3610 Exercises

Ex 1
In this problem, $A=<script type="math/tex; mode=display">\begin{pmatrix}1&1\\-4&1\end{pmatrix}$ and we are interested in the equation $\boldsymbol{u'}=A\boldsymbol{u}$.
(a) Find a fundamental matrix $\Phi(t)$ for $A$.
(b) Find the exponential matrix $e^{At}$.
(c) Find the solution to $\boldsymbol{u'}=A\boldsymbol{u}$ with $\boldsymbol{u}(0)=<script type="math/tex; mode=display">\begin{bmatrix}1\\2\end{bmatrix}$ .
(d) Find a solution to $\boldsymbol{u'}=A\boldsymbol{u}+<script type="math/tex; mode=display">\begin{bmatrix}5\\10\end{bmatrix}$ . What is the general solution? What is the solution with $\boldsymbol{u}(0)=\boldsymbol{0}$.
Solution
(a) The eigenvalues of $A$ are solutions to $(1-\lambda)^2+4=\lambda^2-2\lambda+5=0$, which are in this case the complex conjugates $1\plusmn 2i$.
Choose one of them, say $\lambda_1=1+2i$, and find an eigenvector $v_1$ for it.
The eigenvectors corresponding to $\lambda_1$ are the vectors in the killed by $A-\lambda_1I=<script type="math/tex; mode=display">\begin{bmatrix}-2i&1\\-4&-2i\end{bmatrix}$ , so we can take, say, $v_1=<script type="math/tex; mode=display">\begin{bmatrix}1\\2i\end{bmatrix}$ .
This means this equation has a complex-valued solution $$\begin{aligned} \boldsymbol{u}(t)&=e^{\lambda_1t}v_1=e^{(1+2i)t}\begin{bmatrix} 1\\2i \end{bmatrix}\\ &=e^t\begin{bmatrix} \cos 2t+i\sin 2t\\ 2i\cos 2t-2\sin 2t \end{bmatrix}\\ &=\begin{bmatrix} e^t\cos 2t\\-2e^t\sin 2t \end{bmatrix}+i\begin{bmatrix} e^t\sin 2t\\2e^t\cos 2t \end{bmatrix} \end{aligned}$$ By taking real and imaginary parts, we get two independent real solutions, and so can read off a fundamental matrix $$\Phi(t)=\begin{bmatrix} e^t\cos 2t&e^t\sin 2t \\-2e^t\sin 2t&2e^t\cos 2t \end{bmatrix}$$ (b) we can compute the exponential matrix for this system from any fundamental matrix $\Phi(t)$ found in the first part by using the formula $$e^{At}=\Phi(t)\Phi(0)^{-1}$$ $$\Phi(0)=\begin{bmatrix} 1&0\\0&2 \end{bmatrix}$$ so $$\Phi(0)^{-1}=\begin{bmatrix} 1&0\\0&1/2 \end{bmatrix}$$ $$\begin{aligned} e^{At}&=\begin{bmatrix} e^t\cos 2t&e^t\sin 2t \\-2e^t\sin 2t&2e^t\cos 2t \end{bmatrix}\begin{bmatrix} 1&0\\0&1/2 \end{bmatrix}\\ &=\begin{bmatrix} e^t\cos 2t&\frac{1}{2}\sin 2t\\ -2e^t\sin 2t&e^t\cos 2t \end{bmatrix} \end{aligned}$$ (c) The general strategy is to find the solution with a given initial condition directly from the exponential matrix: $$\begin{aligned} \boldsymbol{u}(t)&=e^{At}\boldsymbol{u}(0)\\ &=\begin{bmatrix} e^t\cos 2t&\frac{1}{2}\sin 2t\\ -2e^t\sin 2t&e^t\cos 2t \end{bmatrix}\begin{bmatrix} 1\\2 \end{bmatrix}\\ &=\begin{bmatrix} e^t(\cos 2t+\sin 2t)\\2e^t(-\sin 2t+\cos 2t) \end{bmatrix} \end{aligned}$$ (d) We guess a constant solution $$\boldsymbol{u}=\begin{bmatrix} k_1\\k_2 \end{bmatrix}$$ Substituting this into the DE gives $$\begin{bmatrix} 0\\0 \end{bmatrix}=A\begin{bmatrix} k_1\\k_2 \end{bmatrix}+\begin{bmatrix} 5\\10 \end{bmatrix}$$ This implies $$\boldsymbol{u}=-A^{-1}\begin{bmatrix} 5\\10 \end{bmatrix}=-\frac{1}{5}\begin{bmatrix} 1&-1\\4&1 \end{bmatrix}\begin{bmatrix} 5\\10 \end{bmatrix}=\begin{bmatrix} 1\\-6 \end{bmatrix}$$ Since all homogeneous solutions are of the form $e^{At}<script type="math/tex; mode=display">\begin{bmatrix}a\\b\end{bmatrix}$ , the general solution is then given by $$e^{At}\begin{bmatrix}a\\b\end{bmatrix}+\begin{bmatrix} 1\\-6 \end{bmatrix}=\begin{bmatrix} e^t(a\cos 2t+\frac{1}{2}b\sin 2t)+1\\ e^t(-2a\sin 2t+b\cos 2t)-6 \end{bmatrix}$$ for some constants $a, b$.
To find the particular solution with $\boldsymbol{u}(0)=0$, we plug $t = 0$ into this expression, and get that $$\begin{bmatrix} a+1\\b-6 \end{bmatrix}=\begin{bmatrix} 0\\0 \end{bmatrix}$$ so the desired solution is given by the constants $a = -1$ and $b = 6$: $$\begin{bmatrix} e^t(-\cos 2t+3\sin 2t)+1\\ e^t(2\sin 2t+6\cos 2t)-6 \end{bmatrix}$$

Ex 2
Suppose $\boldsymbol{u}_1(t)=<script type="math/tex; mode=display">\begin{bmatrix}1\\1\end{bmatrix}$ (the constant trajectory) and $\boldsymbol{u}_2(t)=<script type="math/tex; mode=display">\begin{bmatrix}e^t\\-e^{t}\end{bmatrix}$ are solutions to the equation $\boldsymbol{u'}=B\boldsymbol{u}$ for some matrix $B$.
(a) What is the general solution? What is the solution $\boldsymbol{u}(t)$ with $\boldsymbol{u}(0)=<script type="math/tex; mode=display">\begin{bmatrix}2\\2\end{bmatrix}$ ? What is the solution with $\boldsymbol{u}(0)=<script type="math/tex; mode=display">\begin{bmatrix}1\\0\end{bmatrix}$ ?
(b) Find a fundamental matrix, and compute the exponential $e^{Bt}$. What is $e^B$?
(c) What are the eigenvalues and eigenvectors of $B$?
(d) What is $B$?
Solution.
(a) The general solution is $$\boldsymbol{u}(t)=a\boldsymbol{u}_1(t)+b\boldsymbol{u}_2(t)=a\begin{bmatrix}1\\1\end{bmatrix}+b\begin{bmatrix}e^t\\-e^{t}\end{bmatrix}$$ We can find the particular solution that satisfies any given initial condition $\boldsymbol{u}(0)$ explicitly by solving the system of equations given by $a\boldsymbol{u}_1(0) + b\boldsymbol{u}_2(0) = \boldsymbol{u}(0)$ for $a$ and $b$.
Here $\boldsymbol{u}_1(0)=<script type="math/tex; mode=display">\begin{bmatrix}1\\1\end{bmatrix}$ and $\boldsymbol{u}_2(0)=<script type="math/tex; mode=display">\begin{bmatrix}1\\-1\end{bmatrix}$ , so the solution with initial condition $\boldsymbol{u}(0)=<script type="math/tex; mode=display">\begin{bmatrix}2\\2\end{bmatrix}$ is the constant trajectory $$2\boldsymbol{u}_1(t)=\begin{bmatrix} 2\\2 \end{bmatrix}$$ and the solution with initial condition $\boldsymbol{u}(0)=<script type="math/tex; mode=display">\begin{bmatrix}1\\0\end{bmatrix}$ is $$\frac{1}{2}\boldsymbol{u}_1(t)+\frac{1}{2}\boldsymbol{u}_2(t)=\begin{bmatrix} 1/2+e^t/2\\1/2-e^t/2 \end{bmatrix}$$ (b) Since the two given solutions are linearly independent, a fundamental matrix is $$\begin{bmatrix} 1&e^t\\1&-e^t \end{bmatrix}$$ Evaluating this matrix at $t = 0$ gives $<script type="math/tex; mode=display">\begin{bmatrix}1&1\\1&-1\end{bmatrix}$ which has inverse $\frac{1}{2}<script type="math/tex; mode=display">\begin{bmatrix}1&1\\1&-1\end{bmatrix}$ , so the exponential matrix is $$\begin{aligned} e^{Bt}&=\begin{bmatrix} 1&e^t\\1&-e^t \end{bmatrix}\frac{1}{2}\begin{bmatrix}1&1\\1&-1\end{bmatrix}\\ &=\frac{1}{2}\begin{bmatrix} 1+e^t&1-e^t\\1-e^t&1+e^t \end{bmatrix} \end{aligned}$$ Evaluating $e^{Bt}$ at $t = 1$ gives us $e^B=\frac{1}{2}<script type="math/tex; mode=display">\begin{bmatrix}1+e&1-e\\1-e&1+e\end{bmatrix}$
(c) $\boldsymbol{u}_1=<script type="math/tex; mode=display">\begin{bmatrix}1\\1\end{bmatrix}$ , so $\lambda_1=0$ and $\boldsymbol{v}_1=<script type="math/tex; mode=display">\begin{bmatrix}1\\1\end{bmatrix}$
$\boldsymbol{u}_2(t)=<script type="math/tex; mode=display">\begin{bmatrix}e^t\\-e^{t}\end{bmatrix}$=e^t $$\begin{bmatrix}1\\-1\end{bmatrix}$$ , so $\lambda_2=1$ and $\boldsymbol{v}_2=<script type="math/tex; mode=display">\begin{bmatrix}1\\-1\end{bmatrix}$
(d) $\boldsymbol{u'}=B\boldsymbol{u}$ From the equation for the first eigenvalue $$\begin{bmatrix} 0\\0 \end{bmatrix}=B\begin{bmatrix} 1\\1 \end{bmatrix}$$ so $B$ has the form $<script type="math/tex; mode=display">\begin{bmatrix}a&-a\\c&-c\end{bmatrix}$
From the equation for the second eigenvalue $$e^t\begin{bmatrix}1\\-1\end{bmatrix}=\begin{bmatrix}a&-a\\c&-c\end{bmatrix}\begin{bmatrix}e^t\\-e^t\end{bmatrix}$$ so $a=1/2$ and $c=-1/2$. Thus, $$B=\begin{bmatrix} 1/2&-1/2\\-1/2&1/2 \end{bmatrix}$$

Ex 3.
Say that a pair of solutions $x_1(t), x_2(t)$ of the equation $mx'' + bx' + kx = 0$ is normalized at $t = 0$ if: $$x_1(0)=1,x_1'=0$$ $$x_2(0)=0,x_2'=1$$ For example, find the normalized pair of solutions to $x''+2x'+2x=0$. Then find $e^{At}$ where $A$ is the companion matrix for the operator $D^2 + 2D + 2I$.
Solution.
$s^2 + 2s + 2 = (s + 1)^2 + 1$ so the roots of the characteristic polynomial are $-1 \plusmn i$.
Basic solutions are given by $$y_1=e^{-t}\cos t \text{ and } y_2=e^{-t}\sin t$$ $$y_1(0)=1, y_1'(0)=-1,y_2(0)=0,y_2'(0)=1$$ So $x_1 = y_1 + y_2$ and $x_2 = y_2$ form a normalized pair of solutions: $$x_1(t)=e^{-t}(\cos t+\sin t) \text{ and } x_2(t)=e^{-t}\sin t$$ From 3304, we know

The corresponding coefficient matrix is $$A=\begin{pmatrix}0&1\\-k&-b\end{pmatrix}$$ The companion matrix is $A=<script type="math/tex; mode=display">\begin{bmatrix}0&1\\-2&-2\end{bmatrix}$ .
Its characteristic polynomial is the same, $\lambda^2+2\lambda+2$, so its eigenvalues are the same, $-1\plusmn i$.
An eigenvector for value $-1+i$ is given by $\boldsymbol{v}_1$ such that $$\begin{bmatrix} -1+i&1\\-2&-1-i \end{bmatrix}\boldsymbol{v}_1=\begin{bmatrix} 0\\0 \end{bmatrix}$$ We can take $\boldsymbol{v}_1=<script type="math/tex; mode=display">\begin{bmatrix}1\\-1+i\end{bmatrix}$ . The corresponding normal mode is $$\begin{aligned} e^{-1+i}\begin{bmatrix} 1\\-1+i \end{bmatrix}&=e^{-t}(\cos t+i\sin t)\begin{bmatrix} 1\\-1+i \end{bmatrix}\\ &=e^{-t}(\cos t\begin{bmatrix} 1\\-1 \end{bmatrix}+i\sin t\begin{bmatrix} 0\\i \end{bmatrix})+e^{-t}(\cos t\begin{bmatrix} 0\\i \end{bmatrix}+i\sin t\begin{bmatrix} 1\\-1 \end{bmatrix})\\ &=e^{-t}(\begin{bmatrix} \cos t\\-\cos t-\sin t \end{bmatrix}+i\begin{bmatrix} \sin t\\\cos t-\sin t \end{bmatrix}) \end{aligned}$$ Thus $$\Phi(t)=\begin{bmatrix} \cos t&\sin t\\-\cos t-\sin t&\cos t-\sin t \end{bmatrix}$$ $$\Phi(0)=\begin{bmatrix} 1&0\\-1&1 \end{bmatrix}$$ $$\Phi(0)^{-1}=\begin{bmatrix} 1&0\\1&1 \end{bmatrix}$$ so $$\begin{aligned} e^{At}&=\Phi(t)\Phi(0)^{-1}\\ &=\begin{bmatrix} \cos t+\sin t&\sin t\\ -2\sin t&\cos t-\sin t \end{bmatrix} \end{aligned}$$

Ex 4.
We have seen that a complex number $z = a + bi$ determines a matrix $A(z)$ in the following way: $A(a + bi) =<script type="math/tex; mode=display">\begin{bmatrix}a&-b\\b&a\end{bmatrix}$ . This matrix represents the operation of multiplication by $z$, in the sense that if $z(x + yi) = v + wi$ then $A(z)<script type="math/tex; mode=display">\begin{bmatrix}x\\y\end{bmatrix}$= $$\begin{bmatrix}v\\w\end{bmatrix}$$ . What is $e^{A(z)t}$? What is $A(e^{zt})$?
Solution.
With $A(z)=<script type="math/tex; mode=display">\begin{bmatrix}a&-b\\b&a\end{bmatrix}$ , $$p_A(\lambda)=(\lambda-a)^2+b^2$$ so the eigenvalues are $a \plusmn bi$. An eigenvector for $\lambda_1 = a + bi$ is given by $\boldsymbol{v}_1$ such that $$\begin{bmatrix} -bi&-b\\b&-bi \end{bmatrix}\boldsymbol{v}_1=\begin{bmatrix} 0\\0 \end{bmatrix}$$ and we can take $\boldsymbol{v}_1 = <script type="math/tex; mode=display">\begin{bmatrix}1\\-i\end{bmatrix}$ . The corresponding normal mode is $$\begin{aligned} e^{(a+bi)t}\begin{bmatrix}1\\-i\end{bmatrix}&=e^{at}(\cos bt + i\sin bt)\begin{bmatrix}1\\-i\end{bmatrix}\\ &=e^{at}(\cos bt\begin{bmatrix} 1\\0 \end{bmatrix}+i\sin bt\begin{bmatrix} 0\\-i \end{bmatrix}+\cos bt\begin{bmatrix} 0\\-i \end{bmatrix}+i\sin bt\begin{bmatrix} 1\\0 \end{bmatrix})\\ &=e^{at}(\begin{bmatrix} \cos bt\\\sin bt \end{bmatrix}+i\begin{bmatrix} \sin bt\\-\cos bt \end{bmatrix}) \end{aligned}$$ So a fundamental matrix is given by $$\Phi(t)=e^{at}\begin{bmatrix} \cos bt&\sin bt\\\sin bt&-\cos bt \end{bmatrix}$$ $$\Phi(0)=\begin{bmatrix} 1&0\\0&-1 \end{bmatrix}$$ $$\Phi(0)^{-1}=\begin{bmatrix} 1&0\\0&-1 \end{bmatrix}$$ So $$e^{At}=\Phi(t)\Phi(0)^{-1}=e^{at}\begin{bmatrix} \cos bt&-\sin bt\\\sin bt&\cos bt \end{bmatrix}$$ $$\begin{aligned} A(e^{zt})&=A(e^{(a+bi)t})=A(e^{at}(\cos bt+i\sin bt))\\ &=\begin{bmatrix} e^{at}\cos bt&-e^{at}\sin bt\\ e^{at}\sin bt&e^{at}\cos bt \end{bmatrix}=e^{at}\begin{bmatrix} \cos bt&-\sin bt\\\sin bt&\cos bt \end{bmatrix}\\ &=e^{At}=e^{A(z)t}=e^{A(a+bi)t} \end{aligned}$$