Some effects of nonlocal diffusion on the solutions of Fisher-KPP equations in disconnected domains

Abstract

The question under study is the existence and uniqueness of a non-trivial bounded steady state of a Fisher-KPP equation involving a fractional Laplacian $(-\Delta )^\alpha$ in a fragmented domain with exterior Dirichlet conditions. Of particular interest here is the rigidity on the steady states entailed by the nonlocal dispersion. Our results also provide criteria on the domain for the subsistence of a species subject to a nonlocal diffusion in a fragmented area. Further effects, such as the continuity of this principal eigenvalue with respect to the distance between two compact patches in the one dimensional case, are presented.

Appendix A Proof of Lemma 1

We provide here the proof of Lemma 1. Before giving the proof, we introduce a new notation:

Notation

For all $\phi \in H^\alpha _0 (\Omega _{1,2,\mu })$ and $i \in \left\{ 1,2 \right\}$, we will denote by $\phi ^i$ the function $\phi$ restricted to the set $\Omega _i$ and extended by 0 outside $\Omega _i$

$$\begin{aligned}\text {i.e. } \ \phi ^i (x) = \phi (x) 1_{\Omega _i}(x).\end{aligned}$$

We also denote by ${\underline{\lambda }}_i$ the principal eigenvalue of $(-\Delta )^\alpha - I$ in $\Omega _i$ with 0 exterior Dirichlet conditions.

Remark 4

For all $\phi \in H^\alpha _0(\Omega _{1,2,\mu })$, we have

$$\begin{aligned}\phi (x) = \phi ^1(x) + \phi ^2 (x) \ \text { and } \ \phi ^i \in H^\alpha _0(\Omega _i) \ \text { for all } \ i \in \left\{ 1,2 \right\} . \end{aligned}$$

Proof

The aim of the proof is to prove that for $\mu$ large enough, there exists $C>0$ such that

$$\begin{aligned}\lambda _\alpha (\Omega _{1,2,\mu }) \ge \min ({\underline{\lambda }}_{1,\alpha }, {\underline{\lambda }}_{2, \alpha }) - \frac{C}{\mu ^{1+2\alpha }}.\end{aligned}$$

The conclusion follows by the intermediate value Theorem. We start from the Rayleigh quotient defining ${\underline{\lambda }}_{i, \alpha }$:

$$\begin{aligned} {\underline{\lambda }}_{i,\alpha }&= \frac{ \int _{{\mathbb {R}}}\int _{\mathbb {R}} \frac{({\underline{\phi }}_{i,\alpha }(x) - {\underline{\phi }}_{i,\alpha }(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy - ({\underline{\phi }}_{i,\alpha }(x))^2 dx }{\int _{\Omega _i} ({\underline{\phi }}_{i,\alpha }(x))^2dx} \\&= \frac{ \int _{\Omega _i}\int _{\mathbb {R}} \frac{({\underline{\phi }}_{i,\alpha }(x) - {\underline{\phi }}_{i,\alpha }(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy - ({\underline{\phi }}_{i,\alpha }(x))^2 dx }{\int _{\Omega _i} ({\underline{\phi }}_{i,\alpha }(x))^2 dx} + \frac{ \int _{{\mathbb {R}} \backslash \Omega _i}\int _{\Omega _i} \frac{({\underline{\phi }}_{i,\alpha }(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy dx }{\int _{\Omega _i} ({\underline{\phi }}_{i,\alpha }(x))^2 dx} \\&= \frac{ \int _{\Omega _i}\int _{\Omega _i} \frac{({\underline{\phi }}_{i,\alpha }(x) - {\underline{\phi }}_{i,\alpha }(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy - ({\underline{\phi }}_{i,\alpha }(x))^2 dx }{\int _{\Omega _i} ({\underline{\phi }}_{i,\alpha }(x)) ^2 dx} + 2 \frac{ \int _{{\mathbb {R}} \backslash \Omega _i}\int _{\Omega _i} \frac{({\underline{\phi }}_{i,\alpha }(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy dx }{\int _{\Omega _i} ({\underline{\phi }}_{i,\alpha }(x))^2 dx}. \end{aligned}$$

We continue in the same way by rewriting $\lambda _{1,2,\mu ,\alpha }$:

$$\begin{aligned} \lambda _\alpha (\Omega _{1,2,\mu }) =&\frac{ \int _{{\mathbb {R}}}\int _{\mathbb {R}} \frac{(\phi _{1,2,\mu ,\alpha }(x) - \phi _{1,2,\mu ,\alpha }(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy - (\phi _{1,2,\mu ,\alpha }(x))^2 dx }{\int _{\Omega _{1,2,\mu }} (\phi _{1,2,\mu ,\alpha }(x))^2dx} \\ =&\frac{\int _{\Omega _{1,2,\mu }}\int _{\mathbb {R}} \frac{(\phi _{1,2,\mu ,\alpha }(x) -\phi _{1,2,\mu ,\alpha }(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy - (\phi _{1,2,\mu ,\alpha }(x))^2 dx }{\int _{\Omega _{1,2,\mu }} (\phi _{1,2,\mu ,\alpha }(x))^2dx}\\&+ \frac{ \int _{{\mathbb {R}} \backslash \Omega _{1,2,\mu }}\int _{\mathbb {R}} \frac{(\phi _{1,2,\mu ,\alpha }(x) - \phi _{1,2,\mu ,\alpha }(y))^2}{\vert x-y\vert ^{1+2\alpha }}dydx}{\int _{\Omega _{1,2,\mu }} (\phi _{1,2,\mu ,\alpha }(x))^2dx}. \end{aligned}$$

Thus, we have found:

$$\begin{aligned} \lambda _\alpha (\Omega _{1,2,\mu })= \frac{1}{\int _{\Omega _{1,2,\mu }} (\phi _{1,2,\mu ,\alpha }(x))^2dx} \left( I_1+I_2 \right) . \end{aligned}$$

(A.1)

We rewrite $I_1$ and $I_2$ in order to involving the expression of ${\underline{\lambda }}_{1,\alpha }$ and ${\underline{\lambda }}_{2,\alpha }$. We begin by rewriting $I_1$:

$$\begin{aligned} I_1&= \int _{\Omega _{1,2,\mu }}\int _{\mathbb {R}} \frac{(\phi _{1,2,\mu ,\alpha }(x) - \phi _{1,2,\mu ,\alpha }(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy - (\phi _{1,2,\mu ,\alpha }(x))^2 dx \\&= \int _{\Omega _1} \int _{\mathbb {R}} \frac{(\phi _{1,2,\mu ,\alpha }(x) - \phi _{1,2,\mu ,\alpha }(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy - (\phi _{1,2,\mu ,\alpha }(x))^2 dx \\&\quad \quad + \int _{\Omega _2} \int _{\mathbb {R}} \frac{(\phi _{1,2,\mu ,\alpha }(x) - \phi _{1,2,\mu ,\alpha }(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy - (\phi _{1,2,\mu ,\alpha }(x))^2 dx \\&= \int _{\Omega _1} \int _{\Omega _1} \frac{(\phi _{1,2,\mu ,\alpha }(x) - \phi _{1,2,\mu ,\alpha }(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy - (\phi _{1,2,\mu ,\alpha }(x))^2 dx \\&\quad \quad + \int _{\Omega _1} \int _{{\mathbb {R}} \backslash \Omega _{1,2,\mu }} \frac{(\phi _{1,2,\mu ,\alpha }(x) )^2}{\vert x-y\vert ^{1+2\alpha }}dy dx \\&\quad \quad + \int _{\Omega _2} \int _{\Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }(x) - \phi _{1,2,\mu ,\alpha }(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy - (\phi _{1,2,\mu ,\alpha }(x))^2 dx \\&\quad \quad + \int _{\Omega _2} \int _{{\mathbb {R}} \backslash \Omega _{1,2,\mu }} \frac{(\phi _{1,2,\mu ,\alpha }(x) )^2}{\vert x-y\vert ^{1+2\alpha }}dydx \\&\quad \quad +2 \int _{\Omega _1} \int _{\Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }(x) - \phi _{1,2,\mu ,\alpha }(y))^2}{\vert x-y\vert ^{1+2\alpha }}dydx \\&= \int _{\Omega _1} \int _{\Omega _1} \frac{(\phi _{1,2,\mu ,\alpha }^1(x) - \phi _{1,2,\mu ,\alpha }^1(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy - (\phi _{1,2,\mu ,\alpha }^1(x))^2 dx \\&\quad \quad + \quad \int _{\Omega _1} \int _{{\mathbb {R}} \backslash \Omega _1} \frac{(\phi _{1,2,\mu ,\alpha }^1(x) )^2}{\vert x-y\vert ^{1+2\alpha }}dydx \\&\quad \quad + \int _{\Omega _2} \int _{\Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }^2(x) - \phi _{1,2,\mu ,\alpha }^2(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy - (\phi _{1,2,\mu ,\alpha }^2(x))^2 dx \\&\quad \quad + \int _{\Omega _2} \int _{{\mathbb {R}} \backslash \Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }^2(x) )^2}{\vert x-y\vert ^{1+2\alpha }}dy dx +2 \int _{\Omega _1} \int _{\Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }(x) - \phi _{1,2,\mu ,\alpha }(y))^2}{\vert x-y\vert ^{1+2\alpha }}dydx \\&\quad \quad - \int _{\Omega _1} \int _{\Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }^1(x))^2}{\vert x-y\vert ^{1+2\alpha }} dydx - \int _{\Omega _2} \int _{\Omega _1} \frac{(\phi _{1,2,\mu ,\alpha }^2(x))^2}{\vert x-y\vert ^{1+2\alpha }} dydx. \end{aligned}$$

Finally, we find that

$$\begin{aligned} \begin{aligned} I_1&= \int _{\Omega _1} \int _{\Omega _1} \frac{(\phi _{1,2,\mu ,\alpha }^1(x) - \phi _{1,2,\mu ,\alpha }^1(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy - (\phi _{1,2,\mu ,\alpha }^1(x))^2 dx \\&\quad \quad + \int _{\Omega _1} \int _{{\mathbb {R}} \backslash \Omega _1} \frac{(\phi _{1,2,\mu ,\alpha }^1(x) )^2}{\vert x-y\vert ^{1+2\alpha }}dydx \\&\quad \quad + \int _{\Omega _2} \int _{\Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }^2(x) - \phi _{1,2,\mu ,\alpha }^2(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy - (\phi _{1,2,\mu ,\alpha }^2(x))^2 dx \\&\quad \quad + \int _{\Omega _2} \int _{{\mathbb {R}} \backslash \Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }^2(x) )^2}{\vert x-y\vert ^{1+2\alpha }}dydx \\&\quad \quad +2 \int _{\Omega _1} \int _{\Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }(x) - \phi _{1,2,\mu ,\alpha }(y))^2}{\vert x-y\vert ^{1+2\alpha }}dydx \\&\quad \quad - \int _{\Omega _2} \int _{\Omega _1} \frac{(\phi _{1,2,\mu ,\alpha }^2(x))^2}{\vert x-y\vert ^{1+2\alpha }} dydx - \int _{\Omega _1} \int _{\Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }^1(x))^2}{\vert x-y\vert ^{1+2\alpha }} dydx. \end{aligned} \end{aligned}$$

(A.2)

With similar computations, we find that for $I_2$:

$$\begin{aligned} \begin{aligned} I_2&= \int _{\Omega _1} \int _{{\mathbb {R}} \backslash \Omega _1} \frac{(\phi _{1,2,\mu ,\alpha }^1(y))^2}{\vert x-y\vert ^{1+2\alpha }}dxdy +\int _{\Omega _2} \int _{{\mathbb {R}} \backslash \Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }^2(y))^2}{\vert x-y\vert ^{1+2\alpha }}dxdy \\&\quad \quad - \int _{\Omega _1} \int _{\Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }^1(y))^2}{\vert x-y\vert ^{1+2\alpha }}dxdy - \int _{\Omega _2} \int _{\Omega _1} \frac{(\phi _{1,2,\mu ,\alpha }^2(y))^2}{\vert x-y\vert ^{1+2\alpha }}dxdy. \end{aligned} \end{aligned}$$

(A.3)

Combining (A.2) and (A.3) in (A.1), we deduce

$$\begin{aligned} \lambda _\alpha (\Omega _{1,2,\mu })&= \frac{2 \int _{\Omega _1} \int _{\Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }(x) -\phi _{1,2,\mu ,\alpha }(y))^2}{\vert x-y\vert ^{1+2\alpha }}dydx }{\int _{\Omega _{1,2,\mu }} (\phi _{1,2,\mu ,\alpha }(x))^2 dx} \\&\quad \quad + \frac{\int _{\Omega _1} \int _{\Omega _1} \frac{(\phi _{1,2,\mu ,\alpha }^1(x) - \phi _{1,2,\mu ,\alpha }^1(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy - (\phi _{1,2,\mu ,\alpha }^1(x))^2 dx + 2 \int _{\Omega _1} \int _{{\mathbb {R}} \backslash \Omega _1} \frac{(\phi _{1,2,\mu ,\alpha }^1(x) )^2}{\vert x-y\vert ^{1+2\alpha }}dydx}{\int _{\Omega _{1,2,\mu }} (\phi _{1,2,\mu ,\alpha }(x))^2 dx}\\&\quad \quad + \frac{\int _{\Omega _2} \int _{\Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }^2(x) - \phi _{1,2,\mu }^2(y))^2}{\vert x-y\vert ^{1+2\alpha }}dy - (\phi _{1,2,\mu ,\alpha }^2(x))^2 dx + 2 \int _{\Omega _2} \int _{{\mathbb {R}} \backslash \Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }^2(x) )^2}{\vert x-y\vert ^{1+2\alpha }}dydx}{\int _{\Omega } (\phi _{1,2,\mu ,\alpha }(x))^2 dx} \\&\quad -\frac{ 2 \int _{\Omega _1} \int _{\Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }^1(x))^2}{\vert x-y\vert ^{1+2\alpha }} dydx + 2\int _{\Omega _2} \int _{\Omega _1} \frac{(\phi _{1,2,\mu ,\alpha }^2(x))^2}{\vert x-y\vert ^{1+2\alpha }} dydx }{\int _{\Omega _{1,2,\mu }} (\phi _{1,2,\mu ,\alpha }(x))^2 dx} \\&\ge \frac{\int _{\Omega _1} (\phi _{1,2,\mu ,\alpha }^1(x))^2 dx }{\int _{\Omega _{1,2,\mu }} (\phi _{1,2,\mu ,\alpha }(x))^2 dx} {\underline{\lambda }}_{1,\alpha } + \frac{\int _{\Omega _2} (\phi _{1,2,\mu ,\alpha }^2(x))^2 dx }{\int _{\Omega _{1,2,\mu }} (\phi _{1,2,\mu ,\alpha }(x))^2 dx} {\underline{\lambda }}_{2,\alpha } \\&\quad -\frac{ 2\int _{\Omega _1} \int _{\Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }^1(x))^2}{\vert x-y\vert ^{1+2\alpha }} dydx + 2\int _{\Omega _2} \int _{\Omega _1} \frac{(\phi _{1,2,\mu ,\alpha }^2(x))^2}{\vert x-y\vert ^{1+2\alpha }} dydx}{\int _{\Omega _{1,2,\mu }} (\phi _{1,2,\mu ,\alpha }(x))^2 dx} \\&\ge \min ({\underline{\lambda }}_{1,\alpha }, {\underline{\lambda }}_{2,\alpha }) -\frac{ 2\int _{\Omega _1} \int _{\Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }^1(x))^2}{\vert x-y\vert ^{1+2\alpha }} dydx + 2\int _{\Omega _2} \int _{\Omega _1} \frac{(\phi _{1,2,\mu ,\alpha }^2(x))^2}{\vert x-y\vert ^{1+2\alpha }} dydx}{\int _{\Omega _{1,2,\mu }} (\phi _{1,2,\mu ,\alpha }(x))^2 dx} . \end{aligned}$$

But, for $i \in \left\{ 1,2 \right\}$, we have for all $x \in \Omega _i$ and $y \in \Omega _{3-i}$

$$\begin{aligned}2\mu < \vert x-y\vert \Rightarrow -\frac{1}{(2\mu )^{1+2\alpha }} \le -\frac{1}{\vert x-y\vert ^{1+2\alpha }}.\end{aligned}$$

We deduce that

$$\begin{aligned} \lambda _\alpha (\Omega _{1,2,\mu })&\ge \min ({\underline{\lambda }}_{1,\alpha }, {\underline{\lambda }}_{2,\alpha }) \\&\qquad -\frac{ 2\int _{\Omega _1} \int _{\Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }^1(x))^2}{\vert x-y\vert ^{1+2\alpha }} dydx + 2\int _{\Omega _2} \int _{\Omega _1} \frac{(\phi _{1,2,\mu ,\alpha }^2(x))^2}{\vert x-y\vert ^{1+2\alpha }} dydx}{\int _{\Omega _{1,2,\mu }} (\phi _{1,2,\mu ,\alpha }(x))^2 dx} \\&\ge \min ({\underline{\lambda }}_{1,\alpha }, {\underline{\lambda }}_{2,\alpha }) \\&\qquad -\frac{ 2\int _{\Omega _1} \int _{\Omega _2} \frac{(\phi _{1,2,\mu ,\alpha }^1(x))^2}{\mu ^{1+2\alpha }} dydx + 2\int _{\Omega _2} \int _{\Omega _1} \frac{(\phi _{1,2,\mu ,\alpha }^2(x))^2}{\mu ^{1+2\alpha }} dydx}{\int _{\Omega _{1,2,\mu }} (\phi _{1,2,\mu ,\alpha }(x))^2 dx} \\&\ge \min ({\underline{\lambda }}_{1,\alpha }, {\underline{\lambda }}_{2,\alpha }) - \frac{4A}{(2\mu )^{1+2\alpha }} \underset{\mu \rightarrow + \infty }{\longrightarrow }\min ({\underline{\lambda }}_{1,\alpha }, {\underline{\lambda }}_{2,\alpha }). \end{aligned}$$

Since $\min ({\underline{\lambda }}_{1,\alpha }, {\underline{\lambda }}_{2,\alpha })>0$, we deduce the existence of $\mu _1>0$ such that for all $\mu > \mu _1$,

$$\begin{aligned}\lambda _\alpha (\Omega _{1,2,\mu }) > 0.\end{aligned}$$

$\square$