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Journal of Applied Mathematics
Volume 2012, Article ID 761248, 19 pages
Research Article

Spectral Approach to Derive the Representation Formulae for Solutions of the Wave Equation

Department of Mathematics, Atilim University, Incek, 06836 Ankara, Turkey

Received 28 September 2011; Accepted 30 January 2012

Academic Editor: Pablo González-Vera

Copyright © 2012 Gusein Sh. Guseinov. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


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 𝑢(𝑥1,,𝑥𝑛,𝑡)=𝑢(𝑥,𝑡) of 𝑛 space variables 𝑥1,,𝑥𝑛 and the time 𝑡 is given by𝜕2𝑢𝜕𝑡2=Δ𝑢,(1.1) where𝜕Δ=2𝜕𝑥21𝜕++2𝜕𝑥2𝑛(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 𝑢(𝑥,𝑡) of (1.1) defined for 𝑥𝑛, 𝑡0 that satisfies (1.1) for 𝑥𝑛, 𝑡>0 and the initial conditions𝑢(𝑥,0)=𝜑(𝑥),𝜕𝑢(𝑥,0)𝜕𝑡=𝜓(𝑥)(𝑥𝑛).(1.3)

If 𝑛=1 and 𝜑𝐶2(), 𝜓𝐶1(), then the classical solution of problem (1.1), (1.3) is given by d’Alembert’s formula 𝑢(𝑥,𝑡)=𝜑(𝑥+𝑡)+𝜑(𝑥𝑡)2+12𝑥+𝑡𝑥𝑡𝜓(𝑦)𝑑𝑦.(1.4)

If 𝑛=2 and 𝜑𝐶3(2), 𝜓𝐶2(2), then the solution of problem (1.1), (1.3) is given by Poisson’s formula 1𝑢(𝑥,𝑡)=2𝜋||||𝑦𝑥<𝑡𝜓(𝑦)𝑑𝑦𝑡2||||𝑦𝑥2+𝜕1𝜕𝑡2𝜋||||𝑦𝑥<𝑡𝜑(𝑦)𝑑𝑦𝑡2||||𝑦𝑥2,(1.5) where 𝑥=(𝑥1,𝑥2), 𝑦=(𝑦1,𝑦2), and |𝑦𝑥|2=(𝑦1𝑥1)2+(𝑦2𝑥2)2.

If 𝑛=3 and 𝜑𝐶3(2), 𝜓𝐶2(2), then the solution of problem (1.1), (1.3) is given by Kirchhoff’s formula 1𝑢(𝑥,𝑡)=4𝜋𝑡||||𝑦𝑥=𝑡𝜓(𝑦)𝑑𝑆𝑦+𝜕1𝜕𝑡4𝜋𝑡|𝑦𝑥|=𝑡𝜑(𝑦)𝑑𝑆𝑦,(1.6) where 𝑥=(𝑥1,𝑥2,𝑥3), 𝑦=(𝑦1,𝑦2,𝑦3), |𝑦𝑥|2=(𝑦1𝑥1)2+(𝑦2𝑥2)2+(𝑦3𝑥3)2, and 𝑑𝑆𝑦 is the surface element of the sphere {𝑦3|𝑦𝑥|=𝑡}.

Passing to an arbitrary 𝑛 let us denote by 𝑢(𝑥,𝑡)=𝑁𝜑(𝑥,𝑡) the solution of the problem𝜕2𝑢𝜕𝑡2=Δ𝑢,𝑥𝑛,𝑡>0,(1.7)𝑢(𝑥,0)=𝜑(𝑥),𝜕𝑢(𝑥,0)𝜕𝑡=0,𝑥𝑛.(1.8) It is easy to see that then the function𝑣(𝑥,𝑡)=𝑡0𝑢(𝑥,𝜏)𝑑𝜏(1.9) is the solution of the problem𝜕2𝑣𝜕𝑡2=Δ𝑣,𝑥𝑛,𝑡>0,(1.10)𝑣(𝑥,0)=0,𝜕𝑣(𝑥,0)𝜕𝑡=𝜑(𝑥),𝑥𝑛.(1.11) Indeed, integrating (1.7) we get𝑡0𝜕2𝑢(𝑥,𝜏)𝜕𝜏2𝑑𝜏=𝑡0Δ𝑢(𝑥,𝜏)𝑑𝜏=Δ𝑡0𝑢(𝑥,𝜏)𝑑𝜏=Δ𝑣(𝑥,𝑡).(1.12) Hence,𝜕𝑢(𝑥,𝑡)𝜕𝑡𝜕𝑢(𝑥,0)𝜕𝑡=Δ𝑣(𝑥,𝑡)or𝜕𝑢(𝑥,𝑡)𝜕𝑡=Δ𝑣(𝑥,𝑡),(1.13) by the second condition in (1.8). On the other hand, from (1.9),𝜕𝑣(𝑥,𝑡)𝜕𝜕𝑡=𝑢(𝑥,𝑡),2𝑣(𝑥,𝑡)𝜕𝑡2=𝜕𝑢(𝑥,𝑡)𝜕𝑡.(1.14) Comparing (1.13) and (1.14), we get (1.10). Besides,𝑣(𝑥,0)=0,𝜕𝑣(𝑥,0)𝜕𝑡=𝑢(𝑥,0)=𝜑(𝑥)(1.15) so that initial conditions in (1.11) are also satisfied.

Consequently, the solution 𝑢(𝑥,𝑡) of problem (1.1), (1.3) is represented in the form𝑢(𝑥,𝑡)=𝑁𝜑(𝑥,𝑡)+𝑡0𝑁𝜓(𝑥,𝜏)𝑑𝜏.(1.16)

It follows that it is sufficient to know an explicit form of the solution 𝑁𝜑(𝑥,𝑡) of problem (1.7), (1.8). It is known [1, 2] that𝑁𝜑(1𝑥,𝑡)=2𝑚+1𝜋𝑚𝜕1𝜕𝑡𝑡𝑚||||𝑦𝑥=𝑡𝜑(𝑦)𝑑𝑆𝑦if𝑛=2𝑚+1,(1.17)𝑁𝜑1(𝑥,𝑡)=2𝑚𝜋𝑚𝜕1𝜕𝑡𝑡𝑚1𝜕𝑑𝑡||||𝑦𝑥<𝑡𝜑(𝑦)𝑑𝑦𝑡2||||𝑦𝑥2if𝑛=2𝑚,(1.18) where 𝑥=(𝑥1,,𝑥𝑛), 𝑦=(𝑦1,,𝑦𝑛), |𝑦𝑥|2=(𝑦1𝑥1)2++(𝑦𝑛𝑥𝑛)2, and 𝑑𝑆𝑦 is the surface element of the sphere {𝑦𝑛|𝑦𝑥|=𝑡}.

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(𝑛) by the differential expression𝜕Δ=2𝜕𝑥21𝜕++2𝜕𝑥2𝑛,𝑥1,,𝑥𝑛𝑛,(2.1) on the domain of definition 𝐷(𝐴)=𝐶0(𝑛) that is the set of all infinitely differentiable functions on 𝑛 with compact support. Let 𝐸𝜇 denote the resolution of the identity (the spectral projection) for A:𝐴𝑓=0𝜇𝑑𝐸𝜇𝑓,𝑓𝐷(𝐴).(2.2) Next, let 𝑔(𝑡) be any infinitely differentiable even function on the axis <𝑡< with compact support and̃𝑔(𝜆)=𝑔(𝑡)𝑒𝑖𝜆𝑡𝑑𝑡(2.3) its Fourier transform. Note that the function ̃𝑔(𝜆) tends to zero as |𝜆|  (𝜆) faster than any negative power of |𝜆|. Consider the operator ̃𝑔(𝐴1/2) defined according to the general theory of self-adjoint operators (see [3]):𝐴̃𝑔1/2𝑓=0̃𝑔𝜇𝑑𝐸𝜇𝑓,𝑓𝐿2(𝑛).(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(𝑡).

Theorem 2.1. The operator ̃𝑔(𝐴1/2) is an integral operator 𝐴̃𝑔1/2𝑓(𝑥)=𝑛𝒦(𝑥,𝑦)𝑓(𝑦)𝑑𝑦,𝑓𝐿2(𝑛).(2.5) Further, there is a smooth function 𝑘(𝑡) defined on the interval 0𝑡< such that ||||𝒦(𝑥,𝑦)=𝑘𝑥𝑦2.(2.6) The function 𝑘(𝑡) depends on the function 𝑔(𝑡) as follows. If one sets 𝑄(𝑡)=𝑔𝑡𝑡,thatis,𝑄2=𝑔(𝑡),0𝑡<,(2.7) then 𝑘(𝑡)=(1)𝑚𝜋𝑚𝑄(𝑚)(𝑡)if𝑛=2𝑚+1,(1)𝑚𝜋𝑚𝑡𝑄(𝑚)(𝑤)𝑤𝑡𝑑𝑤if𝑛=2𝑚,(2.8) where 𝑄(𝑚)(𝑡) denotes the 𝑚th order derivative of 𝑄(𝑡). Further, if supp𝑔(𝑡)(𝑎,𝑎), then supp𝑘(𝑡)[0,𝑎2). For any solution 𝜓(𝑥,𝜆) of the equation Δ𝜓(𝑥,𝜆)=𝜆2𝜓(𝑥,𝜆),(2.9) the equality 𝑛𝑘||||𝑥𝑦2𝜓(𝑦,𝜆)𝑑𝑦=̃𝑔(𝜆)𝜓(𝑥,𝜆)(2.10) holds for 𝜆.

Proof. First we consider the case 𝑛=1. In this case, the statements of the theorem take the following form: 𝑘(𝑡)=𝑄(𝑡)=𝑔(𝑡) for 0𝑡<; the operator ̃𝑔(𝐴1/2) is an integral operator of the form 𝐴̃𝑔1/2𝑓(𝑥)=𝑔(𝑥𝑦)𝑓(𝑦)𝑑𝑦,(2.11) and for any solution 𝜓(𝑥,𝜆) of the equation 𝜓(𝑥,𝜆)=𝜆2𝜓(𝑥,𝜆),(2.12) the equality 𝑔(𝑥𝑦)𝜓(𝑦,𝜆)𝑑𝑦=̃𝑔(𝜆)𝜓(𝑥,𝜆)(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(,) by the operation 𝑑2/𝑑𝑥2 and the operator 𝐴1/2 by the operation 𝑖𝑑/𝑑𝑥. The resolvent 𝑅𝜇=(𝐴𝜇𝐼)1 of the operator 𝐴 has the form 𝑅𝜇𝑖𝑓(𝑥)=2𝜇𝑒𝑖|𝑥𝑦|𝜇𝑓(𝑦)𝑑𝑦,(2.14) while the spectral projection 𝐸𝜇 of the operator 𝐴 has the form (see [3, page 201]) 𝐸𝜇𝑓(𝑥)=sin𝜇(𝑥𝑦)𝜋𝐸(𝑥𝑦)𝑓(𝑦)𝑑𝑦,0𝜇<,𝜇=0for𝜇<0.(2.15) Therefore, 𝐴̃𝑔1/2𝑓(𝑥)=0̃𝑔𝜇𝑑𝐸𝜇=𝑓(𝑥)0̃𝑔𝜇cos𝜇(𝑥𝑦)2𝜋𝜇=𝑓(𝑦)𝑑𝑦𝑑𝜇1𝜋0𝑓̃𝑔(𝜆)cos𝜆(𝑥𝑦)𝑑𝜆(𝑦)𝑑𝑦=𝑔(𝑥𝑦)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 𝑐𝜓(𝑥,𝜆)=1cos𝜆𝑥+𝑐2𝑐sin𝜆𝑥if𝜆0,1+𝑐2𝑥if𝜆=0,(2.17) where 𝑐1 and 𝑐2 are arbitrary constants. Then, we have, for 𝜆0, 𝑔(𝑥𝑦)𝜓(𝑦,𝜆)𝑑𝑦=𝑐1𝑔(𝑥𝑦)cos𝜆𝑦𝑑𝑦+𝑐2𝑔(𝑥𝑦)sin𝜆𝑦𝑑𝑦=𝑐1𝑔(𝑡)cos𝜆(𝑥𝑡)𝑑𝑡+𝑐2𝑔(𝑡)sin𝜆(𝑥𝑡)𝑑𝑡=𝑐1𝑔(𝑡)(cos𝜆𝑥cos𝜆𝑡+sin𝜆𝑥sin𝜆𝑡)𝑑𝑡+𝑐2𝑔(𝑡)(sin𝜆𝑥cos𝜆𝑡sin𝜆𝑡cos𝜆𝑥)𝑑𝑡=𝑐1cos𝜆𝑥𝑔(𝑡)cos𝜆𝑡𝑑𝑡+𝑐2sin𝜆𝑥=𝑐𝑔(𝑡)cos𝜆𝑡𝑑𝑡1cos𝜆𝑥+𝑐2sin𝜆𝑥𝑔(𝑡)cos𝜆𝑡𝑑𝑡=𝜓(𝑥,𝜆)̃𝑔(𝜆),(2.18) where we have used the fact that the function 𝑔(𝑡) is even and therefore 𝑔(𝑡)sin𝜆𝑡𝑑𝑡=0.(2.19) The same result can be obtained similarly for 𝜆=0. Thus, (2.13) is also proved.
Now we consider the case 𝑛2. We shall use the integral representation 𝑅𝜇𝑓(𝑥)=𝑛𝑟(𝑥,𝑦;𝜇)𝑓(𝑦)𝑑𝑦(2.20) of the resolvent 𝑅𝜇=(𝐴𝜇𝐼)1 of the operator 𝐴. As is known [4, Section  13.7, Formula (13.7.2)], 𝑟(𝑥,𝑦;𝜇)=𝑖𝜇(𝑛2)/42(𝑛+2)/2𝜋(𝑛2)/2||||𝑥𝑦(𝑛2)/2𝐻(1)(𝑛2)/2||||𝑥𝑦𝜇,(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 𝑑𝐸𝜇1𝑓(𝑥)=𝑅2𝜋𝑖𝜇+𝑖0𝑅𝜇𝑖0𝑓(𝑥)𝑑𝜇.(2.22) Therefore, from (2.4) it follows that the representation (2.5) holds with 1𝒦(𝑥,𝑦)=2𝜋𝑖0̃𝑔𝜇[]𝑟(𝑥,𝑦;𝜇+𝑖0)𝑟(𝑥,𝑦;𝜇𝑖0)𝑑𝜇.(2.23) Now the representation (2.6), which expresses that 𝒦(𝑥,𝑦) is a function of |𝑥𝑦|2, follows from (2.23) by (2.21).
To prove (2.10) we use (2.23). By virtue of (2.23), 𝑛𝑘||||𝑥𝑦2𝜓(𝑦,𝜆)𝑑𝑦=𝑛𝒦(𝑥,𝑦)𝜓(𝑦,𝜆)𝑑𝑦=lim𝜀+0𝑛12𝜋𝑖0̃𝑔𝜇[]𝑟(𝑥,𝑦;𝜇+𝑖𝜀)𝑟(𝑥,𝑦;𝜇𝑖𝜀)𝑑𝜇𝜓(𝑦,𝜆)𝑑𝑦=𝜓(𝑥,𝜆)lim𝜀+0𝜀𝜋0̃𝑔𝜇𝜇𝜆22+𝜀2𝑑𝜇=𝜓(𝑥,𝜆)̃𝑔(𝜆),(2.24) see Appendix. Here we have used the fact that from (2.9) it follows that 𝜆(Δ𝑧)𝜓(𝑥,𝜆)=2𝜓𝑧(𝑥,𝜆),(2.25) that is, 𝜆𝜓(𝑥,𝜆)=2𝑧(Δ𝑧)1𝜓(𝑥,𝜆),(2.26) and therefore 𝑛1𝑟(𝑥,𝑦;𝑧)𝜓(𝑦,𝜆)𝑑𝑦=𝜆2𝑧𝜓(𝑥,𝜆).(2.27)
Finally, to deduce the explicit formulae (2.7), (2.8), we take 𝜓(𝑥,𝜆)=𝑒𝑖𝜆𝑥1 in (2.10). Then, putting ̃𝑥=(𝑥2,,𝑥𝑛), we can write 𝑛𝑘||𝑥1𝑦1||2+||||̃𝑥̃𝑦2𝑒𝑖𝜆𝑦1𝑑𝑦1𝑑̃𝑦=̃𝑔(𝜆)𝑒𝑖𝜆𝑥1.(2.28) If we set 𝑥1𝑦12=𝑤,(2.29) then the left-hand side of (2.28) equals 𝑛1𝑘||||𝑤+̃𝑥̃𝑦2𝑒𝑑̃𝑦𝑖𝜆𝑦1𝑑𝑦1.(2.30) On the other hand, 𝑛1𝑘||||𝑤+̃𝑥̃𝑦2𝑑̃𝑦=0||̃𝑥̃𝑦||=𝑟𝑘||||𝑤+̃𝑥̃𝑦2=𝑑𝑆𝑑𝑟0𝑘𝑤+𝑟2||̃𝑥̃𝑦||=𝑟𝑑𝑆𝑑𝑟=𝜎𝑛10𝑟𝑛2𝑘𝑤+𝑟2=1𝑑𝑟2𝜎𝑛1𝑤(𝑡𝑤)(𝑛3)/2𝑘(𝑡)𝑑𝑡,(2.31) where 𝜎𝑛=2𝜋𝑛/2Γ(𝑛/2)(2.32) is the surface area of the (𝑛1)-dimensional unit sphere (Γ is the gamma function) and 𝑑𝑆 denotes the surface element of the sphere {̃𝑦𝑛1|̃𝑥̃𝑦|=𝑟}. Therefore, setting 1𝑄(𝑤)=2𝜎𝑛1𝑤(𝑡𝑤)(𝑛3)/2𝑘(𝑡)𝑑𝑡,(2.33) we get that (2.28) takes the form 𝑄(𝑤)𝑒𝑖𝜆𝑦1𝑑𝑦1=̃𝑔(𝜆)𝑒𝑖𝜆𝑥1.(2.34) Substituting here the expression of 𝑤 given in (2.29) and making then the change of variables 𝑥1𝑦1=𝑡, we obtain 𝑄𝑡2𝑒𝑖𝜆𝑡𝑑𝑡=̃𝑔(𝜆)=𝑔(𝑡)𝑒𝑖𝜆𝑡𝑑𝑡.(2.35) Hence (2.7) follows. Further, it is not difficult to check that the formula (2.33) for 𝑛2 is equivalent to (2.8), see Appendix.
Since 𝑔(𝑡) is smooth and has a compact support, it follows from (2.7), (2.8) that the function 𝑘(𝑡) also is smooth and has a compact support; more precisely, if supp𝑔(𝑡)(𝑎,𝑎), then supp𝑘(𝑡)[0,𝑎2). This implies, in particular, convergence of the integral in (2.10) for each fixed 𝑥. The theorem is proved.

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

Consider the Cauchy problem (1.7), (1.8):𝜕2𝑢𝜕𝑡2=Δ𝑢,𝑥𝑛,𝑡>0,(3.1)𝑢(𝑥,0)=𝜑(𝑥),𝜕𝑢(𝑥,0)𝜕𝑡=0,𝑥𝑛,(3.2) where 𝑢=𝑢(𝑥,𝑡), 𝑡0, 𝑥=(𝑥1,,𝑥𝑛)𝑛, 𝜑(𝑥)𝐶0(𝑛).

For 𝜈=(𝜈1,,𝜈𝑛), 𝑥=(𝑥1,,𝑥𝑛)𝑛, let us set|𝜈|2=𝜈21++𝜈2𝑛,(𝜈,𝑥)=𝜈1𝑥1++𝜈𝑛𝑥𝑛.(3.3) Since Δ𝑒𝑖(𝜈,𝑥)=|𝜈|2𝑒𝑖(𝜈,𝑥),(3.4) applying (2.9), (2.10), we get𝑛𝑘||||𝑥𝑦2𝑒𝑖(𝜈,𝑦)𝑑𝑦=̃𝑔(|𝜈|)𝑒𝑖(𝜈,𝑥)(𝜈𝑛).(3.5) Hence, by the inverse Fourier transform formula,𝑘||||𝑥𝑦2=1(2𝜋)𝑛𝑛̃𝑔(|𝜈|)𝑒𝑖(𝜈,𝑥)𝑒𝑖(𝜈,𝑦)𝑑𝜈.(3.6) Multiplying both sides of the last equality by 𝜑(𝑦) and then integrating on 𝑦𝑛, we get𝑛𝑘||||𝑥𝑦21𝜑(𝑦)𝑑𝑦=(2𝜋)𝑛𝑛̃𝑔(|𝜈|)𝑒𝑖(𝜈,𝑥)𝑛𝜑(𝑦)𝑒𝑖(𝜈,𝑦)𝑑𝑦𝑑𝜈.(3.7) Substituting here for ̃𝑔(|𝜈|) its expressioñ𝑔(|𝜈|)=20𝑔(𝑡)cos(|𝜈|𝑡)𝑑𝑡(3.8) and setting1𝑢(𝑥,𝑡)=(2𝜋)𝑛𝑛(cos|𝜈|𝑡)𝑒𝑖(𝜈,𝑥)𝑛𝜑(𝑦)𝑒𝑖(𝜈,𝑦)𝑑𝑦𝑑𝜈,(3.9) we obtain𝑛𝑘||||𝑥𝑦2𝜑(𝑦)𝑑𝑦=20𝑔(𝑡)𝑢(𝑥,𝑡)𝑑𝑡.(3.10) Obviously, the function 𝑢(𝑥,𝑡) 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 form𝑘||||𝑥𝑦2𝜑(𝑦)𝑑𝑦=20𝑔(𝑡)𝑢(𝑥,𝑡)𝑑𝑡(3.11) and from (2.7), (2.8) we have𝑘𝑡2𝑡=𝑄2=𝑔(𝑡).(3.12) Therefore, making the change of variables 𝑦𝑥=𝑡 and taking into account the evenness of the function 𝑔(𝑡), we can write𝑘||||𝑥𝑦2𝜑(𝑦)𝑑𝑦=𝑘𝑡2=𝜑(𝑥+𝑡)𝑑𝑡𝑔(𝑡)𝜑(𝑥+𝑡)𝑑𝑡=0[]𝑔(𝑡)𝜑(𝑥+𝑡)+𝜑(𝑥𝑡)𝑑𝑡.(3.13) Substituting this in the left-hand side of (3.11), we obtain0[]𝑔(𝑡)𝜑(𝑥+𝑡)+𝜑(𝑥𝑡)𝑑𝑡=20𝑔(𝑡)𝑢(𝑥,𝑡)𝑑𝑡.(3.14) Hence, by the arbitrariness of the smooth even function 𝑔(𝑡) with compact support, we get𝑢(𝑥,𝑡)=𝜑(𝑥+𝑡)+𝜑(𝑥𝑡)2.(3.15)

Further assume that 𝑛2. Making the change of variables𝜔𝑦𝑥=𝑡𝜔,0𝑡<,|𝜔|=1,𝜔=1,,𝜔𝑛,𝑑𝑦=𝑡𝑛1𝑑𝑡𝑑𝑆𝜔,(3.16) where 𝑑𝑆𝜔 is the surface element of the unit sphere {𝜔𝑛|𝜔|=1}, we get𝑛𝑘||||𝑥𝑦2𝜑(𝑦)𝑑𝑦=0𝑡𝑛1𝑘𝑡2|𝜔|=1𝜑(𝑥+𝑡𝜔)𝑑𝑆𝜔𝑑𝑡.(3.17) Further, making in the right-hand side of (3.17) the change of variables𝑥+𝑡𝜔=𝑦,𝑑𝑆𝑦=𝑡𝑛1𝑑𝑆𝜔,(3.18) where 𝑑𝑆𝑦 is the surface element of the sphere {𝑦𝑛|𝑦𝑥|=𝑡}, we have𝑡𝑛1|𝜔|=1𝜑(𝑥+𝑡𝜔)𝑑𝑆𝜔=||||𝑦𝑥=𝑡𝜑(𝑦)𝑑𝑆𝑦=𝑃𝜑(𝑥,𝑡).(3.19) Therefore,𝑛𝑘||||𝑥𝑦2𝜑(𝑦)𝑑𝑦=0𝑘𝑡2𝑃𝜑(𝑥,𝑡)𝑑𝑡,(3.20) and (3.10) becomes0𝑘𝑡2𝑃𝜑(𝑥,𝑡)𝑑𝑡=20𝑔(𝑡)𝑢(𝑥,𝑡)𝑑𝑡.(3.21) Consider the cases of odd and even 𝑛 separately.

Let 𝑛=2𝑚+1(𝑚). Then, by (2.8) we have𝑘𝑡2=(1)𝑚𝜋𝑚𝑄(𝑚)𝑡2(3.22) and it follows from (2.7) (by successive differentiation) that𝑄(𝑚)𝑡2=1𝜕2𝑡𝜕𝑡𝑚𝑔(𝑡).(3.23) Therefore,𝑘𝑡2=(1)𝑚2𝑚𝜋𝑚1𝑡𝜕𝜕𝑡𝑚𝑔(𝑡),(3.24) and (3.21) takes the form(1)𝑚2𝑚𝜋𝑚01𝑡𝜕𝜕𝑡𝑚𝑃𝑔(𝑡)𝜑(𝑥,𝑡)𝑑𝑡=20𝑔(𝑡)𝑢(𝑥,𝑡)𝑑𝑡.(3.25) Further, integrating 𝑚 times by parts, we get01𝑡𝜕𝜕𝑡𝑚𝑃𝑔(𝑡)𝜑||(𝑥,𝑡)𝑑𝑡=𝑅(𝑥,𝑡)𝑡=𝑡=0+(1)𝑚0𝜕𝑔(𝑡)1𝜕𝑡𝑡𝑚𝑃𝜑(𝑥,𝑡)𝑑𝑡,(3.26) where 𝑅(𝑥,𝑡)=𝑚𝑘=1(1)𝑘1𝑡1𝑡𝜕𝜕𝑡𝑚𝑘𝜕𝑔(𝑡)1𝜕𝑡𝑡𝑘1𝑃𝜑=(𝑥,𝑡)𝑚𝑘=1(1)𝑘1𝑡1𝑡𝜕𝜕𝑡𝑚𝑘𝜕𝑔(𝑡)1𝜕𝑡𝑡𝑘1𝑡2𝑚|𝜔|=1𝜑(𝑥+𝑡𝜔)𝑑𝑆𝜔.(3.27) Since 𝑔(𝑡) is identically zero for large values of 𝑡, we have from (3.27) that 𝑅(𝑥,)=0. Also, it follows directly from (3.27) that 𝑅(𝑥,0)=0. Therefore, (3.25) becomes12𝑚𝜋𝑚0𝜕𝑔(𝑡)1𝜕𝑡𝑡𝑚𝑃𝜑(𝑥,𝑡)𝑑𝑡=20𝑔(𝑡)𝑢(𝑥,𝑡)𝑑𝑡.(3.28) Since in (3.28) 𝑔(𝑡) is arbitrary smooth even function with compact support, we obtain that1𝑢(𝑥,𝑡)=2𝑚+1𝜋𝑚𝜕1𝜕𝑡𝑡𝑚𝑃𝜑(𝑥,𝑡).(3.29) This coincides with (1.17) by (3.19).

Now let us consider the case 𝑛=2𝑚(𝑚). In this case, by (2.8) we have𝑘𝑟2=(1)𝑚𝜋𝑚𝑟2𝑄(𝑚)(𝑤)𝑤𝑟2(𝑑𝑤=1)𝑚𝜋𝑚𝑟𝑄(𝑚)𝑡22𝑡𝑡2𝑟2𝑑𝑡,(3.30) and therefore0𝑘𝑟2𝑃𝜑(𝑥,𝑟)𝑑𝑟=(1)𝑚𝜋𝑚0𝑟𝑄(𝑚)𝑡22𝑡𝑡2𝑟2𝑃𝑑𝑡𝜑=(𝑥,𝑟)𝑑𝑟(1)𝑚𝜋𝑚0𝑟𝑄(𝑚)𝑡22𝑡𝑡2𝑟2𝑑𝑡|𝑦𝑥|=𝑟𝜑(𝑦)𝑑𝑆𝑦=𝑑𝑟(1)𝑚𝜋𝑚0𝑟𝑄(𝑚)𝑡22𝑡|𝑦𝑥|=𝑟𝜑(𝑦)𝑑𝑆𝑦𝑡2||||𝑦𝑥2=𝑑𝑡𝑑𝑟(1)𝑚𝜋𝑚0𝑄(𝑚)𝑡22𝑡𝑡0|𝑦𝑥|=𝑟𝜑(𝑦)𝑑𝑆𝑦𝑡2||||𝑦𝑥2𝑑𝑟𝑑𝑡.(3.31) Hence, setting𝐻𝜑(𝑥,𝑡)=𝑡0|𝑦𝑥|=𝑟𝜑(𝑦)𝑑𝑆𝑦𝑡2||||𝑦𝑥2𝑑𝑟=||||𝑦𝑥<𝑡𝜑(𝑦)𝑑𝑦𝑡2||||𝑦𝑥2,(3.32) we get0𝑘𝑟2𝑃𝜑((𝑥,𝑟)𝑑𝑟=1)𝑚𝜋𝑚0𝑄(𝑚)𝑡22𝑡𝐻𝜑(𝑥,𝑡)𝑑𝑡.(3.33) Substituting this in the left-hand side of (3.21) (beforehand replacing 𝑡 by 𝑟 in the left side of (3.21)), we obtain(1)𝑚𝜋𝑚0𝑄(𝑚)𝑡22𝑡𝐻𝜑(𝑥,𝑡)𝑑𝑡=20𝑔(𝑡)𝑢(𝑥,𝑡)𝑑𝑡(3.34) or, using (3.23),(1)𝑚2𝑚1𝜋𝑚01𝑡𝜕𝜕𝑡𝑚𝑔(𝑡)𝑡𝐻𝜑(𝑥,𝑡)𝑑𝑡=20𝑔(𝑡)𝑢(𝑥,𝑡)𝑑𝑡.(3.35) Further, integrating 𝑚 times by parts, we get01𝑡𝜕𝜕𝑡𝑚𝑔(𝑡)𝑡𝐻𝜑||(𝑥,𝑡)𝑑𝑡=𝐿(𝑥,𝑡)𝑡=𝑡=0+(1)𝑚0𝜕𝑔(𝑡)1𝜕𝑡𝑡𝑚𝑡𝐻𝜑(𝑥,𝑡)𝑑𝑡,(3.36) where𝐿(𝑥,𝑡)=𝑚𝑘=1(1)𝑘11𝑡𝜕𝜕𝑡𝑚𝑘𝜕𝑔(𝑡)1𝜕𝑡𝑡𝑘1𝑡𝐻𝜑(𝑥,𝑡).(3.37) Since 𝑔(𝑡) is identically zero for large values of 𝑡, we have from (3.37) that 𝐿(𝑥,)=0. Also, using the expression of 𝐻𝜑(𝑥,𝑡),𝐻𝜑(𝑥,𝑡)=𝑡0|𝑦𝑥|=𝑟𝜑(𝑦)𝑑𝑆𝑦𝑡2||||𝑦𝑥2=𝑑𝑟𝑡0𝑟2𝑚1|𝜔|=1𝜑(𝑥+𝑟𝜔)𝑡2𝑟2𝑑𝑆𝜔=𝑑𝑟𝑡0𝑟2𝑚1𝑡2𝑟2|𝜔|=1𝜑(𝑥+𝑟𝜔)𝑑𝑆𝜔=𝑑𝑟𝑡0𝑡2𝜉22𝑚2|𝜔|=1𝜑𝑥+𝑡2𝜉2𝜔𝑑𝑆𝜔𝑑𝜉,(3.38) we can check directly from (3.37) that 𝐿(𝑥,0)=0. Therefore, (3.35) becomes12𝑚1𝜋𝑚0𝜕𝑔(𝑡)1𝜕𝑡𝑡𝑚𝑡𝐻𝜑(𝑥,𝑡)𝑑𝑡=20𝑔(𝑡)𝑢(𝑥,𝑡)𝑑𝑡.(3.39) Since in (3.39) 𝑔(𝑡) is arbitrary smooth even function with compact support, we obtain that1𝑢(𝑥,𝑡)=2𝑚𝜋𝑚𝜕1𝜕𝑡𝑡𝑚𝑡𝐻𝜑=1(𝑥,𝑡)2𝑚𝜋𝑚𝜕1𝜕𝑡𝑡𝑚1𝜕𝐻𝜕𝑡𝜑(𝑥,𝑡).(3.40) This coincides with (1.18) by (3.32).


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 𝑛2 implies (2.8).

Let 𝑛=2𝑚+1, where 𝑚1. Then, since12𝜎2𝑚=122𝜋𝑚Γ=𝜋(𝑚)𝑚(𝑚1)!,(A.1) Equation (2.33) takes the form𝑄(𝑤)=𝜋𝑚𝑤(𝑡𝑤)𝑚1(𝑚1)!𝑘(𝑡)𝑑𝑡.(A.2) Hence applying the differentiation formula𝑑𝑑𝑤𝑤𝐺(𝑡,𝑤)𝑑𝑡=𝐺(𝑤,𝑤)+𝑤𝜕𝐺(𝑡,𝑤)𝜕𝑤𝑑𝑡(A.3) repeatedly, we find𝑄(𝑚)(𝑤)=𝜋𝑚(1)𝑚𝑘(𝑤)(A.4) which gives (2.8) for 𝑛=2𝑚+1.

In the case 𝑛=2𝑚 with 𝑚1, (2.33) takes the form1𝑄(𝑤)=2𝜎2𝑚1𝑤(𝑡𝑤)(2𝑚3)/2𝑘(𝑡)𝑑𝑡.(A.5) Hence,𝑄(𝑚1)1(𝑤)=2𝜎2𝑚1𝑤(1)𝑚12𝑚322𝑚5212(𝑡𝑤)1/2𝑘(𝑡)𝑑𝑡.(A.6) Therefore, taking into account that by virtue of1Γ(𝑥)=(𝑥1)Γ(𝑥1),Γ2=𝜋,(A.7) we have12𝜎2𝑚1=𝜋(2𝑚1)/2=𝜋Γ((2𝑚1)/2)(2𝑚1)/2=𝜋((2𝑚3)/2)((2𝑚5)/2)(1/2)Γ(1/2)𝑚1,((2𝑚3)/2)((2𝑚5)/2)(1/2)(A.8) we get𝑄(𝑚1)(𝑤)=(1)𝑚1𝜋𝑚1𝑤𝑘(𝑡)𝑡𝑤𝑑𝑡.(A.9) In the right-hand side we replace 𝑡 by 𝑢, then divide both sides by 𝑤𝑡 and integrate on 𝑤(𝑡,) to get𝑡𝑄(𝑚1)(𝑤)𝑤𝑡𝑑𝑤=(1)𝑚1𝜋𝑚1𝑡1𝑤𝑡𝑤𝑘(𝑢)𝑢𝑤𝑑𝑢𝑑𝑤=(1)𝑚1𝜋𝑚1𝑡𝑘(𝑢)𝑢𝑡𝑑𝑤(𝑤𝑡)(𝑢𝑤)𝑑𝑢=(1)𝑚1𝜋𝑚𝑡𝑘(𝑢)𝑑𝑢,(A.10) because for any 𝑡<𝑢, using the change of variables 𝑤𝑡=𝜉, we have𝑢𝑡𝑑𝑤(𝑤𝑡)(𝑢𝑤)=20𝑢𝑡𝑑𝜉𝑢𝑡𝜉2𝜉=2arcsin|||||𝑢𝑡𝜉=𝑢𝑡𝜉=0=2arcsin1=𝜋.(A.11) Therefore, differentiating (A.10) with respect to 𝑡, we get𝑘(𝑡)=(1)𝑚𝜋𝑚𝑑𝑑𝑡𝑡𝑄(𝑚1)(𝑤)𝑤𝑡𝑑𝑤=(1)𝑚𝜋𝑚𝑑𝑑𝑡0𝑄(𝑚1)(𝑢+𝑡)𝑢=𝑑𝑢(1)𝑚𝜋𝑚0𝑄(𝑚)(𝑢+𝑡)𝑢𝑑𝑢=(1)𝑚𝜋𝑚𝑡𝑄(𝑚)(𝑤)𝑤𝑡𝑑𝑤.(A.12) Thus, (2.8) is obtained also for 𝑛=2𝑚 with 𝑚1.

(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 formula𝐴̃𝑔(1/2)𝑓=𝛿̃𝑔𝜇𝑑𝐸𝜇𝑓,(A.13) where 𝛿 is an arbitrary positive real number and the integral does not depend on 𝛿>0 (𝐸𝜇 is zero on (,0) because 𝐴 is a positive operator). Therefore, for (2.24) we have to show thatlim𝜀+0𝜀𝜋𝛿̃𝑔𝜇𝜇𝜆22+𝜀2𝑑𝜇=̃𝑔(𝜆),𝜆,(A.14) for any 𝛿>0.

Since for any 𝜀>0𝜀𝜋𝛿1𝜇𝜆22+𝜀2𝜀𝑑𝜇=𝜋𝛿𝜆2𝑑𝑢𝑢2+𝜀2=1𝜋𝜋2+arctan𝛿+𝜆2𝜀,(A.15) we havelim𝜀+0𝜀𝜋𝛿1𝜇𝜆22+𝜀2𝜀𝑑𝜇=1,𝜆,𝜋𝑑𝑢𝑢2+𝜀2=1.(A.16) Given 𝛼>0, we can choose a 𝛽>0 such that|||̃𝑔𝑢+𝜆2|||̃𝑔(𝜆)<𝛼for𝑢Ω=𝑢𝛿𝜆2<𝑢<,|𝑢|<𝛽(A.17) since the function ̃𝑔(𝑧) is continuous for 𝑧=𝜆 (we choose the continuous branch of the square root for which 1=1). Further, we choose a number 𝑀 such that||||||||̃𝑔(𝑧)𝑀forIm𝑧𝐶<,(A.18) for sufficiently large positive number 𝐶. This is possible by (2.3) and the fact that 𝑔(𝑡) has a compact support. Let us set Ω=(𝛿𝜆2,)Ω. Then,|||||𝜀𝜋𝛿̃𝑔𝜇𝜇𝜆22+𝜀2𝜀𝑑𝜇̃𝑔(𝜆)𝜋𝛿1𝜇𝜆22+𝜀2|||||𝜀𝑑𝜇𝜋𝛿𝜆2|||̃𝑔𝑢+𝜆2|||̃𝑔(𝜆)𝑢2+𝜀2=𝜀𝑑𝑢𝜋Ω|||̃𝑔𝑢+𝜆2|||̃𝑔(𝜆)𝑢2+𝜀2𝜀𝑑𝑢+𝜋Ω|||̃𝑔𝑢+𝜆2|||̃𝑔(𝜆)𝑢2+𝜀2𝑑𝑢.(A.19) Further,𝜀𝜋Ω|||̃𝑔𝑢+𝜆2|||̃𝑔(𝜆)𝑢2+𝜀2𝛼𝑑𝑢<𝜋𝜀𝑢2+𝜀2𝜀𝑑𝑢=𝛼,𝜋Ω|||̃𝑔𝑢+𝜆2|||̃𝑔(𝜆)𝑢2+𝜀2𝑑𝑢2𝑀𝜋|𝑢|𝛽𝜀𝑢2+𝜀2=𝑑𝑢4𝑀𝜋𝛽𝜀𝑢2+𝜀2𝑑𝑢=4𝑀𝜋𝜋2𝛽arctan𝜀.(A.20)

For fixed 𝛽, the last expression tends to zero as 𝜀+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)(1/2)2(𝑧)=𝜋𝑧1/2𝑒𝑖𝑧.(A.21)

(4) The difference between operators (𝜕/𝜕𝑡1/𝑡)𝑚 (formulae (1.17), (1.18)) and (1/𝑡𝜕/𝜕𝑡)𝑚 (formula (3.25)) is given by𝜕1𝜕𝑡𝑡𝑚=𝜕1𝜕𝑡𝑡𝜕𝜕𝑡𝑚11𝑡.(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 “method of descent” (see [1]).

(6) Since for supp𝑔(𝑡)(𝑎,𝑎), 𝑎>0, we have supp𝑘(𝑡)[0,𝑎2), and on the left-hand side of (2.10) the integral is taken in fact over the ball {𝑦𝑛|𝑦𝑥|<𝑎}, for fixed 𝑥. Therefore, this integral is finite for each 𝑥𝑛 and any solution 𝜓(𝑥,𝜆) of (2.9). We proved (2.10) for 𝜆. If the solution 𝜓(𝑥,𝜆) is an analytic function of 𝜆, then (2.10) will be held also for complex values of 𝜆 by the uniqueness of analytic continuation.


  1. R. Courant and D. Hilbert, Methods of Mathematical Physics. Vol. II, Interscience Publishers, New York, NY, USA, 1962.
  2. V. I. Smirnov, A Course of Higher Mathematics, Vol. II, Addison-Wesley, Reading, Mass, USA, 1964.
  3. M. S. Birman and M. Z. Solomjak, Spectral Theory of Self-Adjoint Operators in Hilbert Space, Reidel, Dordrecht, The Netherlands, 1987.
  4. E. C. Titchmarsh, Eigenfunction Expansions Associated with Second-Order Differential Equations. Vol. 2, Clarendon Press, Oxford, UK, 1958.