One form of the Gronwall's inequality is that

If $\alpha(x),u(x)$ are non-negative continuous functions on $[0,1]$, and $$\forall x\in [0,1], u(x)\leq C+\int_{0}^{x}[\alpha(s)u(s)+K]ds\;(C,K\geq0),$$ then we have that $u(x)\leq[C+Kx]e^{\int_{0}^{x}\alpha(s)ds}$.

One form of comparison theorem is the following.

Assume that $f(x,y),F(x,y)$ are continuous on a domain $\Omega\supset [0,1]\times\mathbb{R}$ and $f(x,y)<F(x,y)(\forall (x,y)\in\Omega)$, $y=\phi(x),y=\varphi(x)$ are solutions to $y'=f(x,y),y'=F(x,y)$ (respectively), and $\phi(0)=\varphi(0)$. Then we have that $\phi(x)<\varphi(x)\,(\forall x\in(0,1])$.

Are there any interpretations for the Gronwall's inequality in view of the comparison theorem? I am not sure if they have any connections besides the fact that Gronwall's inequality can be used to prove the comparison theorem.

Will someone be kind enough to give some comments on this? Thank you very much!

  • $\begingroup$ Here we elaborated a bit on a "comparison-oriented" generalization of Gronwall's lemma. $\endgroup$ Aug 26, 2012 at 22:28

1 Answer 1


Maybe this is a remark, but I'm not qualified to make remark, so I wrote here. For simplicity, we could consider $K$ to be $0$. Now we are lead to $u(x)\le C+\int_0^x a(s)u(s)ds$, "differentiate" on both sides, $u'(x)\le a(x)u(x)$, now consider another ODE, $u'(x)=a(x)u(x)$, with initial value $u(0)=C$, we could apply comparison principle, right?

  • 4
    $\begingroup$ Differentiate do not preserves inequalities. Take $f\equiv 0$ and $g(x)=-e^{-x}$. Then $f\gt g$, but $f'\lt g'$. $\endgroup$
    – leo
    Apr 25, 2012 at 5:11

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