Survival in a harsh environment
We’ll continue studying the logistic reactiondiffusion 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 noflux 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.