Is it legitimate to use a delta function as a composite function?

Question

I am used to $\int f(x) \delta(x) \, dx = f(0) $ and the idea that this represents the action of a dual vector on $f(x)$ as an element of a vector space of functions, mapping it to the real number $f(0)$. Advanced physics books on electromagnetism use the composite $\delta[g(x)] = \delta(x-r)/g'(r)$, $r$ a root of $g$, and integrate $f(x) \delta[g(x)] \, dx$ in deriving the Lienard-Wiechert potential of a moving charge using a Green function. I am skeptical of the delta function, not actually a function, as a composite function. (The result for the potential is correct - I just don't think the derivation is) I wonder if $g(x)$ represents a change of basis and the integral should be of $ f[g(x)] \delta[g(x)] \, dg$ with the same real number output $f(0)$.

Answer 1

One approach to a definition of the composition $\delta[g(x)]$ is to define the delta function as the distribution obtained by taking the second derivative of the absolute value function, $$\delta(x)=\frac{1}{2}\left(\frac{d}{dx}\right)^2|x|,$$ and then to define the composition with a differentiable function $g(x)$ by $$\delta[g(x)]=\frac{1}{2}\left(\frac{1}{g'(x)}\frac{d}{dx}\right)^2 |g(x)|.$$ This approach to a distribution as a derivative of finite order of a continuous function, and the composition law it entails, is worked out in Introduction to the theory of distributions by J. Campos Ferreira. As explained on page 31, if $g(x)$ has isolated roots at $x=x_n$, $$\delta[g(x)]=\sum_n\frac{|g'(x_n)|}{2g'(x_n)^2}\left(\frac{d}{dx}\right)^2|x-x_n|=\sum_n\frac{1}{|g'(x_n)|}\delta(x-x_n).$$

Answer 2

$\newcommand\de\delta\newcommand\ep\varepsilon$The generalized function "$\de(g(x))$" (better denoted as $\de(g)$) must be defined -- just as its special case, the generalized function "$\de(x)$" (better denoted as $\de$), has to be defined.

The action of the generalized function $\de$ on a (say) continuous function $f$ is $$\de(f):=f(0)=\lim_{\ep\downarrow0}\int dx\,K_\ep(x)f(x),$$ where (say) $K_\ep\ge0$, $K_\ep(x)=0$ for real $x\notin[-\ep,\ep]$, and $\int K_\ep=1$, for each real $\ep>0$. In this sense, $K_\ep\approx\de$ for small $\ep>0$.

So, we may define the action of the generalized function $\de(g)$ on a continuous function $f$ as $$\de(g)(f):=\lim_{\ep\downarrow0}\int dx\,K_\ep(g(x))f(x)$$ whenever this limit exists.

If e.g. the function $g$ has $k$ distinct roots $r_1,\dots,r_k$ and $g$ is continuous everywhere and continuously differentiable in a neighborhood of each root $r_j$ with $g'(r_j)\ne0$ for each $j$, then for each $j$ the restriction of $g$ to a neighborhood of the root $r_j$ has a continuously differentiable inverse $g_j^{-1}$, with $g_j^{-1}(0)=r_j$ and $(g_j^{-1})'(0)=1/g'(r_j)$. So, then $$\de(g)(f)=\sum_{j=1}^k\lim_{\ep\downarrow0} \int dy\,|(g_j^{-1})'(y)|\,K_\ep(y)f(g_j^{-1}(y)) =\sum_{j=1}^k\frac{f(r_j)}{|g'(r_j))|}.$$

When $k=1$ and $g'(r)>0$ for $r=r_1$, then the just described action corresponds to your informal expression $\de(x-r)/g'(r)$.

On the other hand, even when $k=1$ but $g'(r)<0$ for $r=r_1$, your informal expression $\de(x-r)/g'(r)$ does not make sense. E.g., suppose that $g(x)=-x$. Then $k=1$, $r=r_1=0$, and $g'(r)=g'(0)=-1<0$ so that your informal expression $\de(x-r)/g'(r)$ for $\de(g(x))$ becomes $-\de(x)$. However, here $\de(g(x))=\de(-x)=\de(x)$ makes much more sense than $-\de(x)$. Indeed, informally, $$\int dx\,\de(-x)f(x)=\int dx\,\de(x)f(-x)=f(-0)=f(0)=\int dx\,\de(x)f(x).$$ Alternatively, again informally, $\de(-x)\approx K_\ep(-x)=K_\ep(x)\approx\de(x)$ for small $\ep>0$ if the function $K_\ep$ is even.

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The approach described above seems quite natural and, as I have just learned, is well known -- see e.g. formulas (9.2) and (9.3) on p. 22 of Methods of the Theory of Generalized Functions by Vladimirov.

Answer 3

Yes, there is a valid notion of a "composite function". It's unique, meaning that the end result doesn't depend on how you might construct it or approximate it by other functions. I'll refer to Hörmander, The Analysis of Linear Differential Operators Vol. I, Theorem 6.1.2:

Let $X_j \subset \mathbb{R}^{n_j}$, $j = 1,2$, be open sets, and $f: X_1 \to X_2$ a $C^\infty$ map such that $f'(x)$ is surjective for every $x \in X_1$. Then there is a unique continuous linear map $f^* : \mathscr{D}'(X_2) \to \mathscr{D}'(X_1)$ such that $f^* u = u \circ f$ when $u \in C^0(X_2)$. ... [The notation $\mathscr{D}'(X_i)$ denotes the space of distributions on $X_i$.]

In particular, let's say $f$ is a diffeomorphism, and $u$ is a distribution. Their composition will act on smooth, compactly supported test functions following the ordinary change of variables formula: $$ \int_{X} u(f(x)) \phi(x) \, dx = \int_{g(X)} u(y) \phi(f^{-1}(y)) \lvert \det (f^{-1})'(y) \rvert \, dy $$ If $u$ is the Dirac delta, we obtain $\phi(f^{-1}(0)) \lvert \det (f^{-1})'(0) \rvert$. Which is sort of like $\int \frac{\delta(x-r)}{g'(r)} \phi(x) \, dx$ with $g(r)=0$, up to some different notation and assumptions to make the $\lvert \det \rvert$ go away.

(The theorem also explains other expressions that are related to the wave equation, electromagnetism, and retarded time / potentials - expressions like $\delta(c^2 t^2 - |x|^2)$ and $\delta(t - \frac{1}{c}|x|)$. For these examples, however, the composite provided by the theorem is defined only on a proper subset of $\mathbb{R}^{n_1}$ - away from $(t,x)=(0,0)$ and $x=0$, respectively. Trying to extend such a distribution to all of $\mathbb{R}^{n_1}$ is a separate step, not covered by the theorem above.)

Answer 4

Added 2/15/2024: See page 79 of Operational Calculus by van der Pol and Bremmer, where the authors give a particularly straightforward and simple derivation of the composition formula starting with the Heaviside step function. (The book also presents several approximate, or nascent, delta functions--what Kanwal in Generalized Functions essentially calls delta series.)

Original response:

To get a simple feel for this type of identity, consider a nascent, or approximate, delta function rep along with the sifting property of the Heaviside-Dirac delta function.

The sifting property along with the local Taylor series expansion about a zero $g(x_0) =0$ with $g'(x_0) = a \neq 0$

$$g(x) = g(x_0) + g'(x_0)(x-x_0) + g''(x_0)(x-x_0)^2/2! + \cdots$$

$$ = g'(x_0)(x-x_0) + g''(x_0)(x-x_0)^2/2! + \cdots$$

$$ = (x-x_0)[g'(x_0) + g''(x_0)(x-x_0)/2! + \cdots)]$$

allows us to consider, about the isolated zero,

$\delta(g(x)) = \delta[(x-x_0)g'(x_0) + O(x-x_0)^2)] \to \delta(a(x-x_0)).$

Then consider, for $L >0$,

$$\lim_{L \to \infty} \int_{-L/2}^{L/2} e^{-i2 \pi uy}du = \lim_{L \to \infty} \frac{\sin(L\pi y)}{\pi y} = \delta(y)$$

so

$$\delta(ay) = \lim_{L \to \infty} \frac{\sin(L\pi ay)}{\pi ay} =\lim_{L \to \infty} \int_{-L/2}^{L/2} e^{-i2 \pi u ay}du $$

$$=\lim_{L \to \infty} \int_{-|a|L/2}^{|a|L/2} e^{-i2 \pi v y}dv/|a| = \frac{\delta(y)}{|a|}$$

with $v =au$.

(Of course limits are understood to be taken outside an integral for the nascent delta reps.)

A simple illustration of the sifting property at zeros reflected in the nascent sinc function rep is provided by the Desmos graphing app with $f(x) = \frac{\sin(L \pi x)}{\pi x}$ and $g(x) = \cos(\pi x)$. Then with $L = 10$, a plot of $f(g(x))$ superposed on $g(x)$ is

Choosing rather $g(x) = \cos(b x)$, with a sliding $b$, illustrates how the area under each sinc function essentially scales in the limit as $1/|b|$, equal to $1/|g'(x)|$ evaluated at the zeros of $g(x)$.