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# 4.4: Constant Coefficient Homogeneous Systems I

• • Contributed by William F. Trench
• Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University
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## Constant Coefficient Homogenous Systems I

We'll now begin our study of the homogeneous system

\begin{equation}\label{eq:4.4.1}
{\bf y}'=A{\bf y},
\end{equation}

where $$A$$ is an $$n\times n$$ constant matrix. Since $$A$$ is continuous on $$(-\infty,\infty)$$, Theorem $$(4.2.1)$$ implies that all solutions of \eqref{eq:4.4.1} are defined on $$(-\infty,\infty)$$. Therefore, when we speak of solutions of $${\bf y}'=A{\bf y}$$, we'll mean solutions on $$(-\infty,\infty)$$.

In this section we assume that all the eigenvalues of $$A$$ are real and that $$A$$ has a set of $$n$$ linearly independent eigenvectors. In the next two sections we consider the cases where some of the eigenvalues of $$A$$ are complex, or where $$A$$ does not have $$n$$ linearly independent eigenvectors.

In Example $$(4.3.2)$$ we showed that the vector functions

\begin{eqnarray*}
{\bf y}_1 = \left[ \begin{array} \\ -e^{2t} \\ 2e^{2t} \end{array} \right] \quad \mbox{and} \quad {\bf y}_2 = \left[ \begin{array} \\ -e^{-t} \\ e^{-t} \end{array} \right]
\end{eqnarray*}

form a fundamental set of solutions of the system

\begin{equation}\label{eq:4.4.2}
{\bf y}' = \left[ \begin{array} \\ {-4} & {-3} \\ 6 & 5 \end{array} \right] {\bf y},
\end{equation}

but we did not show how we obtained $${\bf y}_1$$ and $${\bf y}_2$$ in the first place. To see how these solutions can be obtained we write \eqref{eq:4.4.2} as

\begin{equation}\label{eq:4.4.3}
\begin{array}{ccc}
y_1'&=&-4y_1-3y_2\\y_2'&=&\phantom{-}6y_1+5y_2\end{array}
\end{equation}

and look for solutions of the form

\begin{equation}\label{eq:4.4.4}
\end{equation}

where $$x_1$$, $$x_2$$, and $$\lambda$$ are constants to be determined. Differentiating \eqref{eq:4.4.4} yields

\begin{eqnarray*}
\end{eqnarray*}

Substituting this and \eqref{eq:4.4.4} into \eqref{eq:4.4.3} and canceling the common factor $$e^{\lambda t}$$ yields

\begin{eqnarray*}
\begin{array}{ccc}-4x_1-3x_2&=&\lambda x_1 \\
6 x_1+5x_2&=&\lambda x_2.\end{array}
\end{eqnarray*}

For a given $$\lambda$$, this is a homogeneous algebraic system, since it can be rewritten as

\begin{equation}\label{eq:4.4.5}
\begin{array}{rcl} (-4-\lambda) x_1-3 x_2&=&0\\
6 x_1+(5-\lambda) x_2&=&0.\end{array}
\end{equation}

The trivial solution $$x_1=x_2=0$$ of this system isn't useful, since it corresponds to the trivial solution $$y_1\equiv y_2\equiv0$$ of \eqref{eq:4.4.3}, which can't be part of a fundamental set of solutions of \eqref{eq:4.4.2}. Therefore we consider only those values of $$\lambda$$ for which \eqref{eq:4.4.5} has nontrivial solutions. These are the values of $$\lambda$$ for which the determinant of \eqref{eq:4.4.5} is zero; that is,

\begin{eqnarray*}
\left|\begin{array}{cc}-4-\lambda&-3\\6&5-\lambda\end{array}\right|&=&
(-4-\lambda)(5-\lambda)+18\\&=&\lambda^2-\lambda-2\\
&=&(\lambda-2)(\lambda+1)=0,
\end{eqnarray*}

which has the solutions $$\lambda_1=2$$ and $$\lambda_2=-1$$.

Taking $$\lambda=2$$ in \eqref{eq:4.4.5} yields

\begin{eqnarray*}
-6 x_1-3 x_2&=&0\\
6 x_1+3 x_2&=&0,
\end{eqnarray*}

which implies that $$x_1=-x_2/2$$, where $$x_2$$ can be chosen arbitrarily. Choosing $$x_2=2$$ yields the solution $$y_1=-e^{2t}$$,
$$y_2=2e^{2t}$$ of \eqref{eq:4.4.3}. We can write this solution in vector form as

\begin{equation}\label{eq:4.4.6}
{\bf y}_1= \left[ \begin{array} \\ -1 \\ {\phantom{-}2} e^{2t} \end{array} \right].
\end{equation}

Taking $$\lambda=-1$$ in \eqref{eq:4.4.5} yields the system

\begin{eqnarray*}
-3 x_1-3 x_2&=&0\\
\phantom{-}6 x_1+6 x_2&=&0,
\end{eqnarray*}

so $$x_1=-x_2$$. Taking |(x_2=1\) here yields the solution $$y_1=-e^{-t}$$, $$y_2=e^{-t}$$ of \eqref{eq:4.4.3}. We can write this solution in vector form as

\begin{equation}\label{eq:4.4.7}
{\bf y}_2= \left[ \begin{array} \\ -1 \\ {\phantom{-}1}e^{-t} \end{array} \right].
\end{equation}

In \eqref{eq:4.4.6} and \eqref{eq:4.4.7} the constant coefficients in the arguments of the exponential functions are the eigenvalues of the coefficient matrix in \eqref{eq:4.4.2}, and the vector coefficients of the exponential functions are associated eigenvectors. This illustrates the next theorem.

### Theorem $$\PageIndex{1}$$

Suppose the $$n\times n$$ constant matrix $$A$$ has $$n$$ real eigenvalues $$\lambda_1,\lambda_2,\ldots,\lambda_n$$ (which need not be distinct) with associated linearly independent eigenvectors $${\bf x}_1,{\bf x}_2,\ldots,{\bf x}_n$$. Then the functions

\begin{eqnarray*}
{\bf y}_1 = {\bf x}_1 e^{\lambda_1 t}, \, {\bf y}_2 = {\bf x}_2 e^{\lambda_2 t}, \, \dots, \, {\bf y}_n = {\bf x}_n e^{\lambda_n t}
\end{eqnarray*}

form a fundamental set of solutions of $${\bf y}'=A{\bf y};$$ that is, the general solution of this system is

\begin{eqnarray*}
{\bf y} = c_1 {\bf x}_1 e^{\lambda_1 t} + c_2 {\bf x}_2 e^{\lambda_2 t} + \cdots + c_n {\bf x}_n e^{\lambda_n t}.
\end{eqnarray*}

Proof

Differentiating $${\bf y}_i={\bf x}_ie^{\lambda_it}$$ and recalling that $$A{\bf x}_i=\lambda_i{\bf x}_i$$ yields

\begin{eqnarray*}
{\bf y}_i' = \lambda_i {\bf x}_i e^{\lambda_i t} = A {\bf x}_i e^{\lambda_i t} = A {\bf y}_i.
\end{eqnarray*}

This shows that $${\bf y}_i$$ is a solution of $${\bf y}'=A{\bf y}$$.

The Wronskian of $$\{{\bf y}_1,{\bf y}_2,\ldots,{\bf y}_n\}$$ is

\begin{eqnarray*}
\left| \begin{array} \\ x_{11} e^{\lambda_1 t} & x_{12} e^{\lambda_2 t} & \cdots & x_{1n} e^{\lambda_n t} \\ x_{21} e^{\lambda_1 t} & x_{22} e^{\lambda_2 t} & \cdots & x_{2n} e^{\lambda_n t} \\ \vdots & \vdots & \ddots & \vdots \\ x_{n1} e^{\lambda_1 t} & x_{n2} e^{\lambda_2 t} & \cdots & x_{nn} e^{\lambda x_n} \end{array} \right| = e^{\lambda_1 t} e^{\lambda_2 t} \cdots e^{\lambda_n t} \left| \begin{array} \\ x_{11} & x_{12} & \cdots & x_{1n} \\ x_{21} & x_{22} & \cdots & x_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ x_{n1} & x_{n2} & \cdots & x_{nn} \end{array} \right|.
\end{eqnarray*}

Since the columns of the determinant on the right are $${\bf x}_1$$, $${\bf x}_2$$, $$\dots$$, $${\bf x}_n$$, which are assumed to be linearly independent, the determinant is nonzero. Therefore Theorem $$(4.3.3)$$ implies that $$\{{\bf y}_1,{\bf y}_2,\ldots,{\bf y}_n\}$$ is a fundamental set of solutions of $${\bf y}'=A{\bf y}$$.

Example $$\PageIndex{1}$$

(a) Find the general solution of

\begin{equation}\label{eq:4.4.8}
{\bf y}'= \left[ \begin{array} \\ 2 & 4 \\ 4 & 2 \end{array} \right] {\bf y}.
\end{equation}

(b) Solve the initial value problem

\begin{equation}\label{eq:4.4.9}
{\bf y}'= \left[ \begin{array} \\ 2 & 4 \\ 4 & 2 \end{array} \right] {\bf y}, \quad {\bf y}(0)= \left[ \begin{array} \\ 5 \\-1 \end{array} \right].
\end{equation}

(a) The characteristic polynomial of the coefficient matrix $$A$$ in \eqref{eq:4.4.8} is

\begin{eqnarray*}
\left|\begin{array}{cc} 2-\lambda&4\\4&2-\lambda\end{array}\right|
&=& (\lambda-2)^2-16\\
&=& (\lambda-2-4)(\lambda-2+4)\\
&=& (\lambda-6)(\lambda+2).
\end{eqnarray*}

Hence, $$\lambda_1=6$$ and $$\lambda_2 =-2$$ are eigenvalues of $$A$$. To obtain the eigenvectors, we must solve the system

\begin{equation}\label{eq:4.4.10}
\left[\begin{array}{cc} 2-\lambda&4\\4&2-\lambda\end{array}\right]
\left[\begin{array}{c} x_1\\x_2\end{array}\right]=
\left[\begin{array}{c} 0\\0\end{array}\right]
\end{equation}

with $$\lambda=6$$ and $$\lambda=-2$$. Setting $$\lambda=6$$ in \eqref{eq:4.4.10} yields

\begin{eqnarray*}
\left[ \begin{array} \\ -4 & 4 \\ 4 & -4 \end{array} \right] \left[ \begin{array} \\ x_1 \\ x_2 \end{array} \right] = \left[ \begin{array} \\ 0 \\ 0 \end{array} \right],
\end{eqnarray*}

which implies that $$x_1=x_2$$. Taking $$x_2=1$$ yields the eigenvector

\begin{eqnarray*}
{\bf x}_1 = \left[ \begin{array} \\ 1 \\ 1 \end{array} \right],
\end{eqnarray*}

so

\begin{eqnarray*}
{\bf y}_1 = \left[ \begin{array} \\ 1 \\ 1 \end{array} \right] e^{6t}
\end{eqnarray*}

is a solution of \eqref{eq:4.4.8}. Setting $$\lambda=-2$$ in \eqref{eq:4.4.10} yields

\begin{eqnarray*}
\left[ \begin{array} \\ 4 & 4 \\ 4 & 4 \end{array} \right] \left[ \begin{array} \\ x_1 \\ x_2 \end{array} \right] = \left[ \begin{array} \\ 0 \\ 0 \end{array} \right],
\end{eqnarray*}

which implies that $$x_1=-x_2$$. Taking $$x_2=1$$ yields the eigenvector

\begin{eqnarray*}
{\bf x}_2 = \left[ \begin{array} \\ -1 \\ 1 \end{array} \right],
\end{eqnarray*}

so

\begin{eqnarray*}
{\bf y}_2 = \left[ \begin{array} \\ -1 \\ 1 \end{array} \right] e^{-2t}
\end{eqnarray*}

is a solution of \eqref{eq:4.4.8}. From Theorem $$(4.4.1)$$, the general solution of \eqref{eq:4.4.8} is

\begin{equation}\label{eq:4.4.11}
{\bf y}=c_1{\bf y}_1+c_2{\bf y}_2=c_1\left[\begin{array}{r}1\\1\end{array}\right]e^{6t}+c_2\left[\begin{array}{r}-1\\1
\end{array}\right]e^{-2t}.
\end{equation}

(b) To satisfy the initial condition in \eqref{eq:4.4.9}, we must choose $$c_1$$ and $$c_2$$ in \eqref{eq:4.4.11} so that

\begin{eqnarray*}
c_1 \left[ \begin{array} \\ 1 \\ 1 \end{array} \right] + c_2 \left[ \begin{array} \\ -1 \\ 1 \end{array} \right] = \left[ \begin{array} \\ 5 \\ -1 \end{array} \right].
\end{eqnarray*}

This is equivalent to the system

\begin{eqnarray*}
c_1-c_2&=&\phantom{-}5\\
c_1+c_2&=&-1,
\end{eqnarray*}

so $$c_1=2, c_2=-3$$. Therefore the solution of \eqref{eq:4.4.9} is

\begin{eqnarray*}
{\bf y} = 2 \left[ \begin{array} \\ 1 \\ 1 \end{array} \right] e^{6t} -3 \left[ \begin{array} \\ -1 \\ 1 \end{array} \right] e^{-2t},
\end{eqnarray*}

or, in terms of components,

\begin{eqnarray*}
y_1 = 2e^{6t} + 3e^{-2t}, \quad y_2 = 2e^{6t} - 3e^{-2t}.
\end{eqnarray*}

Example $$\PageIndex{2}$$

(a) Find the general solution of

\begin{equation}\label{eq:4.4.12}
{\bf y}'=\left[\begin{array}{rrr}3&-1&-1\\-2& 3& 2\\4&-1&-2\end{array}\right]{\bf y}.
\end{equation}

(b) Solve the initial value problem

\begin{equation}\label{eq:4.4.13}
{\bf y}'=\left[\begin{array}{rrr}3&-1&-1\\-2&3&
2\\4&-1&-2\end{array}
-1\\8\end{array}\right].
\end{equation}

(a) The characteristic polynomial of the coefficient matrix $$A$$ in \eqref{eq:4.4.12} is

\begin{eqnarray*}
\left| \begin{array} \\ 3-\lambda & -1 & -1 \\ -2 & 3-\lambda & 2 \\ 4 & -1 & -2-\lambda \end{array} \right| = -(\lambda-2)(\lambda-3)(\lambda+1).
\end{eqnarray*}

Hence, the eigenvalues of $$A$$ are $$\lambda_1=2$$, $$\lambda_2=3$$, and $$\lambda_3=-1$$. To find the eigenvectors, we must solve the system

\begin{equation}\label{eq:4.4.14}
\left[\begin{array}{ccc}3-\lambda&-1&-1\\-2&3-\lambda& 2\\4&-1& -2-\lambda\end{array}\right]\left[\begin{array}{c} x_1\\x_2\\x_3 \end{array} \right]=\left[\begin{array}{r}0\\0\\0\end{array}\right]
\end{equation}

with $$\lambda=2$$, $$3$$, $$-1$$. With $$\lambda=2$$, the augmented matrix of \eqref{eq:4.4.14} is

\begin{eqnarray*}
\left[ \begin{array} \\ 1 & -1 & -1 & \vdots & 0 \\ -2 & 1 & 2 & \vdots & 0 \\ 4 & -1 & -4 & \vdots & 0 \end{array} \right],
\end{eqnarray*}

which is row equivalent to

\begin{eqnarray*}
\left[ \begin{array} \\ 1 & 0 & -1 & \vdots & 0 \\ 0 & 1 & 0 & \vdots & 0 \\ 0 & 0 & 0 & \vdots & 0 \end{array} \right].
\end{eqnarray*}

Hence, $$x_1=x_3$$ and $$x_2=0$$. Taking $$x_3=1$$ yields

\begin{eqnarray*}
{\bf y}_1 = \left[ \begin{array} \\ 1 \\ 0 \\ 1 \end{array} \right] e^{2t}
\end{eqnarray*}

as a solution of \eqref{eq:4.4.12}. With $$\lambda=3$$, the augmented matrix of \eqref{eq:4.4.14} is

\begin{eqnarray*}
\left[ \begin{array} \\ 0 & -1 & -1 & \vdots & 0 \ -2 & 0 & 2 & \vdots & 0 \\ 4 & -1 & -5 & \vdots & 0 \end{array} \right],
\end{eqnarray*}

which is row equivalent to

\begin{eqnarray*}
\left[ \begin{array} \\ 1 & 0 & -1 & \vdots & 0 \\ 0 & 1 & 1 & \vdots & 0 \\ 0 & 0 & 0 & \vdots & 0 \end{array} \right].
\end{eqnarray*}

Hence, $$x_1=x_3$$ and $$x_2=-x_3$$. Taking $$x_3=1$$ yields

\begin{eqnarray*}
{\bf y}_2 = \left[ \begin{array} \\ 1 \\ -1 \\ 1 \end{array} \right] e^{3t}
\end{eqnarray*}

as a solution of \eqref{eq:4.4.12}. With $$\lambda=-1$$, the augmented matrix of \eqref{eq:4.4.14} is

\begin{eqnarray*}
\left[ \begin{array} \\ 4 & -1 & -1 & \vdots & 0 \\-2 & 4 & 2 & \vdots & 0 \\ 4 & -1 & -1 & \vdots & 0 \end{array} \right],
\end{eqnarray*}

which is row equivalent to

\begin{eqnarray*}
\left[ \begin{array} \\ 1 & 0 & -{1 \over 7} & \vdots & 0 \\ 0 & 1 & {3 \over 7} & \vdots & 0 \\ 0 & 0 & 0 & \vdots & 0 \end{array} \right].
\end{eqnarray*}

Hence, $$x_1=x_3/7$$ and $$x_2=-3x_3/7$$. Taking $$x_3=7$$ yields

\begin{eqnarray*}
{\bf y}_3 = \left[ \begin{array} \\ 1 \\ -3 \\ 7 \end{array} \right] e^{-t}
\end{eqnarray*}

as a solution of \eqref{eq:4.4.12}. By Theorem $$(4.4.1)$$, the general solution of \eqref{eq:4.4.12} is

\begin{eqnarray*}
{\bf y} = c_1 \left[ \begin{array} \\ 1 \\ 0 \\ 1 \end{array} \right] e^{2t} + c_2 \left[ \begin{array} \\ 1 \\ -1 \\ 1 \end{array} \right] e^{3t} + c_3 \left[ \begin{array} \\ 1 \\ -3 \\ 7 \end{array} \right] e^{-t},
\end{eqnarray*}

which can also be written as

\begin{equation}\label{eq:4.4.15}
{\bf y}=\left[\begin{array}{crc}e^{2t}&e^{3t}&e^{-t}
\\0&-e^{3t}&
-3e^{-t}\\e^{2t}&e^{3t}&\phantom{-}7e^{-t}\end{array}
\right]\left[\begin{array}{c} c_1\\c_2\\c_3\end{array}\right].
\end{equation}

(b) To satisfy the initial condition in \eqref{eq:4.4.13} we must choose $$c_1$$, $$c_2$$, $$c_3$$ in \eqref{eq:4.4.15} so that

\begin{eqnarray*}
\left[ \begin{array} \\ 1 & 1 & 1 \\ 0 & -1 & -3 \\ 1 & 1 & 7 \end{array} \right] \left[ \begin{array} \\ c_1 \\ c_2 \\ c_3 \end{array} \right] = \left[ \begin{array} \\ 2 \\ -1 \\ 8 \end{array} \right].
\end{eqnarray*}

Solving this system yields $$c_1=3$$, $$c_2=-2$$, $$c_3=1$$. Hence, the solution of \eqref{eq:4.4.13} is

\begin{eqnarray*}
{\bf y}&=&\left[\begin{array}{ccc}e^{2t}&e^{3t}&
e^{-t}\\0&-e^{3t}
&-3e^{-t}\\e^{2t}&e^{3t}&7e^{-t}\end{array}
\right]
\left[\begin{array}{r}3\\-2\\1\end{array}\right]\\
&=&3\left[\begin{array}{r}1\\0\\1\end{array}\right]e^{2t}-2
\left[\begin{array}{r}1\\-1\\1\end{array}\right]
e^{3t}+\left[\begin{array}{r}1\\-3\\7\end{array}
\right]e^{-t}.
\end{eqnarray*}

Example $$\PageIndex{3}$$

Find the general solution of

\begin{equation}\label{eq:4.4.16}
{\bf y}'=\left[\begin{array}{rrr}-3&2&2\\
2&-3&2\\2&2&-3
\end{array}\right]{\bf y}.
\end{equation}

The characteristic polynomial of the coefficient matrix $$A$$ in \eqref{eq:4.4.16} is

\begin{eqnarray*}
\left[ \begin{array} \\ -3-\lambda & 2 & 2 \\ 2 & -3-\lambda & 2 \\ 2 & 2 & -3-\lambda \end{array} \right] = -(\lambda-1)(\lambda+5)^2.
\end{eqnarray*}

Hence, $$\lambda_1=1$$ is an eigenvalue of multiplicity $$1$$, while $$\lambda_2=-5$$ is an eigenvalue of multiplicity $$2$$. Eigenvectors associated with $$\lambda_1=1$$ are solutions of the system with augmented matrix

\begin{eqnarray*}
\left[ \begin{array} \\ -4 & 2 & 2 & \vdots & 0 \\ 2 & -4 & 2 \vdots & 0 \\ 2 & 2 & -4 & \vdots & 0 \end{array} \right],
\end{eqnarray*}

which is row equivalent to

\begin{eqnarray*}
\left[ \begin{array} \\ 1 & 0 & -1 & \vdots & 0 \\ 0 & 1 & -1 & \vdots & 0 \\ 0 & 0 & 0 & \vdots & 0 \end{array} \right].
\end{eqnarray*}

Hence, $$x_1=x_2=x_3$$, and we choose $$x_3=1$$ to obtain the solution

\begin{equation}\label{eq:4.4.17}
{\bf y}_1=\left[\begin{array}{r}1\\1\\1\end{array}\right]e^t
\end{equation}

of \eqref{eq:4.4.16}. Eigenvectors associated with $$\lambda_2=-5$$ are solutions of the system with augmented matrix

\begin{eqnarray*}
\left[ \begin{array} \\ 2 & 2 & 2 & \vdots & 0 \\ 2 & 2 & 2 & \vdots & 0 \\ 2 & 2 & 2 & \vdots & 0 \end{array} \right].
\end{eqnarray*}

Hence, the components of these eigenvectors need only satisfy the single condition

\begin{eqnarray*}
x_1 + x_2 + x_3 = 0.
\end{eqnarray*}

Since there's only one equation here, we can choose $$x_2$$ and $$x_3$$ arbitrarily. We obtain one eigenvector by choosing $$x_2=0$$ and $$x_3=1$$, and another by choosing $$x_2=1$$ and $$x_3=0$$. In both cases $$x_1=-1$$. Therefore

\begin{eqnarray*}
\left[ \begin{array} \\ -1 \\ 0 \\ 1 \end{array} \right] \quad \mbox{and} \ quad \left[ \begin{array} \\ -1 \\ 1 \\ 0 \end{array} \right]
\end{eqnarray*}

are linearly independent eigenvectors associated with $$\lambda_2= -5$$, and the corresponding solutions of \eqref{eq:4.4.16} are

\begin{eqnarray*}
{\bf y}_2 = \left[ \begin{array} \\ -1 \\ 0 \\ 1 \end{array} \right] e^{-5t} \quad \mbox{and} \quad {\bf y}_3 = \left[ \begin{array} \\ -1 \\ 1 \\ 0 \end{array} \right] e^{-5t}.
\end{eqnarray*}

Because of this and \eqref{eq:4.4.17}, Theorem $$(4.4.1)$$ implies that the general solution of \eqref{eq:4.4.16} is

\begin{eqnarray*}
{\bf y} = c_1 \left[ \begin{array} \\ 1 \\ 1 \\ 1 \end{array} \right] e^t + c_2 \left[ \begin{array} \\ -1 \\ 0 \\ 1 \end{array} \right] e^{-5t} + c_3 \left[ \begin{array} \\ -1 \\ 1 \\ 0 \end{array} \right] e^{-5t}.
\end{eqnarray*}

## Geometric Properties of Solutions when $$n=2$$

We'll now consider the geometric properties of solutions of a $$2\times 2$$ constant coefficient system

\begin{equation} \label{eq:4.4.18}
\left[ \begin{array} \\ {y_1'} \\ {y_2'} \end{array} \right] = \left[\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22}
\end{array}\right] \left[ \begin{array} \\ {y_1} \\ {y_2} \end{array} \right].
\end{equation}

It is convenient to think of a $$y_1$$-$$y_2$$ plane," where a point is identified by rectangular coordinates $$(y_1,y_2)$$. If $${\bf y}=\displaystyle{ \left[ \begin{array} \\ {y_1} \\ {y_2} \end{array} \right]}$$ is a non-constant solution of \eqref{eq:4.4.18}, then the point $$(y_1(t),y_2(t))$$ moves along a curve $$C$$ in the $$y_1$$-$$y_2$$ plane as $$t$$ varies from $$-\infty$$ to $$\infty$$. We call $$C$$ the $$\textcolor{blue}{\mbox{trajectory}}$$ of $${\bf y}$$. (We also say that $$C$$ is a trajectory of the system \eqref{eq:4.4.18}.) It's important to note that $$C$$ is the trajectory of infinitely many solutions of \eqref{eq:4.4.18}, since if $$\tau$$ is any real number, then $${\bf y}(t-\tau)$$ is a solution of \eqref{eq:4.4.18} (Exercise $$(4.4E.28)$$ part (b)), and $$(y_1(t-\tau),y_2(t-\tau))$$ also moves along $$C$$ as $$t$$ varies from $$-\infty$$ to $$\infty$$. Moreover, Exercise $$(4.4E.28)$$ part (c) implies that distinct trajectories of \eqref{eq:4.4.18} can't intersect, and that two solutions $${\bf y}_1$$ and $${\bf y}_2$$ of \eqref{eq:4.4.18} have the same trajectory if and only if $${\bf y}_2(t)={\bf y}_1(t-\tau)$$ for some $$\tau$$.

From Exercise $$(4.4E.28)$$ part (a), a trajectory of a nontrivial solution of \eqref{eq:4.4.18} can't contain $$(0,0)$$, which we define to be the trajectory of the trivial solution $${\bf y}\equiv0$$. More generally, if $${\bf y}=\displaystyle{ \left[ \begin{array} \\ {k_1} \\ {k_2} \end{array} \right]} \ne{\bf 0}$$ is a constant solution of \eqref{eq:4.4.18} (which could occur if zero is an eigenvalue of the matrix of \eqref{eq:4.4.18}), we define the trajectory of $${\bf y}$$ to be the single point $$(k_1,k_2)$$.

To be specific, this is the question: What do the trajectories look like, and how are they traversed? In this section we'll answer this question, assuming that the matrix

\begin{eqnarray*}
A = \left[ \begin{array} \\ a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right]
\end{eqnarray*}

of \eqref{eq:4.4.18} has real eigenvalues $$\lambda_1$$ and $$\lambda_2$$ with associated linearly independent eigenvectors $${\bf x}_1$$ and $${\bf x}_2$$. Then the general solution of \eqref{eq:4.4.18} is

\begin{equation} \label{eq:4.4.19}
{\bf y}= c_1{\bf x}_1
e^{\lambda_1 t}+c_2{\bf x}_2e^{\lambda_2 t}.
\end{equation}

We'll consider other situations in the next two sections.

We leave it to you (Exercise $$(4.4E.35)$$) to classify the trajectories of \eqref{eq:4.4.18} if zero is an eigenvalue of $$A$$. We'll confine our attention here to the case where both eigenvalues are nonzero. In this case the simplest situation is where $$\lambda_1=\lambda_2\ne0$$, so \eqref{eq:4.4.19} becomes

\begin{eqnarray*}
{\bf y} = (c_1 {\bf x}_1 + c_2 {\bf x}_2) e^{\lambda_1 t}.
\end{eqnarray*}

Since $${\bf x}_1$$ and $${\bf x}_2$$ are linearly independent, an arbitrary vector $${\bf x}$$ can be written as $${\bf x}=c_1{\bf x}_1+c_2{\bf x}_2$$. Therefore the general solution of \eqref{eq:4.4.18} can be written as $${\bf y}={\bf x}e^{\lambda_1 t}$$ where $${\bf x}$$ is an arbitrary $$2$$-vector, and the trajectories of nontrivial solutions of \eqref{eq:4.4.18} are half-lines through (but not including) the origin. The direction of motion is away from the origin if $$\lambda_1>0$$ (Figure $$4.4.1$$), toward it if $$\lambda_1<0$$ (Figure $$4.4.2$$). (In these and the next figures an arrow through a point indicates the direction of motion along the trajectory through the point.)

### Figure $$4.4.1$$

Trajectories of a $$2\times 2$$ system with a repeated positive ### Figure $$4.4.2$$

Trajectories of a $$2\times 2$$ system with a repeated negative Now suppose $$\lambda_2>\lambda_1$$, and let $$L_1$$ and $$L_2$$ denote lines through the origin parallel to $${\bf x}_1$$ and $${\bf x}_2$$, respectively. By a half-line of $$L_1$$ (or $$L_2$$), we mean either of the rays obtained by removing the origin from $$L_1$$ (or $$L_2$$).

Letting $$c_2=0$$ in \eqref{eq:4.4.19} yields $${\bf y}=c_1{\bf x}_1e^{\lambda_1 t}$$. If $$c_1\ne0$$, the trajectory defined by this solution is a half-line of $$L_1$$. The direction of motion is away from the origin if $$\lambda_1>0$$, toward the origin if $$\lambda_1<0$$. Similarly, the trajectory of $${\bf y}=c_2{\bf x}_2e^{\lambda_2 t}$$ with $$c_2\ne0$$ is a half-line of $$L_2$$.

Henceforth, we assume that $$c_1$$ and $$c_2$$ in \eqref{eq:4.4.19} are both nonzero. In this case, the trajectory of \eqref{eq:4.4.19} can't intersect $$L_1$$ or $$L_2$$, since every point on these lines is on the trajectory of a solution for which either $$c_1=0$$ or $$c_2=0$$. (Remember: distinct trajectories can't intersect!). Therefore the trajectory of \eqref{eq:4.4.19} must lie entirely in one of the four open sectors bounded by $$L_1$$ and $$L_2$$, but do not any point on $$L_1$$ or $$L_2$$. Since the initial point $$(y_1(0),y_2(0))$$ defined by

\begin{eqnarray*}
{\bf y}(0) = c_1 {\bf x}_1 + c_2 {\bf x}_2
\end{eqnarray*}

is on the trajectory, we can determine which sector contains the trajectory from the signs of $$c_1$$ and $$c_2$$, as shown in Figure $$4.4.3$$.

The direction of $${\bf y}(t)$$ in \eqref{eq:4.4.19} is the same as that of

\begin{equation} \label{eq:4.4.20}
e^{-\lambda_2 t}{\bf y}(t)=
c_1{\bf x}_1e^{-(\lambda_2-\lambda_1)t}+c_2{\bf x}_2
\end{equation}

and of

\begin{equation} \label{eq:4.4.21}
e^{-\lambda_1 t}{\bf y}(t)=c_1{\bf
x}_1+c_2{\bf x}_2e^{(\lambda_2-\lambda_1)t}.
\end{equation}

Since the right side of \eqref{eq:4.4.20} approaches $$c_2{\bf x}_2$$ as $$t\to\infty$$, the trajectory is asymptotically parallel to $$L_2$$ as $$t\to\infty$$. Since the right side of \eqref{eq:4.4.21} approaches $$c_1{\bf x}_1$$ as $$t\to-\infty$$, the trajectory is asymptotically parallel to $$L_1$$ as $$t\to-\infty$$.

The shape and direction of traversal of the trajectory of \eqref{eq:4.4.19} depend upon whether $$\lambda_1$$ and $$\lambda_2$$ are both positive, both negative, or of opposite signs. We'll now analyze these three cases.

Henceforth $$\|{\bf u}\|$$ denote the length of the vector $${\bf u}$$.

### Figure $$4.4.3$$

Four open sectors bounded by $$L_1$$ and $$L_2$$ ### Figure $$4.4.4$$

Two positive eigenvalues; motion away from origin ## Case 1: $$\lambda_2>\lambda_1>0$$

Figure $$4.4.4$$ shows some typical trajectories. In this case, $$\lim_{t\to-\infty}\|{\bf y}(t)\|=0$$, so the trajectory is not only asymptotically parallel to $$L_1$$ as $$t\to-\infty$$, but is actually asymptotically tangent to $$L_1$$ at the origin. On the other hand, $$\lim_{t\to\infty}\|{\bf y}(t)\|=\infty$$ and

\begin{eqnarray*}
\lim_{t\to\infty} \left\|{\bf y}(t)-c_2{\bf x}_2 e^{\lambda_2 t} \right\| = \lim_{t\to\infty}\|c_1{\bf x_1} e^{\lambda_1 t} \| = \infty,
\end{eqnarray*}

so, although the trajectory is asymptotically parallel to $$L_2$$ as $$t\to\infty$$, it's not asymptotically tangent to $$L_2$$. The direction of motion along each trajectory is away from the origin.

## Case 2: $$0>\lambda_2>\lambda_1$$

Figure $$4.4.5$$ shows some typical trajectories. In this case, $$\lim_{t\to\infty}\|{\bf y}(t)\|=0$$, so the trajectory is asymptotically tangent to $$L_2$$ at the origin as $$t\to\infty$$. On the other hand, $$\lim_{t\to-\infty}\|{\bf y}(t)\|=\infty$$ and

\begin{eqnarray*}
\lim_{t\to-\infty} \left\|{\bf y}(t)-c_1{\bf x}_1 e^{\lambda_1 t} \right\| = \lim_{t\to-\infty} \| c_2{\bf x}_2 e^{\lambda_2 t} \| = \infty,
\end{eqnarray*}

so, although the trajectory is asymptotically parallel to $$L_1$$ as $$t\to-\infty$$, it's not asymptotically tangent to it. The direction of motion along each trajectory is toward the origin.

### Figure $$4.4.5$$

Two negative eigenvalues; motion toward the origin ### Figure $$4.4.6$$

Eigenvalues of different signs ## Case 3: $$\lambda_2>0>\lambda_1$$

Figure $$4.4.6$$ shows some typical trajectories. In this case,

\begin{eqnarray*}
\lim_{t\to\infty}\|{\bf y}(t)\| = \infty \quad \mbox{and} \quad \lim{|t\to\infty}\left\| {\bf y}(t)-c_2{\bf x}_2 e^{\lambda_2 t} \right\| = \lim_{t\to\infty}\| c_1{\bf x}_1 e^{\lambda_1 t}\| = 0,
\end{eqnarray*}

so the trajectory is asymptotically tangent to $$L_2$$ as $$t\to\infty$$. Similarly,

\begin{eqnarray*}
\lim_{t\to\infty}\|{\bf y}(t)\| = \infty \quad \mbox{and} \quad \lim_{t\to\infty}\left\|{\bf y}(t)-c_1{\bf x}_1 e^{\lambda_1 t} \right\| = \lim_{t\to\infty}\|c_2{\bf x}_2 e^{\lambda_2 t} \| = 0,
\end{eqnarray*}

so the trajectory is asymptotically tangent to $$L_1$$ as $$t\to-\infty$$. The direction of motion is toward the origin on $$L_1$$ and away from the origin on $$L_2$$. The direction of motion along any other trajectory is away from $$L_1$$, toward $$L_2$$.