J. Eur. Opt. Society-Rapid Publ. 21, 11( 2025) 115
Set IV. d 0 ¼ d 3 ¼
25lc þ 8r 2 ¼ 0; 64br þ 15c 2 ¼ 0: ð71Þ
2rðf � 2g þ 3hÞ 2rðf � 2g þ 3hÞ; d 1 ¼�; cðg � 3hÞ cðg � 3hÞ rffiffiffiffiffiffiffiffiffiffiffiffiffi r d 2 ¼ 0; ¼ 2; ð72Þ g � 3h under the constraint conditions
lcðg � 3hÞ 2 þ hr 2 ðf � 2g þ 3hÞ ¼0; 4brðf � 2g þ 3hÞ 2 � 3c 2 ðg � 3hÞðf � g þ hÞ ¼0; ð73Þ
where r( g � 3h) > 0. Applying these outcomes to( 27) and using( 24) and( 25), the following cases of solutions to equation( 5) are extracted.
Case IV1. If f = 4, g = �4( 1 + m 2), h = 4m 2, K( f)= sn 2( f, m), equation( 5) acquires JEF solution of the form
2
31
2
6 pðx; tÞ ¼ � 2rð5m2 þ 3Þ 1 cð4m 2 þ 1Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
7 4
5 e ið / t 1ðnÞ�x a 1 a Þ; 2
1 þ sn � r n
4m 2 þ1
2
31
2
6 qðx; tÞ ¼k � 2rð5m2 þ 3Þ 1 cð4m 2 þ 1Þ pffiffiffiffiffiffiffiffiffiffiffiffiffiffi
7 4
5 e ið / t 2ðnÞ�x a 2 a Þ; 2
1 þ sn � r n
4m 2 þ1 ð74Þ
provided that r < 0 and c > 0. The conditions( 73) reduce to lcð4m 2 þ 1Þ 2 þ r 2 m 2 ð5m 2 þ 3Þ ¼0; 2brð5m 2 þ 3Þ 2 þ 3c 2 ðm 2 þ 1Þð4m 2 þ 1Þ ¼0: ð75Þ
As m approaches 1, solution( 74) transfers to a soliton solution given by
2
31
2
6 pðx; tÞ ¼ � 16r 1
5c pffiffiffiffiffiffi
7 4
5 e ið / t 1ðnÞ�x a 1 a Þ; 2
1 þ tanh � r 5 n
2
31
2
6 qðx; tÞ ¼k � 16r 1
5c pffiffiffiffiffiffi
7 4
5 e ið / t 2ðnÞ�x a 2 a Þ; 2
1 þ tanh n
� r 5 ð76Þ
which delineates bright soliton as presented in Figure 5, where r < 0 and c > 0. From( 75), we catch
25lc þ 8r 2 ¼ 0; 64br þ 15c 2 ¼ 0: ð77Þ
Case IV2. If f = 4( 1 � m 2), g = �4( 1 � 2m 2), h = �4m 2, K( f)= cn 2( f, m), equation( 5) has JEF solution of the form
2 6 pðx; tÞ ¼ � 2rð8m2 � 3Þ 1 cð5m 2 � 1Þ pffiffiffiffiffiffiffiffiffiffi
7 4
5 e ið / t 1ðnÞ�x a 1 a Þ; 2 r
1 þ cn n
5m 2 �1
2
31
2
6 qðx; tÞ ¼k � 2rð8m2 � 3Þ 1 cð5m 2 � 1Þ pffiffiffiffiffiffiffiffiffiffi
7 4
5 e ið / t 2ðnÞ�x a 2 a Þ: 2 r
1 þ cn n
The conditions( 73) become
5m 2 �1
lcð5m 2 � 1Þ 2 þ r 2 m 2 ð8m 2 � 3Þ ¼0; 2brð8m 2 � 3Þ 2 þ 3c 2 ð2m 2 � 1Þð5m 2 � 1Þ ¼0:
3
1 2 ð78Þ
ð79Þ
As m approaches 1, solution( 78) converts to a soliton solution given by
" # 1
pðx; tÞ ¼ � 5r
2
1 � pffiffiffi
2c 1 þ sech 1 2 e ið / t 1ðnÞ�x a 1 a Þ; r
2 n " # ð80Þ
1 qðx; tÞ ¼k � 5r
2
1 � pffiffiffi
2c 1 þ sech 1 2 e ið / t 2ðnÞ�x a 2 a Þ; r
2 n
which depicts dark soliton as displayed in Figure 6, where r > 0 and c < 0. From( 79), we obtain
16lc þ 5r 2 ¼ 0; 25br þ 6c 2 ¼ 0: ð81Þ
Case IV3. If f = 1, g = 2( 1 � 2m 2), h = 1, KðfÞ ¼ sn2 ðf; mÞ
, equation( 5) admits JEF solution of the ð1cnðf; mÞÞ form 2 h n pffiffiffiffiffiffiffiffiffiffiffiffiffiffi oi1 pðx; tÞ ¼ � 8m2 r 1 þ cn 2 � r
2 n e ið / t
1ðnÞ�x a 1 cð4m 2 þ1Þ 4m 2 a Þ;
þ1 h n pffiffiffiffiffiffiffiffiffiffiffiffiffiffi oi1
8m qðx; tÞ ¼k 2 r 1 þ cn 2 � r
2 n e ið / t
2ðnÞ�x a 2 cð4m 2 þ1Þ 4m 2 a Þ; þ1 ð82Þ provided that r < 0 and c > 0. The conditions( 73) turn to
lcð4m 2 þ 1Þ 2 þ 8r 2 m 2 ¼ 0; 64brm 2 þ 3c 2 ð4m 2 þ 1Þ ¼0: ð83Þ
As m approaches 1, solution( 82) changes to a soliton solution given by h n pffiffiffiffiffiffi
oi1 pðx; tÞ ¼ � 8r 1 þ sech 2 � r 2 5c 5 n e ið / t
1ðnÞ�x a 1 a Þ; h n pffiffiffiffiffiffi
oi1 qðx; tÞ ¼k � 8r 1 þ sech 2 � r 2 5c 5 n e ið / t
2ðnÞ�x a 2 a Þ;
ð84Þ
which illustrates bright soliton as shown in Figure 5, where r < 0 and c > 0. From( 83), wefind
25lc þ 8r 2 ¼ 0; 64br þ 15c 2 ¼ 0: ð85Þ