Grönwall's inequality

Statement

Let $a < b$ and let $I$ be an interval of the form $[a, b]$, $[a, b)$ or $[a, \infty)$. Let $I^\circ$ be the interior of $I$. Differential 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 $t \in I^\circ$. Then $$u(t) \le u(a) \exp\left( \int_a^t \beta(s)\, ds \right) \quad \text{for each } t \in I.$$ Integral 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 $$u(t) \le \alpha(t) + \int_a^t \beta(s)\, u(s)\, ds \quad \text{for each } t \in I.$$ Then $$u(t) \le \alpha(t) \exp\left( \int_a^t \beta(s)\, ds \right) \quad \text{for each } t \in I.$$

Remarks

The 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. The integral form does not need $u$ to be differentiable. For constant $\alpha = 0$, the integral form gives $u \le 0$. This is the step that proves uniqueness of solutions. A variant with an additional constant term gives the Grönwall bound, and there is a discrete version.

History

The inequality is named after Thomas Hakon Grönwall. He proved it in 1919, in the integral form with constant $\alpha$ and $\beta$. Richard Bellman proved a slightly more general integral form in 1943.

Explanations

L1 human

If a quantity grows at most at a rate proportional to itself, it grows at most exponentially. The inequality compares the quantity with the solution of the equation $v' = \beta v$. This is the main tool to compare two solutions of a differential equation.