Causal Green function of a damped oscillator ($G(t)=H(t)e^{-kt}\sin (\omega t)/\omega$)

For $L=D_t^2+2k{D}_t+k^2+{\omega }^2$, with $k>0$ and $\omega >0$, the causal Green function is $G(t)=H(t)e^{-kt}\sin (\omega t)/\omega$, where $H$ is the Heaviside step function. It satisfies $LG=\delta (t)$: it is continuous at zero, with a unit jump in its first derivative. Its smooth positive-time kernel $g$ obeys $g(0)=0,g^{\prime }(0)=1$. The zero-frequency limit is $H(t)t{e}^{-kt}$. Shifting $G$ gives the zero-initial-data response to a delayed Dirac delta function.

With the Fourier transform convention $\hat{f}(\xi )=\int f(t)e^{-i\xi t}dt$, the transformed equation gives $\hat{G}(\xi )=[k^2+{\omega }^2-{\xi }^2+2ik\xi {]}^{-1}$. Its poles $\xi =ik\pm \omega$ lie in the upper half of the complex plane. Inverting by contour integration therefore gives zero for negative time and an exponentially damped oscillation for positive time.

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.

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.

Integrating the smooth positive-time impulse kernel gives the zero-data response to the Heaviside step function. For nonzero $\omega$,At zero frequency and nonzero $k$ the limit is $[1-e^{-kt}(1+kt)]/k^2$; when both parameters vanish it is $t^2/2$. The integral representation avoids ambiguous divisions in these limiting cases.

Adding a load $M$ to a spring-supported table shifts its static downward position by $\Delta =Mg/k$. If the load is deposited with negligible impulse and moves with the table, the moving mass is $m+M$. Starting at the former equilibrium with zero velocity, an underdamped response isIts transient envelope decays over $1/\beta$. Critical damping $b=2\sqrt{k(m+M)}$ gives $x=\Delta [1-(1+{\omega }_{0}t)e^{-{\omega }_{0}t}]$. A nonzero impact impulse changes the initial velocity and therefore the transient.

For a step-response kernel with $K(0)=0$, differentiating the start time produces no delta term from the response itself, because its coefficient is zero. The negative shift derivative is the impulse response. More generally an additional $\delta (t-b)K(0)$ term appears if the kernel has a nonzero initial value. The identity holds distributionally as well as for the ordinary smooth pieces.