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\title{Numerical Mathematics II \\ SS 2019}
\author{}

\begin{document}
\maketitle
\tableofcontents
\newpage
\section{Basic Facts on Ordinary Differential Equations}
\begin{definition}

	An ODE of first order in some interval $I\subset \R$ is an equation of the form
\[
y'(t)=f(t,y(t)),~ t \in I
\]
where $y: I \to \C^n ,~~  y\in C^1(I)$ and $f: I\times\C^n\to\C^n$. The order is the highest derivative in the ODE. We call an ODE \textbf{explicit} if we can solve it for $y'$ and \textbf{implicit} otherwise.
\end{definition}

\begin{definition}
An ordinary differential equation of order $n$ is given as
\[
y^{(n)}(t) = f(t,y(t),y'(t),\dots ,y^{(n-1)}(t))
\]
for $t \in I\subset \R$ where $y$ is a $n$-times differentiable function on $I$ and $f:I\times(\C^n)^n\to \C^n$ is a function.

A solution $y$ of an ODE on some $J\subset I$ is a (multiple) continuously differentiable function $y: I\to \C^n$ which solves the ODE
\end{definition}
\begin{remark}
	An ODE of order $n$ can be transferred to an ODE of first order by transformation.
\end{remark}

\begin{definition}
	We call an ODE
 \[
 y^{(n)}(t)=f(t,y(t),y'(t),\dots ,y^{(n-1)}(t)), t \in I
 \]
 an \textbf{initial value problem (IVP)} for $y$ if we have additionally the constraints $y(t_0)=y_0, \dots, y^{(n-1)}(t_0)=y_{n-1}$ for $t_0 \in I$.
\end{definition}

\begin{remark}
An ODE has a swarm of solutions, IVP has specific solutions. The swarm of solutions with all constraints is called general solution.
\end{remark}


\begin{theorem}[Picard-Lindelöf]
	For $t_0\in\R,~ y_0\in\R^n,~a,b>0$ we set \[
	I=[t_0 -a, t_0+a] \und Q=\{z\in\C^n ~|~ \norm{z-y_0}_\infty \leq b\}.\] Let furthermore $F:I\times Q\to\C^n$ be continuous, with bounded components by some constant $R$ and Lipschitz-continuous in the second argument, i.e.
	\[
	\big|F_j(t,u)-F(t,v)\big| \leq L \sum_{k=1}^n |u_k - v_k|, ~~j=1,\dots,n~,~~ t\in I,~~u,v \in Q.
	\]
	Then the IVP $y'(t)=F(t,y(t)),~ y(b)=y_0$ has on $J=[t_0-\alpha, t_0 + \alpha]\subset I$ with $\alpha=\min\{a,\frac b R\}$ exactly one differentiable solution.
\end{theorem}
\begin{proof}
No Proof.
\end{proof}


\begin{remark}
The existence is local around $t_0$.
\end{remark}

\begin{definition}
The system $y'(t)=A(t)y(t)+f(t)$ for some interval $I\subset\R$ with $A(t)=(a_{ij}(t))_{ij}\in\C^{n,n},~ a_{ij}:I\to\C \text{ for } i,j \in \{1,\dots, n\},~n\in\N, ~y: I\to \C^n$ and $f:I\to \C^n$ is a linear system of ODEs.

The function $f$ is called inhomogeneity.

The system is called homogenous if $f=0$ and inhomogeneous otherwise.
\end{definition}

\begin{theorem}
	Let $y_1,y_n$ be two solutions of the homogeneous system
\end{theorem}

$ \int_{\varepsilon}^{\text{deine mudda}}\circ ~d \mu = [\text{abc},	\text{def}]$

\setcounter{subsection}{2}
\subsection{Qualitative Behaviour of ODEs}
\begin{example}
Let us consider the $n$-dimensional non contonomous system of first order
\begin{align*}
	y'(t)&=f(t,y(t))\\
	y(t_0)&=y_0
\end{align*}
where $f: D \to \R^n, ~D\subset I\times \R^n,~ t_0 \in I,~ I\subset \R$. The questions we are dealing with are:
\begin{enumerate}
\item Why only first order?
\item What is the relation between a non-autonomous and an autonomous system?
\end{enumerate}
\end{example}

The reason behind 1. is that any ODE of $n$-th order can be transformed into a $n$-dimensional ODE of first order. Consider the ODE
\[
x^{(n)}=F(t,x(t),x'(t),\dots,x^{(n-1)}(t))
\]
and define a vector $y$ with its components $y_i, ~ i=1,\dots,n$ by
\[
y_i(t)=x^{(i-1)}(t)
\]
and a vector field $f(t,y)$ by
\[
f(t,y) = \Big(t,y_1,y_2,\dots,y_n,F(t,y_1,y_2,\dots,y_n)\Big)^T
\]
Then the ODE of $n$-th order is equivalent to $y'(t)=f(t,y)$.\newline

A system of the form $y'(t)=f(t,y)$ is called an non-autonomous system, a system of the form $y=f(y)$ is called autonomous. We can transform a non-autonomous system to an autonomous system.

Consider the ODE
\[
y'(t)=f(t,y) \und y(t_0) = y_0.
\]
We set
\[
z=\begin{bmatrix} y \\ s \end{bmatrix} \und \hat{f}=\begin{bmatrix} f(s,y) \\ 1 \end{bmatrix}, ~~ s\in \R
\]
Then
\[
z'(t)=\hat{f}(z(t)), ~ z(t_0)=z_0=\begin{bmatrix} y_0\\t_0 \end{bmatrix}
\]
is an autonomous system.

In short, each ODE in $\R^n$ can be transformed to an autonomous ODE in $\R^{n+1}$

\begin{remark}
In the theorem of Picard-Lindelöf the ODE is of the form $f(t,y)$, where $y$ has to be Lipschitz-continuous.

In the autonomous system the right hand side looks like $f(y(t))$, where $t$ and $y$ have to be Lipschitz-continuous.
\end{remark}

\subsubsection*{Analytic Continuation}
"Local solutions can be spread onto a maximum time interval."\newline

\setcounter{theorem}{13}
\begin{definition}[Local Lipschitz]
A function $f: X \to Y$ is local Lipschitz in $x\in X$ if the exists a neighbourhood $U_x \subseteq  X$ around $x$ such that $f|_{U_x}$ is Lipschitz-continuous.
\end{definition}

For $G:=I\times Q$ with $I=[t_0-a, t_0+a], ~ Q=\{z\in\C~|~ \norm{z-y_0}\leq b\}$ with $a,b>0$ the theorem of Picard-Lindelöf gives for local Lipschitz $f$ the existence of a solution $y_0(t)$ of the IVP
\begin{align}
	y'(t)&=f(t,y(t)) \tag{1.4}\\
	y(t_0)&=y_0\notag
\end{align}
on some (small) inverval $I_0=[t_0-a_0,t_0+a_0]$ with $a_0=a>0$.

We will have a look at what happens if we apply the theorem of Picard-Lindelöf on one side of the interval $I_0$. Let now be $t_1:=t_0+a_0$ and $y_1 = y_0(t_1)$. We then have that $(y_1,t_1)\in G$ and according to Picard-Lindelöf we know that the IVP with $y(t_1)=y_1$ has a unique solution $y_1(t)$ on $I_1:=[t_1-a_0,t_1+a_1]$ where $a_1>0$.

Due to the uniqueness of the solution if hold $y_0(t)=y_1(t)$ on $I_0\cap I_1$ we are defining a continuoation of our solution on the greater interval.

It holds
\[
y_{+}(t)=y_0(t) ~\text{for}~ t\in[t_0,t_1]
\]
and
\[
y_{+}(t)=y_1(t) ~\text{for}~ t \in (t_1,t_1+a_1]
\]
analogue for $y_{-}(t)$. Thus there exists a unique solution on the interval $[t_0, t_0+a_0+a_1+\dots]$ if $\sum_{k=0}^{\infty}a_k<\infty$. If $\sum_{k=0}^{\infty}a_k$ diverges, the solution exists globally in forward time.

\begin{remark}
It can happen that $a_n$ can arbitrary small when $(t_k,y_{+}(t_k))$ approaches the boundary of $G$. Then either $\norm{f((t_k),y_{+}(t_k))}$ or the Lipschitz-constant $L$ might get arbitrary large.
\end{remark}

\begin{definition}
Let $f:G \to \R^n$ be continuous and local Lipschitz with respect to $y$ and let $(t_0,y_0)\in G$. Let furthermore $t_{\pm}:=t_{\pm}(t_0,y_0)\in\R$ be defined as
\begin{align*}
t_{+}&=\sup \{\tau>t_0 ~|~ \text{there exists a continuation}~ y_+ \text{ of (1.4) on } [t_0,\tau]\}\\
t_{-}&=~\inf \{\tau>t_0 ~|~ \text{there exists a continuation}~ y_- \text{ of (1.4) on } [t_0,\tau]\}.
\end{align*}
The interval $(t_-,t_+)$ is the largest interval of existence of the IVP with some initial point $y(t_0)=y_0$.

The maximum solution $y(t)$ is
\[
y(t)=\begin{cases} y_+(t) \text{ for } t\in[t_0,t_+) \\ y_-(t) \text{ for } t\in (t_-,t_0].\end{cases}
\]
\end{definition}

\begin{example}
Consider
\[
y'=y^2, ~ y(t_0)=y(0)=1, ~ y(t)=\frac{1}{1-t}.
\]
Then we have $(t_-,t_+)=(-\infty, 1) \text{ or } (1,\infty)$.
\end{example}

\begin{remark}
In case of $t_+<\infty$ the maximum solution approches for $t\to t_+$, it can then happen that $\norm{y(t)}$ is unbounded. This is also called "blow up".
\end{remark}

\subsubsection*{Solutions and Initial Data}
"What is the influence of a perturbation in $f,~y_0$ or $t_0$ on the solution?"\newline

To consider this, we need the following Lemma.
\begin{lemma}[Grönwall-Lemma]
Let $I=[a,b]\subseteq \R$ and $g:I\to\R$ be a continuous function. If
\[
0\leq g(t)\leq \delta+\gamma \int_{a}^{t }g(x)~dx
\]
holds for all $t\in I, ~\delta,\gamma >0$, then it holds
\[
g(t)\leq\delta e^{\gamma(t-a)}.
\]
\end{lemma}
\begin{proof}
We set
\[
\varphi(t)=\delta+\gamma \int_{a}^{t }g(x)~dx.
\]
Then we have
\[
\varphi'(t)=\gamma\cdot g(t) \leq \gamma\varphi(t).
\]
Since
\[
\Big(\varphi\cdot e^{-\gamma t}\Big)' = \varphi'\cdot e^{-\gamma t} + \varphi \cdot (-\gamma) e^{-\gamma t}= e^{-\gamma t}\Big(\varphi'(t)-\gamma\varphi(t)\Big)\leq 0
\]
we have that $\varphi e^{-\gamma t}$ is monotone falling. It thus follows
\[
g(t)\cdot e^{-\gamma t}\leq \varphi(t)\cdot e^{\gamma t} \leq \varphi(a)\cdot e^{-\gamma a}= \delta \cdot e^{-\gamma a}
\]
for all $t\geq a$.
\end{proof}
The Grönwall-Lemma allows us to prove the following theorem.
\begin{theorem}[Dependence on initial data]
Let $D\subset I\times \R^n$ be open, $f:D\to\R^n$ continuous and local Lipschitz with respect to $y$ and $(t_0,y_0) \in D$. If the solution of
\begin{align*}
y'(t)&=f(t,y(t))\\
y(t_0)&=y_0, ~y_0\in\R^n
\end{align*}
exists for all $t\in I=[a,b]$ then for each $\varepsilon>0$ there exists a $\delta>0$ such that \begin{enumerate}[(i)]
\item If $\norm{y_0-z_0}<\delta$ there also exists a solution of
\begin{align*}
z'(t)&=f(t,z(t))\\
z(t_0)&=z_0, ~z_0\in\R^n
\end{align*}
for $t \in I$.
\item It holds
\[
\max_{t\in I}\norm{y(t)-z(t)}<\varepsilon.
\]
\end{enumerate}
\end{theorem}
\begin{proof}
Since $D$ is open, there exists a $\bar\delta>0$ and a compact set
\[
K:=\{(t,z(t))~|~t\in I, ~ \norm{y(t)-z(t)}\leq\bar\delta\}\subset D.
\]
On $K$ the function $f$ is Lipschitz (with respect to $y$) with a Lipschitz-constant $L$. Let now $\delta<\bar\delta$ and $\norm{y_0-z_0}<\delta$. Then for all $t_0,t\in[a,b]$ it holds
\[
\norm{z(t)-y(t)}\leq \delta+L \int_{t_0}^{t }\norm{y(x)-z(x)}~dx.
\]
This can be seen by the integral representation of $y(t)$. Applying Grönwall's Lemma with $\gamma=L$ yields
\begin{equation}
\norm{y(t)-z(t)}\leq\delta\cdot e^{L(t-t_0)} \tag{I.5}
\end{equation}
and by choosing $\delta\leq\bar\delta\cdot e^{L(a-b)}$ it holds $\norm{y(t)-z(t)}\leq \bar\delta$ for all $t\in I$. Thus it holds $(t,z(t))\in K$ for $t\in[a,b]$ and hence we have shown \textit{(i)}.

By choosing $\delta<\varepsilon\cdot e^{L(a-b)}$ it follows \textit{(ii)}.
\end{proof}
\begin{remark}
We have thus shown, that the solution $y(t)$ of the IVP with initial value $y(t_0)=y_0$ depends continuously on the initial data. The solution is often written as $y(t;t_0,y_0,f)$.
\end{remark}
\begin{example}
Let us consider the ODE
\begin{align*}
y'&=\lambda y, ~ \lambda \in \R\\
y(0)&=y_0
\end{align*}
Here we have $L=|\lambda|$. The equation (I.5) gives
\[
|y(t)-z(t)|\leq e^{|\lambda|\cdot t} |y_0-z_0|.
\]
For $\lambda<0$ we know that $|y(t)-z(t)|$ decreases exponentially.
\end{example}
\subsection{Stability and Flow}
\subsubsection*{Vector field}
A solution of an ODE is a function $y:I\to\R^n$ which is differentiable on $I$. Its graph $\{(t,y(t))~|~t\in I\}$ is a differentiable curve in $\R^{n+1}$ also known as \textit{solution curve} or \textit{integral curve}. In each point $(t,y(t))$ the direction of the tangent is given by the $(1,f(t,y(t)))$. In other words, $f$ is assigning a direction to each point.

\subsubsection*{Stability and small perturbations}
Consider
\[
y'(t)=f(t,y(t)), ~ y(t_0)=y_0.
\]
We are now interested in a comparison of different solutions for $t\in[t_0,\infty)$ with respect to the initial condition. We denote the solution by $y(t)=y(t,t_0)$.

Stability: $y(t_0)=\tilde y$ with $\tilde y$ near by $y_0$. The question we are dealing with is "How does $y(t,\tilde y)$ behave in comparison with $y(t,y_0)$?".\newline

Let us consider an autonomous ODE, i.e. an ODE of the form $y'(t)=f(y(t))$.

\begin{definition}[Equilibrium Point]
A point $\bar y\in D\subset\R^n$ is called an equilibrium point of a mapping $f:D\to\R^n$ if $f(\bar y)=0$. The constant solution $y(t)=\bar y$ is the only solution with $y(t_0)=\bar y$.
\end{definition}
\begin{remark}
Other names for equilibrium points are fixed points, equilibria and stationary points.
\end{remark}
\begin{definition}[Stability and asymptotic stability]
An equilibrium point is \textbf{stable} (in the sense of Ljapunov) if for each $\varepsilon>0$ there exists a $\delta>0$ such that for $t\geq t_0$ and for all trajectories $y(t)$ with $\norm{y(t_0)-\bar y}\leq \delta$ it holds that
\[
\norm{y(t)-\bar y}\leq \varepsilon.
\]
An equilibrium point is \textbf{instable} if it is not stable. \newline

\noindent An equilibrium point $\bar y$ is \textbf{asymptotic stable} if there exists a neighbourhood $U_{\bar y}$ of $\bar y$ such that
\[
y(t_0)\in U_{\bar y} \Rightarrow \lim_{t\to\infty} y(t)=\bar y.
\]
In this case $\bar y$ is called a sink.

An equilibrium point $\bar y$ is a spring if for each solution $y(t)$ with $y(t_0)\in U_{\bar y}$ and $y(t_0)\neq \bar y$ there exists a $t_1>t_0$ such that $y(t)\not \in U_{\bar y}$ for all $t\geq t_1$.
\end{definition}
\begin{example}
Consider an ODE in $\R^1$ given by $y'(t)=f(t,y(t))$. The equilibrium point is asymptotic stable if in $U_{\bar y}$ it holds that
\[
f(y)<0 \text{ for } y<\bar y ~~\und~~ f(y)>0 \text{ for } y>\bar y.
\]
\end{example}
\begin{definition}[Stability of solutions]
Let $y(t;y_0)$ be a solution of $y'(t)=f(y(t)), ~y(t_0)=y_0 ~\forall t\geq t_0$. Then the solution is \textbf{stable} if for each $\varepsilon>0$ there exists a $\delta>0$ such that
\[
\norm{y_0-\tilde y_0}\leq \delta \Rightarrow \norm{y(t;y_0)-y(t,\tilde y_0)}<\varepsilon
\]
for all $t>t_0$.
The solution is \textbf{attractive} if there exists a $\delta >0$ such that
\[
\norm{y_0-\tilde y_0}<\delta \Rightarrow \lim_{t\to\infty}\norm{y(t;y_0)-y(t,\tilde y_0)}=0.
\]
The solution is \textbf{asymptotic stable} if its stable and attractive.
\end{definition}





















































































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