1 INTRODUCTION

Consider the planar \(4\)-body problem with masses \(m_{1}=m_{2}=m_{3}=1\) and \(m_{4}>0\). Let \(q_{i}\in\mathbb{R}^{2}\) be the positions of the bodies. Assume that the center of mass is at the origin and introduce Jacobi coordinates

$$z_{1}=q_{2}-q_{1}\qquad z_{2}=q_{3}-\frac{1}{2}(q_{1}+q_{2})\qquad z_{3}=q_{4}-\frac{1}{3}(q_{1}+q_{2}+q_{3}).$$

If \(\zeta_{i}=\dot{z}_{i}\), then the motion is governed by the Euler – Lagrange equations for the Lagrangian \(L=K+U\) where

$$K=\frac{1}{2}(\mu_{1}|\zeta_{1}|^{2}+\mu_{2}|\zeta_{2}|^{2}+\mu_{3}|\zeta_{3}|^{2})$$

is the kinetic energy and

$$U(z)=\sum_{i<j}\frac{m_{i}m_{j}}{r_{ij}}\qquad r_{ij}=|q_{i}-q_{j}|$$

is minus the Newtonian potential energy. The Jacobi mass parameters are

$${\textstyle\mu_{1}=\frac{m_{1}m_{2}}{m_{1}+m_{2}}=\frac{1}{2}\quad\mu_{2}=\frac{(m_{1}+m_{2})m_{3}}{m_{1}+m_{2}+m_{3}}=\frac{2}{3}\quad\mu_{3}=\frac{(m_{1}+m_{2}+m_{3})m_{4}}{m_{1}+m_{2}+m_{3}+m_{4}}=\frac{3m_{4}}{m}},$$

where \(m=m_{1}+m_{2}+m_{3}+m_{4}=3+m_{4}\) and the mutual distances are

$$\begin{matrix}r_{12}=|z_{1}|&r_{13}=|z_{2}+\frac{1}{2}z_{1}|&r_{23}=|z_{2}-\frac{1}{2}z_{1}|\\ r_{14}=|z_{3}+\frac{1}{2}z_{1}+\frac{1}{3}z_{2}|&r_{24}=|z_{3}-\frac{1}{2}z_{1}+\frac{1}{3}z_{2}|&r_{34}=|z_{3}-\frac{2}{3}z_{2}|.\end{matrix}$$

The Euler – Lagrange equations are

$$\displaystyle\begin{aligned} \displaystyle\dot{z}&\displaystyle=\zeta,\\ \displaystyle\dot{\zeta}&\displaystyle=M^{-1}\nabla U(z),\end{aligned}$$

where \(z(t)=\big{(}z_{1}(t),z_{2}(t),z_{3}(t)\big{)}\), \(\zeta(t)=\big{(}\zeta_{1}(t),\zeta_{2}(t),\zeta_{3}(t)\big{)}\) and \(M\) is the Jacobi mass matrix, \(M=\text{diag}(\mu_{1},\mu_{1},\mu_{2},\mu_{2},\mu_{3},\mu_{3})\).

A solution has a total collision at time \(T\) if \(z(t)\rightarrow 0\) as \(t\rightarrow T\). Let

$$I(z)=\mu_{1}|z_{1}|^{2}+\mu_{2}|z_{2}|^{2}+\mu_{3}|z_{3}|^{2}$$

be the moment of inertia with respect to the origin. We can think of \(I(z)\) as the square of a mass norm on \(\mathbb{R}^{6}\) and write \(I(z)=\lVert z\rVert^{2}\). The quantity \(r(z)=\sqrt{I(z)}=\lVert z\rVert\) is a convenient measure of the distance to collision. The unit vector \(\hat{z}=z/\lVert z\rVert\) is the corresponding normalized configuration.

A classical result of Chazy [1, 7] about total collision solutions states that, as \(t\rightarrow T\), \(r(t)\rightarrow 0\) with asymptotic rate proportional to \(|T-t|^{\frac{2}{3}}\). Meanwhile, the normalized configuration \(\hat{z}(t)\) converges to the set of normalized central configurations.

Definition 1

A point \(z\in\mathbb{R}^{6}\) is a central configuration or CC if

$$\nabla U(z)+\lambda Mz=0\vspace{-2mm}$$
(1.1)

for some constant \(\lambda>0\).

A simple example of a CC for our four-body problem is a centered equilateral triangle, that is, a configuration with \(m_{1},m_{2},m_{3}\) at the vertices of an equilateral triangle and \(m_{4}\) at the center.

If we release the masses at a central configuration \(z_{0}\) with initial zero velocity, we obtain a simple example of a total collision solution of the form \(z(t)=f(t)z_{0}\) where \(f(t)>0\) is a scalar function with \(f(t)\rightarrow 0\) as \(t\rightarrow T\). In this case, the normalized configuration \(\hat{z}(t)\) is constant. But there are many other less obvious examples of total collision with nonconstant \(\hat{z}(t)\).

A normalized CC with \(I(z)=\lVert z\rVert=1\) is a critical point of the restriction of the Newtonian potential to the unit sphere, \(S\). Since the planar \(n\)-body problem is rotationally symmetric, any configuration \(z\neq 0\) determines a circle of symmetrical configurations. If \(z\) is a CC, then so are all of the rotated configurations. In other words, there are circles of critical points for \(U(z)\) on \(S\). Imagine passing to the \(4\)-dimensional quotient manifold \(S/SO(2)\) (diffeomorphic to the complex projective space \(\mathbb{CP}(2)\)). Let \([z]\) denote the point of the quotient space determined by \(z\) and \(U([z])\) the function on \(\mathbb{CP}(2)\) induced by \(U\).

Suppose \([z_{0}]\) is a nondegenerate critical point of \(U([z])\) whose Hessian has \(k\) positive eigenvalues. Then it turns out that, after eliminating rotations and fixing the energy, there is a \((k+1)\)-dimensional set of initial conditions leading to total collision with \(\hat{z}(t)\rightarrow z_{0}\). Moreover, the rate of convergence of \(\hat{z}(t)\) to \(z_{0}\) is \(O(|T-t|^{p})\) where \(p>0\) is determined by the eigenvalues. The goal of this paper it to give an example of what can happen for degenerate CCs. In particular, Palmore [6, 5] showed that the centered equilateral triangle in our four-body problem is degenerate when the fourth mass is

$$m_{4}=m_{deg}=\frac{81+64\sqrt{3}}{249}\simeq 0.77048695.$$

The Hessian has 2 positive eigenvalues and also 2 zero eigenvalues. We will show that, for this example, there are total collision solutions with \(\hat{z}(t)\rightarrow z_{0}\) at a slower rate, indeed slower than any power of \(|T-t|\).

Theorem 1

In the four-body problem with \(m_{1}=m_{2}=m_{3}\) and \(m_{4}=m_{deg}\) there exist total collision solutions whose normalized configuration \(\hat{z}(t)\) converges to the centered equilateral CC with asymptotic rate proportional to \(-1/\ln|T-t|\) .

Figure 1 gives an idea of the appearance of normalized configurations mentioned in Theorem 1.

Fig. 1
figure 1

Normalized configurations converging to the centered equilateral triangle (black).

Full size image

Theorem 1 will follow from another result about solutions in a blown-up and rotation-reduced phase space. The modern approach to studying total collision uses McGehee’s blow-up technique which involves introducing the size coordinate \(r\) and a set of coordinates on the unit sphere \(S\) as well as suitable velocity variables and a new timescale \(\tau\) [3, 4]. In these coordinates \(\{r=0\}\) becomes an invariant total collision manifold. There are rest points in the collision manifold associate to the normalized CCs. Orbits which previously experienced total collision as \(t\nearrow T\) now converge to the set of rest points in the collision manifold as \(\tau\rightarrow+\infty\). The rest point corresponding to Palmore’s degenerate CC turns out to have a stable manifold and also a nontrivial center manifold. The solutions in the stable manifold converge to the rest point at an exponential rate in the new timescale, but we will show that there are solutions in the center manifold which converge more slowly.

Theorem 2

There are solutions in the center manifold of the rest point corresponding to Palmore’s degenerate CC which converge to the rest point with asymptotic rate in the new timescale proportional to \(1/\tau\) as \(\tau\rightarrow\infty\) .

The center manifold is entirely contained in the collision manifold \(\{r=0\}\). The total collision solutions with \(r(t)\searrow 0\) needed for Theorem 1 will be solutions in the center-stable manifold asymptotically converging to those in Theorem 2.

2 THE ISOSCELES SUBSYSTEMS

Since our goal is to show existence of special solutions, it suffices to work in a simple, lower-dimensional subsystem. Since \(m_{1}=m_{2}\) the subset of phase space such that masses \(m_{1},m_{2},m_{3}\) form an isosceles triangle with \(m_{4}\) on the symmetric axis is invariant. Fixing the orientation so that the symmetry axis is vertical, we will have

$$\displaystyle\begin{aligned} \displaystyle z_{1}&\displaystyle=(x_{1},0),\quad z_{2}=(0,x_{2}),\quad z_{3}=(0,x_{3}),\\ \displaystyle\zeta_{1}&\displaystyle=(\xi_{1},0),\quad\zeta_{2}=(0,\xi_{2}),\quad\zeta_{3}=(0,\xi_{3}).\end{aligned}$$

Since \(m_{1}=m_{3}\) and \(m_{2}=m_{3}\), there are two more invariant isosceles subsystems. We have a Lagrangian system on \(T(\mathbb{R}^{3}\setminus\Delta)\)

$$L(x,\xi)=\frac{1}{2}(\mu_{1}\xi_{1}^{2}+\mu_{2}\xi_{2}^{2}+\mu_{3}\xi_{3}^{2})+U(x_{1},x_{2},x_{3}),$$

where

$$U(x_{1},x_{2},x_{3})=\frac{1}{|x_{1}|}+\frac{4}{\sqrt{x_{1}^{2}+4x_{2}^{2}}}+\frac{3m_{4}}{|2x_{2}-3x_{3}|}+\frac{12m_{4}}{\sqrt{9x_{1}^{2}+4(x_{2}+3x_{3})^{2}}}$$

and \(\Delta=\{x:r_{ij}=0\text{ for some }i\neq j\}\).

Define a mass inner product on \(\mathbb{R}^{3}\)

$$\langle\!\langle v,w\rangle\!\rangle=\mu_{1}v_{1}w_{1}+\mu_{2}v_{2}w_{2}+\mu_{3}v_{3}w_{3}=\frac{1}{2}v_{1}w_{1}+\frac{2}{3}v_{2}w_{2}+\frac{3m_{4}}{m}v_{3}w_{3}$$

and mass norm \(\lVert v\rVert=\sqrt{\langle\!\langle v,v\rangle\!\rangle}\). To blow-up the total collision, let \(r=\lVert x\rVert\) and define the normalized configuration \(\hat{x}=x/r\). Then we will have

$$\lVert\hat{x}\rVert=1\qquad\langle\!\langle\hat{x},\hat{\xi}\rangle\!\rangle=0,$$

where \(\hat{\xi}=\dot{\hat{x}}\). At this point we have a Lagrangian system on the tangent bundle of \(\mathbb{R}^{+}\times\mathbb{S}^{2}\setminus\Delta\). On the open subset where \(x_{1}\neq 0\) we can introduce local coordinates on the sphere

$$s_{1}=\frac{x_{2}}{x_{1}}\qquad s_{2}=\frac{x_{3}}{x_{1}}$$

and their velocities \(\omega_{i}=\dot{s}_{i}\).

We have \(x=rx_{1}(1,s_{1},s_{2})=rx_{1}(1,s)\) and

$$\xi=\dot{x}=(\rho x_{1}+r\dot{x}_{1})(1,s)+rx_{1}(0,\omega),$$

where \(\rho=\dot{r}\). The spherical constraints give

$$x_{1}^{2}=\frac{1}{\lVert(1,s)\rVert^{2}}\qquad x_{1}\dot{x}_{1}=-\frac{\langle\!\langle(1,s),(0,\omega)\rangle\!\rangle}{\lVert(1,s)\rVert^{4}}.$$

Making these substitutions and eliminating \(x_{1},\dot{x}_{1}\) leads to the blown-up Lagrangian

$$L(r,\rho,s,\omega)=\frac{1}{2}\rho^{2}+\frac{r^{2}}{2}\lVert\omega\rVert^{2}_{S}+\frac{1}{r}V(s),$$

where

$$\lVert\omega\rVert_{S}^{2}=\frac{\lVert(1,s)\rVert^{2}\lVert(0,\omega)\rVert^{2}-\langle\!\langle(1,s),(0,\omega)\rangle\!\rangle^{2}}{\lVert(1,s)\rVert^{4}}.$$

and

$$V(s)=\lVert(1,s)\rVert U(1,s).$$

Explicitly, we have

$$\lVert(1,s)\rVert^{2}=\frac{1}{2}+\frac{2}{3}s_{1}^{2}+\frac{3m_{4}}{m}s_{2}^{2}.$$

We will also use the notation \(F(s,\omega)=\lVert\omega\rVert^{2}_{S}\). It is the local coordinate representation of the square of the norm on the unit sphere induced by the mass norm on \(\mathbb{R}^{3}\). It can be written as a quadratic form \(F=\omega^{T}A(s)\omega\) where \(A(s)\) is the \(2\times 2\) positive-definite symmetric matrix

$$A(s)=\frac{1}{\lVert(1,s)\rVert^{4}}\begin{bmatrix}\frac{m+6m_{4}s_{2}^{2}}{3m}&-\frac{2m_{4}s_{1}s_{2}}{m}\\ -\frac{2m_{4}s_{1}s_{2}}{m}&\frac{m_{4}(3+4s_{1}^{2})}{2m}\end{bmatrix}.$$

We have a Lagrangian system on the tangent bundle of \(\mathbb{R}^{+}\times\mathbb{R}^{2}\setminus\Delta\). The Euler – Lagrange equations are

$$\displaystyle\begin{aligned} \displaystyle\dot{r}&\displaystyle=\rho,\\ \displaystyle\dot{\rho}&\displaystyle=rF(s,w)-\frac{1}{r^{2}}V(s),\\ \displaystyle\dot{s}&\displaystyle=\omega,\\ \displaystyle\dot{\omega}&\displaystyle=\frac{1}{r^{3}}A^{-1}(s)\nabla V(s)-\frac{2\rho\omega}{r}+\frac{1}{2}A^{-1}(s)\nabla F(s,\omega)-A^{-1}(s)\dot{A}(s)\omega,\end{aligned}$$
(2.1)

where \(\nabla\) denotes the Euclidean gradient or partial gradient with respect to \(s\). The total energy of the system is

$$\frac{1}{2}\rho^{2}+\frac{r^{2}}{2}F(s,\omega)-\frac{1}{r}V(s)=h.$$

The last equation in (2.1) follows from the Euler – Lagrange equation \((L_{\omega})^{\cdot}=L_{s}\). We have \(L_{\omega}=r^{2}A(s)\omega\), so

$$(L_{\omega})^{\cdot}=r^{2}A(s)\dot{\omega}+r^{2}\dot{A}(s)\omega+2r\rho\big{(}A(s)\omega\big{)}.$$

We can further expand the last term of \(\dot{\omega}\) using

$$\dot{A}(s)\omega=\big{(}DA(s)(\omega)\big{)}\omega.$$

The equation for \(\dot{\omega}\) can be written more concisely if we make use of gradients and covariant derivatives with respect to the spherical metric. Let \(\tilde{\nabla}=A^{-1}(s)\nabla\) denote the gradient or partial gradient with respect to the spherical metric and let \(\tilde{D}_{t}\) denote the covariant time derivative of a vector field along a curve \(s(t)\). Then the usual formula involving Christoffel symbols is equivalent to

$$\tilde{D}_{t}\omega=\dot{\omega}-\frac{1}{2}\tilde{\nabla}F(s,\omega)+A^{-1}(s)\big{(}DA(s)(\omega)\big{)}\omega$$

and the last Euler – Lagrange equation simplifies to

$$\tilde{D}_{t}\omega=\frac{1}{r^{3}}\tilde{\nabla}V(s)-\frac{2\rho\omega}{r}.$$
(2.2)

3 COLLISION MANIFOLD AND CENTER MANIFOLD

To study orbits converging to total collision we will use McGehee’s blowup method [3, 4]. Namely, introduce rescaled velocity variables \(v=\sqrt{r}\rho,w=r^{\frac{3}{2}}\omega\) and a new time variable \(\tau\) such that \({}^{\prime}=r^{\frac{3}{2}}{}^{\cdot}\). The blown-up differential equations are

$$\displaystyle\begin{aligned} \displaystyle r^{\prime}&\displaystyle=vr,\\ \displaystyle v^{\prime}&\displaystyle=\frac{1}{2}v^{2}+F(s,w)-V(s),\\ \displaystyle s^{\prime}&\displaystyle=w,\\ \displaystyle\tilde{D}_{\tau}w&\displaystyle=\tilde{\nabla}V(s)-\frac{1}{2}vw,\end{aligned}$$
(3.1)

where

$$\tilde{D}_{\tau}w=w^{\prime}-\frac{1}{2}\tilde{\nabla}F(s,w)+A^{-1}(s)\big{(}DA(s)(w)\big{)}w.$$

The energy equation is

$$\frac{1}{2}v^{2}+\frac{1}{2}F(s,w)-V(s)=rh.$$

The collision manifold \(\{r=0\}\) is invariant. The equilibrium points are of the form \(P=(r,v,s,w)=(0,v_{0},s_{0},0)\) where \(\tilde{\nabla}V(s_{0})=0\). The last equation, which is equivalent to \(\nabla V(s)=0\), characterizes the reduced central configurations. If \(z(t)\) is a total collision solution with \(t\nearrow T\), then it can be shown that the corresponding solution \(\gamma(\tau)=\big{(}r(\tau),v(\tau),s(\tau),w(\tau)\big{)}\) converges to the set of equilibrium points in some level set \(v=v_{0}<0\) as \(\tau\rightarrow\infty\) (see the Appendix for a proof). The energy equation shows that \(v_{0}=-\sqrt{2V(s_{0})}\). Since \(r^{\prime}=vr\) and \(v(\tau)\rightarrow v_{0}<0\), it follows that \(r(\tau)\) converges to 0 exponentially. This will turn out to be equivalent to the asymptotic rate \(|T-t|^{\frac{2}{3}}\) in the natural timescale.

The centered equilateral triangle is represented in these coordinates by \(s_{0}=(\sqrt{3}/2,0)\) independent of \(m_{4}\). Let \(P=(r,v,s,w)=(0,v_{0},s_{0},0)\) where \(v_{0}=-\sqrt{2V(s_{0})}\). We want to study solutions \(\gamma(\tau)\) converging to \(P\) as \(\tau\rightarrow\infty\) The linearized differential equations at \(P\) have matrix

$$\begin{bmatrix}\delta r^{\prime}\\ \delta v^{\prime}\\ \delta s^{\prime}\\ \delta w^{\prime}\end{bmatrix}=\begin{bmatrix}v_{0}&0&0&0\\ 0&v_{0}&0&0\\ 0&0&0&I\\ 0&0&D\tilde{\nabla}V(s_{0})&-\frac{1}{2}v_{0}I\end{bmatrix}\begin{bmatrix}\delta r\\ \delta v\\ \delta s\\ \delta w\end{bmatrix}.$$

Since \(\nabla V(s)=0\), it follows that \(D\tilde{\nabla}V(s_{0})=D\big{(}A^{-1}(s)\nabla V(s)\big{)}=A^{-1}(s)D\nabla V(s)\).

The tangent space to the energy manifold is given by \(v_{0}\delta v=h\delta r\), so the upper left \(2\times 2\) block gives rise to a single eigenvalue \(v_{0}<0\). This corresponds to the exponential convergence of \(r(\tau)\) to 0. Let \(B\) be the lower right \(4\times 4\) block, representing the linearized differential equations within the collision manifold. If \(\delta s\) satisfies \(D\tilde{\nabla}V(\sigma_{0})\delta s=c\delta s\), then \((\delta s,\delta w)=(\delta s,\lambda_{\pm}\delta s)\) is an eigenvector of \(B\) with eigenvalue

$$\lambda_{\pm}=\frac{-v_{0}\pm\sqrt{v_{0}^{2}+16c}}{4}.$$

Since \(v_{0}<0\), negative eigenvalues \(c\) for \(D\tilde{\nabla}V(\sigma_{0})\) produce two unstable eigenvalues \(\lambda_{\pm}\). For \(c>0\) we have real eigenvalues \(\lambda_{+}>0\) and \(\lambda_{-}<0\). Finally, a degenerate eigenvalue \(c=0\) leads to \(\lambda_{+}>0\) and \(\lambda_{-}=0\).

The computation of the derivatives \(\tilde{\nabla}V(s)\) and \(D\tilde{\nabla}V(s)\) is rather complicated and was carried out with Mathematica. Since the centered equilateral triangle \(s_{0}=(\sqrt{3}/2,0)\) is a central configuration, we have \(\tilde{\nabla}V(s_{0})=0\) and it turns out that the second derivative simplifies to

$$D\tilde{\nabla}V(s_{0})=\frac{1}{2}\begin{bmatrix}9(1+\sqrt{3}m_{4})&-27\sqrt{3}m_{4}\\ -3\sqrt{3}(3+m_{4})&3\big{(}2+3\sqrt{3}(1+m_{4})\big{)}\end{bmatrix}.$$

The eigenvalues, \(c\), satisfy \(c^{2}+a_{1}c+a_{0}=0\) with

$$a_{1}=-\frac{3}{2}\big{(}5+3\sqrt{3}(1+2m_{4})\big{)},\quad a_{0}=\frac{27}{4}\big{(}2+3\sqrt{3}-(18-5\sqrt{3})m_{4}\big{)}.$$

It has real eigenvalues \(c_{1}(m_{4})>0\) and \(c_{2}(m_{4})\) changing from positive to negative at \(m_{4}=m_{deg}\). The four corresponding eigenvalues of \(B\) are all real with signs \(+,+,-,-\) for \(m_{4}<m_{deg}\) and \(+,+,+,-\) for \(m_{4}>m_{deg}\). Thus, for \(m_{4}\neq m_{deg}\) \(P\) is a hyperbolic rest point. Working in a five-dimensional energy manifold and remembering the eigenvalue \(v_{0}<0\), we have a stable manifold \(W^{s}(P)\) of dimension \(3\) for \(m_{4}<m_{deg}\) and \(2\) for \(m_{4}>m_{deg}\).

When \(m_{4}=m_{deg}\) we have a degenerate rest point and \(B\) has real eigenvalues \(+,+,-,0\). In the energy manifold we have a stable manifold with \({\rm dim}W^{s}(P)=2\). There is also a one-dimensional center manifold \(W^{c}(P)\) and a three-dimensional center-stable manifold \(W^{cs}(P)\) [2, 8]. To prove Theorem 2 we will try to understand the flow on the center manifold.

4 FLOW ON THE CENTER MANIFOLD AND PROOFS OF THE THEOREMS

From the discussion of eigenvalues above, the center subspace is

$$\mathbb{E}^{c}=\{(\delta r,\delta v,\delta s,\delta w)=(0,0,\delta s,0):D\nabla V(s_{0})\delta s=0\}.$$

When \(m_{4}=m_{deg}\) we have

$$D\tilde{\nabla}V(s_{0})=\frac{3}{166}\begin{bmatrix}9(49+9\sqrt{3})&-9(64+27\sqrt{3})\\ -4(16+69\sqrt{3})&2(179+165\sqrt{3})\end{bmatrix}$$

and the one-dimensional kernel is spanned by the vector

$$\alpha=\left(\frac{1}{26}(29+9\sqrt{3}),1\right).$$

The line through \(s_{0}\) in the direction of \(\mathbb{E}^{c}\) is given parametrically by

$$s_{1}(u)=s_{0}+u\alpha=(\sqrt{3}/2,0)+u\left(\frac{1}{26}(29+9\sqrt{3}),1\right),\quad u\in\mathbb{R}.$$

Since the collision manifold is invariant, we can find a center manifold with \(r=0\). Working in an energy manifold, \(W^{c}(P)\) will take the form of a a smooth curve through \(s_{0}\) tangent to \(\mathbb{E}^{c}\):

$$s(u)=s_{1}(u)+f(u),\qquad w(u)=g(u),\qquad v(u)=-\sqrt{2V\big{(}s(u)\big{)}-\lVert w(u)\rVert_{S}^{2}},$$

where \(f(u),g(u)\) are \(O(u^{2})\). Let \(u^{\prime}=\phi(u)\) denote the induced differential equation on \(W^{c}\). We have

$$s^{\prime}=\big{(}\alpha+Df(u)\big{)}u^{\prime}=\big{(}\alpha+Df(u)\big{)}\phi(u)=w=g(u).$$

It follows that \(\phi(u)=O(u^{2})\) and \(w=\alpha\phi(u)+O(|u|^{3})\).

Let \(\phi(u)=ku^{2}+O(|u|^{3})\). To determine \(k\), we will use the differential equation for \(w^{\prime}\). We have

$$w^{\prime}=Dg(u)u^{\prime}=Dg(u)\phi(u)=O(|u|^{3}).$$

All of the other terms in the covariant derivative \(\tilde{D}_{\tau}w\) are quadratic in \(w\), so \(\tilde{D}_{\tau}w=O(|u|^{3})\). It follows that the terms of order \(u^{2}\) on the right-hand side of the last equation of (3.1) must sum to zero. We have

$$-\frac{1}{2}v(u)w(u)=-\frac{1}{2}v_{0}ku^{2}\alpha+O(|u|^{3}).$$

Actually, we will focus on the inner product of the right-hand terms with \(\alpha\),

$$\langle\!\langle\alpha,-\frac{1}{2}v(u)w(u)\rangle\!\rangle=-\frac{1}{2}v_{0}ku^{2}\lVert\alpha\rVert^{2}+O(|u|^{3}).$$

The Taylor expansion of the remaining term is

$$\displaystyle\begin{aligned} \displaystyle\tilde{\nabla}V\big{(}s(u)\big{)}&\displaystyle=\tilde{\nabla}V\big{(}s_{0}+u\alpha+f(u)\big{)}\\ &\displaystyle=uD\tilde{\nabla}V(s_{0})\alpha+D\tilde{\nabla}V(s_{0})f(u)+u^{2}D^{2}\tilde{\nabla}V(s_{0})(\alpha,\alpha)+O(|u|^{3}).\end{aligned}$$

Since \(\alpha\) spans the kernel of \(D\tilde{\nabla}V(s_{0})\), the first term vanishes. Write the Taylor series of \(f(u)\) as \(f(u)=u^{2}(a_{1}\alpha+a_{2}\beta)+O(|u|^{3})\) where \(\beta\) is an eigenvector for the nonzero eigenvalue \(c_{1}\) of \(D\tilde{\nabla}V(s_{0})\). Then

$$D\tilde{\nabla}V(s_{0})f(u)=a_{2}c_{2}u^{2}\beta+O(|u|^{3}).$$

Since \(\beta\) is orthogonal to \(\alpha\), taking the inner product with \(\alpha\) gives \(\langle\!\langle\alpha,D\tilde{\nabla}V(s_{0})f(u)\rangle\!\rangle=O(|u|^{3})\). Since the \(\alpha\) components of the \(u^{2}\) terms on the right-hand side are to vanish, we have

$$\frac{1}{2}v_{0}k\lVert\alpha\rVert^{2}=\langle\!\langle\alpha,D^{2}\tilde{\nabla}V(s_{0})(\alpha,\alpha)\rangle\!\rangle.$$

The expression on the right side can be evaluated in terms of the one-dimensional Taylor series expansion of \(V(s)\) along the center subspace. Let \(W(u)=V(s_{0}+u\alpha)\). Then since \(s_{0}\) is a CC and \(\alpha\) is in the kernel of \(D\tilde{\nabla}V(s_{0})\), the first two derivatives of \(W\) at \(u=0\) vanish and we have

$$W(u)=V(s_{0})+\frac{1}{6}W^{\prime\prime\prime}(0)u^{3}+O(u^{4}),\qquad W^{\prime\prime\prime}(0)=\langle\!\langle\alpha,D^{2}\tilde{\nabla}V(s_{0})(\alpha,\alpha)\rangle\!\rangle.$$

Mathematica finds

$$V(s_{0})=\frac{9}{83}(49+9\sqrt{3}),\qquad W^{\prime\prime\prime}(0)=\frac{226557+188323\sqrt{3}}{8788}.$$

Finally, since \(v_{0}=-\sqrt{2V(s_{0})}\), we have

$$k=\frac{2W^{\prime\prime\prime}(0)}{v_{0}\lVert\alpha\rVert^{2}}=-\frac{\sqrt{6(37224915239-21141777681\sqrt{3}}}{793\sqrt{13}}\simeq-21.0944471.$$

Proof (of Theorem 2 )

In the isosceles subsystem, the center manifold of Palmore’s degenerate rest point has dimension one and the differential equation on it has the form \(u^{\prime}=\phi(u)=ku^{2}+O(|u|^{3})\) with \(k<0\). It follows that for initial conditions \(u_{0}>0\) and sufficiently close to the origin, we have \(u(\tau)\rightarrow 0\) as \(\tau\rightarrow\infty\). More precisely, write \(u^{\prime}=\phi(u)=-|k|u^{2}\big{(}1+g(u)\big{)}\) where \(g(u)=O(|u|)\). If \(v=1/u\) we have \(v^{\prime}=|k|\big{(}1+h(v)\big{)}\) where \(h(v)=g(1/v)\). L’Hospital’s rule shows that \(v(t)/(|k|\tau)\rightarrow 1\) as \(\tau\rightarrow\infty\), which means \(u(\tau)\rightarrow 0\) with asymptotic rate \(1/(|k|\tau)\) as claimed. Figure 1 shows normalized configurations of the form \(s_{1}(u)\) for \(u=0,0.1,0.2,0.3\).     \(\square\)

Theorem 1 will follow by using the center-stable foliation as well as a comparison of the blown-up time, \(\tau\), with the usual time, \(t\).

Proof (of Theorem 1 )

Consider a total collision solution converging to the centered equilateral triangle rest point \(P\) as \(t\nearrow T\). By definition \(t(\tau)\) satisfies

$$t^{\prime}(\tau)=r^{\frac{3}{2}}(\tau).$$

Since \(r^{\prime}=vr\) and since \(v(\tau)\rightarrow v_{0}<0\) as \(\tau\rightarrow\infty\), we have

$$\lim\frac{r^{\frac{3}{2}}}{T-t}=\lim\frac{\frac{3}{2}\sqrt{r}r^{\prime}}{-t^{\prime}}=\lim-\frac{3}{2}v(\tau)=\frac{3}{2}|v_{0}|.\vspace{-1mm}$$

This leads to Chazy’s estimate that \(r(t)\) is asymptotically proportional to \(|T-t|^{\frac{2}{3}}\).

By Theorem 2, there are solutions \(\gamma_{c}(\tau)\) in the center manifold of \(P\) such that \(s(\tau)\rightarrow s_{0}\) with asymptotic rate \(1/(|k|\tau)\). From the center-stable foliation, we see that for each such solution there is a two-dimensional set of solutions \(\gamma(\tau)\) in the center stable manifold with \(|\gamma(\tau)-\gamma_{c}(\tau)|\rightarrow 0\) exponentially. Moreover, since one of the two stable eigenvectors at \(s_{0}\) has \(\delta r\neq 0\), it follows that an open subset of the center-stable manifold consists of solutions with \(r(\tau)>0\), that is, genuine total collision solutions. Along these solutions we also have \(s(\tau)\rightarrow s_{0}\) with asymptotic rate \(1/(|k|\tau)\). To complete the proof, it remains to show that \(-1/\ln\big{(}T-t(\tau)\big{)}\) is asymptotically proportional to \(1/\tau\). Using L’Hospital’s rule one more time gives

$$\lim\frac{-\ln\big{(}T-t(\tau)\big{)}}{\tau}=\lim\frac{t^{\prime}}{T-t}=\lim\frac{r^{\frac{3}{2}}}{T-\tau}=\frac{3}{2}|v_{0}|$$

as in the last paragraph.     \(\square\)