CoV18

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Find the Euler equations for $J(y,z)=\int_a^b\left[ y''z' + xyz'' + z'''y^2\right] dx\,$

For a functional of several functions, create a system of equations using the formula

$L_{y_i} - \frac{d}{dx}L_{y^\prime_i} + \frac{d^2}{dx^2} L_{y^{\prime\prime}_i} - \ldots = 0$

So

$\begin{cases} L_{y} - \frac{d}{dx}L_{y^\prime} + \frac{d^2}{dx^2}L_{y^{\prime\prime}} = 0\\ L_{z} - \frac{d}{dx}L_{z^\prime} + \frac{d^2}{dx^2}L_{z^{\prime\prime}} - \frac{d^3}{dx^3}L_{z^{\prime\prime\prime}} = 0 \end{cases}$

$\begin{cases} xz^{\prime\prime} + 2z^{\prime\prime\prime}y - \frac{d}{dx}(0) + \frac{d^2}{dx^2}(z^\prime) = 0\\ 0 - \frac{d}{dx}(y^{\prime\prime}) + \frac{d^2}{dx^2}(xy) - \frac{d^3}{dx^3}(y^2) = 0 \end{cases}$

$\begin{cases} xz^{\prime\prime} + 2z^{\prime\prime\prime}y + z^{\prime\prime\prime} = 0\\ -y^{\prime\prime\prime} + \frac{d}{dx}(y + xy^\prime) - \frac{d^2}{dx^2}(2yy^\prime) = 0 \end{cases}$

$\begin{cases} xz^{\prime\prime} + (2y + 1)z^{\prime\prime\prime} = 0\\ -y^{\prime\prime\prime} + 2y^\prime + xy^{\prime\prime} - \frac{d}{dx}(2y^{\prime 2} + 2yy^{\prime\prime}) = 0 \end{cases}$

$\begin{cases} xz^{\prime\prime} + (2y + 1)z^{\prime\prime\prime} = 0\\ -y^{\prime\prime\prime} + 2y^\prime + xy^{\prime\prime} - 6y^{\prime}y^{\prime\prime} - 2yy^{\prime\prime\prime} = 0 \end{cases}$

$\begin{cases} xz^{\prime\prime} + (2y + 1)z^{\prime\prime\prime} = 0\\ 2y^\prime + xy^{\prime\prime} - 6y^{\prime}y^{\prime\prime} - (2y + 1)y^{\prime\prime\prime} = 0 \end{cases}$

Which are the Euler equations for the functional.

Main Page : Calculus of Variations

CoV18

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