216 Notes
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\documentclass{article}
\usepackage[a4paper]{geometry}
\begin{document}
\title{Course 216: Ordinary Differential Equations}
\author{Notes by Chris Blair}
\date{}
\maketitle
\vspace{-2.5em}
\begin{center}
\begin{minipage}{170pt}
\scriptsize These notes cover the ODEs course given in 2007-2008 by Dr.
John Stalker.
\end{minipage}
\end{center}
\tableofcontents
\section*{Terminology}
\paragraph{Scalar equation} A single ODE.
\paragraph{System of equations} Several ODEs.
\paragraph{Order} The order of an ODE is the order of the highest
derivative appearing in it.
\paragraph{Linear / Non-linear} A linear ODE is an ODE that is linear,
etc.
\paragraph{Homogeneous / Inhomogeneous} Homogeneous means no constant
terms present. Inhomogeneous means constant terms are present.
\paragraph{Invariants}
An invariant of a system of ODEs is a function of the dependent and
independent variables and their derivatives which is constant for any
solution of the equation. They can be used to place bounds on solutions.
\part{Solving Linear ODEs}
\section{Reduction of Order}
\begin{itemize}
\item Any higher order ODE or system of ODEs can be reduced to a system of
first order ODEs by introducing new variables to replace the derivatives in
the original equation/system.
\item For example, the third order equation
\[c_1 x'''(t) + c_2 x''(t) + c_3 x'(t) + c_4 x(t) = 0 \]
can be reduced to a first order system using the following set of
substitutions:
\[ x_1 = x, \, \, x_2 = x', \, \, x_3 = x'' \]
giving:
\[x_1' = x_2, \, \, x_2' = x_3, \,\, x_3' = - \frac{c_4}{c_1} x_1 -
\frac{c_3}{c_1} x_2 - \frac{c_2}{c_1} x_3 \]
We can write this in matrix form:
\[ \left( \begin{array}{c}x_1' \\ x_2' \\ x_3' \end{array} \right) =
\left(\begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\- \frac{c_4}{c_1} & -
\frac{c_3}{c_1} & - \frac{c_2}{c_1} \end{array} \right)
\left(\begin{array}{c}x_1 \\ x_2 \\ x_3 \end{array} \right) \]
\item Hence, any ODE or system of ODEs can be written in the following
matrix form:
\[ \vec{x}' (t) = A(t) \vec{x} (t) \]
which has solution:
\[\vec{x}(t) = \mbox{exp} (tA) \, \vec{x} (0) \]
\end{itemize}
\section{Computing Matrix Exponentials}
\begin{itemize}
\item The exponential of the matrix $tA$ is given by:
\[\mbox{exp} (tA) = \sum_{n=0}^{\infty} \frac{1}{n!} t^n A^n \]
\item For a diagonal matrix,
\[ \exp \left( \begin{array}{cccc} a & 0 & \dots & 0 \\ 0 & b & \dots & 0
\\ \vdots & & \ddots & \\ 0 & \dots & 0 & n \\ \end{array} \right) =
\left( \begin{array}{cccc} \exp(a) & 0 & \dots & 0 \\ 0 & \exp(b) & \dots
& 0 \\ \vdots & & \ddots & \\ 0 & \dots & 0 & \exp(n) \\ \end{array}
\right) \]
\item Given two matrices $A$ and $B$ then
\[\mbox{exp} (A+B) = \mbox{exp} (A) \mbox{exp} (B) \]
if $AB = BA$. Note that any scalar multiple of the identity commutes with
all matrices.
\item \bf 2 by 2 Matrices \normalfont
\[ A = \left( \begin{array}{cc} a & b \\ c & d \\ \end{array} \right) =
\left( \begin{array}{cc} \frac{a+d}{2} & 0 \\ 0 &\frac{a + d}{2} \\
\end{array} \right) + \left( \begin{array}{cc} \frac{a-d}{2} & b \\ c &
\frac{d-a}{2} \\ \end{array} \right) = B + C \]
and we have $BC = CB$ so that $\exp(B+C) = \exp B \, \exp C$. Letting $\mu
= \frac{a+d}{2}$, we then have
\[ \exp(tA) = \exp(tB)\exp(tC) \]
\[\Rightarrow \exp(tA) = \left( \begin{array}{cc} \exp(\mu t) & 0 \\ 0 &
\exp(\mu t) \\ \end{array} \right)\exp(tC) \]
Now, the discriminant $\Delta$ of $A$ is
\[ \Delta = \Big(\mbox{tr}A\Big)^2 - 4 \, \mbox{det} A \]
and $C^2 = \frac{\Delta}{4} I$. This leads to three cases:
i) $\Delta = 0$, then
\[\exp(tC) = I + t C \]
ii) $\Delta < 0$, then
\[\exp(tC) = \cos \Big(\frac{t \sqrt{- \Delta}}{2}\Big) I + \frac{\sin
(\frac{t \sqrt{- \Delta}}{2})}{\frac{ \sqrt{- \Delta}}{2}} C \]
iii) $\Delta > 0$, then
\[\exp(tC) = \cosh \Big(\frac{t \sqrt{\Delta}}{2}\Big) I + \frac{\sinh
(\frac{t \sqrt{\Delta}}{2})}{\frac{ \sqrt{\Delta}}{2}} C \]
\item \bf $n \times n$ Matrices \normalfont
Every $n$ by $n$ matrix $A$ is similar to its Jordan form $J$, which can
be written as the sum of a diagonal and a nilpotent matrix, $J = D + N$.
We have
\[A = P J P^{-1} \]
\[\Rightarrow \exp(tA) = P \, \exp(tJ) \, P^{-1} \]
\[\Rightarrow \exp(tA) = P \, \exp(tD) \,\exp(tN) \, P^{-1} \]
The Jordan form $J$ has the eigenvalues of $A$ on the diagonal, and some
ones below the diagonal, depending on whether the eigenvalues are
distinct. The columns of the matrix $P$ are the eigenvectors of $A$. The
entries of $P$ can also be found once you know $J$, using $AP = PJ$.
The exponential of the nilpotent matrix $N$ is computed directly using the
exponential formula.
Note that in the case of a higher order scalar equation, we only need the
first row of $P$, as we are just looking for $x(t)$.
\end{itemize}
\section{Higher Order Scalar ODEs}
\begin{itemize}
\item Consider a higher order scalar ODE,
\[ c_n \frac{d^n x}{d t^n} + \dots + c_2 \frac{d^2 x}{d t^2} + c_1 \frac{d
x}{dt} + c_0 x = 0 \]
which we can write as
\[p \left(\frac{d}{dt} \right) x = 0\]
where $p$ is the polynomial
\[p(s) = c_n s^n + \dots + c_2 s^2 + c_1 s + c_0 = 0\]
which has roots $\lambda_i$.
\item A basis for the solution space is then
\[ \Big\lbrace \exp(\lambda_1 t), t\, \exp(\lambda_1 t), \dots , t^{r_1 -
1} \exp(\lambda_1 t), \dots , \exp(\lambda_k t), \dots , t^{r_k - 1}
\exp(\lambda_k t) \Big\rbrace \]
where the $\lambda_i$ are the individual roots of the equation and $r_i$
is the multiplicity of the $i^{th}$ root.
\item In the inhomogeneous case, we have $p(\frac{d}{dt}) x = f$, and have
the special case where $f$ itself satisfies some differential equation
$q(\frac{d}{dt}) f = 0$. Hence
\[ q\bigg(\frac{d}{dt}\bigg) p\bigg(\frac{d}{dt}\bigg) x = 0 \]
and we can form a basis for the solution space using the roots of $r(s) =
q(s) p(s)$. It is then possible to evaluate the coefficients of the
particular solution to the inhomogeneous equation by evaluating
$p(\frac{d}{dt}) x = f$
\end{itemize}
\section{Non-constant Coefficients}
\begin{itemize}
\item \bf Homogeneous Scalar Equations \normalfont
The homogeneous equation
\[ x'(t) = a(t) x(t) \]
has unique solution:
\[x(t) = x(0) \, \exp \left( \int_0^t a(s) ds \right) \]
\item \bf Inhomogeneous Scalar Equations \normalfont
The inhomogeneous equation
\[ x'(t) = a(t) x(t) + f(t) \]
has unique solution:
\[x(t) = x(0) \, \exp \left( \int_0^t a(s) ds \right) + \int_0^t \, \exp
\left( \int_s^t a(r)dr \right) f(s) ds \]
\item \bf Systems \normalfont
The equation
\[\vec{x}'(t) = A(t) \vec{x}(t) + \vec{f} (t) \]
has unique solution:
\[ \vec{x}(t) = W(t) \vec{x}(0) + \int_0^t W(t) W^{-1}(s) \vec{f}(s) ds \]
where $W(t)$ satisfies the matrix initial value problem
\[W'(t) = A(t) W(t), \, \, \,W(0) = I \]
\end{itemize}
\section{Method of Wronski}
\begin{itemize}
\item Consider a second order scalar linear homogeneous ODE:
\begin{equation} \label{a} p(t) x''(t) + q(t) x'(t) + r(t) x(t) = 0
\end{equation}
which has a two-dimensional solution space.
\item We define
\[w(t) = x_1 (t) x_2'(t) - x_1'(t) x_2(t) \]
giving
\[p(t) w'(t) + q(t) w(t) = x_1(t) \Big[ p(t) x_2''(t) + q(t) x_2'(t) +
r(t) x_2(t)\Big] - x_2 (t) \Big[p(t) x_1''(t) + q(t) x_1'(t) + r(t)
x_1(t)\Big] \]
so if $x_1, x_2$ solve ($\ref{a}$) then $w(t)$ solves
\begin{equation} \label{b} p(t) w'(t) + q(t) w(t) = 0 \end{equation}
\item Hence, if we have $x_1$ a solution to ($\ref{a}$) and $w$ a solution
to ($\ref{b}$), we can then find $x_2$ such that $x_2$ is a solution to
($\ref{a}$), and is linearly independent to $x_1$.
\item Then, given ($\ref{a}$) and $x_1$:
\[ w(t) = w(0) \, \exp \left( - \int_0^t \frac{q(s)}{p(s)} ds \right)\]
and as $\frac{d}{dt} \left(\frac{x_2(t)}{x_1(t)}\right) =
\frac{w(t)}{x_1^2(t)}$,
\[\frac{x_2(t)}{x_1(t)} = \frac{x_2(0)}{x_1(0)} + \int_0^t
\frac{w(s)}{x_1(s)^2} ds \]
\item The general solution is then any linear combination of $x_1$ and
$x_2$:
\[ x(t) = c_1 x(t) + c_2 x_2 (t) \]
\end{itemize}
\part{Stability}
\section{Non-linear ODEs}
\begin{itemize}
\item \bf Non-linear ODEs \normalfont
A non-linear ODE is of the form
\[\vec{x}' (t) = \vec{F} \Big( \vec{x}(t),t\Big) \]
\item \bf Autonomous Systems \normalfont
An autonomous system is of the form
\[\vec{x}^{\prime} (t) = \vec{F} \Big( \vec{x}(t)\Big) \]
\end{itemize}
\section{Equilibria and Stability}
\begin{itemize}
\item \bf Equilibria \normalfont
An equilibrium of an autonomous system $\vec{x}^{\prime}(t) = \vec{F}
\Big( \vec{x}(t)\Big)$ is a $\vec{c}$ such that
\[\vec{F} (\vec{c}) = 0\]
i.e. the equilibria of a system are the zeros of $\vec{F}$.
\item \bf Stability \normalfont
An equilibrium $\vec{c}$ is said to be stable if $\forall \, \varepsilon >
0$, $\exists \, \delta > 0$ such that if
\[ ||\, \vec{x}(0) - \vec{c}\,|| \leq \delta \]
then
\[ || \, \vec{x} (t) - \vec{c}\, || \leq \varepsilon \]
for all positive $t$.
\item \bf Asymptotic Stability \normalfont
An equilibrium $\vec{c}$ is said to be asymptotically stable if $\exists
\, \delta > 0$ such that
\[ || \, \vec{x}(0) - \vec{c} \, || \leq \delta \Rightarrow \lim_{t
\rightarrow \infty} \vec{x}(t) = \vec{c} \]
\item \bf Strict Stability \normalfont
An equilibrium $\vec{c}$ is said to be strictly stable if it is both
stable and asymptotically stable.
\item \bf Stability and Invariants \normalfont
If $\vec{c}$ is an equilibrium of an autonomous system and $E$ is a
continuously differentiable invariant of the system which has a strict
local minimum at $\vec{c}$, then $\vec{c}$ is stable but not
asymptotically stable.
\item \bf Stability of Linear Constant Coefficient First Order Systems
\normalfont
These are systems
\[\vec{x}' (t) = A \vec{x}(t) \]
with solution
\[\vec{x}(t) = \exp (tA) \vec{x}(0) = P \exp (tJ) P^{-1} \vec{x}(0) \]
$\vec{0}$ is always an equilibrium, and each equilibrium is
stable/asymptotically stable if and only if $\vec{0}$ is
stable/asymptotically stable.
We can determine the stability of the system by considering the real parts
of the eigenvalues of $A$:
\vspace{1em}
\renewcommand\arraystretch{1.25}
\begin{tabular}{|l|l|l|}
\hline
\bf Real Parts & \bf Stable & \bf Asymptotically Stable \\ \hline
all $<0$ & Yes & Yes \\ \hline
all $\leq 0$,& Yes & No \\
\scriptsize geometric multiplicity = algebraic multiplicity for all
imaginary eigenvalues & & \\ \hline
all $\leq 0$, & No & No \\
\scriptsize geometric multiplicity $<$ algebraic multiplicity for some
imaginary eigenvalue & & \\ \hline
some $>0$ & No & No \\ \hline
\end{tabular}
\vspace{1em}
In the 2 by 2 case, then if trace $A<0$ and det $A\geq0$, then $\vec{0}$
is strictly stable. If trace $A\leq0$ and det $A\geq 0$ then $\vec{0}$ is
stable. Otherwise it is not stable or asymptotically stable.
In the scalar high order case where $p(\frac{d}{dt})x = 0$, $p(s)$ a
polynomial, if all roots of $p(s) = 0$ have negative real parts, then we
have strict stability. If all roots have non-positive real parts, and all
imaginary roots have multiplicity one, then we have stability but not
strict stability. Otherwise, neither stability nor asymptotic stability.
\end{itemize}
\section{Linearisation}
\begin{itemize}
\item The linearisation of an autonomous system $\vec{x}'(t) = \vec{F}
\Big(\vec{x}(t)\Big)$ about an equilibrium $\vec{c}$ is the matrix $A$
defined by
\[a_{jk} = \frac{\partial F_j}{\partial x_k} (\vec{c}) \]
\item If all eigenvalues of $A$ have negative real parts, then $\vec{c}$
is strictly stable.
\item If some eigenvalue of $A$ has positive real part, then $\vec{c}$ is
neither stable nor asymptotically stable.
\item Otherwise, we learn nothing.
\end{itemize}
\section{Method of Lyapunov}
\begin{itemize}
\item \bf Lyapunov Function \normalfont
A Lyapunov function for the equilibrium $\vec{c}$ of an autonomous system
is a continuously differentiable function $V$ with a strict local minimum
at $\vec{c}$ such that
\[ \sum_j \frac{\partial V}{\partial x_j} F_j \leq 0 \]
\item \bf Strict Lyapunov Function \normalfont
A strict Lyapunov function is a Lyapunov function satisfying
\[ \sum_j \frac{\partial V}{\partial x_j} F_j \leq - r \Big[ V(\vec{x})
- V(\vec{c}) \Big] \]
for some positive $r$.
\item An equilibrium $\vec{c}$ is stable if it admits a Lyapunov function,
and strictly stable if it admits a strict Lyapunov function.
\end{itemize}
\end{document}
