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(i) The representation formula for gi(��) = gi(x, t, ��), i = one,two, three.We rewrite g1(��) as followsg1(��)=��|��|��Nu��(x,t)�Ŧ���g1(0)+��1��|��|��Ng��[1](x,t)�Ŧ¡�g1(0)+H1,(22)in whichg1(0)=g1(x,t)=u0(x,t),?|��|=0,g��[1](x,t)=u��(x,t),?1��|��|��N.(23)It is similar to g2(��), g3(��); we writeg2(��)=��|��|��Nu��x(x,t)�Ŧ���g2(0)+��1��|��|��Ng��[2](x,t)�Ŧ¡�g2(0)+H2,g3(��)=��|��|��Nu��t(x,t)�Ŧ���g3(0)+��1��|��|��Ng��[3](x,t)�Ŧ¡�g3(0)+H3,(24)whereg2(0)=g2(x,t)=u0x(x,t),g3(0)=g3(x,t)=u0t(x,t),?|��|=0,g��[2](x,t)=u��x(x,t),g��[3](x,t)=u��t(x,t),?1��|��|��N.(25)For Valnemulin HCl eachfixed t +: we have the next.(ii) The representation formula for gi(��) = gi(t, ��), i = four,five, six.

Writesellekchem(26)(27)Similarly,g5(��)=||��|��|��Nu��x(t)�Ŧ�||2=��|��|��N?��|��|��N?u��x(t),u��x(t)?�Ŧ�+��=||u0x(t)||2+��1��|��|��NG��[ux(t)]�Ŧ�?+||��||N+1RN(one)[ux,��]��g5(0)+��1��|��|��Ng��[5](t)�Ŧ�+R5��g5(0)+H5��g5(0)+P5+R5,(28)during which(29)(thirty)whereg6(0)=||u0t(t)||two,g��[6](t)=G��[ut(t)]=��|��|��N?��|��|��N?�Ʀ�+��=��?u��t(t),u��t(t)?,???????????1��|��|��N,R6=||��||N+1RN(1)[ut,��]=��N+1��|��|��2NG��[ut(t)]�Ŧ�.(31)Then(22), (24), (26), (28), and (thirty) imply thatgi(��)?gi(0)=Hi=Pi+Ri=��1��|��|��Ng��[i]�Ŧ�+Ri,Ri=0,?i=1,2,3,??Rj=O(||��||N+1),?j=4,five,six.(32)We also need the next lemmathose.Lemma 4 ��For all �� = (��1,��, ��n) +n, |�� | �� one, thenH��=H1��1H2��2H3��3H4��4H5��5H6��6=��|��|��|��|��N����[��,N,��(1),��,��(6)]�Ŧ�+O(||��||N+1),(33)where����[��,N,��(one),��,��(6)]?=�Ʀ�1��|��1|�ܦ�1N,?��6��|��6|�ܦ�6N�Ʀ�1+?+��6=��TN(��1)[��(one)]��1?TN(��6)[��(six)]��6,��(i)=(�Ҧ�(i)),??�Ҧ�(i)=g��[i],?1��|��|��N,??i=1,two,��,six.

(34)Proof of Lemma four ��We haveH��=H1��1H2��2H3��3H4��4H5��5H6��6=P1��1P2��2P3��3(P4+R4)��4(P5+R5)��5(P6+R6)��6=P1��1P2��2P3��3P4��4P5��5P6��6+P1��1P2��2P3��3?��[(P4+R4)��4(P5+R5)��5(P6+R6)��6?P4��4P5��5P6��6].(35)Note that Rj = O(||��||N+1), j = four,5, six, so(P4+R4)��4(P5+R5)��5(P6+R6)��6?P4��4P5��5P6��6?=(P4+R4)��4(P5+R5)��5(P6+R6)��6???P4��4(P5+R5)��5(P6+R6)��6??+P4��4(P5+R5)��5(P6+R6)��6?P4��4P5��5(P6+R6)��6??+P4��4P5��5(P6+R6)��6?P4��4P5��5P6��6?=[(P4+R4)��4?P4��4](P5+R5)��5(P6+R6)��6??+P4��4[(P5+R5)��5?P5��5](P6+R6)��6??+P4��4P5��5[(P6+R6)��6?P6��6]?=R4[��j=0��4?1(P4+R4)jP4��4?one?j]??��(P5+R5)��5(P6+R6)��6??+R5[��j=0��5?one(P5+R5)jP5��5?one?j]P4��4(P6+R6)��6??+R6[��j=0��6?one(P6+R6)jP6��6?1?j]P4��4P5��5=||��||N+1RN(one)[H,��,��]=O(||��||N+1).

(36)Combining (35) and (36) yieldsH��=H1��1H2��2H3��3H4��4H5��5H6��6=P1��1P2��2P3��3(P4+R4)��4(P5+R5)��5(P6+R6)��6=P1��1P2��2P3��3P4��4P5��5P6��6?+||��||N+1RN(one)[H,��,��]=(��1��|��|��Ng��(1)�Ŧ�)��1?(��1��|��|��Ng��(6)�Ŧ�)��6?+||��||N+1RN(one)[H,��,��]=��|��|��|��|��N����[��,N,��(one),��,��(6)]�Ŧ�?+��N+1��|��|��|��|N����[��,N,��(1),��,��(6)]�Ŧ�?+||��||N+1RN(one)[H,��,��]�ԡ�|��|��|��|��N����[��,N,��(one),��,��(six)]�Ŧ�?+||��||N+1RN(two)[H,��,��],(37)the place||��||N+1RN(two)[H,��,��]??=��N+1��|��|��|��|N����[��,N,��(one),��,��(six)]�Ŧ�???+||��||N+1RN(1)[H,��,��].(38)Lemma four is proved.