maisvendoo August 10, 2015 at 02:09

The Lyapunov function method in the problem of the Janibekov effect

Introduction

This article is not related to the series “The Magic of Tensor Algebra” , but is brought to life by publications from it. Carelessly clicking on the links in the search engine came across a discussion of one of his articles on the Janibekov effect, and drew attention to the fair remark that the study of the stability of the Janibekov nut in a first approximation does not give an unambiguous answer to the question of at what parameters the motion will be stable. This is so, since the roots of the characteristic polynomial, when rotating around the axis with the smallest and greatest moment of inertia, are purely imaginary, their real part is zero. Under such conditions, it is impossible to answer the question whether the movement will be stable without additional research.

The McCullagh interpretation is probably the simplest explanation for the Janibekov effect.

Such a study can be performed using the Lyapunov function method (the second or direct Lyapunov method). And in order to finally close the issue with the Janibekov nut, I decided to write this note.

1. Differential equations of perturbed motion. Again.

Let there be a system, in the general case of nonlinear differential equations of motion of some mechanical system

$\ frac {d \ mathbf y} {dt} = \ mathbf F (t, \, \ mathbf y)$

where $\ mathbf y = \ begin {bmatrix} y_1 && y_2 && \ cdots && y_n \ end {bmatrix} ^ T$ is the column vector of the system state variables; $\ mathbf F (t, \, \ mathbf y)$ Is a nonlinear vector function.

The solution of system (1) $\ mathbf y (t) = \ mathbf y_0 (t)$ gives the so-called unperturbed motion . In fact, this is a usual, steady-state mode of movement of the system under the action of forces applied to it. We set some perturbation determined by the vector of $\ mathbf x (t)$ deviations from the unperturbed motion, i.e.

$\ mathbf y (t) = \ mathbf y_0 (t) + \ mathbf x (t)$

Substituting (3) into (1), we obtain

$\ frac {d \ mathbf y_0} {dt} + \ frac {d \ mathbf x} {dt} = \ mathbf F (t, \, \ mathbf y_0 + \ mathbf x)$

Subtract (1) from (4)

$\ frac {d \ mathbf x} {dt} = \ mathbf F (t, \, \ mathbf y_0 + \ mathbf x) - \ mathbf F (t, \, \ mathbf y_0)$

$\ frac {d \ mathbf x} {dt} = \ mathbf G (t, \, \ mathbf x)$

where $\ mathbf G (t, \, \ mathbf x) = \ mathbf F (t, \, \ mathbf y_0 + \ mathbf x) - \ mathbf F (t, \, \ mathbf y_0)$ , and the resulting equation is called the perturbed motion equation , the trivial solution of which $x_1 = x_2 = ... = x_n = 0$ corresponds to the unperturbed motion of the system.

In our case, we restrict ourselves to considering an autonomous system, where the right-hand side is clearly time-independent

$\ frac {d \ mathbf x} {dt} = \ mathbf G (\ mathbf x)$

2. The tricky function V ( x ) is a candidate for the Lyapunov function

Consider some scalar function

$V = V (\ mathbf x) = V (x_1, \, x_2, \, ..., \, x_n)$

defined in some neighborhood of the origin, such that

$| x_i | <h, \ quad i = \ overline {1, n}$

where $h$ is some, rather small, positive number.

Function (6) is called sign-definite if in region (7) it takes values of only one sign (only positive or only negative), and is equal to zero only at the origin (for $x_1 = x_2 = ... = x_n = 0$ )

Function (6) is called sign-constant if in region (7) it takes values of only one particular sign, but can vanish at $x_1 ^ 2 + x_2 ^ 2 + ... + x_n ^ 2 \ ne 0$ .

We calculate the total time derivative of function (6). Since $x_i = x_i (t), \ quad i = \ overline {1, n}$ , by the definition of the total derivative, we obtain

$\ frac {dV} {dt} = \ sum_ {i = 1} ^ n \ frac {\ partial V} {\ partial x_i} \, \ dot x_i$

which, taking into account equation (5), is equivalent to the relation

$\ frac {dV} {dt} = \ sum_ {i = 1} ^ n \ frac {\ partial V} {\ partial x_i} \, G_i (x_1, \, x_2, \, ..., \, x_n)$

Function (8) is called the total time derivative of function (6), compiled by virtue of equation (5).

3. Lyapunov stability theorems

The two paragraphs above are written in the dry mathematical language of definitions, and probably not otherwise. Add some more formal math, formulating

Lyapunov stability theorem

If for the system of equations (5) there exists a sign-definite function $V (x_1, \, x_2, \, ..., \, x_n)$ (Lyapunov function), the total time derivative of which, compiled by virtue of system (5), is a constant function of the sign opposite to V , or identically equal to zero, then the stationary point of system (5) is $x_1 = x_2 = ... = x_n = 0$ stable

By the resting point of system (5) here we mean its trivial solution corresponding to the unperturbed motion of the mechanical system under consideration. Roughly speaking, according to the stated theorem, one should choose a function $V (x_1, \, x_2, \, ..., \, x_n)$ satisfying the properties indicated in the condition of the theorem. If it satisfies these properties, then it is called the Lyapunov function, and if such a function (at least one!) Exists, then the established mode of motion of the mechanical system under consideration will be stable.

However, in this theorem we are not talking about the asymptoticstability, that is, the nature of the movement of the system in which its perturbed movement will tend to the initial steady state. Here, stable is also understood as such a movement in which the system will fluctuate in the vicinity of the initial steady state, but will never return to it. The asymptotic stability condition will be more stringent

Lyapunov's asymptotic stability theorem

If for a system of equations (5) there exists a sign-definite function $V (x_1, \, x_2, \, ..., \, x_n)$ (Lyapunov function), the total time derivative of which, compiled by virtue of system (5), is a sign-definite function of the sign opposite to V , then the stationary point of system (5) is $x_1 = x_2 = ... = x_n = 0$ asymptotically stable

An asymptotically stable system, after perturbation, will tend to return to the steady state mode of motion, that is, the solution of system (5) will converge to the origin $x_i = 0, \ quad i = \ overline {1, n}$ .

These theorems provide a way to study the stability of linear and nonlinear mechanical systems, more general than the study in a first approximation.

Another question is how to find the Lyapunov function satisfying equation (5) and the requirements of the theorems. Mathematics still does not know a definite answer to this question. There are a number of works entirely devoted to this issue, for example, the book by E. A. Barabashin, “Lyapunov Functions” . For most linear systems, one can look for Lyapunov functions in the form of quadratic forms, for example, for a third-order system this function can be such

$V = x_1 ^ 2 + x_2 ^ 2 + x_3 ^ 2$

this function is definitely positive, and in an arbitrarily large neighborhood of the system’s stationary point. Or such a function

$V = x_1 ^ 2 + x_2 ^ 2 + 2 \, x_1 \, x_2 + x_3 ^ 2$

will be sign-constant, positive, because it $V = (x_1 + x_2) ^ 2 + x_3 ^ 2$ can be equal to zero both at the point of rest of the system $x_1 = x_2 = x_3 = 0$ and at a point that satisfies the condition $x_3 = 0, \ quad x_1 = -x_2$ .

In the case of conservative mechanical systems, the Lyapunov function can be the total mechanical energy of the system, which, in the absence of dissipation, is constant (sign-constant) and also the time derivative equal to zero - it is a constant. And this function follows from the system of equations of motion, because it is one of its integrals.

In the case of the Janibekov nut, I took the idea from A. P. Markeev's book Theoretical Mechanics as a very elegant solution . This solution has been slightly revised and expanded by me to be in the context of previously written articles.

4. Integrals of the motion of the Janibekov nut

We obtain the first two integrals of motion, relying on the system of equations given in the tensor cycle . We will operate with tensor relations so as not to lose hold. So, the equation of rotation of the nut around the center of mass has the form

$I _ {\, j} ^ {\, i} \, \ dot \ omega ^ {j} + \ varepsilon ^ {\, ijk} \, \ omega _ {\, j} \, g _ {\, kl} \, I _ {\, p} ^ {\, l} \, \ omega ^ {\, p} = 0$

let's move in this equation to the MCD vector

$\ dot L ^ {\, i} + \ varepsilon ^ {\, ijk} \, \ omega _ {\, j} \, L _ {\, k} = 0$

Multiply equation (10) scalarly by twice the MCD vector

$2 \, L _ {\, i} \, \ frac {dL ^ {\, i}} {dt} + 2 \, \ varepsilon ^ {\, ijk} \, L _ {\, i} \, L _ {\ , k} \, \ omega _ {\, j} = 0$

It is easy to see that in the second term (11) the convolution $\ varepsilon ^ {\, ijk} \, L _ {\, i} \, L _ {\, k} = 0$ , and in the first, is the derivative of the square of the MCD module. We transform equation (11) and integrate it

$\ begin {align *} & \ frac {d} {dt} \ left (L ^ {\, 2} \ right) = 0 \\ & L ^ {\, 2} = \ rm const \ end {align *}$

$I_x ^ {\, 2} \, \ omega_x ^ 2 + I_y ^ {\, 2} \, \ omega_y ^ 2 + I_z ^ {\, 2} \, \ omega_z ^ 2 = \ rm const$

Expression (12) is the first integral of motion expressing the constancy of the MCD module of the nut under consideration. To obtain another first integral of motion, we multiply (9) scalarly by the angular velocity vector

$\ omega _ {\, i} \, I _ {\, j} ^ {\, i} \, \ dot \ omega ^ {j} + \ varepsilon ^ {\, ijk} \, \ omega _ {\, i} \ , \ omega _ {\, j} \, g _ {\, kl} \, I _ {\, p} ^ {\, l} \, \ omega ^ {\, p} = 0$

after which, suddenly, we find in the second term the convolution $\ varepsilon ^ {\, ijk} \, \ omega _ {\, i} \, \ omega _ {\, j}$ equal to zero, getting the equation

$\ omega _ {\, i} \, I _ {\, j} ^ {\, i} \, \ dot \ omega ^ {j} = 0$

Recall that we have already seen something similar before . After all, the kinetic energy of the body in its rotation relative to the center of mass is

$T = \ frac {1} {2} \, \ omega_i \, I_j ^ i \, \ omega ^ {\, j}$

and if we differentiate it in time, we get

$\ frac {dT} {dt} = \ frac {1} {2} \, \ dot \ omega_i \, I_j ^ i \, \ omega ^ {\, j} + \ frac {1} {2} \, \ omega_i \, I_j ^ i \, \ dot \ omega ^ {\, j} = \ omega_i \, I_j ^ i \, \ dot \ omega ^ {\, j}$

Accordingly, we can rewrite equation (13) and integrate it

$\ begin {align *} & \ frac {d} {dt} \, \ left (\ frac {1} {2} \, \ omega_i \, I_j ^ i \, \ omega ^ {\, j} \ right) = 0 \\ & \ frac {1} {2} \, \ omega_i \, I_j ^ i \, \ omega ^ {\, j} = \ rm const \ end {align *}$

Given that multiplying a constant by a two does not change its “constancy”, we can finally write down the first integral in component form (taking into account the Cartesian basis!)

$I_x \, \ omega_x ^ 2 + I_y \, \ omega_y ^ 2 + I_z \, \ omega_z ^ 2 = \ rm const$

Expression (14) expresses the constancy of the kinetic energy of rotation of the nut around the center of mass. It remains to go to expressions (12) and (14) to dimensionless moments of inertia $i_y = \ frac {I_y} {I_x}, \ quad i_z = \ frac {I_z} {I_x}$

$\ begin {align *} & \ omega_x ^ 2 + i_y ^ {\, 2} \, \ omega_y ^ 2 + i_z ^ {\, 2} \, \ omega_z ^ 2 = \ rm const \\ & \ omega_x ^ 2 + i_y \, \ omega_y ^ 2 + i_z \, \ omega_z ^ 2 = \ rm const \ end {align *}$

The obtained equations are the first integrals of motion that we use to construct the Lyapunov function

4. Construction of the Lyapunov function from the integrals of motion

The method of constructing the Lyapunov function from equations of the form (15) is called the Chetaev method of integral connectives and suggests that the indicated function can be sought in the form of a bunch of integrals of motion of the form

$V = \ lambda_1 \, U_1 + \ lambda_2 \, U_2 + ... + \ lambda_k \, U_k + \ mu_1 \, U_1 ^ 2 + \ mu_2 \, U_2 ^ 2 + ... + \ mu_k \, U_k ^ 2$

where $U_1, ..., U_k$ are the first integrals of the equations of perturbed motion; $\ lambda_1, ..., \ lambda_k$ and $\ mu_1, ..., \ mu_k$ - indefinite constants, the selection of which can make function (16) definitely positive, satisfying Lyapunov's stability theorem.

The unperturbed rotation of the nut occurs around an axis $x$ with a constant angular velocity $\ omega$ . We will outrage this movement, giving the angular velocity a small increment $\ Delta \ vec \ omega$ , and rewrite expressions (15)

$\ begin {align *} & (\ omega + \ Delta \ omega_x) ^ 2 + i_y ^ {\, 2} \, \ Delta \ omega_y ^ 2 + i_z ^ {\, 2} \, \ Delta \ omega_z ^ 2 = \ rm const \\ & (\ omega + \ Delta \ omega_x) ^ 2 + i_y \, \ Delta \ omega_y ^ 2 + i_z \, \ Delta \ omega_z ^ 2 = \ rm const \ end {align *}$

$\begin{align*} &\omega^2 + 2\omega \, \Delta\omega_x + \Delta\omega_x^2 + i_y^{\,2} \, \Delta\omega_y^2 + i_z^{\,2} \, \Delta\omega_z^2 = \rm const \\ &\omega^2 + 2\omega \, \Delta\omega_x + \Delta\omega_x^2 + i_y \, \Delta\omega_y^2 + i_z \, \Delta\omega_z^2 = \rm const \end{align*}$

With the steady rotation of the nut with a constant angular velocity, the constant $\omega^2$ can be subtracted from both sides of the resulting equations, having received in their left side the functions

$\begin{align*} &U_1 = \Delta\omega_x^2 + i_y^{\,2} \, \Delta\omega_y^2 + i_z^{\,2} \, \Delta\omega_z^2 + 2\omega \, \Delta\omega_x \\ &U_2 = \Delta\omega_x^2 + i_y \, \Delta\omega_y^2 + i_z \, \Delta\omega_z^2 + 2\omega \, \Delta\omega_x \end{align*}$

Lyapunov function will have the form

$V = U_1^2 + U_2^2$

Based on equations (15), it is clear that $\frac{dV}{dt} = 0$ , therefore, there will be no talk of asymptotic stability. But, based on Lyapunov's theorem, it is necessary to verify that function (18) is definitely positive. From the expressions (18) and (17) it is clear that its values are positive for any $\Delta\omega_x$ , $\Delta\omega_y$ and $\Delta\omega_z$ . Now we show that (18) vanishes only at the rest point of the system $\Delta\omega_x = \Delta\omega_y = \Delta\omega_z = 0$ . Expression (18) is equal to zero exclusively in the case

$U_1 = 0, \quad U_2 = 0$

Subtract the second from the first equation of system (19)

$U_1 - U_2 = i_y \left( 1 - i_y \right) \, \Delta\omega_y^2 + i_z \left( 1 - i_z \right) \, \Delta\omega_z^2 = 0$

If $i_y, \, i_z < 1$ (the moment of inertia around which the nut rotates the largest ), or $i_y, \, i_z > 1$ (the moment of inertia around which the nut rotates the smallest ), then equality (20) will be valid only when $\Delta\omega_y = \Delta\omega_z = 0$ . We take this fact into account and add equations (19)

$U_1 + U_2 = 2 \, \Delta\omega_x^2 + 4\,\omega \, \Delta\omega_x = 2\,\Delta\omega_x \left(\Delta\omega_x + 2\,\omega \right) = 0$

Equation (21) is valid for $\Delta\omega_x = 0$ and for $\Delta\omega_x = - 2 \, \omega$ . But, as we assume $|\Delta\omega_x| \ll 2\,\omega$ , function (18) will be equal to zero exclusively at the point of rest of the system $\Delta\omega_x = \Delta\omega_y = \Delta\omega_z = 0$ .

Thus, the rotation of the nut around the axis with the smallest and greatest moment of inertia will be stable according to Lyapunov.

However, I hasten to note that with $i_y > 1, \quad i_z < 1$ , or $i_y < 1, \quad i_z > 1$ i.e., when the moment of inertia about the axis around which the rotation takes place is intermediate between the maximum and minimum values, function (18) can no longer be called definite positively, because the terms in (20) will have different signs. But it is absolutely impossible to say that the movement will be unstable. A peculiarity of Lyapunov's stability theorems is that they declare a stability condition, but do not declare the opposite. Motion instability will have to be proven separately.

5. Instability of rotation of the Janibekov nut

We formulate the definition

A domain is a $v > 0$ region of a neighborhood $|x_i| < h, \quad i = \overline{1,n}$ where a $v(x_1, x_2,...,x_n)$ condition is fulfilled for some function $v(x_1, x_2,...,x_n) > 0$ , and at the boundary of the region $v = 0$ and the rest point of the system belongs to this boundary.

and the theorem

Chetaev instability theorem

If the differential equations of the perturbed motion (5) are such that there exists a function $v(x_1, x_2,...,x_n)$ such that in an arbitrarily small neighborhood
$|x_i| < h, \quad i = \overline{1,n}$

there is a region $v > 0$ , and at all points of this region the derivative $\dot v$ takes positive values by virtue of equations (5), then the unperturbed motion is unstable.

The function $v(x_1, x_2,...,x_n)$ referred to in the theorem is called the Chetaev function . Now consider again our nut, the equations of rotation of which look like this (taking into account the work in Cartesian coordinates related to the body and the dimensionless moments of inertia introduced by us)

$\begin{align*} &\dot\omega_x = \left(i_y - i_z) \, \omega_y \, \omega_z \\ &\dot\omega_y = \frac{i_z - 1}{i_y} \, \omega_x \, \omega_z \\ &\dot\omega_z = \frac{1 - i_y}{i_z} \, \omega_x \, \omega_y \end{align*}$

Considering that initially rotation occurs at a constant angular velocity $\omega$ around the axis $x$ , we construct the equations of perturbed motion. We assume that $\omega > 0$ - this can always be achieved by choosing the axes of our own coordinate system.

$\begin{align*} &\Delta\dot\omega_x = \left(i_y - i_z) \, \Delta\omega_y \, \Delta\omega_z \\ &\Delta\dot\omega_y = \frac{i_z - 1}{i_y} \, (\omega + \Delta\omega_x) \, \Delta\omega_z \\ &\Delta\dot\omega_z = \frac{1 - i_y}{i_z} \, (\omega + \Delta\omega_x) \, \Delta\omega_y \end{align*}$

We construct the Chetaev function

$v = \Delta\omega_y \, \Delta\omega_z$

The rest point of the system lies on the boundary $v > 0$ , and function (23) is positive for $\Delta\omega_y, \, \Delta\omega_z > 0$ . The time derivative of (23) by virtue of (22) has the form

$\dot v = \Delta\dot\omega_y \, \Delta\omega_z + \Delta\omega_y \, \Delta\dot\omega_z = (\omega + \Delta\omega_x) \, \left(\frac{i_z - 1}{i_y} \, \Delta\omega_z^2 + \frac{1 - i_y}{i_z} \, \Delta\omega_y^2 \right)$

Due to the fact that $\omega > 0, \quad \omega \gg |\Delta\omega_x|$ , as well as under the condition that the nut rotates around the average moment of inertia, so that $i_z > 1, \quad i_y < 1$ , that is $I_z > I_x > I_y$ , the derivative (24) is positive in the region $v > 0$ , which means that the motion will be unstable.

If, as in the case we initially considered $I_y > I_x > I_z$ , or $i_z < 1, \quad i_y > 1$ ,, then as the Chetaev function we choose

$v = -\Delta\omega_y \, \Delta\omega_z$

Then the region $v > 0$ corresponds to the condition $\Delta\omega_y, \, \Delta\omega_z < 0$ , the stationary point of the system also lies on its boundary, and the derivative (25) equal to

$\dot v = -\Delta\dot\omega_y \, \Delta\omega_z - \Delta\omega_y \, \Delta\dot\omega_z = (\omega + \Delta\omega_x) \, \left(\frac{1 - i_z}{i_y} \, \Delta\omega_z^2 + \frac{i_y - 1}{i_z} \, \Delta\omega_y^2 \right)$

will also be positive. The movement will be unstable.

Conclusion

This article is an addition to the article on the stability of movement of the Janibekov nut . The main material is taken from the above literature, as well as the Math Help Planet website . The author's contribution to this article is a phased detailed discussion of the second Lyapunov method using an example of a specific problem. In addition, a little more extensively than in the book of Markeyev , the question of the instability of motion in relation to various versions of the relationship between the moments of inertia of the nut is considered.

Thus, I believe that I corrected a defect related to the incompleteness of the question about the causes of the Janibekov effect. And at the same time, he himself studied in more detail the second Lyapunov method.

I thank the readers for their attention!

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