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

\centerline{\sc nonlinear systems}
\medskip

Let~$E\subseteq\R^n$ be an open subset of~$\R^n$ and let $C(E)$ denote the
vector space of continuous functions of the form~$E\to\R^{n}$.  Given~$f\in
C(E)$, we are now interested in solutions to the differential equation
\begin{equation}
  x'=f(x).
  \label{eqn: 06wed1}
\end{equation}

The function~$f$ is a vector field in $\R^{n}$ defined on~$E$.
We have just finished studying the linear case of this problem, i.e., in
which~$f(x)=Ax$ for some~$A\in M_n(\R^n)$ and are now particularly interested in
the case where~$f$ is no longer a linear function.

A \emph{solution} to equation~\eqref{eqn: 06wed1} on an interval~$I$ is a
function~$x\colon I\to E\subseteq\R^n$ such that
\[
  x'(t)=f(x(t))
\]
for all~$t\in I$.  Given~$t_0\in I$ with~$x(t_0)=x_0\in E$, we say the solution
satisfies the {\em initial value problem}
\begin{align*}
  x'&= f(x)\\
  x(t_0)&= x_0
\end{align*}
on~$I$.
\medskip

\textbf{Example.} Consider the (non-linear) system
\begin{align*}
  x' &= x^2-y\\
  y' &= xy
\end{align*}
with initial value $(x(0),y(0))=(0.5,1)$.  So in this case, the relevant vector
field is $f(x,y)=(x^{2}-y,xy)$. Here is a plot of the vector field and the solution
to the initial value problem:
\begin{center}
  \begin{tikzpicture}
    \node at (0,0) {\includegraphics[height=2in]{vectorfield06wed.png}};
  \end{tikzpicture}
\end{center}
Note that this system displays behavior one would not see in the linear case.
\medskip


The systems we are studying are called {\em autonomous} since~$f$ is a
function of~$x\in\R^n$ and not~$t$.  However, a {\em nonautonomous} system
\[
  x'=g(x,t)
\]
can be converted to an autonomous system by letting~$x_{n+1}:=t$
and~$x'_{n+1}=1$.
\medskip

{\bf Goals.}  Our first main goal is to find conditions under which the initial
value problem for equation~\eqref{eqn: 06wed1} has a unique solution.  After
that, we'll discuss how solutions change if~$f$ changes a small amount and
discuss the size of the interval on which a solution exists.
\medskip

{\bf New behavior.} In the linear case,~$x'=Ax$ and~$x(0)=x_0$, we saw that
there is always a unique solution.  That's no longer generally true in the
nonlinear case.  For instance, the following initial value problem
\begin{align*}
  x'&= 3x^{2/3}\\ 
  x(0)&= 0
\end{align*}
has two solutions:~$x(t)=0$ and~$x(t)=t^3$.  We'll see that the source of
non-uniqueness here is that~$f(x)=3x^{2/3}$ is not continuously differentiable:
~$f'(x)=2x^{-1/3}$, which is not continuous at~$0$.

Even if~$f'$ is continuous everywhere, the solution may only exist on
subintervals of the real line, again unlike the linear situation.  For example,
consider the system
\begin{align*}
  x'&= x^2\\
  x(0)&= 1.
\end{align*}
The solution is
\[
  x(t)=\frac{1}{1-t}
\]
but is only defined on the interval~$(-\infty,1)$.  The solution blows up
as~$t\to1^{-}$.
\medskip

\textbf{Key idea.} We have solved the initial value problem for
equation~\eqref{eqn: 06wed1} if we can find a continuous function~$x(t)$
satisfying
\[
  x(t)=x_0+\int_{s=0}^tf(x(s))\,ds
\]
for all~$t\in[-a,a]$ for some~$a>0$.

Check: First, by the fundamental theorem of calculus
\[
  x'(t)=(x_0)'+\left(\int_{s=0}^tf(x(s))\,ds\right)'=0+f(x(t))=f(x(t)).
\]
Next,
\[
x(0)=x_0+\int_{s=0}^0f(x(s))\,ds=x_0.
\]
The {\em method of successive approximations} attempts to create a sequence of
functions~$(u_k(t))_{k\geq0}$ converging to a solution:
\begin{align*}
  u_0(t)&:=x_0\\
  u_{k+1}(t)&:=x_0+\int_{s=0}^tf(u_{k}(s))\,ds,\quad \text{for~$k\geq0$}.
\end{align*}
\medskip

{\bf Example.} Consider the initial value problem
\[
  x'=xt,\quad x(0)=1.
\]
This is an autonomous equation, so we first convert it to a nonautonomous system
by letting~$x_1=x$ and~$x_2=t$.  The system becomes
\[
  \left(\begin{array}{c}
      x_1'\\x_2'
  \end{array} \right)
  =
  \left(\begin{array}{c}
      x_1x_2\\1
  \end{array} \right)
  =: f(x_1,x_2)
\]
with initial condition~$x_1(0)=x(0)=1$ and~$x_2(0)=0$ (since~$x_2=t$).  

Apply the method of successive approximations starting with
\[
  u_{0}(t)=\left(\begin{array}{c}
      x_1(0)\\x_2(0)
  \end{array} \right)
  =
  \left(\begin{array}{c}
      1\\0
  \end{array} \right).
\]
We get
\begin{align*}
  u_1(t)
  &= 
  \left(\begin{array}{c}
      1\\0
  \end{array} \right)
  +
  \int_{s=0}^tf(u_0(s))\,ds\\[8pt]
  &= 
  \left(\begin{array}{c}
      1\\0
  \end{array} \right)
  +
  \int_{s=0}^t\left(\begin{array}{c}
      0\\1
  \end{array} \right)\,ds\\[8pt]
  &= 
  \left(\begin{array}{c}
      1\\0
  \end{array} \right)
  +
  \left.\left(\begin{array}{c}
      c\\s
  \end{array} \right)\right|_{s=0}^t\\[8pt]
  &= 
  \left(\begin{array}{c}
      1\\0
  \end{array} \right)
  +
  \left(\begin{array}{c}
      0\\t
  \end{array} \right)
  =
  \left(\begin{array}{c}
      1\\t
  \end{array} \right).
\end{align*}
Next,
\begin{align*}
  u_2(t)
  &= 
  \left(\begin{array}{c}
      1\\0
  \end{array} \right)
  +
  \int_{s=0}^tf(u_1(s))\,ds\\[8pt]
  &= 
  \left(\begin{array}{c}
      1\\0
  \end{array} \right)
  +
  \int_{s=0}^tf(1,s)\,ds\\[8pt]
  &= 
  \left(\begin{array}{c}
      1\\0
  \end{array} \right)
  +
  \int_{s=0}^t\left(\begin{array}{c}
      s\\1
  \end{array} \right)\,ds\\[8pt]
  &= 
  \left(\begin{array}{c}
      1\\0
  \end{array} \right)
  +
  \left.\left(\begin{array}{c}
      s^2/2\\s
  \end{array} \right)\right|_{s=0}^t\\[8pt]
  &= 
  \left(\begin{array}{c}
      1+t^2/2\\t
  \end{array} \right).
\end{align*}
Next,
\begin{align*}
  u_3(t)
  &= 
  \left(\begin{array}{c}
      1\\0
  \end{array} \right)
  +
  \int_{s=0}^tf(u_2(s))\,ds\\[8pt]
  &= 
  \left(\begin{array}{c}
      1\\0
  \end{array} \right)
  +
  \int_{s=0}^tf(1+s^2/2,s)\,ds\\[8pt]
  &= 
  \left(\begin{array}{c}
      1\\0
  \end{array} \right)
  +
  \int_{s=0}^t\left(\begin{array}{c}
      s+s^3/2\\1
  \end{array} \right)\,ds\\[8pt]
  &= 
  \left(\begin{array}{c}
      1\\0
  \end{array} \right)
  +
  \left.\left(\begin{array}{c}
      s^2/2+s^2/(2\cdot 4)\\s
  \end{array} \right)\right|_{s=0}^t\\[8pt]
  &= 
  \left(\begin{array}{c}
      1+t^2/2+t^4/(2\cdot 4)\\t
  \end{array} \right).
\end{align*}
Similarly,
\[
  u_{4}=\left(\begin{array}{c}
      1+t^2/2+t^4/(2\cdot 4)+t^6/(2\cdot4\cdot 6)\\t
  \end{array} \right),
\]
and so on.  Recall that~$x_1=x$, and~$x$ is the function we are trying to
find.  Thus, we are interested in the limit of the first components of
the~$u_{k}$.  The method of successive approximations is delivering
\begin{align*}
  x(t)
&= 1+\frac{t^2}{2}+\frac{t^4}{2\cdot 4}+\frac{t^6}{2\cdot4\cdot 6}
+\frac{t^8}{2\cdot4\cdot 6\cdot 8}+\dots\\[8pt]
&= 1+\frac{t^2}{2}+\frac{t^4}{(1\cdot2)2^2}+\frac{t^6}{(1\cdot2\cdot
3)2^3}+\frac{t^8}{(1\cdot2\cdot 3\cdot 4)2^3}+\dots\\[8pt]
&= 1+\frac{t^2}{2}+\frac{1}{2!}\left( \frac{t^2}{2} \right)^2
+\frac{1}{3!}\left( \frac{t^2}{2} \right)^3
+\frac{1}{4!}\left( \frac{t^2}{2} \right)^4+\dots\\[8pt]
&= e^{t^2/2},
\end{align*}
which converges, and it's easy to check that it satisfies the original initial
value problem:
\[
  x'(t)=\left( e^{t^2/2} \right)'=te^{t^2/2}=x(t)t,
\]
and~$x(0)=1$.

Of course, we could have solved the equation through separation of variables:
\[
x'=xt
\quad\Rightarrow\quad\int\frac{dx}{x}=\int t\,dt
\quad\Rightarrow\quad\ln(x)=t^2/2 + c.
\]
Then~$x(0)=1$ implies~$c=0$.  Exponentiate:
\[
  \ln(x)=t^2/2\quad\Rightarrow\quad x=e^{t^2/2}.
\]
\medskip

{\bf Fixed points.} Consider the operator on functions,~$u\to T(u)$ given by
\[
  T(u)(t):=x_0+\int_{s=0}^tf(u(s))\,ds
\]
In the case we just considered, with~$x_0=(1,0)$ and~$f(x_1,x_2)=(x_1x_2,1)$, the method of
successive iterations produced the function~$u(t)=(e^{t^2/2},t)$.  This
function~$u$ is a {\em fixed point} for the operator~$T$:
\begin{align*}
  T(u(t))
  &= \left(\begin{array}{c}
      1\\0
  \end{array} \right)
  +
  \int_{s=0}^t f(u(s))\,ds\\[8pt]
  &= \left(\begin{array}{c}
      1\\0
  \end{array} \right)
  +
  \int_{s=0}^t f(e^{s^2/2},s)\,ds\\[8pt]
  &= \left(\begin{array}{c}
      1\\0
  \end{array} \right)
  +
  \int_{s=0}^t \left(\begin{array}{c}
      se^{s^2/2}\\1
  \end{array} \right)\,ds\\[8pt]
  &= \left(\begin{array}{c}
      1\\0
  \end{array} \right)
  +
  \left.\left(\begin{array}{c}
      e^{s^2/2}\\s
  \end{array} \right)\right|_{s=0}^t\\[8pt]
  &= \left(\begin{array}{c}
      1\\0
  \end{array} \right)
  +
  \left(\begin{array}{c}
      e^{t^2/2}-1\\t
  \end{array} \right)\\[8pt]
  &= 
  \left(\begin{array}{c}
      e^{t^2/2}\\t
  \end{array} \right)\\[8pt]
  &= u(t).
\end{align*}
\medskip

{\bf Next step.} We have seen that the method of successive approximations
amounts to iterating a operator on a space of functions and converging to a
fixed point for that operator. Our next step is to consider this situation a
little more generally.  Let~$X$ be a space in which convergence makes sense, and
consider a mapping~$T\colon X\to X$.  We would like to know conditions under
which iterates of a point~$x_0\in X$ under~$T$ will converge to a point~$\tilde{x}\in X$ that
is fixed under~$T$, i.e., such that~$T(\tilde{x})=\tilde{x}$.
\end{document}
