# On the reduction of Krasnoselskii’s theorem to Schauder’s theorem

## On the reduction of Krasnoselskii’s theorem to Schauder’s theorem

Applied Mathematics and Computation 250 (2015) 339–351 Contents lists available at ScienceDirect Applied Mathematics and Computation journal homepag...
Applied Mathematics and Computation 250 (2015) 339–351

Contents lists available at ScienceDirect

Applied Mathematics and Computation journal homepage: www.elsevier.com/locate/amc

On the reduction of Krasnoselskii’s theorem to Schauder’s theorem T.A. Burton a,⇑, Bo Zhang b a b

Northwest Research Institute, 732 Caroline St., Port Angeles, WA, United States Department of Mathematics and Computer Science, Fayetteville State University, Fayetteville, NC 28301, United States

a r t i c l e

i n f o

Keywords: Fixed points Krasnoselskii’s theorem on sum of two maps Nonlinear integral equations

a b s t r a c t Krasnoselskii noted that many problems in analysis can be formulated as a mapping which is the sum of a contraction and compact map. He proved a theorem covering such cases which is the union of the contraction mapping principle and Schauder’s second ﬁxed point theorem. In putting the two results together he found it necessary to add a condition which has been difﬁcult to fulﬁll, although a great many problems have been solved using his result and there have been many generalizations and simpliﬁcations of his result. In this paper we point out that when the mapping is deﬁned by an integral plus a contraction term, the integral can generate an equicontinuous map which is independent of the smoothness of the functions. Because of that, it is possible to set up that mapping, not as a sum of contraction and compact map, but as a continuous map on a compact convex subset of a normed space. An application of Schauder’s ﬁrst ﬁxed point theorem will then yield a ﬁxed point without any reference to that difﬁcult condition of Krasnoselskii. Finite and inﬁnite intervals are handled separately. For the class of problems considered, application is parallel to the much simpler Brouwer ﬁxed point theorem. Ó 2014 Elsevier Inc. All rights reserved.

1. Introduction Throughout applied mathematics we see real world problems modeled by various differential equations which are often inverted as integral equations deﬁning natural mappings of certain sets in a Banach space into themselves. In order to get a ﬁxed point solving the differential equation we are frequently faced with severe compactness problems, particularly when we need a solution on an entire interval ½0; 1Þ. Indeed, there is a myriad of real world problems modeled by fractional differential equations which naturally invert as integral equations with singular kernels. These integral equations deﬁne a natural mapping which invites either Schauder’s or Krasnoselskii’s ﬁxed point theorem. Frequently the mapping is continuous and we can locate a convex set mapped into itself. Our task is only beginning as we consider compactness questions and the mixing of contraction and compact maps. Here we arrive at the objective of this project. We show that if the kernel and its coefﬁcient function satisfy reasonable conditions, then there is a natural equicontinuity condition on the part of the mapping generated by the integral. We then restrict our mapping to a convex set in the Banach space for which that equicontinuity holds even for the contraction part of the mapping. This allows us to use Schauder’s ﬁrst ﬁxed point theorem to get a ﬁxed point.

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Here is the advantage. In Krasnoselskii’s theorem there is a complicated condition (item (i) in the theorem below) which ties the contraction mapping to the compact mapping. By the above process we avoid that complication and get the ﬁxed point directly from Schauder’s theorem. We now look at the details. Krasnoselskii  studied an old paper of Schauder on elliptic partial differential equations and deduced a working principle which we formalize as follows: The inversion of a perturbed differential operator yields the sum of a contraction and compact map. Accordingly, he offered the following result to facilitate treatment of that sum. Theorem 1.1 (Krasnoselskii). Let ðS; k  kÞ be a Banach space, M a closed, convex, nonempty subset of S. Suppose that A; B : M ! S such that

ðiÞ x; y 2 M ) Ax þ By 2 M: A is continuous and

ðiiÞ AM resides in a compact set; ðiiiÞ B is a contraction with constant a < 1. Then 9y 2 M with Ay þ By ¼ y. It is clear from (ii) and (iii) that it is intended to be a combination of Schauder’s second ﬁxed point theorem and the contraction mapping principle. As such it would seem to be exactly what is needed in so many problems in differential and integral equations. But the marriage of the two principles takes place in (i) and that has been very challenging in so many standard problems from applied mathematics. We addressed item (i) in . A very nice summary of selected results on Krasnoselskii’s theorem up through 2007 is found in . Other recent results are found in [1,10,11,5]. It is very convenient to ﬁnd Krasnoselskii’s result and proof in , as well as two forms of Schauder’s ﬁxed point theorem used here. To focus on the need for such a result we consider a neutral functional differential equation. Example. Consider the scalar equation

x0 ðtÞ ¼ xðtÞ þ a

d xðt  hÞ þ uðt; xðtÞÞ dt

with a continuous initial function w : ½h; 0 ! R in which, for simplicity in this presentation, we ask that wð0Þ ¼ awðhÞ. By grouping terms and integrating we obtain

xðtÞ ¼ axðt  hÞ þ

Z

t

eðtsÞ ½axðs  hÞ þ uðs; xðsÞÞds:

0

A full treatment using Krasnoselskii’s ﬁxed point theorem is found in [2, pp. 180–184]. The ﬁrst term, axðt  hÞ, does not smooth but the integral smooths in a most remarkable way. When x is restricted to any given bounded set in BC with a bound of a ﬁxed number K, then the integral maps that set into an equicontinuous set where the equicontinuity is completely independent of the behavior of x. This allows us to place an equicontinuity condition on the mapping set so that the integral equation maps that set into itself. The fact that the contraction term does not smooth causes no trouble at all. We then apply Schauder’s ﬁrst ﬁxed point theorem and obtain a bounded and continuous solution on any interval ½0; T. The victory is that in applying the result condition (i) of Krasnoselskii’s theorem is completely avoided. If the mapping set is essentially a ball then the work holds for 0 6 t < 1 in a weighted space. An integral equation with a mild singularity has a natural induced equicontinuity which can be of prime importance in ﬁxed point theory. We will consider two essentially different forms:

xðtÞ ¼ Vðt; xðtÞÞ þ

Z

t

Rðt  sÞuðt; s; xðsÞÞds

ð1aÞ

Rðt  sÞuðt; s; xðÞÞds;

ð1bÞ

0

and

xðtÞ ¼ f ðt; xðÞÞ þ

Z

t

0

where V and f are of a nature to generate a contraction while R will generate a compact map. For example, we may ﬁnd a closed convex nonempty set M in the Banach space ðBC; k  kÞ of bounded continuous functions / : ½0; 1Þ ! R with the supremum norm with the property that / 2 M and for P deﬁned by either

ðP/ÞðtÞ ¼ Vðt; /ðtÞÞ þ

Z

t

Rðt  sÞuðt; s; /ðsÞÞds

ð2aÞ

0

or

ðP/ÞðtÞ ¼ f ðt; /ðÞÞ þ

Z

t

Rðt  sÞuðt; s; /ðÞÞds

0

with a given initial function w : ½h; 0 ! R;; we have P : M ! M.

ð2bÞ

T.A. Burton, B. Zhang / Applied Mathematics and Computation 250 (2015) 339–351

341

Immediately we think of Krasnoselskii’s ﬁxed point theorem with all its beneﬁts, together with real challenges. But there is a more direct way if R has mild singularities of the following form. Assume that for the ﬁxed set M there is a positive constant, q, with 0 < q < 1 so that for / 2 M then

0 6 Rðt  sÞ 6 ðt  sÞq1 :

ð3Þ

There are many sources for such problems. We can then show (see  and Theorem 6.1) that, independently of the particular / 2 M, the integral

ðL/ÞðtÞ :¼

Z

t

Rðt  sÞuðt; s; /ðsÞÞds

ð4Þ

0

satisﬁes

jL/ðtÞ  L/ðsÞj 6 HðtÞjt  sjq ;

ð5Þ

where HðtÞ is an increasing function. That is, LM is an equicontinuous set. In fact, let t P 0 be ﬁxed. For a given  > 0, we ﬁnd d > 0 corresponding to the  and t in the deﬁnition of equicontinuity by HðtÞdq <  or 1=q

d < ð=HðtÞÞ

ð6Þ

:

Next, suppose there is a b < 1 so that x; y 2 R implies that

jVðt; xÞ  Vðt; yÞj 6 bjx  yj:

ð7aÞ

Or, suppose there is an a < 1 and h > 0 so that for t P h,

jf ðt; /ðÞÞ  f ðt; wðÞÞj 6 aj/t  wt j½h;0

ð7bÞ

½h;0

for /; w 2 M, where j/t  wt j ¼ suph6h60 j/ðt þ hÞ  wðt þ hÞj. The process is now clear. We ask that we add to M the property that all functions satisfy the equicontinuity condition (6) in a certain way so that / 2 M implies P/ 2 M. 2. The integral equation without a delay We are going to continue to use the kernel in the stated form so that the reader can see clearly the exactness of the equicontinuity on which the entire process depends. However, with care one can do the same for a more general equation. For example, Garcia–Falset [10, p. 1746, Item 4] considers the integral equation

xðtÞ ¼ gðt; xÞ þ

Z

t

Fðt  s; s; uðsÞÞds

0

and obtains a bounded mapping set M in which there is a contraction condition on g and a relation

kFðt; s; xÞ  Fðh; s; xÞk 6 SGðjt  hjÞ where G is a continuous function with Gð0Þ ¼ 0 and S a constant depending on a bound on the functions in M. The reader is then left to carry out computations parallel to those which we provide below as a template for such work. Let T > 0 and ðBC; k  kÞ be the Banach space of bounded continuous functions / : ½0; T ! R with the supremum norm. We consider an integral equation of the form (1a)

xðtÞ ¼ Vðt; xðtÞÞ þ

Z

t

Rðt  sÞuðt; s; xðsÞÞds:

0

The following assumptions will be used. (i) R : ð0; 1Þ ! ½0; 1Þ is continuous, decreasing, and Rðt  sÞ 6 ðt  sÞq1 ; 0 < q < 1. (ii) u : ½0; T  ½0; T  R ! R is continuous. (iii) There is a closed, bounded, convex, nonempty set M  BC with the following properties: (a) There are positive J; S such that / 2 M and 0 6 s 6 t 6 T implies that juðt; s; /ðsÞÞj 6 S and

juðt; s; /ðsÞÞ  uðs; s; /ðsÞÞj 6 Jjt  sjq : (b) V : ½0; T  R ! R is continuous and there is a positive b with b < 1 such that / 2 M and t; s 2 ½0; T implies that

jVðt; /ðtÞÞ  Vðt; /ðsÞÞj 6 bj/ðtÞ  /ðsÞj: We will proceed with a view of constructing a nonempty compact convex subset M  of M which is mapped into itself by P deﬁned in (2a) in such a way that Schauder’s ﬁrst ﬁxed point theorem can be applied to the restriction mapping P : M  ! M  . The resulting ﬁxed point is then trivially a ﬁxed point of the original mapping P : M ! M.

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Theorem 2.1. Let (i)–(iii) hold. Suppose that the mapping P deﬁned by / 2 M implies that

ðP/ÞðtÞ ¼ Vðt; /ðtÞÞ þ

Z

t

Rðt  sÞuðt; s; /ðsÞÞds;

06t6T

0

maps M ! M. Then P has a ﬁxed point. Proof. Since M is nonempty, we choose a ﬁxed / 2 M and deﬁne

M ¼ f/ 2 M : j/ðtÞ  /ðsÞj 6 xðt; sÞ; 8t; s 2 ½0; Tg; where

xðt; sÞ ¼

 1   j/ ðtÞ  / ðsÞj þ HðtÞjt  sjq þ V  ðt; sÞ 1b

Rt with V  ðt; sÞ ¼ supjxj6D jVðt; xÞ  Vðs; xÞj; HðtÞ ¼ J 0 RðsÞds þ ð2S=qÞ and D the bound of M. We now see that M  is nonempty    since / 2 M . We shall show that M is a compact convex subset of M. First, we note that M is convex so if /; g 2 M and 0 6 k 6 1 then k/ þ ð1  kÞg 2 M. Thus, if /; g 2 M  then

jk½/ðt 1 Þ  /ðt 2 Þ þ ð1  kÞ½gðt1 Þ  gðt 2 Þj 6 xðt 1 ; t2 Þ; 

so M is also convex. In fact, M  is a compact subset of M. To see this, let f/n g denote a sequence in M  . As the sequence is uniformly bounded and equicontinuous on ½0; T, by the Ascoli–Arzela theorem there is a subsequence f/nk g with a limit / residing in the closed set M. If t1 ; t2 2 ½0; T then

j/ðt 1 Þ  /ðt 2 Þj 6 j/ðt1 Þ  /nk ðt 1 Þj þ j/nk ðt1 Þ  /nk ðt 2 Þj þ j/nk ðt2 Þ  /ðt2 Þj: The ﬁrst and last terms on the right-hand-side tend to zero as k ! 1, while the middle term is less than xðt1 ; t2 Þ so that for large k the left-hand term is less than xðt 1 ; t 2 Þ. Thus, / 2 M , and so M  is a compact subset of M. Next, we show that / 2 M implies that P/ 2 M . Certainly, P/ 2 M. By Theorem 6.1, we see (5) holds on ½0; T. Now, for t; s 2 ½0; T, by (iii) and Theorem 6.1 we obtain

jðP/ÞðtÞ  ðP/ÞðsÞj 6 jVðt; /ðtÞÞ  Vðs; /ðsÞÞj þ jL/ðtÞ  L/ðsÞj 6 jVðt; /ðtÞÞ  Vðt; /ðsÞÞj þ jVðt; /ðsÞÞ  Vðs; /ðsÞÞj þ jL/ðtÞ  L/ðsÞj 6 bj/ðtÞ  /ðsÞj þ V  ðt; sÞ þ HðtÞjt  sjq 6 bxðt; sÞ þ ð1  bÞxðt; sÞ ¼ xðt; sÞ; so P : M ! M  . It is rather routine to show that P is continuous. Let

l

Z

 > 0 be given. First, for =2, ﬁnd l > 0 so that

T

RðsÞds < =2:

0

Now M is bounded by a number D and uðt; s; xÞ is uniformly continuous for 0 6 s 6 t 6 T and jxj 6 D so for the l > 0 there is the d1 > 0 of uniform continuity so that k/  gk < d1 implies that juðt; s; /ðsÞÞ  uðt; s; gðsÞÞj < l. Also, since Vðt; xÞ is uniformly continuous for 0 6 t 6 T and jxj 6 D, there exists d2 > 0 such that k/  gk < d2 implies that

jVðt; /ðtÞÞ  Vðt; gðtÞÞj < =2 for all t 2 ½0; T: Take d ¼ min½d1 ; d2 . Then k/  gk < d implies that

e ð/Þ  V e ðgÞk þ l kðP/Þ  ðPgÞk 6 k V

Z

T

RðsÞds 6 ð=2Þ þ ð=2Þ;

0

e ð/ÞðtÞ ¼ Vðt; /ðtÞÞ, as required. where V Thus, P is a continuous map of a compact convex nonempty set M  into itself so, by Schauder’s ﬁrst ﬁxed point theorem, P has a ﬁxed point in M   M. h 2.1. The case for T ¼ 1 For the case of 0 6 t 6 T just covered we allowed M to be any closed bounded convex nonempty set in BC. When we pass to ½0; 1Þ there is a large change since M  may no longer be a compact subset of M in BC even if it satisﬁes the equicontinuity condition. In this case M must be essentially a ball in order for M to be closed and M  compact in a weighted space being considered here. Theorem 2.2. Let (i)–(iii) hold with T ¼ 1. Suppose there are continuous functions v ; w : ½0; 1Þ ! R with v ðtÞ < wðtÞ for t P 0. Let

M ¼ f/ 2 BCjv ðtÞ 6 /ðtÞ 6 wðtÞg: If the mapping P of Theorem 2.1 maps M into M, then P has a ﬁxed point.

T.A. Burton, B. Zhang / Applied Mathematics and Computation 250 (2015) 339–351

343

Proof. Let g : ½0; 1Þ ! ½1; 1Þ be continuous with g 2" 1. Then ðW; j  jg Þ is the Banach space of continuous functions / : ½0; 1Þ ! R for which

j/jg ¼: sup

06t<1

j/ðtÞj < 1: gðtÞ

Since g is an arbitrary continuous strictly increasing function with gðtÞ ! 1 as t ! 1, we may choose g so that

lim

t!1

v ðtÞ gðtÞ

¼ 0 and lim

t!1

wðtÞ ¼ 0: gðtÞ

ð8Þ

We see that v and w may be unbounded, but the functions in M restricted to any compact interval ½0; T are uniformly bounded. We deﬁne M  as in Theorem 2.1 with

xðt; sÞ ¼

 1   j/ ðtÞ  / ðsÞj þ HðtÞjt  sjq þ V  ðt; sÞ 1b

for t; s P 0. Note that M  is a closed convex nonempty subset of ðW; j  jg Þ and the work in the proof of Theorem 2.1 shows that P : M  ! M  . Let f/n g be a sequence in M  and use Ascoli’s theorem and the diagonalization process to show that there is a subsequence f/nk g converging to some / 2 BC uniformly on compact subsets of ½0; 1Þ and / 2 M. Moreover, j/nk  /jg ! 0 as k ! 1 since (8) holds. The proof of Theorem 2.1 shows that / satisﬁes the equicontinuity property so / 2 M . Therefore M  is a compact subset of ðW; j  jg Þ. An argument similar to that in  shows that P is continuous in the g-norm on M. Applying Schauder’s ﬁrst ﬁxed point theorem to P : M  ! M  in ðW; j  jg Þ, we obtain that there exists a point / 2 M  with P/ ¼ /. This completes the proof. h Remark 1. If b ¼ 1 in (iii)-(b), then the argument in the proof of theorems above fails. However, we may still be able to establish the existence of a ﬁxed point for P under additional assumptions on V. The process goes as follows. We ﬁrst prove the existence of an e-ﬁxed point of P; that is, for each e > 0, there exists xe 2 M such that

kPxe  xe k < e: Next, we apply the approximation method to obtain a ﬁxed point of P. This will be demonstrated in Theorem 2.3. Let I denote the identity map and ðI  PÞðMÞ denote the range of I  P on M. Theorem 2.3. Let (i)–(iii) hold with b ¼ 1 and T ¼ 1, and let M be deﬁned in Theorem 2.2. Suppose that (iv) the set ðI  PÞðMÞ is closed in BC. If the mapping P of Theorem 2.1 maps M into M, then P has a ﬁxed point. ~ 2 M and deﬁne, for any positive integer n, a mapping Pn by / 2 M implies Proof. Since M is nonempty, we choose a ﬁxed /

ðP n /ÞðtÞ ¼

      Z t 1 1~ 1 1~ 1 ¼ 1 ðP/ÞðtÞ þ /ðtÞ Vðt; /ðtÞÞ þ /ðtÞ þ 1 1 Rðt  sÞuðt; s; /ðsÞÞds: n n n n n 0

~ for Since P/ 2 M and M is convex, we have Pn / 2 M and thus, Pn ðMÞ  M. Letting V n ðt; xÞ ¼ ð1  1=nÞVðt; xÞ þ ð1=nÞ/ðtÞ n ¼ 2; 3; . . ., we see for / 2 M that

jV n ðt; /ðtÞÞ  V n ðt; /ðsÞÞj ¼



1

   1 1 jVðt; /ðtÞÞ  Vðt; /ðsÞÞj 6 1  j/ðtÞ  /ðsÞj ¼: bj/ðtÞ  /ðsÞj n n

for t; s P 0. Thus, conditions (i)–(iii) are satisﬁed with Vðt; xÞ; uðt; s; xÞ: replaced by V n ðt; xÞ; ð1  1=nÞuðt; s; xÞ, respectively. By Theorem 2.2, there is a point /n 2 M such that P n /n ¼ /n ; that is,

 Z t 1 /n ðtÞ ¼ V n ðt; /n ðtÞÞ þ 1  Rðt  sÞuðt; s; /n ðsÞÞds: n 0 We also see from

ðP/n ÞðtÞ  /n ðtÞ ¼

  Z t 1 1~ Vðt; /n ðtÞÞ þ Rðt  sÞuðt; s; /n ðsÞÞds  /ðtÞ n n 0

that P has an e-ﬁxed point in M for each

e > 0 since

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T.A. Burton, B. Zhang / Applied Mathematics and Computation 250 (2015) 339–351

kP/n  /n k 6

l n

;

~ Let G ¼ I  P. Then l ¼ supfkPuk : u 2 Mg þ k/k.

where

kGð/n Þk ¼ kP/n  /n k ! 0 as n ! 1 and so 0 2 GðMÞ ¼ GðMÞ since GðMÞ is closed in BC. Thus, 9/ 2 M such that P/  / ¼ 0. This completes the proof.

h

e : M ! BC by V e ð/ÞðtÞ ¼ Vðt; /ðtÞÞ for all t P 0 and / 2 M. We now deﬁne a mapping V Corollary 1. Let (i)–(iii) hold with b ¼ 1 and T ¼ 1, and let

M ¼ f/ 2 BC : k/k 6 Kg for a constant K > 0. Suppose that (v) suptP0 jVðt; xÞ  Vðt; yÞj < jx  yj for all x; y 2 ½K; K with x – y. If the mapping P of Theorem 2.1 maps M into M, then P has a ﬁxed point.

Proof. We only need to show that (iv) holds. To this end, let wn 2 ðI  PÞðMÞ with kwn  wk ! 0 as n ! 1 for some w 2 BC. We shall show that w 2 ðI  PÞðMÞ. Let /n 2 M with wn ¼ ðI  PÞ/n for n ¼ 1; 2; . . .. We write ðI  PÞ/n ðtÞ ¼ wn ðtÞ as

/n ðtÞ  Vðt; /n ðtÞÞ ¼ yn ðtÞ þ wn ðtÞ;

ð9Þ

where

yn ðtÞ ¼

Z

t

Rðt  sÞuðt; s; /n ðsÞÞds:

0

Since V is continuous on ½0; 1Þ  ½K; K and PðMÞ  M, we see that the sequence fyn g is uniformly bounded and equicontinuous on any compact subset of ½0; 1Þ by (iii)-(a) and Theorem 6.1. Thus, by the Ascoli–Arzela theorem, there is a subsequence fynk g converging to some y 2 BC uniformly on any closed bounded interval ½0; T. Since (v) holds, we have by e Þ1 is continuous on ðI  V e ÞðMÞ. It now follows from (9) that Theorem 6.2 that ðI  V

h i e Þ1 y þ w ðtÞ: /nk ðtÞ ¼ ðI  V nk nk

ð10Þ

eÞ This implies that f/nk g converges to a function / 2 M uniformly on ½0; T. Since ðI  V ting k ! 1 in (10) we obtain

1

e ÞðMÞ, by letis continuous on ðI  V

1

e Þ ½y þ wðtÞ for t 2 ½0; T: /ðtÞ ¼ ðI  V

ð11Þ

Taking the limit in

Z

ynk ðtÞ ¼

t

0

Rðt  sÞuðt; s; /nk ðsÞÞds;

we also obtain

yðtÞ ¼

Z

t

Rðt  sÞuðt; s; /ðsÞÞds:

ð12Þ

0

Combining (11) and (12), we see that wðtÞ ¼ ðI  PÞ/ðtÞ for all t P 0. Thus, w 2 ðI  PÞðMÞ, and the proof is complete. h Example. Consider the fractional differential equation of Caputo type c

Dq ðx  jðxÞÞ ¼ aðtÞx3 ðtÞ þ Gðt; xðtÞÞ;

with a : ½0; 1Þ ! R;

xð0Þ ¼ x0 ;

0 < q < 1;

j : R ! R; G : ½0; 1Þ  R ! R; continuous. See  for background and deﬁnitions. Suppose that

(~i) aðtÞ is bounded on ½0; 1Þ, e jGðt; xÞj 6 bðtÞjxj3 for jxj 6 1 and t P 0, ( ii) e aðtÞ  bðtÞ P d for all t P 0 and a constant d > 0. ( iii) f ( iv ) 9c > 0 such that j : ½c; c ! R is nondecreasing, odd with x  jðxÞ  x3 increasing on ½0; c, Then the zero solution of (13) is stable.

ð13Þ

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T.A. Burton, B. Zhang / Applied Mathematics and Computation 250 (2015) 339–351

Proof. Choose a constant g > 0 with suptP0 aðtÞ < g and deﬁne

CðtÞ ¼

g CðqÞ

tq1 :

Then the resolvent R satisﬁes

RðtÞ ¼ CðtÞ 

Z

t

Cðt  sÞRðsÞds:

0

This resolvent R is completely monotone on ð0; 1Þ. Moreover,

0 6 RðtÞ 6 CðtÞ;

tRðtÞ ! 0 as t ! 1;

and

Z

1

RðsÞds ¼ 1:

0

If we write (13) as c

Dq ðx  jðxÞÞ ¼ aðtÞx3 ðtÞ þ Gðt; xðtÞÞ ¼ g½x  jðxÞ þ g½x  jðxÞ  x3  þ ½g  aðtÞx3 þ Gðt; xðtÞÞ;

then the solution xðtÞ satisﬁes

xðtÞ ¼ zðtÞ þ jðxÞ þ

Z

t

Rðt  sÞ½xðsÞ  jðxÞ  x3 ðsÞds þ

0

¼: jðxÞ þ zðtÞ þ

Z

Z

t

  Z t aðsÞ 3 1 x ðsÞds þ Rðt  sÞ 1  Rðt  sÞ Gðs; xðsÞÞds

g

0

0

g

t

Rðt  sÞuðs; xðsÞÞds ¼: ðPxÞðtÞ;

0

Rt where zðtÞ ¼ ðx0  jðx0 ÞÞð1  0 RðsÞdsÞ. pﬃﬃﬃ Let 0 < e < c. We may assume that c < 3=3; g P 1, and 0 < d < 1 so that de3 =g < e. Now let jx0 j < de3 =g and deﬁne

M ¼ f/ 2 BC : k/k 6 eg: For the mapping deﬁned above, we can show that P : M ! M. To see this, we observe that r  jðrÞ  r3 is odd and increase Apply ( ii) e and (f iii) to obtain ing on ½0; e by ( iv).

jðPxÞðtÞj 6 jzðtÞj þ jðeÞ þ ðe  jðeÞ  e3 Þ þ e3

Z

t

  aðsÞ jbðsÞj ds 6 jx0 j þ ðe  e3 Þ þ e3 ð1  d=gÞ < e; Rðt  sÞ 1  þ

0

g

g

3

if jx0 j < de =g. Since x  jðxÞ  x3 is increasing on ½0; c, we see that x  jðxÞ is strictly increasing on ½c; c with

jjðxÞ  jðyÞj < jx  yj for all x; y 2 ½c; c with x – y. We now readily verify that all conditions of Corollary 1 are satisﬁed with Vðt; xÞ ¼ jðxÞ þ zðtÞ, and so, P has a ﬁxed point x 2 M which is the solution of (13). Since e > 0 is arbitrary, this proves that the solution x ¼ 0 of (13) is stable. h Critique Notice that conditions (i)–(iii) are simply deﬁning the sets and functions. Theorem 2.1 asks only that the investigator show that P : M ! M and P is continuous. Krasnoselskii’s theorem has been replaced by Schauder’s theorem for certain Banach spaces. 3. The integral equation with a delay Let T > 0 and ðBC; k  kÞ be the Banach space of bounded continuous functions / : ½0; T ! R with the supremum norm. We consider an integral equation with a given continuous initial function w : ½h; 0 ! R of the form

xðtÞ ¼ f ðxðt  hÞÞ þ

Z

t

Rðt  sÞuðs; xðsÞ; xðs  hÞÞds þ FðtÞ;

tP0

0

with xðtÞ ¼ wðtÞ for h 6 t 6 0. The following assumptions will be used. (i) R : ð0; 1Þ ! ½0; 1Þ is continuous, decreasing, and Rðt  sÞ 6 ðt  sÞq1 ; 0 < q < 1. (ii) u : ½0; T  R2 ! R and F : ½0; T ! R are continuous. (iii) There is a closed, bounded, convex, nonempty set M  BC with the following properties: (a) / 2 M implies that /ð0Þ ¼ wð0Þ. (b) There is a positive S such that / 2 M and 0 6 t 6 T implies that juðt; /ðtÞ; /ðt  hÞÞj 6 S. (c) f : R ! R is continuous and there is an a < 1 such that / 2 M and t; s 2 ½0; T implies

ð14Þ

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jf ð/ðtÞÞ  f ð/ðsÞÞj 6 aj/ðtÞ  /ðsÞj: It is understood that /ðsÞ ¼ wðsÞ; 8s 2 ½h; 0, for all / 2 M. We also extend the domain of any function g : ½a; b ! R: to R: by assigning gðsÞ ¼ gðaÞ for s 6 a and gðsÞ ¼ gðbÞ for s P b. For a real-valued function u : R ! R, we set

jut  us j½a;b ¼ sup juðt þ sÞ  uðs þ sÞj a6s6b

for all t; s 2 R. Finally, we point out that functions f ; F in (14) may depend on w so that wð0Þ ¼ f ðwðhÞÞ þ Fð0Þ. Theorem 3.1. Let (i)–(iii) hold. Suppose that the mapping P deﬁned by / 2 M implies that

ðP/ÞðtÞ ¼ f ð/ðt  hÞÞ þ

Z

t

Rðt  sÞuðs; /ðsÞ; /ðs  hÞÞds þ FðtÞ;

06t6T

0

maps M ! M. Then P has a ﬁxed point. Proof. Let / 2 M be ﬁxed and deﬁne

M ¼ f/ 2 M : j/ðtÞ  /ðsÞj 6 hðt; sÞ; 8t; s 2 Xg; where

X ¼ fðt; sÞ : t; s 2 ½0; h or t; s 2 ½h; Tg and

hðt; sÞ ¼

  1 2S j/t  /s jð1;0 þ jt  sjq þ jF t  F s jð1;0 þ supjf ðwðt þ sÞÞ  f ðwðs þ sÞÞj; 1a q s60

for all t; s 2 ½0; T. It is clear that hðt; sÞ ! 0 as jt  sj ! 0. This implies that M  is uniformly bounded and equicontinuous on ½0; T. Since / 2 M  , we see that M  is a compact convex nonempty subset of M. We now claim that hðt  h; s  hÞ 6 hðt; sÞ: for all t; s 2 ½h; T. In fact, we have

  1 2S j/th  /sh jð1;0 þ jt  sjq þ jF th  F sh jð1;0 þ supjf ðwðt  h þ sÞÞ  f ðwðs  h þ sÞÞj 1a q s60   1 2S q ð1;h   ð1;h þ sup jf ðwðt þ rÞÞ  f ðwðs þ rÞÞj j/t  /s j þ jt  sj þ jF t  F s j ¼ 1a q r6h   1 2S j/t  /s jð1;0 þ jt  sjq þ jF t  F s jð1;0 þ supjf ðwðt þ sÞÞ  f ðwðs þ sÞÞj ¼ hðt; sÞ: 6 1a q s60

hðt  h; s  hÞ ¼

Next, we show that / 2 M  implies that P/ 2 M  . Certainly, P/ 2 M. We may assume h < T and still denote the integral term in (14) by L/ðtÞ so that (5) holds with HðtÞ ¼ 2S=q. Now, for t; s 2 ½h; T, we have

jðP/ÞðtÞ  ðP/ÞðsÞj 6 jf ð/ðt  hÞÞ  f ð/ðs  hÞÞj þ jL/ðtÞ  L/ðsÞj þ jFðtÞ  FðsÞj 6 aj/ðt  hÞ  /ðs  hÞj þ

2S 2S jt  sjq þ jFðtÞ  FðsÞj 6 ahðt  h; s  hÞ þ jt  sjq þ jF t  F s jð1;0 q q

6 ahðt; sÞ þ ð1  aÞhðt; sÞ ¼ hðt; sÞ: For t; s 2 ½0; h, we observe that

jf ð/ðt  hÞÞ  f ð/ðs  hÞÞj ¼ jf ðwðt  hÞÞ  f ðwðs  hÞÞj and so

jðP/ÞðtÞ  ðP/ÞðsÞj 6 jf ðwðt  hÞÞ  f ðwðs  hÞÞj þ jL/ðtÞ  L/ðsÞj þ jFðtÞ  FðsÞj 6 jf ðwðt  hÞÞ  f ðwðs  hÞÞj þ

2S jt  sjq þ jFðtÞ  FðsÞj 6 hðt; sÞ: q

This implies that P : M  ! M  . The rest of the proof is exactly as in the proof of Theorem 2.1. h The extension to the interval ½0; 1Þ is exactly as before. 4. A fractional integral equation There is an interesting and important paper  dealing with the scalar integral equation

xðtÞ ¼ gðt; xðtÞÞ þ

f ðt; xðtÞÞ CðqÞ

Z 0

t

v ðt; s; xðsÞÞ ðt  sÞ1q

ds;

0 < q < 1;

ð15Þ

T.A. Burton, B. Zhang / Applied Mathematics and Computation 250 (2015) 339–351

347

which is studied by means of measures of noncompactness and a ﬁxed point theorem of Darbo. It is important because it models a number of real-world problems with a view to showing that at least one solution exists and that any other solution converges to it at inﬁnity. All of the conditions are introduced in Lipschitz form and from these are obtained rapid decay of the functions involved. It is interesting from our point of view because there are no sign conditions on any of the functions. Everything is controlled by the Lipschitz conditions and the decay of functions. We wish to show that we can avoid the measures of noncompactness and one of the Lipschitz conditions (but not the implied decay), and use Theorem 2.2 to obtain some elementary solutions. Indeed, the point of our results is that they are elementary and, in fact, fairly close to Brouwer’s theorem. The idea is to state the conditions in , denoted by ðhi Þ, and then state the consequences which we will use and denote these as ðci Þ. ðh1 Þ g : Rþ  R ! R is continuous, gðt; 0Þ is bounded, and g  ¼ suptP0 jgðt; 0Þj. Also, there is a continuous function ‘ðtÞ with

jgðt; xÞ  gðt; yÞj 6 ‘ðtÞjx  yj for x; y 2 R and t P 0. ðc1 Þ We will use ðh1 Þ, but will not ask for g  (see ðc5 Þ). ðh2 Þ f : Rþ  R ! R is continuous and there is a continuous function m : Rþ ! Rþ such that

jf ðt; xÞ  f ðt; yÞj 6 mðtÞjx  yj for x; y 2 R and t P 0. ðc2 Þ We will use ðh2 Þ, but will not ask for mðtÞnðtÞtq ! 0 as t ! 1. ðh3 Þ v : Rþ  Rþ  R ! R is continuous. There exists a continuous functions n : Rþ ! Rþ and a continuous nondecreasing function U : Rþ ! Rþ with Uð0Þ ¼ 0 so that for all t; s 2 Rþ with t P s we have

jv ðt; s; xÞ  v ðt; s; yÞj 6 nðtÞUðjx  yjÞ: Also, there is a function ðc3 Þ We will use

v  ðtÞ ¼ maxfjv ðt; s; 0Þj : 0 6 s 6 tg.

jv ðt; s; xÞj 6 v  ðtÞ þ nðtÞUðjxjÞ: Now, everything is gathered. ðh4 Þ The functions /; w; n; g : Rþ ! Rþ deﬁned by

/ðtÞ ¼ mðtÞnðtÞtq wðtÞ ¼ mðtÞv  ðtÞtq nðtÞ ¼ nðtÞjf ðt; 0Þjt q

gðtÞ ¼ v  ðtÞjf ðt; 0Þjtq are all bounded on Rþ and limt!1 /ðtÞ ¼ limt!1 nðtÞ ¼ 0. ðc4 Þ We will use the notation of ðh4 Þ, but will not ask boundedness or limit conditions of these functions. ðh5 Þ There exists a positive solution r0 of the inequality

ð‘ r þ g  ÞCðq þ 1Þ þ ½/ rUðrÞ þ w r þ n UðrÞ þ g  6 r Cðq þ 1Þ and ‘ Cðq þ 1Þ þ / Uðr 0 Þ þ w < Cðq þ 1Þ, where ‘ ¼ supf‘ðtÞ : t P 0g; / ¼ supf/ðtÞ : t P 0g; w ¼ supfwðtÞ : t P 0g; n ¼ supfnðtÞ : t P 0g and g ¼ supfgðtÞ : t P 0g. ðc5 Þ We will improve ðh5 Þ by asking that there exists a continuous function r : ½0; 1Þ ! ð0; 1Þ and a b < 1 such that for all tP0

‘ðtÞ þ ½wðtÞ þ /ðtÞUðr  ðtÞÞ

1

Cðq þ 1Þ

6b

and

cðtÞ þ ‘ðtÞr þ ½wðtÞr þ /ðtÞrUðr Þ þ nðtÞUðr Þ

1

Cðq þ 1Þ

6r

where r ¼ rðtÞ; r  ðtÞ ¼ sup06s6t rðsÞ; cðtÞ ¼ jgðt; 0Þj þ gðtÞ=Cðq þ 1Þ. Under the assumptions ðh1 Þ–ðh5 Þ it is shown in [9, p. 77] that there is at least one solution of the integral equation; moreover, if there are other solutions, then they converge to the given solution. This is proved using the full conditions ðh1 Þ—ðh5 Þ and methods of measures of noncompactness. Our purpose here is to show that with our theorems there is an elementary proof using only ðc1 Þ—ðc5 Þ to show that there is a solution x of (15) with jxðtÞj 6 rðtÞ without the Lipschitz condition on v or the limit conditions in ðh4 Þ.

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Theorem 4.1. Let ðc1 Þ—ðc5 Þ hold. Then the integral equation (15) has a solution x with jxðtÞj 6 rðtÞ for all t P 0. Proof. As in [9, p. 79], the stated conditions are sufﬁcient to show that if

M ¼ f/ 2 BC : j/ðtÞj 6 rðtÞg; then the natural mapping deﬁned by the integral equation maps M into itself. The proof of Theorems 2.1 and 2.2 will establish continuity of the map. The appendix will show the required equicontinuity and other technical details. h

5. A neutral delay equation 5.1. Neutral equations A large body of literature can be found concerning applications of neutral differential equations by simply putting ‘‘epidemics and neutral differential equations’’ into a search engine. The basic and heuristic idea of a neutral equation is that the rate of change, x0 ðtÞ, is inﬂuenced not only by a ‘‘position’’ of x in space and time, together with forces acting on x as we would deduce from Newton’s Second Law of Motion, but it is also inﬂuenced by the recent rate of change of x. Epidemics and other problems in mathematical biology are widely studied by neutral differential equations. See, for example, Gopalsamy , Gopalsamy and Zhang , Kuang [13–15]. Investigators have given heuristic arguments to support their use in describing biological phenomena and much of this is formalized in the ﬁnal chapter of each of the books by Gopalsamy  and Kuang . There are many other problems treated by neutral differential equations and our main contribution here is a quick sketch of the way to express a problem of the form

x0 ðtÞ ¼

d f ðxðt  hÞÞ  uðt; xðtÞ; xðt  hÞÞ; dt

xðtÞ ¼ wðtÞ;

h 6 t 6 0;

as a problem readily attacked by the extension of Theorem 3.1 to ½0; 1Þ. Here, h is a positive constant, u is continuous and bounded for x bounded, and f satisﬁes a contraction condition. The function w is a given continuous initial function and we want a solution for 0 6 t < 1. If we were to simply integrate that equation to obtain an integral equation deﬁning a mapping, then almost everything in our theory would fail. Instead we employ a form of a ‘‘linearization trick’’. Let J be a positive constant to be determined, subtract and add JxðtÞ to obtain

x0 ðtÞ ¼ JxðtÞ þ JxðtÞ þ

d f ðxðt  hÞÞ  uðt; xðtÞ; xðt  hÞÞ: dt

Take all the terms except x0 ðtÞ ¼ JxðtÞ as an inhomogeneous term and use the variation of parameters formula to write

xðtÞ ¼ wð0ÞeJt þ

Z

t

eJðtsÞ ½JxðsÞ þ

0

d f ðxðs  hÞÞ  uðs; xðsÞ; xðs  hÞÞds: ds

Integration by parts of the derivative term in the integral yields

xðtÞ ¼ wð0ÞeJt þ f ðxðt  hÞÞ  eJt f ðwðhÞÞ þ

Z

t

eJðtsÞ ½JxðsÞ  Jf ðxðs  hÞÞ  uðs; xðsÞ; xðs  hÞds:

0

The positive constant, J, can be chosen at will to facilitate the construction of a self-mapping set M of the type required for our theory. This equation deﬁnes a mapping which is well-suited to Theorem 3.1 and its extension to ½0; 1Þ. We do not need to satisfy the difﬁcult condition (i) of Krasnoselskii’s theorem. Details for a self-mapping set can be found in . The details are quite lengthy and will not be repeated here. Appendix A Theorem 6.1. Let u : ½0; 1Þ  ½0; 1Þ  R ! R be continuous, and let R : ð0; 1Þ ! ½0; 1Þ be continuous, decreasing, and Rðt  sÞ 6 Dðt  sÞq1 with 0 < q < 1 and D > 0. Then there is a continuous increasing function H so that if t; s P 0, if x 2 BC with juðt; s; xðsÞÞj 6 K and

juðt; s; xðsÞÞ  uðs; s; xðsÞÞj 6 Jjt  sjq ; then

Z t Z s L :¼ Rðt  sÞuðt; s; xðsÞÞds  Rðs  sÞuðs; s; xðsÞÞds 6 HðtÞjt  sjq ; 0

where HðtÞ ¼ 2KD=q þ J

0

Rt 0

RðsÞds.

ð16Þ

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Proof. Note that since RðtÞ is decreasing and there is a constant D with 0 6 RðtÞ 6 Dt q1 we have, for 0 6 s 6 t, that

Z t Z s jRðt  sÞ  Rðs  sÞjjuðt; s; xðsÞÞjds þ jRðt  sÞjjuðt; s; xðsÞÞjds þ jRðs  sÞjjuðt; s; xðsÞÞ  uðs; s; xðsÞÞjds 0 s 0 Z s Z t Z s 6 K½Rðs  sÞ  Rðt  sÞds þ K Rðt  sÞds þ jRðs  sÞjdsJjt  sjq 0 s 0 Z s Z s Z t Z s ¼K Rðs  sÞds  K Rðt  sÞds þ K Rðt  sÞds þ RðsÞdsJjt  sjq 0 0 s 0 Z s Z t Z t Z s ¼K RðsÞds  K RðsÞds þ K Rðt  sÞds þ RðsÞdsJjt  sjq 0 ts s 0 Z s Z t Z t Z s ¼K RðsÞds  K RðsÞds þ 2K Rðt  sÞds þ RðsÞdsJjt  sjq 0 0 s 0 Z t Z s the sum of the first two terms is negative 6 2DK ðt  sÞq1 ds þ RðsÞdsJjt  sjq s 0   Z s Z s ¼ 2ðKD=qÞðt  sÞq jts þ RðsÞdsJjt  sjq ¼ 2KD=q þ J RðsÞds jt  sjq 0 0   Z t 6 2KD=q þ J RðsÞds jt  sjq ¼: HðtÞjt  sjq :

L6

Z

s

0

The same equality holds if 0 6 t 6 s. This completes the proof.

h

Theorem 6.2. Let (i)–(iii) hold with b ¼ 1 and T ¼ 1, and let

M ¼ f/ 2 BC : k/k 6 Kg for a constant K > 0. Suppose that (v) suptP0 jVðt; xÞ  Vðt; yÞj < jx  yj for all x; y 2 ½K; K with x: ¼ y. e Þ1 is continuous on ðI  V e ÞðMÞ, where ð V e /ÞðtÞ ¼ Vðt; /ðtÞÞ for / 2 M. Then ðI  V

1

1

e Þ is one to one, and hence the inverse ðI  V e Þ exists. We now show that ðI  V e Þ is Proof. Since (v) holds, we see that ðI  V e ÞðMÞ. To this end, let fy g be a sequence in ðI  V e ÞðMÞ with ky  y k ! 0 as n ! 1 for a function continuous on ðI  V n n e ÞðMÞ. We need to show that y 2 ðI  V 1

1

e Þ y ! ðI  V e Þ y as n ! 1: ðI  V n 1

1

e Þ y and x ¼ ðI  V e Þ y . Then ðI  V e Þxn ¼ y and ðI  V e Þx ¼ y . Suppose that xn 9x . Then there exists an Set xn ¼ ðI  V n n e0 > 0 and a subsequence fxnk g of fxn g such that kxnk  x k P e0 for all k ¼ 1; 2; . . .. Now choose tk 2 ½0; 1Þ with

jxnk ðtk Þ  x ðt k Þj P e0 =2: Observe that the function suptP0 jVðt; xÞ  Vðt; yÞj is continuous on the compact set X ¼ ½K; K  ½K; K. By (v), we have

sup jxyjPe0 =2

suptP0 jVðt; xÞ  Vðt; yÞj : x; y 2 ½K; K ¼ d < 1: jx  yj

and therefore

jVðtk ; xnk ðt k ÞÞ  Vðtk ; x ðt k ÞÞj 6 djxnk ðtk Þ  x ðt k Þj: We now have

e xn Þðt k Þ  ðx  V e x Þðt k Þj P jxn ðtk Þ  x ðt k Þj  jð V e xn Þðt k Þ  ð V e x Þðtk Þj jynk ðt k Þ  y ðtk Þj ¼ jðxn  V k k P jxnk ðtk Þ  x ðt k Þj  djxnk ðt k Þ  x ðt k Þj ¼ ð1  dÞjxnk ðtk Þ  x ðt k Þj P ð1  dÞe0 =2 > 0: This yields

ð1  dÞe0 =2 6 jynk ðt k Þ  y ðt k Þj 6 kynk  y k ! 0 as n ! 1 e Þ1 is continuous on ðI  V e ÞðMÞ. This completes the proof. h a contradiction. So we obtain xn ! x as n ! 1, and thus ðI  V Proof of Theorem 4.1. Let BC be the Banach space of bounded continuous functions / : ½0; 1Þ ! R with the supremum norm k  k. To simplify notations, we set

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g  ðt; sÞ ¼ sup jgðt; yÞ  gðs; yÞj; jyj6rðsÞ 

f ðt; sÞ ¼ sup jf ðt; yÞ  f ðs; yÞj jyj6rðsÞ

and

Iðt; xðÞÞ ¼

1 CðqÞ

Z

t

v ðt; s; xðsÞÞ ðt  sÞ1q

0

ds:

We deﬁne

M ¼ fx 2 BC : jxðtÞj 6 rðtÞ; 8t P 0g where rðtÞ is given in (c5). Now let P be the natural mapping deﬁned by the integral equation (15). We will show that P maps M into itself. To see this, letting x 2 M, we have

jðPxÞðtÞj ¼ jgðt; xðtÞÞ þ f ðt; xðtÞÞIðt; xðÞÞj 6 jgðt; xðtÞÞ  gðt; 0Þj þ jgðt; 0Þj þ jf ðt; xðtÞÞ  f ðt; 0ÞjjIðt; xðÞÞj þ jf ðt; 0ÞjjIðt; xðÞÞj 6 ‘ðtÞjxðtÞj þ jgðt; 0Þj þ mðtÞjxðtÞjjIðt; xðÞÞj þ jf ðt; 0ÞjjIðt; xðÞÞj ðuse ð16Þ belowÞ 6 ‘ðtÞrðtÞ þ jgðt; 0Þj þ mðtÞrðtÞ½v  ðtÞ þ nðtÞUðr  ðtÞÞt q

1

Cðq þ 1Þ

¼ cðtÞ þ ‘ðtÞrðtÞ þ ½wðtÞrðtÞ þ /ðtÞrðtÞUðr  ðtÞÞ þ nðtÞUðr  ðtÞÞ

þ jf ðt; 0Þj½v  ðtÞ þ nðtÞUðr ðtÞÞt q 1

Cðq þ 1Þ

1

Cðq þ 1Þ

6 rðtÞ

by applying ðc1 Þ—ðc3 Þ and ðc5 Þ. Next, we deﬁne M  by

M ¼ fx 2 M : jxðtÞ  xðsÞj 6 mðt; sÞ; 8t; s P 0g where mðt; sÞ is a continuous function to be deﬁned below with mðt; sÞ ! 0 as jt  sj ! 0. We want to show that jðPxÞðtÞ  ðPxÞðsÞj 6 mðt; sÞ for / 2 M. To this end, we proceed to estimate the terms on the right-hand side of (15) for x 2 M. By (c3), we have

jIðt; xðÞÞj 6

1 CðqÞ

Z

0

t

jv ðt; s; xðsÞÞj ðt  sÞ

1q

6 ½v  ðtÞ þ nðtÞUðr  ðtÞÞ

ds 6 1

1 CðqÞ Z t

CðqÞ

0

Z

v  ðtÞ þ nðtÞUðjxðsÞjÞ

t

ðt  sÞ1q

0

1 1q

ðt  sÞ

ds

ds ¼ ½v  ðtÞ þ nðtÞUðr  ðtÞÞtq

1

Cðq þ 1Þ

¼: J  ðtÞ

ð16Þ

and

jf ðt; xðtÞÞj 6 jf ðt; xðtÞÞ  f ðt; 0Þj þ jf ðt; 0Þj 6 mðtÞjxðtÞj þ jf ðt; 0Þj 6 mðtÞrðtÞ þ jf ðt; 0Þj ¼: ^f ðtÞ:

ð17Þ

From (c1), we have

jgðt; xðtÞÞ  gðs; xðsÞÞj 6 jgðt; xðtÞÞ  gðt; xðsÞÞj þ jgðt; xðsÞÞ  gðs; xðsÞÞj 6 ‘ðtÞjxðtÞ  xðsÞj þ g  ðt; sÞ:

ð18Þ

Apply (c2) and the estimate in (16) to obtain

jf ðt; xðtÞÞ  f ðs; xðsÞÞj jIðt; xðÞÞj 6 jf ðt; xðtÞÞ  f ðt; xðsÞÞj jIðt; xðÞÞj þ jf ðt; xðsÞÞ  f ðs; xðsÞÞj jIðt; xðÞÞj 

6 mðtÞjxðtÞ  xðsÞjJ ðtÞ þ f ðt; sÞJ  ðtÞ ¼ ½wðtÞ þ /ðtÞUðr  ðtÞÞ

1



Cðq þ 1Þ

jxðtÞ  xðsÞj þ f ðt; sÞJ  ðtÞ:

ð19Þ

An argument similar to that in the proof of Theorem 6.1 yields, for 0 6 s 6 t, that

jIðt; xðÞÞ  Iðs; xðÞÞj 6 ½v  ðtÞ þ nðtÞUðr  ðtÞÞ

2

Cðq þ 1Þ

jt  sjq þ

1

Cðq þ 1Þ

tq v  ðt; sÞ ¼: I ðt; sÞ

where v  ðt; sÞ ¼ supjyj6rðsÞ jv ðt; s; yÞ  v ðs; s; yÞj for 0 6 s 6 s 6 t. Taking into account (17), we get

jIðt; xðÞÞ  Iðs; xðÞÞjjf ðs; xðsÞÞj 6 I ðt; sÞ^f ðsÞ:

ð20Þ

Combine (18)–(20) to obtain

jðPxÞðtÞ  ðPxÞðsÞj 6 ½‘ðtÞCðq þ 1Þ þ wðtÞ þ /ðtÞUðr  ðtÞÞ

1

Cðq þ 1Þ

 jxðtÞ  xðsÞj þ g  ðt; sÞ þ f ðt; sÞJ  ðtÞ þ I ðt; sÞ^f ðsÞ

ðassigning the last three terms as ð1  bÞmðt; sÞÞ 6 bjxðtÞ  xðsÞj þ ð1  bÞmðt; sÞ 6 bmðt; sÞ þ ð1  bÞmðt; sÞ ¼ mðt; sÞ;

T.A. Burton, B. Zhang / Applied Mathematics and Computation 250 (2015) 339–351

351

where b is given in (c5). Thus, P : M  ! M  . The rest of the proof follows that of Theorem 2.2 so P has a ﬁxed point in M which is a solution of (15). The proof is complete. h References                 

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