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Assume we have a time-invariant linear ODE that arose from a non-linear ODE by linearizing at some point $U_1=(u_1,v_1,w_1)$ and that has the following form: $$ \dot{\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}}=\underbrace{\begin{pmatrix}0 & 1 & 0\\-f'(u_1) & -c & 1\\0 & 0 & 0\end{pmatrix}}_{=:A}\cdot\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}, $$ where $f'(u_1)<0, c>0$. Then the matrix has eigenvalues $$ \lambda_1=-\frac{c}{2}-\sqrt{\frac{c^2}{4}-f'(u_1)},~~~~~\lambda_2=-\frac{c}{2}+\sqrt{\frac{c^2}{4}-f'(u_1)},~~~~~\lambda_3=0. $$ Since $f'(u_1)<0$, $\lambda_1<0$ and $\lambda_2>0$.

In terms of stability theory, this means that unstable subspace, stable subspace and centre subspace are non-empty. Moreover, as far as I know, the centre subspace here is stable (in the Lyapunov sense), too. So we can say that we have one unstable subspace and two stable ones. Let these be spans of the eigenvectors $X_1$ (unstable), $X_2$ and $X_3$.

Now, project this equation down to the unit sphere $S^2$ by defining $$ s(t)=\frac{x(t)}{\lvert x(t)\rvert}. $$ Then, by the chain rule, we get $$ \dot{s}(t)=h(s(t)),~~~\text{where }h(s)=(A-s^TAs\cdot I)s. $$

Now, there are some statements that I do not understand.

(1.) By homogeneity, the original equation induces a differential equation on $S^2$ obtained by identifying opposite points on $S^2$.

I do not understand this claim, in particular how the new ODE looks like and what it does on the sphere.

(2.) The new ODE on $S^2$ has two saddle points, two attracting points and two repelling points. These come from the eigenspaces. Let $X_2$ be the eigenvector that gives the saddle. Set $C=\text{span}\left\{X_2,X_3\right\}\cap S^2$.

I do not understand where the two saddle points, two attracting points and two saddle points come from and how they are connected with the unstable subspace, stable subspace and centre subspace of the original ODE.

(3.) For the new ODE on $S^2$, the two points contained in $\text{span}\left\{X_1\right\}\cap S^2$ are attractors.

I do not understand why these two points are attractors for the new ODE on the sphere. That might be connected with my problem that I do not understand what the ODE on the sphere does and how it works on the sphere.

Maybe you can explain me this three statements.

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  • $\begingroup$ what is A in equation for $\dot{s}$? $\endgroup$ – nonlinearism Nov 9 '15 at 0:38
  • $\begingroup$ @nonlinearism $A$ is the matrix appearing in the original ODE system. Sorry, I forgot to mention that but I now added it. $\endgroup$ – M. Meyer Nov 9 '15 at 9:57
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After a discussion with the author of the OP, I write some explanations in the case where the "projection on the unit sphere" means intersection subspaces with it and renormalizion of vectors.

$(1)$ Let $s(t):=\frac{x(t)}{|x(t)|}$. Indeed, $|s(t)|=1$ (thus $s:\mathbb{R}\rightarrow\mathbb{S}^2$) and we have \begin{align*} \frac{\mathrm{d}s}{\mathrm{d}t} & = \frac{1}{|x(t)|}\frac{\mathrm{d}x}{\mathrm{d}t}-\frac{x(t)}{|x(t)|^3}\left(x(t)\cdot\frac{\mathrm{d}x}{\mathrm{d}t}\right) \\ & = \frac{1}{|x(t)|}Ax(t)-\frac{x(t)}{|x(t)|^3}\left(x(t)\cdot Ax(t)\right) \\ & = A\frac{x(t)}{|x(t)|}-\frac{x(t)}{|x(t)|}\left(\frac{x(t)^T}{|x(t)|}A\frac{x(t)}{|x(t)|}\right) \\ & = As(t)-(s(t)^TAs(t))s(t) \\ & = \left(1-s(t)^TAs(t)\right)As(t) \\ & = B(s(t))s(t) \end{align*} with $$B(s(t))=\left(1-s(t)^TAs(t)\right)A=\left(1-s_2\left(s_1(1-f'(u_2))-s_2+s_3\right)\right)A$$ if we write $s(t)=(s_1,s_2,s_3)(t)$. This is the ODE induces by the original one on the unit sphere. Not sure why they want to identify opposite points (as we discussed, identifying opposite points on the sphere is quotienting it by a relation of equivalence, and then they must define norm and ODE in the quotient...).

$(2)$ Let $(\lambda_k,X_k)_{1\leq k\leq3}$ the eigen pairs of the matrix $A$, that is $$AX_k=\lambda_kX_k\quad\quad\quad\forall1\leq k\leq3.$$ Then, for all $1\leq k\leq3$, we have \begin{align*} \frac{\mathrm{d}}{\mathrm{d}t}\left(\frac{X_k(t)}{|X_k(t)|}\right) & = B\left(\frac{X_k(t)}{|X_k(t)|}\right)\frac{X_k(t)}{|X_k(t)|} \\ & = \left(1-\frac{X_k(t)^T}{|X_k(t)|}A\frac{X_k(t)}{|X_k(t)|}\right)A\frac{X_k(t)}{|X_k(t)|} \\ & = \left(1-\frac{X_k(t)^T}{|X_k(t)|}\lambda_k\frac{X_k(t)}{|X_k(t)|}\right)\lambda_k\frac{X_k(t)}{|X_k(t)|} \\ & = \left(1-\lambda_k\frac{|X_k(t)|^2}{|X_k(t)|^2}\right)\lambda_k\frac{X_k(t)}{|X_k(t)|} \\ & = \left(1-\lambda_k\right)\lambda_k\frac{X_k(t)}{|X_k(t)|} \\ & = \Lambda_k\frac{X_k(t)}{|X_k(t)|} \end{align*} with $\Lambda_k=\left(1-\lambda_k\right)\lambda_k$. This shows that there are three new eigen pairs, namely $(\Lambda_k,\frac{X_k}{|X_k|})_{1\leq k\leq3}$. Now, following your notations, $\Lambda_3=0$ because $\lambda_3=0$, so $\frac{X_3}{|X_3|}$ is the eigenvector that gives the saddle (they call it $X_3$). Next, as $\lambda_1<0$, $\Lambda_1<0$ also. Finally, $\Lambda_2$ can have any sign, according to $\lambda_2\leq\frac{1}{2}$ or $\lambda_2>\frac{1}{2}$.

$(3)$ To see where are eigenspaces on the sphere (provided that I understand the question), we identify the eigenspaces in $\mathbb{R}^3$ spanned by the eigenvectors $\left(\frac{X_k}{|X_k|}\right)_{1\leq k\leq3}$ (these are the same that those spanned by $\left(X_k\right)_{1\leq k\leq3}$), and then we intersect them with the sphere. Lines and planes passing through the center of the sphere then becomes respectively points and great circles (or longitudes). For example, $C:=\mathrm{span}{X_1}\cap\mathbb{S}^2$ is the union of the two intersection points of $\mathbb{S}^2$ and the line spanned by $X_1$. Since $\Lambda_1<0$, they are attractors. Similaraly, the intersection $\mathrm{span}{X_2,X_3}\cap\mathbb{S}^2$ is the intersection of $\mathbb{S}^2$ with the vector plane spanned by $X_2$ and $X_3$, so it is a great circle passing through the projections of $X_2$ and $X_3$ on the sphere (it is a geodesic of the sphere).

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  • $\begingroup$ First of all: Thanks! Then one question: How do you know by $\Lambda_1 <0$ that the two points on the sphere that come from $\text{span}X_1\cap S^2$ are attractors? $\endgroup$ – M. Meyer Nov 11 '15 at 12:24
  • $\begingroup$ Suppose $\Lambda_1<0$ AND if $\Lambda_2<0$ (I was thinking to it when I wrote my answer but not precised it) ; the integral curves $(x_1(t),x_2(t),x_3(t))$ of the induced ODE satisfy $x_2=C|x_1|^{\Lambda_2/\Lambda_1}$ with $C\in\mathbb{R}$. So integral curves go to the origin (implied that the origin is the critical point $U_1$ here). To see what become these curve once projected on the sphere, the best is to draw a picture (in a plan $x_3=\mathrm{constant}$). But anyway, projection are coninuous, so the limit is unchanged after the projection... $\endgroup$ – Nicolas Nov 11 '15 at 14:59
  • $\begingroup$ I am sorry to not being very clear here, but it becomes hard to explain myself. Writing explanations concerning dynamical systems is really difficult ! But there is something that you need to keep in mind : my answer only is available if we consider the induced ODE as an ODE in $\mathbb{R}^3$, and then we project eigenspaces on the sphere and deal with the dynamic as we do in $\mathbb{R}^3$. Not sure if it was the original goal of the question, but lacking of background, I can not say anything else! You may read carefully your book to collect more clues on the meaning of this problem. $\endgroup$ – Nicolas Nov 11 '15 at 15:04
  • $\begingroup$ Dont we have to identify opposite points since otherwise we would have no uniqueness of solutions? $\endgroup$ – M. Meyer Nov 23 '15 at 17:54
  • $\begingroup$ @M.Meyer It was not clear to me if we were asked to solve the ODE in the quotient space. If it is, then my answer needs to be reviewed, of course. We probably can deal with ODE in quotient space after providing it with a (induced) norm. I have not really time to see it for now ; you may check what was the "true" problem in your book - what authors did ? - and then precise your question, since nobody answered it except me (an I am not sure if I have properly do it). $\endgroup$ – Nicolas Nov 23 '15 at 21:54

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