2. Structure of Arbitrary Function of the Laplace Operator

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Journal of Applied Mathematics Volume 2012, Article ID 761248,19pages doi:10.1155/2012/761248

Research Article

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

Gusein Sh. Guseinov

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

Correspondence should be addressed to Gusein Sh. Guseinov,guseinov@atilim.edu.tr Received 28 September 2011; Accepted 30 January 2012

Academic Editor: Pablo Gonz´alez-Vera

Copyrightq2012 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 Kirchhoﬀrepresentation formulae in low space dimensions.

1. Introduction

The wave equation for a functionux1, . . . , x_n, t ux, tofnspace variablesx₁, . . . , x_nand the timetis given by

∂²u

∂t² Δu, 1.1

where

Δ ∂²

∂x²₁ · · · ∂²

∂x_n² 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

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probleminitial value problemone asks for a solutionux, tof 1.1 defined forx ∈ Rⁿ, t≥0 that satisfies1.1forx∈Rⁿ,t >0 and the initial conditions

ux,0 ϕx, ∂ux,0

∂t ψx x∈Rⁿ. 1.3

Ifn1 andϕ∈C²R,ψ∈C¹R, then the classical solution of problem1.1,1.3is given by d’Alembert’s formula

ux, t ϕxt ϕx−t

2 1

2 _xt

x−tψ y

dy. 1.4

Ifn2 andϕ∈C³R²,ψ ∈C²R², then the solution of problem1.1,1.3is given by Poisson’s formula

ux, t 1 2π

|^y−x|^<t ψ

y dy

t²−y−x² ∂

∂t

⎡

⎢⎣ 1 2π

|^y−x|^<t ϕ

y dy

t²−y−x²

⎤

⎥⎦, 1.5

wherex x1, x2,y y1, y2, and|y−x|² y1−x1² y2−x2².

Ifn3 andϕ∈C³R²,ψ ∈C²R², then the solution of problem1.1,1.3is given by Kirchhoﬀ’s formula

ux, t 1 4πt

|^y−x|^tψ y

dS_y ∂

∂t 1

4πt

|y−x|tϕ y

dS_y

, 1.6

wherex x1, x2, x3,y y1, y2, y3,|y−x|² y1−x1² y2−x2² y3−x3², anddSyis the surface element of the sphere{y∈R³:|y−x|t}.

Passing to an arbitrarynlet us denote byux, t Nϕx, tthe solution of the problem

∂²u

∂t² Δu, x∈Rⁿ, t >0, 1.7

ux,0 ϕx, ∂ux,0

∂t 0, x∈Rⁿ. 1.8

It is easy to see that then the function

vx, t _t

0

ux, τdτ 1.9

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is the solution of the problem

∂²v

∂t² Δv, x∈Rⁿ, t >0, 1.10

vx,0 0, ∂vx,0

∂t ϕx, x∈Rⁿ. 1.11

Indeed, integrating1.7we get _t

0

∂²ux, τ

∂τ² dτ _t

0

Δux, τdτ Δ _t

0

ux, τdτ Δvx, t. 1.12

Hence,

∂ux, t

∂t −∂ux,0

∂t Δvx, t or ∂ux, t

∂t Δvx, t, 1.13 by the second condition in1.8. On the other hand, from1.9,

∂vx, t

∂t ux, t, ∂²vx, t

∂t² ∂ux, t

∂t . 1.14

Comparing1.13and1.14, we get1.10. Besides, vx,0 0, ∂vx,0

∂t ux,0 ϕx 1.15

so that initial conditions in1.11are also satisfied.

Consequently, the solutionux, tof problem1.1,1.3is represented in the form ux, t N_ϕx, t

_t

0

N_ψx, τdτ. 1.16

It follows that it is suﬃcient to know an explicit form of the solutionN_ϕx, tof problem1.7, 1.8. It is known 1,2that

N_ϕx, t 1 2^m1π^m

∂

∂t 1

t _m

|^y−x|^tϕ y

dS_y if n2m1, 1.17

N_ϕx, t 1 2^mπ^m

∂

∂t 1 t

_m−1

∂ dt

|^y−x|^<t ϕ

y dy

t²−y−x² ifn2m, 1.18

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wherex x1, . . . , xn,y y1, . . . , yn,|y−x|² y1−x1²· · · yn−xn², anddSyis the surface element of the sphere{y∈Rⁿ:|y−x|t}.

In the present paper, we give a new proof of formulae1.17,1.18for the solution of problem1.7,1.8. 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. InSection 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 inSection 3to 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

LetAbe the self-adjoint positive operator obtained as the closure of the symmetric operator Adetermined in the Hilbert spaceL²Rⁿby the diﬀerential expression

−Δ − ∂²

∂x²₁ · · · ∂²

∂x²_n

, x1, . . . , xn∈Rⁿ, 2.1

on the domain of definition DA C^∞₀ Rⁿthat is the set of all infinitely diﬀerentiable functions on Rⁿ with compact support. Let E_μ denote the resolution of the identity the spectral projectionfor A:

Af _∞

0

μdEμf, f ∈DA. 2.2

Next, letgt be any infinitely diﬀerentiable even function on the axis−∞ < t < ∞with compact support and

gλ

_∞

−∞gte^iλtdt 2.3

its Fourier transform. Note that the functiongλ tends to zero as|λ| → ∞λ∈Rfaster than any negative power of|λ|. Consider the operatorgA ^1/2defined according to the general theory of self-adjoint operatorssee 3:

g

A^1/2 f

_∞

0

g

μ

dE_μf, f ∈L²Rⁿ. 2.4

The following theorem describes the structure of the operatorgA ^1/2showing that it is an integral operator and giving an explicit formula for its kernel in terms of the function gt.

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Theorem 2.1. The operatorgA ^1/2is an integral operator

g

A^1/2 fx

RⁿK x, y

f y

dy, f∈L²Rⁿ. 2.5

Further, there is a smooth functionktdefined on the interval 0≤t <∞such that

K x, y

kx−y²

. 2.6

The functionktdepends on the functiongtas follows. If one sets Qt g√

t

, that is, Q t²

gt, 0≤t <∞, 2.7

then

kt

⎧⎪

⎪⎪

⎨

⎪⎪

⎪⎩

−1^m

π^m Q^mt if n2m1,

−1^m π^m

_∞

t

Q√^mw

w−t dw if n2m,

2.8

where Q^mt denotes the mth order derivative of Qt. Further, if suppgt ⊂ −a, a, then suppkt⊂ 0, a². For any solutionψx, λof the equation

−Δψx, λ λ²ψx, λ, 2.9

the equality

Rⁿkx−y² ψ

y, λ

dygλψx, λ 2.10

holds forλ∈R.

Proof. First we consider the casen 1. In this case, the statements of the theorem take the following form: kt Qt g√

t for 0 ≤ t < ∞; the operator gA ^1/2 is an integral operator of the form

g

A^1/2 fx

_∞

−∞g x−y

f y

dy, 2.11

and for any solutionψx, λof the equation

−ψx, λ λ²ψx, λ, 2.12

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the equality

_∞

−∞g x−y

ψ y, λ

dygλψx, λ 2.13

holds.

To prove the last statements note that, in the casen1, the operatorAis generated in the Hilbert spaceL²−∞,∞by the operation−d²/dx²and the operatorA^1/2by the operation id/dx. The resolventRμ A−μI⁻¹of the operatorAhas the form

R_μfx i 2√μ

_∞

−∞e^{i|x−y|√μ}f y

dy, 2.14

while the spectral projectionEμof the operatorAhas the formsee 3, page 201

Eμfx _∞

−∞

sin√μ x−y π

x−y f y

dy, 0≤μ <∞,

Eμ0 for μ <0.

2.15

Therefore,

g

A^1/2 fx

_∞

0

g

μ

dE_μfx

_∞

0

g

μ_∞

−∞

cos√μ x−y 2π√μ f

y dy

dμ

_∞

−∞

1 π

_∞

0

gλcosλ x−y

dλ

f y

dy _∞

−∞g x−y

f y

dy,

2.16

where we have used the inversion formula for the Fourier cosine transform. Therefore,2.11 is proved. To prove2.13note that the general solution of2.12is

ψx, λ

⎧⎨

⎩

c1cosλxc2sinλx ifλ /0, c1c2x ifλ0,

2.17

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wherec1andc2are arbitrary constants. Then, we have, forλ /0, _∞

−∞g x−y

ψ y, λ

dyc₁ _∞

−∞g x−y

cosλy dyc₂ _∞

−∞g x−y

sinλy dy c1

_∞

−∞gtcosλx−tdtc2

_∞

−∞gtsinλx−tdt c1

_∞

−∞gtcosλxcosλtsinλxsinλtdt c₂

_∞

−∞gtsinλxcosλt−sinλtcosλxdt c1cosλx

_∞

−∞gtcosλt dtc2sinλx _∞

−∞gtcosλt dt c₁cosλxc₂sinλx

_∞

−∞gtcosλt dtψx, λgλ,

2.18

where we have used the fact that the functiongtis even and therefore _∞

−∞gtsinλt dt0. 2.19

The same result can be obtained similarly forλ0. Thus,2.13is also proved.

Now we consider the casen≥2. We shall use the integral representation

Rμfx

Rⁿr x, y;μ

f y

dy 2.20

of the resolventR_μ A−μI⁻¹ of the operatorA. As is known 4, Section 13.7, Formula 13.7.2,

r x, y;μ

iμ^n−2/4

2^n2/2π^n−2/2x−y^n−2/2H_n−2/2¹ x−yμ

, 2.21

whereH_ν¹zis the Hankel function of the first kind of orderν. Next, according to the general spectral theory of self-adjoint operators 3, page 150, Formula11, we have

dE_μfx 1 2πi

R_μi0−R_μ−i0

fxdμ. 2.22

Therefore, from2.4it follows that the representation2.5holds with K

x, y 1

2πi _∞

0

g

μ r

x, y;μi0

−r

x, y;μ−i0 dμ. 2.23

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Now the representation2.6, which expresses thatKx, yis a function of|x−y|², follows from2.23by2.21.

To prove2.10we use2.23. By virtue of2.23,

Rⁿkx−y² ψ

y, λ dy

RⁿK x, y

ψ y, λ

dy

lim

ε→0

Rⁿ

1 2πi

_∞

0

g

μ r

x, y;μiε

−r

x, y;μ−iε dμ

ψ y, λ

dy

ψx, λlim

ε→0

ε π

_∞

0

g√μ μ−λ²₂

ε²dμψx, λgλ,

2.24

see Appendix. Here we have used the fact that from2.9it follows that

−Δ−zψx, λ λ²−z

ψx, λ, 2.25

that is,

ψx, λ λ²−z

−Δ−z⁻¹ψx, λ, 2.26

and therefore

Rⁿr x, y;z

ψ y, λ

dy 1

λ²−zψx, λ. 2.27

Finally, to deduce the explicit formulae2.7,2.8, we takeψx, λ e^iλx¹ in2.10.

Then, puttingx x2, . . . , x_n, we can write

Rⁿkx₁−y₁²x−y²

e^iλy¹dy₁dygλe^iλx¹. 2.28

If we set

x₁−y₁₂

w, 2.29

then the left-hand side of2.28equals _∞

−∞

Rⁿ⁻¹k

wx−y² dy

e^iλy¹dy₁. 2.30

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On the other hand,

Rⁿ⁻¹k

wx−y² dy

_∞

0

|^x− ^y|^rk

wx−y² dS

dr

_∞

0

k

wr²

|^x− ^y|^rdS

drσ_n−1 _∞

0

rⁿ⁻²k wr²

dr

1 2σ_n−1

_∞

w

t−w^n−3/2ktdt,

2.31

where

σ_n 2π^n/2

Γn/2 2.32

is the surface area of then−1-dimensional unit sphereΓis the gamma functionanddS denotes the surface element of the sphere{y∈Rⁿ⁻¹:|x−y| r}. Therefore, setting

Qw 1

2σ_n−1 _∞

w

t−w^n−3/2ktdt, 2.33

we get that2.28takes the form _∞

−∞Qwe^iλy¹dy₁gλe^iλx¹. 2.34 Substituting here the expression ofwgiven in2.29and making then the change of variables x1−y1t, we obtain

_∞

−∞Q t²

e^iλtdtgλ _∞

−∞gte^iλtdt. 2.35

Hence2.7follows. Further, it is not diﬃcult to check that the formula2.33forn ≥ 2 is equivalent to2.8, see Appendix.

Sincegtis smooth and has a compact support, it follows from2.7,2.8that the functionktalso is smooth and has a compact support; more precisely, if suppgt⊂−a, a, then suppkt ⊂ 0, a². This implies, in particular, convergence of the integral in2.10for each fixedx. The theorem is proved.

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3. Derivation of Formulae 1.17, 1.18

Consider the Cauchy problem1.7,1.8:

∂²u

∂t² Δu, x∈Rⁿ, t >0, 3.1

ux,0 ϕx, ∂ux,0

∂t 0, x∈Rⁿ, 3.2

whereuux, t,t≥0,x x1, . . . , x_n∈Rⁿ,ϕx∈C^∞₀Rⁿ. Forν ν1, . . . , νn,x x1, . . . , xn∈Rⁿ, let us set

|ν|²ν²₁· · ·ν_n², ν, x ν1x1· · ·νnxn. 3.3

Since

−Δe^iν,x|ν|²e^iν,x, 3.4

applying2.9,2.10, we get

Rⁿkx−y²

e^iν,ydyg|ν|e ^iν,x ν∈Rⁿ. 3.5

Hence, by the inverse Fourier transform formula,

kx−y² 1

2πⁿ

Rⁿg|ν|e ^iν,xe^−iν,ydν. 3.6 Multiplying both sides of the last equality byϕyand then integrating ony∈Rⁿ, we get

Rⁿkx−y² ϕ

y

dy 1

2πⁿ

Rⁿg|ν|e^iν,x

!

Rⁿϕ y

e^−iν,ydy

"

dν. 3.7

Substituting here forg|ν| its expression

g|ν| 2 _∞

0

gtcos|ν|tdt 3.8

and setting

ux, t 1 2πⁿ

Rⁿcos|ν|te^iν,x

!

Rⁿϕ y

e^−iν,ydy

"

dν, 3.9

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we obtain

Rⁿkx−y² ϕ

y dy2

_∞

0

gtux, tdt. 3.10

Obviously, the functionux, tdefined by3.9is the solution of problem3.1,3.2. Next we will transform the left-hand side of3.10usingTheorem 2.1.

First we consider the casen1. In this case,3.10takes the form _∞

−∞kx−y² ϕ

y dy2

_∞

0

gtux, tdt 3.11

and from2.7,2.8we have

k t²

Q t²

gt. 3.12

Therefore, making the change of variablesy−xtand taking into account the evenness of the functiongt, we can write

_∞

−∞kx−y² ϕ

y dy

_∞

−∞k t²

ϕxtdt

_∞

−∞gtϕxtdt _∞

0

gt

ϕxt ϕx−t dt.

3.13

Substituting this in the left-hand side of3.11, we obtain _∞

0

gt

ϕxt ϕx−t dt2 _∞

0

gtux, tdt. 3.14

Hence, by the arbitrariness of the smooth even functiongtwith compact support, we get

ux, t ϕxt ϕx−t

2 . 3.15

Further assume thatn≥2. Making the change of variables

y−xtω, 0≤t <∞, |ω|1, ω ω1, . . . , ωn, dytⁿ⁻¹dt dSω, 3.16 wheredSωis the surface element of the unit sphere{ω∈Rⁿ:|ω|1}, we get

Rⁿkx−y² ϕ

y dy

_∞

0

tⁿ⁻¹k t²

|ω|1ϕxtωdSω

dt. 3.17

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Further, making in the right-hand side of3.17the change of variables

xtωy, dS_ytⁿ⁻¹dS_ω, 3.18

wheredSyis the surface element of the sphere{y∈Rⁿ:|y−x|t}, we have

tⁿ⁻¹

|ω|1ϕxtωdSω

|^y−x|^tϕ y

dSy:Pϕx, t. 3.19

Therefore,

Rⁿkx−y² ϕ

y dy

_∞

0

k t²

P_ϕx, tdt, 3.20

and3.10becomes

_∞

0

k t²

P_ϕx, tdt2 _∞

0

gtux, tdt. 3.21

Consider the cases of odd and evennseparately.

Letn2m1m∈N. Then, by2.8we have

k t²

−1^m π^m Q^m

t²

3.22

and it follows from2.7 by successive diﬀerentiationthat

Q^m t²

1

2t

∂

∂t m

gt. 3.23

Therefore,

k t²

−1^m 2^mπ^m

1 t

∂

∂t m

gt, 3.24

and3.21takes the form

−1^m 2^mπ^m

_∞

0

1 t

∂

∂t m

gt

Pϕx, tdt2 _∞

0

gtux, tdt. 3.25

Further, integratingmtimes by parts, we get _∞

0

1 t

∂

∂t _m

gt

P_ϕx, tdt Rx, t|^t∞_t0 −1^m _∞

0

gt ∂

∂t 1 t

_m

P_ϕx, tdt, 3.26

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where

Rx, t #^m

k1

−1^k−1 t

1 t

∂

∂t _m−k

gt∂

∂t 1 t

_k−1 P_ϕx, t

#^m

k1

−1^k−1 t

1 t

∂

∂t _m−k

gt∂

∂t 1 t

_k−1 t^2m

|ω|1ϕxtωdSω.

3.27

Sincegtis identically zero for large values oft, we have from3.27thatRx,∞ 0. Also, it follows directly from3.27thatRx,0 0. Therefore,3.25becomes

1 2^mπ^m

_∞

0

gt ∂

∂t 1 t

_m

Pϕx, tdt2 _∞

0

gtux, tdt. 3.28

Since in3.28gtis arbitrary smooth even function with compact support, we obtain that ux, t 1

2^m1π^m ∂

∂t 1

t m

P_ϕx, t. 3.29

This coincides with1.17by3.19.

Now let us consider the casen2mm∈N. In this case, by2.8we have

k r²

−1^m π^m

_∞

r²

Q√^mw

w−r²dw −1^m π^m

_∞

r

Q^m t²

√ 2t

t²−r² dt, 3.30 and therefore

_∞

0

k r²

Pϕx, rdr −1^m π^m

_∞

0

_∞

r

Q^m t²

√ 2t

t²−r² dt

Pϕx, rdr

−1^m π^m

_∞

0

_∞

r

Q^m t²

√ 2t

t²−r² dt

|y−x|rϕ y

dSy

dr

−1^m π^m

_∞

0

⎧⎪

⎨

⎪⎩ _∞

r

Q^m t²

2t

⎡

⎢⎣

|y−x|r

ϕ y

dSy

t²−y−x²

⎤

⎥⎦dt

⎫⎪

⎬

⎪⎭dr

−1^m π^m

_∞

0

Q^m t²

2t

⎧⎪

⎨

⎪⎩ _t

0

⎡

⎢⎣

|y−x|r

ϕ y

dS_y

t²−y−x²

⎤

⎥⎦dr

⎫⎪

⎬

⎪⎭dt.

3.31

Hence, setting

H_ϕx, t: _t

0

⎡

⎢⎣

|y−x|r

ϕ y

dS_y

t²−y−x²

⎤

⎥⎦dr

|^y−x|^<t ϕ

y dy

t²−y−x², 3.32

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we get

_∞

0

k r²

P_ϕx, rdr −1^m π^m

_∞

0

Q^m t²

2tH_ϕx, tdt. 3.33

Substituting this in the left-hand side of3.21 beforehand replacingtbyrin the left side of 3.21, we obtain

−1^m π^m

_∞

0

Q^m t²

2tH_ϕx, tdt2 _∞

0

gtux, tdt 3.34

or, using3.23,

−1^m 2^m−1π^m

_∞

0

1 t

∂

∂t _m

gt

tH_ϕx, tdt2 _∞

0

gtux, tdt. 3.35

Further, integratingmtimes by parts, we get _∞

0

1 t

∂

∂t _m

gt

tHϕx, tdt Lx, t|^t∞_t0 −1^m _∞

0

gt ∂

∂t 1 t

_m

tHϕx, tdt, 3.36

where

Lx, t #^m

k1

−1^k−11 t

∂

∂t _m−k

gt∂

∂t 1

t _k−1

tH_ϕx, t. 3.37

Sincegtis identically zero for large values oft, we have from3.37thatLx,∞ 0. Also, using the expression ofH_ϕx, t,

Hϕx, t _t

0

⎡

⎢⎣

|y−x|r

ϕ y

dSy

t²−y−x²

⎤

⎥⎦dr

_t

0

r^2m−1

|ω|1

ϕxrω

√t²−r² dS_ω

dr

_t

0

r^2m−1

√t²−r²

|ω|1ϕxrωdSω

dr

_t

0

t²−ξ²_2m−2

|ω|1ϕ

x

t²−ξ²ω

dSω

dξ,

3.38

we can check directly from3.37thatLx,0 0. Therefore,3.35becomes 1

2^m−1π^m _∞

0

gt ∂

∂t 1

t _m

tH_ϕx, tdt2 _∞

0

gtux, tdt. 3.39

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Since in3.39gtis arbitrary smooth even function with compact support, we obtain that ux, t 1

2^mπ^m ∂

∂t 1 t

_m

tHϕx, t 1

2^mπ^m ∂

∂t 1 t

_m−1

∂

∂tHϕx, t.

3.40

This coincides with1.18by3.32.

Appendix

For reader’s convenience, in this section we give some explanation of several points in the paper.

1Let us show how2.33forn≥2 implies2.8.

Letn2m1, wherem≥1. Then, since 1

2σ2m 1 2

2π^m

Γm π^m

m−1!, A.1 Equation2.33takes the form

Qw π^m _∞

w

t−w^m−1

m−1! ktdt. A.2

Hence applying the diﬀerentiation formula d

dw _∞

w

Gt, wdt−Gw, w _∞

w

∂Gt, w

∂w dt A.3

repeatedly, we find

Q^mw π^m−1^mkw A.4

which gives2.8forn2m1.

In the casen2mwithm≥1,2.33takes the form

Qw 1

2σ_2m−1 _∞

w

t−w^2m−3/2ktdt. A.5

Hence,

Q^m−1w 1 2σ_2m−1

_∞

w

−1^m−12m−3 2

2m−5 2 · · ·1

2t−w^−1/2ktdt. A.6

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Therefore, taking into account that by virtue of Γx x−1Γx−1, Γ

1 2

√

π, A.7

we have 1

2σ_2m−1 π^2m−1/2

Γ2m−1/2 π^2m−1/2

2m−3/22m−5/2· · ·1/2Γ1/2 π^m−1

2m−3/22m−5/2· · ·1/2,

A.8

we get

Q^m−1w −1^m−1π^m−1 _∞

w

√kt

t−wdt. A.9

In the right-hand side we replacetbyu, then divide both sides by√

w−tand integrate on w∈t,∞to get

_∞

t

Q√^m−1w

w−t dw −1^m−1π^m−1 _∞

t

√ 1 w−t

_∞

w

√ku u−wdu

dw

−1^m−1π^m−1 _∞

t

ku _u

t

dw

w−tu−w

du

−1^m−1π^m _∞

t

kudu,

A.10

because for anyt < u, using the change of variables√

w−tξ, we have _u

t

dw

w−tu−w 2 ^√_u−t

0

dξ

u−t−ξ² 2 arcsin ξ

√u−t ^ξ

√u−t

ξ0 2 arcsin 1π.

A.11

Therefore, diﬀerentiatingA.10with respect tot, we get

kt −1^m π^m

d dt

_∞

t

Q√^m−1w

w−t dw −1^m π^m

d dt

_∞

0

Q^m−1√ut

u du

−1^m π^m

_∞

0

Q^mut

√u du −1^m π^m

_∞

t

Q^mw

√w−t dw.

A.12

Thus,2.8is obtained also forn2mwithm≥1.

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2Here we explain2.24. Note that since the spectrum of the operatorAis 0,∞ zero is included into the spectrum, the spectral representation formula2.4 should be understood in the sense of the formula

g

A^1/2 f

_∞

−δg μ

dE_μf, A.13

whereδis an arbitrary positive real number and the integral does not depend onδ >0Eμis zero on−∞,0becauseAis a positive operator. Therefore, for2.24we have to show that

εlim→0

ε π

_∞

−δ

g√μ

μ−λ²2ε²dμgλ, λ∈R, A.14

for anyδ >0.

Since for anyε >0

ε π

_∞

−δ

1

μ−λ²₂

ε²dμ ε π

_∞

−δ−λ²

du u²ε² 1

π π

2 arctanδλ² ε

, A.15

we have

εlim→0

ε π

_∞

−δ

1

μ−λ²2ε²dμ1, λ∈R, ε

π _∞

−∞

du u²ε² 1.

A.16

Givenα >0, we can choose aβ >0 such that g

uλ²

−gλ < α foru∈Ω '

u:−δ−λ²< u <∞,|u|< β(

A.17

since the functiongz is continuous forzλwe choose the continuous branch of the square root for which√

11. Further, we choose a numberMsuch that

gz≤M for|Imz| ≤C <∞, A.18

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for suﬃciently large positive numberC. This is possible by2.3and the fact thatgthas a compact support. Let us setΩ −δ−λ²,∞\Ω. Then,

ε π

_∞

−δ

g√μ

μ−λ²2ε²dμ−gλ ε π

_∞

−δ

1

μ−λ²2ε²dμ

≤ ε π

_∞

−δ−λ²

g√ uλ²

−gλ u²ε² du ε

π

Ω

g√ uλ²

−gλ u²ε² du ε

π

Ω

g√ uλ²

−gλ u²ε² du.

A.19

Further,

ε π

Ω

g√ uλ²

−gλ u²ε² du < α

π _∞

−∞

ε

u²ε²duα, ε

π

Ω

g√ uλ²

−gλ

u²ε² du≤ 2M π

|u|≥β

ε u²ε²du 4M

π _∞

β

ε

u²ε²du 4M π

π

2 −arctanβ ε

.

A.20

For fixedβ, the last expression tends to zero asε → 0; hence, and byA.16,A.19, andA.20we getA.14.

3The formula2.14follows from2.21forn1 noting that

H_−1/2¹ z 2

πz 1/2

e^iz. A.21

4 The diﬀerence between operators ∂/∂t 1/t^m formulae 1.17, 1.18 and 1/t ∂/∂t^mformula3.25is given by

∂

∂t 1

t _m

∂

∂t 1

t

∂

∂t _m−1

1

t. A.22

5The explicit formula for the solution of the wave equation in the caseneven can be derived from the casenodd by a known computation called the “method of descent”see 1.

6Since for suppgt ⊂ −a, a,a > 0, we have suppkt ⊂ 0, a², and on the left- hand side of2.10the integral is taken in fact over the ball{y∈Rⁿ:|y−x|< a}, for fixedx.

Therefore, this integral is finite for eachx∈Rⁿand any solutionψx, λof2.9. We proved 2.10forλ∈R. If the solutionψx, λis an analytic function ofλ∈C, then2.10will be held also for complex values ofλby the uniqueness of analytic continuation.

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References

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.

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2. Structure of Arbitrary Function of the Laplace Operator

Research Article

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

Gusein Sh. Guseinov

1. Introduction

2. Structure of Arbitrary Function of the Laplace Operator

3. Derivation of Formulae 1.17, 1.18

Appendix

References

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