# Snail Problem

A friend mentioned this problem a while back, and while I was able to use a computer to get an accurate numerical solution, I really want to find an analytical solution using math. The problem is that it will require differential equations, which I no longer remember. Edit: I haven’t yet been able to solve this problem, but I’m preserving my initial attempts below. See Part 2 where I made more progress.

The problem is this: a snail starts crawling along a twig, from one end towards the other. The snail can crawl at 1 cm/d. But the twig is growing at 2 cm/d. The twist is that the twig is adding material along its entire length, so the snail will be carried forward a bit, too, depending on where it is. Will the snail ever reach the end? Let’s let L0 be the initial length of the twig.

Let’s say the snail crawls at 1 cm/d and the twig grows at 2 cm/d. The length of the twig at time t would be

$\displaystyle L(t) = L_0 + 2t$.

For any point along the twig, the velocity will be proportional to the distance along the length of the twig, so at a distance x

$\displaystyle v_t(x, t) = \frac{x}{L_0 +2t} \cdot 2 = \frac{2x}{L_0 + 2t}$

Let’s double-check this: the beginning of the twig is $x = 0$:

$\displaystyle v_t(0,t) = \frac{2 \cdot 0}{L_0 + 2t} = 0$.

And at the end of the twig, $x = L_0 + 2t$:

$\displaystyle v_t(L_0+2t,t) = \frac{2(L_0+2t)}{L_0+2t} = 2$.

The velocity of the snail will include both its movement along the twig (at 1 cm/d) and the growth of the twig. It will be therefore

$\displaystyle v_s(x,t) = \frac{2x}{L_0+2t} + 1$

Since the velocity is the derivative of its position,

$\displaystyle \frac{dx}{dt} = \frac{2x}{L_0 + 2t} + 1$

and we want to know if and when $x(t) = L(t)$. In other words, solve for t:

$\displaystyle x(t) = L_0 + 2t$

Taking the derivative with respect to t,

$\displaystyle \frac{d}{dt} x(t) = \frac{d}{dt} (L_0 + 2t)$

$\displaystyle \frac{dx}{dt} = 2$

Note that this makes sense, at the time t when the snail reaches the end, … wait, no it doesn’t, that’s not correct. Since the functions are only equal at a specific time t, I can’t differentiate them and expect them to be equal.

Going back:

$\displaystyle \frac{dx}{dt} = \frac{2x}{L_0 + 2t} + 1$

Is this an linear differential equation? I hope so, since I’m hoping that will make it easier to solve (though at this point, I don’t remember how to solve linear or nonlinear differential equations.

$\displaystyle \frac{dx}{dt} - \frac{2}{L_0 + 2t} \cdot x = 1$.

I think I got it into the correct form. I don’t yet know what to do from here, so I’ll read on and also refresh my memory of differentiation rules.

I actually don’t think we’ll need complex differential equation techniques. We should be able to just integrate this entire equation with respect to t. I think I see why linear differential equations are easier to solve.

$\displaystyle \int ( \frac{dx}{dt} - \frac{2}{L_0 + 2t} \cdot x )\,dt = \int 1\,dt$

By the sum rule,

$\displaystyle \int \frac{dx}{dt} \,dt - \int \frac{2}{L_0+2t} \cdot x\,dt = \int 1\,dt$.

I spoke too soon. The first term is easy to integrate and the last is trivial, but I don’t know how to integrate the second.

$\displaystyle x + c_1 - \int \frac{2}{L_0+2t} \cdot x\,dt = t + c_2$

I can rearrange the terms and combine the constants

$\displaystyle \int \frac{2}{L_0+2t} \cdot x\,dt = x - t + c_3$

and I’m now stuck.

OK, I think we might be able to use integration by parts, using the formula

$\displaystyle \int u\,dv = uv - \int v\,du$.

If I set

$\displaystyle u =\frac{2}{L_0+2t}$

and set

$\displaystyle dv = x\,dt$

then to find v, we need to integrate:

$\displaystyle v = \int x\,dt$

$\displaystyle v = \frac{x^2}{2} + c_4$.

And to find du, we need to take the derivative of u. I’m just going to use the product rule, since I’m rusty at this and I don’t remember the quotient rule.

$\displaystyle du = \frac{d}{dt}\frac{2}{2t+L_0} = \frac{d}{dt}[2\cdot (2t+L_0)^{-1}]$

$\displaystyle du = \left(\frac{d}{dt}2\right) \cdot \frac{1}{2t+L_0} + 2 \cdot \frac{d}{dt}(2t+L_0)^{-1}$

$\displaystyle du = 0 + 2 \cdot (-1)\cdot(2t+L_0)^{-2}\cdot 2$

I’m going to keep going, but I just realized that I am not going to be able to integrate $\int v\,du$.

$\displaystyle du = -\frac{4}{(2t+L_0)^2}$

Shoot.

$\displaystyle uv - \int v\,du = \frac{2}{2t+L_0}\cdot \left(\frac{x^2}{4}+c_4 \right) + \int \left(\frac{x^2}{4}+c_4 \right) \left[\frac{4}{(2t+L_0)^2}\right]\,dt$.

This is a disaster.

I will have to study this more and return to this problem later.