SOLVABILITY OF A QUADRATIC INTEGRAL EQUATION OF FREDHOLM TYPE IN HÖLDER SPACES

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1 Electronic Journal of Differential Equations, Vol. 214 (214), No. 31, pp ISSN: URL: or ftp ejde.math.txstate.edu SOLVABILITY OF A QUADRATIC INTEGRAL EQUATION OF FREDHOLM TYPE IN HÖLDER SPACES JOSEFA CABALLERO, MOHAMED ABDALLA DARWISH, KISHIN SADARANGANI Abstract. In this article, we prove the existence of solutions of a quadratic integral equation of Fredholm type with a modified argument, in the space of functions satisfying a Hölder condition. Our main tool is the classical Schauder fixed point theorem. 1. Introduction Differential equations with a modified arguments arise in a wide variety of scientific and technical applications, including the modelling of problems from the natural and social sciences such as physics, biological and economics sciences. A special class of these differential equations have linear modifications of their arguments, and have been studied by several authors, see [1] [9] and their references. The aim of this article is to investigate the existence of solutions of the following integral equation of Fredholm type with a modified argument, x(t) = p(t) + x(t) k(t, τ) x(r(τ)) dτ, t [, 1]. (1.1) Our solutions are placed in the space of functions satisfying the Hölder condition. A sufficient condition for the relative compactness in these spaces and the classical Schauder fixed point theorem are the main tools in our study. 2. Preliminaries Our starting point in this section is to introduce the space of functions satisfying the Hölder condition and some properties in this space. These properties can be found in [2]. Let [a, b] be a closed interval in R, by C[a, b] we denote the space of continuous functions on [a, b] equipped with the supremum norm; i.e., x = sup{ x(t) : t [a, b]} for x C[a, b]. For < α 1 fixed, by H α [a, b] we will denote the space of the real functions x defined on [a, b] and satisfying the Hölder condition; that is, those functions x for which there exists a constant Hx α such that x(t) x(s) Hx α, (2.1) for all t, s [a, b]. It is easily proved that H α [a, b] is a linear subspace of C[a, b]. 2 Mathematics Subject Classification. 4G1, 4M99, 47H9. Key words and phrases. Fredholm; Hölder condition; Schauder fixed point theorem. c 214 Texas State University - San Marcos. Submitted October 31, 213. Published January 27,

2 2 J. CABALLERO, M. A. DARWISH, K. SADARANGANI EJDE-214/31 In the sequel, for x H α [a, b], by Hx α we will denote the least possible constant for which inequality (2.1) is satisfied. More precisely, we put H α x = sup { x(t) x(s) : t [a, b], t s }. (2.2) The spaces H α [a, b] with < α 1 can be equipped with the norm x α = x(a) + sup { x(t) x(s) : t [a, b], t s }, for x H α [a, b]. In [2], the authors proved that (H α [a, b], α ) with < α 1 is a Banach space. The following lemmas appear in [2]. Lemma 2.1. For x H α [a, b] with < α 1, the following inequality is satisfied Lemma 2.2. For < α < γ 1, we have x max ( 1, (b a) α) x α. (2.3) H γ [a, b] H α [a, b] C[a, b]. (2.4) Moreover, for x H γ [a, b] the following inequality holds x α max ( 1, (b a) γ α) x γ. (2.) Now, we present the following sufficient condition for relative compactness in the spaces H α [a, b] with < α 1 which appears in Example 6 of [2] and it is an important result for our study. Theorem 2.3. Suppose that < α < β 1 and that A is a bounded subset in H β [a, b] (this means that x β M for certain constant M >, for any x A) then A is a relatively compact subset of H α [a, b]. 3. Main results In this section, we will study the solvability of (1.1) in the Hölder spaces. We will use the following assumptions: (i) p H β [, 1], < β 1. (ii) k : [, 1] [, 1] R is a continuous function such that it satisfies the Hölder condition with exponent β with respect to the first variable, that is, there exists a constant K β such that k(t, τ) k(s, τ) K β, for any t, s, τ [, 1]. (iii) r : [, 1] [, 1] is a measurable function. (iv) The following inequality is satisfied p β (2K + K β ) < 1 4, where the constant K is defined by which exists by (ii). K = sup { k(t, τ) dτ : t [, 1] },

3 EJDE-214/31 SOLVABILITY OF A QUADRATIC INTEGRAL EQUATION 3 Theorem 3.1. Under assumptions (i) (iv), Equation (1.1) has at least one solution belonging to the space H α [, 1], where α is arbitrarily fixed number satisfying < α < β. Proof. Consider the operator T defined on H β [, 1] by (T x)(t) = p(t) + x(t) k(t, τ) x(r(τ)) dτ, t [, 1]. In the sequel, we will prove that T transforms the space H β [, 1] into itself. In fact, we take x H β [, 1] and t, s [, 1] with t s. Then, by assumptions (i) and (ii), we obtain (T x)(t) (T x)(s) p(t) + x(t) = k(t, τ) x(r(τ)) dτ p(s) x(s) k(s, τ) x(r(τ)) dτ p(t) p(s) x(t) + k(t, τ) x(r(τ)) dτ x(s) k(t, τ) x(r(τ)) dτ 1 x(s) + k(t, τ) x(r(τ)) dτ x(s) k(s, τ) x(r(τ)) dτ p(t) p(s) x(t) x(s) + + x(s) k(t, τ) k(s, τ) x(r(τ)) dτ p(t) p(s) x(t) x(s) + x k(t, τ) x(r(τ)) dτ + x x k(t, τ) k(s, τ) dτ k(t, τ) dτ p(t) p(s) + K x x(t) x(s) + x 2 H β p + K x H β x + K β x 2. K β dτ By Lemma 2.1, since x x β and, as H β x x β, we infer that (T x)(t) (T x)(s) H β p + (K + K β ) x 2 β.

4 4 J. CABALLERO, M. A. DARWISH, K. SADARANGANI EJDE-214/31 Therefore, T x β = (T x)() + sup { (T x)(t) (T x)(s) : t, s [, 1], t s } (T x)() + H β p + (K + K β ) x 2 β p() + x() k(, τ) x(r(τ)) dτ + H β p + (K + K β ) x 2 β p β + x x k(, τ) dτ + (K + K β ) x 2 β (3.1) p β + K x 2 β + (K + K β ) x 2 β = p β + (2K + K β ) x 2 β <. This proves that the operator T maps H β [, 1] into itself. Taking into account that the inequality p β + (2K + K β )r 2 < r is satisfied for values between the numbers and r 1 = p β (2K + K β ) 2(2K + K β ) r 2 = p β (2K + K β ) 2(2K + K β ) which are positive by assumption (iv), consequently, from (3.1) it follows that T transforms the ball Br β = {x H β [, 1] : x β r } into itself, for any r [r 1, r 2 ]; i.e., T : Br β Br β, where r 1 r r 2. By Theorem 2.3, we have that the set Br β is relatively compact in H α [, 1] for any < α < β 1. Moreover, we can prove that Br β is a compact subset in H α [, 1] for any < α < β 1 (see Appendix). Next, we will prove that the operator T is continuous on Br β, where in Br β we consider the induced norm by α, where < α < β 1. To do this, we fix x Br β and ε >. Suppose that y Br β and x y α δ, where δ is a positive ε number such that δ < 2(2K+3K β )r. Then, for any t, s [, 1] with t s, we have [(T x)(t) (T y)(t)] [(T x)(s) (T y)(s)] [ x(t) = k(t, τ) x(r(τ)) dτ y(t) k(t, τ) y(r(τ)) dτ] [ x(s) k(s, τ) x(r(τ)) dτ y(s) k(s, τ) y(r(τ)) dτ] [ x(t) k(t, τ) x(r(τ)) dτ y(t) k(t, τ) x(r(τ)) dτ] [ y(t) + k(t, τ) x(r(τ)) dτ y(t) k(t, τ) y(r(τ)) dτ]

5 EJDE-214/31 SOLVABILITY OF A QUADRATIC INTEGRAL EQUATION [ x(s) k(s, τ) x(r(τ)) dτ y(s) k(s, τ) x(r(τ)) dτ] [ y(s) k(s, τ) x(r(τ)) dτ y(s) k(s, τ) y(r(τ)) dτ] 1 (x(t) = y(t)) (x(s) y(s)) k(t, τ) x(r(τ)) dτ + y(t) k(s, τ) x(r(τ)) dτ y(s) 1 { (x(t) y(t)) (x(s) y(s)) + x(s) y(s) + y(t) (k(t, τ) k(s, τ)) x(r(τ)) dτ k(t, τ)(x(r(τ) y(r(τ))) dτ y(s) (x(t) y(t)) (x(s) y(s)) x k(t, τ)(x(r(τ) y(r(τ))) dτ k(s, τ) (x(r(τ)) y(r(τ))) dτ k(t, τ) x(r(τ)) dτ k(t, τ) dτ + [ (x(s) y(s)) (x() y()) + x() y() ] x k(t, τ) k(s, τ) dτ + 1 y(t) k(t, τ)(x(r(τ) y(r(τ))) dτ y(s) 1 y(s) + y(s) K x y α x + k(t, τ)(x(r(τ) y(r(τ))) dτ k(s, τ)(x(r(τ) y(r(τ))) dτ sup p,q [,1] (x(p) y(p)) (x(q) y(q)) K β x dτ + x() y() x y(s) x(s) + + y(s) k(t, τ) x(r(τ) y(r(τ)) dτ } k(s, τ)(x(r(τ) y(r(τ))) dτ k(t, τ)(x(r(τ) y(r(τ))) dτ k(t, τ) k(s, τ) x(r(τ) y(r(τ)) dτ K x x y α + x K β α { (x(p) y(p)) (x(q) y(q)) sup p,q [,1], p q p q α p q α} + K β x β α x() y() + KH α y x y + y x y K β dτ K β dτ K x β x y α + 2K β x β x y α + K y α x y α + K β y α x y α

6 6 J. CABALLERO, M. A. DARWISH, K. SADARANGANI EJDE-214/31 ( K x β + 2K β x β + K y α + K β y α ) x y α. Since y α y β (see, Lemma 2.2) and x, y B β r, from the above inequality we infer that [(T x)(t) (T y)(t)] [(T x)(s) (T y)(s)] (2Kr + 3K β r ) x y α (2Kr + 3K β r )δ < ε 2. (3.2) On the other hand, (T x)() (T y)() = x() x() + x() x() + (x() y()) k(, τ) x(r(τ)) dτ y() k(, τ) x(r(τ)) dτ x() k(, τ) y(r(τ)) dτ y() k(, τ)y(r(τ)) dτ k(, τ)(x(r(τ)) y(r(τ))) dτ k(, τ) y(r(τ))) dτ K x x y + K y x y α K x β x y α + K y β x y α k(, τ) y(r(τ)) dτ k(, τ) y(r(τ)) dτ From (3.2) and (3.3), it follows that T x T y = (T x)() (T y)() 2Kr x y α < 2Kr δ < ε 2. (3.3) + sup { ((T x)(t) (T y)(t)) ((T x)(s) (T y)(s)) : t, s [, 1], t s } < ε 2 + ε 2 = ε. This proves that the operator T is continuous at the point x Br δ for the norm α. Since Br δ is compact in H α [, 1], applying the classical Schauder fixed point theorem we obtain the desired result. 4. Example To present an example illustrating our result we need some previous results. Definition 4.1. A function f : R + R + is said to be subadditive if f(x + y) f(x) + f(y) for any x, y R +. Lemma 4.2. Suppose that f : R + R + is subadditive and y x then f(x) f(y) f(x y). Proof. Since f(x) = f(x y + y) f(x y) + f(y) the result follows.

7 EJDE-214/31 SOLVABILITY OF A QUADRATIC INTEGRAL EQUATION 7 Remark 4.3. From Lemma 4.2, we infer that if f : R + R + is subadditive then f(x) f(y) f( x y ) for any x, y R +. Lemma 4.4. Let f : R + R + be a concave function with f() =. Then f is subadditive. Proof. For x, y R + and, since f is concave and f() =, we have ( x f(x) = f (x + y) + y ) x + y x + y x f(x + y) + y x + y x + y f() = x f(x + y) x + y and ( x f(y) = f x + y + x f()f(x + y) + y x + y = y f(x + y). x + y y ) (x + y) x + y f(x + y) x + y Adding these inequalities, we obtain f(x) + f(y) x f(x + y) + y f(x + y) = f(x + y). x + y x + y This completes the proof. Remark 4.. Let f : R + R + be the function defined by f(x) = p x, where p > 1. Since this function is concave (because f (x) for x > ) and f() =, Lemma 4.4 says us that f is subadditive. By Remark 4.3, we have for any x, y R +. f(x) f(y) = p x p y p x y Example 4.6. Let us consider the quadratic integral equation x(t) = arctan 4 q sin t + ˆq + x(t) mt2 + τ x ( τ ) dτ, τ + 1 t [, 1], (4.1) where, q, ˆq and m are nonnegative constants. Notice that (4.1) is a particular case of (1.1), where p(t) = arctan q sin t + ˆq, k(t, τ) = 4 mt 2 + τ and r(τ) = τ τ+1. In what follows, we will prove that assumptions (i)-(iv) of Theorem 3.1 are satisfied. Since the inverse tangent function is concave (because its second derivative is nonpositive) and its value at zero is zero, taking into account Remarks 4.3 and 4., we have p(t) p(s) = arctan q sin t + ˆq arctan q sin s + ˆq ( arctan q sin t + ˆq q sin s + ˆq ) q sin t + ˆq q sin s + ˆq q sin t sin s q t s 1/,

8 8 J. CABALLERO, M. A. DARWISH, K. SADARANGANI EJDE-214/31 where we have used that arctan x x for x and sin x sin y x y for any x, y R. This says that p H 1 [, 1] and, moreover, H1/ p = q. Therefore, assumptions (i) of Theorem 3.1 is satisfied. Note that p 1 = p() + sup { p(t) p(s) : t, s [, 1], t s } t s 1/ arctan ˆq + H 1/ p = arctan ˆq + q. Since for any t, s, τ [, 1], we have (see, Remark 4.) k(t, τ) k(s, τ) = 4 mt 2 + τ 4 ms 2 + τ 4 mt 2 ms 2 = 4 m 4 t 2 s 2 = 4 m 4 t + s 4 t s 4 m 4 2 t s 1/4 = 4 m 4 2 t s 1/2 t s 1/ 4 2m t s 1/, assumption (ii) of Theorem 3.1 is satisfied with K β = K 1 = 4 2m. It is clear that r(τ) = τ τ+1 satisfies assumption (iii). In our case, the constant K is given by K = sup { k(t, τ) dτ : t [, 1] } = sup { 4 mt2 + τ dτ : t [, 1] } = sup { 4 ( 4 (mt2 + 1) 4 ) } m t 1 = 4 ( 4 (m + 1) 4 ) m. Therefore, the inequality appearing in assumption (iv) takes the form ( p 1 (2K + K β) = arctan ˆq + )( 8 [ q 4 (m + 1) 4 m ] + 4 2m) < 1 4. It is easily seen that the above inequality is satisfied when, for example, ˆq =, q = 1 2 and m = 1. Therefore, using Theorem 3.1, we infer that (4.1) for ˆq =, 2 q = 1 2 and m = 1 has at least one solution in the space H 2 α [, 1] with < α < 1/. Note that in (4.1), we can take as r(τ) a particular functions such as r(τ) = {e τ }, where { } denotes the fractional part. Remark 4.7. Note that any solution x(t) of (1.1), i.e., x(t) = p(t) + x(t) k(t, τ) x(r(τ)) dτ, t [, 1],

9 EJDE-214/31 SOLVABILITY OF A QUADRATIC INTEGRAL EQUATION 9 satisfies that its zeroes are also zeroes of p(t). From this, we infer that if p(t) for any t [, 1] then x(t) for any t [, 1]. By Bolzano s theorem, this means that the solution x(t) of Eq(1.1) does not change of sign on [, 1] when p(t) for any t [, 1]. These questions seem to be interesting from a practical standpoint.. Appendix Suppose that < α < β 1 and by Br β we denote the ball centered at θ and radius r in the space H β [a, b]; i.e., Br β = {x H β [a, b] : x β r}. Then Br β is a compact subset in the space H α [a, b]. In fact, by Theorem 2.3, since Br β is a bounded subset in H β [a, b], Br β is a relatively compact subset of H α [a, b]. In the sequel, we will prove that Br β is a closed subset of H α [a, b]. Suppose that (x n ) Br β We have to prove that x Br β. α and x n x with x Hα [a, b]. α Since x n x, for ε > given we can find n N such that x x α ε for any n n, or, equivalently, x n (a) x(a) + sup { (x n (t) x(t)) (x n (s) x(s)) : t, s [a, b], t s } < ε, (.1) for any n n. Particularly, this implies that x n (a) x(a). Moreover, if in (.1) we put s = a then we get sup { (x n (t) x(t)) (x n (a) x(a)) t a α : t, s [a, b], t a } < ε, for any n n. This says that (x n (t) x(t)) (x n (a) x(a)) < ε t a α, for any n n and for any t [a, b]. (.2) Therefore, for any n n and any t [a, b] by (.1) and (.2), we have x n (t) x(t) (x n (t) x(t)) (x n (a) x(a)) + x n (a) x(a) From this, it follows that < ε(t a) α + ε = ε(1 + (b a) α ). x n x. (.3) Next, we will prove that x B β r. In fact, as (x n ) B β r H β [a, b], we have that x n (t) x n (s) r for any t, s [a, b] with t s. Consequently, x n (t) x n (s) r for any t, s [a, b]. Letting n and taking into account (.3), we obtain for any t, s [a, b]. Therefore, x(t) x(s) r x(t) x(s) r

10 1 J. CABALLERO, M. A. DARWISH, K. SADARANGANI EJDE-214/31 for any t, s [a, b] with t s, and this means that x B β r. This completes the proof. References [1] C. Bacoţiu; Volterra-Fredholm nonlinear systems with modified argument via weakly Picard operators theory, Carpath. J. Math. 24(2) (28), [2] J. Banaś, R. Nalepa; On the space of functions with growths tempered by a modulus of continuity and its applications, J. Func. Spac. Appl., (213), Article ID 82437, 13 pages. [3] M. Benchohra, M.A. Darwish; On unique solvability of quadratic integral equations with linear modification of the argument, Miskolc Math. Notes 1(1) (29), 3 1. [4] J. Caballero, B. López, K. Sadarangani; Existence of nondecreasing and continuous solutions of an integral equation with linear modification of the argument, Acta Math. Sin. (English Series) 23 (27), [] M. Dobriţoiu; Analysis of a nonlinear integral equation with modified argument from physics, Int. J. Math. Models and Meth. Appl. Sci. 3(2) (28), [6] T. Kato, J.B. Mcleod; The functional-differential equation y (x) = ay(λx)+by(x), Bull. Amer. Math. Soc., 77 (1971), [7] M. Lauran; Existence results for some differential equations with deviating argument, Filomat 2(2) (211), [8] V. Mureşan; A functional-integral equation with linear modification of the argument, via weakly Picard operators, Fixed Point Theory, 9(1) (28), [9] V. Mureşan; A Fredholm-Volterra integro-differential equation with linear modification of the argument, J. Appl. Math., 3(2) (21), Josefa Caballero Departamento de Matemáticas, Universidad de Las Palmas de Gran Canaria, Campus de Tafira Baja, 317 Las Palmas de Gran Canaria, Spain address: fefi@dma.ulpgc.es Mohamed Abdalla Darwish Department of Mathematics, Sciences Faculty for Girls, King Abdulaziz University, Jeddah, Saudi Arabia. Department of Mathematics, Faculty of Science, Damanhour University, Damanhour, Egypt address: dr.madarwish@gmail.com Kishin Sadarangani Departamento de Matemáticas, Universidad de Las Palmas de Gran Canaria, Campus de Tafira Baja, 317 Las Palmas de Gran Canaria, Spain address: ksadaran@dma.ulpgc.es