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Mathematical Biology

grandy

Member
Dec 26, 2012
73
Consider the single species population model defined by
dR/dt = [gR/(R+R_m)] - dR, t>0,
where g,R_m and d are all positive parameters and R(0) =R_0

(a) Describe the biological meaning of each term in the equation.

(b) Determine the steady-states of the system and discuss any constraints on the model parameters for the model to admit biologically meaningful solutions.

(c) Determine the steady-state stability and discuss any variation in this with respect to the model parameter values.

=>
a) gR represents the exponential growth of population
dR represents the exponential decay of population
g is the growth rate
d is the decay rate
what does R and R_m represent?
how can I define this term gR/(R+R_m)?
what is (R+R_m)? does it affect the gR for the grow?

b)
In single steady -state system, dR/dt =0
[gR/(R+R_m)] - dR =0
gR -dR(R+R_m) =0
R[g -d(R+R_m)] =0
either R=0
OR g -d(R+R_m)= 0
g- dR_m =0 (R=0)
R_m = g/d
R* = g/d (R_m = R*)
SO we have : (R*1, R*2) =(0, g/d)
is my R*1, R*2 correct. I am concern about R*2 because not sure about if I am allow to do _m = R*.
I am not sure constraint on the model parameters to admit biologically meaning solutions?

c)
to determine steady-state stability
let f(R) = [gR/(R+R_m)] - dR
df/dR = g ln(R+R_m) -d .
My differentiation may be wrong and don't know the term R_m while differentiating with respect to R.
I really don't know after that. and I know my answer is still incomplete.
please help me.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
Consider the single species population model defined by
dR/dt = [gR/(R+R_m)] - dR, t>0,
where g,R_m and d are all positive parameters and R(0) =R_0

(a) Describe the biological meaning of each term in the equation.

(b) Determine the steady-states of the system and discuss any constraints on the model parameters for the model to admit biologically meaningful solutions.

(c) Determine the steady-state stability and discuss any variation in this with respect to the model parameter values.

=>
a) gR represents the exponential growth of population
dR represents the exponential decay of population
g is the growth rate
d is the decay rate
what does R and R_m represent?
how can I define this term gR/(R+R_m)?
what is (R+R_m)? does it affect the gR for the grow?
I would agree that the term $-dR$ represents an exponential decay. However,
the $\displaystyle \frac{gR}{R+R_m}$ is not exponential growth - exponential growth would look like a plain $gR$. It might be helpful to rewrite this term:
$$\frac{gR}{R+R_m}=g \frac{R}{R+R_m}=g \left[ \frac{R+R_m-R_m}{R+R_m}\right]
=g \left[ 1- \frac{R_m}{R+R_m}\right]=g- \frac{gR_m}{R+R_m}.$$
So your DE would then be
$$ \frac{dR}{dt}=g-dR- \frac{gR_m}{R+R_m}.$$

So, in looking at this, I would agree that $g$ is a growth term; however, it's not an exponential growth term, but linear. This could be a steady influx of population from outside, at a constant rate. The $-gR_m / (R+R_m)$ term represents something that always negatively affects the population, but is worse when the population is smaller, and not so bad when the population is larger. I don't know what this could correspond to. If I think of the pilgrims coming over to the US, then this term could be something like division of labor. When you have a very small population, there is a "critical mass" of people you need in order to sustain growth. This could be one way to think of this weird fraction. $R_m$ is some parameter associated with the "division of labor term". $d$, I would agree, represents a death rate, and this one is exponential, because it's multiplied by $R$. So the more population you have, the worse this term gets.

b)
In single steady -state system, dR/dt =0
Correct.

[gR/(R+R_m)] - dR =0
gR -dR(R+R_m) =0
R[g -d(R+R_m)] =0
either R=0
OR g -d(R+R_m)= 0
Good.

g- dR_m =0 (R=0)
This move is not valid. You said before that EITHER $R=0$ OR $g-d(R+R_m)=0$. You can't suddenly that to BOTH of them are zero simultaneously. Try this:
\begin{align*}
g-d(R+R_m)&=0 \\
g&=d(R+R_m) \\
\frac{g}{d} &=R+R_m \\
\frac{g}{d}-R_m&=R.
\end{align*}
So your other steady-state solution is $g/d-R_m$.

R_m = g/d
R* = g/d (R_m = R*)
SO we have : (R*1, R*2) =(0, g/d)
is my R*1, R*2 correct. I am concern about R*2 because not sure about if I am allow to do _m = R*.
I am not sure constraint on the model parameters to admit biologically meaning solutions?
I am not savvy enough to answer this question, either.

c)
to determine steady-state stability
let f(R) = [gR/(R+R_m)] - dR
df/dR = g ln(R+R_m) -d .
My differentiation may be wrong
It is. You "integrated" one term, and differentiated the other! Differentiating with respect to $R$ is fine - you'll get a "feel" for $d^2R/dt^2$ that way. I get:
\begin{align*}
\frac{d}{dR} \left[ g-dR- \frac{gR_m}{R+R_m}\right]&=
-d-g R_m \frac{d}{dR}(R+R_m)^{-1} \\
&=-d+\frac{gR_m}{(R+R_m)^2}.
\end{align*}

and don't know the term R_m while differentiating with respect to R.
I really don't know after that. and I know my answer is still incomplete.
please help me.
Well, for each steady state you found, plug into the second derivative here, and see if $df/dR$ is positive or negative. One will be stable, the other unstable.
 

grandy

Member
Dec 26, 2012
73
a)
Can I define the first term of RHS of equation (gR/(R+R_m)) be logistic growth at a rate g with carrying capacity (R+R_m) ?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
a)
Can I define the first term of RHS of equation (gR/(R+R_m)) be logistic growth at a rate g with carrying capacity (R+R_m) ?
No, because the logistic equation would have the $R+R_m$ in the numerator, not the denominator. It would also look more like a difference: $R_m-R$, where $R_m$ is your "carrying capacity". The logistic equation is
$$\frac{dN}{dt}= \frac{r N (K-N)}{K},$$
where $K$ is the carrying capacity, and $r$ is the Malthus parameter.
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
On the other hand, if you did a series expansion of $1/(R+R_m),$ you would get
$$ \frac{1}{R+R_m} \approx \frac{R_m-R}{R_{m}^{2}}+O(R^2).$$
That does look like a logistic term with carrying capacity $R_m$. Now it's not multiplying $R$, like it normally would, but that may be ok.
 

grandy

Member
Dec 26, 2012
73
b)
so my other steady state is
g-d(R+R_m) =0
R= g/d -R_m
(R*1, R*2)= (0, g/d -R_m)

From the non zero steady-state we note that g>d, otherwise R*2 <0, which is not possible because we can not have negative population density. if this does occur in practice then R*1 is the only possible steady-state where the population dies out.

Did I answer the discussion of any constraints on the model parameters for the model to admit bilogogically meaningful solutions? Do I need to talk about the term -R_m which is constant parameter I guess
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,193
b)
so my other steady state is
g-d(R+R_m) =0
R= g/d -R_m
(R*1, R*2)= (0, g/d -R_m)

From the non zero steady-state we note that g>d,
Check your algebra here! You need
\begin{align*}
\frac{g}{d}-R_m &\ge 0 \\
\frac{g}{d} &\ge R_m \\
g &\ge d R_m.
\end{align*}

otherwise R*2 <0, which is not possible because we can not have negative population density. if this does occur in practice then R*1 is the only possible steady-state where the population dies out.

Did I answer the discussion of any constraints on the model parameters for the model to admit bilogogically meaningful solutions?
Well, there might be other things you could say (I don't know if there are or not), but this is certainly a good start.

Do I need to talk about the term -R_m which is constant parameter I guess
You've talked about it some already, in conjunction with $g$ and $d$.
 

grandy

Member
Dec 26, 2012
73
a)
I can rearrange the model:
dR/dt = [gR/(R+R_m)] - dR
dR/dt = gR - dR (R+R_m)

gR represents the exponential growth of population
dR represents the exponential decay of population
g is the growth rate
d is the decay rate
R is a variable and R_m is a Malthus parameter

The second term of RHS of equation - dR (R+R_m) be logistic decay at a rate d with carrying capacity R_m.
or carrying capacity (R+R_m)
Is it right thing to do?