keynote:lesson11
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| 故有，$\left\{ {\begin{array}\ {x=c_1sint,} \\ {y=-c_1cost+c_2} \\ \end{array} } \right.$\\ | 故有，$\left\{ {\begin{array}\ {x=c_1sint,} \\ {y=-c_1cost+c_2} \\ \end{array} } \right.$\\ | ||
| 消去参数t，得，$x^2+(y-c_2)^2=c_1^2$.\\ | 消去参数t，得，$x^2+(y-c_2)^2=c_1^2$.\\ | ||
| - | 这表示圆心在纵轴上的一簇圆，其中，$c_1$,$c_2$由边界条件确定。 | + | 这表示圆心在纵轴上的一簇圆，其中，$c_1$,$c_2$由边界条件确定。 \\ |
| - | <note important>edited by 杨冰(0921060)2010/05/22 </note> | + | <note important>edited by 杨冰(0921060)2010/05/22 </note> \\ |
| ∗**例7.** 求最速降线问题的解，即求 \\ $T=\int_{0}^{a} {\frac{\sqrt {1+y'^2}}{\sqrt{2gy}}}dx$,$y(0)=0$,$y(a)=b$,的极值曲线。\\ | ∗**例7.** 求最速降线问题的解，即求 \\ $T=\int_{0}^{a} {\frac{\sqrt {1+y'^2}}{\sqrt{2gy}}}dx$,$y(0)=0$,$y(a)=b$,的极值曲线。\\ | ||
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| $dx=\frac{dy}{y'}=\frac{\frac{c_1}{2}sint}{cot\frac{t}{2}}=\frac{c_1}{2}(1-cost)dt.$\\ | $dx=\frac{dy}{y'}=\frac{\frac{c_1}{2}sint}{cot\frac{t}{2}}=\frac{c_1}{2}(1-cost)dt.$\\ | ||
| 故\\ | 故\\ | ||
| - | $ \left\{ {\begin{array} \{x=\frac{c_1}{2}(t-sint)+c_2,} \\ {y=\frac{c_1}{2}(1-\frac{c_1}{2}(1-cost}. \\ \end{array} } \right.$ \\ | + | $\left\{ {\begin{array}\ {x=\frac{c_1}{2}(t-sint)+c_2,} \\ {y= \frac{c_1}{2}\frac{c_1}{2}(1-cost)}. \\ \end{array} } \right.$\\ |
| - | 由t=0时，x=0,得c_2=0,而c_1由y(a)=b定出。于是，最速降线问题的解为\\ | + | 由t=0时，x=0,得$c_2=0$,而$c_1$由y(a)=b定出。于是，最速降线问题的解为\\ |
| - | $\left\{ {\begin{array} \{x=\frac{c_1}{2}(t-sint),} \\ {y=\frac{c_1}{2}(1-\frac{c_1}{2}(1-cost}.\end{array} } \right.$ \\ | + | $\left\{ {\begin{array}\ {x=\frac{c_1}{2}(t-sint),} \\ {y= \frac{c_1}{2}\frac{c_1}{2}(1-cost)}. \\ \end{array} } \right.$\\ |
| - | 摆线。这是一条过A，B两点的摆线。 | + | 摆线。这是一条过A，B两点的摆线。\\ |
| - | <note important>edited by 谭小球(0921062)2010/06/10</note> | + | ∗**例8.** 求最小旋转面面积问题的解，即求泛函 \\ |
| + | $S[y]=2π\int_{x_0}^{x_1}{y\sqrt{1+y'^2}}dx$,$y(x_0)=y_0$,$y(x_1)=y_1$ \\ 的极值曲线。 \\ | ||
| + | 解：Euler方程具有初积分$y'F_{y'}=c$，即 \\ | ||
| + | $2πy \frac{y'^2}{\sqrt{1+y'^2}}-2πy\sqrt{1+y'^2}=c$ \\ | ||
| + | 化简得，$y=c\sqrt{1+y'^2}.$ \\ | ||
| + | 令$y'=sinht$，代入上式得 \\ | ||
| + | $y=c_1cosht,$ \\ | ||
| + | $dx=\frac{dy}{y'}=\frac{c_1sinht}{sinht}dt=c_1dt.$ \\ | ||
| + | 于是，所求曲面是由平面曲线 \\ | ||
| + | $\left\{ {\begin{array}\ {x={c_1}t-c_2,} \\ {y= c_1cosht}. \\ \end{array} } \right.$\\ | ||
| + | 绕OX轴旋转而成的，消去参数t，得 \\ | ||
| + | $y=c_1cosh\frac{x-c_2}{c_1},$ \\ | ||
| + | 其中，$c_1$,$c_2$由$y(x_0)=y_0$,$y(x_1)=y_1$确定，因此，最小曲面是悬链面。 \\ | ||
| + | ====**2.2. 含有多个未知函数的变分问题**==== \\ | ||
| + | ★含两个未知函数的泛函极值的必要条件 \\ | ||
| + | ☆问题： \\ | ||
| + | $J[y(x),z(x)]=\int_{x_0}^{x_1}{F(x,y,z,y',z')}dx,$ (11) \\ | ||
| + | 其中，F关于所含变量具有二阶连续领导，容许曲线$y(x)$,$z(x)$∈$C^2[x_0,x_1]$,且满足边界条件： \\ | ||
| + | $\left\{ {\begin{array}\ {y(x_0)=y_0, y(x_1)=y_1,} \\ {z(x_0)=z_0, z(x_1)=z_1}. \\ \end{array} } \right.$ \\ | ||
| + | ☆极值的必要条件：使泛函$J[y(x),z(x)]=\int_{x_0}^{x_1}{F(x,y,z,y',z')}dx$达到极值的必要条件是，y(x),z(x)满足Eular方程组，即： \\ | ||
| + | $\left\{ {\begin{array}\ {F_y-\frac{d}{dx}F_{y'}=0,} \\ {F_z-\frac{d}{dx}F_{z'}=0}. \\ \end{array} } \right.$ (12)\\ | ||
| + | |||
| + | ☆例1.求泛函$J[y,z]=\int_{0}^{\frac{π}{2}{(y'^2+z'^2+2yz)}}dx$满足边界条件： \\ | ||
| + | $\left\{ {\begin{array}\ {y(x_0)=0, y(\frac{π}{2})=1,} \\ {z(x_0)=0, z(\frac{π}{2})=-1}. \\ \end{array} } \right.$ \\ | ||
| + | 的极值曲线。 \\ | ||
| + | 解：$F=y'^2+z'^2+2yz$,故Eular方程组为：\\ | ||
| + | $\left\{ {\begin{array}\ {2z-\frac{d}{dx}(2y')=0,} \\ {2y-\frac{d}{dx}(2z')=0}. \\ \end{array} } \right.$\\ | ||
| + | 得： \\ | ||
| + | $\left\{ {\begin{array}\ {z-y^{(2)}=0,} \\ {y-z^{(2)}=0}. \\ \end{array} } \right.$\\ | ||
| + | 由此方程组消去z,得$y^{(4)}-y=0$，其通解为：\\ | ||
| + | $\left\{ {\begin{array}\ {y=c_1e^x+c_2e^{-x}+c_3cosx+c_4sinx,} \\ {z=y''=c_1e^x+c_2e^{-x}-c_3cosx-c_4sinx}. \\ \end{array} } \right.$\\ | ||
| + | 由边界条件可确定$c_1=c_2=c_3=0$,$c_4=1$,故，所求极值曲线为： \\ | ||
| + | $\left\{ {\begin{array}\ {y=sinx,} \\ {z=-sinx}. \\ \end{array} } \right.$\\ | ||
| + | |||
| + | <note> edited by 谭小球(0921062)2010/06/10</note> | ||
| + | ★例2 求泛函$J[y(x),z(x)]=\int_{x_0}^{x_1}{F(y',z')}dx,$ 的极值曲线，其中假设 F_{y'y'}-F^2_{y'z'} ≠0 \\ | ||
| + | 解：因为F_y=F_z=0,故Euler方程组为 \\ | ||
| + | $\left\{ {\begin{array}\ {\frac{d}{dx}(F_{y'})=0,} \\ {\frac{d}{dx}(F_{z'})=0}. \\ \end{array} } \right.$\\ | ||
| + | 得：\\ | ||
| + | $\left\{ {\begin{array}\ {F_{y'y'}y"+F_{y'z'}z"=0,} \\ {F_{y'z'}y"+F_{z'z'}z"=0}. \\ \end{array} } \right.$\\ | ||
| + | 根据假设条件，该方程只有零解y"=0,z"=0,即 \\ | ||
| + | $\left\{ {\begin{array}\ {y=c_1x+c_2,} \\ {z=c_3x+c_4}. \\ \end{array} } \right.$\\ | ||
| + | 是所求的极值曲线。\\ | ||
| + | ·含n(n>2)个未知函数的泛函极值的必要条件\\ | ||
| + | 泛函$J=\int_{x_0}^{x_1}{F(x,y_1,y_2,…,y_n,y'_1,y'_2,…y'_n)}dx,$在满足边界条件 \\ | ||
| + | $\left\{ {\begin{array}\ {y_i(x_0)=y_{i0},} \\ {y_i(x_1)=y_{i1}},i=1,2,…n, \\ \end{array} } \right.$\\ | ||
| + | 下，取得极值的必要条件是y_1(x),y_2(x),…,y_n(x),满足Euler方程组\\ | ||
| + | F_{yi}-\frac{d}{dx}(F_{y'i})=0,i=1,2,…n. \\ | ||
| + | 上述方程组通解中的常数，可以由所给的边界条件确定。\\ | ||
| + | <note> edited by 杨立春(10921048)2010/06/26</note> | ||
keynote/lesson11.txt · Last modified: 2023/08/19 21:02 (external edit)
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