Differential Equations

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Method of Lyapunov Functions

Solved Problems

Example 1.

Investigate the stability of the zero solution of the system

\[\frac{{dx}}{{dt}} = - 2x,\; \frac{{dy}}{{dt}} = x - y.\]

Solution.

This system is a linear homogeneous system with constant coefficients. We take as a Lyapunov function the quadratic form

\[V\left( \mathbf{X} \right) = V\left( {x,y} \right) = a{x^2} + b{y^2},\]

where the coefficients \(a, b\) are to be determined.

Obviously, the function \(V\left( {x,y} \right)\) is positive everywhere except at the origin, where it is zero. We calculate the total derivative of the function \(V\left( {x,y} \right):\)

\[\frac{{dV}}{{dt}} = \frac{{\partial V}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial V}}{{\partial y}}\frac{{dy}}{{dt}} = 2ax\left( { - 2x} \right) + 2by\left( {x - y} \right) = - 4a{x^2} + 2bxy - 2b{y^2} = - 2b\left( {\frac{{4a}}{{2b}}{x^2} - xy + {y^2}} \right) = - 2b\left( {\frac{{2a}}{b}{x^2} - xy + {y^2}} \right).\]

The expression in the brackets can be converted to a square of the difference if the following condition is satisfied:

\[\frac{{2a}}{b} = \frac{1}{4}\;\;\text{or}\;\;8a = b.\]

We can take any suitable combination, for example, we set \(a = 1,\) \(b = 8.\) Then the derivative becomes

\[\frac{{dV}}{{dt}} = - 16\left( {\frac{{{x^2}}}{4} - xy + {y^2}} \right) = - 16{\left( {\frac{x}{2} - y} \right)^2} \lt 0.\]

Thus, for the given system, there is a Lyapunov function, and its derivative is negative everywhere except at the origin. Hence, the zero solution of the system is asymptotically stable (stable node).

Example 2.

Investigate the stability of the zero solution of the system

\[\frac{{dx}}{{dt}} = y,\; \frac{{dy}}{{dt}} = - x.\]

Solution.

Note that the first approximation method is not applicable for this system, since the zero solution is a "center" (that is the system is not rough):

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

We use the method of Lyapunov functions for the stability analysis. Let the function \(V\left( \mathbf{X} \right)\) have the form

\[V\left( \mathbf{X} \right) = V\left( {x,y} \right) = {x^2} + {y^2}.\]

We calculate the derivative of the function \(V\left( \mathbf{X} \right)\) by virtue of the system:

\[\frac{{dV}}{{dt}} = \frac{{\partial V}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial V}}{{\partial y}}\frac{{dy}}{{dt}} = 2x \cdot y + 2y \cdot \left( { - x} \right) \equiv 0.\]

Thus, the derivative is identically zero. Hence, the function \(V\left( \mathbf{X} \right)\) is a Lyapunov function and the zero solution of the system is stable in the sense of Lyapunov. The condition of asymptotic stability is not satisfied (for this, the derivative \({\frac{{dV}}{{dt}}}\) must be negative).

Example 3.

Investigate the stability of the zero solution of the nonlinear system

\[\frac{{dx}}{{dt}} = - x{y^2},\; \frac{{dy}}{{dt}} = -y - y{x^2}.\]

Solution.

First we determine the Jacobian of the system at the point \(\left( {0,0} \right):\)

\[J = {\left. {\left[ {\begin{array}{*{20}{c}} {\frac{{\partial {f_1}}}{{\partial x}}}&{\frac{{\partial {f_1}}}{{\partial y}}}\\ {\frac{{\partial {f_2}}}{{\partial x}}}&{\frac{{\partial {f_2}}}{{\partial y}}} \end{array}} \right]} \right|_{\substack{ x = 0\\ y = 0}}} = {\left. {\left[ {\begin{array}{*{20}{c}} { - {y^2}}&{ - 2xy}\\ { - 2xy}&{ - 1 - {x^2}} \end{array}} \right]} \right|_{\substack{ x = 0\\ y = 0}}} = \left[ {\begin{array}{*{20}{c}} 0&0\\ 0&{ - 1} \end{array}} \right].\]

Find the eigenvalues of this matrix:

\[\det \left( {J - \lambda I} \right) = 0,\;\; \Rightarrow \left[ {\begin{array}{*{20}{c}} { - \lambda }&0\\ 0&{ - 1 - \lambda } \end{array}} \right] = 0,\;\; \Rightarrow \lambda \left( {1 + \lambda } \right) = 0,\;\; \Rightarrow {\lambda _1} = 0,\;{\lambda _2} = - 1.\]

Since one of the eigenvalues is equal to zero, the first approximation method is inapplicable here.

Let's see what results can be obtained using a Lyapunov function. We choose as a Lyapunov function the quadratic form

\[V\left( \mathbf{X} \right) = V\left( {x,y} \right) = {x^2} + {y^2},\]

which is positive definite everywhere except at the origin. Calculate the total derivative:

\[\frac{{dV}}{{dt}} = \frac{{\partial V}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial V}}{{\partial y}}\frac{{dy}}{{dt}} = 2x\left( { - x{y^2}} \right) + 2y\left( {-y-y{x^2}} \right) = - 2{x^2}{y^2} - 2{y^2} - 2{x^2}{y^2} = - 4{x^2}{y^2} - 2{y^2} = -2{y^2}{\left({2{x^2}+1}\right)}.\]

The derivative \(\frac{{dV}}{{dt}}\) is negative everywhere except the points \(\left( {c,0} \right), c \in \mathbb{R},\) where it is equal to zero. This means that the zero solution is stable (in the sense of Lyapunov).

Example 4.

Investigate the stability of the zero solution of the system using the method of Lyapunov functions.

\[\frac{{dx}}{{dt}} = y - 2x,\; \frac{{dy}}{{dt}} = 2x - y - {x^3}.\]

Solution.

As a Lyapunov candidate function we choose a function of the form

\[V\left( \mathbf{X} \right) = V\left( {x,y} \right) = {\left( {x + y} \right)^2} + \frac{{{x^4}}}{2}.\]

Obviously, this function is positive definite everywhere except at the origin, where it is zero. We calculate its derivative (by virtue of the system):

\[\frac{{dV}}{{dt}} = \frac{{\partial V}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial V}}{{\partial y}}\frac{{dy}}{{dt}} = \left( {2x + 2y + 2{x^3}} \right)\left( {y - 2x} \right) + \left( {2x + 2y} \right)\left( {2x - y - {x^3}} \right) = \cancel{\left( {2x + 2y} \right)\left( {y - 2x} \right)} + 2{x^3}\left( {y - 2x} \right) - \cancel{\left( {2x + 2y} \right)\left( {y - 2x} \right)} - {x^3}\left( {2x + 2y} \right) = \cancel{2{x^3}y} - 4{x^4} - 2{x^4} - \cancel{2{x^3}y} = - 6{x^4} \le 0.\]

As one can see, the derivative is negative definite everywhere except at \(\left( {0,0} \right).\) Then the zero solution is asymptotically stable.

Note that the first approximation method is inapplicable here because one of the eigenvalues is zero:

\[J = {\left. {\left[ {\begin{array}{*{20}{c}} {\frac{{\partial {f_1}}}{{\partial x}}}&{\frac{{\partial {f_1}}}{{\partial y}}}\\ {\frac{{\partial {f_2}}}{{\partial x}}}&{\frac{{\partial {f_2}}}{{\partial y}}} \end{array}} \right]} \right|_{\substack{ x = 0\\ y = 0}}} = \left[ {\begin{array}{*{20}{c}} {-2}&{1}\\ 2&{-1} \end{array}} \right];\]
\[\det \left( {J - \lambda I} \right) = 0,\;\; \Rightarrow \left| {\begin{array}{*{20}{c}} { - 2 - \lambda }&1\\ 2&{ - 1 - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow \left( { - 2 - \lambda } \right)\left( { - 1 - \lambda } \right) - 2 = 0,\;\; \Rightarrow \cancel{2} + \lambda + 2\lambda + {\lambda ^2} - \cancel{2} = 0,\;\; \Rightarrow {\lambda ^2} + 3\lambda = 0,\;\; \Rightarrow \lambda \left( {\lambda + 3} \right) = 0,\;\; \Rightarrow {\lambda _1} = 0,\;{\lambda _2} = - 3.\]

Example 5.

Using a Lyapunov function, investigate the stability of the zero solution of the system

\[\frac{{dx}}{{dt}} = x + 3y,\; \frac{{dy}}{{dt}} = 2x.\]

Solution.

Take as a function \(V\left( \mathbf{X} \right)\) the following function:

\[V\left( \mathbf{X} \right) = V\left( {x,y} \right) = 2{x^2} - 3{y^2}.\]

The choice of the coefficients will be clear from the subsequent transformations. We calculate the total derivative of the function \(V\left( \mathbf{X} \right)\) by virtue of this system:

\[\frac{{dV}}{{dt}} = \frac{{\partial V}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial V}}{{\partial y}}\frac{{dy}}{{dt}} = 4x\left( {x + 3y} \right) - 6y\left( {2x} \right) = 4{x^2} + \cancel{12xy} - \cancel{12xy} = 4{x^2} \gt 0.\]

Thus, the derivative \({\frac{{dV}}{{dt}}}\) is positive definite everywhere except at the origin.

On the other hand, we can find points that are arbitrarily close to zero, in which the function \(V\left( \mathbf{X} \right)\) will also be positive. Such points, for example, are located on the \(x\)-axis at \(y = 0.\)

As it can be seen, the conditions of the Lyapunov instability theorem are met. Therefore, the zero solution of the system is unstable.

Example 6.

Investigate the stability of the zero solution of the system

\[\frac{{dx}}{{dt}} = {x^3} + y,\; \frac{{dy}}{{dt}} = x + {y^3}.\]

Solution.

As it follows from the RHS (right-hand sides) of the equations, the derivatives \({\frac{{dx}}{{dt}}},\) \({\frac{{dy}}{{dt}}}\) will increase for the points in the first quadrant of the phase plane (\(x \gt 0,\) \(y \gt 0\)). It can therefore be assumed that the system is unstable. We use the Chetaev theorem to prove this.

Let the function \(V\left( \mathbf{X} \right)\) have the form

\[V\left( \mathbf{X} \right) = V\left( {x,y} \right) = {x^2} - {y^2}.\]

This function is positive definite in the subdomain \({U_1},\) in which the inequality \(\left| x \right| \gt \left| y \right|\) holds (see Figure \(4\)).

An example of application of Chetaev instability theorem
Figure 4.

We calculate the derivative \({\frac{{dV}}{{dt}}}\) by virtue of the system and define its sign in the subdomain \({U_1}.\)

\[\frac{{dV}}{{dt}} = \frac{{\partial V}}{{\partial x}}\frac{{dx}}{{dt}} + \frac{{\partial V}}{{\partial y}}\frac{{dy}}{{dt}} = 2x\left( {{x^3} + y} \right) - 2y\left( {x + {y^3}} \right) = 2{x^4} + \cancel{2xy} - \cancel{2xy} - 2{y^4} = 2\left( {{x^4} - {y^4}} \right).\]

One can see that the derivative \({\frac{{dV}}{{dt}}}\) is also positive definite in the subdomain \({U_1}\) defined by the relationship \(\left| x \right| \gt \left| y \right|.\) Besides that, the function \(V\left( \mathbf{X} \right)\) is zero on the boundary of \({U_1}\) including the point \(\left( {0,0} \right).\) Thus, all conditions of the Chetaev theorem are fulfilled. Therefore, the zero solution of the system is unstable.

Calculating the eigenvalues of the Jacobian of the linearized system, we see that the zero equilibrium point is a saddle:

\[ J = {\left. {\left[ {\begin{array}{*{20}{c}} {\frac{{\partial {f_1}}}{{\partial x}}}&{\frac{{\partial {f_1}}}{{\partial y}}}\\ {\frac{{\partial {f_2}}}{{\partial x}}}&{\frac{{\partial {f_2}}}{{\partial y}}} \end{array}} \right]} \right|_{\substack{ x = 0\\ y = 0}}} = {\left. {\left[ {\begin{array}{*{20}{c}} {3{x^2}}&1\\ 1&{3{y^2}} \end{array}} \right]} \right|_{\substack{ x = 0\\ y = 0}}} = \left[ {\begin{array}{*{20}{c}} 0&1\\ 1&0 \end{array}} \right];\]
\[ \det \left( {J - \lambda I} \right) = 0,\;\; \Rightarrow \left| {\begin{array}{*{20}{c}} { - \lambda }&1\\ 1&{ - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow {\lambda ^2} - 1 = 0,\;\; \Rightarrow {\lambda ^2} = 1,\;\; \Rightarrow {\lambda _{1,2}} = \pm 1.\]
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