Abstract

Using spectral properties of the Laplace operator and some structural formula for rapidly decreasing functions of the Laplace operator, we offer a novel method to derive explicit formulae for solutions to the Cauchy problem for classical wave equation in arbitrary dimensions. Among them are the well-known d'Alembert, Poisson, and Kirchhoff representation formulae in low space dimensions.

1. Introduction

The wave equation for a function 𝑒(π‘₯1,…,π‘₯𝑛,𝑑)=𝑒(π‘₯,𝑑) of 𝑛 space variables π‘₯1,…,π‘₯𝑛 and the time 𝑑 is given byπœ•2π‘’πœ•π‘‘2=Δ𝑒,(1.1) whereπœ•Ξ”=2πœ•π‘₯21πœ•+β‹―+2πœ•π‘₯2𝑛(1.2) is the Laplacian. The wave equation is encountered often in applications. For 𝑛=1 the equation can represent sound waves in pipes or vibrations of strings, for 𝑛=2 waves on the surface of water, for 𝑛=3 waves in acoustics or optics. Therefore, formulae that give the solution of the Cauchy problem in explicit form are of great significance. In the Cauchy problem (initial value problem) one asks for a solution 𝑒(π‘₯,𝑑) of (1.1) defined for π‘₯βˆˆβ„π‘›, 𝑑β‰₯0 that satisfies (1.1) for π‘₯βˆˆβ„π‘›, 𝑑>0 and the initial conditions𝑒(π‘₯,0)=πœ‘(π‘₯),πœ•π‘’(π‘₯,0)πœ•π‘‘=πœ“(π‘₯)(π‘₯βˆˆβ„π‘›).(1.3)

If 𝑛=1 and πœ‘βˆˆπΆ2(ℝ), πœ“βˆˆπΆ1(ℝ), then the classical solution of problem (1.1), (1.3) is given by d’Alembert’s formula 𝑒(π‘₯,𝑑)=πœ‘(π‘₯+𝑑)+πœ‘(π‘₯βˆ’π‘‘)2+12ξ€œπ‘₯+𝑑π‘₯βˆ’π‘‘πœ“(𝑦)𝑑𝑦.(1.4)

If 𝑛=2 and πœ‘βˆˆπΆ3(ℝ2), πœ“βˆˆπΆ2(ℝ2), then the solution of problem (1.1), (1.3) is given by Poisson’s formula 1𝑒(π‘₯,𝑑)=ξ€œ2πœ‹||||π‘¦βˆ’π‘₯<π‘‘πœ“(𝑦)𝑑𝑦𝑑2βˆ’||||π‘¦βˆ’π‘₯2+πœ•βŽ‘βŽ’βŽ’βŽ’βŽ£1πœ•π‘‘ξ€œ2πœ‹||||π‘¦βˆ’π‘₯<π‘‘πœ‘(𝑦)𝑑𝑦𝑑2βˆ’||||π‘¦βˆ’π‘₯2⎀βŽ₯βŽ₯βŽ₯⎦,(1.5) where π‘₯=(π‘₯1,π‘₯2), 𝑦=(𝑦1,𝑦2), and |π‘¦βˆ’π‘₯|2=(𝑦1βˆ’π‘₯1)2+(𝑦2βˆ’π‘₯2)2.

If 𝑛=3 and πœ‘βˆˆπΆ3(ℝ2), πœ“βˆˆπΆ2(ℝ2), then the solution of problem (1.1), (1.3) is given by Kirchhoff’s formula 1𝑒(π‘₯,𝑑)=ξ€œ4πœ‹π‘‘||||π‘¦βˆ’π‘₯=π‘‘πœ“(𝑦)𝑑𝑆𝑦+πœ•ξ‚Έ1πœ•π‘‘ξ€œ4πœ‹π‘‘|π‘¦βˆ’π‘₯|=π‘‘πœ‘(𝑦)𝑑𝑆𝑦,(1.6) where π‘₯=(π‘₯1,π‘₯2,π‘₯3), 𝑦=(𝑦1,𝑦2,𝑦3), |π‘¦βˆ’π‘₯|2=(𝑦1βˆ’π‘₯1)2+(𝑦2βˆ’π‘₯2)2+(𝑦3βˆ’π‘₯3)2, and 𝑑𝑆𝑦 is the surface element of the sphere {π‘¦βˆˆβ„3∢|π‘¦βˆ’π‘₯|=𝑑}.

Passing to an arbitrary 𝑛 let us denote by 𝑒(π‘₯,𝑑)=π‘πœ‘(π‘₯,𝑑) the solution of the problemπœ•2π‘’πœ•π‘‘2=Δ𝑒,π‘₯βˆˆβ„π‘›,𝑑>0,(1.7)𝑒(π‘₯,0)=πœ‘(π‘₯),πœ•π‘’(π‘₯,0)πœ•π‘‘=0,π‘₯βˆˆβ„π‘›.(1.8) It is easy to see that then the functionξ€œπ‘£(π‘₯,𝑑)=𝑑0𝑒(π‘₯,𝜏)π‘‘πœ(1.9) is the solution of the problemπœ•2π‘£πœ•π‘‘2=Δ𝑣,π‘₯βˆˆβ„π‘›,𝑑>0,(1.10)𝑣(π‘₯,0)=0,πœ•π‘£(π‘₯,0)πœ•π‘‘=πœ‘(π‘₯),π‘₯βˆˆβ„π‘›.(1.11) Indeed, integrating (1.7) we getξ€œπ‘‘0πœ•2𝑒(π‘₯,𝜏)πœ•πœ2ξ€œπ‘‘πœ=𝑑0ξ€œΞ”π‘’(π‘₯,𝜏)π‘‘πœ=Δ𝑑0𝑒(π‘₯,𝜏)π‘‘πœ=Δ𝑣(π‘₯,𝑑).(1.12) Hence,πœ•π‘’(π‘₯,𝑑)βˆ’πœ•π‘‘πœ•π‘’(π‘₯,0)πœ•π‘‘=Δ𝑣(π‘₯,𝑑)orπœ•π‘’(π‘₯,𝑑)πœ•π‘‘=Δ𝑣(π‘₯,𝑑),(1.13) by the second condition in (1.8). On the other hand, from (1.9),πœ•π‘£(π‘₯,𝑑)πœ•πœ•π‘‘=𝑒(π‘₯,𝑑),2𝑣(π‘₯,𝑑)πœ•π‘‘2=πœ•π‘’(π‘₯,𝑑)πœ•π‘‘.(1.14) Comparing (1.13) and (1.14), we get (1.10). Besides,𝑣(π‘₯,0)=0,πœ•π‘£(π‘₯,0)πœ•π‘‘=𝑒(π‘₯,0)=πœ‘(π‘₯)(1.15) so that initial conditions in (1.11) are also satisfied.

Consequently, the solution 𝑒(π‘₯,𝑑) of problem (1.1), (1.3) is represented in the form𝑒(π‘₯,𝑑)=π‘πœ‘(ξ€œπ‘₯,𝑑)+𝑑0π‘πœ“(π‘₯,𝜏)π‘‘πœ.(1.16)

It follows that it is sufficient to know an explicit form of the solution π‘πœ‘(π‘₯,𝑑) of problem (1.7), (1.8). It is known [1, 2] thatπ‘πœ‘(1π‘₯,𝑑)=2π‘š+1πœ‹π‘šξ‚€πœ•1πœ•π‘‘π‘‘ξ‚π‘šξ€œ||||π‘¦βˆ’π‘₯=π‘‘πœ‘(𝑦)𝑑𝑆𝑦if𝑛=2π‘š+1,(1.17)π‘πœ‘1(π‘₯,𝑑)=2π‘šπœ‹π‘šξ‚€πœ•1πœ•π‘‘π‘‘ξ‚π‘šβˆ’1πœ•ξ€œπ‘‘π‘‘||||π‘¦βˆ’π‘₯<π‘‘πœ‘(𝑦)𝑑𝑦𝑑2βˆ’||||π‘¦βˆ’π‘₯2if𝑛=2π‘š,(1.18) where π‘₯=(π‘₯1,…,π‘₯𝑛), 𝑦=(𝑦1,…,𝑦𝑛), |π‘¦βˆ’π‘₯|2=(𝑦1βˆ’π‘₯1)2+β‹―+(π‘¦π‘›βˆ’π‘₯𝑛)2, and 𝑑𝑆𝑦 is the surface element of the sphere {π‘¦βˆˆβ„π‘›βˆΆ|π‘¦βˆ’π‘₯|=𝑑}.

In the present paper, we give a new proof of formulae (1.17), (1.18) for the solution of problem (1.7), (1.8). Our method of the proof is based on the spectral theory of the Laplace operator. We hope that such a method may be useful also in some other cases of the equation and space.

The paper consists, besides this introductory section, of three sections. In Section 2, we describe the structure of arbitrary rapidly decreasing function of the Laplace operator, showing that it is an integral operator and giving an explicit formula for its kernel. Next we use these results in Section 3 to derive the explicit representation formulae for the classical solution to the initial value problem for the wave equation in arbitrary dimensions. The final Section  is an appendix and contains some explanation of several points in the paper.

2. Structure of Arbitrary Function of the Laplace Operator

Let 𝐴 be the self-adjoint positive operator obtained as the closure of the symmetric operator 𝐴′ determined in the Hilbert space 𝐿2(ℝ𝑛) by the differential expressionξƒ©πœ•βˆ’Ξ”=βˆ’2πœ•π‘₯21πœ•+β‹…β‹…β‹…+2πœ•π‘₯2𝑛ξƒͺ,ξ€·π‘₯1,…,π‘₯π‘›ξ€Έβˆˆβ„π‘›,(2.1) on the domain of definition 𝐷(π΄ξ…ž)=𝐢∞0(ℝ𝑛) that is the set of all infinitely differentiable functions on ℝ𝑛 with compact support. Let πΈπœ‡ denote the resolution of the identity (the spectral projection) for A:ξ€œπ΄π‘“=∞0πœ‡π‘‘πΈπœ‡π‘“,π‘“βˆˆπ·(𝐴).(2.2) Next, let 𝑔(𝑑) be any infinitely differentiable even function on the axis βˆ’βˆž<𝑑<∞ with compact support andξ€œΜƒπ‘”(πœ†)=βˆžβˆ’βˆžπ‘”(𝑑)π‘’π‘–πœ†π‘‘π‘‘π‘‘(2.3) its Fourier transform. Note that the function ̃𝑔(πœ†) tends to zero as |πœ†|β†’βˆžβ€‰β€‰(πœ†βˆˆβ„) faster than any negative power of |πœ†|. Consider the operator ̃𝑔(𝐴1/2) defined according to the general theory of self-adjoint operators (see [3]):𝐴̃𝑔1/2ξ€Έξ€œπ‘“=∞0ξ€·βˆšΜƒπ‘”πœ‡ξ€Έπ‘‘πΈπœ‡π‘“,π‘“βˆˆπΏ2(ℝ𝑛).(2.4)

The following theorem describes the structure of the operator ̃𝑔(𝐴1/2) showing that it is an integral operator and giving an explicit formula for its kernel in terms of the function 𝑔(𝑑).

Theorem 2.1. The operator ̃𝑔(𝐴1/2) is an integral operator 𝐴̃𝑔1/2ξ€Έξ€œπ‘“(π‘₯)=ℝ𝑛𝒦(π‘₯,𝑦)𝑓(𝑦)𝑑𝑦,π‘“βˆˆπΏ2(ℝ𝑛).(2.5) Further, there is a smooth function π‘˜(𝑑) defined on the interval 0≀𝑑<∞ such that ξ‚€||||𝒦(π‘₯,𝑦)=π‘˜π‘₯βˆ’π‘¦2.(2.6) The function π‘˜(𝑑) depends on the function 𝑔(𝑑) as follows. If one sets ξ‚€βˆšπ‘„(𝑑)=𝑔𝑑𝑑,thatis,𝑄2ξ€Έ=𝑔(𝑑),0≀𝑑<∞,(2.7) then π‘˜βŽ§βŽͺβŽͺ⎨βŽͺβŽͺ⎩(𝑑)=(βˆ’1)π‘šπœ‹π‘šπ‘„(π‘š)(𝑑)if𝑛=2π‘š+1,(βˆ’1)π‘šπœ‹π‘šξ€œβˆžπ‘‘π‘„(π‘š)(𝑀)βˆšπ‘€βˆ’π‘‘π‘‘π‘€if𝑛=2π‘š,(2.8) where 𝑄(π‘š)(𝑑) denotes the π‘šth order derivative of 𝑄(𝑑). Further, if supp𝑔(𝑑)βŠ‚(βˆ’π‘Ž,π‘Ž), then suppπ‘˜(𝑑)βŠ‚[0,π‘Ž2). For any solution πœ“(π‘₯,πœ†) of the equation βˆ’Ξ”πœ“(π‘₯,πœ†)=πœ†2πœ“(π‘₯,πœ†),(2.9) the equality ξ€œβ„π‘›π‘˜ξ‚€||||π‘₯βˆ’π‘¦2ξ‚πœ“(𝑦,πœ†)𝑑𝑦=̃𝑔(πœ†)πœ“(π‘₯,πœ†)(2.10) holds for πœ†βˆˆβ„.

Proof. First we consider the case 𝑛=1. In this case, the statements of the theorem take the following form: βˆšπ‘˜(𝑑)=𝑄(𝑑)=𝑔(𝑑) for 0≀𝑑<∞; the operator ̃𝑔(𝐴1/2) is an integral operator of the form 𝐴̃𝑔1/2ξ€Έξ€œπ‘“(π‘₯)=βˆžβˆ’βˆžπ‘”(π‘₯βˆ’π‘¦)𝑓(𝑦)𝑑𝑦,(2.11) and for any solution πœ“(π‘₯,πœ†) of the equation βˆ’πœ“β€²ξ…ž(π‘₯,πœ†)=πœ†2πœ“(π‘₯,πœ†),(2.12) the equality ξ€œβˆžβˆ’βˆžπ‘”(π‘₯βˆ’π‘¦)πœ“(𝑦,πœ†)𝑑𝑦=̃𝑔(πœ†)πœ“(π‘₯,πœ†)(2.13) holds.
To prove the last statements note that, in the case 𝑛=1, the operator 𝐴 is generated in the Hilbert space 𝐿2(βˆ’βˆž,∞) by the operation βˆ’π‘‘2/𝑑π‘₯2 and the operator 𝐴1/2 by the operation 𝑖𝑑/𝑑π‘₯. The resolvent π‘…πœ‡=(π΄βˆ’πœ‡πΌ)βˆ’1 of the operator 𝐴 has the form π‘…πœ‡π‘–π‘“(π‘₯)=2βˆšπœ‡ξ€œβˆžβˆ’βˆžπ‘’βˆšπ‘–|π‘₯βˆ’π‘¦|πœ‡π‘“(𝑦)𝑑𝑦,(2.14) while the spectral projection πΈπœ‡ of the operator 𝐴 has the form (see [3, page 201]) πΈπœ‡ξ€œπ‘“(π‘₯)=βˆžβˆ’βˆžβˆšsinπœ‡(π‘₯βˆ’π‘¦)πœ‹πΈ(π‘₯βˆ’π‘¦)𝑓(𝑦)𝑑𝑦,0β‰€πœ‡<∞,πœ‡=0forπœ‡<0.(2.15) Therefore, 𝐴̃𝑔1/2ξ€Έξ€œπ‘“(π‘₯)=∞0ξ€·βˆšΜƒπ‘”πœ‡ξ€Έπ‘‘πΈπœ‡=ξ€œπ‘“(π‘₯)∞0ξ€·βˆšΜƒπ‘”πœ‡ξ€Έξƒ―ξ€œβˆžβˆ’βˆžβˆšcosπœ‡(π‘₯βˆ’π‘¦)√2πœ‹πœ‡ξƒ°=ξ€œπ‘“(𝑦)π‘‘π‘¦π‘‘πœ‡βˆžβˆ’βˆžξ‚»1πœ‹ξ€œβˆž0ξ‚Όπ‘“ξ€œΜƒπ‘”(πœ†)cosπœ†(π‘₯βˆ’π‘¦)π‘‘πœ†(𝑦)𝑑𝑦=βˆžβˆ’βˆžπ‘”(π‘₯βˆ’π‘¦)𝑓(𝑦)𝑑𝑦,(2.16) where we have used the inversion formula for the Fourier cosine transform. Therefore, (2.11) is proved. To prove (2.13) note that the general solution of (2.12) is ξ‚»π‘πœ“(π‘₯,πœ†)=1cosπœ†π‘₯+𝑐2𝑐sinπœ†π‘₯ifπœ†β‰ 0,1+𝑐2π‘₯ifπœ†=0,(2.17) where 𝑐1 and 𝑐2 are arbitrary constants. Then, we have, for πœ†β‰ 0, ξ€œβˆžβˆ’βˆžπ‘”(π‘₯βˆ’π‘¦)πœ“(𝑦,πœ†)𝑑𝑦=𝑐1ξ€œβˆžβˆ’βˆžπ‘”(π‘₯βˆ’π‘¦)cosπœ†π‘¦π‘‘π‘¦+𝑐2ξ€œβˆžβˆ’βˆžπ‘”(π‘₯βˆ’π‘¦)sinπœ†π‘¦π‘‘π‘¦=𝑐1ξ€œβˆžβˆ’βˆžπ‘”(𝑑)cosπœ†(π‘₯βˆ’π‘‘)𝑑𝑑+𝑐2ξ€œβˆžβˆ’βˆžπ‘”(𝑑)sinπœ†(π‘₯βˆ’π‘‘)𝑑𝑑=𝑐1ξ€œβˆžβˆ’βˆžπ‘”(𝑑)(cosπœ†π‘₯cosπœ†π‘‘+sinπœ†π‘₯sinπœ†π‘‘)𝑑𝑑+𝑐2ξ€œβˆžβˆ’βˆžπ‘”(𝑑)(sinπœ†π‘₯cosπœ†π‘‘βˆ’sinπœ†π‘‘cosπœ†π‘₯)𝑑𝑑=𝑐1ξ€œcosπœ†π‘₯βˆžβˆ’βˆžπ‘”(𝑑)cosπœ†π‘‘π‘‘π‘‘+𝑐2ξ€œsinπœ†π‘₯βˆžβˆ’βˆž=𝑐𝑔(𝑑)cosπœ†π‘‘π‘‘π‘‘1cosπœ†π‘₯+𝑐2ξ€Έξ€œsinπœ†π‘₯βˆžβˆ’βˆžπ‘”(𝑑)cosπœ†π‘‘π‘‘π‘‘=πœ“(π‘₯,πœ†)̃𝑔(πœ†),(2.18) where we have used the fact that the function 𝑔(𝑑) is even and therefore ξ€œβˆžβˆ’βˆžπ‘”(𝑑)sinπœ†π‘‘π‘‘π‘‘=0.(2.19) The same result can be obtained similarly for πœ†=0. Thus, (2.13) is also proved.
Now we consider the case 𝑛β‰₯2. We shall use the integral representation π‘…πœ‡ξ€œπ‘“(π‘₯)=β„π‘›π‘Ÿ(π‘₯,𝑦;πœ‡)𝑓(𝑦)𝑑𝑦(2.20) of the resolvent π‘…πœ‡=(π΄βˆ’πœ‡πΌ)βˆ’1 of the operator 𝐴. As is known [4, Section  13.7, Formula (13.7.2)], π‘Ÿ(π‘₯,𝑦;πœ‡)=π‘–πœ‡(π‘›βˆ’2)/42(𝑛+2)/2πœ‹(π‘›βˆ’2)/2||||π‘₯βˆ’π‘¦(π‘›βˆ’2)/2𝐻(1)(π‘›βˆ’2)/2ξ€·||||√π‘₯βˆ’π‘¦πœ‡ξ€Έ,(2.21) where 𝐻𝜈(1)(𝑧) is the Hankel function of the first kind of order 𝜈. Next, according to the general spectral theory of self-adjoint operators [3, page 150, Formula (11)], we have π‘‘πΈπœ‡1𝑓(π‘₯)=𝑅2πœ‹π‘–πœ‡+𝑖0βˆ’π‘…πœ‡βˆ’π‘–0𝑓(π‘₯)π‘‘πœ‡.(2.22) Therefore, from (2.4) it follows that the representation (2.5) holds with 1𝒦(π‘₯,𝑦)=ξ€œ2πœ‹π‘–βˆž0ξ€·βˆšΜƒπ‘”πœ‡ξ€Έ[]π‘Ÿ(π‘₯,𝑦;πœ‡+𝑖0)βˆ’π‘Ÿ(π‘₯,𝑦;πœ‡βˆ’π‘–0)π‘‘πœ‡.(2.23) Now the representation (2.6), which expresses that 𝒦(π‘₯,𝑦) is a function of |π‘₯βˆ’π‘¦|2, follows from (2.23) by (2.21).
To prove (2.10) we use (2.23). By virtue of (2.23), ξ€œβ„π‘›π‘˜ξ‚€||||π‘₯βˆ’π‘¦2ξ‚ξ€œπœ“(𝑦,πœ†)𝑑𝑦=ℝ𝑛𝒦(π‘₯,𝑦)πœ“(𝑦,πœ†)𝑑𝑦=limπœ€β†’+0ξ€œβ„π‘›ξ‚»1ξ€œ2πœ‹π‘–βˆž0ξ€·βˆšΜƒπ‘”πœ‡ξ€Έ[]ξ‚Όπ‘Ÿ(π‘₯,𝑦;πœ‡+π‘–πœ€)βˆ’π‘Ÿ(π‘₯,𝑦;πœ‡βˆ’π‘–πœ€)π‘‘πœ‡πœ“(𝑦,πœ†)𝑑𝑦=πœ“(π‘₯,πœ†)limπœ€β†’+0πœ€πœ‹ξ€œβˆž0ξ€·βˆšΜƒπ‘”πœ‡ξ€Έξ€·πœ‡βˆ’πœ†2ξ€Έ2+πœ€2π‘‘πœ‡=πœ“(π‘₯,πœ†)̃𝑔(πœ†),(2.24) see Appendix. Here we have used the fact that from (2.9) it follows that ξ€·πœ†(βˆ’Ξ”βˆ’π‘§)πœ“(π‘₯,πœ†)=2ξ€Έπœ“βˆ’π‘§(π‘₯,πœ†),(2.25) that is, ξ€·πœ†πœ“(π‘₯,πœ†)=2ξ€Έβˆ’π‘§(βˆ’Ξ”βˆ’π‘§)βˆ’1πœ“(π‘₯,πœ†),(2.26) and therefore ξ€œβ„π‘›1π‘Ÿ(π‘₯,𝑦;𝑧)πœ“(𝑦,πœ†)𝑑𝑦=πœ†2βˆ’π‘§πœ“(π‘₯,πœ†).(2.27)
Finally, to deduce the explicit formulae (2.7), (2.8), we take πœ“(π‘₯,πœ†)=π‘’π‘–πœ†π‘₯1 in (2.10). Then, putting Μƒπ‘₯=(π‘₯2,…,π‘₯𝑛), we can write ξ€œβ„π‘›π‘˜ξ‚€||π‘₯1βˆ’π‘¦1||2+||||Μƒπ‘₯βˆ’Μƒπ‘¦2ξ‚π‘’π‘–πœ†π‘¦1𝑑𝑦1𝑑̃𝑦=̃𝑔(πœ†)π‘’π‘–πœ†π‘₯1.(2.28) If we set ξ€·π‘₯1βˆ’π‘¦1ξ€Έ2=𝑀,(2.29) then the left-hand side of (2.28) equals ξ€œβˆžβˆ’βˆžξ‚»ξ€œβ„π‘›βˆ’1π‘˜ξ‚€||||𝑀+Μƒπ‘₯βˆ’Μƒπ‘¦2ξ‚ξ‚Όπ‘’π‘‘Μƒπ‘¦π‘–πœ†π‘¦1𝑑𝑦1.(2.30) On the other hand, ξ€œβ„π‘›βˆ’1π‘˜ξ‚€||||𝑀+Μƒπ‘₯βˆ’Μƒπ‘¦2ξ‚ξ€œπ‘‘Μƒπ‘¦=∞0ξƒ―ξ€œ||Μƒπ‘₯βˆ’Μƒπ‘¦||=π‘Ÿπ‘˜ξ‚€||||𝑀+Μƒπ‘₯βˆ’Μƒπ‘¦2=ξ€œπ‘‘π‘†π‘‘π‘Ÿβˆž0π‘˜ξ€·π‘€+π‘Ÿ2ξ€Έξƒ―ξ€œ||Μƒπ‘₯βˆ’Μƒπ‘¦||=π‘Ÿξƒ°π‘‘π‘†π‘‘π‘Ÿ=πœŽπ‘›βˆ’1ξ€œβˆž0π‘Ÿπ‘›βˆ’2π‘˜ξ€·π‘€+π‘Ÿ2ξ€Έ=1π‘‘π‘Ÿ2πœŽπ‘›βˆ’1ξ€œβˆžπ‘€(π‘‘βˆ’π‘€)(π‘›βˆ’3)/2π‘˜(𝑑)𝑑𝑑,(2.31) where πœŽπ‘›=2πœ‹π‘›/2Ξ“(𝑛/2)(2.32) is the surface area of the (π‘›βˆ’1)-dimensional unit sphere (Ξ“ is the gamma function) and 𝑑𝑆 denotes the surface element of the sphere {Μƒπ‘¦βˆˆβ„π‘›βˆ’1∢|Μƒπ‘₯βˆ’Μƒπ‘¦|=π‘Ÿ}. Therefore, setting 1𝑄(𝑀)=2πœŽπ‘›βˆ’1ξ€œβˆžπ‘€(π‘‘βˆ’π‘€)(π‘›βˆ’3)/2π‘˜(𝑑)𝑑𝑑,(2.33) we get that (2.28) takes the form ξ€œβˆžβˆ’βˆžπ‘„(𝑀)π‘’π‘–πœ†π‘¦1𝑑𝑦1=̃𝑔(πœ†)π‘’π‘–πœ†π‘₯1.(2.34) Substituting here the expression of 𝑀 given in (2.29) and making then the change of variables π‘₯1βˆ’π‘¦1=𝑑, we obtain ξ€œβˆžβˆ’βˆžπ‘„ξ€·π‘‘2ξ€Έπ‘’π‘–πœ†π‘‘ξ€œπ‘‘π‘‘=̃𝑔(πœ†)=βˆžβˆ’βˆžπ‘”(𝑑)π‘’π‘–πœ†π‘‘π‘‘π‘‘.(2.35) Hence (2.7) follows. Further, it is not difficult to check that the formula (2.33) for 𝑛β‰₯2 is equivalent to (2.8), see Appendix.
Since 𝑔(𝑑) is smooth and has a compact support, it follows from (2.7), (2.8) that the function π‘˜(𝑑) also is smooth and has a compact support; more precisely, if supp𝑔(𝑑)βŠ‚(βˆ’π‘Ž,π‘Ž), then suppπ‘˜(𝑑)βŠ‚[0,π‘Ž2). This implies, in particular, convergence of the integral in (2.10) for each fixed π‘₯. The theorem is proved.

3. Derivation of Formulae (1.17), (1.18)

Consider the Cauchy problem (1.7), (1.8):πœ•2π‘’πœ•π‘‘2=Δ𝑒,π‘₯βˆˆβ„π‘›,𝑑>0,(3.1)𝑒(π‘₯,0)=πœ‘(π‘₯),πœ•π‘’(π‘₯,0)πœ•π‘‘=0,π‘₯βˆˆβ„π‘›,(3.2) where 𝑒=𝑒(π‘₯,𝑑), 𝑑β‰₯0, π‘₯=(π‘₯1,…,π‘₯𝑛)βˆˆβ„π‘›, πœ‘(π‘₯)∈𝐢∞0(ℝ𝑛).

For 𝜈=(𝜈1,…,πœˆπ‘›), π‘₯=(π‘₯1,…,π‘₯𝑛)βˆˆβ„π‘›, let us set|𝜈|2=𝜈21+β‹―+𝜈2𝑛,(𝜈,π‘₯)=𝜈1π‘₯1+β‹―+πœˆπ‘›π‘₯𝑛.(3.3) Since βˆ’Ξ”π‘’π‘–(𝜈,π‘₯)=|𝜈|2𝑒𝑖(𝜈,π‘₯),(3.4) applying (2.9), (2.10), we getξ€œβ„π‘›π‘˜ξ‚€||||π‘₯βˆ’π‘¦2𝑒𝑖(𝜈,𝑦)𝑑𝑦=̃𝑔(|𝜈|)𝑒𝑖(𝜈,π‘₯)(πœˆβˆˆβ„π‘›).(3.5) Hence, by the inverse Fourier transform formula,π‘˜ξ‚€||||π‘₯βˆ’π‘¦2=1(2πœ‹)π‘›ξ€œβ„π‘›Μƒπ‘”(|𝜈|)𝑒𝑖(𝜈,π‘₯)π‘’βˆ’π‘–(𝜈,𝑦)π‘‘πœˆ.(3.6) Multiplying both sides of the last equality by πœ‘(𝑦) and then integrating on π‘¦βˆˆβ„π‘›, we getξ€œβ„π‘›π‘˜ξ‚€||||π‘₯βˆ’π‘¦21πœ‘(𝑦)𝑑𝑦=(2πœ‹)π‘›ξ€œβ„π‘›Μƒπ‘”(|𝜈|)𝑒𝑖(𝜈,π‘₯)ξ‚Έξ€œβ„π‘›πœ‘(𝑦)π‘’βˆ’π‘–(𝜈,𝑦)ξ‚Ήπ‘‘π‘¦π‘‘πœˆ.(3.7) Substituting here for ̃𝑔(|𝜈|) its expressionξ€œΜƒπ‘”(|𝜈|)=2∞0𝑔(𝑑)cos(|𝜈|𝑑)𝑑𝑑(3.8) and setting1𝑒(π‘₯,𝑑)=(2πœ‹)π‘›ξ€œβ„π‘›(cos|𝜈|𝑑)𝑒𝑖(𝜈,π‘₯)ξ‚Έξ€œβ„π‘›πœ‘(𝑦)π‘’βˆ’π‘–(𝜈,𝑦)ξ‚Ήπ‘‘π‘¦π‘‘πœˆ,(3.9) we obtainξ€œβ„π‘›π‘˜ξ‚€||||π‘₯βˆ’π‘¦2ξ‚ξ€œπœ‘(𝑦)𝑑𝑦=2∞0𝑔(𝑑)𝑒(π‘₯,𝑑)𝑑𝑑.(3.10) Obviously, the function 𝑒(π‘₯,𝑑) defined by (3.9) is the solution of problem (3.1), (3.2). Next we will transform the left-hand side of (3.10) using Theorem 2.1.

First we consider the case 𝑛=1. In this case, (3.10) takes the formξ€œβˆžβˆ’βˆžπ‘˜ξ‚€||||π‘₯βˆ’π‘¦2ξ‚ξ€œπœ‘(𝑦)𝑑𝑦=2∞0𝑔(𝑑)𝑒(π‘₯,𝑑)𝑑𝑑(3.11) and from (2.7), (2.8) we haveπ‘˜ξ€·π‘‘2𝑑=𝑄2ξ€Έ=𝑔(𝑑).(3.12) Therefore, making the change of variables π‘¦βˆ’π‘₯=𝑑 and taking into account the evenness of the function 𝑔(𝑑), we can writeξ€œβˆžβˆ’βˆžπ‘˜ξ‚€||||π‘₯βˆ’π‘¦2ξ‚ξ€œπœ‘(𝑦)𝑑𝑦=βˆžβˆ’βˆžπ‘˜ξ€·π‘‘2ξ€Έ=ξ€œπœ‘(π‘₯+𝑑)π‘‘π‘‘βˆžβˆ’βˆžξ€œπ‘”(𝑑)πœ‘(π‘₯+𝑑)𝑑𝑑=∞0[]𝑔(𝑑)πœ‘(π‘₯+𝑑)+πœ‘(π‘₯βˆ’π‘‘)𝑑𝑑.(3.13) Substituting this in the left-hand side of (3.11), we obtainξ€œβˆž0[]ξ€œπ‘”(𝑑)πœ‘(π‘₯+𝑑)+πœ‘(π‘₯βˆ’π‘‘)𝑑𝑑=2∞0𝑔(𝑑)𝑒(π‘₯,𝑑)𝑑𝑑.(3.14) Hence, by the arbitrariness of the smooth even function 𝑔(𝑑) with compact support, we get𝑒(π‘₯,𝑑)=πœ‘(π‘₯+𝑑)+πœ‘(π‘₯βˆ’π‘‘)2.(3.15)

Further assume that 𝑛β‰₯2. Making the change of variablesξ€·πœ”π‘¦βˆ’π‘₯=π‘‘πœ”,0≀𝑑<∞,|πœ”|=1,πœ”=1,…,πœ”π‘›ξ€Έ,𝑑𝑦=π‘‘π‘›βˆ’1π‘‘π‘‘π‘‘π‘†πœ”,(3.16) where π‘‘π‘†πœ” is the surface element of the unit sphere {πœ”βˆˆβ„π‘›βˆΆ|πœ”|=1}, we getξ€œβ„π‘›π‘˜ξ‚€||||π‘₯βˆ’π‘¦2ξ‚ξ€œπœ‘(𝑦)𝑑𝑦=∞0π‘‘π‘›βˆ’1π‘˜ξ€·π‘‘2ξ€Έξ‚»ξ€œ|πœ”|=1πœ‘(π‘₯+π‘‘πœ”)π‘‘π‘†πœ”ξ‚Όπ‘‘π‘‘.(3.17) Further, making in the right-hand side of (3.17) the change of variablesπ‘₯+π‘‘πœ”=𝑦,𝑑𝑆𝑦=π‘‘π‘›βˆ’1π‘‘π‘†πœ”,(3.18) where 𝑑𝑆𝑦 is the surface element of the sphere {π‘¦βˆˆβ„π‘›βˆΆ|π‘¦βˆ’π‘₯|=𝑑}, we haveπ‘‘π‘›βˆ’1ξ€œ|πœ”|=1πœ‘(π‘₯+π‘‘πœ”)π‘‘π‘†πœ”=ξ€œ||||π‘¦βˆ’π‘₯=π‘‘πœ‘(𝑦)𝑑𝑆𝑦=βˆΆπ‘ƒπœ‘(π‘₯,𝑑).(3.19) Therefore,ξ€œβ„π‘›π‘˜ξ‚€||||π‘₯βˆ’π‘¦2ξ‚ξ€œπœ‘(𝑦)𝑑𝑦=∞0π‘˜ξ€·π‘‘2ξ€Έπ‘ƒπœ‘(π‘₯,𝑑)𝑑𝑑,(3.20) and (3.10) becomesξ€œβˆž0π‘˜ξ€·π‘‘2ξ€Έπ‘ƒπœ‘ξ€œ(π‘₯,𝑑)𝑑𝑑=2∞0𝑔(𝑑)𝑒(π‘₯,𝑑)𝑑𝑑.(3.21) Consider the cases of odd and even 𝑛 separately.

Let 𝑛=2π‘š+1(π‘šβˆˆβ„•). Then, by (2.8) we haveπ‘˜ξ€·π‘‘2ξ€Έ=(βˆ’1)π‘šπœ‹π‘šπ‘„(π‘š)𝑑2ξ€Έ(3.22) and it follows from (2.7) (by successive differentiation) that𝑄(π‘š)𝑑2ξ€Έ=ξ‚€1πœ•2π‘‘ξ‚πœ•π‘‘π‘šπ‘”(𝑑).(3.23) Therefore,π‘˜ξ€·π‘‘2ξ€Έ=(βˆ’1)π‘š2π‘šπœ‹π‘šξ‚€1π‘‘πœ•ξ‚πœ•π‘‘π‘šπ‘”(𝑑),(3.24) and (3.21) takes the form(βˆ’1)π‘š2π‘šπœ‹π‘šξ€œβˆž0ξ‚»ξ‚€1π‘‘πœ•ξ‚πœ•π‘‘π‘šξ‚Όπ‘ƒπ‘”(𝑑)πœ‘ξ€œ(π‘₯,𝑑)𝑑𝑑=2∞0𝑔(𝑑)𝑒(π‘₯,𝑑)𝑑𝑑.(3.25) Further, integrating π‘š times by parts, we getξ€œβˆž0ξ‚»ξ‚€1π‘‘πœ•ξ‚πœ•π‘‘π‘šξ‚Όπ‘ƒπ‘”(𝑑)πœ‘||(π‘₯,𝑑)𝑑𝑑=𝑅(π‘₯,𝑑)𝑑=βˆžπ‘‘=0+(βˆ’1)π‘šξ€œβˆž0ξ‚€πœ•π‘”(𝑑)1πœ•π‘‘π‘‘ξ‚π‘šπ‘ƒπœ‘(π‘₯,𝑑)𝑑𝑑,(3.26) where 𝑅(π‘₯,𝑑)=π‘šξ“π‘˜=1(βˆ’1)π‘˜βˆ’1𝑑1π‘‘πœ•ξ‚πœ•π‘‘π‘šβˆ’π‘˜ξ‚Όξ‚€πœ•π‘”(𝑑)1πœ•π‘‘π‘‘ξ‚π‘˜βˆ’1π‘ƒπœ‘=(π‘₯,𝑑)π‘šξ“π‘˜=1(βˆ’1)π‘˜βˆ’1𝑑1π‘‘πœ•ξ‚πœ•π‘‘π‘šβˆ’π‘˜ξ‚Όξ‚€πœ•π‘”(𝑑)1πœ•π‘‘π‘‘ξ‚π‘˜βˆ’1𝑑2π‘šξ€œ|πœ”|=1πœ‘(π‘₯+π‘‘πœ”)π‘‘π‘†πœ”.(3.27) Since 𝑔(𝑑) is identically zero for large values of 𝑑, we have from (3.27) that 𝑅(π‘₯,∞)=0. Also, it follows directly from (3.27) that 𝑅(π‘₯,0)=0. Therefore, (3.25) becomes12π‘šπœ‹π‘šξ€œβˆž0ξ‚€πœ•π‘”(𝑑)1πœ•π‘‘π‘‘ξ‚π‘šπ‘ƒπœ‘ξ€œ(π‘₯,𝑑)𝑑𝑑=2∞0𝑔(𝑑)𝑒(π‘₯,𝑑)𝑑𝑑.(3.28) Since in (3.28) 𝑔(𝑑) is arbitrary smooth even function with compact support, we obtain that1𝑒(π‘₯,𝑑)=2π‘š+1πœ‹π‘šξ‚€πœ•1πœ•π‘‘π‘‘ξ‚π‘šπ‘ƒπœ‘(π‘₯,𝑑).(3.29) This coincides with (1.17) by (3.19).

Now let us consider the case 𝑛=2π‘š(π‘šβˆˆβ„•). In this case, by (2.8) we haveπ‘˜ξ€·π‘Ÿ2ξ€Έ=(βˆ’1)π‘šπœ‹π‘šξ€œβˆžπ‘Ÿ2𝑄(π‘š)(𝑀)βˆšπ‘€βˆ’π‘Ÿ2(𝑑𝑀=βˆ’1)π‘šπœ‹π‘šξ€œβˆžπ‘Ÿπ‘„(π‘š)𝑑2ξ€Έ2π‘‘βˆšπ‘‘2βˆ’π‘Ÿ2𝑑𝑑,(3.30) and thereforeξ€œβˆž0π‘˜ξ€·π‘Ÿ2ξ€Έπ‘ƒπœ‘(π‘₯,π‘Ÿ)π‘‘π‘Ÿ=(βˆ’1)π‘šπœ‹π‘šξ€œβˆž0ξƒ―ξ€œβˆžπ‘Ÿπ‘„(π‘š)𝑑2ξ€Έ2π‘‘βˆšπ‘‘2βˆ’π‘Ÿ2ξƒ°π‘ƒπ‘‘π‘‘πœ‘=(π‘₯,π‘Ÿ)π‘‘π‘Ÿ(βˆ’1)π‘šπœ‹π‘šξ€œβˆž0ξƒ―ξ€œβˆžπ‘Ÿπ‘„(π‘š)𝑑2ξ€Έ2π‘‘βˆšπ‘‘2βˆ’π‘Ÿ2ξƒ°ξ‚»ξ€œπ‘‘π‘‘|π‘¦βˆ’π‘₯|=π‘Ÿπœ‘(𝑦)𝑑𝑆𝑦=π‘‘π‘Ÿ(βˆ’1)π‘šπœ‹π‘šξ€œβˆž0⎧βŽͺ⎨βŽͺβŽ©ξ€œβˆžπ‘Ÿπ‘„(π‘š)𝑑2ξ€ΈβŽ‘βŽ’βŽ’βŽ’βŽ£ξ€œ2𝑑|π‘¦βˆ’π‘₯|=π‘Ÿπœ‘(𝑦)𝑑𝑆𝑦𝑑2βˆ’||||π‘¦βˆ’π‘₯2⎀βŽ₯βŽ₯βŽ₯⎦⎫βŽͺ⎬βŽͺ⎭=π‘‘π‘‘π‘‘π‘Ÿ(βˆ’1)π‘šπœ‹π‘šξ€œβˆž0𝑄(π‘š)𝑑2ξ€ΈβŽ§βŽͺ⎨βŽͺβŽ©ξ€œ2𝑑𝑑0βŽ‘βŽ’βŽ’βŽ’βŽ£ξ€œ|π‘¦βˆ’π‘₯|=π‘Ÿπœ‘(𝑦)𝑑𝑆𝑦𝑑2βˆ’||||π‘¦βˆ’π‘₯2⎀βŽ₯βŽ₯βŽ₯⎦⎫βŽͺ⎬βŽͺβŽ­π‘‘π‘Ÿπ‘‘π‘‘.(3.31) Hence, settingπ»πœ‘ξ€œ(π‘₯,𝑑)∢=𝑑0βŽ‘βŽ’βŽ’βŽ’βŽ£ξ€œ|π‘¦βˆ’π‘₯|=π‘Ÿπœ‘(𝑦)𝑑𝑆𝑦𝑑2βˆ’||||π‘¦βˆ’π‘₯2⎀βŽ₯βŽ₯βŽ₯βŽ¦ξ€œπ‘‘π‘Ÿ=||||π‘¦βˆ’π‘₯<π‘‘πœ‘(𝑦)𝑑𝑦𝑑2βˆ’||||π‘¦βˆ’π‘₯2,(3.32) we getξ€œβˆž0π‘˜ξ€·π‘Ÿ2ξ€Έπ‘ƒπœ‘((π‘₯,π‘Ÿ)π‘‘π‘Ÿ=βˆ’1)π‘šπœ‹π‘šξ€œβˆž0𝑄(π‘š)𝑑2ξ€Έ2π‘‘π»πœ‘(π‘₯,𝑑)𝑑𝑑.(3.33) Substituting this in the left-hand side of (3.21) (beforehand replacing 𝑑 by π‘Ÿ in the left side of (3.21)), we obtain(βˆ’1)π‘šπœ‹π‘šξ€œβˆž0𝑄(π‘š)𝑑2ξ€Έ2π‘‘π»πœ‘ξ€œ(π‘₯,𝑑)𝑑𝑑=2∞0𝑔(𝑑)𝑒(π‘₯,𝑑)𝑑𝑑(3.34) or, using (3.23),(βˆ’1)π‘š2π‘šβˆ’1πœ‹π‘šξ€œβˆž0ξ‚»ξ‚€1π‘‘πœ•ξ‚πœ•π‘‘π‘šξ‚Όπ‘”(𝑑)π‘‘π»πœ‘ξ€œ(π‘₯,𝑑)𝑑𝑑=2∞0𝑔(𝑑)𝑒(π‘₯,𝑑)𝑑𝑑.(3.35) Further, integrating π‘š times by parts, we getξ€œβˆž0ξ‚»ξ‚€1π‘‘πœ•ξ‚πœ•π‘‘π‘šξ‚Όπ‘”(𝑑)π‘‘π»πœ‘||(π‘₯,𝑑)𝑑𝑑=𝐿(π‘₯,𝑑)𝑑=βˆžπ‘‘=0+(βˆ’1)π‘šξ€œβˆž0ξ‚€πœ•π‘”(𝑑)1πœ•π‘‘π‘‘ξ‚π‘šπ‘‘π»πœ‘(π‘₯,𝑑)𝑑𝑑,(3.36) where𝐿(π‘₯,𝑑)=π‘šξ“π‘˜=1(βˆ’1)π‘˜βˆ’1ξ‚»ξ‚€1π‘‘πœ•ξ‚πœ•π‘‘π‘šβˆ’π‘˜ξ‚Όξ‚€πœ•π‘”(𝑑)1πœ•π‘‘π‘‘ξ‚π‘˜βˆ’1π‘‘π»πœ‘(π‘₯,𝑑).(3.37) Since 𝑔(𝑑) is identically zero for large values of 𝑑, we have from (3.37) that 𝐿(π‘₯,∞)=0. Also, using the expression of π»πœ‘(π‘₯,𝑑),π»πœ‘ξ€œ(π‘₯,𝑑)=𝑑0βŽ‘βŽ’βŽ’βŽ’βŽ£ξ€œ|π‘¦βˆ’π‘₯|=π‘Ÿπœ‘(𝑦)𝑑𝑆𝑦𝑑2βˆ’||||π‘¦βˆ’π‘₯2⎀βŽ₯βŽ₯βŽ₯⎦=ξ€œπ‘‘π‘Ÿπ‘‘0π‘Ÿ2π‘šβˆ’1ξƒ¬ξ€œ|πœ”|=1πœ‘(π‘₯+π‘Ÿπœ”)βˆšπ‘‘2βˆ’π‘Ÿ2π‘‘π‘†πœ”ξƒ­=ξ€œπ‘‘π‘Ÿπ‘‘0π‘Ÿ2π‘šβˆ’1βˆšπ‘‘2βˆ’π‘Ÿ2ξ‚Έξ€œ|πœ”|=1πœ‘(π‘₯+π‘Ÿπœ”)π‘‘π‘†πœ”ξ‚Ή=ξ€œπ‘‘π‘Ÿπ‘‘0𝑑2βˆ’πœ‰2ξ€Έ2π‘šβˆ’2ξ‚Έξ€œ|πœ”|=1πœ‘ξ‚΅ξ”π‘₯+𝑑2βˆ’πœ‰2πœ”ξ‚Άπ‘‘π‘†πœ”ξ‚Ήπ‘‘πœ‰,(3.38) we can check directly from (3.37) that 𝐿(π‘₯,0)=0. Therefore, (3.35) becomes12π‘šβˆ’1πœ‹π‘šξ€œβˆž0ξ‚€πœ•π‘”(𝑑)1πœ•π‘‘π‘‘ξ‚π‘šπ‘‘π»πœ‘ξ€œ(π‘₯,𝑑)𝑑𝑑=2∞0𝑔(𝑑)𝑒(π‘₯,𝑑)𝑑𝑑.(3.39) Since in (3.39) 𝑔(𝑑) is arbitrary smooth even function with compact support, we obtain that1𝑒(π‘₯,𝑑)=2π‘šπœ‹π‘šξ‚€πœ•1πœ•π‘‘π‘‘ξ‚π‘šπ‘‘π»πœ‘=1(π‘₯,𝑑)2π‘šπœ‹π‘šξ‚€πœ•1πœ•π‘‘π‘‘ξ‚π‘šβˆ’1πœ•π»πœ•π‘‘πœ‘(π‘₯,𝑑).(3.40) This coincides with (1.18) by (3.32).

Appendix

For reader’s convenience, in this section we give some explanation of several points in the paper.(1)Let us show how (2.33) for 𝑛β‰₯2 implies (2.8).

Let 𝑛=2π‘š+1, where π‘šβ‰₯1. Then, since12𝜎2π‘š=122πœ‹π‘šΞ“=πœ‹(π‘š)π‘š(π‘šβˆ’1)!,(A.1) Equation (2.33) takes the form𝑄(𝑀)=πœ‹π‘šξ€œβˆžπ‘€(π‘‘βˆ’π‘€)π‘šβˆ’1(π‘šβˆ’1)!π‘˜(𝑑)𝑑𝑑.(A.2) Hence applying the differentiation formulaπ‘‘ξ€œπ‘‘π‘€βˆžπ‘€ξ€œπΊ(𝑑,𝑀)𝑑𝑑=βˆ’πΊ(𝑀,𝑀)+βˆžπ‘€πœ•πΊ(𝑑,𝑀)πœ•π‘€π‘‘π‘‘(A.3) repeatedly, we find𝑄(π‘š)(𝑀)=πœ‹π‘š(βˆ’1)π‘šπ‘˜(𝑀)(A.4) which gives (2.8) for 𝑛=2π‘š+1.

In the case 𝑛=2π‘š with π‘šβ‰₯1, (2.33) takes the form1𝑄(𝑀)=2𝜎2π‘šβˆ’1ξ€œβˆžπ‘€(π‘‘βˆ’π‘€)(2π‘šβˆ’3)/2π‘˜(𝑑)𝑑𝑑.(A.5) Hence,𝑄(π‘šβˆ’1)1(𝑀)=2𝜎2π‘šβˆ’1ξ€œβˆžπ‘€(βˆ’1)π‘šβˆ’12π‘šβˆ’322π‘šβˆ’521β‹…β‹…β‹…2(π‘‘βˆ’π‘€)βˆ’1/2π‘˜(𝑑)𝑑𝑑.(A.6) Therefore, taking into account that by virtue ofξ‚€1Ξ“(π‘₯)=(π‘₯βˆ’1)Ξ“(π‘₯βˆ’1),Ξ“2=βˆšπœ‹,(A.7) we have12𝜎2π‘šβˆ’1=πœ‹(2π‘šβˆ’1)/2=πœ‹Ξ“((2π‘šβˆ’1)/2)(2π‘šβˆ’1)/2=πœ‹((2π‘šβˆ’3)/2)((2π‘šβˆ’5)/2)β‹…β‹…β‹…(1/2)Ξ“(1/2)π‘šβˆ’1,((2π‘šβˆ’3)/2)((2π‘šβˆ’5)/2)β‹…β‹…β‹…(1/2)(A.8) we get𝑄(π‘šβˆ’1)(𝑀)=(βˆ’1)π‘šβˆ’1πœ‹π‘šβˆ’1ξ€œβˆžπ‘€π‘˜(𝑑)βˆšπ‘‘βˆ’π‘€π‘‘π‘‘.(A.9) In the right-hand side we replace 𝑑 by 𝑒, then divide both sides by βˆšπ‘€βˆ’π‘‘ and integrate on π‘€βˆˆ(𝑑,∞) to getξ€œβˆžπ‘‘π‘„(π‘šβˆ’1)(𝑀)βˆšπ‘€βˆ’π‘‘π‘‘π‘€=(βˆ’1)π‘šβˆ’1πœ‹π‘šβˆ’1ξ€œβˆžπ‘‘1βˆšξƒ―ξ€œπ‘€βˆ’π‘‘βˆžπ‘€π‘˜(𝑒)βˆšξƒ°π‘’βˆ’π‘€π‘‘π‘’π‘‘π‘€=(βˆ’1)π‘šβˆ’1πœ‹π‘šβˆ’1ξ€œβˆžπ‘‘ξƒ―ξ€œπ‘˜(𝑒)π‘’π‘‘π‘‘π‘€βˆšξƒ°(π‘€βˆ’π‘‘)(π‘’βˆ’π‘€)𝑑𝑒=(βˆ’1)π‘šβˆ’1πœ‹π‘šξ€œβˆžπ‘‘π‘˜(𝑒)𝑑𝑒,(A.10) because for any 𝑑<𝑒, using the change of variables βˆšπ‘€βˆ’π‘‘=πœ‰, we haveξ€œπ‘’π‘‘π‘‘π‘€βˆšξ€œ(π‘€βˆ’π‘‘)(π‘’βˆ’π‘€)=2√0π‘’βˆ’π‘‘π‘‘πœ‰βˆšπ‘’βˆ’π‘‘βˆ’πœ‰2πœ‰=2arcsin√|||||π‘’βˆ’π‘‘βˆšπœ‰=π‘’βˆ’π‘‘πœ‰=0=2arcsin1=πœ‹.(A.11) Therefore, differentiating (A.10) with respect to 𝑑, we getπ‘˜(𝑑)=(βˆ’1)π‘šπœ‹π‘šπ‘‘ξ€œπ‘‘π‘‘βˆžπ‘‘π‘„(π‘šβˆ’1)(𝑀)βˆšπ‘€βˆ’π‘‘π‘‘π‘€=(βˆ’1)π‘šπœ‹π‘šπ‘‘ξ€œπ‘‘π‘‘βˆž0𝑄(π‘šβˆ’1)(𝑒+𝑑)βˆšπ‘’=𝑑𝑒(βˆ’1)π‘šπœ‹π‘šξ€œβˆž0𝑄(π‘š)(𝑒+𝑑)βˆšπ‘’π‘‘π‘’=(βˆ’1)π‘šπœ‹π‘šξ€œβˆžπ‘‘π‘„(π‘š)(𝑀)βˆšπ‘€βˆ’π‘‘π‘‘π‘€.(A.12) Thus, (2.8) is obtained also for 𝑛=2π‘š with π‘šβ‰₯1.

(2) Here we explain (2.24). Note that since the spectrum of the operator 𝐴 is [0,∞) (zero is included into the spectrum), the spectral representation formula (2.4) should be understood in the sense of the formula𝐴̃𝑔(1/2)ξ€Έξ€œπ‘“=βˆžβˆ’π›Ώξ€·βˆšΜƒπ‘”πœ‡ξ€Έπ‘‘πΈπœ‡π‘“,(A.13) where 𝛿 is an arbitrary positive real number and the integral does not depend on 𝛿>0 (πΈπœ‡ is zero on (βˆ’βˆž,0) because 𝐴 is a positive operator). Therefore, for (2.24) we have to show thatlimπœ€β†’+0πœ€πœ‹ξ€œβˆžβˆ’π›Ώξ€·βˆšΜƒπ‘”πœ‡ξ€Έξ€·πœ‡βˆ’πœ†2ξ€Έ2+πœ€2π‘‘πœ‡=̃𝑔(πœ†),πœ†βˆˆβ„,(A.14) for any 𝛿>0.

Since for any πœ€>0πœ€πœ‹ξ€œβˆžβˆ’π›Ώ1ξ€·πœ‡βˆ’πœ†2ξ€Έ2+πœ€2πœ€π‘‘πœ‡=πœ‹ξ€œβˆžβˆ’π›Ώβˆ’πœ†2𝑑𝑒𝑒2+πœ€2=1πœ‹ξ‚΅πœ‹2+arctan𝛿+πœ†2πœ€ξ‚Ά,(A.15) we havelimπœ€β†’+0πœ€πœ‹ξ€œβˆžβˆ’π›Ώ1ξ€·πœ‡βˆ’πœ†2ξ€Έ2+πœ€2πœ€π‘‘πœ‡=1,πœ†βˆˆβ„,πœ‹ξ€œβˆžβˆ’βˆžπ‘‘π‘’π‘’2+πœ€2=1.(A.16) Given 𝛼>0, we can choose a 𝛽>0 such that|||ξ‚€βˆšΜƒπ‘”π‘’+πœ†2|||ξ€½βˆ’Μƒπ‘”(πœ†)<𝛼forπ‘’βˆˆΞ©=π‘’βˆΆβˆ’π›Ώβˆ’πœ†2ξ€Ύ<𝑒<∞,|𝑒|<𝛽(A.17) since the function ̃𝑔(𝑧) is continuous for 𝑧=πœ† (we choose the continuous branch of the square root for which √1=1). Further, we choose a number 𝑀 such that||||||||̃𝑔(𝑧)≀𝑀forIm𝑧≀𝐢<∞,(A.18) for sufficiently large positive number 𝐢. This is possible by (2.3) and the fact that 𝑔(𝑑) has a compact support. Let us set Ξ©β€²=(βˆ’π›Ώβˆ’πœ†2,∞)⧡Ω. Then,|||||πœ€πœ‹ξ€œβˆžβˆ’π›Ώξ€·βˆšΜƒπ‘”πœ‡ξ€Έξ€·πœ‡βˆ’πœ†2ξ€Έ2+πœ€2πœ€π‘‘πœ‡βˆ’Μƒπ‘”(πœ†)πœ‹ξ€œβˆžβˆ’π›Ώ1ξ€·πœ‡βˆ’πœ†2ξ€Έ2+πœ€2|||||β‰€πœ€π‘‘πœ‡πœ‹ξ€œβˆžβˆ’π›Ώβˆ’πœ†2|||ξ‚€βˆšΜƒπ‘”π‘’+πœ†2|||βˆ’Μƒπ‘”(πœ†)𝑒2+πœ€2=πœ€π‘‘π‘’πœ‹ξ€œΞ©|||ξ‚€βˆšΜƒπ‘”π‘’+πœ†2|||βˆ’Μƒπ‘”(πœ†)𝑒2+πœ€2πœ€π‘‘π‘’+πœ‹ξ€œΞ©β€²|||ξ‚€βˆšΜƒπ‘”π‘’+πœ†2|||βˆ’Μƒπ‘”(πœ†)𝑒2+πœ€2𝑑𝑒.(A.19) Further,πœ€πœ‹ξ€œΞ©|||ξ‚€βˆšΜƒπ‘”π‘’+πœ†2|||βˆ’Μƒπ‘”(πœ†)𝑒2+πœ€2𝛼𝑑𝑒<πœ‹ξ€œβˆžβˆ’βˆžπœ€π‘’2+πœ€2πœ€π‘‘π‘’=𝛼,πœ‹ξ€œΞ©β€²|||ξ‚€βˆšΜƒπ‘”π‘’+πœ†2|||βˆ’Μƒπ‘”(πœ†)𝑒2+πœ€2𝑑𝑒≀2π‘€πœ‹ξ€œ|𝑒|β‰₯π›½πœ€π‘’2+πœ€2=𝑑𝑒4π‘€πœ‹ξ€œβˆžπ›½πœ€π‘’2+πœ€2𝑑𝑒=4π‘€πœ‹ξ‚΅πœ‹2π›½βˆ’arctanπœ€ξ‚Ά.(A.20)

For fixed 𝛽, the last expression tends to zero as πœ€β†’+0; hence, and by (A.16), (A.19), and (A.20) we get (A.14).

(3) The formula (2.14) follows from (2.21) for 𝑛=1 noting that𝐻(1)βˆ’(1/2)ξ‚€2(𝑧)=ξ‚πœ‹π‘§1/2𝑒𝑖𝑧.(A.21)

(4) The difference between operators (πœ•/πœ•π‘‘1/𝑑)π‘š (formulae (1.17), (1.18)) and (1/π‘‘πœ•/πœ•π‘‘)π‘š (formula (3.25)) is given byξ‚€πœ•1πœ•π‘‘π‘‘ξ‚π‘š=πœ•ξ‚€1πœ•π‘‘π‘‘πœ•ξ‚πœ•π‘‘π‘šβˆ’11𝑑.(A.22)

(5) The explicit formula for the solution of the wave equation in the case 𝑛 even can be derived from the case 𝑛 odd by a known computation called the β€œmethod of descent” (see [1]).

(6) Since for supp𝑔(𝑑)βŠ‚(βˆ’π‘Ž,π‘Ž), π‘Ž>0, we have suppπ‘˜(𝑑)βŠ‚[0,π‘Ž2), and on the left-hand side of (2.10) the integral is taken in fact over the ball {π‘¦βˆˆβ„π‘›βˆΆ|π‘¦βˆ’π‘₯|<π‘Ž}, for fixed π‘₯. Therefore, this integral is finite for each π‘₯βˆˆβ„π‘› and any solution πœ“(π‘₯,πœ†) of (2.9). We proved (2.10) for πœ†βˆˆβ„. If the solution πœ“(π‘₯,πœ†) is an analytic function of πœ†βˆˆβ„‚, then (2.10) will be held also for complex values of πœ† by the uniqueness of analytic continuation.