Solution. \(\min u_0' = \min \cos x = -1\), at \(x = \pi \), so \(t_b = 1\). The characteristic from \(x_0 = \pi \) has speed \(\sin \pi = 0\), so the shock forms at \(x = \pi \), \(t = 1\), and stays there by symmetry: \(u\) is odd about \(\pi \), so \(u_l = -u_r\) and \(s = 0\).

Solution. Characteristics from \(x_0 \in [0, 1]\) have speed \(1 - x_0\) and are \(x = x_0 + (1 - x_0)t\); all of them meet at \(x = 1\), \(t = 1\), so \(t_b = 1\). For \(t < 1\), in the ramp, \(x = x_0 + (1 - x_0)t\) gives \(x_0 = (x - t)/(1 - t)\) and \(u = 1 - x_0 = (1 - x)/(1 - t)\) for \(t \leq x \leq 1\); \(u = 1\) for \(x < t\); \(u = 0\) for \(x > 1\). At \(t = 1\) a shock forms at \(x = 1\) joining \(u_l = 1\) to \(u_r = 0\); it moves at speed \(\frac 12\), so \(x = 1 + \frac 12(t - 1)\) for \(t \geq 1\).

Solution. \(u_t + uu_x = -\frac {x}{t^2} + \frac xt\cdot \frac 1t = 0\). For the discontinuous solution, \(a(u_l) = u_l < \frac 12(u_l + u_r) < u_r = a(u_r)\), so \(a(u_l) < s\) and \(s < a(u_r)\), the reverse of the Lax condition: characteristics leave the shock on both sides.

Solution. \(q(\rho _{\max }) = 0\), so \[ s = -\frac {q(\rho _l)}{\rho _{\max } - \rho _l} = -\frac {v_{\max }\rho _l(1 - \rho _l/\rho _{\max })}{\rho _{\max }(1 - \rho _l/\rho _{\max })} = -v_{\max }\frac {\rho _l}{\rho _{\max }} . \] Here \(s = -33.3\) km/h, so after one minute the back of the queue is \(33.3/60 = 0.556\) km behind the light, and the queue holds \(0.556 \times 150 = 83.3\) cars. As a check by conservation: \(0.556 \times 50 = 27.8\) cars were already on that stretch when the light turned red, and \(q(\rho _l)/60 = 3333/60 = 55.6\) more arrived during the minute, \(83.3\) in all.

Solution. \(s = (0 - 1)/(0 - 1) = 1\). With \(a(u) = 3u^2\), \(a(u_l) = 3 > 1 > 0 = a(u_r)\): the Lax condition holds, and the shock is admissible.