Power-type estimates for a nonlinear fractional differential equation

Nonlinear Analysis 62 (2005) 1025–1036www.elsevier.com/locate/na

Power-type estimates for a nonlinear fractionaldifferential equation

Khaled M. Furati, Nasser-eddine Tatar∗Department of Mathematical Sciences, King Fahd University of Petroleum and Minerals,

Dhahran 31261, Saudi Arabia

Received 28 October 2004; accepted 13 April 2005

Abstract

In thiswork,weareconcernedwitha family of nonlinearordinarydifferential equationsof fractionalorder. It is proved that solutions of these equations with weighted initial data exist globally and decayas a power function.� 2005 Elsevier Ltd. All rights reserved.

MSC:26A33 (primary); 34C11; 34D05; 45D05; 45E10

Keywords:Asymptotic behavior; Fractional derivative; Riemann–Liouville integral; Singular kernel; WeightedCauchy-type problem

1. Introduction

We consider the weighted Cauchy-type problem{D�u(t)= f (t, u), t >0,t1−�u(t)|t=0 = b, (1)

whereD� is the fractional derivative (in the sense ofRiemann–Liouville) of order 0< �<1,f is a continuous nonlinear function andb ∈ R∗ (the set of all real numbers except 0).The local existence and the uniqueness for this problem (withf (t, u) = f (u) = us ,

0<s <1/(1− �)) have been investigated by Delbosco and Rodino in[8]. This has been

∗ Corresponding author. Tel.: +96638604654; fax: +96638602340.E-mail addresses:kmfurati@kfupm.edu.sa(K.M. Furati), tatarn@kfupm.edu.sa(N.-e. Tatar).

0362-546X/$ - see front matter� 2005 Elsevier Ltd. All rights reserved.doi:10.1016/j.na.2005.04.010

1026 K.M. Furati, N.-e. Tatar / Nonlinear Analysis 62 (2005) 1025–1036

established in a weighted space of continuous functions. This problem has also been studiedbyKilbaset al. in[14] for complex valuesof�.Global existenceanduniquenesswereprovedin the space of summable functions on an interval.There are a large number of papers in the literature dealingwith the equation in (1) subject

to the initial condition(I1−� u)(0)=b, whereI1−� is the fractional integral (in the sense ofRiemann–Liouville), see Barrett[3], Campos[7], Diethelem and Ford[9], Fujiwara[10],Kilbas et al.[15], Kilbas and Saigo[17], Tazali[36] and Zhang[37] to cite but a few (seealso Kilbas and Trujillo[19] and the references therein). Whenf (t, u) is linear, very oftenexplicit solutions can be found usingmethods from the ordinary differential equation theorysuch as Laplace transform, operational calculus, decompositionmethod and other methods.Again, we refer the reader to the survey paper by Kilbas and Trujillo[19] for a discussionon results in this regard.Many phenomena in various fields of science and engineering can be described by frac-

tional differential equations. Indeed, we can find numerous applications in viscoelasticity,electrochemistry, control, porous media, electromagnetic, etc. (see Oldham and Spanier[31], Podlubny[32] and many other references in[19]).In this work, we investigate the behavior of solutions for problem (1) with certain nonlin-

earities. Using the equivalence of the fractional differential problemwith the correspondingVolterra integral equation, we prove that solutions exist globally and are bounded (in abso-lute values) byCt�−1 for some positive constantC. Therefore, as time increases, solutions“decay” as a power-type function. To this end we combine and modify certain techniquesdue to Medved[23,24], Kirane and Tatar[20,21], Mazouzi and Tatar[22] and Tatar[35].We mention here that in[17] Kilbas and Saigo showed that if(� + �)/(1− m)> � − 1,then the equation

(D�y)(x)= ax�ym(x), 0<x <d <∞, m ∈ R, m �= 1

has at least one solution given by

y(x)= [�(� + 1)/a�(� − � + 1)]1/(1−m)x�,

where� = (� + �)/(1−m).The plan of our paper is as follows. In the next section, we prepare some material

needed to prove our results. Section 3 is devoted to our main results on the power-typedecay. In the last section, we provide an example and discuss how we can apply ourresults to it.

2. Preliminaries

In this section, we present some definitions, lemmas and notation which will be used inour theorems.

Definition 1. The Riemann–Liouville fractional integral of order�>0 of a Lebesgue-measurable functionf : R+ → R is defined by

I�f (t)= 1

�(�)

∫ t

0(t − s)�−1f (s)ds,

provided that the integral exists.

K.M. Furati, N.-e. Tatar / Nonlinear Analysis 62 (2005) 1025–1036 1027

Definition 2. The fractional derivative (in thesenseofRiemann–Liouville) of order 0<�<1of a continuous functionf : R+ → R is defined as the left inverse of the fractional integralof f:

D�f (t)= d

dt(I1−�f )(t).

That is

D�f (t)= 1

�(1− �)d

dt

∫ t

0(t − s)−�f (s)ds,

provided that the right side exists.

The reader is referred to[26,31–33] for more on fractional integrals and fractionalderivatives.

Theorem 1 (Samkoet al.[33, Corollary toTheorem3.6, p. 69]). The fractional integrationoperator I� is bounded fromLp(0, h) into h�−(1/p)([0, h]) (the little Hölder space) if0<1/p< �<1+ 1/p.

Forh>0, we define the space

C0r ([0, h]) :=

{v ∈ C0((0, h]) : lim

t→0+ trv(t) exists and is finite

}.

Here,C0((0, h]) is the usual space of continuous functions on(0, h]. It turns out that thespaceC0

r ([0, h]) endowed with the norm‖v‖r := max

0� t�htr |v(t)|

is a Banach space. Then, we define the space

C�1−�([0, h]) := {v ∈ C0

1−�([0, h]) : there existc ∈ R andv∗ ∈ C01−�([0, h])

such thatv(t)= ct�−1 + I �v∗(t)}.

Proposition 1. If �> 12, then the spaceC

�1−�([0, h]) endowed with the norm

‖v‖1−�,� := ‖v‖1−� + ‖D�v‖1−�

is a Banach space.

Proof. Let {un}∞n=1 be a Cauchy sequence in the spaceC�1−�([0, h]). Then, clearly{un}∞n=1

and{D�un}∞n=1 are Cauchy sequences inC01−�([0, h]), which is complete. Therefore,un

andD�un converge to someu andw in C01−�([0, h]), respectively. We need to prove that

w =D�u. As �> 12 andD

�un converges tow in C01−�([0, h]), then using the embedding

C01−�([0, h]) ⊂ Lp(0, h) for 1�p< 1

1− �

1028 K.M. Furati, N.-e. Tatar / Nonlinear Analysis 62 (2005) 1025–1036

and Theorem 1, we can deduce that

I �D�un → I �w asn→ ∞ uniformly on [0, h].Hence, from

un(t)= cnt�−1 + I �u∗n(t) for somecn ∈ R, n= 1,2, . . . ,

we have

un(t)− cnt�−1 → I�w(t) asn→ ∞ uniformly on [0, h].Sinceun(t) converges tou(t) (for each fixedt ∈ [0, h]) we see thatcnt�−1 → ct�−1 asn→ ∞. Therefore, limt→0 t

1−�u(t)= c andw =D�u. �

We will also need the following well-known lemmas:

Lemma 1. For all �>0 and�>− 1,we have∫ t

0(t − s)�−1s� ds = �(�)�(� + 1)

�(� + � + 1)t�+�, t�0.

The next lemma can be found in[25] or [20].

Lemma 2. If �, �, �>0, then for anyt >0,we have

t1−�∫ t

0(t − s)�−1s�−1e−�s ds�C,

where C is a positive constant independent of t. In fact,

C =max{1,21−�}�(�)(1+ �/�)�−�.

3. Power-type decay

A local existence result has been proved in[8] provided thatf is continuous on(0,1]×Randt�f (t, u) is a continuous function[0,1] ×R for some� such that 0��<s <1. Then,a uniqueness result was established whenf satisfies a Lipschitz-type condition. For theweighted problem withf (t, u)=f (u) a Lipschitz continuous function andf (0)=0, it hasbeen shown that there exists a unique solution.In [11], the present authors established local existence results for more general (including

nonlocal) nonlinearities provided thatt1−� f (t, u) is continuous with respect tou.If a solution remains bounded by a continuous function, then one can extend it by con-

tinuity to some maximal interval. In particular, uniform boundedness by a constant allowsus to extend solutions for all time. In this case, we will have global existence in time. Wewill assume the following hypotheses on the nonlinearityf (t, u):(F) f (t, u) is continuous onR+ × R and is such that

|f (t, u)|� t�e−�t(t)|u|m, ��0, m>1, �>0,

where(t) is a continuous function onR+.

K.M. Furati, N.-e. Tatar / Nonlinear Analysis 62 (2005) 1025–1036 1029

Proposition 2. Let u ∈ C�1−�([0, h]) with �> 1

2. Suppose further that� − (m − 1)(1−�)>0.Then problem(1) and its associated integral equation

u(t)= bt�−1 + 1

�(�)

∫ t

0(t − s)�−1f (s, u(s))ds (2)

are equivalent.

Proof. Let us first prove the necessity. Ifu is a solution of (1) inC�1−�([0, h]), then applying

the operatorI� to both sides of (1) we clearly haveu∗ = f (t, u(t)) andu(t)= ct�−1 + I �f (t, u(t))

for all t in (0, h] and somec ∈ R. If we denote the right-hand side of this relation byT u(t),then we can check that it is inC�

1−�([0, h]). That is, thatTmapsC�1−�([0, h]) into itself.

Indeed, by definition ofT and the assumptions onf , we have

‖T u(t)‖1−�,�

= maxt∈[0,h] t

1−�|T u(t)| + maxt∈[0,h] t

1−�|f (t, u(t))|

� |c| + ‖‖∞�(�)

maxt∈[0,h] t

1−�∫ t

0(t − s)�−1s�−m(1−�)e−�ssm(1−�)|u(s)|m ds

+ ‖‖∞ maxt∈[0,h] t

�−(m−1)(1−�)tm(1−�)|u(t)|m.

Next by the definition of‖.‖1−� and Lemma 1, we find

‖T u(t)‖1−�,�� |c| + C‖u‖m1−�‖‖∞�(�)

h1+�−m(1−�) + ‖‖∞‖u‖m1−�h�−(m−1)(1−�)

for some positive constantC (it is in fact the constant which appears from the applicationof Lemma 1). Moreover, from the above argument observe that

limt→0

t1−�I �f (t, u(t))= C̃ limt→0

t1+�−m(1−�) = 0.

Observe here that we could have used Lemma 2 instead of Lemma 1 to prove thatTmapsC�1−�([0, h]) into itself. We preferred Lemma 1 to have this last limit vanish.Therefore,

c = limt→0

t1−�u(t)= b,

and with the help of the formula at the beginning of the proof we see thatu is a solution ofthe corresponding integral equation (2).For sufficiency, applyingD� to both sides of (2), using the facts thatD�I �v(t)=v(t) and

D�t�−1=0, we entail thatu satisfies the equation in (1). Since limt→0 t1−�I �f (t, u(t))=0,

we obtain the initial conditiont1−�u(t)|t=0 = b. �

We shall study problem (1) through its associated integral equation (2). In this respect,we would like to draw the attention of the reader to the works of Askhabov, Bushell,

1030 K.M. Furati, N.-e. Tatar / Nonlinear Analysis 62 (2005) 1025–1036

Okrasinski, Karapetyants, Kilbas, Saigo, Samko, Nestell, Chandehari and Schneider in[1,4,5,12,13,16,18,27–30,34]for related results on integral equations with convolutionsinvolving power nonlinearities.Let p andq be conjugate exponents, i.e.pq = p + q and�1 := 1+ p[� − (1− �)m],

�2 := 1+ p(� − 1). Observe that if� − (m− 1)(1− �)>0 andq >1/�, then�1>0 and�2>0. We denote byL the positive real number

L := �(�)21+m−�|b|m−1

(2m

m− 1

)1/q[

(�p)�1

�(�1)(1+ �1/�2)

]1/p

.

Theorem 2. Suppose thatf (t, u) satisfies(F), � − (m − 1)(1 − �)>0 and �> 12. If

‖‖q <L for someq >1/�, then there exists a positive constant C such that|u(t)|�Ct�−1,t >0.

Proof. Let us consider the Volterra integral equation (2) associated to problem (1):

u(t)= bt�−1 + 1

�(�)

∫ t

0(t − s)�−1f (s, u(s))ds. (3)

Multiplying both sides of (3) byt1−� and using the assumption(F) onf , we get

t1−�|u(t)|� |b| + t1−�

�(�)

∫ t

0(t − s)�−1s�e−�s(s)|u(s)|m ds. (4)

Denoting byv(t) the left-hand side of (4), we see that

v(t)� |b| + t1−�

�(�)

∫ t

0(t − s)�−1s�−(1−�)me−�s(s)vm ds. (5)

Using the Hölder inequality withp andq, we find

∫ t

0(t−s)�−1s�−(1−�)me−�s(s)vm ds�

(∫ t

0(t−s)p(�−1)sp[�−(1−�)m]e−�ps ds

)1/p

×(∫ t

0q(s)vqm(s)ds

)1/q

.

Since�1>0 and�2>0, we may apply Lemma 2 to get

∫ t

0(t − s)�−1s�−(1−�)m(s)vm(s)ds�C1t�−1

(∫ t

0q(s)vqm(s)ds

)1/q

, (6)

whereC1 is the constant appearing in Lemma 2 corresponding to the present exponents.That isCp1 = 2p(1−�)�(�1)(1+ �1/�2)/(�p)�1. From (5) and (6)

v(t)� |b| + C̃1(∫ t

0q(s)vqm(s)ds

)1/q

, (7)

K.M. Furati, N.-e. Tatar / Nonlinear Analysis 62 (2005) 1025–1036 1031

whereC̃1 = C1/�(�). Raising both sides of (7) to the powerq and using the inequality(a + b)�2−1(a + b)

for a�0, b�0 and>1, we get

vq(t)�2q−1[|b|q + C̃q1

∫ t

0q(s)vqm(s)ds

]. (8)

If we designate byw(t) the right-hand side of (8), we easily see thatw(0) = 2q−1|b|qand by differentiation

w′(t)= 2q−1C̃q1q(t)vqm(t)�0.

From (8) we havevq(t)�w(t). This implies that

w′(t)�2q−1C̃q1q(t)wm(t). (9)

An integration of (9) leads to

wm−1(t)� wm−1(0)

1− 2q−1C̃q1 (m− 1)wm−1(0)

∫ t0 q(s)ds

as long as

2(q−1)mC̃q1 (m− 1)|b|q(m−1)

∫ t

0q(s)ds <1. (10)

Indeed, asw(t) does not vanish and is positive, dividing (9) bywm(t) and integrating bothsides we find

w1−m(t)1−m − w1−m(0)

1−m �2q−1C̃q1

∫ t

0q(s)ds.

Multiplying by 1−m and noticing that it is negative, we obtain

w1−m(t)�w1−m(0)− (m− 1)2q−1C̃q1

∫ t

0q(s)ds.

In particular, if‖‖q <L/2 thenwm−1(t)�C∗1w

m−1(0) for some positive constantC∗1.

Therefore,w(t) and thereafterv(t) is bounded. This yields|u(t)|�Ct�−1. Notice here thatwe could have used the Gronwall–Bihari inequality (see[2], for instance) to (8) directly.The proof is complete. �

In the next theorem, we shall see that for solutions of the integral equation (2) one canenlarge the class of functionsf (t, u) but at the expense of the decay rate.We will show thatsolutions of the integral equation (2) withf (t, u) satisfying(F)∗ f (t, u) is continuous onR+ × R and is such that|f (t, u)|� t�(t)|u|m, ��0,

m>1, where(t) is a continuous function onR+decay ast−� with �<1− �.

1032 K.M. Furati, N.-e. Tatar / Nonlinear Analysis 62 (2005) 1025–1036

For short notation, let us denote by�̃1 := 1+p(�−�m), �2 := 1+p(�−1) (as before),�3 := � + � − �(m− 1)− 1/q, �4 := � − (1− �) and

K := �(�)

[(m− 1)2m(q−1)+1]1/q |b|m−1a�3+(m−1)�4

[�(�̃1 + �2)

�(�̃1)�(�2)

]1/p

.

Theorem 3. Assume that�> 12, f (t, u) satisfies(F)

∗ and ‖‖q <K for someq >1/�.Suppose further that� + 1/p<m(1− �). Then, there existsC >0 and0< �<1− � suchthat any solution of(2) exists globally and satisfies

|u(t)|�Ct−�, t�a >0.

Proof. Let u be a solution to the integral equation (2). By the assumption(F)∗ we have

|u(t)|� |b|t�−1 + 1

�(�)

∫ t

0(t − s)�−1s�(s)|u(s)|m ds.

If we multiply both sides of this relation byt� with �<1− � and setv(t) := t�|u(t)|, thenwe obtain for allt�a, wherea is any (possibly very small) positive constant,

v(t)� |b|a�−(1−�) + t�

�(�)

∫ t

0(t − s)�−1s�−�m(s)vm(s)ds. (11)

Using the Hölder inequality and Lemma 1, we infer that∫ t

0(t − s)�−1s�−�m(s)vm(s)ds

�(∫ t

0(t − s)p(�−1)sp(�−�m) ds

)1/p(∫ t

0q(s)vqm(s)ds

)1/q

. (12)

Taking0< �<(1/m)(�+1/p) (notehere that according toourassumption�+1/p<m(1−�), we have(1/m)(� + 1/p)<1− �), it is clear that�̃1>0 and sinceq >1/� we have�2>0. Therefore, we can apply Lemma 1 to get∫ t

0(t − s)p(�−1)sp(�−�m) ds�C2t1+p(�−�m+�−1), (13)

where

C2 := �(�̃1)�(�2)/�(�̃1 + �2).

Taking relations (12) and (13) into account in (11), we obtain

v(t)� |b|a�4 + 1

�(�)C1/p2 t�3

(∫ t

0q(s)vqm(s)ds

)1/q

, t�a >0.

K.M. Furati, N.-e. Tatar / Nonlinear Analysis 62 (2005) 1025–1036 1033

Therefore,

vq(t)�2q−1(

|b|qaq�4 + C3tq�3∫ t

0q(s)vqm(s)ds

), t�a >0,

whereC3 := Cq/p2 /�q(�).Choosing now0< �<(1/m)(�+1/p) in such away that�3=�+�−�(m−1)−1/q <0,

i.e.

0<1

m− 1

(� + � − 1

q

)< �<

1

m

(� + 1

p

)

(observe that this possible again by our assumption� + 1/p<m(1− �)), we find

vq(t)�2q−1(

|b|qaq�4 + C4∫ t

0q(s)vqm(s)ds

), t�a >0, (14)

whereC4 := C3aq�3. Letw(t) denote the right-hand side of (14). It is clear thatw(0) =2q−1|b|qaq�4 and by differentiation

w′(t)= 2q−1C4q(t)vqm(t)�2q−1C4q(t)wm(t). (15)

An integration of (15) gives

wm−1(t)� (2q−1|b|qaq�4)m−1

1− (m− 1)2m(q−1)|b|q(m−1)aq(m−1)�4C4∫ t0 q(s)ds

and thereforew(t) is (uniformly) bounded provided that∫ +∞

0q(s)ds <1/(m− 1)2m(q−1)aq(m−1)�4|b|q(m−1)C4.

We deduce the uniform boundedness of|u(t)| by the power-type functiont−� for t�a >0.The proof is complete. �

Remark 1. Our results can be generalized to nondecreasing functionsw(x) (such thatw(t�−1x)�h(t)w(x) for some functionh(t)) instead of power functionsxm. It suffices touse the Gronwall–Bellman inequality in Butler and Rogers[6] for instance.

Remark 2. The result in Theorem 3 concerns solutions of the integral equation (2) onlybecause for solutions of problem (1) the assumption�− (m−1)(1−�)>0 will contradictthe conditions� + 1/p<m(1− �) andq >1/�.

4. Example

In this section, we present an example and discuss how our results apply. Let us considerthe problem{

D�u(t)= tm−1e−t�(t)|u|m, t >0,t1−�u(t)|t=0 = b. (16)

1034 K.M. Furati, N.-e. Tatar / Nonlinear Analysis 62 (2005) 1025–1036

Here, asmwe can take any numberm>1 andb �= 0 any real number. The function� iscontinuous onR+. Suppose also that12 < �<1. It is clear that�−(m−1)(1−�)=m−1−(m− 1)(1− �)>0. If u(t) is a local solution of problem (16) in the spaceC�

1−�([0, h̃]) forsome small̃h>0, then, according to Proposition 2,u(t) is a solution of the integral equation

u(t)= bt�−1 + 1

�(�)

∫ t

0(t − s)�−1sm−1e−s�(s)|u(s)|m ds. (17)

Now, assuming further that� verifies

‖�‖q < �(�)21+m−�|b|m−1

(2m

m− 1

)1/q[

p�1

�(�1)(1+ �1/�2)

]1/p

,

where�1 := 1+ p[�m − 1], �2 := 1+ p(� − 1) for someq >1/� and 1/p + 1/q = 1,Theorem 2 allows us to derive thatu(t) exists everywhere (as a solution of (17)) and that|u(t)|�Ct�−1, t >0 for some positive constantC. Moreover, applying the operatorD� toboth sides of (17) we see that

t1−�|D�(t)|� tm−�e−�t |�(t)||u(t)|m�Cm|�(t)|t�(m−1).

Since�(t) is a continuous function, this means thatu(t) is in C�1−�([0, h]) for anyh>0.

Consequently,u(t) is a global solution (with power-type decay) for problem (16) (seeProposition 2 again).

Acknowledgements

Theauthorswould like to thank the referee for his/her valuable comments and suggestionswhich improved the original version of this work. The authors are also very grateful forthe financial support and facilities provided by King Fahd University of Petroleum andMinerals.

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