Another way:

mat = {{6/10, 1/2}, {-(7/40), 6/5}};
vel = D[MatrixPower[mat, t] . {x, y}, t] /. t -> 0
StreamPlot[vel, {x, -2, 2}, {y, -2, 2}]

This is probably the nicest way to see the overall behavior of the system.

ParametricPlot[] is evaluated before the c's are replaced by numbers. You need to do the replacement first.

If f and r are scalars, then you're plotting a scalar, which ParametricPlot[] won't do.

If f and r are ordered pairs, then you're plotting an ordered pair, and you should see a plot, if the first point is addressed.

You can also use Unevaluated[] to address the first point:

Unevaluated[
  ParametricPlot[Subscript[c, 1]*(0.95^k*{10, 7} . {f, r}) + 
         Subscript[c, 2]*(0.85^k*{2, 1} . {f, r}), {k, 1, 10}]] /. 
   {Subscript[c, 1] -> 1, Subscript[c, 2] -> 1, f -> {1, -1}, 
     r -> {-3, 2}}

Oops, update: Just noticed the blackboard. I think one problem, which is addressed incorrectly in the 2nd and 3rd points, is that you should not be using . (Dot[]) in your code. The blackboard as scalar multiplication (* or Times[]), not vector-vector multiplication nor matrix-vector multiplication.

I am trying to guess what f and r may be:

With[{c1 = 1, c2 = 1, f = 1, r = 1},
 Plot[c1*0.95^k {10, 7} . {f, r} +
   c2*0.85^k {2, 1} . {f, r}, {k, 1, 10}]]

Hi Gianluca;

I discovered that the key to getting what I wanted was using starting values that reflected the Eigenvalues. In the case of the attached Notebook, I used the exact Eigenvalues as the initial values.

I found it interesting that both your method and Eric's method produced exactly the same answer even though you used the MatrixPower function and Eric used Eigenvalues and Eigenvectors in his calculations.

Thanks so much, I not only obtained the answer, but learned a lot in the process.

Mitch Sandlin

Attachments:

PredPrayPlot.nb

If we start with few foxes and many rabbits, fow a while the rabbits will increase. It will take time before the foxes catch up and they will both go extinct. Here is an example:

mat = {{6/10, 1/2}, {-(7/40), 6/5}};
f[t_] = MatrixPower[mat, t];
ParametricPlot[f[t] . {0, 1}, {t, 0, 30},
  PlotRange -> {{0, 2.1}, {0, 2}},
  AxesLabel -> {"foxes", "rabbits"}] /. Line -> Arrow

Using discrete steps:

mat = {{6/10, 1/2}, {-(7/40), 6/5}};
f[t_] = MatrixPower[mat, t];
pts = Table[f[t] . {0, 1}, {t, 0, 30}];
Graphics[{Arrowheads[{.05, .05, .05, .05}], Arrow[pts], 
  PointSize[Medium], Point[pts]}, PlotRange -> {{0, 2.1}, {0, 2}}, 
 Axes -> True, AxesLabel -> {"foxes", "rabbits"}]

Your initial picture only shows the regime where both decrease.

Hi Eric;

I find your response extremely interesting and am still studying your code to get a better understanding of your calculations. Additionally, I find your style of coding excellent since it is easy to read and understand.

Using your function "populationFn[6.5]" you calculated an approximate population of 81 Foxes and 61 Rabbits, and I was curious how you picked the time integral of 6.5.

Furthermore, the graph doesn't seem to model the expected outcome of the populations. For example, since both of the Eigenvalues are less than 1, one would expect both the Foxes and Rabbits to go extinct which the plot and the function do not demonstrate. Actually, it is my understanding that the populations should start decreasing immediately, as displayed in the graph that I attached in my original notebook, with the Rabbits going extinct first and then followed by the Foxes.

Since I am new at this and therefore could be incorrect, please review and give me your thoughts. In any event, I am a lot closer now thanks to you help.

Thanks Again,

Mitch Sandlin