10.4: Constant Coefficient Homogeneous Systems I

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We’ll now begin our study of the homogeneous system

$$\label{eq:10.4.1} {\bf y}'=A{\bf y},$$

where $A$ is an $n\times n$ constant matrix. Since $A$ is continuous on $(-\infty,\infty)$, Theorem 10.2.1 implies that all solutions of Equation $\ref{eq:10.4.1}$ 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 10.3.2 we showed that the vector functions

$${\bf y}_1=\twocol {-e^{2t}}{2e^{2t}}\quad \text{and} \quad {\bf y}_2=\twocol{-e^{-t}}{e^{-t}}\nonumber$$

form a fundamental set of solutions of the system

$$\label{eq:10.4.2} {\bf y}'=\left[\begin{array}{cc} {-4}&{-3}\\[4pt]{6}&{5}\end{array} \right] {\bf y},$$

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 Equation $\ref{eq:10.4.2}$ as

$$\label{eq:10.4.3} \begin{array}{ccc} y_1'&=-4y_1-3y_2\\[4pt]y_2'&=\phantom{-}6y_1+5y_2\end{array}$$

and look for solutions of the form

$$\label{eq:10.4.4} y_1=x_1e^{\lambda t}\quad \text{and} \quad y_2=x_2e^{\lambda t},$$

where $x_1$, $x_2$, and $\lambda$ are constants to be determined. Differentiating Equation $\ref{eq:10.4.4}$ yields

$$y_1'=\lambda x_1e^{\lambda t}\quad\mbox{ and }\quad y_2'=\lambda x_2e^{\lambda t}.\nonumber$$

Substituting this and Equation $\ref{eq:10.4.4}$ into Equation $\ref{eq:10.4.3}$ and canceling the common factor $e^{\lambda t}$ yields

$$\begin{array}{ccc}-4x_1-3x_2&=&\lambda x_1 \\[4pt] 6 x_1+5x_2&=&\lambda x_2.\end{array}\nonumber$$

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

$$\label{eq:10.4.5} \begin{array}{rcl} (-4-\lambda) x_1-3 x_2&=&0\\[4pt] 6 x_1+(5-\lambda) x_2&=&0.\end{array}$$

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 Equation $\ref{eq:10.4.3}$, which can’t be part of a fundamental set of solutions of Equation $\ref{eq:10.4.2}$. Therefore we consider only those values of $\lambda$ for which Equation $\ref{eq:10.4.5}$ has nontrivial solutions. These are the values of $\lambda$ for which the determinant of Equation $\ref{eq:10.4.5}$ is zero; that is,

$$\begin{aligned} \left|\begin{array}{cc}-4-\lambda&-3\\[4pt]6&5-\lambda\end{array}\right|&= (-4-\lambda)(5-\lambda)+18\\[4pt]&=\lambda^2-\lambda-2\\[4pt] &=(\lambda-2)(\lambda+1)=0,\end{aligned}\nonumber$$

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

Taking $\lambda=2$ in Equation $\ref{eq:10.4.5}$ yields

$$\begin{aligned} -6 x_1-3 x_2&=0\\[4pt] 6 x_1+3 x_2&=0,\end{aligned}\nonumber$$

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 Equation $\ref{eq:10.4.3}$. We can write this solution in vector form as

$$\label{eq:10.4.6} {\bf y}_1=\twocol {-1}{\phantom{-}2} e^{2t}.$$

Taking $\lambda=-1$ in Equation $\ref{eq:10.4.5}$ yields the system

$$\begin{aligned} -3 x_1-3 x_2&=0\\[4pt] \phantom{-}6 x_1+6 x_2&=0,\end{aligned}\nonumber$$

so $x_1=-x_2$. Taking $x_2=1$ here yields the solution $y_1=-e^{-t}$, $y_2=e^{-t}$ of Equation $\ref{eq:10.4.3}$. We can write this solution in vector form as

$$\label{eq:10.4.7} {\bf y}_2=\twocol{-1}{\phantom{-}1}e^{-t}.$$

In Equation $\ref{eq:10.4.6}$ and Equation $\ref{eq:10.4.7}$ the constant coefficients in the arguments of the exponential functions are the eigenvalues of the coefficient matrix in Equation $\ref{eq:10.4.2}$, and the vector coefficients of the exponential functions are associated eigenvectors. This illustrates the next theorem.

Theorem 10.4.1

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

$${\bf y}_1={\bf x}_1e^{\lambda_1 t},\, {\bf y}_2={\bf x}_2e^{\lambda_2 t},\, \dots,\, {\bf y}_n={\bf x}_ne^{\lambda_n t} \nonumber$$

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

$${\bf y}=c_1{\bf x}_1e^{\lambda_1 t}+c_2{\bf x}_2e^{\lambda_2 t} +\cdots+c_n{\bf x}_ne^{\lambda_n t}. \nonumber$$

Proof

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

$${\bf y}_i'=\lambda_i{\bf x}_ie^{\lambda_it}=A{\bf x}_ie^{\lambda_it} =A{\bf y}_i. \nonumber$$

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

$$\left|\begin{array}{cccc} x_{11}e^{\lambda_1 t}& x_{12}e^{\lambda_2 t}&\cdots& x_{1n}e^{\lambda_n t}\\[4pt] x_{21}e^{\lambda_1 t}& x_{22}e^{\lambda_2 t}&\cdots& x_{2n}e^{\lambda_n t}\\[4pt]\vdots&\vdots&\ddots&\vdots\\[4pt] x_{n1}e^{\lambda_1 t}& x_{n2}e^{\lambda_2 t}&\cdots& x_{nn}e^{\lambda x_n t}\end{array}\right| =e^{\lambda_1 t}e^{\lambda_2 t}\cdots e^{\lambda_n t} \left|\begin{array}{cccc} x_{11}&x_{12}&\cdots&x_{1n}\cr x_{21}&x_{22}&\cdots&x_{2n}\cr \vdots&\vdots&\ddots&\vdots\cr x_{n1}&x_{n2}&\cdots&x_{nn}\cr \end{array}\right|.\nonumber$$

Since the columns of the determinant on the right are ${\bf x}_1$, ${\bf x}_2$, …, ${\bf x}_n$, which are assumed to be linearly independent, the determinant is nonzero. Therefore Theorem 10.3.3 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 10.4.1

1. Find the general solution of $$\label{eq:10.4.8} {\bf y}'=\left[\begin{array}{cc}{2}&{4}\\[4pt]{4}&{2}\end{array} \right] {\bf y}.$$
2. Solve the initial value problem $$\label{eq:10.4.9} {\bf y}'=\left[\begin{array}{cc}{2}&{4}\\[4pt]{4}&{2}\end{array} \right] {\bf y},\quad{\bf y}(0)=\left[\begin{array}{r}5 \\[4pt]-1 \end{array}\right].$$

Solution a

The characteristic polynomial of the coefficient matrix $A$ in Equation $\ref{eq:10.4.8}$ is

$$\begin{aligned} \left|\begin{array}{cc} 2-\lambda&4\\[4pt]4&2-\lambda\end{array}\right| &= (\lambda-2)^2-16\\[4pt] &= (\lambda-2-4)(\lambda-2+4)\\[4pt] &= (\lambda-6)(\lambda+2).\end{aligned}\nonumber$$

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

$$\label{eq:10.4.10} \left[\begin{array}{cc} 2-\lambda&4\\[4pt]4&2-\lambda\end{array}\right] \left[\begin{array}{c} x_1\\[4pt]x_2\end{array}\right]= \left[\begin{array}{c} 0\\[4pt]0\end{array}\right]$$

with $\lambda=6$ and $\lambda=-2$. Setting $\lambda=6$ in Equation $\ref{eq:10.4.10}$ yields

$$\left[\begin{array}{rr}-4&4\\[4pt]4&-4 \end{array}\right]\left[\begin{array}{c} x_1\\[4pt]x_2\end{array}\right]=\left[\begin{array}{c} 0\\[4pt]0\end{array} \right],\nonumber$$

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

$${\bf x}_1=\left[\begin{array}{c} 1\\[4pt]1\end{array}\right],\nonumber$$

so

$${\bf y}_1=\left[\begin{array}{c} 1\\[4pt]1\end{array}\right]e^{6t}\nonumber$$

is a solution of Equation $\ref{eq:10.4.8}$. Setting $\lambda=-2$ in Equation $\ref{eq:10.4.10}$ yields

$$\left[\begin{array}{rr} 4&4\\[4pt]4&4\end{array}\right] \left[\begin{array}{c} x_1\\[4pt]x_2 \end{array}\right]=\left[\begin{array}{c} 0\\[4pt]0\end{array}\right],\nonumber$$

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

$${\bf x}_2=\left[\begin{array}{r}-1\\[4pt]1\end{array}\right],\nonumber$$

so

$${\bf y}_2=\left[\begin{array}{r}-1\\[4pt]1\end{array} \right]e^{-2t}\nonumber$$

is a solution of Equation $\ref{eq:10.4.8}$. From Theorem 10.4.1 , the general solution of Equation $\ref{eq:10.4.8}$ is

$$\label{eq:10.4.11} {\bf y}=c_1{\bf y}_1+c_2{\bf y}_2=c_1\left[\begin{array}{r}1\\[4pt]1 \end{array}\right]e^{6t}+c_2\left[\begin{array}{r}-1\\[4pt]1 \end{array}\right]e^{-2t}.$$

Solution b

To satisfy the initial condition in Equation $\ref{eq:10.4.9}$, we must choose $c_1$ and $c_2$ in Equation $\ref{eq:10.4.11}$ so that

$$c_1\left[\begin{array}{r}1\\[4pt]1\end{array}\right]+c_2\left[ \begin{array}{r}-1\\[4pt] 1\end{array}\right]=\left[\begin{array}{r}5\\[4pt]-1 \end{array}\right].\nonumber$$

This is equivalent to the system

$$\begin{aligned} c_1-c_2&=\phantom{-}5\\[4pt] c_1+c_2&=-1,\end{aligned}\nonumber$$

so $c_1=2, c_2=-3$. Therefore the solution of Equation $\ref{eq:10.4.9}$ is

$${\bf y}=2\left[\begin{array}{r}1\\[4pt]1\end{array}\right]e^{6t}-3 \left[\begin{array}{r}-1\\[4pt]1\end{array}\right]e^{-2t},\nonumber$$

or, in terms of components,

$$y_1=2e^{6t}+3e^{-2t},\quad y_2=2e^{6t}-3e^{-2t}.\nonumber$$

Example 10.4.2

1. Find the general solution of $$\label{eq:10.4.12} {\bf y}'=\left[\begin{array}{rrr}3&-1&-1\\[4pt]-2& 3& 2\\[4pt]4&-1&-2\end{array}\right]{\bf y}.$$
2. Solve the initial value problem $$\label{eq:10.4.13} {\bf y}'=\left[\begin{array}{rrr}3&-1&-1\\[4pt]-2&3& 2\\[4pt]4&-1&-2\end{array} \right]{\bf y},\quad{\bf y}(0)=\left[\begin{array}{r}2\\[4pt] -1\\[4pt]8\end{array}\right].$$

Solution a

The characteristic polynomial of the coefficient matrix $A$ in Equation $\ref{eq:10.4.12}$ is

$$\left|\begin{array}{ccc}3-\lambda&-1&-1\\[4pt]-2&3-\lambda& 2\\[4pt]4 &-1&-2-\lambda\end{array}\right|=-(\lambda-2)(\lambda-3)(\lambda+1).\nonumber$$

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

$$\label{eq:10.4.14} \left[\begin{array}{ccc}3-\lambda&-1&-1\\[4pt]-2&3-\lambda& 2\\[4pt]4&-1& -2-\lambda\end{array}\right]\left[\begin{array}{c} x_1\\[4pt]x_2\\[4pt]x_3 \end{array} \right]=\left[\begin{array}{r}0\\[4pt]0\\[4pt]0\end{array}\right]$$

with $\lambda=2$, $3$, $-1$. With $\lambda=2$, the augmented matrix of Equation $\ref{eq:10.4.14}$ is

$$\left[\begin{array}{rrrcr} 1&-1&-1&\vdots&0\\[4pt]-2& 1&2&\vdots&0\\[4pt]4&-1&-4&\vdots&0 \end{array}\right],\nonumber$$

which is row equivalent to

$$\left[\begin{array}{rrrcr} 1&0&-1&\vdots&0\\[4pt]0&1&0& \vdots&0\\[4pt]0&0&0&\vdots&0\end{array}\right].\nonumber$$

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

$${\bf y}_1=\left[\begin{array}{rrr}1\\[4pt]0\\[4pt]1\end{array}\right]e^{2t}\nonumber$$

as a solution of Equation $\ref{eq:10.4.12}$. With $\lambda=3$, the augmented matrix of Equation $\ref{eq:10.4.14}$ is

$$\left[\begin{array}{rrrcr}0&-1&-1&\vdots&0\\[4pt]-2& 0& 2&\vdots&0\\[4pt]4&-1&-5&\vdots&0 \end{array}\right],\nonumber$$

which is row equivalent to

$$\left[\begin{array}{rrrcr} 1&0&-1&\vdots&0\\[4pt]0&1&1& \vdots&0\\[4pt]0&0&0&\vdots&0\end{array}\right].\nonumber$$

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

$${\bf y}_2=\left[\begin{array}{r}1\\[4pt]-1\\[4pt]1\end{array} \right]e^{3t}\nonumber$$

as a solution of Equation $\ref{eq:10.4.12}$. With $\lambda=-1$, the augmented matrix of Equation $\ref{eq:10.4.14}$ is

$$\left[\begin{array}{rrrcr} 4&-1&-1&\vdots&0\\[4pt]-2&4& 2&\vdots&0\\[4pt]4&-1&-1&\vdots&0 \end{array}\right],\nonumber$$

which is row equivalent to

$$\left[\begin{array}{rrrcr} 1&0&-{1\over 7}&\vdots&0\\[4pt]0&1& {3\over 7}&\vdots&0\\[4pt]0&0&0&\vdots&0\end{array}\right].\nonumber$$

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

$${\bf y}_3=\left[\begin{array}{r}1\\[4pt]-3\\[4pt]7\end{array} \right]e^{-t}\nonumber$$

as a solution of Equation $\ref{eq:10.4.12}$. By Theorem 10.4.1 , the general solution of Equation $\ref{eq:10.4.12}$ is

$${\bf y}=c_1\left[\begin{array}{r}1\\[4pt]0\\[4pt]1\end{array}\right]e^{2t} +c_2\left[\begin{array}{r}1\\[4pt]-1\\[4pt]1\end{array}\right] e^{3t}+c_3 \left[\begin{array}{r}1\\[4pt]-3\\[4pt]7\end{array}\right]e^{-t},\nonumber$$

which can also be written as

$$\label{eq:10.4.15} {\bf y}=\left[\begin{array}{crc}e^{2t}&e^{3t}&e^{-t} \\[4pt]0&-e^{3t}& -3e^{-t}\\[4pt]e^{2t}&e^{3t}&\phantom{-}7e^{-t}\end{array} \right]\left[\begin{array}{c} c_1\\[4pt]c_2\\[4pt]c_3\end{array}\right].$$

Solution b

To satisfy the initial condition in Equation $\ref{eq:10.4.13}$ we must choose $c_1$, $c_2$, $c_3$ in Equation $\ref{eq:10.4.15}$ so that

$$\left[\begin{array}{rrr}1&1&1\\[4pt]0&-1&-3\\[4pt] 1&1&7\end{array}\right] \left[\begin{array}{c} c_1\\[4pt]c_2\\[4pt]c_3\end{array}\right]= \left[\begin{array}{r}2\\[4pt]-1\\[4pt]8\end{array}\right].\nonumber$$

Solving this system yields $c_1=3$, $c_2=-2$, $c_3=1$. Hence, the solution of Equation $\ref{eq:10.4.13}$ is

$$\begin{aligned} {\bf y}&=\left[\begin{array}{ccc}e^{2t}&e^{3t}& e^{-t}\\[4pt]0&-e^{3t} &-3e^{-t}\\[4pt]e^{2t}&e^{3t}&7e^{-t}\end{array} \right] \left[\begin{array}{r}3\\[4pt]-2\\[4pt]1\end{array}\right]\\[4pt] &=3\left[\begin{array}{r}1\\[4pt]0\\[4pt]1\end{array}\right]e^{2t}-2 \left[\begin{array}{r}1\\[4pt]-1\\[4pt]1\end{array}\right] e^{3t}+\left[\begin{array}{r}1\\[4pt]-3\\[4pt]7\end{array} \right]e^{-t}.\end{aligned}\nonumber$$

Example 10.4.3

Find the general solution of

$$\label{eq:10.4.16} {\bf y}'=\left[\begin{array}{rrr}-3&2&2\\[4pt] 2&-3&2\\[4pt]2&2&-3 \end{array}\right]{\bf y}.$$

Solution

The characteristic polynomial of the coefficient matrix $A$ in Equation $\ref{eq:10.4.16}$ is

$$\left|\begin{array}{ccc}-3-\lambda&2&2\\[4pt]2&-3-\lambda&2\\[4pt]2&2 &-3-\lambda\end{array}\right|=-(\lambda-1)(\lambda+5)^2.\nonumber$$

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

$$\left[\begin{array}{rrrcr}-4&2&2&\vdots&0\\[4pt] 2 &-4&2&\vdots&0\\[4pt]2&2&-4& \vdots&0\end{array}\right],\nonumber$$

which is row equivalent to

$$\left[\begin{array}{rrrcr} 1&0&-1&\vdots& 0\\[4pt]0&1&-1 &\vdots& 0 \\[4pt]0&0&0&\vdots&0\end{array}\right].\nonumber$$

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

$$\label{eq:10.4.17} {\bf y}_1=\left[\begin{array}{r}1\\[4pt]1\\[4pt]1\end{array}\right]e^t$$

of Equation $\ref{eq:10.4.16}$. Eigenvectors associated with $\lambda_2=-5$ are solutions of the system with augmented matrix

$$\left[\begin{array}{rrrcr} 2&2&2&\vdots&0\\[4pt]2&2&2&\vdots&0 \\[4pt]2&2&2&\vdots&0\end{array}\right].\nonumber$$

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

$$x_1+x_2+x_3=0.\nonumber$$

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

$$\left[\begin{array}{r}-1\\[4pt]0\\[4pt]1\end{array}\right]\quad \mbox{ and }\quad\left[\begin{array}{r}-1\\[4pt]1\\[4pt]0 \end{array}\right]\nonumber$$

are linearly independent eigenvectors associated with $\lambda_2= -5$, and the corresponding solutions of Equation $\ref{eq:10.4.16}$ are

$${\bf y}_2=\left[\begin{array}{r}-1\\[4pt]0\\[4pt]1\end{array} \right]e^{-5t}\quad \mbox{ and }\quad{\bf y}_3=\left[\begin{array}{r}-1\\[4pt]1\\[4pt] 0\end{array}\right]e^{-5t}.\nonumber$$

Because of this and Equation $\ref{eq:10.4.17}$, Theorem 10.4.1 implies that the general solution of Equation $\ref{eq:10.4.16}$ is

$${\bf y}=c_1\left[\begin{array}{r}1\\[4pt]1\\[4pt] 1\end{array}\right]e^t+c_2 \left[\begin{array}{r}-1\\[4pt]0\\[4pt]1\end{array}\right] e^{-5t}+c_3\left[\begin{array}{r}-1\\[4pt]1\\[4pt]0\end{array} \right]e^{-5t}.\nonumber$$

Geometric Properties of Solutions when n=2

We’ll now consider the geometric properties of solutions of a $2\times2$ constant coefficient system

$$\label{eq:10.4.18} \twocol{y_1'}{y_2'}=\left[\begin{array}{cc}a_{11}&a_{12}\\[4pt]a_{21}&a_{22} \end{array}\right]\twocol{y_1}{y_2}.$$

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 ${\bf y}=\left[\begin{array}{c}{y_{1}}\\[4pt]{y_{2}}\end{array} \right]$ is a non-constant solution of Equation $\ref{eq:10.4.18}$, 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 trajectory of ${\bf y}$. (We also say that $C$ is a trajectory of the system Equation $\ref{eq:10.4.18}$.) I’s important to note that $C$ is the trajectory of infinitely many solutions of Equation $\ref{eq:10.4.18}$, since if $\tau$ is any real number, then ${\bf y}(t-\tau)$ is a solution of Equation $\ref{eq:10.4.18}$, and $(y_1(t-\tau),y_2(t-\tau))$ also moves along $C$ as $t$ varies from $-\infty$ to $\infty$. Moreover, Exercise 10.4.28c implies that distinct trajectories of Equation $\ref{eq:10.4.18}$ can’t intersect, and that two solutions ${\bf y}_1$ and ${\bf y}_2$ of Equation $\ref{eq:10.4.18}$ have the same trajectory if and only if ${\bf y}_2(t)={\bf y}_1(t-\tau)$ for some $\tau$.

From Exercise 10.4.28a, a trajectory of a nontrivial solution of Equation $\ref{eq:10.4.18}$ can’t contain $(0,0)$, which we define to be the trajectory of the trivial solution ${\bf y}\equiv0$. More generally, if ${\bf y}=\left[\begin{array}{c}{k_{1}}\\[4pt]{k_{2}}\end{array} \right] \ne{\bf 0}$ is a constant solution of Equation $\ref{eq:10.4.18}$ (which could occur if zero is an eigenvalue of the matrix of Equation $\ref{eq:10.4.18}$), 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

$$A=\left[\begin{array}{cc}a_{11}&a_{12}\\[4pt]a_{21}&a_{22} \end{array}\right]\nonumber$$

of Equation $\ref{eq:10.4.18}$ 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 Equation $\ref{eq:10.4.18}$ is

$$\label{eq:10.4.19} {\bf y}= c_1{\bf x}_1 e^{\lambda_1 t}+c_2{\bf x}_2e^{\lambda_2 t}.$$

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

We leave it to you to classify the trajectories of Equation $\ref{eq:10.4.18}$ 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 Equation $\ref{eq:10.4.19}$ becomes

$${\bf y}=(c_1{\bf x}_1+c_2{\bf x}_2)e^{\lambda_1 t}.\nonumber$$

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 Equation $\ref{eq:10.4.18}$ 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 Equation $\ref{eq:10.4.18}$ are half-lines through (but not including) the origin. The direction of motion is away from the origin if $\lambda_1>0$ (Figure 10.4.1 ), toward it if $\lambda_1<0$ (Figure 10.4.2 ). (In these and the next figures an arrow through a point indicates the direction of motion along the trajectory through the point.)

Figure 10.4.1 (left): Trajectories of a $2\times 2$ system with a repeated positive eigenvalue and Figure 10.4.2 (right): Trajectories of a $2\times 2$ system with a repeated negative eigenvalue.

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 Equation $\ref{eq:10.4.19}$ 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 Equation $\ref{eq:10.4.19}$ are both nonzero. In this case, the trajectory of Equation $\ref{eq:10.4.19}$ 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 Equation $\ref{eq:10.4.19}$ 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

$${\bf y}(0)=c_1{\bf x}_1+c_2{\bf x}_2\nonumber$$

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 10.4.3 .

Figure 10.4.3 : Four open sectors bounded by $L_{1}$ and $L_{2}$.

The direction of ${\bf y}(t)$ in Equation $\ref{eq:10.4.19}$ is the same as that of

$$\label{eq:10.4.20} e^{-\lambda_2 t}{\bf y}(t)= c_1{\bf x}_1e^{-(\lambda_2-\lambda_1)t}+c_2{\bf x}_2$$

and of

$$\label{eq:10.4.21} e^{-\lambda_1 t}{\bf y}(t)=c_1{\bf x}_1+c_2{\bf x}_2e^{(\lambda_2-\lambda_1)t}.$$

Since the right side of Equation $\ref{eq:10.4.20}$ 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 Equation $\ref{eq:10.4.21}$ 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 Equation $\ref{eq:10.4.19}$ 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 ${\bf u}$.

Case 1: λ₂ > λ₁

Figure 10.4.4 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

$$\lim_{t\to\infty}\left\|{\bf y}(t)-c_2{\bf x}_2e^{\lambda_2 t}\right\|=\lim_{t\to\infty}\|c_1{\bf x_1}e^{\lambda_1t}\|=\infty,\nonumber$$

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

Figure 10.4.4 : Two positive eigenvalues; motion away from origin.

Case 2: 0 > λ₂ > λ₁

Figure 10.4.5 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

$$\lim_{t\to-\infty}\left\|{\bf y}(t)-c_1{\bf x}_1e^{\lambda_1 t}\right\|=\lim_{t\to-\infty}\|c_2{\bf x}_2e^{\lambda_2t}\|=\infty,\nonumber$$

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

Figure 10.4.5 : Two negative eigenvalues; motion toward the origin.

Case 3: λ₂ > 0 > λ₁

Figure 10.4.6 shows some typical trajectories. In this case,

$$\lim_{t\to\infty}\|{\bf y}(t)\|=\infty \quad \text{and} \quad \lim_{t\to\infty}\left\|{\bf y}(t)-c_2{\bf x}_2e^{\lambda_2 t}\right\|=\lim_{t\to\infty}\|c_1{\bf x}_1e^{\lambda_1t}\|=0,\nonumber$$

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

$$\lim_{t\to-\infty}\|{\bf y}(t)\|=\infty \quad \text{and} \quad \lim_{t\to-\infty}\left\|{\bf y}(t)-c_1{\bf x}_1e^{\lambda_1 t}\right\|=\lim_{t\to-\infty}\|c_2{\bf x}_2e^{\lambda_2t}\|=0,\nonumber$$

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 $L_2$.

Figure 10.4.6 : Eigenvalues of different signs.