Heat kernel

Remarks

Check for $n = 1$: $\partial_t \Phi = \Phi \cdot \big( -\tfrac{1}{2t} + \tfrac{x^2}{4t^2} \big)$ and $\partial_x \Phi = -\tfrac{x}{2t} \Phi$, so $\partial_x^2 \Phi = \Phi \cdot \big( \tfrac{x^2}{4t^2} - \tfrac{1}{2t} \big)$, which is the same. For bounded continuous $g$, the function $u(x, t) = \int_{\R^n} \Phi(x - y, t) g(y) \, dy$ solves the heat equation for $t > 0$ and tends to $g(x)$ as $t \to 0^+$. For each $t$, $\Phi(\cdot, t)$ is the density of a normal distribution with mean $0$ and covariance $2t$ times the identity. Non-example: $p(x, t) = (2 \pi t)^{-n/2} e^{-\abs{x}^2/(2t)}$ is not the heat kernel of $\partial_t = \Delta$; it solves $\partial_t p = \tfrac12 \Delta p$ and is the transition density of Brownian motion. More generally, a domain with a boundary condition, or a Riemannian manifold, has a heat kernel $K(x, y, t)$. It is the function such that $u(x, t) = \int K(x, y, t) g(y) \, dy$ solves the heat equation with that boundary condition and initial value $g$.