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  "body": "\n## Definition\n\nLet $U \\subseteq \\R^n$ be open, and let $a = (a_1, \\dots, a_n) \\colon U \\times \\R \\to \\R^n$ and $c \\colon U \\times \\R \\to \\R$ be $C^1$ functions.\nConsider the first-order [[quasilinear equation|7X9N9H3RK6]]\n$$\\sum_{i=1}^n a_i(x, u)\\, \\partial_i u = c(x, u)$$\nfor an unknown $C^1$ function $u$ on $U$, where $\\partial_i u$ is the [[partial derivative|09EGC6F8PT]] in $x_i$.\nIts *characteristic equations* are the [[ordinary differential equations|0NT64JV96A]]\n$$\\frac{dx}{ds} = a(x, z), \\qquad \\frac{dz}{ds} = c(x, z)$$\nfor a curve $s \\mapsto (x(s), z(s))$ in $U \\times \\R$; the curves $s \\mapsto x(s)$ are the *characteristics*.\nThe *method of characteristics* solves the equation with data $u = g$ on a hypersurface $\\Gamma \\subseteq U$ as follows.\n\n1. For each $y \\in \\Gamma$, solve the characteristic equations with [[initial values|0FXDW992DS]] $x(0) = y$ and $z(0) = g(y)$.\n2. Invert the map $(y, s) \\mapsto x(y, s)$ near $\\Gamma$, which is possible when $\\Gamma$ is non-characteristic, that is, $a(y, g(y))$ is not tangent to $\\Gamma$.\n3. Set $u(x(y, s)) = z(y, s)$.\n\n## Remarks\n\nThe 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 $z' = \\sum_i a_i\\, \\partial_i u = c(x, z)$.\nSo the graph of a solution is a union of solutions of the characteristic system.\nFor 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 $Du$.\n\nExample: for $u_t + u\\, u_x = 0$ in the variables $(x, t)$ with $u(x, 0) = g(x)$, the characteristic equations are $\\dot t = 1$, $\\dot x = z$, $\\dot z = 0$.\nSo $z$ is constant and the characteristics are the lines $x = y + g(y) t$, which gives the implicit solution $u = g(x - ut)$.\nCheck: 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 $u_t + u u_x = 0$.\n\nNon-example: data given on a characteristic curve.\nTake $\\partial_1 u = 0$ on $\\R^2$ with data $u(x_1, 0) = g(x_1)$ on $\\Gamma = \\set{x_2 = 0}$.\nHere $a = (1, 0)$ is tangent to $\\Gamma$, so step 2 fails.\nIndeed no solution exists unless $g$ is constant, and if $g$ is constant there are infinitely many, such as $u = g + h(x_2)$ with $h(0) = 0$.\n",
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