Differential Equations

Systems of Equations

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Basic Concepts of Stability Theory

Solved Problems

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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 region (shaded in green), where the solution of the dynamic system is asymptotically stable
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.

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