finite difference method for the resolution

NATIONAL UNIVERSITY OF RWANDA

FACULTY OF SCIENCES

DEPARTEMENT OF APPLIED MATHEMATICS

ACADEMIC YEAR: 2009

FINITE DIFFERENCE METHOD FOR THE RESOLUTION

OF SOME PARTIAL DIFFERENTIAL EQUATIONS:

A MATLAB APROACH.

Project submitted to obtain Bachelors degree in Applied Mathematics

Presented by: BIZIMUNGU Anaclet & NDAHIMANA RUKUNDO Janvier

Supervisor: Dr NTAGANDA Jean Marie

Huye, October 2009

i

Dedication

To the Almighty God, for his salvation,

To my Darling NYIRAGASIGWA Marie,

To my parents,

To my dear sisters and brothers,

And to all my friends and relatives.

BIZIMUNGU Anaclet

Dedication

To the Almighty God, for his salvation,

To the memory of my late parents,

To my dear sisters and brother,

To all my friends and relatives,

And to the member of the group (GBU).

NDAHIMANA RUKUNDO Janvier

ii

Acknowledgement The work like this is not a fruit of two individuals. Many people were contributed to the realization of it.

We express our sincere gratitude to Dr. NTAGANDA J. Marie for having accepted to supervise this project. His guidance and a great number counseling he provided us contributed to the successful completion of this work.

We would like to express our sincere thanks to all lectures of the faculty of science, especially those of Applied Mathematics department and those of Physic department which contributed to our formation, specially the Dean of faculty of science, Dr. NDAHAYO Fidele, the Head of Applied Mathematics department, Dr. MINANI Froduald, lectures, Dr. MAHARA Isidore, Dr. LYAMBABAJE Alexandre and Dr. SAFARI Bonfils.

Our sincere gratitude goes to our parents, brothers, and sisters especially the family of NDAYISENGA Dogratias, the family of NTAMUKUNZI Joshua and uncle CYIZA Prosper deserve praise for the pain and sacrifices endured during our education.

All members of our family and friend, GBU-UNR, who in one way or another, contributed to our social and academic success we acknowledge your cares equally find our sincere thanks.

Our thanks are also expressed to the Government of Rwanda, the National University of Rwanda, especially the department of Applied Mathematics that availed us skills for the accomplishment of this work.

We would not forget to appreciate the company and friendship from our classmates, roommates especially those who contributed to the completion of this work.

May the Almighty God bless you all.

BIZIMUNGU Anaclet NDAHIMANA RUKUNDO Janvier

iii

Abstract The present work named Finite difference method for the resolution of some partial differential equations, is focused on the resolution of partial differential equation of the second degree.

To establish this work we have first present and classify the partial differential equations. And follow we had present and describe the finite difference method. After we had applied those methods for the numerical resolution of some partial differential equations.

The implementation was done using Matlab language. The result shows that the finite difference method is very efficient for the resolution of partial differential equations, the reason why it is useful for scientists such us engineers, physician...

Key words: Partial differential equations, finite difference method, forward difference, backward difference, centre difference, Euler method, explicit method, implicit method, and Crank-Nicolson method.

.

iv

Contents Dedication.i Acknowledgement...ii Abstract.......iii Content....iv

List of figures..........vi

Introduction......1 1 Presentation of partial differential equations ......................................................................3 1.1 Basic concepts and definitions........3 1.2 Classification of second order equations.....................................................................5 1.3 The canonical form ....8

1.3.1 Hyperbolic type .......8 1.3.2 Parabolic type ........10

1.3.3 Elliptic type ..11

2 Presentation of .finite methods.13 2.1 Grid lines...13 2.2 Finite different schemes ...14 2.3 First order partial derivative......15 2.3.1 Central partial difference formula .....15 2.3.2 Forward difference formula....15 2.3.3 Backward Difference Formula...16 2.4 First order partition derivative in two independent variables ......16 2.4.1 Central partial difference formulas in two independent variables.........16 2.4.2 Forward difference formulas in two independent variables .............18 2.4.3 Backward difference formulas in two independent variables ...........18 2.5 The second order partial derivative.......18 2.6 Partial derivative in two independent variables....18 2.6.1 Central difference formulas in two independent variables.............19

2.6.2 Forward difference schemes in two independent variables ..........................19 2.6.3 Backward difference schemes in two independent variables .......19 2.7 Some schemes that use a Finite Difference Method ........20 2.7.1 Explicit method .....20

v

2.7.2 Implicit method .....21 2.7.3 The Lax-Wendroff method .......25

3 Numerical solution of three forms of partial differential equations......26 3.1 Parabolic equations ......26 3.1.1 Forward difference (Explicitly scheme) ...27 3.1.2 The classical implicit method ...30 3.1.3 The Crank-Nicolson method .....33 3.2 Hyperbolic equations ...36 3.2.1 Implicit scheme .....37

3.2.2 Explicit scheme (Forward difference) ..38 3.3 Elliptic equation .......42

Conclusion.............................................................................................................................49 Bibliography..........................................................................................................................50

vi

List of figures 1.1 Coordinate system....6 2.1 Grid lines....14 2.2 The slop of the chord AB ......15 2.3 Forward derivative ....16 2.4 Backward derivative.......17 2.5 The classical explicit scheme.....22 2.6 The classical implicit scheme ...23 2.7 The Crank Nicolson scheme..24

3.1 Dimensional rod of length L..27 3.2 The MATLAB function for the resolution of parabolic equation in case of heat equation using the forward difference method...29 3.3 Forward difference method for example 3.1......30 3.4 A three dimensional representation of table given by the figure (3.3)...31 3.5 The first part of Matlab function for the resolution of parabolic equation in case of heat equation using the Crank-Nicolson method...34 3.6 The second part of Matlab function for the resolution of parabolic equation in case of heat equation using the Crank-Nicolson method....35 3.7 Crank-Nicolson method for example 3.3 ..36 3.8 A three dimensional representation of table given by the figure 3.7.... 37 3.9 Vibrating string..38 3.10 The MATLAB program for the resolution of the hyperbolic equation using a three level explicit method....40 3.11 Explicit three level difference method for example 3.4...42 3.12 A three dimensional representation of the table given by the figure 3.11...43 3.13 Stead state heat flow .. 44 3.14 The computational molecule for Laplace equation..45 3.15 The MATLAB program for the resolution of the Laplaces equation.....46 3.16 Numbering system for internal grid points..47 3.17 Approximate solution for example 3.6........48 3.18 A three dimensional representation of figure 3.17...48

Introduction

We are in the generation, where, the Communication Information Technology (CIT)

completely challenges the world. According to Economy Development Poverty Reduc-

tion Strategy (EDPRS), Rwanda as a country with low natural resources, need to es-

tablish an Information Technology (IT) based economy. It is why we prefer to develop

an IT education. It is so important for all scientists to be aware to the IT.

In Applied Mathematics we also need to excel in that technology, so that we can

well solve some complex problems. So there is a need to introduce IT into mathe-

matics courses. The use of MATLAB is integrated in such manner that is takes the

interested students and professionals, through this tool to understand the subject even

better. Note that, IT uses methods that have significant advantages over other classical

methods, because of its simplicity of analysis and computer codes, in solving problems

with complex geometries. For example Partial difference Equations (PDEs). PDEs are

equations that describe natural phenomena, and are encountered in a variety of appli-

cations in many fields, including continuum mechanics, potential theory, physics, and

geophysics, chemistry, and biology geology and mathematics economics. Note that those

problems have no relevance, if there are not solutions for them and only few of them

have analytic solutions. Also we must recognize that people who will use PDEs are not

all mathematician, so its necessary for us to develop them other methods, where they

will use computer programs to solve those problems. Therefore we need to establish

some methods that use IT to approach the given problem so that we can easily solve

them. The methods that can use IT are known as numerical methods which are tech-

niques of writing the given problem in the language that the computer can understand.

Here we are interested in resolution of some PDEs; the problem is that the classical

methods for the resolution of PDEs present many challengers such us: much calculus;

long time to get the solution; very difficult to draw the graph solution of problem with

complex geometries; and many of the PDEs which result from engineering problems can-

not be readily solved by analytical methods. Consequently, knowledge of the methods

of obtaining numerical solutions of PDEs is important to the modern engineer. As the

2domain of PDEs is very large, our subject will concern with the solution of the second

order PDEs; specifically our study will carry out the resolution of the three prominent

classical equations of mathematical physics namely: the wave equation, the diffusion

equation, and the Laplaces equation.

Therefore, numerical solution of these equations is the only recourse. Its why in our

work we present, and describe the finite different methods (FDMs), which is a numerical

technique for the resolution of PDEs, and then we apply them to solve some PDEs.

Our work is organized as follows: The first chapter presents the PDEs especially

the classification of second order PDEs in the three principle classical equations of

mathematical physics. In the second chapter describe the FDMs. In this chapter we

will start by analyzing the first and the second order partial derivatives, and give the

central difference scheme, the forward and backward schemes; then we present some

schemes that use finite difference methods. In the last chapter we apply those methods

to discretize the three prominent classical PDEs. In this chapter, we give some numerical

examples for these PDEs.

Chapter 1

Presentation of partial differential

equations

In this chapter we first introduce certain basic concepts and definitions. The rest of

the chapter will be devoted to classification of linear second order differential equations

with constant coefficients into the three principal types and the corresponding canonical

forms will be discussed.

1.1 Basic concepts and definitions

A Partial Differential Equation (PDE) is an equation that involves an unknown function

of two or more independent variables and certain partial derivatives of the unknown

function. More precisely let U denote a function of the n 2 independent variablesx1, ..., xn. Then the relation of the form

F (x1..., xn, U, Ux1,...Uxn , Ux1x1 , ..., Uxnxn) = 0, (1.1)

Where F is a function of its arguments is a partial differential equation. The general

linear second order PDE in one dependent variable U may be written asn

i,j=1

Ai,jUxixj +n

i,=1

BiUxi + FU = G, (1.2)

in which we assume Aij = Aji And Aij, Bi, F and G are real valued functions defined

in some region of the space (x1, x2, ..., xn). Here we shall be concerned with second

order PDE in one dependent variable U and two independents variables x and y hence

equation (1.2) can be put in the form

AUxx +BUxy + CUyy +DUx + EUy + FU = G, (1.3)

1.1 Basic concepts and definitions 4

where the coefficients A,B,C,D,E, F and G are functions of x and y in some domain

of the xy-plane.

Definition 1.1

As an ordinary differential equation the highest-order derivative appearing in a PDE

is called the order of the equation. A PDE is said to be the second order if its order is

two. A PDE in function of U is said to be linear if it is at most of first degree in U. For

detail we can refer in [9].

It is impossible to obtain the general solution of equation (1.3) except in some special

case:

aUxx + bUxy + cUyy + dUx + eUy + fU = 0, (1.4)

where a, b, c, d, e and f are constants such that a, b and c are not zero simultaneously.

The lineal second order PDE with constant coefficients may be written as

AUxx +BUxy + CUyy +DUx + EUy + FU = G, (1.5)

where A,B,C,D,E, F and G are constants such that A,B and C are not zero simul-

taneously. In particular we shall limit our discussion of second order equations to the

three prominent classical equations of Mathematical physics, namely

2U

t2

2U

x2= f (x, t) , (1.6)

U

t

2U

x2= g (x, t) , (1.7)

2U

y2+2U

x2= h (x, y) , (1.8)

these equations are respectively the basic PDE of wave propagation, heat equation and

potential theory. Not only are these equation of fundamental importance in many

branches of physics but they are also serve as a prototypes for the three principal types

of second-order PDEs. The kind of problems that can be solved for each of those equa-

tions and the property of the corresponding solutions are generally typical of what we

can be expected in the general case where more than two independent variables are

involved. For more detail we can refer in [1], [9].

1.2 Classification of second order equations 5

1.2 Classification of second order equations

The study of linear second-order PDE is often facilitating by recognition of the type of

the differential equation in question. For the depending on the type of the equation it

is frequently possible by mean of a coordinate transformation to reduce the equation

to one of the three canonical forms. These canonical forms correspond to differential

simple forms in which the second-order derivatives terms can appear in question. More

over the type of PDE play a decisive role in determining the kind of auxiliary conditions

that can considered with the equation so that the resolution of the problem has a unique

solution.

In this part we shall consider the classification of linear second-order PDE of equation

(1.3). In an attempt to simplify the form of second-order terms of this equation we

introduce new variables and . Consider a curve (see the figure 1.1) defined by

(x, y) = 0,

for convenience consider adjacent curves of the form

(x, y) = const,

where various values of the constant are chosen. Introduce also a family of curves

(x, y) = const,

in such a way that we can use the values of and as a local coordinate system (1.1).

In particular we require that the Jacobean

J

(,

x, y

)= xy yx,

must be equal to zero.

We now think of W as a function of and . At any point on , we know from

the prescribed data what the rate of change of is in each of the tangential and normal

directions, and so in any direction. Thus we can calculate W and W at each point

of since W and W are known functions of on , we can differentiate with respect

to to obtain values of W,W, ...,W,W, ...on . To determine W on we must

however use the equation (1.5), [1].

When the Jacobean J

(,

x, y

)is zero, its possible to write and by means of the

linear transformation = x+ y,

= x+ y.(1.9)


Curves (x,y)=const.

Curves (x,y)=const.

Figure 1.1: Coordinate system.

The coefficient , , and are constant, which will be determined subject to the

condition

6= 0.Writing

U(x, y) =W (, ) ,

we have by chain rule

Ux = W + W,

Uy = W + W,

Uxx = 2W + 2W +

2W,

Uxy = W + ( + )W + W,

Uyy = 2W + 2W +

2W.

(1.10)

Substituting these values in equation (1.5) and collecting similar terms we have

aW + bW + cW + dW + eW + fW = g (, ) , (1.11)

Where

a = A2 + 2B + C2, (1.12)


b = A +B ( ) + C, (1.13)

c = A2 +B + C2, (1.14)

d = D + E,

e = D + E,

f = F,

and

g = G.

Using (1.12), (1.13) and (1.14) it is easy to verify that

b2 ac = (B2 AC)( )2, (1.15)

Now since the constant , , and are at our disposal we shall chose them in

such a way that at least one of the second order terms in (1.11) drop out, as matter of

facts, we shall show that according to B2 AC > 0, B2 AC = 0 and B2 AC < 0.The constant , , and can be chosen such that:

From (1.12) we have, a = c = 0,

From (1.13) we have, b = c = 0,

From (1.14) we obtain, a = c and b = 0.

Thus simplifying the equation to one of the three canonical forms we have:

W = H(W,W,W, , ),

W W = H(W,W,W, , ),(1.16)

W = H(W,W,W, , ), (1.17)

W +W = H(W,W,W, , ), (1.18)

where H is a function of the new dependent variable W and his first partial derivatives,

and the new independents variables and .

Definition 1.2

We say that the PDE (1.5) is

1.3 The canonical form. 8

Hyperbolic type, if, B2 AC > 0

Parabolic type, if, B2 AC = 0 and

Elliptic type, if, B2 AC < 0

We see in the above classification that only the coefficients of the second order terms

matter. The quantity B2 AC is called the discriminate of the equation. In view ofthe formula (1.15) we see that the sign of the discriminant remain unaltered under the

change of variables (1.16) , (1.17) and (1.18). Thus the type of equation (1.5) is invariant

with respect to the linear transformation (1.9).For detail we can refer in [6]

1.3 The canonical form.

We now consider the problem of choosing the constant , , and in (1.9) so that the

equation (1.11) will reduce to one of the canonical form (1.16), (1.17) and (1.18).

1.3.1 Hyperbolic type

The PDE is hyperbolic type if B2 AC > 0, in this case we shall show that (1.9) canbe chosen such that the coefficient a and c in (1.11) vanish while b 6= 0 we thereforeconsider the equations (1.12) and (1.13), if a = c = 0 then of necessity b 6= 0 and thusequation (1.5) readily put in the canonical form (1.16) by dividing out by 2B. In this

case the transformation (1.9) may be chosen as the identity transformation = x, = y

with = = 1 and = = 0 suppose that A 6= 0 and = 0 or = 0, then from (1.12)and (1.13) we would have = 0 or = 0. In either case the condition 6= 0 isviolated. Therefore and cannot be zero and hence equation (1.12) and (1.13) can

be written as

A

(

)2+ 2B

(

)+ C = 0, (1.19)

and

A(

)2+ 2B

(

)+ C = 0. (1.20)

Thus in order that the coefficient a and c will vanish we shall choose , , and such

that

and

are roots of the quadratic equation

Am2 + 2Bm+ C = 0. (1.21)


Since B2 AC > 0 equation (1.21) has two distinct real roots

m1 =B (B2 AC) 12

A,

and

m2 =B + (B2 AC) 12

A.

Let us choose

= B + (B2 + AC) 12 , = A = B (B2 + AC) 12 ,

and

= A.

Then

= 2(B2 AC) 12 > 0,and the coefficient a and c in equation (1.11) drop out. From (1.12)-(1.14) we see that

b = A +B( ) + C,

b = 2A(B2 AC) 6= 0,therefore when equation (1.5) is of hyperbolic type the transformation

=[B + (B2 + AC) 12

]x+ Ay,

=[B (B2 + AC) 12

]x+ Ay,

will reduce the equation to the form

W = D1U + E1U + F1U +G1(, ), (1.22)

where D1, E1 and F1 are constants. Here since the coefficient are constants the lower

order terms are expressed explicitly. This form is called the first canonical form of the

hyperbolic equation. With result from canonical form (1.16) the case A = 0, C 6= 0 canbe treated in very similar way. If we further introduce the variables and defined by

the equations

= +

2,

= 2

,(1.23)


the canonical form becomes

W W = D1U + E 1U + F 1U +G1(, ), (1.24)

where D1, E1 and F

1 are constants. This form is called the second canonical form of

the hyperbolic equation. This alternative form can of cause be achieved directly from

equation (1.5) by means of the transformations

= Bx+ Ay, = (B2 AC) 12x, (1.25)

and dividing out the result by the coefficient A(B2AC) which is different from zero.

1.3.2 Parabolic type

The PDE is parabolic type if B2AC = 0, in this case it is clear that the coefficients Aand C cannot be both zero. Suppose that A 6= 0, the case C = 0 can be treated in similarmanner. Then the quadratic equation (1.21) has only one distinct root m = B

A. If we

can choose = B, = A and let and be any number such that say = 1and = 0 then from (1.12)-(1.14) it follow, that a = A, b = 0 and c = 0. Here we have

used the fact that B2 AC = 0 thus in the parabolic case the transformation

= x,

= Bx Ay,

reduce equation (1.5) to the form

W = D2U + E2U + F2U +G2(, ), (1.26)

where D2, E2 and F2 are constants, which result in the canonical form (1.17). In term

of the characteristic equation

Ad2y 2Bdxdy + Cd2x = 0,

we conclude that in the parabolic case there is only one family of characteristic given

by

Bx Ay = const.


1.3.3 Elliptic type

The PDE is elliptic type if B2 AC < 0, here it is clear that neither A nor C can bezero. Following the discussion in hyperbolic case, we see that since B2 AC < 0 thequadratic equation (1.21) has complex roots

m1 =B + i(AC B2) 12

A,

and

m2 =B i(AC B2) 12

A.

The choice

= B i(AC B2) 12 , = = A,

and

=

with

= B + i(AC B2) 12 ,would then make the coefficients a and c of equation (1.11) vanish,with

b = 2A(AC B2) > 0.

Thus under the change of variables

=[B + i(AC B2) 12

]x+ Ay, (1.27)

=[B i(AC B2) 12

]x+ Ay,

equation (1.5) would be reduce to the form

W = D3U + E3U + F3U +G3(, ), (1.28)

where D3, E3 and F3 are constant. However from (1.27) we see that and are complex

variables such that = . In order that we could have the canonical form in real

variables, we further make the change of variables (1.23) then, as is easy verified

W =1

4(W +W), (1.29)


so that (1.28) reduce to the canonical form

W +W = D3U + E

3U + F

3U +G

3(

, ), (1.30)

where D3, E3 and F

3 are constants. After dividing through by A(AC B2). if we

substitute and from (1.27) in (1.28), we obtain the transformation (1.25), writing

and again in place of and the transformation (1.25) corresponding to the choice

= B,

= A,

= (AC B2) 12 ,and

= 0,

in (1.9) with this choice, it is easily verified from (1.11) that

a = c = (AC B2),

and b = 0. Therefore, under the change of variables

= Bx+ Ay, = (B2 AC) 12x, (1.31)

equation (1.5) can be reduced to the desired canonical form (1.14). We introduce

since B2 AC < 0 equation (1.21) has no real integral curves. This means that anequation (1.5) of the elliptic type has no real characteristic. For detail we can refer in

[1] and [9].

Chapter 2

Presentation of finite methods

Among different numerical techniques for solving PDEs and initial and boundary prob-

lems, the difference finite methods (FDMs) are widely used. These methods are derived

from the truncated Taylors series where a given PDE and boundary and initial con-

ditions are replaced by set of algebraic equations that are then solved by varies well

known numerical techniques. Those methods have significant advantages over other

methods because of its simplicity of analysis and computer codes in solving problems

with complex geometric. [5].

We will discus different schemes for first and second order partial derivatives, and

then apply them to discretize the boundary and initial values problems of the second

order PDEs.

2.1 Grid lines

Let us consider in case of a function U (x, t) of two independent variables x and t, we

partition the x-axes into interval of length h and t-axes into interval of length k. The

(x, t)-plane is divided into equal rectangles of area hk by the gild lines (see figure 2.1)

parallel to 0t, defined by

xi = ih, i = 0,1,2..., (2.1)and by the gild lines parallel to 0x defined by

tj = jk, j = 0, j = 1, j = 2.... (2.2)We will use the following notation

Up = U(xi, tj) = U (ih, jk) = Uij, i = 0, ..., n; j = 0, ...,m,

denote the value of the function U (x, t) at mesh point P (ih, jk) .

2.2 Finite different schemes 14

k h

P

(i,j) (i+1,j) (i-1,j)

(i,j-1)

(i,j+1) P (ih,jk)

X 0

U

Figure 2.1: Grid lines.

2.2 Finite different schemes

Definition 2.1

If U is an open in Rn and K. N we denote by Ck (u) the space of all functions on Upossessing continuous partial derivatives of order k and we set Cc (E) = 1 Ck (U) .Furthermore, for only E Rn we denote by Cc (E) the space of all C functions onRn whose support in compact and contained in E. If E = Rn, we shall usually omit itin naming function spaces thus Ck = Ck(Rn), Cc = Cc (Rn). [3]

Let us consider a single valued, finite function U (x) that belongs to the class C(Rn)then by the Taylors Theorem:

U (x+ h) U (x) + hU (x) +h2

2!U (X) +

h3

3!U (x) + ..., (2.3)

U (x h) U (x) hU (x) + h2

2!U (X) h

3

3!U (x) + ..., (2.4)

where h is a small step around the point x.

2.3 First order partial derivative 15

2.3 First order partial derivative

In first order partial derivatives, we will discuss about three difference formula which

are: central partial difference formula, forward and backward difference formula

2.3.1 Central partial difference formula

The subtraction of (2.3) and (2.4) gives:

U (x) U (x+ h) U(x h)

2h, (2.5)

the approximation relation (2.5) is known as the first-order central partial difference

formula. The figure (2.2) show the geometrical interpretation of the formula (2.5) as

the slops of the chord AB.

U(x-h)

h h

U(x)

U(x+h)

A

P

B

X

U

0

Figure 2.2: The slop of the chord AB.

2.3.2 Forward difference formula

After some minor algebraic manipulation of the relation (2.3) we get

U (x) U(x+ h) U(x)

h, (2.6)

2.4 First order partition derivative in two independents variables 16

which known as forward difference formula because the step h was taken forwardly from

the point x. The interpretation is given by the figure (2.3)

A

B

X

U

0

x x+h

h

U(x+h)

U(x)

U(x+h)-U(x)

U

t

T

Figure 2.3: Forward derivative.

2.3.3 Backward Difference Formula

From the relation (2.4) the previous consideration permit to obtain

U (x) U(x) U(x+ h)

h, (2.7)

which known as backward difference formula, because the step h is taken in the backward

sense from the point x. For geometrical interpretation see the figure (2.4)

2.4 First order partition derivative in two indepen-

dents variables

2.4.1 Central partial difference formulas in two independents

variables

Let us consider the uniform grid

rh,k = {(xi, tj), where xi = ih, yj = jk, i = 0, ..., n; j = 0, ...,m} ,

2.4 First order partition derivative in two independents variables 17

A

B

X

U

0

x-h x h

U(x)

U(x-h)

U(x)-U(x-h)

U

T

t

Figure 2.4: Backward derivative.

give by the figure (2.1). In this case we have the relation (2.1) and (2.2).

Using the relation (2.5) and consider that i = 0, ..., n 1; and, j = 0, ...,m we have

U

x=U((i+ 1)h.jk) U((i 1)h, jk)

h,

U

x=U ji+1 U ji1

h,

(2.8)

where U ji = U(xi, tj) similarly, we have for i = 0, ..., n; and, j = 0, ...,m 1

U

t=U(ih.(j + 1)k) U(ih, (j 1)k)

h,

U

t=U ji+1 U ji1

h.

(2.9)

The relation (2.8) and (2.9) are known as the first-order order central partial differ-

ence formula in two independent variables.

2.5 The second order partial derivative 18

2.4.2 Forward difference formulas in two independents vari-

ables

From the relation (2.6) the previous consideration permit to obtain

U

x=U ji+1 U ji

h,

U

t=U j+1i U ji

k,

(2.10)

which known as forward difference formulas in two independent variables. The geomet-

rical interpretation of forward difference is show in the figure (2.3).

2.4.3 Backward difference formulas in two independents vari-

ables

Using the relation (2.7) the some previous argument permit to obtain

U

x=U ji U ji1

h,

U

t=U ji U j1i

k.

(2.11)

The relation (2.11) is known as backward difference formula in two independent vari-

ables. The figure (2.3).illustrate the geometrical interpretation of backward difference.

2.5 The second order partial derivative

The addition of the relation(2.3) and (2.4) gives

U(x+ h) + U(x h) 2U(x) + h2

U(x). (2.12)

With a minor algebraic manipulation, the relation (2.12) could be written in the

form: U(x)

U(x+ h) 2U(x) + U(x h)h2

, (2.13)

which known as the second order central partial difference formula.

2.6 Partial derivative in two independent variables

Let us consider in the case of function U(x, t) in two independent variables x and t, in

view of formula (2.13) we have central partial difference formula, forward and backward

difference formula.

2.6 Partial derivative in two independent variables 19

2.6.1 Central difference formulas in two independents variables

The Central difference formulas in two independent variables is characterized by the

following relations,

2U

x2=U ji+1 2U ji + U ji1

h2; i = 1, 2...n 1; j = 1, 2...m,

2U

t2=U j+1i 2U ji + U j1i

k2; j = 1, 2...m 1; i = 1, 2...n,

2U

xt=U j+1i+1 U j1i+1 U j+1i1 + U i1j1

4hk, i = 1, 2...n 1, j = 1, 2...m 1.

(2.14)

2.6.2 Forward difference schemes in two independents vari-

ables

When the step h and k are taken in the forward sense, from the relation (2.14) we will

obtain the following formulas:

2U

x2=U ji+2 2U ji+1 + U ji

h2; i = 0, 1, 2...n 2; j = 1, 2...m,

2U

t2=U j+2i 2U j+1i + U ji

k2; j = 0, 1, 2...m 2; i = 1, 2...n,

2U

xt=U j+2i+2 U ji+2 U j+2i + U ij

4hk, i = 0, 1, 2...n 2, j = 1, 2...m 2.

(2.15)

2.6.3 Backward difference schemes in two independents vari-

ables

Let us take the step h and k in the backward sense, then the relation (2.14) permit to

obtain the following formulas [5]

2U

x2=U ji 2U ji1 + U ji2

h2; i = 1, 2...n 2, j = 1, 2...m,

2U

t2=U ji 2U j1i + U j2i

h2; j = 1, 2...m 2; i = 1, 2...n,

2U

xt=U ji U j2i U ji2 + U i2j2

4hk; i = 1, 2...n 2, j = 1, 2...m 2.

(2.16)

2.7 Some schemes that use a Finite Difference Method 20

2.7 Some schemes that use a Finite Difference Method

2.7.1 Explicit method

Euler method

To establish the Euler schemes, we begin by partition the t-axis; it means that we choice

some points: t0, t1, t2, ...tn, tn+1... such that

0 = t0 < t1 < t2 < ...tn < tn+1....

Suppose that

hn = tn+1 tn.

We can approximate U (tn) byU(tn+1) U(tn)

hn.

Forward Euler scheme Let us consider the notation Un = U(tn). The forward Euler

scheme enables us to calculate Un+1 from Un, and it is so possible to determine succes-

sively U1, U2, U2,... from U0. Forward Euler scheme is an explicit scheme because it

is use to approximate the solution Un+1 depending on the values of Un

Un+1 = Un + hnf(Un, tn), n = 0, 1, 2..., (2.17)

where f is a function of Un and tn.

The Euler method are used to approximate the first derivative term of U [7]

Un+1 Unhn

= f(Un, tn), n = 0, 1, 2.... (2.18)

The classical explicit method

The classical explicit method enables us to approximate the solution

U j+1i = f(Uji1, U

ji , U

ji+1), i = 0, 1, 2..., j = 0, 1, 2..., (2.19)

where f is the function of U ji1, Uji , and U

ji+1.

In schematic form of the forward difference method (2.5) the solution at every point

(i, j + 1) on the (j + 1)th time level is expressed in term of the solution values at the

points (i 1, j); (i, j) and (i + 1, j) of previous time level, such method is called anexplicit method.


One important approach to approximate the solution to PDE is FDMs. In order

to approximate the solution for our problem, a network of grid point is established

throughout the rectangular region

R = {(x, t) |0 x L, 0 t T} .

As show in partition R by dividing the intervals [0, L] into n equal subintervals, each of

length

h =L

n,

and the interval [0, T ] into the m equal subintervals each of the length

k =T

m.

The corresponding point of the interval [0, L] and [0, T ] are denoted by

xi, i = 0, 1...n,

and

tj j = 0, 1...m,

respectively. The points P (xi, tj) are called the mesh or grid points and are defined by

xi = hi, i = 0, 1...n,

tj = kj, j = 0, 1...m.

The approximated solution U(x, t) at the mesh point (xi, tj) is denoted by Uji and the

true solution is denoted by U(xi, tj).

To obtain the classical explicit difference formula we approximate each partial deriv-

ative of U by the central partial difference formula, the forward difference formula, or

the backward difference formula [5].

The geometrical interpretation of classical explicit method is show in figure 2.5.

2.7.2 Implicit method

In order to establish the implicit difference schemes we proceed in the same manner as

we use to approximate the explicit difference schemes, where the solution U j+1i of the

(j + 1)th time level must depend on the values, U ji1, Uji and U

j

i+1 of U at previous level

time but the implicit difference schemes cannot directly approximate U j+1i depending


P

(i,j) (i+1,j) (i-1,j)

(i,j+1)

X 0

U

k h h

Figure 2.5: The classical explicit scheme.

on the values, U ji1, Uji and U

ji+1 of U at previous time level. The implicit difference

formulas has the following form

f(U j+1i , Uji1, U

ji , U

ji+1) = 0, i = 0, 1, 2..., j = 0, 1, 2..., (2.20)

where f designates the function of U j+1i , Uji1, U

ji , and U

ji+1.

Euler method

Backward Euler scheme This scheme can be expressed from the following relation

[7]

Un+1 Unhn

= f(Un+1, tn+1), n = 0, 1, 2.... (2.21)

The backward Euler scheme is an implicit scheme because it cannot directly approx-

imate Un+1 depending on Un, if is not trivial. It is easy to see that from (2.21) we

have

Un+1 hnf(Un+1, tn+1) = Un, n = 0, 1, 2.... (2.22)


The classical implicit method

The classical implicit method is used to solve PDE. For this, we approximate each

partial derivative of U by the central partial difference formula, the forward difference

formula, the backward difference formula and after we substitute those derivatives in

the given PDE; and with some minor algebraic manipulation, we express the solution

U ji depending on the values Uj+1i1 , U

j+1i and U

j+1

i+1 of U at the following time level. We

obtain the following implicit difference formula.

f(U j+1i1 , Uj+1i , U

j+1i+1 ) = U

ji i = 0, 1, 2..., j = 0, 1, 2.... (2.23)

where f designates the function of U j+1i1 , Uj+1i and U

j+1i+1 .

The solution value at point (i, j + 1) on the (j + 1)th time level is depend on the

solution values at the neighboring points on the same level and the one point on the

jth time level. which are obtained implicitly [5]. The figure 2.6 shows the geometrical

interpretation of classical implicit backward scheme.

P

(i,j) (i+1,j) (i-1,j)

(i,j+1)

X 0

U

k h h

Figure 2.6: The classical implicit scheme.

Crank-Nicolson method

It is show that this method is indispensable in PDE resolution [5]


To use the Crank-Nicolson method we approximate the first derivative of U by the

forward difference and the second derivative of U by the average of the centered difference

at the time steps (j+ i) and j and we substitute them in the given PDE. using algebraic

theories, the solution value at any point (i, j+1) on the (j+1)th time level is depending

on the solution value at the neighboring point on the same level and three points on the

jth time level. Since values at the (j +1)th time level are obtained implicit, the method

is called an implicit method. For geometrical interpretation see the figure (2.7)

P

(i,j) (i+1,j) (i-1,j)

(i,j+1)

X 0

U

k h h

Figure 2.7: The Crank Nicolson scheme.

This method is unconditionally stable, and sometime it can be written in matrix

form [5].

For example if we have the Crank-Nicolson expression

aU j+1i1 + bUj+1i + cU

j+1i+1 = a

U ji1 + bU ji + c

U ji+1, i = 0, 1, 2..., j = 0, 1, 2..., (2.24)

the matrix form of (2.24) can be written (2.25)

AU (j+1) = BU (j), j = 0, 1, 2..., (2.25)

where

U (j+1) =[U

(j+1)1 , U

(j+1)2 , ..., U

(j+1)n1

]Tj = 0, 1, 2...,


A =

b c 0 0a b c

. . ....

0. . . . . . . . . 0

.... . . a b c

0 0 a b

, B =

b c 0 0a b c

. . ....

0. . . . . . . . . 0

.... . . a b c

0 0 a b

The tridiagonal matrix A is positive definite and strictly dominant. therefore A is

nonsingular and the system of equation (2.24) has a unique solution [5].

2.7.3 The Lax-Wendroff method

The Lax-Wendroff method, named after Peter Lax and Burton Wendroff, is a numerical

method for the solution of hyperbolic partial differential equations. It is based on finite

differences. Suppose one has an equation of the following form:

U (x, t)

t=V (U (x, t))

x, (2.26)

where x and t are independent variables, and the initial state, U(x, 0) is given.

The first step in the LaxWendroff method calculates values for U(x, t) at half time

steps, tn+1/2 and half grid points xi+1/2. In the second step values at tn+1 are calculated

using the data for tn and tn+1/2.

First (Lax) step:

Un+ 1

2

i+ 12

Uni Uni+12

12t

=V ni+i V ni

x, (2.27)

Second step:

Un+1i Unit

=V

n+ 12

i+ 12

V n+12

i 12

x. (2.28)

This method can be further applied to some systems of partial differential equations.

For detail we can refer from [2] .

Chapter 3

Numerical solution of three forms of

partial differential equations

In this chapter, we present some numerical solution for solving the three prominent

classical equations of mathematical physics namely:

1. The wave equation,

2. The diffusion equation,

3. The Laplaces equation,

presented in chapter 1, using some formulas described in chapter 2.

3.1 Parabolic equations

We begin our discussion with one-dimensional PDEs of parabolic type form (1.7). which

is diffusion equation given by the relation

Ut = Uxx, 0 < x < L, 0 < t T, (3.1)

subject to the boundary conditions

U(0, t) = 0, 0 < t T,U(L, t) = 0, 0 < t T, (3.2)

and the initial condition

U(x, 0) = f(x), 0 < x < L. (3.3)

3.1 Parabolic equations 27

Here U(x, t) denote the Temperature at any time t along a thin rod of length L in which

heat is following as illustrated in figure 3.1. We assume that the rod is of homogeneous

material and has a cross-section at area A that is constant throughout the length of the

rod. The rod is laterally insulted along its entire length. The constant is determined

by the thermal properties of the material and is measure of its ability to conduct heat

or denotes the thermal diffusivity. For the grid (xi, tj) = (ih, jk), we will discuss the

following three finite difference schemes.

X=0 X=L

A: Area

Figure 3.1: Dimensional rod of length L.

3.1.1 Forward difference

To present this scheme, two formulas from chapter 2 are used in this context. The central

difference formula (2.14) for approximating Uxx(xi, tj) and the forward difference formula

(2.10) for approximating Ut(xi, tj) by using these schemes the equation 3.1 gives

U j+1i U jik

= U ji+1 2U ji + U ji1

h2, (3.4)

if we set = k

h2and solve U j+1i we will obtain the explicit difference formula

U j+1i = (1 2)U ji + (U ji+1 + U ji1) i = 1, .., .n 1, J = 0, ...,m 1, (3.5)


known as the forward difference or classical explicit method. In schematic form of

forward difference method, the equation (3.5) is show in figure 2.5 the solution at every

point (i, j + 1)of the (i, j + 1)th time is expressed in terms of the solution values at the

points (i 1, j), (i, j) and (i+ 1, j) of the previous time level.The value of the initial condition

U(xi, 0) = f(xi) i = 1, ..., n

are used in the equation (3.5) to find the values of U1i for i = 1, ..., n 1. The boundaryconditions

U(0, tj) = U(L, tj) = 0 (3.6)

imply that

U j0 = Ujn = 0 j = 1, ...,m. (3.7)

Once the approximations U1i , U2i , ..., Umi can be obtained in a similar manner. If n and

m are small values, for example n = 4, m = 3 the values of

U ij i = 1, ..., n; m = 1, ...,m

can be easily found manually. But if n and m are great numbers, it will be diffi-

cult to compute the values of U ij at every grid point; Here it is necessary to compute

the numerical solution using the programming languages like FORTRAN, C++,Maple,

Mathematica, etc. In our work the implementation is made using Matlab.

Example 3.1

The MATLAB function for the resolution of parabolic equation in case of heat equa-

tion using forward difference method is given by the figure 3.2

We illustrate the forward difference method by solving the heat equation

Uxx = Ut , 0 < x < 1, t > 0,

subject to the initial and boundary conditions

U(x, 0) = sin(pix), 0 x 1,U(0, t) = U(1, t) = 0, t 0,

the exact solution to this problem is

U(x, t) = epi2t sin(pit).

JARUKNoteU(x,t)=exp(-(pi)^2t)sin(pix)

not t in place of x

JARUKNote

JARUKNoteMarked set by JARUK

JARUKNoteUnmarked set by JARUK


function heat(f,c1,c2,L,T,h,k,alpha) % Solve the heat equation % with I.C. u(x,0)=f(x) % and B.C. u(0,t)=c1 and u(L,t)=c2. % using the Forward-Difference method n=L/h;m=T/k; lambda=alpha*k/(h^2) z=0:h:L; disp('______________________________________________') fprintf(' t x = ') fprintf('%4.2f ',z) fprintf('\n') disp('____________________________________________ _') fprintf('% 5.4f ',0) % Compute the values of u at t=0 for i=1:n+1 u(i)=feval(f,(i-1)*h); fprintf('%10.6f ',u(i)) end fprintf('\n') % Compute the values of u t=jk, k=1,2,...,m for j=1:m t=j*k; fprintf('% 5.4f ',t) for i=1:n+1 if (i==1) y(i)=c1; elseif (i==n+1) y(i)=c2; else y(i)=(1-2*lambda)*u(i)+lambda*(u(i+1)+u(i-1)); end; fprintf('%10.6f ',y(i)) end; fprintf('\n') u=y; end;

Figure 3.2: The MATLAB function for the resolution of parabolic equation in case of

heat equation using the forward difference method.

The solution will be approximated first with T = 0.025, h = 0.1 and k = 0.0025 so that

= kh2

= 0.25 n = Lh= 10 and m = T

k= 10.

The approximation of U at t = 0.0025 for i = 1, 2 are found in this manner; we

first replace the value of into equation (3.5) we get the simplified difference equation

U j+1i = 0, 5Uji + 0.25(U

ji+1 + U

ji1) i = 1, .., 9; j = 0, ..., 9. (3.8)

The value of initial condition

U0i = sin(pixi), i = 1, .., 10,

are used in equation (3.5) to find the value of U ji for i = 1, ..., n, using the boundary

condition

U j0 = Uj9 = 0 j = 0, ..., 10.


For i = 1, 2 and j = 0, the approximation of U are

U1i = 0, 5U0i + 0.25(U0i+1 + U0i1) i = 1, .., 9;

U11 =1

4(U00 + 2U01 + U02 ) =

1

4[sin(0.0) + 2 sin(0.1pi) + sin(0.2pi)] = 0.301455,

U12 =1

4(U01 + 2U02 + U03 ) =

1

4[sin(0.1pi) + 2 sin(0.2pi) + sin(0.3pi)] = 0.573401.

If we continue in this manner,we get the result show in table of the figure 3.3,which are

obtained using the MATLAB function heat given in the figure 3.2. A three dimensional

representation of table given by the figure 3.3 is shown in figure 3.4.

t \ x 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00

0.0000 0.000000 0.309017 0.587785 0.809017 0.951057 1.000000 0.951057 0.809017 0.587785 0.309017 0.000000

0.0025 0.000000 0.301455 0.573401 0.789219 0.927783 0.975528 0.927783 0.789219 0.573401 0.301455 0.000000

0.0050 0.000000 0.294078 0.559369 0.769905 0.905078 0.951655 0.905078 0.769905 0.559369 0.294078 0.000000

0.0075 0.000000 0.286881 0.545680 0.751064 0.882929 0.928367 0.882929 0.751064 0.545680 0.286881 0.000000

0.0100 0.000000 0.279861 0.532327 0.732685 0.861322 0.905648 0.861322 0.732685 0.532327 0.279861 0.000000

0.0125 0.000000 0.273012 0.519300 0.714755 0.840244 0.883485 0.840244 0.714755 0.519300 0.273012 0.000000

0.0150 0.000000 0.266331 0.506591 0.697263 0.819682 0.861865 0.819682 0.697263 0.506591 0.266331 0.000000

0.0175 0.000000 0.259813 0.494194 0.680200 0.799623 0.840773 0.799623 0.680200 0.494194 0.259813 0.000000

0.0200 0.000000 0.253455 0.482100 0.663554 0.780055 0.820198 0.780055 0.663554 0.482100 0.253455 0.000000

0.0225 0.000000 0.247253 0.470303 0.647316 0.760966 0.800127 0.760966 0.647316 0.470303 0.247253 0.000000

0.0250 0.000000 0.241202 0.458793 0.631475 0.742343 0.780546 0.742343 0.631475 0.458793 0.241202 0.000000

Figure 3.3: Forward difference method for example 3.1 .

3.1.2 The classical implicit method

To apply the classical implicit method, the finite difference equation of this method

is obtained by replacing Uxx(xi, tj) in the equation (3.1) with the centered difference

formula (2.14) at the time steps j+1, and Ut(xi, tj) with the forward difference formula

(2.10), we obtainU j+1i U ji

k=

U j+1i+1 2U j+1i + U j+1i1h2

, (3.9)


00.2

0.40.6

0.81

00.005

0.010.015

0.020.025

0

0.2

0.4

0.6

0.8

1

Figure 3.4: A three dimensional representation of table given by the figure 3.3.

if we set = k

h2for we obtain the implicit difference formula

U j+1i1 + (1 + 2)U j+1i U j+1i+1 = U ji i = 1, ..., n 1, j = 0, ...,m 1, (3.10)

known as the classical implicit method. In schematic form, the equation (3.10) is shown

in figure 2.6 The solution value at any point (i, j + 1) on the (j + 1)th time level is

depending on the solution values at the neighboring points on the same level and one

point on the jth time level. Since values at the (j+1)th time level are obtained implicitly,

the method is called the implicit method. The matrix form of the classical implicit

method is

AU (j+1) = U (j), j = 0, 1, 2... (3.11)

where

U (j+1) =[U

(j+1)1 , U

(j+1)2 , ..., U

(j+1)n1

]T,

A =

2(1 + ) 0 0 2(1 + ) . . . ...0

. . . . . . 0... 2(1 + ) 0 0 2(1 + )

.


Example 3.2

We illustrate the forward difference method by solving the heat equation

Uxx = Ut, 0 < x < 0.5, t > 0,


U(x, 0) = cos(pix), 0 x 0.5,U(0, t) = U(0.5, t) = 0, t 0.

The solution will be approximated with T = 0.025, h = 0.1 and k = 0.0025 so that

= k

h2= 0.25 n = 5 and m = 10, the approximation of U at t = 0.0025 are found in

this manner. We first replace the value of into equation (3.10) we get the simplified

difference equation

0.25U j+1i1 + (1.5)U j+1i 0.25U j+1i+1 = U ji i = 1, .., 4; j = 0, ..., 9.At the first step t = k is given by the solution of the tridiagonal system

1.5 0.25 0 00.25 1.5 0.25 00 0.25 1.5 0.250 0 0.25 1.5

U11U12U13U14

=U01U02U03U04

,the initial conditions

U(xi, 0) = cos(pixi), i = 1, ..., 4,

are used to find the value of U0i for i = 1, ..., 4 and our system becomes1.5 0.25 0 0

0.25 1.5 0.25 00 0.25 1.5 0.250 0 0.25 1.5

U11U12U13U14

=0.99998498

0.99993992

0.99986485

0.999975974

.The solution of the tridiagonal system is

U(xi, 0, 0025) = [0.82756606, 0.965456227, 0.965411405, 0.827552586]t, i = 1, .., 4

and the other values of U(xi, 0, 005), ..., U(xi, 0, 05) are found in the similar manner.

For simplifying many iterations it is important to compute the values of U ij for

i = 1, ..., n;

and

j = 1, ...,m

using the programming languages [4].


3.1.3 The Crank-Nicolson method

The Crank-Nicolson method, is based on the finite difference equation. This equation is

obtained by replacing Uxx(xi, tj) of the equation (3.1) with the average of the centered

difference formula (2.14) at the time steps j + 1, and j, then Ut(xi, tj) with the forward

difference formula (2.10), which gives

U j+1i U jik

=

2

[U ji+1 2U ji + U ji1

h2+U j+1i+1 2U j+1i + U j+1i1

h2i = 1, ..., n 1

], (3.12)

if we set = k

h2as before, equation (3.12) can be written as

U j+1i1 +2 (1 + 2)U j+1i U j+1i+1 = U ji1+2 (1 2)U ji +U ji+1 i = 1, ..., n1, (3.13)known as the Crank-Nicolson method. In schematic form, the equation (3.13) is shown

in figure 2.7. The solution value at any point (i, j + 1) on the (j + 1)th time level

is depending on the solution values at the neighboring points on the same level and

three points on the jth time level. Since values at the (j + 1)th time level are obtained

implicitly, the method is called the implicit method.

The matrix form of the Crank-Nicolson method is

AU (j+1) = BU (j), i = 0, 1, 2... (3.14)

where

U (j+1) =[U

(j+1)1 , U

(j+1)2 , ..., U

(j+1)n1

]T

A =

2(1 + ) 0 0 2(1 + ) . . . ...0

. . . . . . 0... 2(1 + ) 0 0 2(1 + )

,

B =

2(1 ) 0 0 2(1 ) . . . ...0

. . . . . . 0... 2(1 ) 0 0 2(1 )

The MATLAB function for the resolution of parabolic equation in case of heat equation

using the Crank-Nicolson method is is given in the figure 3.5 and the figure 3.6


function heat_crank(f,c1,c2,L,T,h,k,alpha) % Solve the heat equation with I.C. u(x,0)=f(x) and B.C. u(0,t)=c1, and %u(L,t)=c2, using Crank-Nicolson method. n=L/h;m=T/k; lambda=alpha*k/(h^2) z=0:h:L; disp('______________________________________________') fprintf(' t x = ') fprintf('%4.2f ',z) fprintf('\n') disp('______________________________________________') fprintf('% 4.2f ',0) % Compute the values of u at t=0 for i=1:n+1 u(i)=feval(f,(i-1)*h); fprintf('%10.6f ',u(i)) end fprintf('\n') for i=2:n if (i~=n) % Compute the subdiagonal of A. a(i)=-lambda; end % Compute the main diagonal of A. b(i)=2*(1+lambda); if (i~=n) % Compute the superdiagonal of A. c(i)=-lambda; end end

Figure 3.5: The first part of Matlab function for the resolution of parabolic equation in

case of heat equation using the Crank-Nicolson method.

Example 3.3

Solve the heat equation

Uxx = Ut , 0 < x < 1, 0 < t < 0.5,


U(x, 0) = sin(pix), 0 x 1,U(0, t) = U(1, t) = 0, 0 < t < 0.5.

We choose h = 0.2, k = 0.05, so that = 1.25, n = 5, and m = 10 by replacing the value

of into equation (3.13) we get the simplified difference equation

1.25U j+1i1 + 4.5U j+1i 1.25U j+1i+1 = 1.25U ji1 + 0.5U ji + 1.25U ji+1 i = 1, ..., 4.


% save the vecot b in bb. bb=b; for j=1:m t=j*k; fprintf('% 4.2f ',t) % Compute the coefficient vector Bu. for i=2:n d(i)=lambda*u(i-1)+2*(1-lambda)*u(i)+lambda*u(i+1); end %Boundary conditions y(n+1)=c1; y(1)=c2; % Solve the tridiagonal system Au=Bu. for i=3:n ymult=a(i-1)/bb(i-1); bb(i)=bb(i)-ymult*c(i-1); d(i)=d(i)-ymult*d(i-1); end y(n)=d(n)/bb(n); for i=n-1:-1:2 y(i)=(d(i)-c(i)*y(i+1))/bb(i); end % Write the ouput. for i=1:n+1 fprintf('%10.6f ',y(i)) end fprintf('\n') u=y; bb=b; end

Figure 3.6: The second part of Matlab function function for the resolution of parabolic

equation in case of heat equation using the Crank-Nicolson method.

At the first time step t = k, U1i is given by the solution of the tridiagonal system4.5 1.25 0 0

1.25 4.5 1.25 00 1.25 4.5 1.250 0 1.25 4.5

U11U12U13U14

=

0.5U01 + 1.25U021.25U01 0.5U02 + 1.25U031.25U02 0.5U03 + 1.25U04

1.25U03 0.5U04

=

0.894928

1.448024

1.448024

0.894928

where

U0i = sin(piih).

The solution to the tridiagonal system is

U(xi, 0.05) = [0.36122840, 0.58447983, 0.58447983, 0.36122840]T , i = 1, 2, 3, 4.

Using the MATLAB function heat crank, we obtain the solution shown in table given by

the figure 3.7 and three dimensional representation of this is shown in figure 3.8

3.2 Hyperbolic equations 36

t \ x 0.00 0.20 0.40 0.60 0.80 1.00

0.00 0.000000 0.587785 0.951057 0.951057 0.587785 0.000000

0.05 0.000000 0.361228 0.584480 0.584480 0.361228 0.000000

0.10 0.000000 0.221996 0.359197 0.359197 0.221996 0.000000

0.15 0.000000 0.136430 0.220748 0.220748 0.136430 0.000000

0.20 0.000000 0.083844 0.135662 0.135662 0.083844 0.000000

0.25 0.000000 0.051527 0.083372 0.083372 0.051527 0.000000

0.30 0.000000 0.031666 0.051237 0.051237 0.031666 0.000000

0.35 0.000000 0.019461 0.031488 0.031488 0.019461 0.000000

0.40 0.000000 0.011960 0.019351 0.019351 0.011960 0.000000

0.45 0.000000 0.007350 0.011893 0.011893 0.007350 0.000000

0.50 0.000000 0.004517 0.007309 0.007309 0.004517 0.000000

Figure 3.7: Crank-Nicolson method for example 3.3.

3.2 Hyperbolic equations

In this section we will discuss methods for the numerical solutions of the on-dimension

PDEs of hyperbolic type. Consider the wave equation given by form of equation (1.6)

that is

Utt = 2Uxx, 0 < x < L, t > 0, (3.15)

subject to the boundary conditions

U(0, t) = 0, t > 0,

U(L, t) = 0, t > 0,(3.16)

and the initial conditions

U(x, 0) = f(x), 0 < x < L,

Ut(x, 0) = g(x), 0 < x < L.(3.17)

Here U(x, t) denotes the vertical displacement of a uniform, perfectly flexible string

of a contact density that is tightly stretched between two fixed points, 0 and L, we

assume that the equilibrium position of the string is horizontal, with the string aligned

along the x-axis as shown in figure 3.9. Suppose that the string is plucked at time t = 0


00.2

0.40.6

0.81

00.1

0.20.3

0.40.5

0

0.2

0.4

0.6

0.8

1

t

x

Figure 3.8: A three dimensional representation of table given by the figure 3.7.

causing the string to vibrate. Our problem is to determine the vertical displacement

U(x, t) of a point x at time t. We assume that the horizontal displacement is so small

relative to the vertical displacement as to be negligible we also assume that the maximum

displacement of each point on the string is small in comparison with the length L of the

string.

We assume that the region R = {(x, t)|0 x L, 0 t T} to be subdividedinto rectangles as show in the figure 2.1. To derive a difference equation for the solution

we start by replacing the space derivative by the center difference formula but more

frequently used schemes are the explicit schemes and the implicit schemes of Crank

Nicolson.

3.2.1 Implicit scheme

In the Crank-Nicolson scheme, we start by replacing the space derivative Uxx(xi, tj) by

the average of the central difference formula at the time steps j + 1 and j 1; and thetime derivative Utt(xi, tj) by the central difference formula. the result is

U j+1i 2U ji + U j1ik2

=2

2

[U j+1i+1 2U j+1i + U j+1i1

h2+U j1i+1 2U j1i + U j1i1

h2

]. (3.18)


Point on the string

U: displacement

0

Y

X

Figure 3.9: Vibrating string.

If we set = k

h2and rearrange the order of the terms, we obtain the following implicit

three level difference formula

2U j+1i+1 + 2(1 2)U j+1i 2U j+1i1 = (1 + 2)U j1i + 2(U j1i1 + U j1i+1 ) 4U ji ,i = 1, .., .n 1 j = 0, ...,m 1,

(3.19)

the equation (3.14) is show in figure 2.7.

The solution value at any point (i, j + 1) on the (i, j + 1)th time level is depending

on the solution values at the neighboring points on the same level, and the three points

on the (i, j 1)thtime level and also the one point on the jth time level. Since values at(i, j + 1)thtime level are obtained implicitly, the method is an implicit method.

3.2.2 Explicit scheme

We start by replacing the space derivative by the center difference formula and the time

derivative by the center difference formula (2.14) into (3.15) gives the difference equation

U j1i 2U ji + U j1ik2

= 2

(U ji+1 2U ji + U ji1

h2

). (3.20)


If we set = k

h2and rearrange the order of the terms, we obtain the explicit three

level difference formula

U j+1i = 2(1 2)U ji + 2(U ji1 +U ji+1)U j1i , i = 1, ..., n 1 j = 0, ...,m 1. (3.21)

In schematic form the equation (3.21) is shown in figure 2.5. The solution at every point

(i, j + 1) on the (j + 1)th time level is expressed in term of the solution values at the

points (i 1, j), (i, j), (i + 1, j) and (i, j 1) of the two previous time levels. Theexplicit formula (3.21) has a stability problem and it can be shown that the method is

stable if 0 < 1 [4].The right-hand side of equation (3.21) show that to calculate the entry U j+1i on the

(j + 1)th time level, we must use entries from the jth and (j 1)th time levels. thisrequirement creates a problem at the beginning because we only know the very first

row from the initial condition U0i = f(xi). To obtain the second row corresponding to

U1i , the second initial condition Ut(x, 0) = g(x) may be used as follows. If we use the

forward difference approximation

Ut(xi, 0) U(xi, t1) U(xi, 0)k

, (3.22)

we get a finite difference equation that gives an approximation for the second row. To

get a better approximation, we consider the Taylor series expansion of U(x, t) about the

point(xi, 0).

U(xi, k) U(xi, 0) + kUt(xi, 0) + k2

2Utt(xi, 0). (3.23)

Assuming that the second derivative of f(x) exists, we have the result obtained from

the equation (3.15) and (3.17)

Utt(xi, 0) = 2Uxx(xi, 0) =

2f (xi), (3.24)

substituting (3.24) into (3.23) and using the initial condition (3.17)

Ut(xi, 0) = g(xi)

U(xi, 0) = f(xi)(3.25)

we have

U(xi, k) f(xi) + kg(xi) + k2

22f (xi). (3.26)


Finally by replacing f (xi), in the last equation, by the central difference formula, we

get a difference formula for the numerical approximation in second row

U1i f(xi) + kg(xi) +k22

2h2[f(xi1) 2f(xi) + f(xi+1)] ,

U1i (1 2)f(xi) + kg(xi) +2

2[f(xi1) + f(xi+1)] ,

(3.27)

where

=k

h.

The MATLAB program for the resolution of the hyperbolic equation using a three level

explicit method is given by the following figure 3.10

function hyperbolic(f,g,c1,c2,L,T,h,k,alpha) % Solve the hyperbolic equation % with u(x,0)=f(x), ut(x,0)=g(x). % and u(0,t)=c1 and u(L,t)=c2, % using a three level explicit method. n=L/h;m=T/k; lambda=alpha*k/h z=0:h:L; disp('_______________________________________________') fprintf(' t x = ') fprintf('%4.2f ',z) fprintf('\n') disp('________________________________________________') fprintf('% 4.2f ',0) % Compute the values of u at t=0 for i=1:n+1 u0(i)=feval(f,(i-1)*h); fprintf('%10.6f ',u0(i)) end fprintf('\n') fprintf('% 4.2f %10.6f ',k,c1) %Compute the values of u at t=k for i=1:n-1 u1(i+1)=(1-lambda^2)*feval(f,i*h)+lambda^2/2*(feval(f,(i+1)*h)... +feval(f,(i-1)*h))+k*feval(g,i*h); fprintf('%10.6f ',u1(i+1)) end fprintf('%10.6f ',c2) fprintf('\n') %Compute the values of u at t>k for j=2:m t=j*k; fprintf('% 4.2f ',t) %Boundary condition u1(1)=c1; ui(1)=c1; u1(n+1)=c2; ui(n+1)=c2; fprintf('%10.6f ',ui(1)) for i=2:n ui(i)=2*(1-lambda^2)*u1(i)+lambda^2*(u1(i+1)... +u1(i-1))-u0(i); fprintf('%10.6f ',ui(i)) end fprintf('%10.6f ',ui(n+1)) fprintf('\n') u0=u1; u1=ui; end

Figure 3.10: The MATLAB program for the resolution of the hyperbolic equation using

a three level explicit method.


Example 3.4

Using the explicit scheme (Forward difference), approximate the solution of the wave

equation

16Uxx = Ut, 0 < x < 1, t > 0,

subject to the conditions

U(x, 0) = sin(pix), 0 x 1,Ut(x, 0) = 0, 0 x 1,U(0, t) = U(1, t) = 0, t > 0.

The solution will be approximated with T = 0.5, h = 0.2, and k = 0.05, so that

= k

h2= 4

0.05

0.2= 1.0,

n = 5, and m = 10. The approximation of U at t = 0.05 for i = 1, 2, 3, 4 are as follow

U j0 = Uj5 = 0, j = 0, ..., 10

and the initial conditions give

U j0 = sin(pi0.2i), i = 0, ..., 5.

From equation (3.27) we obtain

U1i =12[f(0.2(i+ 1)) + f(0.2(i 1))] + 0.05g(0.2i),

= 0.5{sin[pi0.2(i+ 1)] + sin[pi0.2(i 1)]}, i = 1, ..., 4.Hence

U(xi, 0, 05) = [0.47552826, 0.76942088, 0.769420, 0.47552826]T , i = 1, 2, 3, 4.

For t = 2k = 0.1,we get the difference equation

U2i = U1i+1 + U1i1 U0i , i = 1, 2, 3, 4

which implies that

U(xi, 0, 1) = [0.18163563, 0.29389263, 0.29389263, 0.18163563]T , i = 1, 2, 3, 4

Using the MATLAB function hyperbolic with .T = 0.5, h = 0.2, and k = 0.05 we obtain

the solution shown in table given by the figure 3.11 . A three dimensional representation

of this table is shown in figure 3.12.

3.3 Elliptic equation 42

t \ x 0.00 0.20 0.40 0.60 0.80 1.00

0.00 0.000000 0.587785 0.951057 0.951057 0.587785 0.000000

0.05 0.000000 0.475528 0.769421 0.769421 0.475528 0.000000

0.10 0.000000 0.181636 0.293893 0.293893 0.181636 0.000000

0.15 0.000000 -0.181636 -0.293893 -0.293893 -0.181636 0.000000

0.20 0.000000 -0.475528 -0.769421 -0.769421 -0.475528 0.000000

0.25 0.000000 -0.587785 -0.951057 -0.951057 -0.587785 0.000000

0.30 0.000000 -0.475528 -0.769421 -0.769421 -0.475528 0.000000

0.35 0.000000 -0.181636 -0.293893 -0.293893 -0.181636 0.000000

0.40 0.000000 0.181636 0.293893 0.293893 0.181636 0.000000

0.45 0.000000 0.475528 0.769421 0.769421 0.475528 0.000000

0.50 0.000000 0.587785 0.951057 0.951057 0.587785 0.000000

Figure 3.11: Explicit three level difference method for example 3.4.

3.3 Elliptic equation

Our discussion on elliptic equations will focus on the formulation of the finite difference

equation for the two dimensional Laplace equation given by equation

Uxx + Uyy = 0, (3.28)

on rectangular domain

R = {(x, y)| a < x < b, c < y < d},

and subject to the Dirichlet boundary condition

U(x, c) = f1(x), U(x, c) = f2(x), a x b,U(a, y) = g1(x), U(b, y) = g2(y), c x d.

(3.29)

Equation (3.29) arises in the study of steady-state or time-independent solutions of

heat equations. Because these solution do not depend on time; initial conditions are

irrelevant and only boundary conditions are specified. other applications include the

static displacement U(x, y) of the stretched membrane fastened in space along the

boundary of a region; the electrostatic and gravitational potentials in certain force fields;

and in fluid mechanics for an ideal fluid. A problem of particular interest is rectangular


00.2

0.40.6

0.81

00.1

0.20.3

0.40.5-1

-0.5

0

0.5

1

tx

u

Figure 3.12: A three dimensional representation of the table given by the figure 3.11.

region R, as illustrated in the figure 3.13. This situation is modelled with Laplaces

equation in R with the boundary condition (3.29)

To solve the Laplaces equation by finite difference methods, the region R is parti-

tioned into a grid consisting of n m rectangles with sides h and k. The mesh pointsare given by

xi = a+ ih, i = 1, .., .n,

yj = c+ jk, J = 0, ...,m.(3.30)

By using central difference approximation for the spatial derivatives, the finite difference

equation for equation (3.29) is

U ji+1 2U ji + U ji1h2

+U j+1i 2U ji + U j1i

k2= 0, i = 1, ..., n 1, j = 0, ...,m 1, (3.31)

and the boundary conditions are

U(xi, y0) = f1(xi), i = 1, .., .n 1,U(xi, ym) = f2(xi), i = 1, .., .n 1,U(x0, yj) = g3(yj), j = 1, .., .m 1,U(xn, yj) = g3(yj), j = 1, .., .m 1,

(3.32)

after it is desirable to set h = k and in this case the equation (3.31) becomes

4U ji + U j1i + U j+1i + U ji1 + U ji+1 = 0. (3.33)


Uxx+Uyy=0

f2(x)

f1(x)

g1(x)

g2(x)

X

Y

a b

c

d

0

Figure 3.13: Stead state heat flow.

The computational molecule for the equation (3.33) is shown in the figure 3.14

The equation (3.31) with the boundary conditions (3.32) form an (n 1)(m 1)(n1)(m1) linear system of equations. The coefficient matrix of the system is sparse,but is not banded in quite the same way as we have come to expect of sparse matrices.

Instead, the matrix is striped with a tridiagonal band along the main diagonal, but with

two additional bands displaced from the main diagonal by a considerable amount.

For general use, iterative techniques often represent the best approach to the solution

of such system of equations. However, if the number of equations is not too large, a direct

solution of such systems is practical. We now illustrate the solution of the Laplaces

equation when n = m = 4.

The MATLAB program for the resolution of the Laplaces equation is given by the

figure 3.15

Example 3.5

Find an approximate solution to the steady state heat flow equation

Uxx + Uyy = 0, 0 < x < 2, 0 < y < 2,


P

(i,j) h

(i,j+1) P (ih,jk)

X 0

U

(i+1,j) (i-1,j) h

k

Figure 3.14: The computational molecule for Laplace equation.

with the initial and boundary conditions

U(x, 0) = 300, U(x, 2) = 100, 0 x 2,U(0, y) = 200, U(2, y) = 200, 0 y 2,

using h = k = 0.5.

To set up the linear system, the nine interior grid point are labelled row by row from

V1 to V9 starting from the left-top corner point as shown in figure 3.16. The resulting

system is

4 1 0 1 0 0 0 0 01 4 1 0 1 0 0 0 00 1 4 0 0 1 0 0 01 0 0 4 1 0 1 0 00 1 0 1 4 1 0 1 00 0 1 0 1 4 0 0 10 0 0 1 0 0 4 1 00 0 0 0 1 0 1 4 10 0 0 0 0 1 0 1 4

V1

V2

V3

V4

V5

V6

V7

V8

V9

=

U0,3 + U1,4 = 300U2,4 = 100

U4,3 + U3,4 = 400U0,2 = 200

0

U4,2 = 200U0,4 U1,0 = 500

U2,0 = 300U3,0 + U4,1 = 400


function laplace(f1,f2,g1,g2,a,n,itmax,tol) % Solve the laplace equation in a square % with B.C. u(x,0)=f1(x), u(x,a)=f2(x) % and u(0,y)=g1(y) and u(a,y)=g2(y). % using the five point difference method. % Gauss-Seidel method is used to solve the linear system. h=a/n; z=0:h:a; h disp('__________________________________________________') fprintf(' u= x\\y ') fprintf('%4.2f ',z) fprintf('\n') disp('__________________________________________________') for i=1:n+1 u(i,1)=feval(f1,(i-1)*h); u(i,n+1)=feval(f2,(i-1)*h); u(1,i)=feval(g1,(i-1)*h); u(n+1,i)=feval(g2,(i-1)*h); end iter=0; err=tol+1; while (err>tol)&(iter


x

2

V2 V1 V3

V7

V8 U(2,y)=200y

U(0,y)=200

2

V4

U(x,2)=100

U2,4=300

U4,2 =200

U4,1=100

V6

V9

V5

U(x,0)=300

x X

x Y

x 0

Figure 3.16: Numbering system for internal grid points.

show in table given by the figure3.17; and a three dimensional representation of this is

shown in figure 3.18.


x\y 0.00 0.35 0.70 1.05 1.40 1.75 2.09 2.44 2.79 3.14

0.00 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

0.35 0.0000 0.0108 0.0228 0.0377 0.0570 0.0833 0.1196 0.1703 0.2416 0.3420

0.70 0.0000 0.0202 0.0429 0.0708 0.1072 0.1566 0.2248 0.3201 0.4541 0.6428

1.05 0.0000 0.0273 0.0579 0.0954 0.1445 0.2109 0.3029 0.4313 0.6118 0.8660

1.40 0.0000 0.0310 0.0658 0.1085 0.1643 0.2399 0.3444 0.4905 0.6957 0.9848

1.75 0.0000 0.0310 0.0658 0.1085 0.1643 0.2399 0.3444 0.4905 0.6957 0.9848

2.09 0.0000 0.0273 0.0579 0.0954 0.1445 0.2109 0.3029 0.4313 0.6118 0.8660

2.44 0.0000 0.0202 0.0429 0.0708 0.1072 0.1566 0.2248 0.3201 0.4541 0.6428

2.79 0.0000 0.0108 0.0228 0.0377 0.0571 0.0833 0.1196 0.1703 0.2416 0.3420

3.14 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000 0.0000

iter =

35

Figure 3.17: Approximate solution for example 3.6.

00.5

11.5

22.5

33.5

0

1

2

3

40

0.2

0.4

0.6

0.8

1

xy

u

Figure 3.18: A three dimensional representation of figure 3.17.

Conclusion

Our work is interested in numerical resolution of PDEs. The finite difference method

is used. The implementation has been made using Matlab language. The result show

that the finite difference method has significant advantages to solve PDEs.

For that reason, we recommend scientists who interact with partial differential equa-

tions, especially engineers and physicians to adopter those methods.

Bibliography

[1] G. F. Carrier and C. E. Pearson, Partial Differential Equations and technique, Aca-

demic Press, New York San Francisco London, 1976

[2] http://en.wikipedia.org/wiki/Lax-Wendroff method.

[3] B. F. Gerald, Real analysis, Washington, 1984.

[4] E. Isaacson and H. B. Keller,Analysis of Numerical Methods, Wiley, New York,

2ndEdition, 1990.

[5] P. K. Kythe, P. Puri, M. R. Schaferkotter, Partial Differential Equations and Bound-

ary value problems with mathematica, A CRC Press Company, Boca Raton London

New York Washington, D.C. 2nd Edition, 2003.

[6] P. Prasad & R. Ravindran, Partial Differential Equations, New age international(p)

limited publication Bangole, 1985.

[7] J. Rappaz & M. Picasso, Introduction a` lanalyse numerique, Presses polytechnique

et universitaires romandes, CH-1015 Lausane, 2004.

[8] D. U. von Rosenberg, Method for the numerical solution of partial differential

eqEdited by R. Bellman, American Elsevier Publishing Company, Inc.New York,

1969.

[9] E. C.Young, Partial Differential Equations an itroduction, Allyn and Bacon inc,

Boston, 1972.

Presentation of partial differential equationsBasic concepts and definitionsClassification of second order equationsThe canonical form.Hyperbolic typeParabolic typeElliptic type

Presentation of finite methodsGrid linesFinite different schemesFirst order partial derivativeCentral partial difference formulaForward difference formulaBackward Difference Formula

First order partition derivative in two independents variablesCentral partial difference formulas in two independents variablesForward difference formulas in two independents variablesBackward difference formulas in two independents variables

The second order partial derivativePartial derivative in two independent variablesCentral difference formulas in two independents variablesForward difference schemes in two independents variablesBackward difference schemes in two independents variables

Some schemes that use a Finite Difference MethodExplicit methodImplicit methodThe Lax-Wendroff method

Numerical solution of three forms of partial differential equationsParabolic equationsForward differenceThe classical implicit methodThe Crank-Nicolson method

Hyperbolic equationsImplicit schemeExplicit scheme

Elliptic equation

finite difference method for the resolution

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