Routh-Hurwitz Criterion

Solved Problems

Click or tap a problem to see the 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

${x^{IV}} + x^{\prime\prime\prime} + \alpha x^{\prime\prime} + \beta x' + x = 0.$

Example 4

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

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

Example 1.

Investigate the stability of the zero solution of the equation

$x^{\prime\prime\prime} + 6x^{\prime\prime} + 3x' + 2x = 0.$

Solution.

We write the auxiliary equation:

${\lambda ^3} + 6{\lambda ^2} + 3\lambda + 2 = 0.$

The coefficients $${a_i}$$ are

${a_0} = 1,\;\; {a_1} = 6,\;\; {a_2} = 3,\;\; {a_3} = 2.$

Form the Hurwitz matrix

$\left[ {\begin{array}{*{20}{c}} {{a_1}}&{{a_0}}&0\\ {{a_3}}&{{a_2}}&{{a_1}}\\ 0&{{a_4}}&{{a_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 6&1&0\\ 2&3&6\\ 0&0&2 \end{array}} \right]$

and find all its principal minors:

${\Delta _1} = 6 \gt 0,\;\; {\Delta _2} = \left| {\begin{array}{*{20}{c}} 6&1\\ 2&3 \end{array}} \right| = 18 - 2 = 16 \gt 0,\;\; {\Delta _3} = {a_3}{\Delta _2} = 2 \cdot 16 = 32 \gt 0.$

As it can be seen, all minors are positive. Therefore, the zero solution of the equation is asymptotically stable.

Example 2.

Investigate the stability of the zero solution of the differential equation

${x^{IV}} + 2x^{\prime\prime\prime} + 4x^{\prime\prime} + 7x' + 3x = 0.$

Solution.

The auxiliary equation is given by

${\lambda ^4} + 2{\lambda ^3} + 4{\lambda ^2} + 7\lambda + 3 = 0.$

The coefficients $${a_i}$$ in this equation are

${a_0} = 1,\;\; {a_1} = 2,\;\; {a_2} = 4,\;\; {a_3} = 7,\;\; {a_4} = 3.$

We write the Hurwitz matrix and find its principal diagonal minors:

$\left[ {\begin{array}{*{20}{c}} {{a_1}}&{{a_0}}&0&0\\ {{a_3}}&{{a_2}}&{{a_1}}&{{a_0}}\\ 0&{{a_4}}&{{a_3}}&{{a_2}}\\ 0&0&0&{{a_4}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 2&1&0&0\\ 7&7&2&1\\ 0&3&7&4\\ 0&0&0&3 \end{array}} \right];$
${\Delta _1} = 2 \gt 0,\;\; {\Delta _2} = \left| {\begin{array}{*{20}{c}} 2&1\\ 7&4 \end{array}} \right| = 8 - 7 = 1 \gt 0,\;\; {\Delta _3} = \left| {\begin{array}{*{20}{c}} 2&1&0\\ 7&4&2\\ 0&3&7 \end{array}} \right| = - 2 \cdot \left| {\begin{array}{*{20}{c}} 2&1\\ 0&3 \end{array}} \right| + 7 \cdot \left| {\begin{array}{*{20}{c}} 2&1\\ 7&4 \end{array}} \right| = - 12 + 7 = - 5 \lt 0.$

In the course of calculations, we've got a negative minor $${\Delta _3} \lt 0.$$ This means that the zero solution is unstable.

Example 3.

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

${x^{IV}} + x^{\prime\prime\prime} + \alpha x^{\prime\prime} + \beta x' + x = 0.$

Solution.

The auxiliary equation can be written as

${\lambda ^4} + {\lambda ^3} + \alpha {\lambda ^2} + \beta \lambda + 1 = 0.$

The coefficients have the following values:

${a_0} = 1,\;\; {a_1} = 1,\;\; {a_2} = \alpha,\;\; {a_3} = \beta,\;\; {a_4} = 1.$

Form the Hurwitz matrix

$\left[ {\begin{array}{*{20}{c}} {{a_1}}&{{a_0}}&0&0\\ {{a_3}}&{{a_2}}&{{a_1}}&{{a_0}}\\ 0&{{a_4}}&{{a_3}}&{{a_2}}\\ 0&0&0&{{a_4}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} 1&1&0&0\\ \beta&\alpha&1&1\\ 0&1&\beta&\alpha\\ 0&0&0&1 \end{array}} \right]$

and write the Routh-Hurwitz stability condition:

${\Delta _1} = 1 \gt 0,$
${\Delta _2} = \left| {\begin{array}{*{20}{c}} 1&1\\ \beta &\alpha \end{array}} \right| = \alpha - \beta \gt 0,$
${\Delta _3} = \left| {\begin{array}{*{20}{c}} 1&1&0\\ \beta &\alpha &1\\ 0&1&\beta \end{array}} \right| = \left| {\begin{array}{*{20}{c}} \alpha &1\\ 1&\beta \end{array}} \right| - \beta \left| {\begin{array}{*{20}{c}} 1&0\\ 0&\beta \end{array}} \right| = \alpha \beta - 1 - {\beta ^2} = \beta \left( {\alpha - \beta } \right) - 1 \gt 0,$
${\Delta _4} = {a_4}{\Delta _3} \gt 0,\;\; \Rightarrow {a_4} = 1 \gt 0.$

Thus, the system of inequalities that describes the stability region is as follows:

$\left\{ \begin{array}{l} {\Delta _2} \gt 0\\ {\Delta _3} \gt 0 \end{array} \right.,\;\; \Rightarrow \left\{ {\begin{array}{*{20}{l}} {\alpha - \beta \gt 0}\\ {\beta \left( {\alpha - \beta } \right) \gt 1} \end{array}} \right..$

Let us first consider the second inequality. We solve it with respect to $$\alpha.$$ Here, there are two cases:

1. If $$\beta \gt 0,$$ then dividing both sides of the inequality by $$\beta,$$ we obtain:
$\alpha - \beta \gt \frac{1}{\beta },\;\; \Rightarrow \alpha \gt \beta + \frac{1}{\beta };$
2. If $$\beta \lt 0,$$ then we have:
$\alpha - \beta \lt \frac{1}{\beta },\;\; \Rightarrow \alpha \lt \beta + \frac{1}{\beta }.$

Obviously, the case $$\beta = 0$$ does not satisfy this inequality.

So, the solution of the inequality are two infinite regions, which are symmetric about the origin and bounded by the curve $$\alpha = \beta + {\frac{1}{\beta }}$$ (Figure $$1$$).

The first inequality $$\alpha \gt \beta$$ is satisfied by the set of points above the line $$\alpha = \beta.$$ As a result, the solution of the whole system is the only region $$\alpha \gt \beta + {\frac{1}{\beta }}$$ in the upper right quadrant (with $$\beta \gt 0,$$ $$\alpha \gt 0$$). In Figure $$1,$$ this area is shown in green. This region of the plane $$\left( {\beta ,\alpha } \right)$$ is a region of asymptotic stability of the original equation.

Example 4.

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

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

Solution.

The auxiliary equation is given by

$\left| {\begin{array}{*{20}{c}} {\alpha - \lambda }&\beta &0\\ 1&{ - \lambda }&{ - 1}\\ { - 1}&1&{ - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow \left( {\alpha - \lambda } \right)\left| {\begin{array}{*{20}{r}} { - \lambda }&{ - 1}\\ 1&{ - \lambda } \end{array}} \right| - \beta \left| {\begin{array}{*{20}{r}} 1&{ - 1}\\ { - 1}&{ - \lambda } \end{array}} \right| = 0,\;\; \Rightarrow \left( {\alpha - \lambda } \right)\left( {{\lambda ^2} + 1} \right) - \beta \left( { - \lambda - 1} \right) = 0,\;\; \Rightarrow \alpha {\lambda ^2} - {\lambda ^3} + \alpha - \lambda + \beta \lambda + \beta = 0,\;\; \Rightarrow {\lambda ^3} - \alpha {\lambda ^2} + \left( {1 - \beta } \right)\lambda - \alpha - \beta = 0.$

Here, the coefficients $${a_i}$$ have the following values:

${a_0} = 1,\;\; {a_1} = - \alpha ,\;\; {a_2} = 1 - \beta ,\;\; {a_3} = - \alpha - \beta .$

First we determine at which values of $$\alpha$$ and $$\beta$$ the necessary condition of the stability holds:

${a_i} \gt 0,\;\;i = 0,1,2,3.$

We obtain the following system of inequalities:

$\left\{ \begin{array}{l} - \alpha \gt 0\\ 1 - \beta \gt 0\\ - \alpha - \beta \gt 0 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} \alpha \lt 0\\ \beta \lt 1\\ \beta \lt - \alpha \end{array} \right..$

It is convenient to draw the solution to this system in the plane $$\left( {\alpha ,\beta } \right).$$ Figure $$2$$ shows the straight lines $$\alpha = 0,$$ $$\beta = 1,$$ $$\beta = -\alpha$$ and the regions corresponding to the solution of each inequality. The intersection of these areas is marked in yellow.

$\left\{ \begin{array}{l} - \alpha \gt 0\\ 1 - \beta \gt 0\\ - \alpha - \beta \gt 0 \end{array} \right.,\;\; \Rightarrow \left\{ \begin{array}{l} \alpha \lt 0\\ \beta \lt 1\\ \beta \lt - \alpha \end{array} \right..$

Now consider the sufficient conditions of the stability. Write the Hurwitz matrix:

$\left[ {\begin{array}{*{20}{c}} {{a_1}}&{{a_0}}&0\\ {{a_3}}&{{a_2}}&{{a_1}}\\ 0&0&{{a_3}} \end{array}} \right] = \left[ {\begin{array}{*{20}{c}} { - \alpha }&1&0\\ { - \alpha - \beta }&{1 - \beta }&{ - \alpha }\\ 0&0&{ - \alpha - \beta } \end{array}} \right].$

The principal diagonal minors $${\Delta _i}$$ are given by

${\Delta _1} = - \alpha ,$
${\Delta _2} = \left| {\begin{array}{*{20}{c}} { - \alpha }&1\\ { - \alpha - \beta }&{1 - \beta } \end{array}} \right| = \alpha \left( {\beta - 1} \right) + \alpha + \beta = \alpha \beta - \cancel{\alpha} + \cancel{\alpha} + \beta = \left( {\alpha + 1} \right)\beta ,$
${\Delta _3} = {a_3}{\Delta _2} = \left( { - \alpha - \beta } \right)\left( {\alpha + 1} \right)\beta .$

According to the Routh-Hurwitz stability criterion, the following inequalities must be satisfied:

${\Delta _1} = - \alpha \gt 0,\;\;\; {\Delta _2} = \left( {\alpha + 1} \right)\beta \gt 0,\;\;\; {a_3} \gt 0.$

From these inequalities, the additional one (compared with the necessary conditions) is the only inequality

${\Delta _2} \gt 0\;\;\;\text{or}\;\;\; \left( {\alpha + 1} \right)\beta \gt 0.$

Its solution includes two cases:

1. $$\beta \gt 0,\;\;$$ $$\Rightarrow \alpha + 1 \gt 0\;\;$$ $$\text{or}\;\;\alpha \gt - 1;$$
2. $$\beta \lt 0,\;\;$$ $$\Rightarrow \alpha + 1 \lt 0\;\;$$ $$\text{or}\;\;\alpha \lt - 1.$$

These two cases are represented by two quadrants bounded by the lines $$\alpha = -1,$$ $$\beta = 0$$ (They are depicted in purple in Figure $$2$$).

The region of stability of the system is bounded by the values of the parameters $$\alpha, \beta,$$ at which both necessary and sufficient conditions of the asymptotic stability are satisfied. This region consists of the left lower quadrant bounded by the lines $$\alpha = -1,$$ $$\beta = 0,$$ and the right triangle with vertices $$\left( {0,0} \right),$$ $$\left( {-1,0} \right),$$ $$\left( {-1,1} \right).$$ The resulting stability region is shown in the figure in green.