by Nebulae
Last Updated August 12, 2018 21:20 PM

Show that the first order system described by

$$\frac{dx}{dt} = f(x) = x(x^2-1)$$

is terminating for the initial condition $x(0) = x_0 = 2$ by deriving a finite upper bound for the time needed to reach $\infty$.

$\displaystyle t_1-t_0 = \int_{x_{0}}^{x_1} \frac{1}{f(x)}\,{dx} $ so time needed to reach $x_1$ in this case is given by

$\displaystyle t_1 = \int_2^{x_1} \frac{1}{x(x^2-1)}\,{dx} \le \int_2^{x_1} \frac{1}{x(x+1)}\,{dx} = \log\left|\frac{x_1}{x_1+1}\right|- \log\left(\frac{2}{3}\right) \to \log\left(\frac{3}{2}\right)$ as $x_1 \to \infty$.

Where the inequality comes from $x^2-1 \ge x+1$ for $x \in [2, \infty)$. So $t_1 \le \log(3/2)$. Hence the motion is terminating. Is this right? In my notes the answer they get for the upper bound is $1/2$. I know upper bounds are not unique, but I'm not sure if my solution is correct or not. Thanks.

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