MATH 308 Lecture 32

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Lecture Notes

Quiz over Ch. 7.5 Wednesday

Section 7.5

Exercise 5

Find the general solution of the given system of equations:

$X^{\prime }=\left[\begin{array}{ccc}3& 2& 4\\ 2& 0& 2\\ 4& 2& 3\end{array}\right]X$

Eigenvalues: $\lambda \in \{8,-1,-1\}$

Eigenvectors:

1. ${\lambda }_1=8:⟨2,1,2⟩$
2. ${\lambda }_2=-1:⟨\alpha ,-2\alpha -2\beta ,\beta ⟩=\alpha ⟨1,-2,0⟩+\beta ⟨0,-2,1⟩$ (note we have two eigenvectors)

Corresponding Solutions:

1. $X_1={\mathrm{e}}^{8t}\left[\begin{array}{c}2\\ 1\\ 2\end{array}\right]$
2. $X_2={\mathrm{e}}^{-t}\left[\begin{array}{c}1\\ -2\\ 0\end{array}\right]$
3. $X_3={\mathrm{e}}^{-t}\left[\begin{array}{c}0\\ -2\\ 1\end{array}\right]$

Check Linear Independence (redundant vs. complete solutions):

$W\{X_1,X_2,X_3\}=\left|\begin{array}{ccc}2{\mathrm{e}}^{8t}& {\mathrm{e}}^{-t}& 0\\ {\mathrm{e}}^{8t}& -2{\mathrm{e}}^{-t}& -2{\mathrm{e}}^{-t}\\ 2{\mathrm{e}}^{8t}& 0& {\mathrm{e}}^{-t}\end{array}\right|=-9{\mathrm{e}}^{8t}\equiv ̸0$

Therefore, our solutions $X_1$, $X_2$, and $X_3$ are linearly independent; so the general solution is

$X(t)=c_1{\mathrm{e}}^{8t}\left[\begin{array}{c}2\\ 1\\ 2\end{array}\right]+c_2{\mathrm{e}}^{-t}\left[\begin{array}{c}1\\ -2\\ 0\end{array}\right]+c_3{\mathrm{e}}^{-t}\left[\begin{array}{c}0\\ -2\\ 1\end{array}\right]$

Section 7.6

Lecture Notes

Given the systems:

$\begin{array}{rl}{X}^{\prime }& =\left[\begin{array}{cc}3& -2\\ 4& -1\end{array}\right]X\\ X^{\prime }& =\left[\begin{array}{cc}-1& -4\\ 1& -1\end{array}\right]X\\ \end{array}$

Previously, we would find the eigenvalues and eigenvectors and the general solution is of the form:

$X(t)=\sum _{i=0}^{\left|\lambda \right|}{c}_i{\mathrm{e}}^{{\lambda }_i}{\overrightarrow{v}}_i$

where $\lambda$ represents each eigenvalue and $\overrightarrow{v}$ represents the corresponding eigenvector.

What about matrices with imaginary eigenvalues?

$\begin{array}{rl}\left|\begin{array}{cc}3-\lambda & -2\\ 4& -1-\lambda \end{array}\right|& =0\\ {\lambda }^2-2\lambda +5& =0\\ \lambda & =1\pm 2i\end{array}$

1. $\lambda =1+2i:⟨1,1-i⟩$
2. We don't need to consider the second eigenvalue since they are complex conjugates

Recall from Chapter 3 that a root $a+bi$ has real-valued solution $c_1{\mathrm{e}}^{at}\cos bt+c_2{\mathrm{e}}^{at}\sin bt$. Let's do the same here:

for $X={\mathrm{e}}^t{\mathrm{e}}^{2it}\left[\begin{array}{c}1\\ 1-i\end{array}\right]$, we get:

${\mathrm{e}}^t(\cos 2t+i\sin 2t)(\left[\begin{array}{c}1\\ 1\end{array}\right]+i\left[\begin{array}{c}0\\ -1\end{array}\right])$

Expanding this gives

$X(t)=\underset{P}{\underbrace{({\mathrm{e}}^t\cos 2t\left[\begin{array}{c}1\\ 1\end{array}\right]-{\mathrm{e}}^t\sin 2t\left[\begin{array}{c}0\\ -1\end{array}\right])}}+i\underset{Q}{\underbrace{({\mathrm{e}}^t\sin 2t\left[\begin{array}{c}1\\ 1\end{array}\right]+{\mathrm{e}}^t\cos 2t\left[\begin{array}{c}0\\ -1\end{array}\right])}}$

The real part of this solution $P$ and the imaginary part of this solution $Q$ are each solutions to the original system, and they are always linearly independent

Theorem

Given a system of differential equations

$X^{\prime }(t)=AX(t)$

If $r_1=\lambda +i\mu$ and $r_2=\lambda -i\mu$ is a pair of complex conjugate eigenvalues and $\xi =\overrightarrow{a}\pm i\overrightarrow{b}$ is the corresponding pair of eigenvectors, then the vectors

$\begin{array}{rl}\overrightarrow{u}& ={\mathrm{e}}^{\lambda t}(\overrightarrow{a}\cos \mu t-\overrightarrow{b}\sin \mu t)\\ \overrightarrow{v}& ={\mathrm{e}}^{\lambda t}(\overrightarrow{a}\sin \mu t+\overrightarrow{b}\cos \mu t)\end{array}$

are real valued solutions of the system

Example

$X^{\prime }=\left[\begin{array}{cc}-1& -4\\ 1& -1\end{array}\right]X$

Eigenvalue: $\lambda =-1+2i$

Eigenvector: $\left[\begin{array}{c}2i\\ 1\end{array}\right]=\left[\begin{array}{c}0\\ 1\end{array}\right]+i\left[\begin{array}{c}2\\ 0\end{array}\right]$

Apply the theorem:

$\begin{array}{rl}{X}_1& ={\mathrm{e}}^{-t}\cos 2t\left[\begin{array}{c}0\\ 1\end{array}\right]-{\mathrm{e}}^{-t}\sin 2t\left[\begin{array}{c}2\\ 0\end{array}\right]\\ X_2& ={\mathrm{e}}^{-t}\cos 2t\left[\begin{array}{c}2\\ 0\end{array}\right]+{\mathrm{e}}^{-t}\sin 2t\left[\begin{array}{c}0\\ 1\end{array}\right]\end{array}$