Solution. Try \(u = (Ar + B/r)\cos \theta \), harmonic since \(r^{\pm 1}\cos \theta \) are. Then \(A + B = 0\) and \(2A + B/2 = 1\), so \(A = \frac 23\), \(B = -\frac 23\), and \(u = \frac 23\bigl (r - \frac 1r\bigr )\cos \theta \).

Solution. The recurrence \((x^mJ_m)' = x^mJ_{m - 1}\) with \(m = 1\) gives \((xJ_1(x))' = xJ_0(x)\), so \(\int _0^1J_0(jr)\,r\,dr = j^{-2}\int _0^jxJ_0(x)\,dx = j^{-2}\cdot jJ_1(j)\). Dividing by the norm \(\frac 12J_1(j)^2\) gives \(c_n = 2/(j_{0n}J_1(j_{0n}))\).

Solution. \(\cos \theta = P_1(\cos \theta )\). Inside, \(u = rP_1 = r\cos \theta = z\), a uniform field. Outside, \(u = r^{-2}\cos \theta \), the field of a dipole. Both equal \(\cos \theta \) at \(r = 1\).

Solution. For radial \(u\), \(\Delta u = \frac 1r(ru)_{rr}\), so \(v = ru\) satisfies \(v_t = v_{rr}\) with \(v(0, t) = v(1, t) = 0\) and \(v(r, 0) = r\). The sine coefficients of \(r\) are \(2\int _0^1r\sin n\pi r\,dr = 2(-1)^{n + 1}/n\pi \), so \[ u = \sum _{n \geq 1}\frac {2(-1)^{n + 1}}{n\pi }e^{-n^2\pi ^2t}\,\frac {\sin n\pi r}r , \] and since \(\sin (n\pi r)/r \to n\pi \), the centre temperature is \(u(0, t) = 2\sum _{n \geq 1}(-1)^{n + 1}e^{-n^2\pi ^2t}\).

Solution. A blow at the centre is symmetric about it, so it excites only the radially symmetric modes \((0, n)\), whose frequency ratios \(1, 2.295, 3.598, \ldots \) are far from harmonic and give a dull thud. Striking off centre excites the modes with \(m \geq 1\) as well; the \((1, 1)\), \((2, 1)\), \((3, 1)\) modes have ratios close to \(2 : 3 : 4\) once the air loading of a kettledrum shifts them, a harmonic series on a missing fundamental, which gives the timpani its pitch.

Solution. \(t = T_0^2/\pi \kappa G^2\) is inversely proportional to \(\kappa \) and to \(G^2\): doubling \(\kappa \) halves the age to \(48\) million years, and halving \(G\) multiplies it by \(4\), to \(384\) million years. The estimate is very sensitive to the gradient, which is one reason for Kelvin’s wide range.