Boundary Element Communications, 13(1), (2002)

Tutorial 4:
FUNDAMENTAL SOLUTIONS:
I-SIMPLE AND COMPOUND OPERATORS
YOUSSEF F. RASHED
Dept. of Structural Engineering, Cairo University, Giza Egypt
youssef@eng.cu.edu.eg

Summary and objectives

In the tutorial 3, we presented other examples on the derivation of the boundary integral equation in the direct
form. Mainly, elasticity and plate in bending problems were discussed. In this tutorial, we will discuss the
definitions and the methods of derivation of fundamental solutions. The use of such solution within the
boundary element method was discussed in the former tutorial. A table presents the commonly used forms of
fundamental solution is given. Also a method based on simple analogy to the algebraic partial fraction is
discussed to decompose compound differential operators. In the next tutorial, we will continue discussing
how to set up the fundamental solutions for complex matrix operators.

1 Definitions
The fundamental solution can be defined in the most simple way as the response due to unit source in an
infinite problem. For example when we say Uij(ξ,x) is the fundamental solution for displacements in
elasticity problems that means: Uij(ξ,x) is the displacement at point x in the direction j due to unit point load
applied at ξ in the i direction. It can be seen that Uij(ξ,x) is a kernel between two-points. From Betti
reciprocal theory it is clear that Uij(ξ,x)=Uji(ξ,x).

Mathematically the fundamental solution of a problem is the solution of the governing differential
equation when the Dirac delta is acting as a forcing term, appears of the right hand side [1]. It has
to be noted that no boundary conditions is forced to simulate the infinite nature of the problem. In
other words, it is the particular solution of the problem corresponding to the Dirac delta
distribution. Provided that the Dirac delta posses the singular nature, the fundamental solution is
also singular. The name “fundamental” came from the fact that it is the solution of the most
“fundamental” problem in mechanics, which deals with a unit source in an infinite body. It is
called also the principal solution; the singular solution or the free-space Green’s function. From
this definitions, the fundamental solution can be defined as follows:

LU (ξ, x) = −δ(ξ , x) (1)

where L is a scalar differential operator and δ(ξ,x) is the Paul Dirac delta, in which is the ξ source
point and the x is a field point. If L is a matrix operator, equation (1) can be re-written as follows:
LijU*kj (ξ, x) = −δ(ξ, x)δ ki (2)

It has to be noted that no boundary conditions are enforced in both equations (1) and (2). In this
tutorial we will discuss the derivation of the fundamental solution in equation (1) whereas the
derivation of fundamental solutions for matrix operators as that of equation (2) will be discussed
in the next tutorial.

2 Useful properties of the Paul Dirac delta

The following properties of the Paul Dirac delta could be used in the derivation of the
fundamental solution [1]:

⎧∞ when ξ = x
δ(ξ, x) = ⎨ (3)
⎩0 when ξ ≠ x

lim
Ω →0 ∫ δ(ξ, x)dΩ = 1
Ω
(4)

∫ δ(ξ, x)F(x)dΩ = F(ξ)

Ω
(5)

in which Ω is an arbitrary domain. For simplicity, it will be chosen as either a circle or sphere for
two- and three-dimensional problems, respectively [2]. It has to be noted that we already have
made use of the third property (equation (5)) in the former two tutorials. In this tutorial the second
property (equation (4)) will help in the steps of the fundamental solution derivation.

3 Methods of derivation
Derivation of fundamental solutions is a lengthy task for difficult operators. However, in many
cases, this could be a systematic procedure. The general technique for deriving the fundamental
solution is to use integral transforms, such as, Fourier, Laplace or Hankel transforms [1]. Such a
technique involves complicated mathematics and has very sophisticated procedures. Therefore it
will be covered in latter tutorial as an advanced topic. In the next section, we will demonstrate the
derivation of the fundamental solution for well-known simple operators such as the Laplacian. The
procedures will be described for both two- and three-dimensional problems. In section 5, a table
will be presented to summarize the fundamental solutions for commonly used simple operators.
Then in section 6, a simple analogy to the algebraic partial fraction technique will be used to
decompose compound operators to the simple forms presented in section 5.
4 Fundamental solutions using integration
In this section we will demonstrate a technique based on direct integration and making use of the
properties of the Dirac delta to construct the fundamental solution. The starting point of this
technique is to solve the following homogeneous equation:

LU (ξ , x) = 0 (6)

This equation can be solved using any simple technique in the calculus such as, direct integration
in polar coordinate, separation of variables, variation of parameters or using complex variable
transformation (for the case of two dimensional problems only). Some constants will appear due
to the integration procedures. In order to obtain such constants, we can make use of the Dirac
Delta property in equation (4) after combining it with equation (1), to give:

lim
Ω →0 ∫ LU(ξ, x)dΩ = −1
Ω
(7)

As the domain Ω can be chosen arbitrary, one can present it as a small circle (sphere for the three-
dimensional case) of a radius ε, for simplicity. Then the limit in equation (7) will be performed as
εÆ0. It has to be noted that the integration in equation (7) could be performed easily by
transforming it to the boundary of the circle using the integration by parts procedures presented in
tutorial 2. More details about this method is presented by Rahman in Ref. [2]. The following
examples will demonstrate that idea.

4.1 Laplace operator in two-dimension

Consider the Laplace equation in two-dimension:

∇ 2 U(ξ, x) = −δ(ξ, x) (8)

The first step to construct the fundamental solution is to solve the following homogeneous
equation:

∇ 2 U(ξ, x) = 0 (9)

Or in polar coordinate it could be expressed as follows:

1∂⎛ ∂⎞
⎜ r ⎟ U (ξ, x) = 0 (10)
r ∂r ⎝ ∂r ⎠

By integration, one can obtain:

U (ξ , x) = a + b ln r (11)
In order to obtain the constants a and b in equation (11) we will make use of equation (7), to give:
lim
Ω →0 ∫
Ω
∇ 2 U(ξ, x)dΩ = −1 (12)

or

lim
Ω→0 ∫ ∇ (a + b ln r )dΩ = −1
Ω
2
(13)

Applying the integration by parts (or Green’s second identity), it gives:

∂
lim
ε →0 ∫
Γ
∂n
(a + b ln r )dΓ = −1 (14)

Use the polar coordinate notation, where dΓ = rdθ , one can obtain:
θ= 2π
∂
lim
ε→0
θ=0
∫ ∂n
(a + b ln ε )dΓ = −1 (15)

∂ε 1
Noting that for a circle = 1 , then we can obtain: a is an arbitrary constant and b = . The
∂n 2π
final form of the fundamental solution can be written as:

1
U (ξ , x ) = a + ln r (16)
2π

It has to be noted that, in carrying out the integration in the former example, the following
alternative transformation could be used [3]:

z = x1 + i x 2 and z = x1 − i x 2 (17)

and it is easy to prove that:

z+z z−z
x1 = and x2 = (18)
2 2i

4.2 Laplace operator in the three-dimension

Similar to the two-dimensional case, Equation (9) can be re-written as:

1 ∂⎛ 2 ∂⎞
⎜r ⎟ U (ξ, x) = 0 (19)
r 2 ∂r ⎝ ∂r ⎠

By integration one can obtain:

 b
U (ξ, x) = a + (20)
r

By satisfying the condition in equation (7), one can obtain:

γ = 2π θ= 2π
∂ ⎛ b⎞ 2
lim
ε →0 ∫ ∫
γ =0 θ=0
⎜a +
∂n ⎝
⎟ε dθdγ = −1
ε⎠
(21)

∂ε
Noting that for a sphere = 1 and dΓ = r 2dθdγ , then we can easy obtain: a is arbitrary
∂n
1
constant and b = . Then the final form of the fundamental solution can be written as:
4π
1
U (ξ, x) = a + (22)
4πr

5 Common fundamental solution

The most commonly fundamental solutions are used as basics for many problems in computational
mechanics are presented in Table 1 for one-, two- and three-dimensional case [4].

Table 1: Fundamental solutions for most commonly used operators.

Equation One-dimensional Two-dimensional Three-dimensional

Laplace x 1 1
U=− U = − ln r U=−
∇ U = −δ(ξ, x)
2
2 2π 4πr
Helmholtz 1 1 1 − iλ r
(∇ + λ )U = −δ(ξ, x)
2 2 U = − sin(λ x )
2λ
U = − H (1) (λr )
4i
U=−
4πr
e
Modified Helmholtz 1 1 1 λr
(∇ − λ )U = −δ(ξ, x)
2 2 U = − sin( −iλ x )
2λ
U = − K 0 (λr )
2π
U=−
4πr
e
Bi-harmonic 1
U = − r 2 ln r
∇ 4 U = −δ(ξ, x) 8π

Where H(1) and K0 are Hankel and Bessel functions respectively.

6 Partial fraction analogy

In the former section, we have demonstrated simple procedures to derive the fundamental solution
for simple operators such as the Laplacian. In this section we will present a technique based on an
analogy to the algebraic partial fraction to decompose compound operator to simple operators of
the forms given in Table 1. In order to demonstrate this method, we will consider the following
examples:

6.1 Example 1:
Consider if we want to construct the fundamental solution of the following compound operator:

(∇ 2 − a 2 )(∇ 2 − b 2 ) U = −δ(ξ, x) (23)

From Table 1, we can obtain the following fundamental solutions U1 and U2, where:

(∇ 2 − a 2 ) U1 = −δ(ξ, x) (24)

and

(∇ 2 − b 2 ) U 2 = −δ(ξ, x) (25)

In order to obtain the fundamental solution U in terms of U1 and U2, we try to make an analogy to
the partial fraction theory, as follows:

− δ(ξ, x)
U= (26)
(∇ − a 2 )(∇ 2 − b 2 )
2

− δ(ξ, x) − δ(ξ, x)
= 2 + 2 (27)
(a − b )(∇ − a ) (b − a 2 )(∇ 2 − b 2 )
2 2 2

U U
= 2 1 2 + 2 2 2 (28)
(a − b ) ( b − a )

6.2 Example 2:
The following example will be considered to make the idea more clear. Consider that we want to
construct the fundamental solution of the following operator:

∇ 2 (∇ 2 − a 2 ) U = −δ(ξ, x) (29)

From Table 1, we can obtain fundamental solutions U1 and U2 where:

∇ 2 U1 = −δ(ξ, x) (30)

and

(∇ 2 − a 2 ) U 2 = −δ(ξ, x) (31)
Following the same analogy of the previous example, one can write:

− δ(ξ, x)
U= (32)
∇ (∇ 2 − a 2 )
2

− δ(ξ, x) − δ(ξ, x)
= + (33)
− a 2∇ 2 a 2 (∇ 2 − a 2 )
U U
= 12 + 22 (34)
−a a
As it can be seen that, the partial fraction analogy helps in decomposing compound scalar
operators to well-known simple operators.

7 Conclusions
In this tutorial, we have demonstrated simple procedures for deriving the fundamental solutions
for well know and commonly used differential operators in computational mechanics. We also
covered a technique based on an analogy to the simple concept of algebraic partial fractions to
obtain the fundamental solution for compound operators consist of product of the well-know
simple operators. In the next tutorial we will cover the use of Hörmander method to decouple
complex operators. We will give many examples to show the step-by-step derivation of the
fundamental solution kernels.

8 Exercise:
1- Using the method described in section 4, obtain the fundamental solution for the
modified Helmholtz equation.
2- Use the partial fraction analogy, find the fundamental solution for the following operator:
( )
∇ 4 ∇ 2 − a 2 U = −δ(ξ, x) (35)

9 Solution of the exercise in tutorial 3

In order to derive the direct boundary integral equation for the shear-deformable plate bending
resting on the two-parameter Pasternak foundation model, the following identity can be written:

∫ [(M
Ω
αβ ,β −Q α )Uα + (Qα ,α +q − K f u 3 + G f ∇ 2u 3 )U3 ]dΩ = 0 (36)

or

∫ [(M αβ ,β −Q α )Uα + (Qα ,α +q )U3 ]dΩ + ∫ (− K f u 3 + G f ∇ 2u 3 )U3 dΩ = 0 (37)

Ω Ω

The first integral will lead to the same results as described in Tutorial 3. Now, we will consider the
second integral:
∫ (− K u
Ω
f 3 )
+ G f ∇ 2 u 3 U 3 dΩ =
∫ − K u U dΩ + ∫ (G ∇ u )U dΩ
Ω
f 3 3
Ω
f
2
3 3

=
∫ − K u U dΩ + ∫ (G ∇ u )U dΩ
Ω
f 3 3
Ω
f
2
3 3
(38)

The first integral on the right hand side will remains as it is, whereas the second integral can be
decomposed using the integration by parts procedures (Green’s second identity) twice (in similar
manner as we used before for Laplace equation), to give:

∫ (G ∇ u )U
Ω
f
2
3 3
∫ (
dΩ = G f ∇ 2 U3 u 3 dΩ
Ω
)

∫
+ G f U3u 3 ,n dΓ − G f U3 ,n u 3 dΓ
Γ
∫
Γ
(39)

Then the final integral equation can be written as:

∫ (− U p + P u )dΓ − ∫ G U u , dΓ +∫ G U u ,
Γ
ki i ki i
Γ
f 3 3 n
Γ
f 3 3 n dΓ

−
∫ (M kαβ ,β −Q kα )u αdΩ + ∫ (Qkα ,α −k f U3 − G f ∇2U3 )u3dΩ = 0
Ω Ω
(40)
And the fundamental solution cam be obtained from:
M kαβ ,β (ξ, x) − Q kα (ξ, x) = −δ(ξ, x)δ kα (41)

Q kα ,α −k f U3 − G f ∇ 2 U3 = −δ(ξ, x)δk 3 (42)

Making use of the properties of the Dirac Delta, one can write:

u i (ξ ) +
∫ P (ξ, x)u (x)dΓ(x) + G ∫ U
Γ(x)
ki i f
Γ(x)
i3 , n (ξ, x)u 3 (x)dΓ(x)

=
∫U
Γ(x)
ki (ξ , x) p i ( x)dΓ ( x) + G f
∫U
Γ(x)
i 3 (ξ , x) u 3 , n ( x)dΓ ( x) (43)

Which is the required boundary integral equation. For more details on the derivation, the reader
can consult Ref. [5].
References and Further Reading
[1] Kythe, P.K., Fundamental Solutions for Differential Operators and Applications, Birkhäuser
Press, Berlin, Germany, 1996.

[2] Rahman, M., Mathematical Methods with Applications, WIT Press, Southampton UK, Boston,
USA, 2000.

[3] Katsikadelis, J.T., The Analysis of Plates on Elastic Foundation by the Boundary Integral
Equation Method, Ph.D. Thesis, Polytechnic Institute of New York, 1982.

[4] Brebbia, C.A., Fundamentals of Boundary Elements, In New Developments in Boundary

Element Methods, Proceeding of the 2nd International Seminar, University of Southampton, CMP
Press, Southampton, UK, 1980.

[5] Rashed, Y.F., Boundary Elements Formulations for Thick Plates, WIT Press, Southampton,
UK, 2000.