Electronic Journal of Differential Equations, Vol. 2024 (2024), No. 78, pp. 1–32. ISSN: 1072-6691. URL: https://ejde.math.txstate.edu, https://ejde.math.unt.edu DOI: 10.58997/ejde.2024.78 SOLVABILITY OF TRANSMISSION PROBLEMS WITH GENERALIZED DIFFUSION EQUATION IN Lp-SPACES ALEXANDRE THOREL Abstract. We study a transmission problem in population dynamics between two juxtaposed habitats. In each habitat, we consider a generalized diffusion equation composed by the Laplace operator and a btocould be negative or null. Using semigroup theory and functional calculus, we give some relation between coefficients to obtain the existence and uniqueness of a classical solution in Lp- spaces. 1. Introduction In this work, using analytic semigroups theory, we study a transmission problem for a coupled system of generalized diffusion equations in Lp-spaces, with p ∈ (1,+∞). We denote by generalized diffusion equation, an equation of the form k∆2u− l∆u = g, with k, l ∈ R and g given. This equation is obtained using the Landau-Ginzburg free energy functional, we refer to [8, 29] for more details. This work is a natural continuation of the works done in [19, 35]. Here, we investigate the influence of the Laplace operator and the biharmonic term in the diffusion. In population dynam- ics, the Laplace operator model the short range diffusion whereas the biharmonic operator represents the long range diffusion. Thus, generalized diffusion is a linear combination of these two operators. Usually, in most models k, l > 0, but in many works for instance [8, 16, 27, 28, 29], the authors explain that the biharmonic term plays a stabilizing role if k > 0 and a destabilizing role when k < 0. This is why, in the present paper, we consider k ∈ R \ {0} and l ∈ R. Many works have treated generalized diffusion equations and transmission prob- lems associated to it. For instance, we refer to [3, 6, 8, 10, 20, 21, 22, 23, 28, 29, 30], for the study of such an equation in population dynamics and to [11, 13, 19, 35] for transmission problems associated with it. Note that [19, 35], consider applications in population dynamics wheras [11, 13], consider applications in plate theory. We define Ω = Ω− ∪ Ω+, the n-dimensional area, n ∈ N \ {0, 1}, constituted by the two juxtaposed habitats Ω− := (a, γ) × ω and Ω+ := (γ, b) × ω with their 2020 Mathematics Subject Classification. 35B65, 35J48, 35R20, 47A60, 47D06. Key words and phrases. Operational differential equations; functional calculus; analytic semigroups; BIP operators; maximal regularity. ©2024. This work is licensed under a CC BY 4.0 license. Submitted April 4, 2024. Published November 30, 2024. 1 2 A. THOREL EJDE-2024/78 interface Γ = {γ} × ω, where a, γ, b ∈ R with a < γ < b and ω being a bounded domain of Rn−1. We study the transmission problem k+∆ 2u+ − l+∆u+ = g+, in Ω+ k−∆ 2u− − l−∆u− = g−, in Ω− . (1.1) where k± ∈ R \ {0}, l± ∈ R, u± ∈ Ω± are population density and g± ∈ Lp(Ω±) are given. Note that the case k±, l± > 0 has been already treated in [19] and the case l± = 0 with k± > 0 has been already treated in [35]. Thus, in the present article, the most important new results concern the other cases. Indeed, as in the two previous cases, the key point of this article is the inversion of determinant operator but unlike the two previous cases, the writing of this determinant operator by the functional calculus must take into account the sign of the constants l+ k+ and l− k− . Therefore it is necessary to detail each case and for each of these cases, we give conditions between the different constants k± and l± which make it possible to invert this determinant operator. Here, we denote by (x, y) the spatial variables with x ∈ (a, b) and y ∈ ω. The above equations are supplemented by 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 (1.2)  u−(a, y) = φ− 1 (y), y ∈ ω u+(b, y) = φ+ 1 (y), y ∈ ω ∂u− ∂x (a, y) = φ− 2 (y), y ∈ ω ∂u+ ∂x (b, y) = φ+ 2 (y), y ∈ ω, (1.3) where φ± 1 and φ± 2 are given in suitable spaces, and u− = u+ on Γ ∂u− ∂x = ∂u+ ∂x on Γ k−∆u− = k+∆u+ on Γ ∂ ∂x (k−∆u− − l−u−) = ∂ ∂x (k+∆u+ − l+u+) on Γ. (1.4) In (1.2)-(1.3), the boundary conditions on the two first lines of (1.2) means that the individuals could not lie on the boundaries (a, b) × ∂ω, because, for instance, they die or the edge is impassable. The boundary conditions on the two second lines of (1.2) mean that there is no dispersal in the normal direction. It follows that the dispersal vanishes on (a, b)× ∂ω. In (1.3), the population density and the flux are given, for instance on {a} × ω and on {b} × ω. This signifies that the habitats are not isolated. Then, in (TCpde), the two first transmission conditions mean the EJDE-2024/78 SOLVABILITY OF TRANSMISSION PROBLEMS 3 continuity of the density and its flux at the interface, while the two second express, in some sense, the continuity of the dispersal and its flux at the interface Γ. This article is organized as follows. In section 2, we give our operational problem. Section 3 is devoted to some recall on BIP operators and real interpolation spaces. In section 4, we give our assumptions and main results. Then, in section 5, we state some preliminary results that will be useful to prove our main result. Finally, section 6, which is composed of three parts, is devoted to the proof of our main result. 2. Operational formulation Since ∆u = ∂2u ∂x2 +∆yu, we set D(A0) := {ψ ∈W 2,p(ω) : ψ = 0 on ∂ω} ∀ψ ∈ D(A0), A0ψ = ∆yψ. (2.1) Thus, using operator A0, it follows that each equation of (1.1) becomes u (4) ± (x) + (2A0 − l± k± I)u′′±(x) + (A2 0 − l± k± A0)u±(x) = f±(x), where u±(x) := u±(x, ·), f±(x) := g±(x, ·)/k± and f− ∈ Lp(a, γ;Lp(ω)), f+ ∈ Lp(γ, b;Lp(ω)) with p ∈ (1,+∞). Moreover, since the boundary conditions (1.2) are included in the domain of A0, problem (1.1)-(1.2)-(1.3)-(1.4) becomes u (4) + (x) + (2A0 − l+ k+ I)u′′+(x) + (A2 0 − l+ k+ A0)u+(x) = f+(x), for a.e. x ∈ (γ, b) u (4) − (x) + (2A0 − l− k− I)u′′−(x) + (A2 0 − l− k− A0)u−(x) = f−(x), for a.e. x ∈ (a, γ) u−(a) = φ− 1 , u+(b) = φ+ 1 u′−(a) = φ− 2 , u′+(b) = φ+ 2 u−(γ) = u+(γ) u′−(γ) = u′+(γ) k−u ′′ −(γ) + k−A0u−(γ) = k+u ′′ +(γ) + k+A0u+(γ) k−u (3) − (γ) + k−A0u ′ −(γ)− l−u ′ −(γ) = k+u (3) + (γ) + k+A0u ′ +(γ)− l+u ′ +(γ), Then, we will consider a more general case using (A,D(A)), instead of (A0, D(A0)), with −A a BIP operator of angle θA ∈ (0, π) on a UMD space X, see below for the definitions of BIP operator and UMD spaces, and f ∈ Lp(a, b;X). More precisely, setting r± = l± k± , we study the transmission problem (2.2)–(2.4):{ u (4) + (x) + (2A− r+ I)u ′′ +(x) + (A2 − r+A)u+(x) = f+(x), x ∈ (γ, b) u (4) − (x) + (2A− r− I)u ′′ −(x) + (A2 − r−A)u−(x) = f−(x), x ∈ (a, γ) (2.2){ u−(a) = φ− 1 , u+(b) = φ+ 1 u′−(a) = φ− 2 , u′+(b) = φ+ 2 (2.3) 4 A. THOREL EJDE-2024/78 u−(γ) = u+(γ) u′−(γ) = u′+(γ) k−u ′′ −(γ) + k−Au−(γ) = k+u ′′ +(γ) + k+Au+(γ) k−u (3) − (γ) + k−Au ′ −(γ)− l−u ′ −(γ) = k+u (3) + (γ) + k+Au ′ +(γ)− l+u ′ +(γ). (2.4) The transmission conditions (2.4) will be divided into u−(γ) = u+(γ) u′−(γ) = u′+(γ), (2.5) and k−u ′′ −(γ) + k−Au−(γ) = k+u ′′ +(γ) + k+Au+(γ) k−u (3) − (γ) + k−Au ′ −(γ)− l−u ′ −(γ) = k+u (3) + (γ) + k+Au ′ +(γ)− l+u ′ +(γ). (2.6) Note that (2.6) is well defined, see Lemma 3.8 below. We will search a classical solution of problem (2.2)-(2.3)-(2.4), that is a solution u such that u+ := u|(γ,b) ∈W 4,p(γ, b;X) ∩ Lp(γ, b;D(A2)), u′′+ ∈ Lp(γ, b;D(A)), u− := u|(a,γ) ∈W 4,p(a, γ;X) ∩ Lp(a, γ;D(A2)), u′′− ∈ Lp(a, γ;D(A)), (2.7) and which satisfies (2.2)-(2.3)-(2.4). Note that such a solution is not W 4,p(a, b;X) but uniquely W 4,p(a, γ;X) in Ω− and W 4,p(γ, b;X) in Ω+. 3. Definitions and prerequisites 3.1. The class of Bounded Imaginary Powers of operators. Definition 3.1. A Banach space X is a UMD space if for all p ∈ (1,+∞), the Hilbert transform is bounded from Lp(R, X) into itself (see [4] and [5]). Definition 3.2. A closed linear operator T1 is called sectorial of angle α ∈ [0, π) if (i) σ(T1) ⊂ Sα, (ii) for all α′ ∈ (α, π), sup { ∥λ(λ I − T1) −1∥L(X) : λ ∈ C \ Sα′ } < +∞, where Sα := { {z ∈ C : z ̸= 0 and | arg(z)| < α} if α ∈ (0, π), (0,+∞) if α = 0, (3.1) see [17, p. 19]. Remark 3.3. From [18, p. 342] we know that any injective sectorial operator T1 admits imaginary powers T is 1 , s ∈ R, but, in general, T is 1 is not bounded. Definition 3.4. Let θ ∈ [0, π). We denote by BIP(X, θ), the class of sectorial injective operators T2 such that (i) D(T2) = R(T2) = X, (ii) for all s ∈ R, T is 2 ∈ L(X), (iii) there exists C ≥ 1 such that for all s ∈ R, ∥T is 2 ∥L(X) ≤ Ce|s|θ, see [31, p. 430]. EJDE-2024/78 SOLVABILITY OF TRANSMISSION PROBLEMS 5 3.2. Interpolation spaces. Here we recall some properties about real interpola- tion spaces in particular cases. Definition 3.5. Let T3 : D(T3) ⊂ X −→ X be a linear operator such that (0,+∞) ⊂ ρ(T3) and ∃ C > 0 : ∀t > 0, ∥t(T3 − tI)-1∥L(X) ⩽ C. (3.2) Let k ∈ N \ {0}, θ ∈ (0, 1) and q ∈ [1,+∞]. We will use the real interpolation spaces (D(T k 3 ), X)θ,q = (X,D(T k 3 ))1-θ,q, defined, for instance, in [24, 25]. In particular, for k = 1, we have the characterization (D(T3), X)θ,q := { ψ ∈ X : t 7−→ t1-θ∥T3(T3 − tI)-1ψ∥X ∈ Lq ∗(0,+∞) } , where Lq ∗(0,+∞) is given by Lq ∗(0,+∞;C) := { f ∈ Lq(0,+∞) : (∫ +∞ 0 |f(t)|q dt t )1/q < +∞ } , for q ∈ [1,+∞), and for q = +∞, by L∞ ∗ (0,+∞;C) := sup t∈(0,+∞) |f(t)|, see [9, p. 325], or [15, p. 665, Teorema 3], or [36, section 1.14], where this space is de- noted by (X,D(T3))1-θ,q. Note that we can also characterize the space (D(T3), X)θ,q taking into account the Osservazione, [15, p. 666]. We set also, for each k ∈ N \ {0}, (D(T3), X)k+θ,q := { ψ ∈ D(T k 3 ) : T k 3 ψ ∈ (D(T3), X)θ,q } , (X,D(T3))k+θ,q := { ψ ∈ D(T k 3 ) : T k 3 ψ ∈ (X,D(T3))θ,q } , see [26, definition 3.2, p. 64]. Remark 3.6. The general situation of the real interpolation space (X0, X1)θ,q with X0, X1 two Banach spaces such that X0 ↪→ X1, is described in [24]. Remark 3.7. Note that for T3 satisfying (3.2), T k 3 is closed for each k ∈ N \ {0} since ρ(T3) ̸= ∅; consequently, if kθ < 1, we have (D(T k 3 ), X)θ,q = (X,D(T k 3 ))1−θ,q = (X,D(T3))k−kθ,q = (D(T3), X)(k−1)+kθ,q ⊂ D(T k−1 3 ). For more details see [25, (2.1.13), p. 43], or [15, p. 676, Thm. 6]. Lemma 3.8 ([15]). Let T3 be a linear operator satisfying (3.2). Let u such that u ∈Wn,p(a1, b1;X) ∩ Lp(a1, b1;D(T k 3 )), where a1, b1 ∈ R with a1 < b1, n, k ∈ N \ {0} and p ∈ (1,+∞). Then for any j ∈ N satisfying the Poulsen condition 0 < 1 p + j < n and s ∈ {a1, b1}, we have u(j)(s) ∈ (D(T k 3 ), X) j n+ 1 np ,p . This result is proved in [15, p. 678, Thm. 2]. 6 A. THOREL EJDE-2024/78 4. Assumptions and statement of results 4.1. Hypotheses. In the sequel, r+, r− ∈ R, k+k− > 0 and A denotes a closed linear operator in X. We assume the following hypotheses: (H1) X is a UMD space, (H2) [min(r+, r−, 0),+∞) ⊂ ρ(A), (H3) −A ∈ BIP(X, θA) for some θA ∈ [0, π), (H4) −A ∈ Sect(0). Remark 4.1. (1) Because of (H2), if at least one parameter r+ or r− is negative or null, then 0 ∈ ρ(A). (2) Operator A0, defined by (2.1), satisfies all the previous hypotheses with X = Lq(ω), q ∈ (1,+∞) and r± ∈ ρ(A0). From [33, Proposition 3, p. 207], X satisfies (H1) and taking A0 + r±I in [14, Theorem 9.15, p. 241 and Lemma 9.17, p. 242], we deduce that A0 satisfies (H2). Moreover, (H3) is satisfied for every θA ∈ [0, π), from [32, Theorem C, p. 166-167]. Finally, (H4) is satisfied thanks to [17, section 8.3, p. 232]. (3) In the scalar case, to solve each equation of (1.1), we need to solve the characteristic equations χ4 + (2A− r±)χ 2 + (A2 − r±A) = 0, thus, in our operational case, we consider the following operators L− := − √ −A+ r−I, L+ := − √ −A+ r+I and M := − √ −A. (4.1) By (H2) and (H3), operators −A, −A+ r−I and −A+ r+I are sectorial operators, so the existence of L−, L+ and M is ensured, see for instance [17, e), p. 25], and [1, Theorem 2.3, p. 69]. (4) From [17, Proposition 3.1.9, p. 65], we have D(L−) = D(L+) = D(M). Thus, for n,m ∈ N and m ⩽ n D(Ln ±) = D(Mn) = D(Lm ±M n−m) = D(MmLn−m ± ). (5) From [31, Theorem 3, p. 437] and [1, Theorem 2.3, p. 69], assumptions (H2) and (H3) imply that −A + r± I ∈ BIP(X, θA) and from [17, Proposition 3.2.1, e), p. 71], that −L−,−L+,−M ∈ BIP(X, θA/2). Moreover, from [31, Theorem 4, p. 441], we get −(L− +M),−(L+ +M) ∈ BIP(X, θA/2 + ε), for all ε ∈ (0, π/2− θA/2). Since we have 0 < θA/2 < π/2, then by [31, Theorem 2, p. 437], we deduce that L−, L+,M , L−+M and L++M generate bounded analytic semigroups (exL−)x⩾0, (exL+)x⩾0, (e xM )x⩾0, (e x(L−+M))x⩾0 and (ex(L++M))x⩾0. (6) Using the Dore-Venni sums theorem, see [12], we deduce from (H1), (H2) and (H3) that 0 ∈ ρ(M) ∩ ρ(L−) ∩ ρ(L+) ∩ ρ(L+ +M) ∩ ρ(L− +M). (7) From (4.1), we deduce that ∀ψ ∈ D(M2), (L2 + −M2)ψ = r+ψ and (L2 − −M2)ψ = r−ψ. (4.2) and that for all ψ ∈ D(M) it holds (L+ −M)ψ = r+(L+ +M)−1ψ, (L− −M)ψ = r−(L− +M)−1ψ. (4.3) EJDE-2024/78 SOLVABILITY OF TRANSMISSION PROBLEMS 7 4.2. Main results. To solve our operational problem (2.2)–(2.4), we introduce two problems: u (4) + (x) + (2A− r+ I)u ′′ +(x) + (A2 − r+A)u+(x) = f+(x), for a.e. x ∈ (γ, b) u+(γ) = ψ1, u+(b) = φ+ 1 u′+(γ) = ψ2, u′+(b) = φ+ 2 . (4.4) and u (4) − (x) + (2A− r− I)u ′′ −(x) + (A2 − r−A)u−(x) = f−(x), for a.e. x ∈ (a, γ) u−(a) = φ− 1 , u−(γ) = ψ1 u′−(a) = φ− 2 , u′−(γ) = ψ2. (4.5) Remark 4.2. Recall that u is a classical solution of (2.2)–(2.4) if and only if there exist ψ1, ψ2 ∈ X such that (i) u− is a classical solution of (4.5), (ii) u+ is a classical solution of (4.4), (iii) u− and u+ satisfy (2.6). Therefore, our aim is to state that there exists a unique couple (ψ1, ψ2) which satisfies (i), (ii) and (iii). Theorem 4.3. Let f− ∈ Lp(a, γ;X) and f+ ∈ Lp(γ, b;X) with p ∈ (1,+∞). Assume that (H1)–(H4) hold and k+k− > 0. Thus (1) when r+, r− ∈ R \ {0}, • if r+ > 0 and r− > 0, • if r+ < 0 and r− < 0, such that (l+ − l−)(k+ − k−) ⩾ 0, • if r+ > 0 and r− < 0, such that −6l−k+ + l+k+ + l−k− ⩾ 0. • if r+ < 0 and r− > 0, such that −6l+k− + l+k+ + l−k− ⩾ 0, (2) when r+ ∈ R \ {0} and r− = 0 with k− k+ ⩽ 2, • if r+ > 0 is such that r+ ⩾ ( √ t+ 1 + √ t)2 t2 k2+ 4k2− , for t ∈ ( 0, 1 r+∥A−1∥L(X) ) fixed, • if r+ < 0 is such that r+ ⩽ − 27k2 + 64k2 − , (3) when r+ = 0 and r− ∈ R \ {0} with k+ k− ⩽ 2, • if r− > 0 is such that r− ⩾ ( √ t+ 1 + √ t)2 t2 k2− 4k2+ , for t ∈ ( 0, 1 r−∥A−1∥L(X) ) fixed, • if r− < 0 is such that r− ⩽ − 27k2 − 64k2 + , then, there exists a unique classical solution u, of the transmission problem (2.2)– (2.4) if and only if φ+ 1 , φ − 1 ∈ (D(A), X)1+ 1 2p ,p and φ+ 2 , φ − 2 ∈ (D(A), X)1+ 1 2+ 1 2p ,p . (4.6) 8 A. THOREL EJDE-2024/78 Remark 4.4. Since the third case is the symmetric of the second one, replacing k+ by k− and l+ by l−, the proof if exactly the same. Thus, we omit it. Remark 4.5. If A = A0, to satisfy the first condition set in the second or the third case of Theorem 4.3, since ∥A−1∥L(X) ⩾ 1 Cω , where Cω is the Poincaré constant, it suffices to take ω sufficiently small because the more ω is small, the more Cω is large, see for instance [7, Corollary 2.2, p. 95 and Corollary 2.3, p. 96], or [2, Remark 3, p. 15]. As a consequence of the previous Theorem, we state the following corollary. Corollary 4.6. Let n ⩾ 2, f+ ∈ Lp(Ω+) and f− ∈ Lp(Ω−) with p ∈ (1,+∞) and p > n. Assume that ω is a bounded open set of Rn−1 with C2-boundary. Let k+, k−, l+ > 0 and l− < 0 with k+ = k−. Then, there exists a unique solution u of (Ppde), such that we have u− ∈W 4,p(Ω−) and u+ ∈W 4,p(Ω+), if and only if φ± 1 , φ ± 2 ∈W 2,p(ω) ∩W 1,p 0 (ω) ∆φ± 1 ,∈W 2− 1 p ,p(ω) ∩W 1,p 0 (ω) ∆φ± 2 ∈W 1− 1 p ,p(ω) ∩W 1,p 0 (ω). The proof is quite similar to the one stated in [19, Corollary 1, p. 2941], or in [20, Corollary 2.7, p. 357]. Thus we omit it. 5. Preliminary results In the sequel, we set c = γ − a > 0 and d = b− γ > 0. From Remark 4.2, to solve problem (2.2)–(2.4) we must first study problems (4.4) and (4.5). To this end, we need the following invertibility result obtained in [20] and [34]. Lemma 5.1 ([20] and [34]). Assume that (H1)–(H4) hold. Then operators U±, V± ∈ L(X) defined by U+ := { I − ed(L++M) − 1 r+ (L+ +M)2 ( edM − edL+ ) , if r+ ∈ R \ {0} I − e2dM + 2dMedM , if r+ = 0 V+ := { I − ed(L++M) + 1 r+ (L+ +M)2 ( edM − edL+ ) , if r− ∈ R \ {0} I − e2dM − 2dMedM , if r+ = 0 U− := { I − ec(L−+M) − 1 r− (L− +M)2 ( ecM − ecL− ) , if r− ∈ R \ {0} I − e2cM + 2cMecM , if r− = 0 V− := { I − ec(L−+M) + 1 r− (L− +M)2 ( ecM − ecL− ) , if r− ∈ R \ {0} I − e2cM − 2cMecM , if r− = 0, (5.1) are invertible with bounded inverses. From Remark 4.1, statement 4, U± and V± are well defined. For a detailed proof, see [20, Proposition 5.4 with k = r±] and [34, Proposition 4.5, p. 645]. 5.1. Transmission system. EJDE-2024/78 SOLVABILITY OF TRANSMISSION PROBLEMS 9 5.1.1. First case. Assume that r± ∈ R \ {0}. We set P+ 1 = k+(L+ +M) ( U−1 + (I + edM )(I − edL+) + V −1 + (I − edM )(I + edL+) ) P+ 2 = k+(L+ +M) ( U−1 + (I − edM )(I − edL+) + V −1 + (I + edM )(I + edL+) ) P+ 3 = k+(L+ +M)L+ ( U−1 + (I + edM )(I + edL+) + V −1 + (I − edM )(I − edL+) ) , (5.2) and similarly P− 1 = k−(L− +M) ( U−1 − (I + ecM )(I − ecL−) + V −1 − (I − ecM )(I + ecL−) ) P− 2 = k−(L− +M) ( U−1 − (I − ecM )(I − ecL−) + V −1 − (I + ecM )(I + ecL−) ) P− 3 = k−(L− +M)L− ( U−1 − (I + ecM )(I + ecL−) + V −1 − (I − ecM )(I − ecL−) ) . (5.3) Moreover, we note that S1 = k+(L+ +M) ( U−1 + (I − edL+)φ̃2 + + V −1 + (I + edL+)φ̃4 + ) − k−(L− +M) ( U−1 − (I − ecL−)φ̃2 − + V −1 − (I + ecL−)φ̃4 −) , (5.4) and S2 = −k+(L+ +M) ( U−1 + (I + edM )φ̃1 + + V −1 + (I − edM )φ̃3 + ) − k−(L− +M) ( U−1 − (I + ecM )φ̃1 − + V −1 − (I − ecM )φ̃3 −)− 2M−1R1, (5.5) with R1 = −k+F ′′′ + (γ) + k+M 2F ′ +(γ) + l+F ′ +(γ) + k−F ′′′ − (γ) − k−M 2F ′ −(γ)− l−F ′ −(γ), (5.6) where F+ is the unique classical solution of the problem u (4) + (x) + (2A− r+ I)u ′′ +(x) + (A2 − r+A)u+(x) = f+(x), for a.e. x ∈ (γ, b) u+(γ) = u+(b) = u′′+(γ) = u′′+(b) = 0, (5.7) and F− is the unique classical solution of the problem u (4) − (x) + (2A− r− I)u ′′ −(x) + (A2 − r−A)u−(x) = f−(x), for a.e. x ∈ (a, γ) u−(a) = u−(γ) = u′′−(a) = u′′+(γ) = 0. (5.8) For an explicit representation formula of the solution of both previous problems, we refer to [20, Theorem 2.2, p. 355-356]. . Remark 5.2. Since F± is a classical solution of (5.7), respectively (5.8), from Lemma 3.8, it follows that, for j = 0, 1, 2, 3 and s = a, γ or b F (j) ± (s) ∈ (D(M), X)3−j+ 1 p ,p . Now, with our notations, we recall a useful result of [19, Theorem 4.6, p. 2945]. This result is proved for r± > 0 but clearly remains true for r± < 0. Theorem 5.3 ([19]). Let f− ∈ Lp(a, γ;X) and f+ ∈ Lp(γ, b;X), with p ∈ (1,+∞). Assume that (H1), (H2), (H3) and (H4) hold. Then, the transmission problem 10 A. THOREL EJDE-2024/78 (2.2)–(2.4) has a unique classical solution if and only if the data φ+ 1 , φ − 1 , φ + 2 , φ − 2 satisfy (4.6) andthe system( P− 1 − P+ 1 ) Mψ1 + ( P+ 2 + P− 2 ) ψ2 = S1( P+ 3 + P− 3 ) ψ1 + ( P− 1 − P+ 1 ) ψ2 = S2, (5.9) has a unique solution (ψ1, ψ2) such that (ψ1, ψ2) ∈ (D(A), X)1+ 1 2p ,p × (D(A), X)1+ 1 2+ 1 2p ,p . (5.10) 5.1.2. Second case. Now, assume that r− = 0; then l− = 0 and the transmission conditions (2.6) becomes k−u ′′ −(γ) + k−Au−(γ) = k+u ′′ +(γ) + k+Au+(γ) k−u (3) − (γ) + k−Au ′ −(γ) = k+u (3) + (γ) + k+Au ′ +(γ)− l+u ′ +(γ). (5.11) Our aim here is to establish a similar result to the previous one. To this end, we set Q− 1 = k− ( U−1 − + V −1 − )( I − e2cM ) Q− 2 = k− ( U−1 − ( I − ecM )2 + V −1 − ( I + ecM )2) Q− 3 = k− ( U−1 − ( I + ecM )2 + V −1 − ( I − ecM )2) , (5.12) Moreover, we note that S3 = 2k−M ( U−1 − ( I − ecM ) φ̃2 − + V −1 − ( I + ecM ) φ̃4 −) − k+(L+ +M) ( U−1 + ( I − edL+ ) φ̃2 + + V −1 + ( I + edL+ ) φ̃4 + ) , (5.13) and S4 = −2k−M ( U−1 − ( I + ecM ) φ̃− 2 + V −1 − ( I − ecM ) φ̃− 4 ) − k+(L+ +M) ( U−1 + ( I + edM ) φ̃+ 1 + V −1 + ( I − edM ) φ̃+ 3 ) + 2M−1R2, (5.14) with R2 = −k−F̃ ′′′ − (γ) + k−M 2F̃ ′ −(γ) + k+F ′′′ + (γ)− k+M 2F ′ +(γ)− l+F ′ +(γ), (5.15) where F̃− is the classical solution of the problem u (4) − (x) + 2Au′′−(x) +A2u−(x) = f−(x), a.e. x ∈ (a, γ) u−(a) = u−(γ) = u′′−(a) = u′′−(γ) = 0. (5.16) Remark 5.4. Since F̃− is a classical solution of (5.8), as in Remark 5.2, from Lemma 3.8, it follows that, for j = 0, 1, 2, 3 and s = a, γ or b F̃ (j) − (s) ∈ (D(M), X)3−j+ 1 p ,p . Theorem 5.5. Let f− ∈ Lp(a, γ;X) and f+ ∈ Lp(γ, b;X), with p ∈ (1,+∞). As- sume that (H1)–(H4) hold. Then problem (2.2)–(2.4) has a unique classical solution if and only if the data φ+ 1 , φ − 1 , φ + 2 , φ − 2 satisfy (4.6) and the system( P+ 1 − 2MQ− 1 ) Mψ1 − ( P+ 2 + 2MQ− 2 ) ψ2 = S3( P+ 3 + 2MQ− 3 ) ψ1 + ( 2MQ− 1 − P+ 1 ) ψ2 = S4, (5.17) has a unique solution (ψ1, ψ2) satisfying (5.10). EJDE-2024/78 SOLVABILITY OF TRANSMISSION PROBLEMS 11 Proof. We follow the same steps than the proof of Theorem 4.6, p. 2945 in [19], we only point out the key points. From [20], Theorem 2.5, statement 2, there exists a unique classical solution u+ of (4.4) if and only if φ+ 1 , ψ1 ∈ (D(A), X)1+ 1 2p ,p and φ+ 2 , ψ2 ∈ (D(A), X)1+ 1 2+ 1 2p ,p . (5.18) Recall that, from Remark 3.7, we have (D(A), X)1+ 1 2p ,p = (D(M), X)3+ 1 p ,p , (D(A), X)1+ 1 2+ 1 2p ,p = (D(M), X)2+ 1 p ,p . (5.19) This solution is explicitly given in [19, Proposition 2, p. 2943-2944], from which we deduce that k+ ( u′′+(γ)−M2u+(γ) ) = l+ ( I − edL+ ) α+ 2 + l+ ( I + edL+ ) α+ 4 , and k+ ( u (3) + (γ)−M2u′+(γ) ) − l+u ′ +(γ) = −l+M ( I + edM ) α+ 1 − l+M ( I − edM ) α+ 3 + k+F ′′′ + (γ)− k+M 2F ′ +(γ)− l+F ′ +(γ), where α+ 1 = 1 2r+ (L+ +M)U−1 + [ L+(I + edL+)ψ1 − (I − edL+)ψ2 + φ̃+ 1 ] α+ 2 = − 1 2r+ (L+ +M)U−1 + [ M(I + edM )ψ1 − (I − edM )ψ2 + φ̃+ 2 ] α+ 3 = 1 2r+ (L+ +M)V −1 + [ L+(I − edL+)ψ1 − (I + edL+)ψ2 + φ̃+ 3 ] α+ 4 = − 1 2r+ (L+ +M)V −1 + [ M(I − edM )ψ1 − (I + edM )ψ2 + φ̃+ 4 ] , (5.20) with φ̃1 + = −L+ ( I + edL+ ) φ+ 1 + ( I − edL+ ) ( F ′ +(b) + F ′ +(γ)− φ+ 2 ) φ̃2 + = −M ( I + edM ) φ+ 1 + ( I − edM ) ( F ′ +(b) + F ′ +(γ)− φ+ 2 ) φ̃3 + = L+ ( I − edL+ ) φ+ 1 − ( I + edL+ ) ( F ′ +(b)− F ′ +(γ)− φ+ 2 ) φ̃4 + =M ( I − edM ) φ+ 1 − ( I + edM ) ( F ′ +(b)− F ′ +(γ)− φ+ 2 ) , (5.21) and F+ is the unique classical solution of problem (5.7). In the same way, from [34, Theorem 2.8, statement 2, p. 637], there exists a unique classical solution u− of problem (4.5) if and only if φ− 1 , ψ1 ∈ (D(A), X)1+ 1 2p ,p and φ− 2 , ψ2 ∈ (D(A), X)1+ 1 2+ 1 2p ,p . (5.22) Note that from (5.19), we have φ− 1 , ψ1 ∈ (D(M), X)3+ 1 p ,p and φ− 2 , ψ2 ∈ (D(M), X)2+ 1 p ,p . Moreover, this solution, given in [34, Proposition 4.1, p. 640], is explicitly written in [35, Proposition 4.2], from which it follows that k− ( u′′−(γ)−M2u−(γ) ) = −k− ( 2M ( I − ecM ) α− 2 − 2M ( I + ecM ) α− 4 ) , and k− ( u (3) − (γ)−M2u′−(γ) ) = k− ( 2M2 ( I + ecM ) α− 2 − 2M2 ( I − ecM ) α− 4 ) 12 A. THOREL EJDE-2024/78 + k−F (3) − (γ)− k−M 2F ′ −(γ), where α− 1 := −1 2 U−1 − [( I + (I + cM) ecM ) ψ1 − cecMψ2 + φ̃− 1 ] α− 2 := 1 2 U−1 − [( I + ecM ) Mψ1 + ( I − ecM ) ψ2 + φ̃− 2 ] α− 3 := 1 2 V −1 − [( I − (I + cM) ecM ) ψ1 + cecMψ2 + φ̃− 3 ] α− 4 := −1 2 V −1 − [( I − ecM ) Mψ1 + ( I + ecM ) ψ2 + φ̃− 4 ] , (5.23) with φ̃1 − := − ( I + ecM ) φ− 1 − cecM ( Mφ− 1 + φ− 2 − F̃ ′ −(a)− F̃ ′ −(γ) ) φ̃2 − := −M ( I + ecM ) φ− 1 + ( I − ecM ) ( φ− 2 − F̃ ′ −(a)− F̃ ′ −(γ) ) φ̃3 − := ( I − ecM ) φ− 1 − cecM ( Mφ− 1 + φ− 2 − F̃ ′ −(a) + F̃ ′ −(γ) ) φ̃4 − :=M ( I − ecM ) φ− 1 − ( I + ecM ) ( φ− 2 − F̃ ′ −(a) + F̃ ′ −(γ) ) . (5.24) Note that from (5.18), (5.19), (5.20) and (5.21), respectively to (5.19), (5.22), (5.23) and (5.24), we deduce that α± i ∈ D(M), for i = 1, 2, 3, 4 and α− 2 , α − 4 ∈ D(M2). Thus, system (5.11), given by (5.11), reads −2k−M ( I − ecM ) α− 2 − ( I + ecM ) α− 4 = l+ ( I − edL+ ) α+ 2 + ( I + edL+ ) α+ 4 ) 2k−M 2 ( I + ecM ) α− 2 − ( I − ecM ) α− 4 = −l+M ( I + edM ) α+ 1 + ( I − edM ) α+ 3 +R2, where R2 is given by (5.15). Thus, the previous system gives − 2k−U −1 − M ( I − ecM ) [( I + ecM ) Mψ1 + ( I − ecM ) ψ2 + φ̃− 2 ] − 2k−V −1 − M ( I + ecM ) [( I − ecM ) Mψ1 + ( I + ecM ) ψ2 + φ̃− 4 ] + k+(L+ +M)U−1 + ( I − edL+ ) [ M(I + edM )ψ1 − (I − edM )ψ2 + φ̃+ 2 ] + k+(L+ +M)V −1 + ( I + edL+ ) [ M(I − edM )ψ1 − (I + edM )ψ2 + φ̃+ 4 ] = 0 and 2k−MU−1 − ( I + ecM ) [( I + ecM ) Mψ1 + ( I − ecM ) ψ2 + φ̃− 2 ] + 2k−MV −1 − ( I − ecM ) [( I − ecM ) Mψ1 + ( I + ecM ) ψ2 + φ̃− 4 ] + k+(L+ +M)U−1 + ( I + edM ) [ L+(I + edL+)ψ1 − (I − edL+)ψ2 + φ̃+ 1 ] + k+(L+ +M)V −1 + ( I − edM ) [ L+(I − edL+)ψ1 − (I + edL+)ψ2 + φ̃+ 3 ] = 2M−1R2, Finally, using (5.2), (5.12), (5.13), (5.14) and (5.15), we obtain that the previous system writes as system (5.17). Conversely, if we assume that (4.6) holds and system (5.17) has a unique solution (ψ1, ψ2) satisfying (5.10), then considering u± the unique classical solution of (P±), we obtain that u is the unique classical solution of (2.2)–(2.4). □ EJDE-2024/78 SOLVABILITY OF TRANSMISSION PROBLEMS 13 5.2. Functional calculus. In this section, by using functional calculus, we rewrite the operators defined in (5.1), (5.2), (5.3) and (5.12), to invert the determinant operator of system (5.9) and system (5.17). To this end, we recall some classical notation. For θ ∈ (0, π), we denote by H(Sθ) the space of holomorphic functions on Sθ (defined by (3.1)) with values in C. Moreover, we consider the subspace of H(Sθ): E∞(Sθ) := { f ∈ H(Sθ) : f = O(|z|−s) (|z| → +∞) for some s > 0 } . In other words, E∞(Sθ) is the space of polynomially decreasing holomorphic func- tions at +∞. Let T be an invertible sectorial operator of angle θT ∈ (0, π). If f ∈ E∞(Sθ), with θ ∈ (θT , π), then we can define, by functional calculus, f(T ) ∈ L(X), see [17, p. 45]. Then, we recall a useful result from [20, Lemma 5.3, p. 370]. Lemma 5.6 ([20]). Let P be an invertible sectorial operator in X with angle θ, for all θ ∈ (0, π). Let G ∈ H(Sθ), for some θ ∈ (0, π), such that (i) 1−G ∈ E∞(Sθ), (ii) G(x) ̸= 0 for any x ∈ R+ \ {0}. Then, G(P ) ∈ L(X), is invertible with bounded inverse. Let r ∈ R, rm = max(−r, 0), δ > 0 and z ∈ C \ (−∞, rm]. We set uδ,r(z) = { 1− e−δ( √ z+r+ √ z) − 1 r ( √ z + r + √ z)2 ( e−δ √ z − e−δ √ z+r ) , if r ∈ R \ {0} 1− e−2δ √ z − 2δ √ ze−δ √ z, if r = 0 vδ,r(z) = { 1− e−δ( √ z+r+ √ z) + 1 r ( √ z + r + √ z)2 ( e−δ √ z − e−δ √ z+r ) , if r ∈ R \ {0} 1− e−2δ √ z + 2δ √ ze−δ √ z, if r = 0. When uδ,r(z) ̸= 0, vδ,r(z) ̸= 0, we note that fδ,r,1(z) =  (√ z + r + √ z )√ z + ru−1 δ,r (z) ( 1 + e−δ √ z )( 1 + e−δ √ z+r ) + (√ z + r + √ z )√ z + rv−1 d,r(z) ( 1− e−δ √ z )( 1− e−δ √ z+r ) , if r ∈ R \ {0}( u−1 δ,0(z) + v−1 δ,0 (z) )( 1− e−2δ √ z ) , if r = 0, fδ,r,2(z) =  − (√ z + r + √ z ) u−1 δ,r (z) ( 1 + e−δ √ z )( 1− e−δ √ z+r ) − (√ z + r + √ z ) v−1 δ,r (z) ( 1− e−δ √ z )( 1 + e−δ √ z+r ) , if r ∈ R \ {0} u−1 δ,0(z) ( 1− e−δ √ z )2 + v−1 δ,0 (z) ( 1 + e−δ √ z )2 , if r = 0, fδ,r,3(z) =  − (√ z + r + √ z ) u−1 δ,r (z) ( 1− e−δ √ z )( 1− e−δ √ z+r ) − (√ z + r + √ z ) v−1 δ,r (z) ( 1 + e−δ √ z )( 1 + e−δ √ z+r ) , if r ∈ R \ {0} u−1 δ,0(z) ( 1 + e−δ √ z )2 + v−1 δ,0 (z) ( 1− e−δ √ z )2 , if r = 0 14 A. THOREL EJDE-2024/78 Remark 5.7. Note that, from (H2) and (H3), if r± ̸= 0, we have uc,r−(−A) = U−, ud,r+(−A) = U+, vc,r−(−A) = V−, vd,r+(−A) = V+, k−fc,r−,1(−A) = P− 1 , k+fd,r+,1(−A) = P+ 1 , k−fc,r−,2(−A) = P− 2 , k+fd,r+,2(−A) = P+ 2 , k−fc,r−,3(−A) = P− 3 , k+fd,r+,3(−A) = P+ 3 , and if r− = 0, we obtain uc,0(−A) = U−, vc,0(−A) = V−, k−fc,0,1(−A) = Q− 1 , k−fc,0,2(−A) = Q− 2 , k−fc,0,3(−A) = Q− 3 . Remark 5.8. Let δ > 0, r ∈ R and x ∈ (rm,+∞). Then, when r = 0, we have 1− e−2δ √ x ± 2δ √ xe−δ √ x = 2e−δ √ x ( sinh(δ √ x)± δ √ x ) > 0, and from [20, Lemma 5.2, p. 369] it clear that uδ,r(x) > 0 and vδ,r(x) > 0. Thus, when r ̸= 0, we deduce that fδ,r,1(x) > 0 and fδ,r,2(x), fδ,r,3(x) < 0, and when r = 0, we obtain fδ,0,1(x), fδ,0,2(x), fδ,0,3(x) > 0. Moreover, for z ∈ C \ (−∞, rm] and r ∈ R \ {0}, we define gδ,r(z) =− √ z + r (( 1− e−2δ( √ z+r+ √ z) )2 − 1 r2 ( √ z + r + √ z)4 ( e−2δ √ z − e−2δ √ z+r )2 ) + √ z (( 1− e−δ( √ z+r+ √ z) )2 + 1 r ( √ z + r + √ z)2 ( e−δ √ z − e−δ √ z+r )2 )2 , and for r = 0, we set gδ,0(z) = ( 1 + √ z ) ( 1− e−2δ √ z )4 + 4 ( 1− e−2δ √ z )2 e−2δ √ z − 16 δ2z e−4δ √ z. Lemma 5.9. Let δ > 0 and x ∈ (rm,+∞). Thus, if r ∈ R \ {0}, then gδ,r(x) < 0 and if r = 0, then gδ,0(x) > 0. Proof. Let δ > 0 and r ∈ R \ {0}. For all x ∈ (rm,+∞), from [19], Lemma 4.4, p. 2950, we obtain the result. Now, consider that r = 0. Then gδ,0(x) = (1 + √ x) ( 1− e−2δ √ x )4 + 4e−2δ √ x (( 1− e−2δ √ x )2 − 4 δ2x e−2δ √ x ) = (1 + √ x) ( 1− e−2δ √ x )4 + 4e−2δ √ x ( 1− e−2δ √ x − 2δ √ x e−δ √ x )( 1− e−2δ √ x + 2 δ √ x e−δ √ x ) . Since δ, x > 0, we have 1− e−2δ √ x − 2δ √ x e−δ √ x = e−δ √ x ( eδ √ x − e−δ √ x − 2δ √ x ) = 2e−δ √ x ( sinh(δ √ x)− δ √ x ) > 0. Finally, we deduce that gδ,0 > 0. □ EJDE-2024/78 SOLVABILITY OF TRANSMISSION PROBLEMS 15 6. Proof of the main results In both cases, assume that problem (2.2)–(2.4) has a unique classical solution; thus, from Theorem refTh syst trans, respectively Theorem refTh syst trans M, (4.6) holds. Conversely, assume that (4.6) holds, then due to Theorem refTh syst trans, respectively Theorem refTh syst trans M, we have to prove that system (5.9), respectively system (5.17), has a unique solution such that (5.10) holds. The proof is divided in three parts for both cases. First, we will make explicit, in the first case, the determinant of system (5.9) and in the second case, the deter- minant of system (5.17). Then, in the two cases, we will show the uniqueness of the solution. To this end, we will invert the determinant thanks to functional calculus. Finally, we will prove, in all cases, that ψ1 and ψ2 have the expected regularity. 6.1. Calculus of the determinant. 6.1.1. First case. Here, we consider r+, r− ∈ R\{0}. We have to make explicit the determinant of system (5.9) that we recall here( P− 1 − P+ 1 ) Mψ1 + ( P+ 2 + P− 2 ) ψ2 = S1( P+ 3 + P− 3 ) ψ1 + ( P− 1 − P+ 1 ) ψ2 = S2. We write the previous system as a matrix equation Λ1Ψ = S, where Λ1 = (( P− 1 − P+ 1 ) M ( P+ 2 + P− 2 )( P+ 3 + P− 3 ) ( P− 1 − P+ 1 )) , Ψ = ( ψ1 ψ2 ) , S = ( S1 S2 ) . To solve system (5.9), we will study the determinant det(Λ1) :=M ( P− 1 − P+ 1 )2 − ( P+ 2 + P− 2 ) ( P+ 3 + P− 3 ) , of the matrix Λ1. Thus, we set det(Λ1) = D+ 1 +D− 1 +D2, (6.1) where D+ 1 =M ( P+ 1 )2 − P+ 3 P + 2 D− 1 =M ( P− 1 )2 − P− 3 P − 2 D2 = −P+ 3 P − 2 − P− 3 P + 2 − 2MP+ 1 P − 1 . Then, we recall the result [19, Lemma 5.1, p. 2953] describing the determinant. Lemma 6.1 ([19]). We have (1) D+ 1 = −4k2+(L+ +M)2U−2 + V −2 + D+, with D+ = L+ (( I − e2d(L++M) )2 − 1 r2+ (L+ +M)4 ( e2dM − e2dL+ )2) −M (( I − ed(L++M) )2 + 1 r2+ (L+ +M)2 ( edM − edL+ )2)2 . (2) D− 1 = −4k2−(L− +M)2U−2 − V −2 − D−, with D− = L− (( I − e2c(L−+M) )2 − 1 r2− (L− +M)4 ( e2cM − e2cL− )2) −M (( I − ec(L−+M) )2 + 1 r2− (L− +M)2 ( ecM − ecL− )2)2 . 16 A. THOREL EJDE-2024/78 6.1.2. Second case. Here, we consider r+ ∈ R \ {0} and r− = 0. As previously, we make explicit the determinant of system (5.17) that we recall here( P+ 1 − 2MQ− 1 ) Mψ1 − ( P+ 2 + 2MQ− 2 ) ψ2 = S3( P+ 3 + 2MQ− 3 ) ψ1 + ( 2MQ− 1 − P+ 1 ) ψ2 = S4, We write this system as a matrix equation Λ2Ψ = S̃, where Λ2 = (( P+ 1 − 2MQ− 1 ) M − ( P+ 2 + 2MQ− 2 )( P+ 3 + 2MQ− 3 ) ( 2MQ− 1 − P+ 1 ) ) , Ψ = ( ψ1 ψ2 ) , S̃ = ( S3 S4 ) . To solve system (5.9), we will study the determinant det(Λ2) := −M ( P+ 1 − 2MQ− 1 )2 + ( P+ 3 + 2MQ− 3 ) ( P+ 2 + 2MQ− 2 ) , of the matrix Λ2. We set det(Λ2) = D+ 3 +D− 3 +D4, (6.2) where D+ 3 = P+ 2 P + 3 −M ( P+ 1 )2 and D− 3 = 4M2 ( Q− 2 Q − 3 −M ( Q− 1 )2 ) , with D4 = 2M ( P+ 3 Q − 2 + P+ 2 Q − 3 +MP+ 1 Q − 1 ) . Lemma 6.2. We have (1) D+ 3 = k2+(L+ +M)2U−2 + V −2 + D+ 0 , with D+ 0 = L+ (( I − e2d(L++M) )2 − 1 r2+ (L+ +M)4 ( e2dM − e2dL+ )2) −M (( I − ed(L++M) )2 + 1 r2+ (L+ +M)2 ( edM − edL+ )2)2 . (2) D− 3 = 16k2−M 2U−2 − V −2 − D− 0 , with D− 0 = (I −M) ( I − e2cM )4 + 4 ( I − e2cM )2 e2cM − 16c2M2e4cM . Proof. (1) We have P+ 2 P + 3 = k2+(L+ +M)2L+U −2 + V −2 + D′ +, where D′ + = ( U2 + + V 2 + ) (I − e2dM )(I − e2dL+) + U+V+ ( (I + edM )2(I + edL+)2 + (I − edM )2(I − edL+)2 ) = ( U2 + + V 2 + ) [( I + ed(L++M) )2 − ( edM + edL+ )2] + 2U+V+ [( I + ed(L++M) )2 + ( edM + edL+ )2] = (U+ + V+) 2 ( I + ed(L++M) )2 − (V+ − U+) 2 ( edM + edL+ )2 . Moreover, from (5.1), we obtain that U+ + V+ = 2 ( I − ed(L++M) ) , V+ − U+ = 2 r+ (L+ +M)2 ( edM − edL+ ) . (6.3) EJDE-2024/78 SOLVABILITY OF TRANSMISSION PROBLEMS 17 Then D′ + = 4 ( I − ed(L++M) )2 ( I + ed(L++M) )2 − 4 r2+ (L+ +M)4 ( edM − edL+ )2 ( edM + edL+ )2 = 4 ( I − e2d(L++M) )2 − 4 r2+ (L+ +M)4 ( e2dM − e2dL+ )2 . Furthermore, M ( P+ 1 )2 = k2+(L+ +M)2MU−2 + V −2 + D′′ +, where D′′ + = ( V+(I + edM )(I − edL+) + U+(I − edM )(I + edL+) )2 = [ (U+ + V+) ( I − ed(L++M) ) + (V+ − U+) ( edM − edL+ )]2 , and from (6.3), it follows that D′′ + = [ 2 ( I − ed(L++M) )2 + 2 r+ (L+ +M)2 ( edM − edL+ )2]2 . Finally, we deduce that D+ 3 = P+ 2 P + 3 −M ( P+ 1 )2 = k2+(L+ +M)2U−2 + V −2 + ( L+D ′ + −MD′′ + ) , and setting D+ 0 = L+D ′ + −MD′′ +, we obtain the expected result. (2) We have Q− 2 Q − 3 = k2−U −2 − V −2 − D′ −, where D′ − = ( V− ( I − ecM )2 + U− ( I + ecM )2)( V− ( I + ecM )2 + U− ( I − ecM )2) = ( U2 − + V 2 − ) ( I − e2cM )2 + 2U−V− ( I − e2cM )2 + 16U−V−e 2cM , and M ( Q− 1 )2 = k2−MU−2 − V −2 − (U− + V−) 2 ( I − e2cM )2 = k2−U −2 − V −2 − [ M ( U2 − + V 2 − ) ( I − e2cM )2 + 2MU−V− ( I − e2cM )2] . Thus Q− 2 Q − 3 −M ( Q− 1 )2 = k2−U −2 − V −2 − D′′ −, where D′′ − = (I −M) ( U2 − + V 2 − ) ( I − e2cM )2 + 2(I −M)U−V− ( I − e2cM )2 + 16U−V−e 2cM = (I −M) (U− + V−) 2 ( I − e2cM )2 + 16U−V−e 2cM . Moreover, from (5.1), we obtain that U− + V− = 2 ( I − e2cM ) and U−V− = ( I − e2cM )2 − 4c2M2e2cM . Then D′′ − = 4(I −M) ( I − e2cM )4 + 16 ( I − e2cM )2 e2cM − 64c2M2e4cM . 18 A. THOREL EJDE-2024/78 Therefore, D− 3 = 4M2 ( Q− 2 Q − 3 −M ( Q− 1 )2) = 16k2−M 2U−2 − V −2 − D− 0 , where D− 0 = 1 4 D ′′ −. □ 6.2. Inversion of the determinant. 6.2.1. First case. Here, we consider r+, r− ∈ R \ {0}. Let r = max(−r+,−r−, 0) ⩾ 0. By using functional calculus, we prove that the determinant of system (5.9), given by (6.1), is invertible with bounded inverse. From Lemma 6.1 and the definition of D2, we obtain: D+ 1 = g+1 (−A), D− 1 = g−1 (−A) and D2 = g2(−A), where, for z ∈ C \ R−, we have set g+1 (z) = 4k2+( √ z + r+ + √ z)2u−2 d,r+ (z)v−2 d,r+ (z)gd,r+(z) g−1 (z) = 4k2−( √ z + r− + √ z)2u−2 c,r−(z)v −2 c,r−(z)gc,r−(z) g2(z) = k+fd,r+,1(z)k−fc,r−,3(z) + k−fc,r−,1(z)k+fd,r+,3(z) − 2 √ z k+fd,r+,2(z)k−fc,r−,2(z), with uδ,r, vδ,r, gδ,r and fδ,r,i the complex functions defined in section 5.2. Thus det(Λ1) = D+ 1 +D− 1 +D2 = f1(−A), (6.4) with f1 = g+1 + g−1 + g2. Note that, for some θ ∈ (0, π), we have f1 ∈ H(Sθ) and because of Remark 5.8 and Lemma 5.9, for x > 0, we have f1(x) = g+1 (x) + g−1 (x) + g2(x) < 0. (6.5) Let C1, C2 be two linear operators in X. We denote by C1 ∼ C2 the equality C1 = C2 + Σ, where Σ is a finite sum of terms of type kLl +L m −M neαL+eβL−eδM , where k ∈ R; l,m, n ∈ N; α, β, δ ∈ R+ with α+ β+ δ ̸= 0. Note that Σ is a regular term in the sense Σ ∈ L(X) with Σ(X) ⊂ D(M∞) := ∩k⩾0D(Mk). Since U± ∼ I, V± ∼ I, by setting W = U−U+V−V+ ∼ I we deduce that WP+ 1 ∼ 2k+(L+ +M), WP− 1 ∼ 2k−(L− +M) WP+ 2 ∼ 2k+(L+ +M), WP− 2 ∼ 2k−(L− +M) WP+ 3 ∼ 2k+(L+ +M)L+, WP− 3 ∼ 2k−(L− +M)L−. Thus W 2 det(Λ1) =M ( WP+ 1 )2 − ( WP+ 2 WP+ 3 ) +M ( WP− 1 )2 − ( WP− 2 WP− 3 ) − ( WP− 2 WP+ 3 +WP+ 2 WP− 3 + 2MWP+ 1 WP− 1 ) ∼ −4k2+(L+ +M)2(L+ −M)− 4k2−(L− +M)2(L− −M) − 4k+k−(L+ +M)(L− +M)(L+ + L− + 2M). From (4.3), we have −W 2 det(Λ1) ∼ 4k2+r+(L+ +M) + 4k2−r−(L− +M) + 4k+k−(L+ +M)(L− +M)(L+ + L− + 2M) ∼ 4k+l+(L+ +M) + 4k−l−(L− +M) EJDE-2024/78 SOLVABILITY OF TRANSMISSION PROBLEMS 19 + 4k+k−(L+ +M)(L− +M)(L+ + L− + 2M) Hence, we note that B1 = 4k+l+(L++M)+4k−l−(L−+M)+4k+k−(L++M)(L−+M)(L++L−+2M). Thus, we obtain det(Λ1) = −W−2 ( B1 + ∑ j∈J kjL lj +L mj − MnjeαjL+eβjL−eδjM ) , (6.6) where J is a finite set and for each j ∈ J : kj ∈ R, lj ,mj , nj ∈ N, αj , βj , δj ∈ R+ with αj + βj + δj ̸= 0. We set B2 = I + l+ k− (L− +M)−1(L+ +L− +2M)−1 + l− k+ (L+ +M)−1(L+ +L− +2M)−1 such that B1 = 4k+k−(L+ +M)(L− +M)(L+ + L− + 2M)B2. Proposition 6.3. Assume that (H1)–(H4) hold and k+k− > 0. If one of the following assumptions holds • l+ k− > 0 and l− k+ > 0, • l+ k− < 0 and l− k+ < 0, such that (l+ − l−)(k+ − k−) ⩾ 0, (6.7) • l+ k− > 0 and l− k+ < 0, such that −6l−k+ + l+k+ + l−k− ⩾ 0, (6.8) • l+ k− < 0 and l− k+ > 0, such that −6l+k− + l+k+ + l−k− ⩾ 0, (6.9) then, b2(x) > 0, for x > r ⩾ 0 and the operator B1, defined above, is invertible with bounded inverse. Remark 6.4. Since k+k− > 0, we have the following equivalences l+ k− > 0 ⇐⇒ r+ > 0 and l− k+ > 0 ⇐⇒ r− > 0. Proof. From (H2) and (H3), since k+k− ̸= 0, it is clear that 0 ∈ ρ (4k+k−(L+ +M)(L− +M)(L+ + L− + 2M)) . Thus, it remains to prove that 0 ∈ ρ(B2). To this end, we use Lemma 5.6. Let z ∈ C \ (−∞, r]. We set b2(z) = 1 + l+ k− 1 ( √ z + r− + √ z)( √ z + r+ + √ z + r− + 2 √ z) + l− k+ 1 ( √ z + r+ + √ z)( √ z + r+ + √ z + r− + 2 √ z) , (6.10) hence b2(−A) = B2. Then, for all x > r ⩾ 0, it follows that b2(x) = 1 + l+ k− 1 ( √ x+ r− + √ x)( √ x+ r+ + √ x+ r− + 2 √ x) + l− k+ 1 ( √ x+ r+ + √ x)( √ x+ r+ + √ x+ r− + 2 √ x) 20 A. THOREL EJDE-2024/78 Our aim is to prove that b2(x) > 0, for all x > r. To this end, we set y = x− r > 0, hence b2(y + r) = 1 + l+ k− 1 ( √ y + r + r− + √ y + r)( √ y + r + r+ + √ y + r + r− + 2 √ y + r) + l− k+ 1 ( √ y + r + r+ + √ y + r)( √ y + r + r+ + √ y + r + r− + 2 √ y + r) = 1 + 1 ( √ y + r + r+ + √ y + r + r− + 2 √ y + r) b3(y), where b3(y) = l+ k− ( √ y + r + r− + √ y + r) + l− k+ ( √ y + r + r+ + √ y + r) . Then b′3(y) = − l+ k− ( 1 2 √ y+r+r− + 1 2 √ y+r ) ( √ y + r + r− + √ y + r)2 + − l− k+ ( 1 2 √ y+r+r+ + 1 2 √ y+r ) ( √ y + r + r+ + √ y + r)2 and b′2(y + r) = − ( 1 2 √ y+r+r+ + 1 2 √ y+r+r− + 1√ y+r ) ( √ y + r + r+ + √ y + r + r− + 2 √ y + r)2 b3(y) + 1 ( √ y + r + r+ + √ y + r + r− + 2 √ y + r) b′3(y), Now, we have to study the following fourth cases. (1) If l+ k− > 0 and l− k+ > 0, then it is clear that b3 > 0 and b2 > 0. (2) If l+ k− < 0 and l− k+ < 0, then b′3 > 0 and b′2 > 0. Thus b2(y+ r) > b2(r) where b2(r) = 1 + 1 ( √ r + r+ + 2 √ r) b3(0) > 1 + 1 2 √ r b3(0), with b3(0) = l+ k−√ r + r− + √ r + l− k+√ r + r+ + √ r . Since √ r + r+ > 0 and √ r + r− > 0, it follows that l+ k−√ r + r− + √ r > l+ k−√ r and l− k+√ r + r+ + √ r > l− k+√ r , hence b3(0) > 1√ r ( l+ k− + l− k+ ) . Thus, we obtain b2(r) > 1 + 1 2r ( l+ k− + l− k+ ) . Moreover, we have 1 + 1 2r ( l+ k− + l− k+ ) ⩾ 0 ⇐⇒ 2r ⩾ − ( l+ k− + l− k+ ) , EJDE-2024/78 SOLVABILITY OF TRANSMISSION PROBLEMS 21 where − ( l+ k− + l− k+ ) = −r− ( l+ l− + k− k+ ) , if r = −r− −r+ ( k+ k− + l− l+ ) , if r = −r+. Thus, we obtain that 2r ⩾ − ( l+ k− + l− k+ ) ⇐⇒ { l+k+ + l−k− − 2l−k+ ⩾ 0, if r = −r− l+k+ + l−k− − 2l+k− ⩾ 0, if r = −r+. Furthermore, since k+k− > 0, if r = −r−, it follows that − l− k− ⩾ − l+ k+ , hence −l−k+ ⩾ −l+k− and if r = −r+, then − l+ k+ ⩾ − l− k− , hence −l+k− ⩾ −l−k+. It follows that l+k+ + l−k− − 2l−k+ ⩾ l+k+ + l−k− − l+k− − l−k+, if r = −r− l+k+ + l−k− − 2l+k− ⩾ l+k+ + l−k− − l+k− − l−k+, if r = −r+. Finally, if (6.7) holds, then we obtain b2 > 0. (3) If l+ k− > 0 and l− k+ < 0, then since k+k− > 0, we have l+ k− > 0 ⇐⇒ l+ k+ k+ k− k2− > 0 ⇐⇒ r+ > 0 and l− k+ < 0 ⇐⇒ l− k− k− k+ k2+ < 0 ⇐⇒ r− < 0. Thus r = −r− > 0, r+ > 0 and b3(y) = l+ k− ( √ y + √ y + r) + l− k+ ( √ y + r + r+ + √ y + r) . Since l+ k− > 0 and √ y < √ y + r it follows that l+ k−√ y + √ y + r > l+ k− 2 √ y + r . In the same way, since l− k+ < 0 and √ y + r + r+ > √ y + r, we deduce that l− k+√ y + r + r+ + √ y + r > l− k+ 2 √ y + r , hence b3(y) > 1 2 √ y + r ( l+ k− + l− k+ ) . If l+ k− + l− k+ > 0, then b3 > 0 and b2 > 0. If l+ k− + l− k+ < 0, then we have b2(y + r) = 1 + 1 ( √ y+r+r++ √ y+2 √ y+r) b3(y) > 1 + 1 3 √ y+r b3(y) > 1 + 1 6(y+r) ( l+ k− + l− k+ ) . Moreover, we have 1 + 1 6(y + r) ( l+ k− + l− k+ ) ⩾ 0 ⇐⇒ 6(y + r) + l+ k− + l− k+ ⩾ 0. 22 A. THOREL EJDE-2024/78 It is obvious that 6(y + r) + ( l+ k− + l− k+ ) ⩾ 6r + l+ k− + l− k+ , thus, since k+k− > 0 and here r = −r− = − l− k− , we deduce that the previous inequality becomes −6 l− k− + l+ k− + l− k+ ⩾ 0 ⇐⇒ −6l−k+ + l+k+ + l−k− ⩾ 0. Finally, since k+k− > 0, if (6.8) holds, then b2 > 0. (4) If l+ k− < 0 and l− k+ > 0, then here r = −r+ and in the same way than previously, if (6.9) holds, then b2 > 0. Since r = max(−r+,−r−, 0) ⩾ 0, from (H2) and (H3), we deduce that operator −A− rI ∈ BIP(X, θA) with 0 ∈ ρ(−A− rI). Thus, considering b̃2(z) = b2(z + r), with z + r ∈ C \ R−, it follows that b̃2(−A − rI) = B2. Moreover, for a given θ ∈ (0, π), it is clear that 1 − b2, 1 − b̃2 ∈ E∞. Finally, applying Lemma 5.6 with G = b̃2 and P = −A− rI, we deduce the result. □ From (6.6) and Proposition 6.3, it follows that det(Λ1) = −W−2B1F1, (6.11) where F1 = I + ∑ j∈J kjB −1 1 L lj +L mj − MnjeαjL+eβjL−eδjM . (6.12) For z ∈ C \ (−∞, r], we set b1(z) = −4k+k−( √ z + r+ + √ z)( √ z + r− + √ z) (√ z + r+ + √ z + r− + 2 √ z ) b2(z), (6.13) where b2 is given by (6.10) and f̃1(z) = 1 + ∑ j∈J kjb1(z) −1 ( − √ z + r+ )lj (−√ z + r− )mj (−√ z )nj × e−αj √ z+r+e−βj √ z+r−e−δj √ z. Then, from (H2) and (H3), we have B1 = b1(−A) and F1 = f̃1(−A). Moreover, from (6.4) and (6.11), we obtain f1(−A) = det(Λ1) = −W−2B1f̃1(−A). Note that f1(z) = −u−2 d,r+ (z)v−2 d,r+ (z)u−2 c,r−(z)v −2 c,r−(z)b1(z)f̃1(z). (6.14) Proposition 6.5. Assume that (H1)–(H4) hold and k+k− > 0. Thus • if r+ > 0 and r− > 0, • if r+ < 0 and r− < 0, such that (6.7) holds, • if r+ > 0 and r− < 0, such that (6.8) holds, • if r+ < 0 and r− > 0, such that (6.9) holds. Then F1 ∈ L(X), given by (6.12), is invertible with bounded inverse. EJDE-2024/78 SOLVABILITY OF TRANSMISSION PROBLEMS 23 Proof. For a given θ ∈ (0, π), we have f1, f̃1 ∈ H(Sθ). Moreover, for z ∈ C\(−∞, r], since kjb −1 1 (z) ( − √ z + r+ )lj (−√ z + r− )mj (−√ z )nj , for all j ∈ J, are polynomial functions, we deduce that 1− f̃1 ∈ E∞(Sθ). From (6.5), Proposition 6.3 and Remark 6.4, we know that f1 < 0 and b2 > 0 on (r,+∞). Since ud,r+ , uc,r− , vd,r+ , vc,r− > 0 on (r,+∞) and due to (6.13) and (6.14), we deduce that f̃1 < 0 on (r,+∞). Therefore, noting that ˜fr,1(z) = f̃1(z+ r) and applying Lemma 5.6 with G = f̃1 and operator P = −A− rI, we deduce that operator F1 = ˜fr,1(−A− rI) = f̃1(−A) is invertible with bounded inverse. □ This result finally leads us to state the following main result of this section. Proposition 6.6. Assume that (H1)–(H4) hold and k+k− > 0. Thus • if r+ > 0 and r− > 0, • if r+ < 0 and r− < 0, such that (6.7) holds, • if r+ > 0 and r− < 0, such that (6.8) holds, • if r+ < 0 and r− > 0, such that (6.9) holds. Then det(Λ1) is invertible with bounded inverse. Proof. From (6.11), Propositions 6.3 and 6.5, it follows that det(Λ1) = −W−2B1F1, is invertible with bounded inverse. □ 6.2.2. Second case. Let r+ ∈ R\{0} and r− = 0. In the same way than previously, using functional calculus, we prove that the determinant of system (5.17), given by (6.2), is invertible with bounded inverse. Due to Lemma 6.2, and the definition of D4, we obtain: D+ 3 = g+3 (−A), D− 3 = g−3 (−A) and D4 = g4(−A), where, for z ∈ C \ R−, we have set g+3 (z) = 4k2+( √ z + r+ + √ z)2u−2 d,r+ (z)v−2 d,r+ (z)gd,r+(z) g−3 (z) = 16k2− z u −2 c,0(z)v −2 c,0(z)gc,0(z) g4(z) = −2 √ z ( k+fd,r+,3(z)k−fc,0,2(z) + k+fd,r+,2(z)k−fc,0,3(z) ) + 2zk+fd,r+,1(z)k−fc,0,1(z), with uδ,r, vδ,r, gδ,r and fδ,r,i the complex functions defined in section 5.2. Thus det(Λ2) = D+ 3 +D− 3 +D4 = f2(−A), (6.15) with f2 = g+3 + g−3 + g4. Note that, for some θ ∈ (0, π), we have f2 ∈ H(Sθ) and by Remark 5.8 and Lemma 5.9, for x > max(−r+, 0), we have f2(x) = g+3 (x) + g−3 (x) + g4(x), (6.16) where g+3 < 0 and g−3 , g4 > 0. Lemma 6.7. Let k+k− > 0. Then • if r+ > 0 such that r+ ⩾ (√ t+ 1 + √ t )2 t2 k2+ 4k2− , for t > 0 fixed. (6.17) 24 A. THOREL EJDE-2024/78 for all x ⩾ tr+, we have f2(x) > 0. • if r+ < 0 such that r+ ⩽ − 27k2+ 64k2− , (6.18) for all x ⩾ −r+, we have f2(x) > 0. Proof. From (6.16), we deduce f2(x) ⩾ g+3 (x) + g4(x) ⩾ g+3 (x) + 2k+k−xfd,r+,1(x)fc,0,1(x). Let r = max(−r+, 0). For x ∈ (r,+∞), setting y = x− r > 0 and noting h1(y) = g+3 (y + r) + 2k+k− (y + r) fd,r+,1(y + r)fc,0,1(y + r), it follows that h1(y) = 4k2+ ( √ y + r + r+ + √ y + r)2 u2d,r+(y + r)v2d,r+(y + r) gd,r+(y + r) + 2k+k−(y + r)fd,r+,1(y + r)fc,0,1(y + r). Since 0 > gd,r+(y + r) ⩾ − √ y + r + r+ ( 1− e−2d( √ y+r+r++ √ y+r) )2 , we have h1(y) ⩾ 4 ( √ y + r + r+ + √ y + r) √ y + r + r+ u2d,r+(y + r)v2d,r+(y + r)uc,0(y + r)vc,0(y + r) h2(y), where h2(y) = −k2+( √ y + r + r+ + √ y + r) ( 1− e−2d( √ y+r+r++ √ y+r) )2 × uc,0(y + r)vc,0(y + r) + k+k− (y + r) ( 1− e−2c √ y+r )2 vd,r+(y + r) × ( 1 + e−d √ y+r )( 1 + e−d √ y+r+r+ ) + k+k−(y + r) ( 1− e−2c √ y+r )2 ud,r+(y + r) ( 1− e−d √ y+r )( 1− e−d √ y+r+r+ ) ⩾ −k2+( √ y + r + r+ + √ y + r) ( 1− e−2d( √ y+r+r++ √ y+r) )2( 1− e−2c √ y+r )2 + k+k−(y + r) ( 1− e−2c √ y+r )2 h3(y), with h3(y) = vd,r+(y + r) ( 1 + e−d √ y+r )( 1 + e−d √ y+r+r+ ) + ud,r+(y + r) ( 1− e−d √ y+r )( 1− e−d √ y+r+r+ ) and h3(y) = 2vd,r+(y + r) ( 1 + e−d( √ y+r+r++ √ y+r) ) + 2ud,r+(y + r) ( e−d √ y+r + e−d √ y+r+r+ ) = 2 ( 1− e−d( √ y+r+r++ √ y+r) )( 1 + e−d( √ y+r+r++ √ y+r) ) + 2 ( √ y + r + r+ + √ y + r)2 r+ ( e−d √ y+r − e−d √ y+r+r+ )( e−d √ y+r + e−d √ y+r+r+ ) EJDE-2024/78 SOLVABILITY OF TRANSMISSION PROBLEMS 25 = 2 ( 1− e−2d( √ y+r+r++ √ y+r) ) + 2 ( √ y + r + r+ + √ y + r)2 r+ ( e−2d √ y+r − e−2d √ y+r+r+ ) . Moreover, for all y > 0, since e−2d √ y+r − e−2d √ y+r+r+ r+ > 0, for r+ ∈ R \ {0}, we deduce that h3(y) > 2 ( 1− e−2d( √ y+r+r++ √ y+r) ) and h2(y) > −k2+( √ y + r + r+ + √ y + r) ( 1− e−2d( √ y+r+r++ √ y+r) )2( 1− e−2c √ y+r )2 + 2k+k− (y + r) ( 1− e−2c √ y+r )2( 1− e−2d( √ y+r+r++ √ y+r) ) > −k2+( √ y + r + r+ + √ y + r) ( 1− e−2d( √ y+r+r++ √ y+r) )( 1− e−2c √ y+r )2 + 2k+k−(y + r) ( 1− e−2d( √ y+r+r++ √ y+r) )( 1− e−2c √ y+r )2 > ( 1− e−2d( √ y+r+r++ √ y+r) )( 1− e−2c √ y+r )2 h4(y), where h4(y) = 2k+k− (y + r)− k2+( √ y + r + r+ + √ y + r). Thus h4(y) ⩾ 0 ⇐⇒ y + r√ y + r + r+ + √ y + r ⩾ k2+ 2k+k− . (6.19) We set h5(y) = y + r√ y + r + r+ + √ y + r , (6.20) hence h′5(y) = ( 1√ y + r + r+ + √ y + r )( 1− 1 2 √ y + r y + r + r+ ) . (1) If r+ < 0, then r = −r+ and y + r y + r + r+ = y + r y , moreover h′5(y) ⩾ 0 ⇐⇒ 1− 1 2 √ y + r y ⩾ 0 ⇐⇒ 4 ⩾ y + r y ⇐⇒ y ⩾ r 3 . Thus, h5(y) ⩾ h5 (r 3 ) = 4r 3√ r 3 + 2 √ r 3 = 4 3 √ r 3 > 0. This yields that for all y ⩾ 0, we have h5(y) ⩾ 4 3 √ r 3 > 0. Therefore, from (6.19) and (6.20), we deduce that h4(y) ⩾ 0 ⇐⇒ h5(y) ⩾ k2+ 2k+k− ⇐⇒ 4 3 √ r 3 ⩾ k+ 2k− , 26 A. THOREL EJDE-2024/78 hence, since r = −r+ > 0, we obtain 4 3 √ r 3 ⩾ k+ 2k− ⇐⇒ √ r 3 ⩾ 3k+ 8k− ⇐⇒ r 3 ⩾ 9k2+ 64k2− ⇐⇒ −r+ ⩾ 27k2+ 64k2− . (2) If r+ > 0, then r = 0 and y + r y + r + r+ = y y + r+ < 1, hence h′5 > 0 and h5 is an increasing function. Thus, from (6.20), since r = 0, it follows that h5(y) = y√ y + r+ + √ y , then h5(y) ⩾ k2+ 2k+k− ⇐⇒ y√ y + r+ + √ y ⩾ k+ 2k− . Moreover, for t > 0 fixed, we have h5(tr+) ⩾ k+ 2k− ⇐⇒ tr+√ (t+ 1)r+ + √ tr+ ⩾ k+ 2k− ⇐⇒ t√ t+ 1 + √ t √ r+ ⩾ k+ 2k− , hence √ r+ ⩾ √ t+ 1 + √ t t k+ 2k− ⇐⇒ r+ ⩾ (√ t+ 1 + √ t )2 t2 k2+ 4k2− . Finally, if r+ > 0 such that (6.17) holds, then since y = x, for all x ⩾ tr+, we have h2(x) > 0, h1(x) > 0 and f2(x) > 0. Moreover, r+ < 0 such that (6.18) holds, then for all y > 0, we have h2(y) > 0, h1(y) > 0 and since y = x + r+, for all x > −r+, it follows that f2(x) > 0. □ Therefore, as in the first case, since we have U± ∼ I and V± ∼ I, then by setting W = U−U+V−V+ ∼ I, we deduce that WP+ 1 ∼ 2k+(L+ +M), WQ− 1 ∼ 2k−I WP+ 2 ∼ 2k+(L+ +M), WQ− 2 ∼ 2k−I WP+ 3 ∼ 2k+(L+ +M)L+, WQ− 3 ∼ 2k−I. Thus W 2 det(Λ2) = ( WP+ 2 WP+ 3 −M ( WP+ 1 )2) + 4M2 ( WQ− 2 WQ− 3 −M ( WQ− 1 )2) + 2M ( WP+ 3 WQ− 2 +WP+ 2 WQ− 3 +MWP+ 1 WQ− 1 ) ∼ 4k2+(L+ +M)2(L+ −M) + 16k2−M 2(I −M) + 8k+k−(L+ +M)M(L+ +M + I). From (4.3), we have W 2 det(Λ2) ∼ 4k2+r+(L+ +M) + 16k2−M 2(I −M) + 8k+k−(L+ +M)M(L+ +M + I) ∼ 4k+l+(L+ +M) + 16k2−M 2(I −M) + 8k+k−(L+ +M)M(L+ +M + I). Hence, B3 = 4k+l+(L+ +M) + 16k2−M 2(I −M) + 8k+k−(L+ +M)M(L+ +M + I). EJDE-2024/78 SOLVABILITY OF TRANSMISSION PROBLEMS 27 Thus, det(Λ2) =W−2 ( B3 + ∑ j∈J kjL lj +M mjeαjL+eβjM ) , (6.21) where J is a finite set and for each j ∈ J : kj ∈ R, lj ,mj ∈ N, αj , βj ∈ R+ with αj + βj ̸= 0. Proposition 6.8. Assume that (H1)–(H3) hold and k+k− > 0. If k− k+ ⩽ 2, then 0 ∈ ρ ( 8k+k−(L+ +M)2M − 16k2−M 3 ) . Proof. Since k+k− > 0, we have k− k+ > 0 and 8k+k−(L+ +M)2M − 16k2−M 3 = 8k+k−M [ L2 + + 2L+M +M2 − 2 k− k+ M2 ] . From Remark 4.1, statement 5 and [31, Corollary 3, p. 444], we deduce that L2 +, 2L+M, M2 ∈ BIP(X, θA). Thus, if k− k+ ⩽ 1, then L2 + + 2L+M +M2 − 2 k− k+ M2 = L2 + − k− k+ M2 + 2L+M + ( 1− k− k+ ) M2. Moreover, for all ψ ∈ D(M2) = D(A), from (4.2), we have( L2 + − k− k+ M2 ) ψ = [ − ( 1− k− k+ ) A+ r+I ] ψ, and from [31, Theorem 3, p. 437] and [1, Theorem 2.3, p. 69], assumptions (H2) and (H3) imply that − ( 1− k− k+ ) A+ r+I ∈ BIP(X, θA), and L2 + − k− k+ M2 + 2L+M + ( 1− k− k+ ) M2 ∈ BIP(X, θA + ε), for all ε ∈ (0, π − θA). Moreover, since 0 ∈ ρ(L+M), we deduce from [31, remark at the end of p. 445] that 0 ∈ ρ ( L2 + − k− k+ M2 + 2L+M + ( 1− k− k+ ) M2 ) . Therefore, since 0 ∈ ρ(M) and k+k− > 0, it follows that 0 ∈ ρ ( 8k+k−(L+ +M)2M − 16k2−M 3 ) . In the same way, if 1 < k− k+ ⩽ 2, then L2 + + 2L+M +M2 − 2 k− k+ M2 = L2 + −M2 + 2L+M − 2 (k− k+ − 1 ) M2 +M2 −M2, hence, for all ψ ∈ D(M2) = D(A), from (4.2), we obtain( L2 + + 2L+M +M2 − 2 k− k+ M2 ) ψ = r+ψ + 2M ( L+ − (k− k+ − 1 ) M ) ψ. (6.22) Moreover,( L+ − (k− k+ − 1 ) M ) ψ = ( L+ + (k− k+ − 1 ) M )−1( L2 + − (k− k+ − 1 )2 M2 ) ψ, 28 A. THOREL EJDE-2024/78 and from [31, Theorem 3, p. 437] and [1, Theorem 2.3, p. 69], assumptions (H2) and (H3) imply that( L2 + − (k− k+ − 1 )2 M2 ) = − ( 2− k− k+ ) A+ r+I ∈ BIP(X, θA). (6.23) Finally, from (H2), (H3), (6.22), (6.23) and [31, Theorem 3, p. 437], we deduce that r+ψ + 2M ( L+ − (k− k+ − 1 ) M ) ∈ BIP(X, θA), and 0 ∈ ρ ( r+ψ + 2M ( L+ − (k− k+ − 1 ) M )) . Therefore, since 0 ∈ ρ(M) and k+k− > 0, it follows that 0 ∈ ρ ( 8k+k−(L+ +M)2M − 16k2−M 3 ) . □ We set B4 = I + 4k+l+(L+ +M) ( 8k+k−(L+ +M)2M − 16k2−M 3 )−1 + 16k2−M 2 ( 8k+k−(L+ +M)2M − 16k2−M 3 )−1 + 8k+k−(L+ +M)M ( 8k+k−(L+ +M)2M − 16k2−M 3 )−1 . (6.24) Therefore, B3 = ( 8k+k−(L+ +M)2M − 16k2−M 3 ) B4. Moreover, from (6.21) and noting that B5 = 8k+k−(L+ +M)2M − 16k2−M 3, we have det(Λ2) =W−2B5F2, (6.25) where F2 = B4 + ∑ j∈J kjB −1 5 L lj +M mjeαjL+eβjM . (6.26) Now, for z ∈ C \ [max(−r+, 0),+∞), we set f̃2(z) = b4(z) + ∑ j∈J kjb5(z) −1 √ z + r+ lj√ z mj e−αj √ z+r+e−βj √ z, where b3(z) = b4(z)b5(z), with b4(z) = 1 + 4k+l+( √ z + r+ + √ z) ( 8k+k−( √ z + r+ + √ z)2 √ z − 16k2− √ z 3 )−1 + 16k2−z ( −8k+k−( √ z + r+ + √ z)2 √ z + 16k2− √ z 3 )−1 + 8k+k−( √ z + r+ + √ z) √ z ( 8k+k−( √ z + r+ + √ z)2 √ z − 16k2− √ z 3 )−1 , and b5(z) = −8k+k−( √ z + r+ + √ z)2 √ z + 16k2− √ z 3 . Then f̃2(−A) = F2, b3(−A) = B3, b4(−A) = B4 and b5(−A) = B5. Thus, from (6.15) and (6.25), we deduce that f2(z) = u−2 d,r+ (z)v−2 d,r+ (z)u−2 c,0(z)v −2 c,0(z)b5(z)f̃2(z). (6.27) Proposition 6.9. Assume that (H1)–(H4) hold and k+k− > 0 with k− k+ ⩽ 2. Thus EJDE-2024/78 SOLVABILITY OF TRANSMISSION PROBLEMS 29 • if r+ > 0 such that r+ ⩾ (√ t+ 1 + √ t )2 t2 k2+ 4k2− , for t ∈ ( 0, 1 r+∥A−1∥L(X) ) fixed, (6.28) • if r+ < 0 such that (6.18) holds. Then F2, given by (6.26), is invertible with bounded inverse. Proof. From Proposition 6.8 and (6.26), we deduce that F2 is well defined. First assume that r+ > 0 such that (6.17) holds. Then, from Lemma 6.7 and (6.27), it follows that f2 does not vanish on (tr+,+∞), for t > 0 fixed, which involves that u−2 d,r+ , v−2 d,r+ , u−2 c,0, v −2 c,0 , b5 and f̃2 do not vanish on (tr+,+∞), for t > 0 fixed. Moreover, by (H2), there exists R = 1 ∥A−1∥L(X) > 0 such that B(0, R) ⊂ ρ(A). Therefore, setting f̃tr+,2(z) = f̃2(z + tr+), with t ∈ ( 0, 1 r+∥A−1∥L(X) ) fixed and applying Lemma 5.6 where we have set G = f̃tr+,2 and operator P = −A− tr+I ∈ BIP (X, θA) (due to (H2) and (H3)), we deduce that operator F2 = f̃tr+,2(−A − tr+I) = f̃2(−A) is invertible with bounded inverse. Now, assume that r+ < 0 such that (6.18) holds. Then f̃2 does not vanish on (−r+,+∞). Moreover, from (H2) and (H3), we have −A + r+I ∈ BIP (X, θA). It follows that F2 = f̃−r+,2(−A + r+I) = f̃2(−A) is invertible with bounded inverse. □ This result finally leads us to state the following main result of this section. Proposition 6.10. Assume that (H1)–(H4) hold and k+k− > 0 with k− k+ ⩽ 2. Thus • if r+ > 0 such that (6.28) holds, • if r+ < 0 such that (6.18) holds, then det(Λ2) is invertible with bounded inverse. Proof. From (6.25), Propositions 6.8 and 6.9, we obtain that det(Λ2) =W−2B5F2, is invertible with bounded inverse. □ 6.3. Regularity. 6.3.1. First case. Here, we consider r+, r− ∈ R \ {0}. From Theorem refTh syst trans, we have to prove that system (5.9) has a unique solution (ψ1, ψ2) satisfying (5.10). The existence and uniqueness of this solution is ensured by Proposition 6.6, so we have ψ1 = ( P− 1 − P+ 1 ) [det(Λ1)] −1 S1 − ( P+ 2 + P− 2 ) [det(Λ1)] −1S2 ψ2 = − ( P+ 3 − P− 3 ) [det(Λ1)] −1 S1 +M ( P− 1 − P+ 1 ) [det(Λ1)] −1S2. (6.29) Now, we have to study the regularity of [det(Λ1)] −1 . Since, in this case, the deter- minant det(Λ1) is the same than the one in [19]. From [19, Lemma 5.3, p. 2958] we deduce that there exists Rdet(Λ1) ∈ L(X) such that Rdet(Λ1)(X) ⊂ D(M), [det(Λ1)] −1 = N−1 +N−1Rdet(Λ1), where N = 4k+k−(L−+M)(L++M)(L++L−+2M). Then, the rest of the proof is similar to the one given in [19, section 5.3]. Therefore, from (5.4) and (5.5), it follows that S1, S2 ∈ (D(M), X)1+ 1 p ,p and thus [det(Λ)] −1 S1, [det(Λ)] −1 S2 ∈ (D(M), X)4+ 1 p ,p . (6.30) 30 A. THOREL EJDE-2024/78 Moreover, from (6.29), we have ψ1 = −2 (k+(L+ +M)− k−(L− +M)) [det(Λ1)] −1 S1 + 2 (k+(L+ +M)− k−(L− +M)) [det(Λ1)] −1 S2 + S̃1 ψ2 = −2 (k+(L+ +M)L+ + k−(L− +M)L−) [det(Λ1)] −1 S1 − 2 (k+(L+ +M)− k−(L− +M)) [det(Λ1)] −1 S2 + S̃2, (6.31) where S̃1, S̃2 ∈ D(M∞). Finally, from (5.19), (6.30) and (6.31), we obtain ψ1 ∈ (D(M), X)3+ 1 p ,p = (D(A), X)1+ 1 2p ,p ψ2 ∈ (D(M), X)2+ 1 p ,p = (D(A), X)1+ 1 2+ 1 2p ,p . 6.3.2. Second case. Here, we consider r+ ∈ R \ {0} and r− = 0. From Theorem refTh syst trans M, we have to prove that (5.17) has a unique solution (ψ1, ψ2) satis- fying (5.10). The existence and uniqueness of this solution is ensured by Proposition 6.10, so we have ψ1 = ( 2MQ− 1 − P+ 1 ) [det(Λ2)] −1 S3 + ( P+ 2 + 2MQ− 2 ) [det(Λ2)] −1 S4 ψ2 = − ( P+ 3 + 2MQ− 3 ) [det(Λ2)] −1 S3 +M ( P+ 1 − 2MQ− 1 ) [det(Λ2)] −1S4. (6.32) Now, we have to study the regularity of [det(Λ2)] −1. From (5.1), (6.24), (6.25), (6.26) and [20, Lemma 5.1, p. 365], we deduce that there exists Rdet(Λ2) ∈ L(X) such that Rdet(Λ2)(X) ⊂ D(M), [det(Λ2)] −1 = B−1 5 +B−1 5 Rdet(Λ2), where we recall that B5 = 8k+k−(L+ +M)2M − 16k2−M 3. Moreover, from (4.6), (5.19), (5.21) and (5.24), we have φ̃+ 1 , φ̃ − 2 , φ̃ + 2 , φ̃ + 3 , φ̃ − 4 , φ̃ + 4 ∈ (D(M), X)2+ 1 p ,p and φ̃− 1 , φ̃ − 3 ∈ (D(M), X)3+ 1 p ,p . Thus, from (5.13), (5.14), (5.15), Remarks 5.2 and 5.4, we deduce that R2 ∈ (D(M), X) 1 p ,p and S3, S4 ∈ (D(M), X)1+ 1 p ,p , which implies that [det(Λ2)] −1 S3, [det(Λ2)] −1 S4 ∈ (D(M), X)4+ 1 p ,p . (6.33) Moreover, from (5.2), (5.12), (6.32) and [20, Lemma 5.1, p. 365], we have ψ1 = −2 (k+(L+ +M)− 2k−M) [det(Λ2)] −1 S3 + 2 (k+(L+ +M) + 2k−M) [det(Λ2)] −1 S4 + S̃3 ψ2 = −2 (k+(L+ +M)L+ + 2k−M) [det(Λ2)] −1 S3 + 2M (k+(L+ +M)− 2k−M) [det(Λ2)] −1 S4 + S̃4, (6.34) where S̃3, S̃4 ∈ D(M∞). 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Alexandre Thorel Université Le Havre Normandie, Normandie Univ, LMAH UR 3821, 76600 Le Havre, France Email address: alexandre.thorel@univ-lehavre.fr 1. Introduction 2. Operational formulation 3. Definitions and prerequisites 3.1. The class of Bounded Imaginary Powers of operators 3.2. Interpolation spaces 4. Assumptions and statement of results 4.1. Hypotheses 4.2. Main results 5. Preliminary results 5.1. Transmission system 5.2. Functional calculus 6. Proof of the main results 6.1. Calculus of the determinant 6.2. Inversion of the determinant 6.3. Regularity Acknowledgments References