Bifurcation of limit cycles, classification of centers and isochronicity for a class of non-analytic quintic systems

Abstract

Inspired by Llibre and Vallls (in J. Math. Anal. Appl. 357:427–437, 2009), the conditions of center and isochronous center at the origin for a class of non-analytic quintic systems are studied in this paper. By a transformation, we first transform the systems into analytic systems, then sufficient and necessary conditions for the origin of the systems being a center are obtained. The fact that 11 limit circles could be bifurcated is proved. A complete classification of the sufficient and necessary conditions is given for the origin of the systems being an isochronous center.

Appendices

Appendix A

The recursive formulas to compute singular point quantities at the origin of system (16):

c[0,0]=1;

when (k=j>0) or k<0, or j<0,

c[k,j]=0;

else

$$\begin{aligned} c[k,j]&=\frac{1}{2(j-k)}\bigl(-a_{31}jc[-2+k,-1+j]\\ &\quad{}- b_{22}jc[-2+k,-1+j]\\ &\quad {}+a_{31}kc[-2+k,-1+j]\\ &\quad{}+b_{22}kc[-2+k,-1+j]\\ &\quad {}+2a_{31}rc[-2+k,-1+j]\\ &\quad{}- 2b_{22}rc[-2+k,-1+j]\\ &\quad {}-b_{22}jrc[-2+k,-1+j]\\ &\quad{}+a_{31}jrc[-2+k,-1+j]\\ &\quad {}+a_{31}krc[-2+k,-1+j]\\ &\quad{}- b_{22}krc[-2+k,-1+j]\\ &\quad {}-a_{30}jc[-2+k,j]\\ &\quad{}- b_{12}jc[-2+k,j]+a_{30}kc[-2+k,j]\\ &\quad{}+2a_{30}rc[-2+k,j]+b_{12}kc[-2+k,j]\\ &\quad{}- 2b_{12}rc[-2+k,j]+a_{30}jrc[-2+k,j]\\ &\quad{}- b_{12}jrc[-2+k,j]+a_{30}krc[-2+k,j]\\ &\quad{}- b_{12}krc[-2+k,j]\\ &\quad {}-a_{22}jc[-1+k,-2+j]\\ &\quad{}- b_{31}jc[-1+k,-2+j]\\ &\quad {}+a_{22}kc[-1+k,-2+j]\\ &\quad{}+b_{31}kc[-1+k,-2+j]\\ &\quad {}-2b_{31}rc[-1+k,-2+j]\\ &\quad{}+a_{22}jrc[-1+k,-2+j]\\ &\quad {}-b_{31}jrc[-1+k,-2+j]\\ &\quad{}+a_{22}krc[-1+k,-2+j]\\ &\quad {}-b_{31}krc[-1+k,-2+j]\\ &\quad{}- a_{21}jc[-1+k,-1+j]\\ &\quad {}-b_{21}jc[-1+k,-1+j]\\ &\quad{}+a_{21}kc[-1+k,-1+j]\\ &\quad {}+b_{21}kc[-1+k,-1+j]\\ &\quad{}+2a_{21}rc[-1+k,-1+j]\\ &\quad {}-2b_{21}rc[-1+k,-1+j]\\ &\quad{}+a_{21}jrc[-1+k,-1+j]\\ &\quad {}-b_{21}jrc[-1+k,-1+j]\\ &\quad{}+a_{21}krc[-1+k,-1+j]\\ &\quad {}-b_{21}krc[-1+k,-1+j]\\ &\quad{}- a_{12}jc[k,-2+j]-b_{30}jc[k,-2+j]\\ &\quad{}+a_{12}kc[k,-2+j]+b_{30}kc[k,-2+j]\\ &\quad{}+2a_{12}rc[k,-2+j]-2b_{30}rc[k,-2+j]\\ &\quad{}+a_{12}jrc[k,-2+j]-b_{30}jrc[k,-2+j]\\ &\quad{}+a_{12}krc[k,-2+j]-b_{30}krc[k,-2+j]\\ &\quad{}+2a_{22}rc[-1+k,-2+j]\bigr), \end{aligned}$$

$$\begin{aligned} \mu_k&=r\bigl(a_{31}c[-2+k,-1+k]\\ &\quad {}-b_{22}c[-2+k,-1+k]\\ &\quad{}- b_{12}c[-2+k,k]+a_{30}c[-2+k,k]\\ &\quad{}+a_{22}c[-1+k,-2+k]\\ &\quad {}-b_{31}c[-1+k,-2+k]\\ &\quad{}+a_{21}c[-1+k,-1+k]\\ &\quad {}-b_{21}c[-1+k,-1+k]\\ &\quad{}+a_{12}c[k,-2+k]-b_{30}c[k,-2+k]\bigr). \end{aligned}$$

Appendix B

The recursive formulas to compute period constants of the origin of system (16):

c′[1,0]=d′[1,0]=1;c′[0,1]=d′[0,1]=0;

if k<0 or j<0 or (j>0 and k=j+1) then c′[k,j]=0,d′[k,j]=0;

else

$$\begin{aligned} c'[k,j]&=\frac{1}{j+1-k} \bigl(\bigl(-b_{22}(-1+j) \\ &\quad{}+a_{31}(-2+k)\bigr)e[-2+k,-1+j] \\ &\quad {}+\bigl(-b_{12}j +a_{30}(-2+k)\bigr)e[-2+k,j] \\ &\quad {}+\bigl(-b_{31}(-2+j) \\ &\quad{}+a_{22}(-1+k)\bigr)e[-1+k,-2+j] \\ &\quad {}+\bigl(-b_{21}(-1+j) \\ &\quad{}+a_{21}(-1+k)\bigr)e[-1+k,-1+j] \\ &\quad {}+\bigl(-b_{30}(-2+j) +a_{12}k\bigr)e[k,-2+j] \bigr), \end{aligned}$$

$$\begin{aligned} d'[k,j]&=\frac{1}{j+1-k}\bigl(\bigl(-a_{22}(-1+j) \\ &\quad{}+b_{31}(-2+k)\bigr)d[-2+k,-1+j] \\ &\quad {}+\bigl(-a_{12}j +b_{30}(-2+k)\bigr)d[-2+k,j] \\ &\quad {}+\bigl(-a_{31}(-2+j) \\ &\quad{}+b_{22}(-1+k)\bigr)d[-1+k,-2+j] \\ &\quad {}+\bigl(-a_{21}(-1+j) \\ &\quad{}+b_{21}(-1+k)\bigr)d[-1+k,-1+j] \\ &\quad{}+\bigl(-a_{30}(-2+j) +b_{12}k\bigr)d[k,-2+j]\bigr), \end{aligned}$$

$$\begin{aligned} p'[j]&=\bigl(a_{31}(-1+j) \\ &\quad {}-b_{22}(-1+j)\bigr)e[-1+j,-1+j] \\ &\quad{}+\bigl(a_{30}(-1+j)-b_{12}j\bigr)e[-1+j,j] \\ &\quad{}+\bigl(-b_{31}(-2+j)+a_{22}j\bigr)e[j,-2+j] \\ &\quad{}+\bigl(-b_{21}(-1+j)+a_{21}j\bigr)e[j,-1+j] \\ &\quad{}+\bigl(-b_{30}(-2+j) \\ &\quad {}+a_{12}(1+j)\bigr)e[1+j,-2+j], \\ q'[j]&=\bigl(-a_{22}(-1+j) \\ &\quad {}+b_{31}(-1+j)\bigr)d[-1+j,-1+j] \\ &\quad{}+\bigl(b_{30}(-1+j)-a_{12}j\bigr)d[-1+j,j] \\ &\quad{}+\bigl(-a_{31}(-2+j)+b_{22}j\bigr)d[j,-2+j] \\ &\quad{}+\bigl(-a_{21}(-1+j)+b_{21}j\bigr)d[j,-1+j] \\ &\quad{}+\bigl(-a_{30}(-2+j) \\ &\quad {}+b_{12}(1+j)\bigr)d[1+j,-2+j]. \\ &\tau_m=p[m]+q[m]=p'[m]+q'[m]. \end{aligned}$$

Appendix C

The first eight point quantities at the origin of system (16) are as follows:

$$\begin{aligned} & u_1 =(a_{11}-b_{11})r,\\ & u_2=(a_{22}-b_{22})r,\\ & u_3=-(a_{12}a_{21}-b_{12}b_{21})r,\\ & u_4=-(a_{04}a_{40}-b_{04}b_{40})r. \end{aligned}$$

Case 1 When a ₂₁ b ₂₁ a ₄₀ b ₄₀≠0, let a ₁₂=pb ₂₁,b ₁₂=pa ₂₁,a ₀₄=qb ₄₀,b ₀₄=qa ₄₀, then

$$u_5=\bigl(a_{31}b_{21}^2-a_{21}^2b_{31} \bigr) (-1+p)r(1+p-2r+2pr). $$

Subcase 1.1 Let $(-1+p)r(1+p-2r+2pr)\neq 0, a_{31}=ka_{21}^{2}, b_{31}=k b_{21}^{2}$,

$$u_6=-\bigl(a_{40}b_{21}^4-a_{21}^4b_{40}\bigr)k^2(-1+q)r. $$

If k≠0,q=1,

$$\begin{aligned} u_7& =-\bigl(a_{40}b_{21}^4-a_{21}^4b_{40} \bigr) (-1+p)k r(3+3p-8r+8pr), \\ u_8& =\frac{24(a_{40}b_{21}^4-a_{21}^4b_{40})r^2(54b_{11}k+288b_{11}k r+35r^2+384b_{11}kr^2)}{(3+8r)^4}, \end{aligned}$$

$$\begin{aligned} u_9& =-\frac {96(a_{40}b_{21}^4-a_{21}^4b_{40})r^2}{5(3+8r)^4}\bigl(-135b_{22}k \\ &\quad {} +72a_{21}b_{21}k^2-720b_{22}kr+384a_{21}b_{21}k^2r \\ &\quad {} -175b_{11}r^2-960b_{22}kr^2+512a_{21}b_{21}k^2r^2 \bigr), \end{aligned}$$

$$\begin{aligned} u_{10}&=\frac {7}{(114303+8r)^4}\bigl(94608a_{21}b_{21}k^3 \\ &\quad {}+1009152a_{21}b_{21}k^3r+4036608a_{21}b_{21}k^3r^2 \\ &\quad {}+7176192a_{21}b_{21}k^3r^3+111125r^4 \\ &\quad {}+4784128a_{21}b_{21}k^3r^4 \bigr), \\ u_{11}&=-\frac {4(a_{40}b_{21}^4-a_{21}^4b_{40})r^2}{417195(3+8r)^6}\bigl(4379546232a_{40}b_{40}k \\ &\quad {}+46715159808a_{40}b_{40}kr \\ &\quad {}+186860639232a_{40}b_{40}kr^2 \\ &\quad {}+332196691968a_{40}b_{40}kr^3 \\ &\quad {}-11814392845a_{21}b_{21}r^4 \\ &\quad {}+221464461312a_{40}b_{40}kr^4\bigr). \end{aligned}$$

If k=0,q≠1,

$$\begin{aligned} u_7&=0, \\ u_8& =-\frac{1}{48}\bigl(a_{40}b_{21}^4-a_{21}^4b_{40} \bigr)r(1+p-r+pr) \\ &\quad {}\times(1+p-5r+5pr) (3+3p-7r+7pr) \\ &\quad {}\times(3-5p+5q-3pq+r-pr-qr+pqr), \\ u_9& =\frac {(-8+3m)b_{11}}{12(1+mr)^3}\bigl(a_{40}b_{21}^4-a_{21}^4b_{40} \bigr) \\ &\quad {}\times\bigl(53-55m+8m^2\bigr)r^4 \\ &\quad {}\times(3-5p+5q-3pq+r-pr-qr+pqr), \end{aligned}$$

$$\begin{aligned} u_{10}&=\frac {b_{22}(a_{40}b_{21}^4-a_{21}^4b_{40})(-3+m)(35-31m+4m^2)r^4}{2(1+mr)^3} \\ &\quad {}\times(3-5p+5q-3pq+r-pr-qr+pqr), \\ u_{11}&=\frac{a_{21}b_{21}(-a_{40}b_{21}^4+a_{21}^4b_{40})}{30(1+mr)^5} \\ &\quad {}\times\bigl(65870-97508m+31371m^2+1898m^3 \\ &\quad {}-1271m^4+60m^5\bigr)\times r^6 \\ &\quad {}\times(3-5p+5q-3pq+r-pr-qr+pqr). \end{aligned}$$

Subcase 1.2 Let $a_{31}b_{21}^{2}\neq a_{21}^{2}b_{31}, p=1$,

$$\begin{aligned} & u_6=- \bigl(a_{40}b_{31}^2-a_{31}^2b_{40} \bigr) (-1+q)r, \\ & \mathrm{let}~a_{31}^2=ka_{40},b_{31}^2=kb_{40}, \\ & u_7=3 \bigl(a_{40}b_{21}^2b_{31}-a_{21}^2a_{31}b_{40} \bigr) (-1+q)r, \\ & \mathrm{let}~a_{31}=na_{40}b_{21}^2,b_{31}=na_{21}^2b_{40}, \\ & u_8=- \bigl(a_{40}b_{21}^4-a_{21}^4b_{40} \bigr) (-1+q)r. \end{aligned}$$

Case 2 When a ₂₁ b ₂₁≠0,a ₄₀=b ₄₀=0, let a ₁₂=pb ₂₁,b ₁₂=pa ₂₁, then

$$u_5=\bigl(a_{31}b_{21}^2-a_{21}^2b_{31} \bigr) (-1+p)r(1+p-2r+2pr). $$

Subcase 2.1 Let $a_{31}=ka_{21}^{2}, b_{31}=kb_{21}^{2}, p\neq1$

$$\begin{aligned} u_6&= \bigl(a_{04}a_{21}^4-b_{04}b_{21}^4 \bigr)k^2r, \\ u_7&=0, \\ u_8&=\frac{1}{48} \bigl(a_{04}a_{21}^4-b_{04}b_{21}^4 \bigr)r(5-3p-r+pr) \\ &\quad {}\times (1+p-r+pr) (1+p-5r+5pr) \\ &\quad {}\times(3+3p-7r+7pr), \\ u_9&=-\frac{b_{11}(a_{04}a_{21}^4-b_{04}b_{21}^4)(-8+3m)}{12(1+mr)^3} \\ &\quad {}\times \bigl(53-55m+8m^2 \bigr)r^4(5-3p-r+pr), \end{aligned}$$

$$\begin{aligned} u_{10}&=-\frac{(a_{04}a_{21}^4-b_{04}b_{21}^4)b_{22}(-3+m)}{2(1+mr)^3} \\ &\quad {}\times \bigl(35-31m+4m^2 \bigr)r^4(5-3p-r+pr), \\ u_{11}&=\frac {a_{21}b_{21}(a_{04}a_{21}^4-b_{04}b_{21}^4)r^6(5-3p-r+pr)}{30(1+mr)^5} \\ &\quad {}\times \bigl(65870-97508m+31371m^2+1898m^3 \\ &\quad {}-1271m^4+60m^5 \bigr); \end{aligned}$$

where m=1,5 or $\frac{7}{3}$.

Subcase 2.2 Let $a_{31}b_{21}^{2}\neq a_{21}^{2}b_{31}, p=1$

$$\begin{aligned} & u_6=\bigl(a_{04}a_{31}^2-b_{04}b_{31}^2 \bigr)r, \\ & \mathrm{let}~a_{04}=kb_{31}^2,b_{04}=ka_{31}^2 \\ & u_7=3 a_{31}b_{31}\bigl(a_{31}b_{21}^2-a_{21}^2 b_{31}\bigr)kr. \end{aligned}$$

Case 3 When a ₂₁=b ₂₁=0,a ₄₀ b ₄₀≠0, let a ₀₄=qb ₄₀,b ₀₄=qa ₄₀, then

$$u_5=\bigl(a_{12}^2a_{31}-b_{12}^2 b_{31}\bigr)r(1+2r). $$

Subcase 3.1 When a ₁₂ b ₁₂≠0, let $a_{31}=pb_{12}^{2}, b_{31}=pa_{12}^{2}$, then

$$u_6=-\bigl(a_{12}^4a_{40}-b_{12}^4b_{40} \bigr)p^2(-1+q)r. $$

If p=0,q≠1,

$$\begin{aligned} & u_7=0, \\ & u_8=-\frac{1}{48} \bigl(a_{12}^4a_{40}-b_{12}^4b_{40} \bigr)r(1+r) (1+5r) \\ & \hphantom{u_8=} {}\times(3+7r) (-5 - 3 q - r + q r), \\ &u_9=\frac{1}{96}b_{11} \bigl(a_{12}^4a_{40}-b_{12}^4b_{40} \bigr)r(3+8 r) \\ & \hphantom{u_8=} {}\times(-5-3q-r+qr) \bigl(8+55r + 53 r^2 \bigr), \\ & u_{10}=\frac{1}{16}b_{22} \bigl(a_{12}^4a_{40}-b_{12}^4b_{40} \bigr)r(1 + 3 r) \\ & \hphantom{u_{10}=} {}\times(-5 - 3 q - r +q r) \bigl(4 + 31 r + 35 r^2 \bigr), \\ & u_{11}=\frac{1}{960}a_{12}b_{12} \bigl(a_{12}^4a_{40}-b_{12}^4b_{40} \bigr)\\ & \hphantom{u_{10}=} {}\times r(-5-3q-r+qr) \\ & \hphantom{u_{10}=} {}\times \bigl(-60-1271r-1898r^2+31371r^3 \\ &\hphantom{u_{10}=} {}+97508 r^4 + 65870 r^5 \bigr); \end{aligned}$$

If q=1,p≠0,

$$\begin{aligned} & u_7=-\bigl(a_{12}^4a_{40}-b_{12}^4b_{40} \bigr)pr(3+8r), \\ & u_8=\frac{3(a_{12}^4a_{40}-b_{12}^4b_{40})(35 + 1536 b_{11}p)}{8192}, \end{aligned}$$

$$\begin{aligned} & u_9=-\frac{3(a_{12}^4a_{40}-b_{12}^4b_{40})(-175b_{11}- 3840 b_{22}p+2048 a_{12}b_{12} p^2)}{10240}, \end{aligned}$$

$$\begin{aligned} & u_{10}=\frac{(a_{12}^4a_{40}-b_{12}^4b_{40})(1143 b_{22}+208 a_{12}b_{12} p)}{16384}. \end{aligned}$$

Subcase 3.2 When a ₁₂=b ₁₂=0, then

$$u_6=-\bigl(a_{40}b_{31}^2-a_{31}^2b_{40} \bigr) (-1 + q)r. $$

Case 4 When a ₂₁=b ₂₁=a ₄₀=b ₄₀=0, then

$$u_5=\bigl(a_{12}^2a_{31}-b_{12}^2 b_{31}\bigr)r(1+2r). $$

Subcase 4.1 When a ₁₂ b ₁₂≠0, let $a_{31}=pb_{12}^{2}$, $b_{31}=pa_{12}^{2}$, then

$$\begin{aligned} & u_6=-\bigl(a_{12}^4b_{04}-a_{04}b_{12}^4 \bigr)p^2 r, \\ & u_7=0, \\ & u_8=-\frac{1}{48}\bigl(a_{12}^4b_{04}-a_{04}b_{12}^4 \bigr) (-3 + r) r \\ &\hphantom{u_8}\quad {}\times(1 + r) (1 + 5 r) (3 + 7 r), \\ & u_9=\frac{1}{3360}b_{11}\bigl(a_{12}^4b_{04}-a_{04}b_{12}^4 \bigr) (-3 + r) \\ &\hphantom{u_8}\quad {}\times r\bigl(432 + 2585 r + 3203 r^2\bigr), \\ & u_{10}=-\frac{1}{16}\bigl(a_{12}^4b_{04}-a_{04}b_{12}^4 \bigr)b_{22}(-3+r)r \\ & \hphantom{u_{10}}\quad \times\bigl(5 + 32 r + 43 r^2 \bigr), \\ & u_{11}=-\frac{1}{39200}a_{12}b_{12} \bigl(a_{12}^4b_{04}-a_{04}b_{12}^4 \bigr) \\ &\hphantom{u_{11}}\quad \times\bigl(32397 + 242990 r + 382093 r^2\bigr). \end{aligned}$$

Subcase 4.2 When a ₁₂=b ₁₂=0, then

$$u_6=\bigl(a_{31}^2a_{04}-b_{04}b_{31}^2 \bigr)r. $$