Page 114 - IJOCTA-15-3

U. Arora, S. Singh, V. Vijayakumar, A. Shukla / IJOCTA, Vol.15, No.3, pp.483-492 (2025)
Assume that P be the set of all functions By using this admissible control function, the
z ∈ L 2 (I, Y) determined on I with z(0) = y 0 , state equation (4)-(5) becomes,
c
ν
z(b) = y 1 and the fractional derivative D z ex- c D y(ϱ) =Ay(ϱ) + D z(ϱ) − Az(ϱ)
ν
ν
c
t
ists almost everywhere for ν ∈ (0, 1]. Noting that ϱ ϱ

Z ϱ
P be the set of all feasible trajectories for (1)-(2). − G ϱ, z(ϱ), g(ϱ, α, z(α))dα
0
Definition 6. The system (1)-(2) is called T-
ϱ

Z
controllable provided that for every z ∈ P, ∃ u ∈ + G ϱ, y(ϱ), g(ϱ, α, y(α))dα
L 2 (I, V) such that y(ϱ) fulfills y(ϱ) = z(ϱ) almost 0
everywhere on I. y(0) =y 0 .
Clearly, T-controllability ⇒ Complete con- Let us make the substitution w(ϱ) = y(ϱ) − z(ϱ),
trollability. Then for w(ϱ), we have the subsequent fractional
We show that the trajectory controllability of system
(1)-(2) in the following two cases: Z ϱ
c ν
Case-I: If the control appears linearly: D w(ϱ) =Aw(ϱ) + G ϱ, y(ϱ), g(ϱ, α, y(α))dα
ϱ
0
Let us take a look at the linear control sys- Z ϱ
tem, that is, B(ϱ, v(ϱ)) = b(ϱ)v(ϱ), where b : IR − G ϱ, z(ϱ), g(ϱ, α, z(α))dα ,
and u : IV, then the system (1)-(2) becomes 0
c ν w(0) = 0.
D y(ϱ) =Ay(ϱ) + b(ϱ)v(ϱ)
ϱ
ϱ
Z It follows that the solution of the above system
+ G ϱ, y(ϱ), g(ϱ, α, y(α))dα , with w(0) = 0 is described as
0 Z ϱ
ϱ ∈ I, ν ∈ (0, 1], w(ϱ) = (ϱ − α) ν−1 ˆ
S ν (ϱ − α)
(4) 0
Z α
y(0) = y 0 . (5)
G α, y(α), g(α, β, y(β))dβ
The sufficient conditions to prove the trajectory 0
Z α
controllability of (4)-(5) are as follows:
Assumptions [I] [38] − G α, z(α), g(α, β, z(β))dβ dα.
0
(1) A generates a strongly continuous semi- Thus,
group {S(ϱ) : ϱ ≥ 0} on Y. ν Z ϱ
Mν b
(2) b(ϱ) by no means disappear on I. ∥w(ϱ)∥ ≤ δ 1 ∥y(α) − z(α)∥
(3) G is Lipschitz continuous w.r.t. (i) and Γ(1 + ν) ν 0
α

Z
(ii) argument, that is, ∃ δ 1 > 0 and δ 2 > 0
+ δ 2
g(α, β, x(β))dβ
such that
0
α
Z
∥G(ϱ, y 1 , y 1 )−G(ϱ, x 2 , y 2 )∥ ≤ δ 1 ∥y 1 −x 2 ∥+δ 2 ∥y 1 −y 2 ∥, − g(α, β, z(β))dβ
dα

for all y 1 , x 2 , y 1 , y 2 ∈ Y, ϱ ∈ I. 0 ν Z ϱ
1
(4) g is L -Lipschitz continuous w.r.t. third ≤ Mb (δ 1 ∥y(α)
argument, that is, ∃ γ > 0, ∋, Γ(1 + ν) 0
Z ϱ − z(α)∥ + δ 2 γ∥y(α) − z(α)∥)dα.
∥g(ϱ, α, y(α)) − g(ϱ, α, z(α))∥dα ≤ γ∥y(ϱ)
0 That is,
−z(ϱ)∥, y, z ∈ P, (ϱ, α) ∈ △. Mb ν
∥y(ϱ) − z(ϱ)∥ ≤ (δ 1 + δ 2 γ)
Using these hypotheses, we are able to build Γ(1 + ν)
a control function explicitly to verify T- Z ϱ
controllability of (4)-(5). For showing this, we ∥y(α) − z(α)∥dα.
0
continue like this:
Hence by ‘Gronwall’s inequality’, we get
The existence and uniqueness discussion for
(4)-(5) may be easily proved with the help of Lip- ∥y(ϱ) − z(ϱ)∥ = 0.
schitz continuity of functions G and h, for each Thus, y(ϱ) = z(ϱ), ∀ ϱ ∈ I and which concludes
control v ∈ L 2 (I, V). the T-controllability of (4)-(5).
Assume that z(ϱ) be the considered trajectory Case-II: If the control appears nonlinearly
in P. We extract admissible control v(ϱ) from in (1)-(2):
equation (4)-(5) by In this case, we need the additional assump-
ϱ tions on B, h, and G to prove the trajectory con-
Z
c D z(ϱ) − Az(ϱ) − G ϱ, z(ϱ), g(ϱ, α, z(α))dα
ν
ϱ trollability of (1)-(2) given as:
v(ϱ) = 0 .
b(ϱ) Assumptions [II].
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