0
$\begingroup$

I know that for a 2nd order linear differential equation system, there are 3 possible scenarios: over-damped, critically damped and underdamped. For the underdamped case the solutions are of the form: $e^{-\alpha t}(Acos(\omega_d t) + Bsin(\omega_d t))$

I am interested in a solution of the form $e^{-\alpha t^2}Acos(\omega_d t)$ i.e., I want the oscillations to die at quadratic rate.

Is there a corresponding differential equation that can generate this kind of behavior?

Note: I asked this question before. The answer I got was a trivially constructed linear time-varying system. I am interested in a more compact and physically driven representation.

Improve this question
$\endgroup$
3
  • $\begingroup$ Please define "compact" and "physically driven"; that said, the oscillatory solutions take the natural form $C e^{- \alpha t + i \omega_d t}$. One should expect a quadratic version to have form $$ e^{(-\alpha + i \omega_d) t^2}$$ which would make it $\omega_d t^2$ in the $\cos$ rather than linear. $\endgroup$
    – Willie Wong
    Commented Mar 14, 2016 at 17:39
  • $\begingroup$ @WillieWong by physically driven I am looking for some physical analogy, i.e. maybe a nonlinear damping element. $\endgroup$
    – jkt
    Commented Mar 14, 2016 at 17:44
  • $\begingroup$ I did not want to express the solution in polar form but yeah the oscillation is of the form $Ce^{-\alpha t + j \omega_d t}$. I do not however wish the cosine term wrt $\omega_d t^2$. I am strictly looking for standard sinusoid oscillations amplitudes of which decay according to the rate $e^{-\alpha t^2}$, i.e. quadratic rate instead of a linear one. $\endgroup$
    – jkt
    Commented Mar 14, 2016 at 17:47

1 Answer 1

Reset to default
2
$\begingroup$

I don't know about "physically driven", but you could try

$$\ddot{u} + 4 t \dot{u} + (4 t^2 + 3) u = 0$$

If you want something autonomous, you might make it

$$ \eqalign{\ddot{u} + 4 v \dot{u} + (4 v^2 + 3) u &= 0 \cr \dot{v} & = 1\cr}$$

which can be made into a single nonlinear d.e., but it will be messy.

Improve this answer
$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .