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Mathematical Digression: Euler-Lagrange Equation

Suppose we want to minimize the function:  
 A = \int_a^b \mathcal{L}\bigl(M(x),M'(x)\bigr)\,dx\end{displaymath} (4)
with M' =dM/dx, and

M(a)=M_a,\quad M(b)=M_b\end{displaymath}

(This is called Lagrange problem).

Standard calculus: if we want to minimize y(x), we consider a small deviation:

\delta y = y(x+\delta x) -y (x)\end{displaymath}

At the minimum $\delta y=0$

In this situation we have a functional $\mathcal{L}$ that depends on the function M(x). Same idea: consider a deviation:

M(x)+\delta M(x)\end{displaymath}

and calculate $\delta A$
 \psfrag{dM}{$M(x)+\delta M(x)$}
We have:

\delta A = \int_a^b \left[\mathcal{L}\bigl(M+\delta M,M'+\delta
 M'\bigr) - \mathcal{L}\bigl(M,M'\bigr)\right]\,dx = 0\end{displaymath}


\mathcal{L}\bigl(M+\delta M,M'+\delta M'\bigr) -
 M}\delta M + \frac{\partial\mathcal{L}}{\partial M'}\delta M'\end{displaymath}


\int_a^b \left[\frac{\partial\mathcal{L}}{\partial
 ...frac{\partial\mathcal{L}}{\partial M'}\delta
 M'\right]\,dx = 0\end{displaymath}

Second term: integrating by parts:

\int_a^b \frac{\partial\mathcal{L}}{\partial M'}\delta M'\,d...
 ...eft(\frac{\partial\mathcal{L}}{\partial M'}\right)\delta M\,dx \end{displaymath}


\int_a^b \left[\frac{\partial\mathcal{L}}{\partial
 M}- \fra...
 ...ial\mathcal{L}}{\partial M'}\right)\right]\delta
 M(x)\,dx = 0 \end{displaymath}

This is possible only if  
 M}- \frac{\partial}{\partial
 x}\left(\frac{\partial\mathcal{L}}{\partial M'}\right) =0\end{displaymath} (5)
This is Euler-Lagrange equation, well known in Classical Mechanics.

© 1997 Boris Veytsman and Michael Kotelyanskii
Thu Oct 16 20:58:44 EDT 1997