# Snail Problem, Part 2

I read through some more of my college differential equations textbook and I think I am ready to further tackle the snail problem (see yesterday’s post). In short, a snail crawls along a twig of length L0 at 1 cm/d, and the twig grows 2 cm/d. Will the snail reach the end, and when?

I got as far as this differential equation, which I believe is correct:

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

I read about a technique to solve linear first-order differential equations last night, so I’m going to try it here. First we put this into standard form:

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

Since we already have an x’ and an x, if only they were multiplied by another function μ and μ’ respectively, we would have an inverse of the product rule (μx)’ = (μx’ + μ’x). Of course we can’t multiply different parts of the equation by different functions; we’ll have to multiply the entire equation by the same function μ(t). This will put a μ term in front of the x’, so we have to select a μ that, when multiplied by the function in the second term, will result in μ’. Let’s take a detour to find this function.

$\displaystyle \mu' = \mu\left( -\frac{2}{2t + L_0} \right)$

This is a separable differential equation; we can get the μ terms on one side and the t terms on the other. (Incidentally, I don’t think we could have done that with the original equation, or I would have.)

$\displaystyle \frac{d\mu}{dt} = \mu\left( -\frac{2}{2t + L_0} \right)$

$\displaystyle \frac{1}{\mu}\,d\mu = -\frac{2}{2t+L_0}\,dt$

I just realized we divided both sides by μ. I hope that was “allowed.” Of course we would have lost any solutions where μ = 0, though I’m not sure of the significance of that. In any case, I can’t stop now. Let’s integrate:

$\displaystyle \int \frac{1}{\mu}\,d\mu = \int \left( -\frac{2}{2t+L_0} \right) \,dt$

The left side is easy; it’s a standard rule.

$\displaystyle \ln |\mu|= \int \left( -\frac{2}{2t+L_0} \right) \,dt$

The right side is essentially the same; we need to use the inverse of the chain rule.

$\displaystyle \ln |\mu| = - \int \left( \frac{1}{2t+L_0} \right) (2) \,dt$

$\displaystyle \ln |\mu| = - \ln | 2t+L_0 | + c_1$

Exponentiating both sides yields

$\displaystyle e^{ \ln |\mu| } = e^{- \ln | 2t+L_0 | + c_1}$

$\displaystyle |\mu| = {e^{c_1}} {\left( e^{ln |2t+L_0|} \right)}^{-1}$

Setting $c_2 := e^{c_1}$

$\displaystyle |\mu| = c_2 {|2t+L_0|}^{-1}$

$\displaystyle \mu = \pm \frac{c_2}{2t+L_0}$

I think we can absorb the ± into the constant (though I’m not positive).

$\displaystyle \mu = \frac{c_3}{2t+L_0}$

Actually, we need a solution, not all solutions, so I’ll set the constant equal to 1. So dropping the ± was fine.

$\displaystyle \mu = {(2t+L_0)}^{-1}$

Armed with μ, we should now be ready to return to our earlier work. We’ll take our previous equation, in standard form,

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

and multiply both sides by the function μ:

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

We should now be able to a use the inverse of the product rule.

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

Integrating both sides gives

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

We’ll need to use the inverse of the chain rule again, so let’s set it up.

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

$\displaystyle \frac{x}{2t+L_0} = \frac{1}{2} \left(\ln |2t+L_0| + c_4\right)$

$\displaystyle \frac{x}{2t+L_0} = \frac{ \ln |2t+L_0|}{2} + c_5$

Multiplying both sides by 2tL0 gives

$\displaystyle x = \left( \frac{\ln |2t+L_0|}{2} + c_5 \right) (2t+L_0)$

Wow, we already solved for x. I wasn’t expecting it so soon. The solution is a little more complex than I’d expected, and I suspect I made a mistake somewhere. However, we should keep going. Let’s find the constant. The snail starts at position x = 0 at time t = 0, so

$\displaystyle 0 = \left( \frac{\ln |0+L_0|}{2} + c_5 \right) (0 + L_0)$

L0 is positive, so we don’t need the absolute value.

$\displaystyle 0 = \left( \frac{\ln L_0}{2} + c_5 \right)L_0$

$\displaystyle 0 = \frac{L_0 \ln L_0}{2} + c_5 L_0$

$\displaystyle c_5 L_0 = - \frac{L_0 \ln L_0}{2}$

$\displaystyle c_5 = - \frac{\ln L_0}{2}$

This is actually promising. Substituting c5 back into the equation we solved for x,

$\displaystyle x = \left( \frac{\ln |2t+L_0|}{2} + c_5 \right) (2t+L_0)$

$\displaystyle x = \left( \frac{\ln |2t+L_0|}{2} - \frac{\ln L_0}{2} \right) (2t+L_0)$

$\displaystyle x = \left( \frac{\ln |2t+L_0| - \ln L_0}{2} \right) (2t+L_0)$

I want to get rid of the absolute value. I think we can. The natural logarithm is only defined for positive numbers (at least without using complex numbers), so the absolute value allows our function to be defined for all nonzero arguments (just like $1/x$). If I remove the absolute value function, then the logarithm and therefore the equation will not be defined for 2t + L0 < 0, or when t < –L0/2. Since negative time is meaningless in this example (the snail started at time t = 0), I don’t mind discarding those times. (As an aside, I haven’t kept the units, but the 2 in the denominator is actually 2 cm/d — the speed at which the twig grows. The time –L0/2 corresponds to the time at which the twig had length zero. Clearly it would be meaningless in this problem to consider times before this.)

Back to the problem:

$\displaystyle x = \left[ \frac{\ln (2t+L_0) - \ln L_0}{2} \right] (2t+L_0)$

$\displaystyle x = [\ln (2t+L_0) - \ln L_0] \cdot \frac{2t+L_0}{2}$

$\displaystyle x = \ln \frac{2t+L_0}{L_0} \cdot \left(t + \frac{L_0}{2}\right)$

$\displaystyle x = \left(t+\frac{L_0}{2}\right) \ln \left(\frac{2t}{L_0} + 1\right)$

And that’s it! That is the equation for the distance the snail has crawled, x, at any given time t. We haven’t actually answered the question of if the snail will reach the end and when that would be, but we can save that for the next post.

# 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.