Accelerated front propagation for monostable equations with nonlocal diffusion: multidimensional case

Abstract

We describe acceleration of the front propagation for solutions to a class of monostable nonlinear equations with a nonlocal diffusion in ${{\mathbb {R}}^d}$, $d\ge 1$. We show that the acceleration takes place if either the diffusion kernel or the initial condition has ‘regular’ heavy tails in ${{\mathbb {R}}^d}$ (in particular, decays slower than exponentially). Under general assumptions which can be verified for particular models, we present sharp estimates for the time-space zone which separates the region of convergence to the unstable zero solution with the region of convergence to the stable positive constant solution. We show the variety of different possible rates of the propagation starting from a little bit faster than a linear one up to the exponential rate. The paper generalizes to the case $d>1$ our results for the case $d=1$ obtained early in Finkelshtein and Tkachov (Appl Anal 98(4):756–780, 2019).

Appendix

Proof of Lemma 2.5

Firstly, we note that (1.7) implies (A5). Let G be defined by (2.4), i.e., for $0\le u\in E$ and $x\in {{\mathbb {R}}^d}$,

$$\begin{aligned} (Gu)(x) = \beta - \alpha \frac{f\bigl (u(x)\bigr )}{u(x)} - (1-\alpha ) \beta \bigl (1-(a^-*u)(x)\bigr )^k, \end{aligned}$$

where $\frac{f(s)}{s}:=\beta$ for $s=0$. Then it is straightforward to check that (A2)–(A3) and (A6)–(A8) hold. We are going to prove that there exists $p\ge 0$ such that, for any $v,w\in E_1^+$ with $v\le w$,

$$\begin{aligned} p(w-v)&+ a*(w-v)\nonumber \\&\ge (w-v)Gw + v(Gw-Gv) + \rho \mathbb {1}_{B_\rho (0)}*(w-v). \end{aligned}$$

(Ap.1)

Note that (Ap.1) evidently implies (A4). Next, (A10) will follow from (Ap.1) if we choose any $\delta <\rho$ with $\delta <\rho$ and any $b\in C^\infty ({{\mathbb {R}}^d})\cap L^\infty ({{\mathbb {R}}^d})$, such that $a -\rho \mathbb {1}_{B_\rho (0)} \le b \le a - \delta \mathbb {1}_{B_\delta (0)}$.

By (1.5), there exists a Lipschitz constant $K>0$, such that

$$\begin{aligned} Gw-Gv&=\alpha \Bigr (\frac{f(v)}{v}-\frac{f(w)}{w}\Bigr )+ (1-\alpha )\beta \Bigl ((1-a^-*v )^k - (1-a^-*w)^k \Bigr )\nonumber \\&\le \alpha K (w-v)+(1- \alpha )\beta k a^-*(w-v), \end{aligned}$$

(Ap.2)

where we used an elementary inequality $q^k-r^k\le k(q-r)$ for $0\le r\le q\le 1$. Multiplying both parts of (Ap.2) on $0\le v\le 1$ and using (1.7), we get

$$\begin{aligned} v(Gw-Gv)\le \alpha K (w-v)+ a*(w-v) - \rho \mathbb {1}_{B_\rho (0)}*(w-v). \end{aligned}$$

Finally, by (A2), $(w-v)Gw\le \beta (w-v)$, and therefore, the inequality (Ap.1) holds with $p:= \beta +\alpha K>0$. $\square$

Lemma A.1

Let$\lambda >1$and let$b:{\mathbb {R}}_+\rightarrow {\mathbb {R}}_+$be defined, for larges, as follows

$$\begin{aligned} b(s)=\exp \Bigl (-\frac{s}{(\log s)^\lambda }\Bigr ). \end{aligned}$$

Let$\beta >0$, and define, for larget, the function$\eta (t):=b^{-1}\bigl (e^{-\beta t}\bigr )$. Then

$$\begin{aligned} \eta (t)\sim \beta t (\log t)^\lambda , \quad t\rightarrow \infty . \end{aligned}$$

(Ap.3)

Proof

The equation $b(s)=e^{-\beta t}$ yields $s(\log s)^{-\lambda }=\beta t$. Making substitution $s=e^{\tau }$, one easily gets

$$\begin{aligned} -\frac{\tau }{\lambda } e^{-\frac{\tau }{\lambda }}=-\frac{1}{\lambda (\beta t)^{\frac{1}{\lambda }}}. \end{aligned}$$

Since $s>e^\lambda$ implies $-\frac{\tau }{\lambda }<-1$ and assuming t big enough, to ensure that $-\frac{1}{\lambda (\beta t)^{\frac{1}{\lambda }}}>-\frac{1}{e}$, one has that the solution to the latter equation can be given in terms of the negative real branch $W_{-1}$ of Lambert W-function, that is the function such that $W_{-1}(\nu )\exp (W_{-1}(\nu ))=\nu$, $W_{-1}(\nu )<-1$, $\nu \in (-e^{-1},0)$. Namely, one gets $-\frac{\tau }{\lambda }=W_{-1}\bigl (-\lambda ^{-1}(\beta t)^{-\frac{1}{\lambda }}\bigr ),$ and, therefore

$$\begin{aligned} \eta (t)=\exp \biggl ( -\lambda W_{-1}\Bigl (-\frac{1}{\lambda (\beta t)^{\frac{1}{\lambda }}}\Bigr ) \biggr ). \end{aligned}$$

However, $\exp (-W_{-1}(\nu )) =\nu ^{-1}W_{-1}(\nu )$, therefore,

$$\begin{aligned} \exp (-\lambda W_{-1}(\nu )) =(-\nu )^{-\lambda }(-W_{-1}(\nu ))^\lambda , \end{aligned}$$

i.e.

$$\begin{aligned} \eta (t)=\lambda ^\lambda \beta t\biggl ( -W_{-1}\Bigl (-\frac{1}{\lambda (\beta t)^{\frac{1}{\lambda }}}\Bigr ) \biggr )^\lambda , \quad t>\frac{1}{\beta }\Bigl (\frac{e}{\lambda }\Bigr )^\lambda . \end{aligned}$$

It is well-known that $W_{-1}(\nu )\sim \log (-\nu )$, $\nu \rightarrow 0-$. This yields (Ap.3). $\square$

Lemma A.2

Let a function$X(t)\rightarrow \infty$, $t\rightarrow \infty$, be such that, forlarget,

$$\begin{aligned} \int _{X(t)}^\infty \int _{X(t)}^\infty b(|y|)\,dy_1\,dy_2=e^{-\beta t}, \qquad |y|=\sqrt{y_1^2+y_2^2}, \end{aligned}$$

(Ap.4)

where$\beta >0$and$b:{\mathbb {R}}_+\rightarrow {\mathbb {R}}_+$is a decreasing at$\infty$function, such that$\int _{{\mathbb {R}}_+} b(r)r\,dr<\infty$. Consider the following functions

$$\begin{aligned} c(x):=\frac{\pi }{2}\int _{\sqrt{2}x}^\infty b(r)r\,dr, \qquad \mu (t):=c^{-1}\bigl (e^{-\beta t}\bigr ) \end{aligned}$$

(Ap.5)

for largexandt. Then, for any$\varepsilon \in (0,1)$and larget,

$$\begin{aligned} \mu (t)\ge X(t) \ge \frac{1}{2}\mu (t-\varepsilon t). \end{aligned}$$

Proof

Rewriting the set $\{(y_1,y_2)\in {\mathbb {R}}^2\mid y_1\ge X(t), y_2\ge X(t)\}$ for $X(t)>0$ in polar coordinates, we obtain from (Ap.4) that, for large t,

$$\begin{aligned} e^{-\beta t}&=\int _{\sqrt{2}X(t)}^\infty \int _{\arcsin \frac{X(t)}{r}}^{\arccos \frac{X(t)}{r}} b(r)r\,dr =\int _{\sqrt{2}X(t)}^\infty \biggl (\frac{\pi }{2}-2\arcsin \frac{X(t)}{r}\biggr ) b(r)r\,dr\\&=X(t)^2\int _{\sqrt{2}}^\infty \biggl (\frac{\pi }{2}-2\arcsin \frac{1}{s}\biggr ) b\bigl (X(t)s\bigr )\,s\,ds. \end{aligned}$$

Therefore, for any $\delta >0$,

$$\begin{aligned} c\bigl ( X(t)\bigr )\ge e^{-\beta t}&\ge X(t)^2\int _{\sqrt{2}+\delta }^\infty \biggl (\frac{\pi }{2}-2\arcsin \frac{1}{s}\biggr ) b\bigl (X(t)s\bigr )\,s\,ds\nonumber \\&\ge f(\delta ) X(t)^2 \int _{\sqrt{2}+\delta }^\infty b\bigl (X(t)s\bigr )\,s\,ds=\frac{2}{\pi } f(\delta ) c\biggl (\frac{\sqrt{2}+\delta }{\sqrt{2}}X(t) \biggr ), \end{aligned}$$

(Ap.6)

where

$$\begin{aligned} f(\delta ):=\frac{\pi }{2}-2\arcsin \frac{1}{\sqrt{2}+\delta }\in \Bigl (0,\frac{\pi }{2}\Bigr ), \quad \delta >0, \end{aligned}$$

is an increasing function. Since c(x) is decreasing, we obtain from (Ap.6) that

$$\begin{aligned} c^{-1}(e^{-\beta t})\ge X(t) \ge \frac{\sqrt{2}}{\sqrt{2}+\delta }c^{-1}\biggl (\frac{\pi }{2f(\delta )}e^{-\beta t}\biggr ). \end{aligned}$$

(Ap.7)

Set $\lambda =\frac{\sqrt{2}}{\sqrt{2}-1}>1$. Choose $\delta >0$ such that

$$\begin{aligned} f(\delta )=\frac{\pi }{2\lambda }<\frac{\pi }{2}, \end{aligned}$$

then

$$\begin{aligned} \frac{\sqrt{2}}{\sqrt{2}+\delta }=\sqrt{2}\sin \biggl (\frac{\pi }{4}\Bigl (1-\frac{1}{\lambda }\Bigr )\biggr )>\frac{1}{\sqrt{2}}\Bigl (1-\frac{1}{\lambda }\Bigr )=\frac{1}{2}, \end{aligned}$$

where we used the inequality $\sin x>\frac{2}{\pi }x$ for $0<x<\frac{\pi }{2}$. Then (Ap.7) implies

$$\begin{aligned} \mu (t)\ge X(t) \ge \frac{1}{2}c^{-1}\bigl (\lambda e^{-\beta t}\bigr ). \end{aligned}$$

Take finally an $\varepsilon \in (0,1)$ and assume that t is big enough to ensure that $e^{\varepsilon \beta t}>\lambda$. Since $c^{-1}(x)$ is a decreasing function, one gets the statement. $\square$

Remark A.3

Let (1.17)–(1.18) holds. Then, by Theorem 2.9, (1.9)–(1.10) hold with $\Lambda (t)=\Lambda (t,c)$ given by (1.11) where $c(x_1,x_2)=\int _{x_1}^\infty \int _{x_2}^\infty b(|y|)\,dy_1\,dy_2$, cf. (1.16), and $b\in {\mathcal {E}}_{d}$ is log-equivalent to $e^{-\sqrt{s}}$, $s>0$. Take $b(s)=\frac{1}{\pi } s^{-\frac{3}{2}}e^{-\sqrt{s}}$ for large s. By [32, Corollary 3.1], $b\in {\mathcal {E}}_{d}$. Let $X(t):=X_1(t)=X_2(t)$ describe the motion of the boundary of $\Lambda (t)$ in the diagonal direction in (1.16). Then, by Lemma A.2, we have, cf. (Ap.5),

$$\begin{aligned} c(x)=\frac{\pi }{2}\frac{1}{\pi } \int _{\sqrt{2}x}^\infty \frac{1}{\sqrt{r}}\exp \bigl (-\sqrt{r}\bigr )\,dr =\exp \bigl (-\root 4 \of {2}\sqrt{x}\bigr ). \end{aligned}$$

Then, by (Ap.5), $\mu (t)=\frac{\beta ^2}{\sqrt{2}}t^2$. Therefore, by Lemma A.2, for any $\varepsilon \in (0,1)$ and large t, (1.19) holds.