A two-dimensional pulse has a tail inside the cone

The value at the centre: setting $x=x_0$ in [F1] gives the displayed formula $u(x_0,t)=\frac{1}{2\pi c}{\int }_{B_{ct}(x_0)}(c^2{t}^2-\mathrm{\mid }y-x_0{\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}{u}_1(y)dy$, the integrand being defined and continuous on the open disk because $\mathrm{\mid }y-x_0\mathrm{\mid }<ct$ there.

Positivity: if $t>r\mathrm{/}c$ then ${\bar{B}}_r(x_0)\subseteq B_{ct}(x_0)$, and on that closed support ball $\mathrm{\mid }y-x_0\mathrm{\mid }\le r<ct$ gives $\sqrt{c^2{t}^2-\mathrm{\mid }y-x_0{\mathrm{\mid }}^2}\le ct$, hence the weight $(c^2{t}^2-\mathrm{\mid }y-x_0{\mathrm{\mid }}^2{)}^{-1\mathrm{/}2}\ge 1\mathrm{/}(ct)>0$. Since $u_1\ge 0$ is continuous and nonzero, it is positive on a nonempty open subset of $B_r(x_0)$, so [F3] gives ${\int }_{B_{ct}(x_0)}{u}_1(y)dy>0$ and the displayed integral is at least $(2\pi c{)}^{-1}(ct{)}^{-1}{\int }_{B_{ct}(x_0)}{u}_1(y)dy>0$; this shows that the centre still sees a positive displacement at every time after the front $\mathrm{\mid }x-x_0\mathrm{\mid }=ct$ has passed beyond the support, that is, for $t>r\mathrm{/}c$.

The tail is carried by the interior: for $t>r\mathrm{/}c$ the data are supported strictly inside $B_{ct}(x_0)$ and vanish near its boundary, yet step 2.1 gives $u(x_0,t)>0$. This is the interior tail and illustrates the failure of sphere-only dependence in [F2].