Calculating the functional derivative of an implicit functional

Question

We want to calculate the functional derivative of the following wrt $\rho$: $$F=\int X dx$$ where $X$ is implicitly defined as: $$X=\frac{1}{1+\overline \rho(x) \int \rho(x) X dx}$$ and $\overline \rho$ is the result of the following convolution: $$\overline \rho(x) = \int \rho(x^\prime)H(\sigma^2 - (x-x^\prime)^2)dx^\prime$$ where $H$ is the Heaviside step function: $$\begin{cases} 1,& x\geq 0\\ 0,& x < 0\end{cases}$$ How can I calculate the functional derivative $\frac{\delta F}{\delta \rho}$ analytically?

Note that I can solve this numerically by solving iteratively for X and then computing the functional derivative numerically. But I'm looking for the analytical expression.

Answer 1

Given the implicit operator equation you have at hand, there's no hope to find an analytical expression for the sought for functional derivative: let's see why.
First of all, we cannot calculate directly $\frac{\delta F}{\delta \rho}$ since, as correctly stated in the OP, $F$ depends explicitly only on the variable $X$, which is defined by $\rho$ trough an implicit operator equation. Thus the functional derivative we need to calculate is really $$ \frac{\delta F(X(\rho))}{\delta \rho}=\frac{\delta F\circ X}{\delta \rho}=\lim_{\varepsilon\to 0}\frac{F\circ X(\rho+\varepsilon \rho_0)-F\circ X(\rho)}{\varepsilon} =\frac{\delta F}{\delta X} \Big(\frac{\delta X}{\delta \rho}\Big)(\rho_0)\label{1}\tag{1} $$ where

• $\varepsilon\rho_0$ is the variation of the function $\rho$,
• $X$ is the solution of the implicit operator equation given above.

Said that, we nevertheless see that we do not really need to solve the implicit operator equation above which we write as $$ G(\rho, X)=0 $$ where $$ G(\rho, X)=\left[{1+\overline \rho(x) \int \rho(x) X dx}\right]X-1 $$ and $$ \overline \rho(x) = \int \rho(x^\prime)H(\sigma^2 - (x-x^\prime)^2)dx^\prime $$ Indeed, we can apply the implicit function theorem for functions between Banach spaces (in a formal but fully justifiable way) and obtain the following operator relations $$ \begin{split} \frac{\delta G(\rho, X)}{\delta \rho}\delta\rho +\frac{\delta G(\rho, X)}{\delta X}\delta X =0 & \iff \frac{\delta G(\rho, X)}{\delta X}\delta X=- \frac{\delta G(\rho, X)}{\delta \rho}\delta\rho \\ &\iff \delta X = -\left[\frac{\delta G(\rho, X)}{\delta X}\right]^{-1} \frac{\delta G(\rho, X)}{\delta \rho}\delta\rho\\ &\iff \frac{\delta X}{\delta\rho} = -\left[\frac{\delta G(\rho, X)}{\delta X}\right]^{-1} \frac{\delta G(\rho, X)}{\delta \rho} \end{split}\label{2}\tag{2} $$ and since we know that $$ \lim_{\varepsilon\to 0}\frac{F(X+\varepsilon X_0)-F(X)}{\varepsilon} =\frac{\delta F}{\delta X}= \int X_0dx $$ where $\varepsilon X_0=\delta X$ is the variation of the function $X$, has a very simple expression we may think we obtained the sough for result. But this is not the case, as the operator $\frac{\delta G(\rho, X)}{\delta X}$ in the equations \eqref{2} is invertible but has not a simple analytic expression. Indeed, $$ \frac{\delta G(\rho, X)}{\delta X}\delta X = \left[{1+\overline \rho(x) \int \rho(x) X dx}\right]\delta X + \left[\color {red}{X\cdot{\overline \rho(x) \int \rho(x) \delta X dx}}\right] $$ The part in red makes the operator at the right side of the above equation a linear integral operator of the third kind applied to the variation $\delta X$, i.e. $$ \frac{\delta G(\rho, X)}{\delta X}\delta X = a(x)\delta X+ b(x)\int c(x) \delta X\operatorname{d\!}x $$ and surely has not an explicit analytic inverse $\left[\frac{\delta G(\rho, X)}{\delta X}\right]^{-1}$, needed for an explicit calculation of $ \frac{\delta X}{\delta\rho}$ in \eqref{2} and \eqref{1}.