Solvability of a Maximum Quadratic Integral Equation of Arbitrary Orders

Advances in Dynamical Systems and Applications ISSN 973-5321, Volume 11, Number 2, pp. 93 14 (216) http://campus.mst.edu/aa Solvability of a Maximum Quadratic Integral Equation of Arbitrary Orders Mohamed Abdalla Darwish King Abdulaziz University Sciences Faculty for Girls Department of Mathematics Jeddah, Saudi Arabia dr.madarwish@gmail.com Johnny Henderson Baylor University Department of Mathematics Waco, Texas 76798-7328, USA Johnny Henderson@baylor.edu Kishin Sadarangani Universidad de Las Palmas de Gran Canaria Departamento de Matemáticas Campus de Tafira Baja, 3517 Las Palmas de Gran Canaria, Spain ksadaran@dma.ulpgc.es Abstract We investigate a new quadratic integral equation of arbitrary orders with maximum and prove an existence result for it. We will use a fixed point theorem due to Darbo as well as the monotonicity measure of noncompactness due to Banaś and Olszowy to prove that our equation has at least one solution in C[, 1] which is monotonic on [, 1]. AMS Subject Classifications: 45G1, 47H9, 45M99. Keywor: Fractional, monotonic solutions, monotonicity measure of noncompactness, quadratic integral equation, Darbo theorem. Received August 24, 216; Accepted November 27, 216 Communicated by Martin Bohner

94 M. A. Darwish, J. Henderson and K. Sadarangani 1 Introduction In several papers, among them [1,11], the authors studied differential and integral equations with maximum. In [6 9] Darwish et al. studied fractional integral equations with supremum. Also, in [4, 5], Caballero et al. studied the Volterra quadratic integral equations with supremum. They showed that these equations have monotonic solutions in the space C[, 1]. Darwish [7] generalized and extended the Caballero et al. [4] results to the case of quadratic fractional integral equations with supremum. In this paper we will study the fractional quadratic integral equation with maximum y(t) = f(t) (T y)(t) t ϕ (s)κ(t, s) max [,σ(s)] y(τ) (ϕ(t) ϕ(s)) 1β, t J = [, 1], < β < 1, (1.1) where f, ϕ : J R, T : C(J) C(J), σ : J J and κ : J J R. By using the monotonicity measure of noncompactness due to Banaś and Olszowy [3] as well as the Darbo fixed point theorem, we prove the existence of monotonic solutions to (1.1) in C[, 1]. Now, we assume that (E, ) is a real Banach space. We denote by B(x, r) the closed ball centred at x with radius r and B r B(θ, r), where θ is a zero element of E. We let X E. The closure and convex closure of X are denoted by X and ConvX, respectively. The symbols X Y and λy are using for the usual algebraic operators on sets and M E and N E stand for the families defined by M E = {A E : A, A is bounded} and N E = {B M E : B is relatively compact}, respectively. Definition 1.1 (See [2]). A function µ : M E [, ) is called a measure of noncompactness in E if the following conditions: hold. 1 {X M E : µ(x) = } = kerµ N E, 2 if X Y, then µ(x) µ(y ), 3 µ(x) = µ(x) = µ(convx), 4 µ(λx (1 λ)y ) λµ(x) (1 λ)µ(y ), λ 1 and 5 if (X n ) is a sequence of closed subsets of M E with X n X n1 (n = 1, 2, 3,...) and lim n µ(x n ) = then X = n=1x n, We will establish our result in the Banach space C(J) of all defined, real and continuous functions on J [, 1] with standard norm y = max{ y(τ) : τ J}. Next, we define the measure of noncompactness related to monotonicity in C(J); see [2, 3]. Let Y C(J) be a bounded set. For y Y and ε, the modulus of continuity of the function y, denoted by ω(y, ε), is defined by ω(y, ε) = sup{ y(t) y(s) : t, s J, t s ε}.

A Maximum Quadratic Integral Equation 95 Moreover, we let and Define and d(y) = ω(y, ε) = sup{ω(y, ε) : y Y } ω (Y ) = lim ε ω(y, ε). sup ( y(t) y(s) [y(t) y(s)]) t,s J, s t d(y ) = sup d(y). y Y Notice that all functions in Y are nondecreasing on J if and only if d(y ) =. Now, we define the map µ on M C(J) as µ(y ) = d(y ) ω (Y ). Clearly, µ satisfies all conditions in Definition 3, and therefore, it is a measure of noncompactness in C(J) [3]. Definition 1.2. Let P : M E be a continuous mapping, where M E. Suppose that P maps bounded sets onto bounded sets. Let Y be any bounded subset of M with µ(py ) αµ(y ), α, then P is called verify the Darbo condition with respect to a measure of noncompactness µ. In the case α < 1, the operator P is said to be a contraction with respect to µ. Theorem 1.3 (See [1]). Let Ω E be a closed, bounded and convex set. If P : Ω Ω is a continuous contraction mapping with respect to µ, then P has a fixed point in Ω. We will need the following two lemmas in order to prove our results [4]. Lemma 1.4. Let r : J J be a continuous function and y C(J). If, for t J, then F y C(J). (F y)(t) = max [,σ(t)] y(τ), Lemma 1.5. Let (y n ) be a sequence in C(J) and y C(J). If (y n ) converges to y C(J), then (F y n ) converges uniformly to F y uniformly on J.

96 M. A. Darwish, J. Henderson and K. Sadarangani 2 Main Theorem Let us consider the following assumptions: (a 1 ) f C(J). Moreover, f is nondecreasing and nonnegative on J. (a 2 ) The operator T : C(J) C(J) is continuous and satisfies the Darbo condition with a constant c for the measure of noncompactness µ. Moreover, T y if y. (a 3 ) There exist constants a, b such that (T y)(t) a b y y C(J), t J. (a 4 ) The function ϕ : J R is C 1 (J) and nondecreasing. (a 5 ) The function κ : J J R is continuous on J J and nondecreasing t and s separately. Moreover, κ = κ(t, s). sup (t,s) J J (a 6 ) The function σ : J J is nondecreasing and continuous on J. (a 7 ) r > such that and f κ r (a br ) (ϕ(1) ϕ()) β r (2.1) ck r < (ϕ(1) ϕ())β. Now, we define two operators K and F on C(J) as follows and (Ky)(t) = 1 t ϕ (s)κ(t, s) max [,σ(s)] y(τ) (2.2) (ϕ(t) ϕ(s)) 1β (Fy)(t) = f(t) (T y)(t) (Ky)(t), (2.3) respectively. Solving (1.1) is equivalent to find a fixed point of the operator F. Under the above assumptions, we will prove the following theorem. Theorem 2.1. Assume the assumptions (a 1 ) (a 7 ) are satisfied. Then (1.1) has at least one solution y C(J) which is nondecreasing on J. Proof. First, we claim that the operator F transforms C(J) into itself. For this, it is sufficient to show that if y C(J), then Ky C(J). Let y C(J) and t 1, t 2 J (t 1 t 2 ) such that t 2 t 1 ε for fixed ε >, then we have (Ky)(t 2 ) (Ky)(t 1 )

A Maximum Quadratic Integral Equation 97 = 1 ϕ (s)κ(t 2, s) max [,σ(s)] y(τ) ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) (ϕ(t 1 ) ϕ(s)) 1β 1 ϕ (s)κ(t 2, s) max [,σ(s)] y(τ) ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) 1 ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) 1 ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) (ϕ(t 1 ) ϕ(s)) 1β 1 1 ϕ (s) κ(t 2, s) κ(t 1, s) max [,σ(s)] y(τ) ϕ (s) κ(t 1, s) max [,σ(s)] y(τ) 1 t 1 ϕ (s) (ϕ(t 2 ) ϕ(s)) β1 (ϕ(t 1 ) ϕ(s)) β1 max y ω κ(ε,.) where we used { t 1 ϕ (s) (ϕ(t 2 ) ϕ(s)) κ y 1β [,σ(s)] ϕ (s)[(ϕ(t 1 ) ϕ(s)) β1 (ϕ(t 2 ) ϕ(s)) β1 ] } ϕ (s) (ϕ(t 2 ) ϕ(s)) 1β κ(t 1, s) y(τ) = y ω κ(ε,.)(ϕ(t 2 ) ϕ()) β κ y [ (ϕ(t 1 ) ϕ()) β (ϕ(t 2 ) ϕ()) β 2(ϕ(t 2 ) ϕ(t 1 )) β] y ω κ(ε,.)(ϕ(t 2 ) ϕ()) β 2κ y (ϕ(t 2) ϕ(t 1 )) β y ω κ(ε,.)(ϕ(1) ϕ()) β 2κ y [ω(ϕ, ε)]β, (2.4) ω κ (ε,.) = sup κ(t, s) κ(τ, s) t, τ J, tτ ε

98 M. A. Darwish, J. Henderson and K. Sadarangani and the fact that ϕ(t 1 ) ϕ() ϕ(t 2 ) ϕ(). Notice that, since the function κ is uniformly continuous on J J and the function ϕ is continuous on J, then when ε, we have that ω κ (ε,.) and ω(ϕ, ε). Therefore, Ky C(J) and consequently, Fy C(J). Now, for t J, we have Hence (Fy)(t) (T y)(t) f(t) f a b y t t ϕ (s)κ(t, s) max [,σ(s)] y(τ) (ϕ(t) ϕ(s)) 1β ϕ (s) κ(t, s) (ϕ(t) ϕ(s)) 1β f (a b y )κ y (ϕ(t) ϕ()) β. Fy f (a b y )κ y (ϕ(1) ϕ()) β. By assumption (a 7 ), if y r, we get Fy f (a br )κ r (ϕ(1) ϕ()) β r. max y(τ) [,σ(s)] Therefore, F maps B r into itself. Next, we consider the operator F on the set B r = {y B r : y(t), t J}. It is clear that B r is closed, convex and bounded. By these facts and our assumptions, we obtain F maps B r into itself. In what follows, we will show that F is continuous on B r. For this, let (y n ) be a sequence in B r such that y n y and we will show that Fy n Fy. We have, for t J, (Fy n )(t) (Fy)(t) = (T y n )(t) t ϕ (s)κ(t, s) max [,σ(s)] y n (τ) (ϕ(t) ϕ(s)) 1β (T y)(t) t ϕ (s)κ(t, s) max [,σ(s)] y(τ) (ϕ(t) ϕ(s)) 1β (T y n )(t) t ϕ (s)κ(t, s) max [,σ(s)] y n (τ) (ϕ(t) ϕ(s)) 1β (T y)(t) t ϕ (s)κ(t, s) max [,σ(s)] y n (τ) (ϕ(t) ϕ(s)) 1β (T y)(t) t ϕ (s)κ(t, s) max [,σ(s)] y n (τ) (ϕ(t) ϕ(s)) 1β

A Maximum Quadratic Integral Equation 99 (T y)(t) t (T y n)(t) (T y)(t) (T y)(t) t By applying Lemma 1.5, we get ϕ (s)κ(t, s) max [,σ(s)] y(τ) (ϕ(t) ϕ(s)) 1β t ϕ (s) κ(t, s) (ϕ(t) ϕ(s)) 1β ϕ (s) κ(t, s) max [,σ(s)] y n (τ) (ϕ(t) ϕ(s)) 1β max y n(τ) max [,σ(s)] Fy n Fy κ r (ϕ(1) ϕ()) β T y n T y By the continuity of T, n 1 N such that Also, n 2 N such that T y n T y y n y [,σ(s)] y(τ). κ (a br )(ϕ(1) ϕ()) β y n y. (2.5) ε 2κ r (ϕ(1) ϕ()) β, n n 1. ε 2κ (a br )(ϕ(1) ϕ()) β, n n 2. Now, take n max{n 1, n 2 }, then (2.5) gives us that Fy n Fy ε. This shows that F is continuous in B r. Next, let Y B r be a nonempty set. Let us choose y Y and t 1, t 2 J with t 2 t 1 ε for fixed ε >. Since no generality will loss, we will assume that t 2 t 1. Then, by using our assumptions and (2.4), we obtain (Fy)(t 2 ) (Fy)(t 1 ) f(t 2 ) f(t 1 ) (T y)(t 2 )(Ky)(t 2 ) (T y)(t 2 )(Ky)(t 1 ) (T y)(t 2 )(Ky)(t 1 ) (T y)(t 1 )(Ky)(t 1 ) ω(f, ε) (T y)(t 2 ) (Ky)(t 2 ) (Ky)(t 1 ) (T y)(t 2 ) (T y)(t 1 ) (Ky)(t 1 ) (a b y ) y [ ω(f, ε) ωκ (ε,.)(ϕ(1) ϕ()) β 2κ (ω(ϕ, ε)) β] ω(t y, ε) y κ (ϕ(t 1 ) ϕ()) β ω(f, ε) r (a br ) [ ωκ (ε,.)(ϕ(1) ϕ()) β 2κ (ω(ϕ, ε)) β]

1 M. A. Darwish, J. Henderson and K. Sadarangani Hence, κ r (ϕ(1) ϕ()) β ω(t y, ε). ω(fy, ε) ω(f, ε) r (a br ) Consequently, κ r (ϕ(1) ϕ()) β ω(t y, ε). ω(fy, ε) ω(f, ε) r (a br ) κ r (ϕ(1) ϕ()) β ω(t Y, ε). [ ωκ (ε,.)(ϕ(1) ϕ()) β 2κ (ω(ϕ, ε)) β] [ ωκ (ε,.)(ϕ(1) ϕ()) β 2κ (ω(ϕ, ε)) β] The uniform continuity of the function κ on J J and the continuity of the functions f and ϕ on J, implies the last inequality becomes ω (FY ) κ r (ϕ(1) ϕ()) β ω (T Y ). (2.6) In the next step, fix arbitrary y Y and t 1, t 2 J with t 2 > t 1. Then, by our assumptions, we have (Fy)(t 2 ) (Fy)(t 1 ) [(Fy)(t 2 ) (Fy)(t 1 )] = f(t 2) (T y)(t 2) ϕ (s)κ(t 2, s) max [,σ(s)] y(τ) f(t 1 ) (T y)(t 1) ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) (ϕ(t 1 ) ϕ(s)) 1β [ f(t 2 ) (T y)(t 2) ϕ (s)κ(t 2, s) max [,σ(s)] y(τ) f(t 1 ) (T y)(t ] 1) ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) (ϕ(t 1 ) ϕ(s)) 1β { f(t 2 ) f(t 1 ) [f(t 2 ) f(t 1 )]} (T y)(t 2 ) t2 ϕ (s)κ(t 2, s) max [,σ(s)] y(τ) (T y)(t 1) ϕ (s)κ(t 2, s) max [,σ(s)] y(τ) (T y)(t 1 ) t2 ϕ (s)κ(t 2, s) max [,σ(s)] y(τ)

A Maximum Quadratic Integral Equation 11 But = = (T y)(t 1) {[ (T y)(t2 ) (T y)(t 1) [ (T y)(t1 ) (T x)(t 1) ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) (ϕ(t 1 ) ϕ(s)) 1β ϕ (s)κ(t 2, s) max [,σ(s)] y(τ) ] ϕ (s)κ(t 2, s) max [,σ(s)] y(τ) ϕ (s)κ(t 2, s) max [,σ(s)] y(τ) ]} ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) (ϕ(t 1 ) ϕ(s)) 1β { (T y)(t 2) (T y)(t 1 ) [(T y)(t 2 ) (T y)(t 1 )]} ϕ (s)κ(t 2, s) max [,σ(s)] y(τ) (T y)(t { 1) t2 ϕ (s)κ(t 2, s) max [,σ(s)] y(τ) ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) (ϕ(t 1 ) ϕ(s)) 1β [ ϕ (s)κ(t 2, s) max [,σ(s)] y(τ) ]} ϕ (s)κ(t 1, s) max [,σ(s)] y(τ). (2.7) (ϕ(t 1 ) ϕ(s)) 1β ϕ (s)κ(t 2, s) max [,σ(s)] y(τ) ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) (ϕ(t 1 ) ϕ(s)) 1β ϕ (s)κ(t 2, s) max [,σ(s)] y(τ) t2 ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) t1 ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) t1 ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) (ϕ(t 1 ) ϕ(s)) 1β ϕ (s)(κ(t 2, s) κ(t 1, s)) max [,σ(s)] y(τ) ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) t 1 t1 κ(t 1, s)[(ϕ(t 2 ) ϕ(s)) β1 (ϕ(t 1 ) ϕ(s)) β1 ] max y(τ). [,σ(s)]

12 M. A. Darwish, J. Henderson and K. Sadarangani Since κ(t 2, s) κ(t 1, s) (κ(t, s) is nondecreasing with respect to t), we have and, since ϕ (s)(κ(t 2, s) κ(t 1, s)) max [,σ(s)] y(τ) (2.8) 1 (ϕ(t 2 ) ϕ(s)) 1 1β (ϕ(t 1 ) ϕ(s)) for s [, t 1) then 1β ϕ (s)κ(t 1, s)[(ϕ(t 2 ) ϕ(s)) β1 (ϕ(t 1 ) ϕ(s)) β1 ] max t 1 t1 t 1 ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) [,σ(s)] y(τ) ϕ (s)κ(t 1, t 1 )[(ϕ(t 2 ) ϕ(s)) β1 (ϕ(t 1 ) ϕ(s)) β1 ] max [,σ(t 1 )] y(τ) ϕ (s)κ(t 1, t 1 ) max [,σ(t1)] y(τ) = κ(t 1, t 1 ) max [,σ(t 1 )] y(τ) [ ϕ (s) = κ(t 1, t 1 ) (ϕ(t 2) ϕ()) β (ϕ(t 1 ) ϕ()) β β. Finally, (2.8) and (2.9) imply that max y(τ) [,σ(t 1 )] ] ϕ (s) (ϕ(t 1 ) ϕ(s)) 1β (2.9) ϕ (s)κ(t 2, s) max [,σ(s)] y(τ) The above inequality and (2.7) lea us to ϕ (s)κ(t 1, s) max [,σ(s)] y(τ) (ϕ(t 1 ) ϕ(s)) 1β. Thus, and therefore, (Fy)(t 2 ) (Fy)(t 1 ) [(Fy)(t 2 ) (Fy)(t 1 )] (T y)(t 2) (T y)(t 1 ) [(T y)(t 2 ) (T y)(t 1 )] ϕ (s)κ(t 2, s) max [,σ(s)] y(τ) κ r (ϕ(1) ϕ()) β d(t y). d(fy) κ r (ϕ(1) ϕ()) β d(t y) d(fy ) κ r (ϕ(1) ϕ()) β d(t Y ). (2.1)

A Maximum Quadratic Integral Equation 13 Finally, (2.6) and (2.1) give us that or ω (FY ) d(fy ) κ r (ϕ(1) ϕ()) β (ω (FY ) d(t Y )) µ(fy ) r κ (ϕ(1) ϕ()) β µ(t Y ) κ cr (ϕ(1) ϕ()) β µ(y ). κ r c Since < (ϕ(1) ϕ())β, F is a contraction operator with respect to µ. Finally, by Theorem 1.3, F has at least one fixed point, or equivalently, (1.1) has at least one nondecreasing solution in B r. This finishes our proof. Next, we present the following numerical example in order to illustrate our results. Example 2.2. Let us consider the following integral equation with maximum y(t) = arctan t y(t) t t2 s 2 max [,ln(s1)] y(τ) 5Γ(1/2) 2 s 1, t J. (2.11) t 1 s 1 Notice that (2.11) is a particular case of (1.1), where f(t) = arctan t, (T y)(t) = y(t)/5, β = 1/2, ϕ(s) = s 1, κ(t, s) = t 2 s 2 and σ(t) = ln(t 1). It is not difficult to see that assumptions (a 1 ), (a 2 ), (a 3 ), (a 4 ), (a 5 ) and (a 6 ) are verified with f = π/4, c = 1/5, a =, b = 1/5 and κ = 2. Now, the inequality (2.1) in assumption (a 7 ) takes the expression π 4 which is satisfied by r = 1. Moreover, cκ r = 2 2 1 r 2 r 5Γ(3/2) 2 5Γ(3/2) =.32 < (ϕ(1) ϕ()) β = 1 2 1 = 1.56. Therefore, by Theorem 2.1, (2.11) has at least one continuous and nondecreasing solution which is located in the ball B 1. Acknowledgements The first author s permanent address is Department of Mathematics, Faculty of Science, Damanhour University, Damanhour, Egypt.

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