Murray J.D. Mathematical Biology: I. An Introduction

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246 7. Biological Oscillators and Switches

Figure 7.15. Compartment model for the control of testosterone production in the male. The hypothalmus

secretes luteinising hormone release hormone (LHRH), denoted by R(t), which controls the release of lutein-

ising hormone (LH), denoted by L(t), by the pituitary which controls the production of testosterone, T (t),by

the gonads. The dashed line denotes the feedback control to the hypothalmus from the testes.

where b

, b

, g

and g

are positive parameters and the negative feedback function

f (T ) is a positive monotonic decreasing function of T . At this stage we do not need a

speciﬁc form for f (T ) although we might reasonably take it to be typically of the form

A/(K + T

) as in the prototype feedback model (7.3).

As we mentioned, the blood level of testosterone in men oscillates in time. Exper-

iments in which the natural state is disturbed have also been carried out; see the brief

surveys given by Smith (1980) and Cartwright and Husain (1986). Our interest here is

mainly related to the observed periodic ﬂuctuations.

From the analysis in Section 7.2, or simply by inspection, we know that a positive

steady state R

, L

and T

exists for the model (7.43). With the speciﬁc form (7.3)

for f (T ) oscillations exist for a Hill coefﬁcient m ≥ 8 (Exercise 4), which we noted

before is an unrealistically high ﬁgure. We can modify the speciﬁc form of f (T ) so

that periodic solutions exist but this is essentially the same as choosing the form in

(7.3) with m ≥ 8. We assume therefore that the feedback function f (T ) is such that

the steady state is always stable. Thus we must modify the model to include more of

the physiology. If we consider the actual process that is taking place there must be a

delay between production of the hormone at one level and its effect on the production

of the hormone it stimulates, simply because of their spatial separation and the fact that

the hormones are transported by circulating blood. Accordingly W. R. Smith and J. D.

Murray in the early 1980’s suggested a simple delay model based on the modiﬁcation

to the system (7.3) with m = 1, similar to the delay control model suggested by Murray

(1977), in which the production of testosterone is delayed. Although it is reasonable

to consider a delay in each hormone’s production they incorporated them all in the T -

equation so as to be able to investigate the system analytically and hence get an intuitive

feel for the effect of delay on the system. They took, in place of (7.43),

= f (T ) − b

= g

R − b

= g

L(t −τ)− b

(7.44)

7.6 Testosterone Secretion and Chemical Castration 247

where τ is a delay associated with the blood circulation time in the body. The steady

state is again (R

, L

, T

) determined by

, R

, f (T

) −

= 0, (7.45)

whichalwaysexistsif f (0)>0and f (T ) is a monotonic decreasing function. If we

now investigate the stability of the steady state by writing

x = R − R

, y = L − L

, z = T − T

(7.46)

the linearised system from (7.44) is

= f



)z − b

= g

x −b

= g

y(t −τ)−b

(7.47)

Now look for solutions in the form









= A exp [λt], (7.48)

where A is a constant vector. On substitution into (7.47) we have

+aλ

+bλ + c +de

−λτ

= 0,

a = b

, b = b

c = b

, d =−f



> 0.

(7.49)

We now want to determine the conditions for the steady state to be linearly unstable;

that is, we require the conditions on a, b, c, d and τ such that there are solutions of

(7.49) with Re λ>0.

We know that with τ = 0thesteadystate(R

, L

, T

) is stable; that is, if τ = 0in

(7.49), Re λ<0. Using the Routh–Hurwitz conditions on the cubic in λ given by (7.49)

with τ = 0 this means that a, b, c and d necessarily satisfy

a > 0, c +d > 0, ab − c −d > 0. (7.50)

We know that delay can be destabilising so we now try to determine the critical delay

> 0, in terms of a, b, c and d so that a solution with Re λ>0 exists for τ>τ

.We

determine the conditions by considering (7.49) as a complex variable mapping problem.

From the analysis on transcendental equations in Chapter 1 we know that all so-

lutions of (7.49) have Re λ bounded above. The critical τ

is that value of τ such that

248 7. Biological Oscillators and Switches

Figure 7.16. (a) λ-plane. (b) w-plane.

Re λ = 0, that is, the bifurcation between stability and instability in the solutions (7.48).

Consider the transformation from the λ-plane to the w-plane deﬁned by

w = λ

+aλ

+bλ + c +de

−λτ

,τ>0. (7.51)

Setting λ = µ +iν this gives

w =[µ

−3µν

+a(µ

−ν

) +bµ + c +de

−µτ

cos ντ]

+i[−ν

+3µ

ν +2aµν +bν −de

−µτ

sin ντ].

(7.52)

We wish to ﬁnd the conditions on a, b, c, d and τ such that w = 0 has solutions µ>0;

the bifurcation state is µ = 0.

Consider the contour in the λ-plane consisting of the imaginary axis and a semi-

circle of inﬁnite radius as shown schematically in Figure 7.16(a). Without any delay

τ = 0 and we know that in this case Re λ<0sow as a function of λ in (7.51) does

not pass through the origin in the w-plane; that is, the map in the w-plane of AGH A

in Figure 7.16(a) does not pass through w = 0. Consider ﬁrst τ = 0 and the mapping

(7.52). AG in Figure 7.16(a), and on which µ = 0, is mapped by

w =[(c −aν

) + d]+i[bν −ν

] (7.53)

onto A



in Figure 7.16(b) with D



(= c + d + i0) in the w-plane corresponding to

D(= 0+i0) in the λ-plane. The domain V is mapped into V



; the hatches in Figure 7.16

point into the respective domains. The points A, B, C, D, E, F and G in Figure 7.16(a)

are mapped onto their primed equivalents as

7.6 Testosterone Secretion and Chemical Castration 249

A (∞e

iπ/2

) A



(∞e

3iπ/2

)

B (

√

iπ/2

) B



((ab − c − d)e

iπ

)





c +d



1/2

iπ/2









c +d



1/2



b −

c +d



iπ/2



D (0) D



(c +d)





c +d



1/2

−iπ/2









c +d



1/2



b −

c +d



−iπ/2



F (

√

−iπ/2

) F



((ab − c − d)e

−iπ

)

G (∞e

−iπ/2

) G



(∞e

−3iπ/2

(7.54)

As the semi-circle GHAis traversed λ moves from ∞e

−iπ/2

to ∞e

iπ/2

and so w(∼ λ

)

moves from ∞e

−3iπ/2

, namely, G



,to∞, namely, H



, and then to ∞e

3iπ/2

,thatis,A



all as shown in Figure 7.16(b).

Now consider the mapping in the form (7.52) with τ>0. The line AG in Fig-

ure 7.16(a) has µ = 0 as before, so it now maps onto

w =[(c −aν

) +d cos ντ]+i[bν −ν

−d sin ντ]. (7.55)

The effect of the trigonometric terms is simply to add oscillations to the line



as shown schematically in Figure 7.17(a) which is now the map of

Figure 7.17. Map in the w-plane of the contour in Figure 7.16(a) in the λ-plane under the transformation

(7.55): (a) τ<τ

;(b) τ>τ

. In the latter case, the map encloses the origin.

250 7. Biological Oscillators and Switches

ABCDEFGHA under (7.55). Now let τ increase. As soon as the transformed curve

passes through the origin in the w-plane this gives the critical τ

.Forτ>τ

, the map-

ping is schematically as shown in Figure 7.17(b) and the origin in the w-plane is now

enclosed in V



, that is, the transformation of V under (7.55).

If we now traverse the boundary of V



and compute the change in arg w we imme-

diately get the number of roots of w(λ) deﬁned by (7.51). It helps to refer to both Fig-

ures 7.17(a) and (b). Let us start at A



where arg w = 3π/2. On reaching B



,argw = π

and so on, giving

Point A



(via H



)

arg w

3π

2π

5π

3π

5π

+3π



11π



Thus the change in arg w is 4π which implies two roots with Re λ>0 in the domain

V ; the roots are complex conjugates.

Let us now obtain expressions for the critical τ

such that the curve A



just passes through the origin in the w-plane. The bifurcation value we are

interested in for Re λ is of course µ = 0 so we require the value of τ such that w = 0

in (7.55), namely,

c −aν

+d cos ντ = 0, bν − ν

−d sin ντ = 0, (7.56)

from which we get

cot ντ =

aν

−c

ν(b −ν

)

⇒ ν = ν(τ). (7.57)

If we plot each side of (7.57) as a function of ν, as shown schematically in Figure 7.18

where for illustration we have taken

√

b <π/τ, we see that there is always a solution,

Figure 7.18. Schematic graphical solution ν of (7.57) in

the case where

√

b <π/τ.

7.6 Testosterone Secretion and Chemical Castration 251

given by the intersection of the curves, such that

0 <ν(τ)<

, 0 <ν(τ)<

√

b. (7.58)

Furthermore, the solution ν(τ) as a function of τ satisﬁes

ν(τ

)<ν(τ

) if τ

>τ

√

b >π/τthen (7.58) still holds but another solution exists with π/τ < ν(τ) <

√

Consider now the solution ν(τ) which satisﬁes (7.58). A solution of the simultane-

ous equations (7.56) must satisfy (7.57) and one of (7.56), which to be speciﬁc we take

here to be the second. With ν(τ) the solution of (7.57) satisfying (7.58), the second of

(7.56) gives

d =

[b −ν

(τ)]ν(τ)

sin(ν(τ)τ )

. (7.59)

If τ is such that this equation cannot be satisﬁed with the corresponding ν(τ) then no

solution can exist with Re λ>0. We can not get an analytical solution of (7.57) for

ν(τ) but an indication of how the solution behaves is easily obtained for τ large and

small. For the case in (7.58), we have from (7.57),

ν(τ) =

√

b −(ab − c)

√

+ O(τ

) for 0 <τ  1 (7.60)

and the second of (7.56) becomes

bν(τ) − ν

(τ) −d sin[ν(τ)τ]≈(ab − c)τ

√

b −dτ

√

b +···

= τ(ab − c − d)

√

b + O(τ

)

> 0

from conditions (7.50). Thus for small enough τ no solution exists with Re λ>0. Now

let τ increase until a solution to (7.59) can be satisﬁed: this determines the critical τ

the solution (7.58) of (7.57) which also satisﬁes (7.59). The procedure is to obtain ν(τ)

from (7.57) and then the value τ

which satisﬁes (7.59).

Now suppose τ  1. From (7.57) we get

ν(τ) ∼

−

πb

cτ

+ O





for τ  1,

and now the second of (7.56) gives

bν − ν

−d sin[ν(τ)τ]∼

bπ

−

bdπ

cτ

+ O





< 0ifd > c.

Thus there is a range of τ>τ

> 0 such that a solution λ exists with Re λ>0ifd > c.

252 7. Biological Oscillators and Switches

An approximate range for ν(τ) can be found by noting that with ν<

√

b <π/τ,

the speciﬁc case we are considering and which is sketched in Figure 7.18,

g(ν) = c −aν

+d cos ντ

is monotonic decreasing in 0 ≤ ντ < π.Furthermore,





c +d



1/2



=−d + d cos





c +d



1/2



< 0, g(0) = c +d > 0

which implies that

0 <ν<



c +d



1/2

√

b <

We have thus demonstrated that there is a critical delay τ

such that the steady state

, L

, T

) is linearly unstable by growing oscillations. Since the model system (7.44)

has a conﬁned set we might thus expect limit cycle periodic solutions to be generated;

this is indeed what happens when the parameters are chosen so that the steady state is

linearly unstable.

The model proposed by Cartwright and Husain (1986) is based on further exper-

imental results and includes further delays in the production of each of R, L and T .

They also incorporate feedback by LH as well as T . Their model is necessarily more

complex. Analytical results such as we have derived above, even if possible, would nec-

essarily be much more complicated. Numerical simulations of their model system with

reasonable parameter values show stable periodic solutions in all of R, L and T .They

also carry out mathematical ‘experiments’ which mimic certain laboratory experiments,

with encouraging results.

Chemical Castration

Returning to the effect of drugs, such as Lupron, mentioned above, they effect chemical

castration by blocking the production of the hormone LH produced by the pituitary.

That is, in the model system (7.44), g

= 0. In this case, the governing equation for

L(t), that is, the concentration of LH, is uncoupled from the other equations and L → 0

with time, which in turn implies from the T -equation in (7.44) that T → 0 with time,

which is the equivalent of castration. This castration procedure could be used to replace

the widely used surgical methods currently used by veterinarians on farm and domestic

animals. Such a vaccine has been developed by Carelli et al. (1982).

More recent work related to chemical castration by Ferro and Stimson (1996) and

Ferro et al. (1995) is based on gonadotrophin releasing hormone (GnRH) in the form

of a GnRH-releasing vaccine which blocks the production of the leutinising hormone

LH; it has a possible application to human sex-hormone-dependent disorders. It is also

a potential treatment for oestrogen-dependent breast cancer (Ferro and Stimson 1998).

Before leaving the subject of interfering with testosterone production, it is relevant

to mention that there has been much research trying to develop a contraceptive ‘pill’

Exercises 253

for men. One avenue was to use artiﬁcial hormones to try and fool the body’s system

into shutting down the production of testosterone. In the past, for example, overload-

ing the body with testosterone shut down the pituitary gland’s production of LH. The

pituitary gland also produces another hormone (not included in our simple model), the

follicle-stimulating hormone, FSH, which promotes sperm formation by inducing them

to multiply. LH also helps this process. If there is no LH and low levels of testosterone

in the testes no sperm are produced. The problem is that if there is a high level of

testosterone in the blood there are some unpleasant side effects such as irritability, acne

and weight gain. Recently Anderson et al. (1997) developed a combination of an arti-

ﬁcial form of progesterone (a major ingredient in the female contraceptive pill) and a

slow-release testosterone pill which has less serious side effects. The testosterone pill

is placed under the skin and provides a sufﬁcient supply of testosterone for the body’s

needs for some months without requiring the testes to produce testosterone. Apparently

men involved in the clinical trials have greatly enjoyed participating.

Another approach is being developed (Ferro and Stimson 1998) which is totally

different, namely, to eliminate FSH entirely. It is a vaccine which produces antibodies

which attach to FSH and makes it inactive. This has not yet reached human trials—only

in rats—but if it works it will require about one treatment a year. In spite of the obvious

beneﬁts and the potentially enormous proﬁts for companies which develop such male

contraceptive procedures

few have so far become involved, in part, apparently, because

of fear of litigation in the U.S.A.

Exercises

1 The ‘Brusselator’ reaction mechanism proposed by Prigogene and Lefever (1968) is

→ X, B + X

→ Y + D, 2X

→ 3X, X

→ E,

where the ks are the rate constants, and the reactant concentrations of A and B are

kept constant. Write down the governing differential equation system for the con-

centrations of X and Y and nondimensionalise the equations so that they become

dτ

= 1 −(b +1)u +au

dτ

= bu −au

where u and v correspond to X and Y , τ = k

t, a = k

and b = k

B/k

Determine the positive steady state and show that there is a bifurcation value b =

= 1 + a at which the steady state becomes unstable in a Hopf bifurcation way.

Hence show that in the vicinity of b = b

there is a limit cycle periodic solution with

period 2π/

√

2 In the reaction mechanism du/dt = f(u) the kinetics f is curl free, that is, curl

f(u) =

0. This implies that f can be written in terms of a gradient of a potential F(u);that

The World Health Organisation reported in 1994 that in 70% of couples it is the women who are respon-

sible for contraception.

254 7. Biological Oscillators and Switches

is,

curl

f(u) = 0 ⇒ f(u) =∇

F(u).

The model system becomes

= f(u) =∇

F(u),

which is called a gradient system. By supposing that u(t) is a periodic solution with

period T , show, by considering



t+T





ds,

that a gradient system cannot have periodic solutions.

3 In the feedback control system governed by

= f (u

) −k

= u

r−1

−k

, r = 2, 3,...n

the feedback function is given by

(i) f (u) =

a + u

1 +u

,(ii) f (u) =

1 +u

where a and m are positive constants. Determine which of these represents a positive

feedback control and which a negative feedback control. Determine the steady states

and hence show that with positive feedback multi-steady states are possible while if

f (u) represents negative feedback there is only a unique steady state.

4 Consider the negative feedback mechanism

1 +u

−k

= u

i−1

−k

, i = 2, 3.

(i) For the case m = 1, show that a conﬁned set B is given by a rectangular box

whose sides are bounded by the u

= 0, i = 1, 2, 3axesandU

, i = 1, 2, 3,

where

1 + U

< k

Exercises 255

and hence determine U

, i = 1, 2, 3. For the case m = 1 show that U

is given

by the appropriate solution of the equation

m+1

−

= 0.

(ii) Prove, using the Routh–Hurwitz conditions (see Appendix B), that the model

with m = 1 cannot have limit cycle periodic solutions.

(iii) Prove that limit cycle solutions are possible if m > 8. [At one stage in the

analysis for this case you will need to use the general inequality

≥





1/2

≥ (k

)

1/3

5 Sketch the null clines for the system (Exercise 1):

dτ

= 1 −(b +1)u +au

v = f (u,v),

dτ

= bu −au

v = g(u,v).

Note the signs of f and g in the (u,v) phase plane and ﬁnd a conﬁned set (not a

trivial exercise) enclosing the steady state. Determine the (a, b) parameter domain

where the system has periodic solutions.

6 Consider the dimensionless activator (u)–inhibitor (v) system represented by

= a − bu +

= f (u,v),

= u

−v = g(u,v),

where a, b(> 0) are parameters. Sketch the null clines, append the signs of f and g,

and examine the signs in the stability matrix for the steady state. Is there a conﬁned

set? Show that the (a, b) parameter space in which u and v may exhibit periodic

behaviour is bounded by the curve

b =

1 −a

−1,

and hence sketch the domain in which the system could have periodic solutions.

Consider the modiﬁed system in which there is inhibition by u.Inthis

case u

/v is replaced by u

/[v(1 + Ku

)] in the u equation, where K(> 0) is the

inhibition parameter. Sketch the null clines. Show that the boundary curve in (a, b)

space for the domain in which periodic solutions may exist is given parametrically

b =

(1 + Ku

)

−1,

a =

(1 + Ku

)

−u

−

(1 + Ku

)

b ≥ 0