FIRST ORDER INTEGRO-DIFFERENTIAL EQUATIONS IN BANACH ALGEBRAS INVOLVING CARATHEODORY AND DISCONTINUOUS NONLINEARITIES

In this paper some existence theorems for the rst order dieren tial equations in Banach algebras is proved under the mixed generalized Lipschitz, Carath eodory and monotonicity conditions.

By a solution of IGDE (1.1) we mean a function x ∈ AC 1 (J, R) such that (i) the function t → x f (t, x) is continuous for each x ∈ R, and (ii) x satisfies the equations in (1.1), where, AC 1 (J, R) is the space of absolutely continuous real-valued functions on J.
In recent years, the topic of nonlinear differential equations in Banach algebras is received the attention of several authors and at present, there is a considerable literature available in this direction.See Dhage and O'Regan [5] Dhage et.al. [6] and and the references therein.In this paper we deal with the second order ordinary differential equations in Banach algebras and discuss the existence results under mixed Lipschitz and Carathéodory EJQTDE, 2005 No. 21, p. 1 conditions.We will employ the fixed point theorems of Dhage [2,3,4] for proving our main existence results.The nonlinear differential equation as well as the existence results of this are new to the literature on the theory of ordinary differential equations.
Our method of study is to convert the IGDE (1.1) into equivalent integral equation and apply the fixed point theorems of Dhage [2,3,4] under suitable conditions on the nonlinearities f and g.In the following section we shall give some preliminaries needed in the sequel.

Auxiliary Results
Let X be a Banach algebra with norm • .A mapping A : X → X is called D-Lipschitz if there exists a continuous nondecreasing function ψ : R + → R + satisfying for all x, y ∈ X with ψ(0) = 0.In the special case when ψ(r) = αr (α > 0), A is called a Lipchitz with a Lipschitz constant α.In particular, if α < 1, A is called a contraction with a contraction constant α.Further, if ψ(r) < r for all r > 0, then A is called a nonlinear contraction on X.Sometimes we call the function ψ a D-function for convenience.An operator T : X → X is called compact if T (S) is a compact subset of X for any S ⊂ X. Similarly T : X → X is called totally bounded if T maps a bounded subset of X into the relatively compact subset of X. Finally T : X → X is called completely continuous operator if it is continuous and totally bounded operator on X.It is clear that every compact operator is totally bounded, but the converse may not be true.The nonlinear alternative of Schaefer type recently proved by Dhage [4] is embodied in the following theorem.
It is known that Theorem 2.1 is useful for proving the existence theorems for the integral equations of mixed type.See [2] and the references therein.The method is commonly known as priori bound method for the nonlinear equations.See, for example, Dugundji and Granas [7], Zeidler [12] and the references therein.
An interesting corollary to Theorem 2.
where 0 is the zero element of X.A cone K is called to be positive if (iv) K • K ⊆ K, where "•" is a multiplication composition in X.We introduce an order relation ≤ in X as follows.Let x, y ∈ X.Then x ≤ y if and only if y −x ∈ K.A cone K is called to be normal if the norm • is monotone increasing on K.It is known that if the cone K is normal in X, then every order-bounded set in X is norm-bounded.The details of cones and their properties appear in Guo and Lakshmikantham [9].
We equip the space AC 1 (J, R) with the order relation ≤ with the help of the cone defined by It is well known that the cone K is positive and normal in AC 1 (J, R).As a result of positivity of the cone K in AC 1 (J, R) we have: For any a, b We use the following fixed point theorem of Dhage [3] for proving the existence of extremal solutions of the IGDE (1.1) under certain monotonicity conditions.

Existence Theory
Let M (J, R) and B(J, R) respectively denote the spaces of measurable and bounded realvalued functions on J. Let C(J, R), be the space of all continuous real-valued functions on J. Define a norm • in C(J, R) by EJQTDE, 2005 No. 21, p. 4 Clearly C(J, R) becomes a Banach algebra with this norm and the multiplication "•" defined by (xy)(t) = x(t)y(t) for all t ∈ J.By L 1 (J, R) we denote the set of Lebesgue integrable functions on J and the norm • in L 1 (J, R) is defined by We need the following lemma in the sequel.
if and only if is solution of the integral equation (in short IE) Proof.The proof is simple and omit the details.
We need the following definition in the sequel.
) is measurable for each x ∈ R, and (ii) x → β(t, x) is continuous almost everywhere for t ∈ J.
Again a Carathódory function β(t, x) is called L 1 -Carathéodory if (iii) for each real number r > 0 there exists a function h r ∈ L 1 (J, R) such that Finally a Carathéodory function For convenience, the function h is referred to as a bound function of β.
We will need the following hypotheses in the sequel.
where Proof.By Lemma 3.1, the IGDE (1.1) is equivalent to integral equation Set X = C(J, R).Define the two mappings A and B on X by and Obviously A and B define the operators A, B : X → X.Then the IGDE (1.1) is equivalent to the operator equation We shall show that the operators A and B satisfy all the hypotheses of Corollary 2.1.We first show that A is a Lipschitz on X.Let x, y ∈ X.Then by (H 1 ), Taking the supremum over t we obtain Ax − Ay ≤ k x − y for all x, y ∈ X.So A is a Lipschitz on X with a Lipschitz constant k .Next we show that B is completely continuous on X.Using the standard arguments as in Granas et al. [8], it is shown that B is a continuous operator on X.Let S be a bounded set in X.We shall show that B(X) is a uniformly bounded and equicontinuous set in X.Since g(t, x(t)) is L 1 X -Carathéodory, we have Taking the supremum over t, we obtain Bx ≤ M for all x ∈ S, where M = x 0 f (0,x 0 ) + h L 1 T.This shows that B(X) is a uniformly bounded set in X.Now we show that B(X) is an equicontinuous set.Let t, τ ∈ J. Then for any x ∈ X we have by (3.6), Hence B(X) is an equi-continuous set and consequently B(X) is relatively compact by Arzelà-Ascoli theorem.As a result B is a compact and continuous operator on X.Thus all the conditions of Theorem 2.1 are satisfied and a direct application of it yields that either the conclusion (i) or the conclusion (ii) holds.We show that the conclusion (ii) is not possible.Let x ∈ X be any solution to IGDE (1.1).Then we have, for any λ ∈ (0, 1), Put u(t) = sup s∈[0,t] |x(s)|, for t ∈ J. Then we have |x(t)| ≤ u(t) for all t ∈ J, and so, there is a point t * ∈ [0, t] such that u(t) = |x(t * )|.From (3.9) it follows that where, Then u(t) ≤ w(t) and a direct differentiation of w(t) yields A change of variables in the above integral gives that Now an application of mean value theorem yields that there is a constant M > 0 such that w(t) ≤ M for all t ∈ J.This further implies that for all t ∈ J. Thus the conclusion (ii) of Corollary 2.1 does not hold.Therefore the operator equation AxBx = x and consequently the IGDE (1.1) has a solution on J.This completes the proof.

Existence of Extremal Solutions
We need the following definitions in the sequel.

Carathéodory case
We consider the following set of assumptions: (B 1 ) g is Carathéodory.
(B 2 ) The functions f (t, x) and g(t, x) are nondecreasing in x and y almost everywhere for t ∈ J.
(B 3 ) The IGDE (1.1) has a lower solution u and an upper solution v on J with u ≤ v.
(B 4 ) The function : J → R defined by Proof.Now IGDE (1.1) is equivalent to IE (3.4) on J. Let X = C(J, R) and define an order relation " ≤ " by the cone K given by (2.2).Clearly K is a normal cone in X. Define two operators A and B on X by (3.5) and (3.5) respectively.Then IE (1.1) is transformed into an operator equation Ax(t)Bx(t) = x(t) in a Banach algebra X.Notice that (B 1 ) implies A, B : [u, v] → K. Since the cone K in X is normal, [u, v] is a norm bounded set in X.Now it is shown, as in the proof of Theorem 3.1, that A is a Lipschitz with a Lipschitz constant α and B is completely continuous operator on [u, v].Again the hypothesis (B 2 ) implies that A and B are nondecreasing on [u, v].To see this, let x, y ∈ [u, v] be such that x ≤ y.Then by (B 2 ), Similarly, Since αM ≤ k ( x 0 f (0,x 0 ) +T L 1 ) < 1, we apply Theorem 4.1 to the operator equation Ax Bx = x to yield that the IGDE (1.1) has a minimal and a maximal positive solution on J.This completes the proof.

Discontinuous case
We need the following definition in the sequel.For convenience, the function h is referred to as a bound function of β.
We consider the following hypotheses in the sequel.Proof.Now IGDE (1.1) is equivalent to IE (3.4) on J. Let X = C(J, R) and define an order relation " ≤ " by the cone K given by (2.2).Clearly K is a normal cone in X. Define two operators A and B on X by (3.5) and (3.6) respectively.Then FIE (1.1) is transformed into an operator equation Ax(t) Bx(t) = x(t) in a Banach algebra X.Notice that (B 0 ) implies A, B : [u, v] → K. Since the cone K in X is normal, [u, v] is a norm bounded set in X.
Step I : Next we show that A is completely continuous on [a, b].Now the cone K in X is normal, so the order interval [a, b] is norm-bounded.Hence there exists a constant r > 0 such that x ≤ r for all x ∈ [a, b].As f is continuous on compact J × [−r, r], it attains its maximum, say M .Therefore for any subset S of [a, b] we have This shows that A(S) is an equi-continuous set in X.Now an application of Arzelà-Ascoli theorem yields that A is a completely continuous operator on [a, b].
Step II : Next we show that B is totally bounded operator on [a, b].To finish, we shall show that B(S) is uniformly bounded ad equi-continuous set in X for any subset S of [a, b].Since the cone K in X is normal, the order interval [a, b] is norm-bounded.Hence there is a real number r > 0 such that x ≤ r for all x ∈ [a, b].Let y ∈ B(S) be arbitrary.Then, for some x ∈ S. By hypothesis (B 2 ) one has Taking the supremum over t,

1 Introduction
Let R denote the real line.Given a closed and bounded interval J = [0, T ] in R, consider the first order functional integro-differential equation ( in short IGDE) , x(s)) ds, a.e.t ∈ J x(0) = x 0 ∈ R,

Theorem 2 . 1 (
Dhage[4]) Let X be a Banach algebra and let A, B : X → X be two operators satisfying (a) A is a D-Lipschitz with a D-function ψ, (b) B is compact and continuous, and (c) M ψ(r) < r whenever r > 0, where M = B(X) = sup{ Bx : x ∈ X}.Then either (i) the equation λAx Bx = x has a solution for λ = 1, or

Corollary 2 . 1
Let X be a Banach algebra and let A, B : X → X be two operators satisfying (a) A is Lipschitz with a Lipschitz constant α, (b) B is compact and continuous, and (c) αM < 1 , where M = B(X) := sup{ Bx : x ∈ X}.Then either (i) the equation λAx Bx = x has a solution for λ = 1, or

Theorem 2 . 2 (
Dhage [3]).Let K be a cone in a Banach algebra X and let a, b ∈ X. Suppose that A, B : [a, b] → K are two operators such that (a) A is Lipschitz with a Lipschitz constant α, (b) B is completely continuous, (c) Ax Bx ∈ [a, b] for each x ∈ [a, b], and EJQTDE, 2005 No. 21, p. 3 (d) A and B are nondecreasing.Further if the cone K is positive and normal, then the operator equation Ax Bx = x has a least and a greatest positive solution in [a, b], whenever αM < 1, where M = B([a, b]) := sup{ Bx : x ∈ [a, b]}.

Theorem 2 . 3 (Remark 2 . 1
Dhage [4]).Let K be a cone in a Banach algebra X and let a, b ∈ X. Suppose that A, B : [a, b] → K are two operators such that (a) A is completely continuous, (b) B is totally bounded, (c) Ax By ∈ [a, b] for each x, y ∈ [a, b], and (d) B is nondecreasing.Further if the cone K is positive and normal, then the operator equation Ax Bx = x has a least and a greatest positive solution in [a, b].Theorem 2.4 (Dhage [4]).Let K be a cone in a Banach algebra X and let a, b ∈ X. Suppose that A, B : [a, b] → K are two operators such that (a) A is Lipschitz with a Lipschitz constant α, (b) B is totally bounded, (c) Ax By ∈ [a, b] for each x, y ∈ [a, b], and (d) B is nondecreasing.Further if the cone K is positive and normal, then the operator equation Ax Bx = x has least and a greatest positive solution in [a, b], whenever αM < 1, where M = B([a, b]) := sup{ Bx : x ∈ [a, b]}.Note that hypothesis (c) of Theorems 2.2, 2.3, and 2.4 holds if the operators A and B are monotone increasing and there exist elements a and b in X such that a ≤ Aa Ba and Ab Bb ≤ b.

(C 1 )
The function f is continuous on J × R.(C 2 ) There is a function k ∈ B(J, R) such that k(t) > 0, a.e.t ∈ I and |f (t, x) − f (t, y)| ≤ k(t) |x − y| , a.e.t ∈ J for all x, y ∈ R. (C 3 ) The function f (t, x) is nondecreasing in x almost everywhere for t ∈ J.(C 4 ) The function g is Chandrabhan.

0 ( 0 ( 0 (
, 2005 No. 21, p. 13 which shows that B(S) is a uniformly bounded set in X. Similarly let t, τ ∈ J. Then for any y ∈ B(S), |y(t) − y(τ )| ≤ t 0 (t − s)g(s, x(s)) ds − τ 0 (τ − s)g(s, x(s)) ds≤ t t − s)g(s, x(s)) ds − t 0 (τ − s)g(s, x(s)) ds + t τ − s)g(s, x(s)) ds − τ 0 (τ − s)g(s, x(s)) ds ≤ t t − τ )g(s, x(s)) ds + t τ (τ − s)g(s, x(s)) ds ≤ T 0 |t − τ ||g(s, x(s))| ds + T t τ |g(s, x(s))| ds ≤ T 0 |t − τ | (s) ds + T t τ (s) ds ≤ |t − τ | L 1 + |p(t) − p(τ )|where p(t) = T t 0 (s) ds.Since the function p is continuous on compact interval J, it is uniformly continuous , and therefore we have|y(t) − y(τ )| → 0 as t → τ,for all y ∈ B(S).Hence B(S) is an equi-continuous set in X.Thus B is totally bounded in view of Arzelà-Ascoli theorem.Thus all the conditions of Theorem 2.3 are satisfied and hence an application of it yields that the IGDE (1.1) has a maximal and a minimal solution on J. Theorem 4.3 Suppose that the assumptions (B 0 ),(B 3 ) and (C 1 )-(C 4 ) hold.Further if k x 0 f (0,x 0 ) + T L 1 < 1, and is given in Remark 4.1, then IGDE (1.1) has a minimal and a maximal positive solution on J.Proof.Now IGDE (1.1) is equivalent to IE (3.4) on J. Let X = C(J, R) and define an order relation " ≤ " by the cone K given by (2.2).Clearly K is a normal cone in X. Define two operators A and B on X by (3.5) and (3.6) respectively.Then FIE (1.1) is transformed into an operator equationAx(t) Bx(t) = x(t) in a Banach algebra X.Notice that (B 0 ) implies A, B : [u, v] → K. Since the cone K in X is normal, [u, v]is a norm bounded set in X.Now it can be shown as in the proofs of Theorem 3.1 and Theorem 2.4 that the operator A is a Lipschitz with a Lipschitz constant α = k and B is totally bounded with M = B([u, v]) = x 0 f (0,x 0 ) +T L 1 .respectively.Since αM = k x 0 f (0,x 0 ) +T L 1 < 1, the desired conclusion follows by an application of Theorem 2.4.This completes the proof.EJQTDE, 2005 No. 21, p. 14 EJQTDE, 2005No.21, p. 5 (H 1 ) The function f : J × R → R is continuous and there exists a function k ∈ B(J, R) such that k(t) > 0, a.e.t ∈ J and|f (t, x) − f (t, y)| ≤ k(t) |x − y| , a.e.t ∈ J for all x, y ∈ R. (H 2 ) The function g is L 1 X -Carathéodory with bound function h.(H 3 ) There exists a continuous and nondecreasing function Ω : [0, ∞) → (0, ∞) and a function γ ∈ L 1 (J, R) such that γ(t) > 0, a.e.t ∈ J and |g(t, x)| ≤ γ(t)Ω |x| , a.e.t ∈ J, for all x ∈ R. Theorem 3.1 Assume that the hypotheses (H 1 )-(H 3 ) hold.Suppose that Definition 4.2 A solution x M of the IGDE (1.1) is said to be maximal if for any other solution x to IGDE(1.1) one has x(t) ≤ x M (t), for all t ∈ J. Again a solution x m of the IGDE (1.1) is said to be minimal if x m (t) ≤ x(t), for all t ∈ J, where x is any solution of the IGDE (1.1) on J.
R) is called a lower solution of the IGDE (1.1) on t 0 g(s, u(s)) ds, a.e.t ∈ J, and u(0) ≤ x 0 .Again a function v ∈ AC 1 (J, R) is called an upper solution of the IGDE (1.1) on J if t 0 g(s, u(s)) ds, a.e.t ∈ J, and v(0) ≥ x 0 .EJQTDE, 2005 No. 21, p. 9 ∈ J.So A and B are nondecreasing operators on [u, v].Again Lemma 4.1 and hypothesis (B 3 ) implies that x) is measurable for each x ∈ R, and(ii) x → β(t, x) is nondecreasing almost everywhere for t ∈ J. Again a Chandrabhan function β(t, x) is called L 1 -Chandrabhan if (iii) for each real number r > 0 there exists a function h r ∈ L 1 (J, R) such that |β(t, x)| ≤ h r (t), a.e.t ∈ Jfor all x ∈ R with |x| ≤ r.