# Inverse dynamics control: Proof of asymptotic stability of error system

The inverse dynamics control in robotic applications yields the error system $$$$\ddot{\mathbf{e}} + \mathbf{K}_1 \dot{\mathbf{e}} + \mathbf{K}_0 {\mathbf{e}} = \mathbf{0}$$$$ or rewritten as ODE-system $$\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!

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: $$\mathcal{M} = \{\mathbf{0}\}$$.
• 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. Commented Oct 14, 2021 at 13:53
• @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). Commented Oct 14, 2021 at 17:37