GENERALIZED LINEAR MODELS FOR POPULATION DYNAMICS IN TWO JUXTAPOSED HABITATS

. In this work we introduce a generalized linear model regulating the spread of population displayed in a d -dimensional spatial region Ω of R d constituted by two juxtaposed habitats having a common interface Γ. This model is described by an operator L of fourth order combining the Laplace and Biharmonic operators under some natural boundary and transmission conditions. We then invert explicitly this operator in L p -spaces using the H ∞ -calculus and the Dore-Venni sums theory. This main result will lead us in a later work to study the nature of the semigroup generated by L which is important for the study of the complete nonlinear generalized diﬀusion equation associated to it.

1. Introduction. The partial differential equations play a natural role in population dynamics, in particular in the reaction-diffusion models which are derived from the well known Fick's law.
An important problem in population ecology is the effect of environmental changes on the growth and diffusion of the species in areas made up of various habitats. In this situation and in order to understand how populations interact between the habitats, it is necessary to have spatially explicit models incorporating individual behaviour at different boundaries and interfaces of the habitats.
If u(t, .) denotes the population density, the classical Fickian equations in each habitat for these models are typically of the form where F is the nonlinear growth interaction and l is the positive coefficient diffusion (which can be variable).
The variety and the complexity of the habitats and the individuals are not well modeled by spatial effects to be simply Fickian diffusion (as, for example, models of cell motion). An approach based on a the Landau-Ginzburg free energy functional and on the variational derivative consider the more generalized following diffusion equation for growth and dispersal in a population ∂u ∂t where k is generally positive and l is a number which can be negative, see [4], p. 238. In this paper, we only consider the case when k and l are positive, but our techniques can be extended to k, l ∈ R \ {0}, satisfying some conditions, this will be done in a forthcoming paper. Now consider the d-area Ω = Ω − ∪Ω + constituted by the two juxtaposed habitats with their interface Γ = {γ} × ω, where a, γ, b ∈ R with a < γ < b and ω being an open bounded regular set of R d−1 . Consider the following linear stationary dispersal equations where u = u − in Ω − u + in Ω + and f = f − in Ω − f + in Ω + , with f given in L p (a, b; L p (ω)) = L p (Ω), and k ± , l ± are positive numbers. The spatial variables will be denoted by (x, y), x ∈ (a, b) and y ∈ ω. The above equations will be considered under the following boundary and transmission conditions u − (x, ζ) = 0, x ∈ (a, γ), ζ ∈ ∂ω u + (x, ζ) = 0, x ∈ (γ, b), ζ ∈ ∂ω ∆u − (x, ζ) = 0, x ∈ (a, γ), ζ ∈ ∂ω ∆u + (x, ζ) = 0, x ∈ (γ, b), ζ ∈ ∂ω (2) u − (a, y) = ϕ − 1 (y), u + (b, y) = ϕ + 1 (y), y ∈ ω ∂u − ∂x (a, y) = ϕ − 2 (y), ∂u + ∂x (b, y) = ϕ + 2 (y), y ∈ ω, (ϕ ± 1 and ϕ ± 2 will be given in appropriated spaces) and Now, define the following homogeneous dispersal linear operator      D (L) = u ∈ L p (Ω) : ∆u ± , ∆ 2 u ± ∈ L p (Ω ± ) and u ± satisfies (BC 0 ) and (T C pde ) where (BC 0 ) corresponds to (BC pde ) with ϕ + 1 = ϕ − 1 = ϕ + 2 = ϕ − 2 = 0. Therefore, in this work, we will focus ourselves on proving essentially the invertibility of L; this study will be very useful to analyze the following spectral equation in order to characterize the nature of the semigroup generated by L. On the other hand the same techniques used here will apply for this analysis. We know the importance of this property in the study of the generalized diffusion complete equation quoted above.
Let us comment on the boundary and transmission conditions. The two first boundary conditions of (1) in (BC pde ) simply mean that the individuals die when they reach on the other parts of the boundaries (a, b) × ∂ω (which means that we have an inhospitable border); the next two others of (1) mean that there is no dispersal in the normal direction. We deduce that the dispersal vanishes on (a, b) × ∂ω, that is ∆u − = 0 on (a, γ) × ∂ω and ∆u + = 0 on (γ, b) × ∂ω.
In (2) of (BC pde ), the population density and the flux are given, for instance on {a} × ω and on {b} × ω. This signifies that the habitats are not segregated.
In (T C pde ), the two first transmission conditions mean the continuity of the density and its flux at the interface, while the two second express the continuity of the dispersal and its flux (in some sense) at Γ.
We can consider more realistic transmission conditions with the noncontinuity of the density and the flux but including the continuity of the generalized dispersal: This situation requires to work in spaces built on the continuous functions. We will consider this case in a future work. Note that, when we consider different types of habitats, the response of individuals at the interface is important for the overall movement behaviour.
In many works, a generalized diffusion model is considered. Let us quote a number of them.
In [4] and in [18], the authors have presented in one dimensional case a nonlinear model with spatial structure characterized by a fourth order operator in only one habitat. They used essentially a Landau-Ginzburg free energy functional.
We were essentially inspired by these works to deduce a linear d-dimensional model set in two bounded juxtaposed cylindrical habitats which requires necessarily boundary and transmission conditions. We will then base ourselves on similar techniques to those used in the works of [7] and [14].
The paper is organized as follows. First, in section 2, we present the PDE transmission problem (P pde ) and with the help of operator A 0 defined below, we give its operational writing. We will then study problem (P) with a general operator A instead of A 0 .
Then, in section 3, we recall what is a BIP operator, we precise our notations about interpolation spaces, we set our hypotheses and their consequences. We explain how to solve our problem (P) by introducing two auxiliary problems (P − ) and (P + ). We then present our main result in Theorem 3.3. As a consequence of this theorem, we obtain the Corollary 1 which states existence and uniqueness of the solution of problem (EQ pde ) − (BC pde ) − (T C pde ) quoted above.
In section 4, we give technical results which help us to prove our main result. In Proposition 1 and in Proposition 2 we solve problems (P − ) and (P + ) provided that the data are in some real interpolation spaces. We establish (see Theorem 4.2) a useful technical result which allows us to prove Theorem 3.3. Then, we show some technical lemmas which lead us to apply functional calculus.
Section 5 is devoted to the proof of Theorem 3.3. This section is composed of three parts: in the first part, we use Theorem 4.2 to explicit the determinant of the transmission system. In the second part, we inverse the determinant of the transmission system using functional calculus. Finally, in the last part, we show that the general transmission problem has a unique classical solution by establishing the regularity of this solution.
2. Operational formulation. Consider now the problem Let us define A 0 , the Laplace operator in R d−1 , d ∈ N \ {0, 1}, as follows Thus, using operator A 0 , problem (P pde ) becomes Then, we will consider a generalization of this problem with (−A, D(−A)), instead of (−A 0 , D(−A 0 )), a BIP operator of angle θ ∈ (0, π) on a UMD space X, see section 3 below for the definitions of BIP operator and UMD spaces, and f ∈ L p (a, b; X).
More precisely, we study the following transmission problem (P): The transmission conditions (T C) will be divided into Note that (T C2) is well defined in virtue of Lemma 3.2, see Section 3.2 below. We will search a classical solution of problem (P), that is a solution u such that u − := u |(a,γ) ∈ W 4,p (a, γ; X) ∩ L p (a, γ; D(A 2 )), u − ∈ L p (a, γ; D(A)), and which satisfies (EQ) − (BC) − (T C).

3.
Assumptions, consequences and statement of results.
It is known that any injective sectorial operator T 1 admits imaginary powers T is 1 , s ∈ R, but, in general, T is 1 is not bounded, see [12], p. 342. Let θ ∈ [0, π). We denote by BIP(X, θ), see [19], p. 430, the class of sectorial injective operators T 1 such that In this case, D(T 1 ) ∩ R(T 1 ) = X, see [11], proof of Proposition 3.2.1, c), p. 71. We will use the well-known Dore-Venni theorem, see [6] and its generalization in [19], which needs to consider a UMD space X. Recall that a Banach space X is a UMD space if and only if for some p ∈ (1, +∞) and thus for all p, the Hilbert transform is bounded from L p (R, X) into itself.
Note that for an operator We have the two following lemmas.

3.3.
Hypotheses. In all the sequel, k + , k − , l + , l − ∈ R + \ {0}, A denotes a closed linear operator in X and we set We assume the following hypotheses: Note that (H 4 ) means that −A is a sectorial operator of any angle θ ∈ (0, π). Let us give some consequences of our assumptions.

From (H 2 ) and (H
. Thus, assumptions (H 1 ), (H 2 ) and (H 3 ) lead us to apply the Dore-Venni theorem, see [6], to obtain 0 ∈ ρ(L + + M ) and 0 ∈ ρ(L − + M ). 6. It follows from (5) that and also 3.5. The main results. To solve problem (P), we introduce two problems: So our goal is to prove that there exists a unique couple (ψ 1 , ψ 2 ) which satisfies (i), (ii) and (iii). This will lead us to obtain our main result.
1. The proof of Theorem 3.3 uses operators L − , L + , M and also interpolation spaces (D(M ), X) 3−j+ 1 p ,p , j = 0, 1, 2, 3. But from Lemma 3.1, we get 2. We can generalize this Theorem by considering a transmission problem between n habitats, with n ∈ N \ {0}. It suffices to use Theorem 3.3 on the two first habitats and then apply it on the transmission problem between the second and the third habitat to solve the problem with n = 3. By recurrence, we obtain the result.
As consequence of Theorem 3.3, we deduce some results for problem (P pde ) under some necessary boundary conditions. Let us consider the case A = A 0 (other cases can be treated).

Preliminary results.
In all the sequel, we set Applying Remark 1, we will solve problem (P) by studying first problems (P − ) and (P + ). To this end, we need the following invertibility result obtained in [13].
are invertible with bounded inverse.
All these exponentials are well defined, see statement 4 of section 3.4. For a detailed proof, see [13], Proposition 5.4 with k = r − or k = r + . Moreover where and F − is the unique classical solution of problem Proof. From Remark 3. In the previous proposition, due to (13), (15) and (16), we have α − i ∈ D(M ), for i = 1, 2, 3, 4. Moreover, since F − is a classical solution of (17), by Lemma 3.2, we deduce that, for j = 0, 1, 2, 3 and s = a or γ where and F + is the unique classical solution of problem u (4) Proof. From  [13], where u, L, f , a, F 0,f , k, ϕ 1 , ϕ 2 , ϕ 3 , ϕ 4 are respectively replaced by u + , L + , f + , γ, F + , r + , ψ 1 , ϕ + 1 , ψ 2 , ϕ + 2 . Remark 4. In the previous proposition, due to (18), (20) and (21), we have α + i ∈ D(M ), for i = 1, 2, 3, 4. Moreover, since F + is a classical solution of (22), by Lemma 3.2, we deduce that, for j = 0, 1, 2, 3 and s = γ or b F (j) The transmission system. This section is devoted to the proof of Theorem 4.2 stated below, which gives the link between problem (P) and the following system The coefficients are given by and similarly The second members are with R 1 given by and Then, the transmission problem (P ) has a unique classical solution if and only if the data ϕ + 1 , ϕ − 1 , ϕ + 2 , ϕ − 2 satisfy (8) and system (23) has a unique solution (ψ 1 , ψ 2 ) such that Proof. First, we assume that (P) admits a unique classical solution u. Setting we get that u − (respectively u + ) is the classical solution of (P − ) (respectively (P + )). So, applying Proposition 1 (respectively Proposition 2), we obtain (8) and also (29). It remains to prove that (ψ 1 , ψ 2 ) satisfies (23). To this end we use (T C2) satisfied by u, that is To make explicit this system, we use the expression of u + given in (19). It follows, for x ∈ (γ, b) Then, in virtue of Lemma 3.2, we have Furthermore, from (6), we obtain it follows that Note that, from Remark 4 and Lemma 3.2, all the terms in the previous equalities are justified. By the same arguments and using again (6) and also (22), we have Then, we obtain As previously, for u − , we use (14) and get, for x ∈ (a, γ) Then, in virtue of Lemma 3.2, we have hence, due to (6), we have Then, we obtain Furthermore, from (6), we have Then, we deduce the following equality: Note that, from Remark 3 and Lemma 3.2, all the terms in the previous equalities are justified. It follows, from (30) and (31), that system (T C2) becomes where, R 1 is given by (27). Thus But, from (20), (21), (15) and (16), we have So, using (24) and (25), the first line of system (32) writes where S 1 is given by (26). By the same way, we have Furthermore, since s = √ x and t = √ x + r, we have r = t 2 − s 2 = (t + s)(t − s). It follows Then, we have Moreover, from [13], Lemma 5.2, we have Then, from (36), we obtain Then, from (36), we obtain thatg δ,r (x) < 0. Finally, we have g δ,r (x) < (s + t) e −δs − e −δt 2g δ,r (x) < 0, from which we deduce that h δ,r (x) < 0. Conversely, if (8) holds, due to Theorem 4.2, it suffices to prove that system (23) has a unique solution such that (29) holds. The proof is divided in three parts. First, we will make explicit the determinant of system (23). Then, we will show the uniqueness of the solution, to this end, we will inverse the determinant with the help of functional calculus. Finally, we will prove that ψ 1 and ψ 2 have the expected regularity.
5.1. Calculus of the determinant. Now we have to make explicit the determinant. Recall system (23) We write the previous system as a matrix equation ΛΨ = S, where To solve system (23), we will study the determinant det(Λ) : of the matrix Λ. We develop it to obtain where and D 2 = P + 1 P − 3 + P − 1 P + 3 + 2M P + 2 P − 2 . Using section 3.4, we obtain that , which justify the equality in (37). In the sequel, we precise the terms D + 1 and D − 1 . Lemma 5.1. We have Proof.

We have
Moreover Then Using again (38), we obtain Finally, since we have The result is similarly obtained by replacing respectively d, k + and r + by c, k − and r − in the proof above.