A limit cycle in a planar system encloses three fixed points. If two of them are saddles (index -1 each), what must be true about the third?
AIt must be a saddle as well — saddles attract limit cycles
BIt must have index +3 — to make the total index equal +1. But standard fixed points have index ±1, so this configuration is impossible
CIt must have index +1 (a node, spiral, or center) — but three fixed points with indices -1, -1, +1 sum to -1, not +1. So this configuration is impossible with only these three
DIt could be any type of fixed point — the index theorem places no constraints on enclosed fixed points
The index of a limit cycle is +1, and it must equal the sum of indices of enclosed fixed points. Two saddles contribute -2, and a single node/spiral contributes +1, giving a total of -1 ≠ +1. This configuration is impossible. You would need at least two nodes/spirals (contributing +2) plus two saddles (-2) to reach the required sum of +1... but that requires a fourth fixed point. Index theory thus constrains which phase portrait configurations can support limit cycles.
Question 2 True / False
The index of an isolated fixed point can be determined from the eigenvalues of the Jacobian.
TTrue
FFalse
Answer: True
For a hyperbolic fixed point (no zero or purely imaginary eigenvalues), the index is determined by the sign of the determinant of the Jacobian. If det(Df) > 0, the index is +1 (node, spiral, center). If det(Df) < 0, the index is -1 (saddle). This follows because the index counts the net rotation of the vector field, which is related to how many times the field's direction cycles as you go around the point. Two negative eigenvalues (stable node): arrows all point inward, rotating once → index +1. One positive, one negative (saddle): the field reverses on two axes → index -1.
Question 3 Short Answer
On a sphere, the sum of the indices of all fixed points of a smooth vector field must equal 2 (the Euler characteristic of the sphere). What does this imply about combing a hairy ball?
Think about your answer, then reveal below.
Model answer: Any smooth tangent vector field on a sphere must have fixed points whose indices sum to 2. Since index ±1 are the generic values, the minimum number of fixed points is one (with index +2, a dipole — though this is non-generic) or two (each with index +1, like the north and south poles of a rotational flow). It is impossible to have a smooth, nonvanishing tangent vector field on a sphere — this is the hairy ball theorem. You can't comb a hairy ball flat without creating at least one cowlick (a point where the vector field vanishes).
The Poincare-Hopf theorem generalizes index theory to arbitrary closed surfaces: the sum of indices equals the Euler characteristic χ. For a sphere χ = 2, for a torus χ = 0 (so a vector field with no fixed points IS possible on a torus — you can comb a hairy donut). This connects dynamical systems to topology in a deep way: the global topology of the space constrains what flows are possible on it.
Question 4 Multiple Choice
A student draws a phase portrait with two stable nodes and nothing else (no saddle points). Can this be correct for a flow on the plane?
AYes — two stable nodes can coexist with all trajectories going to one or the other
BNo — two stable nodes (each index +1) require at least one saddle (index -1) between them to separate their basins of attraction
CYes, but only if the system is non-autonomous
DNo — a planar system can have at most one stable fixed point
On the plane (not a closed surface), the index theorem doesn't require a global sum constraint. However, the topological structure of basins of attraction forces a saddle between any two stable nodes — the basins must be separated by a boundary, and that boundary must contain a saddle point (or extend to infinity in a specific way). Generically, two attractors in the plane require a saddle between them whose stable manifold forms the separatrix. This is a practical consequence of the flow topology, even though the formal index theorem on the (non-compact) plane is more subtle.