PATA TYPE FIXED POINT THEOREMS OF MULTIVALUED OPERATORS IN ORDERED METRIC SPACES WITH APPLICATIONS TO HYPERBOLIC DIFFERENTIAL INCLUSIONS

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U.P.B. Sci. Bull., Series A, Vol. 78, Iss. 4, 216 ISSN 1223-727 PATA TYPE FIXED POINT THEOREMS OF MULTIVALUED OPERATORS IN ORDERED METRIC SPACES WITH APPLICATIONS TO HYPERBOLIC DIFFERENTIAL INCLUSIONS Samad Mohseni Kolagar 1, Maryam Ramezani 2, Madjid Eshaghi 3 The purpose of this paper is to present some Pata type fixed point theorems for multivalued mappings on ordered complete metric spaces. Moreover, as an application of our main theorem, we give an existence theorem for the solution of a hyperbolic differential inclusion problem. Keywords: Fixed point, Ordered metric space, Multivalued mapping, Hyperbolic differential inclusion. 1. Introduction and preliminaries In 1922, the Polish mathematician Stefan Banach proved a theorem which ensures, under appropriate conditions, the existence and uniqueness of a fixed point. His result is called Banach s fixed point theorem or the Banach contraction principle[1]. Nadler [2] in 1969, proved a set-valued extension of the Banach s contraction principle in complete metric spaces. Afterward many fixed point theorems have been proved by various authors as generalization of the Nadler s theorem where the nature of contractive mapping is weakened along with some additional requirements, see for instance [3, 4, 5, 6, 7, 8, 9, 1]. Ran and Reurings [11] established the existence of unique fixed point for the monotone single valued mapping in partially ordered metric spaces. Their result was further extended in [12, 13, 4, 14, 15, 16, 17]. Recently, V. Pata [18] improve the Banach principal. In fact, Pata extended the Banach contraction principle with weaker hypotheses than those of the Banach contraction principle with an explicit estimate of the convergence rate. In this paper, using the idea of Pata, we prove some fixed point theorems on ordered complete metric spaces. As an application, we also obtain conditions which guarantee the existence of a solution for hyperbolic differential inclusions problem. Let (M, d) be a metric space. Then 2 M is the class of all nonempty subsets of M and for A, B 2 M, let H d (A, B) = max{sup b B d(b, A), sup d(a, B)}, a A 1 Department of Mathematics, University of Mazandaran, Babolsar, Iran, e-mail: mohseni.samad@gmail.com 2 Department of Mathematics, University of Bojnord, Bojnord, Iran, e-mail: m.ramezani@ub.ac.ir 3 Department of Mathematics, Semnan University, P.O. Box 35195-363, Semnan, Iran, e-mail: madjid.eshaghi@gmail.com 21

22 Samad Mohseni Kolagar, Maryam Ramezani, Madjid Eshaghi where d(a, B) = inf d(a, b), b B then H d is called the Hausdorff-Pompeiu functional induced by d. Definition 1.1. Let (X, ) be a partially ordered set and x, y X. Elements x and y are said to be comparable elements of X if either x y or y x. Definition 1.2. [19] Let M be a metric space. approximative if the multivalued mapping A subset N M is said to be has nonempty values. Ϝ N (x) = {y N : d(x, y) = d(n, x)}, x M, Definition 1.3. [19] The multivalued mapping T : M 2 M is said to have approximative values, AV for short, if T x is approximative for each x M. Definition 1.4. [19] The multivalued mapping T : M 2 M is said to have comparable approximative values, CAV for short, if T has approximative values and, for each z M, there exists y Ϝ T z (x) such that y is comparable to z. Definition 1.5. [19] The multivalued mapping T : M 2 M is said to have upper comparable approximative values, UCAV, for short(resp.lower comparable approximative values, LCAV for short) if T has approximative values and, for each z M, there exists y Ϝ T z (x) such that y z(resp.y z). Definition 1.6. [19] For two subset X, Y of M, we denote X r Y if for each x X there exists y Y such that x y and X Y if each x X and each y Y imply that x y. Definition 1.7. [19] A multivalued mapping T : M 2 M is said to be r-nondecreasing(rnonincreasing) if x y implies that T x r T y(t y r T x) for all x, y M. T is said to be r-monotone if T is r-nondecreasing or r-nonincreasing. The notation of nondecreasing(nonincreasing) is similarly defind by writing instead of the notation r. 2. Main Results Let (M, d, ) be a partial ordered complete metric space. The following hypothesis in M(which appear in [2]) will be applied: (H 1 ) If {x n } is a non-decreasing(resp. non-increasing) sequence in M such that x n x, then x n x(resp.x n x) for all n N. For a metric space (M, d), Selecting an arbitrary x M we denote x = d(x, x ) for all x M.

Pata type fixed point theorems of multivalued operators in ordered metric spaces 23 Let ψ : [, 1] [, ) is an increasing function vanishing with continuity at zero. Also consider the vanishing sequence depending on α 1, w n (α) = ( α n n )α ψ( α k ). Theorem 2.1. Let M be an ordered complete metric space and satisfy (H 1 ). Let Λ, α 1 and β [, α] be a fixed constants. Suppose that the multivalued map T : M 2 M has UCAV and the inequality H d (T x, T y) (1 ϵ)d(x, y) + Λϵ α ψ(ϵ)[1 + x + y ] β, (1) is satisfied for every ϵ [, 1] and for all x, y M with x and y comparable. Then T has a fixed point x T x. Furthermore, for some positive constant K Λ(1 + 4 x ) β. d(x, T x n 1 ) Kw n (α), (2) Proof. Given x M, if x T x, proof is complete. Moreover from the fact that T x has UCAV it follows there exsists x 1 T x with x 1 x and x 1 x such that d(x, x 1 ) = inf d(x, x ) = d(t x, x ). x T x Continuing in this way obtain that, there exists x n+1 T x n with x n+1 x n and x n+1 x n such that Moreover, therefore, d(x n, x n+1 ) = d(t x n, x n ), n = 1, 2,.... d(t x n, x n ) sup d(t x n, x) H d (T x n, T x n 1 ), x T x n 1 d(x n, x n+1 ) H d (T x n 1, T x n ) for n = 2, 3,.... Furthermore for n = 1, 2,... we set C n = x n = d(x n, x ). Since (1) is true for every ε [, 1], setting ε =, we have the following relations d(x n+1, x n ) H d (T x n, T x n 1 ) d(x n, x n 1 ) H d (T x n 1, T x n ) By triangle inequality we have. d(x n 1, x n ) H d (T x 1, T x ) d(x 1, x ) = C 1. (3) d(x n+1, x ) d(x n+1, x 1 ) + d(x 1, x ), (4) and d(x n, x ) d(x n, x n+1 ) + d(x n+1, x ). (5)

24 Samad Mohseni Kolagar, Maryam Ramezani, Madjid Eshaghi Now, using (3), (4) and (5) we have C n d(x n+1, x 1 ) + 2C 1 H d (T x n, T x ) + 2C 1. (6) So for α β, there exist real numbers E, D > such that C n (1 ϵ)c n + Λϵ α ψ(ϵ)[1 + C n ] β + 2C 1 (1 ϵ)c n + Eϵ α ψ(ϵ)c α n + D. (7) Accordingly, εc n Eε α ψ(ε)c α n + D, which holds by hypothesis for any ε [, 1] taken for each n N. If there is a subsequence C nk, then the choice ε nk = min(1, 1+D C nk ), leads to the following contradiction 1 E(1 + D) α ψ(ε nk ) as n k. Then the sequence {C n } n=1 is bounded. Now, we prove that sequence {x n } is Cauchy sequence: d(x n+m+1, x n+1 ) H d (T x n+m, T x n ) (1 ϵ)d(x n+m, x n )+Λϵ α ψ(ϵ)[ x n+m + x n ] β. For fixed m, set K = sup Λ[1 + 2c n ] β, (8) n N and ε = 1 ( n n+1 )α α n+1. So (n + 1) α d(x n+m+1, x n+1 ) n α d(x n+m, x n ) + Kα α α ψ( n + 1 ). Setting r n := n α d(x n+m, x n ), we have r n+1 r n + Kα α α ψ( n + 1 ) r n 1 + Kα α ψ( α n ) + α Kαα ψ( n + 1 )... Therefore n+1 r + Kα α ψ( α n+1 k ) = Kαα ψ( α k ). d(x n+m, x n ) K( α n )α n ψ( α k ) = Kw n(α). (9) Taking limits as n, we get d(x n+m, x n ). This implies that {x n } is Cauchy sequence in X. Since X is a complete metric space, there exists x M such that

Pata type fixed point theorems of multivalued operators in ordered metric spaces 25 lim n x n = x. Using (9) we have Again d(x, T x n 1 ) d(x, x n ) = lim m d(x n+m, x n ) Kw n (α). d(x, T x ) = lim n d(x n 1, T x ) lim n H d(t x n, T x ) (1 ϵ)d(x n, x ) + Λϵ α ψ(ϵ)[1 + x n + x ] β. Since the contractive condition (1) holds for any real constant ε [, 1], we can replace ε, for each n N, by a sequence [, 1] ε n as n. Then by letting ε = ε n as n, we have d(x, T x ) (1 ϵ n )d(x n, x ) + Λϵ α nψ(ϵ n )[1 + x n + x ] β. So d(x, T x ) = and x T x. Also, the convergence rate estimate stated in (2) is achieved from the following relations d(x, x n ) H d (T x, T x n 1 ) d(x, x n 1 ) H d (T x, T x n 2 ) d(x, x n 2 ). H d (T x, T x ) d(x, x ) = x, which implies that C n = d(x n, x ) d(x n, x ) + d(x, x ) x + x = 2 x. From the last inequality and (8) we have K Λ(1 + 4 x ) β. Theorem 2.2. Let M be an ordered complete metric space and satisfy (H 1 ). Let Λ, α 1 and β [, α] be a fixed constants. Suppose that the multivalued map T : M 2 M has LCAV and the inequality H d (T x, T y) (1 ϵ)d(x, y) + Λϵ α ψ(ϵ)[1 + x + y ] β, is satisfied for every ϵ [, 1] and for all x, y M with x and y comparable. Then T has a fixed point x T x. Furthermore, d(x, T x n 1 ) Kw n (α), for some positive constant K Λ(1 + 4 x ) β. Proof. Given x M, if x T x, proof is complete. Moreover from the fact that T x has LCAV it follows there exsists x 1 T x with x 1 x and x 1 x such that d(x, x 1 ) = inf d(x, x ) = d(t x, x ). x T x Continuing in this way obtain that, there exists x n+1 T x n with x n+1 x n and x n+1 x n such that d(x n, x n+1 ) = d(t x n, x n ), n = 1, 2,....

26 Samad Mohseni Kolagar, Maryam Ramezani, Madjid Eshaghi Moreover, therefore, d(t x n, x n ) sup d(t x n, x) H d (T x n, T x n 1 ), x T x n 1 d(x n, x n+1 ) H d (T x n 1, T x n ) for n = 2, 3,.... Furthermore for n = 1, 2,... we set C n = x n = d(x n, x ). Since (1) is true for every ε [, 1], setting ε =, we have the following relations d(x n+1, x n ) H d (T x n, T x n 1 ) d(x n, x n 1 ) H d (T x n 1, T x n ) By triangle inequality we have. d(x n 1, x n ) H d (T x 1, T x ) d(x 1, x ) = C 1. (1) d(x n+1, x ) d(x n+1, x 1 ) + d(x 1, x ), (11) and d(x n, x ) d(x n, x n+1 ) + d(x n+1, x ). (12) Now, using (1), (11) and (12) we have C n d(x n+1, x 1 ) + 2C 1 H d (T x n, T x ) + 2C 1. (13) So for α β, there exist real numbers E, D > such that C n (1 ϵ)c n + Λϵ α ψ(ϵ)[1 + C n ] β + 2C 1 (1 ϵ)c n + Eϵ α ψ(ϵ)c α n + D. (14) Accordingly, εc n Eε α ψ(ε)c α n + D, which holds by hypothesis for any ε [, 1] taken for each n N. If there is a subsequence C nk, then the choice ε nk = min(1, 1+D C nk ), leads to the following contradiction 1 E(1 + D) α ψ(ε nk ) as n k. Then the sequence {C n } n=1 is bounded. Now, we prove that sequence {x n } is Cauchy sequence: d(x n+m+1, x n+1 ) H d (T x n+m, T x n ) (1 ϵ)d(x n+m, x n )+Λϵ α ψ(ϵ)[ x n+m + x n ] β.

Pata type fixed point theorems of multivalued operators in ordered metric spaces 27 For fixed m, set K = sup Λ[1 + 2c n ] β, (15) n N and ε = 1 ( n n+1 )α α n+1. So (n + 1) α d(x n+m+1, x n+1 ) n α d(x n+m, x n ) + Kα α α ψ( n + 1 ). Setting r n := n α d(x n+m, x n ), we have r n+1 r n + Kα α α ψ( n + 1 ) r n 1 + Kα α ψ( α n ) + α Kαα ψ( n + 1 )... Therefore n+1 r + Kα α ψ( α n+1 k ) = Kαα ψ( α k ). d(x n+m, x n ) K( α n )α n ψ( α k ) = Kw n(α). (16) Taking limits as n, we get d(x n+m, x n ). This implies that {x n } is Cauchy sequence in X. Since X is a complete metric space, there exists x M such that lim n x n = x. Using (16) we have Again d(x, T x n 1 ) d(x, x n ) = lim m d(x n+m, x n ) Kw n (α). d(x, T x ) = lim n d(x n 1, T x ) lim n H d(t x n, T x ) (1 ϵ)d(x n, x ) + Λϵ α ψ(ϵ)[1 + x n + x ] β. Since the contractive condition (1) holds for any real constant ε [, 1], we can replace ε, for each n N, by a sequence [, 1] ε n as n. Then by letting ε = ε n as n, we have d(x, T x ) (1 ϵ n )d(x n, x ) + Λϵ α nψ(ϵ n )[1 + x n + x ] β. So d(x, T x ) = and x T x. Also, the convergence rate estimate stated in (2) is achieved from the following relations d(x, x n ) H d (T x, T x n 1 ) d(x, x n 1 ) H d (T x, T x n 2 ). d(x, x n 2 ) H d (T x, T x ) d(x, x ) = x,

28 Samad Mohseni Kolagar, Maryam Ramezani, Madjid Eshaghi which implies that C n = d(x n, x ) d(x n, x ) + d(x, x ) x + x = 2 x. From the last inequality and (15) we have K Λ(1 + 4 x ) β. Theorem 2.3. Let M be an ordered complete metric space and satisfy (H 1 ). Let Λ, α 1 and β [, α] be a fixed constants. Suppose that the multivalued map T : M 2 M has AV, is non-decreasing and the inequality H d (T x, T y) (1 ϵ)d(x, y) + Λϵ α ψ(ϵ)[1 + x + y ] β, is satisfied for every ϵ [, 1] and for all x, y M with x and y comparable. If there exists x M such that {x } T x, then T has a fixed point x T x. Furthermore, d(x, T x n 1 ) Kw n (α), for some positive constant K Λ(1 + 4 x ) β. Proof. The first by using hypothesis of this theorem, we construct sequence {x n }. If x T x, proof is complete. Moreover, according hypothesis, for any x T x, we have x x. Since T has AV, there exists x 1 T x with x 1 x and d(x, x 1 ) = d(t x, x ). Continuing in this way obtain that, there exists x n+1 T x n with x n+1 x n and x n+1 x n such that Moreover, therefore, d(x n, x n+1 ) = d(t x n, x n ), n = 1, 2,.... d(t x n, x n ) sup d(t x n, x) H d (T x n, T x n 1 ), x T x n 1 d(x n, x n+1 ) H d (T x n 1, T x n ) for n = 2, 3,.... Furthermore for n = 1, 2,... we set C n = x n = d(x n, x ). Since (1) is true for every ε [, 1], setting ε =, we have the following relations d(x n+1, x n ) H d (T x n, T x n 1 ) d(x n, x n 1 ) H d (T x n 1, T x n ) By triangle inequality we have. d(x n 1, x n ) H d (T x 1, T x ) d(x 1, x ) = C 1. (17) d(x n+1, x ) d(x n+1, x 1 ) + d(x 1, x ), (18) and d(x n, x ) d(x n, x n+1 ) + d(x n+1, x ). (19)

Pata type fixed point theorems of multivalued operators in ordered metric spaces 29 Now, using (17), (18) and (19) we have C n d(x n+1, x 1 ) + 2C 1 H d (T x n, T x ) + 2C 1. (2) So for α β, there exist real numbers E, D > such that C n (1 ϵ)c n + Λϵ α ψ(ϵ)[1 + C n ] β + 2C 1 (1 ϵ)c n + Eϵ α ψ(ϵ)c α n + D. (21) Accordingly, εc n Eε α ψ(ε)c α n + D, which holds by hypothesis for any ε [, 1] taken for each n N. If there is a subsequence C nk, then the choice ε nk = min(1, 1+D C nk ), leads to the following contradiction 1 E(1 + D) α ψ(ε nk ) as n k. Then the sequence {C n } n=1 is bounded. Now, we prove that sequence {x n } is Cauchy sequence: d(x n+m+1, x n+1 ) H d (T x n+m, T x n ) (1 ϵ)d(x n+m, x n )+Λϵ α ψ(ϵ)[ x n+m + x n ] β. For fixed m, set K = sup Λ[1 + 2c n ] β, (22) n N and ε = 1 ( n n+1 )α α n+1. So (n + 1) α d(x n+m+1, x n+1 ) n α d(x n+m, x n ) + Kα α α ψ( n + 1 ). Setting r n := n α d(x n+m, x n ), we have r n+1 r n + Kα α α ψ( n + 1 ) r n 1 + Kα α ψ( α n ) + α Kαα ψ( n + 1 )... Therefore n+1 r + Kα α ψ( α n+1 k ) = Kαα ψ( α k ). d(x n+m, x n ) K( α n )α n ψ( α k ) = Kw n(α). (23) Taking limits as n, we get d(x n+m, x n ). This implies that {x n } is Cauchy sequence in X. Since X is a complete metric space, there exists x M such that

3 Samad Mohseni Kolagar, Maryam Ramezani, Madjid Eshaghi lim n x n = x. Using (23) we have Again d(x, T x n 1 ) d(x, x n ) = lim m d(x n+m, x n ) Kw n (α). d(x, T x ) = lim n d(x n 1, T x ) lim n H d(t x n, T x ) (1 ϵ)d(x n, x ) + Λϵ α ψ(ϵ)[1 + x n + x ] β. Since the contractive condition (1) holds for any real constant ε [, 1], we can replace ε, for each n N, by a sequence [, 1] ε n as n. Then by letting ε = ε n as n, we have d(x, T x ) (1 ϵ n )d(x n, x ) + Λϵ α nψ(ϵ n )[1 + x n + x ] β. So d(x, T x ) = and x T x. Also, the convergence rate estimate stated in (2) is achieved from the following relations d(x, x n ) H d (T x, T x n 1 ) d(x, x n 1 ) H d (T x, T x n 2 ) d(x, x n 2 ). H d (T x, T x ) d(x, x ) = x, which implies that C n = d(x n, x ) d(x n, x ) + d(x, x ) x + x = 2 x. From the last inequality and (22) we have K Λ(1 + 4 x ) β. 3. Applications In this section, as an application of the result of previous section, is concerned with the existence of solutions for hyperbolic differential inclusions 2 u(t, x) F (t, x, u(t, x)) a.e.(t, x) J a J b, t x (24) u(t, ) = ξ(t), u(, x) = η(x), t J a, x J b, where J a = [, a], J b = [, b], F : J a J b R 2 R is multivalued mapping satisfying some hypotheses which will be specified later, ξ C(J a, R), η C(J b, R). In the present section, we will prove the existence of solutions for problem (24) based on Theorems 2.1. First of all, we introduce notations which are used throughout this section. C(J a J b, R) is the Banach space consisting of all continuous functions from J a J b into R with the norm u = sup{ u(t, x) : (t, x) J a J b } for u C(J a J b, R).

Pata type fixed point theorems of multivalued operators in ordered metric spaces 31 For u, v C(J a J b, R) we define that u v if and only if u(t, x) v(t, x) for each (t, x) J a J b. Let P = {u : u C(J a J b, R), u } and 2 u H = {u C(J a J b, R) : t x exists for each (t, x) J a J b }. L 1 (J a J b, R) stands for the Banach space consisting of measurable functions u : J a J b R which are Lebesgue integrable normed by u L = a b u(t, x) dxdt for u L 1 (J a J b, R). Let M : J a J b R 2 R be a multivalued map with nonempty values. For each u C(J a J b, R) define the set of selections of M by S M,u = {v L 1 (J a J b, R) : v(t, x) M(t, x, u(t, x)) a.e. (t, x) J a J b }, and assign to M the multivalued operator M : C(J a J b, R) 2 L1 (J a J b,r) by letting M(u) = {w L 1 (J a J b, R) : w(t, x) M(t, x, u(t, x)), (t, x) J a J b }. The operator M is called the Niemytsky operator associated with M in the light of some of the current literature. In order to state and verify our results, we need the continuous map L : L 1 (J a J b, R) C(J a J b, R) defined by As an application of results included in Theorem 2.1, we first unify the following Lu(t, x) = t x u(s, τ)dsdτ. Theorem 3.1. Suppose that the multivalued function F : J a J b R 2 R satisfies the following conditions: (L1) F (t, x, u) is compact subset for all (t, x, u) J a J b C(J a J b, R). Moreover, S F,u is nonempty for each u C(J a J b, R). (L2) For any u, v C(J a J b, R), if u, v are comparable then H d (F (t, x, u(t, x)), F (t, x, v(t, x))) l(t, x) 2 u(t, x) v(t, x), for almost each (t, x) J a J b, where l L 1 (J a J b, R) with l L 1. (L3) For each u C(J a J b, R) one has {u(t, x) ξ(t) η(x) + ξ()} Lv(t, x) for (t, x) J a J b and v S F,u. Then the problem (24) has at least a solution u C(J a J b, R). Proof. Let M C(J a J b, R) be a complete metric space. Then M satisfies the condition (H1). It is clear that problem (24) is equivalent to the integral inclusion { t x u(t, x) h C(J a J b, R) : h(t, x) = ξ(t)+η(x) ξ()+ v(s, τ)dsdτ, v S F,u }. Define the multivalued map A : M 2 M by (Au)(t, x) = {h C(J a J b, R) : h(t, x) = ξ(t) + η(x) ξ() + Lv(t, x), v S F,u }. Clearly, the multivalued map A is well defined on in view of hypothesis (L1). We shall show that A satisfy all conditions of Theorem 2.1. We first show that A has compact values. It then suffices to prove that the composition map LoS F has

32 Samad Mohseni Kolagar, Maryam Ramezani, Madjid Eshaghi compact values. Let u C(J a J b, R) be arbitrary and let {u n } be a sequence in S F,u. Then by the definition of S F,u, u n (t, x) F (t, x, u(t, x)) a.e for all (t, x) J a J b. Since F (t, x, u(t, x)) is compact, there exists a convergent subsequence of {u n (t, x)} (without loss of generality, we may assume it is {u n (t, x)} itself) that converges in measure to some v(t, x) F (t, x, u(t, x)) a.e for (t, x) J a J b. Now the continuity of L guarantees that Lu n (t, x) Lv(t, x) pointwise on J a J b provided n. We shall prove that the convergence is uniform. To this end, we show that {Lu n } is an equicontinuous sequence. Let t 1, t 2 J a with t 1 < t 2 and x 1, x 2 J b with x 1 < x 2, then t2 x2 t1 x1 Lu n (t 2, x 2 ) Lu n (t 1, x 1 ) u n (s, τ)dsdτ u n (s, τ)dsdτ t1 x2 x 1 u n (s, τ) dsdτ + t2 x2 t 1 u n (s, τ) dsdτ. Note that u n L 1 (J a J b, R), we infer that the right-hand side of the above expression tends to zero as t 2 t 1 and x 2 x 1. Hence, {Lu n } is equicontinuous and has a uniformly convergent subsequence by virtue of Arzela-Ascoli theorem. Obviously, this convergence is Lv and Lv LoS F,u. This shows that LoS F,u is compact. Therefore, A has compact values. As a result of this and (L3), we obtain that A has UCAV. Next we show that A satisfies relation (1). Let u, v K are comparable, say, u v. Let h 1 Au. Then there exists v 1 S F,u such that h 1 (t, x) = ξ(t) + η(x) ξ() + t x From (L2) there exists w F (t, x, v(t, x)) such that Define the multivalued map U by v 1 (s, τ)dsdτ for (t, x) J a J b. v 1 (t, x) w l(t, x) 2 (v(t, x) u(t, x)). U(t, x) = {w R : v 1 (t, x) w l(t, x) 2 (v(t, x) u(t, x))}. Then the multivalued map V (t, x) = U(t, x) S F,v has nonempty values and is measurable (see [21]). Then there exists a function v 2 which is a measurable selection for V. Clearly, v 2 (t, x) F (t, x, v(t, x)) for each (t, x) J a J b satisfying v 1 (t, x) v 2 (t, x) l(t, x) 2 (v(t, x) u(t, x)). Let us define for each (t, x) J a J b, It follows that h 2 Av and h 2 (t, x) = ξ(t) + η(x) ξ() + h 2 (t, x) h 1 (t, x) t x t x v 2 (s, τ)dsdτ. v 2 (s, τ) v 1 (s, τ) dsdτ.

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