aggierabz2006zw

Answered

2022-07-07

Uniform continuity of heat equation with ${L}^{1}$ data
I came across the following remark in my reading.
Remark. If the initial data is in ${L}^{1}$, then heat equation solutions approach 0 within the sup-norm. That is, if K(x,t) is the heat kernel and $f\in {L}^{1}\left({\mathbb{R}}^{n}\right)$, then
$\underset{t\to \mathrm{\infty }}{lim}{\int }_{{\mathbb{R}}^{n}}K\left(x-y,t\right)f\left(y\right)dy=0,$
uniformly.
I can picture this working if f is uniformly continuous in ${\mathbb{R}}^{n}$ and $\underset{x\to \mathrm{\infty }}{lim}f\left(x\right)=0$. Does $f\in {L}^{1}\left({\mathbb{R}}^{n}\right)$ imply this somehow, and I'm not seeing this? I'm relatively new to ${L}^{p}$-spaces.
If so, then for $ϵ>0$, we can find some radius ${r}_{ϵ}>0$ such that |$|f\left(y\right)|<ϵ/2$.
Then we can bound the heat kernel, and use the fact that, $\int K\left(x-y,t\right)dy=1$ to conclude, more or less.

Answer & Explanation

Alisa Jacobs

Expert

2022-07-08Added 13 answers

This is quite straightforward from the definition. In n dimensions, the heat kernel is given by
$K\left(x-y,t\right)=\frac{1}{\left(4\pi t{\right)}^{n/2}}{e}^{-|x-y{|}^{2}/4t}.$
As ${e}^{-x}\le 1$ for $x\ge 0$ this implies that
$|K\left(x-y,t\right)|\le \left(4\pi t{\right)}^{-n/2}$ and therefore
$|{\int }_{{\mathbb{R}}^{n}}K\left(x-y,t\right)f\left(y\right)dy|\le \left(4\pi t{\right)}^{-n/2}{\int }_{{\mathbb{R}}^{n}}|f\left(y\right)|dy,$
which converges to 0 as $t\to \mathrm{\infty }$ uniformly in x, as soon as f is in ${L}^{1}\left({\mathbb{R}}^{n}\right)$.

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