亨特 - 薩克斯頓方程

亨特 - 薩克斯頓方程(Hunter–Saxton equation)是一個模擬向列型液晶中波動傳播的非線性偏微分方程：

$$(u_t + u u_x)_x = \frac{1}{2} \, u_x^2$$

## 解析解

亨特 - 薩克斯頓方程有解析解：

$$u(x,t)=\frac{1}{1-t/2}*(x-t)+1$$

此外，Maple軟體包 TWSolution 給出多個行波解：

tanh 展開法

$$f1 := u(x, t) = -1.7045454545454545454+1.1363636363636363636*(-(1/4)*(6.336*ln(tanh(2.3+.88*x+1.5*t)-1)-6.336*ln(tanh(2.3+.88*x+1.5*t)+1)+9.2928)^(1/3)-(1/4*I)*sqrt(3)*(6.336*ln(tanh(2.3+.88*x+1.5*t)-1)-6.336*ln(tanh(2.3+.88*x+1.5*t)+1)+9.2928)^(1/3))^2$$

$$f2 := u(x, t) = -1.7045454545454545454+1.1363636363636363636*(-(1/4)*(6.336*ln(tanh(2.3+.88*x+1.5*t)-1)-6.336*ln(tanh(2.3+.88*x+1.5*t)+1)+9.2928)^(1/3)+(1/4*I)*sqrt(3)*(6.336*ln(tanh(2.3+.88*x+1.5*t)-1)-6.336*ln(tanh(2.3+.88*x+1.5*t)+1)+9.2928)^(1/3))^2$$

$$u(x, t) = -1.7045454545454545454 + 0.28409090909090909090 (6.336 <pre> ln(tanh(2.3 + 0.88 x + 1.5 t) - 1) - 6.336 ln(tanh(2.3 + 0.88 x + 1.5 t) + 1) + 9.2928)^(2/3)$$

其中 f1、f2 是重解，因此，只有兩個獨立的行波解。

sech arctan 展開法

$$f[1] := -.20000000000000000000+(1/12)*(-(36*(1.3*arctan(1/sqrt(sech(2+3*x+.6*t)^2-1))*sqrt(sech(2+3*x+.6*t)+1)*sqrt(sech(2+3*x+.6*t)-1)-2.6*sqrt(sech(2+3*x+.6*t)^2-1)))/sqrt(sech(2+3*x+.6*t)^2-1))^(2/3)$$

$$f[2] := -.20000000000000000000+(1/3)*(-(1/4)*(-(36*(1.3*arctan(1/sqrt(sech(2+3*x+.6*t)^2-1))*sqrt(sech(2+3*x+.6*t)+1)*sqrt(sech(2+3*x+.6*t)-1)-2.6*sqrt(sech(2+3*x+.6*t)^2-1)))/sqrt(sech(2+3*x+.6*t)^2-1))^(1/3)-(1/4*I)*sqrt(3)*(-(36*(1.3*arctan(1/sqrt(sech(2+3*x+.6*t)^2-1))*sqrt(sech(2+3*x+.6*t)+1)*sqrt(sech(2+3*x+.6*t)-1)-2.6*sqrt(sech(2+3*x+.6*t)^2-1)))/sqrt(sech(2+3*x+.6*t)^2-1))^(1/3))^2$$ $$f[3] := -.20000000000000000000+(1/3)*(-(1/4)*(-(36*(1.3*arctan(1/sqrt(sech(2+3*x+.6*t)^2-1))*sqrt(sech(2+3*x+.6*t)+1)*sqrt(sech(2+3*x+.6*t)-1)-2.6*sqrt(sech(2+3*x+.6*t)^2-1)))/sqrt(sech(2+3*x+.6*t)^2-1))^(1/3)+(1/4*I)*sqrt(3)*(-(36*(1.3*arctan(1/sqrt(sech(2+3*x+.6*t)^2-1))*sqrt(sech(2+3*x+.6*t)+1)*sqrt(sech(2+3*x+.6*t)-1)-2.6*sqrt(sech(2+3*x+.6*t)^2-1)))/sqrt(sech(2+3*x+.6*t)^2-1))^(1/3))^2$$ $$f[4] := -.20000000000000000000+(1/3)*(-(1/4)*(187.2+140.4*x+28.08*t)^(1/3)-(1/4*I)*sqrt(3)*(187.2+140.4*x+28.08*t)^(1/3))^2$$ $$f[5] := -.20000000000000000000+(1/3)*(-(1/4)*(187.2+140.4*x+28.08*t)^(1/3)+(1/4*I)*sqrt(3)*(187.2+140.4*x+28.08*t)^(1/3))^2$$ $$f[6] := -.20000000000000000000+(1/12)*(187.2+140.4*x+28.08*t)^(2/3)$$ $$<f[7] := -1.6666666666666666667*_C6+1.6666666666666666667*(-(1/4)*(9.36*Intat(1/sqrt(4*_a^4-5*_a^2+1), _a = JacobiSN(3+.6*x+_C6*t, 2))+18.72)^(1/3)-(1/4*I)*sqrt(3)*(9.36*Intat(1/sqrt(4*_a^4-5*_a^2+1), _a = JacobiSN(3+.6*x+_C6*t, 2))+18.72)^(1/3))^2$$ $$f[8] := -1.6666666666666666667*_C6+1.6666666666666666667*(-(1/4)*(9.36*Intat(1/sqrt(4*_a^4-5*_a^2+1), _a = JacobiSN(3+.6*x+_C6*t, 2))+18.72)^(1/3)+(1/4*I)*sqrt(3)*(9.36*Intat(1/sqrt(4*_a^4-5*_a^2+1), _a = JacobiSN(3+.6*x+_C6*t, 2))+18.72)^(1/3))^2$$ $$f[9] := -1.6666666666666666667*_C6+.41666666666666666668*(9.36*Intat(1/sqrt(4*_a^4-5*_a^2+1), _a = JacobiSN(3+.6*x+_C6*t, 2))+18.72)^(2/3)$$

綜合展開

$$p[12] := -1.0714+.71429*(-.50821*((1.1*ln(cos(1.3+1.4*x+1.5*t)+sqrt(cos(1.3+1.4*x+1.5*t)^2-1.))*sqrt((cos(1.3+1.4*x+1.5*t)-1.)*(cos(1.3+1.4*x+1.5*t)+1.))+1.2*sqrt(cos(1.3+1.4*x+1.5*t)+1.)*sqrt(cos(1.3+1.4*x+1.5*t)-1.))*(1.+sqrt(1/((cos(1.3+1.4*x+1.5*t)-1.)*(cos(1.3+1.4*x+1.5*t)+1.)))*sqrt(cos(1.3+1.4*x+1.5*t)-1.)*sqrt(cos(1.3+1.4*x+1.5*t)+1.))/(sqrt(cos(1.3+1.4*x+1.5*t)+1.)*sqrt(cos(1.3+1.4*x+1.5*t)-1.)))^(1/3)-(.88026*I)*((1.1*ln(cos(1.3+1.4*x+1.5*t)+sqrt(cos(1.3+1.4*x+1.5*t)^2-1.))*sqrt((cos(1.3+1.4*x+1.5*t)-1.)*(cos(1.3+1.4*x+1.5*t)+1.))+1.2*sqrt(cos(1.3+1.4*x+1.5*t)+1.)*sqrt(cos(1.3+1.4*x+1.5*t)-1.))*(1.+sqrt(1/((cos(1.3+1.4*x+1.5*t)-1.)*(cos(1.3+1.4*x+1.5*t)+1.)))*sqrt(cos(1.3+1.4*x+1.5*t)-1.)*sqrt(cos(1.3+1.4*x+1.5*t)+1.))/(sqrt(cos(1.3+1.4*x+1.5*t)+1.)*sqrt(cos(1.3+1.4*x+1.5*t)-1.)))^(1/3))^2$$

$$p[13] := -1.0714+.71429*(-.50821*((1.1*ln(cos(1.3+1.4*x+1.5*t)+sqrt(cos(1.3+1.4*x+1.5*t)^2-1.))*sqrt((cos(1.3+1.4*x+1.5*t)-1.)*(cos(1.3+1.4*x+1.5*t)+1.))+1.2*sqrt(cos(1.3+1.4*x+1.5*t)+1.)*sqrt(cos(1.3+1.4*x+1.5*t)-1.))*(1.+sqrt(1/((cos(1.3+1.4*x+1.5*t)-1.)*(cos(1.3+1.4*x+1.5*t)+1.)))*sqrt(cos(1.3+1.4*x+1.5*t)-1.)*sqrt(cos(1.3+1.4*x+1.5*t)+1.))/(sqrt(cos(1.3+1.4*x+1.5*t)+1.)*sqrt(cos(1.3+1.4*x+1.5*t)-1.)))^(1/3)+(.88026*I)*((1.1*ln(cos(1.3+1.4*x+1.5*t)+sqrt(cos(1.3+1.4*x+1.5*t)^2-1.))*sqrt((cos(1.3+1.4*x+1.5*t)-1.)*(cos(1.3+1.4*x+1.5*t)+1.))+1.2*sqrt(cos(1.3+1.4*x+1.5*t)+1.)*sqrt(cos(1.3+1.4*x+1.5*t)-1.))*(1.+sqrt(1/((cos(1.3+1.4*x+1.5*t)-1.)*(cos(1.3+1.4*x+1.5*t)+1.)))*sqrt(cos(1.3+1.4*x+1.5*t)-1.)*sqrt(cos(1.3+1.4*x+1.5*t)+1.))/(sqrt(cos(1.3+1.4*x+1.5*t)+1.)*sqrt(cos(1.3+1.4*x+1.5*t)-1.)))^(1/3))^2$$

$$p[16] := -1.0714+.71429*(-.50821*((1.1*ln(sin(1.3+1.4*x+1.5*t)+sqrt(sin(1.3+1.4*x+1.5*t)^2-1.))*sqrt((sin(1.3+1.4*x+1.5*t)-1.)*(sin(1.3+1.4*x+1.5*t)+1.))+1.2*sqrt(sin(1.3+1.4*x+1.5*t)+1.)*sqrt(sin(1.3+1.4*x+1.5*t)-1.))*(1.+sqrt(1/((sin(1.3+1.4*x+1.5*t)-1.)*(sin(1.3+1.4*x+1.5*t)+1.)))*sqrt(sin(1.3+1.4*x+1.5*t)-1.)*sqrt(sin(1.3+1.4*x+1.5*t)+1.))/(sqrt(sin(1.3+1.4*x+1.5*t)+1.)*sqrt(sin(1.3+1.4*x+1.5*t)-1.)))^(1/3)-(.88026*I)*((1.1*ln(sin(1.3+1.4*x+1.5*t)+sqrt(sin(1.3+1.4*x+1.5*t)^2-1.))*sqrt((sin(1.3+1.4*x+1.5*t)-1.)*(sin(1.3+1.4*x+1.5*t)+1.))+1.2*sqrt(sin(1.3+1.4*x+1.5*t)+1.)*sqrt(sin(1.3+1.4*x+1.5*t)-1.))*(1.+sqrt(1/((sin(1.3+1.4*x+1.5*t)-1.)*(sin(1.3+1.4*x+1.5*t)+1.)))*sqrt(sin(1.3+1.4*x+1.5*t)-1.)*sqrt(sin(1.3+1.4*x+1.5*t)+1.))/(sqrt(sin(1.3+1.4*x+1.5*t)+1.)*sqrt(sin(1.3+1.4*x+1.5*t)-1.)))^(1/3))^2$$

$$p[17] := -1.0714+.71429*(-.50821*((1.1*ln(sin(1.3+1.4*x+1.5*t)+sqrt(sin(1.3+1.4*x+1.5*t)^2-1.))*sqrt((sin(1.3+1.4*x+1.5*t)-1.)*(sin(1.3+1.4*x+1.5*t)+1.))+1.2*sqrt(sin(1.3+1.4*x+1.5*t)+1.)*sqrt(sin(1.3+1.4*x+1.5*t)-1.))*(1.+sqrt(1/((sin(1.3+1.4*x+1.5*t)-1.)*(sin(1.3+1.4*x+1.5*t)+1.)))*sqrt(sin(1.3+1.4*x+1.5*t)-1.)*sqrt(sin(1.3+1.4*x+1.5*t)+1.))/(sqrt(sin(1.3+1.4*x+1.5*t)+1.)*sqrt(sin(1.3+1.4*x+1.5*t)-1.)))^(1/3)+(.88026*I)*((1.1*ln(sin(1.3+1.4*x+1.5*t)+sqrt(sin(1.3+1.4*x+1.5*t)^2-1.))*sqrt((sin(1.3+1.4*x+1.5*t)-1.)*(sin(1.3+1.4*x+1.5*t)+1.))+1.2*sqrt(sin(1.3+1.4*x+1.5*t)+1.)*sqrt(sin(1.3+1.4*x+1.5*t)-1.))*(1.+sqrt(1/((sin(1.3+1.4*x+1.5*t)-1.)*(sin(1.3+1.4*x+1.5*t)+1.)))*sqrt(sin(1.3+1.4*x+1.5*t)-1.)*sqrt(sin(1.3+1.4*x+1.5*t)+1.))/(sqrt(sin(1.3+1.4*x+1.5*t)+1.)*sqrt(sin(1.3+1.4*x+1.5*t)-1.)))^(1/3))^2$$

$$p[20] := -1.0714+.71429*(-.64030*((-1.1*arctan(1/sqrt(sec(1.3+1.4*x+1.5*t)^2-1.))*sqrt(sec(1.3+1.4*x+1.5*t)-1.)*sqrt(sec(1.3+1.4*x+1.5*t)+1.)+1.2*sqrt(sec(1.3+1.4*x+1.5*t)^2-1.))/sqrt(sec(1.3+1.4*x+1.5*t)^2-1.))^(1/3)-(1.1091*I)*((-1.1*arctan(1/sqrt(sec(1.3+1.4*x+1.5*t)^2-1.))*sqrt(sec(1.3+1.4*x+1.5*t)-1.)*sqrt(sec(1.3+1.4*x+1.5*t)+1.)+1.2*sqrt(sec(1.3+1.4*x+1.5*t)^2-1.))/sqrt(sec(1.3+1.4*x+1.5*t)^2-1.))^(1/3))^2$$

$$p[21] := -1.0714+.71429*(-.64030*((-1.1*arctan(1/sqrt(sec(1.3+1.4*x+1.5*t)^2-1.))*sqrt(sec(1.3+1.4*x+1.5*t)-1.)*sqrt(sec(1.3+1.4*x+1.5*t)+1.)+1.2*sqrt(sec(1.3+1.4*x+1.5*t)^2-1.))/sqrt(sec(1.3+1.4*x+1.5*t)^2-1.))^(1/3)+(1.1091*I)*((-1.1*arctan(1/sqrt(sec(1.3+1.4*x+1.5*t)^2-1.))*sqrt(sec(1.3+1.4*x+1.5*t)-1.)*sqrt(sec(1.3+1.4*x+1.5*t)+1.)+1.2*sqrt(sec(1.3+1.4*x+1.5*t)^2-1.))/sqrt(sec(1.3+1.4*x+1.5*t)^2-1.))^(1/3))^2$$

$$p[29] := -1.0714+.73794*((1.1*ln(cos(1.3+1.4*x+1.5*t)+sqrt(cos(1.3+1.4*x+1.5*t)^2-1.))*sqrt((cos(1.3+1.4*x+1.5*t)-1.)*(cos(1.3+1.4*x+1.5*t)+1.))+1.2*sqrt(cos(1.3+1.4*x+1.5*t)+1.)*sqrt(cos(1.3+1.4*x+1.5*t)-1.))*(1.+sqrt(1/((cos(1.3+1.4*x+1.5*t)-1.)*(cos(1.3+1.4*x+1.5*t)+1.)))*sqrt(cos(1.3+1.4*x+1.5*t)-1.)*sqrt(cos(1.3+1.4*x+1.5*t)+1.))/(sqrt(cos(1.3+1.4*x+1.5*t)+1.)*sqrt(cos(1.3+1.4*x+1.5*t)-1.)))^(2/3)$$

$$p[31] := -1.0714+.73794*((1.1*ln(sin(1.3+1.4*x+1.5*t)+sqrt(sin(1.3+1.4*x+1.5*t)^2-1.))*sqrt((sin(1.3+1.4*x+1.5*t)-1.)*(sin(1.3+1.4*x+1.5*t)+1.))+1.2*sqrt(sin(1.3+1.4*x+1.5*t)+1.)*sqrt(sin(1.3+1.4*x+1.5*t)-1.))*(1.+sqrt(1/((sin(1.3+1.4*x+1.5*t)-1.)*(sin(1.3+1.4*x+1.5*t)+1.)))*sqrt(sin(1.3+1.4*x+1.5*t)-1.)*sqrt(sin(1.3+1.4*x+1.5*t)+1.))/(sqrt(sin(1.3+1.4*x+1.5*t)+1.)*sqrt(sin(1.3+1.4*x+1.5*t)-1.)))^(2/3)$$

$$p[32] := -1.0714+1.1714*((-1.1*arctan(1/sqrt(csc(1.3+1.4*x+1.5*t)^2-1.))*sqrt(csc(1.3+1.4*x+1.5*t)-1.)*sqrt(csc(1.3+1.4*x+1.5*t)+1.)+1.2*sqrt(csc(1.3+1.4*x+1.5*t)^2-1.))/sqrt(csc(1.3+1.4*x+1.5*t)^2-1.))^(2/3)$$

$$p[33] := -1.0714+1.1714*((-1.1*arctan(1/sqrt(sec(1.3+1.4*x+1.5*t)^2-1.))*sqrt(sec(1.3+1.4*x+1.5*t)-1.)*sqrt(sec(1.3+1.4*x+1.5*t)+1.)+1.2*sqrt(sec(1.3+1.4*x+1.5*t)^2-1.))/sqrt(sec(1.3+1.4*x+1.5*t)^2-1.))^(2/3)$$

$$<p[34] := -1.0714+1.1714*((-1.1*arctan(1/sqrt(sech(1.3+1.4*x+1.5*t)^2-1.))*sqrt(sech(1.3+1.4*x+1.5*t)-1.)*sqrt(sech(1.3+1.4*x+1.5*t)+1.)+1.2*sqrt(sech(1.3+1.4*x+1.5*t)^2-1.))/sqrt(sech(1.3+1.4*x+1.5*t)^2-1.))^(2/3)$$

雅可比橢圓函數展開

## 參考文獻

- *谷超豪 《孤立子理論中的達布變換及其幾何應用》上海科學技術出版社
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- Dongming Wang, Elimination Practice,Imperial College Press 2004
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- George Articolo Partial Differential Equations & Boundary Value Problems with Maple V Academic Press 1998 ISBN 9780120644759