Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

What's a measure valued solution of a PDE?

For instance the Fokker-Planck equation \begin{align} \partial_t\mu_t+\sum_i\partial_i(b_i\mu_t)-\frac{1}{2}\sum_{ij}\partial_{ij}(a_{ij}\mu_t)=0 \end{align} it says that for a measure $\mu=\mu_(t,x)=\mu_t(x)$ being a solution of the above equation means \begin{align} \frac{d}{dt}\int_{\mathbb{R}^N}\phi(x)d\mu_t(x)=\int_{\mathbb{R}^N}\left(\sum_ib_i(t,x)\partial_i\phi(x)+\frac{1}{2}\sum_{ij}a_{ij}(t,x)\partial_{ij}\phi(x)\right)d\mu_t(x) \end{align} How do we get this? Also if compare this with the weak formulation, What's the connection between a measure valued solution and a distributional solution?

Many thanks!

share|cite|improve this question
What is $\phi(x)$? Is it a test-function or something else? – Pantelis Sopasakis Aug 28 '12 at 10:20
up vote 2 down vote accepted

Here's some motivation for the definition: If we have a (positive) classical solution f(t,x) define the measure $\mu(t,x)$ on $\mathbb{R}^N$ by $\mu(t,x)(A) = \int_{A} f(t,x)dx$, so that $d\mu_t = f(t,x)dx$. Multiplying by a test function $\phi(x),$ differentiating in $t$ and integrating by parts gives us exactly the above formula.

Measure-valued solutions are convenient to use in many situations, as far as I can tell because they simplify the existence theory for PDE. For another example, one can define measure-valued solutions to the Monge-Ampere equation $\det D^2u = \nu$ by requiring that $u$ is a convex function with $|Du(A)| = \nu(A)$, where $Du(A)$ is the collection of slopes of supporting hyperplanes to $u$ in the set $A$. In this setting, the motivation for the definition is the area formula, $|Du(A)| = \int_{A} \det D^2u.$

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.