# Basic Concepts of Stability Theory

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Using the definition of Lyapunov stability, show that the zero solution is stable.

$\frac{{dx}}{{dt}} = - x - y,\;\;\frac{{dy}}{{dt}} = - x + y.$

### Example 2

Investigate the stability of the zero solution of the system, the general solution of which is given by

$\left\{ \begin{array}{l} x\left( t \right) = 3{C_1} + {C_2}{e^{ - t}}\\ y\left( t \right) = 2{C_1}{t^2}{e^{ - t}} - {C_2}\cos t \end{array} \right..$

### Example 3

Determine the values of the parameters $$a, b$$ for which the zero solution of the system is asymptotically stable.

$\frac{{dx}}{{dt}} = ax + y,\;\;\frac{{dy}}{{dt}} = x + by$

### Example 1.

Using the definition of Lyapunov stability, show that the zero solution is stable.

$\frac{{dx}}{{dt}} = - x - y,\;\;\frac{{dy}}{{dt}} = - x + y.$

Solution.

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

$A = \left[ {\begin{array}{*{20}{r}} { - 1}&{ - 1}\\ 1&1 \end{array}} \right],\;\; \det \left( {A - \lambda I} \right) = 0,\;\; \Rightarrow \left| {\begin{array}{*{20}{c}} { - 1 - \lambda }&{ - 1}\\ 1&{1 - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow {\left( { - 1 - \lambda } \right)^2} + 1 = 0,\;\; \Rightarrow {\lambda ^2} + 2\lambda + 2 = 0,\;\; \Rightarrow {\lambda _{1,2}} = - 1 \pm i.$

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

$\left( {A - {\lambda _1}I} \right){\mathbf{V}_1} = \mathbf{0},\;\Rightarrow \left[ {\begin{array}{*{20}{c}} { - 1 - \left( { - 1 + i} \right)}&{ - 1}\\ 1&{ - 1 - \left( { - 1 + i} \right)} \end{array}} \right] \left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow \left[ {\begin{array}{*{20}{c}} { - i}&{ - 1}\\ 1&{ - i} \end{array}} \right]\left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \mathbf{0},\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} { - i{V_{11}} - {V_{21}} = 0}\\ {{V_{11}} - i{V_{21}} = 0} \end{array}} \right..$

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

${V_{11}} = i{V_{21}} = it.$

Hence, the eigenvector $${\mathbf{V}_1}$$ is

${\mathbf{V}_1} = \left[ {\begin{array}{*{20}{c}} {{V_{11}}}\\ {{V_{21}}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} {it}\\ t \end{array}} \right] = t\left[ {\begin{array}{*{20}{c}} i\\ 1 \end{array}} \right] \sim \left[ {\begin{array}{*{20}{c}} i\\ 1 \end{array}} \right].$

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

${\mathbf{X}_1}\left( t \right) = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right] = {e^{{\lambda _1}t}}{\mathbf{V}_1} = {e^{\left( { - 1 + i} \right)t}}\left[ {\begin{array}{*{20}{c}} i\\ 1 \end{array}} \right].$

We expand the exponential function by Euler's formula:

${e^{\left( { - 1 + i} \right)t}} = {e^{ - t}}{e^{it}} = {e^{ - t}}\left( {\cos t + i\sin t} \right).$

As a result, we have

${\mathbf{X}_1}\left( t \right) = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right] = {e^{ - t}}\left( {\cos t + i\sin t} \right)\left[ {\begin{array}{*{20}{c}} i\\ 1 \end{array}} \right] = {e^{ - t}}\left[ {\begin{array}{*{20}{c}} {\left( {\cos t + i\sin t} \right)i}\\ {\cos t + i\sin t} \end{array}} \right] = {e^{ - t}}\left[ {\begin{array}{*{20}{c}} { - \sin t + i\cos t}\\ {\cos t + i\sin t} \end{array}} \right] = {e^{ - t}}\left[ {\begin{array}{*{20}{c}} { - \sin t}\\ {\cos t} \end{array}} \right] + i{e^{ - t}}\left[ {\begin{array}{*{20}{c}} {\cos t}\\ {\sin t} \end{array}} \right].$

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

$\text{Re}\left[ {{\mathbf{X}_1}\left( t \right)} \right] = {e^{ - t}}\left[ {\begin{array}{*{20}{r}} { - \sin t}\\ {\cos t} \end{array}} \right],\;\; \text{Im}\left[ {{\mathbf{X}_1}\left( t \right)} \right] = {e^{ - t}}\left[ {\begin{array}{*{20}{c}} {\cos t}\\ {\sin t} \end{array}} \right].$

Hence, the general solution is expressed by the formula

$\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}} x\\ y \end{array}} \right] = {C_1}{e^{ - t}}\left[ {\begin{array}{*{20}{r}} { - \sin t}\\ {\cos t} \end{array}} \right] + {C_2}{e^{ - t}}\left[ {\begin{array}{*{20}{c}} {\cos t}\\ {\sin t} \end{array}} \right]$

or

$\left\{ \begin{array}{l} x\left( t \right) = - {C_1}{e^{ - t}}\sin t + {C_2}{e^{ - t}}\cos t\\ y\left( t \right) = {C_1}{e^{ - t}}\cos t + {C_2}{e^{ - t}}\sin t \end{array} \right..$

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

$\left\{ \begin{array}{l} x\left( 0 \right) = {C_2} = {x_0}\\ y\left( 0 \right) = {C_1} = {y_0} \end{array} \right..$

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

$\left\{ \begin{array}{l} x\left( t \right) = - {y_0}{e^{ - t}}\sin t + {x_0}{e^{ - t}}\cos t\\ y\left( t \right) = {y_0}{e^{ - t}}\cos t + {x_0}{e^{ - t}}\sin t \end{array} \right..$

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

$\left| {\mathbf{X}\left( 0 \right) - \boldsymbol{\varphi} \left( 0 \right)} \right| \lt \delta,$

the relationship

$\left| {\mathbf{X}\left( t \right) - \boldsymbol{\varphi} \left( t \right)} \right| \lt \varepsilon$

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:

$\left\| {\mathbf{X}\left( 0 \right) - \boldsymbol{\varphi} \left( 0 \right)} \right\| = \sqrt {{{\left| {{x_0}} \right|}^2} + {{\left| {{y_0}} \right|}^2}} \le \left| {{x_0}} \right| + \left| {{y_0}} \right| = \frac{\delta }{2} + \frac{\delta }{2} = \delta .$

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:

$\left\| {\mathbf{X}\left( t \right) - \boldsymbol{\varphi} \left( t \right)} \right\| = {e^{ - t}}\sqrt {{{\left| { - {y_0}\sin t + {x_0}\cos t} \right|}^2} + {{\left| {{y_0}\cos t + {x_0}\sin t} \right|}^2}} \le {e^{ - t}}\sqrt {{{\left( {\left| {{y_0}} \right| + \left| {{x_0}} \right|} \right)}^2} + {{\left( {\left| {{y_0}} \right| + \left| {{x_0}} \right|} \right)}^2}} = {e^{ - t}}\sqrt {2{{\left( {\left| {{y_0}} \right| + \left| {{x_0}} \right|} \right)}^2}} = {e^{ - t}}\sqrt {2{{\left( {\frac{\delta }{2} + \frac{\delta }{2}} \right)}^2}} = {e^{ - t}}\sqrt {2{\delta ^2}} = \sqrt 2 {e^{ - t}}\delta \le \sqrt 2 \delta = \varepsilon .$

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:

$\lim\limits_{t \to \infty } \left| {\mathbf{X}\left( t \right) - \boldsymbol{\varphi} \left( t \right)} \right| = \lim\limits_{t \to \infty } \left( {\sqrt 2 {e^{ - t}}\delta } \right) = 0,$

that is the system is also asymptotically stable.

### Example 2.

Investigate the stability of the zero solution of the system, the general solution of which is given by

$\left\{ \begin{array}{l} x\left( t \right) = 3{C_1} + {C_2}{e^{ - t}}\\ y\left( t \right) = 2{C_1}{t^2}{e^{ - t}} - {C_2}\cos t \end{array} \right..$

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 $${x_0},{y_0}:$$

$\left\{ \begin{array}{l} x\left( 0 \right) = 3{C_1} = {x_0}\\ y\left( 0 \right) = - {C_2} = {y_0} \end{array} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {{C_1} = \frac{{{x_0}}}{3}}\\ {{C_2} = - {y_0}} \end{array}} \right..$

Then

$\left\{ \begin{array}{l} x\left( t \right) = {x_0} - {y_0}{e^{ - t}}\\ y\left( t \right) = \frac{2}{3}{x_0}{t^2}{e^{ - t}} + {y_0}\cos t \end{array} \right..$

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

$\left| {x\left( t \right)} \right| \lt \varepsilon ,\;\;\left| {y\left( t \right)} \right| \lt \varepsilon ,$

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:

$\left| {x\left( 0 \right)} \right| = \left| {{x_0}} \right| \lt \delta \left( \varepsilon \right),\;\; \left| {y\left( 0 \right)} \right| = \left| {{y_0}} \right| \lt \delta \left( \varepsilon \right).$

The result is

$\left| {x\left( t \right)} \right| = \left| {{x_0} - {y_0}{e^{ - t}}} \right| \le \left| {{x_0}} \right| + \left| {{y_0}} \right| \lt \varepsilon ,$
$\left| {y\left( t \right)} \right| = \left| {\frac{2}{3}{x_0}{t^2}{e^{ - t}} + {y_0}\cos t} \right| \le \frac{2}{3}\left| {{x_0}} \right|{t^2}{e^{ - t}} + \left| {{y_0}} \right| \lt \varepsilon .$

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

$g'\left( t \right) = {\left( {{t^2}{e^{ - t}}} \right)^\prime } = 2t{e^{ - t}} - {t^2}{e^{ - t}} = \left( {2t - {t^2}} \right){e^{ - t}} = t\left( {2 - t} \right){e^{ - t}},\;\; \Rightarrow g'\left( t \right) = 0\;\;\text{at}\;\;t = 0,2.$

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

${g_{\max }} = g\left( {t = 2} \right) = {2^2}{e^{ - 2}} \approx 0.54 \lt 1.$

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

$\left| {y\left( t \right)} \right| \le \frac{2}{3}\left| {{x_0}} \right| + \left| {{y_0}} \right| \lt \varepsilon .$

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

$\left| {{x_0}} \right| \lt \delta = \frac{\varepsilon }{2}\;\;\text{and}\;\; \left| {{y_0}} \right| \lt \delta = \frac{\varepsilon }{2},$

the inequalities

$\left| {x\left( t \right)} \right| = \left| {{x_0}} \right| + \;\left| {{y_0}} \right| \lt \frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \varepsilon ,\;\; \left| {y\left( t \right)} \right| \le \frac{2}{3}\left| {{x_0}} \right| + \left| {{y_0}} \right| \lt \frac{2}{3} \cdot \frac{\varepsilon }{2} + \frac{\varepsilon }{2} = \frac{{5\varepsilon }}{6} \lt \varepsilon .$

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 $$t \to \infty.$$

### Example 3.

Determine the values of the parameters $$a, b$$ for which the zero solution of the system is asymptotically stable.

$\frac{{dx}}{{dt}} = ax + y,\;\;\frac{{dy}}{{dt}} = x + by$

Solution.

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

$A = \left[ {\begin{array}{*{20}{c}} a&1\\ 1&b \end{array}} \right],\;\; \det \left( {A - \lambda I} \right) = 0,\;\; \Rightarrow \left| {\begin{array}{*{20}{c}} {a - \lambda }&1\\ 1&{b - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow \left( {a - \lambda } \right)\left( {b - \lambda } \right) - 1 = 0,\;\; \Rightarrow {\lambda ^2} - \left( {a + b} \right)\lambda + ab - 1 = 0.$

Solve the resulting quadratic equation with the parameters $$a, b.$$

$D = {\left( {a + b} \right)^2} - 4\left( {ab - 1} \right) = {a^2} + 2ab + {b^2} - 4ab + 4 = {a^2} - 2ab + {b^2} + 4 = {\left( {a - b} \right)^2} + 4 \gt 0.$

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

${\lambda _{1,2}} = \frac{{a + b \pm \sqrt {{{\left( {a - b} \right)}^2} + 4} }}{2}.$

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):

$\left\{ \begin{array}{l} {\lambda _1} \lt 0\\ {\lambda _2} \lt 0 \end{array} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{c}} {a + b + \sqrt {{{\left( {a - b} \right)}^2} + 4} \lt 0}\\ {a + b - \sqrt {{{\left( {a - b} \right)}^2} + 4} \lt 0} \end{array}} \right..$

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

$\sqrt {{{\left( {a - b} \right)}^2} + 4} \gt a + b$

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

We solve the first inequality:

$\left\{ \begin{array}{l} \sqrt {{{\left( {a - b} \right)}^2} + 4} \lt - \left( {a + b} \right)\\ a + b \lt 0 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} {\left( {a - b} \right)^2} + 4 \lt {\left( {a + b} \right)^2}\\ a + b \lt 0 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} {\left( {a + b} \right)^2} - {\left( {a + b} \right)^2} \gt 4\\ a + b \lt 0 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} \left( {a + b - a + b} \right)\left( {a + b + a - b} \right) \gt 4\\ a + b \lt 0 \end{array} \right.\;\; \Rightarrow \left\{ \begin{array}{l} 4ab \gt 4\\ a + b \lt 0 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} ab \gt 1\\ a + b \lt 0 \end{array} \right..$

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.