02 - Gronwall's Inequality

Gronwall's Inequality

Suppose $A\ge 0$ and $B\ge 0$ are constants and $v$ is a non-negative continious function with

$$v(x)\le B+A\left|{\int }_a^{x}v(s)ds\right|$$

for all $x$ in an interval $[a-h_1,a+h_2]$ and $h_{1,2}\ge 0$ then

$$v(x)\le B{e}^{A|x-a|}\text{ for all }x\in [a-h_1,a+h_2]x$$

Note: need modulus for case where $x\le a$

Proof

For $x\ge a$ let $V(x)={\int }_a^{x}v(s)ds$ so $V^{\prime }(x)=v(x)$
As $x\ge a$ and $v\ge 0$ then $V(x)\ge 0$ and $V(a)=0$ so

$$V^{\prime }(x)\le B+AV(x)$$

By multiplying it through by integrating factor $e^{-}Ax$ then

$$\begin{array}{rl}(V^{\prime }(x)-AV(x))e^{-Ax}& \le B{e}^{-Ax}\\ \frac{d}{dx}(V(x)e^{-Ax})& \le B{e}^{-Ax}\\ V(x)e^{-Ax}-V(a)e^{-Aa}& \le {\int }_a^{x}B{e}^{-As}ds=\frac{B}{A}(e^{-Aa}-e^{-Ax})\\ V(x)& \le \frac{B}{A}(e^{A(x-a)}-1)\end{array}$$

Using the assumption again then

$$v(x)\le B+A{\int }_a^{x}v(s)ds=B+AV(x)\le B+A\cdot \frac{B}{A}(e^{A(x-a)}-1)=B{e}^{A(x-a)}$$

Similar case for $x\le a$