FRACTIONAL ORDER IMPULSIVE PARTIAL HYPERBOLIC DIFFERENTIAL INCLUSIONS
WITH VARIABLE TIMES Sa¨ıd Abbas
Laboratoire de Math´ematiques, Universit´e de Sa¨ıda B.P. 138, 20000, Sa¨ıda, Alg´erie
e-mail: abbasmsaid@yahoo.fr
Mouffak Benchohra
Laboratoire de Math´ematiques, Universit´e de Sidi Bel-Abb`es B.P. 89, 22000, Sidi Bel-Abb`es, Alg´erie
e-mail: benchohra@univ-sba.dz
and
Lech G´orniewicz∗
Institute of Mathematics, Kazimierz Wielki University Weyssenhoffa 11, 85–072 Bydgoszcz, Poland
J. Schauder Center for Nonlinear Studies
University of Nicolaus Copernicus, 87–100 Toru´n, Poland e-mail: lego@ukw.edu.pl
Abstract
This paper deals with the existence of solutions to some classes of partial impulsive hyperbolic differential inclusions with variable times involving the Caputo fractional derivative. Our works will be consid-ered by using the nonlinear alternative of Leray-Schauder type. Keywords and phrases: impulsive functional differential inclusions, fractional order, solution, left-sided mixed Riemann-Liouville integral, Caputo fractional-order derivative, variable times, fixed point. 2010 Mathematics Subject Classification: 26A33, 34A60.
∗
1. Introduction
The paper deals with the existence of solutions for the following impulsive partial fractional order initial value problem (IVP for short), for the system
(1)
(cDr
0u)(x, y) ∈ F (x, y, u(x, y)), where (x, y) ∈ J,
x6= xk(u(x, y)), k = 1, . . . , m,
(2) u(x
+, y) = I
k(u(x, y)), where (x, y) ∈ J,
x= xk(u(x, y)), k = 1, . . . , m,
(3) u(x, 0) = ϕ(x), u(0, y) = ψ(y), x ∈ [0, a], y ∈ [0, b],
where J = [0, a] × [0, b], a, b > 0, cD0r is the fractional Caputo derivative of order r = (r1, r2) ∈ (0, 1] × (0, 1], 0 = x0 < x1 <· · · < xm < xm+1 = a,
F : J × Rn → P(Rn) is a compact valued multivalued map, P(Rn) is the
family of all subsets of Rn, xk: Rn→ R, Ik : Rn → Rn, k= 1, 2, . . . , m are
given functions and ϕ : [0, a] → Rn, ψ : [0, b] → Rnare absolutely continuous
functions with ϕ(0) = ψ(0).
Next we consider the following initial value problem for impulsive partial neutral functional differential inclusions
(4)
cDr
0[u(x, y) − g(x, y, u(x, y))] ∈ F (x, y, u(x, y)), where (x, y) ∈ J,
x6= xk(u(x, y)), k = 1, . . . , m,
(5) u(x
+, y) = I
k(u(x, y)), where (x, y) ∈ J,
x= xk(u(x, y)), k = 1, . . . , m,
(6) u(x, 0) = ϕ(x), u(0, y) = ψ(y), x ∈ [0, a], y ∈ [0, b],
where F, ϕ, ψ, xk, Ik; k = 1, 2, . . . , m are as in problem (1)–(3) and g : J ×
Rn→ Rnis a given function.
of fractional order (see [15, 17, 18, 21, 27]). In recent years, several quali-tative results for ordinary and partial fractional differential equations have been obtained; see the monographs of Kilbas et al. [25], Miller and Ross [28], Samko [30], the papers of Abbas and Benchohra [1, 2], Agarwal et al. [7], Belarbi et al. [8], Belmekki et al. [9], Benchohra et al. [10, 11, 13], Diethelm [15, 16], Kilbas and Marzan [23], Mainardi [27], Podlubny [29], Vityuk and Golushkov [34], and the references therein. The problem of existence of so-lutions to Cauchy-type problems for ordinary differential equations of frac-tional order in spaces of integrable functions was studied in numerous works (see [22, 32]), a similar problem in spaces of continuous functions was studied in [33].
The theory of impulsive integer order differential equations has become important in some mathematical models of real processes and phenomena studied in physics, chemical technology, population dynamics, biotechnology and economics. There has been a significant development in impulse theory in recent years, especially in the area of impulsive differential equations and inclusions with fixed moments; see the monographs of Benchohra et al. [12], Lakshmikantham et al. [26], and Samoilenko and Perestyuk [31], and the references therein. Recently in [3, 4, 6], we have considered some classes of hyperbolic differential equations involving the Caputo fractional derivative and impulses at fixed time. The theory of impulsive differential equations with variable time is relatively less developed due to the difficulties created by the state-dependent impulses [5].
This paper initiates the study of fractional order hyperbolic differential inclusions at variable times. Our existence results will be obtained using the nonlinear alternative of Leray-Schauder type.
2. Preliminaries
In this section, we introduce notations, definitions, and preliminary facts from multivalued analysis which are used throughout this paper. By L1(J, Rn) we denote the space of Lebesgue-integrable functions w : J → Rn
with the norm
kwk1 = Z a 0 Z b 0 kw(x, y)kdydx,
Let L∞
(J, Rn) be the Banach space of measurable functions w : J → Rn
which are bounded, equipped with the norm
kwkL∞ = inf{c > 0 : kw(x, y)k ≤ c, a.e. (x, y) ∈ J},
and AC(J, Rn) be the space of absolutely continuous functions from J into Rn.
Let a1 ∈ [0, a], z+ = (a+1,0) ∈ J, Jz = [a1, a] × [0, b], r1, r2 > 0 and
r= (r1, r2). For w ∈ L1(Jz, Rn), the expression
(Izr+w)(x, y) = 1 Γ(r1)Γ(r2) Z x a+1 Z y 0 (x − s)r1−1(y − t)r2−1w(s, t)dtds is called the left-sided mixed Riemann-Liouville integral of order r, where Γ(.) is the (Euler’s) Gamma function defined by Γ(ξ) =R∞
0 tξ−1e
−tdt, ξ >0.
Denote by D2
xy := ∂
2
∂x∂y the mixed second order partial derivative.
Definition 2.1 ([34]). For w ∈ L1(J
z, Rn) where D2xyf is Lebesque
inte-grable on [xk, xk+1]×[0, b], k = 0, . . . , m, the Caputo fractional-order
deriva-tive of order r is defined by the expression (cDr
z+w)(x, y) = (I
1−r
z+ D2xyw)(x, y). In the definition above by 1 − r we mean (1 − r1,1 − r2) ∈ (0, 1] × (0, 1].
Let (X, k · k) be a Banach space. Let Pcl(X) = {Y ∈ P(X) : Y closed},
Pb(X) = {Y ∈ P(X) : Y bounded}, Pcp(X) = {Y ∈ P(X) : Y compact}
and Pcp,c(X) = {Y ∈ P(X) : Y compact and convex}. A multi-valued map
G : X → 2X has convex (closed) values if G(x) is convex (closed) for all
x ∈ X. G is bounded on bounded sets if G(B) is bounded in X for each bounded set B of X, i.e., supx∈B{sup{kyk : y ∈ G(x)}} < ∞. G is called upper semi-continuous (u.s.c.) on X if for each x0 ∈ X the set G(x0) is a
nonempty, closed subset of X, and if for each open set N of X containing G(x0), there exists an open neighborhood M of x0 such that G(M ) ⊆ N . G
is lower semi-continuous (l.s.c.) if the set {x ∈ X : G(x) ∩ A 6= ∅} is open for any open subset A ⊆ X. G is said to be completely continuous if G(B) is relatively compact for every bounded subset B ⊆ X. If the multi-valued G is completely continuous with nonempty compact values, then G is u.s.c. if and only if G has a closed graph (i.e., xn→ x∗, yn→ y∗, yn∈ G(xn) imply
y∗ ∈ G(x∗)). G has a fixed point if there is x ∈ X such that x ∈ G(x).
Definition 2.2. A multi-valued map F : J × Rn → P(Rn) is said to be
(i) (x, y) 7→ F (x, y, u) is measurable for each u ∈ Rn;
(ii) u 7→ F (x, y, u) is upper semi-continuous for almost each (x, y) ∈ J; (iii) For each q > 0, there exists φq∈ L1(J, R+) such that
kF (x, y, u)kP = sup{kvk : v ∈ F (x, y, u)} ≤ φq(x, y)
for all kuk ≤ q and for almost each (x, y) ∈ J. F is said to be Carath´eodory if (i) and (ii) hold.
Let (X, d) be a metric space induced from the normed space (X, k · k). Consider Hd: P(X) × P(X) −→ R+∪ {∞} given by Hd(A, B) = max sup a∈A d(a, B), sup b∈B d(A, b) ,
where d(A, b) = infa∈Ad(a, b), d(a, B) = infb∈Bd(a, b). Then, (Pb,cl(X), Hd)
is a metric space and (Pcl(X), Hd) is a generalized metric space (see [24]).
For more details on multi-valued maps we refer the reader to the books of Deimling [14], G´orniewicz [20], Kisielewicz [24].
For a function u defined on J we define the set
SF,u= {f ∈ L1(J, Rn) : f (x, y) ∈ F (x, y, u(x, y)) for a.e. (x, y) ∈ J},
which is known as the set of selection functions.
Theorem 2.3 [19] (Nonlinear alternative of Leray-Schauder type). Let X be a Banach space and C a nonempty convex subset of X. Let U be a nonempty open subset of C with 0 ∈ U and T : U → Pcp,c(C) be an upper
semicontinuous and compact operator. Then, either
(a) T has fixed points, or
(b) there exist u ∈ ∂U and λ ∈ [0, 1] with u ∈ λT (u).
To define the solutions to problem (1)–(3), we shall consider the space Ω =u : J → Rn: there exist 0 = x
0 < x1< x2 < ... < xm< xm+1 = a
such that xk= xk(u(xk, .)), and u(x−k, .), u(x+k, .) exist with
where Jk := (xk, xk+1] × [0, b]. This set is a Banach space with the norm
kukΩ = max{kukk, k = 0, . . . , m},
where uk is the restriction of u to Jk, k = 0, . . . , m.
3. Existence of solutions
In what follows, we will assume that F is an L1-Carath´eodory function. Let
us start by defining what we mean by a solution to problem (1)–(3). Set
J′ := J\{(x1, y), . . . , (xm, y), y ∈ [0, b]}.
Definition 3.1. A function u ∈ Ω ∩ ∪mk=1AC(Jk, Rn) whose r-derivative
exists on J′
is said to be a solution to (1)–(3) if there exists a function f(x, y) ∈ F (x, y, u(x, y)) such that u satisfies (cDr
0u)(x, y) = f (x, y) on J ′
and conditions (2), (3) are satisfied.
Let h, w ∈ C([xk, xk+1] × [0, b], Rn), zk = (xk,0), and
µk(x, y) = u(x, 0) + u(x+k, y) − u(x+k,0), k = 0, . . . , m.
For the existence of solutions to problem (1)–(3), we need the following lemma:
Lemma 3.2 [3]. A function u ∈ AC([xk, xk+1] × [0, b], Rn); k = 0, . . . , m is
a solution of the differential equation
(cDzrku)(x, y) = h(x, y); (x, y) ∈ [xk, xk+1] × [0, b],
if and only if u(x, y) satisfies
(7) u(x, y) = µk(x, y) + (Izrkh)(x, y); (x, y) ∈ [xk, xk+1] × [0, b].
Corollary 3.3. A function u∈ AC([xk, xk+1] × [0, b], Rn); k = 0, . . . , m is
a solution of the differential equation
cDr
if and only if u(x, y) satisfies
(8) u(x, y) = µk(x, y)+w(x, y)−w(x, 0)−w(xk, y)+w(xk,0)+(Izrkh)(x, y), for all (x, y) ∈ [xk, xk+1] × [0, b].
In what follows, set µ(x, y) := µ0(x, y); (x, y) ∈ J.
Lemma 3.4. Let 0 < r1, r2 ≤ 1 and let h : J → Rn be continuous. A
function u is a solution of the fractional integral equation
(9) u(x, y) = µ(x, y) + Γ(r 1 1)Γ(r2) Rx 0 Ry 0(x − s)r1 −1(y − t)r2−1h(s, t)dtds; if (x, y) ∈ [0, x1] × [0, b],
ϕ(x) + Ik(u(xk, y)) − Ik(u(xk,0))
+Γ(r 1 1)Γ(r2) Rx xk Ry 0(x − s)r1 −1 (y − t)r2−1h(s, t)dtds; if (x, y) ∈ (xk, xk+1] × [0, b], k = 1, . . . , m,
if and only if u is a solution of the fractional IVP
cDru(x, y) = h(x, y), (x, y) ∈ J′
, (10)
u(x+k, y) = Ik(u(xk, y)), k= 1, . . . , m.
(11)
P roof. Assume u satisfies (10)–(11). If (x, y) ∈ [0, x1] × [0, b], then cDru(x, y) = h(x, y). Lemma 3.2 implies u(x, y) = µ(x, y) + 1 Γ(r1)Γ(r2) Z x 0 Z y 0 (x − s)r1−1(y − t)r2−1h(s, t)dtds.
If (x, y) ∈ (x1, x2] × [0, b], then Lemma 3.2 implies
+ 1 Γ(r1)Γ(r2) Z x x1 Z y 0 (x − s)r1−1(y − t)r2−1h(s, t)dtds
= ϕ(x) + I1(u(x1, y)) − I1(u(x1,0))
+ 1 Γ(r1)Γ(r2) Z x x1 Z y 0 (x − s)r1−1(y − t)r2−1h(s, t)dtds.
If (x, y) ∈ (x2, x3] × [0, b], then from Lemma 3.2 we get
u(x, y) = µ2(x, y) + 1 Γ(r1)Γ(r2) Z x x2 Z y 0 (x − s)r1−1(y − t)r2−1h(s, t)dtds = ϕ(x) + u(x+2, y) − u(x+2,0) + 1 Γ(r1)Γ(r2) Z x x2 Z y 0 (x − s)r1−1(y − t)r2−1h(s, t)dtds = ϕ(x) + I2(u(x2, y)) − I2(u(x2,0))
+ 1 Γ(r1)Γ(r2) Z x x2 Z y 0 (x − s)r1−1(y − t)r2−1h(s, t)dtds.
If (x, y) ∈ (xk, xk+1] × [0, b], then again from Lemma 3.2 we get (9).
Conversely, assume that u satisfies the impulsive fractional integral equation (9). If (x, y) ∈ [0, x1] × [0, b] and using the fact thatcDr is the left
inverse of Ir, we get
cDru(x, y) = h(x, y), for each (x, y) ∈ [0, x
1] × [0, b].
If (x, y) ∈ [xk, xk+1) × [0, b], k = 1, . . . , m and using the fact thatcDrC= 0,
where C is a constant, we get
cDru(x, y) = h(x, y), for each (x, y) ∈ [x
k, xk+1) × [0, b].
Also, we can easily show that
Corollary 3.5. Let 0 < r1, r2 ≤ 1 and let h, w : J → Rn be continuous. A
function u is a solution of the fractional integral equation
(12) u(x, y) = µ(x, y) + w(x, y) − w(x, 0) − w(0, y) + w(0, 0) +Γ(r 1 1)Γ(r2) Rx 0 Ry 0(x − s)r1 −1 (y − t)r2−1h(s, t)dtds; if (x, y) ∈ [0, x1] × [0, b], ϕ(x) + w(x, y) − w(x, 0) − w(xk, y) + w(xk,0)
+Ik(u(xk, y)) − Ik(u(xk,0))
+Γ(r 1 1)Γ(r2) Rx xk Ry 0(x − s)r1 −1(y − t)r2−1h(s, t)dtds; if (x, y) ∈ (xk, xk+1] × [0, b], k = 1, . . . , m,
if and only if u is a solution of the fractional IVP
cDr(u(x, y) − w(x, y)) = h(x, y), (x, y) ∈ J′
, (13)
u(x+k, y) = Ik(u(xk, y)), k= 1, . . . , m.
(14)
Now, we are concerned with the existence result for problem (1)–(3). The following hypotheses will be assumed hereafter.
(H1) The function xk∈ C1(Rn, R) for k = 1, . . . , m. Moreover,
0 = x0(u) < x1(u) < . . . < xm(u) < xm+1(u) = a, for all u ∈ Rn,
(H2) there exist constants ck, dk>0 such that
kIk(u)k ≤ ckkuk + dk, for each u ∈ Rn, k= 1, . . . , m,
(H3) there exist a continuous nondecreasing function δ : [0, ∞) → (0, ∞), and p ∈ L∞
(J, R+) such that
kF (x, y, u)kP ≤ p(x, y)δ(kuk) a.e. (x, y) ∈ J, and each u ∈ Rn,
(H4) there exists l ∈ L∞
(J, R+); k = 1, . . . , m, such that
and
d(0, F (x, y, 0)) ≤ l(x, y), a.e. (x, y) ∈ Jk, k= 0, . . . , m,
(H5) for all (s, t, u) ∈ J × Rn, there exists f ∈ S
F,u such that
x′k(u)[ϕ′(s) + r1− 1 Γ(r1)Γ(r2) Z s xk Z t 0 (s − θ)r1−2(t − η)r2−1f(θ, η)dηdθ] 6= 1; k= 1, . . . , m, (H6) for all u ∈ Rn, x
k(Ik(u)) ≤ xk(u) < xk+1(Ik(u)) for k = 1, . . . , m,
(H7) there exists M > 0 such that
min ( M kµk∞+ p∗ar1br2δ(M ) Γ(r1+ 1)Γ(r2+ 1) , (15) M kϕk∞+ 2ckM+ 2dk+ p∗ar1br2δ(M ) Γ(r1+ 1)Γ(r2+ 1) ; k = 1, . . . , m ) >1, where p∗ = kpkL∞.
Theorem 3.6. Assume that hypotheses (H1)–(H7) hold. Then the initial-value problem (1)–(3) has at least one solution on J.
The proof of this theorem will be given in several steps. Step 1. Consider the problem
(cD0ru)(x, y) ∈ F (x, y, u(x, y)), where (x, y) ∈ J, (16)
u(x, 0) = ϕ(x), u(0, y) = ψ(y), x ∈ [0, a], y ∈ [0, b]. (17)
Clearly, the fixed points of N are solutions to (16)–(17). We shall show that the operator N is completely continuous. The proof will be given in several Claims.
Claim 1. N(u) is convex for each u ∈ C(J, Rn)).
Indeed, if h1, h2 belong to N (u), then there exist f1, f2 ∈ SF,u such that
for each (x, y) ∈ J we have hi(x, y) = µ(x, y)+ 1 Γ(r1)Γ(r2) Z x 0 Z y 0 (x−s)r1−1(y−t)r2−1f i(s, t)dtds, i = 1, 2.
Let 0 ≤ d ≤ 1. Then, for each (x, y) ∈ J we have (dh1+ (1 − d)h2)(x, y) = µ(x, y) + 1 Γ(r1)Γ(r2) Z x 0 Z y 0 (x − s)r1−1(y − t)r2−1 × [f1(s, t) + (1 − d)f2(s, t)]dtds.
Since SF,u is convex (because F has convex values), then for each (x, y) ∈ J,
dh1+ (1 − d)h2∈ N (u).
Claim 2. N maps bounded sets into bounded sets in C(J, Rn).
Indeed, it is sufficient to show that for any q > 0 there exists a positive constant ℓ such that for each u ∈ Bq = {u ∈ C(J, Rn) : kuk∞≤ q} we have
kN (u)k ≤ ℓ. Let u ∈ Bq and h ∈ N (u), then there exists f ∈ SF,u such that
then we obtain
khk ≤ kµk∞+
ar1br2p∗δ(q) Γ(r1+ 1)Γ(r2+ 1)
:= ℓ.
Claim 3. N maps bounded sets into equicontinuous sets of C(J, Rn).
Let (τ1, y1), (τ2, y2) ∈ J, τ1 < τ2 and y1 < y2, Bq be the bounded set
of C(J, Rn) as in Claim 2. Let u ∈ B
q and h ∈ N (u). Then, there exists
f ∈ SF,u such that for each (x, y) ∈ J we have
h(x, y) = µ(x, y) + 1 Γ(r1)Γ(r2) Z x 0 Z y 0 (x − s)r1−1(y − t)r2−1f(s, t)dtds. Then, for each (x, y) ∈ J we have
≤ kµ(τ1, y1) − µ(τ2, y2)k + p ∗ δ(q) Γ(r1+ 1)Γ(r2+ 1) [2yr2 2 (τ2− τ1)r1+ 2τ2r1(y2− y1)r2 + τr1 1 y r2 1 − τ r1 2 y r2 2 − 2(τ2− τ1)r1(y2− y1)r2].
As τ1 −→ τ2and y1 −→ y2, the right-hand side of the above inequality tends
to zero. As a consequence of Claims 1 to 3 together and the Arzela-Ascoli theorem, we can conclude that N : C(J, Rn) → P(C(J, Rn)) is a completely
continuous multi-valued operator. Claim 4. N has a closed graph.
Let un → u∗, hn ∈ N (un) and hn → h∗. We need to show that h∗ ∈
N(u∗). hn ∈ N (un) means that there exists fn ∈ SF,un such that for each (x, y) ∈ J, hn(x, y) = µ(x, y) + 1 Γ(r1)Γ(r2) Z x 0 Z y 0 (x − s)r1−1(y − t)r2−1f n(s, t)dtds.
We must show that there exists f∗∈ SF,u∗ such that for each (x, y) ∈ J, h∗(x, y) = µ(x, y) + 1 Γ(r1)Γ(r2) Z x 0 Z y 0 (x − s)r1−1(y − t)r2−1f ∗(s, t)dtds.
Since F (x, y, ·) is upper semicontinuous, then for every ε > 0, there exist n0(ǫ) ≥ 0 such that for every n ≥ n0 we have
fn(x, y) ∈ F (x, y, un(x,y)) ⊂ F (x, y, u∗(x,y)) + εB(0, 1), a.e. (x, y) ∈ J.
Since F (., ., .) has compact values, then there exists a subsequence fnm such that
fnm(·, ·) → f∗(·, ·) as m → ∞ and
f∗(x, y) ∈ F (x, y, u∗(x,y)), a.e. (x, y) ∈ J.
For every w ∈ F (x, y, u∗(x,y)) we have
Then,
kfnm(x, y) − f∗(x, y)k ≤ d(fnm(x, y), F (x, y, u∗(x,y))).
By an analogous relation, obtained by interchanging the roles of fnm and f∗, it follows that
kfnm(x, y) − u∗(x, y)k ≤ Hd(F (x, y, un(x,y)), F (x, y, u∗(x,y))) ≤ l(x, y)kun− u∗k∞.
Let l∗
:= klkL∞,then by (H4) we obtain for each (x, y) ∈ J, khn(x,y)− h∗(x,y)k ≤ 1 Γ(r1)Γ(r2) Z x 0 Z y 0 (x − s)r1−1(y − t)r2−1kf m(s, t) − f∗(s, t)kdtds ≤ kunm− u∗k∞ Γ(r1)Γ(r2) Z x 0 Z y 0 (x − s)r1−1(y − t)r2−1l(s, t)dtds ≤ a r1br2l∗ku nm− u∗k∞ Γ(r1+ 1)Γ(r2+ 1) . Hence, khnm− h∗k∞ ≤ ar1br2l∗ku nm− u∗k∞ Γ(r1+ 1)Γ(r2+ 1) → 0 as m → ∞.
Claim 5. A priori bounds on solutions.
Let u be a possible solution to problem (1)–(3). Then, there exists f ∈ SF,u such that for each (x, y) ∈ J,
This implies that for each (x, y) ∈ J we have kuk∞ kµk∞+ p∗ar1br2δ(kuk ∞) Γ(r1+ 1)Γ(r2+ 1) <1.
Then, by condition (15), there exists M such that kuk∞6= M. Let
U = {u ∈ C(J, Rn) : kuk∞< M}.
The operator N : U → P(C(J, Rn)) is upper semicontinuous and completely
continuous. From the choice of U , there is no u ∈ ∂U such that u ∈ λN (u) for some λ ∈ (0, 1). As a consequence of the nonlinear alternative of Leray-Schauder type (Theorem 2.3), we deduce that N has a fixed point which is a solution to (16)–(17). Denote this solution by u1. Define the function
rk,1(x, y) = xk(u1(x, y)) − x, for x ≥ 0, y ≥ 0.
Hypothesis (H1) implies that rk,1(0, 0) 6= 0 for k = 1, . . . , m.
If rk,1(x, y) 6= 0 on J for k = 1, . . . , m; i.e.,
x6= xk(u1(x, y)), on J for k = 1, . . . , m,
then u1 is a solution to problem (1)–(3).
It remains to consider the case when r1,1(x, y) = 0 for some (x, y) ∈ J.
Now, since r1,1(0, 0) 6= 0 and r1,1 is continuous, there exists x1 >0, y1 >0
such that r1,1(x1, y1) = 0, and r1,1(x, y) 6= 0, for all (x, y) ∈ [0, x1) × [0, y1).
Thus, by (H1) we have
r1,1(x1, y1) = 0 and r1,1(x, y) 6= 0, for all (x, y) ∈ [0, x1) × [0, y1] ∪ (y1, b].
Suppose that there exist (¯x,y) ∈ [0, x¯ 1) × [0, y1] ∪ (y1, b] such that
r1,1(¯x,y) = 0. The function r¯ 1,1 attains a maximum at some point (s, t) ∈
[0, x1) × [0, b]. Since
(cD0ru1)(x, y) ∈ F (x, y, u1(x, y)), for (x, y) ∈ J,
Hence ∂u1(x, y) ∂x exists, and ∂r1,1(s, t) ∂x = x ′ 1(u1(s, t)) ∂u1(s, t) ∂x − 1 = 0. Since ∂u1(x, y) ∂x = ϕ ′ (x) + r1− 1 Γ(r1)Γ(r2) Z x 0 Z y 0 (x − s)r1−2(y − t)r2−1f(s, t)dtds, then x′1(u1(s, t))[ϕ′(s) + r1− 1 Γ(r1)Γ(r2) Z s 0 Z t 0 (s − θ)r1−2(t − η)r2−1f(θ, η)dηdθ] = 1,
witch contradicts (H5). From (H1) we have
rk,1(x, y) 6= 0 for all (x, y) ∈ [0, x1) × [0, b] and k = 1, . . . , m.
Step 2. In what follows, set
Xk := [xk, a] × [0, b]; k = 1, . . . , m.
Consider now the problem
(cD0ru)(x, y) ∈ F (x, y, u(x, y)), where (x, y) ∈ X1,
(18)
u(x+1, y) = I1(u1(x1, y)).
(19)
Consider the operator N1: C(X1, Rn) → P(C(X1, Rn)) defined as
N1(u) = h∈ C(X1, Rn) : h(x, y) = ϕ(x)+I1(u1(x1, y))−I1(u1(x1,0)) +Γ(r 1 1)Γ(r2) Rx x1 Ry 0(x − s)r1 −1 (y − t)r2−1 ×f (s, t)dtds; (x, y) ∈ X1, f ∈ SF,u. As in Step 1 we can show that N1 is upper semicontinuous and completely
there exists f ∈ SF,u such that for each (x, y) ∈ X1,
ku(x, y)k ≤ kϕ(x)k + kI1(u1(x1, y))k + kI1(u1(x1,0))k
+ 1 Γ(r1)Γ(r2) Z x x1 Z y 0 (x − s)r1−1(y − t)r2−1kf (s, t)kdtds ≤ kϕk∞+ 2c1kuk + 2d1 + 1 Γ(r1)Γ(r2) Z x 0 Z y 0 (x − s)r1−1(y − t)r2−1p(s, t)δ(kuk)dtds. Then, kuk∞ kϕk∞+ 2c1kuk∞+ 2d1+ p∗ar1br2δ(kuk ∞) Γ(r1+ 1)Γ(r2+ 1) <1.
Then by condition (15), there exists M′
such that kuk∞6= M′.Let
U′ = {u ∈ C(X1, Rn) : kuk∞< M′}.
The operator N1 : U ′
→ P(C(X1, Rn)) is upper semicontinuous and
com-pletely continuous. From the choice of U′
, there is no u ∈ ∂U′
such that u∈ λN1(u) for some λ ∈ (0, 1). As a consequence of the nonlinear
alterna-tive of Leray-Schauder type (Theorem 2.3), we deduce that N1 has a fixed
point which is a solution to (18)–(19). Denote this solution by u2. Define
rk,2(x, y) = xk(u2(x, y)) − x, for (x, y) ∈ X1.
If rk,2(x, y) 6= 0 on (x1, a] × [0, b] and for all k = 1, . . . , m, then
u(x, y) =(u1(x, y), if (x, y) ∈ [0, x1) × [0, b], u2(x, y), if (x, y) ∈ [x1, a] × [0, b],
is a solution to problem (1)–(3). It remains to consider the case where r2,2(x, y) = 0, for some (x, y) ∈ (x1, a] × [0, b]. By (H6), we have
r2,2(x+1, y1) = x2(u2(x+1, y1) − x1
= x2(I1(u1(x1, y1))) − x1
> x1(u1(x1, y1)) − x1
Since r2,2 is continuous, there exist x2 > x1, y2 > y1 such that r2,2(x2, y2)
= 0, and r2,2(x, y) 6= 0 for all (x, y) ∈ (x1, x2) × [0, b].
It is clear by (H1) that
rk,2(x, y) 6= 0 for all (x, y) ∈ (x1, x2)] × [0, b], k = 2, . . . , m.
Now suppose that there are (s, t) ∈ (x1, x2) × [0, b] such that r1,2(s, t) = 0.
From (H6) it follows that
r1,2(x+1, y1) = x1(u2(x+1, y1) − x1
= x1(I1(u1(x1, y1))) − x1
≤ x1(u1(x1, y1)) − x1
= r1,1(x1, y1) = 0.
Thus, r1,2 attains a nonnegative maximum at some point (s1, t1) ∈ (x1, a) ×
[0, x2) ∪ (x2, b]. Since
(cD0ru2)(x, y) ∈ F (x, y, u2(x, y)), for (x, y) ∈ X1,
Therefore, x′1(u2(s1, t1)) ϕ′(s1) + r1− 1 Γ(r1)Γ(r2) Z s1 x1 Z t1 0 (s1− θ)r1−2(t1− η)r2−1f(θ, η)dηdθ = 1, which contradicts (H5).
Step 3. We continue this process and take into account that um+1 := u
X
m is a solution to the problem
(cD0ru)(x, y) ∈ F (x, y, u(x, y)), a.e. (x, y) ∈ (xm, a] × [0, b],
u(x+m, y) = Im(um−1(xm, y)).
The solution u to problem (1)–(3) is then defined by
u(x, y) = u1(x, y), if (x, y) ∈ [0, x1] × [0, b], u2(x, y), if (x, y) ∈ (x1, x2] × [0, b], . . . um+1(x, y), if (x, y) ∈ (xm, a] × [0, b].
Now we present (without proof), by using Corollaries 3.3 and 3.5, an ex-istence result as an extension of the result presented in Theorem 3.6 to problem (4)–(6).
Definition 3.7. A function u ∈ Ω ∩ ∪m
k=1AC(Jk, Rn) whose r-derivative
exists on J′
is said to be a solution to (4)–(6) if there exists a function f(x, y) ∈ F (x, y, u(x, y)) such that u satisfies (cDr0u)(x, y) = f (x, y) on J′
and conditions (5), (6) are satisfied.
Theorem 3.8. Assume (H1)–(H4), (H6) and the following conditions (H8) The function g is nonnegative and completely continuous and there
exist constants 0 ≤ l1<1, l2 ≥ 0 such that
kg(x, y, u)k ≤ l1kuk + l2; (x, y) ∈ J, u ∈ Rn.
(H9) For all (s, t, u) ∈ J × Rn there exists f∈ S
x′k(u)[ϕ′(s) +∂g(s, t, u(s, t)) ∂x − ∂g(s, 0, u(s, 0)) ∂x + r1− 1 Γ(r1)Γ(r2) Z s xk Z t 0 (s − θ)r1−2(t − η)r2−1f(θ, η)dηdθ] 6= 1; k = 1, . . . , m.
(H10) There exists a number M′ >
0 such that min ( M′ kµk∞+ 4l1M′+ 4l2+ p∗ar1br2δ(M′) Γ(r1+ 1)Γ(r2+ 1) , (20) M′ kϕk∞+ (2ck+ 4l1)M′+ 2dk+ 4l2+ p∗ar1br2δ(M′) Γ(r1+ 1)Γ(r2+ 1) ; k = 1, . . . , m ) >1,
hold. Then, the initial-value problem (4)–(6) has at least one solution on J.
4. An example
As an application of our results we consider the following impulsive partial hyperbolic functional differential inclusions of the form
(21)
(cD0ru)(x, y) ∈ F (x, y, u(x, y)),
where (x, y) ∈ J, x 6= xk(u(x, y)); k = 1, . . . , m,
(22) u(x+k, y) = Ik(u(xk, y)), y ∈ [0, 1], k = 1, . . . , m,
(23) u(x, 0) = x, u(0, y) = y2, x∈ [0, 1], y ∈ [0, 1],
where J = [0, 1] × [0, 1], r = (r1, r2), 0 < r1, r2 ≤ 1, xk(u) = 1 − 2k 1
(1+u2); k= 1, . . . , m, and for each u ∈ R we have xk+1(Ik(u)) > xk(u); k = 1, . . . , m,
and also there exist constants ck, dk>0 such that
Let u ∈ R, then we have xk+1(u) − xk(u) =
1
2k+1(1 + u2) >0; k = 1, ..., m.
Hence, 0 < x1(u) < x2(u) < . . . < xm(u) < 1 for each u ∈ R. Set
F(x, y, u(x, y)) = {u ∈ R : f1(x, y, u(x, y)) ≤ u ≤ f2(x, y, u(x, y))},
where f1, f2 : [0, 1] × [0, 1] × R → R. We assume that for each (x, y) ∈
J, f1(x, y, .) is lower semi-continuous (i.e., the set {z ∈ R : f1(x, y, z) > ν}
is open for each ν ∈ R), and assume that for each (x, y) ∈ J, f2(x, y, .) is
upper semi-continuous (i.e., the set {z ∈ R : f2(x, y, z) < ν} is open for
each ν ∈ R). Assume that there are p ∈ C(J, R+) and δ : [0, ∞) → (0, ∞)
continuous and nondecreasing such that
max(|f1(x, y, z)|, |f2(x, y, z)|) ≤ p(x, y)δ(|z|), for a.e. (x, y) ∈ J and z ∈ R.
It is clear that F is compact and convex valued, and it is upper semi-continuous (see [14]). Moreover, assume that conditions (H5) and (H7) are satisfied. Since all conditions of Theorem 3.6 are satisfied, problem (21)–(23) has at least one solution on [0, 1] × [0, 1].
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