Method of characteristics

Remarks

The key fact is that if $u$ is a $C^1$ solution and $x(s)$ solves $x' = a(x, u(x))$, then $z(s) = u(x(s))$ satisfies $z' = \sum_i a_i\, \partial_i u = c(x, z)$. So the graph of a solution is a union of solutions of the characteristic system. For a fully nonlinear equation $F(Du, u, x) = 0$ the method uses the larger system $\dot x = D_p F$, $\dot z = D_p F \cdot p$, $\dot p = -D_x F - D_z F\, p$, where $p$ stands for $Du$.

Example: for $u_t + u\, u_x = 0$ in the variables $(x, t)$ with $u(x, 0) = g(x)$, the characteristic equations are $\dot t = 1$, $\dot x = z$, $\dot z = 0$. So $z$ is constant and the characteristics are the lines $x = y + g(y) t$, which gives the implicit solution $u = g(x - ut)$. Check: differentiating $u = g(x - ut)$ gives $u_t = -g'\,(u + t u_t)$ and $u_x = g'\,(1 - t u_x)$, so $u_t = -u g'/(1 + t g')$ and $u_x = g'/(1 + t g')$, and $u_t + u u_x = 0$.

Non-example: data given on a characteristic curve. Take $\partial_1 u = 0$ on $\R^2$ with data $u(x_1, 0) = g(x_1)$ on $\Gamma = \set{x_2 = 0}$. Here $a = (1, 0)$ is tangent to $\Gamma$, so step 2 fails. Indeed no solution exists unless $g$ is constant, and if $g$ is constant there are infinitely many, such as $u = g + h(x_2)$ with $h(0) = 0$.