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Critical damping is the repeated-root case of the viscous damping model: its two characteristic roots coincide at $r=-{\omega }_0$. The solution with arbitrary initial conditions isThe extra factor $t$ is essential: one exponential function alone cannot fit arbitrary initial displacement and velocity when the characteristic equation has a double polynomial root. Substitution, or taking the limit of the underdamped formula as its angular frequency tends to zero, gives the displayed expression.

The precise fastest-return statement is this: hold $m$, $k$, $x_0>0$ and $v_0=0$ fixed, and compare all $\zeta \ge 1$. Every response decreases positively toward equilibrium; the critically damped responseis smaller than every overdamped response at each $t>0$. It therefore reaches and stays within any specified band $|x|\le ϵx_0$, $0<ϵ<1$, sooner than those responses. Exact arrival at rest takes infinite time. This is a comparison among nonoscillatory regimes for these particular initial conditions, not a universal minimum of every possible settling-time criterion.

One short comparison argument uses $q=\partial X/\partial \zeta$. Its initial conditions are zero and $q^{\prime \prime }+2\zeta q^{\prime }+q=-2X^{\prime }$. For release from rest, $X^{\prime }<0$ at $\tau >0$; the causal Green function of a damped oscillator is positive at positive time for $\zeta \ge 1$. Its convolution with $-2X^{\prime }$ gives $q>0$: increasing damping beyond the critical value increases the remaining displacement.

There is no repeated oscillation at critical damping, but there can be overshoot once. For $x_0>0$, the factor in square brackets crosses zero precisely when $v_0<-{\omega }_0{x}_0$, atFor example, $v_0=-2{\omega }_0{x}_0$ gives $x=x_0(1-{\omega }_{0}t)e^{-{\omega }_{0}t}$, crossing equilibrium at $t=1/{\omega }_0$ as shown above. An overdamped response can also cross once for suitable initial conditions. Conversely, specially chosen initial conditions can remove its slow exponential decay entirely. Thus the generic slow-root argument must not be applied to a response whose slow-root coefficient is zero.

For $x^{\prime \prime }+2\gamma x^{\prime }+({\gamma }^2+{\mu }^2)x=g(t)$ with $\gamma ,\mu >0$, zero initial data give $x(t)={\mu }^{-1}{\int }_0^t{e}^{-\gamma (t-s)}\sin (\mu (t-s))g(s)ds$. If $|g(s)|\le C{e}^{-as}$ for $a>0$, the causal Green function of a damped oscillator givesBoth bounds tend to zero. A decaying homogeneous transient can be added for arbitrary finite initial data. The coincident decay rates require the extra polynomial factor and cannot be handled by dividing by zero in the first expression.

For $\beta >0$, $0<\gamma <{\omega }_0$ and zero prehistory, the equation $y^{\prime \prime }+2\gamma y^{\prime }+{\omega }_0^{2}y=F{e}^{-\beta t}H(t)$ hasOne obtains it by convolution with the causal Green function of a damped oscillator. Alternatively, a particular exponential has amplitude $F/[(\beta -\gamma {)}^2+{\Omega }^2]$; the two homogeneous terms enforce $y(0)=y^{\prime }(0)=0$. These continuity conditions avoid adding impulsive forcing at switch-on. The forcing may have either sign; its Fourier transform is $-iF/(\omega -i\beta )$ with the negative-exponential transform convention.

The step response of a damped oscillator is the time integral of its causal Green function of a damped oscillator. Put $s=t-b$ and define $K(s)={\int }_0^s{e}^{-ku}\sin (\omega u)/\omega du$ for $s\ge 0$. Then $K(0)=K^{\prime }(0)=0$ and its oscillator equation has constant right side $1$. Integrating explicitly givesThis formula applies when $\omega \ne 0$. It is a convolution of the causal impulse response with the shifted Heaviside step function, so the response is zero before the forcing starts and has the stated zero initial data.

The integral definition also handles the degenerate parameters without division ambiguities. For $\omega =0,k\ne 0$,for $\omega =k=0$, $y_3=H(s)s^2/2$. These are the continuous zero-frequency limits of the step response.

Before the impulse the zero initial data force $y_2=0$. An impulse modeled by the Dirac delta function requires continuity of $y_2$: a jump in $y_2$ would introduce an unwanted derivative of a delta in $y_2^{\prime \prime }$. Integrating the equation across $t=a$ then gives the velocity jump $y_2^{\prime }(a+)-y_2^{\prime }(a-)=1$.

The causal Green function of a damped oscillator isThe jump conditions give the causal impulse responseHere $H$ is the Heaviside step function. For $\omega =0$, interpret the quotient continuously as $t-a$, giving $H(t-a)(t-a)e^{-k(t-a)}$. Both initial values are zero because $a>0$.

Take $x\ge 0$ and a source position $\xi >0$. Before the source, the zero initial data require $G=0$. After it, set $G=e^{-2(x-\xi )}h(x-\xi )$; the homogeneous equation becomes $h^{\prime \prime }+a^{2}h=0$. Continuity of $G$ and a unit jump in $G_x$ at $x=\xi$ give $h(0)=0$, $h^{\prime }(0)=1$. Thus the causal Green function of a damped oscillator isHere $H$ is the Heaviside step function. For $a=0$, take the continuous limit $\sin (as)/a\to s$, giving $G=H(x-\xi )(x-\xi )e^{-2(x-\xi )}$. The formula is also unchanged when a nonzero $a$ changes sign. Its continuity removes a delta derivative, and its unit first-derivative jump supplies the required Dirac delta function.

For the forced equation use $a=2$. The zero-initial-data Green function representation isThe sine-product identity reduces this to elementary exponential-trigonometric integrals. Evaluating givesFor an independent check, the steady particular part is $\sin 2x/20-\cos 2x/10$, while the decaying homogeneous part has coefficients $1/10$ and $1/20$ that impose $y(0)=y^{\prime }(0)=0$.

Use the zero initial displacement and velocity given in the PDF. The characteristic roots of the homogeneous equation are $-1\pm i\mu$. Its causal Green function of a damped oscillator iswhere $H$ is the Heaviside step function. The positive-time kernel has $G(0+)=0$ and $G^{\prime }(0+)=1$, so its derivative jump generates a unit impulse. The zero-data forced solution is consequently the causal convolutionDifferentiating once gives $x^{\prime }(0)=0$; differentiating twice supplies the endpoint term $g(t)$ and verifies the differential equation. No homogeneous transient is required with these initial data.

Use the Fourier transform convention $\tilde{y}(\omega )={\int }_{\mathbb{R}}y(t)e^{-i\omega t}dt$. For a transformable response with the necessary boundary terms vanishing, integration by parts givesTransforming the differential equation therefore gives $(5-{\omega }^2+2i\omega )\tilde{y}=\tilde{f}$, soThe identities also hold for tempered distributions. The denominator factors as $(\omega -i-2)(\omega -i+2)$; its two poles are in the upper half-plane, corresponding under inverse factor $e^{i\omega t}$ to the causal Green function of a damped oscillator.

The frequency-domain formula selects the Fourier-transformable response. Decay of the force alone would not remove arbitrary homogeneous solutions, which grow towards negative time and do not have an ordinary bilateral Fourier transform.

The result of part (i) vanishes identically on the open interval $t<0$, so its first and second derivatives also vanish there. Substituting into the equation of motion immediately givesThus the force has no negative-time prehistory, consistent with the causal Green function of a damped oscillator. In fact, direct inversion or the positive-time substitution above identifies $f(t)=-e^{-2t}H(t)$; the given numerator $+i$ corresponds to a negative exponential force with this Fourier convention. Here $H$ is the Heaviside step function; its value exactly at the jump does not affect the transform.

For $t>0$, close the inverse-transform contour in the upper half-plane. There $|e^{i\omega t}|\le 1$; the rational factor is $O(|\omega {|}^{-2})$, so the arc integral vanishes. For $\Omega >\kappa$, write $\beta =\sqrt{{\Omega }^2-{\kappa }^2}>0$. The two simple poles are $i\kappa \pm \beta$, and the residue theorem yieldsThis causal Green function of a damped oscillator has damped oscillations with envelope $e^{-\kappa t}$ and tends to zero as $t\to \infty$.

For the zero-data initial value problem, choose a causal Green function satisfying $L_{x}G(x,\xi )=\delta (x-\xi )$ and $G=0$ for $x<\xi$. Continuity at $x=\xi$ prevents a derivative of the Dirac delta distribution, while integration across the source givesFor $x>\xi$, part (i) gives the shifted homogeneous solutions. The zero value at the join removes the cosine term, and the unit derivative jump fixes the sine coefficient. Therefore the causal Green function of a damped oscillator isEquivalently, multiply the first expression by the Heaviside step function $H(x-\xi )$. The required solution isIts initial value and first derivative vanish: the interval has zero length at $x=0$ and the kernel vanishes at its upper limit. Differentiating twice shows that the upper-limit contribution to $y^{\prime \prime }$ is precisely $g(x)$, while the remaining homogeneous kernel terms cancel in $Ly$.

Let $\tau =t-t^{\prime }$ and $\Omega =\sqrt{k^2+m^2-{\lambda }^2}>0$. First take $\lambda =e\gamma >0$. Since $|e^{-i\omega \tau }|=e^{\tau \mathrm{Im}\omega }$, close the contour below for $\tau >0$, and above for $\tau <0$. On the chosen semicircle the exponential is bounded and the denominator grows as $R^2$, so the arc integral tends to zero. The upper contour encloses no pole, giving $G=0$ for $t<t^{\prime }$. The lower contour is clockwise, hence the real-axis integral divided by $2\pi$ is $-i$ times the sum of residues. Since $D^{\prime }({\omega }_{+})=-2\Omega$ and $D^{\prime }({\omega }_{-})=2\Omega$, that sum isThusThis is the causal Green function of a damped oscillator. Its value is continuous through $\tau =0$, and its derivative jumps from zero to one; integrating the differential equation through zero verifies the delta-source normalization and the residue sign. The limits are $H(\tau )\tau e^{-\lambda \tau }$ at critical damping and $H(\tau )e^{-\lambda \tau }\sinh (\kappa \tau )/\kappa$ in the overdamped regime.

For completeness, if the unspecified charge has $e<0$, the same real-axis integral instead encloses upper poles for $\tau <0$ and gives $-H(-\tau )e^{-\lambda \tau }\sin (\Omega \tau )/\Omega$, an advanced Green function. A causal growing response would require an inverse-Laplace contour above those poles, not the stated real-axis Fourier integral. At $e=0$ the real-axis integral needs a pole prescription; shifting both poles below it gives the retarded zero-damping limit. These distinctions are necessary because $\gamma >0$ alone does not fix the sign of damping.