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\begin{document}
\centerline{Math 322 lecture for Monday, Week 5}
\bigskip

\centerline{\sc Stability}
\medskip

Let $A\in M_n(\C)$.  For each eigenvalue $\lambda\in\C$, the {\em generalized
eigenspace} for $\lambda$ is
\[
V_{\lambda}=\{v\in\C^n:(A-\lambda I)^kv=0\ \mbox{for some $k>0$}\}.
\]
We can choose bases $\mathcal{B}_{\lambda}$ for the
generalized eigenspaces resulting in a basis
$\mathcal{B}=\cup_{\lambda}\mathcal{B}_{\lambda}$ for $\C^n$ with respect to
which the matrix $A$ attains its Jordan form. Define the {\em stable}, {\em
center}, and {\em unstable} spaces for $A$ respectively by
\begin{align*}
  E^s&=
  \mathrm{Span}\,\displaystyle\cup_{\lambda:\mathrm{Re}(\lambda)<0}\mathcal{B}_{\lambda}\\[5pt]
  E^c&=
  \mathrm{Span}\,\displaystyle\cup_{\lambda:\mathrm{Re}(\lambda)=0}\mathcal{B}_{\lambda}\\[5pt]
  E^u&=
  \mathrm{Span}\,\displaystyle\cup_{\lambda:\mathrm{Re}(\lambda)>0}\mathcal{B}_{\lambda}.
\end{align*}
Since $\mathcal{B}$ is a basis, we can write
\[
\C^n=E^s\oplus E^c\oplus E^u,
\]
i.e., every $v\in\C^n$ can be written uniquely as $v=v_s+v_c+v_u$ where
$v_s\in E^s$, $v_c\in E^c$, and $v_u\in E^u$.

If~$A$ is a real matrix and we are working over the real numbers, then define
the (real) {\em stable}, {\em center}, and {\em unstable}  spaces for~$A$ by
intersecting each of~$E^s$,~$E^c$, and~$E^u$ with~$\R^n$.  Not that if~$A$ is
real, is nonreal eigenvalues will occur in conjugate pairs~$a\pm bi$, and the
conjugates have the same real part.  We can also adjust the
basis~$\mathcal{B}$ so that with respect to~$\mathcal{B}$, the matrix~$A$ has
its real Jordan form.

If $L:F^n\to F^n$ is a linear function and $W\subseteq F^n$, we say that $W$ is
{\em invariant under $L$} if $L(W)\subseteq W$.  If $M$ is the matrix
representing $L$, we similarly say that $W$ is invariant under~$M$ if $Mw\in W$
for all $w\in W$.
\bigskip

\textbf{Proposition.}
Each generalized eigenspace, the stable, center, and unstable spaces are
invariant under $A$ and under $e^{At}$ for all $t\in\R$.
\begin{proof}
 Staring at the Jordan form for $A$ and its exponential makes this result
 obvious, but we will give a formal proof.  First consider the action of $A$.
 Fix an eigenvalue~$\lambda$ for $A$ and consider the corresponding generalized
 eigenspace $V_{\lambda}$. Let $v\in V_{\lambda}$.  To show that~$Av\in
 V_{\lambda}$, we first let $w=(A-\lambda\, I)v$.
 We claim that $w\in V_{\lambda}$.  To see this, take~$k>0$ such that
 $(A-\lambda\,I)^kv=0$.  Then $(A-\lambda\, I)^{k-1}w=0$ (in the special case where~$k=1$, we
 have~$(A-\lambda\, I)^{0}w=(A-\lambda\, I)v =w=0\in V_{\lambda}$).  Since
 $v,w\in V_{\lambda}$ and~$V_{\lambda}$ is a subspace,
 \[
 Av=\lambda\,v+w\in V_{\lambda}.
 \]
 This shows that $V_{\lambda}$ is invariant under $A$.  Now since each of the
 stable, center, and unstable spaces is formed by taking the linear span of
 bases for certain generalized eigenspaces, it follows that each of these is
 invariant under $A$.  It follows that they are invariant under~$e^{At}$ by
 homework.  
\end{proof}

Thus, let $x(t)$ be the solution to the initial value problem $x'=Ax$,
$x(0)=x_0$, i.e., let $x(t)=e^{At}x_0$.  It follows that if $x_0\in E^s$, then
$x(t)\in E^s$ for all $t$.  The solution never leaves the stable space.
Similarly, a solution starting in the center or the unstable space never leaves
that space.  Further, from the Jordan form, one sees that
\begin{align*}
  x_0\in E^s\setminus\{0\}&\Longrightarrow\lim_{t\to\infty}x(t)=0\qquad\mbox{and}\quad 
 \lim_{t\to-\infty}|x(t)|=\infty\\
x_0\in E^u\setminus\{0\}&\Longrightarrow
\lim_{t\to\infty}|x(t)|=\infty\quad\mbox{and}\quad\lim_{t\to-\infty}x(t)=0.
\end{align*}
In particular, if all eigenvalues of $A$ have negative real part, then all
solutions, no matter what the initial condition, are drawn into the origin.  If
all eigenvalues have positive real part, all solutions with non-zero initial
condition will eventually leave any fixed compact set.
\bigskip

\centerline{\sc linear systems in~$\R^3$}
\medskip

\textbf{Linear systems in~$\R^3$.}  Let~$A\in M_3(\R)$. Then~$A$ either has three
real eigenvalues (counting multiplicities) or it has a single real eigenvalue
and pair of conjugate nonreal eigenvalues.  Therefore, the possibilities for the
Jordan form and for the solutions to~$x'=Ax$ up to a linear change of
coordinates are:

{\bf I.} $u,v,w\in\R$:
\[
J=
\left(\begin{array}{ccc}
  u&0&0\\
  0&v&0\\
  0&0&w
\end{array} \right)
\qquad
x(t) = e^{Jt}x_0=
\left(\begin{array}{ccc}
    e^{ut}&0&0\\
    0&e^{vt}&0\\
    0&0&e^{wt}
\end{array} \right)x_0.
\]
The behavior of the various trajectories will depend on the signs of~$u,v,w$,
with saddle-like behavior if they don't all have the same sign.

{\bf II.} $u,v\in\R$:
\[
J=
\left(\begin{array}{ccc}
  u&1&0\\
  0&u&0\\
  0&0&v
\end{array} \right)
\qquad
x(t) = e^{Jt}x_0=
\left(\begin{array}{ccc}
    e^{ut}&te^{ut}&0\\
    0&e^{ut}&0\\
    0&0&e^{vt}
\end{array} \right)x_0.
\]
{\bf III.} $u\in\R$:
\[
J=
\left(\begin{array}{ccc}
  u&1&0\\
  0&u&1\\
  0&0&u
\end{array} \right)
\qquad
x(t) = e^{Jt}x_0=
\left(\begin{array}{ccc}
    e^{ut}&te^{ut}&\frac{t^2}{2}e^{ut}\\
    0&e^{ut}&te^{ut}\\
    0&0&e^{ut}
\end{array} \right)x_0.
\]
{\bf IV.} $a,b,u\in\R$ and~$b\neq0$:
\[
J=
\left(\begin{array}{ccc}
  a&-b&0\\
  b&a&0\\
  0&0&u
\end{array} \right)
\qquad
x(t) = e^{Jt}x_0=
\left(\begin{array}{ccc}
    e^{at}\cos(bt)&-e^{at}\sin(bt)&0\\
    e^{at}\sin(bt)&e^{at}\cos(bt)&0\\
    0&0&e^{ut}
\end{array} \right)x_0.
\]
An interesting special case is where~$a=0$.

We will take a look at examples of all of these in class.
\bigskip

\centerline{\sc nonhomogeneous systems}
\medskip

\textbf{Proposition.}
Let~$A\in M_n(F)$ and consider the system
\[
  x'(t)=Ax(t) +b(t)
\]
where~$t\mapsto b(t)\in F^n$ is continuous.  The solution with initial
condition~$x_0$ is
\[
  x(t)=e^{At}x_0+e^{At}\int_{s=0}^te^{-As}b(s)\,ds.
\]
The solution is unique.
\begin{proof}
 Defining~$x(t)$ as above, use the product rule and the fundamental
 theorem of calculus to see
 \begin{align*}
   x'(t)
   &= \left( e^{At}x_0 \right)'+\left( e^{At}
   \right)'\int_{s=0}^te^{-As}b(s)\,ds+e^{At}\left( \int_{s=0}^te^{-As}b(s)\,ds
   \right)'\\
 &= Ae^{At}x_0+Ae^{At}\int_{s=0}^te^{-As}b(s)\,ds+e^{At}e^{-At}b(t)\\
 &= Ae^{At}x_0+Ae^{At}\int_{s=0}^te^{-As}b(s)\,ds+e^{At}e^{-At}b(t)\\
 &= A\left(e^{At}x_0+e^{At}\int_{s=0}^te^{-As}b(s)\,ds\right)+b(t)\\
 &= Ax(t)+b(t).
 \end{align*}
 Uniqueness of the solution will be a homework problem.
\end{proof}
\end{document}
