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we know $u$ (if it is a solution to the wave equation in $\mathbb{R}^3$) decays as $1/t$ as $t$ goes to $\infty$. this comes easily from spherical means. but how do we know this is the maximum possible rate of decay?

flag	asked May 9 2010 at 19:15 chris 35●2

Harald Hanche-Olsen · Accepted Answer · 2010-05-09 21:14:59Z UTC

0

Let $\bar u(t,r)$ be the spherical mean over the sphere of radius $r$, centered at (say) the origin. Then $r\bar u(t,r)$ satisfies the one-dimensional wave equation. If $u$ has compact support initially, then eventually $r\bar u(t,r)$ is an outward traveling wave. This wave has constant shape, and so $\bar u(t,r)$ decays as $1/r$.

answered May 9 2010 at 21:14

Harald Hanche-Olsen
5,750●1●13●34

	why does it have a constant shape? – chris May 9 2010 at 23:19
	That's the “wave” in “wave equation”. The general solution of $v_{tt}-v_{rr}=0$ is $f(r-t)+g(r+t)$ for functions $f$, $g$. After the $g$ contribution has left the region $r>0$ , only $f(r-t)$ remains. I.e., constant shape. – Harald Hanche-Olsen May 10 2010 at 0:12
	oh i wasn't thinking at all =) thanks, that cleared it up a lot – chris May 10 2010 at 0:22
	i am being a bit dense here, why does decay @ $1/r$ imply decay @ $1/t$? – chris May 10 2010 at 3:48
	The wave travels outward at speed 1. Hence, for large $t$, $r$ and $t$ are asymptotically equal where $u\ne0$. – Harald Hanche-Olsen May 10 2010 at 3:58

Willie Wong · Answer 2 · 2010-05-10 10:50:45Z UTC

Harald has already shown that the $1/t$ decay is the best "general decay rate" for, say, smooth data with compact support. I claim that a slightly stronger result can hold: if you have smooth data with compact support, and the evolution is such that there exists a constant $C$ such that your solution decays faster than $C/t^{1+\epsilon}$, your initial data must be trivial.

Harald's answer already tells you that, after fixing an origin, the spherical mean of the solution cannot decay faster than $1/t$, unless it vanishes. So if the solution has our supposed decay, the spherical mean around any point is 0. This is equivalent to saying that the spherical Radon transform of your initial data is 0. So if your data is in a sufficiently nice class (smooth with compact support will do), you can invert the spherical Radon transform and get that your initial data must be 0.

Another way to see this is to take the inverse scattering point of view. Look at the fundamental solution of the wave equation. If you multiply it by $t$ and formally take the limit as $t=r\to \infty$, and consider the function $f(x,e) = \lim t u(t,x + re)$ (where $e$ is a unit vector), you see that $f$ is the Radon transform of the data. For data with compact support, the $\limsup$ and $\liminf$ are well-defined (by the general decary rate of $1/t$), and they both go to 0 if the $u$ decays any faster. So taking the inverse transform you have that the data must be 0 also.

maximum decay rate

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