Dynamical Systems on the Sphere | 3 Nov. 2019 |

In this post we start a bit with control of systems living on compact metric spaces. This is interesting since we have to rethink what we want, xtoinfty cannot even occur. Let a great arc (circle) mathcal{G} on $mathbf{S}^{n-1}:={xin mathbf{R}^n;:;|x|_2=1}$ be parametrized by the normal of a hyperplane $mathcal{H}_a:={xin mathbf{R}^n;:;langle a , xrangle =0}$ through , i.e., $mathcal{G}=mathcal{H}_acap mathbf{S}^{n-1}$ , is the axis of rotation for .

To start, consider a linear dynamical systems dot{x}=Ax , x(0)=x_0 , for which the solution is given by $x(t)=mathrm{exp}(At)x_0$ . We can map this solution to the sphere via x(t) mapsto x(t)/|x|_2=:y(t) . The corresponding vector field for y(t) is given by

$begin{array}{ll} dot{y} &= f(y) &= frac{dot{x}}{|x|_2}+x frac{partial}{partial t}left(frac{1}{|x|_2} right) &= frac{dot{x}}{|x|_2}-x frac{1}{|x|_2^3}x^{top}dot{x} &= A frac{x}{|x|_2} - frac{x}{|x|_2} frac{x^{top}}{|x|_2}Afrac{x}{|x|_2} &= left(A-(y^{top}Ay) I_nright) y. end{array}$

The beauty is the following, to find the equilibria, consider $(A-(y^{top}Ay)I_n)y = 0$ . Of course, y=0 is not valid since $0notin mathbf{S}^{n-1}$ . However, take any eigenvector of , then v/|v|_2 is still an eigenvector, plus it lives on $mathbf{S}^{n-1}$ . If we plug such a scaled (from now on consider all eigenvectors to live on $mathbf{S}^{n-1}$ ) into f(y) = 0 we get (A-lambda I_n)v=0 , which is rather cool. The eigenvectors of are (at least) the equilibria of . Note, if f(v)=0 , then so is f(-v)=0 , each eigenvector gives rise to two equilibria. We say at least, since if mathrm{ker}(A) is non-trivial, then for any $yin mathrm{ker}(A)subset mathbf{R}^n$ , we get f(y)=0 .

Let lambda(A)subset mathbf{R} , then what can be said about the stability of the equilibrium points? Here, we must look at $T_vmathbf{S}^{n-1}$ , where the -dimensional system is locally a n-1 -dimensional linear system. To do this, we start by computing the standard Jacobian of , which is given by

$Df(y) = A-I_n (y^{top}Ay) - 2 yy^{top}A.$

As said before, we do not have a -dimensional system. To that end, we assume Ain mathsf{Sym}(n,mathbf{R}) with diagonalization $A=QLambda Q^{top}$ (otherwise peform a Gram-Schmidt procedure) and perform a change of basis $Df'=Q^{top}DfQ$ . Then if we are interested in the qualitative stability properties of v_i (the $i^{mathrm{th}}$ eigenvector of ), we can simply look at the eigenvalues of the matrix resulting from removing the $i^{mathrm{th}}$ row and column in $Df'|_{v_i}$ , denoted $Df'|_{T_{v_i}mathbf{S}^{n-1}}$ .

We do two examples. In both figures are the initial conditions indicated with a dot while the equilibrium points are stars.

First, let

$A_1 = left[ begin{array}{lll} 1 & 0 & 0 0 & -1 & 0 0 & 0 & -1 end{array}right]$

Clearly, for A_1 we have Q=I_3 and let (v_1,v_2,v_3) be (e_1,e_2,e_3) . Then using the procedure from before, we find that pm v_1 is stable (attracting), while pm (v_2,v_3) are locally unstable in one direction while in the other (along the great arc). Hence, this example gives rise to the two stable poles.

Secondly, we consider

$A_2 = left[ begin{array}{lll} 1 & 2 & 0 2 & -10 & 0 0 & 0 & 0.1 end{array}right]$

To assess stability, we compute $Df'|_{T_{v_i}mathbf{S}^{2}}$ for i=1,2,3 and v_1=(-0.17,0.98,0) , v_2=(-0.98,-0.17,0) , v_3=(0,0,1) , lambda=(-10.3,1.3,0.1) :

$Df'|_{T_{v_1}mathbf{S}^{2}} = left[ begin{array}{ll} 11.7 & 0 0 & 10.4 end{array}right],quad Df'|_{T_{v_2}mathbf{S}^{2}} = left[ begin{array}{ll} -11.7 & 0 0 & -1.2 end{array}right],quad Df'|_{T_{v_3}mathbf{S}^{2}} = left[ begin{array}{ll} -10.4 & 0 0 & 1.2 end{array}right].$

Recall that tangent space basis is in line with the eigenvectors plus that the qualitative behaviour around v_1 equals that around -v_1 . This explains the attracting great arc parametrized by $mathcal{H}_{v_1}$ . Note, although the arc is attracting, f(y) does not vanish on the complete arc, we are locally unstable along v_2 in $T_{v_3}mathbf{S}^{2}$ and therefore converge to $pm v_2 in mathcal{H}_{v_1}cap mathbf{S}^2$ .

Now, a valid question is, why do stable equilibrium points (for dot{x}=Ax on $mathbf{R}^3$ ) become unstable on the sphere? For our example, we see that the full (minus ) line: all cv_1 s.t. |c|in mathbf{R} collapses to just two points: pm v_1 . Not just any two points, but two points surrounded by instability, hence the positive definite Jacobian. In a similar fashion it follows that the unstable asymptotes pm infty (v_2,v_3) now relate to (finite!) attracting equilibria pm(v_2,v_3) . The fact that v_2 is attracting from all directions while v_3 relates to a saddle follows from the corresponding eigenvalues, lambda_2>lambda_3 . From the figure you can take away that $mathrm{span}(v_2,v_3)$ is the attracting plane, resulting in $H_{v_1}cap mathbf{S}^2$ being an attracting great arc (since v_1perp (v_2,v_3) ).

We observe that by construction it seems impossible to have 1 stable pole since the equilibrium points come in pairs. You might say, sure, but these vector fields where created by using linear systems. Perhaps there is some wild differential equation which does the trick? The answer is yes, we can however not remove the equilibrium points completely, we cannot comb a hairy ball.