Propagation or extinction in bistable equations: The non-monotone role of initial fragmentation

Figure 1.  Evolution of the numerical solution $ u(t, x) $ of (1), with $ N = 1 $, $ f(s) = s(1-s)(s-\theta) $, and $ \theta = 0.4 $, starting with an initial condition $ u_0(x) = \mathbb{1}_E(x) $ with (left) $ E = E_1: = (-L/2, L/2) $ and (right) $ E = E_2: = \bigcup_{x\in{\mathbb{Z}} \cap [-k, k]}\Big[\frac{x}{z}+\Big(\!\!-\!\frac{\alpha}{2z}, \frac{\alpha}{2z}\Big)\Big] $. The curves correspond to the solution for $ t = 0 $, $ t = 0.05 $ and for $ t $ ranging from 4 to 80 with a step size of 4. The gradient color goes from blue for the earliest times to red for the latest times. In the left panel, $ \lambda(E_1) = L = 4.55 $ and $ \delta_1(E_1) = 0 $; in the right panel $ \alpha = 3/4 $, $ z = 2.16 $, $ k = 6 $, $ \lambda(E_2) = 4.52 $ and $ \delta_1(E_2) = 0.23 $. We observe here that, though $ z $ is not that large and $ \lambda(E_2) $ is even smaller than $ \lambda(E_1) $, fragmentation (right panel) improves invasion success. We also note that $ u(t, x)\approx \alpha \mathbb{1}_{B_R} $ with $ R = k/z+\alpha/(2 \, z) $ when $ t\ll 1 $. The Matlab code used for the computations is available at http://doi.org/10.17605/OSF.IO/ZM479

Figure 2.  The two sets $ E_1 $ (in red) and $ E_2 $ (in blue) have the same Lebesgue measure $ \lambda(E_1) = \lambda(E_2) $. Invasion occurs for (1) with initial condition $ \mathbb{1}_{E_1} $ but not with $ \mathbb{1}_{E_2} $. To understand why $ 0<\delta_1(E_2)<\delta_1(E_1) $ for all $ |x| $ large enough, observe that the value of $ \max_{B\in\mathcal{B}, \, \lambda(B) = \lambda(E_1)}\lambda(E_1\cap B) $ corresponds in dimension $ N = 2 $ to the measure $ \lambda(E_1\cap \widetilde {B}) $ of the intersection between $ E_1 $ and the disk $ \widetilde {B} $ inside the dashed circle. For $ |x| $ large enough, the value of $ \max_{B\in\mathcal{B}, \, \lambda(B) = \lambda(E_2)}\lambda(E_2\cap B) $ simply corresponds to the measure of $ \lambda(E_2\cap \widetilde {B}) $, and is higher than $ \lambda(E_1\cap \widetilde {B}) = \max_{B\in\mathcal{B}, \, \lambda(B) = \lambda(E_1)}\lambda(E_1\cap B) $, hence $ \delta_1(E_2)<\delta_1(E_1) $. The details are provided below