Linear Normal Form Planar Systems

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There are three linear systems of ordinary differential equations that we now solve explicitly using matrix exponentials. Remarkably, in a sense to be made precise, these are the only linear planar systems. The three systems are listed in Table ??.


name	equations	closed form solution

(a)	$\dot {X} = \mattwo {\lambda _1}{0}{0}{\lambda _2} X$	$X(t) = \mattwo {e^{\lambda _1 t}}{0}{0}{e^{\lambda _2 t}}X_0$

(b)	$\dot {X}=\mattwo {\sigma }{-\tau }{\tau }{\sigma }X$	$X(t) = e^{\sigma t} \mattwo {\cos (\tau t)}{-\sin (\tau t)}{\sin (\tau t)}{\cos (\tau t)}X_0$

(c)	$\dot {X} = \mattwo {\lambda _1}{1}{0}{\lambda _1}$	$X(t) = e^{\lambda _1 t}\mattwo {1}{t}{0}{1}X_0$

Table 1: Solutions to normal form ODEs with $X(0)=X_0$ .

The verification of Table ??(a) follows from (??), but it just reproduces earlier work in Section ?? where we considered uncoupled systems of two ordinary differential equations. To verify the solutions to (b) and (c), we need to prove:

Let $A$ and $B$ be two $n\times n$ matrices such that $\begin{equation} \label{e:AB=BA} AB = BA. \end{equation}$ Then $e^{A+B} = e^A e^B.$

Proof: Note that (??) implies that
$\begin{eqnarray} A^kB & = & BA^k \label{e:AkB=BAk}\\ e^{tA}B & = & Be^{tA}. \label{e:etAB=BetA} \end{eqnarray}$
Identity (??) is verified when $k=2$ using associativity of matrix multiplication, as follows $A^2B = AAB = ABA = BAA = BA^2.$ The argument for general $k$ is identical. Identity (??) follows directly from (??) and the power series definition of matrix exponentials (??).
We use Theorem ?? to complete the proof of this proposition. Recall that $X(t) = e^{t(A+B)}X_0$ is the unique solution to the initial value problem
$\begin{eqnarray*} \frac{dX}{dt} & = & (A+B)X \\ \\ X(0) & = & X_0. \end{eqnarray*}$
We claim that $Y(t) = e^{tA}e^{tB}X_0$ is another solution to this equation. Certainly $Y(0)=X_0$ . It follows from (??) that $\frac {d}{dt}e^{tA} = Ae^{tA} \AND \frac {d}{dt}e^{tB} = Be^{tB}.$ Thus the product rule together with (??) imply that
$\begin{eqnarray*} \frac{dY}{dt} & = & \left(Ae^{tA}\right)e^{tB}X_0 + e^{tA}\left(Be^{tB}\right)X_0 \\ & = & (A+B) e^{tA} e^{tB} X_0\\ & = & (A+B)Y(t). \end{eqnarray*}$
Thus $\frac {dY}{dt} = (A+B)Y,$ and $Y(t)=X(t)$ . Since $X_0$ is arbitrary it follows that $e^{t(A+B)} = e^{tA}e^{tB}.$ Evaluating at $t=1$ yields the desired result. $\blacksquare$

Verification of Table ??(b)

We begin by noting that the $2\times 2$ matrix $C$ in (b) is $C = \mattwo {\sigma }{-\tau }{\tau }{\sigma } = \sigma I_2 + \tau J,$ where $J= \mattwo {0}{-1}{1}{0}.$ Since $I_2J=JI_2$ , it follows from Proposition ?? that $e^{tC} = e^{(\sigma t)I_2} e^{(\tau t)J}.$ Thus (??) and (??) imply

$\begin{equation} \label{e:exprotation} e^{tC} = e^{\sigma t} \mattwo{\cos(\tau t)}{-\sin(\tau t)}{\sin(\tau t)}{\cos(\tau t)}, \end{equation}$ and (b) is verified.

Verification of Table ??(c)

To determine the solutions to Table ??(c), observe that $C = \mattwoc {\lambda _1}{1}{0}{\lambda _1} = \lambda _1 I_2 + N,$ where $N = \mattwo {0}{1}{0}{0}.$ Since $I_2N=NI_2$ , Proposition ?? implies

$\begin{equation} \label{e:expshear} e^{tC} = e^{(t\lambda_1)I_2}e^{tN} = e^{t\lambda}\mattwo{1}{t}{0}{1} \end{equation}$ by (??) and (??).

Summary

The normal form matrices in Table ?? are characterized by the number of linearly independent real eigenvectors. We summarize this information in Table ??. We show, in Section ??, that any planar linear system of ODEs can be solved just by noting how many independent eigenvectors the corresponding matrix has; general solutions are found by transforming the equations into one of the three types of equations listed in Table ??.


Matrix	Number of Real Eigenvectors	Reference

$\mattwoc {\lambda _1}{0}{0}{\lambda _2}$	two linearly independent	Section ??

$\mattwo {\sigma }{-\tau }{\tau }{\sigma }$	none	Chapter ??, (??)

$\mattwoc {\lambda _1}{1}{0}{\lambda _1}$	one linearly independent	Lemma ??

Table 2: Number of linearly independent real eigenvectors.

Exercises

Solve the initial value problem $\begin {array}{rcr} \dot {x} & = & 2x + 3y \\ \dot {y} & = & -3x + 2y \end {array}$ where $x(0) = 1 \AND y(0) = -2$ .

Solve the initial value problem $\begin {array}{rcr} \dot {x} & = & -2x + y \\ \dot {y} & = & -2y \end {array}$ where $x(0) = 4 \AND y(0) = -1$ .

Let $A$ be an $n\times n$ matrix such that $A^3=0$ . Compute $e^{tC}$ where $C=2I_n+A$ .

Use pplane5 to plot phase plane portraits for each of the three types of linear systems (a), (b) and (c) in Table ??. Based on this computer exploration answer the following questions:

If a solution to that system spirals about the origin, is the system of differential equations of type (a), (b) or (c)?
How many eigendirections are there for equations of type (c)?
Let $(x(t),y(t))$ be a solution to one of these three types of systems and suppose that $y(t)$ oscillates up and down infinitely often. Then $(x(t),y(t))$ is a solution for which type of system?