Solvability of a Maximum Quadratic Integral Equation of Arbitrary Orders


 Κόσμος Ζέρβας
 11 ημέρες πριν
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1 Advances in Dynamical Systems and Applications ISSN , Volume 11, Number 2, pp (216) 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 Johnny Henderson Baylor University Department of Mathematics Waco, Texas , USA Johnny Kishin Sadarangani Universidad de Las Palmas de Gran Canaria Departamento de Matemáticas Campus de Tafira Baja, 3517 Las Palmas de Gran Canaria, Spain 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
2 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}.
3 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.
4 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 )
5 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
6 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
7 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]
8 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)
9 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)]
10 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)
11 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 = r 2 r 5Γ(3/2) 2 5Γ(3/2) =.32 < (ϕ(1) ϕ()) β = = 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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