J. Eur. Opt. Society-Rapid Publ. 21, 11( 2025) 113 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2r pðx; tÞ ¼m � sn � r n e ið / t
1ðnÞ�x a 1 cðm 2 þ 1Þ m 2 a Þ; þ 1 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2r qðx; tÞ ¼km � sn � r n e ið / t
2ðnÞ�x a 2 cðm 2 þ 1Þ m 2 a Þ; þ 1 ð50Þ
provided that r < 0 and c > 0. The conditions( 45) reduce to lcðm 2 þ 1Þ 2 þ r 2 m 2 ¼ 0; b ¼ 0: ð51Þ
As m approaches 1, solution( 50) reduces to a soliton solution given by rffiffiffiffiffiffiffi pðx; tÞ ¼ � r rffiffiffiffiffiffiffi tanh � r n e ið / t 1ðnÞ�x a 1 a Þ; c
2 rffiffiffiffiffiffiffi qðx; tÞ ¼k � r rffiffiffiffiffiffiffi tanh � r ð52Þ n e ið / t 2ðnÞ�x a 2 a Þ; c
2
which presents dark soliton as shown in Figure 1, where r < 0 and c > 0. From( 51), we arrive at
4lc þ r 2 ¼ 0; b ¼ 0: ð53Þ
Case II3. If f = 1, g = 2( 1 � 2m 2), h = 1, KðfÞ ¼ sn2 ðf; mÞ
, equation( 5) gains JEF solution of the ð1cnðf; mÞÞ 2 form See Equation( 54) at the bottom of this page p where n ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffi m 2 � 1. The conditions( 45) reduce to
lcðm 2 þ n 2 � 6mnÞ 2 � 4r 2 mnðm � nÞ 2 ¼ 0; b ¼ 0: ð55Þ
As m approaches 1, solution( 54) degeneratestoasoliton solution given by
pðx; tÞ ¼ � 4r p
sech 2 ffiffiffi 1
ð r nÞ p c 1 þ sech 2 ffiffiffi
2 e ið / t 1ðnÞ�x a 1 a Þ; ð r nÞ qðx; tÞ ¼k � 4r p
sech 2 ffiffiffi 1
ð r nÞ p c 1 þ sech 2 ffiffiffi
2 e ið / t 2ðnÞ�x a 2 a Þ; ð r nÞ ð56Þ
which describes bright soliton as shown in Figure 2, where r > 0 and c < 0. From( 55), we arrive at
l ¼ 0; b ¼ 0: ð57Þ
Set III. 2rð3f � 2g þ hÞ d 0 ¼ d 2 ¼ 0; d 1 ¼; cðg � 3f Þ rffiffiffiffiffiffiffiffiffiffiffiffiffi 2rð3f � 2g þ hÞ r d 3 ¼�; ¼ 2; ð58Þ cðg � 3f Þ g � 3f under the constraint conditions
lcðg � 3f Þ 2 þ f r 2 ð3f � 2g þ hÞ ¼0; 4brð3f � 2g þ hÞ 2 � 3c 2 ðg � 3f Þðf � g þ hÞ ¼0; ð59Þ
where r( g � 3f) > 0. Applying these outcomes to( 27) and using( 24) and( 25), the following cases of solutions to equation( 5) are extracted.
Case III1. If f = 4, g = �4( 1 + m 2), h = 4m 2, K( f)= sn 2( f, m), equation( 5) attains JEF solution of the form
2 pffiffiffiffiffiffiffiffiffiffiffiffiffi 3 2 1 pðx; tÞ ¼ � 2rð3m2 þ 5Þ sn � r
2
6 n m 2 þ4 cðm 2 þ 4Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi
7 4
5 e ið / t 1ðnÞ�x a 1 a Þ; 2
1 þ sn � r n
m 2 þ4
2 pffiffiffiffiffiffiffiffiffiffiffiffiffi 3 2 1 qðx; tÞ ¼k � 2rð3m2 þ 5Þ sn � r
2
6 n m 2 þ4 cðm 2 þ 4Þ pffiffiffiffiffiffiffiffiffiffiffiffiffi
7 4
5 e ið / t 2ðnÞ�x a 2 a Þ; 2
1 þ sn � r n m 2 þ4 ð60Þ provided that r < 0andc > 0. The conditions( 59) reduce to lcðm 2 þ 4Þ 2 þ r 2 ð3m 2 þ 5Þ ¼0;
2brð3m 2 þ 5Þ 2 þ 3c 2 ðm 2 þ 4Þðm 2 þ 1Þ ¼0: ð61Þ
As m approaches 1, solution( 60) transforms to a soliton solution given by
2 pffiffiffiffiffiffi
3 2 1 pðx; tÞ ¼ � 16r tanh � r 2
5 n 6
5c pffiffiffiffiffiffi
7 4
5 e ið / t 1ðnÞ�x a 1 a Þ; 2
1 þ tanh � r 5 n 2 pffiffiffiffiffiffi
3 2 1 qðx; tÞ ¼k � 16r tanh � r 2
6
5 n 5c pffiffiffiffiffiffi
7 4
5 e ið / t 2ðnÞ�x a 2 a Þ; 2
1 þ tanh n
� r 5 ð62Þ
which represents dark soliton as exhibited in Figure 3, where r < 0 and c > 0. From( 61), we arrive at
2 n pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
o31 4r 1 �ðm 2 2
r
� mnÞ 1 þ cn 2 n m pðx; tÞ ¼ 2 þn 2 �6mn 4
cðm 2 þ n 2 � 6mnÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
5 e ið / t 1ðnÞ�x a 1 a Þ; r
1 þ cn 2 n m 2 þn 2 �6mn
2 n pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
o31 4r 1 �ðm 2 2
r
� mnÞ 1 þ cn 2 n m qðx; tÞ ¼k 2 þn 2 �6mn 4
cðm 2 þ n 2 � 6mnÞ pffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
5 e ið / t 2ðnÞ�x a 2 a Þ; r
1 þ cn 2 n m 2 þn 2 �6mn ð54Þ