A Method for Solving the Three-dimensional Wave
Equation
A. S. Yakimov
Tomsk State University, Tomsk, 634034, Russia *Corresponding Author: [email protected]
Copyright © 2013 Horizon Research Publishing All rights reserved.
Abstract
On the basis of the Laplace integral transform, locally one-dimensional scheme of cleavage and quasi-linearization method to obtain an approximate analytical solution of the three-dimensional nonlinear hyperbolic equation of second order. The assessment of the accuracy of analytical formulas when compared with the exact solution of the first boundary value problem and numerical solution by a known method.Keywords
Analytical Solution, Three-dimensional Telegraph Equation, Laplace Integral Transform1. Introduction
In mathematical modeling of heat and mass transfer [1], the heat transfer in high-frequency processes [2], vibrations [3] and so on. there is a problem the solution of telegraph type an equation [1-3]. If the solution of a nonlinear parabolic equation [1] there are a number of analytical techniques (reviewed in [4]), the exact analytical solutions are obtained for the linear one-dimensional (in the absence of a source) [5,6] or multi-dimensional [3] an equation of the telegraph type. However, in practice most often of interest to the solution of nonlinear boundary-value problems [1-4,7].
For one-dimensional solutions of nonlinear ordinary differential equations in [8] proposed a method of quasi-linearization. With this method, a decision along the nonlinear problem is reduced to solving a sequence of linear problems, which is essentially a development of the well-known Newton's method and its generalized variant proposed by L. V. Kantorovich [9]. Otherwise, the quasi-linearization − is the application of a nonlinear functional generated by the nonlinear boundary value problem, the Newton-Kantorovich.
In the numerical solution of problems of mathematical physics were effective splitting methods [10,11]. In particular, the locally one-dimensional scheme cleavage [10] proposed to solve the multi-dimensional heat equation in combination with the analytical (constant coefficients) and
numerical methods
The purpose of article − with the help of the locally one-dimensional scheme splitting [11], quasi-linearization [8,9] and the Laplace integral transform [12] to find an approximate analytical solution of the nonlinear three-dimensional hyperbolic heat conduction equation in a finite region and to assess the accuracy of analytical formulas.
2. Statement of the Problem and the
Algorithm of Method
Suppose you want to find a solution to a hyperbolic equation of the second order [1,3], with sources
t
T
T
C
t
T
z
∂
∂
+
∂
∂
−
(
)
2 2
1 =
3
1 2
1 ( ) ( ) ( , )
k j
j j j j
T T
A T Y T A T A x t
x x x
=
∂ ∂ ∂
+ + +
∂ ∂ ∂
∑
(1)in the parallelepiped
Q x
:[
=
( , , ), 0
x x x
1 2 3<
x
j<
L
j ,0
<
L
j< ∞ =
,
j
1, 2,3]
, Q Q= +Г, Q Qt = × < ≤[0 t t0],Г − boundary surface of the domain Q with initial conditions
),
(
|
0p
1x
T
t==
∂
T
t
t 0=
p
2(
x
)
∂
= ,
x
=
(
x
1,
x
2,
x
3)
(2)and to simplify further calculations with the boundary condition of the first kind
,
|
Г=
Ψ
T
Ψ
≠
const
, (3)where
A
1=
const
,z
=
c
2/
A
н,Y
j=
C T w T
( ) ( )
j , j =– the thermal conductivity,
A
н – the thermal conductivity at the initial temperature, W/(m⋅K); T – the temperature, K;j
x
, j = 1, 2, 3 – axis of the Descartes coordinate system, m; t – time, s;L
j, j = 1, 2, 3 – length of the sides of the parallelepiped, m.According to [1], the equation (1) is obtained assuming that
C
(
T
)
,Y
j(
T
)
, j = 1, 2, 3 do not explicitly depend on time t andw
j<<
1
m/s, j = 1, 2, 3 disregard mixedderivative
∑
=
∂
∂
∂
3
1 2
j j
t
x
jT
w
as compared∂
2T
/
∂
t
2 to the left of equation (1). Furthermore, in equation (1) on the right side are absent form mixed derivatives∑
≠=
∂
∂
∂
∂
3
, 1
)
(
ij
j j
x
iT
T
A
x
and summand −τ
r(
∂
A
2/
∂
t
)
,2
/
c
r
=
χ
τ
(τ
r − relaxation time, χ− coefficient thermal diffusivitym
2/
s
), whose value for the times considered below:t
>
100
×
τ
r (τ
r≈
10
−9 s) is negligible.We will always assume:
1. Problem (1) − (3) has a unique solution
T
(
x
,
t
)
, which continuously inQ
t and has continuous derivatives2 2 2
2
/
,
/
,
/
,
/
t
T
t
T
x
jT
x
jT
∂
∂
∂
∂
∂
∂
∂
∂
j = 1, 2, 3. 2.Performed following conditions:
A
(
T
)
≥
l
1>
0
,0
)
(
T
≥
l
2>
C
,z
≥
l
3>
0
,l
i=
const
, i = 1, 2, 3,2 1
,
p
p
− set to continuous function inQ
, а C,A
,Y
j,2
A
, j = 1, 2, 3 − continuous functions inQ
t. The coefficientsC
(
T
)
,Y
j(
T
)
, j = 1, 2, 3 in the general case can be non-linearly dependent on the solution of problem [1]. 3. ViewA
(
T
)
is defined by the following formula (16);Ψ
− given continuous function on the boundary Г for0
<
t
≤
t
0, having bounded partial derivatives of the first order.Applicable the locally one-dimensional scheme splitting of the equations (1) − (3) on the differential level [11] and introduce the superscripts (1), (2), (3) to denote the solution of the intermediate stages, as well as ξ there is the direction of wave decision and η there is direction the solution of the parabolic part equation (1). Then we have
2 1 1
) 1 ( ) 0 (
1 2
) 1 ( 2
1
(
)
A
x
T
T
A
x
t
T
z
+
ξσ
∂
∂
∂
∂
ξ
=
∂
∂
ξξ ξ
− ,
*
0
<
t
<
t
, (4))
(
|
0 1) 1
(
p
x
T
ξ t==
, 0 2(
)
) 1 (
x
p
t
T
t
=
∂
∂
=
ξ ,
j j
L
x
<
<
0
,j = 1, 2, 3,
)
,
,
(
* 2 31 0 ) 1 (
1
G
t
x
x
T
ξ x==
,)
,
,
,
(
* 1 2 32 )
1 (
1
1
G
t
L
x
x
T
ξ x=L=
, (5)
∂
∂
∂
∂
η
=
∂
∂
ηξ η
ξ
1 ) 1 ( ) 1 ( 1 )
1 ( ) 1
(
)
(
)
(
x
T
T
A
x
t
T
T
C
++
∂
∂
ηξ
1 ) 1 ( ) 1 ( 1
(
)
x
T
T
Y
(6)+
σ
1[
A
1(
T
η(1))
k+
η
A
2]
,0
<
t
<
t
*,)
,
(
)
,
0
(
(1) *) 1
(
x
T
t
x
T
η=
ξ ,T
η(1) x1=0=
G
1,2 )
1 (
1
1
G
T
η x=L=
; (7)2 2 2
) 2 ( ) 1 (
2 2
) 2 ( 2
1
(
)
A
x
T
T
A
x
t
T
z
+
ξσ
∂
∂
∂
∂
ξ
=
∂
∂
ξη ξ
− ,
*
0
<
t
<
t
, (8)),
,
(
)
,
0
(
(1) *) 2
(
x
T
t
x
T
ξ=
ηt
x
t
T
t
x
T
∂
∂
=
∂
∂
ξ(2)(
0
,
)
η(1)(
*,
)
,)
,
,
(
* 1 31 0 ) 2 (
2
Q
t
x
x
T
ξ x==
,)
,
,
,
(
* 1 2 32 )
2 (
2
2
Q
t
x
L
x
T
ξ x =L=
, (9)
∂
∂
∂
∂
η
=
∂
∂
ηξ η
ξ
2 ) 2 ( ) 2 (
2 )
2 ( ) 2
(
)
(
)
(
x
T
T
A
x
t
T
T
C
++
∂
∂
η ξ2 ) 2 ( ) 2 ( 2
(
)
x
T
T
Y
(10)+
σ
2[
A
1(
T
η(2))
k+
η
A
2]
,0
<
t
<
t
*,)
,
(
)
,
0
(
(2) *) 2
(
x
T
t
x
T
η=
ξ ,T
η(2) x2=0=
Q
1,2 )
2 (
2
2
Q
T
η x=L=
; (11)2 3 3
) 3 ( ) 2 (
3 2
) 3 ( 2
1
(
)
A
x
T
T
A
x
t
T
z
+
ξσ
∂
∂
∂
∂
ξ
=
∂
∂
ξη ξ
−
,
*
0
<
t
<
t
, (12)),
,
(
)
,
0
(
(2) *) 3
(
x
T
t
x
T
ξ=
ηt
x
t
T
t
x
T
∂
∂
=
∂
)
,
,
(
* 1 21 0 ) 3 (
3
D
t
x
x
T
ξ x==
,)
,
,
,
(
* 1 2 32 )
3 (
3
3
D
t
x
x
L
T
ξ x=L=
, (13)
∂
∂
∂
∂
η
=
∂
∂
ηξ η
ξ
3 ) 3 ( ) 3 (
3 )
3 ( ) 3
(
)
(
)
(
x
T
T
A
x
t
T
T
C
++
∂
∂
ηξ
3 ) 3 ( ) 3 ( 3
(
)
x
T
T
Y
(14)+
σ
3[
A
1(
T
η(3))
k+
η
A
2]
,0
<
t
<
t
*,)
,
(
)
,
0
(
(3) *) 3
(
x
T
t
x
T
η=
ξ ,T
η(3) x3=0=
D
1,2 )
3 (
3
3
D
T
η x =L=
, (15)where ξ + η = 1,
σ
1+
σ
2+
σ
3=
1
. WhenC
(
T
)
≠
0
solved the system of equations (4) − (15) for the telegraph equation: the presence of friction (conductive medium). WhenC
(
T
)
=
0
,A
1=
0
,η
=
0
solved the system ofequations (4), (5), (8), (9), (12), (13) for the wave equation: the lack of friction (decaying environment). Indexes: н − the initial value, * − characteristic value, 0 at the bottom − the ultimate value, r− value of relaxation, ξ− bottom part of the wave equation (1), η− below the parabolic component of the equation (1).
We are talking about the next model, for example, the conductive-convective heat transfer for the hyperbolic heat equation at
A
1=
A
2=
0
. First, in the first stage is turned off by the conductive-convective heat transfer coordinate directionsx
2,
x
3, that is consider the problem (4) − (7). Then att
=
t
* we get a temperature distribution)
,
(
* ) 1 (t
x
T
. Taking her by the initialing, turn off the conductive-convective heat transfer on directionsx
1,
x
3 and solve the problem (8) − (11), then att
=
t
* we have the distributionT
(2)(
t
*,
x
)
. We take it for an initial temperature, turn off the conductive-convective heat transfer on directionsx
1,
x
2 and consider the problem (12) − (15). Then in momentt
=
t
* we finally get the temperature)
,
(
*) 3 (
t
x
T
, which coincides with the true valueT
(
t
*,
x
)
.According to this model, the process the conductive-convective heat transfer "stretched" in time and takes place during the time of the gap 3
t
* [10], and instead*
t
. Such an approach to solve multi-dimensional equations of partial differential equations with constant coefficients is proposed and justified in [10,11]. For the wave equation (C
(
T
)
=
0
,A
1=
0
,η
=
0
) are excluded by thecoordinate directions
x
1,
x
2,
x
3 – wave velocity, shear elasticity of the medium and so on.But before that, to the system (4) – (15) must apply Kirchhoff transformation [4] and quasi-linearization [8,9] to obtain the differential equations with constant coefficients, which can be solved with Laplace integral transform [12].
In the future, to use the inversion formula
A
(
T
)
in (1) taken in the form)
(
T
A
=N
T
m, m ≥ 0,N
>
0
,N
=
const
.(16) We use Kirchhoff transformation [4]dT
A
T
A
v
=
T∫
0 н
)
(
. (17)
Then, taking into account the relations [4]:
T
T
A
A
∇
∂
∂
=
∇
,t
T
A
A
t
v
∂
∂
=
∂
∂
н
,
T
A
A
v
=
∇
∇
н
, (18)
get out (4) – (18)
φ
=
T
s/
v
,φ
=
sA
н/
N
, s = m + 1, (19)2 1 2 1
) 1 ( 2 ) 1 (
н
a
x
v
t
v
A
A
t
b
+
σ
∂
∂
=
∂
∂
∂
∂
ξ ξ ,*
0
<
t
<
t
, (20)),
(
|
0 1) 1
(
p
x
T
ξ t==
(1) 0p
2(
x
)
t
T
t
=
∂
∂
= ξ
,
0
<
x
j<
L
j, j = 1, 2, 3, (21)1 0 ) 1 (
1
g
v
ξ x==
,v
ξ(1) x1=L1=
g
2, (22)+
∂
∂
+
∂
∂
=
∂
∂
η η η1 ) 1 ( 1 2 1 ) 1 ( 2 ) 1 (
1
x
v
r
x
v
t
v
c
(23)+
σ
1[
a
1(
T
η(1))
k+
a
2]
,0
<
t
<
t
*,)
,
(
)
,
0
(
(1) *) 1
(
x
v
t
x
v
η=
ξ ,v
η(1) x1=0=
g
1,2 )
1 (
1
1
g
v
η x=L=
; (24)2 2 2 2 ) 2 ( 2 ) 2 (
н
a
x
v
t
v
A
A
t
b
+
σ
∂
∂
=
∂
∂
∂
∂
ξ ξ,
0
<
t
<
t
*,(25)),
,
(
)
,
0
(
(1) *) 2
(
x
T
t
x
T
ξ=
ηt
x
t
T
t
x
T
∂
∂
=
∂
∂
ξ(2)(
0
,
)
η(1)(
*,
)
,1 0 ) 2 (
2
q
v
ξ x==
,v
ξ(2) x2=L2=
q
2,+
∂
∂
+
∂
∂
=
∂
∂
η η η2 ) 2 (
2 2 2 ) 2 ( 2 ) 2 (
2
x
v
r
x
v
t
+
σ
2[
a
1(
T
η(2))
k+
a
2]
,0
<
t
<
t
*,)
,
(
)
,
0
(
(2) *) 2
(
x
v
t
x
v
η=
ξ ,v
η(2) x2=0=
q
1,2 )
2 (
2
2
q
v
η x=L=
;2 3 2 3 ) 3 ( 2 ) 3 (
н
a
x
v
t
v
A
A
t
b
+
σ
∂
∂
=
∂
∂
∂
∂
ξ ξ ,*
0
<
t
<
t
,(26)),
,
(
)
,
0
(
(2) *) 3
(
x
T
t
x
T
ξ=
ηt x t T t
x T
∂ ∂ = ∂
∂ ξ(3)(0, ) η(2)(*, ),
1 0 ) 3 (
3
d
v
ξ x==
,v
ξ(3) x3=L3=
d
2,+
∂
∂
+
∂
∂
=
∂
∂
η ξ η3 ) 3 ( 3 2 3 ) 3 ( 2 ) 3 (
3
x
v
r
x
v
t
v
c
+
σ
3[
a
1(
T
η(3))
k+
a
2]
,0
<
t
<
t
*,)
,
(
)
,
0
(
(3) *) 3
(
x
v
t
x
v
η=
ξ ,v
η(3) x3=0=
d
1,2 )
3 (
3
3
d
v
η x =L=
,where
c
C
(
T
(i))
/(
A
)
i
=
ξη
,r
i=
Y
i/(
η
A
)
, i = 1, 2, 3,)
/(
1
c
2b
=
ξ
, a1=A1/(ηAн), a2=A2/Aн,g
i=
G
is/
φ
,φ
=
s/
i i
Q
q
,=
s/
φ
i i
D
d
, i = 1, 2.This intermediate value in directions:
T
ξ(j)(
t
*,
x
)
,)
,
(
*) (
t
x
T
jη , j = 1, 2, 3 determined from (19) the inversion
formulas
s j
j
v
T
( )(
( ))
1/ξ
ξ
=
φ
,T
η(j)=
(
φ
v
η(j))
1/s, j = 1, 2, 3. (27)Then the final solution of the problem (1) – (3) is write
)
,
(
)
,
(
t
*x
T
(3)t
*x
T
=
η . (28)Note that the range of the independent variables and the type of boundary conditions do not change under the transformation Kirchhoff (17), and in the presence of the inversion formula boundary conditions of the first kind of goes to the Dirichlet condition.
Our purpose to receive a solution of the nonlinear boundary value problem, if it exists, as the limit of a sequence of solutions of linear boundary value problems. To do this, we use the results [7-9]. Assume further that all the coordinate directions in space are equivalent. Let
const
) 1 (
=
v
there is an initial approximation [as an initial approximation to take forv
(1) the first formula of (21) with the first equation in (19)]. For simplicity, consider the case of a quasi-one dimension analysis and sequence)
,
(
t
x
v
n defined by the recurrence relation [8] (dot and barat the top indicate the partial derivative with respect to time
and space)
n n n n n n n
v f v v v
f v v f y v
′ ∂
∂ ′ − ′ + ∂
∂ − + = ∂ ∂
+ +
+ ( ) ( )
1 1
21 2
+
n n n
v
v
f
v
∂
∂
−
+
)
(
1 , (29))
,
,
(
v
nv
nv
nf
f
=
′
,v
н=
v
n+1(
0
,
x
)
,v
n+1|
Г=
Ψ
s/
φ
,
n = 0, 1, 2,…, (30) where y there is the any coordinate from
x
j, j = 1, 2, 3 in (29). Then fory
=
x
1 the rest coordinates in (29)j j
L
x
<
<
0
, j = 2, 3 change parametrically. The remaining coordinates the preparation of expressions (29) and (30) occurs circular replacement index if instead substitute y respectivelyx
2,
x
3 . Note that in solving the three-dimensional boundary value problem (20) – (26), while in the first coordinatex
1 in the direction of an initialiteration acts
v
n, the subsequent iterationv
n+1 is obtainedfrom the final expression
v
n+1(
t
*,
x
)
=
v
(3)(
t
*,
x
)
[seecomment below to formulas (56) – (61)]. Then in a quasi-one dimension of equation (29), (30) can be rewritten in the coordinate
x
1 [7]:) 0 ( 1 ) 0 ( ) 1 ( ) 0 (1 ) 0 ( ) 1 ( 1 2 1 ) 1 ( 2
)
(
)
(
v
f
v
v
v
f
v
v
f
x
v
′
∂
∂
′
−
′
+
∂
∂
−
+
=
∂
∂
+
(
(1) (0))
(10)v
f
v
v
∂
∂
−
(31))
,
,
(
(0) (0) (0) 11
f
v
v
v
f
=
′
,v
(1)(
0
,
x
)
=
v
(0)(
t
*,
x
)
,н ) 0 (
v
v
=
,,
|
0 1) 1 (
1
g
v
x==
v
(1)|
x1=L1=
g
2.(32)Expressions similar to (29), (30) can be written in other coordinate directions
x
2,
x
3 . In particular, the second coordinate directionx
2 is necessary in (31), (32) everywhere to replace the top and bottom indexes (1) and 1 on (2) and 2, and the top index (0) on (1). Thus for the entry condition in the second co-ordinate directionx
2 have)
,
(
)
,
0
(
(1) *) 2
(
x
v
t
x
v
=
.Each function
v
n+1 in (29), (30) in the case of quasi-onedimension or
v
(1) in (31), (32) is solution of the linear equation, which is a very important feature of this algorithm. The algorithm follows from the approximation method of Newton-Kantorovich [9] in the functional space.2 )
(
a
R
i=
σ
iξ ,
)
(
/
1 / 2) 1 ( ) 1
(
r
v
x
a
Z
a
v
c
f
k si i i i i i i
i
=
η−−
∂
η−∂
−
σ
+
,)
(
(−1)η
φ
=
ii
v
Z
,c
i(
v
η(i−1))
=
∂
f
i/
∂
v
η(i−1) ,) 1 (
/
−η
′
∂
∂
−
=
ii
i
f
v
r
,)
/(
н /
1
kZ
A
Z
N
a
ik s i ii
=
σ
Φ
,Φ
i=
−
∂
f
i/
∂
v
η(i−1),)]
/(
1
[
( 1) н/ 1 )
(
a
Z
v
kA
Z
N
R
k s i ii i
i −
η
η
=
σ
−
+σ
ia
2,) ( ) ( )
( i i
i
i
c
v
R
h
η=
η−
η , i = 1, 2, 3. (33)Obtain a quasi-one dimension solution of the problem (31), (32) in coordinate direction
x
1 , using equation (20) − (22)1 2 1 ) 1 ( 2
h x
v =
∂
∂ ξ , (1) (1)
н
1 ξ − ξ
∂ ∂ ∂
∂
= R
t v A A t b
h ,
0
<
t
<
t
* , (34))
(
|
0 1) 1
(
p
x
T
ξ t==
, 0 2(
)
) 1 (
x
p
t
T
t
=
∂
∂
= ξ
, (35)
1 0 ) 1 (
1
g
v
ξ x==
,v
ξ(1) x1=L1=
g
2. (36)Laplace integral transform is applicable to the differential equation (34), except for derivative on
x
1 and replacing it with a linear expression relative to the image of the desired function. In what follows we consider functions for which the Laplace integral transform is absolutely convergent. The real part of the complex numberp
=
α
+
i
β
,i
=
−
1
is positive, that is Re p > 0. We denote the image in big lettersV
(1),
H
1. It is assumed that in calculating the image coordinatesx
i , i = 1, 2, 3, we operate with the functions of the analytic continuation to the valuesx
i>
L
ion that law, which they are defined in the interval(
0
,
L
i)
, i = 1, 2, 3.It is believed that the solution
v
(1)(
t
,
x
)
and its derivatives in equation (34), satisfy the conditions for the existence of Laplace integral transform onx
1 , and thedegree of growth on
x
1 the functionv
(1) and its derivatives do not depend from t,x
2,
x
3. For simplicity, we omit the index calculations * at t at the bottom of the index ξat
v
ξ(1), as well as the index of (1) at the top and at thebottom at
v
ξ(1),
V
(1),
h
1,
H
1 , indicating1 3 2 ) 1
(
(
t
,
0
,
x
,
x
)
/
x
v
∂
∂
ξ =∂
g
1/
∂
x
1, then we have from (34) [12]:)
,
,
,
(
2 32
V
t
p
x
x
p
−pg
1 −1 1
x
g
∂
∂
=
H
(
t
,
p
,
x
2,
x
3)
,j j
L
x
<
<
0
, j = 2, 3,)
,
,
,
(
t
p
x
2x
3V
= 1 1/
2 1 2p
H
p
x
g
p
g
+
∂
∂
+
. (37)
Using the inverse Laplace integral transform [12]:
1 2 1
(
1
/
)
L
−p
=
x
, −=
∫
10 1
[
(
)
/
]
(
)
L
x
dy
y
h
p
p
H
, to restorethe original для
v
(
t
,
x
)
from (37) [12])
,
(
t
x
v
=+
∫
−
∂
∂
+
10 1 1 1 1
1
(
)
(
)
x
dy
y
h
y
x
x
g
x
g
. (38)The derivative
∂
g
1/
∂
x
1 in (38) we find, using the second boundary condition of (36)2
g
=+
∫
−
∂
∂
+
10 1 1
1 1
1
(
)
(
)
L
dy
y
h
y
L
x
g
L
g
. (39)Therefore, finding
∂
g
1/
∂
x
1 in (39) and substituting it into (38), we obtain1
)
,
(
t
x
g
v
=
+
−
−
−
∫
10 1 1 2 1
1
(
)
(
)
L
dy
y
h
y
L
g
g
L
x
+(40)+
∫
−
1
0
1
)
(
)
(
x
dy
y
h
y
x
.Transform the expression on the right-hand side of (40) so as to get rid of the integral with variable upper limit. Then, by introducing the Green's function
E
1(
x
1,
y
)
[7,8]
≤
≤
−
≤
≤
−
=
,
,
/
)
(
,
0
,
/
)
(
)
,
(
1 1
1 1
1
1 1
1 1 1
1
x
L
y
L
x
y
L
x
y
L
x
L
y
y
x
E
(41)expression (40) is rewritten in the return of the upper index (1) and lower ξ, noting that b, A is clearly not dependent on
1
x
, unlikeR
ξ(1) in (33)) 1 (
0 1 1 ) 1 (
н 1
(
,
)
ξ
ξ
+
∂
∂
∂
∂
∫
E
x
y
dy
v
t
v
A
A
t
b
L = (42)= 1
0
) 1 ( 1 1 1
2 1 1 1
1
)
,
(
)
(
g
g
E
x
y
R
dy
F
L
x
g
L=
+
−
+
∫
ξ .We apply again Kirchhoff transformation (17) to return to the original variable T in equation (4)
dv
A
A
T
=
∫
v0
Transform the left side of equation (42), using (43)
t
v
A
A
t
T
∂
∂
=
∂
∂
н,
∂
∂
∂
∂
=
∂
∂
t
v
A
A
t
t
T
н2 2
,
v
=
T
s/
φ
,N
sA
н/
=
φ
, s = m + 1.Since the second term on the left-hand side of equation (42) takes the form
v
ξ(1)=
(
T
ξ(1))
s/
φ
, then it is necessary toreapply the quasi-linearization (31), (32), then we have
) 1 ( ) 0 ( 1 2
) 1 ( 2
)
(
ξ ξξ
φ
+
∂
∂
T
T
B
s
t
T
m= (44)
=
[
Y
1−1F
1+
(
s
/
φ
−
1
)
B
1(
T
ξ(0))
s]
=
P
1,∫
=
10 1 1
1
(
,
)
L
dy
y
x
E
b
Y
,где, 1
1 1
=
Y
−B
, B1>0,T
ξ(0)=
p
1(
x
)
,E
1(
x
1,
y
)
>
0
из (41). We use the idea of Doetsch [13] on the applicability of Laplace integral transform to the partial differential equation as many times as its dimension. Then, with the initial conditions (35), we obtain [7,12] of (44))
,
(
* ) 1 (t
x
T
ξ =s
1(
t
*)
p
1(
x
)
+
s
2(
t
*)
p
2(
x
)
+∫
*−
τ
τ
τ
0
1 *
2
(
)
(
,
)
t
d
x
P
t
s
, (45))
(
sin
)
(
* 11 1 *2
t
t
s
=
γ
−γ
,)]
(
sin[
)
(
* 11 1 *2
t
−
τ
=
γ
−γ
t
−
τ
s
, (46))
(
cos
)
(
* 1 *1
t
t
s
=
γ
,s
1(
t
*−
τ
)
=
co
s[
γ
1(
t
*−
τ
)]
,1 1
=
β
γ
,β
1=
sB
1(
T
ξ(0))
m/
φ
,t
x
t
T
∂
∂
ξ(1)(
*,
)
=
s
1(
t
*)
p
2(
x
)
−
γ
12s
2(
t
*)
p
1(
x
)
+ (47)+
∫
−
τ
τ
τ
*
0
1 *
1
(
)
(
,
)
t
d
x
P
t
s
.Now we obtain an analytical solution of the problem of the intermediate (23), (24), using the notation (33)
1 ) 1 ( 1 ) 1 ( ) 1 ( 1 2 1 ) 1 ( 2
x
v
r
h
v
x
v
∂
∂
−
=
Φ
+
∂
∂
ηη η η
,
0
<
t
<
t
*, (48))
,
(
)
,
0
(
(1) *) 1
(
x
v
t
x
v
η=
ξ ,v
η(1) x1=0=
g
1, 2 )1 (
1
1
g
v
η x=L=
.Note that
Φ
1 of (33) does not depend explicitly ofx
1in (48) [
Φ
1 you can always ask at the bottom of iteration n, knowing the valueT
ξ(1) of the initial andsubsequent times from (45) and
v
ξ(1) from (19)]. Forsimplicity, we omit the index calculations * at t the bottom of the index η at
v
η(1), and as well as the index of(1) at the top and at the bottom at
v
η(1),
V
(1),
h
η(1),
H
1.Then, as in (34) − (41) we have
)
,
,
,
(
2 32
V
t
p
x
x
p
−pg
1 −1 1
x
g
∂
∂
+
Φ
1V
(
t
,
p
,
x
2,
x
3)
+
r
1[
pV
(
t
,
p
,
x
2,
x
3)
−
g
1]
=H
(
t
,
p
,
x
2,
x
3)
,j j
L
x
<
<
0
, j = 2, 3,]
)
[(
)
/
(
)
(
)
(
2 1 2 1
1 1 1 1 2 1
2 1
b
p
b
H
x
g
g
b
b
p
g
p
V
+
δ
+
+
∂
∂
+
δ
+
+
δ
+
δ
+
=
, (49)where
δ
=
r
1/
2
,b
1=
Φ
1−
δ
2 .Using the inverse Laplace integral transform [12]:
)
cos(
)]
/(
[
L
−1p
p
2+
b
12=
b
1x
1 atb
12=
Φ
1−
δ
2>
0
,)]
/(
[
L
21 2 1
p
p
−
b
− =
cosh(
)
1 1
x
b
atb
12<
0
,)
exp(
]
)
[(
L
−1p
+
δ
−1=
−
δ
x
1 ,∫
=
− 1
0 1
[
(
)
/
]
(
)
L
H
p
p
xh
y
dy
, to restore the original for)
,
(
t
x
v
of (49) [12])
,
(
t
x
v
=exp(
−
δ
x
1)
}
/
)
(
)]
(
)
(
[
{
g
1u
1x
1+
δ
u
2x
1+
u
2x
1∂
g
1∂
x
1 + (50)+
exp(
−
δ
x
1)
∫
δ
−
10
1
2
(
)
(
)
)
exp(
x
dy
y
h
y
x
u
y
,j j
L
x
<
<
0
, j = 2, 3;)
(
sin
)
(
1 1 11 1
2
x
b
b
x
u
=
− ,)]
(
(
sin[
)
(
1 11 1 12
x
y
b
b
x
y
u
−
=
−−
, (51))
(
cos
)
(
1 1 11
x
b
x
u
=
,b
12=
Φ
1−
δ
2>
0
;)
(
cosh
)
(
1 1 11
x
b
x
u
=
,u
2(
x
1)
=
b
1−1sinh
(
b
1x
1)
,)]
(
(
sinh[
)
(
1 11 1 12
x
y
b
b
x
y
u
−
=
−−
, 21
