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edited Apr 29, 2012 at 17:06

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Assume $x_1(0) < x_2(0) < x_3(0) < x_4(0)$. Note that $\dfrac{dx_2}{dt}$ and $\dfrac{dx_4}{dt}$ are positive and the other two are negative. So for $t > 0$ (and before the collision), $x_1 < x_1(0) < x_2(0) < x_2 < x_3 < x_3(0) < x_4(0) < x_4$. Now $ \dfrac{dx_4}{dt} \le \dfrac{m_4}{(x_4(0)-x_3(0))^3}$, call$\dfrac{dx_4}{dt} \le \dfrac{m_4}{(x_4-x_3(0))^3}$. Solving the differential equation obtained by making this $A_4$an equality, and similarlywe find that $\dfrac{dx_1}{dt} \ge - A_1 = -\dfrac{m_1}{(x_2(0) - x_1(0))^3}$$x_4(t) \le x_3(0) + ((x_4(0) - x_3(0))^4 + 4 m_4 t)^{1/4}$. Call the right side $B_4(t)$. SoSimilarly $$ \dfrac{dx_2}{dt} > \frac{m_2}{(x_3(0) - x_1(0) + A_1 t)(x_4(0) + A_4 t - x_2(0))(x_3 - x_2)}$$$x_1(t) \ge B_1(t) = x_2(0) - ((x_2(0) - x_1(0))^4 + 4 m_1 t)^{1/4}$.

So $$ \dfrac{dx_2}{dt} > \frac{m_2}{(x_3(0) - B_1(t))(B_4(t) - x_2(0))(x_3 - x_2)}$$ $$ \dfrac{dx_3}{dt} < \frac{m_3}{(x_3(0) - x_1(0) + A_1 t)(x_4(0) + A_4 t - x_2(0))(x_2 - x_3)}$$$$ \dfrac{dx_3}{dt} < \frac{m_3}{(x_3(0) - B_1(t))(B_4(t) - x_2(0))(x_2 - x_3)}$$ $$ (x_3 - x_2) \dfrac{d}{dt} (x_3 - x_2) < - \dfrac{m_3+m_2}{(x_3(0) - x_1(0) + A_1 t)(x_4(0) + A_4 t - x_2(0))} $$$$ (x_3 - x_2) \dfrac{d}{dt} (x_3 - x_2) < - \dfrac{m_3+m_2}{(x_3(0) - B_1(t))(B_4(t) - x_2(0))} $$ Thus a collision will occur by time $T$ if $$ \int_0^T \dfrac{ dt}{(x_3(0) - x_1(0) + A_1 t)(x_4(0) + A_4 t - x_2(0))} > \frac{(x_3(0)-x_2(0))^2}{2(m_2 + m_3)} $$$$ \int_0^T \dfrac{ dt}{(x_3(0) -B_1(t))(B_4(t) - x_2(0))} > \frac{(x_3(0)-x_2(0))^2}{2(m_2 + m_3)} $$ The integral of the left side from $0$ to $\infty$ is finiteinfinite, but doesn't depend on $m_2$ or $m_3$. So certainly if $m_2$ or $m_3$ is large enoughsince (depending on$x_3(0) - B_1(t)$ and $m_1$,$B_4(t) - x_2(0)$ only grow like $m_2$ and the$t^{1/4}$ as $x_j(0)$)$t \to \infty$.

So there will always be a collision will occur in finite time.

Assume $x_1(0) < x_2(0) < x_3(0) < x_4(0)$. Note that $\dfrac{dx_2}{dt}$ and $\dfrac{dx_4}{dt}$ are positive and the other two are negative. So for $t > 0$ (and before the collision), $x_1 < x_1(0) < x_2(0) < x_2 < x_3 < x_3(0) < x_4(0) < x_4$. Now $ \dfrac{dx_4}{dt} \le \dfrac{m_4}{(x_4(0)-x_3(0))^3}$, call this $A_4$, and similarly $\dfrac{dx_1}{dt} \ge - A_1 = -\dfrac{m_1}{(x_2(0) - x_1(0))^3}$. So $$ \dfrac{dx_2}{dt} > \frac{m_2}{(x_3(0) - x_1(0) + A_1 t)(x_4(0) + A_4 t - x_2(0))(x_3 - x_2)}$$ $$ \dfrac{dx_3}{dt} < \frac{m_3}{(x_3(0) - x_1(0) + A_1 t)(x_4(0) + A_4 t - x_2(0))(x_2 - x_3)}$$ $$ (x_3 - x_2) \dfrac{d}{dt} (x_3 - x_2) < - \dfrac{m_3+m_2}{(x_3(0) - x_1(0) + A_1 t)(x_4(0) + A_4 t - x_2(0))} $$ Thus a collision will occur by time $T$ if $$ \int_0^T \dfrac{ dt}{(x_3(0) - x_1(0) + A_1 t)(x_4(0) + A_4 t - x_2(0))} > \frac{(x_3(0)-x_2(0))^2}{2(m_2 + m_3)} $$ The integral of the left side from $0$ to $\infty$ is finite, but doesn't depend on $m_2$ or $m_3$. So certainly if $m_2$ or $m_3$ is large enough (depending on $m_1$, $m_2$ and the $x_j(0)$) a collision will occur in finite time.

So $$ \dfrac{dx_2}{dt} > \frac{m_2}{(x_3(0) - B_1(t))(B_4(t) - x_2(0))(x_3 - x_2)}$$ $$ \dfrac{dx_3}{dt} < \frac{m_3}{(x_3(0) - B_1(t))(B_4(t) - x_2(0))(x_2 - x_3)}$$ $$ (x_3 - x_2) \dfrac{d}{dt} (x_3 - x_2) < - \dfrac{m_3+m_2}{(x_3(0) - B_1(t))(B_4(t) - x_2(0))} $$ Thus a collision will occur by time $T$ if $$ \int_0^T \dfrac{ dt}{(x_3(0) -B_1(t))(B_4(t) - x_2(0))} > \frac{(x_3(0)-x_2(0))^2}{2(m_2 + m_3)} $$ The integral of the left side from $0$ to $\infty$ is infinite, since $x_3(0) - B_1(t)$ and $B_4(t) - x_2(0)$ only grow like $t^{1/4}$ as $t \to \infty$.

So there will always be a collision in finite time.

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created Apr 29, 2012 at 16:20

Robert Israel

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Assume $x_1(0) < x_2(0) < x_3(0) < x_4(0)$. Note that $\dfrac{dx_2}{dt}$ and $\dfrac{dx_4}{dt}$ are positive and the other two are negative. So for $t > 0$ (and before the collision), $x_1 < x_1(0) < x_2(0) < x_2 < x_3 < x_3(0) < x_4(0) < x_4$. Now $ \dfrac{dx_4}{dt} \le \dfrac{m_4}{(x_4(0)-x_3(0))^3}$, call this $A_4$, and similarly $\dfrac{dx_1}{dt} \ge - A_1 = -\dfrac{m_1}{(x_2(0) - x_1(0))^3}$. So $$ \dfrac{dx_2}{dt} > \frac{m_2}{(x_3(0) - x_1(0) + A_1 t)(x_4(0) + A_4 t - x_2(0))(x_3 - x_2)}$$ $$ \dfrac{dx_3}{dt} < \frac{m_3}{(x_3(0) - x_1(0) + A_1 t)(x_4(0) + A_4 t - x_2(0))(x_2 - x_3)}$$ $$ (x_3 - x_2) \dfrac{d}{dt} (x_3 - x_2) < - \dfrac{m_3+m_2}{(x_3(0) - x_1(0) + A_1 t)(x_4(0) + A_4 t - x_2(0))} $$ Thus a collision will occur by time $T$ if $$ \int_0^T \dfrac{ dt}{(x_3(0) - x_1(0) + A_1 t)(x_4(0) + A_4 t - x_2(0))} > \frac{(x_3(0)-x_2(0))^2}{2(m_2 + m_3)} $$ The integral of the left side from $0$ to $\infty$ is finite, but doesn't depend on $m_2$ or $m_3$. So certainly if $m_2$ or $m_3$ is large enough (depending on $m_1$, $m_2$ and the $x_j(0)$) a collision will occur in finite time.