I'm using Arnol'd's book on ordinary differetial equations. I'm having difficulty with problems after the example of small oscillations of a spherical pendulum.

We have the system of differential equations $\dot{x}_1=x_2$, $\dot{x}_2=-x_1$, $\dot{x}_3=x_4$, $\dot{x}_4=-x_3$.

Problem 1. Prove that the phase curves of this field lie on the three-dimensional spheres $x_1^2+\cdots+x_4^2=const$.

I know how to do this one because the phase curves are defined by $x_1^2+x^2=c_1$ and $x_3^2+x_4^2=c_2$.

Problem 2. Prove that the phase curves are great circle of these spheres.

What is the definition of a great circle of $S^3$? I only know the definition of great circles of $S^2$.

Problem 3. Prove that the set of all phase curves on each three-dimensional sphere itself form a two-dimensional sphere.

First I need to figure out what the author means here. Does he mean that on each three-dimensional sphere, the union of all phase curves form a two-dimensional sphere? This is not true as the union is the whole three-dimensional sphere. Or does he mean that on each three-dimensional sphere, the set itself is a two-dimensional sphere when equipped with a topology in a natural way? This does not seem true because this set can be indexed by one single parameter $c_1$.

Thanks in advance for your help!


In my question I made the wrong assumption that the phase curve of the system is the product of phase curves of two smaller systems that describe the motion in two directions. In fact, for a fixed $c>0$, the set $F_c$ of phase curves that lie on the sphere $x_1^2+\cdots+x_4^2=c$ is $$\{im(f_{\phi,c_1}) | f_{\phi,c_1}:[0,2\pi]\rightarrow \mathbb{R^4}, t\mapsto (\sqrt{c_1}\sin(t),\sqrt{c_1}\cos(t),\sqrt{c-c_1}\sin(t+\phi),\sqrt{c-c_1}\cos(t+\phi)), 0\leq c_1\leq c,0\leq\phi<2\pi \}.$$

For fixed $c_1$ and $\phi$, the components $f_{\phi,c_1}^3$ and $f_{\phi,c_1}^4$ are linear combinations of $f_{\phi,c_1}^1$ and $f_{\phi,c_1}^2$. This gives us two linear equations that define a plane whose intersection with the sphere $x_1^2+\cdots+x_4^2=c$ is $im(f_{\phi,c_1})$. As this plane passes through the center of the sphere, the intersection is a great circle.

If we define the distance between two elements of $F_c$ to be their Hausdorff distance, we see that $F_c$ is the suspension $$([0,2\pi[\times [0,c]) / \{(\phi_1,0)\equiv (\phi_2,0) \mbox{ and }(\phi_1,c)\equiv (\phi_2,c) \mbox{ for all } \phi_1,\phi_2\in [0,2\pi[ \} $$, thus a two-dimensional sphere.



$$ x_1 \dot x_1 = x_1 x_2\\ x_2\dot x_2 = -x_2 x_1\\ x_3\dot x_3 = x_3 x_4\\ x_4\dot x_4 = - x_4 x_3 $$

we get

$$ x_1\dot x_1+x_2\dot x_2+x_3\dot x_3+x_4\dot x_4 = 0\Rightarrow x_1^2+x_2^2+x_3^2+x_4^2=C_0 $$


Note that in the same way

$$ x_1^2+x_2^2 = C_1\\ x_3^2+x_4^2 = C_2 $$


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