Harmonic oscillator boundary condition issue for an impulse force

Question

I am trying to solve an equation of an underdamped harmonic oscillator with a damping, and I get a weird boundary condition that perplexes me. Let me precise the issue, the equation is : \begin{equation} \ddot{y} + 2\gamma \omega_0 \dot{y} + \omega^2_0 y = f(t) \label{eq:harmonic_osc} \end{equation} And I am also interested in the case where the force $f(t)$ is an impulse as following: $$ f(t) = A \delta(t)$$

By the applying the Laplace transform to the both sides of the equation, we get:

\begin{equation} s^2 y(s) - \dot{y}(t = 0 )+ 2\gamma \omega_0 s y(s) + \omega^2_0 y(s) = f(s) \end{equation} where $y(t = 0 ) = 0 $, thanks to the boundary condition that we impose. As for the $\dot{y}(t=0)$ we can get it from the integrating the harmonic oscillator equation:

\begin{equation} \int \limits_{-t_2}^t \, dt \left(\ddot{y} + 2\gamma \omega_0 \dot{y} + \omega^2_0 y \right)= \int \limits_{-t_2}^t \, dt \, f(t), \end{equation} where all the values for time $-t_2$ are equal to zero by the definition of the Laplace transform and, thus we get that $\dot{y}(t=0) = A \int \limits_{-t_2}^0 \, dt \, \delta(t) = A/2$

so:

$$ y(s) = \frac{A}{s^2 + 2 \gamma \omega_0 s + \omega^2_0} $$

Next we can calculate the inverse Laplace transform: $$ y(t) = \frac{1}{2 \pi i }\int \limits^{c+i\infty}_{c-i\infty} y(s) e^{st} \,ds$$

This integral is being calculated by finding residues and etc, so I get that :

$$ y(t) = \frac{A e^{-\gamma \omega_0 t}}{\tilde{\omega}_{0} } \sin{\tilde{\omega}_{0} t}$$ where $\tilde{\omega}_0 = \omega_0 \sqrt{1-\gamma^2}$

Now we see that $y(0) = 0$, however as for the first derivative of $y(t)$ we have:

$$ \dot{y} (t) = \frac{A} {\tilde{\omega}_0} (-\gamma \omega_0 e^{-\gamma \omega_0 t} \sin{\tilde{\omega}_{0} t} + e^{-\gamma \omega_0 t} \cos{\tilde{\omega}_{0} t}) $$ and the $\dot{y}(0) = A,$ which is different from the initial assumption where $\dot{y}(0) = A/2$, could someone please tell me where is a catch? It seems that I have checked a lot of things but I could not find a problem.

Answer 1

When solving differential equation with distribution, you must use the distribution derivation to be consistent. This restrict the boundary conditions.

It comes for the first derivative:

$$\dot{y}=\{\dot{y}\}+(y(0^{+})-y(0^{-})) \delta (t)$$

Where:

• $\{\dot{y}\}$ is the usual derivative.
• $y(0^{+})=\lim_{t \to 0^{+}}y(t)$
• $ y(0^{-})= \lim_{t \to 0^{-}}y(t)$

Applying again this result to: $\ddot{{y}}$, one gets:

$$\ddot{y}=\{\ddot{y}\}+(\dot{y}(0^{+})-\dot{y}(0^{-}))~\delta (t)+(y(0^{+})-y(0^{-}))~\dot{\delta} (t)$$

The substitution into your differential equation gives:

$$\{\ddot{y}\} + \big(\dot{y}(0^{+})-\dot{y}(0^{-})\big) ~\delta (t)+ \big(y(0^{+})-y(0^{-})\big) ~\dot{\delta} (t)+ 2\gamma \omega_0 \{\dot{y}\} +2\gamma \omega_0~ \big(y(0^{+})-y(0^{-})\big) ~\delta (t)+ \omega^2_0 \{y\}=A~\delta (t)$$

Regular functions cannot be equal to distributions, delta distribution can only be equal to delta distribution and the same goes for the derivative. So we get:

• Regular function: $\{\ddot{y}\} + 2\gamma \omega_0 \{\dot{y}\} + \omega^2_0 \{y\}=0$

• $\dot{\delta} (t)$: $~~~~~~y(0^{+})-y(0^{-})=0$

• $\delta (t)$: $~~~~~~\dot{y}(0^{+})-\dot{y}(0^{-})=A$

So the function y(t) is continuous at t=0 but its first derivative is discontinuous.

$$\begin{cases}y(t)=0 & t \leq 0\\\dot{y}(0^{+})=A \end{cases}$$

Your boundary condition: $\dot{y}(0^{+})= \frac{A}{2}~$ is false. Further more, the continuity of y(t) at t=0 is the only consistent boundary condition.

Answer 2

So, the Laplace transform of a second derivative is $ L(\ddot{y})(s) = s^2 y(s) - sy(t=0) - \dot{y}(t=0) $ so your equation reads

$ s^2 y(s) - s y(0) - \dot{y}(0) + 2\gamma \omega_0 (s y(s) - y(0)) + \omega_0^2 y(s)=f(s)$

so you need also a boundary condition for $ y(t=0)$ and the equation becomes

$ y(s) = \frac{sy(0)+A+\dot{y}(0)-2\gamma \omega_0^2 y(0)}{s^2 +2\gamma \omega_0 s + \omega_0^2}$

This is a very ugly inverse Laplace transform. I personally would really advise to work with Fourier transform (and Greens functions). Sorry, that I couldn't do the calculations because I have no time, but I hope it could help...