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The inverse dynamics control in robotic applications yields the error system \begin{equation} \ddot{\mathbf{e}} + \mathbf{K}_1 \dot{\mathbf{e}} + \mathbf{K}_0 {\mathbf{e}} = \mathbf{0} \end{equation} or rewritten as ODE-system $$ \frac{d}{dt} \begin{bmatrix} \mathbf{e} \\ \dot{\mathbf{e}} \end{bmatrix} = \underbrace{ \begin{bmatrix} \mathbf{0} & \mathbf{I}\\ -\mathbf{K}_0 & -\mathbf{K}_1 \end{bmatrix} }_{\mathbf{A}} \begin{bmatrix} \mathbf{e} \\ \dot{\mathbf{e}} \end{bmatrix} \text{ .} $$ To prove asymptotic stability of the error system, A has to be a Hurwitz-Matrix.

Literature says that it is sufficient for $\mathbf{K}_0$ and $\mathbf{K}_1$ to be positive definite to guarantee asymptotic stability of the error system.

During the proof the assumption of $\mathbf{K}_0 = diag\{k_{0,1}, \dots, k_{0,n} \}$ and $\mathbf{K}_1 = diag\{k_{1,1}, \dots, k_{1,n} \}$ was made. This in return yields a decoupled system $$ \frac{d}{dt} \begin{bmatrix} {e_j} \\ \dot{{e}}_j \end{bmatrix} = \underbrace{ \begin{bmatrix} {0} & 1\\ -{k}_{0,j} & -{k}_{1,j} \end{bmatrix} }_{\mathbf{A}_j} \begin{bmatrix} {e}_j \\ \dot{{e}_j} \end{bmatrix} \text{ .} $$ for $j=1, \dots, n$ with the characteristic polynomial of $\mathbf{A}_j$ $$ p_j(s) = s^2 + k_{1,j} s + k_{0,j} $$ which is a Hurwitz polynomial for $k_{1,j} > 0$ and $k_{0,j} > 0$, hence guarantees the asymptotic stability of the decoupled system.

$\textbf{BUT}$ I have not found a proof which relies only on the assumption of $\mathbf{K}_0$ and $\mathbf{K}_1$ to be positive definite.

I tried solving the Lyapunov equation $$ \mathbf{A}^{T} \mathbf{P} + \mathbf{P} \mathbf{A} + \mathbf{Q} = \mathbf{0} $$ with the positive definite matrices $\mathbf{P}$ respectively $\mathbf{Q}$ without much luck.

How can I proof asymptotic stability of the above-mentioned error system with only the assumption of $\mathbf{K}_0$ and $\mathbf{K}_1$ to be positive definite?

Any help would be much appreciated!

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A related post lead me to the answer: Help! Lyapunov proof for calculated torque control with friction term for robot

The Lyapunov function $$ V(\mathbf{e}, \dot{\mathbf{e}}) = \frac{1}{2} \dot{\mathbf{e}}^{T} \dot{\mathbf{e}} + \frac{1}{2} {\mathbf{e}}^{T} \mathbf{K}_0{\mathbf{e}} $$ yields $V(\mathbf{e}, \dot{\mathbf{e}}) > 0$ in case $\mathbf{K}_0$ is positive definite and \begin{align} \dot{V}(\mathbf{e}, \dot{\mathbf{e}}) &= \dot{\mathbf{e}}^{T} \ddot{\mathbf{e}} + \dot{\mathbf{e}}^{T} \mathbf{K}_0 \mathbf{e} \\ &= - \dot{\mathbf{e}}^{T} \mathbf{K}_1 \dot{\mathbf{e}} \end{align} yields $\dot{V}(\mathbf{e}, \dot{\mathbf{e}}) \leq 0 $ in case $\mathbf{K}_1$ is positive definite.

According to LaSalle's invariance principle the error system is asymptotic stable as the largest invariant set $\mathcal{M} \subseteq \left\{\mathbf{e}, \dot{\mathbf{e}} \in \mathbb{R}^n ~|~ \dot{V}(\mathbf{e}, \dot{\mathbf{e}}) = \mathbf{0}\right\}$ is the origin itself: $\mathcal{M} = \{\mathbf{0}\} $.

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    $\begingroup$ I think $\dot{V} \leq 0$ only because you have $\mathbf{e}$ and $\dot{\mathbf{e}}$ as states so $V$ and $\dot{V}$ are functions of $\mathbf{e}$ and $\dot{\mathbf{e}}$. So $\dot{V}(\mathbf{e}, \dot{\mathbf{e}})$ should be negative semi definite. $\endgroup$
    – SampleTime
    Commented Oct 14, 2021 at 13:53
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    $\begingroup$ @SampleTime but in this case one can apply LaSalle to finish the proof. $\endgroup$ Commented Oct 14, 2021 at 15:11
  • $\begingroup$ @KwinvanderVeen True. Alternatively there should exist a different $V$ such that $\dot{V}$ is actually negative definite. For scalar error $e$ the following should work: $V(e, \dot{e}) = \frac{1}{2}(k_0^2 + k_0 + k_1^2) e^2 + k_1 e \dot{e} + \frac{1}{2}(k_0 + 1) \dot{e}^2$ which leads to $\dot{V}(e, \dot{e}) = -k_0 k_1 (e^2 + \dot{e}^2)$ which is negative definite. This should probably also work if $\mathbb{e}$ is not a scalar (though I didn't check that). $\endgroup$
    – SampleTime
    Commented Oct 14, 2021 at 17:37
  • $\begingroup$ Thanks, I finished the proof with the invariance principle. $\endgroup$
    – Gerald
    Commented Oct 14, 2021 at 18:37

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