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  "body": "\n## Definition\n\nLet $n \\ge 1$ and let $b \\in \\R^n$ be a fixed vector.\nThe *transport equation* with velocity $b$ is the [[partial differential equation|GC5BDZ6QBC]]\n$$\\partial_t u + b \\cdot \\nabla u = 0$$\nfor $u \\colon \\R^n \\times (0, \\infty) \\to \\R$.\nHere $\\nabla u$ is the [[gradient|0ABK9V9Y64]] of $u$ in the space variable $x$.\nThe *initial value problem* adds the condition $u(x, 0) = g(x)$ for a given function $g \\colon \\R^n \\to \\R$.\n\n## Remarks\n\nFor $g \\in C^1(\\R^n)$ the initial value problem has the unique $C^1$ solution $u(x, t) = g(x - tb)$.\nThe reason is that $u$ is constant along each line $s \\mapsto (x + sb, t + s)$, since the derivative of $u$ along it is $\\partial_t u + b \\cdot \\nabla u = 0$.\nThe inhomogeneous equation $\\partial_t u + b \\cdot \\nabla u = f$ has the solution $u(x, t) = g(x - tb) + \\int_0^t f(x + (s - t) b, s)\\, ds$.\nWith a variable velocity field $b(x, t)$ one also calls $\\partial_t u + b \\cdot \\nabla u = 0$ a transport equation, and $\\partial_t u + \\nabla \\cdot (u\\, b) = 0$ the continuity equation; the two agree when $\\nabla \\cdot b = 0$.\n\nExample: $n = 1$, $b = 2$, $g(x) = e^{-x^2}$, and $u(x, t) = e^{-(x - 2t)^2}$.\nCheck: $\\partial_t u = 4(x - 2t) u$ and $\\partial_x u = -2(x - 2t) u$, so $\\partial_t u + 2 \\partial_x u = 0$, and $u(x, 0) = g(x)$.\n\nNon-example: $u(x, t) = e^{-t} \\sin x$ with $n = 1$ is not a solution for any $b$.\nIndeed $\\partial_t u + b\\, \\partial_x u = e^{-t}(-\\sin x + b \\cos x)$, which is not identically zero.\n",
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