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Correction to: On Stacked Planar Central Configurations with Five Bodies when One Body is Removed

  • Antonio Carlos Fernandes
  • Luis Fernando Mello
Correction

1 Correction to: Qual. Theory Dyn. Syst. (2013) 12:293–303  https://doi.org/10.1007/s12346-012-0084-y

The final part of the proof of Lemma 2.7 in [3] is not correct as pointed out in [1]. The correct part, after Eq. (12), is the following one.

Suppose that \(r_5 \ne (0, 0)\). Then, from Eq. (12), we have \(d_6^3 \lambda + M = 0\), which is equivalent to
$$\begin{aligned} \frac{\lambda }{M}+R=0, \end{aligned}$$
where \(R=d_6^{-3}\), for every mass \(m_5\). The last equation can be written as
$$\begin{aligned} R-\frac{\sum _{1\le i< j\le 5}\frac{m_im_j}{r_{ij}}}{\sum _{1\le i < j\le 5}m_im_jr_{ij}^2}=0, \end{aligned}$$
which is equivalent to
$$\begin{aligned} R\sum _{1\le i< j\le 5}m_im_jr_{ij}^2-\sum _{1\le i < j\le 5}\frac{m_im_j}{r_{ij}}=0. \end{aligned}$$
Separating \(m_5\) in the last equation, we have
$$\begin{aligned} m_5\left( R\sum _{i=1}^4m_id_6^2-\sum _{i=1}^4\frac{m_i}{d_6}\right) +R\sum _{1\le i< j\le 4}m_im_jr_{ij}^2-\sum _{1\le i < j\le 4}\frac{m_im_j}{r_{ij}}=0. \end{aligned}$$
In the last equation, the factor multiplying \(m_5\) is null, then we get
$$\begin{aligned} R\sum _{1\le i< j\le 4}m_im_jr_{ij}^2-\sum _{1\le i < j\le 4}\frac{m_im_j}{r_{ij}}=0, \end{aligned}$$
which is equivalent to
$$\begin{aligned} R-\frac{\sum _{1\le i< j\le 4}\frac{m_im_j}{r_{ij}}}{\sum _{1\le i < j\le 4}m_im_jr_{ij}^2}=0. \end{aligned}$$
Note that the quotient in the last equation is the Lagrange multiplier of the four co-circular bodies divided by \(\mathcal {M}=m_1+m_2+m_3+m_4\). Let \(\lambda _4\) be this Lagrange multiplier. So the last equation can be written as
$$\begin{aligned} R+\frac{\lambda _4}{\mathcal {M}}=0. \end{aligned}$$
(13)
Since the four co-circular bodies are in a convex central configuration with positions disposed counterclockwise, the following inequalities must hold (see for instance [2] p. 349)
$$\begin{aligned} R_{13}, R_{24}< -\frac{\lambda _4}{\mathcal {M}}<R_{12},R_{14}, R_{23}, R_{34}. \end{aligned}$$
On the other hand, using the perpendicular bisector theorem, we see that in a co-circular central configuration the center of the circle belongs to the interior of the convex hull of the quadrilateral, see [2]. Thus, from the geometry of a quadrilateral inscribed in the circle of radius \(d_6\) with the center of the circle inside the convex hull of the quadrilateral, at least one side is greater than \(\sqrt{2}d_6\). Thus, Eq. (13) is never satisfied, because the four co-circular bodies are in a convex central configuration. Therefore, in order to satisfy (12) we must have \(r_5=(0,0)=\mathcal {C}\) and the proof is complete.

References

  1. 1.
    Chen, K.C., Hsiao, J.S.: Strictly convex central configurations of the planar five-body problem. Trans. Am. Math. Soc. 370, 1907–1924 (2018)MathSciNetCrossRefzbMATHGoogle Scholar
  2. 2.
    Cors, J.M., Roberts, G.: Four-body co-circular central configurations. Nonlinearity 25, 343–370 (2012)MathSciNetCrossRefzbMATHGoogle Scholar
  3. 3.
    Fernandes, A.C., Mello, L.F.: On stacked planar central configurations with five bodies when one body is removed. Qual. Theory Dyn. Syst. 12, 293–303 (2013)MathSciNetCrossRefzbMATHGoogle Scholar

Copyright information

© Springer International Publishing AG, part of Springer Nature 2018

Authors and Affiliations

  • Antonio Carlos Fernandes
    • 1
  • Luis Fernando Mello
    • 1
  1. 1.Instituto de Matemática e ComputaçãoUniversidade Federal de ItajubáItajubáBrazil

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