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I am trying to show what is asked in question 3.18. After using the hint I have

$z(t)\leq k_1+k_2\int_{t_0}^t z(\tau)d\tau+k_3\int_{t_0}^t e^{\alpha (\tau-t_0)}d\tau \\ \quad \,= k_1+\dfrac{k_3}{\alpha}\big[e^{\alpha(t-t_0)}-1\big]+k_2\int_{t_0}^t z(\tau)d\tau\\$

then using the Gronwall-Bellman inequality I get

$z(t)\leq k_1e^{k_2(t-t_0)}+\dfrac{k_3}{\alpha}\big[e^{(\alpha+k_2)(t-t_0)}-e^{k_2(t-t_0)}\big]$

or equivalently

$y(t)\leq k_1e^{-(\alpha-k_2)(t-t_0)}+\dfrac{k_3}{\alpha}\big[e^{k_2(t-t_0)}-e^{-(\alpha-k_2)(t-t_0)}\big]$, which is not quite what the book has: $y(t)\leq k_1e^{-(\alpha-k_2)(t-t_0)}+\dfrac{k_3}{\alpha-k_2}\big[1-e^{-(\alpha-k_2)(t-t_0)}\big]$.

I am not sure if I have made an error in application of Gronwalls inequality, or something else entirely.

If anyone can give me a hint as to which direction to go I would greatly appreciate it.

problem 3.18

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  • $\begingroup$ What is the name and author of the textbook? $\endgroup$ – Gordon Sep 12 '16 at 15:07
  • $\begingroup$ It's Nonlinear Systems by Khalil, 3rd edition (international version maybe?), but the version I have doesn't have an appendix, which is where the Gronwall inequality should be in the regular version. So I had to look up the inequality on wiki $\endgroup$ – Thomas Kirven Sep 12 '16 at 20:41
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As I do not know the specific form of the Gronall-Bellman inequality in your textbook, I provide a direct proof below.

Let \begin{align*} v(t) = \int_{t_0}^t e^{-\alpha(t-\tau)}(k_2 y(\tau) + k_3) d\tau. \end{align*} Then \begin{align*} y(t) \le v(t) + k_1e^{-\alpha(t-t_0)}. \end{align*} Moreover, \begin{align*} \frac{d\left(v(t)e^{(\alpha-k_2)(t-t_0)}\right)}{dt} &= (\alpha-k_2) e^{(\alpha-k_2)(t-t_0)}v(t) + \frac{d\left(v(t)\right)}{dt}e^{(\alpha-k_2)(t-t_0)}\\ &=(\alpha-k_2) e^{(\alpha-k_2)(t-t_0)}v(t) \\ &\qquad + e^{(\alpha-k_2)(t-t_0)}\left(k_2y(t)+k_3 -\alpha \int_{t_0}^t e^{-\alpha(t-\tau)}(k_2 y(\tau) + k_3) d\tau\right) \\ &=(\alpha-k_2) e^{(\alpha-k_2)(t-t_0)}v(t) + e^{(\alpha-k_2)(t-t_0)}\left(k_2y(t)+k_3 -\alpha v(t)\right)\\ &\le (\alpha-k_2) e^{(\alpha-k_2)(t-t_0)}v(t) \\ &\qquad + e^{(\alpha-k_2)(t-t_0)}\left(k_2v(t)+k_3 -\alpha v(t) + k_2 k_1e^{-\alpha(t-t_0)}\right)\\ &= k_3 e^{(\alpha-k_2)(t-t_0)} + k_2 k_1 e^{-k_2(t-t_0)}. \end{align*} That is, \begin{align*} \frac{d\left(v(t)e^{(\alpha-k_2)(t-t_0)} - \frac{k_3}{\alpha-k_2}e^{(\alpha-k_2)(t-t_0)} + k_1 e^{-k_2(t-t_0)}\right)}{dt} \le 0. \end{align*} In other words, $$(v(t)e^{(\alpha-k_2)(t-t_0)} - \frac{k_3}{\alpha-k_2}e^{(\alpha-k_2)(t-t_0)} + k_1 e^{-k_2(t-t_0)}$$ is a decreasing function. Therefore, \begin{align*} v(t)e^{(\alpha-k_2)(t-t_0)} - \frac{k_3}{\alpha-k_2}e^{(\alpha-k_2)(t-t_0)} + k_1 e^{-k_2(t-t_0)} &\le - \frac{k_3}{\alpha-k_2} + k_1. \end{align*} Then, \begin{align*} v(t) \le k_1 e^{-(\alpha-k_2)(t-t_0)}+ \frac{k_3}{\alpha-k_2}\left( 1- e^{-(\alpha-k_2)(t-t_0)}\right) - k_1e^{-\alpha(t-t_0)} \end{align*} Finally, \begin{align*} y(t) &\le v(t) + k_1e^{-\alpha(t-t_0)}\\ &\le k_1 e^{-(\alpha-k_2)(t-t_0)}+ \frac{k_3}{\alpha-k_2}\left( 1- e^{-(\alpha-k_2)(t-t_0)}\right). \end{align*}

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