Lehner G. Electromagnetic Field Theory for Engineers and Physicists

Подождите немного. Документ загружается.

384 Time Dependent Problems I (Quasi Stationary Approximation)

θ-functions that allow one to transform poorly converging series into well

converging series.

Now, we return to the above announced proof of (6.167) and (6.168), where it

was claimed that

(6.177)

It may be sufficient to prove one of the two cases. Let us pick the case . This

equation meets the requirements to apply the residue theorem to find the inverse

Laplace transform (see Sect. 6.3). According to (6.60), we need the residues for

(6.178)

Using

(6.179)

one might as well write

(6.180)

The zeros of the denominator are at

(6.181)

However, notice that there is no pole for , i.e. , as the nominator also

vanishes with and the limit has a finite value. The

poles are therefore at

(6.182)

or in terms of p

Fig. 6.25

t 0=

x 0= xd=

t ∞=

L 2 nπξ()sin nπξ'()sin n

τ–[]exp

n 1=

∞

∑







pξ[]sinh p 1 ξ'–()[]sinh

pp[]sinh

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

for ξ' ξ >

pξ'[]sinh p 1 ξ–()[]sinh

pp[]sinh

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

for ξ' ξ <











ξ' ξ>

p() pτ()exp

pξ[]sinh p 1 ξ'–()[]sinh

pp[]sinh

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

pτ()exp=

z()sinh iiz()sin–=

p() pτ()exp i–

ipξ[]sin ip1 ξ'–()[]sin

pip[]sin

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

pτ()exp=

ip nπ=

n 0=

0= f

p() pτ()exp[]

p 0→

ip nπ, = n 1≥

6.6 Field Diffusion in a Plane Plate 385

(6.183)

These are poles of order 1. There are no other poles. Its residues are

This requires l’Hospital’s rule (6.63), by which we find

(6.184)

and its residue is

(6.185)

Summing all these residues from through totals to exactly what

was claimed in (6.177).

6.6.3 Impact of Boundary Conditions

Now we study the impact of the boundary conditions at and (this

relates to and , respectively). The first two parts from (6.165)

responsible for the boundary conditions are

(6.186)

The convolution theorem shall be used to find the solution in the time domain, if

we are able to find the inverse Laplace transform of the two functions

and

Both have poles of order 1 at

(6.187)

(6.188)

Again, there is no pole for because both functions are finite there. Now one

needs to find the residues for

– π

, = n 1≥

i– nπξ()sin nπ 1 ξ'–()[]sin n

τ–[]exp

nπ

------

ip[]sin

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

+()

–→

lim=

ip[]sin

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

–→

lim

2 p

----------

ip[]cos

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

–→

lim

2nπ

nπ()cos

--------------------–==

2– nπξ()sin nπ()sin nπξ'()cos nπ()cos nπξ'()sin–[]n

τ–[]exp

nπ()cos

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

2 nπξ()sin nπξ'()sin n

τ–[]exp=

n 1= n ∞=

ξ 0= ξ 1=

x 0= xd=

ξ p,()f

p()

p 1 ξ–()[]sinh

p[]sinh

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

p()

pξ[]sinh

p[]sinh

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

p 1 ξ–()[]sinh

p[]sinh

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

ip1 ξ–()[]sin

ip[]sin

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

pξ[]sinh

p[]sinh

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

ipξ[]sin

ip[]sin

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

ip nπ, = n 1≥

– π

, = n 1≥

386 Time Dependent Problems I (Quasi Stationary Approximation)

(6.189)

and

(6.190)

For the first case it is

and by (6.184)

(6.191)

In like manner, one finds for the second case:

(6.192)

The residues of (6.191) belong to the function of (6.189) and the residues of

(6.192) belong to the function (6.190). This enables one to write:

(6.193)

. (6.194)

In principle, one of the formulas is sufficient because they are equivalent since

Using this, the convolution theorem (6.54), and eq. (6.186), i.e. for

, yields

ip1 ξ–()[]sin pτ()exp

ip[]sin

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

ipξ[]sin pτ()exp

ip[]sin

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

nπ 1 ξ–()[]sin n

τ–[]exp

ip[]sin

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

+()

–→

lim=

2– πnnπ 1 ξ–()[]sin n

τ–[]exp

nπ()cos

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

2– πnnπ()sin nπξ()cos nπ()cos nπξ()sin–[]n

τ–[]exp

nπ()cos

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

2nπ nπξ()sin n

τ–[]exp=

nπξ()sin n

τ–[]exp

ip[]sin

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

+()

–→

lim=

2– πnnπξ()sin n

τ–[]exp

nπ()cos

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

2nπ 1–()

nπξ()sin n

τ–[]exp–=

L 2nπ nπξ()sin n

τ–[]exp

n 1=

∞

∑







p 1 ξ–()[]sinh

p[]sinh

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

L 2– nπ 1–()

nπξ()sin n

τ–[]exp

n 1=

∞

∑







pξ[]sinh

p[]sinh

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

nπ 1 ξ–()[]sin nπ()sin nπξ()cos nπ()cos nπξ()sin–1–()

– nπξ()sin==

h ξ() 0=

6.6 Field Diffusion in a Plane Plate 387

or slightly rewritten

(6.195)

This solves the problem for arbitrary boundary conditions. The solution of the

overall problem is the result of adding all contributions resulting from the initial

condition (6.169) and all contributions due to the just calculated boundary

conditions (6.195).

It is time for a simple example. The boundary conditions shall be

(6.196)

(6.197)

Their Laplace transform is

(6.198)

, (6.199)

and therefore with (6.186)

(6.200)

This problem can be solved by the inverse transform of (6.200) as well as by

applying (6.195). We shall do both. First, the inverse transform of (6.200) is carried

out in a similar way as in the previous example. The difference is essentially that

now, there is also a pole at and all the other residues have an additional

factor in the denominator of . The residue of

at the pole is

and the overall result is therefore

ξτ,() f

() 2nπ nπξ()sin n

ττ

–()–[]τ

dexp

n 1=

∞

∑

∫

() 2nπ 1–()

nπξ()sin n

ττ

–()–[]τ

dexp

n 1=

∞

∑

∫

–

ξτ,()2 nπ nπξ()sin n

τ–[]exp

n 1=

∞

∑

() 1–()

()–[]n

[]τ

.dexp

∫

⋅

τ() 0=

τ()







= for

τ 0 ≥

τ 0 .<

p() 0=

p()

------=

ξ p,()

------

pξ[]sinh

p[]sinh

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

n 1≥()–=

------

pξ[]sinh

p[]sinh

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

pτ()exp

ξ=

388 Time Dependent Problems I (Quasi Stationary Approximation)

(6.201)

Now, taking the other approach, using eq. (6.195) as staring point, then

This is the same result as before, where it is to be noted that

(6.202)

The proof can be provided by means of eqs. (3.118) and (3.120). We conclude that

both methods produce the same result. There is another way to understand this. For

the limit of large times, it is

(6.203)

(6.204)

The consequence is that since the boundary conditions (6.196) and (6.197) are time

independent, that for large times the field has to be time independent. This means

that is must be

(6.205)

Its general solution is

(6.206)

It follows from (6.196) that

ξτ,()B

ξ 2B

1–()

nπξ()sin n

τ–[]exp

nπ

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

n 1=

∞

∑

ξτ,()2 nπ nπξ()sin n

τ–[]exp

n 1=

∞

∑

[]τ

1–()

–[]dexp

∫

⋅

nπ nπξ()sin n

τ–[]

τ[]exp 1–

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

1–()

–[]exp

n 1=

∞

∑

ξτ,()2B

1–()

nπξ()sin

nπ

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

–

n 1=

∞

∑

1–()

nπξ()sin n

τ[]exp

nπ

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

n 1=

∞

∑

ξτ,()B

ξ 2B

1–()

nπξ()sin n

τ[]exp

nπ

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

n 1=

∞

∑

ξ 21–()

nπξ()sin

nπ

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

–

n 1=

∞

∑

= for -1<ξ +1<

ξτ,()

τ∞→

lim B

ξ=

xt,()

t ∞→

lim B

---

∂

x() 0=

x() ax b+=

6.7 The Cylindrical Diffusion Problem 389

(6.207)

and from (6.196) follows that

(6.208)

which leads to the solution (6.204). The complete solution (6.201) describes how

the steady state (6.204) is gradually achieved. This is illustrated qualitatively in

Fig. 6.26. If one considers the current, for which one can write:

(6.209)

Initially, there is a surface current

(6.210)

which gradually spreads out in order to uniformly occupy the entire available space

in its final state, the steady state:

(6.211)

The overall current per unit length remains unchanged

(6.212)

It is fixed by the boundary conditions chosen in this example.

6.7 The Cylindrical Diffusion Problem

6.7.1 The Basic Formulas

We will now focus on problems that are important for practical applications, but

nevertheless represent very simple specific cases, such as the skin effect in a

b 0=

------=

Fig. 6.26

t 0=

x 0= xd=

t ∞=

xt,()

x∂

∂

xt,()–

------

x∂

∂

xt,()–==

------

δ xd–()–=

---------–=

- g

x()xd

∫

------–==

390 Time Dependent Problems I (Quasi Stationary Approximation)

cylindrical wire. The wire considered, is to be infinite in length and rotationally

symmetric (i.e. a circular) cylinder. The magnetic field which surrounds it, or is

present inside depends only on r. Consequently, all derivatives of ϕ and z have to

vanish. The requirement that B has to be source free leads under these

circumstances to the condition that the radial component has to vanish.

Otherwise, it would diverge at the axis. Eq. (3.32) gives

(6.213)

from which it follows that

(6.214)

Therefore,

(6.215)

The diffusion equation

(6.216)

for the assumptions made here and with (5.14) becomes

(6.217)

. (6.218)

Thus, we have to solve different equations for and . Subsequently, we will

discuss the field components separately. To simplify, we will limit our discussion to

a solid, circular cylinder of radius . We further simplify the problem by

assuming a vanishing initial field,

(6.219)

It shall be noted that the general initial value problem with an arbitrary initial field

can be solved without particular difficulties, for example, by expanding the initial

field into a Fourier-Bessel series. The boundary conditions for our problem are

(6.220)

. (6.221)

We Introduce again the dimensionless variables

(6.222)

and

(6.223)

The equations for and read now:

B∇•

---

r∂

∂

()0==

const.

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

B 0 B

rt,()B

rt,(),,〈〉=

∇

B µκ

t∂

∂B

---

r∂

∂

r∂

∂

-----–





rt,() µκ

t∂

∂

rt,()=

---

r∂

∂

r∂

∂





rt,() µκ

t∂

∂

rt,()=

B rt,()[]

to=

B r 0,()0==

B rt,()[]

B r

t,()f t()==

B rt,()[]

r 0=

B 0 t,() finite==

xrr

⁄=

µκr

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

6.7 The Cylindrical Diffusion Problem 391

(6.224)

, (6.225)

with the initial condition

(6.226)

and the boundary conditions

(6.227)

. (6.228)

This is the form for which eqs. (6.224) and (6.225) have to be solved – now, with

the conditions (6.226) through (6.228). We will first solve for the longitudinal field

, and thereafter for the azimuthal field .

6.7.2 The longitudinal Field B

Consider an infinitely long cylinder in a space where a uniform magnetic field is

created, that is, the field is oriented parallel to the cylinder axis:

(6.229)

If there is no field initially in the cylinder, then one has to solve the above specified

problem. Instead of immediately solving for , we introduce its Laplace

transformed field and then, from (6.225) with the initial condition

(6.226), we obtain the equation

(6.230)

and the boundary conditions in the p-domain

(6.231)

. (6.232)

This needs to be solved. Previously, in eq. (3.164), we had already met the Bessel

differential equation:

(6.233)

Substituting

(6.234)

in eq. (6.230) gives the new equation

(6.235)

---

x∂

∂

x∂

∂

-----–





x τ,()

τ∂

∂

x τ,()=

---

x∂

∂

x∂

∂





x τ,()

τ∂

∂

x τ,()=

B x 0,()0=

B 1 τ,()f τ()=

B 0 τ,()finite=

τ() f

τ()=

x τ,()

xp,()

---

x∂

∂

x∂

∂

p–





xp,()0=

1 p,()f

p()=

0 p,()finite=

---

ξ∂

∂

ξ∂

∂

-------–





+ Z

ξ() 0=

ξ xi p=

---

ξ∂

∂

ξ∂

∂

1+ B

ξ p,()0=

392 Time Dependent Problems I (Quasi Stationary Approximation)

This is Bessel’s differential equation with index or order zero. Therefore

(6.236)

where A and B may depend on p but not on x. Because of the boundary conditions

(6.232)

(6.237)

and because of the boundary condition (6.231), the other constant has to be

(6.238)

Therefore, the Laplace transformed field is

(6.239)

We can obtain the general solution for arbitrary boundary conditions in the time

domain from this formula, if we are able to find the inverse transform for

. This is possible by means of the residue theorem. Notice the

broad analogy to the problem of Sect. 6.6.3.

For the inverse transform, we need the residues of the function

(6.240)

This has an infinite number of poles of order 1. If we use the notation introduced in

Sect. 3.7.3.3, eq. (3.209) to address the zeros of by , then it is for the poles

(6.241)

(6.242)

The residues of the function in (6.240) are

Using the identity

gives

xp,()AJ

xi p()BN

xi p()+=

B 0=

p()

ip()

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

xp,()f

p()

xi p()

ip()

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

xi p()J

ip()⁄

xi p()

ip()

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

pτ()exp

ip λ

–=

x() λ

– τ()exp

ip()

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

p λ

–→

lim p λ

+()=

x() λ

– τ()exp

2 p

----------

J '

ip()

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

p λ

–→

lim .=

J '

–=

6.7 The Cylindrical Diffusion Problem 393

(6.243)

and therefore, we have the transform

(6.244)

Applying the convolution theorem to eq. (6.239) yields

(6.245)

The result is in the form of a Fourier-Bessel series according to (3.213). Its

coefficients are time dependent functions:

(6.246)

The structure of eq. (6.245) is very similar to the corresponding structure of the

plane equation (6.195).

The current in the cylinder is

(6.247)

It is remarkable that this is not a Fourier-Bessel series, but rather a so-called Dini

series, which we will not discuss further. Details for further study to the various

types of series involving Bessel functions, including Dini series, can be found in

[7].

As an example, let us consider the simple boundary condition

(6.248)

(6.249)

Then one writes the field with (6.239)

2λ

x() λ

– τ()exp

()

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

2λ

x() λ

– τ()exp

()

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

n 1=

∞

∑

xi p()

ip()

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

x τ,() f

()

2λ

x() λ

– ττ

–()[]exp

()

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

n 1=

∞

∑

∫

x τ,()

2λ

x() λ

– τ[]exp

()

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

() λ

()exp τ

∫

n 1=

∞

∑

τ()

2λ

– τ()exp

()

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

() λ

()exp τ

∫

r()

r∂

∂H

–

------

r∂

∂B

–

-----------

x∂

∂B

–== =

2λ

x() λ

– τ()exp

()

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

() λ

()exp τ

∫

n 1=

∞

∑

τ() B

p() B

p⁄=