参数曲面上测地线方程及求解
已知参数曲面\(x=f(u,v),y=g(u,v),z=h(u,v)\)及曲面上的两点\((x_1,y_1,z_1),(x_2,y_2,z_2) \)，如何求解经过已知两点的测地线方程及解析式渐近表达？例如：对于双椭圆曲面台\(x=(ka_1+(1-k)a_2)\cos(t)+kx_0,y=(kb_1+(1-k)b_2)\sin(t)+ky_0,z=kh_1-(1-k)h_2\) 及曲面上的两点\((x_1,y_1,z_1),(x_2,y_2,z_2) \) ，如何求解经过已知两点的测地线方程及解析式渐近表达？其中\(k,t\)为变量，\(a_1,a_2,b_1,b_2,h_1,h_2,x_0,y_0\)为已知常数
对于隐函数z=f(x,y)的曲面方程，其测地线有下面经典的结果：
\(\frac{d^2y}{dx^2}[(1+(\frac{\partial f}{\partial x})^2+(\frac{\partial f}{\partial y})^2]=[\frac{\partial^2 f}{\partial x^2}+2\frac{\partial^2 f}{\partial x \partial y}+\frac{\partial^2 f}{\partial y^2}(\frac{dy}{dx})^2](\frac{\partial f}{\partial x}\frac{dy}{dx}-\frac{\partial f}{\partial y})\)
关于参数的曲面方程的测地线的微分方程如何？如何求解其解析解的渐近表达？ 试着使用隐函数求导，把z看成函数,x,y看成隐变量。1#和2#都是用f表示不同意义，容易造成混淆，建议2#中f用z替换，这样就不容易混淆了。
然后对于1#中表达式分别对x,y求偏导，比如第一式子分别对x,y求偏导得到
\(\begin{cases}1=\frac{\partial f}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial x}\\ 0=\frac{\partial f}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial f}{\partial v}\frac{\partial v}{\partial y}\end{cases}\)
同样对于第二表达式也对x,y求偏导得到
\(\begin{cases}0=\frac{\partial g}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial g}{\partial v}\frac{\partial v}{\partial x}\\ 1=\frac{\partial g}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial g}{\partial v}\frac{\partial v}{\partial y}\end{cases}\)
然后四个式子解方程，可以解得\(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial v}{\partial x},\frac{\partial v}{\partial y}\)。
此后我们对第三式子也分别对x,y求偏导得到
\(\begin{cases}\frac{\partial z}{\partial x}=\frac{\partial h}{\partial u}\frac{\partial u}{\partial x}+\frac{\partial h}{\partial v}\frac{\partial v}{\partial x}\\ \frac{\partial z}{\partial y}=\frac{\partial h}{\partial u}\frac{\partial u}{\partial y}+\frac{\partial h}{\partial v}\frac{\partial v}{\partial y}\end{cases}\)
将上面\(\frac{\partial u}{\partial x},\frac{\partial u}{\partial y},\frac{\partial v}{\partial x},\frac{\partial v}{\partial y}\)结果带入。
于是我们得到z关于x,y的一阶偏导数。由于2#中还需要计算z关于x,y的二阶偏导，那么还需要对上面一阶偏导继续计算下去，这个计算量看上去有点大。
最后将这些计算结果带入2#的公式就应该可以了
没记错的话（微分几何教程），测地线方程的经典表达就是参数化的 终于找到了相关的结果:
设C是曲面上过点P的弧长正则曲线：\( r(s)=r(u(s),v(s))\),测地曲率为
\(k_g=[\varGamma_{11}^2 (\frac{du}{ds})^3+(2 \varGamma_{12}^2-\varGamma_{11}^1)(\frac{du}{ds})^2(\frac{dv}{ds})+(\varGamma_{22}^2-2\varGamma_{12}^1)(\frac{du}{ds})(\frac{dv}{ds})^2-\varGamma_{22}^1 (\frac{dv}{ds})^3+\frac{du}{ds}\frac{d^2v}{ds^2}-\frac{d^2u}{ds^2}\frac{dv}{ds}]\sqrt{EG-F^2}\),
其中\(E,F,G\)为曲面的第一基本形式的系数。
当\(k_g=0\)即为测地线
其中
\(\varGamma_{11}^1=\frac{GE_u-2FF_u+FE_v}{2(EG-F^2)}\)
\(\varGamma_{11}^2=\frac{2EF_u-EE_v+FE_u}{2(EG-F^2)}\)
\(\varGamma_{12}^1=\frac{GE_v-FG_u}{2(EG-F^2)}\)
\(\varGamma_{12}^2=\frac{EG_u-FE_v}{2(EG-F^2)}\)
\(\varGamma_{22}^1=\frac{2GF_v-GG_u+FG_v}{2(EG-F^2)}\)
\(\varGamma_{22}^2=\frac{EG_v-2FF_v+FG_u}{2(EG-F^2)}\)
最后得到一个二阶常微分方程组，为了降低阶次，可以转化为下面的一阶常微分方程组：
\(\frac{du}{ds}=p\)
\(\frac{dv}{ds}=q\)
\(\frac{dp}{ds}=-\varGamma_{11}^1 p^2-2\varGamma_{12}^1 pq -\varGamma_{22}^1 q^2\)
\(\frac{dq}{ds}=-\varGamma_{11}^2 p^2-2\varGamma_{12}^2 pq -\varGamma_{22}^2 q^2\)
请各位大神利用上面的结论，给出1# 第2问的解答~~ 或许利用下面的刘维尔公式计算更方便一些：
对于(2)的问题
\(\)
参数条件：
\(a_1=50,b_1=60,a_2=80,b_2=90,x_0=30,y_0=50,t_0=\frac{\pi}{6},h_0=500,h_1=500\)
可以计算得到第一，第二基本形式系数：
\(E=\left( -600k+1700 \right) \left( \cos ^2\left( t \right) \right) +900\left( k-\frac{8}{3} \right) ^2,F=\left( -1500k-300\sin \!\left( t \right) +4500 \right) \cos \!\left( t \right) +\left( 900k-2400 \right) \sin \!\left( t \right) ,G=1004300-3000\sin \!\left( t \right) -1800\cos \!\left( t \right)\)
\(L=\frac{-900000k^2+5100000k-7200000}{D},M=\frac{150000\sin \!\left( 2t \right)}{D},N=0\)
\(D^2=90000\left( \cos ^4\left( t \right) \right) +\left( 540000k-1620000 \right) \left( \cos ^3\left( t \right) \right) +\left( \left( 900000k-2400000 \right) \sin \!\left( t \right) -1440000k^2-593400000k+1692730000 \right) \left( \cos ^2\left( t \right) \right) +2700000\left( \sin \!\left( t \right) -\frac{3}{5} \right) \left( k-\frac{8}{3} \right) \left( k-3 \right) \cos \!\left( t \right) -2700000\left( k-\frac{8}{3} \right) ^2\left( \sin \!\left( t \right) -\frac{5017}{15} \right)\)
有哪位大神可以 继续简化得到最终结果？（若能得测地线弧长s关于t的微分方程或者渐近表达式更好）
请利用数学软件给出下面微分方程的解（渐近解k(t)）及绘出k(t)的图像？
\(c_0(k(t),t)\frac{d^2 {k(t)}}{dt^2}+c_1(k(t),t)(\frac{d {k(t)}}{dt})^3+c_2(k(t),t)({\frac{d {k(t)}}{dt}})^2+c_3(k(t),t){\frac{d {k(t)}}{dt}}+c_4(k(t),t)
=0\)
其中方程系数满足：
c_0(k(t),t)=180000*cos(t)^4 + (1080000*k(t) - 3240000)*cos(t)^3 + ((1800000*k(t) - 4800000)*sin(t) - 2880000*k(t)^2 - 1186800000*k(t)+ 3385460000)*cos(t)^2 + 5400000*(sin(t) - 3/5)*(k(t) - 8/3)*(k(t) - 3)*cos(t) - 5400000*(sin(t) - 5017/15)*(k(t) - 8/3)^2
c_1(k(t),t)=1080000*sin(t)*(-(2*cos(t)^2*k(t)^2)/3 + (k(t)/2 - 3/2)*cos(t) + (-(34*k(t))/9 + 140/27)*sin(t)^2 + ((5*k(t))/6 - 20/9)*sin(t) + k(t)^3 - (49*k(t)^2)/6 + (51*k(t))/2 - 24)*cos(t)
c_2(k(t),t)=(1080000*k(t) - 3060000)*cos(t)^4 + (17820000*k(t) - 23940000)*cos(t)^3 + ((-5400000*k(t)^2 + 31500000*k(t) - 45000000)*sin(t) - 3960000*k(t) - 590340000)*cos(t)^2 + (3240000*sin(t)^2*k(t)^2 + (5400000*k(t) - 15300000)*sin(t) - 21060000*k(t)+ 33660000)*cos(t) - 5400000*(k(t) - 8/3)*(sin(t) - 5017/15)
c_3(k(t),t)=(8100000*k(t) + 360000*sin(t) - 23400000)*cos(t)^3 + ((-4860000*k(t) + 13500000)*sin(t) - 5400000*(k(t) - 8/3)*(k(t) - 3))*cos(t)^2 + ((-2880000*k(t)^2 + 2428680000*k(t) - 6858400000)*sin(t) + 2700000*k(t)^2 - 23400000*k(t)+ 45000000)*cos(t) - 1620000*(sin(t) - 5/3)*(k(t) - 8/3)*(k(t)- 3)
c_4(k(t),t)=1620000*cos(t)^3 + (2880000*k(t)+ 2700000*sin(t) - 1214340000)*cos(t)^2 + ((-5400000*k(t) + 15300000)*sin(t) + 1807740000*k(t) - 4821180000)*cos(t) + (3012900000*k(t)- 9040500000)*sin(t) - 10620000*k(t)+ 633900000
\(c_0(k(t),t)=180000\cos(t)^4 + (1080000k(t)- 3240000)\cos(t)^3 + ((1800000k - 4800000)\sin(t) - 2880000k(t)^2 - 1186800000k(t)+ 3385460000)\cos(t)^2 + 5400000(\sin(t) - \frac{3}{5})(k(t)- \frac{8}{3})(k(t) - 3)\cos(t) - 5400000(\sin(t) - \frac{5017}{15})(k(t) -\frac{8}{3})^2\)
\(c_1(k(t),t)=1080000\frac{\sin(t)(-(2\cos(t)^2 k(t)^2)}{3} + (\frac{k(t)}{2} - \frac{3}{2})\cos(t) + (-\frac{34k(t)}{9} + \frac{140}{27})\sin(t)^2 + (\frac{5k(t)}{6}- \frac{20}{9})\sin(t) + k(t)^3 - \frac{49k(t)^2}{6} + \frac{51k(t)}{2} - 24)\cos(t)\)
\(c_2(k(t),t)=(1080000k(t) - 3060000)\cos(t)^4 + (17820000k(t) - 23940000)\cos(t)^3 + ((-5400000k^2 + 31500000k(t) - 45000000)\sin(t) - 3960000k(t)- 590340000)\cos(t)^2 + (3240000\sin(t)^2 k(t)^2 + (5400000k(t)- 15300000)\sin(t) - 21060000k(t)+ 33660000)\cos(t) - 5400000(k(t)- \frac{8}{3}(\sin(t) - \frac{5017}{15})\)
\(c_3(k(t),t)=(8100000k(t) + 360000\sin(t) - 23400000)\cos(t)^3 + ((-4860000k(t)+ 13500000)\sin(t) - 5400000(k(t)- \frac{8}{3})(k(t)- 3))\cos(t)^2 + ((-2880000k(t)^2 + 2428680000k(t)- 6858400000)\sin(t) + 2700000k(t)^2 - 23400000k(t) + 45000000)\cos(t) - 1620000(\sin(t) - \frac{5}{3})(k(t)- \frac{8}{3})(k(t) - 3)\)
\(c_4(k(t),t)=1620000\cos(t)^3 + (2880000k(t) + 2700000\sin(t) - 1214340000)\cos(t)^2 + ((-5400000k(t)+ 15300000)\sin(t) + 1807740000k(t)- 4821180000)\cos(t) + (3012900000k(t)- 9040500000)sin(t) - 10620000k(t) + 633900000\)
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