Bounding Lyapunov functionals

Question

In a calculation that I'm doing, I end up with the following derivative \begin{equation} V^{\prime}(t)=-a[x^2(t)+x^2(t-h)]+2b(t)x(t)x(t-h), \label{eqn1} \end{equation} where $V(t)$ is a functional, $a>0$, $h>0$, $t\geq t_0$, and $b(t)$ is a continuous function. I want to show that $$ V^{\prime}(t)\leq -\nu [x^2(t)+x^2(t-h)], $$ where $\displaystyle \nu=a-\sup_{t\geq t_0}|b(t)|$. What's the easiest strategy to prove the inequality? It seems easy, but I'm not so sure anymore.

Answer 1

The idea is to use the inequality $2AB\le A^2 + B^2$, which can be seen by expanding the inequality $(A-B)^2\ge 0$. Hence \begin{align*} V'(t) & = -a\left[x^2(t) + x^2(t-h)\right] + b(t)\left[2x(t)x(t-h)\right] \\ & \le -a\left[x^2(t) + x^2(t-h)\right] + b(t)\left[x^2(t) + x^2(t-h)\right] \\ & = \left[-a + b(t)\right]\left[x^2(t) + x^2(t-h)\right] \\ & \le \left[-a + \sup_{t\ge t_0}|b(t)|\right]\left[x^2(t) + x^2(t-h)\right] \\ & = -\left[a - \sup_{t\ge t_0}|b(t)|\right]\left[x^2(t) + x^2(t-h)\right]. \end{align*}

Answer 2

This problem has nothing to do with the assertion that $V(t)$ is a Lyapunov function; rather, the result follows from simple algebraic manipulation of the given expressions, to wit:

Let

$y(t) = x(t - h); \tag 1$

then

$V'(t) = -a[x^2(t) + x^2(t - h)] + 2b(t) x(t) x(t - h) \tag 2$

is equivalent to

$V'(t) = -a[x^2(t) + y^2(t)] + 2b(t) x(t) y(t); \tag 3$

now with

$\nu = \displaystyle a - \sup_{t \ge t_0} \vert b(t) \vert, \tag 4$

we have

$-\nu[x^2(t) + y^2(t)] = \displaystyle (\sup_{t \ge t_0} \vert b(t) \vert - a)[x^2(t) + y^2(t)]$ $= -a[x^2(t) + y^2(t)] + \displaystyle \sup_{t \ge t_0} \vert b(t) \vert [x^2(t) + y^2(t)]; \tag 5$

we observe that

$b(t) \le \vert b(t) \vert \le \displaystyle \sup_{t \ge t_0} \vert b(t) \vert; \; -b(t) \le \vert b(t) \vert \le \displaystyle \sup_{t \ge t_0} \vert b(t) \vert; \tag 6$

also,

$0 \le (x(t) - y(t))^2 = x^2(t) - 2x(t)y(t) + y^2(t), \tag 7$

whence

$2x(t)y(t) \le x^2(t) + y^2(t); \tag 8$

now if $0 \le b(t)$, then

$2b(t)x(t)y(t) \le b(t)[x^2(t) + y^2(t)] \le \displaystyle \sup_{t \ge t_0} \vert b(t) \vert [x^2(t) + y^2(t)], \tag 9$

so that

$V'(t) = -a[x^2(t) + y^2(t)] + 2b(t) x(t) y(t)$ $\le -a[x^2(t) + y^2(t)] + \displaystyle \sup_{t \ge t_0} \vert b(t) \vert [x^2(t) + y^2(t)]; \tag{10}$

if $b(t) < 0$, we use

$0 \le (x(t) + y(t))^2 = x^2(t) + 2x(t) y(t) + y^2(t) \tag{11}$

from which it follows that

$-2x(t) y(t) \le x^2(t) + y^2(t), \tag{12}$

or

$2x(t) y(t) \ge -[x^2(t) + y^2(t)], \tag{13}$

and then

$2b(t)x(t)y(t) \le -b(t) [x^2(t) + y^2(t)] \le \displaystyle \sup_{t \ge t_0} \vert b(t) \vert [x^2(t) + y^2(t)], \tag{14}$

and again we have

$V'(t) = -a[x^2(t) + y^2(t)] + 2b(t) x(t) y(t)$ $\le -a[x^2(t) + y^2(t)] + \displaystyle \sup_{t \ge t_0} \vert b(t) \vert [x^2(t) + y^2(t)]; \tag{15}$

now re-substituting $x(t - h)$ for $y(t)$ we see that

$V'(t) \le -a[x^2(t) + x^2(t - h)] + \displaystyle \sup_{t \ge t_0} \vert b(t) \vert [x^2(t) + x^2(t - h)] \tag{16}$

in all cases.