# Preface

This section studies some first order nonlinear ordinary differential equations describing the time evolution (or “motion”) of those hamiltonian systems provided with a first integral linking implicitly both variables to a motion constant. An application has been performed on the Lotka--Volterra predator-prey system, turning to a strongly nonlinear differential equation in the phase variables.

Introduction to Linear Algebra with Mathematica

# Predator-Prey Equations

Some situations require more than one differential equation to model a particular situation. We might use a system of differential equations to model two interacting species, say where one species preys on the other. For example, we can model how the population of Canadian lynx (lynx Canadians) interacts with a the population of snowshoe hare (lepus americanis) (see https://www.youtube.com/watch?v=ZWucOrSOdCs).

A good historical data are known for the populations of the lynx and snowshoe hare from the Hudson Bay Company. This large Canadian retail company, which owns and operates a large number of retail stores in North America and Europe, including Galeria Kaufhof, Hudson's Bay, Home Outfitters, Lord & Taylor, and Saks Fifth Avenue, was originally founded in 1670 as a fur trading company. The Hudson Bay Company kept accurate records on the number of lynx pelts that were bought from trappers from 1821 to 1940. The company noticed that the number of pelts varied from year to year and that the number of lynx pelts reached a peak about every ten years. The ten year cycle for lynx can be best understood using a system of differential equations.

The primary prey for the Canadian lynx is the snowshoe hare. We will denote the population of hares by H(t) and the population of lynx by L(t), where t is the time measured in years. We will make the following assumptions for our predator-prey model.

• If no lynx are present, we will assume that the hares reproduce at a rate proportional to their population and are not affected by overcrowding. That is, the hare population will grow exponentially,
$\frac{{\text d}H}{{\text d}t} = a\,H \qquad (\mbox{where a is a constant}).$
• Since the lynx prey on the hares, we can argue that the rate at which the hares are consumed by the lynx is proportional to the rate at which the hares and lynx interact. Thus, the equation that predicts the rate of change of the hare population becomes
$\frac{{\text d}H}{{\text d}t} = a\,H -b\, H\, L ,\qquad (\mbox{where a, b are constants}).$
We are thinking of H L as the number of possible interactions between the lynx and the hare populations.
• If there is no food, the lynx population will decline at a rate proportional to itself,
$\frac{{\text d}L}{{\text d}t} = \delta\,L \qquad (\mbox{where \delta is a constant}).$
• The lynx receive benefit from the hare population. The rate at which lynx are born is proportional to the number of hares that are eaten, and this is proportional to the rate at which the hares and lynx interact. Consequently, the growth rate of the lynx population can be described by
$\frac{{\text d}L}{{\text d}t} = c\,H\,L -\delta\, L ,\qquad (\mbox{where c, \delta are constants}).$
We now have a system of differential equations that describe how the two populations interact,
$\dot{H} = a\,H -b\, H\, L ,\qquad \dot{L} = c\,H\,L -\delta\, L ,$
where we use the dot notation to represent the derivatives with respect to time t.
The system of differential equation used to model predator-prey interactions
$\begin{split} \dot{x} = x \left( a - \alpha y \right) , \\ \dot{y} = y \left( -\delta + bx \right) , \end{split}$
is usually referred to as Lotka--Volterra model. Here x is the number of prey (for example, rabbits); y is the number of some predator (for example, foxes); $$\dot{x} \ \mbox{ and} \ \dot{y}$$ represent the instantaneous growth rates of the two populations; t represents time; a, α, δ, and b are positive real parameters describing the interaction of the two species.

The Lotka--Volterra system of equations is an example of a Kolmogorov model, which is a more general framework that can model the dynamics of ecological systems with predator-prey interactions, competition, disease, and mutualism. The equations which model the struggle for existence of two species (prey and predators) bear the name of two scientists: Lotka (1880--1949) and Volterra (1860--1940). They lived in different countries, had distinct professional and life trajectories, but they are linked together by their interest and results in mathematical modeling.

The predator–prey model was initially proposed by Alfred J. Lotka in the theory of autocatalytic chemical reactions in 1910. Lotka was born in Lemberg, Austria-Hungary, but his parents immigrated to the US. In 1925, he utilized the equations to analyze predator-prey interactions. Lotka published almost a hundred articles on various themes in chemistry, physics, epidemiology or biology, about half of them being devoted to population issues. He also wrote six books.

The same set of equations was published in 1926 by Vito Volterra, a mathematician and physicist, who had become interested in mathematical biology because of the impact by the marine biologist Umberto D'Ancona (1896--1964). Vito Volterra was born in Ancona, then part of the Papal States, into a very poor Jewish family. He attended the University of Pisa, where he became professor of rational mechanics in 1883. His most famous work was done on integral equations. In 1892, he became professor of mechanics at the University of Turin and then, in 1900, professor of mathematical physics at the University of Rome La Sapienza. His daughter Luisa married Umberto D’Ancona.

The predator-prey system of equations was later extended by many researchers, including C. S. Holling, Arditi--Ginzburg model, Rosenzweig-McArthur model, and some others.

The critical points of the Lotka--Volterra system of equations are the solutions of the algebraic equations

$x \left( a - \alpha y \right) =0 , \qquad y \left( -\delta + bx \right) = 0,$
namely (0,0) and (δ/b, a/α). Next we try to examine the local behavior of solutions near each critical point using linearization approach. Neglecting the nonlinear terms, we obtain the corresponding linear system near the origin:
$\frac{\text d}{{\text d}t} \begin{pmatrix} x \\ y \end{pmatrix} = \begin{bmatrix} a & 0 \\ 0 & -\delta \end{bmatrix}\begin{pmatrix} x \\ y \end{pmatrix} .$
The eigenvalues of the above system of equations are λ1 = a > 0 and λ2 = - δ < 0. So the origin is a saddle point, unstable. At another critical point, linearization yields
$\frac{\text d}{{\text d}t} \begin{pmatrix} x- \delta / b \\ y - a/\alpha \end{pmatrix} = \begin{bmatrix} 0 & -\frac{\alpha \delta}{b} \\ \frac{ab}{\alpha} & 0 \end{bmatrix}\begin{pmatrix} x- \delta / b \\ y- a/\alpha \end{pmatrix} .$
The eigenvalues of the above system of equations are pure complex conjugate:
$\lambda = \pm {\bf j} \sqrt{a\delta} .$
Therefore, the Lotka--Volterra system of equation is hyperbolic at the second critical point and linearization is inconclusive. So we need to find another approach. It is not hard to verify that the energy function
$E = bx + \alpha y - \delta \,\ln x - a\,\ln y + \mbox{constant}$
is a Lyapunov function for the predator-prey model. Indeed, its derivative with respect to the system of differential equation is
$\dot{E} = b\,\dot{x} + \alpha\,\dot{y} - \frac{\delta}{x}\,\dot{x} - \frac{a}{y}\,\dot{y} = \left( a - \alpha\,y \right) \left( b\,x - \delta \right) + \left( -\delta + b\,x \right) \left( -a + \alpha\,y \right) \equiv 0.$
Therefore, the stationary point (δ/b, a/α) is a stable critical point.

We may try to find the general solution of the Lotka--Volterra system of equations. From both equations, we get

$\frac{\dot{x}}{x} = a- \alpha \,y \qquad \mbox{and} \qquad \frac{\dot{y}}{y} = - \delta + b\,x .$
Multiplying the first relation by (-δ + bx) and the second one by (a - αy), we obtain
$\frac{\dot{x}}{x} \left( - \delta + b\,x \right) = \left( a- \alpha \,y \right) \left( - \delta + b\,x \right) = \frac{\dot{y}}{y} \left( a- \alpha \,y \right) .$
So we separated variables
$\frac{\dot{x}}{x} \left( - \delta + b\,x \right) = \frac{\dot{y}}{y} \left( a- \alpha \,y \right) ,$
which upon integration yields the energy integral.
Example:
 s = NDSolve[{H'[t] == 0.4*H[t] - 0.01*H[t]*L[t], L'[t] == -0.3*L[t] + 0.005*H[t]*L[t], H[0] == 70, L[0] == 50}, {H, L}, {t, 0, 100}] a = Plot[Evaluate[H[t] /. s], {t, 0, 100}, PlotStyle -> {Thick, Red}, PlotRange -> {20, 85}, PlotLabels -> Placed[{"L(t)"}, Above], PlotLegends -> {"L(t)"}] b = Plot[Evaluate[L[t] /. s], {t, 0, 100}, PlotStyle -> {Thick, Blue}, PlotRange -> {20, 80}, PlotLabels -> Placed[{"H(t)"}, Below], PlotLegends -> {"H(t)"}] Show[{a, b}, PlotRange -> {20, 85}]

Notice that the predator population, L, begins to grow and reaches a peak after the prey population, H reaches its peak. As the prey population declines, the predator population also declines. Once the predator population is smaller, the prey population has a chance to recover that the cycle begins again.

f[x_, y_] = {x - 2 x y, -.5 y + x y};
Print["X' = ", MatrixForm[f[x, y]]]
Out[4]=
X' = (x-2 x y
-0.5 y+x y )
Print["x' = ", f[x, y][[1]], ", y' = ", f[x, y][[2]]]
Out[5]=
x' = x-2 x y, y' = -0.5 y+x y

 Show[VectorPlot[f[x, y], {x, 0, 2}, {y, 0, 2}], Frame -> True, BaseStyle -> {FontFamily -> "Times", FontSize -> 14}]
 Show[VectorPlot[ f[x, y]/(10^-8 + Norm[f[x, y]]), {x, 0, 2}, {y, 0, 2}], Frame -> True, BaseStyle -> {FontFamily -> "Times", FontSize -> 14}]

 p2 = Show[ ParametricPlot[ Table[{x[t], y[t]} /. sol[[j]], {j, Length[ic]}], {t, 0, T}, PlotRange -> {{0, 4}, {0, 3}}], Table[Graphics[{Arrowheads[.03], Arrow[{ic[[j]], ic[[j]] + .01 f[ic[[j, 1]], ic[[j, 2]] ] }]}], {j, Length[ic]}], AxesLabel -> {"x", "y"}, BaseStyle -> {FontFamily -> "Times", FontSize -> 14}, PlotLabel -> "Phase portrait" ]

 Plot[{x[t] /. sol[[1]], y[t] /. sol[[1]]}, {t, 0, T}, PlotStyle -> {{Thickness[0.01], RGBColor[1, 0, 0]}, {Thickness[0.01], RGBColor[0, 1, 0]}}, AxesOrigin -> {0, 0}, AxesLabel -> {"t", "x(t),y(t)"}, BaseStyle -> {FontFamily -> "Times", FontSize -> 14}, PlotLabel -> "Time courses: Red - x(t), Green - y(t)"]

we can place two plots sude by side:

a = 2.2; b = 0.03; c = 1.4; d = 0.02;
eq1 = r'[t] == a*r[t] - b*r[t]*f[t]
eq2 = f'[t] == -c*f[t] + d*f[t]*r[t]
sol = NDSolve[{eq1, eq2, r[0] == 150, f[0] == 10}, {r, f}, {t, 0, 30}]
gr1 = Plot[Evaluate[{r[t], f[t]} /. sol], {t, 0, 30}, PlotStyle -> {{Thick, Hue[0.7]}, { Dashing[{0.03}], Hue[1]}}, Ticks -> {{{0.01, 0}, {10, "time"}, 20}, {500, {700, "number"}, 1000}}, LabelStyle -> {FontFamily -> "Times", FontSize -> 16}, PlotLabel -> "solid blue: rabbits, dashed red: foxes"];
gr2 = ParametricPlot[Evaluate[{r[t], f[t]} /. sol], {t, 0, 30}, PlotStyle -> {Thick, Hue[0.3]}, Ticks -> {{{0.01, 0}, 500, {700, "rabbits"}, 1000}, {500, {700, "foxes"}, 1000}}, LabelStyle -> {FontFamily -> "Times", FontSize -> 16}];
sp = StreamPlot[{2.2 r - 0.03 f r, -1.4 f + 0.02` f r}, {r, 0, 200}, {f, 0, 150}, StreamScale -> Large]
GraphicsRow[{gr1, gr2, sp}]
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Example: Consider the system of autonomous equations

$\dot{x} = x^2 -3\,xy, \qquad \dot{y} = 2xy-y^2 .$
First, we plot the direction field using StreamPLot command:

 StreamPlot[{x^2 - 3*x*y, 2*x*y - y^2}, {x, -3, 3}, {y, -3, 3}]
Then we can use NDSolve function to solve the given system numerically subject to some initial conditions:

 deq1 = x'[t] == x[t]^2 - 3*x[t]*y[t]; deq2 = y'[t] == 2*x[t]*y[t] - y[t]^2; soln = NDSolve[{deq1, deq2, x[0] == 1, y[0] == 1}, {x[t], y[t]}, {t, -10, 10}]; Next, we plot the corresponding solutions curve with the command x = soln[[1, 1, 2]]; y = soln[[1, 2, 2]]; ParametricPlot[{x, y}, {t, -10, 10}, PlotStyle -> Thick]
We can plot as many such solutions curves as we want, and then display them simultaneously. For example, the following command creates a list of solution curves corresponding to the initial conditions $$x(0) = n/10, \quad y(0) =n/10$$ for $$n=-4,-3,\ldots , 3,4.$$

 curve = Table[n, {n, -4, 4}]; curvep = Table[n, {n, 0, 4}]; Do[Clear[x, y]; soln = NDSolve[{deq1, deq2, x[0] == n/10, y[0] == n/10}, {x[t], y[t]}, {t, -10, 10}]; x = soln[[1, 1, 2]]; y = soln[[1, 2, 2]]; curve[[n]] = ParametricPlot[{x, y}, {t, -10, 10}, PlotStyle -> Thick], {n, -4, 4}]; Show[curve[[-4]], curve[[-3]], curve[[-2]], curve[[-1]], curve[[1]], curve[[2]], curve[[3]], curve[[4]], AspectRatio -> 1]
To display solutions in the first quadrant, we type:

 Do[Clear[x, y]; soln = NDSolve[{deq1, deq2, x[0] == n/10, y[0] == n/10}, {x[t], y[t]}, {t, -10, 10}]; x = soln[[1, 1, 2]]; y = soln[[1, 2, 2]]; curvep[[n]] = ParametricPlot[{x, y}, {t, -10, 10}, PlotStyle -> Thick], {n, 0, 4}]; Show[curvep[[0]], curvep[[1]], curvep[[2]], curvep[[3]], curvep[[4]], AspectRatio -> 1]
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Example: Consider the following competition of two species for a single nutrient:
$\begin{split} \dot{S} = \gamma\,S(t) \left[ 1 - S(t) /K \right] - \frac{m_1}{y_1} \,\frac{x_1 S(t)}{a_1 + S(t)} - \frac{m_2}{y_2} \,\frac{x_2 S(t)}{a_2 + S(t)} , \\ \dot{x}_1 = \frac{m_1 x_1 (t)\,S(t)}{a_1 + S(t)} - \delta_1 x_1 (t) , \\ \dot{x}_2 = \frac{m_2 x_2 (t)\,S(t)}{a_2 + S(t)} - \delta_2 x_1 (t) , \end{split}$
under the assumptions
$S(0) = S_0 > 0, \quad x_1 (0) = x_{10} > 0, \quad x_2 (0) = x_{20} > 0.$
Here xi is the number of the i-th predator at time t, S(t) is the number of the prey at time t, mi is the maximum growth (birth) rate of the i-th predator, δi is the death rate for the i-th predator, yi is the yield factor for the i-th predator, which is the prey density at which the functional response of the predator is half maximal. The parameters γ and K are the intrinsic rate of increase and the carrying capacity or the prey population, respectively.
Theorem: Let
1. $$\displaystyle 0 < a_1 \delta_1 < K \left( m_1 - \delta_1 \right)$$ and
2. $$\displaystyle a_2 \delta_2 > K \left( m_2 - \delta_2 \right)$$ or $$b_2 \le 1 .$$
If $$\displaystyle K < a_1 + \frac{2a_1 \delta_1}{m_1 - \delta_1} ,$$ then
$\begin{split} \lim_{t\to\infty} S(t) &= S^{\ast} = \frac{a_1 \delta_1}{m_1 - \delta_1} , \\ \lim_{t\to\infty} x_1 (t) &= x_1^{\ast} = \frac{\gamma \left( 1-S^{\ast} /K \right) \left( a_1 + S^{\ast} \right)}{m_1 / y_1} , \\ \lim_{t\to\infty} x_2 (t) &= 0. \end{split}$
The coexistence of two predators is possible although the problem remains open analytically.     ■
Example: Consider the following predator-prey system with Beddington--DeAngelis functional response (see complete mathematical analysis in the article of D.T.Dimitrov and H.V. Kojouharov, Complete mathematical analysis of predator–prey models with linear prey growth and Beddington–DeAngelis functional response, Applied Mathematics and Computation, 162, (2), 523--538, 2005)
$\dot{x} = x - \frac{a\,xy}{1+x+y} , \qquad \dot{y} = \frac{b\,xy}{1+x+y} -c\,y ,$
where the constants are a = 6, b = 7.5, and c = 5.

Mathematical analysis of the Beddington--DeAngelis system shows that there exist two equilibria (0,0) and $$\left( \frac{ac}{ab - b -ac} , \frac{b}{ab - b -ac} \right) = (4,1) ,$$ being globally stable in the interior of the first quadrant. The eigenvalues of the Jacobian matrix at the origin are $$\lambda_1 =1 \quad\mbox{and}\quad \lambda_2 =5 ,$$ and the eigenvalues of the Jacobian at the point (4,1) are given by $$- \frac{1}{12} \pm {\bf j} \, \frac{\sqrt{119}}{12} .$$

StreamDensityPlot[{x-x*y/(1+x+y), 2*x*y/(1+x+y) -3*y}, {x,-3,3}, {y,-3,3}, StreamColorFunction->"Rainbow"]
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