Basic Concepts of Stability Theory

Solved Problems

Solution.

First we find the general solution. The eigenvalues \({\lambda _i}\) of the coefficient matrix \(A\) are

Determine the eigenvector \({\mathbf{V}_1} = {\left( {{V_{11}},{V_{21}}} \right)^T}\) for the eigenvalue \({\lambda _1} = - 1 + i:\)

The resulting equations are linearly dependent. Therefore, by setting \({V_{21}} = t,\) we find from the second equation:

Hence, the eigenvector \({\mathbf{V}_1}\) is

Then the solution \({\mathbf{X}_1}\left( t \right),\) corresponding to the complex eigenvalue \({\lambda _1} = - 1 + i\) is given by

We expand the exponential function by Euler's formula:

As a result, we have

This shows that the real and imaginary parts of the solution are equal:

Hence, the general solution is expressed by the formula

or

Assuming that at the initial time \(t = 0\) the system is at the point \(\left( {{x_0},{y_0}} \right),\) we obtain:

Taking this into account, the solution can be written as follows:

The trajectory of this solution goes from the point \(\left( {{x_0},{y_0}} \right).\)

Now we study the stability of the zero solution, which we denote as \(\boldsymbol{\varphi} \left( t \right) \equiv 0.\) According to the definition of Lyapunov stability, we introduce a number \(\varepsilon \gt 0\) and find the corresponding number \(\delta = \delta \left( \varepsilon \right) \gt 0\) such that if

the relationship

will hold for all \(t \ge 0.\)

Suppose that the perturbed solution \(\mathbf{X}\left( t \right)\) at the initial moment has the coordinates \(\left( {{x_0},{y_0}} \right).\) Assuming that the deviation from zero for each of the coordinates does not exceed \(\frac{\delta }{2}\) and applying the triangle inequality, we can write:

We have used here the usual Euclidean metric as the norm.

Next, we establish a relationship between the numbers \(\delta\) and \(\varepsilon.\) Substituting the known expressions for the solution \(\mathbf{X}\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right),\) we get:

So, if we set \(\delta = {\frac{\varepsilon }{{\sqrt 2 }}},\) all perturbed trajectories emanating from the point \(\left( {{x_0},{y_0}} \right),\) provided that \(\left| {{x_0}} \right| \lt {\frac{\delta }{2}},\;\left| {{y_0}} \right| \lt {\frac{\delta }{2}},\) will remain in the tube with radius \(\varepsilon.\) Thus, the system is stable in the sense of Lyapunov.

Note that a stronger condition is satisfied here:

that is the system is also asymptotically stable.

Solution.

Let the initial conditions be given as \(x\left( 0 \right) = {x_0},y\left( 0 \right) = {y_0}.\) Express the general solution in terms of the coordinates \({x_0},{y_0}:\)

Then

Suppose that the solutions \(x\left( t \right), y\left( t \right)\) for all values \(t \ge 0\) satisfy the relations

where \(\varepsilon\) is a positive number. According to the definition of Lyapunov stability, we try to choose a number \(\delta \left( \varepsilon \right)\) depending on \(\varepsilon\) such that the following inequalities hold:

The result is

We take into account that the function \(g\left( t \right) = {t^2}{e^{ - t}}\) is bounded. Indeed,

At \(t = 2,\) the function \(g\left( t \right) = {t^2}{e^{ - t}}\) has a maximum equal

Then the inequality for \(\left| {y\left( t \right)} \right|\) can be written as

If we now choose \(\delta = {\frac{\varepsilon }{2}},\) so that

the inequalities

are obviously satisfied. Hence, the zero solution of the given system of equations is stable.

It follows from the formulas for \({x\left( t \right)},{y\left( t \right)} \) that the system is not asymptotically stable, since the values of \({x\left( t \right)},{y\left( t \right)} \) do not tend to zero as \(t \to \infty.\)

Solution.

We calculate the eigenvalues \({\lambda _i}\) of the coefficient matrix \(A:\)

Solve the resulting quadratic equation with the parameters \(a, b.\)

As it can be seen, the discriminant is always positive. Therefore, the eigenvalues are real numbers and are defined by

We find the set of values of the numbers \(a, b\) at which the eigenvalues \({\lambda _1},\) \({\lambda _2}\) are negative (this means that the system is asymptotically stable):

By adding the two inequalities, we get \(a + b \lt 0.\) In this case, the second inequality

holds for all \(a, b,\) satisfying \(a + b \lt 0.\)

We solve the first inequality:

The solutions of both elementary inequalities are shown graphically in Figure \(5.\)

The common solution is the region (shaded in green) below the hyperbola \(ab = 1\) in the left half-plane. For all values of \(a, b\) from this region, the solution of the system is asymptotically stable.