One-dimensional conservative and dissipative models — Examples

Declare the harmonic model of The harmonic (linear restoring) model: a declared linear force law with parameter $k>0$ with the data $m=1\mathrm{k}\mathrm{g},k=4\mathrm{N}\mathrm{/}\mathrm{m}=4\mathrm{k}\mathrm{g}\mathrm{/}{\mathrm{s}}^2,$ time window $\mathbb{R}$ and configuration domain $\mathbb{R}$, and the datum at $t_0=0$ $x_0=0\mathrm{m},v_0=1\mathrm{m}\mathrm{/}\mathrm{s}\mathrm{.}$ The claim checked below is that the curve $x_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}(t_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}})=\frac{1}{2}\sin (2t_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\mathrm{/}(1\mathrm{s}))\mathrm{m}(t_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\in \mathbb{R})$ is the unique maximal motion of the model with this datum, that its least positive period is $\pi$ seconds, that its energy is the constant $\frac{1}{2}$ joule, and that its phase curve is the ellipse $4x^2+v^2=1$ in the numerical SI phase coordinates $x:=x_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\mathrm{/}(1\mathrm{m})$ and $v:=v_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\mathrm{/}(1\mathrm{m}\mathrm{/}\mathrm{s})$ of The phase plane of a one-dimensional model: phase curves, the energy function and its level sets, and turning points. All coefficient calculations below use these numerical coordinates and $t:=t_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\mathrm{/}(1\mathrm{s})$; equivalently the physical ellipse is $4(x_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\mathrm{/}(1\mathrm{m}){)}^2+(v_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\mathrm{/}(1\mathrm{m}\mathrm{/}\mathrm{s}){)}^2=1$.

• What the check exhibits. The data produce the motion $\frac{1}{2}\sin (2t)$ with exact period $\pi$ s, energy $\frac{1}{2}$ J and phase ellipse $4x^2+v^2=1$; the check exercises clause 1-4 of The motions of the harmonic model are the sinusoids $A\cos (\omega (t-t_0))+B\sin (\omega (t-t_0))$: amplitude and phase, the least period, the conserved energy $k{R}^2\mathrm{/}2$ and the elliptical phase curves and the conservation and level-set statements of Existence, uniqueness and conservation of energy for a one-dimensional conservative model: phase curves lie in energy level sets and the velocity is determined by the position and energy up to sign on one motion.

• Scope. The values $m=1$ kg and $k=4$ N/m are declared data of this computation; no measured value is asserted, and no real spring is claimed to obey the linear law in any regime.

Declare the damped oscillator model of The damped oscillator model: a harmonic restoring force with a phenomenological linear drag $-c\dot{x}$, $c\ge 0$ with $m=1\mathrm{k}\mathrm{g},k=1\mathrm{N}\mathrm{/}\mathrm{m},$ and the common datum at $t_0=0$ $x_0=1\mathrm{m},v_0=0\mathrm{m}\mathrm{/}\mathrm{s},$ for each of the three drag coefficients $c=1,c=2,c=5(\mathrm{k}\mathrm{g}\mathrm{/}\mathrm{s})\mathrm{.}$ In displayed numerical formulas, $t:=t_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\mathrm{/}(1\mathrm{s})$, $x:=x_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\mathrm{/}(1\mathrm{m})$, and $x^{\mathrm{\prime }}(t)$ is the numerical velocity in metres per second; model parameters and rates are their SI numerical values. Energy values are numerical joules, so the numerical energy identity is $E^{\mathrm{\prime }}(t)=-c{x}^{\mathrm{\prime }}(t{)}^2$ with these conventions. The claims checked below are that the three displayed curves are the unique maximal motions with this datum in the three cases, that all three have initial energy $\frac{1}{2}$ J and satisfy $E^{\mathrm{\prime }}(t)=-c{x}^{\mathrm{\prime }}(t{)}^2$, and that the decay bounds of Strictly damped oscillator: explicit decay bounds, convergence to rest and asymptotic stability of the origin hold with the displayed constants:

$\begin{array}{rl}{\displaystyle c=1:}& {\displaystyle x(t)=e^{-t\mathrm{/}2}(\cos \frac{\sqrt{3}t}{2}+\frac{1}{\sqrt{3}}\sin \frac{\sqrt{3}t}{2}),}\\ {\displaystyle c=2:}& {\displaystyle x(t)=(1+t)e^{-t},}\\ {\displaystyle c=5:}& {\displaystyle x(t)=\frac{5+\sqrt{21}}{2\sqrt{21}}{e}^{\frac{-5+\sqrt{21}}{2}t}+\frac{\sqrt{21}-5}{2\sqrt{21}}{e}^{\frac{-5-\sqrt{21}}{2}t}\mathrm{.}}\end{array}$

• The three regimes. The discriminant test $\mathrm{\Delta }<0$, $\mathrm{\Delta }=0$, $\mathrm{\Delta }>0$ separates the decaying oscillation ($\omega =\sqrt{3}\mathrm{/}2$, quasi-period $4\pi \mathrm{/}\sqrt{3}$ s), the critical case ($x=(1+t)e^{-t}$, no oscillation) and the overdamped case (two decaying exponentials with rates $(-5\pm \sqrt{21})\mathrm{/}2$). The numerical evaluations above are decimal approximations of the explicit expressions, quoted to display the order of magnitude; the exact inequalities are the displayed bounds.

• Scope. The data $m=k=1$, $c=1,2,5$ are declared model data of this check; no measured value is asserted and no real damper is claimed to obey the linear drag law in any regime.

Declare the sinusoidally driven oscillator model of The sinusoidally driven oscillator model: a prescribed external force $f_0\cos (\mathrm{\Omega }t)$ added to the damped oscillator with $m=1\mathrm{k}\mathrm{g},k=1\mathrm{N}\mathrm{/}\mathrm{m},f_0=1\mathrm{N},c=\frac{1}{5}\mathrm{k}\mathrm{g}\mathrm{/}\mathrm{s},\mathrm{\Omega }=1{\mathrm{s}}^{-1},$ so that the physical natural frequency is $\sqrt{k\mathrm{/}m}=1{\mathrm{s}}^{-1}$. In the displayed calculations, $t$ denotes the numerical SI time $t_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\mathrm{/}(1\mathrm{s})$, and position formulas give numerical values in metres. Unless a physical unit is explicitly attached, parameters in those calculations also mean SI numerical values: $m,k,c,f_0$ are normalized by $1\mathrm{k}\mathrm{g}$, $1\mathrm{N}\mathrm{/}\mathrm{m}$, $1\mathrm{k}\mathrm{g}\mathrm{/}\mathrm{s}$ and $1\mathrm{N}$, respectively, and $\mathrm{\Omega },{\omega }_0,{\omega }_d,\gamma$ mean physical rates divided by $1{\mathrm{s}}^{-1}$. Thus the numerical drive is $f(t)=\cos t$, and derivatives of numerical position with respect to $t$ give numerical velocity in metres per second. The claims checked below are:

1. the steady response is $x_p(t)=5\sin t$ metres, a motion on $\mathbb{R}$ with amplitude $A(1)=5$ m and phase lag $\delta =\pi \mathrm{/}2$;
2. the amplitude function $A(\mathrm{\Omega })=1\mathrm{/}\sqrt{(1-{\mathrm{\Omega }}^2{)}^2+{\mathrm{\Omega }}^2\mathrm{/}25}$ attains its maximum at ${\mathrm{\Omega }}_{\ast }=\sqrt{0.98}=\sqrt{49\mathrm{/}50}$ with the value $A({\mathrm{\Omega }}_{\ast })=10\mathrm{/}\sqrt{3.96}$, larger than $A(1)=5$;
3. the motion with datum $(0,0,0)$ is $x(t)=5\sin t-\frac{5}{{\omega }_d}{e}^{-t\mathrm{/}10}\sin ({\omega }_{d}t),{\omega }_d=\sqrt{0.99}=\sqrt{99}\mathrm{/}10,$ which is bounded on $[0,\mathrm{\infty })$ and satisfies $x(t)-x_p(t)\to 0$ and $x^{\mathrm{\prime }}(t)-x_p^{\mathrm{\prime }}(t)\to 0$ as $t\to +\mathrm{\infty }$;
4. at $c=0$ the model with the same drive at $\mathrm{\Omega }=1$ has the unbounded motion $t\sin t\mathrm{/}2$, while at $\mathrm{\Omega }=2$ every motion is bounded with amplitude $1\mathrm{/}3$ m for the steady response.

• What the check exhibits. At the drive frequency equal to the natural frequency of the undamped model, the damped model's steady amplitude is $5$ m, and the peak of the amplitude curve is at ${\mathrm{\Omega }}_{\ast }=\sqrt{0.98}$ with value $10\mathrm{/}\sqrt{3.96}\approx 5.025$ m; the transient from rest dies with rate $1\mathrm{/}10$ s${}^{-1}$. The dimensionless ratio $A({\omega }_0)\mathrm{/}A(0)=5$ is ${\omega }_0\mathrm{/}(2\gamma )$ in the notation of the source's quality factor. All decimal numbers here are approximations of the exact expressions displayed.

• Scope. The data $m=k=f_0=1$, $c=1\mathrm{/}5$, $\mathrm{\Omega }=1$ are declared model data of this check; no measured value is asserted and no apparatus is claimed to be described by the model.

Adopt the ideal planar pendulum angle-chart equation as an effective physical model, using the mathematical constant-metric chart of The mechanical first-order field, Lagrangian and energy of a configuration-dependent kinetic metric in a chart. This angular chart model is not an instance of the one-dimensional translational model class of The one-dimensional force models: data, motions, phase system, potential presentation, energy and equilibria: $\theta$ is a dimensionless angle and its generalized inertia has dimension mass times length squared. The chart data are configuration domain $\mathbb{R}$, coordinate $\theta$ (the angle from the downward vertical, in radians), effective mass parameter $m_{\mathrm{e}\mathrm{f}}:=m{l}^2>0,\text{force law}F(\theta ):=-mgl\sin \theta ,$ and constants $m,l,g>0$ of dimensions mass, length and length per time squared (Dimensions and SI unit conventions of the non-relativistic classical mechanics framework). The force $F$ is a generalized torque. Its equation of motion is $m{l}^2{\theta }^{\mathrm{\prime }\mathrm{\prime }}=-mgl\sin \theta ,\text{equivalently}{\theta }^{\mathrm{\prime }\mathrm{\prime }}=-\frac{g}{l}\sin \theta ,$ the recorded source's point-pendulum equation (Eq. (1.18) there) expressed with the effective inertia $m_{\mathrm{e}\mathrm{f}}=m{l}^2$ and generalized torque $F=-mgl\sin \theta$. The potential $U=-mgl\cos \theta$ is an energy-valued scalar function of the dimensionless angle. The stability and linearization calculations use the autonomous scalar equation as a mathematical chart system with constant coefficient $M=m_{\mathrm{e}\mathrm{f}}$. All phase norms mean the Euclidean norm of the numerical pair $(\theta ,(1\mathrm{s}){\theta }^{\mathrm{\prime }})$; the normalized time variable used in the finite-time estimate is $s=t_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\mathrm{/}(1\mathrm{s})$. The claims checked below are:

1. the generalized torque is $-U^{\mathrm{\prime }}(\theta )$ for $U(\theta )=-mgl\cos \theta$, and $\theta =0$ is a strict local minimum;
2. the hanging rest point $(\theta ,{\theta }^{\mathrm{\prime }})=(0,0)$ is Lyapunov stable, and the linearised angular equation is $m_{\mathrm{e}\mathrm{f}}{\theta }^{\mathrm{\prime }\mathrm{\prime }}+mgl\theta =0$ with frequency $\omega =\sqrt{g\mathrm{/}l}$;
3. at $g=1\mathrm{m}\mathrm{/}{\mathrm{s}}^2$, $l=1\mathrm{m}$, use $s=t_{\mathrm{p}\mathrm{h}\mathrm{y}\mathrm{s}}\mathrm{/}(1\mathrm{s})$ and $z=(\theta ,d\theta \mathrm{/}ds)$. For every initial phase vector $z_0$ with $\mathrm{\parallel }{z}_0{\mathrm{\parallel }}_2=\delta =1\mathrm{/}100$, radius $r=1\mathrm{/}2$ and normalized horizon $T=1$, write $t$ for the numerical time $s$ in the bound. Then the difference between the exact motion and the harmonic motion with the same datum obeys $\parallel z(t)-y(t){\parallel }_2\le \frac{t{e}^{3t}}{12\cdot {10}^4}(0\le t\le 1),\parallel z(1)-y(1){\parallel }_2\le \frac{e^3}{12\cdot {10}^4}\approx 1.67\cdot {10}^{-4},$ the source-supported bound of The pendulum field near the downward equilibrium: bounded force derivative, cubic remainder and the finite-time small-angle error bound;
4. the quadratic energy of the linearised angular equation, $E_{\mathrm{q}\mathrm{u}\mathrm{a}\mathrm{d}}:=\frac{1}{2}{m}_{\mathrm{e}\mathrm{f}}{\theta }^{\mathrm{\prime }2}+\frac{1}{2}mgl{\theta }^2$, is not conserved along exact motions, with $\frac{d}{dt}[\frac{1}{2}{m}_{\mathrm{e}\mathrm{f}}{\theta }^{\mathrm{\prime }2}+\frac{1}{2}mgl{\theta }^2]=mgl{\theta }^{\mathrm{\prime }}(\theta -\sin \theta ),$ which is positive at the phase point $(\theta ,{\theta }^{\mathrm{\prime }})=(\pi \mathrm{/}2,1{\mathrm{s}}^{-1})$.