# Variation of a functional

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1. Feb 19, 2016

### Whitehole

1. The problem statement, all variables and given/known data
Consider the functional $S(a,b) = \int_0^∞ r(1-b)a' \, dr$ of two functions $a(r)$ and $b(r)$ (with $a' = \frac{da}{dr}$). Find the $a(r)$ and $b(r)$ that extremize $S$, with boundary conditions $a(∞) = b(∞) = 1$.

2. Relevant equations

3. The attempt at a solution
I know how to find $b(r)$, my problem is $a(r)$. This is what I've done,

$δS = \int_0^∞ r(1-η)a' \, dr = \int_0^∞ ra' \, dr + \int_0^∞ ra'η \, dr$
where $η$ is the variation in $b$.

Can I say that since both terms should be $0$ so for the right term, since $η$ is arbitrary $a' = 0$ which implies the left term is also zero?

2. Feb 19, 2016

### Staff: Mentor

I could be way off-base here, but isn't $dS = \frac{\partial S}{\partial a} \cdot da + \frac{\partial S}{\partial b} \cdot db$?
The two partial derivatives above are the partials of S(a, b), one with respect to a and the other with respect to b.

3. Feb 19, 2016

### Orodruin

Staff Emeritus
$S$ is a functional, not a function. In the expression, $a$ and $b$ are not numbers, they are functions. As such the derivatives should be functional derivatives and not partial ones. Still, the variation with respect to $b$ is not done correctly.

4. Feb 19, 2016

### Whitehole

Should it be $δb(r) = (b(r + η(r)) - b(r)) = b'(r)η(r)$? So that $δS = \int_0^∞ r(1-b'η)a' \, dr$?

5. Feb 20, 2016

### Orodruin

Staff Emeritus
No, I suggest you start from the definition of the functional derivative

6. Feb 20, 2016

### Whitehole

$S(a,b) = \int_0^∞ f(b, a', r) \, dr = \int_0^∞ r(1-b)a' \, dr, \quad b = b(r) + αη(r)$

$\frac{dS}{dα} = \int_0^∞ \frac{∂f(b(r) + αη(r),~ a(r),~ r)}{∂α}~dr = \int_0^∞ r(-η)a'(r)~dr = -~ \int_0^∞ ra'(r)η(r)~dr = 0$

This implies $a'(r) = 0$

7. Feb 20, 2016

### Orodruin

Staff Emeritus
Right, and therefore ...

8. Feb 20, 2016

### Whitehole

$a = constant$ but from the boundary condition, $a(∞) = 1$, so $a(r) = 1$. Thanks!!!