144
G. Yu. Levi et al.
u
(1)
4 = u
(0)
4 ,
−λ
(1)
3 u
(1)
4,3 + λ
(0)
3 u
(0)
4,3 = 0,
(I)
(10.15)
−λ
(1)
3 u
(1)
4,3 = 0,
−λ
(0)
3 u
(0)
4,3 = 0.
(II)
(10.16)
Here c
(n)
2 = c
∗(n)
13 + c
∗(n)
1331 , the index after the comma denotes the partial derivatives
at the corresponding coordinates.
To solve the boundary problem, we apply a one-dimensional Fourier transform
along the coordinate x 1 . Thus, we rewrite all the functions in the form [Levi M.O.
et al., 2017]:
u j , σ i j
=
U j , , i j
e
−iαx 1 .
The equations of motion and heat conduction in the Fourier images are rewritten:
−α
2 c
(n)∗
1111 U
(n)
1 + c
(n)∗
3113 U
(n)
1
+ ω
2 U
(n)
1 − iαc
(n)
2 U
(n)
3
+ iαβ
(n)∗
11 U
(n)
4 = 0,
−iαc
(n)
2 U
(n)
1
− α
2 c
(n)∗
1331 U
(n)
3 + ω
2 U
(n)
3 + c
(n)∗
3333 U
(n)
3
− β
(n)∗
33 U
(n)
4
= 0, (10.17)
−α
2 U
(n)
4 + λ
(n)
33 U
(n)
4
+ iωθ
(n)
1 U
(n)
4 + iωE
(n)∗
−iαβ
(n)∗
11 U
(n)
1 + β
(n)∗
33 U
(n)
3
= 0,
Here are the notation: f
= d f
dx 3 , f = d
2 f
dx
2
3 . Solutions of Eqs. (10.17)
will be sought in the form [Levi G.Yu. et al., 2015]:
U
(1)
1 (α, x 3 , ω) = −iα
3
k=1
f
(1)
1k
C k shσ
(1)
k x 3 + C k+3 chσ
(1)
k x 3
,
U
(1)
3 (α, x 3 , ω) =
3
k=1
f
(1)
3k
C k chσ
(1)
k x 3 + C k+3 shσ
(1)
k x 3
,
(10.18)
U
(1)
4 (α, x 3 , ω) =
3
k=1
f
(1)
4k
C k shσ
(1)
k x 3 + C k+3 chσ
(1)
k x 3
, −h ≤ x 3 ≤ 0;
U
(0)
1 (α, x 3 , ω) = −iα
3
k=1
f
(0)
1k C k+6 e
σ
(0)
k x 3 ,
U
(0)
3 (α, x 3 , ω) =
3
k=1
f
(0)
3k C k+6 e
σ
(0)
k x 3 ,
(10.19)
U
(0)
4 (α, x 3 , ω) =
3
k=1
f
(0)
4k C k+6 e
σ
(0)
k x 3 , x 3 ≤ −h.