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Spreading of Two Competing Species in Advective Environment Governed by Free Boundaries with a Given Moving Boundary

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Abstract

In this paper, we study a free boundary problem of two competing species in the left-shifting environment. This model may be used to describe the interaction of the spreading phenomena of two competing species over a one dimensional habitat influenced by an external effect such as the effect of global warming. Here, we assume that only the habitat of inferior competitor is eroded away by the left moving boundary at constant speed c and we consider the how its spreading influences to the spreading of the superior competitor. We prove, as \(c_{2}^{\ast }<c<c_{1}^{\ast }\), a trichotomy result: (i) vanishing, (ii) spreading, or (iii) transition for inferior competitor influenced by an advection term caused by the left-shifting boundary and vanishing for superior competitor while both species go extinct in the long run as \(c_{2}^{\ast }<c_{1}^{\ast }<c\). This extends the result of Matsuzawa (Commun. Pure Appl. Anal. 17, 1821–1852, 2018) in two aspects: the model is considered for the two competing species and it takes into account the influence of the drift term caused by the the effect of left-shifting boundary.

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Acknowledgments

We are very grateful to the anonymous referees for careful reading and helpful suggestions which led to an improvement of our original manuscript. This work is supported by Vietnam National Foundation for Science and Technology Development (NAFOSTED) under grant 101.02-2018.312. Part of this work had been done while H-H. Vo was visiting Vietnam Institute for Advanced Study in Mathematics (VIASM), whose hospitality has been acknowledged.

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Correspondence to Hoang-Hung Vo.

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Appendix A

Appendix A

Let us prove the claim (1.10). When the spreading happens for v, i.e., \(\lim _{t\to \infty }g(t) = \infty \), we will show that

$$ \lim\limits_{t\to\infty}\frac{g(t)}{t}=c_{2}^{\ast}. $$

First, we recall some known results that are used in the proof of (1.10).

Lemma A.1 ([7, Theorem 3.3] and [11, Theorem 1.2])

Let (w, h) be a solution of

$$ \left\{\begin{array}{rll} {w}_{t} &= d_{2}{w}_{xx} + {w}(1 - c_{2}{w}), &\quad 0<x< g(t),~t>0, \\ {w}_{x}(0, t) &= 0,~{w}(g(t), t)=0, &\quad t>0,\\ g^{\prime}(t) &= -\mu_{2} {w}_{x}(g(t), t),&\quad t>0,\\ g(0) &= g_{0},~{w}(x,0) = {w}_{0}(x),&\quad 0 < x < g_{0}, \end{array}\right. $$
(A.1)

where g0 > 0, w0C2[0, g0] and \({w}_{0}(x) > 0 = {w}^{\prime }_{0}(0) = 0 = {w}_{0}(g_{0})\) for x ∈ [0, g0). Then the followings hold:

  1. (i)

    (Spreading-vanishing dichotomy) Either

    $$ \lim\limits_{t\to\infty}g(t) = \infty, \quad \lim\limits_{t\to\infty} {w}(x, t) = \frac{1}{c_{2}} $$

    uniformly in any bounded subset of \([0, \infty )\) or

    $$ \lim\limits_{t\to\infty}g(t) \leq \frac{\pi}{2}\sqrt{d},\quad \lim\limits_{t\to\infty}\|{w}(\cdot, t)\|_{C([0, g(t)])}=0. $$
  2. (ii)

    When \(\lim _{t\to \infty }g(t) = \infty \), \(g(t)/t \to c_{2}^{\ast }\) as \(t \to \infty \) and

    $$ \lim\limits_{t\to\infty}\sup\limits_{x\in [0, g(t)]}\left|{w}(x, t) - U_{c_{2}^{\ast}}(g(t) - x)\right| = 0, $$

    where \(c_{2}^{\ast }\) and \(U_{c_{2}^{\ast }}\) are defined in Proposition 4.1 of [7].

Lemma A.2 ([7, Lemma 3.5])

Assume that \(\overline {g} \in C^{1}[0, T]\) and \(\overline {{w}} \in C(\overline {D^{\overline {g}}}_ T)\bigcap C^{2,1}(D^{\overline {g}}_{T})\), where \(D^{\overline {g}}_{T} := \{(x, t)\in \mathbb {R}^{2} : 0 < x < D^{\overline {g}}_{T} (t), 0 < t \leq T\}\) and

$$ \left\{\begin{array}{rll} \overline{{w}}_{t} &\geq d_{2}\overline{{w}}_{xx} + \overline{{w}}(1 - c_{2}\overline{{w}}), &\quad 0<x<\overline{g}(t),~t>0, \\ \overline{{w}}_{x}(0, t) &\leq 0,~\overline{{w}}\left( \overline{g}(t), t\right)=0, &\quad t>0,\\ \overline{g}^{\prime}(t) &\geq -\mu_{2} \overline{{w}}_{x}\left( \overline{g}(t), t\right),&\quad t>0. \end{array}\right. $$

If \(g_{0} \leq \overline {g}(0)\) and \({w}_{0}(x)\leq \overline {{w}}(x, 0)\) for all x ∈ [0, g0], then the solution (w, g) of (A.1) satisfies \(g(t) \leq \overline {g}(t)\) for all t ∈ (0, T] and \({w}(x, t)\leq \overline {{w}}(x, t)\) for 0 ≤ xg(t), 0 ≤ tT.

Remark A.3

We call \((\overline {{w}}, \overline {g})\) defined in Lemma A.2 a super solution of (A.1). A subsolution can be defined if we reverse all the inequalities in Lemma A.2.

Now, let us prove the statement (1.10). It is clear that (v, g) in the second equation of (1.1) forms a subsolution of

$$ \left\{\begin{array}{rll} \overline{{w}}_{t} &= d_{2}\overline{{w}}_{xx} + \overline{{w}}(1- c_{2}\overline{{w}}), &\quad t>0,~0<x<\overline{g}(t), \\ \overline{{w}}_{x}(t,0) &=\overline{{w}}(t,\overline{g}(t))=0, &\quad t>0, \\ \overline{g}^{\prime}(t) &= -\mu_{2} \overline{{w}}_{x}(t, \overline{g}(t)),&\quad t>0,\\ \overline{g}(0) &= g_{0},~\overline{{w}}(0, x) = \overline{{w}}_{0}(x),&\quad 0 \leq x \leq g_{0}. \end{array}\right. $$

By Lemma A.2, we have \(\overline {g}(t)\geq g(t)\) for all t, which implies \(\overline {g}(t) = \infty \). Therefore, it follows Lemma A.1, we see that \(\frac {\overline {g}(t)}{t}\to c_{2}^{\ast }(d_{2}, \mu _{2})=c_{2}^{\ast }\) as \(t\to \infty \). As a consequence, we have

$$ \limsup\limits_{t\to\infty}\frac{g(t)}{t}\leq c_{2}^{\ast}. $$
(A.2)

Now, it remains to prove that \(\liminf _{t\to \infty }\frac {g(t)}{t}\geq c_{2}^{\ast }\). To do so, it suffices to show that for any \(\widetilde {c}_{2}\in (0, c_{2}^{\ast })\), one has

$$ \liminf\limits_{t\to\infty}\frac{g(t)}{t}\geq \widetilde{c}_{2}. $$

To derive this lower estimate, we take 0 < ε ≪ 1 small enough and T(ε) ≫ 1 in such a way that

$$ \begin{array}{@{}rcl@{}} u(t,x) &\leq& \frac{1}{b_{1}} + \varepsilon \quad\text{ for all }~x\in [0, \infty)\text{ and }t\geq T(\varepsilon), \end{array} $$
(A.3)
$$ \begin{array}{@{}rcl@{}} g(T(\varepsilon)) &>& \frac{\pi}{2}\sqrt{\frac{d_{2}}{1-b_{2}\left( \frac{1}{b_{1}}+\varepsilon\right)}}, \end{array} $$
(A.4)

where (A.4) can be derived by using b1 > b2 in (1.2). It then follows from (A.3) that (v, g) forms a super solution of

$$ \left\{\begin{array}{rll} \underline{{w}}_{t} &= d_{2}\underline{{w}}_{xx} + \underline{{w}}\left( 1- b_{2}\left( \frac{1}{b_{1}}+\varepsilon\right) - c_{2}\underline{{w}} \right), &\quad t>T(\varepsilon),~0<x< \underline{g}(t), \\ \underline{{w}}_{x}(t,0) &= \underline{{w}}(t,g(t))=0, &\quad t>T(\varepsilon), \\ \underline{g}^{\prime}(t) &= -\mu_{2} \underline{{w}}_{x}\left( t, \underline{g}(t)\right),&\quad t>T(\varepsilon),\\ \underline{{w}}(T(\varepsilon), x) &= {v}(x, T(\varepsilon)),&\quad 0 < x < \underline{g}(T(\varepsilon)). \end{array}\right. $$

Using (A.4), we see that \(\underline {g}(t) = \infty \). From Lemma A.2, we get

$$ \begin{array}{@{}rcl@{}} \frac{\underline{g}(t)}{t} &\to& \widetilde{c}_{2} (\varepsilon) := c_{2}^{\ast}\left( 1-b_{2}\left( \frac{1}{b_{1}}+\varepsilon\right), d_{2}, \mu_{2}\right)\quad\text{ as }~ t\to\infty,\\ \underline{{w}} &\geq& V_{\widetilde{c}_{2} (\varepsilon)}\left( \underline{g}(t)-x\right) - \varepsilon\quad\text{ for }~x\in \left[0, \underline{g}(t)\right]~\text{ and }~t\gg1, \end{array} $$

where \(V_{\widetilde {c}_{2} (\varepsilon )}\) is defined in Proposition 4.1 of [7]. By Lemma A.2, one derives

$$ \liminf\limits_{t\to\infty}\frac{g(t)}{t}\geq \widetilde{c}_{2} (\varepsilon). $$
(A.5)

Thus, by taking ε → 0 and incorporating with (A.2), (A.5), we obtain the desired equality (1.10).

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Nguyen, TH., Trong, D.D. & Vo, HH. Spreading of Two Competing Species in Advective Environment Governed by Free Boundaries with a Given Moving Boundary. Vietnam J. Math. 49, 1199–1225 (2021). https://doi.org/10.1007/s10013-020-00457-5

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