We get Equations (12) and (13) for 1 . In what follows, r and are monotonic functions on [ two , ), where two 1 . Take into account the case when r 0, 0 for two . Therefore, for s t two , r (s) (s) r implies that(s)r , r (s)that may be,s(s) r Td . r Due to the fact r is nonincreasing, there C6 Ceramide manufacturer exists a constant C 0 such that r s -C for 2 . Consequently, (s) – C T rd) . For s , it follows that ( 0 – CR for two . Clearly, (i ) CR(i ), i N. So, z F CR and hence z – CR F . Thinking about z – CR 0 we’ve F 0, a contradiction. So, z – CR 0 implies that z CR F F . In addition, z(i ) F (i ), i N. Consequently, Equations (14) and (15) lower to r r ( i ) G ( a) r ( – ) ( i ) G ( a ) r ( i – )( – ) G F ( – 0 (i – ) k G F (i – for three 2 , = i , i N. Integrating the last inequality from three to ( three ), we come across r G ( a) r ( – )( – )tt-3 i r ( i )( i )- G ( a)that’s,three i r ( i – )( i – ) Q G F ( – d 0,Q G F ( – d3 i Hk G F (i – – r – r G ( a) r ( – ) G ( a) r ( – ) ( – ) ( – )t- 1 G ( a) r implies that 1 1 G ( a) r Q G F ( – d three i Hk G F (i – -( ).Further integration on the above inequality, we get that 1 G ( a)u1 r Q G F ( – d 3 i Hk G F (i – d- =- ( three ) u 3 u 3 i u 3 i u( i )[ ( i 0) -(i – 0)]3 i u( i 0).Symmetry 2021, 13,9 ofSince is monotonic and bounded, hence,1 r Q G F ( – d Hk G F (i – i =d ,which contradicts to (H16). The rest of your proof follows in the proof Theorem 1. This completes the proof of your theorem. Theorem 5. Assume that (H1), (H4), (H5), (H9)H12), (H15) and (H18)H21) hold and -1 p 0, R . Then each and every resolution of (S) is oscillatory. Proof. For contrary, let u be a nonoscillatory solution of (S). Then preceding as in the proof of the Theorem 2, we acquire and r are monotonic on [ two , ). If 0 and r 0 for three 2 , then we make use of the same sort of (Z)-Semaxanib Purity & Documentation argument as in Theorem two to obtain that u is bounded, that’s, lim exists. Clearly, z 0. So, -z – F , and therefore, -z F – . So, for- u ( – ) p ( ) u ( – ) z ( ) – F – ( ).Consequently, u( – F – ( – , four 3 and Equations (12) and (13) yield r r ( i ) q G F – ( – 0, = i , i N (i ) h(i ) G F – (i – 0, i Nfor four . Integrating the preceding impulsive program from 4 to , we obtainq G F – ( – d four i h ( i ) G F – ( i – ) -r ( )( ),that is certainly, 1 r q G F – ( – d 4 i h ( i ) G F – ( i – )-( ).From additional integration of the final inequality, we find1 r q G F – ( – d four i h ( i ) G F – ( i – )d which contradicts (H19). If 0 and r 0 for 3 , then following Theorem four, we find z F CR F and z 0, that is, u F . The rest in the proof follows from the proof of Theorem two. Thus, the theorem is proved. Theorem six. Take into account – -b p -1, R . Assume that (H1), (H4), (H5), (H9)H12), (H15), (H20) and (H21)H23) hold. Then every bounded resolution of (S) is oscillatory. Proof. The proof of your theorem follows the proof of Theorem 5. four. Adequate Circumstances for Nonoscillation This section deals together with the existence of optimistic options to show that the IDS (S) has positive resolution. nonincreasing. Theorem 7. Look at p C (R , [-1, 0]) and assume that (H1) holds. If (H24) holds, then the IDS (S) includes a optimistic answer.Symmetry 2021, 13,10 ofProof. (i) Contemplate -1 -b p 0, R where b 0. For (H24), we are able to obtain a = max{, such thatT1 r qd h(i ) d i =1-b . 10G (1)We consider the set M = u : u C ([ – , ), R), u = 0 for [ – , ] and and defin.
