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Decay integral solutions for neutral fractional differential equations with infinite delays

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Page 1: Decay integral solutions for neutral fractional differential equations with infinite delays

Research Article

Received 12 January 2014 Published online in Wiley Online Library

(wileyonlinelibrary.com) DOI: 10.1002/mma.3172MOS subject classification: 34A08; 47H08; 47H10

Decay integral solutions for neutral fractionaldifferential equations with infinite delays

Nguyen Thanh Anh and Tran Dinh Ke*†

Communicated by M. Kirane

Our aim in this work is to find decay integral solutions for a class of neutral fractional differential equations in Banachspaces involving unbounded delays. By constructing a suitable measure of noncompactness on the space of solutions andestablishing new estimates for fractional resolvent operators, we prove the existence of a compact set of decay integralsolutions to the mentioned problem. Copyright © 2014 John Wiley & Sons, Ltd.

Keywords: resolvent operator; decay solution; fractional neutral differential equation; infinite delay; condensing map; fixed point;measure of noncompactness

1. Introduction

Let .X , k � k/ be a Banach space. We consider the Cauchy problem for neutral fractional differential equations with infinite delays of thefollowing form

D˛0 Œx.t/ � h .t, xt/� D Ax.t/C f .t, x.t/, xt/ , t > 0, (1)

x.t/ D '.t/, t � 0, (2)

where D˛0 , 0 < ˛ � 1 is the Caputo derivative of order ˛, A is the infinitesimal generator of an analytic semigroup on X , ' 2 B withB being an admissible phase space that will be specified later, h : Œ0,1/ � B ! X , f : Œ0,1/ � X � B ! X are given functions, andxt : .�1, 0�! X is the history of the state function defined by xt.�/ D x.tC �/ for � 2 .�1, 0�.

In recent years, abstract neutral functional differential equations similar to (1) have been studied extensively. In most cases, thequestion of existence of integral solutions has been considered. Let us refer to some recent works, without being exhaustive withquotations: for ˛ D 1, we cite [1,2] for neutral functional differential equations with finite delays and [3] for a case of unbounded delays.Furthermore, in the case ˛ 2 .0, 1/, we refer the reader to [4] for equation with finite delays and [5, 6] for infinite delay ones. Althoughproblem (1)–(2) has been an increasingly interesting subject, no attempt has been made to find its decay solutions with explicit decayrates, up to our knowledge. This is the main motivation for our study.

In investigation of stability for differential equations, Burton and Furumochi [7, 8] introduced a new approach that deploys the fixedpoint theory to search for solutions lying in a stable subset of the space of state functions. They noticed that this method helps inovercoming difficulties that appeared while using the Lyapunov method to deal with differential equations containing unboundeddelays or unbounded parts. We make use of this idea to prove the existence of decay integral solutions to (1)–(2), but now with thequestion of determining decay rate of solutions. For this purpose, we will set up some features:

� deriving new estimates for fractional resolvent operators,� constructing a satisfactory space of solutions and find on this space a regular measure of noncompactness (MNC).

These features enable us to utilize the fixed point theory for condensing maps. Consequently, we obtain the existence of decay integralsolutions x with kx.t/k D O .t�˛/ as t!1. Because the case ˛ 2 .0, 1/ is more involved than the case ˛ D 1, we focus on the formerone and make a note that our technique can be applied to the latter case with the same manner.

Department of Mathematics, Hanoi National University of Education, 136 Xuan Thuy, Cau Giay, Hanoi, Vietnam* Correspondence to: Tran Dinh Ke, Department of Mathematics, Hanoi National University of Education, 136 Xuan Thuy, Cau Giay, Hanoi, Vietnam.† E-mail: [email protected]

Copyright © 2014 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2014

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N.T. ANH AND T.D. KE

Our paper is organized as follows. In the next section, we recall some notions and facts concerning the fractional calculus, includingthe representation for integral solutions of (1)–(2). In addition, we construct a regular MNC on BC.0,1; X/ (the space of continuousbounded functions from .0,1/ to X) and give a fixed point principle, which will be used for our purpose. For the completeness, inSection 3, we prove the global existence for (1)–(2) on interval .�1, T� for each T > 0, under some regular conditions imposed onthe nonlinearities h and f . Section 4 is devoted to showing that decay integral solutions with the certain decay rate exist if someadditional conditions for h, f and the phase space B are added. In the last section, we give an example involving neutral functionalpartial differential equations in unbounded domain to illustrate the obtained results.

2. Preliminaries

2.1. Fractional calculus

Let L1.0, T ; X/ be the space of integrable functions on Œ0, T� in the sense of Bochner.

Definition 2.1The fractional integral of order ˛ > 0 of a function f 2 L1.0, T ; X/ is defined by

I˛0 f .t/ D1

�.˛/

Z t

0.t � s/˛�1f .s/ds,

where � is the Gamma function, provided the integral converges.

Definition 2.2For a function f 2 CN.Œ0, T�; X/, the Caputo fractional derivative of order ˛ 2 .N � 1, N� is defined by

D˛0 f .t/ D

8̂<:̂

1

�.N � ˛/

Z t

0.t � s/N�˛�1f .N/.s/ds, if ˛ 2 .N � 1, N/,

f .N/.t/, if ˛ D N.

It should be noted that there are some notions of fractional derivatives, in which the Riemann–Liouville and Caputo definitionshave been used widely. Many application problems, expressed by differential equations of fractional order, require initial conditionsrelated to u.0/, u0.0/, and so on, and the Caputo fractional derivative satisfies these demands. For u 2 CN.Œ0, T�; X/, we have thefollowing formulas

D˛0 I˛0 u.t/ D u.t/,

I˛0 D˛0 u.t/ D u.t/ �N�1XkD0

u.k/.0/

kŠtk .

Let fS.t/, t � 0g be the analytic semigroup generated by A and fS˛.t/, P˛.t/ : t � 0g be the family of operators on X defined by

S˛.t/z D

Z 10

�˛.�/S�

t˛��

zd� , (3)

P˛.t/z D ˛

Z 10

��˛.�/S�

t˛��

zd� , z 2 X , (4)

where �˛ is a probability density function defined on .0,1/, that is, �˛.�/ � 0 andR1

0 �˛.�/d� D 1. Moreover, �˛ has the expression

�˛.�/ D1

˛��1� 1

˛ ˛

���

�, (5)

˛.�/ D1

1XnD1

.�1/n�1��˛n�1 �.n˛ C 1/

nŠsin.n�˛/. (6)

By the same arguments as in [4], we have the following formula derived from (1) and (2):

x.t/ D h .t, xt/C S˛.t/ Œx.0/ � h .0, x0/�C

Z t

0.t � s/˛�1AP˛.t � s/h .s, xs/ dsC

Z t

0.t � s/˛�1P˛.t � s/f .s, x.s/, xs/ ds. (7)

Copyright © 2014 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2014

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N.T. ANH AND T.D. KE

Proposition 2.1We have the following properties:

1. S˛.�/ and P˛.�/ are norm continuous, that is, t 7! S˛.t/ and t 7! P˛.t/ are continuous for t > 0;2. If S.�/ is a compact semigroup, then S˛.t/ and P˛.t/ are compact for t > 0.

For the proof of the last proposition, see [9].Let p > 1

˛. We define the operator Q˛ : Lp.0, T ; X/! C.Œ0, T�; X/ as follows:

Q˛.f /.t/ D

Z t

0.t � s/˛�1P˛.t � s/f .s/ds. (8)

Using Proposition 2.1, we obtain the following.

Proposition 2.2The operator Q˛ defined by (8) maps any bounded set in Lp.0, T ; X/ into an equicontinuous one in C.Œ0, T�; X/.

ProofBecause P˛ is norm continuous, the operator

ˆ.t, s/ D .t � s/˛�1P˛.t � s/

satisfies the assumption of [10, Lemma 1]. It follows that Q˛.�/ is equicontinuous for each bounded set � � Lp.0, T ; X/. The proof iscomplete.

Now, we prove some important estimates for S˛ and P˛ in the case 0 2 �.A/. In this case, we have (see [11])

kS.t/kL.X/ � Me�at , t � 0,���.�A/ˇS.t/���L.X/� Cˇt�ˇe�at ,ˇ > 0,

where a, M, Cˇ are positive constants and k � kL.X/ is the operator norm in L.X/, the space of bounded linear operators on X .For a given 0 < ˛ < 1, ı > �1, let us introduce the function F˛,ı : Œ0,1/! Œ0,1/ defined by

F˛,ı.s/ D

Z 10

e�s��ı�˛.�/d� , s � 0, (9)

where �˛ is defined by (5) and (6).

Lemma 2.3There exists a constant Dı such that

0 � F˛,ı.s/ �Dı

s1Cı,8s > 0. (10)

ProofIt is known that (see, for example, [12, (F.15), p. 241])

�˛.�/ D1

2� i

ZHa

e����˛

˛�1d , � � 0,

where Ha (Hankel path) is a loop in the complex plane, which starts and ends at �1 and encircles the circular disk j j � �1=˛ . Thus,we have

F˛,ı.s/ D

Z 10

e�s��ı�˛.�/d� D

Z 10

e�s��ı�

1

2� i

ZHa

e����˛

˛�1d

�d�

D1

2� i

ZHa

e�˛�1

Z 10

�ıe��.sC�˛/d�d D

�.1C ı/

2� i

ZHa

e�˛�1

.˛ C s/ıC1d . (11)

Here, we have used the indentity (see, for example, [12, (A.21)])

Z 10

�ıe��.sC�˛/d� D

�.1C ı/

.˛ C s/ıC1. (12)

Copyright © 2014 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2014

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N.T. ANH AND T.D. KE

Because 0 < ˛ < 1, we can take a real number satisfying

2< < min

� �2˛

,��

. (13)

By repeating the arguments in [13, Section 1.1.6], we can transform the contour Ha in (11) to obtain the following representation

F˛,ı.s/ D�.1C ı/

2� i

Z�.1,�/

e�˛�1

.˛ C s/ıC1d , (14)

where by �.�,'/ .� > 0, 0 < ' � �/, we denote the contour consisting of three parts of the complex plane as follows:

1. arg D �', j j � �;2. j j D �, �' � arg � ';3. arg D ', j j � �.

The contour is traced so that arg is non-decreasing. Now, by substituting D 1=˛ in (14), we arrive at the following integralrepresentation

F˛,ı.s/ D�.1C ı/

2�˛i

Z�.1,˛�/

exp� 1=˛

�. C s/ıC1

d . (15)

It is noted that �.1,˛/ is contained in right half of the complex plane due to ˛ < �2 by (13). Thus we see that

j C sj D j � .�s/j � j � sj D s,8 2 �.1,˛/.

Therefore, from (15) we have

jF˛,ı.s/j ��.1C ı/

2�˛s1Cı

Z�.1,˛�/

ˇ̌̌exp

� 1=˛

�ˇ̌̌d ,8s > 0. (16)

Consequently, we obtain the assertion of the lemma with

Dı D�.1C ı/

2�˛

Z�.1,˛�/

ˇ̌̌exp

� 1=˛

�ˇ̌̌d .

Here it is noted that

1

2�˛

Z�.1,˛�/

ˇ̌̌exp

�ˇ̌̌d < C1,

since for such that arg D ˛, j j � 1 the following holds:

ˇ̌̌exp

�ˇ̌̌D exp

�j j

1˛ cos

�,

where cos < 0 due to (13).

Lemma 2.4For any fixed t � 0, S˛.t/ and P˛.t/ are linear bounded operators, moreover,

kS˛.t/kL.X/ � M min

�1,

D0

at˛

�(17)

and

kP˛.t/kL.X/ � ˛M min

�1

�.1C ˛/,

D1

a2t2˛

�, (18)

for all t > 0, where D0, D1 are constants defined in Lemma 2.3.

Copyright © 2014 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2014

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N.T. ANH AND T.D. KE

ProofIt is well-known that (see, e.g., [4, Lemma 3.2])

kS˛.t/kL.X/ � M, kP˛.t/kL.X/ �˛M

�.1C ˛/,8t � 0. (19)

By (3) and (10) we have

kS˛.t/kL.X/ �

Z 10

�˛.�/��S�� t˛

���L.X/ d�

� M

Z 10

�˛.�/e�at˛�d� D MF˛,0

�at˛

��

MD0

at˛(20)

and

kP˛.t/kL.X/ � ˛

Z 10

��˛.�/��S�� t˛

���L.X/ d�

� M

Z 10

˛��˛.�/e�at˛�d� D ˛MF˛,1

�at˛

��˛MD1

a2t2˛, (21)

for all t > 0. Thus (17) and (18) follow from (19), (20) and (21).

Lemma 2.5For any x 2 X , ˇ 2 .0, 1/ and � 2 .0, 1�, we have

� AP˛.t/x D .�A/1�ˇP˛.t/.�A/ˇx,8t � 0, (22)

and ��.�A/�P˛.t/��L.X/ � min

�˛C�t˛�

.�.2 � �/

�.1C ˛.1 � �//,˛C�D1��

a2��t2˛

�,8t > 0. (23)

ProofIn Lemma 3.5 of [4], it was proved that the equality (22) holds and

��.�A/�P˛.t/��L.X/ �

˛C�t˛�

.�.2 � �/

�.1C ˛.1 � �//for all t > 0. (24)

Now by (4) and (10), we obtain

��.�A/�P˛.t/��L.X/ � ˛

Z 10

��˛.�/��.�A/�S

�t˛�

���L.X/ d�

�˛C�t˛�

Z 10

�1���˛.�/e�at˛�d�

D˛C�t˛�

F˛,1��

�at˛

��˛C�t˛�

.D1��

.at˛/2�� D˛C�D1��

a2��t2˛, (25)

for all t > 0. The proof is complete.

2.2. Phase space

Let .B, j � jB/ be a semi-normed linear space, consisting of functions mapping .�1, 0� into a Banach space X . The definition of a phasespace B, introduced in [14], is based on the following axioms stating that, if a function v : .�1, T C � ! X is such that vjŒ� ,TC�� 2

C.Œ , T C �; X/ and v� 2 B, then

(B1) vt 2 B for t 2 Œ , T C �;(B2) the function t 7! vt is continuous on Œ , T C �;(B3) jvtjB � K.t � / supfkv.s/k : � s � tg CM.t � /jv� jB , where K , M : Œ0,1/! Œ0,1/, K is continuous, M is locally bounded,

and they are independent of v.

Let us give some examples of phase spaces. The first one is given by

C� D

� 2 C..�1, 0�; X/ : lim

�!�1e���.�/ exists in X

, (26)

Copyright © 2014 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2014

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N.T. ANH AND T.D. KE

where � is a positive number. This phase space satisfies (B1)-(B3) with

K.t/ D 1, M.t/ D e�� t , (27)

and it is a Banach space with the norm

j�j� D sup��0

e��k�.�/k.

Regarding another typical example, suppose that 1 � p < C1, 0 � r < C1 and g : .�1,�r� ! R is nonnegative,Borel measurable function on .�1,�r/. Let CLp

g is a class of functions ' : .�1, 0� ! X such that ' is continuous on Œ�r, 0� andg.�/k'.�/kp 2 L1.�1,�r/. A seminorm in CLp

g is given by

j'jCLpgD sup�r���0

fk'.�/kg C

�Z �r

�1

g.�/k'.�/kpd�

� 1p

. (28)

Assume further that

Z �r

sg.�/d� < C1, for every s 2 .�1,�r/ and (29)

g.sC �/ � G.s/g.�/ for s � 0 and � 2 .�1,�r/, (30)

where G : .�1, 0� ! RC is locally bounded. We know from [15] that if (29)-(30) hold true, then CLpg satisfies (B1)-(B3). Moreover, one

can take

K.t/ D

(1 for 0 � t � r,

1C R�r�t g.�/d�

�1=pfor t > r;

(31)

M.t/ D

8̂̂<ˆ̂:

maxn

1C R�r�r�t g.�/d�

�1=p, G.�t/1=p

ofor 0 � t � r,

max

hR�t�r�t g.�/d�

i1=p, G.�t/1=p

for t > r.

(32)

For more examples of phase spaces, see [15].

2.3. Measure of noncompactness and condensing operators

Let E be a Banach space. Denote

P.E/ D fB � E : B ¤ ;g,

B.E/ D fB 2 P.E/ : B is bounded g.

We will use the following definition of MNC given in [16].

Definition 2.3A function ˇ : B.E/! RC is called MNC in E if

ˇ .co �/ D ˇ.�/ for every � 2 B.E/,

where co � is the closure of the convex hull of�. An MNC ˇ is called

(i) monotone if�0,�1 2 B.E/, �0 � �1 implies ˇ .�0/ � ˇ.�1/;(ii) nonsingular if ˇ.fag [�/ D ˇ.�/ for any a 2 E,� 2 B.E/;

(iii) invariant with respect to union with compact set if ˇ.K [�/ D ˇ.�/ for every relatively compact set K � E and� 2 B.E/;(iv) algebraically semi-additive if ˇ.�0 C�1/ � ˇ.�0/C ˇ.�1/ for any�0,�1 2 B.E/;(v) regular if ˇ.�/ D 0 is equivalent to the relative compactness of�.

An important example of MNC is the Hausdorff MNC �.�/, which is defined as follows

�.�/ D inff" : � has a finite "-net g.

Copyright © 2014 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2014

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N.T. ANH AND T.D. KE

For T > 0, it is known that the Hausdorff MNC on C .Œ0, T�;Rn/ is given by (see [17])

�T .D/ D1

2limı!0

supx2D

maxt,s2Œ0,T�,jt�sj<ı

kx.t/ � x.s/k. (33)

The last measure can be seen as the modulus of equicontinuity of a subset in C .Œ0, T�;Rn/. In C.Œ0, T�; X/ with X being infinitedimensional, there is no such formulation as (33). However, if D � C.Œ0, T�; X/ is an equicontinuous set then

�T .D/ D supt2Œ0,T�

�.D.t//, (34)

here � is the Hausdorff MNC in X .Consider the space BC

�RC; X

�of bounded continuous functions on Œ0,1/ taking values on X . Denote by �T the restriction operator

on this space, that is �T .x/ is the restriction of x on Œ0, T�. Then

�1.D/ D supT>0

�T .�T .D// , D � BC�RC; X

�, (35)

is an MNC. One can check that this MNC satisfies all properties given in Definition 2.3, but regularity. Indeed, we will testify this claimby choosing the sequence ffkg � BC

�RC;R

�as follows

fk.t/ D

8̂̂<ˆ̂:

0, t 62 Œk, kC 1�,

2t � 2k, t 2

k, kC 12

�,

�2tC 2kC 2, t 2

kC 12 , kC 1

�.

Then it is obvious that f�T .fk/g is compact (converging to 0 in C.Œ0, T�;R/) for any T > 0. However, one sees that

supt�0jfk.t/ � fl.t/j D 1 for k ¤ l,

and then ffkg is not a Cauchy sequence in BC�RC;R

�. This fact tells us that �T .�T .ffkg/// D 0 for any T > 0 and then �1.ffkg/ D 0,

but ffkg is non-compact.We also make use of the following MNCs on BC

�RC; X

�(see [18] for the case X D Rn):

dT .D/ D supx2D

supt�Tkx.t/k, (36)

d1.D/ D limT!1

dT .D/, (37)

��.D/ D �1.D/C d1.D/. (38)

The regularity of �� will be proved in the following lemma.

Lemma 2.6The MNC �� defined by (38) is regular on BC

�RC; X

�.

ProofLet D � BC

�RC; X

�be a bounded set such that ��.D/ D 0. We show that D is relatively compact. Let PBC

�RC; X

�be the space of

piecewise continuous and bounded functions on RC, taking values in X . This is a Banach space with the norm

kxkPBC D supt�0kx.t/k,

and contains BC�RC; X

�as a closed subspace.

For � > 0, since d1.D/ D 0 one can choose T > 0 such that supt�T kx.t/k < 2 ,8x 2 D. This means that

kx � �T .x/kPBC <�

2,8x 2 D,

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N.T. ANH AND T.D. KE

here �T .x/ agrees with a function in PBC�RC; X

�by the following manner

�T .x/ D

8<:

x.t/, t 2 Œ0, T�,

0, t > T .

Now since �T .D/ is a compact set in C.Œ0, T�; X/, we can write

�T .D/ �N[

iD1

BT

�xi ,�

2

�(39)

where xi 2 C.Œ0, T�; X/, i D 1, : : : , N, the notation BT .x, r/ stands for the ball in C.Œ0, T�; X/ centered at x with radius r. Define

Oxi.t/ D

8<:

xi.t/, t 2 Œ0, T�,

0, t > T ,

then˚Oxi

�N

iD1belong to PBC

�RC; X

�. We assert that

D �N[

iD1

B1�Oxi ,�

2

�,

here B1.x, r/ is the ball in PBC�RC; X

�with center x and radius r. Indeed, let x 2 D then by (39), there is a number k 2 f1, : : : , Ng

such that

k�T .x/ � xkkC.Œ0,T�;X/ <�

2.

This implies that ���T .x/ � Oxk

��PBC

<�

2.

Then

kx � OxkkPBC � kx � �T .x/kPBC C k�T .x/ � OxkkPBC

<�

2C�

2D �.

Thus x 2 B1�Oxk , �

�. We have D �

NSiD1

B1�Oxi , �

�, and hence D is relatively compact in PBC

�RC; X

�. Since BC

�RC; X

�and PBC

�RC; X

�have the same norm, we conclude that D is a relatively compact set in BC

�RC; X

�. The proof is complete.

In the sequel, we need some basic MNC estimates. Recall that based on the Hausdorff MNC � in X , one can define the sequential MNC�0 as follows:

�0.�/ D supf�.D/ : D 2 �.�/g, (40)

where�.�/ is the collection of all at-most-countable subsets of� (see [17]). We know that

1

2�.�/ � �0.�/ � �.�/, (41)

for all bounded set� � X . Then the following property is evident.

Proposition 2.7Let � be the Hausdorff MNC in X and� � X be a bounded set. Then for every � > 0, there exists a sequence fxng � � such that

�.�/ � 2� .fxng/C �.

We have the following estimate, whose proof can be found in [16].

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N.T. ANH AND T.D. KE

Proposition 2.8If fwng � L1.0, T ; X/ such that

kwn.t/k � �.t/, for a.e. t 2 Œ0, T�,

for some � 2 L1.0, T/, then we have

�Z t

0wn.s/ds

�� 2

Z t

0� .fwn.s/g/ ds

for t 2 Œ0, T�.

Using Proposition 2.7 and 2.8, we get

Proposition 2.9Let D � L1.0, T ; X/ such that

(1) k�.t/k � �.t/, for all � 2 D and for a.e. t 2 Œ0, T�,(2) �.D.t// � q.t/ for a.e. t 2 Œ0, T�,

where �, q 2 L1.0, T/. Then

�Z t

0D.s/ds

�� 4

Z t

0q.s/ds,

here

Z t

0D.s/ds D

Z t

0�.s/ds : � 2 D

.

ProofFor � > 0, there exists a sequence �n 2 D such that

�Z t

0D.s/ds

�� 2�

�Z t

0�n.s/ds

�C �,

thanks to Proposition 2.7. Applying Proposition 2.8 for the last expression, we have

�Z t

0D.s/ds

�� 4

Z t

0� .f�n.s/g/ dsC �

� 4

Z t

0q.s/dsC �.

Since � is arbitrary, we get the conclusion of the proposition as desired.

To end this section, we recall a fixed point principle for condensing maps that will be used in the next section.

Definition 2.4A continuous map F : Z E ! E is said to be condensing with respect to an MNC ˇ (ˇ-condensing) if for any bounded set � � Z,the relation

ˇ.�/ � ˇ.F.�//

implies the relative compactness of�.

Let ˇ be a monotone nonsingular MNC in E. The application of the topological degree theory for condensing maps (see, e.g., [16, 17])yields the following fixed point principle.

Theorem 2.10[16, Corollary 3.3.1] Let M be a bounded convex closed subset of E and let F : M!M be a ˇ-condensing map. Then Fix.F/ :D fx DF.x/g is a non-empty compact set.

3. Existence result

In formulation of problem (1)-(2), we assume that

(A) A is the infinitesimal generator of an analytic semigroup fS.t/g such that 0 2 �.A/ and

kS.t/kL.X/ � Me�at , t � 0, (42)

where a, M are positive constants.

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N.T. ANH AND T.D. KE

(B) The phase space B verifies (B1)-(B3).(H) h : Œ0, T� � B! X is a continuous function such that

k.�A/ˇŒh.t,�/ � h.t, /�k � �.t/j� � jB ,

for all t 2 Œ0, T�, and �, 2 B, here ˇ 2 .0, 1/ and � is a continuous function.(F) f : Œ0, T� � X � B! X is a function satisfying that

(1) t 7! f .t, v, w/ is measurable for each .v, w/ 2 X � B and .v, w/ 7! f .t, v, w/ is continuous for a.e. t 2 .0, T/;(2) there exists a function m 2 Lp.0, T/, p > 1

˛, such that

kf .t, v, w/k � m.t/.kvk C jwjB/, 8v 2 X , w 2 B,

and for a.e. t 2 .0, T/;(3) if the semigroup S.�/ generated by A is non-compact then for any bounded sets B � X , C � B, we have

�.f .t, B, C// � k.t/

"�.B/C sup

��0�.C.�//

#,

for a.e. t 2 .0, T/, where k 2 Lp.0, T/ is a nonnegative function.

Remark 3.1If assumption (A) holds, then one can define the operator .�A/ˇ for ˇ 2 .0, 1/ as follows (see [11]):

.�A/ˇz D

�1

�.ˇ/

Z 10

tˇ�1etAzdt

��1

, z 2 X .

Motivated by (7), we have the following definition.

Definition 3.1A function x : .�1, T�! X is said to be an integral solution of problem (1)-(2) iff x.t/ D '.t/ for t � 0 and

x.t/ D h .t, xt/C S˛.t/Œ'.0/ � h.0,'/�C

Z t

0.t � s/˛�1AP˛.t � s/h.s, xs/ds

C

Z t

0.t � s/˛�1P˛.t � s/f .s, x.s/, xs/ds, for t > 0.

For ' 2 B and y 2 C.Œ0, T�; X/, we denote the function yŒ'� : .�1, T�! X as follows:

yŒ'�.t/ D

8<:

y.t/ for t 2 Œ0, T�,

'.t/ for t < 0.

Put

C' D fy 2 C.Œ0, T�; X/ : y.0/ D '.0/g.

Then C' is a closed subspace of C.Œ0, T�; X/with the norm

kykC D supt2Œ0,T�

ky.t/k.

For given ' 2 B, we define the solution operator † : C' ! C' as follows

†.x/.t/ D h .t, xŒ'�t/C S˛.t/ Œ'.0/ � h.0,'/�C

Z t

0.t � s/˛�1AP˛.t � s/h .s, xŒ'�s/ dsC

Z t

0.t � s/˛�1P˛.t � s/f .s, x.s/, xŒ'�s/ds. (43)

It is obvious that if x is a fixed point of† then xŒ'� is an integral solution of (1)-(2) on .�1, T�. Due to the continuity of h and f , it is easilyproved that† is continuous.

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N.T. ANH AND T.D. KE

Denote

�T D supt2Œ0,T�

�.t/, KT D supt2Œ0,T�

K.t/, MT D supt2Œ0,T�

M.t/;

ƒT D supt2Œ0,T�

Z t

0.t � s/˛�1

���.�A/1�ˇP˛.t � s/���L.X/

K.s/�.s/ds;

‚T D

(0, if S.�/ is compact ,

8 supt2Œ0,T�

R t0 .t � s/˛�1kP˛.t � s/kL.X/k.s/ds, otherwise ;

‡T D supt2Œ0,T�

Z t

0.t � s/˛�1kP˛.t � s/kL.X/Œ1C K.s/�m.s/ds.

Lemma 3.1Let (A), (B), (H) and (F) hold. Then

�T .†.D// �

��T KT

���.�A/�ˇ���L.X/CƒT C‚T

��T .D/,

for any bounded set D � C' .

ProofLet

†1.x/.t/ D h .t, xŒ'�t/C S˛.t/Œ'.0/ � h.0,'/�,

†2.x/.t/ D

Z t

0.t � s/˛�1AP˛.t � s/h.s, xŒ'�s/ds,

†3.x/.t/ D

Z t

0.t � s/˛�1P˛.t � s/f .s, x.s/, xŒ'�s/ds,

and DŒ'� D fyŒ'� : y 2 Dg. We first give an estimate for†1. Let x1, x2 2 D, then

k†1.x1/.t/ �†1.x2/.t/k D kh.t, x1Œ'�t/ � h.t, x2Œ'�t/k

D���.�A/�ˇ.�A/ˇ Œh .t, x1Œ'�t/ � h .t, x2Œ'�t/�

�������.�A/�ˇ

���L.X/

�.t/ jx1Œ'�t � x2Œ'�tjB

����.�A/�ˇ

���L.X/

K.t/�.t/ sup0�s�t

kx1.s/ � x2.s/k ,

thanks to (B3) and the fact that x1Œ'�0 D x2Œ'�0 D '. Then

k†1.x1/ �†1.x2/kC � �T KT

���.�A/�ˇ���L.X/� kx1 � x2kC .

The last inequality ensures that

�T .†1.D// � �T KT

���.�A/�ˇ���L.X/� �T .D/. (44)

Regarding†2, we have

k†2.x1/.t/ �†2.x2/.t/k D

����Z t

0.t � s/˛�1.�A/1�ˇP˛.t � s/.�A/ˇ Œh .s, x1Œ'�s/ � h .s, x2Œ'�/� ds

�����

Z t

0.t � s/˛�1

���.�A/1�ˇP˛.t � s/���L.X/

K.s/�.s/ supr2Œ0,s�

kx1.r/ � x2.r/k ds.

Thus

k†2.x1/ �†2.x2/kC �

"sup

t2Œ0,T�

Z t

0.t � s/˛�1

���.�A/1�ˇP˛.t � s/���L.X/

K.s/�.s/ds

#� kx1 � x2kC .

This implies that

�T .†2.D// � ƒT � �T .D/. (45)

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N.T. ANH AND T.D. KE

For †3, we see that ft 7! f .t, x.t/, xŒ'�t/ : x 2 Dg is a bounded set in Lp.0, T ; X/. Then according to Proposition 2.2, †3.D/ is anequicontinuous set. Therefore

�T .†3.D// D supt2Œ0,T�

� .†3.D/.t//

� 4 supt2Œ0,T�

Z t

0.t � s/˛�1� .P˛.t � s/f .s, D.s/, DŒ'�s// ds,

(46)

thanks to Proposition 2.9. If S.�/ is a compact semigroup then by Lemma 2.1, P˛.t � s/ is compact for s < t. This deduces that� .P˛.t � s/f .s, D.s/, DŒ'�s// D 0 for a.e. s 2 .0, t/ and hence �T .†3.D// D 0. In the opposite case, we have

� .P˛.t � s/f .s, D.s/, DŒ'�s// � kP˛.t � s/kL.X/k.s/

"�.D.s//C sup

��0�.DŒ'�.sC �//

#

� kP˛.t � s/kL.X/k.s/

"�.D.s//C sup

r2Œ0,s��.D.r//

#

� 2kP˛.t � s/kL.X/k.s/ supr2Œ0,s�

�.D.r//

� 2kP˛.t � s/kL.X/k.s/ � �T .D/,

(47)

here we use the fact that DŒ'�.r/ D f'.r/g for r � 0. Putting (47) into (46), we obtain

�T .†3.D// � ‚T � �T .D/. (48)

Now combining (44), (45) and (48), we arrive at

�T .†.D// �

��T KT

���.�A/�ˇ���L.X/CƒT C‚T

��T .D/.

The proof is complete.

We are in a position to state the main result of this section.

Theorem 3.2Let the assumptions of Lemma 3.1 hold. Then problem (1)-(2) has at least one integral solution on .�1, T� provided that

�T KT

���.�A/�ˇ���L.X/CƒT C‚T < 1, (49)

�T KT

���.�A/�ˇ���L.X/CƒT C ‡T < 1. (50)

ProofBy (49), the solution operator † is �T -condensing. Indeed, if D � C' is a bounded set such that �T .D/ � �T .†.D//, then Lemma 3.1ensures that

�T .D/ � �T .†.D// � `�T .D/,

for ` D �T KT

��.�A/�ˇ��L.X/ CƒT C‚T < 1. Therefore �T .D/ D 0 and then D is relatively compact.

According to Theorem 2.10, it remains to show that one can find R > 0 verifying the inclusion†.BR/ � BR, where BR is the ball in C'centered at origin with radius R. Assume to the contrary that for each n 2 N , there exists xn 2 C' with kxnkC � n but k†.xn/kC > n.Then we have

k†.xn/.t/k ����.�A/�ˇ

���L.X/

���.�A/ˇh.t, xnŒ'�t/���C kS˛.t/Œ'.0/ � h.0,'/�k

C

Z t

0.t � s/˛�1

���.�A/1�ˇP˛.t � s/���L.X/

���.�A/ˇh.s, xnŒ'�s/��� ds

C

Z t

0.t � s/˛�1kP˛.t � s/kL.X/ � kf .s, xn.s/, xnŒ'�s/kds

� �.t/���.�A/�ˇ

���L.X/jxnŒ'�tjB C

���.�A/�ˇ���L.X/kh.t, 0/k C kS˛.t/Œ'.0/ � h.0,'/�k

C

Z t

0.t � s/˛�1k.�A/1�ˇP˛.t � s/kL.X/ Œ�.s/jxnŒ'�sjB C kh.s, 0/k� ds

C

Z t

0.t � s/˛�1kP˛.t � s/kL.X/m.s/ Œkxn.s/k C jxnŒ'�sjB� ds,

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N.T. ANH AND T.D. KE

thanks to the assumptions (H) and (F).Noting that

jxnŒ'�sjB � K.s/ supr2Œ0,s�

kxn.r/k CM.s/j'jB

� K.s/kxnkC CMT j'jB

� K.s/nCMT j'jB ,

we have

k†.xn/.t/k � �.t/K.t/���.�A/�ˇ

���L.X/� n

C n

Z t

0.t � s/˛�1

���.�A/1�ˇP˛.t � s/���L.X/

K.s/�.s/ds

C n

Z t

0.t � s/˛�1kP˛.t � s/kL.X/Œ1C K.s/�m.s/dsCW.t/,

with W.t/ being independent of xn. This implies that

1 <k†.xn/kC

n� �T KT

���.�A/�ˇ���L.X/

C supt2Œ0,T�

Z t

0.t � s/˛�1

���.�A/1�ˇP˛.t � s/���L.X/

K.s/�.s/ds

C supt2Œ0,T�

Z t

0.t � s/˛�1kP˛.t � s/kL.X/Œ1C K.s/�m.s/ds

Csupt2Œ0,T� W.t/

n

D �T KT

���.�A/�ˇ���L.X/CƒT C ‡T C

supt2Œ0,T� W.t/

n.

Passing the last relation into limits as n!1, we get a contradiction to (50), which completes the proof.

4. Existence of decay integral solutions

In this section, we consider the solution operator† on the following space:

BC�' D

y 2 C.Œ0,C1/; X/ : y.0/ D '.0/ and sup

t�0t�ky.t/k <1

,

where � is a positive number chosen later. This space is endowed with the supremum norm

kykBC D supt�0ky.t/k,

and it becomes a closed subspace of

BC D fy 2 C.Œ0,C1/; X/ : kykBC <1g .

On BC�' we make use of MNC �� given by (38). We will prove that† keeps BC�' invariant, i.e.†�

BC�'�� BC�' , and† is ��-condensing

on BC�' . To this end, we have to replace (B), (H) and (F) by stronger ones. Specifically, we assume that for � � ˛:

(B*) The phase space B satisfies (B) with K 2 BC�RC;RC

�and M being such that t�M.t/ D O.1/ as t!1.

(H*) The function h verifies (H) for all T > 0 and for � 2 BC�RC,RC

�. In addition, h.t, 0/ D 0 for all t 2 RC.

(F*) f satisfies (F) for all T > 0.

Put

�1 D supt�0

�.t/, K1 D supt�0

K.t/, and M1 D supt�0

M.t/.

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N.T. ANH AND T.D. KE

Lemma 4.1Let (A), (B*), (H*) and (F*) hold. Then†

�BC�'

�� BC�' for all � � ˛.

ProofLet x 2 BC�' with kxkBC D r > 0. Then we have

jxŒ'�tjB � K.t/ sups2Œ0,t�

kx.s/k CM.t/j'jB

� K1rCM1j'jB , 8t � 0.

On the other hand

jxŒ'�tjB � K

�t

2

�sup

s2Œ t2 ,t�kx.s/k CM

�t

2

�jxŒ'� t

2jB

� K1 sups2Œ t

2 ,t�kx.s/k CM

�t

2

�.K1rCM1j'jB/.

Thus

t� jxŒ'�tjB � K1t� sups2Œ t

2 ,t�kx.s/k C t�M

�t

2

�.K1rCM1j'jB/

� K12� sups2Œ t

2 ,t�s�kx.s/k C 2�

�t

2

��M

�t

2

�.K1rCM1j'jB/

D O.1/ as t!1,

(51)

thanks to (B*). We show that t�k†.x/.t/k D O.1/ as t ! 1. Take the decomposition of † as in Lemma 3.1: †.x/ D †1.x/C †2.x/C†3.x/, where

†1.x/.t/ D h .t, xŒ'�t/C S˛.t/ Œ'.0/ � h.0,'/� ,

†2.x/.t/ D

Z t

0.t � s/˛�1AP˛.t � s/h .s, xŒ'�s/ ds,

†3.x/.t/ D

Z t

0.t � s/˛�1P˛.t � s/f .s, x.s/, xŒ'�s/ ds.

We have

t�k†1.x/.t/k � �.t/���.�A/�ˇ

���L.X/

t� jxŒ'�tjB C t�kS˛.t/kL.X/ .k'.0/k C kh.0,'/k/ .

This implies

t�k†1.x/.t/k D O.1/ as t!1, (52)

thanks to (H*), (51) and estimate (17) that kS˛.t/kL.X/ D O .t�˛/ as t!1.For†2, one has

k†2.x/.t/k �

Z t

0.t � s/˛�1

���.�A/1�ˇP˛.t � s/���L.X/����.�A/ˇh .s, xŒ'�s/

��� ds

Z t

0.t � s/˛�1

���.�A/1�ˇP˛.t � s/���L.X/

�.s/ jxŒ'�sjB ds

D

Z t2

0C

Z t�1

t2

C

Z t

t�1

!.t � s/˛�1

���.�A/1�ˇP˛.t � s/���L.X/

�.s/jxŒ'�sjBds

D I1.t/C I2.t/C I3.t/.

Using estimate (23) that��.�A/1�ˇP˛.t/

��L.X/ D O

�t�2˛

�as t!1, we have

I1.t/ � C

Z t2

0.t � s/˛�1.t � s/�2˛�.s/jxŒ'�sjBds

� C

Z t2

0.t � s/�˛�1ds D

C .2˛ � 1/

˛t�˛ ,

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N.T. ANH AND T.D. KE

thanks to the boundedness of �.s/ and jxŒ'�sjB , here we use a generic constant C which may be changed from line to line. Since � � ˛,the last estimate deduces that

t� I1.t/ D O.1/ as t!1. (53)

Considering I2.t/, we get

t� I2.t/ � Ct�Z t�1

t2

.t � s/˛�1.t � s/�2˛jxŒ'�sjBds

� Ct�Z t�1

t2

s�� .t � s/�˛�1s� jxŒ'�sjBds

� C

Z t�1

t2

.t � s/�˛�1ds DC

˛

�1 �

�t

2

��˛�,

thanks to the fact that s�� � 2� t�� and s� jxŒ'�sjB is bounded. Thus

t� I2.t/ D O.1/ as t!1. (54)

Now deploying estimate (23) that��.�A/1�ˇP˛.t/

��L.X/ D O

�t�˛.1�ˇ/

�, we have

t� I3.t/ � Ct�Z t

t�1.t � s/˛�1.t � s/˛ˇ�˛�.s/jxŒ'�sjBds

� Ct�Z t

t�1s�� .t � s/˛ˇ�1s� jxŒ'�sjBds

� C

�t

t � 1

�� Z t

t�1.t � s/˛ˇ�1ds D

C

˛ˇ

�t

t � 1

��,

here we use the boundedness of s� jxŒ'�sjB and the fact s � t � 1. The last estimate leads to

t� I3.t/ D O.1/ as t!1. (55)

From (53)-(55), we obtain

t�k†2.x/.t/k D O.1/ as t!1. (56)

It remains to deal with†3. Proceeding similarly, we have

k†3.x/.t/k �

Z t

0.t � s/˛�1kP˛.t � s/kL.X/m.s/ Œkx.s/k C jxŒ'�sjB� ds

D

Z t2

0C

Z t�1

t2

C

Z t

t�1

!.t � s/˛�1kP˛.t � s/kL.X/m.s/ Œkx.s/k C jxŒ'�sjB� ds

D I4.t/C I5.t/C I6.t/.

Using estimate (18) for P˛ that kP˛.t/kL.X/ D O�

t�2˛�

as t!1, we get

I4.t/ D

Z t2

0.t � s/˛�1kP˛.t � s/kL.X/m.s/ Œkx.s/k C jxŒ'�sjB� ds

� C

Z t2

0.t � s/�˛�1m.s/ds,

thanks to the boundedness of kx.s/k C jxŒ'�sjB . Using the Hölder inequality, one has

I4.t/ � CkmkLp.RC/

"Z t2

0.t � s/�

.˛C1/pp�1 ds

# p�1p

D CkmkLp.RC/

�p � 1

˛pC 1

� p�1p �

2˛pC1

p�1 � 1� p�1

pt�

˛pC1p .

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N.T. ANH AND T.D. KE

Since � � ˛ < ˛pC1p we get

t� I4.t/ � Ct��˛pC1

p D o.1/ as t!1. (57)

For I5.t/, using estimate (18) again, we see that

t� I5.t/ � Ct�Z t�1

t2

.t � s/�˛�1m.s/ Œkx.s/k C jxŒ'�sjB� ds

� Ct�Z t�1

t2

s�� .t � s/�˛�1m.s/

s�kx.s/k C s� jxŒ'�sjB�

ds

� C

Z t�1

t2

.t � s/�˛�1m.s/ds,

thanks to the fact that s � t2 . Then the Hölder inequality gives

t� I5.t/ � CkmkLp.RC/

241 �

�t

2

�� ˛pC1p�1

35

p�1p

D O.1/ as t!1. (58)

Now for I6.t/, we have

t� I6.t/ � Ct�Z t

t�1.t � s/˛�1m.s/ Œkx.s/k C jxŒ'�sjB� ds

D Ct�Z t

t�1s�� .t � s/˛�1m.s/

s�kx.s/k C s� jxŒ'�sjB

�ds

� C

�t

t � 1

�� Z t

t�1.t � s/˛�1m.s/ds,

thanks to the fact that kP˛.t � s/kL.X/ and s�kx.s/k C s� jxŒ'�sjB are bounded. Then using the Hölder inequality again, we get

t� I6.t/ � C

�t

t � 1

��kmkLp.RC/

�Z t

t�1.t � s/

.˛�1/pp�1 ds

� p�1p

D C

�t

t � 1

��kmkLp.RC/

�p � 1

˛p � 1

� p�1p

D O.1/ as t!1. (59)

Therefore, it follows from (57)-(59) that

t�k†3.x/.t/k D O.1/ as t!1. (60)

Finally, combining (52), (56) and (60) yields

t�k†.x/.t/k D O.1/ as t!1.

The proof is complete.

Now denote

ƒ1 D supt�0

Z t

0.t � s/˛�1

���.�A/1�ˇP˛.t � s/���L.X/

K.s/�.s/ds; (61)

‚1 D

(0, if S.�/ is compact ,

8 supt�0

R t0 .t � s/˛�1kP˛.t � s/kL.X/k.s/ds, otherwise ;

(62)

‡1 D supt�0

Z t

0.t � s/˛�1kP˛.t � s/kL.X/Œ1C K.s/�m.s/ds. (63)

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N.T. ANH AND T.D. KE

We will prove the ��-condensing property for†.

Lemma 4.2Let (A), (B*), (H*) and (F*) hold. Then† is ��-condensing on BC�' with 0 < � � ˛, provided that

�1K1���.�A/�ˇ

���L.X/Cƒ1 C‚1 < 1. (64)

ProofLet D � BC�' be a bounded set. We first show that d1.†.D// D 0. Indeed, taking r > 0 such that kxkBC � r for all x 2 D, we have

t�k†.x/.t/k D O.1/ as t!1,

according to the proof of Lemma 4.1. This means

k†.x/.t/k � Ct�� , 8x 2 D,

for all large t. Equivalently, for a large T , one has dT .†.D// � CT�� . Then

d1.†.D// D limT!1

dT .†.D// D 0. (65)

Now arguing as in Lemma 3.1, we have

�T .�T .†.D/// �

��T KT

���.�A/�ˇ���L.X/CƒT C‚T

��T .�T .D//

��1K1

���.�A/�ˇ���L.X/Cƒ1 C‚1

��T .�T .D// .

Then

�1.†.D// D supT>0

�T .�T .†.D///

��1K1

���.�A/�ˇ���L.X/Cƒ1 C‚1

��1.D/. (66)

It follows from (65)-(66) that

��.†.D// D �1.†.D//C d1.†.D//

��1K1

���.�A/�ˇ���L.X/Cƒ1 C‚1

���.D/.

Now if ��.D/ � ��.†.D// then

��.D/ �

��1K1

���.�A/�ˇ���L.X/Cƒ1 C‚1

���.D/.

This implies that ��.D/ D 0 according to condition (64). Since �� is regular due to Lemma 2.6, we get that D is relatively compact. Theproof is complete.

The following theorem is our main result.

Theorem 4.3Let (A), (B*), (H*) and (F*) hold. Then problem (1)-(2) has a compact set of integral solutions in BC

�RC; X

�satisfying kx.t/k D O .t�˛/

as t!C1, provided that

�1K1���.�A/�ˇ

���L.X/Cƒ1 C‚1 < 1, (67)

�1K1���.�A/�ˇ

���L.X/Cƒ1 C ‡1 < 1. (68)

ProofBy Lemma 4.2, condition (67) ensures that† is ��-condensing on BC˛' . On the other hand, by condition (68) and the arguments in theproof of Theorem 3.2, one can find R > 0 such that†.BR/ � BR where BR is the ball in BC˛' with center at origin and radius R. ApplyingTheorem 2.10, we obtain the conclusion as desired.

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N.T. ANH AND T.D. KE

We end this section by giving simple estimates forƒ1,‚1 and‡1 defined by (61)-(63). We have

ƒ1 �˛C1�ˇ�.1C ˇ/

�.1C ˛ˇ/supt�0

Z t

0.t � s/˛ˇ�1K.s/�.s/ds

�C1�ˇ�.ˇ/

�.˛ˇ/K1 sup

t�0

Z t

0.t � s/˛ˇ�1�.s/ds, (69)

thanks to estimate (23). Assume that �.t/ � b�1Ct˛ˇ

for some b� > 0, then

Z t

0.t � s/˛ˇ�1�.s/ds � b�

Z t=2

0

.t � s/˛ˇ�1

1C s˛ˇdsC b�

Z t

t=2

.t � s/˛ˇ�1

1C s˛ˇds

� b�

�t

2

�˛ˇ�1 Z t=2

0

ds

1C s˛ˇC

b�

1C�

t2

�˛ˇZ t

t=2.t � s/˛ˇ�1ds

� b�

�.t/C

1

˛ˇ

�,

where .t/ D�

t2

�˛ˇ�1 R t=20

ds1Cs˛ˇ

. Noting that limt!0

.t/ D 0 and limt!1

.t/ D 12.1�˛ˇ/ , we get that 1 D supt�0 .t/ <1. Thus

Z t

0.t � s/˛ˇ�1�.s/ds � b�

�1 C

1

˛ˇ

�.

Plugging the last inequality in (69), we have

ƒ1 �C1�ˇ�.ˇ/

�.˛ˇ/K1b�

�1 C

1

˛ˇ

�.

Similarly, if k.t/ � bk1Ct˛ and m.t/ � bm

1Ct˛ for some bk , bm > 0, then we have

‚1 �8Mbk

�.˛/

��1 C

1

˛

�, ‡1 �

Mbm

�.˛/.1C K1/

��1 C

1

˛

�,

where

�1 D supt�0

�t

2

�˛�1 Z t=2

0

ds

1C s˛<1.

5. An example

In this section, we apply our abstract results to the following problem with functional partial differential equations:

@12

t

hu.t, x/ � Qh .t, x, ut/

iD .Dx/ u.t, x/C Qf .t, x, u.t, x/, ut/ , x 2 Rn, t > 0, (70)

u.s, x/ D '.s, x/, x 2 Rn, s � 0, (71)

where @˛t .˛ > 0/ stands for the Caputo derivative of order ˛ with respect to t,

Qh.t, x, ut/ D

Z t

�1

ZRn

K.x, y/�.s � t, y/�.t, u.s � t, y//dyds,

Qf .t, x, u.t, x/, ut/ D Qf1.t, x, u.t, x//

C .t, x/

Z 0

�1

ZRn�.� , y/Qf2.y, u.tC � , y//dyd� .

The notation .Dx/ stands for the partial differential operator defined by

.Dx/ DXj˛j�m

a˛D˛x ,

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N.T. ANH AND T.D. KE

where a˛ 2 C and

D˛x D

�@

@x1

�˛1�@

@x2

�˛2

: : :

�@

@xn

�˛n

,

for a multi-index ˛ D .˛1,˛2, : : : ,˛n/ 2 Nn with length j˛j D ˛1 C ˛2 C : : :C ˛n.The symbol of .Dx/ is given by

.�/ DXj˛j�m

a˛.i�/˛ D

Xj˛j�m

a˛.i�1/˛1.i�2/

˛2 : : : .i�n/˛n .

The operator A D .Dx/ is assumed to be elliptic, i.e. the principal part m.�/ DPj˛jDm a˛.i�/˛ of its symbol satisfies

m.�/ ¤ 0 for all 0 ¤ � 2 Rn.

Let X D L2 .Rn/ , D.A/ D Hm .Rn/ and B D CL2g (g and r will be specified in the sequel). In this example, we assume that

(H1) ! D sup2Rn Re .�/ < 0;(H2) there is ı 2 .0, �2 � such that .Rn/ � Cn†ıC �

2, where†ıC�

2D˚� 2 C : jarg�j < ı C �

2

�nf0g.

Under (H1)-(H2), .A, D.A// generates a bounded analytic semigroup fS.t/gt�0 on X (see [19, Theorem 5.15]). Furthermore, adopting theresult in [20, Theorem 2.2] (for the case p D 2, B D I), we have

kS.t/kL.X/ � Ce!t ,8t > 0,

where C is a positive constant and ! is given in (H1).Now we give the description for the nonlinearities Qh and Qf :

(H3) K : Rn �Rn ! R, � : .�1, 0� �Rn ! R and � : RC �R! R such that

(a)p� .Dx/K.x, y/ 2 L2

�Rn; L2 .Rn/

�;

(b) � is continuous, j�.� , x/j � Ceh0� for some C > 0, 0 < h0 < 1,8x 2 Rn;(c) � is continuous and Lipschitzian with respect to second argument, i.e. j�.t, z1/� �.t, z2/j � Q�.t/jz1 � z2j, for all z1, z2 2 R,

where Q� is a continuous function;

(H4) Qf1 : RC �Rn �R! R, : RC �Rn ! R, � : .�1, 0� �Rn ! R and Qf2 : Rn �R! R such that

(a) Qf1 is a continuous function such that Qf1.t, x, 0/ D 0 and

ˇ̌̌Qf1 .t, x, z1/ � Qf1 .t, x, z2/

ˇ̌̌� m1.t/ jz1 � z2j

for all x 2 Rn, z1, z2 2 R, here m1 2 Lp�RC

�with p > 2;

(b) 2 Lp�RC; L2 .Rn/

�;

(c) � is continuous and satisfies j�.t, x/j � C�e�0t for all t 2 .�1, 0�, x 2 Rn, here �0 � h0;(d) Qf2 is continuous and jQf2.y, z/j � `.y/jzj for ` 2 L2 .Rn/.

Let h : RC � B! X , f : RC � X � B! X such that

h.t, w/.x/ D

Z 0

�1

ZRn

K.x, y/�.s, y/g.t, w.s, y//ds

f .t, v, w/.x/ D f1.t, v/.x/C f2.t, w/.x/,

where

f1.t, v/.x/ D Qf1.t, x, v.x//

f2.t, w/.x/ D .t, x/

Z 0

�1

ZRn�.� , y/Qf2.y, w.� , y//dyd� .

Copyright © 2014 John Wiley & Sons, Ltd. Math. Meth. Appl. Sci. 2014

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N.T. ANH AND T.D. KE

Let K0.x, y/ Dp� .Dx/K.x, y/, then we have

���p� .Dx/ Œh.t, w1/ � h.t, w2/����2

X

D

ZRn

�Z 0

�1

ZRn

K0.x, y/�.s, y/Œ�.t, w1.s, y// � �.t, w2.s, y//�dyds

�2

dx

� C2 Q�

2.t/

ZRn

�Z 0

�1

eh0s

ZRn

K0.x, y/jw1.s, y/ � w2.s, y/jdyds

�2

dx

� C2 Q�

2.t/

ZRn

�Z 0

�1

eh0skK0.x, �/kX � kw1.s, �/ � w2.s, �/kX ds

�2

dx

D C2 Q�

2.t/

ZRnkK0.x, �/k2

X

�Z 0

�1

eh0skw1.s, �/ � w2.s, �/kX ds

�2

dx,

thanks to the Hölder inequality. Since

�Z 0

�1

eh0skw1.s, �/ � w2.s, �/kX ds

�2

D

�Z 0

�1

e12 h0se

12 h0skw1.s, �/ � w2.s, �/kX ds

�2

�1

h0

Z 0

�1

eh0skw1.s, �/ � w2.s, �/k2X ds. (72)

Then

kp� .Dx/ Œh.t, w1/ � h.t, w2/� k

2X

� C2 Q�

2.t/kK0k

2L2.Rn ;X/

h0

Z 0

�1

eh0s kw1.s, �/ � w2.s, �/k2X ds

D C2 Q�

2.t/kK0k

2L2.Rn ;X/

h0

�Z 0

�rC

Z �r

�1

�eh0skw1.s, �/ � w2.s, �/k2

X ds

� C2 Q�

2.t/kK0k

2L2.Rn ;X/

h0

�1 � e�h0r

h0kw1 � w2k

2C.Œ�r,0�;X/ C

Z �r

�1

eh0s kw1.s, �/ � w2.s, �/k2X ds

�.

Choose B D CL2g (described in (28)) with r D � 1

h0ln.1 � h0/ and g.s/ D eh0s. Then the last inequality means that

���p� .Dx/ Œh .t, w1/ � h .t, w2/����

X�

Cp

h0kK0kL2.Rn ;X/ Q�.t/jw1 � w2jB .

Accordingly, h satisfies (H) with �.t/ D Cph0kK0kL2.Rn ;X/ Q�.t/.

Concerning f1, we have

kf1 .t, v1/ � f1 .t, v2/kX � m1.t/kv1 � v2kX .

This implies

� .f1.t, V// � m1.t/�.V/, for all bounded set V � X . (73)

Regarding f2, we observe that

kf2.t, w/k2X D k.t, �/k2

X

�Z 0

�1

ZRn�.� , y/Qf2.y, w.� , y//dyd�

�2

� k.t, �/k2X C2�

�Z 0

�1

e�0s

ZRn`.y/jw.s, y/jdyds

�2

� k.t, �/k2X C2�k`k

2X

�Z 0

�1

e�0skw.s, �/kX ds

�2

�1

�0k.t, �/k2

X C2�k`k

2X

Z 0

�1

eh0skw.s, �/k2X ds

D1

�0k.t, �/k2

X C2�k`k

2X

�Z 0

�rC

Z �r

�1

�eh0skw.s, �/k2

X ds

�1

�0k.t, �/k2

X C2�k`k

2X

�kwk2

C.Œ�r,0�;X/ C

Z �r

�1

eh0skw.s, �/k2X ds

�,

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N.T. ANH AND T.D. KE

thanks to the Hölder inequality and the argument as in (72), here we take into account the hypothesis �0 � h0. Then we have

kf2.t, w/kX �C�p�0k.t, �/kXk`kX jwjB . (74)

On the other hand, for any bounded set W � B, we see that

f2.t, W/ � f�.t, �/ : � 2 Rg,

that is, f2.t, W/ lies in an one dimensional subspace of X. Hence

�.f2.t, W// D 0. (75)

It follows from (73) and (75) that

�.f .t, V , W// � � .f1.t, V//C �.f2.t, W// � m1.t/�.V/.

Thus f fulfills (F) with k.t/ D m1.t/ and m.t/ D maxfm1.t/,C�p�0k.t, �/kXk`kXg.

As far as the phase space is concerned, for B D CL2g with r D � 1

h0ln.1� h0/ and g.s/ D eh0s, one sees that (29)-(30) are satisfied with

G.s/ D g.s/. Then B verifies (B1)-(B3) with

K.t/ D

(1, 0 � t � r,

1C 1ph0

pe�h0r � e�h0t , t > r.

M.t/ D

8̂<:̂

max

e�

12 h0t , 1C

qe�h0 r

h0

�1 � e�h0t

�, 0 � t � r,

e�12 h0t , t > r,

thanks to the expressions of K and M in (31)-(32). This implies B satisfies (B*).By the above description for (70)-(71), we can apply Theorem 3.2 and 4.3 to get the existence of integral solutions on .�1, T� for all

T > 0 as well as the existence of decay solutions with decay rate described byp

tku.t, �/kX D O.1/ as t!1.

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