# Cleaning up the Great Lakes - Differential equation systems

## Problem Details

The idea of the problem is to find out how long it would take to flush the Great Lakes of pollution. They're set up as a series of five tanks and you are given inflow rates of clean water, inflow rates from the other tanks, and outflow rates.

For our model, we make the following assumptions:

1. The volume of each lake remains constant.
2. The flow rates are constant throughout the year.
3. When a liquid enters the lake, perfect mixing occurs and the pollutants are uniformly distributed.
4. Pollutants are dissolved in the water and enter or leave by inflow or outflow of solution.

Before using this model to obtain estimates on the cleanup times for the lakes, we consider some simpler models:

• (a) Use the outflow rates given in the figure to determine the time it would take to “drain” each lake. This gives a lower bound on how long it would take to remove all the pollutants.

• (b) A better estimate is obtained by assuming that each lake is a separate tank with only clean water flowing in. Use this approach to determine how long it would take the pollution level in each lake to be reduced to 50% of its original level. How long would it take to reduce the pollution to 5% of its original level?

• (c) Finally, to take into account the fact that pollution from one lake flows into the next lake in the chain, use the entire multiple compartment model given in the figure to determine when the pollution level in each lake has been reduced to 50% of its original level, assuming pollution has ceased (that is, inflows not from a lake are clean water). Assume that all the lakes initially have the same pollution concentration $p$. How long would it take for the pollution to be reduced to 5% of its original level?

## Solution so far:

$\frac{dA}{dt}$ = rate in-rate out where $A$ is the amount of pollution at time $t$.

First I wrote equations for each lake. Rather than using $A$ as my variable, I used the first letter of each lake (with $n$ for Ontario) to stand for the amount of pollution in the given lake at time $t$. This gives...

$$\frac{ds}{dt} = \frac{-15s}{2900}$$ $$\frac{dm}{dt} = \frac{-38m}{1180}$$ $$\frac{dh}{dt} = \frac{15s}{2900} + \frac{38m}{1180} - \frac{68h}{850}$$ $$\frac{de}{dt} = \frac{68h}{850} - \frac{85e}{116}$$ $$\frac{dn}{dt} = \frac{85e}{116} - \frac{99n}{393}$$

Rearranging and pulling out the differential operator leads to the following system:

$$\left(D + \frac{38}{1180}\right)[m] = 0$$ $$\left(D - \frac{68}{850}\right)[h] + \frac{38m}{1180} - \frac{15s}{2900}=0$$ $$\left(D + \frac{15}{2900}\right)[s] = 0$$ $$\left(D + \frac{85}{116}\right)[e] - \frac{68h}{850} = 0$$ $$\left(D + \frac{99}{393}\right)[n] - \frac{85e}{116} = 0$$

This is where I'm stuck. I have this system of five equations with five variables. It seems like it should be fairly straightforward to solve from here, but I can't figure out what to do next.

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By the way, this problem comes from Nagle, Saff, and Snider, "Fundamentals of Differential Equations", chapter 5, group project E. – Teepeemm Apr 8 '15 at 19:40

$(D + \frac{38}{1180})[m] = 0$

$(D - \frac{68}{850})[h] + \frac{38m}{1180} - \frac{15s}{2900}=0$

$(D + \frac{15}{2900})[s] = 0$

$(D + \frac{85}{116})[e] - \frac{68h}{850} = 0$

$(D + \frac{99}{393})[n] - \frac{85e}{116} = 0$

Just for the sake of brevity , I have converted all the fractions into decimal form ie

$\frac{38}{1180} = 0.0322$

$\frac{68}{850} = 0.0800$

$\frac{15}{2900} = 0.0051$

$\frac{85}{116} = 0.7327$

$\frac{99}{393} = 0.2519$

I am going to use the Laplace method for converting these differential equations into algebraic equations , thereby we will solve the equations and get the Laplace transforms of $S,H,M,E$ and $N$ and then take the inverse Laplace transforms of these to get the final equations of pollution for these lakes . Now $s(0) = h(0) = m(0) = e(0) = n(0) = p$ , since the initial conditions are same for all the lakes .

Therfore using Laplace notation we have ,

$sS - s(0) + 0.0051S = 0$

Therfore $S = \frac{p}{s+.0051}$

Similarly we have $M = \frac{p}{s+0.0322}$

Now for (2) differential equation we have

$(D - \frac{68}{850})[h] + \frac{38m}{1180} - \frac{15s}{2900}=0$

Applying Laplace we will have

$H = \frac{-.0322p}{(s+.0322)(s+.08)}\frac{.0051p}{(s+.0051)(s+.08)}+\frac{p}{(s+.08)}$

Similarly , you can plug in this value of $H$ in the remaining differential equations ( along with the value of $M$ and $S$ found previously to find the Laplace expressions for $E$ and $N$ , and then take the Laplace inverse of all and you will get the equations for all lakes .

Now for getting $5\%$ time, just put $0.05p$ and equate it to get $t$ for each case .

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So to get the 5% time I do 0.05p and for 50% it's 0.5p? – Contourette. Dec 9 '13 at 10:03
Yes you have to put $0.05p$ for 5% and $0.5p$ for 50% . – AbKDs Dec 10 '13 at 0:44
I haven't studied Laplace yet so could you give me an idea of Paul Safier's answer ? I've commented to his solution. – Contourette. Dec 10 '13 at 9:49
I will definitely give a try :) ! – AbKDs Dec 10 '13 at 10:52
@Contourette , I realised the equations that you have written and we have subsequently solved are for part (c) . Actually in part (b) says to just assume separate tanks , ie no pollution comes in only it goes out . Say for Ontario the equations will be $\frac{dn}{dt} = \frac{-99}{393}n$ ie no incoming flow of pollution (only outgoing pollution) . So (b) part is very straightforward . The solutions me and Paul Safier have written are for part (c) . – AbKDs Dec 10 '13 at 14:50

Now that you have your system of differential equations defined, you can start solving them. The equations for $s(t)$ and $m(t)$ are not coupled so you can solve those first. All equations have the initial condition of $p$ amount of polutant. After you solve for $s(t)$ and $m(t)$ (and let me know if this is a problem) you can plug in those to the equations for $h(t)$. After you have $h(t)$ you can get $e(t)$ and finally $n(t)$. Then for each equation plug in $0.05 p$ for the concentration and solve for $t$ to determine the time it takes for the concentration in each lake to drop to 5% of the original concentration.

For example, the solution of $s(t)$ is: $$s(t)=p e^{-\frac{15}{2900}t}.$$ Setting $s=0.05p$ and solving for time gives $t=579$. Do the same for the others.

Does this help any?

Paul Safier

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Okay so I have some uncertainties while solving. As you said I computed the $s(t)$ and $m(t)$ and replaced in $h(t)$. Though when computing $h(t)$ I got really large numbers and I'm not sure if it's correct. Then another question I have is about b). So to compute the time to reduce to 5% you said it's $0.05p$ so that means for 50% is $0.5p$ ? And for the question at c), I don't really understand what it asks. I mean, the input of a lake is equal to the output ( taking into consideration the clean water) so I have no idea what it asks.I edited the requirements adding last 2 lines to c. Thanks – Contourette. Dec 7 '13 at 12:58
A reason you are getting huge values for $h$ might be because you wrote the equation wrong the second time. It should be $(D+\frac{68}{850})[h] = \text{stuff}$. – Stephen Montgomery-Smith Dec 12 '13 at 2:00