Property implies finite propagation speed

Question

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!

Answer 1

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.