
% extra credit ch8 2250, S2015
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 {\bf Name~}\underline{~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~}\hfill
 {Class Time~}\underline{~~~~~~~~}\newline


\begin{center}
\bf Math 2250 Extra Credit Problems \\
\bf Chapter 8 \\
{\small\bf S2015}
\end{center}

{\bf Submitted work}. Please submit one stapled package with this sheet
on top. Kindly check-mark the problems submitted and
label the paper \fbox{\bf Extra Credit}. Label each solved problem with
its corresponding problem number, e.g., \fbox{Xc10.3-20}.

%05 Dec, 8.1: \quad \Y{4}, \Y{12}, \Y{38}
%08 Dec, 8.2: \quad \Y{4}, \Y{19} \\
\Prob{Xc8.1-4}{(Fundamental Matrix)} Find a fundamental matrix $\Phi(t)$ by each of the following methods. Report $e^{At}=\Phi(t)\Phi(0)^{-1}$, using one of the answers for $\Phi$.
$$\.u'=A\.u,\quad A=\twobytwo{2 & -5}{0 & 1},\quad \.u=\colvector{x(t)\\ y(t)}.$$
\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.
\item[\BF{(b)}] {\em Eigenanalysis method}. Find the eigenpairs $(\lambda_1,\.v_1), (\lambda_2,\.v_2)$ of $A$.
Let $\Phi$ have columns $e^{\lambda_1 t}\.v_1$, $e^{\lambda_2 t}\.v_2$.
\end{itemize}

\Prob{Xc8.1-12}{(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 from Putzer's formula $e^{At}$ 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{Xc8.1-38}{(Laplace's 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}
\Prob{Xc8.2-4}{(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 technology 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{Xc8.2-19}{(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}
{\bf End of extra credit problems chapter 8.}

\end{document}
