I have been studying the wave equation in $\mathbb{R}^n$ for the cases $n=1,2$ and $3$. In the three cases, working all over $\mathbb{R}^n$. That is:
$u_{tt}(x,t)=c^2 \Delta_{x} u(x,t)$ for $x \in \mathbb{R}^n$ and $t>0$
$u(x,0)=g(x)$ for $x \in \mathbb{R}^n$
$u_{t}(x,0)=h(x)$ for $x \in \mathbb{R}^n$
where $g$ and $h$ are given initial conditions.
The math part goes like this:
Study the case $n=1$ and derive d'Alembert's formula for the solution.
Applying the method of spherical means reduce the cases $n=2,3$ to the case $n=1$.
Solve the case $n=3$ and obtain Kirchhoff's formula for the solution.
Reduce the case $n=2$ to the case $n=3$ and obtain Poisson's formula for the solution.
For the cases $n=1$ and $n=2$ the solution $u$ at $(x,t)$ depends on the values of $g$ and $h$ over the entire, closed ball $B_{ct}(x)$. That is, the value of a wave that behaves according to the wave equation can be predicted at the point $x$ and at time $t$ knowing its values over the entire ball $B_{ct}(x)$ at time $0$.
However, for the case $n=3$ the solution $u$ at $(x,t)$ depends only on the values of $g$ and $h$ over the boundary of the ball $B_{ct}(x)$. That is, the value of a wave that behaves according to the wave equation can be predicted at the point $x$ end at time $t$ knowing only its values over $\partial B_{ct}(x)$ at time $0$.
I've read that the different behavior for the case $n=3$ has something to do with Huygens' principle. Anyway, I didn't get it.
My question is: is there a physical argument that allows one to understand, or even predict, why is it that the case $n=3$ depends only on the boundary of the ball around $x$ but the same doesn't hold for the cases $n=1,2$?
