# How to take 'non-local' functional derivatives?

I am currently in the process of getting into linear response theory in general, and I have often met functional derivatives of the type

$$\frac{\partial J[f(x)]}{\partial f(y)} = \chi(x,y).$$

I've came across these things mostly in the context of defining non-local response functions. The physical picture behind this definition is often something like "how does the quantity $$J$$ change at $$x$$ by changing $$f$$ infinitesimally at location $$y$$". Some ideas include

• Eq. 11 of this publication, describing non-local polarizabilities and susceptibilities of molecules and materials
• In many-body perturbation theory, I think it is used to express two-point functions by taking a functional derivative with the functional depending on one, and the test function depending on the other coordinate. For an example with Hedin's formalism, Eq. 11 is here
• A seemingly general quantity called "fundamental response tensor", found for example Eq. 3.60 here

Now I only had some basic introduction to functional analysis in my studies - mostly in classical mechanics, the Euler-Lagrange equations and such. But there, and basically any other places I've looked, I see functionals being defined for example as

$$J[f] = \int F(x,f,f') \mathrm{d} x$$

and their derivatives using the limits of expressions such as

$$J[f+h] = \int \left[ F(x,f,f') + F(x,f+h,f'+h')x\right] \mathrm{d} x$$

corresponding to taking a derivative in form of

$$\frac{\partial J[f(x)]}{\partial f(x)}$$

As you can see, the main difference here is the variable in which the variation is defined. I just can't wrap my head around the fact that if the functional $$J[f(x)]$$ has admissable functions defined on the domain $$x$$, then how can I take any kind of variation using functions defined on the domain $$y$$? Could someone clear these things up, maybe provide me with a mathematical example or a detailed derivation somewhere from physics that shows how such variations are to be taken?

This not always explained particularly well, but the idea is that $$J$$ is a functional, which means that it eats a function and returns a number. In particular, it is not strictly correct to write $$J[f(x)]$$, since $$f(x)$$ is the number obtained by evaluating the function $$f$$ at the point $$x$$. One should instead write $$J[f]$$.

Given a second function $$g$$, the quantity $$\left.\frac{d}{d\epsilon}J[f+\epsilon \cdot g]\right|_{\epsilon = 0}$$ is a generalization of the directional derivative which applies to functionals. In essence, it measures the rate at which $$J$$ changes when we change its input by a tiny amount in the "direction" of $$g$$.

It is sometimes possible to write this derivative in the following form:

$$\left.\frac{d}{d\epsilon}J[f+\epsilon \cdot g]\right|_{\epsilon = 0} = \int d^nx \ M\bigg(x,f(x),f'(x),\ldots\bigg) g(x)$$

In such cases, we call $$M(x,f(x),f'(x),\ldots)$$ the functional derivative of $$J$$ with respect to $$f$$, denoted $$\frac{\delta J}{\delta f}$$. Since this $$M$$ is inside the integral, we should expect $$\frac{\delta J}{\delta f}$$ to be a function of $$x$$; we might therefore choose the notation $$\frac{\delta J}{\delta f}(x)$$ or (more commonly) $$\frac{\delta J}{\delta f(x)}$$.

If we have a functional of the form $$J[f] = \int dx'\ \mu(x,x') f(x')$$, it's important to note that this functional is itself a function of $$x$$. In that spirit, it should really be labeled $$J_x[f]$$ or $$J[f;x]$$, as for each value of $$x$$, $$J_x$$ is a functional.

With that slight ambiguity cleared up, the functional derivative as defined above is trivial to evaluate, and the result is $$\frac{\delta J_x}{\delta f(x')} = \mu(x,x')$$

Let me start by mentioning that the definition you mention from classical mechanics is the definition of a functional derivative. J. Murray's answer already covers this, so let me try and make some additional comments I think you might find helpful. Your notation is leading you somewhat astray. The function $$J[f]=\int F(x,f,f^\prime,f^{\prime\prime},\ldots)dx$$ in classical mechanics is a function of $$f(x)$$ at all points $$x$$, not just at one specific point, so it's really not correct to write $$\frac{\partial J[f(x)]}{\partial f(x)}$$, but rather the more standard notation $$\frac{\delta J[f]}{\delta f(x)}$$.

Since you mention linear response theory, let me try and describe things in a way I suspect will adapt easily to that area. Suppose we have some input function $$f$$ and the output of some process is given by $$J[f](x)=\int_{-\infty}^xG(x-y)f(y)dy,$$ so really we are just thinking about a convolution out to some position $$x$$. This function $$G$$ I will treat as given and, in particular, independent of $$f$$. This is not the most general thing we could write down, but it's very good to understand.

Looking at this $$J$$ I've written down, something you can note right away is that, while it's a function of $$x$$, it depends on the values of $$f$$ at more than just that one point. In principle, it could depend on all the values of $$f$$ below $$x$$. As a result, it would be entirely misleading to write $$J[f(x)]$$.

Next, we can compute the functional derivative by writing $$J[f+\eta\delta f]=J[f]+\eta\int_{-\infty}^x G(x-y)\delta f(y)dy+\mathcal{O}(\eta^2)$$ for an arbitrary function $$\delta f$$. By the definition of the functional derivative, it follows that we should rewrite the above as $$J[f+\eta\delta f]=J[f]+\eta\int_{-\infty}^\infty \theta(x-y) G(x-y)\delta f(y)dy+\mathcal{O}(\eta^2)$$ where $$\theta(x)$$ is the Heaviside step function and we should now identify $$\frac{\delta J[f](x)}{\delta f(y)}=\theta(x-y)G(x-y).$$ Depending on how much linear response theory you've seen it may be satisfying that the response function is essentially the transfer function of the convolution for the kind of in-out system I defined by $$J$$.

As a final point, let me mention that there is a notion of a functional Taylor series expansion (without commenting on the convergence, or lack thereof, of such objects). These commonly appear in quantum field theories, particularly in relation to effective theories. The idea is to write a functional $$J[f]$$ as $$J[f](x)=J_0(x)+\int dy_1J_1(x,y_1)f(y_1)+\iint dy_1dy_2 J_2(x,y_1,y_2)f(y_1)f(y_2)+\cdots.$$ The coefficients here will be related to the functional derivatives of $$J[f]$$, just like with the Taylor series.

• Thank you for the explanation. Indeed I am working with an expression that you have in your response theory explanation. However, I think I have the 'inverse' problem on hand: I want to find the 'Green function' of a system, and for that I want to evaluate the 'mixed' functional derivative. Hence I was asking the question.
– user112876
Feb 3 at 14:26
• @Ezze Taking a mixed derivative is simply taking another derivative, so once you know how to do one derivative, you can always do more. Feb 3 at 18:36
• @Ezze But also, I have given the method for calculating the system response function. It is the first derivative evaluated at zero fields. Feb 3 at 18:40
• @Richard_Myers I am not quite sure I can follow... By saying you gave a way to calculate the response function, you basically mean Eq. 4 of your answer, right? I took it as a 'proof' of the connection between the mixed derivative and the response function, and I am glad that I see the connection now. However, I am still uncertain how to actually do the derivative itself. Do you maybe have a reference somewhere where this derivative is taken? By 'mixed' derivative I've meant 'non-local', eg. delta J[f](x)/delta f(y), not the second derivative.
– user112876
Feb 4 at 10:05
• In functional analysis, the same language used for normal derivatives is standard, hence my confusion about mixed derivatives. Anyway, Eq. 4 was only meant to be an example of a calculation which I thought would be relevant to you. Eq. 2 is where I did the calculation. As I have done, evaluate the functional at $f+\delta f$, expand to first order in $\delta f$, and from this expansion identify the derivative as the coefficient of the $\delta f$. Feb 4 at 19:21