Differential Equations, Vol. 40, No. 4, 2004, pp. 491–501. Translated from Differentsial’nye Uravneniya, Vol. 40, No. 4, 2004, pp. 455–464. c 2004 by Borovskikh. Original Russian Text Copyright
ORDINARY DIFFERENTIAL EQUATIONS
Investigation of Relaxation Oscillations with the Use of Constructive Nonstandard Analysis: II A. V. Borovskikh Voronezh State University, Voronezh, Russia Received December 16, 2002
This paper is a continuation of [1], where we introduced the set HR of hyperreals, which contains ordinary reals as well as infinitesimals and infinite quantities. Below we use the representation of an element of HR by a series in powers of infinitesimals for the computation of slow and rapid motions in relaxation oscillations. Technically, there is nothing unusual except that the time, as well as the phase variables, is expanded in powers of a small parameter. This is formally possible, but the expansion of time becomes meaningful only if the time continuum is represented not simply as R1 but as an infinite system of “embedded” numerical lines, so that each point of an axis “contains” the subsequent axis. This model of continuum, as well as R, is an idealization of the real property that all real bodies consist of sufficiently many sufficiently small particles; in the idealization, we proceed to an ideal body consisting of infinitely many infinitely small particles. The difference between these models is in the assumption of divisibility and the existence of an internal structure of “atoms.” If they are indivisible, then we obtain the model R. But if they have their own internal structure and consist of smaller particles, then the idealization results in a model like HR. In what follows, we study relaxation oscillations for a single van der Pol equation and for a pair of van der Pol equations. These results were obtained together with I.V. Sereda and E.A. Kakunina. 1. LOADED VAN DER POL EQUATION The classical van der Pol equation [2–4] ε¨ x + x˙ x2 − 1 + x = a
(a ≥ 0)
(1)
3 −x to the Lienard system, with a constant load can be reduced by the substitution u = εx+(1/3)x ˙ which we represented for convenience via differentials: du = [a − x]dt. (2) ε dx = u + x − (1/3)x3 dt,
We represent each of the quantities occurring in the equation (including the constant and time) as series in powers of ε : x = x0 + x1 ε + x2 ε2 + · · · , t = t0 + t1 ε + t2 ε2 + · · · ,
u = u0 + u1 ε + u2 ε2 + · · · , a = a0 + a1 ε + a2 ε2 + · · · ,
and substitute them together with the differentials into system (2). The matching of coefficients of likes powers of ε gives two systems of equations 0 = u0 + x0 − (1/3)x30 dt0 , dx0 = u0 + x0 − (1/3)x30 dt1 + u1 + x1 − x20 x1 dt0 , (3) dx1 = u0 + x0 − (1/3)x30 dt2 + u1 + x1 − x20 x1 dt1 + u2 + x2 − x0 x21 − x20 x2 dt0 , ............................................................... c 2004 MAIK “Nauka/Interperiodica” 0012-2661/04/4004-0491
492
BOROVSKIKH
du0 = [a0 − x0 ] dt0 , du1 = [a0 − x0 ] dt1 + [a1 − x1 ] dt0 ,
(4)
du2 = [a0 − x0 ] dt2 + [a1 − x1 ] dt1 + [a2 − x2 ] dt0 , ................................................
These two systems readily permit one to separate the slow and rapid motions and construct a complete picture of motion on the plane (x0 , u0 ). 2. SLOW MOTIONS If dt0 = 0 (slow motions), then from the first equation in system (3), we obtain u0 = (1/3)x30 −x0 ; i.e., the slow trajectory is a cubic parabola. The substitution of this expression into the first equation in (4) results in the relation dx0 =
a0 − x0 dt0 , x20 − 1
(5)
and the comparison of the resulting relation with the second equation in (3) (where the first term on the right-hand sides proves to be zero) shows that u1 depends on x1 (and obviously on x0 ) : u1 + x1 − x20 x1 =
a0 − x0 . x20 − 1
(6)
By substituting the function (6) into the second equation in (4) and by comparing the result with the third equation in (3), for u2 , we obtain the formula u2 + x2 −
x0 x21
−
x20 x2
a1 − x1 a0 − x0 a0 − x0 − 2x0 x1 + = 2 2 2 x0 − 1 x0 − 1 (x20 − 1)
(7)
and so on. By proceeding in this way, we obtain all expressions for ui via xj for j ≤ i and hence a closed-form expression for u via x. Relation (5) permits one to determine the direction of motion on the curve. By assuming that one always has dt0 ≥ 0, we find that x0 is increasing (motion to the right) for x0 ∈ (−∞, −1) ∪ (min {a0 , 1} , max {a0 , 1}) and decreasing (motion to the left) for x0 ∈ (−1, min {a0 , 1}) ∪ (max {a0 , 1} , +∞) .
3. RAPID MOTIONS If dt0 = 0 (rapid motions), then from the first equation of system (4), we obtain du0 = 0; i.e., u0 = const, and rapid trajectories prove to be horizontal lines. In this case, the second equations in systems (3) and (4) specify a relationship between the variables u1 , t1 , and x0 . Moreover, from the second equation in system (3) (in which the second term on the right-hand side proves to be zero) for the case in which dt1 > 0, we find that the motion from the left to the right (dx0 > 0) takes place above the graph of the function u0 = (1/3)x30 − x0 , and the motion from the right to the left (dx0 < 0) takes place below the graph. Once the rapid and slow trajectories are known, one can readily construct the phase portrait of the system and use it to show that breakdown points (points of passage from slow to rapid motion) can occur only within the decreasing middle part of a slow trajectory. Further, one readily sees that the breakdown point is always (−1, 2/3) [and the point (1, −2/3) if a0 < 1]; the motion from these points along slow trajectories is impossible. DIFFERENTIAL EQUATIONS
Vol. 40
No. 4
2004
INVESTIGATION OF RELAXATION OSCILLATIONS . . . II
493
Here one can readily trace that if a0 > 1, then, after finitely many (≤ 3) changes of the motion type (from rapid motion to slow motion and vice versa), all motions enter the point (a0 , a30 /3 − a0 ) (from which no trajectory issues), and if a0 < 1, then all motions exit to a cycle consisting of parts of two exterior branches of a parabola and horizontal lines u0 = ±2/3. Therefore, the relaxation cycle proves to be a trajectory of system (2) for infinitesimal ε. 4. “DUCKS” The case a0 = 1 is the special case related to the well-known “duck hunt.” Unlike the case a0 = 1, in which, for an arbitrary given point on the plane, the trajectory issuing from it is uniquely determined, here we have a bifurcation: from the point (1, −2/3), one can move along a slow trajectory as well as a rapid trajectory; moreover, the possibility of switching to rapid trajectories is preserved on the entire interior part of the cubic parabola. Since we have analyzed only the initial terms in the expansions of x and u, to clarify the presence of a bifurcation, we should analyze higher terms as well. For slow motions with a0 = 1, formulas (5)–(7) imply that 1 dt0 , x0 + 1 1 , u1 + x1 − x20 x1 = − x0 + 1 a1 − x1 1 1 2 2 + 2x0 x1 + u2 + x2 − x0 x1 − x0 x2 = 2 2 2 x0 − 1 (x0 + 1) (x0 − 1) (x0 + 1) dx0 = −
(5 ) (6 )
3
x1 a1 + 1/ (x0 + 1) + = 2, x20 − 1 (x0 + 1)
(7 )
and it follows from the last formula that the motion through the point (1, −2/3) along a slow trajectory is impossible even for a1 = −1/8; for x0 = 1, the denominator of the first term on the right-hand side in (7 ) is zero [the same happens in (6) for a0 = 1; thus the passage from a slow trajectory to a rapid trajectory actually occurs “before” the point x0 = 1]. Following Section 2 and using the fourth equation in system (3), from the third equation of system (4), we obtain 1 1 x20 + 5x0 + 12 u3 = x3 x20 − 1 + 2x0 x1 x2 + x31 − x21 3 + x1 5 3 (x0 + 1) 4 (x0 + 1) 1 3x2 + 14x0 + 31 1 + 2 a2 + 0 − x2 , 6 x0 + 1 x0 − 1 8 (x0 + 1) which implies that the motion through the point (1, −2/3) along a slow trajectory can hold only if a2 = −3/32. Further, by continuing to compute the extendibility conditions for higher-order terms, we find that the trajectory of a solution can remain on a slow curve for x0 = 1 only if a takes the unique value 3 1 (8) a = 1 − ε − ε2 + · · · 8 32 In the classical theory, the possibility to continue the motion on an “unstable” part of a slow trajectory is related to the appearance of a family of trajectories referred to as “ducks” [5–11]. One can readily see that such a family appears in the model HR due to the nonuniqueness of a solution of the equation for the parameter value given by (8). The parameter value thus obtained exactly coincides with the result in [9]. 5. A PAIR OF COUPLED VAN DER POL EQUATIONS Let us now consider a system of van der Pol equations describing a two-contour generator [12]: ˙ ε¨ y + C y˙ + Dy = Ex + k 1 − x2 x˙ (9) ε¨ x + Ax˙ + Bx = y + 1 − x2 x, DIFFERENTIAL EQUATIONS
Vol. 40
No. 4
2004
494
BOROVSKIKH
(where, in view of the physical properties of the system, A > 0, B > 0, C > 0, D > 0, BD − E > 0, and the inequality A < 1 is a necessary condition for the existence of oscillations). By setting u = εx˙ + Ax − (x − (1/3)x3 ) and v = εy˙ + Cy − k (x − (1/3)x3 ), we reduce it to a system similar to (2) : du = [−Bx + y]dt, ε dx = u − Ax + x − (1/3)x3 dt, (10) 3 dv = [Ex − Dy]dt. ε dy = v − Cy + k x − (1/3)x dt, By substituting the variables x, y, u, v, and t expanded in powers of ε into system (10): x = x0 + x1 ε + x2 ε2 + · · · , v = v0 + v1 ε + v2 ε2 + · · · ,
y = y0 + y1 ε + y2 ε2 + · · · , t = t0 + t1 ε + t2 ε2 + · · · ,
u = u0 + u1 ε + u2 ε2 + · · · ,
and by matching the coefficients of like powers of ε, we obtain four systems 0 = u0 − Ax0 + x0 − (1/3)x30 dt0 , dx0 = u0 − Ax0 + x0 − (1/3)x30 dt1 + u1 − Ax1 + x1 − x20 x1 dt0 , dx1 = u0 − Ax0 + x0 − (1/3)x30 dt2 + u1 − Ax1 + x1 − x20 x1 dt1 + u2 − Ax2 + x2 − x20 x2 − x0 x21 dt0 , du0 = [−Bx0 + y0 ] dt0 ,
(11)
du1 = [−Bx0 + y0 ] dt1 + [−Bx1 + y1 ] dt0 ,
(12)
du2 = [−Bx0 + y0 ] dt2 + [−Bx1 + y1 ] dt1 + [−Bx2 + y2 ] dt0 , 0 = v0 − Cy0 + k x0 − (1/3)x30 dt0 , dy0 = v0 − Cy0 + k x0 − (1/3)x30 dt1 + v1 − Cy1 + k x1 − x20 x1 dt0 , dy1 = v0 − Cy0 + k x0 − (1/3)x30 dt2 + v1 − Cy1 + k x1 − x20 x1 dt1 + v2 − Cy2 + k x2 − x20 x2 − x0 x21 dt0 , dv0 = [Ex0 − Dy0 ] dt0 ,
(13)
dv1 = [Ex0 − Dy0 ] dt1 + [Ex1 − Dy1 ] dt0 ,
(14)
dv2 = [Ex0 − Dy0 ] dt2 + [Ex1 − Dy1 ] dt1 + [Ex2 − Dy2 ] dt0 . Just as in the previous case, we have two cases: either dt0 > 0, and we have slow motions, or dt0 = 0, and we have rapid trajectories. 6. SLOW MOTIONS If dt0 > 0, then from the first two equations in systems (11) and (12), we obtain explicit formulas for u0 and v0 : v0 = Cy0 − k x0 − (1/3)x30 . (15) u0 = Ax0 − x0 − (1/3)x30 , By substituting them into the first equations in systems (12) and (14), we obtain dx0 = y0 − Bx0 , A − 1 − x20 dt0 dx0 dy0 − k 1 − x20 = Ex0 − Dy0 , C dt0 dt0
(16) (17)
and the division of one of these equations by the other implies the equation Ex0 − Dy0 Cdy0 − k (1 − x20 ) dx0 = 2 [A − (1 − x0 )] dx0 −Bx0 + y0 DIFFERENTIAL EQUATIONS
Vol. 40
No. 4
2004
INVESTIGATION OF RELAXATION OSCILLATIONS . . . II
495
Fig. 1. Monotonicity domains and slow trajectories.
or C
(E − BD)x0 dy0 . = k 1 − x20 + A − 1 − x20 −D + dx0 −Bx0 + y0
(18)
This is a differential equation describing [together with (15)] slow trajectories. Since it contains only the variables x0 and y0 , it is convenient to represent these trajectories in the projection on the plane (x0 , y0 ). lines First, we note the following: it follows √ from (16) that the plane (x0 , y0 ) can be divided by the ∗ ∗ 2 y0 = Bx0 and x0 = ±x (where x = 1 − A is the positive root of the equation [A − (1 − x0 )] = 0) into six domains (which are numbered by Roman numerals I–V I, see Fig. 1) inside each of which the motion is monotone with respect to x0 . To refine the picture, we compute the vector field on the boundaries of these domains. On the line y0 = Bx0 , we have dx0 /dt0 = 0, C dy0 /dt0 = (E − BD)x0 . Therefore, the field is directed vertically downwards (since BD − E > 0) in the right half-plane and vertically upwards in the left half-plane. however, it follows from (18) that On the lines x0 = ±x∗ , both derivatives become infinite; 2 their ratio tends to a finite limit equal to k 1 − (x∗ ) /C = kA/C. Therefore, all trajectories intersect the vertical lines x0 = ±x∗ at the same angle, and the direction of motion near the point of intersection is determined by the motion in the corresponding domain. 7. STRUCTURE OF THE FAMILY OF SLOW MOTIONS All motions issuing from domains I and V I reach one of two rays (referred to as breakdown rays), either the ray x0 = x∗ , y0 ≤ Bx∗ , or the ray x0 = −x∗ , y0 ≥ −Bx∗ , in finite time t0 [which follows from (16)]. In finite time t0 , trajectories lying in domains III and IV reach the line y0 = Bx0 and pass into the domains V I and I, respectively. Indeed, if a trajectory would always remain above the line (in the case of domain III), then, by (18), the derivative dy0 /dx0 would be bounded, i.e., vertical asymptotes would be absent, and, by (16), x0 would be monotone increasing at a positive ray and hence would range over all values from the initial one to +∞. On the other hand, if relation (17) would be reduced to the form C
dx0 d (y0 − Bx0 ) + BC − k 1 − x20 = (E − BD)x0 − D (y0 − Bx0 ) , dt0 dt0
and dx0 /dt0 would be substituted from (16), then we would obtain BC − Ak d (y0 − Bx0 ) + (E − BD)x0 , = − (y0 − Bx0 ) D + k + C dt0 A − (1 − x20 )
(19)
which would imply that the derivative d (y0 − Bx0 ) /dt0 would be necessarily negative (since BD − E > 0 and the first term is negative for large values of x0 ) and would be separated away DIFFERENTIAL EQUATIONS
Vol. 40
No. 4
2004
496
BOROVSKIKH
from zero for sufficiently large positive values of x0 and for y0 − Bx0 ≥ 0. Therefore, the difference y0 − Bx0 attains the zero value in finite time, and hence the trajectory intersects the line y0 = Bx0 . It is quite obvious that on the left-hand side (in which x0 < 0) of domain II and on the righthand side of domain V , trajectories also reach breakdown rays [in finite time by virtue of (16) again], and from the right-hand side of the domain II (respectively, from the left-hand side of the domain V ), they can get only in the two previous domains, either by passing through the ordinate axis or after the intersection of the line y0 = Bx0 . The fact that the exit time is finite follows from (16) in the first case and from (17) in the second one. Note that there is an exceptional case in which a trajectory passes through the origin. The point (0, 0) is a singular point of system (16), (17), it is a saddle, its matrix in the coordinates (x0 , z0 = y0 − Bx0 ) has the form
0 1/(A − 1) , (20) (E − BD)/C −[D + k + (BC − Ak)/(A − 1)]/C and the product of the eigenvalues is equal to −(E − BD)/[C(A − 1)] < 0 (since C > 0, A < 1, and BD − E > 0). It also follows from (20) that the trajectories entering the point (the eigenvalue is negative) have a tangent with positive inclination angle, i.e., passing through the right half of domain II and the left half of domain V , and trajectories issuing from this point have a tangent with negative inclination angle, i.e., passing through the left half of domain II and the right half of domain V . Finally, note that the point (0, 0) is attained in infinite time t0 ; thus, this trajectory is the only exceptional case in which there is no oscillation process. But below, in the investigation of rapid trajectories, we show that, first, domains II and V contain unstable slow trajectories, and second, in general, the motion along slow trajectories occurs in the domain x∗ < |x0 | < 2x∗ ; thus, after attaining breakdown rays, the trajectories cannot return into domains II and V . Some typical trajectories on the plane (x0 , y0 ), as well as trajectories entering the origin, are shown in Fig. 1. 8. SINGULAR POINTS OF THE EQUATION OF SLOW MOTIONS The last detail in the constructed picture is the behavior of trajectories near two more singular points (other than the origin) of Eq. (18), namely, the points ± (x∗ , Bx∗ ). These points are specified by the fact that if there are slow trajectories entering some of them, then there are trajectories issuing from it “on the other side.” If the entering trajectories are stable, then the outgoing trajectories are unstable, and we have a possibility to continue the motion on the unstable part of a slow trajectory, which, as was mentioned above, leads to duck-type trajectories. Just as in the case of Eq. (17), to analyze these singular points, we reduce Eq. (18) to an equation for x0 and z0 = y0 − Bx0 : C
(E − BD)x0 − Dz0 dz0 = −BC + k 1 − x20 + A − 1 − x20 , dx0 z0
(18 )
for which the singular points in question become (±x∗ , 0). The matrix of these singular points (the common matrix for both points) has the form
0 C , (21) 2(E − BD)(1 − A) Ak − BC its characteristic polynomial is λ2 + (BC − Ak)λ + 2C(BD − E)(1 − A), and therefore, a singular point is either a node, a focus, or a center. The case of a center is the simplest one: here trajectories do not enter the singular point. In the case of a focus, the integral curve of Eq. (18) reaches the singular point; however, the trajectory of system (16), (17) does not coincide with the entire integral curve; it terminates at the intersection of this curve with the line x0 = x∗ , more precisely, with the breakdown ray on which the passage to rapid motion takes place. DIFFERENTIAL EQUATIONS
Vol. 40
No. 4
2004
INVESTIGATION OF RELAXATION OSCILLATIONS . . . II
497
Fig. 2. The case in which λ1,2 > 0: a, the separatrix of the point (x∗ , Bx∗ ) lies in domain II as well as in domain V ; b, the separatrix of the point (x∗ , Bx∗ ) entirely lies in domain V ; the separatrices of the point (0, 0) are shown by the double line.
Fig. 3. The case in which λ1,2 < 0 : a, b, domains III and V I (a, the separatrix entirely lies in domain V I; b, the separatrix passes from domain III in domain V I); c, d, domains II and V (c, the separatrix entirely lies in domain II; d, the separatrix passes from domain II in domain V ).
But in the case of a node, we have effects mentioned at the beginning of this section. Consider this case in more detail. By (21), (C, λ) is the eigenvector corresponding to the eigenvalue λ; therefore, λ with the minimum absolute value corresponds to the line nearest to the line y0 = Bx0 . If λ1,2 > 0 (i.e., Ak > BC), then both lines [for the point (x∗ , Bx∗ )] lie in domains V and III; if λ1,2 < 0, then they lie in domains II and V I. (The singular point is a center for Ak = BC and is a focus for close values of Ak and BC; thus, the two cases under consideration are “separated” from each other.) Let us consider each of these cases. In the first case, the phase portrait of slow trajectories has the form shown in Fig. 2. The parts of domains III and V I in which all trajectories issue from the singular point is bounded by the curve issuing from the singular point along the line “contracting” the node; this curve is referred to as the separatrix curve of the corresponding singular point. The part of domains II and V in which all trajectories enter the singular point is also bounded by a separatrix curve (see Fig. 2a); if this curve entirely lies in the domain V , then this part is also bounded by the separatrices of the saddle (0, 0) (see Fig. 2b). Note that, in this case, the passage through the singular points occurs from the unstable part of a slow curve to the stable one; thus there are no duck cycles in this case. Just as in the previous case, in the second case λ1,2 < 0, the phase portrait of slow trajectories has the form shown in Fig. 3. The parts of domains III and V I in which all trajectories enter the singular point are bounded by the separatrix, which can entirely lie in domain V I (and then this domain is semibounded; see Fig. 3a) or can pass into it from domain III (then this part is bounded, see Fig. 3b). But in domains II and V , the part in which trajectories issue from the singular point is bounded either by the separatrix of this singular point and by the separatrices of the saddle (0, 0) (if this separatrix entirely lies in domain II, see Fig. 3c) or only by the separatrix of this singular point (if the separatrix passes from domain II into domain V , see Fig. 3d). It is the case in which a trajectory can pass from a stable part to an unstable one and can generate a duck-type solution, which becomes a duck-type cycle provided that the entire slowrapid trajectory is closed. DIFFERENTIAL EQUATIONS
Vol. 40
No. 4
2004
498
BOROVSKIKH
9. RAPID MOTIONS In the case dt0 = 0, we obtain equations of rapid motions: du0 = dv0 = 0, 1 3 1 3 dx0 = u0 − Ax0 + x0 − x0 dt1 , dy0 = v0 − Cy0 + k x0 − x0 dt1 , 3 3
(22)
which implies the equation of a slow trajectory in the form u0 = const,
dy0 v0 − Cy0 + k (x0 − (1/3)x30 ) . = dx0 u0 − Ax0 + (x0 − (1/3)x30 )
v0 = const,
(23)
Let (¯ x0 , y¯0 , u¯0 , v¯0 ) be an arbitrary point. Rapid trajectories passing through this point are ¯0 and v0 = v¯0 , which satisfy the initial condition x0 (0) = x ¯0 , solutions of system (22) with u0 = u y0 (0) = y¯0 for t1 = 0. To clarify the behavior of solutions of this system, we note that the first equation in (22), together with the equation du0 = 0, forms exactly a system determining rapid motions for the van der Pol equation (with a = 0), which was considered in the preceding section; therefore, the projection of rapid trajectories on the plane (x0 , u0 ) coincides with the rapid trajectories of the single van der Pol equation (see Section 3). Therefore, the motion on any rapid trajectory is monotone with respect to x0 , is bounded, and terminates on the axis u0 = Ax0 − (x0 − (1/3)x30 ); i.e., by the first equation in (22), it attains that curve in infinite time ˆ0 (its absolute value always exceeds x∗ ). t1 , i.e., as t1 → ∞. We denote the limit value of x0 by x Then ˆ0 − (1/3)ˆ x30 = u (24) ¯0 . Aˆ x0 − x By using the function x0 (t1 ) found from the first equation in (22), which is continuous, bounded, and monotone and tends to the limit x ˆ0 as t1 → +∞, from the second equation (which is a linear nonhomogeneous equation), one can find the function y0 (t1 ) : −Ct1
y0 (t1 ) = y¯0 e
C(s−t1 )
+
1 3 v¯0 − k x0 (s) − x0 (s) ds, 3
t1 e
0
which tends to the limit yˆ0 as t1 → +∞, and this limit is equal to ˆ0 − (1/3)ˆ x30 . C yˆ0 = v¯0 − k x
(25)
¯0 , v¯0 ) of this Thus, each solution of system (22) is bounded and tends to the equilibrium (ˆ x0 , yˆ0 , u system as t1 → +∞, which, by (24) and (25), satisfies Eq. (15), i.e., lies on the manifold of slow trajectories. 10. STABLE AND UNSTABLE PARTS OF SLOW TRAJECTORIES ¯0 , v¯0 ) be a point of a slow trajectory. By (15), Now let (¯ x0 , y¯0 , u x0 − x ¯0 − (1/3)¯ x30 , u ¯0 = A¯
¯0 − (1/3)¯ x30 . v¯0 = C y¯0 − k x
¯0 Let us consider rapid trajectories passing through this point. They satisfy system (22) with u0 = u for v0 = v¯0 , and the point (¯ x0 , y¯0 ) is a singular point of this system. At this point, the first-order approximation matrix has the form
0 −A + (1 − x ¯20 ) , (26) −C k (1 − x ¯20 ) x0 | < x∗ ). Therefore, and the singular point is either a stable node (for |¯ x0 | > x∗ ) or a saddle (for |¯ ∗ slow trajectories are stable for |x0 | > x (the passage to rapid trajectories is impossible) and DIFFERENTIAL EQUATIONS
Vol. 40
No. 4
2004
INVESTIGATION OF RELAXATION OSCILLATIONS . . . II
499
unstable for |x0 | < x∗ ; moreover, a breakdown takes place in the direction of the vector (±1, 0, 0, 0), i.e., towards a grow or a decay of x0 . 11. BREAKDOWN RAYS One should separately consider the case in which x ¯0 = x∗ and a rapid trajectory issues from a point lying on a breakdown ray. In this case, the matrix (26) of the singular point is degenerate, and to clarify the behavior of rapid trajectories near this point, we need a more detailed analysis. Since
3 3 2 v¯0 = C y¯0 − k x∗ − (1/3) (x∗ ) , A = 1 − (x∗ ) , u ¯0 = Ax∗ − x∗ − (1/3) (x∗ ) , it follows that Eq. (23) acquires the form dy0 −C (y0 − y¯0 ) + Ak (x0 − x∗ ) +k = 2 dx0 −(1/3) (x0 − x∗ ) (x0 + 2x∗ ) or
A [y0 − y¯0 − (Ak/C) (x0 − x∗ )] d [y0 − y¯0 − (Ak/C) (x0 − x∗ )] +k 1− . =C 2 dx0 C (1/3) (x0 − x∗ ) (x0 + 2x∗ )
This is a linear equation, which can be integrated explicitly, and its solution has the form x0 C/(3x∗2 ) A 1 s − x∗ M +k 1− ds exp C s + 2x∗ x∗ (s − x∗ ) Ak b , (x0 − x∗ ) + y0 = y¯0 + C/(3x∗2 ) C 1 x0 − x∗ exp x0 + 2x∗ x∗ (x0 − x∗ )
(27)
(28)
where b is an arbitrary value and M is an arbitrary constant. It follows from (28) that if x0 → x∗ +0 (and b should be chosen larger than x∗ ), then all solutions tend to y¯0 : the limit of the second term is zero, and the l’Hˆopital rule applied to the ratio indicates the limit equal to the limit of the expression −k(1 − A/C)(3C)−1 (x0 − x∗ ) (x0 + 2x∗ ) , 2
which is zero. In this case, the difference between two distinct solutions has the infinitesimal order x0 → x∗ , it is determined by the exponential factor exp (−1/[x∗ (x0 − x∗ )]). To the left of the point x∗ (where it is convenient to set b = x∗ ), solutions tend to infinity as x0 → x∗ − 0 (which is determined by an exponential factor), the unique exception is a solution (with M = 0) whose opital rule to the ratio on the right-hand side in (28), limit is y¯0 : in this case, by applying the l’Hˆ we obtain the limit of the same expression −k(1 − A/C)(3C)−1 (x0 − x∗ ) (x0 + 2x∗ ) , 2
which is equal to zero (see Fig. 4). Since, in the case under consideration, the first equation in (22) 2 has the form dx0 = −(1/3) (x0 − x∗ ) (x0 + 2x∗ ) dt1 , it follows that the motion is directed from the right to the left for x0 > −2x∗ . To complete the investigation of Eq. (27), we note that the second singular point (−2x∗ , y¯0 − 3Akx∗ /C) of this equation is a stable node (which was mentioned above), and thus all trajectories enter it; moreover, the trajectory issuing from the point (x∗ , y¯0 ) enters the point (−2x∗ , y¯0 − 3Akx∗ /C) and either is tangent to the line y0 − y¯0 + 3Akx∗ /C = H (x0 + 2x∗ ) 2
opital rule again; H = (3 − 4A)/(3 − 3A − C)] for C ≥ 3 (x∗ ) = 3(1 − A) [here we have used the l’Hˆ or is tangent to the vertical axis x = −2x∗ for C < 3(1 − A) [in which case the integral occurring in the numerator is convergent; therefore, the entire ratio admits the asymptotic representation C/3(x∗ )2 O(1) as x → −2x∗ ]. (x + 2x∗ ) DIFFERENTIAL EQUATIONS
Vol. 40
No. 4
2004
500
BOROVSKIKH
Fig. 4. Trajectories of rapid motions issuing from a slow trajectory for x ¯0 = x∗ .
In conclusion of this section, we note that all above-represented considerations can be reproduced word for word for x ¯0 = −x∗ as well; in this case, the motion on a trajectory issuing from (−x∗ , y¯0 ) takes place in the increasing direction of x0 and is terminated at the point (2x∗ , y¯0 + 3Akx∗ /C). 12. CONCLUSIONS I. On the initial segment of the motion, from an arbitrary point (x0 , y0 , u0 , v0 ), one can reach the manifold of slow trajectories determined by Eq. (15) along a rapid trajectory [except for two trajectories, those are separatrices of the point (0, 0)]; moreover, the terminal value of x0 necessarily satisfies the inequality |x0 | > x∗ . II. If |x0 | > x∗ , then slow trajectories are stable: no rapid trajectory issues from them; therefore, for |x0 | > x∗ , the motion always takes place on a slow trajectory. If |x0 | ≤ x∗ , then slow trajectories are unstable. If |x0 | = x∗ , then exactly one rapid trajectory issues from each slow trajectory and attains (exactly once) the manifold of slow trajectories for |x0 | = 2x∗ ; the passage from the point (±x∗ , y0 , u0 , v0 ) [where u0 and v0 are given by (15)] takes place to the point (∓2x∗ , y0 ∓ 3Akx∗ /C). If |x0 | < x∗ , and then exactly two rapid trajectories achieving the manifold of slow trajectories issue from each point of a slow trajectory. The part of the manifold of slow motions (15) satisfying the condition |x0 | > x∗ is said to be stable, and the part on which |x0 | ≤ x∗ is said to be unstable. III. All slow trajectories are terminated on two rays referred to as breakdown rays, namely, x0 = x∗ ,
y0 ≤ Bx∗ ,
x0 = −x∗ ,
y0 ≥ Bx0 .
A guaranteed passage to a rapid motion takes place at points of these rays (except for their vertices). Definition. A trajectory is said to be generic if its slow parts do not pass through any of the points (0, 0) and ± (x∗ , Bx∗ ). Theorem. Each generic trajectory corresponds to relaxation oscillations with permanent switches between rapid and slow motions. Trajectories passing through the origin do not induce oscillations : they approach the origin in infinite time. Trajectories passing through the points ± (x∗ , Bx∗ ) also correspond to relaxation oscillations, but if (BC − Ak)2 − 8C(BD − E)(1 − A) > 0,
BC − Ak > 0,
then duck trajectories can appear. Therefore, the use of the method of time expansion in powers of ε based on the concept of the mathematical continuum HR has permitted one to completely describe rapid and slow trajectories in a very complicated problem, determine possible schemes of motion, and derive conditions providing the motions on the unstable part of the slow manifold. The results can be used to find cycles, investigate their stability, and compute the period [but in this case, HR should be supplemented by “logarithmic” infinite quantities ln(1/ε)]. DIFFERENTIAL EQUATIONS
Vol. 40
No. 4
2004
INVESTIGATION OF RELAXATION OSCILLATIONS . . . II
501
REFERENCES 1. Borovskikh, A.V., Differents. Uravn., 2004, vol. 40, no. 3, pp. 291–298. 2. Van der Pol, B., Nelineinaya teoriya elektricheskikh kolebanii (Nonlinear Theory of Electric Oscillations), Moscow, 1935. 3. Dorodnitsyn, A.A., Prikl. Mat. Mekh., 1947, vol. 11, no. 3, pp. 313–328. 4. Mishchenko, E.F. and Rozov, N.Kh., Differentsial’nye uravneniya s malym parametrom i relaksatsionnye kolebaniya (Differential Equations with a Small Parameter and Relaxation Oscillations), Moscow, 1975. 5. Benoit, E., Callot, J.L., Diener, F., Diener, M., Collect. math., 1981, vol. 31, no. 1–3, pp. 37–119. 6. Vasil’eva, A.B. and Butuzov, V.F., Asimptoticheskie razlozheniya reshenii singulyarno vozmushchennykh uravnenii (Asymptotic Expansions of Solutions of Singularly Perturbed Equations), Moscow, 1973. 7. Lutz, R. and Sari, T., Theory and Applications of Singular Perturbations: Proc. Conf. Oberwolfach, Aug. 16–22, 1981 , Berlin, 1982. 8. Troesch, A., Theory and Applications of Singular Perturbations: Proc. Conf. Oberwolfach, Aug. 16–22, 1981 , Berlin, 1982. 9. Cartier, P., Uspekhi Mat. Nauk , 1984, vol. 39, no. 2, pp. 57–76. 10. Zvonkin, A.K. and Shubin, M.A., Uspekhi Mat. Nauk , 1984, vol. 39, no. 2, pp. 77–127. 11. Shchepakina, E. and Sobolev, V., Nonlinear Anal., 2001, vol. 44, pp. 897–908. 12. Neprintsev, V.I., Zadorozhnij, V.G., and Lobanov, S.N., in Euromech—2nd European Nonlinear Oscillations Conference, Prague, 1996, pp. 157–160.
DIFFERENTIAL EQUATIONS
Vol. 40
No. 4
2004