Techniques and methods for obtaining solutions to different kind of Ordinary Differential Equations is investigated in Scilab. The approach is based on solving different kind of Ordinary Differential Equations with different method some which are user defined for example Euler's method and other which are ready made in Scilab for example Runge-Kutta, Fehlberg's runge-Kutta, Adams-Bashforth and Stiff (belonging to stiff category problems). Emphasis is placed on mathematical justifiation of the approach. Time required to complete a task and step size for desired accuracy of solution are the main concern and basis of comparison between methods. On the basis of this approach problem and difficulties in Scilab are observed and suggestions are made for their remedies.

1. Kishor Vaigyanik Protsahan Yojana
(Department of Science and Technology, Government of India)
June 11, 2009
Summer Camp Report on
Numerical Solution to Ordinary Dierential
Equations in Scilab
by
Rahul Kumar Soni
(rahul@ismu.ac.in)
Department of Fuel and Mineral Engineering
Indian School of Mines University Dhanbad
under the mentorship of
Professor Kannan M. Moudgalya
(kannan@iitb.ac.in)
Department of Chemical Engineering
Indian Institute of Technology Bombay, India
1

2. Certicate
This is to certify that Mr. Rahul Kumar Soni, a stu-
dent of Department of Fuel and Mineral Engineering from
Indian School of Mines University Dhanbad has done his
Kishor Vaigyanik Protsahan Yojana summer camp project
at Department of Chemical Engineering of Indian Insti-
tute of Technology Bombay from May 11, 2009 to June
10, 2009 under my guidance.
The project work entitled Numerical Solution to Ordi-
nary Dierential Equations in Scilab embodies the origi-
nal work done by Mr. Rahul Kumar Soni during his sum-
mer camp.
Signature
Professor Kannan M. Moudgalya
Department of Chemical Engineering
Indian Institute of Technology Bombay
2

3. Acknowledgement
My sincere thanks to Prof. Kannan Moudgalya for having given
me this opportunity to work on such a lucrative project. I wish to
thank the KVPY cell for having provided a great work environment.
This project would not have been complete without the guidance and
timely inputs from Prof. G K Srinivasan ( Professor, Department
of Mathematics, IIT Bombay), Prof. V K Gupta (Tata Steel Chair
Professor at Department of Fuel and Mineral Engineering, ISMU),
Dr. Nikkam Suresh(Asso. Prof. and HOD, Department of Fuel and
Mineral Engineering, ISMU), Dr. Biswajit Paul (Department of En-
vironment Science and Engineering, ISMU), Dr. S Bhattacharya
(Asso. Prof., Department of Fuel and Mineral Engineering, ISMU),
Dr. N K Singh(Asso. Prof. and HOW, Department of Mechani-
cal Egineering, ISMU), Ms. Inderpreet Arora (M.Tech Student, IIT
Bombay), Ms. Sandhya Sourirajan (B.E. student, Coimbtore Insti-
tute of Technology), all teachers and my friends.
I am thankful to the entire work force at ERTS lab, CDEEP and
Chemical Department of IIT Bombay for the part they played in
making the period of work a joyful experience.
Last but not the least, I thank to Indian Institute of Science
Banglore (the Organizing Institute) and Department of Science and
Technology, Government of India (the sponsoring agency) for their
support and encouragement.
3

4. Abbreviations and meaning of symbols
ODE: ordinary dierential equation
IVP: initial value problem
BVP: boundary value problem
rk: Runge-Kutta 4th order method
rkf: Fehlberg's Runge-Kutta order 4 and 5 method (RKF45)
LT: Laplace Transform
ILT: Inverse Laplace Transforms
//**S: Shows the start of the part of program executed online in Scilab
window.
//**E: Shows the end of the part of program executed online in Scilab win-
dow.
//##S: Shows start of the writer's contribution to the program or in the
matter taken from references or
complete work of writer.
//##E: Shows end of the writer's contribution to the program or in the
matter taken from references or
complete work of writer.
Machine congurations
Simulation has done in two machines of following congurations
1. Notepad
Microsoft Windows Vista Home Premium, Intel(R) Core(TM)2 Duo
CPU 1.67GHz 32 bit operating system, 2GB RAM, 160GB HDD.
2. Lab PC
Microsoft Windows XP Professional version 2002 SP2, Intel(R) Pen-
tium(R) D CPU 3.40GHz, 1GB RAM, 160GB HDD.
Disclaimer
1. All the programs written in the report are tested in Scilab and
produced gure are attached in the report, so in any case not proper
working of program is might be due to printing mistake.
2. For purpose of contracting space in the report; position of some
plots/gures have changed, however they are correct.
3. In some places in programs '...' this notation has used at last of
the lines. Notation is for continuation purpose, in running program
this may replace by continuing next line.
4

5. Contents
1 Scilab 8
1.1 SCILAB HISTORY . . . . . . . . . . . . . . . . . . . 8
1.2 HOW TO OBTAIN ? . . . . . . . . . . . . . . . . . 8
1.3 INSTALLATION REQUIREMENTS . . . . . . . 8
2 Method involve in solution to ODE 9
2.1 Analytical Solutions of Ordinary Dierential Equa-
tions (ODEs) . . . . . . . . . . . . . . . . . . . . . . 9
2.1.1 Introduction to Dierent kind of ODEs and
their analytical Solution . . . . . . . . . . . . 9
2.2 Numerical methods for solution of ODEs available in
Scilab . . . . . . . . . . . . . . . . . . . . . . . . . . 12
2.2.1 Euler's method (Non-Sti ): Solution of ODEs
of non-sti type by the basic denition of dier-
entiation and by fourier transform expansion. . . 12
2.2.2 Scilab ODE Solvers . . . . . . . . . . . . . . 12
3 Some Useful Commands regarding to ODE solutions: 13
4 Stability of a solution and Sti/Non-sti problems 13
4.1 Stability . . . . . . . . . . . . . . . . . . . . . . . . . 13
4.2 Sti and Non-sti problems . . . . . . . . . . . . . . 14
5 Scilab Programs for solving ODEs and their exam-
ples 16
5.1 Solution to Simple ODE explicit in terms of indepen-
dent variable . . . . . . . . . . . . . . . . . . . . . . 16
5.2 Solution by Euler method . . . . . . . . . . . . . . . 18
5.3 A simple ODE of type dy/dt = f(t,y) . . . . . . . . . 20
5.4 Sti type problem and its solution by other methods
comparing with sti method . . . . . . . . . . . . . . 21
5.5 Solution to Linear rst order linear system of ODEs 25
5.6 Solution to the system of Linear ODEs with help of
Laplace Transform . . . . . . . . . . . . . . . . . . . 26
5.7 Solution to a Second Order ODE with constant coef-
cients . . . . . . . . . . . . . . . . . . . . . . . . . . 28
5.8 Solution to Higher order non-homogeneous ODEs with
constant coecients . . . . . . . . . . . . . . . . . . 33
5

6. 6 Miscellaneous Problems 36
6.1 Predator-Prey model . . . . . . . . . . . . . . . . . . 36
6.1.1 Predator-Prey problem solution by Euler's method 38
6.1.2 Predator-Prey problem solution by Scilab ODE
Solver . . . . . . . . . . . . . . . . . . . . . . 42
6.2 Lorenz ow and Lorenz equation . . . . . . . . . . . 44
6.3 Curve tting with Scilab for Data analysis purpose . 46
6.4 Solution to ODE where analytical is solution is dicult 51
6.4.1 Solution of Example 11 by Scialb ode solver . 51
6.5 Modeling and Simulation of a Pressure wave generator 52
7 Demerits and diculties in Scilab and suggestion for
improvement 55
6

7. Abstract
Techniques and methods for obtaining solutions to dierent kind
of Ordinary Dierential Equations is investigated in Scilab. The
approach is based on solving dierent kind of Ordinary Dieren-
tial Equations with dierent method some which are user dened
for example Euler's method and other which are ready made in
Scilab for example Runge-Kutta, Fehlberg's runge-Kutta, Adams-
Bashforth and Sti (belonging to sti category problems). Empha-
sis is placed on mathematical justication of the approach. Time
required to complete a task and step size for desired accuracy of so-
lution are the main concern and basis of comparison between meth-
ods. On the basis of this approach problem and diculties in Scilab
are observed and suggestions are made for their remedies.
Preamble
In mathematics, an ordinary dierential equation (or ODE) is
a relation that contains functions of only one independent variable,
and one or more of its derivatives with respect to that variable. Sir
Isaac Newton has rst developed the concept of ordinary Dierential
Equations for a series solution in the eld of Astronomy and from
that time ODE has wide applications in Industry, scientic work,
Economics, Ecology, Geology and many other elds including bio-
logical sciences. ODE always gives the relation among independent
variables, dependent variable and their derivatives which are di-
cult to analyze. A better analysis of these equations requires their
solution in terms of independent and dependent variables. Various
techniques are available for the solution of these ODEs in dierent
software packages like Mathematica, Maple, Matlab, Scilab etc. This
report mainly deals with the solution of ODEs in Scilab by dierent
methods like Adams-Bashforth, Runge-kutta , Fehlberg's Runge-
Kutta , Discrete time simulation and nally Sti type of equations
along with examples, comparison of their solutions (error involved,
smoothness of curve).
7

8. Aim
Aim of this paper is to examine the apparent trend in
Simulation of Ordinary Dierential Equation using Scilab
by various methods, their applicability and nally ana-
lyzing the method with optimize solution for a various
category of problems.
1 Scilab
1.1 SCILAB HISTORY
It is an open source software package developed at INRIA (France), for system
control and signal processing applications. It also features a wide variety of
tools for various Engineering and Mathematical applications. It was introduced
as an Open source alternative to MATLAB. It is also a vector and matrix based
program. Scilab introduced Scicos equivalent to Simulink. It has constantly
undergone vital changes ever since its inception in 1994.
1.2 HOW TO OBTAIN ?
It can be freely downloaded at the home page(http://www.scilab.org/). The site
oers the latest version for all the prominent Operating systems. The Windows
version is a 83.6 MB zip le ( Scilab 5.1.1). The user is free to modify the source
which is also readily available at the same link. The package also includes a
simulator called SCICOS which is an open source alternative to SIMULINK of
MatLab. The minimal package does not include subject specic toolboxes and
are to be downloaded separately.
1.3 INSTALLATION REQUIREMENTS
The Scilab installation le has to be unzipped into a folder from which it can
be easily installed into the desired location by double clicking the installation
le. In case of Linux, the setup le in binary version is a 15.6 MB .tar le which
has to be extracted into a directory. Open the Linux terminal and then move
to the directory into which the SCILAB .tar le has been extracted. Typing
the make command from the SCILAB directory produces an executable le,
which can be invoked by typing scilab .The standard package occupies only
130 MB of disk space upon installation.
8

9. 2 Method involve in solution to ODE
2.1 Analytical Solutions of Ordinary Dierential Equa-
tions (ODEs)
(Preparation done by attending Continuing Education Program-
CEP02 on Ordinary Dierential Equations at Indian Institute of
Technology Bombay held from May 11,09 to May 22,09.)
2.1.1 Introduction to Dierent kind of ODEs and their analytical
Solution
1. Simple linear ordinary dierential equation
dy
= ax + bxy
dt
dx
= cy + dxy
dt
2. Clairut's ODE
dy dy
y=x + f( )
dx dx
3. Singular and General solution, General solution as Envelope of Singular
solution.
4. Curl, Divergence, Laplacian of a vector, Convex and Concave domain.
5. Exact or Total Dierential Equation
M dx + N dy = 0
6. Necessary and Sucient condition for Exact or Total Dierential Equa-
tion.
∂M ∂N
=
∂y ∂x
∂f (x, y)
M=
∂x
∂f (x, y)
N=
∂y
7. Along with vector eld Mˆ + N ˆ
i j must be conservative in nature.
8. Linear rst order ODE and its solution by operation with Integrating
factor.
dy
+ P (x)y = Q(x)
dx
9

10. 9. Bernoulli's Equation and its solution.
dy
+ P (x)y = Q(x)y n
dx
10. Orthogonal Trajectories.
11. Solution of second order Non-homogeneous dierential Equation by method
of variation of parameters.
12. Higher order dierential Equation with variable coecients.
n−1
Dn y + aj (x)Dj y = 0
j=0
Calculation of Wronskian.
13. Homogeneous Linear Dierential Equation with constant coecients
n−1
Dn y + aj Dj y = 0
j=0
In cases of Real distinct, Real repeative and complex roots.
14. Non-homogeneous Linear Dierential Equation with constant coecients
n−1
Dn y + aj Dj y = R(x)
j=0
Solution of such problems require Annihilators, which are
T able1
R(x) Annihilator
xk eax (D − a)k+1
xk eax sinbx {(D − a)2 + b2 }k+1
xk eax cosbx {(D − a)2 + b2 }k+1
15. Improper Integrals (1st and 2nd kind).
16. Riemann Integrals and its properties, theorems, tests.
17. Functional function.
f (x + 1) = x(x)
and
f (1) = 1
10

11. 18. Gamma function and its properties.
∞
Γ(a) = exp(−t)ta−1 dt
0
19. Beta function and its properties.
1
β(a, b) = xa−1 (1 − x)b−1 dx
0
20. Euler's formula
π
Γ(a)Γ(a − 1) =
sin(πa)
21. Exponential type function, Necessary and Sucient condition for Laplace
Transform.
22. Laplace Transform and its properties.
∞
F (s) = L{f (t)} = exp(−st)f (t)dt
0
23. Shift theorem, Uniqueness Theorem for Laplace Transform.
24. Inverse Laplace Transform , properties, theorems.
L−1 {F (s)} = f (t)
25. Solution of ODE, system of linear dierential equation by Laplace Trans-
form method.
26. Application of Laplace Transform in calculating Integrals.
27. Convolution theorem.
H(s) = F (s) ∗ G(s)
then
1
h(t) = (f ∗ g)(t) = f (τ )g(τ − t)dτ
0
28. Homogeneous Linear Dierential Equation with constant coecients and
their dierent combinations as:
3 Legendre Equation
3 Tchebychev Equation
3 Airy Equation
3 Hermits Equation
3 Laguerre Equation
3 Bessel's Equation
3 Hypergeometric Equation
3 Jacobi's Equation.
11

12. 2.2 Numerical methods for solution of ODEs available in
Scilab
2.2.1 Euler's method (Non-Sti): Solution of ODEs of non-sti type
by the basic denition of dierentiation and by fourier transform
expansion.
2.2.2 Scilab ODE Solvers
Although there are other ode solvers like ode, dassl, dassrt, odedc are available
in Scilab but we will consider only 'ode' solver here.
General call syntax of Scilab ODE solver 'ode'
y = ode( type , y0 , t0 , t, f (t, y))
where
y is a function of t and dy/dt = f(t,y)
y0 and t0 are initial condition as y(to ) = y0
t is the range for the solution of ODE
type is a character string given in a single quote and direct machine for any
specic type of numerical solution method. This methods and corresponding
string are as follows;
3 '(empty input)': Default Lsoda method uses non-sti procedure initially and
then uses Sti backward dierentiation method , if required.
3 'adams': Solves Non-sti problems by Adams-Bashforth method.
3 'rk': Uses Runge-Kutta 4th order method to solve non-sti problem.
3 For non-sti and mildly sti problems, uses Fehlberg's Runge-Kutta pair
'rkf ':
of order 4 and 5 (RKF45) method, should not use when high accuracy is
desired.
3 'x':Similar to Fehlberg's Runge Kutta method, with easier user interface (for
non-sti problems).
3 'root':Use default lsoda method and gives roots of the solution (for non-sti
problems).
3 'discrete':For Discrete time simulation, can solve ODE at discrete points (for
non-sti problems).
3 'sti ': Only method which can deal with to deal with sti problems.
Note: However, sti ode solver is particularly for sti problems but other can
also solve them under depression of intervals for the whole range of t or for the
segment where solution is unstable, same is explained in an example later also.
12

13. 3 Some Useful Commands regarding to ODE so-
lutions:
3 ODE solvers: ode, dassl, dassrt, odedc
3 Polynomials: poly, roots, coe, horner, clean, freq
3 Linear systems: syslin
3 Programming: function, de, argn, for, if, end, while, select,warning, er-
ror, break, return
3 Comparison symbols: ==, =, , =, (and),| (or)
3 Execution of a le: exec
3 Character strings: string, part, evstr, execstr
3 Graphics: plot, xset, driver, plot2d, xgrid, locate, plot3d, Graphics
3 Interconnected dynamic systems: scicos
Furthermore, Scilab provides several facilities for nonlinear calculations. Numer-
ical simulation of systems of dierential equations is made by the ode primitive.
Many solvers are available, mostly from odepack, for solving sti or non-sti
systems. Implicit systems can be solved by dassl. There is a number of optional
arguments available for solving ode's (tolerance parameters, jacobian, order of
approximation, time steps etc). For ode solvers, these parameters are set by the
global variable %ODEOPTIONS. Minimizing non linear functions is done the
optim function. Several algorithms (including non dierentiable optimization)
are available. Codes are from INRIA's modulopt library. Enter help optim for
more a more detailed description.
(Reference: http://www.scilab.org/doc/intro/node51.html)
4 Stability of a solution and Sti/Non-sti prob-
lems
4.1 Stability
Let us take dy/dt = ky, k C solution of this is in the form of y = exp(kt) , this
approaches y→ 0 as t→ ∞ when Re(k)0. If numerical solution exhibit this
nature then called A-stable solution and don't have stability problems.
While solving above equation by Runge-Kutta method we get yn = ϕ(h, k)yn
n
or yn = {ϕ(h, k)} y0 , for a stable solution |ϕ(h, k)| must be 1.
13

14. 4.2 Sti and Non-sti problems
What's a sti equation? No, precise denition exist. Operationally
dY
= f (x, y)
dx
Y (x0 ) = Y0
is 'Sti ' if its solution by some methods requires (perhaps in a portion of an
interval) a signicant depression of the step size in order to avoid instabilities.
Take an example of
dy
= −15y
dt
solution of this by Euler's method shows followoing plots for dierent values of
steps of t (i.e. h). As h goes smaller solution tends to stable condition.
Although a sti problem can be solve by general methods of numerical
solution of ODE with a precision of step increment.
Problems other than sti therefore those not show instability while solve by
general methods of Solution of ODE are Non-sti problems, also these don't
requires more precision on step increment.
Another example of sti equation is following linear combination of dierential
equations as:
14

15. and the ODE is
dY
= AY
dt
solution of this ODE for two values of h; h1 = 2.702703 × 10−2 and
−2
h2 = 2.73972 × 10 is
(References: http://en.wikipedia.org/wiki/stiff_equation,http://
www.physics.arizona.edu/~restrepo/475B/Notes/source/node16.html)
15

16. 5 Scilab Programs for solving ODEs and their
examples
5.1 Solution to Simple ODE explicit in terms of indepen-
dent variable
dy
= f (x)
dx
Above can be easily solve by basic integration method like Summation of
elemental rectangular area, Trapezoidal rule, Simpon's 1/3 rd etc.
Here an examples present to show integration by Summation of elemental
rectangular area. Here area between xi and xi+1 can be calculated in three ways
a. h*f(xi )
f (xi )+f (xi+1 )
b. h*(
2 )
c. h*f(xi+1 )
Above three are called Lower sum, Middle sum and Upper sum respectively.
Program 1:
//##S
function [I]=integration(stype,a,b,n,f )
//stype is the way of integration may be 'U','M' 'L' for upper, middle
lower sum
//a,b is the range of x for integration, n=no. of intervals, f be the function
dened online
h=(b-a)/n;
xset('window',1)
if (stype'L')(stype'M')(stype'U') then
error('stype must be L,M,U as string ');
abort;
end
if stype=='L' then x=a j=1
elseif stype=='M' then x=a+(h/2) j=2
else x=a+h j=3
end
A=zeros(1,n)
X=zeros(1,n)
Y=zeros(1,n)
for i=1:n
X(1,i)=x
y=feval(x,f )
Y(1,i)=y
16

17. A(1,i)=h*y
x=x+h
end
subplot(2,2,1)
plot(X,Y)
title('x,y function plot')
I=sum(A)
B=zeros(1,n);
for i=1:n
B(1,i)=sum(A(1,1:i))
end
subplot(2,2,2)
plot(X,B) title('Integrated curve for '+stype)
endfunction
//##E
The function name above is given as integration and calling syntax is
integration(stype,a,b,n,f ), here stype is the type of integration (Lower, Middle
or Upper sum), a and b are limit of integration, n is the no. of intervals for
acccuracy of result.
Example 1:
Let
dy
= x2 + x + 3
dx
//**S
//##S
de('[z]=f(x)','z=x^2+x+3');
getf('integration.sci');
integration('L',0,5,10,f )
ans =
68.41875
integration('M',0,5,10,f )
ans =
69.165625
integration('M',0,5,10,f )
ans =
69.91875
//**E
//##E
Three answer shown above are for Lower, Middle and Upper sum respetively,
let them call L-ans, M-ans and U-ans.
Note:
17

18. L − ans M − ans U − ans
and from the exact solution we have answer as 69.166667, which is closest to
the M-ans. Why? This can be explained by gure.
Fig 1.4 shows the plot of integrated curve of given problem, there are three
curves for each sum and so make the curve line thick.
Fig 1
Note: However, above program is correct to get answer and individual plots
but given Fig 1 is not produce by above program. Some more programming is
required for the same.
5.2 Solution by Euler method
as we know of the ODE is
dy
= g(t, y)
dt
then
g(y + h) = g(y) + hg(t, y)
18

19. or
yn+1 = yn + hg(tn , yn )
Following above program for solution by Euler's method can be written as:
Program 2:
function [t,y]=Euler(t0,y0,tn,h,g)
//Euler 1st order method for solving ODE
//dy/dx=g(t,y)
//t0 and tn are the range of t and h is the interval bet
ymaxallowed=1e+100; //Actually it is abs(ymaxallowed)
t=t0:h:tn; y=zeros(t); n=length(y); y(1)=y0;
for j=1:n-1
y(j+1)=y(j)+h*g(t(j),y(j));
if abs(y(j+1))ymaxallowed then
disp('Euler-warning: underow or overow')
n=j; t=t(1,1:n); y=y(1,1:n);
break;
end;
end;
endfunction;
(Reference: numerical and statistical methods with SCILAB
for science and engineering, vol. 1, gilberto e. urroz)
Example 2:
dy 1
//Let the function is
dt = f (t, y) = 2(1+y) with initial condition y(0)=-1
,from t:2 to 10
//**S
//##S
t0=1; y0=0;
de('[z]=g(t,y)','z=1./(2*(1+y))');
getf('Euler.sci') //calling earlier dened Euler program
Euler(t0,y0,tn,h,g);
plot(t,y)
xtitle('Fig 2-Solution plot to Example 1 by Euler method','t','y')
//##E
//##S
19

20. Fig 2
5.3 A simple ODE of type dy/dt = f(t,y)
Example 3: Let us solve the problem of Example 2 again
//**S
//##S
t0=1; y0=0; //initial condition
de('[z]=f(t,y)','z=1./(2*(1+y))')//dening function
t=1:1:10;
y=ode(y0,t0,t,f );
plot(t,y)
xtitle('Fig 1-Solution plot to Example 1','t','y')
//**E
//##S
It can be observe that the output plots between dependent and independent
variable are similar for both the method therefore by Euler's method and Scilab
ODE solver.
It is dicult to distinguish them with such simple problems. Their perfor-
mance may be distinguish in case of complicated problems or in system of ODEs
with large no. of iterations.
20

21. Fig 3
5.4 Sti type problem and its solution by other methods
comparing with sti method
What is Sti type of problem?
Example 4:
Let us take an example dy/dx=-1000(y-x)+2001 (whose exact solution is
y1 (x)=-exp(-1000x) and y2 (x)=x+2 )
We will plot a particular solution y=y1(x)+y2(x)=-exp(-1000x)+x+2 of given
ODE
Note: In particular solution rst part goes zero rapidly while second part
produces a straight line
Program 4:
//##S
de('[y]=f(x)','y=-exp(-1000.*x)+x+2')
//PHASE 1: Plotting exact solution for small values of x
x1=linspace(0,0.01,100);
y1=feval(x1,f );
xset('window',0)
clf
subplot(4,1,1)
plot(x1,y1)
21

22. xtitle('Fig 4.1:Plot for exact solution for small values of x','x','y')
//PHASE 2:Plotting exact solution for small values of x
x2=linspace(0,1,100);
y2=feval(x2,f );
subplot(4,1,2)
plot(x2,y2)
xtitle('Fig 4.2:Plot for exact solution for comparatively large values of x','x','y')
//PHASE 3:Now we will see what happen when we plot the curves for
dierent numerical solutions
de('[Dy]=g(x,y)','Dy=-1000.*(y-x)+2001')
//Call 'Euler' program for Numerical solution by Euler method
getf('Euler.sci')
//Plot the curve for dierent increment as 0.0005,0.0010,0.0020,0.0025
//For initial condition y(0)=1
y0=1; x0=0;
h1=0.0005; h2=0.0010; h3=0.0020; h4=0.0025;
[x1,y1]=Euler(x0,y0,0.01,h1,g);
[x2,y2]=Euler(x0,y0,0.01,h2,g);
[x3,y3]=Euler(x0,y0,0.01,h3,g);
[x4,y4]=Euler(x0,y0,0.01,h4,g);
ymin=min([y1 y2 y3 y4]);
ymax=max([y1 y2 y3 y4]);
disp(ymin)
disp(ymax)
//Let us dene the axis properties
rect=[0 -4 0.01 6];
subplot(4,1,3)
//plots to give discrete curve with dierent signs
plot2d(x1,y1,-1,'011',,rect)
plot2d(x2,y2,-2,'011',,rect)
plot2d(x3,y3,-3,'011',,rect)
plot2d(x4,y4,-4,'011',,rect)
//plots to give continuous curves
plot2d(x1,y1,1,'011','h=0.0005',rect)
plot2d(x2,y2,5,'011','h=0.0010',rect)
plot2d(x3,y3,3,'011','h=0.0020',rect)
plot2d(x4,y4,2,'011','h=0.0025',rect)
title('Fig 4.3:Plots by Euler method for dierent value of h (increment)')
//Here corresponding curves for dierent colors are as follows
//h=0.0005, color=Black
//h=0.0010, color=Red
//h=0.0020, color=Pale Green
//h=0.0025, color=Blue
//PHASE 4:Now Let us see what happen when we plot the above curves
with sti ODE solver of Scilab
subplot(4,1,4)
22

23. rect=[0 1 0.01 2.2]
x1=x0:h1:0.01;
x2=x0:h2:0.01;
x3=x0:h3:0.01;
x4=x0:h4:0.01;
y1=ode(y0,x0,x1,g);
y2=ode(y0,x0,x2,g);
y3=ode(y0,x0,x3,g);
y4=ode(y0,x0,x4,g);
plot2d(x1,y1,1,'011','h=0.0005',rect)
plot2d(x2,y2,5,'011','h=0.0010',rect)
plot2d(x3,y3,3,'011','h=0.0020',rect)
plot2d(x4,y4,2,'011','h=0.0025',rect)
title('Fig 4.4:Plot for dierent h (x-increment) by sti solver of Scilab')
//From the plot it is obvious that sti ODE solver of Scilab stabilizes solution
for x-increment=0.0005,0.0010,0.0020,0.0025
//##E
Answer: - 3.0525, 5.3825
Explanation: However, program itself explains everything but still detailed
explanation is as follows:
Fig 4.1 is the plot of exact solution {y(x) = −exp(−1000x)andy(x) = x + 2}
for the smaller values of x.
Fig 4.2 gives the same solution plot for comparatively larger values of x. It is
obvious from the plots that solution faces sudden changes between 0.0 and
0.01.
Fig 4.3 gives the solution of ODE by Euler's Method at dierent values of
x-increment (h), for those curves are shown with dierent colors. Answer given
after execution of program is the minimum and maximum value of y
respectively, comes in the solution of ODE by Euler's method (just to check
degree of instability).
Color and their corresponding (h) curves are as follows:
Table 2
Value of regular x-increment Color of curve
h=0.0005 Black
h=0.0010 Red
h=0.0020 Pale Green
h=0.0025 Blue
Apparently as value of interval increases (h) instability of curve increases.
23

24. Fig 4.4 comes after solution of given ODE by SCILAB Sti ODE solver at all
previously dened values of h (colors of curve are corresponds to values of h as
similar to Fig 4.3).
It is apparent from the plots that sti solver stabilize the solution but in
better way for higher value of h.
Fig 4
(Reference: numerical and statistical method with SCILAB
for science and engineering, Volume 1, gilaberto e. urroz)
24

25. 5.5 Solution to Linear rst order linear system of ODEs
This is example gives the idea behind how to solve system of ODE with a single
ode solver
Example 5:
Let us take system of ODEs as
dy1
= y2 + x
dx
and
dy2
= −y1 + y2
dx
with initial condition as
y1 (0) = 1;y2 (0) = 2;between 0 to 2 with regular increment of 0.1
Program 5:
//**S
de('[w]=f(x,y)',['f1=y(2)+x';'f2=-y(1)+y(2)';'w=[f1;f2]'])
x0=0; Dx=0.1; xn=2;
y0=[1;2];
x=[x0:Dx:xn];
y = ode(y0,x0,x,f );
plot2d([x',x'],[y(1,:)',y(2,:)'],[1,-1],'111','y1@y2',[0 -3 2 4])
xtitle('ode solution to a system of ODEs in example 3','x','f(x)')
pause
//**E
In the Fig 5 continuous plot gives the solution y1 and the plot which is shown
in discrete pattern shows solution y2 .
25

26. Fig 5
(Reference: Youngstown state University,http://www.eng.ysu.
edu/~jalam/engr6924s07/sessions/session27/session27.pdf)
5.6 Solution to the system of Linear ODEs with help of
Laplace Transform
Let a system is as
dY
= A ∗ Y + g(t)
dt
where A is a n×n matrix if n no. of ODEs are there.
Taking Laplace Transform at both side
dY
L{ } = A ∗ L{Y } + L{g(t)}
dt
sL{Y } − Y0 = A ∗ L{Y } + L{g(t)}
(sI − A)L{Y } = Y0 + L{g(t)}
L{Y } = (sI − A)−1 Y0 + (sI − A)−1 L{g(t)}
26

27. Taking Inverse Laplace Transform at both side
Y = L−1{ (sI − A)−1 }Y0 + L−1 {(sI − A)−1 L{g(t)}}
Program 6: Following program has been made in MatLab.
//##S
function [Y]=LT(A,x0)
[m,n]=size(A);
if mn
error('matrix must of square type')
elseif mn
error('matrix must of square type')
end
syms s;
S=s*eye(2,2)-A;
Si=inv(S);
IL=ilaplace(Si);
syms t;
syms C1 C2;
ParticularSolutions=IL*x0'
GeneralSolutions=IL*[C1; C2]
//##E
Example 6: Let us consider the following problem
dy1
= 4y1 + 2y2
dt
dy2
= −2y2
dt
for initial conditions as y1 (0) =0 and y2 (0) =1 and 0t20
This problem can be re-written as
dY
=A∗Y
dt
where
4 2
A=
0 −2
//**S
//##S
A=[4 2;0 -2];
x0=[0 1];
LT(A,x0)
//##E
//**E
27

28. Answers:
ParticularSolutions =
y1=2/3*exp(t)*sinh(3*t)
y2=exp(-2*t)
GeneralSolutions =
y1=exp(4*t)*C1+2/3*exp(t)*sinh(3*t)*C2
y2=exp(-2*t)*C2
5.7 Solution to a Second Order ODE with constant coef-
cients
This type of ODE is discussed here, because it comes frequently in dierent
disciplines of Engineering especially in Electrical, Mechanical and Civil
Engineering.
Let us take an example of spring-mass system as shown
Let m be the mass, k be the spring constant and b be the damping constant
which appear due to frictional losses (may be heat losses in case of Electrical
LCR circuits).
And forces corresponding to above are
F = −kx
F = −bv
28

29. where x and v are instant position and velocity of mass respectively.
From Newton's 2nd law
d2 x
−kx − bv = m
dt2
dx d2 x
−kx − b =m 2
dt dt
d2 x k b dx
2
+ x+ =0
dt m m dt
b k
D2 + D+ =0
m m
which is a quadratic equation whose roots may be real distinct, real repeated
and complex roots. A particular type of roots gives the relation among m, b
and k or inequalities among them which is important because a particular type
of root decides the characteristic of system and its functioning, this can be
seen as follows:
Program to solve above ODE
Program 7:
//##S
function []=dampedoscillation(m,B,k,t0,x0,v0,t1,t2)
//m=mass, b=damping constant, k=spring constant, t0=initial time...
//at which xo,vo are given,
//t=time range for plot,
//F1=-kx, F2=-bv
//Di. Equation: (D^2+(b/m)D+(k/m))x=0
t=linspace(t1,t2,1000);
D=poly([(k/m) (B/m) 1],'D','coe ')
R=roots(D)
disp('roots')
disp(R)
a=R(1,1)
b=R(2,1)
//PHASE 1: For distinct real roots
if imag(a)==0 ab then
M=[exp(a*t0) exp(b*t0);a*exp(a*t0) b*exp(b*t0)][x0 v0]'
//A B are coecients in the solution x=Aexp(R1*t)+Bexp(R2*t)
A=M(1,1)
29

30. B=M(2,1)
x=(A*exp(a*t))+(B*exp(b*t))
v=(a*A*exp(a*t))+(b*B*exp(b*t))
a=(a*a*A*exp(a*t))+(b*b*B*exp(b*t))
clf
subplot(2,2,1)
plot(t,x)
title('x-t curve')
subplot(2,2,2)
plot(t,v)
title('v-t curve')
subplot(2,2,3)
plot(t,a)
title('a-t curve')
subplot(2,2,4)
plot(t,x,t,v,t,a)
title('x-t, v-t a-t curve at one place')
//PHASE 2: For repeated real roots
elseif a==b then
M=(exp(a*t0))*[1 t0;a (a*t0)+1][x0 v0]' //A B are coecients ...
//in the solution x=Aexp(R1*t)+Bexp(R2*t)
A=M(1,1)
B=M(2,1)
x=(A*exp(a*t))+(B*t.*exp(a*t))
v=(a*A*exp(a*t))+(B*(a*t+1).*exp(b*t)) a=(a*a*A*exp(a*t))+
...(a*B*(a*t+2).*exp(b*t))
clf
subplot(2,2,1)
plot(t,x)
title('x-t curve')
subplot(2,2,2)
plot(t,v)
title('v-t curve')
subplot(2,2,3)
plot(t,a)
title('a-t curve')
subplot(2,2,4)
plot(t,x,t,v,t,a)
title('x-t, v-t a-t curve at one place')
//PHASE 3: For complex roots
else
b=imag(a)
b=abs(b)
a=real(a)
c=(a*cos(b*t0))-(b*sin(b*t0))
d=(a*sin(b*t0))+(b*cos(b*t0))
30

31. M=(((exp(a*t0)))*[cos(b*t0) sin(b*t0);c d])[x0; v0]
//A B are coecients in the solution
x=Aexp(R1*t)+Bexp(R2*t)
A=M(1,1)
B=M(2,1)
x=((A*cos(b*t))+(B*sin(b*t))).*exp(a*t)
v=(exp(a*t)).*(((a*A+b*B)*cos(b*t))+((a*B-b*A)*sin(b*t)))
a=(exp(a*t)).*((((a*a*A+b*a*B)*cos(b*t))+((a*a*B-b*a*A)*sin(b*a*t)))+...
...(((-b*a*A-b*b*B)*sin(b*t))+((a*b*B-b*b*A)*cos(b*a*t))))
// Note that in all above three equation 'a' i.e. the real part of the roots
has the role for damping similarly 'b' i.e. the imaginary part of roots...
//has the role for frequency or say angular frequency=b
clf
subplot(2,2,1)
plot(t,x)
title('x-t curve')
subplot(2,2,2)
plot(t,v)
title('v-t curve')
subplot(2,2,3)
plot(t,a)
title('a-t curve')
subplot(2,2,4)
plot(t,x,t,v,t,a)
title('x-t, v-t a-t curve at one place')
end
xset('window',5)
clf subplot(3,1,1)
plot(x,v)
title('v(vertical axis)-x(horizontal axis)')
subplot(3,1,2)
plot(v,a)
title('a(vertical axis)-v(horizontal axis)')
subplot(3,1,3) plot(x,a)
title('a(vertical axis)-x(horizontal axis)')
endfunction;
//##E
Now we will see characteristics of system by their response plots
for three cases
3 Real distinct roots
3 Real repeated roots
3 Complex roots
31

32. Fig 6
Explanation: To plot above gures following of m, b and k have been taken
Table 3
Root type m b k Roots
Real distinct 2 4 1 -0.2928,-1.7071
Real repeated 2 4 2 -1,-1
Complex 2 4 50 -1±4.8989795 i
32

33. Assume Fig 6 a matrix of 4×3 then the rst row plots the position vs time
(x-t), velocity vs time (v-t) and acceleration vs time (a-t). Second, third and
forth rows gives velocity vs position (v-x), acceleration vs velocity (a-v) and
acceleration vs position (a-x) plots respectively.
Conclusions from the plot:
3 If damping constant is not zero then all motion decays.
3 In case of both real root cases motion is only decaying not oscillatory.
3 To make motion oscillatory roots must be complex which concludes
b2 4mk
this gives an inequlity between damping constant, mass and spring
constant.
3 In the complex root case when damping constant b= 0, oscillatory motion
goes on reducing which is obvious from all last plots of rst, second and
third rows.
3 In case of oscillatory motion acceleration is always proportionate to posi-
tion of massive body.
5.8 Solution to Higher order non-homogeneous ODEs with
constant coecients
n−1
Dn y + aj Dj y = R(x)
j=0
or
n−1
Dn y = − aj Dj y + R(x)
j=0
Example 7: Let the problem is
d4 y d3 y d2 y dy x2
+3 3 −2 2 +5 +y =
dx4 dx dx dx 2
or
d4 y d3 y d2 y dy x2
4
= −3 3 + 2 2 − 5 −y+
dx dx dx dx 2
for initial conditions at x=0
D3 y = −1, D2 y = 0, Dy = −1,y = 1
33

34. This problem can be re-written as
0.5x2
       
u3 (x) −3 2 −5 −1 u3 (x)
d  u2 (x)   1
 = 0 0 0   u2 (x)
∗
  0 
+ 
dx  u1 (x)   0 1 0 0   u1 (x)   0 
y(x) 0 0 1 0 y(x) 0
where ui (x)are intermediate function of x.
for this initial condition matrix can be written as
 
−1
 0 
u(0) =  
 −1 
1
Now above problem can be easily solve by Scilab ode solver for 0x10.
Let us also check here is sti ode solver is required for a non-sti problem or
already stabilize problem.
Program 8:
// D^4(y)+3D^3(y)-2D^2(y)+5D(y)+y=(x^2)/2, 0t10
clf
t0=0;
v0=[-1 0 -1 1]';
A=[-3 2 -5 -1;1 0 0 0;0 1 0 0;0 0 1 0];
t=t0:0.5:10;
de('[Dv]=f(x,v)','Dv=A*[v(1);v(2);v(3);v(4)]+[(x^2)/2;0;0;0]')
v=ode(v0,t0,t,f );
subplot(5,1,1)
plot(t,v(4,:))
xgrid(5)
xtitle('solution to the 4th order non-homogeneous ODE ...
(constant coecient), Abscissa-time',,'y')
subplot(5,1,2)
plot(t,v(3,:))
xgrid(5)
xtitle('solution to the 4th order non-homogeneous ODE...
(constant coecient), Abscissa-time',,'Dy')
subplot(5,1,3)
plot(t,v(2,:))
xgrid(5)
xtitle('solution to the 4th order non-homogeneous ODE...
(constant coecient), Abscissa-time',,'(D^2)y')
subplot(5,1,4)
plot(t,v(1,:))
34

35. xgrid(5)
xtitle('solution to the 4th order non-homogeneous ODE...
(constant coecient), Abscissa-time',,'(D^3)y')
yprim4=-3*v(1,:)+2*v(2,:)-5*v(3,:)-v(4,:)+(t.^2)/2;
subplot(5,1,5)
plot(t,yprim4)
xgrid(5)
xtitle('solution to the 4th order non-homogeneous ODE...
(constant coecient), Abscissa-time',,'(D^4)y')
pause
solution to the 4th order non−homogeneous ODE (constant coefficient), Abscissa−time
120
100
80
60
y
40
20
0
−20
0 1 2 3 4 5 6 7 8 9 10
solution to the 4th order non−homogeneous ODE (constant coefficient), Abscissa−time
100
80
60
Dy
40
20
0
−20
0 1 2 3 4 5 6 7 8 9 10
solution to the 4th order non−homogeneous ODE (constant coefficient), Abscissa−time
60
40
20
(D^2)y
0
−20
−40
−60
−80
0 1 2 3 4 5 6 7 8 9 10
solution to the 4th order non−homogeneous ODE (constant coefficient), Abscissa−time
40
20
0
(D^3)y
−20
−40
−60
−80
−100
−120
−140
−160
0 1 2 3 4 5 6 7 8 9 10
solution to the 4th order non−homogeneous ODE (constant coefficient), Abscissa−time
40
20
0
(D^4)y
−20
−40
−60
−80
−100
−120
0 1 2 3 4 5 6 7 8 9 10
Fig 7
(Reference: numerical and statistical method with SCILAB
for science and engineering, Volume 1, gilaberto e. urroz)
35

36. 6 Miscellaneous Problems
6.1 Predator-Prey model
Model describes the interaction of two species in an eco-system when they are
isolated from others. Here we will discuss growth rate of Fish and Shark when
they don't interact with other species (assumed). In this situation rate
changes will follow relations given below
Growth rate of sh=Rate at which sh born-Rate at which sh are eaten by
sharks
Growth rate of shark=Rate at which sh turned into sharks-Rate at which
shark die without sh
Now
Rate at which sh born∝F
Rate at which sh are eaten by sharks∝FS
Rate at which sh turned into sharks∝FS
Rate at which shark die without sh∝S
where F and S are instant population of Fish and Shark respectively. So, we
have
dF
= αF − βF S (1)
dt
dS
= βF S − γS (2)
dt
with the initial conditions
F (0) = F0
S(0) = S0
where
F0 and S0 are initial population at time t=0 (relatively).
α :growth rate of sh in absence of shark (1/years)
γ :death rate of sharks in the absence of their prey i.e. sh (1/years)
β :death rate per encounter of sh with sharks (1/sharks/years)
36

37. :eciency of turning predated sh into shark (shark/sh)
From above equations, it is apparent that
1. In absence of shark
dF
= αF
dt
F = F0 eαt (3)
Thus population of sh increases exponentially.
2. In absence of sh
dS
= −γS
dt
S = S0 e−γt (4)
Thus population of sh decreases exponentially.
Non-dimensionalization:
Non-dimensionalization of equations is a process to make all terms
dimensionless. Benets of this are described later
If we nondimensionalize according to
F
F∗ =
F0
S
S∗ =
S0
t∗ = αt
putting this in equations 1 and 2, we get
dF ∗ βS0 ∗ ∗
∗
= F∗ − F S
dt α
dS ∗ βF0 ∗ ∗ γ ∗
= F S − S
dt∗ α α
if we write
βS0
a=
α
βF0
b=
α
37

38. γ
c=
α
Then equations can be re-writtem as
dF ∗
= F ∗ − aF ∗ S ∗ (5)
dt∗
dS ∗
= bF ∗ S ∗ − cS ∗ (6)
dt∗
with initial conditions
F ∗ (0) = 1
S ∗ (0) = 1
So, benets of non-dimensionalization
Equation becomes easy by reducing no. of undetermined constants.
Initial condition are not undetermined for solution of this ODE.
6.1.1 Predator-Prey problem solution by Euler's method
Program 9: Program for the Solution of the above problem.
//##S
function []=PP(Alpha,Beta,Gamma,Epsilon,t,Dt,F0,S0)
//Note: t must be in years (absolute, not by transformation) and...
//Dt must be in terms of months
a=Beta*S0/Alpha;
b=Epsilon*Beta*F0/Alpha;
c=Gamma/Alpha;
t=t*Alpha;
Dt=(1/52)*Dt;
//Since Dt is given in terms of weeks, so converting it to years
Dt=Alpha*Dt;
t=0:Dt:t; //t here is the vector from 0 to t multiplied by Alpha
//We have {dF^{*}}/{dt^{*}}=F^{*}-aF^{*}S^{*} and...
//{dS^{*}}/{dt^{*}}=bF^{*}S^{*}-cS^{*} as transformed equations
//Above equations can be re-written as dy1/dt=y1-a*y1*y2 and...
//dy2/dt=b*y1*y2-c*y2
//Now the RHS of both above equations are assumed as f1 and f2
d=size(t); e=d(1,2);
f1=zeros(1,e);
38

39. f2=zeros(1,e);
y1=zeros(1,e+1);
y2=zeros(1,e+1);
y1(1,1)=1;
y2(1,1)=1;
f1(1,1)=y1(1,1)-a*y1(1,1)*y2(1,1);
f2(1,1)=b*y1(1,1)*y2(1,1)-c*y2(1,1);
for i=1:e
f1(1,i+1)=y1(1,i)-a*y1(1,i)*y2(1,i);
f2(1,i+1)=b*y1(1,i)*y2(1,i)-c*y2(1,i);
y1(1,i+1)=y1(1,i)+Dt*f1(1,i);
y2(1,i+1)=y2(1,i)+Dt*f2(1,i);
end
y1max=max(y1);
y2max=max(y2);
y1min=min(y1);
y2min=min(y2);
Fmax=y1max*F0
Smax=y2max*S0
Fmin=y1min*F0
Smin=y2min*S0
disp(Fmin, Fmax, Smin, Smax)
disp(Fmin)
disp(Fmax)
disp(Smin)
disp(Smax)
if y1min y2min then
p=-y1min
else
p=-y2min
end
if y1max y2max then
q=y1max+0.1*y1max
else
q=y2max+0.1*y2max
end
r=(max(t))/Alpha;
rect=[0 p r q];
t=t/Alpha;
subplot(2,1,1)
plot2d(t,y2(1,1:e),5,'111',,rect)
plot2d(t,y1(1,1:e),11,'111',,rect)
xtitle('Fish and Shark population Vs time','time',...
'Shark(S/S0-Red),Fish(F/F0-Blue)')
subplot(2,1,2)
plot(y1(1,1:e),y2(1,1:e))
39

40. xtitle('Shark Vs Fish relative population plot',...
'Fish(F/F0)','Shark(S/S0)')
endfunction
//##E
Example 8: Following are the results come from the solution by Euler's
method for given values of
α = 0.7, β = 0.007, γ = 0.5, = 0.3, t = 50, F0 = 200, S0 = 50
Table 4
Time step(weeks) Fmin Fmin Smin Smax
1.00e+00 58 654 30 226
5.00e-01 79 543 40 206
2.50e-01 90 501 45 191
1.25e-01 95 483 47 184
6.25e-02 98 474 48 181
3.13e-02 99 470 45 180
1.56e-02 100 468 49 179
7.81e-03 100 467 49 178
Conclusion:
1. As steps goes smaller, answers tends to stabilized values.
2. It is aapprent from the plots that as time steps goes smaller solution tends
to cyclic solution with invariable cycle amplitudes.
Here, Fish and Shark relative population plot with time and their phase
portrait for three values of time steps 1.00e+00, 6.25e-02 and 7.81e-03 are
given.
40

41. Fig 8
However by Euler' method solution of Predator-Prey problems are possible
but for accurate result (i.e. cyclic curve Population Vs time, should be of
constant amplitudes or Predator Vs Prey population plot should be thin in
nature) time step should be suciently less and while doing such simulation, it
takes a longer time than usual and sometimes due to limited stack size not
gives the desired result.
(Reference: Stanford University, http://fluid.stanford.edu/
~finger/teaching/numerical_methods_02/tutorials/tutorial2.pdf)
41

42. 6.1.2 Predator-Prey problem solution by Scilab ODE Solver
Program 10:
//##S
function []=PPS(stype,Alpha,Beta,Gamma,Epsilon,t,Dt,F0,S0) //Here stype
is the method by which...
//solution is going to be taken and it may be 'adams', 'rk', 'rkf ', 'sti ' etc.
a=Beta*S0/Alpha;
b=Epsilon*Beta*F0/Alpha;
c=Gamma/Alpha;
t=t*Alpha;
Dt=(Dt/52)*Alpha;
t=0:Dt:t; y0=[1;1];
t0=0;
de('[w]=f(x,y)',['f1=y(1)-a*y(1).*y(2)';'f2=b*y(1).*y(2)-c*y(2)';'w=[f1;f2]'])
y=ode('stype',y0,t0,t,f );
t=t/Alpha;
subplot(2,1,1)
plot2d([t',t'],[y(1,:)',y(2,:)'],[11,5],'111','Fish@Shark',[0 -0.5 50 5])
title('Fish and Shark population Vs time')
subplot(2,1,2)
plot(y(1,:),y(2,:))
xtitle('Shark Vs Fish relative population plot','Fish(F/F0)','Shark(S/S0)')
endfunction
//##E
Above Predator-Prey problem has been also solved by adams, rk, rkf , sti
method and conclusions after them are as follows:
1. All these methods are able to give ne results even for a time step of 1 week.
2. Problem is not of sti type because for a same time step solution by
method other than sti method and solution by sti method are similar.
Solution is already stabilize in nature so there is no need for sti solution.
3. It has been observed that for larger time steps (for eg: 1 year) plots are non
smooth and similar for all method, one of this i.e. by rk method is given in Fig
10.
4. Plots in Fig 9 are given for adams, rk, rkf and sti method with time step
of 1 week.
42

43. Fig 9.1
Fig 9.2
43

44. Fig 10
6.2 Lorenz ow and Lorenz equation
The Lorenz equation was published in 1963 by a meteorologist and
mathematician from MIT called Edward N. Lorenz.This is a model for some of
the unpredictable behavior which we normally associate with the weather.
The Lorenz equation is commonly dened as three coupled ordinary
dierential equation like
dx
= σ(y − x)
dt
dy
= x(τ − z) − y
dt
dz
= xy − βz
dt
where
44

45. where the three parameter σ, τ , β are positive and are called the Prandtl
number, the Rayleigh number, and a physical proportion, respectively. It is
important to note that the x, y , z are not spacial coordinate. The x is
proportional to the intensity of the convective motion, while y is proportional
to the temperature dierence between the ascending and descending currents,
similar signs of x and y denoting that warm uid is rising and cold uid is
descending. The variable z is proportional to the distortion of vertical
temperature prole from linearity, a positive value indicating that the
strongest gradients occur near the boundaries.
Program 11:
//##S
function []=Lorenz(stype,Sigma,Tou,Beta,x0,y0,z0)
//Let for convenience xset('window',4)
clf
a=Sigma;
b=Tou;
c=Beta;
de('[w]=f(t,u)','w=[a*(u(2)-u(1)); u(1).*(b-u(3))-u(2); u(1).*u(2)-c*u(3)]');
t0=0;
t=0:0.01:50;
u0=[x0 y0 z0]';
u=ode(stype,u0,t0,t,f );
subplot(3,1,1)
plot(t,u(1,:),,'t','x')
subplot(3,1,2)
plot(t,u(2,:),,'t','y')
subplot(3,1,3)
plot(t,u(3,:),,'t','z')
xset('window',2)
clf
plot3d3(u(1,:),u(2,:),u(3,:))
title('Phase portrait of x,y,z')
endfunction
//##E
Example 9: Lorenz Equation is famous for its drastic dependency on initial
conditions and parameters σ, τ, β .
We will take here three examples by changing parameters to see their eect.
1. σ = 10, τ = 28 and β = 8/3 (which is the classical example).The initial
condition of the system is (x0 , y0 , z0 )= (3,15,1).
2. σ = 5, τ = 28 and β = 8/3 (which is the classical example).The initial
condition of the system is (x0 , y0 , z0 )= (3,15,1).
45

46. 3. σ = 10, τ = 100 and β = 8/3 (which is the classical example).The initial
condition of the system is (x0 , y0 , z0 )= (3,0,1).
Conclusions:
1. For all rk, rkf, adams solution are similar.
2. Fig 11 shows how solution changes drastically with variation in parameter
and initial conditions.
Fig 11
(Reference: http://planetmath.org/encyclopedia/lorenzEquation.html)
6.3 Curve tting with Scilab for Data analysis purpose
Curve tting is nding a curve which has the best t to a series of data points
and possibly other constraints. This section is an introduction to both inter-
polation (where an exact t to constraints is expected) and regression analysis.
Both are sometimes used for extrapolation. Regression analysis allows for an
approximate t by minimizing the dierence between the data points and the
curve. Technique to t set of data in a curve (or polynomial) is based on min-
imizing their sum of square of dierence between experimental data and data
provided by considered polynomial.
46

47. Here a program has been made without using Scilab readymade curve tting
functions. Program can t up to ve curve at a time and shoes their plot at
one place for analysis purpose. Function curvetting given below requires
three input these are vector of independent variable, vector of experimental
data and a vector which may have size up to ve and contains degree of
polynomials on which curve has to be tted.
As the result of executing program it gives the coecient of individual terms
of a polynomial. Program also gives the error plot for each tted curve and
value of sum of squares of dierence between experimental and value provided
by plot.
y = a0 + a1 x + ax2 + a3 x3 + a4 x4 ........
2
Program 12:
//##S
//let k be the degree of the polynomial
//Fundamental equation A*a=B
function []=curvetting(x,y,D)
xset('window',1);
clf
//let us rst dene how many elements are there in D
p=size(D);
q=p(1,2);
//##PHASE1: Getting A
subplot(6,2,1);
plot(x,y)
xgrid(#)
title('DATA CURVE')
m=size(x);
//n actually gives the no. of elements in the vector x or y n=m(1,2);
for l=1:q // starting of this for loop
k=D(1,l)
A=zeros(k+1,k+1);
for j=1:k+1
for i=1:k+1
A(i,j)=sum(x.^(i+j-2))
end
end
//##PHSAE2:Getting B
B=zeros((k+1),1);
for i=1:k+1;
B(i,1)=sum(y.*(x.^(i-1)))
end
//##PHASE3:Getting 'a' the coecient matrix
47

48. //'a' is the desired coecient matrix shows a0,a1,a2...
// in the tting polynomial
a=(AB)';
disp(a)
//##PHASE4:Ploting curve by dierent solutions
//c d be the rst last value of vector x
c=x(1,1);
d=x(1,n);
x1=linspace(c,d,50);
//getting the smooth value of x
x2=x1';
//getting transpose of x1
//we have made x1 because to get a smooth curve of resultant polynomial
x3=zeros(50,k+1);
// an assumed matrix for further operation
//we going to make x3 a matrix having rst row as...
//1,x2.^2,x2.^3.... for rst value of x2, similarly second row...
//will be for second value of x2, so in the above manner we...
//will get a matrix having its row as 1,x2.^2,x2.^3....
//now if any of the row of this matrix multiply with 'a'...
// suppose then give a vector then sum of that vector...
//element will be the corresponding y2(or say y3 here) for x2
x3(:,1)=1;
x3(:,2)=x1';
for i=1:k+1;
x3(:,i)=x2.^(i-1)
end
y3=zeros(50,1);
for i=1:50
y3(i,1)=sum(a.*x3(i,:))
end
//dening y4 as calculated from result polynomial according to 'a' matrix
subplot(6,2,(2*(l-1)+3))
plot(x2,y3)
xgrid(3)
title('CURVE FOR DEGREE='+string(k));
//PHASE5:To get the value of y corresponding to given x for resultant poly-
nomial
x4=zeros(n,k+1);
for i=1:k+1;
x4(:,i)=x'.^(i-1)
end
y4=zeros(n,1);
for i=1:n
y4(i,1)=sum(a.*x4(i,:))
end
48

49. //PHASE6:To get the error curve for corresponding plot,...
//i.e. plot between (ye-yr) x. where ye is the experimental... //
value and yr is the real value
z=y'-y4;
z1=z'.^2;
z2=sum(z1);
z3=sqrt(z1);
b=[k,z2];
subplot(6,2,2*(l+1))
plot(x,z3)
xgrid(3)
title('SUM OF SQUARE OF DEVIATIONS FOR THIS ERROR CURVE='+string(z2));
ylabel('DEGREE OF CURVE='+string(k));
end // end of rst for loop
endfunction;
//##E
Example 10.1: Let us try to t the known data for a circle of radius 5 centered
at origin.
Producing data
//**S
de('[y]=f(x)','y=sqrt(25-x.^2)')
x=-5:0.1:5;
y=feval(x,f );
//**E
Now tting the generated data for polynomial of degrees 1,2,3,4, and 5 by
using the program 12.
It is apparent from the plots second and third degree curves are more or less
representing the circle.
The answer came from program are as follows:
Degree Coecients Sum of square of deviations
1 3.9,-1.6D-17 142.87233
2 5.2,1.0D-16,-0.2 7.0324985
3 5.2,-2.5D-17,-0.2,8.5D-18 7.0324985
4 5.0,2.0D-16,-0.1,-3.0D-18,-0.01 1.8451398
5 5.0,-6.0,-0.1,1.1D-15,-0.01,-4.0D-17 1.8451398
49

50. Fig 12.1
From the above plot conclusion can be made that while approaching to
higher degree plots although sum of square of deviation (sum of least squares)
reduces but along with this smoothness of the curve is also reduces, which are
opposite to each other for a desired result in industry. So one should go with
the optimize result.
Example 10.2:
Similar thing has been done a Gaussian curve for σ = 0.9 ,xmean = 5and
thus the Fig 12.2 came out concludes that no tted curve able to represent the
Gaussian curve.
50

51. DATA CURVE
0.45
0.40
0.35
0.30
0.25
0.20
0.15
0.10
DEGREE OF CURVE=5 OF CURVE=4 OF CURVE=3 OF CURVE=2 OF CURVE=1
0.05
0.00
−10 −5 0 5 10 15 20
CURVE FOR DEGREE=1 SUM OF SQUARE OF DEVIATIONS FOR THIS ERROR CURVE=2.8021607
1.0 0.45
0.8 0.40
0.6 0.35
0.4 0.30
0.2 0.25
0.0 0.20
−0.2 0.15
−0.4
DEGREE
−0.6 0.10
−0.8 0.05
−1.0 0.00
−10 −5 0 5 10 15 20 −10 −5 0 5 10 15 20
CURVE FOR DEGREE=2 SUM OF SQUARE OF DEVIATIONS FOR THIS ERROR CURVE=2.3957277
0.08 0.40
0.06 0.35
0.04 0.30
0.02 0.25
0.00 0.20
−0.02 0.15
DEGREE
0.10
−0.04 0.05
−0.06 0.00
−10 −5 0 5 10 15 20 −10 −5 0 5 10 15 20
CURVE FOR DEGREE=3 SUM OF SQUARE OF DEVIATIONS FOR THIS ERROR CURVE=2.3957277
0.08 0.40
0.06 0.35
0.04 0.30
0.02 0.25
0.00 0.20
−0.02 0.15
DEGREE
0.10
−0.04 0.05
−0.06 0.00
−10 −5 0 5 10 15 20 −10 −5 0 5 10 15 20
CURVE FOR DEGREE=4 SUM OF SQUARE OF DEVIATIONS FOR THIS ERROR CURVE=3.0490262
0.03 0.45
0.02 0.40
0.01 0.35
0.00 0.30
−0.01 0.25
−0.02 0.20
0.15
DEGREE
−0.03 0.10
−0.04 0.05
−0.05 0.00
−10 −5 0 5 10 15 20 −10 −5 0 5 10 15 20
CURVE FOR DEGREE=5 SUM OF SQUARE OF DEVIATIONS FOR THIS ERROR CURVE=3.1182013
0.015 0.45
0.010 0.40
0.005 0.35
0.30
0.000 0.25
−0.005 0.20
−0.010 0.15
−0.015 0.10
0.05
−0.020 0.00
−10 −5 0 5 10 15 20 −10 −5 0 5 10 15 20
Fig 12.2
6.4 Solution to ODE where analytical is solution is di-
cult
Example 11:
Consider a problem
dy 2 + 18t + 68t2 + 180t3 + 250t4 + 250t5
=
dt y2
for initial condition y(t=1)=0, and 0t24.
Here we will solve this by Scilab ODE solver.
6.4.1 Solution of Example 11 by Scialb ode solver
//**S
//##S
de('[z]=g(t,y)','z=(2+18*t+68*t^2+180*t^3+250*t^4+250*t^4)/(y^2)')
t0=0; y0=1; tn=24; h=0.1; t=0:0.1:24;
y=ode(y0,t0,t,g); //We have already dene f(t,y) in previous solu-
tion.
plot(t,y)
xtitle('Solution plot to Example 11','t','y')
51

52. //##E
//**E
Solution plot to Example 11
1400
1200
1000
800
y
600
400
200
0
0 5 10 15 20 25
t
Fig 13
6.5 Modeling and Simulation of a Pressure wave generator
(One of the KVPY mentorship holder work on the same project in the Refrig-
eration Lab at Department of Mechanical Engineering of IIT Bombay during
KVPY summer camp 2009. We have model the system and simulate for opti-
mize result).
since there are spring attached (not shown) in the system for restoring pur-
pose and it has found that cylinders also behave like spring, so if we have have
combined spring constant as k, then natural frequency of the system
w0 = k/m
where m is the mass of the moving part of the system.
In this derivation damping forces due to friction and others are assumed to
be zero.
Since a current carrying coil is suspended in the magnetic eld, so a Lorentz
force appear as
52

53. 2πrnBi0
Fl = (L − x)sin(wt)
m
where
r =radius of coil
n =no. of turns per unit length of coil
B =Magnetic eld produce by Magnetic core
i0 =Amplitude of supplied AC current to coil
L =Length of the coil inside magnetic core at steady or mean state
x =Position of the moving part (or piston) at any time
w =Angular frequency of the AC current
while balancing these force we have the dierential equation as
d2 x 2 Bi0
+ w0 x = (L − x)sin(wt)
dt2 m
or
d2 x 2 2πrnBi0 2πrnBi0
+ {w0 + sin(wt)}x = Lsin(wt)
dt2 m m
Let
2
a = w0
2πrnBi0
b=
m
So we have equations as
d2 x
+ {a + bsin(wt)}x = bLsin(wt)
dt2
Above things are for ideal conditions, in practical we have one more term in
the equation as
d2 x b
+ {a + bsin(wt)}x = (2L + hg )sin(wt)
dt2 2
where hg is the height of air gap between coil and magnetic core.
To solve this dierential equations in Scilab we can re-write them as
dx1
= x2
dt
and
dx2 b
= −{a + bsin(wt)}x1 + (2L + hg )sin(wt)
dt 2
Following program have been made to solve above dierential equations
53

54. Program 13:
//##S
function []=mech1(n,r,B,w,i0,m,v0,t0,tn,Dt,hg,k,w0,L)
if length(hu)length(w0) then
error('length of w0 and lv must be same')
abort;
end
t=t0:Dt:tn;
p=length(w0);
a=(w0)^2; b=B*(i0)*n*2*%pi*r/m;
de('[z]=f(t,x)','z=[x(2);-(a+b*sin(w*t))*x(1)+(b/2)*(hg+2*L)*sin(w*t)]')
t0=[0; 0];
x=ode([0 v0]',0,t,f );
xset('window',k)
clf
subplot(2,1,1)
plot(t,x(1,:))
xtitle('Position Vs time plot','time','Position(x)')
subplot(2,1,2)
plot2d(x(1,:)',x(2,:)')
xtitle('Phase plot between velocity and position','Position','Velocity')
endfunction
##E
In the program v0 is the initial condition for the velocity, t0 and tn are start
and end time for solution, Dt is the time increment for time range, k is the
gure window no. on which we want the current solution plot.
For ideal conditions of manufacturing the solution plot should have sinusoidal
pattern, which depends upon all input parameteres.
Simulation for the above has been done for following parameters (all values
are given in their corresponding SI units) to approach ideal conditions and the
corresponding solutions are plotted in Fig 13:
−1
(a)n=12000m ,r=0.02m,B=0.5T,w=314rad/sec,
i0 = 2amp,m=0.03kg,v0 =0.25m/sec,t0 =0sec,tn =25sec,
Dt==0.05sec,hg =0.01m,k=0,w0 =250rad/sec,L=0.1m.
−1
(b)n=12000m ,r=0.02m,B=0.5T,w=314rad/sec,
i0 = 2amp,m=0.03kg,v0 =0.25m/sec,t0 =0sec,tn =25sec
,Dt==0.05sec,hg =0.01m,k=1,w0 =350rad/sec,L=0.1m.
−1
(c)n=12000m ,r=0.02m,B=0.5T,w=314rad/sec,
i0 = 2amp,m=0.03kg,v0 =0.25m/sec,t0 =0sec,tn =100sec
,Dt==0.05sec,hg =0.01m,k=2,w0 =330rad/sec,L=0.15m.
54

55. Fig 14
Fig 13 clearly explains how changes in parameters can change the solution.
7 Demerits and diculties in Scilab and sugges-
tion for improvement
(Scilab has been compared with MatLab for same problems and tasks. How-
ever, it is found that Scilab has some feature better than Matlab but here only
demerits and diculties found are shown for its development).
1. It is limited to the developer community and open source users. It needs
advertisement and implementation in various disciplines of Engineering
and Science for better feedback and development.
2. The implementation of some eld especially studied ODE is quite simple
and readymade functions are available. The program crashes after 4-5
function implementations or when no. of iterations are large. This has
been especially seen during implementation of a Chemical Engineering
ODE of sti type.
3. Scilab has a very poor help browser as compared to its competitive soft-
ware packages. While searching by some string, mostly Scilab only looks
for matching strings in the topic title only and not inside the topic also.
The help contains neither a categorical listing of functions nor an exten-
sive search. It does not feature tutorials or a Getting Started tool. The
online help available is also limited and not many tutorials are available.
55

56. 4. Scilab has week GUI in command, Editor and other windows while com-
paring with Matlab. While error for something wrong in Editor window,
Scilab only shows at which line it is and not shows the column position,
for complicated functions sometimes it becomes dicult to x it. Simi-
lar case is with brackets in the Editor window, at any moment it doesn't
shows desired animation for paired brackets.
5. Limited no. of toolboxes availability (if available then dicult to access)
lacks Scilab. For example a basic toolbox for computation of Laplace and
Inverse Laplace Transform which comes frequently in Electrical, Mechani-
cal, Chemical Engineering disciplines. The spectrum of topics dealt with,
even in the contributors page is not large.
6. Many times it crash or hang or shows 'stack size exceeded' for large no.
of iterations.
7. A good timer function is not available for comparative studies. Example:
comparison of Runge-Kutta, Fehlberg's Runge-Kutta, Adams-Bashforth
and Sti methods for the solution of ODEs.
8. Scilab don't have symbolic solution environment that means readymade
functions like sin, log, sinh etc. which reduces its capacity to solve sym-
bolic equations. Dicult to dene functions with symbolic terms, in the
manner we dene polynomial by 'poly' command. Example
−− y = poly([2, 43, 4], x, coef f )
−− derivat(y)
−− ans = 43 + 8x
Now how to derivate this
y = cosx + eax lnx
similar thing is for integration.
9. Frequently while loading les it hangs and request for unnecessary time.
10. While looking for help regarding plot or scicos related commands, help
browser doesn't show visual graphic example like plots and scicos model.
56

57. Summary and Conclusion
The purpose of the present investigation was to study an optimize
approach for solution of a particular category of ordinary dierential
equation. However, almost all type of ordinary dierential equation
can solve by all given methods but our concern is on two points rst
is the simulation time for numerical solution and the second is the
time step required for a desired accuracy of solution. In general for
a particular method time taken for computation increases as time
step (or may be independent variable step in some cases) reduces.
Comparative study has done by keeping one of this constant and
other as variable. It is found that although Euler's method can
solve problems but simulation time required by it is usually more or
say time step should be less in this case, especially in case of sti
category problem it takes large time as compared to other methods.
On, other hands in this study methods other than Euler's method
all methods are found to more or less similar (Note: Higher Engi-
neering problems are not taken in this study they may distinguish
these methods) until unless problem is not of sti type. In the sti
category problem it is possible to solve them other methods but sim-
ulation time are very high for a desired accuracy which is found to
be highest in case of Euler's method, Sti solver stands best here
in terms of simulation time, time step and performance of machine.
Meanwhile time of above work some other task related to main work
done for example curve tting, solution of ode by Laplace Transform
Technique which concludes that Scilab generates interactive plots at
one place for a better analysis but it should be equipped with some
essential toolboxes. The main conclusion of the approach is that
Scilab is capable to lead everyone but needs a sincere and successive
development for this achievement. Also Scilab is capable to solve dif-
ferent kind of ordinary dierential equations and their system which
regularly comes in industry. The important consequence is being
the Scilab free of cost and open source so that can be use easily and
sometimes it is faster than other software packages.
57

58. References
1. http://www.physics.arizona.edu/~restrepo/475B/Notes/source/node16.html,
http://en.wikipedia.org/wiki/sti_equation
2. numerical and statistical methods with SCILAB for science and
engineering, vol. 1, gilberto e. urroz
3. Youngstown state University, http://www.eng.ysu.edu/~jalam/
engr6924s07/sessions/session27/session27.pdf
4. Stanford University, http://uid.stanford.edu/~nger/teaching/
numerical_methods_02/tutorials/tutorial2.pdf
5. http://planetmath.org/encyclopedia/lorenzEquation.html
58