Classifying Equilibria for a 2x2 Matrix

The Equilibrium Point of a 2x2 Matrix calculator computes the equilibrium point of a system of differential equations.

INSTRUCTIONS: Enter the following:

• (A)  This is the 2x2 matrix

Equilibrium Point: The calculator returns the equilibrium point or set of points for the 2x2 matrix.

2x2 Matrix Calculators :

• To compute the Characteristic Polynomial of a 3x3 matrix,CLICK HERE.
• To compute the Trace of a 2x2 Matrix, CLICK HERE.
• To compute the Determinant of a 2x2 Matrix, CLICK HERE.
• To compute the Inverse of a 2x2 Matrix, CLICK HERE.
• To compute the Eigenvalues of a 2x2 Matrix, CLICK HERE.
• To compute the Eigenvalues and Eigenvectors of a 2x2 Matrix, CLICK HERE.
• To multiply a 2x2 matrix by a scalar, CLICK HERE.
• To compute the Characteristic Polynomial of a 2x2 Matrix, CLICK HERE.

The Math

The equilibrium point of a system of differential equations is a point or set of points at which the system is unchanging. That is the point where `(dY)/(dt)=0`.

For a linear system of equations, the origin is always an equilibrium point, though there may be others. Consider the linear system
     `(dY)/(dt)=AY`
where `A` is a 2x2 matrix. Then the equilibrium point `Y_0` is the point where
     `(dY)/(dt)=AY_0 =  ((a,b),(d,c))*((x_0),(y_0))= ((ax_0+by_0),(cx_0+dy_0))=((0),(0))`
This is equivalent to the pair of linear equations
     `ax_0+by_0 =0`
     `cx_0+dy_0 =0`
So `(x_0,y_0)=(0,0)` is a solution to the system and therefore an equilibrium point. 

Any other equilibrium point must also satisfy these equations, so suppose `a!=0` and rewrite the first equation
     `x_0=-(b)/(a) y_0`
Then the second equation becomes
     `c(-(b)/(a) y_0)+dy_0=0`, or
     `(ad-bc)y_0=0`.
So a solution to the system must be where either `y_0=0` or `ad-bc=0`. When `y_0=0`, we know that `x_0=0`, so the linear system has non-trivial solutions (that is, solutions besides `(0,0)`) when `ad-bc=0`. Notice that `ad-bc` is the determinant of our 2x2 matrix, so we can say that a linear system of differential equations always has the origin as an equilibrium, and has other equilibria only when the determinant of the corresponding matrix is `0`).

Classification of Equilibria

Just as important as the coordinates of the equilibria is the behavior around them. We want to know what our system of equations looks like, and we can understand their general shape based on what happens around these equilibrium points. This behavior is determined by the eigenvalues of our matrix.

For real, non-zero, distinct eigenvalues, there are three classifications of equilibrium points. If `0<lambda_1<lambda_2`, the equilibrium is classified as a "source". As `t -> infty`, all solutions to the system of equations tend away from the equilibrium point between the straight line solutions. So as time goes on everything moves further and further away from the origin.

If `lambda_1<lambda_2<0`, the equilibrium is classified as a "sink". As `t-> infty`, all solutions to the system of equations tend toward the equilibrium point between the straight line solutions. That is, as time goes on everything eventually ends up near the origin.

If `lambda_1<0<lambda_2`, the equilibrium point is classified as a "saddle". In a saddle, the solutions approach the origin in the direction of one of the straight line solutions, and tend away from the origin in the direction of the other. 

Real repeated eigenvalues and eigenvalues equal to `0` behave differently from the non-zero, real, distinct cases. With repeated eigenvalues, we have just one straight line solution instead of two. We still have that if `lambda<0` it is a sink and if `lambda>0` it is a source, but the phase portraits of these systems look different from the distinct eigenvalue case. The solutions appear to spiral around the origin, but can never cross the one straight line solution.

If `lambda_1=0`, we again have different behavior. Notice that if `lambda_1=0`, it must be that the determinant of our matrix is also `0`. (For the simplest way to see this, recall that `prod lambda_i = det`, so if some `lambda_i=0` then the determinant is necessarily `0`.) So we know from above that the system should have multiple equilibrium points. If we find the eigenvectors `vec(v)_1` and `vec(v)_2` corresponding to the two eigenvalues (with `lambda_1=0`), the equilibrium points lie along the `vec(v)_1`, and the solutions approach the equilibrium along lines parallel to `vec(v)_2`.

For complex eigenvalues (of the form `lambda=a+bi`), we have 3 cases based on the real part of the eigenvalue (`a`). Recall that in our 2x2 case, our eigenvalues are complex conjugates of each other (that is, `lambda_1=a+bi` and `lambda_2=a-bi`), so the real part is the same for either of them and we need only consider one.) For `a>0`, the origin is a "spiral source". For complex eigenvalues there are no straight line solutions as in the real case, so the solutions can simply spiral out from the origin.

For `a<0`, we have a spiral sink. Similar to the "spiral source", the solutions now spiral into the origin as `t-> infty`.

For `a=0`, the equilibrium is classified as a "center". In a center, the solutions neither approach the equilibrium point nor tend away from it, but follow constant elliptical orbits around the equilibrium point.

Sources

Blanchard, Paul, Robert L. Devaney, and Glen R. Hall. Differential Equations. 3rd ed. Belmont, CA: Thomson Brooks/Cole, 2006. Print. 

See Also

https://youtu.be/bOreOaAjDno