Physics Challenge I: The Raindrop solved by mfb and voko

[mfb]

I don't see the issue. t=0 is just the point where the second equation is satisfied.

With
$$r(t) = (a+bt)^2$$
i get $\dot{r}(0)=2ba$ and $r(0)=a^2$
Apart from the special case of r(0)=0, I can find a,b for every pair of $(r,\dot{r})$ (assuming both are positive, to stay in the physical range).

 

[jostpuur]

You should find a and b such that the following three condition are satisfied:

(1) r(0) is what we want.

(2) r'(0) is what we want.

(3) r(t) satisfies the ODE.

If you choose a and b so that (1) and (2) hold, the third one will probably not.

 

[lugita15]

It seems that a DE

<br /> \ddot{r}(t) = \alpha - \beta\frac{\dot{r}(t)^2}{r(t)}<br />

is relevant for this problem.

If you want to solve that differential equation, you can start by substituting v = dr/dt, so d^2 r/dt^2 = dv/dt = dv/dr * dr/dt = v* dv/dr. So we have a first-order differential equation of the form v*dv/dx = a - bv^2/r.

 

[jostpuur]

Actually I had a two component first order ODE in the beginning, and I was hoping to make things easier by eliminating the other component Now I'm not sure which way would turn out easier...

Here's a summary of equations: First, the physics must be used to justify the following three equations:

<br /> m(t) = \frac{4\pi \mu r(t)^3}{3}<br />
<br /> \dot{m}(t) = \rho\pi r(t)^2 v(t)<br />
<br /> \dot{m}(t)v(t) + m(t)\dot{v}(t) = gm(t)<br />

If m(t) is eliminated, we get the following two equations:

<br /> \dot{v}(t) = g - \frac{3\rho}{4\mu}\frac{v(t)^2}{r(t)}<br />
<br /> \dot{r}(t) = \frac{\rho}{4\mu} v(t)<br />

Which form a two component first order ODE, and which should have a unique solution as long as r(t)\neq 0.

If v(t) is eliminated from this, we get a second order DE for r(t) that has the form I was asking about.

 

[mfb]

You should find a and b such that the following three condition are satisfied:

(1) r(0) is what we want.

(2) r'(0) is what we want.

(3) r(t) satisfies the ODE.

If you choose a and b so that (1) and (2) hold, the third one will probably not.

Hmm...

$r(t) = (a+bt)^2$
$r'(t) = 2b(a+bt)$
$r''(t) = 2b$
$$2b^2=g-3\frac{r'^2}{r} = g-12b^2$$
Good point, b is fixed by the differential equation. Actually, your solutions are all solutions which start with a zero size somewhere, as "a" is a shift in time (by -a/b).

To find the other solutions, we can set r'(0)=0 and look at r(0) > 0.
r''(0)=g and r''(t) approaches g/7 for large t.

 

[jostpuur]

Forgetting all physics and merely focusing on mathematics, the problem is this ODE:

<br /> \left(\begin{array}{c}<br /> \dot{x}_1(t) \\ \dot{x}_2(t) \\<br /> \end{array}\right)<br /> = \left(\begin{array}{c}<br /> \alpha - \beta\frac{x_1(t)^2}{x_2(t)} \\ \gamma x_1(t) \\<br /> \end{array}\right)<br />

Considering the style in which voko solved the stuff, I think there's is a signifigant chance that a nice formula for all solutions doesn't necessarily exist. However, it could be that a formula exists for a graph of the solutions. That means that we don't seek a mapping t\mapsto (x_1(t),x_2(t)), but instead a mapping x_1\mapsto \mathcal{X}_2(x_1) such that the solution is of the form t\mapsto (x_1(t),\mathcal{X}_2(x_1(t))). This lead me to the one component first order ODE

<br /> f&#039;(x) = \frac{Cxf(x)}{Af(x)-Bx^2}<br />

which seems like some progress, but still was too difficult for me to solve in a moment.

 

[jostpuur]

Initial value f(0)=0 has a simple solution

<br /> f(x) = \frac{2B+C}{2A} x^2<br />

hm.. what are the rest?

 

[jostpuur]

I assumed

<br /> f(x) = a_0 + a_1x + a_2x^2 + a_3x^3 +\cdots<br />

with a_0\neq 0 and got

<br /> f(x) = a_0 + \frac{C}{2A}x^2 + \frac{BC}{4A^2a_0}x^4 + \frac{BC}{6A^3a_0^2}\Big(B-\frac{C}{2}\Big)x^6 + \cdots<br />

Doesn't look recognizable.

Altought this seems to hint that there could exist a function x\mapsto G_{B,C}(x) such that

<br /> f(x) = f(0) + f(0)G_{B,C}\Big(\frac{x^2}{Af(0)}\Big)<br />

...

 

[jostpuur]

The attempt

<br /> f(x) = \frac{2B+C}{2A}x^2 + g(x)<br />

works if g satisfies the DE

<br /> g&#039;(x) = -\frac{2Bxg(x)}{Ag(x) + \frac{C}{2}x^2}<br />

which is similar to the original DE for f, but with some key differences. If we assume that the constants are positive, the new DE has property g(x)&gt;0\implies g&#039;(x)&lt;0 for all x&gt;0. So it seems reasonable that g will probably approach zero if it starts at some positive value. With very large x approximation

<br /> g&#039;(x) \approx -\frac{4B}{Cx}g(x)<br />

should hold, which implies that there is signifigant chance for the approximation

<br /> g(x) = D x^{-\frac{4B}{C}} + o(?)\quad\quad\quad\textrm{when}\;x\to\infty<br />

with some constant D and with some small error term.

 

[jostpuur]

In thread how to prove asymptotic properties I finally managed to give a rigorous proof that the solutions for f with all initial values f(0)&gt;0 will have the form

<br /> f(x) = \frac{2B+C}{2A}x^2 + O\Big(x^{-\frac{4B}{C}+\epsilon}\Big)<br />

in the limit x\to\infty.

 

[jostpuur]

I believe that the solution which won this challenge contains a mistake. The rate of mass increase does not obey a formula

<br /> \dot{m}(t) = \pi\rho r(t)^2v(t)<br />as is believed by everyone, but instead a formula

<br /> \dot{m}(t) = \pi\rho r(t)^2\big(v(t)+2\dot{r}(t)\big)<br />

which follows by taking into account the way in which the lower boundary of the water drop moves downwards due to the increase in the drop size.

 

[mfb]

That's the approximation which has been discussed several times now.

 

[jostpuur]

Most comments seem to imply that the contribution from increase in size would be so small that the final differential equation (which was used in solution) would be correct. As if the correction was of infinitesimal order dt^2 or something like that. Also, the discussion must have been insufficient, since the correct differential equation has not been mentioned.

I checked that with the correct differential equation the result is still the same, that acceleration converges to \frac{g}{7}. A result, which should be seen as interesting.

 

[jostpuur]

A different problem: (Heavily related to the original challenge.)

A free planet moves with constant velocity in space. Then it encounters a massive cloud of dust, which it starts to absorb to itself on its path. The question is what happens. This could be slightly easier than the accelerating water drop actually, but it's still not trivial.

It's obvious that the planet will slow down and increase in size. But for example, it is not obvious at all if it will come to halt at some point (converge to some point), or continue infinitely far with always slower speeds.

I solved this (or I believe I did), and I proved that the velocity will slow down with rate v\sim t^{-\frac{3}{4}}. Therefore the traveled distance will be \sim t^{\frac{1}{4}}, and the planet will not stop due to the cloud.

 

[mfb]

For a planet, gravity would certainly be relevant. What was wrong with the raindrop (we can do the experiment at the ISS)?

 

[jostpuur]

Yes to be precise we should take into account how the planet affects the dust cloud, but then on the other hand, we also should take into account the air resistance with the water drop, so this is merely mathematics anyway, and these are interesting math problems.

Also, it might seem strange to not make the approximation v+2\dot{r}\approx v considering that we are making other approximations anyway. Still it would be preferrable that we know what approximations we are making. One point of view is that we should make approximations only if they make the calculations essentially easier. Keeping v+2\dot{r} doesn't lead to much added difficulty in the end.

 

[jostpuur]

Forgetting all physics and merely focusing on mathematics, the problem is this ODE:

<br /> \left(\begin{array}{c}<br /> \dot{x}_1(t) \\ \dot{x}_2(t) \\<br /> \end{array}\right)<br /> = \left(\begin{array}{c}<br /> \alpha - \beta\frac{x_1(t)^2}{x_2(t)} \\ \gamma x_1(t) \\<br /> \end{array}\right)<br />

Considering the style in which voko solved the stuff, I think there's is a signifigant chance that a nice formula for all solutions doesn't necessarily exist. However, it could be that a formula exists for a graph of the solutions. That means that we don't seek a mapping t\mapsto (x_1(t),x_2(t)), but instead a mapping x_1\mapsto \mathcal{X}_2(x_1) such that the solution is of the form t\mapsto (x_1(t),\mathcal{X}_2(x_1(t))).

I just understood that this choice was a mistake. The attempt must be to find a mapping x_2\mapsto\mathcal{X}_1(x_2) such that the solution is of the form t\mapsto (\mathcal{X}_1(x_2(t)),x_2(t)). This way the we end up with the same formulas that were in voko's post.