universitÀ degli studi di bologna … · 2008-03-11 · universitÀ degli studi di bologna ......

UNIVERSITÀ DEGLI STUDI DI BOLOGNA ——————————————————

DIPARTIMENTO DI INGEGNERIA ELETTRICA Viale Risorgimento n°2 - 40136 BOLOGNA (ITALIA)

ANALYTICAL SOLUTIONS

FOR THE CURRENT DISTRIBUTION

IN A RUTHERFORD CABLE WITH N STRANDS

M. Fabbri, M. Breschi

Abstract −−−− The geometrical properties of the auto/mutual induction coefficient matrix among the strands of a Rutherford cable are investigated. The solution for the general linear case is

given. The comparison with a known two-strand solution is carried out. The convergence

analysis is done and the regime solution is evaluated. For the non-linear case a diagonal sys-

tem is obtained and solved under the assumption of “mean” resistance. Finally, the flux ramp

case and the perfect conductor cases are fully described.

RAPPORTO INTERNO DIEUB – APRILE 2001

DIE/UNIBO – Aprile 2001 1 di 37

Index

1. Introduction .................................................................................................................... pp. 1

2. The Solution for the Linear Case .................................................................................... pp. 5

3. Comparison with the Krempasky-Schmidt solution ....................................................... pp. 9

4. Convergence Analysis .................................................................................................... pp. 13

5. Regime Solution.............................................................................................................. pp. 16

6. Non-linear Case............................................................................................................... pp. 17

7. Flux Ramp....................................................................................................................... pp. 19

8. Perfect Conductor Case (r = 0) ....................................................................................... pp. 27

References ........................................................................................................................... pp. 32

Appendix A – Forced current distribution in a 3-strand cable............................................ pp. 33

Appendix B – 3-strand cable excited as in the Krempasky-Schmidt case.......................... pp. 35

1. Introduction

Consider a square N × N circulant symmetric matrix [A], defined by the following two properties:

1- ai,1 = ai−1,N, ai,j = ai−1,j−1, with i, j = 2, … , N;

2- ai,j = aj,i, with i, j = 1, … , N.

Assume that N is even, in particular N = 2p. In this case p+1 different elements exist in the matrix [A]

(in fact, consider the firs row of [A], for which a1,j = a1,N+2−j with j = 2, … , N. There exist p−1 pairs and

the unpaired elements a11 and a1,p+1 ha to be added). Therefore the matrix [A] can be written as follows:

[ ]

=

−+

−−

−−+

−−+

+−−

−−

+−

+

11121p,1p,11p,1141312

12112p,11p,1p,1151413

1p,12p,1111213p,11p,1p,1

p,11p,11211121p,1p,11p,1

1p,1p,11312112p,11p,1p,1

1415p,11p,12p,1111213

13141p,1p,11p,1121112

1213p,11p,1p,1131211

aaaaaaaa

aaaaaaaa

aaaaaaaa

aaaaaaaa

aaaaaaaa

aaaaaaaa

aaaaaaaa

aaaaaaaa

A

LL

LL

MMOMMMOMMM

LL

LL

LL

MMOMMMOMMM

LL

LL

LL

The eigenvalues of [A] can be obtained from the N-th complex roots of the unity 1:

π==ω k2jN e1 ⇒ p/jkN/k2jk ee ππ ==ω , with k = p, p−1, …, 1, 0, −1, …, −(p−1)

The eigenvalues are given with k = p, p−1, …, 1, 0, −1, …, −(p−1), by:


( )

( ) ( )

( ) ( ) ( )

( ) ( ) ( )

( ) ( )k1p,1

p

2s

s,111

p

2

p/k1j,1

k1p,1

p

2s

p/k1sjs,111

p

2

p/k1p2j2p2,1

k1p,1

p

2s

p/k1sjs,111

p2

2ps

p/k1sjs,1

jk1p,1

p

2s

p/k1sjs,111

N

1s

p/k1sjs,1

N

1s

1sks,1

1NkN1

1sks,1k1211k

1ap

k1scosa2a

ea1aeaa

ea1aeaa

eaeaeaa

eaaaaaa

−+π−+=

=+−++=

=+−++=

=+++=

==ω=ω++ω++ω+=λ

+=

=σ

πσ−σ+

=

π−

=σ

πσ−+σ−++

=

π−

+=

π−π+

=

π−

=

π−

=

−−−

∑

∑∑

∑∑

∑∑

∑∑KK

The equation above shows that λq = λ−q, with q = 1, … , p−1. Thus, p+1 different eigenvalues of [A] ex-

ist: the algebraic multiplicity of λ0 and λp is equal to 1 (and therefore also their geometric multiplicity

and the dimension of the corresponding subspace are equal to 1); the algebraic multiplicity of λq, with q

= 1, … , p−1, is equal to 2. To determine the existence of the associated spectral basis it’s sufficient to

demonstrate that the geometric multiplicity of λq, with q = 1, … , p−1, is equal to 2, determining two dif-

ferent eigenvectors.

It is simple to show that the vector bk (∈ CN), with k = p, p−1, …, 1, 0, defined by:

1Nkk

Tk ,,,1 −ωω= Lb

is an eigenvector of [A] (the index T denotes the transpose operator). In fact, we have:

[ ] [ ][ ]

[ ][ ]

kk

k1N

k

kk

k

112k13

1k12

1Nk

2Nk1N,111

1NkN1k

1NkN,1k1211

112N

k131N

k121N

k

2Nk1N,111

1kN1k

1NkN,1k1211

1Nk11k1312

1Nk1N,1k11N1

1NkN,1k1211

k

aaa

aaa

aaa

aaa

aaa

aaa

aaa

aaa

aaa

A

b

b

λ=

λω

λωλ

=

++ω+ωω

ω+++ωωω++ω+

=

=

++ω+ωω

ω+++ωωω++ω+

=

ω++ω+

ω++ω+ω++ω+

=

−−

−−

−

−

+−+−−

−−

−

−

−

−−

−

M

K

M

K

K

K

M

K

K

K

M

K

K

Note that of all the bk (∈ CN), with k = p, p−1, …, 1, 0, only b0 and bp belong to RN

:

1,1,,1,1 T0 L=b 1,1,,1,1 T

p −−= Lb

Concerning bk, with k = p−1, …, 1, to construct a real spectral basis the following vectors are defined:

( )

π−π=ωωℜ= −

p

q1p2cos,,

p

qcos,1,,,1 1N

qqTq LLb

( )

π−π=ωωℑ= −

−p

q1p2sin,,

p

qsin,0,,,1 1N

qqT

q LLb


with q = 1, … , p−1. The eigenvectors bq and b-q, associated to λq, are orthogonal. In fact, defining

Bq = bq + j b-q, it can be found that:

( )1N2q

2qq

Tq 1 −ω++ω+= LBB ⇒ 1 N2

qqTqq

Tq

2q −ω+=ω BBBB ⇒ ( ) 0 1 q

Tq

2q =−ω BB

Thus, for p ≠ q ≠ 0, we get:

( ) ( )qT

qqTqq

Tqq

Tqq

Tq j0 bbbbbbbb −−−− ++−==BB ⇒

=

=

−

−

0qTq

2

q

2

q

bb

bb

Finally, the real spectral basis bk, with k = p, p−1, …, 1, 0, −1, …, −(p−1) is orthogonal. In fact, recall-

ing that 2:

( )

−=∑−

= y2

1sin

y2

Nsin

y2

1Ncoskycos

1N

0k

( )

−=∑−

= y2

1sin

y2

Nsin

y2

1Nsinkysin

1N

0k

It results that, with s ≠ q (and thus s ± q ≠ ±N):

( ) ( )

( )( ) ( )( )( )

( )( ) ( )( )( ) 0

N

qssin

qssin

N

qs1Ncos

2

1

N

qssin

qssin

N

qs1Ncos

2

1

N

qs2kcos

2

1

N

qs2kcos

2

1

N

q2kcos

N

s2kcos

1N

0k

1N

0k

1N

0k

qTs

=

π−π−

π−−+

π+π+

π+−=

=

π−+

π+=

π

π= ∑∑ ∑−

=

−

=

−

=

bb

( ) ( )

( )( ) ( )( )( )

( )( ) ( )( )( ) 0

N

sqsin

sqsin

N

sq1Nsin

2

1

N

qssin

qssin

N

qs1Nsin

2

1

N

sq2ksin

2

1

N

sq2ksin

2

1

N

q2ksin

N

s2kcos

1N

0k

1N

0k

1N

0k

qTs

=

π−π−

π−−+

π+π+

π+−=

=

π−+

π+=

π

π= ∑∑ ∑−

=

−

=

−

=−bb

( ) ( )

( )( ) ( )( )( )

( )( ) ( )( )( ) 0

N

qssin

qssin

N

qs1Ncos

2

1

N

qssin

qssin

N

qs1Ncos

2

1

N

qs2kcos

2

1

N

qs2kcos

2

1

N

q2ksin

N

s2ksin

1N

0k

1N

0k

1N

0k

qT

s

=

π+π+

π+−+

π−π−

π−−=

=

π+−

π−=

π

π= ∑∑ ∑−

=

−

=

−

=−− bb

To obtain an orthonormal spectral basis, note that: 1) ||b0||2 = ||bp||2 = N; 2) ||bq||2 + ||b−q||2 = N; 3) ||bq||2 =

||b−q||2; and thus ||bq||2 = ||b−q||2 = N/2 = p. therefore, it’s possible to define the orthonormal spectral basis,

as follows:

=

N

1,

N

1,,

N

1,

N

1 T

0 Lb

−−=

N

1,

N

1,,

N

1,

N

1 T

p Lb


( )

π−π=

p

q1p2cos

p

1,,

p

qcos

p

1,

p

1 T

q Lb

( )

π−π=−

p

q1p2sin

p

1,,

p

qsin

p

1,0 T

q Lb

with q = 1, … , p−1. Since any spectral basis is complete, it’s possible to decompose uniquely any given

vector u ∈ RN as follows:

( )( )∑

−−=

=p

1pk

Tkk ubbu

and thus: [ ] ( )( )∑

−−=

λ=p

1pk

Tkkk A ubbu

Finally note that:

1) given a square N × N circulant symmetric positive-definite matrix [M];

2) given a square N × N circulant symmetric [G] such that ∑=

−=N

2k

k111 gg ;

Thus we have that:

1 - the eigenvalues of [M] are positive and given by:

( ) ( )k1p,1

p

2s

s,111k 1mp

k1scosm2m −+π−+=λ +

=∑

with k = p, p−1, …, 1, 0, −1, …, −(p−1). Thus [M] bk = λk bk;

2 - the eigenvalues of [G] are negative (except for one that is null) e given by:

( ) ( )[ ]k1p,1

p

2s

s,1k 11gp

k1scos1g2 −−−

π−−−=γ +=∑

with k = p, p−1, …, 1, 0, −1, …, −(p−1) (therefore, in particular γ0 = 0). Thus [G] bk = γk bk;

3 - the orthonormal spectral basis bk, with k = p, p−1, …, 1, 0, −1, …, −(p−1), it’s the same for [M] and

[G]:

=

N

1,

N

1,,

N

1,

N

1 T

0 Lb

−−=

N

1,

N

1,,

N

1,

N

1 T

p Lb

( )

π−π=

p

q1p2cos

p

1,,

p

qcos

p

1,

p

1 T

q Lb ( )

π−π=−

p

q1p2sin

p

1,,

p

qsin

p

1,0 T

q Lb

The analysis for N odd is really equivalent, but the formulas for the eigenvalues and for the orthonormal

spectral basis are different. Assuming that N = 2p + 1, it can be found that:

1) the eigenvalues of [M] are again positive and given by:


( )∑

+

= +π−+=λ

1p

2s

s,111k1p2

k1s2cosm2m

where the index k spans the set p, p−1, …, 1, 0, −1, …, −p.

2) the eigenvalues of g are again negative and given by:

( )∑

+

=

+−−−=γ

1

2

,112

π12cos12

p

s

skp

ksg

where the index k spans the set p, p−1, …, 1, 0, −1, …, −p.

3) the orthonormal spectral basis b is defined as follows:

=

N

1,

N

1,,

N

1,

N

1 T

0 Lb

+π

++π

++=

1p2

q2p2cos

1p2

2,,

1p2

q2cos

1p2

2,

1p2

2 T

q Lb

+π

++π

+=−

1p2

q2sin

1p2

2,,

1p2

q2sin

1p2

2,0 T

q 2p Lb

with q = 1, … , p.

2. The Solution for the Linear Case

Assume that the currents flowing in the N-strand Rutherford cable are described by the following para-

bolic system:

(P)

[ ][ ] ( ) [ ][ ] ( ) ( ) [ ] ( )

( ) ( ) ( )

( ) ( )( )

==

=α====

=∂∂++

∂∂

αα

x0t,x

N,,1con,N

tIt,Lxit,0xi

t,xGt,xx

t,xRGt,xt

MG

0

2

2

ii

vi

ii

L and i, v ∈ RN

with [R] = r [I], where [I] is the identity matrix, r is constant and [M] and [G] are constant-coefficient

matrixes with the above described properties. The boundary conditions can be synthetically written as (o)

:

(o)

Note that projecting (P) on b0, it results:

(P0)

( ) ( ) ( ) ( )

( ) ( ) ( )

==

====

=∂∂

x0t,x

N

tIt,Lxt,0x

0t,xx

0T0

T0

T0

T0

T02

2

ibib

ibib

ib

⇒ ( ) ( )N

tIt,x

T0 =ib ⇒

( ) ( ) ( )N

0tIx

0T0

==ib (compatibility condition)


( ) ( ) ( )0

N

tIt,Lxt,0x bii ====

Defining the current variations as:

( ) ( ) ( )0

N

tIt,xt,x bii −=δ

the problem can be restated as (note that [G] [M] b0 = [G] λ0 b0 = γ0 λ0 b0 = 0):

(P')

[ ][ ] ( ) [ ] ( ) ( ) [ ] ( )

( ) ( )( ) ( )( ) ( )

=−==δ

==δ==δ

=∂

δ∂+δ+∂δ∂

00

2

2

N

0tIx0t,x

0t,Lxt,0x

t,xGt,xx

t,xrGt,xt

MG

bii

ii

vi

ii

Utilizing the trigonometric basis, orthogonal in [0, L], sin(nπx/L)n, with n∈N, the following series are

defined:

( ) ( )

π=∑∞

= L

xnsintt,x

1n

nFF ⇔ ( ) ( ) ξ

πξξ= ∫ dL

nsint,

L

2t

L

0

n FF .

and then the problem can be restated as (note that the boundary conditions are essential to obtain

( ) n

2L

0

2

2

L

nd

L

nsint,

xL

2i

i δ

π−=ξ

πξξ∂

δ∂∫ ):

(P'n)

[ ][ ] ( ) [ ] ( ) ( ) [ ] ( )

( ) ( ) ( ) ( )[ ]

−−π

=−==δ

=δ

π−δ+δ

n0

0nn

nn

2

nn

11n

2

N

0tI0t

tGtL

ntrGt

dt

dMG

bii

viii

, with n∈N

Utilizing the orthonormal spectral basis bk, with k = p, p−1, …, 1, 0, −1, … −(p−1), defined above, we

get:

( )( )

( )tt k,n

p

1pk

kn η=δ ∑−−=

bi ⇔ ( ) ( )tt nTkk,n ib δ=η ( )

( )( )tt k,n

p

1pk

kn ν= ∑−−=

bv ⇔ ( ) ( )tt nTkk,n vb=ν

and then the problem can be restated as (with n∈N):

(P'n,k)( ) ( ) ( ) ( )

( ) ( )

==η

νγ=η

π−ηγ+η

λγ

0n

Tkk,n

k,nkk,n

2

k,nkk,n

kk

0t

ttL

ntrt

dt

d

ib

, with k = p, p−1, …, 1, −1, … −(p−1)

Note that k ≠ 0, in fact ηn,0 = 0 as a consequence of ( ) 0t,xT0 =δib . The problem (P'n,k) is directly solv-

able:

( )

( ) ( )

==η

ν=η

πγ

−+η

λ

0n

Tkk,n

k,nk,n

2

k

k,nk

0t

tL

n1r

dt

d

ib

⇒


⇒ ( ) ( )( ) ( )( )

ττνλ

+=η

πγ

−λ

τ−−

πγ

−λ−

∫ de1

et

2

kk

2

kk L

n1r

tt

0

k,nk

L

n1r

t

0n

Tkk,n ib

Reconstructing the solution, we get:

( ) ( ) ( )( )∑ ∑

≠−−=

∞

=

πη+=p

0k1pk 1n

k,nk0L

xnsint

N

tIt,x bbi ⇒

( ) ( ) ( )( )( )

( )( )( )

( )∑ ∑∫

∑ ∑

≠−−=

∞

=

πγ

−λ

τ−−

≠−−=

∞

=

πγ

−λ−

πτλ

τ+

+

π+=

p

0k1pk 1n

L

n1r

t

nTk

k

k

t

0

p

0k1pk 1n

L

n1r

t

0n

Tkk0

L

xnsined

L

xnsine

N

tIt,x

2

kk

2

kk

vbb

ibbbi

⇒

( ) ( ) ( )( )( )( )

( )( )( )

( )∑ ∑∫∫

∑ ∑∫

≠−−=

∞

=

πγ

−λ

τ−−

≠−−=

∞

=

πγ

−λ−

πξ

πτξλ

τξ+

+

πξ

πξξ+=

p

0k1pk 1n

L

n1r

t

Tk

k

k

t

0

L

0

p

0k1pk 1n

L

n1r

t

0Tkk

L

0

0

L

nsin

L

xnsine,dd

L

2

L

nsin

L

xnsined

L

2

N

tIt,x

2

kk

2

kk

vbb

ibbbi

⇒

( ) ( ) ( )( )( ) ( )( )

( )( ) ( )( )∑∫∫

∑∫

≠−−=

≠−−=

τ−ξΓτξλ

τξ+

+ξΓξξ+=

p

0k1pk

kTk

k

k

t

0

L

0

p

0k1pk

k0T

kk

L

0

0

t,,x,ddL

2

t,,xdL

2

N

tIt,x

vbb

ibbbi

Recalling that ( ) ( ) ( ) ( )[ ]212121 coscos2

1sinsin ω+ω−ω−ω=ωω and the definition of the elliptic theta

function ϑ3 3(oo)

: ( ) ∑∞

=

+=ϑ1n

n3 nu2cosq21q,u

2

, the Green functions can be written as (ooo)

:

(oo) the fundamental propriety of the function ϑ3 (u, q) is: ( ) ( )q,u

u4

1q,u

qq

2

32

3

∂ϑ∂

−=∂ϑ∂

(ooo) Note that: ( ) ( ) ( ) 0t,,x

xt,,xrt,,x

t 2

k2

kkk

kk =ξ∂

Γ∂+ξΓγ+ξ

∂Γ∂

λγ , with k = p, p−1, …, 1, −1, … −(p−1).


( )

( ) ( )

ξ+πϑ−

ξ−πϑ=

ξ+π−

ξ−π=

=

πξ

π=ξΓ

πγλ

πγλ

λ−

∞

=

πγλ

λ−

∞

=

πγλ

λ−

∞

=

πγ

−λ−

∑∑

∑

2

kk

2

kkk

2

kk

2

k

2

kk

2

k

2

kk

L

t

3L

t

3

rt

1n

L

tnrt

1n

L

tnrt

1n

L

n1r

t

k

e,L2

xe,

L2

x

4

e

L

xncose

2

e

L

xncose

2

e

L

nsin

L

xnsinet;,x

with k = p, p−1, …, 1, −1, … −(p−1). Therefore the Green matrixes are given by:

( )( )[ ] ( )( )∑

≠−−=

ξΓ=ξp

0k1pk

Tkkk

0 t,,xt,,xK bb ( )[ ] ( )( )∑

≠−−= λ

ξΓ=ξp

0k1pk k

Tkk

k t,,xt,,xKbb

And the solution of (P) results to be:

( ) ( ) ( )( )[ ] ( )( ) ( )[ ] ( )τξτ−ξτξ+ξξξ+= ∫∫∫ ,t,,xKddL

2t,,xKd

L

2

N

tIt,x

t

0

L

0

00

L

0

0 vibi

Note that this solution of (P) results to be invariant by addition to the sources i(0)

(x) and v(x, t) of terms

proportional to a b0. In fact:

( )( )[ ] ( )( ) ( ) ( )( )[ ] ( )( ) ( ) ( )( )[ ] ( )( )[ ] ( ) ( )ξξ=ξξ+ξξ=ξ+ξξ 000

0000

00 ,xK,xKv,xKi,xK ibibi

( )[ ] ( ) ( ) ( )[ ] ( ) ( ) ( )[ ] ( )[ ] ( )τξτξ=τξτξ+τξτξ=τξ+τξτξ ,,,xK,,xK,v,,,xK,v,,,xK 00 vbvbv

This means that the solution does not change if in the first integral i(0)

(x) is replaced by δi(0)(x) and if the

reference node for the voltages is changed.

Note that the calculation of the integration kernels [K(0)

] and [K] may cause convergence difficulties,

since the function Γk approaches the Dirac δ distribution, for t going to zero:

( ) ( )ξ−δ=

πξ

π=ξΓ ∑∞

=→x

2

L

L

nsin

L

xnsint;,xlim

1n

k0t

Thus, if i(0)

(x) and v(x, t) are differentiable with respect to x, it’s better to utilize an alternative form of

the solution, obtained through an integration by parts. Defining the new function Γk(oooo)

:

( ) ( ) ∑∫∞

=

πγ

−λ−

ξ

πξ

ππ

=ξ′ξ′Γ=ξΓ1n

2

L

n1r

t

0

kkL2

nsin

L

xnsin

n

eL2dt;,xt;,x

2

kk

(oooo) Note that Γk satisfies the same equations of Γk: ( ) ( ) ( ) 0t,,x

xt,,xrt,,x

t 2

k2

kkk

kk =ξ∂

Γ∂+ξΓγ+ξ

∂Γ∂

λγ


with k = p, p−1, …, 1, −1, … −(p−1), and the new integration kernels [K(0)

] and [K]:

( )( )[ ] ( )( )∑

≠−−=

ξΓ=ξp

0k1pk

Tkkk

0t,,xt,,xK bb ( )[ ] ( )

( )∑

≠−−= λ

ξΓ=ξp

0k1pk k

Tkk

k t,,xt,,xKbb

The solution of (P) can be written as:

( ) ( ) ( )( )[ ] ( )( ) ( )[ ] ( )

( ) ( )( )[ ] ( )( ) ( )( )[ ] ( )( )

( )[ ] ( ) ( )[ ] ( )τξξ∂

∂τ−ξτξ−

τξτ−ξτ+

+ξξ∂

∂ξξ−

ξξ+=

=τξτ−ξξ∂

∂τξ+ξξξ∂

∂ξ+=

∫∫∫

∫

∫∫∫

=ξ

=ξ

=ξ

=ξ

,t,,xKddL

2,t,,xKd

L

2

t,,xKdL

2t,,xK

L

2

N

tI

,t,,xKddL

2t,,xKd

L

2

N

tIt,x

t

0

L

0

L

0

t

0

00

L

0

L

0

000

t

0

L

0

00L

0

0

vv

iib

vibi

and finally:

( ) ( ) ( )( )[ ] ( )( ) ( )( )[ ] ( )( )

( )[ ] ( ) ( )[ ] ( )τξξ∂

∂τ−ξτξ−ττ−τ+

+ξξ∂

∂ξξ−+=

∫∫∫

∫

,t,,xKddL

2,Lt,L,xKd

L

2

t,,xKdL

2Lt,L,xK

L

2

N

tIt,x

t

0

L

0

t

0

00

L

0

000

vv

iibi

Note that the calculation of the integration kernels [K(0)

] and [K] is no more cause of convergence diffi-

culties, since the function Γk approaches the Heaviside step function, for t going to zero:

( ) ( )

ξ>ξ<

=−ξ=

πξ

ππ

=ξΓ ∑∞

=→ x,0

x,1

2

LxU

2

L

L2

nsin

L

xnsin

n

L2t;,xlim

1n

2k

0t

In the particular case in which i(0)

(x) has equal components and v(x, t) is time-independent, the solution

further simplifies, defining:

( ) ( ) ∑∫∞

=

πγ

−λ−

πξ

π

−

πγ

−

λ=ττξΓ=ξΓ

1n

L

n1r

t

2

k

k

t

0

kkL

nsin

L

xnsine1

L

n1r

d;,xt;,x*

2

kk

with k = p, p−1, …, 1, −1, … −(p−1); and the integration kernel [K*]:

( )[ ] ( )( )∑

≠−−= λ

ξΓ=ξp

0k1pk k

Tkk

k t,,xt,,x*K *bb

the solution of (P) can be written as:


( ) ( ) ( )[ ] ( )ξξξ+= ∫ vbi t,,x*KdL

2

N

tIt,x

L

0

0

3. Comparison with the Krempasky-Schmidt solution

The solution found above is now compared with a known one for a two-strand cable 4. With the con-

ditions:

1- I(t) = 0;

2- r = 0,

3- even number of periods,

4- i(0)

(x) = 0,

5- time-independent excitation spatially bounded excitation to an interval in the center of the cable.

Hypotheses 3 and 5 lead to write:

( )

−+

ξ−δ+δ−−ξδ

Φ=ξ1

1)

22

L(U)

22

L(U

2/&v

where U is the Heaviside step function. For two strand, N = 2, p = 1; thus, given

[ ]

=

1112

1211

mm

mmM [ ]

−++−

=1212

1212

gg

ggG

1 - the eigenvalues of [M] are positive and given by:

12110 mm +=λ ; 12111 mm −=λ

2 - the eigenvalues of [G] are negative (except for one that is null) e given by:

00 =γ ; 121 g2−=γ

3 - the orthonormal spectral basis bk, with k = 1, 0, it’s the same for [M] and [G]:

=

2

1,

2

1 T

0b

−=

2

1,

2

1 T

1b

To reproduce the nomenclature of 4 the following variables are defined:

L1 = 2 (m11 − m12), G1 = g12, α = π w/(2w + δ), L = 2w + δ,

2

112 2

w2GL

4

δ+π

=τ

(Note that π / L = α / w, L / w = π / α and (πδ / 2L) + α = π / 2)

Thus we get: ( ) 2)22

L(U)

22

L(U

21bv ξ−δ+δ−−ξ

δΦ=ξ&

. When the simplified solution is utilized, we

have:


( )

∑

∑

∞

=

τ−

∞

=

πγ

−λ−

πξ

π−πλ

=

=

πξ

π

−

πγ

−

λ=ξΓ

1n

/tn

22

2112

1n

L

n1t

2

1

11

L

nsin

L

xnsine1

n

1Lg2

L

nsin

L

xnsine1

L

n1t;,x*

2

2

11

(Note that

( ) τ=

δ+π=

δ+π=

π−

=

πγ

−λ

1

w2

2

GL

1

4w2GL

1

Lg2

1

mm

1

L

112

11

2

2

2

11

2

121211

2

11

)

Therefore the kernel becomes:

( )[ ] ( ) ( )

( ) ( )

+−−+

πξ

π−π

=

=

+−−+

λξΓ=

+−−+

λξΓ=

=

−

−λξΓ

=λ

ξΓ=ξ

∑∞

=

τ−

11

11

L

nsin

L

xnsine1

n

1Lg

11

11

2

t,,x*

2/12/1

2/12/1t,,x*

2

1,

2

1

2

1

2

1

t,,x*t,,x*t,,x*K

1n

/tn

22

212

1

1

1

1

1

1

1

T11

1

2

bb

and the solution can be written as:

( ) ( )[ ] ( )

( )

( )

( )

( )

( )

−+

πξξ

π−δπ

Φ

=

−+

+−−+

πξξ

π−δπ

Φ=

=ξ

+−−+

πξξ

π−π

=

=ξξξ=

∑ ∫

∑ ∫

∑ ∫

∫

∞

=

δ+

δ−

τ−

∞

=

δ+

δ−

τ−

∞

=

τ−

2

2

L

nsind

L

xnsine1

n

1Lg

1

1

11

11

L

nsind

L

xnsine1

n

1Lg

11

11

L

nsind

L

xnsine1

n

1Lg2

t,,x*KdL

2t,x

1n

2/L

2/L

/tn

22

12

1n

2/L

2/L

/tn

22

12

1n

L

0

/tn

22

12

L

0

2

2

2

&

&

v

vi

Since i2 = − i1, only i1(x, t) is considered:

( ) ( )

( )

( )

( )

( )

( )

∑ ∫

∑ ∫

∑ ∫

∞

=

δ+

δ−

τ−

∞

=

δ+

δ−

τ−

∞

=

δ+

δ−

τ−

πξξδ

α−πα

=

=

πξξ

α−δπ

=

=

πξξ

π−δπ

Φ=

1n

2/L

2/L

/tn

2m

1n

2/L

2/L

/tn

22

m

1n

2/L

2/L

/tn

22

11

L

nsind

1

w

xnsine1

n

1I

4

L

nsind

w

xnsine1

n

1

w

LI4

L

nsind

L

xnsine1

n

1LG2t,xi

2

2

2&


where 2/wGI 1m Φ= & . The last integral can be made explicit, as follows:

( )

( )

πδ

ππδ

=

=

δ+π−

δ−ππδ

=

πξξδ ∫

δ+

δ−

L2

nsin

2

nsin

n

L2

L2

Lncos

L2

Lncos

n

L

L

nsind

12/L

2/L

Note that, since sin (nπ/2) = 0 for even n, it’s possible to restrict la summation to the odd n only. More-

over, note that

( ) ( )α

πδ+α

πδ=

α+πδ=

πnsin

L2

ncosncos

L2

nsinn

L2

nsin

2

nsin

Thus, assuming δ << L :

( )

( )( ) ( ) ( )α≅

α

πδ+α

πδ

πδ

πδ

=

πξξδ ∫

δ+

δ−

nsinnsinL2

ncosncos

L2

nsin

L2

n

L2

nsin

L

nsind

12/L

2/L

Note that this corresponds to a first order approximation of the original integral:

( )

( )( )α=δ

δ+π

δ≅

δ+πξξ

δ=

πξξδ ∫∫

δ+δ+

δ−

nsinw2

wnsin

1

w2

nsind

1

L

nsind

1w

w

2/L

2/L

and finally:

( ) ( )∑∞

=

τ− α

α−πα

=

nodd1n

/tn

2m1 nsinw

xnsine1

n

1I

4t,xi

2

If the external field ramp is stopped at time t1 and the field is kept constant, the external voltage drops to

zero. In this case the induced currents start decaying. In order to prove this, we start noting that, since

the external voltage drop to zero for t > t1, the external voltage is time dependent and it can found multi-

plying v(ξ) by U( t1 − t). Thus, the solution can be written as follows:

( ) ( ) ( ) ( )τ−ξτ−ξτξ= ∫∫ 1

t

0

L

0

tUt,,xddL

2t,x vKi

Therefore for t > t1 this equation gives

( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )ξ−ξξ−ξξξ=

=ξ

′ξ′ξ=

=ξτ−ξτξ=

∫∫

∫∫

∫∫

−

vKvK

vK

vKi

1*

L

0

*

L

0

t

tt

L

0

t

0

L

0

tt,,xdL

2t,,xd

L

2

t,,xtddL

2

t,,xddL

2t,x

1

1


These last two integrals are the same previously solved. Thus, we can evaluate directly them as follows:

( ) ( )

( ) ( )

( ) ( ) 1

nodd1n

/t/tt

2m

1

nodd1n

/tt

2m

1

nodd1n

/t

2m

nsinw

xnsinee

n

1I

4

nsinw

xnsine1

n

1I

4

nsinw

xnsine1

n

1I

4t,x

nn1

n1

n

b

b

bi

∑

∑

∑

∞

=

τ−τ−−

∞

=

τ−−

∞

=

τ−

α

α−πα

=

=α

α−πα

−

+α

α−πα

=

Therefore, factorizing out the exponential, it can be seen that each component under the sum decays with

its corresponding time constant τn. Thus, the strand currents are given by:

( ) ( ) ( ) 1/)tt(

n1n

/t

2mn1n1 ensin

w

xnsine1

n

1I

4t,x bi

α

α−πα

= τ−−∞

=

τ−∑odd

4. Convergence Analysis

The calculation of the function Γk may cause convergence difficulties, since the summation terms are

oscillating.

( )

( ) ( )

ξ+πϑ−

ξ−πϑ=

ξ+π−

ξ−π=

=

πξ

π=ξΓ

πγλ

πγλ

λ−

∞

=

πγλ

λ−

∞

=

πγλ

λ−

∞

=

πγ

−λ−

∑∑

∑

2

kk

2

kkk

2

kk

2

k

2

kk

2

k

2

kk

L

t

3L

t

3

rt

1n

L

tnrt

1n

L

tnrt

1n

L

n1r

t

k

e,L2

xe,

L2

x

4

e

L

xncose

2

e

L

xncose

2

e

L

nsin

L

xnsinet;,x

It can be demonstrated 5 that the elliptic theta function ϑ3 admit the following representation (where

the summation terms are non-oscillating):

( )( )

∑∑∞

−∞=

π−−∞

−∞=

−− π==ϑn

s

nu

n

sns3

2

2

es

nu2cosee,u

Thus the function Γk admits the following alternative representation:


( )

∑

∑

∑

∞

−∞=

−ξ+γλ+λ−

−ξ−γλ+λ−

∞

−∞=

−ξ+γλ

−ξ−γλλ−

∞

−∞=

π−ξ+ππ

γλ

π−ξ−ππ

γλλ−

πγλ

πγλ

λ−

−πγλ

−=

=

−πγλ

−=

=

−πγλ−=

=

ξ+πϑ−

ξ−πϑ=ξΓ

n

nL2

x

t

rtnL

2

x

t

rt

kk

n

nL2

x

tnL

2

x

tkk

rt

n

nL2

x

tLn

L2

x

tL

kk

rt

L

t

3L

t

3

rt

k

2kk

k

2kk

k

2kk

2kk

k

2

2kk2

2

2kk2

k

2

kk

2

kkk

eet4

L

eet4

eL

eet4

eL

e,L2

xe,

L2

x

4

et;,x

Moreover, for what concerns Γk*, we get, after defining ττ=θ⇒τ=θ d

d2 :

( ) ( )

∑ ∫∫

∑∫

∑∫∫

∞

−∞=

−ξ+θ

γλ+λ

θ−

−ξ−θ

γλ+λ

θ−

∞

−∞=

−ξ+θ

γλ+λ

θ−

−ξ−θ

γλ+λ

θ−

∞

−∞=

−ξ+τγλ+

λτ−

−ξ−τγλ+

λτ−

θπγλ

−−θπγλ

−=

=

−θπγλ

−=

=τ

−πτ

γλ−=ττξΓ=ξΓ

n

t

0

nL2

xr

kk

t

0

nL2

xr

kk

n

nL2

xrnL

2

xrt

0

kk

n

nL2

xrnL

2

xr

kk

t

0

t

0

kk

de2

Lde

2

L

eed2

L

dee4

Ld;,xt;,x*

2

2kk

k

22

2kk

k

2

2

2kk

k

22

2kk

k

2

2kk

k

2kk

k

Consider now the summation term:

θπγλ−=θ

πγλ− ∫∫

θγλ−

−λ

θ−γ−−θ

γλ+λ

θ−

dee2

Lde

2

Lt

0

Qr

rQ2kk

t

0

Qr

kk

2

kk

kk

2

2kk

k

2

Defining r

Qr4

y

r2

yQry k

kk

2k

kkk

γ−λ+

λ+

λ=θ⇒

θγλ−−

λθ= we have that:

dyrQ4y

y1

r2

1d

k2

k

γ−++

λ=θ

Therefore:

dyrQ4y

y1ee

r4

Lde

2

L

k2

Qt

rt

yrQ2kk

t

0

Qr

kk

kk

k2

k

2

2kk

k

2

γ−++

πγ

−λ

=θπγλ

− ∫∫

γλ−−λ

∞−

−γ−−θγλ+

λθ−


Defining the functions ( ) dye2

xerf

x

0

y2

∫−

π= and erfc (x) = 1 − erf (x), we have:

( )

γ−++

γ−+−

+

γλ−−λ

π+∞−π−πγ−λ=θ

πγλ−

∫∫

∫γλ−−

λ −∞− −

γ−−θγλ+

λθ−

Qt

rt

0 k2

y

0 k2

y

kk

k

rQ2kk

t

0

Qr

kk

kk

k 22

k

2

2kk

k

2

dyrQ4y

yedy

rQ4y

ye

Qt

rterf

2erf

2e

r4

Lde

2

L

Defining ydywdwrQ4yw k22 =⇒γ−+= we have:

γ−−

γ−+π=

==γ−+

∫∫γ−+

γ−

−γ−−

kk2

rQ4X

rQ4

wrQ4X

0 k2

y

rQ4erfrQ4Xerf2

dweedyrQ4y

yek

2

k

2k

2

Thus:

γλ−+

λ−

γλ−−

λ−

γ−

λ=

=

γλ−+

λ+−

+

γλ−−

λ+

γ−

λ=θ

πγλ

−

γ−γ−−γ−−

γ−γ−

γ−−θγλ+

λθ−

∫

Qt

rterfceQ

t

rterfcee2

r8

L

Qt

rterfee

Qt

rterf1e

r8

Lde

2

L

kk

k

rQ2kk

k

rQ2rQ2kk

kk

k

rQ4rQ4

kk

k

rQ2kk

t

0

Qr

kk

kkk

kk

k

2

2kk

k

2

and finally:

( )

−ξ+γλ−

+λ

+

−ξ+γλ−

−λ

+

+

−ξ−γλ−

+λ

−

−ξ−γλ−

−λ

−

+−γ−λ=ξΓ

γ−−ξ+γ−−ξ+−

γ−−ξ−γ−−ξ−−

γ−−ξ+−∞

−∞=

γ−−ξ−−∑

nL2

x

t

rterfcenL

2

x

t

rterfce

nL2

x

t

rterfcenL

2

x

t

rterfce

e2e2r8

Lt;,x*

kk

k

rnL2xkk

k

rnL2x

kk

k

rnL2xkk

k

rnL2x

rnL2x

n

rnL2xkkk

kk

kk

kk

Note that the first two terms are time-independent and, therefore, they represent the regime solution.

Thus:


( ) ( )

−ξ+γλ−

+λ

+

−ξ+γλ−

−λ

+

+

−ξ−γλ−

+λ

−

−ξ−γλ−

−λ

−

γ−

λ+∞ξΓ=ξΓ

γ−−ξ+γ−−ξ+−

γ−−ξ−γ−−ξ−−

∞

−∞=∑

nL2

x

t

rterfcenL

2

x

t

rterfce

nL2

x

t

rterfcenL

2

x

t

rterfce

r8

L;,x*t;,x*

kk

k

rnL2xkk

k

rnL2x

kk

k

rnL2xkk

k

rnL2x

n

kkkk

kk

kk

Note that the function Γk* has two (unessential) singularity points for t → 0 and for t → +∞; in fact

erfc(−∞) → 2 and erfc(+∞) → 0.

5. Regime Solution

Note that since ( ) 0t;,x*lim k0t

=ξΓ→

it’s possible to evaluate easily the regime value of Γk*. In fact:

( )

∑

∑ ∫∫

∑∫

∞

−∞=

−ξ+γλ

λ−=

−ξ−γλ

λ−=

∞

−∞=

∞

−ξ+τγλ+

λτ−

∞

−ξ−τγλ+

λτ−

∞

−∞=

−ξ+τγλ+

λτ−

−ξ−τγλ+

λτ−∞

−

πγλ

−=

=

ττ

−ττπ

γλ−=

=τ

−πτ

γλ−=∞ξΓ

n

nL2

x

t

rs

nL2

x

t

rs

kk

n 0

nL2

xr

0

nL2

xr

kk

n

nL2

xrnL

2

xr

kk

0

k

t

e

t

e

4

L

de

de

4

L

dee4

L;,x*

2kk

k

2kk

k

2kk

k

2kk

k

2kk

k

2kk

k

LL

Recalling that ast4/a

s est

e −− π=

L 6, it follows:

( )

∑

∑

∞

−∞=

−ξ+γ−−−ξ−γ−−

∞

−∞=

−ξ+γλ

λ−=

−ξ−γλ

λ−=

−

γ−

λ=

=

−

πγλ

−=∞ξΓ

n

nL2

xr2nL

2

xr2

kk

n

nL2

x

t

rs

nL2

x

t

rs

kkk

kk

2kk

k

2kk

k

eer4

L

t

e

t

e

4

L;,x* LL

Since 0 < x < L and 0 < ξ < L, it’s simple to show that, separating the summation on the positive and

negative indexes, we have


( )

=

−+−+

−

γ−

λ=∞ξΓ

∑∑

∑∑

∞

=

−ξ+γ−∞

=

−ξ−γ−

ξ+γ−−ξ−γ−−

−

−∞=

−ξ+γ−−−

−∞=

−ξ−γ−−

1n

nL2

xr2

1n

nL2

xr2

2

xr2

2

xr2

1

n

nL2

xr21

n

nL2

xr2

kkk

kkkk

kk

eeee

eer4

L;,x*

( ) ( )

( ) ( ) ( )

( ) ( ) ( )

−+

−

γ−ξ+−γ−ξ−γ−

λ=

=

−

+−+

−

γ

−λ

=

γ−ξ+−γ−ξ−−

γ−

∞

=

γ−−γ−ξ+−∞

=

γ−−γ−ξ−−γ−ξ+−γ−ξ−−

∞

=

γ−−γ−ξ+−∞

=

γ−−γ−ξ−−

∑∑

∑∑

2

ee

1e

rxcoshrxcosh

r2

L

eeeeee

eeeer4

L

kk

k

kkkkkk

kkkk

rxrx

rL2

kkkk

1n

nrL2rx

1n

nrL2rxrxrx

1n

nrL2rx

1n

nrL2rxkk

where it has been used the fact that

1q

1

1q

1n

n

−=∑

∞

=

. Finally:

( )( ) ( ) ( )

−

γ−ξγ−−−γ

−λ=∞ξΓγ−


1e

rsinhrxsinh

4

ee

rL;,x*

k

kk

rL2

kkrxrx

kkk

6. Non-linear Case

If the resistance of the strand is a known function of the current, i.e.

( )( )t,xirr jj = , with j = 1, …, N

the problem (P) can still reduced to a diagonal form. Assigning

[ ]( )∑

−−=

λ=p

1pk

TkkkM bb [ ]

( )∑

−−=

γ=p

1pk

TkkkG bb [ ] ∑

=

=N

1j

TjjjrR ee

where ej denote the unit vectors of the Cartesian basis, we have (with k ≠ 0):

[ ][ ] ( ) [ ][ ] ( ) ( ) [ ] ( )

( )[ ] [ ][ ]( )

( )[ ] ( )[ ] ( )[ ]

γ−∂∂+γ+

∂∂λγ=

=−∂∂++

∂∂=

∑ ∑−−= =

t,xt,xx

t,xrt,xt

t,xGt,xx

t,xRGt,xt

MG0

Tkk

Tk2

2Th

p

1ph

N

1j

hTjj

Tkjk

Tkkkk

2

2

vbibibbeebibb

vi

ii

and the following system of N−1 equations (k = −(p−1), …,−1, 1, …, p) is obtained:


(Pk)

( )[ ] ( )[ ] ( )[ ] [ ][ ]( )

( )[ ]( ) ( )( ) ( )( )

==

====

−=∂∂

γ+

∂∂λ ∑ ∑

−−= =

x0t,x

0t,Lxt,0x

t,xrt,xt,xx

1t,x

t

0Tk

Tk

Tk

Tk

Th

p

1ph

N

1j

hTjj

Tkj

Tk

Tk2

2

k

Tkk

ibib

ibib

ibbeebvbibib

Nevertheless, the equations are still coupled since ( )( ) [ ] ( )[ ]( )

== ∑

−−=

p

1ph

Thh

Tjjj t,xrt,xirr ibbe , with j = 1,

…, N. Anyway, note that for k = 0, we get:

( )[ ]( ) ( ) ( )

( ) ( )( ) ( )

====

====

=∂∂

N

0tIx0t,x

N

tIt,Lxt,0x

0t,xx

0T0

T0

T0

T0

T02

2

ibib

ibib

ib

with the usual solution: ( ) ( )N

tIt,xT

0 =ib . Thus, the solution of problem (P) results to be:

( ) ( ) ( )( )( )∑

≠−−=

+=p

0k1pk

Tkk0 t,x

N

tIt,x ibbbi

Finally, in the particular case for which

( )( )tIrrj = , with j = 1, …, N

it’s still possible to solve exactly the problem (P). In fact, with the same symbols of §2:

( )( ) ( )

( ) ( )

==η

ν=η

πγ

−+η

λ

0n

Tkk,n

k,nk,n

2

k

k,nk

0t

tL

n1tIr

dt

d

ib

⇒

⇒ ( ) ( )( ) ( )( )( )

( )( ) ( )

τ∫

τνλ

+∫

=η

πγλτ−+′′

λ−

πγλ

+′′λ−

τ∫ de1

et

2

kk

t

k

2

kk

t

0k L

nttdtIr

1t

0

k,nk

L

nttdtIr

1

0n

Tkk,n ib

Thus the solution, formally, is unchanged, provided that the function Γk is defined as:

( )( )( )

( )( )

ξ+πϑ−

ξ−πϑ

∫

=

=

πξ

π∫=τξΓ

πγλ

πγλ

′′λ−

∞

=

πγλ

+′′λ−

+τ

τ

+τ

τ∑

2

kk

2

kk

t

k

2

kk

t

k

L

t

3L

t

3

tdtIr1

1n

L

nttdtIr

1

k

e,L2

xe,

L2

x

4

e

L

nsin

L

xnsine,t;,x


Therefore the Green matrixes are expresses by:

( )( )[ ] ( )( )∑

≠−−=

ξΓ=ξp

0k1pk

Tkkk

0 0,t,,x0,t,,xK bb ( )[ ] ( )( )∑

≠−−= λ

τξΓ=τξp

0k1pk k

Tkk

k ,t,,x,t,,xKbb

and the solution of (P) can be written as:

( ) ( ) ( )( )[ ] ( )( ) ( )[ ] ( )τξττ−ξτξ+ξξξ+= ∫∫∫ ,,t,,xKddL

20,t,,xKd

L

2

N

tIt,x

t

0

L

0

00

L

0

0 vibi

7. Flux Ramp

In the particular case in which i(0)

(x) has equal components and v(x, t) is given by:

( ) ( )

><

=0

0

tt if ,0

tt if ,x*t,x

vv

the solution of (P) can be written as:

( ) ( ) ( )[ ] ( )τξτ−ξτξ+= ∫∫ ,t,,xKddL

2

N

tIt,x

t

0

L

0

0 vbi

that can be specialized in two cases:

I) t < t0

( )[ ] ( ) ( )[ ] ( ) ( )[ ] ( )ξξξ=ξ

τ−ξτξ=τξτ−ξτξ ∫∫∫∫∫ *t,,x*Kd*t,,xKdd,t,,xKdd

L

0

t

0

L

0

t

0

L

0

vvv

II) t > t0

( )[ ] ( ) ( )[ ] ( ) ( )[ ] ( )

( )[ ] ( ) ( )[ ] ( )ξ−ξξ−ξξξ=

=ξ

′ξ′ξ=τξτ−ξτξ=τξτ−ξτξ

∫∫

∫∫∫∫∫∫−

*tt,,x*Kd*t,,x*Kd

*t,,xKtdd,t,,xKdd,t,,xKdd

0

L

0

L

0

t

tt

L

0

t

0

L

0

t

0

L

0 0

0

vv

vvv

thus, after same manipulation, we obtain:

I) t < t0

( ) ( ) ( )[ ] ( )ξξξ+= ∫ *t,,x*KdL

2

N

tIt,x

L

0

0 vbi

II) t > t0

( ) ( ) ( )[ ] ( ) ( )[ ] ( )ξ−ξξ−ξξξ+= ∫∫ *tt,,x*KdL

2*t,,x*Kd

L

2

N

tIt,x 0

L

0

L

0

0 vvbi

In particular, the last integral can be written as:


( )[ ] ( ) ( ) ( )000

0

L

0

tt,xN

ttI*tt,,x*Kd

L

2 −+−−=ξ−ξξ∫ ibv

and therefore:

( ) ( ) ( ) ( )[ ] ( ) ( )0

L

0

00 tt,x*t,,x*Kd

L

2

N

ttItIt,x −−ξξξ+−+= ∫ ivbi

N.B. In the particular case in which v* is a piecewise constant function, we get:

( ) ( )∑=

+=M

1m

1mm*m x,x;xUx* vv

(with x1 = 0 and xM+1 = L) and where:

( )

><<

<=

bx fi,0

bxa fi,1

ax fi,0

b,a;xU

We have:

I) t < t0

( ) ( ) ( )[ ]∑=

++=M

1m

*m1mm0 x,x;t,x**K2

N

tIt,x vbi

II) t > t0

( ) ( ) ( )[ ] ( )[ ]∑∑=

+=

+ −−+=M

1m

*m1mm0

M

1m

*m1mm0 x,x;tt,x**K2x,x;t,x**K2

N

tIt,x vvbi

where (for 0 ≤ a < b ≤ L):

( )[ ] ( )[ ] ( )( )

( )( )∑∑∫∫

≠−−=

≠−−= λ

Γ=λ

ξΓξ=ξξ=p

0k1pk k

Tkk**

k

p

0k1pk k

Tkk*

k

b

a

b

ab,a;t,xt,,xd

L

1t,,x*Kd

L

1b,a;t,x**K

bbbb

( ) ( )t,,xdL

1b,a;t,x *

k

b

a

**k ξΓξ=Γ ∫

From the definition, we get:


( ) ( )

−ξ+γλ−

+λ

+

−ξ+γλ−

−λ

+

+

−ξ−γλ−

+λ

−

−ξ−γλ−

−λ

−

γ−

λ+∞ξΓ=ξΓ

γ−−ξ+γ−−ξ+−

γ−−ξ−γ−−ξ−−

∞

−∞=∑

nL2

x

t

rterfcenL

2

x

t

rterfce

nL2

x

t

rterfcenL

2

x

t

rterfce

r8

L;,x*t;,x*

kk

k

rnL2xkk

k

rnL2x

kk

k

rnL2xkk

k

rnL2x

n

kkkk

kk

kk

( )( ) ( ) ( )

−


−λ=∞ξΓγ−


1e

rsinhrxsinh

4

ee

rL;,x*

k

kk

rL2

kkrxrx

kkk

Therefore:

( ) ( )( ) ( ) ( )

−


−λξ=

=∞ξΓξ=∞Γ

γ−


∫

∫

1e

rsinhrxsinh

4

ee

rd

;,xdL

1b,a;,x

k

kk

rL2

kkrxrx

kk

b

a

*k

b

a

**k

Assuming ( ) krxs γ−−ξ=− , ( ) krxs γ−+ξ=+ , krs γ−ξ= and thus krddsdsds γ−ξ=== +− ,

it result:

( )( )( )

( )( ) ( ) ( )

( )

( )

( )

( )

( ) ( )[ ]

( ) ( )

( ) ( )[ ] b

akrL2

kk

b

a

rxkb

a

rxk

rbs

rasrL2

kk

rxbs

rxas

sk

rxbs

rxas

sk

rb

rarL2

kkrxb

rxa

skrxb

rxa

sk**k

rcosh1e

rxsinh

r

er4

e1xsgnr4

scosh1e

rxsinh

r

er4

e1)tsgn(r4

dsssinh1e

rxsinh

rdse

r4dse

r4b,a;,x

k

kk

k

kk

k

k

k

k

k

kk

k

k

k

k

=ξ=ξγ−

=ξ

=ξ

γ−+ξ−=ξ

=ξ

γ−−ξ−

γ−=γ−=γ−

γ−+=

γ−+=

−γ−−=

γ−−=

−−

γ−

γ−+

γ−

γ−+

γ−++−γ−−

γ−−−−

γ−ξ−

γ−λ−

+

λ

+

−−ξ

λ=

−

γ−λ−

+

λ

+

−

λ=

−

γ−λ−

λ−

λ=∞Γ

+

+

+

−

−

−

+−

∫∫∫

and finally (o)

:

( )( ) ( ) ( ) ( )

b

a

rL2

kk

rxrx

k**k

1e

rcoshrxsinh

4

ee1xsgn

rb,a;,x

k

kk

=ξ

=ξ

γ−

γ−+ξ−γ−−ξ−

−

γ−ξγ−−

+

−−ξλ

=∞Γ

Regarding the transient term, it is:

(o) Note that ( )[ ] qu

0u

uq

0

ue1)usgn(due

==

−− −=∫


( ) ( )

−ξ+γλ−

+λ

+

−ξ+γλ−

−λ

+

+

−ξ−γλ−

+λ

−

−ξ−γλ−

−λ

−

ξγ

−λ

+∞Γ=Γ

γ−−ξ+γ−−ξ+−

γ−−ξ−γ−−ξ−−

∞

−∞=∫∑

nL2

x

t

rterfcenL

2

x

t

rterfce

nL2

x

t

rterfcenL

2

x

t

rterfce

dr8

b,a;,xb,a;t,x

kk

k

rnL2xkk

k

rnL2x

kk

k

rnL2xkk

k

rnL2x

b

an

kk**k

**k

kk

kk

Assuming ( ) krnL2xw γ−−ξ−=− , ( ) krnL2xw γ−−ξ+=+ and thus krddwdw γ−ξ==− +− ,

it result:

( ) ( )

( )( )

( )( )

λ+

λ+

λ−

λ+

+

λ+

λ+

λ−

λ

λ=∞Γ−Γ

++−γ−−+

γ−−++

−−−γ−−−

γ−−−−

∞

−∞=

++

−−

∫

∫

∑

wrt2

1rterfcew

rt2

1rterfcedw

wrt2

1rterfcew

rt2

1rterfcedw

r8b,a;,xb,a;t,x

k

k

wk

k

wrnL2bx

rnL2ax

k

k

wk

k

wrnL2bx

rnL2ax

n

k**k

**k

k

k

k

k

Substitution of the relations

( )qu

0u

2

1

uq

0

uuerfeu

2

1erfce

2

1erfc)usgn(duu

2

1erfce

2=

=

Ω−−−

Ω+

Ω−Ω

−

Ω=

Ω−Ω∫

( )qu

0u

2

1

uq

0

uuerfeu

2

1erfce

2

1erfc)usgn(duu

2

1erfce

2=

=

Ω−

Ω−

Ω+Ω

−

Ω−=

Ω+Ω∫

(where rt2

1 kλ=Ω ) leads to:


( ) ( )

( )

( )( )

( )

( )

( )( )

( ) k

k

k

k

k

k

k

k

rnL2bxw

rnL2axw

k

rt

k

k

w

k

k

rt

k

k

w

k

rnL2bxw

rnL2axw

k

rt

k

k

w

k

k

rt

k

k

w

k

n

k**k

**k

wrt2

1erfew

rt2

1rterfce

rterfcwsgn

wrt2

1erfew

rt2

1rterfce

rterfcwsgn

wrt2

1erfew

rt2

1rterfce

rterfcwsgn

wrt2

1erfew

rt2

1rterfce

rterfcwsgn

r8b,a;,xb,a;t,x

γ−−+=

γ−−+=

+λ−

++

+λ−

+−+

γ−−−=

γ−−−=

−λ−

−−

−λ−

−−−

∞

−∞=

+

+

+

+

−

−

−

−

λ−

λ+

λ−

λ−

+

λ+

λ−

λ−

λ

λ−

λ+

λ−

λ−

+

λ+

λ−

λ−

λ

λ=∞Γ−Γ ∑

Simplifying, it result:

( ) ( )

( )

( )

( )

( )

( )

( ) k

k

k

k

k

k

rnL2bxw

rnL2axw

k

rt

k

k

wk

k

w

rnL2bxw

rnL2axw

k

rt

k

k

wk

k

w

n

k

**k

**k

wrt2

1erfe2

wrt2

1rterfcew

rt2

1rterfcewsgn

wrt2

1erfe2

wrt2

1rterfcew

rt2

1rterfcewsgn

r8

b,a;,xb,a;t,x

γ−−−=

γ−−−=

+λ−

+−++

γ−−−=

γ−−−=

−λ−

−−−−∞

−∞=

+

+

++

−

−

−−

λ+

+

λ−

λ−

λ+

λ+

+

λ+

+

λ−

λ−

λ+

λλ

=

=∞Γ−Γ

∑

Substituting:


( ) ( )

( )

( )

b

a

kk

rt

kk

k

rnL2x

kk

k

rnL2x

kk

rt

kk

k

rnL2x

kk

k

rnL2x

n

k**k

**k

nL2xt2

1erfe2nL2x

t2

1rterfce

nL2xt2

1rterfcenL2xsgn

nL2xt2

1erfe2nL2x

t2

1rterfce

nL2xt2

1rterfcenL2xsgn

r8b,a;,xb,a;t,x

kk

k

kk

k

=ξ

=ξ

λ−

γ−−ξ+−

γ−−ξ+

λ−

γ−−ξ−−

γ−−ξ−

∞

−∞=

−ξ+

γλ−+

−ξ+

γλ−−

λ−

+

−ξ+

γλ−+

λ−ξ++

+

−ξ−

γλ−+

−ξ−

γλ−−

λ−

+

−ξ−

γλ−+

λ−ξ−

λ+∞Γ=Γ ∑

Taking into account that for convergence regions it’s convenient to substitute erf = 1 − erfc, we get:

( ) ( ) ( ) ( )[ ]

( )

( )

b

a

kk

rt

kk

k

rnL2x

kk

k

rnL2x

kk

rt

kk

k

rnL2x

kk

k

rnL2x

n

k

n

ba

rt

k**k

**k

nL2xt2

1erfce2nL2x

t2

1rterfce

nL2xt2

1rterfcenL2xsgn

nL2xt2

1erfce2nL2x

t2

1rterfce

nL2xt2

1rterfcenL2xsgn

r8

nL2xsgnnL2xsgner4

b,a;,xb,a;t,x

kk

k

kk

k

k

=ξ

=ξ

λ−

γ−−ξ+−

γ−−ξ+

λ−

γ−−ξ−−

γ−−ξ−

∞

−∞=

∞

−∞=

=ξ=ξ

λ−

−ξ+

γλ−−

−ξ+

γλ−−

λ−

+

−ξ+

γλ−+

λ−ξ++

+

−ξ−

γλ−−

−ξ−

γλ−−

λ−

+

−ξ−

γλ−+

λ−ξ−

λ+

−ξ++−ξ−λ

+∞Γ=Γ

∑

∑

Breaking the first series in positive and negative terms, we have:

( ) ( )[ ] ( ) ( )[ ]

( ) ( ) ( ) ( )[ ]

( )[ ] ( )[ ] ba

rt

kba

rt

k

1n

ba

rt

k

ba

rt

k

n

ba

rt

k

xsgner4

1xsgner4

nL2xsgnnL2xsgnnL2xsgnnL2xsgner4

xsgnxsgner4

nL2xsgnnL2xsgner4

kk

k

kk

=ξ=ξ

λ−

=ξ=ξ

λ−

∞

=

=ξ=ξ

λ−

=ξ=ξ

λ−∞

−∞=

=ξ=ξ

λ−

ξ−λ

=+ξ−λ

=

=+ξ+++ξ−+−ξ++−ξ−λ

+

+ξ++ξ−λ

=−ξ++−ξ−λ

∑

∑

Note that, being n > 1, 0 < x < L and 0 < ξ < L, it result x − ξ − 2nL < 0, x + ξ − 2nL < 0, x − ξ + 2nL >

0, x + ξ + 2nL > 0 and thus the sign series is null. Substitution leads to:


( ) ( ) ( )[ ]

( )

( )

b

a

kk

rt

kk

k

rnL2x

kk

k

rnL2x

kk

rt

kk

k

rnL2x

kk

k

rnL2x

n

kba

rt

k**k

**k

nL2xt2

1erfce2nL2x

t2

1rterfce

nL2xt2

1rterfcenL2xsgn

nL2xt2

1erfce2nL2x

t2

1rterfce

nL2xt2

1rterfcenL2xsgn

r8xsgne

r4b,a;,xb,a;t,x

kk

k

kk

k

k

=ξ

=ξ

λ−

γ−−ξ+−

γ−−ξ+

λ−

γ−−ξ−−

γ−−ξ−

∞

−∞=

=ξ=ξ

λ−

−ξ+

γλ−−

−ξ+

γλ−−

λ−

+

−ξ+

γλ−+

λ−ξ++

+

−ξ−

γλ−−

−ξ−

γλ−−

λ−

+

−ξ−

γλ−+

λ−ξ−

λ+ξ−

λ+∞Γ=Γ ∑

Note that the erfc functions in the series do not tend a 0, for increasing n, since the argument is negative

and decreasing. In order to improve the convergence the following substitution can be done: erfc(−x) = 1

− erf(−x) = 1 + erf(x) = 2 − erfc(x). The remaining series can be treated separately:

( ) ( ) ( )[ ]

( ) ( )

( )

( )

b

a

kk

rt

kk

k

rnL2x

kk

k

rnL2x

kk

rt

kk

k

rnL2x

kk

k

rnL2x

n

k

n

b

a

rnL2xrnL2xk

ba

rt

k**k

**k

nL2xt2

1erfce2nL2x

t2

1rterfce

nL2xt2

1rterfcenL2xsgn

nL2xt2

1erfce2nL2x

t2

1rterfce

nL2xt2

1rterfcenL2xsgn

r8

enL2xsgnenL2xsgnr4

xsgner4

b,a;,xb,a;t,x

kk

k

kk

k

kk

k

=ξ

=ξ

λ−

γ−−ξ+−

γ−−ξ+

λ−

γ−−ξ−−

γ−−ξ−

∞

−∞=

∞

−∞=

=ξ

=ξ

γ−−ξ+−γ−−ξ−−

=ξ=ξ

λ−

−ξ+

γλ−−

−ξ+

γλ−+

λ−+

+

−ξ+

γλ−+

λ−ξ++

+

−ξ−

γλ−−

−ξ−

γλ−+

λ−+

+

−ξ−

γλ−+

λ−ξ−

λ+

+

−ξ++−ξ−

λ−

+ξ−λ

+∞Γ=Γ

∑

∑

In fact:


( ) ( )

( ) ( )

( ) ( )

( ) ( ) =+ξ+++ξ−+

+−ξ++−ξ−

λ−

ξ++ξ−

λ−=

=

−ξ++−ξ−

λ−

γ−+ξ+−γ−+ξ−−

γ−−ξ+−γ−−ξ−−

∞

=


∞

−∞=

γ−−ξ+−γ−−ξ−−

∑

∑

kk

kk

kk

kk

rnL2xrnL2x

rnL2xrnL2x

1n

krxrxk

n

rnL2xrnL2xk

enL2xsgnenL2xsgn

enL2xsgnenL2xsgn

r4exsgnexsgn

r4

enL2xsgnenL2xsgnr4

( ) ( )

( ) ( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( ) ( )

( ) ( )

( ) ( ) ( ) ( )1e

rcoshrxsinh

reexsgn

r4

rcoshe2rcoshe2

1e

1

r4eexsgn

r4

eeee

er4

eexsgnr4

eeee

r4eexsgn

r4

k

kk

kk

k

kk

kkkk

kkk

kkkk

kk

rL2

kkkrxrxk

krx

krx

rL2

krxrxk

rxrxrxrx

1n

rnL2krxrxk

rnL2xrnL2xrnL2xrnL2x

1n

krxrxk

−

γ−ξγ−λ+

+ξ−

λ−=

=γ−ξ+

γ−ξ−

−

λ−

+ξ−

λ−=

=++

−−

λ−

+ξ−

λ−=

=++

−−

λ−

+ξ−

λ−=

γ−γ−ξ+−γ−ξ−−

γ−−γ−

γ−γ−ξ+−γ−ξ−−

γ−ξ+−γ−ξ−−γ−ξ+γ−ξ−

∞

=

γ−−γ−ξ+−γ−ξ−−

γ−+ξ+−γ−+ξ−−γ−−ξ+γ−−ξ−

∞

=


∑

∑

Therefore, finally:

( ) ( )

( ) ( ) ( ) ( )

( )

( )

b

a

kk

rt

kk

k

rnL2x

kk

k

rnL2x

kk

rt

kk

k

rnL2x

kk

k

rnL2x

n

k

b

a

rL2

kkkrxk

rt

rxk

**k

**k

nL2xt2

1erfce2nL2x

t2

1rterfce

nL2xt2

1rterfcenL2xsgn

nL2xt2

1erfce2nL2x

t2

1rterfce

nL2xt2

1rterfcenL2xsgn

r8

1e

rcoshrxsinh

re

r4eexsgn

r4

b,a;,xb,a;t,x

kk

k

kk

k

k

kkk

=ξ

=ξ

λ−

γ−−ξ+−

γ−−ξ+

λ−

γ−−ξ−−

γ−−ξ−∞

−∞=

=ξ

=ξ

γ−γ−ξ+−λ

−γ−ξ−−

−ξ+

γλ−−

−ξ+

γλ−+

λ−+

+

−ξ+

γλ−+

λ−ξ++

+

−ξ−

γλ−−

−ξ−

γλ−+

λ−+

+

−ξ−

γλ−+

λ−ξ−

λ+

+

−

γ−ξγ−λ+

λ−

−ξ−

λ−

+∞Γ=Γ

∑


8. Perfect Conductor Case (r = 0)

In the particular case in which r → 0, it’s possible to specialize the form of Γ* and of Γ** as follow:

( )( ) ( ) ( )

( )

( )

ξ−

ξ−−ξ+γλ−=

−γ−+γ−ξγ−

−γ−ξ++−γ−ξ−−γ

−λ=

−


−λ=∞ξΓ

→

γ−


L2

x

4

xxL

1rL21

rrx

4

rx1rx1

rL

1e

rsinhrxsinh

4

ee

rL;,x*

kk

k

kkkkkk

0r

rL2

kkrxrx

kkk

k

kk

Thus:

( ) ( )

ξ−

ξ−−ξ+γλ−=∞ξΓ

→

L

x

2

xx

2

L;,x* kk

0r

k

Moreover:

( ) ( )

−ξ+γλ−

+λ

+

−ξ+γλ−

−λ

+

+

−ξ−γλ−

+λ

−

−ξ−γλ−

−λ

−

γ−

λ=∞ξΓ−ξΓ

γ−−ξ+γ−−ξ+−

γ−−ξ−γ−−ξ−−

∞

−∞=∑

nL2

x

t

rterfcenL

2

x

t

rterfce

nL2

x

t

rterfcenL

2

x

t

rterfce

r8

L;,x*t;,x*

kk

k

rnL2xkk

k

rnL2x

kk

k

rnL2xkk

k

rnL2x

n

kkkk

kk

kk

thus

( ) ( )

( )

( ) +

λπ−

−ξ−γλ−

−γ−−ξ−−−

+

λπ−

−ξ−γλ−

+γ−−ξ−+−

γ−

λ=∞ξΓ−ξΓ

−ξ−γλ

−ξ−γλ

∞

−∞=

→

∑

k

nL2

x

tkkk

k

nL2

x

tkkk

n

kk0r

kk

rte

2nL

2

x

terf1rnL2x1

rte

2nL

2

x

terf1rnL2x1

r8

L;,x*t;,x*

2kk

2kk

( )

( )

λπ−

−ξ+γλ−

−γ−−ξ+++

+

λπ−

−ξ+γλ−

+γ−−ξ+−+

−ξ+γλ

−ξ+γλ

k

nL2

x

tkkk

k

nL2

x

tkkk

rte

2nL

2

x

terf1rnL2x1

rte

2nL

2

x

terf1rnL2x1

2kk

2kk


−γλ−π

+

+

−ξ+γλ−

−ξ+−

−ξ−γλ−

−ξ−

γ−γ

−λ

=

−ξ+γλ

−ξ−γλ

∞

−∞=∑

2kk

2kk nL

2

x

tnL

2

x

t

kk

kkkk

n

kkk

eet2

nL2

x

terfnL2xnL

2

x

terfnL2x

r2r8

L

and finally:

( ) ( )

−γλ−π

+

+

−ξ+

γλ−−ξ+−

−ξ−

γλ−−ξ−

λγ−=∞ξΓ−ξΓ

−ξ+γλ

−ξ−γλ

∞

−∞=

→

∑

2kk

2kk nL

2

x

tnL

2

x

t

kk

kkkk

n

kk0r

kk

eet2

nL2xt2

1erfnL2xnL2x

t2

1erfnL2x

4

L;,x*t;,x*

Note that while for the last term of this series the convergence is assured for n → ±∞, the sum of the first

two terms is semi-convergent since goes to ±2ξ for n → ±∞. In order to improve the convergence for n

→ ±∞ the following the operations are performed:

1° the substitution erf(x) = 1 − erfc(x)

2° the separation of the series from − ∞ to +∞ in two series (changing sign to the negative n):

( ) ( ) [ ]

−γλ−

πλγ

−

−ξ+

γλ−−ξ++

−ξ−

γλ−−ξ−−

λγ−−ξ+−−ξ−

λγ−=∞ξΓ−ξΓ

−ξ+γλ

−ξ−γλ∞

−∞=

∞

−∞=

∞

−∞=

→

∑

∑∑

2kk

2kk nL

2

x

tnL

2

x

t

kkn

kk

kkkk

n

kk

n

kk0r

kk

eet2

4

L

nL2xt2

1erfcnL2xnL2x

t2

1erfcnL2x

4

LnL2xnL2x

4

L;,x*t;,x*

[ ]

∑

∑

∞

=

∞

=

+ξ+

γλ−+ξ++

+ξ−

γλ−+ξ−−

λγ−

+

ξ+

γλ−ξ+−

ξ−

γλ−ξ−

λγ+

+

+ξ+−+ξ−+−ξ+−−ξ−+ξ+−ξ−λγ

−=

1n

kkkkkk

kkkkkk

1n

kk

nL2xt2

1erfcnL2xnL2x

t2

1erfcnL2x

4

L

xt2

1erfcxx

t2

1erfcx

4

L

nL2xnL2xnL2xnL2xxx4

L


−γλ−

πλγ

−

−ξ+

γλ−−ξ++

−ξ−

γλ−−ξ−−

−ξ+γλ

−ξ−γλ∞

−∞=∑

2kk

2kk nL

2

x

tnL

2

x

t

kkn

kk

kkkk

eet2

4

L

nL2xt2

1erfcnL2xnL2x

t2

1erfcnL2x

Note that, being n > 1, 0 < x < L and 0 < ξ < L it result x − ξ − 2nL < 0, x + ξ − 2nL < 0, x − ξ + 2nL >

0, x + ξ + 2nL > 0 and thus the first series is null. Moreover, the separation of the last series, evidencing

the zero-th order terms e simplifying, gives:

( ) ( )

γλ−π−

−ξ+

γλ−−ξ++

+γλ−π

+

−ξ−

γλ−−ξ−−

+γλ−π

−

+ξ+

γλ−+ξ++

+γλ−π

+

+ξ−

γλ−+ξ−−

λγ−

+

γλ−π−

ξ+

γλ−ξ+−

+γλ−π

+

ξ−

γλ−ξ−

λγ−=

=∞ξΓ−ξΓ

−ξ+γλ

−ξ−γλ

+ξ+γλ

+ξ−γλ∞

=

ξ+γλ

ξ−γλ

→

∑

2kk

2kk

2kk

2kk

2kk

2kk

nL2

x

t

kk

kk

nL2

x

t

kk

kk

nL2

x

t

kk

kk

nL2

x

t

kk1n

kkkk

2

x

t

kk

kk

2

x

t

kk

kkkk

0r

kk

et2

nL2xt2

1erfcnL2x

et2

nL2xt2

1erfcnL2x

et2

nL2xt2

1erfcnL2x

et2

nL2xt2

1erfcnL2x

4

L

et2

xt2

1erfx

et2

xt2

1erfx

4

L

;,x*t;,x*

It’s now possible to eliminate all the modules thanks to the symmetry of the erf function and being n > 1,

0 < x < L and 0 < ξ < L, so that it result x − ξ − 2nL < 0, x + ξ − 2nL < 0, x − ξ + 2nL > 0, x + ξ + 2nL >

0 e x + ξ > 0:


( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

γλ−π−

−ξ+

γλ−−−ξ+−

+γλ−π

+

−ξ−

γλ−−−ξ−+

+γλ−π

−

+ξ+

γλ−+ξ++

+γλ−π

+

+ξ−

γλ−+ξ−−

λγ−

+

γλ−π−

ξ+

γλ−ξ+−

+γλ−π

+

ξ−

γλ−ξ−

λγ−=

=∞ξΓ−ξΓ

−ξ+γλ

−ξ−γλ

+ξ+γλ

+ξ−γλ∞

=

ξ+γλ

ξ−γλ

→

∑

2kk

2kk

2kk

2kk

2kk

2kk

nL2

x

t

kk

kk

nL2

x

t

kk

kk

nL2

x

t

kk

kk

nL2

x

t

kk1n

kkkk

2

x

t

kk

kk

2

x

t

kk

kkkk

0r

kk

et2

nL2xt2

1erfcnL2x

et2

nL2xt2

1erfcnL2x

et2

nL2xt2

1erfcnL2x

et2

nL2xt2

1erfcnL2x

4

L

et2

xt2

1erfx

et2

xt2

1erfx

4

L

;,x*t;,x*

Regarding Γ**, from the definition it result:

( ) ( )t,,xdL

1b,a;t,x *

k

b

a

**k ξΓξ=Γ ∫

thus (*)

, assigning ( )x−ξ=σ and ξ=σ dd , it result:

( ) ( )( )( )

σσ−

ξ−ξ+γλ−=∞Γ ∫

−

−

=ξ

=ξ

→ xb

xa

b

a

22kk

0r**

k d2

1

L2

x

4

x

2b,a;,x

and finally:

( ) ( ) ( )( ) b

a

222kk

0r**

kL

x

2

xxsgnx

4b,a;,x

=ξ

=ξ

→

ξ−ξ−−ξ−ξ+γλ−=∞Γ

for what concerns the transient term, assigning ( )nL2xt2

1Lx

kk θ+ξ−θγλ−

=ω− ,

( )nL2xt2

1Lx

kk θ+ξ+θγλ−

=ω+ , with θx,θL = 0, ±1 and ξγλ−

=ω=ω− +− dt2

1dd kk it result:

(*)Note that

qu

0u

2q

0 2

u)usgn(duu

=

=

=∫


( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

( ) ( )

ω

λγπ−ωωω

λγ+

+ωλγπ

−ωωωλγ

+

+ωλγπ

+ωωωλγ

−

+ωλγπ

+ωωω

λγ−

λγ−

+

ω

λγπ+ωωω

λγ+

+ωλγπ

+

ωωω

λγλγ

−

=∞Γ−Γ

+ω−

+++

−ω−

−−−

+ω−

+++

−ω−

∞

=−−−

+ω−

+++

−ω−

−−−

→

∫∫

∫∫

∫∫

∫∑ ∫

∫∫

∫∫

+

+

++

+

−

−

−−

−

+

+

++

+

−

−

−−

−

+

+

++

+

−

−

−−

−

det22

derfct4

det22

derfct4

det22

derfct4

det22

derfct4

4

det22

derft4

det22

derft4

4

b,a;,xb,a;t,x

b

akk

b

akk

b

akk

b

akk

b

akk

b

akk

b

akk1n

b

akk

kk

b

akk

b

akk

b

akk

b

akk

kk

0r**

k**

k

2

2

2

2

2

2

(

(

(

(

(

(

(

(

)

)

)

)

)

)

)

)

&&&&&&

&&&&&&

where

( )axt2

1a kk −γλ−=− , ( )bx

t2

1b kk −γλ−=− ,

( )axt2

1a kk +γλ−=+ , ( )bx

t2

1b kk +γλ−=+

( )nL2axt2

1a kk +−

γλ−=−

), ( )nL2bx

t2

1b kk +−

γλ−=−

),

( )nL2axt2

1a kk ++

γλ−=+

), ( )nL2bx

t2

1b kk ++

γλ−=+

)

( )nL2axt2

1a kk −+

γλ−−=−

(, ( )nL2bx

t2

1b kk −+

γλ−−=−

(,

( )nL2axt2

1a kk −−

γλ−−=+

(, ( )nL2bx

t2

1b kk −−

γλ−−=+

(

Simplifying e taking into account the relations

( )qu

0u

u2q

0

uq

0

22

eu

)u(erf2

1udue

2duuerfu2

=

=

−−

π+

+=π

+ ∫∫

( )qu

0u

u2q

0

uq

0

22

eu

)u(erfc2

1udue

2duuerfcu2

=

=

−−

π−

+=π

− ∫∫

we get:


( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

( ) ( ) ( ) ( ) ( ) ( )

πω

−ω

+ω+

πω

−ω

+ω+

+

πω

−ω

+ω

−

πω

−ω

+ω−−

+

πω

+ω

+ω+

πω

+ω

+ω−

=∞Γ−Γ

−

−

−+

+

+

+

+

+−

−

−

+

+

+−

−

−

ω−−−−

ω−+++

ω−+++

∞

=

ω−−−−

ω−+++

ω−−−−

→

∑

b

a

2b

a

2

b

a

2

1n

b

a

2

b

a

2

b

a

2

0r**

k**

k

22

22

22

eerfc2

1eerfc

2

1

eerfc2

1eerfc

2

1

2

t

eerf2

1eerf

2

1

2

t

b,a;,xb,a;t,x

(

(

(

(

)

)

)

)

&&&&&&

and finally:

( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

( ) ( ) ( ) ( )b

a

nL2xt4kkkk2kk

nL2xt4kkkk2kk

nL2xt4kkkk2kk

nL2xt4kkkk2kk

1n

b

a

xt4kkkk2kk

xt4kkkk2kk

0r**

k**

k

2kk

2kk

2kk

2kk

2kk

2kk

enL2xt

nL2xt2

1erfcnL2x

t21

enL2xt

nL2xt2

1erfcnL2x

t21

enL2xt

nL2xt2

1erfcnL2x

t21

enL2xt

nL2xt2

1erfcnL2x

t21

4

tex

tx

t2

1erfx

t21

ext

xt2

1erfx

t21

4

t

b,a;,xb,a;t,x

=ξ

=ξ

−ξ+γλ

−ξ−γλ

+ξ+γλ

+ξ−γλ

∞

=

=ξ

=ξ

ξ+γλ

ξ−γλ

→

−ξ+

πγλ−

+

−ξ+

γλ−−

−ξ+γλ

−+

+−ξ−π

γλ−+

−ξ−

γλ−−

−ξ−γλ

−+

++ξ+π

γλ−+

+ξ+

γλ−

+ξ+γλ

−−

+

+ξ−

πγλ−

+

+ξ−

γλ−

+ξ−γλ

−−

−

ξ+

πγλ−

+

ξ+

γλ−

ξ+γλ

−+

+

ξ−

πγλ−

+

ξ−

γλ−

ξ−γλ

−−

=∞Γ−Γ

∑

References

1 I.S. Gradshteyn, I.M. Ryzhik, Tables of Integrals, Series, and Products, Academic Press: London, 1980, p.

1112.

2 I.S. Gradshteyn, I.M. Ryzhik, Tables of Integrals, Series, and Products, Academic Press, London, 1980, p.

29.


921.

4 L. Krempasky, C. Schmidt, “Theory of Supercurrents and their influence on field quality and stability of su-

perconducting magnets”, J. Appl. Phys. 78 (9) 1995

5 R. Courant, D. Hilbert, Method of Mathematical Physics, Interscience, NY,1962, vol.2, p. 200; vol.1, p. 75.


921.


APPENDIX A – FORCED CURRENT DISTRIBUTION IN A 3-STRAND CABLE

Consider, as shown in the figure, a 3-strand cable with an insulated input strand. Moreover, assume that

the flowing current is equally distributed on the two input strand and on the three output strands.

I(t)/3

I(t)/3

I(t)/3

I(t)/2

I(t)/2

The longitudinal resistance (r) ant the external voltage (v) are assumed to be zero. Consequently, the cur-

rents flowing in the 3-strand cable are described by the following system:

(P)

[ ][ ] ( ) ( )

( )( ) ( ) ( ) ( )

( ) ( ) ( ) ( )

======

======

==

=∂∂+

∂∂

3

tIt,Lxit,Lxit,Lxi

2

tIt,0xit,0xi,0t,0xi

00t,x

0t,xx

t,xt

MG

321

321

2

2

i

ii

and i, v ∈ R3

with I(0) = 0 and where [M] and [G] are the following constant-coefficient circulant symmetric ma-

trixes:

[ ]

=

111212

121112

121211

mmm

mmm

mmm

M [ ]

−−

−=

121212

121212

121212

g2gg

gg2g

ggg2

G

Defining the orthonormal spectral basis bk, with k = 1, 0, −1, that it’s the same for [M] and [G]:

=1

1

1

3

1 0b

−−=

2/1

2/1

1

3

2 1b

−=−

1

1

0

2

1 1b

the matrixes [M] and [G] can be written as

[ ] T111

T111

T000M −−−λ+λ+λ= bbbbbb [ ] T

111T111

T000G −−−γ+γ+γ= bbbbbb

with eigenvalues given by

12110 m2m +=λ 00 =γ

121111 mm −=λ=λ − 1211 g3−=γ=γ −

Decomposing i as follows

( ) ( ) ( ) ( ) 111100 t,xt,xt,xt,x −−η+η+η= bbbi

Problem (P) is simply divided in the following three problems:

(P0)

( )

( ) ( ) ( ) ( ) ( )

==η==η==η

=∂

η∂

3

tIt,Lx,

3

tIt,0x,00t,x

0t,xx

000

2

02


(P−1) ( ) ( )

( ) ( ) ( )

==η==η==η

=∂

η∂+∂η∂λγ

−−−

−−−−

0t,Lx,0t,0x,00t,x

0t,xx

t,xt

111

2

12

111

(P1)

( ) ( )

( ) ( ) ( ) ( )

==η−==η==η

=∂

η∂+

∂η∂

λγ

0t,Lx,6

tIt,0x,00t,x

0t,xx

t,xt

111

2

12

111

Problems (P0) and (P−1) have the simple solutions

( ) ( )3

tIt,x0 =η ( ) 0t,x1 =η−

While problem (P1) requires a different approach. Defining

( ) ( ) ( )

−+η=ϕL

x1

6

tIt,xt,x 1

gives

(Pϕ) ( ) ( ) ( )

( ) ( ) ( )

==η==η==η

γ=∂

ϕ∂+∂ϕ∂λγ

0t,Lx,0t,0x,00t,x

t,xut,xx

t,xt

111

12

2

11

which has been already treated and gives (having defined ( ) ( )

−′

λ=L

x1

6

tIt,xu 1 where the prime de-

notes differentiation with respect to the argument)

( ) ( ) ( )τ−ξΓλ

τξτξ=ϕ ∫∫ t,,x,u

ddL

2t,x 1

1

t

0

L

0

Thus

( ) ( ) ( ) ( )τ−ξΓ

ξ−τ′τξ+

−−=η ∫∫ t,,xL

16

Idd

L

2

L

x1

6

tIt,x 1

t

0

L

0

1

and the currents are

( ) ( ) ( ) ( ) ( ) 11

t

0

L

0

0 t,,xL

16

Idd

L

2

L

x1

6

tI

3

tIt,x bbi

τ−ξΓ

ξ−τ′τξ+

−−+= ∫∫

In particular, for a current ramp

( ) ( )

><<

<=′⇒

><<

<=

0

000

00

000

tt fi,0

tt0 fi,t/I

0 tfi,0

tI

tt fi,I

tt0 fi,t/tI

0 tfi,0

tI

this becomes


( ) ( )

( ) ( )

>

−ξΓ−ξΓ

ξ−ξα+

−−+

<<

ξΓ

ξ−ξα+

−α−+α

<

=

∫

∫

010*1

*1

L

0

00

0

01*1

L

0

0

tt fi,tt,,xt,,xL

1dL

2

6L

x1

6

I

3

I

tt0 fi ,t,,xL

1dL

2

6L

x1

6

t

3

t

0 tfi,0

t,x

bb

bbi

where α = I0/t0 is the ramp rate.

APPENDIX B – 3-STRAND CABLE EXCITED AS IN THE KREMPASKY-SCHMIDT CASE

Consider a 3-strand cable with the conditions:

1. I(t) = 0;

2. r = 0,

3. even number of periods,

4. i(0)(x) = 0,

5. time-independent excitation spatially bounded to an interval in the center of the cable.

Hypotheses 3 and 5 lead to write:

( ) 111 )()22

L(U)

22

L(U

2/

1

1

0

)22

L(U)

22

L(U

2/−−− ξν=ξ−δ+δ−−ξ

δΦ=

−+ξ−δ+δ−−ξ

δΦ=ξ bbv

&&

Consequently, the currents flowing in the 3-strand cable are described by the following system:

(P) [ ][ ] ( ) ( ) [ ] ( )

( ) ( ) ( )

======

=∂∂+

∂∂

0t,Lxt,0x0t,x

xGt,xx

t,xt

MG2

2

iii

vii

Decomposing i as follows

( ) ( ) ( ) ( ) 111100 t,xt,xt,xt,x −−η+η+η= bbbi

Problem (P) is simply divided in the following three problems:

(P0) ( )

( ) ( ) ( )

==η==η==η

=∂

η∂

0t,Lxt,0x0t,x

0t,xx

000

2

02

(P1) ( ) ( )

( ) ( ) ( )

==η==η==η

=∂

η∂+∂η∂λγ

0t,Lxt,0x0t,x

0t,xx

t,xt

111

2

12

111


(P−1) ( ) ( ) ( )

( ) ( ) ( )

==η==η==η

νγ=∂

η∂+∂η∂λγ

−−−

−−−−

−−

0t,Lxt,0x0t,x

xt,xx

t,xt

111

112

12

111

Problems (P0) and (P1) have the simple solutions

( ) 0t,x0 =η ( ) 0t,x1 =η

While problem (P−1) can be solved as usual. Note that the currents becomes

( ) ( ) ( ) ( ) ( )t,xit,xi,0t,xit,xt,x 32111 −==⇒η= −− bi

Thus, since

( ) ( ) ( )t,,xdL

2t,x *

11

1

L

0

1 ξΓλ

ξνξ=η

−

−− ∫

The current i2 is given by

( ) ( ) ( )( )

( )( )

δ+δ−Γδλ

Φ=ξΓξδλΦ=ξΓ

λξν

ξ= −−

−

δ+

δ−−−

−

− ∫∫ 2

L,

2

L;t,xt,,xd

Lt,,x

2d

L

2t,xi **

11

*1

2/L

2/L1

*1

1

1

L

0

2

&&

By using the same notation of Krempasky-Schmidt (but note that the numerical values are different

since the eigenvalues for the 3-strand case are different from the 2-strand case) he same solution can be

obtained:

( )

∑

∑

∞

=

τ−

∞

=

πγ

−λ−

−

−−

πξ

π−πλ

=

=

πξ

π

−

πγ

−

λ=ξΓ −−

1n

/tn

22

2112

1n

L

n1t

2

1

11

L

nsin

L

xnsine1

n

1Lg3

L

nsin

L

xnsine1

L

n1t;,x*

2

2

11

and finally

( ) ( )∑∞

=

τ− α

α−πα

=

nodd1n

/tn

2m2 nsinw

xnsine1

n

1I

6t,xi

2

universitÀ degli studi di bologna … · 2008-03-11 · universitÀ degli studi di bologna ......

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