請問單擺勢能函數問題...

2010-02-11 9:21 am

有哪位高手可以解以下問題
微分方程：
x''=x-3/2x^2
(1)證明量函數 H(x,x'):=1/2(x' ^2-x^2+x^3) 沿任何軌跡量不變！
(2)繪此系統之相位圖。找出其平衡點，及其穩定形式。
(3)找出相平面上之Separatrix曲線。

更新1:

抱歉，有亂碼，重貼部分方程式：微分方程： x''=x-3/2x^2 (1)證明量函數 H(x,x'):=1/2(x' ^2 -x^2+x^3) 沿任何軌跡量不變！ ................ 其餘同

回答 (1)

天助

2010-02-20 6:44 am

✔ 最佳答案

參考一下:
(1)x"=x- 1.5x^2, then x'x"= xx'- 1.5x^2 x'
integrate w.r.t. time t, then 0.5(x')^2 = 0.5x^2 - 0.5x^3+C
thus, 0.5[(x')^2-x^2+x^3]= C independent on time t
(2) (a)phase diagram is the famliy of y^2-x^2+x^3= 2C, C is any real number
(b) critical point (node)
x"=x-1.5x^3 then
{ x'=y
{ y'=x- 1.5x^2 = -1.5(x- 2/3)^2 - (x- 2/3)
thus, critical point are (x,y)=(0,0), (2/3, 0)
case 1: node(0,0)
the linear form of the system is
{ x'=y
{ y'=x
eigenvalue of the system are 1,-1, so, the node(0,0) is a saddle point
(unstable)
case 2: node(2/3, 0)
the linear form of the system is
{ X'= Y
{ Y'= -X , where X=x-2/3, Y=y
eigenvalue are i, -i, so, the node (2/3, 0) is a center point(stable)
(3)Separatrix
the separatrix of the system is y^2-x^2+x^3=0, shown as:

圖片參考：http://imgcld.yimg.com/8/n/AD04686329/o/161002110070113872639760.jpg

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