On the Green function of an orthotropic clamped plate in a half-plane

First, we calculate, in a heuristic manner, the Green function of an orthotropic plate in a half-plane which is clamped along the boundary. We then justify the solution and generalize our approach to operators of the form (Q(∂′)-a2∂n2)(Q(∂′)-b2∂n2)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$(Q(\partial ')-a^2\partial _n^2)(Q(\partial ')-b^2\partial _n^2)$$\end{document} (where ∂′=(∂1,⋯,∂n-1)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\partial '=(\partial _1,\dots ,\partial _{n-1})$$\end{document} and a>0,b>0,a≠b)\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$a>0,b>0,a\ne b)$$\end{document} with respect to Dirichlet boundary conditions at xn=0.\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$x_n=0.$$\end{document} The Green function Gξ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$G_\xi $$\end{document} is represented by a linear combination of fundamental solutions Ec\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$E^c$$\end{document} of Q(∂′)(Q(∂′)-c2∂n2),\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$Q(\partial ')(Q(\partial ')-c^2\partial _n^2),$$\end{document}c∈{a,b},\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$c\in \{a,b\},$$\end{document} that are shifted to the source point ξ,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$\xi ,$$\end{document} to the mirror point -ξ,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$-\xi ,$$\end{document} and to the two additional points -abξ\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$-\frac{a}{b}\xi $$\end{document} and -baξ,\documentclass[12pt]{minimal} \usepackage{amsmath} \usepackage{wasysym} \usepackage{amsfonts} \usepackage{amssymb} \usepackage{amsbsy} \usepackage{mathrsfs} \usepackage{upgreek} \setlength{\oddsidemargin}{-69pt} \begin{document}$$-\frac{b}{a}\xi ,$$\end{document} respectively.


Introduction and notation
We will derive in this study an explicit formula for the deflection G (x) of a semi-infinite orthotropic plate in H = {x ∈ ℝ 2 ; x 2 > 0}, which is clamped along the boundary x 2 = 0 and loaded by a unit point force at = (0, 2 ) with 2 > 0.
In the case of the orthotropic plate operator, the deduction of G from E in (1.3) is more complicated. In Section 2, we shall derive the Green functions of the operators in a heuristic manner by partial Fourier transform with respect to x 1 . The correctness and the uniqueness of the Green functions under appropriate conditions will be investigated in Sections 3 and 4 more generally for operators of the form P( ) = (Q( � ) − a 2 2 n )(Q( � ) − b 2 2 n ). Therein, we shall also provide further examples. Let us introduce some notation. ℕ and ℕ 0 denote the sets of positive and of nonnegative integers, respectively. We consider as differentiation symbols and we denote by P( ) linear partial differential operators In some examples, we set = ( t , 1 , … , n ) and P( ) is then an operator in the n + 1 variables t, x 1 , … , x n . We employ the standard notation for the distribution spaces D ′ , S ′ , the dual spaces of the spaces D, S of "test functions" and of "rapidly decreasing functions," respectively, see [13,15]. In order to display the active variable in a distribution, say x ∈ ℝ n , we use notation as T(x) or T ∈ D � (ℝ n x ). For the evaluation of a distribution T on a test function , we use angle brackets, i.e., ⟨ , T⟩.
The Heaviside function is denoted by Y, see [15, p. 36], and we set The function z ↦ z can be analytically continued in S � (ℝ 1 ) and thus yields an entire function see [4,Equs. (3.1), (3.2), pp. 314, 315], [8, (3.2.17), p. 73]. Note that ( z ) � = z−1 , z ∈ ℂ, and −m = (m−1) , m ∈ ℕ. We write for the delta distribution with support in 0 i.e., ⟨ , ⟩ = (0) for ∈ D(ℝ n ). We use the Fourier transform F in the form this being extended to S ′ by continuity. (Herein and also elsewhere, the Euclidean inner product ( , x) ↦ x is simply expressed by juxtaposition.) For the partial Fourier transforms of a distribution T ∈ S � (ℝ m+n xy ) with respect to x ∈ ℝ m or y ∈ ℝ n , respectively, we use the notation F x T and F y T, respectively.

3
The restriction of a distribution T ∈ D � (ℝ n ) to an open set H ⊂ ℝ n will be denoted by T| H . Similarly, we write T| x n = for the restriction of T to the hyperplane {x ∈ ℝ n ; x n = }, > 0, if the distribution T continuously depends on x n , i.e., if it belongs to the subspace of D � (H), H = {x ∈ ℝ n ; x n > 0}, constituted by the continuous mappings According to [8,Thm. 4.4.8,p. 115], T continuously depends on x n if it solves a linear partial differential equation of order m with constant coefficients and with a non-vanishing coefficient of m n .
2 Green functions of (@ 2 1 + a 2 @ 2 2 )(@ 2 1 + b 2 @ 2 2 ) and of @ 4 1 + 2(1 − 2 2 )@ 2 1 @ 2 2 + @ 4 2 We shall first determine the Green function g (x) = g (x), > 0, x > 0, of the ordinary differential operator with respect to Dirichlet boundary conditions at x = 0. Hence, g fulfills The uniquely determined temperate fundamental solution of ( − a 2 d 2 ∕dx 2 ) is given by and this easily yields that the linear combination i.e., shows that g continuously depends on also near = 0 and that In the next step, we consider the operator We are going to derive the Green function G (x), = (0, 2 ), 2 > 0, in the half-space H = {x ∈ ℝ 2 ; x 2 > 0} subject to Dirichlet boundary conditions at the border line x 2 = 0. Hence, G fulfills Note that (2.6) does not determine G uniquely since, e.g., for each solution G , the distributions G + cx 2 2 , c ∈ ℂ, also fulfill (2.6). As we shall show in Section 4, G becomes uniquely determined if we add to (2.6) the growth condition Upon a partial Fourier transform with respect to x 1 , we obtain the ordinary differential operator in (2.1) as an operator in d∕dx 2 with being the square of the transformed variable of x 1 . Hence, we conclude, at least heuristically, that G (x) = F −1 We next observe that is, at first sight, not well defined at x 1 = 0, but that the linear combination of these functions in g We can therefore evaluate the inverse Fourier transform of g 2 1 (x 2 ) by the finite part at z = −3 of the meromorphic distribution-valued function which has simple poles for z ∈ −ℕ.
For Re z > −1 and fixed x 2 ≠ 0, S a z (x) is an integrable function of x 1 . Hence, we obtain, for Re z > −1 and x 2 ≠ 0, the following: z as a locally integrable function in ℝ 2 if |Re z| < 1. By analytic continuation, this generally holds for Re z < 1 outside the poles of Γ(z + 1), i.e., if z ∉ −ℕ. Thus, we obtain We now use [12, Prop. 1.6.3, p. 28], and Upon summing up the six terms which constitute g , the second-order polynomials with the factor (3) cancel out. Furthermore, for Re z > 0, we have , and this implies, by analytic continuation, that Let us eventually observe that, here and similarly in the following sections, dilation by the factor a implies E a = a −1 E 1 (x 1 , x 2 ∕a). Thus, we arrive at the following proposition.
, with respect to Dirichlet boundary conditions at x 2 = 0, i.e., let G be determined by the conditions (2.6) and (2.7).
In order to derive the Green function (of the Dirichlet problem) for the orthotropic plate operator in (1.1), let us represent G slightly more explicitly in the case b = 1 a . We assume that a > 1 and we set = 1 2 (a − b). This implies a 2 + b 2 = 2 + 4 2 , Of course, the last part of G , i.e., is the most laborious one. It gives We now obtain the Green function of the orthotropic plate operator P( ) in (1.1) by con- The result is the following.

Proposition 2.2 The Green function
This settles the first term in formula (2.12). For the second one, observe that where or − is added in the first two formula lines if the complex number x 2 1 + (bx 2 + a 2 ) 2 belongs to the second or to the third quadrant, respectively. The remaining two terms in Eq. (2.12) are treated similarly. We therefore obtain for G (x) the representation which is enunciated in the proposition.
The three conditions in (2.6) then hold automatically by analytic continuation. Condition (2.7) can be checked directly. Finally, the uniqueness of G (x) under these conditions follows from a reasoning which is similar to that in the proof of Proposition 4.1 below. This completes the proof.

Quasi-hyperbolic and non-vanishing symbol operators
Let us generalize now the context of Section 2 and consider operators of the form where Q( � ) is an operator in the n − 1 variables x � = (x 1 , … , x n−1 ) and P( ) either is quasihyperbolic or has a non-vanishing symbol P(ix). Let us first recall the notion of "quasi-hyperbolicity", see [  . This fundamental solution is then given by the formula and it holds supp E ⊂ {x ∈ ℝ n ; x ⋅ N ≥ 0} and e − xN E ∈ S � (ℝ n ) for each ≥ 0 . Note that operators P( ) whose symbols P(ix) do not vanish are formally contained in the above definition if we allow for N = 0.
In the next proposition, we shall state uniqueness conditions and we shall give a formula for the Green function G (x) of the Dirichlet problem in the half-space H = {x ∈ ℝ n ; x n > 0} for quasi-hyperbolic and for non-vanishing symbol operators P( ) as in (3.1). We assume that Q( � ) is an operator in the n − 1 variables x � = (x 1 , … , x n−1 ), that N � ∈ ℝ n−1 and we set N = (N � , 0). Then, P( ) as in (3.1) is quasi-hyperbolic in direction N (in case N ≠ 0) or has a non-vanishing symbol P(ix) (in case N = 0) if and only if, for some 0 ∈ ℝ, the following condition is satisfied: be a linear partial differential operator with constant coefficients in n − 1 variables such that condition (3.4) is satisfied. Let P( ) be defined as in

Then, the Green function G ∈ D � (H) of P( ) fulfilling is uniquely determined and given by
[Note that the restrictions of k n T to the subspaces {x ∈ ℝ n ; x n = }, > 0, are well defined due to (i) and [8,Thm. 4.4.8,p. 115]. Condition (iii) requires that these restrictions belong to S � (ℝ n−1 ) and converge therein to 0.] Proof The statement in the proposition corresponds to Lemma 3.2 and Proposition 3.7 in [14], where the operator (Q( � ) − 2 n ) m , m ∈ ℕ, is investigated. We therefore only outline the proof here and refer to [14] for details.
Since the values of the polynomial (x � ) = Q(ix � + N � ) lie in the set ℂ ⧵ (−∞, 0] due to condition (3.4), we can choose the root √ (x � ) such the Re √ (x � ) > 0. If G , G are two solutions of (3.5), then we consider the difference S = G −G and the partial Fourier transform with respect to x ′ It fulfills ( (x � ) − a 2 2 n )( (x � ) − b 2 2 n )S = 0 in D � (H) and accordingly we obtain the representation with distributions A , B ∈ D � (ℝ n−1 ), = ±. From F x � (T −T ) ∈ S � (ℝ n ) , we infer that A + and B + must vanish, and from the boundary condition (iii) in (3.5), we then conclude that also A − = B − = 0, i.e., S = 0 and G =G . Therefore, G is uniquely determined.
In order to show that G defined by formula (3.6) satisfies the conditions in (3.5), let E denote the uniquely determined fundamental solution of P( ) such that e − N � x � E is temperate for ≥ 0 . Then, E satisfies Furthermore, P( )E a = P( )E b = 0 in H. This implies that G satisfies conditions (i) and (ii) in (3.5).
Finally, condition (iii) in (3.5) follows from the calculation leading to formula (2.4), which corresponds to the representation (3.6) in the one-dimensional case. We observe that the definition in , c ∈ {a, b}. Using the method of parameter integration, see [13,Chap. 3], we have E c = c −2 ∫ c 2 0 E d with E being the "forward" fundamental solution of ( 2 t − Δ n−1 − 2 n ) 2 , i.e., the one with support in the half-space If z is defined as in (1.6), then the forward fundamental solution E 1 of ( 2 t − Δ n ) 2 can be written as according to [ function (t, x) ↦ t 2 − |x| 2 is submersive outside the origin and that the composition with z defined in (1.6) thus is meaningful outside the origin. Furthermore, the resulting distribution continuously depends on t ∈ ℝ, and it can therefore by multiplied by Y(t), see also [12,Prop. 2.4.2,p. 56], and the ensuing remark.) E.g., let us evaluate the integral in (3.8) in the case of n = 3 spatial dimensions. Due to 0 = Y and setting x � = (x 1 , we obtain see also [17,Ex. 5,p. 27]. Therefore, formula (3.6) yields the following Green function for the operator P( ) in (3.7) with n = 3 ∶ Example 3.4 By means of Proposition 3.2 and using analytic continuation, we can also derive the Green function for the Cauchy-Dirichlet problem of the equation of transverse vibrations of a semi-infinite clamped beam, see [11]. The fundamental solution E c referring to formula (3.3) for the operator t ( t − c 2 2 x ), c > 0, is given by Hence, formula (3.6) in Proposition 3.2 yields the following for the Green function G (0, ) , > 0, of the operator We now use analytic continuation as in Section 2 and set a = Then, x and formula (3.9) yields in accordance with [11, p. 239]. (3.9) Remark 3.5 Our faithful reader, who has followed us up to this point, might wonder how the representation of G in (3.6) would look in the case of a product of m > 2 factors, i.e., if P( = ∏ m j=1 (Q( � ) − a 2 j 2 n ) for pairwise different positive numbers a j . Since we will not use the corresponding formula, we just mention the result. For c > 0, let E c denote the funda- A technically relevant application of formula (3.10) would consist in the derivation of an explicit formula for the Green function of a product P 1 ( )P 2 ( ) of two orthotropic plate operators P 1 ( ), P 2 ( ) as in (1.1). This product operator describes the deflection of an orthotropic cylindrical shell, see [3,Equs. (14) and (22), pp. 738, 739].

Operators with symbols that are positive outside the origin
In this final section, let us treat operators P( ) in ℝ n such that P(ix) > 0 for x ∈ ℝ n ⧵ {0}. As has been shown in [1], the analytic distribution-valued function has a meromorphic extension S to the whole complex plane, see also [13,Prop. 2.3.1,p. 134]. For simplicity, we shall write P(ix) z =S(z) if S is analytic in z ∈ ℂ and P(ix) z 0 = Pf z=z 0S (z) if z 0 ∈ ℂ is a pole of S . (Here, Pf stands for the finite part of a meromorphic distribution-valued function, compare [12].) Since P(ix) ⋅ P(ix) z = P(ix) z+1 holds by analytic continuation for each z ∈ ℂ, E ∶= F −1 (P(ix) −1 ) yields a temperate fundamental solution of P( ).
Then, the Green function G ∈ D � (H) of P( ) fulfilling is uniquely determined and given by formula (3.6).
Proof In order to prove the uniqueness of G under the four conditions in (4.4), we proceed as in the corresponding Lemma 4.1 in [14] and we refer to it for details. As in the proof of Proposition 3.2, we consider the difference S = G −G of two solutions of (4.4) and its partial Fourier transform Condition (i) in (4.4) implies that the support of S is contained in the half-axis {(0, … , 0, x n ) ∈ ℝ n ; x n > 0}. More precisely, we obtain the representation for l ∈ ℕ 0 and polynomials q in one variable. Hence, S is a polynomial. Condition (iii) in (4.4) implies S = x 2 n ⋅ R for another polynomial and condition (iv) then yields R = 0, i.e., G =G .
For the verification of the representation of G in formula (3.6), we consider again S a z as in (4.1). Since we obtain, by analytic continuation, that vanishes in H and is independent of a. This implies that P( )E a = P( )E b = 0 in H and that . On the other hand, the explicit representations yield and hence, for x > 0, > 0, Due to the symmetry in the left-hand side of (4.7) with respect to x and , we can interchange x and on the right-hand side of (4.7). From we then obtain Due to g (x) = A + B and taking into account the estimate for A above, we arrive at the inequality in (4.5). This completes the proof. According to Proposition 4.1 above, G (x) can be expressed by formula (3.6) if E a denotes the fundamental solution of Δ n−1 (Δ n−1 + a 2 2 n ) which is given by with x � = (x 1 , … , x n−1 ). Due to E a = a −1 E 1 (x � , x n ∕a), we can restrict ourselves to calculating the fundamental solution E 1 of Δ n−1 Δ n .