CoV17

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Find the extremals of $J(y) = \int_0^1(yy'+y''^2)dx, y(0)=0, y'(0)=1, y(1)=2, y'(1)=4\,$

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

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

So

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

Solving for the constants

$y(0) = 0\,$	$\Longrightarrow c_4 = 0$
$y^\prime(0) = 1\,$	$\Longrightarrow c_3 = 1$
$y(1) = 2\,$	$\Longrightarrow \frac{c_1}{6} + \frac{c_2}{2} + 1 + 0 = 2$
$y^\prime(1) = 4\,$	$\Longrightarrow \frac{c_1}{2} + c_2 + 1 = 4$

Thus, solving the system of equations

$\begin{cases} \frac{c_1}{6} + \frac{c_2}{2} = 1\\ \frac{c_1}{2} + c_2 = 3 \end{cases}$

$\begin{cases} c_1 + 3c_2 = 6\\ c_1 + 2c_2 = 6 \end{cases}$

$\Longrightarrow c_2 = 0, c_1 = 6$

Therefore, $y = x^3 + x\,$ is an extremal.

Main Page : Calculus of Variations

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