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<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 !



Stabilize the vector field of $y' = f (y) - \gamma H^T(HH^T)^{-1}h( y ) $ of ODE $y' = f(y)$
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.


