A. G A S U L L and R. P R O H E N S (Barcelona)
EFFECTIVE COMPUTATION
OF THE FIRST LYAPUNOV QUANTITIES FOR A PLANAR DIFFERENTIAL EQUATION
Abstract. We take advantage of the complex structure to compute in a short way and without using any computer algebra system the Lyapunov quantities V 3 and V 5 for a general smooth planar system.
1. Introduction. Consider the differential equation ( ˙x, ˙y) = (f (x, y), g(x, y)), (x, y) ∈ R 2 , in the plane where f and g are analytic functions satisfying f (0, 0) = g(0, 0) = 0. It is well known that when the origin is a non-hyperbolic critical point of focus type the study of its stability can be reduced to the computation of the so called Lyapunov quantities, V 2k+1 , k = 1, 2, . . . ; see [ALGM] for more details. By making a linear change of coordinates and a rescaling of the time variable if necessary, the planar differential equation can be written as
(1) ˙z = F (z, z) = iz +
X ∞ k=2
F k (z, z),
where z = x + iy = Re(z) + i Im(z), and F k is a complex homogeneous polynomial of degree k.
In this paper we make some modifications in the standard techniques ex- plained in [ALGM] to obtain the Lyapunov quantities. These modifications simplify their effective computation. The main idea is to keep the complex structure of (1) during all the process.
In Section 2 we give some preliminary results and in Section 3 we prove:
1991 Mathematics Subject Classification: 34C25, 58F14.
Key words and phrases : stability, weak focus, Lyapunov quantities.
Research of the first author partially supported by a DGICYT grant, number PB93- 0860.
[243]
Theorem A. Consider the differential equation (1). Set F 2 (z, z) = Az 2 + Bzz + Cz 2 ,
F 3 (z, z) = Dz 3 + Ez 2 z + F zz 2 + Gz 3 ,
F 4 (z, z) = Hz 4 + Iz 3 z + Jz 2 z 2 + Kzz 3 + Lz 4 ,
F 5 (z, z) = M z 5 + N z 4 z + Oz 3 z 2 + P z 2 z 3 + Qzz 4 + Rz 5 . Then the first Lyapunov quantities of (1) are:
V 3 = 2π[Re(E) − Im(AB)], (i)
V 5 = π
3 [6 Re(O) + Im(3E 2 − 6DF + 6AI (ii)
− 12BI − 6BJ − 8CH − 2CK)
+ Re(−8CCE + 4ACF + 6ABF + 6BCF − 12B 2 D − 4ACD
− 6ABD + 10BCD + 4ACG + 2BCG) + Im(6AB 2 C + 3A 2 B 2 − 4A 2 BC + 4B 3 C)].
The above result already appears in [CGMM, FLLL, G, GW, HW], but the proof that we present is shorter and does not use any computer algebra system.
2. Preliminary results. We briefly recall the definition of the Lya- punov constants.
In the (r, θ)-polar coordinates zz = r 2 , θ = arctan Im(z) Re(z) , (1) is converted
into dr
dθ = Re[zF (z, z)]/r Im[zF (z, z)]/r 2
z=reiθ
, or equivalently, for r small enough,
(2) dr
dθ =
P ∞
k=2 r k Re(S k (θ)) 1 + P ∞
k=2 r k−1 Im(S k (θ)) = X ∞ k=2
R k (θ)r k ,
where S k (θ) = zF k (z, z)| z=eiθ = e −iθ F k (e iθ , e −iθ ), R 2 (θ) = Re(S 2 (θ)) and (3) R k (θ) = Re(S k (θ)) −
k−2 X
j=1
R k−j (θ) Im(S j+1 (θ)) for k ≥ 3.
Denote by r(θ, s) the solution of (2) which takes the value s at θ = 0.
Consider
(4) r(θ, s) − s = X ∞ k=2
u k (θ)s k , where u k (0) = 0 for k ≥ 2.
Then the stability of the origin of (1) is given by the sign of the first non-zero
value u k (2π). It is well known that the corresponding k is odd (see [ALGM,
p. 243]).
Assume that u k (2π) = 0 for k = 1, . . . , 2m and u 2m+1 (2π) 6= 0. Then the mth Lyapunov quantity is defined by V 2m+1 = u 2m+1 (2π).
The next result is inspired by [AL] and it allows us to compute the first values u k (2π). In the sequel, we use the notation e f = e f (θ) = (f ) ∼ (θ) =
T
θ
0 f (s) ds.
Proposition 1. Given (2), the functions u i (θ), i = 2, 3, 4, 5, involved in its solution (4) are
u 2 (θ) = e R 2 ,
u 3 (θ) = ( e R 2 ) 2 + e R 3 ,
u 4 (θ) = ( e R 2 ) 3 + 2 e R 2 R e 3 + g R e 2 R 3 + e R 4 ,
u 5 (θ) = ( e R 2 ) 4 + 3( e R 2 ) 2 R e 3 + ( e R g 2 ) 2 R 3 + 2 e R 2 R e g 2 R 3
+ 3 2 ( e R 3 ) 2 + 2 e R 2 R e 4 + 2 g R 4 R e 2 + e R 5 . P r o o f. Direct substitution gives
X ∞ k=2
R k (θ)[r(θ, s)] k = X ∞ k=2
u ′ k (θ)s k .
By using the expression for a power series raised to some power (see [GR], for instance), whenever k ≥ 2, we have
u ′ k (θ) = X k m=2
R m (θ) X
M
m
a 1 . . . a k−1
u a 22(θ)u a 33(θ) . . . u a k−1k−1(θ)
(θ) . . . u a k−1k−1(θ)
, where M = {(a 1 , . . . , a k−1 ) ∈ N k−1 : a 1 + . . . + a k−1 = m, a 1 + . . . + (k − 1)a k−1 = k}. Then the proof follows from judicious integration. As an example we prove the expression for u 4 (θ). By using the previous formula we have
u 4 (θ) =
θ
\
0
(R 2 (Ψ )(2u 3 (Ψ ) + u 2 2 (Ψ )) + R 3 (Ψ )3u 2 (Ψ ) + R 4 (Ψ )) dΨ.
We obtain the desired result from the last expression, by substituting the values of u 2 (Ψ ) and u 3 (Ψ ) and integrating, as follows:
u 4 (θ) =
θ
\
0
[R 2 (Ψ )(2 e R 3 (Ψ ) + 3( e R 2 (Ψ )) 2 ) + R 3 (Ψ )3 e R 2 (Ψ ) + R 4 (Ψ )] dΨ
= ( e R 2 (θ)) 3 + 2
θ
\
0
[ e R 3 (Ψ )R 2 (Ψ ) + e R 2 (Ψ )R 3 (Ψ )] dΨ +
θ
\
0
R e 2 (Ψ )R 3 (Ψ ) dΨ + e R 4 (θ)
= ( e R 2 ) 3 + 2 e R 2 R e 3 + g R e 2 R 3 + e R 4 .
Corollary 2. The first Lyapunov quantities of (1) are V 3 = e R 3 (2π),
V 5 = ( R 3 g ( e R 2 ) 2 + 2 g R 4 R e 2 + e R 5 )(2π), where the functions R i (θ) are defined by
R 2 = Re S 2 ,
R 3 = Re S 3 − Re S 2 Im S 2 ,
R 4 = Re S 4 − Re S 3 Im S 2 + Re S 2 (Im S 2 ) 2 − Re S 2 Im S 3 , R 5 = Re S 5 − Re S 4 Im S 2 − Re S 2 Im S 4 + 2 Re S 2 Im S 2 Im S 3
− Re S 3 Im S 3 + Re S 3 (Im S 2 ) 2 − Re S 2 (Im S 2 ) 3 , and S k (θ) = e −iθ F k (e iθ , e −iθ ).
P r o o f. From the fact that u 2 (2π) = 0 and using Proposition 1, we have e R 2 (2π) = 0. Hence, the result on V 3 follows by using Proposition 1.
Assuming that V 3 = 0, we get u 4 (2π) = 0 and from Proposition 1, again, we get the desired result on V 5 . On the other hand, the expression of R k
when k = 2, 3, 4 and 5 follows directly from (3). As an example we prove the expression for R 4 . From (3) we have
R 4 = Re S 4 − X 2 j=1
R 4−j (θ) Im S j +1 (θ)
= Re S 4 − (Re S 3 − Re S 2 Im S 2 ) Im S 2 − Re S 2 Im S 3 , which gives the expected value of R 4 .
We now recall the following formulas that will be frequently used in the sequel:
(5)
2 Re α Re β = Re[αβ + αβ], 2 Im α Im β = Re[−αβ + αβ],
2 Re α Im β = Im[αβ + αβ], α, β ∈ C.
3. Proof of Theorem A. Firstly we will express the Lyapunov quan- tities of (1) in terms of the trigonometric polynomials S k .
Proposition 3. The first two Lyapunov quantities of system (1) are V 3 = Re
2π
\0
S 3 (Ψ ) dΨ − 1 2 Im
2π
\0
S 2 2 (Ψ ) dΨ,
V 5 = Re
2π
\0
S 5 (Ψ ) dΨ
− Im
2π
\
0
T 2 (Ψ )(S 4 (Ψ ) + S 4 (Ψ )) + S 2 (Ψ )S 4 (Ψ ) + 1 2 S 3 2 (Ψ ) dΨ
+ 1 4 Re
2π
\
0
S 3 (Ψ )[(S 2 (Ψ ) + S 2 (Ψ )) 2 − (T 2 (Ψ ) − T 2 (Ψ ) + 2S 2 (Ψ )) 2 ] dΨ
+ 1 8 Im
2π
\0
S 2 2 (Ψ )[T 2 (Ψ ) − T 2 (Ψ ) + S 2 (Ψ ) − S 2 (Ψ )] 2 dΨ, where S k (ψ) = e −iψ F k (e iψ , e −iψ ) and T 2 (Ψ ) = −i[ e S 2 (Ψ ) − 2π 1
T
2π
0 S e 2 (θ)dθ]
P r o o f. By using Corollary 2 and formulas (5) we get the expression for V 3 .
To obtain V 5 we recall that by Corollary 2,
V 5 = (R 3 ( e R 2 ) 2 + 2R 4 R e 2 + R 5 ) ∼ (2π).
In order to simplify the calculations of V 5 we define, for any real number v, V 5 (v) = (R 3 ( e R 2 + v) 2 + 2R 4 ( e R 2 + v) + R 5 ) ∼ (2π).
By using the fact that V 3 = 0 ( e R 3 (2π) = 0) and also that u 4 (2π) = 0 (( e R 4 + g R e 2 R 3 )(2π) = 0), it turns out that V 5 (v) ≡ V 5 . Therefore we can choose any v for computing V 5 . We choose it such that
R e 2 + v = Re( e S 2 + v) = Re(iT 2 ) = − Im(T 2 ).
Hence V 5 =
2π
\
0
(R 3 (θ)(Im(T 2 (θ))) 2 − 2R 4 (θ) Im(T 2 (θ)) + R 5 (θ)) dθ.
To get a more suitable expression for the integrated function we again use Corollary 2, obtaining
(Re S 3 − Re S 2 Im S 2 )(Im T 2 ) 2
− 2(Re S 4 − Re S 3 Im S 2 + Re S 2 (Im S 2 ) 2 − Re S 2 Im S 3 ) Im T 2
+ Re S 5 − Re S 4 Im S 2 − Re S 2 Im S 4 + 2 Re S 2 Im S 2 Im S 3
− Re S 3 Im S 3 + Re S 3 (Im S 2 ) 2 − Re S 2 (Im S 2 ) 3 .
Collecting terms taking into account the number of factors they have, we get
Re S 5 − 2 Re S 4 Im T 2 − Re S 4 Im S 2 − Re S 2 Im S 4 − Re S 3 Im S 3
+ Re S 3 (Im T 2 ) 2 + 2 Re S 3 Im S 2 Im T 2 + 2 Re S 2 Im S 3 Im T 2
+ 2 Re S 2 Im S 2 Im S 3 + Re S 3 (Im S 2 ) 2
− Re S 2 Im S 2 (Im T 2 ) 2 − 2 Re S 2 (Im S 2 ) 2 Im T 2 − Re S 2 (Im S 2 ) 3 .
Afterwards we will apply iteratively the formulas (5) to arrive at the final expression of V 5 .
Firstly we consider the terms with one, two and three factors. The unique term with exactly one factor is Re S 5 , and its integral appears in the expression of V 5 . With exactly two factors we have
−2 Re S 4 Im T 2 − Re S 4 Im S 2 − Re S 2 Im S 4 − Re S 3 Im S 3 . The use of formulas (5) gives
− Im
T 2 (S 4 + S 4 ) + S 2 S 4 + 1 2 S 3 2 , which is the result that appears in the expression of V 5 .
We have the following terms with exactly three factors:
Re S 3 (Im T 2 ) 2 + 2 Re S 3 Im S 2 Im T 2 + 2 Re S 2 Im S 3 Im T 2
+ 2 Re S 2 Im S 2 Im S 3 + Re S 3 (Im S 2 ) 2 . Transforming this expression term after term by applying formulas (5), we have
Re S 3 (Im T 2 ) 2 = − 1 4 Re(S 3 (T 2 − T 2 ) 2 ),
2 Re S 3 Im S 2 Im T 2 + 2 Re S 2 Im S 3 Im T 2 = − Re(S 2 S 3 (T 2 − T 2 )), 2 Re S 2 Im S 2 Im S 3 + Re S 3 (Im S 2 ) 2 = 1 4 Re(S 3 [(S 2 + S 2 ) 2 − 4S 2 2 ]).
Integrating the sum of the last three expressions we obtain the corresponding term that appears in V 5 .
The computations involving the terms with four factors are tedious but straightforward and we omit them.
As a consequence of the previous proposition we can prove our main result.
P r o o f o f T h e o r e m A. If we express S 2 (θ), S 3 (θ), S 4 (θ), S 5 (θ) and T 2 (θ) in terms of the coefficients of the differential equation we get
S 2 (θ) = Ae iθ + Be −iθ + Ce − 3iθ , S 3 (θ) = De 2iθ + E + F e − 2iθ + Ge − 4iθ ,
S 4 (θ) = He 3iθ + Ie iθ + Je −iθ + Ke − 3iθ + Le − 5iθ ,
S 5 (θ) = M e 4iθ + N e 2iθ + O + P e − 2iθ + Qe − 4iθ + Re − 6iθ , T 2 (θ) = −Ae iθ + Be − iθ + C
3 e − 3iθ .
To compute V 3 , from Proposition 3, we need to calculate
Re
2π
\