Simple Functional Derivatives

[ryanwilk]

Homework Statement

Hey, can I just check these functional derivatives?:

1) $$\frac{\delta F[g]}{\delta g(y)}$$ where $$F[g] = \int dx \left[ \frac{1}{\sqrt{1+(g'(x))^2}} - 2g(x) + 5 \right]\>.$$

2) $$\frac{\delta F[a,b,g]}{\delta g(y)}$$ where $$F[a,b,g] = \int d^4x \left[ A(\partial_{\mu} g(x))a(x)b(x) + Bg^3(x) \right]\>.$$

Homework Equations

$$\frac{\delta F[g]}{\delta g(y)} = \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ F[g(x) +\epsilon\delta(x-y)] - F[g(x)] \right]\>.$$

The Attempt at a Solution

The answers I get are 1) -2 and 2) 3Bg²(y).

Thanks in advance!

 

[Mute]

Homework Statement

Hey, can I just check these functional derivatives?:

1) $$\frac{\delta F[g]}{\delta g(y)}$$ where $$F[g] = \int dx \left[ \frac{1}{\sqrt{1+(g'(x))^2}} - 2g(x) + 5 \right]\>.$$

2) $$\frac{\delta F[a,b,g]}{\delta g(y)}$$ where $$F[a,b,g] = \int d^4x \left[ A(\partial_{\mu} g(x))a(x)b(x) + Bg^3(x) \right]\>.$$

Homework Equations

$$\frac{\delta F[g]}{\delta g(y)} = \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ F[g(x) +\epsilon\delta(x-y)] - F[g(x)] \right]\>.$$

The Attempt at a Solution

The answers I get are 1) -2 and 2) 3Bg²(y).

Thanks in advance!

(Just realized this is in the completely wrong forum... =/)

How did you get your answers? I don't think they are correct. It looks like you have assumed that $\delta g'(x)/\delta g(y) = 0$. This is not true; you need to use $\delta g'(x)/\delta g(y) = \delta'(x-y)$, the derivative of the dirac delta function.

 

[ryanwilk]

Oh. Yes, I did assume that. I'll have another go.

 

[ryanwilk]

So, for (2),

$$F[a,b,g] = \int d^4x \left[ A(\partial_{\mu} g(x))a(x)b(x) + Bg^3(x) \right]\>.$$

$$\implies \frac{\delta F[a,b,g]}{\delta g(y)} = \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ \int d^4x\>[A(\partial_{\mu}(g(x)+\epsilon \delta(x-y)))a(x)b(x) - A(\partial_{\mu}g(x))a(x)b(x) + B{(g(x)+\epsilon \delta(x-y))}^3-Bg^3(x)] \right]$$

$$= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[\int d^4x\>A\epsilon\>\partial_{\mu}(\delta(x-y))\>a(x)b(x) + B(3g^2(x)\epsilon\delta(x-y)+3g(x)\epsilon^2\delta^2(x-y)+\epsilon^3\delta^3(x-y)) \right]$$

$$= \left[\int d^4x\>A\>\partial_{\mu}(\delta(x-y))\>a(x)b(x) + B(3g^2(x)\delta(x-y)) \right]\>.$$

Then apparently $$\int \delta'(x-y)f(x) = - \int \delta(x-y)f'(x)$$ so,

$$\frac{\delta F[a,b,g]}{\delta g(y)} = 3Bg^2(y) - A\partial_{\mu}[a(y)b(y)]\>.$$

And similarly, (1) gives -2g''(y)-2?

 

[ryanwilk]

Does this seem correct? =/

 

[Mute]

So, for (2),

$$F[a,b,g] = \int d^4x \left[ A(\partial_{\mu} g(x))a(x)b(x) + Bg^3(x) \right]\>.$$

$$\implies \frac{\delta F[a,b,g]}{\delta g(y)} = \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[ \int d^4x\>[A(\partial_{\mu}(g(x)+\epsilon \delta(x-y)))a(x)b(x) - A(\partial_{\mu}g(x))a(x)b(x) + B{(g(x)+\epsilon \delta(x-y))}^3-Bg^3(x)] \right]$$

$$= \lim_{\epsilon \to 0} \frac{1}{\epsilon} \left[\int d^4x\>A\epsilon\>\partial_{\mu}(\delta(x-y))\>a(x)b(x) + B(3g^2(x)\epsilon\delta(x-y)+3g(x)\epsilon^2\delta^2(x-y)+\epsilon^3\delta^3(x-y)) \right]$$

$$= \left[\int d^4x\>A\>\partial_{\mu}(\delta(x-y))\>a(x)b(x) + B(3g^2(x)\delta(x-y)) \right]\>.$$

Then apparently $$\int \delta'(x-y)f(x) = - \int \delta(x-y)f'(x)$$ so,

$$\frac{\delta F[a,b,g]}{\delta g(y)} = 3Bg^2(y) - A\partial_{\mu}[a(y)b(y)]\>.$$

And similarly, (1) gives -2g''(y)-2?

Sorry for late reply. I've been travelling. Your (2) looks superficially good - the problem I see is that you have a $\partial_\mu$, but there is no indication that the index is being summed over, so either you have a dangling index or you're missing another term with a $\mu$ index (which could change the result - e.g., if the derivative were really $\partial^\mu\partial_\mu$, then this is a second derivative, so when you apply the dirac derivative rule twice the minus signs cancel out).

Your (1) still looks incorrect - the first term should be $-\frac{d}{dx}\left[g'(x)/(1+(g'(x))^2)^{3/2}\right]$, evaluated at x = y, which I don't think reduces to $-2g''(y)$.