*This is crosspost from math.stackexchange http://math.stackexchange.com/questions/311213/heat-kernel-asymptotics-on-manifold-with-boundary where the question did not yield any answer*

On a closed Riemannian manifold $M$, the heat kernel $k_t(x, y)$ of the Laplace-Beltrami operator (or more general of any generalized symmetric Laplace-type operator acting on sections of a vector bundle) admits an asymptotic expansion of the form $$ k_t(x, y) \sim \exp\left( -\frac{1}{4t}d(x, y)^2\right) \sum_{j=0}^\infty t^j \Phi_j(x, y) $$ where $d(x, y)$ denotes the Riemannian distance and $\Phi_j$ are appropriate smooth functions, not depending on $t$. This is meant in the sense that for each $N \in \mathbb{N}$, there exists a constant $C>0$ such that for all $x, y \in M$, we have $$ \left| k_t(x, y) - \chi(x, y)\exp\left( -\frac{1}{4t}d(x, y)^2\right) \sum_{j=0}^Nt^j \Phi_j(x, y) \right| < C t^{N+1}$$ where $\chi(x, y)$ is an appropriate cutoff function that is $\equiv 1$ near the diagonal.

In the case that $M$ is still compact but has a boundary, in many books there can be found an asymptotic expansion of the trace, but I could not find an asymptotic expansion of the kernel itself, uniform on $M \times M$. Is there such an expansion?

**Remark 1:** When $M$ has a totally geodesic boundary, such an expansion is easy to get with help of the "Riemannian double". But of course, this case is quite unlikely (it is not even fulfilled for domains in Euclidean space).

**Remark 2:** I did not specify any boundary conditions, but one can assume that we are in the simplest case, i.e. the Laplace-Beltrami operator acting on functions with either Dirichlet or Neumann boundary conditions, whichever you like.