Owen Lewis
Math 5110
Homework #4
11/20/2015
1 Read Chapters 1 and 2 in the textbook.
Just read. This one isn’t hard. If you haven’t done it already, please do in the future. It
will provide valuable supplemental information to what was covered in lecture.
2 A semi-toxic chemical is ingested by an animal and enters its bloodstream at a constant rate D. It distributes within the body, passing from the blood to soft tissue to bones
with (positive) rate constants indicated in the figure. It may be removed from the blood
(and excreted as urine) and the soft tissue (through sweat) at rates u and s respectively.
A model of this type was proposed by Batsschelet, et. al in 1979 to study the kinetics of
lead in the human body. Let x1 , x2 , and x3 represent the concentration of the chemical in
the three resevoirs. From the diagram below, we may write the following equation for
the evolution of chemical concentration in the blood,
dx1
= D − ux1 − k12 x1 + k21 x2 .
dt
(1)
(a) Assuming that exchange between all three compartments is linear (as in the above
equation), write equations for the evolution of x2 and x3 .
(b) Find the steady-state concentration of chemicals throughout the body.
(c) Do you think that this steady state is stable? How would you find out?
1
(a) Top define the rate of change of concentration in each compartment, we simply sum
the flux in (or out) through each channel. This gives the equations
dx2
= k12 x1 − k21 x2 + k32 x3 − k23 x2 − sx2 ,
dt
dx3
= −k32 x3 + k23 x2 .
dt
(b) The steady state is defined to be a value ( x1∗ , x2∗ , x3∗ ) that results in all three time derivatives being zero. Algebraically, this may be written as
0 = D − ux1∗ − k12 x1∗ + k21 x2∗ ,
0 = k12 x1∗ − k21 x2∗ + k32 x3∗ − k23 x2∗ − sx2∗ ,
0 = −k32 x3∗ + k23 x2∗ .
We may use the third equation to solve for x3∗ as a function of x2∗
x3∗ =
k23 ∗
x .
k32 2
We can then substitute this into the second equation, eliminating x3∗ , and solve for x2∗ as a
function of x1∗ ,
k12
x2∗ =
x∗ .
s + k21 1
This allows us to eliminate x2∗ from the first equation, and directly solve for x1∗ ,
x1∗ =
D (k21 + s)
.
uk21 + us + k12 s
We can now readily evaluate x1∗ and x3∗
Dk12
,
uk21 + us + k12 s
Dk12 k23
x3∗ =
.
k32 (uk21 + us + k12 s)
x2∗ =
(c) First, notice that the equations which describe this system are nearly linear. The only
non-linear term is the additive constant D which appears in the first equation. Second,
notice that all of the coefficients in front of x1 are negative in the first equation. Similarly,
all of the coefficients in front of x2 are negative in the second equation, and all of the
coefficients in front of x3 are negative in the third equation. This implies a behavior like
exponential decay for all three variables. Based on this, I would guess that the fixed point
is stable. However, to test this, I would first form the Jacobian matrix


−u − k12
k21
0
−k21 − k23 − s k32  ,
J |( x∗ ,x∗ ,x∗ ) =  k12
1 2 3
0
k23
−k32
and calculate its eigenvalues. If they are all negative, the equilibrium is stable.
2
3 In 1971 J.S. Griffith proposed the following model for the interaction of messenger
RNA and protein. The concentration of messenger RNA is given by M and the concentration of the protein is given by by E. These concentrations evolve according to the
equations
aKEm
dM
=
− bM
dt
1 + KEm
dE
= cM − gE.
dt
(2)
(3)
(a) Rescale these equations to arrive at the non-dimensional set of equations
d M̃
Ẽm
− α M̃
=
dt
1 + Ẽm
d Ẽ
= M̃ − β Ẽ.
dt
(4)
(5)
Give the non-dimensional parameters α and β in terms of the dimensional parameters a, K, b, c, and g. For the rest of this problem, analyze the non-dimensional
equation.
(b) Find one steady state, and show that there is another steady state that is defined by
an algebraic equation. Now set m = 1. What condition must the non-dimensional
parameters satisfy for the second equilibrium to exist and be positive (physical).
Draw a phase-plane diagram for both cases where this equilibrium does and does
not exist. What happens to the concentrations M̃ and Ẽ?
(c) Now set m = 2. Find an algebraic equation that defines the steady state protein
concentration. Show that for certain conditions on α and β, there are two positive
physical equilibrium and if this condition is broken, there are none. Draw the
phase diagram for each case. What happens to the concentrations M̃ and Ẽ?
(a) We begin by addressing the denominator of the first equation. Because it is added to
one (which is dimensionless), the quantity KEm must be dimensionless. From here, it is
relatively easy to determine the units of the various parameters in the problem.
Variable or Parameter
[ M]
[ E]
[t]
[K ]
[ a]
[b]
[c]
[ g]
Units
Concentration of MRNA
Concentration of protein
Time
(Concentration of protein)−m
Concentration of MRNA /Time
1/Time
Concentration of protein/(Time · Concentration of MRNA)
1/Time
3
Now, we have some potential candidates for scales with which to non-dimensionalize
the system. However, we will address that issue later. For now, let us define our nondimensional variables with the relations
M = M M̃,
E = E Ẽ,
t = Tτ,
where M̃, Ẽ and τ are non-dimensional variables, while M, E, and T are dimensional
constants with units of concentration of MRNA, concentration of protein, and time. If we
substitute these expressions into the dimensional equations and rearrange, we arrive at
the following equations:
m
d M̃
aT KE Ẽm
=
− bT M̃
m
dτ
M 1 + KE Ẽm
cTM
d Ẽ
=
M̃ − gT Ẽ.
dτ
E
We now set the unknown scale for protein concentration to E = K −1/m . This reduces our
equation to
d M̃
aT Ẽm
=
− bT M̃
dτ
M 1 + Ẽm
cTM
d Ẽ
= −1/m M̃ − gT Ẽ.
dτ
K
Now, notice that each of the groups of parameters is non-dimensional. We are now free
to choose our MRNA and time scales in any reasonable manner. To eliminate the nondimensional groups in front of the first term of each equation, we must satisfy the following two algebraic equations
aT
=1
M
cTM
= 1.
K −1/m
These equations can be solved for T and M to find that
r
a
M=
cK1/m
r
1
T=
.
acK1/m
Plugging these back into the non-dimensional
equations easy gives us the system (4) &
q
q
2
2
g
(5), where α = acKb1/m , and β = acK1/m .
4
(b) Taking the non-dimensional form of the system, we set both derivatives to zero to
arrive at the set of equations that define the equilibria ( E∗ , M∗ ).
E∗m
− αM∗
1 + E∗m
0 = M∗ − βE∗ .
0=
(6)
(7)
We can solve the second equation to get M∗ = βE∗ . This will be useful later, as anytime
we find E∗ , we will immediately know M∗ . Substituting this into the first equation, we
have
E∗m
0=
− αβE∗ .
1 + E∗m
By inspection, E∗ = 0 clearly satisfies this relationship. Therefore we have one equilibrium, ( E∗ , M∗ ) = (0, 0) the trivial equilibrium. Now, if E∗ 6= 0, we may safely divide
through by E∗ and arrive at the relationship.
0=
Which can be rewritten as
E∗m−1
− αβ,
1 + E∗m
0 = E∗m−1 − αβ (1 + E∗m ) .
This defines any other equilibria that the system may have.
We now set m = 1 and solve for
1 − αβ
.
E∗ =
αβ
This implies that there is an equilibrium at
E∗ =
1 − αβ
,
αβ
M∗ =
β (1 − αβ)
.
αβ
(8)
This equilibrium is physical assuming that 1 − αβ ≥ 0, or equivalently when αβ ≤ 1.
In Figure 1, we show the behavior of the system in both cases. First, we show the phase
plane when α = 1 and β = 1/2. In this case, the trivial equilibrium is a saddle point, while
the non-trivial equilibrium is a stable node. All physical (positive) solutions eventually
settle to the non-trivial steady state where there is some finite amount of MRNA and the
protein. Second, we show the phase plane when α = 1 and β = 6/5. Now the non-trivial
steady state has passed through zero and become unphysical. A transcritical bifurcation
has occurred. The trivial steady state is now stable and all trajectories decay to zero. No
MRNA or protein exists.
(c) From part (b), we know that any non-trivial equilibria are defined by
0 = E∗m−1 − αβ (1 + E∗m ) .
5
a=1
b = 0.5
a=1
b = 1.2
M ’ = E/(1 + E) − a M
E’=M−bE
4
4
3.5
3.5
3
3
2.5
2.5
2
2
E
E
M ’ = E/(1 + E) − a M
E’=M−bE
1.5
1.5
1
1
0.5
0.5
0
0
0
0.5
1
1.5
2
M
2.5
3
3.5
4
0
0.5
1
1.5
2
M
2.5
3
3.5
4
Figure 1: Phase plane showing null-clines and a few sample solutions for both cases. E
null-clines are shown in orange. M null-clines are shown in pink. Solution trajectories are
shown in blue. The first plot was generated with α = 1 and β = 1/2. The second panel
was generated with α = 1 and β = 6/5.
When m = 2, this results in quadratic polynomial, which we can solve via the quadratic
formula. We therefore have
p
1 ± 1 − 4α2 β2
.
E∗ =
2αβ
These E∗ values are associated with
M∗ =
1±
p
1 − 4α2 β2
.
2α
Clearly, the will be physical if 1 − 4α2 β2 ≥ 0. This can be rewritten as αβ ≤ 1/2. Clearly,
if this condition is not met, then the roots expressed above are imaginary, and do not
correspond to equilibria in the phase plane.
In Figure 2 we show the phase plane of the system where non-trivial steady states do
and do not exist. First, we show the behavior of the system when α = 9/10 and β = 1/2.
Because αβ < 1/2, two non-trivial equilibria exist. One is a saddle point and one is a
stable node. In contrast to the case explored in (b), the trivial equilibrium is stable in
this case. Therefore, if the initial concentrations are above a threshold (defined by the
separatrix of the saddle point), the system will go to a positive stable equilibrium. If the
initial concentration is below this threshold, both species will decay to zero. In the second
case, we show the behavior when α = 11/10 and β = 1/2. Because αβ > 1/2, the nontrivial equilibria are gone. A saddle node bifurcation has occurred. Now the MRNA and
protein decay to zero regardless of starting concentration.
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M ’ = E2/(1 + E2) − a M
E’=M−bE
a = 0.9
b = 0.5
5
5
4.5
4.5
4
4
3.5
3.5
3
3
2.5
2.5
E
E
M ’ = E2/(1 + E2) − a M
E’=M−bE
2
2
1.5
1.5
1
1
0.5
0.5
0
a = 1.1
b = 0.5
0
0
0.5
1
1.5
2
M
2.5
3
3.5
4
0
0.5
1
1.5
2
M
2.5
3
3.5
4
Figure 2: Phase plane showing null-clines and a few sample solutions for both cases. E
null-clines are shown in orange. M null-clines are shown in pink. Solution trajectories
are shown in blue. The stable and unstable orbits of the saddle point in the first panel are
shown in green. The first plot was generated with α = 9/10 and β = 1/2. The second
panel was generated with α = 11/10 and β = 1/2.
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