Let $u(x, t)$ be a (non-negative, bounded) function on $\mathbb{R}^{n}\times [0, +\infty)$ and suppose that $u$ satisfies some time-independent PDE, e.g. $\partial\_{t}u=\Delta\_{p}u$. Let us assume that $u$ has the following property: if there exists a ball $B(x\_{0}, r\_{0})$ in $\mathbb{R}^{n}$ such that $$B(x\_{0}, r\_{0})\cap \text{supp}~u(\cdot, 0)=\emptyset,$$ then there is $t\_{0}=t\_{0}(r\_{0})$ so that for all $0\leq t<t\_{0}$, $$B(x\_{0}, r\_{0}/2)\cap \text{supp}~u(\cdot, t)=\emptyset.$$


Does this property imply that, if we suppose that $\text{supp}~u(\cdot, 0)$ is compact, there exists $T=T(r)>0$ and an increasing, non-negative function $$r:[0, T)\to [0, \infty)$$ so that for any $0\leq t<T$, $$\text{supp}~u(\cdot, t)\subset U\_{r(t)},$$ where $U\_{r}=\{x\in \mathbb{R}^{n}:\text{dist}(x, \text{supp}~u(\cdot, 0))\leq r\}$?
In other words, the latter property says that $u$ has finite propagation speed.


Does anyone have a reference for this?


Any help is appreciated!



Property implies finite propagation speed
As noted by the OP in a comment, $u(x, t)$ is assumed to be a (non-negative, bounded) function on $\mathbb{R}^{n}\times [0, +\infty)$ which solves a time-homogenous PDE.
``Time homogenous'' means that if $u(x,t)$ is a solution, then $\tilde{u}(x,t):=u(x,s+t)$ is also a solution for every $s \ge 0$. In particular, this holds for equations of the form $u\_t=F(u,\nabla u, \nabla^2u)$, and the equation is of this form.


Indeed, suppose $u(x,t)$ solves $$u\_t(x,t)=F\bigl(u(x,t),\nabla u(x,t), \nabla^2 u (x,t)\bigr) \,.$$
Then
$$\tilde{u}\_t(x,t)=u\_t(x,s+t)=F\bigl(u(x,t),\nabla u(x,s+t), \nabla^2 u (x,s+t)\bigr)$$ $$=F\bigl(\tilde{u}(x,t),\nabla\tilde{u}(x,t), \nabla^2 \tilde{u}(x,t)\bigr) \,.$$


Thus the hypothesis in the problem can be rewritten in the following form:


For every $s\ge 0$, if there exists a ball $B(x\_{0}, r\_{0})$ in $\mathbb{R}^{n}$ such that $$B(x\_{0}, r\_{0})\cap \text{supp}~u(\cdot, s)=\emptyset,$$ then there is $t\_{0}=t\_{0}(r\_{0})$ so that for all $0\leq t<t\_{0}$, $$B(x\_{0}, r\_{0}/2)\cap \text{supp}~u(\cdot, s+ t)=\emptyset.$$


Since supp $u(\cdot,0)$ is compact, it is contained in a closed ball $\bar{B}(0,R)$.


**Claim:** If for some $s\ge 0$ and $R\_1>0$, we have supp $u(\cdot,s) \subset \bar{B}(0,R\_1)$, then for all $t\in [0,t\_0/2]$ we have supp $u(\cdot,s+t) \subset \bar{B}(0,R\_1+r\_0/2)$.


Proof: It suffices to verify that every $z \notin \bar{B}(0,R\_1+r\_0/2)$ satisfies $z \notin \: \text{supp} \: u(\cdot,s+t)$ for every $t\in [0,t\_0/2]$.


If $|z| \ge R\_1+r\_0$, , let $z\_1=z$.


If $|z|<R\_1+r\_0$, let $z\_1:= \displaystyle(R\_1+r\_0)\frac{z}{|z|},$
and observe that in this case,
$$ z\_1-z =(R\_1+r\_0-|z|)\frac{z}{|z|} \quad \text{so} \quad |z\_1-z|=R\_1+r\_0-|z| \in (0, r\_0/2)\,.$$
In both cases, we have
$$B(z\_1, r\_{0})\cap \text{supp}~u(\cdot, s)=\emptyset\, ,$$
so for all $t\in [0,t\_0/2]$,
$$B(z\_1, r\_{0}/2)\cap \text{supp}~u(\cdot, s+t)=\emptyset\, .$$
In particular, $ z \notin \: \text{supp} \: u(\cdot,s+t)$.


$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$


The claim implies, by induction on $k$, that
$$\forall k \ge 1, \quad \forall t\in [(k-1)t\_0/2,\, kt\_0/2), 
\quad \text{supp} \:u(\cdot,t) \subset \bar{B}(0, R+kr\_0/2) \tag{\*} \,. $$


as required.


