A variation of Gronwall's lemma

TL;DR

This paper presents a modified version of Gronwall's lemma, providing a new mathematical inequality useful for analyzing differential equations and related fields.

Contribution

It introduces a novel variation of Gronwall's lemma, expanding the toolkit for mathematical analysis in differential equations.

Findings

Proves a new form of Gronwall's inequality

Provides potential applications in differential equations

Enhances existing mathematical analysis methods

Abstract

We prove a variation of Gronwall's lemma.

Equations

$$\forall u,t0\le u\le tg(t)-g(u)\le {\int }_u^t-\lambda g(s)+Cds$$

$$\forall t\ge 0g(t)\le \frac{C}{\lambda }+{\left[g(0)-\frac{C}{\lambda }\right]}^{+}{e}^{-\lambda t}$$

$$\forall t\ge 0h(t)=g(0)e^{-\lambda t}$$

$$\forall u,t0\le u\le th(t)-h(u)=-{\int }_u^t\lambda h(s)ds$$

$$m=\inf \{a^{\prime }<a|\forall t\in ]a^{\prime },a[g(t)>h(t)\}$$

$$\forall t\in ]m,a[g(t)>h(t)$$

$$\forall t\ge mg(t)\le \varphi (t)$$

$${\left(\frac{\varphi }{h}\right)}^{\prime }=\frac{{\varphi }^{\prime }h-h^{\prime }\varphi }{h^2}=\frac{-\lambda gh+\lambda h\varphi }{h^2}=\frac{\lambda h(\varphi -g)}{h^2}\ge 0$$

$$\forall t\in ]m,a[\varphi (t)\ge h(t)$$

$$\forall t\in ]m,a[{\int }_m^{t}g(s)ds\le {\int }_m^{t}h(s)ds$$

$${\int }_m^{t_0}g(s)ds>{\int }_m^{t_0}h(s)ds$$

$$m=\inf \{a^{\prime }<a|\forall t\in ]a^{\prime },a[g(t)>0\}$$

$$\forall t\in ]m,a[g(t)>0$$

$$g(t_0)\le -\lambda {\int }_m^{t_0}g(s)ds\le 0$$

$$\forall u,t0\le u\le t\hat{g}(t)-\hat{g}(u)=g(t)-g(u)\le {\int }_u^t-\lambda g(s)+Cds=-{\int }_u^t\lambda \hat{g}(s)ds$$

$$\forall t\ge 0\hat{g}(t)\le [\hat{g}(0){]}^{+}{e}^{-\lambda t}$$

Full text

The formulation and proof of the classical Gronwall’s lemma can be found in [1]. We prove here a variation of this lemma, which we were not able to find in the literature. The main difference from usual versions of Gronwall’s lemma is that $-\lambda$ is negative.

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Then

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where $[\cdot]^{+}=\max(0,\cdot)$.

Proof

Case 1 : $C=0$, $g(0)>0$.

Define $h(t)$ by

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Remark that $h$ is positive with $h(0)=g(0)$, and satisfies (1) where the inequality has been replaced by an equality

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Consider now the set $S=\{t\geq 0\ |\ g(t)>h(t)\}$. If $S=\emptyset$ then the lemma holds true. Assume by contradiction that $S\neq\emptyset$. In this case, consider an element $a\in S$. One has by definition $g(a)>h(a)$. Since $g(0)=h(0)$, one also has $a>0$. Consider now

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By continuity of $g$ and $h$ and by the fact that $g(0)=h(0)$, one has $g(m)=h(m)$. One thus also has $m<a$ and

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Consider now $\phi(t)=g(m)-\lambda\int_{m}^{t}g(s)ds$. Equation (1) implies that

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In order to compare $\phi(t)$ and $h(t)$ for $t\in]m,a[$, let us differentiate the ratio $\phi(t)/h(t)$.

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Thus $\phi(t)/h(t)$ is increasing for $t\in]m,a[$. Since $\phi(m)/h(m)=1$, one can conclude that

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which implies, by definition of $\phi$ and $h$, that

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Choose now a $t_{0}\in]m,a[$. Then one has by (3)

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which clearly contradicts (4).

Case 2 : $C=0$, $g(0)\leq 0$

Consider the set $S=\{t\geq 0\ |\ g(t)>0\}$. If $S=\emptyset$ then the lemma holds true. Assume by contradiction that $S\neq\emptyset$. In this case, consider an element $a\in S$. One has by definition $g(a)>0$. Since $g(0)\leq 0$, one also has $a>0$. Consider now

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By continuity of $g$ and by the fact that $g(0)\leq 0$, one has $g(m)=0$. One thus also has $m<a$ and

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Choose now a $t_{0}\in]m,a[$. Equation (1) implies that

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which clearly contradicts (5).

Case 3 : $C\neq 0$

Define $\hat{g}=g-C/\lambda$. One has

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Thus $\hat{g}$ satisfies the conditions of Case 1 or Case 2, and as a consequence

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The conclusion of the lemma follows by replacing $\hat{g}$ by $g-C/\lambda$ in the above equation. $\Box$

Acknowledgments

The author would like to thank N. Tabareau and J.-J. Slotine for their helpful comments and V. Valmorin for having pointed out an error in an earlier version of the manuscript.