MathOverflow is a question and answer site for professional mathematicians. 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

Consider the heat equation $u_t=\Delta u$ with Neumann boundary condition and initial condition $u(x,0)=u^0(x)$ in a bounded domain $\Omega$ with smooth boundary. Is this true:

Any solution $u(x,t)\in W^{2,p}$ of the equation can be written as $$u(x,t)=k(x,t)\star u^0(x)$$ where $k$ is a green function (depends on $\Omega$).

share|cite|improve this question
Since the heat equation with the given condition has a unique solution and the convolution form is one of the solutions, the answer is in affirmative. – Uday Feb 21 '12 at 8:58
Are you expecting a formula like $u(x,t) = \int_\Omega k(x-y,t)u_0(y) dy$ or something like $u(x,t) = \int_\Omega k(x,y,t)u_0(y)dy$ ? I don't think something like the first formula can be true. For small $t$, such a $k$ would have to like the fundamental solution of the heat equation. At the same time, $u(x,t) \approx u_0(x)$ for small $t$. Near the boundary of $\Omega$, this cannot be possible. – Hans Engler Mar 7 '12 at 1:55
This is possible inside the domain. For the Dirichlet problem that is indeed so (the question is copied from the formulation in the Krylov's book). The case of the Neumann condition must not be utterly different - one should study the existing literature to have a precise formulation. – Anatoly Kochubei Mar 19 '12 at 5:31

There exists a theory of Green functions for general parabolic boundary value problems which covers the case you are interested in, in particular papers by Eidelman, Ivasishen, Solonnikov. For references see

S. D. Eidelman and N. V. Zhitarashu, Parabolic boundary value problems. Basel: Birkhäuser (1998).

Unfortunately, most of the papers on this subject are available only in Russian.

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.