A question about the derivation of the Euler-Lagrange equation

[planck42]

In the book Mathematical Methods for Engineers and Scientists 3, the derivation of the Euler-Lagrange equation starts roughly along the lines of this:

In order to minimize the functional I=\int_{x_1}^{x_2}{f(x,y,y')dx}, one should define two families of functions Y(x) and Y'(x), where Y(x) is y(x)+{\alpha}{\eta}(x) and Y'(x) is the x-derivative of y(x), \alpha is an arbitrary constant coefficient of {\eta}(x), which is an arbitrary function that must be zero at x_1 and x_2.

These steps aren't difficult to understand; it allows for any possible function that could render the functional stationary while maintaining the boundary conditions.

The minimum value of I should occur when \frac{dI}{d\alpha} is zero at {\alpha}=0

Whoa! Why are we dealing in \alpha's all of a sudden? How does this minimize I and not the derivative with respect to x? Does working it out with x always produce a trivial solution? I need help with this one step!

 

[Office_Shredder]

I isn't a function of x (x is a dummy variable inside the integral, not a variable that you put into I).

On the other hand I DOES change as alpha changes (and as eta changes, but we're not going to worry about that)

 

[planck42]

Makes sense. However, I still have one question that I'll attempt to answer: why does {\alpha}=0 produce an extreme, and not, for instance, {\alpha}=1? My guess is that the value of \alpha is irrelevant; any value of \alpha can make the first derivative of I with respect to \alpha vanish. Hence the simplest choice, {\alpha}=0.

 

[Galileo]

The point is that y(x) is a function that minimizes (or extremizes) the integral.
So in y(x)+{\alpha}{\eta}(x)[/itex], the y(x) is the sought after solution and {\alpha}{\eta}(x) is an arbitrary deviation from that solution.<br /> Therefore, the minimum occurs at {\alpha}=0