Lehner G. Electromagnetic Field Theory for Engineers and Physicists

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514 Numerical Methods

Next, one calculates the integral at the outside as well as on the inside boundary, or

the difference between the two integrals. One then finds the value of discontinuity

to be

From eq. (8.13), in complete analogy, while recognizing the changed sign, one sees

that the discontinuities of the integral is

Obviously, these discontinuities are local properties of each individual surface

element. For instance, suppose that is true only for one particular surface

area element. Therefore, the results apply in general to an arbitrary surface.

Suppose is given on the entire surface and if the coordinate system is

chosen appropriately, then the boundary values can be expanded in terms of

spherical harmonics:

(8.28)

For we use the Ansatz

(8.29)

which allows to determine by means of eq. (8.11) and eq. (8.26), as well as

the orthogonality relation eq. (3.300). One finds

(8.30)

This solves the problem. One can now use eq. (8.11), together with eq. (8.29) and

eq. (8.30) to write the solution for the interior of the sphere

(8.31)

which was expected based on eq. (8.28). The purpose of this example was to

demonstrate that the integral relation eq. (8.11) is suitable to calculate the correct

result.

In similar manner, one may use eq. (8.14) and the Ansatz

(8.32)

σ r'()

4πε

------------

nm–()!

nm+()!

--------------------

2 δ

–()P

θcos()P

θ'cos()m ϕϕ'–()[]cos

m 0=

∑

n 0=

∞

∑∫

–

2n 1+

---------------

θ'sin dθ'dϕ'⋅

σ r'()

------------

δθ θ'–()δϕϕ'–()dθ'dϕ'

∫

–

σ r'()

------------–==

τ r'()

-----------

δθ θ'–()δϕϕ'–()dθ'dϕ'

∫

τ r'()

-----------=

σ 0≠

φ r()

φ r() B

θcos()mϕ()cos

m 0=

∑

n 0=

∞

∑

σ r

()

σ r

() C

cos()mϕ

()cos

m 0=

∑

n 0=

∞

∑

σ r

()

2n 1+()ε

------------------------------------=

φ r() B

----





θcos()mϕ()cos

m 0=

∑

n 0=

∞

∑

τ r

() D

cos()mϕ

()cos

m 0=

∑

n 0=

∞

∑

8.2 Basics of Potential Theory 515

to determine . Then

(8.33)

which solves the problem as well. Again, one finds the potential (8.31).

Finally, with the Ansatz

(8.34)

we may base the analysis on eq. (8.2), where we let . One obtains

(8.35)

and using eq. (8.1) gives again the potential (8.31).

Of course, this approach is applicable to arbitrarily shaped surfaces. If

analytical solutions do not exist, then the problem can be solved numerically. To do

this we partition the surface (the “boundary”) into small surface elements

(“boundary elements”). This leads to the boundary value method, which we will

discuss in Sect. 8.8.

8.2.3 Mean Value Theorems of Potential Theory

The so-called mean value theorems of potential theory are a consequence of

Kirchhoff’s theorems. They illustrate important relations in a clear and intuitive

manner. Consider a spherically shaped area, centered at the origin with radius R

that does not contain charges ( ). Then by eq. (8.1)

Because there are no charges, one has

Furthermore

This gives

(8.36)

i.e., the potential at the center of a charge-free sphere is equal to the mean value of

the potential on the surface of the sphere. We might as well consider the entire

volume of the sphere and apply the mean value theorem onto all spherical shells

that are contained within the sphere and obtain

2n 1+()ε

n 1+

------------------------------------–=

∂φ

∂n

------





cos()mϕ

()cos

m 0=

∑

n 0=

∞

∑

C 2π=

----

∇

ϕ 0=

ϕ 0()

4π

------

---

∫

n'∂

∂

ϕ r'()dA'

4π

------

ϕ r'()

n'∂

∂ 1

------ dA'

∫

–=

n'∂

∂

ϕ r'()

∫

dA'0=

n'∂

∂ 1

------

---------





–

------–==

ϕ 0()

4πR

-------------

ϕ r'()A'd

∫

ϕ〈〉

516 Numerical Methods

(8.37)

This applies in much the same way to two-dimensional spherical regions. Eq. (8.6)

for the center of a circle reads:

Using

and

gives

(8.38)

and for the entire area of the circle

(8.39)

There are analogous statements for the one-dimensional case

which has only a linear solution that is

This case yields

(8.40)

and

(8.41)

The conclusion is, that the solution of the Laplace equation in one-, two-, and three

dimensions has the property that its values at the center of a line, circle, sphere are

equal to the mean value of the corresponding values on its boundary, which is the

average of the line, the entire circle, or the entire sphere. We will encounter this

fact again when we will discuss the method of finite differences.

ϕ 0()

4πR

-------------

ϕ r'()τ'd

∫

ϕ〈〉

ϕ 0()

Rln

2π

---------

∂ϕ r'()

∂n'

---------------

ds'

∫

–

2π

------

ϕ r'()

n'∂

∂

r'ln ds'

∫

∂ϕ r'()

∂n'

---------------

ds'

∫

n'∂

∂

r'ln

------





---==

ϕ 0()

2πR

----------

ϕ r'()s'd

∫

ϕ〈〉

ϕ 0()

πR

----------

ϕ r'()A'd

∫

ϕ〈〉

∇

ϕ x() 0=

ϕ ABx+=

ab+

------------





ϕ a() ϕb()+

------------------------------

ab+

------------





ba–

------------

ϕ x()xd

∫

8.3 Boundary Value Problems as Variational Problems 517

8.3 Boundary Value Problems as Variational Problems

8.3.1 Variational Integrals and Euler’s Equations

Many problems that are formulated with differential equations can be reduced to

variational problems. This is also true for some problems of the electromagnetic

field theory. We already encountered one such example when studying

electromagnetic waves inside a cylindrical wave guide (Sect. 7.11). The general

task in Variational Calculus is to find the function (or functions) which constitutes

an extremum of its dependent integral, which is called functional, meaning a

function of functions. Consider the following integral that spans a region in space:

(8.42)

Initially, we require that vanishes on the boundary. Later, we will

consider different boundary conditions. represent the partial derivatives

of u with respect to . Let u be the function that gives an extremum to the

integral. We define a differing, so-called reference function

(8.43)

where f represents a mostly arbitrary function, which vanishes on the boundary and

its derivative is continuous. The reference function satisfies the same boundary

conditions as u. If we replace u in I(u) by , then we obtain a function that depends

on ε and whose derivative with respect to ε vanishes for , i.e., for .

This gives

(8.44)

and

whereby

Using the same procedure for and one obtains

The second integral vanishes because of Gauss’ integral theorem and because of

the boundary condition for f (i.e. on the boundary). Since f is an arbitrary

function, the integrand of the first integral has to vanish as its integral must do so

and therefore:

Iu() Fxyzuu

,,,,,,()τd

∫

extremum==

u uxyz),,()=

xyz,,

uuεf+=

ε 0= uu=

I ε() Fxyzu εf+ u

εf

+ u

εf

+ u

εf

+,,, , , ,()τd

∫

∂I ε()

∂ε

-------------

ε 0=

+++()τd

∫

0==

∂F

∂u

--------

∂f

∂x

-----

∂x

-----

∂F

∂u

--------





∂

∂u

∂x

---------------





–==

∂

∂u

∂x

---------------–

∂

∂u

∂y

---------------–

∂

∂u

∂z

--------------–





f τd

∫

∂x

-----

∂F

∂u

--------





∂y

-----

∂F

∂u

--------





∂z

-----

∂F

∂u

--------









f τd

∫

+0=

f 0=

518 Numerical Methods

(8.45)

This is the so-called Euler differential equation (also known as Euler-Lagrange

equation). Its solution is the solution of the variational problem given by eq. (8.42)

and vice versa.

As an example, consider the following integral

(8.46)

The corresponding Euler-Lagrange equation is

(8.47)

which as a special case contains Laplace’s equation ( ), Poisson’s

equation ( ), and the Helmholtz equation ( ). In case of the

Laplace equation, the integral

possesses the very interesting property to assume an extremum (a minimum,

specifically). This also minimizes the integral

(8.48)

This expresses the remarkable theorem that the electric field in a charge-free region

arranges itself such that the electrostatic energy it contains is minimized

(compatible with the boundary conditions).

The reader is reminded, that we have started with a Dirichlet problem and

then introduced competing reference functions which had to satisfy the boundary

conditions, who’s task was to facilitate minimization of the functional. Therefore,

not every arbitrary solution of the corresponding Euler-Lagrange equation is a

suitable solution, but only the one which satisfies the Dirichlet boundary

conditions. The next question is how to proceed when the boundary condition is

given in a different form. This question leads to the need to distinguish between

essential boundary conditions and natural boundary conditions . Suppose the

boundary is A, partitioned into the parts and . Let the essential Dirichlet

boundary condition

(8.49)

apply to part and the natural boundary condition

(8.50)

shall apply on the remaining part of the boundary . For , this condition

reduces to the Neumann boundary condition (in case of or , these

boundary conditions are called homogeneous, otherwise they are inhomogeneous).

∂F

∂u

------

∂

∂x

-----

∂F

∂u

--------

–

∂

∂y

-----

∂F

∂u

--------

–

∂

∂z

-----

∂F

∂u

--------

–0 =

IFτd

∫

∂u

∂x

------





∂u

∂y

------





∂u

∂z

------





2u r()g r()– u

r()h r()–++ τ d

∫

∇

ughu++ 0 =

g 0 h, 0==

h 0= g 0 h, const.==

∇u()

τd

∫

τd

∫

-----

∇u()

τd

∫

-----------

τd

∫

ubr()=

∂u

∂n

------ d r()u+ e r()=

d 0=

b 0= e 0=

8.3 Boundary Value Problems as Variational Problems 519

The distinction between essential and natural boundary conditions is based on the

fact that the reference functions used in the variation must satisfy the essential

boundary conditions, but do not need to satisfy the natural boundary conditions. In

order to satisfy the natural boundary conditions, the variational integral has to be

supplemented by a boundary integral, i.e., eq. (8.46) is replaced by

(8.51)

Notice that the boundary integral applies to the surface only, that is, the part

where the natural boundary conditions (8.50) apply. We will skip the proof of this,

which can be found for example, at Davies [17]. If and , then the

boundary integral is obsolete, which means that in case of Neumann boundary

conditions, only the volume integral remains. Its variation automatically results in

the solution which satisfies the homogeneous Neumann boundary conditions.

Unlike the Dirichlet boundary conditions, these do not require the choice of

suitable reference functions to ensure a solution. We will demonstrate this

difference by means of simple examples.

The reader shall be reminded that the Neumann boundary condition may not

be prescribed entirely arbitrarily. For the inner (but not the outer) Neumann

problem it has to be

(8.52)

that is, the electric flux has to be compatible with the total charge inside the region

(this is irrelevant for the outer problem because there, the electric flux is assumed

to extend to infinity and the flux through the imaginary surface at infinity may take

any value).

One can avoid the need to choose the reference functions in case of the

Dirichlet boundary condition and start with arbitrary functions when

supplementing the variational integral by appropriate supplemental conditions and

consider these by means of Lagrange parameters (Lagrange multipliers). We will

illustrate this with an example.

The variational integrals are very well suited to obtain both an exact, as well

as an approximative solution of the corresponding problem. Oftentimes,

astonishingly accurate results can be obtained even with relatively simple means.

They also provide starting points for the method of Finite Elements (Sect. 8.7).

A practically important method to obtain approximate solutions of the

variation problems is the so-called Ritz method also known as Rayleigh-Ritz

method. It is based on writing the solution as a series of linear independent

functions in the following form:

(8.53)

I ∇u()

2u r()g r()– u

r()h r()–[]τd

∫

d r()u

r() 2u r()e r()–[]A .d

∫

d 0= e 0=

DdA•

∫

∂ϕ

∂n

------

∫

– Q==

i 1=

∑

520 Numerical Methods

Substitute this series into the variational integral and let the derivatives of the

resulting function with respect to be zero:

(8.54)

If the functions form a complete basis within the region where the functions u

are defined, then this allows to obtain the exact solution.

8.3.2 Examples

8.3.2.1 Poisson’s Equation

As our first example to demonstrate the variational calculation, consider the one-

dimensional Poisson equation

(8.55)

with its general solution

(8.56)

Initially, the boundary conditions shall be left undefined. Later we will solve the

problem under the following boundary conditions

a) First, the problem shall be solved by means of the variational integral for the

Dirichlet boundary condition

(8.57)

The series solution eq. (8.53) shall have the form

(8.58)

It must satisfy the essential boundary conditions. This requires that

(8.59)

Next, the coefficients C and D have to be chosen such that the variational inte-

gral is minimized:

(8.60)

Calculating this integral and letting its derivative with respect to C and D be

zero gives

(8.61)

and

(8.62)

… c

,,() c

∂Ic

… c

,,()

∂c

---------------------------------0,= i 1 … n,,=

∇

ugx()– abx+ ,== 0x 1≤≤

uABx

---

++ +=

u 0() γ ,= u 1() δ=

ux() ABxCx

++ +=

ux() γ δ γ– C– D–()xCx

+++=

Iu' x()[]

2ux() abx+()⋅+{}xd

∫

ICD,()==

---

,= D

---=

ux() γ δ γ–

---–





---

+++=

8.3 Boundary Value Problems as Variational Problems 521

This represents the exact solution because the statement (8.58) is flexible

enough to contain this as a special case.

b) The same problem can be solved in a different manner. The boundary condi-

tions are introduced here as supplementary conditions when varying the inte-

gral and considered by means of the Lagrange multiplier λ and µ. Then with

(8.58), the integral gives

(8.63)

The supplemental conditions that correspond to the boundary conditions are

(8.64)

This makes the functional which we have to vary

(8.65)

The next step is to take the derivative with respect to , and µ:

(8.66)

Eliminating λ and µ gives

(8.67)

which, again, is the solution eq. (8.62).

c) Now, we will cover Neumann’s boundary value problem. Because of

eq. (8.52), the Neumann boundary conditions have to be chosen such that

(8.68)

This means, we require that

(8.69)

Now, by eq. (8.51), we have to add the expression

IABCD,,,()B

---

2BC 2BD 3CD+++ + +=

2 aA

---

bD+++ +++++

A γ–0 ,= ABCDδ–+++ 0=

FABCD,,,()IABCD,,,()λA γ–()µABCDδ–+++()++=

ABCDλ,,,,

∂F ∂A⁄ 2abλµ+++ 0==

∂F ∂B⁄ 2B 2C 2Da

---

b µ+++++ 0==

∂F ∂C⁄ 2B

---

C 3D

---

b µ+ + +++ 0==

∂F ∂D⁄ 2B 3C

------

---

b µ++ +++ 0==

∂F ∂λ⁄ A γ–0==

∂F ∂µ⁄ ABCDδ–+++ 0==











A γ ,= B δγ–

---–

---

,–= C

---

,= D

---=

u'1() u'0()–

2ab+

---------------=

u'1() β ,= u'0() β

2ab+

---------------–=

522 Numerical Methods

(8.70)

to the variational integral. Then from the functional

(8.71)

we obtain the equations

(8.72)

The first of these four equations is identically satisfied. The reason is that the

Neumann boundary conditions were chosen in the allowed manner. Had we

not considered eq. (8.68), that condition would have to be imposed posteri-

orly. The remaining three equations give

(8.73)

while A may be chosen arbitrarily (since in case of a Neumann problem u is

only specified up to a constant). Therefore:

(8.74)

This represents the exact solution of the problem. In contrast to the Dirichlet

boundary conditions, the Neumann boundary conditions had to be supple-

mented by the boundary integral (where in this one-dimensional case, this

consists of only two terms). Omitting the boundary integral in an attempt to

satisfy the Neumann boundary conditions by means of a reference function

leads to an incorrect result. The reason is that the wrong quantity is mini-

mized. The reader is encouraged to try this and thereby verify the claim.

Thus far, in above examples, we have found the exact solution. Now, consider a

simplified Ansatz, which only allows to solve by approximation:

(8.75)

d) When considering the Dirichlet boundary value problem with the boundary

conditions eq. (8.57) we have to require

2 u

∂u

∂n

------

∫

–2uu'[]

–=

2A β

2ab+

---------------–





2 ABCD+++()β–=

GABCD,,,()IABCD,,,()2A β

2ab+

---------------–





2 ABCD+++()β–+=

∂G ∂A⁄ 2ab2β 2ab+()–2β–++ 0==

∂G ∂B⁄ 2B 2C 2Da

---

b 2β–++++ 0==

∂G ∂C⁄ 2B

---

C 3D

---

b 2β–++++ 0==

∂G ∂D⁄ 2B 3C

------

---

b 2β–++ ++ 0==











B β

2ab+

---------------

,–= C

---

,= D

---=

ux() A β

2ab+

---------------–





---

+++=

ux() ABxCx

++=

8.3 Boundary Value Problems as Variational Problems 523

(8.76)

The variational integral obtained yields

and therefore

(8.77)

e) Considering the Neumann boundary value problem with the boundary condi-

tions eq. (8.69) gives the first three of the eqs. (8.72) when and thus

, , ,

(8.78)

and

(8.79)

This solution neither satisfies the Poisson equation nor the boundary condi-

tions exactly, but only approximately.

f) To satisfy the boundary conditions of the Neumann problem exactly, requires

one to find suitable reference functions. This requires a consideration of the

additional boundary integral for the variation, because otherwise, the solution

would be incorrect. For the present case, this leads to the trial function

(8.80)

There are no more variational parameters left. The parameter A is arbitrary

and the variation does not depend on it. When comparing the two approxima-

tions (8.79) and (8.80) with the exact solution (8.74), then we observe that the

approximation (8.79) more closely resembles the exact solution and therefore

can be regarded as the better of the two approximations. This can be general-

ized. When starting from the same Ansatz, then the case where the Neumann

boundary conditions are considered exactly, is generally the less accurate

approximation. This is still the case, even if variational parameters remain

after the Neumann boundary conditions were applied. A better approach to

determine the variational parameters is, to leave this to the variation integral,

which defines the approximation and yields an optimized compromise for the

solution.

8.3.2.2 Helmholtz Equation

As our second example, we shall employ the Helmholtz equation, which describes

for example, the waves in wave guides and was discussed in sects. 7.7 through

7.11. The Helmholtz equation is

ux() γ δ γ– C–()xCx

++=

2ab+

---------------=

ux() γ δ γ–

2ab+

---------------–





2ab+

---------------

++=

D 0=

A arbitrary= B β

12a 7b+

----------------------–= C

2ab+

---------------=

ux() A β

12a 7b+

----------------------–





2ab+

---------------

++=

ux() A β

2ab+

---------------–





2ab+

---------------

++=