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planck42

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In the book

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!

*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!

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