Pond Pollution Consider three ponds connected by streams, as in Figure 4. The first pond has a pollution source, which spreads via the connecting streams to the other ponds. The plan is to determine the amount of pollutant in each pond. f(t) 1 Figure 4. Three ponds 1, 2, 3 of volumes V1, V2, V3 connected by streams. The pollution source f(t) is in pond 1. Assume the following. Symbol f(t) is the pollutant flow rate into pond 1 (lb/min). Symbols fi, f2, f3 denote the pollutant flow rates out of ponds 1, 2, 3, respectively (gal/min). It is assumed that the pollutant is well-mixed in each pond. The three ponds have volumes V1, V2, V3 (gal), which remain con- stant. Symbols x1(t), 2(t), 3(t) denote the amount (lbs) of pollutant in ponds 1, 2, 3, respectively. 11.1 Examples of Systems 525 The pollutant flux is the flow rate times the pollutant concentration, e.g., pond 1 is emptied with flux f₁ times x₁(t)/V₁. A compartment analysis is summarized in the following diagram. f(t) fixi/V₁ 12 f3x3/V3 f2x2/V2 23 Figure 5. Pond diagram. The compartment diagram represents the three-pond pollution problem of Figure 4. The diagram plus compartment analysis gives the following differential equations. x₁(t) x3(t) − x1(t) + f(t), x2(t) = x(t) = *1(t)-2(1), #50-500 For a specific numerical example, take fi/V = 0.001, 1 ≤ i ≤3, and let f(t) 0.125 lb/min for the first 48 hours (2880 minutes), thereafter f(t) = 0. We expect due to uniform mixing that after a long time there will be (0.125) (2880) 360 pounds of pollutant uniformly deposited, which is 120 pounds per pond. Initially, 1(0)=2(0)=3(0) = 0, if the ponds were pristine. The specialized problem for the first 48 hours is x(t) =0.001 x3 (t) 0.00121(t) + 0.125, x2(t) =0.0011(t) 0.001 x2(t), x's(t) 0.001 2(t) 0.001 23(t), 1(0)=2(0)=3(0) = 0. The solution to this system is 21(t) = e 2000 125√3 9 √3t 125 125 sin COS + 2000 3 2000 3 24 125 √√3t 125√3 √√3t t x3(t) =e 2000 COS + sin 3 2000 9 2000 24 ⅰ)) + 옯-뜸. 125 3 After 48 hours elapse, the approximate pollutant amounts in pounds are x1 (2880) 162.30, x2(2880) = 119.61, x3 (2880) = 78.08. It should be remarked that the system above is altered by replacing 0.125 by zero, in order to predict the state of the ponds after 48 hours. The 526 Systems of Differential Equations corresponding homogeneous system has an equilibrium solution x₁(t): = x2(t) = x3(t) = 120. This constant solution is the limit at infinity of the solution to the homogeneous system, using the initial values x1(0)≈ 162.30, x2 (0) 119.61, x3(0) ≈ 78.08.