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

Abstract

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 $(Q(\partial ')-a^2\partial _n^2)(Q(\partial ')-b^2\partial _n^2)$ (where $\partial '=(\partial _1,\dots ,\partial _{n-1})$ and $a>0,b>0,a\ne b)$ with respect to Dirichlet boundary conditions at $x_n=0.$ The Green function $G_\xi$ is represented by a linear combination of fundamental solutions $E^c$ of $Q(\partial ')(Q(\partial ')-c^2\partial _n^2),$ $c\in \{a,b\},$ that are shifted to the source point $\xi ,$ to the mirror point $-\xi ,$ and to the two additional points $-\frac{a}{b}\xi$ and $-\frac{b}{a}\xi ,$ respectively.

1 Introduction and notation

We will derive in this study an explicit formula for the deflection $G_\xi (x)$ of a semi-infinite orthotropic plate in $H=\{x\in {\mathbb {R}}^2;\, x_2>0\},$ which is clamped along the boundary $x_2=0$ and loaded by a unit point force at $\xi =(0,\xi _2)$ with $\xi _2>0.$

Hence, $G_\xi$ is the Green function of the differential operator

$$\begin{aligned} P(\partial )&=\partial _1^4+2(1-2\epsilon ^2)\partial _1^2\partial _2^2+\partial _2^4\\&=(\partial _1^2+2\epsilon \partial _1\partial _2+\partial _2^2)(\partial _1^2-2\epsilon \partial _1\partial _2+\partial _2^2),\quad 0<\epsilon <1,\end{aligned} $$

(1.1)

with respect to Dirichlet boundary conditions at $x_2=0,$ i.e., $G_\xi$ is the unique solution of $P(\partial ) G_\xi =\delta (x-\xi ) \text { in }H$ satisfying $\lim _{\epsilon \searrow 0}\partial _2^k G_\xi |_{x_2=\epsilon }=0,$ $k=0,1,$ and some growth condition for $x_2\rightarrow \infty .$ The parameter $\epsilon$ characterizes the orthotropy of the plate.

If $\epsilon =0,$ then $P(\partial )$ coincides with the bi-harmonic operator $\Delta _2^2,$ which is the operator of isotropic plates. The Green function $G_\xi ^0$ of isotropic plates was derived in 1901 by J. H. Michell, see [10, p. 225, last line], [6, Equ. (633), p. 233]. In our notation, it is given by $G^0_\xi (\xi )=|\xi |^2/(4\pi )$ and

$$\begin{aligned} G^0_\xi (x)=\frac{1}{16\pi }\,|x-\xi |^2\log \left (\frac{|x-\xi |^2}{|x+\xi |^2}\right )+\frac{x\xi }{4\pi },\quad x_2>0,\ x\ne \xi . \end{aligned}$$

(1.2)

A fundamental solution (also called singularity function) E of $P(\partial )$ in (1.1) is known since 1959 at least, see [16, Equ. (B9), p. 11], [17, p. 44], [13, Ex. 5.2.4, p. 351]. It reads

$$\begin{aligned} E(x)&=\frac{1}{32\pi \sqrt{1-\epsilon ^2}}\bigl [\bigl (x_1^2+\tfrac{2}{\epsilon }\,x_1x_2+x_2^2\bigr )\log (x_1^2+2\epsilon x_1x_2+x_2^2)\,+ \\&\quad +\bigl (x_1^2-\tfrac{2}{\epsilon }\,x_1x_2+x_2^2\bigr )\log (x_1^2-2\epsilon x_1x_2+x_2^2)\bigr ]\\&\quad +\frac{x_1^2-x_2^2}{16\pi \epsilon }\,\arctan \left (\frac{\epsilon }{\sqrt{1-\epsilon ^2}}\frac{x_1^2-x_2^2}{x_1^2+x_2^2}\right), \quad 0<\epsilon <1, \end{aligned}$$

(1.3)

and the limit $\epsilon \searrow 0$ yields, up to the bi-harmonic polynomial $\frac{|x|^2}{16\pi },$ the well-known fundamental solution $E^0$ of the isotropic plate operator $\Delta _2^2,$ i.e.,

$$\begin{aligned} E^0(x)=\frac{|x|^2}{16\pi }\,\log |x|^2. \end{aligned}$$

(1.4)

Let us note, in parentheses, that formula (1.3) also furnishes, by linear transformations, a fundamental solution E of the fourth-order operator $\nabla ^TA\nabla \cdot \nabla ^TB\nabla$ in ${\mathbb {R}}^2,$ if $A,B\in {\mathbb {R}}^{2\times 2}$ are linearly independent symmetric positive definite matrices. The result is

$$\begin{aligned} E&=\,\frac{1}{8\pi \alpha \sqrt{\det A}}\cdot x^T[{\text {tr}}(BA^{\text {ad}})\cdot A^{\text {ad}}-2(\det A)\cdot B^{\text {ad}}]x\cdot \log (x^T A^{\text {ad}}x) \\&+\frac{1}{8\pi \alpha \sqrt{\det B}}\cdot x^T[{\text {tr}}(AB^{\text {ad}})\cdot B^{\text {ad}}-2(\det B)\cdot A^{\text {ad}}]x\cdot \log (x^T B^{\text {ad}}x) \\&+\frac{\beta (x)}{2\pi \alpha }\cdot \arctan \left (\frac{\beta (x)}{\sqrt{\det A}\cdot x^T B^{\text {ad}}x+\sqrt{\det B}\cdot x^T A^{\text {ad}}x}\right) \end{aligned}$$

(1.5)

where $x=\left( {\begin{array}{c}x_1\\ x_2\end{array}}\right) ,$ $x^T=(x_1,x_2),$ $A^{\text {ad}}=(\det A)\cdot A^{-1}$ and $\alpha ={\text {tr}}^2(BA^{\text {ad}})-4\det A\det B$ is positive since B is not a multiple of A and

$$\begin{aligned} \beta (x)=\sqrt{{\text {tr}}(BA^{\text {ad}})x^T A^{\text {ad}}x\cdot x^T B^{\text {ad}}x-\det B\cdot (x^T A^{\text {ad}}x)^2 -\det A\cdot (x^T B^{\text {ad}}x)^2}. \end{aligned}$$

Let us observe, furthermore, that we can derive the Green function $G^0_\xi$ in formula (1.2) from the fundamental solution $E^0$ in (1.4) by the ansatz

$$\begin{aligned} G_\xi ^0=\frac{|x-\xi |^2}{16\pi }\bigl [\log (|x-\xi |^2)-\log (|x+\xi |^2)\bigr ]+Z \end{aligned}$$

where Z fulfills the conditions $\Delta _2^2Z=0$ and $Z|_{x_2=0}=0,$ $\partial _2Z|_{x_2=0}=\frac{\xi _2}{4\pi }.$ This yields $Z=x\xi /(4\pi ).$

In the case of the orthotropic plate operator, the deduction of $G_\xi$ from E in (1.3) is more complicated. In Section 2, we shall derive the Green functions of the operators $(\partial _1^2+a^2\partial _2^2)(\partial _1^2+b^2\partial _2^2),$ $a>0, b>0,a\ne b,$ and $\partial _1^4+2(1-2\epsilon ^2)\partial _1^2\partial _2^2+\partial _2^4,$ $0<\epsilon <1,$ 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 $P(\partial )=(Q(\partial ')-a^2\partial _n^2)(Q(\partial ')-b^2\partial _n^2).$ Therein, we shall also provide further examples.

Let us introduce some notation. ${\mathbb {N}}$ and ${\mathbb {N}}_0$ denote the sets of positive and of nonnegative integers, respectively. We consider as differentiation symbols

$$\begin{aligned} \partial _t:=\frac{\partial }{\partial t},\ \partial _j:=\frac{\partial }{\partial x_j},\ \partial :=(\partial _1,\dots ,\partial _n), \, \Delta _n:=\partial _1^2+\dots +\partial _n^2 \end{aligned}$$

and we denote by $P(\partial )$ linear partial differential operators $\sum _{|\alpha |\le m} c_\alpha \partial ^\alpha$ with constant coefficients $c_{\alpha }\in {\mathbb {C}}$ for $\alpha \in {\mathbb {N}}_0^n,$ $|\alpha |=\alpha _1+\dots +\alpha _n.$ In some examples, we set $\partial =(\partial _t,\partial _1,\dots ,\partial _n)$ and $P(\partial )$ is then an operator in the $n+1$ variables $t,x_1,\dots ,x_n.$

We employ the standard notation for the distribution spaces ${\mathcal {D}}',\,{\mathcal {S}}',$ the dual spaces of the spaces ${\mathcal {D}},\,{\mathcal {S}}$ of “test functions” and of “rapidly decreasing functions,” respectively, see [13, 15]. In order to display the active variable in a distribution, say $x\in {\mathbb {R}}^n,$ we use notation as T(x) or $T\in {\mathcal {D}}'({\mathbb {R}}^n_x).$ For the evaluation of a distribution T on a test function $\phi ,$ we use angle brackets, i.e., $\langle \phi ,T\rangle .$

The Heaviside function is denoted by Y,  see [15, p. 36], and we set

$$\begin{aligned} \chi ^z(t)=\frac{Y(t)t^z}{\Gamma (z+1)}\in L^1_{\text {loc}}({\mathbb {R}}^1_t)\quad \text {for }z\in {\mathbb {C}}\text { with } \text {Re}\,z>-1. \end{aligned}$$

(1.6)

The function $z\mapsto \chi ^z$ can be analytically continued in ${\mathcal {S}}'({\mathbb {R}}^1)$ and thus yields an entire function

$$\begin{aligned} \chi :{\mathbb {C}}\longrightarrow {\mathcal {S}}'({\mathbb {R}}^1):z\longmapsto \chi ^z, \end{aligned}$$

see [4, Equs. (3.1), (3.2), pp. 314, 315], [8, (3.2.17), p. 73]. Note that $(\chi ^z)'=\chi ^{z-1},$ $z\in {\mathbb {C}},$ and $\chi ^{-m}=\delta ^{(m-1)},$ $m\in {\mathbb {N}}.$ We write $\delta$ for the delta distribution with support in 0 i.e., $\langle \phi ,\delta \rangle =\phi (0)$ for $\phi \in {\mathcal {D}}({\mathbb {R}}^n).$

The pull-back $h^*T=T\circ h\in {\mathcal {D}}'(\Omega )$ of a distribution T in one variable t with respect to a submersive $C^\infty$ function $h:\Omega \rightarrow {\mathbb {R}},\ \Omega \subset {\mathbb {R}}^n\text { open},$ is defined as in [5, Equ. (7.2.4/5), p. 82] or in [13, Def. 1.2.12, p. 19], i.e.,

$$\begin{aligned} \langle \phi ,h^*T\rangle =\Big \langle \frac{\text {d}}{\text {d}t}\biggl (\int _\Omega Y(t-h(x)) \phi (x)\,\text {d}x\biggr ), T\Big \rangle ,\quad \phi \in {\mathcal {D}}(\Omega ). \end{aligned}$$

We use the Fourier transform ${\mathcal {F}}$ in the form

$$\begin{aligned} ({\mathcal {F}}\phi )(\xi ):= \int _{{\mathbb {R}}^n}\text {e}^{-\text {i}\xi x}\phi (x)\,\text {d} x,\qquad \phi \in {\mathcal {S}}({\mathbb {R}}^n), \end{aligned}$$

this being extended to ${\mathcal {S}}'$ by continuity. (Herein and also elsewhere, the Euclidean inner product $(\xi ,x)\mapsto \xi x$ is simply expressed by juxtaposition.) For the partial Fourier transforms of a distribution $T\in {\mathcal {S}}'({\mathbb {R}}^{m+n}_{xy})$ with respect to $x\in {\mathbb {R}}^m$ or $y\in {\mathbb {R}}^n,$ respectively, we use the notation ${\mathcal {F}}_xT$ and ${\mathcal {F}}_yT,$ respectively.

The restriction of a distribution $T\in {\mathcal {D}}'({\mathbb {R}}^n)$ to an open set $H\subset {\mathbb {R}}^n$ will be denoted by $T|_H.$ Similarly, we write $T|_{x_n=\epsilon }$ for the restriction of T to the hyperplane $\{x\in {\mathbb {R}}^n;\,x_n=\epsilon \},$ $\epsilon >0,$ if the distribution T continuously depends on $x_n,$ i.e., if it belongs to the subspace of ${\mathcal {D}}'(H),$ $H=\{x\in {\mathbb {R}}^n;\, x_n>0\},$ constituted by the continuous mappings

$$\begin{aligned} (0,\infty )\longrightarrow {\mathcal {D}}'({\mathbb {R}}^{n-1}_{x'}):x_n\longmapsto T_{x_n}(x'). \end{aligned}$$

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 $\partial _n^m.$

2 Green functions of $(\partial _1^2+a^2\partial _2^2)(\partial _1^2+b^2\partial _2^2)$ and of $\partial _1^4+2(1-2\epsilon ^2)\partial _1^2\partial _2^2+\partial _2^4$

We shall first determine the Green function $g_\xi (x)=g_\xi ^\lambda (x),$ $\xi >0,$ $x>0,$ of the ordinary differential operator

$$\begin{aligned} p\left( \tfrac{\text {d}}{\text {d}x}\right) =\left( \lambda -a^2\tfrac{{\text {d}}^2}{\text {d} x^2}\right) \left( \lambda -b^2\tfrac{{\text {d}}^2}{\text {d} x^2}\right) , \quad a>0,b>0,\lambda >0, a\ne b, \end{aligned}$$

(2.1)

with respect to Dirichlet boundary conditions at $x=0.$ Hence, $g_\xi$ fulfills

$$\begin{aligned}&\text {(i)}\ p(\tfrac{\text {d}}{\text {d}x}) g_\xi =\delta (x-\xi )\text { for }x>0, \\&\text {(ii)}\ \exists T_\xi \in {\mathcal {S}}'({\mathbb {R}}):g_\xi =T_\xi |_{(0,\infty )} ,\\&\text {(iii)}\ \lim _{\epsilon \searrow 0}g_\xi ^{(k)}(\epsilon )=0\text { for }k=0,1. \end{aligned}$$

(2.2)

The uniquely determined temperate fundamental solution of $\lambda (\lambda -a^2{\text {d}}^2/\text {d} x^2)$ is given by

$$\begin{aligned} E^a(x)=E^{a,\lambda }(x)=\frac{1}{2a \lambda ^{3/2}}\, \text {e}^{-\sqrt{\lambda } |x|/a},\quad x\in {\mathbb {R}}, \end{aligned}$$

(2.3)

and this easily yields that the linear combination $E=(a^2 E^a-b^2 E^b)/(a^2-b^2)$ coincides with the temperate fundamental solution of $p(\tfrac{\text {d}}{\text {d}x}).$ In fact,

$$\begin{aligned} p(\tfrac{\text {d}}{\text {d}x})E=\frac{1}{a^2-b^2}\Bigl [\frac{a^2}{\lambda }\Bigl (\lambda -b^2\frac{\text {d}^2}{\text {d}x^2}\Bigr )\delta - \frac{b^2}{\lambda }\Bigl (\lambda -a^2\frac{\text {d}^2}{\text {d}x^2}\Bigr )\delta \Bigr ]=\delta . \end{aligned}$$

If we use the ansatz

$$\begin{aligned} g_\xi (x)=E(x-\xi )+c_1 E^a(x+\xi )+c_2 E^b(x+\xi )+c_3 E^a(x+\tfrac{a\xi }{b})+c_4 E^b(x+\tfrac{b\xi }{a}), \end{aligned}$$

then conditions (i) and (ii) in (2.2) are clearly fulfilled. Now note that $E^a(\frac{a\xi }{b})=\frac{b}{a} E^b(\xi )$ and $E^{a\prime }(\frac{a\xi }{b})=\frac{b^2}{a^2} E^{b\prime }(\xi ).$ Therefore, the remaining condition (iii) in (2.2) yields

$$\begin{aligned} 0=g_\xi (0)=\,&\tfrac{a^2}{a^2-b^2} E^a(\xi )-\tfrac{b^2}{a^2-b^2} E^b(\xi )\\&+c_1 E^a(\xi )+c_2E^b(\xi )+c_3\tfrac{b}{a} E^b(\xi )+c_4\tfrac{a}{b} E^a(\xi ), \end{aligned} $$

i.e.,

$$\begin{aligned} c_1+\frac{a}{b} c_4=\frac{a^2}{b^2-a^2},\ c_2+\frac{b}{a} c_3=\frac{b^2}{a^2-b^2}, \end{aligned}$$

and

$$\begin{aligned} 0=g_\xi '(0)=\,&-\tfrac{a^2}{a^2-b^2} E^{a\prime }(\xi )+\tfrac{b^2}{a^2-b^2} E^{b\prime }(\xi )\\&+c_1 E^{a\prime }(\xi )+c_2E^{b\prime }(\xi )+c_3\tfrac{b^2}{a^2}E^{b\prime }(\xi )+c_4\tfrac{a^2}{b^2} E^{a\prime }(\xi ), \end{aligned} $$

i.e.,

$$\begin{aligned} c_1+\frac{a^2}{b^2} c_4=\frac{a^2}{a^2-b^2},\ c_2+\frac{b^2}{a^2} c_3=\frac{b^2}{b^2-a^2}. \end{aligned}$$

These linear equations for $c_1,\dots ,c_4$ have the solutions

$$\begin{aligned} c_1=-\frac{a^2}{(a-b)^2}=\frac{a^2}{b^2} c_2,\quad c_3=\frac{2a^2b}{(a-b)^2(a+b)}=\frac{a}{b} c_4, \end{aligned}$$

and hence,

$$\begin{aligned} g_\xi (x)&=g_\xi ^\lambda (x)=F^{a,b,\lambda }_\xi (x)+F^{b,a,\lambda }_\xi (x),\quad x>0,\\ F^{a,b,\lambda }_{\xi }(x)&=\frac{a^2}{a^2-b^2} E^{a,\lambda }(x-\xi )-\frac{a^2}{(a-b)^2} E^{a,\lambda }(x+\xi )\\&\quad +\frac{2a^2b}{(a-b)^2(a+b)} E^{a,\lambda }\Bigl (x+\frac{a\xi }{b}\Bigr ). \end{aligned}$$

(2.4)

We finally observe that the expansion into a power series with respect to $\sqrt{\lambda }$ shows that $g_\xi ^\lambda$ continuously depends on $\lambda$ also near $\lambda =0$ and that

$$\begin{aligned} \lim _{\lambda \searrow 0} g_\xi ^\lambda (x)=\frac{1}{12 a^2b^2}\bigl [|x-\xi |^3-x^3+3x^2\xi +3x\xi ^2-\xi ^3\bigr ]. \end{aligned}$$

(2.5)

In the next step, we consider the operator

$$\begin{aligned} P(\partial )=(\partial _1^2+a^2\partial _2^2)(\partial _1^2+b^2\partial _2^2),\quad a>0,b>0, a\ne b. \end{aligned}$$

We are going to derive the Green function $G_\xi (x),$ $\xi =(0,\xi _2),$ $\xi _2>0,$ in the half-space $H=\{x\in {\mathbb {R}}^2;\, x_2>0\}$ subject to Dirichlet boundary conditions at the border line $x_2=0.$ Hence, $G_\xi$ fulfills

$$\begin{aligned} &\text {(i)}\ P(\partial ) G_\xi =\delta (x-\xi )\text { for }x_2>0, \\&\text {(ii)}\ \exists T_\xi \in {\mathcal {S}}'({\mathbb {R}}^2):G_\xi =T_\xi |_H ,\\&\text {(iii)}\ \lim _{\epsilon \searrow 0}\partial _2^kG_\xi |_{x_2=\epsilon }=0\text { in }{\mathcal {S}}'({\mathbb {R}}^1_{x_1})\text { for }k=0,1. \end{aligned}$$

(2.6)

Note that (2.6) does not determine $G_\xi$ uniquely since, e.g., for each solution $G_\xi$, the distributions $G_\xi +cx_2^2,$ $c\in {\mathbb {C}},$ also fulfill (2.6). As we shall show in Section 4, $G_\xi$ becomes uniquely determined if we add to (2.6) the growth condition

$$\begin{aligned} \text {(iv)}\ \lim _{N\rightarrow \infty }N^{-2}G_\xi |_{x_2=N}=0\text { in }{\mathcal {S}}'({\mathbb {R}}^1_{x_1}). \end{aligned}$$

(2.7)

Upon a partial Fourier transform with respect to $x_1,$ we obtain the ordinary differential operator in (2.1) as an operator in $\text {d}/\text {d}x_2$ with $\lambda$ being the square of the transformed variable of $x_1.$ Hence, we conclude, at least heuristically, that $G_\xi (x)={\mathcal {F}}^{-1}_{x_1}g_{\xi _2}^{x_1^2}(x_2).$

We next observe that

$$\begin{aligned} E^{a,x_1^2}(x_2)=\frac{1}{2a|x_1|^3}\,\text {e}^{-|x_1x_2|/a} \end{aligned}$$

is, at first sight, not well defined at $x_1=0,$ but that the linear combination of these functions in $g_{\xi _2}^{x_1^2}(x_2)$ is continuous at $x_1=0$ due to Eq. (2.5). We can therefore evaluate the inverse Fourier transform of $g_{\xi _2}^{x_1^2}(x_2)$ by replacing $E^{a,x_1^2}(x_2)$ by the finite part at $z=-3$ of the meromorphic distribution-valued function

$$\begin{aligned} z\longmapsto \frac{1}{2a}|x_1|^z \text {e}^{-|x_1x_2|/a}=S^a_z(x)\in {\mathcal {S}}'({\mathbb {R}}^2), \end{aligned}$$

which has simple poles for $z\in -{\mathbb {N}}.$

For $\text {Re}\,z>-1$ and fixed $x_2\ne 0,$ $S^a_z(x)$ is an integrable function of $x_1.$ Hence, we obtain, for $\text {Re}\,z>-1$ and $x_2\ne 0,$ the following:

$$\begin{aligned}{\mathcal {F}}^{-1}_{x_1} S^a_z&=\frac{1}{4\pi a}\int _{-\infty }^\infty \text {e}^{\text {i} tx_1-|x_2 t|/a}|t|^z\,\text {d}t \\&=\frac{\Gamma (z+1)}{4\pi a}\,\Bigl [\Bigl (\frac{|x_2|}{a}+\text {i}x_1\Bigr )^{-z-1}+\Bigl (\frac{|x_2|}{a}-\text {i}x_1\Bigr )^{-z-1}\Bigr ], \end{aligned}$$

(2.8)

see [7, Equ. 3.381.4], [9, p. 103]. Furthermore, ${\mathbb {R}}\rightarrow {\mathcal {S}}'({\mathbb {R}}^1_{x_1}):x_2\mapsto S^a_z(x)$ is continuous if $\text {Re}\,z>-1.$ Therefore, the result in (2.8) represents ${\mathcal {F}}^{-1}_{x_1} S^a_z$ as a locally integrable function in ${\mathbb {R}}^2$ if $|\text {Re}\,z|<1.$ By analytic continuation, this generally holds for $\text {Re}\, z<1$ outside the poles of $\Gamma (z+1),$ i.e., if $z\not \in -{\mathbb {N}}.$ Thus, we obtain

$$\begin{aligned} {\mathcal {F}}^{-1}_{x_1}(\begin{array}{c} \\ \mathrm{Pf}\\ {z=-3}\\ \end{array} S^a_z)=\begin{array}{c} \\ \mathrm{Pf}\\ {z=-3}\\ \end{array} ({\mathcal {F}}^{-1}_{x_1} S^a_z) =\frac{1}{2\pi a}\text {Re}\Bigl [ \begin{array}{c} \\ \mathrm{Pf}\\ {z=-3}\\ \end{array} \Gamma (z+1)\Bigl (\frac{|x_2|}{a}+\text {i}x_1\Bigr )^{-z-1}\Bigr ]. \end{aligned}$$

We now use [12, Prop. 1.6.3, p. 28], and ${\text {Res}}_{z=-2}\Gamma (z)=\frac{1}{2},$ ${\text {Pf}}_{z=-2}\Gamma (z)=\frac{1}{2}\psi (3)$ in order to conclude that

$$\begin{aligned} {\mathcal {F}}^{-1}_{x_1}(\begin{array}{c} \\ \mathrm{Pf}\\ {z=-3}\\ \end{array} S^a_z)&=\frac{1}{2\pi a}\text {Re}\Bigl [ \begin{array}{c} \\ \mathrm{Pf}\\ {z=-3}\\ \end{array} \Gamma (z+1)\cdot \Bigl (\frac{|x_2|}{a}+\text {i}x_1\Bigr )^2\\&\quad +\begin{array}{c} \\ \mathrm{Res}\\ {z=-3}\\ \end{array} \Gamma (z+1)\cdot \frac{\text {d}}{\text {d}z}\Bigl (\frac{|x_2|}{a}+ \text {i}x_1\Bigr )^{-z-1}\Big |_{z=-3}\Bigr ]\\&=\frac{1}{2\pi a}\text {Re}\Bigl [ \frac{1}{2}\psi (3)\Bigl (\frac{|x_2|}{a}+\text {i}x_1\Bigr )^2-\frac{1}{2} \Bigl (\frac{|x_2|}{a}+\text {i}x_1\Bigr )^2 \log \Bigl (\frac{|x_2|}{a}+\text {i}x_1\Bigr )\Bigr ] \\&=-\frac{\psi (3)}{4\pi a} \Bigl (x_1^2-\frac{x_2^2}{a^2}\Bigr )+\frac{1}{8\pi a}\Bigl (x_1^2-\frac{x_2^2}{a^2}\Bigr ) \log \Bigl (x_1^2+\frac{x_2^2}{a^2}\Bigr )+\frac{x_1x_2}{2\pi a^2}\arctan \Bigl (\frac{ax_1}{x_2}\Bigr ). \end{aligned} $$

Upon summing up the six terms which constitute $g_{\xi _2}^{x_1^2}(x_2),$ the second-order polynomials with the factor $\psi (3)$ cancel out. Furthermore, for $\text {Re}\, z>0,$ we have $x_1^2(x_1^2-a^2\partial _2^2)S^a_z=|x_1|^{z+3}\delta (x_2),$ and this implies, by analytic continuation, that $E^a={\mathcal {F}}^{-1}_{x_1}({\text {Pf}}_{z=-3} S^a_z)$ is a fundamental solution of $\partial _1^2(\partial _1^2+a^2\partial _2^2),$ also compare [13, Equ. (3.1.15), p. 194]. 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.

Proposition 2.1

For $c>0,$ let $E^c$ denote the following fundamental solution of $\partial _1^2(\partial _1^2+c^2\partial _2^2):$

$$\begin{aligned} E^c(x)=\frac{1}{8\pi c}\Bigl (x_1^2-\frac{x_2^2}{c^2}\Bigr ) \log \Bigl (x_1^2+\frac{x_2^2}{c^2}\Bigr )+\frac{x_1x_2}{2\pi c^2}\arctan \Bigl (\frac{cx_1}{x_2}\Bigr ). \end{aligned}$$

(2.9)

Let $G_\xi (x),$ $x_2>0,$ $\xi =(0,\xi _2),$ $\xi _2>0$ be the Green function of the operator $P(\partial )=(\partial _1^2+a^2\partial _2^2)(\partial _1^2+b^2\partial _2^2),$ $a>0,b>0,a\ne b,$ with respect to Dirichlet boundary conditions at $x_2=0,$ i.e., let $G_\xi$ be determined by the conditions (2.6) and (2.7).

Then $G_\xi (x)= F^{a,b}_\xi (x)+F^{b,a}_\xi (x),$ $x_1\in {\mathbb {R}},\ x_2>0,$ where

$$\begin{aligned} F^{a,b}_\xi (x)&=\frac{a^2}{a^2-b^2} E^a(x-\xi )-\frac{a^2}{(a-b)^2} E^a(x+\xi )\\&\quad +\frac{2a^2b}{(a-b)^2(a+b)} E^a\Bigl (x+\frac{a\xi }{b}\Bigr ). \end{aligned}$$

In order to derive the Green function (of the Dirichlet problem) for the orthotropic plate operator in (1.1), let us represent $G_\xi$ slightly more explicitly in the case $b=\frac{1}{a}.$ We assume that $a>1$ and we set $\mu =\frac{1}{2}(a-b).$ This implies $a^2+b^2=2+4\mu ^2,$ $\begin{matrix} a\\ b\end{matrix}\Big \}=\pm \mu +\sqrt{1+\mu ^2}$ and $P(\partial )=\partial _1^4+2(1+2\mu ^2)\partial _1^2\partial _2^2+\partial _2^4.$ Then, the fundamental solution $E=(a^2E^a-b^2 E^b)/(a^2-b^2)$ of $P(\partial )$ takes the form

$$\begin{aligned} E&=\frac{1}{8\pi (a^2-b^2)}\Bigl [(ax_1^2-bx_2^2)\log (x_1^2+b^2x_2^2)-(bx_1^2-ax_2^2)\log (x_1^2+a^2x_2^2)\\&\quad +4x_1x_2\arctan \Bigl (\frac{ax_1}{x_2}\Bigr )-4x_1x_2\arctan \Bigl (\frac{bx_1}{x_2}\Bigr )\Bigr ]\\&=\frac{1}{8\pi (a^2-b^2)}\Bigl [\frac{a-b}{2}|x|^2\log P(x)+\frac{a+b}{2}(x_1^2-x_2^2)\log \Bigl (\frac{x_1^2+b^2x_2^2}{x_1^2+a^2x_2^2}\Bigr )\\&\quad +4x_1x_2\arctan \Bigl (\frac{(a-b) x_1x_2}{|x|^2}\Bigr )\Bigr ]. \end{aligned}$$

Hence, we obtain

$$\begin{aligned} E&=\frac{|x|^2}{32\pi \sqrt{1+\mu ^2}}\log P(x)+\frac{x_1^2-x_2^2}{32\pi \mu }\log \left (\frac{x_1^2+b^2x_2^2}{x_1^2+a^2x_2^2}\right ) \\&\quad +\frac{x_1x_2}{8\pi \mu \sqrt{1+\mu ^2}}\arctan\left(\frac{2\mu x_1x_2}{|x|^2}\right). \end{aligned}$$

(2.10)

Similarly, the term $H=(a^2 E^a+b^2 E^b)/(a-b)^2$ yields

$$\begin{aligned} H&=\frac{\sqrt{1+\mu ^2}(x_1^2-x_2^2)}{32\pi \mu ^2}\log P(x)+\frac{|x|^2}{32\pi \mu }\log \left (\frac{x_1^2+b^2x_2^2}{x_1^2+a^2x_2^2}\right) \\&+\frac{x_1x_2}{8\pi \mu ^2}\Bigl [\arctan \left(\frac{2\sqrt{1+\mu ^2} x_1x_2}{x_2^2-x_1^2}\right )+\pi Y(x_1^2-x_2^2){\text {sign}}(x_1x_2)\Bigr ]. \end{aligned}$$

(2.11)

Of course, the last part of $G_\xi ,$ i.e.,

$$\begin{aligned} J=\frac{2ab}{(a-b)^2(a+b)}\bigl [a E^a(x+\tfrac{a\xi }{b})+b E^b(x+\tfrac{b\xi }{a})\bigr ], \end{aligned}$$

is the most laborious one. It gives

$$\begin{aligned} J&=\frac{x_1^2-(x_2+\xi _2)^2-2\mu ^2(x_2^2+\xi _2^2)}{32\pi \mu ^2\sqrt{1+\mu ^2}}\\&\quad \times \log \bigl [\bigl (x_1^2+(bx_2+a\xi _2)^2\bigr ) \bigl (x_1^2+(ax_2+b\xi _2)^2\bigr )\bigr ]\\&\quad +\frac{x_2^2-\xi _2^2}{16\pi \mu }\log \Bigl (\frac{x_1^2+(bx_2+a\xi _2)^2}{x_1^2+(ax_2+b\xi _2)^2}\Bigr )\\&\quad +\frac{x_1(x_2+\xi _2)}{8\pi \mu ^2}\biggl [\arctan \Bigl (\frac{2\sqrt{1+\mu ^2}x_1(x_2+\xi _2)}{(x_2+\xi _2)^2+ 4\mu ^2x_2\xi _2-x_1^2}\Bigr )\\&\quad +\pi Y\bigl (x_1^2-(x_2+\xi _2)^2-4\mu ^2x_2\xi _2\bigr ){\text {sign}}(x_1(x_2+\xi _2))\biggr ]\\&\quad +\frac{x_1(\xi _2-x_2)}{8\pi \mu \sqrt{1+\mu ^2}} \arctan \left (\frac{2\mu x_1(x_2-\xi _2)}{(x_2+\xi _2)^2+ 4\mu ^2x_2\xi _2+x_1^2}\right). \end{aligned}$$

(2.12)

We now obtain the Green function of the orthotropic plate operator $P(\partial )$ in (1.1) by continuing $G_\xi (x)=E(x-\xi )-H(x+\xi )+J(x,\xi )$ analytically with respect to $\mu ,$ i.e., we set $\mu =\text {i} \epsilon ,$ $0<\epsilon <1.$ The result is the following.

Proposition 2.2

The Green function $G_\xi (x),$ $x_2>0,$ $\xi =(0,\xi _2),$ $\xi _2>0,$ of the operator of the orthotropic plate $P(\partial )=\partial _1^4+2(1-2\epsilon ^2)\partial _1^2\partial _2^2+\partial _2^4,$ $0<\epsilon <1,$ with respect to Dirichlet boundary conditions at $x_2=0,$ see (2.6) and (2.7), is given by $G_\xi (x)={\tilde{G}}_\xi (x)+{\tilde{G}}_\xi (-x_1,x_2)$ where ${\tilde{G}}_\xi (x)=f_1(x-\xi )+f_2(x+\xi )+f_3(x,\xi )$ and

$$\begin{aligned} f_1(x)=&\ \frac{\epsilon |x|^2+2x_1x_2}{32\pi \epsilon \sqrt{1-\epsilon ^2}}\log (|x|^2+2\epsilon x_1x_2)+\frac{x_2^2-x_1^2}{16\pi \epsilon } \arctan \left(\frac{\epsilon |x_2|+x_1}{|x_2|\sqrt{1-\epsilon ^2}}\right ), \\ f_2(x)=&\ \frac{\sqrt{1-\epsilon ^2}(x_1^2-x_2^2)}{32\pi \epsilon ^2}\log (|x|^2+2\epsilon x_1x_2)+ \\&\quad +\frac{\epsilon |x|^2+2x_1x_2}{16\pi \epsilon ^2} \arctan \left(\frac{\epsilon x_2+x_1}{x_2\sqrt{1-\epsilon ^2}}\right ), \\ f_3(x,\xi )=&\ \frac{(x_2+\xi _2)^2-2\epsilon ^2(x_2^2+\xi _2^2)-x_1^2-2\epsilon x_1(x_2-\xi _2)}{32\pi \epsilon ^2\sqrt{1-\epsilon ^2}} \times \\&\times \log [x_1^2+(x_2+\xi _2)^2-4\epsilon ^2x_2\xi _2+2\epsilon x_1(x_2-\xi _2)]\\&-\frac{\epsilon (x_2^2-\xi _2^2)+x_1(x_2+\xi _2)}{8\pi \epsilon ^2} \arctan \left(\frac{\epsilon (x_2-\xi _2)+x_1}{(x_2+\xi _2)\sqrt{1-\epsilon ^2}}\right ). \end{aligned}$$

Proof

If we replace $\mu$ by $\text {i}\epsilon ,$ then $\begin{matrix} a\\ b\end{matrix}\Big \}=\pm \text {i}\epsilon +\sqrt{1-\epsilon ^2}$ are conjugate complex numbers of modulus one. Hence, in the second term of E in formula (2.10), the logarithm is purely imaginary and given by

$$\begin{aligned} \log \Bigl (\frac{x_1^2+b^2x_2^2}{x_1^2+a^2x_2^2}\Bigr )&= 2\text {i}\biggl [\arctan \Bigl (\frac{\text {Im}(b^2)x_2^2}{x_1^2+\text {Re}(b^2)x_2^2}\Bigr )- \pi Y\bigl (-x_1^2-\text {Re}(b^2) x_2^2\bigr )\biggr ]\\&=-2\text {i}\biggl [\arctan \Bigl (\frac{2\epsilon \sqrt{1-\epsilon ^2} x_2^2}{|x|^2-2\epsilon ^2x_2^2}\Bigr ) +\pi Y(2\epsilon ^2x_2^2-|x|^2)\biggr ]\\&=-2\text {i}\biggl [\arctan \Bigl (\frac{\epsilon |x_2|+x_1}{|x_2|\sqrt{1-\epsilon ^2}}\Bigr )+\arctan \Bigl (\frac{\epsilon |x_2|-x_1}{|x_2|\sqrt{1-\epsilon ^2}}\Bigr )\biggr ]. \end{aligned}$$

Furthermore, the arctangent in Eq. (2.10) yields

$$\begin{aligned} \arctan \Bigl (\frac{2\text {i}\epsilon x_1x_2}{|x|^2}\Bigr )=\frac{\text {i}}{2}\log \Bigl (\frac{|x|^2+2\epsilon x_1x_2}{|x|^2-2\epsilon x_1x_2}\Bigr ). \end{aligned}$$

The analytic continuation of H in formula (2.11) is similar.

Let us yet explain the analytic continuation of formula (2.12). The polynomial

$$\begin{aligned} Q(x)=\bigl (x_1^2+(bx_2+a\xi _2)^2\bigr )\bigl (x_1^2+(ax_2+b\xi _2)^2\bigr ) \end{aligned}$$

can be factored as $Q(x)=q(x)q(-x_1,x_2)$ where

$$\begin{aligned} q(x)&=\bigl (x_1+\text {i}(bx_2+a\xi _2)\bigr )\bigl (x_1-\text {i}(ax_2+b\xi _2)\bigr )\\&=x_1^2+(x_2+\xi _2)^2-4\epsilon ^2x_2\xi _2+2\epsilon x_1(x_2-\xi _2). \end{aligned}$$

This settles the first term in formula (2.12). For the second one, observe that

$$\begin{aligned} &\log \Bigl (\frac{x_1^2+(bx_2+a\xi _2)^2}{x_1^2+(ax_2+b\xi _2)^2}\Bigr ) = 2\text {i}\biggl [\arctan \Bigl (\frac{\text {Im}\bigl ((bx_2+a\xi _2)^2\bigr )}{x_1^2+\text {Re}\bigl ((bx_2+a\xi _2)^2\bigr )}\Bigr ) +\biggl \{\begin{aligned}&0\\ \pm&\pi \end{aligned}\biggr \}\biggr ]\\&\quad =-2\text {i}\biggl [\arctan \Bigl (\frac{2\epsilon \sqrt{1-\epsilon ^2}( x_2^2-\xi _2^2)}{x_1^2+(x_2+\xi _2)^2-2\epsilon ^2(x_2^2+\xi _2^2)}\Bigr ) +\biggl \{\begin{aligned}&0\\ \mp&\pi \end{aligned}\biggr \}\biggr ]\\&\quad =-2\text {i}\biggl [\arctan \Bigl (\frac{\epsilon (x_2-\xi _2)+x_1}{(x_2+\xi _2)\sqrt{1-\epsilon ^2}}\Bigr ) +\arctan \Bigl (\frac{\epsilon (x_2-\xi _2)-x_1}{(x_2+\xi _2)\sqrt{1-\epsilon ^2}}\Bigr )\biggr ], \end{aligned}$$

where $\pi$ or $-\pi$ is added in the first two formula lines if the complex number $x_1^2+(bx_2+a\xi _2)^2$ 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 $G_\xi (x)$ 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 $G_\xi (x)$ under these conditions follows from a reasoning which is similar to that in the proof of Proposition 4.1 below. This completes the proof. $\square$

Remark 2.3

If $f_1$ is as in Proposition 2.2, then $f_1(x)+f_1(-x_1,x_2)$ coincides with the fundamental solution E(x) in Eq. (1.3) of the orthotropic plate operator (1.1) up to the second-order polynomial $\frac{\arcsin \epsilon }{16\pi \epsilon }(x_1^2-x_2^2).$

3 Quasi-hyperbolic and non-vanishing symbol operators

Let us generalize now the context of Section 2 and consider operators of the form

$$\begin{aligned} P(\partial )=(Q(\partial ')-a^2\partial _n^2)(Q(\partial ')-b^2\partial _n^2),\quad a>0,b>0,a\ne b, \end{aligned}$$

(3.1)

where $Q(\partial ')$ is an operator in the $n-1$ variables $x'=(x_1,\dots ,x_{n-1})$ and $P(\partial )$ either is quasi-hyperbolic or has a non-vanishing symbol $P(\text {i} x).$

Let us first recall the notion of “quasi-hyperbolicity”, see [13, Def. and Prop. 2.4.13, p. 162].

Definition 3.1

The operator $P(\partial )=P(\partial _1,\dots ,\partial _n)=\sum _{|\alpha |\le m} c_\alpha \partial ^\alpha ,$ $c_\alpha \in {\mathbb {C}},$ $\alpha \in {\mathbb {N}}_0^n,$ is called quasi-hyperbolic in the direction $N\in {\mathbb {R}}^n\setminus \{0\}$ if and only if there exists $\sigma _0\in {\mathbb {R}}$ such that

$$\begin{aligned} \forall \sigma >\sigma _0:\forall x\in {\mathbb {R}}^n:P(\text {i} x+\sigma N)\ne 0. \end{aligned}$$

(3.2)

As shown in [13, Def. and Prop. 2.4.13, p. 162], for a quasi-hyperbolic operator $P(\partial ),$ there exists a uniquely determined fundamental solution E of $P(\partial )$ fulfilling $\text {e}^{-\sigma Nx}E\in {\mathcal {S}}'({\mathbb {R}}^n)$ for some $\sigma$ larger than the constant $\sigma _0$ in (3.2). This fundamental solution is then given by the formula

$$\begin{aligned} E=\text {e}^{\sigma Nx}{\mathcal {F}}^{-1}\bigl (P(\text {i}x+\sigma N)^{-1}\bigr ),\quad \sigma >\sigma _0, \end{aligned}$$

(3.3)

and it holds $\text {supp}\,E\subset \{x\in {\mathbb {R}}^n;\, x\cdot N\ge 0\}$ and $\text {e}^{-\sigma xN}E\in {\mathcal {S}}'({\mathbb {R}}^n)$ for each $\sigma \ge \sigma _0.$ Note that operators $P(\partial )$ whose symbols $P(\text {i}x)$ 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 $G_\xi (x)$ of the Dirichlet problem in the half-space $H=\{x\in {\mathbb {R}}^n;\,x_n>0\}$ for quasi-hyperbolic and for non-vanishing symbol operators $P(\partial )$ as in (3.1). We assume that $Q(\partial ')$ is an operator in the $n-1$ variables $x'=(x_1,\dots ,x_{n-1}),$ that $N'\in {\mathbb {R}}^{n-1}$ and we set $N=(N',0).$ Then, $P(\partial )$ as in (3.1) is quasi-hyperbolic in direction N (in case $N\ne 0)$ or has a non-vanishing symbol $P(\text {i}x)$ (in case $N=0)$ if and only if, for some $\sigma _0\in {\mathbb {R}},$ the following condition is satisfied:

$$\begin{aligned} \forall \sigma >\sigma _0:\forall x=(x',x_n)\in {\mathbb {R}}^n:Q(\text {i} x'+\sigma N')+x_n^2\ne 0. \end{aligned}$$

(3.4)

Proposition 3.2

Let $N'\in {\mathbb {R}}^{n-1},$ $\sigma _0\in {\mathbb {R}},$ $H=\{x\in {\mathbb {R}}^n;\,x_n>0\},$ $\xi =(0,\dots ,0,\xi _n)\in {\mathbb {R}}^n,$ $\xi _n>0.$ Let $Q(\partial ')=Q(\partial _1,\dots ,\partial _{n-1})$ be a linear partial differential operator with constant coefficients in $n-1$ variables such that condition (3.4) is satisfied. Let $P(\partial )$ be defined as in (3.1). For $c>0,$ let $E^c=c^{-1} E^1(x',x_n/c)$ be the uniquely determined fundamental solution of $Q(\partial ')\bigl (Q(\partial ')-c^2\partial _n^2\bigr )$ such that $\text {e}^{-\sigma N'x'}E^c$ is temperate for $\sigma \ge \sigma _0.$

Then, the Green function $G_\xi \in {\mathcal {D}}'(H)$ of $P(\partial )$ fulfilling

$$\begin{aligned} &\text {(i)}\ P(\partial ) G_\xi =\delta (x-\xi )\text { in }{\mathcal {D}}'(H), \\&\text {(ii)}\ \exists \sigma >\sigma _0:\exists T_\xi \in {\mathcal {S}}'({\mathbb {R}}^n):\text {e}^{-\sigma N'x'}G_\xi =T_\xi |_H,\\&\text {(iii)}\ \lim _{\epsilon \searrow 0}\partial _n^k T_\xi |_{x_n=\epsilon }=0\text { in }{\mathcal {S}}'({\mathbb {R}}^{n-1}) \text { for }T_\xi \text { as in {(b)} and } k=0,1 \end{aligned}$$

(3.5)

is uniquely determined and given by

$$\begin{aligned} G_\xi (x)&=F^{a,b}_\xi (x)+F^{b,a}_\xi (x),\quad x\in H,\\ F^{a,b}_\xi (x)&=\frac{a^2}{a^2-b^2} E^a(x-\xi )-\frac{a^2}{(a-b)^2} E^a(x+\xi )\\&\quad +\frac{2a^2b}{(a-b)^2(a+b)} E^a\Bigl (x+\frac{a\xi }{b}\Bigr ). \end{aligned}$$

(3.6)

[Note that the restrictions of $\partial _n^kT_\xi$ to the subspaces $\{x\in {\mathbb {R}}^n;\ x_n=\epsilon \},$ $\epsilon >0,$ are well defined due to (i) and [8, Thm. 4.4.8, p. 115]. Condition (iii) requires that these restrictions belong to ${\mathcal {S}}'({\mathbb {R}}^{n-1})$ 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 $(Q(\partial ')-\partial _n^2)^m,$ $m\in {\mathbb {N}},$ is investigated. We therefore only outline the proof here and refer to [14] for details.

Since the values of the polynomial $\lambda (x')=Q(\text {i}x'+\sigma N')$ lie in the set ${\mathbb {C}}\setminus (-\infty ,0]$ due to condition (3.4), we can choose the root $\sqrt{\lambda (x')}$ such the $\text {Re}\,\sqrt{\lambda (x')}>0.$ If $G_\xi ,$ ${\tilde{G}}_\xi$ are two solutions of (3.5), then we consider the difference $S=G_\xi -{\tilde{G}}_\xi$ and the partial Fourier transform with respect to $x'$

$$\begin{aligned} {\tilde{S}}={\mathcal {F}}_{x'}(\text {e}^{-\sigma N'x'}S)={\mathcal {F}}_{x'}(T_\xi -{\tilde{T}}_\xi )|_H\in {\mathcal {D}}'(H). \end{aligned}$$

It fulfills $(\lambda (x')-a^2\partial _n^2)(\lambda (x')-b^2\partial _n^2){\tilde{S}}=0$ in ${\mathcal {D}}'(H)$ and accordingly we obtain the representation

$$\begin{aligned} {\tilde{S}}=\sum _{\epsilon =\pm } A_\epsilon (x')\text {e}^{\epsilon x_n \sqrt{\lambda (x')}/a}+ B_\epsilon (x')\text {e}^{\epsilon x_n \sqrt{\lambda (x')}/b} \end{aligned}$$

with distributions $A_\epsilon ,B_\epsilon \in {\mathcal {D}}'({\mathbb {R}}^{n-1}),$ $\epsilon =\pm .$ From ${\mathcal {F}}_{x'}(T_\xi -{\tilde{T}}_\xi )\in {\mathcal {S}}'({\mathbb {R}}^n)$, 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 $G_\xi ={\tilde{G}}_\xi .$ Therefore, $G_\xi$ is uniquely determined.

In order to show that $G_\xi$ defined by formula (3.6) satisfies the conditions in (3.5), let E denote the uniquely determined fundamental solution of $P(\partial )$ such that $\text {e}^{-\sigma N'x'}E$ is temperate for $\sigma \ge \sigma _0.$ Then, E satisfies

$$\begin{aligned} E&=\text {e}^{\sigma N'x'} {\mathcal {F}}^{-1}\bigl ((\lambda (x')+a^2 x_n^2)^{-1}(\lambda (x')+b^2 x_n^2)^{-1}\bigr )\\&=\frac{\text {e}^{\sigma N'x'}}{a^2-b^2} {\mathcal {F}}^{-1} \Bigl (\frac{a^2}{\lambda (x')(\lambda (x')+a^2 x_n^2)}-\frac{b^2}{\lambda (x')(\lambda (x')+b^2 x_n^2)}\Bigr )\\&=\frac{a^2 E^a-b^2 E^b}{a^2-b^2}. \end{aligned}$$

Furthermore, $P(\partial ) E^a=P(\partial ) E^b=0$ in H. This implies that $G_\xi$ 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 (2.3) and the ensuing calculations are equally valid for $\lambda \in {\mathbb {C}}\setminus (-\infty ,0]$ if we choose $\sqrt{\lambda }$ with positive real part. $\square$

Example 3.3

Let us calculate the Green function $G_{(0,\xi )}(t,x)$ where $\xi =(0,\dots ,0,\xi _n)\in {\mathbb {R}}^n,$ $\xi _n>0,$ for the product of wave operators

$$\begin{aligned} P(\partial )=(\partial _t^2-\Delta _{n-1}-a^2\partial _n^2)(\partial _t^2-\Delta _{n-1}-b^2\partial _n^2),\quad a>0,b>0,a\ne b, \end{aligned}$$

(3.7)

with respect to Dirichlet conditions at the hyperplane $x_n=0$ and vanishing Cauchy data at $t=0.$ According to Proposition 3.2, we can express $G_{(0,\xi )}$ through a combination of shifted fundamental solutions $E^c$ of $(\partial _t^2-\Delta _{n-1})(\partial _t^2-\Delta _{n-1}-c^2\partial _3^2),$ $c\in \{a,b\}.$

Using the method of parameter integration, see [13, Chap. 3], we have $E^c=c^{-2}\int _0^{c^2} E_\mu \text {d}\mu$ with $E_\mu$ being the “forward” fundamental solution of $(\partial _t^2-\Delta _{n-1}-\mu \partial _n^2)^2,$ i.e., the one with support in the half-space $\{(t,x)\in {\mathbb {R}}^{n+1};\,t\ge 0,x\in {\mathbb {R}}^n\}.$

If $\chi ^z$ is defined as in (1.6), then the forward fundamental solution $E_1$ of $(\partial _t^2-\Delta _n)^2$ can be written as

$$\begin{aligned} E_1(t,x)=\frac{Y(t)\chi ^{2-(n+1)/2}(t^2-|x|^2)}{8\pi ^{(n-1)/2}} \end{aligned}$$

according to [4, Lemma 4.2, p. 317], [8, Section 6.2], [13, Equs. (2.3.11), (2.3.12), pp. 141, 142]. (Note that the function $(t,x)\mapsto t^2-|x|^2$ is submersive outside the origin and that the composition with $\chi ^z$ defined in (1.6) thus is meaningful outside the origin. Furthermore, the resulting distribution continuously depends on $t\in {\mathbb {R}},$ and it can therefore by multiplied by Y(t),  see also [12, Prop. 2.4.2, p. 56], and the ensuing remark.)

Hence,

$$\begin{aligned} E^c=\frac{Y(t)}{8\pi ^{(n-1)/2}c^2}\int _0^{c^2} \chi ^{2-(n+1)/2}\Bigl (t^2-|x'|^2-\frac{x_n^2}{\mu }\Bigr )\frac{\text {d}\mu }{\sqrt{\mu }}. \end{aligned}$$

(3.8)

E.g., let us evaluate the integral in (3.8) in the case of $n=3$ spatial dimensions. Due to $\chi ^0=Y$ and setting $x'=(x_1,x_2),$ $R=\sqrt{t^2-|x'|^2},$ $t_+=\chi ^1(t)=Y(t)t,$ we obtain

$$\begin{aligned} E^c(t,x)=\frac{Y(t)}{8\pi c^2}\int _0^{c^2} Y(t^2-|x'|^2-x_3^2/\mu )\frac{\text {d}\mu }{\sqrt{\mu }} =\frac{Y(t-|x'|)}{4\pi c^2R}(cR-|x_3|)_+, \end{aligned}$$

see also [17, Ex. 5, p. 27]. Therefore, formula (3.6) yields the following Green function for the operator $P(\partial )$ in (3.7) with $n=3:$

$$\begin{aligned} G_{(0,\xi )}(t,x)&=\frac{Y(t-|x'|)}{4\pi R}\Bigl [\frac{(aR-|x_3-\xi _3|)_+-(bR-|x_3-\xi _3|)_+}{a^2-b^2}\\&\quad -\frac{(aR-x_3-\xi _3)_++(bR-x_3-\xi _3)_+}{(a-b)^2}\\&\quad +2\frac{(abR-ax_3-b\xi _3)_++(abR-bx_3-a\xi _3)_+}{(a-b)^2(a+b)}\Bigr ]. \end{aligned}$$

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 $\partial _t(\partial _t-c^2\partial _x^2),$ $c>0,$ is given by

$$\begin{aligned} E^c=\frac{Y(t)}{2c\sqrt{\pi }}\int _0^t\text {e}^{-x^2/(4c^2\tau )}\frac{\text {d}\tau }{\sqrt{\tau }}. \end{aligned}$$

Hence, formula (3.6) in Proposition 3.2 yields the following for the Green function $G_{(0,\xi )},$ $\xi >0,$ of the operator

$$\begin{aligned} P(\partial )=&(\partial _t-a^2\partial _x^2)(\partial _t-a^{-2}\partial _x^2)=\partial _t^2+\partial _x^4-(a^2+a^{-2})\partial _t\partial _x^2,\quad a>0:\\ G_{(0,\xi )}(t,x)=&\frac{Y(t)}{2\sqrt{\pi }}\int _0^t\Bigl \{\frac{1}{a^2-a^{-2}}\Bigl [a\,\text {e}^{-(x-\xi )^2/(4 a^2\tau )} -a^{-1}\text {e}^{-a^2(x-\xi )^2/(4\tau )}\Bigr ] \\&-\frac{1}{(a-a^{-1})^2}\Bigl [a\,\text {e}^{-(x+\xi )^2/(4 a^2\tau )} +a^{-1}\text {e}^{-a^2(x+\xi )^2/(4\tau )}\Bigr ]\\&+\frac{2}{(a-a^{-1})^2(a+a^{-1})}\Bigl [\text {e}^{-(ax+\xi /a)^2/(4\tau )}+\text {e}^{-(a\xi +x/a)^2/(4\tau )}\Bigr ]\Bigr \} \frac{\text {d}\tau }{\sqrt{\tau }}. \end{aligned}$$

(3.9)

We now use analytic continuation as in Section 2 and set $a=\sqrt{\text {i}}=\frac{1+\text {i}}{\sqrt{2}}.$ Then, $P(\partial )=\partial _t^2+\partial _x^4$ and formula (3.9) yields

$$\begin{aligned} G_{(0,\xi )}(t,x)&=\frac{Y(t)}{2\sqrt{\pi }}\int _0^t\Bigl \{\sin \Bigl [\frac{(x-\xi )^2}{4\tau }+\frac{\pi }{4}\Bigr ] -\sin \Bigl [\frac{(x+\xi )^2}{4\tau }-\frac{\pi }{4}\Bigr ]\\&\quad -\sqrt{2}\,\text {e}^{-x\xi /(2\tau )} \cos \Bigl [\frac{x^2-\xi ^2}{4\tau }\Bigr ]\Bigr \}\frac{\text {d}\tau }{\sqrt{\tau }}. \end{aligned}$$

in accordance with [11, p. 239].

Remark 3.5

Our faithful reader, who has followed us up to this point, might wonder how the representation of $G_\xi$ in (3.6) would look in the case of a product of $m>2$ factors, i.e., if $P(\partial =\prod _{j=1}^m (Q(\partial ')-a_j^2\partial _n^2)$ 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 fundamental solution of $Q( \partial ')^{m-1}(Q(\partial ')-c^2\partial _n^2)$ such that $\text {e}^{-\sigma N'x'}E$ is temperate for $\sigma \ge \sigma _0.$ Then, $G_\xi (x)=\sum _{j=1}^m F^j_\xi (x)$ where

$$\begin{aligned} F^j_\xi (x)&=\biggl (\prod _{k\ne j}\frac{a_j^2}{a_j^2-a_k^2}\biggr )E^{a_j}(x-\xi )- \biggl (\prod _{k\ne j}\Bigl (\frac{a_j}{a_j-a_k}\Bigr )^2\biggr )E^{a_j}(x+\xi ) \\&\quad +2\sum _{k\ne j} \biggl (\prod _{l\ne j,l\ne k}\frac{a_ja_k}{(a_j-a_l)(a_k-a_l}\biggr ) \frac{a_j^2a_k E^{a_j}(x+a_j\xi /a_k)}{(a_j-a_k)^2(a_j+a_k)}. \end{aligned}$$

(3.10)

A technically relevant application of formula (3.10) would consist in the derivation of an explicit formula for the Green function of a product $P_1(\partial )P_2(\partial )$ of two orthotropic plate operators $P_1(\partial ),P_2(\partial )$ as in (1.1). This product operator describes the deflection of an orthotropic cylindrical shell, see [3, Equs. (14) and (22), pp. 738, 739].

4 Operators with symbols that are positive outside the origin

In this final section, let us treat operators $P(\partial )$ in ${\mathbb {R}}^n$ such that $P(\text {i}x)>0$ for $x\in {\mathbb {R}}^n\setminus \{0\}.$ As has been shown in [1], the analytic distribution-valued function

$$\begin{aligned} S:\{z\in {\mathbb {C}};\,\text {Re}\,z>0\}\longrightarrow {\mathcal {S}}'({\mathbb {R}}^n):z\longmapsto P(\text {i}x)^z \end{aligned}$$

has a meromorphic extension ${\tilde{S}}$ to the whole complex plane, see also [13, Prop. 2.3.1, p. 134]. For simplicity, we shall write $P(\text {i}x)^z={\tilde{S}}(z)$ if ${\tilde{S}}$ is analytic in $z\in {\mathbb {C}}$ and $P(\text {i}x)^{z_0}={\text {Pf}}_{z=z_0}{\tilde{S}}(z)$ if $z_0\in {\mathbb {C}}$ is a pole of ${\tilde{S}}.$ (Here, ${\text {Pf}}$ stands for the finite part of a meromorphic distribution-valued function, compare [12].) Since $P(\text {i}x)\cdot P(\text {i}x)^z=P(\text {i}x)^{z+1}$ holds by analytic continuation for each $z\in {\mathbb {C}},$ $E:={\mathcal {F}}^{-1}(P(\text {i}x)^{-1})$ yields a temperate fundamental solution of $P(\partial ).$

In particular, let $Q(\partial ')$ be an operator in $n-1$ variables $x'=(x_1,\dots ,x_{n-1})$ such that $Q(\text {i}x')>0$ for $x'\in {\mathbb {R}}^{n-1}\setminus \{0\}.$ Then, $z\mapsto Q(\text {i}x')^{z/2}$ is meromorphic. Upon expanding the exponential function into a power series, this implies that, for $a>0,$ the holomorphic function

$$\begin{aligned} S^a:\{z\in {\mathbb {C}};\,\text {Re}\,z>0\}\longrightarrow {\mathcal {S}}'({\mathbb {R}}^n):z\longmapsto S_z^a=\frac{1}{2a} Q(\text {i}x')^{z/2} \text {e}^{-\sqrt{Q(\text {i} x')}|x_n|/a} \end{aligned}$$

(4.1)

can be extended as a meromorphic function to ${\mathbb {C}}.$ From

$$\begin{aligned} Q(\text {i} x')(Q(\text {i} x')-a^2\partial _n^2) S^a_z=Q(\text {i} x')^{(z+3)/2}\otimes \delta (x_n),\quad \text {Re}\, z>0, \end{aligned}$$

we conclude, by analytic continuation, that

$$\begin{aligned} E^a:= {\mathcal {F}}^{-1}_{x'}\left(\begin{array}{c} \\ \mathrm{Pf}\\ {z=-3}\\ \end{array} S^a_z\right) \end{aligned}$$

(4.2)

is a fundamental solution of $Q(\partial ')(Q(\partial ')-a^2\partial _n^2).$

In the next proposition, we shall generalize Proposition 2.1 to operators of the form

$$\begin{aligned} P(\partial )=(Q(\partial ')-a^2\partial _n^2)(Q(\partial ')-b^2\partial _n^2),\quad a>0,b>0,a\ne b. \end{aligned}$$

(4.3)

Proposition 4.1

Let $Q(\partial ')=Q(\partial _1,\dots ,\partial _{n-1})$ be a linear partial differential operator with constant coefficients in $n-1$ variables such that $Q(\text {i}x')>0$ for $x'\in {\mathbb {R}}^{n-1}\setminus \{0\}.$ Let $H=\{x\in {\mathbb {R}}^n;\,x_n>0\},$ $\xi =(0,\dots ,0,\xi _n)\in {\mathbb {R}}^n,$ $\xi _n>0,$ and $P(\partial )$ be as in (4.3). For $c>0,$ let $E^c$ be the fundamental solution of $Q(\partial ')\bigl (Q(\partial ')-c^2\partial _n^2\bigr )$ defined in (4.2).

Then, the Green function $G_\xi \in {\mathcal {D}}'(H)$ of $P(\partial )$ fulfilling

$$\begin{aligned}&\text {(i)}\ P(\partial ) G_\xi =\delta (x-\xi )\text { in }{\mathcal {D}}'(H), \\&\text {(ii)}\ \exists T_\xi \in {\mathcal {S}}'({\mathbb {R}}^n):G_\xi =T_\xi |_H,\\&\text {(iii)}\ \lim _{\epsilon \searrow 0}\partial _n^k G_\xi |_{x_n=\epsilon }=0\text { in }{\mathcal {S}}'({\mathbb {R}}^{n-1}) \text { for } k=0,1,\\&\text {(iv)}\ \lim _{N\rightarrow \infty }N^{-2}G_\xi |_{x_n=N}=0\text { in }{\mathcal {S}}'({\mathbb {R}}^{n-1}) \end{aligned}$$

(4.4)

is uniquely determined and given by formula (3.6).

Proof

In order to prove the uniqueness of $G_\xi$ 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 $S=G_\xi -{\tilde{G}}_\xi$ of two solutions of (4.4) and its partial Fourier transform

$$\begin{aligned} {\tilde{S}}={\mathcal {F}}_{x'} S={\mathcal {F}}_{x'}( T_\xi -{\tilde{T}}_\xi )|_H\in {\mathcal {D}}'(H). \end{aligned}$$

Condition (i) in (4.4) implies that the support of ${\tilde{S}}$ is contained in the half-axis $\{(0,\dots ,0,x_n)\in {\mathbb {R}}^n;\ x_n>0\}.$ More precisely, we obtain the representation

$$\begin{aligned} {\tilde{S}}=\sum _{\alpha \in {\mathbb {N}}_0^{n-1},\,|\alpha |\le l} \partial ^\alpha \delta (x')\otimes q_\alpha (x_n) \end{aligned}$$

for $l\in {\mathbb {N}}_0$ and polynomials $q_\alpha$ in one variable. Hence, S is a polynomial. Condition (iii) in (4.4) implies $S=x_n^2\cdot R$ for another polynomial and condition (iv) then yields $R=0,$ i.e., $G_\xi ={\tilde{G}}_\xi .$

For the verification of the representation of $G_\xi$ in formula (3.6), we consider again $S^a_z$ as in (4.1). Since

$$\begin{aligned} (Q(\text {i} x')-a^2\partial _n^2)S^a_z=Q(\text {i} x')^{(z+1)/2}\otimes \delta (x_n),\quad \text {Re}\,z>0, \end{aligned}$$

we obtain, by analytic continuation, that

$$\begin{aligned} V=(Q(\partial ')-a^2\partial _n^2)E^a={\mathcal {F}}^{-1}_{x'}\bigl (\begin{array}{c} \\ \mathrm{Pf}\\ {z=-1}\\ \end{array} Q(\text {i}x')^z\bigr )\otimes \delta (x_n) \end{aligned}$$

vanishes in H and is independent of a. This implies that $P(\partial )E^a=P(\partial )E^b=0$ in H and that $E=(a^2 E^a-b^2 E^b)/(a^2-b^2)\in {\mathcal {S}}'({\mathbb {R}}^n)$ is a fundamental solution of $P(\partial )$ due to

$$\begin{aligned} P(\partial )E=\frac{a^2}{a^2-b^2}(\delta -b^2\partial _n^2 V)-\frac{b^2}{a^2-b^2}(\delta -a^2\partial _n^2 V)=\delta . \end{aligned}$$

Hence, conditions (i) and (ii) in (4.4) are fulfilled for $G_\xi$ as in (3.6).

By construction, the partial Fourier transform ${\mathcal {F}}_{x'}G_\xi =({\mathcal {F}}_{x'} T_\xi )|_H$ coincides, for fixed $x'\in {\mathbb {R}}^{n-1},$ with the one-dimensional Green function $g_{\xi _n}^{Q(\text {i}x')}(x_n)$ given in (2.4) for $Q(\text {i}x')>0$ and in (2.5) for $Q(\text {i}x')=0.$ The following lemma shows that $g_{\xi _n}^\lambda (\epsilon )$ and $N^{-2}g_{\xi _n}^\lambda (N)$ converge to 0 for $\epsilon \searrow 0$ and $N\rightarrow \infty$ uniformly with respect to $\lambda \in [0,\infty ).$ The same assertion holds for $(g_{\xi _n}^\lambda )'(\epsilon )$, and this can be shown by inspection of the explicit representation of $(g_{\xi _n}^\lambda )'(\epsilon )$ by exponential functions, i.e., $(g_\eta ^\lambda )'(\epsilon )=h^{a,b,\lambda }_\eta (\epsilon )+h^{b,a,\lambda }_\eta (\epsilon )$ with

$$\begin{aligned} h^{a,b,\lambda }_\eta (\epsilon )=\frac{\text {e}^{-\sqrt{\lambda }\eta /a}}{\lambda (a-b)^2(a+b)}\Bigl [a\cosh \Bigl (\frac{\sqrt{\lambda }\epsilon }{a}\Bigr ) -a\,\text {e}^{-\sqrt{\lambda }\epsilon /b}-b\sinh \Bigl (\frac{\sqrt{\lambda }\epsilon }{a}\Bigr )\Bigr ] \end{aligned}$$

for $0<\epsilon <\eta .$ This implies the conditions (iii) and (iv) in (4.4) and completes the proof. $\square$

Lemma 4.2

For $a>0,b>0,\lambda>0,\xi >0,a\ne b,$ let $g_\xi ^\lambda$ be the one-dimensional Green function of the operator in (2.1), i.e., the function satisfying the conditions in (2.2) and given by formula (2.4).

Then $g_\xi ^\lambda$ fulfills the estimate

$$\begin{aligned} |g_\xi ^\lambda (x)|\le \frac{x\xi \min (x,\xi )}{a^2b^2},\quad x>0. \end{aligned}$$

(4.5)

Proof

As before, let us denote by E the temperate fundamental solution of the operator $p(\text {d}/\text {d}x)$ in (2.1). Then, the Green function can be represented by

$$\begin{aligned} g_\xi ^\lambda (x)=E(x-\xi )-E(x+\xi )-4ab(a+b)\sqrt{\lambda }\, E'(x)E'(\xi ),\quad x>0. \end{aligned}$$

(4.6)

In fact, the function on the right-hand side of formula (4.6) fulfills all the conditions in (2.2). (In parentheses, let us observe that, indeed, formula (4.6) yields a simpler and more symmetric representation of $g_\xi ^\lambda$ than (2.4). But this comes at the expense of the fact that the third term in (4.6), i.e., $\sqrt{\lambda }\, E'(x)E'(\xi ),$ is multiplicative. Hence, a partial Fourier transform with respect to the variables $x_2,\dots ,x_n$ results in a convolution and cannot be immediately expressed by fundamental solutions as in formula (3.6).)

Furthermore, let $F^t$ denote the temperate fundamental solution of $(\lambda -t\text {d}^2/\text {d}x^2)^2,$ $t>0,$ and write $h^{\lambda ,t}_\xi$ for the Green function of $(\lambda -t\text {d}^2/\text {d}x^2)^2,$ i.e.,

$$\begin{aligned} h^{\lambda ,t}_\xi (x)=\lim _{b\rightarrow a=\sqrt{t}} g_\xi ^\lambda (x)=F^t(x-\xi )-F^t(x+\xi )-8\sqrt{\lambda }\, t^{3/2} F^{t\prime }(x)F^{t\prime }(\xi ). \end{aligned}$$

By parameter integration, see [13, Chapter 3], we have $E=(b^2-a^2)^{-1}\int _{a^2}^{b^2} F^t\text {d}t$ and hence

$$\begin{aligned} g_\xi ^\lambda (x)=\frac{1}{b^2-a^2}\int _{a^2}^{b^2}\bigl [h^{\lambda ,t}_\xi (x)+&8\sqrt{\lambda }\, t^{3/2}F^{t\prime }(x) F^{t\prime }(\xi )\bigr ]\text {d}t \\&-4ab(a+b)\sqrt{\lambda }\, E'(x)E'(\xi ). \end{aligned}$$

We now refer to [14, Eqns. (2.7), (2.10)] which imply that

$$\begin{aligned} 0\le h^{\lambda ,t}_\xi (x)\le h^{0,t}_\xi (x)=\frac{1}{6t^2}\left\{ \begin{aligned} x^2(3\xi -x)\,&:0<x\le \xi \\ \xi ^2(3x-\xi )\,&:x\ge \xi \end{aligned} \right\} \le \frac{x\xi \min (x,\xi )}{2 t^2}. \end{aligned}$$

Therefore, $0\le A:=(b^2-a^2)^{-1}\int _{a^2}^{b^2}h^{\lambda ,t}_\xi (x)\text {d}t\le \frac{1}{2} a^{-2}b^{-2}x\xi \min (x,\xi ).$

On the other hand, the explicit representations

$$\begin{aligned} F^t(x)=\frac{\text {e}^{-|x|\sqrt{\lambda /t}}}{4\lambda ^{3/2}\sqrt{t}} \biggl (1+|x|\sqrt{\frac{\lambda }{t}}\biggr ),\quad F^{t\prime }(x)=-\frac{x \text {e}^{-|x|\sqrt{\lambda /t}}}{4\sqrt{\lambda }\, t^{3/2}} \end{aligned}$$

yield

$$\begin{aligned} E'(x)=\frac{1}{b^2-a^2}\int _{a^2}^{b^2} F^{t\prime }(x)\text {d}t=\frac{x}{2\sqrt{\lambda }(b^2-a^2)} \int _{1/a}^{1/b} \text {e}^{-\sqrt{\lambda }|x| u}\text {d}u \end{aligned}$$

and hence, for $x>0,\xi >0,$

$$\begin{aligned} B&:=\frac{8\sqrt{\lambda }}{b^2-a^2} \int _{a^2}^{b^2} t^{3/2} F^{t\prime }(x) F^{t\prime }(\xi )\text {d}t -4ab(a+b)\sqrt{\lambda }\, E'(x)E'(\xi )\\&=\frac{ ab(a+b)x\xi }{(b^2-a^2)^2\sqrt{\lambda }} \int _{1/a}^{1/b}\int _{1/a}^{1/b} \text {e}^{-\sqrt{\lambda }x u} \bigl ( \text {e}^{-\sqrt{\lambda }\xi u}-\text {e}^{-\sqrt{\lambda }\xi v}\bigr )\text {d}u\text {d}v. \end{aligned}$$

(4.7)

Due to the symmetry in the left-hand side of (4.7) with respect to x and $\xi ,$ we can interchange x and $\xi$ on the right-hand side of (4.7). From

$$\begin{aligned} \big |\text {e}^{-\sqrt{\lambda }x u}\big |< 1,\ \big |\text {e}^{-\sqrt{\lambda }\xi u}-\text {e}^{-\sqrt{\lambda }\xi v}\big |\le \sqrt{\lambda }\xi |u-v|,\quad u>0,v>0, \end{aligned}$$

we then obtain

$$\begin{aligned} |B|\le \frac{|a-b|x\xi }{3a^2b^2(a+b)}\,\min (x,\xi )\le \frac{1}{3a^2b^2}\,x\xi \min (x,\xi ). \end{aligned}$$

Due to $g_\xi ^\lambda (x)=A+B$ and taking into account the estimate for A above, we arrive at the inequality in (4.5). This completes the proof. $\square$

Example 4.3

In this final example, let us generalize Proposition 2.1 to n dimensions and calculate the Green function $G_\xi (x)$ for the operator

$$\begin{aligned} P(\partial )=(\Delta _{n-1}+a^2\partial _n^2)(\Delta _{n-1}+b^2\partial _n^2),\quad a>0,b>0,a\ne b, \end{aligned}$$

with respect to Dirichlet boundary conditions at the border $x_n=0.$

According to Proposition 4.1 above, $G_\xi (x)$ can be expressed by formula (3.6) if $E^a$ denotes the fundamental solution of $\Delta _{n-1}(\Delta _{n-1}+a^2\partial _n^2)$ which is given by

$$\begin{aligned} E^a= \begin{array}{c} \\ \mathrm{Pf}\\ {z=-3}\\ \end{array} \bigl ({\mathcal {F}}^{-1}_{x'} S^a_z\bigr ),\quad S_z^a=\frac{|x'|^z}{2a}\text {e}^{-|x'||x_n|/a} \end{aligned}$$

with $x'=(x_1,\dots ,x_{n-1}).$ Due to $E^a=a^{-1}E^1(x',x_n/a),$ we can restrict ourselves to calculating the fundamental solution $E^1$ of $\Delta _{n-1}\Delta _n.$

In order to evaluate ${\mathcal {F}}^{-1}_{x'} S^1_z,$ we use the Poisson–Bochner formula, see [15, Equ. (VII,7;22), p. 259], [13, Equ. (1.6.14), p. 97]. This yields

$$\begin{aligned} {\mathcal {F}}^{-1}_{x'} S^1_z=\frac{|x'|^{(3-n)/2}}{2(2\pi )^{(n-1)/2}}\int _0^\infty \rho ^{z+(n-1)/2} J_{(n-3)/2}(\rho |x'|)\text {e}^{-\rho |x_n|}\text {d}\rho \end{aligned}$$

for $x_n\ne 0$ and $\text {Re}\,z>1-n.$ Note that $S^1_z\in L^1({\mathbb {R}}^{n-1}_{x'})$ for $x_n\ne 0$ and $\text {Re}\, z>1-n,$ and that $S^1_z$ continuously depends on $x_n$ for $\text {Re}\,z>1-n.$ An appeal to [7, Equ. 6.621.1] furnishes

$$\begin{aligned}&{\mathcal {F}}^{-1}_{x'} S^1_z=\frac{ \Gamma (z+n-1)}{(4\pi )^{(n-1)/2}\Gamma (\frac{n-1}{2})} |x|^{-(z+n-1)} \nonumber \\&\quad \times {}_2F_1\Bigl (\frac{z+n-1}{2},-\frac{z+1}{2};\frac{n-1}{2};\frac{|x'|^2}{|x|^2}\Bigr ),\quad \text {Re}\,z>1-n. \end{aligned}$$

(4.8)

We observe that the hypergeometric function in (4.8) can also be expressed by the Legendre function, see [7, Equ. 6.621.1].

If $n\ge 5,$ then $S^1_z$ is analytic near $z=-3$ and thus

$$\begin{aligned} E^1(x)={\mathcal {F}}^{-1}_{x'}S^1_{-3}=\frac{(n-5)!|x|^{4-n}}{(4\pi )^{(n-1)/2} \Gamma (\frac{n-1}{2})} \,{}_2F_1\Bigl (1,\frac{n}{2}-2;\frac{n-1}{2};\frac{|x'|^2}{|x|^2}\Bigr ). \end{aligned}$$

(4.9)

In particular, if $n\ge 5$ is odd, i.e., if $n=1+2m,$ $m\ge 2,$ then

$$\begin{aligned} {}_2F_1(1,m-\tfrac{3}{2};m;u)=-\frac{2(m-1)!\sqrt{\pi }}{\Gamma (m-\frac{3}{2})u^{m-1}}\biggl [\sqrt{1-u}-\sum _{k=0}^{m-2}\left( {\begin{array}{c}1/2\\ k\end{array}}\right) (-u)^k \biggr ] \end{aligned}$$

according to [2, Equ. 7.3.1.123], and hence,

$$\begin{aligned} E^1(x)=\frac{(m-2)!}{8\pi ^m |x'|^{2(m-1)}}\biggl [-|x_n|+|x|\sum _{k=0}^{m-2}\left( {\begin{array}{c}1/2\\ k\end{array}}\right) \Bigl (-\frac{|x'|^2}{|x|^2}\Bigr )^k\biggr ]. \end{aligned}$$

If, instead, $n=2m$ is even with $m\ge 3,$ then we can use [7, Equ. 9.137.17] and [2, Equ. 7.3.1.133], which furnish

$$\begin{aligned} {}_2F_1(1,m-2;m-\tfrac{1}{2};u)&=(2m-3)\,{}_2F_1(1,m-2;m-\tfrac{3}{2};u) \\&\quad -(2m-4)\,{}_2F_1(1,m-1;m-\tfrac{1}{2};u) \\&=\frac{2(\frac{1}{2})_{m-1}(u-1)}{(m-3)! u^{m-1}}\biggl [2\sqrt{\frac{u}{1-u}}\arctan \sqrt{\frac{u}{1-u}}- \sum _{k=1}^{m-3}\frac{(k-1)! u^k}{(\frac{1}{2})_k}\biggr ]+\frac{2m-3}{u} \end{aligned}$$

where $(\frac{1}{2})_k=\frac{1}{2}\cdot \frac{3}{2}\cdots (\frac{1}{2}+k-1)$ is an instance of Pochhammer’s symbol. From formula (4.9), we then obtain

$$\begin{aligned} E^1(x)&=-\frac{\Gamma (m-\frac{3}{2}) x_n}{4\pi ^{1/2+m}|x'|^{2m-3}} \arctan \Bigl (\frac{|x'|}{x_n}\Bigr )\\&\quad +\frac{\Gamma (m-\frac{3}{2}) x_n^2}{8\pi ^{1/2+m}}\sum _{k=1}^{m-3} \frac{(k-1)! |x'|^{2(k+1-m)}}{(\frac{1}{2})_k|x|^{2k}} +\frac{(m-3)!}{8\pi ^m|x|^{2m-6}|x'|^2}. \end{aligned}$$

Let us finally treat the cases $n=3$ and $n=4$ where $S^1_z$ has a pole at $z=-3.$ (The case $n=2$ was the content of Proposition 2.1.) If $n=4,$ then formula (4.8) furnishes

$$\begin{aligned} {\mathcal {F}}^{-1}_{x'} S^1_{z-3}&=\frac{\Gamma (z)}{4\pi ^2} |x|^{-z} {}_2F_1\Bigl (\frac{z}{2},1-\frac{z}{2};\frac{3}{2};\frac{|x'|^2}{ |x|^2}\Bigr )\\&\quad =\frac{\Gamma (z-1)}{4\pi ^2|x'|} |x|^{1-z}\sin \biggl ((z-1) \arcsin \Bigl (\frac{|x'|}{|x|}\Bigr )\biggr ) \end{aligned}$$

due to [2, Equ. 7.3.1.91]. Here $x'=(x_1,x_2,x_3).$ Using ${\text {Res}}_{z=0}\Gamma (z-1)=-1$ and ${\text {Pf}}_{z=0}\Gamma (z-1)=-\psi (2)$ and $E^a=a^{-1}({\text {Pf}}_{z=0}{\mathcal {F}}^{-1}_{x'} S^1_{z-3})(x',x_4/a)$, we obtain

$$\begin{aligned} E^a(x)=\frac{1}{4\pi ^2}\Bigl [\frac{\psi (2)}{a}-\frac{1}{2a}\log \Bigl (|x'|^2+\frac{x_4^2}{a^2}\Bigr ) -\frac{x_4}{a^2|x'|}\arctan \Bigl (\frac{a|x'|}{x_4}\Bigr )\Bigr ]. \end{aligned}$$

(4.10)

Note that the terms corresponding to the constant $\psi (2)/a$ in formula (4.10) cancel in the linear combination of fundamental solutions making up the Green function $G_\xi (x)$ according to formula (3.6).

For $n=3,$ the easiest derivation consists in descending from the five-dimensional case by integration with respect to the $(x_3,x_4)$-plane and by renaming afterwards $x_5$ as $x_3.$ When regularizing the corresponding integral, this furnishes

$$\begin{aligned} E^1(x)=\frac{1}{4\pi }\bigl [|x_3|\log (|x|+|x_3|)-|x|\bigr ]. \end{aligned}$$