Kevorkian Cole Multiple Scale Singular Perturbation Methods

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oint Source in Biological Cell (from Cole, Peskoff, Barclion, and Eisenbe

J

Kevorkian J D Cole

Multiple Scale and Singular

Perturbation Methods

With

83

Illustrations

Springer



oint Source in Biological Cell (from Cole, Peskoff, Barclion, and Eisenb

J Kevorkian

Department of Applied Mathematics

University

of

Washington

Seattle,

WA

98195

USA

ditors

J.E. Marsden

Control and Dynamical Systems, 104-44

California Institute of Technology

Pasadena, CA 91125

USA

J.D. Cole

Department of Mathematical Sciences

Rensselaer Polytechnic Institute

Troy, NY

12181

USA

L Sirovich

Division of Applied Mathematics

Brown University

Providence, RI 02912

USA

Mathematics Subject Classification (1991): 34E 10, 35B20, 76Bxx

Library of Congress Cataloging-in-Publication

Data

Kevorkian, J.

Multiple scale

and

singular

perturbation

methods/

J.

Kevorkian,

J.D. Cole.

p. em. - (Applied mathematical sciences; v 114)

Includes bibliographical references

and

index.

ISBN 0-387-94202-5 (hardcover:alk. paper)

I

Differential equations- Numerical solutions. 2. Differential

equations-Asymptotic

theory. 3.

Perturbation

(Mathematics)

I

Cole, Julian D. II. Title.

Ill.

Series: Applied mathemati cal

sciences (Springer-Verlag New York Inc.); v 114.

QA1.A647 vol. 114

[QA371]

510

s-dc20

[515' .35] 95-49951

Printed

on

acid-free paper.

© 1996 Springer-Verlag New York, Inc.

All rights reserved. This work may not be translated or copied in whole or in part without the

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SPIN

10424264



oint Source in Biological Cell (from Cole, Peskoff, Barclion, and Eisenb218

3.

Limit Process Expansions for Partial Differential Equations

fi Re) 1/[log ljRe)] as Re 0. In particular, when successive terms of

the inner expansions are constructed, each satisfies the same Stokes equations

3.3.193). The nonlinear effects appear only explicitly in the outer equation and

outer expansion. Thus, the nonlinearity indicates the existence

of

terms,

fi

Re)

fi

2

Re) log

2

~ e )

,

3.3.227)

so that u

1

, v

1

,

p

1

) satisfy nonhomogeneous Oseen equations. Of course, terms

of

intermediate order satisfying the homogeneous Oseen equations may appear

between

u,

u

1

) to complete the matching. For the incompressible case, it turns

out that the outer expansion includes the inner expansion and that a uniformly valid

solution is found from the outer expansion with a boundary condition satisfied on

r

Re/2.

Such a result cannot be expected in the more general compressible

case.

A much more sophisticated version

of

this problem and the general problem

of

low Re flow appears in [3.12], [3.14], and [3.17].

Independently

of

the above work, similar results were also obtained in [3.35].

3 3 5 Potential Induced by a Point Source of Current in

the Interior of a Biological Cell

Certain boundary-value problems become singular, in the perturbation sense, be

cause the solution fails to exist for a limiting value of a parameter. For example,

if a heat source

is

turned on and maintained inside a finite conducting body that

is

imperfectly insulated at its surface, a steady temperature will

be

reached.

f

the

insulation

is

made more and more perfect, the body will heat up more quickly and

in the limiting case

of

perfect insulation, a steady state is never reached.

An analogous problem that occurs

in

electrophysiology is discussed in this

section. In certain experiments, in order to measure passive electrical properties,

a microelectrode

is

used to introduce a point source

of

current into a cell, and the

potential

is

measured at another point. The analysis

of

this experiment depends on

the theoretical treatment sketched

in

the following.

Problem formulation

The model for the cell

is

a finite body

of

characteristic dimension a enclosed by a

membrane

of

thickness

8,

surrounded by a perfectly conducting external medium

constant potential). The more general case

of

finite external conductivity can be

worked out by similar methods and appears

in

[3.32]. The geometry and coordinate

system are shown in Figure 3.3.5.

The conductivities

of

the cell interior and membrane are a; and

a

mhos/em),

respectively. The membrane thickness and conductivity are considered to approach

zero individually

in

such a way that the ratio

a

18

the surface conductivity, remains

finite. For a typical cell used in physiological experiments,

8

=

1

o

6

cm and

a

=

1o

3

to 5 x lo

2

cm. The membrane is also assumed to have a surface




3.3. Singular Boundary Problems 219

FIGURE 3.3.5. Coordinate System for Spherical Cell

capacitance em ' 111 farad/cm

2

•

Typical values are rm = 3 x 10 IO mhos/em,

cr; = 7 x lo-

3

mhos/em, and the basic dimensionless small parameter of the

problem

is

crma

to-3

€

=

<

cr;

Let (

)

temporarily denote quantities with physical dimensions, and assume

that a point source

of

current (

4.rr

amps) at

r =

R

is

turned on in a quiescent

system. The current density J amps/cm

2

is then given by

div · J

=

4.rr8(r -

R )H(t ),

(3.3.228a)

where

8 is

the Dirac delta function res presenting a point source in three dimensions,

and

is

the Heaviside step function:

H(t )

=

l

Ohm s law

is

t < 0

t

>

0.

J = -cr;grad V , V = potential(volts).

Thus, V obeys the Laplace equation

4.rr

Ll V = 8(r -

R )H(t ).

CI

(3.3.228b)



Point Source in Biological Cell (from Cole, Peskoff, Barclion, and Eisenbe


av1 l

-

an

= o + aVo/3t

if

ao E)

«

1/E,

if

ao E)

= 0\. ( .

(3.3.236b)

It

is clear that the second case in (3.3.236b) is the only one to have a solution, so

we choose

a

0

= E I

for simplicity, and

V

1

satisfies

av1

avo

V

0

+ - on the cell surface.

at

n

(3.3.237)

Thus,az = E a 3 = E

2

,etc.,and-aVkf3n =

vk 1

+aVk-1/at,k = 2,3, . . .

on the cell surface.

The solution for

V

0

is thus uniform inside the cell:

o = fo t).

(3.3.238)

As is typical

in

singular boundary-value problems, this is all the information that

can be obtained from a study

of

V

0

.

In order to find out more about /o t), it is

necessary to use a solvability condition derived from the equation and boundary

condition for

V

1•

Integration

of

(3.3.235) and the use

of

Gauss s theorem gives

d

3

r

=

infinitesimal volume element)

1

,

/1

av1

V I d · r = 4 r r = . dS.

cell volume cell surface

dn

(3.3.239)

Using the boundary condition (3.3.237), this gives

{

dfo t) + fo(t))

dS

= 4rr,

J ell surface d t

I.e.,

dfo

7r

- fo=4-,

dt

A

(3.3.240)

where

A

is the surface area

of

the cell membrane. The solution for

f

0

t)

is

4rr

1

o

= fo

= - + aoe

A

Thus, if we assume that the potential is initially zero, we have

4rr

1

V

0

r, t = - (I - e · ).

A

(3.3.241)

(3.3.242)

This result shows that the cell builds up to a large uniform potential independent

of cell shape but dependent on cell surface area

A.

Next. we consider the problem for

V

1

using the result for

\l l

to write the boundary

condition (3.3.237) in the form

4rr

an A

(3.3.243)

A corresponding solvability condition derived from the problem for

V

2

exists

for the pqtential V

1

:

.

':l Vzd

3

r =

0

= '

{

a z d S.

lcell volume lcell surface

an




222

3.


Thus,

Vt

dS = 0.

1

av1

cell surface t

(3.3.244)

We can effectively split V

into a steady-state part, which is a characteristic function

G

1

for the domain, and a transient part. Let

where

Vt

r, t =

Gt r)

+ ft r,

t ,

~ t

= - 47r8 r -

R

47r

A

f

r{ GtdS

= 0.

J

ell surface

(3.3.245)

(3.3.246a)

(3.3.246b)

(3.3.246c)

The condition (3.3.246c) serves to define the arbitrary constant that would exist

for G

1

otherwise. Correspondingly, we have the problem for f

1

:

=

0,

o

_ =

n

(3.3.247)

I

oft

)

ft

dS

= 0.

cell surface

t

Again, we see that, because of the equations and boundary conditions, f

f

t), and it follows that

(3.3.248)

Now the representation for

V

is

Vt

=

Gt r, R)

+

a1e-r

(3.3.249)

and a new difficulty comes to light. The initial condition V

= 0 cannot be satisfied

so that this particular limit process expansion

E

---+ 0, t fixed)

is

not initially valid.

nitially valid expansion

In

order to construct an initially valid expansion,

we

must take into account that

there

is

another important short time scale in the problem, and that av jot in the

boundary condition (3.3.232) can be large. However, since the time

is

short, the

potential has not yet had time to reach a large value. A consistent expansion that

keeps the time derivative term in the boundary condition

is

V r, t; E) =

v

r, t*

+ Ev2(r, t* + ... ,

(3.3.250)





where

t* =

t j E. The limit process associated with this is,

of

course, E 0,

r,

t*

fixed. The following sequence of problems results:

~ v = 4rro r R)H t*),

OV]

- on the cell surface,

r•

v

1

= 0 at

t

= 0 on the cell surface.

= 0.

OV2 OV2

v

1

on the cell surface,

an at

v

2

= 0 at

t

= 0 on the cell surface.

3.3.25la)

3.3.25lb)

3.3.25lc)

3.3.252a)

3.3.252b)

3.3.252c)

Some indication

of

the general form that the solution must have can be obtained

by integration over the cell volume. From 3.3.251), we have

fort*

> 0

~ v

r

= -4rr = - dS

Iii

3

I OV]

cell volume cell surface

an

/I

V]dS.

at*

cell surface

3.3.253)

Thus, using the initial condition, we find

/

r

f

v

1

dS = 4rrr*. 3.3.254)

J

ell surface

Part

of

v

1

must increase linearly with t . The following decomposition of v

1

is

suggested:

Vt r, t*)

=

u

1

r,

r*)

h r)t*,

3.3.255)

where

u

1

r,

t*) is a potential that does not grow with time. The boundary condit ion

3.3.251 b) becomes

au 8h

t -

an an

This implies that

h

an

au

-

h

on the cell surface.

r•

3.3.256)

=

0 on the cell surface

3.3.257)

since u

1

does not grow with time. Since ~ = 0 inside the cell, it follows that

h r) =

const. 3.3.258)

This constant can be evaluated from the integral condition 3.3.254)

4rr

h r)

=

- A = cell surface area. 3.3.259)

A

Thus, we have

• 4rr •

v

1

r,t*)

= u

1

r , t )

t

A

3.3.260)



oint Source in Biological Cell (from Cole, Peskoff, Barclion, and Eisenb224 3. Limit Process Expansions for Partial Differential Equations

Now,

u

1

satisfies the problem

= 4rr8 r R)H t*),

aul

aul 4rr

- = - - - - on the cell surface,

an at A

u

1

=

0 at

t

=

0 on the cell surface.

3.3.261a)

3.3.26lb)

3.3.261c)

It is clear that as

u

1

approaches a steady state, it will tend toG

1

,

the characteristic

function for the cell, defined by 3.3.246). To get some idea how this approach

might take place, let us assume that the geometry is such that the characteristic

function can be represented by a separation-of-variables type

of

expansion:

Gt

r, R) = L ckfdPt)1/lk(P2, PJ).

k

3.3.262)

Here, p

1

r)

= Pc = const. defines the cell surface.Thus p

1

is a coordinate normal

to the surface;

{>2

P3 are coordinates in the surface. The function fk p

1

satisfies

an equation

of

the form

d Ki dfk

2

- 'AkKtfk

=

0, Ku pt) > 0,

dpt Kt dpt

3.3.263)

where

Af

is a separation constant. Typically,

p

1

=

0

is a singular point inside the

cell and, depending on the type of expansion, a delta function may appear on the

right-hand side of 3.3.263). In any case, the energy integral

1

c { d Ki d fk )

2

l

k - - - - 'AkKtfk dpt

=

0

o

dpt Kt dpt

implies that

/k dfkjdpt) > 0

at the cell surface, Kif Kt)fddfkfdpt)

0

as

Pt

0) or

dfk

2

- (Pc) = J.LkfdPc)

on

the cell surface.

dpt

Now we

can

try to expand

u

1

in terms

of

the same functions

Ut r,

t)

= Gt r, R) L

ak(t*)fk(pt)1/tdP2. p3),

k

3.3.264)

3.3.265)

where ak 0) = q so that the singularity at r = R is removed

just

at t = 0.

But for all

t

>

0, it remains.

The

boundary condit ion 3.3.261 b) then becomes

fort* > 0

3.3.266)

or

using 3.3.264)

3.3.267)





l

where

f lZ

= ,:

at

the cell surface. Therefore,

(3.3.268)

The preceding calculations can be regarded as symbolic, but they are verified in

detail for the explicit case

of

a sphere. In summary, for t* > 0,

'

-

; t• JT *

v = G1

r,

R)

- ~ k

1

JdpJ)1/fk(pz, P3)

+ -

t .

k

A

(3.3.269)

In a similar way, the form

of

the short-time correction potential

v

2

can also be

found. In the calculation

of

v

1

only the term corresponding to membrane capac

itance remains in the boundary condition. Now, for v

2

, the effect of membrane

resistance appears, since (3.3.252b) implies that

a z a 1J

*

-. d S = 0 = : ; Vzd S

- U

IdS +

4Jr

t .

cell surface an d

t

cell surface cell surface

3.3.270)

A

suitable

decomposition

for

v

2

is

t*z

vz r, t*) =

hz r) T + uz r, t*),

3.3.271)

where u

2

does not grow with t*.

Then

the

boundary

condition (3.3.252b) is

l i z *2

l

o

2 t

ouz

*

- - - =

hz r)t

+ u

1

+

an 2 an

7r * auz

t +

A at*

Again, iJhzjan = 0, h

2

= const. =

- 4n A.

The resulting problem for

u

2

is

t:J.uz

=

0,

uz = 0 at t* = 0.

Thus,

a representation for u

2

is sought:

uz

=

L

bk t*)fdPI )1/fk. pz, PJ).

k

The boundary condition (3.3.272b) then gives

so that

MatchinR

d

hk

2 •

- + khk = - cd l - e-

1

11

) . bk O) = 0

dt*

bA(t*) = - ck

2

( l - e-

1

';

1

·)

+

Ct.J*e

_,,;

1

••

Jlk

(3.3.272a)

(3.3.272c)

3.3.273)

3.3.274)

3.3.275)

Now we can discuss the matching between the long-time and short-time expansions

in terms

of t,

1

=

t /7J(E), where

7

belongs to an appropriate subc lass oft: «

7

«

1.




226 3. Limit Process Expansions for Partial Differential Equations

Thus, for

t

fixed t*

=

17t

jE)

oo and

t =

7t

0. We have, for the

long-time expansion,

V = G1 r, R =

G1

r, R

a1

1 - 7t

... ,

and for the short-time expansion, neglecting transcendentally small terms,

4.rr 7f ]

v1

= G1 r,

R

- - -

A E

2 2

Ck

4.rr

t

-v2

=

2

/k Pt)t/Jk />2, P3)

-A -

2

2

·

k I Lk E

Comparing

1

/t:) Vo V

1

with v

1

EV2, we see that the term 4.rrI A)17t

matches,

that G

1

matches, and that we must choose

at 0.

The term 4.rr/A)

7

2

t;j2 also matches, and neglected terms are

0 17

3

E

2

.

Therefore, the overlap domain for the matching

to 0 E)

is

E I

log

E I << 7 « E

1

1

3

•

The voltage response at a typical point, as indicated by this theory, is given in

Figure 3.3.6. For further discussion, see also [3.30].

Transmembrane

Potential

/

Short Time)

y

/

- - - -

- - - - - - - - - -/

_

-

= .: ;.: ;:..; ;;:. ;;

/

6

Jo

/

/

\ _ ~ V

~ V L o n g Time)

Time t)

FIGURE 3.3.6. Matching of Short-Time and Long-Time Expansions





3.3.6 Green s Function; Infinite Cylindrical Cell

A number of problems in biology require the solution of Laplace s equation with

a boundary condition that describes the properties of the membrane surrounding a

biological cell, separating the interior from the exterior, and buffering the internal

environment from external disturbance. The membrane serves as an electrical

buffer because its resistivity

is

much greater than the resistivity

of

the cell interior.

The membrane boundary condition, therefore, contains a small parameter E, the

ratio of the internal resistance to the membrane resistance

in

appropriate units (see

Sec. 3.3.5).

Problem formulation

Here we consider a problem that arises when the electrical properties of very long

cylindrical cells are investigated by the application of current to the interior of the

cell from a microelectrode, a glass micropipette filled with conducting salt solution.

The potential in the interior

of

the cell obeys Laplace s equation. The boundary

condition is that the normal derivative

of

the potential at the inside surface

of

the

membrane (proportional to the normal component

of

current) is proportional to

the potential difference across the membrane.

If

the microelectrode is considered

a point source of current, the solution to the problem is the Green s function for

the electric potential

in

a cylinder with a membrane boundary condition.

The same method applies with other boundary conditions or source distributions

in

a part of the cell near the origin. There are also, of course, analogies with

other problems for the Laplace equation, for example, steady heat conduction or

incompressible flow. A more detailed discussion of this problem appears in [3.31]

and in references therein. Another approach by classical analysis appears in [3.29].

The problem for determining the potential V x, r, }; E) may be written,

in

cylindrical coordinates,

1 1 1

- rVr)r

+

2 VoH + Vu = 8 x 8 r -

R)8 0),

r r r

(3.3.276a)

Vr X, 1

} ;E)+

EV x, 1

} ;E )=

0,

(3.3.276b)

V

±oo,

r, })

=

0. (3.3.276c)

When

E

is small, the boundary condition at

r =

1 in (3.3.276b) implies that the

current

flow

will be predominantly in the axial direction, i.e., only a small fraction

of

the local current, O E), crosses the membrane in an axial distance

of

0 1).

We

are tempted to try to find an expansion

in

the small parameter

E, in

which the

leading term is the potential forE =

0.

Denoting this term by V

1

(x,

r, 0 ,

we see

from (3.3.276a, b) that V

1

satisfies

I 1 1

- r V,,)r +

2

V .

+ V, .. = 8 x)8 r - R)8 })

r r r

v,, x,

1

}) =

0.

(3.3.277)




3.


The boundary condition at

r =

1 implies that no current crosses the membrane;

all the current

is

confined to the interior

of

the cell. Consequently, V

1

must contain

a part that is linearly decreasing with increasing

x 1

This would lead to a potential

V

1

-+

-oo as

x I

-+ oo, making it impossible to satisfy the boundary condition

V = 0 at

x

I = oo.

To

avoid this divergence, any expansion that contains V

1

can

be valid only over a limited range of

x,

designated the near field, which contains

the source point. At large distances from the source, we must look for another,

far-field, expansion.

We expect that as

E

-+ 0, the region

of

validity

of

any near-field expansion of

which V

1

is a part becomes larger.

If

there is a linearly decaying potential over a

large distance, and the potential approaches zero as

x

I

-+ oo, then the potential

at x = 0 must be very large, i.e., V (0, r,

)) -+

oo as

E

-+ 0. Clearly, V1 must

be 0 (

1

and cannot be the leading term in the expansion. The leading term can be

found by matching to the far-field solution, and therefore we first solve the far-field

problem.

Far-field

expansion

In the far field, a long distance from the source, current flow is predominantly in

the axial direction. Since only a small fraction of the current within the cell, at

any value of

x,

leaks out of the cylinder in an axial distance

of 0 1),

the variation

in the x direction will be slow. We therefore, for convenience in ordering the far

field expansion, write the far-field potential in terms

of

a new slow variable

x

Denoting the potential in the far field by W we write the following expansion:

V = W x,

r, ) ;E )=

o E)W

0

x,

r,

))

+ {

1

E) W

1

{X, r,

))

+

...

, (3.3.278)

where the slow variable is defined by

x =

a E)x,

for a

a E) «

I to be defined. Thus,

(3.3.279)

~ = 0 =

oa

2

Wo,, +

{1a

2

W1 +

...

+

o ~ r W o

+

{ l ~ r W I

+

...

,

w h e r e ~ ~ = l jr) ajar) ra;ar) +

l r

2

) a

2

;ae

2

)

is

the transverse Laplacian,

and on the boundary,

r

= I , we have

Wr

+ EW = 0 =

oWo,

+{I

W1,

+ ... + E{oWo + E{1 W1 +

. . . .

The corresponding approximating sequence of problems is

~ r W o = 0

Wo, x, 1 e) = o

Wo(±oo, ·. 8) = 0

(3.3.280)



oint Source in Biological Cell (from Cole, Peskoff, Barclion, and Eisenbe3.3. Singular Boundary Problems 229

W1, x, 1, 0) = - WoCX, 1, B

W

1

(±oo,

r,

;1) = 0,

where we have set

E(o

=

I

to

obtain the surface boundary condition in (3.3.281).

Writing an expansion for a (E) in the form

a E)

= O o(E) + a

1

E)+ 0 2(E) + . . . ,

where the a; E) are an asymptotic sequence, we further set

o a ~ = I

to obtain (3.3.281). Thus

ao = ,JE.

(3.3.281)

(3.3.282)

We

could take a

=

a

0

with a

1

= a

2

= . . . =

0 and still obtain a sequence of

problems of increasing order in E. It will be seen later, however, that we would not

be able to maintain uniform validity of the asymptotic expansion for W at large

x

Assuming

a (E)

to have the more general form makes it possible to obtain a

uniform expansion. Continuing this procedure with E I

= (

2

, a

1

=

a

1

Ea

0

, etc.,

we find

(3.3.283)

and

(3.3.284)

W

3

(±oo,

r,

0 = 0,

where the a; are unknown constants. Thus, the far-field expansion

of

the potential

is

taken in the form

....._ - - 2 -

W x,

r 8;

E)= {o E)[Wo x,

r

8)

+ EW

1

x,

r

0)

+ E W

x,

r ;1) +

. . .

],

(3.3.285)

where the axial coordinate variable

is

(3.3.286)

So far, (o E). the order of the leading term

in

the W expansion,

is

unknown. It

will be determined by matching to the near field. The constants a

1

a

2

in the



oint Source in Biological Cell (from Cole, Peskoff, Barclion, and Eisenb230 3. Limit Process Expansions for Partial Differential Equations

expansion of x which couple different orders of W

in

the sequence of problems,

will be determined by requiring uniform validity of the W expansion for large

values ofx.

We now solve the sequence of problems. The solution to the first problem is

independent

of

r and }. Thus, we have

Wo x, r,

}) =

F(x),

3.3.287)

where F(x) is an as yet arbitrary function

ofx.

We must go to the second problem

to determine its functional form.

From

3.3.281)

and

3.3.287),

we obtain

1 1

- (rWJ, )+ w ~ , '

=

-F (x),

r r

wl,(x, 1

e ) =

-F x) , 3.3.288)

W

±oo,

r,

})

=

0.

Since the inhomogeneous term in the equation and the boundary condition at

r

= 1 are both independent

of

}, clearly,

W

1

is independent of }. Examining

3.3.283)- 3.3.284), the same reasoning then implies that W

2

, W

3

,

•.• are all

independent of(}.

Integrating

3.3.288),

we obtain, for the solution that is bounded at

r = 0,

2

-

r _

-

W1 x, r)

4

F (x) + G(x), 3.3.289)

where G(x) is an arbitrary function of x that cannot be determined until we go to

the next problem for

W

2•

Substituting the result 3.3.289) in the

r

= I boundary condition yields

F - 2F = 0,

3.3.290)

3.3.29I)

where A is a constant to be determined by matching to the near field.

Continuing in the same way, we find

I 2 ,

II

- rWz,)

=

r

- 4al) f - G

3.3.292)

r

Wz, X, I = F/2 - G

Wz ±oo, r)

= 0

and then

-

r4

2 I

)

-

Wz x, r) = J6 r a

1

F

4

c H(x).

3.3.293)





Substituting the expression for W

2

and Fin the

r

= 1 boundary condition yields

G - 2G = -4A a

1

+

~ e J2I

7

1 3.3.294)

The right-hand side is a homogeneous solution

of

the equation. Therefore, the

particular solution contains a term proportional to

x

imes exp

-.J21xl).

f

such

a term appears

in

G, then for lxl =

O E-

1

) , the expansion will not be valid

uniformly in

x To

avoid this, we require the right-hand side of 3.3.294) to vanish;

this occurs if

1

C¥] = .

8

3.3.295)

It is now clear why we could not assume the simple relation

x

= JEx but required

the more general form. The freedom to choose

az,

. . . allows us to force all

of the

x

dependence of

W

into exp -v Zx), eliminating nonuniformities

in

the

expansion.

The solution to 3.3.294)

is

thus

G x)

= Be-./2

171

•

3.3.296)

where

x =

y E 1 - i + .. .) x.

3.3.297)

The constant

B

will be determined by matching to the near field.

Thus,

W

1

, the second term in the far-field expansion see 3.2.289)), is

W

x, r =

- Ar

2

+ B

e J2

7

1

3.3.298)

We continue the same procedure and calculate z = 5E

2

/384.

The order

of

the expansion E) and the three constants A, B, C) are to be

found by matching to the near-field expansion. We introduce the matching variable

xTJ

= xry E)

3.3.299)

for a class

of T

E) contained in

JE « T

E)

«

1 and we take the limit E 0

with xT fixed. Under this limit, the near-field coordinate x = x Jja oo and the

far-field coordinate

- ..jE E

5

2 )

X

=

ry Xa 1 -

S

+

384

E - . . . 0.

Because

of

the simple way

x

nters the expansion, it is appropriate to express the

far field in terms of

x

as x 0 and to compare directly with the near field as

~

00

The far field has the expansion

W [xvlf I

-i + -

..

}

r; ]




232

3.


16

3.3.300)

Near-field expansion

Next, we consider the sequence

of

near-field problems implied by the far-field

behavior.

In

the vicinity of the point source, the potential is a rather complex function of

position, and there is no simple mathematical representation in terms ofelementary

functions as there

is

in the far field. The potential has a singularity at the source

point; the current diverges from this point, half going toward

x

=

+oo

and half

toward x

= -oo.

Close to the source, the lines

of

current flow are diverging

outward, equally in all directions. Those lines directed toward the membrane must

curve to avoid the membrane as, again, only a small fraction of the local current

leaves the cylinder. As the current flows down the cylinder, the lines become

predominantly in the axial direction, and the potential joins smoothly onto the

far-field potential.

In terms of the asymptotic expansions representing the near and far fields, this

behavior requires that the near-field expansion increase in powers of ,JE so it can

join the expansion 3.3.300) of the far field. Furthermore, in accordance with the

earlier arguments, which concluded that the 0 I) term in the near field has a linear

dependence on

lx

I as

lx

I --- oo, we see that the second term in 3.3.300) must be

0

1

in order to match the near field. Consequently,

1

o E)

= ./E.

The near-field expansion must be

of

the form

V x,

r,

8;

E =

E-

1

/

2

V

0

x,

r, 8 + V

x,

r,

8)

+ E

1

/

2

V2 x, r, 8 + EV3 x, r,

8

+ . . . .

Thus, we obtain the following sequence of near-field problems:

ll Vo =

0

V

0

, x,

1 8 = 0; Vo x, r,

8 ---+

A as lxl---+ oo,

3.3.301)

3.3.302)

3.3.303)



oint Source in Biological Cell (from Cole, Peskoff, Barclion, and Eisenb3.3. Singular Boundary Problems 233

1

V

1

=

o x)o r -

R)8 })

r

V

1

, x, 1, })

= o

3.3.304)

~ z = 0

Vz, x, 1, })

=

-Vo x, 1, })

V,(x,

r,

9 -+ A(x - r:) + B, as lxl -+ oo,

3.3.305)

V3, x,

1, })

=

1

x, 1, })

s x I

oo,

3.3.306)

V4, x, 1, }) = -Vz x, 1, });

+B (x -

r: +Caslxl -+ oo,

3.3.307)

~ s

= 0

Vs, x, 1,

}) =

-V3 x, 1, })

Vs,(x,

r,

9) -+ h lx l

[

-

3

4

+

x

8

+

x;

-

(

1 x2

rz)

]

B

g -

3

+ Z -

C

s x oo.

3.3.308)

The delta function source appears in the V

1

problem, consistent with the linear

decrease with x as x I oo. All other orders of the potential are source-free.



oint Source in Biological Cell (from Cole, Peskoff, Barclion, and Eisenb234 3 Limit Process Expansions for Partial Differential Equations

Each even odd) order problem except for the first two)

is

coupled to the preced

ing even odd) order problem via the boundary condition on the x = I surface. The

physical interpretation

of

this coupling is that the current crossing the membrane

in the nth problem

is

proportional to the membrane potential in the n - 2)nd

problem. The even order problems are coupled to the odd order problems by their

asymptotic behavior as

lxl

oo, i.e., the constants

A,

B,

C, appear in both

even and odd order problems.

It should be noted thatthe V

1

,

V

3

,

terms alone are sufficient to satisfy 3.3.276)

at small x. It

is

only from considerations of behavior for large

lx

I required of the

far-field potential, that we conclude that

V

0

V

2

are even necessary. These

terms are thus known as switchback terms.

By direct substitution

of

the

lx I

oo asymptotic forms

of V

0

Vz, and

V

4

in

the respective equations and boundary conditions 3.3.303, 305, 307), it is seen

that the lx I - oo forms are the solutions valid for all x. This part

of

the near field

is

thus completely contained in the far field:

Vo

=A

3.3.309)

V

2

=A (x -

r ~

+ B

3.3.310)

v =A

_ x

+

:•

- r ; + r +

~ :

+ B (x -

r ~

+c.

3.3.311)

Now we evaluate the constant

A.

Integrating 3.3.304) over the large volume of

the cylinder between - x and x, lxl oo, and using the divergence theorem, we

obtain

-1

= lim { rr

d(} {

1

rdr

fx

dx L\ V

JxJ-- oo

Jo

o -x

= lim

{zrr d(} t

rdr[V1, x,

r, 0

- V1, -x

r,

0)]

lxJ-- oo o Jo

= -27 { Aj../2,

where

in

accordance with the

r =

1 boundary condition, the integral over the

surface

of

the cylinder

is

zero, leaving only the integral over the disks at ±x.

The last equality follows from substitution

of

the asymptotic behavior

of

V

1

,

as

lxl

oo.

The problem 3.3.304) for V

1

is

now definite.

In

order to solve the problem, it

is

convenient to decompose the near-field potential V

1

into two terms:

3.3.312)



oint Source in Biological Cell (from Cole, Peskoff, Barclion, and Eisenb3.3. Singular Boundary Problems 235

We obtain the following problem for

<l>

1

:

tl<l>1 = -8(x) [ 8(r -

R)8(())

-

~

,

<l>I,(x, 1,8) =

0,

(3.3.313)

<1>1 (±oo, r,

))

= 0.

The right-hand side of (3.3.312) consists of a unit point source at (0, R

0

plus a

uniform distribution of sinks in the

x = 0

plane.

The

net current source for <1>

1

is

zero, i.e., all the current that enters the cylinder at the point 0, R 0) is removed

uniformly in the cross section 0,

r,

8). Unlike the problem for V

1

, which contains

unit current flowing outward as lx

I

--+

oo,

the problem for <1>

1

contains no current

flow as lx

I --+ oo.

The

boundary-value problem may be solved by Fourier transformation in the()

and x coordinates. Defining the double Fourier transform of

<l>

1

by

1/1; \k,

r

=

J

0

2

rr

dee-i

11

f : XJ dx cos(kx)<l>

1

(x, r, fJ ,

>

1

(x, r,

f = J ~ X J dk

cos(kx)

.L:- )0

e;

1111

1/J; >(k, r ,

(3.3.314)

noting that

< >

1

is even in x and

},

we see that the problem (3.3.313) becomes, in

Fourier transform space,

) )

Ill>

2 i l l )

I

-

(rl/1

1

)r

-

k

+ -

2

1 1

1

= 8(r -

R)

+ 28o

11

,

r r r

where

2mr

[ -1 )

- I J .

The solution is

28 K' k)

,J,

1

111

>(k, r - I (kr) - " - (kR)

'/' k2 II l,;(k) II

+

I

11

(kR)I

11

(kr),

0

:S

r

S

R,

K

11

(kr) l

11

(kR),

R :S r :S

l.

(3.3.315)

Taking the inverse transform, we obtain

V

1

(x, r, fJ + -

1

-

(x

2

+ r

2

+ R

2

-

2rRcosfJ

112

)

2n

4n

I

I I I I I 1 Jk

k [ K;, k) k

I

k

- 2 . L e l cos( ·x - -1

11

·R

11

r

2n l l = -oo

0

l,;(k)

2

8o11 ]

+

k2

.

(3.3.316)




3.


The integral over

k

can be replaced by an equivalent sum by considering the

integral in 3.3.316) as a portion

of

a contour integral, so that

ln

A.nJ

R)

ln

A.nJ

r)

X

AnJ

:r:-

-

1 l}(Ans)

'

3.3.317)

where A.

s

is

the

sth

zero of 1 ; A.) excluding the one at A.

=

0.

We

can see that as

x --+ oo, V

1

--+ - x /2:rr plus terms that are exponentially small in 3.3.317).

We now tum to the V

3

problem and evaluate the constant B. Integrating the

Laplacian in 3.3.306) over the volume

of

a large cylinder extending from - x to

x and using the divergence theorem, we have

0 =

Jim

X

dX t rdr {ZIT de/:).. V3

lxl---.oo

x

Jo Jo

=

lim [ X.

dx {ZIT de

V3r (x

I

e) + 2

t rdr {

2

JT de V3, (x r, e ] .

111---.oo _ Jo Jo Jo

(3.3.3I8)

Using the boundary condit ion and the transform for V

1

,

we see that the first integral

becomes

x

· 2rr

-

{

dx

1

de V

1

(x, I,

e)

=x

2

-

dx

dk cos kx)l/fi

0

J k, I)

foo

1oo

;rr

00

0

2

,/,(0) 0

= X -

f )

, I).

3.3.319)

From 3.3.315), we obtain, using the Wronskian

of

1

and K

and the power series

expansion of 1

k),

1/J O) O,

I ) =

l i m - 2_ + fo(kR)) =

R z -

~ .

I

k--+0 k

2

kft(k)

2 2

3.3.320)

Using the asymptotic form for large x for from 3.3.306), we see that the

second integral in 3.3.3I8) becomes

1 rdr

t

d82-/2 [A -

x

2

+ r;)-

B]

3.3.321)





Thus, from (3.3.318),

B = .Ji ( _ R1 ) .

4Jr

8 2

(3.3.322)

As a consequence

of

(3.3.322),

W

1

and V

2

depend on

R

the distance from the

source to the axis of the cylinder, whereas lower-order terms do not.

Having evaluated

A

and

B

we have now obtained the near field and far field up

to terms of

O(E

1

1

2

,

i.e., we have obtained Vo

Vt,

V2 Wo and W1. These terms

represent that part of the potential that is numerically significant in a physiological

experiment: all higher-order terms are too small to detect anywhere in a cylindrical

cell. In [3.31

]

the calculations are carried out further to find the constant C.

The leading terms in the far-field expansion and in the near-field expansion are

each of order

E- I 1

1

. In

the near field, the leading term is a constant. Thus, near

the point source, the interior of the cylinder is raised to a large, constant potential,

relative to the zero potential at infinity. The physical basis for the large potential

is that the membrane permits only a small fraction

of

the current to leave the

cylinder per unit length. Consequently, most of the current flows a long distance

before getting out, and a large potential drop

is

required to force this current down

the cylinder. The existence of this large constant potential, and its magnitude of

0 E

I1

2

, could only be deduced from considerations

of

the far field.

The leading term in the far field decays as

exp -.Ji€1xl).

Consequently, to

lowest order,

1

e of the current leaves the cylinder in a distance

of 1

.J2€. The

corresponding potential required to drive a current this distance is of O E-

1

1

2

,

which is the physical basis for the order of the large potential in the near field.

The precise numerical values of the leading terms for V and W were determined

by requiring in the limit lx I -+ oo, x - 0, that the two terms be identical to

the lowest order in

E.

In the far field, i.e.,

x

=

x

JE

1

-

E

18

+ ... )

-

oo, the

potential is seen

to

approach zero exponentially.

The leading term in the far-field expansion is independent of

r

and

e

Thus, to

the lowest order, the far-field current is distributed uniformly over the circular cross

section of the cylinder. The leading term is the known result of one-dimensional

cable theory, equation 14 in [3.39]. The high-order terms are all independent

of

the

polar angle. They do, however, depend on the radial coordinate,

r.

The dependence

is

in the form

of

a polynomial in r

2

, the degree

of

the polynomial increasing by one

in

each successive term.

We

also see that the higher-order terms also depend on R

the radial distance between the source and the axis of the cylinder. The potential

is seen to be symmetric with respect to an interchange of

r

and

R.

This must be

so because the potential is the Green s function (with source at x

=

0,

e

=

0) for

the cylindrical problem.

The solution obtained here is close to the solution derived by Barcilon, Cole,

and Eisenberg [3.2], using multiple scaling.

The result of the multiple scale analysis differs from our result only because

[3.2] contains a sign error and a secular term V 3) which has not been removed. If

these errors are corrected, and the infinite sum over Bessel functions is written in




3. Limit Process Expansions for Partial Differential Equations

closed form, the multiple scale result, the expansion

of

the exact solution, and the

present results are identical.

3.3.

7

Whispering Gallery Modes

The propagation

of

sound waves with little attenuation inside a curved surface

is

called the whispering gallery effect. Rayleigh commented on this effect in [3.36]

and said that the effect is easily observed in the dome

of

St. Paul s, London. He

noted that the explanation

of

the Astronomer Royal (Airy), that rays from one

pole converge toward the opposite pole, was not correct since the effect can be felt

all around the circumference. A simple ray explanation was offered, but a more

complete theory based on solutions of the wave equation was given in [3.37]. The

idea is to examine the normal modes and identify those that can support waves

along the surface; these might be excited by a source of disturbance. Rayleigh

then studies the planar case and shows that the desired effect can be seen

in

high

frequency modes, whose wavelength is considerably shorter than the characteristic

radius

of

the circle. Analytically, the result depends on the asymptotic behavior

of

Bessel functions of nearly equal argument and order.

A similar result is worked out here for the interior of a spherical dome by

using boundary-layer ideas. Only axisymmetric modes are considered. The wave

equation of acoustics for the velocity potential is

a

2

J> 2

a c ~

I

a

2

J> cot 0 a I> I

a

2

J>

- -

+ -

+ - -

0.

ar

2

r ar r

2

ao

2

r

2

ao c

2

at2

The velocity and density perturbations are, as usual,

q =

grad< >,

o

o

(3.3.323)

(3.3.324a)

(3.3.324b)

where o is the ambient density, c =

Jy

Pol o is the speed

of

sound, o is the

ambient pressure, and

y

is the ratio of specific heats.

The coordinates are the usual spherical polar coordinates

x

=

r cos 0 y

=

r sin() cos 1/J z

=

r sin() sin 1/J

(3.3.325)

where ) is the pole angle, and

1/J

is the azimuth angle. The radius of the sphere is

taken to be a. The boundary condition at the wall, r

=

a is that the radial velocity

is

zero

a I>

ar a, ), t = 0,

(3.3.326)

and the solution should be regular at r = 0.

A point source at()

= 0

ould produce an axisymmetric wave field (

=

0 .

Since these modes propagate almost unattenuated along the ) direction but are

confined to a thin region close t r = a we can assume the following form for

Kevorkian Cole Multiple Scale Singular Perturbation Methods

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