On The Maximum Number Of Limit Cycles Of Generalized Polynomial Liénard Di¤erential Systems Via Averaging Theory

On The Maximum Number Of Limit Cycles Of Generalized Polynomial Liénard Di¤erential Systems Via Averaging Theory

Amel Boulfoul

, Nawal Mellahi

Received 31 March 2019

Abstract

In this paper, we apply the averaging theory of …rst and second order for studying the limit cycles of generalized polynomial Liénard systems of the form

_

x=y l(x)y; y_= x f(x) g(x)y h(x)y² p(x)y³;

wherel(x) = l1(x) + ²l2(x); f(x) = f1(x) + ²f2(x); g(x) = g1(x) + ²g2(x); h(x) = h1(x) + ²h2(x) andp(x) = p1(x) + ²p2(x)wherelk(x) has degreem,fk(x),gk(x),hk(x)andpk(x) have degreenfor eachk= 1;2;and is a small parameter.

1 Introduction

Polynomial Liénard systems occur as models or at least as simpli…cations of models in many domains in science. According to Smale [16], they are also a good starting point to try to solve the second part of Hilbert’s 16th problem, which asks for the maximal number of limit cycles that a polynomial planar vector

…elds can have, depending on the degree of the system [8]. Recall that a limit cycle of a planar polynomial di¤erential system is a periodic orbit of the system isolated in the set of all periodic orbits of the system.

The notion of a center of real planar polynomial di¤erential system is an isolated equilibrium point having a neighborhood such that all the orbits of this neighborhood are periodic orbits with the unique exception of the singular point, which is de…ned by Poincaré in [14]. A classical way of producing limit cycles is by perturbing a system which has a center. The techniques used for studying the limit cycles that can bifurcate from the periodic orbits of a center are: Poincaré return map [2], Abelian integrals or Melnikov integrals (note that for systems in the plane the two notions are equivalent) [7], inverse integrating factor [5] and averaging theory ([3,15]). In the plane at same order all these techniques produce the same results, but the computations can change with the di¤erent technique.

In this paper, using averaging theory we want to study the number of limit cycles which bifurcate from the periodic orbits of the linear centerx_ =y;y_= xof the more generalized Liénard polynomial di¤erential

system (

_

x=y l(x)y;

_

y= x f(x) g(x)y h(x)y² p(x)y³; (1)

where l(x) = l1(x) + ²l2(x); f(x) = f1(x) + ²f2(x); g(x) = g1(x) + ²g2(x); h(x) = h1(x) + ²h2(x) and p(x) = p1(x) + ²p2(x) where lk(x) has degree m, fk(x), gk(x), hk(x) and pk(x) have degree n for eachk= 1;2;and is a small parameter. Note that system (1) was been studied in [4] whenl(x) = 0 and p(x) = 0.

There are some results concerning the maximum number of limit cycles bifurcating from the linear center _

x=y;y_= xof generalized polynomial Liénard di¤erential systems using averaging theory.

Mathematics Sub ject Classi…cations: 34C29, 34C25, 47H11.

yDepartment of Mathematics, 20 august 1955 University, El Hadaiek 21000, Skikda, Algeria

zDepartment of Applied Mathematics, 20 august 1955 University, El Hadaiek 21000, Skikda, Algeria

167

In [10], Llibre et al. studied how many limit cycles of system x_ =y;y_ = g(x) f(x)y can bifurcate from the periodic orbits of the linear center x_ =y;y_ = xusing the averaging theory of orderk for k= 1;2;3;where f(x)andg(x)are polynomials in the variable xof degrees nandmrespectively.

In [11], the authors studied using the averaging theory of …rst and second order the following system (x_ =y (g₁₁(x) +f₁₁(x)y) ²(g₁₂(x) +f₁₂(x)y);

_

y= x (g21(x) +f21(x)y) ²(g22(x) +f22(x)y); (2) whereg1i; f1i; g2i; f2ihave degreel; k; mandnrespectively for eachi= 1;2;and is a small parameter.

In [1], Alavez-Ramirez et al. studied the polynomial di¤erential system (x_ =y g₁₁(x) ²g₁₂(x);

_

y= x (g21(x) +f21(x)y) ²(g22(x) +f22(x)y); (3) where g1i; g2i; f2i have degreel; mandnrespectively for eachi= 1;2;and is a small parameter.

In [12,13], the authors studied using the averaging theory of …rst second and third order the number of limit cycles of the polynomial di¤erential system

(x_ =y f₁(x)y;

_

y= x g2(x) f2(x)y; (4)

where f₁(x) = f₁₁(x) + ²f₁₂(x) + ³f₁₃(x); g₂(x) = g₂₁(x) + ²g₂₂(x) + ³g₂₃(x) and f₂(x) = f₂₁(x) + ²f₂₂(x) + ³f₂₃(x) where f_1i, f_2i and g_2i have degree l; n and m respectively for each i= 1;2;3;and is a small parameter.

In what follows we present our main results.

Theorem 1 Forj j>0su¢ ciently small, the maximum number of limit cycles of the generalized polynomial Liénard di¤ erential systems (1) bifurcating from the periodic orbits of linear centerx_ =y;y_= xusing the averaging theory of …rst order is

1= max n+ 2

2 :

The proof of Theorem1 is given in Section 3.

Theorem 2 Forj j>0su¢ ciently small, the maximum number of limit cycles of the generalized polynomial Liénard di¤ erential systems (1) bifurcating from the periodic orbits of linear centerx_ =y;y_= xusing the averaging theory of second order is

2= max n+ 1; n 1

2 + m 1

2 + 2 : The proof of Theorem2 is given in Section 4.

In [4] it has been shown that there exist generalized Liénard systems (1) withl(x) =p(x) = 0, having at least nlimit cycles. The result in Theorem2 improves this lower estimate ( ₂> n): Note that we do not know if the upper bound ₁ (respectively ₂) for the number of limit cycles of the polynomial di¤erential system (1), which bifurcate from the periodic solutions of the linear centerx_ =y;y_= xusing the averaging theory of …rst order (respectively second order) is reached. In Section 5 we prove that the upper bound is reached for 1 n 9 in the …rst order averaging theory and for1 n 4;1 m 4 in the second order averaging theory (see Examples1and2).

In Section 2, we introduce the averaging theory of …rst and second order.

2 Averaging Theory of First and Second Order

We consider the di¤erential system _

x= F₁(t; x) + ²F₂(t; x) + ³R(t; x; ); (5) where F₁; F₂ : R D ! R; R : R D ( _f; _f) ! R are continuous functions, T-periodic in the …rst variable, andD is an open subset ofRⁿ:Assume that the following hypotheses (i), (ii) hold.

(i) F1(t; :)2C²(D); F2(t; :)2C¹(D) :for all :t2R;:F1; F2; R are locally Lipschitz with respect tox, and R is twice di¤erentiable with respect to .

We de…neF_k0:D!Rfork= 1;2as

F₁₀(z) = 1 T

Z T 0

F₁(s; z)ds;

F₂₀(z) = 1 T

Z T 0

[D_zF₁(s; z)y₁(s; z) +F₂(s; z)]ds;

where

y1(s; z) = Z s

0

F1(t; z)dt:

(ii) For V D an open and bounded set and for each 2 ( f; f)nf0g; there exists a 2 V sach that F10(a) + F20(a) = 0anddB(F10+ F20; V; a)6= 0. The expressiondB(F10+ F20; V; a)6= 0means that the Brouwer degree (see [3]) of the function F10+ F20 : V ! Rat the …xed point a is not zero. A su¢ cient condition for the inequality to be true is that the Jacobian of the functionF10+ F20atais not zero.

Then, forj j>0su¢ ciently small there exists aT-periodic solution'(:; )of the equation (5) such that '(0; )!awhen !0.

- IfF10is not identically zero, then the zeros ofF10+ F20are mainly the zeros ofF10for su¢ ciently small.

In this case the previous result provides the averaging theory of …rst order.

- IfF10 is identically zero andF20 is not identically zero, then the zeros ofF10+ F20 are mainly the zeros ofF20 for su¢ ciently small. In this case the previous result provides the averaging theory of second order.

For a general introduction to averaging theory see ([9,17]).

3 Proof of Theorem 1

We shall need the …rst-order averaging theory to prove Theorem1. We write system (1) in polar coordinates (r; )wherex=rcos ; y=rsin ; r >0:If we take

l1(x) = Xm i=0

ei;1xⁱ, f1(x) = Xn i=0

ai;1xⁱ, g1(x) = Xn i=0

bi;1xⁱ, h1(x) = Xn i=0

ci;1xⁱ, p1(x) = Xn i=0

di;1xⁱ;

the system (1) becomes 8>

>>

<

>>

>: _ r=

Pn i=0

Ai;1( ; r)rⁱ+ Pm i=0

ei;1Ri+1( )rⁱ⁺¹ +O( ²);

_ = 1 _r( Pn i=0

D_i;1( ; r)rⁱ Pm i=0

e_i;1T_i( )rⁱ⁺¹) +O( ²);

(6)

where

Ai;1(r; ) = ai;1Ri( ) +bi;1Ti( )r+ci;1Si( )r²+di;1Ui( )r³; Rj( ) = cos^j sin ;

T_j( ) = cos^j sin² =cos^j cos^j+2 ;

Sj( ) = cos^j sin³ =cos^j sin cos^j+2 sin ; U_j( ) = cos^j sin⁴ =cos^j 2cos^j+2 +cos^j+4 ;

Di;1(r; ) = ai;1cosⁱ⁺¹ +bi;1Ri+1( )r+ci;1Ti+1( )r²+di;1Si+1( )r³: Now taking as the new independent variable, system (6) becomes

dr d =r_

_ = F1(r; ) +O(²); (7)

where

F1(r; ) = Xn i=0

Ai;1( ; r)rⁱ+ Xm i=0

ei;1Ri+1( )rⁱ⁺¹: Therefore, from Section 2 we must study the simple positive zeros of the function

F10(r) = 1 2

Xn i=0

rⁱ(ai;1Ji(2 ) +bi;1rI~i(2 ) +ci;1r²J~i(2 ) +di;1r³I~~i(2 ))

+ Xm i=0

ei;1Ji+1(2 )rⁱ⁺¹

!

;

where

Jk(2 ) = Z 2

0

Rk( )d ; I~k(2 ) = Z 2

0

Tk( )d ; J~k(2 ) = Z 2

0

Sk( )d ; I~~k(2 ) = Z 2

0

Uk( )d : To calculate the exact expression ofF10(r), we use the following integrals (see for more details [6]):

Ji( ) = Z

0

cosⁱtsintdt= 1

i+ 1 1 cosⁱ⁺¹ ; (8)

J~i( ) = Z

0

cosⁱtsin³tdt= 2 (i+ 1) (i+ 3)

1

i+ 1cosⁱ⁺¹ + 1

i+ 3cosⁱ⁺³ ; (9)

Ji(2 ) = ~Ji(2 ) = 0; (10)

Ii( ) = Z

0

cosⁱtdt= 8>

><

>>

: Pk

l=0 k;lsin(2l+ 1) ifi= 2k+ 1;

k+ Pk

l=1 k;lsin(2l ) ifi= 2k;

(11)

where

i= 1

2²ⁱ 2i

i , _k;l= 1 2²ⁱ

2i+ 1 i l

1

2l+ 1, _k;l= 2i i+ 1

1 l:

I_i(2 ) = 0 ifi= 2k+ 1;

k

2^{k 1}k! ifi= 2k;

where _k= 3 5 (2k 1); _k+1= (2k+ 1) _k. And I2k+2(2 ) = 2k+ 1

2k+ 2I2k(2 ); (12)

I~i(2 ) =Ii(2 ) Ii+2(2 ); (13)

~~

Ii(2 ) =Ii(2 ) 2Ii+2(2 ) +Ii+2(2 ): (14) And thus we have

F₁₀(r) =r

[ⁿ₂]

X

i=0

r^{2i i}

2ⁱ⁺¹(i+ 1)! b_2i;1+d_2i;1r² 3

2(i+ 2) : (15)

From (15) the functionF₁₀(r)has at most[ⁿ⁺²₂ ]simple positive zeros.

4 Proof of Theorem 2

For proving Theorem2we shall use the second order averaging theory. If we write lk(x) =

Xm i=0

ei;kxⁱ; fk(x) = Xn i=0

ai;kxⁱ; gk(x) = Xn i=0

bi;kxⁱ; hk(x) = Xn i=0

ci;kxⁱ; pk(x) = Xn i=0

di;kxⁱ;

for eachk= 1;2:Then system (1) in polar coordinates(r; ); r >0becomes 8>

>>

>>

><

>>

>>

>>

: _

r = (Pn

i=0A_i;1( ; r)rⁱ+Pm

i=0e_i;1R_i+1( )rⁱ⁺¹) ²(Pn

i=0A_i;2( ; r)rⁱ +Pm

i=0ei;2Ri+1( )rⁱ⁺¹) +O( ³);

_ = 1 _r(Pn

i=0D_i;1( ; r)rⁱ Pm

i=0e_i;1T_i( )rⁱ⁺¹) _r²(Pn

i=0D_i;2( ; r)rⁱ Pm

i=0ei;2Ti( )rⁱ⁺¹) +O( ³);

(16)

where

Ai;j( ; r) =ai;jRi( ) +bi;jTi( )r+ci;jSi( )r²+di;jUi( )r³; j= 1;2;

D_i;j( ; r) =a_i;jcosⁱ⁺¹ +b_i;jR_i+1( )r+c_i;jT_i+1( )r²+d_i;jS_i+1( )r³; j= 1;2:

Taking as the new independent variable system (16) writes dr

d = r_

= F1(r; ) + ²F2(r; ) +O( ³); (17) where

F1(r; ) = Xn i=0

Ai;1( ; r)rⁱ+ Xm i=0

ei;1Ri+1( )rⁱ⁺¹; (18)

F2(r; ) = Xn i=0

Ai;2( ; r)rⁱ+ Xm i=0

ei;2Ri+1( )rⁱ⁺¹ 1

r Xn i=0

A_i;1( ; r)rⁱ+ Xm i=0

e_i;1R_i+1( )rⁱ⁺¹

! _n X

i=0

D_i;1( ; r)rⁱ Xm i=0

e_i;1T_i( )rⁱ⁺¹

! : (19)

In order to apply the averaging theory of second order, F₁₀(r) must be identically zero. From (15), F₁₀ is identically zero if and only if 8

>>

<

>>

:

b0=d_2[ⁿ

2] = 0;

b2i =_2i³₁d2i 2; 1 i [ⁿ₂];

b2i =_2i³₁d2i 2; 1 i [ⁿ₂]:

(20)

In the next Proposition we obtain the value ofF1(r; )whenF10(r) 0:

Proposition 1 If F10(r) 0; then

F1(r; ) = Xn i=0

rⁱ ai;1Ri( ) +ci;1r²Si( ) +

[Xⁿ₂¹] i=0

r²ⁱ⁺² b2i+1;1T2i+1( ) +d2i+1;1r²U2i+1( )

+ Xm i=0

ei;1rⁱ⁺¹Ri+1( ) +

[ⁿ₂]

X

i=0

d2i 2r²ⁱ⁺¹ T2i 2

2i+ 2

2i 1T2i( ) : (21)

Proof. Since F10(r) 0;we have the relation (20). If the relationsUk( ) =Tk( ) Tk+2( ) and (20) are substituted in the expression (18) we obtain (21).

Now we continue our proof for Theorem2. If we derive the expression of (21) with respect tor, we obtain dF₁(r; )

dr =

Xn i=0

r^{i 1}(ia_i;1R_i( ) + (i+ 2)c_i;1r²S_i( ))

+

[Xⁿ₂¹] i=0

r²ⁱ⁺¹ (2i+ 2)b2i+1;1T2i+1( ) + (2i+ 4)d2i+1;1r²U2i+1( ) +

Xm i=0

e_i;1(i+ 1)rⁱR_i+1( )

+

[ⁿ₂]

X

i=0

(2i+ 1)d2i 2r²ⁱ T2i 2

2i+ 2

2i 1T2i( ) : (22)

In the next Proposition we obtainy(r ) =R

0 F(r; s)ds.

Proposition 2 We have

y(r; ) = Xn i=0

rⁱ a_i;1J_i( ) +c_i;1r²J~_i( ) +

[ⁿ₂¹]

X

i=0

r²ⁱ⁺² b_2i+1;1 Xi+1 l=0

~_i;lsin(2l+ 1)

+d2i+1;1

Xi+2 l=0

~~_i;lsin(2l+ 1) r²

! +

Xm i=0

ei;1rⁱ⁺¹Ji+1( )

+

[ⁿ₂]

X

i=0

d_{2i 2}r²ⁱ⁺¹ Xi+1

l=0

~i;lsin(2l ); (23)

where

~_i;l= (

i;l i+1;l;: 0 l i;

i+1;i+1;: l=i+ 1: ; ~~_i;l= 8>

<

>:

i;l 2 _i+1;l+ _i+2;l;: 0 l i;

2 _i+1;i+1+ _i+2;i+1;: l=i+ 1;

i+2;i+2;: l=i+ 2;

~i;l= 8>

>>

<

>>

>:

i 1;l 4i+1

2i 1 i;l+²ⁱ⁺²_{2i 1 i+1;l}; 0 l i 1;

4i+1

2i 1 i;i+²ⁱ⁺²_{2i 1 i+1;i}; l=i;

2i+2

2i 1 i+1;i+1; l=i+ 1:

Proof. Simplifying by using the relationsI~k( ) =Ik( ) Ik+2( )and I~~k( ) =Ik( ) 2Ik+2( ) +Ik+4( ).

Taking into account that

i 1

4i+ 1

2i 1 ⁱ +2i+ 2

2i 1 ⁱ⁺¹ = 0;

and using the integrals (8), (9) and (11), we obtain (23). Thus, the Proposition ‡lows.

Now we determine the corresponding function F₂₀(r) = 1

2 (II(r) +III(r)); (24)

where

II(r) = Z 2

0

@F1(r; )

@r :y(r; )d and III(r) = Z 2

0

F2(r; )d : First we shall compute the integralsII(r). Using the expressions (22) and (23), We have

II(r) = Xn i=0

Xn i=0

r^{i+j 1}M_i;j¹ + Xn i=0

[ⁿ₂¹]

X

k=0

r^i+2k+1M_i;j² + Xn i=0

Xm j=0

r^i+jM_i;j³

+ Xn i=0

[ⁿ₂]

X

k=0

r^i+2kM_i;j⁴ +

[ⁿ₂¹]

X

s=0 [ⁿ₂¹]

X

k=0

r^2s+2k+3M_i;j⁵ +

[ⁿ₂¹]

X

k=0

Xm i=0

r^i+2k+2M_i;j⁶

+

[ⁿ₂]

X

s=0 [Xⁿ₂¹]

k=0

r^2s+2k+2M_i;j⁷ + Xm i=0

Xm j=0

r^i+j+1M_i;j⁸ +

[ⁿ₂]

X

k=0

Xm i=0

r^2k+i+1M_i;j⁹

+

[ⁿ₂]

X

s=0 [ⁿ₂]

X

k=0

r^2s+2k+1M_i;j¹⁰; (25)

where

M_i;j¹ = ia_i;1a_j;1 Z 2

0

R_i( )J_j( )d +ia_i;1c_j;1r² Z 2

0

R_i( ) ~J_j( )d +(i+ 2)c_i;1a_j;1

Z 2 0

S_i( )J_j( )d + (i+ 2)c_i;1c_j;1r⁴ Z 2

0

S_i( ) ~J_j( )d ;

M_i;j² = ai;1b2k+1;1 i

k+1X

l=0

~_k;l Z 2

0

Ri( ) sin(2l+ 1) d + (2k+ 2) Z 2

0

T2k+1( )

Ji( )d ) +ai;1d2k+1;1r² i

k+2X

l=0

~~_k;l Z 2

0

Ri( ) sin(2l+ 1) d + (2k+ 4) Z 2

0

U2k+1( )Ji( )d +ci;1b2k+1;1r² (i+ 2)

k+1X

l=0

~_k;l Z 2

0

Si( ) sin(2l+ 1)

d + (2k+ 2) Z 2

0

T_2k+1( ) ~J_i( )d +c_i;1d_2k+1;1r⁴ (i+ 2)

k+2X

l=0

~~_k;l Z 2

0

S_i( ) sin(2l+ 1) d + (2k+ 4) Z 2

0

U_2k+1( ) ~J_i( )d ;

M_i;j³ = ai;1ej;1 i Z 2

0

Ri( )Jj+1( )d + (j+ 1) Z 2

0

Rj+1( )Ji( )d +ci;1ej;1r² (i+ 2)

Z 2 0

S_i( )J_j+1( )d + (j+ 1) Z 2

0

R_j+1( ) ~J_i( )d ;

M_i;j⁴ = a_i;1d_{2k 2;1} i

k+1X

l=1

~k;l

Z 2 0

R_i( ) sin(2l )d + (2k+ 1) Z 2

0

T_{2k 2}

( )Ji( )d 2k+ 2 2k 1

Z 2 0

T2k( )Ji( )d ) +ci;1d2k 2;1r²((i+ 2)

k+1X

l=1

~k;l

Z 2 0

Si( ) sin(2l )d + (2k+ 1) Z 2

0

T2k 2( ) ~Ji( )d 2k+ 2

2k 1 Z 2

0

T_2k( ) ~J_i( )d ;

M_i;j⁵ = (2k+ 2) b_2s+1;1b_2k+1 Xs+1

l=0

~_s;l Z 2

0

T_2k+1( ) sin(2l+ 1) d +d_2s+1;1b_2k+1r²

s+2X

l=0

~~_s;l Z 2

0

T_2k+1( ) sin(2l+ 1) d

!

+ (2k+ 4) b_2s+1;1d_2k+1r² Xs+1 l=0

~_s;l Z 2

0

U2k+1( ) sin(2l+ 1) d +d2s+1;1d2k+1r⁴

s+2X

l=0

~~_s;l Z 2

0

U2k+1( ) sin(2l+ 1) d );

M_i;j⁶ = e_i;1b_2k+1 (2k+ 2) Z 2

0

T_2k+1( )J_i+1( )d + (j+ 1)

k+1X

l=0

~_k;l Z 2

0

R_i+1( ) sin(2l+ 1) )d +ei;1d2k+1r² (2k+ 4)

Z 2 0

U2k+1( )Ji+1( )d + (i+ 1)

k+2X

l=0

~~_k;l Z 2

0

Ri+1( ) sin(2l+ 1)

! d ;

M_i;j⁷ = d2s 2;1b2k+1 (2k+ 2) Xs+1

l=0

~s;l

Z 2 0

T2k+1( ) sin(2l )d + (2s+ 1)

k+1X

l=0

~_k;l Z 2

0

T2s 2( ) sin(2l+ 1) d 2s+ 2 2s 1

Z 2 0

T2s( ) sin((2l+ 1) )d +d_{2s 2;1}d_2k+1r² (2k+ 4)

Xs+1 l=0

~s;l

Z 2 0

U_2k+1( ) sin(2l )d + (2s+ 1)

k+2X

l=0

~~_k;l Z 2

0

T_{2s 2}( ) sin(2l+ 1) d 2s+ 2 2s 1

Z 2 0

T_2s( ) sin(2l+ 1) )d ) ;

M_i;j⁸ = (j+ 1)ei;1ej;1

Z 2 0

Rj+1( )Ji+1( )d ;

M_i;j⁹ = ei;1d2k 2

"

(i+ 1)

k+1X

l=0

~k;l

Z 2 0

Ri+1( ) sin(2l )d + (2k+ 1) Z 2

0

T2k 2( ) 2k+ 2

2k 1 Z 2

0

T_2k( )J_i+1( )d ))

#

;

and

M_i;j¹⁰ = (2s+ 1)d_{2s 2;1}d_{2k 2;1}

k+1X

l=0

~k;l

h Z ²

0

T_{2s 2}( ) sin(2l )d 2s+ 2 2s 1

Z 2 0

T_2s( ) sin(2l )d )

i :

In the next propositions we obtain some results of the integrals on the right hand part of (25).

Proposition 3 Fori; j2Nand 2Rthe following statements hold.

(a) cosⁱRj( ) =Ri+j( );

(b) cosⁱTj( ) =Ri( )Rj( ) =Ti+j( );

(c) cosⁱS_j( ) =R_i( )T_j( ) =S_i+j( );

(d) cosⁱU_j( ) =R_i( )S_j( ) =T_i( )T_j( ) =U_i+j( );

(e) Ri( )Uj( ) =Ti( )Sj( ) =Si+j( ) Si+j+2( );

(f ) Ti( )Uj( ) =Si( )Sj( ) =Ui+j( ) Ui+j+2( );

(g) S_i( )U_j( ) =S_i+j( ) 2S_j+j+2( ) +S_i+j+4( ):

Proof. By using the elemental trigonometric relations we obtain easily the previous equalities.

Corollary 1 Fori; j2N, the following statements hold.

(a)

Z 2 0

cosⁱ Rj( )d = Z 2

0

cosⁱ Sj( )d = Z 2

0

Ri( )Tj( )d

= Z 2

0

Ri( )Uj( )d Z 2

0

Ti( )Sj( )d = Z 2

0

Si( )Uj( )d = 0:

(b) Z 2

0

cosⁱTj( )d = Z 2

0

Ri( )Rj( )d = ~Ii+j(2 ) = 0; ifi+j is odd, I~_2k(2 ); ifi+j is even.

(c)

Z 2 0

cosⁱUj( )d = Z 2

0

Ri( )Sj( )d = Z 2

0

Ti( )Tj( )d

= I~~_i+j(2 ) =

( 0; if i+j is odd,

~~

I_2k(2 ); if i+j is even.

(d)

Z 2 0

Ti( )Uj( )d = Z 2

0

Si( )Sj( )d =I~~i+j(2 ) I~~i+j+2(2 )

=

( 0; if i+j is odd,

~~

I2k(2 ) I~~2k+2(2 ); if i+j is even.

Proof. Using Proposition 3 and using (10) for all i; j 2 N; we obtain (a). Again from Proposition 3 and using (13) and (14) we obtain (b), (c) and (d). Hence the Corollary follows.

Proposition 4 Fori; j2Nand 2Rthe following statements hold.

(a) R_i( )J_j( ) = _j+1¹ (R_i( ) R_i+j+1( )):

(c) Ri( ) ~Jj( ) = _(j+1)(j+3)^{2Ri( ) Ri+j+1}_j+1^{( )}+^Ri+j+3_j+3^{( )}: (e) T_i( )J_j( ) =_j+1¹ (T_i( ) T_i+j+1( )):

(g) Ti( ) ~Jj( ) =_(j+1)(j+3)^{2Ti( ) Ti+j+1}_j+1^{( )} +^Ti+j+3_j+3^{( )}: (i) Si( )Jj( ) = _j+1¹ (Si( ) Si+j+1( )):

(k) S_i( ) ~J_j( ) = _(j+1)(j+3)^{2Si( ) Si+j+1}_j+1^{( )} +^Si+j+3_j+3^{( )}: (m) U_i( )~j_j( ) = _j+1¹ (U_i( ) U_i+j+1( )):

(o) Ui( ) ~Ij( ) = _(j+1)(j+3)^{2Ui( ) Ui+j+1}_j+1^{( )} +^Ui+j+3_j+3^{( )}:

Proof. The Proposition follows if we use the integrals (8) and (9) and Proposition3. In the next Corollary

we use Z 2

0

cosⁱ( ) sin^j( ) sin((2l+ 1) )d 6= 0; ifieven andj odd, Z 2

0

cosⁱ( ) sin^j( ) sin((2l+ 1) )d = 8<

:

0; ifiodd orj even,

Ck;l; ifi= 2k,j= 1 andl 0, Kk;l; ifi= 2k,j= 3 andl 0,

(26) whereC_k;l; K_k;l are non-zero constants.

Z 2 0

cosⁱ( ) sin^j( ) sin(2l )d 6= 0; ifiandj odd, Z 2

0

cosⁱ( ) sin^j( ) sin(2l )d = 8<

:

0; ifiodd orj even,

C~k;l; ifi= 2k+ 1,j= 1 andl 0, K~k;l; ifi= 2k+ 1,j= 3 andl 0,

(27)

whereC~k;l;K~k;l are non-zero constants.

Corollary 2 Fori; j2N, the following statements hold.

(a)

Z 2 0

T_i( ) sin(2l+ 1) d = Z 2

0

T_i( ) sin(2l )d = Z 2

0

U_i( ) sin(2l+ 1) d

= Z 2

0

U_i( ) sin(2l )d = 0:

(b) R2

0 Ri( ) sin(2l+ 1) d = Ci;l: (c) R2

0 Ri( ) sin(2l )d = C~i;l: (d) R2

0 Si( ) sin(2l+ 1) d = Ki;l: (e) R2

0 Si( ) sin(2l )d = K~i;l:

Proof. The Corollary follows if we use the integrals (26) and (27).

Corollary 3 Fori; j2N the following statements hold.

(a) Z 2

0

R_i( )J_j( )d = Z 2

0

R_i( ) ~J_j( )d = Z 2

0

S_i( )J_j( )d = Z 2

0

S_i( ) ~J_j( )d = 0:

(b)

Z 2 0

T_i( )J_j( )d = 8>

>>

<

>>

>:

1

2s+1I~2k(2 ); if i= 2k; j= 2s;

1

2s+2( ~I_2k(2 ) I~_2k+2s+2(2 )); ifi= 2k; j= 2s+ 1;

1

2s+2I~2k+2s+2(2 ); ifi= 2k+ 1; j= 2s;

0; ifi= 2k+ 1; j= 2s+ 1:

(c)

Z 2 0

T_i( ) ~J_j( )d

= 8>

>>

>>

>>

<

>>

>>

>>

>:

2

(2s+1)(2s+3)I~_2k(2 ); if i= 2k; j= 2s,

2

(2s+2)(2s+4)I~2k(2 ) _(2s+2)² I~2k+2s+2(2 ) +_(2s+4)² I~2k+2s+4(2 ); if i= 2k; j= 2s+ 1;

2

(2s+1)I~2k+2s+2(2 ) +_(2s+3)² I~2k+2s+4(2 ); if i= 2k+ 1; j = 2s;

0; if i= 2k+ 1; j = 2s+ 1:

(d)

Z 2 0

Ui( )Jj( )d = 8>

>>

>>

>>

<

>>

>>

>>

>:

1 2s+1

~~

I2k(2 ); if i= 2k; j = 2s,

1

2s+2(I~~_2k(2 ) I~~_2k+2s+2(2 )); if i= 2k; j = 2s+ 1,

1 2s+2

~~

I2k+2s+2(2 ); if i= 2k+ 1; j= 2s,

0; if i= 2k+ 1; j= 2s+ 1.

(e)

Z 2 0

Ui( ) ~Jj( )d

= 8>

>>

>>

>>

<

>>

>>

>>

>:

2 (2s+1)(2s+3)

~~

I2k(2 ); if i= 2k; j = 2s,

2 (2s+2)(2s+4)

~~

I2k(2 ) _(2s+2)² I~~2k+2s+2(2 ) +_(2s+4)² I~~2k+2s+4(2 ); if i= 2k; j = 2s+ 1,

2 (2s+1)

~~

I2k+2s+2(2 ) + _(2s+3)² I~~2k+2s+4(2 ); if i= 2k+ 1; j= 2s,

0; if i= 2k+ 1; j= 2s+ 1.

Proof. Using Proposition 4 and using (10) for all i; j 2N; we obtain (a). Again from Proposition 4 and using (13) and (14) one has (b), (c), (d) and (e) and the Corollary follows.

Now we can compute the integral II(r).

Lemma 1 The integral II(r)is given by

II(r) =

[ⁿ₂]

X

s=0 [ⁿ₂¹]

X

k=0

r^2s+2k+1M_2s;2k+1² +

[ⁿ₂¹]

X

s=0 [ⁿ₂]

X

k=0

r^2s+2k+1M_2s+1;2k⁴

+

[Xⁿ₂¹] k=0

[^mX₂¹] s=0

r^2s+2k+3M_2s+1;2k+1⁶ +

[ⁿ₂]

X

k=0 [^m₂]

X

s=0

r^2k+2s+1M_2s;2k⁹ ; (28) where

M_2s;2k+1² = a2s;1b2k+1;1(2s

k+1X

l=0

~_k;l Cs;l

(k+ 1) k+s+1

2^k+s(2s+ 1)(k+s+ 2)!) +a2s;1d2k+1;1r² (2s

k+2X

l=0

~~_k;l Ci;l

3(k+ 2) k+s+1

2^k+s+1(2s+ 1)(k+s+ 3)!) +c2s;1b2k+1;1r²((2s+ 2)

k+1X

l=0

~_k;l Ks;l

(k+ 1)(4k+ 10s+ 15) k+s+1

2^k+s+1(2s+ 1)(2s+ 3)(k+s+ 3)!) +c2s;1d2k+1;1r⁴ ((2s+ 2)

k+2X

l=0

~~_k;l K2s;l

3(k+ 2)(4k+ 14s+ 21) k+s+1

2^k+s+2(2s+ 1)(2s+ 3)(k+s+ 4)!); (29)

M_2s+1;2k⁴ = a_2s+1;1d_{2k 2;1}((2s+ 1)

k+1X

l=1

~k;l C~_s;l+ 3(2k+ 1) _k+s 2^k+s+1(2k 1)(k+s+ 2)!) +c_2s+1;1d_{2k 2;1}r²((2s+ 3)

k+1X

l=1

~k;l K~_s;l+ 15(2k+ 1) k+s

2^k+s+2(2k 1)(k+s+ 3)!); (30)

M_2s+1;2k+1⁶ = e2s+1;1b2k+1( (k+ 1) k+s+2

2^k+s+1(2s+ 3)(k+s+ 3)!+ (2s+ 2)

k+1X

l=0

~_k;l Cs+1;l) +e2s+1;1d2k+1r²( 3(k+ 2) _k+s+2

2^k+s+2(2s+ 3)(k+s+ 4)!

+(2s+ 2)

k+1X

l=0

~~_k;l Cs+1;l); (31)

M_2s;2k⁹ =e_2s;1d_{2k 2} (2s+ 1)

k+1X

l=0

~k;l C~_s;l+ 3(2k+ 1) 2^k+s+1(2k 1)

!

k+s

(k+s+ 2)! (32)

Proof. Using the expression (25) and the results of Corollaries2and3, we have

M_2s+1;2k+1⁵ =M_2s;2k+1⁷ =M_2s;2k¹⁰ = 0for alls, k2N:

M_i;j¹ =M_i;j³ =M_i;j⁸ = 0for alli,j 2N:

If i= 2s and by substituting in (25), we obtain M_2s;2k+1² =M_2s;2k⁹ = 0and also the expressions (30) and (31).

Finally ifi= 2s+ 1and by substituting in (25) we obtain M_2s+1;2k⁴ =M_2s+1;2k+1⁶ = 0;(29) and (32).

The proof of the Lemma is completed.

In order to complete the computation ofF20(r)we must determine the functionIII(r)of the expression ofF20(r)given in (24). In the next Proposition we obtain the value ofF2(r; )whenF10(r) 0:

Proposition 5 If F10(r) 0; then

F₂(r; ) = Xn i=0

rⁱ a_i;2R_i( ) +b_i;2rT_i( ) +c_i;2r²S_i( ) +d_i;2r³U_2i+1( )

+ Xm i=0

ei;2rⁱ⁺¹Ri+1( ) 1 r

" _n X

i=0

rⁱ(ai;1Ri( ) +ci;1r²Si( )) +

[Xⁿ₂¹] k=0

r^2k+2

(b2k+1;1T2k+1( ) +d2k+1;1r²U2k+1( )) + Xm i=0

ei;1rⁱ⁺¹Ri+1( )

+

[ⁿ₂]

X

k=0

d2k 2r²ⁱ⁺¹ T2k 2( ) 2k+ 2 2k 1T2k( )

# " _n X

i=0

rⁱ(ai;1cosⁱ⁺¹( )

+ci;1r²Ti+1( )) +

[Xⁿ₂¹] k=0

r^2k+2(b2k+1;1R2k+2( )d2k+1;1r²S2k+2( ))

+ Xm i=0

e_i;1rⁱ⁺¹T_i( ) +

[ⁿ₂]

X

k=0

d_{2k 2}r^2k+1 R_{2k 1} 2k+ 2

2k 1R_2k+1( )

#

: (33)

Proof. If F10(r) 0; from (15) we have that b0 = d_2[ⁿ

2] = 0; and b2i = _2i³₁d2i 2: By substituting the relationUk( ) =Tk( ) Tk+2( )in (19) we obtain (33).

Now we shall compute the integralsIII(r):Using the expression of (33) we have III(r) =

Xn i=0

rⁱ(ai;2

Z 2 0

Ri( )d +bi;2r Z 2

0

Ti( )d +ci;2r² Z 2

0

Si( )d

+d_i;2r³ Z 2

0

U_i( )d ) + Xn

i=0

Xn i=0

r^{i+j 1}M~_i;j¹ + Xn i=0

[ⁿ₂¹]

X

k=0

r^i+2k+1M~_i;j²

+ Xn i=0

Xm j=0

r^i+jM~_i;j³ + Xn i=0

[ⁿ₂]

X

s=0

r^i+2sM~_i;j⁴ +

[ⁿ₂¹]

X

s=0 [ⁿ₂¹]

X

k=0

r^2s+2k+3M~_i;j⁵

+

[ⁿ₂¹]

X

k=0

Xm i=0

r^i+2k+2M~_i;j⁶ +

[ⁿ₂]

X

s=0 [ⁿ₂¹]

X

k=0

r^2s+2k+2M~_i;j⁷ + Xm i=0

Xm j=0

r^i+j+1M~_i;j⁸

+

[ⁿ₂]

X

s=0

Xm i=0

r^2s+i+1M~_i;j⁹ +

[ⁿ₂]

X

s=0 [ⁿ₂]

X

k=0

r^2s+2k+1M~_i;j¹⁰; (34)

where

M~_i;j¹ = ai;1aj;1

Z 2 0

Ri( ) cos^j+1 d +ai;1cj;1r² Z 2

0

Ri( )Tj+1( )d +ci;1aj;1r² Z 2

0

Si( ) cos^j+1 d +ci;1cj;1r⁴ Z 2

0

Si( )Tj+1( )d ;

M~_i;j² = ai;1b2k+1;1( Z 2

0

T2k+1( ) cosⁱ⁺¹ d + Z 2

0

R2k+2( )Ri( )d ) +ai;1d2k+1;1r² (

Z 2 0

U2k+1( ) cosⁱ⁺¹ d + Z 2

0

S2k+2( )Ri( )d ) +ci;1b2k+1;1r² (

Z 2 0

T2k+1( )Ti( )d + Z 2

0

R2k+2( )Si( )d ) +ci;1d2k+1;1r⁴ (

Z 2 0

U2k+1( )Ti( )d+ Z 2

0

S2k+2( )Si( )d );

M~_i;j³ = a_i;1e_j;1( Z 2

0

R_i( )T_j( )d + Z 2

0

R_j+1( ) cosⁱ⁺¹ d ) +c_i;1e_j;1r² (

Z 2 0

S_i( )T_j( )d + Z 2

0

R_j+1( )T_i( )d );

M~_i;j⁴ = ai;1d2k 2;1( Z 2

0

Ri( )R2k 1( )d + Z 2

0

T2k 2( ) cosⁱ⁺¹ d 2k+ 2 2k 1 (

Z 2 0

Ri( )R2k+1( )d + Z 2

0

T2k( ) cosⁱ⁺¹ d )) +ci;1d2k 2;1r² (

Z 2 0

Si( )R2k 1( )d + Z 2

0

T2k 2( )Ti( )d 2k+ 2 2k 1

Z 2 0

Si( )R2k+1( )d 2k+ 2

2k 1 Z 2

0

T2k( )Ti( )d );

M~_i;j⁵ = b2s+1;1b2k+1

Z 2 0

T2s+1( )R2k+2( )d +d2s+1;1b2k+1r² Z 2

0

U2s+1( ) R2k+2( )d +b2s+1;1d2k+1r²

Z 2 0

T2s+1( )S2k+2( )d +d2s+1;1d2k+1

r⁴ Z 2

0

U_2s+1( )S_2k+2( )d ;

M~_i;j⁶ = ei;1b2k+1

Z 2 0

R2k+2( )Ri+1( )d + Z 2

0

Ti( )T2k+1( )d +ei;1d2k+1r²

Z 2 0

S2k+2( )Ri+1( )d + Z 2

0

Ti( )U2k+1( )d ;

M~_i;j⁷ = d2s 2;1b2k+1

Z 2 0

T2s 2( )R2k+2( )d + Z 2

0

R2s 1( )T2k+1( )d 2s+ 2

2s 1 Z 2

0

T_2s( )R_2k+2( )d + Z 2

0

R_2s+1( )T_2k+1( )d +d2s 2;1d2k+1r²

Z 2 0

T2s 2( )S2k+2( )d + Z 2

0

R2s 1( )U2k+1( )d 2s+ 2

2s 1 Z 2

0

T_2s( )S_2k+2( )d + Z 2

0

R_2s+1( )U_2k+1( )d ; M~_i;j⁸ =ei;1ej;1

Z 2 0

Rj+1( )Ti( )d ; M~_i;j⁹ = e_i;1d_{2k 2}

Z 2 0

R_i+1( )R_{2k 1}( )d + Z 2

0

T_{2k 2}( )T_i( )d 2k+ 2

2k 1 Z 2

0

R2k+1( )Ri+1( )d + Z 2

0

T2k( )Ti( )d ;

M~_i;j¹⁰ = d2s 2;1d2k 2;1

Z 2 0

T2s 2( )R2k 1( )d 2s+ 2 2s 1 Z 2

0

T_{2s 2}( )R_2k+1( )d + Z 2

0

T_2s( )R_{2k 1}( )d +(2s+ 2)²

(2s 1)² Z 2

0

T2s( )R2k+1( )d :

In the next propositions we obtain some results on the integrals of the right hand part of (34).

Proposition 6 If F10(r) 0; then III(r) =

[ⁿ₂]

X

s=0

r^{2s+1 s}

2^s(s+ 1)! b2s;2+ 3

2(s+ 2)d2s;2r² +

[ⁿ₂]

X

s=0 [ⁿ₂¹]

X

k=0

r^2s+2k+1 M~_2s;2k+1²

+

[ⁿ₂¹]

X

s=0 [ⁿ₂]

X

k=0

r^2s+2k+1M~_2s+1;2k⁴ +

[ⁿ₂¹]

X

k=0 [^m₂¹]

X

s=0

r^2s+2k+3 M~_2s+1;2k+1⁶

+

[ⁿ₂]

X

k=0 [^m₂]

X

s=0

r^2k+2s+1M~_2s;2k⁹ ; (35)