Stability and instability

Definition 2

 The point (a,b) is a critical point of

$$\dot x_1=F(x_1,x_2), \quad \dot x_2=G(x_1,x_2)$$

if

$$F(a,b)=0, \quad G(a,b)=0.$$

A critical point x_c of $\dot x=f(x)$ is stable if given any R>0, there exists an r>0 such that every solution $x=\phi(t)$ which satisfies $\vert\vert\phi(0)-x_c\vert\vert<r$ exists and satisfies $\vert\vert\phi(t)-x_c\vert\vert<R$, $r\leq R$, for all $t\ge 0$.

A critical point x_c is asymptotically stable if

(i) it is stable and

(ii) there is a C>0 such that if $\phi (0)$ satisfies $\vert\vert\phi(0)-x_c\vert\vert<C$, then $\lim_{t\to\infty} \phi(t)=x_c$.

Example 6

 The difference between stability and asymptotic stability is that stability merely requires solutions which start close to x_c to remain close to x_c; they do not have to approach x_c. For instance, the equation $\dot x =0$ has solutions x=constant in 1-D. A critical point is x_c=0. If you pick any R around x_c=0, then any solution with |x(0)|<R also satisfies |x(t)|<R for all t>0, so that the solution remains within a distance R of the critical point. However, this solution never approaches 0 as $t\to\infty$.

Another example, in 2D, is if the trajectories are closed orbits
around x_c. Solutions which start close to x_c will remain there (stable) but not approach x_c as $t\to\infty$. In conservative systems, solutions are never asymptotically stable because there is no energy dissipation, so that a critical point can be stable but not asymptotically stable.

For linear systems with constant coefficients, there is a very simple criterion for stability.

Theorem 2

 The solution X=0 of the system $\dot X=AX$ is stable if all eigenvalues of A are in the closed left half plane and, in addition, all eigenvalues on the imaginary axis are semisimple, i.e. their algebraic and geometric multiplicities are equal. It is asymptotically stable if all eigenvalues are in the open left half plane.

The proof follows immediately from the representation of the solution in terms of exponentials (or powers of t times exponentials). We note that $t^n\exp(\sigma t)\to 0$ if the real part of $\sigma$ is negative. If $\sigma$ is purely imaginary, then $t^n\exp(\sigma t)$ grows unbounded if n>0 and has constant modulus if n=0.

Two-dimensional homogeneous constant coefficient system.

We consider the case of an autonomous system in the plane. In the following, we shall visualize the types of qualitative behavior which can arise. The system we consider is

$${dX\over dt}=AX,\quad X=\pmatrix{x_1\cr x_2}.$$

An example is

$$A=\pmatrix{1 &-1\cr 2 & -2}.$$

The eigenvalues of A are $\sigma=0,-1$.The fact that 0 is an eigenvalue means that

$$A\pmatrix{c\cr c}=0$$

and the critical points consist of a subspace spanned by the vector $\pmatrix{1\cr 1}$.In the x₁-x₂ plane, there is a line of critical points. To consider isolated critical points, we require that the matrix A not have 0 as an eigenvalue: det $A \ne 0$. Then dX/dt=0 implies that X=A^-1(0)=0 is the only isolated critical point. In this case, we distinguish between 4 kinds of critical points.

1. Center. Trajectories are closed. This case occurs if the eigenvalues of A are purely imaginary.

 

Figure 1.2: Phase vector diagram for $\dot x_1=x_2$, $\dot x_2=-x_1$ (example of a center).  

2. Saddle point. This case occurs if the eigenvalues are real and have different signs. There is one line along which the trajectory moves towards zero and another line along which it moves away from 0.

 

Figure 1.3: Phase vector diagram for $\dot x_1=x_2$, $\dot x_2=x_1$ (example of a saddle).  

3. Spiral point. For this case, the eigenvalues of A are complex conjugates. Solutions approach zero in an inward spiral if the real part of the eigenvalues is negative, and they move away from zero in an outward spiral if the real part of the eigenvalues is positive.

 

Figure 1.4: Phase vector diagram for $\dot x_1=-x_1+2x_2$, $\dot x_2=-x_2-2x_1$ (example of a stable spiral).  

4. Node. Here the eigenvalues of A are real and of equal sign, Say the eigenvalues are $\sigma_1<\sigma_2<0$. Let $\xi_1$ and $\xi_2$ be the corresponding eigenvectors. Then the line spanned by $\xi_1$ is invariant, and the solution approaches to zero along that line. The same is true for the line spanned by $\xi_2$. A general initial condition is a superposition $c_1\xi_1+c_2\xi_2$. The corresponding solution is $c_1\xi_1\exp(\sigma_1 t)+c_2 \xi_2\exp(\sigma_2t)$. Since $\exp(\sigma_1t)<<\exp(\sigma_2t)$ for large t, the term $c_2\xi_2\sigma_2t$ is dominant. Hence all solutions except those which are multiples of $\xi_1$ will approach the origin along the $\xi_2$-direction. A similar picture applies if there are two positive eigenvalues; of course the trajectories move away from the origin in that case.

 

Figure 1.5: Phase vector diagram for $\dot x_1=-2x_1$, $\dot x_2=-x_2$ (example of a stable node).