Difference between revisions of "Scilab/C4/ODE-Applications/English-timed"
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| − | | Welcome to the Spoken Tutorial on '''Solving ODEs using Scilab ode | + | | Welcome to the Spoken Tutorial on '''Solving ODEs using Scilab ode function''' |
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| − | |The '''ode''' function is an '''ordinary differential equation''' | + | |The '''ode''' function is an '''ordinary differential equation solver'''. |
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| − | || Here '''y zero''' is the initial | + | || Here '''y zero''' is the initial conditions of the '''ODEs''', |
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|01:36 | |01:36 | ||
| − | |'''theta zero''' is equal to '''pi by four''' and | + | |'''theta of zero''' is equal to '''pi by four''' and |
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|02:03 | |02:03 | ||
| − | |''' l equal to zero point five meter''' is the | + | |''' l equal to zero point five meter''' is the length of the '''pendulum.''' |
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| − | |Open ''' | + | |Open '''Pendulum dot sci''' on '''Scilab editor.''' |
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|03:03 | |03:03 | ||
| − | |Then we call the '''ode | + | |Then we call the '''ode''' function with arguments '''y zero, t zero, t''' and the function '''Pendulum.''' |
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| 03:12 | | 03:12 | ||
| − | | The solution to the | + | | The solution to the equation will be a '''matrix''' with two '''rows.''' |
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| − | |Let us solve '''Van der Pol equation''' using the '''ode | + | |Let us solve '''Van der Pol equation''' using the '''ode''' function. |
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| − | |The initial | + | |The initial conditions are '''v of two equal to one''' and '''v dash of two equal to zero. ''' |
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| − | | Switch to '''Scilab editor''' and open ''' | + | | Switch to '''Scilab editor''' and open '''Vander pol dot sci.''' |
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| − | | Then we define the '''function | + | | Then we define the '''function Vander pol''' and construct a system of '''first order ODEs.''' |
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| − | | Then we call '''ode | + | | Then we call '''ode''' function and solve the system of '''equations.''' |
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| − | | Save and execute the file ''' | + | | Save and execute the file '''Vander pol dot sci.''' |
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| − | |The initial conditions are '''x one zero | + | |The initial conditions are '''x one zero equal to minus ten''', '''x two zero equal to ten''' and '''x three zero equal to twenty five.''' |
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| − | |Let '''sigma''' be equal to '''ten, r''' be equal to '''twenty eight and b''' equal to '''eight by three.''' | + | |Let '''sigma''' be equal to '''ten, r''' be equal to '''twenty eight''' and '''b''' equal to '''eight by three.''' |
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| − | |We define the '''function Lorenz''' and then define the given constants '''sigma, r and b.''' | + | |We define the '''function Lorenz''' and then define the given constants '''sigma, r''' and '''b.''' |
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| − | |Then we call the '''ode | + | |Then we call the '''ode''' function to solve the '''Lorenz system of equations.''' |
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Revision as of 11:17, 3 March 2015
| Time | Narration |
| 00:01 | Dear Friends, |
| 00:02 | Welcome to the Spoken Tutorial on Solving ODEs using Scilab ode function |
| 00:09 | At the end of this tutorial, you will learn how to: |
| 00:12 | Use Scilab ode function |
| 00:15 | Solve typical examples of ODEs and |
| 00:18 | Plot the solution. |
| 00:21 | The typical examples will be: |
| 00:24 | * Motion of simple pendulum |
| 00:26 | * Van der Pol equation |
| 00:28 | * and Lorenz system. |
| 00:30 | To record this tutorial, I am using |
| 00:33 | Ubuntu 12.04 as the operating system |
| 00:36 | and Scilab 5.3.3 version. |
| 00:40 | To practice this tutorial, a learner should have basic knowledge of Scilab |
| 00:45 | and should know how to solve ODEs. |
| 00:48 | To learn Scilab, please refer to the relevant tutorials available on the Spoken Tutorial website. |
| 00:56 | The ode function is an ordinary differential equation solver. |
| 01:01 | The syntax is y equal to ode within parenthesis y zero, t zero, t and f. |
| 01:10 | Here y zero is the initial conditions of the ODEs, |
| 01:15 | t zero is the initial time, |
| 01:17 | t is the time range, |
| 01:19 | and f is the function. |
| 01:22 | Consider the motion of simple pendulum. |
| 01:25 | Let theta t be the angle made by the pendulum with the vertical at time t. |
| 01:33 | We are given the initial conditions- |
| 01:36 | theta of zero is equal to pi by four and |
| 01:39 | theta dash of zero is equal to zero. |
| 01:44 | Then the position of the pendulum is given by: |
| 01:47 | theta double dash t minus g by l into sin of theta t equal to zero. |
| 01:56 | Here g equal to 9.8 m per second square is the acceleration due to gravity and |
| 02:03 | l equal to zero point five meter is the length of the pendulum. |
| 02:11 | For the given initial conditions, we have to solve the ODE within the time range zero less than equal to t less than equal to five. |
| 02:22 | We also have to plot the solution. |
| 02:25 | Let us look at the code for solving this problem. |
| 02:29 | Open Pendulum dot sci on Scilab editor. |
| 02:34 | The first line of the code defines the initial conditions of the ODE. |
| 02:40 | Then we define the intial time value. And we provide the time range. |
| 02:46 | Next, we convert the given equation to a system of first order ODEs. |
| 02:52 | We substitute the values of g and l . |
| 02:56 | Here we take y to be the given variable theta and y dash to be theta dash. |
| 03:03 | Then we call the ode function with arguments y zero, t zero, t and the function Pendulum. |
| 03:12 | The solution to the equation will be a matrix with two rows. |
| 03:17 | The first row will contain the values of y in the given time range. |
| 03:21 | The second row will contain the values of y dash within the time range. |
| 03:27 | Hence we plot both the rows with respect to time. |
| 03:31 | Save and execute the file Pendulum dot sci. |
| 03:37 | The plot shows how the values of y and y dash vary with time. |
| 03:44 | Switch to Scilab console. |
| 03:47 | If you want to see the values of y, type y on the console and press Enter. |
| 03:54 | The values of y and y dash are displayed. |
| 03:58 | Let us solve Van der Pol equation using the ode function. |
| 04:03 | We are given the equation - |
| 04:06 | v double dash of t plus epsilon into v of t square minus one into v dash of t plus v of t equal to zero. |
| 04:20 | The initial conditions are v of two equal to one and v dash of two equal to zero. |
| 04:28 | Assume epsilon is equal to zero point eight nine seven. |
| 04:32 | We have to find the solution within the time range two less than t less than ten and then plot the solution. |
| 04:42 | Let us look at the code for Van der Pol equation. |
| 04:47 | Switch to Scilab editor and open Vander pol dot sci. |
| 04:53 | We define the initial conditions of the ODEs and time and then define the time range. |
| 05:01 | Since the inital time value is given as two, we start the time range at two. |
| 05:07 | Then we define the function Vander pol and construct a system of first order ODEs. |
| 05:15 | We substitute the value of epsilon with zero point eight nine seven. |
| 05:21 | Here y refers to the voltage v. |
| 05:25 | Then we call ode function and solve the system of equations. |
| 05:30 | Finally we plot y and y dash versus t. |
| 05:35 | Save and execute the file Vander pol dot sci. |
| 05:41 | The plot showing voltage versus time is shown. |
| 05:45 | Let's move onto Lorenz system of equations. |
| 05:50 | The Lorenz system of equations is given by: |
| 05:53 | x one dash equal to sigma into x two minus x one, |
| 06:00 | x two dash equal to one plus r minus x three into x one minus x two and |
| 06:08 | x three dash equal to x one into x two minus b into x three. |
| 06:16 | The initial conditions are x one zero equal to minus ten, x two zero equal to ten and x three zero equal to twenty five. |
| 06:29 | Let sigma be equal to ten, r be equal to twenty eight and b equal to eight by three. |
| 06:37 | Switch to Scilab editor and open Lorenz dot sci. |
| 06:44 | We start by defining the initial conditions of the ODEs. |
| 06:48 | Since there are three different ODEs, there are three initial conditions. |
| 06:54 | Then we define the inital time condition and next the time range. |
| 07:00 | We define the function Lorenz and then define the given constants sigma, r and b. |
| 07:08 | Then we define the first order ODEs. |
| 07:12 | Then we call the ode function to solve the Lorenz system of equations. |
| 07:18 | We equate the solution to x. |
| 07:21 | Then we plot x one, x two and x three versus time. |
| 07:28 | Save and execute the file Lorenz dot sci. |
| 07:33 | The plot of x one, x two and x three versus time is shown. |
| 07:39 | Let us summarize this tutorial. |
| 07:41 | In this tutorial we have learnt to develop Scilab code to solve an ODE using Scilab ode function. |
| 07:50 | Then we have learnt to plot the solution. |
| 07:53 | Watch the video available at the link shown below. |
| 07:56 | It summarizes the Spoken Tutorial project. |
| 07:59 | If you do not have good bandwidth, you can download and watch it. |
| 08:04 | The spoken tutorial project Team: |
| 08:06 | Conducts workshops using spoken tutorials. |
| 08:09 | Gives certificates to those who pass an online test. |
| 08:13 | For more details, please write to contact@spoken-tutorial.org. |
| 08:20 | Spoken Tutorial Project is a part of the Talk to a Teacher project. |
| 08:23 | It is supported by the National Mission on Eduction through ICT, MHRD, Government of India. |
| 08:31 | More information on this mission is available at the link shown below. |
| 08:36 | This is Ashwini Patil, signing off. |
| 08:38 | Thank you for joining. |
