雙重sine-Gordon方程

雙重sine-Gordon方程是一個非線性偏微分方程，它在諸如鐵磁物質研究、帶電密度波、液晶研究、氦自旋波、超短光脈衝等工程物理系領域有廣泛的運用。雙重sine-Gordon方程形式如下：

$u_{xt}=asin(u)+bsin(2u)$

變換

作變換

$u=2*arctan(v)$

經簡化後，上列方程化為

$2*v_{xt}+2*v_{xt}*v^{2}-4*v_{t}*v*v_{x}=2*v*(a+a*v^{2}+2*b-2*b*v^{2})$

中間解

此v方程有雅可比橢圓函數解：

${v=_{C}5*JacobiCN(_{C}2+_{C}3*x-(a*_{C}5^{2}-2*b*_{C}5^{2}+a+2*b)*t/(_{C}3*(_{C}5^{2}+1)),{\sqrt {(}}(a*_{C}5^{4}+2*a*_{C}5^{2}+a+2*b-2*b*_{C}5^{4})*(a*_{C}5^{2}-2*b*_{C}5^{2}+a))*_{C}5/(a*_{C}5^{4}+2*a*_{C}5^{2}+a+2*b-2*b*_{C}5^{4}))}$

${v=_{C}5*JacobiDN(_{C}2+_{C}3*x-_{C}5^{2}*(a*_{C}5^{2}-2*b*_{C}5^{2}+a)*t/(_{C}3*(_{C}5^{4}+2*_{C}5^{2}+1)),{\sqrt {(}}(a*_{C}5^{4}+2*a*_{C}5^{2}+a+2*b-2*b*_{C}5^{4})*(a*_{C}5^{2}-2*b*_{C}5^{2}+a))/((a*_{C}5^{2}-2*b*_{C}5^{2}+a)*_{C}5))}$

${v=_{C}5*JacobiNC(_{C}2+_{C}3*x+(a*_{C}5^{2}-2*b*_{C}5^{2}+a+2*b)*t/(_{C}3*(_{C}5^{2}+1)),{\sqrt {(}}(a*_{C}5^{4}+2*a*_{C}5^{2}+a+2*b-2*b*_{C}5^{4})*(a*_{C}5^{2}+a+2*b))/(a*_{C}5^{4}+2*a*_{C}5^{2}+a+2*b-2*b*_{C}5^{4}))}$

${v=_{C}5*JacobiND(_{C}2+_{C}3*x+(a*_{C}5^{2}+a+2*b)*t/(_{C}3*(_{C}5^{4}+2*_{C}5^{2}+1)),{\sqrt {(}}(a*_{C}5^{4}+2*a*_{C}5^{2}+a+2*b-2*b*_{C}5^{4})*(a*_{C}5^{2}+a+2*b))/(a*_{C}5^{2}+a+2*b))}$

${v=_{C}5*JacobiNS(_{C}2+_{C}3*x+_{C}5^{2}*(a*_{C}5^{2}-2*b*_{C}5^{2}+a)*t/(_{C}3*(_{C}5^{4}+2*_{C}5^{2}+1)),{\sqrt {(}}-(a*_{C}5^{2}-2*b*_{C}5^{2}+a)*(a*_{C}5^{2}+a+2*b))/((a*_{C}5^{2}-2*b*_{C}5^{2}+a)*_{C}5))}$

…………………………

以及雙曲函數解

${v={\sqrt {(}}(a-2*b)*a)*sinh(_{C}1+_{C}2*x-(a-2*b)*t/_{C}2)/(a-2*b)}$

${v={\sqrt {(}}-(a-2*b)*a)*cosh(_{C}1+_{C}2*x-(a-2*b)*t/_{C}2)/(a-2*b)}$

${v={\sqrt {(}}-(a-2*b)*(a+2*b))*tanh(_{C}1+_{C}2*x+(1/8)*(a^{2}-4*b^{2})*t/(_{C}2*b))/(a-2*b)}$

${v=-{\sqrt {(}}a*(a+2*b))*csch(_{C}1+_{C}2*x+(a+2*b)*t/_{C}2)/a}$

………………

和三角函數解

${v={\sqrt {(}}(a-2*b)*(a+2*b))*cot(_{C}1+_{C}2*x-(1/8)*(a^{2}-4*b^{2})*t/(_{C}2*b))/(a-2*b)}$

${v={\sqrt {(}}(a-2*b)*(a+2*b))*tan(_{C}1+_{C}2*x-(1/8)*(a^{2}-4*b^{2})*t/(_{C}2*b))/(a-2*b)}$

${v={\sqrt {(}}-(a-2*b)*a)*cos(_{C}1+_{C}2*x+(a-2*b)*t/_{C}2)/(a-2*b)}$

${v={\sqrt {(}}-(a-2*b)*a)*sin(_{C}1+_{C}2*x+(a-2*b)*t/_{C}2)/(a-2*b)}$

……

行波解

作反變換

$u=tan(v/2)$即得原來sine-Gordon方程的行波解：

${\begin{aligned}*2arctan(1.5JacobiCN(-1.2-1.3x+0.17751479289940828402t,1.0741723110591493207I))\\*2arctan(1.5JacobiDN(1.2+1.3x+0.20482476103777878925t,0.93094933625126274463I))\\*2arctan(1.5JacobiNC(1.2+1.3x+0.17751479289940828402t,1.4675987714106856141))\\2arctan(1.5JacobiND(1.2+1.3x+0.38233955393718707328t,0.68138514386924689225))\\*-2arctan(1.5JacobiNS(-1.2-1.3x+0.20482476103777878925t,1.3662601021279464511))\\*-2arctan(1.5JacobiSN(-1.2-1.3x+0.38233955393718707328t,0.73192505471139988450))\\*-(2*I)*arctanh(sinh(-15.1+1.2*x+.83333333333333333333*t))\\*2*arctan(sin(15.1-1.2*x+.83333333333333333333*t))\\*2*arctan(sqrt(3)*coth(15.1-1.2*x+.31250000000000000000*t))\\\end{aligned}}$

行波圖