
% extra credit ch5 2280, S2015
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\begin{document}
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 {\bf Name~}\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\hfill
 {Class Time~}\underline{~~~~~~~~}\newline


\begin{center}
\bf Math 2280 Extra Credit Problems \\
\bf Chapter 5 \\
{\small\bf S2015}
\end{center}

{\bf Submitted work}. Please submit one stapled package per chapter.
Kindly label problems \fbox{\bf Extra Credit}. Label each problem with
its corresponding problem number, e.g., \fbox{Xc5.1-8}. Please attach
this printed sheet to simplify your work.

\Prob{Xc5.0-1}{(Eigenpairs of a Matrix $A$)}
 ~\newline
 (a) Let $A=\twobytwo{9 & -10}{2 & 0}$. Find the eigenpairs of $A$. Then report
 eigenpair
packages $P$ and $D$ such that $AP=PD$.


(b) Let
$A=\threebythree{5 & -6 & 3}{6 & -7 & 3}{6 & -6 & 2}$. Find the
eigenpairs of $A$. Then report eigenpair packages $P$ and $D$ such
that $AP=PD$.
% A1:=matrix([[9,-10],[2,0]]);eigenvects(A1);
% [4, 1, {vector([2, 1])}], [5, 1, {vector([5/2, 1])}]

(c) Let
$A=\twobytwo{0 & -6}{24 & 0}$. Find the complex eigenpairs of $A$. Then
report eigenpair packages $P$ and $D$ such that $AP=PD$.
% A:=matrix([[0,-6],[24,0]]);eigenvects(A);

(d) Let $A=\threebythree{2 & -2 & 1}{2 & -2 & 1}{2 & -2 & 1}$. Find the
 eigenpairs of $A$. Then report eigenpair packages $P$ and $D$ such that
 $AP=PD$. Check the answer by hand, expanding both products $AP$ and
 $PD$, finally showing equality.
% G:=matrix([[2 , -2 ,1],[2 , -2 , 1],[2 , -2 , 1]]);eigenvects(G);
% [1, 1, {vector([1, 1, 1])}], [0, 2, {vector([1, 0, -2]), vector([0, 1, 2])}]

\Prob{Xc5.0-36}{(Eigenvalues of band matrices)}
 Find the eigenvalues of the matrix $A$ below without the aid of
 computers.
$$
 A = \left(\begin{array}{rrrrrr}
 1 & 2 & 0 & 0 & 0 & 0 \\
 2 & 1 & 2 & 0 & 0 & 0 \\
 0 & 2 & 1 & 2 & 0 & 0 \\
 0 & 0 & 2 & 1 & 2 & 0 \\
 0 & 0 & 0 & 2 & 1 & 2 \\
 0 & 0 & 0 & 0 & 0 & 1
 \end{array}
 \right)
$$
%B:=matrix([
%  [1 , 2 , 0 , 0 , 0 , 0],
%  [2 , 1 , 2 , 0 , 0 , 0],
%  [0 , 2 , 1 , 2 , 0 , 0] ,
%  [0 , 0 , 2 , 1 , 2 , 0] ,
%  [0 , 0 , 0 , 2 , 1 , 2] ,
%  [0 , 0 , 0 , 0 , 0 , 1]]);
% eigenvals(B);
% -1, 3, 1+2*sqrt(3), 1-2*sqrt(3), 1, 1



\Prob{Xc5.0-18}{(Fourier's model for a $3\times 3$)}
 Assume Fourier's model for a certain matrix $A$:
$$
 A\left( c_1\colvector{1\\ 0 \\ -2}+c_2\colvector{1\\ 1\\
 0}+c_3\colvector{0\\ 0\\ 1}\right) =
 3c_1\colvector{1\\ 0 \\ -2}+c_2\colvector{1\\ 1\\
 0}+c_3\colvector{0\\ 0\\ 1}.
$$
Fourier's model (the above equation) is assumed valid for all constants $c_1,c_2,c_3$. Find $A$ explicitly from $AP=PD$. Check your answer by finding the
eigenpairs of $A$.
% C:=matrix([[3 , -2 ,0],[0 , 1 , 0],[-4 , 4 , 1]]);eigenvects(C);
% [3, 1, {vector([1, 0, -2])}], [1, 2, {vector([1, 1, 0]), vector([0, 0, 1])}]

\Prob{Xc5.0-28}{(Eigenpairs and Diagonalization of a $4\times 4$)}
 Determine the eigenpairs of $A$ below. If diagonalizable, then
 report eigenpair packages $P$ and $D$ such that $AP=PD$.
$$
 A = \left(\begin{array}{rrrrrr}
 1 & 2 & 0 & 0 \\
 2 & 1 & 2 & 0 \\
 0 & 2 & 1 & 2 \\
 0 & 0 & 0 & 13
 \end{array}
 \right)
$$
%F:=matrix([
%  [1 , 2 , 0 , 0],
%  [2 , 1 , 2 , 0],
%  [0 , 2 , 1 , 2] ,
%  [0 , 0 , 0 , 3]]);
% eigenvals(F);
% [1, 1, {vector([-1, 0, 1, 0])}], [1+2*sqrt(2), 1, {vector([1, sqrt(2), 1, 0])}],
% [1-2*sqrt(2), 1, {vector([1, -sqrt(2), 1, 0])}],[3, 1, {vector([1, 1, 0, -1])}]
\newpage
\Prob{Xc5.1-14}{(Particular solution)}

 (a) Find the constants $c_1$, $c_2$ in the general solution
 $$ \.x(t)=c_1e^{3t}\colvector{1\\ -1} + c_2e^{5t}\colvector{1\\
 -3}$$
 satisfying the initial conditions $x_1(0)=4$, $x_2(0)=-1$.

  (b) Find the matrix $A$ in the equation $\.x'=A\.x$. Use the
  formula $AP=PD$ and Fourier's model for $A$, which is given
  implicitly in (a) above, and explicitly as
$$ A(c_1\.v_1+c_2\.v_2) = c_1\lambda_1 \.v_1 + c_2\lambda_2 \.v_2$$
where $c_1$, $c_2$ are arbitrary constants and
$(\lambda_1,\.v_1)$,
$(\lambda_2,\.v_2)$ are the eigenpairs of the $2\times 2$ matrix $A$.

\Prob{Xc5.2-8}{(Eigenanalysis method $2\times 2$)}

 (a) Find $\lambda_1$, $\lambda_2$, $\.v_1$, $\.v_2$ in Fourier's
 model
 $A\left(c_1\.v_1+c_2\.v_2\right)=c_1\lambda_1\.v_1+c_2\lambda_2\.v_2$
 for
 $$A=\twobytwo{3&-4}{4&3}.$$

 (b) Display the general solution of $\.x'=A\.x$.

\Prob{Xc5.2-20}{(Eigenanalysis method $3\times 3$)}

(a) Find $\lambda_1$, $\lambda_2$, $\lambda_3$, $\.v_1$, $\.v_2$,
$\.v_3$ in Fourier's
 model
 $A\left(c_1\.v_1+c_2\.v_2+c_3\.v_3\right)=c_1\lambda_1\.v_1+c_2\lambda_2\.v_2+c_3\lambda_3\.v_3$
 for
 $$A=\threebythree{2&1&-1}{-4&-3&-1}{4&4&2}.$$

 (b) Display the general solution of $\.x'=A\.x$.

\Prob{Xc5.2-30}{(Brine Tanks)}
 Consider two brine tanks satisfying the equations
 $$
 x_1'(t) = -k_1x_1+k_2x_2,\quad x_2'=k_1x_1-k_2x_2.
 $$
 Assume $r=10$ gallons per minute, $k_1=r/V_1$, $k_2=r/V_2$, $x_1(0)=30$ and $x_2(0)=0$. Let the tanks have volumes
 $V_1=50$ and $V_2=25$ gallons. Solve for $x_1(t)$ and $x_2(t)$.


\Prob{Xc5.2-40}{(Eigenanalysis method $4\times 4$)}
 Display (a) Fourier's model and (b) the general solution of
 $\.x'=A\.x$ for the $4\times 4$ matrix
 $$A=\left(\begin{array}{rrrr}
  2 & 0 & 0 & 0 \\
   -21 & -5 & -27 & -9 \\
    0 & 0 & 5 & 0 \\
     0 & 0 & -16 & -4
  \end{array}\right).
 $$
%A:=matrix([[ 2 , 0 , 0 , 0],
%    [-21 , -5 , -27 , -9],
%    [ 0 , 0 , 5 , 0],
%    [  0 , 0 , -16 , -4]]);
%
% [5, 1, {vector([0, -11/10, 1, -16/9])}],
% [2, 1, {vector([1, -3, 0, 0])}],
% [-5, 1, {vector([0, 1, 0, 0])}],
% [-4, 1, {vector([0, -9, 0, 1])}]

\Prob{Xc5.5-4}{(Fundamental Matrix)}
This problem is double credit, to match the effort required. Consider the $2\times 2$ vector-matrix differential equation
$$\.u'=A\.u,\quad A=\twobytwo{2 & -5}{0 & 1},\quad \.u=\colvector{x(t)\\ y(t)}.$$
Complete all parts below.
\begin{itemize}
\item[\BF{(a)}] {\em Cayley-Hamilton method}. Compute the characteristic
equation $\det(A-\lambda I)=0$. Find two atoms from the roots of this
equation. Then $x(t)$ is a linear combination of these atoms. The first
equation $x'=2x-5y$ can be solved for $y$ to find the second answer.
Construct a fundamental matrix $\Phi$ from these scalar answers.
\item[\BF{(b)}] {\em Eigenanalysis method}. Find the eigenpairs $(\lambda_1,\.v_1), (\lambda_2,\.v_2)$ of $A$.
Let $\Phi$ have columns $e^{\lambda_1 t}\.v_1$, $e^{\lambda_2 t}\.v_2$.
Explain why $\Phi$ is a fundamental matrix.
\item[\BF{(c)}] {\em Putzer's formula}.
Find $e^{At}$ from the formula
$$e^{At} = e^{\lambda_1 t}T + \frac{e^{\lambda_1 t}-e^{\lambda_2
t}}{\lambda_1 - \lambda_2}(A-\lambda_1 I).$$
If the eigenvalues are complex, then $e^{At}$ is the real part of the
right side. If $\lambda_1=\lambda_2$, then $e^{At}$ is the limit of the
right side as $\lambda_2 \to \lambda_1$ (use L'Hopital's rule).
\item[\BF{(d)}] Report $e^{At}=\Phi(t)\Phi(0)^{-1}$,
using the answer for $\Phi$ from part (a) or (b). Check your answer
against the one in part (c).
\end{itemize}

\Prob{Xc5.5-12}{(Putzer's Method)}
The exponential matrix $e^{At}$ can be found in the $2\times 2$ case from Putzer's formula
$$e^{At} = e^{\lambda_1 t}I+\dd\frac{e^{\lambda_1 t}-e^{\lambda_2 t}}{\lambda_1 - \lambda_2}(A - \lambda_1 I).$$
If the roots $\lambda_1$, $\lambda_2$ of $\det(A-\lambda I)=0$ are equal, then compute the Newton
quotient factor by L'Hopital's rule, limiting $\lambda_2 \to \lambda_1$ [$\lambda_1$, $t$ fixed].
If the roots are complex, then take the real part of the right side of the equation.

\smallskip
Compute $e^{At}$ from Putzer's formula for the following cases.
\begin{itemize}
\item[\BF{(a)}] $A=\twobytwo{1 & 0}{0 & 2}$. Answer $e^{At}=\twobytwo{e^t & 0}{0 & e^{2t}}$.
\item[\BF{(b)}] $A=\twobytwo{2 & 1}{1 & 2}$.
\item[\BF{(c)}] $A=\twobytwo{0 & 1}{-1 & -2}$.
\item[\BF{(d)}] $A=\twobytwo{2 & -5}{4 & -2}$.
\end{itemize}

\Prob{Xc5.5-38}{(Laplace's Resolvent Method)}
The exponential matrix $e^{At}$ can be found from the Laplace resolvent formula
for the problem $\Phi'=A\Phi$, $\Phi(0)=I$:
$$\L(\Phi(t))=(sI-A)^{-1}\Phi(0)=(sI-A)^{-1}.$$
For example, $A=\twobytwo{1 & 0}{0 & 2}$ gives $\L(e^{At})=\twobytwo{s-1 & 0}{0 & s-2}^{-1}=\twobytwo{\frac{1}{s-1} & 0 }{0 & \frac{1}{s-2}}=\twobytwo{\L(e^t) & 0 }{0 & \L(e^{2t})}$,
which implies $e^{At}=\twobytwo{e^t & 0 }{0 & e^{2t}}$.

\smallskip
Compute $\Phi(t)=e^{At}$ using the resolvent formula for the following cases.
\begin{itemize}
\item[\BF{(a)}] $A=\twobytwo{2 & 1}{1 & 2}$.
\item[\BF{(b)}] $A=\twobytwo{0 & 1}{-1 & -2}$.
\item[\BF{(c)}] $A=\twobytwo{2 & -5}{4 & -2}$.
\end{itemize}
\newpage
\Prob{Xc5.6-4}{(Variation of Parameters)}
Use the variation of parameters formula $\.u_p(t)=e^{At}\int e^{-At}\.f(t)dt$ to find
a particular solution of the given system. Please use {\tt maple} to do the indicated integration,
following the example below.

\begin{itemize}
\item[\BF{(a)}] $\.u'=\twobytwo{2 & 1}{1 & 2}\.u+\colvector{1\\ 2}$.
\item[\BF{(b)}] $\.u'=\twobytwo{0 & 1}{-1 & -2}\.u+\colvector{e^t\\ 1}$.
\end{itemize}

\BF{Example}: Solve for $\.u_p(t)$: $\.u'=\twobytwo{0 & 1}{1 & 0}\.u+\colvector{1\\ 0}$.
\begin{verbatim}
 with(LinearAlgebra):
 A:=Matrix([[0,1],[1,0]]);
 f:=t->Vector([1,0]);
 expAt:=t->MatrixExponential(A,t);
 integral:=Map(g->int(g,t),expAt(-t).f(t));
 up:=simplify(expAt(t).integral);
\end{verbatim}

\Prob{Xc5.6-19}{(Initial Value Problem)}
Solve the given initial value problem using a computer algebra system. Follow
the example given below.

\begin{itemize}
\item[\BF{(a)}] $\.u'=\twobytwo{2 & 1}{1 & 2}\.u+\colvector{1\\ 2}$, $\.u(0)=\colvector{0\\ 0}$.
\item[\BF{(b)}] $\.u'=\twobytwo{0 & 1}{-1 & -2}\.u+\colvector{e^t\\ 1}$, $\.u(0)=\colvector{1\\ 0}$.
\end{itemize}

\BF{Example}: Solve for $\.u(t)$: $\.u'=\twobytwo{0 & 1}{1 & 0}\.u+\colvector{-1\\ 0}$, $\.u(0)=\colvector{-1\\ 0}$. The answer is $\.u=\colvector{-e^{-t}\\ e^{-t}-1}$.
\begin{verbatim}
 with(LinearAlgebra):
 A:=Matrix([[0,1],[1,0]]);
 f:=t->Vector([1,0]);
 expAt:=t->MatrixExponential(A,t);
 integral:=Map(g->int(g,t=0..t),expAt(-t).f(t));
 up:=unapply(expAt(t).integral,t):
 u0:=Vector([-1,0]);
 uh:=t->expAt(t).(u0-up(0));
 u:=simplify(uh(t)+up(t));
\end{verbatim}


 \vfill {\bf End of extra credit problems chapter 5.}




\end{document}
