4. Points of Equilibrium of Linear Systems
Definition. The ODE
x'(t) = f(x(t)), x, f:Rn → Rn
has solution x(t) = a if f(a) = 0. a is called a point of equilibrium of the above vector differential equation (that is, system of scalar equations).
Example. Linear systems. Let A be an n×n matrix. Then dx/dt = A x has a point of equilibrium at the origin. More generally, dx/dt = A (x - c) has a point of equilibrium at c.
Two dimensional case. A 2×2 matrix can be put into diagonal form in the case of distinct eigenvalues, or upper triangular form in the case of a single eigenvalue. For real A, complex eigenvalues appear with their conjugates. We also have the relations
trace (A) = sum of eigenvalues,
determinant (A) = product of eigenvalues.
The following table summarises the tyes of possible equilibrium points:
| eigenvalue λ | fixed point |
| λ1 < λ2 < 0 | stable node |
| λ1 > λ2 > 0 | stable node |
| λ1 < 0 < λ2 | saddle point |
| λ1,2 = - α ± iβ | stable spiral |
| λ1,2 = α ± iβ | unstable spiral |
| λ1,2 = ± iω | elliptic fixed point |
| λ1 = λ2 < 0, diagonal | stable star |
| λ1 = λ2 > 0, diagonal | unstable star |
| λ1 = λ2 < 0, upper triangular | stable proper node |
| λ1 = λ2 > 0, upper triangular | unstable proper node |
The solution trajectories are sketched below.
Higher dimensions; the Jordan decomposition. In the case of higher dimensional linear systems, the Jordan decomposition of matrix or linear transformation enables us to study the behaviour of the system dx/dt = T x. If T is a linear transformation, we set null (T) = {x| Tx = 0}.
Theorem. Let T be a linear transformation on the finite dimensional vector space V over the algebraically closed field F, and let the scalars c1, ... , ck be the distinct eigenvalues of T. Then there exist numbers ri, for 1 ≤ i ≤ k, such that
V = null (T - c1)r1 ⊕ ... ⊕ null (T - ck)rk
is a direct sum decomposition of V into subspaces invariant under T.
[This is Theorem 6 of http://math.princeton.edu/nelson/217/jordan.pdf.]
Definition. A linear transformation N is nilpotent if Nr = 0 for some r.
The decomposition. For T as above, the restriction of T - ciI to null (T - ciI)ri is nilpotent. Thus we can write
T|null (T - ciI)ri = ciI + Ni
where Ni is nilpotent. Taking the direct sum over all nullspaces,
T = S + N , N = N1 ⊕ ... ⊕ Nk
with S diagonalisable (or semisimple) and N nilpotent. Further SN = NS. it is also shown in the cited pdf file that a nilpotent matrix can be put into triangular form with 0 on the diagonal.
Remark. If T = S + N as above, dx/dt = T x has solution x(t) = exp (tT) x(0). Since SN = NS, we have exp (tT) = exp (tS) exp (tN); further we note that exp (diag (a1, ..., an)) = diag (ea1, ..., ean), and that the exponential of a nilpotent matrix truncates to a polynomial. This facilitates the computation of the solution.