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Statistics, Ningxia University, In this paper, we study Riemann boundary-value problem for doubly-periodic Yinchuan, P.R. China Full list of author information is bianalytic functions. By the decomposition of doubly-periodic polyanalytic functions, available at the end of the article the problem is transformed into two equivalent and independent Riemann boundary-value problems of doubly-periodic analytic functions, which has been discussed according to growth order of functions at the origin by Jianke Lu. Finally, we obtain the explicit expression of solutions and the conditions of solvability for the doubly-periodic bianalytic functions. MSC: 30G30; 45E05 Keywords: Doubly-periodic bianalytic function; Riemann boundary-value problem; Growth order; Canonical function; Freedom 1 Introduction An extension of analytic function leads to polyanalytic function, which is usually deﬁned as solutions of simple complex partial diﬀerential equation ∂ f =0,where ∂ is the classical z ¯ z ¯ Cauchy–Riemann operator ∂ = 1/2[∂/∂x + i(∂/∂y)]. Polyanalytic function stemmed from z ¯ planar elasticity problems and was ﬁrst investigated by Kolossov in 1908. A good overview of polyanalytic function is included in Balk’s excellent monograph [1]orthe literature [2]. Recently, various boundary-value problems (BVP) of polyanalytic functions and other functions determined by the general partial diﬀerential equations have been widely in- vestigated by Begehr, Schmersau, Hile, Vanegas, Kumar, Jinyuan Du, Yufeng Wang, Ying Wang, Zhihua Du and others (see, for example, [1–21]). The general partial diﬀerential equations include the inhomogeneous polyanalytic equation [5], the higher order Poisson equation [6], and polyharmonic equations [7, 8]. Riemann BVP of single-periodic polyanalytic functions has been investigated [10, 11]. Analogously, a Riemann BVP of rotation-invariant polyanalytic functions has been dis- cussed in [20]. Actually, a single-periodic polyanalytic function is deﬁned by a translation- invariant group T = τ : τ (z)= z + nω, n ∈ Z with ω ∈ C \{0}, n n which is generated by two elements {τ , τ } with τ (z)= z ± ω. Generally speaking, the 1 –1 ±1 single-periodic polyanalytic function is translation-invariant under the group T ,and the © The Author(s) 2018. This article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, pro- vided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made. Han et al. Boundary Value Problems (2018) 2018:88 Page 2 of 20 rotation-invariant polyanalytic function is invariant under a rotation group 2kπi R = ρ : ρ (z)= e z, k = 0,1,2,..., m –1 . k k In general, single-periodic polyanalytic function and rotation-invariant polyanalytic func- tion are automorphic. In 1935, Natanzon ﬁrst made use of doubly-periodic bianalytic function to deal with a problem on stresses deriving from a stretched plate. In 1957, Erwe also studied other classes of doubly-periodic polyanalytic functions. In 1982, Pokazeev further considered a general form of doubly-periodic polyanalytic functions. A concise history of investiga- tion of doubly-periodic polyanalytic function has been introduced in the literature [2]. Doubly-periodic polyanalytic function is also automorphic and is determined by another translation-invariant group G = τ : τ (z)= z +2kω +2ω , k, ∈ Z with Im(ω /ω ) =0. k, k, 1 2 2 1 Investigation of Riemann BVP for this kind of functions is a spontaneous thing. For the compact Riemann surfaces of ﬁnite genus, the classical BVP of analytic func- tions was discussed in [22, 23]. However, for the very important doubly-periodic prob- lem, it is essential to have an eﬀective method of solution which has been systematically investigated by Jianke Lu [24]. Later, BVP of automorphic analytic functions has been ﬁrst discussed by Gakhov and Chibrikova [25, 26]. Up to now, Riemann BVP for doubly-periodic polyanalytic function has not been well- posed and systematically investigated. In this article, our main objective is to set up the theory of doubly-periodic bianalytic function. The way to solve this problem is the conver- sion method used in [16]. Riemann BVP for doubly-periodic polyanalytic functions will be presented in the forthcoming paper. This article is organized as follows. In Sect. 2, we give a decomposition of doubly- periodic polyanalytic functions, which will be used to solve BVPs of doubly-periodic bi- analytic functions. It is worth mentioning that the decomposition obtained here is dis- tinct from the classical decomposition described in [2]. In Sect. 3, the growth order of doubly-periodic polyanalytic functions at the origin is deﬁned. To pose the reasonable BVPs of doubly-periodic bianalytic functions, the deﬁnition of growth order at the origin is needed. In the classical monographs [24, 25], the growth order of doubly-periodic func- tions is not explicitly deﬁned. In Sect. 4, Riemann BVP of doubly-periodic bianalytic func- tions is presented. The solutions and conditions of solvability of this kind of problem are obtained by Jianke Lu [24]. By the decomposition of doubly-periodic bianalytic functions, the problem is transformed into two independent Riemann-type BVPs of doubly-periodic analytic functions. Finally, the solution is explicitly expressed as an integral representation. 2 Doubly-periodic polyanalytic functions Without loss of generality, we always assume that Im(ω /ω ) > 0 in the following. The 2 1 parallelogram with vertices ω + ω ,–ω + ω ,–ω – ω ,and ω – ω is denoted by S ,which 1 2 1 2 1 2 1 2 0 is usually called the fundamental cell. Obviously, the origin is the center of the fundamental cell S . 0 Han et al. Boundary Value Problems (2018) 2018:88 Page 3 of 20 The classical Weierstrass’s ζ -function is deﬁned by 1 1 1 z ζ (z)= + + + , (2.1) z z – k k (k,)=(0,0) with =2kω +2ω ,and k, l are integers. Clearly, k 1 2 ζ (z +2ω )= ζ (z)+2η , j =1,2, j j where η = ζ (ω ) satisﬁes the relation j j 2η ω –2η ω = πi. (2.2) 1 2 2 1 Let φ(z)= z ¯ – λz – δζ (z), (2.3) with 2i 2i λ = (w η – w η ), δ =– (w w – w w ). (2.4) 1 2 2 1 1 2 2 1 π π By simple computation, one has φ(z +2ω )= z +2ω – λ(z +2ω )– δζ (z +2ω )= φ(z), z =0. j j j j This implies that φ is a doubly-periodic bianalytic function. If the open set on the complex plane C satisﬁes the condition z +2kω +2ω ∈ for 1 2 ∀z ∈ , ∀k, ∈ Z,then is called a doubly-periodic open set with periods 2ω ,2ω .Sim- 1 2 ilar to the deﬁnition of single-periodic polyanalytic function in [10], we give the following deﬁnition. Deﬁnition 2.1 Suppose f to be a polyanalytic function [1]oforder n on ,where is a doubly-periodic open set with periods 2ω ,2ω .If 1 2 f (z +2ω )= f (z), ∀z ∈ , j = 1, 2, (2.5) then we say that f is a doubly-periodic polyanalytic function of order n with periods 2ω ,2ω on , or simply doubly-periodic polyanalytic function. The collection of all the 1 2 doubly-periodic polyanalytic functions on is denoted by DPH ( ). By Deﬁnition 2.1, DPH ( ) is just a set of all the doubly-periodic analytic functions on the doubly-periodic open set . The function f ∈ DPH ( ) is called doubly-periodic bianalytic function. DPH ( ) is a subset of the collection of polyanalytic functions on denoted by H ( )= {f : ∂ f (z)=0, z ∈ }.Equation(2.5)isequivalentto z ¯ k k ∂ f (z +2ω )= ∂ f (z), ∀z ∈ for k = 0,1,..., n –1 and j = 1, 2. (2.6) z ¯ z ¯ Han et al. Boundary Value Problems (2018) 2018:88 Page 4 of 20 Now we introduce the subset of DPH ( ) as follows: DPH ( ;2ω ,2ω )= f ∈ DPH ( ): f (z +2ω )= f (z), ∀z ∈ for j =1,2 , n 1 2 n j which is an object of investigation in the following. Finally, one arrives at the decomposition of doubly-periodic polyanalytic functions, used to solve Riemann BVP of doubly-periodic bianalytic functions in the sequel. Theorem 2.1 Let be a doubly-periodic open set with periods 2ω ,2ω and 0/ ∈ . Then 1 2 DPH ( ;2ω ,2ω )= DPH ( ;2ω ,2ω ) n 1 2 1 1 2 ⊕ z – λz – δζ (z) DPH ( ;2ω ,2ω ) 1 1 2 n–1 ⊕ ··· ⊕ z – λz – δζ (z) DPH ( ;2ω ,2ω ), (2.7) 1 1 2 j j with [z – λz – δζ (z)] DPH ( ;2ω ,2ω )= {[z – λz – δζ (z)] f (z): f ∈ DPH ( ;2ω ,2ω )} for 1 1 2 1 1 2 j = 0,1,..., n –1, where ζ is deﬁned by (2.1), and λ, δ are given by (2.4). Proof We only need to verify the relation ⊆ by induction. Obviously, if n =1,the theorem is straightforward. Suppose that the relation DPH ( ;2ω ,2ω ) ⊆ DPH ( ;2ω ,2ω ) n–1 1 2 1 1 2 ⊕ z – λz – δζ (z) DPH ( ;2ω ,2ω ) 1 1 2 n–2 ⊕ ··· ⊕ z – λz – δζ (z) DPH ( ;2ω ,2ω ) 1 1 2 is valid. Next one has to verify DPH ( ;2ω ,2ω ) ⊆ DPH ( ;2ω ,2ω ) n 1 2 1 1 2 ⊕ z – λz – δζ (z) DPH ( ;2ω ,2ω ) 1 1 2 n–1 ⊕ ··· ⊕ z – λz – δζ (z) DPH ( ;2ω ,2ω ). (2.8) 1 1 2 Let f ∈ DPH ( ;2ω ,2ω ). Then ∂ f ∈ DPH ( ;2ω ,2ω ). And hence, by the induc- n 1 2 z ¯ n–1 1 2 tive hypothesis, there exist g (z) ∈ DPH ( ;2ω ,2ω ), j = 0,1,..., n –2, such that j 1 1 2 n–2 [z – λz – δζ (z)] ∂ f (z)= g (z)+ g (z), z ∈ . (2.9) z ¯ 0 j j! j=1 Setting n–2 j+1 [z – λz – δζ (z)] h(z)= f (z)– z – λz – δζ (z) g (z)– g (z), z ∈ , (2.10) 0 j (j +1)! j=1 one has, by (2.9), n–2 [z – λz – δζ (z)] ∂ h(z)= ∂ f (z)– g (z)– g (z)=0, z ∈ . z ¯ z ¯ 0 j j! j=1 Han et al. Boundary Value Problems (2018) 2018:88 Page 5 of 20 Therefore, h ∈ DPH ( ;2ω ,2ω ). And (2.10)isrewritten as 1 1 2 n–2 j+1 [z – λz – δζ (z)] f (z)= h(z)+ z – λz – δζ (z) g (z)+ g (z), z ∈ , 0 j (j +1)! j=1 which implies that (2.8) remains true. Theorem 2.1 indicates that f ∈ DPH ( ;2ω ,2ω ) admits a unique decomposition n 1 2 n–1 f (z)= z – λz – δζ (z) f (z), z ∈ , (2.11) j=0 where f ∈ DPH ( ;2ω ,2ω ) is called j-component of f with respect to the base {[z – λz – j 1 1 2 δζ (z)] : j = 0,1,2,..., n –1}. Specially, g ∈ DPH ( ;2ω ,2ω ) has theuniqueexpansion 2 1 2 g(z)= g (z)+ z – λz – δζ (z) g (z), z ∈ , (2.12) 1 2 with g ∈ DPH ( ;2ω ,2ω ), j =1,2. j 1 1 2 3 Growth order of doubly-periodic bianalytic functions First, the following deﬁnition is analogous to that in [10]. This is just the deﬁnition of growth order for the general polyanalytic function at the origin [15]. Deﬁnition 3.1 Suppose to be a doubly-periodic open set with periods 2ω ,2ω ,0 ∈ 1 2 and f ∈ DPH ( ;2ω ,2ω ). If there exists an integer m such that n 1 2 lim sup z f (z) = α, α ∈ (0, +∞), z∈ ,z→0 then we say that f possesses order m at the origin, denoted by Ord(f ,0) = m.If lim sup z f (z) =+∞ for any m ∈ Z, z∈ ,z→0 then we say that f has order +∞ at the origin, denoted by Ord(f ,0) = +∞. We assume Ord(f ,0) = –∞ if and only if f =0. Now, one has the following result needed in the sequel. Lemma 3.1 Suppose to be a doubly-periodic open set with periods 2ω ,2ω ,0 ∈ and 1 2 f ∈ DPH ( ;2ω ,2ω ). At the deleted neighbor of the origin, one has 1 1 2 (j) f (z)= c (z)+ c ζ (z), (3.1) 0 j j=1 where ζ is deﬁned by (2.1), the order of c (z) ∈ DPH ( ;2ω ,2ω ) is not more than 1 and 0 1 1 2 c ∈ C, j = 1,2,... . j Han et al. Boundary Value Problems (2018) 2018:88 Page 6 of 20 Proof Since 0 ∈ and f ∈ DPH ( ;2ω ,2ω ), at the deleted neighbor of the origin, one 1 1 2 has Laurent’s expansion +∞ 1 1 j–1 f (z)= d , d = f (t)t dt, (3.2) j j z 2πi |t|=r j=–∞ for suﬃciently small r >0. Let 1 1 1 z ζ (z)= + ζ (z), ζ (z)= + + , (3.3) 0 0 z z – k k (k,)=(0,0) where ζ (0) = 0 and ζ (z) is analytic at the origin. Thus one gets 0 0 (–1) (j) (j) + ζ (z)= + ζ (z), j ∈ Z , j+1 which is equivalent to (j–1) j–1 (j–1) + =(–1) ζ (z)– ζ (z) , j ∈ Z . (3.4) Inserting (3.4)into(3.2), we get j (j) f (z)= d(z)+ d (–1) ζ (z), j+1 j=1 with +∞ ∞ (j) j j d(z)= d z – d (–1) ζ (z) ∈ DPH ( ;2ω ,2ω ). –j j+1 1 1 2 j=–1 j=1 The order of d(z) is obviously not more than 1. This completes the proof. Corollary 3.1 Suppose that f ∈ DPH (C;2ω ,2ω ) possesses a uniquely possible singular 1 1 2 point z =0 in the fundamental cell S . Then one has (j) f (z)= c + c ζ (z), z ∈ S \{0}, (3.5) 0 j 0 j=1 where ζ is deﬁned by (2.1), and c ∈ C, j = 0,1,2,... . Proof By Lemma 3.1, at the deleted neighbor of the origin, one has (j) f (z)= c (z)+ c ζ (z), 0 j j=1 where ζ is deﬁned by (2.1), the order of c (z) ∈ DPH ( ;2ω ,2ω ) is not more than 1 and 0 1 1 2 c ∈ C, j = 1,2,... . And c (z) ∈ DPH (C;2ω ,2ω )implies that c (z) is an elliptic function. j 0 1 1 2 0 By Liouville’s theorem, c (z) is a constant function. 0 Han et al. Boundary Value Problems (2018) 2018:88 Page 7 of 20 Lemma 3.2 Suppose to be a doubly-periodic open set with periods 2ω ,2ω ,0 ∈ and 1 2 f ∈ DPH ( ;2ω ,2ω ). If Ord(f ,0) ≤ m, then Ord(f ,0) ≤ m +2 and Ord(f ,0) ≤ m +1, 2 1 2 1 2 where f is j-component of f . Proof First, this theorem is veriﬁed under m >1. By Theorem 2.1,there exist f , f ∈ 1 2 DPH ( ;2ω ,2ω )such that f (z)= f (z)+ φ(z)f (z). This leads to 1 1 2 1 2 f (z)= f (z)– λz + δζ (z) f (z) + zf (z) 1 2 2 ∞ ∞ j k = a z + z b z , (3.6) j k j=–∞ k=–∞ near the origin, where a , b are constants. j k and ≥ m.Wechoosesuﬃcientlysmall r >0 such that D = {z : |z| < r}⊆ . Let ∈ Z By (3.6), one has ∞ ∞ 1 1 r j k f (t)t dt = a t + b t t dt j k 2πi 2πi t |t|=r |t|=r j=–∞ k=–∞ 1 a 1 b –(+1) – = dt + r dt 2πi t 2πi t |t|=r |t|=r = a + r b . (3.7) –(+1) – Now Ord(f ,0) ≤ m implies that there exist M >0, r >0 such that f (z) ≤ , |z| < r . |z| We assume 0 < r < r , and one has the estimation +1–m f (t)t dt ≤ Mr . (3.8) 2πi |t|=r Combining (3.7)with(3.8), we get 2 +1–m |a | + r |b |≤ Mr –(+1) – for suﬃciently small r >0. Let r → 0 ,and onehas a =0, ≥ m. –(+1) This leads to Ord f – λz + δζ (z) f ,0 ≤ m. 1 2 Therefore, Ord(f ,0) = Ord(zf ,0) + 1 ≤ max Ord(f ,0), Ord f – λz + δζ (z) f ,0 +1 ≤ m +1, 2 2 1 2 Han et al. Boundary Value Problems (2018) 2018:88 Page 8 of 20 and Ord(f ,0) ≤ max Ord(f ,0), Ord(φf ,0) ≤ m +2, 1 2 where φ is deﬁned by (2.3). Finally, if m ≤ 1, by Corollary 3.1 in [15], similar to the discussion above, it is not diﬃcult to know that the conclusion remains true. In what follows,weneed theoperators L [f ]= d , j ∈ Z, (3.9) j j where d is the j-coeﬃcient of Laurent’s expansion of the function f deﬁned in (3.2). Theorem 3.1 Suppose to be a doubly-periodic open set with periods 2ω ,2ω ,0 ∈ 1 2 and f ∈ DPH ( ;2ω ,2ω ). Then Ord(f ,0) ≤ mif and only if 2 1 2 Ord(f ,0) ≤ m +3– j, j = 1, 2, (3.10) and L [f ]= δL [f ], j = m, m + 1, (3.11) j+1 1 j 2 where L is the operator deﬁned by (3.9), and δ is given in (2.4). Proof First, we assume m > 3. And we prove the necessity. By Theorem 2.1,there exist f , f ∈ DPH ( ;2ω ,2ω )such that f (z)= f (z)+ φ(z)f (z). By Lemma 3.2, Ord(f ,0) ≤ m 1 2 1 1 2 1 2 implies Ord(f ,0) ≤ m +2 and Ord(f ,0) ≤ m + 1. Also by Lemma 3.1, 1 2 m+1 (j) f (z)= c (z)+ c ζ (z), (3.12) 1 0 j j=1 (j) f (z)= d (z)+ d ζ (z). (3.13) 2 0 j j=1 Inserting (3.12)and (3.13)intothe expression f (z)= f (z)+ φ(z)f (z), one easily gets 1 2 m+1 (k) f (z)= c (z)+ c ζ (z) 0 k k=1 (j) + z – λz – δζ (z) d (z)+ d ζ (z) . (3.14) 0 j j=1 Thus, Ord(f ,0) ≤ m,where f givenin(3.14), if and only if (k+1) (k) P.P c ζ (z)– δd ζ (z)ζ (z), 0 =0, k = m –1, m, (3.15) k+1 k Han et al. Boundary Value Problems (2018) 2018:88 Page 9 of 20 where P.P(h, 0) denotes the principal part of h.Equation(3.15)isequivalentto k+1 d = , k = m –1, m. Inserting this expression into (3.14), one has m–1 (k) (m) (m+1) f (z)= c (z)+ c ζ (z)+ c ζ (z)+ c ζ (z) 0 k m m+1 k=1 m–2 c c m m+1 (j) (m–1) (m) + z – λz – δζ (z) d (z)+ d ζ (z)+ ζ (z)+ ζ (z) , (3.16) 0 j δ δ j=1 where c (z), d (z) ∈ DPH ( ;2ω ,2ω ), Ord(c ,0) ≤ 1, Ord(d ,0) ≤ 1, and d , c ∈ C. 0 0 1 1 2 0 0 j k Therefore, (3.11)istrue. The suﬃciency is obvious. Finally, if m ≤ 3, analogously to the discussion above, the conclusion is also true. This completes the proof of this theorem. 4 Riemann BVP for doubly-periodic analytic functions In this section, we will give the solutions and conditions of solvability for Riemann BVP for doubly-periodic analytic functions, which was investigated in detail by Jianke Lu [24]. Let L be a closed smooth Jordan curve, oriented counterclockwise. The fundamen- + – tal cell S is divided into two domains denoted by S and S ,respectively. Withoutloss 0 0 + + of generality, we always assume 0 ∈ S .Let L =2kω +2ω + L for k, ∈ Z, S = k, 1 2 0 + – (2kω +2ω + S ), S = C \ S ,and we assume that L has the same orientation 1 2 k, k,∈Z 0 as L for every k, ∈ Z. For the convenience, we set L = L . 0 k, k,∈Z Now, our problem is to ﬁnd a sectionally doubly-periodic analytic function (z) ∈ DPH ( ;2ω ,2ω ) satisfying a boundary condition and a growth condition 1 1 2 ⎨ + – (t)= G(t) (t)+ g(t), t ∈ L, (4.1) Ord(,0) ≤ m, where the given Hölder-continuous functions G, g satisfy G(t+2ω )= G(t), g(t+2ω )= g(t), j j j =1,2 and G(t) =0, t ∈ L. This problem is simply called DR problem. Introduce the function σ (z)σ (z – ω – ω ) 1 2 μ(z)= , (4.2) σ (z – ω )σ (z – ω ) 1 2 with z z z σ (z)= z 1– exp + . (4.3) 2kω +2ω 2kω +2ω 2kω +2ω 1 2 1 2 1 2 (k,)=(0,0) Then μ is an elliptic function with periods 2ω ,2ω and Ord(μ,0) = –1. Let 1 2 κ = G(t) , (4.4) 2π Han et al. Boundary Value Problems (2018) 2018:88 Page 10 of 20 which is called the index, and –κ G = log μ (t)G(t) dt. (4.5) 2πi Without loss of generality, we assume G ∈/ L in the following. The solutions and condi- tions of solvability of DR problem (4.1)are presentedintwo cases. 4.1 The case G =2kω +2ω for some k, ∈ Z ∗ 1 2 In this case, G =0 (mod 2ω ,2ω ). Let ∗ 1 2 ⎨ κ (z) + μ (z)h(z)e , z ∈ S , X(z)= (4.6) (z) – h(z)e , z ∈ S , where –κ + – (z)= log μ (t)G(t) ζ (t – z)dt, z ∈ S ∪ S (4.7) 2πi and h(z)= exp 2(kη + η )z . (4.8) 1 2 In (4.8), η = ζ (ω ), j =1,2. X deﬁned by (4.6) is called the canonical function which pos- j j sesses the following ﬁve properties: (1) X ∈ DPH ( ;2ω ,2ω ); 1 1 2 + – (2) X (t)= G(t)X (t), t ∈ L; (3) X (t) ∈ H(L); (4) X(z) =0, z =2(pω + qω ) with p, q ∈ Z and X (t) =0 for t ∈ L; 1 2 (5) X(z) has a pole of order –κ at the origin, or say Ord(X,0) = –κ. If the function Y also satisﬁes ﬁve properties from (1) to (5) above, then there exists C ∈ C such that Y (z)= CX(z), where X is given by (4.6). For convenience, we introduce the set of elliptic functions of order k with an exclusive singular point z = 0 as follows: (k) {c + c ζ (z)+ ··· + c ζ (z): c ∈ C, j = 0,1,..., k}, k >0, ⎪ 0 1 k j (ζ)= (4.9) C, k =0, {0}, k <0. Now we state the results obtained by Jianke Lu. Theorem 4.1 Under this case, two subcases arise: (1) When κ + m >0, DR problem (4.1) is solvable and its solution can be expressed as X(z) g(t) (z)= ζ (t – z)+ ζ (z) dt 2πi X (t) + X(z)p (z), p ∈ (ζ ), (4.10) κ+m–1 κ+m–1 κ+m–1 where (ζ ) is deﬁned by (4.9) and X is given by (4.6). κ+m–1 Han et al. Boundary Value Problems (2018) 2018:88 Page 11 of 20 (2) When κ + m ≤ 0, if and only if 1 g(t) (k) ζ (t)dt =0, k = –1,0,1,2,...,–κ – m – 1, (4.11) 2πi X (t) DR problem (4.1) is solvable and its solution can be written as X(z) g(t) (z)= ζ (t – z)– ζ (t) dt + CX(z), C ∈ (ζ ). (4.12) κ+m 2πi X (t) (–1) (0) We assume ζ (t)= ζ (t)=1 in (4.11). In general, the freedom of solutions is κ + m. 4.2 The case G =2kω +2ω for any k, ∈ Z ∗ 1 2 In this case, there exists G ∈ S such that G = G (mod 2ω ,2ω ), and G =0. Let X be 0 0 ∗ 0 1 2 0 deﬁned by (4.6), where σ (z) h(z)= . (4.13) σ (z – G ) At this time, X is also the canonical function which satisﬁes four properties from (1) to (4) in Sect. 4.1,and (5 ) X(z) has a pole of order –κ –1 at the origin, precisely Ord(X,0) = –κ –1. The following result is also obtained by Jianke Lu. X used in the following theorem is deﬁned by (4.6)with(4.13). Theorem 4.2 Under this case, two subcases arise: (1) When κ + m +1 ≥ 0, if and only if the condition of solvability 1 g(t) (k) ζ (t)dt =0, k = 0,1,2,...,–κ – m – 1, (4.14) 2πi X (t) is fulﬁlled, DR problem (4.1) is solvable and its solution can be expressed as X(z) g(t) (z)= ζ (t – z)+ ζ (z)– ζ (t – G )– ζ (G ) dt 0 0 2πi X (t) + X(z) p (z)– p (G ) , (4.15) κ+m κ+m 0 (0) (j) with p ∈ (ζ ). We assume ζ (t)=1 and ζ (t)=0 for j <0 in (4.14). κ+m κ+m (2) When κ + m +1 < 0, if and only if 1 g(t) ζ (t – G )– ζ (t) dt = 0 (4.16) 2πi X (t) and 1 g(t) (k) ζ (t)dt =0, k = 0,1,2,...,–κ – m – 2, (4.17) 2πi X (t) are satisﬁed, DR problem (4.1) is solvable and its solution can be written as (4.15). (0) We assume ζ (t)=1 in (4.17). In general, the freedom of solutions is κ + m. Han et al. Boundary Value Problems (2018) 2018:88 Page 12 of 20 5 Riemann BVP for doubly-periodic bianalytic functions In this section, we consider the following Riemann BVP for doubly-periodic bianalytic + – functions with the same factor: ﬁnd a function V ∈ DPH (S ∪ S ;2ω ,2ω )satisfying 2 1 2 two Riemann-type boundary conditions and a growth condition + – V (t)= G(t)V (t)+ g (t), t ∈ L, ⎪ 1 + – (5.1) (∂ V) (t)= G(t)(∂ V) (t)+ g (t), t ∈ L, z ¯ z ¯ 2 Ord(V,0) ≤ m, where the given boundary datum G and g , j = 1, 2, are Hölder-continuous on every curve L and G(t) =0, t ∈ L. In addition, G(t +2ω )= G(t), g (t +2ω )= g(t), g (t +2ω )= g(t) k, j 1 j 2 j for j =1,2 and t ∈ L. This problem is simply called DBR problem. + – Since V ∈ DPH (S ∪S ;2ω ,2ω ), by Theorem 2.1 or (2.12), one has the decomposition 2 1 2 + – V(z)= V (z)+ z – λz – δζ (z) V (z), z ∈ S ∪ S , (5.2) 1 2 where V ∈ DPH ( ;2ω ,2ω )for j =1,2. j 1 1 2 Now, we will prove that DBR problem (5.1) can be transformed to two independent DR problem (5.3)and DR problem (5.4) as follows: m+2 m+1 ⎨ + – V (t)= G(t)V (t)+ g ˜ (t), t ∈ L, 1 1 (5.3) Ord(V ,0) ≤ m +2, and ⎨ + – V (t)= G(t)V (t)+ g (t), t ∈ L, 2 2 (5.4) Ord(V ,0) ≤ m +1, where g ˜ (t)= g (t)– t – λt – δζ (t) g (t). (5.5) 1 1 2 The solutions and conditions of solvability for those two problems have been presented in the preceding section. Lemma 5.1 Let V , V , V be given in (5.2). Then V is the solution of DBR problem (5.1) if 1 2 m and only if V , V are respectively the solutions of DR problem (5.3) and DR problem 1 2 m+2 m+1 (5.4) satisfying the relation L [V ]= δL [V ], j = m, m + 1, (5.6) j+1 1 j 2 where L is the operator deﬁned by (3.9), and δ is given in (2.4). Proof Suppose that V , V are respectively the solutions of DR problem (5.3)and 1 2 m+2 DR problem (5.4) satisfying relation (5.6). By Theorem 3.1 and (5.6), Ord(V ,0) ≤ m+2 m+1 1 Han et al. Boundary Value Problems (2018) 2018:88 Page 13 of 20 and Ord(V ,0) ≤ m +1 lead to Ord(V,0) ≤ m. (5.7) On the other hand, one has + + + V (t)= V (t)+ φ(t)V (t) 1 2 – – – = G(t) V (t)+ φ(t)V (t) + g (t)= G(t)V (t)+ g (t), (5.8) 1 1 1 2 and + + – (∂ V) (t)= ∂ (V + φV ) (t)= V (t)= G(t)V (t) z ¯ z ¯ 1 2 2 2 = ∂ (V + φV ) (t)= G(t)(∂ V) (t)+ g (t). (5.9) z ¯ 1 2 z ¯ 2 Combining (5.7), (5.8)with(5.9), V is just a solution of DBR problem (5.1). Conversely, if V is the solution of DBR problem (5.1), obviously boundary conditions in (5.3)and (5.4) are valid. By Theorem 3.1, Ord(V,0) ≤ m implies Ord(V ,0) ≤ m +2 and Ord(V ,0) ≤ m + 1, and the validity of relation (5.6). This completes the proof. Analogously to the preceding section, we will discuss DBR problem (5.1)intwo cases according to G =0 (mod 2ω ,2ω )or G =0 (mod 2ω ,2ω ). ∗ 1 2 ∗ 1 2 5.1 The case G =2kω +2ω for some k, ∈ Z ∗ 1 2 In this case, G =0 (mod 2ω ,2ω ). And we will discuss DBR problem (5.1)inthree sub- ∗ 1 2 m cases. Theorem 5.1 If κ + m +1 > 0, DBR problem (5.1) is solvable and its solution can be expressed as V(z)= X(z)W[g , g ](z)+ X(z) p (z)+ φ(z)q (z) , (5.10) 1 2 κ+m+1 κ+m with L [p ]= δL [q ], j = κ + m, κ + m + 1, (5.11) j+1 j+1 j j where p ∈ (ζ ), q ∈ (ζ ) and κ+m+1 κ+m+1 κ+m κ+m 1 g (t) W[g , g ](z)= ζ (t – z)+ ζ (z) dt 1 2 2πi X (t) 1 g (t) + φ(z)– φ(t) ζ (t – z)+ ζ (z) dt. (5.12) 2πi X (t) Proof By Theorem 4.1, the solution of DR problem (5.3)can be expressedas m+2 X(z) g (t) V (z)= ζ (t – z)+ ζ (z) dt 2πi X (t) + X(z)p (z), p ∈ (ζ ), (5.13) κ+m+1 κ+m+1 κ+m+1 Han et al. Boundary Value Problems (2018) 2018:88 Page 14 of 20 where (ζ)is deﬁned by (4.9)and X is given by (4.6). By Theorem 4.1,the solution κ+m+1 of DR problem (5.4)can be expressedas m+1 X(z) g (t) V (z)= ζ (t – z)+ ζ (z) dt + X(z)p (z), p ∈ (ζ ). (5.14) 2 κ+m κ+m κ+m 2πi X (t) According to Lemma 5.1, inserting (5.13)and (5.14)into(5.2), one gets X(z) g (t)– φ(t)g (t) 1 2 V(z)= ζ (t – z)+ ζ (z) dt + X(z)p (z) κ+m+1 2πi X (t) φ(z)X(z) g (t) + ζ (t – z)+ ζ (z) dt + φ(z)X(z)p (z) κ+m 2πi X (t) X(z) g (t)+[φ(z)– φ(t)]g (t) 1 2 = ζ (t – z)+ ζ (z) dt 2πi X (t) + X(z) p (z)+ φ(z)p (z) , κ+m+1 κ+m which leads to (5.10). At the same time, (5.6) is reduced to (5.11). This completes the proof. Remark 5.1 Under this case, combining (5.10)with(5.11), the solution of DBR problem (5.1)can be rewrittenas (κ+m) (κ+m+1) V(z)= X(z)W[g , g ](z)+ X(z) p (z)+ c ζ (z)+ c ζ (z) 1 2 κ+m–1 κ+m κ+m+1 c c κ+m κ+m+1 (κ+m–1) (κ+m) + φ(z) q (z)+ ζ (z)+ ζ (z) , κ+m–2 δ δ where p ∈ (ζ ), q ∈ (ζ ), c ∈ C and c ∈ C. κ+m–1 κ+m–1 κ+m–2 κ+m–2 κ+m κ+m+1 Theorem 5.2 If κ + m +1 = 0, if and only if 1 g (t) dt = 0, (5.15) 2πi X (t) DBR problem (5.1) is solvable and its solution can be represented as V(z)= X(z)W[g , g ](z)+ X(z) δ + φ(z) , C ∈ C, (5.16) 1 2 with L W[g , g ] = δL W[g , g ] , (5.17) –1 1 2 –2 1 2 0 1 where {W[g , g ]} is j-component of W[g , g ] deﬁned by 1 2 j 1 2 φ(z) g (t) W[g , g ](z)= W[g , g ](z)– ζ (z)+ ζ (t) dt. (5.18) 1 2 1 2 2πi X (t) 0 Han et al. Boundary Value Problems (2018) 2018:88 Page 15 of 20 Proof By Theorem 4.1, the solution of DR problem (5.3)can be expressedas m+2 X(z) g (t) V (z)= ζ (t – z)+ ζ (z) dt + X(z)C , C ∈ C. (5.19) 1 1 1 2πi X (t) By Theorem 4.1, if and only if the condition of solvability (5.15) is fulﬁlled, the solution of DR problem (5.4)can be expressedas m+1 X(z) g (t) V (z)= ζ (t – z)– ζ (t) dt + X(z)C , C ∈ C. (5.20) 2 2 2 2πi X (t) By Lemma 5.1, putting (5.19)and (5.20)into(5.2), one easily gets 1 g (t)– φ(t)g (t) 1 2 V(z)= X(z) ζ (t – z)+ ζ (z) dt 2πi X (t) φ(z) g (t) + ζ (t – z)– ζ (t) dt 2πi X (t) + X(z) C + φ(z)C , (5.21) 1 2 with C , C ∈ C. Also by Lemma 5.1, V given by (5.21) is the solution of DBR problem 1 2 m (5.1)ifand only if C = δC and (5.17) are satisﬁed. And hence the proof of the theorem is 1 2 completed. Theorem 5.3 If κ + m +1 < 0, if and only if 1 g (t)– φ(t)g (t) 1 2 (k) ζ (t)dt =0, k = –1,0,1,2,...,–κ – m – 3, (5.22) 2πi X (t) and 1 g (t) (k) ζ (t)dt =0, k = –1,0,1,2,...,–κ – m – 2, (5.23) 2πi X (t) DBR problem (5.1) is solvable and its solution can be represented as V(z)= X(z) W[g , g ](z)+ C , (5.24) 1 2 with C = δL [{W[g , g ]} ], κ+m+1 1 2 1 (5.25) L [{W[g , g ]} ]= δL [{W[g , g ]} ], κ+m+1 1 2 0 κ+m 1 2 1 where {W[g , g ]} is j-component of W[g , g ] deﬁned by 1 2 j 1 2 1 g (t) W[g , g ](z)= ζ (t – z)– ζ (t) dt 1 2 2πi X (t) 1 g (t) + φ(z)– φ(t) ζ (t – z)– ζ (t) dt. (5.26) 2πi X (t) 0 Han et al. Boundary Value Problems (2018) 2018:88 Page 16 of 20 Proof By Theorem 4.1, if and only if the conditions of solvability (5.22) are fulﬁlled, the solution of DR problem (5.3)can be expressedas m+2 X(z) g (t)– φ(t)g (t) 1 2 V (z)= ζ (t – z)– ζ (t) dt 2πi X (t) + CX(z), C ∈ (ζ ). (5.27) κ+m+2 Analogously, by Theorem 4.1, if and only if the conditions of solvability (5.23) are satisﬁed, the solution of DR problem (5.4)can be writtenas m+2 X(z) g (t) V (z)= ζ (t – z)– ζ (t) dt. (5.28) 2πi X (t) Therefore, by Lemma 5.1, if and only if the conditions of solvability (5.22)and (5.23)are fulﬁlled, the solution of DBR problem (5.1)can be expressedas X(z) g (t)– φ(t)g (t) 1 2 V(z)= ζ (t – z)– ζ (t) dt + X(z)C 2πi X (t) φ(z)X(z) g (t) + ζ (t – z)– ζ (t) dt 2πi X (t) X(z) g (t)+[φ(z)– φ(t)]g (t) 1 2 = ζ (t – z)– ζ (t) dt 2πi X (t) + X(z)C, C ∈ (ζ ), κ+m+2 satisfying relation (5.25). This completes the proof. 5.2 The case G =2kω +2ω for any k, ∈ Z ∗ 1 2 In this case, there exists G ∈ S such that G = G (mod 2ω ,2ω ), and G =0. We will 0 0 ∗ 0 1 2 0 investigate the problem in four subcases. Theorem 5.4 If κ + m +2 ≥ 0, DBR problem (5.1) is solvable and its solution can be expressed as V(z)= X(z) W[g , g ](z)– W[g , g ](G ) 1 2 1 2 0 + X(z) p (z)– p (G ) + φ(z) q (z)– q (G ) , (5.29) κ+m+2 κ+m+2 0 κ+m+1 κ+m+1 0 with L [p ]= δL [q ], j = κ + m +1, κ + m + 2, (5.30) j+1 j+1 j j and p ∈ (ζ ), q ∈ (ζ ), where X is thesameasthatinSect. 4.2 and κ+m+1 κ+m+1 κ+m κ+m W[g , g ](z) is given in (5.12). 1 2 Proof By Theorem 4.2, the solution of DR problem (5.3)can be expressedas m+2 X(z) g (t)– φ(t)g (t) 1 2 V (z)= ζ (t – z)+ ζ (z)– ζ (t – G )– ζ (G ) dt 1 0 0 2πi X (t) + X(z) p (z)– p (G ) , (5.31) κ+m+2 κ+m+2 0 Han et al. Boundary Value Problems (2018) 2018:88 Page 17 of 20 with p ∈ (ζ ), where (ζ)is deﬁned by (4.9)and X is given by (4.6)with κ+m+2 κ+m+2 κ+m+2 (4.13). By Theorem 4.2, the solution of DR problem (5.4)can be representedas m+1 X(z) g (t) V (z)= ζ (t – z)+ ζ (z)– ζ (t – G )– ζ (G ) dt 2 0 0 2πi X (t) + X(z) p (z)– p (G ) , (5.32) κ+m+1 κ+m+1 0 with p ∈ (ζ ). Observe κ+m+1 κ+m+1 W[g , g ](z)– W[g , g ](G ) 1 2 1 2 0 1 g (t)– φ(t)g (t) 1 2 = ζ (t – z)+ ζ (z)– ζ (t – G )– ζ (G ) dt 0 0 2πi X (t) φ(z) g (t) + ζ (t – z)+ ζ (z)– ζ (t – G )– ζ (G ) . 0 0 2πi X (t) Inserting (5.31)and (5.32)into(5.2), one easily gets expression (5.29). By Lemma 5.1,(5.29) is the solution of DBR problem (5.1)ifand only if (5.30)issatisﬁed. Theorem 5.5 If κ + m +2 = –1, if and only if 1 g (t) dt = 0, (5.33) 2πi X (t) DBR problem (5.1) is solvable and its solution can be expressed as V(z)= X(z) W[g , g ](z)– W[g , g ](G ) , (5.34) 1 2 1 2 0 with L W[g , g ] = δL W[g , g ] , j = –1, 0, (5.35) j+1 1 2 j 1 2 0 1 where W[g , g ](z) is given by (5.12). 1 2 Proof By Theorem 4.2, the solution of DR problem (5.3)can be expressedas m+2 X(z) g (t)– φ(t)g (t) 1 2 V (z)= ζ (t – z)+ ζ (z)– ζ (t – G )– ζ (G ) dt. (5.36) 1 0 0 2πi X (t) By Theorem 4.2, if and only if the condition of solvability (5.33) is fulﬁlled, the solution of DR problem (5.4)can be writtenas m+1 X(z) g (t) V (z)= ζ (t – z)– ζ (t – G ) dt. (5.37) 2 0 2πi X (t) By (5.33), expression (5.37)can be rewrittenas X(z) g (t) V (z)= ζ (t – z)+ ζ (z)– ζ (t – G )– ζ (G ) dt. 2 0 0 2πi X (t) 0 Han et al. Boundary Value Problems (2018) 2018:88 Page 18 of 20 Thus, by Lemma 5.1, similar to the preceding discussion, the desired conclusion is ob- tained. Theorem 5.6 If κ + m +2 < –1, if and only if 1 g (t)– φ(t)g (t) 1 2 ζ (t – G )– ζ (t) dt = 0, (5.38) 2πi X (t) 1 g (t) ζ (t – G )– ζ (t) dt = 0, (5.39) 2πi X (t) 1 g (t)– φ(t)g (t) 1 2 (k) ζ (t)dt =0, k = 0,1,2,...,–κ – m – 4 (5.40) 2πi X (t) and 1 g (t) (k) ζ (t)dt =0, k = 0,1,2,...,–κ – m – 3, (5.41) 2πi X (t) DBR problem (5.1) is solvable and its solution can be expressed as V(z)= X(z) W[g , g ](z)– W[g , g ](G ) , (5.42) 1 2 1 2 0 with L W[g , g ] = δL W[g , g ] , j = κ + m +1, κ + m + 2, (5.43) j+1 1 2 j 1 2 0 1 where W[g , g ](z) is given by (5.12). 1 2 Proof By Theorem 4.2, if and only if conditions (5.38)and (5.40) are fulﬁlled, the solution of DR problem (5.3)can be expressedas m+2 X(z) g (t)– φ(t)g (t) 1 2 V (z)= ζ (t – z)– ζ (t – G ) dt. (5.44) 1 0 2πi X (t) By Theorem 4.2, if and only if the conditions of solvability (5.39)and (5.41) are fulﬁlled, the solution of DR problem (5.4)can be writtenas m+1 X(z) g (t) V (z)= ζ (t – z)– ζ (t – G ) dt. (5.45) 2 0 2πi X (t) And hence, analogously to the preceding discussion, the desired conclusion is obtained. Remark 5.2 To sum up the discussion above, the freedom of solutions of DBR problem (5.1)is 2(κ + m)+1. 6Conclusion In this article, we deﬁne doubly-periodic polyanalytic functions and growth order of doubly-periodic polyanalytic functions at the origin. Riemann BVP of doubly-periodic bianalytic functions is presented. The problem is transformed into two independent Han et al. Boundary Value Problems (2018) 2018:88 Page 19 of 20 Riemann-type BVPs of doubly-periodic analytic functions. Finally, the solution is explic- itly expressed as the integral representation. Boundary value problems are always related with the theory of elasticity (see, for ex- ample, [24, 27, 28]). If the stresses and the elastic region are doubly periodic, BVPs of doubly-periodic functions can be applied to the theory of planar elasticity. Furthermore, the number and the shape of cracks in the so-called fundamental periodic parallelogram described in Sect. 2 could be arbitrary. In some sense, the results obtained here could contribute to the investigation of planar elasticity of doubly-periodic functions. Acknowledgements The authors would like to thank the referees for their valuable suggestions which helped to improve this work. Funding This research is supported by Major Innovation Projects for Building First-class Universities in China’s Western Region (ZKZD2017009). Availability of data and materials Not applicable. Competing interests The authors declare that they have no competing interests. Consent for publication Not applicable. Authors’ contributions HH carried out theoretical calculation, participated in the design of the study, and drafted the manuscript. HL conceived of the study and participated in the design of the study. YW participated in its design and helped to draft the manuscript. All authors read and approved the ﬁnal manuscript. Author details 1 2 School of Mathematics and Statistics, Ningxia University, Yinchuan, P.R. China. Department of Mathematics, Tianjin University of Technology and Education, Tianjin, P.R. China. School of Mathematics and Statistics, Wuhan University, Wuhan, P.R. China. Publisher’s Note Springer Nature remains neutral with regard to jurisdictional claims in published maps and institutional aﬃliations. Received: 26 November 2017 Accepted: 23 May 2018 References 1. Balk, M.B.: Polyanalytic Functions. Akademie Verlag, Berlin (2001) 2. Gonchar, A.A., Havin, V.P., Nikolski, N.K. (eds.): Complex Analysis I: Entire and Meromorphic Functions, Polyanalytic Functions and Their Generalizations. Springer, Berlin (1997) 3. Begehr, H., Schmersau, D.: The Schwarz problem for polyanalytic functions. Z. Anal. Anwend. 24(2), 341–351 (2005) 4. Begehr, H., Hile, G.N.: A hierarchy of integral operators. Rocky Mt. J. Math. 27, 669–706 (1997) 5. 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Boundary Value Problems – Springer Journals
Published: May 31, 2018
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