30
D. Parra-Guevara and Y.N. Skiba
D
φ∇ · (μ∇φ) dr =
∂ D
φμ
∂φ
∂n
d S −
D
μ |∇φ|
2 dr,
D
φ∇ · φ s dr =
1
2
∂ D
φφ s · n d S.
Finally, dividing each integral over ∂ D into the four integrals over S T , S + , S −
and S B , and applying the conditions (2.6)–(2.9) and (2.12), we get
(Aφ, φ) =
D
σ φ
2 dr +
D
μ |∇φ|
2 dr +
S T
ζ φ
2 k · n d S
+
1
2
S +
U n φ
2 d S −
S −
U n φ
2 d S +
S T
v s φ
2 k · n d S −
S B
v s φ
2 k · n d S
(2.15)
Since U n < 0 in S − , k · n > 0 at S T and k · n < 0 at S B , Eq.(2.15) can be
rewritten as
(Aφ, φ) =
D
σ φ
2 dr +
D
μ |∇φ|
2 dr +
S T
ζ φ
2 k · n d S
+
1
2
⎧
⎪ ⎨
⎪ ⎩
S + ∪S −
|U n | φ
2 d S +
S T ∪S B
v s φ
2
|k · n| d S
⎫
⎪ ⎬
⎪ ⎭
.
Thus, operator A is positive semidefinite: (Aφ, φ) ≥ 0.
Taking the inner product of every term of Eq. (2.4) with φ we obtain
∂φ
∂t
, φ
= ( f, φ) − (Aφ, φ),
f (r, t) =
N
i=1
Q i (t)δ(r − r i ).
Using the condition (Aφ, φ) ≥ 0 and the Schwarz inequality [17], the last equation
implies the inequality
φ,
∂φ
∂t
≤ ≤φ f ,
φ =
(φ, φ).
Further,
φ,
∂φ
∂t
=
1
2
∂
∂t
φ
2
= φ
∂
∂t
φ