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 $\dot{y}(t=0) = A \int \limits_{-t_2}^0 \, dt \, \delta(t) = A/2$


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

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.


2 Answers 2


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:

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


  • $\{\dot{y}\}$ 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.

  • $\begingroup$ Thank you, so you going to calcuate the inverse Laplace transform from which of the follwing formulas, my problem might come from here: $$ y(s) = \frac{f(s) + \dot{y}(t =+ 0)}{s^2 + 2 \gamma \omega_0 s + \omega^2_0}$$ $$ y(s) = \frac{\dot{y}(t = +0)}{s^2 + 2 \gamma \omega_0 s + \omega^2_0}$$ $\endgroup$ Commented Sep 11, 2022 at 16:52
  • $\begingroup$ where $f(s)$ is a delta function Laplace transform in general sense (If it is rigorous to say that) $\endgroup$ Commented Sep 11, 2022 at 16:55
  • $\begingroup$ Sorry, I forgot to add a question mark in the first comment, StackExchange does not let me do that any longer. I just want to ask you if you are going to use one of proposed $y(s)$ in order to calculate the $y(t)$ $\endgroup$ Commented Sep 11, 2022 at 17:00
  • $\begingroup$ @ Pierre What are you asking. I do not understand. $\endgroup$
    – Shaktyai
    Commented Sep 11, 2022 at 18:48

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...

  • $\begingroup$ Thank you. The usage of a Laplace transform is motivated by a control theory. I am reading references on this topic. The engineers do use Laplace transform a lot in this topic and generally speaking the transfert function is a ratio of output value to the input value, and it is given in Laplace s-space. So the Laplace transform is mandatory. $\endgroup$ Commented Sep 11, 2022 at 16:23
  • $\begingroup$ Also, I have seen that in a lot of problems with differential equations when there are higher derivatives up to a value (n). It is usually considered that all the following values are equal to 0: $y^{(n-1)}(t=0) = 0$ $y^{(n-2)}(t=0) = 0 $... $\endgroup$ Commented Sep 11, 2022 at 16:24
  • $\begingroup$ Another thing, I am not sure that Fourier transform is used for impulses, I might be wrong, but I have not seen any exemples. $\endgroup$ Commented Sep 11, 2022 at 16:24
  • $\begingroup$ Of course Green function method could work. However it seems that is is necessary to know the Green function for every differential equation and with Laplace transform method it is not necessary $\endgroup$ Commented Sep 11, 2022 at 16:32

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