[cylinder kernel of R2 in polar coordinates]
We have:
2πT (t, r, r0 ) =
=
=
t
(1)
3
(t2 + |r − r0 |2 ) 2
t
3
(t2 + (r cos θ − r0 cos θ0 )2 + (r sin θ − r0 sin θ0 )2 ) 2
t
3
(t2 + r2 + r02 − 2rr0 cos(θ − θ0 )) 2
∂ T̄
= 2π
∂t
(2)
(3)
(4)
1
where 2π T̄ = ((t2 + r2 + r02 − 2rr0 cos(θ − θ0 ))− 2
Consider:
∂ 2T
1 ∂ 2T
∂ 2T
1 ∂T
+
+
+
=0
∂t2
∂r2
r ∂r
r2 ∂θ2
We assume T (θ + θ1 ) = T (θ) and
1
T (0, r, r0 ) = δ(r − r0 )δ(θ − θ0 )
r
Expand T in Fourier sum in θ:
T (t, r, θ) =
∞
X
inθ( 2π
)
θ
e
1
n=−∞
Z θ1
Tn (t, r) =
1
θ1
Tn (t, r)
−inθ( 2π
)
θ
e
1
T (t, r, θ)
(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) =
δ(r − r0 )δ(θ − θ0 )
dθe
θ1 0
r
1 −inθ0 ( 2π
)1
θ1
=
e
δ(r − r0 )
θ1
r
Try Tsep (t, r) = T (t)R(r). Let λ =
2nπ
.
θ1
(10)
(11)
Then
1
λ2
T 00 R + T R00 + T R0 − 2 T R = 0
r
r
T 00 R00 1 R0 λ2
+
+
− 2 =0
T
R
rR
r
(12)
(13)
Let
T 00
R00 1 R0 λ2
=
+
− 2
T
R
rR
r
2
= −ω
−
(14)
(15)
So that
T = e−ωt
λ2
1 0
2
00
R + R + ω R− 2 R=0
r
r
(16)
(17)
The solution of (17) 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 ∞
Tn (t, r) =
ωdω T̃ (ω)J|λ| (ωr)e−ωt
(18)
0
When t = 0, from (11), we have
0
e−iλθ δ(r − r0 )
Tn (0, r) =
= P (r)
θ1 r
(19)
and from (18):
∞
Z
ωdω T̃ (ω)J|λ| (ωr)
Tn (0, r) =
(20)
0
So one can solve for T̃ (ω):
∞
Z
T̃ (ω) =
rdrJ|λ| (ωr)P (r)
(21)
0
1 ∞
0
drJ|λ| (ωr)e−iλθ δ(r − r0 )
θ1 0
0
e−iλθ
J|λ| (ωr0 )
=
θ1
Z
=
(22)
(23)
From this one can get
Z
∞
T (t, r, θ) =
ωdω
0
T (t, r, θ, r0 , θ0 ) =
1
θ1
Z
∞
∞
X
T̃n (ω)J|λ| (ωr)einθ e−ωt
n=−∞
∞
X
ωdω
0
n=−∞
0
eiλ(θ−θ ) e−ωt J|λ| (ωr)J|λ| (ωr0 )
(24)
(25)
The formula for T̄ is the same with ωdω replaced by −dω. From Gradshteyn–Ryzhik,
(6.612.3), we get a Legendre function of the second kind:
∞
1 X iλ(θ−θ0 ) 1
√ Q|λ|− 1 (cosh u0 )
T (t, r, θ, r , θ ) =
e
2
θ1 n=−∞
rr0
0
2
2
02
0
(26)
where cosh u0 ≡ t +r2rr+r
. If we follow equation (2.17) and (2.18) of Smith, we’ll get the
0
same final answer as tft3d.