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:

$X'={\begin{bmatrix}3&2&4\\2&0&2\\4&2&3\end{bmatrix}}\,X$

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

Eigenvectors:

1. $\lambda _{1}=8:\left\langle 2,1,2\right\rangle$
2. $\lambda _{2}=-1:\left\langle \alpha ,-2\alpha -2\beta ,\beta \right\rangle =\alpha \left\langle 1,-2,0\right\rangle +\beta \left\langle 0,-2,1\right\rangle$ (note we have two eigenvectors)

Corresponding Solutions:

1. $X_{1}=\mathrm {e} ^{8t}{\begin{bmatrix}2\\1\\2\end{bmatrix}}$
2. $X_{2}=\mathrm {e} ^{-t}{\begin{bmatrix}1\\-2\\0\end{bmatrix}}$
3. $X_{3}=\mathrm {e} ^{-t}{\begin{bmatrix}0\\-2\\1\end{bmatrix}}$

Check Linear Independence (redundant vs. complete solutions):

$W\{X_{1},X_{2},X_{3}\}={\begin{vmatrix}2\mathrm {e} ^{8t}&\mathrm {e} ^{-t}&0\\\mathrm {e} ^{8t}&-2\mathrm {e} ^{-t}&-2\mathrm {e} ^{-t}\\2\mathrm {e} ^{8t}&0&\mathrm {e} ^{-t}\end{vmatrix}}=-9\mathrm {e} ^{8t}\not \equiv 0$

Therefore, our solutions $X_{1}$, $X_{2}$, and $X_{3}$ are linearly independent; so the general solution is

$X(t)=c_{1}\,\mathrm {e} ^{8t}\,{\begin{bmatrix}2\\1\\2\end{bmatrix}}+c_{2}\mathrm {e} ^{-t}\,{\begin{bmatrix}1\\-2\\0\end{bmatrix}}+c_{3}\mathrm {e} ^{-t}\,{\begin{bmatrix}0\\-2\\1\end{bmatrix}}$

Section 7.6

Lecture Notes

Given the systems:

${\begin{aligned}X'&={\begin{bmatrix}3&-2\\4&-1\end{bmatrix}}\,X\\X'&={\begin{bmatrix}-1&-4\\1&-1\end{bmatrix}}\,X\\\end{aligned}}$

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

$X(t)=\sum _{i=0}^{\left|\lambda \right|}c_{i}\,\mathrm {e} ^{\lambda _{i}}\,{\vec {v}}_{i}$

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

What about matrices with imaginary eigenvalues?

${\begin{aligned}{\begin{vmatrix}3-\lambda &-2\\4&-1-\lambda \end{vmatrix}}&=0\\\lambda ^{2}-2\lambda +5&=0\\\lambda &=1\pm 2i\end{aligned}}$

1. $\lambda =1+2i:\left\langle 1,1-i\right\rangle$
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 $c_{1}\,\mathrm {e} ^{a\,t}\,\cos {b\,t}+c_{2}\,\mathrm {e} ^{a\,t}\,\sin {b\,t}$. Let's do the same here:

for $X=\mathrm {e} ^{t}\,\mathrm {e} ^{2i\,t}\,{\begin{bmatrix}1\\1-i\end{bmatrix}}$, we get:

$\mathrm {e} ^{t}\,\left(\cos {2t}+i\,\sin {2t}\right)\left({\begin{bmatrix}1\\1\end{bmatrix}}+i{\begin{bmatrix}0\\-1\end{bmatrix}}\right)$

Expanding this gives

$X(t)=\underbrace {\left(\mathrm {e} ^{t}\,\cos {2t}\,{\begin{bmatrix}1\\1\end{bmatrix}}-\mathrm {e} ^{t}\,\sin {2t}\,{\begin{bmatrix}0\\-1\end{bmatrix}}\right)} _{P}+i\,\underbrace {\left(\mathrm {e} ^{t}\,\sin {2t}\,{\begin{bmatrix}1\\1\end{bmatrix}}+\mathrm {e} ^{t}\,\cos {2t}\,{\begin{bmatrix}0\\-1\end{bmatrix}}\right)} _{Q}$

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

$X'(t)=A\,X(t)$

If $r_{1}=\lambda +i\,\mu$ and $r_{2}=\lambda -i\,\mu$ is a pair of complex conjugate eigenvalues and $\xi ={\vec {a}}\pm i\,{\vec {b}}$ is the corresponding pair of eigenvectors, then the vectors

${\begin{aligned}{\vec {u}}&=\mathrm {e} ^{\lambda \,t}\,\left({\vec {a}}\,\cos {\mu \,t}-{\vec {b}}\,\sin {\mu \,t}\right)\\{\vec {v}}&=\mathrm {e} ^{\lambda \,t}\,\left({\vec {a}}\,\sin {\mu \,t}+{\vec {b}}\,\cos {\mu \,t}\right)\end{aligned}}$

are real valued solutions of the system

Example

$X'={\begin{bmatrix}-1&-4\\1&-1\end{bmatrix}}\,X$

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

Eigenvector: ${\begin{bmatrix}2i\\1\end{bmatrix}}={\begin{bmatrix}0\\1\end{bmatrix}}+i{\begin{bmatrix}2\\0\end{bmatrix}}$

Apply the theorem:

${\begin{aligned}X_{1}&=\mathrm {e} ^{-t}\,\cos {2t}\,{\begin{bmatrix}0\\1\end{bmatrix}}-\mathrm {e} ^{-t}\,\sin {2t}\,{\begin{bmatrix}2\\0\end{bmatrix}}\\X_{2}&=\mathrm {e} ^{-t}\,\cos {2t}\,{\begin{bmatrix}2\\0\end{bmatrix}}+\mathrm {e} ^{-t}\,\sin {2t}\,{\begin{bmatrix}0\\1\end{bmatrix}}\end{aligned}}$