A question about the derivation of the Euler-Lagrange equation

  • Thread starter planck42
  • Start date
  • #1
82
0
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 [tex]I=\int_{x_1}^{x_2}{f(x,y,y')dx}[/tex], one should define two families of functions Y(x) and Y'(x), where Y(x) is [tex]y(x)+{\alpha}{\eta}(x)[/tex] and Y'(x) is the x-derivative of y(x), [tex]\alpha[/tex] is an arbitrary constant coefficient of [tex]{\eta}(x)[/tex], which is an arbitrary function that must be zero at [tex]x_1[/tex] and [tex]x_2[/tex].

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 [tex]\frac{dI}{d\alpha}[/tex] is zero at [tex]{\alpha}=0[/tex]

Whoa! Why are we dealing in [tex]\alpha[/tex]'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!
 
Last edited:

Answers and Replies

  • #2
Office_Shredder
Staff Emeritus
Science Advisor
Gold Member
4,420
505
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)
 
  • #3
82
0
Makes sense. However, I still have one question that I'll attempt to answer: why does [tex]{\alpha}=0[/tex] produce an extreme, and not, for instance, [tex]{\alpha}=1[/tex]? My guess is that the value of [tex]\alpha[/tex] is irrelevant; any value of [tex]\alpha[/tex] can make the first derivative of I with respect to [tex]\alpha[/tex] vanish. Hence the simplest choice, [tex]{\alpha}=0[/tex].
 
  • #4
Galileo
Science Advisor
Homework Helper
1,989
6
The point is that [itex]y(x)[/itex] is a function that minimizes (or extremizes) the integral.
So in [tex]y(x)+{\alpha}{\eta}(x)[/itex], the [tex]y(x)[/tex] is the sought after solution and [tex]{\alpha}{\eta}(x)[/tex] is an arbitrary deviation from that solution.
Therefore, the minimum occurs at [tex]{\alpha}=0[/tex]
 

Related Threads on A question about the derivation of the Euler-Lagrange equation

Replies
9
Views
3K
Replies
5
Views
959
Replies
1
Views
2K
  • Last Post
Replies
1
Views
2K
  • Last Post
Replies
11
Views
3K
  • Last Post
Replies
4
Views
5K
Replies
4
Views
876
  • Last Post
Replies
2
Views
1K
Replies
6
Views
3K
Top