# Calculus of variations changing variables

1. Apr 9, 2015

### bobred

1. The problem statement, all variables and given/known data
Hi
I am given the functional

I am asked to show that if and with an appropriate value for that

2. Relevant equations

3. 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?

2. Apr 9, 2015

### Orodruin

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

3. Apr 9, 2015

### 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.

4. Apr 9, 2015

### Orodruin

Staff Emeritus
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 ...).

5. Apr 9, 2015

### PeroK

Yes, didn't notice that.

6. Apr 22, 2015

### 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

7. Apr 22, 2015

### Orodruin

Staff Emeritus
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 $2f(x)$.

8. Apr 22, 2015

### bobred

Yes, thanks for the clarification.
James

9. Apr 22, 2015

### Ray Vickson

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

10. Apr 22, 2015

### Orodruin

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

11. Apr 22, 2015

### Ray Vickson

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.