mth3060303m9.html

3.3 Phase Planes for Linear Systems with Real Eigenvalues

October 21, 1998

Every linear system with non-zero real eigenvalues has an isolated stationary point at the origin. Looking at the phase plane, just as we looked at the phase line for a single autonomous equation, we can classify the singularity as a sink, a source, or (new to higher dimensions) a saddle. We'll start with the new type of singularity.

> with(linalg):
with(DEtools):

Warning, new definition for norm

Warning, new definition for trace

Warning, new definition for adjoint

Saddles

The isolated singularity is a saddle if one eigenvalue is positive and one is negative.

Example 1

Here is a simple example. The coefficient matrix is diagonal.

> A:=matrix(2,2,[[-2,0],[0,3]]);

> phaseportrait({D(x)(t)=-2*x(t),D(y)(t)=3*y(t)},
[x(t),y(t)],
t=-.5..0.5,
[[x(0)=1,y(0)=1],[x(0)=1,y(0)=-1],
[x(0)=-1,y(0)=1],[x(0)=-1,y(0)=-1]]);

Example 2

Here the matrix is not diagonal.

> B:=matrix([[-41, -45], [36, 40]]);
eigenvalues(B);
eigenvectors(B);

> phaseportrait({D(x)(t)=-41*x(t)-45*y(t),D(y)(t)=36*x(t)+40*y(t)},
[x(t),y(t)],
t=-1..1,
[[x(0)=0.5,y(0)=0.5],[x(0)=1,y(0)=-0.9],
[x(0)=-1,y(0)=0.9],[x(0)=-0.5,y(0)=-0.5],
[x(0)=-1,y(0)=1],[x(0)=-5/4,y(0)=1],
[x(0)=1,y(0)=-1],[x(0)=5/4,y(0)=-1]],
x=-2..2,y=-2..2);

The outside curves look rather straight, but they are not.

> dsolve({D(x)(t)=-41*x(t)-45*y(t),D(y)(t)=36*x(t)+40*y(t),
x(0)=0.5,y(0)=0.5},{x(t),y(t)});

This is not a straight line. Let's find its slope .

> slope:=simplify(diff(-4.000000000*exp(-5*t)+4.500000000*exp(4*t),t)/
diff(-4.500000000*exp(4*t)+5.000000000*exp(-5*t),t));

This is not a constant.

> plot(slope,t=-1..1);

Note that the slope doesn't change much, though.

Sinks

The isolated singularity is a sink if both eigenvalues are negative.

Example 1

Once again, we start with a diagonal example.

> C:=matrix(2,2,[[-1,0],[0,-3]]);

> eigenvectors(C);

> phaseportrait({D(x)(t)=-x(t),D(y)(t)=-3*y(t)},
[x(t),y(t)],
t=-.5..5,
[[x(0)=1,y(0)=1],[x(0)=1,y(0)=-1],
[x(0)=-1,y(0)=1],[x(0)=-1,y(0)=-1],
[x(0)=-0.5,y(0)=2],[x(0)=-0.5,y(0)=-2],
[x(0)=0.5,y(0)=2],[x(0)=0.5,y(0)=-2]]);

Notice that all solutions seem to approach the origin along the -axis. Indeed, the general solution is as follows.

> dsolve({D(x)(t)=-x(t),D(y)(t)=-3*y(t)},
[x(t),y(t)]);

${x(t) = exp(-t)*_C2, y(t) = exp(-3*t)*_C1}$

Thus as , and , and the ratio is given by

> simplify(C1*exp(-3*t)/(C2*exp(-t)));

which goes to zero as .
This is a general phenomenon. The solutions will approach the origin along the line spanned by the eignevector whose eigenvalue has smaller magnitude.

Example 2

This is not a diagonal example, but the solutions still approach the origin along the line spanned by the eigenvector whose eigenvalue has smaller magnitude.

> E:=matrix(2,2,[[-1,1],[-2,-4]]);
eigenvectors(E);

> phaseportrait({D(x)(t)=-1*x(t)+1*y(t),D(y)(t)=-2*x(t)-4*y(t)},
[x(t),y(t)],
t=-1..5,
[[x(0)=0.5,y(0)=0.5],[x(0)=0,y(0)=-1],
[x(0)=0,y(0)=1],[x(0)=-0.5,y(0)=-0.5],
[x(0)=-1,y(0)=2],[x(0)=-1,y(0)=1],
[x(0)=1,y(0)=-2],[x(0)=1,y(0)=-1],
[x(0)=1,y(0)=-1.5],[x(0)=-1,y(0)=1.5]],
x=-2..2,y=-2..2,
stepsize=0.1);

Sources

The isolated singularity is a source if both eigenvalues are positive.
This is just like a sink except the arrows go the other direction. The pictures in the next two example should look very familiar if you have looked at the sink examples.

Example 1

Once again, we start with a diagonal example.

> F:=matrix(2,2,[[1,0],[0,3]]);

> eigenvectors(F);

> phaseportrait({D(x)(t)=x(t),D(y)(t)=3*y(t)},
[x(t),y(t)],
t=-5..0.5,
[[x(0)=1,y(0)=1],[x(0)=1,y(0)=-1],
[x(0)=-1,y(0)=1],[x(0)=-1,y(0)=-1],
[x(0)=-0.5,y(0)=2],[x(0)=-0.5,y(0)=-2],
[x(0)=0.5,y(0)=2],[x(0)=0.5,y(0)=-2]]);

Notice that all solutions seem to leave the origin along the -axis. Indeed, the general solution is as follows.

> dsolve({D(x)(t)=x(t),D(y)(t)=3*y(t)},
[x(t),y(t)]);

${y(t) = exp(3*t)*_C1, x(t) = exp(t)*_C2}$

Thus as , and , and the ratio is given by

> simplify(C1*exp(3*t)/(C2*exp(t)));

which goes to zero as .
This is a general phenomenon. The solutions will leave the origin along the line spanned by the eignevector whose eigenvalue has smaller magnitude.

Example 2

This is not a diagonal example, but the solutions still leave the origin along the line spanned by the eigenvector whose eigenvalue has smaller magnitude.

> E:=matrix(2,2,[[1,-1],[2,4]]);
eigenvectors(E);

> phaseportrait({D(x)(t)=1*x(t)-1*y(t),D(y)(t)=2*x(t)+4*y(t)},
[x(t),y(t)],
t=-5..1,
[[x(0)=0.5,y(0)=0.5],[x(0)=0,y(0)=-1],
[x(0)=0,y(0)=1],[x(0)=-0.5,y(0)=-0.5],
[x(0)=-1,y(0)=2],[x(0)=-1,y(0)=1],
[x(0)=1,y(0)=-2],[x(0)=1,y(0)=-1],
[x(0)=1,y(0)=-1.5],[x(0)=-1,y(0)=1.5]],
x=-2..2,y=-2..2,
stepsize=0.1);

Stability

Suppose P = (x0, y0) is an isolated singularity of a system

.

We say that is stable if any solution starting sufficiently close to stays close to , that is, more exactly,
for all , , there is a , , such that for any with dist( , ) < , the solution with initial value satisfies dist( , ) < for all , .

We say that is asymptotically stable if it is stable and any solution starting sufficiently close to approaches as , that is, more exactly, there is a , , such that for any with dist( , ) < , the solution with initial value satisfies as .

Note that if is asymptotically stable, it is stable.

We say that is unstable if it is not stable.

Any sink is asymptotically stable. Indeed, as we shall see, any system whose eigenvalues have negative real part has an asymptotically stable stationary point.

Both saddles and sources are unstable. Indeed, if any eigenvalue has positive real part, will be unstable.

The system

whose solutions are circles around the origin, provides an example of a system that is stable but not asymptotically stable.