# Functional derivative of a functional that depends on antiderivative

It is well known how to calculate functional derivative if a functional depends of the function and it's derivatives (Euler-Lagrange rule): $\mathcal{L}=\int F(x,\dot{x},t)dt$. There is also straight-forward generalization for a functional that depend on higher order function derivatives.

From now on I will use $x$ instead of $t$ and will denote functions by small english letters (e.g. $f$, $\phi$. etc.).

In my problem, a functional depends on the function's antiderivative:

$$\mathcal{L}=\int dx \int_{-1}^x f(x')dx'.$$

How can I calculate ${\delta \mathcal{L}}/{\delta f}$ ? I tried to do it by definition:

$$\int \frac{\delta \mathcal{L}}{\delta f} \phi(x) dx = \left[\frac{dF[f+\epsilon\phi]}{d\epsilon}\right]_{\epsilon=0}.$$ However, after simplifying the right hand side I found out that the result cannot be factorized to $\int (...)\phi(x)dx$.

Denoting $F=\int_{-1}^xf(x')dx'$, I was also trying to use the chain rule, but did not succeed.

In my actual problem I need to minimize $\mathcal{L}=\int dx\int \exp(-f(x'))dx'$.

Thanks,

Mikhail

Edit: Sorry, I should have been more specific from the beginning. The actual problem is find the functional gradient w.r.t. $f(x)$ of the following functional:

$$\mathcal{L} = \int_{-1}^1 dx \exp\left[\int_{-1}^x f(x')dx'\right]\beta(x).$$

Note, that the second integral is entirely inside of the exponent. $\beta(x)$ does not depend on f(x).

• You could set $g(x)=\int_{-1}^x f(x') dx'$ and then find $g$. Then $f(x)=g'(x)$. (But you say you tried that). For your concrete problem, there is probably no minimizer. (Hint: Consider $f(x)=n$ and let $n\to\infty$). – Kusma Aug 6 '18 at 15:07
• Hi Kusma, I just did not quite understand the chain rule for functional derivative. I am interested how can I solve problems like this one using the tools of functional calculus – Mikhail Genkin Aug 6 '18 at 15:17
• I don't see that you need a chain rule here. If we have $\mathscr{L}=\int L(f,\int f(x')dx',x)dx$, we can set $F=\int f(x')dx'$ then $\mathscr{L}=\int L(F',F,x)dx$. We can forget for the moment how $F$ was defined, solve for $F$ using functional derivative and then recover $f=F'$. – Kusma Aug 6 '18 at 15:55
• @Kusma, No, the way you describe will result in $\delta \mathcal{L} /\delta F$ that described in terms of $f$. It is different from $\delta \mathcal{L} /\delta f$. The relanshionship is manifested in the chain rule – Mikhail Genkin Aug 7 '18 at 13:55

## 2 Answers

1. OP first considers a functional $S\equiv {\cal L}$ of the form \begin{align}S[f]~=&~\int_a^b\!\mathrm{d}x \int_a^x\!\mathrm{d}x^{\prime} ~g\circ f(x^{\prime}) \cr ~=&~\iint_{[a,b]^2}\!\mathrm{d}x ~\mathrm{d}x^{\prime} ~\theta(x-x^{\prime})~g\circ f(x^{\prime}) \cr ~=&~ (b-x^{\prime}) \int_a^b\!\mathrm{d}x^{\prime}~g\circ f(x^{\prime}), \end{align} \tag{1a} where $\theta$ is the Heaviside step function and $g$ is a fixed function. (In OP's examples $g(y)=y$ and $g(y)=e^{-y}$ and $a=-1$.)

The functional/variational derivative is then $$\forall x^{\prime}~\in~[a,b]:~~\frac{\delta S[f]}{\delta f(x^{\prime})} ~=~ (b-x^{\prime}) ~g^{\prime}\circ f(x^{\prime}) .\tag{1b}$$

2. In an edit OP considers a functional $S\equiv {\cal L}$ of the form $$S[f]~=~\int_a^b\!\mathrm{d}x ~g(x,F(x)), \qquad F(x)~:=~\int_a^x\!\mathrm{d}x^{\prime}~f(x^{\prime}) ~=~ \int_a^b\!\mathrm{d}x^{\prime}~\theta(x-x^{\prime})~f(x^{\prime}). \tag{2a}$$ Let $g_F=\frac{\partial g}{\partial F}$ denote the partial derivative wrt. the second argument of the $g$-function. The functional/variational derivative is then $$\forall x^{\prime}~\in~[a,b]:~~\frac{\delta S[f]}{\delta f(x^{\prime})} ~=~ \int_a^b\!\mathrm{d}x~\theta(x-x^{\prime})~g_F(x,F(x)) ~=~ \int_{x^{\prime}}^b\!\mathrm{d}x~g_F(x,F(x)).\tag{2b}$$

• Hi, thanks for the response. You are right: for this simplified problem. I am sorry, I should have been more specific from the beginning. Please see the edit. – Mikhail Genkin Aug 7 '18 at 14:26
• I updated the answer. – Qmechanic Aug 7 '18 at 14:52
• Yeah, I actually got similar result. It looks correct. Can you explain how Heaviside function turned out to be outside of $g$ function in your answer (originally it was inside the function $g(x,F(x))$) ? – Mikhail Genkin Aug 7 '18 at 15:04

I am not convinced that calculus of variations will help you find a minimiser $f$ for your problem, as no minimiser exists. Consider $$\mathcal{L}[f]=\int_{-1}^1 \int_{-1}^x \exp(-f(x'))dx'dx$$ Clearly $\mathcal{L}[f]\ge 0$ for all functions $f$. However, choosing $f_n\equiv \ln n$ (constant function), we have $$\mathcal{L}[f_n]=\int_{-1}^1 \int_{-1}^x \frac1n dx' dx = \frac1n\int_{-1}^1 (x+1)dx = \frac2n,$$ which tends to $0$ for $n\to\infty$. So if there is a minimiser, the value of the minimum must be zero. Hence we must have $\int_{-1}^x \exp(-f(x'))dx'=0$ for almost every $x$, which implies $\exp(-f(y))=0$ for almost every $y$. The exponential is never zero, so no such function exists.

• _Well, I omitted another function. My actual problem $\int_-1^1 \int_-1^x exp(-f(x')dx' ) \beta(x) dx$. Function $\beta(x)$ is does not depend on $f$. Note, that $dx'$ is also in the exponent. Sorry, I should have been more specific – Mikhail Genkin Aug 7 '18 at 14:20
• Please see the edit, you are actually absolutely right – Mikhail Genkin Aug 7 '18 at 14:24