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Controllability of Fractional Chapter | 8  239


              ρ,E:Φðx ρ ÞðtÞ: 2
                 5:E q2p ðAt q2p Þx 0 2At q2p E q2p;q2p11 ðAt q2p Þx 0 1tE q2p;2 ðAt q2p Þx 0 0
                   ð t                           ð t
                 1  ðt2sÞ q21  E q2p;q ðAðt2sÞ q2p ÞBu ρ ðsÞds1  ðt2sÞ q21  E q2p;q ðAðt2sÞ q2p Þv ρ ðsÞds
                    0                             0
                  ð  t                    ð  s
                        q21          q2p                    2
                 1 ðt2sÞ  E q2p;q ðAðt2sÞ  Þ  Gðθ;x ρ ðθÞÞdwðθÞ ds:
                   0                      0
                                2
                 #6:E q2p ðAt  q2p Þx 0 : 16:At  q2p E q2p;q2p11 ðAt q2p Þx 0 : 2
                                        ð t
                                  2
                 16:tE q2p;2 ðAt q2p Þx : 16E: ðt2sÞ q21 E q2p;q ðAðt2sÞ q2p ÞBu ρ ðsÞds: 2
                                0
                                0
                                         0
                      ð t
                            q21          q2p      2
                 16E: ðt2sÞ   E q2p;q ðAðt2sÞ  Þv ρ ðsÞds:
                       0
                      ð t                     ð s
                            q21          q2p                   2
                 16E: ðt2sÞ   E q2p;q ðAðt2sÞ  Þ  Gðθ;x ρ ðθÞÞdwðθÞ ds:
                       0                      0
                                   T 2q  2  2    2     2
                 #6n 1 16n 2 16n 3 136  n :B: :B : l E:x 1 : 1n 1 1n 2 1n 3
                                       4
                                   q 2
                   T 2q       T  2q                    2      T  2q
                 1    n 4 ϕ ðtÞ1  Tn 4 M 1 M G 11 supE:x ρ ðtÞ:  16  n 4 ϕ ðtÞ
                                                                   ρ
                         ρ
                                      b
                   q 2        q 2            tAJ              q 2
                    T 2q                     2
                 16    Tn 4 M 1 M G 11 supE:x ρ ðtÞ:
                            b
                    q 2            tAJ
                     0                   1
                           T  2q  2  2    2    T 2q  2  2    2    2
                 #6n 1 116    n :B: :B : l A  136  n :B: :B : lE:x 1 :
                     @
                               4
                                                   4
                            q 2                q 2
                     0                   1     0                   1
                           T  2q   2    2            T  2q   2    2
                               2
                                                         2
                 16n 2 116    n :B: :B : l A  16n 3 116  n :B: :B : l A
                                                         4
                                               @
                     @
                               4
                            q 2                       q 2
                            0                   1    0                  1
                    T  2q         T  2q   2    2           T  2q  2    2
                                                              2
                                      2
                 16    n 4 ϕ ðtÞ 116  n :B: :B : l A  16116  n :B: :B : l A
                          ρ
                                                              4
                                                     @
                                      4
                            @
                    q 2            q 2                     q 2
                   T 2q                     2
                 3    Tn 4 M 1 M G 11 supE:x ρ ðtÞ:  :
                           b
                   q 2             tAJ
                Dividing both sides of the above inequality by ρ and taking the limit as
             ρ-N; one can get
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