Chapter 2_603

8/7/2019 Chapter 2_603

1/28

CHAPTER 2

Finite Difference Equations,

Continuous Approach

Introduction

Finite Difference Equation for the Diffusivity Equation (1-D)

Types and Methods of Solution of F.D.E.

Accuracy of Solutions

Implicit vs. Explicit F.D.E.

Stability of Convergence of Finite Difference Equation(Von Neumann Analysis)

8/7/2019 Chapter 2_603

2/28

F.D.E. - Continuous Approach

1

Finite Difference Equations, Continuous Approach

Introduction

It is not possible to develop analytical solutions for many problems. The most common

reasons are the heterogeneity and irregular shapes of reservoirs. Another common reason, as in

the case for gases, is the nonlinearity of the equations (properties changing with pressure).

Reservoir simulation is a viable method for solving such problems and has become the standard

method for solving reservoir flow problems in both field offices and in research environments.

Reservoir simulation has become a common tool in reservoir engineering. Reservoir

simulation is used to solve research-type problems, such as analysis of enhanced oil recovery,

horizontal wells, well stimulation techniques, etc. More common applications, however, are thesimulation of actual wells and actual reservoirs. Since the late 1960's the technology and

computer hardware have advanced enormously and the cost and speed of doing reservoir

simulation have been improved dramatically.

The valuable aspect of reservoir simulation for field problems is the ability to make

production forecasts. Usually a number of different simulated cases are compared for the

purpose of making decisions. Drilling locations, production rates, injection plans, etc. are factors

that may vary from case to case.

The simulator data is taken from available information such as well logs, core analysis,

geological description, pressure data, and production data. In many cases, the well or wells havealready produced. For these cases, the past production period is simulated for the purpose of

"history matching". The input data usually must be adjusted in order to match the past

performance. This calibration process provides a more accurate reservoir "model" to be used

for forecasts.

In this chapter, we introduce the finite difference approach as an alternative to analytical

solutions. This approach is usually called reservoir simulation. The theory and practice of

reservoir simulation are covered in several publications.

8/7/2019 Chapter 2_603

3/28


2

Finite Difference Approach for The Diffusivity Equation (1-D)

We can start with the diffusivity equation in one dimension:

2

2

p

x=

c

0.00633k

p

t

(1)

This applies, of course, to the flow of a liquid with a small compressibility. We will

discuss gas flow later.

In order to solve problems which involve this equation, the finite difference method can

be used. This equation is "discretized" into the following finite difference form

i-1

n+1

i

n+1

i+1

n+1

2

i

n+1

i

np - 2 p + p

( x )=

c

0.00633 k

p - p

t

(2)

The n superscript indicates the "old" time level. All of the unknowns have already been

solved at the n time level. The n+1 superscript indicates the "new" time level. We want to solve

for these unknown values at the new time level.

Eq. 2 is called an implicit finite difference equation since it involves more than one

unknown. Three unknowns, pi-1n+1, pi

n+1, and pi+1n+1 occur because we chose the n+1 time level

to discretize the left-hand side of the equation. A template of this finite difference equation is

shown in

Fig. 1.

8/7/2019 Chapter 2_603

4/28


3

t

x

t

p p p

P(x+x)

x+xx

P(x)

x

P

Fig. 1 - Finite difference template, Fig. 2 - Illustration of Taylor showing pressures

involved in the finite series analysis.

difference equation.

We know the value ofp at the n time level and we are trying to determine the values ofp

at the n+1 time level.

The solution procedure for the finite difference equations is as follows:

(1) specify values for all pi0 at t = 0. (this is the initial condition)

(2) solve for all pin+1 for timestep 1.

(3) repeat this procedure for a sequence of timesteps, using the values from the previous

timestep as the "old" n values.

(4) stop when the desired time has been simulated.

It would have been possible to discretize the left-hand side of Eq. 1 at the n time level. In

that case, we would have only had one unknown, pin+1. This method would be called an explicit

finite difference equation. This would be easier to solve, but is of no practical use because it is

unstable (oscillates or blows up) for practical timestep sizes.

Theoretical basis, Taylor series.

Our justification for approximating Eq.1 with Eq.2 is based on Taylor series analysis.

Consider the function p(x) in Fig. 2. Suppose we know the value ofp(x) at the point x. Alsosuppose we know all the derivatives ofp(x) at the point x. If we want to approximate the value

ofp(x+x) at the point x+x, we can do this with a Taylor series as follows,

8/7/2019 Chapter 2_603

5/28


4

p(x+ x) = p(x)+ xp (x)+x

2!p (x)+

x

3!p (x)+ +

x

n!p (x)+

2 3 nn

(3)

where pn is the n th derivative ofp. This is an infinite series which is theoretically exact for an

infinite number of terms. However, if we truncate the series after n terms, then we introduce a

truncation error, et, (the remaining terms which are not included). This truncation error is

t

n+1 n

e =p(x ) x

(n+1)!+

( ) +

1

(4)

and is equal to

t

n+1 n

e =p(x + ) x

(n+1)!(0 x)

( )

+

1

(5)

It should be remembered that the function p(x) and all its derivatives must be continuousthroughout the interval under consideration. (In fact, we often deal with discontinuities in both

time and space. That limits the Taylor series analysis. Still we will continue our discussion

about using Taylor series for developing our finite difference equations.)

We will now simplify our notation by using pi rather than p(x), pi+1 rather than p(x+x),

and pi-1 rather than p(x-x). We can expand our Taylor series in either direction as follows:

i+1 i

2 2

2

3

3

4

4p = p + x

p

x

+x

2!

p

x

+x

3!

p

x

+x

4!

p

x

+

( ) ( ) ( )3 4

(6)

i-1 i

2

2

3

3

4

4p = p - x

p

x+

x

2!

p

x-

x

3!

p

x+

x

4!

p

x-

( ) ( ) ( )2 3 4

(7)

Notice that the right-hand side now uses partial derivatives (evaluated at xi) since p is a

function of both x and t, p(x,t).

We have a couple of choices of approximations to p/x:

8/7/2019 Chapter 2_603

6/28


5

Forward difference (from Eq. 6):

p

x

p - p

x

i+1 i

(8)

Backward difference (from Eq. 7):

)( xx

p-p

x

p 1-ii +

(9)

These expressions are both (x) (pronounced "first order correct") since the first term

truncated has x in it. The terms with higher powers ofx are called "higher order terms".

We can also develop an approximation for 2p/x2 by combining Eqs. 6 and 7. This gives us

Central difference (from Eqs. 6 and 7):

)( 2x)x(

p+p2-p

x

p2

1+ii1-i

2

2

+

(10)

This expression is used for the left hand side of Eq.1. It is (x2) (second order correct)

since the x terms cancel and the first term truncated contains x2.

We apply our Taylor series in terms oftto find the finite difference expression for p/t.

In this case, we will use a backward difference (in the negative tdirection from tn+1

). We will

simplify our notation to use superscripts to indicate the time level. Again, n is the "old" time

level for which we have a complete solution and know all variables and properties. n+1 is the

new time level for which the solution is unknown. With this notation, our finite difference

approximations to the 1-D diffusivity equation is

),( txt

p-p

k0.00633

c=

)x(

p+p2-pn

i

1+n

i

2

1+n

1+i

1+n

i

1+n

1-i +

(11)

Eq. 2 is the basic finite difference equation for the 1-D diffusivity equation. We say that

it is (x2,t). It must be solved for all the new pressures, p

n+1, simultaneously. Once these

new pressures are solved, then they become the old pressures for the next timestep. In thismanner, solutions to Eq. 2 are solved in a time sequence for as many timesteps as required.

Now we will turn our attention to the initial conditions and boundary conditions which

complete the specification of the problem.

8/7/2019 Chapter 2_603

7/28


6

Initial conditions. Initial conditions are required to begin the timestep sequence. For the initial

conditions, n=0, a value is specified for each pressure. The most common initial condition is to

specify each pressure to be equal to a constant, such as

i

0

initialp = p i = 1,.., IMAX (11)

where IMAXis the total number of points in the x direction. This typifies a well or reservoir at

initial conditions of no fluid flow.

Boundary conditions. For a 1-D problem, we need to specify equations other than Eq. 2 at the

left boundary and the right boundary. Usually, either a fixed pressure (mathematically called the

Dirichlet condition) or a fixed first derivative (the Neumann condition) are specified, as follows:

Fixed pressure:

p1n+1

= C1 (12)

or,

pIMAXn+1

= C2 (13)

Fixed first derivative:

2 13

p - p

x = C (14)

or,

IMAX IMAX -14

p - p

x= C

(15)

where C1, C2, C3, and C4 are values specified by the problem. Although these values may

change with time, they are held constant during the timestep.

We now will discuss how the solution is obtained for a timestep.

8/7/2019 Chapter 2_603

8/28


7

Types and Methods of Solution of F.D.E.

The pressure solution - a matrix problem. We can now state all of the equations that are to be

solved simultaneously for a particular timestep. Let us take a case where the boundary

conditions are fixed pressure. For the following equations, the unknown pressures are on the left

hand side and the right hand side has known values.

The left boundary:

1 1p = C

The interior points, i = 2,...,IMAX-1:

- p + 2+c

0.00633k

x

t p - p =

c

0.00633k

x

t pi-1

n+1 t

i

n+1

i+1

n+1

i

n ( ) ( )

2 2

(17)

The right boundary:

IMAX 2p = C (18)

Eqs. 16-18 comprise a system of simultaneous equations with IMAX unknowns and

IMAXequations. All of these equations are linear with respect to the unknown pressures. That

is, there are no pressure squared, etc. All of the pressure terms have coefficients which do not

depend on pressure.

As an example, let us consider a problem where IMAX = 5. For this problem, we have

five unknown pressures and five equations. Three of these equations are for interior points,

represented by the finite difference equation, Eq. 2. The first and last equations are for the

boundary conditions.

This set of equations can be represented by a matrix equation which can simply be

written as

A p = d (19)

where A is a coefficient matrix, and p and dare column vectors. The matrix equation, Eq. 19, can

be shown as

8/7/2019 Chapter 2_603

9/28


8

b c

a b c

a b c

a b c

a b

p

p

p

p

p

d

d

d

d

d

1 1

2 2 2

3 3 3

4 4 4

5 5

1

2

3

4

5

1

2

3

4

5

=

(20)

The rows of the matrix represent equations and the columns represent unknowns. Only

the non-zero elements of the matrix are shown in Eq. 20. Notice that the non-zero elements

follow a diagonal trend, lying in three adjacent diagonals. This is called a tri-diagonal matrix

and is characteristic of the matrix form of Eq. 2. Since matrix A is mostly zeros for large

problems, it is not necessary to store all of its elements in the computer. The values ofa, b, c,

and d are stored as arrays and are computed as follows, for a problem with fixed boundary

pressures:

The left boundary:

b1 = 1

c1 = 0

d1 = C1

The interior points: i = 2, IMAX-1

ai = -1

bi = 2 + (c)/(0.00633k) ((x)2/t)

ci = -1

di = (c)/(0.00633k) ((x)2/t) pin

The right boundary:

aIMAX= 0

bIMAX= 1

dIMAX= C2

Notice that a1 and cIMAXare not used because they fall outside the matrix.

Tri-diagonal solution - Thomas algorithm. The tri-diagonal matrix of Eq. 20 is a form whichcommonly occurs in engineering problems and has a very efficient solution procedure called the

Thomas Algorithm. This algorithm is similar to Gaussian elimination in that it has a forward

sweep and a backward sweep. The algorithm can be summarized as follows:

8/7/2019 Chapter 2_603

10/28


9

Forward sweep:

w1 = b1

g1 = d1/w1

wi = bi - ai ci-1 / wi-1gi = (di - ai gi-1) / wi

carried out with i increasing, and

Backward sweep:

pIMAX= gIMAX

pi = gi - ci pi+1 / wi

carried out with i decreasing.

After this computation is completed, we have now solved for the new pressures and are

ready to go to the next timestep.

Example 2.1 1-D Simulation

Problem. Solve the following initial/boundary value problem with the method just discussed:

permeability = 10 md

viscosity = 10 cpcompressibility = 20 x 10-6 psi-1

length = 400 ft

porosity = 0.20

Initial condition:

p(x,0) = 1,000 psi

Boundary conditions:

p(0,t) = 0 psi

p(400,t) = 1,000 psi

8/7/2019 Chapter 2_603

11/28


10

Model parameters:

x = 100 ft (IMAX = 5)

t = 5 days

Solution.

This problem is solved by writing a FORTRAN program. The following program is

specifically written to solve this problem and output the pressure results each timestep.

Boundary Conditions Example.

p1 = 0

p5 = 1,000

1 01 1

1 1

1 1

0 1

0

1000

2

3

4

1

2

3

4

5

2

3

4

=

b

b

b

p

p

p

p

p

d

d

d

b2 = b3 = b4 = 2 +

d2 = p2n

d3 = p3n

d4 = p4

n

where: = (c)/(0.00633k) ((x)2/t)

8/7/2019 Chapter 2_603

12/28


11

c===================================================== 1dsim

c Program for solving diffusivity equation with fixed

c boundary pressures

DIMENSION a(40),b(40),c(40),d(40),p(40),pn(40)

WRITE(*,*) 'imax, delt'

READ (*,*) imax, delt

pinit = 1000.

xlength = 400.

c1 = 0.c2 = 1000.

delx = xlength / (imax-1)

phi = 0.20

perm = 10.

visc = 10.

comp = 20.E-6

time = 80.

alpha = (phi * visc * comp) / (0.00633 * perm) * (delx**2 / delt)

c ... initial conditions

DO 10 i = 1, imax

p(i) = pinit

pn(i) = pinit

10 CONTINUE

c ... interior gridblocks

DO 20 i = 2, imax-1

a(i) = -1.b(i) = 2. + alpha

c(i) = -1.

20 CONTINUE

c ... boundary conditions

b(1) = 1.

c(1) = 0.

d(1) = c1

a(imax) = 0.

b(imax) = 1.

d(imax) = c2

c ... taking timesteps

t = 0

30 CONTINUE

t = t + delt

DO 40 i = 2, imax-1

d(i) = alpha * pn(i)

40 CONTINUE

c

CALL thomas(a,b,c,d,p,imax)

WRITE(*,800) t, (p(i),i = 1,imax)

DO 50 i = 1, imax

pn(i) = p(i)

50 CONTINUE

IF ( t .LT. time ) GO TO 30

800 FORMAT( 200F8.1)

END

8/7/2019 Chapter 2_603

13/28


12

c==================================================== thomas

SUBROUTINE thomas(a,b,c,d,p,n)

c this is a solution of tridiagonal system of equations

c a,b,c,d : coefficients of the equation,

c a(i) * x(i-1) + b(i) * x(i) + c(i) * x(i+1) = d(i)

c p = solution vector

c n = number of unknowns

DIMENSION a(*),b(*),c(*),d(*),p(*),w(101),g(101)

w(1) = b(1)g(1) = d(1) / w(1)

DO 1 i = 2,n

w(i) = b(i) - a(i) * c(i-1) / w(i-1)

g(i) = (d(i) - a(i) * g(i-1)) / w(i)

1 CONTINUE

p(n) = g(n)

DO 2 i = n-1,1,-1

p(i) = g(i) - c(i) * p(i+1) / w(i)

2 CONTINUE

RETURN

END

8/7/2019 Chapter 2_603

14/28


13

The numerical results are shown in Fig. 3 Although results are only computed at discrete

points, it is customary to draw straight lines between points. The straight lines, rather than a

smoothed curve, emphasize the discrete nature of the solution and clearly shows where the

discrete points are. The solutions in Fig. 3 show that the early solution is transient but

approaches steady state at 80 days.

0 100 200 300 400

X (ft)

0

200

400

600

800

1000p(x,t) (psi)

t = 5 days10

2030

80

0 100 200 300 400

X (ft)

0

200

400

600

800

1000p(x,5) (psi)

. t = 5 days

. t = 2.5 days

t = 1.25 da s. t = 0.625. t = 0.3125. t = 0.15625

Fig. 3 - Pressure profiles Fig. 4 - Comparison of pressure

for Example 1. profiles at t = 5 days for differenttimestep sizes.

8/7/2019 Chapter 2_603

15/28


14

Accuracy of solutions

Now the question occurs: "how accurate are the finite difference solutions". We can

compare with an exact analytical solution, if it is available. However, we do not have an exact

analytical solution for many problems. We need to develop a methodology for analyzing the

accuracy of finite difference solutions - at least to have confidence that they are sufficientlyaccurate for the purposes of our objectives.

The accuracy of the solutions is related to the truncation error in our Taylor series

analysis. We could do a complicated analysis of the Taylor series approximations, but we would

be analyzing the accuracy of the finite difference equation (it's approximation to the partial

differential equation), not the accuracy of the solution.

A more practical approach is to vary the values oftand x and analyze the comparative

behavior of our solutions. Let us use Example 1 to illustrate this numerical experimentation.

First, we will vary the value oftand observe the results. tis usually much easier to change

than x. Fig. 4 shows such a comparison at t= 5 days. It can be seen that as tbecomes smaller,the solution converges.

The value ofp at x = 100 ft and t= 5 days, p(100,5), is then plotted in Fig. 5 to see more

detail. As t approaches zero, p(100,5) seems to approach the correct analytical solution of

573.34 psi. Fig. 5 also shows other values for x and they all approach the exact solution as t

approaches zero. On the other hand, Fig. 6 shows the effect ofx on the solution with various

values oft. It is seen that p(100,5) does not approach the correct solution as x approaches

zero unless tis sufficiently small. For this particular problem, we can conclude the small tis

much more important than small x. In general, however, we should realize that complete

numerical experiments may have to be carried out to have confidence in our results.

Fig. 7 shows our first solution ofp(x,5) using t= 5 days and x = 100 ft, compared with

our improved values oft= 1.25 days and X= 25 ft. It is easy to see the improved accuracy

and smoothness of the solution. However, this solution is still higher than the analytical solution

at x = 100 ft (the analytical solution is 573.34 psi).

Fig. 8 shows a comparison of the two numerical solutions on a time plot. The same

general behavior is true as on the x profile plot - the smaller values of x and t give more

accurate results.

For this problem, we may not require great accuracy. We may decide from Figs. 5 and2.6 that values oft= 1.25 days and x = 25 ft are sufficiently small to give good results for

practical purposes. This gives an error of about 35 psi for p(100,5) which seems high. However,

if we look at Fig. 7, it looks like these values give an acceptable result. (The error at x = 100 is

8/7/2019 Chapter 2_603

16/28


15

about the highest in the x profile). This selection oft and x is a judgement, according to thepracticality for the problem being solved.

0 1 2 3 4 5

. t (days)

550

575

600

625

650

675

700p(100,5) (psi)

x = 12.5 ft

0 20 40 60 80 100

. x (ft)

550

575

600

625

650

675

700p(100,5) (psi)

t = 5 days

t = 1.25 days

t = 0.15625 days

t = 0.01 days

x ft=625.

x = 100 ft

x = 50 ft

Fig. 5 - Effect of timestep size on Fig. 6 - Effect of grid spacing on

pressures at x = 100 feet and t = 5 days pressures at x = 100 feet and t = 5 days

0 100 200 300 400

X (ft

0

200

400

600

800

1000p(x,5) (psi)

x = 25 ft , t = 1.25 days

x=100 ft , t =5 days

Analytical Solution

0 20 40 60 80Time (days)

0

200

400

600

800

1000p(100,t) (psi)

x = 100 ft , t = 5 days

x = 25 ft ,t = 1.25 days

Analytical Solution

Fig. 7 - Comparison of pressure Fig. 8 - Comparison of pressure

profiles at t = 5 days vs. time at x = 100 feet

8/7/2019 Chapter 2_603

17/28


16

On large problems encountered in reservoir simulation, there is a cost involved in taking

small values oftand space grids (x, y, and z). It is also recognized that much of the field

data is somewhat inaccurate and uncertain. Therefore, something less then complete accuracy

may be acceptable in order to reduce cost and simplify data preparation.

Although we have found practical values of t and x for this particular problem, weshould not expect these values to apply to other problems. The trial and error procedure for

determining practical values ofx and tis called a sensitivity analysis and is commonly done

for practical problems when the question of accuracy arises. Most of the time the correct

analytical solution is not known, so converging results are taken to be converging on the correct

solution.

It should also be pointed out that the best solution may not always be achieved by t

approaching zero. For certain other problems, such as water displacing oil (the Buckley-Leverett

problem), it is known that the best value of t depends on the value of x and the physical

properties of the problem6. Therefore, we should be cautious in accepting that small values oft

always give the best results.

Other Taylor Series Approximations.The following is a table that shows forward, backward,

and central difference expressions for first and second derivatives.

pi = p(x)

pi+1 = p(x+x)pi+2 = p(x+2x)

Difference Expressionpi-3 pi-2 pi-1 pi pi+1 pi+2 pi+3

Forward Difference

x p ' (x) -1 1

(x)2 p ''(x) 1 -2 1

Backward Difference

x p ' (x) -1 1 + O (x)

(x)2 p ''(x) 1 -2 1

Central Difference

2 x p ' (x) -1 0 1 + (x)2

(x)2p ''(x) 1 -2 1

Higher Order Difference Expressions

8/7/2019 Chapter 2_603

18/28


17

Forward Difference

2x p ' (x) -3 4 -1 + (x)2

(x)2 p ''(x) 2 -5 4 -1

Backward Difference

2x p ' (x) 1 -4 3 + (x)2

(x)2

p ''(x) -1 4 -5 2

Central Difference

12x p ' (x) 1 -8 0 8 -1 + (x)4

12 (x)2 p ''(x) -1 16 -30 16 -1

8/7/2019 Chapter 2_603

19/28


18

Implicit vs. Explicit F.D.E.

2

2

p

x=

p

t

(21)

)x(+)x(

p+p2-p=

x

p 22

1+ii1-i

2

2

(22)

where

x)+x*xx-(xt)(x*,x

p

12

)x(

-=)x( 4

422

(23)

t)(+t

p-p=

t

pn

i

1+n

i

(24)

where

t)+tt(t)t(x,x

p

2

t-=t)(

2

2

** (25)

Explicit FDE. The solution is only stable when (t) < (x)2/2.

1

( x )( p - 2 p + p ) =

1

t( p - p )

2 i-1

n

i

n

i+1

n

i

n+1

i

n

(26)

Implicit FDE. The solution is unconditionally stable for any t.

1

( x )( p - 2 p + p ) =

1

t( p - p )

2 i-1

n+1

i

n+1

i+1

n+1

i

n+1

i

n

(27)

Crank-Nicolson Method (C-N). This is a half explicit/half implicit method

8/7/2019 Chapter 2_603

20/28


19

1

2

1

( x )( p - 2 p + p ) +

1

2

1

( x )( p - 2 p + p ) =

1

t( p - p )

2 i-1

n

i

n

i+1

n

2 i-1

n+1

i

n+1

i+1

n+1

i

n+1

i

n

(28)

Truncation Error : ( (x)2 , (t)2 )

General Implicit Methods. This is also a mixed explicit/implicit method, but is more flexible

than the C-N method.

(1 - )1

( x )( p - 2 p + p ) +

1

( x )( p - 2 p + p ) =

1

t( p - p )

2 i-1

n

i

n

i+1

n

2 i-1

n+1

i

n+1

i+1

n+1

i

n+1

i

n

(29)

Weighting factor : ( 0 1 )

= 1 ( Implicit)

= 0 ( Explicit)

= 1/2 ( C-N)

Truncation Error: ( (x)2 , (t) ) (generally)

Only the implicit method is unconditionally stable (for any timestep size). The explicit

method is stable only for sufficiently small timesteps, which are generally impractical. Althoughthe Crank-Nicholson method has higher order accuracy, it is on the borderline of instability and

is conditionally stable for practical purposes. Therefore, the implicit method is the standard in

reservoir simulation. If more accuracy is required, smaller timesteps are used.

8/7/2019 Chapter 2_603

21/28


20

Stability and Convergence of Finite Difference Equations (Von Neumann

Analysis )

Consider the equation

2

2

p

x=

p

t

(21)

The initial conditions can be represented by a Fourier series

==

1)0,(

m

isxmeaxf

(30)

But we can look at an arbitrary element and analyze the stability of its behavior. We can ignore

am and just represent the element as

e=p(x,0)isx (31)

At time t, we can represent the solution as

(t)e=p(x,t)isx (32)

Since the solution is separable (a function ofx times a function oft).

Time period operator, E.

We will represent one time period as an operator acting on the previous time period:

(t)E=t)+(t (33)

Analytical solution.

A solution can be written in the form:

ee=t)p(x,ts-isx

2

(34)

which means that

e=(t)ts-

2

(35)

8/7/2019 Chapter 2_603

22/28


21

Then the solution can be expanded as follows,

]...+3!

t)s(-

2!

t)s(+ts-1[e=p(x,t)

2 32 2

2isx (36)

but

-s -s t -s t -s t2 (t+ t) 2 2 2

e = e e = E e (37)

So

e=Ets-

2 (38)

analytical2

22 32

E = 1 - s t +( s t)

2!-

( s t)

3!+ ...

(39)

Note that for any (positive) value oft, 0

8/7/2019 Chapter 2_603

23/28


22

So,

(t)-)t+(t=(t)1]-)x(s[)x(

t22

cos (44)

(t + t ) = (1 + 2t

( x )[ (s x ) - 1] ) (t)

2

cos (45)

)2

(sin21)cos( 2x

sxs

= (46)

(t))])2

x(s[

)x(

t4-(1=t)+(t 2

2

sin (47)

Then, the value ofEis

)]2

x(s[

)x(

t4-1=E

2

2explicit

sin (48)

For small values ofx,

22sin xsxs (49)

)2

x)(s

2

x(s=)

2

x(s 2

0x

sinlim (50)

)2

x(s)

2

x(s

)x(

t4-1=E 2explicit

(51)

So,

ts-1=E2

explicit (52)

8/7/2019 Chapter 2_603

24/28


23

Which does converge to Eanalyticalas tbecomes small.

But since we can write the initial conditions as

p0

= eisx

and after n timesteps

p(x,t) = Enp(x,0)

n isx

2

2 np = e (1 - 4t

( x )[ (s

x

2)] )

sin (53)

The quantity, Emust be -1 E1, otherwise changes in p will grow instead of decayingand errors will grow instead of decaying. So, in order for the absolute value ofEto be less than

1, the condition for stability is

t

( x )

Chapter 2_603

Documents

Transcript of Chapter 2_603