We’ll continue studying the logistic reaction-diffusion model,

$\pd{u}{t}=D\nabla^2 u+u\left(1-\frac{u}{K}\right),$

but now we consider different boundary conditions.

• Load the interactive simulation. The initial conditions leads to population growth in select areas of the domain, which eventually reach a large density and begin spreading. As the system initially has no-flux boundary conditions, the population will reach carrying capacity everywhere.

• Now change the boundary conditions to “Dirichlet”, taking the constant value to be $0$. This will affect the solution near the domain edges, but not the interior, as the diffusion is small.

• Now increase the diffusion coefficient by deleting zeroes from the value of $D$. This will increase the region over which the boundary conditions affect the structure of the solution.

• One can show that the positive equilibrium exists and is globally stable if and only if $$D < \frac{L^2}{2\pi^2} \approx 5.066,$$ as $L=10$ in our simulation. For diffusion coefficients larger than this value, the extinction equilibrium, $u=0$, is stable. Try simulating the system at $D=4$, and $D=6$, clicking if needed to introduce some additional population into the domain.

Note: You can check the condition more precisely by looking at $D$ values closer to the boundary, but the timescale to reach equilibrium will be long, and the amplitude of $u$ will become very small.

• The analytical condition above does not depend on the carrying capacity $K$. Set $K=1000$, and simulate the system at $D=4$ and $D=6$ in this case to confirm that $K$ will not change the boundary of where the populations persists, though it will change the structure of the solution when $u>0$. Again you may need to click to introduce some population into the domain. With this value of $K$, it is easier to see the solution’s behaviour near the critical diffusion threshold. The values $D=5$ and $D=5.2$ are good choices, for example.