Method of Lyapunov functions in the problem of the Janibekov effect

Introduction

This article is not related to the cycle “Magic of tensor algebra” , but is brought to life by publications from it. Carelessly clicking on the links in a search engine, I came across a discussion of one of my articles on the Janibekov effect, and drew attention to a fair observation that the study of the sustainability of the Janibekov nut does not give an unambiguous answer to the question of which parameters the movement will be sustainable. 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 sustainable without additional research.

Interpretation of Mac-Kullag - probably the simplest explanation of the Janibekov effect

Such a study can be performed using the Lyapunov function method (the second or the direct Lyapunov method). And in order to finally close the issue with the nut of Janibekov, 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 a certain mechanical system
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$\ frac {d \ mathbf y} {dt} = \ mathbf F (t, \, \ mathbf y)$

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

System Solution (1) $\ mathbf y (t) = \ mathbf y_0 (t)$ gives the so-called undisturbed movement . In fact, this is a normal, steady-state mode of motion of the system under the action of forces applied to it. Set some perturbation defined by the vector $\ mathbf x (t)$ deviations from 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 undisturbed motion of the system.

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

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

2. Simple function V ( x ) - candidate for 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

$| x_i | & lt; h, \ quad i = \ overline {1, n}$

Where $h$ - some, rather small, positive number.

The function (6) is called sign-defined if in the domain (7) it takes values of only one sign (only positive or only negative), and is equal to zero only at the origin of coordinates (with $x_1 = x_2 = ... = x_n = 0$ )

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

Calculate the total derivative of function (6) with respect to time. Because $x_i = x_i (t), \ quad i = \ overline {1, n}$ , by the definition of the total derivative we get

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

that, taking into account equation (5), is equivalent to

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

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

3. Lyapunov stability theorems

Two paragraphs, which are higher, are written in a dry mathematical language of definitions, and it probably cannot be otherwise. Add some more formal mathematics by formulating

Lyapunov theorem on stability

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

The point of rest of the system (5) here is understood as its trivial solution, corresponding to the unperturbed motion of the mechanical system under consideration. Roughly speaking, according to the formulated theorem, the function should be chosen $V (x_1, \, x_2, \, ..., \, x_n)$ satisfying the properties specified in the condition of the theorem. If it satisfies these properties, then it is called a Lyapunov function, and if such a function (at least one!) Exists, then the steady state of motion of the considered mechanical system will be stable.

However, in this theorem we are not talking about asymptotic stability, that is, the nature of the motion of the system, in which its disturbed motion will tend to the original steady state. Stable here is understood as such a movement in which the system will fluctuate in the neighborhood of the original steady state, but will never return to it. The condition of asymptotic stability will be more stringent.

Lyapunov theorem on asymptotic stability

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

An asymptotically stable system, after a perturbation, will tend to return to the steady state 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 first approximation study.

Another question is how to find a Lyapunov function that satisfies equation (5) and the requirements of theorems. Mathematics does not yet know an unambiguous 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, you can search for Lyapunov functions in the form of quadratic forms, for example, for a third-order system, this function can be

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

this function is definitely positive, moreover, in an arbitrarily large neighborhood of the point of rest of the system. Or such a function

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

will be permanent, positive, for $V = (x_1 + x_2) ^ 2 + x_3 ^ 2$ may be zero as in the resting point of the system $x_1 = x_2 = x_3 = 0$ , and at the 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 a constant (sign-consistent) and also a time derivative equal to zero — it is a constant. And this function follows from the system of equations of motion, for it is one of its integrals.

In the case of the nut of Janibekov, I took an idea from the book of A.P. Markeev “Theoretical Mechanics” as a very elegant solution. This decision is somewhat revised and expanded by me to be in the context of previously written articles.

4. Integrals of the movement of the nut Dzhanibekov

We obtain the first two integrals of motion, based on the system of equations given in the tensor cycle . We will operate with tensor relations in order not to lose grip. So, the equation of rotation of the nut around the center of mass is

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

let's move in this equation to the vector MKD

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

Multiply the equation (10) by a scalar on the doubled MKD 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) convolution $\ varepsilon ^ {\, ijk} \, L _ {\, i} \, L _ {\, k} = 0$ , and in the first - the derivative of the square of the MKD module. Transform equation (11) and integrate it

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

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

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

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

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

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

Recall, after all, something similar we have already seen earlier . 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 by 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

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

Considering that multiplying a constant by a two does not change its “constancy”, we can finally write the first integral in component form (given 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 in expressions (12) and (14) to dimensionless moments of inertia $i_y = \ frac {I_y} {I_x}, \ quad i_z = \ frac {I_z} {I_x}$

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

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 a 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 as a bundle 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 $U_1, ..., U_k$ - 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 the Lyapunov stability theorem.

Unperturbed nut rotation occurs around the axis $x$ with constant angular velocity $\ omega$ . We stir up this movement, giving the angular velocity a small increment $\ Delta \ vec \ omega$ , and rewrite the expression (15)

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

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

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

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

Lyapunov function will have the form

$V = U_1 ^ 2 + U_2 ^ 2$

Based on equations (15) it is clear that $\ frac {dV} {dt} = 0$ , it means there will be no asymptotic speech stability. But, proceeding from Lyapunov's theorem, it is necessary to make sure that function (18) is definitely positive. From 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 will show that (18) vanishes only at the point of rest of the system $\ Delta \ omega_x = \ Delta \ omega_y = \ Delta \ omega_z = 0$ . Expression (18) is zero only in the case

$U_1 = 0, \ quad U_2 = 0$

From the first equation of system (19) we subtract the second

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

If a $i_y, \, i_z & lt; one$ (the moment of inertia around which the nut rotates the greatest ), or $i_y, \, i_z & gt; one$ (the moment of inertia, around which the nut rotates the smallest ), then equality (20) will be valid only in the case 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) holds for $\ Delta \ omega_x = 0$ and at $\ Delta \ omega_x = - 2 \, \ omega$ . But, since we assume $| \ Delta \ omega_x | \ ll 2 \, \ omega$ , function (18) will be zero only 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 Lyapunov stable.

However, I hasten to note that with $i_y & gt; 1, \ quad i_z & lt; one$ , or $i_y & lt; 1, \ quad i_z & gt; one$ that is, when the moment of inertia about the axis around which rotation takes place is intermediate between the maximum and minimum value, function (18) cannot already be called definite positive, because the terms in (20) will have different signs. But it cannot be said at all that the movement will be unstable. A specific feature of Lyapunov stability theorems is that they declare a stability condition, but do not declare the opposite. The instability of the movement will have to be proved separately.

5. The instability of the rotation of the nut Dzhanibekov

We formulate the definition

Area $v & gt; 0$ we will call any region of the neighborhood $| x_i | & lt; h, \ quad i = \ overline {1, n}$ where for some function $v (x_1, x_2, ..., x_n)$ condition is met $v (x_1, x_2, ..., x_n) & gt; 0$ , and on the border of the region $v = 0$ and the point of rest of the system belongs to this boundary.

and theorem

Chetaev's instability theorem

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

there is an area $v & gt; 0$ and in all points of this domain the derivative $\ dot v$ by virtue of equations (5) takes positive values, the unperturbed motion is unstable.

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

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

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

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

Construct the Chetaev function

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

The rest point of the system lies on the border $v & gt; 0$ and function (23) is positive with $\ Delta \ omega_y, \, \ Delta \ omega_z & gt; 0$ . 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 $\ omega & gt; 0, \ quad \ omega \ gg | \ Delta \ omega_x |$ , and also provided that the nut is rotated around the average moment of inertia, so that $i_z & gt; 1, \ quad i_y & lt; one$ , i.e $I_z & gt; I_x & gt; I_y$ , derivative (24) is positive in the region $v & gt; 0$ so the movement will be unstable.

If, as in the case we initially considered, $I_y & gt; I_x & gt; I_z$ , or $i_z & lt; 1, \ quad i_y & gt; one$ , then as a function of Chetaev we choose

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

Then the area $v & gt; 0$ matches the condition $\ Delta \ omega_y, \, \ Delta \ omega_z & lt; 0$ , the point of rest 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. Movement will be unstable.

Conclusion

This article is an addition to the article on the stability of the movement of the nut Dzhanibekov . The main material is taken from the above references, as well as the site Math Help Planet . The author's contribution to this article is a phased detailed consideration of the second Lyapunov method on the example of a specific task. In addition, a little more detailed than in the book by Markeev , the question of the instability of motion with respect to different variants of the relationship between the moments of inertia of the nut was considered.

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

Thank you for your attention!

Source: https://habr.com/ru/post/264419/

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