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This question has been asked here but there is no answer: https://math.stackexchange.com/questions/1585400/stabilize-the-vector-field-of-y-f-y-hthht-1h-y-of-ode-y

Consider autonomous ODE $y' = f(y)\quad (1)$ which has an invariant set $M$ defined by the equations $$h (y) = 0 \qquad (2) $$ i.e., assuming that the initial conditions satisfy $h (y (0)) = 0$, the solution of the ODE satisties $h ( y ( t )) = 0$ for all later times $t\geq 0$. Defined the Jacobian matrix ($y,h\in \mathbb{R}^n$) $$H(y)=\frac{\partial h}{\partial y}$$ and assume that it has full row rank for all $t$ (in particular, there are no more equations in (2) than in (1)). Next we stabilize the vector field, replacing the autonomous (1) by $$y' = f (y) - \gamma H^T(HH^T)^{-1}h( y )\quad(3) $$ Show that if there is a constant $\gamma_0$ such that $$|Hf(y)|_2 \leq \gamma_0 |h(y)|_2$$ for all $y$ in the neighborhood of the invariant set $M$ then $M$ becomes asymptotically stable, i.e. $|h(y(t))|$ decreases in $t$ for trajectories of (3) starting near $M$ , provided that $\gamma\geq \gamma_0$.

I have no clue to prove the claim. Can anyone help me? Thank you in advance !

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  • $\begingroup$ Perhaps an exercise from "Computer Methods for Ordinary Differential Equations and Differential-Algebraic Equations" by Ascher and Petzold? $\endgroup$
    – John B
    Commented Dec 25, 2015 at 18:11
  • $\begingroup$ @JohnB do you know the proof? $\endgroup$
    – David Li
    Commented Dec 25, 2015 at 20:20
  • $\begingroup$ Hint: $Hy'=Hf(y)-\gamma h(y)$. $\endgroup$
    – John B
    Commented Dec 25, 2015 at 21:00
  • $\begingroup$ @JohnB Thank you for your hint. Does $|H(f(y))| \leq \gamma_0 |h(y)|$ implies there exits $\gamma_1$ s.t $H(f(y)) \leq \gamma_ 1 h(y)$? If it's true, from your hint I get $Hy' \leq (\gamma_1-\gamma)h(y) => y(h(t))' \leq (\gamma_1-\gamma)h(y) $. This implies that h(y(t)) decreasing in t for trajectories of (3). Am I right? $\endgroup$
    – David Li
    Commented Dec 26, 2015 at 11:54
  • $\begingroup$ Not really, these may be vectors. See my answer. $\endgroup$
    – John B
    Commented Dec 26, 2015 at 17:07

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Notice that $$ \begin{split}\frac12\frac{d}{dt} \|h(y(t))\|^2 &=\frac12\frac{d}{dt}\bigl\langle h(y(t)),h(y(t))\bigr\rangle\\ &=\bigl\langle H(y(t))y'(t),h(y(t))\bigr\rangle\\ &=\bigl\langle H(y(t))f(y(t))−\gamma h(y(t)),h(y(t))\bigr\rangle\\ &=\bigl\langle H(y(t))f(y(t)),h(y(t))\bigr\rangle−\gamma \|h(y(t))\|^2\\ &\le \|H(y(t))f(y(t))\|\cdot\|h(y(t))\|−\gamma \|h(y(t))\|^2\\ &\le \gamma_0 \|h(y(t))\|^2−\gamma \|h(y(t))\|^2. \end{split}$$ The rest is also standard.

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