I've just started to learn mathematical physics, and I read Stone and Goldbart's Mathematics for Physics. But right at the beginning when they introduce the functional derivative, I couldn't understand their explanation:

1.2.1. The functional derivative:

We restrict ourselves to expressions of the form

$$J[y] = \int_{x_1}^{x_2} f[ \, x, y, y',y'',...y^{(n)} ] \, dx $$

Consider a functional $J=\int f dx$ in which f depends only on $x, y$ and $y'$. Make a change $y(x) \rightarrow y(x) + \epsilon \eta (x)$, where $\epsilon$ is a (small) $x$-independent constant. The resultant change in J is: $$\begin{align} J[y+\epsilon \eta]-J[y] &=\int_{x_1}^{x_2}\{f(x,y+\epsilon \eta,y'+\epsilon \eta')-f(x,y,y')\}dx \\ &= \int_{x_1}^{x_2} \{\epsilon \eta\frac{\partial f}{\partial y}+\epsilon\frac{d\eta}{dx}\frac{\partial f}{\partial y'}+O(\epsilon^2)\}\\ &= ......(\text{this part I understand})\end{align} $$

I can't really wrap my head around this. What is that $\eta(x)$ that suddenly showed up? How did they manage from the first line to the second line in the above expression? And why is there a big-O at the end? (If you don't have the book you can actually find on page 16 right here, the next part is just an integration by part). I'm honestly clueless.

It'd be awesome if someone can explain this to me please?

  • 2
    $\begingroup$ "Make a change $y(x) \rightarrow y(x) + \epsilon \eta (x)$, where $\epsilon$ is a (small) $x$-indipendent constant." $\eta$ is just some random (well-behaved) function. And the big-O is because you are expanding in powers of $\epsilon$; it just means "some terms of order $\epsilon^2$". $\endgroup$
    – valerio
    Nov 24, 2016 at 12:25
  • 3
    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Nov 24, 2016 at 12:32
  • $\begingroup$ @Qmechanic I think so. There is no physics involved. $\endgroup$
    – valerio
    Nov 24, 2016 at 13:01

1 Answer 1


$\eta(x)$ is some random well-behaved function you add to $y$ (it is the "variation").

Now, for a single-variable function, you have (Taylor series)

$$f(x+x_0) \simeq f(x) + x_0 \partial_x f(x_0)$$

For a function of three variables, we have

$$f(x+x_0,y+y_0,z+z_0) \simeq f(x,y,z)+x_0 \partial_xf(x_0,y_0,z_0)+y_0 \partial_yf(x_0,y_0,z_0)+z_0 \partial_zf(x_0,y_0,z_0)$$

Applying this result to your integrand, you obtain

$$f(x,y+\epsilon \eta, y'+\epsilon \eta') \simeq f(x,y,y')+\epsilon \eta \ \partial_y f (x,y,y')+ \epsilon \eta' \ \partial_{y'}f(x,y,y')$$

The other terms in the expansion contain multiplicative factors $\epsilon^2,\epsilon^3,\epsilon^4 \dots$, so we just write

$$f(x,y+\epsilon \eta, y'+\epsilon \eta') = f(x,y,y')+\epsilon \eta \ \partial_y f (x,y,y')+ \epsilon \eta' \ \partial_{y'}f(x,y,y')+O(\epsilon^2)$$

meaning that there is an additional part containin terms $t_k$ every one of which satisfies

$$\lim_{\epsilon \to 0} \left| \frac{t_k}{\epsilon^2} \right| < \infty$$

Plugging this result in the integral, you obtain the final result.


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