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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 '12 at 8:58
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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.

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Probably yes, if I interpret your question correctly. The answer is based on the theory of operator semigroups.

First of all, let us consider the case $p=2$: Then the solution is given by a semigroup on $L^2(\Omega)$ generated by an operator associated with a quadratic form: it is known that the semigroup is ultracontractive (and in particular has a kernel $k$ that is of class $L^\infty(\Omega\times \Omega)$ with respect to space) if and only if the form domain $H^1(\Omega)$ is continuously embedded in some $L^p(\Omega)$ space for $p$ large enough, which luckily does hold by virtue of the Sobolev embedding theorem.

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