Phase curve of $\ddot{x}=-x,\ddot{y}=-y$

Rewriting the ODE in the title, we get $$\dot{x_1}=x_2,\dot{x_2}=-x_1, \dot{x_3}=x_4, \dot{x_4}=-x_3.$$ It is easy to show that $$x_1=A\cos(t)+B\sin(t),x_2=B\cos(t)-A\sin(t),\\x_3=C\cos(t)+D\sin(t),x_4=D\cos(t)-C\sin(t).$$ In Ordinary Differential Equations by V. I. Arnold, there are several problems on this equation. For example, it can be shown that each phase curve is on a 3-sphere, and is a great circle of that.

The next problem is more difficult: show that the phase curves on a given 3-sphere form a 2-sphere.

I attempt to find out which 3-dimensional subspace the 2-sphere lies in, but in vain.

And the last question is about the linking number. Since 3-sphere can be regarded as $\mathbb R^3\cup \{\infty\}$, a partition of 3-sphere into circles determines a partition of $\mathbb R^3$ into circles and nonclosed circles. Show that any two of the circles of this partition are linked with linking number 1.

I am not sure how to visualize this partition, and what the circles stand for.

• @Amzoti Yes, and I have edited it. Mar 11 '15 at 14:39
• I'm not an expert on this, but it seems to be connected with the so called Hopf fibration; see here: en.wikipedia.org/wiki/Hopf_fibration Mar 11 '15 at 15:00
• It seems that every property you are interested in derives from the fact that each solution of this differential system stays on a surface $$[x_1^2+x_2^2=a^2,\,x_3^2+x_4^2=b^2],$$ which, in turn, is included in the 3-sphere $$x_1^2+x_2^2+x_3^2+x_4^2=r^2$$ for some suitable $r$.
– Did
Mar 13 '15 at 15:02
• The two-sphere (the set of orbits) turns out not to be embedded in the three-sphere. It may help to introduce complex coordinates $z_{1} = x_{1} + ix_{2}$, $z_{2} = x_{3} + ix_{4}$; the flow of your system is $$(t, z_{1}, z_{2}) \mapsto (e^{it}z_{1}, e^{it}z_{2}).$$The three-sphere is invariant (see also Did's comment), and the quotient of $S^{3}$ by this circle action is the Hopf map (see Christian's comment); the quotient space is the set of complex lines through the origin of $\mathbf{C}^{2}$. Mar 13 '15 at 16:04
• By not rewriting the ODE we find that: $$x=A\cos(t)+B\sin(t)\\ y=C\cos(t)+D\sin(t)$$ Solving for $\cos(t)$ and $\sin(t)$ and summing the squares of these then yields a Lisajous Ellipse, degenerated eventually (if $AD-BC=0$): $$\left(\frac{Cx-Ay}{AD-BC}\right)^2+\left(\frac{Dx-By}{AD-BC}\right)^2 = 1$$ That's all "interesting" I can see. What more is there to be said? Mar 14 '15 at 9:44


The Riemann Sphere: For present purposes, the Riemann sphere is the holomorphic curve obtained by taking two copies $U_{0}$ and $U_{1}$ of the complex line, with respective coordinates $z$ and $w$, and identifying $w$ in $U_{1}$ with $z = 1/w$ in $U_{0}$. The origin ($w = 0$) in $U_{1}$ is regarded as the point at infinity in $U_{0}$, and vice versa.

The Complex Projective Line: If $z_{0}$ and $w_{0}$ are complex numbers, not both zero, the complex line they determine is the set of (complex) scalar multiples of $(z_{0}, w_{0})$. Throughout, the term "line" refers to a complex line through the origin in $\Cpx^{2}$. (A complex line is an oriented, real $2$-plane. However, most real $2$-planes through the origin of $\Cpx^{2}$ are not complex lines.)

Every non-zero vector in $\Cpx^{2}$ lies on a unique line. Two non-zero vectors $(z_{1}, w_{1})$ and $(z_{2}, w_{2})$ lie on the same line if and only if there exists a complex number $\lambda$ (necessarily non-zero) such that $$(z_{2}, w_{2}) = \lambda (z_{1}, w_{1}).$$

The set lines is, by definition, the complex projective line, $\Cpx\Proj^{1}$. This space acquires the structure of a holomorphic curve, equivalent to the Riemann sphere $\Cpx \cup \{\infty\}$, as follows: A complex number $z$ in $U_{0}$ corresponds to the line through $(z, 1)$. A complex number $w$ in $U_{1}$ corresponds to the line through $(1, w)$. If $w \neq 0$ and $z = 1/w$, then $$(1, w) = w(1/w, 1) = w(z, 1) \sim (z, 1);$$ that is, each non-zero number $w$ in $U_{1}$ is identified with $z = 1/w$ in $U_{0}$. This is precisely the gluing that defines the Riemann sphere.

The Hopf Fibration: Define a (holomorphic) mapping $\Pi:\Cpx^{2} \setminus\{(0, 0\} \to \Cpx\Proj^{1}$ by sending each (non-zero) pair $(z, w)$ to the line through $(z, w)$. The restriction of $\Pi$ to the $3$-sphere $$S^{3} = \{(z, w) : |z|^{2} + |w|^{2} = 1\}$$ is the Hopf map $\pi:S^{3} \to \Cpx\Proj^{1}$. The preimage of each point is the intersection of $S^{3}$ with a line, a great circle called a Hopf fibre. Since $\pi$ induces a bijection between Hopf fibres and points of $\Cpx\Proj^{1}$, the set of Hopf fibres is a $2$-sphere.

To see that distinct fibres of the Hopf map link once inside $S^{3}$, note that if $L_{1}$ and $L_{2}$ are distinct lines, then $\Cpx^{2} = L_{1} \oplus L_{2}$. Projection to the second summand is surjective, and the Hopf fibre in $L_{2}$ "winds around" $L_{1}$ (since this circle winds around the origin in $L_{2}$).

Consider the first-order linear system $$\dot{x}_{1} = x_{2},\quad \dot{x}_{2} = -x_{1},\quad \dot{x}_{3} = x_{4},\quad \dot{x}_{4} = -x_{3}.$$ Introducing complex coordinates $z = x_{1} + ix_{2}$ and $w = x_{3} + ix_{4}$ (and therefore identifying $\mathbf{R}^{4}$ with $\Cpx^{2}$, the preceding system becomes $$\dot{z} = iz,\quad \dot{w} = iw. \tag{1}$$ The solution of (1) with initial conditions $z(0) = z_{0}$, $w(0) = w_{0}$, is $$\left[\begin{array}{@{}cc@{}} z(t) \\ w(t) \\ \end{array}\right] = e^{it} \left[\begin{array}{@{}cc@{}} z_{0} \\ w_{0} \\ \end{array}\right] = \left[\begin{array}{@{}cc@{}} e^{it} & 0 \\ 0 & e^{it} \\ \end{array}\right] \left[\begin{array}{@{}cc@{}} z_{0} \\ w_{0} \\ \end{array}\right]. \tag{2}$$

Since scalar multiplication maps each line to itself, each solution curve (2) is the Hopf fibre in the line determined by the unit vector $(z_{0}, w_{0})$. Since solutions of (1) are precisely Hopf fibres, the set of solutions is a $2$-sphere, and distinct solution curves link once in $S^{3}$.