Example with

> A := matrix( [ [ 2, 4 ], [ 3, 6 ] ] );

>

> charpoly( A, lambda );

> eigenvals( A );

Since zero is an eigenvalue, there will be a line of equilibrium solutions for this system. Since the other eigenvalue is positive, all other solutions will converge (along straight lines) to an equilibrium solution - not necessarily the origin.

>

> eigenvects( A );

>

The straight-line solutions are:

> Y1 := exp(8*t) * matrix( 2, 1, [ 2, 3 ] );

> Y2 := exp(0*t) * matrix( 2, 1, [ -2, 1 ] );

>

> ode1 := diff( x(t), t ) = 2*x(t) + 4*y(t);

> ode2 := diff( y(t), t ) = 3*x(t) + 6*y(t);

> MODEL := { ode1, ode2 }:

> VARS := { x(t), y(t) }:

> DOMAIN := t=0..1:

> RANGE := x=-2..2, y=-2..2:

> dirPLOT := DEplot( MODEL, VARS, DOMAIN, RANGE, arrows=MEDIUM ):

> dirPLOT;

>

> IC := [ [ x(0)=-0.8, y(0)=0.8 ], [ x(0)=-1.2, y(0)=0.2 ],
[ x(0)=1.2, y(0)=-0.3 ], [ x(0)=0.8, y(0)=-0.8 ],
[ x(0)=-1.8, y(0)=1.3 ], [ x(0)=-2.2, y(0)=0.7 ],
[ x(0)=2.2, y(0)=-0.7 ], [ x(0)=1.8, y(0)=-1.3 ] ]:

> solPLOT := DEplot( MODEL, VARS, DOMAIN, RANGE, IC, arrows=NONE, stepsize=0.1, linecolor=GREEN ):

> #solPLOT;

>

> ICline := [ [ x(0)=0.2, y(0)=0.3], [ x(0)=-0.2, y(0)=-0.3 ],
[ x(0)=-2, y(0)=1 ], [ x(0)=2, y(0)=-1 ],
[ x(0)=-1, y(0)=0.5 ], [ x(0)=1, y(0)=-0.5 ] ]:

> linePLOT := DEplot( MODEL, VARS, DOMAIN, RANGE, ICline, arrows=NONE, linecolor=BLUE ):

> #linePLOT;

>

> display( [ dirPLOT, solPLOT, linePLOT ], title=`Example of Sink` );

Notice how the solutions travel along paths that are parallel to the eigenvector for the non-zero eigenvalue and the line of equilibrium solutions conicides with the direction of the zero eigenvalue.

>