0
$\begingroup$

corresponding paper

enter image description here enter image description here enter image description here

why it said that $-\lambda \frac{|\dot{x}|\dot{x}}{2}$ is a small second order dissipating term ?

For a linear second order system, the velocity is a dissipating term, because I know the equation is derived from the mass-damper-spring phsyical system.

I am new to ordinary differential equation, the first derivative state term always plays a dissipating role no matter its power and its sign in a second order system?

enter image description here

$\endgroup$

1 Answer 1

2
$\begingroup$

The word dissipation here is referring to the energy dissipated as heat in the canonical mass-spring-damper system. Usually this term is related to $\dot{x}$ in the form of $\ddot{x} = -b\dot{x}$, and in many cases we call this damping - but it is essentially just friction. Now, if we were concerned with our system's velocity and not position, we should clearly see that the above system is stable (assuming $b \gt 0$).

This is the simple linear case, but what happens if we do not care about small velocities, but want to heavily "dampen" large velocities. In this case, we could look for damping of the form $\ddot{x} = -b\dot{x}^{2}$. But what happens when this system starts out with a negative velocity? Clearly it is unstable. So, we will have to look for a different method of damping.

Consider the system described by $\ddot{x} = -b\ sgn(\dot{x})\dot{x}^2$. Whenever our velocity is negative, acceleration will be positive. Whenever velocity is positive, acceleration will be negative. This is now a stable system.

Finally, consider this system's behavior close to the origin. Our damping force decreases exponentially as we approach a velocity of zero. As this is the case, it makes sense to consider this a small dissipating term - as it will not dissipate all of our systems energy in a finite amount of time.

As a final note, let $b = \frac{\lambda}{2}$ and note that $|\dot{x}|\dot{x} = sgn(\dot{x})\dot{x}^{2}$.

$\endgroup$
1
  • $\begingroup$ Very nice explanation. $\endgroup$ Commented May 26, 2022 at 12:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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