Calc of variations, minimizing functionals question

[Coffee_]

Consider the following problem:

$A$ is a functional (some integral operator to be more specific) of a (complex) function $F$.

We want to minimize $A[F]$ wrt. to a constraint $B[F]=\int (|F|²)=N$

If I read around online I find that in general such extremization problems are done by minimizing:

$A[F]-µ(B[F]-N)$

Where $µ$ is a Lagrange multiplier.

My physics book does it slightly differently however, it doesn't include the $-µN$ term in the minimization.

They just say that it's alright to minimize $A[F]-µB[F]$ for a fixed $µ$.

Why can they use a slightly different way or extremizing a functional under a constraint than the general method I seem to find online?

 

[RUber]

N is an arbitrary constant, right?
So it should be acceptable to use N=0.

 

[Coffee_]

No, this is a physics book on Bose Einstein condensates and N is the number of particles in the system. I am seriously confused now because now I've started looking more online for general explanations of this integral constraint minimization and some DO include the ''µN'' part and some don't.

Example where they don't : http://www.mpri.lsu.edu/textbook/Chapter8-b.htm (see section integral constraints)

Example where they DO: http://liberzon.csl.illinois.edu/teaching/cvoc/node38.html (see eq 2.50)

 

[fresh_42]

Why isn't $A[F]-µC[F] = A[F]-µ(B[F]-N)$ with $C = B[F] - N$ the same optimization problem?

 

[Coffee_]

Why isn't $A[F]-µC[F] = A[F]-µ(B[F]-N)$ with $C = B[F] - N$ the same optimization problem?

It would be if you indeed correctly switch to $C[F]$ which they don't, they explicitly do minimize $A[F]-µB[F]$ but I think I understand it now.

What they do is do not vary µ during the minimization such that $µN$ is just a constant which can be neglected. They will then find a solution $F(µ,x)$ which would need to be plugged into $B[F(x,µ)]=N$ to fix $µ$.

The difference here is that if you'd minimize $A[F]-µ(B[F]-N)$ wrt. to both $µ$ and $F$ you'd get the correct $µ$ and $F$ simultaneously. Am I correct?

 

[fresh_42]

As far as I can see (draw some lines ...) you are right as long as there are no other constraints. If there were, the minimal $A[F]$ could have led you to a solution $F(μ,x)$ where the "plug-in" $B[F(μ,x)] = N$ is already out of the allowed area. This cannot happen if you couple it beforehand.