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?
