EXISTENCE OF LIMIT CYCLES
FOR COUPLED VAN DER POL EQUATIONS
NORIMICHI HIRANO AND SLAWOMIR RYBICKI†
Abstract. In this paper, we consider the existence of limit cycles of coupled van der Pol
equations by using S 1 −degree theory due to Dylawerski et al., see [3].
1. Introduction
In this paper, we consider the existence of limit cycles of coupled van der Pol equations of the
form

2

 ü1 + ε1 (u21 − a1 )u̇1 + c11 u1 + c12 u2 + c13 u3 + . . . + c1n un = 0,

ü2 + ε2 (u2 − a2 )u̇2 + c21 u2 + c22 u2 + c23 u3 + . . . + c2n un = 0,
.
..



ün + εn (u2n − an )u̇n + cn1 un + cn2 u2 + cn3 u3 + . . . + cnn un = 0,
where n ≥ 1, εi > 0 for each i = 1, · · · , n, and cij ∈ R for all 1 ≤ i, j ≤ n.
The coupled van der Pol equation has been studied as a model of self-excited systems. This
kind of systems appears in a wide variety of mechanical, electronical and biological systems. A
limit cycle is a nontrivial periodic solution of the autonomous system above. The existence of a
limit cycle of one dimensional van der Pol equation
ü + ε(u2 − 1)u̇ + u = 0
(1.1)
is well known. The proof of the existence of the limit cycle is based on the Poincaré-Bendixson
theorem(cf. [6], [15]). In contrast to one dimensional case, the existence of limit cycles for
coupled van der Pol equations is not yet established except some restrictive cases(cf. [17]). In
the present paper, we prove the existence of limit cycles for coupled van der Pol equations by
using S 1 -degree theory, see [3]. To avoid unnecessary complexity, we restrict ourselves to the
case n = 2. That is we consider the problem
(
ü1 + ε1 (u21 − 1)u̇1 + u1 + c2 u2 = 0,
(P)
ü2 + ε2 (u22 − 1)u̇2 + c1 u1 + u2 = 0,
Our argument below remains valid for the case that n > 2. We impose that following condition
on c1 and c2 :
c1 · c2 ∈ (0, 1) ∪ (1, +∞)
(A)
We can now state our main results:
Date: January 12, 2006.
1991 Mathematics Subject Classification. Primary 05C38, 15A15; Secondary 05A15, 15A18.
Key words and phrases. Limit cycles, von der Pol system, S 1 -degree.
†
Research supported by the State Committee for Scientific Research grant No. 5 PO3A 026 20.
1
NORIMICHI HIRANO AND SLAWOMIR RYBICKI†
2
1
, there exist ε1 , ε2 > 0 such that problem (P) has a
√
1 + c1 c2
nontrivial periodic solution u ∈ C 2 (R, R2 ) with period 2πα.
Theorem 1.1. For any α > p
Theorem 1.2. There exists C > 0 such that for any (ε1 , ε2 ) ∈ R+ × R+ such that
max {ε1 , ε2 } < C, problem (P) possesses a nontrivial periodic solution u ∈ C 2 (R, R2 ) with period
2πα for some α > 1.
For the convenience of the reader we have included some references related to the equivariant
degree methods.
The first degree theory for admissible S 1 −equivariant gradient maps, which is a rational
number, is due to Dancer [1]. Degree theories for S 1 −equivariant maps have been defined in
[3], [5], [7], [8], [13]. The class of S 1 −equivariant orthogonal maps have been introduced in
1
[16]. The first degree
! theory for S −equivariant orthogonal maps, which is an element of the
∞
M
group Z ⊕
Z , is due to the second author [16]. Degree theories for abelian actions have
i=1
been defined by Ize and Vignoli in [9], [11], [12]. A definition of degree theory for equivariant
orthogonal maps (symmetries of any compact abelian Lie group are admitted) is due to Ize and
Vignoli [12]. Finally, degree theory for G-equivariant gradient maps, where G is any compact
Lie group, is due to Gȩba [4].
2. Preliminaries
We denote by < ·, · >2 the scalar product of L2 ([0, 2π], R2 ). Define
Hper = {v : R → R2 : v is absolutely continuous, < v̇, v̇ >2 < ∞ and v(t) = v(t + 2π) ∀ t ∈ R},
and scalar products < ·, · >Hper : Hper × Hper → R as follows
< w, v >Hper =< w, v >2 + < ẇ, v̇ >2 .
Let S 1 = {z ∈ C : | z |= 1} = {ei·θ : θ ∈ R} be the group of complex numbers of module 1 with
an action given by multiplication of complex numbers. For any fixed m ∈ N we denote by Zm a
cyclic group of order m and define homomorphism ρm : S 1 → GL(2, R) as follows
cos(mθ) − sin(mθ)
i·θ
ρm e
=
.
sin(mθ)
cos(mθ)
Of course R[1, m] := (R2 , ρm ) is a two-dimensional representation of the group S 1 . We will denote
by R[k, m] the direct sum of k copies of representation R[1, m] and by R[k, 0] k−dimensional
trivial representation of the group S 1 . Define action ρ : S 1 × Hper → Hper of the group S 1 as
follows
ρ(ei·θ , v(t)) = v(t + θ)
(2.1)
In the following fact we collect some well known properties of the space Hper .
Fact 2.1. Under the above assumptions:
(1) (Hper , < ·, · >Hper ) is a separable Hilbert space,
(2) (Hper , < ·, · >Hper ) is an orthogonal representation of the group S 1 with S 1 −action given
by (2.1),
VAN DER POL SYSTEM
(3) Hper = cl
∞
M
3
!
R[2, n] .
n=0
is a closed subgroup of S 1 , one can consider Hper as a Z2 −space. Let (Hperp )Z2
⊥
denote the set of fixed points of the action of the group Z2 on Hper . Moreover, by (Hper )Z2
Since Z2 ⊂
S1
we denote the orthogonal complement of (Hper )Z2 in Hper .
Define
H = {v : R → R2 : v is absolutely continuous, < v̇, v̇ >2 < ∞ and v(t) = −v(π + t) ∀ t ∈ R}.
In the following fact we collect some well known properties of the space H.
Fact 2.2. Under the above assumptions:
⊥
(1) H = (Hperp )Z2
,
(2) (H, < ·, · >H ) is a separable Hilbert space,
(3) (H, < ·, · >H ) is an orthogonal representation of the group S 1 with S 1 −action given by
the restriction of (2.1), !
∞
M
(4) H = cl
R[2, 2n − 1] .
n=1
Let v = (v1 , v2 ) be a periodic solution of (P) with period 2πα for some α > 1. Then by putting
t = ατ and u(τ ) = (u1 (τ ), u2 (τ )) = (v1 (ατ ), v2 (ατ )), we find that u = (u1 , u2 ) ∈ H is a 2π−
periodic solution of problem
(
ü1 + ε1 α(u21 − 1)u̇1 + α2 (u1 + c2 u2 ) = 0,
(2.2)
ü2 + ε2 α(u22 − 1)u̇2 + α2 (c1 u1 + u2 ) = 0,
Here we put
F (u) =
ε1 ( 31 u31 − u1 )
!
1 c2
c1 1
,
A=
.
ε2 ( 13 u32 − u2 )
In this way we have converted problem (P) of finding of periodic solutions of any period into
1−parameter problem (2.2) of finding of periodic solutions of a fixed period 2π.
Lemma 2.1. Let Hilbert space H and operator F will be defined as above. Then,
(1) F : H → H is well-defined continuous operator,
(2) for any w ∈ H the following holds true:
(a) w(t) = w(t + 2π), for any t ∈ R,
Z 2π
w(t)dt = 0,
(b)
0
Z t
(c)
w(τ )dτ ∈ Hper ,
0Z
Z t
t
(d) <
w(τ )dτ, (cos(2nt), cos(2nt)) >Hper =<
w(τ )dτ, (sin(2nt), sin(2nt)) >Hper =
0
0, for any Zn ∈ N.
t
(3) if w ∈ H, then
F (w(τ ))dτ ∈ R[2, 0] ⊕ H.
0
0
Proof. The easy proof is left to the reader.
NORIMICHI HIRANO AND SLAWOMIR RYBICKI†
4
Notice that (2.2) can be rewritten as
d
F (u) + α2 Au = 0.
(2.3)
dt
We will find a solution u = (u1 , u2 ) ∈ H of (2.2). We denote by L1 : H → H the inverse of the
mapping u → −ü for u ∈ H. That is u = L1 (−ü) for u ∈ H. We also put Lu = L1 Au for each
u ∈ H. Denote by σ(L) the spectrum of L. If µ ∈ σ(L) then V (µ) denotes the eigenspace of L
corresponding to the eigenvalue µ. Notice that to describe eigenvalues and eigenspaces of the
operator L it is enough to consider equation ü + µ1 Au = 0.
√
[
1 ± c1 c2
µ±
=
. MoreFact 2.3. Suppose that c1 c2 ∈ (0, 1) ∪ (1, +∞). Then, σ(L) =
n
(2n − 1)2
n∈N
over,
(1) V (µ±
n ) ≈ R[1, 2n − 1],
!
∞
M
−
+
(2) H = cl
V (µn ) ⊕ V (µn ) ,
ü + α
i=1
−
+
−
+
−
(3) if c1 c2 ∈ (0, 1), then µ+
1 > µ1 > µ2 > µ2 > . . . > µn > µn > . . . > 0,
+
+
−
+
(4) if c1 c2 > 1, then µ1 > µ2 > . . . > µn > . . . > 0 and µn < 0 for any n ∈ N.
Let m, M ∈ R be such that 0 < m < M. Let η : R → [0, 1] be a smooth function such that

m2

0, if ktk ≤ 2 ,
η(t) =

2

1, if ktk ≥ M2 ,
2
m M2
0
and that η (t) > 0 for t ∈
,
. Moreover, define a smooth S 1 −equivariant function
2
2
θ : H → [0, 1] by the following formula
kuk2
θ(u) = η
(2.4)
2
Denote by π : R[2, 0] ⊕ H → H the S 1 −equivariant orthogonal projection.
Since Lemma 2.1, for each α > 0 and δ ∈ [0, 1], we define a mapping G(·, α, δ) : H → H by
Z t
G (v, α, δ) = −δαπ
F (v(τ ))dτ + α2 θ(v)L(v)
(2.5)
0
Then each solution u ∈ H of problem G(u, α, δ) = u for some (α, δ) ∈ R+ × R+ satisfies
d
F (u) + α2 θ(u)Au = 0
dt
We will also consider the following family of differential equations
ü + δα
(2.6)
d
F (u) + α2 Au = 0
(2.7)
dt
Below we formulate and prove three technical lemmas which we will apply in the next sections.
ü + δα
Lemma 2.2. There exists a monotone increasing function m(·) : R+ → R+ such that for each
δ ∈ (0, 1], α ∈ R+ and each solution u ∈ H of problem (2.7) inequality kuk ≤ m(α) holds.
VAN DER POL SYSTEM
5
+
Proof. Fix α ∈ R and δ ∈ (0, 1]. Let u ∈ H be a solution of (2.7). Multiplying (2.7) by u̇ and
integrating over [0, 2π], we find that

Z 2π
Z 2π

2
2
2


u̇1 u2 = 0,
(u1 − 1)u̇1 + α c2
 δαε1
0
0
Z 2π
Z 2π


2
2
2

u1 u̇2 = 0.
(u2 − 1)u̇2 + α c1
 δαε2
0
0
Then we have that
2π
Z
(u21
c1 ε1
−
1)u̇21
Z
Z
c1 ε1
2π
u21 u̇21
(u22 − 1)u̇22 = 0
(2.8)
0
0
That is
2π
+ c2 ε2
Z
2π
+ c2 ε2
0
u22 u̇22
2π
Z
u̇21
= c1 ε1
0
Z
2π
+ c2 ε2
0
0
Since c1 c2 > 0, it follows that there exists C1 > 0 such that
Z 2π
Z 2π
2 2
2 2
u1 u̇1 + u2 u̇2 ≤ C1
u̇21 + u̇22
(2.9)
0
0
On the other hand by the Schwartz inequality, we have
Z t
√ Z
2
|u1 (t)| ≤ 2
|u1 u̇1 | dt ≤ 2 2π(
s1
and
Z
2
u̇22 .
t
|u2 (t)| ≤ 2
2π
|u1 |2 |u̇1 |2 )1/2
0
√ Z
|u2 u̇2 | dt ≤ 2 2π(
s2
2π
|u2 |2 |u̇2 |2 )1/2
0
for all t ∈ [0, 2π], where s1 , s2 ∈ [0, 2π] satisfy u1 (s1 ) = u2 (s2 ) = 0. It then follows from (2.9)
and the inequalities above that there exists C2 > 0 such that
|u(t)|2 ≤ C2 |u̇|2
for all t ∈ [0, 2π]
(2.10)
We next multiply (2.7) by u and integrate over [0, 2π]. Then we have that
|u̇|22 = α2 hAu, ui ≤ α2 kAk |u|22 ,
2
where kAk
p denotes the operator norm of matrix A. Then by (2.10), we have |u|2 ≤ 2πC2 |u̇|2 ≤
2πC2 α kAk |u|2 . It then follows from the inequality above that |u|2 ≤ 2πC2 α kAk and then
p
p
p
+
|u̇|2 ≤ 2πC2 α2 kAk. Then by putting m(α) = 2πC2 kAkα(1 + kAkα) for each α ∈ R , we
reach to the assertion.
Lemma 2.3. For each α0 > 0, there exists δ1 (α0 ) > 0 such that the problem (2.7) has no
nontrivial solution in H for each α ∈ (0, α0 ) and δ > δ1 (α0 ).
Proof. Fix δ > 0, α0 > 0 and α ∈ (0, α0 ). Let u ∈ H be a solution of (2.7). Multiplying (2.7) by
u̇ and integrating over [0, 2π] we find that
Z 2π
Z 2π
2
2
c1 ε1
(u1 − 1)u̇1 + c2 ε2
(u22 − 1)u̇22 = 0.
0
0
Since u(t) = −u(t + π) and the above, there exists t0 ∈ [0, 2π] such that u1 (t0 ) = 1 or u2 (t0 ) = 1.
Without loss of the generality one can assume that u1 (t0 ) = 1 holds. We also may assume that
u̇1 (t0 ) ≤ 0. Suppose that u̇1 (t0 ) > 0. Since u̇1 (t0 ) > 0, u1 (t0 ) = 1 and u1 (t0 + π) = −1, there
NORIMICHI HIRANO AND SLAWOMIR RYBICKI†
6
exists t00 > t0 such that u1 (t00 ) = 1 and u̇1 (t00 ) ≤ 0. Then we assume that u̇1 (t0 ) ≤ 0. We next
integrate (2.7) over [t0 − π, t0 ]. Then notice that u̇(t) = −u̇(t − π) and u(t) = −u(t − π), we have
Z t0
1 3
4
2
2u̇1 (t0 ) + 2αδ1
(u1 (t) + c2 u2 (t))dt.
u (t0 ) − u1 (t0 ) = 2u̇1 (t0 ) − αδ1 = −α
3 1
3
t0 −π
¿From the above and Lemma 2.2, by the Schwartz inequality we obtain the following
Z
2
α2 t0
0 ≥ u̇1 (t0 ) = αδ1 −
(u1 (t) + c2 u2 (t)) dt ≥
3
2 t0 −π
Z
2
α0 t0
≥α
δ1 −
(| u1 (t) | + | c2 || u2 (t) |) dt ≥
3
2 t0 −π
Z t0
2
α0
(| u1 (t) | + | u2 (t) |) dt ≥
≥α
δ1 −
max{1, | c2 |}
3
2
t0 −π
1/2 !!
Z t0
1/2 Z t0
√
2
α0
2
2
≥α
+
| u2 (t) | dt
≥
δ1 −
max{1, | c2 |} π
| u1 (t) | dt
3
2
t0 −π
t0 −π
√
√
2
2
0
0
0
≥α
δ1 − α max{1, | c2 |} πkuk ≥ α
δ1 − α max{1, | c2 |} πm(α ) .
3
3
Therefore to complete the proof it is enough to put
√
3 0
δ1 (α0 ) =
α max{1, | c2 |} πm(α0 ).
21
1
∈ R+ \σ(L), there exists δ2 (α) > 0 such that there exists no nontrivial
α2
solution of (2.7) in H for all δ ∈ (0, δ2 (α)).
Lemma 2.4. For each
1
∈ R+ \ σ(L). Suppose contrary to our claim that there exists a sequence
α2
{(un , δn )} ⊂ H × R+ such that lim δn = 0 and each un is a solution of (2.7) with δ = δn .
Proof. Fix
n→∞
Then by Lemma 2.2, sequence {un } is bounded in H. Therefore we may assume that un converges to u ∈ H weakly in H and strongly in L2 ([0, 2π]; R2 ). Then since sup |un (t)| ≥ 1 for
t∈[0,2π]
1
each n ≥ 1, we find that u 6≡ 0. Also one can see that u satisfies ü+α2 Au = 0. That is Lu = 2 u.
α
1
/ σ(L), this is a contradiction. Then the assertion holds.
Since 2 ∈
α
3. S 1 -degree
Denote by Γ0 the free abelian group generated by N and let Γ = Z2 ⊕ Γ0 . Then γ ∈ Γ
means γ = {γr } , where γ0 ∈ Z2 and γr ∈ Z for r ∈ N. Let V be a Hilbert space which is
a representation of S 1 . For each proper subgroup Q of S 1 and each S 1 -equivariant subset X
of V, we denote X Q the subset of fixed points of Q in X. For each U ⊂ V ⊕ R and each S 1
equivariant compact mapping f : U → V, we define, by using the fact that there is a one-to-one
correspondence between N and the proper, closed subgroups Q of S 1 , Deg(I − f, U ) = {γr } ∈ Γ
by γ0 = degS 1 (I − f, U ) and γr = degZr (I − f, U ), r ∈ N (cf. [3] and [16]). Next theorem has
been formulated and proved in [3] and describe properties of S 1 -degree.
VAN DER POL SYSTEM
7
Theorem 3.3 ([3]). Let V be a Hilbert space which is a representation of S 1 , U be an open
bounded, invariant subset of V ⊕ R and f : U → V is a compact S 1 −mapping such that
(I − f )(∂U ) ⊂ V \ {0} . Then there exists a Γ−valued function Deg(I − f, U ) called S 1 −degree,
satisfying the following properties:
(a) if degQ (I − f, U ) 6= 0, then (I − f )−1 (0) ∩ U Q 6= φ,
(b) if U0 ⊂ U is open, invariant and (I − f )−1 (0) ∩ U ⊂ U0 , then
Deg(I − f, U ) = Deg(I − f, U0 );
(c) if h : cl(U ) × [0, 1] → V is an S 1 −equivariant homotopy of compact mappings such that
(I − h)(∂U × [0, 1]) ⊂ V \ {0} . Then
Deg(I − h0 , U ) = Deg(I − h1 , U ).
To apply S 1 −degree theory to our problem, we need some definitions. Since Lemma 2.1, we
define bounded operator E : H → H as follows


Z t
v1 dt 
 ε1


0
E(v) = π 
for each v = (v1 , v2 ) ∈ H.

Z t


ε2
v2 dt
0
For each α > 0 and δ ∈ [0, 1], we define a mapping H(·, ·, α, δ) : H ⊕ R → H by
H(u, λ, α, δ) = G(u, α, δ) + λαE(u).
It is easy to see that H(·, ·, α, δ) is an S 1 -equivariant compact mapping. One can see that if
u ∈ H satisfies u = H(·, ·, α, δ) for (α, δ) ∈ R+ × R+ then
ü + δα
or in equivalent form
(
d
d2
F (u) + α2 θ(u)Au = λα 2 E(u)
dt
dt
ü1 + δε1 α(u21 − 1)u̇1 + α2 θ(u)(u1 + c2 u2 ) = ε1 αλu̇1
ü2 + δε2 α(u22 − 1)u̇,2 +α2 θ(u)(c1 u1 + u2 ) = ε2 αλu̇2 ,
(3.1)
(3.2)
Lemma 3.5. Let α, δ > 0 and λr∈ R. Suppose that u = (u1 , u2 ) ∈ H is a nontrivial solution of
δ
(3.1). Then λ + δ > 0 and w =
u is a solution of equation
λ+δ
d
(3.3)
ẅ + (λ + δ)α F (w) + α2 θ(u)Aw = 0
dt
Proof. Let u ∈ H be a nontrivial solution of (3.1). First of all we will show that λ + δ > 0.
Notice that (3.2) can be represented in the following form
(
ü1 + ε1 α(δu21 − (λ + δ))u̇1 + α2 θ(u)(u1 + c2 u2 ) = 0,
(3.4)
ü2 + ε2 α(δu22 − (λ + δ))u̇2 + α2 θ(u)(c1 u1 + u2 ) = 0,
Multiplying (3.4) by (c1 u̇1 , c2 u̇2 ) and integrating over [0, 2π] we obtain the following
c1 ε1 (u21 − (λ + δ))u̇21 + c2 2 (u22 − (λ + δ))u̇22 = 0.
r
λ+δ
Since u is a nonzero function, we obtain λ + δ > 0. Putting u =
w in (3.4) we find that
δ
w satisfies (3.3).
NORIMICHI HIRANO AND SLAWOMIR RYBICKI†
8
Lemma 3.6. Let α > 0 be such that
1
1
∈
/ σ(L) and n0 = max{µ+
n > 2 }. Then,
2
n∈N
α
α
(1) if c1 c2 ∈ (0, 1), then

0,










2,






degQ (Id − H(·, ·, α, 0), U ) = 2,








1,








0,
Q = S1,
Q = Z2m−1 for m ∈ {1, . . . , n0 − 1},
Q = Z2n0 −1 and µ−
n0 >
1
,
α2
Q = Z2n0 −1 and µ−
n0 <
1
,
α2
otherwise,
where U = {u ∈ H : m < kuk < M } × [−1, 1],
(2) if c1 c2 > 1, then

1

0, Q = S ,





degQ (Id − H(·, ·, α, 0), U ) = 1, Q = Z2m−1 for m ∈ {1, . . . , n0 },






0, otherwise,
where U = {u ∈ H : m < kuk < M } × [−1, 1].
1
. The assertion is a slight
α2
modification of Corollary 4.7 of [16] and the proof is basically the same as that of Theorem
4.2 of [16]. Then we just show the sketch of the proof. From the definition H(u, λ, α, 0) =
α2 θ(u)Lu + λαE(u). Then one can see that u = H(u, λ, α, 0) if and only if u = α2 θ(u)Lu.
1
/ σ(L), 0 < θ(u) < 1 and therefore u ∈ U. It is easy to verify that the set
Moreover, since 2 ∈
α
2
u ∈ U : u = α θ(u)Lu consists of a finite number of S 1 −orbits and is defined as follows
Proof. 1. Without loss of the generality one can assume that µ−
n0 >
2
u ∈ U : u = α θ(u)Lu =
n0
[
[
i=1 ν∈{−,+}
u∈V
(µνi )
:
µνi θ(u)
1
= 2
α
.
Let Ui± ⊂ cl Ui± ⊂ U, i = 1, . . . , n0 , be open, disjoint, S 1 −invariant sets such that
1
±
±
u ∈ V (µi ) : µi θ(u) = 2 ⊂ Ui± , i = 1, . . . , n0 .
α
Therefore,
Deg(Id − H(·, ·, α, 0), U ) =
n0
X
X
i=1 ν∈{−,+}
Deg(Id − H(·, ·, α, 0), Ui± ).
VAN DER POL SYSTEM
9
Fix i0 ∈ {1, . . . , n0 } and ν0 ∈ {−, +}. What is left is to show that


1, if Q = Z2i0 −1 ,
ν0
degQ (Id − H(·, ·, α, 0), Ui0 ) =


0, otherwise.
⊥ ×[−1, 1] such that α2 θ(v0 ) =
Fix (v0 , 0, 0) ∈ Uiν00 ×[−1, 1] ⊂ H×[−1, 1] = V µνi00 ⊕ V µνi00
1
. It is easy to show that D(u−H(u, λ, α, 0))(v0 , 0, 0) is a surjection. Finally applying Theorem
µνi00
6.7 (a) of [3] we complete the proof.
2. The proof is literally the same as the proof of (1).
4. Proof of Theorems
1
Proof of Theorem 1.1. Fix α > p
and M > m(α). We also choose m > 0 so small
√
1 + c1 c2
that kuk∞ < 1 for each u ∈ H with kuk ≤ m. Choose λ0 and δ0 such that λ0 − δ0 > δ1 (α) and,
δ0 m2
(4.1)
[0, λ0 ] ⊂ {λ : λ ≥ δ1 (α)} ∪ λ : 0 ≤ λ ≤
m(α)2
Let U ⊂ H ⊕ R be a open set defined by
U = {v ∈ H : m < kvk < M } × (−λ0 , λ0 ).
Then the boundary ∂U of U has the form ∂U = B1 ∪ B2 ∪ B3 , where
B1 = {v ∈ H : kvk = m} × [−λ0 , λ0 ],
B2 = {v ∈ H : kvk = M } × [−λ0 , λ0 ],
B3 = {v ∈ H : m < kvk < M } × {−λ0 , λ0 } .
We put
S1 = {(u, λ) ∈ cl(U ) : u = H(u, λ, α, δ)
for some δ ∈ [0, δ0 ]} .
Then we claim that S1 ∩ (B1 ∪ B3 ) = φ. Suppose contrary to our claim that Let (u, λ) ∈ S1 ∩ B1 .
Then θ(u) = 0 by the definition. Then by (3.1), we have
d
d2
F (u) = αλ 2 E(u).
dt
dt
Multiplying (4.2) by u and integrating over [0, 2π], we find by the periodicity of u that
Z 2π
Z 2π
2
u̇1 =
u̇22 = 0,
ü + δα
0
(4.2)
0
which implies u ≡ 0. This contradicts that (u, λ) ∈ B1 ⊂ (H \ {0}) × [−λ0 , λ0 ]. Suppose that
(u, λ) ∈ S1 ∩ B3 . That is we assume that λ = ±λ0 and m < kuk ≤ M. Fix λ = −λ0 . Since
δ − λ0 < 0, from Lemma 3.5 it follows that u = 0, a contradiction.
Suppose now that λ = λ0 .
r
δ + λ0
δ
Since (3.2) holds with λ = λ0 , putting δ̃ = p
and w =
u we have that
δ
+
λ0
θ(u)
(
e 1e
ẅ1 + δε
a(w12 − 1)ẇ1 + e
a2 (w1 + c2 w2 ) = 0,
(4.3)
e 2e
ẅ2 + δε
a(w22 − 1)ẇ2 + e
a2 (c1 w1 + w2 ) = 0,
NORIMICHI HIRANO AND SLAWOMIR RYBICKI†
10
δ + λ0
θ(u) ≤ α and δe = p
≥ δ1 (α). Therefore putting in Lemma 2.3
θ(u)
α = α̃, δ = δ̃, α0 = α we obtain that (4.3) can not hold. Notice that we have just shown that
S1 ∩ (B1 ∪ B3 ) = ∅.
Notice that if (u, λ) ∈ S1 ∩ B2 , then taking into account that θ(u) = 1, we obtain
Notice that α
e = α
p
d
F (w) + α2 Aw = 0.
dt
That is we have a solution of (P) with period 2πα and with ε1 , ε2 replaced with
((δ + λ)ε1 , (δ + λ)ε2 ). We next define a homotopy I : H × [0, 1] → H of S 1 − equivariant compact
mappings by
Z t
Fs (v)dt + a2 θ(v)Lv + αλE(u)
I(u, λ, s) = −π δ0 α
ẅ + (δ + λ)α
0
where
Fs (u) =
ε1 ( 31 u31 − (1 − s)u1 )
!
ε2 ( 31 u32 − (1 − s)u2 )
for (u, s) ∈ H × [0, 1].
It is easy to verify that I is a homotopy of S 1 −equivariant compact mappings. We put
S2 = {(u, λ) ∈ U : I(u, λ, s) = u
for some s ∈ [0, 1]} .
Then one can see that (u, λ) ∈ S2 if and only if
(
ü1 + δ0 ε1 α(u21 − 1 + s)u̇1 + α2 θ(u)(u1 + c2 u2 ) = ε1 αλu̇1 ,
.
ü2 + δ0 ε2 α(u22 − 1 + s)u2 + α2 θ(u)(c1 u1 + u2 ) = ε2 αλu̇2 ,
(4.4)
for some s ∈ [0, 1]. Repeating reasoning given above we can show that S2 ∩ (B1 ∪ B3 ) = φ.
It also follows that if (u, λ) ∈ B2 ∩ S2 , then (P) has a solution with period 2πα. Therefore to
complete the proof it is enough to show that (S1 ∪ S2 ) ∩ B2 6= ∅.
We next show that there is no element (u, λ) ∈ cl(U ) which satisfies that I(u, λ, 1) = u. Suppose
contrary that there exists (u, λ) ∈ cl(U ) satisfying I(u, λ, 1) = u. Repeating
reasoning given in
r
δ0
the proof of Lemma 3.5 we show that λ > 0. Then by putting w =
u we have that,
λ

λα


ẅ
+
ε1 e
a(w12 − 1)ẇ1 + e
a2 (w1 + c2 w2 ) = 0,
 1
α
e
(4.5)

λα

2
2
 ẅ2 +
ε2 e
a(w2 − 1)ẇ2 + e
a (c1 w1 + w2 ) = 0,
α
e
p
p
λα
λ
where α
e = α θ(u). If λ > δ1 (α), then δ =
= p
> δ1 (α). Since α
e = α θ(u) < α,
α̃
θ(u)
2
2
from Lemma
r 2.3 it follows that (4.5) can not hold. On the other hand, if δ0 m > λm(α) , then
δ0
kwk ≥ kuk
> m(α). Then by Lemma 2.2, we have that (4.5) can not hold. Thus we find that
λ
I(u, λ, 1) 6= u for all (u, λ) ∈ cl(U ). We now define a homotopy of mapping Ψ : U × [0, 1] → H
by

 H(u, λ, α, 2sδ0 ) for (u, λ, s) ∈ U × [0, 21 ],
Ψ(u, λ, s) =

I(u, λ, 2s − 1) for (u, λ, s) ∈ U × [ 12 , 1].
VAN DER POL SYSTEM
11
1
1
, µ+
and
√
1 >
α2
1 + c1 c2
Lemma 3.6, we obtain Deg(I − Ψ(·, ·, 0), U ) 6= 0. Now suppose that u 6= Ψ(u, λ, s) for all
u ∈ ∂U and s ∈ [0, 1]. Then by the homotopy invariance of S 1 −degree we find that Deg(I −
Ψ(·, ·, 1), U ) 6= 0. Then by property (1) of Theorem 3, we have that there exists (u, λ) ∈ U
satisfying u = Ψ(u, λ, 1). This contradicts to the observation above. Therefore we have that
there exists (u, λ) ∈ ∂U such that u = Ψ(u, λ, s) for some s ∈ [0, 1). On the other hand, we have,
from the argument above, that there is no solution (u, λ) of problem u = Ψ(u, λ, s) on B1 ∪ B3
for all s ∈ [0, 1]. Thus we obtain that there exists a solution (u, λ) of problem u = Ψ(u, λ, s) on
B2 for some s ∈ [0, 1]. Then by the argument above, we obtain a solution of (P) with period
2πα.
Then since Ψ(u, λ, 0) = H(u, λ, α, 0) = Lu + αλE(u). Since α > p
1 1
+
,
, where µ = µ−
Proof of Theorem 1.2. We choose
∈
1 when c1 c2 < 1 and µ = µ2
µ
µ+
1
when c1 c2 > 1. Then by Lemma 2.4 there exists δ2 (α0 ) > 0 such that the problem (2.7) with
α = α0 has no solution for δ ∈ (0, δ2 (α0 )]. We will show that for any δ ∈ (0, δ2 (α0 )), the problem
(2.7) has a solution for some α ∈ (0, α0 ). Suppose contrary that there exists δ1 ∈ (0, δ2 (α0 )) such
that problem (2.7) has no solution with δ = δ1 for any α ∈ (0, α0 ). Then we claim that there
exists ρ ∈ (0, δ1 ) such that δ1 + ρ < δ2 (a0 ) and for any δ ∈ [δ1 − ρ, δ1 + ρ], problem (2.7) has no
+
solution for any α ∈ (0, α0 ). Suppose that there exists a sequence {(un , δn , αn )} ∈ H×R ×(0, α0 )
such that lim δn = δ1 and each un is a solution of problem (2.7) with δ = δn and α = αn . By
α02
n→∞
Lemma 2.2 sequence {un } is bounded in H. Therefore there exists a subsequence {uni } of {un }
such that uni * u ∈ H weakly as i → ∞ and lim αni = α ∈ (0, α]. One can see from (2.8) with
i→∞
u = (u1 , u2 ) replaced by un = (u1n , u2,n ) that either | u1n (t) |≥ 1 | or | u2n (t) |≥ 1 for some
t ∈ [0, 2π]. That is u is a nontrivial
solution od (2.7) with δ = δ1 , a contradiction.
r
ρ
> m(α0 ). Put
Choose M > 0 such that M
ρ + δ1
U = {v ∈ H : m < kvk < M } × (−δ1 , δ1 )
B1 = {v ∈ H : kvk = m} × [−δ1 , δ1 ],
B2 = {v ∈ H : kvk = M } × [−δ1 , δ1 ],
B3 = {v ∈ H : m < kvk < M } × {−δ1 , δ1 } .
We first see that there is no nontrivial solution u of problem u = H(u, λ, α0 , (1 − s)ρ) for any
s ∈ [0, 1]. That is there is no nontrivial solution of
ü + (1 − s)ρα0
d2
d
F (u) + α02 Au = λα0 2 E(u)
dt
dt
(4.6)
Suppose that u ∈ U is a solution of problem (4.6). If s = 1, then u is a solution of
ü − λα0
d2
E(u) + α02 Au = 0
dt2
(4.7)
Then multiplying (4.7) by u̇ and integrating over [0, 2π], we obtain u = 0, a contradiction.
Suppose now that s < 1. In case (1 − s)ρ + λ ≤ 0, problem (4.6) has no nontrivial solution by
12
NORIMICHI HIRANO AND SLAWOMIR RYBICKI†
s
Lemma 3.5. If (1 − s)ρ + λ > 0, then w =
(1 − s)ρ
u is a solution of
(1 − s)ρ + λ
d
F (w) + α02 w = 0,
dt
where δ = λ + (1 − s)ρ. Since δ ≤ δ1 + ρ < δ2 (α0 ), this contradicts to the definition of δ2 (α0 ).
e :
Thus we find that (4.6) has no solution in U. Define a homotopy of compact mappings H
U × [0, 1] → H by
ẅ + δα0
e
H(u,
λ, s) = H(u, λ, α0 ((1 − s) + sθ(u)), (1 − s)ρ),
for (u, λ) ∈ U
and s ∈ [0, 1],
e
where θ : H → [0, 1] is given by (2.4). One can see that H(u,
λ, s) = H(u, λ, α0 , ρ) for (u, λ) ∈ U.
e
We will see that there exists no solution u = H(u, λ, s) on ∂U for any s ∈ [0, 1]. Suppose,
e
contrary to our claim that u = H(u,
λ, s) for some u ∈ ∂U and s ∈ [0, 1]. Suppose that u ∈ B1 .
Then since θ(u) = 1, u satisfies (4.6), a s
contradiction. We next suppose that u ∈ B2 . Then
putting α = α0 ((1 − s) + sθ(u)) and w =
(1 − s)ρ
u we have that w is a solution of
(1 − s)ρ + λ
d
F (w) + α2 w = 0
dt
where δ = (1 − s)ρ + λ. Since kuk = M, we have that
s
s
(1 − s)ρ
(1 − s)ρ
kuk =
> m(α0 ).
kwk =
(1 − s)ρ + λ
(1 − s)ρ + λ
ẅ + δα
(4.8)
On the other hand, we have by Lemma 2.2 that kwk < m(α) ≤ m(α0 ), a contradiction. Finally
suppose that u ∈ B3 . If λ = −δ1 , then we reach a contradiction by Lemma 3.5. Suppose that λ =
δ1 . Then δ = (1 − s)ρ + δ1 ∈ [δ1 − ρ, δ1 + ρ]. Therefore by the assumption, (4.8) has no nontrivial
solution. Summing up, we have shown that u ∈
/ ∂U. Therefore by the homotopy invariance of
1
e ·, 0), U ) = Deg(I − H(·,
e ·, 1), U ).
degree for S −equivariant maps we obtain that Deg(I − H(·,
e
e
Since I − H(·, ·, 0) 6= 0 on U by the same argument we have that Deg(I − H(·, ·, 0), U ) =
e ·, 1) = H(·, ·, α0 , 0), by Lemma 3.6 we obtain that
0. On the other hand, noting that H(·,
e ·, 1), U ) 6= 0, a contradiction. Which completes the proof.
Deg(I − H(·,
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Department of Mathematics, Graduate School of Environment and Information Sciences, Yokohama National University, 156 Tokiwadai, Hodogayaku, Yokohama 240, Japan
E-mail address: [email protected]
Faculty of Mathematics and Computer Science, Nicolaus Copernicus University, PL-87-100
Toruń, ul. Chopina 12/18, Poland
E-mail address: [email protected]