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Consider the piecewise constant function $\psi:I=[-1,1] \rightarrow \mathbb{R}$ given by

$$\psi(x) = \begin{cases} \psi_1 & x \leq 0, \ \psi_2 & x > 0 \end{cases}$$

for some constants $\psi_1, \psi_2 \in \mathbb{R}$. I would like to evaluate the integral

$$\int_I \delta(x) \psi(x) dx$$

where $\delta$ is the Dirac delta distribution centered at $x=0$. Of course, if we think about distributions in the usual way then you might say that $\delta(\psi) = \langle \delta, \psi \rangle = \psi(0) = \psi_1.$ But then the result depends on a fairly arbitrary choice when defining $\psi$: should the left- or right- half of the interval be closed? This question doesn't seem to have a meaningful answer when dealing with a problem that arises from a physical system (say).

Instead, consider a family of distributions $\phi_\epsilon(x)$ such that

  • $\phi_\epsilon(x)=\phi_\epsilon(-x)$,
  • $\int_I \phi_\epsilon(x)=1$ for all $\epsilon$, and
  • $\lim_{\epsilon \rightarrow 0} \phi_\epsilon = \delta$,

i.e., any family of even distributions with unit mass that approaches the Dirac delta distribution as $\epsilon$ approaches zero. (For instance, you could use the family of Gaussians $\phi_\epsilon(x) = \frac{1}{\epsilon\sqrt{\pi}}e^{-x^2/\epsilon^2}$.)

I can now think of my integral as

$$\lim_{\epsilon \rightarrow 0} \int_I \phi_\epsilon(x) \psi(x) dx.$$

For some $\epsilon > 0$, the integral inside the limit can be expressed as

$$u(\epsilon) = \psi_1 \int_{-1}^0 \phi_\epsilon(x) dx + \psi_2 \int_0^1 \phi_\epsilon(x) dx = \frac{1}{2}\left( \psi_1 + \psi_2 \right) = \bar{\psi},$$

where $\bar{\psi}$ is the mean of the constant values. Can we say, then, that $\lim_{\epsilon \rightarrow 0} u(\epsilon) = \bar{\psi}$? It would seem so: for any $\mu > 0$ there exists an $\epsilon_0$ such that $\epsilon < \epsilon_0$ implies $|u(\epsilon) - \bar{\psi}|<\mu$ (namely, $\epsilon$ is any positive constant!). But clearly I've got a problem somewhere, because $\bar{\psi} \ne \psi_1$, i.e., this result does not agree with my earlier interpretation.

So what's the right thing to do here? The latter answer ($\bar{\psi}$) agrees more with my "physical" intuition (because it's invariant with respect to deciding which half-interval is open), but I'm concerned about rigor.

Edit: Since the problem as stated is not well-posed ($\delta$ cannot be evaluated on discontinuous functions), let me give some motivation. Imagine that I have a pair of piecewise linear functions $f,g:I^2 \rightarrow \mathbb{R}$, which are again discontinuous only at $x=0$. I would like to integrate the wedge product of $df$ and $dg$ over the domain:

$$\int_{I^2} df \wedge dg = \int_I \int_I \frac{\partial f}{\partial x} \frac{\partial g}{\partial y} - \frac{\partial f}{\partial y}\frac{\partial g}{\partial x} dx dy.$$

Consider just the first term $(\partial f/\partial x)(\partial g/\partial y)$ and consider just the inner integral $\int_I \cdot dx$. We now have (almost) the original problem: $\partial f/\partial x$ can be thought of as a $\delta$ (plus a piecewise constant), and $\partial g/\partial y$ is simply piecewise constant along the $x$-direction.

So, the problem could be restated as: how do I integrate the wedge product $df \wedge dg$ of piecewise linear 0-forms $f$ and $g$ defined over a planar region? Formally this problem may again be ill-posed, yet it is a real problem that comes up in the context of finite element analysis where basis functions are nonsmooth or even discontinuous.

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It does not make sense to evaluate δ at a function which is not continuous at zero. Distributions, like δ, are certain functions defined on certain vector spaces (Wikipedia should have some of the details; Rudin's book on functional analysis is a most efficient way to get them in textbook form) and none of those on which δ is defined include your discontinuous function –  Mariano Suárez-Alvarez Aug 31 '10 at 15:49
    
Thanks -- fair enough. I've added some motivation to the question in the hopes that it might lead to a better-formulated question. –  user1612 Aug 31 '10 at 16:14
    
While taking an average is usually valid, you should be careful. Anyone interested should read Griffiths, David, and Stephen Walborn. "Dirac deltas and discontinuous functions." American Journal of Physics 67 (1999): 446-446. –  user147869 May 4 at 23:37

2 Answers 2

up vote 4 down vote accepted

The solution you have suggested is a perfectly good one. The only problem is that you are venturing outside the bounds of conventional Distribution theory. Consequently, the onus is on you to prove whatever properties of your definition that you use.

Actually, the Heaviside step function is more usually treated as a distribution itself. Bracewell's book, "The Fourier Transform and its Applications" considers the problem of defining a value at the discontinuity and concludes that it mostly doesn't matter.

What you are doing seems somewhat similar to the idea of density in physics. I'd be surprised if the physicists or the electrical engineers (signal processing) people hadn't already confronted this issue. However, after looking around a little bit I am unable to find anything specific.

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It seems your question contains its own answer. $\delta$ is not defined as evaluation-at-a-point except for continuous functions. (The inner product, strictly speaking, is defined in $L^p$ spaces of equivalence classes of functions and the "value" of such a class at a point is undefined unless you limit the evaluation to continuous functions.) Your second method is close to one standard method of creating (and calculating with) $\delta$ as a weak limit of integrals. (Another standard method, based on integration by parts, requires the trial function to be differentiable, so applying that method won't work here.) Thus, you have clearly demonstrated why evaluate-at-a-point is an unsuitable definition of $\delta$ for discontinuous functions.

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