3. Limit cycles of the Li´ enard equation
3.3 The existence of the limit cycle
The basic theme of this paper is the search for a simple robust model of periodic behaviour. The last two theorems show us how the system explodes on one condition and converges on another condition. It would be a nice procedure to show that the system tends to explode near the origin and tends to converge far from the origin. In this case the Poincar´e-Bendixson theorem would trivially affirm the existence of the limit cycle. Unfortunately the set of divergence
E = (x, y)|d
dt(x2+y2)≥0 ={(x, y)|xF(x)≤0} (3.20)
is unbounded. So we must do some extra work to confirm the necessary conditions for the Poincar´e-Bendixson theorem.
Let us consider the Li´enard equation
¨
x+f(x)x˙ +x= 0 (3.21)
where f ∈C0(IR). In the Li´enard plane the equation has the equivalent form
˙
x=y−F(x)
˙ y=−x (3.22)
where F(x) = 0xf(s)ds.
Throughout this section we assume that there exists numbers β1 <α1 <0<α2 <
β2 and ε>0 such that the generating function F fulfills the following properties:
F(x)<−ε, if x∈]− ∞,β1[, F(x)>0, if x∈]α1,0[, F(x)<0, if x∈]0,α2[, F(x)>ε, if x∈]β2,∞[.
(3.23)
We will further use the following notations:
M1 = max{F(x)|x∈[β1,0]}, (3.24)
m1 = min{F(x)|x∈[β1,0]}, (3.25)
M2 = max{F(x)|x∈[0,β2]}, (3.26)
m2 = min{F(x)|x∈[0,β2]}. (3.27)
These numbers are well defined because the function F is continuous on the closed intervals [β1,0] and [0,β2]. The constraints for the graph of the function are drawn in Figure 15.
Next we will construct an orbit from a point on the positive y-axis to a point on the negative y-axis. In the construction we will subsequently show that the orbit will reach some target infinite time.
Let G ⊂ IR2 be a bounded connected region such that the velocity of motion v =
√x2 +y2is positive inGand the time derivativesx andy do not change their signs in G. Let the diameter ofGbed(G) = supp,q∈Gd(p, q) = supp,q∈G (xp−xq)2+ (yp−yq)2 Because the setGis bounded and closed, it is compact. Thus there exists a positive number usuch that v≥u >0 in G. Let p= (xp, yp)∈G and q= (xq, yq) =ϕ(p, tq)∈
M1 M2
m2 m1
−ε ε
α1
α2 β1
β2
Figure 15. The bounds of the graphy =F(x).
p
q ds G
| |dx
| |dy
Figure 16. An arc of the orbit γ+(p) in a bounded region G.
γ+(p). If ϕ(p, t)∈G,∀t∈[0, tq], then the length of the arc fromptoq along the orbit γ+(p) is
q p =
q
γ+,pds ≤
q
γ+,p(|dx|+|dy|) =|xq−xp|+|yq−yp| (3.28)
But now
2d(G)≥|xq−xp|+|yq−yp| ≥ qp ≥utq
(3.29)
⇒ tq ≤ 2d(G)
u ,∀q∈G.
(3.30)
If t >2d(G)/u, then ϕt(p) ∈/ G and so the orbit γ+(p) intersects the boundary ∂G of the bounded regionG in a finite time.
In the subsequent reasoning we will use the following rule:
LetG⊂IR2 be a bounded connected region. Let p∈∂Gbe a boundary point such that the positive orbit γ+(p) goes insideG. If the velocity of motion is positive in G (i.e. the critical point is not in G) and the time derivatives x and y do not change their signs in G, then the positive orbit γ+(p) reaches the boundary ∂G in finite time.
Next we consider the positive orbit γ+ under the flow (3.22)generated by F with the initial point p0 = (0, y0), y0 > 0 on the positive y-axis. At p0 the derivatives are y = 0 and x =y0−F(x0)>0, and so the direction of the motion atp0 is to the right (the direction of the positive x-axis). Just after the initial point y =−x < 0, and so the direction of the motion is to the right and down (see Figure 14). So the orbit must intersect in finite time either the line x = β2 or the line y = M2. Let us denote the point of intersection by p1 = (x1, y1). Then either x1 = β2 and y1 ≥ M2 or y1 = M2
andx1 ≤β2. If ρ0 andρ1 are the corresponding distances from the origin, then we can estimate the value of y1 by the line-integral
ρ21−ρ20 = p1
γ+,p0
d(ρ2) = p1
γ+,p0
2F(x)dy (3.31)
≥ 2M2(y1 −y0)
⇒ x21+y12−y02 ≥2M2(y1 −y0) (3.32)
⇔ y1−y0 ≥ −β22 y1+y0−2M2
(3.33)
So by choosing a y02 big enough so that y21 > M22 we can assure y0 > y1 > M2
(3.34)
from which we conclude that p1 lies on the line x=β2 over the curve y=F(x).
After the point p1 as far as the orbit is over the curvey =F(x) we have x >β2
0< F(x)< y ⇒
y =−x <0 x =y−F(x)>0
d(ρ2)/dt=−2xF(x)<0 (3.35)
Because the distance ρ is decreasing, the orbit must in finite time intersect the curve y=F(x) at a point which we denote by p2 (see Figure 18).
After the point p2 as far as x > β2 we have
x >β2
y < F(x) F(x)>0
⇒
y =−x <0 x =y−F(x)<0 d(ρ2)/dt=−2xF(x)<0 (3.36)
M2
m2
ε
β1 x =
x y
p0
p1
G
y = F x( )
Figure 17. The first arc p0 →p1, of the orbit γ+(p0).
M2
m2
ε
β1 x =
x yp
0 p
1
p2
arcρ = ρ1
G y = F x( )
Figure 18. The first two arcs p0 →p1→ p2, of the orbit γ+(p0).
M2
m2
ε
β1 x =
x y
p0
p1
p2
arcρ = ρ2 p3
G y = F x( )
Figure 19. The first three arcs p0→p1 →p2 →p3 of the orbit γ+(p0).
Because the distance ρis decreasing, the orbit must infinite time intersect the line x=β2 at a point which we denote by p3 (see Figure 19).
After the point p3 as far as x > 0 we have
x < β2
y < F(x) F(x)≥m2
⇒
y =−x <0 x =y−F(x)<0
d(ρ2) = 2F(x)dy≤2m2dy (3.37)
Letp= (x, y) be a point of the orbit γ+(p0) after the point p3. Then ρ2−ρ23 =
p γ+,p3
d(ρ2)≤2m2(y−y3) (3.38)
⇔ y2+x2−(β22+y32)≤2m2(y−y3) (3.39)
⇔ (y−y3)(y+y3−2m2)≤β22−x2 ≤β22 (3.40)
⇔ y−y3 ≥ −β22 2m2−y−y3
(3.41)
From the inequality (3.41) it follows thatyis bounded and from (3.37) we then conclude that the orbit after the point p3 intersects in finite time the positive y-axis at a point which we denote by p4 (see Figure 20).
Theorem 3.3.1 In the construction above, the initial point p0 can be chosen in such a way that y42 < y02.
Proof. If |y4| is bounded when y0 → ∞ then the proof is trivial. Let us suppose y4 → −∞when y0 → ∞. Let ρi be the distance of the pointpi from the origin. Then
ρ21−ρ20 =
p1 p0
2F(x)dy≤2m2(y1−y0) (3.42)
M2
m2
ε β1
x =
x y
p0
p1
p2
p3
y = ymin p4
G y = F x( )
Figure 20. The arcs p0→p1 →p2 →p3 →p4 of the orbit γ+(p0).
ρ23−ρ21 =
p3 p1
2F(x)dy≤2ε(y3−y1) (3.43)
ρ24−ρ23 =
p4 p3
2F(x)dy≤2m2(y4−y3) (3.44)
Combining these we have
ρ24−ρ20 ≤2m2(y4−y3)
>0
+ 2ε(y3−y1)
<0
+ 2m2(y1−y0)
>0
(3.45)
From (3.33) we see that y1 − y0 → 0 as y0 → ∞ and from (3.41) we see that 0 ≥ y4 −y3 ≥ −β2/(2m2 − y4 − y3) → 0 as y0 → ∞. So according to (3.45) the differenceρ24 −ρ20 becomes negative when y0 becomes large enough. 2
Now we can finally formulate the proper proof of the existence of the limit cycle.
Theorem 3.3.2 Let F be a continuous function such that there exist numbers β1 <
α1 <0<α2 <β2 and ε>0 with the following properties F(x)<−ε, if x∈]− ∞,β1[, F(x)>0, if x∈]α1,0[, F(x)<0, if x∈]0,α2[, F(x)>ε, if x∈]β2,∞[.
Then the dynamical system
˙
x=y−F(x)
˙ y=−x has a limit cycle around the origin.
W Γ1
Γ2
Γ3
Γ4 Dr
Figure 21. The invariant set W.
Proof. According to the previous theorem we can construct an arcΓ1 of an orbit with the initial point p0 = (0, y0), y0 >0 and the end point p1 = (0, y1), y1 <0 such that
|y1|<|y0|. In a similar way we can construct an arcΓ2 of another orbit with the initial point p2 = (0, y2), y2 <0 and the end point p3 = (0, y3), y3 >0 such that |y3|< |y2|. We can always choose the initial points so thaty0 > y3 >0> y1 > y2. We connect the two arcs with line intervalsΓ3 =p3p0 andΓ4 =p1p2. ThenΩ=Γ1+Γ4+Γ2+Γ3 is a closed Jordan curve and a boundary of a simply connected regionG (i.e. ∂G=Ω).
Letr= 12min{|α1|,α2}and letDr ={(x, y)|x2+y2 ≤r2}be a closedr-disk. Then the bounded region W = G−Dr, with the boundary ∂W = Ω−∂Dr has no critical points.
Because two orbits cannot intersect, no orbit can penetrate Γ1 or Γ2. On the line interval Γ3 there are y = 0 and x > 0. So any orbit penetrating Γ3 must go from outside to inside of G. In a similar way we see that any orbit penetrating Γ4 must go from outside to inside ofG. So G is a positively invariant set.
From the theorem (3.2.1) we see that any orbit penetrating ∂Dr must go from out-side to inout-side W. So W is an positively invariant set. Let p be a point on the line intervalΓ3 =p3p0 and letγ+ be a positive orbit with the initial pointp. Thenγ+⊂W is bounded and contains no critical point. It follows from the Poincar´e-Bendixson
the-orem (2.3.8) that W contains a limit cycle. 2
2
-2
-2 2
Figure 22. The limit cycle generated by the function F(x) = 1+ee3x3x(x2−x).
Corollary 3.3.3 Let F be a C1-function such that F (0)< 0, limx→∞F(x) > 0 and limx→−∞F(x)<0, then the dynamical system
˙
x=y−F(x)
˙ y=−x has a limit cycle around the origin.
2
It should be noticed that the conditions in the theorem above are sufficient but not necessary for the existence of the limit cycle. For example the dynamical system generated by the function
F(x) = e3x
1 +e3x(x2−x) (3.46)
has a unique limit cycle although the generating function does not fulfil the conditions for the theorems before. The limit cycle of this counterexample is drawn in Figure 22.