Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We know that the Dirac delta can be loosely thought of as a function on the real line which is zero everywhere except at the origin, where it is infinite, $$ \delta(x)=\begin{cases}+\infty, &x=0\\ 0, &x\neq 0 \end{cases} $$ and which is also constrained to satisfy the identity $$ \int_{-\infty}^{\infty}\delta(x)dx=1 $$ However, this is merely a heuristic characterization. The Dirac delta function is rigorously defined either as a distribution or as a measure.

Here comes the problem. I've seen many PDEs that contain delta functions, which are not functions at all, in the traditional sense. For example the following:

Stokes equations with singularly forced: $$ \begin{align} -\nabla p+\mu\nabla^2{\bf u}+{\bf g}\delta({\bf x}-{\bf x}_0) &= 0,\\ \nabla\cdot {\bf u} &= 0 \end{align} $$ where ${\bf g}=(g_i)_{1\leq i\leq 3}\in{\mathbb R}^3$ is an arbitrary constant vector, ${\bf x}_0$ is an arbitrary point in the domain, and $\delta$ is the three-dimensional Dirac delta function.

Here are my questions:

  1. For the first equation, the first two terms, $-\nabla p$, $\mu\nabla^2{\bf u}$, are functions in the traditional sense, how should I understand "a traditional sense function plus a distribution/measure"?
  2. Generally, what does a PDE mean when it contains a measure or distribution?
share|cite|improve this question
up vote 8 down vote accepted

When you see a differential equation that uses distributions, like the Dirac delta distribution, then you must understand that "equality" is always meant in the sense of distributions.

Since distributions are elements of the dual spaces of various functional spaces, then they are actually linear functionals that act on these spaces. The Dirac delta distribution is an element of $\mathcal{D}'$, the dual space of all smooth functions. Hence, $\delta$ acts on a smooth function as $\langle\delta,f\rangle=f(0)$, for all $f\in C^{\infty}$ (notice that this action is linear).

So, when you see a differential equation like $u''-u=\delta$, then you must interpret it as saying $\langle u''-u,f\rangle=\langle \delta,f\rangle = f(0)$. Note here that $u$ may indeed be a function, and not a distribution, but you can always identify a function with its linear functional $T_u$, defined by $\langle T_u,f \rangle=\int_{\mathbb{R}^n}uf d\mathbf{x}$ (which is again a linear action).

Therefore you would say $u''-u=\delta$ if $$\langle u''-u,f\rangle=\langle u'',f\rangle-\langle u,f \rangle$$ $$=\langle T_{u''},f \rangle-\langle T_u,f \rangle$$ $$=\int_{\mathbb{R}}u''fdx-\int_{\mathbb{R}}ufdx$$ $$=f(0)$$ $$=\langle\delta,f\rangle,$$

for all $f\in C^{\infty}.$

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.