Discussion and Examples

We have shown that the SSD in the case

g(x)

=

gx

and

b(x)

= bxl,

C ₀₀ (-l)lIak"e-�

y(x)

=

x

2

!

_(ak - 1 ) _{... (ab'} - 1 ) '

where a =

�

and

{ (k)l

_•

_{-1 )}

1 ...

a

C =

(k)l

1

• , t =

�, m

E 1+ , t #

�,m

E 1+. (3.62) (3.63) (3.M)

Continuity of C

In order to show that C is a continuous function of le, we need to prove that

lim

_K(�,a)r(l

_ =

(ak - 1)(a2k

. .

(a(lII-l)k - 1 ) K(�,l) loga.

(3.65)

1 __ -1

m.

Now using (3.42) and separating out the one factor which becomes zero as le _

m-I

, we write

00

K(ak, a)

=

(1 -ai-h)

n

(1 -

ai-kit).

II=I,llPt

(3.66)

Similarly, we separate out the factor in

r(1 - i)

which causes difficulties as le _

m-I ,

by applying the iterative relationship

r(x - 1)

=

��l m

times, giving

r ( 1 _ !. -

_{t) -}

_{m - I (1 - i)(2 - i) . . . «m - 1) - i> "} r(l - i + m)

Substituting these expansions into _the left hand side of(3.65) we have

lim K(ak a)r(l - !.)

-

lim

{ r(l - i

ai-kit)

1 __ -1 ' le

-

1 __ -1

(1 - )(2 - . . . «m - 1) -

m -

(3.67)

(3.68)

Now the limit of the expression in square brackets can be found by a single application of

l'Hopital's rule,

(3.69)

m

Substituting le =

m-I

into the rest of (3.68) then gives

Now

lim

K(ak, a)r(l - !.)

= aI-�) log a

1 ... 111-1 le

(-1

)",-1

(m - 1 )(m -

2)

. . . 1 m

(3.70)

(3.71)

so after expanding the left-hand product then changing the summation variable in the

right-hand product to get it into the form required by (3.42) we get

00

n

(1 -

_aI-�) = (-1

j-l(� - 1

)(�

- 1)

. . .

(a(lII-l)k - l)K(ak,l ).

(3.72)

1I=I,llPt

Substitution ofthis expression into (3.70) now gives (3.65), so C is a continuous function of le as expected.

Moments of �he SSD

When k is an integer, we can obtain the kth moment of the SSD about the origin, Ilk =

fooo xky(x)dx,

directly from the integral condition (3.27) as

1

Pt. = a

(a - 1)"

(3.73)

An iterative expression for moments beyond the kth is obtainable by applying a Mellin Transform to (3.20): multiplying through by .%"'-1 , where

m

is an integer, and integrating

over all

x,

remembering that

y(x)

=

x-2Z(x)

gives

1000 ,t"-lZ'(x)dx

= -ap",+k + a

1000 ak,t"+A:[x-2Z(ax)]dt.

(3.74)

Integration by parts on the left and the substitution z = ax on the right then leads to (3.75)

so we have the iterative equation

(m - 1)

_(3.76)

provided

m

> 1. This enables us to deduce all the moments of the SSD once we have found the kth moment by using (3.73) and all other moments up to the (t + 1 )th by integration of

(3.63) directly.

For example, in the particular case t =

1,

the mean is

1

_(3.77)

Jl = a

(a - 1)

and the second moment about the origin

(Jl2

=

fO'ry(x)dx = fooo Z(x)dt)

can be obtained by

integrating (3.25) term by term to give

1

Jl2 =

a2(a

- 1 ) log a ·

We can then apply (3.76) repeatedly to obtain a"(

a

-1 ) log

a

-

(3.78)

1 .2 ... . 8 x - >- .6 .4 .2 .5 2.5 x

Figure 3.1 : SSD probability density functions y(x) for g(x) = gx and hex)

=

hxk for a range of values of le. The variables a and

a

have been fixed at

1

and

2

respectively.

An Example

Figure 3.1 shows the shape of the SSD function y(x) with a = 1 and

a = 2

for a range of values of /c. Consideration of the differential equation formed in the extreme cases shows that as le -- 0, the SSD y(x) approaches the Dirac delta function 6(x), and as /c __ 00 it

approaches (Koch and Schaechter [30)) the function

y(x)

= {

11 a < 11 x <.Q

o , OtherwISe.

3.4 A Minimum Size for Division: The Case

b(x)

=

bxkH(x

-Xl

)

This case is the same as that discussed in the previous section, except that there is a minimum size

Xl

> 0 below which cells do not divide. Substituting

b(x) = bxkH(x - Xl )'

k

> 0, (where H is the Heavyside or unit step function and

b

and

Xl

are constants) into equation

(3.14)

we get

Z'(X) = xf-l ( -aH(x - XI )Z(X)

+

a�H(ax -XI)Z(ax» ,

(3.81 )

where as in Section 3.3 we have made the substitution

a =

�.

In this case we choose to couple (3.81 ) with the integral condition

(3.16)

as this simplifies to a condition dependent only on

Z(x)

for

X

�

Xl,

namely

100

-2

1

;q xf

Z(x)dt = a(a - 1

r (3.82)

It is convenient to solve the equation in three regions starting with the region

X

>

Xl .

We

let

{ ZI(X),X

�

Xl

(region

1 )

Z(x) =

_Z:a

_(x)

, � �

X

�

Xl

(region 2)

Zs(x),x

� � (region 3)

where continuity considerations require

ZI (XI) = Z:a(XI )

and Zs(�)

= Z:a(�).

3.4.1 Region 1, x �

Xl

In this region, (3.81 ) becomes (3.20), so the solution is

00

( -1 )

" aD

!!!{t

ZI (X) =

C ,,

� _{(a-t - l ) .}

.

. (aG

'lb find the constant C, we substitute

Z(x) = Zl (x)

into (3.82) above, giving

c

100 [I

( -1

)"

ab'

xl-2

-

/l�

l

dx _

1

;q ,,=0

(a-t - 1) ... (aG - 1 )

e

- a(a - 1) "

(3.83)

(3.84)

(3.85)

This time there is no difficulty in immediately interchanging the integral and summation operations as the integration starts at

Xl

> 0 and therefore avoids the singularity at

X =

0

when le � 1 . Making the change of variable z =

a�

in each integral leads to

•

Hence we can write

(le)

1 1

c - -- a

(a - l)S'

where

and r(c,x) =

f%oo

r-le-'ds is an incomplete gamma (unction.

In the particular case le = 1 , (3.87) reduces to c =

_a(a

le _{- 1 )S}

where

and El is the exponential integral.

8.4.2 Region 2, � :::; x < Xl (3.86) (3.87) (3.88) (3.89) (3.90)

In this region we have H(x -Xl ) = 0 and H( c:u -Xl ) = 1 80 we may immediately integrate (3.81 ) to obtain

(3.91 )

This integration can be carried out term by term, using the substitution

z = a�,

to give

00 (-1 'f

[ _*

�(x) =

C,.!O

(at - 1 )

... (abt

- 1 ) e - e ,

where the constant C is the same as that in the expression for Zl (x) in (3.84).

3.4.3 Region 3, x < � In this region the (3.81 ) becomes

�(X) =

0

so as we have

Z3(�)

= ZJ(�) =

0

from (3.92) the solution is simply

Za(X) =

0

as expected.

3.4.4 Discussion and EXamples

(3.93)

(3.94)

Combining the results above and noting Z(x) = ry(x), the SSD in the case g(x) = gx and

b(x) = bxkH(x -Xl ) is given by (3.95) where (k) l

1

c- -

_{- a} (a - l )S (3.96) and (3.97)

Figure 3.2 shows this piecewise function for y(x) for a range of mjnjmum sizes for cell

division xlt with k = 2, a =

�

=

1,

and a = 2. The discontinuity in the slope ofy(x) at x = Xl

is a natural consequence of the step change in the probability of cell division b(x) (from

0

to

b�) at this size. The form ofy(x) for small values of Xl approaches the form given in Fig. 2.1 for k = 2.

1 .2 ... .8 x ... _>. .6 .4 .2 00 _{.5 1 .5 2} x

Figure 3.2: SSD probability density functions y(x) for g(x) = gx and b(x) = bxkH(x -Xl),

with k = 2, a = 1 ,

a

= 2, and a range of values of Xl

In document Steady size distributions in cell populations : a thesis presented in partial fulfilment of the requirements for the degree of Doctor Philosophy in Mathematics at Massey University (Page 61-68)