{"id":25139,"date":"2025-09-16T15:19:09","date_gmt":"2025-09-16T15:19:09","guid":{"rendered":"https:\/\/www.newsbeep.com\/ie\/25139\/"},"modified":"2025-09-16T15:19:09","modified_gmt":"2025-09-16T15:19:09","slug":"analytical-construction-and-visualization-of-nonlinear-waves-in-the-21-dimensional-kadomtsev-petviashvili-sawada-kotera-ramani-equation-with-stability-analysis","status":"publish","type":"post","link":"https:\/\/www.newsbeep.com\/ie\/25139\/","title":{"rendered":"Analytical construction and visualization of nonlinear waves in the (2+1) dimensional Kadomtsev-Petviashvili-Sawada-Kotera-Ramani equation with stability analysis"},"content":{"rendered":"

The steps of the improved modified extended Tanh method that were explained in the preceding section will be applied in this section to determine the exact solution for Eq. (1<\/a>).<\/p>\n

$$\\begin{aligned} u(x,y,t)=z(\\xi ), ~ \\xi =x+y+c t. \\end{aligned}$$<\/p>\n

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

Substitute Eq. (7<\/a>) into the original PDE Eq. (1<\/a>) to obtain this ODE equation:<\/p>\n

$$\\begin{aligned} & (6+90 z) (z’)^2+(c+\\sigma +6z+45z^2) z”+15(z”)^{2}+30z’z^{(3)}+z^{(4)}+15zz^{(4)}+z^{(6)}=0. \\end{aligned}$$<\/p>\n

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

After applying the principle of balance between the highest-order linear term \\(z^{(6)}\\) with nonlinear term \\(zz^{(4)}\\) or \\(z’z”’\\) in Eq. (8<\/a>), we obtain \\(l=2\\) using the equation \\((l + 6 = 2 l + 4)\\). Equation (8<\/a>)\u2019s solution can therefore be expressed as follows:<\/p>\n

$$\\begin{aligned} z(\\xi )=a_0+a_1 \\phi (\\xi )+a_2 \\phi ^2(\\xi )+\\dfrac{b_1}{\\phi (\\xi )}+\\dfrac{b_2}{\\phi ^2(\\xi )}. \\end{aligned}$$<\/p>\n

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

Case 1: \\(d_0=d_1=d_3=0\\)<\/p>\n

Result (1)<\/p>\n

$$\\begin{aligned} a_0= & \\frac{1}{15} \\left( -20 d_2-1\\right) ,~d_4=-\\frac{a_2}{4},~a_1=b_1=b_2=0,~d_2=\\frac{\\sqrt{-5 c-5 \\sigma +1}}{4 \\sqrt{5}}.\\nonumber \\\\ u(x,y,t)= & \\frac{1}{15} \\left( -\\sqrt{5} \\sqrt{-5 c-5 \\sigma +1}+3 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1} ~\\text {sech}^2\\left( \\frac{1}{2} \\root 4 \\of {-c-\\sigma +\\frac{1}{5}} (c t+x+y)\\right) -1\\right) ,\\nonumber \\\\ & -5 c-5 \\sigma +1>0. \\end{aligned}$$<\/p>\n

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

This is a representation of a bright soliton solution.<\/p>\n

Result (2)<\/p>\n

$$\\begin{aligned} a_0= & \\frac{1}{15} \\left( -20 d_2-1\\right) ,~d_4=-\\frac{a_2}{4},~a_1=b_1=b_2=0,~d_2=-\\frac{\\sqrt{-5 c-5 \\sigma +1}}{4 \\sqrt{5}}.\\nonumber \\\\ u(x,y,t)= & \\frac{1}{15} \\left( \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1}-3 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1}~ \\sec ^2\\left( \\frac{1}{2} \\root 4 \\of {-c-\\sigma +\\frac{1}{5}} (c t+x+y)\\right) -1\\right) ,\\nonumber \\\\ & -5 c-5 \\sigma +1>0. \\end{aligned}$$<\/p>\n

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

This is a representation of a singular periodic solution.<\/p>\n

Case 2: \\(d_1=d_3=0,~d_0=\\frac{d_2^2}{4d_4}\\)<\/p>\n

Result (1)<\/p>\n

$$\\begin{aligned} a_0= & \\frac{1}{15} \\left( -20 d_2-1\\right) ,~d_4=-\\frac{a_2}{4},~a_1=b_1=b_2=0,~d_2=\\frac{\\sqrt{-5 c-5 \\sigma +1}}{2 \\sqrt{5}}.\\nonumber \\\\ u(x,y,t)= & \\frac{1}{15} \\left( -2 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1}-3 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1} ~\\tan ^2\\left( \\frac{1}{2} \\root 4 \\of {-c-\\sigma +\\frac{1}{5}} (c t+x+y)\\right) -1\\right) ,\\nonumber \\\\ & -5 c-5 \\sigma +1>0. \\end{aligned}$$<\/p>\n

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

This is a representation of a singular periodic solution.<\/p>\n

Result (2)<\/p>\n

$$\\begin{aligned} a_0= & \\frac{1}{15} \\left( -20 d_2-1\\right) ,~d_4=-\\frac{a_2}{4},~a_1=b_1=b_2=0,~d_2=-\\frac{\\sqrt{-5 c-5 \\sigma +1}}{2 \\sqrt{5}}.\\nonumber \\\\ u(x,y,t)= & \\frac{1}{15} \\left( 2 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1}-3 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1} ~\\tanh ^2\\left( \\frac{1}{2} \\root 4 \\of {-c-\\sigma +\\frac{1}{5}} (c t+x+y)\\right) -1\\right) ,\\nonumber \\\\ & -5 c-5 \\sigma +1>0. \\end{aligned}$$<\/p>\n

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

This is a representation of a hyperbolic solution.<\/p>\n

Result (3)<\/p>\n

$$\\begin{aligned} a_0= & \\frac{1}{15} \\left( -20 d_2-1\\right) ,~a_1=b_1=a_2=0,~d_4=-\\frac{d_2^2}{b_2},~d_2=\\frac{\\sqrt{-5 c-5 \\sigma +1}}{2 \\sqrt{5}}.\\nonumber \\\\ u(x,y,t)= & \\frac{1}{15} \\left( -2 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1}-3 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1}~ \\cot ^2\\left( \\frac{1}{2} \\root 4 \\of {-c-\\sigma +\\frac{1}{5}} (c t+x+y)\\right) -1\\right) ,\\nonumber \\\\ & -5 c-5 \\sigma +1>0. \\end{aligned}$$<\/p>\n

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

This is a representation of a singular periodic solution.<\/p>\n

Result (4)<\/p>\n

$$\\begin{aligned} a_0= & \\frac{1}{15} \\left( -20 d_2-1\\right) ,~a_1=b_1=a_2=0,~d_4=-\\frac{d_2^2}{b_2},~d_2=-\\frac{\\sqrt{-5 c-5 \\sigma +1}}{2 \\sqrt{5}}.\\nonumber \\\\ u(x,y,t)= & \\frac{1}{15} \\left( -2 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1}-3 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1}~ \\coth ^2\\left( \\frac{1}{2} \\root 4 \\of {-c-\\sigma +\\frac{1}{5}} (c t+x+y)\\right) -1\\right) ,\\nonumber \\\\ & -5 c-5 \\sigma +1>0. \\end{aligned}$$<\/p>\n

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

This is a representation of a singular soliton solution.<\/p>\n

Result (5)<\/p>\n

$$\\begin{aligned} a_0= & \\frac{1}{15} \\left( -20 d_2-1\\right) ,~d_4=-\\frac{a_2}{4},~b_2=-\\frac{d_2^2}{d_4},~d_2=\\frac{\\sqrt{-5 c-5 \\sigma +1}}{8 \\sqrt{5}},~a_1=b_1=0.\\nonumber \\\\ u(x,y,t)= & \\frac{1}{60} \\Bigg (-2 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1}-3 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1} ~\\tan ^2\\left( \\frac{1}{4} \\root 4 \\of {-c-\\sigma +\\frac{1}{5}} (c t+x+y)\\right) \\nonumber \\\\ & -3 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1}~ \\cot ^2\\left( \\frac{1}{4} \\root 4 \\of {-c-\\sigma +\\frac{1}{5}} (c t+x+y)\\right) -4\\Bigg ),\\quad \\quad \\quad -5 c-5 \\sigma +1>0. \\end{aligned}$$<\/p>\n

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

This is a representation of a singular periodic solution.<\/p>\n

Result (6)<\/p>\n

$$\\begin{aligned} a_0= & \\frac{1}{15} \\left( -20 d_2-1\\right) ,~d_4=-\\frac{a_2}{4},~b_2=-\\frac{d_2^2}{d_4},~d_2=-\\frac{\\sqrt{-5 c-5 \\sigma +1}}{8 \\sqrt{5}},~a_1=b_1=0.\\nonumber \\\\ u(x,y,t)= & \\frac{1}{60} \\Bigg (-2 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1}-3 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1} ~\\tanh ^2\\left( \\frac{1}{4} \\root 4 \\of {-c-\\sigma +\\frac{1}{5}} (c t+x+y)\\right) \\nonumber \\\\ & -3 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1} ~\\coth ^2\\left( \\frac{1}{4} \\root 4 \\of {-c-\\sigma +\\frac{1}{5}} (c t+x+y)\\right) -4\\Bigg ),\\quad \\quad \\quad -5 c-5 \\sigma +1>0. \\end{aligned}$$<\/p>\n

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

This is a representation of a singular soliton solution.<\/p>\n

Case (3): \\(d_1=d_3=d_4=0\\)<\/p>\n

Result (1)<\/p>\n

$$\\begin{aligned} a_0= & \\frac{1}{15} \\left( -20 d_2-1\\right) ,~a_1=b_1=a_2=0,~d_0=-\\frac{b_2}{4},~d_2=\\frac{\\sqrt{-5 c-5 \\sigma +1}}{4 \\sqrt{5}}.\\nonumber \\\\ u(x,y,t)= & \\frac{1}{15} \\left( -\\sqrt{5} \\sqrt{-5 c-5 \\sigma +1}-3 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1} ~\\text {csch}^2\\left( \\frac{1}{2} \\root 4 \\of {-c-\\sigma +\\frac{1}{5}} (c t+x+y)\\right) -1\\right) ,\\nonumber \\\\ & -5 c-5 \\sigma +1>0. \\end{aligned}$$<\/p>\n

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

This is a representation of a singular soliton solution.<\/p>\n

Result (2)<\/p>\n

$$\\begin{aligned} a_0= & \\frac{1}{15} \\left( -20 d_2-1\\right) ,~a_1=b_1=a_2=0,~d_0=-\\frac{b_2}{4},~d_2=-\\frac{\\sqrt{-5 c-5 \\sigma +1}}{4 \\sqrt{5}}.\\nonumber \\\\ u(x,y,t)= & \\frac{1}{15} \\left( -\\sqrt{5} \\sqrt{-5 c-5 \\sigma +1}-3 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1}~ \\csc ^2\\left( \\frac{1}{2} \\root 4 \\of {-c-\\sigma +\\frac{1}{5}} (c t+x+y)\\right) -1\\right) ,\\nonumber \\\\ & -5 c-5 \\sigma +1>0. \\end{aligned}$$<\/p>\n

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

This is a representation of a singular periodic solution.<\/p>\n

Case (4): \\(d_0=d_1=0\\),\u00a0\\(d_{4}=\\dfrac{d_3^2}{4d_2}\\)<\/p>\n

Result (1)<\/p>\n

$$\\begin{aligned} a_0= & \\frac{1}{15} \\left( -5 d_2-1\\right) ,~a_2=-\\frac{d_3^2}{d_2},~d_3=-\\frac{a_1}{2},~d_2=\\frac{\\sqrt{-5 c-5 \\sigma +1}}{\\sqrt{5}},~b_1=b_2=0.\\nonumber \\\\ u(x,y,t)= & \\frac{1}{15} \\left( 2 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1}-3 \\sqrt{5} \\sqrt{-5 c-5 \\sigma +1}~\\tanh ^2\\left( \\frac{1}{2} \\root 4 \\of {-c-\\sigma +\\frac{1}{5}} (c t+x+y)\\right) -1\\right) ,\\nonumber \\\\ & -5 c-5 \\sigma +1>0. \\end{aligned}$$<\/p>\n

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

This is a representation of a hyperbolic solution.<\/p>\n","protected":false},"excerpt":{"rendered":"The steps of the improved modified extended Tanh method that were explained in the preceding section will be…\n","protected":false},"author":2,"featured_media":25140,"comment_status":"","ping_status":"","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[24],"tags":[21376,2026,21375,61,60,21373,21372,2027,21374,4968,248,82],"class_list":{"0":"post-25139","1":"post","2":"type-post","3":"status-publish","4":"format-standard","5":"has-post-thumbnail","7":"category-physics","8":"tag-graphical-analysis","9":"tag-humanities-and-social-sciences","10":"tag-hyperbolic-wave-structures","11":"tag-ie","12":"tag-ireland","13":"tag-kpskr-equation","14":"tag-mathematics-and-computing","15":"tag-multidisciplinary","16":"tag-nonlinear-wave-propagation","17":"tag-optics-and-photonics","18":"tag-physics","19":"tag-science"},"_links":{"self":[{"href":"https:\/\/www.newsbeep.com\/ie\/wp-json\/wp\/v2\/posts\/25139","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=25139"}],"version-history":[{"count":0,"href":"https:\/\/www.newsbeep.com\/ie\/wp-json\/wp\/v2\/posts\/25139\/revisions"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.newsbeep.com\/ie\/wp-json\/wp\/v2\/media\/25140"}],"wp:attachment":[{"href":"https:\/\/www.newsbeep.com\/ie\/wp-json\/wp\/v2\/media?parent=25139"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.newsbeep.com\/ie\/wp-json\/wp\/v2\/categories?post=25139"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.newsbeep.com\/ie\/wp-json\/wp\/v2\/tags?post=25139"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}