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% Author: FOSSEE

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% Title page

\title[]{ODEs \& Finding Roots}

\author[FOSSEE] {FOSSEE}

\institute[IIT Bombay] {Department of Aerospace Engineering\\IIT Bombay}

\date[] {31, October 2009\\Day 1, Session 6}

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\begin{document}

\begin{frame}

  \maketitle

\end{frame}

%% \begin{frame}

%%   \frametitle{Outline}

%%   \tableofcontents

%%   % You might wish to add the option [pausesections]

%% \end{frame}

\section{ODEs}

\begin{frame}[fragile]

\frametitle{ODE Integration}

We shall use the simple ODE of a simple pendulum. 

\begin{equation*}

\ddot{\theta} = -\frac{g}{L}sin(\theta)

\end{equation*}

\begin{itemize}

\item This equation can be written as a system of two first order ODEs

\end{itemize}

\begin{align}

\dot{\theta} &= \omega \\

\dot{\omega} &= -\frac{g}{L}sin(\theta) \\

 \text{At}\ t &= 0 : \nonumber \\

 \theta = \theta_0\quad & \&\quad  \omega = 0 \nonumber

\end{align}

\end{frame}

\begin{frame}[fragile]

\frametitle{Solving ODEs using SciPy}

\begin{itemize}

\item We use the \typ{odeint} function from scipy to do the integration

\item Define a function as below

\end{itemize}

\begin{lstlisting}

In []: def pend_int(unknown, t, p):

  ....     theta, omega = unknown

  ....     g, L = p

  ....     f=[omega, -(g/L)*sin(theta)]

  ....     return f

  ....

\end{lstlisting}

\end{frame}

\begin{frame}[fragile]

\frametitle{Solving ODEs using SciPy \ldots}

\begin{itemize}

\item \typ{t} is the time variable \\ 

\item \typ{p} has the constants \\

\item \typ{initial} has the initial values

\end{itemize}

\begin{lstlisting}

In []: t = linspace(0, 10, 101)

In []: p=(-9.81, 0.2)

In []: initial = [10*2*pi/360, 0]

\end{lstlisting}

\end{frame}

\begin{frame}[fragile]

\frametitle{Solving ODEs using SciPy \ldots}

\begin{small}

  \typ{In []: from scipy.integrate import odeint}

\end{small}

\begin{lstlisting}

In []: pend_sol = odeint(pend_int, 

                         initial,t, 

                         args=(p,))

\end{lstlisting}

\end{frame}

\section{Finding Roots}

\begin{frame}[fragile]

\frametitle{Roots of $f(x)=0$}

\begin{itemize}

\item Roots --- values of $x$ satisfying $f(x)=0$

\item $f(x)=0$ may have real or complex roots

\item Presently, let's look at real roots

\end{itemize}

\end{frame}

\begin{frame}[fragile]

\frametitle{Initial Estimates}

\begin{itemize}

\end{itemize}

\begin{enumerate}

\item Check for change of signs of $f(x)$ in the given interval

\item A root lies in the interval where the sign change occurs

\end{enumerate}

\end{frame}

\begin{frame}[fragile]

\frametitle{Initial Estimates \ldots}

\begin{lstlisting}

  In []: def our_f(x):

   ....:     return cos(x)-x**2

   ....: 

  In []: x = linspace(-pi/2, pi/2, 11)

\end{lstlisting}

\begin{itemize}

\item Get the intervals of x, where sign changes occur

\end{itemize}

\end{frame}

%% \begin{frame}[fragile]

%% \frametitle{Initial Estimates \ldots}

%% \begin{lstlisting}

%% In []: pos = y[:-1]*y[1:] < 0

%% In []: rpos = zeros(x.shape, dtype=bool)

%% In []: rpos[:-1] = pos

%% In []: rpos[1:] += pos

%% In []: rts = x[rpos]

%% \end{lstlisting}

%% \end{frame}

\begin{frame}[fragile]

\frametitle{Fixed Point Method}

\begin{enumerate}

\item Convert $f(x)=0$ to the form $x=g(x)$

\item Start with an initial value of $x$ and iterate successively

\item $x_{n+1}=g(x_n)$ and $x_0$ is our initial guess

\item Iterate until $x_{n+1}-x_n \le tolerance$

\end{enumerate}

\end{frame}

%% \begin{frame}[fragile]

%% \frametitle{Fixed Point \dots}

%% \begin{lstlisting}

%% In []: def our_g(x):

%%  ....:     return sqrt(cos(x))

%%  ....: 

%% In []: tolerance = 1e-5

%% In []: while abs(x1-x0)>tolerance:

%%  ....:     x0 = x1

%%  ....:     x1 = our_g(x1)

%%  ....:   

%% In []: x0

%% \end{lstlisting}

%% \end{frame}

\begin{frame}[fragile]

\frametitle{Bisection Method}

\begin{enumerate}

\item Start with an interval $(a, b)$ within which a root exists

\item $f(a)\cdot f(b) < 0$

\item Bisect the interval. $c = \frac{a+b}{2}$

\item Change the interval to $(a, c)$ if $f(a)\cdot f(c) < 0$

\item Else, change the interval to $(c, b)$

\item Go back to 3 until $(b-a) \le$ tolerance

\end{enumerate}

\end{frame}

%% \begin{frame}[fragile]

%% \frametitle{Bisection \dots}

%% \begin{lstlisting}

%% In []: tolerance = 1e-5

%% In []: a = -pi/2

%% In []: b = 0

%% In []: while b-a > tolerance:

%%  ....:     c = (a+b)/2

%%  ....:     if our_f(a)*our_f(c) < 0:

%%  ....:         b = c

%%  ....:     else:

%%  ....:         a = c

%%  ....:         

%% \end{lstlisting}

%% \end{frame}

\begin{frame}[fragile]

\frametitle{Newton Raphson Method}

\begin{enumerate}

\item Start with an initial guess of x for the root

\item $\Delta x = -f(x)/f^{'}(x)$

\item $ x \leftarrow x + \Delta x$

\item Go back to 2 until $|\Delta x| \le$ tolerance

\end{enumerate}

\end{frame}

%% \begin{frame}[fragile]

%% \frametitle{Newton Raphson \dots}

%% \begin{lstlisting}

%% In []: def our_df(x):

%%  ....:     return -sin(x)-2*x

%%  ....: 

%% In []: delx = 10*tolerance

%% In []: while delx > tolerance:

%%  ....:     delx = -our_f(x)/our_df(x)

%%  ....:     x = x + delx

%%  ....:     

%%  ....:     

%% \end{lstlisting}

%% \end{frame}

\begin{frame}[fragile]

\frametitle{Newton Raphson \ldots}

\begin{itemize}

\item What if $f^{'}(x) = 0$?

\item Can you write a better version of the Newton Raphson?

\item What if the differential is not easy to calculate?

\item Look at Secant Method

\end{itemize}

\end{frame}

\begin{frame}[fragile]

\frametitle{Scipy Methods - \typ{roots}}

\begin{itemize}

\item Calculates the roots of polynomials

\item Array of coefficients is the only parameter

\end{itemize}

\begin{lstlisting}

  In []: coeffs = [1, 6, 13]

  In []: roots(coeffs)

\end{lstlisting}

\end{frame}

\begin{frame}[fragile]

\frametitle{Scipy Methods - \typ{fsolve}}

\begin{small}

\begin{lstlisting}

  In []: from scipy.optimize import fsolve

\end{lstlisting}

\end{small}

\begin{itemize}

\item Finds the roots of a system of non-linear equations

\item Input arguments - Function and initial estimate

\item Returns the solution

\end{itemize}

\begin{lstlisting}

  In []: fsolve(our_f, -pi/2)

\end{lstlisting}

\end{frame}

\begin{frame}[fragile]

\frametitle{Scipy Methods \dots}

\begin{lstlisting}

In []: from scipy.optimize import fixed_point

In []: from scipy.optimize import bisect

In []: from scipy.optimize import newton

\end{frame}

\end{document}