[cylinder kernel of R3 in polar coordinates]
We have:
t
(t2 + |r − r0 |2 )2
π 2 T (t, r, r0 ) =
=
(t2
+
∂ T̄
= π2
∂t
r2
+
r02
(1)
−
2rr0
t2
+
r2
+
(2)
(3)
where
−2π 2 T̄ =
t
cos(θ − θ0 ) + (z − z 0 )2 )2
r02
−
1
cos(θ − θ0 ) + (z − z 0 )2
2rr0
(4)
Consider:
1 ∂ 2T
∂ 2T
∂ 2T
1 ∂T
∂ 2T
+
+
+
+
=0
∂t2
∂r2
r ∂r
r2 ∂θ2
∂z 2
We assume T (θ + θ1 ) = T (θ) and
1
T (0, r, r0 ) = δ(r − r0 )δ(θ − θ0 )δ(z − z 0 )
r
Expand T in Fourier sum in θ:
T (t, r, θ, z) =
∞
X
n=−∞
Z θ1
Tn (t, r, z) =
1
θ1
inθ( 2π
)
θ
e
1
Tn (t, r, z)
−inθ( 2π
)
θ
e
1
T (t, r, θ, z)
(5)
(6)
(7)
(8)
0
Hence
∂ 2 Tn ∂ 2 Tn 1 ∂Tn n2
+
+
− 2
∂t2
∂r2
r ∂r
r
2π
θ1
2
Tn = 0
(9)
From (6) and (8), take θ0 = 0 WLOG, we obtain
Z
1 θ1
−inθ( 2π
)1
θ1
Tn (0, r, z) =
dθe
δ(r − r0 )δ(θ − θ0 )δ(z − z 0 )
θ1 0
r
1 −inθ0 ( 2π
)1
θ1
=
e
δ(r − r0 )δ(z − z 0 )
θ1
r
Try Tsep (t, r) = T (t)R(r)Z(z). Let λ =
2nπ
.
θ1
00
00
(11)
Then
1
λ2
0
T RZ + T R Z + T RZ + T R Z − 2 T R = 0
r
r
T 00 R00 Z 00 1 R0 λ2
+
+
+
− 2 =0
T
R
Z
rR
r
00
(10)
(12)
(13)
Let
−
T 00 Z 00
R00 1 R0 λ2
−
=
+
− 2
T
Z
R
rR
r
2
= −ω
(14)
(15)
So that
0
T = e−ω t
Z = eikz
1 0
λ2
02
00
2
R + R + ((ω + k )R − 2 )R = 0
r
r
(16)
(17)
(18)
0
This implies T 00 = (ω 2 + k 2 )T . Let ω 2 = ω 2 + k 2 . The solution of (18) has the form
of Bessel functions J|λ| (ωr) (and not Y|λ| (ωr) because of the minimal irregularity at r = 0).
Now we have
Z ∞
Z ∞
0
Tn (t, r, z) =
ωdω
dk T̃ (ω, k)J|λ| (ωr)e−ω t eikz
(19)
−∞
0
When t = 0, from (11), we have
0
e−iλθ δ(r − r0 )δ(z − z 0 )
= P (r, z)
Tn (0, r, z) =
θ1 r
(20)
and from (19):
∞
Z
Z
∞
dk T̃ (ω, k)J|λ| (ωr)eikz
ωdω
Tn (0, r, z) =
(21)
−∞
0
So one can solve for T̃ (ω, k):
Z ∞ Z ∞
1
T̃ (ω, k) =
rdrJ|λ| (ωr)P (r, z)e−ikz
dz
2π −∞
Z ∞ 0
1
0
0
=
drJ|λ| (ωr)e−iλθ −ikz δ(r − r0 )
2πθ1 0
0
e−iλθ
0
=
J|λ| (ωr0 )e−ikz
2πθ1
(22)
(23)
(24)
From this one can get
Z
∞
T (t, r, θ, z) =
Z
dk
−∞
0
1
T (t, r, θ, z, r , θ , z ) =
2πθ1
0
0
∞
ωdω
0
∞ Z
X
n=−∞
∞
X
e
(25)
n=−∞
∞
Z
∞
ωdω
−∞
0
iλ(θ−θ0 ) −ω 0 t ik(z−z 0 )
×e
0
T̃n (ω)J|λ| (ωr)einθ e−ω t eikz
e
dkJ|λ| (ωr)J|λ| (ωr0 )
(26)
The formula for T̄ is the same with ωdω replaced by − dω
= −dω(ω 2 + k 2 )−1/2 . So
ω0
Z
∞
1 X iλ(θ−θ0 ) ∞
ωdωJ|λ| (ωr)J|λ| (ωr0 )
T (t, r, θ, z, r , θ , z ) =
e
2πθ1 n=−∞
0
Z ∞
1
2
2
0
dke−(ω +k ) 2 t eik(z−z )
×
0
0
0
(27)
−∞
R∞
p
p
2 + x2 ] cos ax √ dx
exp[−β
=
K
(γ
γ
a2 + β 2 ),
0
0
γ 2 +x2
p
then the k integral becomes 2K0 (ωζ), where ζ = t2 + (z − z 0 )2 . Hence we have
If we use (3.961.2) of Gradshteyn–Rhyzhik,
Z
∞
1 X iλ(θ−θ0 ) ∞
e
ωdωJ|λ| (ωr)J|λ| (ωr0 )K0 (ωζ)
T̄ (t, r, θ, z, r , θ , z ) = −
πθ1 n=−∞
0
0
0
0
(28)
The integral over ω may be found in (6.522.3) of Gradshteyn–Rhyzhik, giving us the
Fourier-series representation of T̄ :
|λ|
∞
r2 − r1
1 X iλ(θ−θ0 ) 1
e
T̄ (t, r, θ, z, r , θ , z ) = −
πθ1 n=−∞
r1 r2 r2 + r1
0
0
0
(29)
p
p
where r1 ≡ (r − r0 )2 + ζ 2 and r2 ≡ (r + r0 )2 + ζ 2 .
−r1
Let rr22 +r
≡ e−u . If we follow the steps illustrated in Smith, we get the final closed form
1
of T̄ :
∞
X
1
0
T̄ (t, r, θ, z, r , θ , z ) = −
e−|λ|u+iλ(θ−θ )
0
2πθ1 rr sinh u n=−∞
0
0
0
sinh( 2π
u)
1
θ1
= −
2π
0
2πθ1 rr sinh u cosh( θ1 u) − cos 2π
(θ − θ0 )
θ1
(30)
(31)
When θ1 = 2π,
1
1
2
0
4π rr cosh u − cos(θ − θ0 )
1
= − 2 2
02
0
2π (r + r − 2rr cos(θ − θ0 ) + (z − z 0 )2 + (t − t0 )2 )
T̄ (t, r, θ, z, r0 , θ0 , z 0 ) = −
which agrees with (4).
(32)
(33)