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  "body": "\n## Definition\n\nLet $U \\subseteq \\R^m$ be a nonempty set, let $T \u003e 0$, and let $f \\colon \\R^n \\times U \\to \\R^n$, $L \\colon \\R^n \\times U \\to \\R$ and $g \\colon \\R^n \\to \\R$ be continuous.\nConsider the [[control system|ZPDTVZQN5K]] $x' = f(x, u)$ with running cost $L$ and terminal cost $g$.\nThe *Hamilton–Jacobi–Bellman equation* of this problem is the partial differential equation\n$$\\frac{\\partial V}{\\partial t}(t, x) + \\min_{u \\in U} \\left\\{ \\nabla_x V(t, x) \\cdot f(x, u) + L(x, u) \\right\\} = 0, \\qquad (t, x) \\in (0, T) \\times \\R^n,$$\nwith terminal condition $V(T, x) = g(x)$, for an unknown function $V \\colon [0, T] \\times \\R^n \\to \\R$.\nWhen the minimum is not attained, it is replaced by the infimum.\n\n## Remarks\n\nIts intended solution is the value function $V(t, x) = \\inf_u \\left( \\int_t^T L(x(s), u(s)) \\, ds + g(x(T)) \\right)$, where the infimum is over admissible controls on $[t, T]$ and $x$ is the trajectory with $x(t) = x$.\nIf the value function is continuously differentiable, it satisfies the equation.\nUnder standard hypotheses, such as $U$ compact and $f$, $L$, $g$ bounded and Lipschitz, it is the unique bounded uniformly continuous viscosity solution.\nExample: for $x' = u$, $U = \\R$, $L = u^2 / 2$ and $g(x) = x^2 / 2$, the minimum over $u$ is $-\\frac{1}{2} (\\partial_x V)^2$, and $V(t, x) = \\dfrac{x^2}{2 (1 + T - t)}$ solves the equation.\nIndeed $\\partial_t V = \\dfrac{x^2}{2 (1 + T - t)^2}$ and $\\frac{1}{2} (\\partial_x V)^2 = \\dfrac{x^2}{2 (1 + T - t)^2}$, and $V(T, x) = x^2 / 2$.\nNon-example: $W(t, x) = x^2 / 2$ satisfies the terminal condition but not the equation, since $\\partial_t W - \\frac{1}{2} (\\partial_x W)^2 = -x^2 / 2 \\ne 0$ for $x \\ne 0$.\nFor infinite-horizon problems with discount rate $\\rho \u003e 0$ the equation becomes stationary: $\\rho V(x) = \\min_{u \\in U} \\{ \\nabla V(x) \\cdot f(x, u) + L(x, u) \\}$.\n",
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