Recall the Kirchhoff's
$$ u(x, t)=\mathrel{\int\!\!\!\!\!\!-}_{\partial B(x, t)} t h(y)+D g(y) \cdot(y-x)+g(y) \mathrm{d} \sigma_{y}, \quad x \in \mathbb{R}^{3}, t>0 $$ and Poisson's fomulas $$ u(x, t)=\frac{1}{2} \mathrel{\int\!\!\!\!\!\!-}_{B(x, t)} \frac{\operatorname{tg}(y)+t^{2} h(y)+t D g(y) \cdot(y-x)}{\left(t^{2}-|x-y|^{2}\right)^{1 / 2}} \mathrm{~d} y, \quad t>0, x \in \mathbb{R}^{2} . $$
What can you say about the behavior of $u(x, t)$ when $t \rightarrow \infty$, if the initial data $g$ and $h$ are compactly supported?
My attempt:
For Poisson's:
Let's assume that the initial data has compact support $\Omega$, where $\Omega$ is connected and regular enough. $\forall x_{0} \notin \Omega$ and $d_{1}=\operatorname{dist}\left(x_{0}, \Omega\right)>0, d_{2}=\max \left\{\operatorname{dist}\left(x_{0}, x\right): x \in \Omega\right\}$. In 2-D, the possible nonzero point of $u\left(x_{0}, t\right)$ must be the half line $\left[d_{1},+\infty\right)$.
For Kirchhoff's:
$$ u(x, t)=\frac{1}{2}\left[\partial_{t}\left(\frac{t^{2}}{\left|B_{t}(x)\right|} \int_{B_{t}(x)} \frac{g(y)}{\left(t^{2}-|y-x|^{2}\right)^{\frac{1}{2}}} d y\right)+\left(\frac{t^{2}}{\left|B_{t}(x)\right|} \int_{B_{t}(x)} \frac{h(y)}{\left(t^{2}-|y-x|^{2}\right)^{\frac{1}{2}}} d y\right)\right] $$
The fraction $t^{2} /\left|B_{t}(x)\right|$ simplifies to $1 / \pi$, and it should become clear that the second integral is $O(1 / t)$ while the first is even $O\left(1 / t^{2}\right)$, as the differentiation brings the power down.