Massless limit of a dynamical system

[muzialis]

Hello All,

from considering the equations (apologies for the poor notation, primes denote differentiation)

G = mv' + p (v)

and

G' = a - v

(where v = v(t), for the interested is the crack speed for a genrelized crack dynamics model, m and a are constants, p a function)

one obtains the system


G' = a - v
v' = (1/m) (G - p (v))


Now I paper I found considers the function dG / dV, dividing the top eqautionby the lower one, and this represent the slope of integral curves in the phase plane, as

dG / dV = m (a - v) / (G- p(v))

Now it is said, considering the massless limit one gets

dG(G- p(v)) = 0.

This is my first question. I see how this comes out, but I am wondering, from the original equation dG / dV one mght be tempted to say tha the massless limit is dG / dV = 0, which is different from dG(G- p(v)) = 0.


Second point: should one not be able to get to the massless limit by ignoring the term mv' (inertial term) from the start?

If I try i do not recover the relationship dG(G- p(v)) = 0

Any help would be the most appreciated

thanks

 

[jvicens]

Hello there,

Thank you for sharing your equations and questions. It seems like you are working on a fascinating problem in generalized crack dynamics. Let me address your questions one by one:

1. The massless limit in the equation dG/dV = m(a-v)/(G-p(v)) comes from the fact that for a massless object, the term mv' becomes zero. This means that the massless limit is actually dG/dV = 0, which is the same as dG(G-p(v)) = 0. So, you are correct in saying that the massless limit in this case is dG(G-p(v)) = 0, not dG/dV = 0.

2. Ignoring the inertial term (mv') from the start will not necessarily lead to the relationship dG(G-p(v)) = 0. This is because the inertial term is still present in the original equations, and ignoring it from the start means that you are not considering the full dynamics of the system. In order to get the relationship dG(G-p(v)) = 0, you need to consider the full system of equations and then take the massless limit.

I hope this helps clarify your questions. Keep up the great work on your research!