Theorem 9.
Let
{
u
ɛ
}
be a sequence of solutions in
W
2
1
(
0
,
T
;
H
0
1
(
Ω
)
,
L
2
(
Ω
)
)
to (29). Then it holds that
(31)
u
ɛ
(
x
,
t
)
⟶
u
(
x
,
t
)
i
n
L
2
(
Ω
T
)
,
u
ɛ
(
x
,
t
)
⇀
u
(
x
,
t
)
i
n
L
2
(
0
,
T
;
H
0
1
(
Ω
)
)
,
∇
u
ɛ
(
x
,
t
)
⇀
n
+
1
,
m
+
1
∇
u
(
x
,
t
)
+
∑
j
=
1
n
∇
y
j
u
j
(
x
,
t
,
y
j
,
s
m
)
,
where
u
∈
W
2
1
(
0
,
T
;
H
0
1
(
Ω
)
,
L
2
(
Ω
)
)
is the unique solution to
(32)
∂
t
u
(
x
,
t
)

∇
·
(
b
(
x
,
t
)
∇
u
(
x
,
t
)
)
=
f
(
x
,
t
)
i
n
Ω
T
,
u
(
x
,
t
)
=
0
o
n
∂
Ω
×
(
0
,
T
)
,
u
(
x
,
0
)
=
u
0
(
x
)
i
n
Ω
with
(33)
b
(
x
,
t
)
∇
u
(
x
,
t
)
=
∫
𝒴
n
,
m
a
(
y
n
,
s
m
)
×
(
∇
u
(
x
,
t
)
+
∑
j
=
1
n
∇
y
j
u
j
(
x
,
t
,
y
j
,
s
m
)
)
d
y
n
d
s
m
.
Here
u
1
∈
L
2
(
Ω
T
×
S
m
;
H
♯
1
(
Y
1
)
/
ℝ
)
and
u
j
∈
L
2
(
Ω
T
×
𝒴
j

1
,
m
;
H
♯
1
(
Y
j
)
/
ℝ
)
,
j
=
2
,
…
,
n
, are the unique solutions to the system of local problems
(34)
ρ
i
∂
s
m

d
i
u
i
(
x
,
t
,
y
i
,
s
m
)

∇
y
i
·
∫
S
m

d
i
+
1
⋯
∫
S
m
∫
Y
i
+
1
⋯
∫
Y
n
a
(
y
n
,
s
m
)
×
(
∇
u
(
x
,
t
)
+
∑
j
=
1
n
∇
y
j
u
j
(
x
,
t
,
y
j
,
s
m
)
)
×
d
y
n
⋯
d
y
i
+
1
d
s
m
⋯
d
s
m

d
i
+
1
=
0
,
for
i
=
1
,
…
,
n
, where
u
i
is independent of
s
m

d
i
+
1
,
…
,
s
m
.
Proof of Theorem <xref reftype="statement" rid="thm10">9</xref>.
Since
{
u
ɛ
}
is bounded in
W
2
1
(
0
,
T
;
H
0
1
(
Ω
)
,
L
2
(
Ω
)
)
and the lists of scales are jointly wellseparated, we can apply Theorem 4 and obtain that, up to a subsequence,
(35)
u
ɛ
(
x
,
t
)
⟶
u
(
x
,
t
)
in
L
2
(
Ω
T
)
,
u
ɛ
(
x
,
t
)
⇀
u
(
x
,
t
)
in
L
2
(
0
,
T
;
H
0
1
(
Ω
)
)
,
∇
u
ɛ
(
x
,
t
)
⇀
n
+
1
,
m
+
1
∇
u
(
x
,
t
)
+
∑
j
=
1
n
∇
y
j
u
j
(
x
,
t
,
y
j
,
s
m
)
,
where
u
∈
W
2
1
(
0
,
T
;
H
0
1
(
Ω
)
,
L
2
(
Ω
)
)
,
u
1
∈
L
2
(
Ω
T
×
S
m
;
H
♯
1
(
Y
1
)
/
ℝ
)
, and
u
j
∈
L
2
(
Ω
T
×
𝒴
j

1
,
m
;
H
♯
1
(
Y
j
)
/
ℝ
)
,
j
=
2
,
…
,
n
.
To obtain the homogenized problem, we introduce the weak form
(36)
∫
Ω
T

u
ɛ
(
x
,
t
)
v
(
x
)
∂
t
c
(
t
)
222
+
a
(
x
ε
q
1
,
…
,
x
ε
q
n
,
t
ε
r
1
,
…
,
t
ε
r
m
)
∇
u
ɛ
(
x
,
t
)
·
∇
v
(
x
)
c
(
t
)
d
x
d
t
=
∫
Ω
T
f
(
x
,
t
)
v
(
x
)
c
(
t
)
d
x
d
t
of (29) where
v
∈
H
0
1
(
Ω
)
and
c
∈
D
(
0
,
T
)
, and letting
ɛ
→
0
, we get using Theorem 4
(37)
∫
Ω
T

u
(
x
,
t
)
v
(
x
)
∂
t
c
(
t
)
+
∫
𝒴
n
,
m
a
(
y
n
,
s
m
)
(
∇
u
(
x
,
t
)
+
∑
j
=
1
n
∇
y
j
u
j
(
x
,
t
,
y
j
,
s
m
)
)
2222222
·
∇
v
(
x
)
c
(
t
)
d
y
n
d
s
m
d
x
d
t
=
∫
Ω
T
f
(
x
,
t
)
v
(
x
)
c
(
t
)
d
x
d
t
.
We proceed by deriving the system of local problems (34) and the independencies of the local temporal variables. Fix
i
=
1
,
…
,
n
and choose
(38)
v
(
x
)
=
ε
p
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
+
1
(
x
ε
q
i
)
,
p
>
0
,
c
(
t
)
=
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
c
λ
+
1
(
t
ε
r
λ
)
,
λ
=
1
,
…
,
m
with
v
1
∈
D
(
Ω
)
,
v
j
∈
C
♯
∞
(
Y
j

1
)
for
j
=
2
,
…
,
i
,
v
i
+
1
∈
C
♯
∞
(
Y
i
)
/
ℝ
,
c
1
∈
D
(
0
,
T
)
and
c
l
∈
C
♯
∞
(
S
l

1
)
for
l
=
2
,
…
,
λ
+
1
. Here
p
and
λ
will be fixed later. Using this choice of test functions in (36), we have
(39)
∫
Ω
T

u
ɛ
(
x
,
t
)
ε
p
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
+
1
(
x
ε
q
i
)
×
(
∂
t
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
c
λ
+
1
(
t
ε
r
λ
)
+
∑
l
=
2
λ
+
1
ε

r
l

1
c
1
(
t
)
×
c
2
(
t
ε
r
1
)
⋯
∂
s
l

1
c
l
(
t
ε
r
l

1
)
⋯
c
λ
+
1
(
t
ε
r
λ
)
)
+
a
(
x
ε
q
1
,
…
,
x
ε
q
n
,
t
ε
r
1
,
…
,
t
ε
r
m
)
∇
u
ɛ
(
x
,
t
)
·
(
ε
p
∇
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
+
1
(
x
ε
q
i
)
+
∑
j
=
2
i
+
1
ε
p

q
j

1
v
1
(
x
)
×
v
2
(
x
ε
q
1
)
⋯
∇
y
j

1
v
j
(
x
ε
q
j

1
)
⋯
v
i
+
1
(
x
ε
q
i
)
)
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
c
λ
+
1
(
t
ε
r
λ
)
d
x
d
t
=
∫
Ω
T
f
(
x
,
t
)
ε
p
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
+
1
(
x
ε
q
i
)
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
c
λ
+
1
(
t
ε
r
λ
)
d
x
d
t
,
where, for
l
=
2
and
l
=
λ
+
1
, the interpretation should be that the partial derivative acts on
c
2
and
c
λ
+
1
, respectively, and where the
j
=
2
and
j
=
i
+
1
terms are defined analogously. We let
ɛ
→
0
and using Theorem 4, we obtain
(40)
lim
ɛ
→
0
∫
Ω
T

u
ɛ
(
x
,
t
)
ε
p
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
+
1
(
x
ε
q
i
)
×
∑
l
=
2
λ
+
1
ε

r
l

1
c
1
(
t
)
c
2
(
t
ε
r
1
)
222222222222
⋯
∂
s
l

1
c
l
(
t
ε
r
l

1
)
⋯
c
λ
+
1
(
t
ε
r
λ
)
+
a
(
x
ε
q
1
,
…
,
x
ε
q
n
,
t
ε
r
1
,
…
,
t
ε
r
m
)
∇
u
ɛ
(
x
,
t
)
·
∑
j
=
2
i
+
1
ε
p

q
j

1
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
∇
y
j

1
v
j
(
x
ε
q
j

1
)
⋯
v
i
+
1
(
x
ε
q
i
)
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
c
λ
+
1
(
t
ε
r
λ
)
d
x
d
t
=
0
,
and extracting a factor
ε

q
i
in the first term, we get
(41)
lim
ɛ
→
0
∫
Ω
T

ε

q
i
u
ɛ
(
x
,
t
)
×
∑
l
=
2
λ
+
1
ε
p
+
q
i

r
l

1
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
+
1
(
x
ε
q
i
)
22222222222
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
∂
s
l

1
c
l
(
t
ε
r
l

1
)
⋯
c
λ
+
1
(
t
ε
r
λ
)
+
a
(
x
ε
q
1
,
…
,
x
ε
q
n
,
t
ε
r
1
,
…
,
t
ε
r
m
)
∇
u
ɛ
(
x
,
t
)
·
∑
j
=
2
i
+
1
ε
p

q
j

1
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
∇
y
j

1
×
v
j
(
x
ε
q
j

1
)
⋯
v
i
+
1
(
x
ε
q
i
)
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
c
λ
+
1
(
t
ε
r
λ
)
d
x
d
t
=
0
.
Suppose that
p
+
q
i

r
λ
≥
0
and
p

q
i
≥
0
(which also guarantees that
p
>
0
as required above); then, by Theorems 7 and 4, we have left
(42)
lim
ɛ
→
0
∫
Ω
T

ε

q
i
u
ɛ
(
x
,
t
)
ε
p
+
q
i

r
λ
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
+
1
(
x
ε
q
i
)
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
∂
s
λ
c
λ
+
1
(
t
ε
r
λ
)
+
a
(
x
ε
q
1
,
…
,
x
ε
q
n
,
t
ε
r
1
,
…
,
t
ε
r
m
)
∇
u
ɛ
(
x
,
t
)
·
ε
p

q
i
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
(
x
ε
q
i

1
)
∇
y
i
v
i
+
1
(
x
ε
q
i
)
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
c
λ
+
1
(
t
ε
r
λ
)
d
x
d
t
=
0
,
which is the point of departure for deriving the local problems and the independency.
We distinguish four different cases where
ρ
i
is either zero (nonresonance) or one (resonance) and
d
i
is either zero or positive.
Case 1. Consider
ρ
i
=
0
and
d
i
=
0
. We choose
λ
=
m
and
p
=
q
i
. This means that
p
+
q
i

r
λ
=
2
q
i

r
m
>
0
since
d
i
=
ρ
i
=
0
and
p

q
i
=
q
i

q
i
=
0
. This implies that (42) is valid. We get
(43)
lim
ɛ
→
0
∫
Ω
T

ε

q
i
u
ɛ
(
x
,
t
)
ε
2
q
i

r
m
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
+
1
(
x
ε
q
i
)
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
∂
s
m
c
m
+
1
(
t
ε
r
m
)
+
a
(
x
ε
q
1
,
…
,
x
ε
q
n
,
t
ε
r
1
,
…
,
t
ε
r
m
)
∇
u
ɛ
(
x
,
t
)
·
ε
0
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
(
x
ε
q
i

1
)
∇
y
i
v
i
+
1
(
x
ε
q
i
)
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
c
m
+
1
(
t
ε
r
m
)
d
x
d
t
=
0
,
where we let
ɛ
→
0
and obtain by means of Theorems 7 and 4
(44)
∫
Ω
T
∫
𝒴
n
,
m
a
(
y
n
,
s
m
)
(
∇
u
(
x
,
t
)
+
∑
j
=
1
n
∇
y
j
u
j
(
x
,
t
,
y
j
,
s
m
)
)
·
v
1
(
x
)
v
2
(
y
1
)
⋯
v
i
(
y
i

1
)
∇
y
i
v
i
+
1
(
y
i
)
c
1
(
t
)
×
c
2
(
s
1
)
⋯
c
m
+
1
(
s
m
)
d
y
n
d
s
m
d
x
d
t
=
0
.
By the Variational Lemma, we have
(45)
∫
Y
i
⋯
∫
Y
n
a
(
y
n
,
s
m
)
(
∇
u
(
x
,
t
)
+
∑
j
=
1
n
∇
y
j
u
j
(
x
,
t
,
y
j
,
s
m
)
)
22222222
·
∇
y
i
v
i
+
1
(
y
i
)
d
y
n
⋯
d
y
i
=
0
,
a.e. in
Ω
T
×
S
m
×
Y
1
×
⋯
×
Y
i

1
for all
v
i
+
1
∈
C
♯
∞
(
Y
i
)
/
ℝ
and by density for all
v
i
+
1
∈
H
♯
1
(
Y
i
)
/
ℝ
. This is the weak form of the local problem in this case. In what follows Theorems 7 and 4, the variational lemma and the density argument are used in a corresponding way.
Case 2. Consider
ρ
i
=
1
and
d
i
=
0
. We again choose
λ
=
m
and
p
=
q
i
. We then have
p
+
q
i

r
λ
=
2
q
i

r
m
=
0
since
d
i
=
0
and
ρ
i
=
1
and
p

q
i
=
q
i

q
i
=
0
which implies that we may again use (42). We get
(46)
lim
ɛ
→
0
∫
Ω
T

ε

q
i
u
ɛ
(
x
,
t
)
ε
0
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
+
1
(
x
ε
q
i
)
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
∂
s
m
c
m
+
1
(
t
ε
r
m
)
+
a
(
x
ε
q
1
,
…
,
x
ε
q
n
,
t
ε
r
1
,
…
,
t
ε
r
m
)
∇
u
ɛ
(
x
,
t
)
·
ε
0
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
(
x
ε
q
i

1
)
∇
y
i
v
i
+
1
(
x
ε
q
i
)
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
c
m
+
1
(
t
ε
r
m
)
d
x
d
t
=
0
and, passing to the limit,
(47)
∫
Ω
T
∫
𝒴
n
,
m

u
i
(
x
,
t
,
y
i
,
s
m
)
v
1
(
x
)
v
2
(
y
1
)
⋯
v
i
+
1
(
y
i
)
2222222
×
c
1
(
t
)
c
2
(
s
1
)
⋯
∂
s
m
c
m
+
1
(
s
m
)
2222222
+
a
(
y
n
,
s
m
)
(
∇
u
(
x
,
t
)
+
∑
j
=
1
n
∇
y
j
u
j
(
x
,
t
,
y
j
,
s
m
)
)
2222222
·
v
1
(
x
)
v
2
(
y
1
)
⋯
v
i
(
y
i

1
)
∇
y
i
v
i
+
1
(
y
i
)
c
1
(
t
)
2222222
×
c
2
(
s
1
)
⋯
c
m
+
1
(
s
m
)
d
y
n
d
s
m
d
x
d
t
=
0
.
By the variational lemma
(48)
∫
S
m
∫
Y
i
⋯
∫
Y
n

u
i
(
x
,
t
,
y
i
,
s
m
)
v
i
+
1
(
y
i
)
∂
s
m
c
m
+
1
(
s
m
)
+
a
(
y
n
,
s
m
)
(
∇
u
(
x
,
t
)
+
∑
j
=
1
n
∇
y
j
u
j
(
x
,
t
,
y
j
,
s
m
)
)
·
∇
y
i
v
i
+
1
(
y
i
)
c
m
+
1
(
s
m
)
d
y
n
⋯
d
y
i
d
s
m
=
0
a.e. for all
v
i
+
1
∈
H
♯
1
(
Y
i
)
/
ℝ
and
c
m
+
1
∈
C
♯
∞
(
S
m
)
, which is the weak form of the local problem in this second case.
Case 3. Consider
ρ
i
=
0
and
d
i
>
0
. Let
λ
be fixed and successively be
m
,
…
,
m

d
i
+
1
. Choose
p
=
r
λ

q
i
which immediately yields that
p
+
q
i

r
λ
=
0
. Furthermore,
p

q
i
=
r
λ

2
q
i
>
0
by the restriction of
λ
and the definition of
d
i
.
Thus we have from (42)
(49)
lim
ɛ
→
0
∫
Ω
T

ε

q
i
u
ɛ
(
x
,
t
)
ε
0
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
+
1
(
x
ε
q
i
)
222222
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
∂
s
λ
c
λ
+
1
(
t
ε
r
λ
)
222222
+
a
(
x
ε
q
1
,
…
,
x
ε
q
n
,
t
ε
r
1
,
…
,
t
ε
r
m
)
∇
u
ɛ
(
x
,
t
)
222222
·
ε
r
λ

2
q
i
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
(
x
ε
q
i

1
)
∇
y
i
v
i
+
1
(
x
ε
q
i
)
222222
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
c
λ
+
1
(
t
ε
r
λ
)
d
x
d
t
=
0
.
We let
ɛ
tend to zero and obtain
(50)
∫
Ω
T
∫
𝒴
i
,
λ

u
i
(
x
,
t
,
y
i
,
s
λ
)
v
1
(
x
)
v
2
(
y
1
)
⋯
v
i
+
1
(
y
i
)
222222
×
c
1
(
t
)
c
2
(
s
1
)
⋯
∂
s
λ
c
λ
+
1
(
s
λ
)
d
y
i
d
s
λ
d
x
d
t
=
0
and we have left
(51)
∫
S
λ

u
i
(
x
,
t
,
y
i
,
s
λ
)
∂
s
λ
c
λ
+
1
(
s
λ
)
d
s
λ
=
0
,
a.e. for all
c
λ
+
1
∈
C
♯
∞
(
S
λ
)
. This means that
u
i
is independent of
s
λ
; thus,
u
i
does not depend on
s
m

d
i
+
1
,
…
,
s
m
. Next we choose
p
=
q
i
and
λ
=
m

d
i
. We have
p
+
q
i

r
λ
=
2
q
i

r
m

d
i
>
0
and
p

q
i
=
0
and we may again use (42). We have
(52)
lim
ɛ
→
0
∫
Ω
T

ε

q
i
u
ɛ
(
x
,
t
)
ε
2
q
i

r
m

d
i
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
+
1
(
x
ε
q
i
)
222222
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
∂
s
m

d
i
c
m

d
i
+
1
(
t
ε
r
m

d
i
)
222222
+
a
(
x
ε
q
1
,
…
,
x
ε
q
n
,
t
ε
r
1
,
…
,
t
ε
r
m
)
∇
u
ɛ
(
x
,
t
)
222222
·
ε
0
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
(
x
ε
q
i

1
)
∇
y
i
v
i
+
1
(
x
ε
q
i
)
222222
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
c
m

d
i
+
1
(
t
ε
r
m

d
i
)
d
x
d
t
=
0
,
where a passage to the limit yields
(53)
∫
Ω
T
∫
𝒴
n
,
m
a
(
y
n
,
s
m
)
(
∇
u
(
x
,
t
)
+
∑
j
=
1
n
∇
y
j
u
j
(
x
,
t
,
y
j
,
s
m

d
i
)
)
22222222
·
v
1
(
x
)
v
2
(
y
1
)
⋯
v
i
(
y
i

1
)
∇
y
i
v
i
+
1
(
y
i
)
22222222
×
c
1
(
t
)
c
2
(
s
1
)
⋯
c
m

d
i
+
1
(
s
m

d
i
)
d
y
n
d
s
m
d
x
d
t
=
0
,
and finally
(54)
∫
S
m

d
i
+
1
⋯
∫
S
m
∫
Y
i
⋯
∫
Y
n
a
(
y
n
,
s
m
)
×
(
∇
u
(
x
,
t
)
+
∑
j
=
1
n
∇
y
j
u
j
(
x
,
t
,
y
j
,
s
m

d
i
)
)
·
∇
y
i
v
i
+
1
(
y
i
)
d
y
n
⋯
d
y
i
d
s
m
⋯
d
s
m

d
i
+
1
=
0
,
a.e. for all
v
i
+
1
∈
H
♯
1
(
Y
i
)
/
ℝ
, which is the weak form of the local problem.
Case 4. Consider
ρ
i
=
1
and
d
i
>
0
. Let
λ
be fixed and successively be
m
,
…
,
m

d
i
+
1
. Choose
p
=
r
λ

q
i
directly implying that
p
+
q
i

r
λ
=
0
. Moreover,
p

q
i
=
r
λ

2
q
i
>
0
by the restriction of
λ
and the definition of
d
i
and
ρ
i
. Hence using (42), we obtain
(55)
lim
ɛ
→
0
∫
Ω
T

ε

q
i
u
ɛ
(
x
,
t
)
ε
0
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
+
1
(
x
ε
q
i
)
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
∂
s
λ
c
λ
+
1
(
t
ε
r
λ
)
+
a
(
x
ε
q
1
,
…
,
x
ε
q
n
,
t
ε
r
1
,
…
,
t
ε
r
m
)
∇
u
ɛ
(
x
,
t
)
·
ε
r
λ

2
q
i
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
(
x
ε
q
i

1
)
∇
y
i
v
i
+
1
(
x
ε
q
i
)
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
c
λ
+
1
(
t
ε
r
λ
)
d
x
d
t
=
0
.
Passing to the limit, we get
(56)
∫
Ω
T
∫
𝒴
i
,
λ

u
i
(
x
,
t
,
y
i
,
s
λ
)
v
1
(
x
)
v
2
(
y
1
)
⋯
v
i
+
1
(
y
i
)
22222222
×
c
1
(
t
)
c
2
(
s
1
)
⋯
∂
s
λ
c
λ
+
1
(
s
λ
)
d
y
i
d
s
λ
d
x
d
t
=
0
.
That is,
(57)
∫
S
λ

u
i
(
x
,
t
,
y
i
,
s
λ
)
∂
s
λ
c
λ
+
1
(
s
λ
)
d
s
λ
=
0
a.e. for all
c
λ
+
1
∈
C
♯
∞
(
S
λ
)
, and hence
u
i
is independent of
s
λ
. Next we choose
p
=
q
i
and
λ
=
m

d
i
in (42). Thus we have
p
+
q
i

r
λ
=
2
q
i

r
m

d
i
=
0
and
p

q
i
=
0
and we get
(58)
lim
ɛ
→
0
∫
Ω
T

ε

q
i
u
ɛ
(
x
,
t
)
ε
0
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
+
1
(
x
ε
q
i
)
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
∂
s
m

d
i
c
m

d
i
+
1
(
t
ε
r
m

d
i
)
+
a
(
x
ε
q
1
,
…
,
x
ε
q
n
,
t
ε
r
1
,
…
,
t
ε
r
m
)
∇
u
ɛ
(
x
,
t
)
·
ε
0
v
1
(
x
)
v
2
(
x
ε
q
1
)
⋯
v
i
(
x
ε
q
i

1
)
∇
y
i
v
i
+
1
(
x
ε
q
i
)
×
c
1
(
t
)
c
2
(
t
ε
r
1
)
⋯
c
m

d
i
+
1
(
t
ε
r
m

d
i
)
d
x
d
t
=
0
.
We let
ɛ
go to zero obtaining
(59)
∫
Ω
T
∫
𝒴
n
,
m

u
i
(
x
,
t
,
y
i
,
s
m

d
i
)
v
1
(
x
)
v
2
(
y
1
)
⋯
v
i
+
1
(
y
i
)
2222222
×
c
1
(
t
)
c
2
(
s
1
)
⋯
∂
s
m

d
i
c
m

d
i
+
1
(
s
m

d
i
)
2222222
+
a
(
y
n
,
s
m
)
(
∇
u
(
x
,
t
)
+
∑
j
=
1
n
∇
y
j
u
j
(
x
,
t
,
y
j
,
s
m

d
i
)
)
2222222
·
v
1
(
x
)
v
2
(
y
1
)
⋯
v
i
(
y
i

1
)
∇
y
i
v
i
+
1
(
y
i
)
2222222
×
c
1
(
t
)
c
2
(
s
1
)
⋯
c
m

d
i
+
1
(
s
m

d
i
)
d
y
n
d
s
m
d
x
d
t
=
0
and finally we arrive at
(60)
∫
S
m

d
i
⋯
∫
S
m
∫
Y
i
⋯
∫
Y
n

u
i
(
x
,
t
,
y
i
,
s
m

d
i
)
v
i
+
1
2222
×
(
y
i
)
∂
s
m

d
i
c
m

d
i
+
1
(
s
m

d
i
)
+
a
(
y
n
,
s
m
)
(
∇
u
(
x
,
t
)
+
∑
j
=
1
n
∇
y
j
u
j
(
x
,
t
,
y
j
,
s
m

d
i
)
)
·
∇
y
i
v
i
+
1
(
y
i
)
c
m

d
i
+
1
(
s
m

d
i
)
d
y
n
⋯
d
y
i
d
s
m
⋯
d
s
m

d
i
=
0
a.e. for all
v
i
+
1
∈
H
♯
1
(
Y
i
)
/
ℝ
and
c
m

d
i
+
1
∈
C
♯
∞
(
S
m

d
i
)
, the weak form of the local problem.