author | Puneeth Chaganti <punchagan@gmail.com> |
Sat, 01 May 2010 16:30:45 +0530 | |
changeset 396 | 502d4d151fa2 |
parent 341 | 7ae88b9da553 |
child 438 | 8af5dfa5432b |
permissions | -rwxr-xr-x |
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\documentclass[12pt]{article} |
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\title{Solving Equations \& ODEs} |
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\author{FOSSEE} |
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\usepackage{listings} |
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\lstset{language=Python, |
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basicstyle=\ttfamily, |
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commentstyle=\itshape\bfseries, |
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showstringspaces=false, |
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} |
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\newcommand{\typ}[1]{\lstinline{#1}} |
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\usepackage[english]{babel} |
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\usepackage[latin1]{inputenc} |
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\usepackage{times} |
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\usepackage[T1]{fontenc} |
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\usepackage{ae,aecompl} |
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\usepackage{mathpazo,courier,euler} |
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\usepackage[scaled=.95]{helvet} |
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\usepackage{amsmath} |
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\begin{document} |
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\date{} |
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\vspace{-1in} |
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\begin{center} |
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\LARGE{Solving Equations \& ODEs}\\ |
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\large{FOSSEE} |
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\end{center} |
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\section{Solving linear equations} |
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Consider following sets of equations:\\ |
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\begin{align*} |
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3x + 2y - z & = 1 \\ |
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2x - 2y + 4z & = -2 \\ |
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-x + \frac{1}{2}y -z & = 0 |
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\end{align*}\\ |
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The matrix representation is:\\ |
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\begin{center} |
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$A*x = B$ |
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\end{center} |
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Where A is coefficient matrix(in this case 3x3)\\ |
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B is constant matrix(1x3)\\ |
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x is the required solution.\\ |
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\begin{lstlisting} |
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In []: A = array([[3,2,-1], [2,-2,4], [-1, 0.5, -1]]) |
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In []: B = array([[1], [-2], [0]]) |
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In []: x = solve(A, B) |
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\end{lstlisting} |
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Solve the equation $A x = B$ for $x$.\\ |
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To check whether solution is correct try this: |
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\begin{lstlisting} |
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In []: Ax = dot(A,x) #Matrix multiplication of A and x(LHS) |
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In []: allclose(Ax, B) |
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Out[]: True |
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\end{lstlisting} |
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\typ{allclose} Returns \typ{True} if two arrays(in above case Ax and B) are element-wise equal within a tolerance. |
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\newpage |
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\section{Finding roots} |
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\subsection{Polynomials} |
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\begin{center} |
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$x^2+6x+13=0$ |
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\end{center} |
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to find roots, pylab provides \typ{roots} function. |
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\begin{lstlisting} |
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In []: coeffs = [1, 6, 13] #list of all coefficients |
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In []: roots(coeffs) |
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\end{lstlisting} |
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\subsection{functions} |
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Functions can be defined and used by following syntax: |
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\begin{lstlisting} |
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def func_name(arg1, arg2): |
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#function code |
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return ret_value |
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\end{lstlisting} |
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A simple example can be |
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\begin{lstlisting} |
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def expression(x): |
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y = x*sin(x) |
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return y |
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\end{lstlisting} |
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Above function when called with a argument, will return $xsin(x)$ value for that argument. |
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\begin{lstlisting} |
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In [95]: expression(pi/2) |
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Out[95]: 1.5707963267948966 |
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|
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In [96]: expression(pi/3) |
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Out[96]: 0.90689968211710881 |
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\end{lstlisting} |
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\subsection{Roots of non-linear equations} |
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For Finding the roots of a non linear equation(defined as $f(x)=0$), around a starting estimate we use \typ{fsolve}:\\ |
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\typ{In []: from scipy.optimize import fsolve}\\ |
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\typ{fsolve} is not a part of \typ{pylab}, instead is a function in the \textbf{optimize} module of \textbf{scipy}, and hence we \textbf{import} it.\\ |
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%\typ{fsolve} takes first argument as name of function, which evaluates $f(x)$, whose roots one wants to find. And second argument is starting estimate, around which roots are found. |
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For illustration, we want to find roots of equation: |
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\begin{center} |
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$f(x)=sin(x)+cos(x)^2$ |
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\end{center} |
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So just like we did above, we define a function: |
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\begin{lstlisting} |
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In []: def f(x): |
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....: return sin(x)+cos(x)**2 |
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....: |
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99 |
\end{lstlisting} |
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Now to find roots of this non linear equation, around a initial estimate value, say 0, we use \typ{fsolve} in following way: |
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101 |
\begin{lstlisting} |
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In []: fsolve(f, 0) #arguments are function name and estimate |
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103 |
Out[]: -0.66623943249251527 |
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104 |
\end{lstlisting} |
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105 |
|
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\section{ODE} |
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107 |
|
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We wish to solve an (a system of) Ordinary Differential Equation. For this purpose, we shall use \typ{odeint}.\\ |
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\typ{In []: from scipy.integrate import odeint}\\ |
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\typ{odeint} is a function in the \textbf{integrate} module of \textbf{scipy}.\\ |
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As an illustration, let us solve the ODE below:\\ |
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$\frac{dy}{dt} = ky(L-y)$, L = 25000, k = 0.00003, y(0) = 250\\ |
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We define a function (as below) that takes $y$ and time as arguments and returns the right hand side of the ODE. |
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\begin{lstlisting} |
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In []: def f(y, t): |
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.... k, L = 0.00003, 25000 |
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.... return k*y*(L-y) |
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.... |
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\end{lstlisting} |
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Next we define the time over which we wish to solve the ODE. We also note the initial conditions given to us. |
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\begin{lstlisting} |
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In []: t = linspace(0, 12, 61) |
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In []: y0 = 250 |
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\end{lstlisting} |
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To solve the ODE, we call the \typ{odeint} function with three arguments - the function \typ{f}, initial conditions and the time vector. |
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\begin{lstlisting} |
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In []: y = odeint(f, y0, t) |
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\end{lstlisting} |
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Note: To solve a system of ODEs, we need to change the function to return the right hand side of all the equations and the system and the pass the required number of initial conditions to the \typ{odeint} function. |
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\section{Links and References} |
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\begin{itemize} |
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\item Documentation for Numpy and Scipy is available at:\\ http://docs.scipy.org/doc/ |
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\item For "recipes" or worked examples of commonly-done tasks in SciPy explore: \\ http://www.scipy.org/Cookbook/ |
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\end{itemize} |
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\end{document} |