Mathematical Digression: Euler-Lagrange Equation

Suppose we want to minimize the function:  

$$A = \int_a^b \mathcal{L}\bigl(M(x),M'(x)\bigr)\,dx$$ (4)

with M' =dM/dx, and

$$M(a)=M_a,\quad M(b)=M_b$$

(This is called Lagrange problem).

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

$$\delta y = y(x+\delta x) -y (x)$$

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:

$$M(x)+\delta M(x)$$

and calculate $\delta A$

We have:

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

Expansion:

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

and

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

Second term: integrating by parts:

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

Result:

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

This is possible only if  

$$\frac{\partial\mathcal{L}}{\partial M}- \frac{\partial}{\partial x}\left(\frac{\partial\mathcal{L}}{\partial M'}\right) =0$$ (5)

This is Euler-Lagrange equation, well known in Classical Mechanics.