# Heat Equation with positive initial condition with compact support

Consider $u(x,t)$, the solution to the initial value problem $$u_t = ∆u \ \ \ \ with \ x ∈ \mathbb{R^n},t> 0$$ $$u(x,0) = g(x)$$ Assume that $g ≥ 0$ is a continuous function with compact support. Let $D := diam(supp(g))$ and $d(x) := dist(x,supp(g))$. Show that for all $x ∈ \mathbb{R^n},t> 0$ $$u(x,t) > \frac{{||g||}_{1}}{(4πt)^{n/2}} \exp\biggl[-\frac{D^2+d(x)^2}{2t}\biggl]$$ My attempt at understanding the problem --> I know that the formula I need to prove is some sort of form of the fundamental solution to the heat equation where we consider the case of $g \ge0$ in a bounded domain (because $g$ has compact support). Then, we also expect $u \ge0$ for $t \ge0$. Additionally, I know I must use the "infinite speed of propagation" fact along with maximum/minimum principle for the heat equation. I am just not really sure how to combine these ideas to get started. Thank you for your help!

• You have the following with a Brownian motion : $u(x,t)=\mathbb{E}[g(x+W_t)]$. With an integral form : $u(x,t)=\int_{\mathbb{R}^n} g(x+y)\frac{1}{(2\pi t)^{n/2}}\exp(-\frac{y^T y}{2t})dy$ (I'm not sure about the constants 2 but it should approximately look like this) – Bertrand R Oct 19 '13 at 17:12
• Thanks for the help. From here, how do you think I should proceed then? (by the way, no worries about the constants!) – johnsteck Oct 19 '13 at 18:32
• Typographical suggestion: use \operatorname{dist} to get proper font and spacing: $\operatorname{dist}(x,\operatorname{supp} g)$. – Post No Bulls Dec 19 '13 at 5:28

You know that $$u(x,t)= \frac{1}{(4\pi t)^{n/2}} \int_{\mathbb R^n} e^{-(x-y)^2/(4t)} g(y)\,dy$$ How can you get a lower bound in terms of $\int g$? The standard way is to replace the other factor $e^{-(x-y)^2/(4t)}$ by its lower bound on the support of $g$, and pull out this bound out of the integral, as a constant factor. When $y\in\operatorname{supp} g$, we have $|x-y|\le D+d$, hence
$$e^{-(x-y)^2/(4t)} \ge e^{-(D+d)^2/(4t)}$$ The rest should not be hard.