# Variational derivative of function with respect to its derivative [closed]

What is $$\frac{\delta f(t)}{\delta \dot{f}(t)}~?$$

Where $\dot{f}(t) = df/dt$.

## closed as off-topic by sammy gerbil, peterh, AccidentalFourierTransform, Kyle Kanos, garypFeb 26 '17 at 22:29

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• could you provide some context please? – ZeroTheHero Feb 25 '17 at 3:32
• It is purely a math question ,so better to ask in mathematics stack exchange – Lapmid Feb 25 '17 at 4:00
• @SherlockHolmes yeah and all possible answers are purely math too. – ZeroTheHero Feb 25 '17 at 6:05
• Related question by OP: physics.stackexchange.com/q/263261/2451 – Qmechanic Feb 25 '17 at 7:36
• – Qmechanic Feb 25 '17 at 7:47

The definition of the functional derivative of a functional $I[g]$ is the distribution $\frac{\delta I}{\delta g}(\tau)$ such that $$\left\langle \frac{\delta I}{\delta g}, h\right\rangle := \frac{d}{d\alpha}\bigg\rvert_{\alpha=0} I[g+ \alpha h]$$ for every test function $h$. In our case, assuming to deal with functions which suitably vanish before reaching $\pm \infty$, $$I[g] = \int_{-\infty}^t g(x)dx$$ so that $$I[\dot{f}]= f(t)$$ as requested. Going on with the procedure $$\left\langle \frac{\delta I}{\delta g}, h\right\rangle = \frac{d}{d\alpha}|_{\alpha=0} \int_{-\infty}^t(g(\tau)+ \alpha h(\tau)) d\tau = \int_{-\infty}^t h(\tau) dx = \int_{-\infty}^{+\infty} \theta(t-\tau)h(\tau) d\tau$$ where $\theta(\tau)=1$ for $\tau\geq 0$ and $\theta(\tau)=0$ for $\tau<0$ and so $$\frac{\delta f(t)}{\delta \dot{f}}(\tau) = \frac{\delta I}{\delta g}(\tau)= \theta(t-\tau)$$

• Just a LaTeX tip: you can use \big, \bigg and so forth to have a larger vertical line, \rvert. (See edit.) – JamalS Feb 25 '17 at 9:54
• Thanks. I usually use $\left.$ $\right|$ but I did not exploit them here. – Valter Moretti Feb 25 '17 at 9:56
• Great explanation, thanks! I guess it makes physical sense too - varying at $\tau$ you only expect any effect at $t\geq\tau$ – smörkex Feb 26 '17 at 3:43

The important thing to keep in mind is that a functional derivative is more like a gradient than an ordinary derivative. The reason that this is an important consideration is because, practically, we always specify functions with (possibly infinite) lists of numbers, be they: Taylor series coefficients, continued fraction constants, a list of constant values (approximating with boxcars), a list of points (connect the dots), Fourier series coefficients, or etc.

The important part of this consideration is that the function's derivative doesn't carry any information about a constant vertical offset. Thus, because any function of the form $f(t) + c$ has the same derivative, $\dot{f}(t)$, the functional derivative in the question will not be defined in the "direction" that corresponds to the degree of freedom defined by $c$.

In equations, let \begin{align} g(t) &\equiv \dot{f}(t) \Rightarrow \\ f(t) - f(t_0) & = \int_{t_0}^t g(t') \operatorname{d} t'\end{align} From there: \begin{align} \frac{\delta f(t)}{ \delta \dot{f}(\tau)} - \frac{\delta f(t_0)}{ \delta \dot{f}(\tau)} & = \frac{\delta \int_{t_0}^t g(t') \operatorname{d}t'}{ \delta g(\tau)} \\ & = \int_{t_0}^t \delta(t' - \tau) \operatorname{d}t' \\ & = \Theta(t-\tau) \, \Theta(\tau - t_0) - \Theta(t_0 - \tau)\, \Theta(\tau - t). \end{align}

This now satisfies: $$\frac{\partial}{\partial t} \left(\frac{\delta f(t)}{\delta \dot{f}(\tau)}\right) = \delta(t - \tau),$$ as expected. Because $\dot{f}(t)$ doesn't carry any information about the vertical offset of $f(t)$, only differences of the functional derivative, like above, are well defined.

If the space of functions is limited to those that satisfy $\lim_{t\rightarrow -\infty} f(t) = 0$, then we can take $t_0\rightarrow -\infty$ to get the expression from Valter Moretti's answer.

• Since the choice of $t_0$ is arbitrary, your calculation seems to suggest that this variation $\frac{\delta f(t)}{\delta \dot{f}(t)}$ is not well defined. – taper Feb 25 '17 at 4:15
• @taper I have now addressed that, and you're right, only differences in that functional derivative are well defined. – Sean E. Lake Feb 25 '17 at 19:35
• Thanks for this insight. So combining with Valter Moretti's answer, the full solution is then $\delta f(t) / \delta \dot{f}(\tau) = \theta(t-\tau) + c(\tau)$. But what kinds of initial conditions would determine $c(\tau)$ - it seems like for this problem these should be context-independent of what $f$ actually is. One natural condition seems to be $\delta f(a) / \delta \dot{f}(b) = 0$ where $b>a$. Then $0 = \delta f(a) / \delta \dot{f}(b) = \theta(a-b) + c(b) = c(b)$, so the full solution is just $\delta f(t) / \delta \dot{f}(\tau) = \theta(t-\tau)$ - is this true? – smörkex Feb 26 '17 at 3:53
• I would disagree with that. Looking at your other questions, related to Euler-Lagrange equations, this isn't relevant anyway. There are two main ways to do that problem. First, variational derivative of the action w.r.t. $x(t)$ witch uses chain rules and $\frac{\delta x(t)}{\delta x(t')} = \delta(t-t')$. The second is to use partial derivatives in which $x$, $\dot{x}$, $\ddot{x}$, etc are all treated as independent variables. – Sean E. Lake Feb 26 '17 at 5:59
• @Kurt In either case, the result is: \begin{align} \frac{\delta S[x]}{\delta x(t)} &= \int \left(\frac{\partial L}{\partial x} \delta(t'-t) + \frac{\partial L}{\partial \dot{x}} \delta'(t'-t) + \frac{\partial L}{\partial \ddot{x}} \delta''(t'-t) + \ldots \right) \operatorname{d}t' \\ & = \sum_{n=0}^\infty (-1)^n \frac{\operatorname{d}^n}{\operatorname{d}t^n} \left(\frac{\partial L}{\partial \frac{\mathrm{d}^n\, x}{\mathrm{d}\, t^n}}\right) \end{align} – Sean E. Lake Feb 26 '17 at 6:03