最美情侣中文字幕电影,在线麻豆精品传媒,在线网站高清黄,久久黄色视频

歡迎光臨散文網(wǎng) 會(huì)員登陸 & 注冊(cè)

最速降線求法的簡(jiǎn)介

2022-07-13 13:04 作者:現(xiàn)代微積分  | 我要投稿

此篇文章主要科普下最速降線的求解方法。之所以說(shuō)是科普是因?yàn)閮?nèi)容門檻有些高)有一半以上都是借助于《泛函分析》這本書(shū)的內(nèi)容以及我查閱的諸多文獻(xiàn)的,所以就當(dāng)大部分是“復(fù)述”好了[doge]。(我花費(fèi)了一上午才大致看明白“泛函”和“變分”的基本概念,看來(lái)探索真理任重道遠(yuǎn)呀)

如圖,一小球(視為質(zhì)點(diǎn))由A點(diǎn)靜止釋放,沿一光滑路徑運(yùn)動(dòng)至點(diǎn)C,不計(jì)一切阻力,則取何種路徑時(shí)運(yùn)動(dòng)時(shí)間最短?

當(dāng)路徑y(tǒng)=y(x)改變時(shí),對(duì)應(yīng)的時(shí)間t也會(huì)改變,因此路徑可以映射到時(shí)間上,換言之時(shí)間是關(guān)于路徑的函數(shù)t=t(y)

以A為坐標(biāo)原點(diǎn),建立左手系(圖如所示)

設(shè)路徑所在曲線為y=y(x)

根據(jù)功能關(guān)系,有:mgy%3D%5Cfrac%7B1%7D%7B2%7D%20mv%5E2

解得:v%3D%5Csqrt%7B2gy%7D

v%3D%5Cfrac%7Bds%7D%7Bdt%7D%20得:

dt%3D%5Cfrac%7Bds%7D%7Bv%7D%20%3D%5Cfrac%7B%5Csqrt%7B(dx)%5E2%2B(dy)%5E2%7D%20%7D%7B%5Csqrt%7B2gy%7D%20%7D%20%3D%5Cfrac%7B%5Csqrt%7B1%2B(y')%5E2%7D%20dx%7D%7B%5Csqrt%7B2gy%7D%20%7D%20

(ps:ds%3D%5Csqrt%7B(dx)%5E2%2B(dy)%5E2%7D%20為弧微分,即微小的一段弧可近似用直線替代,用勾股定理求之)

取其在區(qū)間[x?,x?]上的積分得:

T%3D%5Cint_%7Bx_1%7D%5E%7Bx_2%7D%5Cfrac%7B%5Csqrt%7B1%2B(y')%5E2%7D%7D%7B%5Csqrt%7B2gy%7D%20%7D%20dx%20

選取不同的y(曲線),時(shí)間t也會(huì)不同

輸入一個(gè)函數(shù)y,則輸出一個(gè)時(shí)間t,這是一種由函數(shù)空間到數(shù)域的映射,稱之為一種泛函

若我們需要找到一個(gè)y(曲線),使得輸出的時(shí)間t最小,也就是要找到這個(gè)泛函極值

求解該問(wèn)題則需要運(yùn)用變分法

由上式可知,T是關(guān)于x,y,y'的函數(shù),即:

T%3D%5Cint_%7Bx_1%7D%5E%7Bx_2%7D%20F(x%2Cy%2Cy')dx

設(shè)y(曲線)取得關(guān)于T的泛函極小值(默認(rèn)其一階導(dǎo)存在)

我們對(duì)這一曲線附加一微小的擾動(dòng)(且兩個(gè)端點(diǎn)處微擾值為0,即定端變分),則曲線形式發(fā)生輕微的改變使得T略有增大。當(dāng)微擾→0時(shí),總時(shí)間的改變→0,記作:

%5Cdelta%20T%3DY(y%2B%5Cdelta%20y)-T(y)

ps:符號(hào)δ和微分dx的d思想類似,但后者的微擾對(duì)應(yīng)為一個(gè)數(shù)x,而前者的微擾對(duì)應(yīng)為一族函數(shù),因此為了區(qū)別此便將其記作δ

上式展開(kāi)得:

%5Cdelta%20T%3D%5Cint_%7Bx_1%7D%5E%7Bx_2%7D%5BF(x%2Cy%2B%5Cdelta%20y%2Cy'%2B%5Cdelta%20y')%20-F(x%2Cy%2Cy')%5Ddx

F(x%2Cy%2B%5Cdelta%20y%2Cy'%2B%5Cdelta%20y')%3DF(x%2Cy%2Cy')%2B%5Cfrac%7B%5Cpartial%20F(x%2Cy%2Cy')%7D%7B%5Cpartial%20y%7D%20%5Cdelta%20y%2B%5Cfrac%7B%5Cpartial%20F(x%2Cy%2Cy')%7D%7B%5Cpartial%20y'%7D%20%5Cdelta%20y'

%3DF(x%2Cy%2Cy')%2B%5Cfrac%7B%5Cpartial%20F(x%2Cy%2Cy')%7D%7B%5Cpartial%20y%7D%20%5Cdelta%20y%2B%5Cfrac%7B%5Cpartial%20F(x%2Cy%2Cy')%7D%7B%5Cpartial%20y'%7D%20(%5Cdelta%20y)'

取后兩項(xiàng)積分得微擾的累計(jì)時(shí)間δT

%5Cdelta%20T%3D%5Cint_%7Bx_1%7D%5E%7Bx_2%7D(%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D%20%5Cdelta%20y%2B%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D%20(%5Cdelta%20y)')dx

考慮到端點(diǎn)處變分值為0,對(duì)上式第2項(xiàng)采用分部積分得:

%20%5Cint_%7Bx_1%7D%5E%7Bx_2%7D%20%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D(%5Cdelta%20y)'dx%3D%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D%5Cdelta%20y%7C%5E%7Bx_2%7D_%7Bx_1%7D%20-%5Cint_%7Bx_1%7D%5E%7Bx_2%7D(%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D%20%5Cdelta%20y)dx%20%3D-%5Cint_%7Bx_1%7D%5E%7Bx_2%7D(%5Cfrac%7Bd%7D%7Bdx%7D%20%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D%20%5Cdelta%20y)dx%20

%5Cdelta%20T%3D%5Cint_%7Bx_1%7D%5E%7Bx_2%7D%20(%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D-%5Cfrac%7Bd%7D%7Bdx%7D%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D)%20%5Cdelta%20y%20%20dx

令δT→0,則可讓T收斂于所設(shè)泛函極值

%5Cdelta%20T%3D%5Cint_%7Bx_1%7D%5E%7Bx_2%7D%20(%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D-%5Cfrac%7Bd%7D%7Bdx%7D%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D)%20%5Cdelta%20y%20%20dx%3D0

由于δy為任取的一階微小量,則上述積分恒為0需滿足括號(hào)內(nèi)恒為0,即:

%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y%7D-%5Cfrac%7Bd%7D%7Bdx%7D%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D%3D0

上式即著名的E-L方程(歐拉-拉格朗日方程)

43下面用E-L方程求解原題的最佳解泛函(即最速曲線)

據(jù)表達(dá)式F(x%2Cy%2Cy')%3D%5Cfrac%7B%5Csqrt%7B1%2B(y')%5E2%7D%20%7D%7B%5Csqrt%7B2gy%7D%20%7D%20可知,F(xiàn)是不含變量x的函數(shù)

%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20x%7D%3D0,由E-L方程可得,泛函T出現(xiàn)極值時(shí),有:

F-y'%5Cfrac%7B%5Cpartial%20F%7D%7B%5Cpartial%20y'%7D%20%3DC(C為常數(shù))

代入F表達(dá)式整理得:

%5Cbegin%7Barray%7D%0A%5C%5C%5Cfrac%7B%5Csqrt%7B1%2B(y')%5E2%7D%20%7D%7B%5Csqrt%7B2gy%7D%20%7D-y'%5Cfrac%7B%5Cfrac%7B1%7D%7B2%7D%5B1%2B(y')%5E2%5D%5E%7B-%5Cfrac%7B1%7D%7B2%7D%7D%20%5Ccdot%202y'%7D%7B%5Csqrt%7B2gy%7D%20%7D%20%0A%5C%5C%3D%5Cfrac%7B%5Csqrt%7B1%2B(y')%5E2%7D%20%7D%7B%5Csqrt%7B2gy%7D%20%7D-%5Cfrac%7B(y')%5E2%7D%7B%5Csqrt%7B2gy%7D%5Ccdot%20%5Csqrt%7B1%2B(y')%5E2%7D%20%20%7D%20%0A%5C%5C%3D%5Cfrac%7B1%7D%7B%5Csqrt%7B2gy%7D%5Ccdot%20%5Csqrt%7B1%2B(y')%5E2%7D%20%20%7D%20%0A%5C%5C%3DC%0A%5Cend%7Barray%7D

y(1%2B(y')%5E2)%3D%5Cfrac%7B1%7D%7B2gC%5E2%7D%3Dk%20

y'%3D%5Cfrac%7Bdy%7D%7Bdx%7D%3D%5Csqrt%7B%5Cfrac%7Bk-y%7D%7By%7D%20%7D%20%20

接下來(lái)就是求解微分方程了

分離變量得:%5Csqrt%7B%5Cfrac%7Bk-y%7D%7By%7D%20%7D%20%20dy%3Ddx

換元令y%3D%5Cfrac%7Bk%7D%7B2%7D%20(1-cos%5Ctheta%20),則

dy%3D%5Cfrac%7Bk%7D%7B2%7D%20sin%5Ctheta%20d%5Ctheta%20

代入得:%5Csqrt%7B%5Cfrac%7B%5Cfrac%7Bk%7D%7B2%7D(1-cos%5Ctheta%20)%20%7D%7Bk-%5Cfrac%7Bk%7D%7B2%7D%20(1-cos%5Ctheta%20)%7D%20%7D%20%20dy%3Ddx

%5Csqrt%7B%5Cfrac%7B1-cos%5Ctheta%20%7D%7B1%2Bcos%5Ctheta%20%7D%20%7D%20%5Ccdot%20%5Csqrt%7B%5Cfrac%7B1-cos%5Ctheta%20%7D%7B1-cos%5Ctheta%20%7D%20%7D%5Ccdot%20%5Cfrac%7Bk%7D%7B2%7Dsin%5Ctheta%20d%5Ctheta%20%3Ddx

dx%3D%5Cfrac%7Bk%7D%7B2%7D%20(1-cos%5Ctheta%20)d%5Ctheta%20

兩邊積分可得:

x%3D%5Cfrac%7Bk%7D%7B2%7D%20(%5Ctheta%20-sin%5Ctheta%20)%2BC

質(zhì)點(diǎn)在初始位置時(shí),y%3D%5Cfrac%7Bk%7D%7B2%7D%20(1-cos%5Ctheta%20)%3D0,θ=0

此時(shí)x=0,解得C=0

得軌跡參數(shù)方程:

%5Cleft%5C%7B%5Cbegin%7Bmatrix%7D%20%0Ax%3D%5Clambda%20(%5Ctheta%20-sin%5Ctheta%20)%5C%5Cy%3D%5Clambda%20(1-cos%5Ctheta%20)%0A%5Cend%7Bmatrix%7D%5Cright.%20

(其中θ為參數(shù))

此參數(shù)方程與擺線的參數(shù)方程相同,因此最速降線也是一條擺線

求出了該最佳泛函解,我們可以利用光學(xué)中的“光線傳播時(shí)間最短”(費(fèi)馬原理)以及斯涅爾定律(光的折射定律),通過(guò)取空介微元結(jié)合擺線的性質(zhì)加以驗(yàn)證。由于篇幅原因,驗(yàn)證的解析這里就不多贅述了。

數(shù)學(xué)家們的游戲希望有朝一日眼前的你也能參與其中[doge]!

最速降線求法的簡(jiǎn)介的評(píng)論 (共 條)

分享到微博請(qǐng)遵守國(guó)家法律
财经| 万盛区| 西城区| 衡东县| 临泽县| 东港市| 英超| 仪征市| 枣阳市| 屏东市| 开原市| 阿拉尔市| 秦皇岛市| 三都| 阿勒泰市| 伊宁市| 鄯善县| 共和县| 岗巴县| 铜川市| 萨迦县| 获嘉县| 潜山县| 沙河市| 江门市| 广昌县| 虎林市| 东乌| 松潘县| 肥城市| 溧水县| 曲周县| 渑池县| 原平市| 邓州市| 阜宁县| 正定县| 五常市| 蒙自县| 通河县| 黎平县|