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Controllability of Fractional Chapter | 8  237


                One needs to show that Φ satisfies all the conditions of Lemma 8.4.For
             the sake of convenience, subdivide the proof into four steps.
                Step 1. For each xAB; Φ is convex.
                In fact, if Ψ 1 ; Ψ 2 AΦðxÞ; then for each tAJ; there exists v 1 ; v 2 AN F;x such
             that
              Ψ i ðtÞ5E q2p ðAt q2p Þx 0 2At q2p E q2p;q2p11 ðAt q2p Þx 0 1tE q2p;2 ðAt q2p Þx 0
                                                                0
                                            "
                     t
                    ð
                   1    t2s   q21  E q2p;q ðAðt2sÞ q2p  ÞB B E q2p;q ðA ðT2sÞ q2p ÞW 21


                     0
                     (
                      ðEx 1 2E q2p ðAT q2p Þx 0 1AT q2p E q2p;q2p11 ðAT q2p Þx 0 2TE q2p;2 ðAT q2p Þx 0
                                                                            0
                   ð T
                          q21          q2p
                 2 ðT2sÞ    E q2p;q ðAðT2sÞ  Þ
                    0
                                           )#
                          ð s                       ð t
                 3 v i ðsÞ1  Gðθ;xðθÞÞdwðθÞ ds  ðsÞds1 ðt2sÞ q21 E q2p;q ðAðt2sÞ q2p Þ
                           0                         0
                          ð s
                 3 v i ðsÞ1  Gðθ;xðθÞÞdwðθÞ ds; i51;2:
                           0
                Let 0 # λ # 1; then for each tAJ; one can have


              ½λΨ 1 1ð12λÞΨ 2 ŠðtÞ5E q2p ðAt q2p Þx 0 2At q2p E q2p;q2p11 ðAt q2p Þx 0 1tE q2p;2 ðAt q2p Þx 0 0
                                                      "
                                ð  t

                              1   ðt2sÞ q21  E q2p;q ðAðt2sÞ q2p ÞB B E q2p;q ðA ðT2sÞ q2p ÞW 21

                                 0
                                 (
                                  Ex 1 2E q2p ðAT  q2p Þx 0
                              1AT q2p E q2p;q2p11 ðAT q2p Þx 0 2TE q2p;2 ðAT q2p Þx 0 0

                                ð  T
                              2   ðT2sÞ q21  E q2p;q ðAðT2sÞ q2p Þ
                                 0
                                "                                # )#
                                                    ð s
                              3 ðλv 1 ðsÞ1ð12λÞv 2 ðsÞÞ1  Gðθ;xðθÞÞdwðθÞ ds  ðsÞds
                                                     0
                                ð t
                              1   ðt2sÞ q21 E q2p;q ðAðt2sÞ q2p Þ
                                 0

                                                   ð s
                              3 ðλv 1 ðsÞ1ð12λÞv 2 ðsÞÞ1  Gðθ;xðθÞÞdwðθÞ ds:
                                                    0
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