MATH 308 Lecture 34

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Lecture Notes

Section 7.8

To solve a system with 1 eigenvalue with multiplicity 2:

Find space of eigenvectors for eigenvalue $\lambda$.

If there are two linearly independent eigenvectors ${\vec {v}}_{1}$ and ${\vec {v}}_{2}$, then

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

are two linearly independent solutions

If there is no pair of linearly independent eigenvectors, find an eigenvector ${\vec {v}}_{1}$, and then find a second (generalized) eigenvector satisfying

$\left(A-\lambda \,I\right)\,{\vec {v}}_{2}={\vec {v}}_{1}$

Then

$X_{1}=\mathrm {e} ^{\lambda \,t}\,{\vec {v}}_{1}$ and $X_{2}=\mathrm {e} ^{\lambda \,t}\,\left(t\,{\vec {v}}_{1}+{\vec {v}}_{2}\right)$

Exercise 2

Find general solution to $X'={\begin{bmatrix}4&-2\\8&-4\end{bmatrix}}\,X$.

Eigenvalues: $\lambda \in \{0,0\}$

Eigenvectors: $V={\begin{bmatrix}1\\2\end{bmatrix}}$

Therefore, the only solution we have so far is $X_{1}=\mathrm {e} ^{0t}\,{\begin{bmatrix}1\\2\end{bmatrix}}={\begin{bmatrix}1\\2\end{bmatrix}}$

Let's find another solution in the form

$X_{2}=t\,\mathrm {e} ^{0t}\,{\begin{bmatrix}1\\2\end{bmatrix}}+\mathrm {e} ^{0t}\,{\vec {v}}_{2}$

Where

${\vec {v}}_{1}={\begin{bmatrix}1\\2\end{bmatrix}}=(A-0I)\,{\vec {v}}_{2}$

We find ${\vec {v}}_{2}={\begin{bmatrix}{\frac {1}{2}}\,\alpha +{\frac {1}{2}}\\\alpha \end{bmatrix}}$

Let $\alpha =1$, and we get the vector ${\vec {v}}_{2}={\begin{bmatrix}1\\1\end{bmatrix}}$

Therefore, $X_{2}=t\,{\begin{bmatrix}1\\2\end{bmatrix}}+{\begin{bmatrix}1\\1\end{bmatrix}}$

and the general solution is

$X=c_{1}\,{\begin{bmatrix}1\\2\end{bmatrix}}+c_{2}\left(t\,{\begin{bmatrix}1\\2\end{bmatrix}}+{\begin{bmatrix}1\\1\end{bmatrix}}\right)$

Note: We could have taken any value for $\alpha$. The resulting general solution will look different, but the constants $c_{1}$ and $c_{2}$ will change to compensate.

Exercise 3

Find particular solution to initioal value problem $X'={\begin{bmatrix}1&-4\\4&-7\end{bmatrix}}\,X$, where $X(0)={\begin{bmatrix}3\\2\end{bmatrix}}$

Eigenvalues: $\lambda \in \{-3,-3\}$

Eigenvector: ${\vec {v}}_{1}=\left\langle 1,1\right\rangle$

Solve $\left(A+3I\right)\,{\vec {v}}_{2}=\left\langle 1,1\right\rangle$ for $v_{2}=\left\langle {\frac {1}{4}}+\alpha ,\alpha \right\rangle$

Let $\alpha ={\frac {3}{4}}$, then ${\vec {v}}_{2}=\left\langle 1,{\frac {3}{4}}\right\rangle$

Therefore, our general solution is

$X=c_{1}\,\mathrm {e} ^{-3t}\,{\begin{bmatrix}1\\1\end{bmatrix}}+c_{2}\,\mathrm {e} ^{-3t}\,\left(t\,{\begin{bmatrix}1\\1\end{bmatrix}}+{\begin{bmatrix}1\\{\frac {3}{4}}\end{bmatrix}}\right)$

Plug in $t=0$ and solve for ${\vec {c}}$:

${\begin{bmatrix}3\\2\end{bmatrix}}=c_{1}\,{\begin{bmatrix}1\\1\end{bmatrix}}+c_{2}\,{\begin{bmatrix}1\\{\frac {3}{4}}\end{bmatrix}}$

${\vec {c}}=\left\langle -1,4\right\rangle$

So our particular solution is

$X=-1\mathrm {e} ^{-3t}\,{\begin{bmatrix}1\\1\end{bmatrix}}+4\mathrm {e} ^{-3t}\,\left(t\,{\begin{bmatrix}1\\1\end{bmatrix}}+{\begin{bmatrix}1\\{\frac {3}{4}}\end{bmatrix}}\right)$