Comparison theorem

Remarks

Proof idea: suppose $v(t_1) > u(t_1)$ and let $s$ be the last time before $t_1$ with $v(s) \le u(s)$, so $v(s) = u(s)$. Just after $s$ the difference $w = v - u$ is positive and satisfies $w' \le f(t, v) - f(t, u) \le K w$ by the Lipschitz condition. Then $w(t) e^{-K(t - s)}$ is non-increasing and $w(s) = 0$, so $w \le 0$ there, a contradiction; this is the differential form of Grönwall's inequality. Example: $v(t) = 0$ satisfies $v' \le 1 + v^2$, so the solution $u = \tan t$ of $u' = 1 + u^2$, $u(0) = 0$, is $\ge 0$ on $[0, \pi/2)$; with $v(t) = t$, which satisfies $v' = 1 \le 1 + v^2$, it gives $\tan t \ge t$ there. The Lipschitz hypothesis is needed: for $f(t, x) = 2\sqrt{\abs{x}}$, the solution $u = 0$ and the function $v = t^2$ satisfy $v' = 2\sqrt{v}$ and $v(0) = u(0)$, but $v > u$ for $t > 0$. Without Lipschitz conditions the conclusion holds when $u$ is replaced by the maximal solution of the initial value problem.

Going deeper

For a student

Comparison arguments give a priori bounds, blow-up times and global existence results for scalar equations. For systems, one applies the theorem to a scalar quantity such as $\norm{x(t)}^2$ or a Lyapunov function. The Bihari–LaSalle inequality and Grönwall's inequality are comparison results in integral form. The Sturm comparison theorem is a different result, about zeros of solutions of second-order linear equations.