Proof. Write \(\sigma (x_0) = a(u_0(x_0))\), the speed of the characteristic from \(x_0\). The map \(x_0 \mapsto X(x_0, t) = x_0 + \sigma (x_0)t\) has derivative \(1 + \sigma '(x_0)t\), which is positive for all \(x_0\) exactly when \(t < t_b\); it is then strictly increasing and, since \(\sigma \) is bounded, onto \(\mathbb R\). So each \((x, t)\) lies on exactly one characteristic, with foot \(x_0(x, t)\), and \(u(x, t) = u_0(x_0(x, t))\) is \(C^1\) by the inverse function theorem. Differentiating, \[ u_x = \frac {u_0'(x_0)}{1 + \sigma '(x_0)t} , \] and along the characteristic where \(\sigma '\) attains (or approaches) its infimum, the denominator tends to \(0\) as \(t \to t_b\) while the numerator does not (since \(\sigma ' = a'(u_0)u_0'\)), so \(u_x \to \infty \). At \(t_b\) two characteristics carrying different values meet, and \(u\) would have to take both values at one point. □