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An explicit weakly damped oscillator: the averaging constants and the energy-decay bound over five thousand seconds (about 796 undamped periods)

Physical scope

A worked numerical instance of the weak-damping averaging estimate: the model m x'' + epsilon gamma x' + k x = 0 with m = 1 kg, k = 1 kg/s^2, epsilon = 10^-4 and gamma = 1 kg/s, initial rotating-frame amplitude zeta_0 = 1, ball radii R = 1/2 and rho = 1/4 and horizon parameter A = 1/2, verifying the smallness and trapping hypotheses with the exact constants M = 3 sqrt2/2, L = sqrt2, K = 2 pi (3 sqrt2 + 6) exp(sqrt2/2) and concluding that the mechanical energy tracks the averaged exponential exp(-t/10^4) within 2.63 x 10^-2 on the interval [0,5000]. The numbers are exact-arithmetic evaluations of the model formulas, not measurements; no apparatus, no observed decay and no empirical statement is present.

Example

Adopt the damped oscillator model of The damped oscillator model: a harmonic restoring force with a phenomenological linear drag $- c \dot{x}$ , $c \geq 0$ with data

$m = 1 kg, k = 1 {kg/s}^{2}, ε = 10^{- 4}, γ = 1 kg/s,$

so that $c = ε γ = 10^{- 4}$ kg/s and $ω_{0} = \sqrt{k / m} = 1$ rad/s; the angles and frequencies are read under the convention of Dimensions and SI unit conventions of the non-relativistic classical mechanics framework. Fix amplitude unit $a_{*} = 1 m s^{- 1 / 2}$ and time unit $1 s$ . Take $ζ_{0} = a_{*}$ , $R = a_{*} / 2$ , $ρ = a_{*} / 4$ and $A = (1 s) / 2$ . Below these symbols, $t$ and the constants in coefficient calculations mean their numerical values in these fixed units. In particular $M$ is in $a_{*} / s$ , $L$ in $s^{- 1}$ , $B, K$ in $a_{*}$ and $B_{1}$ is dimensionless. Every energy exponential below is multiplied by $E_{0} = 1 J$ when read as a physical quantity, and the horizon $[0, 5000]$ means $[0, 5000 s]$ . For $x > 0$ , write $\ln x : = \log x$ , where $\log : (0, \infty) \to R$ is the inverse of the real exponential (The natural logarithm as the inverse of the exponential function). Then:

the averaging data of Averaging for a weakly damped oscillator: the rotating-frame amplitude, the averaged exponential energy decay and the order- $ε$ bound on the $O (1 / ε)$ time scale specialize to $P = 2 π$ , $F (t, ζ) = - i e^{i t} (\cos t Im ζ - \sin t Re ζ)$ and $\overset{ˉ}{F} (ζ) = - ζ / 2$ , with averaged solution $\overset{ˉ}{ζ} (t) = e^{- ε t / 2} = e^{- t / (2 \cdot 10^{4})}$ ;
the constants of the averaging estimate are $M = \frac{3 \sqrt{2}}{2}, L = \sqrt{2}, B = 6 \sqrt{2} π, B_{1} = 4 \sqrt{2} π, K = 2 π (3 \sqrt{2} + 6) e^{\sqrt{2} / 2},$ so that $K \approx 1.30522 \times 10^{2}, ε K \approx 1.30522 \times 10^{- 2} < 0.25 = ρ;$
the trapping hypothesis holds: $∣ ζ_{0} ∣ (1 - e^{- γ A / (2 m)}) = 1 - e^{- 1 / 4} \approx 0.22120 < R - ρ = 0.25$ , and also $1 / 4 = γ A / (2 m) \leq \ln (4 / 3) \approx 0.28768$ ;
consequently the motion with datum $x (0) = \sqrt{2}$ m, $x^{'} (0) = 0$ — which has $ζ (0) = 1$ and mechanical energy $E_{0} = m ω_{0} ∣ ζ_{0} ∣^{2} = 1$ J — satisfies $∣ E (t) - e^{- t / 10^{4}} ∣ \leq m ω_{0} ε K (2 ∣ ζ_{0} ∣ + ε K) \approx 2.62748 \times 10^{- 2} J (0 \leq t \leq 5000),$ the interval $[0, 5000]$ being $[0, A / ε]$ and containing $5000 / (2 π) \approx 795.8$ oscillation periods of the undamped frequency;
at the end of the horizon the averaged energy is $e^{- 1 / 2} \approx 0.60653$ J, so the estimate bounds the deviation of the exact energy from the averaged exponential decay by about $4.3 %$ of its value there.

Facts & Assumptions

Given: The model data $m = 1$ , $k = 1$ , $ε = 10^{- 4}$ , $γ = 1$ ; the numbers $ζ_{0} = 1$ , $R = 1 / 2$ , $ρ = 1 / 4$ , $A = 1 / 2$ ; the constants $M, L, B, B_{1}, K$ of the Example.

[F1]

Averaging theorem (Averaging for a weakly damped oscillator: the rotating-frame amplitude, the averaged exponential energy decay and the order- $ε$ bound on the $O (1 / ε)$ time scale clauses 1-4): for the model with data $(m, ε γ, k)$ the rotating-frame amplitude maps motions bijectively onto solutions of $ζ^{'} = ε F (t, ζ)$ ; $F$ is $P = 2 π / ω_{0}$ -periodic with average $\overset{ˉ}{F} (ζ) = - γ ζ / (2 m)$ ; under the hypotheses $ε K < ρ$ and the trapping condition, the estimate $∥ ζ (t) - \overset{ˉ}{ζ} (t) ∥_{2} \leq ε K$ and the energy estimate $∣ E (t) - E_{0} e^{- ε γ t / m} ∣ \leq m ω_{0} ε K (2 ∣ ζ_{0} ∣ + ε K)$ hold on $[0, A / ε]$ , with $M = (γ / m) \sqrt{2} (∣ ζ_{0} ∣ + R)$ , $L = (γ / m) \sqrt{2}$ , $B = 2 M P$ , $B_{1} = 2 L P$ and $K = e^{L A} [B + (L B + B_{1} M) A]$ .

[F2]

Data specialization (Periodic averaging data: the average field, the oscillatory part and the periodic deviation, Real and imaginary parts, complex conjugation, and modulus): with $ω_{0} = 1$ and $γ / m = 1$ the field of the theorem reads $F (t, ζ) = - i e^{i t} (\cos t Im ζ - \sin t Re ζ)$ and $P = 2 π$ ; $∣ ζ_{0} ∣ = 1$ .

[F3]

Elementary arithmetic (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The derivatives of sine and cosine are cosine and minus sine): the displayed decimal values are evaluations of the exact expressions, obtained by direct arithmetic from $\sqrt{2} \approx 1.41421356$ , $π \approx 3.14159265$ and $e^{\sqrt{2} / 2} \approx 2.02811498$ ; in particular the datum with $ζ_{0} = 1$ corresponds to $x (0) = \sqrt{2}$ m, $x^{'} (0) = 0$ and $E_{0} = m ω_{0} ∣ ζ_{0} ∣^{2} = 1$ J.

[F4]

Exact numerical inequalities: $\sqrt{2} < 3 / 2$ by squaring and $2 < e < 3$ by The elementary numerical bound $2 < e < 3$ . For $x \in [1 / 4, 3 / 4]$ , $x (1 - x) \geq 3 / 16$ and $1 + x^{2} \leq 25 / 16$ , so $\frac{x^{4} (1 - x)^{4}}{1 + x^{2}} \geq \frac{16}{25} (\frac{3}{16})^{4} > 0.$ Polynomial division gives $\frac{x^{4} (1 - x)^{4}}{1 + x^{2}} = x^{6} - 4 x^{5} + 5 x^{4} - 4 x^{2} + 4 - \frac{4}{1 + x^{2}} .$ By Principal arctangent: derivative, integral, power series, and the Gregory–Leibniz series, $\int_{0}^{1} d t / (1 + t^{2}) = \arctan 1$ . Also $\arctan 1 = π / 4$ : the principal inverse tangent of The principal inverse tangent $\arctan : R \to (- π / 2, π / 2)$ inverts $\tan$ on $(- π / 2, π / 2)$ ; the cofunction identity of Cofunction, supplementary, quarter-turn, and reflection identities for the six trigonometric functions at $x = π / 4$ gives $\sin (π / 4) = \cos (π / 4)$ , hence $\tan (π / 4) = 1$ by Tangent, cotangent, secant, and cosecant on their exact natural domains. Termwise integration of the polynomial using For a natural $n \geq 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $ι (n) x^{n - 1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \geq 1$ the function $x \mapsto x^{- n}$ is differentiable at every $x \neq 0$ with derivative $- ι (n) x^{- n - 1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term and Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ now gives $\int_{0}^{1} \frac{x^{4} (1 - x)^{4}}{1 + x^{2}} d x = \frac{22}{7} - 4 \arctan 1 = \frac{22}{7} - π .$ The integrand is nonnegative on $[0, 1]$ , and its integral on $[1 / 4, 3 / 4]$ is at least $\frac{1}{2} \cdot \frac{16}{25} (\frac{3}{16})^{4} > 0$ by If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ , so the full integral is positive and $π < 22 / 7$ . Also, for $x > 0$ , write $\ln x : = \log x$ as in The natural logarithm as the inverse of the exponential function. Then $\ln (4 / 3) = \int_{1}^{4 / 3} d t / t \geq \frac{3}{4} \cdot \frac{1}{3} = \frac{1}{4}$ by The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t and integral monotonicity. For $u > 0$ , the fundamental theorem and exponential derivative give $1 - e^{- u} = \int_{0}^{u} e^{- s} d s < u$ : on $[0, u / 2]$ the integrand is at most $1$ , and on $[u / 2, u]$ it is at most $e^{- u / 2} < 1$ by strict increase of the exponential (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ , The exponential function is smooth and $(\exp)^{'} = \exp$ , The exponential function is strictly increasing, If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ ). Finally, $e^{- 1 / 2} > 1 / 3$ follows from $e^{1 / 2} < e < 3$ and $e^{- x} = 1 / e^{x}$ (The elementary numerical bound $2 < e < 3$ , The exponential function is strictly increasing, The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ ).

Verification

technique · direct; specialize the theorem's data, evaluate the constants, check the two explicit hypotheses, and state the conclusion

1.1F1F2algebra

Specialization: with $m = k = γ = 1$ and $ε = 10^{- 4}$ the quantities of [F1] become $ω_{0} = 1$ , $P = 2 π$ , $\overset{ˉ}{F} (ζ) = - ζ / 2$ and $\overset{ˉ}{ζ} (t) = e^{- ε t / 2} = e^{- t / (2 \cdot 10^{4})}$ ; the field is the one displayed in clause 1.

1.2F1F4givenalgebra

Constants: substituting $∣ ζ_{0} ∣ = 1$ , $R = 1 / 2$ , $γ / m = 1$ gives $M = \sqrt{2} \cdot \frac{3}{2} = \frac{3 \sqrt{2}}{2}$ and $L = \sqrt{2}$ , hence $B = 2 M P = 6 \sqrt{2} π$ and $B_{1} = 2 L P = 4 \sqrt{2} π$ . Then $L B = \sqrt{2} \cdot 6 \sqrt{2} π = 12 π$ and $B_{1} M = 4 \sqrt{2} π \cdot \frac{3 \sqrt{2}}{2} = 12 π$ , so $K = e^{\sqrt{2} / 2} [6 \sqrt{2} π + \frac{1}{2} (12 π + 12 π)] = 2 π (3 \sqrt{2} + 6) e^{\sqrt{2} / 2},$ and by [F4] $K < 198 < 200$ : $\sqrt{2} < 3 / 2$ and $e^{\sqrt{2} / 2} < e < 3$ give $K < 2 π \cdot (3 \cdot \frac{3}{2} + 6) \cdot 3 = 63 π < 63 \cdot 22 / 7 = 198$ . This is clause 2.

2.1F1F4step 1.2givenalgebra

Hypotheses: $ε K < 198 / 10^{4} < 1 / 50 < 1 / 4 = ρ$ by step 1.2, so the smallness condition holds; at the full horizon $A / ε = 5000$ the trapping quantity is $∣ ζ_{0} ∣ (1 - e^{- γ A / (2 m)}) = 1 - e^{- 1 / 4} < 1 / 4 = R - ρ$ by [F4]. The sufficient logarithmic condition also holds, since $γ A / (2 m) = 1 / 4 \leq \ln (4 / 3)$ by [F4]. Hence the averaged solution stays in the required ball. This is clause 3.

3.1F1F3F4step 1.1step 1.2step 2.1algebra

Conclusion: applying [F1] with the verified hypotheses and the horizon $A / ε = 5000$ gives $∥ ζ (t) - \overset{ˉ}{ζ} (t) ∥_{2} \leq ε K < 1 / 50$ for $0 \leq t \leq 5000$ , hence the energy bound $∣ E (t) - e^{- t / 10^{4}} ∣ \leq ε K (2 + ε K) < \frac{101}{2500} < \frac{1}{24}$ J; the datum corresponds to $x (0) = \sqrt{2}$ m, $x^{'} (0) = 0$ , $E_{0} = 1$ J, and the horizon contains more than $795$ periods of the undamped frequency, because $5000 > 1590 π$ and [F4] gives $1590 π < 1590 \cdot \frac{22}{7} = \frac{34980}{7} < 5000$ ; at $t = 5000$ the averaged energy is $e^{- 1 / 2} > 1 / 3$ by [F4], so the bound is below one eighth of it. This is clauses 4 and 5.

4.1step 1.1step 1.2step 2.1step 3.1∎

Clauses 1-5 are verified by exact computation from the formulas of the averaging theorem, with the rational and elementary estimates above in place of decimal arithmetic; no measured quantity, apparatus or empirical claim is present.

Remarks

Decimal orientation values. On the stated instance, $K \approx 1.30522 \times 10^{2}$ , $ε K \approx 1.30522 \times 10^{- 2}$ and the energy bound is approximately $2.63 \times 10^{- 2}$ J; the averaged energy at $t = 5000$ is approximately $0.60653$ J. These decimals are evaluations of the exact expressions of the proof and play no role in it.
Reading the constant. The dominating factor in $K$ is $e^{L A} = e^{\sqrt{2} / 2}$ , below $3$ , because the horizon parameter $A = 1 / 2$ is small: the estimate is useful for horizons of order one in the slow time $A$ , and $K$ grows like $A e^{L A}$ with $A$ . Taking $A$ larger lengthens the interval $[0, A / ε]$ but increases both $K$ and the required smallness margin.
The finite check is arithmetic, not empirical. Every number above is computed from the model formulas; nothing in the example depends on an experiment, and the estimate is a conditional bound for the declared model.

An explicit weakly damped oscillator: the averaging constants and the energy-decay bound over five thousand seconds (about 796 undamped periods)

Example

Facts & Assumptions

Verification

Remarks

Depends on

Dependency tree · 2 levels

Sources