{"id":171783,"date":"2025-12-02T22:30:15","date_gmt":"2025-12-02T22:30:15","guid":{"rendered":"https:\/\/www.newsbeep.com\/ie\/171783\/"},"modified":"2025-12-02T22:30:15","modified_gmt":"2025-12-02T22:30:15","slug":"soliton-dynamics-and-stability-in-resonant-nonlinear-schrodinger-systems-with-cubic-quintic-effects-via-enhanced-modified-extended-tanh-function-method","status":"publish","type":"post","link":"https:\/\/www.newsbeep.com\/ie\/171783\/","title":{"rendered":"Soliton dynamics and stability in resonant nonlinear Schr\u00f6dinger systems with cubic quintic effects via enhanced modified extended tanh function method"},"content":{"rendered":"

Our goal in this part is to get the next form of solutions for Eq. (1<\/a>):<\/p>\n

$$\\begin{aligned} R(x,y,z,t)=R(\\xi ) e^{i(\\omega ~t-\\nu _1~ x-\\nu _2~ y-\\nu _3~ z+\\theta )},~ \\xi =\\eta _1~x+\\eta _2~y+\\eta _3~z-\\gamma ~t, \\end{aligned}$$<\/p>\n

\n (7)\n <\/p>\n

in which \\(\\gamma\\) denotes the velocity of the solitonn along the x, y, and z axes, while the wave numbers are symbolized by \\(\\nu _1\\), \\(\\nu _2\\), and \\(\\nu _3\\). Additionally, \\(\\theta\\) stands for phase constant, and \\(\\omega\\) represents the soliton frequency. Substituting by Eq. (7<\/a>) into Eq. (1<\/a>) then separating into real and imaginary parts, we have:<\/p>\n

The real component:<\/p>\n

$$\\begin{aligned} & s_1~ R(\\xi )^{1-n}+s_2~ R(\\xi )^{1-2n}+s_3~ R(\\xi )^{1-3n} +s_4~ R(\\xi )^{1-4n}+w_1~ R(\\xi )^{1+n}+w_2~ R(\\xi )^{1+2n}+w_3~ R(\\xi )^{1+3n}\\nonumber \\\\ & \\quad +w_4~ R(\\xi )^{1+4n}+(-\\omega +m_1~ \\nu _1^2+2m_4~ \\nu _1 ~ \\nu _2-2m_6~\\nu _3~\\nu _2+m_2~ \\nu _2^2-2m_5~\\nu _3~\\nu _1+m_3~ \\nu _3^2)~R(\\xi )\\nonumber \\\\ & \\quad -R”(\\xi )~(m_1~ \\eta _1^2+2m_4~ \\eta _1~\\eta _2-2m_6~\\eta _3~\\eta _2+m_2~\\eta _2^2-2m_5~\\eta _1~\\eta _3 +m_3~\\eta _3^2)=0. \\end{aligned}$$<\/p>\n

\n (8)\n <\/p>\n

The imaginary component:<\/p>\n

$$\\begin{aligned} & \\gamma -2m_1~\\nu _1~\\eta _1+2m_6~\\nu _2~\\eta _3+2m_5~\\nu _3~\\eta _1+2m_6~\\nu _1~\\eta _2-2m_2~\\nu _2~\\eta _2\\nonumber \\\\ & \\quad -2m_4~(\\nu _2~\\eta _1+\\nu _1~\\eta _2)+2m_5~\\nu _1~\\eta _3-2m_3~\\nu _3~\\eta _3=0. \\end{aligned}$$<\/p>\n

\n (9)\n <\/p>\n

Equalizing the coefficient of Eq. (9<\/a>) to zero provides<\/p>\n

$$\\begin{aligned} \\gamma =2\\eta ~(m_1~\\nu _1+m_4~\\nu _2-m_5~\\nu _3)+2\\eta _2~(-m_6~\\nu _1+m_2~\\nu _2+m_4~\\nu _2)+2\\eta _3~(m_3~\\nu _3-m_5~\\nu _1-m_6~\\nu _2). \\end{aligned}$$<\/p>\n

\n (10)\n <\/p>\n

To implement the IMETF scheme on the real portion, the balance principle has to applied to get the integer k, by comparing the highest-order linear derivative term with the leading-order nonlinear term.<\/p>\n

According to Eq. (9<\/a>), the chain rule applied to the function \\(R(\\xi ) = \\sum _{i=-N}^{N} \\alpha _{i} W(\\xi )^{i}\\) results in terms with dominant behavior of order \\(\\mathbf {W^{N+2}}\\) from the second derivative \\(R”\\). Conversely, terms of leading order \\(\\mathbf {W^{2N}}\\) are produced by the nonlinear term \\(\\mathbf {R^{1+4n}}\\).<\/p>\n

By equating the dominant powers of \\(W(\\xi )\\) in both terms, we get:<\/p>\n

$$\\begin{aligned} k+2=(1+4n)~k\\quad \\implies \\quad k=\\frac{1}{2n}. \\end{aligned}$$<\/p>\n

, so the following transformation can be applied on the real part to get an integer value for k:<\/p>\n

$$\\begin{aligned} R(\\xi ) =U(\\xi )^{\\frac{1}{2n}}. \\end{aligned}$$<\/p>\n

\n (11)\n <\/p>\n

Eq. (8) becomes:<\/p>\n

$$\\begin{aligned} & 4n^2~s_4+4n^2~s_2~U(\\xi )+4n^2~(\\Pi _2-\\omega )~U(\\xi )^2+4n^2~w_2~U(\\xi )^3+4n^2~w_4~U(\\xi )^4+(1-2n)~\\Pi _1~U'(\\xi )^2\\nonumber \\\\ & \\quad +2n~\\Pi _1~U”(\\xi )~U(\\xi )=0, \\end{aligned}$$<\/p>\n

\n (12)\n <\/p>\n

Where<\/p>\n

$$\\begin{aligned} & \\Pi _1=-(m_1~\\eta _1^2+2m_4~\\eta _1~\\eta _2+m_2~\\eta _2^2-2m_5~\\eta _1~\\eta _3-2m_6~\\eta _2~\\eta _3+m_3~\\eta _3^2), \\end{aligned}$$<\/p>\n

\n (13)\n <\/p>\n

$$\\begin{aligned} & \\Pi _2= m_1~\\nu _1^2+2m_4~\\nu _1~\\nu _2+m_2~\\nu _2^2-2m_5~\\nu _1~\\nu _3-2m_6~\\nu _2~\\nu _3+m_3~\\nu _3^2. \\end{aligned}$$<\/p>\n

\n (14)\n <\/p>\n

Balancing \\(U(\\xi )^4\\) with \\(U”(\\xi )~U(\\xi )\\), we get \\(k=1\\) by the balancing procedure that was previously mentioned, \\(4k=k+2+k\\). According to the proposed approach, the solution of Eq.(12<\/a>) will be declared into the next form<\/p>\n

$$\\begin{aligned} U(\\xi )=h_0+h_1 \\rho (\\xi )+j_1 \\rho (\\xi )^{-1}. \\end{aligned}$$<\/p>\n

\n (15)\n <\/p>\n

Substituting by Eq. (15<\/a>) and its Riccati equation Eq. (6<\/a>) into Eq. (12<\/a>) then equating all the coefficients of \\(\\rho ^i\\) to zero, a system of non-linear equations will be resulted that can be resolved using mathematica packages. The results will be as follow:<\/p>\n

Case 1. \\(q_0=q_1=q_3=0\\)<\/p>\n

Result(1)<\/p>\n

$$q_2=\\frac{q_4~(\\omega +2n~\\omega +6w_4~h_0^2-\\Pi _2 -2n\\Pi _2 )}{w_4~h_1^2},~ q_4=\\frac{-4n^2~w_4~h_1^2}{(1+2n)\\Pi _1},~ j_1=0, ~h_0 =\\frac{-(1+2n)w_2}{4(1+n)w_4},$$<\/p>\n

$${-22em} s_2=\\frac{2(-1+n)~w_4~h_0~(2q_4~h_0^2+q_2~h_1^2)}{(1+2n)~q_4}.$$<\/p>\n

The next bright soliton solution can then be elevated<\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\left[ -\\frac{(1+2n)w_2}{4(1+n)w_4}+\\sqrt{\\frac{\\frac{3(1+2n)^2~w_2^2}{8(1+n)^2~w_4}+(1+2n)(\\omega -\\Pi _2)}{w_4}}\\right. \\nonumber \\\\ & \\left. \\quad \\operatorname {sech}\\left( 2(\\eta _1~x+\\eta _2~y+\\eta _3~z-\\gamma ~t) \\sqrt{\\frac{n^2\\left( \\frac{3(1+2n)^2~w_2^2}{8(1+n)^2~w_4}+(1+2n)(\\omega -\\Pi _2)\\right) }{(1+2n)~\\Pi _1}}\\right) \\right] ^{\\frac{1}{2n}}\\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1 x-\\nu _2 y-\\nu _3 z)} \\nonumber \\\\ & ,{\\frac{\\frac{3(1+2n)^2~w_2^2}{8(1+n)^2~w_4}+(1+2n)(\\omega -\\Pi _2)}{w_4}>0,~ \\frac{n^2\\left( \\frac{3(1+2n)^2~w_2^2}{8(1+n)^2~w_4}+(1+2n)(\\omega -\\Pi _2)\\right) }{(1+2n)~\\Pi _1}>0.} \\end{aligned}$$<\/p>\n

\n (16)\n <\/p>\n

The next singular periodic solution can then be elevated<\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\left[ \\frac{-(1+2n)~w_2}{4(1+n)~w_4}+\\sqrt{\\frac{\\frac{3(1+2n)^2~w_2^2}{8(1+n)^2~w_4}+(1+2n)(\\omega -\\Pi _2)}{w_4}}\\right. \\nonumber \\\\ & \\left. \\times \\sec \\left( (\\eta _1~x+\\eta _2~y+\\eta _3~z-\\gamma ~t)~\\sqrt{\\frac{-n^2\\left( \\frac{3(1+2n)^2~w_2^2}{8(1+n)^2~w_4} +(1+2n)(\\omega -\\Pi _2)\\right) }{(1+2n)~\\Pi _1}}\\right) \\right] ^{\\frac{1}{2n}}\\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1~ x-\\nu _2~y-\\nu _3~z)},\\nonumber \\\\ & {\\frac{\\frac{3(1+2n)^2~w_2^2}{8(1+n)^2~w_4}+(1+2n)(\\omega -\\Pi _2)}{w_4}>0,~\\frac{n^2 \\left( \\frac{3(1+2n)^2~w_2^2}{8(1+n)^2~w_4}+(1+2n)(\\omega -\\Pi _2)\\right) }{(1+2n)~\\Pi _1}<0.} \\end{aligned}$$<\/p>\n

\n (17)\n <\/p>\n

Case 2. \\(q_1=q_3=0\\)<\/p>\n

Result(1)<\/p>\n

$$h_1=0,~h_0=\\frac{-(1+2n)~w_2}{4~(1+n)~w_4},~q_2=\\frac{n^2~((-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2)}{2~(1+n)^2~(-1-n+2~n^2)~w_2~w_4~\\Pi _1},~q_4=\\frac{-(1+2n)~q_2^2~\\Pi _1}{16~n^2~w_4~j_1^2}.$$<\/p>\n

The next singular soliton is then elevated<\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\frac{1}{4}\\left[ -\\frac{(1+2n)w_2}{(1+n)w_4}\\right. \\nonumber \\\\ & \\left. + \\frac{\\coth \\left( \\frac{1}{2}(\\eta _1 x+\\eta _2 y+\\eta _3 z-\\gamma t)\\sqrt{-\\frac{n^2((-1+n)(1+2n)^2w_2^3+16(1+n)^3s_2w_4^2)}{(1+n)^2(-1-n+2n^2)w_2w_4\\Pi _1}}\\right) }{\\sqrt{\\frac{(-1+n)(1+n)^2w_2w_4^2}{(-1+n)(1+2n)^2w_2^3+16(1+n)^3s_2w_4^2}}}\\right] ^{\\frac{1}{2n}}\\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1 x-\\nu _2 y-\\nu _3 z)}, \\quad q_0=\\frac{q_2^2}{4q_4} {,~\\frac{n^2((-1+n)(1+2n)^2w_2^3+16(1+n)^3s_2w_4^2)}{(1+n)^2(-1-n+2n^2)w_2w_4\\Pi _1}<0}\\nonumber \\\\ & {,~\\frac{(-1+n)(1+n)^2w_2w_4^2}{(-1+n)(1+2n)^2w_2^3+16(1+n)^3s_2w_4^2}>0.} \\end{aligned}$$<\/p>\n

\n (18)\n <\/p>\n

The next singular periodic solution is then elevated<\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\frac{1}{4}~\\left[ -\\frac{(1+2n)~w_2}{(1+n)~w_4}+\\frac{\\cot \\left[ \\frac{1}{2}~(\\eta _1~x+\\eta _2~y+\\eta _3~z-\\gamma ~t)~\\sqrt{\\frac{n^2~((-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2)}{(1+n)^2~(-1-n+2~n^2)~w_2~w_4~\\Pi _1}}\\right] }{\\sqrt{-\\frac{(-1+n)~(1+n)^2~w_2~w_4^2}{(-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2}}}\\right] ^{\\frac{1}{2n}}\\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1 x-\\nu _2 y-\\nu _3 z)},~q_0=\\frac{q_2^2}{4~q_4} {,~\\frac{n^2~((-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2)}{(1+n)^2~(-1-n+2~n^2)~w_2~w_4~\\Pi _1}>0}\\nonumber \\\\ & {,~\\frac{(-1+n)~(1+n)^2~w_2~w_4^2}{(-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2}<0.} \\end{aligned}$$<\/p>\n

\n (19)\n <\/p>\n

The next Jacobi elliptic solutions can then be elevated<\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\frac{1}{8}~\\left[ -~\\frac{2~(1+2~n)~w_2}{(1+n)~w_4}+\\right. \\nonumber \\\\ & \\left. \\frac{\\sqrt{2}~ j_1}{{{\\,\\textrm{cn}\\,}}\\left[ \\frac{(\\eta _1~x+\\eta _2~y+\\eta _3~z-\\gamma ~t)~\\sqrt{\\frac{n^2~\\left( (-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2\\right) }{(1+n)^2~(-1-n+2~n^2)~(-1+2p^2)~w_2~w_4~\\Pi _1}}}{\\sqrt{2}},p\\right] ~\\sqrt{\\frac{(-1+n)~(1+n)^2~p^2~w_2~w_4^2~j_1}{(-1+2p^2)~\\left( (-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2\\right) }}}\\right] ^{\\frac{1}{2n}}\\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1 x-\\nu _2 y-\\nu _3 z)},~q_0=\\frac{q_2^2~p^2~(1-p^2)}{q_4~(2p^2-1)^2} {,~\\frac{n^2~\\left( (-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2\\right) }{(1+n)^2~(-1-n+2~n^2)~(-1+2p^2)~w_2~w_4~\\Pi _1}>0}\\nonumber \\\\ & {,~\\frac{(-1+n)~(1+n)^2~p^2~w_2~w_4^2~j_1}{(-1+2p^2)~\\left( (-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2\\right) }>0.} \\end{aligned}$$<\/p>\n

\n (20)\n <\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\frac{1}{8}~\\left[ -\\frac{2~(1+2~n)~w_2}{(1+n)~w_4}+\\frac{j_1}{{{\\,\\textrm{dn}\\,}}\\left[ \\frac{(\\eta _1~x+\\eta _2~y+\\eta _3~z-\\gamma ~t)~\\sqrt{-\\frac{n^2~((-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2)}{(1+n)^2~(-1-n+2~n^2)~(-2+p^2)~w_2~w_4~\\Pi _1}}}{\\sqrt{2}},p\\right] }\\right. \\nonumber \\\\ & \\left. \\sqrt{-\\frac{(-1+n)^2~(1+n)^4~(1+2n)~p^2~w_2^2~w_4^3~j_1^2~\\Pi _1}{n^2~(-2+p^2)((-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2)^2}}\\right] ^{\\frac{1}{2n}}\\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1 x-\\nu _2 y-\\nu _3 z)},~q_0=\\frac{q_2^2~(1-p^2)}{q_4~(2-p^2)^2}\\nonumber \\\\ & {,~\\frac{n^2~((-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2)}{(1+n)^2~(-1-n+2~n^2)~(-2+p^2)~w_2~w_4~\\Pi _1}<0}\\nonumber \\\\ & {,~\\frac{(-1+n)^2~(1+n)^4~(1+2n)~p^2~w_2^2~w_4^3~j_1^2~\\Pi _1}{n^2~(-2+p^2)((-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2)^2}<0.} \\end{aligned}$$<\/p>\n

\n (21)\n <\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\frac{1}{8}~\\left[ -\\frac{2~(1+2~n)~w_2}{(1+n)~w_4}+\\frac{\\sqrt{2}~j_1}{{{\\,\\textrm{sn}\\,}}\\left[ \\frac{(\\eta _1~x+\\eta _2~y+\\eta _3~z-\\gamma ~t)~\\sqrt{-\\frac{n^2~((-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2)}{(1+n)^2~(-1-n+2~n^2)~(1+p^2)~w_2~w_4~\\Pi _1}}}{\\sqrt{2}},p\\right] }\\right. \\nonumber \\\\ & \\left. \\sqrt{\\frac{(-1+n)~(1+n)^2~p^2~w_2~w_4^2~j_1^2}{(1+p^2)((-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2)}}\\right] ^{\\frac{1}{2n}}\\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1 x-\\nu _2 y-\\nu _3 z)},~q_0=\\frac{q_2^2~p^2}{q_4~(p^2+1)^2}\\nonumber \\\\ & {,~\\frac{n^2~((-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2)}{(1+n)^2~(-1-n+2~n^2)~(1+p^2)~w_2~w_4~\\Pi _1}<0}\\nonumber \\\\ & {,~\\frac{(-1+n)~(1+n)^2~p^2~w_2~w_4^2~j_1^2}{(1+p^2)((-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2)}>0} \\end{aligned}$$<\/p>\n

\n (22)\n <\/p>\n

Result(2)<\/p>\n

$$j_1=0,~q_4=\\frac{-4~n^2~w_4~h_1^2}{(1+2n)~\\Pi _1},~h_0=\\frac{-(1+2n)~w_2}{4~(1+n)~w_4},~q_2=\\frac{q_4~(s_2+2~n~s_2+4~w_4~h_0^3-4~n~w_4~h_0^3}{2~(-1+n)~w_4~h_0~h_1^2}.$$<\/p>\n

The next dark soliton solution is then elevated<\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\left[ \\frac{-(1+2n)~w_2}{4~(1+n)~w_4}+\\sqrt{\\frac{(1+n)~\\left( (s_2+2~n~s_2-\\frac{(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}+\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}\\right) }{(-1+n)~(1+2n)~w_2}} \\right. \\nonumber \\\\ & \\left. \\tanh \\left( 2~(\\eta _1~x+\\eta _2~y+\\eta _3~z-\\gamma ~t)~\\sqrt{-\\frac{n^2~(1+n)~\\left( s_2+2~n~s_2-\\frac{(1+2n)^3~w_2~3}{16~(1+n)^3~w_4^2}+\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}\\right) ~w_4}{(-1+n)~(1+2n)~w_2~\\Pi _1}}\\right) \\right] ^{\\frac{1}{2n}} \\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1 x-\\nu _2 y-\\nu _3 z)},~q_0=\\frac{q_2^2}{4~q_4} {,~\\frac{(1+n)~\\left( (s_2+2~n~s_2-\\frac{(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}+\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}\\right) }{(-1+n)~(1+2n)~w_2}>0}\\nonumber \\\\ & \\quad {,~\\frac{n^2~(1+n)~\\left( s_2+2~n~s_2-\\frac{(1+2n)^3~w_2~3}{16~(1+n)^3~w_4^2}+\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}\\right) ~w_4}{(-1+n)~(1+2n)~w_2~\\Pi _1}<0.} \\end{aligned}$$<\/p>\n

\n (23)\n <\/p>\n

The next singular periodic solution is then elevated<\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\left[ \\frac{-(1+2n)~w_2}{4~(1+n)~w_4}+\\sqrt{-\\frac{(1+n)~\\left( (s_2+2~n~s_2-\\frac{(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}+\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}\\right) }{(-1+n)~(1+2n)~w_2}} \\right. \\nonumber \\\\ & \\left. \\tan \\left( 2~(\\eta _1~x+\\eta _2~y+\\eta _3~z-\\gamma ~t)~\\sqrt{\\frac{n^2~(1+n)~\\left( s_2+2~n~s_2-\\frac{(1+2n)^3~w_2~3}{16~(1+n)^3~w_4^2}+\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}\\right) ~w_4}{(-1+n)~(1+2n)~w_2~\\Pi _1}}\\right) \\right] ^{\\frac{1}{2n}} \\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1 x-\\nu _2 y-\\nu _3 z)},~q_0=\\frac{q_2^2}{4~q_4}{,~\\frac{(1+n)~\\left( (s_2+2~n~s_2-\\frac{(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}+\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}\\right) }{(-1+n)~(1+2n)~w_2}<0}\\nonumber \\\\ & {,~\\frac{n^2~(1+n)~\\left( s_2+2~n~s_2-\\frac{(1+2n)^3~w_2~3}{16~(1+n)^3~w_4^2}+\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}\\right) ~w_4}{(-1+n)~(1+2n)~w_2~\\Pi _1}>0.} \\end{aligned}$$<\/p>\n

\n (24)\n <\/p>\n

The next Jacobi elliptic solutions can then be elevated<\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\left[ \\frac{-(1+2n)~w_2}{4~(1+n)~w_4}+\\sqrt{2}~{{\\,\\textrm{cn}\\,}}\\left[ 2~\\sqrt{2}~(\\eta _1~x+\\eta _2~y+\\eta _3~z-\\gamma ~t)\\right. \\right. \\nonumber \\\\ & \\left. \\left. \\sqrt{\\frac{n^2~(1+n)~\\left( s_2+2~n~s_2-\\frac{(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}+\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}\\right) ~w_4}{(-1+n)~(1+2~n)^2~(-1+2~p^2)~w_2~\\Pi _1}},p\\right] \\right. \\nonumber \\\\ & \\left. \\sqrt{\\frac{(1+n)~p^2~\\left( (s_2+2~n~s_2-\\frac{(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}+\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}\\right) }{(-1+n)~(1+2n)~(-1+2p^2)~w_2~h_1^2}}~h_1 \\right] ^{\\frac{1}{2n}} \\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1 x-\\nu _2 y-\\nu _3 z)}, ~q_0=\\frac{q_2^2~p^2~(1-p^2)}{q_4~(2p^2-1)^2}{,~\\frac{n^2~(1+n)~\\left( s_2+2~n~s_2-\\frac{(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}+\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}\\right) ~w_4}{(-1+n)~(1+2~n)^2~(-1+2~p^2)~w_2~\\Pi _1}>0}\\nonumber \\\\ & {,~\\frac{(1+n)~p^2~\\left( (s_2+2~n~s_2-\\frac{(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}+\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}\\right) }{(-1+n)~(1+2n)~(-1+2p^2)~w_2~h_1^2}>0.} \\end{aligned}$$<\/p>\n

\n (25)\n <\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\left[ \\frac{-(1+2n)~w_2}{4~(1+n)~w_4}+\\frac{1}{2}~{{\\,\\textrm{dn}\\,}}\\left[ 2~\\sqrt{2}~(\\eta _1~x+\\eta _2~y+\\eta _3~z-\\gamma ~t)\\right. \\right. \\nonumber \\\\ & \\left. \\left. \\sqrt{\\frac{n^2~(1+n)~\\left( s_2+2~n~s_2-\\frac{(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}+\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}\\right) ~w_4}{(-1+n)~(1+2~n)^2~(2-p^2)~w_2~\\Pi _1}},p\\right] \\right. \\nonumber \\\\ & \\left. \\sqrt{\\frac{(1+2n)~p^2~\\Pi _1}{n^2~(2-p^2)~w_4~h_1^2}}~h_1\\right] ^{\\frac{1}{2n}} \\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1 x-\\nu _2 y-\\nu _3 z)}, ~q_0=\\frac{q_2^2~(1-p^2)}{q_4~(2-p^2)^2} {,~\\frac{n^2~(1+n)~\\left( s_2+2~n~s_2-\\frac{(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2} +\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}\\right) ~w_4}{(-1+n)~(1+2~n)^2~(2-p^2)~w_2~\\Pi _1}>0}\\nonumber \\\\ & {\\frac{(1+2n)~p^2~\\Pi _1}{n^2~(2-p^2)~w_4~h_1^2}>0} \\end{aligned}$$<\/p>\n

\n (26)\n <\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\left[ \\frac{-(1+2n)~w_2}{4~(1+n)~w_4}+\\sqrt{2}~{{\\,\\textrm{sn}\\,}}\\left[ 2~\\sqrt{2}~(\\eta _1~x+\\eta _2~y+\\eta _3~z-\\gamma ~t)\\right. \\right. \\nonumber \\\\ & \\left. \\left. \\sqrt{\\frac{-n^2~(1+n)~\\left( s_2+2~n~s_2-\\frac{(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}+\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}\\right) ~w_4}{(-1+n)~(1+2~n)^2~(1+p^2)~w_2~\\Pi _1}},p\\right] \\right. \\nonumber \\\\ & \\left. \\sqrt{\\frac{(1+n)~p^2~\\left( (s_2+2~n~s_2-\\frac{(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}+\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2} \\right) }{(-1+n)~(1+2n)~(1+p^2)~w_2~h_1^2}}~h_1\\right] ^{\\frac{1}{2n}} \\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1 x-\\nu _2 y-\\nu _3 z)},~q_0=\\frac{q_2^2~p^2}{q_4~(p^2+1)^2} {,~\\frac{n^2~(1+n)~\\left( s_2+2~n~s_2-\\frac{(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}+\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}\\right) ~w_4}{(-1+n)~(1+2~n)^2~(1+p^2)~w_2~\\Pi _1}<0}\\nonumber \\\\ & {,~\\frac{(1+n)~p^2~\\left( (s_2+2~n~s_2-\\frac{(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}+\\frac{n~(1+2n)^3~w_2^3}{16~(1+n)^3~w_4^2}\\right) }{(-1+n)~(1+2n)~(1+p^2)~w_2~h_1^2}>0.} \\end{aligned}$$<\/p>\n

\n (27)\n <\/p>\n

Case 3. \\(q_2=q_4=0\\), \\(q_0\\ne 0\\), \\(q_1\\ne 0\\), \\(q_3>0.\\)<\/p>\n

Result(1)<\/p>\n

$$q_0=\\frac{-4n^2~w_4~j_1^2}{(1+2n)~\\Pi _1},~ q_1=\\frac{4~q_0~h_0}{j_1}-\\frac{4n^2~w_2~j_1}{(1+n)~\\Pi _1},~ q_3=\\frac{h_0^2~(-3~q_1~j_1+4~q_0~h_0)}{j_1^3}+\\frac{4n^2~s_2}{(-1+n)~j_1~\\Pi _1},$$<\/p>\n

$$h_0=\\frac{1}{12}~\\left( \\frac{6-3(1+2n)~w_2}{(1+n)~c_4}+\\sqrt{\\frac{\\left( -\\frac{3}{1+n}\\right) ^2~w_2^2+24(1+2n)~w_4~(\\omega -\\Pi _2 )}{w_4^2}}\\right) .$$<\/p>\n

The next Weierstrass elliptic doubly periodic function is elevated for Eq. (1<\/a>)<\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\left[ \\frac{1}{12}~\\left( \\frac{-3~(1+2n)~w_2}{(1+n)~w_4}+\\tau \\right) +j_1~\/~\\wp \\left[ \\frac{1}{12\\sqrt{3}}~(\\eta _1~x+\\eta _2~y+\\eta _3~z-\\gamma ~t)~\\sqrt{\\frac{n^2}{j_1~\\Pi _1}~\\delta }\\right. \\right. \\nonumber \\\\ & \\left. \\left. ,~\\{\\left( 576~w_4~j_1^2~\\tau \\right) ~\/~((1+2n)~\\delta ~)~,~(1728~w_4~j_1^2)~\/~((1+2n)~\\delta ~)\\}~\\right] +\\right. \\nonumber \\\\ & \\left. h_1~ \\wp \\left[ \\frac{1}{12\\sqrt{3}}~(\\eta _1~x+\\eta _2~y+\\eta _3~z-\\gamma ~t)~\\sqrt{\\frac{n^2}{j_1~\\Pi _1}~\\delta }~,\\right. \\right. \\nonumber \\\\ & \\left. \\left. \\{\\left( 576~w_4~j_1^2~\\tau \\right) ~\/~((1+2n)~\\delta ~)~,~(1728~w_4~j_1^2)~\/~((1+2n)~\\delta ~)\\}~\\right] \\right] ^{\\frac{1}{2n}}\\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1 x-\\nu _2 y-\\nu _3 z)}, q_0\\ne 0, ~q_1\\ne 0{,~\\Pi _1>0.}, \\end{aligned}$$<\/p>\n

\n (28)\n <\/p>\n

where<\/p>\n

$$\\tau = \\sqrt{\\frac{\\left( 6-\\frac{3}{1+n}\\right) ^2~w_2^2+24(1+2n)~w_4~(\\omega -\\Pi _2 )}{w_4^2}}$$<\/p>\n

$$\\delta = \\left( \\frac{432~s_2}{-1+n}+\\frac{1}{(1+n)^3~(1+2n)~w_4^2}~\\left( -3~(1+2n)~w_2+(1+n)~w_4~\\tau \\right) ^2~(3~(1+2n)~w_2+2~(1+n)~w_4~\\tau )~\\right) .$$<\/p>\n

Case 4. \\(q_0=q_1=q_2=0\\)<\/p>\n

Result(1)<\/p>\n

$$\\begin{aligned} Q&= (1+n-2n^2)^3~w_2^3 \\\\&\\quad +8\\left( -4~(-1+n)^2~(1+n)^3~(1+2n)~s_2~w_4^2\\right. \\\\&\\quad \\left. +\\sqrt{(-1+n)^4~(1+n)^3~(1+2n)^2~s_2~w_4^2((-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2)} \\right) . \\end{aligned}$$<\/p>\n

$$\\begin{aligned} h_0= & (w_2^2+2~n~w_2^2-3~n^2~w_2^2-4~n^3~w_2^2+4~n^4~w_2^2\\\\ & +w_2~Q^{1\/3}+n~w_2~Q^{1\/3}-2~n^2~w_2~Q^{1\/3}+\\frac{Q^{2\/3}}{8~(-1+n)~(1+n)~w_4~Q^{1\/3}}, \\end{aligned}$$<\/p>\n

$$j_1=0,~q_4=-\\frac{4~n^2~w_4~h_1^2}{(1+2n)~\\Pi _1}, q_3=-\\frac{q_4~(s_2+2~n~s_2+4~w_4~h_0^3-4~n~w_4~h_0^3)}{3~(-1+n)~w_4~h_0^2~h_1}.$$<\/p>\n

The next rational solution is elevated<\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\left[ \\frac{A}{-8(1-n^2)w_4 Q^{1\/3}} \\right. \\nonumber \\\\ & \\left. + \\frac{1}{12288 (1+2n) A^2 \\Pi _1 \\left( \\frac{-8(1+n)n^2 Q^{1\/3} (s_2(1+2n)-A)^3}{(1-n^2)^2} + \\frac{n A^3}{128(1-n^2)^3 w_4^2 Q} \\right) (\\eta _1 x+\\eta _2 y+\\eta _3 z-\\gamma t)^2} + \\frac{16 n^2 w_4 h_1^2}{\\Pi _1(1+2n)} \\right] ^{\\frac{1}{2n}}\\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1 x-\\nu _2 y-\\nu _3 z)},~q_4 \\ne 0 {,(-1+n)^4~(1+n)^3~(1+2n)^2~ s_2~w_4^2((-1+n)~(1+2n)^2~w_2^3+16~(1+n)^3~s_2~w_4^2)>0}. \\end{aligned}$$<\/p>\n

\n (29)\n <\/p>\n

The next exponential functions can be elevated<\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\left[ 3 (-1+n) \\left( A\\right) ^3+2 e^{\\frac{64 (-1+n) n^2 (1+n)^2 w_4^2 Q^{2\/3} \\left( s_2+2 n~ s_2-\\frac{\\left( A\\right) ^3}{B}\\right) ~ s_1~ (\\eta _1 x+\\eta _2 y+\\eta _3 z-\\gamma t)}{3 (1+2 n)~ \\left( A\\right) ^2 ~\\Pi _1 \\sqrt{\\frac{n^2~ w_4 ~s_1^2}{\\Pi _1+2 n ~\\Pi _1}}}} \\left( 128 \\left( 1-n^2\\right) ^3~ s_2 ~w_4^2~Q+\\right. \\right. \\nonumber \\\\ & \\left. \\left. n ~\\left( 1-n^2\\right) ^3~ s_2~ w_4^2 ~Q-\\left( A\\right) ^3+n~ \\left( A\\right) ^3\\right) \/\\left( 24 (-1+n)^2~ (1+n) ~w_4~ Q^{1\/3} ~\\left( A\\right) ^2 \\right) \\right] ^{\\frac{1}{2n}} \\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1 x-\\nu _2 y-\\nu _3 z)},~q_4< 0,~w_2<0,~w_4<0{,~n<0,~s_2<0}, \\end{aligned}$$<\/p>\n

\n (30)\n <\/p>\n

where:<\/p>\n

$$A = w_2^2~(1 + 2n – 3n^2 – 4n^3 + 4n^4) + w_2 ~Q^{1\/3}~(1 + n – 2n^2) + Q^{2\/3},$$<\/p>\n

$$\\begin{aligned} B&= 128(1 – n^2)^3~ w_4^2 \\left( -(1 + n – 2n^2)^3 ~w_2^3 + 32(-1 + n)^2~ (1 + n)^3 ~(1 + 2n)~ s_2~ w_4^2 \\right. \\\\&\\quad \\left. – 8 \\sqrt{(-1 + n)^4~ (1 + n)^3 ~(1 + 2n)^2~ s_2~ w_4^2~ \\left( (-1 + n)~(1 + 2n)^2~ w_2^3 + 16(1 + n)^3 ~s_2~ w_4^2 \\right) } \\right) . \\end{aligned}$$<\/p>\n

Case 5. \\(q_3=q_4=0\\)<\/p>\n

Result(1)<\/p>\n

$$h_1=0,~q_0=-\\frac{4 n^2~ w_4~ j_1^2}{(1+2 n) ~\\Pi _1},~q_2=\\frac{n^2~ \\left( (1+2 n)^2~ w_2^3+\\frac{16 (1+n)^3~ s_2~ w_4^2}{-1+n}\\right) }{2 (1+n)^2 ~(1+2 n)~ w_2~ w_4~\\Pi _1},~h_0=-\\frac{(1+2 n)~ w_2}{4 (1+n)~ w_4}.$$<\/p>\n

The next singular periodic solution is elevated<\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\left[ \\frac{1}{4}~ \\left( -\\frac{(1+2 n)~ w_2}{(1+n)~ w_4}+\\frac{\\csc \\left( \\frac{(\\eta _1 x+\\eta _2 y+\\eta _3 z-\\gamma t) ~\\sqrt{-\\frac{n^2 \\left( (1+2 n)^2~ w_2^3+\\frac{16 (1+n)^3~s_2~ w_4^2}{-1+n}\\right) }{(1+n)^2 ~(1+2 n)~ w_2~ w_4 ~\\Pi _1}}}{\\sqrt{2}}\\right) }{\\sqrt{\\frac{(-1+n)~ (1+n)^2 ~w_2 ~w_4^2}{\\left( -2-6 n+8 n^3\\right) ~ w_2^3+32 (1+n)^3 ~s_2 ~w_4^2}}}\\right) \\right] ^{\\frac{1}{2n}} \\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1 x-\\nu _2 y-\\nu _3 z)},~q_1=0 {,~\\frac{n^2 \\left( (1+2 n)^2~ w_2^3+\\frac{16 (1+n)^3~s_2~ w_4^2}{-1+n}\\right) }{(1+n)^2 ~(1+2 n)~ w_2~ w_4 ~\\Pi _1}<0,~\\frac{(-1+n)~ (1+n)^2 ~w_2 ~w_4^2}{\\left( -2-6 n+8 n^3\\right) ~ w_2^3+32 (1+n)^3 ~s_2 ~w_4^2}>0} \\end{aligned}$$<\/p>\n

\n (31)\n <\/p>\n

The next singular soliton is elevated<\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\left[ \\frac{1}{4} ~\\left( -\\frac{(1+2 n)~ w_2}{(1+n)~ w_4}+\\frac{\\sqrt{2} {{\\,\\textrm{csch}\\,}}\\left( \\frac{(\\eta _1 x+\\eta _2 y+\\eta _3 z-\\gamma t)~ \\sqrt{\\frac{n^2 ~\\left( (1+2 n)^2~ w_2^3+\\frac{16 (1+n)^3~ s_2~ w_4^2}{-1+n}\\right) }{(1+n)^2~ (1+2 n)~ w_2~ w_4 ~\\Pi _1}}}{\\sqrt{2}}\\right] }{\\sqrt{-\\frac{(1+n)^2~w_2~ w_4^2 }{(1+2 n)^2 ~w_2^3+\\frac{16 (1+n)^3 ~s_2 w_4^2}{-1+n}}}}\\right) \\right] ^{\\frac{1}{2n}} \\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1 x-\\nu _2 y-\\nu _3 z)},~q_1=0{,~\\frac{n^2 ~\\left( (1+2 n)^2~ w_2^3+\\frac{16 (1+n)^3~ s_2~ w_4^2}{-1+n}\\right) }{(1+n)^2~ (1+2 n)~ w_2~ w_4 ~\\Pi _1}>0} {,~\\frac{(1+n)^2~w_2~ w_4^2 }{(1+2 n)^2 ~w_2^3+\\frac{16 (1+n)^3 ~s_2 w_4^2}{-1+n}}<0} \\end{aligned}$$<\/p>\n

\n (32)\n <\/p>\n

Case 6. \\(q_0=q_1=0, ~q_4>0\\)<\/p>\n

Result(1)<\/p>\n

$$j_1=0, h_0=\\frac{h_1 ~\\left( 4 n^2~ w_2~ h_1+q_3 ~\\Pi _1+n~ q_3 ~\\Pi _1\\right) }{4 (1+n) ~q_4~ \\Pi _1}, q_2=\\frac{4 n^2~ \\omega ~ h_1^2-6 q_4 ~h_0^2~ \\Pi _1+3 q_3 ~h_0~ h_1~ \\Pi _1-4 n^2~ h_1^2 ~\\Pi _2}{h_1^2~ \\Pi _1},$$<\/p>\n

$$q_4=-\\frac{4 n^2 ~w_4~ h_1^2}{(1+2 n)~ \\Pi _1},~q_3=2\\sqrt{q_2~ q_4}.$$<\/p>\n

The next singular periodic solution is elevated The next dark soliton solution is elevated<\/p>\n

$$\\begin{aligned} R(x,y,z,t)= & \\left[ \\frac{1}{16 n ~\\sqrt{w_4} ~\\sqrt{q_4}~ \\sqrt{\\Pi _1}} ~ \\sqrt{-1-2 n}~ \\left( \\frac{4 ~ n ~\\sqrt{-1-2 n}~ w_2 ~\\sqrt{q_4} \\sqrt{\\Pi _1}}{(1+n) ~\\sqrt{w_4}}+2 q_3 ~\\Pi _1+q_4 ~\\Pi _1 \\right. \\right. \\nonumber \\\\ & \\left. \\left. \\sqrt{\\frac{24 n^2 ~(1+2 n)~ w_2^2~ q_4+2 (1+n)^2~ w_4 ~(3 q_3^2 ~\\Pi _1+32 n^2 ~q_4 (\\omega -\\Pi _2))}{(1+n)^2~ w_4~ q_4^2~ \\Pi _1}} \\left( 1 \\right. \\right. \\right. \\nonumber \\\\ & \\left. \\left. \\left. +\\tanh \\left( \\frac{(\\eta _1 x+\\eta _2 y+\\eta _3 z-\\gamma t)~ \\sqrt{\\frac{\\frac{12 n^2 ~(1+2 n)~ w_2^2}{(1+n)^2~ w_4}+\\frac{3 q_3^2~ \\Pi _1}{q_4}+32 n^2~ (\\omega -\\Pi _2)}{\\Pi _1}}}{4 \\sqrt{2}}\\right) \\right) \\right) \\right] ^{\\frac{1}{2n}} \\nonumber \\\\ & \\times e^{i(t \\omega +\\theta -\\nu _1 x-\\nu _2 y-\\nu _3 z)}, ~w_4<0,\\Pi _1>0 {,~n<0}\\nonumber \\\\ & {,~\\frac{24 n^2 ~(1+2 n)~ w_2^2~ q_4+2 (1+n)^2~ w_4 ~(3 q_3^2 ~\\Pi _1+32 n^2 ~q_4 (\\omega -\\Pi _2))}{(1+n)^2~ w_4~ q_4^2~ \\Pi _1}>0}\\nonumber \\\\ & {\\frac{\\frac{12 n^2 ~(1+2 n)~ w_2^2}{(1+n)^2~ w_4}+\\frac{3 q_3^2~ \\Pi _1}{q_4}+32 n^2~ (\\omega -\\Pi _2)}{\\Pi _1}>0.} \\end{aligned}$$<\/p>\n

\n (33)\n <\/p>\n","protected":false},"excerpt":{"rendered":"Our goal in this part is to get the next form of solutions for Eq. (1): $$\\begin{aligned} R(x,y,z,t)=R(\\xi…\n","protected":false},"author":2,"featured_media":171784,"comment_status":"","ping_status":"","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[24],"tags":[94806,94805,2026,61,60,21372,2027,4968,248,82,94804,94807],"class_list":{"0":"post-171783","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-physics","8":"tag-analytical-wave-localization","9":"tag-higher-order-dispersive-media","10":"tag-humanities-and-social-sciences","11":"tag-ie","12":"tag-ireland","13":"tag-mathematics-and-computing","14":"tag-multidisciplinary","15":"tag-optics-and-photonics","16":"tag-physics","17":"tag-science","18":"tag-soliton-behavior","19":"tag-stability-analysis"},"_links":{"self":[{"href":"https:\/\/www.newsbeep.com\/ie\/wp-json\/wp\/v2\/posts\/171783","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.newsbeep.com\/ie\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.newsbeep.com\/ie\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.newsbeep.com\/ie\/wp-json\/wp\/v2\/users\/2"}],"replies":[{"embeddable":true,"href":"https:\/\/www.newsbeep.com\/ie\/wp-json\/wp\/v2\/comments?post=171783"}],"version-history":[{"count":0,"href":"https:\/\/www.newsbeep.com\/ie\/wp-json\/wp\/v2\/posts\/171783\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.newsbeep.com\/ie\/wp-json\/wp\/v2\/media\/171784"}],"wp:attachment":[{"href":"https:\/\/www.newsbeep.com\/ie\/wp-json\/wp\/v2\/media?parent=171783"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.newsbeep.com\/ie\/wp-json\/wp\/v2\/categories?post=171783"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.newsbeep.com\/ie\/wp-json\/wp\/v2\/tags?post=171783"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}