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Massless limit of a dynamical system

  1. Dec 2, 2011 #1
    Hello All,

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

    G = mv' + p (v)


    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

  2. jcsd
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