) ge 0

For any 119906 isin 119870 from (6) and Lemma 7(i) we have

max119905isin[01]

119879 (119906 (119905)) = max119905isin[01]

(120582int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904 + 120583Φ (119905))

le 120582int

1

0

119866 (1 119904) 119891 (119904 119906 (119904)) 119889119904

+ 120583 (ℎ (119906 (0)) minus 119892 (119906 (1)))

(10)

Hence

119879119906 le 120582int

1

0

119866 (1 119904) 119891 (119904 119906 (119904)) 119889119904 + 120583 (ℎ (119906 (0)) minus 119892 (119906 (1)))

(11)



min120591le119905le1

119879 (119906 (119905)) = min120591le119905le1

(120582int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904

+ 120583 (ℎ (119906 (0)) 119905 minus 119892 (119906 (1)) 120601 (119905)) )

ge

1205912

3

(120582int

1

0

119866 (1 119904) 119891 (119904 119906 (119904)) 119889119904

+120583 (ℎ (119906 (0)) minus 119892 (119906 (1))) )

ge

1205912

3

119879119906

(12)







119903=

119906 isin 119870 | 119906 = 119903


119903 119870) = 0


119903 119870) = 1




119872 = max(119905119904)isin[01]times[01]

119866 (119905 119904) gt 0 (13)

Let 119902 gt 0 set

119868 (119902) = 119872 max119906isin119870119906=119902

(int

1

0

119891 (119904 119906 (119904)) 119889119904) gt 0 (14)


1199031

= 119906 isin 119870 | 119906 le 1199031 Choose

120590 gt 0 so that

120590 le

1199031

2119868 (1199031)

forall119906 isin 1205971198701199031

120590 (ℎ (119906 (0)) minus 119892 (119906 (1))) le

1199031

2

forall119906 isin 1205971198701199031

(15)



119879119906 (119905) le 120582119872int

1

0

119891 (119904 119906 (119904)) 119889119904

+ 120583 (ℎ (119906 (0)) minus 119892 (119906 (1)))

le 120590119872int

1

0

119891 (119904 119906 (119904)) 119889119904 + 120590 (ℎ (119906 (0)) minus 119892 (119906 (1)))

le 120590119868 (1199031) +

1199031

2

le 1199031

(16)

which implies that

119879119906 le 1199031= 119906 119906 isin 120597119870

1199031

(17)


119894 (1198791198701199031

119870) = 1 (18)


such that

119891 (119905 119906) ge 120578119906 forall119906 ge 119901 (19)


120582(

1205912

3

)

2

120578int

1

120591

119866 (1 119904) 119889119904 ge 1 forall120591 isin (0 1) (20)


1 and set

1198701199032

= 119906 isin 119870 | 119906 le 1199032 If 119906 isin 120597119870

1199032

then

min120591le119905le1

119906 (119905) ge

1205912

3

1199032gt 119901 (21)


min119905isin[1205911]

119879 (119906 (119905)) ge min119905isin[1205911]

(120582int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904)

ge 120582

1205912

3

int

1

0

119866 (1 119904) 119891 (119904 119906 (119904)) 119889119904

ge 120582

1205912

3

int

1

0

119866 (1 119904) 120578119906 (119904) 119889119904

ge 120582

1205912

3

120578int

1

120591

119866 (1 119904) 119906 (119904) 119889119904

ge 120582(

1205912

3

)

2

120578int

1

120591

119866 (1 119904) 119889119904 119906

ge 119906

(22)

which implies that

119879 (119906) ge 119906 119906 isin 1205971198701199032

(23)



119894 (1198791198701199032

119870) = 0 (24)


119894 (1198791198701199032

1198701199031

119870) = 119894 (1198791198701199032

119870) minus 119894 (1198791198701199031

119870) = minus1 (25)


1198701199031



119891 (119905 119906) ge 119888119906 forall119906 ge 0 (26)


120582(

1205912

3

)

2

119888 int

1

120591

119866 (1 119904) 119889119904 gt 1 forall120591 isin (0 1) (27)


min119905isin[1205911]

119906 (119905) ge min119905isin[1205911]

(120582int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904)

ge 120582

1205912

3

int

1

120591

119866 (1 119904) 119888119906 (119904) 119889119904

ge 120582(

1205912

3

)

2

119888 int

1

120591

119866 (1 119904) 119889119904 119906

gt 119906

(28)




1199061015840101584010158401015840

(119905) = 120582 (1199063

(119905) (3 + sin (119906 (119905))) + 1199052 + 1) 119905 isin (0 1)

119906 (0) = 0 1199061015840

(0) = 120583 (2[119906 (0)]2+ 1)

11990610158401015840

(1) = 0 119906101584010158401015840

(1) = 120583 ([minus (119906 (1))]12

minus 1)

(29)


119891 (119905 119906) = 1199063

(119905) (3 + sin (119906 (119905))) + 1199052 + 1

(119905 119906) isin [0 1] times [0infin)

ℎ (119904) = 21199042+ 1 119892 (119904) = minus119904


(30)


If 1199031= 4 (where 119903


119906 isin 1205971198701199031

then 11990312119868(119902) = 217Φ(119905) le 36 and take 120590 = 120



119879 (119906) ge 119906 forall119906 isin 1205971198701199032

(31)


1198701199031



119891 (119905 119906) ge 3119906 forall119906 ge 0 (32)




Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











min120591le119905le1

119879 (119906 (119905)) = min120591le119905le1

(120582int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904

+ 120583 (ℎ (119906 (0)) 119905 minus 119892 (119906 (1)) 120601 (119905)) )

ge

1205912

3

(120582int

1

0

119866 (1 119904) 119891 (119904 119906 (119904)) 119889119904

+120583 (ℎ (119906 (0)) minus 119892 (119906 (1))) )

ge

1205912

3

119879119906

(12)







119903=

119906 isin 119870 | 119906 = 119903


119903 119870) = 0


119903 119870) = 1




119872 = max(119905119904)isin[01]times[01]

119866 (119905 119904) gt 0 (13)

Let 119902 gt 0 set

119868 (119902) = 119872 max119906isin119870119906=119902

(int

1

0

119891 (119904 119906 (119904)) 119889119904) gt 0 (14)


1199031

= 119906 isin 119870 | 119906 le 1199031 Choose

120590 gt 0 so that

120590 le

1199031

2119868 (1199031)

forall119906 isin 1205971198701199031

120590 (ℎ (119906 (0)) minus 119892 (119906 (1))) le

1199031

2

forall119906 isin 1205971198701199031

(15)



119879119906 (119905) le 120582119872int

1

0

119891 (119904 119906 (119904)) 119889119904

+ 120583 (ℎ (119906 (0)) minus 119892 (119906 (1)))

le 120590119872int

1

0

119891 (119904 119906 (119904)) 119889119904 + 120590 (ℎ (119906 (0)) minus 119892 (119906 (1)))

le 120590119868 (1199031) +

1199031

2

le 1199031

(16)

which implies that

119879119906 le 1199031= 119906 119906 isin 120597119870

1199031

(17)


119894 (1198791198701199031

119870) = 1 (18)


such that

119891 (119905 119906) ge 120578119906 forall119906 ge 119901 (19)


120582(

1205912

3

)

2

120578int

1

120591

119866 (1 119904) 119889119904 ge 1 forall120591 isin (0 1) (20)


1 and set

1198701199032

= 119906 isin 119870 | 119906 le 1199032 If 119906 isin 120597119870

1199032

then

min120591le119905le1

119906 (119905) ge

1205912

3

1199032gt 119901 (21)


min119905isin[1205911]

119879 (119906 (119905)) ge min119905isin[1205911]

(120582int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904)

ge 120582

1205912

3

int

1

0

119866 (1 119904) 119891 (119904 119906 (119904)) 119889119904

ge 120582

1205912

3

int

1

0

119866 (1 119904) 120578119906 (119904) 119889119904

ge 120582

1205912

3

120578int

1

120591

119866 (1 119904) 119906 (119904) 119889119904

ge 120582(

1205912

3

)

2

120578int

1

120591

119866 (1 119904) 119889119904 119906

ge 119906

(22)

which implies that

119879 (119906) ge 119906 119906 isin 1205971198701199032

(23)



119894 (1198791198701199032

119870) = 0 (24)


119894 (1198791198701199032

1198701199031

119870) = 119894 (1198791198701199032

119870) minus 119894 (1198791198701199031

119870) = minus1 (25)


1198701199031



119891 (119905 119906) ge 119888119906 forall119906 ge 0 (26)


120582(

1205912

3

)

2

119888 int

1

120591

119866 (1 119904) 119889119904 gt 1 forall120591 isin (0 1) (27)


min119905isin[1205911]

119906 (119905) ge min119905isin[1205911]

(120582int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904)

ge 120582

1205912

3

int

1

120591

119866 (1 119904) 119888119906 (119904) 119889119904

ge 120582(

1205912

3

)

2

119888 int

1

120591

119866 (1 119904) 119889119904 119906

gt 119906

(28)




1199061015840101584010158401015840

(119905) = 120582 (1199063

(119905) (3 + sin (119906 (119905))) + 1199052 + 1) 119905 isin (0 1)

119906 (0) = 0 1199061015840

(0) = 120583 (2[119906 (0)]2+ 1)

11990610158401015840

(1) = 0 119906101584010158401015840

(1) = 120583 ([minus (119906 (1))]12

minus 1)

(29)


119891 (119905 119906) = 1199063

(119905) (3 + sin (119906 (119905))) + 1199052 + 1

(119905 119906) isin [0 1] times [0infin)

ℎ (119904) = 21199042+ 1 119892 (119904) = minus119904


(30)


If 1199031= 4 (where 119903


119906 isin 1205971198701199031

then 11990312119868(119902) = 217Φ(119905) le 36 and take 120590 = 120



119879 (119906) ge 119906 forall119906 isin 1205971198701199032

(31)


1198701199031



119891 (119905 119906) ge 3119906 forall119906 ge 0 (32)




Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra











119894 (1198791198701199032

119870) = 0 (24)


119894 (1198791198701199032

1198701199031

119870) = 119894 (1198791198701199032

119870) minus 119894 (1198791198701199031

119870) = minus1 (25)


1198701199031



119891 (119905 119906) ge 119888119906 forall119906 ge 0 (26)


120582(

1205912

3

)

2

119888 int

1

120591

119866 (1 119904) 119889119904 gt 1 forall120591 isin (0 1) (27)


min119905isin[1205911]

119906 (119905) ge min119905isin[1205911]

(120582int

1

0

119866 (119905 119904) 119891 (119904 119906 (119904)) 119889119904)

ge 120582

1205912

3

int

1

120591

119866 (1 119904) 119888119906 (119904) 119889119904

ge 120582(

1205912

3

)

2

119888 int

1

120591

119866 (1 119904) 119889119904 119906

gt 119906

(28)




1199061015840101584010158401015840

(119905) = 120582 (1199063

(119905) (3 + sin (119906 (119905))) + 1199052 + 1) 119905 isin (0 1)

119906 (0) = 0 1199061015840

(0) = 120583 (2[119906 (0)]2+ 1)

11990610158401015840

(1) = 0 119906101584010158401015840

(1) = 120583 ([minus (119906 (1))]12

minus 1)

(29)


119891 (119905 119906) = 1199063

(119905) (3 + sin (119906 (119905))) + 1199052 + 1

(119905 119906) isin [0 1] times [0infin)

ℎ (119904) = 21199042+ 1 119892 (119904) = minus119904


(30)


If 1199031= 4 (where 119903


119906 isin 1205971198701199031

then 11990312119868(119902) = 217Φ(119905) le 36 and take 120590 = 120



119879 (119906) ge 119906 forall119906 isin 1205971198701199032

(31)


1198701199031



119891 (119905 119906) ge 3119906 forall119906 ge 0 (32)




Acknowledgments


References



























Volume 2014




Journal of











Journal of


Function Spaces






Algebra

























Volume 2014




Journal of











Journal of


Function Spaces






Algebra









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