Novi Sad J. Math. Vol. 46, No. 2, 2016, 45-53
ON A TWO-VARIABLES FRACTIONAL PARTIAL DIFFERENTIAL INCLUSION VIA RIEMANN-LIOUVILLE DERIVATIVE S. Etemad1 and Sh. Rezapour23 Abstract. We investigate the existence a solution to a two-variable fractional partial differential inclusion via Riemann-Liouville derivative. Also, we provide an example to illustrate our main result. AMS Mathematics Subject Classification (2010): 26A33; 34A00; 34A08 Key words and phrases: boundary value problem; endpoint; fixed point; fractional partial derivative; two-variables fractional partial differential inclusion
1.
Introduction
There are many published works about fractional partial differential equations by using the notions of delay or time-fractional (see for example, [1], [2], [9] and [10]). It is interesting to work on two variables fractional partial differential equations (see, for example, [3], [4], [6] and [11]). Let θ = (0, 0) and α = (α1 , α2 ) where 0 < α1 , α2 ≤ 1. Also, put Ja × Jb = [0, a] × [0, b] where a and b are positive constants. The Riemann-Liouville fractional partial integral of u ∈ L1 (Ja × Jb ) is defined by ∫ x∫ y 1 α (Iθ u)(x, y) = (x − s)α1 −1 (y − t)α2 −1 u(s, t)dtds Γ(α1 )Γ(α2 ) 0 0 whenever the integral exists (see, for example, [14], [15] and [16]). The RiemannLiouville partial derivative of fractional order α for a function u ∈ L1 (Ja × Jb ) is defined by ∫ x∫ y ∂2 (x − s)−α1 (y − t)−α2 2 (Dθα u)(x, y) = Dxy (Iθ1−α u)(x, y) = dtds ∂x∂y 0 0 Γ(1 − α1 )Γ(1 − α2 ) (see for more details [14], [15] and [16]). Note that (Iθα Iθβ u)(x, y) = (Iθα+β u)(x, y) whenever β = (β1 , β2 ) > 0 ([16]). 1 Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, Iran e-mail:
[email protected] 2 Department of Mathematics, Azarbaijan Shahid Madani University, Tabriz, Iran e-mail:
[email protected] 3 Corresponding author
46
S. Etemad, Sh. Rezapour
Let (X, d) be a metric space, P(X) the class of all nonempty subsets of X, Pcl (X) the class of all closed subsets of X, Pbd (X) the class of all bounded subsets of X, Pcp (X) the class of all compact subsets of X and Pcv (X) the class of all convex subsets of X. A multi-valued map F : Ja × Jb → Pcl (R) is measurable whenever the function (x, y) 7→ d(w, F (x, y)) = inf{∥w − v∥ : v ∈ F (x, y)} is measurable for all w ∈ R, where Ja × Jb = [0, a] × [0, b] ([12]). Also, the Pompeiu-Hausdorff metric Hd : P(X) × P(X) → [0, ∞) is defined by Hd (A, B) = max{sup d(a, B), sup d(A, b)}, a∈A
b∈B
where d(A, b) = inf a∈A d(a, b) ([7]). Then (Pcl,bd (X), Hd ) is a metric space and (Pcl (X), Hd ) is a generalized metric space ([7]). Recall that a multifunction F : X → P(X) is said to be a contraction if there exists k ∈ (0, 1) such that Hd (F (u), F (v)) ≤ kd(u, v) for all u, v ∈ X ([13]). An element u ∈ X is called endpoint of the multifunction F : X → P(X) whenever F u = {u} ([8]). We say that the multifunction F has an approximate endpoint property whenever inf u∈X supw∈F u d(u, w) = 0 ([8]). A real-valued function f on R is called upper semi-continuous whenever lim supn→∞ f (λn ) ≤ f (λ) for all sequence {λn }n≥1 with λn → λ. In this paper, using the main idea of [5], [6] and [17], we investigate the existence of solutions for the two-variables fractional partial differential inclusion (Dθα u)(x, y) ∈ F (x, y, u(x, y)),
(1.1)
with the partial integral boundary value conditions (1.2)
(Iθ1−α u)(x, 0) = λ1 φ(x), (Iθ1−α u)(0, y) = λ2 γ(y),
where Dθα denotes the Riemann-Liouville fractional partial derivative of order α, (x, y) ∈ Ja × Jb , 0 < αi ≤ 1, λi ∈ R+ (i = 1, 2) and F : Ja × Jb × R → P(R) is a compact valued multi-valued map. Here, the functions φ : Ja → R and γ : Jb → R are absolutely continuous with φ(0) = γ(0) = 0. We need the following endpoint result. Theorem 1.1. ([8]) Suppose that (X, d) is a complete metric space, ψ : [0, ∞) → [0, ∞) is an upper semi-continuous function such that ψ(t) < t and lim inf t→∞ (t − ψ(t)) > 0 for all t > 0 and T : X → CB(X) is a multifunction such that Hd (T x, T y) ≤ ψ(d(x, y)) for all x, y ∈ X. Then T has a unique endpoint if and only if T has approximate endpoint property.
2.
Main results
Now we are ready to state and prove our main results. First, we give the following key result. Lemma 2.1. Let f ∈ L(Ja × Jb ) and α = (α1 , α2 ) ∈ (0, 1] × (0, 1]. Then the continuous function u0 ∈ L(Ja × Jb ) is a solution for the fractional partial differential equation (2.1)
Dθα u(x, y) = f (x, y)
On a two-variables fractional partial differential inclusion...
47
with boundary conditions (Iθ1−α u)(x, 0) = λ1 φ(x) and (Iθ1−α u)(0, y) = λ2 γ(y) if and only if u0 is a solution for the fractional integral equation (2.2) u(x, y) =
λ2 xα1 −1 α2 λ1 y α2 −1 α1 (Iθ γ)(y) ˙ + (I φ)(x) ˙ + (Iθα f )(x, y). Γ(α1 ) Γ(α2 ) θ
Proof. Let u0 be a solution for the fractional partial differential equation (2.1). 2 Then, we have Dxy (Iθ1−α u0 )(x, y) = f (x, y) and so (Iθ1−α u0 )(x, y) − (Iθ1−α u0 )(x, 0) − (Iθ1−α u0 )(0, y) + (Iθ1−α u0 )(0, 0) = (Iθ1 f )(x, y). Using the boundary conditions we get (Iθ1−α u0 )(x, y) − λ1 φ(x) − λ2 γ(y) + λ1 φ(0) = (Iθ1 f )(x, y) and so (Iθ1−α u0 )(x,(y) − (Iθ1 f )(x, y) = λ1 φ(x) +)λ2 γ(y). Since (Iθ0 u0 )(x, y) =
u0 (x, y), we obtain Iθ1−α (u0 (x, y)−(Iθα f )(x, y)) (x, y) = λ1 φ(x)+λ2 γ(y). On the other hand, we have [ ] ( ) Iθα Iθ1−α (u0 (x, y) − (Iθα f )(x, y))(x, y) (x, y) = Iθα λ1 φ(x) + λ2 γ(y)
and so ( ) ( ) Iθ1 u0 (x, y) − (Iθα f )(x, y) (x, y) = Iθα λ1 φ(x) + λ2 γ(y) = (Iθα p)(x, y). (2.3) But we have
∫ x∫ y λ1 (x − s)α1 −1 (y − t)α2 −1 φ(s)dtds Γ(α1 )Γ(α2 ) 0 0 ∫ x∫ y λ2 (x − s)α1 −1 (y − t)α2 −1 γ(t)dtds. + Γ(α1 )Γ(α2 ) 0 0 ∫ x ∫ y Since φ(x) = φ(s)ds ˙ and γ(y) = γ(t)dt, ˙ we obtain (Iθα p)(x, y) =
0
(Iθα p)(x, y)
= + = + = + =
λ1 Γ(α1 )Γ(α2 )
∫
x
∫
0 y
(x − s) 0
0
α1 −1
(y − t)
α2 −1
(∫ 0
s
) φ(τ ˙ )dτ dtds
∫ x∫ y (∫ t ) λ2 α1 −1 α2 −1 (x − s) (y − t) γ(τ ˙ )dτ dtds Γ(α1 )Γ(α2 ) 0 0 0 ∫ x ∫ s λ1 y α2 (x − s)α1 −1 ( φ(τ ˙ )dτ )ds Γ(α1 )Γ(α2 + 1) 0 0 ∫ y ∫ t λ2 xα1 (y − t)α2 −1 ( γ(τ ˙ )dτ )dt Γ(α1 + 1)Γ(α2 ) 0 0 ∫ ∫ x( s ) λ1 y α2 1 (s − τ )α1 −1 φ(τ ˙ )dτ ds Γ(α2 + 1) 0 Γ(α1 ) 0 ∫ y( ∫ t ) λ2 xα1 1 (t − τ )α2 −1 γ(τ ˙ )dτ dt Γ(α1 + 1) 0 Γ(α2 ) 0 ∫ x ∫ y λ1 y α2 λ2 xα1 α1 (I φ)(s)ds (I α2 γ)(t)dt. ˙ + ˙ Γ(α2 + 1) 0 θ Γ(α1 + 1) 0 θ
48
S. Etemad, Sh. Rezapour ∫ x Since (Iθα1 φ)(x) ˙ ∈ L(Ja ) and (Iθα2 γ)(y) ˙ ∈ L(Jb ), the functions (Iθα1 φ)(s)ds ˙ 0 ∫ y 2 and (Iθα2 γ)(t)dt ˙ are absolutely continuous and so there exists Dxy (Iθα p)(x, y) 0
2 for almost all (x, y) ∈ Ja × Jb . By applying the operator Dxy on both sides of [ ( ) ] [ ] 2 1 α 2 α (2.3), we get Dxy Iθ u0 (x, y) − (Iθ f )(x, y) (x, y) = Dxy (Iθ p)(x, y) . Thus,
u0 (x, y) − (Iθα f )(x, y) ∫ y ] [ λ y α2 ∫ x λ2 xα1 1 ˙ + ˙ (Iθα1 φ)(s)ds (Iθα2 γ)(t)dt Γ(α2 + 1) 0 Γ(α1 + 1) 0 [ λ α y α2 −1 ∫ x ] λ 2 x α1 1 2 = Dx (Iθα2 γ)(y) ˙ (Iθα1 φ)(s)ds ˙ + Γ(α2 + 1) 0 Γ(α1 + 1)
2 = Dxy
=
λ1 y α2 −1 α1 λ2 xα1 −1 α2 (Iθ φ)(x) ˙ + (I γ)(y). ˙ Γ(α2 ) Γ(α1 ) θ
λ2 xα1 −1 α2 λ1 y α2 −1 α1 (Iθ φ)(x) ˙ + (I γ)(y) ˙ + (Iθα f )(x, y). This Γ(α2 ) Γ(α1 ) θ shows that u0 is a solution of the fractional integral equation (2.2). Now, let u0 be a solution for the fractional integral equation (2.2). Then, (Iθ1−α u0 )(x, y) = [ λ y α2 −1 ] λ2 xα1 −1 α2 ˙ 1 Iθ1−α (Iθα1 φ)(x)+ ˙ (Iθ ψ)(y) (x, y)+(Iθ1 f )(x, y). On the other Γ(α2 ) Γ(α ) ∫ 1 1 Γ(z)Γ(w) hand by using B(z, w) = (1 − x)w−1 xz−1 dx = , we get Γ(z + w) 0 Hence, u0 (x, y) =
Iθ1−α ∫
∫
[ λ xα1 −1 ] 2 (Iθα2 γ)(y) ˙ (x, y) Γ(α1 )
) (x − s)−α1 (y − t)−α2 ( sα1 −1 α2 (Iθ γ)(t) ˙ dtds Γ(1 − α1 )Γ(1 − α2 ) Γ(α1 ) 0 0 ∫ x∫ y λ2 = (x − s)−α1 sα1 −1 (y − t)−α2 (Iθα2 γ)(t)dtds ˙ Γ(α1 )Γ(1 − α1 )Γ(1 − α2 ) 0 0 ∫ x ∫ y ( ) 1 λ2 (x − s)−α1 sα1 −1 ds (y − t)−α2 (Iθα2 γ)(t)dt ˙ = Γ(α1 )Γ(1 − α1 ) 0 Γ(1 − α2 ) 0 ( ) λ2 B(α1 , 1 − α1 ) 1−α2 α2 = Iθ (Iθ γ)(t) ˙ (y) Γ(α1 )Γ(1 − α1 ) x
y
= λ2
λ2 (I 1 γ)(y) ˙ = λ2 (γ(y) − γ(0)) = λ2 γ(y) Γ(1) θ [ λ y α2 −1 ] 1 and similarly Iθ1−α (Iθα1 φ)(x) ˙ (x, y) = λ1 φ(x). Thus, Γ(α2 ) =
(2.3)
(Iθ1−α u0 )(x, y) = λ2 γ(y) + λ1 φ(x) + (Iθ1 f )(x, y).
On a two-variables fractional partial differential inclusion...
49
2 By applying the operator Dxy on both sides of (2.3), we obtain
[ ] [ ] 2 2 Dxy (Iθ1−α u0 )(x, y) = Dxy λ2 γ(y) + λ1 φ(x) + (Iθ1 f )(x, y) and so (Dθα u0 )(x, y) = f (x, y). By using (2.3), we get (Iθ1−α u0 )(x, 0) = λ2 γ(0) + λ1 φ(x) + (Iθ1 f )(x, 0) = λ1 φ(x) and (Iθ1−α u0 )(0, y) = λ2 γ(y) + λ1 φ(0) + (Iθ1 f )(0, y) = λ2 γ(y). This completes the proof. Consider the Banach space X = C(Ja × Jb , R) endowed with the norm ∥u∥ = sup(x,y)∈Ja ×Jb |u(t)|. For u ∈ X, define the set of selections of F by SF,u := {v ∈ L1 (Ja ×Jb , R) : v(x, y) ∈ F (x, y, u(x, y)) for almost all (x, y) ∈ Ja ×Jb }.
It has been proved that SF,u ̸= ∅ for all u ∈ C(Ja × Jb , X) ([11]). We say that u ∈ X is a solution for the boundary value problem (1.1)-(1.2) whenever it satisfies the boundary value conditions (1.2) and also there is a function v ∈ L1 (Ja × Jb , R) such that v(x, y) ∈ F (x, y, u(x, y)) for all (x, y) ∈ Ja × Jb and λ2 xα1 −1 α2 λ1 y α2 −1 α1 u(x, y) = (Iθ γ)(y) ˙ + (I φ)(x) ˙ Γ(α1 ) Γ(α2 ) θ ∫ x∫ y 1 + (x − s)α1 −1 (y − t)α2 −1 v(s, t)dtds Γ(α1 )Γ(α2 ) 0 0 for almost all (x, y) ∈ Ja × Jb . Define the multifunction N : X → P(X) by N (u) = {h ∈ X : h(x, y) 1 + Γ(α1 )Γ(α2 )
∫
x 0
∫
y
λ2 xα1 −1 α2 λ1 y α2 −1 α1 (Iθ γ)(y) ˙ + (I φ)(x) ˙ Γ(α1 ) Γ(α2 ) θ
(x − s)α1 −1 (y − t)α2 −1 v(s, t)dtds} for all (x, y) ∈ Ja × Jb }.
0
Here, we provide our main result. Theorem 2.2. Suppose that ψ : [0, ∞) → [0, ∞) is a nondecreasing upper semi-continuous map such that lim inf t→∞ (t − ψ(t)) > 0 and ψ(t) < t for all t > 0, F : Ja × Jb × R → Pcp (R) is an integrable bounded multifunction such that F (·, ·, u) : Ja × Jb → Pcp (R) is measurable for all u ∈ R. Assume that there exits m ∈ C(Ja × Jb , [0, ∞)) such that Hd (F (x, y, u) − F (x, y, u′ )) ≤
1 m(x, y)ψ(|u − u′ |) Λ1
{ for all (x, y) ∈ Ja ×Jb and u, u′ ∈ R, where Λ1 = ∥m∥
} aα1 bα2 . If Γ(α1 + 1)Γ(α2 + 1) the multifunction N has the approximate endpoint property, then the fractional partial differential inclusion problem (1.1)-(1.2) has a solution.
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S. Etemad, Sh. Rezapour
Proof. First, we prove that the multifunction N has at least one endpoint. Let u ∈ X. Since the multivalued map (x, y) 7→ F (x, y, u(x, y)) is measurable and is closed-value, it has measurable selection and so SF,u is nonempty. Let {pn }n≥1 be a sequence in N (u) with pn → p. For each n, choose vn ∈ SF,un such that λ2 xα1 −1 α2 λ1 y α2 −1 α1 pn (x, y) = (Iθ γ)(y) ˙ + (I φ)(x) ˙ Γ(α1 ) Γ(α2 ) θ ∫ x∫ y 1 + (x − s)α1 −1 (y − t)α2 −1 vn (s, t)dtds Γ(α1 )Γ(α2 ) 0 0 for all (x, y) ∈ Ja × Jb . Since the operator F is compact, the sequence {vn }n≥1 has a subsequence converging to some v ∈ L1 (Ja × Jb ). We denote this subsequence again by {vn }n≥1 . It is easy to see that v ∈ SF,u and p(x, y) =
+
λ1 y α2 −1 α1 λ2 xα1 −1 α2 (Iθ γ)(y) ˙ + (I φ)(x) ˙ Γ(α1 ) Γ(α2 ) θ
1 Γ(α1 )Γ(α2 )
∫
x
∫
0
y
(x − s)α1 −1 (y − t)α2 −1 v(s, t)dtds
0
for all (x, y) ∈ Ja × Jb . This shows that p ∈ N (u) and so N (u) is closed. Note that, N (u) is bounded because F has a compact values. Now, we show that Hd (N (u), N (w)) ≤ ψ(∥u − w∥) for all u, w ∈ X. Let u, w ∈ X and h1 ∈ N (w). Choose v1 ∈ SF,w such that h1 (x, y) = 1 + Γ(α1 )Γ(α2 )
λ2 xα1 −1 α2 λ1 y α2 −1 α1 (Iθ γ)(y) ˙ + (I φ)(x) ˙ Γ(α1 ) Γ(α2 ) θ ∫
x
∫
0
y
(x − s)α1 −1 (y − t)α2 −1 v1 (s, t)dtds
0
for almost all (x, y) ∈ Ja × Jb . By using the hypothesis, we have Hd (F (x, y, u(x, y)) − F (x, y, w(x, y))) ≤
1 m(x, y)ψ(|u(x, y) − w(x, y)|) Λ1
and so we can choose z ∈ F (x, y, u(x, y)) such that |v1 (x, y) − z| ≤
1 m(x, y)ψ(|u(x, y) − w(x, y)|). Λ1
Define the multivalued map U : Ja × Jb → P(R) by U (x, y) = {z ∈ R : |v1 (x, y) − z| ≤
1 m(x, y)ψ(|u(x, y) − w(x, y)|). Λ1
( ) 1 mψ |u − w| are measurable, the multifunction U (·, ·) ∩ Λ1 F (·, ·, u(·, ·)) is measurable. Hence, there exists v2 (x, y) ∈ F (x, y, u(x, y)) such Since v1 and η =
On a two-variables fractional partial differential inclusion...
51
1 m(x, y)ψ(|u(x, y) − w(x, y)|). Now, consider the Λ1 element h2 ∈ N (u) defined by
that |v1 (x, y) − v2 (x, y)| ≤
h2 (x, y) = +
1 Γ(α1 )Γ(α2 )
λ2 xα1 −1 α2 λ1 y α2 −1 α1 (Iθ γ)(y) ˙ + (I φ)(x) ˙ Γ(α1 ) Γ(α2 ) θ ∫ x∫ y (x − s)α1 −1 (y − t)α2 −1 v2 (s, t)dtds 0
0
for all (x, y) ∈ Ja × Jb . Put sup(x,y)∈Ja ×Jb |m(x, y)| = ∥m∥. Then, we have λ xα1 −1 λ1 y α2 −1 α1 2 |h1 (x, y) − h2 (x, y)| ≤ (Iθα2 γ)(y) ˙ + (I φ)(x) ˙ Γ(α1 ) Γ(α2 ) θ ∫ x∫ y 1 + (x − s)α1 −1 (y − t)α2 −1 v1 (s, t)dtds Γ(α1 )Γ(α2 ) 0 0 λ1 y α2 −1 α1 λ2 xα1 −1 α2 (Iθ γ)(y) ˙ − (I φ)(x) ˙ Γ(α1 ) Γ(α2 ) θ ∫ x∫ y 1 − (x − s)α1 −1 (y − t)α2 −1 v2 (s, t)dtds Γ(α1 )Γ(α2 ) 0 0 ∫ x∫ y 1 ≤ (x − s)α1 −1 (y − t)α2 −1 |v1 (s, t) − v2 (s, t)|dtds Γ(α1 )Γ(α2 ) 0 0 { } 1 x α1 y α2 ≤ ∥m∥ψ(∥u − w∥) Λ1 Γ(α1 + 1)Γ(α2 + 1) −
=
Λ1 ψ(∥u − w∥) = ψ(∥u − w∥) Λ1
and so ∥h1 − h2 ∥ = sup(x,y)∈Ja ×Jb |h1 (x, y) − h2 (x, y)| = ψ(∥u − w∥). Hence, Hd (N (u), N (w)) ≤ ψ(∥u − w∥) for all u, w ∈ X. Since the multifunction N has approximate endpoint property according to Theorem 1.1, there exists u∗ ∈ X such that N (u∗ ) = {u∗ }. It is easy to check that u∗ is a solution for the fractional partial differential inclusion problem (1.1)-(1.2). For illustration of our main result, we give the following example. Example 2.3. Consider the fractional partial differential inclusion [ ] 0.03xy| sin u(x, y)| α Dθ u(x, y) ∈ 0, 1 + | sin u(x, y)| with boundary value conditions (Iθ1−α u)(x, 0) = 0.1(ex −1) and (Iθ1−α u)(0, y) = 0.01y 2 , where (x, y) ∈ [0, 1] × [0, 1]. Let α = (α1 , α2 ) with α1 , α2 ∈ (0, 1], λ1 = 0.1 and λ2[= 0.01. Define the]multifunction F : [0, 1] × [0, 1] × R → P(R) 0.03xy| sin z(t)| . If m : [0, 1] × [0, 1] → [0, ∞) is defined by by F (x, y, z) = 0, 1 + | sin z(t)|
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S. Etemad, Sh. Rezapour
3 t 3 xy, then ∥m∥ = . Consider the map ψ(t) = . It is clear that 100 100 2 ψ is nondecreasing, upper semi-continuous on (0, 1], lim inf t→∞ (t − ψ(t)) > 0 1 and ψ(t) < t for all t > 0. Since Γ(αi + 1) < , we get 2 { } ( ) α1 α2 a b 3 1 Λ1 = ∥m∥ = < 0.12. Γ(α1 + 1)Γ(α2 + 1) 100 Γ(α1 + 1)Γ(α2 + 1) m(x, y) =
One can easily check that Hd (F (x, y, u1 ) − F (x, y, u2 )) ≤
( ) 1 m(x, y)ψ |u1 − u2 | . Λ1
Put X = CR ([0, 1] × [0, 1]). Define N : X → P(X) by N (u) = {h ∈ X : there exists v ∈ SF,u such that h(x, y) = w(x, y) for all (x, y) ∈ [0, 1] × [0, 1]}, where
0.1y α2 −1 α1 0.01xα1 −1 α2 (Iθ γ)(y) ˙ + (Iθ φ)(x) ˙ Γ(α1 ) Γ(α2 ) ∫ x∫ y 1 + (x − s)α1 −1 (y − t)α2 −1 v(s, t)dtds. Γ(α1 )Γ(α2 ) 0 0 w(x, y) =
Since supu∈N (0) ∥u∥ = 0, inf u∈X sups∈N (u) ∥u − s∥ = 0 and so N has the approximate endpoint property. Now by using Theorem 2.2, we conclude that the above fractional partial differential inclusion problem has a solution.
Acknowledgement Research of the authors was supported by Azarbaijan Shahid Madani University.
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Received by the editors December 14, 2015 First published online July 13, 2016
