Abstract
In this paper, we give the algebraic conditions that a configuration of 5 points in the plane must satisfy in order to be the configuration of zeros of a polynomial isochronous vector field. We use the obtained results to analyze configurations having some of its zeros satisfying some particular geometric conditions.
1. Introduction
We start defining an isochronous vector field, and we express its general associated -form, with its respective residues.
An isochronous vector field is as a complex polynomial vector field on whose zeros are all isochronous centers. A center is isochronous if the periods of the trajectories surrounding it are constant.
Let be a complex polynomial vector field on of degree , nonidentically zero, as follows:where the coefficients can be calculated by Vieta’s formulas, in particular . An isochronous vector field is characterized by their associated 1-form:which has a unique zero at infinity of multiplicity and simple poles with nonzero pure imaginary residues. For , the residue of at iswhere the hat means that the factor is omitted (see [1]).
The following well-known result characterizes the polynomial isochronous vector fields.
Theorem 1 (see [1, 2]). Let be a complex polynomial vector field on of degree defined as in (1); then, the following statements are equivalent:(a) has isochronous centers(b)The zeros of satisfy(c)Their residues satisfy
Here, is the set of the pure imaginary complex numbers different from zero.
Next, we characterize the polynomial isochronous vector fields over the Riemann spheres in terms of the quotients of its residues.
Definition 1. We say that the collection of zeros is an isochronous configuration if there exists a rotation such that the vector field is isochronous.
If the residues belong to the line for some , then the configuration is isochronous. In short, the configuration is isochronous if and only iffor all with . Here, is the set of real numbers different from zero.
We can associate to each isochronous vector field a weighted -tree in the following way. The vertices correspond to the zeros of , and two vertices are connected with an edge if the basins of the corresponding centers are adjacent and the weights are the periods [1, 3]. We know that each embedded -tree (without weights) is realized by an isochronous vector field and that if the phase portraits of two different isochronous vector fields are topologically equivalent, then they have the same embedded -tree (see [3]).
A topological classification of the isochronous vector fields of degree 2 can be found in [1, 4–6]. In [1], the authors characterize the isochronous vector fields of degree in terms of the shape of the configuration of zeros when , and they give partial results for when the zeros present some symmetries. The known results for and are summarized in the next section. Another reference about polynomial vector fields is given in [7] and rational vector field on the Riemann sphere in [8]. Finally, more general vector fields are studied in [9, 10].
The aim of this paper is to characterize the isochronous vector fields of degree 5 in terms of the configurations of zeroes without imposing any condition of symmetry.
If a polynomial vector field over the Riemann sphere is given, we can change the coordinates in such a way that has a zero at 0 and another zero at 1 because an affine transformation over the Riemann space does not change the main characteristics of a polynomial vector field, in particular the condition of to be isochronous, [1, 3]. Then, if the complex polynomial vector field has degree 5, it has 8 free real parameters: six for the zeros and two for the main coefficient. Assuming that the position of two zeros is fixed, it will be proved that there can be up to seven different three-parameter families of isochronous configurations, see Theorem 6. Notice that the family of 5-degree isochronous vector fields has only three real parameters. There are still too much parameters to give a result providing all the possible shapes of the zeros of the isochronous configurations for . Nevertheless, we can reduce the parameter space by fixing either the position or the shape of some zeros. At the end of the paper, we will analyze the shape of the isochronous configurations in some particular cases where the zeros present some symmetries and we will complete some of the cases studied in [1].
The paper is structured as follows. In the next section, we give a summary of the known results on isochronous vector fields of degree 5. After that, in the following section, we describe how to solve the set of equation (6) for an arbitrary vector field of degree 5. Finally, in the last section, we give explicitly the isochronous configurations in some particular cases: when four zeros are at the vertices of either a parallelogram or an isosceles trapezoid, when two zeros are on the line orthogonal to the line passing through other two zeros, and when three zeros are at the vertices of an equilateral triangle. We give numerical examples of some isochronous configurations with three zeros at the vertices of two isosceles triangles when , , and and when , , and . Finally, we give examples of isochronous configurations where the zeros do not satisfy any symmetry. We also analyze the phase portrait of the isochronous vector fields associated to the configurations. In particular, we find either the star or fork topology (Figure 1) in all cases except when we have all the zeros in the line. We have not found examples of the line topology when the zeros are not all in a line.
(a)
(b)
(c)
2. Some Known Results
In this section, we summarize some known results on isochronous configurations of vector fields of degree given in [1].
Theorem 2 (see [1]). A polynomial vector field of degree 4 is isochronous if and only if the zeros , , , and are either in a line or three are in the vertices of a triangle and one is at its orthocenter.
The phase portrait of the isochronous vector fields of degree 5 can have as many different topologies as 5-trees, so they can have three different topologies (see Figure 1), namely, the star topology, the fork topology, and the line topology.
The next theorem summarizes the results of isochronous configurations that are valid for vector fields of degree .
Theorem 3 (see [1]). The following statements hold.(a)For each , if the zeros , are in a line, then is isochronous and his phase portrait has the line topology(b)For each , if the zeros , are at the vertices of a regular polygon and is at its center, then is isochronous and his phase portrait has the star topology(c)For each , there exist isochronous vector fields with the zeros in a line and the zeros and in new line orthogonal to the previous one In addition, for , the following statements hold.(d)If , , , , and , then is isochronous if and only if (e)For , if are at the vertices of a rhombus and is at its center and if the residues of satisfy , , and , then is isochronous; moreover, its phase portrait has the star topology(f)Assume that is isochronous and , , and are in the bisector line of the segment with endpoints and (i)If is in the convex hull of , , , and and the residues of satisfy and , then is isochronous; moreover, the phase portrait of has the star topology(ii)If and are in the convex hull of , , and and the residues of satisfy and , then is isochronous; moreover, its phase portrait has the fork topology
Although we are not interested in physical applications in this paper, we mention some of them studied in [11]. For example, let ; then, the th terms may be regarded as the force with which a fixed mass (or electric charge) at repeals (attracts if ) a movable unit mass (or charge) at , being the law of repulsion, the inverse distance law. Equivalent interpretation can be made in terms of masses (or charges) repelling according to the inverse-square laws. Actually, in Theorem (3.1) in [10], it says that the zeros of with all real are the points of equilibrium in the field of force due to the systems of masses (point charges) at the fixed points repelling a movable unit mass at according to the inverse distance law. Note that the expression has the same form as the 1-form given in (2) associated to an isochronous vector field , and it can be written as , with given in (3).
Still another interpretation in is that each term is the vector velocity in a two-dimensional flow of an incompressible fluid due to a source of strength at (sink if ).
3. Characterization of the Isochronous Vector Fields of Degree 5
As previously said, without loss of generality, we can consider that the position of two of the zeros of are fixed to 0 and 1.
Assume that and , and let , , and . From (6), the configuration is isochronous if and only if for all with . Note that we only are interested in configurations with for . We define
The denominators of the functions are defined when for so we can drop them. Let be the numerator of the factorization of for all . Then, is isochronous if and only if is a solution of the set of polynomial equations:
Next, we will provide information about the solutions of (8). We only are interested in solutions of (8) satisfying for . In what follows, these solutions will be called valid solutions.
The first two equations of system (8) can be written aswhere
Systems of the form (9) will appear often in our analysis. We give the solution of this kind of systems in the next section, considering a generic system as (11).
Many solutions of (8) satisfy that for some , these solutions correspond to isochronous configurations with at least three zeros aligned. In order to simplify our computations, this case is treated separately, at Theorem 4. From now on, the solutions of (8) with for all that are valid solutions will be called admissible solutions.
3.1. The Resolution of Systems of the Form (9)
Let us consider a generic system of the form (9):
Note that the two equations of (11) correspond to the equations of two circles (eventually degenerated to a line) passing through the point . So, is always a solution of (11).
After tedious but not difficult computations, we see that when and are not simultaneously zero, then the solutions of system (11) are and with
Solving system and , we get the following conditions: , , and . Then, we can prove the following result.
Lemma 1. Let , , and . Then, the following statements hold.(a) is a solution of (11) for all , , , , and (b)If neither of conditions , , and is satisfied, then the solutions of (11) are and as given in (12)(c)If condition is satisfied, then the solutions of (11) are the solutions of equation (d)If condition is not satisfied and condition is satisfied, then the solution of (11) is when and when (e)If condition is not satisfied and condition is satisfied, then either the second equation is identically 0 or the two equations of (11) are linearly dependent
3.2. Isochronous Configurations with Three Zeros in a Line
Without loss of generality, we can assume that the zeros that are aligned are , , and , so we assume that .
We substitute into equations and , and we get the equations:where we define and with
The solutions with and are not valid because they correspond to and , respectively. Now, we analyze the solutions of system and . This is a system of the form (11); hence, from Lemma 1, if conditions with are not satisfied, then the nontrivial solution of system and is
Condition does not provide valid solutions because it is satisfied when either or (i.e., when either or ). Condition is satisfied when ; under this condition, ; so, from Lemma 1, the solution of system and is with . Finally, condition is satisfied either when , which does not provide a valid solution, or when . This last case satisfies condition , so it has already been studied.
In short, system and with have only two valid solutions:
By substituting into equation , we get
The solutions and of this equation are not valid, and the solutions of the last factor of the equation arewhen and either , , or when (or equivalently when ). It is easy to check that the solutions with do not provide valid solutions. On the contrary, the solution with given in (18) and the solution satisfy all equations . Therefore, they provide isochronous configurations. In short, we have proved the following theorem.
Theorem 4. If the zeros , , and are in a line, then the configuration is isochronous if and only if it satisfies one of the following statements:(a), , , , and are in a line(b), , and are in the bisector line of the segment with endpoints and In particular, if , , , , and , then is isochronous if and only if one of the following statements holds:(c)(d), , and
Theorem 4 completes in some sense the results in Theorem 3.
From now on, we only will consider solutions of system (8) with .
3.3. Admissible Solutions of System (8)
Now we analyze the solutions of (9) that provide solutions of (8) with for and , that is, those provide admissible solutions. System (9) is of the form (11) with , , , , , and so we will apply Lemma 1. Here, we use the notation:
In order to simplify our computations, we will use resultants’ theory in some cases. Next, we summarize the basic properties of the resultants.
Let and be two polynomials in the variable with leading coefficient one. Let , , be the roots of and , , be the roots of . The resultant of and , , is the expression formed by the product of all the differences , and , see for instance [12, 13]. The main property of the resultant is that if and have a common solution, then necessarily .
Let now and be polynomials in the variables . These polynomials can be considered as polynomials in with polynomial coefficients in ; then, the resultant with respect to , , is a new polynomial in the variable with the following property. If and have a common solution , then and similarly for the variable .
3.3.1. Solutions of (9) Satisfying Condition
System and (condition ) can be written in the form (11) with , , and and can be solved by applying Lemma 1 again. Conditions for provide solutions with either or . The solution in Lemma 1 becomes (i.e., ). Therefore, neither of the solutions of (9) satisfying condition can provide valid solutions of (8).
3.3.2. Solutions of (9) Satisfying Condition
System and (condition ) can be written again in the form (11) with , , , , and . All conditions in Lemma 1 provide solutions with . Therefore, conditions cannot provide admissible solutions of (8).
The solution in Lemma 1 becomes
We substitute this solution into equation , and we get an equation equivalent to
Clearly, neither of the solutions of this equation can provide admissible solutions of (8). In short, neither of the solutions of (9) satisfying condition can provide valid solutions of (8).
3.3.3. Admissible Solutions of (8) Satisfying Condition
Finally, we analyze the solutions of system and (condition ). System and cannot be written in the form (11). We will analyze the solution of this system by using the properties of resultants.
We compute the resultant of and with respect to , and we getwhere
By the properties of the resultant, we know that if is a solution of system and , then the coordinates satisfy equation . Thus, in order to solve system and , it is sufficient to find the solutions of that satisfy system and .
Clearly, the first four factors of do not provide admissible solutions of (8). Thus, we only will consider solutions with either or .
Case :
We substitute into equation , and we get
The first two factors of this equation do no provide admissible solutions of (8). From the last factor of the equation, we have
We substitute into equation , and we get the following solutions:
The first solution corresponds to , the second and the third imply , and the fourth corresponds to . Therefore, the unique solutions of system and that can provide admissible solutions of (8) are
Notice that is defined for and when either or .
We substitute and into (8). Since condition is satisfied, equations and are linearly dependent, so we will work with equations and instead of equations and . If and , then system and can be written aswhere
Here,where the upper sign corresponds to the positive value and the lower sign corresponds to the negative value . Notice that, in order to obtain the expressions and , we have substituted into system and not only and but also , , and .
Next, we analyze the solutions of (28) providing admissible solutions of (8). The factors and do not provide solutions in the domain of definition of , and the factor provides solutions with . We analyze the solutions of and by applying again Lemma 1. Conditions and in Lemma 1 give solutions with either or , and condition gives solutions with either , , or . The solution given by in Lemma 1 becomesand it corresponds to . Therefore, the factor does not provide admissible solutions of (8).
Case :
From equation , we getwhere
Note that is defined when and .
Solutions with : solving equation with respect to , we get and where
The domain of definition of and is the set
On the contrary, analyzing the functions and , we can see that when , when , when either or , when , and when . Thus, the solutions and are defined only when . Therefore, the denominator of is zero when either or . The solution does not provide admissible solutions of system (8).
Now we analyze the solutions of (8) with . In this case, is identically 0. By substituting into equation and solving the resulting equation, we get the solutions and where
These solutions also satisfy equation . We substitute the solution with and into equations and , and we obtain a system of equations of the form (9) in the variables where the coefficients depend on . We solve this system by applying Lemma 1 as we have done in the previous cases, and we see that it has two unique solutions, one satisfying and the other one satisfying . The same occurs with the solution and . Analyzing the cases and in a similar way, we get three unique solutions, which satisfy , , and , respectively. Therefore, system (8) has no valid solutions when .
Solutions with : assume that . We substitute into equation , and we solve the resulting equation obtaining in this way the solutions:
Note that the last solution is defined only when .
First, we analyze the solution . It is easy to check that equation is always satisfied when . We substitute this solution into equations and , and we getwhere
Here,
The first factor of equation (38) gives solutions with , the second one gives solutions with , and the third factor does not give real solutions. The fourth factor of the first equation of (38) gives solutions with , and the fourth factor of the second equation gives solutions with . Therefore, neither of them provides valid solutions.
We solve system and by applying Lemma 1 again, and we get the solutions and withwhen conditions , , and are not satisfied. Conditions and provide solutions with , and the condition provides solutions with either , , , , or . Therefore, these conditions cannot provide admissible solutions of (8). Finally, it is easy to check that, for all , the equations are satisfied when and . In short, this solution will provide admissible solutions when is such that for and .
Now, we consider the solution (assuming that ), and we proceed in a similar way. First, we substitute the solution into equation , and we get the following equation:where
Clearly, the second factor of equation does not provide real solutions, the first and fourth factors provide solutions with , the third provides solutions with , and the fifth provides solutions with . Therefore, the unique factor that can provide admissible solutions of (8) is .
It is not difficult to check that
So, if and , then , and this is the case we have just been studied. Furthermore, if , then equation has only the two solutions and . Since the solution is defined when , we do not need to consider this case separately.
3.3.4. Admissible Solutions of (8) that Do Not Satisfy Any Condition
From Lemma 1, when neither of conditions is satisfied, the solutions of system and are and withwhere is defined in Lemma 1 and , , , , and are defined in (9). We substitute the solution into equations for , we factorize the resulting equations, and we drop the denominators obtaining a new system of polynomial equations:where the function is identically zero. Next, we analyze the solutions of (46).
The factorization of consists of two factors:where
Then, the solutions of (46) must satisfy either or . We see that factor is common to all the functions for . Therefore, all the solutions of equation that do not satisfy conditions with provide isochronous configurations with .
It is easy to check that . Therefore, all the solutions of equation satisfy condition , and the solution is not defined in this case.
Remark 1. Equation has at most seven different solutions with when , at most six different solutions with when and , and solution and when .
Indeed, if , then is a polynomial of degree 7 in the variable ; therefore, there could exist up to seven different real solutions of . The coefficient of degree 7 of , , is equal to zero when either or . Here, we are not interested in solutions with . So, becomes a polynomial of degree 6 with at most 6 solutions when . If , then the coefficient of degree 6 of iswhich becomes zero when . But if and , then .
We note that the solution with , , and always satisfies equation , but in this case, the solution is not defined because (or equivalently, because conditions are satisfied for some ). In short, we have proved the following theorem.
Theorem 5. Let , , , , and . Then, we can have the following families of solutions of (8) with for providing isochronous configurations with for .(a)Up to seven different three-parameter families of solutions with (see (45)) and satisfying equation when (b)Up to six different two-parameter families of solutions with and satisfying equation when and (c)The two-parameter family of solutions with , , and (d)The two-parameter family with and , see (32), (37), and (41)
Notice that the solution is defined only for solutions of that do not satisfy any of conditions , , and defined in (19). The solution in (d) is defined when .
Theorem 4 gives a complete description of the isochronous configurations with for some , and Theorem 5 gives a complete description of the isochronous configurations with for all . Hence, we have a complete description of all the isochronous configurations of the vector fields of degree 5, which is summarized in Theorem 6.
We can see that all the zeros given in Theorem 4 satisfy equation . The zeros given by Theorem 4 (c) do not provide solutions with because in this case, . This does not happen with the zeros given by Theorem 4 (d). Moreover, we can see that if and , then . Hence, statement (d) of Theorem 4 can be included in either statement (a), (b), or (c) of Theorem 5 if we do not consider the assumption for all .
Theorem 6. Let , , , , and . The only solutions of (8) with for providing isochronous configurations are the following.(a)(b) (see (45)) and is a solution of equation that does not satisfy any of conditions defined in (19)(c) and , see (32), (37), and (41)
Remark 2. Using the symmetries of the configuration, it is not difficult to see that if , , , , and are an isochronous configuration, then so is the configuration , , , , and , the configuration , , , , and the configuration , . So, it is sufficient to analyze the configurations with any .
4. Isochronous Configurations with = in Some Particular Cases
In the previous section, more precisely in Theorem 6, we have given a complete description of the isochronous configurations with . Here, we will analyze the isochronous configurations for some particular configurations of zeros.
4.1. Four Zeros at the Vertices of a Parallelogram
Without loss of generality, we can assume that the segment joining and is an edge of the parallelogram; then, the remaining vertices of the parallelogram can be taken as and with . The assumption is not restrictive by Remark 2.
Since , all isochronous configurations of this type are given by Theorem 5. First, we see that the solution with and does not provide isochronous configurations with four zeros at the vertices of a parallelogram. Indeed, if the configuration of is a parallelogram, then satisfies equations and , and this system of equations does not have real solutions.
Now we substitute into equation , and we get
This equation has a unique real solution with , . By substituting this solution into the expression of , we get
In short, we have a unique one parameter family of configurations with four zeros at the vertices of a parallelogram and . It is given by the solution
It is easy to check that, in this family of configurations, , , , and are at the vertices of a rhombus (degenerated to a square when ) and is at its center. These configurations correspond to the ones given by Theorem 3(e), so its phase portrait has the star topology.
In short, we have proved the following result, which in some sense can be thought as a generalization of the statement (b) of Theorem 3.
Lemma 2. Assume that the configuration is isochronous. The following statements hold.(a)If the zeros , , , and are at the vertices of a parallelogram, then this parallelogram is a rhombus and is at the center of the rhombus.(b)There exist a unique configuration with four zeros at the vertices of a given rhombus. Assuming that the vertices of the rhombus are , , , and , the configuration satisfies Notice that the rhombus is a square when .(c)The phase portrait associated to the configuration has the star topology.
4.2. Four Zeros at the Vertices of an Isosceles Trapezoid
Without loss of generality, we can assume that the segment joining and is an edge of the isosceles trapezoid; then, the remaining vertices of the trapezoid can be taken as and . As in the previous case, here we have assumed that , and this is not restrictive by Remark 2.
As in the previous section, since , all isochronous configurations of this type are given by Theorem 5. First, we analyze the solution with and . By imposing that the configuration is an isosceles trapezoid, we get the following equations:
Solving this system of equations, we get the solutionswhich is not a valid solution because .
Now we substitute into equation , and we get
The solution with is not possible because it corresponds to . The last factor in (56) provides a unique real solution with , and the solutionwhere
Substituting this solution into , we get
So, we have a unique one parameter family of isosceles trapezoid configurations which is given bywhich are defined for . It is not difficult to check that for all and that when . In short, we have proved the following result.
Lemma 3. Assume that the configuration is isochronous. The following statements hold.(a)If the zeros , , , and are at the vertices of an isosceles trapezoid, then is at the interior of the trapezoid on its axis of symmetry.(b)There exist a unique configuration with four zeros at the vertices of a given isosceles trapezoid. Assuming that the vertices of the trapezoid are , , , and , the configuration satisfies with given in (57) and with given in (59).
We have plotted the phase portrait of the isochronous vector fields associated to the configurations given by Lemma 3(b) for many values of . After analyzing the obtained results, we conjecture that all the isochronous configurations given by Lemma 3(b) have a phase portrait with the star topology.
4.3. Two Zeros on the Line Orthogonal to the Line Passing through Other Two Zeros
In Theorem 4, we have proved that if , , and are in a line , then and are either in the same line or is the bisector line of the segment with endpoints and . Now we prove the following more generic result.
Lemma 4. Assume that the configuration is isochronous. If and are in a line and and are in a line orthogonal to and , then either is in and is the bisector line of the segment with endpoints and or is in and is the bisector line of the segment with endpoints and .
Proof. Assume that , , , and with . This is not restrictive by Remark 2. Let be the line passing through and and be the line passing through and .
Since , all the isochronous configurations of this type are given by Theorem 5. The equation evaluated at becomesWe note that the solution of (61) with does not provide valid solutions of (8) because it corresponds to . If , then condition is satisfied, and therefore, the solution of (8) with is not defined. The solutions of (61) and provide valid solutions of (8) with which are given byrespectively. In the first solution, is on and is the bisector line of the segment with endpoints and . In the second solution, is on and is the bisector line of the segment with endpoints and . Notice that the solutions given in (62) correspond to the configuration given in Theorem 3(d).
Finally, we analyze the solutions of (8) with and . When , equation is equivalent toWe can see that all the solutions of the second factor provide solutions of (8) with which are not valid. The solution provides the following solution:In this solution, is on and is the bisector line of the segment with endpoints and . This completes the proof.
We note that the first solution in (62) and the solution in (64) coincide when .
4.4. Three Zeros at the Vertices of an Equilateral Triangle
Let , , and be at the vertices of an equilateral triangle. Without loss of generality, we can assume that ; then, the positions of the zeros are , , , , and .
Since , all the isochronous configurations of this type are given by Theorem 5. By substituting into equation and solving the resulting equation, we get the following solutions:
The first three solutions provide the following valid solutions of (8) with :
The last solution provides solutions of (8) satisfying condition ; thus, the solution is not defined in this case. Finally, the solutions of (8) with and are not defined when because .
Analyzing the shape of the solutions of (66), we see that, in the first solution of (66), the zeros and are on the median of the triangle that passes through the vertex , in the second solution of (66), the zeros and are on the median passing through the vertex , and in the third solution, and are on the median passing through the vertex . We note that when , the first solution of (66) is a square of vertices , , , and and at its center.
We have proved the following result.
Lemma 5. Assume that the configuration is isochronous. The following statements hold.(a)If , , and are at the vertices of an equilateral triangle, then and are in one of the medians of the triangle, and therefore, , , , , and satisfy statements of Lemma 4.(b)There exists three different one-parameter families of configurations with three zeros at the vertices of a given equilateral triangle and two zeros in a median of the triangle, one for each median. If the vertices of the equilateral triangle are , , and , then these three one-parameter families of configurations are given by and with , given in (66).
We have plotted the phase portrait of the isochronous vector fields associated to the configurations given by Lemma 5(b) for many values of the variable that acts as a parameter. After analyzing the obtained results, we conjecture the following.(i)If and are given by the first solution in (66), then the phase portrait of the isochronous configuration has star topology when and fork topology when (ii)If and are given by the second solution in (66), then the phase portrait of the isochronous configuration has star topology when and fork topology when (iii)If and are given by the third solution in (66), then the phase portrait of the isochronous configuration has fork topology when and star topology when
4.5. Three Zeros at the Vertices of an Isosceles Triangle
Let , , and be at the vertices of an isosceles triangle. Without loss of generality, we can assume that the positions of the zeros are , , , , and with .
Since , all the isochronous configurations of this type are given by Theorem 5. From Theorem 5(d), the solutions of (5) with and become
From Theorem 5(c), we obtain the two-parameter family of solutions
This solution corresponds to the configuration given by Theorem 3(d). We note that the solutions (67) and (68) coincide when
From Theorem 5(b), we know the existence of up to six different two-parameter families of solutions with , , and satisfying . The exact number of such families will depend on the values of and . Unfortunately, equation for the isosceles triangle configurations cannot be solved explicitly. We can solve it numerically by setting the values of the parameters and . Hence, we have proved the following result.
Lemma 6. Assume that is isochronous and that , , and are at the vertices of an isosceles triangle. Let , , and with be the vertices of the isosceles triangle. Then, the following statements hold.(a)There exist up to six different two-parameter families of isochronous configurations with , , and satisfying (b)There exists the two-parameter family given by (68)(c)There exists the additional one-parameter family given by (67)We note that configurations given in statements (b) and (c) coincide when
To give some examples of how the families given by Lemma 6 are, now we will analyze numerically the families of solutions of with for two particular values of . We have chosen , which provides a zero near (i.e., the configuration with at the midpoint of the line segment joining and ), and , which provides a zero near (i.e., the configuration with , , and at the vertices of an equilateral triangle).
First, fixing the value of , we find the values of where the number of solutions can change by solving numerically the following system of polynomial equations:
Let for denote the values of corresponding to the solutions of (1), and let and . For all , we find the number of solutions of equation for a value by solving numerically the polynomial equation . We note that due to the symmetry of the configuration, it is sufficient to consider values of . Finally, to understand how the families of solutions are related, we compute numerically the solutions of for values of near the bifurcation values for . Here, we only give the solution at a value for all and at for all . These are the results that we have obtained.
Case:
Equation with has the following families of solutions.(i)Two one-parameter families of solutions when with and (ii)Three solutions when (iii)Four one parameter families of solutions when with (iv)Five solutions when (v)Six one-parameter families of solutions when with (vi)Four one-parameter families of solutions when (vii)Six one-parameter families of solutions when with (viii)Five solutions when (ix)Four solutions when with
Now, we give the isochronous configurations , , , , and at a given value in each interval with and at with . We will use the following notation: we denote the families of solutions by for , and the notation denotes a solution from which we bifurcate the two families and .(i)If , then the two solutions of provide the isochronous configurations with The phase portrait associated to the isochronous vector field given by has the fork topology, and the one given by has the star topology.(ii)If , then the solutions of provide the isochronous configurations with The phase portraits associated to the isochronous vector fields given by and have the fork topology, and the one given by has the star topology.(iii)If , then the four solutions of provide the isochronous configurations with Note that, from the family given in (ii), bifurcate two families of solutions that we denote by and . The phase portraits associated to the isochronous vector fields given by , , and have the fork topology, and the one given by has the star topology.(iv)If , then the five solutions of provide the isochronous configurations with The phase portraits associated to the isochronous vector fields given by , , , and have the fork topology, and the one given by has the star topology.(v)If , then the six solutions of provide the isochronous configurations with The phase portraits associated to the isochronous vector fields given by , , , , and have the fork topology, and the one given by has the star topology.(vi)If , then the four solutions of provide the three isochronous configurations with The fourth solution of is , and it does not provide an isochronous configuration. At this solution, the three families of solutions of corresponding to , , and coincide. The phase portraits associated to the isochronous vector fields given by and have the fork topology, and the one given by has the star topology.(vii)If , then the six solutions of provide the isochronous configurations with The phase portraits associated to the isochronous vector fields given by , , , , and have the fork topology, and the one given by has the star topology.(viii)If , then the five solutions of provide the isochronous configurations with The phase portraits associated to the isochronous vector fields given by , , , and have the fork topology, and the one given by has the star topology.(ix)If , then the four solutions of provide the isochronous configurations with
The phase portraits associated to the isochronous vector fields given by , , and have the fork topology, and the one given by has the star topology.
We note that when , the solution families and tend to a solution of with . This solution does not provide a solution of (8) because , and therefore, is not defined. Moreover, along the solutions’ families and when . On the contrary, the solution families and tend to a solution of with . This solution does not provide a valid solution because . In this case, and along the solutions’ families and , respectively, when .
Case: :
Equation with has the following families of solutions.(i)Two one-parameter families of solutions when with and (ii)Three solutions when (iii)Four one-parameter families of solutions when with (iv)Two solutions when (v)Four one-parameter families of solutions when with
Now we give the isochronous configurations , , , , and at a given value in each interval with and at with . As above, with denotes the families of solutions and denotes a solution from which we bifurcate the two families and .(i)If , then two solutions of provide the isochronous configurations with The phase portrait associated to the isochronous vector field given by has the fork topology, and the one given by has the star topology.(ii)If , then the solutions of provide the isochronous configurations with The phase portrait associated to the isochronous vector field given by has the fork topology, and the ones given by and have the star topology.(iii)If , then the four isochronous configurations are given by Note that, from the family given in (ii), bifurcate two families of solutions that we denote by and . The phase portrait associated to the isochronous vector field given by has the fork topology, and the ones given by , , and have the star topology.(iv)If , then the two solutions of provide a unique isochronous configuration with The other solution of is , and it does not provide isochronous configurations. At this solution, the three families of solutions of corresponding to , , and coincide. The phase portrait associated to the isochronous vector field given by has the star topology.(v)If , then the four solutions of provide a unique isochronous configuration with
The phase portrait associated to the isochronous vector field given by has the fork topology, and the ones given by , , and have the star topology.
We note that when , the solution families and tend to a solution of with . This solution does not provide a solution of (8) because , and therefore, is not defined. Moreover, along the solutions’ families and when . The solution families and tend to the following solutions, which provide isochronous configurations:
The phase portraits associated to the isochronous vector fields given by and have the star topology.
4.6. More Examples
In the previous sections, we have studied particular cases where the configuration of some of the zeros is symmetric. By proceeding as in the last section, we could analyze the isochronous configurations when , , and are in an arbitrary triangle obtaining in this way isochronous configurations having no symmetries. For example, if , , and , then we obtain five isochronous configurations that are given by
In Figure 2, we show the phase portraits of the isochronous vector fields associated to these configurations with their corresponding 5-trees. We observe that the phase portrait of the first and the third configurations have the star topology, while the other three have the fork topology. Note that, in the second configuration, three zeros are in the bisector line of the other two, and this does not happen in the fourth and the fifth.
(a)
(b)
(c)
(d)
(e)
For the moment, we have not been able to find examples of isochronous vector fields having a phase portrait with line topology whose zeros are not aligned. We conjecture that there are no isochronous vector fields having a phase portrait with line topology whose zeros are not aligned.
Conflicts of Interest
The authors declare that they have no conflicts of interest.
Acknowledgments
This work was partially supported by Tecnologico de Monterrey. The second author was partially supported by the Ministerio de Ciencia, Innovación y Universidades, Agencia Estatal de Investigación grants MTM2016-77278-P (FEDER) and PID2019-104658GB-I00 (FEDER).