Solution. The solution is radial, and \(V = ru\) solves the one-dimensional equation with \(V(r, 0) = 0\) and \(V_t(r, 0) = re^{-r^2}\), which is odd. By d’Alembert, \[ V = \frac 1{2c}\int _{r - ct}^{r + ct}se^{-s^2}\,ds = \frac 1{4c}\Bigl (e^{-(r - ct)^2} - e^{-(r + ct)^2}\Bigr ) , \] so \(u = V/r\). Letting \(r \to 0\), \(u(0, t) = \frac 1{4c}\cdot 4ct\,e^{-c^2t^2} = te^{-c^2t^2}\), which agrees with Kirchhoff: \(tM_g(0, ct) = te^{-c^2t^2}\).

Solution. Put the centre of the ball at the origin and the point at \(x = (R, 0, 0)\). A point of the sphere is \(y = x + \rho \omega \) with \(|y|^2 = R^2 + \rho ^2 + 2R\rho \omega _1\), and it lies in the ball when \(\omega _1 < (a^2 - R^2 - \rho ^2)/2R\rho \). Since \(\omega _1\) is uniformly distributed on \([-1, 1]\) for \(\omega \) uniform on the sphere (Archimedes’ hat-box theorem), the fraction is \(\frac 12\bigl (1 + (a^2 - R^2 - \rho ^2)/2R\rho \bigr ) = \bigl (a^2 - (R - \rho )^2\bigr )/4R\rho \).

Solution. \(t - \sqrt {t^2 - a^2} = a^2/(t + \sqrt {t^2 - a^2}) \sim a^2/2t\). For large \(t\) the whole disc is near the centre of the circle of radius \(t\), where the weight \(1/(2\pi \sqrt {t^2 - \rho ^2}) \approx 1/2\pi t\), so \(u \approx \pi a^2/2\pi t\): the area struck times \(1/2\pi t\).

Solution. \(u = \frac 1{4\pi c^2}\int _{|y - x| \leq ct}\frac {dy}{|x - y|} = \frac 1{4\pi c^2}\int _0^{ct}\frac {4\pi r^2}{r}\,dr = \frac {1}{c^2}\cdot \frac {c^2t^2}{2} = \frac {t^2}{2}\), which plainly solves \(u_{tt} = 1\), \(\Delta u = 0\).

Solution. With \(h = \sqrt {c^2t^2 - (y_1 - x_1)^2}\), the disc \(|y - x| < ct\) is \(|y_1 - x_1| < ct\), \(|y_2 - x_2| < h\), and \[ u = \frac 1{2\pi c}\int _{x_1 - ct}^{x_1 + ct}g(y_1)\int _{-h}^{h}\frac {dy_2}{\sqrt {h^2 - y_2^2}}\,dy_1 = \frac 1{2\pi c}\int _{x_1 - ct}^{x_1 + ct}\pi g(y_1)\,dy_1 = \frac 1{2c}\int _{x_1 - ct}^{x_1 + ct}g . \]

Solution. The source is not a point: the stroke is a long line of sources, all at different distances. Each point of the channel is heard sharply, at time distance\(/c\). The nearest point is \(2\) km away and the farthest \(\sqrt {2^2 + 3^2} = 3.61\) km, so the thunder lasts from \(2000/343 = 5.8\) s to \(3606/343 = 10.5\) s, about \(4.7\) s. (Echoes and the branching of the channel lengthen it further.)