Lecture 18 - Numerical Lecture 18 - Numerical Differentiation Differentiation
CVEN 302
July 19, 2002
Lecture’s GoalsLecture’s Goals
• Finite Difference Schemes
• Taylor Series Expansion for Differentiation
• Basic Numerical Integration– Trapezoidal Rule– Simpson’s Rule– Midpoint– Gaussian Quadrature
dvdt
v
ta
tdt
dva
v
t
b
avdty
y
DifferentiationDifferentiation IntegrationIntegration
Calculus - Numerical MethodsCalculus - Numerical Methods
Numerical DifferentiationNumerical Differentiation
• Estimate the derivatives (slope, curvature, etc.) of a function by using the function values at only a set of discrete points
• Ordinary differential equation (ODE)
• Partial differential equation (PDE)
Numerical DifferentiationNumerical Differentiation
• Represent the function by Taylor polynomials or Lagrange interpolation
• Evaluate the derivatives of the interpolation polynomial at selected nodal points
Numerical DifferentiationNumerical Differentiation A Taylor series or Lagrange interpolation of points can be used to find the derivatives. The Taylor series expansion is defined as:
!3!2
!3!2
0
30i
0
20i
00i0i
0i
xx
3
33
xx
2
22
xx0i
000
xfxx
xfxx
xfxxxfxf
xxx
dx
fdx
dx
fdx
dx
dfxxfxf
First Derivative at a Point
)x(f
i-2 i-1 i i+1 i+2
Numerical DifferentiationNumerical Differentiation
Use the Taylor series expansion to represent three points about single location:
3 !2
2
1 !2
i
2i1i
ii1ii1-i
ii
i
2i1i
ii1ii1i
xfxx
xfxxxfxf
xfxf
xfxx
xfxxxfxf
Numerical DifferentiationNumerical Differentiation
Assume that the data points are “equally spaced” and the equations can be written as:
3 !3
!2
2
1 !3
!2
i
3
i
2
ii1-i
ii
i
3
i
2
ii1i
xfx
xfx
xfxxfxf
xfxf
xfx
xfx
xfxxfxf
Forward DifferentiationForward Differentiation
For a forward first derivative, subtract eqn[2] from eqn[1]:
Rearrange the equation:
!3
!2 i
3
i
2
ii1i
xfx
xfx
xfxxfxf
!3
!2 i
3
i
2
i1ii
xfx
xfx
xfxfxfx
2
i 1 ii i i
2! 3!
f x f x x xf x f x f x
x
Forward DifferentiationForward Differentiation
As the x gets smaller the error will get smaller
The error is defined as:
Error i1i
i
x
xfxfxf
!3
!2
Error i
2
i
xfx
xfx
Backward DifferentiationBackward Differentiation
Subtract eqn[3] from eqn[2]:
The error is defined as:
Error 1-ii
i
x
xfxfxf
!3
!2
Error i
2
i
xfx
xfx
Central DifferentiationCentral Differentiation
Subtract eqn[3] from eqn[1]:
The error is defined as:
Error
21-i1i
i
x
xfxfxf
!3
Error i
2
xf
x
Differential ErrorDifferential Error
Notice that the errors of the forward and backward 1st derivative of the equations have an error of the order of O(x) and the central differentiation has an error of order O(x2). The central difference has an better accuracy and lower error that the others. This can be improved by using more terms to model the first derivative.
Numerical DifferentiationNumerical DifferentiationIf you want to improve the accuracy and decrease the error you will need to eliminate terms :
=
0
=
0
!3
8
!2
4 2 *
!3
!2
*
*
i
3
i
2
ii2-i
i
3
i
2
ii1-i
ii
xfx
xfx
xfxxfxfC
xfx
xfx
xfxxfxfB
xfxfA
0
Higher Order Errors in Higher Order Errors in DifferentiationDifferentiation
The terms become :
The terms become A=-3, B= 4 and C=-1
xOE 2
43 22-i1-iii
x
xfxfxfxf
04
2# 2
0
CB
CB
CBA
Higher Order 1st Derivative)x(f
i-2 i-1 i i+1 i+2
Parabolic curve
Backward difference xOE 2
43 22-i1-iii
x
xfxfxfxf
Forward difference xOE 2
43 2i1i2ii
x
xfxfxfxf
Higher Order DerivativesHigher Order Derivatives
To find higher derivatives, use the Taylor series expansions of term and eliminate the terms from the sum of equations. To improve the error in the problem add additional terms.
22ndnd Derivative of the Function Derivative of the Function
It will require three terms to get a central 2nd derivative of discrete set of data.
=
0
=
0
=
#
!3
!2
*
*
!3
!2
*
i
3
i
2
ii1-i
ii
i
3
i
2
ii1i
xfx
xfx
xfxxfxfC
xfxfB
xfx
xfx
xfxxfxfA
2nd Order Central Difference2nd Order Central DifferenceThe terms become :
The terms become A=1,B=-2 and C=1. Therefore
2#
0
0
CA
CA
CBA
xOE 2 2
21-ii1i
i
x
xfxfxfxf
Lagrange DifferentiationLagrange Differentiation
Another form of differentiation is to use the Lagrange interpolation between three points. The values can be determine for unevenly spaced points. Given:
3
1323
212
3212
311
3121
32
332211
yxxxx
xxxxy
xxxx
xxxxy
xxxx
xxxx
yxLyxLyxLxL
Lagrange DifferentiationLagrange Differentiation
Differentiate the Lagrange interpolation
Assume a constant spacing
31323
212
3212
31
13121
32
22
2
yxxxx
xxxy
xxxx
xxx
yxxxx
xxxxLxf
3221
2231
1232
2
22
2
2y
x
xxxy
x
xxxy
x
xxxxf
Lagrange DifferentiationLagrange Differentiation
Differentiate the Lagrange interpolation
Various locations
3221
2231
1232
2
22
2
2y
x
xxxy
x
xxxy
x
xxxxf
x
yyyy
x
xxxy
x
xxxy
x
xxxxf
2
43
2
22
2
2 32132
21122
31112
3211
x
yyy
x
xxxy
x
xxxy
x
xxxxf
22
22
2
2 1332
21222
31212
3222
x
yyyy
x
xxxy
x
xxxy
x
xxxxf
2
34
2
22
2
2 32132
21322
31312
3233
Lagrange DifferentiationLagrange Differentiation
To find a higher order derivative from the Lagrange interpolation for a three point Lagrange
3221
2231
1232
2
22
2
2y
x
xxxy
x
xxxy
x
xxxxf
2
321322212
2121
x
yyyy
xy
xy
xxf
Take the derivative
Partial DerivativesPartial Derivatives• Straightforward extension of one-
dimensional formula
(i-2, j) (i-1, j) (i, j) (i+1, j) (i+2, j)
(i, j+1)
(i, j+2)
(i, j-1)
(i, j-2)
(i+1, j+1)(i-1, j+1)
(i-1, j-1) (i+1, j-1)
(i-2, j) (i-1, j) (i, j) (i+1, j) (i+2, j)
(i, j+1)
(i, j+2)
(i, j-1)
(i, j-2)
(i+1, j+1)(i-1, j+1)
(i-1, j-1) (i+1, j-1)
121 h
1
x
uu
101 h2
1
x
u u
22
2
xx
x
Partial DerivativesPartial DerivativesLaplacian Operator
1
|
141
|
1
h
1
uuu
2
yyxx2
i-1 i i+1
j+1
j
j-1
Partial DerivativesPartial Derivatives
i-1 i i+1
j+1
j
j-1
Mixed Derivative
101
|||
000
|||
101
h4
1u
2xy
Bi-harmonic OperatorBi-harmonic Operator
i-1 i i+1
j+1
j
j-1
i-2 i+2j-2
j+2
1
|
282
|||
182081
|||
282
|
1
h4
1
x
u
y x
u2
x
uu
4
4
4
22
2
4
44
Richardson ExtrapolationRichardson ExtrapolationThis technique uses the concept of variable grid sizes to reduce the error. The technique uses a simple method for eliminating the error. Consider a second order central difference technique. Write the equation in the form:
xaxa
2 42
212
1-ii1ii
x
xfxfxfxf
Richardson ExtrapolationRichardson ExtrapolationThe central difference can be defined as
i 1 i i-1 2 4i 1 22
2 a a
f x f x f xf x x x
x
2 4
i 1 2
2 4
i 1 2
A a a
A a a 2 2 2
f x A x x x
x x xf x A
Write the equation with different grid sizes
Richardson ExtrapolationRichardson Extrapolation
Expand the terms:
2 4i 1 2
2 4
i 1 2
A a a 1
A a a 22 4 16
f x A x x x
x x xf x A
Richardson ExtrapolationRichardson ExtrapolationMultiply eqn [2] by 4 and subtract eqn [1] from it.
2 4
1 2
2 41 2
4A 4 4a 4a 2 4 16
A a a
x x xA
A x x x
4
23A 4 12a 2 16
x xA A x
Richardson ExtrapolationRichardson ExtrapolationThe equation can be rewritten as:
It can be rewritten in the form
4
2
4 2
A a 3 4
xA A x
x
4 61 2A b b B x x x
Richardson ExtrapolationRichardson ExtrapolationThe technique can be extrapolated to include the higher order error elimination by using a finer grid.
6
16 2
A 15
xB B x
O x
Richardson Extrapolation Richardson Extrapolation ExampleExample
The function is given:
Find the first derivative at x=1.25 using a central difference scheme and h = 0.25. The exact solution:
842 23 xxxxf
6875.3425.1425.13443 22 xxxf
Richardson Extrapolation Richardson Extrapolation ExampleExample
The data points are:
The derivatives using central difference
x f(x)1 -5
1.125 -4.6071.25 -4.1721.375 -3.6821.5 -3.125
7032.3
25.0
6074.46816.3
125.02
125.1375.125.1
75.35.0
5125.3
25.02
0.15.125.1
fff
fff
Richardson Extrapolation Richardson Extrapolation ExampleExample
The results of the central difference scheme are:
The Richardson Extrapolation uses these results to find a better solution
%425.0Error 7032.325.1
%69.1Error 75.325.1
f
f
4
4 3.7032 3.7521.25 3.6876 Error 0.003%
3 3
xA A x
f
SummarySummary
• Finite Difference Techniques
– Taylor Series
– Lagrange Polynomials
• Error Calculation
SummarySummary
• Finite Difference Techniques
– Forward Difference Scheme
– Backward Difference Scheme
– Central Difference Scheme
Homework
• Check the Homework webpage