Rigid Body Nutation. Euler’s equations describe the motion of the principal-axis components of the angular velocity of a freely rotating rigid body (such as a space station), as seen by an observer rotating with the body (the astronauts, for example). This motion is called nutation. If the angular velocity components are denoted by x, y, and z, then an example of Euler’s equations is the three-dimensional autonomous system

\(\begin{array}{l}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = yz}}\\\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = - 2xz}}\\\frac{{{\bf{dz}}}}{{{\bf{dt}}}}{\bf{ = xy}}\end{array}\)

The trajectory of a solution x(t),y(t), z(t) to these equations is the curve generated by the points (x(t), y(t), z(t) ) in xyz-phase space as t varies over an interval I.

(a) Show that each trajectory of this system lies on the surface of a (possibly degenerate) sphere centered at the origin (0, 0, 0).[Hint: Compute\(\frac{{\bf{d}}}{{{\bf{dt}}}}{\bf{(}}{{\bf{x}}^{\bf{2}}}{\bf{ + }}{{\bf{y}}^{\bf{2}}}{\bf{ + }}{{\bf{z}}^{\bf{2}}}{\bf{)}}\)What does this say about the magnitude of the angular velocity vector?

(b) Find all the critical points of the system, i.e., all points\({\bf{(}}{{\bf{x}}_{\bf{o}}}{\bf{,}}{{\bf{y}}_{\bf{o}}}{\bf{,}}{{\bf{z}}_{\bf{o}}}{\bf{)}}\) such that \({\bf{x(t) = }}{{\bf{x}}_{\bf{o}}}{\bf{,y(t) = }}{{\bf{y}}_{\bf{o}}}{\bf{,z(t) = }}{{\bf{z}}_{\bf{o}}}\) is a solution. For such solutions, the angular velocity vector remains constant in the body system.

(c) Show that the trajectories of the system lie along the intersection of a sphere and an elliptic cylinder of the form\({{\bf{y}}^{\bf{2}}}{\bf{ + 2}}{{\bf{x}}^{\bf{2}}}{\bf{ = C}}\) for some constant C. [Hint: Consider the expression for dy/dx implied by Euler’s equations.]

(d) Using the results of parts (b) and (c), argue that the trajectories of this system are closed curves. What does this say about the corresponding solutions?

(e) Figure 5.19 displays some typical trajectories for this system. Discuss the stability of the three critical points indicated on the positive axes.

Now,

\(\begin{array}{c}\frac{{\bf{d}}}{{{\bf{dt}}}}{\bf{(}}{{\bf{x}}^{\bf{2}}}{\bf{ + }}{{\bf{y}}^{\bf{2}}}{\bf{ + }}{{\bf{z}}^{\bf{2}}}{\bf{) = 2x}}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ + 2y}}\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ + 2z}}\frac{{{\bf{dz}}}}{{{\bf{dt}}}}\\{\bf{ = 2x}}{\bf{.yz + 2y( - 2xz) + 2z(xy)}}\\{\bf{ = 2xyz - 4xyz + 2xyz}}\\{\bf{ = 0}}\end{array}\)

This means that \({\bf{(}}{{\bf{x}}^{\bf{2}}}{\bf{ + }}{{\bf{y}}^{\bf{2}}}{\bf{ + }}{{\bf{z}}^{\bf{2}}}{\bf{)}}\)is a constant, which is true only if trajectories (x, y, z) lie on a sphere centered at the origin.

The magnitude of the angular velocity vector is given as \(\sqrt {{{\bf{x}}^{\bf{2}}}{\bf{ + }}{{\bf{y}}^{\bf{2}}}{\bf{ + }}{{\bf{z}}^{\bf{2}}}} \). But \({\bf{(}}{{\bf{x}}^{\bf{2}}}{\bf{ + }}{{\bf{y}}^{\bf{2}}}{\bf{ + }}{{\bf{z}}^{\bf{2}}}{\bf{)}}\)is a constant, so the magnitude of the angular velocity vector is a constant.

\(\begin{array}{c}\frac{{{\bf{dx}}}}{{{\bf{dt}}}}{\bf{ = yz = 0}}\\\frac{{{\bf{dy}}}}{{{\bf{dt}}}}{\bf{ = - 2xz = 0}}\\\frac{{{\bf{dz}}}}{{{\bf{dt}}}}{\bf{ = xy = 0}}\end{array}\)

If both \({\bf{y = 0}}\) and\({\bf{z = 0}}\), I can take any x and the equation will be satisfied. Therefore, all points of the form (x, 0, 0) are critical points, for all real numbers x.

If both\({\bf{x = 0}}\), and\({\bf{z = 0}}\), I can take any y and the equation will be satisfied. Therefore, all points of the form (0, y, 0) are critical points, for all real numbers y.

If both\({\bf{x = 0}}\), and\({\bf{y = 0}}\), I can take any y and the equation will be satisfied. Therefore, all points of the form (0, 0, z) are critical points, for all real numbers z.

From part (a) trajectories lie on a sphere.

Consider

\(\begin{array}{c}\frac{{{\bf{dy}}}}{{{\bf{dx}}}}{\bf{ = }}\frac{{{\bf{ - 2xz}}}}{{{\bf{yz}}}}\\{\bf{ = }}\frac{{{\bf{ - 2x}}}}{{\bf{y}}}\\{\bf{ydy = - 2xdx}}\\\int {{\bf{ydy = }}\int {{\bf{ - 2xdx}}} } \\{{\bf{y}}^{\bf{2}}}{\bf{ = - 2}}{{\bf{x}}^{\bf{2}}}{\bf{ + C}}\\{{\bf{y}}^{\bf{2}}}{\bf{ + 2}}{{\bf{x}}^{\bf{2}}}{\bf{ = C}}\end{array}\)

This is the result.

I know that critical points are on the axis and the trajectories lie on the intersections of an elliptic cylinder and a sphere. The intersection of these curves will always make a closed on sphere. Since the trajectories are closed curves, the corresponding solutions are periodic.

From the image, it is clear that the trajectories move from the critical points on the y axis, so this critical point is unstable. As for the critical points on the x and z axis, they are stable.

Therefore, the critical points on the x and z axis are stable and the critical points on the y axis are unstable.