Calculus of variations changing variables

[bobred]

Homework Statement

Hi
I am given the functional

I am asked to show that if

and with an appropriate value for

that

Homework Equations

[/B]

The Attempt at a Solution

So I get

If I set

then I get

I think that it is correct but what about the factor of 2?

 

[Orodruin]

Your work looks correct. Unless you do another variable transformation, you will not get rid of the 2.

 

[PeroK]

I imagine that since you are trying to find stationary values of this integral, they are not affected by a constant multiple, so you can drop the factor. I think it should be -2, but that's a minor point.

 

[Orodruin]

I think it should be -2, but that's a minor point.

This would depend on the order of the integration limits (note that $a$ was the lower integration limit in $x$ but $A$ is the upper integration limit in $u$ - of course I am just making the arbitrary inference that $a$ corresponds to $A$ here ...).

 

[PeroK]

This would depend on the order of the integration limits (note that $a$ was the lower integration limit in $x$ but $A$ is the upper integration limit in $u$ - of course I am just making the arbitrary inference that $a$ corresponds to $A$ here ...).

Yes, didn't notice that.

 

[bobred]

Thanks guys.
I may have missed this in my notes PeroK but if we are trying to find stationary points of a functional constant multiples can be ignored?
James

 

[Orodruin]

but if we are trying to find stationary points of a functional constant multiples can be ignored?

Yes, but there is nothing strange about this. It works this way for functions as well, if $f(x)$ has the stationary point $x=2$, then so does $2f(x)$.

 

[bobred]

Yes, thanks for the clarification.
James

 

[Ray Vickson]

Thanks guys.
I may have missed this in my notes PeroK but if we are trying to find stationary points of a functional constant multiples can be ignored?
James

Positive multiples can be ignored, but omitting negative multiples changes the direction of optimization.

 

[Orodruin]

Positive multiples can be ignored, but omitting negative multiples changes the direction of optimization.

But it does not change the fact that the point is stationary. Just exchanges minima for maxima.

 

[Ray Vickson]

But it does not change the fact that the point is stationary. Just exchanges minima for maxima.

Of course, but that is another issue. It certainly changes the second-order tests, so that instead of looking to see if the Hessian is positive-definite, we look instead to see if it is negative-definite. However, when I taught this stuff I recommended that students just "memorize" the conditions for a minimum, then switch the sign of the objective if the problem was a maximization; that eliminates the need for a whole raft of special cases, etc.