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Lecture notes

Next:Introduction

Boundary Element Methods

Lothar Gaul and Matthias Fischer

IntroductionBE Formulation of Laplace's Equation

Weak formulation of the differential equationTransformation on the boundary

Fundamental solution as weighting functionBoundary integral equation of the 2-D problemPreparative example for the limit processCalculation of the limit

Discretisation of the boundaryThe collocation methodExample: Laplace problem of heat transfer

Numerical solution with the collocation methodAnalytical solution

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Computation of solution in the domainCalculation of Dirichlet variable in the domainCalculation of flux in the domain

BE formulation of Poisson's equationCalculation of domain integrals by integration of cellsCalculation of domain integrals by transformation into a boundary integralCalculation of the unknown boundary variables

Orthotropic constitutive behaviour in the domain

Indirect calculation of diagonal elements inConcentrated source termsSubstructure techniqueExample: Orthotropic heat transfer and subregion couplingFundamental solutions

Laplace equationFundamental solution of the 2D Laplace equationFundamental solution of the 3D Laplace equation

Helmholtz equationsFundamental solution of the 3D Helmholtz equation

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Introduction

Next:BE Formulation of Laplace's EquationUp:Boundary Element MethodsPrevious:Boundary Element Methods

Introduction

This manuscript accompanies the lecture Boundary Element Methods in Statics and Dynamics. However, thematerial presented on the web cannot include all the aspects that are discussed in the class.

Focus point of the manuscript is the derivation of the standard boundary element method for Laplace's equation.Starting from the differential equation, the BEM is formulated step by step. Simple examples are calculated andcompared to analytical solutions. The handling of domain integrals in the BEM is discussed on the example ofPoisson's equation. Some advanced techniques and the derivation of selected fundamental solutions conclude

the manuscript.

The lecture covers additional important aspects of boundary elements. For example the application of themethod to elastostatics and elastodynamics as well as to acoustics. Furthermore advanced formulations such asthe Dual Reciprocity BEM and variational BEM are presented.

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Page 2 of 2BE Formulation of Laplace's Equation


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The collocation methodExample: Laplace problem of heat transfer

Numerical solution with the collocation method

Analytical solution

Computation of solution in the domainCalculation of Dirichlet variable in the domainCalculation of flux in the domain

Next:Weak formulation of the differential equationUp:Boundary Element MethodsPrevious:Introduction


Page 2 of 2BE Formulation of Laplace s Equation


Page 1 of 2Weak formulation of the differential equation


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Weak formulation of the differential equation

Next:Transformation on the boundaryUp:BE Formulation of Laplace's EquationPrevious:BE Formulation of Laplace's Equation

Weak formulation of the differential equation

Starting point for the boundary element formulation is the weighted residual (or weak) statement of the differentialequation. For Laplace's equation , it is given by

with a test function.

Next:Transformation on the boundaryUp:BE Formulation of Laplace's EquationPrevious:BE Formulation of Laplace's Equation

(1)

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Page 1 of 2Transformation on the boundary


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Transformation on the boundary

Next:Fundamental solution as weighting functionUp:BE Formulation of Laplace's EquationPrevious:Weak formulation of the differential equation

Transformation on the boundary

This step corresponds in 1-D to the partial integration of the differential operator. It requires the application ofspecial integral theorems depending on the problem dimension. These theorems reduce domain integrals inboundary integrals. This is different from the 1-D case where integrals reduce to scalar quantities. In the followingparagraphs the transformation on the boundary is treated for 2-D and 3-D by adopting Green's integral theoremin the plane and in space.

The transformation of the differential operator to the boundary is done by applying Green's theorem twice to theweighted residual statement. In index notation this reads

(2)


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Next:Fundamental solution as weighting functionUp:BE Formulation of Laplace's EquationPrevious:Weak formulation of the differential equation


gy


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Fundamental solution as weighting function

Next:Boundary integral equation of the 2-D problemUp:BE Formulation of Laplace's EquationPrevious:Transformation on the boundary

Fundamental solution as weighting function

The fundamental solution is the Green's function for the unbounded space and solves the differential equation

The minus sign of the Dirac distribution is introduced for convenience such that the obtained system matricesbecome positive. In 2-D the fundamental solution (s.f. Appendix 9) is given by:

with the common abbreviation of the Euclidean distance . The 3-D case leads to (s.f.

(3)

(4)

(5)

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Appendix 9)

The functions and are denoted as single layer and double layer potentials, respectively. After selecting

, Eq. (2) and associated with the sifting property of the Dirac distribution lead to

where

The common notations for the field point or receiver point (marked by the vector ) and for the load point or

source point (marked by the vector ) have been used. It has to be noticed, that the definition (9) of deviates

from the physical definition of the heat flux vector

In an actual heat transfer problem, physical constants such as the heat conductivity need to be taken into

account.

(6)

(7)

(8)

(9)

(10)




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Other than in the 1-D case, higher dimensional problems lead to integrals over the boundary of the domain ,or - to be more precise - over the boundary consisting of all field points on .

By recalling all steps necessary to derive Eq. (8), one recognizes, that the weighted residual statement Eq. (1)does not lead to an approximation. The question arises, whether an exact solution of Eq. (8) is an exact solutionof Laplace's equation as well. This seems not to be the case, since the weighted residual statement allows forlocal errors in the domain but averages them to zero by domain integration. An exact solution of Eq. (8) whichfulfills the integral pointwise represents a weighting with infinitely many linearly independent test functions in theresidual statement (1). It follows that statement (1) is only fulfilled if the differential equation is satisfiedidentically. Thus, the exact solution of (8) is an exact solution of the corresponding differential equation as well.

Next:Boundary integral equation of the 2-D problemUp:BE Formulation of Laplace's EquationPrevious:Transformation on the boundary



Page 1 of 2Boundar inte ral e uation of the 2-D roblem


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Boundary integral equation of the 2-D problem

Next:Preparative example for the limit processUp:BE Formulation of Laplace's EquationPrevious:Fundamental solution as weighting function

Boundary integral equation of the 2-D problemIf the load point in Eq. (8) moves on the boundary, only boundary data are present in Eq. (8). This equation is

called a boundary integral equation. Calculation of the integrals by boundary discretisation leads to algebraicequations for solving unknown boundary data in terms of known boundary data. The singularities of fundamentalsolutions requires careful analysis when the load point is accommodated on the boundary.

Subsections

Preparative example for the limit processCalculation of the limit

Discretisation of the boundary

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Preparative example for the limit process

Next:Calculation of the limitUp:Boundary integral equation of the 2-D problemPrevious:Boundary integral equation of the 2-D problem

Preparative example for the limit process

If the integral

is solved as shown, the correct result is obtained by chance but the integration 'with closed eyes' incorporates

the singularity which is improper. The singularity at becomes obvious when the integral is

considered.

If one approaches the singularity in Eq. (11) from both sides by the small quantities and , one obtains

(11)

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Page 3 of 3Preparative example for the limit process


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In stress calculations and special BEM formulations such as the hybrid BEM, so called hyper singularities areencountered (s.f. Part III).

Next:Calculation of the limitUp:Boundary integral equation of the 2-D problemPrevious:Boundary integral equation of the 2-D problem

Table 1: Classification of singularities

Type Property Example

1. weak singularityIntegral is finite atsingularity

2. strong singularityInterpretation as Cauchy

Principal Value



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Calculation of the limit

Next:Discretisation of the boundaryUp:Boundary integral equation of the 2-D problemPrevious:Preparative example for the limit process

Calculation of the limit

To locate the load point on the boundary, we first adjust the boundary such that it contains the point inside acircle of radius according to Fig. 1

Thus the point is inside the domain and Eq. (8) is still valid.

(16)

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The integration along the -circle is parameterized by

(s.f. Fig. 2). Furthermore holds

Figure 1: Boundaryextension by a circle

(17)

(18)




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The limit value by taking Eq. (8) on the boundary can now be calculated. For we get by Eq. (4)

In the limit, the first integral is weakly singular. With Eqs (17, 18) and l'Hospital's rule, the last integral in Eq. (19)results in a vanishing contribution

Figure 2: Geometry for accommodating the load point on theboundary

(19)




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With Eq. (5), the integral leads to

The first integral in Eq. (21) is a strongly singular integral calculated by Cauchy's Principal Value. The secondintegral leads to

Summarizing these results and inserting in Eq. (8) leads to the integral equation

(20)

(21)

(22)




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respectively

The factor is called boundary factor and denotes the fraction of which is inside

Solving the integrals in Eq. (24) analytically is only possible for special cases. For a numerical integration theboundary is divided in segments with the interpolation of boundary data by piecewise continuous functions suchas polynomials. This approach is called discretisation.

Next:Discretisation of the boundaryUp:Boundary integral equation of the 2-D problemPrevious:Preparative example for the limit process

(23)

(24)

(25)



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For an approximation of the geometry, the boundary of the domain is divided in boundary elements(Fig. 3). Every element has one or more nodes. At node of element the value of is and the

value of is . Shape functions describe the spatial distribution on the element. With nodes in element

, the shape of and are interpolated by

Figure 3: Discretisation of the

boundary

(26)




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respectively. Or in matrix notation by

where and are 1 x M row vectors and is a M x 1 column vector. The simplest shape functions areconstant and linear shape functions.

Constant shape function

Only one node exists per element, the values of and are constant throughout the element and have

the value at the node. This means and

and

and (27)

and (28)




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Linear shape function

Figure 4: Constant element shape function and localcoordinate



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with the local coordinate . The shape functions are depicted in Fig. 6.

Discretisation of Eq. (24) in 2-D leads to

(30)

Figure 6: Linear shape functions




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The nodal values and are constants and can be brought outside the integrals

If constant elements are used, the node is usually located in the middle of the perfectly flat element (s.f. Fig. 4).Therefore, and Eq. (25) lead to

The vector is perpendicular to if load point and field point are located on the same element, and

therefore

(31)

(32)

(33)




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This simplifies the calculation and makes the numerical implementation easier.

Next:The collocation methodUp:Boundary integral equation of the 2-D problemPrevious:Calculation of the limit

(34)



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The collocation method

Next:Example: Laplace problem of heat transferUp:BE Formulation of Laplace's EquationPrevious:Discretisation of the boundary

The collocation method

The collocation method allows to calculate the unknown boundary data from Eq. (32). The simplest approach isto establish a system of equations with as many unknowns as equations.

The principle of collocation means to locate the load point sequentially at all nodes of the discretisation such thatthe domain variable at the load point coincides with the nodal value. Because linear and higher order

polynomial shape functions lead to nodes which belong to more than one element, it is worthwhile to introduce aglobal node numbering ( ) which does not depend on the element.

If the load point is located on the first global node the first equation of the system reads

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The notation means the sum of integrals contributed from those elements which contain the global

node, where is the corresponding shape function. The Eq. (35) is given in matrix notation by

where ( ) denotes the element which contains the boundary term .

By collocating the load point with the nodes to the additional equations of the system (37) are obtained

(35)

(36)




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and read in matrix notation

The diagonal elements of the matrices and contain singular integrals because the distance

vanishes at the nodes. All other matrix elements contain regular integrals. Since both vectors and in

Eq. (38) contain known as well as unknown boundary data, it is necessary to rewrite the equations with allunknowns appearing in a vector on one side

A systematic way of doing this and solving the system of equations is demonstrated with a simple example whichcan be calculated by hand.

Next:Example: Laplace problem of heat transferUp:BE Formulation of Laplace's EquationPrevious:Discretisation of the boundary

(37)

(38)

(39)



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Example: Laplace problem of heat transfer

Next:Numerical solution with the collocation methodUp:BE Formulation of Laplace's EquationPrevious:The collocation method

Example: Laplace problem of heat transfer

Now, Laplace's equation of heat transfer is considered in a 2-D rectangular domain with aspect ratio 1:2 asdepicted in Fig. 7. At the horizontal boundary lines, the temperatures are prescribed. At the vertical

boundary lines the heat flux is given.

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The remaining boundary values , and , are unknown for a discretisation with four elements. The

numerical results are afterwards compared to the analytical solution.

Subsections

Figure 7: Example: Heat transfer in

rectangular domain




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Numerical solution with the collocation methodAnalytical solution

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Next:Analytical solutionUp:Example: Laplace problem of heat transferPrevious:Example: Laplace problem of heat transfer


For simplicity, constant elements are chosen in the example (i.e. , , . Each element has

only one node located in the middle. If Eq. (32) is written for the load point , one obtains

(40)

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Four equations for the four unknown boundary values are obtained if takes the values 1 to 4 and is located

at the four nodes sequentially. The elements of the matrices are the integrals in Eq. (40) for different values ofand . In matrix notation these equations are

Calculation of matrix elements and : ,

(41)




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Fig. 8 shows:

Figure 8: Calculation of matrix elementsand




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Inserting leads to

and

(42)




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Spatial isotropy of the problem at hand leads to

These symmetries do nothold in general for BEM.

and : ,

(43)

and (44)




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Fig. 9 shows:

Figure 9: Calculation of matrix

elements and




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Inserting leads to

and

(45)




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Spatial isotropy leads to

and : and

The geometry for calculating and is obtained the same way

(46)

and (47)




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Inserting leads to

and

(48)




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and by virtue of symmetry

and : and

The geometry for calculating and is obtained the same way

(49)

and (50)




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Inserting leads to

and

(51)




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and by virtue of symmetry

Diagonal terms:According to Eq. (34) the main diagonal of matrix vanishes

(52)

and (53)

(54)




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For one obtains

And by inserting

The integrand in Eq. (55) is weakly singular, but the integral exists. With (s.f. Eq. (15))

(55)




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is obtained and because of symmetry

holds. The same way follows

and

(56)

(57)

(58)


(59)



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Assembling and solving the system of equations

Rewriting of Eq. (41) in index notation and summation of the terms leads to

Plugging the matrix elements and the known boundary data leads to

(59)

(60)

(61)




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and after rewriting the equations with all unknown boundary data appearing on the left side

Solving the equations leads to

In terms of physics, the flux has to be multiplied by in order to obtain the heat flux (s.f. Eq. (8) and

(62)

(63)


Eq. (10)). After multiplication the nodal value is positive and is negative. This means that the heat flux



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through element 1 has the direction of the outward normal vector and the heat flux through element 3 has thedirection opposite to the outward normal vector.

After having used the crudest form of discretisation, a finer boundary mesh with six constant elements of length 1is used. This discretisation still allows the calculation by hand. The matrix entries are calculated as shown for thefour element mesh. The system of equations is obtained as

Rearranging of known and unknown nodal data in the equations and solving the system of equations leads to thesolution

(64)

(65)




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where the nodes 1 and 4 of the six element discretisation coincide with the nodes 1 and 3 of the four element

discretisation.

Next:Analytical solutionUp:Example: Laplace problem of heat transferPrevious:Example: Laplace problem of heat transfer




Page 1 of 3Analytical solution


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Analytical solution

Next:Computation of solution in the domainUp:Example: Laplace problem of heat transferPrevious:Numerical solution with the collocation method

Analytical solution

The 2-D Laplacian is the field equation of the heat flux problem

The flux in direction vanishes on the boundaries and , . On the boundaries and

the gradient of vanishes in direction. This leads to the conclusion that the seeked solution is

independent of and a trial function with unknown coefficients and depends linearly on

(66)

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Fitting the boundary conditions

leads to the constants

The analytical solution is thus given by

The comparison between analytical and numerical solution shows, that for the Dirichlet variable even a coarse

(67)

(68)

and (69)

and (70)


discretisation leads to good accuracy. The larger error of the Neumann variable for the four element

discretisation can be explained by the fact that the differentiated quantity requires finer discretisation because



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discretisation can be explained by the fact that the differentiated quantity requires finer discretisation becauseintegration smoothes while differentiation creates roughness. The finer discretisation by six elements already

shows a considerable improvement of accuracy.

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Computation of solution in the domain

Next:Calculation of Dirichlet variable in the domainUp:BE Formulation of Laplace's EquationPrevious:Analytical solution

Computation of solution in the domainThe solution of unknown data in the domain can only be obtained after the data on the boundary have been

calculated. The load point is placed where the domain data shall be calculated. Integration of Eq. (32) along

the boundary with a vector connecting each boundary point with the interior load point gives the value of the

field variables. The boundary factor is chosen according to Eq. (25). The boundary data is completelyknown and consist of given boundary conditions and the values that were calculated using the collocationmethod.

Subsections

Calculation of Dirichlet variable in the domainCalculation of flux in the domain

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Next: Calculation of Dirichlet variable in the domain

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Next:Calculation of Dirichlet variable in the domainUp:BE Formulation of Laplace's EquationPrevious:Analytical solution




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Calculation of Dirichlet variable in the domain

Next:Calculation of flux in the domainUp:Computation of solution in the domainPrevious:Computation of solution in the domain

Calculation of Dirichlet variable in the domain

The calculation of in the domain is demonstrated on the 2-D example. Rewriting Eq. (40) for the load point in

the domain leads to

(71)

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with the boundary factor according to Eq. (25) and the nodal data and . For the solution of the

domain variable all boundary integrals need to be evaluated

Page 2 of 2Calculation of Dirichlet variable in the domain


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domain variable all boundary integrals need to be evaluated.

Next:Calculation of flux in the domainUp:Computation of solution in the domainPrevious:Computation of solution in the domain




Page 1 of 3Calculation of flux in the domain


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Calculation of flux in the domain

Next:BE formulation of Poisson's equationUp:Computation of solution in the domainPrevious:Calculation of Dirichlet variable in the domain

Calculation of flux in the domain

For the calculation of the flux, it is necessary to calculate the gradient of at the load point . This

leads to both flux coordinates

With a matrix notation of scalar products, Eq. (71) leads to

and

and (72)

(73)

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Exchanging differentiation and integration, the derivatives with respect to are obtained as

and

The derivatives of the matrix entries of and with respect to are

and

After this, the corresponding integrals need to be solved and lead to the flux at the load point in the domain.

(75)

(76)

(77)

(78)


N t BE f l ti f P i ' ti



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Next:BE formulation of Poisson's equationUp:Computation of solution in the domain

Previous:Calculation of Dirichlet variable in the domain




H | L t t | E i | I tit t f A li d d E i t l M h i

Page 1 of 4BE formulation of Poisson's equation


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BE formulation of Poisson's equation

Next:Calculation of domain integrals by integration of cellsUp:Boundary Element MethodsPrevious:Calculation of flux in the domain

Boundary element formulation ofPoisson's equation

Poisson's equation with a non-homogeneous term

describes for example the local heat conduction with sources in the domain or torsion of non-circular crosssections. The weighted residue statement

in (79)

(80)

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and the inverse form with Green's theorem lead to the presence of a domain integral



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With the domain integral is given in 2-D by

A mean to calculate the domain integral is to discretise the domain into integration cells and then

using subsequent numerical integrations. The cells look like a finite element mesh. However, the procedure hasan essential difference because there are no unknowns in the domain. The cells are used as integration regionsover which analytical or Gaussian quadrature is performed.

Discretisation of the boundary with elements and of the domain with cells leads to (s.f. Eq. (32))

(81)

(82)

(83)




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The result is a system of equations

and after separating given boundary variables from unknown boundary variables

By adding the vectors and , the system of equations allows to calculate the vector containing the

unknown boundary variables.

Subsections

Calculation of domain integrals by integration of cellsCalculation of domain integrals by transformation into a boundary integral

Calculation of the unknown boundary variables

Next:Calculation of domain integrals by integration of cellsUp:Boundary Element MethodsPrevious:Calculation of flux in the domain

(84)

(85)





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Page 1 of 4Calculation of domain integrals by integration of cells


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o e| ectu e otes| e c ses | st tute o pp ed a d pe e ta ec a cs


Calculation of domain integrals by integration of cells

Next:Calculation of domain integrals by transformation into a boundary integralUp:BE formulation of Poisson's equationPrevious:BE formulation of Poisson's equation

Calculation of domain integrals by integration of cells

The example of Laplace's equation in a rectangular domain is now modified such that Poisson's equation (79)holds with const. If the complete domain is taken as integration cell and the boundary is dicretised

with four constant elements, Eq. (83) leads to

with

(86)

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The integration is carried out

Symmetry results in

The result for is obtained by

(87)

(88)




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Symmetry results in

Next:Calculation of domain integrals by transformation into a boundary integralUp:BE formulation of Poisson's equationPrevious:BE formulation of Poisson's equation

(89)

(90)





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B d El M h d

Page 1 of 7Calculation of domain integrals by transformation into a boundary integral


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Calculation of domain integrals by transformation into a boundary integral

Next:Calculation of the unknown boundary variablesUp:BE formulation of Poisson's equationPrevious:Calculation of domain integrals by integration of cells

Calculation of domain integrals by transformation into a boundary integral

If is a harmonic function, that is, if it satisfies , the domain integral may be transformed into a

boundary integral.

After introducing a function defined by

and using Green's theorem

along with , one arrives at

(91)

(92)

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with the fundamental solution of Laplace's equation. A so called higher order fundamental solution can becalculated from Eq. (91). With its directional derivative all terms in the boundary representation Eq. (93)

are known.

Determination of

With the 2-D fundamental solution, Eq. (91) is given by

In polar coordinates , Eq. (94) is expressed by

If is assumed to have no dependence, Eq. (96) remains

A first integration

where (94)

(95)

(96)


(97)



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and a second integration lead to

The choice of results in

Determination of

The directional derivative is executed by

(98)

(99)

(100)




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With Eq. (93), can be obtained from

The simple example where const leads to

The boundary integrals from to can be split into sub-integrals corresponding to each boundary element.

This means e.g. for

For an element which contains the load point as well as the field point, orthogonality leads to

(101)

(102)

(103)

(104)

(105)


The second integral leads to



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and because of symmetry follows

The result for is

so that at the end is obtained as

(106)

(107)

(108)

(109)


The result for contains



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The integral is now given by

and because of the problem, symmetry holds

The integral leads to

(110)

(111)

(112)

(113)


and finally

(114)



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is obtained. One realizes that this result is identical with the one calculated by domain integration. The presentedapproach allows to transform the domain integral onto the boundary But it has to be noticed that the approachonly applies for special functions of . For more general distributions other methods are available such as the

Multiple Reciprocity Method[#!nowak!#] which represents an extension of the approach presented in thischapter, or the Dual Reciprocity Method[#!drm!#] with a slightly different approach.

Next:Calculation of the unknown boundary variablesUp:BE formulation of Poisson's equationPrevious:Calculation of domain integrals by integration of cells






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y


Next:Orthotropic constitutive behaviour in the domainUp:BE formulation of Poisson's equationPrevious:Calculation of domain integrals by transformation into a boundary integral


Inserting the results for in Eq. (85) leads to

As compared to the inhomogeneous set of equations (62), another known vector is added on the right hand side.Solving Eq. (115) for a fixed value of leads to the unknown boundary variables. The calculation of the domainvariables proceeds analogue as shown for Laplace's equation.

Next:Orthotropic constitutive behaviour in the domain

(115)

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Page 2 of 2


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Page 1 of 6Orthotropic constitutive behaviour in the domain


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Orthotropic constitutive behaviour in the domain

Next:Indirect calculation of diagonal elements inUp:Boundary Element MethodsPrevious:Calculation of the unknown boundary variables

Orthotropic constitutive behaviour

in the domainIn an anisotropic domain the constitutive parameters depend on the direction. Orthotropic heat transfer, e.g. withcoordinates and in the direction of orthotropy, is associated with conduction coefficients and .

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The modified Fourier heat conduction equation in an orthotropic domain reads in index notation

where the brackets around the index exclude summation. In 2-D, this leads to

Figure 10: Direction of orthotropy

(116)

(117)




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and

Eq. (116) leads to the orthotropic heat transfer equation

Stationary heat transfer along with homogeneous orthotropic constants and lead to

(118)

(119)

(120)


The associated fundamental solution is obtained from



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by invoking a coordinate transformation

such that the left hand side leads to the ordinary Laplacian operator

(121)

and (122)

(123)


Theorem 2 The -distribution has the property



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With this property, Eq. (123) is given by

and leads to the already known 2D fundamental solution (s.f. Eq. (4))

(124)

(125)


where

d (126)



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can be transformed back to the physical space using Eq. (122). The following calculation is now handled in amanner analogous to the case const.

Next:Indirect calculation of diagonal elements inUp:Boundary Element MethodsPrevious:Calculation of the unknown boundary variables

and (126)






Page 1 of 3Indirect calculation of diagonal elements in


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Indirect calculation of diagonal elements in

Next:Concentrated source termsUp:Boundary Element MethodsPrevious:Orthotropic constitutive behaviour in the domain

Indirect calculation of diagonal elements in matrix

from physical considerations

Calculation of diagonal elements of matrix requires to determine the fractional boundary coefficients by

integration. Different from the coefficient for a boundary point on a constant element, the determination is more

complex for more complex elements. In the following, a simple way is discussed in which the diagonal elements

can be computed regardless of the element complexity.

The most simple solution to be described by the system matrices is a uniformly constant temperature on theboundary. In this homogeneous case, there is no flux in the domain or on the boundary. With these boundaryconditions and an arbitrary constant , the vectors and are given by

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and (127)



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Eq. (60) leads to

with the singular matrix . The sum of terms in any row of must vanish. This leads to the diagonal elements

of by the negative sum of the off diagonal elements

As the matrix entries of in Eq. (61) show, the sum of the entries in a row does as well vanish when theboundary are determined explicitly. Both procedures lead to the same result.

(128)

(129)


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Page 1 of 2Concentrated source terms


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Concentrated source terms

Next:Substructure techniqueUp:Boundary Element MethodsPrevious:Indirect calculation of diagonal elements in

Concentrated source terms

In the presence of concentrated source terms in the domain, the volume integral

can be simplified for

(130)

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(131)

Page 2 of 2Concentrated source terms


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By inserting in Eq. (130):

Next:Substructure techniqueUp:Boundary Element MethodsPrevious:Indirect calculation of diagonal elements in

(132)






Page 1 of 7Substructure technique


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Substructure technique

Next:Example: Orthotropic heat transfer and subregion couplingUp:Boundary Element MethodsPrevious:Concentrated source terms

Substructure technique

So far, only homogeneous domains have been treated in which the constitutive properties do not vary. Domainswith piecewise non-homogeneity are now subdivided into homogeneous separate subregions. Afterwards theformulations of the distinct regions are coupled by a substructure technique.

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Fig. 11 illustrates homogeneous subregions 1 and 2 with different constitutive parameters. According to Eq. (84),the formulation for subdomain 1 is

Figure 11: Substructure technique: Division of non-homogeneousdomain in piecewise homogeneous subregions


(133)



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and for subdomain 2 is

Compatibility of at the interface ( compatibility)

as well as compatibility of ( compatibility)

(134)

(135)

(136)


lead to a coupled system of equations which can be solved by two methods.



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Method 1

After rearranging Eq. (133) for subregion 1

and Eq. (134) for subregion 2

(137)

(138)


the coupling with and compatibility according to the constraints in Eq. (135) and Eq. (136), respectively,leads to



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No inversion is necessary for setting up the coupled equations and the system matrix is banded which is anadvantage for the numerical treatment. Another advantage is that all unknowns in the interface, and , are

obtained at once.

Method 2

Multiplication of Eq. (84) with the inverse gives

(139)

(140)


with matrix and vector .

The application to subregion 1 is



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and correspondingly to subregion 2

Coupling with the constraints in Eq. (135) and Eq. (136) leads to

(141)

(142)

(143)




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As compared to Eq. (139), the smaller set of equations is an advantage. But this is obtained at the cost of aninversion of and the necessity to calculate the flux in the interface from Eq. (141) or Eq. (142) after Eq. (143)has been solved.

Next:Example: Orthotropic heat transfer and subregion couplingUp:Boundary Element MethodsPrevious:Concentrated source terms






E l O th t i h t t f d b i li

Page 1 of 9Example: Orthotropic heat transfer and subregion coupling


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Example: Orthotropic heat transfer and subregion coupling

Next:Fundamental solutions

Up:Boundary Element MethodsPrevious:Substructure technique

Coupling of an orthotropic and

an isotropic subregionFor illustrating the methods outlined in the preceding chapters, the coupling of an orthotropic and an isotropicsubregion is treated. The matrices from the example in Chapter 2.6 for the isotropic subregion are used. Fig. 12shows the rectangular subdomains with aspect ratio 2:1 which are discretised by 6 constant elements,respectively. Subregion 1 is isotropic and subregion 2 is orthotropic with conduction coefficients and

.

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The elements of matrices and of Eq. (64) in Chapter 2.6.1 define the system of equations for subregion 1.After rearranging, one obtains

Figure 12: Example: Coupling of isotropic and orthotropic subregions




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The calculation of the system matrices for subregion 2 is demonstrated for an example of the matrix and anelement of the matrix . According to Eq. (122), new variables are introduced

Calculation of matrix elements and ( )

(144)

and (145)




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According to Fig. 13 the following relations hold. The variables and are replaced in the correspondingexpressions

Figure 13: Calculation of matrix elements H78 and G78




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Inserting leads to

(146)


and



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The remaining elements can be calculated the same way. The following system of equations is obtained

(147)




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Coupling of Eq. (144) and Eq. (148) according to either method 1 in Eq. (139) or method 2 in Eq. (143) along withEq. (141) or Eq. (142) leads to sets of equations from which the unknown boundary variables and the interfacevariables can be solved.

The solution is given by:

(148)

(149)


Analytical solution

The field equations for the analytical solution are



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where , in subregion 1 and , in subregion 2. According to the boundary and

compatibility conditions, the variables at the nodes of the discretisation are obtained as

(150)

(151)


The crude discretisation of Fig. 12 should be taken into consideration when the numerical results in Eq. (149) arecompared to the analytical results in Eq. (151). The Dirichlet data lead to very good accuracy while the

Neumann data or fluxes show reasonable approximations.



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Next:Fundamental solutionsUp:Boundary Element MethodsPrevious:Substructure technique






Fundamental solutions

Page 1 of 2Fundamental solutions


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Next:Laplace equation

Up:Boundary Element MethodsPrevious:Example: Orthotropic heat transfer and subregion coupling

Fundamental solutions

In this section fundamental solutions are derived for potential problems. Potential problems are scalar fieldproblems, thus the fundamental solution consists of a scalar function relating the effect of a source term at theload-point to its influence point . This point is usually called field-point.

Subsections

Laplace equation

Fundamental solution of the 2D Laplace equationFundamental solution of the 3D Laplace equation

Helmholtz equationsFundamental solution of the 3D Helmholtz equation

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Next:Laplace equationUp:Boundary Element MethodsPrevious:Example: Orthotropic heat transfer and subregion coupling


Page 2 of 2Fundamental solutions


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Laplace equation

Page 1 of 2Laplace equation


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Next:Fundamental solution of the 2D Laplace equationUp:Fundamental solutionsPrevious:Fundamental solutions

Laplace equationThe fundamental solution of the Laplace equation is a solution of the equation

Note that is the distance between the load- and field-point. This implies that a fundamental solution

is a symmetric function

(152)

(153)

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The derivation of the fundamental solution of the Laplace equation in 2D and 3D is carried out here as anexample for the general appraoch.

Subsections

Page 2 of 2Laplace equation


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Fundamental solution of the 2D Laplace equationFundamental solution of the 3D Laplace equation

Next:Fundamental solution of the 2D Laplace equationUp:Fundamental solutionsPrevious:Fundamental solutions






Fundamental solution of the 2D Laplace equation

Page 1 of 4Fundamental solution of the 2D Laplace equation


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Next:Fundamental solution of the 3D Laplace equation

Up:Laplace equationPrevious:Laplace equation


For simplicity the load-point is shifted in the origin. The derivation starts out by transforming the Laplace operatorto polar coordinates

The excitation with the Dirac impulse is radial-symmetric and, since we are dealing with an infinite problem, thereare no disturbances from the boundary, it is implied that the fundamental solution is radial-symmetric, too. Thusthe last term in Eq. (154) vanishes. The Dirac impulse in polar coordinates is stated as .

Hence, a way to solve for is to integrate Eq. (154). This yields

(154)

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with the integration constants and . It will be shown in the next section that for Eq. (155) being a

valid solution of Eq. (152). The constant introduces the notion of a constant potential. It is arbitrary and is

generally set to zero.

Verification of the impulse condition

The validity of a fundamental solution can be verified by evaluating the impulse condition. This condition carriesout the integral over the partial differential equation over an arbitrary volume enclosing the Dirac impulse

Application of Gauss' theorem transforms the volume integral on the left to a surface integral

(155)

(156)


(157)



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Since is radial-symmetric, the gradient is also a pure function of the radius. In polar coordinates this reads as

Moreover, the outward normal on a circle is defined in polar coordinates as

Choosing the surface as a circle of arbitrary radius leads to the impulse condition

(158)

(159)

(160)


Since depends only on and is constant on a specific circle , the impulse condition is reformulated as



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which proves that Eq. (155) is indeed a valid fundamental solution of Eq. (152). Note that this condition alsoimplies the must be zero as stated before because otherwise the terms would not cancel to -1.

Next:Fundamental solution of the 3D Laplace equationUp:Laplace equationPrevious:Laplace equation

(161)









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Next:Helmholtz equations

Up:Laplace equationPrevious:Fundamental solution of the 2D Laplace equation


In this case a transformation on spherical coordinates is carried out

Assuming radial symmetry yields

(162)

(163)

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The Dirac impulse in spherical coordinates is

(164)



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since

and by equivalence

(164)

(165)

(166)


The integration of the Laplace equation yields for the 3D case



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Again, the impulse condition will show that . As before is arbitrary and is set to zero for convenience.

Verification of the impulse condition

The derivation is analogous to the 2D case. The integral over the partial differential equation is transformed tothe boundary. Since the solution is radial symmetric the gradient has only a component in the radial direction.

A sphere is chosen as arbitrary enclosing surface in the 3D case. The normal vector is a unit vector in sphericalcoordinates

(167)

(168)


With this the impulse condition is

(169)



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p

Again, only depends on and thus is constant on a sphere of constant radius. It follows

As in the 2D case, this shows that must be set to zero so that fulfills this equation.

The solutions for the potential and the flux as the normal derivative of the potential in 2D and 3D are

(170)

(171)


summarized in Table 2.

Table 2: Fundamental solutions of

the Laplace equation

2D 3D



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Next:Helmholtz equationsUp:Laplace equationPrevious:Fundamental solution of the 2D Laplace equation






Helmholtz equations

Page 1 of 2Helmholtz equations


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Next:Fundamental solution of the 3D Helmholtz equationUp:Fundamental solutionsPrevious:Fundamental solution of the 3D Laplace equation

Helmholtz equationsA fundamental solution for the Helmholtz equation is derived by solving

As in the case of the Laplace equation, the function is scalar.

Subsections

Fundamental solution of the 3D Helmholtz equation

(172)

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Page 2 of 2Helmholtz equations


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Page 1 of 5Fundamental solution of the 3D Helmholtz equation


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Next:About this document ...

Up:Helmholtz equationsPrevious:Helmholtz equations


Transformation on spherical coordinates and taking radial symmetry into account yields

A solution is obtained by choosing

(173)

(174)

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where the term with the function is singular for while the term with the function remains regular.To verify the impulse condition the behavior for small is considered. A series expansion of the cosine-function

shows that the singularity behavior is , which after comparison to Eq. (167) yields



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g y p q ( ) y

The sine-term is again a homogeneous solution and does not contribute to the Dirac impulse. Hence, this second

constant is free and is adjusted such that the ansatz in Eq. (174) fulfills the Sommerfeld condition Eq. ( ) for the3D case.

This yields

(175)

(176)


and the complete solution is

(177)



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For small the behavior is

This implies that also in 3D the fundamental solution behaves for small or like the fundamental solution of

the Laplace equation.

The flux of the 3D solution is

(178)

(179)


The solutions for the potential and the flux as the normal derivative of the potential in 2D and 3D are

summarized in Table 3.

(180)



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Next:About this document ...Up:Helmholtz equationsPrevious:Helmholtz equations

Table 3: Fundamental solutions of the Helmholtz equation

2D 3D





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