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\begin{document}
\centerline{Math 322 lecture for Monday, Week 8}
\bigskip

\centerline{\sc stable manifold theorem}
\medskip

{\bf Review of stable, unstable, and center subspaces.} Consider the linear
system~$x'=Ax$ for some~$A\in M_n(\R)$.  Suppose that the generalized
eigenvectors and their corresponding eigenvalues for~$A$ are~$u_j+iv_j$
and~$\lambda_j=a_j+ib_j$, respectively, for~$j=1,\dots,n$.  Thus, putting these vectors as
columns in a matrix~$P$, we have~$P^{-1}AP=J$ where~$J$ is the Jordan form
of~$A$.  The generalized eigenvectors~$u_j+iv_j$ for which~$b_j\neq0$ come in
conjugate pairs since~$A$ is a real matrix.   Then the {\bf stable, unstable,
and center subspaces} for the system are, respectively, 
\begin{align*}
  E^{s}&:=\Span\left\{ u_j,v_j:a_j<0 \right\} \\ 
  E^{u}&:=\Span\left\{ u_j,v_j:a_j>0 \right\} \\ 
  E^{c}&:=\Span\left\{ u_j,v_j:a_j=0 \right\}.
\end{align*} 
Recall that up to a change of coordinates, the solution to the system
is~$e^{Jt}$ and that for a Jordan block corresponding to~$\lambda_j=a_j+ib_j$,
we can factor out~$e^{\lambda_jt}=e^{a_jt}(\cos(b_jt)+i\sin(b_jt))$, leaving a
matrix that is polynomial in~$t$:
\[
  e^{J_{\ell}(\lambda_j)t}
  = 
  e^{\lambda_j t}
  \left(\begin{array}{cccccc}
      1&t&\frac{t^2}{2!}&\dots&\dots&\frac{t^{\ell-1}}{(\ell-1)!}\\[4pt]
      0& 1&t&\dots&\dots&\frac{t^{\ell-2}}{(\ell-2)!}\\[4pt]
      0& 0& 1&\dots&\dots&\frac{t^{\ell-3}}{(\ell-3)!}\\[4pt]
      &\ddots & &\vdots&\vdots&\vdots\\[4pt]
      0&\dots&\dots&0&1&t\\[4pt]
      0&\dots&\dots&\dots&0&1\\[4pt]
  \end{array} \right).
\]
Thus, it is the signs of the~$a_j$ that
determine the long-term behavior of the system.
\medskip

\textbf{Theorem.} (Stable manifold theorem.)  Let~$E\subseteq\R^n$ and
let~$f\in C^1(E)$.  Suppose that~$f(0)=0$ and that~$Df_0$ has~$k$ eigenvalues
with negative real part and~$n-k$ eigenvalues with positive real part.
Let~$\phi$ be the flow for the system~$x'=f(x)$.  Then there exists a
~$k$-dimensional differentiable manifold~$S$ tangent to the stable
subspace~$E^s$ of the linearized system~$x'=Df_0(x)$ at~$0$ and there exists
an~$(n-k)$-dimensional differentiable manifold~$U$ tangent to the unstable
space~$E^u$ of the linearized system.  Further
\[
  \lim_{t\to\infty}\phi_t(p)=0
\]
for any~$p\in S$ and
\[
  \lim_{t\to-\infty}\phi(p)=0
\]
for any~$p\in U$.
\medskip

{\bf Remark.} To apply this theorem to an arbitrary equilibrium point~$x_0$,
make the change of coordinates~$x\mapsto x-x_0$, find the stable and unstable
manifolds at the origin, and translate back~$x\mapsto x+x_0$.

{\bf Example.} The system
\begin{align*}
  x'&= -x-y^2\\
  y'&= y+x^2
\end{align*}
has an equilibrium point at the origin.  The Jacobian
for~$f(x,y)=(-x-y^2,y+x^2)$ is
\[
  Jf(x,y)=
  \left(\begin{array}{cc}
      -1&-2y\\
      2x&1
  \end{array} \right).
\]  
Therefore,
\[
  Jf(0,0)=
  \left(\begin{array}{cc}
      -1&0\\
      0&1
  \end{array} \right)
\]
and the linearized system is
\begin{align*}
  x'&= -x\\
  y'&= y.
\end{align*}
(The linearized system in this case is easy to read off of the original system
  in this case since the equilibrium point is the origin and~$f$ has components
that are polynomials since~$f$ is it's own Taylor expansion at the origin.)

The main thing that concerns us, though, is that the eigenvalues
for~$Df_{(0,0)}$ are~$\pm1$.  The eigenspace for~$-1$ is spanned by~$(1,0)$,
i.e., the~$x$-axis, and
the eigenspace for~$1$ is spanned by~$(0,1)$, the~$y$-axis.  So a stable manifold for our
original system should be tangent to the~$x$-axis and an unstable manifold
should be tangent to the~$y$-axis at the origin.  Here is a picture of the flow of the vector
field~$f$:
\begin{center}
  \includegraphics[height=2in]{08mon1.png}
\end{center}
From the picture, we can see that the vector field is not tangent to the stable
and unstable spaces for the linearized system at the origin everywhere.  The stable manifold theorem is a statement
about what is happening locally, very close to the equilibrium point.  Below, we
zoom in on the origin:
\begin{center}
  \begin{tikzpicture}
    \node at (-5,0) {\includegraphics[height=1.5in]{08mon2.png}};
    \node at (0,0) {\includegraphics[height=1.5in]{08mon3.png}};
    \node at (5,0) {\includegraphics[height=1.5in]{08mon4.png}};
  \end{tikzpicture}
\end{center}

\begin{proof}[Sketch of proof of the stable manifold theorem]  The proof of the stable
 manifold theorem, like the proof of the fundamental existence and uniqueness
 theorem can be done by the method of successive approximations.  
 
 We start with some ``pre-processing'':  As mentioned above, if the equilibrium
 point~$x_0$ is not the origin, first replace~$x$ by~$x-x_0$.  Suppose that has
 been done.  Second, write
 \[
   x' = f(x) = Jf(0)x+\left( f(x)-Jf(0)x \right).
 \]
 Defining~$F(x):=f(x)-Jf(0)x$, our system becomes
 \[
   .
 \]
 Third, choose an~$n\times n$ real matrix~$P$ such that
 \[
   P^{-1}Jf(0)P
   =
   \left(\begin{array}{cc}
       A&0\\
     0&B
   \end{array} \right)
 \]
 where~$A$ has~$k$ eigenvalues with negative real parts and~$B$ has~$n-k$
 eigenvalues with positive real parts.  Finally, make the change of
 variables~$y=P^{-1}x$.  Then
 \begin{align*}
   x' = Jf(0)x+F(x)
   &\quad\Rightarrow\quad Py'=Jf(0)Py+F(Py)\\[5pt]
   &\quad\Rightarrow\quad y'=P^{-1}Jf(0)Py+P^{-1}F(Py).
 \end{align*}
 Define~$G(y)=P^{-1}F(Py)$ to get the system
 \begin{equation}
 y' = 
 \left(\begin{array}{cc}
     A&0\\0&B
 \end{array} \right)y
 +G(y).
   \label{eqn:08mon}
 \end{equation}
 Each step we've made is reversible.  So solving this system is equivalent to
 solving the original system.
 
 We now find stable and unstable manifolds through the method of successive
 approximations.  Define
 \[
   U(t) := 
   \left(\begin{array}{cc}
       e^{At}&0\\0&0
   \end{array} \right)
   \quad\text{and}\quad 
   V(t) := 
   \left(\begin{array}{cc}
       0&0\\0&e^{Bt}
   \end{array} \right)
 \]
 so that
 \[
   e^{\left(\begin{smallmatrix}
       A&0\\0&B
 \end{smallmatrix} \right)t}=U(t)+V(t).
 \]
 For~$t\in\R$ and~$a\in\R^n$, define an operator~$T$ on $\R^{n}$-valued functions~$u$ with
 domain in a region near the origin in~$\R\times \R^{n}$ by
 \begin{equation}
   (Tu)(t,a):=
   U(t)a+\int_{s=0}^tU(t-s)G(u(s,a))\,ds-\int_{s=t}^{\infty}V(t-s)G(u(s,a))\,ds.
   \label{eqn:08mon_approx}
 \end{equation}
 Now use the method of successive approximations starting
 with 
 \[
   u^{(0)}(t,a)=0\in\R^{n}.
 \]
 Calculations like those we did for the proof of the fundamental existence and
 uniqueness theorem show the approximations~$u^{(m)}(t,a)$ converge to a
 fixed point~$u(t)$ of~$T$ for~$t$ in a small interval about the origin and
 for~$a$ restricted to a sufficiently small neighborhood of the origin
 in~$\R^n$.  
 
 A stable manifold for equation~\eqref{eqn:08mon} is given as the set of points
 \[
   \left( a_1,\dots,a_k,u_{k+1}(0,a_1,\dots,a_k,0,\dots,0),\dots, u_{n}(0,a_1,\dots,a_k,0,\dots,0)\right)
 \]
 as~$(a_1,\dots,a_k)$ varies in a neighborhood of the origin in~$\R^k$.  We get a
 stable manifold for the original system by applying~$P$ to these points,
 since~$y=P^{-1}x$, then
   translating back~$x\mapsto x+x_0$, if the original equilibrium point was not
 the origin.
 
 To find an unstable manifold, replace~$t$ by~$-t$ to get the system
 \[
 y' = 
 -\left(\begin{array}{cc}
     A&0\\0&B
 \end{array} \right)y
 -G(y),
 \]
 since~$(y(-t))'=-y'(-t)$.  However, now note that~$-A$ has~$k$ positive
 eigenvalues and~$-B$ has~$n-k$ negative eigenvalues, so to apply the above
 argument, we need to swap
 coordinates~$\phi\colon y\mapsto(y_{k+1},\dots,y_n,y_1,\dots,y_k)$ to get the
 system
 \[
   (\phi(y))' = 
 \left(\begin{array}{cc}
     -B&0\\0&-A
 \end{array} \right)\phi(y)
 -G(\phi(y)),
 \]
 apply the method of successive approximations, then swap back by applying~$\phi^{-1}$ to the
 points in the resulting manifold.
\end{proof}

As evidence for the reasonableness of the method presented in the sketch above
suppose that~$y(t)$ is a solution to equation~\eqref{eqn:08mon_approx} with (i) initial
condition~$y(0)$ close to~$0$ and such that (ii) $y(t)$ is bounded
as~$t\to\infty$.  We will show that~$y$ must satisfy equation~\eqref{eqn:08mon_approx} (and thus
will be a fixed point of the iterative process). Let
\[
M := 
\left(\begin{array}{cc}
    A&0\\
    0&B
\end{array} \right)
\]
so that the system becomes
\[
  y' = My+G(y).
\]
Then
\allowdisplaybreaks
\begin{align*}
   y' = My+G(y)
   &\quad\Rightarrow\quad e^{-Mt}y'=e^{-Mt}My+e^{-Mt}G(y)\\[5pt]
   &\quad\Rightarrow\quad e^{-Mt}y'=Me^{-Mt}y+e^{-Mt}G(y)\\[5pt]
   &\quad\Rightarrow\quad e^{-Mt}y'-Me^{-Mt}y=e^{-Mt}G(y)\\[5pt]
   &\quad\Rightarrow\quad (e^{-Mt}y)'=e^{-Mt}G(y)\\[5pt]
   &\quad\Rightarrow\quad \int_{s=0}^{t}(e^{-Ms}y(s))'\,ds=\int_{s=0}^te^{-Ms}G(y(s)\,ds\\[5pt]
     &\quad\Rightarrow\quad
     e^{-Mt}y(t)-y(0)=\int_{s=0}^te^{-Ms}G(y(s))\,ds\\[5pt]
   &\quad\Rightarrow\quad
   y(t)-e^{Mt}y(0)=\int_{s=0}^te^{M(t-s)}G(y(s))\,ds\\[5pt]
   &\quad\Rightarrow\quad y(t)=e^{Mt}y(0)+\int_{s=0}^te^{M(t-s)}G(y(s))\,ds\\[10pt]
   &\quad\Rightarrow\quad y(t)= (U(t)+V(t)) y(0)\\[8pt]
   &\qquad\qquad\qquad+\int_{s=0}^t (U(t-s)+V(t-s)) G(y(s))\,ds\\[8pt]
   &\quad\Rightarrow\quad y(t)= (U(t)+V(t))y(0)+\int_{s=0}^t U(t-s)G(y(s))\,ds\\[8pt]
 &\qquad\qquad\qquad+\int_{s=0}^{\infty}V(t-s)) G(y(s))\,ds
 -\int_{s=t}^{\infty}V(t-s)) G(y(s))\,ds\\[8pt]
\end{align*}
To see that the integrals here are all bounded, first note that since~$y$ is
bounded as $t\to\infty$ (by assumption) and~$G$ is continuous, we have that~$G(y(s))$ is 
bounded as $s\to\infty$.  Next note that since the real part of the eigenvalues of~$B$ are
positive,~$V(t-s)$ is bounded as~$s\to\infty$ (recall that
$V(t-s)=\left(\begin{smallmatrix}0&0\\0&e^{B(t-s)}\end{smallmatrix}\right)$).
Continuing,
\begin{align*}
y(t)&=  (U(t)+V(t))y(0)+\int_{s=0}^t U(t-s)G(y(s))\,ds\\[8pt]
 &\qquad\qquad\qquad+\int_{s=0}^{\infty}V(t-s)) G(y(s))\,ds
 -\int_{s=t}^{\infty}V(t-s)) G(y(s))\,ds\\[8pt]
&\Rightarrow
y(t)=
U(t)y(0)+V(t)\left(y(0)+\int_{s=0}^{\infty}V(-s)G(y(s))\,ds\right)\qquad(\star)\\[8pt]
 &\qquad\qquad\qquad+\int_{s=0}^t U(t-s)G(y(s))\,ds
 -\int_{s=t}^{\infty}V(t-s)) G(y(s))\,ds\\[8pt]
\end{align*}
Consider the above equation.  On the left, we have~$y(t)$, which is bounded
as~$t\to\infty$.  One the right, considering the eigenvalues of~$A$
and~$B$ we see that first, third, and fourth summands are bounded
as~$t\to\infty$.  This implies that
\[
V(t)\left(y(0)+\int_{s=0}^{\infty}V(-s)G(y(s))\,ds\right)
\]
is bounded as~$t\to\infty$.  But recall that
\[
  V(t)=
  \left(\begin{array}{cc}
      0&0\\
      0&e^{Bt}
  \end{array} \right)
\]
where~$B$ has~$n-k$ eigenvalues, each with positive real parts.  Since
\[
  y(0)+\int_{s=0}^{\infty}V(-s)G(y(s))\,ds
\]
is bounded (in fact, constant), this means that
\[
V(t)\left(y(0)+\int_{s=0}^{\infty}V(-s)G(y(s))\,ds\right) =0
\]
(Note that the above equation is a product of two matrices.  So we cannot
conclude that either of the factors is the zero matrix.)
From equation~$(\star)$, above, it follows that
\[
y(t)= U(t)y(0)+\int_{s=0}^t U(t-s)G(y(s))\,ds -\int_{s=t}^{\infty}V(t-s))
G(y(s))\,ds,
\]
as we wanted to show.
\end{document}
