CE 30125 - Lecture 14
p. 14.1
LECTURE 14
NUMERICAL INTEGRATION
• Find
or
• Often integration is required. However the form of may be such that analyticalintegration would be very difficult or impossible. Use numerical integration techniques.
• Finite element (FE) methods are based on integrating errors over a domain. Typically weuse numerical integrators.
• Numerical integration methods are developed by integrating interpolating polyno-mials.
I f x xd
a
b
=
I f x y yd
u x
v x
xd
a
b
=
f x
CE 30125 - Lecture 14
p. 14.2
Trapezoidal Rule
• Trapezoidal rule uses a first degree Lagrange approximating polynomial ( , nodes, linear interpolation).
76
• Define the linear interpolating function
• Establish the integration rule by computing
N 1=N 1+ 2=
f0
f1
x0 x1
f(x)
g(x)
g x fo
x1 x–
x1 xo–---------------- f1
x xo–
x1 xo–---------------- +=
I f x xd
xo
x1
g x e x + xd
xo
x1
= =
CE 30125 - Lecture 14
p. 14.3
• Trapezoidal Rule
I g x x E+d
xo
x1
=
I fo
x1 x–
x1 xo–---------------- f1
x xo–
x1 xo–---------------- + x E+d
xo
x1
=
I fo
x1xx
2
2-----–
x1 xo–-------------------
f1
x2
2----- xox–
x1 xo–-------------------
+
x1
xo
E+=
I fo
x21
x21
2-------–
x1 xo–---------------------
f1
x21
2------- xox1–
x1 xo–------------------------
fo
x1xo
xo2
2-----–
x1 xo–----------------------
f1
xo2
2----- xo
2–
x1 xo–----------------
E+––+=
Ix1 xo–
2---------------- fo f1+ E+=
CE 30125 - Lecture 14
p. 14.4
• Trapezoidal Rule integrates the area of the trapezoid between the two data or interpola-tion points.
• Evaluating the error for trapezoidal rule.
• The error is dependent on the integral of the difference .However integrating the dependent error approximation for the interpolatingfunction does not work out in general since is a function of !
• We must express in terms of a series of terms expanded about in order to
evaluate correctly.
• An alternative strategy is to evaluate
by developing Taylor series expansions for , and .
• We do note that as E
E e x f x g x –=
x
e x xo
E e x xd
xo
x1
=
E f x xx1 xo–
2---------------- fo f1+ –d
xo
x1
=
f x fo f1
x1 xo– h=
CE 30125 - Lecture 14
p. 14.5
Evaluation of the Error for Trapezoidal Rule
Evaluation of the error by integrating e(x)
• We note that
• However
• Thus
• Recall that for Lagrange interpolation was expressed as:
E I g x xd
xo
xN
– E f x x g x xd
xo
xN
–d
xo
xN
e x f x g x – e x xd
xo
xN
f x x g x xd
xo
xN
–d
xo
xN
=
E e x xd
xo
xN
=
e x
e x x xo– x x1– x xN–
N 1+ !---------------------------------------------------------------f
N 1+ =
CE 30125 - Lecture 14
p. 14.6
• Notes
• Procedure applies to higher order integration rules as well.
• In general is a function of
• Neglecting the dependence of , can lead to incorrect results. e.g. for Simpson’s
rule you will integrate out the dependent term and the result would be !
• A way you can apply is to take as a series of terms! Then we will
always get the correct answer!
Evaluation of the error for Trapezoidal Rule by Taylor Series expansion
x
x 13--- E 0=
E e x xd
xo
x1
= e x
E Ix1 xo–
2---------------- fo f1+ –=
E f x xo
x1
dxx1 xo–
2---------------- fo f1+ –=
CE 30125 - Lecture 14
p. 14.7
• Let’s now develop Taylor series expansions for , and about 77
• In general Taylor series expansion about :
• Now evaluate using the Taylor series
• However since
f x fo f1 xxo x1+
2----------------=
xx0 x1
h
x
x
f x f x x x– f 1 x x x– 2
2!------------------ f 2 x O x x– 3+ + +=
fo f xo =
fo f x xo x– f 1 x xo x– 2
2---------------------f 2 x O xo x– 3+ + +=
xo x– h2---–=
fo f x h2--- f 1 x –
h2
8-----f 2 x O h 3+ +=
CE 30125 - Lecture 14
p. 14.8
• Similarly
• Let’s substitute in for , and into the expression for
f1 f x h2--- f 1 x h2
8-----f 2 x O h 3+ + +=
f x fo f1 E
E f x x x– f 1 x x x– 22!
------------------f 2 x O x x– 3+ + + xd
xo
x1
=
x1 x0–
2---------------- – f x h
2--- f 1 x –
h2
8-----f 2 x O h 3+ + f x h
2--- f 1 x h2
8-----f 2 x O h 3+ + + +
E f x x x x– 2
2------------------f 1 x x x– 3
6------------------f 2 x O x x– 4+ + +
x1
xo
=
x1 xo–
2--------------------- 2f x h2
4-----f 2 x O h 3+ +–
CE 30125 - Lecture 14
p. 14.9
• Notes
• Higher order terms have been truncated in this error expression.
• This integration will be exact only for linear.
• However it is third order accurate in
• Error evaluation procedure using T.S. applies to higher order methods as well
E f x x1 xo– x1 x– 2
2---------------------f 1 x
xo x– 2
2---------------------f 1 x –+=
x1 x– 3
6---------------------f 2 x
xo x– 3
6---------------------f 2 x O x1 x– 4 O xo x– 4+ +–+
x1 xo– – f x x1 xo–
8---------------------h2f 2 x –
x1 xo– 2
---------------------O h 3+
Eh2
8----- f 1 x h2
8-----f 1 x –
h3
48------f 2 x h3
48------f 2 x O h 4 h3
8-----f 2 x O h 4+–+ + +=
E h3
12------ f 2 x –=
f x =
h
CE 30125 - Lecture 14
p. 14.10
Extended Trapezoidal Rule
• Apply trapezoidal rule to multiple “sub-intervals” 78
• Integrate each sub-interval with trapezoidal rule and sum
• Split into equispaced sub-intervals with
• Compute I as:
a
f0f1 f2
x0 x1 x2
fi
fN
bxNxi
x
f(x)
a b N hb a–
N------------=
I f x xd
a
b
f x xd
xi
xi 1+
i 0=
N 1–
= =
CE 30125 - Lecture 14
p. 14.11
where
.
.
.
• Thus extended trapezoidal rule can be expressed as:
where
Ix1 xo–
2---------------- fo f1+
x2 x1–
2---------------- f1 f2+
xN xN 1––
2------------------------ fN 1– fN+ E a b ++ ++=
Ih2--- f0 2f1 2f2 2fN 1– fN+ + + + + E a b +=
fo f a =
f1 f a h+ =
f2 f a 2h+ =
fi f a ih+ =
Ih2--- f a f b 2 f a ih+
i 1=
N 1–
+ += Nb a–
h------------=
CE 30125 - Lecture 14
p. 14.12
• Error is simply the sum of the individual errors:
where = the average within each sub-interval
• Defining the average of the second derivatives
• Thus
• Error is 2nd order over the interval
• Thus the error over the interval decreases as .
• The slope of error vs. on a log-log plot is 2.
E a b - 1
12------h
3f
2 xi
i 1=
N
=
xi x
E a b 112------– b a– h2 1
N---- f 2 xi
i 1=
N
=
f 2 1N---- f 2 xi
i 1=
N
E a b 112------ b a– h2f 2 –=
a b
h2
h
CE 30125 - Lecture 14
p. 14.13
Romberg Integration
• Uses extended trapezoidal rule with two or more different integration point to integra-tion point spacings (in this case equal to the sub-interval spacing), , in conjunctionwith the general form of the error in order to compute one or more terms in the serieswhich represents the error.
• This will then result in a higher order estimate of the integrand.
• More importantly, it will allow us to easily derive an error estimate for the numericalintegrations based on the results using the different grid spacings.
• Consider
where
the exact integrand,
the approximate integral with integration point to integration point spacing
the associated error.
h
I Ih E a b h–+=
I
Ih h
E a b h–
Ihh2--- f a f b 2 f a ih+
i 1=
b a–h
------------ 1–
+ +
CE 30125 - Lecture 14
p. 14.14
• In general the form of the error term if we had worked out more terms in the error series.
• Notes
• The coefficient
• In general, C, D, E etc. are functions of the average of the various derivatives of over the interval of interest.
• These coefficients are not dependent on the spacing .
• Also we do not worry about the exact form of these coefficients.
• As far as we are concerned, they are unknown constant coefficients over the interval.
E a b h– Ch2 Dh4 Eh6 O h 8+ + +=
C 1
12------ b a– f 2 –=
f
h
a b
CE 30125 - Lecture 14
p. 14.15
• Thus the integral
• Unknowns: = the exact integral; = the coefficients of the error term.
• Knowns: = the approximation to the integral; = the integration pointspacing.
• We must generate equations to solve for some of the unknowns
• Solve for and . This will improve the accuracy of to !
• Two unknowns must have two equations use two different integration point to inte-gration point spacings.
I Ih Ch2 Dh4 Eh6 O h 8+ + + +=
I C D E
Ih h
I C I O h 4
I Ih1Ch2
1 Dh41 Eh6
1 O h1 8+ + + +=
I Ih2Ch2
2 Dh42 Eh6
2 O h2 8+ + + +=
CE 30125 - Lecture 14
p. 14.16
• We now have two equations and can therefore solve for 2 unknowns.
• = the exact integral is unknown: = the leading coefficient of the error term isunknown.
• We can solve for and .
• We can not solve for , , ... and the other coefficients since we do not have enoughequations!
• We must select and such that is divided into an integer number of sub-inter-vals. Let
• We compute the approximation to the integral twice.
I C
I C
D E
h1 h2 a b
h1 2h*=
h2 h* base interval = =
2h* I2h*
h* Ih*
CE 30125 - Lecture 14
p. 14.17
• Thus
• Two equations and 2 unknowns. Thus we can solve for both and .
• Therefore if you have 2 second order accurate approximations to
You can extrapolate a 4th order accurate approximation using the above formula.
I I2h*4Ch2
* 16Dh4* 64Eh6
* O h*
8+ + + +=
I Ih*Ch2
*Dh4
*Eh6
*O h
* 8+ + + +=
I C
I– I2h*– 4Ch2
*16Dh4
*–– 64Eh6
*– O h
* 8+=
4I 4 Ih*4Ch2
* 4Dh4* 4Eh6
* O h*
8+ + + +=
I4 Ih*
I2h*–
3------------------------- 4Dh4
* 20Eh6*–– O h
* 8+=
I
Ih* using h*
I2h* using 2h*
CE 30125 - Lecture 14
p. 14.18
• More importantly, we can estimate the errors for both the coarse and the fine integrationpoint solutions simply by solving for using the 2 simultaneous equations
• Thus the estimated error associated with the coarse integration point spacing solution,using the coarse and fine integration point spacing solutions is,
• The estimated error associated with the fine integration point spacing solution, using thecoarse and fine integration point spacing solutions is,
C
CIh*
I2h*–
3h*
2--------------------- 5Dh
*
2– O h
* 4+=
2h* h*
E a b 2h*–43--- Ih*
I2h*– O h
* 4+=
2h* h*
E a b h*–13--- Ih*
I2h*– O h
* 4+=
CE 30125 - Lecture 14
p. 14.19
Example
• Consider:
• Integrating exactly
• Let’s integrate numerically
• Apply extended trapezoidal rule using:
• (using one interval of 8)
• (using two intervals of 4)
• Apply the Romberg integration rule we derived when two integral estimates wereobtained using intervals 2h* and h* to obtain a fourth order estimate for the integral
• Estimate the errors associated with the extended trapezoidal rule results
I5x4
8-------- 4x3– 2x 1+ + xd
0
8
=
I 72=
f x 5x4
8-------- 4x3– 2x 1+ += a 0= b 8=
h 2h* 8= =
h h* 4= =
CE 30125 - Lecture 14
p. 14.20
• Applying (using one interval of 8)
• Since the index runs from 1 to 0, we do not evaluate the summation term. Thus
h 2h* 8= =
I2h*
2h*
2-------- f a f b 2 f 0 i2h*+
i 1=
8 0–2h*
------------ 1–
+ +=
I2h*4 f 0 f 8 2 f 0 i2h*+
i 1=
0
+ +=
i
I2h*= 4 1 529+
I2h*2120=
CE 30125 - Lecture 14
p. 14.21
• Applying (using two intervals of 4)
h h* 4= =
Ih*
h*
2----- f a f b 2 f a ih+
i 1=
8 0–h*
------------ 1–
+ +=
Ih*2 f 0 f 8 2 f 0 4i+
i 1=
1
+ +=
Ih*2 f 0 f 8 2f 4 + + =
Ih*2 1 529 2 87+ + =
Ih*712=
CE 30125 - Lecture 14
p. 14.22
• We can obtain an accurate answer using the trapezoidal rule results,
and
• We can also estimate the error associated with the two trapezoidal rule results,
and
• Let’s estimate the error for the trapezoidal rule result with
• Note that the actual error for the trapezoidal rule results with
O h* 4 O h* 2 I2h*
Ih*
I4 Ih*
I2h*–
3------------------------ O h* 4+=
I4 712 2120–
3------------------------------------ O h* 4+=
I 242.6667 O h* 4+=
O h* 2
I2h*Ih*
h 2h* 8= =
E a b 2h*– estimated–43--- Ih*
I2h*– O h
* 4+
43--- 712 2120– 1877.33–= = =
h 2h* 8= =
CE 30125 - Lecture 14
p. 14.23
• Let’s estimate the error for the trapezoidal rule results with
• Note that the actual error for the trapezoidal rule results with equals
Romberg Integration Using 3 Estimates of the Integral
• Let’s consider using three estimates on
E a b 2h*– actual– I I2h*– 72 2120– 2048.–= = =
h h* 4= =
E a b h*– estimated–13--- Ih*
I2h*– O h
* 4+
13--- 712 2120– 469.33–= = =
h h* 4= =
E a b h*– actual– I Ih*– 72 712– 640.–= = =
I
I Ih1Ch2
1 Dh41 Eh6
1 O h1 8+ + + +=
I Ih2Ch2
2 Dh42 Eh6
2 O h2 8+ + + +=
CE 30125 - Lecture 14
p. 14.24
• Three equations we can solve for three unknowns: Solve for = exact integral and and = coefficients of the first two terms in the error series!
• Therefore we can now derive an accurate approximation to
• Apply integration point spacings: , and .
• Estimates of the integral are related to the exact integral, I, as:
• We can solve for the unknowns , and !
I Ih3Ch2
3 Dh43 Eh6
3 O h3 8+ + + +=
I CD
O h 6 I
h1 2h*= h2 h*= h3
h*
2-----=
I I2h*4Ch2
* 16Dh4* 64Eh6
* O h* 8+ + + +=
I Ih*Ch2
* Dh4* Eh6
* O h* 8+ + + +=
I Ih*2-----
C4----h2
*D16------h4
*E64------h6
* O h* 8+ + + +=
I C D
CE 30125 - Lecture 14
p. 14.25
SUMMARY OF LECTURE 14
• Trapezoidal rule is simply applying linear interpolation between two points and inte-grating the approximating polynomial.
• Error for Trapezoidal Rule
• The error can be determined by computing if is expressed in series
form.
• The error can also be determined by Taylor series expansions of the integrationformula and the exact integral.
• Extended trapezoidal rule applies piecewise linear approximations and sums up indi-vidual integrals.
• Error for extended trapezoidal rule is obtained simply by adding errors over all sub-intervals
I145------ 64 Ih*
2-----
20 Ih*I2h*
+– E h* 6 O h* 8+ +=
e x xd
xo
x1
e x
CE 30125 - Lecture 14
p. 14.26
• Romberg Integration
• Uses trapezoidal rule with different intervals.
• Extrapolates a better answer by estimating the error.
• This can be a much more efficient process than increasing the number of intervals.
• Romberg Integration can be applied to any of the integration methods we will develop