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There is a wonderful roller that transports mass from A to B and there is a pile of sand with weight $W$ in point A and we want to transport it to B.

Wonderfulness of roller comes from this property that it has:

  • When the weight of the mass that is on the roller be $W$ then speed of its transporting from left to right is $f(W)$ (meters per second) that $f$ is a function that is fixed for this roller (distance between A and B is a constant like $L$.).

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The main problem in the most general formulation is this:

  • What is the minimum time that we can transport this mass from A to B? And how we should insert masses along the time on the roller, that the minimum time occur?

Maybe solving the problem in the above general formulation is a far goal to reach. But I have thought on the problem in some certain cases and there are some results as below:

  • If $f(W)=\frac{1}{W}$ then the measure of the mass that we can move one meter from left to right in $1$ second is fixed and every case is optimum.
  • For some functions like $f(W) = w^{- \alpha}$ that $\alpha$ is a positive constant, solving problem is elementary with writing some inequalities.
  • If the optimum case occurs in $T$ second then if the weight of the mass that is on the roller in time $t$ be $w(t)$ then we have this two bellow equations ($M$ is the weight of all masses that we want to transport):

$$\int_{0}^{T} w(t)\times f(w(t)) \ dt = ML$$

$$w(0)=w(T)=0$$

Finally, some other questions about this problem:

  • Can connect this problem to Mass-Transportation Problems that are famous and well-studied in Mathematics?
  • Can solve the problem in some general cases like decreasing or convex functions?
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  • $\begingroup$ Your second equation simplifies to $w(0)=w(T)$. Don't you want to add $w(0)=w(T)=0$, so there is no mass on the roller at the beginning and the end of transportation? $\endgroup$
    – Alexandre Eremenko
    Commented Mar 27, 2015 at 2:11
  • $\begingroup$ ok! that's true. my purpose from writing that equation was this: sum of all wight that comes to the ruller is equal with sum of all weight that goes out from it, and know I changed it with the equation $w(0)=w(T)=0$. AlexandreEremenko $\endgroup$
    – Mohammad Ghiasi
    Commented Mar 27, 2015 at 6:03
  • $\begingroup$ Still I'm afraid that your equations are incorrect. From your equations it is clear that $T$ can be arbitrarily small, which is absurd. $\endgroup$
    – Alexandre Eremenko
    Commented Mar 28, 2015 at 18:08
  • $\begingroup$ You can have $w>0$ and $T$ with the property that $w(0)=w(T)=0$, $T$ arbitrarily small, and still integral $\int_0^T wf(w)dt$ is arbitrary, if for example $f(w)=1/\sqrt{w}$. $\endgroup$
    – Alexandre Eremenko
    Commented Mar 28, 2015 at 18:10
  • $\begingroup$ the equation is correct and is a necessary condition for every solution but it isn't sufficient (every $w(t)$ that satisfy the equation can not be a solution). @AlexandreEremenko $\endgroup$
    – Mohammad Ghiasi
    Commented Mar 28, 2015 at 19:32

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