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Trajectory controllability of integro-differential system of fractional orders in Hilbert spaces
                By using this control, (16)-(17) becomes,         In this case, trajectory controllability of the
                                                              system (11)-(12) can be verified by the same ap-
              c  ν              c  ν
               D y(ϱ) =Ay(ϱ) + D z(ϱ) − Az(ϱ)                 proach discussed in Theorem 1 using the Lips-
                                   ϱ
                 ϱ
                                        ϱ                     chitz continuity of G and h and some additional
                                    Z
                         − G ϱ, z(ϱ),    g(ϱ, α, z(α))dα
                                                              assumptions on B, g, and f given in Assumptions
                                       0
                                        ϱ                     [II] of Section 2.
                                    Z
                         + G ϱ, y(ϱ),    g(ϱ, α, y(α))dα ,
                                       0                      4. Examples
                         ν ∈ (1, 2]
                                ′
                  y(0) =y 0 and y (0) = z 0 .                 In this section, we present numerical examples to
                                                              support and illustrate the theoretical results.
            Fixing w(ϱ) = y(ϱ) − z(ϱ), one can get                Example 1.       Assume that the follow-
              c  ν                                            ing nonlinear integrodifferential system with
               D w(ϱ) =Aw(ϱ)
                 ϱ
                                        ϱ
                                                            controlb(ϱ, v) = v|v|.
                                      Z
                                                                 ν
                         + G ϱ, y(ϱ),    g(ϱ, α, y(α))dα      c D y(ϱ) =a(ϱ)y(ϱ) + B(ϱ, v(ϱ))
                                                                 ϱ
                                       0                                              Z  ϱ
                                        ϱ                               + sin y(ϱ) + 3     x(α)dα , 0 < ν ≤ 1,
                                     Z
                         − G ϱ, z(ϱ),    g(ϱ, α, z(α))dα ,                               0
                                       0
                                                                                                         (18)
                         ν ∈ (1, 2],
                                                                                y(0) = y 0 .             (19)
                                ′
                  w(0) =0 and w (0) = 0.                          The control B(ϱ, v) is continuous and coer-
            By using the semigroup theory, the solution of the  cive. It is easy to show that G and h fulfill all the
            above equation may be given by                    requirements of the Theorem 1. Therefore, the
                                                              system (18) is T-controllable.
                       Z  ϱ
                w(ϱ) =    P ν (ϱ − α) G α, y(α),                  Example 2. Assume that Ω = (0, 1) be the
                        0                                     bounded domain in R   n  with smooth boundary
                         α                                    ∂Ω. Assume that the subsequent fractional sys-
                       Z
                           g(α, β, y(β))dβ  − G α, z(α),
                        0                                     tem
                                                                           2
                       Z  α                                   c  ν      ∂ y            1    2
                           g(α, β, z(β))dβ  dα                  D y(ϱ) =  ∂ϱ 2  + v(y, ϱ) + [sin y(ϱ)
                                                                  ϱ
                                                                                         2
                        0
                                                                          + sin y(ϱ)] in Ω × (0, b), 0 < ν ≤ 1,
            Thus,
                                                                                                         (20)
                      Mb ν−1  Z  ϱ
            ∥w(ϱ)∥ ≤              δ 1 ∥y(α) − z(α)∥
                       Γ(ν)   0                                              z(y, 0) = 0 in Ω,           (21)
                           α
                        
 Z

                   + δ 2  
  g(α, β, x(β))dβ
                        
                                               z(y, ϱ) = 0 in ∂Ω × (0, b).      (22)
                          0
                        α                                                              2           2
                      Z
                                        
                         Let us define A : L (0, 1) → L (0, 1) by
                   −     g(α, β, z(β))dβ  
  dα                       ′′
                                        
                     Aw = w , where
                       0
                                                                                   ′
                      Mb ν−1  Z  ϱ                            D(A) = {w ∈ Y : w, w are absolutely continuous ,
                    ≤           (δ 1 ∥x(s) − z(s)∥ + δ 2 γ∥x(α)
                       Γ(ν)   0
                   − z(α)∥)dα.
            Hence,                                                           w(0) = w(1) = 0}
                                                              and
                                    Mb ν−1
                    ∥y(ϱ) − z(ϱ)∥ ≤       (δ 1 + δ 2 γ)                 +∞
                                     Γ(ν)                               X    2
                                                                  Aw =      k < w, w k > w k ,  w ∈ D(A),
                                  ϱ
                                Z
                                   ∥y(α) − z(α)∥dα.                     k=1
                                 0                                                1/2
                                                              where w k (α) = (2/π)  sin kα,  k = 1, 2, 3 · · · is
            Using “Grownwall’s inequality,” we get            the orthogonal set of eigenfunctions of A. Here,
                                                                                       2
                                                              < w, w k > stands for the L inner product. Addi-
                           ∥y(ϱ) − z(ϱ)∥ = 0.
                                                              tionally, A generates a strongly continuous semi-
            Therefore, y(ϱ) = z(ϱ) for all ϱ ∈ I. This proves  group {S(ϱ), ϱ ≥ 0} in Y presented as
            T-controllability of (16)-(17).                            +∞
                                                                       X
            Case-II: If the control appears nonlinearly        S(ϱ)w =     exp(−k ϱ) < w, w k > w k ,  w ∈ Y,
                                                                                  2
            in (11)-(12):
                                                                       k=1
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