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  "body": "\n## Definition\n\nLet $\\Omega \\subseteq \\R^3$ be an open set, let $T \u003e 0$, let $\\nu \u003e 0$ be a constant, and let $f \\colon \\Omega \\times (0, T) \\to \\R^3$ be a given force.\nThe *incompressible Navier–Stokes equations* are the system\n$$\\partial_t u + (u \\cdot \\nabla) u = \\nu \\Delta u - \\nabla p + f, \\qquad \\nabla \\cdot u = 0,$$\nfor an unknown [[vector field|0299VCW1TH]] $u \\colon \\Omega \\times (0, T) \\to \\R^3$, the velocity, and an unknown function $p \\colon \\Omega \\times (0, T) \\to \\R$, the pressure.\nHere $\\nabla p$ is the [[gradient|0ABK9V9Y64]] of $p$ in $x$, and $\\nabla \\cdot u$ is the [[divergence|0MWV6ZTK6S]] of $u$.\nThe [[Laplacian|0E0B6RA8K0]] $\\Delta$ acts on each component, and $(u \\cdot \\nabla) u$ has components $\\sum_{j=1}^3 u_j \\partial_j u_i$.\nThe constant $\\nu$ is the kinematic viscosity, and the density is normalised to $1$.\n\n## Remarks\n\nExample: on $\\Omega = \\R^3$ with $f = 0$, the shear flow $u(x, y, z, t) = (y, 0, 0)$ with constant pressure is a solution.\nIndeed $\\partial_t u = 0$, $(u \\cdot \\nabla) u = y \\, \\partial_x u = 0$, $\\Delta u = 0$, $\\nabla p = 0$ and $\\nabla \\cdot u = 0$.\n\nNon-example: $u = (x, 0, 0)$ with any pressure is not a solution, because $\\nabla \\cdot u = 1 \\ne 0$.\n\nThe same equations are studied in two dimensions and on the torus.\nFor a compressible fluid the density $\\rho$ becomes an unknown, with the continuity equation $\\partial_t \\rho + \\nabla \\cdot (\\rho u) = 0$ in place of $\\nabla \\cdot u = 0$.\nSetting $\\nu = 0$ gives the Euler equations of an ideal fluid.\n",
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