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J. Eur. Opt. Society-Rapid Publ. 21, 11( 2025)
Figure 1. Intensity profile of dark soliton given by solutions( 34),( 42) and( 52) with parameter values a 1 = a 2 = b 1 = b 2 = c 1 = c 2 = x 1 = 0.5, m = 1.25, k = 1.5.( a) 3D plot of dark soliton;( b) | p | 2 with distinct values of a;( c) | q | 2 with distinct values of a;( d) Chirping profile.
Figure 2. Intensity profile of bright soliton given by solutions( 38),( 48) and( 56) with parameter values a 1 = 0.93, a 2 = b 2 = �0.5, b 1 = c 1 = c 2 = 0.5, x 1 = 1.5, m = 0.75, k = 1.5.( a) 3D plot of bright soliton;( b) | p | 2 with distinct values of a;( c) | q | 2 with distinct values of a;( d) Chirping profile.
" pðx; tÞ ¼ � r �pffiffiffiffiffiffiffiffiffi
# 1 2
1 � sech �2rn �pffiffiffiffiffiffiffiffiffi
e ið / t 1ðnÞ�x a 1 a Þ; c 1 þ sech �2rn
" qðx; tÞ ¼k � r �pffiffiffiffiffiffiffiffiffi
# 1 2
1 � sech �2rn �pffiffiffiffiffiffiffiffiffi
e ið / t 2ðnÞ�x a 2 a Þ; c 1 þ sech �2rn ð42Þ
which characterizes dark soliton as plotted in Figure 1, where r < 0 and c > 0. From( 41), we arrive at
Set II.
4lc
þ r 2 ¼ 0; b ¼ 0: ð43Þ
2rðg þ sÞ d 0 ¼� cðg þ 3sÞ; d 1 ¼� 4rh
cðg þ 3sÞ; d 2 ¼ d 3 ¼ 0; sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
¼ 2 � 2r; ð44Þ g þ 3s
under the constraint conditions
lcðg þ 3sÞ 2 þ 2sr 2 ðg þ sÞ ¼0; b ¼ 0; ð45Þ p where s ¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
g 2 � 4fh and r( g + 3s) < 0. Applying these outcomes to( 27) and using( 24) and( 25), the following cases of solutions to equation( 5) are extracted.
Case II1. 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 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2r pðx; tÞ ¼ dn � r n e ið / t
1ðnÞ�x a 1 cðm 2 � 2Þ m 2 a Þ; � 2 sffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi rffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi
2r qðx; tÞ ¼k dn � r n e ið / t
2ðnÞ�x a 2 cðm 2 � 2Þ m 2 a Þ: � 2
where r > 0 and c < 0. The conditions( 45) reduce to
ð46Þ
lcðm 2 � 2Þ 2 þ r 2 ðm 2 � 1Þ ¼0; b ¼ 0: ð47Þ
As m approaches 1, solution( 46) changes to a soliton solution given by sffiffiffiffiffiffiffiffiffiffi pðx; tÞ ¼ � 2r �pffiffiffi
sech r n e ið / 1 ðnÞ�x t a 1 a Þ; c sffiffiffiffiffiffiffiffiffiffi ð48Þ qðx; tÞ ¼k � 2r �pffiffiffi
sech r n e ið / 2 ðnÞ�x t a 2 a Þ; c
which illustrates bright soliton as exhibited in Figure 2, where r > 0 and c < 0. From( 47), we arrive at
l ¼ 0; b ¼ 0: ð49Þ
Case II2. 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