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I'm struggling with yet another question for the classical heat equation in the whole space $R^d$. This question seems basic at first sight, but I think it is nontrivial in the end so here it is.

The setting:

Let $E\subset R^d$ be a closed set with positive finite Lebesgue measure $0<|E|<\infty$ (no smoothness assumptions at all so $E$ may be very ugly), and consider a nonnegative function $0\leq u_0\in L^1\cap L^p(R^d)$ for some $p>1$ with support $supp(u_0)=E$. Let $\Gamma_t(z)=\frac{1}{(4\pi t)^{d/2}}e^{-\frac{|z|^2}{4t}}$ be the usual heat kernel, and $u(t,x)=[\Gamma_t * u_0](x)$ the distributional solution of the heat equation $\partial_t u=\Delta u$ in $R^d$ with initial datum $u_0$. We know that $|u(t)|_{L^p(R^d)}\leq |u_0|_{L^p(R^d)}$ and $|u(t)|_{L^1}=|u_0|_{L^1}$ for all $t\geq 0$, and by the minimum principle there is instantaneous invasion $u(t,x)>0$ for any $t>0$ and all $x\in R^d$.

My question:

Is it true (and how to show) that the variation of the $L^p$ norm at time $t=0$ is concentrated on the initial support $E=supp(u_0)$? More precisely, do we have $$ \lim\limits_{t\downarrow 0}\frac{|u(t)|_{L^p(E^c)}}{t}=\lim\limits_{t\downarrow 0}\frac{1}{t}\left(\int\limits_{E^c} u(t)^p\right)^{1/p}\overset{?}{=}0\hspace{3cm} (1) $$ (I'm writing here the complement $E^c=R^d\setminus E$).

What I believe/tried:

I have the intuition that it should be true, since for any $x\notin E$ and $r>0$ we have $$ dist(x,E)\geq r>0\quad\Rightarrow\quad 0\leq u(t,x)\leq C\frac{e^{-r^2/4t}}{t^{d/2}}.\hspace{3cm} (2) $$ so that $u(t,x)$ is exponentially slowly growing in time outside of the initial support. With this estimate (2) it is easy to show rigorously that, as long as we step $r>0$ away from $E=supp(u_0)$ then the $L^p$ variation is zero. By that I mean that, if we replace $E^c$ by "the complement $(E+B_r)^c$ of an $r$-neighborhood of $E$" in $(1)$, then the result is true (here $r>0$ is fixed). Roughly, my question is equivalent to: What happens if $r=0$ (or $r\downarrow 0$?) Obviously one would expect $\lim\limits_{t\to 0}(\lim\limits_{r\to 0}...)\overset{?}{=}\lim\limits_{r\to 0}(\lim\limits_{t\to 0}...)=\lim\limits_{r\to 0} (0)=0$, but the (optimal) estimate (2) degenerates when $r\to 0$ so a priori it is not legitimate to swap the limits in the first equality.

Of course I'm aware that there may (or may not) be boundary effects localized on $\partial E$, which is another way to put my question (the limit $r\downarrow$ really means that I'm approaching the boundary $\partial E$). Let me point out that I only have $L^1\cap L^p$ regularity of the initial datum $u_0$. I'm typically interested in the case where $u_0$ "jumps" from $u_0\geq M>0$ in it's support $E$ to zero in $E^c=supp(u_0)^c$, so at time $t=0$ the "time derivative" $\partial_t u|_{t=0}=\Delta u_0\notin L^1_{loc}$ has a "bad" contribution on $\partial E$. This distribution $\Delta u_0$ typically involves the normal derivative of test functions integrated on the boundary, provided the normal derivative even makes sense (I'm not assuming that $E$ is smooth!)

Thank you in advance!

share|improve this question
    
I just realized that my question can be summarized as follows: For initial datum $0\leq u_0\in L^1\cap L^p(R^d)$ with initial support $supp(u_0)=E$ of finite measure, is it true that the restriction $t\mapsto u(t)|_{E^c}$ of $u(t,x)$ to the complement of the initial support $E^c$ is $\mathcal{C}^1$ from $[0,\infty)$ to $L^p(E^c)$ and that the time derivative at time $t=0$ is zero? –  leo monsaingeon Oct 17 '13 at 11:30
    
I am starting to believe that my statement $\lim\limits_{t\searrow 0}\frac{1}{t}|u(t)|_{L^p(E^c)}=0$ is too much to ask for, since this really means differentiability at time $t=0$ in $L^p(E^c)$ with zero derivative. Instead, can we get an upper bound on how fast $|u(t)|_{L^p(E^c)}$ grows for small $t>0$? Maybe some estimate $|u(t)|_{L^p(E^c)}\leq Ct^{\alpha}$ with $\alpha<1$? but this may be again too much to ask for. –  leo monsaingeon Oct 21 '13 at 13:36

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