Question

2. We are concerned about the pollution in a long lake with a dam at one end. We can model the lake as a one-dimensional medium on a domain 2€ (0, 100), with pollution density u(, t). A stream flows into and out of the lake, bringing in pollution at a density of v = 3/10 and removing pollution at a rate of a = 2/5 per unit density. The lake contains bacteria that break down the pollution at a rate of a = 1/40 and the pollution diffuses within the lake with diffusion coefficient D = 1/10. The PDE to describe the pollution level in the lake is U = Dur-au u(0,t) = V (100, 1) = u(100, 1) What is the equation for the pollution level in terms of re [0, 100 at equilibrium? Provide a sketch of the solution.

Answer 1

Solution At equilibrium, the pollution density doesn't change with time, hence is only a function of position vcx). The derivative Up Of u with respect to time is 280. and the fpost per'n becomes. OD - au 2 1 of u since the coefticeint of u on the RHS is the the solution to this PDF are given by Mouw ) - AVD & Be ex - Bez The and equation in the question states that UO) V bence A-B zv - A= Btv The expression for u becomes ux) = (B+V) ex - Bellax The derivative of u is uxl) - Į LB +v) et + + Belt

Then, from the 3rd ean Uze(100) ţ (B+v)+ Re 50 BA - (Btu) - ape 50 - Į (B+V) 0001B 100 - L (Btude – LB 2B(1&- 12) - (2+2) - ( -a velo since the value of 100 is very lara to &, we can ignore the and term in the Parantheses in the LHS to set Bld - 1 h) e a 11 av 100 100 = -V B becomes u UX) - vetz uia) - 0.32 let)

This at equilibriu is of the this pollution eab level eah is shown A graph 0.25 los of 10 20 30 40 50 60 70 80 90 Thanking you.