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

k_{1}(t, s) f

s,

Z t 0

k (s, θ) f_{1}(θ, x (φ (θ))) dθ

ds, t ≥ 0 and

x(t) = g(t) + Z t

0

k_{1}(t, s) f_{1}

s,

Z s 0

k_{2}(s, θ) f_{2}(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, f_{1}, f_{2} which stated
in [5],[6] and prove the existence of integrable solution x ∈ L_{1}[0, 1] of the integral
equation of Volterra-Fredholm type

(1) x(t) = Z t

0

k_{1}(t, s) f_{1}

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 ∈ L_{1}[0, 1] for the initial value problem of fractional order

(2)

dz
dt = f_{1}

t,R1

0 k (t, s) f s, D^{β}z (s)
ds

, t ∈ [0, 1] , β ∈ (0, 1]

z (0) = 0.

Denote by L_{1}= L_{1}[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 L_{1}into itself if and only if |f (t, x)| ≤ a(t) + b |x| for all t ∈ [0, 1]

and x ∈ R. Here a belongs to L_{1}and 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] − D_{ε}) = meas.D^{c}_{ε}< 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, f_{1}, k, k_{1}. 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

I_{a}^{α}f (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

D_{a}^{α}g (t) = I_{a}^{1−α}Dg (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 ∈ L_{1}and α, β ∈ (0, 1], then

I_{a}^{α}I_{a}^{β}f (t) = I_{a}^{β}I_{a}^{α}f (t) = I_{a}^{α+β}f (t) .
Furthermore,

(I_{a}^{α})^{n}f (t) = I_{a}^{nα}f (t) , n = 1, 2, 3, · · · .
(II) Let α, β ∈ (0, 1] and f, Df ∈ L_{1}, then

(i) DI_{a}^{α}f (t) = I_{a}^{α}Df (t) , when f (a) = 0.

(ii) D^{α}_{a}I_{a}^{β}f (t) = I_{a}^{β−α}f (t) , α < β.

(iii) D^{α}_{a}I_{a}^{β}f (t) = D_{a}^{α−β}f (t) , α > β, f (a) = 0.

(iv) I_{a}^{α}D^{α}_{a}f (t) = f (t) − f (a) .

(III) The operator I_{a}^{α} maps L_{1}into 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, f_{1}: [0, 1] × R → R satisfy the Carathéodory conditions. There exist func-
tions a and a_{1}that belong to L_{1}and positive constants b and b_{1}, such that

|f (t, u)| ≤ a(t) + b |u| , and

|f1(t, u)| ≤ a_{1}(t) + b_{1}|u| ,
hold for (t, u) ∈ [0, 1] × R.

(V2) k_{1}, k : [0, 1] × [0, 1] → R_{+} are measurable with respect to both variables.

Let K_{1}be the Volterra operator associated with the kernal k_{1}while K is the
Fredholm operator associated with k. Both K_{1}and K map L_{1} into itself.

(V3) φ : [0, 1] → [0, 1] is increasing, absolutely continuous and there exists a con-
stant M > 0 such that φ^{0}(t) ≥ M for almost all t ∈ [0, 1] .

(V4) ^{b}^{1}^{bkK}_{M}^{1}^{kkKk} < 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 x_{0}that belongs to L_{1}, such that

(3) x_{0}(t) =
Z t

0

k_{1}(t, s)

a_{1}(s) + b_{1}
Z 1

0

k (s, θ) (a(θ) + bx_{0}(φ (θ))) dθ

ds.

Proof Set

h (t) = Z t

0

k_{1}(t, s)

a_{1}(s) + b_{1}
Z 1

0

k (s, θ) a(θ)dθ

ds
Clearly h ∈ L_{1}as it is evident from our assumption (V 2) . Let
B_{r}= {x : x ∈ L_{1}, kxk ≤ r} with r = _{1−b} ^{khk}

1bkK1kkKk/M. Next, we consider the oper- ator A defined by

(Ax) (t) = Z t

0

k_{1}(t, s)

a_{1}(s) + b_{1}
Z 1

0

k (s, θ) (a(θ) + bx (φ (θ))) dθ

ds.

It is evident that A maps L_{1}into itself continuously.

For x ∈ B_{r}, we have

kAxk = Z 1

0

|Ax(t)| dt =

= Z 1

0

Z t 0

k_{1}(t, s)

a_{1}(s) + b_{1}
Z 1

0

k (s, θ) (a(θ) + bx (φ (θ))) dθ

ds

dt

≤ Z 1

0

Z t 0

k_{1}(t, s)

a_{1}(s) + b_{1}
Z 1

0

k (s, θ) a(θ)dθ

ds

dt + +

Z 1 0

Z t 0

b_{1}bk_{1}(t, s)

Z 1 0

k (s, θ) x (φ (θ)) dθ

ds

dt

≤ khk + Z 1

0

Z t 0

b_{1}bk_{1}(t, s)

Z 1 0

k (s, θ) |x (φ (θ))| dθ

ds

dt

≤ khk + b_{1}b kK_{1}k kKk
Z 1

0

|x (φ (s))| ds

≤ khk + b_{1}b kK_{1}k kKk
Z 1

0

|x(φ (s))| φ^{0}(s)
M ds

≤ khk +b_{1}b

M kK_{1}k kKk
Z φ(1)

φ(0)

|x(v)| dv.

Hence

kAxk ≤ khk +b_{1}b

M kK1k kKk Z 1

0

|x(v)| dv

≤ khk +b_{1}b

M kK_{1}k kKk kxk ≤ khk + b

M kK_{1}k kKk r

≤ khk +b_{1}b

M kK_{1}k kKk khk

1 −b_{1}b kK_{1}k kKk
M

= r.

This proves that A maps B_{r} into B_{r}.

Let B_{r}^{+}= {x : x ∈ B_{r}, x (t) ≥ 0 a.e.} , then we have A (B^{+}_{r}) ⊂ B^{+}_{r} and it can be
shown that B_{r}^{+}is a closed subset of L_{1}and consequently we get that B_{r}^{+}is a complete
metric space. Next we show that A is a contraction in B_{r}^{+}. Let x_{1}, x_{2} ∈ B_{r}^{+} we
have

kAx1− Ax2k =

= Z 1

0

Z t 0

k_{1}(t, s)

Z 1 0

k (s, θ) b_{1}b (x_{1}(φ (θ)) − x_{2}(φ (θ))) dθ

ds

dt

≤ Z 1

0

Z t 0

k_{1}(t, s)

Z 1 0

k (s, θ) b_{1}b (|x_{1}(φ (θ)) − x_{2}(φ (θ))|) dθ

dsdt

≤ kK1k kKk Z 1

0

b_{1}b (|x_{1}(φ (θ)) − x_{2}(φ (θ))|) φ^{0}(θ)
M dθ

≤ kK_{1}k kKk
Z 1

0

b_{1}b

M (|x_{1}(v) − x_{2}(v)|) dv

≤ b_{1}b

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 x_{0}that belongs

to L_{1}and 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 ∈ L_{1}.

Proof Assume that x0the zero element of L_{1}that solves equation (3), and let

y(t) = Z t

0

k_{1}(t, s) f_{1}

s,

Z 1 0

k (s, θ) f (θ, x_{0}(φ (θ))) dθ

ds, t ∈ [0, 1] , hence

|y(t)| ≤ Z t

0

k_{1}(t, s)

a_{1}(s) + b_{1}
Z 1

0

k (s, θ) (a(θ) + bx_{0}(φ (θ))) dθ

ds

|y(t)| ≤ x_{0}(t) ,

where the last equality follows from Lemma 3.1 and hence y = 0. Hence x_{0} solves
equation (1). Now, assume that x_{0} is not the zero element of L_{1}and consider the
set

Q = {y : y ∈ L_{1}, |y(t)| ≤ x_{0}(t) , a.e.} ,

this set Q can be shown to be nonempty, closed, bounded and convex in L_{1}.
Now let H be the nonlinear operator associated with equation (1) and defined
by

Hx(t) = Z t

0

k_{1}(t, s) f_{1}

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 L_{1}into itself continuously. In
what follows, we will prove that H : Q → Q. For this purpose, let x ∈ Q, then

|Hx(t)| =

Rt

0k_{1}(t, s) f_{1}
s,R1

0k (s, θ) f (θ, x (φ (θ))) dθ ds

≤Rt

0k_{1}(t, s)

a_{1}(s) + b_{1}R1

0k (s, θ) (a (θ) + b |x (φ (θ))|) dθ ds

≤Rt
0k_{1}(t, s)

a_{1}(s) + b_{1}R1

0k (s, θ) (a (θ) + bx_{0}(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 L_{1}. To achieve this we assume that {y_{n}} is a sequence in Q.

Theorem 1.2 imply that there is a closed subset D_{ε}of [0, 1] with meas.D_{ε}^{c}< ε such
that all the restrictions x_{0}|_{D}

ε, k_{1}|_{D}

ε×[0,1]and k|_{D}

ε×[0,1] are uniformly continuous.

In what follows we show that {Hy_{n}} is an equicontinuous on Dε, for that let ε_{1}> 0
be given. Since k_{1}|_{D}

ε×[0,1]is uniformly continuous, there exists δ > 0 such that for
t^{0}, t^{00} ∈ D_{ε}, |t^{00}− t^{0}| < δ we have

|k_{1}(t^{00}, s) − k_{1}(t^{0}, s)| < ε_{1}
2γ,

where γ = ka_{1}k + b_{1}kKk kak + ^{bb}_{M}^{1}kKk kx_{0}k. Furthermore, the function k_{1}(t, s)
attains its maximum say C on D_{ε}× [0, 1]. The integral

Rt^{00}
t^{0} f_{1}

s,R1

0k (s, θ) f (θ, x_{0}(φ (θ))) dθ

ds can be made less than _{2C}^{ε}^{1}, for any t^{0}, t^{00} ∈
D_{ε}, with |t^{00}− t^{0}| < δ, because of the property of the intgrand. Finally, for t^{0},
t^{00} ∈ D_{ε}, |t^{00}− t^{0}| < δ we get

|Hy_{n}(t^{00}) − Hy_{n}(t^{0})| ≤

≤
Z t^{0}

0

|k_{1}(t^{00}, s) − k_{1}(t^{0}, s)|

·

|a_{1}(s)| + b_{1}
Z 1

0

k (s, θ) |a (θ) + b |x_{0}(φ (θ))|| dθ

ds +

+

Z t^{00}
t^{0}

k_{1}(t^{00}, s) f_{1}

s,

Z 1 0

k (s, θ) f (θ, x_{0}(φ (θ))) dθ

ds

≤ ε_{1}

2γγ + Cε_{1}
2C = ε_{1.}

Also, the sequence {Hy_{n}} is a sequence of uniformly bounded functions, since

|Hyn(t)| =

=

Z t 0

k_{1}(t, s) f_{1}

s,

Z 1 0

k (s, θ) f (θ, y_{n}(φ (θ))) dθds

, t ∈ D_{ε}

≤ Z t

0

k_{1}(t, s)

a_{1}(s) + b_{1}
Z 1

0

k (s, θ) (a (θ) + b |y_{n}(φ (θ))|) dθ

ds

≤ Z t

0

k_{1}(t, s)

a_{1}(s) + b_{1}
Z 1

0

k (s, θ) (a (θ) + bx_{0}(φ (θ))) dθ

ds

≤ x_{0}(t) , ∀n.

Hence

kHynk_{C(D}

ε)≤ kx0k_{C(D}

ε), ∀n.

and consequently the sequence {Hy_{n}} is a sequence of uniformly bounded and
equicontinuous functions on D_{ε}. Hence, in view of Ascoli-Arz´ela theorem we deduce
that {H (y_{n})} is relatively compact subset of C (D_{ε}) . In fact there exists a sequence
{D_{k}} of closed subsets of [0, 1] with meas.D_{k}^{c} → 0 such that {H (y_{n})} is relatively
compact subset in each C (D_{k}). This implies that {Hy_{n}} has a convergent subse-
quence in each C (D_{k}) k = 1, 2, · · · (We call this subsequence again {H (y_{n})}). But
C (D_{k}) is a complete metric space, hence this subsequence is a Cauchy sequence in
each C (D_{k}), k = 1, 2, · · · . That is for any given > 0 and arbitrary large m, n
kHy_{m}− Hy_{n}k_{C(D}

k) < . Next we show that {Hy_{n}} is convergent in L_{1}. For that
purpose we verify that {Hy_{n}} is a Cauchy sequence in L_{1}.

Let η > 0 be given, we will show that there exists N (η) such that
kHy_{m}− Hy_{n}k_{L}

1< η, whenever m, n > N (η) . Given η > 0, there is a δ > 0 such
that meas.D_{δ} < δ implies

Z

Dδ

x_{0}(s) ds < η
4.
Choose k^{∗}∈ N with meas.D^{c}_{k}∗ < δ and

1

Z

0

|(Hy_{m}) (t) − (Hy_{n}) (t)| dt =

= Z

D^{c}_{k∗}

|(Hy_{m}) (t) − (Hy_{n}) (t)| dt +
Z

Dk∗

|(Hy_{m}) (t) − (Hy_{n}) (t)| dt

≤ Z

D^{c}_{k∗}

{|(Hy_{m}) (t)| + |(Hy_{n}) (t)|} dt + kHy_{m}− Hy_{n}k_{C(D}

k∗)

Z

Dk∗

dt

≤ η

4+η

4 + kHy_{m}− Hy_{n}k_{C(D}

k∗)(1 − δ)

≤ η

2+ kHy_{m}− Hy_{n}k_{C(D}

k∗),

choose N (η) such that m, n > N (η) we have kHy_{m}− Hy_{n}k_{C(D}

k∗)<^{η}_{2}. Then

1

Z

0

|(Hy_{m}) (t) − (Hy_{n}) (t)| dt <η
2+η

2 = η.

This means that {H (y_{n})} is a Cauchy sequence in the space L_{1}and hence the
set H(Q) is relatively compact in L_{1}. Then H has at least one fixed point . This
proves that there exists at least one x ∈ L_{1}that solves equation (1). _{}

3.1. Convolution kernel. Assume that ˜k : [0, 1] → R_{+} is an integrable
function. For an arbitrary function x ∈ L_{1}set

(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 L_{1}into
itself continuously.

Now, consider the following additional condition
(V5) ˜k : [0, 1] → R_{+}and ˜k ∈ L_{1}.

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 x_{0}that belongs to L_{1}, such that

x_{0}(t) =
Z t

0

˜k (t − s)

a_{1}(s) + b_{1}
Z 1

0

k (s, θ) (a(θ) + bx_{0}(φ (θ))) dθ

ds.

Furthermore, the equation (4) x(t) =

Z t 0

˜k (t − s) f_{1}

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) f_{1}

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) = I^{1−β}f_{1}

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 ^{b}_{Γ(2−β)}^{1}^{bkKk}< 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) = I^{β}x (t) ,

then we get

z (t) = I^{β}x (t) =

= I^{β}I^{1−β}f_{1}

t,

Z 1 0

k (t, θ) f (θ, x (θ)) dθ

= I^{1}f_{1}

t,

Z 1 0

k (t, θ) f (θ, x (θ)) dθ

= I^{1}f_{1}

t,

Z 1 0

k (t, θ) f (θ, x (θ)) dθ

= Z t

0

f_{1}

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 D^{β}z (t) is defined.

Furthermore

D^{β}z (t) = I^{1−β}d

dtz (t) = I^{1−β}d

dt I^{β}x (t)

= I^{1−β}d
dtI^{β}

I^{1−β}f_{1}

t,

Z 1 0

k (t, θ) f (θ, x (θ)) dθ

= d

dtI^{1−β}I^{β}I^{1−β}f_{1}

t,

Z 1 0

k (t, θ) f (θ, x (θ)) dθ

= d

dtI^{1}I^{1−β}f_{1}

t,

Z 1 0

k (t, θ) f (θ, x (θ)) dθ

= d
dtI^{1}

I^{1−β}f_{1}

t,

Z 1 0

k (t, θ) f (θ, x (θ)) dθ

= x (t) .

From equation (7)

dz dt = d

dtI^{β}x (t) .
Substituting from equation (6), we get

dz

dt = d

dtI^{β}I^{1−β}f_{1}

t,

Z 1 0

k (t, θ) f θ, D^{β}z (θ) dθ

= d

dtI^{1}f_{1}

t,

Z 1 0

k (t, θ) f θ, D^{β}z (θ) dθ

= f_{1}

t,

Z 1 0

k (t, θ) f θ, D^{β}z (θ) dθ

.

Which proves that if x (t) is any solution of equation (6), then z (t) = I^{β}x (t) is a
solution of the initial value problem (2). Hence our proof is complete. _{}

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J. Stecenko, Integral equations, Noordhoff Leyden 1975.

A. M. A. El-Sayed

Faculty of Science, Alexandria University Alexandria, Egypt

E-mail: [email protected] Nagwa Sherif

Faculty of Science, Suez Canal University Ismailia, Egypt

E-mail: [email protected] I. A. Ibrahim

Faculty of Science, Suez Canal University Ismailia, Egypt

E-mail: [email protected]

(Received: 26.07.05)