Er, t) 0 from Lemma 5. Hence, F and F 0 0 , implies that F is nondecreasing, F :[ ,0 0]F0 ],0.This, with each other withF[ 0,andFfor any[ 0,F0 ].In consequence, F [ ,00][ 0,0]andmax,:= d.Let be an mis in [ , 0 ]. Then, F n 0 and F d. For any n 0 0 (t,) i [-d, d], there exists a positive constant L0 such that |V (t,)| L0 . Then, for any t1 , t2 i with t1 t2 , we get 1 t tt |F (t) – F (| =( (t) -)-1 V (,)dt-1 t1 t t( ( -) t-V (,)d1 ( (t) -)–1 V (,)d – ( ( -) t1 t1 t( (t) -)-1 V (,)dL0 t ( (t) – (t1)) – ( ( – (t1)) ( 1)- ( (t) – ( (t) – , t tFractal Fract. 2021, five,7 ofwhich converges to zero as t. Let us observe that for t, , t t t |F (t) – F (| 0, when t. Therefore, F n is equicontinuous on all . So, F is reasonably compact on [ , t 0 Therefore, the Arzel Ascoli theorem implies that F is compact on [ , 0 ], and so,0 0 ].FnF,converges. Alternatively, Theorem 1 implies that the sequence from the F -iteration of and 0 converges towards the least as well as the greatest fixed points and of F , respectively. 0 This, in turn, implies that difficulty (three) has extremal solutions , [ , 0 ], which can be 0 obtained using the corresponding iterative sequences and( n) defined in (17) and (18), n respectively. Additionally, we havenn0.This completes the proof. 4. Some Relevant Examples Example 1. Contemplate the following difficulty: 1 C D four , (t) = 2t – 0 where = 1 1 three 2 1 i, = (0, 1], = 0, = (0, 1), = , t1 = 0, t2 = 1, 4 two 4 3 two V (t,) = 2t -2 (t) 1,t i := [0, 1], (21)(0) =3 2 4 I0 (two) 1, 3and V : i R R is offered by 1,for t i, R. We take (t) = 0 as the reduced solution and 0 (t) = 1 2t as the upper remedy 0 of challenge (21), and we take 0 for t i. So, ( H1) of Theorem two holds. Now, we take into account 0 3 instances for : 1 (t) = t, two (t) = 2t , 3 (t) = t. Note that 1 (t) = t and three (t) = t give the Caputo and Katugampola (for = 0.5) derivatives within this example. Together with the data offered in Aztreonam custom synthesis Tables 1 and two, we are able to see from assumption ( H3) that ( – ( a)) = ( 1)three four 13 4 3 two three four three 2 3 four 32- (0)1( ( ))210.690988 1,1(t) = t.=23 -210.648610 1, (t) = 2t .(t) =0.764704 1,t.Fractal Fract. 2021, five,eight of1 ( t)Tables 1 and two show these results. One can see the 2D line plots of (t) and n (t) for the n = t Caputo derivative, 2 (t) = 2t , along with the three (t) = t Katugampola Nitrocefin Purity derivative (for = 0.five) in Figure 1a,b. Furthermore, assumption ( H2) is clearly happy.nTable 1. Numerical outcomes offor (t) = t, 2t ,t in Example 1.n (t) = 2tn 1 two three 4 5 six 7 eight 9 ten 11 12 13 14 15 16t 0.00000 0.06250 0.12500 0.18750 0.25000 0.31250 0.37500 0.43750 0.50000 0.56250 0.62500 0.68750 0.75000 0.81250 0.87500 0.93750 1.n(t)=t(t) =t6.16427 4.28485 six.94358 3.08762 7.27414 1.44744 6.79598 1.96779 7.91490 3.00121 0.40120 six.91232 1.12735 6.93050 two.52051 five.59257 1.1.42292 1.48618 2.16988 1.32280 two.49829 0.99029 two.59341 1.26042 three.16083 0.16361 0.71785 three.19814 0.60722 3.44249 0.25975 3.21189 1.9.07593 six.19447 10.07951 4.40385 ten.38117 2.03591 9.54727 two.73167 10.95387 four.18637 0.62015 9.34005 1.66134 9.21394 three.59983 7.26441 1.Table two. Numerical final results offor (t) = t, 2t ,t in Example 1.nn 1 two three 4 five 6 7 8 9 10 11 12 13 14 15 16t 0.00000 0.06250 0.12500 0.18750 0.25000 0.31250 0.37500 0.43750 0.50000 0.56250 0.62500 0.68750 0.75000 0.81250 0.87500 0.93750 1.(t) = t 06.1643 06.7849 04.4538 11.6960 99.2723 51.8545 102 13.8157 106 97.9661 1012 49.2583 1026 12.4534 1054 79.5978 10108 32.5184 10218 NaN NaN NaN NaN NaN(t) = 2t 01.4229 01.9923 01.4135 02.3487 01.1392 02.5625 01.0813 02.7994 03.4989 1012 88.4591 1038 05.6540.