THE METHOD OF UPPER AND LOWER SOLUTIONS FOR PARTIAL HYPERBOLIC FRACTIONAL ORDER
DIFFERENTIAL INCLUSIONS WITH IMPULSES
Sa¨ ıd Abbas
Laboratoire de Math´ ematiques Universit´ e de Sa¨ıda B.P. 138, 20000, Sa¨ıda, Alg´ erie e-mail: abbas said dz@yahoo.fr
and
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
Abstract
In this paper we use the upper and lower solutions method to inves- tigate the existence of solutions of a class of impulsive partial hyper- bolic differential inclusions at fixed moments of impulse involving the Caputo fractional derivative. These results are obtained upon suitable fixed point theorems.
Keywords and phrases: impulsive hyperbolic differential inclusion, fractional order, upper solution, lower solution, left-sided mixed Riemann-Liouville integral, Caputo fractional-order derivative, fixed point.
2000 Mathematics Subject Classification: 26A33, 34A60.
∗
Corresponding author.
1. Introduction
This paper deals with the existence of solutions to impulsive fractional order initial value problems (IV P for short) for the system
(1) (
cD
0ru)(x, y) ∈ F (x, y, u(x, y)), if (x, y) ∈ J, x 6= x
k, k = 1, . . . , m,
(2) u(x
+k, y) = u(x
−k, y) + I
k(u(x
−k, y)), if (x, y) ∈ J, k = 1, . . . , m,
(3) u(x, 0) = ϕ(x), u(0, y) = ψ(y), (x, y) ∈ J,
where ϕ(0) = ψ(0), J = [0, a] × [0, b], a, b > 0,
cD
0ris the fractional Caputo derivative of order r = (r
1, r
2) ∈ (0, 1] × (0, 1], 0 = x
0< x
1< · · · < x
m<
x
m+1= a, F : J × R
n→ P(R
n) is a compact valued multivalued map, P(R
n) is the family of all subsets of R
n, I
k: R
n→ R
n, k = 1, . . . , m, ϕ : [0, a] → R
n, ψ : [0, b] → R
nare given functions. Here u(x
+k, y) and u(x
−k, y) denote the right and left limits of u(x, y) at x = x
k, respectively.
It is well known that differential equations of fractional order play an im- portant role in describing some real world problems. They are useful for solv- ing some problems in physics, mechanics, viscoelasticity, electrochemistry, control, porous media, electromagnetic, etc. (see [12, 14, 15, 18, 29, 30, 33]).
The theory of differential equations of fractional order has recently received a lot of attention and now constitutes a significant branch of nonlinear anal- ysis. Numerous research papers and monographs devoted to fractional dif- ferential equations have appeared in the monographs of Kilbas [24], Laksh- mikantham et al. [26], Miller and Ross [31], Samko [35], the papers of Abbas and Benchohra [1, 2], Agarwal et al. [3], Belarbi et al. [4], Benchohra et al. [5, 6, 8], Diethelm [12, 13], Kilbas et al. [22], Kilbas and Marzan [23], Mainardi [29], Podlubny [34], Semenchuk [36], Vityuk [38], Vityuk and Go- lushkov [39], Yu and Gao [40], Zhang [41].
The method of upper and lower solutions plays an important role in the investigation of solutions for differential equations and inclusions. See the monographs by Benchohra et al. [7], Heikkila and Lakshmikantham [17], Ladde et al. [28] and the references therein.
In this paper, we present an existence result for problem (1)–(3) by
means of the concept of upper and lower solutions combined with fixed
point theorem of Bohnnenblust-Karlin. This paper initiates the application
of the method of upper and lower solutions for impulsive partial hyperbolic
differential inclusions with Caputo fractional derivative and fixed moments of impulse. The present results extend those considered with integer order derivative [7, 10, 20, 21, 27, 32] and those with fractional derivative and without impulses [23].
2. Preliminaries
In this section we introduce notations, definitions, and preliminary facts which are used throughout this paper. Let k · k denote a suitable complete norm on R
n.
By L
1(J, R
n) we denote the space of Lebesgue-integrable functions f : J → R
nwith the norm
kf (x, y)k
1= Z
a0
Z
b 0kf (x, y)kdydx.
Let L
∞(J, R
n) be the Banach space of measurable functions f : J → R
nwhich are bounded, equipped with the norm
kf k
L∞= inf{c > 0 : kf (x, y))k ≤ c, a.e. (x, y) ∈ J}.
Let a
1∈ [0, a], z
+= (a
+1, 0) ∈ J, J
z= [a
1, a] × [0, b], r
1, r
2> 0 and r = (r
1, r
2). For f ∈ L
1(J
z, R
n), the expression
(I
zr+f)(x, y) = 1 Γ(r
1)Γ(r
2)
Z
x a+1Z
y 0(x − s)
r1−1(y − t)
r2−1f (s, t)dtds,
where Γ(·) is the Euler gamma function, is called the left-sided mixed Riemann-Liouville integral of order r.
Definition 2.1 ([39]). For f ∈ L
1(J
z, R
n), the Caputo fractional-order derivative of order r is defined by the expression (
cD
zr+f )(x, y) = (I
z1−r+ ∂2∂x∂y
f )(x, y).
We also need some properties of set-valued maps. Let (X, k · k) be a Banach
space. Let P
cl(X) = {Y ∈ P(X) : Y closed}, P
b(X) = {Y ∈ P(X) :
Y bounded}, P
cp(X) = {Y ∈ P(X) : Y compact} and P
cp,c(X) = {Y ∈
P(X) : Y compact and convex}.
Definition 2.2. A multivalued map G : X → P (X) is convex (closed) val- ued if G(x) is convex (closed) for all x ∈ X. G is bounded on bounded sets if G(B) = ∪
x∈BG(x) is bounded in X for all B ∈ P
b(X) (i.e., sup
x∈B{sup{|y| : y ∈ G(x)}} < ∞).
A multivalued map G : X → P (X) is called upper semi-continuous (u.s.c.) on X if for each x
0∈ X, the set G(x
0) is a nonempty closed subset of X, and if for each open set N of X containing G(x
0), there exists an open neighborhood N
0of x
0such that G(N
0) ⊆ N. G is said to be completely continuous if G(B) is relatively compact for every B ∈ P
b(X). G has a fixed point if there is x ∈ X such that x ∈ G(x). The fixed point set of the multivalued operator G will be denoted by F ixG.
A multivalued map G : J → P
cl(R
n) is said to be measurable if for every u ∈ R
n, the function
(x, y) 7−→ d(u, G(x, y)) = inf{ku − zk : z ∈ G(x, y)}
is measurable.
Definition 2.3. A multivalued map F : J × R
n→ P(R
n) is said to be L
1-Carath´eodory if
(i) (x, y) 7−→ F (x, y, u) is measurable for each u ∈ R
n;
(ii) u 7−→ F (x, y, u) is upper semicontinuous for a.e. (x, y) ∈ J.
(iii) for each q > 0, there exists ϕ
q∈ L
1(J, R
+) such that kF (x, y, u)k
P= sup{kf k : f ∈ F (x, y, u)} ≤ ϕ
q(x, y)
for all kuk ≤ q and a.e. (x, y) ∈ J.
F is said to be Carath´eodory if (i) and (ii) hold.
For each u ∈ C(J, R
n), define the set of selections of F by
S
F,u= {f ∈ L
1(J, R
n) : f (x, y) ∈ F (x, y, u(x, y)) a.e. (x, y) ∈ J}.
Let (X, d) be a metric space induced from the normed space (X, k · k).
Consider H
d: P(X) × P(X) −→ R
+∪ {∞} given by H
d(A, B) = max n
sup
a∈A
d(a, B), sup
b∈B
d(A, b) o
,
where d(A, b) = inf
a∈Ad(a, b), d(a, B) = inf
b∈Bd(a, b). Then (P
b,cl(X), H
d) is a metric space and (P
cl(X), H
d) is a generalized metric space (see [25]).
For more details on multi-valued maps we refer the reader to the books of Deimling [11], G´orniewicz [16], Hu and Papageorgiou [19] and Tolstonogov [37].
Lemma 2.4 (Bohnenblust-Karlin [9]). Let X be a Banach space and K ∈ P
cl,c(X) and suppose that the operator G : K → P
cl,c(K) is upper semicon- tinuous and the set G(K) is relatively compact in X. Then G has a fixed point in K.
3. Main Result We set
J
k:= (x
k, x
k+1] × (0, b].
To define the solutions of problems (1)–(3), we shall consider the space P C (J, R
n) = u : J → R
n: u ∈ C(J
k, R
n); k = 1, . . . , m, and there exist
u(x
−k, y) and u(x
+k, y); k = 1, . . . , m, with u(x
−k, y) = u(x
k, y) . This set is a Banach space with the norm
kuk
P C= sup
(x,y)∈J
ku(x, y)k.
Set J
0:= J\{(x
1, y), . . . , (x
m, y), y ∈ [0, b]}.
Definition 3.1. A function u ∈ P C(J, R
n) T S
mk=0
AC((x
k, x
k+1) × [0, b], R
n) whose r-derivative exists on J
0is said to be a solution of (1)–(3) if there exists a function f ∈ L
1(J, R
n) with f (x, y) ∈ F (x, y, u(x, y)) such that u satisfies (
cD
0ru)(x, y) = f (x, y) on J
0and conditions (2), (3) are satisfied.
Let z, ¯ z ∈ C(J, R
n) be such that
z(x, y) = (z
1(x, y), z
2(x, y), . . . , z
n(x, y)), (x, y) ∈ J, and
¯
z(x, y) = (¯ z
1(x, y), ¯ z
2(x, y), . . . , ¯ z
n(x, y)), (x, y) ∈ J.
The notation z ≤ ¯ z means that
z
i(x, y) ≤ ¯ z
i(x, y), i = 1, . . . , n.
Definition 3.2. A function v ∈ P C(J, R
n) T S
mk=0
AC((x
k, x
k+1)×[0, b], R
n) is said to be a lower solution of (1)–(3) if there exists a function f
1∈ L
1(J, IR
n) with f
1(x, y) ∈ F (x, y, v(x, y)) such that v satisfies
(
cD
r0v)(x, y) ≤ f
1(x, y, v(x, y)), v(x, 0) ≤ ϕ(x), v(0, y) ≤ ψ(y) on J
0, v(x
+k, y) ≤ v(x
−k, y) + I
k(v(x
−k, y)), if (x, y) ∈ J; k = 1, . . . , m, v(x, 0) ≤ ϕ(x), v(0, y) ≤ ψ(y) on J, and v(0, 0) ≤ ϕ(0).
A function w ∈ P C(J, R
n) T S
mk=0
AC((x
k, x
k+1) × [0, b], R
n) is said to be an upper solution of (1)–(3) if there exists a function f
2∈ L
1(J, IR
n) with f
2(x, y) ∈ F (x, y, w(x, y)) such that w satisfies
(
cD
r0w)(x, y) ≥ f
2(x, y, w(x, y)), w(x, 0) ≥ ϕ(x), w(0, y) ≥ ψ(y) on J
0, w(x
+k, y) ≥ w(x
−k, y) + I
k(w(x
−k, y)), if (x, y) ∈ J; k = 1, . . . , m, w(x, 0) ≥ ϕ(x), w(0, y) ≥ ψ(y) on J, and w(0, 0) ≥ ϕ(0).
Let h ∈ C([x
k, x
k+1] × [0, b], R
n), z
k= (x
k, 0), and
µ
k(x, y) = u(x, 0) + u(x
+k, y) − u(x
+k, 0), k = 0, . . . , m.
For the existence of solutions for problem (1)–(3), we need the following lemma:
Lemma 3.3. A function u ∈ AC([x
k, x
k+1] × [0, b], R
n); k = 0, . . . , m is a solution of the differential equation
(
cD
zrku)(x, y) = h(x, y); (x, y) ∈ [x
k, x
k+1] × [0, b], if and only if u(x, y) satisfies
(4) u(x, y) = µ
k(x, y) + (I
zrkh)(x, y); (x, y) ∈ [x
k, x
k+1] × [0, b].
P roof. Let u(x, y) be a solution of (
cD
zrku)(x, y) = h(x, y); (x, y) ∈ [x
k, x
k+1] × [0, b]. Then, taking into account the definition of the derivative (
cD
rzk+
u)(x, y), we have I
1−rzk+
(D
2xyu)(x, y) = h(x, y).
Hence, we obtain I
zr+k
(I
z1−rkD
xy2u)(x, y) = (I
zr+ kh)(x, y), then
I
1z+k
D
2xyu(x, y) = (I
zr+ kh)(x, y).
Since I
z1+k
(D
2xyu)(x, y) = u(x, y) − u(x, 0) − u(x
+k, y) + u(x
+k, 0), we have
u(x, y) = µ
k(x, y) + (I
zr+ kh)(x, y).
Now let u(x, y) satisfy (4). It is clear that u(x, y) satisfies (
cD
r0u)(x, y) = h(x, y), on [x
k, x
k+1] × [0, b].
In the following we set
µ
0(x, y) := µ(x, y); (x, y) ∈ J.
Lemma 3.4. Let 0 < r
1, r
2≤ 1 and let h : J → R
nbe continuous. A function u is a solution of the fractional integral equation
(5)
u(x, y) =
µ(x, y) +
Γ(r1)Γ(r1 2)R
x 0R
y0
(x − s)
r1−1(y − t)
r2−1h(s, t)dtds;
if (x, y) ∈ [0, x
1] × [0, b], µ(x, y) + P
ki=1
(I
i(u(x
−i, y)) − I
i(u(x
−i, 0))) +
Γ(r 11)Γ(r2)
P
k i=1R
xixi−1
R
y0
(x
i− s)
r1−1(y − t)
r2−1h(s, t)dtds +
Γ(r 11)Γ(r2)
R
x xkR
y0
(x − s)
r1−1(y − t)
r2−1h(s, t)dtds;
if (x, y) ∈ (x
k, x
k+1] × [0, b], k = 1, . . . , m,
if and only if u is a solution of the fractional IVP
c
D
ru(x, y) = h(x, y), (x, y) ∈ J
0, (6)
u(x
+k, y) = u(x
−k, y) + I
k(u(x
−k, y)), k = 1, . . . , m.
(7)
P roof. Assume that u satisfies (6)–(7). If (x, y) ∈ [0, x
1] × [0, b], then
c
D
ru(x, y) = h(x, y).
Lemma 3.3 implies
u(x, y) = µ(x, y) + 1 Γ(r
1)Γ(r
2)
Z
x 0Z
y 0(x − s)
r1−1(y − t)
r2−1h(s, t)dtds.
If (x, y) ∈ (x
1, x
2] × [0, b], then Lemma 3.3 implies
u(x, y) = µ
1(x, y) + 1 Γ(r
1)Γ(r
2)
Z
x x1Z
y 0(x − s)
r1−1(y − t)
r2−1h(s, t)dtds
= ϕ(x) + u(x
+1, y) − u(x
+1, 0)
+ 1
Γ(r
1)Γ(r
2) Z
xx1
Z
y 0(x − s)
r1−1(y − t)
r2−1h(s, t)dtds
= ϕ(x) + u(x
−1, y) − u(x
−1, 0) + I
1(u(x
−1, y)) − I
1(u(x
−1, 0))
+ 1
Γ(r
1)Γ(r
2) Z
xx1
Z
y 0(x − s)
r1−1(y − t)
r2−1h(s, t)dtds
= ϕ(x) + u(x
1, y) − u(x
1, 0) + I
1(u(x
−1, y)) − I
1(u(x
−1, 0))
+ 1
Γ(r
1)Γ(r
2) Z
xx1
Z
y 0(x − s)
r1−1(y − t)
r2−1h(s, t)dtds
= µ(x, y) + I
1(u(x
−1, y)) − I
1(u(x
−1, 0))
+ 1
Γ(r
1)Γ(r
2) Z
x10
Z
y 0(x
1− s)
r1−1(y − t)
r2−1h(s, t)dtds
+ 1
Γ(r
1)Γ(r
2) Z
xx1
Z
y 0(x − s)
r1−1(y − t)
r2−1h(s, t)dtds.
If (x, y) ∈ (x
2, x
3] × [0, b], then from Lemma 3.3 we get
u(x, y) = µ
2(x, y) + 1 Γ(r
1)Γ(r
2)
Z
x x2Z
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(r
1)Γ(r
2) Z
xx2
Z
y 0(x − s)
r1−1(y − t)
r2−1h(s, t)dtds
= ϕ(x) + u(x
−2, y) − u(x
−2, 0) + I
2(u(x
−2, y)) − I
2(u(x
−2, 0))
+ 1
Γ(r
1)Γ(r
2) Z
xx2
Z
y 0(x − s)
r1−1(y − t)
r2−1h(s, t)dtds
= ϕ(x) + u(x
2, y) − u(x
2, 0) + I
2(u(x
−2, y)) − I
2(u(x
−2, 0))
+ 1
Γ(r
1)Γ(r
2) Z
xx2
Z
y 0(x − s)
r1−1(y − t)
r2−1h(s, t)dtds
= µ(x, y) + I
2(u(x
−2, y))−I
2(u(x
−2, 0))+I
1(u(x
−1, y))−I
1(u(x
−1, 0))
+ 1
Γ(r
1)Γ(r
2) Z
x10
Z
y 0(x
1− s)
r1−1(y − t)
r2−1h(s, t)dtds
+ 1
Γ(r
1)Γ(r
2) Z
x2x1
Z
y 0(x
2− s)
r1−1(y − t)
r2−1h(s, t)dtds
+ 1
Γ(r
1)Γ(r
2) Z
xx2
Z
y 0(x − s)
r1−1(y − t)
r2−1h(s, t)dtds.
If (x, y) ∈ (x
k, x
k+1] × [0, b], then again from Lemma 3.3 we get (3.4).
Conversely, assume that u satisfies the impulsive fractional integral equation (3.4). If (x, y) ∈ [0, x
1] × [0, b] and using the fact that
cD
ris the left inverse of I
rwe get
c
D
ru(x, y) = h(x, y), for each (x, y) ∈ [0, x
1] × [0, b].
If (x, y) ∈ [x
k, x
k+1) × [0, b], k = 1, . . . , m and using the fact that
cD
rC = 0, where C is a constant, we get
c
D
ru(x, y) = h(x, y), for each (x, y) ∈ [x
k, x
k+1) × [0, b].
Also, we can easily show that
u(x
+k, y) = u(x
−k, y) + I
k(u(x
−k, y)), y ∈ [0, b], k = 1, . . . , m.
For the study of problem (1)–(3), we first list the following hypotheses:
(H1) F : J × R
n−→ P
cp,c(R
n) is L
1-Carath´eodory.
(H2) There exists l ∈ L
∞(J, R
+) such that
H
d(F (x, y, u), F (x, y, u)) ≤ l(x, y)ku − uk for every u, u ∈ R
n, and
d(0, F (x, y, 0)) ≤ l(x, y), a.e. (x, y) ∈ J.
(H3) There exist v and w ∈ P C(J, R
n) T AC((x
k, x
k+1) × [0, b], R
n), k = 0, . . . , m, lower and upper solutions for problem (1)–(3) such that v(x, y) ≤ w(x, y) for each (x, y) ∈ J.
(H4) For each y ∈ [0, b], we have v(x
+k, y) ≤ min
u∈[v(x−k,y),w(x−k,y)]
I
k(u)
≤ max
u∈[v(x−k,y),w(x−k,y)]
I
k(u) ≤ w(x
+k, y), k = 1, . . . , m.
Theorem 3.5. Assume that hypotheses (H1)–(H4) hold. Then problem (1)–
(3) has at least one solution u such that
v(x, y) ≤ u(x, y) ≤ w(x, y) for all (x, y) ∈ J.
P roof. Transform problem (1)–(3) into a fixed point problem. Consider the following modified problem
(8) (
cD
r0u)(x, y) ∈ F (x, y, g(u(x, y))), if x 6= x
k, (x, y) ∈ J, k = 1, . . . , m,
(9) u(x
+k, y) = u(x
−k, y)+I
k(g(x
−k, y, u(x
−k, y))), if y ∈ [0, b]; k = 1, . . . , m,
(10) u(x, 0) = ϕ(x), u(0, y) = ψ(y), (x, y) ∈ J,
where g : P C(J, R
n) −→ P C(J, R
n) is the truncation operator defined by
(gu)(x, y) =
v(x, y), u(x, y) < v(x, y)
u(x, y), v(x, y) ≤ u(x, y) ≤ w(x, y) w(x, y), u(x, y) > w(x, y).
A solution to (8)–(10) is a fixed point of the operator G : P C(J, R
n) −→
P(P C(J, R
n)) defined by:
G(u) =
h ∈ P C(J, R
n) : h(x, y) = µ(x, y) + P
0<xk<x
(I
k(g(x
−k, y, u(x
−k, y))) − I
k(g(x
−k, 0, u(x
−k, 0)))) +
Γ(r 11)Γ(r2)
P
0<xk<x
R
xkxk−1
R
y0
(x
k− s)
r1−1(y − t)
r2−1f (s, t)dtds +
Γ(r 11)Γ(r2)
R
x xkR
y0
(x − s)
r1−1(y − t)
r2−1f(s, t)dtds, where
f ∈ ˜ S
F,g(u)1= f ∈ S
F,g(u)1: f (x, y) ≥ f
1(x, y) on A
1and f (x, y) ≤ f
2(x, y) on A
2, A
1= {(x, y) ∈ J : u(x, y) < v(x, y) ≤ w(x, y)}, A
2= {(x, y) ∈ J : v(x, y) ≤ w(x, y) < u(x, y)}, and
S
F,g(u)1= {f ∈ L
1(J, R
n) : f (x, y) ∈ F (x, y, g(u(x, y))), for (x, y) ∈ J}.
Remark 3.6.
(A) For each u ∈ P C(J, R
n), the set ˜ S
F,g(u)is nonempty. In fact, (H
1) implies that there exists f
3∈ S
F,g(u), so we set
f = f
1χ
A1+ f
2χ
A2+ f
3χ
A3,
where χ
Aiis the characteristic function of A
i; i = 1, 2, and A
3= {(x, y) ∈ J : v(x, y) ≤ u(x, y) ≤ w(x, y)}.
Then, by decomposability, f ∈ ˜ S
F,g(u).
(B) By the definition of g it is clear that F (., ., g(u)(., .)) is an L
1-Cara- th´eodory multi-valued map with compact convex values and there exists φ
1∈ L
∞(J, R
+) such that
kF (x, y, g(u(x, y)))k
P≤ φ
1(x, y) for a.e. (x, y) ∈ J and u ∈ R
n. (C) By the definition of g and from (H4), we have
v(x
+k, y) ≤ I
k(g(x
k, y, u(x
k, y))) ≤ w(x
+k, y); y ∈ [0, b]; k = 1, . . . , m.
Set
φ
∗1:= kφ
1k
L∞,
η = kµk
∞+ 2
m
X
k=1
y∈[0,b]
max (kv(x
+k, y)k, kw(x
+k, y)k) + 2a
r1b
r2φ
∗1Γ(r
1+ 1)Γ(r
2+ 1) , and
D = {u ∈ P C(J, R
n) : kuk
P C≤ η}.
Clearly, D is a closed convex subset of P C(J, R
n) and G maps D into D.
We shall show that D satisfies the assumptions of Lemma 2.4. The proof will be given in several steps.
Step 1. G(u) is convex for each u ∈ D.
Indeed, if h
1, h
2belong to G(u), then there exist f
1, f
2∈ ˜ S
F,g(u)1such that for each (x, y) ∈ J we have
h
i(u)(x, y) = µ(x, y) + X
0<xk<x
(I
k(g(x
−k, y, u(x
−k, y))) − I
k(g(x
−k, 0, u(x
−k, 0))))
+ 1
Γ(r
1)Γ(r
2) X
0<xk<x
Z
xkxk−1
Z
y 0(x
k− s)
r1−1(y − t)
r2−1f
i(s, t)dtds
+ 1
Γ(r
1)Γ(r
2) Z
xxk
Z
y 0(x − s)
r1−1(y − t)
r2−1f
i(s, t)dtds.
Let 0 ≤ ξ ≤ 1. Then, for each (x, y) ∈ J, we have (ξh
1+ (1− ξ)h
2)(x, y) = µ(x, y) + X
0<xk<x
(I
k(g(x
−k, y, u(x
−k, y))))
− X
0<xk<x
(I
k(g(x
−k, 0, u(x
−k, 0))))
+ 1
Γ(r
1)Γ(r
2) X
0<xk<x
Z
xkxk−1
Z
y 0(x
k− s)
r1−1(y− t)
r2−1× [ξf
1(s, t) + (1 − ξ)f
2(s, t)] dtds
+ 1
Γ(r
1)Γ(r
2) Z
xxk
Z
y 0(x − s)
r1−1(y − t)
r2−1× [ξf
1(s, t) + (1 − ξ)f
2(s, t)]dtds.
Since ˜ S
F,g(u)1is convex (because F has convex values), we have ξh
1+ (1 − ξ)h
2∈ G(u).
Step 2. G(D) is bounded.
This is clear since G(D) ⊂ D and D is bounded.
Step 3. G(D) is equicontinuous.
Let (τ
1, y
1), (τ
2, y
2) ∈ J, τ
1< τ
2and y
1< y
2, let u ∈ D and h ∈ G(u), then there exists f ∈ ˜ S
F,g(u)1such that for each (x, y) ∈ J we have
kh(u)(τ
2, y
2) − h(u)(τ
1, y
1)k
≤ kµ(τ
1, y
1) − µ(τ
2, y
2)k +
m
X
k=1
(kI
k(g(x
−k, y
1, u(x
−k, y
1)))
− I
k(g(x
−k, y
2, u(x
−k, y
2)))k)
+ 1
Γ(r
1)Γ(r
2)
m
X
k=1
Z
xkxk−1
Z
y10
(x
k− s)
r1−1[(y
2− t)
r2−1− (y
1− t)
r2−1]
× kf (s, t)kdtds
+ 1
Γ(r
1)Γ(r
2)
m
X
k=1
Z
xkxk−1
Z
y2y1
(x
k− s)
r1−1(y
2− t)
r2−1kf (s, t)kdtds
+ 1 Γ(r
1)Γ(r
2)
Z
τ10
Z
y10
[(τ
2− s)
r1−1(y
2− t)
r2−1− (τ
1− s)
r1−1(y
1− t)
r2−1]
× kf (s, t)kdtds
+ 1
Γ(r
1)Γ(r
2) Z
τ2τ1
Z
y2y1
(τ
2− s)
r1−1(y
2− t)
r2−1kf (s, t)kdtds
≤ kµ(τ
1, y
1) − µ(τ
2, y
2)k +
m
X
k=1
(kI
k(g(x
−k, y
1, u(x
−k, y
1))) − I
k(g(x
−k, y
2, u(x
−k, y
2)))k)
+ φ
∗1Γ(r
1)Γ(r
2)
m
X
k=1
Z
xkxk−1
Z
y10
(x
k− s)
r1−1[(y
2− t)
r2−1− (y
1− t)
r2−1]dtds
+ φ
∗1Γ(r
1)Γ(r
2)
m
X
k=1
Z
xkxk−1
Z
y2y1
(x
k− s)
r1−1(y
2− t)
r2−1dtds
+ φ
∗1Γ(r
1)Γ(r
2)
Z
τ10
Z
y10
[(τ
2− s)
r1−1(y
2− t)
r2−1− (τ
1− s)
r1−1(y
1− t)
r2−1]dtds
+ φ
∗1Γ(r
1)Γ(r
2)
Z
τ2τ1
Z
y2y1
(τ
2− s)
r1−1(y
2− t)
r2−1dtds.
As τ
1−→ τ
2and y
1−→ y
2, the right-hand side of the above inequality tends to zero. As a consequence of Steps 1 to 3 together with the Arzel´a-Ascoli theorem, we can conclude that G : D −→ P(D) is compact.
Step 4. G has a closed graph.
Let u
n→ u
∗, h
n∈ G(u
n) and h
n→ h
∗. We need to show that h
∗∈ G(u
∗).
h
n∈ G(u
n) means that there exists f
n∈ ˜ S
F,u1 nsuch that for each (x, y) ∈ J, h
n(x, y) = µ(x, y) + X
0<xk<x
(I
k(g(x
−k, y, u
n(x
−k, y)))− I
k(g(x
−k, 0, u
n(x
−k, 0))))
+ 1
Γ(r
1)Γ(r
2) X
0<xk<x
Z
xkxk−1
Z
y 0(x
k− s)
r1−1(y − t)
r2−1f
n(s, t)dtds
+ 1
Γ(r
1)Γ(r
2) Z
xxk
Z
y 0(x − s)
r1−1(y − t)
r2−1f
n(s, t)dtds.
We must show that there exists f
∗∈ ˜ S
F,u1 ∗such that for each (x, y) ∈ J,
h
∗(x, y) = µ(x, y) + X
0<xk<x
(I
k(g(x
−k, y, u
∗(x
−k, y)))− I
k(g(x
−k, 0, u
∗(x
−k, 0))))
+ 1
Γ(r
1)Γ(r
2) X
0<xk<x
Z
xkxk−1
Z
y 0(x
k− s)
r1−1(y − t)
r2−1f
∗(s, t)dtds
+ 1
Γ(r
1)Γ(r
2) Z
xxk
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 n
0() ≥ 0 such that for every n ≥ n
0, we have
f
n(x, y) ∈ F (x, y, g(u
n(x, y))) ⊂ F (x, y, g(u
∗(x, y))) + εB(0, 1), a.e. (x, y) ∈ J.
Since F (., ., .) has compact values, then there exists a subsequence f
nmsuch that
f
nm(·, ·) → f
∗(·, ·) as m → ∞ and
f
∗(x, y) ∈ F (x, y, g(u
∗(x, y))), a.e. (x, y) ∈ J.
For every w(x, y) ∈ F (x, y, g(u
∗(x, y))), we have
kf
nm(x, y) − u
∗(x, y)k ≤ kf
nm(x, y) − w(x, y)k + kw(x, y) − f
∗(x, y)k.
Then,
kf
nm(x, y) − f
∗(x, y)k ≤ d(f
nm(x, y), F (x, y, g(u
∗(x, y))).
By an analogous relation, obtained by interchanging the roles of f
nmand f
∗, it follows that
kf
nm(x, y) − u
∗(x, y)k ≤ H
d(F (x, y, g(u
n(x, y))), F (x, y, g(u
∗(x, y))))
≤ l(x, y)ku
n− u
∗k
∞.
Then by (H2), we have for each (x, y) ∈ J, kh
n(x, y) − h
∗(x, y)k ≤
≤
m
X
k=1
kI
k(g(x
−k, y, u
nm(x
−k, y))) − I
k(g(x
−k, y, u
∗(x
−k, y)))k
+
m
X
k=1
kI
k(g(x
−k, 0, u
nm(x
−k, 0))) − I
k(g(x
−k, 0, u
∗(x
−k, 0)))k
+ 1
Γ(r
1)Γ(r
2) X
x1<xk<x
Z
xkxk−1
Z
y 0(x
k− s)
r1−1(y − t)
r2−1× kf
nm(s, t) − f
∗(s, t)kdtds
+ 1
Γ(r
1)Γ(r
2) Z
xxk
Z
y 0(x − s)
r1−1(y − t)
r2−1× kf
nm(s, t) − f
∗(s, t)kdtds
≤
m
X
k=1
kI
k(g(x
−k, y, u
nm(x
−k, y))) − I
k(g(x
−k, y, u
∗(x
−k, y)))k
+
m
X
k=1
kI
k(g(x
−k, 0, u
nm(x
−k, 0))) − I
k(g(x
−k, 0, u
∗(x
−k, 0)))k
+ 2ku
nm− u
∗k
∞Γ(r
1)Γ(r
2) Z
x0
Z
y 0(x − s)
r1−1(y − t)
r2−1l(s, t)dtds
≤
m
X
k=1
kI
k(g(x
−k, y, u
nm(x
−k, y))) − I
k(g(x
−k, y, u
∗(x
−k, y)))k
+
m
X
k=1
kI
k(g(x
−k, 0, u
nm(x
−k, 0))) − I
k(g(x
−k, 0, u
∗(x
−k, 0)))k
+ 2l
∗a
r1b
r2Γ(r
1+ 1)Γ(r
2+ 1) ku
nm− u
∗k
∞, where
l
∗:= klk
L∞. Hence,
kh
nm− h
∗k
∞→ 0 as m → ∞.
Step 5. The solution u of (8)–(10) satisfies
v(x, y) ≤ u(x, y) ≤ w(x, y) for all (x, y) ∈ J.
Let u be the above solution to (8)–(10).
We prove that
u(x, y) ≤ w(x, y) for all (x, y) ∈ J.
Assume that u − w attains a positive maximum on [x
+k, x
−k+1] × [0, b] at (x
k, y) ∈ [x
+k, x
−k+1] × [0, b] for some k = 0, . . . , m; that is,
(u − w)(x
k, y) = max{u(x, y) − w(x, y) : (x, y) ∈ [x
+k, x
−k+1] × [0, b]} > 0, for some k = 0, . . . , m.
We distinguish the following cases.
Case 1. If (x
k, y) ∈ (x
+k, x
−k+1)×[0, b] there exists (x
∗k, y
∗) ∈ (x
+k, x
−k+1)×
[0, b] such that
(11) u(x
∗k, y
∗) − w(x
∗k, y
∗) ≤ 0, and
(12) u(x, y) − w(x, y) > 0, for all (x, y) ∈ (x
∗k, x
k] × [y
∗, b].
By the definition of h, one has
c
D
ru(x, y) ∈ F (x, y, w(x, y)) for all (x, y) ∈ [x
∗k, x
k] × [y
∗, b].
An integration on [x
∗k, x] × [y
∗, y] for each (x, y) ∈ [x
∗k, x
k] × [y
∗, b] yields
(13)
u(x, y) − u(x
∗k, y
∗)
= 1
Γ(r
1)Γ(r
2) Z
xx∗k
