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Strictly damped homogeneous oscillator with $k > 0$ : every solution tends to zero

Statement

Let $m > 0$ , $c > 0$ and $k > 0$ , let $x : R \to R$ be a solution of the homogeneous model $m x^{''} + c x^{'} + k x = 0$ with time domain $I = R$ and $f \equiv 0$ (The damped driven oscillator model: second-order equation, phase-plane system and characteristic data, The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data), and write $Δ = c^{2} - 4 m k$ , $γ = c / (2 m)$ , $λ_{\pm} = (- c \pm \sqrt{Δ}) / (2 m)$ in case $Δ > 0$ and $λ = - c / (2 m)$ in case $Δ = 0$ . Then $x (t) \to 0$ as $t \to + \infty$ (Limits at $+ \infty$ and $- \infty$ , and infinite limits at a point); more precisely, with the coefficients $A, B$ of the solution in the notation of The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data:

if $Δ > 0$ : $λ_{+} < 0$ and $∣ x (t) ∣ \leq (∣ A ∣ + ∣ B ∣) e^{λ_{+} t}$ for all $t \geq 0$ ;
if $Δ = 0$ : $λ < 0$ and $∣ x (t) ∣ \leq (∣ A ∣ + ∣ B ∣ t) e^{λ t}$ for all $t \geq 0$ ;
if $Δ < 0$ : $γ > 0$ and $∣ x (t) ∣ \leq (∣ A ∣ + ∣ B ∣) e^{- γ t}$ for all $t \geq 0$ .

Facts & Assumptions

Given: $m > 0$ , $c > 0$ , $k > 0$ , a solution $x$ of the homogeneous model, and the notation $Δ, γ, λ_{\pm}, λ$ of the statement.

[F1]

The solution is one of the three explicit families of The homogeneous damped oscillator: solution classification by the discriminant and uniqueness for initial data, on all of $R$ , with the coefficients $A, B$ of the relevant case. [given]

[F2]

Signs: if $Δ > 0$ then $λ_{+} = \frac{- c + \sqrt{Δ}}{2 m} < \frac{- c + c}{2 m} = 0$ because $k > 0$ gives $Δ < c^{2}$ and hence $\sqrt{Δ} < c$ ; if $Δ = 0$ then $λ = - c / (2 m) < 0$ ; and $γ = c / (2 m) > 0$ . [F1, algebra]

[F3]

Ranges: $∣ \cos u ∣ \leq 1$ and $∣ \sin u ∣ \leq 1$ for every real $u$ (Signs, monotonicity intervals, and ranges of sine and cosine).

[F4]

Exponential limits: $\exp (u) \to 0$ as $u \to - \infty$ and $\exp (u) \to + \infty$ as $u \to + \infty$ (The exponential tends to $+ \infty$ at $+ \infty$ and to $0$ at $- \infty$ ); consequently for $λ < 0$ , $\exp (λ t) \to 0$ as $t \to + \infty$ . The reciprocal identity $e^{λ t} = 1 / e^{- λ t}$ and positivity hold by The exponential is positive and satisfies $\exp (- x) = 1 / \exp (x)$ . For $t > 0$ and $λ < 0$ , the exponential series has nonnegative terms, so $\exp (- λ t) \geq (- λ t)^{2} / 2$ and hence $0 \leq t e^{λ t} \leq \frac{2}{λ^{2} t}$ (The real exponential function and the number $e$ by a power series, A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum).

[F5]

Convergence to $0$ at $+ \infty$ is the quantified assertion of Limits at $+ \infty$ and $- \infty$ , and infinite limits at a point: for every $ε > 0$ there is $T$ with $∣ u (t) ∣ < ε$ for all $t > T$ .

[F6]

The real exponential is strictly increasing (The exponential function is strictly increasing), so $u \leq v$ implies $e^{u} \leq e^{v}$ .

Proof

technique · cases; read the three explicit solution families off the classification theorem, bound them with the sign and range facts, and take the limits

1.1F1F3F6

Bounds for the three families. [assume-case 1, assume-case 2, assume-case 3] By [F1], in case 1 the solution is $x (t) = A e^{λ_{+} t} + B e^{λ_{-} t}$ with $λ_{-} < λ_{+}$ ; since $t \geq 0$ and $λ_{-} < λ_{+}$ give $λ_{-} t \leq λ_{+} t$ , [F6] gives $e^{λ_{-} t} \leq e^{λ_{+} t}$ , the triangle inequality yields $∣ x (t) ∣ \leq (∣ A ∣ + ∣ B ∣) e^{λ_{+} t}$ . In case 2, $∣ x (t) ∣ = ∣ A + B t ∣ e^{λ t} \leq (∣ A ∣ + ∣ B ∣ t) e^{λ t}$ . In case 3, $∣ x (t) ∣ = e^{- γ t} ∣ A \cos (ω t) + B \sin (ω t) ∣ \leq (∣ A ∣ + ∣ B ∣) e^{- γ t}$ by [F3].

2.1step 1.1F2F4F5

The three majorants tend to zero at $+ \infty$ . By [F2] the exponents $λ_{+}, λ, - γ$ are negative, so $e^{λ_{+} t} \to 0$ and $e^{- γ t} \to 0$ by [F4]. For the middle case, [F4] gives $0 \leq t e^{λ t} \leq 2 / (λ^{2} t) \to 0$ : given $ε > 0$ , every $t > 2 / (λ^{2} ε)$ has $t e^{λ t} < ε$ ; hence $t e^{λ t} \to 0$ and therefore $(∣ A ∣ + ∣ B ∣ t) e^{λ t} = ∣ A ∣ e^{λ t} + ∣ B ∣ t e^{λ t} \to 0$ .

3.1cases-exhaustive: the three discriminant casesstep 1.1step 2.1F5∎

Since $∣ x (t) ∣$ is bounded above in each case by the corresponding majorant, and all three majorants tend to $0$ , [F5] gives $x (t) \to 0$ as $t \to + \infty$ in each case; the estimates are those displayed in clause 1.

Strictly damped homogeneous oscillator with $k > 0$ : every solution tends to zero

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Strictly damped homogeneous oscillator with k>0k>0: every solution tends to zero

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources

Strictly damped homogeneous oscillator with $k > 0$ : every solution tends to zero