Periodic solutions of a generalized Sitnikov problem

Abstract

In this paper, we study the problem of determining whether a global family of even periodic solutions of a generalized Sitnikov problem, which emerges from a circular generalized Sitnikov problem, continues for all values of eccentricity in [0, 1) or ends in the equilibrium.

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Appendix

Appendix

In this section, we enunciate and demonstrate some technical lemmas, which deal with Taylor series of \(\sqrt{q_\lambda (u,e)}\) and \(\int _{0}^{\pi }\sqrt{q_\lambda (u,e)}\mathrm{d}u\) and are used to demonstrate other results of the article.

Before enunciating the first lemma, it should be noted the definition of combinatory number, \(\begin{pmatrix} \gamma \\ n \end{pmatrix}:=\frac{\prod _{i=1}^{n}(\gamma -i+1)}{n!}\), where \(\gamma \in {\mathbb {R}}\) and \(n\in {\mathbb {N}}\).

Lemma 6

Let \(q_\lambda (u,e)\) be the function defined in (16). Then, for \(u \in (0,\pi )\), we have that

$$\begin{aligned} \sqrt{q_\lambda (u,e)}=\sqrt{8\lambda }\sum _{l=0}^{\infty }c_{\lambda ,l}(u)e^{l}, \end{aligned}$$

where

$$\begin{aligned}&c_{\lambda ,l}(u)=\sum _{j=0}^{\left[ \frac{l}{2}\right] }\sum _{k=0}^{l-2j}\begin{pmatrix} \frac{1}{2}-j\\ k \end{pmatrix}b_\lambda ^k(u)\cos ^{l-k-2j}(u)\begin{pmatrix} \frac{1}{2}\\ j \end{pmatrix}a_\lambda ^{j}(u), \end{aligned}$$
(32)

with \(a_\lambda (u)=\frac{\cos ^2(u)-3}{32\lambda }\), \(b_\lambda (u)=-\frac{32\lambda -2}{32\lambda }\cos (u)\).

Proof

Let us first observe that

$$\begin{aligned} \sqrt{q_\lambda (u,e)}= & {} \frac{1}{2}\frac{\sqrt{32\lambda -(32\lambda -2)\cos (u)e+(\cos ^2(u)-3)e^2}}{1-e\cos (u)}\\= & {} \sqrt{8\lambda }\frac{\sqrt{1+b_\lambda (u)e+a_\lambda (u)e^2}}{1-e\cos (u)} \end{aligned}$$

where \(a_\lambda (u)=\frac{\cos ^2(u)-3}{32\lambda }\) and \(b_\lambda (u)=-\frac{32\lambda -2}{32\lambda }\cos (u)\). As \(\sqrt{1+x}=\sum _{i=0}^{\infty }\begin{pmatrix} \frac{1}{2}\\ i \end{pmatrix}x^i\) for all \(-1<x< 1\), we have that

$$\begin{aligned} \sqrt{q_\lambda (u,e)}=\sqrt{8\lambda }f_\lambda (u,e)\sum _{n=0}^{\infty }\cos ^n(u)e^n, \end{aligned}$$

where \(f_\lambda (u,e):=\sum _{n=0}^{\infty }\begin{pmatrix} \frac{1}{2}\\ n \end{pmatrix}e^n(b_\lambda (u)+ea_\lambda (u))^n\).

Since \(\sqrt{1+x}\) is analytic in \((-1,1)\) and \(b_\lambda (u)e+a_\lambda (u)e^2\) is analytic, as a function of e in \({\mathbb {R}}\), we have that, for every \(u\in (0,\pi )\), \(\sqrt{1+b_\lambda (u)e+a_\lambda (u)e^2}=f_\lambda (u,e)\) is analytic, as a function of e, in \((-1,1)\). Then, for \(e\in [0,1)\)

$$\begin{aligned} f_\lambda (u,e)=\sum _{n=0}^{\infty }\frac{\partial ^nf_\lambda (u,0)}{\partial e^n}\frac{e^n}{n!}. \end{aligned}$$

Let us calculate \(\frac{\partial ^nf_\lambda (u,0)}{\partial e^n}\).

$$\begin{aligned} \frac{\partial ^nf_\lambda (u,0)}{\partial e^n}= & {} \sum _{m=\left[ \frac{n+1}{2}\right] }^{n}\begin{pmatrix} \frac{1}{2}\\ m \end{pmatrix}\frac{\partial ^n}{\partial e^n}(e^m(b_\lambda (u)+ea_\lambda (u))^m)\Big |_{e=0}\\= & {} \sum _{m=\left[ \frac{n+1}{2}\right] }^{n}\begin{pmatrix} \frac{1}{2}\\ m \end{pmatrix}\sum _{i=0}^{n} \begin{pmatrix} n\\ i \end{pmatrix}\left( \frac{\partial ^ie^m}{\partial e^i}\frac{\partial ^{n-i}}{\partial e^{n-i}}(b_\lambda (u)+ea_\lambda (u))^m\right) \bigg |_{e=0}\\= & {} \sum _{m=\left[ \frac{n+1}{2}\right] }^{n}\begin{pmatrix} \frac{1}{2}\\ m \end{pmatrix} \begin{pmatrix} n\\ m \end{pmatrix}\left( \frac{\partial ^me^m}{\partial e^m}\frac{\partial ^{n-m}}{\partial e^{n-m}}(b_\lambda (u)+ea_\lambda (u))^m\right) \bigg |_{e=0}\\= & {} \sum _{m=\left[ \frac{n+1}{2}\right] }^{n}\begin{pmatrix} \frac{1}{2}\\ m \end{pmatrix} \begin{pmatrix} n\\ m \end{pmatrix}\frac{(m!)^2}{(2m-n)!}b_\lambda ^{2m-n}(u)a_\lambda ^{n-m}(u). \end{aligned}$$

Now,

$$\begin{aligned} \sqrt{q_\lambda (u,e)}=\sqrt{8\lambda }\sum _{l=0}^{\infty }c_{\lambda ,l}(u)e^l, \end{aligned}$$

where

$$\begin{aligned} c_{\lambda ,l}(u)=\sum _{n=0}^{l}\frac{\partial ^nf_\lambda (u,0)}{\partial e^n}\frac{\cos ^{l-n}(u)}{n!}. \end{aligned}$$

Considering \(j=n-m\), and \(k=n-2j\), we have that

$$\begin{aligned} c_{\lambda ,l}(u)= & {} \sum _{n=0}^{l}\sum _{m=\left[ \frac{n+1}{2}\right] }^{n}\begin{pmatrix} \frac{1}{2}\\ m \end{pmatrix}\begin{pmatrix} n\\ m \end{pmatrix}\frac{(m!)^2}{(2m-n)!}\frac{b_\lambda ^{2m-n}(u)a_\lambda ^{n-m}(u)}{n!}\cos ^{l-n}(u)\nonumber \\= & {} \sum _{n=0}^{l}\sum _{m=\left[ \frac{n+1}{2}\right] }^{n} \left[ \prod _{s=1}^{m}\left( \frac{1}{2}-s+1\right) \right] \frac{b_\lambda ^{2m-n}(u)}{(2m-n)!}\frac{a_\lambda ^{n-m}(u)}{(n-m)!}\cos ^{l-n}(u)\nonumber \\= & {} \sum _{j=0}^{\left[ \frac{l}{2}\right] }\sum _{n=2j}^{l}\prod _{s=1}^{n-j}\left( \frac{1}{2}-s+1\right) \frac{b_\lambda ^{n-2j}(u)}{(n-2j)!}\cos ^{l-n}(u)\frac{a_\lambda ^j(u)}{j!}\nonumber \\= & {} \sum _{j=0}^{\left[ \frac{l}{2}\right] }\sum _{k=0}^{l-2j}\prod _{s=j+1}^{k+j}\left( \frac{1}{2}-s+1\right) \frac{b_\lambda ^k(u)}{k!}\cos ^{l-k-2j}(u)\prod _{i=1}^{j}\left( \frac{1}{2}-i+1\right) \frac{a_\lambda ^{j}(u)}{j!}\nonumber \\= & {} \sum _{j=0}^{\left[ \frac{l}{2}\right] }\sum _{k=0}^{l-2j}\begin{pmatrix} \frac{1}{2}-j\\ k \end{pmatrix}b_\lambda ^k(u)\cos ^{l-k-2j}(u)\begin{pmatrix} \frac{1}{2}\\ j \end{pmatrix}a_\lambda ^{j}(u). \end{aligned}$$
(33)

\(\square \)

Lemma 7

Let \(q_\lambda (u,e)\) be the function defined in (16). Then

$$\begin{aligned} \int _{0}^{\pi }\sqrt{q_\lambda (u,e)}\mathrm{d}u=\sqrt{8\lambda }\sum _{\tau =0}^{\infty }\int _{0}^{\pi }c_{\lambda ,2\tau }(u)\mathrm{d}ue^{2\tau }. \end{aligned}$$

Proof

Note first that \(\begin{pmatrix} \frac{1}{2}\\ k \end{pmatrix}(-1)^k<0\) if \(k\ge 1\), and \(\begin{pmatrix} \frac{1}{2}-j\\ k \end{pmatrix}(-1)^k>0\) if \(j\ge 1\) and \(k\ge 0\). Since the function \(-1<\frac{a_\lambda (u)}{1+b_\lambda (u)} < 0\) for \(u\in (0,\pi )\), and \(\sum _{k=0}^{\infty }\begin{pmatrix} \gamma \\ k \end{pmatrix}\beta ^k=(1+\beta )^{\gamma }\) for \(\gamma \in {\mathbb {R}}\) and \(|\beta |<1\), if \(u\in (0,\pi )\), we have that

$$\begin{aligned}&|c_{\lambda ,l}(u)|\\&\quad \le 1+ \sum _{k=1}^{l}\left| \begin{pmatrix} \frac{1}{2}\\ k \end{pmatrix}b_\lambda ^k(u)\right| \left| \cos ^{l-k}(u)\right| +\sum _{j=1}^{\left[ \frac{l}{2}\right] }\sum _{k=0}^{l-2j}\left| \begin{pmatrix} \frac{1}{2}-j\\ k \end{pmatrix}b_\lambda ^k(u)\right| \left| \cos ^{l-k-2j}(u)\begin{pmatrix} \frac{1}{2}\\ j \end{pmatrix}a_\lambda ^{j}(u)\right| \\&\quad \le 1-\sum _{k=1}^{\infty }\begin{pmatrix} \frac{1}{2}\\ k \end{pmatrix}b_\lambda ^k(u)-\sum _{j=1}^{\infty }\sum _{k=0}^{\infty }\begin{pmatrix} \frac{1}{2}-j\\ k \end{pmatrix}b_\lambda ^k(u)\begin{pmatrix} \frac{1}{2}\\ j \end{pmatrix}a_\lambda ^{j}(u)\\&\quad =2-\sqrt{1+b_\lambda (u)}\sum _{j=0}^{\infty }\begin{pmatrix} \frac{1}{2}\\ j \end{pmatrix}\left( \frac{a_\lambda (u)}{1+b_\lambda (u)}\right) ^j\\&\quad \le 2. \end{aligned}$$

Thus, by the dominated convergence theorem we have that

$$\begin{aligned} \int _{0}^{\pi }\sqrt{q_\lambda (u,e)}\mathrm{d}u=\sqrt{8\lambda }\sum _{l=0}^{\infty }\int _{0}^{\pi }c_{\lambda ,l}(u)\mathrm{d}ue^l, \end{aligned}$$

for every \(0\le e<1\). Taking into account the expression of \(c_{\lambda ,l}(u)\), if l is an odd number then \(\int _{0}^{\pi }c_{\lambda ,l}(u)\mathrm{d}u=0\). Therefore,

$$\begin{aligned} \int _{0}^{\pi }\sqrt{q_\lambda (u,e)}\mathrm{d}u=\sqrt{8\lambda }\sum _{\tau =0}^{\infty }\int _{0}^{\pi }c_{\lambda ,2\tau }(u)\mathrm{d}ue^{2\tau }. \end{aligned}$$

\(\square \)

Remark 5

From (33) in Lemma 6, if we define \(h_\tau (u,\omega ):=\cos ^{2\tau -2\omega }(u)(3-\cos ^2(u))^\omega \), we have that

$$\begin{aligned}&c_{\lambda ,2\tau }(u) =\sum _{n=0}^{2\tau }\sum _{m=\left[ \frac{n+1}{2}\right] }^{n}\begin{pmatrix} \frac{1}{2}\\ m \end{pmatrix}\begin{pmatrix} n\\ m \end{pmatrix}\frac{(m!)^2}{(2m-n)!}\frac{b_\lambda ^{2m-n}(u)a_\lambda ^{n-m}(u)}{n!}\cos ^{2\tau -n}(u)\\&\quad =\sum _{n=0}^{2\tau }\sum _{m=\left[ \frac{n+1}{2}\right] }^{n}\left[ \prod _{s=1}^{m}\left( \frac{1}{2}-s+1\right) \right] \frac{(-1)^m}{(n-m)!}\frac{1}{(2m-n)!}\frac{(32\lambda -2)^{2m-n}}{(32\lambda )^m} h_{\tau }(u,n-m). \end{aligned}$$

Now, taking into account that \(\int _{0}^{\pi }\cos ^i(x)\mathrm{d}x=\frac{i-1}{i}\int _{0}^{\pi }\cos ^{i-2}(x)\mathrm{d}x\) for \(i\ge 2\), if \(\tau \ge n-m\ge 0\) then

$$\begin{aligned}&\int _{0}^{\pi } h_{\tau }(u,n-m)\mathrm{d}u\\&\quad =\sum _{i=0}^{n-m}\begin{pmatrix} n-m\\ i \end{pmatrix}3^{i}(-1)^{n-m-i}\int _{0}^{\pi }\cos ^{2(\tau -i)}(u)\mathrm{d}u\\&\quad =\sum _{i=0}^{n-m}\begin{pmatrix} n-m\\ i \end{pmatrix}3^{i}(-1)^{n-m-i}\prod _{j=1}^{\tau -i}\frac{2\tau -2i-2j+1}{2\tau -2i-2j+2}\pi \\&\quad =:{\bar{g}}(\tau ,n-m)\pi . \end{aligned}$$

Hence,

$$\begin{aligned}&\int _{0}^{\pi }c_{\lambda ,2\tau }(u)\mathrm{d}u\nonumber \\= & {} \sum _{n=0}^{2\tau }\sum _{m=\left[ \frac{n+1}{2}\right] }^{n}\left[ \prod _{s=1}^{m}\left( \frac{1}{2}-s+1\right) \right] \frac{(-1)^m}{(n-m)!}\frac{1}{(2m-n)!}\frac{(32\lambda -2)^{2m-n}}{(32\lambda )^m}{\bar{g}}(\tau ,n-m)\pi \nonumber \\= & {} : g_\lambda (\tau )\pi . \end{aligned}$$
(34)

Note that \(\int _{0}^{\pi }c_{\lambda ,2\tau }(u)\mathrm{d}u\) is always a product between a number \(g_\lambda (\tau )\) and \(\pi \), so we can calculate with a symbolic computation software the expression of \(g_\lambda (\tau )\) without error. Thus, for example, we can verify that

$$\begin{aligned}&\int _{0}^{\pi }c_{\lambda ,0}(u)\mathrm{d}u=\pi \\&\quad \int _{0}^{\pi }c_{\lambda ,2}(u)\mathrm{d}u=\frac{768 \lambda ^{2} - 64 \lambda - 1}{4096 \lambda ^{2}}\pi \\&\quad \int _{0}^{\pi }c_{\lambda ,4}(u)\mathrm{d}u=\frac{3 \left( 2293760 \lambda ^{4} - 163840 \lambda ^{3} - 4608 \lambda ^{2} - 128 \lambda - 5\right) }{67108864 \lambda ^{4}}\pi . \end{aligned}$$

In general it seems to be that \(\int _{0}^{\pi }c_{\lambda ,2\tau }(u)\mathrm{d}u>0\) for all \(\lambda \ge \frac{1}{8}\) and \(\tau \in {\mathbb {N}}_0\).

The code used in the Python library “Sympy” with which we calculate \(\int _{0}^{\pi }c_{\lambda ,2\tau }(u)\mathrm{d}u\) is described below

$$\begin{aligned}&\text {def conv(m,n):}\\&\quad \text {return factorial(m)/factorial(m-n)/factorial(n)}\\&\quad \text { def intcos(s):}\\&\quad \text {return prod(Rational(2*s-2*j+1,2*s-2*j+2) for j in range(1,s+1))}\\&\quad \text {def bargt(t,s):}\\&\quad \text {return sum((-1)**(s-i)*3**(i)*conv(s,i)*intcos(t-i)\ for i in}\\&\quad \text {range(s+1))}\\&\quad \text {def gt(tau,lam):}\\&\quad \text {A=sum(sum((-1)**m*prod(Rational(1,2)-s+1}\\&\quad \text {for s in range(1,m+1))*1/factorial(n-m)/factorial(2*m-n)}\\&\quad \text {*(32*lam-2)**(2*m-n)/(32*lam)**m*bargt(tau,n-m) for m in}\\&\quad \text {range(floor(Rational(n+1,2)),n+1)) for n in range(2*tau+1))}\\&\quad \text {return A} \end{aligned}$$

The function \(\text {gt(tau,lam)}\) calculates \(g_\lambda (\tau )\) with error zero.

Lemma 8

Let \(c_{1,l}(u)\) be the coefficient defined in (32). Then

$$\begin{aligned} \int _{0}^{\pi }c_{1,2\tau }(u)\mathrm{d}u\ge -\frac{\pi }{4}. \end{aligned}$$

for all \(\tau \in {\mathbb {N}}_0\).

Proof

For \(u\in (0,\frac{\pi }{2})\), we have that

$$\begin{aligned}&c_{1,2\tau }(u)\nonumber \\&\quad =\sum _{k=0}^{2\tau }\begin{pmatrix} \frac{1}{2}\\ k \end{pmatrix}b_1^k(u)\cos ^{2\tau -k}(u)+\sum _{j=1}^{\tau }\sum _{k=0}^{2\tau -2j}\begin{pmatrix} \frac{1}{2}-j\\ k \end{pmatrix}b_1^k(u)\cos ^{2\tau -k-2j}(u)\begin{pmatrix} \frac{1}{2}\\ j \end{pmatrix}a_1^{j}(u)\nonumber \\&\quad =\cos ^{2\tau }(u)\sum _{k=0}^{2\tau }\begin{pmatrix} \frac{1}{2}\\ k \end{pmatrix}\left( -\frac{15}{16}\right) ^k+\sum _{j=1}^{\tau }\sum _{k=0}^{2\tau -2j}\begin{pmatrix} \frac{1}{2}-j\\ k \end{pmatrix}b_1^k(u)\cos ^{2\tau -k-2j}(u)\begin{pmatrix} \frac{1}{2}\\ j \end{pmatrix}a_1^{j}(u)\nonumber \\&\quad \ge \cos ^{2\tau }(u)\sqrt{1-\frac{15}{16}}+\sum _{j=1}^{\tau }\left[ \sum _{k=0}^{2\tau -2j}\begin{pmatrix} \frac{1}{2}-j\\ k \end{pmatrix}b_1^k(u)\right] \begin{pmatrix} \frac{1}{2}\\ j \end{pmatrix}a_1^{j}(u)\nonumber \\&\quad =(1+b_1(u))^{\frac{1}{2}}\sum _{j=1}^{\infty }\begin{pmatrix} \frac{1}{2}\\ j \end{pmatrix}\left( \frac{a_1(u)}{1+b_1(u)}\right) ^{j}\nonumber \\&\quad =(1+b_1(u))^{\frac{1}{2}}\left( -1+\sqrt{1+\frac{a_1(u)}{1+b_1(u)}}\right) \nonumber \\&\quad =\frac{a_1(u)}{(1+b_1(u))^{\frac{1}{2}}}\frac{1}{1+\sqrt{1-\frac{-a_1(u)}{1+b_1(u)}}}. \end{aligned}$$
(35)

Since the function \(\frac{a_1(u)}{(1+b_1(u))^{\frac{1}{2}}}\) is negative and increasing in \((0,\frac{\pi }{2})\) (because its derivative is equal to \( \frac{ \left( 45 \cos ^{2}{\left( u \right) } - 64 \cos {\left( u \right) } + 45\right) \sin {\left( u \right) }}{16 \left( 16 -15 \cos {\left( u \right) }\right) ^{\frac{3}{2}}} \), which is positive for all \(u\in (0,\frac{\pi }{2})\)), and the function \(\frac{-a_1(u)}{1+b_1(u)}\) is positive and decreasing in \((0,\frac{\pi }{2})\) (since its derivative \(- \frac{\left( 15 \cos ^{2}{\left( u \right) } - 32 \cos {\left( u \right) } + 45\right) \sin {\left( u \right) }}{2 \left( 15 \cos {\left( u \right) } - 16\right) ^{2}}\) is negative for \(u\in (0,\frac{\pi }{2})\)), we have that the last expression in the chain of inequalities mentioned above is increasing. Then, it reaches its minimum in \(\left[ 0,\frac{\pi }{2}\right] \) at \(u=0\). Hence \(c_{1,2\tau }(u)\ge c_{1,2\tau }(0)= - \frac{1}{4}\) for every \(u\in (0,\frac{\pi }{2})\). Therefore

$$\begin{aligned} \int _{0}^{\pi }c_{1,2\tau }(u)\mathrm{d}u=2\int _{0}^{\frac{\pi }{2}}c_{1,2\tau }(u)\mathrm{d}u\ge - \frac{\pi }{4}. \end{aligned}$$

\(\square \)

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Beltritti, G. Periodic solutions of a generalized Sitnikov problem. Celest Mech Dyn Astr 133, 6 (2021). https://doi.org/10.1007/s10569-021-10005-z

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Keywords

  • Generalized Sitnikov problem
  • Even periodic solutions
  • Global families
  • Continuation from circular Sitnikov problem