Proof. From each point \(\tau \) of \(\Gamma \), solve the characteristic ODEs \(x' = a\), \(y' = b\) with \((x, y)(0) = (x_0(\tau ), y_0(\tau ))\), and the ODE \(u' = f - cu\) along it with \(u(0) = u_0(\tau )\). By the existence, uniqueness and differentiable dependence theorems for ODEs, this defines a \(C^1\) map \((\tau , s) \mapsto (x, y)\) and a function \(u(\tau , s)\). At \(s = 0\) the Jacobian of the map has columns \((x_0', y_0')\) and \((a, b)\), so its determinant is non-zero by the non-characteristic condition, and by the inverse function theorem the map is a \(C^1\) diffeomorphism near the point. Then \(u(x, y) = u(\tau (x, y), s(x, y))\) is a \(C^1\) function satisfying the equation (by Theorem 1.3.2, read backwards) and the data. Any solution satisfies the same ODEs along the characteristics, so it coincides with this one.

If \(\Gamma \) is a characteristic, the data \(u_0\) must themselves satisfy the ODE \(u' = f - cu\) along \(\Gamma \); if they do not, there is no solution, and if they do, the values off \(\Gamma \) are not determined, since nothing links \(\Gamma \) to the neighbouring characteristics, and any smooth choice of data on a transverse curve gives a solution. □