In this article, we state and prove Gronwall lemma and give some of its applications. In fact, we will use this lemma for the stability of the solution to the differential equations. In particular, the Lyapunov stability of nonlinear systems.

## The proof of Gronwall lemma

Many proofs of known theorems in mathematics are based on Gronwall’s lemma. In fact, this lemma is used to prove the uniqueness of the solution of differential equations, as well as the stability of equilibrium points.

**Lemma:** Let $\varphi:[a,b]\to [0,+\infty)$ be a continuous function and assume that there exist two real constants $\alpha,\beta\ge 0$ such that for any $c\in [a,b]$, \begin{align*}\varphi(t)\le \alpha+\beta\left|\int^t_c \varphi(s)ds\right|,\qquad \forall t\in [a,b].\end{align*}Then \begin{align*}\varphi(t)\le \alpha e^{\beta |t-c|},\quad\forall t\in [a,b].\end{align*}

**Proof: **For simplicity, we suppose that $t\ge c$. We select\begin{align*}F(t)= \alpha e^{\beta |t-c|},\quad\forall t\in [a,b]. \end{align*}Then $F$ is a $C^1$ function and $F'(t)=\beta \varphi(t)\le \beta F(t)$. By multiplying $e^{-\beta t}$ on the both sides of this inequality, we obtain \begin{align*} e^{-\beta t} F'(t)- e^{-\beta t} \beta F(t)\le 0.\end{align*} Hence, \begin{align*} \frac{d}{ds} \left(e^{-s\beta}F(s)\right)\le 0.\end{align*} By integrating between $c$ and $t$, the result then follows.

**An other version of the lemma:** Let $\varphi,\psi:[a,b]\to [0,+\infty)$ be a continuous function and assume that there exist two real constants $\alpha,\beta\ge 0$ such that for any $c\in [a,b]$, \begin{align*}\varphi(t)\le \alpha+\beta\left|\int^t_c \psi(s)\varphi(s)ds\right|,\qquad \forall t\in [a,b].\end{align*}Then \begin{align*}\varphi(t)\le \alpha \exp\left(\beta\left|\int^t_c \psi(s)ds\right|\right),\quad\forall t\in [a,b].\end{align*}

## Applications to the stability of the solution of differential equations

In this paragraph, we give an application of the Gronwall lemma to the stability of nonlinear differential equations. In fact, let $\lambda>0$ be a real number, and consider the following semilinear Cauchy problem \begin{align*}\tag{CP} \begin{cases}\dot{u}(t)=-\lambda u(t)+\arctan(t^2 u(t)^5) \frac{\sin{u(t)}}{1+2 u(t)^2},&t\ge 0,\cr u(0)=\frac{1}{2}.\end{cases}\end{align*} If we select \begin{align*}f(t,x)=\lambda x+\arctan(t^2 x^5)\frac{\sin(x)}{1+2x^2},\quad (t,x)\in \mathbb{R}^+\times \mathbb{R},\end{align*} then our equation has the following standard form \begin{align*}\tag{CP} \begin{cases}\dot{u}(t)=f(t,u(t)),& t\ge 0,\cr u(0)=\frac{1}{2}.\end{cases}\end{align*} Clearly $f$ is a $C^1$ function, then locally Lipschitz function. Thus by the Cauchy-Lipschitz theorem, the Cauchy problem (CP) admits a unique maximal solution $u:[0,T)\to \mathbb{R}$. This solution is global, i.e. defined on all $[0,+\infty)$ because $|f(t,x)|\le \lambda |x|+1$ for any $(t,x)\in\mathbb{R}^+\times \mathbb{R}$.

By computing the derivative $(e^{\lambda t}u(t))’$ and using the expression of $\dot{u}(t)$ in (CP), we can prove that \begin{align*}u(t)=\frac{1}{2}e^{-\lambda t}+\int^t_0 e^{-\lambda (t-s)}\arctan(t^2 u(s)^5) \frac{\sin{u(s)}}{1+2 u(s)^2}.\end{align*} Now using the facts that $|\arctan(y)|\le \frac{\pi}{2},$ $|\sin(y)|\le |y|$ and $\frac{1}{1+y^2}\le 1,$ for any $y\in\mathbb{R}$, we obtain \begin{align*} |u(t)|\le \frac{1}{2}e^{-\lambda t}+\frac{\pi}{2} e^{-\lambda t}\int^t_0 e^{\lambda s}|u(s)|ds,\quad t\ge 0.\end{align*} It follows that \begin{align*} e^{\lambda t}|u(t)|\le \frac{1}{2}+\frac{\pi}{2}\int^t_0 e^{\lambda s}|u(s)|ds,\quad t\ge 0.\end{align*} Now by using the Gronwall lemma we have \begin{align*} e^{\lambda t}|u(t)|\le \frac{1}{2}e^{\frac{\pi}{2}t}.\end{align*} Hence $$ |u(t)|\le \frac{1}{2}e^{-\left(\lambda-\frac{\pi}{2}\right)t},\quad \forall t\ge 0.$$ We deduce that the solution, of the equilibrium point $0$, is exponentially stable if $\lambda>\frac{\pi}{2}$.