Understanding functional derivative

In summary, the functional ## J[f] = \int[f(y)]^pφ(y)\, dy ## has a functional derivative with respect to ## f(x) ## given by:$$ \frac {δJ[f]} {δf(x)} = \lim_{ε \rightarrow 0} \frac 1 ε \left[ \int[f(y) + εδ(y-x)]^pφ(y)\, dy - \int [f(y)]^pφ(y)\, dy\right] $$$$ = p[f(x)]^{p-1}φ(x) $$
  • #1
Bishamonten
17
1

Homework Statement



"The functional ## J[f] = \int [f(y)]^pφ(y)\, dy ## has a functional derivative with respect to ## f(x) ## given by:

$$ \frac {δJ[f]} {δf(x)} = \lim_{ε \rightarrow 0} \frac 1 ε \left[ \int[f(y) + εδ(y-x)]^pφ(y)\, dy - \int [f(y)]^pφ(y)\, dy\right] $$

$$ = p[f(x)]^{p-1}φ(x) $$

Homework Equations



Limit definition of a functional derivative:

$$ \frac {δF} {δf(x)} = \lim_{ε \rightarrow 0} \frac {F[f(x') + εδ(x-x')] - F[f(x')]} ε$$

The Attempt at a Solution


From line 1:

$$ = \lim_{ε \rightarrow 0} \frac 1 ε \left[ \int[f(y) + εδ(y-x)]^pφ(y)\, dy - \int [f(y)]^pφ(y)\, dy\right] $$


$$ = \lim_{ε \rightarrow 0} \frac 1 ε \left[ \int \sum_{k=0}^p
\begin{pmatrix}
p \\
k \\
\end{pmatrix}
[f(y)]^{p-k}[εδ(y-x)]^kφ(y)\, dy - \int [f(y)]^pφ(y)\, dy\right] $$

$$ \lim_{ε \rightarrow 0} \frac 1 ε \left[ \int \sum_{k=0}^p \frac {p!} {k!(p-k)!} [f(y)]^{p-k}[εδ(y-x)]^kφ(y)\, dy - \int [f(y)]^pφ(y)\, dy\right] $$



This comes straight out of Lancaster and Blundell's "Quantum Field Theory for the Gifted Amateur", and is in the first chapter where they are introducing Lagrangians, Fermat's principle of least time, and Hamilton's principle of least action. This is simply them demonstrating functional derivatives, only, I wanted to see how they arrived from the limit definition of the functional ## J[f] ## to ## p[f(x)]^{p-1}φ(x) ##.

I wanted to evaluate the limit myself, through first evaluating the integral. I saw the power of the sum expression and thought ok, maybe I should introduce the binomial theorem, but I feel like I'm only making a mess of things and am not arriving any closer at ## p[f(x)]^{p-1}φ(x) ##, may someone let me know if I'm going at this in the right direction, and if not, any pointers, or if I'm simply making this more complicated than it should be?

 
Physics news on Phys.org
  • #2
Which terms in your binomial expansion survive the limit ##\varepsilon \to 0##?
 
  • Like
Likes Bishamonten
  • #3
\
Orodruin said:
Which terms in your binomial expansion survive the limit ##\varepsilon \to 0##?
Oooh, ok I don't know why I didn't put it in its expanded form earlier. Maybe it was my lack of confidence talking :DD Thank yo so much for responding!

So when I expand the term out, the 1st term in the series matches ## -\int [f(y)]^pφ(y)dy ##, except it's positive, thus cancelling. Every other term in the series besides the 2nd term ## \frac {p!} {(p-1)!} [f(y)]^{p-1}δ(y-x)φ(y) ## zeroes out when evaluating the limit because they will all continue to have the kth power of epsilon multiplying it, whereas the 2nd term has it's epsilon cancel the one distributed, and of course ## \frac {p!} {(p-1)!} ## is just p.

So here's what I got so far, I am a bit uncomfortable with the dirac delta function argument I make in this last step however, as you'll see:

$$ p\int[f(y)]^{p-1}φ(y)δ(y-x)dy $$

$$ [f(y)]^{p-1}φ(y) ≡ g(y) $$

$$ = p\int g(y)δ(y-x)dy = g(x) = p[f(x)]^{p-1}φ(x) $$

Is that a proper use of the indefinite integral of this delta function?
 
  • #4
That is the definition of how the delta function works, so yes, it is perfectly fine.
 
  • Like
Likes Bishamonten
  • #5
Orodruin said:
That is the definition of how the delta function works, so yes, it is perfectly fine.

Thank you very much for the help. We touched up on the dirac delta function during the last part of our differential equations class, but looks like I need to review how it works a bit. I remember that this was the definition of how it works, but the intuition for why it works that way didn't survive me unfortunately.

Also, I wish to ask another question while I'm at it, related to tensors, covariant and contravariant vectors. May I post it here, or should I start a new thread?
 
  • #6
Bishamonten said:
I remember that this was the definition of how it works, but the intuition for why it works that way didn't survive me unfortunately.
It works like that because it is defined to work like that. For more in depth on what it really is you would need to study distributions (it really is a distribution and not a function), in particular in terms of derivatives and how it relates to differential equations.

Bishamonten said:
Also, I wish to ask another question while I'm at it, related to tensors, covariant and contravariant vectors. May I post it here, or should I start a new thread?
You should start a new thread. If it is a problem like this one, post it in the appropriate homework forum. If it is a more general question, you can post it in one of the technical forums.
 
  • Like
Likes Bishamonten
  • #7
Once again, thank you for all your concise help Orodruin, I've no doubt I'll be needing it again soon enough :smile:
 
  • #8
Sorry, post made by mistake
 

What is a functional derivative?

A functional derivative is a mathematical tool used in calculus of variations to find the rate of change of a functional with respect to a function. It is similar to the ordinary derivative, but instead of finding the rate of change of a function at a specific point, it finds the rate of change of a functional for a whole class of functions.

How is a functional derivative calculated?

The functional derivative is calculated by taking the derivative of the functional with respect to the function, and then evaluating it at a specific function. This involves finding the partial derivatives of the function and then integrating them over the domain of the function.

What is the practical use of functional derivatives?

Functional derivatives have various practical applications, such as in physics, engineering, and economics. They are used to solve problems involving optimization, finding extremum values, and minimizing functional quantities.

What is the relationship between functional derivatives and ordinary derivatives?

Functional derivatives and ordinary derivatives are closely related but have some key differences. While ordinary derivatives deal with functions of one variable, functional derivatives deal with functions of multiple variables. Additionally, ordinary derivatives give the rate of change of a function at a specific point, while functional derivatives give the rate of change of a functional over a whole class of functions.

Are there any limitations to using functional derivatives?

Like any mathematical tool, functional derivatives have limitations. They can only be applied to differentiable functions, and not all functions are differentiable. Additionally, the domain of the function must be well-defined and finite for the functional derivative to be calculated.

Similar threads

Replies
1
Views
806
  • Advanced Physics Homework Help
Replies
4
Views
462
  • Advanced Physics Homework Help
Replies
1
Views
690
  • Advanced Physics Homework Help
Replies
8
Views
805
  • Advanced Physics Homework Help
Replies
19
Views
830
  • Advanced Physics Homework Help
Replies
3
Views
687
  • Advanced Physics Homework Help
Replies
2
Views
829
  • Advanced Physics Homework Help
Replies
10
Views
582
  • Advanced Physics Homework Help
Replies
19
Views
945
  • Advanced Physics Homework Help
Replies
3
Views
505
Back
Top