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Also,
L u 1 , ∂
∂ Ω =
∂ Ω = ∂
∂ Ω
L ∂
∂
L W ϕ , ∂
u 1
∂
W ϕ
∂
∂
∂
W ϕ
∂
W ϕ
∂
t ,
=
0
,
n
n
t
∂
t
n
)= ∂
∂
u 1 (
M
,
0
t W ϕ (
M
,
0
)= ϕ (
M
) ,
2
∂
∂
)= ∂
a 2
t u 1 (
M
,
0
t 2 W ϕ (
M
,
0
)=
Δ
W ϕ (
M
,
0
)=
0
.
∂
Hence, u 1 = ∂
∂
also satisfies the boundary and initial conditions of
(2.1), so that it is indeed the solution of (2.1).
t W ϕ (
M
,
t
)
2. As W f τ (
M
,
t
− τ )
satisfies
⎧
⎨
2 W f τ
∂
∂
a 2
=
W f τ ,
Ω × (
, + ∞ ) ,
Δ
0
t 2
L W f τ , ∂
∂ Ω =
W f τ
∂
,
0
n
⎩
t = τ =
W f τ t = τ =
∂
∂
0
,
t W f τ
f
(
M
, τ ) ,
then
∂
∂
t
t
2 u 3
∂
∂
= ∂
∂
a 2
a 2
t 2 −
Δ
u 3
W f τ (
M
,
t
− τ )
d
τ
−
Δ
W f τ (
M
,
t
− τ )
d
τ
t
t
0
0
t
W f τ τ = t
a 2 t
= ∂
∂
∂
W f τ
∂
τ +
−
W f τ (
,
− τ )
d
0 Δ
M
t
d
τ
t
t
0
τ = t −
t
a 2 t
2 W f τ
∂
∂
τ + ∂
W f τ
∂
=
d
0 Δ
W f τ (
M
,
t
− τ )
d
τ
t 2
t
0
∂
d
t
2 W f τ
∂
a 2
=
−
Δ
W f τ
τ +
f
(
M
,
t
)=
f
(
M
,
t
) .
t 2
0
Therefore, u 3 satisfies the equation of (2.3).
Also,
L u 3 , ∂
∂ Ω =
∂ Ω
L t
0
t
u 3
∂
∂
∂
W f τ (
M
,
t
− τ )
d
τ ,
W f τ (
M
,
t
− τ )
d
τ
n
n
0
L W f τ , ∂
∂ Ω
t
W f τ
∂
=
d
τ =
0
,
n
0
t = 0 =
W f τ τ = t t = 0 =
t
∂
u 3
∂
∂
W f τ
∂
u 3 | t = 0 =
0
,
d
τ +
0
.
t
t
0
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