PROJECT REPORT

IAS-NASI-INSA

Summer Research Fellowship Programme 2014

Spectral methods for solving Differential Equations

Rajesh Aggarwal Faculty of Engineering and Technology

Gurukula Kangri Vishwavidyalaya Haridwar

Under the guidance of

Dr. Pravin Kumar Gupta Department of Earth Sciences

Indian Institute of Technology, Roorkee Uttarakhand

NAME OF SRF:

Rajesh Aggarwal

REGISTRATION NUMBER:

MATS605

INSTITUTE WHERE WORKING:

Indian Institute of

Technology, Roorkee

DATE OF JOINING THE PROJECT:

11/06/2014

DATE OF COMPLETION OF THE PROJECT: 06/08/2014

NAME OF THE GUIDE:

Dr. Pravin Kr. Gupta

PROJECT TITLE:

Spectral methods for

solving Differential

equations

Dr. Pravin Kumar Gupta Rajesh Aggarwal

Department of Earth Sciences MATS605

Indian Institute of Technology Date:06/08/2014

Roorkee, Uttarakhand Date: 06/08/2014

Dr. Pravin Kumar Gupta

Department of Earth Sciences, Indian Institute of Technology Roorkee, Uttarakhand E-mail: pkgptfes@iitr.ernet.in

CERTIFICATE

This is to certify that the project entitled “Spectral methods for solving Differential Equations” is a bonafide work carried out by Rajesh Aggarwal, Second Year B.Tech in Electrical Engineering , Faculty of

Engineering and Technology, Gurukula Kangri Vishwavidyalaya, Haridwar at Indian Institute of Technology, Roorkee under my guidance, during the period of 11th June 2014 to 06th August 2014.

(Dr. Pravin Kumar Gupta)

Acknowledgement

I would like to thank Dr. Pravin Kumar Gupta for the continuous guidance

and patience. Being my 1st summer project I had lot of things to learn and

the amount of knowledge and experience I have gained from him is

matchless.

I owe my sincere thanks to him.

I thank Mr. Rahul Dehiya for his guidance and overall for making it a great learning experience.

I also thank IAS-NASI-INSA for their support throughout the programme

and for giving me this great opportunity. It would have not been possible

without this programme.

Rajesh Aggarwal

Table of Contents

1. Introduction...................................................................... 6

1.1. Analytical methods for solving Differential Equation (D.E.)

……………………………………………………………….……..………7 1.2. Numerical Methods……………………………………………....8

1.2.1. Conventional finite difference method (CFDM)…...……8 1.2.2. Spectral finite difference method (SFDM)…………………9

2. Methods ........................................................................ 10

2.1. Methods for solving D.E by CFDM…………….……………11

2.1.1. Matlab code for solving D.E by CFDM………………………12

2.2. Methods for solving D.E. by SFDM…………………………14 2.2.1. Matlab code for solving D.E by SFDM………………………15 2.2.2. Matlab code for function interpolant_raj………………..16

2.3. Solving Problem on Half-space with different number of grid points …………………………………………………………………..18

2.3.1. Conventional finite difference method…………………....18 2.3.1.1. Matlab code for CFDM………………………………………18

2.3.2. Spectral finite difference method………………….….………20 2.3.2.1. Matlab code for SFDM…………………………….…………20 2.3.2.2. Matlab code for function points_raj_finite………..23 2.3.2.3. Matlab code for function earths_k…………………….24

2.4. Layered earth Problem……………………...………….………25 2.4.1. Conventional finite difference method………….….……..25

2.4.1.1. Matlab code for CFDM………………………….…………..25 2.4.2. Spectral finite difference method…………………..…………27

2.4.2.1. Matlab code for SFDM…………………………..…………..27

3. Conclusion………….......................................................….30

4. References..................................................................... 32

6

Introduction

7

1.1. Analytical methods for solving differential equations:

Analytical solutions are exact solutions of differential equations.

It satisfies differential equations and boundary conditions.

For example: For specific problem of Newton laws of cooling we have differential

equations associated with it as:

2 25dy

ydt

With initial condition 0 40y

This problem can be easily solved analytically and give exact results having solution

as:

12

(25 )

dy

dt y

=> 2 ,

(25 )

dydt

y

=> 2 ,

(25 )

dydt

y

=> 0( 1) ln 25 2y t C

=> ln 25 2y t C

=> 2 225 t C t Cy e e e

=> 225 C ty e e

=> 2 225 25 ,C t ty e e Ae

On putting initial conditions we get

225 15 ty e

But all real physical system do not provide such simple differential equation

And it is tough task to solve complicated differential equations with analytical methods.

So, we need another methods that can be used to solve differential equations

associated with physical system.

8

1.2. Numerical Methods:

Numerical methods provide numerical approximation to the solutions of differential

equations.

Numerical methods do not provide exact solutions of differential equations. They give

approximation to the results.

We will deal with finite difference method methods for solving differential equations:

(1). Conventional finite difference method (CFDM)

(2). Spectral finite difference method (SFDM)

1.2.1. Conventional finite difference method: In conventional finite difference method we convert our differential equation to difference equation and then solve it to get our desired result. Differential equations can be converted to difference equation by replacing first and second prime by respective difference formula.

The derivative of a function f at a point x is defined by the limit

0

( ) ( )( ) lim

h

f x h f xf x

h

And, double derivative of a function f at a point x is defined by the limit

20

( ) 2 ( ) ( )( ) lim

h

f x h f x f x hf x

h

The above two formula can be derived by considering Taylor series of function f.

2( ) ( )( ) ( ) ( )

1! 2!

f a f af a h f a h h O h

Where ( )O h denotes higher order terms or by simply neglecting all higher order terms

we can write

( )( ) ( )

1!

f af a h f a h

Or,

( ) ( )( )

f a h f af a

h

9

1.2.2 Spectral finite difference methods:

When data are smooth Spectral finite difference methods can be used to solve

ordinary differential equations and partial differential equations with high accuracy in

simple domain.

Spectral finite difference methods can achieve high accuracy then conventional finite

difference methods (spectral finite difference methods can achieve accuracy up to ten

digits while finite difference can up to 2 digits with the same number of grid points).

In spectral finite difference methods we write the differential equations as the sum of

basic functions and find there coefficient such that the resultant function can satisfy

the differential equations.

For example: To solve any differential equation we can take a function

( ) ( )n n

n

f x a x

Such that ( )f x satisfy ( )j jf x u

Where ju is corresponding value of function at jx

Here we are free to choose ( )n x , ( )n x can be either

(i) trigonometric,

(ii) exponentials or

(iii) Simple combinations of polynomials.

10

Methods

11

2.1. Methods for solving differential equations by CFDM:

Let the given differential equation be:

22

20

d yk y

dx

………………………….. <01>

With boundary conditions (0) 1y & (1) ky e and k=1

To solve this equation by conventional finite difference method we need to convert this differential equations to difference equations using:

0

( ) ( )( ) lim

h

f x h f xf x

h

, And

20

( ) 2 ( ) ( )( ) lim

h

f x h f x f x hf x

h

Or say 1 1

2

2( ) i i i

i

y y yy x

h

where

1 1( )i iy y x (value of y at1ix ),

ix denotes the grid

points where we have to evaluate the value of differential equation and denotes space

between these grid points.

Then our differential equation will become:

=> 21 1

2

20i i i

i

y y yk y

h

Or, 2 2

1 1(2 ) 0i i iy k h y y --------------------- <02>

Let we have to solve this differential equation in interval [0, 1], here we can discretise

our data in as many interval we want by choosing appropriate value of h .

And get points as ix ih where i =0, 1, 2, ………

Let h =0.25 then from equation 02 we get set of algebraic equations as:

2 2

2 1 0(2 (0.25) ) 0y k y y

2 2

3 2 1(2 (0.25) ) 0y k y y

2 2

4 3 2(2 (0.25) ) 0y k y y

2 2

5 4 3(2 (0.25) ) 0y k y y

Now writing these set of equations in matrix form as:

Ay B

And then after putting boundary conditions and solving the matrix equation for y we

get our desired result.

12

2.1.1 Matlab codes for solving this differential equations

clc; clear; A=input('enter lower limit'); B=input('enter upper limit'); h=input('enter value of h'); K=input('enter value of k'); P=(B-A)/h+1; for i=0:P-1 Q(i+1)=A+i*h; end for i=1:P+1; E(i,i)=1; E(i,i+1)=-2; E(i,i+2)=1; end E=E./(h^2); E=E(1:P,2:P+1); disp('Enter boundary conditions'); a=input('Enter initial condition'); b=input('Enter final condition'); % solving differential equation starts here R=E; I=eye(P);L=zeros(P,P); for k=1:P L(k,k)=(K^2).*I(k,k); end m=R-L; j=1; p=P; y(j)=a; y(p)=b; e=zeros(1,P); c=zeros(1,P); d=zeros(1,P); for i=1:P; e(i)=-1*(m(i,j)*y(j)); end for i=1:P; d(i)=-1*(m(i,p)*y(p)); end for i=1:P; c(i)=e(i)+d(i); end c=c(2:P-1); c=c'; q=m; q=q(2:P-1,2:P-1); l=q\c; O=zeros(1,P); for i=1:P if i==1 O(1)=y(j); elseif i==P O(P)=y(p); else O(i)=l(i-1); end end plot (Q,abs(O));

13

Table 1: solution of differential equations 22

20

d yk y

dx

by CFDM

Graph 1: Solution of given differential equation by conventional finite difference

method

x Calculated

y

Exact y Error

0 1.0000 1.0000 0

0.25 1.2847 1.2840 0.0006803

0.50 1.6497 1.6487 0.0009844

0.75 2.1178 2.1170 0.0008121

1.0 2.7183 2.7183 0

0

0.5

1

1.5

2

2.5

3

0 0.2 0.4 0.6 0.8 1 1.2

y-va

lue

x-value

solution of

14

2.2. Methods for solving differential equations by SFDM:

WE can solve the same problem by spectral finite difference methods:

Our differential equation is:

22

20

d yk y

dx

………………………….. <01>

With boundary conditions (0) 1y & (1) ky e and k=1

To solve the given equation by spectral finite difference methods we have to choose

an interpolant ( ) ( )i iy x L x y such that it satisfy the differential equation.

Then we can write ( ) ( )i iy x L x y and ( ) ( )i iy x L x y

Now on putting value of ( )y x and ( )y x in equation (01) we get a matrix in form:

Ay B .

And then after putting boundary conditions and solving the matrix equation for y we

get our desired result.

15

2.2.1 Matlab codes for solving this differential equations by SFDM:

clc; clear; A=input('enter lower limit'); B=input('enter upper limit'); h=input('enter value of h'); K=input('enter value of k'); P=(B-A)/h+1; for i=0:P-1 Q(i+1)=A+i*h; end E=interpolant_raj(Q); % this function provides D-matrix disp('Enter boundary conditions'); a=input('Enter initial condition'); b=input('Enter final condition'); % solving differential equation starts here R=E^(2); I=eye(P);L=zeros(P,P); for k=1:P L(k,k)=(K^2).*I(k,k); end m=R-L; j=1; p=P; y(j)=a; y(p)=b; e=zeros(1,P); c=zeros(1,P); d=zeros(1,P); for i=1:P; e(i)=-1*(m(i,j)*y(j)); end for i=1:P; d(i)=-1*(m(i,p)*y(p)); end for i=1:P; c(i)=e(i)+d(i); end c=c(2:P-1); c=c'; q=m; q=q(2:P-1,2:P-1); l=q\c; O=zeros(1,P); for i=1:P if i==1 O(1)=y(j); elseif i==P O(P)=y(p); else O(i)=l(i-1); end end plot (Q,O);

16

2.2.2 Matlab code for function interpolant_raj.

function E=interpolant_raj(T) % T is vector containing points where we

have to interpolate % E is D matrix x=T; s=x; [~, n]=size(x); A=zeros(1,n); D=zeros(1,n); B=ones(1,n); E=zeros(n,n); k=1; for j=1:n

for i=1:n if i~=j A(k)=A(k)+1/(s(j)-x(i)); B(k)=B(k).*(s(j)-x(i)); D(k)=B(k).*A(k); E(k,k)=A(k); end

end k=k+1; end

G=ones(n,n); k=1; if n>2 for j=1:n l=1:n; l(:,j)=[]; m=1; for v=1:n-1 z=l(v); for i=1:n if i~=j && i~=z G(k,m)=G(k,m).*(s(j)-x(i)); E(k,z)=G(k,m)./B(z); end

end m=m+1; end k=k+1; end else

for j=1:n l=1:n; l(:,j)=[]; m=1; for v=1:n-1 z=l(v); for i=1:n E(k,z)=G(k,m)./B(z); end m=m+1; end k=k+1; end end

17

Table 2: solution of differential equations 22

20

d yk y

dx

by SFDM

Graph 2: Solution of given differential equation by SFDM

X Calculated y Exact y Error

0 1.000000000000000 1.000000000000000 0

0.25 1.283972703420206

1.284025416687741

0.00005271

0.50 1.648733207743882

1.648721270700128

-0.00001193

0.75 2.117079044491258

2.117000016612675

-0.00007902

1.0 2.718281828459046

2.718281828459046

0

0

0.5

1

1.5

2

2.5

3

0 0.2 0.4 0.6 0.8 1 1.2

y-va

lues

x-values

Solution of D.E.

18

2.3. Solving Problem on Half-space with different number of grid

points and comparing results:

Let our differential equation be 2

2

20

d Ek E

dx this differential equation is arises due

to decay of electric field when it pass through earth surface of different conductivity.

Here 2k j where = earth conductivity, = Permeability having value,

= 74 10 and = angular frequency.

2.3.1. Conventional finite difference method:

For solving this differential equation by conventional finite difference method we will

first need to discretise the interval [0, 3000] into number of subinterval. Let h be gap

between two consecutive subintervals and N denotes number of subintervals.

2.3.1.1 Matlab code to check what happens when we increase number of grid

points (N) in CFDM.

Clc clear n=input('enter number of earth layer'); H=zeros(1,n); N=zeros(1,n); for i=1:n H(i)=input('Enter Depth of layers');

end M=input('enter frequency'); M=M*8*(pi)^2*(1e-07); S=zeros(1,n); for i=1:n S(i)=input('Enter sigma for each layer'); end disp('Enter boundary conditions'); a=input('Enter initial condition'); b=input('Enter final condition'); %h=input('enter h'); for N=3:3:30

h=H/N ;

[Q P]=points_raj_finite(n,h,H,N);

K=earths_k(n,M,S,N,Q,P); for i=1:P+1; E(i,i)=1; E(i,i+1)=-2; E(i,i+2)=1; end E=E./(h^2);

19

E=E(1:P,2:P+1);

% solving differential equation starts here R=E; I=eye(P);L=zeros(P,P); for k=1:P L(k,k)=(K(k)).*I(k,k); end m=R+(L*1i); j=1; p=P; y(j)=a; y(p)=b; e=zeros(1,P); c=zeros(1,P); d=zeros(1,P); for i=1:P; e(i)=-1*(m(i,j)*y(j)); end for i=1:P; d(i)=-1*(m(i,p)*y(p)); end for i=1:P; c(i)=e(i)+d(i); end c=c(2:P-1); c=c'; q=m; q=q(2:P-1,2:P-1); l=q\c; O=zeros(1,P); for i=1:P if i==1 O(1)=y(j); elseif i==P O(P)=y(p); else O(i)=l(i-1); end end T=(exp(1i*Q*(K(1)*1i)^(1/2))); J=abs(T); error = norm(J-abs(O),inf); loglog(N,error,'marker','*'); hold on end grid on, xlabel N, ylabel error title('Convergence of conventional finite difference method')

For:

Number of earth layer,(n)=1 Depth of layer=3000 Frequency=1 Sigma=1 Enter boundary conditions Initial condition (0)E =1

Final condition (3000)E =0

20

Graph 3: Variation in error as we increase number of grid points in conventional finite

difference method

2.3.2. Spectral finite difference method:

For solving this differential equation by spectral finite difference method we will first

discretise the interval [0, 3000] into number of subinterval. Let h be gap between two

consecutive subintervals and N denotes number of subintervals.

2.3.2.1 Matlab code to check what happens when we increase number of grid

points (N) in SFDM.

Clc clear n=input('enter number of earth layer'); H=zeros(1,n); N=zeros(1,n); for i=1:n H(i)=input('Enter Depth of layers');

end M=input('enter frequency'); M=M*8*(pi)^2*(1e-07); S=zeros(1,n); for i=1:n S(i)=input('Enter sigma for each layer');

21

end disp('Enter boundary conditions'); a=input('Enter initial condition'); b=input('Enter final condition'); %h=input('enter h'); for N=3:3:30

h=H/N ;

[Q P]=points_raj_finite(n,h,H,N);

K=earths_k(n,M,S,N,Q,P); E=interpolant_raj(Q);

% solving differential equation starts here R=E^2; I=eye(P);L=zeros(P,P); for k=1:P L(k,k)=(K(k)).*I(k,k); end m=R+(L*1i); j=1; p=P; y(j)=a; y(p)=b; e=zeros(1,P); c=zeros(1,P); d=zeros(1,P); for i=1:P; e(i)=-1*(m(i,j)*y(j)); end for i=1:P; d(i)=-1*(m(i,p)*y(p)); end for i=1:P; c(i)=e(i)+d(i); end c=c(2:P-1); c=c'; q=m; q=q(2:P-1,2:P-1); l=q\c; O=zeros(1,P); for i=1:P if i==1 O(1)=y(j); elseif i==P O(P)=y(p); else O(i)=l(i-1); end end T=(exp(1i*Q*(K(1)*1i)^(1/2))); J=abs(T); error = norm(J-abs(O),inf); loglog(N,error,'marker','*'); hold on end grid on, xlabel N, ylabel error title('Convergence of spectral finite difference method')

22

For:

Number of earth layer,(n)=1 Depth of layer=3000 Frequency=1 Sigma=1 Enter boundary conditions Initial condition (0)E =1

Final condition (3000)E =0

Graph 4: Variation in error as we increase number of grid points in SFDM

23

2.3.2.2 Matlab code for function points_raj_finite(n,h,H,N):

function [Q P]=points_raj_finite(n,h,H,N) % n is scalar having number of earth layer % H is vector containing Depth of each layers from the end of last one % N is vector containing number of points we want to calculate in each % layer % Q is vector containing coordinate of interpolated points % P is scalar having total number of points y=zeros(n,N(n));

for i=1:N(1) y(1,i)=i*h;

end

for i=1:(n-1) a=0; for l=1:i a=a+H(l); end for k=1:N(i+1)+1

y(i+1,k)=a+(k)*h;

end end P=sum(N)+1; k=2; Q(1)=0; for j=1:n for i=1:N(j) Q(k)=y(j,i); k=k+1; end end

24

2.3.2.3 Matlab code for function earths_k:

function K=earths_k(n,M,S,N,Q,P) j=1;K=zeros(1,P); % N is vector containing number of points we want to calculate in each % layer % n is scalar having number of earth layer % Q is vector containing coordinate of interpolated points % P is scalar having total number of points % K is vector having value of K for each corresponding points in Q % M Define (i*w*miu) here (scalar product of j,omega,miu) % S define sigma (vector(1,n) sigma for different layer) for i=1:N(1) K(i)=M*S(j); end h=N(1); a=2; for j=1:n-1 v=h+a; h=h+N(j+1); for i=v:h K(i)=M*S(j+1); end end a=1; for j=1:n-1

K(N(j)+a)=M*((S(j)*(Q(N(j)+a)-Q(N(j)+a-1))+S(j+1)*(Q(N(j)+a+1)-

Q(N(j)+a)))/(Q(N(j)+a)-Q(N(j)+a-1)+Q(N(j)+a+1)-Q(N(j)+a))); a=a+N(j); end

Comparison of error of CFDM and SFDM:

From Graph 3 and Graph 4 we can conclude that Spectral finite difference method

provide error of approx. 310 with only 9 grid points while conventional finite difference

method with 20 grid points and more.

25

2.4. Layered earth Problem:

Solving nonlinear differential equations arises due to Layered earth of different

conductivity by numerical methods:

Let us consider one example to understand this method.

Let our differential equation be 2

2

20

d Ek E

dx this differential equation is arises due to

decay of electric field when it pass through earth surface of different conductivity.

Here 2k j where = earth conductivity, = Permeability, having value

= 74 10 and = angular frequency

In this differential equation since for different earth layer is different or say is

function of x therefore 2k will be function of x .

We will solve this differential equation by conventional finite difference method and

spectral finite difference method and compare the results.

2.4.1. Conventional finite difference method:

For solving this differential equation by conventional finite difference method we will

first need to discretise the interval [0, 3000] into number of subinterval. Let h be gap

between two consecutive subintervals and N denotes number of subintervals.

2.4.1.1 Matlab code for solving this differential equation by CFDM:

clc clear n=input('enter number of earth layer'); H=zeros(1,n); N=zeros(1,n); h=input('enter h'); for i=1:n H(i)=input('Enter Depth of layers'); N(i)=H(i)/h ; end [Q P]=points_raj_finite(n,h,H,N); M=input('enter frequency'); M=M*8*(pi)^2*(1e-07); S=zeros(1,n); for i=1:n S(i)=input('Enter sigma for each layer'); end K=earths_k(n,M,S,N,Q,P); for i=1:P+1; E(i,i)=1;

26

E(i,i+1)=-2; E(i,i+2)=1; end E=E./(h^2); E=E(1:P,2:P+1); disp('Enter boundary conditions'); a=input('Enter initial condition'); b=input('Enter final condition'); % solving differential equation starts here R=E; I=eye(P);L=zeros(P,P); for k=1:P L(k,k)=(K(k)).*I(k,k); end m=R+(L*1i); j=1; p=P; y(j)=a; y(p)=b; e=zeros(1,P); c=zeros(1,P); d=zeros(1,P); for i=1:P; e(i)=-1*(m(i,j)*y(j)); end for i=1:P; d(i)=-1*(m(i,p)*y(p)); end for i=1:P; c(i)=e(i)+d(i); end c=c(2:P-1); c=c'; q=m; q=q(2:P-1,2:P-1); l=q\c; O=zeros(1,P); for i=1:P if i==1 O(1)=y(j); elseif i==P O(P)=y(p); else O(i)=l(i-1); end end plot (Q,abs(O),'marker','*'); grid on, xlabel Depth-from-earth, ylabel Electric-field title(['Calulation using finite difference D-Matrix for h=',num2str(h)])

27

2.4.2. Spectral finite difference method:

For solving this differential equation by Spectral finite difference method we will first

need to discretise the interval [0, 3000] into number of subinterval. Let h be gap

between two consecutive subintervals and N denotes number of subintervals.

2.4.2.1 Matlab code for solving this differential equation by SFDM:

Clc; clear; n=input('enter number of earth layer'); H=zeros(1,n); N=zeros(1,n); h=input('enter h'); for i=1:n H(i)=input('Enter Depth of layers'); N(i)=H(i)/h ; end [Q P]=points_raj_finite(n,h,H,N); M=input('enter frequency'); M=M*8*(pi)^2*(1e-07); S=zeros(1,n); for i=1:n S(i)=input('Enter sigma for each layer'); end K=earths_k(n,M,S,N,Q,P);

E=interpolant_raj(Q); disp('Enter boundary conditions'); a=input('Enter initial condition'); b=input('Enter final condition'); % solving differential equation starts here R=E^2; I=eye(P);L=zeros(P,P); for k=1:P L(k,k)=(K(k)).*I(k,k); end m=R+(L*1i); j=1; p=P; c=zeros(1,P); d=zeros(1,P); y(j)=a; y(p)=b; e=zeros(1,P); for i=1:P; e(i)=-1*(m(i,j)*y(j)); end for i=1:P; d(i)=-1*(m(i,p)*y(p)); end for i=1:P; c(i)=e(i)+d(i); end c=c(2:P-1); c=c'; q=m; q=q(2:P-1,2:P-1); l=q\c; O=zeros(1,P); for i=1:P if i==1 O(1)=y(j); elseif i==P O(P)=y(p); else O(i)=l(i-1); end end plot (Q,abs(O),'marker','*'); grid on, xlabel Depth-from-earth, ylabel Electric-field title(['Calulation using Spectral finite difference method D-Matrix for

h=',num2str(h)])

28

For:

Number of earth layer,(n)=2 Depth of 1st layer=1500 Depth of 2nd layer=1500 Frequency=1

Sigma 1 =1 and

2 =0.8

Enter boundary conditions Initial condition (0)E =1

Final condition (3000)E =0

Comparing Accuracy of conventional finite difference and spectral finite difference

method with different number of grid points

Conventional finite difference method Spectral conventional finite difference method

N=10 N=30 N=50 N=10

x E x E x E x E

0

300 600

900 1200

1500 1800

2100 2400

2700 3000

1

0.5421 0.2935

0.1589 0.0871

0.0494 0.0289

0.0173 0.0102

0.0049 0

0

100 200

300 400

500 600

700 800

900 1000

1100 1200

1300 1400

1500 1600

1700 1800

1900 2000

2100 2200

2300 2400

1

0.8193 0.6712

0.5499 0.4504

0.3688 0.302

0.2472 0.2023

0.1657 0.1358

0.1115 0.0919

0.0761 0.0633

0.0529 0.0443

0.0372 0.0313

0.0264 0.0223

0.019 0.0161

0.0136 0.0114

0 60

120 180

240 300

360 420

480 540

600 660

720 780

840 900

960 1020

1080 1140

1200 1260

1320 1380

1440

1

0.8875 0.7877

0.699 0.6204

0.5505 0.4885

0.4335 0.3846

0.3412 0.3027

0.2685 0.2382

0.2112 0.1874

0.1663 0.1476

0.131 0.1165

0.1036 0.0923

0.0824 0.0737

0.066 0.0593

0

300 600

900 1200

1500 1800

2100 2400

2700 3000

1

0.5557 0.3053

0.1678 0.0937

0.0546 0.0327

0.0188 0.0083

0.0034 0

29

2500 2600

2700 2800

2900 3000

0.0094 0.0074

0.0055 0.0037

0.0018 0

1500

1560 1620

1680 1740

1800 1860

1920 1980

2040 2100

2160 2220

2280 2340

2400 2460

2520 2580

2640 2700

2760 2820

2880 2940

3000

0.0533 0.0479

0.0431 0.0388

0.035 0.0315

0.0285 0.0257

0.0233 0.0211

0.0191 0.0173

0.0157 0.0142

0.0128 0.0115

0.0103 0.009

0.0079 0.0067

0.0056 0.0045

0.0034 0.0022

0.0011 0

Table 3: Solution of differential equation

22

20

d Ek E

dx with different number of grid points (N).

From the above table we can conclude that spectral finite difference methods give

better results with less number of grid points while conventional finite difference

method give the same result with at least 3 to 5 times number of grid points that

required by spectral finite difference methods.

30

Conclusion

31

[01]. All differential equations associated with real physical problems are not easily

solved with analytical methods so we need numerical methods to solve them.

[02]. We have compared the results of two numerical methods to solve our differential

equations:

(01). Conventional finite difference method

(02). Spectral finite difference method

[03]. Spectral finite difference methods give better results than conventional finite

difference method.

[04]. Spectral finite difference methods require less number of grid points as compared

conventional finite difference method to solve a problem with the same accuracy.

32

References

33

[01] Trefethen, L.N,(2000), Spectral methods in Matlab, SIAM publication.

[02]. Abramowitz and Stegun, (1972), Handbook of Mathematical Function, National

Bureau of Standards

[03]. Gilberto E. Urroz,( 2004), Numerical Solution to Ordinary Differential Equations

[04]. http://en.wikipedia.org/wiki/Finite_difference_method

[05]. http://www.engr.sjsu.edu/trhsu/Chapter%209%20Intro%20to%20FDM.pdf

[06]. http://en.wikipedia.org/wiki/Spectral_method

[07]. http://en.wikipedia.org/wiki/Finite_difference

[08]. http://www.csc.kth.se/utbildning/kth/kurser/DN2266/matmod11/Fourier3_2p.pdf

[09]. maths-people.anu.edu.au/~mike/NQinODE.pdf

[10]. tutorial.math.lamar.edu/Classes/DE/BoundaryValueProblem.aspx

[11].http://www.cdeep.iitb.ac.in/nptel/Mechanical/Heat%20and%20Mass%20Transfer

/Conduction/Module%205/main/5.1.html

[12]. en.wikipedia.org/wiki/Differential_equation

[13].https://www.whitman.edu/mathematics/multivariable/multivariable_17_Differential_Equations.pdf [14]. F. N. van de Vosse P. D. Minev ,Spectral element methods: theory and applications (w3.pppl.gov/m3d/reference/se5_nov_28_07.pdf)

[15]. Katarina Gustavsson, Spectral differentiation (www.math.nus.edu.sg/~matgkv/Lecture5.pdf) [16]. A.N. Malyshev,On the spectral differentiation (www.ii.uib.no/~sasha/INF263/spectral.pdf)

[17]. Lu Yixin , Introduction to Spectral conventional finite difference methods, September 26, 2007 [18]. ms.mcmaster.ca/~bprotas/CES712a/diff_02_4.pdf