Timeline for Noninvariance for a specific nonlinear oscillator
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Oct 26, 2015 at 19:21 | comment | added | Frits Veerman | Right, that's essentially the same. Transforming the system might streamline the argument a bit though, but that aspect depends more on personal taste. | |
| Oct 26, 2015 at 13:23 | comment | added | user45183 | But in the end your argument is essentially the same as using the equations derived by differentiating $y_2 + z_2$, right? It only differs in that you are transforming the system slightly beforehand. | |
| Oct 26, 2015 at 12:31 | comment | added | Frits Veerman | And you're of course absolutely right that the relation $c = \pm \sqrt{a(8-a)}$ defines a circle: my bad. | |
| Oct 26, 2015 at 12:25 | comment | added | Frits Veerman | Addendum to the previous: the relations will also be satisfied by the other equilibria of the system: $(8,0,0,0)$ and $(4,0,\pm 4,0)$. Anyway, there is no nontrivial solution for which $y_2(t) \equiv 0$ for all $t$. | |
| Oct 26, 2015 at 12:15 | comment | added | Frits Veerman | Furthermore, for the situation $y_2(t) \neq 0$ not to occur, we want $\dot{y}_2 = 0$ for all time $t$, i.e. that for all time, we have $w_1(t)^2 = y_1(t)(8-y_1(t))$. Taking the time derivative of this identity and using the dynamical system, you obtain a new nontrivial relation between $y_{1,2}(t),w_{1,2}(t)$. Repetition of this process then would yield a number of relations between the dynamical variables which can only be satisfied when $y_{1,2}(t) \equiv 0$ and $w_{1,2} \equiv 0$. | |
| Oct 26, 2015 at 12:11 | comment | added | user45183 | But $-8a + a^2 + c^2 = 0$ gives you a circle. Complete the square in $a$ to get $(a-4)^2 + c^2 = 4^2$. | |
| Oct 26, 2015 at 12:06 | comment | added | Frits Veerman | Well, not really a circle, but yes: there are choices for $a,c$ such that $\dot{y}_2(0)$ is indeed zero: choose $c = \pm \sqrt{a(8-a)}$, i.e. $x_1(0) = 2 \pm \sqrt{8-(z_1(0)-2)^2}$. | |
| Oct 26, 2015 at 11:40 | comment | added | user45183 | Thanks for your detailed answer. I have one question: You say that $\dot{y}_2(0) = \dots \ne$ for general $a,c$, but looking closer at it, the set for which $\dot{y}_2(0) = 0$ has infinitely many solutions, namely the set of $(a,c)$ defines a circle centered at 4 with radius 4. But maybe $x_1 + z_1 = a$ and $x_1 - z_1 = c$ does not have a solution for such $(a,c)$? | |
| Oct 26, 2015 at 11:05 | history | answered | Frits Veerman | CC BY-SA 3.0 |