Introduction to 2-Dim Gaussian Quadrature over a rectangle
Section 4A:
Introduction to iterated 2-dimensional Gaussian Quadrature over
a rectangle
We are now going to use the work on 1-dimensional quadrature to
develop a rule for 2-dimensional quadrature. First recall Gaussian
Quadrature for an interval [a, b]: For example, with n = 3, we get
4 points in [a, b], x0, x1, x2, x3, and 4 positive weights w0, w1,
w2, w3.
In general, for a degree 2n+1 rule, we have n+1 points, xi, and
weights, wi, and the quadrature rule is then Q(f) = w0f(x0) + .. +
wnf(xn) (with the nodes in [a, b] and the weights positive). For a
continuous function f(x) on [a, b], we then approximate ([a,b]
f(x)dx ( Q(f). We know that we get exact results if f is any
polynomial with
deg f ( 2n+1.
Now we consider a function f(x, y) defined on a rectangular
region R ( [a, b] x [c, d]:
We want to estimate I = ((R f(x, y)dx dy. In Calculus we learned
that this can be computed as an iterated integral:
I = ([c,d] ([a,b] f(x, y)dx dy,
or I = ([a,b] ([c,d] f(x, y)dy dx.
For example, let n = 3 and consider a polynomial f(x, y) with
degrees of x and y separately at most 2n+1 (= 7). We next apply
Gaussian Quadrature on [a, b] with n = 3, to find x0, .., xn in [a,
b], and positive weights w0, , wn. Then we apply Gaussian
Quadrature on [c, d] with n = 3 and find y0, .., yn in [c, d], and
positive weights v0, , vn.
Now we define our 2-dimensional quadrature rule by the following
formula:
Q[f] = (0(i(n (0(j(n wi vj f(xi, yj).
For example, for n = 1 we get:
Q[f] = w0 v0 f(x0, y0) + w0 v1 f(x0, y1) + w1 v0 f(x1, y0) + w1
v1 f(x1, y1).
We claim that Q[f] = ((R f(x, y)dx dy if the degree of f in x (
2n+1 and if the degree of f in y ( 2n+1. We will show this for n =
1. Suppose f(x, y) has degree ( 3 in x and y separately.
Let q(y) = f(x0, y), a polynomial in y of degree at most 3.
Therefore,
v0 q(y0) + v1 q(y1) = ([c,d] q(y) dy .
(A)
Now let r(y) = f(x1, y), a polynomial in y of degree at most 3.
Therefore,
v0 r(y0) + v1 r(y1) = ([c,d] r(y) dy .
(B)
Now let p(x) = ([c,d] f(x, y) dy, a polynomial in x of degree (
2n+1, so
([a,b] p(x) dx = w0 p(x0) + w1 p(x1)
(C).
Now Q[f] = w0 v0 f(x0, y0) + w0 v1 f(x0, y1) + w1 v0 f(x1, y0) +
w1 v1 f(x1, y1)
= w0 (v0 f(x0, y0) + v1 f(x0, y1)) + w1 (v0 f(x1, y0) + w1 v1
f(x1, y1))
= w0 (v0 q(y0) + v1 q(y1)) + w1 (v0 r(y0) + v1 r(y1))
= w0 p(x0) + w1 p(x1)(by (A) & (B))
= ([a,b] p(x) dx
(by (C))
= ([a,b] ([c,d] f(x, y)dy dx
= ((R f(x, y)dy dx.
For n = 1, this completes the proof that Q is a quadrature rule
with exact results for polynomials f(x, y) with degrees at most
2n+1 in x and y separately. For n > 1, the same kind of proof
works. This kind of rule is called 2-dimensional iterated Gaussian
Quadrature. The same kind of rule can be developed for triple
integrals or in higher dimensions. Notice that with n = 1 we have a
rule with 4 nodes which is exact for any polynomial with degrees at
most 3 in x and y separately, so the total degree of this
polynomial is at most 6. But this rule is not exact for all
polynomials of degree 6, which include x6, x5 y, x4 y2, x3 y3, x2
y4, x y5, y6. To obtain a rule that is exact for all polynomial of
degree 6 requires 10 points [2]. The minimum number of nodes
required for exact results up to a given degree > 6 is
unknown.
We conclude this section with Mathematica code that implements
this approach to 2-dimensional iterated Gaussian Quadrature. Recall
the code for 1-dimensional Gaussian Quadrature:
The Gaussian nodes are x[0], , x[n] and the weights are
weights[1], , weights[n+1]. We now use the 1-dimensional rule
separately in the x-direction and the y-direction to compute the
following 2-dimensional iterated Gaussian quadrature rule.
The variable II holds estimate of ([a,b] ([c,d] f(x, y)dy dx
based on 2-dimensional iterated Gaussian quadrature. In the next
section we illustrate this method in several examples.
Section 4B:
Experiments with iterated 2-dimensional Gaussian Quadrature for
rectangular regions
In these experiments we are trying to find the smallest n which
gives a Gaussian relative error less than 1% in the 2-dimensional
Gaussian Quadrature approximation. The error is based on a
comparison with the Mathematica approximation, which we consider to
be the
"correct" value.
Throughout these experiments, we use 100-digit precision in
order to get accurate results for the matrix inversion in the
Gaussian code (Section 4A). Final results are only shown to around
6 significant digits (Mathematica's default precision).
Example 1:
f(x) = 3x2y2 + 4x3y 2xy2+ 3x2+ 2y2 6xy +3x + 2y +1,
a = 1, b = 3, c = 2, d = 5.
The graph for this function is:
We see that N ( n = 1 satisfies our objective:
MathematicaIterated Gaussian
Gaussian
Relative Error
N=11608.1608.1.41402*10-16
The maximum degree in each variable separately for the above
function is ( 3, so n = 1 should give exact results. Therefore, the
answer for the Gaussian relative error should be 0, but because of
roundoff error, it gave us 1.41402*10-16.
Example 2:
f(x) = x4 + 3x3y + 7y4 6x2y2 +2x 4y +6,
a = 1, b = 3, c = 2, d = 5.
The graph for this function is:
We were able to get a relative error less than 0.01 on our first
attempt (with n = 1) as you can see in the table. Since the maximum
degree in each variable separately is 4, n = 2 should give exact
results, which the table shows (to within machine roundoff
error).
MathematicaIterated Gaussian
Gaussian
Relative Error
N = 17383.67364.160.00263196
N = 27383.602.46355*10-16
Example 3:
f(x,y) = ((x2+y2),
a = 1, b = 6, c = 4, d = 6.
The graph for this function is:
n = 1 gave us a good Gaussian relative error, but we included n
= 2 for confirmation.
MathematicaIterated GaussianGaussian Relative Error
N = 162.276962.26920.000124067
N = 262.27735.80135x10-6
Example 4:
f(x,y) = (x+y)/(x-y),
a = 1, b = 5, c = 1/4, d = 1/2
The graph for this function is:
n = 1 did not give us a Gaussian relative error less than 1%, so
we used n = 2 in order to obtain a better Gaussian relative error.
Next, n = 3 confirms the approximation.
MathematicaIterated GaussianGaussian Relative Error
N = 11.380161.35770.0162743
N = 21.37490.00381203
N = 31.378940.000878947
Example 5a:
f(x,y) = Sin(x2y + yx2),
a = 0, b = 3, c = 1, d = 3.
The graph for this function is:
The method seemed to start giving us positive results up to N =
2, but it broke down when it reached n = 3, since the relative
error suddenly increased. In addition, the method had to reach n =
6 in order to give us a Gaussian relative error less than 1%, but
as n increases further the relative error does not stabilize. This
difficulty is caused by the extremely irregular graph: small errors
in measuring (x, y) will cause big errors in measuring f(x, y).
MathematicaIterated GaussianGaussian Relative Error
N = 10.6459021.528521.36648
N = 20.5936420.0809097
N = 3-0.3333411.51609
N = 41.305961.02192
N = 50.7526820.165318
N = 60.64520.00108682
N = 70.9264950.43442
N = 80.3871340.400631
N = 200.6490140.00481717
Example 5b:
f(x,y) = Sin(x2y + yx2),
a = 0, b = 1, c = 0, d = 1.
We use the same function as before, but we move the region
closer to the origin, where the graph is more regular.
n = 1 did not satisfy Gaussian relative error less than 1%.
However, n = 2 and n = 3 showed small errors.
MathematicaIterated GaussianGaussian Relative Error
N = 10.289550.2950640.0190411
N = 20.2894470.000424248
N = 30.2895440.0000218553
In conclusion, when can compare the tables for Examples 5a and
5b. We can see that for the intervals a = 0, b = 3, c = 1, and d =
3, where the graph shows more variations, the results started to
break down. It is difficult to integrate f(x, y) in this region. On
the other hand, with the intervals a = 0, b = 1, c = 0, and d = 1,
the graph is smoother and the method quickly gave us a Gaussian
relative error less than 1%.
