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  "aliases": [
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  "body": "\n## Statement\n\nLet $a \u003c b$ and let $I$ be an [[interval|0Q1D85B8G6]] of the form $[a, b]$, $[a, b)$ or $[a, \\infty)$. Let $I^\\circ$ be the interior of $I$.\nDifferential form: let $\\beta, u : I \\to \\mathbb{R}$ be continuous, let $u$ be differentiable on $I^\\circ$, and assume $u'(t) \\le \\beta(t)\\, u(t)$ for each $t \\in I^\\circ$.\nThen\n$$u(t) \\le u(a) \\exp\\left( \\int_a^t \\beta(s)\\, ds \\right) \\quad \\text{for each } t \\in I.$$\nIntegral form: let $\\beta, u : I \\to \\mathbb{R}$ be continuous with $\\beta \\ge 0$, let $\\alpha : I \\to \\mathbb{R}$ be non-decreasing, and assume\n$$u(t) \\le \\alpha(t) + \\int_a^t \\beta(s)\\, u(s)\\, ds \\quad \\text{for each } t \\in I.$$\nThen\n$$u(t) \\le \\alpha(t) \\exp\\left( \\int_a^t \\beta(s)\\, ds \\right) \\quad \\text{for each } t \\in I.$$\n\n## Remarks\n\nThe differential form needs no assumption on the signs of $\\beta$ and $u$. Its proof shows that $u(t) \\exp\\left(-\\int_a^t \\beta\\right)$ is non-increasing.\nThe integral form does not need $u$ to be differentiable.\nFor constant $\\alpha = 0$, the integral form gives $u \\le 0$. This is the step that proves [[uniqueness of solutions|025PKW8AMR]].\nA variant with an additional constant term gives the [[Grönwall bound|0FRX8JD7QH]], and there is a [[discrete version|0NTM3J5MFE]].\n\n## History\n\nThe inequality is named after Thomas Hakon Grönwall. He proved it in 1919, in the integral form with constant $\\alpha$ and $\\beta$.\nRichard Bellman proved a slightly more general integral form in 1943.\n\n## Explanations\n\n### L1 human\n\nIf a quantity grows at most at a rate proportional to itself, it grows at most exponentially.\nThe inequality compares the quantity with the solution of the equation $v' = \\beta v$.\nThis is the main tool to compare two solutions of a differential equation.\n",
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