autarkaw – Page 22 – Numerical Methods Guy

How do I solve an initial value ODE in MATLAB?

Many students ask me how do I do this or that in MATLAB. So I thought why not have a small series of my next few blogs do that. In this blog, I show you how to solve an initial value ordinary differential equation.

The MATLAB program link is here.
The HTML version of the MATLAB program is here.
DO NOT COPY AND PASTE THE PROGRAM BELOW BECAUSE THE SINGLE QUOTES DO NOT TRANSLATE TO THE CORRECT SINGLE QUOTES IN MATLAB EDITOR. DOWNLOAD THE MATLAB PROGRAM INSTEAD

%% HOW DO I DO THAT IN MATLAB SERIES?
% In this series, I am answering questions that students have asked
% me about MATLAB. Most of the questions relate to a mathematical
% procedure.

%% TOPIC
% How do I solve an initial value ordinary differential equation?

%% SUMMARY

% Language : Matlab 2008a;
% Authors : Autar Kaw;
% Mfile available at
% http://nm.mathforcollege.com/blog/ode_initial.m;
% Last Revised : May 14, 2009;
% Abstract: This program shows you how to solve an
% initial value ordinary differential equation.
clc
clear all

%% INTRODUCTION

disp(‘ABSTRACT’)
disp(‘   This program shows you how to solve’)
disp(‘   an initial value ordinary differential equation’)
disp(‘ ‘)
disp(‘AUTHOR’)
disp(‘   Autar K Kaw of http://autarkaw.wordpress.com’)
disp(‘ ‘)
disp(‘MFILE SOURCE’)
disp(‘   http://nm.mathforcollege.com/blog/ode_initial.m’)
disp(‘ ‘)
disp(‘LAST REVISED’)
disp(‘   May 14, 2009’)
disp(‘ ‘)

%% INPUTS
% Solve the ordinary differential equation 3y”+5y’+7y=11exp(-x)
% Define x as a symbol
syms x
%The ODE
ode_eqn=’3*D2y+5*Dy+7*y=11*exp(-13*x)’;
% The initial conditions
iv_1=’Dy(0)=17′;
iv_2=’y(0)=19′;
% The value at which y is sought at
xval=23.0;
%% DISPLAYING INPUTS

disp(‘INPUTS’)
func=[‘ The ODE to be solved is ‘ ode_eqn];
disp(func)
iv_explain=[‘ The initial conditions are ‘ iv_1 ‘ ‘ iv_2];
disp(iv_explain)
fprintf(‘ The value of y is sought at x=%g’,xval)
disp(‘ ‘)

%% THE CODE

% Finding the solution of the ordinary differential equation
soln=dsolve(ode_eqn,iv_1,iv_2,’x’);
% vpa below uses variable-precision arithmetic (VPA) to compute each
% element of soln to 5 decimal digits of accuracy
soln=vpa(soln,5);

%% DISPLAYING OUTPUTS
disp(‘ ‘)
disp(‘OUTPUTS’)
output=[‘ The solution to the ODE is ‘ char(soln)];
disp(output)
value=subs(soln,x,xval);
fprintf(‘ The value of y at x=%g is %g’,xval,value)
disp(‘ ‘)

This post is brought to you by Holistic Numerical Methods: Numerical Methods for the STEM undergraduate at http://nm.mathforcollege.com, the textbook on Numerical Methods with Applications available from the lulu storefront, and the YouTube video lectures available at http://nm.mathforcollege.com/videos and http://www.youtube.com/numericalmethodsguy

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How do I solve a nonlinear equation that needs to be setup in MATLAB?

Many students ask me how do I do this or that in MATLAB. So I thought why not have a small series of my next few blogs do that. In this blog, I show you how to solve a nonlinear equation that needs to be set up.

For example to find the depth ‘x’ to which a ball is floating in water is based on the following cubic equation
4*R^3*S=3*x^2*(R-x/3)
where
R= radius of ball
S= specific gravity of ball
So how do we set this up if S and R are input values?

The MATLAB program link is here.

The HTML version of the MATLAB program is here.

%% HOW DO I DO THAT IN MATLAB SERIES?
% In this series, I am answering questions that students have asked
% me about MATLAB. Most of the questions relate to a mathematical
% procedure.

%% TOPIC
% How do I solve a nonlinear equation if I need to set it up?

%% SUMMARY

% Language : Matlab 2008a;
% Authors : Autar Kaw;
% Mfile available at
% http://nm.mathforcollege.com/blog/integration.m;
% Last Revised : March 28, 2009;
% Abstract: This program shows you how to solve a nonlinear equation
% that needs to set up as opposed that is just given to you.
clc
clear all

%% INTRODUCTION

disp(‘ABSTRACT’)
disp(‘   This program shows you how to solve’)
disp(‘   a nonlinear equation that needs to be setup’)
disp(‘ ‘)
disp(‘AUTHOR’)
disp(‘   Autar K Kaw of http://autarkaw.wordpress.com’)
disp(‘ ‘)
disp(‘MFILE SOURCE’)
disp(‘   http://nm.mathforcollege.com/blog/nonlinearequation.m’)
disp(‘ ‘)
disp(‘LAST REVISED’)
disp(‘   April 17, 2009’)
disp(‘ ‘)

%% INPUTS
% Solve the nonlinear equation where you need to set up the equation
% For example to find the depth ‘x’ to which a ball is floating in water
% is based on the following cubic equation
% 4*R^3*S=3*x^2*(R-x/3)
% R= radius of ball
% S= specific gravity of ball
% So how do we set this up if S and R are input values

S=0.6
R=0.055
%% DISPLAYING INPUTS
disp(‘INPUTS’)
func=[‘ The equation to be solved is 4*R^3*S=3*x^2*(R-x/3)’];
disp(func)
disp(‘ ‘)

%% THE CODE
% Define x as a symbol
syms x
% Setting up the equation
C1=4*R^3*S
C2=3
f=[num2str(C1) ‘-3*x^2*(‘ num2str(R) ‘-x/3)’]
% Finding the solution of the nonlinear equation
soln=solve(f,x);
solnvalue=double(soln);

%% DISPLAYING OUTPUTS

disp(‘OUTPUTS’)
for i=1:1:length(solnvalue)
fprintf(‘\nThe solution# %g is %g’,i,solnvalue(i))
end
disp(‘ ‘)

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How do I solve a nonlinear equation in MATLAB?

Many students ask me how do I do this or that in MATLAB. So I thought why not have a small series of my next few blogs do that. In this blog, I show you how to solve a nonlinear equation.

The MATLAB program link is here.

The HTML version of the MATLAB program is here.

%% HOW DO I DO THAT IN MATLAB SERIES?
% In this series, I am answering questions that students have asked
% me about MATLAB. Most of the questions relate to a mathematical
% procedure.

%% TOPIC
% How do I solve a nonlinear equation?

%% SUMMARY

%% INTRODUCTION

disp(‘ABSTRACT’)
disp(‘   This program shows you how to solve’)
disp(‘   a nonlinear equation’)
disp(‘ ‘)
disp(‘AUTHOR’)
disp(‘   Autar K Kaw of http://autarkaw.wordpress.com’)
disp(‘ ‘)
disp(‘MFILE SOURCE’)
disp(‘   http://nm.mathforcollege.com/blog/nonlinearequation.m’)
disp(‘ ‘)
disp(‘LAST REVISED’)
disp(‘   April 11, 2009’)
disp(‘ ‘)

%% INPUTS
% Solve the nonlinear equation x^3-15*x^2+47*x-33=0
% Define x as a symbol
syms x
% Assigning the fleft hand side o the equation f(x)=0
f=x^3-15*x^2+47*x-33;
%% DISPLAYING INPUTS

disp(‘INPUTS’)
func=[‘ The equation to be solved is ‘ char(f), ‘=0’];
disp(func)
disp(‘ ‘)

%% THE CODE

% Finding the solution of the nonlinear equation
soln=solve(f,x);
solnvalue=double(soln);

%% DISPLAYING OUTPUTS

disp(‘OUTPUTS’)
for i=1:1:length(solnvalue)
fprintf(‘\nThe solution# %g is %g’,i,solnvalue(i))
end
disp(‘ ‘)

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How do I integrate a discrete function in MATLAB?

Many students ask me how do I do this or that in MATLAB. So I thought why not have a small series of my next few blogs do that. In this blog, I show you how to integrate a discrete function.

The MATLAB program link is here.

The HTML version of the MATLAB program is here.

_____________________________________________________

%% HOW DO I DO THAT IN MATLAB SERIES?
% In this series, I am answering questions that students have asked
% me about MATLAB. Most of the questions relate to a mathematical
% procedure.

%% TOPIC
% How do I integrate a discrete function? Three cases of data are
% discussed.

%% SUMMARY

% Language : MATLAB 2008a;
% Authors : Autar Kaw;
% Mfile available at
% http://nm.mathforcollege.com/blog/integrationdiscrete.m;
% Last Revised : April 3, 2009;
% Abstract: This program shows you how to integrate a given discrete function.

clc
clear all

%% INTRODUCTION

disp(‘ABSTRACT’)
disp(‘   This program shows you how to integrate’)
disp(‘   a discrete function’)
disp(‘ ‘)
disp(‘AUTHOR’)
disp(‘   Autar K Kaw of http://autarkaw.wordpress.com’)
disp(‘ ‘)
disp(‘MFILE SOURCE’)
disp(‘   http://nm.mathforcollege.com/blog/integrationdiscrete.m’)
disp(‘ ‘)
disp(‘LAST REVISED’)
disp(‘   April 3, 2009’)
disp(‘ ‘)

%% CASE 1

%% INPUTS

% Integrate the discrete function y from x=1 to 6.5
% with y vs x data given as (1,2), (2,7), (4,16), (6.5,18)
% Defining the x-array
x=[1 2 4 6.5];
% Defining the y-array
y=[2 7 16 18];

%% DISPLAYING INPUTS
disp(‘____________________________________’)
disp(‘CASE#1’)
disp(‘LOWER LIMIT AND UPPER LIMITS OF INTEGRATION MATCH x(1) AND x(LAST)’)
disp(‘ ‘)
disp(‘INPUTS’)
disp(‘The x-data is’)
x
disp(‘The y-data is’)
y
fprintf(‘ Lower limit of integration, a= %g’,x(1))
fprintf(‘\n Upper limit of integration, b= %g’,x(length(x)))
disp(‘ ‘)

%% THE CODE

intvalue=trapz(x,y);

%% DISPLAYING OUTPUTS

disp(‘OUTPUTS’)
fprintf(‘ Value of integral is = %g’,intvalue)
disp(‘ ‘)
disp(‘___________________________________________’)

%% CASE 2

%% INPUTS

% Integrate the discrete function y from x=3 to 6
% with y vs x data given as (1,2), (2,7), (4,16), (6.5,18)
% Defining the x-array
x=[1 2 4 6.5];
% Defining the y-array
y=[2 7 16 18];
% Lower limit of integration, a
a=3;
% Upper limit of integration, b
b=6;
%% DISPLAYING INPUTS

disp(‘CASE#2’)
disp(‘LOWER LIMIT AND UPPER LIMITS OF INTEGRATION DO not MATCH x(1) AND x(LAST)’)
disp(‘ ‘)
disp(‘INPUTS’)
disp(‘The x-data is’)
x
disp(‘The y-data is’)
y
fprintf(‘ Lower limit of integration, a= %g’,a)
fprintf(‘\n Upper limit of integration, b= %g’,b)
% Choose how many divisions you want for splining from a to b
n=1000;
fprintf(‘\n Number of subdivisions used for splining = %g’,n)
disp(‘ ‘)
disp(‘ ‘)

%% THE CODE

xx=a:(b-a)/n:b;
% Using spline to approximate the curve from x(1) to x(last)
yy=spline(x,y,xx);
intvalue=trapz(xx,yy);

%% DISPLAYING OUTPUTS

disp(‘OUTPUTS’)
fprintf(‘ Value of integral is = %g’,intvalue)
disp(‘ ‘)
disp(‘___________________________________________’)
%% CASE 3

%% INPUTS

% Integrate the discrete function y from x=1 to 6.5
% with y vs x data given as (1,2), (4,16), (2,7), (6.5,18)
% The x-data is not in ascending order
% Defining the x-array
x=[1 4 2 6.5];
% Defining the y-array
y=[2 16 7 18];
% Lower limit of integration, a
a=3;
% Upper limit of integration, b
b=6;
%% DISPLAYING INPUTS

disp(‘CASE#3’)
disp(‘LOWER LIMIT AND UPPER LIMITS OF INTEGRATION DO not MATCH x(1) AND x(LAST) ‘)
disp(‘AND X-DATA IS NOT IN ASCENDING OR DESCENDING ORDER’)
disp(‘ ‘)
disp(‘INPUTS’)
disp(‘The x-data is’)
x
disp(‘The y-data is’)
y
fprintf(‘ Lower limit of integration, a= %g’,a)
fprintf(‘\n Upper limit of integration, b= %g’,b)
% Choose how many divisions you want for splining from a to b
n=1000;
fprintf(‘\n Number of subdivisions used for splining = %g’,n)
disp(‘ ‘)
disp(‘ ‘)

%% THE CODE
[x,so] = sort(x); % so is the sort order
y = y(so); % y data is now in same order as x data
xx=a:(b-a)/n:b;
% Using spline to approximate the curve from x(1) to x(last)
yy=spline(x,y,xx);
intvalue=trapz(xx,yy);

%% DISPLAYING OUTPUTS

disp(‘OUTPUTS’)
fprintf(‘ Value of integral is = %g’,intvalue)
disp(‘ ‘)

____________________________________________________________

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How do I integrate a continuous function in MATLAB

Many students ask me how do I do this or that in MATLAB. So I thought why not have a small series of my next few blogs do that. In this blog I show you how to integrate a continuous function.

The MATLAB program link is here.

The HTML version of the MATLAB program is here.

___________________________________________

%% HOW DO I DO THAT IN MATLAB SERIES?
% In this series, I am answering questions that students have asked
% me about MATLAB. Most of the questions relate to a mathematical
% procedure.

%% TOPIC
% How do I integrate a continuous function?

%% SUMMARY

%% INTRODUCTION

disp(‘ABSTRACT’)
disp(‘   This program shows you how to integrate’)
disp(‘   a given function ‘)
disp(‘ ‘)
disp(‘AUTHOR’)
disp(‘   Autar K Kaw of http://autarkaw.wordpress.com’)
disp(‘ ‘)
disp(‘MFILE SOURCE’)
disp(‘   http://nm.mathforcollege.com/blog/integration.m’)
disp(‘ ‘)
disp(‘LAST REVISED’)
disp(‘   March 29, 2009’)
disp(‘ ‘)

%% INPUTS

% Integrate exp(x)*sin(3*x) from x=2.0 to 8.7
% Define x as a symbol
syms x
% Assigning the function to be differentiated
y=exp(x)*sin(3*x);
% Assigning the lower limit
a=2.0;
% Assigning the upper limit
b=8.7;

%% DISPLAYING INPUTS

disp(‘INPUTS’)
func=[‘ The function is to be integrated is ‘ char(y)];
disp(func)
fprintf(‘ Lower limit of integration, a= %g’,a)
fprintf(‘\n Upper limit of integration, b= %g’,b)
disp(‘ ‘)
disp(‘ ‘)

%% THE CODE

% Finding the integral using the int command
% Argument 1 is the function to be integrated
% Argument 2 is the variable with respect to which the
% function is to be integrated – the dummy variable
% Argument 3 is the lower limit of integration
% Argument 4 is the upper imit of integration
intvalue=int(y,x,a,b);
intvalue=double(intvalue);

%% DISPLAYING OUTPUTS

disp(‘OUTPUTS’)
fprintf(‘ Value of integral is = %g’,intvalue)
disp(‘ ‘)

_________________________________________________________

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How do I differentiate in MATLAB?

Many students ask me how do I do this or that in MATLAB. So I thought why not have a small series of my next few blogs do that. In this blog, I show you how to differentiate a function.

The MATLAB program link is here.

The HTML version of the MATLAB program is here.

___________________________________________

%% HOW DO I DO THAT IN MATLAB SERIES?
% In this series, I am answering questions that students have asked
% me about MATLAB. Most of the questions relate to a mathematical
% procedure.

%% TOPIC
% How do I differentiate a function?

%% SUMMARY
% Language : Matlab 2008a
% Authors : Autar Kaw
% Mfile available at
% http://nm.mathforcollege.com/blog/differentiation.m
% Last Revised : March 21, 2009
% Abstract: This program shows you how to differentiate a given function

%% INTRODUCTION
clc
clear all
disp(‘ABSTRACT’)
disp(‘   This program shows you how to differentiate’)
disp(‘   a given function and then find its value’)
disp(‘   at a given point’)
disp(‘ ‘)
disp(‘AUTHOR’)
disp(‘   Autar K Kaw of http://autarkaw.wordpress.com’)
disp(‘ ‘)
disp(‘MFILE SOURCE’)
disp(‘   http://nm.mathforcollege.com/blog/differentiation.m’)
disp(‘ ‘)
disp(‘LAST REVISED’)
disp(‘   March 21, 2009’)
disp(‘ ‘)

%% INPUTS

% Differentiate 7 exp(3*x) once and find the value of the
% first derivative at x=0.5
% Define x as a symbol
syms x
% Defining the function to be differentiated
y=7*exp(3*x);
% Defining the point where you want to find the derivative
xx=0.5;

%% DISPLAYING INPUTS
disp(‘INPUTS’)
func=[‘ The function is to be differentiated is ‘ char(y)];
disp(func)
fprintf(‘ Value of x where you want to find the derivative, x= %g’,xx)
disp(‘ ‘)
disp(‘ ‘)

%% THE CODE
% Finding the derivative using the diff command
% Argument 1 is the function to be differentiated
% Argument 2 is the variable with respect to which the
% function is to be differentiated – the independent variable
% Argument 3 is the order of derivative
dydx=diff(y,x,1);
% subs command substitues the value of x
dydx_val=subs(dydx,x,xx);
%% DISPLAYING OUTPUTS
disp(‘OUTPUTS’)
derivative_func=[‘ The derivative of function ‘ char(y) ‘ is ‘ char(dydx)];
disp(derivative_func)
fprintf(‘ Value of dydx at x=%g is =%g’,xx,dydx_val)
disp(‘ ‘)

______________________________________________________________________

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Numerical Methods YouTube Video Progress

Since January 2009, I have been videotaping numerical methods course lectures in the Educational Outreach studio of University of South Florida, Tampa. Videos are made in 10-minute segments, not just because that is the limit of the length of YouTube videos, but because we firmly believe in making our resources pedagogically neutral. Ten minute videos allow an instructor to pick and choose what, when and how he or she wants the students to learn.

But some people have asked me – Why YouTube – why not put the same 10-minute videos on your own website. We do have the links to the YouTube videos on our own website, but there are many good reasons to go the YouTube route. The list of reasons below may be obvious to some, while others may not have thought about some of them.

1. Use YouTube’s storage space for the videos.

2. Use YouTube’s compression technology to make the videos stream faster on slow connections.

3. Use the power of Google and YouTube to tag and search the videos.

4. Use YouTube’s bandwidth as opposed to that of my school. My school’s IT department most probably would start screaming when the downloads pick up pace.

5. Use YouTube’s editing facilities to add annotations, links, and play lists.

6. Use the ubiquity of YouTube to reach a large audience.

7. Simple one stop process to let others embed the videos on their website.

8. Have open discussion on the videos via comments.

9. Get the videos rated so that we can judge their quality.

10. Use the “insight” tool of YouTube to analyze who is watching the videos.

______________________________________________________________________

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MATLAB code for the efficient automatic integrator

In the previous post, we discussed why doubling the number of segments in the automatic integrator based on multiple-segment trapezoidal rule is more efficient than increasing the number of segments one at a time. But this advantage involves having to store the individual function values from previous calculations and then having to retrieve them properly. This drawback was circumvented very efficiently by using the formula derived in another previous post where there is no need to store individual function values.

The matlab file for finding a definite integral by directly using the multiple segment trapezoidal rule from this post is given here (matlab file, html file), while the matlab file that uses the more efficient formula from this post is given here (matlab file, html file). Here are the inputs to the programs.

% a = Lower limit of integration
% b = Upper limit of integration
% nmax = Maximum number of segments
% tolerance = pre-specified tolerance in percentage
% f = inline function as integrand

a=5.3;
b=10.7;
nmax=200000;
tolerance=0.000005;
f=inline(‘exp(x)*sin(2*x)’)

We ran both the program on a PC and found that the more efficient algorithm (51 seconds) ran in half the time as the other one (82 seconds). This is expected, as only n function evaluations are made for 2n-segments rule with the efficient formula, while 2n+1 functions evaluations are made for the original formula.

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An efficient formula for an automatic integrator based on trapezoidal rule

In the previous post, we discussed why doubling the number of segments in the automatic integrator based on multiple-segment trapezoidal rule is more efficient than increasing the number of segments one at a time. But this advantage involves having to store the individual function values from previous calculations and then having to retrieve them properly. This drawback can be circumvented very efficiently as explained below. What you will see is that there is no need to store individual function values.