Periodic Solutions of Some Classes of One Dimensional Non-autonomous Equation

In this paper, the periodic solutions of a certain one-dimensional differential equation are investigated for the first order cubic non-autonomous equation. The method used here is the bifurcation of periodic solutions from a fine focus z = 0. We aimed to find the maximum number of periodic solutions into which a given solution can bifurcate under perturbation of the coefficients. For classes C_{3, 8}, C_{4, 3}, C_{7, 5}, C_{7, 6}, eight periodic multiplicities have been found. To investigate the multiplicity >9, the formula for the focal value was not available in the literature. We also succeeded in constructing the formula for η₁₀. By implementing our newly developed formula, we are able to get multiplicity ten for classes C_{7, 3}, C_{9, 1}, which is the highest known to date. A perturbation method has been properly established for making the maximal number of limit cycles for each class. Some examples are also presented to show the implementation of the newly developed method. By considering all of these facts, it can be concluded that the presented methods are new, authentic, and novel.

1. Introduction

On August 8, 1900, David Hilbert presented a set of mathematical problems [1] to the Second International Congress of Mathematicians in Paris. The sixteenth problem he posed was titled the Problem of the Topology of Algebraic Curves and Surfaces. It is stated in two parts. In the first part, Hilbert suggested a thorough investigation of the relative positions of the separate branches of algebraic curves in nth-order vector fields, which is in the area of real algebraic geometry. In the second part, Hilbert asked for a search for the upper bound of the number of limit cycles and their relative locations in polynomial vector fields of order n. This part of the problem is related to ordinary differential equations and dynamical systems. Generally, this part of the problem is what is usually meant when talking about Hilbert's 16th problem.

Limit cycle theory takes a central role in Hilbert's 16th problem. Studying the number of limit cycles for differential equations is the most difficult part of the problem. The phenomenon of the limit cycle was first discovered and introduced by Poincaré in his four-part article, Integral curves defined by differential equations [2–5] published between 1881 and 1886.

At that time, Poincaré also noticed the close relationship between the study of limit cycles and the solutions of the global structural problems of a family of integral curves of differential equations. His work was later extended by Bendixson to the well-known Poincaré-Bendixson theorem [6] on the limit set of trajectories of dynamical systems in a bounded region. The driving force behind the study of limit cycle theory was the invention of the triode vacuum tube, which was able to produce stable self-excited oscillations of constant amplitude. It was noted that this kind of oscillation phenomenon could not be described by linear differential equations. At the end of the 1920's Van der Pol [7] developed a differential equation to describe the oscillations of constant amplitude in a triode vacuum tube. Limit cycles are common solutions for all types of dynamical systems. They model systems that exhibit self-sustained oscillations. In other words, these systems oscillate even in the absence of external periodic forcing. For a practical example, consider a specific Holling-Tanner predator-prey model [8]. This model appears to match very well with what happens for many predator-prey species in the natural world, for example, house sparrows and sparrow hawks in Europe, muskrat and mink in Central North America, and white-tailed deer and wolf in Ontario, Canada.

Other examples of self-excited oscillation are the beating of a heart, rhythms in body temperature, hormone secretion, chemical reactions that oscillate spontaneously, and vibrations in bridges and airplane wings. Due to the wide occurrence of limit cycles in science and technology, limit cycle theory has also been extensively studied by physicists, and more recently by chemists, biologists, and economists [9–16].

We consider the differential equation of the form

$\begin{array}{ll}\stackrel{·}{z}=\gamma (t)z^3+\delta (t)z^2+\upsilon (t)z& (1)\end{array}$

where independent variable t and coefficients γ, δ, υ are real-valued functions but z ∈ ℂ. To find the maximum count of periodic solutions we use the complexified form of the equation (1); for more details, see [17–20]. Also consider that ∃ β ∈ ℝ such that:

$\begin{array}{l}z(\beta )=z(0).\end{array}$

These solutions are periodic, even if γ, δ, and υ are not themselves periodic. Our fundamental focus is to get the maximum number of periodic solutions of any class of the form (1) in which a solution may bifurcate by perturbing the coefficients. Neto [21] states that for Equation (1), until some coefficients are restricted, we are unable to have an upper bound for the number of focal values. The number of periodic solutions depends upon the multiplicity of the solutions z = 0. The multiplicity of z = 0, as solution of (1) is also multiplicity of z = 0; as a zero of the following displacement function

$\begin{array}{ll}p:r\to z(\beta ,0,r)-r& (2)\end{array}$

described in complex function theory. For z = 0, the means of computing multiplicity (μ) is explained in Alwash and Llyod [22], but for the sake of ease, we explained it briefly here. We write $z(t,0,r)=\sum _{i=1}^{\infty }{a}_i(t)r^i$ for 0 ≤ t ≤ β where also r lies in neighborhood of z = 0, and use it in equation (2); for more detail, see [21, 23–26]. This provides a differential equation for a_μ(t) with some starting conditions a₁(0) = 1 and a_μ(0) = 0 for i > 1. Therefore

$\begin{array}{ll}p(r)=(a_1(\alpha )-1)r+{\displaystyle \sum _{i=2}^{\infty }}{a}_i(\beta )r^i& (3)\end{array}$

The multiplicity (μ) is “μ > 1” if

$\begin{array}{l}{a}_1(\beta )=1\\ a_2(\beta )=a_3(\beta )=...=a_{\mu -1}(\beta )=0.\end{array}$

However, a_μ(β) ≠ 0. When a₁(β) = 1 and a_μ(β) = 0, ∀ μ > 1, the origin is center. We can observe from Equation (1) that $\stackrel{·}{a_1}(t)=a_1(t)\upsilon (t),$ where a₁(t) is defined as

$\begin{array}{l}{a}_1(t)=e^{{\int }_0^t\upsilon (s)ds}.\end{array}$

In this way, μ > 1 iff

$\begin{array}{ll}{\displaystyle {\int }_0^t}\upsilon (s)ds=0& (4)\end{array}$

because a₁(t) = 1. We are especially interested in the situation where z = 0 has multiple solutions, so we consider that (4) holds. By using the following transformation

$\begin{array}{l}\xi =z{e}^{-{\int }_0^t\upsilon (s)ds}.\end{array}$

(1) takes the form

$\begin{array}{ll}\stackrel{·}{\xi }=\hat{\gamma }(t){\xi }^3+\hat{\delta }(t){\xi }^3& (5)\end{array}$

where $\hat{\gamma }(t)=\gamma (t)e^{2{\int }_0^t\upsilon (s)ds}$ and

$\begin{array}{l}\hat{\delta }(t)=\delta (t)e^{2{\int }_0^t\upsilon (s)ds}.\end{array}$

We can see that $\hat{\gamma }$ and $\hat{\delta }$ are periodic if γ, δ, and υ are periodic. By using Lemma (2.6) in Alwash and Llyod [22], consider multiplicity of z = 0 as a periodic solution of (1). If, for equation (1), μ > 1, then the multiplicity of ξ = 0 as a periodic solution of (5) is also μ. So we consider that υ(t) $\widetilde{=}$ 0 in (1). As a result, equation (1) takes the form

$\begin{array}{ll}\stackrel{·}{z}=\gamma (t)z^3+\delta (t)z^2& (6)\end{array}$

where γ and δ may be polynomials (i) in t (ii) in cost and sint (trigonometric functions). The functions a_i(t), for i > 1 are calculated by utilizing the relation

$\begin{array}{ll}\stackrel{·}{a_i}=\gamma {\displaystyle \sum _{\begin{array}{l}j+k+l=i\\ j,k,l\ge 1\end{array}}}{a}_j{a}_k{a}_l+\delta {\displaystyle \sum _{\stackrel{j+k=i}{j,k\ge 1}}{a}_j}{a}_k& (7)\end{array}$

with a₁(t) = 1. Calculation of these functions is tough because of the integration by parts used in it. Assume that η_i= a_i(β); at that point, μ = i if η₁ = 1 and η_k = 0 for 2 ≤ k ≤ i − 2 but η_i ≠ 0. These ${\eta }_{i^{´}s}$ are known as focal values. For i ≤ 8 functions a_i(t) and η_i are given in Alwash and Llyod [22]. For i = 9 N, Yasmin calculated a₉(t) and η₉ in [27]. For i = 10 we have calculated a₁₀(t) and η₁₀ in Nawaz [28], also presented in theorem 2.1 and 2.2.

In section 2, we write formulas with which we can calculate the highest focal value, and we also implement stopping criteria defined in Alwash and Llyod [22]. Some required conditions and the method of perturbation are described in section 3. Section 2 and 3 are mainly concerned with the calculation of focal values, which we will utilize in section 4. In section 4, we consider polynomial coefficients for equation (6) and calculate the focal values. Some examples are given in section 5. In the last section, 6, we make discussions and conclusions.

2. Calculation of Focal Values η₁₀

In the following theorem (2.1), some functions a₂, a₃, ..., a₁₀ are given that are obtained from Equation (7) and are helpful in calculating the periodic solutions.

Theorem 2.1. For the equation (7), conclusive functions a₂, a₃, ..., a₈ are given in Alwash and Llyod [22], and a₉, a₁₀ are described below:

$\begin{array}{l}{a}_9={\overline{\delta }}^8+7{\overline{\delta }}^6\overline{\gamma }+\overline{{\overline{\delta }}^6\gamma }+6{\overline{\delta }}^5\overline{\overline{\delta }\gamma }+2\overline{{\overline{\delta }}^5\gamma }\overline{\delta }+5{\overline{\delta }}^4\overline{{\overline{\delta }}^2\gamma }\\ +3\overline{{\overline{\delta }}^4\gamma }\overline{\gamma }+3\overline{{\overline{\delta }}^3\gamma }{\overline{\delta }}^2+5\overline{{\overline{\delta }}^4\gamma \overline{\gamma }}+\frac{39}{2}{\overline{\delta }}^4{\overline{\gamma }}^2-2{\overline{\delta }}^3\overline{\delta {\overline{\gamma }}^2}\\ +24{\overline{\delta }}^3\overline{\overline{\delta }\gamma }\overline{\gamma }+6\overline{{\overline{\delta }}^3\gamma \overline{\gamma }}\overline{\delta }-10\overline{{\overline{\delta }}^3\gamma \overline{\overline{\delta }\gamma }}+12\overline{\delta }\overline{\gamma }\overline{{\overline{\delta }}^3\gamma }\\ +4\overline{\overline{\gamma }\delta \overline{{\overline{\delta }}^3\gamma }}+4\overline{{\overline{\delta }}^3\gamma }{\overline{\delta }}^3+\frac{43}{6}{\overline{\delta }}^2{\overline{\gamma }}^3+4\overline{{\overline{\gamma }}^3\delta \overline{\delta }}+4{\overline{\delta }}^2\overline{\gamma }\overline{\gamma {\overline{\delta }}^2}\\ -10\overline{\delta \overline{\delta }\overline{\gamma }\overline{\gamma {\overline{\delta }}^2}}+\frac{15}{2}{\overline{\gamma }}^2\overline{{\overline{\delta }}^2\gamma }+2{\overline{\delta }}^2\overline{{\overline{\delta }}^2\gamma }-4\overline{{\overline{\delta }}^3\gamma }{\overline{\delta }}^3+\frac{43}{6}{\overline{\delta }}^2{\overline{\gamma }}^3\\ +4\overline{{\overline{\gamma }}^3\delta \overline{\delta }}+4{\overline{\delta }}^2\overline{\gamma }\overline{\gamma {\overline{\delta }}^2}-10\overline{\delta \overline{\delta }\overline{\gamma }\overline{\gamma {\overline{\delta }}^2}}+\frac{15}{2}{\overline{\gamma }}^2\overline{{\overline{\delta }}^2\gamma }\\ +2{\overline{\delta }}^2\overline{{\overline{\delta }}^2\gamma }-2{\overline{\delta }}^4\overline{\gamma }+8\overline{\overline{\gamma }\delta {\overline{\delta }}^3}+2\overline{\delta }\overline{{\overline{\delta }}^2\gamma \overline{\overline{\delta }\gamma }}\\ +26\overline{\overline{\delta }\gamma {\overline{\delta }}^2\gamma }\overline{\delta }+6\overline{{\overline{\delta }}^2\gamma }\overline{\gamma }-6\overline{{\overline{\delta }}^2\gamma \overline{\gamma }}+12{\overline{\delta }}^2\overline{\overline{\delta }\gamma \overline{\gamma }}\\ +16\overline{{\overline{\delta }}^2\gamma }\delta \overline{\delta }\overline{\gamma }-16\overline{{\overline{\delta }}^3\gamma \overline{\overline{\delta }\gamma }}+9{\overline{\delta }}^2\overline{{(\overline{\delta }\gamma )}^2}+9\overline{{(\overline{\delta }\gamma )}^2}\overline{\gamma }-\\ \overline{\delta {\overline{\gamma }}^3}\overline{\delta }+\frac{35}{8}{\overline{\gamma }}^4-6\overline{\delta }\overline{\gamma }\overline{\delta {\overline{\gamma }}^2}+8\overline{\delta {\overline{\delta }}^4\gamma \overline{\gamma }}-2\overline{\gamma \overline{\delta }\overline{{\overline{\delta }}^4\gamma }}+\frac{1}{2}{\overline{\delta }}^4\overline{\overline{\delta }\gamma }\\ +2\overline{\delta }\overline{\overline{\delta }\gamma }\overline{{\overline{\delta }}^3\gamma }+\overline{\delta \overline{\overline{\delta }\gamma }\overline{\delta {\overline{\gamma }}^2}}+\overline{\delta }{\left(\overline{{\overline{\delta }}^2\gamma }\right)}^2.\end{array}$

and

$\begin{array}{l}{a}_{10}={\overline{\delta }}^9-\frac{23}{2}\overline{{\overline{\delta }}^7\gamma }-\frac{1235}{6}\overline{{\overline{\delta }}^5\overline{\gamma }\gamma }+3\overline{{\overline{\delta }}^5\gamma }\overline{\gamma }+111\overline{\gamma }{\overline{\delta }}^4\overline{\overline{\delta }\gamma }\\ -444\overline{\overline{\gamma }\delta {\overline{\delta }}^3\overline{\overline{\delta }\gamma }}+20\overline{\gamma }\overline{\delta }\overline{{\overline{\delta }}^4\gamma }-12\overline{\overline{\gamma }\delta \overline{{\overline{\delta }}^4\gamma }}+\frac{214}{3}\overline{\gamma }{\overline{\delta }}^3\overline{{\overline{\delta }}^2\gamma }\\ \begin{array}{l}+3\gamma {\overline{\delta }}^7-160\overline{\overline{\gamma }\delta {\overline{\delta }}^2\overline{{\overline{\delta }}^2\gamma }}+\frac{15}{2}{\overline{\gamma }}^2\overline{{\overline{\delta }}^3\gamma }-\frac{970}{3}\overline{\gamma {\overline{\gamma }}^2{\overline{\delta }}^3}\\ +30\overline{\gamma }{\overline{\delta }}^2\overline{{\overline{\delta }}^3\gamma }-68\overline{\overline{\gamma }\delta \overline{\delta }\overline{{\overline{\delta }}^3\gamma }}+9\overline{\gamma }\overline{{\overline{\delta }}^3\gamma \overline{\gamma }}+\frac{1015}{9}{\overline{\gamma }}^3{\overline{\delta }}^3\\ -237\overline{\delta {\overline{\delta }}^2{\overline{\gamma }}^3}+8\overline{\gamma }{\overline{\delta }}^7-\frac{11}{2}\overline{\gamma }{\overline{\delta }}^2\overline{\delta {\overline{\gamma }}^2}+26\overline{\overline{\gamma }\delta \overline{\delta }\overline{\delta {\overline{\gamma }}^2}}\\ +\frac{319}{2}{\overline{\gamma }}^2{\overline{\delta }}^2\overline{\overline{\delta }\gamma }-174\overline{\delta \overline{\delta }{\overline{\gamma }}^2\overline{\overline{\delta }\gamma }}-90\overline{\overline{\gamma }\gamma \overline{\delta }\overline{{\overline{\delta }}^2\gamma }}+24\overline{\gamma }\overline{{\overline{\delta }}^2\gamma \delta \overline{\gamma }}\\ +40\overline{\gamma }\overline{\delta }\overline{\gamma \overline{\gamma }{\overline{\delta }}^2}-24\overline{\gamma }\delta \overline{\gamma \overline{\gamma }{\overline{\delta }}^2}+3\overline{\gamma }\overline{{\overline{\delta }}^2\gamma \overline{\overline{\delta }\gamma }}-154\overline{\overline{\gamma }\gamma {\overline{\delta }}^{{}^2}\overline{\overline{\delta }\gamma }}\\ -24\overline{\gamma {\overline{\gamma }}^2{\overline{\delta }}^2\delta }+70\overline{\gamma }\overline{\delta }{\left(\overline{\overline{\delta }\gamma }\right)}^2+42\overline{\overline{\gamma }\delta {\left(\overline{\overline{\delta }\gamma }\right)}^2}-70\overline{\overline{\gamma }{\overline{\delta }}^3{\gamma }^2}\\ -\frac{3}{2}\overline{\gamma }\overline{\delta {\overline{\gamma }}^3}-21\overline{\delta {\overline{\gamma }}^4}+\overline{\delta \overline{\overline{\delta }\gamma }\overline{\delta {\overline{\gamma }}^2}}-\frac{15}{4}{\overline{\gamma }}^2\overline{\delta {\overline{\gamma }}^2}+\frac{169}{4}{\overline{\gamma }}^4\overline{\delta }\\ +24\gamma {\overline{\gamma }}^2\overline{\delta }\overline{{\overline{\delta }}^2\gamma }-24\overline{{\overline{\gamma }}^2\delta \overline{{\overline{\delta }}^2\gamma }}+10{\overline{\gamma }}^3\overline{\overline{\delta }\gamma }+\frac{9}{2}{\overline{\delta }}^4\overline{{\overline{\delta }}^3\gamma }\\ -74\overline{\gamma {\overline{\gamma }}^3\overline{\delta }}+8\overline{\delta }\overline{\delta \overline{\delta }{\overline{\gamma }}^3}-5\overline{\gamma {\overline{\delta }}^6}-15\overline{{\overline{\delta }}^5{\gamma }^2}\\ +\frac{34}{3}\overline{\gamma }{\overline{\delta }}^3\overline{{\overline{\delta }}^2\gamma }+2\overline{\delta }\overline{{\overline{\delta }}^6\gamma }+7{\overline{\delta }}^6\overline{\overline{\delta }\gamma }+6{\overline{\delta }}^5\overline{{\overline{\delta }}^2\gamma }-6\overline{\gamma \overline{\delta }\overline{{\overline{\delta }}^4\gamma }}\\ +2{\overline{\delta }}^3\overline{{\overline{\delta }}^3\gamma }+10\overline{\delta }\overline{\gamma \overline{\gamma }{\overline{\delta }}^4}+26{\overline{\gamma }}^2{\overline{\delta }}^5-\frac{5}{2}{\overline{\delta }}^4\overline{\delta {\overline{\gamma }}^2}+\frac{5}{2}\overline{\delta {\overline{\delta }}^4{\overline{\gamma }}^2}\\ +\frac{73}{2}\overline{\gamma }{\overline{\delta }}^4\overline{\overline{\delta }\gamma }-\frac{127}{2}\overline{{\overline{\delta }}^4\gamma \overline{\overline{\delta }\gamma }}+9{\overline{\delta }}^2\overline{\gamma \overline{\gamma }{\overline{\delta }}^3}-20\overline{\delta }\overline{{\overline{\delta }}^3\gamma \overline{\overline{\delta }\gamma }}\\ +19\gamma {\overline{\gamma }}^2\overline{{\overline{\delta }}^3\gamma }-21\overline{{\overline{\delta }}^2\gamma \overline{{\overline{\delta }}^3\gamma }}+8\overline{\delta }\overline{\overline{\gamma }\delta \overline{{\overline{\delta }}^3\gamma }}-\frac{160}{3}\overline{\gamma {\overline{\delta }}^3\overline{{\overline{\delta }}^2\gamma }}\\ -\frac{4}{5}\overline{\gamma }{\overline{\delta }}^5+32\overline{{\overline{\delta }}^4\gamma \overline{\delta \overline{\gamma }}}-20\overline{\delta }\overline{\overline{\gamma }\overline{\delta }\delta \overline{{\overline{\delta }}^2\gamma }}+24\overline{\delta }{\overline{\gamma }}^2\overline{{\overline{\delta }}^2\gamma }\\ +\frac{4}{3}{\overline{\delta }}^3\overline{{\overline{\delta }}^2\gamma }-\frac{31}{30}\overline{\gamma {\overline{\delta }}^5}+16\overline{\delta }\overline{{\overline{\delta }}^3\overline{\gamma }\delta }-16\overline{\overline{\gamma }\delta {\overline{\delta }}^4}+\frac{13}{2}\overline{{\overline{\delta }}^2\gamma \overline{\delta {\gamma }^2}}\\ +3{\overline{\delta }}^2\overline{{\overline{\delta }}^2\gamma \overline{\overline{\delta }\gamma }}+42{\overline{\delta }}^2\overline{{\overline{\delta }}^2\gamma }\overline{\overline{\delta }\gamma }+12\overline{\gamma }\overline{\delta }\overline{{\overline{\delta }}^2\gamma }-12\overline{\gamma \overline{\delta }\overline{{\overline{\delta }}^2\gamma }}\\ -12\overline{\delta }\overline{{\overline{\delta }}^2\overline{\gamma }\gamma }+12{\overline{\delta }}^3\overline{{\overline{\delta }}^2\overline{\gamma }\gamma }+32\overline{{\delta }^2\overline{\gamma }\overline{\delta }\overline{{\overline{\delta }}^2\gamma }}-32\overline{\delta }\overline{{\overline{\delta }}^3\gamma \overline{\delta \overline{\gamma }}}\\ +14{\overline{\delta }}^3{\left(\overline{\overline{\delta }\gamma }\right)}^2-28\gamma \overline{\delta }{\left(\overline{\overline{\delta }\gamma }\right)}^2-\frac{3}{2}{\overline{\delta }}^2\overline{\delta {\overline{\gamma }}^3}+\frac{1}{2}{\overline{\delta }}^4\overline{\overline{\delta }\gamma }\\ +12\overline{\delta }\overline{\delta {\overline{\gamma }}^2}\overline{\delta \overline{\gamma }}-8\overline{\delta {\overline{\delta }}^4\gamma \overline{\gamma }}-2\overline{\gamma \overline{\delta }\overline{{\overline{\delta }}^4\gamma }}+2\overline{\delta }\overline{\overline{\delta }\gamma }\overline{{\overline{\delta }}^3\gamma }+\overline{\delta }{\left(\overline{{\overline{\delta }}^2\gamma }\right)}^2\\ -36\overline{\left({\overline{\gamma }}^2\delta \overline{\delta }\overline{\left(\delta \overline{\gamma }\right)}\right)}-48\overline{\delta }\overline{\left({\overline{\gamma }}^2\delta \overline{\left(\overline{\delta }\gamma \right)}\right)}-16\overline{\left(\gamma \overline{\delta }\overline{\left(\overline{\gamma }{\overline{\delta }}^2\gamma \right)}\right)}\\ -8\overline{\delta }\overline{\left(\delta \overline{\gamma }\overline{\left(\delta {\overline{\gamma }}^2\right)}\right)}+8{\overline{\delta }}^2\overline{\left({\overline{\delta }}^2\gamma \delta \overline{\gamma }\right)}.\end{array}\end{array}$

By using these functions, we obtained the next theorem, 2, which enables us to find the maximum multiplicity in which the integral is like $\int \gamma (t)\overline{\delta (t)}dt$; bar “−” shows that integral $\overline{\delta (t)}={\int }_0^t\delta (t)dt$ is definite.

Theorem 2.2. The solution z = 0 of (6) has a multiplicity k, wherever 2 ≤ k ≤ 10 iff η_n = 0 for 2 ≤ n ≤ k − 1 and η_n ≠ 0 where

${\eta }_2={\int }_0^{\beta }\delta$

${\eta }_3={\int }_0^{\beta }\gamma$

${\eta }_4={\int }_0^{\beta }\gamma \overline{\delta }$

${\eta }_5={\int }_0^{\beta }\gamma {\overline{\delta }}^2$

${\eta }_6={\int }_0^{\beta }\gamma {\overline{\delta }}^3-\frac{1}{2}{\overline{\gamma }}^2\delta$

${\eta }_7={\int }_0^{\beta }\gamma {\overline{\delta }}^4+2\gamma {\overline{\delta }}^2\overline{\gamma }$

${\eta }_8={\int }_0^{\beta }\gamma {\overline{\delta }}^5+3\gamma {\overline{\delta }}^3\overline{\gamma }+\gamma {\overline{\delta }}^2\overline{\overline{\delta }\gamma }-\frac{1}{2}{\overline{\gamma }}^3\delta$

${\eta }_9={\int }_0^{\beta }\gamma {\overline{\delta }}^6-5\gamma {\overline{\delta }}^4\overline{\gamma }-2{\overline{\delta }}^3\overline{\delta \overline{\gamma }}+20\overline{\delta {\overline{\gamma }}^2}+2\overline{\delta \overline{\gamma }}\delta {\overline{\gamma }}^2$

and

${\eta }_{10}={\int }_0^{\beta }\gamma {\overline{\delta }}^7-\frac{1235}{6}\gamma \overline{\gamma }{\overline{\delta }}^5-\frac{970}{3}\gamma {\overline{\gamma }}^2{\overline{\delta }}^3-237\delta {\overline{\delta }}^2{\overline{\gamma }}^3-24\gamma {\overline{\gamma }}^2\delta {\overline{\delta }}^2-$$70\overline{\gamma }{\overline{\delta }}^3{\gamma }^2-21{\overline{\gamma }}^4\delta -74\gamma {\overline{\gamma }}^3\overline{\delta }+\frac{5}{2}{\overline{\gamma }}^2\delta {\overline{\delta }}^4+32{\overline{\delta }}^4\gamma \overline{\delta \overline{\gamma }}-16\delta {\overline{\delta }}^4\overline{\gamma }-15{\overline{\delta }}^5{\gamma }^2$ $-36\delta \overline{\delta }{\overline{\gamma }}^2\overline{\delta \overline{\gamma }}-8\delta {\overline{\delta }}^4\gamma \overline{\gamma }.$

In theorem (2.1) some functions a₂, a₃, ..., a₁₀ are given that are obtained from Equation (7) and are helpful in calculating the periodic solutions. As future work, one can calculate a maximum multiplicity >10 by firstly generalizing theorem 2.1 and 2.2. This should be calculated by substituting the value of i > 10 into Equation (7).

3. Conditions for the Center and Method of Perturbation

In this section, we describe some conditions for the center. From theorem 2.2, we find the maximum value μ for different classes of equations. We have to stop calculating multiplicity η_k. We need some conditions that assure that there is no need to proceed further with η_k. For this, we require some conditions that are sufficient for z = 0 as a center. The conditions are given in theorem 3 and corollary 4.

Theorem 3.1. Consider that there are continuous functions f, g defined on I = σ([0, α]) and differentiable function σ with σ(α) = σ(0) such that

$\begin{array}{l}\gamma (t)=f(\sigma (t))\stackrel{·}{\sigma }\end{array}$

$\begin{array}{l}\delta (t)=g(\sigma (t))\stackrel{·}{\sigma },\end{array}$

then the origin is a center for (6).

Corollary 3.2. Consider that if any δ or γ is identically 0 and the other has mean value zero. The origin is a center.

For more detail, see [17, 19, 20]. After determining the maximum multiplicity μ, we now have to make a series of perturbations of the coefficients, every one of which results in one periodic solution coming out of origin.

For this, suppose the equation of the form given below:

$\begin{array}{ll}\stackrel{·}{z}=\gamma (t)z^3+\delta (t)z^2& (8)\end{array}$

having multiplicity μ = j (suppose). Let U be in the region near 0 in the complex plane containing no periodic solution except z = 0. From theorem (2.4) in Alwash and Llyod [22], the initial point that is contained in U remained fixed concerning a total number of periodic solutions. With the condition that perturbations of the coefficients considered are small enough, our goal is to get η₂ = η₃ = ... = η_j−2 = 0 but η_j−1 ≠ 0 by perturbing and making suitable choices of γ and δ, if possible. Obviously the most effective solutions in U and ψ are zero solutions while we get periodic solution ψ(t), where ψ(0) ∈ U as a non-trivial solution. By considering the underlying fact that the complex solutions always appear in conjugate pairs, we can say that ψ is real. Now, let U₁ and V₁ be the neighborhood of zero and ψ, respectively, such that V₁ ∪ U₁ ⊂ U and V₁ ∩ U₁ = γ. The periodic solutions around V₁ and U₁ are preserved when we make a small perturbation in the coefficients. By applying the same procedure as above, our choice is to perturb the coefficients such that η_k = 0 for k = 2, 3, ..., j − 3, but η_j ≠ 0. So that we get μ = j − 2. By applying that procedure, we get two non-trivial real periodic solutions and the zero solution is of multiplicity j − 2. Continuously, in this way, we end up with Equation (8) with μ = 2 and j − 2 being distinct non-trivial (other than zero) real periodic solutions.

4. Polynomial Coefficients for Some Classes

Let C_i,j indicate the class of the shape (6) in which the degree of γ is i and δ is j and these are polynomial in “t” only. We consider some classes C_7,3, C_7,5, C_7,6, C_3,8, C_4,3, C_9,1 and will evaluate the maximum multiplicity; for more classes, see [19, 20, 28]. These are described below in the form of theorems as:

Theorem 4.1. Let C_7,3 be class of equation of the form

$\begin{array}{ll}\stackrel{·}{z}=\gamma (t)z^3+\delta (t)z^2.& (9)\end{array}$

with

$\begin{array}{l}\gamma (t)=a+b(2t)+c{(2t)}^2+d{(2t)}^3+e{(2t)}^4+h{(2t)}^7,\\ \\ \begin{array}{l}\delta (t)=i+l{(2t)}^3.\end{array}\end{array}$

where the degree of γ(t) is 7 and δ(t) is 3. Then, μ_max(C_7,3) ≥ 10.

Proof. By using theorem 2.2, we calculate

$\begin{array}{ll}{\eta }_2=i+2l,& (10)\end{array}$

$\begin{array}{ll}{\eta }_3=a+b+\frac{4}{3}c+2d+\frac{16}{5}e+16h.& (11)\end{array}$

Thus, multiplicity of z = 0 is μ = 2 if η₂ ≠ 0, and multiplicity μ = 3 if η₂ = 0 but η₃ ≠ 0. If η₂ = η₃ = 0, then from (10) and (11), we take

$\begin{array}{ll}i=-2l,& (12)\end{array}$

and

$\begin{array}{ll}a=-b-c\frac{4}{3}-2d-\frac{16}{5}e-16h.& (13)\end{array}$

Now, by using (12) and (13), γ(t) and δ(t) take the form of:

$\begin{array}{l}\gamma (t)=b(2t-1)+c[{\left(2t\right)}^2-\frac{4}{3}]+d[{\left(2t\right)}^3-2]+e[{\left(2t\right)}^4\\ -\frac{16}{5}]+h[{\left(2t\right)}^7-16],\end{array}$

$\begin{array}{l}\delta (t)=l\left[{(2t)}^3-2\right].\end{array}$

and η₄ is a constant multiple of “l ” given as:

$\begin{array}{l}{\eta }_4=-\frac{l(-3920h-224e+90c+105b)}{1575}.\end{array}$

If η₄ = 0 then either l = 0 or

$\begin{array}{ll}h=-\frac{224}{3920}e+\frac{90}{3920}c+\frac{105}{3920}b.& (14)\end{array}$

If l = 0, then δ(t) = 0 and η₃ = 0 shows the mean value of γ(t) is zero. So by corollary 3.2, the origin is a center. Suppose l ≠ 0. If (14) holds, then η₅ is calculated as:

$\begin{array}{l}{\eta }_5=-\frac{8l^2(264e-35c+165b)}{675675}.\end{array}$

If η₅ = 0, then either l = 0 or 264e − 35c + 165b = 0. But l ≠ 0(taken above), so we substitute

$\begin{array}{ll}e=\frac{35}{264}c-\frac{165}{264}b.& (15)\end{array}$

and calculate η₆ as:

$\begin{array}{l}{\eta }_6=\frac{cl(200984c+6416388l^2+534699b)}{81723972330}.\end{array}$

If η₆ = 0, then either c = 0 or

$\begin{array}{ll}c=-\frac{6416388}{200984}{l}^2-\frac{534699}{200984}b,& (16)\end{array}$

because we already take l ≠ 0. If c = 0 then by using (15), γ(t), and δ(t) take the following form:

$\begin{array}{l}\gamma (t)=\left[8t^3-2\right]\left[b(t^4-t)+d\right],\end{array}$

$\begin{array}{l}\delta (t)=l\left[8t^3-2\right].\end{array}$

Let σ(t) = 2t⁴ − 2t; then $\stackrel{·}{\sigma }(t)=8t^3-2$. Also, σ(0) = σ(1). So γ(t) and δ(t) are as follows:

$\begin{array}{l}\gamma (t)=\left[b(t^4-t)+d\right]\stackrel{·}{\sigma },\end{array}$

$\begin{array}{l}\delta (t)=l\stackrel{·}{\sigma }.\end{array}$

By using theorem 3.1 the origin is a center having f(σ)= [b(t⁴ − t) + d] and g(σ) = l. Thus, suppose c ≠ 0. By using (16), we have η₇ as follows:

$\begin{array}{l}{\eta }_7=\frac{491l^2(12l^2+b)(2060488754705b+22506362768324l^2-757960558278d)}{1336288792941284910}.\end{array}$

Recall that l ≠ 0 (considered above). If η₇ = 0 then either b = −12l² or

$\begin{array}{ll}b=-\frac{22506362768324}{2060488754705}{l}^2-\frac{757960558278}{2060488754705}d.& (17)\end{array}$

If (17) holds, then we find

$\begin{array}{l}{\eta }_8=\frac{491l(307857d-901484l^2)\rho }{170323661397720454173105679513834037327588400000}.\end{array}$

Here

$\begin{array}{ll}\begin{array}{c}\rho =17315692509357951934114134681d^2-\\ 5633881623608845837583322950744d{l}^2\\ -50148902845361498768071379821226736l^4.\end{array}& (18)\end{array}$

Now if η₈ = 0 then either

$\begin{array}{ll}d=-\frac{901484}{307857}{l}^2,& (19)\end{array}$

or because l ≠ 0, ρ ≠ 0. If l ≠ 0, 307857d − 901484l² ≠ 0 but (18) holds, then we have $d=p_i{l}^2.$ For i = 1, 2, with p₁ = 340.28404100, p₂ = −377.123069100, and in each case η₉ is a multiple of l⁷, and l ≠ 0 (taken above). If (19) holds, then we compute η₉ as:

$\begin{array}{l}{\eta }_9=\frac{32l^5(37026759569911l-1736569072760400)}{999670687490475}.\end{array}$

If ${\varkappa }_9=0$ then, as l ≠ 0 considered above gives that l⁵ ≠ 0, we takes value of l as:

$\begin{array}{ll}l=\frac{1736569072760400}{37026759569911}.& (20)\end{array}$

If (20) holds, then we calculate ${\varkappa }_{10}$ as:

$\begin{array}{l}{\varkappa }_{10}=-\frac{\begin{array}{c}59819098508664589134261280679103302616549721\\ 839869947627237537076566683783002143601338\\ 80839216718016650781737345237712655039420363\\ 0821875056577085440000000000000000\end{array}}{\begin{array}{c}4973878997017328732440906864985754494327312\\ 9235904875486944604484243602120619367493715\\ 976441567349747422942013452414709681196758\\ 182064297581227921\end{array}}.\end{array}$

Here, ${\varkappa }_{10}$ is equal to a constant number that is non-zero. Thus, we conclude that the multiplicity of class C_7,3 is 10, i.e., μ_max(C_7,3) ≥ 10.

Theorem 4.2. For equation

$\begin{array}{ll}\stackrel{·}{z}=\gamma (t)z^3+\delta (t)z^2& (21)\end{array}$

With

$\begin{array}{ll}\begin{array}{c}\gamma (t)=-\frac{1802968}{307857}{(\frac{1736569072760400}{37026759569911}+{ϵ}_1)}^2-\frac{163228503184}{41209750941}{ϵ}_2+\frac{656215}{121027}{ϵ}_3\\ -\frac{1080}{539}{ϵ}_4-\frac{16}{7}{ϵ}_5-16{ϵ}_6+{ϵ}_7+2(-12{(\frac{1736569072760400}{37026759569911}+{ϵ}_1)}^2-\\ (\frac{757960558278}{2060488754705}{ϵ}_2+{ϵ}_3)t+4(\frac{8065930672107}{8241955018820}{ϵ}_2-\frac{534699}{200984}{ϵ}_3+{ϵ}_4)t^2\\ +8{(\frac{1736569072760400}{37026759569911}+{ϵ}_1)}^2+{ϵ}_2)t^3+16(\frac{4742797224165}{13187128030112}{ϵ}_2-\frac{224575}{229696}{ϵ}_3\\ +\frac{35}{264}{ϵ}_4+\frac{15}{2}{(\frac{1736569072760400}{37026759569911}+{ϵ}_1)}^2+{ϵ}_5)t^4+128(-\frac{30780466431}{3878567067680}{ϵ}_2+\\ \frac{1699711}{78785728}{ϵ}_3+\frac{199}{12936}{ϵ}_4-{(\frac{1736569072760400}{37026759569911}+{ϵ}_1)}^2-\frac{2}{35}{ϵ}_5+{ϵ}_6)t^7.\end{array}\}& (22)\end{array}$

$\begin{array}{l}\delta (t)=-\frac{3473138145520800}{37026759569911}-2{ϵ}_1+{ϵ}_8\\ \begin{array}{l}+8(\frac{1736569072760400}{37026759569911}+{ϵ}_1)t^3.\end{array}& (23)\end{array}$

Choose ϵ_j for 1 ≤ j ≤ 8 to be non-zero and small as compared to ϵ_j−1. Then (21) has eight distinct non-trivial real periodic solutions.

Proof. If we substitute ϵ_p = 0, ∀ p = 1, 2, ..., 8, and coefficients are as given in Equations (22) and (23). So, the multiplicity of the origin $\varkappa$ is 10. Now, choose ϵ₁ ≠ 0 and ϵ_p = 0 for 2 ≤ p ≤ 8; then it can be easily seen that ${\varkappa }_9$ is a constant multiple of ϵ₁, but ${\varkappa }_2={\varkappa }_3=...={\varkappa }_7={\varkappa }_8=0.$ So, the multiplicity reduces by one and $\varkappa =9.$ For that reason, one periodic solution bifurcates out of the origin. Now, set ϵ₁ ≠ 0, ϵ₂ ≠ 0 and ϵ_p = 0 for 3 ≤ p ≤ 8; then we have ${\varkappa }_p=0$ for p = 2, 3, ..., 7. But ${\varkappa }_8$ results in a form of ϵ₂ with some constant multiple. So, $\varkappa =8.$ Now, set ϵ₁ ≠ 0, ϵ₂ ≠ 0, ϵ₃ ≠ 0 and ϵ_p = 0 for 4 ≤ p ≤ 8; then we have ${\varkappa }_p=0$ for p = 2, 3, ..., 6. But ${\varkappa }_7$ results in a form of ϵ₃ with some constant multiple. If ϵ₂ is sufficient small, then there are two non-trivial real periodic solutions. Further, moving in the present way, we have eight real periodic non-trivial solutions.

Corollary 4.1. For an equation

$\begin{array}{ll}\stackrel{·}{z}=\gamma (t)z^3+\delta (t)z^2+\gamma +\upsilon .& (24)\end{array}$

if γ(t) and δ(t) are as given in theorem 4.1, Equation (24) has ten real periodic solutions if γ and υ are small enough.

Proof. If γ = 0 and υ = 0, μ = 2 then (24) has eight real periodic solutions. If γ ≠ 0 but is small enough, then μ = 1 and by using the same arguments as in the above theorem, there are nine distinct periodic solutions other than 0; z = 0 is another solution. Thus, we have ten real periodic solutions.

Theorem 4.3. For class C_7,5 consider δ(t) = i + n(2t + 1)⁵ and

$\begin{array}{l}\gamma (t)=a+b(2t+1)+c{(2t+1)}^2+f{(2t+1)}^5+h{(2t+1)}^7.\end{array}$

Then μ_max(C_7,5) ≥ 8.

Proof. By using theorem 2.2, we have

$\begin{array}{l}{\eta }_2=\frac{182}{3}n+i,\end{array}$

$\begin{array}{l}{\eta }_3=410h+\frac{182}{3}f+\frac{13}{3}c+2b+a.\end{array}$

Thus, the multiplicity of z = 0 is μ = 2 if η₂ ≠ 0. And multiplicity μ = 3 if η₂ = 0 but η₃ ≠ 0. If η₂ = η₃ = 0, then we calculate η₄ as:

$\begin{array}{l}{\eta }_4=-\frac{n(-114470h+1079c+372b)}{189}.\end{array}$

If η₄ = 0 then either n = 0 or

$\begin{array}{ll}h=\frac{1079}{114470}c+\frac{372}{114470}b.& (25)\end{array}$

If n = 0, then η₃ = 0 shows i = 0; hence, δ(t) = 0 and η₂ = 0 implies that the mean value of γ(t) is zero. By corollary 3.2, the origin is a center. Suppose n ≠ 0. By using (25), we compute η₅ as:

$\begin{array}{l}{\eta }_5=-\frac{8n^2(886519816c+499588553b)}{2109395925}.\end{array}$

If η₅ = 0 then

$\begin{array}{ll}c=-\frac{499588553}{886519816}b.& (26)\end{array}$

because we already take n ≠ 0. If (26) holds, then η₆ is:

$\begin{array}{l}{\eta }_6=\frac{2bn(-379667599239958655624n^2+5813410092109719b)}{815558371657762539807}.\end{array}$

If η₆ = 0 then either b = 0 or

$\begin{array}{ll}b=\frac{379667599239958655624}{5813410092109719}{n}^2,& (27)\end{array}$

because n ≠ 0. If b = 0, then by using (26), (25), γ(t) and δ(t) take the form:

$\begin{array}{l}\gamma (t)=f[{(2t+1)}^5-\frac{182}{3}],\end{array}$

$\begin{array}{l}\delta (t)=n[{(2t+1)}^5-\frac{182}{3}].\end{array}$

Let $\sigma (t)=\frac{1}{2}{(2t+1)}^6-182t$; then $\stackrel{·}{\sigma }(t)=3{(2t+1)}^5-182$. Also, σ(0) = σ(1). So, we can write it as:

$\begin{array}{l}\gamma (t)=\frac{1}{3}f\stackrel{·}{\sigma },\\ \delta (t)=\frac{1}{3}n\stackrel{·}{\sigma }.\end{array}$

Then, by theorem 3.1, the origin is a center with f(σ)= $\frac{1}{3}f$ and $g(\sigma )=\frac{1}{3}n.$ So we take b ≠ 0. If (27) holds, then η₇ is as follows:

$\begin{array}{l}586877587954432n^4(-9888560913986218905316\\ 8639013752599n^2+54255683216749379832543\\ {\eta }_7=\frac{\mathrm{161535450f})}{392738038968759387059090863949933323736325}\end{array}$

Now, if η₇ = 0, recalling that n ≠ 0 then

$\begin{array}{ll}f=\frac{98885609139862189053168639013752599}{54255683216749379832543161535450}{n}^2.& (28)\end{array}$

With holding (28), we have

$\begin{array}{l}{\eta }_8=\frac{2300397726692597332894199628622545916981609410130707279936768}{793823482030814697509595313095537997395261781753311944606875}{n}^7.\end{array}$

Which is a constant multiple of n⁷ and is non-zero because n ≠ 0 (taken above). Thus we conclude that the multiplicity of class C_7,5 is 8, i.e., μ_max(C_7,5) ≥ 8.

Theorem 4.4. For equation

$\begin{array}{ll}\stackrel{·}{z}=\gamma (t)z^3+\delta (t)z^2+{\sigma }_{1}z+{\sigma }_2.& (29)\end{array}$

Let

$\begin{array}{l}\delta (t)=-\frac{182}{3}n+{ϵ}_6+n{(2t+1)}^5,\end{array}$

$\begin{array}{l}\gamma (t)=-v_1+{ϵ}_5+b(2t+1)+c{(2t+1)}^2+f{(2t+1)}^5+h{(2t+1)}^7.\end{array}$

Proof. With

$\begin{array}{l}{v}_1=-410h-\frac{182}{3}f-\frac{13}{3}c-2b,\\ b=\frac{379667599239958655624}{5813410092109719}{n}^2+{ϵ}_2,\\ c=-\frac{19470134860245482491747}{529020318381984429}{n}^2-\frac{499588553}{886519816}{ϵ}_2+{ϵ}_3,\\ f=\frac{98885609139862189053168639013752599}{54255683216749379832543161535450}{n}^2+{ϵ}_1,\end{array}$

and

$\begin{array}{l}h=-\frac{576907546430940497}{4283565331028214}{n}^2-\frac{281257}{136387664}{ϵ}_2+\frac{1079}{114470}{ϵ}_3+{ϵ}_4.\end{array}$

If ϵ_j(1 ≤ j ≤ 6), σ₁ and σ₂ are chosen to be non-zero and also

$\begin{array}{l}\left|{\sigma }_2\right|\ll \left|{\sigma }_1\right|\ll \left|{ϵ}_6\right|\ll \left|{ϵ}_5\right|\ll ...\ll \left|{ϵ}_1\right|.\end{array}$

Then (29) has eight distinct real periodic solutions other than zero.

Theorem 4.5. Let δ(t) = j + p(t − 1)⁶ and

$\begin{array}{l}\gamma (t)=a+c{(t-1)}^2+d{(t-1)}^3+g{(t-1)}^6+h{(t-1)}^7.\end{array}$

For class C_7,6 of the form (9), then μ_max(C_7,6) ≥ 8.

Proof. By using theorem 2.2, we have

$\begin{array}{ll}{\eta }_2=\frac{1}{7}p+j,& (30)\end{array}$

$\begin{array}{ll}{\eta }_3=-\frac{1}{8}h+\frac{1}{7}g-\frac{1}{4}d+\frac{1}{3}c+a.& (31)\end{array}$

Thus, the multiplicity of z = 0 is μ = 2 if η₂ ≠ 0, and multiplicity μ = 3 if η₂ = 0 but η₃ ≠ 0. If η₂ = η₃ = 0, by using values of a & j, , δ(t) and γ(t) are as follows:

$\begin{array}{l}\gamma (t)=c[{(t-1)}^2-\frac{1}{3}]+d[{(t-1)}^3+\frac{1}{4}]+g[{(t-1)}^6-\frac{1}{7}]\\ +h[{(t-1)}^7-\frac{1}{8}],\end{array}$

$\begin{array}{l}\delta (t)=p[{(t-1)}^6-\frac{1}{7}].\end{array}$

Also, we calculate η₄ as:

$\begin{array}{l}{\eta }_4=\frac{p(792c-486d+77h)}{221760}.\end{array}$

If η₄ = 0 then either p = 0 or

$\begin{array}{ll}h=\frac{486}{77}d-\frac{792}{77}c.& (32)\end{array}$

If p = 0, then δ(t) = 0 and η₃ = 0 implies that mean value γ(t) = 0. From corollary 3.2, the origin is a center, so consider p ≠ 0. If (32) holds, then we have η₅ as:

$\begin{array}{l}{\eta }_5=-\frac{p^2(484c-51d)}{54428220}.\end{array}$

If η₅ = 0, then as p ≠ 0 (considered above) implies

$\begin{array}{ll}d=\frac{484}{51}c.& (33)\end{array}$

And by using (33), we calculate η₆ as:

$\begin{array}{l}{\eta }_6=\frac{cp(51382814976p^2+520996995995c)}{115437830545837800}.\end{array}$

If η₆ = 0, then either c = 0 or

$\begin{array}{ll}c=-\frac{51382814976}{520996995995}{p}^2,& (34)\end{array}$

because p ≠ 0. If c = 0 then γ(t) and δ(t) are:

$\begin{array}{l}\gamma (t)=g[{(t-1)}^6-\frac{1}{7}],\\ \delta (t)=p[{(t-1)}^6-\frac{1}{7}].\end{array}$

Let σ(t) = (t − 1)⁷ − t, then $\stackrel{·}{\sigma }(t)=7{(t-1)}^6-1$. Also, σ(0) = σ(1). So it takes new form as:

$\begin{array}{l}\gamma (t)=\frac{g}{7}\stackrel{·}{\sigma },\\ \delta (t)=\frac{p}{7}\stackrel{·}{\sigma }.\end{array}$

By theorem 3.1, having f(σ)= $\frac{g}{7}$ and $g(\sigma )=\frac{p}{7}$, the origin is a center, so take c ≠ 0. Using (34), we have η₇ as:

$\begin{array}{l}{\eta }_7=-\frac{41979424p^4(4135364653107809477799p^2+748144295421365642240g)}{536575324409227144872262909825473125}.\end{array}$

If η₇ = 0, recalling that p ≠ 0 (η₅), then

$\begin{array}{ll}g=-\frac{4135364653107809477799}{748144295421365642240}{p}^2.& (35)\end{array}$

If (35) holds, then we find η₈ as:

$\begin{array}{l}{\eta }_8=\frac{6795652249525465319539097007446902881077050577}{31767297065597743067007681874695840512233681534101600000}{p}^7.\end{array}$

which is a constant multiple of p⁵, and p is also non-zero (as shown above). Thus, we conclude that the multiplicity of class C_7,6 is 8, i.e., μ_max(C_7,6) ≥ 8.

Theorem 4.6. Let C_9,1 be a class of equation of the form (9), with

$\begin{array}{l}\gamma (t)=c+dt+e{t}^2+f{t}^3+k{t}^8+l{t}^9.\end{array}$

$\begin{array}{l}\delta (t)=m+nt.\end{array}$

We then see that μ_max(C_9,1) ≥ 10.

Proof. Using theorem 2.2, we take

$\begin{array}{l}{\eta }_2=m+\frac{1}{2}n,\\ \begin{array}{l}{\eta }_3=c+\frac{1}{2}d+\frac{1}{3}e+\frac{1}{4}f+\frac{1}{9}k+\frac{1}{10}l.\end{array}\end{array}$

Thus, the multiplicity of z = 0 is μ = 2 if η₂ ≠ 0, and the multiplicity μ = 3 if η₂ = 0 but η₃ ≠ 0. If η₂ = η₃ = 0, then by using values of “k” and “a, ” γ(t) and δ(t) are as follows:

$\begin{array}{ll}\gamma (t)=d(t-\frac{1}{2})+e\left(t^2-\frac{1}{3}\right)+f\left(t^3-\frac{1}{4}\right)+k\left(t^8-\frac{1}{9}\right)+l\left(t^9-\frac{1}{10}\right),& (36)\end{array}$

$\begin{array}{ll}\delta (t)=n\left(t-\frac{1}{2}\right).& (37)\end{array}$

Also, we compute η₄, given below as:

$\begin{array}{l}{\eta }_4=\frac{n(108l+112k+99f+66e)}{23760}.\end{array}$

If η₄ = 0 then either n = 0 or

$\begin{array}{ll}l=-\frac{112}{108}k-\frac{99}{108}f-\frac{66}{108}e.& (38)\end{array}$

If n = 0, then δ(t) = 0 and η₃ = 0 gives that the mean value of γ(t) is zero. Thus, the origin is a center from corollary 3.2, so consider n ≠ 0. Now, if (38) holds, then η₅ is as below:

$\begin{array}{l}{\eta }_5=-\frac{n^2(56k+297f+198e)}{7207200}.\end{array}$

If η₅ = 0, then as we already take n ≠ 0 it implies

$\begin{array}{ll}k=-\frac{297}{56}f-\frac{198}{56}e.& (39)\end{array}$

and by using (39), η₆ is:

$\begin{array}{l}{\eta }_6=-\frac{n(2e+3f)(32e-57n^2+29f)}{836559360}.\end{array}$

If η₆ = 0, then as we already consider n ≠ 0 either $f=-\frac{2}{3}e$ or

$\begin{array}{ll}e=\frac{57}{32}{n}^2-\frac{29}{32}f.& (40)\end{array}$

If $f=-\frac{2}{3}e$ then (36) and (37) are of the following form:

$\begin{array}{l}\gamma (t)=\left(t-\frac{1}{2}\right)\left[d+e\left(-\frac{2}{3}{t}^2+\frac{2}{3}t-\frac{1}{3}\right)\right],\end{array}$

$\begin{array}{l}\delta (t)=n\left(t-\frac{1}{2}\right).\end{array}$

Let σ(t) = t² − t; then $\stackrel{·}{\sigma }(t)=2t-1$. Also, σ(0) = σ(1). So we can write

$\begin{array}{l}\gamma (t)=\frac{1}{2}\left[d+c\left(-\frac{2}{3}{t}^2+\frac{2}{3}t\right)\right]\stackrel{·}{\sigma },\end{array}$

$\begin{array}{l}\delta (t)=\frac{n}{2}\stackrel{·}{\sigma }.\end{array}$

Thus, from theorem 3.1, the origin is a center with

$\begin{array}{l}f(\sigma )=\frac{1}{2}\left[d+c\left(-\frac{2}{3}{t}^2+\frac{2}{3}t\right)\right],\end{array}$

and $g(\sigma )=\frac{n}{2},$ so we take $f\ne -\frac{2}{3}e.$ Holding (40), we compute η₇ as:

$\begin{array}{l}{\eta }_7=\frac{19n^2(3n^2+f)(-348307f+1697231n^2+1445136d)}{254514115215360}.\end{array}$

If η₇ = 0, recalling that n ≠ 0, then either f = −3n² or

$\begin{array}{ll}f=\frac{1697231}{348307}{n}^2+\frac{1445136}{348307}d.& (41)\end{array}$

If f = −3n², then

$\begin{array}{l}\gamma (t)=\frac{1}{2}\left[d+n^2(-3t^2+3t)\right]\stackrel{·}{\sigma },\end{array}$

$\begin{array}{l}\delta (t)=\frac{n}{2}\stackrel{·}{\sigma }.\end{array}$

From theorem (3.1), the origin is a center with f(σ)= $\frac{1}{2}\left[d+n^2(-3t^2+3t)\right]$ and $g(\sigma )=\frac{n}{2},$ so consider f ≠ −3n². Using (41), we calculate η₈ as:

$\begin{array}{l}{\eta }_8=\frac{23n(1122d+2129n^2)\zeta }{1029474284594079894479022764851200}.\end{array}$

where ζ = 273879615326996052d² + 1713735341555455508dn² − 132695961322089231627n⁴. Now, if η₈ = 0 then either ζ = 0 or

$\begin{array}{ll}d=-\frac{2129}{1122}{n}^2.& (42)\end{array}$

because n ≠ 0. If Equation (42) ≠ 0, n ≠ 0, but ζ = 0, then $b=r_i{n}^2$ for i = 1, 2, where r₁ = 38.208145460, r₂ = −50.722660920. If (42) holds, but ζ ≠ 0, n ≠ 0, then we compute η₉

$\begin{array}{l}{\eta }_9=-\frac{n^5(5168n+449059)}{2749402656}.\end{array}$

If η₉ = 0, then we substitute $n=-\frac{449059}{5168}$ and calculate η₁₀ as:

$\begin{array}{l}{\eta }_{10}=-\frac{\begin{array}{c}277526056388652430908651014962873\\ 088740929681514867177588679\end{array}}{\begin{array}{c}164738202978811962713659536291595\\ 7451747785441280\end{array}}\end{array}$

That is a non-zero constant number. Thus, we conclude that the multiplicity of class C_9,1 is 10, i.e., μ_max(C_9,1) ≥ 10.

Theorem 4.7. Choose n, k, l with nl ≠ 0. In the equation

$\begin{array}{ll}\stackrel{·}{z}=\gamma (t)z^3+\delta (t)z^2+{\sigma }_{1}z+{\sigma }_2.& (43)\end{array}$

Let

$\begin{array}{l}\delta (t)=\frac{449059}{10336}-\frac{1}{2}{ϵ}_1+{ϵ}_8+(-\frac{449059}{5168}+{ϵ}_1)t,\end{array}$

and

$\begin{array}{l}\gamma (t)=u_1+dt+e{t}^2+f{t}^3+k{t}^8+l{t}^9.\end{array}$

With

$\begin{array}{l}{u}_1=\frac{223}{1122}{(-\frac{449059}{5168}+{ϵ}_1)}^2-\frac{287723}{4179684}{ϵ}_2+\frac{419}{4032}{ϵ}_3\\ -\frac{31}{126}{ϵ}_4-\frac{1}{135}{ϵ}_5-\frac{1}{10}{ϵ}_6+{ϵ}_7,\\ d=-\frac{2129}{1122}{(-\frac{449059}{5168}+{ϵ}_1)}^2+{ϵ}_2,\\ f=-3{(-\frac{449059}{5168}+{ϵ}_1)}^2+\frac{1445136}{348307}{ϵ}_2+{ϵ}_3,\\ e=\frac{9}{2}{(-\frac{449059}{5168}+{ϵ}_1)}^2-\frac{2619309}{696614}{ϵ}_2-\frac{29}{32}{ϵ}_3+{ϵ}_4,\\ k=-\frac{24270543}{2786456}{ϵ}_2-\frac{1881}{896}{ϵ}_3-\frac{99}{28}{ϵ}_4+{ϵ}_5,\end{array}$

and

$\begin{array}{l}l=\frac{31461815}{4179684}{ϵ}_2+\frac{1045}{576}{ϵ}_3+\frac{55}{18}{ϵ}_4-\frac{28}{27}{ϵ}_5+{ϵ}_6.\end{array}$

If ϵ_k(1 ≤ k ≤ 8), σ₁ and σ₂ are chosen to be non-zero such that

$\begin{array}{l}\left|{\sigma }_2\right|\ll \left|{\sigma }_1\right|\ll \left|{ϵ}_8\right|\ll \left|{ϵ}_7\right|\ll ...\ll \left|{ϵ}_1\right|.\end{array}$

then (43) has ten real distinct non-trivial periodic solutions.

Proof. The proof is similar to that in theorem in (4.2), so it is omitted.

It is pertinent to mention that μ_max(C_4,3) ≥ 5 given in Yasmin and Ashraf [20] but we succeeded in increasing its multiplicity from 5 to 8, i.e., μ_max(C_4,3) ≥ 8 by using variable (t) instead of (2t-1).

Theorem 4.8. Let

$\begin{array}{l}\gamma (t)=e+ft+g{t}^2+h{t}^3+i{t}^4,\end{array}$

$\begin{array}{l}\delta (t)=a+d{t}^3.\end{array}$

for the class C_4,3 of form (9). Then we prove that μ_max(C_4,3) ≥ 8.

Proof. From theorem 2.2, we calculate solutions as:

$\begin{array}{l}{\eta }_2=a+\frac{d}{4}.\end{array}$

$\begin{array}{l}{\eta }_3=e+\frac{f}{2}+\frac{g}{3}+\frac{h}{4}+\frac{i}{5}.\end{array}$

Thus, the multiplicity of z = 0 is μ = 2 if η₂ ≠ 0, and multiplicity μ = 3 if η₂ = 0 but η₃ ≠ 0. If η₂ = η₃ = 0, then

$\begin{array}{ll}a=-\frac{d}{4}.& (44)\end{array}$

$\begin{array}{ll}e=-\frac{f}{2}-\frac{g}{3}-\frac{h}{4}-\frac{i}{5}.& (45)\end{array}$

By using (44) and (45), γ(t) and δ(t) are as given below:

$\begin{array}{ll}\gamma (t)=f\left(t-\frac{1}{2}\right)+g\left(t^2-\frac{1}{3}\right)+h\left(t^3-\frac{1}{4}\right)+i\left(t^4-\frac{1}{5}\right),& (46)\end{array}$

$\begin{array}{ll}\delta (t)=d\left(t^3-\frac{1}{4}\right).& (47)\end{array}$

Also we calculate η₄ as:

$\begin{array}{l}{\eta }_4=-\frac{d(-28i+45g+105f)}{25200}.\end{array}$

If η₄ = 0 then either d = 0 or

$\begin{array}{ll}i=\frac{(45g+105f)}{28}.& (48)\end{array}$

If d = 0, then δ(t) = 0, and also η₃ = 0 gives that δ(t) has mean value 0. From corollary 3.2, the origin is a center. Consider d ≠ 0. By using (48), we compute η₅ as:

$\begin{array}{l}{\eta }_5=-\frac{d^2(199g+1617f)}{60540480}.\end{array}$

If η₅ = 0 as d ≠ 0, then

$\begin{array}{ll}f=-\frac{199}{1617}g.& (49)\end{array}$

By using (49), we compute η₆ as:

$\begin{array}{l}{\eta }_6=\frac{gd(78600753d^2+13597688g)}{2050291017815040}.\end{array}$

If η₆ = 0, then either g = 0 or

$\begin{array}{ll}g=-\frac{78600753}{13597688}{d}^2,& (50)\end{array}$

because we already take d ≠ 0. If g = 0, then (46) and (47) can be written as:

$\begin{array}{l}\gamma (t)=h\left(t^3-\frac{1}{4}\right),\end{array}$

$\begin{array}{l}\delta (t)=d\left(t^3-\frac{1}{4}\right).\end{array}$

Consider σ(t) = t⁴ − t; then $\stackrel{·}{\sigma }(t)=4t^3-1$. Also σ(0) = σ(1), so it can written as:

$\begin{array}{l}\gamma (t)=\frac{h}{4}\stackrel{·}{\sigma }(t),\end{array}$

$\begin{array}{l}\delta (t)=\frac{d}{4}\stackrel{·}{\sigma }(t).\end{array}$

By theorem 3.1, the origin is a center with $f(\sigma )=\frac{1}{4}h$ and $g(\sigma )=\frac{1}{4}d$. So take g ≠ 0. If (50) holds, then η₇ is:

$\begin{array}{l}{\eta }_7=-\frac{491d^4(-704698056497d^2+61560932862h)}{102299877970079928320}.\end{array}$

If η₇ = 0, recalling that d ≠ 0, then we take

$\begin{array}{ll}h=\frac{704698056497}{61560932862}{d}^2.& (51)\end{array}$

If (51) holds, then we compute η₈ as:

$\begin{array}{l}{\eta }_8=\frac{466979144516058112634649177}{23384478186490400805647418695680}{d}^7.\end{array}$

where η₈ is a constant multiple of d⁷ and d ≠ 0 (taken above). Thus, we conclude that the multiplicity of class C_4,3 is 8, i.e., μ_max(C_4,3) ≥ 8.

Theorem 4.9. Let C_3,8 be a class of equation of the form (9) with

$\begin{array}{l}\gamma (t)=bt+d{t}^3,\end{array}$

$\begin{array}{l}\delta (t)=ft+h{t}^3+j{t}^5+l{t}^7+m{t}^8.\end{array}$

Then μ_max(C_3,8) ≥ 8 when j = 1.

Proof. First we suppose that j ≠ 1. From theorem 2.2, we see:

$\begin{array}{ll}{\eta }_2=\frac{1}{2}f+\frac{1}{4}h+\frac{1}{6}j+\frac{1}{8}l+\frac{1}{9}m.& (52)\end{array}$

$\begin{array}{ll}{\eta }_3=\frac{1}{2}b+\frac{1}{4}d.& (53)\end{array}$

Thus the multiplicity of z = 0 is μ = 2 if η₂ ≠ 0, and multiplicity μ = 3 if η₂ = 0 but η₃ ≠ 0. If η₂ = η₃ = 0, then from (52) and (53) we take:

$\begin{array}{ll}f=-\frac{1}{2}h-\frac{1}{3}j-\frac{1}{4}l-\frac{2}{9}m,& (54)\end{array}$

and

$\begin{array}{ll}b=-\frac{1}{2}d.& (55)\end{array}$

Now by using (54) and (55), we calculate η₄ as:

$\begin{array}{l}{\eta }_4=-\frac{d(1400m+1287l+858j)}{1235520}.\end{array}$

If η₄ = 0 then either d = 0 or

$\begin{array}{ll}m=-\frac{1287}{1400}l-\frac{858}{1400}j.& (56)\end{array}$

If d = 0, then from (55), γ(t) = 0 and η₃ = 0 gives that the mean value of δ(t) is zero. So by corollary 3.2, the origin is a center. Thus, suppose d ≠ 0. If (56) holds, then η₅ is:

$\begin{array}{l}{\eta }_5=-\frac{d(2j+3l)(30744j+59850h+8041l)}{683726400000}.\end{array}$

If η₅ = 0 then either

$\begin{array}{ll}2j+3l=0,& (57)\end{array}$

or

$\begin{array}{ll}h=-\frac{30744}{59850}j-\frac{8041}{59850}l,& (58)\end{array}$

because we already take d ≠ 0. If (57) holds, then γ(t) and δ(t) are of the form:

$\begin{array}{l}\gamma (t)=d\left(t^3-\frac{t}{2}\right),\end{array}$

$\begin{array}{l}\delta (t)=\left(t^3-\frac{t}{2}\right)\left[h+j\left(\frac{\frac{3}{5600}{t}^2+\frac{88}{525}t}{\frac{3}{700}{t}^5+\frac{2}{3}{t}^4+\frac{3}{1400}{t}^3+\frac{4}{3}{t}^2-\frac{3}{2800}t+\frac{2}{5}}\right)\right].\end{array}$

Let $\sigma (t)=\frac{t^4}{4}-\frac{t^2}{4}$, then $\stackrel{·}{\sigma }(t)=t^3-\frac{t}{2}$. Also σ(0) = σ(1). γ(t) and δ(t) are thus written as:

$\begin{array}{l}\gamma (t)=d\stackrel{·}{\sigma (t)},\end{array}$

$\begin{array}{l}\delta (t)=\left[h+j\left(\frac{\frac{3}{5600}{t}^2+\frac{88}{525}t}{\frac{3}{700}{t}^5+\frac{2}{3}{t}^4+\frac{3}{1400}{t}^3+\frac{4}{3}{t}^2-d32800t+\frac{2}{5}}\right)\right]\stackrel{·}{\sigma }(t).\end{array}$

So by theorem 3.1, the origin is a center with f(σ)= d and

$\begin{array}{l}g(\sigma )=\left[h+j\left(\frac{\frac{3}{5600}{t}^2+\frac{88}{525}t}{\frac{3}{700}{t}^5+\frac{2}{3}{t}^4+\frac{3}{1400}{t}^3+\frac{4}{3}{t}^2-\frac{3}{2800}t+\frac{2}{5}}\right)\right].\end{array}$

Therefore, we suppose that 3l + 2j ≠ 0. Holding (58), we compute η₆, which is a constant multiple of ξ, as:

$\begin{array}{l}{\eta }_6=-d(3l+2j)\xi .\end{array}$

where

$\begin{array}{l}\xi =1532329720524j^2+1273229285332jl\\ +1690381942750100d+25542706569l^2.\end{array}$

Now η₆ = 0 only if ξ = 0, because we have already discussed the possibility of d = 0, (3l + 2j) = 0; in each case, the origin is a center. For ξ = 0, we substitute

$\begin{array}{l}d=-\frac{153232639720524}{169038161950100}{j}^2-\frac{127322239285332}{169038942750100}jl\\ -\frac{25542700956569}{169038161942750100}{l}^2.\end{array}$

and obtain

$\begin{array}{l}{\eta }_7=d\left(3l+2j\right)\left(\text{homogeneous cubic in }j\text{ and }l\right).\end{array}$

and

$\begin{array}{l}{\eta }_8=-d\left(3l+2j\right)\left(\text{homogeneous quartic in }j\text{ and }l\right).\end{array}$

We cannot draw any conclusion looking at η₇ and η₈. Thus, for simplification, we substitute j = 1. Then, η₇ and η₈ becomes

${\eta }_7=\{\begin{array}{c}d\left(3l+2\right)(-\frac{20052663449157741455869750}{437}\\ +\frac{525461424420752097957709100}{4807}l-\\ \frac{26054076800585569433880148475}{302841}{l}^2\\ +\frac{248636375396821626044300}{11}{l}^3).\end{array}$

and

${\eta }_8=\{\begin{array}{c}-d\left(3l+2\right)(-60491257464742335697522\\ 71066018358646532627l+\\ 704985851569387782175507850879295207088l^2\\ -367735177988424214344\\ 94904030251207123117l^3+72607900452360540\\ 07143552578113438l^4).\end{array}$

For multiplicity >6, we have to prove that the cubic in η₇ and quartic in η₈ have no common zero. For this, we suppose that both have a common zero. Then we get a linear relation in j and l, j + kl = 0(say), and for this value of j, η₈ is a constant multiple of dl⁵ ≠ 0. Thus, we conclude that the multiplicity of class C_3,8 is 8, i.e., μ_max(C_3,8) ≥ 8 with j = 1.

Theorem 4.10. Suppose l = α be a real root of the equation

$\begin{array}{l}\{\begin{array}{c}-\frac{20052663449157741455869750}{437}\\ +\frac{525461424420752097957709100}{4807}l-\\ \frac{26054076800585569433880148475}{302841}{l}^2\\ +\frac{248636375396821626044300}{11}{l}^3=0.\end{array}\end{array}$

Choose

$\begin{array}{l}l=\alpha +{ϵ}_1,\\ d=-\frac{5197145561}{573321672577500}{j}^2-\frac{38865152407}{5159895053197500}j(\alpha +{ϵ}_1)-\\ \frac{1110552215503}{734948530185870000}{(\alpha +{ϵ}_1)}^2+{ϵ}_2,\\ h=-\frac{244}{475}j-\frac{8041}{59850}l+{ϵ}_3,\\ m=-\frac{1287}{1400}\alpha -\frac{1287}{1400}{ϵ}_1-\frac{429}{700}j+{ϵ}_4,\\ f=\frac{397}{6650}j+\frac{8041}{119700}-\frac{1}{2}{ϵ}_3-\frac{8}{175}\alpha -\frac{8}{175}{ϵ}_1-\frac{2}{9}{ϵ}_4+{ϵ}_{6,}\\ b=\frac{5195561}{11461500}{j}^2+\frac{3886407}{1035000}j(\alpha +{ϵ}_1)+\frac{11103}{14600}{(\alpha +{ϵ}_1)}^2\\ -\frac{1}{2}{ϵ}_2+{ϵ}_5,\end{array}$

Such that |ϵ₆| ≪ |ϵ₅| ≪ |ϵ₄| ≪ |ϵ₃| ≪ |ϵ₂| ≪ |ϵ₁|. Then equation $\stackrel{·}{z}=\gamma (t)z^3+\delta (t)z^2,$ has six real periodic non-trivial solutions. Where

$\begin{array}{l}\gamma (t)=bt+d{t}^3.\end{array}$

$\begin{array}{l}\delta (t)=ft+h{t}^3+j{t}^5+l{t}^7+m{t}^8.\end{array}$

with j = 1.

Proof. If we put ϵ_j for j as; 1 ≤ j ≤ 6, instead of 1 ≤ j ≤ 8 then the proof is similar to that for theorem (4.2), so it is omitted.

5. Examples

The following examples demonstrate the applicability of our main results.

Example: Consider the differential equation:

$\begin{array}{ll}\frac{dz}{dt}=(e^t)z^3+(\text{cos }t)z^2.& (59)\end{array}$

Here γ(t), δ(t) are transcendental functions, but we use the power series representations by neglecting the terms ‘tⁿ’ for n > 4. Like γ(t) = e^t = a + bt + ct² + dt³ + et⁴, δ(t) = cos t = j − kt² + lt⁴, with a = b = j = 1, $c=k=\frac{1}{2!},d=\frac{1}{3!},e=\frac{1}{4!},$ and then calculate the periodic solutions.

SOLUTION: We substitute k = 0, and by using theorem 2.2, we calculate:

$\begin{array}{ll}{\eta }_2=j+\frac{1}{120}l,& (60)\end{array}$

$\begin{array}{ll}{\eta }_3=a+\frac{1}{2}b+\frac{1}{6}c+\frac{1}{24}d+\frac{1}{120}e.& (61)\end{array}$

Thus, the multiplicity of z = 0 is μ = 2 if η₂ ≠ 0, and multiplicity μ = 3 if η₂ = 0 but η₃ ≠ 0. If η₂ = η₃ = 0, then we take $j=-\frac{1}{120}l,$ and $a=-\frac{1}{2}b-\frac{1}{6}c-\frac{1}{24}d-\frac{1}{120}e.$ By using these values, η₄ is calculated as:

$\begin{array}{l}{\eta }_4=-\frac{l(14d+105c+360b)}{1814400}.\end{array}$

If η₄ = 0 then either l = 0 or

$\begin{array}{ll}d=-\frac{105}{14}c-\frac{360}{14}b.& (62)\end{array}$

If l = 0, then δ(t) = 0 and η₃ = 0 shows that the mean value of γ(t) is zero. So by corollary 3.2, the origin is a center. Suppose l ≠ 0. If (62) holds, we have

$\begin{array}{l}{\eta }_5=-\frac{l^2(28c+325b)}{6054048000}.\end{array}$

If η₅ = 0, then l ≠ 0(taken above), and we substitute $c=-\frac{325}{28}b,$ and calculate η₆ as:

$\begin{array}{l}{\eta }_6=\frac{bl(-21078407l^2+10315069600b)}{1261504744980480000}.\end{array}$

If η₆ = 0, then either b = 0 or

$\begin{array}{ll}b=\frac{21078407}{10315069600}{l}^2,& (63)\end{array}$

because l ≠ 0. If b = 0 then d = c = 0, by using these values, γ(t) and δ(t) takes the following form:

$\begin{array}{l}\gamma (t)=e\left(t^4-\frac{1}{5}\right),\end{array}$

$\begin{array}{l}\delta (t)=l\left(t^4-\frac{1}{5}\right).\end{array}$

Let σ(t) = t⁵ − t, then $\stackrel{·}{\sigma }(t)=5t^4-1$. Also, σ(0) = σ(1). Therefore, we can write $\gamma (t)=\frac{1}{5}e\stackrel{·}{\sigma }$ and $\delta (t)=\frac{1}{5}l\stackrel{·}{\sigma }.$ From theorem 3.1, The origin is a center having f(σ)= $\frac{1}{5}e$ and $g(\sigma )=\frac{1}{5}l.$ Thus, suppose that b ≠ 0. By using (63), we have η₇ as follows:

$\begin{array}{l}{\eta }_7=-\frac{3011201l^4(4904530106070545l^2+18047929302961536e)}{1193751333049572276388321296384000000}.\end{array}$

Recalling that l ≠ 0 (considered above), if η₇ = 0 then

$\begin{array}{ll}e=-\frac{4904530106070545}{18047929302961536}{l}^2.& (64)\end{array}$

If (64) holds, then we find

$\begin{array}{l}{\eta }_8=\frac{37395284143096731929725996189267}{86014498207710495466937428606344400899932160000000}{l}^7.\end{array}$

which is constant multiple of l⁷, and l ≠ 0. Thus we conclude that (59) have eight periodic solutions.

Example: Consider the differential equation:

$\begin{array}{ll}\frac{dz}{dt}=\gamma (t)z^3+\delta (t)z^2.& (65)\end{array}$

With γ(t) =equation of circle = ax² + by² + cx + dy + f, δ(t) =quadratic equation= gx² − h, we calculate the periodic solutions.

SOLUTION: By using theorem 2.2, we calculate:

$\begin{array}{l}{\eta }_2=\frac{g}{3}-h,\end{array}$

$\begin{array}{l}{\eta }_3=\frac{1}{3}a+\frac{1}{3}b+\frac{1}{2}c+\frac{1}{2}d+f.\end{array}$

Thus, the multiplicity of z = 0 is μ = 2 if η₂ ≠ 0, and multiplicity μ = 3 if η₂ = 0 but η₃ ≠ 0. If η₂ = η₃ = 0, then we put $h=\frac{g}{3},$ and $f=-\frac{1}{3}a-\frac{1}{3}b-\frac{1}{2}c-\frac{1}{2}d.$ By using these values, we get

$\begin{array}{l}{\eta }_4=-\frac{1}{360}cg.\end{array}$

If η₄ = 0 then c = g = 0. If g = 0, then δ(t) = 0 and η₃ = 0 shows the mean value of γ(t) is zero. So by corollary 3.2, the origin is a center. Suppose that g ≠ 0. Now by using the value of c = 0, we calculate η₅ = 0. Thus we conclude that (65) have four periodic solutions.

6. Conclusion and Discussion

In this article, periodic solutions are calculated. The solutions satisfying 𝔷(β) = 𝔷(0), are called periodic orbits of the Equation (1). The periodic orbits that are isolated in the set of all periodic orbits are usually called the limit cycle. Periodic solutions are found for algebraic coefficients for various classes by using bifurcation analysis. We examined classes C_3,8, C_4,3, C_7,3, C_7,5, C_7,6,C_9,1. We could only get a maximum multiplicity of 10 by using the classical formulas existing in the literature. We succeeded in developing the formula η₁₀ by which classes C_7,3, and C_9,1 have maximum multiplicity 10, which is the highest known until this time. We also improved some already calculated results of Yasmin and Ashraf [20] for class C_4,3, where μ_max is improved from 5 to 8. A systematic procedure has been established in defining coefficients of higher-order polynomial functions. A perturbation method has been properly established for making the maximal number of limit cycles in section 3, which was used numerically to calculate all the classes mentioned in the article. Some examples are also presented to show the applicability of the method. Since the journey toward solving Hilbert's 16th problem is still far at an end, searching for more limit cycles and raising the general lower bound form could be an effective choice for approaching the problem.

Data Availability Statement

All datasets generated for this study are included in the article/supplementary material.

Author Contributions

SA and AN: conceptualization. SA, AN, and NY: writing original draft. DB and KN: methodology, review, and editing. AG and DB: formal analysis. SA, AG, and KN: software. All authors contributed to the article and approved the submitted version.

Conflict of Interest

The authors declare that the research was conducted in the absence of any commercial or financial relationships that could be construed as a potential conflict of interest.

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