Bihari–LaSalle inequality

Remarks

The function $G$ is continuous and strictly increasing with $G(c) = 0$, so $G^{-1}$ is defined on $[0, G(\infty))$. Proof idea: let $v(t) = c + \int_a^t f(s)\, w(u(s))\, ds$. Then $v \ge c > 0$, $u \le v$, and $v' = f\, w(u) \le f\, w(v)$ since $w$ is non-decreasing. So $\frac{d}{dt} G(v(t)) = v'(t) / w(v(t)) \le f(t)$, and integration gives $G(v(t)) \le \int_a^t f$. For $w(r) = r$ one gets $G(r) = \ln(r/c)$ and the bound $u(t) \le c\, \exp\left(\int_a^t f\right)$, which is Grönwall's inequality with constant $\alpha = c$. For $w(r) = r^2$ and $f = 1$ one gets $u(t) \le c / (1 - c(t - a))$ while $c(t - a) < 1$; the restriction on $t$ matches the blow-up of the solution of $u' = u^2$.

Going deeper

For a student

The inequality is the nonlinear version of the integral form of Grönwall's inequality. It gives a priori bounds for solutions of nonlinear equations and estimates of the time of existence. With $c \to 0$ it gives the Osgood uniqueness criterion, which relaxes the Lipschitz condition in uniqueness theorems.