One of several defining characteristics of our planet is that there is a lot of water. According to [1],
Since life on Earth relies heavily on water, the safety and cleanliness of the water is a natural topic of interest for us.
In this section, we introduce two simple models for contaminant movement (advection) through a well-mixed body of water by considering two cases from our book [7]:
The advection-only case leads to an ODE model and is a good place to start.
What don't ants get sick?
They have anty-bodies.
A common usage of salt water is the production of brine. According to Wikipedia [2],
Determine the amount of salt (the solute) dissolved in a tank of water (the solution) as a function of time.
The input-output compartmental diagram for salt solution is:
From the balance law and compartment diagram, we have
The rate at which salt enters tank is the product of flow rate in and concentration of salt in the incoming mixture:
\[ \left(10 \frac{L}{min}\right) \left(c_{in}(t) \frac{kg}{L}\right) = 10 c_{in}(t) \frac{kg}{min} \]
The rate at which salt leaves tank is the product of flow rate out and concentration of salt in the outgoing mixture is given by
\[ \left( 10 \frac{L}{min}\right) \left(\frac{S(t)}{100} \frac{kg}{L} \right) = \frac{S(t)}{10} \frac{kg}{min} \]
Our IVP will then have the following form:
\[ \frac{dS}{dt}= 10 c_{in}(t) - \frac{S(t)}{10}, \,\, S(0) = s_0 \]
The Normal form of the IVP is
\[ \frac{dS}{dt} + \frac{1}{10}S(t) = 10 c_{in}(t) , \,\, S(0)= s_0 \]
\[ \begin{aligned} \frac{dy}{dt} + p(t)y & = g(t) \\ \mu(t) & = e^{\int p(t)dt } \\ y(t) & = \frac{\int_0^t \mu(s)g(s) ds}{\mu(t)} + \frac{C}{\mu(t)} \end{aligned} \]
Thus for the salt tank IVP, we have
\[ \begin{aligned} \frac{dS}{dt} + \frac{1}{10}S(t) &= 10 c_{in}(t) , \,\, S(0)= s_0 \\ \mu(t) & = e^{\int \frac{1}{10} dt } = e^{\frac{t}{10} } \\ S(t) & = \frac{\int_0^t 10 e^{\frac{s}{10} }c_{in}(s) ds}{e^{\frac{t}{10}}} + \frac{C}{e^{\frac{t}{10}}} \\ S(t) & = e^{-\frac{t}{10} } \int_0^t 10 e^{\frac{s}{10} }c_{in}(s) ds + s_0 e^{-\frac{t}{10} } \end{aligned} \]
Methods of integration are required to solve for \( S(t) \), the specifics of which depend on the form of \( c_{in}(t) \).
Suppose \( c_{in}(t) = c_1 \). Recalling that
\[ \small{ \int e^{Ax }dx = \frac{1}{A}e^x + C } \]
we can solve our IVP as follows:
\[ \small{ \begin{aligned} S_1(t) & = e^{-\frac{t}{10} } \int_0^t 10 e^{\frac{s}{10} }c_{in}(s) ds + s_0 e^{-\frac{t}{10}} \\ & = 10c_1 e^{-\frac{t}{10} } \int_0^t e^{\frac{s}{10} } ds + s_0 e^{-\frac{t}{10}} \\ & = 10c_1 e^{-\frac{t}{10} } \cdot 10\left( e^{\frac{t}{10} } - 1 \right) + s_0 e^{-\frac{t}{10}} \\ & = s_0 e^{-t/10} + 100c_1(1-e^{-\frac{t}{10} }) \end{aligned} } \]
For \( c_{in} \) as below, integration by parts can be used:
\[ \small{ \begin{aligned} c_{in}(t) & = 0.2 - 0.1\sin(t) \\ S_2(t) & = e^{-\frac{t}{10} } \int_0^t 10 e^{\frac{s}{10} }c_{in}(s) ds + s_0 e^{-\frac{t}{10}} \\ & = s_0 e^{-t/10} + 20 + \frac{10}{101}(\sin(t) - 10\cos(t) - 192e^{-t/10}) \end{aligned} } \]
\[ \small{ S(t) = e^{-\frac{t}{10} } \int_0^t 10 e^{\frac{s}{10} }c_{in}(s) ds + s_0 e^{-\frac{t}{10}} } \]
According to the EPA,
Determine the concentration of pollutant in the lake as a function of time, and use it to determine the length of time required for mitigation effects to reduce pollution levels by a specified amount.
The compartment diagram is similar to the salt water tank:
Using the balance law and the compartment diagram, we have:
\[ \small{ \begin{Bmatrix} \mathrm{rate \, of \, change} \\ \mathrm{of \, concentration \, of } \\ \mathrm{pollutant \,in \, lake } \\ \mathrm{at \, time \,} t \end{Bmatrix} = \begin{Bmatrix} \mathrm{rate \, of \, change} \\ \mathrm{of \, concentration \, of } \\ \mathrm{pollutant \, entering \, lake } \\ \mathrm{at \, time \,} t \end{Bmatrix} - \begin{Bmatrix} \mathrm{rate \, of \, change} \\ \mathrm{of \, concentration \, of } \\ \mathrm{pollutant \, leaving \, lake } \\ \mathrm{at \, time \,} t \end{Bmatrix} } \]
The rate at which pollutant enters & leaves lake is product of flow rate in and concentration of pollutant in water flow:
\[ \small{ \begin{aligned} \left(F \frac{m^3}{day}\right) \left(c_{in}(t) \frac{g}{m^3}\right) &= F c_{in} \frac{g}{day} \\ \left(F \frac{m^3}{day}\right) \left(C(t) \frac{g}{m^3}\right) & = F C(t) \frac{g}{day} \end{aligned}} \]
To convert these mass expressions into concentrations, we divide by \( V \) and write our IVP as
\[ \frac{dC}{dt} = \frac{F}{V}\left[c_{in} - C(t)\right], \,\,\, C(0) = c_0 \]
The IVP is given by
\[ \frac{dC}{dt} = \frac{F}{V}\left[c_{in} - C(t)\right], \,\,\, C(0) = c_0 \]
The exact solution can be found using either by the method of integrating factors or by separating variables (see text):
\[ C(t) = c_{in} - (c_{in} - c_0) e^{-Ft/V } \]
Observe that \( C(t) \) tends to the value of the incoming concentration \( c_{in} \) as \( t \) gets large:
\[ \lim_{t \rightarrow \infty} \left(c_{in} - (c_{in} - c_0) e^{-Ft/V }\right) = c_{in} \]
\[ C(t) = c_{in} - (c_{in} - c_0) e^{-Ft/V } = c_0 e^{-Ft/V } \]
Next, let \( C(t) = 0.05c_0 \) and solve for \( t \):
\[ \begin{aligned} C(t) &= c_0 e^{-Ft/V } \\ 0.05c_0 &= c_0 e^{-Ft/V } \\ \ln(0.05) &= - \frac{Ft}{V} \\ t &= - \frac{V}{F}\ln(0.05) \\ t &\cong \frac{3V}{F} \end{aligned} \]
-log(0.05)
[1] 2.995732
\[ V = 4.58 \times 10^{11} m^3,\,\, \,\, F = 4.80 \times 10^8 \frac{m^3}{day}\left(\frac{365 \, day}{year} \right) \]
Thus Lake Erie requires almost 8 years to clear up to within 5% of its current levels; see R computation below:
\[ t \cong \frac{3V}{F} \, \mathrm{years} \]
-log(0.05)*(4.58*10^(11))/(4.8*10^8*365)
[1] 7.831309
\[ V = 1.636 \times 10^{12} m^3,\,\, \,\, F = 5.72 \times 10^8 \frac{m^3}{day}\left(\frac{365 \, day}{year} \right) \]
Thus Lake Ontario requires almost 24 years to clear up to within 5% of its current levels; see R computation below:
\[ t \cong \frac{3V}{F} \, \mathrm{years} \]
-log(0.05)*(1.636*10^(12))/(5.72*10^8*365)
[1] 23.47456
[1] How Much Water is There on Earth?, https://www.usgs.gov/special-topics/water-science-school/science/how-much-water-there-earth, retrieved on 9/2/2022.
[2] Brine, https://en.wikipedia.org/wiki/Brine, retrieved on 9/2/2022.
[3] Water Supply, https://gjcity.org/297/Water-Supply, retrieved on 9/2/2022.
[4] Drought Forces Grand Junction To Dip Into Colorado River, https://www.cbsnews.com/colorado/news/drought-grand-junction-colorado-river/, retrieved on 9/2/2022.
[5] Where This Occurs: Lakes and Rivers, https://www.epa.gov/nutrientpollution/where-occurs-lakes-and-rivers, retrieved on 9/2/2022.
[6] Pollution, https://www.lakescientist.com/pollution/, Sarah Hicks, Kent State University, retrieved on 9/2/2022.
[7] Mathematical Modeling with Case Studies, Barnes and Fulford, CRC Press, 2015.