Qualitative Analysis

Below we perform some qualitative analysis for a differential equation. We find equilibria and determine the eigenvalues (hence behavior) at the equilibria. Consider a modified Lotka-Volterra model, including intraspecies competition.

> MLVx := diff(x(t),t) = 0.2*x(t)*(1 - 0.1*x(t) - 0.2*y(t));
MLVy := diff(y(t),t) = 0.3*y(t)*(0.3*x(t)-1);

We begin by viewing the graph of the model.

> DEplot([MLVx, MLVy], [x(t),y(t)], t=0..50, [[x(0)=1,y(0)=5],[x(0)=1,y(0)=.7]], stepsize=.2,
title=`Lotka-Volterra model`, color = [.3*y(t)*(x(t)-1),x(t)*(1-y(t)),.1],
linecolor=t/2, arrows=MEDIUM, method=rkf45);

We find the equilibria for this model by setting the right hand side of the differential equations equal to zero.

> equil := solve({rhs(MLVx)=0, rhs(MLVy)=0}, {x(t), y(t)});

Linear analysis of a differential equation near an equilibrium is best done by finding the eigenvalues for the linearized system near the equilibrium.

> with(linalg):

Warning, the name adjoint has been redefined

Warning, the protected names norm and trace have been redefined and unprotected

Create a vector for the right hand side of the differential equation.

> A := vector([0.2*X*(1 - 0.1*X - 0.2*Y),
0.3*Y*(0.3*X-1)]);

Find the Jacobian matrix for this vector function.

> J := jacobian(A, [X,Y]);

Analyze the eigenvalues of the Jacobian matrix near the origin (one equilibrium).

> X := 0: Y := 0: J0 := J(0,0);

> eigenvects(J0);

This give a saddle node since the eigenvalues are real and opposite in sign.

> X := rhs(equil[3][1]); Y := rhs(equil[3][2]);

> J1 := J(X,Y);

> eigenvects(J1);

This give a stable node since the eigenvalues are complex with a negative real part.

The phase portrait above shows these properties very clearly.