Page 115 - IJOCTA-15-3

Trajectory controllability of integro-differential system of fractional orders in Hilbert spaces
(1) B, h, and G fulfill Caratheadory condi- y(0) = y 0 . (7)
tions, that is, B(ϱ, ·) : V 7→ Y is continu- We define B 1 (ϱ) by
ous for ϱ ∈ I and B(·, y) : I 7→ Y is mea-
ν
c
B 1 (ϱ) = D z(ϱ) − Az(ϱ)
surable for y ∈ V and g(ϱ, α, ·) : Y 7→ Y ϱ
Z ϱ
is continuous ∀(ϱ, α) ∈ ∆ and g(·, ·, y) :
− G ϱ, z(ϱ), g(ϱ, α, z(α))dα
∆ 7→ Y is measurable ∀y ∈ Y and G ful-
0
fills Caratheadory conditions identical to
for the given trajectory z(ϱ) ∈ P.
h.
Along with this control, (6) can be wriiten as,
(2) B, g, and G satisfy subsequent character-
c ν c ν
istics: D y(ϱ) =Ay(ϱ) + D z(ϱ) − Az(ϱ)
ϱ
ϱ
ϱ
Z
∥B(ϱ, v)∥ Y ≤ l 0 (ϱ) + l 1 ∥v∥ V , ∀ v ∈ V, ϱ ∈ I,
− G ϱ, z(ϱ), g(ϱ, α, z(α))dα
∀ ϱ ∈ I, y ∈ Y, 0
∥g(ϱ, α, y)∥ ≤ m 0 (ϱ) + m 1 ∥y∥ Y
ϱ
Z
∥G(ϱ, y, z)∥ Y ≤ n 0 (ϱ) + n 1 ∥y∥ Y + n 2 ∥z∥ Y . + G ϱ, y(ϱ), g(ϱ, α, y(α))dα ,
0
(3) B fulfills monotonicity and coercivity con- y(0) =y 0 .
ditions, i.e.,
Setting w(ϱ) = y(ϱ) − z(ϱ), one can get
⟨B(ϱ, v)−B(ϱ, w), v−w⟩ ≥ 0 ∀ v, w ∈ V, ϱ ∈ I
ϱ
Z
ν
and c D w(ϱ) =Aw(ϱ) + G ϱ, y(ϱ), g(ϱ, α, y(α))dα
ϱ
⟨B(ϱ, v), v⟩ 0
lim = ∞. Z ϱ
∥u∥→∞ ∥v∥ − G ϱ, z(ϱ), g(ϱ, α, z(α))dα ,
0
Theorem 1. Under assumptions [I] (8)
((i),(iii),(iv)) and [II], the nonlinear system (1)- w(0) = 0. (9)
(2) is T-controllable.
From semigroup theory, the solution of (8)-(9)
Proof. The existence and uniqueness of the non- given by
linear system (1)-(2) can be proved by employing Z ϱ ν−1 ˆ
Lipschitz continuity of G and g for each fixed v w(ϱ) = (ϱ − α) S ν (ϱ − α)
0
and the solution fulfills Z α
ϱ G α, y(α), g(α, β, y(β))dβ
Z
S ν (ϱ − α)
y(ϱ) =S ν (ϱ)y 0 + (ϱ − α) ν−1 ˆ 0
0 Z α
B(α, v(α))dα − G α, z(α), g(α, β, z(β))dβ dα.
ϱ
Z 0
S ν (ϱ − α)
+ (ϱ − α) ν−1 ˆ Now taking norm on both sides and using
0
α
Z Gronwall’s inequality just in the previous case,
G α, y(α), g(α, β, y(β))dβ dα. one can attain
0
∥y(ϱ) − z(ϱ)∥ = 0.
Assume that z ∈ P is the given trajectory with
z(0) = y 0 . Our duty is to check the control v Therefore, y(ϱ) = z(ϱ), ∀ ϱ ∈ I. Therefore,
fulfilling the mild solution of the given system equals the
prescribed trajectory z(ϱ), when the control u is
Z ϱ
S ν (ϱ − α)
z(ϱ) =S ν (ϱ)y 0 + (ϱ − α) ν−1 ˆ given by B 1 (ϱ) = B(ϱ, v(ϱ)). Hence to prove the
0 trajectory controllability, it is enough to extract
B(α, v(α))dα v(ϱ) from B 1 (ϱ). For extracting v(ϱ), we define
Z ϱ N : L 2 (I, V) → L 2 (I, Y) by
+ (ϱ − α) ν−1 ˆ
S ν (ϱ − α)
0 (Nv)(ϱ) = B(ϱ, v(ϱ)). (10)

Z α
G α, z(α), g(α, β, z(β))dβ dα. By referring to the assumption [II(i)-(ii)], we ob-
0 serve that the operator N is well-defined, contin-
To show this, we put B 1 (ϱ) = B(ϱ, v(ϱ)) in (1)- uous, and bounded. Using the assumption[II(iii)],
(2), we get we can say that N is monotone and coercive.
c ν A hemi-continuous monotone map is of the type
D y(ϱ) = Ay(ϱ) + B 1 (ϱ)
ϱ
(M) [44]. Hence, by employing Theorem 3.6.9 of
ϱ 44

Z
+ G ϱ, y(ϱ), g(ϱ, α, y(α))dα , (6) Joshi and Bose, we can say that the nonlinear
0 map N is onto. Therefore, there exists a con-
ϱ ∈ I, ν ∈ (0, 1] trol u fulfilling (10). Also, since u ∈ L 2 (I, V),
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