Heat potential

Remarks

Only the part of $\mu$ with $s < t$ contributes to $\Phi\mu(x, t)$, since $\Phi$ vanishes for nonpositive times. Example: for the unit point mass $\mu = \delta_{(0, 0)}$, the heat potential is $\Phi$ itself, the heat kernel started at time $0$; it solves the heat equation away from $(0, 0)$. Degenerate case: for Lebesgue measure on $\R^n \times \R$, $\Phi\mu(x, t) = \int_{-\infty}^{t} \int_{\R^n} \Phi(x - y, t - s) \, dy \, ds = \int_{-\infty}^t 1 \, ds = \infty$ at every point, so finiteness of a heat potential needs growth conditions on $\mu$. For a bounded function $f$ on $\R^n \times (0, T)$ that is locally Hölder continuous, the function $u(x, t) = \int_0^t \int_{\R^n} \Phi(x - y, t - s) f(y, s) \, dy \, ds$ solves $\partial_t u - \Delta u = f$ with $u(\cdot, 0) = 0$. A measure carried by the lateral boundary of a space-time cylinder gives a single layer heat potential. Using a normal derivative of $\Phi$ instead gives a double layer heat potential. Both are used to solve boundary value problems for the heat equation.