10.5: Constant Coefficient Homogeneous Systems II

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$trench-1995.jpg$
Contributed by William F. Trench
Andrew G. Cowles Distinguished Professor Emeritus (Mathamatics) at Trinity University

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We saw in Section 10.4 that if an $n\times n$ constant matrix $A$ has $n$ real eigenvalues $\lambda_1$, $\lambda_2$, …, $\lambda_n$ (which need not be distinct) with associated linearly independent eigenvectors ${\bf x}_1$, ${\bf x}_2$, …, ${\bf x}_n$, then the general solution of ${\bf y}'=A{\bf y}$ is

\[{\bf y}=c_1{\bf x}_1e^{\lambda_1t}+c_2{\bf x}_2e^{\lambda_2 t} +\cdots+c_n{\bf x}_ne^{\lambda_n t}. \nonumber \]

In this section we consider the case where $A$ has $n$ real eigenvalues, but does not have $n$ linearly independent eigenvectors. It is shown in linear algebra that this occurs if and only if $A$ has at least one eigenvalue of multiplicity $r>1$ such that the associated eigenspace has dimension less than $r$. In this case $A$ is said to be defective. Since it is beyond the scope of this book to give a complete analysis of systems with defective coefficient matrices, we will restrict our attention to some commonly occurring special cases.

Example $\PageIndex{1}$

Show that the system

\[\label{eq:10.5.1} {\bf y}'=\left[\begin{array}{cc}{11}&{-25}\\{4}&{-9}\end{array} \right] {\bf y}\]

does not have a fundamental set of solutions of the form $\{{\bf x}_1e^{\lambda_1t},{\bf x}_2e^{\lambda_2t}\}$, where $\lambda_1$ and $\lambda_2$ are eigenvalues of the coefficient matrix $A$ of Equation \ref{eq:10.5.1} and ${\bf x}_1$, and ${\bf x}_2$ are associated linearly independent eigenvectors.

Solution

The characteristic polynomial of $A$ is

\[\begin{align*} \twochar{11}{-25}4{-9} &=(\lambda-11)(\lambda+9)+100\\ &=\lambda^2-2\lambda+1 \\ &=(\lambda-1)^2.\end{align*}\nonumber\]

Hence, $\lambda=1$ is the only eigenvalue of $A$. The augmented matrix of the system $(A-I){\bf x}={\bf 0}$ is

\[\left[\begin{array}{rrcr}10&-25&\vdots&0\\4& -10&\vdots&0\end{array}\right],\nonumber\]

which is row equivalent to

\[\left[\begin{array}{cccc}{1}&{-\frac{5}{2}}&{\vdots }&{0}\\{0}&{0}&{\vdots }&{0} \end{array} \right] \nonumber\]

Hence, $x_1=5x_2/2$ where $x_2$ is arbitrary. Therefore all eigenvectors of $A$ are scalar multiples of $\bf {x}_1=\left[\begin{array}{c}{5}\\{2}\end{array} \right] $, so $A$ does not have a set of two linearly independent eigenvectors.

From Example $\PageIndex{1}$, we know that all scalar multiples of $\bf {y}_1= \twocol52 e^t$ are solutions of Equation \ref{eq:10.5.1}; however, to find the general solution we must find a second solution $\bf {y}_2$ such that $\{ \bf {y}_1, \bf {y}_2\}$ is linearly independent. Based on your recollection of the procedure for solving a constant coefficient scalar equation

\[ay''+by'+cy=0\nonumber\]

in the case where the characteristic polynomial has a repeated root, you might expect to obtain a second solution of Equation \ref{eq:10.5.1} by multiplying the first solution by $t$. However, this yields ${\bf y}_2=\twocol52 te^t$, which does not work, since

\[{\bf y}_2'=\twocol52(te^t+e^t),\quad \text{while} \quad \left[\begin{array}{cc}{11}&{-25}\\{4}&{-9} \end{array} \right] {\bf y}_2=\twocol52te^t.\nonumber\]

The next theorem shows what to do in this situation.

Theorem $\PageIndex{1}$

Suppose the $n\times n$ matrix $A$ has an eigenvalue $\lambda_1$ of multiplicity $\ge2$ and the associated eigenspace has dimension $1;$ that is$,$ all $\lambda_1$-eigenvectors of $A$ are scalar multiples of an eigenvector ${\bf x}.$ Then there are infinitely many vectors ${\bf u}$ such that

\[\label{eq:10.5.2} (A-\lambda_1I){\bf u}={\bf x}.\]

Moreover$,$ if ${\bf u}$ is any such vector then

\[\label{eq:10.5.3} {\bf y}_1={\bf x}e^{\lambda_1t}\quad\mbox{and }\quad {\bf y}_2={\bf u}e^{\lambda_1t}+{\bf x}te^{\lambda_1t}\]

are linearly independent solutions of ${\bf y}'=A{\bf y}.$

A complete proof of this theorem is beyond the scope of this book. The difficulty is in proving that there’s a vector ${\bf u}$ satisfying Equation \ref{eq:10.5.2}, since $\det(A-\lambda_1I)=0$. We’ll take this without proof and verify the other assertions of the theorem. We already know that ${\bf y}_1$ in Equation \ref{eq:10.5.3} is a solution of ${\bf y}'=A{\bf y}$. To see that ${\bf y}_2$ is also a solution, we compute

\[\begin{align*} {\bf y}_2'-A{\bf y}_2&=\lambda_1{\bf u}e^{\lambda_1t}+{\bf x} e^{\lambda_1t} +\lambda_1{\bf x} te^{\lambda_1t}-A{\bf u}e^{\lambda_1t}-A{\bf x} te^{\lambda_1t}\\ &=(\lambda_1{\bf u}+{\bf x} -A{\bf u})e^{\lambda_1t}+(\lambda_1{\bf x} -A{\bf x} )te^{\lambda_1t}.\end{align*}\nonumber\]

Since $A{\bf x}=\lambda_1{\bf x}$, this can be written as

\[{\bf y}_2'-A{\bf y}_2=- \left((A-\lambda_1I){\bf u}-{\bf x}\right)e^{\lambda_1t},\nonumber\]

and now Equation \ref{eq:10.5.2} implies that ${\bf y}_2'=A{\bf y}_2$. To see that ${\bf y}_1$ and ${\bf y}_2$ are linearly independent, suppose $c_1$ and $c_2$ are constants such that

\[\label{eq:10.5.4} c_1{\bf y}_1+c_2{\bf y}_2=c_1{\bf x}e^{\lambda_1t}+c_2({\bf u}e^{\lambda_1t} +{\bf x}te^{\lambda_1t})={\bf 0}.\]

We must show that $c_1=c_2=0$. Multiplying Equation \ref{eq:10.5.4} by $e^{-\lambda_1t}$ shows that

\[\label{eq:10.5.5} c_1{\bf x}+c_2({\bf u} +{\bf x}t)={\bf 0}.\]

By differentiating this with respect to $t$, we see that $c_2{\bf x}={\bf 0}$, which implies $c_2=0$, because ${\bf x}\ne{\bf 0}$. Substituting $c_2=0$ into Equation \ref{eq:10.5.5} yields $c_1{\bf x}={\bf 0}$, which implies that $c_1=0$, again because ${\bf x}\ne{\bf 0}$

Example $\PageIndex{2}$

Use Theorem $\PageIndex{1}$ to find the general solution of the system

\[\label{eq:10.5.6} {\bf y}'=\left[\begin{array}{cc}{11}&{-25}\\{4}&{-9}\end{array} \right]{\bf y}\]

considered in Example $\PageIndex{1}$.

Solution

In Example $\PageIndex{1}$ we saw that $\lambda_1=1$ is an eigenvalue of multiplicity $2$ of the coefficient matrix $A$ in Equation \ref{eq:10.5.6}, and that all of the eigenvectors of $A$ are multiples of

\[{\bf x}=\twocol52.\nonumber\]

Therefore

\[{\bf y}_1=\twocol52e^t\nonumber\]

is a solution of Equation \ref{eq:10.5.6}. From Theorem $\PageIndex{1}$, a second solution is given by ${\bf y}_2={\bf u}e^t+{\bf x}te^t$, where $(A-I){\bf u}={\bf x}$. The augmented matrix of this system is

\[\left[\begin{array}{rrcr}10&-25&\vdots&5\\4&-10&\vdots&2\end{array}\right],\nonumber\]

which is row equivalent to

\[\left[\begin{array}{rrcr}1&-\frac{5}{2}&\vdots&\frac{1}{2}\\0&0&\vdots&0\end{array}\right],\nonumber\]

Therefore the components of ${\bf u}$ must satisfy

\[u_1-{5\over2}u_2={1\over2},\nonumber\]

where $u_2$ is arbitrary. We choose $u_2=0$, so that $u_1=1/2$ and

\[{\bf u}=\twocol{1\over2}0.\nonumber\]

Thus,

\[{\bf y}_2=\twocol10{e^t\over2}+\twocol52te^t.\nonumber\]

Since ${\bf y}_1$ and ${\bf y}_2$ are linearly independent by Theorem $\PageIndex{1}$, they form a fundamental set of solutions of Equation \ref{eq:10.5.6}. Therefore the general solution of Equation \ref{eq:10.5.6} is

\[{\bf y}=c_1\twocol52e^t+c_2\left(\twocol10{e^t\over2}+\twocol52te^t\right).\nonumber\]

Note that choosing the arbitrary constant $u_2$ to be nonzero is equivalent to adding a scalar multiple of ${\bf y}_1$ to the second solution ${\bf y}_2$ (Exercise 10.5.33).

Example $\PageIndex{3}$

Find the general solution of

\[\label{eq:10.5.7} {\bf y}'=\left[\begin{array}{ccc}{3}&{4}&{-10}\\{2}&{1}&{-2}\\{2}&{2}&{-5}\end{array} \right] {\bf y}.\]

Solution

The characteristic polynomial of the coefficient matrix $A$ in Equation \ref{eq:10.5.7} is

\[\left|\begin{array}{ccc} 3-\lambda & 4 & -10\\ 2 & 1-\lambda & -2\\ 2 & 2 &-5-\lambda\end{array}\right| =- (\lambda-1)(\lambda+1)^2.\nonumber\]

Hence, the eigenvalues are $\lambda_1=1$ with multiplicity $1$ and $\lambda_2=-1$ with multiplicity $2$. Eigenvectors associated with $\lambda_1=1$ must satisfy $(A-I){\bf x}={\bf 0}$. The augmented matrix of this system is

\[\left[\begin{array}{rrrcr} 2 & 4 & -10 &\vdots & 0\\ 2& 0 & -2 &\vdots & 0\\ 2 & 2 & -6 & \vdots & 0\end{array}\right],\nonumber\]

which is row equivalent to

\[\left[\begin{array}{rrrcr} 1 & 0 & -1 &\vdots& 0\\ 0 & 1 & -2 &\vdots& 0\\ 0 & 0 & 0 &\vdots&0\end{array}\right].\nonumber\]

Hence, $x_1 =x_3$ and $x_2 =2 x_3$, where $x_3$ is arbitrary. Choosing $x_3=1$ yields the eigenvector

\[{\bf x}_1=\threecol121.\nonumber\]

Therefore

\[{\bf y}_1 =\threecol121e^t\nonumber\]

is a solution of Equation \ref{eq:10.5.7}. Eigenvectors associated with $\lambda_2 =-1$ satisfy $(A+I){\bf x}={\bf 0}$. The augmented matrix of this system is

\[\left[\begin{array}{rrrcr} 4 & 4 & -10 &\vdots & 0\\ 2 & 2 & -2 & \vdots & 0\\2 & 2 & -4 &\vdots & 0\end{array}\right],\nonumber\]

which is row equivalent to

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

Hence, $x_3=0$ and $x_1 =-x_2$, where $x_2$ is arbitrary. Choosing $x_2=1$ yields the eigenvector

\[{\bf x}_2=\threecol{-1}10,\nonumber\]

\[{\bf y}_2 =\threecol{-1}10e^{-t}\nonumber\]

is a solution of Equation \ref{eq:10.5.7}. Since all the eigenvectors of $A$ associated with $\lambda_2=-1$ are multiples of ${\bf x}_2$, we must now use Theorem $\PageIndex{1}$ to find a third solution of Equation \ref{eq:10.5.7} in the form

\[\label{eq:10.5.8} {\bf y}_3={\bf u}e^{-t}+\threecol{-1}10te^{-t},\]

where ${\bf u}$ is a solution of $(A+I){\bf u=x}_2$. The augmented matrix of this system is

\[\left[\begin{array}{rrrcr} 4 & 4 & -10 &\vdots & -1\\ 2 & 2 & -2 & \vdots & 1\\ 2 & 2 & -4 &\vdots & 0\end{array}\right],\nonumber\]

which is row equivalent to

\[\left[\begin{array}{rrrcr} 1 & 1 & 0 &\vdots& 1\\ 0 & 0 & 1 &\vdots& {1\over2} \\ 0 & 0 & 0 &\vdots&0\end{array}\right].\nonumber\]

Hence, $u_3=1/2$ and $u_1 =1-u_2$, where $u_2$ is arbitrary. Choosing $u_2=0$ yields

\[{\bf u} =\threecol10{1\over2},\nonumber\]

and substituting this into Equation \ref{eq:10.5.8} yields the solution

\[{\bf y}_3=\threecol201{e^{-t}\over2}+\threecol{-1}10te^{-t}\nonumber\]

of Equation \ref{eq:10.5.7}. Since the Wronskian of $\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$ at $t=0$ is

\[\left|\begin{array}{rrr} 1&-1&1\\2&1&0\\1&0&1\over2\end{array}\right|={1\over2},\nonumber\]

$\{{\bf y}_1,{\bf y}_2,{\bf y}_3\}$ is a fundamental set of solutions of Equation \ref{eq:10.5.7}. Therefore the general solution of Equation \ref{eq:10.5.7} is

\[{\bf y}=c_1\threecol121e^t+c_2\threecol{-1}10e^{-t}+c_3\left (\threecol201{e^{-t}\over2}+\threecol{-1}10te^{-t}\right).\nonumber\]

Theorem $\PageIndex{2}$

Suppose the $n\times n$ matrix $A$ has an eigenvalue $\lambda_1$ of multiplicity $\ge 3$ and the associated eigenspace is one–dimensional$;$ that is$,$ all eigenvectors associated with $\lambda_1$ are scalar multiples of the eigenvector ${\bf x}.$ Then there are infinitely many vectors ${\bf u}$ such that

\[\label{eq:10.5.9} (A-\lambda_1I){\bf u}={\bf x},\]

and, if ${\bf u}$ is any such vector$,$ there are infinitely many vectors ${\bf v}$ such that

\[\label{eq:10.5.10} (A-\lambda_1I){\bf v}={\bf u}.\]

If ${\bf u}$ satisfies Equation \ref{eq:10.5.9} and ${\bf v}$ satisfies Equation \ref{eq:10.5.10}, then

\[\begin{aligned} {\bf y}_1 &={\bf x} e^{\lambda_1t},\\ {\bf y}_2&={\bf u}e^{\lambda_1t}+{\bf x} te^{\lambda_1t},\mbox{ and }\\ {\bf y}_3&={\bf v}e^{\lambda_1t}+{\bf u}te^{\lambda_1t}+{\bf x} {t^2e^{\lambda_1t}\over2}\end{aligned}\nonumber\]

are linearly independent solutions of ${\bf y}'=A{\bf y}$.

Again, it is beyond the scope of this book to prove that there are vectors ${\bf u}$ and ${\bf v}$ that satisfy Equation \ref{eq:10.5.9} and Equation \ref{eq:10.5.10}. Theorem $\PageIndex{1}$ implies that ${\bf y}_1$ and ${\bf y}_2$ are solutions of ${\bf y}'=A{\bf y}$. We leave the rest of the proof to you (Exercise 10.5.34).

Example $\PageIndex{4}$

Use Theorem $\PageIndex{2}$ to find the general solution of

\[\label{eq:10.5.11} {\bf y}'=\left[\begin{array}{ccc}1&1&1 \\ 1&3&-1 \\ 0&2&2 \end{array} \right] {\bf y}.\]

Solution:

The characteristic polynomial of the coefficient matrix $A$ in Equation \ref{eq:10.5.11} is

\[\left|\begin{array}{ccc} 1-\lambda & 1 & \phantom{-}1\\ 1 & 3-\lambda & -1\\ 0 & 2 & 2-\lambda\end{array}\right| =-(\lambda-2)^3.\nonumber\]

Hence, $\lambda_1=2$ is an eigenvalue of multiplicity $3$. The associated eigenvectors satisfy $(A-2I){\bf x=0}$. The augmented matrix of this system is

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

which is row equivalent to

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

Hence, $x_1 =x_3$ and $x_2 = 0$, so the eigenvectors are all scalar multiples of

\[{\bf x}_1=\threecol101.\nonumber\]

Therefore

\[{\bf y}_1=\threecol101e^{2t}\nonumber\]

is a solution of Equation \ref{eq:10.5.11}. We now find a second solution of Equation \ref{eq:10.5.11} in the form

\[{\bf y}_2={\bf u}e^{2t}+\threecol101te^{2t},\nonumber\]

where ${\bf u}$ satisfies $(A-2I){\bf u=x}_1$. The augmented matrix of this system is

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

which is row equivalent to

\[\left[\begin{array}{rrrcr} 1 & 0 &- 1 &\vdots& -{1\over2}\\ 0 & 1 & 0 &\vdots& {1\over2}\\ 0 & 0 & 0 &\vdots&0\end{array}\right].\nonumber\]

Letting $u_3=0$ yields $u_1=-1/2$ and $u_2=1/2$; hence,

\[{\bf u}={1\over2}\threecol{-1}10\nonumber\]

and

\[{\bf y}_2=\threecol{-1}10{e^{2t}\over2}+\threecol101te^{2t}\nonumber\]

is a solution of Equation \ref{eq:10.5.11}. We now find a third solution of Equation \ref{eq:10.5.11} in the form

\[{\bf y}_3={\bf v}e^{2t}+\threecol{-1}10{te^{2t}\over2}+\threecol101{t^2e^{2t}\over2}\nonumber\]

where ${\bf v}$ satisfies $(A-2I){\bf v}={\bf u}$. The augmented matrix of this system is

\[\left[\begin{array}{rrrcr} -1 & 1 & 1 &\vdots &-{1\over2}\\ 1& 1 & -1 &\vdots & {1\over2}\\ 0 & 2 & 0 & \vdots & 0\end{array}\right],\nonumber\]

which is row equivalent to

\[\left[\begin{array}{rrrcr} 1 & 0 &- 1 &\vdots& {1\over2}\\ 0 & 1 & 0 &\vdots& 0\\ 0 & 0 & 0 &\vdots&0\end{array}\right].\nonumber\]

Letting $v_3=0$ yields $v_1=1/2$ and $v_2=0$; hence,

\[{\bf v}={1\over2}\threecol100.\nonumber\]

Therefore

\[{\bf y}_3=\threecol100{e^{2t}\over2}+ \threecol{-1}10{te^{2t}\over2}+\threecol101{t^2e^{2t}\over2}\nonumber\]

is a solution of Equation \ref{eq:10.5.11}. Since ${\bf y}_1$, ${\bf y}_2$, and ${\bf y}_3$ are linearly independent by Theorem $\PageIndex{2}$, they form a fundamental set of solutions of Equation \ref{eq:10.5.11}. Therefore the general solution of Equation \ref{eq:10.5.11} is

\[\begin{aligned} {\bf y} = c_{1}\left[\begin{array}{c}{1}\\{0}\\{1}\end{array} \right]e^{2t}+c_{2}\left(\left[ \begin{array}{c}{-1}\\{1}\\{0}\end{array} \right]\frac{e^{2t}}{2}+\left[\begin{array}{c}{1}\\{0}\\{1}\end{array} \right] te^{2t} \right) + c_{3}\left(\left[\begin{array}{c}{1}\\{0}\\{0}\end{array} \right]\frac{e^{2t}}{2}+\left[\begin{array}{c}{-1}\\{1}\\{0}\end{array} \right]\frac{te^{2t}}{2}+\left[\begin{array}{c}{1}\\{0}\\{1}\end{array} \right]\frac{t^{2}e^{2t}}{2} \right) \end{aligned}\nonumber \]

Theorem $\PageIndex{3}$

Suppose the $n\times n$ matrix $A$ has an eigenvalue $\lambda_1$ of multiplicity $\ge 3$ and the associated eigenspace is two–dimensional; that is, all eigenvectors of $A$ associated with $\lambda_1$ are linear combinations of two linearly independent eigenvectors ${\bf x}_1$ and ${\bf x}_2$$.$ Then there are constants $\alpha$ and $\beta$ $($not both zero$)$ such that if

\[\label{eq:10.5.12} {\bf x}_3=\alpha{\bf x}_1+\beta{\bf x}_2,\]

then there are infinitely many vectors ${\bf u}$ such that

\[\label{eq:10.5.13} (A-\lambda_1I){\bf u}={\bf x}_3.\]

If ${\bf u}$ satisfies Equation \ref{eq:10.5.13}, then

\[\label{eq:10.5.14}\begin{array}{ll}{y_{1}}&{=x_{1}e^{\lambda _{1}t}}\\{y_{2}}&{=x_{2}e^{\lambda _{1}t},and}\\{y_{3}}&{ue^{\lambda _{1}t}+x_{3}te^{\lambda _{1}t},} \end{array}\]

are linearly independent solutions of ${\bf y}'=A{\bf y}.$

We omit the proof of this theorem.

Example $\PageIndex{5}$

Use Theorem $\PageIndex{3}$ to find the general solution of

\[\label{eq:10.5.15} {\bf y}'=\left[\begin{array}{ccc}{0}&{0}&{1}\\{-1}&{1}&{1}\\{-1}&{0}&{2}\end{array} \right]{\bf y}.\]

Solution

The characteristic polynomial of the coefficient matrix $A$ in Equation \ref{eq:10.5.15} is

\[\left|\begin{array}{ccc} -\lambda & 0 & 1\\ -1 & 1-\lambda & 1\\ -1 & 0 & 2-\lambda\end{array}\right| =-(\lambda-1)^3.\nonumber\]

Hence, $\lambda_1=1$ is an eigenvalue of multiplicity $3$. The associated eigenvectors satisfy $(A-I){\bf x=0}$. The augmented matrix of this system is

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

which is row equivalent to

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

Hence, $x_1 =x_3$ and $x_2$ is arbitrary, so the eigenvectors are of the form

\[{\bf x}_1=\threecol{x_3}{x_2}{x_3}=x_3\threecol101+x_2\threecol010.\nonumber\]

Therefore the vectors

\[\label{eq:10.5.16} {\bf x}_1 =\threecol101\quad\mbox{and }\quad {\bf x}_2=\threecol010\]

form a basis for the eigenspace, and

\[{\bf y}_1 =\threecol101e^t \quad \text{and} \quad {\bf y}_2=\threecol010e^t\nonumber\]

are linearly independent solutions of Equation \ref{eq:10.5.15}. To find a third linearly independent solution of Equation \ref{eq:10.5.15}, we must find constants $\alpha$ and $\beta$ (not both zero) such that the system

\[\label{eq:10.5.17} (A-I){\bf u}=\alpha{\bf x}_1+\beta{\bf x}_2\]

has a solution ${\bf u}$. The augmented matrix of this system is

\[\left[\begin{array}{rrrcr} -1 & 0 & 1 &\vdots &\alpha\\ -1& 0 & 1 &\vdots &\beta\\ -1 & 0 & 1 & \vdots &\alpha\end{array}\right],\nonumber\]

which is row equivalent to

\[\label{eq:10.5.18} \left[\begin{array}{rrrcr} 1 & 0 &- 1 &\vdots& -\alpha\\ 0 & 0 & 0 &\vdots&\beta-\alpha\\ 0 & 0 & 0 &\vdots&0\end{array} \right].\]

Therefore Equation \ref{eq:10.5.17} has a solution if and only if $\beta=\alpha$, where $\alpha$ is arbitrary. If $\alpha=\beta=1$ then Equation \ref{eq:10.5.12} and Equation \ref{eq:10.5.16} yield

\[{\bf x}_3={\bf x}_1+{\bf x}_2= \threecol101+\threecol010=\threecol111,\nonumber\]

and the augmented matrix Equation \ref{eq:10.5.18} becomes

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

This implies that $u_1=-1+u_3$, while $u_2$ and $u_3$ are arbitrary. Choosing $u_2=u_3=0$ yields

\[{\bf u}=\threecol{-1}00.\nonumber\]

Therefore Equation \ref{eq:10.5.14} implies that

\[{\bf y}_3={\bf u}e^t+{\bf x}_3te^t=\threecol{-1}00e^t+\threecol111te^t\nonumber\]

is a solution of Equation \ref{eq:10.5.15}. Since ${\bf y}_1$, ${\bf y}_2$, and ${\bf y}_3$ are linearly independent by Theorem [thmtype:10.5.3}, they form a fundamental set of solutions for Equation \ref{eq:10.5.15}. Therefore the general solution of Equation \ref{eq:10.5.15} is

\[{\bf y}=c_1\threecol101e^t+c_2\threecol010e^t +c_3\left(\threecol{-1}00e^t+\threecol111te^t\right).\bbox\nonumber\]

Geometric Properties of Solutions when $n=2$

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

\[\label{eq:10.5.19} \twocol{y_1'}{y_2'}=\left[\begin{array}{cc}a_{11}&a_{12}\\a_{21}&a_{22} \end{array}\right]\twocol{y_1}{y_2}\]

under the assumptions of this section; that is, when the matrix

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

has a repeated eigenvalue $\lambda_1$ and the associated eigenspace is one-dimensional. In this case we know from Theorem $\PageIndex{1}$ that the general solution of Equation \ref{eq:10.5.19} is

\[\label{eq:10.5.20} {\bf y}=c_1{\bf x}e^{\lambda_1t}+c_2({\bf u}e^{\lambda_1t}+{\bf x}te^{\lambda_1t}),\]

where ${\bf x}$ is an eigenvector of $A$ and ${\bf u}$ is any one of the infinitely many solutions of

\[\label{eq:10.5.21} (A-\lambda_1I){\bf u}={\bf x}.\]

We assume that $\lambda_1\ne0$.

Figure $\PageIndex{1}$: Positive and negative half-planes.

Let $L$ denote the line through the origin parallel to ${\bf x}$. By a half-line of $L$ we mean either of the rays obtained by removing the origin from $L$. Equation \ref{eq:10.5.20} is a parametric equation of the half-line of $L$ in the direction of ${\bf x}$ if $c_1>0$, or of the half-line of $L$ in the direction of $-{\bf x}$ if $c_1<0$. The origin is the trajectory of the trivial solution ${\bf y}\equiv{\bf 0}$.

Henceforth, we assume that $c_2\ne0$. In this case, the trajectory of Equation \ref{eq:10.5.20} can’t intersect $L$, since every point of $L$ is on a trajectory obtained by setting $c_2=0$. Therefore the trajectory of Equation \ref{eq:10.5.20} must lie entirely in one of the open half-planes bounded by $L$, but does not contain any point on $L$. Since the initial point $(y_1(0),y_2(0))$ defined by ${\bf y}(0)=c_1{\bf x}_1+c_2{\bf u}$ is on the trajectory, we can determine which half-plane contains the trajectory from the sign of $c_2$, as shown in Figure . For convenience we’ll call the half-plane where $c_2>0$ the positive half-plane. Similarly, the-half plane where $c_2<0$ is the negative half-plane. You should convince yourself (Exercise 10.5.35) that even though there are infinitely many vectors ${\bf u}$ that satisfy Equation \ref{eq:10.5.21}, they all define the same positive and negative half-planes. In the figures simply regard ${\bf u}$ as an arrow pointing to the positive half-plane, since wen’t attempted to give ${\bf u}$ its proper length or direction in comparison with ${\bf x}$. For our purposes here, only the relative orientation of ${\bf x}$ and ${\bf u}$ is important; that is, whether the positive half-plane is to the right of an observer facing the direction of ${\bf x}$ (as in Figures $\PageIndex{2}$ and $\PageIndex{5}$), or to the left of the observer (as in Figures $\PageIndex{3}$ and $\PageIndex{4}$).

Multiplying Equation \ref{eq:10.5.20} by $e^{-\lambda_1t}$ yields

\[e^{-\lambda_1t}{\bf y}(t)=c_1{\bf x}+c_2{\bf u}+c_2t {\bf x}.\nonumber\]

Since the last term on the right is dominant when $|t|$ is large, this provides the following information on the direction of ${\bf y}(t)$:

Along trajectories in the positive half-plane ($c_2>0$), the direction of ${\bf y}(t)$ approaches the direction of ${\bf x}$ as $t\to\infty$ and the direction of $-{\bf x}$ as $t\to-\infty$.
Along trajectories in the negative half-plane ($c_2<0$), the direction of ${\bf y}(t)$ approaches the direction of $-{\bf x}$ as $t\to\infty$ and the direction of ${\bf x}$ as $t\to-\infty$.

Since \[\lim_{t\to\infty}\|{\bf y}(t)\|=\infty\quad \text{and} \quad \lim_{t\to-\infty}{\bf y}(t)={\bf 0}\quad \text{if} \quad \lambda_1>0,\nonumber\]

\[\lim_{t-\to\infty}\|{\bf y}(t)\|=\infty \quad \text{and} \quad \lim_{t\to\infty}{\bf y}(t)={\bf 0} \quad \text{if} \quad \lambda_1<0,\nonumber\] there are four possible patterns for the trajectories of Equation \ref{eq:10.5.19}, depending upon the signs of $c_2$ and $\lambda_1$. Figures $\PageIndex{2}$ - $\PageIndex{5}$ illustrate these patterns, and reveal the following principle:

If $\lambda_1$ and $c_2$ have the same sign then the direction of the traectory approaches the direction of $-{\bf x}$ as $\|{\bf y} \|\to0$ and the direction of ${\bf x}$ as $\|{\bf y}\|\to\infty$. If $\lambda_1$ and $c_2$ have opposite signs then the direction of the trajectory approaches the direction of ${\bf x}$ as $\|{\bf y} \|\to0$ and the direction of $-{\bf x}$ as $\|{\bf y}\|\to\infty$.