A mathematically rigorous proof on the decoupling of the plate and disc problem

The three-dimensional problem of linear elasticity (linear 3D problem) can be derived from the elastic and dual potential using the calculus of variations and integration by parts in all three dimensions. For the special geometry of a thin cuboid solid, the application of integration by parts only in two dimensions and the use of Taylor series expansions yields the quasi-two-dimensional problem (quasi-2D problem). We show in a mathematically rigorous way that this problem can be separated into the plate and disc subproblem. Further, we prove that the decoupling behaviour of both subproblems depends on the sparsity scheme of the elasticity tensor. As a consequence of our mathematically rigorous proofs, the (decoupled) subproblems are together equivalent to the linear 3D problem. At the end, we discuss the implications of our results based on an example.

community. Furthermore, we assume a thin rectangular cuboid solid so that the cross section is symmetric with regard to the height. By means of the calculus of variations, integration by parts and Taylor series expansions we arrive at the quasi-two-dimensional problem, which depends only on the two in-plane coordinates. This problem consists of the two-dimensional equilibrium conditions, the Neumann conditions on the midplane boundary and the Dirichlet conditions on the midplane boundary.
We show that the two-dimensional equilibrium conditions, written in the displacements coefficients, decouple into two subproblems in dependence of the sparsity of the elasticity tensor. In this way, we confirm the postulation of [1, Section 5.1.2]. Furthermore, it will turn out that the Neumann and Dirichlet conditions on the midplane boundary reveal the same decoupling behaviour. Altogether, for special material properties (e.g. isotropy) the plate and disc subproblems are decoupled and, therefore, they can be treated separately. To the authors best knowledge, this is the first time that the decoupling of the plate and disc subproblem is proved mathematically.
Another outcome is that the applied loads have to be separated into a symmetric and antisymmetric part to be uniquely allocated with one of the subproblems. In this way, both subproblems can be indeed driven by loads in all directions. As a consequence, the classical definitions of plates and discs are not precise and have to be extended. Moreover, by separating the applied loads, we show that the classical problem "simply supported rectangular plate under constant topside pressure" in fact belongs to both the plate and the disc problem. Of course, this is also valid for other problems, where loads are applied to only one side of the plate.
Last but not least, we show that the assumption of the preservation of the height is only necessary since the wrong load case is considered.

The index notation
The physical quantities of linear 3D theory are tensor fields. Therefore, this paper is based on tensor calculus, which we assume the reader is familiar with. For an introduction, we refer to [7]. For convenience, we will further use a fixed Cartesian coordinate system x = (x i ). Thus, we deal with a positively orientated orthonormal basis and the covariant derivative becomes the partial derivative ∂(•) /∂x i = (•) |i . For vectors and higher-order tensors we use a bold font. The corresponding order can be gained from the context. For scalars, on the other hand, we use a normal font.
Our calculations are mainly performed in the index notation. Our functions can have two indices: j a i . The right lower index i is the tensor index. Latin tensor indices are always from the set {1, 2, 3}, whereas Greek tensor indices are from the set {1, 2}. The index j is an element of N 0 and used for series expansions. Furthermore, also exponents like k of x k 3 are assumed to be elements of N 0 . To avoid confusion, we do not use Einstein notation in this contribution.
In the whole paper, we assume the given functions to be real-analytic, i.e. they are locally given by convergent power series and they are infinitely often differentiable. This assumption, which is not a real drawback for the engineering practice (cf. [15,Section 4.4]), enables us to use Taylor series expansions and Fubini's theorem without checking the regularity.

The three-dimensional theory of linear elasticity
Because the linear 3D theory is commonly known in the engineering community, we only give a short review of the basic assumptions and equations so that the reader is able to comprehend the argumentation in Sects. 3 and 4. For more detail, we refer to [2,7].
The linear 3D theory is limited to small deformations (geometric linearity). Therefore, the undeformed reference configuration and the deformed actual configuration of the elastic body are approximately the same. Further, we do not have to distinguish between the Eulerian or substantial specification and the Lagrangian or spatial specification. So, in the following, every physical quantity is described on the undeformed reference configuration of the elastic body. We denote this configuration with . For the sake of brevity, when we talk about the elastic body in the following, we implicitly mean the undeformed reference configuration of it. Furthermore, we omit all inertia effects in the whole contribution. So, we deal only with the static problem and all quantities are time independent.
The basic equations of the linear 3D theory are the kinematic relations, the kinetic relations (equilibrium conditions) and the constitutive equations (Hooke's law). They are, in order: Here, we used the quantities • ε = (ε i j ) for the symmetric strain tensor field, • u = (u i ) for the displacement vector field, • σ = (σ i j ) for the symmetric stress tensor field, • f = ( f i ) for the volume force vector field, • E = (E i jrs ) for the stiffness tensor field and the abbreviation f.a. for "for all". Note that the stiffness tensor field is only related to the material parameters and must not be confused with the so-called "plate stiffness" K , which depends on both the material and geometric parameters. The stiffness tensor field is homogeneous, because we assume the material to be homogeneous. The stiffness tensor itself is furthermore subject to the following symmetry relations 1 In this contribution, we deal with the mixed boundary value problem. So, the overall boundary of the elastic body ∂ can be composed of both a Neumann boundary ∂ N and a Dirichlet boundary ∂ 0 . For them, the following conditions apply: On ∂ 0 , we prescribe the displacement vector field u 0 = u 0i , and on ∂ N we prescribe the stress vector field g = g i : Here, we used n = (n i ) as outer-unit normal vector field. Considering the symmetries of the strain and stress tensor field, we have 15 Eqs. (1), (2) and (3) for the 15 unknown field functions u, ε and σ . The quantities E, f, u 0 and g are assumed to be given data. A solution (u, ε, σ ) satisfying these 15 equations and the boundary conditions (6) and (7), i.e. the mixed boundary value problem, is called a solution of the problem of the linear 3D theory.
If we insert (1) and (3) in (2) and (7), the mixed boundary value problem can be entirely written in the three displacement functions u i . This simpler representation is called the Navier-Lamé formulation: 3 j,r,s=1 These equations can also be derived by two potentials of the linear 3D theory (cf. Section 3). The first one is the elastic potential (or potential energy) and the second one is the dual potential (or dual energy) where we used the compliance tensor field D = D i jrs . For further information about the elastic and dual potential, especially the separation of them, we refer to [17, 2.5] and [1, 1.3.6].

The anisotropic constitutive law
The following description is based on [7, Section 4] but with the notation of [16]. According to the symmetries (4), the stiffness tensor contains 21 independent components. To avoid setting up a complete fourth-fold tensor with redundant content, we will use an easier representation of Hooke's law. With σ := (σ 11 , σ 22 , σ 33 , σ 12 , σ 23 , σ 31 ) T as the stress vector, ε := (ε 11 , ε 22 , ε 33 , 2ε 12 , 2ε 23 , 2ε 31 ) T as the strain vector and as the stiffness matrix, (3) can be represented by which is called Voigt notation. Note that the underline denotes the corresponding quantities to belong to the Voigt notation and with (•) T we introduced the transposition. As mentioned before, in the most general case the stiffness matrix has 21 independent components (cf. (13)). Such a material is called aeolotropic or triclinic. Instead, if the material possesses exactly one plane of reflection symmetry, it has 13 independent components and is called monoclinic or monotropic. In the case that the x 1 -x 2 plane is this symmetry plane, the corresponding stiffness matrix is given by For a monoclinic material with the x 1 -x 3 plane as symmetry plane, the corresponding matrix is then given by Finally, for a monoclinic material with the x 2 -x 3 plane as symmetry plane, we get If a material possesses two independent axes of rotational symmetry, the material is invariant with respect to any rotation and thus independent of direction. Such a material is called isotropic, and its stiffness matrix is given by where E 1212 = 1 /2 (E 1111 − E 1122 ). It has only two independent components.

The two-dimensional geometry
So far, all equations are valid for a general elastic body . From now on, we will limit ourselves to the treatment of the plate geometry described in Fig. 1. According to (a), the plate is a cuboid solid with the length a in x 1 -direction, the width b in x 2 -direction and the height h in x 3 -direction. Because the thickness dimension is much smaller than the in-plane dimensions we call this continuum quasi-two-dimensional. The midplane A M is parallel to the upper and lower face of the plate (cf. Figure 1) and bisects it in the thickness direction. The origin of our coordinate system is located in a corner of the midplane. We orientate it in such a way that the x 3 axis points to the lower face of the plate and the other axes are parallel to the edges of the midplane. Please note that the coordinate system in Fig. 1b-d is located in the same position as in (a). But, for the sake of clarity, we have placed it next to it. With the measures of Fig. 1a, we are now able to mathematically specify the plate continuum. For this purpose, we use the Cartesian product A × B, which merges the two sets A and B to the set of all ordered pairs (a, b), where a ∈ A and b ∈ B. So, the plate body becomes With the mathematical expression of the midplane we can rewrite (20) to According to Fig. 1b and (5), the boundary of the midplane ∂ A M is given by Here, with 0 and N we introduced the Dirichlet part on the midplane boundary and the Neumann part on the midplane boundary, respectively. These two curves enable us to express the Neumann boundary as and the Dirichlet boundary as This decomposition is also shown in Fig. 1d. Keep in mind that we treat mixed boundary value problems and therefore, Neumann and Dirichlet conditions can appear at the same edge! With regard to the boundary conditions in Sect. 3, we split the Neumann and Dirichlet part on the midplane boundary each into two sets: Accordingly, one set contains the lines with x 1 = const. (e.g. N 1 ) and the second contains the lines with x 2 = const. (e.g. N 2 ). Furthermore, we introduce the indicator functions to be able to clearly formulate the boundary conditions in Sect. 3. With ∃, we denote the existential quantifier, the sets M and N are single tuple and the sets S and T are ordered pairs. The sets S and M are related to each other in the way that the first factor of the Cartesian product of S is a subset of M. Analogously, the second factor of the Cartesian product of T is a subset of N . With these indicator functions, we can define the indicator operators which ensure that the function F(x 1 , x 2 ) is only evaluated at points (x 1 , x 2 ) belonging to the sets S or T . As an example, we assign N 1 = {0, a} × (0, b) and 01 = ∅. Then, we get 3 The quasi-two-dimensional theory

The role of the elastic and dual potential
In the following, we use the two potentials (11) and (12) to derive the equilibrium conditions (8) and the boundary conditions (9) and (10). According to the principle of virtual work, the displacements and stresses adjust themselves so that the potentials (11) and (12) Here, with v = (v i ) and μ = (μ i j ) we introduced the virtual displacements and virtual stresses, respectively. The former have to satisfy the kinematic relations (1) and the Dirichlet boundary conditions (9), i.e. they have to be geometrically compatible. The latter have to satisfy Hooke's law (3), the equilibrium conditions (2), the Neumann boundary conditions (7) and Cauchy's formula (σ ji n j = t i ) (cf. [17, 2.5]). The calculation of the first variations in E pot and E dual is done separately for each summand. The first one is in each case a symmetric bilinear form: According to [14,Section 4.1], the differentiation rules are also valid for the calculus of variations. Consequently, by using Leibniz integral rule (cf. [14, (3.101)]) we obtain for the first variation in B(u, u) In this integral, every summand consists of the product of two strain functions and the factor E i jrs . Thus, with the sum and product rule (cf. [1, Section 10.1]) we arrive further at Here, from the first to the second line we used the first two symmetries of (4) and from the second to the third line we used the remaining one. In the same manner, we get the variation in C(σ , σ ): The two other summands of E pot and E dual are linear forms. With we defined the linear form in u for the elastic potential and the linear form in σ for the dual potential. With the sum rule, the Leibniz integral rule and the product rule, the first variation in F(u) reads and the first variation in G(σ ) follows analogously to So, with (11), (35) and (37) the principle of minimum of potential energy yields Here, from the second to the third row we used integration by parts. As mentioned before, we expect the Dirichlet boundary conditions (9) to hold for v i to arrive at the elastic potential. Consequently, the virtual displacements have to vanish at the Dirichlet boundary, because there the displacements are prescribed. That is why the Dirichlet part in (39) has also vanished. Finally, if we apply the variational lemma (cf. [10]) to (39) we recover the three-dimensional equilibrium equations (8) and the Neumann boundary conditions (10).
With (12), (36) and (38), the principle of maximum of dual energy reveals Here, we have used the inverted Hooke's law to arrive at the third row. From the third to the fourth row, like in (39), we applied integration by parts. Because the equilibrium conditions hold for the dual potential, their first variations with respect to the stresses yield μ i j| j = 0. Consequently, the first term of the second last row of (40) has to vanish. Note that the volume and surface loads are assumed as given data (cf. Section 2.2), and thus, their variation is zero! Also, the third term of (40) has to vanish. This is because, due to the Neumann boundary conditions, the stress vectors are prescribed on the Neumann boundary ∂ N and so, the virtual stress vectors have to be zero. Finally, if we apply the variational lemma to (40), we recover the Dirichlet boundary conditions (9). The problem in terms of the equilibrium conditions (8), the Neumann boundary conditions (10) and Dirichlet boundary conditions (9) is called the three-dimensional problem of linear elasticity (linear 3D problem). Under the third requirement of (5), both the uniqueness and the existence of the solution of the linear 3D problem are guaranteed (cf. [9] and [Theorem 61.D] [20]).

Derivation of the quasi-two-dimensional theory
The derivation of the quasi-two-dimensional problem is similar to the procedure of the previous Section or [ 16,Section 3]. Starting from the application of integration by parts only in x 1 and x 2 -direction yields: Next, we replace the virtual displacements v i by its Taylor series expansions With the plate continuum of Sect. 2.4, especially the partition of the Neumann boundary (24) and the indicator operators (29) and (30), we then arrive at With the stress resultants, the Neumann stress resultants and the load resultants defined by we finally get Note that we introduced the plate parameter c = h / √ 12a, which quantifies the thinness of the plate, in (47). By means of the variational lemma, Eq. (48) gives us the two-dimensional equilibrium conditions and the Neumann conditions on the midplane boundary With the definition of the shift operator we can further simplify (49) to On the other side, applying integration by parts only in x 1 and x 2 -direction to Next, we replace the displacements u i and the Dirichlet displacements u 0i by their Taylor series expansions With the plate continuum of Sect. 2.4, especially the partition of the Dirichlet boundary (25), and the indicator operators (29) and (30), we then arrive at Using the definition of the stress resultants (45), this equation can be simplified to Because all stress resultants are virtual, we recover the virtual form of the two-dimensional equilibrium conditions (cf., (49)) in the first row. We assume that the two-dimensional equilibrium conditions, as well as the three-dimensional equilibrium conditions (cf. [17, Section 2.5]), can be provided for the dual potential.
In order to prove this, we would have to calculate the free energy like the elastic and dual potential [cf. (48) and (59)]. Since the variation in the given load resultants is zero, the virtual form of the two-dimensional equilibrium conditions has to equal zero, too, and thus, the first row has to vanish. Therefore, we achieve By means of the variational lemma, we finally arrive at the Dirichlet conditions on the midplane boundary We call the two-dimensional equilibrium conditions, the Neumann conditions on the midplane boundary and the Dirichlet conditions on the midplane boundary together the quasi-two-dimensional problem. By preceding backwards, the infinite sets of the quasi-two-dimensional problem lead to the first variations in the elastic and dual potential [cf. (41) and (53)]. Thus, the quasi-two-dimensional problem is actually an exact representation of the linear 3D problem and therefore has also an unique and exact solution. This result is analog to that in [16, Section 3] (they have proved that the quasi-one-dimensional problem is equivalent to the linear 3D problem). Schneider, Kienzler and Böhm [18,Problem V] have also shown the equivalence between the linear 3D problem and the quasi-two-dimensional problem but with Fourier series.

The decoupling into a plate and a disc problem
To solve the quasi-two-dimensional problem it is not necessary to calculate every stress resultant, displacement coefficient and load resultant. Rather, we show in this subsection that these quantities can be split into two sets. So, the quasi-two-dimensional problem divides into two subproblems, which are, for special kinds of anisotropy, decoupled. The following preceding is analog to [16, Section 4].

Key observation and notations
At first, we take a look at the two-dimensional equilibrium conditions. According to (48) and (49), they are related to the virtual displacement coefficients and include the load resultants and the stress resultants. To analyse the latter, we need to transform Hooke's law. Therefore, we first take a look at the partial derivatives of the Taylor series expansions of (55), which are gained by differentiating each summand. Due to reordering of the resulting sums, we get the following differentiation rules Thus, differentiation with respect to x 3 leads to a change in the Taylor coefficient n u i . With the shift operator we can express the partial derivatives of u i in one formula: So, by inserting (55) and by using the differentiation rule (67), we can rewrite Hooke's law (3) to Here, with n σ i j , we introduced the stress coefficients. If we insert (68) into the stress resultants (45) and use the shorthand we arrive at Here, according to (69), we can see that we get only a non-vanishing stress resultant, if q and n have the same parity. So, for stress resultants with an even (odd) upper index, we only have to take into account stress coefficients with an even (odd) upper index. By expanding the stress coefficients as follows we can further recognize that, e.g., the two stress resultants 1 m i j and 2 m i j contain both the displacement coefficients 2 u 1 , 2 u 2 and 2 u 3 . Note that the tensor indices i and j are not related to the displacement coefficients! Thus, we see that for general anisotropy the even and odd stress resultants in terms of the displacement coefficients do not decouple. But, if we assume, e.g., isotropic material, the stress resultant 1 m 11 contains only 1 u 1 , 1 u 2 and 2 u 3 and the stress resultant 2 m 11 contains only 2 u 1 , 2 u 2 and 3 u 3 [cf. (18)]. We conclude that the classes o m 11 and e m 11 (with odd (o) and even (e) left upper indices) contain only displacement coefficients from the classes o u α and e u 3 and e u α and o u 3 , respectively. So, both classes are completely decoupled and the decoupling behaviour depends on the sparsity of the stiffness matrix (13). Provided that the classes e m i j and o m i j decouple for each i, j ∈ {1, 2, 3} (we will show the corresponding conditions in Sect. 4.4), the two-dimensional equilibrium conditions in terms of the stress resultants have also to decouple in order to arrive at two independent problems. To show this, we first need some notations, which hopefully make the subsequent explanations more comprehensible. Because we have obviously two different classes for every quantity and the classification depends on the parity of the left upper index, we use the quotient set Z 2 . This set contains exactly the two elements e := {z ∈ Z | z mod 2 = 0} and o := {z ∈ Z | z mod 2 = 1}. Here, Z is the set of all integers and x mod y is the modulo operator, which gives us the remainder of the division of x by y. The element e defines the even integers and the element o defines the odd integers. We have used these elements already above for the stress resultants. As class, we refer to the series coefficients of a specific quantity, whose left upper indices have the same parity, e.g. for the stress resultant n m 12  Further, with t 1 , t 2 we denote unknown elements of Z 2 and also introduce the function [•] Z 2 , which maps integers to their parity class. The latter is defined by [•] Z 2 : Z → Z 2 , z → z mod 2 and, e.g., implicates Last but not least, we define the abstract shift operator K j as an abstract form of the shift operators S j and K j . The definition of this operator reads and therefore, it maps elements from Z 2 to each other. The properties of K j are Here, id Z 2 is the identity on Z 2 and so (a) states that K 1 and K 2 are mapping back to their argument. According to the second property (b), the composition of the same component of the abstract shift operator is the identity. And (c) gives us the commutative property of the abstract shift operator. These properties can easily be checked by inserting both elements of Z 2 . With the definition K i [n] Z 2 T := K i ([n] Z 2 ) T , the abstract shift operator can also be used for classes of quantities (e.g.

The classification of the stress resultants
With the notations from above, we are now able to analyse the two-dimensional equilibrium conditions (52) in an abstract way: In the Following, we connect the virtual displacements with the symbol δ of the variational calculus to avoid confusion between the n v i and n u i . An equation which is gained by the variation in E pot with respect to δ n v i (which is gained by δ n v i ) is given by Hence, we can easily see that an equation class δ t 1 v i potentially contains the stress resultant classes K j t 1 m i j , j ∈ {1, 2, 3}. (Here, we use the word potentially, because some stress resultants might be zero.) As a consequence, the operation K j δ t 1 v i leads to the stress resultant class K j K j t 1 m i j = t 1 m i j . On the other hand, an equation class δ t 1 v j potentially contains the stress resultant classes K i t 1 m ji , i ∈ {1, 2, 3} and so, the operation K i δ t 1 v j leads to the stress resultant class t 1 m ji . Since the stress resultants are symmetric n m i j = n m ji , one specific stress resultant class t 1 m i j can be gained by the equation classes K i δ t 1 v j and K j δ t 1 v i (for i = j). The class t 1 m ii (i = j), on the other hand, is only gained by the class K i δ t 1 v i .
Because the stress resultants with equal tensor indices can be directly assigned to their corresponding equilibrium conditions, we base the following consideration on them. So, let us assume a specific stress resultant from the class t 1 m ii . By the considerations from above, this stress resultant appears in an equation of the class K i δ t 1 v i and the other two stress resultants in that specific equation belong to the classes K i K j t 1 m i j , i = j. Because a stress resultant from the class t 1 m i j can be found in turn in one equation each gained by the two classes K i δ t 1 v j and K j δ t 1 v i , each stress resultant from the classes K j K i t 1 m i j appears in an equation of the classes The same result is also achieved, if we start with t 1 m j j from the equation class K j δ t 1 v j and repeat the argumentation from above. Thus, it has been shown that the stress resultants from the equilibrium conditions, gained by the classes 3 i=1 K i δ t 1 v i , do not appear in equilibrium conditions, gained by the classes So, the equilibrium conditions in terms of the stress resultants always decouple into two problems. Since these problems (later on called: disc and plate problem) are identified by the parity (t 1 ) ∈ Z 2 of the stress resultants with equal tensor indices, we call this parity the problem identifier.

Theorem 1 (Classification of the virtual displacements and stress resultants)
The two-dimensional equilibrium conditions in terms of the stress resultants (52) decouple into two subproblems. We identify each subproblem with one of the two elements of Z 2 :

(e), (o).
The equilibrium conditions of subproblem (t 1 ) ∈ Z 2 are gained precisely by the first variations in the elastic potential with respect to the virtual displacement coefficients from the set 3 i=1 K i δ t 1 v i . Furthermore, the equilibrium conditions of subproblem (t 1 ) are formulated in the stress resultants from the set Aiming to make the abstract proof from above clearer, we now look exemplarily at the subproblem (e). An equilibrium condition for this subproblem can be gained by virtual displacements of the classes 3 i=1 K i δ e v i . We start with K 1 δ e v 1 = δ e v 1 . Here, the equilibrium conditions contain stress resultants from the classes Stress resultants from the class e m 11 (which identifies the problem) can be gained only by virtual displacements from the class δ e v i and thus belong clearly to problem (e). However, stress resultants from the class e m 12 can be gained by virtual displacements from the classes K 1 δ e v 2 = δ e v 2 = K 2 δ e v 2 and K 2 δ e v 1 = δ e v 1 (already analysed). The virtual displacement class δ e v 2 also belongs to problem (e) and, so, stress resultants Table 1 Classification of the displacement coefficients, the virtual displacement coefficients, the load resultants and the volume load coefficients Table 2 Classification of the stress resultants Because of the identical tensor indices, the class e m 22 is clearly related to (e). Stress resultants from the class o m 13 are gained by variation with respect to virtual displacements from the classes  Table 1. Here, instead of δ t 1 v i , we display the displacement classes t 1 u i , because we later on will find the same classification for them (e.g. for isotropy). Furthermore, it will turn out that the load resultants and the volume load coefficients have also the same classification. So, Table 1 is valid for four quantities. The stress resultant classes are classified in Table 2. In both tables, we use the acronyms D (disc problem) and P (plate problem) for the subproblems.
According to (52), an equilibrium condition, which is gained by δ n v i , contains the load resultant n p i . So, for each equilibrium condition the indices of the corresponding virtual displacements and the load resultants are identical. Therefore, in equilibrium conditions, which are gained by virtual displacements from the classes K i δ t 1 v i , there are only load resultants from the classes K i t 1 p i . Accordingly, the classification of the load resultants is the same as for the virtual displacements. It is given in Table 1, if we substitute u i by p i .

The classification of the volume and surface loads
The identification of subproblem (e) as disc problem and subproblem (o) as plate problem results from examining special load cases. In the following, we assume the surface loads to be zero (g i = 0) and allow only for dead weight. Because we restrict ourselves to homogeneous material, the volume loads have to be constant ( f i = const.) for this load case. With dead weight in the midplane direction, we associate the disc problem and with dead weight perpendicular to the midplane we associate the plate problem. So, for the latter we have f 3 = k; k ∈ R and f α = 0. According to the definition of the load resultants (47) and the shorthand (69), we then get From this result, we conclude that we can apply dead weight in x 3 -direction only for load resultants from the class K 3 o p 3 = e p 3 . Therefore, the problem identifier for the plate problem is (o). Otherwise, if we apply dead weight in the midplane direction, we have f 1 = k 1 , f 2 = k 2 for k 1 , k 2 ∈ R and f 3 = 0. In this case, we get Here, we see that dead weight in the midplane direction can only be applied to the subproblem which contains load resultants from the classes K 1 e p 1 = e p 1 and K 2 e p 2 = e p 2 . So the problem identifier for the disc problem has to be (e).
So far we know which load resultant belongs to which subproblem. Now, we want to analyse the load resultants in order to assign the volume and surface loads to the subproblems. We start with the examination of the volume loads and therefore continue neglecting the surface loads. By substituting the volume loads by their infinite Taylor series the load resultants (47) turn into Like in (70), we can easily conclude that for non-vanishing load resultants the parities of n and q have to be equal. This leads to the volume load coefficients having the same classification as the load resultants: K i t 1 f i . So, it can be taken from Table 1 by substituting u i by f i . Exemplarily, we show the Taylor series expansions of the volume loads for the plate problem: In contrast with t 1 f i , the f t 1 i are no classes but functions. Note that the constant coefficient 0 f 3 is that one we used for the dead weight in x 3 -direction above. Accordingly, the other two constant coefficients 0 f 1 and 0 f 2 belong to the disc problem and therefore do not appear in (80).
If we squeeze the plate continuum by applying loads to the upper and lower face of it, we cause deformations which belong to the disc problem. Otherwise, if we bend the plate continuum by applying loads to the upper and lower face of it, we cause deformations which belong to the plate problem. Therefore, the surface loads in (47) also have to be assigned to the two subproblems. But, because the surface loads are applied on the upper and lower face of the plate continuum, they do not have a Taylor series expansion in x 3 -direction. To be able to split the surface loads anyway, we use the geometrical aspect of the parities, namely the symmetry relations.
According to [21, Section 0.2.1], we call a function f , which depends on a variable x, even with respect to x, if is satisfied for all x in the domain D of f and odd with respect to x if is satisfied for all x in the domain D of f . In the first case, we also say that f is axisymmetric with respect to the function axis and in the second case we call f point-symmetric to the origin of the coordinate system. By using these definitions, the Taylor series expansion and the decomposition into a symmetric and antisymmetric part (cf. [3, (2.11)]), we arrive at Theorem 2 (Parity decomposition) A real analytic function is even with respect to x if, and only if, all Taylor coefficients, whose left upper index is odd with respect to x, are zero. Likewise, a real analytic function is odd with respect to x if, and only if, all Taylor coefficients, whose left upper index is even with respect to x, are zero. Furthermore, if the domain D of a function f satisfies the condition "from x ∈ D follows −x ∈ D," then f can be decomposed into a sum of an even function f e and an odd function f o : If we apply this decomposition with regard to the x 3 -direction to the volume loads f i , we get Using the Taylor series expansions of f i , the even functions read and the odd functions are given by So, the parity decomposition leads also to the result (80). If we want to split the surface loads in (47) with the parity decomposition, we have to treat them as two functions with two distinct values at the points x 3 = h /2 and x 3 = − h /2, in each case. In this way, the parity decomposition gives us By introducing the notations , we get the easier representation: So, the even and odd functions of the surface loads on the upper and lower face of the plate simply read Using these functions, we derive from (47) With the binary function η η : which assigns the elements e and o to 0 and 1, respectively, we further can write for a load resultant with an arbitrary parity By this formulation, we have referred the surface load functions g + i and g − i to the classes So, we finally arrive at Theorem 3 (Classification of the volume and surface loads) In problem (t 1 ), the corresponding load resultants, which belong to the set 3 i=1 K i t 1 p i , include volume load coefficients from the set 3 i=1 K i t 1 f i and surface The parts of the volume and surface loads, which drive the problem t 1 , have in x i -direction the parities K i (t 1 ) regarding the x 3 axis. Table 3 Classification of the surface loads Class (t 1 ) The classification of the volume load coefficients and the surface loads of Theorem 3 is given in Tables 1  and 3.
With Theorem 3, we are now able to decompose every surface load on the upper and lower face of the plate and every volume load into the driving forces of the disc and the plate problem (cf. Section 5). Note that these decompositions are not valid for the surface loads on the midplane boundary N (cf. Section 4.5)!

The anisotropic coupling and classification of the displacement coefficients
We have already shown in Theorem 1 that the two-dimensional equilibrium conditions in terms of the stress resultants (49) can be separated into two problems. Now, we will investigate the coupling of these equilibrium conditions in terms of the displacement coefficients. As mentioned before, this coupling depends on the anisotropy of the material.
According to Theorem 1, the equilibrium conditions of subproblem (t 1 ) are formulated in the stress resultants from the set 3 (70), we can easily derive that the stress coefficients belong to the same set. By abstraction of (71) with K j , a stress coefficient from the class t 1 σ i j contains potentially (if all components of the stiffness tensor are nonzero) the displacement coefficients from the set 3 r =1 3 s=1 K s t 1 u r . So, altogether we can state that the equilibrium conditions of subproblem (t 1 ) are formulated in the displacement coefficients of the set 3 i, j,r,s=1 E i jrs = 0 Because this set contains all classes of displacement coefficients, it follows that the equilibrium conditions of both subproblems include each all displacement coefficients, i.e. for an aeolotropic material, where the 21 components of the stiffness tensor are independent and nonzero, both subproblems cannot be treated individually. Consequently, we arrive at one PDE system, written in all displacement coefficients, for the coupled problem. In order to investigate the coupling behaviour depending on the stiffness tensor, we define the effective shift operator If we insert the identity K r K r = id Z 2 into the set (87), apply the commutative property (c) of the abstract shift operator and use the effective shift operator, we can say that the equilibrium conditions of subproblem (t 1 ) are formulated in the displacement coefficients of the set 3 i, j,r,s=1 E i jrs = 0 Keep in mind that, according to (71), a displacement coefficient, which emerges from the index combination (i, j, r, s), is multiplied by the component E i jrs from the stiffness tensor. Therefore, it is obvious to apply the effective shift operator to the index combinations of the stiffness tensor. Because in this way we can recognize the coupling behaviour of the subproblems straight from the anisotropy of the material. So, in Voigt notation, we derive from (13) ⎛ With K, we defined the abstract-stiffness matrix, whose entries are the effective shift operators applied to the indices of the stiffness tensor components at the corresponding position, e.g., for position (1, 1) of K we get with E 1111 the result K eff (1, 1, 1, 1) = K 1 K 1 K 1 K 1 = id Z 2 and for (1, 6) we analogously achieve With K and the results of Sect. 2.3, we are now able to derive from (88) for which kind of anisotropy the equilibrium conditions in terms of the displacement coefficients are decoupled. If is valid for all components of the stiffness matrix or abstract-stiffness matrix, the equilibrium conditions of subproblem (t 1 ) are entirely formulated in the displacement coefficients of the set 3 r =1 K r t 1 u r and, therefore, both subproblems are decoupled. But if, on the contrary, one entry of K is equal to K 3 , both subproblems are coupled. So, with regard to Sect. 2.3 and [7, Section 4] we find that both subproblems are decoupled for isotropic, transversely isotropic and orthotropic material. Furthermore, they are also decoupled for monoclinic material with the x 1 -x 2 plane as symmetry plane. On the other hand, both subproblems are coupled for monoclinic material with either the x 1 -x 3 plane or the x 2 -x 3 plane as symmetry plane [cf. (16) or (17)]. Since there, some components of the stiffness matrix, which are connected with K 3 in the abstract-stiffness matrix, are not zero (e.g. E 2213 ). And of course, as mentioned before, both subproblems are coupled for aeolotropic material.
The previous results lead to Note that the classification of the displacement coefficients, as mentioned before, is given in Table 1. This is of course only valid for that kind of anisotropy which leads to decoupled subproblems (cf. above).

The classification of the boundary conditions
With Theorems 1, 3 and 4, we have shown that, for all materials which have a reflection symmetry with respect to the x 1 -x 2 plane, the two-dimensional equilibrium conditions (49) are decoupled. However, if we want to show the complete decoupling of both subproblems, we also have to treat the Neumann and Dirichlet conditions on the midplane boundary. This task will turn out to be easy with the previous results.
According to Theorem 1, for a subproblem (t 1 ) the Neumann boundary conditions on the midplane boundary (50) can only be gained by virtual displacement coefficients from the set 3 i=1 K i δ t 1 v i . So, the corresponding stress resultants are from the sets 3 i=1 K i t 1 m i1 and 3 i=1 K i t 1 m i2 . Because these two sets are subsets of the set 3 i=1 3 j=1 K i K j t 1 m i j , we still deal only with subproblem (t 1 ). Now, we investigate the RHSs of the conditions (50). With the definition (46) and the Taylor series expansions of the surface loads on the lateral faces we arrive at Here, we can see that we get only nonzero right-hand sides, if the stress resultants have the same parities as the surface loads (q has to equal n). So, it follows that the surface loads from the sets 3 i=1 K i t 1 •x 2 g i belong to subproblem (t 1 ). The Dirichlet conditions on the midplane boundary are gained by δ n m i1 (μ) and δ n m i2 (μ). According to Theorem 4, for a subproblem (t 1 ) and for all materials which have a reflection symmetry with respect to the x 1 -x 2 plane, the displacement coefficients in (60) can only be from the set 3 i=1 K i t 1 u i . Consequently, the Dirichlet displacement coefficients have to be from the set 3 i=1 K i t 1 u 0i and the virtual stress resultants have to belong to the sets 3 The last two sets are the virtual counterparts of the stress resultant sets from the Neumann boundary conditions on the midplane boundary (cf. above). So, they also belong to problem (t 1 ) and therefore the Dirichlet conditions on the midplane boundary decouple, in accordance with the two-dimensional equilibrium conditions, into two subproblems.
In summary, the main result of Sect. 4 is given in: Theorem 5 (Decoupling theorem for the quasi-two-dimensional problem) The quasi-two-dimensional problem is equivalent to the linear 3D problem and therefore has also an unique and exact solution. For all materials which have a reflection symmetry with respect to the x 1 -x 2 plane, it decouples into two independent subproblems, which are characterized by their problem identifier (t 1 ) ∈ Z 2 . The subproblem with the identifier (e) is called the disc problem, and the subproblem with the identifier (o) is called the plate problem. A specific subproblem (t 1 ) is given by the two-dimensional equilibrium conditions and the Neumann boundary conditions on the midplane boundary, both gained by the first variation in the elastic potential with respect to virtual displacement coefficients from the set Furthermore, the subproblem (t 1 ) is given by the Dirichlet boundary conditions on the midplane boundary, which are gained by the first variation in the dual potential with respect to virtual stress resultants from the sets 5 Example: simply supported plate under constant topside pressure A classical problem in continuum mechanics is the simply supported rectangular plate under constant topside pressure (cf. [1,Section 3.2.3.7]). This problem is generally assigned only to the plate problem and its solution is called Navier solution. Now, we want to analyse this problem with the previous findings, i.e. Theorems 2 and 3. Because we only deal with the constant pressure p, which is pointing in positive x 3 -direction (cf. Figure 2) on the upper face of the plate, we arrive at the load case With the decomposition of (84), we get for the even and odd parts of g + 3 and g − 3 the functions According to Table 1 [even (odd) parts of the surface loads appear only in load resultant classes with even (odd) upper indices], the even functions belong to the plate problem and the odd functions belong to the disc problem. With these findings, we arrive at Fig. 2. So, the classical problem of the simply supported rectangular plate under constant topside pressure, in fact, drives the plate problem and the disc problem. Since this classical problem is related to isotropic material, both subproblems can be calculated independently. But only together they correctly describe the stress and the deformation state of the initial problem! In Fig. 2, also the plate and disc deformations are illustrated. Here, we can see that the plate problem leads to a deflection in x 3 -direction and the disc problem leads to a squeezing in x 3 -direction and an elongation in x 1 -direction. According to Theorem 4, we have assigned the corresponding displacement coefficients to these deformations. Thus, for the plate problem we have to consider only the even part of the displacement in x 3 -direction, namely u e 3 . According to Theorem 2, we then have the same displacement on the upper and lower face of the plate. Because the change of height of the plate is given by the difference of these displacements, we conclude Fig. 2 Decomposition of the constant topside pressure into the load cases of the plate (even parts) and the disc problem (odd parts). The corresponding deformation of the plate problem includes, e.g., the constant deflection 0 u 3 . The corresponding deformation of the disc problem includes, e.g., the odd coefficient 1 u 3 , which describes the squeezing of the plate, and the even coefficient 0 u 1 , which characterizes the elongation in x 1 -direction This means that for the pure plate problem, and only for it, the height of the plate is preserved. So, in the classical plate theories of Kirchhoff, Mindlin, Reddy [8,11,12] the assumption of the preservation of the height is only necessary because of the wrong load case! As a short remark, on the basis of Fig. 2 we want to explain the geometrical aspect of the parity decomposition. Here, the load case of the plate problem is even because we have the same amount and direction of the surface load on the upper and lower face of the plate, whereas the load case of the disc problem is odd because the direction of g o− 3 is the opposite of g o+ 3 .

Conclusion
We have assumed a quasi-two-dimensional continuum (cf. Figure 1), which is symmetric with respect to the height. Then, from the fist variations in the elastic and dual potential of the three-dimensional theory of linear elasticity we arrive by the application of integration by parts only in the in-plane directions and the substitution of the corresponding test functions by its Taylor series expansions at the quasi-two-dimensional problem.
We have shown in a mathematically rigorous way that this problem can be separated into a plate and a disc problem. Furthermore, we proved that the decoupling of both subproblems depends on the sparsity scheme of the elasticity tensor. Because of the rigorousness of our proofs, the (decoupled) plate and disc problems are together equivalent to the three-dimensional problem of linear elasticity. It turns out, that for isotropic, transversely isotropic, orthotropic and monoclinic material (with the x 1 -x 2plane as symmetry plane) both subproblems are decoupled. On the contrary, for aeolotropic material and for monoclinic material with either the x 1 -x 3 plane or the x 2 -x 3 plane as symmetry plane we have to treat both subproblems together. This result agrees with [1, Section 5.1.2] and provides a reasoning for it. Additionally, it generalizes the statement [1, Section 5.1.2], because it is valid for all and not only for the shear-rigid theory.
Our proofs further show that the common definition that discs are loaded in direction of their midplanes and plates are loaded transverse to it, is incomplete (cf. Theorem 1). Instead, each subproblem is driven by loads in all directions but only with a specific symmetry in each case, e.g., plates can only be loaded by symmetric loads in x 3 -direction and anti-symmetric loads in the midplane-directions.
According to Theorems 1 and 3, applying loads only on one side of the plate drives in fact both, the plate and the disc problem. We highlighted this result by means of the classical problem of the simply supported rectangular plate under constant topside pressure. So, by treating the pure plate or disc problem there have to be either symmetric or antisymmetric loads with regard to the x 3 -direction.
Because of Theorem 4, the displacement in x 3 -direction of the pure plate problem consists only of even parts. As a consequence, the height of the plate is always preserved and the assumption of the "preservation of the height" in the classical plate theories of Kirchhoff, Mindlin, Reddy [8,11,12] and so on is only necessary because of the wrong load case.
Instead of Taylor series expansions, we can also use, e.g., Fourier series expansions with scaled Legendre polynomials for our proof. The results would be the same; only with Taylor series expansions it is more descriptive. Further, our somewhat long and complex proof should be also possible without reducing the dimension. We strongly assume that the resulting proof would be shorter, less complex and give more insight in the three-dimensional theory of linear elasticity. However, that kind of proof will be object of our ongoing research. Our proof is also extendable to the dynamic case. For this extension, the kinetic potential has to be added to the elastic and dual potential. Furthermore, the displacements, especially the deflection 0 u 3 , are functions of the midplane coordinates and the time. If we choose then the separation approach of Bernoulli, i.e. the displacements are synchronous regarding the time, we get no coupled derivations between time and space so that procedure from above is applicable. Another extension of our proof should include arbitrary cross sections. In that way, the influence of the cross section on the decoupling of the plate and the disc problem would be revealed. Lastly, we like to mention that we have already extended our proof towards inhomogeneous materials (e.g. laminates, sandwiches, functionally graded materials). This proof is among others the subject of a subsequent paper.