How can I put the "3 body problem" mathematically? I'm trying to put the 3 body problem mathematically. But I don't know how. I always get something reasonable, but I get something that is wrong.
 A: *

*Suppose that we have three celestial bodies, called Bodies $ 1 $, $ 2 $ and $ 3 $.

*Let $ \mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3} $ and $ m_{1},m_{2},m_{3} \in \mathbb{R}_{> 0} $ denote the displacement functions and masses of Bodies $ 1 $, $ 2 $ and $ 3 $ respectively.

*Let $ \mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3} \in \mathbb{R}^{3} $ and $ \mathbf{x}_{1},\mathbf{x}_{2},\mathbf{x}_{3} \in \mathbb{R}^{3} $ denote the initial velocities and initial displacements of Bodies $ 1 $, $ 2 $ and $ 3 $ respectively.

*According to Newton's Third Law of Motion and his Law of Universal Gravitation, the vector equation of motion for Body $ i $ is given as
\begin{align}
m_{i} \cdot \mathbf{X}_{i}''
&= \sum_{j \neq i} \frac{G m_{i} m_{j}}{\| \mathbf{X}_{j} - \mathbf{X}_{i} \|^{2}} \cdot \underbrace{\left[ \frac{1}{\| \mathbf{X}_{j} - \mathbf{X}_{i} \|} \cdot (\mathbf{X}_{j} - \mathbf{X}_{i}) \right]}_{\text{Unit vector in the direction $ \mathbf{X}_{j} - \mathbf{X}_{i} $}} \\
&= \sum_{j \neq i} \frac{G m_{i} m_{j}}{\| \mathbf{X}_{j} - \mathbf{X}_{i} \|^{3}} \cdot (\mathbf{X}_{j} - \mathbf{X}_{i}).
\end{align}
Dividing by $ m_{i} $ on both sides of the equation, we obtain
$$
\mathbf{X}_{i}'' = \sum_{j \neq i} \frac{G m_{j}}{\| \mathbf{X}_{j} - \mathbf{X}_{i} \|^{3}} \cdot (\mathbf{X}_{j} - \mathbf{X}_{i}).
$$

*Therefore, the Three-Body Problem is mathematically expressed as the following set of nine equations:

\begin{align}
\forall i \in \{ 1,2,3 \}: \quad
{\mathbf{X}_{i}''}(t) &= \sum_{j \neq i} \frac{G m_{j}}{\| {\mathbf{X}_{j}}(t) - {\mathbf{X}_{i}}(t) \|^{3}} \cdot [{\mathbf{X}_{j}}(t) - {\mathbf{X}_{i}}(t)], \\
{\mathbf{X}_{i}'}(0)  &= \mathbf{v}_{i}, \\
{\mathbf{X}_{i}}(0)   &= \mathbf{x}_{i}.
\end{align}

Note: For certain initial-data sets $ (\mathbf{v}_{1},\mathbf{v}_{2},\mathbf{v}_{3},\mathbf{x}_{1},\mathbf{x}_{2},\mathbf{x}_{3}) \in \mathbb{R}^{18} $, a global solution $ (\mathbf{X}_{1},\mathbf{X}_{2},\mathbf{X}_{3}): \mathbb{R} \to \mathbb{R}^{9} $ does not exist, due to singularities that result from binary or triple collisions among the bodies. However, the collection of all such initial-data sets has Lebesgue measure $ 0 $ in $ \mathbb{R}^{18} $, so these initial-data sets are not generic in the sense of measure.
