Differential Equations

Systems of Equations

Routh-Hurwitz Criterion

Page 1
Theory
Page 2
Problems 1-4

Suppose we are given an \(n\)th order homogeneous system of differential equations with constant coefficients:

\[
{\mathbf{X’}\left( t \right) = A\mathbf{X}\left( t \right),\;\;}\kern-0.3pt
{\mathbf{X}\left( t \right) = \left[ {\begin{array}{*{20}{c}}
{{x_1}\left( t \right)}\\
{{x_2}\left( t \right)}\\
\vdots \\
{{x_n}\left( t \right)}
\end{array}} \right],\;\;}\kern-0.3pt
{{A \text{ = }}\kern0pt{\left[ {\begin{array}{*{20}{c}}
{{a_{11}}}&{{a_{12}}}& \vdots &{{a_{1n}}}\\
{{a_{21}}}&{{a_{22}}}& \vdots &{{a_{2n}}}\\
\cdots & \cdots & \cdots & \cdots \\
{{a_{n1}}}&{{a_{n2}}}& \vdots &{{a_{nn}}}
\end{array}} \right],}}
\]

where \(\mathbf{X}\left( t \right)\) is an \(n\)-dimensional vector containing the unknown functions, \(A\) is a square matrix of size \(n \times n.\)

A nonlinear autonomous system can be reduced to the linear system by performing a linearization around an equilibrium point. Then without loss of generality, we may assume that the equilibrium point is at the origin. It is always possible to reach by choosing a suitable coordinate system.

The stability or instability of the equilibrium state is determined by the signs of the real parts of the eigenvalues of \(A.\) To find the eigenvalues \(\lambda,\) it is necessary to solve the auxiliary equation

\[\det \left( {A – \lambda I} \right) = 0,\]

which is reduced to an algebraic equation of the \(n\)th degree

\[{{a_0}{\lambda ^n} + {a_1}{\lambda ^{n – 1}} }+{ {a_2}{\lambda ^{n – 2}} + \cdots }+{ {a_{n – 1}}\lambda + {a_n} }={ 0.}\]

The roots of this equation can be easily calculated in the case \(n = 2,\) and in some cases when \(n \ge 3.\) In other cases, solving the auxiliary equation can be a difficult problem. Moreover, Norwegian mathematician Niels Henrik Abel \(\left( 1802-1829 \right)\) proved a theorem according to which the general algebraic equation of degree \(n \ge 5\) cannot be solved using four basic arithmetical operations, i.e. there is no formula expressing the roots of the equation through its coefficients in the case \(n \ge 5.\)

In such a situation, methods allowing to determine whether all roots have negative real parts and establish the stability of the system without solving the auxiliary equation itself, are of great importance. One of these methods is the Routh-Hurwitz criterion, which contains the necessary and sufficient conditions for the stability of the system.

Consider again the auxiliary equation

\[{{a_0}{\lambda ^n} + {a_1}{\lambda ^{n – 1}} }+{ {a_2}{\lambda ^{n – 2}} + \cdots }+{ {a_{n – 1}}\lambda + {a_n} }={ 0,}\]

describing the dynamic system. Note that the necessary condition for the stability is satisfied if all the coefficients \({a_i} \gt 0.\) Therefore, we assume that the coefficient \({a_0} \gt 0.\) We write the so-called Hurwitz matrix. It is composed as follows.

The main diagonal of the matrix contains elements \({a_1},\) \({a_2}, \ldots ,\) \({a_n}.\) The first column contains numbers with odd indices \({a_1},\) \({a_3},\) \({a_5}, \ldots \) In each row, the index of each following number (counting from left to right) is \(1\) less than the index of its predecessor. All other coefficients \({a_i}\) with indices greater than \(n\) or less than \(0\) are replaced by zeros. The result is a matrix of kind

The principal diagonal minors \({\Delta _i}\) of the Hurwitz matrix are given by the formulas

\[
{{\Delta _1} = {a_1},\;\;}\kern-0.3pt{{\Delta _2} = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{a_0}}\\
{{a_3}}&{{a_2}}
\end{array}} \right|,\;\;}\kern-0.3pt
{{\Delta _3} = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{a_0}}&0\\
{{a_3}}&{{a_2}}&{{a_1}}\\
{{a_5}}&{{a_4}}&{{a_3}}
\end{array}} \right|,\;\;}\kern-0.3pt
{{{\Delta _n} \text{ = }}\kern0pt{ \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{a_0}}&0& \vdots &0\\
{{a_3}}&{{a_2}}&{{a_1}}& \vdots &0\\
{{a_5}}&{{a_4}}&{{a_3}}& \vdots &0\\
\cdots & \cdots & \cdots & \cdots & \cdots \\
0&0&0& \vdots &{{a_n}}
\end{array}} \right|.}}
\]

We now formulate the Routh-Hurwitz stability criterion: The roots of the auxiliary equation have negative real parts if and only if all the principal diagonal minors of the Hurwitz matrix are positive provided that \({a_0} \gt 0:\) \({\Delta _1} \gt 0,\) \({\Delta _2} \gt 0,\ldots,\) \({\Delta _n} \gt 0.\) As \({\Delta _n} = {a_n}{\Delta _{n – 1}},\) the last inequality can be written as \({a_n} \gt 0.\)

For the most common systems of the \(2\)nd, \(3\)rd and \(4\)th order, we obtain the following stability criteria:

For a second order system, the condition of the stability is given by

\[
{{a_0} \gt 0,\;\;}\kern-0.3pt{{\Delta _1} = {a_1} \gt 0,\;\;}\kern-0.3pt
{{\Delta _2} = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{a_0}}\\
{{a_3}}&{{a_2}}
\end{array}} \right| }={ {a_1}{a_2} \gt 0}
\]

or

\[{{a_0} \gt 0,\;\;}\kern-0.3pt{{a_1} \gt 0,\;\;}\kern-0.3pt{{a_2} \gt 0,}\]

that is, all coefficients of the quadratic characteristic equation must be positive. In other words, for a system of \(2\)nd order, the necessary condition of the stability is also the sufficient one. We emphasize that we consider here the asymptotic stability of the zero solution.

For a \(3\)rd order system, the stability criterion is defined by the inequalities

\[
{{a_0} \gt 0,\;\;}\kern-0.3pt{{\Delta _1} = {a_1} \gt 0,\;\;}\kern-0.3pt
{{{\Delta _2} = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{a_0}}\\
{{a_3}}&{{a_2}}
\end{array}} \right| }
= {{a_1}{a_2} – {a_0}{a_3} \gt 0,\;\;}}\kern-0.3pt
{{\Delta _3} = {a_3} \gt 0.}
\]

or

\[
{{a_0} \gt 0,\;\;}\kern-0.3pt{{a_1} \gt 0,\;\;}\kern-0.3pt
{{a_2} \gt 0,\;\;}\kern-0.3pt
{{a_3} \gt 0,\;\;}\kern-0.3pt
{{a_1}{a_2} – {a_0}{a_3} \gt 0.}
\]

Similarly, for a \(4\)th order system, we obtain the following set of inequalities:

\[
{{a_0} \gt 0,\;\;}\kern-0.3pt{{\Delta _1} = {a_1} \gt 0,\;\;}\kern-0.3pt
{{{\Delta _2} = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{a_0}}\\
{{a_3}}&{{a_2}}
\end{array}} \right| }={ {a_1}{a_2} – {a_0}{a_3} \gt 0,\;\;}}\kern-0.3pt
{{{\Delta _3} = \left| {\begin{array}{*{20}{c}}
{{a_1}}&{{a_0}}&0\\
{{a_3}}&{{a_2}}&{{a_1}}\\
0&{{a_4}}&{{a_3}}
\end{array}} \right| }
= {{a_1}{a_2}{a_3} }-{ a_1^2{a_4} }-{ {a_0}a_3^2 \gt 0,\;\;}}\kern-0.3pt
{{\Delta _4} = {a_4} \gt 0}
\]

or

\[
{{a_i} \gt 0\;\left( {i = 0, \ldots ,4} \right),\;\;\;}\kern-0.3pt
{{a_1}{a_2} – {a_0}{a_3} \gt 0,\;\;\;}\kern-0.3pt
{{{a_1}{a_2}{a_3} }-{ a_1^2{a_4} }-{ {a_0}a_3^2 \gt 0.}}
\]

If all the \(n – 1\) principal minors of the Hurwitz matrix are positive and the \(n\)th order minor is zero: \({\Delta _n} = 0,\) the system is at the boundary of stability. As \({\Delta _n} = {a_n}{\Delta _{n – 1}},\) then there are two options:

  1. The coefficient \({a_n} = 0.\) This corresponds to the case when one of the roots of the auxiliary equation is zero. The system is on the boundary of the aperiodic stability.
  2. The determinant \({\Delta _{n – 1}} = 0.\) In this case, there are two complex conjugate imaginary roots. The system is on the boundary of the oscillatory stability.

The Routh-Hurwitz stability criterion belongs to the family of algebraic criteria. It can be conveniently used to analyze the stability of low order systems. The computational complexity grows significantly with the increase of the order. In such cases, it may be preferable to use other criteria such as the Lienard-Shipart theorem or the Nyquist frequency criterion.

Solved Problems

Click on problem description to see solution.

 Example 1

Investigate the stability of the zero solution of the equation

\[{x^{\prime\prime\prime} + 6x^{\prime\prime} }+{ 3x’ }+{ 2x }={ 0.}\]

 Example 2

Investigate the stability of the zero solution of the differential equation

\[{{x^{IV}} + 2x^{\prime\prime\prime} }+{ 4x^{\prime\prime} + 7x’ }+{ 3x = 0.}\]

 Example 3

For what values of the parameters\(\alpha\) and \(\beta\) the zero solution of the equation

\[{{x^{IV}} + x^{\prime\prime\prime} + \alpha x^{\prime\prime} }+{ \beta x’ + x = 0}\]

is asymptotically stable?

 Example 4

Investigate for which values of the parameters \(\alpha\) and \(\beta\) the zero solution of the system is asymptotically stable:

\[
{\frac{{dx}}{{dt}} = \alpha x + \beta y,\;\;}\kern-0.3pt
{\frac{{dy}}{{dt}} = x – z,\;\;}\kern-0.3pt
{\frac{{dz}}{{dt}} = – x + y.}
\]
Page 1
Theory
Page 2
Problems 1-4