Problem

Solve the system of convection-diffusion-reaction equations on a circular domain (circle.m) with

$$\begin{array}{rcl} f_1(u_1,u_2,x,t) &=& - c u_1 u_2, \\ f_2(u_1,u_2,x,t) &=& - c u_1 u_2 + f(x,t). \end{array}$$ (4)

Take $c = 3$, $a = 0.01$, $b = 0$ and use a time step of size $k = 0.1$ with final time $T = 1$.

As initial conditions, take $u_{10} = 1$ and $u_{20} = 0$ in the domain $\Omega$. This corresponds to a situation where we have a beaker containing substance $A$ (with concentration $1$). There is nothing of substance $B$ in the beaker so to get the reaction going, we need to add $B$ to the system. We do this by taking

$$f(x,t) = \left\{ \begin{array}{l} 2, \quad \mbox{if } \v... ...rt < 0.2, \\ 0, \quad \mbox{otherwise}. \end{array} \right.$$ (5)

This means that we continuously pour substance $B$ into the beaker close to $x = (0.5,0.5)$.

To solve the problem, look at your solver for the bistable equation from computer session E4, or try yourself without looking.

Since the problem is nonlinear, we need to use fixed-point iteration. The procedure is the same as in computer session E4, but since we now have a system of two equations, we need to assemble two vectors and solve two linear systems in each fixed-point iteration.

Note also that both $f_1$ and $f_2$ depend on both $u_1$ and $u_2$, so we need to pass the values of both $u_1$ and $u_2$ (at the right-hand side of the interval) to the assembler. As before, you also need to pass the left-hand side values to the assembler (since the problem is time-dependent).

Check your answer: Compare your solution with the figure below, which shows the solution at time $t = 1$.