Ch 2.5 Lake Pollution Models

• Pollution in our lakes and rivers has become a major problem, particularly over the past 50 years.

• In order to improve this situation in the future, it is necessary to gain a good understanding of the processes involved.

• Some way of predicting how the situation might improve (or decline) as a result of current management practices is vital.

• To this end we need to be able to predict how pollutant amounts or concentrations vary over time and under different management strategies.

• Linear first order ODE

\[ \frac{dS}{dt} = 10 c_{in}(t) - \frac{1}{10}S(t), \,\, S(0)= s_0 \]

• Normal form

\[ \frac{dS}{dt} + \frac{1}{10}S(t) = 10 c_{in}(t) , \,\, S(0)= s_0 \]

• Solve using method of integrating factors (next slide)

• Normal form

\[ \frac{dS}{dt} + \frac{1}{10}S(t) = 10 c_{in}(t) , \,\, S(0)= s_0 \]

• Solve using method of integrating factors

\[ \begin{align*} \mu(t) & = e^{\int \frac{1}{10} dt } = e^{\frac{t}{10} }\\ S(t) & = e^{-\frac{t}{10} } \int 10 e^{\frac{s}{10} }c_{in}(s) ds + s_0 e^{-\frac{t}{10} } \end{align*} \]

• Response due to initial condition goes to zero as t gets large.

\[ \lim_{t \rightarrow \infty} s_0 e^{-\frac{t}{10}} = 0 \]

• Normal form for IVP

\[ \frac{dS}{dt} + \frac{1}{10}S(t) = 10 c_{in}(t) , \,\, S(0)= s_0 \]

• Solution for \( c_{in}(t) = c_1 \):

\[ S(t) = s_0 e^{-t/10} + 100c_1(1-e^{-t/10}) \]

• Solution for \( c_{in}(t) = 0.2 - 0.1\sin(t) \):

\[ S(t) = s_0 e^{-t/10} + 20 + \frac{10}{101}(\sin(t) - 10\cos(t) - 192e^{-t/10}) \]

• \( c_0 \) = initial concentration of pollution in lake.
• Concentration \( c_{in} \) of pollutant in the flow entering lake is constant:

\[ c_{in}\, \frac{g}{m^3} = \mathrm{constant} \]

• Rate \( F \) at which water flows in and out of the lake is constant:

\[ F \, \frac{m^3}{day} = \mathrm{constant} \]

• For the mass \( M(t) \), we have:

\[ M'(t) = F c_{in} - F \frac{M(t)}{V} \]

• Since \( C(t) = M(t)/V \) and \( C'(t) = M'(t)/V \), we have:

\[ C'(t) = \frac{F}{V} c_{in} - \frac{F}{V}C \]

\[ \begin{align*} \frac{dC}{dt} & = \frac{F}{V}c_{in} - \frac{F}{V}C(t) , \,\, C(0)= c_0 \\ & = \frac{F}{V}\left(c_{in} - C(t)\right) \end{align*} \]

• Solve using separation of variables (see text)

\[ \begin{align*} C(t) & = c_{in} - (c_{in} - c_0) e^{-Ft/V } \\ & = c_{in}\left(1 - e^{-Ft/V }\right) + c_0e^{-Ft/V } \end{align*} \]

• Solution

\[ \begin{align*} C(t) & = c_{in} - (c_{in} - c_0) e^{-Ft/V } \\ & = c_{in}\left(1 - e^{-Ft/V }\right) + c_0e^{-Ft/V } \end{align*} \]

• Concentration of \( C(t) \) as t gets large.

\[ \lim_{t \rightarrow \infty} \left(c_{in} - (c_{in} - c_0) e^{-Ft/V }\right) = c_{in} \]

• How long will it take for the lake's pollution level to reach 5% of its initial level if only fresh water flows into the lake?
• From previous slide, with \( c_{in} = 0 \):

\[ C(t) = c_{in} - (c_{in} - c_0) e^{-Ft/V } = c_0 e^{-Ft/V } \]

• Solve for \( t \), and then use \( C = 0.05c_0 \):

\[ t = \frac{V}{F} \ln \left( \frac{C}{c_0} \right) = - \frac{V}{F}\ln(0.05) \cong \frac{3F}{V} \]

• From previous slide with \( C = 0.05c_0, \,c_{in} = 0 \):

\[ t \cong \frac{3F}{V} \]

• For Lake Erie, we can plug in values for \( F \) and \( V \)

\[ t \cong 7.8 \, \mathrm{years} \]

• For Lake Ontario:

\[ t \cong 23.5 \, \mathrm{years} \]