# Overdamped vs underdamped

1. Aug 18, 2015

### LagrangeEuler

In the paper
http://fmc.unizar.es/people/juanjo/papers/falo93.pdf [Broken]

how it is possible to obtain equation of motion which is first order in time

$$\dot{u}_{j}=u_{j+1}+u_{j-1}-2u_j-\frac{K}{2\pi}\sin(2\pi u_j)+F(t)$$
How to obtain this equation of first order?

Last edited by a moderator: May 7, 2017
2. Aug 18, 2015

### Orodruin

Staff Emeritus
It is likely a typo.

3. Aug 18, 2015

### LagrangeEuler

No it is not. You have this in many papers.

4. Aug 18, 2015

### Orodruin

Staff Emeritus
If it is not, there must be additional underlying assumptions, such as a drag force proportional to velocity which is dominating over the acceleration term.

5. Aug 18, 2015

### LagrangeEuler

Could you write this down as relation?

6. Aug 18, 2015

### Orodruin

Staff Emeritus
They even state that the system is overdamped before the equation.

7. Aug 18, 2015

### LagrangeEuler

Yes I know that. But I am trying to figure out what that means, or how them got this equation?

8. Aug 18, 2015

### Orodruin

Staff Emeritus
Are you at all familiar what the term overdamped means?

9. Aug 18, 2015

### LagrangeEuler

Not really.

10. Aug 18, 2015

### Orodruin

Staff Emeritus
So what happens to Newton's second law for particle i if, in addition to the external forces acting on each particle, tell you that there is also a drag force proportional to (but with a negative coefficient) the velocity of the particle? What approximations can you then do if I tell you that this force is generally much larger than the term $ma$ on the right-hand side?

11. Aug 18, 2015

### LagrangeEuler

If I understand you well there is equation with second $\ddot{u}_j$ and first derivative $\dot{u_j}$ such that term $C\dot{u}_j$ is dominant if it is compared with $\ddot{u}_j$. But I am really not sure what is the form of equation before that approximation?

12. Aug 18, 2015

### Orodruin

Staff Emeritus
What does Newton's second law for the ith particle look like?

13. Aug 18, 2015

### LagrangeEuler

$$m_ia=\sum F$$
$$m_ia=-V'(u)+F(t)+F_{el_{i,i+1}}+F_{el_{i,i-1}}$$
where $F_{el}$ is elastic force between the particles. Right?

14. Aug 18, 2015

### Orodruin

Staff Emeritus
You forgot the drag term which we just talked about.

15. Aug 18, 2015

### LagrangeEuler

$$m_ia=\sum F$$
$$m\ddot{u}=-V'(u)+F(t)+F_{el_{i,i+1}}+F_{el_{i,i-1}}-C\dot{u}$$
where $F_{el}$ is elastic force between the particles. Right?
So why I have this drag term?

16. Aug 18, 2015

### Orodruin

Staff Emeritus
Uhmmm, because there is drag in the model? Generally drag results from a resistance to movement due to dissipation of energy to the environment. Think air resistance. The form of the drag will depend on the Reynold's number. The linear drag is usually valid for relatively small velocities. See https://en.wikipedia.org/wiki/Drag_(physics)

17. Aug 18, 2015

### LagrangeEuler

Thanks a lot!

18. Aug 18, 2015

### LagrangeEuler

Now just one thing. Equation should be
$$m\ddot{u}=-V'(u)+F(t)+F_{el_{i,i+1}}+F_{el_{i,i-1}}-C\dot{u}$$
So if $C\dot{u}>>m\ddot{u}$ then
$$C\dot{u}=-V'(u)+F(t)+F_{el_{i,i+1}}+F_{el_{i,i-1}}$$
or equation in paper should be
$$C\dot{u_j}=u_{j+1}+u_{j-1}-2u_j-V'(u_j)+F(t)$$
Why $C=1$?

19. Aug 18, 2015

### Orodruin

Staff Emeritus
You will notice that the equation as it stands in the paper does not make sense dimensionally if t and x have dimensions. However, you can always introduce dimensionless parameters by scaling with a constant of the appropriate dimensions. This can be done in such a way that C becomes equal to one. (Take s = t/C. This gives d/ds = (dt/ds) d/dt = C d/dt.)