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 Calculus of Variations, Euler-Lagrange Equation and Brachistochrone Problem.
10-17-2017, 11:47 PM (This post was last modified: 10-18-2017 12:18 AM by elim.)
Post: #1
 elim Moderator Posts: 582 Joined: Feb 2010 Reputation: 0
Calculus of Variations, Euler-Lagrange Equation and Brachistochrone Problem.
Consider a functional $F:C^1[a,b]\to\mathbb{R},\;y\mapsto F[y]={\small\displaystyle{\int_{\underset{\,}{a}}^b}}f(x,y(x),y'(x))dx$.
Assume $y=\varphi$ minimizes $F$. i.e. $F[\varphi]\le F[y]\;(\forall y\in C^1[a,b]),\;\;f\,$has all
the necessary smoothness and boundedness for the order change of integration
and differentiation below$\underset{\,}{.}$
For $h\in C^1[a,b]$ with $h(a)=h(b)=0$, let $y_{\epsilon}=\underset{\tiny\,}{\varphi}+\epsilon h,\;H(\epsilon)=F[y_{\epsilon}].$ Then we
have $\quad\displaystyle{0 = H'(0) = \int_a^b\bigg({\small\frac{\partial f}{\partial y}}h+{\small\frac{\partial f}{\partial y'}}h'\bigg)dx = \int_a^b h\bigg({\small\frac{\partial f}{\partial y}}-{\small\frac{d}{dx}\frac{\partial f}{\partial y'}}\bigg)dx}$
The last equality is due to $h(\partial[a,b])=\{0\}$ and integration by parts of ${\large\frac{\partial f}{\partial y'}}h'$.
Since $h$ is rather arbitrary on $(a,b)$, we obtain the famous Euler-Lagrange Equation
${\small(\dagger)}\;\;\;\boxed{{\small\frac{\partial f}{\partial y}-\frac{d}{dx}\frac{\partial f}{\partial y'}}=0\;\;\text{or more precisely,}\;\;f_2({\small x,y(x),y'(x)})={\scriptsize\frac{d}{dx}}f_3({\small x,y(x),y'(x)}).}$
10-18-2017, 12:16 PM
Post: #2
 elim Moderator Posts: 582 Joined: Feb 2010 Reputation: 0
Brachistochrone Problem.
The functional for the brachistochrone problem is $F[y]=\frac{1}{\sqrt{2g}}{\small\displaystyle{\int_0^L}}\sqrt{\small\frac{1+(y')^2}{y​}}$
$f(x,y,p)=\sqrt{\frac{1+p^2}{y}}\;(p=y',q = p').$ Then $f_2=-\frac{\sqrt{1+p^2}}{2y\sqrt{y}},\;f_3=\frac{p}{\sqrt{y(1+p^2)}},$
$\frac{d}{dx}f_3=\underset{\,}{\frac{1}{\sqrt{y(1+p^2)}}}\big(\frac{q}{1+p^2}​-\frac{p^2}{2y}\big),\;y''=-\frac{1+p^2}{2y},\;2yy''+1+(y')^2=0,$
$(y+y(y')^2)'=y'+2yy'y''+(y')^3=y'(2yy''+1+(y')^2) = 0,$
$y+y(y')^2=c,\;(y')^{-1}=\sqrt{\frac{y}{c-y}}=:\tan\beta,\;y=c\sin^2\beta = \frac{c}{2}(1-\cos 2\beta),$
$\frac{dy}{d\beta}=c\sin2\beta,\;\frac{dx}{d\beta}=\frac{dx}{dy}\frac{dy}{d\beta​}=2c\sin^2\beta(=2y)=c(1-\cos 2\beta).\;$Therefore
$(x,y)=\frac{c}{2}(2\beta-\sin 2\beta,1-\cos 2\beta)=r(\theta-\sin\theta,1-cos\theta).\;$ ( A cycloid)
To determine $r$, take $(u,v)=\lambda(\cos t,\sin t)\in\Gamma_y\;(\lambda>0< t< \frac{\pi}{2},\;\Gamma_y\,$is
the graph of curve). Then $\frac{t-\sin t}{1-\cos t}=\frac{u}{v}$ has unique solution $t=T\in (0,2\pi)$
And so $r = \frac{v}{1-\cos T}.\quad\square$
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