Hindawi Publishing Corporation
Journal of Applied Mathematics
Volume 2012, Article ID 761248, 19 pages
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 González-Vera
Copyright q 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 ux1 , . . . , xn , t ux, t of n space variables x1 , . . . , xn and
the time t is given by
∂2 u
Δu,
∂t2
1.1
∂2
∂2
·
·
·
∂x12
∂xn2
1.2
where
Δ
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
2
Journal of Applied Mathematics
problem initial value problem one asks for a solution ux, t of 1.1 defined for x ∈ Rn ,
t ≥ 0 that satisfies 1.1 for x ∈ Rn , t > 0 and the initial conditions
ux, 0 ϕx,
∂ux, 0
ψx x ∈ Rn .
∂t
1.3
If n 1 and ϕ ∈ C2 R, ψ ∈ C1 R, then the classical solution of problem 1.1, 1.3 is
given by d’Alembert’s formula
ϕx t ϕx − t 1
ux, t 2
2
xt
ψ y dy.
1.4
x−t
If n 2 and ϕ ∈ C3 R2 , ψ ∈ C2 R2 , then the solution of problem 1.1, 1.3 is given
by Poisson’s formula
⎡
⎤
ψ y dy
ϕ y dy
1
∂⎢ 1
⎥
ux, t 2 ∂t ⎣ 2π y−x <t 2 ⎦,
2π |y−x|<t
| |
t2 − y − x
t2 − y − x
1.5
where x x1 , x2 , y y1 , y2 , and |y − x|2 y1 − x1 2 y2 − x2 2 .
If n 3 and ϕ ∈ C3 R2 , ψ ∈ C2 R2 , then the solution of problem 1.1, 1.3 is given
by Kirchhoff’s formula
1
∂
1
ψ y dSy ϕ y dSy ,
ux, t 4πt |y−x|t
∂t 4πt |y−x|t
1.6
where x x1 , x2 , x3 , y y1 , y2 , y3 , |y − x|2 y1 − x1 2 y2 − x2 2 y3 − x3 2 , and dSy is
the surface element of the sphere {y ∈ R3 : |y − x| t}.
Passing to an arbitrary n let us denote by ux, t Nϕ x, t the solution of the problem
∂2 u
Δu,
∂t2
ux, 0 ϕx,
x ∈ Rn , t > 0,
∂ux, 0
0,
∂t
1.7
x ∈ Rn .
1.8
It is easy to see that then the function
vx, t t
ux, τdτ
0
1.9
Journal of Applied Mathematics
3
is the solution of the problem
∂2 v
Δv,
∂t2
x ∈ Rn , t > 0,
∂vx, 0
ϕx,
∂t
vx, 0 0,
1.10
x ∈ Rn .
1.11
Indeed, integrating 1.7 we get
t
0
∂2 ux, τ
dτ ∂τ 2
t
Δux, τdτ Δ
0
t
ux, τdτ Δvx, t.
1.12
∂ux, t
Δvx, t,
∂t
1.13
0
Hence,
∂ux, t ∂ux, 0
−
Δvx, t
∂t
∂t
or
by the second condition in 1.8. On the other hand, from 1.9,
∂vx, t
ux, t,
∂t
∂2 vx, t ∂ux, t
.
∂t
∂t2
1.14
Comparing 1.13 and 1.14, we get 1.10. Besides,
∂vx, 0
ux, 0 ϕx
∂t
vx, 0 0,
1.15
so that initial conditions in 1.11 are also satisfied.
Consequently, the solution ux, t of problem 1.1, 1.3 is represented in the form
ux, t Nϕ x, t t
1.16
Nψ x, τdτ.
0
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
that
Nϕ x, t 1
2m1 π m
1
Nϕ x, t m m
2 π
∂ 1
∂t t
∂ 1
∂t t
m m−1
|y−x|t
ϕ y dSy
if n 2m 1,
ϕ y dy
∂
2
dt |y−x|<t
t2 − y − x
if n 2m,
1.17
1.18
4
Journal of Applied Mathematics
where x x1 , . . . , xn , y y1 , . . . , yn , |y − x|2 y1 − x1 2 · · · yn − xn 2 , and dSy is the
surface element of the sphere {y ∈ Rn : |y − x| 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 Rn by the differential expression
−Δ −
∂2
∂2
·
·
·
∂x12
∂xn2
,
x1 , . . . , xn ∈ Rn ,
2.1
on the domain of definition DA C0∞ Rn that is the set of all infinitely differentiable
functions on Rn with compact support. Let Eμ denote the resolution of the identity the
spectral projection for A:
Af ∞
μdEμ f,
f ∈ DA.
2.2
0
Next, let gt be any infinitely differentiable even function on the axis −∞ < t < ∞ with
compact support and
g λ ∞
−∞
gteiλt dt
2.3
its Fourier transform. Note that the function g λ tends to zero as |λ| → ∞ λ ∈ R 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
:
∞
g μ dEμ f,
g A1/2 f f ∈ L2 Rn .
2.4
0
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
gt.
Journal of Applied Mathematics
5
Theorem 2.1. The operator g A1/2 is an integral operator
g A1/2 fx Rn
K x, y f y dy,
f ∈ L2 Rn .
2.5
Further, there is a smooth function kt defined on the interval 0 ≤ t < ∞ such that
2 K x, y k x − y .
2.6
The function kt depends on the function gt as follows. If one sets
Qt g
√ t ,
that is, Q t2 gt,
0 ≤ t < ∞,
2.7
then
⎧
−1m m
⎪
⎪
⎪
⎨ π m Q t
kt if n 2m 1,
⎪
⎪
−1m ∞ Qm w
⎪
⎩
dw
√
πm t
w−t
2.8
if n 2m,
where Qm t denotes the mth order derivative of Qt. Further, if supp gt ⊂ −a, a, then
supp kt ⊂ 0, a2 . For any solution ψx, λ of the equation
−Δψx, λ λ2 ψx, λ,
2.9
the equality
Rn
2 k x − y ψ y, λ dy g λψx, λ
2.10
holds for λ ∈ R.
Proof. First we consider the case n √
1. In this case, the statements of the theorem take the
following form: kt Qt g t for 0 ≤ t < ∞; the operator g A1/2 is an integral
operator of the form
g A
1/2
fx ∞
−∞
g x − y f y dy,
2.11
and for any solution ψx, λ of the equation
−ψ x, λ λ2 ψx, λ,
2.12
6
Journal of Applied Mathematics
the equality
∞
−∞
g x − y ψ y, λ dy g λψx, 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 A − μI−1 of the operator A has the form
i
Rμ fx √
2 μ
∞
−∞
ei|x−y|
√
μ
f y dy,
2.14
while the spectral projection Eμ of the operator A has the form see 3, page 201
Eμ fx ∞
−∞
√ sin μ x − y f y dy,
π x−y
0 ≤ μ < ∞,
2.15
Eμ 0 for μ < 0.
Therefore,
∞
g A1/2 fx g μ dEμ fx
0
∞
∞
g μ
0
∞ −∞
1
π
∞
−∞
√ cos μ x − y f y dy dμ
√
2π μ
2.16
∞
g λ cos λ x − y dλ f y dy g x − y f y dy,
−∞
0
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
ψx, λ ⎧
⎨c1 cos λx c2 sin λx
⎩
c1 c2 x
if λ / 0,
if λ 0,
2.17
Journal of Applied Mathematics
7
0,
where c1 and c2 are arbitrary constants. Then, we have, for λ /
∞
−∞
g x − y ψ y, λ dy c1
c1
c1
∞
g x − y cos λy dy c2
−∞
∞
gt cos λx − tdt c2
−∞
∞
∞
−∞
∞
−∞
g x − y sin λy dy
gt sin λx − tdt
gtcos λx cos λt sin λx sin λtdt
−∞
∞
c2
2.18
−∞
c1 cos λx
gtsin λx cos λt − sin λt cos λxdt
∞
−∞
gt cos λt dt c2 sin λx
c1 cos λx c2 sin λx
∞
−∞
∞
−∞
gt cos λt dt
gt cos λt dt ψx, λ
g λ,
where we have used the fact that the function gt is even and therefore
∞
−∞
gt sin λt dt 0.
2.19
The same result can be obtained similarly for λ 0. Thus, 2.13 is also proved.
Now we consider the case n ≥ 2. We shall use the integral representation
Rμ fx Rn
r x, y; μ f y dy
2.20
of the resolvent Rμ A − μI−1 of the operator A. As is known 4, Section 13.7, Formula
13.7.2
,
r x, y; μ iμn−2/4
1
n−2/2 Hn−2/2 x − y μ ,
2n2/2 π n−2/2 x − y
2.21
1
where Hν 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
dEμ fx 1 Rμi0 − Rμ−i0 fxdμ.
2πi
2.22
Therefore, from 2.4 it follows that the representation 2.5 holds with
1
K x, y 2πi
∞
0
g
μ r x, y; μ i0 − r x, y; μ − i0 dμ.
2.23
8
Journal of Applied Mathematics
Now the representation 2.6, which expresses that Kx, y is a function of |x − y|2 , follows
from 2.23 by 2.21.
To prove 2.10 we use 2.23. By virtue of 2.23,
Rn
2 k x − y ψ y, λ dy Rn
K x, y ψ y, λ dy
lim
ε → 0
Rn
1
2πi
∞
g
μ r x, y; μ iε
0
−r x, y; μ − iε dμ ψ y, λ dy
ε
ψx, λ lim
ε → 0 π
∞
0
2.24
√ μ
dμ ψx, λ
g λ,
2
2
ε2
μ−λ
g
see Appendix. Here we have used the fact that from 2.9 it follows that
−Δ − zψx, λ λ2 − z ψx, λ,
2.25
ψx, λ λ2 − z −Δ − z−1 ψx, λ,
2.26
that is,
and therefore
Rn
r x, y; z ψ y, λ dy 1
ψx, λ.
λ2 − z
2.27
Finally, to deduce the explicit formulae 2.7, 2.8, we take ψx, λ eiλx1 in 2.10.
Then, putting x x2 , . . . , xn , we can write
Rn
2 2 k x1 − y1 x − y eiλy1 dy1 dy g λeiλx1 .
2.28
If we set
x1 − y1
2
w,
2.29
2 k w x − y dy eiλy1 dy1 .
2.30
then the left-hand side of 2.28 equals
∞ −∞
Rn−1
Journal of Applied Mathematics
9
On the other hand,
2 k w x − y dy Rn−1
∞ y|r
|x−
0
2 k w x − y dS dr
∞ 2
k wr
0
1
σn−1
2
∞
y|r
|x−
dS dr σn−1
∞
r n−2 k w r 2 dr
0
t − wn−3/2 ktdt,
w
2.31
where
σn 2π n/2
Γn/2
2.32
is the surface area of the n − 1-dimensional unit sphere Γ is the gamma function and dS
denotes the surface element of the sphere {y ∈ Rn−1 : |x − y|
r}. Therefore, setting
Qw 1
σn−1
2
∞
t − wn−3/2 ktdt,
2.33
w
we get that 2.28 takes the form
∞
−∞
Qweiλy1 dy1 g λeiλx1 .
2.34
Substituting here the expression of w given in 2.29 and making then the change of variables
x1 − y1 t, we obtain
∞
−∞
∞
Q t2 eiλt dt g λ gteiλt dt.
−∞
2.35
Hence 2.7 follows. Further, it is not difficult to check that the formula 2.33 for n ≥ 2 is
equivalent to 2.8, see Appendix.
Since gt is smooth and has a compact support, it follows from 2.7, 2.8 that the
function kt also is smooth and has a compact support; more precisely, if supp gt ⊂ −a, a,
then supp kt ⊂ 0, a2 . This implies, in particular, convergence of the integral in 2.10 for
each fixed x. The theorem is proved.
10
Journal of Applied Mathematics
3. Derivation of Formulae 1.17, 1.18
Consider the Cauchy problem 1.7, 1.8:
∂2 u
Δu,
∂t2
x ∈ Rn , t > 0,
∂ux, 0
0,
∂t
ux, 0 ϕx,
3.1
x ∈ Rn ,
3.2
where u ux, t, t ≥ 0, x x1 , . . . , xn ∈ Rn , ϕx ∈ C0∞ Rn .
For ν ν1 , . . . , νn , x x1 , . . . , xn ∈ Rn , let us set
|ν|2 ν12 · · · νn2 ,
ν, x ν1 x1 · · · νn xn .
3.3
Since
−Δeiν,x |ν|2 eiν,x ,
3.4
applying 2.9, 2.10, we get
Rn
2 k x − y eiν,y dy g |ν|eiν,x
ν ∈ Rn .
3.5
Hence, by the inverse Fourier transform formula,
2 k x − y 1
2πn
Rn
g |ν|eiν,x e−iν,y dν.
3.6
Multiplying both sides of the last equality by ϕy and then integrating on y ∈ Rn , we get
Rn
2 k x − y ϕ y dy 1
2πn
!
Rn
g |ν|eiν,x
Rn
"
ϕ y e−iν,y dy dν.
3.7
Substituting here for g |ν| its expression
g |ν| 2
∞
gt cos|ν|tdt
3.8
0
and setting
1
ux, t 2πn
!
Rn
cos|ν|te
iν,x
Rn
"
−iν,y
ϕ y e
dy dν,
3.9
Journal of Applied Mathematics
11
we obtain
Rn
∞
2 k x − y ϕ y dy 2
gtux, tdt.
3.10
0
Obviously, the function ux, 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 form
∞
−∞
∞
2 k x − y ϕ y dy 2
gtux, tdt
3.11
0
and from 2.7, 2.8 we have
k t2 Q t2 gt.
3.12
Therefore, making the change of variables y − x t and taking into account the evenness of
the function gt, we can write
∞
−∞
2 k x − y ϕ y dy ∞
−∞
∞
−∞
k t2 ϕx tdt
gtϕx tdt ∞
gt ϕx t ϕx − t dt.
3.13
0
Substituting this in the left-hand side of 3.11, we obtain
∞
gt ϕx t ϕx − t dt 2
∞
0
3.14
gtux, tdt.
0
Hence, by the arbitrariness of the smooth even function gt with compact support, we get
ux, t ϕx t ϕx − t
.
2
3.15
Further assume that n ≥ 2. Making the change of variables
y − x tω,
0 ≤ t < ∞,
|ω| 1,
ω ω1 , . . . , ωn ,
dy tn−1 dt dSω ,
3.16
where dSω is the surface element of the unit sphere {ω ∈ Rn : |ω| 1}, we get
Rn
∞
2 n−1
2
k x − y ϕ y dy t k t
0
|ω|1
ϕx tωdSω dt.
3.17
12
Journal of Applied Mathematics
Further, making in the right-hand side of 3.17 the change of variables
x tω y,
dSy tn−1 dSω ,
3.18
where dSy is the surface element of the sphere {y ∈ Rn : |y − x| t}, we have
n−1
t
|ω|1
ϕx tωdSω |y−x|t
ϕ y dSy : Pϕ x, t.
3.19
Therefore,
Rn
2 k x − y ϕ y dy ∞ k t2 Pϕ x, tdt,
3.20
0
and 3.10 becomes
∞ ∞
k t2 Pϕ x, tdt 2
gtux, tdt.
0
3.21
0
Consider the cases of odd and even n separately.
Let n 2m 1m ∈ N. Then, by 2.8 we have
−1m
m 2
t
k t2 Q
πm
3.22
and it follows from 2.7 by successive differentiation that
1 ∂ m
Qm t2 gt.
2t ∂t
3.23
−1m 1 ∂ m
k t2 m m
gt,
2 π
t ∂t
3.24
Therefore,
and 3.21 takes the form
−1m
2m π m
∞ 0
1 ∂
t ∂t
m
∞
gt Pϕ x, tdt 2
gtux, tdt.
3.25
0
Further, integrating m times by parts, we get
∞ 0
1 ∂
t ∂t
m
∞
∂ 1 m
t∞
m
gt Pϕ x, tdt Rx, t|t0 −1
gt
Pϕ x, tdt,
∂t t
0
3.26
Journal of Applied Mathematics
13
where
m
#
∂ 1 k−1
1 ∂ m−k
−1k−1
Rx, t gt
Pϕ x, t
t
t ∂t
∂t t
k1
m
#
∂ 1 k−1 2m
1 ∂ m−k
−1k−1
gt
t
ϕx tωdSω .
t
t ∂t
∂t t
|ω|1
k1
3.27
Since gt is identically zero for large values of t, we have from 3.27 that Rx, ∞ 0. Also,
it follows directly from 3.27 that Rx, 0 0. Therefore, 3.25 becomes
1
2m π m
∞
0
∂ 1
gt
∂t t
m
Pϕ x, tdt 2
∞
gtux, tdt.
3.28
0
Since in 3.28 gt is arbitrary smooth even function with compact support, we obtain that
ux, t 1
2m1 π m
∂ 1
∂t t
m
Pϕ x, t.
3.29
This coincides with 1.17 by 3.19.
Now let us consider the case n 2m m ∈ N. In this case, by 2.8 we have
−1m ∞ Qm w
−1m ∞ Qm t2 2t
2
dw dt,
k r √
√
πm
πm
w − r2
t2 − r 2
r2
r
3.30
and therefore
∞ m 2 ∞ Q
t 2t
−1m ∞
2
k r Pϕ x, rdr dt Pϕ x, rdr
√
πm
t2 − r 2
0
0
r
∞ m 2 Q
t 2t
−1m ∞
dt
ϕ y dSy dr
√
πm
t2 − r 2
0
r
|y−x|r
⎧
⎡
⎤ ⎫
⎪
∞
m ∞⎪
⎨
ϕ y dSy ⎥ ⎬
−1
⎢
m 2
dr
t 2t⎣
Q
2 ⎦dt⎪
πm
⎭
⎩ r
0 ⎪
|y−x|r
t2 − y − x
m
−1
πm
∞
0
3.31
⎧ ⎡
⎤ ⎫
⎪
⎨ t ⎪
ϕ y dSy ⎥ ⎬
⎢
dt.
dr
Qm t2 2t
⎣
⎦
⎪
⎪
⎩ 0 |y−x|r t2 − y − x2
⎭
Hence, setting
⎡
t ⎢
Hϕ x, t : ⎣
0
|y−x|r
⎤
ϕ y dSy ⎥
ϕ y dy
2 ⎦dr y−x <t 2 ,
| |
t2 − y − x
t2 − y − x
3.32
14
Journal of Applied Mathematics
we get
∞ −1m ∞ m 2 t 2tHϕ x, tdt.
k r 2 Pϕ x, rdr Q
πm
0
0
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
−1m
πm
∞
Q
m
∞
2
t 2tHϕ x, tdt 2
gtux, tdt
0
3.34
0
or, using 3.23,
−1m
2m−1 π m
∞ 0
1 ∂
t ∂t
m
∞
gt tHϕ x, tdt 2
gtux, tdt.
3.35
0
Further, integrating m times by parts, we get
∞ 0
1 ∂
t ∂t
m
∞
∂ 1 m
m
gt tHϕ x, tdt Lx, t|t∞
gt
tHϕ x, tdt,
−1
t0
∂t t
0
3.36
m
#
∂ 1 k−1
1 ∂ m−k
k−1
gt
tHϕ x, t.
Lx, t −1
t ∂t
∂t t
k1
3.37
where
Since gt is identically zero for large values of t, we have from 3.37 that Lx, ∞ 0. Also,
using the expression of Hϕ x, t,
⎡
t ⎢
Hϕ x, t ⎣
⎤
ϕ y dSy ⎥
2 ⎦dr
0
|y−x|r
t2 − y − x
t
ϕx rω
2m−1
r
dSω dr
√
t2 − r 2
0
|ω|1
t 2m−1 r
ϕx rωdSω dr
√
t2 − r 2 |ω|1
0
t 2m−2 2
2
t −ξ
ϕ x t2 − ξ2 ω dSω dξ,
0
3.38
|ω|1
we can check directly from 3.37 that Lx, 0 0. Therefore, 3.35 becomes
1
2m−1 π m
∞
0
∞
∂ 1 m
gt
tHϕ x, tdt 2
gtux, tdt.
∂t t
0
3.39
Journal of Applied Mathematics
15
Since in 3.39 gt is arbitrary smooth even function with compact support, we obtain that
ux, t 1
2m π m
1
m m
2 π
∂ 1
∂t t
∂ 1
∂t t
m
tHϕ x, t
m−1
∂
Hϕ x, t.
∂t
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 n ≥ 2 implies 2.8.
Let n 2m 1, where m ≥ 1. Then, since
1
πm
1 2π m
σ2m ,
2
2 Γm m − 1!
A.1
Equation 2.33 takes the form
Qw π
m
∞
w
t − wm−1
ktdt.
m − 1!
A.2
Hence applying the differentiation formula
d
dw
∞
Gt, wdt −Gw, w w
∞
w
∂Gt, w
dt
∂w
A.3
repeatedly, we find
Qm w π m −1m kw
A.4
which gives 2.8 for n 2m 1.
In the case n 2m with m ≥ 1, 2.33 takes the form
1
Qw σ2m−1
2
∞
t − w2m−3/2 ktdt.
A.5
w
Hence,
Q
m−1
1
w σ2m−1
2
∞
w
−1m−1
1
2m − 3 2m − 5
· · · t − w−1/2 ktdt.
2
2
2
A.6
16
Journal of Applied Mathematics
Therefore, taking into account that by virtue of
√
1
Γ
π,
2
Γx x − 1Γx − 1,
A.7
we have
1
π 2m−1/2
π 2m−1/2
σ2m−1 2
Γ2m − 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−1
w −1
m−1
π
m−1
∞
w
kt
dt.
√
t−w
In the right-hand side we replace t by u, then divide both sides by
w ∈ t, ∞ to get
∞
t
∞
A.9
√
w − t and integrate on
∞
ku
du dw
√
w−t w u−w
t
u
∞
dw
m−1 m−1
du
−1 π
ku
w − tu − w
t
t
∞
−1m−1 π m
kudu,
Qm−1 w
dw −1m−1 π m−1
√
w−t
√
1
A.10
t
because for any t < u, using the change of variables
u
t
dw
w − tu − w
2
√u−t
√
w − t ξ, we have
dξ
u − t − ξ2
0
√
ξ ξ u−t
2 arcsin √
2 arcsin 1 π.
u − t ξ0
A.11
Therefore, differentiating A.10 with respect to t, we get
−1m d ∞ Qm−1 w
−1m d ∞ Qm−1 u t
kt du
dw √
√
π m dt t
π m dt 0
u
w−t
−1m ∞ Qm u t
−1m ∞ Qm w
du
dw.
√
√
πm
πm
u
w−t
0
t
Thus, 2.8 is obtained also for n 2m with m ≥ 1.
A.12
Journal of Applied Mathematics
17
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
1/2
g A
f
∞
−δ
g
μ dEμ f,
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 that
ε
lim
ε → 0 π
∞
−δ
√ μ
dμ g λ,
2
2
ε2
μ−λ
g
λ ∈ R,
A.14
for any δ > 0.
Since for any ε > 0
ε
π
∞
−δ
1
ε
dμ 2
π
μ − λ2 ε2
∞
−δ−λ2
du
1
2
2
π
u ε
π
δ λ2
arctan
2
ε
,
A.15
we have
ε
lim
ε → 0 π
∞
−δ
1
dμ 1,
ε2
μ−
ε ∞
du
1.
π −∞ u2 ε2
2
λ2
λ ∈ R,
A.16
Given α > 0, we can choose a β > 0 such that
g
u λ2 − g λ < α
'
(
for u ∈ Ω u : −δ − λ2 < 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 ≤ M
for |Im z| ≤ C < ∞,
A.18
18
Journal of Applied Mathematics
for sufficiently large positive number C. This is possible by 2.3 and the fact that gt has a
compact support. Let us set Ω −δ − λ2 , ∞ \ Ω. Then,
√ ε ∞
g μ
ε ∞
1
dμ
−
g
dμ
λ
π −δ μ − λ2 2 ε2
π −δ μ − λ2 2 ε2 √
g u λ2 − g λ
ε ∞
≤
du
π −δ−λ2
u2 ε 2
√
√
g u λ2 − g λ
g u λ2 − g λ
ε
ε
du du.
π Ω
π Ω
u2 ε 2
u2 ε 2
A.19
Further,
√
2 −g
g
u
λ
λ
ε
ε
α ∞
du <
du α,
2
2
2
π Ω
π −∞ u ε2
u ε
√
2 −g
g
u
λ
λ
ε
ε
2M
du ≤
du
2
2
2
π Ω
π |u|≥β u ε2
u ε
β
ε
4M ∞
4M π
−
arctan
.
du
π β u2 ε 2
π
2
ε
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 that
1
H−1/2 z 2
πz
1/2
eiz .
A.21
4 The difference between operators ∂/∂t 1/tm formulae 1.17, 1.18 and
1/t ∂/∂tm formula 3.25 is given by
∂ 1
∂t t
m
∂ 1 ∂ m−1 1
.
∂t t ∂t
t
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 supp gt ⊂ −a, a, a > 0, we have supp kt ⊂ 0, a2 , and on the lefthand side of 2.10 the integral is taken in fact over the ball {y ∈ Rn : |y − x| < a}, for fixed x.
Therefore, this integral is finite for each x ∈ Rn and any solution ψx, λ of 2.9. We proved
2.10 for λ ∈ R. If the solution ψx, λ is an analytic function of λ ∈ C, then 2.10 will be held
also for complex values of λ by the uniqueness of analytic continuation.
Journal of Applied Mathematics
19
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.
Advances in
Operations Research
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Advances in
Decision Sciences
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Mathematical Problems
in Engineering
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Journal of
Algebra
Hindawi Publishing Corporation
http://www.hindawi.com
Probability and Statistics
Volume 2014
The Scientific
World Journal
Hindawi Publishing Corporation
http://www.hindawi.com
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
International Journal of
Differential Equations
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Volume 2014
Submit your manuscripts at
http://www.hindawi.com
International Journal of
Advances in
Combinatorics
Hindawi Publishing Corporation
http://www.hindawi.com
Mathematical Physics
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Journal of
Complex Analysis
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
International
Journal of
Mathematics and
Mathematical
Sciences
Journal of
Hindawi Publishing Corporation
http://www.hindawi.com
Stochastic Analysis
Abstract and
Applied Analysis
Hindawi Publishing Corporation
http://www.hindawi.com
Hindawi Publishing Corporation
http://www.hindawi.com
International Journal of
Mathematics
Volume 2014
Volume 2014
Discrete Dynamics in
Nature and Society
Volume 2014
Volume 2014
Journal of
Journal of
Discrete Mathematics
Journal of
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Applied Mathematics
Journal of
Function Spaces
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Optimization
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014
Hindawi Publishing Corporation
http://www.hindawi.com
Volume 2014