The handout "The Euler Equations" in dvi, ps, pdf formats.
No measurement is perfectly precise. You may think you know the coefficients in some linear ODE, but really they may be slightly different. This will change the eigenvalues slightly. I drew a picture of the complex plane with the eigenvalues corresponding to a center - a conjugate pair of purely imaginary numbers - and then pointed out that a slight change will produce a conjugate pair with small real part: a spiral, either stable or unstable. So in the absence of other structure, usually some sort of symmetry, a center is virtually impossible, and what you should expect is a spiral. Changing the coefficients of an equation giving a spiral, on the other hand, will produce another similar spiral. We say that a center is structurally unstable, a spiral is structurally stable. This use of the word stable is unrelated to its use earlier, as in neutrally or asymptotically stable.
Small changes can make
center into spiral
improper node into spiral or proper node
star node into improper node, spiral, or proper node.
I drew eigenvalue pictures for each. The matrix giving a star node is a multiple of the identity matrix, and it can vary to give an improper node by making one of the off-diagonal entries nonzero, for example.
Centers, improper nodes, star nodes, and all the degenerate examples, are structurally unstable. Spirals, proper nodes, and saddles are structurally stable.
In a similar way, only structurally stable phase portraits should be relied on if they occur as the linear approximation at a critical point of a nonlinear autonomous system. For example, the system
x' = -y + x^3
y' = x
linearizes at (0,0) to an equation whose phase portrait is a center. But compute: (1/2)(d/dt)r^2 = xx' + yy' = x^4. This implies that the trajectories are continually getting farther from the origin: they are spirals, but very very slow ones, not exponential ones of the sort given by linear systems. (In fact pplane5 mistakes them for periodic orbits.)
I got a volunteer to throw a book up and catch it. We noticed that the spin seems steady if it's around the long or the short axis of the book, but wildly unstable about the middle axis. I proposed to explain this using ODEs. The physics is desribed in the (optional) handout on the Euler equations. I explained that at any instant the body is rotating about an axis (assuming it's rotating at all), which we represent by a vector pointing so the rotation satisfies the right hand rule and whose magnitude is the angular speed. This is the angular velocity vector omega, and the first surprise is that omega is not generally constant. We agreed to use a coordinate system with origin at the center of mass. As usual in physics there is a conserved quantity, made by combining velocity with mass distribution. This is the angular momemtum vector l. For physicists I wrote down
l = integral (r x rho r') dV
where r is the position vector, rho is the mass distribution, and x denotes the cross product. Anyway, the information about the mass distribution can be packaged in a symmetric 3x3 matrix L in such a way that
l = L omega
The trouble is that as the object spins, the mass distribution changes if we measure it in the inertial frame. Euler's brilliant idea was to use a frame attached to the spinning body. Then L is fixed but l and omega both change. There are always preferred axes, for which L is diagonal, with diagonal entries I1, I2, I3, the "moments of inertia." Often these axes are obvious - as in the case of the book. The equations of motion that emerge are:
I1 omega1' = (I2 - I3) omega2 omega3
I2 omega2' = (I3 - I1) omega3 omega1
I3 omega3' = (I1 - I2) omega1 omega2
I took as an example I1=2, I2=3, I3=4, so
omega1' = (-1/2) omega2 omega3
omega2' = (2/3) omega3 omega1
omega3' = (-1/4) omega1 omega2
We can reduce this to two dimensions by finding a "constant of motion":
4 omega1 omega1' + 9 omega2 omega2' + 16 omega3 omega3' = 0 so
4(omega1)^2 + 9(omega2)^2 + 16(omega3)^2 = constant:
the vector omega is constrained to lie on an ellipsoid. This number is the square of the length of the angular momentum, so we should have known that it was constant. I drew this (with constant = 1) but scaled it so I drew a sphere.
Then I found the critical points. omega' = 0 exactly when any two of the coordinates are zero. These are along the axes: there are six critical points.
(0,1/3,0): then omega2 is essentially constant, so the linearization is
omega1' = (-1/2)(1/3)omega3
omega3' = (-1/4)(1/3)omega1
This is a saddle, which I drew onto the sphere.
(1/2,0,0): then omega1 is essentially constant, so the linearization is
omega2' = (2/3)(1/2)omega3
omega3' = (-1/4)(1/2)omega1
This gives a center, but as we know we shouldn't trust this. In this case there is in fact more symmetry, in the form of another constant of motion:
2 omega1 omega1' + 3 omega2 omega2' + 4 omega3 omega3' = 0 so
2(omega1)^2 + 3(omega2)^2 + 4(omega3)^2 = constant:
This is another ellipsoid, and the trajectories are constrained to lie on the intersection of the two. This second constant of motion is in fact essentially the (kinetic) energy. I imagined the energy ellipsoid growing, meeting the earlier one at two opposite points, continuing to grow so the intersection forms centers, continuing till it becomes tangent along the omega2-axis in crosses, then continuing to grow till it is intersecting the first ellipsoid in centers around the omege3 axis and then finally leaves it altogether. The result is a beautiful singular foliation of the constant-angular momentum ellipsoid. It shows that the motion exhibits restricted periodic motion around the long and short axes, but wild flips from one direction to the other if you start near the middle axis: just as we observed.
A final comment is that if we introduce some friction into the system, the energy symmetry is broken and we are likely to see spirals after all.