Spreading and Vanishing for a Monostable Reaction–Diffusion Equation with Forced Speed

Abstract

Invasion phenomena for heterogeneous reaction–diffusion equations are contemporary and challenging questions in applied mathematics. In this paper we are interested in the question of spreading for a reaction–diffusion equation when the subdomain where the reaction term is positive is shifting/contracting at a given speed c. This problem arises in particular in the modelling of the impact of climate change on population dynamics. By placing ourselves in the appropriate moving frame, this leads us to consider a reaction–diffusion–advection equation with a heterogeneous in space reaction term, in dimension $N\ge 1$. We investigate the behaviour of the solution u depending on the value of the advection constant c, which typically stands for the velocity of climate change. We find that, when the initial datum is compactly supported, there exists precisely three ranges for c leading to drastically different situations. In the lower speed range the solution always spreads, while in the upper range it always vanishes. More surprisingly, we find that both spreading and vanishing may occur in an intermediate speed range. The threshold between those two outcomes is always sharp, both with respect to c and to the initial condition. We also briefly consider the case of an exponentially decreasing initial condition, where we relate the decreasing rate of the initial condition with the range of values of c such that spreading occurs.

Maximum Principle Lemmas

1.1 An Elliptic Maximum Principle

In our proofs, we use extensively the following lemma:

Lemma A.1

Assume that $c \ge 2\sqrt{g_+ '(0)}$. Let $\underline{\phi } (x,y)$ and $\overline{\phi }(x,y)$ be two nonnegative functions, respectively a sub and a supersolution of (1.1), and assume that they both decay to 0 at $x \rightarrow +\infty$ as

$$\begin{aligned}&o(e^{-\frac{c}{2} x})\quad \left( \text{ if } c > 2 \sqrt{g_+ '(0)} \, \right) ,\\&o((1+\sqrt{x}) e^{-\frac{c}{2} x})\quad \left( \text{ if } c = 2 \sqrt{g_+ '(0)} \, \right) , \end{aligned}$$

where these estimates are understood to be uniform with respect to $y \in \omega$.

Then there exists $X>0$ large enough such that, for any $X ' \ge X$, if $\underline{\phi } (X',\cdot ) \le \overline{\phi } (X',\cdot )$, then either $\underline{\phi } \equiv \overline{\phi }$ or $\underline{\phi } (x,y) < \overline{\phi } (x,y)$ for all $(x,y) \in (X' ,+\infty ) \times \omega$.

Note first that when $c < 2 \sqrt{g_+'(0)}$, then we prove in Sect. 3 that 0 is unstable with respect to compact perturbations so that there does not exist any supersolution decaying to 0 at infinity. Therefore, it is meaningful to only consider the case $c \ge 2 \sqrt{g_+ '(0)}$.

Moreover, if instead of our assumptions, one chooses f to have a negative derivative with respect to its third variable, then the lemma reduces to the usual maximum principle. In particular, one can clearly derive a similar property on the far left of our problem. Because $\partial _s f(x,y,0) >0$ for large positive x and all $y \in \omega$, the situation is rather different on the far right. However, it is still quite standard that the maximum principle remains valid in an appropriate subspace of fast decaying functions, in which 0 is again stable. This is exactly Lemma A.1, whose proof we include below for the sake of completeness.

Proof

Let us first assume that $c > 2 \sqrt{g_+'(0)}$. Let $\varepsilon >0$ be small so that $c > 2 \sqrt{g_+'(0) + 2 \varepsilon }$. We consider $\delta >0$ and $X>0$ such that, for all $x \ge X$, $y\in \omega$ and $s \in [0,\delta ]$, one has

$$\begin{aligned} \partial _s f(x,y,s) \le g_+'(0) + \varepsilon , \end{aligned}$$

and for all $x \ge X$ and $y \in \omega$,

$$\begin{aligned} \underline{\phi } (x,y), \, \overline{\phi } (x,y) \le \delta . \end{aligned}$$

Then $\psi := e^{\frac{c}{2}x } \left( \underline{\phi } - \overline{\phi }\right)$ satisfies, when $x \ge X$:

$$\begin{aligned} \varDelta \psi - \frac{c^2}{4} \psi + e^{\frac{c}{2} x} \left[ f (x,y,\underline{\phi }) - f (x, y,\overline{\phi }) \right] \ge 0. \end{aligned}$$

Proceed by contradiction and assume that $\psi >0$ at some point $(x_0,y_0)$ with $x_0 > X'\ge X$ and $y \in \overline{\omega }$. From our choice of X, we have wherever $\psi >0$ that

$$\begin{aligned} f(x,y,\underline{\phi }) - f (x,y,\overline{\phi }) \le (g_+ ' (0) + \varepsilon ) (\underline{\phi } - \overline{\phi }) < \left( \frac{c^2}{4} - \varepsilon \right) e^{-\frac{c}{2} x} \psi , \end{aligned}$$

hence $\varDelta \psi > \varepsilon \psi$. Noting that $\lim _{x \rightarrow +\infty } \psi (x,y) =0$ (uniformly with respect to y), we can assume without loss of generality that

$$\begin{aligned} \psi (x_0,y_0) = \max _{x \ge X', y \in \omega } \psi (x,y). \end{aligned}$$

If $y_0 \in \omega$, then $\varDelta \psi (x_0,y_0) \le 0$, a contradiction. The case $y_0 \in \partial \omega$ is also ruled out, thanks to Hopf lemma. When $\omega = \mathbb R^{N-1}$, it may be necessary to pass to the limit in a sequence of appropriate shifts of $\psi$. More precisely, choosing a sequence $(x_n, y_n) \in (X',+\infty ) \times \mathbb R^{N-1}$ such that $\psi (x_n,y_n) \rightarrow \max _{x \ge X', y \in \omega } \psi (x,y)$, we can extract a subsequence so that $\psi _n:=\psi (x_n+ \cdot , y_n+ \cdot )$ converges to a function $\psi _\infty$ which reaches a positive maximum at the point (0, 0), and satisfies the parabolic inequality $\varDelta \psi \ge \varepsilon \psi$ on an open neighborhood of (0, 0). We reach the same contradiction.

We conclude that $\underline{\phi } \le \overline{\phi }$ in $(X',+\infty ) \times \omega$. By the strong maximum principle, either $\underline{\phi } \equiv \overline{\phi }$ or this inequality is also strict.

The case $c = 2 \sqrt{g_+ '(0)}$ can be dealt with similarly, by defining instead

$$\begin{aligned} \psi := \frac{e^{\frac{c}{2} x}}{1+\sqrt{x}} (\underline{\phi } - \overline{\phi }), \end{aligned}$$

which satisfies

$$\begin{aligned}&\varDelta \psi + \frac{1}{\sqrt{x}(1+\sqrt{x})} \partial _x \psi + \left( \frac{1}{4x^{3/2}}- \frac{3}{4x} \right) \frac{1}{(1+\sqrt{x})^2} \psi \\&\quad + \frac{e^{\frac{c}{2} x}}{1+\sqrt{x} } \left( f(x,y,\underline{\phi } ) - f (x,y,\overline{\phi }) - g_+'(0) (\underline{\phi }-\overline{\phi }) \right) \ge 0 \end{aligned}$$

for all $x>0$. Because of the critical choice of c, here we use the fact that $\partial _s f(x,y,s)$ converges not only locally uniformly in s to $g_+ ' (s)$ as $x \rightarrow +\infty$, but also in the $C^{0,r}$ topology where $0<r<1$. Thus one can find $\delta >0$, $X>0$ and $\eta >0$ such that for all $x \ge X$, $y \in \omega$ and $s \in [0,\delta ]$, one has

$$\begin{aligned} \partial _s f(x,y,s) \le g_+ ' (0) + \eta s^{r}. \end{aligned}$$

Then, when $\psi >0$ and x large enough, we get

$$\begin{aligned}&\frac{e^{\frac{c}{2} x}}{1+\sqrt{x} } \left( f(x,y,\underline{\phi } ) - f (x,y,\overline{\phi }) - g_+ '(0) (\underline{\phi }-\overline{\phi }) \right) \\&\quad = O\left( \frac{e^{\frac{c}{2} x}}{1+\sqrt{x} } (\underline{\phi }-\overline{\phi })^{1+r}\right) \\&\quad = O \left( (\underline{\phi }-\overline{\phi })^r \psi \right) \\&\quad = o \left( \psi (1+\sqrt{x})^re^{-r\frac{c}{2}x}\right) \\&\quad = o\left( \frac{\psi }{x(1+\sqrt{x})^2} \right) . \end{aligned}$$

Proceeding as above, we assume that there exists $(x_0,y_0)\in \varOmega$ such that $x_0>X'\ge X$ and $\psi (x_0,y_0)>0$ as well as $\lim _{x\rightarrow +\infty } \psi (x,y)=0$ uniformly with respect to y. Then we apply the maximum principle to reach the conclusion. $\square$

The interest of this lemma is that a solution of (1.1) satisfies such exponential bounds when it does not spread: see Proposition 4.1. As we cannot a priori exclude the possibilty that the solution converges to nonstationary solutions as time goes to infinity, we also provide below a parabolic version of this maximum principle.

1.2 A Parabolic Maximum Principle

Lemma A.2

Assume $c \ge 2 \sqrt{g_+'(0)}$. Assume that $\underline{u}$ is a global in time subsolution, and $\overline{u}$ a global in time supersolution, satisfying the exponential estimates from Lemma A.1 as $x\rightarrow +\infty$, uniformly with respect to $t \in \mathbb {R}$ and $y\in \omega$.

Then there exists $X>0$ large enough such that, for any $X' \ge X$, if $\underline{u} (\cdot ,X',\cdot ) \le \overline{u} (\cdot , X',\cdot )$, then $\underline{u} \le \overline{u}$ in $\mathbb {R} \times (X',+\infty )\times \omega$.

Note that the previous elliptic maximum principle lemma can be seen as a particular case. In fact the proof is quite similar and therefore we only briefly sketch it.

Proof

We start with the case $c>2\sqrt{g'_+(0)}$, and define $\varepsilon >0$ such that $c>2\sqrt{g'_+(0)+\varepsilon }$. We use some exponential lift as above, so that $\psi =e^{\frac{c}{2}x} (\underline{u} - \overline{u})$ satisfies

$$\begin{aligned} \partial _t \psi - \varDelta \psi + \frac{c^2}{4} \psi - \frac{ f(x,y,\underline{u}) - f (x,y,\overline{u})}{\underline{u} - \overline{u}} \times \psi \le 0. \end{aligned}$$

Proceeding as above, there exists $X>0$ such that for all $t>0$, $x\ge X$, $y\in \omega$,

$$\begin{aligned} \frac{ f(x,y,\overline{u}) - f (x,y,\underline{u})}{\overline{u} - \underline{u}}<g'_+(0)+\frac{\varepsilon }{2}. \end{aligned}$$

Now choosing $0<\delta <\varepsilon /2$ and $\gamma >0$, the function $w:=\gamma e^{-\delta t}$ is a supersolution of the above linear parabolic problem for all $t\in \mathbb {R}$, $x\ge X$, $y\in \omega$. Noting that $e^{\frac{c}{2} x} \times (\underline{u} - \overline{u})$ is uniformly bounded in space and time, one can choose $\gamma$ so that $\psi < \gamma$ for all time, $x\ge X$ and $y\in \omega$, as well as $\psi (\cdot ,X',\cdot ) \ge 0$ (for some $X'\ge X$ as assumed in Lemma A.2). By a comparison principle, we get that $\psi (t,\cdot ) \le \gamma e^{-\delta s}$ for all $t \in \mathbb {R}$, $s >0$, all $y \in \omega$ and $x \ge X'$. Therefore $\psi \le 0$, and in other words, $\overline{u}- \underline{u} \ge 0$.

The case $c = 2 \sqrt{g'_+ (0)}$ is similar. Posing now $\psi = \frac{e^{\frac{c}{2}x}}{1 + \sqrt{x}} (\underline{u} - \overline{u})$, one may find that $\psi$ is a subsolution of the heat equation on a half-space $(X,+\infty ) \times \omega$, for all times. It also satisfies $\psi (\cdot , +\infty , \cdot )$ and the conclusion of Lemma A.2 follows. $\square$