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- | ===== **2.1. 泛函极值的必要条件与Euler方程** ===== | ||
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- | ·泛函极值的必要条件 | ||
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- | 最简单的泛函的极值问题:\\ | ||
- | $ J[y(x)]= \int_{x_0}^{x_1} {F(x,y,y')dx,} $ (5) \\ | ||
- | 其中,$F∈C^2$,容许曲线$y(x)∈C^2[x_0,x_1]$,且满足边界条件: \\ | ||
- | $y(x_0)=y_0, y(x_1)=y_1.$ (6) \\ | ||
- | 极值的必要条件:使泛函$J[y]=\int_{x_0}^{x_1} {F(x,y,y')dx}$达到极值的必要条件,是y(x)满足Euler方程: \\ | ||
- | $F_y- \frac{d}{dx}F_y=0.$ (7) \\ | ||
- | 证明:构造以α为参数的容许曲线族$\overline {y}(\alpha )=y(x)+\alpha \delta y$,\\ | ||
- | 其中,$\delta y$为宗量y(x)的变分,满足$\delta y|_{x=x_0}=\delta y|_{x=x_1}=0$.\\ | ||
- | 所以,$\overline {y}|_{x=x_0}=y_0,\overline {y}|_{x=x_1}=y_1,\overline {y}\in C^2[x_0,x_1]$.\\ | ||
- | 假设,当α=0时,$\overline {y}=y(x)$是使泛函(18)达到极值的曲线,将$\overline {y}$代入方程(18),得到\\ | ||
- | $\Phi (\alpha )=J[y+\alpha \delta y]=\int_{x_0}^{x_1} {F(x,y+\alpha \delta y,y'+\alpha \delta y')dx.}$\\ | ||
- | 由于$\Phi (\alpha )$在α=0时取得极值,由极值的必要条件,有\\ | ||
- | $\Phi '(0)=\frac{\partial }{\partial \alpha }J[y+\alpha \delta y]|_{\alpha =0}=\int_{x_0}^{x_1} {(F_y\delta y+F_y\delta y')dx}=0.$ (8)\\ | ||
- | 由分部积分法,并注意到$\delta y|_{x=x_0}=\delta y|_{x=x_1}=0$,有\\ | ||
- | $\int_{x_0}^{x_1} {F_{y'}}\delta y'dx=\int_{x_0}^{x_1} {F_{y'}}d(\delta y)=F_{y'}\delta y|_{x_0}^{x_1}-\int_{x_0}^{x_1} {\delta y\frac{d}{dx}F_{y'}dx}=-\int_{x_0}^{x_1} {\delta y\frac{d}{dx}F_{y'}dx}$.\\ | ||
- | 将上式代入方程(8),得到\\ | ||
- | $\Phi '(0)=\delta J=\int_{x_0}^{x_1} {[F_y-\frac{d}{dx}F_{y'}]\delta ydx}=0.$\\ | ||
- | 由基本引理1知,使泛函J[y]达到极值的函数y(x)必满足Euler方程7.\\ | ||
- | 说明:注意到满足Euler方程7仅是泛函取到极值的必要条件,Euler方程7的解曲线仅仅是可能的极值曲线。但是,在实际问题中,往往可以事先确定泛函的极值是否存在。在确定了极值存在的情况下,Euler方程的解曲线就是极值曲线.\\ | ||
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- | ·Euler方程的几种不同形式\\ | ||
- | 为计算简便,Euler方程7,即\\ | ||
- | $F_y-\frac{d}{dx}F_{y'}=0,$\\ | ||
- | 可以改写成如下形式:\\ | ||
- | $F_{y'y'}y''+F_{yy'}y'+F_{xy'}-F_y=0$,(9)\\ | ||
- | 和 \\ | ||
- | $\frac{d}{dx}(F-y'F_{y'})-F_x=0$. (10) \\ | ||
- | ∗**例1.** 求泛函$J[y(x)]=\int_{0}^{1} {(y'^2+12xy)dx}$满足边界条件y(0)=0,y(1)=1的极值曲线.\\ | ||
- | **解:**由于$F(x,y,y')=y'^2+12xy$,其Euler方程为\\ | ||
- | $12x-2y''=0.$\\ | ||
- | 解此二阶常微分方程,得\\ | ||
- | $y=x^3+c_1x+c_2.$\\ | ||
- | 由y(0)=0,y(1)=1得,$c_1=c_2=0$,因此,J[y(x)]的极值曲线为$y=x^{3}$.\\ | ||
- | ∗**例2.** 求泛函$J[y(x)]=\int_{0}^{1} {(y'^2-y^2-2xy)dx}$满足边界条件y(0)=y(1)=0的极值曲线.\\ | ||
- | **解:**$F(x,y,y')=y'^2-y^2-2xy$,其Euler方程为\\ | ||
- | $2y''+2y+2x=0$.\\ | ||
- | 解这个方程,得\\ | ||
- | $y=c_1cosx+c_2sinx-x.$\\ | ||
- | 由边界条件y(0)=y(1)=0,得$c_1=0,c_2=\frac{1}{sin1}$,于是,J[y(x)]的极值曲线为\\ | ||
- | $y=\frac{sinx}{sin1}-x.$\\ | ||
- | --- //[[xiaosai567@gmail.com|黄经州]] 2010/05/21 20:52// \\ | ||
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- | ·几种特殊的Euler方程 \\ | ||
- | *$F=F(x,y)$,F中不含y' \\ | ||
- | 此时,Euler方程为$F_y(x,y)=0$,这是一个普通的函数方程,它可以表示一条或几条曲线,但未必适合边界条件。因此,在一般情况下无解,只有曲线满足边界条件时有解。\\ | ||
- | ∗**例3.** 讨论泛函$J[y(x)]= \int_{x_0}^{x_1} {y^2dx,}$,$y(x_0)=y_0$,$y(x_1)=y_1$的极值曲线。\\ | ||
- | **解:**$F=y^2$,$F_y=2y=0$,所以$y=0$.故当y_0=y_1=0时有极值曲线$y=0$,否则无极值曲线。\\ | ||
- | * $F=P(x,y)+Q(x,y)y'$,F是y'的线性函数。\\ | ||
- | 此时,Euler方程为$P_y=Q_x$.与上述情况一样,这也是一个普通的函数方程。\\ | ||
- | 但是,如果$P_y=Q_x$,则$Pdx+Qdy$是某一函数u(x,y)的全微分, | ||
- | 因而$ J[y(x)]= \int_{x_0}^{x_1} {(P+Qy')dx,} =\int_{x_0,y_0}^{x_1,y_1}{Pdx+Qdy,}= \int_{x_0,y_0}^{x_1,y_1}{du}=u(x_1,y_1)-u(x_0,y_0)=const$\\ | ||
- | 此时变分失去意义。\\ | ||
- | ∗**例4.** 讨论泛函$J[y(x)]= \int_{x_0}^{x_1} {(y+xy')dx}$,$y(x_0)=y_0$,$y(x_1)=y_1$的极值曲线。\\ | ||
- | **解:**此时,Euler方程成为恒等式,变分失去意义。\\ | ||
- | * $F=F(y')$,F只依赖于$y'$\\ | ||
- | 若$y''=0$,则$y=c_1(x)+c_2$是含有两个参数的直线簇。\\ | ||
- | 若y''≠0,则$F_{y'y'}(y')=0$,解之得,$y'=k(常数)$,即,$y=kx+c$.这是含有一个参数的直线簇,它包含在前面的直线簇中。\\ | ||
- | ∗**例5.** 在联接两点$A(x_0,y_0)$,$B(x_1,y_1)$的所有平面曲线中,试求长度最短的曲线。\\ | ||
- | 解:问题可以化为在边界条件$y(x_0)=y_0$,$y(x_1)=y_1$下求泛函\\ | ||
- | $J[y(x)]= \int_{x_0}^{x_1} {\sqrt (1+y'^2)dx}$的极小值。\\ | ||
- | 由于$F=sqrt(1+y'^2)$仅依赖于y',它的极值曲线为直线簇$y=c_1(x)+c_2$,代入边界条件确定$c_1$,$c_2$,得\\ | ||
- | $y=y_0+\frac{y_1-y_0}{x_1-x_0}(x-x_0)$.因此,所求最短曲线就是连接A,B两点的直线。\\ | ||
- | * $F=F(x,y')$,F不含$y$\\ | ||
- | 此时,Euler方程为$\frac{d}{dx}F_{y'}(x,y')=0$,它的积分为$F_{y'}(x,y')=c_1$.这是不含y的一阶常微分方程,解此方程即得可能的极值曲线。\\ | ||
- | ∗**例6.** 求泛函$J[y]=\int_{x_0}^{x_1} {\frac{\sqrt {1+y'^2 }}{x}}dx$,$y(x_0)=y_0$,$y(x_1)=y_1$,的极值曲线。\\ | ||
- | 解:$F=\frac{\sqrt {1+y'^2} }{x}$的Euler方程为\\ | ||
- | $\frac{d}{dx}\frac{y'}{x\sqrt {1+y'2} }=0$\\ | ||
- | 它的初积分为$\frac{y'}{x\sqrt {1+y'^2} }=c,\Rightarrow x=\frac{y'}{c\sqrt {1+y'^2} }$\\ | ||
- | 令$y'=tant$,则:\\ | ||
- | $x=\frac{tant}{c\sqrt {tan^2t} }=\frac{1}{t}sint=c_1sint,c_1=\frac{1}{t}$\\ | ||
- | $dy=y'dx=tant\cdot c_1costdt=c_1sintdt$,\\ | ||
- | $y=-c_1cost+c_2$.\\ | ||
- | 故有,$\left\{ {\begin{array}\ {x=c_1sint,} \\ {y=-c_1cost+c_2} \\ \end{array} } \right.$\\ | ||
- | 消去参数t,得,$x^2+(y-c_2)^2=c_1^2$.\\ | ||
- | 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> | ||
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- | ∗**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。\\ | ||
- | 解:$F=\frac{\sqrt {1+y'^2}}{\sqrt{2gy}}$不含x,故Euler方程具有初始积分$y'F_{y'}=c$,即\\ | ||
- | $y'\frac{y'}{\sqrt{2gy}\sqrt{1+y'^2}}-\frac{\sqrt{1+y'^2}}{\sqrt{2gy}}=c$\\ | ||
- | 化简,得$y(1+y'^2)=c_1.$\\ | ||
- | 令$y'=cot\frac{t}{2}$,则\\ | ||
- | $y=\frac{c_1}{1+y'^2}=c_1sin^2\frac{t}{2}=\frac{c_1}{2}(1-cost)$\\ | ||
- | $dx=\frac{dy}{y'}=\frac{\frac{c_1}{2}sint}{cot\frac{t}{2}}=\frac{c_1}{2}(1-cost)dt.$\\ | ||
- | 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.$ \\ | ||
- | 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\\ | ||
- | $\left\{ {\begin{array} \{x=\frac{c_1}{2}(t-sint),} \\ {y=\frac{c_1}{2}(1-\frac{c_1}{2}(1-cost}.\end{array} } \right.$ \\ | ||
- | 摆线。这是一条过A,B两点的摆线。 | ||
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- | <note important>edited by 谭小球(0921062)2010/06/10</note> | ||