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23 Januari 2017

Persamaan Euler-Lagrange untuk fungsi integran yang mempunyai turunan kedua

Bagaimanakah bentuk persamaan Euler Lagrange jika $F=F(x,y,y',y'')$, yang berarti fungsi $F$ mempunyai variabel yang merupakan turunan kedua?

Bentuk integral yang ingin dicari kondisi stasionernya adalah
\[ I=\underset{x_{1}}{\overset{x_{2}}{\int}}F(x,y,y',y'')dx\]
Kondisi stasioner dipenuhi dengan syarat
\[ \left(\frac{dI}{d\epsilon}\right)_{\epsilon=0}=0 \]
dengan menggunakan kurva variasi yang dinyatakan dengan $Y(x,\epsilon)=y(x)+\epsilon\eta(x)$
dengan $\eta(x_{1})=0$ dan $\eta(x_{2})=0$. Maka diperoleh
\begin{eqnarray*}
Y'(x,\epsilon) & = & y'(x)+\epsilon\eta'(x)\Rightarrow Y'(x,\epsilon)|_{\epsilon=0}=y'(x)\\ Y''(x,\epsilon) & = & y''(x)+\epsilon\eta''(x)\Rightarrow Y''(x,\epsilon)|_{\epsilon=0}=y''(x)
\end{eqnarray*}
\[
\frac{dy}{d\epsilon}=\eta;\frac{dy'}{d\epsilon}=\eta';\frac{dy''}{d\epsilon}=\eta''
\]
Jadi
\begin{eqnarray*}
\left(\frac{dI}{d\epsilon}\right)_{\epsilon=0} & = & \frac{d}{d\epsilon}\underset{x_{1}}{\overset{x_{2}}{\int}}F(x,y,y',y'')dx\\
& = & \underset{x_{1}}{\overset{x_{2}}{\int}}\frac{d}{d\epsilon}\left[F(x,y,y',y'')\right]dx
\end{eqnarray*}
\begin{eqnarray*}
\frac{dF}{d\epsilon} & = & \frac{\partial F}{\partial y}\frac{dy}{d\epsilon}+\frac{\partial F}{\partial y'}\frac{dy'}{d\epsilon}+\frac{\partial F}{\partial y''}\frac{dy''}{d\epsilon}\\
& = & \frac{\partial F}{\partial y}\eta+\frac{\partial F}{\partial y'}\eta'+\frac{\partial F}{\partial y''}\eta''
\end{eqnarray*}
maka
\begin{eqnarray*}
\frac{dI}{d\epsilon} & = & \underset{x_{1}}{\overset{x_{2}}{\int}}\left[\frac{\partial F}{\partial y}\eta+\frac{\partial F}{\partial y'}\eta'+\frac{\partial F}{\partial y''}\eta''\right]dx\\
& = & \underset{x_{1}}{\overset{x_{2}}{\int}}\frac{\partial F}{\partial y}\eta dx+\underset{x_{1}}{\overset{x_{2}}{\int}}\frac{\partial F}{\partial y'}\eta'dx+\underset{x_{1}}{\overset{x_{2}}{\int}}\frac{\partial F}{\partial y''}\eta''dx
\end{eqnarray*}
Tinjau suku kedua yang dapat diselesaikan dengan metode integral parsial
(misalkan $u=\dfrac{\partial F}{\partial y'}\Rightarrow du=\dfrac{d}{dx}\left(\dfrac{\partial F}{\partial y'}\right)dx$ dan $dv=\eta'dx\Rightarrow v=\eta$)
\[
\underset{x_{1}}{\overset{x_{2}}{\int}}\frac{\partial F}{\partial y'}\eta'dx=\dfrac{\partial F}{\partial y'}\eta\Bigr|_{x_{1}}^{x_{2}}-\underset{x_{1}}{\overset{x_{2}}{\int}}\eta\dfrac{d}{dx}\left(\dfrac{\partial F}{\partial y'}\right)dx=-\underset{x_{1}}{\overset{x_{2}}{\int}}\eta\dfrac{d}{dx}\left(\dfrac{\partial F}{\partial y'}\right)dx
\]
Selanjutnya tinjau suku ketiga yang juga dapat diselesaikan dengan metode integral parsial (misalkan $u=\dfrac{\partial F}{\partial y''}\Rightarrow du=\dfrac{d}{dx}\left(\dfrac{\partial F}{\partial y''}\right)dx $ dan $dv=\eta''dx\Rightarrow v=\eta'$)
\begin{eqnarray*}
\underset{x_{1}}{\overset{x_{2}}{\int}}\frac{\partial F}{\partial y''}\eta''dx & = & \dfrac{\partial F}{\partial y''}\eta'\Biggr|_{x_{1}}^{x_{2}}-\underset{x_{1}}{\overset{x_{2}}{\int}}\eta'\dfrac{d}{dx}\left(\dfrac{\partial F}{\partial y''}\right)dx\\
& = & 0-\underset{x_{1}}{\overset{x_{2}}{\int}}\eta'\dfrac{d}{dx}\left(\dfrac{\partial F}{\partial y''}\right)dx\\
& = & -\left[\dfrac{d}{dx}\left(\dfrac{\partial F}{\partial y''}\right)\eta''\Biggr|_{x_{1}}^{x_{2}}-\underset{x_{1}}{\overset{x_{2}}{\int}}\eta\frac{d^{2}}{dx^{2}}\left(\dfrac{\partial F}{\partial y''}\right)dx\right]\\
& = & -\left[0-\underset{x_{1}}{\overset{x_{2}}{\int}}\eta\frac{d^{2}}{dx^{2}}\left(\dfrac{\partial F}{\partial y''}\right)dx\right]=\underset{x_{1}}{\overset{x_{2}}{\int}}\eta\frac{d^{2}}{dx^{2}}\left(\dfrac{\partial F}{\partial y''}\right)dx
\end{eqnarray*}
Maka
\begin{eqnarray*}
\left(\frac{dI}{d\epsilon}\right)\Biggr|_{\epsilon=0}=0 & = & \underset{x_{1}}{\overset{x_{2}}{\int}}\frac{\partial F}{\partial y}\eta dx+\underset{x_{1}}{\overset{x_{2}}{\int}}\frac{\partial F}{\partial y'}\eta'dx+\underset{x_{1}}{\overset{x_{2}}{\int}}\frac{\partial F}{\partial y''}\eta''dx\\
0 & = & \underset{x_{1}}{\overset{x_{2}}{\int}}\frac{\partial F}{\partial y}\eta dx-\underset{x_{1}}{\overset{x_{2}}{\int}}\eta\dfrac{d}{dx}\left(\dfrac{\partial F}{\partial y'}\right)dx+\underset{x_{1}}{\overset{x_{2}}{\int}}\eta\frac{d^{2}}{dx^{2}}\left(\dfrac{\partial F}{\partial y''}\right)dx\\
0 & = & \underset{x_{1}}{\overset{x_{2}}{\int}}\left[\frac{\partial F}{\partial y}-\dfrac{d}{dx}\left(\dfrac{\partial F}{\partial y'}\right)+\frac{d^{2}}{dx^{2}}\left(\dfrac{\partial F}{\partial y''}\right)\right]\eta dx
\end{eqnarray*}
Sehingga akan diperoleh bentuk persamaan Euler Lagrange dalam bentuk
\[\frac{d^{2}}{dx^{2}}\left(\dfrac{\partial F}{\partial y''}\right)-\dfrac{d}{dx}\left(\dfrac{\partial F}{\partial y'}\right)+\frac{\partial F}{\partial y}=0\]

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