Solution. By Theorem 1.2.1, \(u = \sin (x - 3t)\) for the first. For the second, \(u = \sin (x - 3t) + \int _0^ts\,ds = \sin (x - 3t) + \frac {t^2}{2}\): the source raises \(u\) along each characteristic.

Solution. The characteristics solve \(x' = y\), \(y' = -x\): circles \(x^2 + y^2 = \text {const}\). So \(u = F(x^2 + y^2)\), radially symmetric. With \(u(x, 0) = F(x^2) = x^2\), \(F(r) = r\) and \(u = x^2 + y^2\). The data line \(y = 0\), \(x > 0\) crosses each circle exactly once, as Theorem 1.4.1 requires; data on the whole line \(y = 0\) would have to be even in \(x\), or no solution would exist.

Solution. The characteristics are \(x = x_0 + s\), \(y = 2s\), starting from \((x_0, 0)\), along which \(du/ds = -u\), so \(u = f(x_0)e^{-s}\). Inverting, \(s = y/2\) and \(x_0 = x - y/2\), so \(u(x, y) = f(x - y/2)e^{-y/2}\).

Solution. The line \(y = x\) is a characteristic (direction \((1, 1)\)). Along it \(u\) must be constant. Data \(u = x\) are not constant, so there is no solution. Data \(u = 1\) are constant, and \(u = 1 + G(x - y)\) is a solution for every \(C^1\) function \(G\) with \(G(0) = 0\): infinitely many.

Solution. Expanded, \(u_t + xu_x = -u\). Along \(x = x_0e^t\), \(du/dt = -u\), so \(u = f(x_0)e^{-t} = f(xe^{-t})e^{-t}\). Then \(\int u\,dx = e^{-t}\int f(xe^{-t})\,dx = e^{-t}e^t\int f(y)\,dy = \int f\), constant: the conservative form preserves the total amount, spreading it out while lowering its peak. For \(u_t + xu_x = 0\), \(u = f(xe^{-t})\) and \(\int u = e^t\int f\), which grows.

Solution. \(M(a) = \mu a\), so the stable distribution is \(Be^{-\mu a}\), with total \(\int _0^\infty Be^{-\mu a}\,da = B/\mu \). The mean age is \(\int _0^\infty aBe^{-\mu a}\,da/(B/\mu ) = (B/\mu ^2)/(B/\mu ) = 1/\mu \), which is also the life expectancy at birth: with a death rate of \(2\%\) a year, the mean age is \(50\) years.