Show that if \(\mathbf{x}=\mathbf{v}e^{\lambda t}\) is a solution to
\begin{align*}
\mathbf{x}'=A\mathbf{x}
\text{ where }
A=\left[
\begin{array}{rr}
1 & -1 \\
2 & 4
\end{array}
\right]
\end{align*}
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Handout Lessons 25-27, Linear Systems of Differential Equations with Constant Coefficients: The Eigenvalue Method
Textbook Section(s).
This lesson is based on Sections 4.2 and 5.5 of your textbook by Edwards, Penney, and Calvis.
An Educated Guess Solution for a Homogeneous System of Differential Equations with Constant Coefficients.
\begin{align*}
\begin{cases}
x_1'& =x_1-x_2 \\
x_2'& =2x_1+4x_2 \\
\end{cases}
\end{align*}
Which can be written in matrix notation as
\begin{align}
\mathbf{x}'=
\left[
\begin{array}{rr}
1 & -1 \\
2 & 4
\end{array}
\right]
\mathbf{x}
\qquad
\text{ or }
\qquad
\mathbf{x}'=A\mathbf{x}
\text{ where }
A=\left[
\begin{array}{rr}
1 & -1 \\
2 & 4
\end{array}
\right]\tag{✶}
\end{align}
Our understanding of functions and their derivatives might lead us to make an educated guess that \(x_i=v_ie^{\lambda t}\) for some scalars \(v_1\text{,}\) \(v_2\text{,}\) and \(\lambda\text{.}\) Why?
\begin{gather*}
\left[
\begin{array}{r}
x_1 \\
x_2
\end{array}
\right]
=
\left[
\begin{array}{r}
v_1 \\
v_2
\end{array}
\right]
e^{\lambda t}
\qquad
\mathbf{x}=\mathbf{v}e^{\lambda t}
\end{gather*}
Example 161. Discovering \((A-\lambda I)\mathbf{v}=\mathbf{0}\).
We are looking for a set of 2 linearly independent solutions (\(\mathbf{x_1}\) and \(\mathbf{x_2}\)) to
\begin{align*}
\mathbf{x}'=A\mathbf{x}
\text{ where }
A=\left[
\begin{array}{rr}
1 & -1 \\
2 & 4
\end{array}
\right]
\end{align*}
We are hoping to find solutions of the form \(\mathbf{x}=\mathbf{v}e^{\lambda t}\text{.}\) We now know that these solutions have to satisfy
\begin{gather}
(A-\lambda I)\mathbf{v}=\mathbf{0}\tag{#}
\end{gather}
If we are going to keep linear independence, then we need to make sure that \(\mathbf{v} \neq \mathbf{0}\text{.}\) In other words, we seek nontrivial solutions to (#).
Theorem 162.
The system of equations
\begin{gather*}
(A-\lambda I)\mathbf{v}=\mathbf{0}
\end{gather*}
has a nontrivial solution if and only if
\begin{gather*}
\det(A-\lambda I) = 0
\end{gather*}
Example 161 continued: Continuing our efforts to find solutions of the form \(\mathbf{x}=\mathbf{v}e^{\lambda t}\) to
\begin{align*}
\mathbf{x}'=A\mathbf{x}
\text{ where }
A=\left[
\begin{array}{rr}
1 & -1 \\
2 & 4
\end{array}
\right]
\end{align*}
we now know that we seek \(\lambda\) values that satisfy
\begin{gather*}
\det(A-\lambda I)=0.
\end{gather*}
Now that we have found that we require \(\lambda=3\) or \(\lambda=2\text{,}\) we now seek nonzero \(\mathbf{v}\) vectors for each \(\lambda\) value that satisfy
\begin{align*}
(A-\lambda I)\mathbf{v} = \mathbf{0}\\
\left[
\begin{array}{cc}
1-\lambda & -1 \\
2 & 4-\lambda
\end{array}
\right]\mathbf{v} = \mathbf{0}
\end{align*}
-
\(\lambda=3\)We have only equation for variables. This means that we have one degree of . In other words, we can freely choose a value for one of these variables. The other variable will be forced by the value we choose for the first variable.
-
\(\displaystyle \lambda=2\)\begin{align*} \left[ \begin{array}{cc} 1-\lambda & -1 \\ 2 & 4-\lambda \end{array} \right]\mathbf{v} = \mathbf{0} \end{align*}
So we now see that the general solution to
\begin{align*}
\mathbf{x}'=A\mathbf{x}
\text{ where }
A=\left[
\begin{array}{rr}
1 & -1 \\
2 & 4
\end{array}
\right]
\end{align*}
is given by
Eigenvalues, Eigenvectors, and Solutions of Homogeneous Systems of Differential Equations with Constant Coefficients.
Definition 163.
The number \(\lambda\) (real, complex, zero, or nonzero) is an eigenvalue of the matrix \(A\) if
\begin{gather*}
\det(A-\lambda I) = 0
\end{gather*}
If \(\lambda\) is an eigenvalue of \(A\text{,}\) then any nonzero vector \(\mathbf{v}\) that satisfies
\begin{gather*}
A\mathbf{v}=\lambda \mathbf{v}
\end{gather*}
or, equivalently,
\begin{gather*}
(A-\lambda I)\mathbf{v} = \mathbf{0}
\end{gather*}
is called an eigenvector of A associated with the eigenvalue \(\lambda\).
In our very first example, the solutions we found to
\begin{gather*}
\mathbf{x}'=A\mathbf{x}
\end{gather*}
were of the form
\begin{gather*}
\mathbf{v}e^{\lambda t}
\end{gather*}
where \(\lambda\) was and \(\mathbf{v}\) was .
Theorem 164.
If \(\lambda\) is an eigenvalue of the constant coefficient matrix \(A\) and \(\mathbf{v}\) is an eigenvector of \(A\) associated with \(\lambda\text{,}\) then
\begin{gather*}
\mathbf{x}(t)=\mathbf{v}e^{\lambda t}
\end{gather*}
is a solution of the linear homogeneous system of differential equations with constant coefficients
\begin{gather*}
\mathbf{x}'=A\mathbf{x}.
\end{gather*}
Algorithm 165. Eigenvalue Method for \(\mathbf{x}'=A\mathbf{x}\).
Assume that \(A\) is an \(n\times n\) matrix that contains constants.
-
Find \(n\) solutions, \(\lambda_1, \lambda_2, \dots, \lambda_n\) of\begin{gather*} \det(A-\lambda I) = 0. \end{gather*}(There will be \(n\) solutions. Some may not be unique. Some may be complex.)
-
If possible, find \(n\) linearly independent eigenvectors of \(A\text{,}\) \(\mathbf{v_1}, \mathbf{v_2}, \dots \mathbf{v_n}\) associated with \(\lambda_1, \lambda_2, \dots, \lambda_n\text{,}\) respectively.
-
If (2) is possible then\begin{gather*} \mathbf{x_1}=\mathbf{v_1}e^{\lambda_1 t}, \mathbf{x_2}=\mathbf{v_2}e^{\lambda_2 t}, \dots \mathbf{x_n}=\mathbf{v_n}e^{\lambda_n t}, \end{gather*}is a set of \(n\) linearly independent solutions of \(\mathbf{x}'=A\mathbf{x}\) and\begin{gather*} {\bf x}(t)= c_1\mathbf{v_1}e^{\lambda_1 t}+ c_2\mathbf{v_2}e^{\lambda_2 t}+ \dots + c_n\mathbf{v_n}e^{\lambda_n t} \end{gather*}is a general solution of \(\mathbf{x}'=A\mathbf{x}\text{.}\)
Theorem 166.
If \(A_{n\times n}\) has \(n\) distinct real eigenvalues \(\lambda_1, \lambda_2, \dots, \lambda_n\text{,}\) with associated eigenvectors \(\mathbf{v_1}, \mathbf{v_2}, \dots, \mathbf{v_n}\text{,}\) then
\begin{gather*}
\mathbf{x_1}=\mathbf{v_1}e^{\lambda_1 t},
\mathbf{x_2}=\mathbf{v_2}e^{\lambda_2 t},
\dots
\mathbf{x_n}=\mathbf{v_n}e^{\lambda_n t},
\end{gather*}
is a set of \(n\) linearly independent solutions of \(\mathbf{x}'=A\mathbf{x}\text{,}\) and
\begin{gather*}
{\bf x}(t)=
c_1\mathbf{v_1}e^{\lambda_1 t}+
c_2\mathbf{v_2}e^{\lambda_2 t}+
\dots +
c_n\mathbf{v_n}e^{\lambda_n t}
\end{gather*}
is a general solution of \(\mathbf{x}'=A\mathbf{x}\text{.}\)
Example 167.
Let
\begin{align*}
A=\left[
\begin{array}{rrr}
1 & 2 & 2 \\
2 & 0 & 3 \\
2 & 3 & 0
\end{array}
\right]
\end{align*}
-
The characteristic equation of A, \(\det(A-\lambda I)=0\) is \(\lambda^3-\lambda^2-17\lambda-15=0\text{.}\) Find the eigenvalues of \(A\text{.}\)
-
Find an eigenvector \(\mathbf{v_1}\) of \(A\) associated with the eigenvalue \(\lambda_1=-1\text{.}\)\begin{align*} A=\left[ \begin{array}{rrr} 1 & 2 & 2 \\ 2 & 0 & 3 \\ 2 & 3 & 0 \end{array} \right] \end{align*}
-
\(\mathbf{v_2}= \left[ \begin{array}{r} 0 \\ 1 \\ -1 \end{array} \right]\) is an eigenvector of \(A\) associated with the eigenvalue \(\lambda_2=-3\) and \(\mathbf{v_3}= \left[ \begin{array}{r} 1 \\ 1 \\ 1 \end{array} \right]\) is an eigenvector of \(A\) associated with the eigenvalue \(\lambda_3=5\text{.}\) Find a general solution for \(\mathbf{x'}=A\mathbf{x}\text{.}\)
You should look at the Compartmental Analysis subsection of Example 2 about multiple brine tanks in Section 5.2 of your textbook.
Complex Eigenvalues.
Definition 168.
Properties of the Complex Conjugate
-
\(\displaystyle \overline{\alpha+\beta}=\overline{\alpha}+\overline{\beta}\)
-
\(\displaystyle \overline{\alpha-\beta}=\overline{\alpha}-\overline{\beta}\)
-
\(\displaystyle \overline{\alpha\beta}=\overline{\alpha}\overline{\beta}\)
-
\(\overline{\left(\frac{\alpha}{\beta}\right)}= \frac{\overline{\alpha}}{\overline{\beta}}\text{,}\) provide \(\beta \neq 0\)
Theorem 169.
Let \(A \in \mathbb{R}^{n\times n}\text{.}\) If there are real vector functions \(\mathbf{p}(t)\) and \(\mathbf{q}(t)\) such that
\begin{gather*}
\mathbf{v}(t)=\mathbf{p}(t)\pm i\mathbf{q}(t)
\end{gather*}
are solutions to
\begin{gather}
\mathbf{x}'=A\mathbf{x}\tag{#}
\end{gather}
Theorem 170.
Let \(A \in \mathbb{R}^{n\times n}\text{.}\) If \(\alpha+i\beta\) is an eigenvalue of \(A\) with eigenvector \(\mathbf{a}+i\mathbf{b}\text{,}\) then \(\alpha-i\beta\) is an eigenvalue of \(A\) with eigenvector \(\mathbf{a}-i\mathbf{b}\text{.}\)
Example 171. Working with complex eigenvalues.
Find a general solution of \(\mathbf{x}'=A\mathbf{x}\) where
\begin{align*}
A=\left[
\begin{array}{rr}
-1 & 2 \\
-1 & -3
\end{array}
\right]
\end{align*}
NOTE: When \(A\) is a \(2 \times 2\) matrix and \(\lambda\) is an eigenvalue of \(A\text{,}\) then \(A-\lambda I\) will either be
\begin{equation*}
\fillinmath{XXXXXXXXXXXX}
\end{equation*}
or one row of \(A\) will be a scalar multiple of the other row so you can use in \((A-\lambda I)\mathbf{v}=\mathbf{0}\) to find an eigenvector of \(A\) associated with \(\lambda\text{.}\)
Theorem 172.
If \(A\in \mathbb{R}^{n \times n}\) has complex conjugate eigenvalues \(\alpha \pm i \beta\) with corresponding eigenvectors \(\mathbf{a} \pm i \mathbf{b}\text{,}\) then
\begin{gather*}
\mathbf{x_1}=e^{\alpha t}\cos(\beta t)\mathbf{a}-e^{\alpha
t}\sin(\beta t)\mathbf{b}
\end{gather*}
and
\begin{gather*}
\mathbf{x_2}=e^{\alpha t}\sin(\beta t)\mathbf{a}+e^{\alpha
t}\cos(\beta t)\mathbf{b}
\end{gather*}
are two linearly independent real vector solutions of
\begin{gather*}
\mathbf{x}'=A\mathbf{x}.
\end{gather*}
Example 173. Finding general solutions.
Let
\begin{align*}
A=\left[
\begin{array}{rrr}
3 & 0 & 1 \\
9 & -1 & 2 \\
-9 & 4 & -1
\end{array}
\right]
\end{align*}
It can be shown that the eigenvalues of \(A\) are \(3\text{,}\) \(-1+i\) and \(-1-i\text{.}\) Moreover, \(\left[
\begin{array}{r}
4 \\ 9 \\ 0
\end{array}
\right]\) is an eigenvector for \(A\) associated with \(3\) and \(\left[
\begin{array}{r}
1 \\ 2-i \\ -4+i
\end{array}
\right]\) is an eigenvector for \(A\) associated with \(-1+i\text{.}\) Find a general solution for \(\mathbf{x}'=A\mathbf{x}\text{.}\)
Eigenvalues with multiplicity.
So far, we have been able to find general solutions to \(\mathbf{x}'=A\mathbf{x}\) because the eigenvalues have been distinct and, consequently, we have been able to find a set of \(n\) linearly independent eigenvectors. But what happens with the eigenvalues are not distinct? Said another way, what happens when some of the eigenvalues have multiplicity \(>1\text{?}\) Sometimes this is not a problem. Consider the matrix
\begin{align*}
A=\left[
\begin{array}{rr}
5 & 0 \\
0 & 5
\end{array}
\right]
\end{align*}
It is not to hard to see that this matrix has only one eigenvalue, namely \(\lambda=5\) of multiplicity 2, but we can use this eigenvalue to find 2 linearly independent eigenvectors associated with \(\lambda = 5\text{.}\)
But what happens when an eigenvalue \(\lambda\) has multiplicity \(k\text{,}\) but there are fewer than \(k\) linearly independent eigenvectors for \(\lambda\text{,}\) as in the next example.
Example 174. Defective eigenvalues.
Find a general solution for \(\mathbf{x}'=A\mathbf{x}\) where
\begin{align*}
A=\left[
\begin{array}{rr}
-7 & 18 \\
-2 & 5
\end{array}
\right]
\end{align*}
-
At home, you should check that the characteristic equation, \(\det(A-\lambda I) = 0\text{,}\) is\begin{gather*} (\lambda+1)^2=0. \end{gather*}Verify that \((A-(-1)I)\mathbf{v}=(A+I)\mathbf{v}=\mathbf{0}\) has one degree of freedom, and find one solution, \(\mathbf{x_1}\text{,}\) of \(\mathbf{x}'=A\mathbf{x}\text{.}\)
-
We might suspect that we could find a second solution of \(\mathbf{x}'=A\mathbf{x}\) that has the form \(\displaystyle \mathbf{x_2}= \left[\begin{array}{c} u_1 \\ u_2 \end{array}\right]te^{-t}\text{.}\) Show that if \(\mathbf{x_2}\) has this form, then \(\mathbf{x_2} = \displaystyle \left[\begin{array}{c} 0 \\ 0 \end{array}\right]\text{.}\)
Example 175. Continuing Example 174.
-
Show that if \(\mathbf{x_2}\) is a solution of \(\mathbf{x}'=A\mathbf{x}\) that has the form\begin{gather*} \displaystyle \mathbf{x_2}= \left( \left[\begin{array}{c} u_1 \\ u_2 \end{array}\right]t+ \left[\begin{array}{c} w_1 \\ w_2 \end{array}\right]\right)e^{-t}, \end{gather*}then
-
\(\displaystyle (A-(-1)I)\mathbf{u} = (A+I)\mathbf{u}= \mathbf{0}\)
-
\(\displaystyle (A-(-1)I)\mathbf{w} = (A+I)\mathbf{w}= \mathbf{u}\)
-
-
We now seek solutions \(\mathbf{u}\) and \(\mathbf{w}\) to:\begin{gather} (A+I)\mathbf{u} = \mathbf{0}\tag{✶✶}\\ (A+I)\mathbf{w} = \mathbf{u}\tag{✶✶✶} \end{gather}where\begin{align*} A=\left[ \begin{array}{rr} -7 & 18 \\ -2 & 5 \end{array} \right] \qquad \text{and} \qquad A+I=\left[ \begin{array}{rr} -6 & 18 \\ -2 & 6 \end{array} \right] \end{align*}Comments:
-
Notice that (✶✶) says that \(\mathbf{u}\) is an . Earlier, we found \(\displaystyle \mathbf{x_1}=\left[ \begin{array}{c} 3 \\1 \end{array} \right]e^{-t}\) to be one solution of \(\mathbf{x}'=A\mathbf{x}\) because \(\displaystyle \left[ \begin{array}{c} 3 \\1 \end{array} \right]\) was an eigenvector of \(A\) associated with the eigenvalue \(\lambda=-1\text{.}\) Note that \(\displaystyle \left[ \begin{array}{c} 3 \\1 \end{array} \right]\) could be replaced with any other eigenvector of \(A\) associated with \(\lambda=-1\text{.}\)
-
Example 176. Continuing Example 174.
-
So we are trying to find \(\mathbf{u}\) and \(\mathbf{w}\) that satisfy\begin{gather*} (A+I)\mathbf{u} = \mathbf{0} \end{gather*}and\begin{gather*} (A+I)\mathbf{w} = \mathbf{u} \neq \mathbf{0} \end{gather*}\begin{align*} A=\left[ \begin{array}{rr} -7 & 18 \\ -2 & 5 \end{array} \right] \qquad \text{and} \qquad A+I=\left[ \begin{array}{rr} -6 & 18 \\ -2 & 6 \end{array} \right] \end{align*}
-
Find a general solution for\begin{align*} \mathbf{x}'= \left[ \begin{array}{cc} -7 & 18 \\ -2 & 5 \end{array} \right] \mathbf{x} \end{align*}
Definition 177.
Let \(\lambda\) be an eigenvalue of \(A\) of multiplicity \(k>1\text{.}\)
-
If \(\lambda\) has \(k\) linearly independent eigenvectors, then \(\lambda\) is a complete eigenvalue.
-
If \(\lambda\) has only \(p<k\) linearly independent eigenvectors, then \(\lambda\) is a defective eigenvalue and its deficit is \(k-p\text{.}\)
Procedure for Defective Eigenvalues of Multiplicity 2
-
Find an eigenvector \(\mathbf{v_1}\text{:}\)\begin{equation*} (A-\lambda I)\mathbf{v_1} = \mathbf{0}\text{.} \end{equation*}Solve\begin{equation*} (A-\lambda I)\mathbf{v_2} = \mathbf{v_1}\text{.} \end{equation*}
-
Then\begin{gather*} \mathbf{x_1}(t)=\mathbf{v_1}e^{\lambda t} \end{gather*}and\begin{gather*} \mathbf{x_2}(t)=(\mathbf{v_1}t+\mathbf{v_2})e^{\lambda t} \end{gather*}are linearly independent solutions of \(\mathbf{x}'=A\mathbf{x}\text{.}\)
Example 178. Defective eigenvalue in a \(3\times 3\) matrix.
Let
\begin{align*}
A=\left[
\begin{array}{rrr}
-19 & 12 & 84 \\
0 & 5 & 0 \\
-8 & 4 & 33
\end{array}
\right]
\end{align*}
Then the eigenvalues of \(A\) are
-
9 (multiplicity 1) with associated eigenvector \(\mathbf{v}=\left[ \begin{array}{r} 3 \\ 0\\ 1 \end{array} \right]\)
-
and 5 (multiplicity 2).
Find a general solution for \(\mathbf{x}'=A\mathbf{x}\text{.}\)
