Abstract
Insect pests pose a major threat to a balanced ecology as it can threaten local species as well as spread human diseases; thus, making the study of pest control extremely important. In practice, the sterile insect release method (SIRM), where a sterile population is introduced into the wild population with the aim of significantly reducing the growth of the population, has been a popular technique used to control pest invasions. In this work we introduce an integro-differential equation to model the propagation of pests in a heterogeneous environment, where this environment is divided into three regions. In one region SIRM is not used making this environment conducive to propagation of the insects. A second region is the eradication zone where there is an intense release of sterile insects, leading to decay of the population in this region. In the final region we explore two scenarios. In the first case, there is a small release of sterile insects and we prove that if the eradication zone is sufficiently large the pests will not invade. In the second case, when SIRM is not used at all in this region we show that invasions always occur regardless of the size of the eradication zone. Finally, we consider the limiting equation of the integro-differential equation and prove that in this case there is a critical length of the eradication zone which separates propagation from obstruction. Moreover, we provide some upper and lower bound for the critical length.
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Acknowledgments
This was partially supported by the NSF Postdoctoral Fellowship in Mathematical Sciences DMS-1103765. I am grateful to Ricardo Cortez for bringing this application to my attention. I would like to thank Lenya Ryzhik and Henri Beresticky for their helpful discussions.
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Appendix
Appendix
1.1 Proof of Theorem 3
We state a lemma proved in Coville and Dupaigne (1994) that will enable us to construct a solution from an appropriate super-solution and sub-solution.
Lemma 1
[Coville and Dupaigne (1994)] Let \(f\in C^0(\mathbb R) \cap L^2(\mathbb R).\) There exists a unique solution \(v_0\in C^0(\mathbb R) \cap L^2(\mathbb R)\) to
Proof
(Theorem 3) Let
such that \(f_{ l}(u)\) is monostable with roots zero and one, \(f_{ g}(u)\) is negative and monotonically decreasing with \(f_{ g}(0)=0,\) and \(f{ r}(u)\) is bistable with roots: \(0, a,b\) with \(0<a<b<1\). Refer to Fig. 3 for a visual of the reaction terms \(f_L(x,u)\).
Step 1: (Super-solution and sub-solution) The first step is to construct a super-solution and a sub-solution. To construct a super-solution we let \(g(u)\) be a bistable function such that
and with the additional properties that \(g(u) = f_{ r}(u)\) for \(u\in [0,\delta ],\) \(g(u) = f_{ l}(u)\) for \(u\in [1-\delta ,1],\) for some \(\delta >0\) and \(g(u)\ge f_{ g}(u)\) for all \(u\).
Let \(W(x)\) be the stationary wave solution satisfying
Moreover, we choose the translation of the solution so that \(W(0) = 1-\delta .\) Then, if \(L>0\) is sufficiently large so that \(W(x)<\delta \) for all \(x>L\) and we claim that \(W(x)\) is a super-solution. Indeed, one can verify that
In the same spirit, let \(g_1(u)\) be a bistable function such that
and \(g_1(u) = f_{ g}(u)\) for \(u\in [0,\delta ],\) and \(g_1(u)\le f_{ l} \) for all \(u\). Additionally, if \(h(u):=g(u)+u\) we require
Let \(V(x)\) be the stationary traveling wave solution satisfying
such that \(V(0) = \delta \). Then for all \(L\ge 0\) we obtain that
Thus, we have constructed to a super-solution for \(L\) sufficiently large and a sub-solution for all \(L\ge 0\). Note that \(V\in C^2(\mathbb R)\) due to (32) [see Bates et al. (1997)].
Step 2: (Sequence of solutions) We now construct a solution starting with \(V(x)\) and using \(W(x)\) as a barrier. Indeed, let \(u_0(x) = V(x)\) and let \(u_n(x)\) be the solution to
We first show that such \(u_n(x)\) exists. Following Coville and Dupaigne (1994) we choose a \(\phi \in C_c^\infty \) satisfying \(\Vert \phi \Vert _{L^1} =1\) and define
Then, \(v_n=u_n-\varphi \) satisfies
Note, that by our choice of \(u_0\) we know that \(v_0\in C^0\cap L^2(\mathbb R)\) because \(\varphi \equiv 0\) for \(x\) large and \(\varphi \equiv 1\) for \(x\) negative enough. Note that to apply Lemma 1 we need \(f_L(x,v_{0}+\varphi )\in C^0(\mathbb R).\) Since the reaction term \(f_L(x,u)\) is discontinuous we regularize it by convolving with the standard mollifier and denote it by \(f^{\epsilon _1}_L(x,v_{0}+\varphi ),\) with \(\epsilon _1\) being the regularizing parameter. We first find a solution to
Now, for \(x<0\) we have, using the mean value theorem and \(f^{\epsilon _1}_l(1)=0\), that
and for \(x>L\) we have that \(f^{\epsilon _1}_r(0)=0\)
Thus, \(f^{\epsilon _1}_L(x,v_0+\varphi )\in L^2(\mathbb R)\). Finally, the fact that \(\mathcal {J}*\varphi - \varphi \in L^2(\mathbb R)\) follows from Lemma 3.1 in Coville and Dupaigne (1994). Now, applying Lemma 1 gives a solution to (33). However, since the estimates above are independent of the regularizing parameter we can take the limit as \(\epsilon _1\rightarrow \infty \) to obtain a solution to the original problem. We omit the details as the procedure is standard. Repeating this process iteratively gives us a sequence of solutions, \(\left\{ u_n\right\} _{n\in \mathbb {N}}\).
Step 3: (Passing to the limit as \(n\rightarrow \infty \)) From induction and the maximum principle we know that
for all \(n\in \mathbb {N}\). Furthermore, if we define \(w_n = u_n(x-\tau )-u_n(x)\) for some \(\tau > 0\) then \(w_n\) solves
Now, we choose \(\lambda >\Vert f'(u)\Vert \) so that \(-f_L+\lambda \) is non-decreasing and positive; thus, making the right hand side of the above equality non-negative for \(u_1\) given that \(V\) is monotone decreasing. An application of the maximum principle implies that \(w_1\) is always non-negative. Iterating this idea gives that \(u_n(x)\) is non-increasing in \(x\). Invoking, Helly’s lemma we obtain that a subsequence converges to a non-increasing solution \(u(x)\) also satisfying \(V(x)\le u(x)\le W(x).\) Moreover, the dominated convergences implies
Note also that \(\mathcal {J}*u_n \in C^\infty (\mathbb R)\) and
which implies that \(u_n - f_L(x,u_n)\in C^\infty (\mathbb R).\) This gives uniform convergence in \(u_n-f_L(x,u_n)\rightarrow u-f_L(x,u).\) This concludes the existence. We are left to show that the solution must be discontinuous.
Step 4: (Discontinuity) The solution \(u(x)\) satisfies
and since \(\mathcal {J}*u\) is smooth then \(u+f_L(x,u)\) must be smooth. This implies that
Since these functions do not intersect we can denote \(u(0^+)\) to be \(\lim _{x\rightarrow 0^+} u(x)\) (similarly for the other limits) then we have that
With this we conclude the proof. \(\square \)
1.2 Proof of Theorem 6
Proof
Let \(u(x)\) be the solution to (12) with reaction term \(f_L(x,u)\) defined in the hypotheses of the theorem. Since \(s_r=0\) in (11) then \(f_{ r}(x)\) is monostable, which implies that \(f(s)\ge 0\) for \(s\in (0,1)\). For \(x>L\) we have that
with equality only if \(u\equiv 1\). We know that there is a \(\delta >0\) such that \(u(L)=\delta \); hence, we know that \(u(x)\ge \delta \) for \(x>L\). We claim that the limit of the reaction term is zero,
which implies that \(\lim _{x\rightarrow \infty } u(x) = 1\) and from this we can conclude. To prove this claim we let \(\mathcal {J}= \mathcal {J}_\epsilon \) satisfy \(H0-H2\) with \(\epsilon <<1.\) We will approximate the non-local operator by the Laplacian with appropriate diffusion coefficient; indeed (provided \(u\) is sufficiently smooth),
Let \(v\) be the solution to
In fact, \(v(x) = \frac{1-x^2}{2\epsilon ^2}\) is the solution with a maximum value at \(x=0\) and equal to \(\frac{1}{2\epsilon ^2}.\) We define
such that \(u(L)=\delta \) and claim that
Note that if (37) is holds then we are done by the definition of \(\varGamma \). Assume, to get a contradiction, that (37) does not hold, so that there exists an \(y_0>L\) such that
Fix \(R\) such that \(L<R<y_0\) sufficiently close to \(y_0\) so that
Note that \(y_0\) cannot be a minimum of \(u(x)\) by (34). Thus, there exists a \(R<y_1<y_0\) so that \(\delta <u(y_1)<u(y_0)\). Let \(r=R-L\) and define
which satisfies
Note that
Now, consider \(z^\tau (x) := \tau z(x)\) for \(\tau >0\). For \(\tau \) small enough we have that
Let \(\tau ^o\) be the first value where the graph of \(z^{\tau _o}\) touches the graph of \(u(x)\) in \(B(y_1,r)\). Then there exists an \(x_0\) in the interior of \(B(y_1, r)\) (as \(z=0\) on the boundary) where \(\tau ^oz(x_0) = u(x_0)\) and \(z^{\tau ^0}(x)\le u(x)\) for \(x\in B(y_1, r).\) So we have
From this follows that
by the choice of \(R\) and because \(\epsilon <<1\). Now, we define \(w:=\tau ^oz - u\) and we know that \(w\le 0\) for \(x\in B(y_1,r)\) and \(w(x_0) = 0\). Moreover, we know that \(u(x_0) <u(y_0)\) so there exists a neighborhood centered at \(x_0\), \(\mathcal {N}(x_0)\) such that
Thus,
and by (35) provided \(\epsilon \) is sufficiently small
and
for \(x\in \mathcal {N}(x_0)\). This contradicts the fact that \(w(x)\) has a local minimum at \(x=x_0\) and we conclude that \(f(u)\rightarrow 0\) as \(x\rightarrow \infty \) which implies that \(u\rightarrow 1\) as \(x\rightarrow \infty \) in the case when \(\epsilon \) is small. For general \(\epsilon \) we know that the solution is simply a rescaling of the problem in \(\epsilon \). Indeed, if \(u(x,t)\) is a solution to
then \(v(x):=u(\epsilon x)\) is a solution to
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Rodríguez, N. On an integro-differential model for pest control in a heterogeneous environment. J. Math. Biol. 70, 1177–1206 (2015). https://doi.org/10.1007/s00285-014-0793-8
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DOI: https://doi.org/10.1007/s00285-014-0793-8