MATH 308 Lecture 33

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Lecture Notes

Section 7.6

Find general solution

$X'={\begin{bmatrix}1&0&0\\2&1&-2\\3&2&1\end{bmatrix}}\,X$

Eigenvalues: $\lambda \in \{1,1+2i,1-2i\}$

Eigenvectors:

1. $\lambda =1:{\vec {v}}_{1}=\left\langle 2,-3,2\right\rangle$
2. $\lambda =1+2i:{\vec {v}}_{2}=\left\langle 0,i,1\right\rangle =\left\langle 0,0,1\right\rangle +i\,\left\langle 0,1,0\right\rangle$
3. $\lambda =1-2i:{\vec {v}}_{3}={\text{not needed}}$

Solutions

1. $X_{1}=\mathrm {e} ^{t}\,{\begin{bmatrix}2\\-3\\2\end{bmatrix}}$
2. $X_{2}=\mathrm {e} ^{t}\,\left(\cos {2t}\,{\begin{bmatrix}0\\0\\1\end{bmatrix}}-\sin {2t}\,{\begin{bmatrix}0\\1\\0\end{bmatrix}}\right)$
3. $X_{3}=\mathrm {e} ^{t}\,\left(\sin {2t}\,{\begin{bmatrix}0\\0\\1\end{bmatrix}}+\cos {2t}\,{\begin{bmatrix}0\\1\\0\end{bmatrix}}\right)$

General Solution

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

Phase portrait

For $X'={\begin{bmatrix}2&-5\\\alpha &-2\end{bmatrix}}\,X$

1. Determine eigenvalues in terms of $\alpha$
2. Find the critical value or values of α where the qualitative nature of the phase portrait for the system changes.
3. Draw a phase portrait for a value slightly below and for another value slightly above each critical value.

$\lambda =\pm {\sqrt {4-5\alpha }}$

Note that for some values of $\alpha$, the solutions are real (and the limit of the solution as $t$ approaches infinity is infinity) or imaginary (and the limit as $t$ approaches infinity diverges)

• $\alpha =0:\lambda =\pm 2$
• $\alpha =1:\lambda =\pm i$

Critical value of $\alpha$ is when ${\sqrt {4-5\alpha }}=0$: $\alpha ={\frac {4}{5}}$

Section 7.8

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

Only one eigenvalue ($\lambda =1$) with multiplicity 2

Eigenvectors: $\left\langle 2,1\right\rangle$

Therefore one solution $X_{1}=\mathrm {e} ^{t}\,{\begin{bmatrix}2\\1\end{bmatrix}}$

We cannot find another solution since that would require us to find an eigenvector that is linearly independent of what we already found. This is impossible with what we've learned so far.

Recall that when we had a characteristic root with multiplicity 2, we multiplied the original solution by $t$.

However, $X_{2}=t\,\mathrm {e} ^{t}\,{\begin{bmatrix}2\\1\end{bmatrix}}$ is not a solution since $X'$ does not match up with our original equation.

Therefore, we add a second term $\mathrm {e} ^{t}\,{\vec {v}}_{2}$ to "cancel out" the extra term in $X'$:

$X_{2}=t\,\mathrm {e} ^{\lambda \,t}\,{\vec {v}}_{1}+\mathrm {e} ^{\lambda \,t}\,{\vec {v}}_{2}$

Where

• $v_{1}$ is our original eigenvector, and
• ${\vec {v}}_{1}=\left(A-\lambda \,I\right)\,{\vec {v}}_{2}$.

In our example above, ${\begin{bmatrix}2&-4\\1&-2\end{bmatrix}}\,{\begin{bmatrix}x\\y\end{bmatrix}}={\begin{bmatrix}2\\1\end{bmatrix}}$

Therefore ${\vec {v}}_{2}=\left\langle 1,0\right\rangle$