Topic 6b -- Finite-difference approximationsemlab.utep.edu/ee4386_5301cmee/Topic 6b -- Finite-difference... · Types of Finite‐Differences ... Backward Finite‐Difference ... Recall

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Topic 6b –Finite‐Difference Approximations

EE 4386/5301 Computational Methods in EE

Course InstructorDr. Raymond C. RumpfOffice: A‐337Phone: (915) 747‐6958E‐Mail: [email protected]

Outline

• What are finite‐difference approximations?

• Polynomial technique

– Description of the method

– Examples

– Implementation in MATLAB

– Examples using MATLAB

2Topic 6b ‐‐ Finite‐difference approximations

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What are Finite‐Difference Approximations?


What are Finite‐Difference Approximations? (1 of 3)

Slide 4Topic 6b ‐‐ Finite‐difference approximations

Very often in science and engineering we must calculate a derivative.

When we are processing data from measurements or simulations, there may not be an analytical equation to work with symbolically.

Typically, we only know the function at discrete points.

1x 2x 3x

1f

2f3f

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Suppose we wish to numerically calculate the first‐order derivative at x2.

The first‐order derivative is slope. We can estimate the slope as riserunusing information from surrounding points.

3 12

rise

run 2

f ff x

x

1x 2x 3x

1f

2f3f

x

run = 2x

rise

= f 3

–f 1



We can estimate the derivative at the midpoint between data points.

1x 2x 3x

1f

2f3f

2 11.5

f ff x

x

3 22.5

f ff x

x

1.5f

2.5f

The second‐order derivative is the slope of the slope.

2.5 1.52

3 2 2 1

3 2 12

2

f x f xf x

xf f f fx x

xf f f

x

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Our First Two Finite‐Difference Approximations

Topic 6b ‐‐ Finite‐difference approximations Slide 7

3 12 2

f ff x

x

We already derived two finite‐difference approximations

3 2 12 2

2f f ff x

x

General Concept of Finite‐Difference Approximations (1 of 2)


2 2 2 at ,f x y 1 1 1 at ,f x y

3 3 3 at ,f x y

4 4 4 at ,f x y

5 5 5 at ,f x y

6 6 6 at ,f x y

7 7 7 at ,f x y

or at ,a ai i

i ia a

f fx y

x y

Suppose we wish to estimate the function f(x) or one of its derivatives at location (xi,yi).

1 1 2 2 3 3 4 4 5 5 6 6 7 7 or a ai ia a

f fa f a f a f a f a f a f a f

x y

It is always possible to estimate this from a linear sum of the function values at surrounding points.

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General Concept of Finite‐Difference Approximations (2 of 2)


The trick is, how do we calculate the coefficients an?

These are a function of the positions of the points.

1 2 3 4 5 6 7, , , , , ,a a a a a a a 1 1,x y

7 7,x y 6 6,x y

2 2,x y 3 3,x y 4 4,x y

5 5,x y

Same coefficients

Different coefficients

Different coefficients

Finite‐difference coefficients depend only on the relative position of the points. They do not depend on the absolute positions.

Types of Finite‐Differences


2 11df f f

dx x

2f1f

xForward Finite‐Difference

Reaches ahead to use data in the forward direction.

1.5 2 1df f f

dx x

2f1f

xCentral Finite‐Difference

Reaches symmetrically to use data in both directions for highest accuracy.

2 12df f f

dx x

2f1f

xBackward Finite‐Difference

Reaches behind to use data in the backward direction.

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Continuum of Finite‐Difference Approximations (1 of 2)


Continuum of Finite‐Difference Approximations (2 of 2)


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Two Key Considerations


1. The position of the points from which the finite‐difference approximation is calculated. More closely spaced points improves accuracy, but typically leads to more computations.

2. The location of the point where the finite‐difference is being evaluated. We typically want to be as centered as possible for best accuracy.

Polynomial Technique for Deriving Finite‐Difference

Approximations


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Concept of Using Polynomials

Slide 15

We can fit an Nth order polynomial given N+1 points.

20 1 2

NNf x a a x a x a x

After we have performed the curve fit, we can interpolate the function or any of its derivatives from the polynomial.

2 3 40 1 2 3 4

2 3 11 2 3 4

2 22 3 4

33 4

2 3 4

2 6 12 1

6 24 1 2

NN

NN

NN

NN

f x a a x a x a x a x a x

f x a a x a x a x Na x

f x a a x a x N N a x

f x a a x N N N a x

Topic 6b ‐‐ Finite‐difference approximations

Easiest Point for Evaluating f(x)

Slide 16

Recall the equations we will use to evaluate f(x) or one of its derivatives:

2 3 40 1 2 3 4

2 3 11 2 3 4

2 22 3 4

2 3 4

2 6 12 1

NN

NN

NN

f x a a x a x a x a x a x

f x a a x a x a x Na x

f x a a x a x N N a x

0

1

2

3

0

0

0 2

0 6

f a

f a

f a

f a

These are most easily evaluated at x = 0 because the above equations reduce to


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How to Make Any Point Easy

Slide 17

Now suppose we wish to evaluate f(x) or one of its derivatives at the general point x = xfd.

To do this, we shift our x‐axis by xfd before fitting the polynomial.

Recall that the finite‐difference coefficients depend only on the relative position of the points. An offset will not affect their values.

Now we write our polynomial at each shifted point.

21 0 1 1 2 1 1

22 0 1 2 2 2 2

21 0 1 1 2 1 1

NN

NN

NN N N N N

f x a a x a x a x

f x a a x a x a x

f x a a x a x a x

fdn nx x x

In our shifted coordinate system, our finite‐difference is being evaluated at .0x

0 1 2 30 0 0 2 0 6 f a f a f a f a


Four‐Step Procedure to Derive Finite‐Difference Approximations

Slide 18

Step 1 – Identify set of points x1, x2, …, to xN from which to derive a finite‐difference approximation.


20 1 2

NNf x a a x a x a x

Step 3 – Fit shifted points to a polynomial.fdi ix x x

Step 4 – Write finite‐difference approximation directly from one of the derivatives of the polynomial.

0

1

2

0

0

0 2

f a

f a

f a

Step 2 – Shift coordinates so that corresponds to where you wish to approximate the function or one of its derivatives.

0x

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Step 1 – Choose x Coordinates

Identify the x‐coordinates of the points from which you wish to approximate a derivate.

Store these in a column

1

2

1N

x

xx

x


Step 2 – Shift x‐Axis

Shift the function across the x‐axis until corresponds to the point where you wish to approximate the derivative.

20

fd 0a a

a a

d df x x f x

dx dx


0x

Subtract xfd from the column vector [x] to shift coordinates.

1 fd 1

2 fd 2fd

1 fd 1N N

x x x

x x xx x x

x x x

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Step 3 – Fit Points to Polynomial (1 of 3)

Use the column vector to build matrix .

0 1

1 fd 1 fd1 1

2 2 2 fd 2 fd

1 1 1 fd 1 fd

11

1 1

1 1

N

NN

NN

N NN N N N

X x x x

x x x xx x

x x x x x x

x x x x x x


x

Insert 1’s instead of . 0x

X



11 12 1, 1

1 21 22 2, 1

1,1 1,2 1, 1

N

N

N N N N

y y y

y y yY X

y y y

X Invert the matrix .

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0 11 1 12 2 13 3 1, 1 1

1 21 1 22 2 23 3 2, 1 1

2 31 1 32 2 33 3 3, 1 1

1,1 1 1,2 2 1,3 3 1, 1 1

N N

N N

N N

N N N N N N

a y f y f y f y f

a y f y f y f y f

a y f y f y f y f

a y f y f y f y f

Calculate the polynomial coefficients.

11 12 13 1, 10 1

21 22 23 2, 11 21

31 32 33 3, 12 3

1,1 1,2 1,3 1, 1 1

N

N

N

N N N N NN N

y y y ya f

y y y ya f

y y y ya X f Y f a f

y y y ya f

At this point, f1 to fN+1 will be symbolic.

Step 4 – Write Finite‐Difference Approximation

0 11 1 12 2 13 3 1, 1 1

1 21 1 22 2 23 3 2, 1 1

2 31 1 32 2 33 3 3, 1 1

1,1 1 1,2 2 1,3 3 1, 1 1

N N

N N

N N

N N N N N N

a y f y f y f y f

a y f y f y f y f

a y f y f y f y f

a y f y f y f y f

24

0

1

2

22

0

0

0 2

f x a

df x a

dx

df x a

dx


Recall how we interpolate the function or one of it’s derivatives given our polynomial…

11 1 12 2 13 3 1, 1 1

21 1 22 2 23 3 2, 1 1

2

1,1 1 1,2 2 1,3 3 1, 1 12

0

0

0 2 2 2 2

N N

N N

N N N N N

f x y f y f y f y f

df x y f y f y f y f

dx

df x y f y f y f y f

dx

The rows of areY

essentially our finite‐difference coefficients.

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Examples UsingPolynomial Technique


Example #1

26

Derive first‐order and second‐order finite‐difference approximations that span across three points. The approximations should be evaluated at the midpoint.

0

h

x

h

2

0 1 2

2

1

1 0 0

1

h h

X x x x

h h

1

2 2 2

0 1 0

1 2 0 1 2

1 2 1 1 2

Y X h h

h h h

0 1 2 3 20 1

1 31 2 1 1 2 3

2 2 22 3

22

0 1 0 0 1 0

1 2 0 1 2 1 2 0 1 2 2

1 2 1 1 21 2

a f f f fa f

f fa h h f a h f f h f

ha h h h f

a h f

2 2 1 2 31 2 3 2

21 1 2

2

f f fh f h f

h

fd 0 2

fd 3 11

2fd 1 2 3

22 2

2

22

f x a f

df x f fa

dx h

d f x f f fa

dx h


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Example #2

27

Derive first‐order and second‐order finite‐difference approximations that span across three points. The approximations should be evaluated at the first point.

0

2

x h

h

0 1 2 2

2

1 0 0

1

1 2 2

X x x x h h

h h

1

2 2 2

1 0 0

3 2 2 1 2

1 2 1 1 2

Y X h h h

h h h

0 1 2 3 1

0 1

1 2 31 2 1 1 2 3

2 2 22 3

2

1 0 0 1 0 0

1.5 2 0.53 2 2 1 2 3 2 2 1 2

1 2 1 1 21

a f f f fa f

f f fa h h h f a h f h f h f

ha fh h h

a

2 2 2 1 2 31 2 3 2

22 1 1 2

2

f f fh f h f h f

h

fd 0 1

fd 1 2 31

2fd 1 2 3

22 2

1.5 2 0.5

22

f x a f

df x f f fa

dx h

d f x f f fa

dx h


Example #3 – Higher Order Accuracy (1 of 2)

28

Let’s evaluate some derivatives at the midpoint of four discrete points.

3 2

2

2

3 2

h

hx

h

h

2 3

2 3

0 1 2 3

2 3

2 3

3 9 271

2 4 8

12 4 8

12 4 8

3 9 271

2 4 8

h h h

h h h

X x x x xh h h

h h h

1

2 2 2 2

3 3 3 3

1 9 9 1

16 16 16 161 9 9 1

24 8 8 241 1 1 1

4 4 4 41 1 1 1

6 2 2 6

h h h hY X

h h h h

h h h h


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Example #3 – Higher Order Accuracy (2 of 2)

29

The coefficients are then

0 1

1 2

2 32 2 2 2

3 4

3 3 3 3

1 9 9 1

16 16 16 161 9 9 1

24 8 8 241 1 1 1

4 4 4 41 1 1 1

6 2 2 6

Y

a f

a fh h h ha f

h h h ha f

h h h h

0 1 2 3 4

1 1 2 3 4

2 1 2 3 42 2 2 2

3 1 2 3 43 3 3 3

1 9 9 1

16 16 16 161 9 9 1

24 8 8 241 1 1 1

4 4 4 41 1 1 1

6 2 2 6

a f f f f

a f f f fh h h h

a f f f fh h h h

a f f f fh h h h

1 2 3 42.5 0

2.5 1 2 3 41

22.5 1 2 3 4

2 22

9 9

16

27 27

24

22

f f f ff x a

df x f f f fa

dx x

d f x f f f fa

dx x


Implementing the Polynomial Technique

Using MATLAB


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General Form of the Polynomial Fit

Slide 31

We have so far derived finite‐difference approximations symbolically.What if we want 6th‐order accurate finite‐differences? This is unreasonable to do symbolically.

Recall our matrix equation representing the polynomial written at each discrete point. It always had the following form where the w’s were just numerical constants. The h’s were symbolic.

21 012 13 1

22 122 23 2

21 1,2 1,3 1,

1

1

1

NN

NN

NN NN N N N

f aw h w h w h

f aw h w h w h

f aw h w h w h


Factor Out Symbolic Term h

Slide 32

Now, we are able to separate the w terms from the h terms.

12 13 11 0

22 23 22 12

1,2 1,3 1,1

11

1

1

N

N

N N N NN NN

w w wf ah

w w wf ah

w w wf ah


We were able to put numbers to all of these coefficients. This is a fully numerical matrix. It does not contain any symbolic variables.

These are our symbolic variables.

Hint: when 1W X h

So we build [W] by building and pretending h = 1.X

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Solve Matrix Equation for [a]

Slide 33

Solving our matrix equation for [a] gives1

1

12 13 10 1

22 23 21 22

1,2 1,3 1, 1

11 12 13 10

21 22 231

2

11

1

1

1

1

1

1

N

N

N N N NN NN

N

N

N

w w wa fh

w w wa fh

w w wa fh

v v v va hv v v va

h

a

h

1

2 2

1,1 1,2 1,3 1, 1

N

N N N N N N

f

f

v v v v f


1V W

The key aspect here is that [W] will be completely numerical so it is easily inverted using MATLAB. This accommodates large matrices and avoids symbolic manipulation.

Incorporate Symbolic h Again

Slide 34

Next we reincorporate symbolic h by multiplying our matrices.

11 12 13 10 1

21 22 23 21 2

2

1,1 1,2 1,3 1, 1

11 12 13 1

0

21 22 231

1

1

1

1

1 1 1 1

1 1 1 1

N

N

N N N N NN N

N

N

N

v v v va fhv v v va f

hv v v va f

h

v v v va

v v va h h h h

a

1

22

11,1 1,2 1,3 1,

1 1 1 1

N

NN N N N NN N N N

fv f

fv v v v

h h h h


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Extract Polynomial Coefficients

Slide 35

Next, we read off the polynomial coefficients from our matrix equation.

11 12 13 1

0 1

21 22 23 21 2

11,1 1,2 1,3 1,

1 1 1 1

1 1 1 1

1 1 1 1

N

N

N NN N N N NN N N N

v v v va f

v v v va fh h h h

a fv v v v

h h h h


0 11 1 12 2 1 1

21 1 22 2 2 11

1,1 1 1,2 2 1, 1

N N

N N

N N N N NN N

a v f v f v f

v f v f v fa

h

v f v f v fa

h

Write Finite‐Difference Approximations

Slide 36

Last, we write our finite‐difference approximations from the polynomial coefficients.


0 11 1 12 2 1 1

21 1 22 2 2 11

231 1 32 2 3 1

22 22 2

N N

N N

N N

f a v f v f v f

v f v f v fdfa

dx h

v f v f v fd fa

dx h

Staring at these equations long enough, we realize that the vijcoefficients can be determined completely numerically. We just have to remember to divide by h and perhaps multiply the finite‐difference expression by a constant.

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MATLAB Examples


Example #4 – 6th Order Accurate Finite‐Differences (1 of 2)


1. Here we need seven points to calculate seven polynomial coefficients.

3 2 0 2 3T

x h h h h h h

2. To build the [W] matrix, think h = 1 for now.

ˆ 3 2 1 0 1 2 3T

x

0 1 2 3 4 5 6

1 -3 9 -27 81 -243 729

1 -2 4 -8 16 -32 64

1 -1 1 -1 1 -1 1ˆ ˆ ˆ ˆ ˆ ˆ ˆ 1 W x x x x x x x

0 0 0 0 0 0

1 1 1 1 1 1 1

1 2 4 8 16 32 64

1 3 9 27 81 243 729

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Example #4 – 6th Order Accurate Finite‐Differences (2 of 2)


3. Invert [W].

1

-0.0

0.0056 -0.0750 0.7500

167 0.1500 -0.750

0 0 0 1.

0 -0.0000 0.7500 -0.1500 0.01

0000 0 0

67

-

0

V W 0.0208 -0.1667 0.2708 0.0000 -0.2708 0.1667 -0.0208

-0.0069 0.0833 -0.2708 0.3889 -0.2708 0.083

1.3611 0.7500 -0.0750 0.

3 -0.0069

-0.0042 0.0167

0056

-0.0208 -0.0000 0.0208 -0.0167 0.0042

0.0014 -0.0083 0.0208 -0.0278 0.0208 -0.0083 0.0014

4. Write the finite‐difference approximations, remembering to incorporate the symbolic h’s back in.

1 2 3 4 5 6 70

1 2 3 4 5 6 71

21 2 3 4 5 6 7

22 2

0.0167 0.15 0.75 0 0.75 0.15 0.01

0.0056 0.0750 0.7500 1.3611 0.7500 0.075

1

2 2 2 2 2 2 22

0 0.

6

0056

0 0 0 0

7

1 0 0f f f f f f ff a

f f f f f f ffa

x h

f f f f f f ffa

x h

Example #5 – Nonuniform Grid(1 of 3)


x1

= 1.1

x2

= 1.6

x3

= 2.6

x4

= 3.2

x5

= 4.7

x6

= 5.6

x7

= 6.0

x8

= 6.5

x fd

= 4

.4

Derive the finite‐difference equations for first‐and second‐order derivatives at the point xfd.

1. Choose range of points.

2.6 3.2 4.7 5.6T

x

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2. Shift x axis.

fd

2.6 3.2 4.7 5.6 4.4

1.8 1.2 0.3 1.2

T

T

x x x

3. Build matrix.

0 1 2 3

1 1.8 3.24 5.832

1 1.2 1.44 1.728ˆ ˆ ˆ ˆ1 0.3 0.09 0.027

1 1.2 1.44 1.728

X x x x x

X

Note: We are not calling this the [W] matrix because we did not have to factor out a symbolic h term.



4. Invert matrix.

1 0.3810 -1.0833 0.5079 0.194

0.0794 0.1389 -0.6349 0

-0.1143

.416

-0.2

646 0.4630 -0.3527

0.3000 0.9143

0.15

-0

7

4

000

4

3

.1

Y X

5. Write the finite‐difference approximations directly from the rows of .

0 1 2 3 4

1 1 2 3 4

2

2 1 2 3 42

0.3810 1.0833 0.5079

0.1

0.0794 0.1389 0

0.1944

.6349 0.2 2 2 2

143 0.3 0.914

4167

0

2

3 .1f a f f f f

fa f f f f

x

fa f f f f

x

Note: The symbolic variable h did not appear in

so it does not need to be incorporated here. x

X

Y

Documents

Topic 6b -- Finite-difference approximationsemlab.utep.edu/ee4386_5301cmee/Topic 6b -- Finite-difference... · Types of Finite‐Differences ... Backward Finite‐Difference ... Recall