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\maketitle

In this project, we consider logistic models of population growth that have
been modified to account for the fact that the carrying capacity of the
population may change with time.

Recall that the usual logistic differential equation is:

$P^{\prime}=kP(1-\frac{P}{L})$ where $L$ represents the carrying capacity. \
We will now consider $L$ to be a function of time; that is our first attempt
to find a suitable differential equation is:

(1) \ \ \ \ $P^{\prime}=kP(1-\frac{P}{L(t)})$

\ where $L(t)$ is now a function which gives the variational carrying
capacity.$\,\,$ \ We assume that $0<L(t)$ for all $t>0$. \ We can also
assume that $0<P(0)<L(0)$.

\bigskip 

Question One. \ Why is it reasonable to assume that $P(0)<L(0)$?

\bigskip 

Answer: \ because it is reasonable to assume that the initial population is
less than the assumed starting carrying capacity of the system.

\bigskip 

Question Two. \ If $L$ is assumed to be a smooth positive function, given an
initial condition $P(0)=P_{0}$, are we guaranteed to have a unique solution
over our region (in the $t,P$ plane) of interest (which is what region?)? \
Why? \ Note: \ there are both existence and uniqueness considerations here.
\ 

\bigskip 

Answer: \ the region of interest is $P>0,\,t\geq 0.\,\,$If we consider any
rectangle $R$ in the first quadrant of the $(t,P)$ plane, we see that $%
P^{\prime }=f(t,P)=kP(1-\frac{P}{L(t)})$

\ is continuous everywhere (as $L(t)>0$). \ Hence a solution exits. \ The
solution is unique as $\frac{\partial f}{\partial P}=k(1-\frac{P}{L(t)}%
)+kP(1-\frac{1}{L(t)})$ is continuous on the whole $(t,\,P)$ plane. \ 

\bigskip 

Question Three. \ If we assume that $L(t)$ is an increasing function, what
can we say about the relative growth rate $\frac{P^{\prime }}{P}$?

\bigskip 

Answer: \ $\frac{P^{\prime }}{P}=k(1-\frac{P}{L(t)})$. \ Since $k>0$, as $t$
increases, $L$ increases. \ When $P(t)$ is relatively small compared to $L(t)
$, the relative growth rate is approximately $k$. Note that $(1-\frac{P}{L(t)%
})^{\prime }=-(\frac{LP^{\prime }-L^{\prime }P}{L^{2}})=\frac{L^{\prime
}P-LP^{\prime }}{L^{2}}$. \ It is reasonable to assume that initially, $%
P^{\prime }>L^{\prime }$. \ Otherwise, we are not in a constrained growth
situation. \ Hence $LP^{\prime }>L^{\prime }P$ and we expect the relative
growth rate to decrease with time. \ However, the sign of \ $L^{\prime
}P-P^{\prime }L$ determines whether the relative growth rate increases,
decreases or remains constant.

\bigskip 

Question Four. \ \ If \ $L(t_{1})=P(t_{1})$ what can you say about $%
P^{\prime }(t_{1})$? \ If $P^{\prime }(t_{1})=0$, what can you say about $%
L(t_{1})$ and $P(t_{1})$?

\bigskip 

Answer: \ $P^{\prime }=kP(t)(1-\frac{P(t)}{L(t)})$. \ If $%
P(t_{1})=L(t_{1}),\,P^{\prime }(t_{1})=0$. \ If $P^{\prime }(t_{1})=0$, then 
$1-\frac{P(t_{1})}{L(t_{1})}=0$ so $P(t_{1})=L(t_{1})$.

\bigskip 

Question Five. \ Here is a challenge: \ if $L$ is a strictly increasing
smooth function, conclude that $P$($t)<L(t)$ for all $t>0$ (hint: \ what can
you say about $L^{\prime }(t)$)? \ 

\bigskip 

Answer: \ suppose $t_{1}$ is the first $t$ such that $P(t_{1})=L(t_{1}).\,\,$%
Let $F(t)=L(t)-P(t).\,\,F^{\prime }(t)=L^{\prime }(t)-P^{\prime }(t).\,\,$%
Because $L$ is strictly increasing, $F^{\prime }(t_{1})=L^{\prime
}(t_{1})-P^{\prime }(t_{1})=L^{\prime }(t_{1})>0.\,\,$Because $L$ and $P$
are twice differentiable, they have a continuous derivative; hence there is
an interval $(a,b)$ containing $t_{1}$ where $F^{\prime }>0$. \ Therefore $F$
is increasing at $t_{1}$. \ But $F(t_{1})=0,\,$therefore $F$ is negative on $%
(a,\,t_{1})$ which implies that $P>L$ on $(a,\,t_{1})$, which is impossible.

Note: \ there is an alternate argument which uses $P^{\prime\prime}$.

\bigskip 

Question Six. \ Here is a challenge: \ If $L$ is a smooth, strictly
increasing function, conclude that $P$($t)<L(t)$ for all $t>0$ (hint: \ what
can you say about $L^{\prime }(t)$)? \ 

\bigskip 

Answer to Question 6: \ $P^{\prime \prime }(t_{1})=P^{\prime }(t_{1})(1-%
\frac{P(t_{1})}{L(t_{1})})+P(-\frac{L(t_{1})P^{\prime }(t_{1})-L^{\prime
}(t_{1})P(t_{1})}{L^{2}(t_{1})})\,=0+P(t_{1})(-\frac{-L^{\prime
}(t_{1})P(t_{1})}{(L(t_{1}))^{2}})=L^{\prime }(t_{1})(\frac{P(t_{1})}{%
L(t_{1})})^{2}=L^{\prime }(t_{1})$ \ so $P^{\prime \prime }(t_{1})=L^{\prime
}(t_{1})$.

\bigskip 

Question Seven: \ Calculate $P^{\prime \prime \prime }$ and evaluate at $%
t_{1}$ and leave your answer in terms of $L,L^{\prime }$ and $L^{\prime
\prime }$.

\bigskip 

Solution: \ $P=kP(1-P/L),\,P^{\prime }=k(P^{\prime }(1-P/L)+P(1-P/L)^{\prime
}),\,P^{\prime \prime }=k(P^{\prime \prime }(1-P/L)+2P^{\prime
}(1-P/L)^{\prime }+P(1-P/L)^{\prime }$

$P^{\prime\prime\prime}=k(P^{\prime\prime\prime}(1-P/L)+3P^{\prime%
\prime}(1-P/L)^{\prime}+3P^{\prime}(1-P/L)^{\prime\prime}+P(1-P/L)^{^{\prime%
\prime}})\,=$

at ($t=t_1)$ $=k(3P^{\prime\prime}(1-P/L)^{\prime})\,+P(1-P/L)^{\prime%
\prime})=$

$k(3P^{\prime\prime}(\frac{L^{\prime}P-LP^{\prime}}{L^2})+P(\frac{%
L^2(L^{\prime\prime}P+L^{\prime}P^{\prime}-L^{\prime}P^{\prime}-LP^{\prime%
\prime})-2LL^{\prime}(L^{\prime}P-LP^{\prime})}{L^4}))=$

$k(3P^{\prime\prime}L^{\prime}/L)+P/L^4(L^2(L^{\prime\prime}P-LP^{\prime%
\prime})-2LL^{\prime}L^{\prime}P))=$

$k(3(L^{\prime})^2/L+L^{\prime\prime}-L^{\prime}-2(L^{\prime})^2/L)=$

$k(L^{\prime\prime}-L^{\prime}-(L^{\prime})^2/L)$

\bigskip 

Question Eight: \ Tell how we would get $P^{(n)}$. \ 

\bigskip 

Answer: \ We can obtain $P^{(n)}$ merely by starting with an expression for $%
P^{(n-1)}$ and taking the derivative. \ That is, we differentiate both sides
of (1) to obtain $P^{\prime \prime }$, then both sides of ``$P^{\prime
\prime }=$'' to obtain $P^{\prime \prime \prime }$, and so on.

\bigskip 

\begin{center}
\bigskip {\Large Computer Project}
\end{center}

\bigskip 

Consider the following candidates for $L$:

$L_1(t)=L_0+w_1\,t\,\,\,\,\,(w_1>0)$

$L_2(t)=L_0+w_2\,ln(1+t)\,\,\,\,(w_2>0)$

$L_3(t)=L_0+w_3t^2\,\,\,\,(w_3>0)\,\,$

1. \ Use a computer ODE solver to plot direction fields (of the ODE sans the
initial condition) and of \ solutions to (1) using the appropriate carrying
capacity function $L_{i}$ and given the following table of initial
conditions and model parameters:

\begin{tabular}[t]{|l|l|l|l|l|l|}
\hline
$P(0)$ & $L(0)$ & $k$ & $w_1$ & $w_2$ & $w_3$ \\ \hline
4 & 200 & .03 & 1.4 & 15 & .002 \\ \hline
10 & 200 & .03 & 1.4 & 15 & .002 \\ \hline
50 & 200 & .03 & 1.4 & 15 & .002 \\ \hline
4 & 100 & .05 & 1.4 & 15 & .002 \\ \hline
10 & 100 & .05 & 1.4 & 15 & .002 \\ \hline
4 & 200 & .03 & 2.0 & 45 & .004 \\ \hline
\end{tabular}

Example: \ for the first row, you would plot the direction fields and
solutions to three ODE's:

$P^{\prime }=.03$ $P(1-\frac{P}{2001.4+\text{ }t})$, $P^{\prime }=.03P(1-%
\frac{P}{200+15\ln (1+t)})$, \ $P^{\prime }=.03P(1-\frac{P}{200+.002\text{ }%
t^{2}})$, each of which has initial condition $P(0)=4$.

Note: \ you will be plotting the graphs of \ the solutions to 6*3 = 18
different ODE's. \ Don't worry, you won't have to turn in 18 different
graphs.

Write a few paragraphs describing the effects that \ $P(0)$, $L(0),$ $k,\,L_i
$ and $w_i$ have on the solution. \ Include some carefully chosen graphs as
part of your explanation. Note that the first row of data is taken from
studies on the population growth of the United States (where $P$ is in
millions).

\bigskip 

Answer: \ in the case where $L$ grows linearly, $P$ is initially concave up
and then flattens out to linear growth. \ Increasing $w$ increases the slope
of $P$, and increasing $k$ causes quicker initial growth of $P$.

In the case where $L$ grows logarithmically, the graphs of $P$ vary little
from the classical constant $L$ solutions. \ Increasing $k$ increases the
initial rate of growth of $P$ and increasing $w$ causes the point of
diminishing returns to occur later. \ The lower values of $w$ makes the
graph of $P\,$have a ``plateau'' with a slight upward slope.

In the case where $L$ grows quadratically, the shape of $P$ is greatly
affected by $w$ and to a lesser extent by $P(0)$. \ If $w\,$ is large, $P$
starts concave up and remains so; one thinks of a quadratic approximation to
the exponential function. \ If $w$ is smaller, we obtain concavity changes;
at first $P$ is concave up. \ The graph switches concavity past the
classical ``point of diminishing returns'' only to become concave up again.
\ The exception to this is at the $P(0)=50$ case; here the concavity changes
are difficult to discern as the graph of $P$ almost looks linear at first.

\bigskip 

\bigskip 

2. \ Now assume that $L(t)\,=\,200\,+\,1.4t\,+40\,sin(\frac{2\pi t}{50})$,
and that $k=.05$.

Describe the carrying capacity function. \ Use the computer to plot a
direction field and to plot 3 solutions where $P(0)=4,\,P(0)=10,\,P(0)\,=\,50
$. \ Notice that your solution curves for $P$ have local maxima and minima.
\ Comment on what must be happening at such points.

\bigskip 

\bigskip 

One possible answer: \ each solution starts off increasing and concave up. \
The $P(0)=50$ solution then has two slight concavity changes until it
reaches a relative maximum. \ At the relative maximum, $P=L$. \ But, because 
$P$ is concave down at that point, say $t_{a}$ the result of question six
implies that $L^{\prime }<0$; that is $L$ is on a down cycle. \ Similarly,
when $P$ is at a relative minimum, $P=L$ at, say $t_{b}$, but $L$ is
increasing. \ Therefore, on $(t_{a},t_{b})$, $L$ must have passed from
decreasing to increasing and have a local minimum at some intermediate point.

The smaller initial value solutions exhibit similar phenomena, and
eventually converge to the larger initial value solution. \ The ``change of
concavity while $P$ increases'' for the $P(0)=4$ solution is less evident
than it is with the other solutions.

\end{document}
