Toronto Math Forum
APM3462016F => APM346Misc => Topic started by: XinYu Zheng on November 05, 2016, 06:17:23 PM

Here is a problem with a counter intuitive result. Before reading the solution, I strongly encourage to try solving the problem yourself. It is rather fun to work out. This is not a PDE problem, but an ODE one. However, solving the ODE requires some cleverness.
Suppose we start with a rubber rope of rest length $L_0$ with one end attached to the wall. At $t=0$, an ant is placed on the rope at the loose end and begins crawling towards the wall with a speed of $u$ relative to the rope. Meanwhile, the free end of the rope is being pulled away from the wall at a constant speed $v\gg u$, causing the rope to stretch uniformly.
(a) Show that, no matter how large $v$ is compared to $u$, the ant will always reach the wall eventually. Find an explicit expression for the time it takes to do so in terms of $u$, $v$, and $L_0$.
(b) Suppose now that in addition to the rope being pulled away at speed $v$, it is also accelerating with acceleration $a>0$, such that $2aL_0<v^2$. Show that in this case, the ant may or may not reach the wall depending on what $u$ is. Show that there is a lower bound on $u$ in terms of $a$, $L_0$, and $v$ where if the ant crawls any slower it will never reach the wall.
(c) Examine the case where $2aL_0=v^2$. Why is this case interesting?
(d) Examine the case where $2aL_0>v^2$.
(e) It is sometimes said that because the universe is expanding, light from galaxies that are too distant may never reach us. In light of the results of this problem, comment on this statement.
Solutions:
(a) This is a one dimensional problem. Let us place our origin at the location of the wall. Then as a function of $t$, the location of the loose end of the rope is $L(t)=L_0+vt$. Let the location of the ant be $X(t)$. We know that the ant crawls towards the wall with speed $u$ relative to the rope. But at every moment it is also being pulled away from the wall by virtue of standing on the rope because the rope is stretching. Since the rope stretches uniformly, the speed at which the ant is being pulled away is $vX(t)/L(t)$ (ie. it is proportional to the fraction of the rope the ant occupies as measured from the wall). Then we can write the following differential equation describing the ant's motion:
$$X'(t)=u+v\frac{X(t)}{L(t)}$$
Solving this equation can be rather tricky using standard techniques, so we make a change of variables $S(t)=X(t)/L(t)$. Then $X'=S'L+L'S$. Subbing this into the equation we get a major simplification:
$$S'L=u\implies S'=\frac{u}{L}=\frac{u}{L_0+vt}$$
Now this can be solved easily:
$$S(t)=\frac{u}{v}\ln(L_0+vt)+C$$
Note that the initial condition at $t=0$ requires $S(0)=1$, thus $C=\frac{u}{v}\ln L_0+1$, and the final equation becomes
$$S(t)=\frac{u}{v}\ln\left(1+\frac{vt}{L_0}\right)+1$$
Requiring that the ant reaches the wall is equivalent to finding $T$ such that $S(T)=0$. Solving:
$$\frac{u}{v}\ln\left(1+\frac{vt}{L_0}\right)+1=0\implies T=\frac{L_0}{v}(e^{v/u}1)$$
So we can see that the time it takes the ant to reach the wall grows exponentially with the ratio $v/u$, but nevertheless always finite. Just as a comparison, suppose that $v=1km/s$ and $u=1m/s$. Then $T$ will be on the order of $e^{1000}\approx 1.97\times 10^{434}$ seconds, which is about $10^{417}$ times the age of the universe.
(b) This time, we have $L(t)=L_0+vt+at^2/2$. We make the same substitution $S=X/L$ and now the equation reduces to
$$S'(t)=\frac{u}{L_0+vt+at^2/2}$$
Completing the square in the denominator yields
$$S'(t)=\frac{u}{L_0\frac{v^2}{2a}+\frac{a}{2}(t+v/a)^2}=\frac{2uak^2}{1k^2(at+v)^2}$$
Where $k\equiv 1/\sqrt{v^22aL_0}$. The denominator can be factored as a difference of squares and then decomposed into partial fractions. The result after integration will be two logarithms (here combined into one):
$$S(t)=ku\ln\left(\frac{1+k(at+v)}{1k(at+v)}\right)+C$$
Once again, we require $S(0)=1$. Then $C=ku\ln\left(\frac{1kv}{1+kv}\right)+1$ and after some rearranging we obtain
$$S(t)=ku\ln\left(\frac{1+\frac{kat}{1+kv}}{1+\frac{kat}{kv1}}\right)+1$$
Now let us try to solve $S(T)=0$ where $T>0$. Define $d=e^{\frac{1}{ku}}$. Then we have
$$d=\frac{1+\frac{kaT}{1+kv}}{1+\frac{kaT}{kv1}}$$
After some rearranging we arrive at
$$kaT=\frac{(1d)(k^2v^21)}{1kv+d(1+kv)}$$
Note that since $d<1$ and $kv>1$, the numerator is positive. Thus for a positive solution of $T$ to exist we must demand that $1kv+d(1+kv)>0$, which is equivalent to
$$\frac{kv1}{kv+1}<d$$
Now subbing in for $d=e^{\frac{1}{ku}}$ and solving for $u$ we obtain the lower bound
$$u>\frac{1}{k\ln\left(\frac{kv+1}{kv1}\right)}$$
Thus in the case where there is nonzero acceleration satisfying $2aL_0<v^2$, there exists a lower bound on $u$. The ant can no longer make it to the wall all the time.
(c) In the borderline case where $2aL_0=v^2$, the ODE for $S(t)$ reduces to
$$S'(t)=\frac{2au}{(at+v)^2}$$
Which can be solved right away:
$$S(t)=\frac{2u}{at+v}+C$$
Applying the initial condition $S(0)=1$ yields $C=12u/v$ and we have
$$S(t)=2u\left(\frac{1}{at+v}\frac{1}{v}\right)+1$$
Now, solving for $S(T)=0$ yields (after rearranging)
$$\frac{2uv}{2uv}v=aT$$
For positive $T$ solutions, we demand that $\frac{2uv}{2uv}>v$, or $\frac{2u}{2uv}>1$. For this to happen, we only need to demand that the denominator is positive, which yields the lower bound $u>v/2$. This case is particularly interesting because the lower bound only depends on $v$ and nothing else.
(d) Once again, the same story, except this time we define $k\equiv 1/\sqrt{2aL_0v^2}$. Then the ODE for $S(t)$ becomes
$$S'(t)=\frac{2uak^2}{1+k^2(at+v)^2}$$
The solution of this ODE is in the form of arctan:
$$S(t)=2ku\arctan(k(at+v))+C$$
Using initial condition $S(0)=1$ yields $C=2ku\arctan(kv)+1$ and thus we have
$$S(t)=2ku(\arctan(k(at+v))\arctan(kv))+1$$
Note that this function is decreasing with increasing $t$ since $\arctan$ is strictly increasing. Thus to get a positive solution for $S(t)=0$, we can find the threshold by taking $t\to \infty$ and demand $\lim_{t\to\infty}S(t)=0$. Note that $\arctan(k(at+v))\to \pi/2$ in the limit, thus we have
$$0=2ku(\pi/2\arctan(kv))+1\implies u=\frac{1}{2k(\pi/2\arctan(kv))}$$
This is the threshold velocity of crawling. Anything slower than this will cause $S(t)$ to never intersect the $x$ axis, thus the ant will never reach the wall.
(e) In light of the results of the problem, we can say that if the rate of expansion of the universe is constant and that the universe expands uniformly, then light from galaxies will always eventually reach us, regardless of how far away they are or how fast the galaxies are receding. This corresponds to the result of (a), where the light plays the role of the ant and the rubber rope is the fabric of spacetime. However, the results of Hubble tells us that the universe is accelerating in its expansion. From the conclusions of (b), (c), and (d) above, we can see that if this were the case, then it is possible that the light of some galaxies will never reach us. Their existence will never be known to us, nor to any future lifeforms that will ever inhabit our planet.

This is a well known sequence of problems and they are usually solved by introducing $x(t)=X(t)/L(t)$ (as you do) which is the coordinate of the ant in the coordinate system frozen into the rubber band in contrast to the external coordinate system. Compare with
https://en.wikipedia.org/wiki/Lagrangian_and_Eulerian_specification_of_the_flow_field (https://en.wikipedia.org/wiki/Lagrangian_and_Eulerian_specification_of_the_flow_field) in hydrodynamics: $x(t)$ is an Eulerian coordinate and $X(t)$ is a Lagrangian coordinate.