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NUMERICAL SOLUTIONS OF PARTIAL

DIFFERENTIAL EQUATIONS

PRIYA LEKSHMI S

1

PREFACE

Partial differential equations occur in many branches of applied

mathematics. Some of the area of application includes hydrodynamics,

elasticity, quantum mechanics and electromagnetic theory. The analytic

treatment of these equations is rather involved process and requires

application of advanced mathematical methods. This book is intended to

familiarize the reader with the basic concept of partial differential equation

and its numerical solution. Chapter I contain Preliminaries, chapter II

contains Partial Differential Equations of 2nd

order, chapter III contains

Parabolic Equations and chapter IV contains Hyperbolic Equations.

2

INDEX

Sl.No. Content Page

Number

1 Preface 1

2 Introduction 3

3 Chapter I

Preliminaries

4

4 Chapter II

Partial Differential

Equations of second order

6

5 Chapter III

Parabolic Equations

9

6 Chapter IV

Hyperbolic Equations

3

INTRODUCTION

In the language of mathematics, changing entities are called variables and

the rate of change of one variable with respect to another a derivative.

Equations which express a relationship among these variables and their

derivatives are called differential equations. If there are two or more

independent variables present, so that the equations contains partial

derivatives, it is called partial differential equations.

Second order partial differential equations are classified into elliptic,

parabolic and hyperbolic types.

4

CHAPTER-I

PRELIMINARIES

Definition 1.1

Let y=f(x) be a function of x. Let ∆x be the increment in the value of

x and ∆y, the corresponding increment in the value of v so that

y +∆y = f(x+∆x)

y = f(x + ∆x) - f(x)

( ) ( )

⁄

⁄ is called incremental ratio.

Definition 1.2

If f is differentiable at each point of its domain D, then f is said to be

a differentiable function.

Definition 1.3

An equation involving independent and dependent variables and the

derivatives or differentials of one or more dependent variables with respect

to one or more independent variables is called differential equation.

Definition 1.4

A differential equation which involves derivatives with respect to a

single independent variables is known as an ordinary differential equation.

5

Definition 1.5

A differential equation which contains two or more independent

variables and partial derivatives with respect to them is called a partial

differential equation.

Definition 1.6

The order of the highest order derivative involved in a differential

equation is called the order of a differential equation.

Definition 1.7

The degree of a differential equation is the degree of the highest

order derivative present in the equation.

6

CHAPTER-II

PARTIAL DIFFERENTIAL EQUATIONS OF

SECOND ORDER

2:1 - Classification of Second Order Partial Differential Equation

The general second order linear partial differential equations in two

independent variables is of the form.

which can be written as,

Auxx + Buxy + Cuvy + Dux + Euy + Fu = G where A, B. C, D, E, F, G

are functions of x and y

The above equations is said to be elliptic or parabolic or hyperbolic at

point (x,y) in the plane according as B2- 4AC <0 or B

2- 4AC = 0 or

B2 - 4AC >0

It is possible for a second order partial differential equation to be

elliptic in one region, parabolic in another and hyperbolic in some other

region.

For example consider

xuxx + uyy = 0 →(2.1)

Here A=x; B = 0; C =1 so that,

7

B2 - 4A C = - 4x

The equation (2.1) is elliptic if B2 -4AC <0

- 4 x < 0 i f x > 0

Similarly, parabolic If x = 0

And hyperbolic if x < 0

Examples 2:2:1

Classify the equations

(i) uxx + 2uxy + uyy = 0

(ii) x2fxx+(1-y

2)fyy=0

(iii) uxx + 4uxy + (x2 + 4y

2 )uyv = sinxy

Solution: (i) comparing the given equations with the general second order

linear, partial differential equation, we have

A = 1; B = 2; C = 1

Now B2- 4AC = 4 - 4 = 0

The given equation is parabolic

(ii) Here A=x2; B=0; C=1-y

2

Now B2- 4AC = 0 - 4x

2(1-y

2)= 4x

2(y

2-1)

We note that x 2>0 for all x except x = 0 and y

2 -1 < 0 for all y

such that -1 <y< 1

8

B2 -4AC <0 for all x≠0 and - 1 < y < 1

The equations is elliptic in the region given by x ≠0 and - 1 < y < 1

Similarly the equation is hyperbolic in the region given by x≠0 and

y < -1 or y > 1

B2 - 4AC = 4x

2(y

2 -1) = 0

If x = 0 or y2 =1,

The equation is parabolic if x = 0 or y=±1

(iii) uyy + uyy + (x2 + 4y

2 )uyy = sin xy

Here A=1, B=4 and C=x2+4y

2

Now, B2 - 4AC = 16 - 4(x

2 + 4y

2)

= 4[4 - x2 - 4y

2]

The equation is elliptic if 4 - x 2- 4y

2 < 0

ie, if x2+4y

2> 4

if

⁄

⁄

The equation is elliptic in the region outside ellipse

⁄

⁄

It is hyperbolic inside the ellipse

⁄

⁄

It is hyperbolic on the ellipse

⁄

⁄

9

CHAPTER - III

PARABOLIC EQUATIONS

3:1 - Parabolic Equations

We consider the heat conduction equation.

, C being a constant (3:1)

Let the (x,t) plane divided into smaller rectangles by means at the sets of

lines

x=ih,i=0,1,2,…

t=jk, j=0,1,2,…

Using the approximations

And

⁄

Equation (3:1) can be replaced by the finite - difference analogue.

⁄ [ ]

⁄ ( )

Which can be written as.

( )

10

Where ( )⁄

This formula express the unknown function value at the (i,j+l)th

interior point in terms of the known function. Values and hence it is called

the explicit formula, it can be shown that this formula is valid only for

⁄

For ⁄

⁄ ( )

Which is called Bender-Schmidt recurrence relation.

We have used the function values along the jth

row only the

approximation of

Crank and Nicolson proposed a method in 1947 according to which

is replaced by the average of its finite difference approximations on the

jth

and (j+1)th

rows.

11

Thus,

⁄

Hence the equation (5;1) is replaced by

⁄ [ ]

⁄

Which gives on rearranging

( ) ( )

where ( )⁄

On the left side we have three unknowns and on the right side all the

three quantities are known which is an implicit scheme is called Crank-

Nicolson formula and is convergent for all finite values of .

Problem: 3:1:1

Solve the parabolic equation

Given that ux(0,t) = 0

u(2,t) = 1

u(x,0) = 1 if 0 < x < 2 & i > 0

giving the solution for x = 0(0.5)2.0 and t = 0(0.125)0.5 ( t varies from 0 to

0.5 with step 0.125).

12

Solution: Given

a = 1 , Also h = 0.5 , and k = 0.125

Hence ⁄ = 0.5

u2,t = I and u(x,0) =1,

give all values of u as 1 along the last column and first row.

We know that ,

( )

ux(0,t) = 0 implies u i, j = u -1 , j

Thus the initial condition. ux(0,t)= 0 implies that the values of u at different

points along the lines x = k, and x = -k are equal.

In particular

u0.5,0 = u-0.5,0 =1

The other values of ui,j are found by using Schmidt’s equation as

shown below.

13

i

j

-0.5 0 0.5 1.0 1.5 2.0

-1 0 1 2 3 4

0 0 1 1 1 1 1 1

0.125 1 1 1 1 1 1 1

0.250 2 1 1 1 1 1 1

0.375 3 1 1 1 1 1 1

0.50 4 1 1 1 1 1 1

3:2 Schmidt’s Method

Consider a rectangular mesh in the x-t plane with side length h and k

in the x and t directions respectively.

Let us denote the mesh point

(x,y) = (ih,jk) by i , j

Then we have,

=

Substituting in the equation

= 0 , we get ,

( )

14

( )

( )

where =

( )

Now , the boundary conditions ,

u(0,t) = T0

u(1,t) = T1

Can be written in the difference notation as

u0,j = T0

and un,j = T1 where nh =1 and the condition

u(x,0) = f (x) as

ui,0 = f (ih) ; i = 1, 2, 3, …

we choose k in such a way that the coefficient of ui,j in (3:2) is zero

then (3:2) becomes

Coefficient of

( )

=

15

Further

This equation called Schmidt Recurrence equation.

Problem 3:2:1

Using Schmidt’s method solve

given that

u(0,t) = 0 ; u(1,t) = 0 ; u(x,0) =sin πx for 0 < x < 1 and taking h = 1/3 ,

k = 1/36 carry out the computation only for 2 levels ?

Solution :

Hence a = 1 ; since h = 1/3 and k =

1/36 ;

⁄

Since 0 < < 1/2 we can use Schmidt’s explicit formula

( ) for ⁄

⁄ ⁄

⁄

⁄ ( )

The boundary condition u(0,t) = 0 gives all values of ui,j on the t axis and

u(1,t) = 0 , gives all values of ui,j on the line x= 1

16

The initial condition,

u(x,0) = sin x can be written as

u1,0 = sin x

u1,0 = sin( ⁄ )= √

⁄

u2,0 = sin( ⁄ )= √

⁄

Putting j = 0 and i = 1

u1,1 =

⁄ u2,0 + 2u1,0+ u0,0]

= ⁄ √

⁄ (√

⁄ )

=0.65

Putting j = 0 and i = 2

u2,1 =

⁄ u3,0 + 2u2,0+ u1,0]

= ⁄ (√

⁄ ) (√

⁄ )

=0.65

Putting j = 1 and i = 2 respectively , we get

u1,2 =

⁄ u2,1 + 2u1,1+ u0,1]

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= ⁄ ( )

=0.49

U2,2 =

⁄ u3,1 + 2u2,1+ u1,1]

= ⁄ ( )

=0.49

3:3 – Crank-Nicholson Difference Method

In Schmidt’s explicit formula we used the function. Values along the j

th row only in the approximation of uxx as.

=

Crank and Nicholson proposed a method in which uxx is replaced by

the average of it’s finite difference approximation in the

j th

and (j+1) th

rows.

Thus

⁄

For ut we use the forward difference approximation ,

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The one dimensional heat equation reduces to

(

)

Setting

the above equation becomes,

(

)

⁄

⁄ ⁄

⁄

( ) ( ) ( ) ( )

This equation is called Crank-Nicholson Difference Scheme or

Crank-Nicholson Difference Method .

Problem 3:3:1

Solve by Crank- Nicholson’s method

for 0 < x < 1 , t > 0

Given that u(0,t) = 0, u(1,t) = 0 and u(x,0) = 100(x-x2) compute u for one

time step with h = ⁄

Solution : Here a = 1, h= ⁄

Choose k = ah2 = ⁄ , Hence =

⁄ = 1

we can use the simplified formula of Crank- Nicholson for = 1

⁄ [ ] →(3:3)

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Since we have to compute the values of u for only one step of time ‘t’ let the

required values of u be u1,u2,u3

Put j = 0 and i = 1,2, 3, 4 in (3:3)

We get all these values as follows :

Putting j =0 and i = 1,

u1,1= u 1 = ⁄ u0,1 + u2,1+ u0,0+ u2,0]

u 1 = ⁄ 0 + u2+ 0 + 25]

Hence u 1 = ⁄ u2 + 25] →(3:4)

Putting j =0 and i = 2,

u2,1= u 2 = ⁄ u1,1 + u3,1+ u1,0+ u3,0]

u 2 = ⁄ u1 + u3+ 18.75+18.75]

Hence u 2 = ⁄ u1 +u3+ 37.5] →(3:5)

Putting j = 0 and i = 3,

u3,1= u 3 = ⁄ u2,1 + u4,1+ u2,0+ u4,0]

u 3 = ⁄ u2 + 0 + 25 +0]

20

Hence u 3 = ⁄ u2 +25] →(3:6)

From (3:4) & (3:5)

u1 = u3

Hence (5:4) & (5:5) become

u 1 = ⁄ u2 + 25]

u 2 = ⁄ 2u1 + 37.5]

Solving for u1 and u2 we get u1 = 9.82 , u2 = 14.29, u3 = 9.82

Problem 3:3:2

Solve by Schmidt’s method ut = 5uxx with the conditions,

u(0,t) = 0

u(5,t) = 60,

and u(x,0) = {

for 5 times steps having h = 1,

Solution: Given uxx = ⁄ ut →(3:7)

a = ⁄ , take h = 1 , and

K =

⁄ , = ⁄

⁄

21

Hence we can use Schmidt’s recurrence method to solve (3:7)

The boundary condition u(0,t) = 0 gives all entries zero in the first

column. Similarly u(5,t) = 60 gives all entries 60 in the fifth column,

The initial condition,

u(x,0) = {

gives the values of u i,j of the first row, they are

u1,0 =20, u2,0 =40, u3,0 =60,

u4,0 =60, u5,0 =60,

All these above values of u i,j are shown in the following table in bold

letters,

i 0 1 2 3 4 5

j

0 1 2 3 4 5

0 0 0 20 40 60 60 60

0.1 1 0 20 40 50 60 60

0.2 2 0 20 35 50 55 60

0.3 3 0 17.5 35 45 55 60

0.4 4 0 17.5 31.25 45 52.5 60

0.5 5 0 15.625 31.25 41.875 52.5 60

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The other values of the u got from Schmidt’s recurrence formula

⁄

Problem 3:3:3

Using Crank - Nicholson methods

Solve uxx= ut subject to,

( ) ( ) ( )

Taking h = ⁄ & k = ⁄ compute u for one time step only?

Solution:

Here a = 1 , 0 ≤ x ≤ 1 and

t= 0 , ⁄

⁄

⁄

( ⁄ )

Since Crank – Nicholson formula , here,

[ ] [ ]

ie →(3:8)

The initial and boundary values of u , given in the table in bold

letters, and the other values are got from (5:8

Let →(3:9)

23

i

j

0

1

2

3

4

0 0 0 0 0 0

1 0 u1 u2 u3 ⁄

Putting j = 0 and i = 1 in (3:8) we have

u2,1 + u0,1+ u1,1 = u1,0 - u2,0- u0,0

(ie) u2 + 0 – 3u1 = 0 – 0- 0

(ie) u2 = 3u1 (3:10)

Putting j = 0 and i = 2 in (3:8) we have

u3,1 + u1,1-3 u2,1 = u2,0 – u3,0- u1,0

(ie) u3 + 0 – 3u2 = 0

(ie) u1 + u3 = 3u2 (3:11)

Putting j = 0 and i = 3 in (3:8) we have

u4,1 + u2,1-3 u3,1 = u3,0 – u4,0- u2,0

⁄ + u2 – 3u3 = 0

(ie) 3u3 – u2 = ⁄ (3:12)

Solving (3:10) , (3:11) & (3:12) , we get

u1 = 0.00595 , u2 = 0.01789 , u3 = 0.04762

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CHAPTER – IV

HYPERBOLIC EQUATIONS

4:1-Hyperbolic Equations

We consider the equations

utt=c2uxx (4:1)

u (x,0)= f(x) (4:2)

ut(x,0) = (x) (4:3)

u(0,t) = 1(t) (4:4)

u(1,t) = 2(t) (4:5)

For which models the traverse vibrations of a stretched string.

As in the previous cases, we use the following difference approximations

for the derivatives,

⁄ [ ] ( ) → (4:6)

and ⁄ [ ] ( ) → (4:7)

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Where x = ih , i = 0 , 1 , 2,…..

and t =jk, j = 0 , 1 , 2 , ……..

ut(x,t) =

+O(k

2)

Substituting (4:6),(4:7), in (4:1), we obtain

⁄ (ui,j-1 – 2ui,j + ui,j+1) =

⁄ (ui-1,j – 2ui,j + ui+1,j ) → (4:8)

Putting ⍺ = ⁄ in the above and rearranging the terms, we obtain

ui,j+1 = -ui,j-1 + (ui-1,j + ui+1,j) + 2(1- )ui,j → (4:9)

formula (6:9) shows that the function values at the jth and (j-1)

th time

levels are required in order to determine those at the (j+1)th time level. Such

difference schemes are called three level difference schemes compared to

the two level schemes derived in the parabolic case.

By expanding the terms in (4:9) as Tayolr’s series and simplifying it

can be shown that the truncation error in (4:9) is 0( + ) Further ,

formula (4:9) holds good ⍺ < 1, which is the condition for stability.

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There exist implicit finite difference schemes for the equation given by (4:1)

two such schemes are ,

=

⁄ ui+1,j+1 – 2ui,j+1 + ui-1,j+1]

+ ui+1,j-1 – 2ui,j-1 + ui-1,j-1] → (4:10)

and,

=

⁄ ui+1,j+1 – 2ui,j+1 + ui-1,j+1]

+2 ui+1,j – 2ui,j + ui-1,j]

+ ui+1,j-1 , 2ui,j-1 + ui-1,j-1] → (4:11)

Formula (6:10) & (6:11) hold good for all values of ⁄

Problem :4:1:1

Solve the equation

=

Subject to the following conditions

u(0,t) = 0 & u(1,t) = 0, if t < 0

27

and

(0,t) = 0 & u(x,0) = sin3( x), for all x in 0 x 1

solution

u(x,t) = ⁄ sin x cos t - ⁄ sin x cos → (4:12)

we use the explicit formula,

ui,j+1 = -ui,j-1 + (ui-1,j + ui+1,j) + 2(1- )ui,j → (4:13)

where = ⁄ < 1

Let h = 0.25, and k = 0.2

Hence = ⁄ = 0.8

So that the stability condition is satisfied,

Let u i,j = u(ih, jk)

So that the boundary conditions become,

u0,j = 0 → ( )

u4,j = 0 → ( )

ui,0 = sin3( ) → ( )

i = 1, 2, 3, 4

ui,j – ui,-1 = 0 so that

ui,-1 = ui, 1

Substituting the value of = 0.8

Equation (6:13) becomes

ui,j+1 = ui,j-1 + 0.64(ui-1,j + ui+1,j) + 2(0.36)ui,j → (iv)

At the first step j = 0

ui,1 = -ui,-1 + 0.64(ui-1,j + ui+1,j) + 2(0.36)ui,j → (v)

28

ui,1 = 0.32(ui-1,j + ui+1,j) + 0.36ui,j

using (vi)

Hence u1,1 = 0.32(u0,j + u2,0) + 0.36u1,0

= 0.32(0+1) + 0.36(0.357)

= 0.4473

The exact value u(0.25, 0.2) = 0.4838

u2,1 = 0.32[0.3537 + 0.3537] + 0.36(1.0) = 0.5867

Exact value = 0.5296

finally,

u3,1 = 0.32 (1.0+0) + 0.36 (0.3537)

= 0.4473,

exact value = 0.4838

4:2-Wave Equation

One of the most important and typical homogeneous hyperbolic

differential equations is the wave equation.

It is of the form

Where C is the wave speed. This differential equation is used in

many branches of physics and Engineering and is seen in many situations

such as transverse vibrations of a string or membrane, longitudinal

29

vibrations in a bar, propagation of sound waves, electromagnetic waves, sea

waves, elastic waves in solids, and surface waves as in earth quakes.

The solution of a wave equation is called a wave function,

An example for inhomogeneous wave equation is

Where F is a given function of partial variables and time. In physical

problems F represents an external driving force such as gravity force,

Another related equation is,

Where is a real positive constant. This equation is called a wave

equation with damping term, the amplitude of which decreases

exponentially as t decreases.

Definition 4:2:1-The one dimensional wave equation

The one dimensional wave equation is given by,

30

This is a hyperbolic equation

We proceed to develop a method for obtaining numerical solution of

the one – dimensional wave equation.

a2uxx – uu = 0 → (4:14)

Subject to the boundary conditions

u(0,t) = 0 → (4:15)

u(1,t) = 0 → (4:16)

and the initial condition

u(x,0) = f(x) → (4:17)

ut(x,0) = 0 → (4:18)

Replacing the partial derivatives in(4:14) by the difference quotients.

uxx =

utt =

we get a2 = [

-

]

ie,

Taking ⁄ , we get.

( )

( ) ( ) → (4:19)

This scheme is called an explicit scheme for the solution of the wave

equation.

Problem: 4:2:2

Solve 4uxx=utt subject to the conditions

u(0,t) = 0 = u(4,t),

ut(x,0) = 0 and

31

u(x,o) = x (4-x)

Take h = 1, and obtain solution up to 5 time steps ?

Solution:

Here a2 = 4 and h =1, we choose k such that k = ⁄ = 0.5

The boundary conditions,

u(0 ,t) = 0 and u(0 ,t) = 0

give all values 0 in the first and fourth columns.

Now , u(x,0) = x (4-x)

= 0;

= 3;

= 4;

= 3;

= 0;

Now ut(x,0) = 0

⁄

for …

we get,

⁄ [ ]

⁄

32

⁄ [ ]

⁄

⁄ [ ]

⁄

The other values of u1,1 are obtained from the recurrence relation.

And are the given in the table

i

j

0 1 2 3 4

0 1 2 3 4

0 0 0 3 4 3 0

0.5 1 0 2 3 2 0

1.0 2 0 0 0 0 0

1.5 3 0 -2 -3 -2 0

2.0 4

0 -3 -4 -3

0

2.5 5 0 -2

-3 -2 0