Math 325: Differential	Equations    Quiz 4    11/14/97

<<Graphics`PlotField`

Consider the non-linear initial value problem:
    x'[t] == 4x[t] - x[t]^2 - x[t] y[t]
    y'[t] == y - 2x y - y^2
a) Find all the critical points of this system.
b) Find the corrseponding linear system near each critical point and determine its eigenvalues.
c) Discuss the type and stability of each critical point of the original system.

Solution

F[x_,y_] = 4x - x^2 - x y;
G[x_,y_] = y - 2x y - y^2;

a) Find the critical points:

cp = Solve[{F[x,y] == 0, G[x,y] == 0}]

The critical points are thus {-3,7}, {0,0}, {0,1}, and {4,0}.

b) and c) To find the correspoinding linear systems and eigenvalues we need to examine the matrix of partial derivatives evaluated at each critical point.

A = {{D[F[x,y],x], D[F[x,y],y]},
     {D[G[x,y],x], D[G[x,y],y]}};
A//MatrixForm

At the critical point {-3,7}:

A1 = A /. cp[[1]];
A1//MatrixForm

Eigenvalues[A1]

The eigenvalues are complex with negative real part so {-3,7} is an asymptotically stable spiral.
At the point {0,0}:

A2 = A /. cp[[2]];
A2//MatrixForm

The eigenvalues are both positive so {0,0} is an unstable node.
At the point {0,1}:

A3 = A /. cp[[3]];
A3//MatrixForm

The eigenvalues have opposite signs, so {0,1} is an unstable saddle.
At the point {4,0}:

A4 = A /. cp[[4]];
A4//MatrixForm

The eigenvalues are both negative, so {4,0} is an asymptotically stable node.

PlotVectorField[{F[x,y],G[x,y]}, {x,-4,5}, {y,-1,8}]