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## Solution of Heat equation with Neumann BC in an arbitrary domain

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 $\Omega$).

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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 2012 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 2012 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 2012 at 5:31