# Property implies finite propagation speed

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 $$0\leq t, $$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 $$0\leq t, $$\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!

• Is $u(x,t)$ the solution of some time-homogenous PDE? That is, can you conclude from a bound at time $s$ another bound in a smaller ball in the time interval $[s,s+t_0]$? You stated such an assumption only for $s=0$. Jul 19, 2022 at 16:14
• I am sorry, but what do you exactly mean by time-homogenous? Jul 20, 2022 at 14:12
• In a previous comment, that has now unfortunately been deleted, you indicated that your equation has the form $u_t=\Delta_p(u)$, which implies it is time-homogenous. My solution was based on this comment; can you please resurrect it? Jul 20, 2022 at 15:56
• I added this assumption into my question. Jul 21, 2022 at 7:50
• Thank you! That now makes the whole discussion consistent. Jul 21, 2022 at 16:48

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 $$0\leq t, $$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 $$|z|, let $$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.

• Thank you very much for your answer! I am sorry, but what do you exactly mean by time-homogenous? I am not so sure if the equation I consider is in fact time-homogenous... Jul 20, 2022 at 13:38
• 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 $s \ge 0$. Jul 20, 2022 at 16:00
• Why are these equations then time homogeneous? We have $F(u,\nabla u, \nabla^2u)(\cdot, t+s)\neq F(u,\nabla u, \nabla^2u)(\cdot, t)$, or am I missing something? Jul 20, 2022 at 23:08
• @Shaq155 Yes, this equation is time homogeneous. I added an explanation in my answer. I hope it is clearer now. Jul 21, 2022 at 0:52
• Ah ok this is indeed by chain rule since $u_t(x,s+t)=u_{s+t}(x,s+t)$. Thank you! Jul 21, 2022 at 12:19