Lyapunov’s inequality on timescales

The purpose of this work is to establish the timescale version of Lyapunov’s inequality as follows: Let $x\left(t\right)$ be a nontrivial solution of ${\left(r\left(t\right)x^{\Delta }\left(t\right)\right)}^{\Delta }+p\left(t\right)x^{\sigma }\left(t\right)=0\text{on }[a,b]$ satisfying $x\left(a\right)=x\left(b\right)=0$. Then, under suitable conditions on $p$, $r$, $a$ and $b$, we have ${\int }_a^b{p}_{+}\left(t\right)\Delta t\ge \{\begin{array}{cc}\frac{r\left(a\right)}{r\left(b\right)}\frac{b-a}{f\left(d\right)}\text{,}& \text{if }r\text{ is increasing}\text{,}\\ \frac{r\left(b\right)}{r\left(a\right)}\frac{b-a}{f\left(d\right)}\text{,}& \text{if }r\text{ is decreasing}\text{,}\end{array}$ where $p_{+}\left(t\right)=max\{p\left(t\right),0\},f\left(t\right)=(t-a)(b-t)$ and $d\in \mathbb{T}$ satisfies $|\frac{a+b}{2}-d|=min\{|\frac{a+b}{2}-s|\mid s\in [a,b]\cap \mathbb{T}\}$ if $\frac{a+b}{2}\in \mathbb{T}$. Here $\mathbb{T}$ is a timescale (see below).