# 9. Classification of equilibrium points in specific systems.

Again consider the system The equilibrium point obviously is (0,0). Let and be the eigenvalues to the matrix A. We have the following different possibilities:

1) If Let be the eigenvectors belonging to the corresponding eigenvalues and . Then the general solution of the system above is We again have a number of possibilities:
i)   c2=0,
c1>0x(t) is a curve along l1+ (away from origo for increasing t).
ii)
c2=0, c1<0x(t) is a curve along l1 (away from origo for increasing t).
iii) c1=0, c2>0x(t) is a curve along l2+ (away from origo for increasing t).
iv) c1=0, c2<0x(t) is a curve along l2 (away from origo for increasing t). v) neither c1 nor c2 are zero. Then for we have hence for large (in absolute value) negative t. In particular we have that When we have for large positive t. Hence x(t) diverges to infinity with a slope asympotically v2 as t goes to (positive) infinity. This means that (0,0) is an unstable node. The lines defined by the eigenvectors v1 and v2 are called separatrices. The behavior is shown below in the phase portrait. 2) If In the same way as above we realize that (0,0) is a stable node. The phase portrait looks the same but with reversed arrows.

3) If We have the following possibilities:
i)   c2=0, c1>0x(t) is a curve along l1+ (towards origo for increasing t).
ii)
c2=0, c1<0x(t) is a curve along l1 (towards origo for increasing t).
iii) c1=0, c2>0x(t) is a curve along l2+ (away from origo for increasing t).
iv) c1=0, c2<0x(t) is a curve along l2 (away from origo for increasing t).

We have a saddle point. See the phase portrait below. 4) If We have two cases:

a) We have two linearly independent eigenvectors Then we can write the solution where a1 and a2 are arbitrary. This means that every curve is a halfline towards origo. See the phase portrait below. b) We only have one eigenvector Then the solution on the form where the vector satisfies For large t this means that the solution is In this case the phase portrait will look like this. Both these cases are examples of a stable node.

5) If This case is analogous to case 4) above, but with reversed arrows; (0,0) is an unstable node.

6) If By using similar arguments as in Example 9 we realize that the real solutions are Three cases:
i) : Then x(t) and y(t) are periodic with period The point (0,0) is a center.

ii) : The amplitude of x decreases and we thus have a stable spiral.

iii) The amplitude of x increases and we thus have an unstable spiral. Example 10:
Consider the system The coefficient matrix has the characteristic equation and thus the eigenvalues This corresponds to case 3) above and we conclude that (0,0) is an unstable saddle point. The eigenvectors are obtained by solving the linear equation system: Eigenvalue 1: implies that that is, the corresponding eigenvector is Eigenvalue 2: implies that that is, the corresponding eigenvector is The general solution of the system thus is that is The eigenvectors define the directions of the separatrices. Example 11:
Consider the system Here we have the critical point (2,1). We make the change of variables x1=x-2 and y1=y-1 and rewrite the system as By using the result from example 10 above we see that the point (2,1) is an unstable saddle point and that the solution to the original system is We get the phase portrait by taking that from example 10 and move it two steps in the x-direction and one step in the y-direction.
In the same way as in Example 11 we may instead study the more general system The equlibrium point is here (a0,b0). By making the change of variables x1=x-a0 and y1=y-b0 we can transfer the system to the ones studied above with equilibrium point (0,0).