Corrigendum and addendum to

This article refers to:

A note on stability conditions for planar switched systems

[*] A note on stability conditions for planar switched systems, M. Balde, U. Boscain, P. Mason, Int. J. Control, Vol. 82 (10), p. 1882–1888 (2009)

The paper [*] provides a characterisation of the stability properties for two-dimensional switched systems of the form (1) $$\dot{x}\left(t\right)=u\left(t\right)A_{1}x\left(t\right)+(1-u\left(t\right))A_{2}x\left(t\right),u:[0,\infty )\to [0,1].$$(1) Unfortunately, several misprints in the galley proofs remained unnoticed. Namely, many indices in the definitions of τ_i and k were incorrect, while the definition of Δ was missing. In addition, an extra multiplicative factor 2 in the definition of t_i (in the case ${\delta }_{A_i}=0$) remained unnoticed until recently. In this note, we correct the misprints and we provide a full proof in the case S4 (that was not detailed in [*]) in order to correct the multiplicative factor.

Let us recall the notations and the objects that are needed in order to state the stability result of [*] in its correct formulation. In the following by invariant, we mean an object that is invariant with respect to linear coordinate transformations. We denote by $det\left(X\right)$ and tr(X) the determinant and the trace of a matrix X. If $X\in {\mathbb{R}}^{2\times 2}$, the discriminant is defined as $${\delta }_X=\mathrm{tr}{\left(X\right)}^2-4det\left(X\right).$$Given a pair of matrices X, Y, we define the following object: $${\displaystyle \Gamma (X,Y):=\frac{1}{2}(\mathrm{tr}\left(X\right)\mathrm{tr}\left(Y\right)-\mathrm{tr}\left(XY\right)).}$$By means of these quantities, we can define other invariants associated with (1(1) $$\dot{x}\left(t\right)=u\left(t\right)A_{1}x\left(t\right)+(1-u\left(t\right))A_{2}x\left(t\right),u:[0,\infty )\to [0,1].$$(1) ) as follows: $$\begin{array}{ccc}& & {\tau }_i:=\left\{\begin{array}{c}{\displaystyle \frac{\mathrm{tr}\left(A_i\right)}{\sqrt{|{\delta }_{A_i}|}}\text{if}{\delta }_{A_1}\ne 0,{\delta }_{A_2}\ne 0,}\hfill \\ {\displaystyle \frac{\mathrm{tr}\left(A_i\right)}{\sqrt{|{\delta }_{A_j}|}}\text{if}{\delta }_{A_1}{\delta }_{A_2}=0\text{but}{\delta }_{A_j}\ne 0,}\hfill \\ {\displaystyle \frac{\mathrm{tr}\left(A_i\right)}{2}\text{if}{\delta }_{A_1}={\delta }_{A_2}=0,}\hfill \end{array}\right.\hfill \\ & & {\displaystyle k:=\frac{2{\tau }_1{\tau }_2}{\mathrm{tr}\left(A_1\right)\mathrm{tr}\left(A_2\right)}\left(\mathrm{tr}\left(A_1{A}_2\right)-\frac{1}{2}\mathrm{tr}\left(A_1\right)\mathrm{tr}\left(A_2\right)\right),}\hfill \\ & & \Delta :=4(\Gamma {(A_1,A_2)}^2-\Gamma (A_1,A_1)\Gamma (A_2,A_2)).\hfill \\ & & {\displaystyle \mathcal{R}:=\frac{2\Gamma (A_1,A_2)+\sqrt{\Delta }}{2\sqrt{det\left(A_1\right)det\left(A_2\right)}}{e}^{{\tau }_1{t}_1+{\tau }_2{t}_2},\text{where,}\text{for}i=1,2,}\hfill \\ & & t_i:=\left\{\begin{array}{cc}\frac{\pi }{2}-\text{arctan}\frac{\mathrm{tr}\left(A_1\right)\mathrm{tr}\left(A_2\right)(k{\tau }_i+{\tau }_{3-i})}{2{\tau }_1{\tau }_2\sqrt{\Delta }}\hfill & \text{if}{\delta }_{A_i}<0,\hfill \\ \text{arctanh}\frac{2{\tau }_1{\tau }_2\sqrt{\Delta }}{\mathrm{tr}\left(A_1\right)\mathrm{tr}\left(A_2\right)(k{\tau }_i-{\tau }_{3-i})}\hfill & \text{if}{\delta }_{A_i}>0,\hfill \\ \frac{\sqrt{\Delta }}{(\mathrm{tr}\left(A_1{A}_2\right)-\mathrm{tr}\left(A_1\right)\mathrm{tr}\left(A_2\right)/2){\tau }_i}\hfill & \text{if}{\delta }_{A_i}=0.\hfill \end{array}\right.\hfill \end{array}$$Note that the value of t_i in the case ${\delta }_{A_i}=0$ differs from that in [*] by a multiplicative factor 2.

Finally, we define the sign function as follows: $$sign\left(x\right):=\left\{\begin{array}{cc}\hfill +1& \text{if}x>0\text{,}\hfill \\ \hfill 0& \text{if}x=0\text{,}\hfill \\ \hfill -1& \text{if}x<0\text{.}\hfill \end{array}\right.$$

The following theorem is the main result of [*] and is reported here verbatim (the result is correct, provided that the definitions above are adopted). LF means Lyapunov function and GUAS means globally uniformly asymptotically stable.

Theorem 1 ([*]):

We have the following stability conditions for system (1(1) $$\dot{x}\left(t\right)=u\left(t\right)A_{1}x\left(t\right)+(1-u\left(t\right))A_{2}x\left(t\right),u:[0,\infty )\to [0,1].$$(1) )

S1	If $\Gamma \mathrm{tr}(A_1,A_2)>-\sqrt{det\left(A_1\right)det\left(A_2\right)},\text{and}\mathrm{tr}\left(A_1{A}_2\right)>-2\sqrt{det\left(A_1\right)det\left(A_2\right)}$, then the system admits a quadratic LF. If   $-\sqrt{det\left(A_1\right)det\left(A_2\right)}<\Gamma (A_1,A_2)\le \sqrt{det\left(A_1\right)det\left(A_2\right)}$, then the condition $\mathrm{tr}\left(A_1{A}_2\right)>-2\sqrt{det\left(A_1\right)det\left(A_2\right)}$ is automatically satisfied. As a consequence, the system admits a quadratic LF.
S2	If $\Gamma (A_1,A_2)<-\sqrt{det\left(A_1\right)det\left(A_2\right)},$ then the system is unbounded,
S3	If $\Gamma (A_1,A_2)=-\sqrt{det\left(A_1\right)det\left(A_2\right)},$ then the system is uniformly stable but not GUAS,
S4	If $\Gamma (A_1,A_2)>\sqrt{det\left(A_1\right)det\left(A_2\right)},\text{and}\mathrm{tr}\left(A_1{A}_2\right)\le -2\sqrt{det\left(A_1\right)det\left(A_2\right)}$, then the system is GUAS, uniformly stable (but not GUAS) or unbounded, respectively, if $$\mathcal{R}<1,\mathcal{R}=1,\mathcal{R}>1.$$

Remark 1 In the diagonalisable case ${\delta }_{A_1}{\delta }_{A_2}\ne 0$ the parameters τ₁, τ₂, and k are invariant under the transformation (A₁, A₂) → (A₁/α₁, A₂/α₂), for every α₁, α₂ > 0. This is no more true in the nondiagonalisable case. Notice however that in any case the stability conditions of Theorem 1 do not depend on coordinate transformations or on rescalings of the type (A₁, A₂) → (A₁/α₁, A₂/α₂). This is true in particular for the function $\mathcal{R}$.

Remark 2 Condition S1 may be rephrased by saying that the system admits a quadratic LF if and only if the condition $-\sqrt{det\left(A_1\right)det\left(A_2\right)}<\Gamma (A_1,A_2)<\sqrt{det\left(A_1\right)det\left(A_2\right)}+\frac{1}{2}\mathrm{tr}\left(A_1\right)\mathrm{tr}\left(A_2\right)$ is satisfied. Similarly, the preliminary assumptions in S4 are equivalent to the condition $\Gamma (A_1,A_2)\ge \sqrt{det\left(A_1\right)det\left(A_2\right)}+\frac{1}{2}\mathrm{tr}\left(A_1\right)\mathrm{tr}\left(A_2\right)$.

Proof in the case S4. In the case S4, the following result holds.

Lemma 1 ([*]):

If A₁, A₂ satisfy the preliminary assumptions in S4, then, up to a linear change of coordinates and up to multiplication by positive constants, A₁, A₂ assume the normal forms $$A_1=\left(\begin{array}{cc}{\tau }_1& 1\\ sign\left({\delta }_{A_1}\right)& {\tau }_1\end{array}\right),A_2=\left(\begin{array}{cc}{\tau }_2& sign\left({\delta }_{A_2}\right)/F\\ F& {\tau }_2\end{array}\right),$$where $F\in \mathbb{R}$ is such that |F| ≥ 1 and $F+\frac{sign\left({\delta }_{A_1}{\delta }_{A_2}\right)}{F}=2k$.

In the case S4 and assuming that the matrices A₁, A₂ are in normal form, the stability properties of the system are completely determined by the behaviour of special trajectories of the system turning around the origin in clockwise sense, sometimes called worst trajectories: the system is GUAS if and only if such trajectories converge to the origin and it is unstable if and only if they are unbounded (the worst trajectories are formally defined as those trajectories that follow at each time the vector field forming the smallest angle measured in clockwise sense from the exiting radial direction). The switching points of a worst trajectory coincide with its intersections with the two lines defined as the set of zeroes of det(A₁z, A₂z), a quadratic polynomial in $z\in {\mathbb{R}}^2$ (note that in the case S4, the discriminant of det(A₁z, A₂z), which is equal to Δ, is strictly positive). A direct computation shows that the two lines correspond to the directions $$w_{\pm }=\left(\begin{array}{c}1\\ m_{\pm }\end{array}\right)\text{with}{m}_{\pm }=\frac{-F+sign\left({\delta }_{A_1}{\delta }_{A_2}\right)/F\pm \sqrt{\Delta }}{2({\tau }_2-{\tau }_{1}sign\left({\delta }_{A_2}\right)/F)}\text{if}{\tau }_2-{\tau }_{1}sign\left({\delta }_{A_2}\right)/F\ne 0,$$or $$w_{+}=\left(\begin{array}{c}0\\ 1\end{array}\right),w_{-}=\left(\begin{array}{c}1\\ -{\tau }_1\end{array}\right)\text{if}{\tau }_2-{\tau }_{1}sign\left({\delta }_{A_2}\right)/F=0.$$By comparing the values A₁e₂ and A₂e₂, where e₂ = (0, 1)^T, one obtains that:

• if ${\tau }_2-{\tau }_{1}sign\left({\delta }_{A_2}\right)/F<0$, then the worst trajectory starting at e₂ must be initially made of an arc of trajectory of the first vector field reaching the line of direction w₋ (note that in this case, m⁻ > m⁺),

• if ${\tau }_2-{\tau }_{1}sign\left({\delta }_{A_2}\right)/F>0$, then the worst trajectory starting at e₂ must be initially made of an arc of trajectory of the second vector field reaching the line of direction w₊ (in this case, m⁺ > m⁻).

Similarly, in the case ${\tau }_2-{\tau }_{1}sign\left({\delta }_{A_2}\right)/F=0$, the worst trajectory starting at (1, 0)^T must be initially made of an arc of trajectory of the second vector field up to the line of direction w₊.

As a consequence of the observations above, the worst trajectories always coincide with those trajectories following the first vector field from the line of direction w₊ to that of direction w₋, and the second vector field from the line of direction w₋ to that of direction w₊.

By symmetry reasons, in order to determine if the system is GUAS or not, it is enough to compute, for a worst trajectory, the ratio $\mathcal{R}$ between the modulus of the point obtained after half a turn and the modulus of the initial point. If t₁ and t₂ are the time lengths for the arcs corresponding to the first vector field and the second one, respectively, then one has that $$\mathcal{R}=\tilde{\mathcal{R}}{e}^{{\tau }_1{t}_1+{\tau }_2{t}_2},$$where $\tilde{\mathcal{R}}$ is the ratio obtained after half a turn by following the same switching strategy as for the worst trajectory replacing the matrices A₁ and A₂ with $${\tilde{A}}_1=\left(\begin{array}{cc}0& 1\\ sign\left({\delta }_{A_1}\right)& 0\end{array}\right),{\tilde{A}}_2=\left(\begin{array}{cc}0& sign\left({\delta }_{A_2}\right)/F\\ F& 0\end{array}\right).$$

In order to compute $\tilde{\mathcal{R}}$, let us observe that $V_1\left(z\right)=z_2^2-sign\left({\delta }_{A_1}\right)z_1^2$ and $V_2\left(z\right)=F^2{z}_1^2-sign\left({\delta }_{A_2}\right)z_2^2$ are quadratic prime integrals for the differential equations $\dot{z}={\tilde{A}}_{1}z$ and $\dot{z}={\tilde{A}}_{2}z$, respectively. Thus, the trajectory described above corresponds to the application of the first vector field from w₊ to $\sqrt{\frac{V_1\left(w_{+}\right)}{V_1\left(w_{-}\right)}}{w}_{-}$ and then of the second one reaching the point $-\sqrt{\frac{V_2\left(w_{-}\right)}{V_2\left(w_{+}\right)}·\frac{V_1\left(w_{+}\right)}{V_1\left(w_{-}\right)}}{w}_{+}$. By replacing the values w_± obtained above and after simplifications, one gets $$\tilde{\mathcal{R}}=\sqrt{\frac{V_2\left(w_{-}\right)}{V_2\left(w_{+}\right)}·\frac{V_1\left(w_{+}\right)}{V_1\left(w_{-}\right)}}=\frac{2\Gamma (A_1,A_2)+\sqrt{\Delta }}{2\sqrt{det\left(A_1\right)det\left(A_2\right)}}.$$

It remains to compute the switching times. To compute the time length corresponding to the arc of trajectory of the first vector field, let us define m(t) = x₂(t)/x₁(t). Then, it is easy to see that the dynamics of the first subsystem are given by $$\dot{m}\left(t\right)=sign\left({\delta }_{A_1}\right)-m{\left(t\right)}^2.$$As a consequence, the time length for the arc between the lines of directions w₊ and w₋ is equal to $$\begin{array}{cc}\hfill {\int }_{m_{-}}^{+\infty }\frac{1}{m^2-sign\left({\delta }_{A_1}\right)}& dm+{\int }_{-\infty }^{m_{+}}\frac{1}{m^2-sign\left({\delta }_{A_1}\right)}dm\text{if}{\tau }_2-{\tau }_{1}sign\left({\delta }_{A_2}\right)/F\le 0\text{or}\hfill \\ & {\int }_{m_{-}}^{m_{+}}\frac{1}{m^2-sign\left({\delta }_{A_1}\right)}dm\text{if}{\tau }_2-{\tau }_{1}sign\left({\delta }_{A_2}\right)/F>0\text{.}\hfill \end{array}$$Therefore, we get the following:

• If $sign\left({\delta }_{A_1}\right)=-1$ and ${\tau }_2-{\tau }_{1}sign\left({\delta }_{A_2}\right)/F\le 0$, then $t_1=\frac{\pi }{2}-\arctan \left(m_{-}\right)+\arctan \left(m_{+}\right)+\frac{\pi }{2}=\pi +\arctan \frac{m_{+}-m_{-}}{1+m_{-}{m}_{+}}\text{(mod}\pi \text{)}=\pi +\arctan \frac{\sqrt{\Delta }}{2(k{\tau }_1+{\tau }_2)}\text{(mod}\pi \text{)}=\frac{\pi }{2}-\arctan \frac{2(k{\tau }_1+{\tau }_2)}{\sqrt{\Delta }}$. The same result is obtained in the case ${\tau }_2-{\tau }_{1}sign\left({\delta }_{A_2}\right)/F>0$.

• If $sign\left({\delta }_{A_1}\right)=1$, then $t_1=\frac{1}{2}\log \left|\frac{1-m_{+}}{1+m_{+}}\right|-\frac{1}{2}\log \left|\frac{1-m_{-}}{1+m_{-}}\right|=\frac{1}{2}\log \left(\frac{2(k{\tau }_1-{\tau }_2)+\sqrt{\Delta }}{2(k{\tau }_1-{\tau }_2)-\sqrt{\Delta }}\right)=\mathrm{arctanh}\frac{\sqrt{\Delta }}{2(k{\tau }_1-{\tau }_2)}$. Note that the computation relies on the fact that in this case $2(k{\tau }_1-{\tau }_2)>\sqrt{\Delta }$.

• If $sign\left({\delta }_{A_1}\right)=0$, then $t_1=\frac{1}{m_{-}}-\frac{1}{m_{+}}=\frac{\sqrt{\Delta }}{2k{\tau }_1}.$

By symmetry reasons (in particular the transformation y₁ = x₂, y₂ = Fx₁ exchanges the role of A₁ and A₂), the time length t₂ corresponding to the arcs of trajectory of the second vector field is computed by means of the same formula, when one exchanges the values of ${\delta }_{A_1}$ and τ₁ with ${\delta }_{A_2}$ and τ₂, respectively.

The final formula for t_i is then computed by taking into account the rescaling of $\sqrt{\Delta }$ when passing from a general pair of matrices to the corresponding normal forms.

Notes

¹ We thank Prof. Tewfik Sari for pointing out this error.