Periodic trajectory of a linear spring-mass-damper

Question

Given the following equation:

$\ddot y + \frac{1}{4}\dot y + 2y = 0; \space\space\space y(0) = 0, \space \dot y(0) = 2$

Under what condition would the phase plot become closed?

Solving the ivp to obtain the phase-plot shown here:

where: $$y = \frac{16}{\sqrt{127}}e^{-t/8} * \sin(\frac{\sqrt{127}}{8}t)$$ $$\dot y = \frac{2}{127}e^{-t/8}*[127\cos(\frac{\sqrt{127}}{8}t) - \sqrt{127}\sin(\frac{\sqrt{127}}{8}t)]$$

My initial thought would be if the coefficient of $\dot y = 0 $ or if $\dot y = 0$ itself then the phase plot would close. However, I believe this would change the governing equation and isn't necessarily a 'condition'

My question basically is: What needs to be enforced for a linear spring-mass-damper system to display a periodic trajectory? (closed phase-plot)

Any help would be greatly appreciated!

Answer 1

You have a closed system (no external force). As long as there is damping, the energy is dissipated, that is $y^2+(y')^2\to 0$ as $t\to\infty$, and the trajectories will spiral towards the origin. Mathematically, as long as you have a term $ay'$ with $a>0$ in the equation, your solutions will contain a decaying exponential in front of the oscillatory part. The short answer is: it never happens, unless you do not have damping or you have an external force supplying energy.

Answer 2

I decided to address this question in the context of second-order, linear ordinary differential equations with non-constant coefficients, a considerable leap from the specific result requested by our OP Clark. So please bear with me if you choose to read my (admittedly somewhat long-winded) response.

Given the equation

$\ddot y + \gamma(t) \dot y + \kappa(t) y = 0, \tag 1$

where $\gamma$ and $\kappa$ may depend on $t$, we write it as

$\ddot y + \kappa(t) y = - \gamma(t) \dot y, \tag 2$

and multiply through by $\dot y$ to obtain

$\dot y \ddot y + \kappa(t) y \dot y = - \gamma(t) \dot y^2, \tag 3$

or

$\dfrac{1}{2} \dfrac{d}{dt}(\dot y)^2 + \kappa(t) \dfrac{1}{2} \dfrac{d}{dt} (y^2) = - \gamma(t) \dot y^2; \tag 4$

assumming

$\kappa(t) = \kappa = \text{constant}, \tag 5$

(4) may be written in the form

$\dfrac{d}{dt} \left (\dfrac{1}{2} \dot y^2 + \dfrac{1}{2} \kappa y^2 \right ) = -\gamma(t) \dot y^2; \tag 6$

via this equation we may express the net change in the quantity $\frac{1}{2}\dot y^2 + \frac{1}{2} \kappa y^2$ as $t$ traverses any interval $[t_0, t_1]$:

$\left (\dfrac{1}{2} \dot y^2(t_1) + \dfrac{1}{2} \kappa y^2(t_1) \right ) - \left (\dfrac{1}{2} \dot y^2(t_0) + \dfrac{1}{2} \kappa y^2(t_0) \right )$ $= \displaystyle \int_{t_0}^{t_1} \dfrac{d}{ds} \left (\dfrac{1}{2} \dot y^2(s) + \dfrac{1}{2} \kappa y^2 \right ) \; ds = -\displaystyle \int_{t_0}^{t_1} \gamma(s) \dot y^2(s) \; ds; \tag 7$

now if

$\gamma(t) > 0, \tag 8$

we have

$-\displaystyle \int_{t_0}^{t_1} \gamma(t) \dot y^2(s) \; ds < 0 \tag 9$

over any intervql $[t_0, t_1]$ on which

$\dot y(t) \ne 0 \tag{10}$

identically and thus, by (7), $\frac{1}{2}\dot y^2 + \frac{1}{2} \kappa y^2$ decreases over $[t_0, t_1]$; from this it follows that, under the conditions stated here, no non-trivial trajectory of (1) can be closed; for if there were $t_0$, $t_1$ with

$(y(t_0). \dot y(t_0)) = (y(t_1), \dot y(t_1)) \tag{11}$

then we would of course have

$\dfrac{1}{2} \dot y^2(t_1) + \dfrac{1}{2} \kappa y^2(t_1) = \dfrac{1}{2} \dot y^2(t_0) + \dfrac{1}{2} \kappa y^2(t_0), \tag{12}$

precluded by (7).

Note this conclusion applies to any trajectory, and not only the one characterized by

$y(0) = 0, \; \dot y(0) = 2. \tag{13}$

We have seen there can be no non-trivial closed orbit under the condition (8); if on the other hand we take

$\gamma(t) = 0, \tag{14}$

then (1) becomes

$\ddot y + \kappa(t) y = 0, \tag{15}$

and with

$\kappa > 0, \; \text{a constant}, \tag{16}$

this equation has the well-known sinusoidal solutions which are obviously periodic, hence possessed of closed trajectories. Indeed, the period $T$ in this case is $T = 2\pi / \sqrt \kappa$.

Of course, when

$\gamma(t) < 0, \tag{17}$

then (7) shows that $\frac{1}{2}\dot y^2(t) + \frac{1}{2} \kappa y^2(t)$ is increasing with $t$, so periodic trajectories are prohibited here as well.

We observe that $\gamma(t)$ need not be contant for these conclusions to bind, as long as one of (8), (14) or (17) holds.

A Few Further Remarks:

If

$\kappa(t) \ne \text{a constant}, \tag{18}$

and

$\kappa(t) \in C^1, \tag{19}$

then

$\dfrac{d}{dt} \left (\dfrac{1}{2} \dot y^2 + \dfrac{1}{2} \kappa(t) y^2 \right ) - \dfrac{1}{2} \dot \kappa(t) y^2 = -\gamma(t) \dot y^2, \tag{20}$

and thus

$\dfrac{d}{dt} \left (\dfrac{1}{2} \dot y^2 + \dfrac{1}{2} \kappa(t) y^2 \right ) = \dfrac{1}{2} \dot \kappa(t) y^2 -\gamma(t) \dot y^2; \tag{21}$

we may integrate this equation in a manner analogous to (7):

$\left ( \dfrac{1}{2} \dot y^2(t_1) + \dfrac{1}{2} \kappa y^2(t_1) \right ) - \left (\dfrac{1}{2} \dot y^2(t_0) + \dfrac{1}{2} \kappa y^2(t_0) \right )$ $= \displaystyle \int_{t_0}^{t_1} \dfrac{d}{ds} \left ( \dfrac{1}{2} \dot y^2(s) + \dfrac{1}{2} \kappa(s) y^2(s) \right ) \; ds $ $= \displaystyle \int_{t_0}^{t_1} \left (\dfrac{1}{2} \dot \kappa(s) y^2(s) -\gamma(s) \dot y^2(s) \right ) \; ds; \tag{22}$

we see from this equation that we may ensure there are no periodic traectories provided we assume that $\dot \kappa$ and $\gamma$ are of opposite signs:

$\dot \kappa(t) \gamma(t) < 0, \tag{23}$

for in this event the integrand on the right of (22),

$\dfrac{1}{2} \dot \kappa(s) y^2(s) -\gamma(s) \dot y^2(s), \tag{24}$

will be of fixed sign, that of $\dot \kappa(t)$.

So to return to our OP Clark's query, "What needs to be enforced for a linear spring-mass-damper system to display a periodic trajectory?" In accord with (22) it is clear that a necessary condition is that the integral on the right of that equation,

$\displaystyle \int_{t_0}^{t_1} \left (\dfrac{1}{2} \dot \kappa(s) y^2(s) -\gamma(s) \dot y^2(s) \right ) \; ds = 0 \tag{25}$

over the interval $[t_0, t_1]$ on which the periodic orbit is defined; of course, this condition may not be sufficient, since it is concievable that

$\dfrac{1}{2} \dot y^2(t_1) + \dfrac{1}{2} \kappa(t_1) y^2(t_1) = \dfrac{1}{2} \dot y^2(t_0) + \dfrac{1}{2} \kappa(t)_0 y^2(t_0) \tag{26}$

without

$y(t_1) = y(t_0), \; \dot y_1(t_1) = \dot y_1(t_0), \tag{27}$

the condition for periodicity.

Finding sufficient conditions is much more difficult once one has dispensed with the simple case

$\gamma, \kappa = \text{constant}, \; \gamma = 0 \tag{28}$

mentioned ca. (14)-(16); indeed, there have been more than a few book chapters/journal articles devoted to this and similar problems. In this context a word about the "constraint" (23) may be in order: it may be possible, through judicious choice of $\kappa(t)$ and $\gamma(t)$, to cause the sign of (24) to alternate over the interval $[t_0, t_1]$ is such a manner that (25) applies, which allows the necessary condition (26) to remain in force, though of course as we change $\kappa(t)$ and $\gamma(t)$ the solution $y$ will generally change as well, so some care must be taken if such an approach to ensuring (26) is to be attempted. Perhaps further analysis can clarify the relationships between $\gamma(t)$, $\kappa(t)$, $\dot \kappa(t)$ and $y(t)$ which make periodic orbits for (1) not only possible but essential as well.

A Final, Final Note: One cannot help but notice that the two integrals occurring in (22) are each related to certain bilinear mappings defined on appropriate linear spaces of functions on intervals of the form $[t_0, t_1]$. For example, based upon (22), (24) we might define

$S(y_1(s), y_2(s)) = \dfrac{1}{2} \dot \kappa(s) y_1(s)y_2(s) -\gamma(s) \dot y_1(s) \dot y_2(s), \tag{29}$

and

$R(y_1(s), y_2(s)) = \dfrac{d}{ds} \left ( \dfrac{1}{2} \dot y_1(s) \dot y_2(s) + \dfrac{1}{2} \kappa(s) y_1(s) y_2(s) \right ); \tag{30}$

it is easy to see that $S(y_1(s), y_2(s))$ and $R(y_1(s), y_2(s))$ are bilinear in $y_1(s)$ and $y_2(s)$, and that

$S(y(s), y(s)) = \dfrac{1}{2} \dot \kappa(s) y^2(s) -\gamma(s) \dot y^2(s), \tag{31}$

whist

$R(y(s), y(s)) = \dfrac{d}{ds} \left ( \dfrac{1}{2} \dot y(s) \dot y(s) + \dfrac{1}{2} \kappa(s) y(s) y(s) \right ). \tag{32}$

These two equations allow (22) to be written in the form

$\left ( \dfrac{1}{2} \dot y^2(t_1) + \dfrac{1}{2} \kappa y^2(t_1) \right ) - \left (\dfrac{1}{2} \dot y^2(t_0) + \dfrac{1}{2} \kappa y^2(t_0) \right )$ $= \displaystyle \int_{t_0}^{t_1} R(y(s), y(s)) \; ds = \displaystyle \int_{t_0}^{t_1} S(y(s), y(s)) \; ds. \tag{33}$

The introduction of the bilinear maps $R(y_1(s), y_2(s))$ and $S(y_1(s), y_2(s))$ and the integral relation (33), and its connection with periodic orbits, might lead us to suspect that the integrands/intagrals occurring there might play a deeper and more extensive role in understanding the solutions, periodic or not, to the time-varying system (1); but resolving such a question is beyond the scope of this answer, and will be saved for another question.