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https://en.wikipedia.org/wiki/Functional_derivative

Is there a straightforward way to find a functional whose functional derivative has a particular form? That is, is there something like functional integration?

Here's my specific problem. I want to find a functional $F(\rho)$ whose functional derivative is

$$\frac{\delta F}{\delta\rho}(x) = f((g\star\rho)(x)) $$

Here $g\star\rho$ means the convolution of functions $g:\mathbb{R}\rightarrow\mathbb{R}$ and $\rho:\mathbb{R}\rightarrow\mathbb{R}$. $f:\mathbb{R}\rightarrow\mathbb{R}$ is some other (generally nonlinear) function. The functions $f$ and $g$ are given. Feel free to assume $f$ has as many continuous derivatives as you need, and that $f$ and $g$ can be integrated analytically.

If it matters to anyone, this question arises out of a strategy to develop a Lyapunov functional for a nonlinear PDE system. A numerical solution would be good enough for my purposes.

Thanks for any ideas.

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  • $\begingroup$ Typical notation for convolution is $*$ or $\star$. Symbol $\circ$ is used for composition of functions generally. $\endgroup$ – Radost May 8 '19 at 22:36
  • $\begingroup$ This exercise is the only time I have seen a nontrivial "functional integration", as you say. $\endgroup$ – Giuseppe Negro May 8 '19 at 22:49
  • $\begingroup$ For $f$ linear I've managed to get some progress by diagonalising convolution using Fourier transform and then putting this in basis of eigenfunctions of Fourier transform. It gives close to explicit expression for the 'integral' functional. I have no idea if this method is going somewhere for $f$ nonlinear. $\endgroup$ – Radost May 8 '19 at 23:55
  • $\begingroup$ Perhaps a little more tractable question would be to find some nice $C_\infty$ functions s.t. there is no integral at all. $\endgroup$ – Radost May 8 '19 at 23:56
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I begin by adopting the strategy Radost suggests in a comment above: solve the linear case in the Fourier basis. Here's the answer I find. The functional

$$\mathscr{F}(\rho)=\frac{1}{2}(g\star\rho\star\rho_-)(0)+C$$

where $\rho_-(x)\equiv\rho(-x)$ and $C$ is a constant of integration, has functional derivative

$$\frac{\delta\mathscr{F}}{\delta\rho}=g\star\rho$$

I'm going to start with several assumptions. I don't think any of these is essential, but it is easier to give a specific answer to a more specific question (and these happen to match the application I have in mind). First, I will assume that all functions are defined on $\mathbb{R}$, but periodic with period 1. Each such periodic function $f$ has a Fourier series,

$$f(x)=\sum_{k\in2\pi\mathbb{Z}}\tilde{f}_k e^{i k x}$$

Integrals will generally be from 0 to 1 if I don't specify bounds. Thus, I define convolution as

$$(f\star g)(y)\equiv\int_0^1 f(x)g(y-x)dx$$

It is easy to verify the Convolution Theorem with these definitions: $\left(\tilde{f\star g}\right)_k=\tilde{f}_k\tilde{g}_k$. I also define an inner product,

$$\left<f,g\right>\equiv\int f(x)^*g(x) dx$$

using $f^*$ for complex conjugate. $\left<af,bg\right>=a^*b\left<f,g\right>$, and $\left<e^{i k x},e^{i j x}\right>=\delta_{kj}$. The functional derivative can be defined as

$$\lim_{\epsilon\rightarrow 0}\frac{\mathscr{F}(\rho+\epsilon\phi)-\mathscr{F}(\rho)}{\epsilon}=\left<\frac{\delta\mathscr{F}}{\delta\rho}^*,\phi\right>$$

I'm also going to assume that $g$, the kernel of the convolution below, is real and even, $g(x)=g(-x)\in\mathbb{R}$, which implies $\tilde{g}_k=\tilde{g}_{-k}\in\mathbb{R}$.

OK, enough throat-clearing. Now, let's try to find a functional $\mathscr{F}(\rho)$ whose functional derivative is $g\star\rho$. I'll begin by finding $\mathscr{F}(\phi)$, for $\phi(x)=a e^{ikx}+b e^{-ikx}$, $a,b\in\mathbb{C}$. I assume $\mathscr{F}(0)=0$. (This just amounts to the choice of a constant of integration.) I start at $\rho=0$ and integrate $d\mathscr{F}(\alpha\phi)d\alpha$ from $\alpha=0$ to 1. In stepping from $\rho=\alpha\phi$ to $\rho=(\alpha+d\alpha)\phi$, I have

$$ \begin{align} d\mathscr{F}&=\left<g\star(\alpha\phi)^*,\phi d\alpha\right> \\ &=\left<\tilde{g}_k\alpha(a e^{ikx}+b e^{-ikx})^*,(a e^{ikx}+b e^{-ikx})d\alpha\right> \\ &=\tilde{g}_k(2ab)\alpha d\alpha \\ \mathscr{F}(\phi)=\int_{\alpha=0}^1d\mathscr{F}(\alpha) &=2ab\tilde{g}_k\int_0^1\alpha d\alpha \\ &=ab\tilde{g}_k \\ &=\sum_k\frac{1}{2}\tilde{g}_k\tilde{\phi}_k\tilde{\phi}_{-k} \end{align} $$

The factor of $\frac12$ is there because the product $ab$ appears twice in the sum, once for $k$ and once for $-k$. (Also, it is intuitive: this is the same $\frac12$ that appears when you integrate a linear.)

Now, suppose I have $\phi_1(x)=a e^{ikx}+b e^{-ikx}$ and $\phi_2(x)=c e^{ijx}+d e^{-ijx}$, with $j\ne k$. What is $\mathscr{F}(\phi_1+\phi_2)$? The obvious answer, $\tilde{g}_k a b + \tilde{g}_j c d$, is correct: whether you integrate from 0 to $\phi_1$ then from $\phi_1$ to $\phi_1+\phi_2$, or from 0 to $\phi_2$ then from $\phi_2$ to $\phi_1+\phi_2$, or diagonally from 0 to $\phi_1+\phi_2$, because $\phi_1$ and $\phi_2$ are orthogonal. So, this is the answer:

$$\mathscr{F}(\rho)=\sum_k\frac{1}{2}\tilde{g}_k\tilde{\rho}_k\tilde{\rho}_{-k}$$

That's a perfectly fine functional. Numerically, it could easily be evaluated using the FFT. However, it is also possible to write $\mathscr{F}$ in terms of the functions $g$ and $\rho$, as follows. First, define the function $\rho_-$ by $\rho_-(x)\equiv\rho(-x)$. It is easy to show that $\tilde{(\rho_-)}_k=\tilde{\rho}_{-k}$. Thus, by the Convolution Theorem, $\tilde{g}_k\tilde{\rho}_k\tilde{\rho}_{-k}$ is the $k^{th}$ Fourier coefficient of $g\star\rho\star\rho_-$. If you evaluate a function at 0, you get the sum of all its Fourier coefficients. Thus,

$$\mathscr{F}(\rho)=\frac12\left(g\star\rho\star\rho_-\right)(0)$$

I'll point out that $(\rho\star\rho_-)(y)=\int\rho(x)\rho(x-y)dx$ is a familiar object: it is the autocorrelation of $\rho$.

Now let's try a couple of sanity checks. First, let $g=\delta$ (the Dirac delta function). Then $g\star\rho=\rho$. And we know what functional has $\rho$ as its functional derivative: it is just $\int\frac{\rho^2(x)}{2}dx$. That indeed checks out:

$$ \begin{align} \frac12g\star\rho\star\rho_-&=\frac12\delta\star\rho\star\rho_- \\ &=\frac12\rho\star\rho_- \\ (\frac12g\star\rho\star\rho_-)(0)&=\frac12(\rho\star\rho_-)(0) \\ &=\int\frac12\rho(x)\rho(x-0)dx \\ &=\int\frac{\rho^2(x)}{2}dx \space \checkmark \\ \end{align} $$

Now let's test $\mathscr{F}$ in the delta function basis

$$ \begin{align} \mathscr{F}(\rho+\epsilon\delta(x-y))-\mathscr{F}(\rho)&=\frac12\left(g\star\left((\rho+\epsilon\delta(x-y)\star(\rho_-+\epsilon\delta(-x-y))\right)\right)(0)-\frac12\left(g\star\rho\star\rho_-\right)(0) \\ &\approx\frac{\epsilon}{2}\left(g\star\left(\delta(x-y)\star\rho(-x)+\rho(x)\star\delta(-x-y)\right)\right)(0) \\ \frac{\mathscr{F}(\rho+\epsilon\delta(x-y))-\mathscr{F}}{\epsilon}&=\frac12\left(g\star(\rho(y-x)+\rho(y+x))\right)(0) \\ &=\frac12\left(\int\rho(y-x)g(0-x)dx + \int\rho(y+x)g(0-x)dx\right) \\ &=\frac12\left(\int\rho(y-x)g(x)dx + \int\rho(y-u)g(u)du\right) \\ &=(g\star\rho)(y) \space \checkmark \end{align} $$

In the second-to-last step, I used the evenness of $g$ to replace $g(-x)$ with $g(x)$. And I substituted $u=1-x$ in the second integral and swapped the bounds of integration.

OK, so I think that works for the linear case. What about nonlinear? By integrating $d\mathscr{F}(\alpha\rho)$ from $\alpha=0$ to 1 it is easy to see that if $(g\star\rho)^n$ has a functional antiderivative, it must be

$$ \begin{align} \mathscr{F}(\rho) & = \frac{1}{n+1}\sum_k\tilde{\left((g\star\rho)^n\right)}_k\tilde{\rho}_{-k} + C \\ & = \frac{1}{n+1}\left((g\star\rho)^n\star\rho_-\right)(0) + C \\ & = \frac{1}{n+1}\int(\left(g\star\rho)(x)\right)^n\rho(x)dx + C \end{align} $$ There remain two problems (both really the same problem) with this that I have yet to nail down. First, I have only shown that this result is what you get if you integrate radially from 0 to $\rho$. I have not shown that the integral is independent of the path. The nice orthogonality that made this easy for the linear case has no direct counterpart for $(g\star\rho)^n$. Second, although $\left<\left((g\star\rho)^n\right)^*,\epsilon\phi\right>$ gives the correct $O(\epsilon)$ change in $\mathscr{F}(\rho+\epsilon\rho)$ when $\phi=\rho$, it doesn't work for arbitrary $\phi$. That is, $(g\star\rho)^n$ does not have a functional antiderivative for $n>1$.

Here is the functional derivative of the functional $\mathscr{F}$ just described:

$$ \frac{\delta\mathscr{F}}{\delta\rho}=\frac{1}{n+1}\left((g\star\rho)^n\right)+\frac{n}{n+1}\left(\left((g\star\rho)^{n-1}\rho\right)\star g\right) $$

The second term is the problem, of course. The convolution of a product is not the product of the convolutions, so this doesn't reduce to the desired derivative $(g\star\rho)^n$, except in the special cases $g(x)=\delta(x)$ and $n=1$. Nevertheless, some quick numerical experiments suggest that $\mathscr{F}$ might work as an approximate functional antiderivative of $(g\star\rho)^n$.

If we accept that $\mathscr{F}=\frac{1}{n+1}\left((g\star\rho)^n\star\rho_-\right)(0)$ is an adequate approximate functional antiderivative of $(g\star\rho)^n$, it would follow that if $f(\rho)$ has a convergent power series, and if $F(\rho)$ is an antiderivative of $f$, i.e. $F^\prime(\rho)=f(\rho)$, then an approximate functional antiderivative of $f(g\star\rho)$ is $$ \mathscr{F}(\rho) = \left(\frac{F(g\star\rho)}{g\star\rho}\star\rho_-\right)(0) + C $$

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  • $\begingroup$ I've been tracking this question for a while, I'm curious right now. Can you maybe already give the main idea? $\endgroup$ – RMWGNE96 May 13 '19 at 14:19
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    $\begingroup$ It's based on Radost's second comment about using the Fourier basis to find a functional antiderivative for the linear $f$ case. The quick answer is that the functional derivative of $(g\star\rho\star\rho(-x))(0)$ is $g\star\rho$. I think (though I haven't fully nailed this down yet) that a similar answer works for nonlinear $f$. $\endgroup$ – Leon Avery May 13 '19 at 14:24
  • $\begingroup$ Thanks! I wouldn't have a quick idea to prove it, looking forward to see your reasoning. $\endgroup$ – RMWGNE96 May 13 '19 at 14:25
  • $\begingroup$ Oops -- I left out a factor of $\frac{1}{2}$ in the comment above. $\endgroup$ – Leon Avery May 13 '19 at 15:23
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Here's a general recipe for constructing a functional antiderivative.

Lemma: Radial integration.

Let $\mathscr{F}:C^1(\Omega)\rightarrow\mathbb{R}$ be a functional with functional derivative $\delta\mathscr{F}(U)/\delta U(x)$. $\mathscr{F}$ can be computed from its functional derivative as follows, provided the integral exists

\begin{equation} \mathscr{F}(U) = \int_{r=0}^1\left( \int_\Omega \frac{\delta\mathscr{F}(rU)}{\delta U(x)}U(x)d x \right)d r + C, \end{equation}

where $C$ is a constant of integration.

Proof:

Fix a $U:\Omega\rightarrow\mathbb{R}$ and define $\mathscr{f}_U:[0,1]\rightarrow\mathbb{R}$ by

\begin{equation} \mathscr{f}_U(r) = \mathscr{F}(rU) \end{equation}

The derivative of $\mathscr{f}_U$ is

\begin{align} \frac{d\mathscr{f}_U(r)}{dr} &= \lim_{\epsilon\rightarrow 0} \frac{\mathscr{f}_U(r+\epsilon) - \mathscr{f}_U(r)}{\epsilon} \\ &= \lim_{\epsilon\rightarrow 0} \frac{\mathscr{F}\left( (r+\epsilon)U \right) - \mathscr{F}(rU)}{\epsilon} \notag \\ &= \lim_{\epsilon\rightarrow 0} \frac{\mathscr{F}\left( rU+\epsilon U \right) - \mathscr{F}(rU)}{\epsilon} \notag \\ &= \int_\Omega \frac{\delta\mathscr{F}(rU)}{\delta U(x)}U(x)dx, \end{align}

using the definition of the functional derivative in the last step. It follows that $\mathscr{f}_U(r)$ is an antiderivative of the function of $r$ defined by the final integral, i.e.

\begin{equation} \mathscr{f}_U(R) = \int_{r=0}^R\left( \int_\Omega \frac{\delta\mathscr{F}(rU)}{\delta U(x)}U(x)dx \right)d r + C, \end{equation}

provided the integral exists. ($C$ is a constant of integration.) Finally, note that by its definition, $\mathscr{f}_U(1) = \mathscr{F}(U)$ and substitute $R = 1$ to get the stated result. $\Box$

This lemma provides a test for the existence of a functional antiderivative. Given a function $\mathscr{d}(U(x))$, compute the integral $\mathscr{F}(U) = \int_{r=0}^1 \left(\int_\Omega\mathscr{d}(rU)U(x)dx\right)d r$. The function $\mathscr{d}$ may not have a functional antiderivative, but if it does, it must be $\mathscr{F}$ (up to a constant of integration). To complete the test, find the functional derivative of $\mathscr{F}$. If $\delta\mathscr{F}(U)/\delta U(x) = \mathscr{d}(U(x))$, then $\mathscr{d}$ has a functional antiderivative and it is $\mathscr{F}$. If, however, $\delta\mathscr{F}(U)/\delta U(x)\ne\mathscr{d}(U(x))$, no functional antiderivative of $\mathscr{d}$ exists. This test of course depends on the existence of the integral $\int_{r=0}^1 \left(\int_\Omega\mathscr{d}(rU)U(x)dx\right)d r$, but the existence of this integral says nothing, by itself, about whether a functional antiderivative of $\mathscr{d}$ exists.

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