We shall give here a geometrical interpretation of an autonomous system of equations in the form of the phase space of this system. This interpretation differs essentially from the geometrical interpretation of the system of equations in §1 and should, therefore, more correctly be called a kinematic interpretation, since to every solution of its system of equations corresponds not a curve in a space, but the motion of a point along the curve. The kinematic interpretation (the phase space) is in certain respects more expressive than the geometrical (the system of integral curves).
Autonomous systems. A system of ordinary differential equations is called autonomous if it does not explicitly contain the independent variable t (or, as we shall call it, the time). This means that the law of variation of the unknown functions which are described by the system of equations does not change with time, as is usually the case with physical laws. It is very easy to prove that if
x i = f i(t), i = 1, ... , n,
is a solution of a certain autonomous system of equations, then
x i = j i* (t) = j i (t + c), i = 1, ... , n,
where c is a constant, is also a solution of the same autonomous system of equations. We shall carry out the proof of this fact by an example of a normal autonomous system of equations.
| . x i = f i( x 1, ... , x n ), |
(1) |
be an autonomous normal system of nth-order equations and
| . x = f (x), |
its vector notation. The autonomy of system (1) consists of the fact that the functions f i(x1, ... , xn), i = 1, ... , n, are functions of the variables x1, ... , xn and do not depend on the time t. We shall assume that the functions f i(x1, ... , xn), are defined in a certain domain D of the n-dimensional space where x1, ... , xn are the coordinates of a point. We shall assume further that f i(x1, ... , xn) and their first-order partial derivatives are continuous in the domain D. Thus, if
| x i = j i(t), i = 1, ... , n, | (2) |
is a solution of (1), then
| x i = j i* (t) = j i (t + c), i = 1, ... , n, | (3) |
is also a solution of (1). It is evident that if the solution (2) has as a maximal interval of existence the interval m1 < t < m2 then the solution (3) has the maximal interval
m1 – c < t < m2 – c
From the differentiation formula for a composite function, we have the relation
| . . | ||
| j i* (t) = j i (t + c), i = 1, ... , n, | (4) |
Indeed,
|
We shall now prove that (3) is a solution of the system (1). Since (2) is a solution, we have the identities
| . j i(t) = f i(j 1(t), ... , j n (t)), i = 1, ... , n, |
Replacing t in these identities by t + c we obtain
| . j i(t + c) = f i(j 1(t + c), ... , j n (t + c)), i = 1, ... , n, |
Combining this with (4) and (3), we have
| . j i*(t) = |
. j i(t + c) = f i(j 1(t + c), ... , j n (t + c)) = f i(j 1*(t), ... , j n *(t)), i = 1, ... , n, |
We shall now turn to the kinematic interpretation of the solutions of system (1). Formally we shall speak about an interpretation in n-dimensional space, but for the sake of clarity it is reasonable to imagine the case of a plane (n = 2).
(B) To every solution
| x i = j i(t), i = 1, ... , n, | (5) |
of the autonomous system (1) we make correspond the motion of a point in n-dimensional space denned by equations (5), where x1, ... , xn are the coordinates of the point in space and t is the time. In the course of its motion the point describes a curve known as the trajectory of the motion. If we associate with the solution (5) not the process of motion, but the trajectory of the motion of the point, then we shall obtain a less complete picture of the solution, since it is also desirable to indicate the direction of the motion on the trajectory. Thus, if there is another solution
| x i = y i(t), i = 1, ... , n, | (6) |
in addition to (5), then the trajectories corresponding to these solutions either do not intersect in the space or else they coincide. That is, if the trajectories have even one common point, i.e.,
| j i(t1) = y i(t2), i = 1, ... , n, | (7) |
then
| y i(t) = j i(t + c), where c = t1– t2. | (8) |
These last equalities show that the trajectories described by the first and second solutions coincide, but the first solution describes the same trajectory as the second with the time “delay” c. If the point corresponding to the first solution has reached a certain position on the trajectory at instant t + c, then the point corresponding to the second solution has already been in this position at the instant t.
In order to derive (8) from (7), we shall examine the solution
| j i* (t) = j i (t + c), | (9) |
along with (5) [see (A)]. Equations (7), for c = t1– t2, yields the equality
j i* (t2) = j i(t2 + c) = j i(t1) = y i(t2), i = 1, ... , n,
Thus, the solutions (6) and (9) of the system (1) have common initial conditions (namely, their values at the instant t2),and therefore by the uniqueness theorem they must coincide, so that we have
y i(t) = j i* (t) = j i(t + c), i = 1, ... , n,
States of equilibrium and closed trajectories. We pose the question of whether a trajectory representing a solution of the system can intersect itself.
(C) Let
| xi = ji(t), i = 1, ... , n, | (10) |
be a certain solution of the system (1) defined on a maximal interval m1 < t < m2. We shall assume that the equalities
| ji(t1) = ji(t2), i = 1, ... , n, t1 ¹ t2 | (11) |
are valid, where t1 and t2, of course, belong to the interval m1 < t < m2. It then turns out that m1 = –¥, m2 = +¥, [i.e., the maximum interval of existence for the solution (10) is the entire line] and that the following two mutually exclusive cases are possible.
1. For all values of t the equality
j i(t) = a i, i = 1, ... , n,
is valid, where ( a1, a2, ... , an ) is a point of the domain D which does not depend on t. Thus in this case the point (j1(t), ... , jn(t)) actually does not move as t varies but remains fixed. In this case the solution (10) itself with the point ( a1, a2, ... , an ) is called a state of equilibrium of the system (1).
2. There exists a positive number T such that for arbitrary t, the equalities
ji(t + T ) = ji(t), i = 1, ... , n,
are valid, but for |t1– t2| < T and for at least one i = 1, ... , n, the inequality
ji(t1) ¹ ji(t2)
is valid. In this case the solution (10) is called periodic with period T, and the trajectory described by (10) is called a closed trajectory or a cycle.
First of all, we shall show that the maximal interval of existence of solution (10) is the entire straight line. As was noted in proposition (B), the identities
| ji(t + c) = ji(t), i = 1, ... , n, c = t1– t2, | (12) |
follow from equality (11). Since by this equality the interval m1– c < t < m2– c coincides with the interval m1 < t < m2, we have m1 = –¥, m2 = +¥.
Every number c for which (12) is satisfied will be called a period of the solution (10); the set of all periods of the solution (1) is designated by F, which is a certain set of numbers. We shall establish some of its properties. Substituting t – c for t in (12), we obtain ji(t) = ji(t – c) Thus, if c is a period, then –c is also a period. Let us assume that c1 and c2 are periods, i.e., that
ji(t + c1) = ji(t), ji(t + c2) = ji(t), i = 1, ... , n.
Then
ji((t + c2) + c1) = ji(t + c2) = ji(t), i = 1, ... , n.
Thus if c1 and c2 are periods, then c1– c2 is also a period. Let us assume that c1, c2, ..., cm, ... is a sequence of periods which converges to a certain number c0; then we have
ji(t + cm) = ji(t), i = 1, ... , n, m = 1, 2, ... .
Since the functions ji(t) are continuous, then for m ® ¥ we have
ji(t + c0) = ji(t),
i.e., we see that c0 is also a period, because the set F is closed.
Since the number c in (12) is distinct from zero (t1¹ t2), the set F contains numbers distinct from zero. From the properties of the set F which have been established it is easily seen that there exist only two possibilities: (1) the set F coincides with the set of all real numbers; (2) in the set F there is a minimal positive number T, such that F consists of all integer multiples of the number T. Let us prove that there are actually only these two possibilities. Since the set F contains the number –c whenever it contains the number c, and since in F there are numbers different from zero, there are positive numbers in F.
Let us assume that there is no least positive number in F, i.e., that for every positive number e there is a positive period c < e. From the above properties of the set F it follows, since c is a period, that all numbers mc, where m is an integer, are also periods. Since c < e, then for an arbitrary real number c0 it is possible to find an integer m such that |c0 – mc| < e. Thus an arbitrary number c0 is a limit point of the set F, and therefore, ill view of the fact that set F is closed, this set coincides with the set of all real numbers.
Let us now assume that F is not the set of all real numbers. By what has been proved, there then exists in F a least positive number T. Let c be an arbitrary period. Then it is possible to select an integer m such that | c – mT | < T. Let us assume that c ¹ mT; then | c – mT | is a period distinct from zero; but this is impossible since | c – mT | < T. which contradicts the minimal character of the number T. It is thus proved that every number c from F can be written in the form c = mT, where m is an integer.
Now it is easy to verify that, if F is the set of all real numbers, then case (1) occurs, and if F is not the set of all real numbers, then case (2) occurs. Thus proposition (C) is proved.
Proposition (C) can be formulated briefly by saying that there exist three kinds of trajectories: (1) those of the state of equilibrium; (2) periodic trajectories (cycles); and (3) nonintersecting trajectories. It is natural to take the last case as the "most general."
From Theorem 2 it follows that a trajectory representing a solution of the system passes through every point of the domain of definition of the system (1). Thus, the entire domain D is filled with trajectories, and in accordance with (B) these trajectories do not intersect each other in pairs. Those trajectories which do not intersect are of particular interest; they represent either states ot equilibrium or cycles, and are quite important.
This is the kinematic interpretation of solutions of an autonomous system of equations. The system of equations itself also admits a geometric interpretation.
Phase spaces. (D) Since the autonomous system of equations (1) is defined in the domain D, each point (x10, ..., xn0) of the domain D corresponds to a sequence of n numbers, namely the sequence
f 1(x10, ..., xn0), ... , f n(x10, ..., xn0)
These numbers can be thought of as components of a vector f(x10, ..., xn0) in an n-dimensional space emanating from the point (x10, ..., xn0). Thus the autonomous system gives rise to a geometric picture, a vector field defined in the domain D. The vector f(x10, ..., xn0) is defined at every point (x10, ..., xn0) of D, starting from this point. The connection between the geometrical interpretation of the solutions and the geometrical interpretation of the system of equations itself is given by the following. Let (x10, ..., xn0) be an arbitrary point of D. In the geometrical interpretation of the system of equations the vector f(x10, ..., xn0) corresponds to the point from which it starts. Further, by Theorem 2 there exists a solution xi = ji(t) of (1) which satisfies the initial conditions
ji(t0) = xi0 i = 1, ... , n.
According to the kinematic interpretation, the solution xi = ji(t) corresponds in the space to the motion of a point which describes a trajectory which, at the instant t = t0, passes through the point (x10, ..., xn0) in the space. Thus the vector velocity of the point which describes the solution xi = ji(t) at the instant of its passage through the point (x10, ..., xn0) coincides with the vector f(x10, ..., xn0). It is just this coincidence which is expressed by the system of equations (1) for
xi = xi0, i = 1, ... , n, t = t0.
The n-dimensional space, in which solutions of autonomous system (1) are interpreted in the form of trajectories and the autonomous system (1) itself in the form of a vector field, is called the phase space of the system (1). The trajectories in this space are called the phase trajectories, and the vectors f(x10, ..., xn0) are called the phase velocities. The connection between the two interpretations consists in the fact that the velocity of the motion of a point along a trajectory at each instant coincides with the phase velocity given at that point of the space where the moving point is located at that instant.
Let us now examine states of equilibrium from the point of view of phase velocities.
(E) In order that the point ( a1, ..., a n ) of the domain D be a state of equilibrium of the system (1), i.e., that xi = ji(t) be a solution of the system for which
| ji(t) º ai, i = 1, ... , n, | (13) |
it is necessary and sufficient that the phase velocity f( a1, ..., a n ) at ( a1, ..., a n ) be equal to zero. Thus to find all states of equilibrium of (1) it is necessary to solve the system of equations
f i( a1, ..., a n ) = 0, i = 1, ... , n.
This system is not a system of differential equations, but rather a system of finite (or algebraic) equations since it does not include derivatives.
To prove proposition (E), we shall assume that ( a1, ..., a n ) is a state of equilibrium, i.e., that there exists a solution xi = ji(t) for which the relations (13) are satisfied, and we shall substitute this solution into (1). Since the derivative of a constant is zero, the substitution yields
| f i( a1, ..., a n ) = |
d dt |
ji(t) = |
d dt |
a i = 0. |
Thus the phase velocity vector f( a1, ..., a n ) actually vanishes at the point ( a1, ..., a n ). Let us assume that, conversely, the phase velocity vector f( a1, ..., a n ) vanishes at the point ( a1, ..., a n ), i.e., that
f i( a1, ..., a n ) = 0, i = 1, ... , n.
and we shall show that in this case the equalities (13) determine a solution of (1). Substitution gives
| . ji(t) = |
f i( a1, ..., a n ), i = 1, ... , n; |
these equalities are satisfied, since on the left we have tlie derivative of a constant and on the right, zero.
Examples
1. (Compare Example 3 in §2). We shall study the autonomous first-order differential equation
|
(14) |
where f(x) and its derivative are continuous over the entire line P where x is allowed to vary. We shall assume in addition that the zeros of f(x) or, what is the same thing, the states of equilibrium of equation (14) do not have limit points. Under this hypothesis the states of equilibrium divide the straight line P into a system S of intervals. Each interval (a, b) of S has the property that the function f(x) does not vanish on it and that each of its endpoints, a or b, is either a zero of the f unction f(x), or is equal to ±¥. Thus S consists of a finite or countable number of finite intervals and not more than two semi-infinite intervals, or consists only of the infinite interval (–¥, ¥). Let (a, b) be a certain interval of S, x0 a point of this interval, x = j(t) a solution of equation (14) with initial values 0, x0, and m1 < t < m2 the maximal interval of existence of the solution j(t). To be definite we shall assume that f(x0) > 0, so that
| a < j(t) < b for m1 < t < m2, | (15) |
|
(16) |
Further, if either number a or b is finite, then the number m1 or mg, respectively, is infinite. Thus (Fig. 22) every interval (a, b) represents one unique phase trajectory of equation (14).
We shall prove relations (15), (16). From the hypothesis that f(x0) > 0 it follows that the function f(x) is positive on the interval (a, b), and therefore every point of this interval describing a phase trajectory moves from left to right. Thus with increasing t the point j(t) can leave the interval (a, b) only by passing over the right-hand endpoint b. Let us assume that this takes place for some t = t1; then at t = t1; we have j(t1) = b, which creates the impossible situation that the two different trajectories x = j(t) and x = b intersect. In exactly the same way we can prove that the point j{t) cannot leave interval (a, b) for decreasing t. Thus (15) is proved.
Let us now assume that lim t®m2 j(t) = c < b, and let y(t) be a solution of (14) with the initial values 0, c. Since f(c) > 0, then for some negative value of t2 we have y(t2) < c, but this means that two different trajectories j(t) and y(t) intersect, which is impossible.
Thus it is proved that lim t®m2 j(t) = b.
The relation
|
is proved in exactly the same way. (Here some mistypings were fixed: t—m2 was changed to t®m2 and lim = j... changed to lim j... . The subscript 2 due to HTML restrictions is typed here as normal text, so you see m2 instead of m2.)
Let us assume, finally, that b < ¥, and show that under this assumption m2 = ¥. Let us assume the contrary, that is, m2 < ¥. Let us then define a function c(t) by setting c(t) = j(t) for m1 < t < m2, and c(t) = b for t ≥ m2. It is evident that the function c(t) is continuous and satisfies equation (14), but this is impossible since two different trajectories c(t) and x = b would then intersect. This contradiction shows that m2 = +¥. In exactly the same way it can be proved that for a > –¥ we have m1 = –¥.
Let b be an arbitrary state of equilibrium of equation (14) and let (a, b) and (b, c) be the two intervals of S adjoining it (on the left and right, respectively). Each of the intervals (a, b) and (b, c) represents one trajectory. If both points describing the trajectories (a, b) and (b, c) approach (with increasing t) the state of equilibrium b, then the state of equilibrium b is called stable [Fig. 23(a)]. If both points describing the trajectories (a, b) and (b, c) recede from the point b, then the state of equilibrium b is called unstable [Fig. 23(b)]. If along one of the trajectories the point approaches and along the other it recedes, then the state of equilibrium b is called semistable [Fig. 23(c)]. In order that a state of equilibrium b be stable, it is necessary and sufficient that the function f(x) be positive on the interval (a, b) and negative on the interval (b, c). For state of equilibrium b to be unstable, it is necessary and sufficient that the function f(x) be negative on the interval (a, b) and positive on the interval (b, c). For a state of equilibrium b to be semistable, it is necessary and sufficient that the function f(x) have the same sign on both of the intervals (a, b) and (b, c).
Let us assume that |
. f(b) ¹ 0; |
then the sign of function f(x) in the neighborhood of the point b is the same as |
the sign of the quantity |
. f(b) (x – b). |
Hence it follows that for |
. f(b) < 0 |
the state of equilibrium b of equation (14) |
is stable and for |
. f(b) > 0 |
it is unstable. |
2. We shall study the equation
| (17) |
where f(x) is a periodic function with a continuous first derivative. To be definite we shall assume that its period is equal to 2p. Everything said in Example 1 concerning equation (14) remains valid for equation (17) as well, since equation (17) is a particular case of equation (14). However, in order to take into account the specific character of equation (17) [the periodicity of function f(x)], we shall assume that the phase space of equation (17) is not a straight line but a circle K of radius one on which we choose a reference point 0 and a direction of motion (for example, counter-clockwise). To every number x we make correspond the point x of the circle K by marking counterclockwise from the reference point an arc of length x (Fig. 24). Then to all numbers x + 2kp, where k is an integer, there corresponds a unique point x on the circumference. Since
f(x + 2kp) = f(x),
it is possible to set f(x) = f(x), and the function f is then defined on the circumference K. Equation (17) now defines the motion of point x along the circumference K. If x(t) is a certain solution of equation (17), then the point x(t) corresponding to the number x(t) moves along circumference K. If a is a point on K such that f(a) = 0, then there exists a solution x(t) of (17) such that x(t) = a and a is a state of equilibrium of (17). Let us assume for simplicity that the state of equilibrium of (17) on K has no limit points; then there is only a finite number of points or none at all (Fig. 25). States of equilibrium divide the circumference into a finite system S of intervals. If there are no states of equilibrium at all, then the system S contains only one “interval” (the circumference). If there is only one state of equilibrium a, then the system S also contains only one interval, which consists of all points of K with the exception of the point a. In the first case the interval has no endpoints at all; in the second case both its endpoints coincide. Let I be a certain interval of S and x(t) a certain solution of (17) with initial values 0, x0, where x0 is a point of I. The solution x(t) is always defined for all values of t, and the point x(t) belongs to the interval I. If the interval I has endpoints (one or two), then the point traverses I in a fixed direction, the solution x(t) passing once through each point of I. If the interval I coincides with the entire circumference, then after leaving the position x0, the point will return to it after a certain time T, so that x(0) = x(T). In this case the motion x(T) depends periodically with period T on the number t. The numerical solution x(t) of equation (17) corresponding to the motion x(t) satisfies the condition
x(t + T ) = x(t) ± 2p
From this example it is apparent that it is not always appropriate to consider a euclidean coordinate space as the phase space of a system of equations; sometimes it is necessary to consider a more complex geometrical configuration. In Example 3 below we shall encounter this circumstance in a more complex situation than in this example.
3. We shall investigate the system of equations
|
(18) |
where the function f i(x1, x2) is periodic of period 2p in each of its arguments:
f i(x1 + 2kp, x2 + 2lp)= f i(x1, x2), i = 1, 2.
As always, we shall assume that the functions f i(x1, x2), are continuous and have continuous first-order partial derivatives. In view of the periodicity of the functions f i(x1, x2), it is reasonable to assume that the phase space of (18) is not a plane but a more complex geometrical configuration, namely, the surface of a torus or, as it is called, a torus (Fig. 26). We shall describe this surface.
In a three-dimensional euclidean space with cartesian coordinates x, y, z, we take in the xz-plane the circle K with a radius of one and with its center at the point (2, 0, 0). We take as origin on the circumference the point with coordinates (3, 0, 0). Then to every number x1 we make correspond the point; x1 of K (see Example 2). We now rotate the xz-plane about the z-axis in the (x, y, z) space. The surface P described by the circumference K in this rotation is a torus. Let x1 be some point of K. As a result of rotating the xz-plane through the angle x2 measured in radians, the point x1 goes over into a certain point p of the torus P (Fig. 26). If the rotation is made not through the angle x2 but through x2 + 2kp, then we arrive at the same point p on the torus P. Thus the point p on the torus P is denned uniquely by two cyclic coordinates xl, x2, and to each pair of cyclic coordinates x1, x2 corresponds one well-defined point on the torus. Thus we see that the functions f i(x1, x2) can be considered as defined not on a plane, but on the surface of the torus P:
f i(x1, x2 ) = f i(x1, x2 )
Now let x1(t), x2(t) be a certain solution of system (18). If we make the correspondence between the numbers x1(t) and x2(t) and the cyclic coordinates x1(t) and x2(t), we obtain the point x1(t), x2(t) on the torus P. Thus, every solution x1(t), x2(t) of (18) can be represented by the motion of a point on the torus, the law of motion at each instant being defined by that point x1(t), x2(t) of the torus through which the trajectory passes at that instant. This is explained by the fact that functions f i(x1, x2) are defined on the torus. Thus the entire torus P is found to be covered by trajectories, each two of which either do not intersect, or else coincide. In particular, if a trajectory intersects itself, it is then either closed or it is a state of equilibrium.
The representation of phase trajectories of (18) not on a plane but on the surface of a torus reflects the specific property of the system (18) (periodicity of the functions f i) and is a convenient means of studying it.