3/27/08
EP 471  Engineering Problem Solving II
Exercise 14: Numerical Integration: Gaussian Quadrature
In CS310/NEEP271 you probably discussed methods of numerical integration that
involved breaking up a onedimensional spatial domain into uniform intervals of width
Δ
x
, subsequently using the trapezoidal rule, Simpson’s rule, or something similar, to
evaluate the integral.
The problem with this method is that it is not easily extendable to
2D or 3D problems and it provides relatively low order accuracy.
The purpose of this
and the subsequent exercise is to introduce a method that is more accurate and extendable
to 2D or 3D problems, and that is also the foundation of integration in finite element
analysis (FEA).
Along the way, we’ll also answer a question that has occasionally
proved vexing during your construction of Matlab scripts: What is a Jacobian
determinant, and why is it mucking up my programming?
Gaussian integration or “quadrature” differs from other methods in that it doesn’t use
uniform intervals as the basis of integration.
The idea is that by permitting nonuniform
spacing in the evaluation of the integrand, we can introduce another degreeoffreedom
that permits higher order accuracy with fewer terms.
To begin, we consider a “second
order integration” over the interval from
−
1 to +1.
We want to approximate the integral
below with a weighted sum of the integrand evaluated at two locations.
We want to
choose the “best”
a
,
b
,
t
1
and
t
2
so that the approximation provides the highest order
accuracy of the integral.
The nonuniform spacing is evident in that, at this point,
t
1
and
t
2
can be any locations within the domain from
−
1 to +1.
The weighting coefficients
a
and
b
are also not yet known.
These four “degreesoffreedom” should, in principle,
allow us to get an exact representation of the integration of a cubic polynomial, which
also produces four terms.
Our goal then is to represent the integral as the weighted sum
below:
2
1
2
1
1
1
,
,
,
;
)
(
)
(
)
(
t
t
b
a
t
bf
t
af
dt
t
f
+
≅
∫
+
−
to be determined
The conditions that allow us to pick values for the four degreesoffreedom involve an
exact match between the left and right hand sides for certain
f
(
t
).
Suppose we let
f
(
t
) be
t
3
,
t
2
,
t
and 1 respectively.
These generate four equations:
(1)
3
2
3
1
1
1
3
3
0
:
)
(
bt
at
dt
t
t
t
f
+
=
=
=
∫
+
−
2
2
2
1
1
1
2
2
3
2
:
)
(
bt
at
dt
t
t
t
f
+
=
=
=
∫
+
−
(2)
2
1
1
1
0
:
)
(
bt
at
dt
t
t
t
f
+
=
=
=
∫
+
−
(3)
b
a
dt
t
f
+
=
=
=
∫
+
−
2
:
1
)
(
1
1
(4)
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Multiplying the third equation by
, and subtracting from the first, we have:
2
1
t
)
)(
(
)
(
0
0
1
2
1
2
2
2
1
2
3
2
t
t
t
t
bt
t
t
t
b
+
−
=
−
+
=
We can satisfy this equation with either
b
= 0,
t
2
= 0,
t
1
=
t
2
or
t
2
=
−
t
1
.
Only the last of
these options is acceptable, so we choose
t
2
=
−
t
1
.
Equation (3) then tells us that
a
=
b
,
and Equation (4) that
a
=
b
= 1.
Equation (1) is now identically satisfied, so that leaves
Equation (2), which tells us that:
5773
.
0
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 Spring '08
 Witt
 Gaussian quadrature, Jacobian, detJ

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