2D Wave Equation

We can use the solution of the wave equation in three dimensions leads to a solution of the wave equation in two dimensions

We define the corresponding mean: $$\widetilde M t(F)(x) =\frac1{2\pi} \int{|y| \le 1} F(x-ty) (1-|y|^2)^{-1/2}, dy$$
This is not a spherical mean, but is related being a weighted ball average, since it has a weight function .

Let $f,g\in \mathcal{S}(R_2)$, then we would like to solve the problem:

$\mathrm{\Delta }u=\frac{{\partial }^{2}u}{\partial t^2}$ \text{subject to} $u(x,0)=f(x)\text{and}\frac{\partial u}{\partial t}(x,0)=g(x)$

with this in mind we could just extend to with a mute third variable. This is a good approach, the problem is that if it is constant with respect to the third variable, then those extensions aren't in

to solve this problem, let's fix , and consider the function , being , when . Now the functions: $$\tilde f^\flat(x_{1}, x_{2}, x_{3}) = f(x_{1}, x_{2}) \eta(x_{3}) \qquad\text{and}\qquad \tilde g^\flat(x_{1}, x_{2}, x_{3}) = g(x_{1}, x_{2}) \eta(x_{3})$$
We can see that if we have a function , then we can get the neat identity $$M_{t}(H)(x_{1}, x_{2}, 0) = \widetilde M_{t} (h)(x_{1}, x_{2})$$
We can solve the wave equation $$\Delta \tilde u^\flat = \frac{\partial^2 \tilde u^\flat}{\partial t^2} \quad \text{subject to} \quad \tilde u^\flat(x, 0) = \tilde f^\flat(x) \quad \text{and}\quad \frac{\partial u}{\partial t} (x, 0)=\tilde f^\flat(x)$$
Then what we can do know is define the function . We know how to calculate it is: $$\tilde u^\flat(x,t) = \frac{\partial}{\partial t}(t M_{t}(\tilde f^\flat)(x)) + t M_{t}(\tilde g^\flat)(x)$$
We can calculate how would , the solution to the 2D wave equation by applying our identity, getting that $$u(x,t) = \frac{\partial}{\partial t}(t \widetilde M_{t}(f)(x))+t\widetilde M_t{(g)(x)}$$

We see that that in the case of , then the solution at the point depends only on the data at the boundary of the base of the backward light cone will affect the solution. For the case where , we don't see this behaviour, we actually see that the solution at depends on the whole base of the backward light cone.