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  "body": "\n## Statement\n\nLet $n \\ge 1$, let $T \u003e 0$, and let $H \\colon \\R^n \\times \\R^n \\to \\R$ be continuous, written $H(x, p)$.\nAssume that there is a constant $C$ such that, for all $x, y, p, q \\in \\R^n$,\n$$\\abs{H(x, p) - H(x, q)} \\le C \\abs{p - q}, \\qquad \\abs{H(x, p) - H(y, p)} \\le C \\abs{x - y} (1 + \\abs{p}).$$\nLet $g \\colon \\R^n \\to \\R$ be bounded and [[uniformly continuous|0M4JJT780X]].\nConsider the Cauchy problem for the [[Hamilton–Jacobi equation|5TRXQKP7DY]]\n$$u_t + H(x, D_x u) = 0 \\quad \\text{in } \\R^n \\times (0, T), \\qquad u(x, 0) = g(x) \\quad \\text{for } x \\in \\R^n.$$\nThen this problem has at most one solution $u \\colon \\R^n \\times [0, T] \\to \\R$ in the following sense.\nThe function $u$ is bounded and uniformly continuous, satisfies the initial condition, and is a [[viscosity solution|H1ERJ3T75T]] of the equation in $\\R^n \\times (0, T)$.\n\n## Remarks\n\nThe theorem is a consequence of a comparison principle: if $u$ is a bounded uniformly continuous subsolution and $v$ a bounded uniformly continuous supersolution with $u(\\cdot, 0) \\le v(\\cdot, 0)$, then $u \\le v$.\nThe proof doubles the variables.\nOne maximises $u(x, t) - v(y, s) - \\abs{x - y}^2 / \\varepsilon - \\abs{t - s}^2 / \\varepsilon$ over pairs of points, after adding small extra terms that force a maximum to exist on the unbounded domain.\nAt the maximum point the two halves of the penalty serve as test functions for $u$ and for $v$, and the two inequalities of the definition are subtracted.\nThe Lipschitz hypotheses on $H$ make the error terms tend to $0$ with $\\varepsilon$, which gives a contradiction if $u - v$ had a positive supremum.\n\nExample: for $H(x, p) = \\abs{p}$ the hypotheses hold with $C = 1$.\nThe unique bounded uniformly continuous viscosity solution is given by the Hopf–Lax formula $u(x, t) = \\min_{\\abs{y - x} \\le t} g(y)$.\n\nThe viscosity sense is needed.\nFor $H(x, p) = \\abs{p}$ and $g = 0$, both $u = 0$ and $w(x, t) = \\min(0, \\abs{x} - t)$ (with $n = 1$) are Lipschitz and satisfy the equation almost everywhere.\nOnly $u = 0$ is a viscosity solution: at a point $(0, t)$ the function $w - \\varphi$ with $\\varphi(x, s) = -s$ has a local minimum, while $\\varphi_t + \\abs{\\varphi_x} = -1 \u003c 0$.\n\n## Explanations\n\n### L1 human\n\nHamilton–Jacobi equations usually have no smooth solutions, and they have many solutions that hold only almost everywhere.\nViscosity solutions are a way to choose the right one.\nThis theorem says the choice is never ambiguous: given the initial data, there is only one viscosity solution.\n\n### L2 human\n\nUniqueness is what makes viscosity solutions useful in control theory.\nTo show that a value function equals a given formula, it is enough to check that both are viscosity solutions of the same Hamilton–Jacobi–Bellman equation with the same data.\nThe same doubling-of-variables argument gives comparison for stationary equations such as $u + H(x, Du) = 0$, and, with more work, for second-order degenerate elliptic equations.\nExistence comes separately, for instance from the value function of a control problem or from the vanishing viscosity method.\n",
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