First Example

Lets assume that the population has a growth rate of approximately 5 %, so the value of the parameter a is 0.05. Assume about 1 % of the encounters result in deaths, so s = 0.01.
We will let P[t] be the population in hundreds of individuals. Let the units for t be days. Suppose that the initial population is 300 individuals. This means we have P[0]=3.
Lets have Mathematica solve the IVP. We will denote the initial population by Po. (Do not use Pi which is the Mathematica way of denoting the number pi.)

In[1]:=

  
  a=0.05;
  s=0.01;
  Po=3;
  equations={P'[t]==a P[t] - s (P[t])^2,P[0]==Po}
  depvar=P[t];
  indepvar=t;

Out[1]=

                                 2
  {P'[t] == 0.05 P[t] - 0.01 P[t] , P[0] == 3}

In[2]:=

  messysolution=DSolve[equations,depvar,indepvar]

Out[2]=

                    0.05 t
                   E
  {{P[t] -> ----------------------}}
                            0.05 t
            0.133333 + 0.2 E

The cell below defines solution to be the part to the right of the arrow in the messysolution above.

In[3]:=

  solution=messysolution[[1,1,2]];

In[4]:=

  plot1=Plot[solution,{t,0,100},AxesLabel->{indepvar,depvar},
  	PlotStyle->{RGBColor[1.000,0.000,0.000]},
  	AxesOrigin->{0,0},PlotRange->{0,2a/s}];

Is this reasonable? If the initial population is 300, should the population increase as the graph suggests? Look at the differential equation at the begining and convince yourself that P should increase by considering the sign of ist first derivative P'[t].

How will things be different if the initial population is 200? Should it increase or not? Will it attain the levels as with initial value 300? Lets find out.

In[5]:=

  
  Clear[Po,equations,messysolution,solution];
  Po=2;
  equations={P'[t]==a P[t] - s (P[t])^2,P[0]==Po}
  messysolution=DSolve[equations,depvar,indepvar]
  solution=messysolution[[1,1,2]];

Out[5]=

                                 2
  {P'[t] == 0.05 P[t] - 0.01 P[t] , P[0] == 2}

Out[6]=

                  0.05 t
                 E
  {{P[t] -> -----------------}}
                       0.05 t
            0.3 + 0.2 E

In[7]:=

  plot2=Plot[solution,{t,0,100},AxesLabel->{indepvar,depvar},
  	PlotStyle->{RGBColor[0.000,1.000,0.000]},
  	AxesOrigin->{0,0},PlotRange->{0,2a/s}];

We can compare the previous two cases graphically to get an idea. I have given names to the graphs above so we can display them using the command Show.

In[8]:=

  Show[plot1,plot2];

To Do

It seems that after about 100 days the population would be about the same, regardless of the initial condition. What happens if the initial population is more than 500 individuals?

Convince yourself that the same would happen if the initial population is 150. Copy the cell where Po is defined above first, paste it below and change the value of Po. Evaluate the cell after that.

Then copy the cell with the commands for plot2 and paste it below and make the following changes:
Change the name to plot3 and the part with RGBColor[0.000,1.000,0.000] to RGBColor[0.000,0.000,1.000]. This is just to change the color of the plot.
Then evaluate the cell with plot3 and display the 3 graphs by using Show. That is type Show[ plot1, plot2, plot3];

To Do
BEFORE you evaluate the cells below write down below this text what you think should happen to P. Will P increase , will it decrease? After you write the explantation evalute the cells below.

Look at the differential equation at the start and assume that Po is equal to 7 (700 individuals), what would you expect to happen with P this time. (Consider the sign of P'.)

Answer this before evaluating the cells below.


WRITE YOUR TEXT HERE

In[9]:=

  
  Clear[Po,messysolution,solution];
  Po=7;
  equations={P'[t]==a P[t] - s (P[t])^2,P[0]==Po}
  messysolution=DSolve[equations,depvar,indepvar]
  solution=messysolution[[1,1,2]];

Out[9]=

                                 2
  {P'[t] == 0.05 P[t] - 0.01 P[t] , P[0] == 7}

Out[10]=

                     0.05 t
                    E
  {{P[t] -> ------------------------}}
                              0.05 t
            -0.0571429 + 0.2 E

In[11]:=

  plot4=Plot[solution,{t,0,100},AxesLabel->{indepvar,depvar},
  	PlotStyle->{RGBColor[0.502,0.000,0.251]},
  	AxesOrigin->{0,0},PlotRange->{0,2 a/s}];

In[12]:=

  Show[plot1,plot2,plot3,plot4];

  Show::gcomb: An error was encountered in combining the graphics

       objects in Show[-Graphics-, -Graphics-, plot3, -Graphics-].

To Do

What happens if the initial population is 900? Copy the cells used in creating plot 4, make the necessary change in Po and define the new graph as plot5. Evaluate and then have the five graphs displayed below using Show as before.

To Do
Answer the following questions

You should be "convinced" that the population will tend to a certain value, in time, regardless of the initial size of the population. This value is refered to as the carrying capacity of the population.

1) What is that value? What does it appear to be from the graphs.

2) Is there any relationship between the carrying capacity and the values of a and s in the model that you can notice?

3) What is the value of P'[t] if P[t] is equal to the carrying capacity?

4) What does the value of P' in question 3) imply for the size of P?

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