Open access peer-reviewed chapter

Fixed Point Theory Approach to Existence of Solutions with Differential Equations

By Piyachat Borisut, Konrawut Khammahawong and Poom Kumam

Submitted: November 25th 2017Reviewed: January 28th 2018Published: May 23rd 2018

DOI: 10.5772/intechopen.74560

Downloaded: 824

Abstract

In this chapter, we introduce a generalized contractions and prove some fixed point theorems in generalized metric spaces by using the generalized contractions. Moreover, we will apply the fixed point theorems to show the existence and uniqueness of solution to the ordinary difference equation (ODE), Partial difference equation (PDEs) and fractional boundary value problem.

Keywords

  • fixed point
  • contraction
  • generalized contraction
  • differential equation
  • partial differential equation
  • fractional difference equation

1. Introduction

The study of differential equations is a wide field in pure and applied mathematics, chemistry, physics, engineering and biological science. All of these disciplines are concerned with the properties of differential equations of various types. Pure mathematics investigated the existence and uniqueness of solutions, but applied mathematics focuses on the rigorous justification of the methods for approximating solutions. Differential equations play an important role in modeling virtually every physical, technical, or biological process, from celestial motion, to bridge design, to interactions between neurons. Differential equations such as those used to solve real-life problems may not necessarily be directly solvable, i.e. do not have closed form solutions. Instead, solutions can be approximated using numerical methods.

Following the ordinary differential equations with boundary value condition

dnxdtn=ftxdxdtdn1xdtn1

where yx0=0,yx1=c1,,yn1xn1=cn1the positive integer n(the order of the highest derivative). This will be discussed. Existence and uniqueness of solution for initial value problem (IVP).

ut=ftutut0=u0.

Differential equations contains derivatives with respect to two or more variables is called a partial differential equation (PDEs). For example,

A2ux2+B2uxy+C2uy2+Dux+Euy+Fu=G

where uis dependent variable and A,B,C,D,E,Fand Gare function of x,yabove equation is classified according to discriminant B24ACas follows,

  1. Elliptic equation if B24AC<0,

  2. Hyperbolic equation if B24AC>0,

  3. Parabolic equation if B24AC=0.

This will be discussed. Existence of solution for semilinear elliptic equation. Consider a function u:ΩRnRnthat solves,

Δu=fuinΩu=u0on∂Ω

where f:RmRmis a typically nonlinear function. And fractional differential equations. This will be discussed. Fractional differential equations are of two kinds, they are Riemann-Liouville fractional differential equations and Caputo fractional differential equations with boundary value.

cDtαut=But;t>0u0=u0X

where cDtαis the Caputo fractional derivative of order α01, and t0τ, for all τ>0.

The following fractional differential equation will boundary value condition.

D0+αut+ftut=0,0<t<1,1<α2u0=0,u1=01usds,

where f:01×00is a continuous function and D0+αis the standard Riemann-Liouville fractional derivative.

One method for existence and uniqueness of solution of difference equation due to fixed point theory. The primary result in fixed point theory which is known as Banach’s contraction principle was introduced by Banach [1] in 1922.

Theorem 1.1. Let Xdbe a complete metric spaces and T:XXbe a contraction mapping (that is, there exists 0α<1) such that

dTxTyαdxy

for all x,yX, then Thas a unique fixed point.

Since Banach contraction is a very popular and important tool for solving many kinds of mathematics problems, many authors have improved, extended and generalized it (see in [24]) and references therein.

In this chapter, we discuss on the existence and uniqueness of the differential equations by using fixed point theory to approach the solution.

2. Basic results

Throughout the rest of the chapter unless otherwise stated Xdstands for a complete metric space.

2.1. Fixed point

Definition 2.1. Let Xbe a nonempty set and T:XXbe a mapping. A point xXis said to be a fixed point of Tif Tx=x.

Definition 2.2. Let Xdbe a metric space. The mapping T:XXis said to be Lipschitzian if there exists a constant α>0(called Lipschitz constant) such that

dTxTyαdxyforallx,yX.

A mapping Twith a Lipschitz constant α<1is called contraction.

Definition 2.3. Let Fand Xbe normed spaces over the field K, T:FXan operator and cF. We say that Tis continuous at cif for every ε>0there exists δ>0such that TxTc<ϵwhenever xc<δand xF. If Tis continuous at each xF, then Tis said to be continuous on T.

Definition 2.4. Let Xand Ybe normed spaces. The mapping T:XYis said to be completely continuous if TCis a compact subset of Yfor every bounded subset Cof X.

Definition 2.5. Compact operator is a linear operator Lform a Banach space Xto another Banach space Y, such that the image under Lof any bounded subset of Xis a relatively compact subset (has compact closure) of Ysuch an operator is necessarily a bounded operator, and so continuous.

Definition 2.6. A subset Cof a normed linear space Xis said to be convex subset in Xif λx+1λyCfor each x,yCand for each scalar λ01.

Definition 2.7. vis called the αthweak derivative of u

Dαu=v

if

ΩuDαψdx=1αΩvψdx

for all test function ψCcΩ.

Theorem 2.8. (Schauder’s Fixed Point Theorem) Let Xbe a Banach space, MXbe nonempty, convex, bounded, closed and T:MXMbe a compact operator. Then Thas a fixed point.

Lemma 2.9. ref. [5] Given fCRsuch that fta=btrwhere a>0,b>0and r>0are positive constants. Then the map ufuis continuous for LpΩto LprΩfor pmax1rand maps bounded subset of LpΩto bounded subset of LprΩ.

Proof. Form to Jensen’s inequality

a+btrpr2pr1apr+2pr1bprtpC1+tp

where Cis a positive constant depending on a,b,pand ronly, since uLpΩ, we have

ΩfuprdxCabprΩ+Ωupdx<

therefore fuLprΩ. Let unbe a sequence converging to uin LpΩ. There exists a subsequence un, and a function gLpΩsuch that set, un'ux, and un'xgx, almost everywhere. This is sometimes called the generalized DCT, or the partial converse of the DCT, or the Riesz-Fisher Theorem. From the continuity of f, fuxfun'0on Ω\, and

fuxfun'prC1+gxp+fup

where Cis another positive constant depending on a,b,pand ronly, the left-hand-side is independent of n'and is in L1Ω. We can apply the Dominated Convergence Theorem to conclude the

Ωfuxfun'prdx0

or in other words, fuxfun'LprΩ0.Since the limit does not depend on the subsequence this convergence uholds for un. □

Corollary 2.10. ref. [5] Let μ0. Then the map gΔ+μId1gis

  1. continuous as map from L2Ωto H01Ωin other words

    vH01ΩCΩgL2Ω.

  2. compact as map form L2Ωto L2Ω.

Proof. The first part is due to the fact that L2Ωis continuously in H1Ω. The second part follows as Δ+μId1:L2ΩL2Ωcan be viewed as composition of the continuous map Δ+μId1:L2ΩH01Ωand the compact embedding H01ΩL2Ωand as the composition of a compact linear operator a continuous linear operator is again compact.

Theorem 2.11. (Poincare) For p1, there exists a constant C=CΩpsuch that W01,pΩ; uLpΩCuLpΩ:Rn. A key tool to obtain the compactness of the fixed point maps.

2.2. Fuzzy

A fuzzy set in Xis a function with domain X and values in 01. If Ais a fuzzy set on Xand xX,then the functional value Axis called the grade of membership of xin A. The αlevel set of A, denoted by Aαis defined by

Aα=x:Axαifα01,A0=x:Ax>0¯,

where denotes by A¯the closure of the set A. For any Aand Bare subset of Xwe denote by HABthe Huasdorff distance.

Definition 2.12. A fuzzy set Ain a metric linear space is called an approximate quantity if and only if Aαis convex and compact in Xfor each α01and supxXAx=1.

Let I=01and WXIXbe the collection of all approximate in X. For α01,the family WαXis given by {AIX:Aαis nonempty and compact}.

For a metric space Xdwe denoted by VXthe collection of fuzzy sets Ain Xfor which Aαis compact and supAx=1for all α01. Clearly, when Xis a metric linear space WXVX.

Definition 2.13. Let A,BVX,α01.Then

pαAB=infxAα,yBαdxy,DαAB=HAαBα

where His the Hausdorff distance.

Definition 2.14. Let A,BVX.Then Ais said to be more accurate than B(or Bincludes A), denoted by AB, if and only if AxBxfor each xX.

Denote with Φ,the family of nondecreasing function ϕ:0+0+such that n=1ϕnt<for all t>0.

Theorem 2.15. ref. [6] Let Xdbe a complete ordered metric space and T1,T2:XWαXbe two fuzzy mapping satisfying

DαT1xT2yϕMxy+LminpαxT1xpαyT2ypαxT2ypαyT1x

for all comparable element x,yX,where L0and

Mxy=maxdxypα(xT1x)pα(yT2y)12pαxT2y+pα(yT1x).

Also suppose that

  1. if yT1x0α,then y,x0Xare comparable,

  2. if x,yXare comparable, then every uT1xαand every vT2yαare comparable,

  3. if a sequence xnin Xconverges to xXand its consecutive terms are comparable, then xnand xare comparable for all n.

Then there exists a point xXsuch that xαT1xand xαT2x.

Proof. See in [6].

Corollary 2.16. ref. [6] Let Xdbe a complete ordered metric space and T1,T2:XWαXbe two fuzzy mappings satisfying

DαT1xT2yqmaxdxypα(xT1x)pα(yT2y)12[pα(xT2y)+pα(yT1x)]

for all comparable elements x,yX. Also suppose that

  1. if yT1x0α,then y,x0Xare comparable,

  2. if x,yXare comparable, then every uT1xαand every vT2yαare comparable,

  3. if a sequence xnin Xconverges to xXand its consecutive terms are comparable, then xnand xare comparable for all n.

Then there exists a point xXsuch that xαT1xand xαT2x.

2.3. Metric-like space

Definition 2.17. [7] Let Xbe nonempty set and function p:X×XR+be a function satisfying the following condition: for all x,y,zX,

p1pxx=pxy=pyyif and only ifx=y,p2pxxpxy,p3pxx=pyx,p4pxy=pxz+pzypzz.

Then pis called a partial metric on X, so a pair Xpis said to be a partial metric space.

Definition 2.18. [8] A metric-like on nonempty set Xis a function σ:X×XR+. If for all x,y,zX, the following conditions hold:

σ1σxy=0x=y;σ2σxy=σyx;σ3σxy=σxz+σzy.

Then a pair Xσis called a metric-like space.

It is easy to see that a metric space is a partial metric space and each partial metric space is a metric-like space, but the converse is not true but the converse is not true as in the following examples:

Example 2.19. [8] Let X=01and σ:X×XR+be defined by

σxy=2,ifx=y=0,1,otherwise.

Then Xσis a metric-like space, but it is not a partial metric space, cause σ00σ01.

Lemma 2.20. ref. [9] Let Xpbe a partial metric space. Then

  1. xnis a Cauchy sequence in Xpif and only if it is a Cauchy sequence in the metric space Xdp,

  2. Xis complete if and only if the metric space Xdpis complete.

Definition 2.21. [8, 10] Let Xσbe a metric-like space. Then:

  1. A sequence xnin Xconverges to a point xXif limnσxnx=σxx.The sequence xnis said to be σCauchy if limn,mσxnxmexists and is finite. The space Xσis called complete if for every σCauchy sequence in xn, there exists xXsuch that

    limnσxnx=σxx=limn,mσxnxm.

  2. A sequence xnin Xσis said to be a 0σCauchy sequence if limn,mσxnxm=0.The space Xσis said to be 0σcomplete if every 0σCauchy sequence in Xconverges (in τσ) to a point xXsuch that σxx=0.

  3. A mapping T:XXis continuous, if the following limits exist (finite) and

    limnσxnx=σTxx.

Following Wardowski [11], we denote by Fthe family of all function, F:R+Rsatisfying the following conditions:

  1. (F1) Fis strictly increasing on R+,

  2. (F2) for every sequence snin R+,we have limnsn=0if and only if limnFsn=,

  3. (F3) there exists a number k01such that lims0+skFs=0.

Example 2.22. The following function F:R+Rbelongs to F:

  1. Fs=lns,with s>0,

  2. Fs=lns+s,with s>0.

Definition 2.23. [11] Let Xdbe a metric space. A self-mapping Ton Xis called an F-contraction mapping if there exist FFand τR+such that

x,yX,dTxTy>0τ+FdTxTyFdxy.E2.1

Definition 2.24. [12] Let Xσbe a metric-like space. A mapping T:XXis called a generalized Roger Hardy type Fcontraction mapping, if there exist FFand τR+such that

σTxTy>0τ+FσTxTyF(ασxy+βσxTx+γσyTy+ησxTy+δσyTx)E2.2

for all x,yXand α,β,γ,η,δ0with α+β+γ+2η+2δ<1.

Theorem 2.25. ref. [12] Let Xσbe 0σcomplete metric-like spaces and T:XXbe a generalized Roger Hardy type Fcontraction. Then Thas a unique fixed point in X, either Tor Fis continuous.

Proof. See in [12]. □

2.4. Modular metric space

Let Xbe a nonempty set. Throughout this paper, for a function ω:0×X×X0,we write

ωλxy=ωλxy

for all λ>0and x,yX.

Definition 2.26 [13, 14] Let Xbe a nonempty set. A function ω:0×X×X0is called a metric modular on Xif satisfying, for all x,y,zXthe following conditions hold:

  1. ωλxy=0for all λ>0if and only if x=y,

  2. ωλxy=ωλyxfor all λ>0,

  3. ωλ+μxyωλxz+ωμzyfor all λ,μ>0.

If instead of (i) we have only the condition (i’)

ωλxx=0forallλ>0,xX,

then ωis said to be a pseudomodular (metric) on X.A modular metric ωon Xis said to be regular if the following weaker version of (i) is satisfied:

x=yif and only if ωλxy=0for some λ>0.

Note that for a metric (pseudo)modular ωon a set X,and any x,yX,the function λωλxyis nonincreasing on 0.Indeed, if 0<μ<λ,then

ωλxyωλμxx+ωμxy=ωμxy.

Note that every modular metric is regular but converse may not necessarily be true.

Example 2.27. Let X=Rand ωis defined by ωλxy=if λ<1,and ωλxy=xyif λ1,it is easy to verify that ωis regular modular metric but not modular metric.

Definition 2.28. [13, 14] Let Xωbe a (pseudo)modular on X.Fix x0X.The two sets

Xω=Xωx0=xX:ωλxx00asλ

and

Xω=Xωx0=xX:λ=λx>0such thatωλxx0<

are said to be modular spaces (around x0).

Throughout this section we assume that Xωis a modular metric space, Dbe a nonempty subset of Xωand G{Gωis a directed graph with VGω=Dand ΔEGω}.

Definition 2.29. [15, 16] The pair DGωhas Property (A) if for any sequence xnnin D, with xnxas nand xnxn+1EGω,then xnxEGω,for all n.

Definition 2.30. ref. [17] Let FFand GωG.A mapping T:DDis said to be F-Gω-contraction with respect to R:DDif

  1. RxRyEGωTxTyEGωfor all x,yD,i.e. Tpreserves edges w.r.t. R,

  2. there exists a number τ>0such that

ω1TxTy>0τ+Fω1TxTyFω1RxRy

for all x,yDwith RxRyEGω.

Example 2.31. ref. [17] Let FFbe arbitrary. Then every F-contractive mapping w.r.t. Ris an F-Gω-contraction w.r.t. Rfor the graph Gωgiven by VGω=Dand EGω=D×D.

We denote CTRxD:Tx=Rxthe set of all coincidence points of two self-mappings Tand R, defined on D.

Theorem 2.32. ref. [17] Let Xωbe a regular modular metric space with a graph Gω.Assume that D=VGωis a nonempty ω-bounded subset of Xωand the pair DGωhas property (A) and also satisfy ΔM-condition. Let R,T:DDbe two self mappings satisfying the following conditions:

  1. there exists x0Dsuch that Rx0Tx0EGω,

  2. Tis an F-Gω-contraction w.r.t R,

  3. TDRD,

  4. RDis ωcomplete.

Then CRTØ.

Proof. See in [17].□

3. Fixed point approach to the solution of differential equations

Next, we will show a differential equation which solving by fixed point theorem in suitable spaces.

3.1. Ordinary differential equation

Lemma 3.1. ref. [18] uis a solution of ut=ftutsatisfying the initial condition ut0=u0if and only if ut=u0+t0tfsusds.

Proof. Suppose that uis a solution of ut=ftutdefined on an interval Iand satisfying ut0=u0. We integrate both sides of the equation ut=ftutfrom t0to t, where tis in I

t0tusds=t0tfsusdsutut0=t0tfsusds.

Since ut0=u0, we have

ut=u0+t0tfsusds,tI.E3.1

We will show that, conversely, any function which satisfies this integral equation satisfies both the differential equation and the initial condition. Suppose that uis a function defined on an interval Iand satisfies (3.1). Setting t=t0yields ut0=u0, so that usatisfies the initial condition. Next, we note that an integral is always a continuous function, so that a solution of (3.1) is automatically continuous. Since both uand fare continuous, it follows that the integrand fsusis continuous. We may therefore apply the fundamental theorem of calculus to (3.1) and conclude that uis differentiable, and that is ut=ftut. □

The contraction mapping theorem may by used to prove the existence and uniqueness of the initial problem for ordinary differential equations. We consider a first-order of ODEs for a function utthat take value in Rn

u't=ftutE3.2
ut0=u0.E3.3

The function ftutalso take value in Rnand is assumed to be a continuous function of tand a Lipschitz continuous function of uon suitable domain.

Definition 3.2. Suppose that f:I×RnRnwhere Iis on interval in R. We say that ftutis a globally Lipschitz continuous function of uuniformly in t if there is a constant C>0such that

ftuftvCuvE3.4

for all x,yRnand all tI.

The initial value problem can be reformulated as an integral equation.

ut=u0+t0tfsusds.E3.5

By the fundamental theorem of calculus, a continuous solution of (3.5) is a continuously differentiable solution of (3.2). Eq. (3.5) may by written as fixed point equation.

u=Tu

for the map Tdefined by

Tut=u0+t0tfsusds.

Theorem 3.3. ref. [19] Suppose that f:I×RnRnwhere Iis on interval in Rand t0is a point in the interior of I. If ftu, is a continuous function of tuand a globally Lipschitz continuous function of uuniformly in ton I×Rn, then there is a unique continuously differentiable function u:IRnthat satisfies (3.2).

Proof. We will show that Tis a contraction on the space of continuous function defined on a time interval t0tt0+δ, for sufficiently small δ.

Suppose that u,v:t0t0+δRnare two continuous function. Then, form (3.4), (3.5) we estimate,

|TuTv|=supt0tt0+δ|TutTvt|=supt0tt0+δ|t0tfsusfsvsds|supt0tt0+δt0t|fsusfsvs|dssupt0tt0+δC|usvs|t0tds|uv|.

If follow that if δ1cthen Tis contraction on Ct0t0+δ. Therefore, there is a unique solution u:t0t0+δRn.

Let fxybe a continuous real-valued function on ab×cd. The Cauchy initial value problem is to find a continuous differentiable function yon absatisfying the differential equation

dydx=fxy,yx0=y0.E3.6

Consider the Banach space Cabof continuous real-valued functions with supremum norm defined by y=supyx:xab.

Integrating (3.6), that yield an integral equation

yx=y0+x0xftytdt.E3.7

The problem (3.6) is equivalent the problem solving the integral Eq. (3.7).

We define an integral operator T:CabCabby

Tyx=y0+x0xftytdt.

Therefore, a solution of Cauchy initial value problem (3.6) corresponds with a fixed point of T. One may easily check that if Tis contraction, then the problem (3.6) has a unique solution.

Theorem 3.4. ref. [20] Let fxybe a continuous function of Domf=ab×cdsuch that fis Lipschitzian with respect to y, i.e., there exists k>0such that

fxufxvkuvforallu,vcdandforxab.

Suppose x0y0intDomf.Then for sufficiently small h>0,there exists a unique solution of the problem (3.6).

Proof. Let M=supfxy:xyDomfand choose h>0such that

CyCx0hx0+h:yxy0Mh.

Then Cis a closed subset of the complete metric space Cx0hx0+hand hence Cis complete. Note T:CCis a contraction mapping. Indeed, for xx0hx0+hand two continuous functions y1,y2C, we have

Ty1Ty2=x0xfxy1fxy2dtxx0supsx0hx0+hky1sy2skhy1y2.

Therefore, T has a unique fixed point implying that the problem (3.6) has a unique fixed point.

3.2. Ordinary fuzzy differential equation

Now, we consider the existence of solution for the second order nonlinear boundary value problem:

xt=ktxtxt,t0Λ,Λ>0,xt1=x1,xt2=x2,t1,t20ΛE3.8

where k:0Λ×WX×WXWXis a continuous function. This problem is equivalent to the integral equation

xt=t1t2Gtsksxsxsds+βt,t0Λ,E3.9

where the Green’s function Gis given by

Gts=t2tst1t2t1ift1stt2,t2stt1t2t1ift1tst2,

and βtsatisfies β=0,βt1=x1,βt2=x2.Let us recall some properties of Gts,precisely we have

t1t2Gtsdst2t128

and

t1t2Gttsdst2t12.

If necessary, for a more detailed explanation of the background of the problem, the reader can refer to the reference [21, 22]. Here, we will prove our results, by establishing the existence of a common fixed point for pair of integral operators defined as

Tixt=t1t2Gtskisxsxsds+βt,t0Λ,i12E3.10

where k1,k2C0Λ×WX×WXWX,xC10ΛWX,and βC0ΛWX.

Theorem 3.5 ref. [6] Assume that the following conditions are satisfied:

  1. k1,k2:0Λ×WX×WXWXare increasing in its second and third variables,

  2. there exists x0C10ΛWXsuch that, for all t0Λ,we have

    x0tt1t2Gtsk1tx0sx0sds+βt,

    where t1,t20Λ,

  3. there exist constants γ,δ>0such that, for all t0Λ,we have

    k1txtxtk2tytytγxtyt+δxtyt

    for all comparable x,yC10ΛWX,

  4. for γ,δ>0and t1,t20Λwe have

    γt2t128+δt2t12<1,

  5. if x,yC10ΛWXare comparable, then every uT1x1and every vT2y1are comparable.

Then the pair of nonlinear integral equations

xt=t1t2Gtskisxsxsds+βtt0Λ,i12E3.11

has a common solution in C1t1t2WX.

Proof. Consider C=C1t1t2WXwith the metric

Dxy=maxt1tt2γxtyt+δxtyt.

The CDis a complete metric space, which can also be equipped with the partial ordering given by

x,yC,xtytforallt0Λ.

In [23], it is proved that Csatisfies the following condition:

(r) for every nondecreasing sequence xnin Cconvergent to some xC,we have xnxfor all n0.

Let T1,T2:CCbe two integral operators defined by (3.10); clearly, T1,T2are well defined since k1,k2,and βare continuous functions. Now, xis a solution of (3.9) if and only if xis a common fixed point of T1and T2.

By hypothesis (a), T1,T2are increasing and, by hypothesis (b), x0T1x0.Consequently, in view of condition (r), hypothesis (i)-(iii) of Corollary 2.16 hold true.

Next, for all comparable x,yC,From hypothesis (c) we obtain successively

T1xtT2ytt1t2Gtsk1sxsxsk1sysysdsDxyt1t2Gtsdst2t128DxyE3.12

and

T1x'tT2y'tt1t2Gttsk1sxsx'sk1sysy'sdsDxyt1t2Gttsdst2t12Dxy.E3.13

From (3.12) and (3.13), we obtain easily

DT1xT2yγt2t128+δt2t12Dxy.

Consequently, in view of hypothesis (d), the contractive condition (5) is satisfied with

q=γt2t128+δt2t12<1.

Therefore, Corollary 2.16 applied to T1and T2,which have common fixed point xC,that is, xis a common solution of (3.9).□

3.3. Second-order differential equation

Now, we consider the boundary value problem for second order differential equation

xt=ftxt,tI,x0=x1=0,E3.14

where I=01and f:I×RR.is a continuous function.

It is known, and easy to check, that the problem (3.14) is equivalent to the integral equation

xt=01Gtsfsxsds,fortI,E3.15

where Gis the Green’s function define by

Gts=t1sif0ts1s1tif0st1.

That is, if xC2IR, then xis a solution of problem (3.14) iff xis a solution of the integral Eq. (3.15).

Let X=CIbe the space of all continuous functions defined on I. Consider the metric-like σon Xdefine by

σxy=xy+x+yfor allx,yX,

where u=maxt01utfor all uX.

Note that σis also a partial metric on Xand since

dσxy2σxyσxxσyy=2xy.

By Lemma 2.20, hence Xσis complete since the metric space Xis complete.

Theorem 3.6. ref. [12] Suppose the following conditions:

  1. there exist continuous functions p:IR+such that

    fsafsb8psab
    for all sIand a,bR;

  2. there exist continuous functions q:IR+such that

    fsa8qsa
    for all sIand aR;

  3. maxsIps=αλ1<149,which is 0α<17;

  4. maxsIqs=αλ2<149which is 0α<17.

Then problem (3.14) has a unique solution uX=CIR.

Proof. Define the mapping T:XXby

Txt=01Gtsfsxsds,
for all xXand tT.Then the problem (3.14) is equivalent to finding a fixed point uof Tin X. Let x,yX, we obtain
TxtTyt=01Gtsfsxsds01Gtsfsysds01Gtsfsxsf(s,ysds801Gtspsxsysds8αλ1xy01Gtsds=αλ1xy.

In the above equality, we used that for each tI, we have 01Gtsds=t21tand so suptI01Gtsds=18.Therefore,

TxTyαλ1xy.E3.16

Moreover, we have

Txt=01Gtsfsxsds801Gtsqsxsds8αλ2x.

Hence,

Txαλ2x.E3.17

Similar method, we obtain

Tyαλ2y.E3.18

Let eτ=λ1+2λ2<1where τ>0.Form (3.16), (3.17) and (3.18), we obtain

σTxTy=|TxTy|+|Tx|+|Ty|αλ1|xy|+αλ2|x|+αλ2|y|λ1+2λ2α|TxTy|+|Tx|+|Ty|=eτασxy.E3.19

Let β,γ,η,δ>0where β<17,γ<17,η<17,δ<17.It following (3.19), we get

σTxTyeτασxy+βσxTx+γσyTy+ησxTy+δσyTx,E3.20

where α+β+γ+2η+2δ<1. Taking the function F:R+Rin (3.20), where Ft=lnt,which is FF, we get

τ+FσTxTyFασxy+βσxTx+γσyTy+ησxTy+δσyTx.

Therefore all hypothesis of Theorem (2.25) are satisfied, and so Thas a unique fixed point uX, that is, the problem (3.14) has a unique solution uC2I.

3.4. Partial differential equation

Consider the Laplace operator is a second order differential operator in the n-dimensional Euclidean space, defined as the divergence of the gradient f. Thus if fis a twice-differentiable real-valued function, then the Laplacian of fis defined by

Δf=2f=fE3.21

where the latter notations derive from formally writing =x1x2xn. Equivalently, the Laplacian of fthe sum of all the unmixed

Δf=i=0n2fxi2.E3.22

As a second-order differential operator, the Laplace operator maps Ckfunctions to Ck2functions for k2. the expression (3.21)(or equivalently(3.22)) defines an operator Δ:CkRnCk2Rnor more generator Δ:CkΩCk2Ωfor any open set ΩConsider semilinear elliptic equation. Look for a function u:ΩRnRmthat solves

Δu=fuinΩE3.23
u=u0on∂ΩE3.24

where f:RnRmis a typically nonlinear function. Equivalently look for a fixed point of TuΔu01fu.

Theorem 3.7. ref. [5] Let fCRand supxRfx=a<. then (3.23) has a weak solution uH01Ω, i.e.

Ωu∇Φdx=ΩfuΦdx,ΦC0Ω.

Proof. Our strategy is to apply Schauder’s Fixed Point Theorem to the map

T:L2ΩL2ΩuΔ1fu,

where Tis continuous. Lemma (2.9) show that ufuis continuous form L2Ωinto itself. Corollary (2.10) shows that Δ1is continuous form L2Ωinto H01Ω, which is continuously embedded in L2Ω. Find a closed, non-empty bounded convex set such that T:MM. Given uL2Ω, Tusatisfies

ΩTuTudx=ΩfuTudxaΩTuL2ΩE3.25

Cauchy-Schwarz. There fore, using Ponincare’s inequality

TuL2Ω2CΩTuL2Ω2aΩTuL2Ω2.

Thus if we set R=aΩCΩand choose M=u:uL2Ω2R. We have established that T:MM, Tis compact. Using Poincare’s inequality on the right-hand-side in (3.25), we obtain. TuL2Ω2RTuL2Ω. Thus TMu:uH1ΩR, and since the embedding of H1Ωinto L2Ωis compact, Tis compact.□

3.5. A non-homogeneous linear parabolic partial differential equation

We consider the following initial value problem

utxt=uxxxt+Hxtuxtuxxt,<x<,0<tT,ux0=φx0,<x<,E3.26

where His continuous and φassume to be continuously differentiable such that φand φare bounded.

By a solution of the problem (3.26), we mean a function uuxtdefined on R×I, where I0T, satisfying the following conditions:

  1. u,ut,ux,uxxCR×I.{CR×Idenote the space of all continuous real valued functions},

  2. uand uxare bounded in R×I,

  3. utxt=uxxxt+Hxtuxtuxxtfor all xtR×I,

  4. ux0=φxfor all xR.

It is important to note that the differential equation problem (3.26) is equivalent to the following integral equation problem

uxt=kxξtφξ+0tkxξtτHξτuξτuxξτdξdτE3.27

for all xRand 0<tT, where

kxt14πtex24t.

The problem (3.26) admits a solution if and only if the corresponding problem (3.27) has a solution.

Let

ΩuxtuuxCR×Iandu<,

where

usupxR,tIuxt+supxR,tIuxxt.

Obviously, the function ω:R+×Ω×ΩR+given by

ωλuv1λuv=1λduv
is a metric modular on Ω. Clearly, the set Ωωis a complete modular metric space independent of generators.

Theorem 3.8. ref. [17] Consider the problem (3.26) and assume the following:

  1. for c>0with s<cand p<c, the function Fxtspis uniformly Hölder continuous in xand tfor each compact subset of R×I,

  2. there exists a constant cHT+2π12T121q, where q01such that

    01λHxts2p2Hxts1p1cHs2s1+p2p1λ
    for all s1p1,s2p2R×Rwith s1s2and p1p2,

  3. His bounded for bounded sand p.

Then the problem (3.26) admits a solution.

Proof. It is well known that uΩωis a solution (3.26) iff uΩωis a solution integral Eq. (3.27).

Consider the graph Gwith VG=D=Ωωand EG=uvD×D:uxtvxtanduxxtvxxtat eachxtR×I. Clearly EGis partial ordered and DEGsatisfy property (A).

Also, define a mapping Λ:ΩωΩωby

Λuxtkxξtφξ+0tkxξtτHξτuξτuxξτdξdτ
for all xtR×I. Then, finding solution of problem (3.27) is equivalent to the ensuring the existence of fixed point of Λ.

Since uvEG,uxvxEGand hence ΛuΛvEG,ΛuxΛvxEG.

Thus, from the definition of Λand by (ii) we have

1λΛvxtΛuxt1λ0tkxξtτHξτvξτvxξτHξτuξτuxξτdξdτ0tkxξtτcH1λ(vξτu(ξτ)+vx(ξτ)ux(ξτ))dξdτcHωλuvT.E3.28

Similarly, we have

1λΛvxxtΛuxxtcHωλuv0tkxxξtτdξdτ2π12T12cHωλuv.E3.29

Therefore, from (3.28) and (3.29) we have

ωλΛuΛvT+2π12T12cHωλuv

i.e.

ωλΛuΛvqωλuv,q01

i.e.

dΛuΛveτduv,τ>0.

Now, by passing to logarithms, we can write this as

lndΛuΛvlneτduv
τ+lndΛuΛvlnduv.

Now, from example 2.22 (i) and taking T=Λand R=I(Identity map), we deduce that the operator T satisfies all the hypothesis of theorem 2.32.

Therefore, as an application of theorem 2.32 we conclude the existence of uΩωsuch that u=Λuand so uis a solution of the problem 3.26.

3.6. Fractional differential equation

Before we will discuss the source of fractional differential equation.

Cauchy’s formula for repeated integration. Let fbe a continuous function on the real line. Then the nthrepeated integral of fbased at a,

fnx=axaσ1aσ2aσn1fσndσndσ3dσ2dσ1
is given by single integration
fnx=1n1!axxtn1ftdt.

A proof is given by mathematical induction. Since fis continuous, the base case follows from the fundamental theorem of calculus.

ddxf1x=ddxaxftdt=fx

where

f1a=aaftdt=0.

Now, suppose this is true for n, and let us prove it for n+1.

Firstly, using the Leibniz integral rule. Then applying the induction hypothesis

fn+1x=axaσ1aσ2aσnfσn+1dσndσ3dσ2dσ1=ax1n1!aσ1σ1tn1ftdtdσ1=axddσ11n!aσ1σ1tnftdtdσ1=1n!axxtnftdt.

This completes the proof. In fractional calculus, this formula can be used to construct a notion of differintegral, allowing one to differentiate or integrate a fractional number of time. Integrating a fractional number of time with this formula is straightforward, one can use fractional nby interpreting n1!as Γn, that is the Riemann-Liouville integral which is defined by

Iαfx=1Γαaxftxtα1dt.

This also makes sense if a=, with suitable restriction on f. The fundamental relation hold

ddxIα+1fx=IαfxIαIβf=Iα+βfx
the latter of which is semigroup properties. These properties make possible not only the definition of fractional differentiation by taking enough derivative of Iαf. One can define fractional-order derivative of as well by
dαdxαf=dαdxαIααf

where denote the ceilling function. One also obtains a differintegral interpolation between differential and integration by defining

Dxαfx=dαdxαIααfxifα>0fxifα=0Iαfxifα<0.

An alternative fractional derivative was introduced by Caputo in 1967, and produce a derivative that has different properties it produces zero from constant function and more importantly the initial value terms of the Laplace Transform are expressed by means of the value of that function and of its derivative of integer order rather than the derivative of fractional order as in the Riemann-Liouville derivative. The Caputo fractional derivative with base point xis then

cDxαfx=Iααdαdxαfx.

Lemma 3.9. ref. [24] Let u:0Xbe continuous function such that uC0τXfor all τ>0. Then uis a global solution of

cDtαut=But;t>0E3.30
u0=u0XE3.31

if and only if uthe integral equation

ut=u0+1Γα0ttsα1Busds,t0.

Proof. Let τ>0. Since uis a global solution of (3.30), then uC0τX, cDtαuC0τXandt

cDtαut=But,t0τ.

Thus, by applying Itαin both sides of the equality (since cDtαuL10τX) we obtain

ut=u0+ItαBut=u0+1Γα0ttsα1Busds,t0.

Since τ>0was an arbitrary choice, usatisfies the integral equation for all t0, as we wish.

On the other hand, choose τ>0(but arbitrary). By hypothesis, uC0τX, and satisfies the integral equation,

ut=u0+1Γα0ttsα1Busds,t0τ.

Observing also u0=u0and rewriting the equality above, we obtain

ut=u0+ItαBus,t0τ.

Since BusC0τX, we conclude, by cDtαItαft=ftof the fractional integral and derivative property that we can apply cDtαin both sides of the integral equation, obtaining

cDtαut=But,t0τ

what lead us to verify that cDtαuC0τX. Since τ>0was an arbitrary choice, we conclude that the function uis a global solution of (3.30).□

Theorem 3.10. ref. [24] Let α01, BLXand u0Xthen the problem (3.30).

have a unique global solution.

Proof. Choose τ>0. then consider Kτ=uC0τX;u0=u0and operator.

T:KτKτgiven by

Tut=u0+1Γα0tButdt.

We will show that a power (with respect to be composition) of this operator is a contraction and therefore by Banach’s Fixed Point Theorem, Thave a unique fixed point in Kτto this end, observe that for any u,vKτ

TutTvt=1Γα0ttsα1Busvsds1Γα0ttsα1BLXusvs)dsBLXΓαusvs)0ttsα1dstαBLXαΓαusvs)tαΓα+1BLXusvs)tαΓα+1BLXsup0sτusvs).

By iterating this relation, we find that

T2utT2vttαΓα+1BLXsup0sτTusTvs)t2αΓ2α+1BLX2sup0sτusvs)T3utT3vttαΓα+1BLXsup0sτT2usT2vs)t3αΓ3α+1BLX3sup0sτusvs)TnutTnvttnαΓnα+1BLXnsup0sτusvs)

and for an sufficiently large n,the constant in question is less than 1, i.e., there exists a fixed point uKτ. Observe now that τ>0was an arbitrary choice, so we conclude that the fixed point uC0τXfor all τ>0and Lemma (3.9), we obtain the existence and uniqueness of a global solution to the problem (3.30). □

Corollary 3.11. ref. [24] Consider the same hypothesis of theorem (3.10).

  1. Let Untn=0be a sequence of continuous functions Un:0Xgiven by U0t=u0, Un=u0+1Γα0ttsα1BUn1sds, n12.

    Then there exists a continuous function U:0X, such that for any τ>0, we conclude that UnUin C0τX. Moreover, Utis the unique global solution of (3.30).

  2. It holds that

    Ut=k=0tαBku0Γαk+1.

Proof. (i) It follows directly from proof of Theorem (3.10).

(ii) It is trivial that U0t=u0. So we compute, using the gamma function properties, that

U1t=u0+1Γα0ttsα1Bu0sds=u0+tαBu0αΓα=u0+tαBu0Γα+1.

By a simple induction process, we conclude that

Unt=k=0ntαBku0Γαk+1
and therefore
Ut=limnk=0ntαBku0Γαk+1=k=0tαBku0Γαk+1EαtαBu0.

From the above works, we can see a fact, although the fractional boundary value problems have been studied, to the best of our knowledge, there have been a few works using the lower and upper solution method. However, only positive solution are useful for many application, motivated by the above works, we study the existence and uniqueness of positive solution of the following integral boundary value problem.

D0+αut+ftut=0,0<t<1,1<α2E3.32
u0=0,u1=01usds,E3.33

where f:01×00is a continuous function and D0+αis the standard Riemann-Liouville fractional derivative.

We need the following lemmas that will be used to prove our main results.

Lemma 3.12. ref. [25] Let α>0and uC01L01. Then fractional differential equation

D0+αut=0

has

ut=C1tα1+C2tα2++CNtαN,E3.34

CiR,i=1,2,,N,N=α+1as unique solution.

Lemma 3.13. ref. [25] Assume that uC01L01with a fractional derivative of order α>0that belongs to C01L01. Then

I0+αD0+αut=utC1tα1C2tα2CNtαNE3.35

for some CiR,i=1,2,,N,N=α+1.

In the following, we present the Green function of fractional differential equation with integral boundary value condition.

Theorem 3.14. ref. [26] Let 1<α<2, Assume ytC01, then the following equation

D0+αut+yt=0,0<t<1E3.36
u0=0,u1=01usds,E3.37

has a unique solution

ut=01GtsysdsE3.38

where

Gts=t1sα1α1+stsα1α1α1Γαif0st1t1sα1α1+sα1Γαif0ts1.

Proof. We may apply Lemma (3.13) to reduce Eq. (3.36) to an equivalent integral equation

ut=I0+αyt+C1tα1+C2tα2
for some C1,C2R. Therefore, the general solution of (3.36) is

ut=01tsα1Γαysds+C1tα1+C2tα2.E3.39

By u0=0, we can get C2=0. In addition, u1=0ttsα1Γαysds+C1, it follows

C1=01tsα1Γαysds+01usds.E3.40

Take (3.40) into (3.39), we have

ut=01tsα1Γαysds+tα1011sα1Γαysds+tα101usds.E3.41

Let 01usds=A,by (3.41), we can get

01utdt=010ttsα1Γαysdsdt+01tα10t1sα1Γαysdsdt+A01tα1dt=011sααΓαysds+011sα1αΓαysds+Aα=01s1sα1αΓαysds+Aα.

So,

A=αα101s1sα1αΓαysds=01s1sα1α1Γαysds.

Combine with (3.41), we have

ut=0ttsα1Γαysds+tα1011sα1Γαysds+tα101s1sα1α1Γαysds=0ttsα1Γαysds+01t1sα1α1+sα1Γαysds=01t1sα1α1+stsα1α1α1Γαysds+t1[t1sα1α1+sα1Γαysds=01Gtsysds.

This complete the proof.

Remark 3.15. Obviously, the Green function Gtssatisfies the following properties:

  1. Gts>0,t,s01;

  2. Gts2α1Γα;0t,s1.

Theorem 3.16. ref. [26] Assume that function fsatisfies

ftuftvatuvE3.42

where t01,u,v0,a:010is a continuous function. If

01sα1α1+sasds<α1ΓαE3.43

then the Eq. (3.32) has a unique positive solution.

Proof. If Tnis a contraction operator for nsufficiently large, then the Eq. (3.32) has a unique positive solution.

In fact, by the definition of Green function Gts, for u,vP, we have the estimate

TutTvt=01Gtsfsusfsvsds01Gtsasusvsds01[t1sα1α1+sα1Γαasuvdsuvtα1α1Γα011sα1α1+sasds.

Denote K=011sα1α1+sasds, then

TutTvtKtα1α1Γαuv.

Similarly,

T2utT2vt=01GtsfsTusfsTvsds01GtsasTusTvsds01GtsasKsα1α1Γαuvds01Kt1sα1α1+sα12Γ2αassα1uvdsKuvtα1α12Γ2α01sα11sα1α1+sasds=KHtα1α12Γ2αuv

where H=01sα11sα1α1+sasds. By mathematical induction, it follows

TnutTnvtKHn1tα1α1nΓnαuv
by (3.43), for nlarge enough, we have
KHn1tα1α1nΓnα=Kα1ΓαHα1Γαn1<1.

Hence, it holds

TnuTnv<uv,
which implies Tnis a contraction operator for nsufficiently large, then the Eq. (3.32) has a unique positive solution.□

Acknowledgments

This project was supported by the Theoretical and Computational Science (TaCS) Center under Computational and Applied Science for Smart Innovation Cluster (CLASSIC), Faculty of Science, KMUTT. The authors acknowledge the financial support provided by King Mongkut’s University of Technology Thonburi through the KMUTT 55th Anniversary Commemorative Fund”. Furthermore, the second author would like to thank the Research Professional Development Project Under the Science Achievement Scholarship of Thailand (SAST) for financial support.

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Piyachat Borisut, Konrawut Khammahawong and Poom Kumam (May 23rd 2018). Fixed Point Theory Approach to Existence of Solutions with Differential Equations, Differential Equations - Theory and Current Research, Terry E. Moschandreou, IntechOpen, DOI: 10.5772/intechopen.74560. Available from:

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