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OPTIMIZATION OF STABILITY OF A FLEXIBLE MISSILE UNDER FOLLOWER THRUST
Arab. J. Math. https://doi.org/10.1007/s40065-023-00427-y Arabian Journal of Mathematics Bertin Zinsou Stability of a ﬂexible missile and asymptotics of the eigenvalues of fourth order boundary value problems Received: 18 April 2022 / Accepted: 26 March 2023 © The Author(s) 2023 (4) 2 1 Abstract Fourth order problems with the differential equation y − (gy ) = λ y,where g ∈ C [0, a] and a > 0, occur in engineering on stability of elastic rods. They occur as well in aeronautics to describe the stability (4) of a ﬂexible missile. Fourth order Birkhoff regular problems with the differential equation y − (gy ) = λ y and eigenvalue dependent boundary conditions are considered. These problems have quadratic operator representations with non-self-adjoint operators. The ﬁrst four terms of the asymptotics of the eigenvalues of the problems as well as those of the eigenvalues of the problem describing the stability of a ﬂexible missile are evaluated explicitly. Mathematics Subject Classiﬁcation 34L20 · 34B08 · 34B09 1 Introduction Higher order boundary value problems, compared to Sturm–Liouville problems, are less explored. However, they have been attracting a lot of attention. Investigations on eigenvalue parameter for dependent boundary conditions in higher order boundary value problems have seen recent developments, see for example [1– 3,5,8,10–12,14–20,22,23,25–27]. The mathematical model of many of the above problems gives an eigenvalue problem represented by L(λ) = λ M − i λK − A, (1.1) known as operator pencil or operator polynomial, where the boundary condition may depend linearly on the eigenvalue parameter λ in the Hilbert space L (I ) ⊕ C . The operators M, K and A are coefﬁcients of powers of λ, k is the number of boundary conditions depending on λ while I is an interval. Applying separation of variables to the stability of elastic rod problems investigated in [12,14,16,17,20,25– 27], we get fourth order eigenvalue problems with boundary conditions depending on the eigenvalue parameter λ with the differential equation (4) 2 y − (gy ) = λ y (1.2) depending quadratically on λ. Supplementary Information The online version contains supplementary material available at https://doi.org/10.1007/ s40065-023-00427-y. B. Zinsou (B) School of Mathematics, University of the Witwatersrand, Private Bag 3, Wits 2050, Johannesburg, South Africa E-mail: bertin.zinsou@wits.ac.za 123 Arab. J. Math. Other problems represented by the above differential equation are problems describing stability of ﬂexible missiles. Beal [4] has studied the dynamic stability of a ﬂexible missile under an end thrust. He describes the bending characteristics of the vehicle using simple beam theory. Conditions that may be typical of similar conditions existing for an actual missile are obtained. However, they are not quantitatively applicable to a speciﬁc missile. Guran and Ossia [6] have studied the dynamic stability of an uniform free-free rod, representing a ﬂexible missile, subjected to an end thrust by applying ﬁnite difference techniques. They obtain eingenvalue curves and in graphical forms their convergence characteristics. Studying a moving beam subjected to a tangential end force in space, Kirillov and Seyranian, see [7], have obtained its mass distribution with the critical ﬂutter load and they have presented optimal distributions of nonstructural mass with a few switching points. The boundary conditions, of the problems investigated in [4,6,7], are (3) (3) y (0) = y (0) = 0and y (a) = y (a) = 0 (1.3) and their operator polynomial representation is L(λ) = λ M − A, (1.4) in the Hilbert space L (0, a). Recently, Xu, Rong, Xiang, Pan and Yin [24], using numerical calculations, have investigated the dynamic response and stability of a rotating and ﬂexible missile under thrust. Although many studies have been conducted on stability of ﬂexible missiles, no reference can be found on the eigenvalues of these problems. In [16], we have provided necessary and sufﬁcient conditions for the associated operator pencils of classes of boundary value problems to have only self-adjoint operators. In [25], we have derived necessary and sufﬁcient conditions for the associated operator pencils for classes of boundary conditions to be Birkhoff regular. In this paper, we investigate the asymptotics of the eigenvalues of a class of boundary conditions obtained in [25], where only the right end side boundary conditions depend on the eigenvalue parameter. We should observe that the problems under consideration are obtained with more general conditions. Hence, the coefﬁcient operators K and A of the corresponding quadratic operator pencils, see (1.1), are not necessary self-adjoint. We investigate as well the asymptotics of the eigenvalues of the problem describing the stability of a ﬂexible missile. We give a characterization of fourth order Birkhoff regular problems in Sect. 2. In Sect. 3,wepresent the quadratic operator pencil under consideration as well as the boundary conditions that will be investigated. In Sect. 4, we classify the problems under consideration in two different subclasses according to the right endpoint boundary conditions and we derive the eigenvalue asymptotics for g = 0. Since these problems are Birkhoff regular, then the eigenvalues for general g can be derived from those of g = 0 because they are small perturbations of the latter ones. Whence, in Sect. 5, we use the eigenvalue asymptotics for g = 0to provide the ﬁrst four terms of the asymptotics of the eigenvalue of the two relevant classes and we compare the results obtained to those in [17]. Finally in Sect. 6, we give the asymptotics of the eigenvalues of the problem describing the stability of a ﬂexible missile. 2 Fourth order Birkhoff regular problems We consider, on the interval [0, a], the eigenvalue problem (4) 2 y − (gy ) = λ y, (2.1) B (λ)y = 0, j = 1, 2, 3, 4, (2.2) with a > 0, g is a real valued function such that g ∈ C [0, a],and (2.2) are separated boundary conditions which may depend on λ linearly. We make the following assumption [ p ] [q ] j j B (λ)y = y (a ) + iβ λy (a ), (2.3) j j j j with a = 0for j = 1,2and a = a for j = 3, 4. In addition, 0 ≤ q < p ≤ 3, for β ∈ C \{0} while j j j j j β = 0 corresponds to q =−∞, j = 1, 2, 3, 4. j j 123 Arab. J. Math. Note that the following quasi-derivatives [4] (4) [3] (3) [2] [1] [0] y = y − (gy ) , y = y − gy , y = y , y = y , y = y, (2.4) are associated to (2.1), see [13, Deﬁnition 10.2.1]. Observe that λ comes from derivatives with respect to time in the original partial differential equation while the highest space derivative appears in the expression not depending on time derivative. Whence, the boundary conditions satisfying q < p ,for j = 1, 2, 3, 4, are the most relevant. j j We deﬁne { } = s ∈{1, 2, 3, 4}: B (λ) depends on λ , ={1, 2, 3, 4}\ , (2.5) 1 s 0 1 0 a = ∩{1, 2}, = ∩{3, 4}, (2.6) 1 1 1 1 and 0 a = {s ∈{1, 2, 3, 4}: p > −∞}, = ∩{1, 2}, = ∩{3, 4}. (2.7) 0 0 a a Assumption 2.1 By assumption, p for s ∈ , q for j ∈ are distinct and p for s ∈ , q for j ∈ s j s j 1 1 are distinct. [r ] The meaning of Assumption 2.1 is as follows: for any pair (r, a ) the term y (a ) appears at most once in j j the boundary conditions (2.2)and q , p , j = 1, 2, 3, 4 are mutually disjoint. j j Let p , q ∈{0, 1, 2, 3},where p , q are as introduced in Assumption 2.1, j = 1, 2, 3, 4. Choose u so j j j j that u = 0if j = 1,2and u = 1if j = 3, 4. Deﬁne C (r, u), with r = 1, 2, 3, 4,5and u = 0, 1, be the conditions below: C (1, u): p > q + 2, p > q + 2; 1+2u 1+2u 2+2u 2+2u C (2, u): p > q + 2, q + 2 > p ; 1+2u 1+2u 2+2u 2+2u C (3, u): p > q + 2, p = q + 2and β = (−1) , where l = 1, 2; 1+2u 1+2u 2+2u 2+2u 2+2u C (4, u): q + 2 > p , q + 2 > p ; 1+2u 1+2u 2+2u 2+2u C (5, u): q + 2 > p , p = q + 2, 1+2u 1+2u 2+2u 2+2u l+1 (−1) if q − q = 1, 1+2u 2+2u β = 2+2u (−1) if q − q = 3, 1+2u 2+2u where l = 1, 2. Because the boundary conditions (2.2) and the assumptions already made, [25, Theorem 3.4 ] gives. { } Proposition 2.2 The problem (2.1), (2.2) is Birkhoff regular if and only if there are r , r ∈ 1, 2, 3, 4, 5 so 0 1 that C (r , 0) and C (r , 1) hold. 0 1 3 The quadratic operator pencil L Denoting the collection of boundary conditions (2.2)by U , we deﬁne next operators related to U [ p ] [q ] j j U y = y (a ) , r = 0, 1, and V y = β y (a ) , r 1 j j j j ∈ j ∈ r 1 y ∈ W (0, a), (3.1) where W (0, a) is the Sobolev space of order 4 on the interval (0, a). Putting k =| |, we deﬁne in the space L (0, a) ⊕ C the linear operators A(U ), K and M with domains 1 2 D (A(U )) = y = : y ∈ W (0, a), U y = 0 , V y D (K ) = D (M ) = L (0, a) ⊕ C , 123 Arab. J. Math. given by (4) y − (gy ) (A(U )) y = for y ∈ D (A(U )), U y I 0 00 M = and K = with K = diag(β : j ∈ ). 0 j 1 00 0 K It is obvious that M is nonnegative and self-adjoint. In addition, K and M are bounded operators. We associate the operator polynomial L(λ) = λ M − i λK − A(U ), λ ∈ C (3.2) in the space L (0, a) ⊕ C with the problems (2.1), (2.2). A function y satisﬁes (2.1), (2.2) if and only if it satisﬁes L(λ) y = 0. We can conclude that the eigenvalue problem (2.1), (2.2) can be represented by the polynomial operator (3.2). Note that if all the boundary conditions in (2.2) are independent of λ,then V y = 0 and U y = 0, where y ∈ W (0, a). Hence, (3.2) will be reduced to L(λ) = λ M − A(U ), λ ∈ C (3.3) in the space L (0, a). We will investigate the asymptotics of the eigenvalues of the classes of the problems where the boundary conditions at the left endpoint are independent of the parameter λ, while the boundary conditions at the right endpoint may depend on the parameter. For the case β β = 0, we are going to compare the results of our 3 4 investigation to those obtained in the case of self-adjoint problems studied in [17]. The four boundary conditions (2.2)are [ p ] [ p ] 1 2 y (0) = 0, y (0) = 0, (3.4) [ p ] [q ] [ p ] [q ] 3 3 4 4 y (a) + iβ λy (a) = 0, y (a) + iβ λy (a) = 0, 3 4 where 0 ≤ p < p ≤ 3, 0 ≤ q < p ≤ 3, 0 ≤ q < p ≤ 3and 0 < p < p ≤ 3. Therefore, taking 1 2 3 3 4 4 3 4 Assumption 2.1 into account, we will distinguish the following different cases of boundary conditions at the endpoint 0: Case 1 : (p , p ) = (0, 1), Case 2 : (p , p ) = (0, 2), 1 2 1 2 (3.5) Case 3 : (p , p ) = (0, 3), Case 4 : (p , p ) = (1, 2), 1 2 1 2 Case 5 : (p , p ) = (1, 3), Case 6 : (p , p ) = (2, 3). 1 2 1 2 However, the boundary conditions at the right endpoint a will be classiﬁed as (a) Case 1 : (p , q ) = (1, 0) and (p , q ) = (3, 2), 3 3 4 4 (3.6) (a) Case 2 : (p , q ) = (2, 1) and (p , q ) = (3, 0). 3 3 4 4 As we have two sets of boundary conditions at the endpoint a and six sets of boundary conditions at the endpoint 0, then we have 12 sets of boundary conditions in total. We will classify these 12 sets of boundary conditions according to the endpoint a. Hence, we will have two classes of boundary conditions that we will categorise by the pair ( p , q ), j = 3, 4, see (3.6). j j Deﬁne the condition C (2, u): p < q + 2, q + 2 < p , u = 0, 1. Note that the conditions 1+2u 1+2u 2+2u 2+2u C (2, u) and C (2, u), u = 0, 1, are redundant, see [25, page 5]. Hence, for u = 0, 1, any result that is valid for C (2, u), the equivalent result is valid for C (2, u),aswell. Note that the left endpoint boundary conditions satisfy the condition C (1, 0), while the right endpoint (a) (a) boundary conditions satisfy the conditions C (4, 1) for Case 1 and the condition C (2, 1) for Case 2. (a) (a) Whence, the problems are Birkhoff regular for the classes Case 1and Case 2, see Proposition 2.2. We will investigate as well the asymptotics of the eigenvalues of the problem describing the stability of a (3) (3) ﬂexible missile, where the boundary conditions are y (0) = y (0) = 0and y (a) = y (a) = 0. Note that the left endpoint boundary conditions of this problem satisfy the condition C (1, 0), while the right endpoint boundary conditions satisfy the condition C (1, 1). Hence, the problem is Birkhoff regular according to Proposition 2.2. 123 Arab. J. Math. 4 Asymptotics of eigenvalues for the case g = 0 The boundary value problems (2.1), (3.4) with g = 0 is considered in this section. We provide the number of the eigenvalues and their multiplicities. We provide as well the eigenvalue asymptotics for the case g = 0. We consider the canonical fundamental system y , j = 1,..., 4, (4.1) of (2.1). It is analytic with respect to λ on C and [m] y (0) = δ , (4.2) j,m+1 [m] (m) for m = 0,..., 3, when g = 0. Observe that for g = 0, y (0) = y . Putting j j M (λ) = (B (λ)y (·,λ)) , i j i, j =1 the eigenvalues of problems (2.1), (3.4)for g = 0, are those of the analytic matrix function M,ofwhich the corresponding algebraic and geometric multiplicities coincide, see [9, Theorem 3.1.2]. Setting λ = μ and 1 1 y(x,μ) = sinh(μx ) − sin(μx ), 3 3 2μ 2μ it is obvious that (4− j ) y (x,λ) = y (x,μ), j = 1,..., 4. (4.3) Observe that each of the ﬁrst two rows of M (λ) has exactly one entry with 1 and the remaining entries have 0. Whence, for each of the two different classes of boundary conditions, det M (λ) =±φ(μ), with 2 2 B (μ )y (·,μ) B (μ )y (·,μ) 3 σ(1) 3 σ(2) φ(μ) = det . 2 2 B (μ )y (·,μ) B (μ )y (·,μ) 4 σ(1) 4 σ(2) (3, 4) in Case 1,(2, 4) in Case 2,(2, 3) in Case 3, (σ (1), σ (2)) = (4.4) (1, 4) in Case 4, (1, 3) in Case 5, (1, 2) in Case 6. Therefore, 2 2 2 2 φ(μ) = B (μ )y (·,μ)B (μ )y (·,μ) − B (μ )y (·,μ)B (μ )y (·,μ) 3 σ(1) 4 σ(2) 4 σ(1) 3 σ(2) ( p ) (q ) ( p ) (q ) 3 2 3 4 2 4 = y (a) + iβ μ y (a) y (a) + iβ μ y (a) 3 4 σ(1) σ(1) σ(2) σ(2) ( p ) (q ) ( p ) (q ) 4 2 4 3 2 3 − y (a) + iβ μ y (a) y (a) + iβ μ y (a) 4 3 σ(1) σ(1) σ(2) σ(2) ( p ) ( p ) ( p ) ( p ) (q ) ( p ) 3 4 4 3 2 3 4 = y (a)y (a) − y (a)y (a) + i μ β y (a)y (a) σ(1) σ(2) σ(1) σ(2) σ(1) σ(2) ( p ) (q ) ( p ) (q ) (q ) ( p ) 4 3 3 4 4 3 −y (a)y (a) + β y (a)y (a) − y (a)y (a) σ(1) σ(2) σ(1) σ(2) σ(1) σ(2) (q ) (q ) (q ) (q ) 4 4 3 3 4 + β β μ y (a)y (a) − y (a)y (a) . (4.5) 3 4 σ(1) σ(2) σ(1) σ(2) (a) Next, we discuss the asymptotics of the zeros of the problems for each class Case j, j = 1, 2. 123 Arab. J. Math. (a) 4.1 Asymptotics of eigenvalues for g = 0 of the problems of Case 1 (a) It follows from (3.6)and (4.5) that the characteristic functions φ(μ) of the eigenvalue problems of Case 1 are given by: (3) (3) (3) φ(μ) = y (a)y (a) − y (a)y (a) + i μ β y (a)y (a) 3 σ(1) σ(1) σ(2) σ(2) σ(1) σ(2) (3) −y (a)y (a) + β y (a)y (a) − y (a)y (a) σ(2) 4 σ(1) σ(1) σ(2) σ(1) σ(2) + β β μ y (a)y (a) − y (a)y (a) . (4.6) 3 4 σ(2) σ(1) σ(1) σ(2) Each term of φ is a product of a power in μ with the sum of two products of a hyperbolic and a trigonometric functions. The highest μ-power is given by β β μ y (a)y (a) − y (a)y (a) . 3 4 σ(2) σ(1) σ(1) σ(2) Next, we investigate the zeros of φ (μ) = 2μ y (a)y (a) − y (a)y (a) . 0 σ(2) σ(1) σ(1) σ(2) It follows from (4.3)and (4.4) that for these six cases obtained above, we get: Case 1: p = 0, p = 1: 1 2 φ (μ) = μ(cos(μa) sinh(μa) − sin(μa) cosh(μa)). Case 2: p = 0, p = 2: 1 2 φ (μ) =−μ sin(μa) sinh(μa). Case 3: p = 0, p = 3: 1 2 φ (μ) =−μ (cos(μa) sinh(μa) + sin(μa) cosh(μa)). Case 4: p = 1, p = 2: 1 2 φ (μ) =−μ (cos(μa) sinh(μa) + sin(μa) cosh(μa)). Case 5: p = 1, p = 3: 1 2 φ (μ) =−2μ cos(μa) cosh(μa). Case 6: p = 2, p = 3: 1 2 φ (μ) =−μ (cos(μa) sinh(μa) − sin(μa) cosh(μa)). Next, we derive the asymptotic distributions of the zeros of φ (μ), with their number. Lemma 4.1 Case 1: p = 0, p = 1, φ has a zero of multiplicity 4 at 0, exactly one simple zero μ ˜ in each 1 2 0 k 1 π 1 π interval k − , k + ,for k ∈ N, with asymptotics 2 a 2 a μ ˜ = (4k − 3) + o(1), k = 2, 3,..., 4a simple zeros −˜ μ , μ ˜ = i μ ˜ and −i μ ˜ for k = 2, 3,..., and no additional zeros. k −k k k Case 2: p = 0, p = 2, φ has a zero of multiplicity 4 at 0, simple zeros 1 2 0 μ ˜ = (k − 1) , k = 2, 3,..., simple zeros −˜ μ , μ ˜ = i μ ˜ and −i μ ˜ for k = 2, 3,... , and no additional zeros. k −k k k 123 Arab. J. Math. Case 3: p = 0, p = 3, φ has a zero of multiplicity 4 at 0, exactly one simple zero μ ˜ in each interval 1 2 0 k 1 π 1 π k − , k + ,for k ∈ N, with asymptotics 2 a 2 a μ ˜ = (4k − 5) + o(1), k = 2, 3,..., 4a simple zeros −˜ μ , μ ˜ = i μ ˜ and −i μ ˜ for k = 2, 3,... , and no additional zeros. k −k k k Case 4: p = 1, p = 2, φ has a zero of multiplicity 4 at 0, exactly one simple zero μ ˜ in each interval 1 2 0 k 1 π 1 π k − , k + ,for k ∈ N, with asymptotics 2 a 2 a μ ˜ = (4k − 5) + o(1), k = 2, 3,..., 4a simple zeros −˜ μ , μ ˜ = i μ ˜ and −i μ ˜ for k = 2, 3,... , and no additional zeros. k −k k k Case 5: p = 1, p = 3, φ has a zero of multiplicity 4 at 0,simplezeros 1 2 0 μ ˜ = (2k − 1) , k = 2, 3,..., 2a simple zeros −˜ μ , μ ˜ = i μ ˜ and −i μ ˜ ,k = 2, 3,... , and no additional zeros. k −k k k Case 6: p = 2, p = 3, φ has a zero of multiplicity 8 at 0, exactly one simple zero μ ˜ in each interval 1 2 0 k 1 π 1 π k − , k + ,for k ∈ N, with asymptotics 2 a 2 a μ ˜ = (4k − 7) + o(1), k = 3, 4,..., 4a simple zeros −˜ μ , μ ˜ = i μ ˜ and −i μ ˜ for k = 3, 4,..., and no additional zeros. k −k k k Proof The result is evident in Cases 2 and 5. Cases 3 and 4 are identical, while Cases 1 and 6 differ in the factor with the power of μ. We will consider Case 3. The proof for this case is similar to the proof of [27, Lemma 4.1]. The result for Case 1 is similar to that of Case 3 if we replace the trigonometric and hyperbolic functions by their derivatives. Proposition 4.2 For g = 0, there exists k ∈ N such the eigenvalues λ ,k ∈ Z, of the problems (2.1), (3.4), 0 k ˆ ˆ ˆ where (p , q ) = (1, 0) and ( p , q ) = (3, 2),are λ =−λ , λ =ˆ μ for k ≥ k and the μ ˆ have the 3 3 4 4 −k k k 0 k following asymptotic descriptions as k →∞: Case 1: p = 0, p = 1, μ ˆ = (4k − 3) + o(1). 1 2 k 4a Case 2: p = 0, p = 2, μ ˆ = (k − 1) + o(1). 1 2 k Case 3: p = 0, p = 3, μ ˆ = (4k − 5) + o(1). 1 2 k 4a Case 4: p = 1, p = 2, μ ˆ = (4k − 5) + o(1). 1 2 k 4a Case 5: p = 1, p = 3, μ ˆ = (2k − 1) + o(1). 1 2 k 2a Case 6: p = 2, p = 3, μ ˆ = (4k − 7) + o(1). 1 2 k 4a In particular, there is an even number of the pure imaginary eigenvalues in each case. Proof We prove that the zeros of φ are asymptotically close to those of φ . We begin with Case 3. Case 3: We get β β μ 3 4 φ(μ) =− (cos(μa) sinh(μa) + sin(μa) cosh(μa)) 2 2 i μ β i μ β 3 4 − (1 − cos(μa) cosh(μa)) + (1 + cos(μa) cosh(μa)) 2 2 + (cos(μa) sinh(μa) − sin(μa) cosh(μa)). (4.7) 123 Arab. J. Math. Let 2φ(μ) + β β φ (μ) 3 4 0 φ (μ) = . (4.8) φ (μ) The ﬁrst term of φ(μ), up to constant − β β ,is φ (μ). Thus, for μ such φ (μ) = 0, sin(μa) = 0, 3 4 0 0 sinh(μa) = 0, we get 2φ(μ) + β β φ (μ) 1 i (β − β ) 1 3 4 0 4 3 φ (μ) = = φ (μ) μ cos(μa) cosh(μa) tan(μa) + tanh(μa) 1 i (β + β ) 4 3 μ tan(μa) + tanh(μa) 1 2cos(μa) tanh(μa) + 1 − . (4.9) μ sin(μa) + cos(μa) tanh(μa) Fix ε ∈ (0, ) and consider R , k = 2, 3,... , as the boundaries of the squares determined by the k,ε 4a π π vertices (4k − 5) ± ε ± i ε. The squares R do not intersect because ε< . Since tan z =−1 if and only k,ε 4a 2a if z = j π − with j ∈ Z, by the periodicity of tan C (ε) = 2min{|tan(μa) + 1|: μ ∈ R } 1 k,ε is a positive number independent of ε. Since lim tanh(μa) = 1 uniformly in the strip {μ ∈ C : Re μ ≥ |μ|→∞ 1, |Im μ|≤ }, there exist k (ε) ∈ N such that 4a |tan(μa) + tanh(μa)|≥ C (ε) for all μ ∈ R with k > k (ε). 1 k,ε 1 By periodicity, we can ﬁnd a positive number C (ε) such that |cos(μa)| > C (ε) for all μ ∈ R and all k. 2 2 k,ε Note that |cosh(μa)|≥|sinh( μa)|. Hence, there exists k (ε) ≥ k (ε) so that for all μ on the squares R 2 1 k,ε with k ≥ k (ε),weget |φ (μ)| < 1. We conclude from Lemma 4.1 that μ ˜ is inside of R for k > k (ε) and 2 1 k k,ε 2 no additional zero of φ satisﬁes this condition. By deﬁnition of φ in (4.8) and the estimate |φ (μ)| < 1for 0 1 1 all μ on the square R ,wehave k,ε |2φ(μ) + β β φ (mu)| < |φ (μ)|, (4.10) 3 4 0 0 for all μ on the square R . Hence, by Rouché’s theorem, there is one and only one (simple) zero μ ˆ of φ k,ε k in each R for k ≥ k (ε).Since φ (i μ) = φ (μ) and φ (i μ) =−φ (μ) for all μ ∈ C, we can apply the k,ε 2 0 0 1 1 same reasoning to similar squares along the positive imaginary semiaxis. Since φ is an even function, the same estimate can apply to the similar squares along the other two remaining semiaxes. Whence, the zeros of φ are ±ˆ μ , ±ˆ μ for k > k (ε) with the same asymptotic behaviour as the zeros ±˜ μ , ±i μ ˜ of φ as stated in k −k 2 k k 0 Lemma 4.1. π π Next, we provide the estimate of φ on the squares S , k ∈ N, of which vertices are ±k ± ik .For 1 k a a integers k and a real number γ , kπ tan + i γ a = tan(i γ a) = i tanh(γ a) ∈ i R. (4.11) kπ Whence, we get for μ = + i γ , with γ ∈ R and k ∈ Z, |tan(μa)| < 1and |tan(μa) ± 1|≥ 1. (4.12) For μ = x + iy, x , y ∈ R and x = 0, we have (ax +iay) −(ax +iay) e − e tanh(μa) = →±1 (4.13) (ax +iay) −(ax +iay) e + e uniformly in y as x tends to ±∞.Thus,wecan ﬁnd k > 0 so that for k ∈ Z, |k|≥ k ,and γ ∈ R, 1 1 kπ 1 tanh + i γ a − sgn(k)< . (4.14) a 2 123 Arab. J. Math. By (4.12)and (4.14), k ∈ Z, |k|≥ k and γ ∈ R,weget tan(μa) + tanh(μa) ≥ . (4.15) Moreover, we use the estimates kπ cosh + i γ a ≥| sinh(kπ)|, (4.16) kπ cos + i γ a = cosh(γ a) ≥ 1. (4.17) They hold for all integers k and all real numbers γ . Using (4.12), (4.15)–(4.17) and the corresponding estimates where μ is replaced by i μ, we can conclude that there exists k ≥ k such that |φ (μ)| < 1for all μ ∈ S with 1 1 1 k k > k ,where φ is as deﬁned in (4.8). Using the deﬁnition of φ in (4.8) and the estimate |φ (μ)| < 1for all 1 1 1 1 μ ∈ S , then by Rouché’s theorem, the functions φ and φ have the same number of zeros in the square S , k 0 k for all natural numbers k where k ≥ k . The number of zeros of the function φ in the interior of the square S is 4k + 4 while it is 4k + 4(n + 1) 0 k inside the squares S for n ∈ N. We can conclude that for sufﬁciently big k, the only big zeros of φ are k+n those found in Preposition 4.2. In addition, λ =ˆ μ represent all the eigenvalues of the problems (2.1), (3.4), where p = 0, p = 3, ( p , q ) = (1, 0) and ( p , q ) = (3, 2). The non-imaginary eigenvalues appear in pair 1 2 3 3 4 4 ˆ ˆ and can be indexed as λ =−λ . The number of indices left is even. Therefore, we have an even number of −k k pure imaginary eigenvalues. Case 4: The value of σ(1) in this case is equal to that of σ(1) − 1 in Case 3 while the value of σ(2) is equal to the value of σ(2) + 1 in Case 3, see (4.4). Thus, in this case, the function φ is a multiple by a constant factor of that in Case 3. Hence, the results in Cases 3 and 4 are similar. Case 1: The values of σ(1) and σ(2) differ from those in Case 3 by 1. Hence, in this case, the function φ −2 is obtained from that in Case 3 by multiplication by μ and by changing the trigonometric and hyperbolic functions to their derivatives. Hence, the result can be derived from that in Case 3. Case 6: The values of σ(1) and σ(2) are respectively those of σ(1) + 2and σ(2) + 2. Hence, in this case, the function φ can be derived from that in Case 1 by multiplication by μ and by replacing each trigonometric function by its negative. Case 2: We ﬁnd φ(μ) =−β β μ sin(μa) sinh(μa) 3 4 i (β + β )μ 3 4 + (cos(μa) sinh(μa) + sin(μa) cosh(μa)) + cos(μa) cosh(μa). (4.18) Then it follows from (4.8)that 2φ(μ) + β β φ (μ) 3 4 0 φ (μ) = φ (μ) 1 1 = (coth(μa) + cot(μa)) + cot(μa) coth(μa). (4.19) 2μ 2μ π π Using reasoning and estimates as in the proof of Case 3 and substituting μ by μ ± and μ ± i respectively, 2 2 we get the result. Case 5: The values of σ( j ), j = 1, 2 in this case are equal to those of σ( j ) + 1, j = 1, 2 in Case 2. Thus, in this case, the function φ can be obtained from φ in Case 2 by multiplication by μ and by substituting the trigonometric and hyperbolic functions by their derivatives. Using reasoning and estimates similar to those in Case 3, we get the result. 123 Arab. J. Math. (a) 4.2 Asymptotics of eigenvalues for g = 0 of the problems of Case 2 (a) It follows from (3.6)and (4.5) that the characteristic functions φ(μ) of the eigenvalue problems of Case 2 are given by: (3) (3) (3) φ(μ) = y (a)y (a) − y (a)y (a) + i μ β y (a)y (a) σ(1) σ(2) σ(1) σ(2) σ(1) σ(2) (3) −y (a)y (a) + β y (a)y (a) − y (a)y (a) 4 σ(2) σ(1) σ(2) σ(1) σ(2) σ(1) + β β μ y (a)y (a) − y (a)y (a) . (4.20) 3 4 σ(1) σ(2) σ(2) σ(1) (a) The highest μ-powers of the characteristic functions of the problems of Case 2 occur with (3) (3) iβ μ y (a)y (a) − y (a)y (a) . (4.21) σ(1) σ(2) σ(1) σ(2) We investigate the zeros of (3) (3) φ (μ) = 2μ y (a)y (a) − y (a)y (a) . σ(1) σ(2) σ(2) σ(1) (a) It follows from (4.3)and (4.4) that for the six cases of Case 2, we obtain: Case 1: p = 0, p = 1: 1 2 φ (μ) = μ(cos(μa) sinh(μa) + sin(μa) cosh(μa)). Case 2: p = 0, p = 2: 1 2 φ (μ) = μ cos(μa) cosh(μa). Case 3: p = 0, p = 3: 1 2 φ (μ) = μ (cos(μa) sinh(μa) − sin(μa) cosh(μa)). Case 4: p = 1, p = 2: 1 2 φ (μ) = μ (cos(μa) sinh(μa) − sin(μa) cosh(μa)). Case 5: p = 1, p = 3: 1 2 φ (μ) =−2μ sin(μa) sinh(μa). Case 6: p = 2, p = 3: 1 2 φ (μ) =−μ (sin(μa) cosh(μa) + cos(μa) sinh(μa)). (a) Next, we provide the asymptotic distribution of the zeros of the functions φ of the problems of Case 2, with their number. Lemma 4.3 Case 1: p = 0, p = 1, φ has a zero of multiplicity 2 at 0, exactly one simple zero μ ˜ in each 1 2 0 k 1 π 1 π interval k − , k + ,for k ∈ N, with asymptotics 2 a 2 a μ ˜ = (4k − 1) + o(1), k = 1, 2,..., 4a simple zeros −˜ μ , μ ˜ = i μ ˜ and −i μ ˜ for k = 1, 2,..., and no additional zeros. k −k k k Case 2: p = 0, p = 2, φ has a zero of multiplicity 2 at 0,simplezeros 1 2 0 μ ˜ = (2k − 1) , k = 1, 2,..., 2a simple zeros at −˜ μ , μ ˜ = i μ ˜ and −i μ ˜ ,k = 1, 2,... , and no additional zeros. k −k k k 123 Arab. J. Math. Case 3: p = 0,p = 3, φ has a zero of multiplicity 6 at 0, exactly one simple zero μ ˜ in each interval 1 2 0 k 1 π 1 π k − , k + ,for k ∈ N, with asymptotics 2 a 2 a μ ˜ = (4k − 5) + o(1), k = 2, 3,..., 4a simple zeros −˜ μ , μ ˜ = i μ ˜ and −i μ ˜ for k = 3, 4,..., and no additional zeros. k −k k k Case 4: p = 1, p = 2, φ has a zero of multiplicity 6 at 0, exactly one simple zero μ ˜ in each interval 1 2 0 k 1 π 1 π k − , k + ,for k ∈ N, with asymptotics 2 a 2 a μ ˜ = (4k − 5) + o(1), k = 2, 3,..., 4a simple zeros −˜ μ , μ ˜ = i μ ˜ and −i μ ˜ for k = 3, 4,..., and no additional zeros. k −k k k Case 5: p = 1, p = 3, φ has a zero of multiplicity 6 at 0,simplezeros 1 2 0 μ ˜ = (k − 1) , k = 2, 3,..., simple zeros −˜ μ , μ ˜ = i μ ˜ and −i μ ˜ for k = 2, 3,... , and no additional zeros. k −k k k Case 6: p = 2, p = 3, φ has a zero of multiplicity 6 at 0, exactly one simple zero μ ˜ in each interval 1 2 0 k 1 π 1 π k − , k + ,for k ∈ N, with asymptotics 2 a 2 a μ ˜ = (4k − 5) + o(1), k = 2, 3,..., 4a simple zeros −˜ μ , μ ˜ = i μ ˜ and −i μ ˜ for k = 2, 3,... , and no additional zeros. k −k k k Proof The proof is similar to the proof of Lemma 4.1. Proposition 4.4 For g = 0, there exists k ∈ N such the eigenvalues λ ,k ∈ Z, of the problems (2.1), (3.4), 0 k ˆ ˆ ˆ where (p , q ) = (2, 1) and ( p , q ) = (3, 0), are λ =−λ , λ =ˆ μ for k ≥ k and the μ ˆ have the 3 3 4 4 −k k k 0 k following asymptotic descriptions as k →∞: Case 1: p = 0, p = 1, μ ˆ = (4k − 1) + o(1). 1 2 k 4a Case 2: p = 0, p = 2, μ ˆ = (2k − 1) + o(1). 1 2 k 2a Case 3: p = 0, p = 3, μ ˆ = (4k − 5) + o(1). 1 2 k 4a Case 4: p = 1, p = 2, μ ˆ = (4k − 5) + o(1). 1 2 k 4a Case 5: p = 1, p = 3, μ ˆ = (k − 1) + o(1). 1 2 k Case 6: p = 2, p = 3, μ ˆ = (4k − 5) + o(1). 1 2 k 4a In particular, there exists an odd number of the pure imaginary eigenvalues in each case. Proof Case 3: We obtain iβ μ φ(μ) = (cos(μa) sinh(μa) − sin(μa) cosh(μa)) (1 − β β )μ 3 4 − sin(μa) sinh(μa) iβ μ − (sin(μa) cosh(μa) + cos(μa) sinh(μa)). (4.22) All the estimates are similar to those in Case 3 of the proof of Proposition 4.2 and the result can be derived from that in Case 3 of the proof Proposition 4.2. The results in Case 1, Case 4 and Case 6 follow from reasonings respectively similar to those in Case 1, Case 4 and Case 6 of the proof of Proposition 4.2. 123 Arab. J. Math. Case 2: We get φ(μ) = iβ μ cos(μa) cosh(μa) (1 − β β )μ 3 4 + (cos(μa) sinh(μa) − sin(μa) cosh(μa)) − iβ sin(μa) sinh(μa). (4.23) All the estimates are as in Case 2 of the proof of Proposition 4.2 and the result follows from that in Case 2 of the proof of Proposition 4.2. The result in Case 5 follows from reasonings similar to those in Case 5 of the proof of Proposition 4.2. 5 Asymptotics of eigenvalues We consider the canonical fundamental system y , j = 1,..., 4, as deﬁned in (4.1) of the problems (2.1), (3.4). Denoting respectively by D and D the characteristic functions for g = 0 and general g and using the same reasoning as in [16, Section 5], then Proposition 4.2 leads to Proposition 5.1 For g ∈ C [0, a], there exists k ∈ N such the eigenvalues λ ,k ∈ Z of the problem (2.1), 0 k [ p ] [ p ] [3] 1 2 (3.4), where B (λ)y = y (0),B (λ)y = y (0),B (λ)y = y (a) + iβ λy(a),B (λ)y = y (a) + 1 2 3 3 4 ˆ ˆ ˆ iβ λy (a) are λ =−λ , λ =ˆ μ for k ≥ k and the μ ˆ have the following asymptotic descriptions as 4 −k k k 0 k k →∞: Case 1: p = 0, p = 1, μ ˆ = (4k − 3) + o(1). 1 2 k 4a Case 2: p = 0, p = 2, μ ˆ = (k − 1) + o(1). 1 2 k Case 3: p = 0, p = 3, μ ˆ = (4k − 5) + o(1). 1 2 k 4a Case 4: p = 1, p = 2, μ ˆ = (4k − 5) + o(1). 1 2 k 4a Case 5: p = 1, p = 3, μ ˆ = (2k − 1) + o(1). 1 2 k 2a Case 6: p = 2, p = 3, μ ˆ = (4k − 7) + o(1). 1 2 k 4a In particular, the number of the pure imaginary eigenvalues is even in each case. However, Proposition 4.4 leads to Proposition 5.2 For g ∈ C [0, a], there exists k ∈ N such the eigenvalues λ ,k ∈ Z of the problem (2.1), 0 k [ p ] [ p ] [3] 1 2 (3.4), where B (λ)y = y (0),B (λ)y = y (0),B (λ)y = y (a) + iβ λy (a),B (λ)y = y (a) + 1 2 3 3 4 ˆ ˆ ˆ iβ λy(a) are λ =−λ , λ =ˆ μ for k ≥ k and the μ ˆ have the following asymptotic descriptions as 4 −k k k 0 k k →∞: Case 1: p = 0, p = 1, μ ˆ = (4k − 1) + o(1). 1 2 k 4a Case 2: p = 0, p = 2, μ ˆ = (2k − 1) + o(1). 1 2 k 2a Case 3: p = 0, p = 3, μ ˆ = (4k − 5) + o(1). 1 2 k 4a Case 4: p = 1, p = 2, μ ˆ = (4k − 5) + o(1). 1 2 k 4a Case 5: p = 1, p = 3, μ ˆ = (k − 1) + o(1). 1 2 k Case 6: p = 2, p = 3, μ ˆ = (4k − 5) + o(1). 1 2 k 4a In particular, the number of the pure imaginary eigenvalues in each case is odd. 123 Arab. J. Math. (a) We establish, in the sequel, more precise eigenvalue asymptotics respectively of the problems in Case 1and (a) 2 Case 2. Replacing λ with μ , then according to [9, Theorem 8.2.1], the differential equation (2.1) has an asymptotic fundamental system {η ,η ,η ,η } of the form 1 2 3 4 ν−1 ( j ) i μx η (x,μ) = δ (x,μ)e , (5.1) ν, j where d ν−1 ν−1 ν−1 −r i μx −i μx −4+ j δ (x,μ) = (μi ) ϕ (x )e e + o(μ ), (5.2) ν, j r dx r =0 (k) j = 0, 1, 2, 3, where [ ] signiﬁes that expressions of the Leibniz expansion containing a function ϕ with j r dx [3] k > 4 − r are omitted. Because the coefﬁcient of y in (2.1) is zero, we have ϕ (x ) = 1, see [9, (8.2.3)]. Next, we ﬁnd the functions ϕ and ϕ . Note from [9, (8.1.2) and (8.1.3)] that n = 0and l = 4, see [9, 1 2 0 Theorem 8.1.2]. From [9, (8.2.45)], we have T [r ] ϕ = ϕ = ε VQ ε , (5.3) r 1,r 1 4 ( j −1)(k−1) 4 [r ] where ε is the ν-th unit vector in C , V = (i ) ,and Q are 4 × 4 matrices given by [9, (8.2.28), j,k=1 [0] (8.2.33) and (8.2.34)], that is, Q = I , [1] [1] [0] Q − Q = Q = 0, (5.4) 4 4 [2] [2] [1] −2 [0] Q − Q = Q − g εε Q , (5.5) 4 4 4 −1− j T [2] T [2− j ] 0 = ε Q + k εε Q ε (ν = 1, 2, 3, 4), (5.6) 3− j 4 ν j =1 with = diag(1, i, −1, −i ), k =−g , k =−g and ε = (1, 1, 1, 1) . Denote G(x ) = g(t )dt.An 4 1 2 extensive calculation yields 1 1 1 ϕ = G,ϕ = G − g (5.7) 1 2 4 32 8 and thus, 1 1 1 ν−1 −ν+1 −1 ν−1 2 −2 i μx η = 1 + i Gμ + (−1) G − g μ e 4 32 8 ν−1 −2 i μx +{o(μ )} e (5.8) for ν = 1, 2, 3, 4, where {o()} signiﬁes that the estimate is uniform in x. (a) Next, we provide the ﬁrst four terms of the eigenvalue asymptotics of the problems (2.1), (3.4)for Case (a) (a) 1and Case 2 respectively. We start with the problems of Case 1. (a) The characteristic function of (2.1), (3.4) for the problems of Case 1is D(μ) = det(γ exp(ε )) , j,k j,k j,k=1 where k−1 ε = ε = 0,ε = ε = i μa,γ = δ (0,μ), 1,k 2,k 3,k 4,k 1,k k, p γ = δ (0,μ) if p ≤ 2,γ = δ (0,μ) − g(0)δ (0,μ) if p = 3, 2,k k,2 2 2,k k,3 k,1 2 γ = δ (a,μ) + iβ μ δ (a,μ), 3,k k,1 3 k,0 γ = δ (a,μ) − g(a)δ (a,μ) + iβ μ δ (a,μ). 4,k k,3 k,1 4 k,2 123 Arab. J. Math. Note that ω μa D(μ) = ψ (μ)e , (5.9) m=1 where ω = 1 + i, ω =−1 + i, ω =−1 − i, ω = 1 − i, ω = 0. The functions ψ ,...,ψ have the 1 2 3 4 5 1 5 k k−1 k k 0 0 asymptotics c μ + c μ +· · · + c μ + o(μ ). k k−1 k It follows from (5.9)that −ω μa (ω −ω )μa 1 m 1 D (μ) := D(μ)e = ψ (μ) + ψ (μ)e , (5.10) 1 1 m m=2 where ω − ω =−2, ω − ω =−2 − 2i, ω − ω =−2i, ω − ω =−1 − i. 2 1 3 1 4 1 5 1 3π π (ω −ω )μa − sin |μ|a m 3 Thus, for arg μ ∈[− , ],wehave |e |≤ e for m = 2, 3,5and ψ (μ) will absorb 8 8 (ω −ω )μa −s m 1 the expressions ψ (μ)e ,for m = 2, 3, 5, because they are of the form o(μ ) for any integer s. 3π π Hence, for arg μ ∈[− , ], 8 8 (ω −ω )μa −2i μa 4 3 D (μ) = ψ (μ) + ψ (μ)e = ψ (μ) + ψ (μ)e , (5.11) 1 1 4 1 4 where ψ (μ) = γ γ − γ γ γ γ − γ γ , (5.12) 1 13 24 23 14 31 42 32 41 ψ (μ) = γ γ − γ γ γ γ − γ γ . (5.13) 4 12 23 22 13 31 44 34 41 Calculations give 6 5 γ γ − γ γ = 2β β μ + (1 − i )(2β β φ (a) + (β + β ))μ (5.14) 31 42 32 41 3 4 3 4 1 3 4 2 4 4 − 2i (β β φ (a) + (β + β )φ (a) + 1)μ + o(μ ), 3 4 3 4 6 5 γ γ − γ γ = 2β β μ + (1 + i )(2β β φ (a) − (β + β ))μ 31 44 34 41 3 4 3 4 1 3 4 2 4 4 + 2i (β β φ (a) − (β + β )φ (a) + 1)μ + o(μ ). (5.15) 3 4 3 4 1 For the remaining two factors in (5.12)and (5.13) we consider the six different cases. Case 1: p = 0, p = 1. We have for this case 1 2 γ γ − γ γ = (1 − i )μ + o(μ), (5.16) 13 24 23 14 γ γ − γ γ =−(1 + i )μ + o(μ). (5.17) 12 23 22 13 Therefore, 7 6 ψ (μ) = 2(1 − i )β β μ − i (β β G(a) + 2(β + β ))μ 1 3 4 3 4 3 4 2 5 5 − (1 + i )(G (a) − 4β β g(0) + 4(β + β )G(a) + 16)μ + o(μ ), 3 4 3 4 7 6 ψ (μ) =−2(1 + i )β β μ − i (β β G(a) − 2(β + β ))μ (5.18) 4 3 4 3 4 3 4 2 5 5 + (1 − i )(β β G (a) − 4β β g(0) − 4(β + β )G(a) − 16)μ + o(μ ). (5.19) 3 4 3 4 3 4 Case 2: p = 0, p = 2. Here we get 1 2 2 2 γ γ − γ γ =−2μ + o(μ ), (5.20) 13 24 23 14 2 2 γ γ − γ γ = 2μ + o(μ ). (5.21) 12 23 22 13 123 Arab. J. Math. Thus, 8 7 ψ (μ) =−4β β μ − (1 − i )(β β G(a) + 2(β + β ))μ 1 3 4 3 4 3 4 2 6 6 + i (β β G (a) + (β + β )G(a) + 4)μ + o(μ ), (5.22) 3 4 3 4 8 7 ψ (μ) = 4β β μ + (1 + i )(β β G(a) − 2(β + β ))μ 4 3 4 3 4 3 4 2 6 6 + i (β β G (a) + 4(β − β )G(a) + 16)μ + o(μ ). (5.23) 3 4 3 4 Case 3: p = 0, p = 3. We obtain 1 2 3 3 γ γ − γ γ = (1 + i )μ + o(μ ), (5.24) 13 24 23 14 3 3 γ γ − γ γ =−(1 − i )μ + o(μ ). (5.25) 12 23 22 13 Hence, 9 8 ψ (μ) = 2(1 + i )β β μ + (β β G(a) + 2(β + β ))μ 1 3 4 3 4 3 4 2 7 7 + (1 − i )(β β G (a) − 4β β g(0) + 4(β + β )G(a) + 16)μ + o(μ ), (5.26) 3 4 3 4 3 4 9 8 ψ (μ) =−2(1 − i )β β μ − (β β G(a) − 2(β + β ))μ 4 3 4 3 4 3 4 2 7 7 − (1 + i )(β β G (a) − 4β β g(0) − 4(β + β )G(a) + 16)μ + o(μ ). (5.27) 3 4 3 4 3 4 Case 4: p = 1, p = 2. Here we have 1 2 3 3 γ γ − γ γ = (1 + i )μ + o(μ ), (5.28) 13 24 23 14 3 3 γ γ − γ γ =−(1 − i )μ + o(μ ). (5.29) 12 23 22 13 Thus, 9 8 ψ (μ) = 2(1 + i )β β μ + (β β G(a) + 2(β + β ))μ 1 3 4 3 4 3 4 2 7 7 + (1 − i )(β β G (a) − 4β β g(0) + 4(β + β )G(a) + 16)μ + o(μ ). (5.30) 3 4 3 4 3 4 9 8 ψ (μ) =−2(1 − i )β β μ − (β β G(a) − 2(β + β ))μ 4 3 4 3 4 3 4 2 7 7 − (1 + i )(β β G (a) − 4β β g(0) − 4(β + β )G(a) + 16)μ + o(μ ). (5.31) 3 4 3 4 3 4 Case 5: p = 1, p = 3. We get 1 2 4 4 γ γ − γ γ =−2i μ + o(μ ), (5.32) 13 24 23 14 4 4 γ γ − γ γ =−2i μ + o(μ ). (5.33) 12 23 22 13 Therefore, 10 9 ψ (μ) =−4iβ β μ − (1 + i )(β β G(a) + 2(β + β ))μ 1 3 4 3 4 3 4 2 8 8 − (β β G (a) + 4(β + β )G(a) + 16)μ + o(μ ), (5.34) 3 4 3 4 10 9 ψ (μ) =−4iβ β μ + (1 − i )(β β G(a) − 2(β + β ))μ 4 3 4 3 4 3 4 2 8 8 + (β β G (a) − 4(β + β )G(a) + 16)μ + o(μ ). (5.35) 3 4 3 4 Case 6: p = 2, p = 3. We obtain 1 2 5 5 γ γ − γ γ =−(1 − i )μ + o(μ ), (5.36) 13 24 23 14 5 5 γ γ − γ γ = (1 + i )μ + o(μ ). (5.37) 12 23 22 13 123 Arab. J. Math. Hence, 11 10 ψ (μ) =−2(1 − i )β β μ + i (β β G(a) + 2(β + β ))μ 1 3 4 3 4 3 4 2 9 9 + (1 + i )(β β G (a) + 12β β g(0) + 4(β + β )G(a) + 16)μ + o(μ ), (5.38) 3 4 3 4 3 4 11 10 ψ (μ) = 2(1 + i )β β μ + i (β β G(a) − 2(β + β ))μ 4 3 4 3 4 3 4 2 9 9 − (1 − i )(β β G (a) + 12β β g(0) − 4(β + β )G(a) + 16)μ + o(μ ). (5.39) 3 4 3 4 3 4 Proposition 5.1 tells us that the zeros μ of D can be expressed asymptotically as μ = k + τ + o(1) as k k 0 k →∞. To ameliorate the asymptotics, we write −m −n μ = k + τ(k), τ (k) = τ k + o(k ), k = 1, 2,.... (5.40) k m m=0 Since the eigenvalues are symmetric, we just need to ﬁnd asymptotics as k →∞. The number τ is known from Proposition 5.1 and we need to ﬁnd τ and τ . Hence, we will replace (5.40)into D (μ ) = 0and then 1 2 1 k 0 −1 −2 equating the coefﬁcients of k , k and k . Observe that τ τ 1 2 −2i μ a −2iτ(k)a −2i τ a −2 k 0 e = e = e exp −2ia + + o(k ) k k 1 1 −2i τ a 2 2 −2 = e 1 − 2iaτ − 2a τ + 2iaτ + o(k ) , (5.41) 1 2 k k while −1 1 a aτ(k) a a τ −2 = 1 + = − + o(k ). (5.42) 2 2 μ πk kπ kπ k π We know that D (μ ) = 0 can be written as 1 k −γ −γ −2i τ a μ ψ (μ ) + μ ψ (μ )e = 0, (5.43) 1 k 4 k k k where γ is the highest μ-power in ψ (μ) and ψ (μ). Replacing (5.41)and (5.42)into(5.43) and equating the 1 4 0 −1 −2 coefﬁcients of k , k and k we obtain Theorem 5.3 For g ∈ C [0, a], there exists k ∈ N such that the eigenvalues λ ,k ∈ Z of the problem (2.1), 0 k [ p ] [ p ] [3] 1 2 (3.4),where B (y) = y (0),B (y) = y (0),B y = y (a) + iβ λy(a) and B y = y (a) + iβ λy (a) 1 2 3 3 4 4 are λ =−λ , λ = μ for k ≥ k and the μ have the asymptotics −k k k 0 k π τ τ 1 2 −2 μ = k + τ + + + o(k ) k 0 a k k and the numbers τ , τ , τ are as follows: 0 1 2 Case 1: p = 0,p = 1, 1 2 3π 1 G(a) 1 i 1 − β β 3 4 τ =− ,τ = + , 0 1 4a 4 π 2 π β 3 G(a) 1 g(0) 1 a 1 1 2 3 i 1 1 τ = − − + − + + . 2 2 2 2 16 π 4 π 4 π β β 8 π β β β β 3 4 3 4 3 4 Case 2: p = 0,p = 2, 1 2 π 1 G(a) 1 i 1 − β β 3 4 =− = , τ ,τ + 0 1 a 4 π 2 π β 1 G(a) 1 a 1 1 2 1 i 1 1 τ = − + − + + . 2 2 2 4 π 4 π β β β β 2 π β β 3 4 3 4 3 4 123 Arab. J. Math. Case 3: p = 0,p = 3, 1 2 5π 1 β β − 1 i G(a) 3 4 τ =− ,τ = − , 0 1 4a 2 πβ 4 π 5 G(a) 1 ag(0) 1 a 1 1 2 5 i 1 1 τ = + − + − + + . 2 2 2 2 16 π 4 π 4 π β β 8 π β β β β 3 4 3 4 3 4 Case 4: p = 1,p = 2, 1 2 5π 1 β β − 1 i G(a) 3 4 τ =− ,τ = − , 0 1 4a 2 πβ 4 π 5 G(a) 1 ag(0) 1 a 1 1 2 5 i 1 1 τ = − − + − + + . 2 2 2 2 16 π 4 π 4 π β β β β 8 π β β 3 4 3 4 3 4 Case 5: p = 1,p = 3, 1 2 π 1 G(a) 1 i 1 − β β 3 4 τ =− ,τ = + , 0 1 2a 4 π 2 π β 1 G(a) 1 i 1 1 1 a 1 1 2 τ = + + − + − . 2 2 2 8 π 4 π β β 4 π β β β β 3 4 3 4 3 4 Case 6: p = 2,p = 3, 1 2 7π 1 G(a) 1 i 1 − β β 3 4 τ =− ,τ = + , 0 1 4a 4 π 2 π πβ 7 G(a) 3 ag(0) 7 i 1 1 1 a 1 1 2 τ = + + + − + − . 2 2 2 16 π 4 π 8 π β β 4 π β β β β 3 4 3 4 3 4 In particular, the number of the pure imaginary eigenvalues is even in each case. (a) Next, we provide the ﬁrst four terms of the eigenvalue asymptotics of the problems (2.1), (3.4)ofCase 2. (a) The characteristic function of (2.1), (3.4) for the problems of Case 2is D(μ) = det(γ exp(ε )) , j,k j,k j,k=1 where k−1 ε = ε = 0,ε = ε = i μa,γ = δ (0,μ), 1,k 2,k 3,k 4,k 1,k k, p γ = δ (0,μ) if p ≤ 2,γ = δ (0,μ) − g(0)δ (0,μ) if p = 3, 2,k k,2 2 2,k k,3 k,1 2 γ = δ (a,μ) + iβ μ δ (a,μ), 3,k k,2 3 k,1 γ = δ (a,μ) − g(a)δ (a,μ) + iβ μ δ (a,μ). 4,k k,3 k,1 4 k,0 Note that for the calculations of the functions ψ and ψ respectively deﬁned in (5.12)and (5.13) only the 1 4 factors γ γ − γ γ and γ γ − γ γ respectively given in (5.14)and (5.15) will change. Hence, we 31 42 32 41 31 44 34 41 are going to provide these two terms. Calculations yield 1 1 6 5 2 γ γ − γ γ = 2β μ + (1 − i )(β G(a) − 2β β + 2)μ − i (β G (a) 31 42 32 41 3 3 3 4 3 2 8 4 4 + 4(1 − β β )G(a) − 16β )μ + o(μ ), (5.44) 3 4 4 1 1 6 5 2 γ γ − γ γ =−2β μ − (1 + i )(β G(a) + 2β β − 2)μ − i (β G (a) 31 44 34 41 3 3 3 4 3 2 8 4 4 − 4β (1 − β )G(a) − 16)μ + o(μ ). (5.45) 3 4 (a) Using the same method as for Case 1, we get 123 Arab. J. Math. Case 1: p = 0, p = 1. It follows from (5.12), (5.16), (5.44) on one hand and from (5.13), (5.17), (5.45)on 1 2 the other hand that 7 6 2 ψ (μ) = 2(1 − i )β μ − i (β G(a) − 2β β + 2)μ − (1 + i )(β G (a) 1 3 3 3 4 3 5 5 + 4(1 − β β )G(a) − 4β g(0) − 16)μ + o(μ ), (5.46) 3 4 3 7 6 2 ψ (μ) = 2(1 + i )β μ + i (β G(a) + 2β β − 2)μ − (1 − i )(β G (a) 4 3 3 3 4 3 5 5 − 4(1 − β β )G(a) − 4β g(0) − 16β )μ + o(μ ). (5.47) 3 4 3 4 Case 2: p = 0, p = 2. Using (5.12), (5.20), (5.44)and (5.13), (5.21), (5.45), we have 1 2 8 7 ψ (μ) =−4β μ + (1 − i )(β G(a) − 2β β − 2)μ 1 3 3 3 4 2 6 6 − i (β G (a) + 4(1 − β β )G(a) − 16β )μ + o(μ ), (5.48) 3 3 4 4 8 7 ψ (μ) =−4β μ − (1 + i )(β G(a) + 2β β − 2)μ 4 3 3 3 4 2 6 6 − i (β G (a) − 4(1 − β β )G(a) − 16β )μ + o(μ ). (5.49) 3 3 4 4 Case 3: p = 0, p = 3. Putting respectively (5.12), (5.24), (5.44)and (5.13), (5.25), (5.45), together gives 1 2 9 8 2 ψ (μ) = 2(1 + i )β μ + (β G(a) − 2β β + 2)μ + (1 − i )(β G (a) 1 3 3 3 4 3 7 7 + 4(1 − β β )G(a) − 4β g(0) − 16β )μ + o(μ ), (5.50) 3 4 3 4 9 8 2 ψ (μ) = 2(1 − i )β μ + (β G(a) + 2β β − 2)μ + (1 + i )(β G (a) 4 3 3 3 4 3 7 7 − 4(1 − β β )G(a) − β g(0) − 16β )μ + o(μ ). (5.51) 3 4 3 4 Case 4: p = 1, p = 2. The equations (5.12), (5.28), (5.44)and (5.13), (5.29), (5.45), respectively yield 1 2 9 8 2 ψ (μ) = 2(1 + i )β μ + (β G(a) − 2β β + 2)μ + (1 − i )(β G (a) 1 3 3 3 4 3 7 7 + 4(1 − β β )G(a) − 4β g(0) − 16β )μ + o(μ ), (5.52) 3 4 3 4 9 8 2 ψ (μ) = 2(1 − i )β μ + (β G(a) + 2β β − 2)μ + (1 + i )(β G (a) 4 3 3 3 4 3 7 7 − 4(1 − β β )G(a) − β g(0) − 16β )μ + o(μ ). (5.53) 3 4 3 4 Case 5: p = 1, p = 3. It follows from (5.12), (5.32)and (5.44) on one hand and from (5.13), (5.33)and 1 2 from (5.45) on the other hand that 10 9 ψ (μ) =−4iβ μ + (1 + i )(2β β G(a) − β G(a) − 2)μ 1 3 3 4 3 2 8 8 − (β G (a) + 4(1 − β β )G(a) − 16β )μ + o(μ ), (5.54) 3 3 4 4 10 9 ψ (μ) = 4iβ μ − (1 − i )(2β β + β G(a) − 2)μ 4 3 3 4 3 2 8 8 − (β G (a) − 4(1 − β β )G(a) − 16β )μ + o(μ ). (5.55) 3 3 4 4 Case 6: p = 2, p = 3. Using respectively (5.12), (5.36)and (5.44) on one hand and (5.13), (5.37)and (5.45) 1 2 on the other hand, we get 11 10 2 ψ (μ) =−2(1 − i )β μ + i (β G(a) − 2β β + 2)μ + (1 + i )(β G (a) 1 3 3 3 4 3 9 9 + 4(1 − β β )G(a) + 12β g(0) − 16β )μ + o(μ ), (5.56) 3 4 3 4 11 10 2 ψ (μ) =−2(1 + i )β μ − i (β G(a) + 2β β − 2)μ + (1 − i )(β G (a) 4 3 3 3 4 3 9 9 − 4(1 − β β )G(a) + 12β g(0) − 16β )μ + o(μ ). (5.57) 3 4 3 4 123 Arab. J. Math. Using (5.40)–(5.43) and applying to Proposition 5.2 the same reasoning and calculations as for Proposition 5.1, we get Theorem 5.4 For g ∈ C [0, a], there exists k ∈ N such that the eigenvalues λ ,k ∈ Z of the problem (2.1), 0 k [ p ] [ p ] [3] 1 2 (3.4),where B (y) = y (0),B (y) = y (0),B y = y (a) + iβ λy (a) and B y = y (a) + iβ λy(a) 1 2 3 3 4 4 are λ =−λ , λ = μ for k ≥ k and the μ have the asymptotics −k k k 0 k π τ τ 1 2 −2 μ = k + τ + + + o(k ) k 0 a k k and the numbers τ , τ , τ are as follows: 0 1 2 Case 1: p = 0,p = 1, 1 2 π 1 G(a) 1 i 1 1 τ =− ,τ = + − , 0 1 4a 4 π 2 π β β 3 4 2 2 1 G(a) 1 ag(0) 1 a(β β + 2β β + 1) 1 i 1 − β β 3 4 3 4 3 4 τ = − − + . 2 2 2 16 π 4 π 4 π β 8 π β Case 2: p = 0,p = 2, 1 2 π 1 G(a) 1 i 1 1 τ =− ,τ = + − , 0 1 2a 4 π 2 π β β 3 4 2 2 1 G(a) 1 a(β β + 2β β + 1) 1 i 1 − β β 3 4 3 4 3 4 τ = − + . 8 π 4 4 π β π β Case 3: p = 0,p = 3, 1 2 5π i G(a) 1 1 1 τ =− ,τ =− + − , 0 1 4a 4 π 2 β β 4 3 2 2 2 a(β β − 2β β + 1) 1 aG (a) 1 aG(a) β β − 1 5 β β − 1 1 4 3 4 3 3 4 3 4 τ =− + + − 2 2 2 16 π 4 π β 8 πβ 4 π β 3 3 5i G(a) − . 16 π Case 4: p = 1,p = 2, 1 2 5π i G(a) 1 1 1 τ =− ,τ =− + − , 0 1 4a 4 π 2 β β 4 3 2 2 2 1 aG (a) 1 aG(a) β β − 1 5 β β − 1 1 a(β β − 2β β + 1) 4 3 4 3 3 4 3 4 τ =− + + − 2 2 2 16 π 4 π β 8 πβ 4 π β 3 3 5i G(a) − . 16 π Case 5: p = 1,p = 3, 1 2 π 1 G(a) 1 i 1 1 τ =− ,τ = + − , 0 1 a 4 π 2 π β β 3 4 2 2 1 G(a) 1 a(β β + 2β β + 1) 1 i 1 − β β 3 4 3 4 3 4 τ = − + . 2 2 4 π 4 π β 2 π β Case 6: p = 2,p = 3, 1 2 5π 1 G(a) 1 i 1 1 τ =− ,τ = + − , 0 1 4a 4 π 2 π β β 3 4 2 2 5 G(a) 3 ag(0) 1 a(β β + 2β β + 1) 5 i 1 − β β 3 4 3 4 3 4 τ = + − + . 2 2 16 π 4 π 4 8 π πβ π β In particular, in each case, the number of pure imaginary eigenvalues is odd. 123 Arab. J. Math. Remark 5.1 In [17], we have considered the differential equation (2.1) with the boundary terms B (λ)y and B (λ)y at 0 as in this paper. However, only Cases 1,2,5 and 6 of this paper appear in [17]. (a) The boundary terms B (λ)y and B (λ)y considered in Case 1 of this paper differ from those of [17]. 3 4 However, according to the values of τ , we can observe that if β > 0, j = 3, 4, or if β β < 0and β +β ≤ 0, 1 j 3 4 3 4 then the eigenvalues of the operator pencil L(λ) lie on the closed upper half-plane satisfying [17, Proposition 2.3]. (a) The boundary terms B (λ)y and B (λ)y considered in Case 2 of this paper are those of [17]but where 3 4 β > 0and β < 0. We can observe that all eigenvalues of L(λ) lie in the closed upper half-plane in Cases 1, 3 4 2, 5and6if β > 0and β β < 1orif β < 0and β β > 1. However, the eigenvalues in Cases 3 and 4 will 3 3 4 3 3 4 lie in the closed upper half-plane if β > 0and β β > 1or β < 0and β β < 1. 3 3 4 3 3 4 6 Asymptotics of eigenvalues of the problem describing the stability of a ﬂexible missile We consider, in this section, the problem (2.1), (3.4)where β = β = 0, p = p = 2and p = p = 3. 3 4 1 3 2 4 This problem represents the stability of ﬂexible missile problem investigated in [4,6,7]. It follows from (4.3), (4.4)and (4.20) that the characteristic function of the problem for g = 0is: φ(μ) = 2μ [1 − cos(μa) cosh(μa)]. (6.1) Next, we give the asymptotic distributions of the zeros of φ(μ) with their numbers. Lemma 6.1 For g = 0 the function φ has a zero of multiplicity eight at 0, exactly one simple zero in each π 1 π 3 π π interval 2 m , 2 m + and 2 m + ,(2 m + 2) ,for m ∈ N ∪{0} with asymptotics a 2 a 2 a a μ ˜ = (2k − 5) + o(1), k = 3, 4,..., 2a simple zeros at −˜ μ , μ ˜ = i μ ˜ ,-i μ ˜ ,for k = 3, 4,... , and no other zeros. k −k k k Proof The proof of this lemma is similar to the proof of Case 1 of [20, Lemma 3.1]. Proposition 6.2 For g = 0, β = β = 0,p = p = 2 and p = p = 3, there exists k ∈ N such that the 3 4 1 3 2 4 0 eigenvalues λ ,k ∈ Z, numbered with multiplicity, of the problem (2.1), (3.4), can be indexed such that the ˆ ˆ ˆ ˆ eigenvalues λ are real and satisfy λ =−λ .For k > 0, we put λ =ˆ μ ,where the μ ˆ have the following k −k k k k asymptotic description as k →∞: μ ˆ = (2k − 5) + o(1). 2a Note that in this case, there is no perturbed term. Hence, φ (μ) = 0and φ(μ) = φ (μ). 1 0 the characteristic function of (2.1), (3.4), in this case, is D(μ) = det(γ exp(ε )) , j,k j,k j,k=1 where k−1 ε = ε = 0,ε = ε = i μa, 1,k 2,k 3,k 4,k γ = δ (0,μ), γ = δ (0,μ) − g(0)δ (0,μ), 1,k k,2 2,k k,3 k,1 γ = δ (a,μ), γ = δ (a,μ) − g(a)δ (a,μ). 3,k k,2 4,k k,3 k,1 We calculate the functions ψ and ψ respectively deﬁned in (5.12)and (5.13). An extensive calculation gives 1 4 5 3 3 γ γ − γ γ =−(1 − i )μ + (1 + i )g(0)μ + o(μ ), (6.2) 13 24 23 14 5 3 3 γ γ − γ γ = (1 + i )μ − (1 − i )g(0)μ + o(μ ), (6.3) 12 23 22 13 1 1 5 4 2 3 3 γ γ − γ γ = (1 − i )μ + iG(a)μ − (1 + i ) G (a) + 12g(a) μ + o(μ ), (6.4) 31 42 32 41 2 16 1 1 5 4 2 3 γ γ − γ γ = (1 + i )μ + iG(a)μ − (1 − i )(G (a) + 12g(a)) + o(μ ). (6.5) 31 44 34 41 2 16 123 Arab. J. Math. Therefore, it follows from (5.12)and (5.13)that 1 1 10 9 2 8 8 ψ (μ) = 2i μ + (1 + i )G(a)μ + (G (a) + 12(g(0) + g(a))μ + o(μ ), (6.6) 2 8 1 1 10 9 2 8 8 ψ (μ) = 2i μ − (1 − i )G(a)μ + (G (a) + 12(g(0) + g(a))μ + o(μ ). (6.7) 2 8 Using (5.40)–(5.43) and applying to Proposition 6.2 the same reasoning and calculations as for Proposition 5.1, we get Theorem 6.3 For g ∈ C [0, a], there exists k ∈ N such that the eigenvalues λ ,k ∈ Z of the problem 0 k describing the stability of a ﬂexible missile are λ =−λ , λ = μ for k ≥ k and the μ have the −k k k 0 k asymptotics π τ τ 1 2 −2 μ = k + τ + + + o(k ) k 0 a k k and the numbers τ , τ , τ are as follows: 0 1 2 5π 1 G(a) 5 G(a) 1 a τ =− ,τ = ,τ = + (5g(0) + 3g(a)). 0 1 2 2 2 2a 4 π 8 π 4 π In particular, all the eigenvalues are real. Remark 6.1 Note from Lemma 6.1 and the values of τ , τ and τ in Theorem 6.3 that the asymptotics of 0 1 2 the zeros of φ(μ) deﬁned in (6.1) are either real or pure imaginary. Hence, the eigenvalues of the problem describing the stability of a ﬂexible missile are all real. Note as well that according to [16, Theorem 1.2] the problem describing the stability of a ﬂexible missile is self-adjoint and therefore its eigenvalues must necessary be real. Acknowledgements This research was partially supported by a grant from the NRF of South Africa, Grant number 119201. Open Access This article is licensed under a Creative Commons Attribution 4.0 International License, which permits use, sharing, adaptation, distribution and reproduction in any medium or format, as long as you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons licence, and indicate if changes were made. The images or other third party material in this article are included in the article’s Creative Commons licence, unless indicated otherwise in a credit line to the material. 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Arabian Journal of Mathematics – Springer Journals
Published: Dec 1, 2023
Keywords: 34L20; 34B08; 34B09
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