Seria I: PRACE MATEMATYCZNE XLV (2) (2005), 237-247
A. M. A. El-Sayed, Nagwa Sherif, I. A. Ibrahim
On a mixed-type integral equation and fractional-order functional differential equations
Abstract. In this paper we study the existence of solution of a nonlinear integral equation of (mixed type) Volterra-Fredholm type. As an application we prove the existence of solution of an initial value problem of fractional order in the space of Lebesgue integrable functions on the interval [0, 1].
200 Mathematics Subject Classification: 26A33, 45J05, 39B05, 34A60, 34G20, 34K05.
Key words and phrases: Nonlinear functional integral equation, fractional calculus, Volterra operator, mixed type integral equation, Schauder fixed point theorem.
1. Introduction. In [5] and [6] the authors prove the existence of monotonic solution of the following mixed type integral equations
x(t) = Z ∞
0
k1(t, s) f
s,
Z t 0
k (s, θ) f1(θ, x (φ (θ))) dθ
ds, t ≥ 0 and
x(t) = g(t) + Z t
0
k1(t, s) f1
s,
Z s 0
k2(s, θ) f2(θ, x (φ (θ))) dθ
ds,
in the space of Lebesgue integrable functions on the interval [0, ∞) and [0, 1], re- spectively.
In this work we omit the conditions of monotonicity on f, f1, f2 which stated in [5],[6] and prove the existence of integrable solution x ∈ L1[0, 1] of the integral equation of Volterra-Fredholm type
(1) x(t) = Z t
0
k1(t, s) f1
s,
Z 1 0
k (s, θ) f (θ, x (φ (θ))) dθ
ds, t ∈ [0, 1] .
As a consequence of this result we prove that there exists at least one solution z ∈ L1[0, 1] for the initial value problem of fractional order
(2)
dz dt = f1
t,R1
0 k (t, s) f s, Dβz (s) ds
, t ∈ [0, 1] , β ∈ (0, 1]
z (0) = 0.
Denote by L1= L1[0, 1] , the class of Lebesgue integrable functions on the inter- val [0, 1] with the usual norm
kxk = Z 1
0
|x(t)| dt, x ∈ L1.
We assume that f : [0, 1] × R → R satisfies Carathéodory conditions. The function f gives rise to the superposition operator F defined for each measurable function x as in [1],[2],[10] as
(F x) (t) = f (t, x (t)) , t ∈ [0, 1] .
The continuity of superposition operator F is characterized by the following theorem [9].
Theorem 1.1 The superposition operator F generated by the function f maps con- tinuously the space L1into itself if and only if |f (t, x)| ≤ a(t) + b |x| for all t ∈ [0, 1]
and x ∈ R. Here a belongs to L1and b is a nonnegative constant.
The following theorem [13] plays major role in the proof of our main result. We denote by meas.D the Lebesgue measure of the set D.
Theorem 1.2 (Scorza,Dragoni) Let f : [0, 1] × R → R be a function satisfying Carathéodory hypotheses. Then for each ε > 0 there exists a closed subset Dε of [0, 1] with meas. ([0, 1] − Dε) = meas.Dcε< ε and such that f |D
ε×R is continuous.
Finally in our approach to prove the existence result of (1) we will need the following fixed point theorem due to Schauder [8].
Theorem 1.3 Let C be a nonempty, convex, closed, and bounded subset of a Banach space E. Let H : C → C be a completely continuous mapping. Then H has at least one fixed point in C.
In the above theorem by a completely continuous map H : E → E we mean that H is continuous and H(Y ) is relatively compact for every bounded subset Y of E.
Some basic definitions and results from fractional calculus are presented in Sec- tion 2. Our main result is given in Section 3, where we establish the existence of integrable solution of equation (1) under suitable assumptions on f, f1, k, k1. Also several consequences of the main result are discussed.
2. Fractional calculus. The fractional derivative is a singular integro-differential operator and the fractional differential equations (differential equations of fractional orders) are singular integro-differential equations. We give the definition of both differential operator and the integral operator of fractional order.
Definition 2.1 [11] Let f ∈ L1, α ∈ R+. The Riemman-Liouville (R-L) fractional integral of the function f of order α is defined as
Iaαf (t) = Z t
a
(t − s)α−1
Γ (α) f (s) ds, α > 0, a ≤ t ≤ b.
Definition 2.2 [12] Let g be an absolutely continuous function on [a, b]. Then the fractional derivative of order α ∈ (0, 1] of g is defined as
Daαg (t) = Ia1−αDg (t) , D = d dt.
We state here some results concerning the above mentioned operators, that are relevant to our work [4],[11].
(I) Let f ∈ L1and α, β ∈ (0, 1], then
IaαIaβf (t) = IaβIaαf (t) = Iaα+βf (t) . Furthermore,
(Iaα)nf (t) = Ianαf (t) , n = 1, 2, 3, · · · . (II) Let α, β ∈ (0, 1] and f, Df ∈ L1, then
(i) DIaαf (t) = IaαDf (t) , when f (a) = 0.
(ii) DαaIaβf (t) = Iaβ−αf (t) , α < β.
(iii) DαaIaβf (t) = Daα−βf (t) , α > β, f (a) = 0.
(iv) IaαDαaf (t) = f (t) − f (a) .
(III) The operator Iaα maps L1into itself continuously.
3. Main result. In this section we present the existence of integrable solutions of equation (1). For that purpose we impose the following assumptions on the different functions and operators arising in equation (1).
(V1) f, f1: [0, 1] × R → R satisfy the Carathéodory conditions. There exist func- tions a and a1that belong to L1and positive constants b and b1, such that
|f (t, u)| ≤ a(t) + b |u| , and
|f1(t, u)| ≤ a1(t) + b1|u| , hold for (t, u) ∈ [0, 1] × R.
(V2) k1, k : [0, 1] × [0, 1] → R+ are measurable with respect to both variables.
Let K1be the Volterra operator associated with the kernal k1while K is the Fredholm operator associated with k. Both K1and K map L1 into itself.
(V3) φ : [0, 1] → [0, 1] is increasing, absolutely continuous and there exists a con- stant M > 0 such that φ0(t) ≥ M for almost all t ∈ [0, 1] .
(V4) b1bkKM1kkKk < 1.
Next, we prove the following lemma that is essential to our main result.
Lemma 3.1 Let the above conditions (V 1) → (V 4) be satisfied. Then there exists a unique a.e. nonnegative function x0that belongs to L1, such that
(3) x0(t) = Z t
0
k1(t, s)
a1(s) + b1 Z 1
0
k (s, θ) (a(θ) + bx0(φ (θ))) dθ
ds.
Proof Set
h (t) = Z t
0
k1(t, s)
a1(s) + b1 Z 1
0
k (s, θ) a(θ)dθ
ds Clearly h ∈ L1as it is evident from our assumption (V 2) . Let Br= {x : x ∈ L1, kxk ≤ r} with r = 1−b khk
1bkK1kkKk/M. Next, we consider the oper- ator A defined by
(Ax) (t) = Z t
0
k1(t, s)
a1(s) + b1 Z 1
0
k (s, θ) (a(θ) + bx (φ (θ))) dθ
ds.
It is evident that A maps L1into itself continuously.
For x ∈ Br, we have
kAxk = Z 1
0
|Ax(t)| dt =
= Z 1
0
Z t 0
k1(t, s)
a1(s) + b1 Z 1
0
k (s, θ) (a(θ) + bx (φ (θ))) dθ
ds
dt
≤ Z 1
0
Z t 0
k1(t, s)
a1(s) + b1 Z 1
0
k (s, θ) a(θ)dθ
ds
dt + +
Z 1 0
Z t 0
b1bk1(t, s)
Z 1 0
k (s, θ) x (φ (θ)) dθ
ds
dt
≤ khk + Z 1
0
Z t 0
b1bk1(t, s)
Z 1 0
k (s, θ) |x (φ (θ))| dθ
ds
dt
≤ khk + b1b kK1k kKk Z 1
0
|x (φ (s))| ds
≤ khk + b1b kK1k kKk Z 1
0
|x(φ (s))| φ0(s) M ds
≤ khk +b1b
M kK1k kKk Z φ(1)
φ(0)
|x(v)| dv.
Hence
kAxk ≤ khk +b1b
M kK1k kKk Z 1
0
|x(v)| dv
≤ khk +b1b
M kK1k kKk kxk ≤ khk + b
M kK1k kKk r
≤ khk +b1b
M kK1k kKk khk
1 −b1b kK1k kKk M
= r.
This proves that A maps Br into Br.
Let Br+= {x : x ∈ Br, x (t) ≥ 0 a.e.} , then we have A (B+r) ⊂ B+r and it can be shown that Br+is a closed subset of L1and consequently we get that Br+is a complete metric space. Next we show that A is a contraction in Br+. Let x1, x2 ∈ Br+ we have
kAx1− Ax2k =
= Z 1
0
Z t 0
k1(t, s)
Z 1 0
k (s, θ) b1b (x1(φ (θ)) − x2(φ (θ))) dθ
ds
dt
≤ Z 1
0
Z t 0
k1(t, s)
Z 1 0
k (s, θ) b1b (|x1(φ (θ)) − x2(φ (θ))|) dθ
dsdt
≤ kK1k kKk Z 1
0
b1b (|x1(φ (θ)) − x2(φ (θ))|) φ0(θ) M dθ
≤ kK1k kKk Z 1
0
b1b
M (|x1(v) − x2(v)|) dv
≤ b1b
M kK1k kKk kx1− x2k .
It is immediate from condition (V 4) that A is a contraction. Now, the pervious results guarantee the existence of a unique a.e. nonnegative function x0that belongs
to L1and satisfies equation (3).
To prove the existence of a solution to equation (1), we use Lemma 3.1 and an additional condition on k which allows the application of Theorem 1.2
Theorem 3.2 Assume that the conditions (V 1) → (V 4) are satisfied and in addition k satisfies the Carathéodory condition, then equation (1) has at least one solution x ∈ L1.
Proof Assume that x0the zero element of L1that solves equation (3), and let
y(t) = Z t
0
k1(t, s) f1
s,
Z 1 0
k (s, θ) f (θ, x0(φ (θ))) dθ
ds, t ∈ [0, 1] , hence
|y(t)| ≤ Z t
0
k1(t, s)
a1(s) + b1 Z 1
0
k (s, θ) (a(θ) + bx0(φ (θ))) dθ
ds
|y(t)| ≤ x0(t) ,
where the last equality follows from Lemma 3.1 and hence y = 0. Hence x0 solves equation (1). Now, assume that x0 is not the zero element of L1and consider the set
Q = {y : y ∈ L1, |y(t)| ≤ x0(t) , a.e.} ,
this set Q can be shown to be nonempty, closed, bounded and convex in L1. Now let H be the nonlinear operator associated with equation (1) and defined by
Hx(t) = Z t
0
k1(t, s) f1
s,
Z 1 0
k (s, θ) f (θ, x (φ (θ))) dθ
ds, t ∈ [0, 1] .
Based on our assumptions, we can see that H maps L1into itself continuously. In what follows, we will prove that H : Q → Q. For this purpose, let x ∈ Q, then
|Hx(t)| =
Rt
0k1(t, s) f1 s,R1
0k (s, θ) f (θ, x (φ (θ))) dθ ds
≤Rt
0k1(t, s)
a1(s) + b1R1
0k (s, θ) (a (θ) + b |x (φ (θ))|) dθ ds
≤Rt 0k1(t, s)
a1(s) + b1R1
0k (s, θ) (a (θ) + bx0(φ (θ))) dθ
ds
|Hx(t)| ≤ x0(t) , so Hx belongs to Q.
To apply the Schauder fixed point theorem, we need to prove that the set H(Q) is relatively compact in L1. To achieve this we assume that {yn} is a sequence in Q.
Theorem 1.2 imply that there is a closed subset Dεof [0, 1] with meas.Dεc< ε such that all the restrictions x0|D
ε, k1|D
ε×[0,1]and k|D
ε×[0,1] are uniformly continuous.
In what follows we show that {Hyn} is an equicontinuous on Dε, for that let ε1> 0 be given. Since k1|D
ε×[0,1]is uniformly continuous, there exists δ > 0 such that for t0, t00 ∈ Dε, |t00− t0| < δ we have
|k1(t00, s) − k1(t0, s)| < ε1 2γ,
where γ = ka1k + b1kKk kak + bbM1kKk kx0k. Furthermore, the function k1(t, s) attains its maximum say C on Dε× [0, 1]. The integral
Rt00 t0 f1
s,R1
0k (s, θ) f (θ, x0(φ (θ))) dθ
ds can be made less than 2Cε1, for any t0, t00 ∈ Dε, with |t00− t0| < δ, because of the property of the intgrand. Finally, for t0, t00 ∈ Dε, |t00− t0| < δ we get
|Hyn(t00) − Hyn(t0)| ≤
≤ Z t0
0
|k1(t00, s) − k1(t0, s)|
·
|a1(s)| + b1 Z 1
0
k (s, θ) |a (θ) + b |x0(φ (θ))|| dθ
ds +
+
Z t00 t0
k1(t00, s) f1
s,
Z 1 0
k (s, θ) f (θ, x0(φ (θ))) dθ
ds
≤ ε1
2γγ + Cε1 2C = ε1.
Also, the sequence {Hyn} is a sequence of uniformly bounded functions, since
|Hyn(t)| =
=
Z t 0
k1(t, s) f1
s,
Z 1 0
k (s, θ) f (θ, yn(φ (θ))) dθds
, t ∈ Dε
≤ Z t
0
k1(t, s)
a1(s) + b1 Z 1
0
k (s, θ) (a (θ) + b |yn(φ (θ))|) dθ
ds
≤ Z t
0
k1(t, s)
a1(s) + b1 Z 1
0
k (s, θ) (a (θ) + bx0(φ (θ))) dθ
ds
≤ x0(t) , ∀n.
Hence
kHynkC(D
ε)≤ kx0kC(D
ε), ∀n.
and consequently the sequence {Hyn} is a sequence of uniformly bounded and equicontinuous functions on Dε. Hence, in view of Ascoli-Arz´ela theorem we deduce that {H (yn)} is relatively compact subset of C (Dε) . In fact there exists a sequence {Dk} of closed subsets of [0, 1] with meas.Dkc → 0 such that {H (yn)} is relatively compact subset in each C (Dk). This implies that {Hyn} has a convergent subse- quence in each C (Dk) k = 1, 2, · · · (We call this subsequence again {H (yn)}). But C (Dk) is a complete metric space, hence this subsequence is a Cauchy sequence in each C (Dk), k = 1, 2, · · · . That is for any given > 0 and arbitrary large m, n kHym− HynkC(D
k) < . Next we show that {Hyn} is convergent in L1. For that purpose we verify that {Hyn} is a Cauchy sequence in L1.
Let η > 0 be given, we will show that there exists N (η) such that kHym− HynkL
1< η, whenever m, n > N (η) . Given η > 0, there is a δ > 0 such that meas.Dδ < δ implies
Z
Dδ
x0(s) ds < η 4. Choose k∗∈ N with meas.Dck∗ < δ and
1
Z
0
|(Hym) (t) − (Hyn) (t)| dt =
= Z
Dck∗
|(Hym) (t) − (Hyn) (t)| dt + Z
Dk∗
|(Hym) (t) − (Hyn) (t)| dt
≤ Z
Dck∗
{|(Hym) (t)| + |(Hyn) (t)|} dt + kHym− HynkC(D
k∗)
Z
Dk∗
dt
≤ η
4+η
4 + kHym− HynkC(D
k∗)(1 − δ)
≤ η
2+ kHym− HynkC(D
k∗),
choose N (η) such that m, n > N (η) we have kHym− HynkC(D
k∗)<η2. Then
1
Z
0
|(Hym) (t) − (Hyn) (t)| dt <η 2+η
2 = η.
This means that {H (yn)} is a Cauchy sequence in the space L1and hence the set H(Q) is relatively compact in L1. Then H has at least one fixed point . This proves that there exists at least one x ∈ L1that solves equation (1).
3.1. Convolution kernel. Assume that ˜k : [0, 1] → R+ is an integrable function. For an arbitrary function x ∈ L1set
(Kx) (t) = Z t
0
˜k (t − s) x (s) ds, t ∈ [0, 1] .
This operator K is a linear integral operator of convolution type and maps L1into itself continuously.
Now, consider the following additional condition (V5) ˜k : [0, 1] → R+and ˜k ∈ L1.
Then we have the following Corollary
Corollary 3.3 Let the hypotheses (V 1)− (V 5) are satisfied. Then there exists a unique a.e. nonnegative function x0that belongs to L1, such that
x0(t) = Z t
0
˜k (t − s)
a1(s) + b1 Z 1
0
k (s, θ) (a(θ) + bx0(φ (θ))) dθ
ds.
Furthermore, the equation (4) x(t) =
Z t 0
˜k (t − s) f1
s,
Z 1 0
k (s, θ) f (θ, x (φ (θ))) dθ
ds, t ∈ [0, 1] . has at least one integrable solution.
In the next, we prove an existence theorem of equation (2) with the aid of a special form of equation (4).
3.2. Integral equation of fractional order. As a special case of equation (4), we consider
(5) x(t) = Z t
0
(t − s)−β Γ (1 − β) f1
s,
Z 1 0
k (s, θ) f (θ, x (θ)) dθ
ds, t ∈ (0, 1] . Equation (5) is an integral equation of fractional order that can be written in the form
(6) x(t) = I1−βf1
t,
Z 1 0
k (t, θ) f (θ, x (θ)) dθ
, t ∈ (0, 1] , β ∈ (0, 1] . We now prove our main result concerning equation (2).
Theorem 3.4 Let bΓ(2−β)1bkKk< 1 and β ∈ (0, 1] . If assumptions (V 1) − (V 3) and (V 5) are satisfied, then the initial value problem (2) has at least one solution z (t) . Proof Let x (t) be a solution of the integral equation (6). Put
(7) z (t) = Iβx (t) ,
then we get
z (t) = Iβx (t) =
= IβI1−βf1
t,
Z 1 0
k (t, θ) f (θ, x (θ)) dθ
= I1f1
t,
Z 1 0
k (t, θ) f (θ, x (θ)) dθ
= I1f1
t,
Z 1 0
k (t, θ) f (θ, x (θ)) dθ
= Z t
0
f1
s,
Z 1 0
k (s, θ) f (θ, x (θ)) dθ
ds,
with z (0) = 0 and z is absolutely continuous on [0, 1], hence Dβz (t) is defined.
Furthermore
Dβz (t) = I1−βd
dtz (t) = I1−βd
dt Iβx (t)
= I1−βd dtIβ
I1−βf1
t,
Z 1 0
k (t, θ) f (θ, x (θ)) dθ
= d
dtI1−βIβI1−βf1
t,
Z 1 0
k (t, θ) f (θ, x (θ)) dθ
= d
dtI1I1−βf1
t,
Z 1 0
k (t, θ) f (θ, x (θ)) dθ
= d dtI1
I1−βf1
t,
Z 1 0
k (t, θ) f (θ, x (θ)) dθ
= x (t) .
From equation (7)
dz dt = d
dtIβx (t) . Substituting from equation (6), we get
dz
dt = d
dtIβI1−βf1
t,
Z 1 0
k (t, θ) f θ, Dβz (θ) dθ
= d
dtI1f1
t,
Z 1 0
k (t, θ) f θ, Dβz (θ) dθ
= f1
t,
Z 1 0
k (t, θ) f θ, Dβz (θ) dθ
.
Which proves that if x (t) is any solution of equation (6), then z (t) = Iβx (t) is a solution of the initial value problem (2). Hence our proof is complete.
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A. M. A. El-Sayed
Faculty of Science, Alexandria University Alexandria, Egypt
E-mail: amasayed@hotmail.com Nagwa Sherif
Faculty of Science, Suez Canal University Ismailia, Egypt
E-mail: nasherif@yahoo.com I. A. Ibrahim
Faculty of Science, Suez Canal University Ismailia, Egypt
E-mail: iabouelfarag@hotmail.com
(Received: 26.07.05)
