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Proof of Dirichlet stability theorem

  1. Dec 24, 2016 #1
    Hola, I tried to give a proof of this theorem and then check it against the one given by my book(Fasano, Marmi - Analytical Mechanics); I feel like mine seems reasonable and pretty intuitive, but the one on the book is a bit different and I don't really understand it completely, so I'd like to have someone check my work and maybe give me some hint to understand my book's proof.

    1. The problem statement, all variables and given/known data

    Prove Dirichlet stability theorem:
    If [itex]\mathbf{\bar{q}}[/itex] is an isolated minimum of the potential energy, then the corresponding configuration is of stable equilibrium

    2. Relevant equations
    $$ \dfrac{\partial U}{\partial \mathbf{\bar{q}}} = 0$$

    3. The attempt at a solution
    We will consider a holonomic system with fixed smooth costraints.
    Let [itex]\mathbf{\bar{q}} \in R^{n}[/itex] be an isolated minimum of [itex]U = U(\mathbf{q})[/itex], and be it
    $$ U(\mathbf{\bar{q}}) = 0 $$
    For every other point the potential energy will then be positive, and so will the total mechanical energy E = T+U, since T is a positive definite quadratic form in [itex]\mathbf{\dot{q}}[/itex].
    Since E=0 only in [itex](\mathbf{\bar{q}}, \mathbf{\dot{\bar{q}}})[/itex](the velocity in that configuration is zero), I'll consider it to be the origin of my frame of reference;
    E is a continuos function in phase space (subset of [itex]R^{2n}[/itex]) and it is constant throughout the motion since the constraints are fixed.

    Now suppose that [itex]\mathbf{\bar{q}}[/itex] is not a point of stable equilibrium, and consider two balls in phase space:
    $$
    A = \{(\mathbf{q}, \mathbf{\dot{q}}) \in R^{2n} : \vert (\mathbf{q}, \mathbf{\dot{q}}) - (\mathbf{\bar{q}}, \mathbf{\dot{\bar{q}}}) \vert \leq \epsilon \}\\
    B = \{(\mathbf{q}, \mathbf{\dot{q}}) \in R^{2n} : \vert (\mathbf{q}, \mathbf{\dot{q}}) - (\mathbf{\bar{q}}, \mathbf{\dot{\bar{q}}}) \vert \leq \delta \}
    $$
    where [itex]\delta > \epsilon[/itex], therefore [itex]A \subset B[/itex]
    Since the point is unstable there exists a trajectory of E which will start in A, reach B and then get out of it (thus violating Lyapunov stability condition). [this does not look really formal, maybe I'll need to rephrase this somehow]
    If now I compute E over the boundary of B, which is a closed and bounded subset of [itex]R^{2n}[/itex], then for Weierstrass' theorem there exists a minimum value of the mechanical energy
    $$ 0 < \tilde{E} \leq E $$
    To "escape" B, it must be [itex]E>\tilde{E}[/itex], but since E is constant it must have had the same value already in the initial position in A; but if I take [itex]\epsilon \rightarrow 0[/itex], then I narrow A down to the unique point [itex](\mathbf{\bar{q}}, \mathbf{\dot{\bar{q}}})[/itex], where E=0, therefore it cannot be greater than [itex]\tilde{E} > 0[/itex].
    [itex](\mathbf{\bar{q}}, \mathbf{\dot{\bar{q}}})[/itex] must then be a point of stable equilibrium.



    Here, instead, is my book's proof, which is way more compact and less verbose:
    View attachment 110755
    They use V as the potential energy.
    eq. 4.132 refer to $$ \dfrac{\partial U}{\partial \mathbf{\bar{q}}} = 0$$, Theorem 4.1 states the kinetic energy is a positive definite quadratic form, and Definition 4.6 is Lyapunov stability condition.

    It seems to me that the approach is somewhat similar to mine, since they define some neighbourhoods of the equilibrium position and then make considerations about energy levels but I don't understand what is the meaning of
    $$ T(\mathbf{q}, \mathbf{\dot{q}}) \geq a_o\vert\mathbf{\dot{q}}^2\vert$$
    and it seems to be a central point in the proof.

    Hope anyone can help me.
     
  2. jcsd
  3. Dec 29, 2016 #2
    Thanks for the thread! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post? The more details the better.
     
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