Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 259 https://internationalpubls.com Common Fixed Point Solutions for Pair of Mappings Via −− ),,( G Contraction with Admissibility D. Sattemma1*,2*, Dr. S. Vijaya Lakshmi3 1*Research Scholar, Department of mathematics, Osmania University, Hyderabad-500007, Telangana, India 2*Assistant professor, Department of Mathematics, MALD Government Degree College, Gadwal-509125, Telangana, India 3Professor, Department of Mathematics, University College of Science, Osmania University, Hyderabad-500007, Telangana, India (vijayalakshmi.sandra@gmail.com) *Corresponding Author: satya.dyavathi@gmail.com Article History: Received: 28-04-2024 Revised: 17-06-2024 Accepted: 30-06-2024 Abstract: We explain the idea of contraction for a pair of mappings and propose fixed point theorems in the arrangement of G-metric spaces using contraction. Relevant examples are provided. Keywords: BCMP, G-metric space, quasi-metric space, and b-metric. 1. Introduction In 1922, S. Banach [1] presented the well-known Banach contraction mapping theorem (BCMP) for the complete metric space. Several generalizations of BCMP utilizing various contractive conditions in the framework of metric, G-metric space, quasi-metric space, and b-metric space have been published in the literature by a wide spectrum of mathematicians during the last several decades. Khan [2] employed distance mapping to develop a new contractive condition in fixed point theory, hence expanding the BCP into novel configurations. Abodyeh et al. [21] just suggested a new idea, almost-perfect contraction, to create new contractive conditions in order to modify and expand some well-known fixed point theorems. Samet et al. [22] proposed the idea of  - admissibility. Karapinar et al. [23] developed the concept of triangular admissibility. Abdeljawad [24] extended the concept of  - admissibility to a pair of functions. Chary et al. [33] developed the novel idea if rectangular  -G -admissible mapping in 2021, as well as rectangular  - G -admissible concerning another function  in G -metric space. 2. Preliminaries Chary et al. [ 33] have been established the concepts of G − −admissible mapping; G − −admissible with respect to  ; G − ,  − Cauchy;  ,  -complete G -metric space;  ,  − G - continuous; G − −admissible rectangular mapping; G − −admissible rectangular with respect to  . These concepts are used in this work. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 260 https://internationalpubls.com 3. Main Results First, we will define the following terms. Definition. 3.1. Let JH , be two self-mappings on  and }0{: → +R be a function. The pair ),( JH is called G - - admissible if  ,, and 1),,(  imply 1),,(  JJH and 1),,(  HHJ . Definition 3.2. Let JH , be two self-mappings on  and }0{:, → +R be a functions. The pair ),( JH is called G - ),(  - admissibility if  ,, and ),,(),,(   imply ),,(),,(  JJHJJH  an ).,,(),,(  HHJHHJ  Example 3.3. Define H, J from R to R by 2 =H and     − = 0 0 2 2    if if J . Additionally, define }0{:, → +R and  ++= e),,( and  e=),,( then ),( JH is a pair of G - ),(  - admissibility. Proof. Let  ,, such that ),,(),,(   then  ee ++ . So  ++ and hence , are non - negative real numbers. Therefore ),,(),,(),,( 2222222   JJHeeJJH === ++ imply that ),,(),,(  JJHJJH  . Now, if ,0 then ),,(),,(),,( 2222222   HHJeeHHJ === ++ . While if ,0 then ),,(),,(),,( 2222222   HHJeeHHJ ==−= −++− imply ),,(),,(  HHJHHJ  . Definition 3.4. Let G be a metric on . Let H, J be two self-mappings on  ,  be a perfect self - mapping on }0{+R , }0{:, → +R be functions. We say that the pair ),( JH is an G - ),,(  - contraction if there exists )1,0[k such that  ,, and ),,(),,(   imply )1....(........................................))},,(( 3 1 )),,,(( 3 1 )),,,(( 3 1 )),,,(( )),,,(()),,,(()),,,(( )),,,(()),,,(()),,,((max{)),,((     JJGk JJGkJJGkJJGk HHGkHHGkJJGk JJGkHHGkGkJJHG  Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 261 https://internationalpubls.com )2(........................................))},,(( 3 1 )),,,(( 3 1 )),,,(( 3 1 )),,,(( )),,,(()),,,(()),,,(( )),,,(()),,,(()),,,((max{)),,((     JJGk HHGkHHGkJJGk HHGkHHGkJJGk HHGkJJGkGkHHJG  Example.3.5.Define }0{] 4 1 ,0[] 4 1 ,0[] 4 1 ,0[: → +RG by ||||||),,(  −+−+−=G and ] 4 1 ,0[] 4 1 ,0[:, →JH by 2 =H and 4 =J . Also define  on }0{}0{ → ++ RR . By    + = 1 )( and }0{] 4 1 ,0[] 4 1 ,0[] 4 1 ,0[:, → +R by  e=),,( and  ++= e),,( then ),( JH is an G - ),,(  - contraction. Proof. Given        4 1 ,0,,  is such that ),,(),,(   then  ++ ee . Therefore, we conclude that 0==  . Since 4 1  , we have )),,(( 3 1 21 2 3 1 21 2 )2())0,0,(()),,(( 2 2 22       GGJJHG = +  + === and )),,(( 3 1 21 2 3 1 21 2 )2())0,0,(()),,(( 4 4 44       GGHHJG = +  + === . So pair ),( JH is an G - ),,(  - contraction. The main outcome of this study is Theorem. 3.6. On the set , let }0{:, → +R be two functions and H, J be two self- mappings on . Assume there exists a metric G on  such that the following hypothesis hold: Let }0{:, → +R be two functions on , with H and J being two self-mappings. Assume there is a metric G on which the following hypothesis holds: 1. ),( G is an −−G , Complete metric space. 2. H and J are −−G , Continuous. 3. )J H, ( is an G - ),,(  - Contraction. 4. )J H, ( is a pair of −−G),(  admissibility. Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 262 https://internationalpubls.com 5. If  ,, satisfy the condition ),,(),,(   and ),,(),,(   then ),,(),,(   6. There exists 0 Such that ),,(),,( 000000  JHJHHJHJHH  and ),,(),,( 000000  HHJHHHJH  Then both mappings H and J have common fixed point. Proof: In considering premise (6), we begin with 0 in such a way that ),,(),,( 000000  JHJHHJHJHH  and ),,(),,( 000000  HHJHHHJH  . Now, Let 01  H= , 12  J= Then ),,(),,( 110110   and ),,(),,( 001001   . In the view of hypothesis (4), we have ),,(),,(),,(),,( 221110110221  == JJHJJH and ),,(),,(),,(),,( 112001001112  == HHJHHJ Again we put 23  H= , Then, hypothesis (4) implies that ),,(),,(),,(),,( 332221221332  == HHJHHJ and ),,(),,(),,(),,( 223112112223  == JJHJJH putting 34  J= and referring to hypothesis (4), we conclude ),,(),,(),,(),,( 443332332443  == JJHJJH and ),,(),,(),,(),,( 334223223334  == HHJHHJ Proceeding in a similar fashion, we create a sequence }{ n with nn H 212  =+ and 1222 ++ = nn H So that Nnnnnnnn  ++++ ),,(),,( 1111  and Nnnnnnnn  ++ ),,(),,( 11  . Based on premise (5), we observe that Nmnmmnmmn  ,),,(),,(  . If there exists Np so that 122 += pp  then 122 += pp H and hence H has fixed point. Based on premise (1), we have ))},,(( 3 1 )),,,(( 3 1 )),,,(( 3 1 )),,,(( )),,,(()),,,(( )),,,(()),,,(( )),,,(()),,,((max{)),,(( )),,(( 221212122 121212222 2212121212 121212121212 2221212212122 222212 pppppp pppppp pppppp pppppp ppppppppp ppp JJGkJJGk JJGkJJGk HHGkHHGk JJGkJJGk HHGkGkJJHG G       +++ +++ ++++ ++++++ ++++ +++  = Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 263 https://internationalpubls.com )),,(( ))},,(( 3 1 )),,,((max{ ))},,(( 3 1 )),,,(( 3 1 )),,,(( 3 1 )),,,(( )),,,(()),,,(( )),,,(()),,,(( )),,,(()),,,((max{ 222212 222212222212 121212222212 2222212122 22122212 222212222212 222222 +++ ++++++ ++++++ ++++ ++ ++++++    ppp pppppp pppppp pppppp pppppp pppppp pppppp Gk JGkGk GkGk GkGk GkGk GkGk HHGkGk        The last inequality is correct only if 0)),,(( 222212 =+++ pppG  . The properties  and G imply that 2212 ++ = pp  . Hence ppp JH 222  == . Thus, H and J have a common fixed point. If there is a natural number p with 2212 ++ = pp  , then 1212 ++ = pp J and hence J has a fixed point. Based on premise (2), we observe that )),,(( ))},,(( 3 1 )),,,((max{ ))},,(( 3 1 )),,,(( 3 1 )),,,(( 3 1 )),,,(( )),,,(()),,,(( )),,,(()),,,(( )),,,(()),,,((max{)),,(( )),,(( 323222 323222323222 121222222222 222212121222 121222222222 121222121212 121222222212222212 323222 +++ ++++++ ++++++ ++++++ ++++++ ++++++ +++++++++ +++    = ppp pppppp pppppp pppppp pppppp pppppp ppppppppp ppp Gk JGkGk JJGkHHGk HHGkJJGk HHGkHHGk JJGkJJGk HHGkGkHHJG G         The last inequality holds only if 0)),,(( 323222 =+++ pppG  . The properties of  and G imply that 3222 ++ = pp  . Hence 121212 +++ == ppp JH  . Thus we conclude that 12 +p is a common fixed point of H and J . As a result, we determine that H and J have a single fixed point i.e 12 +p . Now, presume that's the case Niii  +1 . For }0{Ni , we get Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 264 https://internationalpubls.com ))},,(( 3 1 )),,,(( 3 1 )),,,(( 3 1 )),,,(( )),,,(()),,,(( )),,,(()),,,(( )),,,(()),,,((max{)),,(( )),,(( 221212122 121212222 2212121212 121212121212 2221212212122 222212 iiiiii iiiiii iiiiii iiiiii iiiiiiiii iii JJGkJJGk JJGkJJGk HHGkHHGk JJGkJJGk HHGkGkJJHG G       +++ +++ ++++ ++++++ ++++ +++  = ))},,(( 3 1 )),,,(( 3 1 )),,,(( 3 1 )),,,(( )),,,(()),,,(( )),,,(()),,,(( )),,,(()),,,((max{ 121212222212 2222212122 121212222212 222212222212 1212212122 ++++++ ++++ ++++++ ++++++ ++++ iiiiii iiiiii iiiiii iiiiii iiiiii GkGk GkGk GkGk GkGk GkGk      ))},,(()),,,((max{ })),,,(( 3 1 )),,,(( 3 1 )),,,(()),,,((max{ 22221212122 22222222212 22221212122 +++++ +++++ +++++   iiiiii iiiiii iiiiii GkGk GkJGk GkGk    Thus if )),,(())},,(()),,,((max{ 22221222221212122 ++++++++ = iiiiiiiii GkGkGk  then )),,(()),,(( 222212222212 ++++++  iiiiii GkG  . Since 1k , condition (1) on  implies that 2212 ++ = ii  , a contradiction, therefore )),,(())},,(()),,,((max{ 1212222221212122 +++++++ = iiiiiiiii GkGkGk  Hence )),,(()),,(( 12122222212 +++++  iiiiii GkG  ………….. (3) Using arguments similar to above, we may show that )),,(()),,(( 221212122 iiiiii GkG  −++  ………….. (4) Combining eqn (3) and (4) together, we reach )),,(()),,(( 111 iiiiii GkG  −++  ………….. (5) By recurring eqn (5) n-times, we deduce Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 265 https://internationalpubls.com )6.(....................)),,(( . . . )),,(( )),,(()),,(( 110 112 2 111    Gk Gk GkG n iii iiiiii    −−− −++ On allowing +→n in eqn (6), we get 0)),,((lim 11 =++ +→ iii n G  ……………. (7) Criterion (2) for the role  indicates that 0),,(lim 11 =++ +→ iii n G  ………….. (8) We seek confirmation that }{ i is a G-Cauchy sequence in . Take Nji , with ij  . We categorize the documentation into four separate instances. Case 1: i is odd integer and j is even integer. Therefore, there exists Nl and an odd integer ‘ h ’ such that 12 += li and hlhij ++=+= 12 . Since ),,(),,( jjijji   we have n ))},,(( 3 1 )),,,(( 3 1 )),,,(( 3 1 )),,,(( )),,,(()),,,(( )),,,(()),,,(( )),,,(()),,,((max{ )),,(()),,(()),,(( 222222 222222 222222 222222 222222 222121212 hlhlhlhlhlhl hlhlllll llhlhlhlhl hlhlhlhlhlhl lllhlhll hlhllhlhlljji JJGkJJGk JJGkJJGk HHGkHHGk JJGkJJGk HHGkGk JJHGGG ++++++ ++ ++++ ++++++ ++ +++++++  ==       ),,(( 3 1 )),,,(( 3 1 )),,,(( 3 1 )),,,(( )),,,(()),,,(( )),,,(()),,,(( )),,,((,)),,((max{ 1212212122 1212212122 1212212122 1212212122 12122 12 2 11 ++++++++++ ++++++ ++++++++ ++++++++++ ++ −+ = ++ hlhlhlhlhlhl hlhlllll llhlhlhlhl hlhlhlhlhlhl lll hl ls sss GkGk GkGk GkGk GkGk GkGk      Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 266 https://internationalpubls.com ),,(( 3 1 )),,,(()),,,(( )),,,((,)),,((max{ 12122 1212212122 12122 12 2 11 ++++ ++++++++ ++ −+ = ++ hlhll llhlhlhlhl lll hl ls sss Gk GkGk GkGk    })),,(( 3 1 ,),,(()),,,(( )),,,((,))),,(((max{ 2 2 11 12 12 1112122 12122 12 2 11    + = ++ −+ += +++++++ ++ −+ = ++ hl ls sss hl ls ssshlhlhl lll hl ls sss Gk GkGk GkGk    })),,(( 3 1 ,),,(()),,,(( )),,,((,)),,((max{ 2 2 11 12 12 1112122 12122 12 2 11    + = ++ −+ += +++++++ ++ −+ = ++ hl ls sss hl ls ssshlhlhl lll hl ls sss Gk GkGk GkGk    )),,(( )),,,(()),,,(( 1 max{ )),,(( )),,,((,))),,(((max{ 222 222110 12 12122 12122 2 11 hlhlhl lll l hlhlhl lll ls sss HHGk HHGkG k k Gk GkGk +++ + +++++ ++ + = ++ −        By allowing +→ji, in the previous solutions and applying argument (7), we obtain 0)),,((lim , = +→ jji ji G  The properties of  imply that 0),,(lim , = +→ jji ji G  ………….. (9) Case 2: i and j are both even integers. Applying the rectangular inequality of the metric ‘G’, we have ),,(),,(),,( 111 jjiiiijji GGG  +++ + Letting +→i and in view of eqn (8) and (9), we get 0),,(lim , = +→ jji ji G  . Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 267 https://internationalpubls.com Case3: i is even integer and j is an odd integer. Applying the rectangular inequality of the metric G, we have ),,(),,(),,(),,( 111111 jjjjjiiiijji GGGG  −−−+++ ++ Letting +→i and considering eqn (8) and (9), we get 0),,(lim , = +→ jji ji G  Case4: i and j are both odd integer, applying the rectangular inequality of the G-metric, we have ),,(),,(),,( 111 jjjjjijji GGG  −−− + On permitting +→i and in view of eqn (8) and (9) 0),,(lim , = +→ jji ji G  After putting everything together, we may decide that 0),,(lim , = +→ jji ji G  Thus, we conclude that }{ i is a G-Cauchy sequence in  then. −−G , Completeness of metric space ),( G ensure that there is  so that  →i using −−G , Continuous of the mappings H and J , we deduce that  HH ii →=+ 212 and  JJ ii →= ++ 1222 . By unique of limit, we obtain  == JH . Thus,  is a fixed point of H and J . EXAMPLE: Define ),0[),0[),0[),0[: +→+++G By    ==  =    if if G ,0 },,,max{ ),,( Let H , J be two self-mappings on ),0[ + defined by  2 2 1 SinH = and  2 4 1 SinJ = . In addition, define the function ),0[),0[: +→+ by    + = 1 )( . Furthermore, we define the function ),0[:, +→ by      = ++ 111,0 ]1,0[,, ),,(     ororif ife And Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 268 https://internationalpubls.com      = 111,1 ]1,0[,, ),,(     ororif ife Then 1.  is a perfect function 2. there exists 0 ),,(),,( 0 2 0 2 00 2 0 2 0  HHHHHH  and ),,(),,( 000 2 000 2  HHHHHH  3. ),( JH is a pair of −−G),(  admissibility 4. H and J are −−G),(  Continuous 5. G, is an −−G , Complete metric space 6. ),( JH is anG - ),,(  - Contraction Proof: It is an easy matter to see eqn’s (1)-(3). Prove eqn (4), let ( i ) be any sequence in ),0[ + such that ),0[ +→i and ),,(),,( 1111 ++++  iiiiii  for all Ni . Thus ]1,0[i for all Ni if  =i for any except a finite number, we infer that  HH i → as +→i . If  i for any save that has an infinite number, we observe that 0= . Hence, 0→i in ( |.|],1,0[ ). Therefore  Hi =→ 0}0,0,sin 2 1 max{ 2 in )),,0([ G+ . That is H is −−G),(  Continuous. To prove (5), let }{ i be a G-Cauchy sequence in )),,0([ G+ such that ),,(),,( 1111 ++++  iiiiii  then ]1,0[i for all Ni if there exists ]1,0[ such that  =i for all but finitely many, then  →i as +→i now suppose the elements of }{ i are distinct for all but finitely many. Given 0 since }{ i is a G-Cauchy sequence in )),,0([ G+ , then there exists Ni 0 such that  },,max{ jji for all 0iij  Therefore,  }0,0,max{ i for 0ii  so, 0→i in )),,0([ G+ Thus, )),,0([ G+ is an −−G , Complete metric space To prove (6), let  ,, such that ),,(),,(   .then ]1,0[,,  so, Communications on Applied Nonlinear Analysis ISSN: 1074-133X Vol 31 No. 8s (2024) 269 https://internationalpubls.com } 4 1 , 4 1 , 2 1 max{1 } 4 1 , 4 1 , 2 1 max{ }) 4 1 , 4 1 , 2 1 (max{ )) 4 1 , 4 1 , 2 1 (()),,(( 222 222 222 222     SinSinSin SinSinSin SinSinSin SinSinSinGJJHG + = = = }1 2 1 , 2 1 ,max{2 } 2 1 , 2 1 ,max{ 222 222   SinSinSin SinSinSin + = )),,(( 5 4 },,max{1 },,max{ 5 4 ,,max{1 },,max{ 5 4 222 222      G SinSinSin SinSinSin =      +        +  ),,(( 15 4 )),,,(( 15 4 )),,,(( 15 4 )),,,(( 5 4 )),,,(( 5 4 )),,,(( 5 4 )),,,(( 5 4 )),,,(( 5 4 )),,,(( 5 4 )),,,(( 5 4 max{     JJGJJG JJGJJGHHG HHGJJGJJG HHGG Similarly, we can show that ))},,(( 15 4 )),,,(( 15 4 )),,,(( 15 4 )),,,(( 5 4 )),,,(( 5 4 )),,,(( 5 4 )),,,(( 5 4 )),,,(( 5 4 )),,,(( 5 4 )),,,(( 5 4 max{)),,((     JJG HHGHHGJJG HHGHHGJJG HHGJJGGHHJG  Hence, H and J Satisfy definition 3.4 for 5 4 =k Therefore, H , J Satisfy all the conditions of theorem 3.6 . 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