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Proof of Gronwall's Inequality 📂Lemmas

Proof of Gronwall's Inequality

Theorem

Let’s assume there are two continuous functions f,w:IRf,w : I \to \mathbb{R} defined in an interval IRI \subset \mathbb{R} that contains the minimum value aRa \in \mathbb{R}. If ww is tI\forall t \in I in w(t)0w(t) \ge 0 and for some constant CRC \in \mathbb{R}, f(t)C+atw(s)f(s)ds,tI f(t) \le C + \int_{a}^{t} w(s) f(s) ds \qquad , \forall t \in I then the following holds. f(t)Cexp(atw(s)ds),tI f(t) \le C \exp \left( \int_{a}^{t} w(s) ds \right) \qquad , \forall t \in I

Explanation

The fact that interval II has a minimum value aa means that II looks like the following with respect to a<ba < b. [a,b] or [a,b) or [a,) [a,b] \text{ or } [a,b) \text{ or } [a,\infty)

Proof 1

α(t)=C+atw(s)f(s)dsβ(t)=Cexp(atw(s)ds) \begin{align*} \alpha (t) =& C + \int_{a}^{t} w(s) f(s) ds \\ \beta (t) =& C \exp \left( \int_{a}^{t} w(s) ds \right) \end{align*}

When we define two functions α,β\alpha, \beta as shown above and suppose γ(t):=α(t)β(t)\displaystyle \gamma (t) := {{ \alpha (t) } \over { \beta (t) }}, then γ(a)=1\gamma (a) = 1 and by the quotient rule of differentiation, γ(t)=α(t)β(t)α(t)β(t)[β(t)]2=w(t)f(t)Cexp(atw(s)ds)(C+atw(s)f(s)ds)w(t)Cexp(atw(s)ds)(Cexp(atw(s)ds))2=w(t)f(t)Catw(s)f(s)dsCexp(atw(s)ds) \begin{align*} & \gamma ' (t) \\ =& {{ \alpha ' (t) \beta (t) - \alpha (t) \beta ' (t) } \over { \left[ \beta (t) \right]^{2} }} \\ =& {{ w(t) f(t) \cdot C \exp \left( \int_{a}^{t} w(s) ds \right) - \left( C + \int_{a}^{t} w(s) f(s) ds \right) \cdot w (t) C \exp \left( \int_{a}^{t} w(s) ds \right) } \over { \left( C \exp \left( \int_{a}^{t} w(s) ds \right) \right)^{2} }} \\ =& w(t) {{ f(t) - C - \int_{a}^{t} w(s) f(s) ds } \over { C \exp \left( \int_{a}^{t} w(s) ds \right) }} \end{align*} it follows. Given that f(t)C+atw(s)f(s)ds\displaystyle f(t) \le C + \int_{a}^{t} w(s) f(s) ds and w(t)0w(t) \ge 0, then γ(t)0\gamma ' (t) \le 0. Therefore, γ(t)\gamma (t) is a non-increasing function, and since γ(t)1\gamma (t) \le 1, the following is established. γ(t)=α(t)β(t)1    α(t)β(t)    C+atw(s)f(s)dsCexp(atw(s)ds)    f(t)C+atw(s)f(s)dsCexp(atw(s)ds)    f(t)Cexp(atw(s)ds) \begin{align*} & \gamma (t) = {{ \alpha (t) } \over { \beta (t) }} \le 1 \\ \implies& \alpha (t) \le \beta (t) \\ \implies& C + \int_{a}^{t} w(s) f(s) ds \le C \exp \left( \int_{a}^{t} w(s) ds \right) \\ \implies& f(t) \le C + \int_{a}^{t} w(s) f(s) ds \le C \exp \left( \int_{a}^{t} w(s) ds \right) \\ \implies& f(t) \le C \exp \left( \int_{a}^{t} w(s) ds \right) \end{align*}