I am trying to solve the following differential equation;

$$\frac{d^2 x}{d t^2}=-\omega^2 x \delta(t-t^\prime).$$

I know this is of the form

$$x(t)= A \sin(\omega t) + B \cos(\omega t).$$

However this delta Dirac function is confusing me. My reasoning is the following; An increase in acceleration at $t^\prime$ results in an increase of constant velocity starting from $t^\prime$ and a linear constant increase in distance so;

$$x(t) = \left[A \sin(\omega t) + B \cos(\omega t)\right] t \Theta(t>t^\prime).$$

With $\Theta$ being the step function. This still feels more like a guess than a straightforward answer, then I am not 100% sure of the first derivative.

Any help would be welcome.

  • 2
    $\begingroup$ This question may be better suited for the Mathematics SE. $\endgroup$ – Davide Morgante Aug 26 at 16:21
  • $\begingroup$ Comment to the post (v3): Are you sure the delta function is multiplied rather than added on the rhs? $\endgroup$ – Qmechanic Aug 26 at 16:58
  • $\begingroup$ @Qmechanic You know I was just thinking the same thing. The equation as stated is not dimensionally consistent. $\endgroup$ – Philip Aug 26 at 17:15
  • $\begingroup$ Okay, you could be right.. The lagrangian that is given is the following: $L=\frac{m \frac{dx}{dt}^2}{2} - \frac{m \omega^2 x^2}{2} \delta (t-t')$. With the Euler Lagrange equations I arrived at the diff eq.. (part of the exercice) $\endgroup$ – Alex Aug 26 at 17:25
  • $\begingroup$ @Alex The Lagrangian isn't dimensionally consistent either! The delta function here has dimensions 1/time. The second term does not have dimensions of energy. Furthermore, your problem is not an oscillator, it looks more like a free particle before and after $t'$, since $\delta(t-t')=0$ for $t\neq t'$. I'm assuming that there's a mistake in the original source. $\endgroup$ – Philip Sep 1 at 8:02

You need to find the solutions for $t> t'$ and for $t> t'$ and then impose the integration constants by the boundary conditions. The problem is actually very similar to that of a delta-function barrier in quantum mechanics, but with zero energy.

The solution with sines and cosines does not belong here, since it comes from the theory of ordinary differential equations with constant coefficients, which is not the case here.

Here is another type of problem where such an equation appears: the field of an infinite charged plane link.

| cite | improve this answer | |

Not the answer you're looking for? Browse other questions tagged or ask your own question.