references. [ 1 ] abramowitz m., stegun i. pp. 149 …978-3-642-84716-5/1.pdf · corrosion and...

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APPENDIX A.1.1

PRIMARY CURRENT DISTRIBUTION ALONG A FREE CATHODE IN PARALLEL WITH AN ANODE AND PERPENDICULAR TO AN INSULATING BOUNDARY.

The primary current distribution around the cathode

presented in figure 1 is to be calculated.

u V ANODE

g

o V

CATHODE q

- fig. 1 -

It is supposed that the anode, the cathode and the in

sulator tend to infinity. The insulating boundary is

replaced using the image method (fig. 2).

z-plane

A g g E

_________ B D __ ~ ____________ __

C C

q q

- fig. 2 -

244

The obtained geometry can be transformed into a half

plane (fig. 3) by means of a Schwartz-Christoffel

transformation.

t-plane

o a b

E' A' B' C' D'

u V o V

- fig. 3 -

Choosing A' the origin and scaling the transformation

such that c = 1, one has

dz S(t - a)(t - b) (1) dt = t(t - 1)2

with z = -q jg for t = a (2a)

z = q jg for t b (2b)

z = _00 jg for t 1 (2c)

z = _00 for t 0 (2d)

z = +00 jg for t +00 (2e)

z = +co for t _co (2f)

Integration of equation (1) yields

S[ab lnt + ( 1 ab)ln(t 1) + (a + b - ab - 1)J z = - - (t 1) -

+ C . 0)

E'

245

The evaluation of the conditions (2a) to (2f) permits

to determine S, a, band C (C is complex) as function

of g and q. One finds:

S = gin (4a)

a.b = 1 (4b)

C -JOg + (a - 1)2 2aIT g,

[ (a-1)(a+1)] In a + 2a - IT -q

g

and finally the following equation is obtained

= £. [1 n t + (a - 1) 2 + (a - 1) 2 ] Z IT aCt - 1) 2a - jg

in which a is defined by the non-linear equa

tion:

In a + (a - 1) (a + 1) 2a

IT -qg

(6)

In the t-plane the field problem is defined as follows.

The potential along the negative part of the real

axis is U volt and zero volt along its positive part.

It is well known that the solution is equal to

w = ¢ + H = 1!.ln t IT (7 )

where ¢ is the streamfunction and W is the potential

function.

As we are interested in the electric field along the

cathode we need [ 15J

or

dw dw dt dz dtOdz

dw Ua(t-1)2 dz = g (t - a)(at - 1)

246

(8 )

by virtue of equation (1), (4b) and the derivative of

equation (7).

Whent takes the values from zero to one all points

of the cathode are obtained with equation (5) and

equation (8) gives the corresponding current density.

247

APPENDIX A.1.2

PRIMARY CURRENT DISTRIBUTION ALONG AN LSHAPED CATHODE.

The primary current distribution along the cathode

presented in fig. 1 is to be calculated.

U V ANODE

/

/

q /

/

/

P

o V CATHODE

- fig. 1 -

It is supposed that the electrodes tend to infinity

and the insulating boundary is replaced using the ima

ge method (fig. 2).

z-plane

A C

B D

- fig. 2 -

E

248

This geometry can be transformed into a half-plane

(fig. 3) by means of a Schwarz-Christoffel transfor

mation.

t-plane

o a c=1 b

E' A' B' C' D' E'

u V o V

- fig. 3 -

Choosing A' the origin and scaling the transformation

such that c = 1, one has

dz S t dt

/(t -

with z -jp

z = -jp

z = 00

z = 00 +

z = _00

z = _00

z = 0

Integration of

- 1

a) (t - b)

for

for

jp for

jp for

jp for

+ jp for

for

equation

t

t = t

t

t

t

t

t

(1)

a

b

+00

+00

0

0

1

yields

(2a)

(2b)

(2c)

(2d)

(2e)

( 2f)

(2g)

249

z = a) (t - b) + 2t - (a + b)] t

~ln[2/ab(t - a)(t - b) + 2ab - (a + b)tJ+ A + Bj • lab t

The conditions (2a) to (2g) permit one to determine S, a,

b, A and B as functions of g and q. One finds:

S = (p + 9) , (4a) 1T

ab = 1 , (4b)

A = -2{p + 9~ln(b - a) , (4c) 1T

B = -q - 2p , (4d)

2/{1 - a){b - 1) t g [ { 9 + 2p) 1T ] (4e) = 2 - (a + b) 2 (p + q)

and finally one has

2(p + 9)ln(b _ a) - (q + 2p)j 1T

in which b = 1/a (6a)

and a is defined by the following non-linear equation:

(6b)

The electric field along the cathode is obtained on

250

the same way as described in appendix A.1.1 and as the

potential in the t-plane is the same, one has (du'e to

equations (1) and (6a)):

dw::: -u I(t - a)(at - 1) dz (p + q) /a(t _ 1)

(7)

When t takes the values from one to infinity, all

points of the cathode are obtained with equation (5)

and equation (7) gives the corresponding current den

sity.

251

APPENDIX A.1. 3

PRIMARY CURRENT DISTRIBUTION ALONG A CATHO

DE BEING IN LINE WITH AN INSULATING BOUNDA

RY.

Using a Schwartz-Christoffel transformation, the pri

mary current distribution along the cathode presented in fig. 1 is calculated.

U V ANODE

p

o V CATHODE ~-------,I"I"'I""J"Ir"'l""T'T""""""'v.

A B

- fig. 1 -

It is supposed that the anode, the cathode and the boundary AB tend to infinity which is allowed when the vertical insulating boundaries are far (>4p) from the singularity. Using the image method the boundary AB can be replaced as represented in fig. 2.

252

A p z-plane

B

c p

- fig. 2 -

This new geometry can be transformed into a half

plane (fig. 3).

t-plane

a=-1 o c

DI AI BI C I

o V u V o V

- fig. 3 -

D

DI

Due to symmetry we choose the orlgln of the t-plane

in BI. The scale is defined by taking a = -1 and one

has

dz dt

St (t + n(t c) ,

with z = 0 for

z = +co pj for

z = +co + pj for

z = -co for

z = -co for

z = -co + pj for

z = _co pj for

Integration of equation

S c [In (t + 1) + z = 1 +

253

t 0

t co

t _co

t = -1 ( t> -1 )

t = c (t <c)

t -1 (t <-1 )

t = c (t>c)

(1 ) yields

cln( t - c) ] + A + jB

(2a)

(2b) (2c)

(2d)

(2e) (2f)

(2g)

By means of the conditions (2a) to (2g), the constants

S, c, A and B can be evaluated and one finds:

S 212. 1T

(4a)

A 0 (4b)

c = 1

B = -p (4d)

After substitution of these values equation (3) beco-

mes

z = J2.[ln(t + 1) + In(t - 1)J - jp . 1T

The potential distribution in the t-plane can easily

be found by superposition. The result is

where $ is the streamfunction and W is the potential

function.

The electric field along the cathode is

dw dz

U pt- (7)

When t takes values from -1 to zero, all points of

the cathode are obtained with equation (5) and equa

tion (7) gives the corresponding current density_

APPENDIX A.2

SOLUTION OF THE POTENTIAL MODEL USING TRIAL

FUNCTIONS SATISFYING THE FIELD EQUATION :

EXAMPLE.

Let the domain and its boundary conditions be as given

in figure 1.

y

b

U=O V U=O V

--+-----------------~-------------.-x a

U=O V

- fig. 1 -

The set of trial functions satisfying all boundary

conditions except those along y = b O~x~a is easily

obtained and a linear combination is

00

~ L Ansin nn~·sinh nn~ n=1 a a

The unknown constants A are calculated by making n

the residual zero along y = b O~x~a

00

L n=1

256

with

= attl ay y=b

= 00

\' X b 7Tn L Ansin n7T--coshn7T--- •

n=1 a a a

The choice of the weight function is free, but using

the same functions as the trial functions namely

sin m7Ti (m = 1,(0), the residual is made orthogonal to

the trial functions (this is called a Galerkin method)

and one has

n=1,2, ••• 00

Unless n1 (-ott') is linear, to find a solution, the

series must be truncated and using only N-terms, an

NxN non-linear system of equations is obtained to

find the An's. When the overpotential function is

linear, say

n - Z J 1 - n -oZ U' ,

the equations are uncoupled and one has

A n

= ....1!z.!i 1 n7T 2 (sinhn7Tb _ oZcoshn7Tb7Tn)

a a a·

n = 1, 2, ••• 00

Since the series is not truncated and complete, an

exact solution is found.

APPENDIX A. 3.1

ANALYTIC INTEGRATION OF INTEGRALS INVOLVED BY THE TWO-DIMENSIONAL BOUNDARY ELEMENT METHOD USING STRAIGHT ELEMENTS.

1. Introduction.

When the boundary of the domain is divided in straight

"segments", using trial functions for the variation

of U and Uf (shape function) in each segment, one ob-

tains the following integrals

t<Pk

2 k aw~~ and k J <Pkw*dr (1) H .. -dr G .. =

lJ an lJ r~ r~

J J

where <Pk are the shape functions for U and UI and w*

is the fundamental solution.

For the Laplace equation in two dimensions

w* = In(1/r)/2n.

r. J

i

t

X,y)

- fig. 1 -

o L/2 1/2

- In two-dimensional problems the most used shape

1 -1 o o

k functions are polynomials and the integrals Hij and

258

k Gij become linear combination of integrals like

and (2)

- Since the elements are straight segments, we have:

- Fig. 2 -

* I:!.X X2 X1

* I:!.Y = Y2 Y1

12 = I:!. X 2 + l:!.y 2

* X X1(1 t) + X2t

* Y = Y1 (1 t) + Y2t (linear shape functions)

* dr = 1dt

* r 1 = (X1 - X) 2 + (Y1 _ Y) 2

259

b 2[(X1-X)~X + (Y1-Y)~YJ/L2

c = 1

* aw* grad w* T an • n

X,y - Y).(~Y/L, -~X/L)·1~

= [(X1 - X)~Y - (Y1 - Y)~XJ._1 2~

a + bt + t 2

Substitution of all this in expressions (2) yields

that all integrals are linear combinations of inte

grals with the form

H J1 t n dt n = 0 (a + bt + t 2 )

and

with n O,1, •• m.

These integrals will be calculated.

2. Integration of Hn.

To calculate the HI s a difference must be made den

pending on the value of the discriminant

260

q = 4a - b 2

= 4r1sin261 ~ 0 •

Depending on the value of the discriminant, the super

script 0 or t is used. If q can take all values, no

superscript is used.

When q equals zero, the three points (X, Y), (X1, Y1)

and (X2,Y2) are colinear and the integrals of Hare n

not to be evaluated since aw*/an equals zero. How

ever in order to calculate the GO,s we can make use n

of the HO n

, s.

2.1 • HO q °

0, n = o.

When q 0 b 2 and a + bt + t 2 (t + £.) 2 a = 4 = 2

HO = f~ dt

° (t + £.) 2 2

HO -1 + 2 0, 0 (5) b b q = n ° 1 + 2'

This integral is +00 when b = -2 and 0, this means

when (X,Y) equal s (X2,Y2) or (X1,Y1).

2.2. HO 1 q = 0, n = 1 •

HO f~ tdt 1 (t + £.) 2

2

HO In b + 2 £.Ho or 1 b 2 °

261

q = 0, n 1 (6)

This integral is also +00 when b = a and -2.

2.3. q 0, n = 2, •••

HO = f01 ~t_n~d~t~ __ n (t + .2.)2

2

n = 2, •••

Using twice integration by parts, one finds (a

1 n - 1 ° aH 2 n- q 0, n = 2,3 ••• (7)

Remark: When b = 0, the integrals are finite and

equal to

q = 0, n = 2, ••• ; b = 0(8)

"'Jhen b = -2, the integral s are infinite

n-1 HO = I r + n~ (n 1)Ho (9) - -n r=1 n - r °

n-1 HO I r + 00 = n r=1 n - r

q = 0, n 2, ••• ; b =-2

262

2.4. H+ q > 0, n = o. 0

H+ f~ dt 0 (a + bt + t 2)

H+ 2 [ 2+b atan ~] (10) = rq atan rq -

0 q > 0, n 0

2.5. H+ 1 q > 0, n = 1 •

H+ = f~ tdt 1 a + bt + t 2

H+ 1 a + b + 1 '£H+ q > 0, 1 (11) = 2'ln n 1 a 2 0

2.6. H+ q > 0, n = 2,3 ••• n

H+ J~ t n dt = n a + bt + t 2

H+ 1 bH+ + > 0 ( 12) = 1 aH 2 q n n - n-1 n-n = 2,.3 •••

3. Integration of Gn •

3.1 • GO q = 0, n = 0,1 ••• , b F -2 and O. n

GO J: tnIn(t + ~)2dt + InL = n + 1 i.1

GO 1 1)[ In(a + b + 1) 0 0 + = 2(n + - 2Hn+2 - bH 1 n n+

2InL ] (13 )

263

(integration by parts and a = b 2 /4)

Problems can arise when b = -2 or O.

3.2. GO : q = 0, n n = o , 1 ... , b = O.

if tnlnt 2 dt + lnL 2 0 n + 1

1 1)[ -2 1 + 2lnL ] 2(n + n + 1

q = 0, b F -2

b F 0

n=0,1 ••.

q=O (14)

n = 0,1 •••

b = 0

Provided that H1 is made finite, equation (13) can be

used due to a = b = 0 and equation (8).

3.3. 0, n 0,1,2 ••. , b -2.

1 -2 n+1 1 lnL = 1 ) r~1 + 0 + 2" (n + r n + 1

1 n+1 1 = 2{n + 1 ) [ -2 L + 2lnL ] r=1 r

Using equation (9) and defining H1 = Ho 0 one has:

-2H n+2 - bH n+1 = -2(Hn +2 - Hn +1 )

n+1 n = -2( L r L r

r) r=1 n + 2 - r r=1 n + 1 -

and

with H1 =

3.4. G+ n

a+ n = it 2 0

G+ -![ = n

H = 0

q >

0

n+1 _1 = -2 L

r=1 r

.

264

0, n = 0,1 .••

tnln(a + bt + t 2 )dt +

t n+1 n + 1ln (a + bt + t 2 )

t n+2

(15)

q = 0, n = 0,1 •••

b = -2

lnL n + 1

1 ]

0

b t n+1 2 t dt -

1 f: dt n + 1 0 a + bt + t 2 n + a + bt + t 2

G+ 1 1)[ In(a + b + 1) - + + = 2{n + 2Hn+2 bHn+1 + n

2lnL ] (16 )

q > 0, n 0,1 .••

4. Conclusion.

Equations (5), ( 6), ( 7 ), ( 8 ), ( 9 ), ( 1 0 ), (11), ( 1 2) ,

(13), (14), (15) and (16) can be put together in the following general recurrence relations:

265

H : 0

1 • q 0 b -2 make H = 0 0

make In(a + b + 1) = 0

2. q 0 b = 0 make Ho 0

make In(a) = 0

3. 0 b F 0, -2 H -1 + g q b 0 1 + b '2

4. q > 0 H = L[ atan (2 + b) 0 ;q ;q

~ln(a + b + 1) - ~ln(a)

1 --~~1 - bH 1 - aH 2 n - n- n-

2lnL ]

Once the G 's and H 's are calculated, the integrals n n

Ie k Hij and Gij are easily obtained.

Defining A = -[ (X1 - X)~Y - (Y1 - Y)~X ]/L22~

and B -L/(2~). we have

- for constant elements:

G]. = BG J.J 0

- for linear- and singularity elements with two nodal

266

points:

1 A(H H ) 1 B(G Gn ) H .. - , G .. = -lJ 0 n lJ 0

2 2 AH BGn H .. Gij = 1 ,2 ••• lJ n n

- for quadratic elements:

1 A(2H 2 3H1 + H ) 1 B(2G2 3G1 + G ) H .. G .. lJ 0 lJ 0

2 A(2H2 - H1 ) 2 B(2G2 - G1 ) H .. G ..

lJ lJ

3 = A(-4H2 + 4H1 ) 3 = B (-4G 2 + 4G1 ) H .. G .. lJ lJ

- higher order shape functions can be obtained in the

same way.

Important remark:

When b is large, soon numerical round-off errors are

involved. Therefore a combination o~ analytic and

numerical integration is necessary. This numerical

integration needs only a few Gauss points.

267

APPENDIX A.3.2

EVALUATION OF INTEGRALS INVOLVED BY THE BOUNDARY ELEMENT METHOD USED TO SOLVE AXISYMHETRIC POTENTIAL PROBLEMS.

For the sake of completeness and referring to fig. 1, all definitions and equations of three-dimensional axisymmetric potential problems are repeated:

observation point (r,e,z)=(x,y,z)

I z

(X,O,Z) (R,O,Z) field point

'I'-..>.......---~- x

- fig. 1 -

Let us introduce the following notations in cartesian and cylindrical coordinates (fig. 1)

- (X,O,Z) or (R,O,Z) the coordinates of the field point,

- (x,y,z) or (r,e,z) the coordinates of the observation point,

268

- R the distance between field and observation point,

- x = f(z) or r = f(z) the equation of the meridional

boundary curve.

The problem being three-dimensional, one has for an

isotropic medium:

w* 1 - = 41TR

with 1R2 (X_X)2 + (y)2 + (Z_Z)2

or 1R2 r2 + R2 -2rRcose + (Z_Z)2

Vw* -=1-[(x-X)T + yT + (z-z)T ] 41T1R 3 x Y z

or

= -=1-[(r-Rcose)T - RsineTe + (z-z)T ] 41T1R 3 r z

(T - f' (z)T )_;=::::1===== r z A + f'(Z)2

aw* = Vw*.T - ~ n - Rcose f' (z) (z_Z)]-;:=::1=====

A + f'(Z)2

Substitution of these expressions in the direct inte

gral equation (2.20) gives:

c.u. = f f21T au f(z);; + f'(Z)2 dedz + l l f(z) 0 an 41TR

f f21T u[f(z) - Rcose - f'(z)(z-z)] f(z)dedz • f(z) 0 41TR 3

269

The meridional boundary curve f(z) is divided into M straight elements. Linear shape functions are in

troduced to approximate the potential and the normal

electric field in each element. With the notation of

fig. 2, it follows that:

- ~ 2r L

f(z) is in parameter form related to the coordinates

of the nodal points:

R2 - R1~ + R1 + R2 r = 2 2

Z2 - Z1~ + Z1 + Z2 z = 2 2

hence f I (z) R2 - R1 11 + fl(Z)2 L Z2 - Z1 , Z2 - Z1

and dz = {Z2 - Z1 ~ d~ 2 .

dr

(R1 ,Z1)

- fig. 2 -

270

Substitution of these equations in equation (1) yields:

with

[H~ . l.J

GLJ {Uj1} l.J U'

j2

-f t1T [(Z2-Z1) (r-Rcose)-(R2-R1) (z-Z)] (1+~)rd8d~ 2041TR3 2 -1 0

(2)

(3)

1. Analytical integration with respect to ~.

1et us define 2a R2 R1

2b R2 + R1

2c = Z2 Z1

2d = Z2 + Z1

Then we can write:

- r a~ + b

- z = c~ + d

- f' (z) = a/c ,

- 12 = 4(a 2 + c 2) ,

- R = /C~2 + B~ + A = Ii ,

with C = a 2 + c 2 = 1 2 /4,

B = 2ab + 2c(d-Z) - 2aRcos8 ,

271

A = b 2 + R2 + (d_Z)2 - 2bRcosS ~

- r - RcosS - f'(z)(z-Z) = (bc - Rc cosS - ad + aZ)/c ,

- 2c:; - 1 = -1 I n r

Equation (2) becomes:

1,2 -1J21T Hij = ~ 0 (bc - Rc cosS - ad + aZ)n'(cosS)dS

with

fi'(cose) = taJ1 ~2d~ + (a + b)J1 ~ + bJ1 ~ -1 xIX -1 xIX -1 xIX

or (after some calculus)

fi'(cosS) = [+a«2B2-4AC)~+2AB)-(atb)2C(B~+2A)+2C(2C~+B)b] cQIX .

+ ~[ln(21GX + 2Cs + B)] 1 • C IC -1

In a similar way equation (3) becomes:

1,2 ICJ21T Gij = 87T 0 g(cosS)dS

with

g(cosS) = tJ1 a~ + (a t b)J1 ~ + bJ1 ~ -1 IX -1 IX -1 IX

or

g(cosS) = [ta(~2C - 3B )IX + (a t b)lX] 1 4C 2 C -1

272

_1 [+a(3B2-4AC)+(a+b)B_b}[ln(21CX+2C~+B)] 1 • /G 8C 2 2C

-1

A further evaluation of the functions g(cos8) and

fi'(cos8) is troublesome, therefore they were coded in

the given form.

Finally we have

1 , 2

H .. lJ

1J21f = 81f 0 fi(cos8)d8 1 J1f 41f 0 fi(cos8)d8

with

fi(cos8)

and

1 , 2

G .• lJ

(bc - Rc cos8 - ad + aZ)fi'(cos8)

/GJ21f A /GJ1f - g(cos8)d8 = - g(cos8)d8 81f 0 41f 0

With the change of the variable of integration x =

cos8, the integrals (4) and (5) transform respective

ly into

1 , 2 -1 t fi(x)dx H .. 41f -1 lJ ;; _ x 2

(6)

and

1 , 2 let g(x)dt G .. = lJ 41f -1 ;; _ x 2

(7)

which can be integrated numerical with a Gauss-Cheby

shev quadrature [ 1J:

1,2 -1 n Hl· J· -4 L w.fi(x.)

1f i=1 l l

273

and

1 , 2 IC n G .. 4 L wi~Hxi) lJ i=1

where w. 1!. and x. 1 n 1

( (2i - 1)1r) cos 2n .

These weights wi and abscissas xi are easily generated.

Remarks:

- [In (2/CX + 2CI; + B) l~ should not be evaluated when

a = b = 0, c = d = 0, (R,Z) = (0,0) and

(R1,Z1) or (R2,Z2) = (0,0).

hThen (R,Z) = (0,0), the distance IR. is constant and

two Gauss-points suffice.

2. Analytical integration with respect to 8.

Let us write equation (3) as

1>2 _ 1J1 (1 + l;)rd8d~ Gij - 2 -1 wA 2 ~

with

w* A LJ27T 1

-d8 47T a R

(8)

wA is the fundamental solution for axisymmetric pro

blems. The distance IR. between the field point (R,Z)

and an observation point (r,z) can be written as (see

fig. 1)

or

274

R2 (r+R)2 - 4rRCOs2~ + (Z_Z)2

with

4rR (r+R)2 + (Z_Z)2

Substitution of equation (9) in (8) yields

or

w* A

w* A

1T

1T[(r+R)2 + (Z_Z)2]1/2°J; (1_k 2::n28)172

(10)

The complete elliptic integral of the first kind K(k)

is recognized with the result

w* A K(k) ( 11)

Differentiating eq. (11) with respect to rand z gives

Clw* Clw* Vw* = _A_r + _A_r

A Clr r Clz z (12 )

with

-k [1 r 2 _R 2 _(Z_Z)2] 2rK(k) + E(k)

(r-R)2 + (Z_Z)2

(13)

and

275

-k [ (z-Z) E(k)]

(r_R)2 + (Z_Z)2 (.14)

where E(k) is the complete elliptic integral of the

second kind.

Equations (11) and (12) can be introduced in (3) and (2) respectively and using the notation as we did for the integration with respect to ~, one obtains:

1,2 H ••

l.J

and

(15)

aw1 (~) ( c -ar

awA* (~) 1-an a)( ~ ~)(a~ + b)d~ •

(16)

These integrals are evaluated numerically with a Gauss

Legendre quadrature.

The functions K(k) and E(k) are calculated with their

polynomial approximations [ 1].

K(k) = 1.3862944 + 0.1119723x + 0.0725296x2 +

(0.5 + 0.1213478x + 0.0288729x2)ln(1/x)

and

E(k) = (1 + 0.463015x + 0.1077812x 2) +

(0.2452727x + 0.0412496x 2)ln(1/x) ,

with x = 1 - k 2 •

APPENDIX A.4

THE GLOBAL NEWTON CONVERGENCE OF THE PO

TENTIAL PROBLEM WITH NON-LINEAR BOUNDARY CONDITIONS.

In general, for an irreducible NxN non-linear system

of equations

{$(X)} = [K(X)]{X} - (F(X)} = 0

with tangential matrix

the global Newton convergence is assured [119], [ 78] when

[KT(X)], the tangential or Jacobian matrix is non-

singular, (1)

- [KT(X)]-1 ~ 0, V XERN component by component, (2)

- {$(l3) - $(X)} ~ (or ~) [KT(X)] {B - X}, V X,B£1EtN

component by component. (3)

In the case considered, after reordering the system of equations in such a way that all unknowns are passed

to the right-hand side, one obtains a system of equations of the form (equation (2.24»

[G*] {X} = [H*] (F(X)) • (4)

As the Laplace equation with mixed boundary conditions

has a unique solution, the role of unknowns and knowns can be inverted and hence the matrices G* and H*,

which depend only on the geometry, are non-singular.

277

Equation (4) can be written as

or

[K]{X} - {F(X)} = 0 ,

with

This matrix is independent on the vector X.

The non-linear part of equation (5) can be written as

follows (equation (2.24)):

(6)

It is observed that F. is a function of only the i-th 1

component X. of the vector X (i=1, ••• ,N). With the 1

terminology used by Ortega and Rheinboldt [ 78J, this

means that Y(X) is diagonal and because the matrix

[KJ is also constant, one says that the system of

equations is "almost linear".

In reality many overpotential relations satisfy over

a wide range the following properties:

* they are continuously differentiable (true on phy

sical grounds);

* they are isotone: for all J1 ~ J 2 , n(J1 ) ~ n(JZ)

278

with J = -oX ;

* they are convex.

More explicitly, under these conditions the functions

Fi(Xi ) take on anode and cathode respectively the form

of figures 1 and 2.

F. J.

v

F. (b) J.

Fi (.)(b-.) I Fi(a)

o

I I -----y--------------------------------I I I

o b

on anode: Fi = V - n1 ~ 0

aF Fi = ax. ~ 0

J.

- fig. 1 -

a

279

---- ---------------------- F. (b) 1

F. (a) 1

o ------------~--------------~----~~ __ x. o 1 b

on cathode: Fi ~ 0

F! ~ 0 1

a

F! (a) (b-a) 1

F.(b) - F.(a) ~ F!(a)(b - a) 111

- fig. 2 -

Remarks: - The overvoltages can never become larger than the

driving potential V, implying that Fi (i=1,N) is non

negative.

_ Overvoltages on electrodes where passivation may

occur, do not correspond with the third condition.

With these properties of [KJ and F, the tangential

matrix becomes

280

* This tangential matrix is non-singular (condition (1 ) ) •

Indeed, the matrix [K] is non-singular and has only real and positive eigenvalues (discretization of a

well-posed Laplace problem). This property holds also when the diagonal of [K] is incremented with only positive constants (Fi ~ 0).

* Inequality (3) can be written as follows:

[KJ {B} -{F(13)} - [K] {A} +{F(A)} ~ ~

{ -F (B) + F (A) } ~

~

{F(13) -F(A)} ~

~

or component by component

~ F1.!(a)(b - a) • ~

[K-F! (A) o .. ]{ B-A} 1. ]. J

-F! (A) o .. {B-A} 1. 1.J

Fi(A)oij{B-A}

This proves that, when the overvoltages are continuous

ly differentiable isotone and convex, inequality (3)

holds also for the system of equations.

* We still have to prove that each component of

[KTJ-1 is non-negative.

This holds when the tangential matrix is an M-matrix

(section 13.3.8 [78]). A matrix B is an M-matrix if

B is invertible, B-1~0 and bij~O for all i,j=1, ••• N,

iFj. This definition is not very useful but there exist more practical characterizations.

281

Because the sum of an M-matrix and a positive diagonal

matrix is also an M-matrix, "it is sufficient" to

prove that only the system matrix [K] is an M-matrix.

This was attempted in vain. Probably, in this case,

M-matrices are a too restrictive subclass of matri

ces with non-negative inverse.

Thus, this part of the theorem remains un-solved.

In practice however, at least after the first itera

tion, a monotonic convergence up to the resolution

of the computer, is always observed when condition

(3) is fulfilled.

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