### Characterization of linear 2D flows

#### Characterization of linear 2D flows

Any linear homogeneous 2D system of ordinary differential equations may be written in the form \[ \left\{ \begin{array}{r c l} x' & = & \alpha x + \beta y \\ y' & = & \gamma x + \delta y \\ \end{array} \right. \] Setting \( x'=y'=0 \) we can easily find that the unique equilibrium point for a homogenous linear system is the origin, i.e. \( E=(0,0) \) for the above system. In the next few sections we will characterize the different types of solutions for various parameters in order to later understand conditions for the orgin to be either a stable equilibrium point or an unstable equilibrium point.

##### growth and decay

In 2 dimensions the simplest linear system is really just 2 one dimensional equations which exhibit exponential growth or decay in each variable. In other words we have the following decoupled system \[ \left\{ \begin{array}{r c l} x' & = & ax \\ y' & = & by \;\;. \\ \end{array} \right. \] for the variables \( x=x(t) \) and \( y=y(t) \) with parameters \(a\) and \(b\). Since each of these equations is "decoupled" from the other we can solve them exactly just as we would for a 1D problem so that \[ x(t) = x(0)e^{at} \] \[ y(t) = y(0)e^{bt}. \]

The Geogebra applet below plots this solution for different values of \(a\), \(b\) and initial condition \( \mathbf{X_0} = (x(0),y(0)) \).

*Linear 2d flow: growth and decay.*

If \( a>0 \) then we see exponential growth along the x-direction and if \( a<0 \) we see exponential decay along the x-direction and similarly the sign of the parameter \(b\) controls growth and decay in the y-direction. This allows us to characterize three cases

exponential growth in both directions | (0,0) unstable equilibrium point |

exponential growth in one direction and exponential decay in the other direction | (0,0) unstable equilibrium point |

exponential decay in both directions | (0,0) stable equilibrium point |

In other words both \(a\) and \(b\) must be negative for the solution to converge to the equilibrium \( (0,0) \).

##### pure oscillation

The system \[ \left\{ \begin{array}{r c l} x' & = & -a^2y \\ y' & = & b^2x \;\;. \\ \end{array} \right. \] exhibits elliptical orbits about the equilibrium \( E= (0,0) \). The paramters above are squared to ensure that the sign of each coefficient does not change as the parameters \(a\) and \(b\) vary. In the geogebra applet below, we again start the dynamics at a location \( \mathbf{X_0} = (x(0),y(0)) \), but in this case will follow periodic orbits about the equilibrium. The green vector \( V \) is the tangent vector to the curve at a given point.

*Linear 2d flow: pure oscillation.*

We classify these elliptic orbits as stable. Even though the solution does not converge to the equilibrium point the solution "stays close."

##### oscillatory growth/decay

\[ \left\{ \begin{array}{r c l} x' & = & -y \\ y' & = & x + (2e)y \;\;. \\ \end{array} \right. \]

*Linear 2d flow: oscillatory growth and decay.*

We can characterise the behaviour of this system as follows:

Oscillation and decay (Spiral in) | \( E= (0,0) \) stable for \( e<0 \) |

Oscillation and growth (Spiral out) | \( E= (0,0) \) unstable for \( e>0 \) |

##### Simplified general linear system

The system \[ \left\{ \begin{array}{r c l} x' & = & -y \\ y' & = & -dx + ty \\ \end{array} \right. \] summarizes all the previous cases together. The two parameters \( d\) and \( t \) are sufficient to create an equivalent system to most any case for the original equations \[ \left\{ \begin{array}{r c l} x' & = & \alpha x + \beta y \\ y' & = & \gamma x + \delta y \\ \end{array} \right. \] with parameters \( \alpha \), \( \beta \), \( \gamma \) and \( \delta \).

*Linear 2d flow: simplified general linear system.*

Can you determine conditions for \(d\) and \(t\) for stability of the equilibrium \(E= (0,0) \)?

#### Linear Algebra

In order to classify the behaviour of a linear flow in the simplest way we will use two ideas from Linear Algebra for a matrix. Any 2x2 matrix, \( A\) can be written as
\[
A = \begin{bmatrix}
a & b \\
c & d \\
\end{bmatrix} \;\;.
\]
We will need to know how to find the *determinant* and the *trace* of such a matrix, both of which return a single number for any given matrix.

##### Trace

The trace of a matrix is simply the sum of the diagonal elements. For the 2x2 matrix \( A\) above we have \[ tr(A) = a + d . \]

##### Determinant

The determinant is a little more complicated than the trace. For the 2x2 matrix \( A\) above we write its determinant as \[ \det(A) = |A| = ad-bc . \] The 2x2 determinant is simply the product of the diagonals minus the product of the off-diagonals.

##### Examples

Find the determinant and trace of the matrices P,Q,R below. \begin{align*} P &= \begin{bmatrix} 1 & 2 \\ 3 & 4 \end{bmatrix} & Q &= \begin{bmatrix} 3 & 0 \\ 1 & -2 \end{bmatrix} & R = \begin{bmatrix} 1 & -1 \\ 3 & 2 \\ \end{bmatrix} \\ \det(P) &= (1)(4)-(2)(3) & \det(Q) &= (3)(-2)-(0)(1) & \det(R) &= (1)(2)-(-1)(3) \\ \det(P) &= -2 & \det(Q) &= -6 & \det(P) &= 5 \\ tr(P) &= 1 + 4 & tr(Q) &= 3 -2 & tr(R) &= 1 + 2 \\ tr(P) &= 5 & tr(Q) &= 1 & tr(P) &= 3 \end{align*}