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\begin{document}
\centerline{Math 322 worksheet for Friday, Week 5}
\bigskip

Consider the~$4$th order linear homogeneous equation with constant coefficients:
\begin{equation}
  y^{(iv)}+a_3y'''+a_2y''+a_1y'+a_0y=0
  \label{eqn: 05friws1}
\end{equation}
with initial condition~$y^{(i)}(0)=b_i$ for~$i=0,1,2,3$.  

Let~$x_i=y^{(i-1)}$ for~$i=1,2,3,4$, and let~$x(t)=(x_1(t),\dots,x_4(t))$.
Use equation~(\ref{eqn: 05friws1}) to create a linear system:
\[
  x' = Ax
\]
with initial condition~$x(0)$. 
\medskip

\textsc{problem 1.} What is the~$4\times 4$ matrix~$A$?  And what is~$x(0)$ in
terms of~$y$?
\vskip3.5in

\textsc{Problem 2.} The solution to the above system is~$x(t)=e^{At}x_0$.
Suppose you have calculated~$e^{At}$.  How do
read off the solution to our original equation~(\ref{eqn: 05friws1})?
\newpage

According to the recipe we learned during the first couple of weeks of
class, to solve equation~\eqref{eqn: 05friws1}, we first consider its
characteristic polynomial $P(x)=x^4+a_3x^3+a_2x^2+a_1x+a_0$.  We would like to
compare $P(x)$ to~$p_A(x):=\det(A-xI_4)$, the characteristic polynomial for the
matrix~$A$.
\medskip

\textsc{problem 3.} Compute~$\det(A-x I_4)$ by first performing the following
column operations (which don't affect the value of the determinant):
add~$x$ times the second column to the first column, then add~$x^2$ times the
third column to the first column, then add~$x^3$ times the fourth column to the
first column.  (i) What is the result?  The first column should consist of
zeros except for the last entry.  (ii) What is this last entry?  (iii) Compute
the determinant by expanding along the first column.  What do you get?  (iv)
What would you get if instead of starting with a~$4$-th degree equation, we
started with an~$n$-th degree equation?
\newpage

Let~$\lambda$ be an eigenvalue for~$A$, and consider the corresponding
eigenspace,
\[
  E_{\lambda}=\left\{ v\in F^4 : Av=\lambda v\right\}.
\]
\medskip

\textsc{problem 4.} Prove that
\[
  E_{\lambda}=\Span\left\{ (1,\lambda,\lambda^2,\lambda^3) \right\}.
\]
Thus,~$\dim E_{\lambda}=1$, i.e., the geometric multiplicity of~$\lambda$
is~$1$.  (Hint: let~$v=(v_{1},\ldots,v_{n})$ and then compare components on both
sides of the equation $Av=\lambda v$.)
\newpage

Suppose that~$A$ has~$k$ distinct eigenvalues~$\lambda_1,\dots,\lambda_k$
over~$\C$ with multiplicities~$m_1,\dots,m_k$, respectively.  So the
characteristic polynomial factors as~$p_A(x)=\prod_{i=1}^k(\lambda_i-x)^{m_i}$. 
\medskip

\textsc{Problem 5.} Why do we know that the Jordan form for~$A$ over the complex numbers is
\begin{center}
  \begin{tikzpicture}
    \node at (0,0) {$\displaystyle 
  \left(\begin{array}{cccc}
      J_{m_1}(\lambda_1)&&&\\
      &J_{m_2}(\lambda_2)&&\\
      &&\ddots&\\
      &&&J_{m_k}(\lambda_k)\\
  \end{array} \right)?
    $};
    \node at (-1.7,-0.5) {\huge 0};
    \node at (1.7,0.5) {\huge 0};
  \end{tikzpicture}
\end{center}
\vskip2.5in

\textsc{Problem 6.} Define the {\em basic functions} for equation~(\ref{eqn:
05friws1}) to be
\[
  \left\{ t^{j}e^{\lambda_i t}: 0\leq j< m_i, 1\leq i\leq k \right\}.
\]
Prove that every solution to equation~(\ref{eqn: 05friws1}) is a linear
combination of these basic functions.
\newpage

We would like to show that each of the basic functions is a solution to
equation~(\ref{eqn: 05friws1}).  Consider the differential operator~$D:=\frac{d}{dt}$.  We can write
equation~(\ref{eqn: 05friws1}) as
\[
  P(D)y=0
\]
where~$P(D)=x^4+a_{3}x^{3}+a_2x^2+a_1x+a_0$.  Further, we know (why?) that 
\[
  P(D)=\prod_{i=1}^k(D-\lambda_i)^{m_i}.
\]
\smallskip

\textsc{Problem 7.}\  
\begin{enumerate}[label={(\alph*)}]
    \item Prove by induction that for every sufficiently differentiable
      function~$f(t)$, we have
      \[
	(D-\lambda)^k(f(t)e^{\lambda t})=e^{\lambda t}D^kf(t)
      \]
      for~$k\geq 0$.
      \vskip3in
\newpage

  \item Show that it follows that
    \[
      P(D)(f(t)e^{\lambda t})=e^{\lambda t}P(D+\lambda)f(t).
    \]
    \vskip2in
  \item Use these results to show that each basic function is a solution to equation~(\ref{eqn: 05friws1}).
\end{enumerate} 
\newpage

Finally, we'd like to show that each solution to equation~(\ref{eqn:
05friws1}) with the given initial condition is a {\em unique} linear combination
of the basic functions.  To do so, list the basic functions in some
order~$f_1,f_2,f_3,f_4$. 
For each~$\alpha=(\alpha_1,\alpha_,\alpha_3,\alpha_4)\in\C^4$, consider the solution
\[
  s_{\alpha}(t)=\alpha_1f_1+\alpha_2 f_2+\alpha_3 f_3+\alpha_4 f_4,
\]
and, in general, define
\begin{align*}
  \phi_{\alpha}\colon\C^4&\to\C^4\\
  \alpha&\mapsto(s(0),s'(0),s''(0),s'''(0)).
\end{align*}
It's clear that~$\phi$ is linear (since differentiation and evaluation
are linear).
\medskip

\textsc{Problem 8.} You have already shown that is~$\phi$ surjective.  How?  Why
does it then follow that~$\phi$ injective?  How does this prove uniqueness?
\end{document}


