CoV19

From Exampleproblems

Jump to: navigation, search

Minimize $J(y)=\int_0^1(1+y''^2)dx, y(0)=0,y'(0)=1,y(1)=1,y'(1)=1\,$

Use the Euler-Lagrange equation for this integrand functional $L\,$

$L_y - \frac{d}{dx}L_{y^\prime} + \frac{d^2}{dx^2}L_{y^{\prime\prime}} = 0$

So

$0 - \frac{d}{dx}(0) + \frac{d^2}{dx^2}(2y^{\prime\prime})$	$= 0\,$
$2y^{\prime\prime\prime\prime}\,$	$= 0\,$
$y^{\prime\prime\prime\prime}\,$	$= 0\,$
$y^{\prime\prime\prime}\,$	$= c_1\,$
$y^{\prime\prime}\,$	$= c_1x + c_2\,$
$y^{\prime}\,$	$= \frac{1}{2}c_1x^2 + c_2x + c_3\,$
$y\,$	$= \frac{1}{6}c_1x^3 + \frac{1}{2}c_2x^2 + c_3x + c_4\,$

Solving for the constants

$y(0) = 0\,$	$\Longrightarrow c_4 = 0$
$y^\prime(0) = 1\,$	$\Longrightarrow c_3 = 1$
$y(1) = 1\,$	$\Longrightarrow \frac{c_1}{6} + \frac{c_2}{2} + 1 + 0 = 1$
$y^\prime(1) = 1\,$	$\Longrightarrow \frac{c_1}{2} + c_2 + 1 = 1$

Thus, solving the system of equations

$\begin{cases} \frac{c_1}{6} + \frac{c_2}{2} = 0\\ \frac{c_1}{2} + c_2 = 0 \end{cases}$

$\begin{cases} c_1 + 3c_2 = 0\\ c_1 + 2c_2 = 0 \end{cases}$

$\Longrightarrow c_2 = 0, c_1 = 0$

Therefore, $y = x\,$ is an extremal.

Main Page : Calculus of Variations

Retrieved from "http://www.exampleproblems.com/wiki/index.php/CoV19"

Views

Personal tools

Create an account or log in

Search

Toolbox

Argan Oil
Natural Skin Care
Organic Skin Care