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\begin{document}
\centerline{Math 322 lecture for Monday, Week 6}
\bigskip

\centerline{\sc $n$-th order linear homogeneous equations revisited}
\medskip

Consider the homogeneous linear equation
\begin{equation}
  y^{(n)}+a_{n-1}y^{(n-1)}+\dots+a_1y'+a_0y=0
  \label{eqn: 05wedeq1}
\end{equation}
with initial condition~$y^{(i)}(0)=b_i$ for~$i=0,1,\dots,n-1$.  Recall the
method of solution introduced during the first two weeks of class.  First we
factor the characteristic polynomial
\[
  P(x)=\prod_{i=1}^k(x-\lambda_i)^{m_i},
\]
where the~$\lambda_i$ are distinct.  We claimed that the most general solution
was an arbitrary linear combination of the basic functions
\[
  e^{\lambda_i t},\ te^{\lambda_i t},\ \dots,\ t^{m_i-1}e^{\lambda_i t}
\]
for~$i=1,\dots,k$.  We also claimed that for each initial condition, there would
be a unique solution.  We now want to justify those claims.
\medskip

Define
\[
  x_1:=y,\ x_2:=y',\ x_3:=y'',\ \dots\ ,\ x_n:=y^{(n-1)}.
\]
We get a corresponding matrix equation
\begin{center}
  \begin{tikzpicture}
    \node at (0,0) 
    {$\displaystyle
  \left(\begin{array}{c}
      x_1'\\x_2'\\x_3'\\\vdots\\x_{n-1}'\\x_n'
  \end{array} \right)
  =
  \left(\begin{array}{cccccc}
     0&1&&&&\\
     &0&1&&&\\
     &&0&1&&\\
     &&&\ddots&\ddots& \\
     &&&&0&1\\
    -a_0&-a_1&-a_2&\dots&-a_{n-2}&-a_{n-1}
  \end{array} \right)
  \left(\begin{array}{c}
      x_1\\x_2\\x_3\\\vdots\\x_{n-1}\\x_n
  \end{array} \right).
    $};
    \node at (-2.5,-0.2) {\huge 0};
    \node at (2.3,0.7) {\huge 0};
  \end{tikzpicture}
\end{center}
Let~$A$ denote the~$n\times n$ matrix above, and we can consider the
differential equation $x'=Ax$ with initial
condition~$x_0=x(0)=(y(0),y'(0),\ldots,y^{(n-1)}(0))=(b_0,b_1,\dots,b_{n-1})$.  The solution is
\[
  x(t)=e^{At}x_0,
\]
and the first component of this solution is~$x_1(t)=y(t)$, the solution to the
original homogeneous system.
\medskip

\textbf{Proposition.}  Let~$P(x)=\sum_{i=0}^na_i x^i$ be the characteristic
polynomial for the linear homogeneous equation~\eqref{eqn: 05wedeq1}.  Then the
characteristic polynomial for~$A$ is
\[
  p_A(x):=\det(A-xI_n)=(-1)^{n+1}P(x).
\]

\begin{proof}
 We have
 \begin{center}
   \begin{tikzpicture}
     \node at (0,0) 
     {$\displaystyle
 A - xI_nA =
   \left(\begin{array}{cccccc}
      -x&1&&&&\\
      &-x&1&&&\\
      &&-x&1&&\\
      &&&\ddots&\ddots& \\
      &&&&-x&1\\
     -a_0&-a_1&-a_2&\dots&-a_{n-2}&-x-a_{n-1}
   \end{array} \right)
 $};
 \node at (-2,-0.2) {\huge 0};
 \node at (3.2,0.7) {\huge 0};
   \end{tikzpicture}
 \end{center}
 Multiply the second column by~$x$ and add it to the first column; then multiply
 the third column by~$x^2$ and add it to the first column; and so on.  This
 does not affect the determinant, and the rows of the first column are all~$0$
 now except the last, which is
 \[
   -a_0-a_1x-a_2x^2-\dots-a_{n-2}x^{n-2}-(x+a_{n-1})x^{n-1}=-P(x).
 \]
 It follows that
 \begin{center}
   \begin{tikzpicture}
     \node at (0,0) 
     {$\displaystyle
       \det(A-xI_n)=
       \det
   \left(\begin{array}{cccccc}
      0&1&&&&\\
      0&-x&1&&&\\
      0&&-x&1&&\\
      \vdots&&&\ddots&\ddots& \\
      0&&&&-x&1\\
      -P(x)&-a_1&-a_2&\dots&-a_{n-2}&-a_{n-1}-x
   \end{array} \right)
 $};
 \node at (-0.1,-0.3) {\huge 0};
 \node at (4.0,0.7) {\huge 0};
   \end{tikzpicture}
 \end{center}
 Expand along the first column to get the result.
\end{proof}

In order to solve the system, we are interested in the Jordan form
for~$A$.  So we think about this next.
\medskip

\textbf{Proposition.} Let~$\lambda$ be an eigenvalue for~$A$.  Then the
corresponding eigenspace is
\[
  E_{\lambda}=\Span\{(1,\lambda,\lambda^2,\dots,\lambda^{n-1})\}
\]
and is, hence, one-dimensional.  So the geometric multiplicity of each
eigenvalue for~$A$ is~$1$.

\begin{proof}
 Suppose that~$Av=\lambda v$ where~$v=(v_1,\dots,v_n)$.  Note that
 \[
 Av = 
   \left(\begin{array}{llllll}
      0&1&&&&\\
      &0&1&&&\\
      &&0&1&&\\
      &&&\ddots&\ddots& \\
      &&&&0&1\\
     -a_0&-a_1&-a_2&\dots&-a_{n-2}&-a_{n-1}
   \end{array} \right)
   \left(\begin{array}{c}
       v_1\\v_2\\v_3\\\vdots\\v_{n-1}\\v_n
   \end{array} \right)
   =
   \lambda
   \left(\begin{array}{c}
       v_1\\v_2\\v_3\\\vdots\\v_{n-1}\\v_n
   \end{array} \right)
   =\lambda v
 \]
 says $v_2=\lambda v_1,\ v_3=\lambda v_2,\ \dots\ ,\ v_n=\lambda v_{n-1}$.  Thus,
 \begin{align*}
   v 
   &=  (v_1,\lambda v_1,\lambda^2v_1,\dots,\lambda^{n-1} v_1)\\
   &=  v_1(1,\lambda,\lambda^2,\dots,\lambda^{n-1}).
 \end{align*}
\end{proof}

\textbf{Corollary.} Suppose that~$A$ has distinct
eigenvalues~$\lambda_1,\dots,\lambda_k$ (over~$\C$) with algebraic
multiplicities,~$m_1,\dots,m_k$, respectively, so the its characteristic
polynomial is
\[
  p_A(x)=\prod_{i=1}^k(\lambda_i-x)^{m_i}.
\]
Then the Jordan form for~$A$ 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).
    $};
    \node at (-1.7,-0.5) {\huge 0};
    \node at (1.7,0.5) {\huge 0};
  \end{tikzpicture}
\end{center}
\begin{proof} This follows immediately from the preceding Proposition.  The diagonal of
  the Jordan form consists of the eigenvalues of~$A$, repeated according to
  multiplicities.  For each Jordan block, there is a corresponding eigenvector
  for~$A$ (and several generalized eigenvectors).  If there where more than one
  Jordan block for a particular eigenvalue~$\lambda$, there would be more
  than one linearly independent eigenvector for~$\lambda$, and we've just seen
  that this cannot happen---each eigenspace has dimension~$1$.
\end{proof}

\textbf{Theorem.}
Suppose the roots for the characteristic
polynomial for equation~\eqref{eqn: 05wedeq1} or, equivalently, the eigenvalues
for~$A$ are~$\lambda_1,\dots,\lambda_k$ with multiplicities~$m_1,\dots,m_k$,
respectively.  Every solution to~equation~\eqref{eqn: 05wedeq1} (with a given
initial condition) is a unique linear combination of the {\em basic functions} 
\begin{equation}
  \left\{ t^{j}e^{\lambda_i t}: 0\leq j< m_i, 1\leq i\leq k \right\},
  \label{eqn: 05wedeq2}
\end{equation}
and each linear combination of these functions is a solution for some initial
condition.
\smallskip

{\bf Proof.} There are three parts to this proof: (i) show each solution is a
linear combination of the basic functions; (ii) show each basic function
satisfies the differential equation~\eqref{eqn: 05wedeq1}; and (iii) show the
basic equations are linearly independent.

(i) The solution to equation~\eqref{eqn: 05wedeq1} is the first component
of~$e^{At}x_0$.  Letting~$P^{-1}AP=J$ be the Jordan form for~$A$, the solution
is
\[
  y(t) = e^{At}x_0=Pe^{Jt}P^{-1},
\]
and hence, is a linear combination of the entries of~$e^{Jt}$.  The result then
follows from the previous corollary recalling that
\[
  e^{J_{m_i}(\lambda_i t)}=
  e^{\lambda_i t}
  \left(\begin{array}{cccccc}
      1&t&\frac{t^2}{2!}&\dots&\dots&\frac{t^{m_i-1}}{(m_i-1)!}\\[4pt]
      0& 1&t&\dots&\dots&\frac{t^{k-2}}{(m_i-2)!}\\[4pt]
      0& 0& 1&\dots&\dots&\frac{t^{k-3}}{(m_i-3)!}\\[4pt]
      &\ddots & &\vdots&\vdots&\vdots\\[4pt]
      0&\dots&\dots&0&1&t\\[4pt]
      0&\dots&\dots&\dots&0&1\\[4pt]
  \end{array} \right).
\]

(ii)  Consider the differential operator~$D:=\frac{d}{dt}$.  We can write
equation~\eqref{eqn: 05wedeq1} as
\[
  P(D)y=0
\]
where~$P(D)=x^n+a_{n-1}x^{n-1}+\dots+a_1x+a_0$.  We are given that
\[
  P(D)=\prod_{i=1}^k(D-\lambda_i)^{m_i}.
\]
That the basic functions satisfy the differential equation~$P(D)y=0$ is left as
homework.  It follows from two facts (which are part of the homework problem):
\begin{enumerate}
  \item $(D-\alpha)(D-\beta)f(t)=(D-\beta)(D-\alpha)f(t)$ for every
    sufficiently differentiable function~$f(t)$ and pair of constants~$\alpha$ and~$\beta$.
  \item $P(D)(f(t)e^{\lambda t})=e^{\lambda t}P(D+\lambda)(f(t))$ for every
    sufficiently differentiable function~$f(t)$ and constant~$\lambda$.
\end{enumerate} 

(iii) For uniqueness, list the~$n$ functions in~\eqref{eqn: 05wedeq2} in some
order~$f_1,\dots,f_n$, and consider the mapping~$\phi\colon\C^n\to\C^n$ defined
as follows: for each~$(\alpha_1,\dots,\alpha_n)\in\C^n$, consider the solution
\[
  s_{\alpha}(t)=\alpha_1f_1+\dots\alpha_n f_n,
\]
and let
\[
  \phi(\alpha_1,\dots,\alpha_n):=(s_{\alpha}(0),s_{\alpha}'(0),\dots,s_{\alpha}^{(n-1)}(0))\in\C^n.
\]
Since taking differentiation and evaluation are both linear operations,~$\phi$
is linear.  It is surjective since we know from part~(ii) that we can find a
solution as a linear combination of~$f_1,\dots,f_n$ for each initial condition.
Since~$\phi$ is linear and has rank~$4$, i.e., $\dim(\im\phi)=4$, the rank-nullity
theorem says that the kernel of~$\phi$ is trivial.  So~$\phi$ is injective. Now
take any two solutions that satisfy the same initial conditions.  Each of these
solutions is a linear combination of the basic functions, so they have the
form~$s_{\alpha}$ and~$s_{\beta}$ for some~$\alpha,\beta\in\C^n$.  Since they
satisfy the same initial condition, we have~$\phi(\alpha)=\phi(\beta)$.
Since~$\phi$ is injective, we have~$\alpha=\beta$. 
\end{document}


