Project
Problem 1
Consider the following equations that describe a remediation process in porous media.
Contaminant concentration (u) is governed by the diffusion equation where the diffusion
coefficient depends on the concentration of the chemical X (v) (that is injected from the
opposite side for remediation purpose) and given by
Du = k0 /(1 + v(x)).
Similarly, the diffusion of the chemical X depends on the concentration of the contaminant,
Dv = k0 /(1 + u(x)).
Consider a square domain where initial concentrations of contaminant (u) and chemical X (v)
as shown in the figure 1. Take k0 to be 0.1 everywhere in D = [−0.5, 0.5]2 except k0 = 0.03
in the rectangle [−0.4, 0.4] × [−0.1, 0.1].
a) Solve coupled equations ut = div(Du (x, v(x, t))∇u), vt = div(DRu (x, u(x, t))∇v).
b) Calculate the average concentration of the contaminant (1/|D| D u(x, t)dx) as a function of time and plot.
c) Compare the average concentration of the contaminant with and without remediation.
Plot.
d) Plot the concentration at time t = 2 with and without remediation.
no flow
u=1
u(t=0)=0
no flow
no flow
no flow
v(t=0)=0
v=1
no flow
Figure 1: Boundary and initial conditions for contaminant and chemical X concentrations
Problem 2
Consider predator-prey model with no flow boundary conditions for both predator and
prey in [0, 1]2 .
∂u
= 0.01∇2 u + u − uv − u2
∂t
∂v
= 0.01∇2 v − v + uv.
∂t
(1)
Assume u(x, t = 0) = 1 in [0.1, 0.2]2 and v(x, t = 0) = 1 in [0.8, 0.9]2 . Solve this equation
using MATLAB up to time 2, and plot both predator and prey population at time instants
t = 0.3, 0.7, 1, 1.5, 2.
1