Cm Math Dynamics — Examples

For fixed $a$, the curve $x(t):=e^{tA}a$ is defined on all of $\mathbb{R}$ and satisfies $x^{\mathrm{\prime }}(t)=A{e}^{tA}a=Ax(t)$ and $x(0)=a$ by [F1]. Hence it is a solution of the initial value problem on $\mathbb{R}$, and by [F2] it is the unique maximal solution through $(0,a)$; therefore $I_a=\mathbb{R}$ and $\mathrm{\Phi }(t,a)=x(t;a)=e^{tA}a$ for every $t\in \mathbb{R}$. As $a$ was arbitrary, $\mathrm{\Omega }=\mathbb{R}\times {\mathbb{R}}^n$.

For fixed $t$, the map $b\mapsto \mathrm{\Phi }(t,b)=e^{tA}b$ is linear, and its derivative at every point is itself: $\mathrm{\Phi }(t,a+h)-\mathrm{\Phi }(t,a)-e^{tA}h=0$, so $D_a\mathrm{\Phi }(t,a)=e^{tA}$ with zero remainder; this holds for every real $t$.

The composition law for the flow follows from [F1]: $\mathrm{\Phi }(s+t,a)=e^{(s+t)A}a=e^{sA}{e}^{tA}a=\mathrm{\Phi }(s,\mathrm{\Phi }(t,a))$, in agreement with the general identity of The local flow is a partial one-parameter group: identity, composition and inverse-time laws.

The function $Y(t):=e^{tA}$ satisfies $Y^{\mathrm{\prime }}(t)=AY(t)$ and $Y(0)=I_n$ by [F1]; by the uniqueness clause of [F3] it is the unique solution of the variational initial value problem on every compact interval containing $0$, and every compact interval is contained together with $0$ in one such interval. In particular the conclusion of the smooth-dependence theorem for the derivative of $a\mapsto \mathrm{\Phi }(t,a)$ is achieved exactly, with no error term.

$S=M^{-1\mathrm{/}2}K{M}^{-1\mathrm{/}2}=K\mathrm{/}m$, and $S{w}_1=\frac{1}{m}\left(\begin{array}{cc}k+c& -c\\ -c& k+c\end{array}\right)\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right)=\frac{1}{\sqrt{2}}\frac{1}{m}\left(\begin{array}{c}k\\ k\end{array}\right)=\frac{k}{m}{w}_1,$ $S{w}_2=\frac{1}{m}\left(\begin{array}{cc}k+c& -c\\ -c& k+c\end{array}\right)\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ -1\end{array}\right)=\frac{1}{\sqrt{2}}\frac{1}{m}\left(\begin{array}{c}k+2c\\ -(k+2c)\end{array}\right)=\frac{k+2c}{m}{w}_2\mathrm{.}$ Since $w_1,w_2$ are orthonormal, $O=(w_1{w}_2)$ is orthogonal and diagonalises $S$ with the displayed eigenvalues; the eigenvalues are non-negative, ${\lambda }_1=k\mathrm{/}m>0$ exactly when $k>0$, and ${\lambda }_2=(k+2c)\mathrm{/}m>0$ exactly when $k+2c>0$.

By [L1] the general solution is $q=M^{-1\mathrm{/}2}Oz$, where each modal coordinate is given by the theorem according to the sign of its eigenvalue: if $k>0$ then $z_1=a_1\cos ({\omega }_{1}t)+b_1\sin ({\omega }_{1}t)$ with ${\omega }_1=\sqrt{{\lambda }_1}$, and if $k=0$ then $z_1=a_1+b_{1}t$; if $k+2c>0$ then $z_2=a_2\cos ({\omega }_{2}t)+b_2\sin ({\omega }_{2}t)$ with ${\omega }_2=\sqrt{{\lambda }_2}$, and if $k=c=0$ then $z_2=a_2+b_{2}t$. The modal energies are constant. The verification of the scalar solution formulas is the one given by the theorem via The derivatives of sine and cosine are cosine and minus sine.

The zero-mode branch $k=0$: then ${\lambda }_1=0$ and the first modal solution is the affine motion $a_1+b_{1}t$ along $w_1$ by [L1], which is unbounded unless $b_1=0$; the second eigenvalue is $2c\mathrm{/}m\ge 0$, with $2c\mathrm{/}m>0$ when $c>0$, so the second mode oscillates with frequency $\sqrt{2c\mathrm{/}m}$ unless also $c=0$, in which case both modes are affine. Thus when $k=0<c$ the only free direction is $\frac{1}{\sqrt{2}}(1,1)$, while when $k=c=0$ both independent directions are free; in either case a zero mode is not an oscillator.

Differentiating the two components of the displayed curve with respect to $t$ using $(\sin {)}^{\mathrm{\prime }}=\cos$, $(\cos {)}^{\mathrm{\prime }}=-\sin$ gives $(-\omega x\sin \omega t+y\cos \omega t,-{\omega }^{2}x\cos \omega t-\omega y\sin \omega t)$, whose first component is the second component of the curve and whose second component is $-{\omega }^2$ times the first; hence the curve solves the phase system $z^{\mathrm{\prime }}=F(z)$ and takes the value $(x,y)$ at $t=0$. It is defined on all of $\mathbb{R}$, so [L1] makes it the unique maximal solution and gives $\mathrm{\Omega }=\mathbb{R}\times {\mathbb{R}}^2$.

For $x\mathrm{\ne }0$ on the section, $\mathrm{\Phi }(t,(x,0))=(x\cos \omega t,-\omega x\sin \omega t)$: the second component vanishes exactly when $\sin \omega t=0$, i.e. at $t=k\pi \mathrm{/}\omega$, $k\in \mathbb{Z}$, by [L3]. For $x>0$ the first positive time at which the orbit meets $\mathrm{\Sigma }$ is $t=\pi \mathrm{/}\omega$, with $\mathrm{\Phi }(\pi \mathrm{/}\omega ,(x,0))=(-x,0)$; therefore $\tau (x,0)=\pi \mathrm{/}\omega$ and $P(x,0)=(-x,0)$ for every $x>0$ close to $1$.

The transversality conditions: with $h(x,y)=y$ we have $Dh=(0,1)$, $F(x_0)=(0,-{\omega }^2)$ and $F(x_1)=(0,{\omega }^2)$, so $Dh(x_0)F(x_0)=-{\omega }^2\mathrm{\ne }0$ and $Dh(x_1)F(x_1)={\omega }^2\mathrm{\ne }0$. By step 2.1 and [L3] the open reference segment has no hit of the closed section $\mathrm{\Sigma }$, which verifies the extra exclusion condition of [L2]. Hence [L2] applies at the reference data $(t_0,x_0)=(\pi \mathrm{/}\omega ,(1,0))$, and its conclusion is realized by the explicit map of step 2.1: $P$ maps a neighbourhood of $(1,0)$ in $\mathrm{\Sigma }$ onto a neighbourhood of $(-1,0)$ in $\mathrm{\Sigma }$, and $Q(y,0)=(-y,0)$ inverts it.

Finally the map $x\mapsto -x$ is a $C^1$ diffeomorphism of $\mathbb{R}$ onto itself, so its restriction to the section is a $C^1$ diffeomorphism between the stated neighbourhoods; this agrees with the general conclusion of the proposition and shows that the local inverse is again given by the reversed crossing, since the same computation with $x<0$ runs the reference orbit backwards.

Since $y_j=x_j$ for $\mathrm{\mid }j\mathrm{\mid }\le 2$, every term of $d(x,y)$ with $\mathrm{\mid }j\mathrm{\mid }\le 2$ vanishes; the only nonzero term is the one with $j=3$, equal to $2^{-3}\mathrm{\mid }{x}_3-y_3\mathrm{\mid }=2^{-3}$. Hence $d(x,y)=1\mathrm{/}8\le 2^{1-2}$, which illustrates the second inclusion of The bilateral shift metric is a metric, cylinders are its basic balls, and the shift space is sequentially compact on a concrete pair.

The displayed sequence $s$ is well defined because the prescription is periodic, it lies in the cylinder $[1,0,1{]}_{-1,1}$ by construction, and ${\sigma }^3(s{)}_k=s_{k+3}=s_k$ for every $k$, so ${\sigma }^3(s)=s$; thus $s$ is a periodic point of $\sigma$ and it lies in the prescribed cylinder, illustrating clause 3 of [L2] on an explicit block.

For $y$ as in item 3, exactly one term of the defining sum of $d(x,y)$ is nonzero, so $d(x,y)=2^{-k}$. Moreover ${\sigma }^k(x{)}_0=x_k\mathrm{\ne }{y}_k={\sigma }^k(y{)}_0$, so the term with $j=0$ in the sum for $d({\sigma }^k(x),{\sigma }^k(y))$ equals $1$ and all terms are nonnegative; hence $d({\sigma }^k(x),{\sigma }^k(y))\ge 1$. This realizes the sensitivity constant $1$ of [L2] with the pair $(x,y)$.

Compute ${\sigma }^n(U)=\{z:z_{-n}=1,z_{1-n}=0\}$ and $V=\{z:z_{-1}=0\}$; the intersection is nonempty exactly when the three coordinate prescriptions are consistent, which they are as soon as no two of the coordinates $-n$, $1-n$, $-1$ coincide with conflicting values. For $n=1$ the prescriptions are $z_{-1}=1$ and $z_{-1}=0$, a contradiction, so $\sigma (U)\cap V=\mathrm{\varnothing }$. For $n\ge 2$: if $n=2$ the coordinates are $-2,-1,-1$ and the prescriptions are $z_{-2}=1$, $z_{-1}=0$, consistent; if $n\ge 3$ the three coordinates are distinct, so a sequence with the prescribed values exists. Hence ${\sigma }^n(U)\cap V\mathrm{\ne }\mathrm{\varnothing }$ for every $n\ge 2$, as claimed, and the mixing condition of [L2] holds on this pair of basic open sets with $N=2$.

Write $A=-I_2+N$ with $N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, so that $N^2=0$ and $N$ commutes with $I_2$. Define $F(t):=e^{-t}(I_2+tN)=e^{-t}\left(\begin{array}{cc}1& t\\ 0& 1\end{array}\right)$; differentiating entrywise with the product rule and $(e^{-t}{)}^{\mathrm{\prime }}=-e^{-t}$ [L4] gives $F^{\mathrm{\prime }}(t)=-e^{-t}(I_2+tN)+e^{-t}N=e^{-t}(-I_2+N-tN)$ and $AF(t)=(-I_2+N)e^{-t}(I_2+tN)=e^{-t}(-I_2+N-tN+t{N}^2)=e^{-t}(-I_2+N-tN)$; since $F(0)=I_2$, [L2] gives $e^{tA}=F(t)$ for all $t$.

For $B=\mathrm{diag}(1\mathrm{/}2,-3\mathrm{/}2)$ define $G(t):=\mathrm{diag}(e^{t\mathrm{/}2},e^{-3t\mathrm{/}2})$; differentiating entrywise gives $G^{\mathrm{\prime }}(t)=\mathrm{diag}(\frac{1}{2}{e}^{t\mathrm{/}2},-\frac{3}{2}{e}^{-3t\mathrm{/}2})=BG(t)$ and $G(0)=I_2$, so [L2] gives $e^{tB}=G(t)$; in particular $e^{tB}{e}_0=e^{t\mathrm{/}2}{e}_0$ and $e^{tB}{e}_1=e^{-3t\mathrm{/}2}{e}_1$. No eigenvalue of $B$ is purely imaginary: if $B_{\mathbb{C}}v=\lambda v$ with $\lambda \in \mathbb{C}$ and $0\mathrm{\ne }v=(v_0,v_1)\in {\mathbb{C}}^2$, the two coordinates give $\frac{1}{2}{v}_0=\lambda v_0$ and $-\frac{3}{2}{v}_1=\lambda v_1$, so $v_0\mathrm{\ne }0$ forces $\lambda =1\mathrm{/}2$ while $v_1\mathrm{\ne }0$ forces $\lambda =-3\mathrm{/}2$. Hence [L3] applies to $B$.

The only eigenvalue of $A$ is $-1$: if $A_{\mathbb{C}}v=\lambda v$ with $\lambda \in \mathbb{C}$ and $0\mathrm{\ne }v=(v_0,v_1)\in {\mathbb{C}}^2$, the second coordinate gives $-v_1=\lambda v_1$ and the first gives $-v_0+v_1=\lambda v_0$. If $v_1=0$ then $v_0\mathrm{\ne }0$ and $\lambda =-1$; if $v_1\mathrm{\ne }0$ then $\lambda =-1$ from the second coordinate. So no eigenvalue of $A$ has real part $0$, and $[L3]$ applies to $A$.

Norm estimate for $A$: put $w(t):=e^{tA}v$, so $w^{\mathrm{\prime }}=Aw$ by step 1.1. For $f(t):=\mathrm{\parallel }w(t){\mathrm{\parallel }}_2^2=w_0(t{)}^2+w_1(t{)}^2$ the product rule gives $f^{\mathrm{\prime }}(t)=2⟨Aw,w{⟩}_1=-2w_0^2+2w_0{w}_1-2w_1^2\le -w_0^2-w_1^2=-f(t),$ because $2w_0{w}_1\le w_0^2+w_1^2$, equivalently $(w_0-w_1{)}^2\ge 0$. Hence $(e^{t}f(t){)}^{\mathrm{\prime }}=e^t(f+f^{\mathrm{\prime }})\le 0$, and the derivative sign criterion [L5] gives $e^{t}f(t)\le f(0)=\mathrm{\parallel }v{\mathrm{\parallel }}_2^2$ for $t\ge 0$. Taking nonnegative square roots yields $\mathrm{\parallel }{e}^{tA}v{\mathrm{\parallel }}_2\le e^{-t\mathrm{/}2}\mathrm{\parallel }v{\mathrm{\parallel }}_2$ for all $v$ and $t\ge 0$, including $v=0$. Thus $C=1$, $c=1\mathrm{/}2$ are admissible constants for $A$ in the sense of [L3].

Identification of the subspaces for $A$: the estimate of step 2.2 holds for every $v\in {\mathbb{R}}^2$, so $e^{tA}v\to 0$ as $t\to +\mathrm{\infty }$ for every $v$ (by $e^{-t\mathrm{/}2}\to 0$ from [L4]); by the characterisation in [L3] the stable subspace is $E_s={\mathbb{R}}^2$, and since ${\mathbb{R}}^2=E_s\oplus E_u$ we get $E_u=\{0\}$.