one‐dimensional finite‐difference method · the finite‐difference method slide 4 the...

9/22/2019

1

Computational Science:Computational Methods in Engineering

One‐DimensionalFinite‐Difference Method

Outline

• Introduction & Problem Setup•Conventional Finite‐Difference Method• Improved Finite‐Difference Method•Matrix Operators• Incorporating Boundary Conditions• Solving ODE’s

2

1

2

9/22/2019

2

Slide 3

Introduction & Problem Setup

The Finite‐Difference Method

Slide 4

The finite‐difference method is a way of obtaining a numerical solution to differential equations. It does not give a symbolic solution.

2

2 0 0 10

0 1, 10 5

d y dy y xdx dxy y

Governing Equation

A y bMatrix Equation

1y A bNumerical Solution

3

4

9/22/2019

3

Functions are Discrete

Slide 5

To obtain a numerical solution using the finite‐difference method, functions are stored as arrays of discrete points.

The function can be resolved more accurately using more points, but the solution will be more computationally intensive to obtain. This is a fundamental tradeoff.

Discrete Functions are Stored as Column Vectors

Slide 6

Discrete functions are stored as a 1D array of numbers in a column vector.

2

2

2

2

0

1 0

d y dy ydx dx

d d ydx dx

0A y y

5

6

9/22/2019

4

Slide 7

ConventionalFinite‐Difference Method

Step 1 – Identify Governing Equation & Boundary Values

Slide 8

2

2 0 0 10d y dy y xdx dx

Governing Equation

0 110 5y

y

Boundary Values

7

8

9/22/2019

5

0y x

Step 2 – Approximate Derivatives with Finite‐Differences (1 of 3)

Slide 9

2

2 0dydx

ddx

yy

First, let the function be discrete.

This allows us to approximate the derivatives with finite‐differences.

2

2y x x y x y x xx

y x y x xx


Slide 10

22

0y x x y x y x x y x y x x

y xx x

It is critical to ensure that each term in the finite‐difference equation exists at the same point.

Exists at x Exists at x - x/2

Exists at x

Exists at x

This is not a healthy or stable formulation because not all of the terms exist at the same point.

9

10

9/22/2019

6


Slide 11

2

2

2

0

20

2

d y dy ydx dx

y x x y x y x x y x x y x xy x

x x

This is the correct finite‐difference equation.

All terms exist at x.

Step 3 – Write Finite‐Difference Equation Using Array Indices

Slide 12

2

2

2

1 1 1 12

0

20

2

2 02

i i i i ii

d y dy ydx dx

y x x y x y x x y x x y x xy x

x x

y y y y y yx x

11

12

9/22/2019

7

Step 4 – Rearrange Finite‐Difference Equation

Slide 13

1 1 1 12

1 1 1 12 2 2

1 12 2 2

2 02

1 2 1 1 1 02 2

1 1 2 1 11 02 2

i i i i ii

i i i i i i

i i i

y y y y y yx x

y y y y y yx x x x x

y y yx x x x x

The finite‐difference equation is rearranged so as to collect the y terms.

Step 5 – Setup Grid

Slide 14

Let’s solve this problem using 21 points.

x1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

x=

0

x=

10

The grid spacing x is then

10 0 0.51 21 1

b ax xxN

x

13

14

9/22/2019

8

Step 6 – Revise Finite‐Difference Equation

Slide 15

1 12 2 2

1 12 2 2

1 1

1 1 2 1 11 02 2

1 1 2 1 11 00.5 2 0.5 0.5 0.5 2 0.5

5 7 3 0

i i i

i i i

i i i

y y yx x x x x

y y y

y y y

Substituting x = 0.5 into our finite‐difference equation gives

Step 7 – Write Finite‐Difference Equation at Each Point on Grid

Slide 16

x1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

1 15 7 3 0i i iy y y

These terms exist outside of the grid.

15

16

9/22/2019

9

y 1=

1

y 21

= 5

Step 8 – Incorporate Boundary Values

Slide 17

x1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

1 15 7 3 0i i iy y y

Step 8 – Write Set of Equations as a Single Matrix Equation

Slide 18

A y b

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

00000000

0

1

000

yyyyyyyyyyyyyyyyyyyyy

5

0000000

17

18

9/22/2019

10

Step 9 – Solve Matrix Equation

Slide 19

11

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

10000000000


0000000005

1.00002.04973.11603.85453.80052.4436-0.6324-5.5482-11.8919-18.5007-23.3484-23.6453-16.25821.473030.533968.7909109.6222141.1336146.6082106.86305.0000

Step 10 – Plot the Result

Slide 20

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21


1.00002.04973.11603.85453.80052.4436-0.6324-5.5482-11.8919-18.5007-23.3484-23.6453-16.25821.473030.533968.7909109.6222141.1336146.6082106.86305.0000

19

20

9/22/2019

11

Slide 21

ImprovedFinite‐Difference Method

Step 1 – Identify Governing Equation & Boundary Values

Slide 22

2

2 0 0 10d y dy y xdx dx

Governing Equation

0 110 5y

y

Boundary Values

21

22

9/22/2019

12

Step 2 – Write Equation in Matrix Form Going Term‐by‐Term

Slide 23

2

2 0d dy x y x y xdx dx

2D D 0x xy y y

Step 3 – Factor Out [y] To Put in Standard Form

Slide 24

2D D 0x x I y

2D D 0x xy y y

2

0 standard form

D Dx x

A y

A I

A = DX2 - DX + I;

23

24

9/22/2019

13

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

0000000000000


00000000

Step 4 – Incorporate Boundary Values (1 of 4)

Slide 25

Start with direct matrix equation.

0A y

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

0000000000000


00000000


Slide 26

Zero out rows in [A]that correspond to the boundary points. For this case, these are the rows for y1 and y21.

25

26

9/22/2019

14

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

0000000000000


00000000


Slide 27

Insert 1’s in the diagonal position of the rows in [A]where zeros were inserted.


Slide 28

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

00000000

0

1

000


5

0000000

Insert boundary values in column vector [b].

27

28

9/22/2019

15

Step 5 – Solve Matrix Equation

Slide 29

11

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

10000000000


0000000005

1.00002.04973.11603.85453.80052.4436-0.6324-5.5482-11.8919-18.5007-23.3484-23.6453-16.25821.473030.533968.7909109.6222141.1336146.6082106.86305.0000

Step 6 – Plot the Result

Slide 30

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21


1.00002.04973.11603.85453.80052.4436-0.6324-5.5482-11.8919-18.5007-23.3484-23.6453-16.25821.473030.533968.7909109.6222141.1336146.6082106.86305.0000

29

30

9/22/2019

16

Slide 31

Matrix Operators

Functions Vs. Operations (1 of 2)

Slide 32

2

2 f x f x f x ga b x xxxx c

x

FunctionsFunctions only appear in a differential equation as the unknown or as the excitation.

unknownexcitation

f x

g x

OperationsEverything else in a differential equation is something that operates on a function.

2

2

, , point-by-point multiplication on

, calculates derivatives of

scales entire

a x b x c x

f x

f xx x

f x

31

32

9/22/2019

17

Functions Vs. Operations (2 of 2)

Slide 33

2x xA D B D Cf f f g

FunctionsFunctions are stored as column vectors.

OperationsOperations are always stored in square matrices. Any linear operation can be put into matrix form.

1 1

2 2

M M

f gf g

f g

f g

11 12 1

21 22 2

1 2

M

M

M M MM

l l ll l l

L

l l l

Point‐by‐Point Multiplication (1 of 2)

Slide 34

Since we are storing our “functions” in vector form, how do we perform a point‐by‐point multiplication using a square matrix?

f xb x B f1x 2x 3x 4x 5x 6x

1f2f

3f 4f 5f 6f

1b2b 3b 4b 5b 6b

1 11 1

2 2

3 3

4 4

2 2

3 3

4

5 5

4

5 5

6 66 6

0 0 0 0 00 0 0 0 00 0 0 0 00 0 0 0 00 0 0 0 00 0 0 0 0

f ff ff

b bb b

b bb b

ff ff ff f

b bb b

f fB B

?

33

34

9/22/2019

18

Point‐by‐Point Multiplication (2 of 2)

Slide 35

1x 2x 3x 4x 5x 6x

1f2f

3f 4f 5f 6f

1b2b 3b 4b 5b 6b

1 11 1

2 2

3 3

4 4

2 2

3 3

4

5 5

4

5 5

6 66 6

0 0 0 0 00 0 0 0 00 0 0 0 00 0 0 0 00 0 0 0 00 0 0 0 0

f ff ff

b bb b

b bb b

ff ff ff f

b bb b

f fB B

Since we are storing our “functions” in vector form, how do we perform a point‐by‐point multiplication using a square matrix?

f xb x B f

First‐Order Partial Derivative (1 of 2)

Slide 36

How can a square matrix be constructed so that when it premultiplies a vector it gives a vector containing the first‐order partial derivative?

xf Dxx f

1x 2x 3x 4x 5x 6x

1f2f

3f 4f 5f 6f

1 2 0

3 1

4

2

3

4

5

2

5 3

6 4

76 5

22222

0 1 0 0 0 01 0 1 0 0 0

0 1 0 1 0 010 0 1 0 1 020 0 0 1 0

21

0 0 0 0 1 0x x

x

D

x

x

x

x

x

f

x

D f

f ff ff ff

fffff

ff f

f f f

?

35

36

9/22/2019

19

First‐Order Partial Derivative (2 of 2)

Slide 37

1x 2x 3x 4x 5x 6x

1f2f

3f 4f 5f 6f

1 2 0

3 1

4

2

3

4

5

2

5 3

6 4

76 5

22222

0 1 0 0 0 01 0 1 0 0 0

0 1 0 1 0 010 0 1 0 1 020 0 0 1 0

21

0 0 0 0 1 0x x

x

D

x

x

x

x

x

f

x

D f

f ff ff ff

fffff

ff f

f f f

xf Dxx f

How can a square matrix be constructed so that when it premultiplies a vector it gives a vector containing the first‐order partial derivative?

Second‐Order Partial Derivative (1 of 2)

Slide 38

How can a square matrix be constructed so that when it premultiplies a vector it gives a vector containing the second‐order partial derivative?

2

22 xDf fx

x

1x 2x 3x 4x 5x 6x

1f2f

3f 4f 5f 6f

2

22 1 0

23 2 1

24 3 2

25 4 3

26 5

2

42

1

2

3

4

5

7 6 56

2 1 0 0 0 01 2 1 0 0 00 1 2 1 0 010 0 1 2 1 00 0 0 1 2 10 0 0 0 1 2

222222

x

x

D f

x

x

x

x

x

x

ffffff

f f ff f ff f ff f ff f ff f f

2xD f

?

37

38

9/22/2019

20

Second‐Order Partial Derivative (2 of 2)

Slide 39

1x 2x 3x 4x 5x 6x

1f2f

3f 4f 5f 6f

2

22 1 0

23 2 1

24 3 2

25 4 3

26 5

2

42

1

2

3

4

5

7 6 56

2 1 0 0 0 01 2 1 0 0 00 1 2 1 0 010 0 1 2 1 00 0 0 1 2 10 0 0 0 1 2

222222

x

x

D f

x

x

x

x

x

x

ffffff

f f ff f ff f ff f ff f ff f f

2xD f

2

22 xDf fx

x

How can a square matrix be constructed so that when it premultiplies a vector it gives a vector containing the second‐order partial derivative?

Why Do We Need Separate Derivative Matrices for First‐ and Second‐Order Derivatives?

Slide 40

It is known that,2

2x x x

Can we just calculate D(2) from D(1)? ?

2 1 1x x xD D D

Yes, but this does not make efficient use of the grid. For a 5‐point, 1D grid, we have

1 1 2

2 2

1 0 1 0 0 2 1 0 0 00 2 0 1 0 1 2 1 0 0

1 1 1 0 2 0 1 0 1 2 1 02 0 1 0 2 0 0 0 1 2 1

0 0 1 0 1 0 0 0 1 2

x x xxx

D D D

This is not as accurate because it calculates the derivative with poorer grid resolution than is available.

This derivate matrix makes optimal use of the available grid resolution.

39

40

9/22/2019

21

USE SPARSE MATRICES!!!!!!!

Slide 41

WARNING !!The derivative operators will be EXTREMELY large matrices.

For a small grid that is just 100200 points:

Total Number of Points: 20,000Size of Derivate Operators: 20,000 20,000Total Elements in Matrices: 400,000,000Memory to Store One Full Matrix: 6 GbMemory to Store One Sparse Matrix: 1 Mb

NEVER AT ANY POINT should you use FULL MATRICES in the finite‐difference method. Not even for intermediate steps. NEVER!

Placing Diagonals into Sparse Matrices in MATLAB

42

M = 6;Z = sparse(M,M);d = ones(M,1);A = spdiags(d,0,Z);

M = 6;Z = sparse(M,M);d = ones(M,1);A = spdiags(-d,-1,Z);A = spdiags(+d,+1,A);

A =[ 0 1 0 0 0 0 ][ -1 0 1 0 0 0 ][ 0 -1 0 1 0 0 ][ 0 0 -1 0 1 0 ][ 0 0 0 -1 0 1 ][ 0 0 0 0 -1 0 ]

A =[ 1 0 0 0 0 0 ][ 0 1 0 0 0 0 ][ 0 0 1 0 0 0 ][ 0 0 0 1 0 0 ][ 0 0 0 0 1 0 ][ 0 0 0 0 0 1 ]

41

42

9/22/2019

22

Slide 43

Incorporating Boundary Conditions

Dirichlet Boundary Conditions(1 of 2)

44

x1x 2x 4x 5x 6x 7x3x1f 2f 4f 5f 6f 7f3f

The simplest boundary condition is to assume all function values outside of the grid are zero.

27 6 7

22

2 0d f f fdx x

21 1 2

2220d f f f

dx x

21 1

22

2i i i id f f f fdx x

43

44

9/22/2019

23

Dirichlet Boundary Conditions(2 of 2)

45

x

7x

6x

4x

3x

2x

1x

5x

x

21 1 2

2220d f f f

dx x

27 6 7

22

2 0d f f fdx x

21 2 32

222f f fd f

dx x

23 2 3 4

22

2d f f f fdx x

23 4 54

22

2f f fd fdx x

25 4 5 6

22

2d f f f fdx x

26 5 6 7

22

2d f f f fdx x

2

3

42

5

7

6

12 1 0 0 0 0 01 2 1 0 0 0 00 1 2 1 0 0 0

1 0 0 1 2 1 0 00 0 0 1 2 1 00 0 0 0 1 2 10 0 0 0 0 1 2

fff

x ff

f

f

2xD

Periodic Boundary Conditions (1 of 2)

46


If the problem is periodic (i.e. keeps repeating), then the value outside of the grid is the same as the value at the opposite side of the grid.

27 6 7

21

2

2d f f fdx x

f

21 21

27

2

2f f fd fdx x

21 1

22

2i i i id f f f fdx x

45

46

9/22/2019

24

Periodic Boundary Conditions(2 of 2)

47

x

7x

6x

4x

3x

2x

1x

5x

x

21 21

27

2

2f f fd fdx x

27 6 7

21

2

2d f f fdx x

f

21 2 32

222f f fd f

dx x

23 2 3 4

22

2d f f f fdx x

23 4 54

22

2f f fd fdx x

25 4 5 6

22

2d f f f fdx x

26 5 6 7

22

2d f f f fdx x

2

7

3

42

5

6

12 1 0 0 0 01 2 1 0 0 0 00 1 2 1 0 0 0

1 0 0 1 2 1 0 00 0 0 1 2 1 00 0 0 0 1 2 1

0 0 0 0 1 2

1

1

fff

x ff

f

f

2xD

Neuman Boundary Conditions(1 of 3)

48


The Neuman boundary condition allows functions to continue linearly off of the grid as if to infinity.

27 7 6 7

2 0df f f d fdx x dx

21 2 1 1

2 0df f f d fdx x dx

21 1 1 1

22

2 2

i i i i i i idf f f d f f f fdx x dx x

47

48

9/22/2019

25


49

x

7x

6x

4x

3x

2x

1x

5x

x

1 2 12 22

df f fdx x

7 7 62 22

df f fdx x

3 12

2f fdf

dx x

3 4 2

2df f fdx x

5 34

2f fdf

dx x

5 6 4

2df f fdx x

6 7 5

2df f fdx x

1

7

2

3

4

5

6

2 2 0 0 0 0 0

0 0 0 0 0 2 2

1 0 1 0 0 0 00 1 0 1 0 0 0

1 0 0 1 0 1 0 02

0 0 0 1 0 1 00 0 0 0 1 0 1

fff

xff

f

f

xD


50

x

7x

6x

4x

3x

2x

1x

5x

x

21

2 0d fdx

27

2 0d fdx

21 2 32

222f f fd f

dx x

23 2 3 4

22

2d f f f fdx x

23 4 54

22

2f f fd fdx x

25 4 5 6

22

2d f f f fdx x

26 5 6 7

22

2d f f f fdx x

2

3

42

1

5

6

7

1 2 1 0 0 0 00 1 2 1 0 0 0

1 0 0 1 2 1 0 00 0 0 1 2 1 00 0 0 0 1 2 1

0 0 0 0 0 0 0

0 0 0 0 0 0 0

ff

f

f

ff

x f

2xD

49

50

9/22/2019

26

High‐Order Boundary Conditions (1 of 2)

51


Here we estimate the derivative at the boundaries using special finite‐difference equations derived specifically for these points.

27 4 5 6 7

2 2

4 5 2d f f f f fdx h

21 2 3 41

2 2

2 5 4f f f fd fdx h

21 1

2 2

2i i i id f f f fdx h

High‐Order Boundary Conditions (2 of 2)

52

x

7x

6x

4x

3x

2x

1x

5x

x

21 2 3 41

22

2 5 4f f f fd fdx x

27 4 5 6 7

224 5 2d f f f f f

dx x

21 2 32

222f f fd f

dx x

23 2 3 4

22

2d f f f fdx x

23 4 54

22

2f f fd fdx x

25 4 5 6

22

2d f f f fdx x

26 5 6 7

22

2d f f f fdx x

4

7

2

6

1

2

3

5

2 5 4 1 0 0 0

0 0 0 1 4 5

1 2 1 0 0 0 00 1 2 1 0 0 0

1 0 0 1 2 1 0 00 0 0 1 2 1 00 0 0 0 1 2

21

fff

x ff

f

f

2xD

51

52

9/22/2019

27

Slide 53

Solving OrdinaryDifferential Equations

Boundary Values

Slide 54

Many differential equations give boundary values, or initial conditions, that must be incorporated into the matrix equation in order to obtain a numerical solution.

For example, a problem may be stated as…

2

2 5 6 0 0 2

0 2 2 0.2

d f x df xf x x

dx dx

f f

Boundary values

53

54

9/22/2019

28

Why are Boundary Values Needed?

Slide 55

To solve the differential equation, it is first converted to matrix form.

2

2 5 6 0d df x f x f xdx dx

2 5 6 0x xD f D f f

Next, this is rearranged into standard form 𝐀𝐱 𝐛. 20 5 6x xA f A D D I

This is not solvable.

1 0 0f A The boundary values have to be incorproated in order to obtain a nontrivial solution. When this is done, the matrix equation will becomes

A f b

Incorporating Boundary Values

Slide 56

It is possible to incorporate the boundary values into the matrix operators that we used to build [A]. However, it is usually easier to incorporate these directly into the final matrix equation.

The steps are:

1. Build direct matrix equation [A][f]=[0].2. For each row in [A] that corresponds to a boundary point:

1. Replace entire row with 0’s.2. Insert a ‘1’ into the diagonal, or pivot, position.3. Place the boundary value in the same row of [b].

55

56

9/22/2019

29

Solve Matrix Equation

Slide 57

This matrix equation is now in the form A f b

This is solved for [f] as follows. 1f A b

For small to moderate size problems, this can be solved directly using LU decomposition, or just backward division in MATLAB.

f = A\b; DO NOT USE f = inv(A)*bFor large problems, iterative methods are preferred, but the conditioning of [A] becomes important and a solution is not guaranteed.

Example (1 of 10)

Slide 58

2

2 5 6 0 0 2

0 2 2 0.2

d f x df xf x x

dx dx

f f

Solve the following ordinary differential equation.

% DEFINE BOUNDARY VALUESxa = 0;xb = 2;fa = 2;fb = 0.2;

57

58

9/22/2019

30

Example (2 of 10)

Slide 59

Step 1 – Formulate the matrix equation [A][f] = [0].

2

2 5 6 0d df x f x f xdx dx

2 5 6 0x xD f D f f

2

0

5 6x x

A f

A D D I

2 5 6 0x xD D I f

Example (3 of 10)

Slide 60

Step 2 – Calculate grid parameters N and x.

Choose N = 11. This is just a first guess!

Calculate x.

2 0 0.21 11 1

b axN

% GRID PARAMETERSNx = 11;dx = (xb - xa)/(Nx - 1);

59

60

9/22/2019

31

Example (4 of 10)

Slide 61

Step 3 – Build Matrix Operators , , .2xD I xD

2

50 25 0 0 0 0 0 0 0 0 025 50 25 0 0 0 0 0 0 0 00 25 50 25 0 0 0 0 0 0 00 0 25 50 25 0 0 0 0 0 00 0 0 25 50 25 0 0 0 0 00 0 0 0 25 50 25 0 0 0 00 0 0 0 0 25 50 25 0 0 00 0 0 0 0 0 25 50 25 0 00 0 0 0 0 0 0 25 50 25 00 0 0 0 0 0 0 0 25 50 250 0 0 0 0 0 0 0 0 25 50

xD

0 2.5 0 0 0 0 0 0 0 0 02.5 0 2.5 0 0 0 0 0 0 0 00 2.5 0 2.5 0 0 0 0 0 0 00 0 2.5 0 2.5 0 0 0 0 0 00 0 0 2.5 0 2.5 0 0 0 0 00 0 0 0 2.5 0 2.5 0 0 0 00 0 0 0 0 2.5 0 2.5 0 0 00 0 0 0 0 0 2.5 0 2.5 0 00 0 0 0 0 0 0 2.5 0 2.5 00 0 0 0 0 0 0 0 2.5 0 2.50 0 0 0 0 0 0 0 0 2.5 0

xD

1 0 0 0 0 0 0 0 0 0 00 1 0 0 0 0 0 0 0 0 00 0 1 0 0 0 0 0 0 0 00 0 0 1 0 0 0 0 0 0 00 0 0 0 1 0 0 0 0 0 00 0 0 0 0 1 0 0 0 0 00 0 0 0 0 0 1 0 0 0 00 0 0 0 0 0 0 1 0 0 00 0 0 0 0 0 0 0 1 0 00 0 0 0 0 0 0 0 0 1 00 0 0 0 0 0 0 0 0 0 1

I

% BUILD MATRIX OPERATORSI = speye(Nx,Nx);[DX,DX2] = fdder1d(Nx,dx);

Example (5 of 10)

Slide 62

Step 4 – Calculate [A] and [b].

2

44 37.5 0 0 0 0 0 0 0 0 012.5 44 37.5 0 0 0 0 0 0 0 0

0 12.5 44 37.5 0 0 0 0 0 0 00 0 12.5 44 37.5 0 0 0 0 0 00 0 0 12.5 44 37.5 0 0 0 0 0

5 6 0 0 0 0 12.5 44 37.5 0 0 0 00 0 0 0 0 12.5 44 37.5 0 0 00 0 0 0 0 0 12.5 44 37.5 0 00 0 0 0 0 0 0 12.5 44 37.5 00 0 0 0 0 0 0 0 12.5 44

x xA D D I

37.5

0 0 0 0 0 0 0 0 0 12.5 44

00000000000

b

% CALCULATE [A] AND [b]A = DX2 + 5*DX + 6*I;b = sparse(Nx,1);

61

62

9/22/2019

32

Example (6 of 10)

Slide 63

Step 5 – Incorporate Boundary Values into [A] and [b].

2

0 0 0 0 0 0 0 0 012.5 44 37.5 0 0 0 0 0 0 0 0

0 12.5 44 37.5 0 0 0 0 0 0 00 0 12.5 44 37.5 0 0 0 0 0 00 0 0 12.5 44 37.5 0 0 0 0 0

5 6 0 0 0 0 12.5 44 37.5 0 0 0 00 0 0 0 0 12.5 44 37.5 0 0 00 0 0 0 0 0 12.5 44 37.5 0 00 0 0 0 0 0 0 12.5 44 37.5 00 0 0 0 0 0 0 0 12.

1

5

0

44 37.50

x xA D D I

0 0 0 0 0 0 0 0 0 1

0000000

2

.2

00

0

b

% INCORPORATE BOUNDARY VALUESA([1 Nx],:) = 0;A(1,1) = 1;A(Nx,Nx) = 1;b([1 Nx]) = [ fa fb ];

Example (7 of 10)

Slide 64

Step 6 – Solve for [f].

1

2.00002.68422.48282.01841.54071.13490.81810.58160.40970.28680.2000

f A b

% SOLVE FOR [f]f = A\b;

63

64

9/22/2019

33

Example (8 of 10)

Slide 65

Step 7 – Check for Convergence

A solution was obtained, but it was all based on a guess for how many points N to use on the grid.

It is critical to check for convergence by increasing the value of N until the changes in the solution are neglibable.

Converged aroundN ≈ 20.

Example (9 of 10)

Slide 66

Step 8 – Obtain the final answer at convergence (i.e. N = 20).

% PLOT RESULTh = plot(x,f,'-ob','LineWidth',2);h2 = get(h,'Parent');set(h2,'LineWidth',2,'FontSize',18);xlabel('$x$','Interpreter','LaTex');ylabel('$f(x)$','Interpreter','LaTex');T = [0 0.5 1 1.5 2];L = {'0' '0.5' '1.0' '1.5' '2.0'};set(gca,'XTick',T,'XTickLabel',L);T = [0.5 1 1.5 2 2.5 3];L = {'0.5' '1.0' '1.5' '2.0' '2.5' '3.0'};set(gca,'YTick',T,'YTickLabel',L);

65

66

9/22/2019

34

Example (10 of 10)

Slide 67

Step 9 – Post process the data.

The finite‐difference method is complete.

Usually, after obtaining a solution, the results are analyzed further.

Code Altogether

Slide 68

% DEFINE BOUNDARY VALUESxa = 0;xb = 2;fa = 2;fb = 0.2;

% GRID PARAMETERSNx = 20;dx = (xb - xa)/(Nx - 1);

% BUILD MATRIX OPERATORSI = speye(Nx,Nx);[DX,DX2] = fdder1d(Nx,dx);

% CALCULATE [A] AND [b]A = DX2 + 5*DX + 6*I;b = sparse(Nx,1);

% INCORPORATE BOUNDARY VALUESA([1 Nx],:) = 0;A(1,1) = 1;A(Nx,Nx) = 1;b([1 Nx]) = [ fa fb ];

% SOLVE PROBLEMf = A\b;

2

2 5 6 0

0 20 2 2 0.2

d f x df xf x

dx dxx

f f

67

68

one‐dimensional finite‐difference method · the finite‐difference method slide 4 the...

Documents