Eduardo D. Sontag, Lecture Notes on Mathematical Biology 5
1 Modeling, Growth, Number of Parameters
1.1 Exponential Growth: Modeling
Let us start by reviewing a subject treated in the basic differential equations course, namely how one
derives differential equations for simple exponential growth,
Suppose that N (t) counts the population of a microorganism in culture, at time t, and write the
increment in a time interval [t, t + h] as “g(N (t), h)”, so that we have:
N (t + h) = N (t) + g(N (t), h) .
(The increment depends on the previous N (t), as well as on the length of the time interval.)
We expand g using a Taylor series to second order:
g(N, h) = a + bN + ch + eN 2 + f h2 + KN h + cubic and higher order terms
(a, b, . . . are some constants). Observe that
g(0, h) ≡ 0 and g(N, 0) ≡ 0 ,
since there is no increment if there is no population or if no time has elapsed. The first condition tells
us that
a + ch + f h2 + . . . ≡ 0 ,
for all h, so a = c = f = 0, and the second condition (check!) says that also b = N = 0.
Thus, we conclude that:
g(N, h) = KN h + cubic and higher order terms.
So, for h and N small:
N (t + h) = N (t) + KN (t)h , (1)
which says that
the increase in population during a (small) time interval
is proportional to the interval length and initial population size.
This means, for example, that if we double the initial population or if we double the interval,
the resulting population is doubled.
Obviously, (1) should not be expected to be true for large h, because of “compounding” effects.
It may or may not be true for large N , as we will discuss later.
We next explore the consequences of assuming Equation (1) holds for all small h>0 and all N .
As usual in applied mathematics, the “proof is in the pudding”:
one makes such an assumption, explores mathematical consequences that follow from it,
and generates predictions to be validated experimentally.
If the predictions pan out, we might want to keep the model.
If they do not, it is back to the drawing board and a new model has to be developed!
,Eduardo D. Sontag, Lecture Notes on Mathematical Biology 6
1.2 Exponential Growth: Math
From our approximation
KN (t)h = N (t + h) − N (t)
we have that
1
KN (t) =
(N (t + h) − N (t))
h
Taking the limit as h → 0, and remembering the definition of derivative, we conclude that the right-
dN
hand side converges to (t). We conclude that N satisfies the following differential equation:
dt
dN
= KN . (2)
dt
We may solve this equation by the method of separation of variables, as follows:
Z Z
dN dN
= Kdt ⇒ = K dt ⇒ ln N = Kt + c .
N N
Evaluating at t = 0, we have ln N0 = c, so that ln(N (t)/N0 ) = Kt. Taking exponentials, we have:
N (t) = N0 eKt (exponential growth: Malthus, 1798)
Bacterial populations tend to growth exponentially, so long as enough nutrients are available.
1.3 Limits to Growth: Modeling
Suppose now there is some number B (the carrying capacity of the environment) so that
populations N > B are not sustainable, i.e.. dN/dt < 0 whenever N = N (t) > B:
dN/dt
0 B N
It is reasonable to pick the simplest function that satisfies the stated requirement;
in this case, a parabola:
dN N
= rN 1 − (for some constant r > 0) (3)
dt B
But there is a different way to obtain the same equation, as follows.
Suppose that the growth rate “K” in Equation (2) depends on availability of a nutrient:
K = K(C) = K(0) + κC + o(C) ≈ κC (using that K(0) = 0)
where C = C(t) denotes the amount of the nutrient, which is depleted in proportion to the population
change: 1
dC dN
= −α = −αKN
dt dt
1
if N (t) counts the number of individuals, this is somewhat unrealistic, as it the ignores depletion of nutrient due to
the growth or individuals once they are born; it is sometimes better to think of N (t) as the total biomass at time t
, Eduardo D. Sontag, Lecture Notes on Mathematical Biology 7
(“20 new individuals formed ⇒ α × 20 less nutrient”). It follows that
d dC dN
(C + αN ) = +α = −αKN + αKN = 0
dt dt dt
and therefore C(t) + αN (t) must be constant, which we call “C 0 ”2
(we use this notation because C(0) + αN (0) ≈ C(0), if the population starts as N (0) ≈ 0).
So K = κC = κ(C0 − αN ), and Equation (2) becomes the same equation as (3), just with different
names of constants:
dN
= κ (C0 − αN ) N
dt
1.4 Logistic Equation: Math
dN N N (B − N )
We solve = rN 1− = r using again the method of separation of variables:
dt B B
Z Z
B dN
= r dt .
N (B − N )
We compute the integral using a partial fractions expansion:
Z Z
1 1 N N ecB
+ dN = r dt ⇒ ln = rt+c ⇒ cert ⇒ N (t) =
= e
N B−N B−N B−N c + e−rt
e
N0 B
⇒ e
c = N0 /(B − N0 ) ⇒ N (t) =
N0 + (B − N0 )e−rt
We can see that there is a B asymptote as t → ∞. Let’s graph with Maple:
with(plots):
f(t):=t->(0.2)/(0.2+0.8*exp(-t)):
p1:=plot(f(t),0..8,0..1.3,tickmarks=[0,2],thickness=3,color=black):
g:=t->1:
p2:=plot(g(t),0..8,tickmarks=[0,2],thickness=2,linestyle=2,color=black):
display(p1,p2);
Gause’s 1934 Experiments
G.F. Gause carried out experiments in 1934, involving Paramecium caudatum and Paramecium aure-
lia, which show clearly logistic growth:
2
this is an example of a “conservation law”, as we’ll discuss later
, Eduardo D. Sontag, Lecture Notes on Mathematical Biology 8
(# individuals and volume of P. caudatum and P. aurelia, cultivated separately, medium changed daily,
25 days.)
1.5 Changing Variables, Rescaling Time
We had this equation for growth under nutrient limitations:
dN
= κ (C0 − αN ) N
dt
which we solved explicitly and graphed for some special values of the parameters C 0 , κ, α.
But how do we know that “qualitatively” the solution “looks the same” for other parameter values?
Can the qualitative behavior of solutions depend upon the actual numbers C 0 , κ, α?
First of all, we notice that we could collect terms as
dN
= ((κC0 ) − (κα)N ) N = Ce0 − α
eN N
dt
(where C
e0 = κC0 and α
e = κα), so that we might as well suppose that κ = 1 (but change α, C 0 ).
But we can do even better and use changes of variables in N and t in order to eliminate the two
remaining parameters!
We will always proceed as follows:
• Write each variable (in this example, N and t) as a product of a new variable and a still-to-be-
determined constant.
• Substitute into the equations, simplify, and collect terms.
• Finally, pick values for the constants so that the equations (in this example, there is only one
differential equation, but in other examples there may be several) have as few remaining param-
eters as possible.
The procedure can be done in many ways (depending on how you collect terms, etc.), so different
people may get different solutions.
1 Modeling, Growth, Number of Parameters
1.1 Exponential Growth: Modeling
Let us start by reviewing a subject treated in the basic differential equations course, namely how one
derives differential equations for simple exponential growth,
Suppose that N (t) counts the population of a microorganism in culture, at time t, and write the
increment in a time interval [t, t + h] as “g(N (t), h)”, so that we have:
N (t + h) = N (t) + g(N (t), h) .
(The increment depends on the previous N (t), as well as on the length of the time interval.)
We expand g using a Taylor series to second order:
g(N, h) = a + bN + ch + eN 2 + f h2 + KN h + cubic and higher order terms
(a, b, . . . are some constants). Observe that
g(0, h) ≡ 0 and g(N, 0) ≡ 0 ,
since there is no increment if there is no population or if no time has elapsed. The first condition tells
us that
a + ch + f h2 + . . . ≡ 0 ,
for all h, so a = c = f = 0, and the second condition (check!) says that also b = N = 0.
Thus, we conclude that:
g(N, h) = KN h + cubic and higher order terms.
So, for h and N small:
N (t + h) = N (t) + KN (t)h , (1)
which says that
the increase in population during a (small) time interval
is proportional to the interval length and initial population size.
This means, for example, that if we double the initial population or if we double the interval,
the resulting population is doubled.
Obviously, (1) should not be expected to be true for large h, because of “compounding” effects.
It may or may not be true for large N , as we will discuss later.
We next explore the consequences of assuming Equation (1) holds for all small h>0 and all N .
As usual in applied mathematics, the “proof is in the pudding”:
one makes such an assumption, explores mathematical consequences that follow from it,
and generates predictions to be validated experimentally.
If the predictions pan out, we might want to keep the model.
If they do not, it is back to the drawing board and a new model has to be developed!
,Eduardo D. Sontag, Lecture Notes on Mathematical Biology 6
1.2 Exponential Growth: Math
From our approximation
KN (t)h = N (t + h) − N (t)
we have that
1
KN (t) =
(N (t + h) − N (t))
h
Taking the limit as h → 0, and remembering the definition of derivative, we conclude that the right-
dN
hand side converges to (t). We conclude that N satisfies the following differential equation:
dt
dN
= KN . (2)
dt
We may solve this equation by the method of separation of variables, as follows:
Z Z
dN dN
= Kdt ⇒ = K dt ⇒ ln N = Kt + c .
N N
Evaluating at t = 0, we have ln N0 = c, so that ln(N (t)/N0 ) = Kt. Taking exponentials, we have:
N (t) = N0 eKt (exponential growth: Malthus, 1798)
Bacterial populations tend to growth exponentially, so long as enough nutrients are available.
1.3 Limits to Growth: Modeling
Suppose now there is some number B (the carrying capacity of the environment) so that
populations N > B are not sustainable, i.e.. dN/dt < 0 whenever N = N (t) > B:
dN/dt
0 B N
It is reasonable to pick the simplest function that satisfies the stated requirement;
in this case, a parabola:
dN N
= rN 1 − (for some constant r > 0) (3)
dt B
But there is a different way to obtain the same equation, as follows.
Suppose that the growth rate “K” in Equation (2) depends on availability of a nutrient:
K = K(C) = K(0) + κC + o(C) ≈ κC (using that K(0) = 0)
where C = C(t) denotes the amount of the nutrient, which is depleted in proportion to the population
change: 1
dC dN
= −α = −αKN
dt dt
1
if N (t) counts the number of individuals, this is somewhat unrealistic, as it the ignores depletion of nutrient due to
the growth or individuals once they are born; it is sometimes better to think of N (t) as the total biomass at time t
, Eduardo D. Sontag, Lecture Notes on Mathematical Biology 7
(“20 new individuals formed ⇒ α × 20 less nutrient”). It follows that
d dC dN
(C + αN ) = +α = −αKN + αKN = 0
dt dt dt
and therefore C(t) + αN (t) must be constant, which we call “C 0 ”2
(we use this notation because C(0) + αN (0) ≈ C(0), if the population starts as N (0) ≈ 0).
So K = κC = κ(C0 − αN ), and Equation (2) becomes the same equation as (3), just with different
names of constants:
dN
= κ (C0 − αN ) N
dt
1.4 Logistic Equation: Math
dN N N (B − N )
We solve = rN 1− = r using again the method of separation of variables:
dt B B
Z Z
B dN
= r dt .
N (B − N )
We compute the integral using a partial fractions expansion:
Z Z
1 1 N N ecB
+ dN = r dt ⇒ ln = rt+c ⇒ cert ⇒ N (t) =
= e
N B−N B−N B−N c + e−rt
e
N0 B
⇒ e
c = N0 /(B − N0 ) ⇒ N (t) =
N0 + (B − N0 )e−rt
We can see that there is a B asymptote as t → ∞. Let’s graph with Maple:
with(plots):
f(t):=t->(0.2)/(0.2+0.8*exp(-t)):
p1:=plot(f(t),0..8,0..1.3,tickmarks=[0,2],thickness=3,color=black):
g:=t->1:
p2:=plot(g(t),0..8,tickmarks=[0,2],thickness=2,linestyle=2,color=black):
display(p1,p2);
Gause’s 1934 Experiments
G.F. Gause carried out experiments in 1934, involving Paramecium caudatum and Paramecium aure-
lia, which show clearly logistic growth:
2
this is an example of a “conservation law”, as we’ll discuss later
, Eduardo D. Sontag, Lecture Notes on Mathematical Biology 8
(# individuals and volume of P. caudatum and P. aurelia, cultivated separately, medium changed daily,
25 days.)
1.5 Changing Variables, Rescaling Time
We had this equation for growth under nutrient limitations:
dN
= κ (C0 − αN ) N
dt
which we solved explicitly and graphed for some special values of the parameters C 0 , κ, α.
But how do we know that “qualitatively” the solution “looks the same” for other parameter values?
Can the qualitative behavior of solutions depend upon the actual numbers C 0 , κ, α?
First of all, we notice that we could collect terms as
dN
= ((κC0 ) − (κα)N ) N = Ce0 − α
eN N
dt
(where C
e0 = κC0 and α
e = κα), so that we might as well suppose that κ = 1 (but change α, C 0 ).
But we can do even better and use changes of variables in N and t in order to eliminate the two
remaining parameters!
We will always proceed as follows:
• Write each variable (in this example, N and t) as a product of a new variable and a still-to-be-
determined constant.
• Substitute into the equations, simplify, and collect terms.
• Finally, pick values for the constants so that the equations (in this example, there is only one
differential equation, but in other examples there may be several) have as few remaining param-
eters as possible.
The procedure can be done in many ways (depending on how you collect terms, etc.), so different
people may get different solutions.
