test problems for the 2d discrete laplace integral operators problems for...9 fortran codes for...
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Test problems for the 2D discrete Laplace Integral Operators
In the development of an appropriate boundary element method1 for the solution of Laplaceβs Equation2 the equation is first reformulated as an integral equation containing (the Laplace) integral operators3. The integral equation is then re-cast as a linear system of equations4 by an integral equation method such as collocation. The integral operators thus replaced by a discrete analogue5, that is as a set of linear equations in which matrices are numerically determined.The vectors that repesent the boundary functions are determined either before or following the solution of the system. In this document, a set of test problems are developed for the module l2lc, which computes the discrete Laplace integral operators in 2D when the boundary is approximated by a set of straight line panels6 and the boundary functions are approximated by a constant on each panel.
The discrete forms of the Laplace integral operators3 are defined as follows5:
{πΏπ}βπ€π
(π) = β« πΊ(π, π)πππβπ€π
{π π}βπ€π
(π) = β«ππΊ(π, π)
πππ πππ
βπ€π
{ππ‘ π}βπ€π
(π; ππ) =π
ππ£πβ« πΊ(π, π)πππβπ€π
{π π}βπ€π
(π; ππ) =π
ππ£πβ«
ππΊ(π, π)
πππ πππ
βπ€π
where πΊ(π, π) = β1
2πln π in two dimensions
In the latter two definitions above the derivative can be taken inside the integral:
{ππ‘ π}βπ€π
(π; ππ) = β«π
ππ£ππΊ(π, π) πππ
βπ€π
and
{π π}βπ€π
(π; ππ) = β« π2πΊ(π, π)
ππ£π πππ πππ
βπ€π
,
1 Boundary Element Method 2 Laplaceβs Equation 3 The Laplace Integral Operators 4 Linear Systems of Equations 5 Discretization of the Laplace Integral Operators 6 Representation of a line by flat panels
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except in the special case for the π operator, when the observation point p lies on the panel βπ€π
β²,
and in this case the evaluation of the discrete form requires special treatment.
Let us now go on to derive geometrical integral expressions for the other operators. In two dimensions we have
ππΊ
ππ= β
1
2π 1
π
π2πΊ
ππ2=
1
2π 1
π2
We also have the following ,
ππ
πππ=
π. ππ
π
ππ
ππ£π= β
π. ππ
π
π2π
ππ£ππππ=
1
π(ππ. ππ +
ππ
ππ£π
ππ
πππ)
Using the chain rule allows us to derive expressions for the kernels of the M and Mt operators:
ππΊ
πππ=
ππΊ
ππ
ππ
πππ=
1
2π
π. ππ
π2 and
ππΊ
ππ£π=
ππΊ
ππ
ππ
ππ£π= β
1
2π
π. ππ
π2 .
Substituting these into the integrals gives the following expressions for the M and Mt operators:
{π π}βπ€πβ²(π) = β«
ππΊ(π, π)
πππ πππ
βπ€πβ²
=1
2πβ«
π.ππ
π2 πππβπ€π
β² and
{ππ‘ π}βπ€πβ²(π; ππ) = β«
ππΊ(π, π)
ππ£π πππ
βπ€πβ²
= β1
2πβ«
π. ππ
π2 πππβπ€π
β² .
(13)
The N operator has second derivatives in its definition. Using the following identity
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π2π
ππ£ππππ= β
1
π(ππ. ππ +
ππ
ππ£π
ππ
πππ),
along with the results above, gives the following,
(15)
π2πΊ
ππ£ππππ=
π
ππ£π(ππΊ
ππ
ππ
πππ) =
π
ππ£π(ππΊ
ππ)
ππ
πππ+
ππΊ
ππ
π2π
ππ£ππππ
=π2πΊ
ππ2
ππ
ππ£π
ππ
πππβ
ππΊ
ππ
1
π(ππ. ππ +
ππ
ππ£π
ππ
πππ)
=1
2π 1
π2 ππ
ππ£π
ππ
πππ+
1
2π 1
π2(ππ. ππ +
ππ
ππ£π
ππ
πππ)
=1
π 1
π2 ππ
ππ£π
ππ
πππ+
1
2π 1
π2ππ. ππ = β
1
π 1
π4 (π. ππ)(π. ππ) +
1
2π 1
π2ππ. ππ.
Substituting the above into the definition of the N operator gives the following result,
note that this is not applicable when the point p lies on the panel βπ€π :
{ππ}βπ€π
(π) = β1
πβ«
1
π4 (π. ππ)(π. ππ)βπ€π
ππ +1
2πππ. ππ β«
1
π2βπ€π
ππ
In the following tests particular values of the disrete Laplace integral operators are found using
vector geometry7 and analytic integration8.
Test(s) 1
Let π = (0,0), and let the edges of the panel be ππ = (1,0) and ππ = (1.5,0) with the
normal to the panel being ππ = (01), as illustrated in the following diagram. The length
of the panel is |πππ
πββ ββ ββ ββ ββ | = 1.
7 Vector Arithmetic and Geometry 8 Integration
ππ p ππ
ππ
0 1 1.5
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{πΏπβ²}βπ€πβ²(π) = β« β
1
2πln(π₯) ππ₯ = β
1
2π[π₯ lnπ₯ β π₯]1
1.5 = β1
2π(1.5ln1.5 β 0.5)
1.5
1
= β0.01722 (5 π. π. ),
For Test 1A , let ππ = (01). Since π = ππββββ β , then for any point π on the panel, π is
perpendicular to ππand ππ and hence {π πβ²}βπ€πβ²(π) and {ππ‘ πβ²}βπ€π
β²(π; ππ) are both zero. Since
ππ
ππ£π= 0 and
ππ
πππ= 0, then {ππβ²}βπ€π
β²(π) =1
2πβ«
1
π2 ππ =1
2πβ«
1
π₯2
1.5
1 ππ₯ = β
1
2π[1
π₯]1
1.5=
β1
2π(
1
1.5β
1
1)=
1
6π .
For Test 1B, let us vary the direction of the vector so that ππ = (β10
). Changing ππ does not
alter {πΏπβ²}βπ€πβ²(π) or {π πβ²}βπ€π
β²(π). Since ππand ππ are perpendicular and ππ
πππ= 0 it follows that
{ππβ²}βπ€πβ²(π) = 0.
{ππ‘ πβ²}βπ€πβ²(π; ππ)=β
1
2πβ«
π.ππ
π2 =βπ€π
β² β1
2πβ«
βπ₯
π₯2
1.5
1 ππ₯ =
1
2πβ«
1
π₯
1.5
1ππ₯ =
1
2π[+ ln(π₯)]1
1.5 =
1
2π[ln(1.5)]1
1.5.
For Test 1B let us vary the direction of the vector so that ππ = (0.60.8
). Changing ππ does not alter
{πΏπβ²}βπ€πβ²(π) or {π πβ²}βπ€π
β²(π). Since ππ. ππ = 0.8 are perpendicular and ππ
πππ= 0 it follows that
{ππβ²}βπ€πβ²(π) = 0.
{ππ‘ πβ²}βπ€πβ²(π; ππ)=β
1
2πβ«
π.ππ
π2 =βπ€π
β² β1
2πβ«
0.6π₯
π₯2
1.5
1 ππ₯ = β
0.6
2πβ«
1
π₯
1.5
1ππ₯ = β
0.6
2π[+ ln(π₯)]1
1.5 =
β0.6
2π[ln(1.5)]1
1.5. {ππ}βπ€πβ²(π) = 0.8 β¨―
1
6π
Test 2
In this test let π = (0,0), ππ = (β10
),ππ = (2,0) and ππ = (2,1), and hence ππ = (β10
).
The points and vectors are illustrated in the following diagram.
ππ p
ππ
2
1
ππ
0
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In order to find the values of the discrete operators, in this case, let y vary from 0 to 1;
from ππ to ππ, so that π = (2, π¦). Hence, we may write π = ππββββ β = (2π¦) and π = |π| =
β4 + π¦2. Substituting these expressions into the definition of the L operator returns the
following result:
{πΏπβ²}βπ€πβ²(π) = β« β
1
2πln (β4 + π¦2) ππ¦ =
1
0
β1
π β« ln(4 + π¦2) ππ¦ =
1
0
β1
π[π¦(ln(4 + π¦2) β 2) + 4 tanβ1 (
π¦
2)]
0
1
= β1
π(ln(5) β 2 + 4 tanβ1 (
1
2))
= β0.116503674 (9 d. p. ) .
For the other operators, we note that π. ππ = β2, π. ππ = β2 and ππ. ππ = 1. Hence
{π π}βπ€πβ²(π) =
1
2πβ«
π.ππ
π2 πππ = β1
πβ«
π
π2 πππ = β1
πβ«
π
4 + π¦2 ππ¦ = β1
2π[tanβ1 (
π¦
2)]
0
11
0βπ€πβ²βπ€π
β²
= β1
2πtanβ1
1
2= β0.073791809 (9 d. p. ) .
Similarly, {ππ‘ π}βπ€πβ²(π; ππ) = β0.073791809 (9 d. p. ) .
{ππ}βπ€πβ²(π) = β
1
πβ«
1
π4 (π.ππ)(π. ππ) ππ +1
2πππ. ππ β«
1
π2 ππ
= β4
πβ«
1
π4 ππ +1
2πβ«
1
π2 ππ = β4
πβ«
1
(4 + π¦2)2
1
0
ππ¦ +1
2πβ«
1
4 + π¦2
1
0
ππ¦
= β1
4π[
2π¦
4 + π¦2]0
1
=1
10π= 0.031830989 (9 d. p. ) .
Test 3
For the final test we consider the special case when the point p lies on the panel.
Formulae9 for the discrete operators are as follows:
{πΏπ}βπ€π
(π) =1
2π[π + π β π lnπ β π ln π],
{π π}βπ€π
(π) = 0 ,
{ππ‘ π}βπ€π
(π; ππ) = 0 and
{ππ}βπ€π
(π; ππ) =1
2 π[1
π+
1
π] ,
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where a is the distance from ππ to p and b is the distance from ππ to p. The derivation of the expression for {πΏπ}
βπ€π(π) πan be derived by integration, as before. The expression for
{ππ}βπ€π
(π) is derived by maintaining the π
ππ£π outside the integral and taking the limiting
value as the point p reached the boundary10.
In order to provide a test problem, let ππ = (1,3) and ππ = (1.3,2.4), and hence ππ =
(β0.80.6
). Let π = (1.15,2.2) and since π lies on the panel ππ = ππ = (β0.80.6
).
The length of the panel is 0.5 and the lengths π = π = 0.25 and hence the non-zero
valued discrete operators are as follows:
{πΏπ}βπ€π
(π) =1
2π[0.25 + 0.25 β 0.25 ln 0.25 β 0.25 ln 0.25] = 0.189895272 (9 d. p. ),
{ππ}βπ€π
(π; ππ) =1
2 π[
1
0.25+
1
0.25] = 1.273239545 (9 d. p. ) ,
Test 4
This set of tests repeat the problem in Test 3. However, when the βl2lcβ routine is called logical
parameters are given the value of βtrueβ if the discrete operator value is required and βfalseβ if the
discrete operator value is not required. In the case where the value of the discrete operator is
not required, a zero is returned. This set of tests goes through every variation of the logical
parameters and verifies that the same result to Test 3 is returned when they are βtrueβ and a
zero is returned when they are βfalseβ.
Test 5
In this test the observation point p is not on the panel. The discrete integral operators are
evaluated for a particular panel. The panel is then split in two and the sum of the discrete
integral operators over the two panels is compared to the initial discrete integral operators. If
they are the same then the test is passed.
Test 6
This test is similar to Test 5, except this time the observation point p is at the centre of the
original panel. The discrete integral operators are again evaluated for a particular panel. The
panel is then split in to three equal parts, with p lying at the centre of the middle panel. The sum
of the discrete integral operators over the three panels is compared to the initial discrete
integral operators. If they are the same then the test is passed.
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Test 7
This test verifies the evaluation of the discrete M and N operators by utilising the equations
{ π1}π (π) + 1 = 0 (π β π·) ,
{ π1}π (π) +1
2 = 0 (π β π, π is smooth at π) ,
{ π1}π (π) = 0 (π β πΈ) .
{ π1}π (π) = 0 (π β π· βͺ π βͺ πΈ) .
where S is a closed boundary and β1β represents the unit function. The boundary is that of the
unit square and the point p takes values of (-0.5,0.5) (in E), (0,0.5) (on S) and (0.5,0.5) (in D).
Test Problem
In the final test the discrete operators are calculated for a closed boundary, for which the
boundary condition and solution is known, and it is shown that two integral equations for the
interior laplace problems are approximately satisfied. The boundary is the unit square divided
into 32 panels.
The unit square domain and the 32 panels.
The simple solution of Laplaceβs equation is assumed in which π = 1 + π¦. The boundary
solution for Ο is clear. It can also be deduced that ππ
ππ= π£ = 0 on the vertical boundaries when
π¦ = 0, 1 and ππ
ππ= π£ = β1 on the lower boundary when π₯ = 0 and
ππ
ππ= π£ = 1 on the upper
boundary when π₯ = 0. This data is fed in to show that they are solutions to the integral
equations {(π +1
2 πΌ) π} (π) = {πΏπ£}(π) and its derivative {ππ}(π) = {(ππ‘ β
1
2 πΌ) π£} (π). Hence
tis verifies all four discrete operators.