2.1 Linear systems in R²

Consider the linear system

 

x' = Ax, (2.1)

where

$$x = \left( \begin{array}{l} x_{1} \\ x_{2} \end{array} \righ... ...} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right) \, .$$

By a linear transformation

x = Py

system (2.1) can be reduced to

 

y' = By , (2.2)

where B has one of the following three forms:

$$\left[ \begin{array}{cc} \lambda_{1} & 0 \\ 0 & \lambda_{2} ... ...\ - \beta & \alpha \end{array} \right] \, , \, \beta > 0 \, ,$$

which are called Jordan canonical forms of A.

Let's begin by describing the phase portraits of system (2.2).

Case I:      $B = \left[ \begin{array}{cc} \lambda_{1} & 0 \\ 0 & \lambda_{2} \end{array} \right]$

          

All solutions are half lines which approach (leave) (0,0).

          

$\left( \frac{y_{1}}{k_{1}} \right)^{\frac{1}{\lambda_{1}}} = \left( \frac{y_{2}}{k_{2}} \right)^{\frac{1}{\lambda_{2}}}$
All solutions starting from $(k_{1}, k_{2}) \neq (0,0)$ are tangent to the y₂-axis as they approach (leave) (0,0).


                     

Case II:      $B = \left[ \begin{array}{cc} \lambda & 1 \\ 0 & \lambda \end{array} \right]$      $\left\{ \begin{array}{l} y_{1} (t) = (k_{1} + tk_{2}) e^{\lambda t} \, , \\ y_{2} (t) = k_{2} e^{\lambda t} \, , \end{array} \right.$     $\left( \begin{array}{l} y_{1} (0) \\ y_{2} (0) \end{array} \right) \, = \, \left( \begin{array}{l} k_{1} \\ k_{2} \end{array} \right)$.
If $\lambda \neq 0$ and $k_{2} \neq 0$, then $y_{1} = y_{2} \left[ \frac{k_{1}}{k_{2}} + \frac{1}{\lambda} \ln \left( \frac{y_{2}}{k_{2}} \right) \right]$.

                     

$$\frac{dy_{2}}{dy_{1}} = \pm \infty \quad \mbox{at} \quad y_{... ...\quad \lim_{t \raro \pm \infty} \frac{dy_{2}}{dy_{1}} = 0 \, .$$

All solutions have vertical tangent lines when they cut the line $y_{2} = - \lambda y_{1}$. All solutions are tangent to the y₁-axis as they approach (leave) (0,0).

Case III:      $B = \left[ \begin{array}{rc} \alpha & \beta \\ - \beta & \alpha \end{array} \right] \, , \quad \beta > 0$.

$$\left\{ \begin{array}{l} y_{1}' = \alpha y_{1} + \beta y_{2} ... ... y_{2}' = - \beta y_{1} + \alpha y_{2} \, . \end{array}\right.$$

Let z = y₁ + i y₂. Then $z' = \ol{\lambda} z$, where $\ol{\lambda} = \alpha - i \beta$.

Thus the solution is

$$z (t) = k e^{\ol{\lambda}t} \, ,\quad k = k_{1} + i k_{2} \, ,$$

i.e. $y_{1} (t) + iy_{2} (t) = (k_{1} + ik_{2}) e^{(\alpha - i \beta)t} = e^{\alpha t} (k_{1} + ik_{2}) (\cos \beta t - i \sin \beta t)$ which yields

$$\left\{ \begin{array}{l} y_{1} (t) = e^{\alpha t} (k_{1} \cos... ...-k_{1} \sin \beta t + k_{2} \cos \beta t) \end{array} \right.$$

         

Remark:     All the phase portraits we have made so far are in terms of the canonical coordinates y. You may wonder what the portraits look like in terms of the original coordinates x. Here are some examples

Example 2.1.1:     Consider the system x' = Ax, where $A = \left( \begin{array}{rl} -4 & 3 \\ -2 & 1 \end{array} \right)$.

$$\mbox{Let} \quad x = Py , \quad \mbox{where} \quad P = \left( \begin{array}{cc} 3 & 1 \\ 2 & 1 \end{array} \right),$$

$P^{-1} = \left( \begin{array}{rr} 1 & -1 \\ -2 & 3 \end{array} \right)$. Then y' = By, where $B= \left( \begin{array}{rr} -2 & 0 \\ 0 & -1 \end{array} \right) = P^{-1} AP$.

       

Example 2.1.1 Simulation

Example 2.1.2:     Consider the system x' = Ax, where $A = \left( \begin{array}{rl} -2 & 2 \\ -1 & 0 \end{array} \right)$. Let

$$x = Py, \quad P = \left( \begin{array}{lr} 1 & -1 \\ 1 & 0 \e... ... \left( \begin{array}{rl} 0 & 1 \\ -1 & 1 \end{array} \right).$$

Then y' = By, $B = \left( \begin{array}{rr} -1 & 1 \\ -1 & -1 \end{array} \right) = P^{-1} AP$.

        

Example 2.1.2 Simulation

Remark:     It should be noted that systems x' = Ax and x' = kAx, k > 0, have identical phase portraits.