Time Frequency Method of Solving One Boundary Value Problem for a Hyperbolic System and Its Application to the Oil Extraction
We consider the boundary value problem, where the motion of the object is described by the two-dimensional linear system of partial differential equations of hyperbolic type where a discontinuity is at a point within the interval that defines the phase coordinate x. Using the method of series and La...
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| Zitieren: | Time Frequency Method of Solving One Boundary Value Problem for a Hyperbolic System and Its Application to the Oil Extraction / F.A. Aliev, N.A. Aliev, A.P. Guliev // Журнал математической физики, анализа, геометрии. — 2016. — Т. 12, № 2. — С. 101-112. — Бібліогр.: 27 назв. — англ. |
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| citation_txt | Time Frequency Method of Solving One Boundary Value Problem for a Hyperbolic System and Its Application to the Oil Extraction / F.A. Aliev, N.A. Aliev, A.P. Guliev // Журнал математической физики, анализа, геометрии. — 2016. — Т. 12, № 2. — С. 101-112. — Бібліогр.: 27 назв. — англ. |
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| description | We consider the boundary value problem, where the motion of the object is described by the two-dimensional linear system of partial differential equations of hyperbolic type where a discontinuity is at a point within the interval that defines the phase coordinate x. Using the method of series and Laplace transformation in time t (time-frequency method), an analytical solution is found for the determination of debit Q(2l, t) and pressure P(2l, t), which can be effective in the calculation of the coefficient of hydraulic resistance in the lift at oil extraction by gas lift method where l is the well depth. For the case where the boundary functions are of exponential form, the formulas for P(2l, t) and Q(2l, t) depending only on t are obtained. It is shown that at constant boundary functions, these formulas allow us to determine the coefficient of hydraulic resistance in the lift of gas lift wells, which determines the change in the dynamics of pollution.
|
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Journal of Mathematical Physics, Analysis, Geometry
2016, vol. 12, No. 2, pp. 101–112
Time Frequency Method of Solving One Boundary
Value Problem for a Hyperbolic System and Its
Application to the Oil Extraction
F.A. Aliev, N.A. Aliev, and A.P. Guliev
Institute of Applied Mathematics, Baku State University
23, Z. Khalilov Str., Baku AZ1148, Azerbaijan
E-mail: f aliev@yahoo.com
Received May 29, 2014, revised November 30, 2015
We consider the boundary value problem, where the motion of the ob-
ject is described by the two-dimensional linear system of partial differential
equations of hyperbolic type where a discontinuity is at a point within the
interval that defines the phase coordinate x. Using the method of series and
Laplace transformation in time t (time-frequency method), an analytical so-
lution is found for the determination of debit Q(2l, t) and pressure P (2l, t),
which can be effective in the calculation of the coefficient of hydraulic re-
sistance in the lift at oil extraction by gas lift method where l is the well
depth. For the case where the boundary functions are of exponential form,
the formulas for P (2l, t) and Q(2l, t) depending only on t are obtained. It
is shown that at constant boundary functions, these formulas allow us to
determine the coefficient of hydraulic resistance in the lift of gas lift wells,
which determines the change in the dynamics of pollution.
Key words: hyperbolic equation, boundary problems, method of series,
Laplace transformation, time-frequency method, gas lift, coefficient of hy-
draulic resistance.
Mathematics Subject Classification 2010: 65M38, 35L02, 35L40, 58J45,
58J90.
Introduction
As known [1–4], only a considerably small part of oil is extracted from the
reservoir by gushing method. The rest of oil is extracted by secondary and tertiary
methods (gas lifting [2–4], subsurface pumping [1, 2], etc.). As considerable
volumes of oil are produced from wells by gas lift method [2, 3, 5–12], researchers
have paid a lot of attention to creating mathematical models of this process
c© F.A. Aliev, N.A. Aliev, and A.P. Guliev, 2016
F.A. Aliev, N.A. Aliev, and A.P. Guliev
lately. However, in [5], the authors, averaging the motion equation (in partial
derivatives) with respect to time t and depth E, solved the general problem of
control and optimization for the corresponding ordinary differential equations.
These simplifications make it possible to obtain “approximate” mathematical
models instead of exact ones. Therefore the authors of [13] applying the difference
schemes [14] and the method of series by the well depth [15] for the motion
equation of gas-liquids mixture in the lift obtained the total number of terms of
the series which defines the solution representation to a given accuracy. In the
present paper, under some restrictions imposed on the boundary conditions [16–
18], an analytical form of the solution of the motion equation in the considered
problem is given. In some partial cases, such a representation (when the volume of
the gas supplied ’shoe’ is constant) allows us to define the analytical expression
of the coefficient of hydraulic resistance in the pump-compressor pipe and the
change of pollution dynamics [19, 20].
1. The Problem Statement and the Method of Series for
Solving a Boundary Value Problem in an Annular Space
Let us consider the boundary value problem for the system of equations of
hyperbolic type
{
−F1
∂P1(x,t)
∂x = ∂Q1(x,t )
∂t + 2a1Q1(x, t),
−F1
∂P1( x,t)
∂t = c1
∂Q1(x,t)
∂x , x > 0, t ∈ R,
(1)
with the boundary conditions
{
P1(0, t) = P10(t),
Q1(0, t) = Q10(t), t ∈ R,
(2)
where P1(x, t) is a pressure, Q1(x, t) is a gas volume,
2a1 =
g
ωc1
+
λωc1
2D1
, (3)
ωc1 is an average velocity of a flow with respect to the length of the annular
pipeline, λ is the coefficient of hydraulic resistances, D1 is the effective diameter
of the annular space, F1 is the area of the cross-section of pump-compressor
pipes, which is constant along the axis x, c1 is the sound speed in gas, g is the
gravity acceleration, P10(t) and Q10(t) are given continuous functions specifying
the initial pressure and the volume of the gas supplied to the “shoe”. Thus, it
is required to find P (x, t) and Q(x, t) which satisfy conditions (1), (2), where
P (x, t) is the pressure and Q(x, t) is the gas volume at each point (x, t) in the
pipe.
102 Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 2
Time Frequency Method of Solving One Boundary Value Problem...
We will search the solution of problem (1)–(3) in the form of [15, 21–23] as
following from [24],
{
P1(x, t) =
∑∞
k=0 P1k(t)xk
k! ,
Q1(x, t) =
∑∞
k=0 Q1k(t)xk
k! , x > 0, t ∈ R,
(4)
which satisfies the boundary condition (2). Differentiating (4) and substituting
it into (1), for the unknown functions P1k(t) and Q1k(t), for k > 0, we have:
P1,2k(t) =
(
d
dt + 2a1
)k P
(k)
10 (t)
ck
1
,
Q1,2k(t) =
(
d
dt + 2a1
)k Q
(k)
10 (t)
ck
1
, k ≥ 0,
(5)
and
P1,2k+1(t) = − (
d
dt + 2a1
)k+1 Q
(k)
10 (t)
F1 ck
1
,
Q1,2k+1(t) = − (
d
dt + 2a1
)k F1P
(k+1)
10 (t)
ck+1
1
, k ≥ 0.
(6)
Introduce the Laplace transforms [25]:
{
P̃10(ρ) =
∫∞
0 e−ρtP10(t)dt,
Q̃10(ρ) =
∫∞
0 e−ρtQ10(t)dt,
(7)
allowing one to recover P10 and Q10 by the the formulas
{
P10(t) = 1
2πi
∫
L1
eρtP̃10(ρ)dρ,
Q10(t) = 1
2πi
∫
L1
eρtQ̃10(ρ)dρ,
(8)
where L1 = ρ : Reρ = σ1 > 0. Then, taking into account (8), in (5) and (6) we
obtain {
P1,2k(t) = 1
2πi
∫
L1
eρt(ρ + 2a1)kρkP̃10(ρ)dρ
ck
1
,
Q1,2k(t) = 1
2πi
∫
L1
eρt(ρ + 2a1)kρkQ̃10(ρ)dρ
ck
1
(9)
and
P1,2k+1(t) = − 1
2πi
∫
L1
eρt(ρ + 2a1)k+1ρkQ̃10(ρ) dρ
F1ck
1
,
Q1,2k+1(t) = − 1
2πi
∫
L1
eρt(ρ + 2a1)kF1ρ
k+1P̃10(ρ) dρ
ck+1
1
.
(10)
Thus, in (9), (10) and (4), we have the following analytic expressions for the
solution of equation (1):
P1(x, t) =
∑∞
k=0 P1,2k(t) x2k
(2k)! +
∑∞
k=0 P1,2k+1(t) x2k+1
(2k+1)!
= 1
2πi
∫
L1
eρt
{
P̃10(ρ)ch
(√
ρ(ρ+2a1)
c1
x
)
− Q̃10(ρ)
F1
√
c1(ρ+2a1)
ρ sh
(√
ρ(ρ+2a1)
c1
x
)}
dρ,
(11)
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 2 103
F.A. Aliev, N.A. Aliev, and A.P. Guliev
Q1(x, t) =
∑∞
k=0 Q1,2k(t) x2k
(2k)! +
∑∞
k=0 Q1,2k+1(t) x2k+1
(2k+1)!
= 1
2πi
∫
L1
eρt
{
Q̃10(ρ)ch
(√
ρ(ρ+2a1)
c1
x
)
− P̃10(ρ) F1
√
ρ
c1(ρ+2a1) sh
(√
ρ(ρ+2a1)
c1
x
)}
dρ,
(12)
which can be formulated as the theorem below.
Theorem 1. If F1, a1, c1 are constant, P10(t) and Q10(t) are infinitely dif-
ferentiable functions, then BVP (1)-(2) has a solution analytically represented in
(11) and (12). The proof of Theorem 1 follows from (11) and (12).
From (11) and (12) at x = l, we define P1(l − 0, t) and Q1(l − 0, t) in the
form:
P1(l − 0, t) = P1(l, t) = 1
2πi
∫
L1
eρt
{
P̃10(ρ)ch(µl)− Q̃10(ρ)
F1
√
c1(ρ+2a1)
ρ sh(µl)
}
dρ,
Q1(l − 0, t) = Q1(l, t) = 1
2πi
∫
L1
eρt
{
Q̃10(ρ)ch(µl)−P̃10(ρ)F1
√
ρ
c1(ρ+2a1) sh(µl)
}
dρ,
(13)
where (µ) =
(√
ρ(ρ+2a1)
c1
l
)
.
2. The Solution of Equation (1) in the Lift x > l
Now we pass from the end of the first stage to the initial position of the second
stage, i.e., let the hyperbolic equation (1) at (l, t) have discontinuity [26],
{
P2(l + 0, t)=P2(l, t) = F 1
δ P1(l − 0, t) + χ1(P1(l − 0, t), α1, α2, α3)P̄ (t) ≡ P20(t),
Q2(l + 0, t)=Q2(l, t) = F 2
δ Q1(l − 0, t) + χ2(Q1(l − 0, t), β1, β2, β3)Q̄(t)≡Q20(t),
(14)
where χ1(·), χ2(·) are continuous functions of their arguments, P̄ (t), Q̄(t) are
given continuous functions that are a possible balance of oil products [9], F i
δ(i =
1, 2), αk, βk(k = 1, 3) are real numbers to be determined by using the corre-
sponding identification problems for creation of gas-liquid mixture in the well
bottom zone.
Now we pass to the second stage. Let us consider the boundary value problem
(1)–(3) for a system of equations of hyperbolic type
{
−F2
∂P2(x,t)
∂x = ∂Q2(x,t )
∂t + 2a2Q2(x, t),
−F2
∂P2( x,t)
∂t = c2
∂Q2(x,t)
∂x , x > l, t ∈ R,
(15)
with the boundary conditions
{
P2(l, t) = P20(t),
Q2(l, t) = Q20(t), t ∈ R,
(16)
104 Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 2
Time Frequency Method of Solving One Boundary Value Problem...
where P20(t) and Q20(t) are given in the form of (14). Then system (15) and
unknowns P2(·), Q2(·) are similar to those of the previous case, and a2, similarly
to k in (3), will be
2a2 =
g
ω2
+
λω2
2D2
. (17)
Analogously to the previous case from Sec. 2, the solution of the boundary
problem (15)–(17) will be sought in the form of [15]:
{
P2(x, t) =
∑∞
k=0 P2k(t)xk
k! ,
Q2(x, t) =
∑∞
k=0 Q2k(t)xk
k! , x > l, t ∈ R,
(18)
for which the boundary conditions (16) are satisfied automatically, and P2k(t)
and Q2k(t) are to be defined for k > 0. If not to repeat the calculations given
above for solving (18), we can get the following expressions:
P2(x, t) = 1
2πi
∫
L2
ert
{
P̃20(r)ch(µx)− Q̃20(r)
F2
√
c2(r+2a2)
r sh(µx)
}
dr,
Q2(x, t) = 1
2πi
∫
L2
ert
{
Q̃20(r)ch(µx)− P̃20(r)F2
√
r
c2(r+2a2) sh(µx)
}
dr,
(19)
where µ =
(√
r(r+2a2)
c2
x,
)
, which, for x = 2l, have the form
P2(2l, t) = 1
2πi
∫
L2
ert
{
P̃20(r)ch(µ2l)− Q̃20(r)
F2
√
c2(r+2a2)
r sh(µ2l)
}
dr,
Q2(2l, t) = 1
2πi
∫
L2
ert
{
Q̃20(r)ch(µ2l)− P̃20(r)F2
(√
r
c2(r+2a2)
)
sh(µ2l)
}
dr,
(20)
where µ =
(√
r(r+2a2)
c2
2l
)
, L2, like L1, is a Laplace line, i.e., if r ∈ L2, then
Rer = σ2 > 0. Thus, the following theorem is proved:
Theorem 2. Under the conditions of Theorem 1, if F2, a2, c2 are con-
stant numbers, P20(t) and Q20(t) are infinitely differentiable functions, then BVP
(15)–(16) has the solutions represented in the analytical form (19).
3. Simplified Scheme
It should be noted that if the boundary functions P0(t), Q0(t) have an expo-
nential form, then the solutions (11), (12) and (19) become much more simplified.
In fact, let {
P10(t) = α1e
ν1t,
Q10(t) = β1e
µ1t, t ∈ R.
(21)
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 2 105
F.A. Aliev, N.A. Aliev, and A.P. Guliev
Then
P̃10(ρ) =
∞∫
0
e−ρtP10(t)dt =
∞∫
0
α1e
ν1t e−ρtdt = α1
∞∫
0
e−(ρ−ν1)tdt,
where the constants α1, β1, ν1, µ1 are given real numbers. If Reρ = σ1 > ν1,
then {
P̃10(ρ) = α1
e−(ρ−ν1)t
−(ρ−ν1)
∣∣∣
∞
t=0
= α1
ρ−ν1
,
Q̃10(ρ) = β1
ρ−µ1
.
(22)
Substituting (22) into (11), (12) and taking into account the analyticity of the
functions ch
(√
ρ(ρ+2a1)
c1
x
)
and
√
c1(ρ+2a1)
ρ sh
(√
ρ(ρ+2a1)
c1
x
)
, via the variable ρ
we find
P1(x, t) = α1e
ν1tch
(√
ν1(ν1+2a1)
c1
x
)
− β1e
µ1t 1
F1
√
c1(µ1+2a1)
µ1
sh
(√
µ1(µ1+2a1)
c1
x
)
,
Q1(x, t) = β1e
µ1tch
(√
µ1(µ1+2a1)
c1
x
)
− α1e
ν1tF1
√
ν1
c1(ν1+2a1) sh
(√
ν1(ν1+2a1)
c1
x
)
.
(23)
Thus, expression (13) finally takes the form
P1(l, t) = α1e
ν1tch
(√
ν1(ν1+2a1)
c1
l
)
− β1e
µ1t 1
F1
√
c1(µ1+2a1)
µ1
sh
(√
µ1(µ1+2a1)
c1
l
)
,
Q1(l, t) = β1e
µ1tch
(√
µ1(µ1+2a1)
c1
l
)
− α1e
ν1tF1
√
c1(ν1+2a1)
ν1
sh
(√
ν1(ν1+2a1)
c1
l
)
.
(24)
Now we suggest that from (14) we have
{
P20(t) = P 2
1 (l, t) · P2,
Q20(t) = −Q2
1(l, t) Q2,
(25)
where P2 > 0 and Q2 > 0 are constant numbers. Then
P̃20(r) =
∫∞
0 e−rtP20(t)dt = P2α
2
1ch
2
(√
ν1(ν1+2a1)
c1
l
)
1
r−2ν1
−2α1β1P2
1
F1
√
c1(µ1+2a1)
µ1
sh
(√
µ1(µ1+2a1)
c1
l
)
ch
(√
ν1(ν1+2a1)
c1
l
)
× 1
r−(ν1+µ1) + P2β
2
1
1
F 2
1
c1(µ1+2a1)
µ1
sh2
(√
µ1(µ1+2a1)
c1
l
)
1
r−2µ1
,
(26)
106 Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 2
Time Frequency Method of Solving One Boundary Value Problem...
Q̃20(r) =
∫∞
0 e−rtQ20(t)dt = −Q2β
2
1ch2
(√
µ1(µ1+2a1)
c1
l
)
1
r−2µ1
+2α1β1Q2F1
√
ν1
c1(ν1+2a1) sh
(√
ν1(ν1+2a1)
c1
l
)
ch
(√
µ1(µ1+2a1)
c1
l
)
1
r−(ν1+µ1)
−Q2α
2
1F
2
1
ν1
c1(µ1+2a1) sh2
(√
ν1(ν1+2a1)
c1
l
)
1
r−2ν1
.
(27)
Here it is supposed that Re r > max {2ν1, (ν1 + µ1), 2µ1} . At last, from (20), we
obtain
P2(2l, t) = P2α
2
1ch
(√
2ν1 2(ν1+a2)
c2
2l
)
ch2
(√
ν1(ν1+2a1)
c1
l
)
e2ν1t
−2α1β1P2
1
F1
√
c1(µ1+2a1)
µ1
sh
(√
µ1(µ1+2a1)
c1
l
)
ch
(√
ν1(ν1+2a1)
c1
l
)
×ch
(√
(ν1+µ1)(ν1+µ1+2a1)
c2
2l
)
e(ν1+µ1)t + P2β
2
1
1
F 2
1
c1(µ1+2a1)
µ1
×sh2
(√
µ1(µ1+2a1)
c1
l
)
ch
(√
2µ1 2(µ1+a2)
c2
2l
)
e2µ1t + Q2β
2
1
1
F2
√
c2(µ1+a2)
µ1
×sh
(√
2µ1 2(µ1+a2)
c2
2l
)
ch2
(√
µ1(µ1+2a1)
c1
l
)
e2µ1t − 2α1β1Q2
F1
F2
×
√
ν1
c1(ν1+2a1) sh
(√
ν1(ν1+2a1)
c1
l
)
ch
(√
µ1(µ1+2a1)
c1
l
)√
c2(ν1+µ1+2a2)
ν1+µ1
×sh
(√
(ν1+µ1)(ν1+µ1+2a2)
c2
2l
)
e(ν1+µ1)t + Q2α
2
1F
2
1
ν1
c1(ν1+2a1)
×sh2
(√
ν1(ν1+2a1)
c1
l
)
1
F2
√
c2(ν1+a2)
ν1
sh
(√
2ν1 2(ν1+a2)
c2
2l
)
e2ν1t,
(28)
Q2(2l, t) = −Q2β
2
1ch2
(√
µ1(µ1+2a1)
c1
l
)
ch
(√
2µ1 2(µ1+a2)
c2
2l
)
e2µ1t
+2α1β1Q2F1
√
ν1
c1(ν1+2a1) sh
(√
ν1(ν1+2a1)
c1
l
)
ch
(√
µ1(µ1+2a1)
c1
l
)
×ch
(√
(ν1+µ1)(ν1+µ1+2a2)
c2
2l
)
e(ν1+µ1)t −Q2α
2
1F
2
1
ν1
c1(ν1+2a1)
×sh2
(√
ν1(ν1+2a1)
c1
l
)
ch
(√
2ν1 2(ν1+a2)
c2
2l
)
e2ν1t − P2α
2
1
×ch2
(√
ν1(ν1+2a1)
c1
l
)
F2
(√
ν1
c2(ν1+a2)
)
sh
(√
2ν1 2(ν1+a2)
c2
2l
)
e2ν1t
+2α1β1P2
1
F1
√
c1(µ1+2a1)
µ1
sh
(√
µ1(µ1+2a1)
c1
l
)
ch
(√
ν1(ν1+2a1)
c1
l
)
×F2
(√
ν1+µ1
c2(ν1+µ1+2a2)
)
sh
(√
(ν1+µ1)(ν1+µ1+2a2)
c2
2l
)
e(ν1+µ1)t
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 2 107
F.A. Aliev, N.A. Aliev, and A.P. Guliev
−P2β
2
1
1
F 2
1
c1(µ1+2a1)
µ1
sh2
(√
µ1(µ1+2a1)
c1
l
)
F2
(√
µ1
c2(µ1+a2)
)
×sh
(√
2µ1 2(µ1+a2)
c2
2l
)
e2µ1t,
(29)
which proves the theorem below.
Theorem 3. If Fi, ai, ci, i = 1, 2, are constant numbers, P10(t) and Q10(t)
have the form of (21), P20(t) and Q20(t) have the form of (25), α1, β1, ν1,
µ1, P2, Q2 are constant numbers, then (28) and (29) hold true for solving BVP
(1)–(3), (14)–(17).
4. Determination of Coefficient of Hydraulic Resistance [19, 20]
Now we use the relationships (28) and (29) to define λ, the coefficient of
hydraulic resistance [19, 20] (CHR), from (16), i.e., to define the pollution dy-
namics in the lift. Note that this problem is essentially the inverse problem to
that considered for parabolic systems in [27].
Let statistical data for the volume of the supplied gas Q̃10(t) from (2) and
debit Q̃(2l, t) be given. Then writing the functional in the form
RQ(λ) =
T∫
0
[
Q(2l, t)− Q̃(2l, t)
]2
dt, (30)
we have to find λ∗ such that the functional R(λ) gets its minimum. Therefore,
we calculate the first variation of the functional (30) and make it equal zero
T∫
0
[
Q(2l, t)− Q̃(2l, t)
]
Q′
λ(2l, t)dt = 0. (31)
Taking into account (29) and (31), for defining the CHR λ we can obtain
a transcendent equation which allows us to find a numerical meaning of the
optimal λ∗.
E x a m p l e. Let us consider BVP (1), (2) (see, [8]), where the boundary
data P10(t) = P0, Q10(t) = Q0 from (2) are constant. While in [15], in the
domain [0, l]× [0, T ],
P (x, t) = P0 − 2
a1
F1
Q0x, Q(x, t) = Q0 (32)
and at the point (l − 0, t),
P (l − 0, t) = P0 − 2
a1
F1
Q0l, Q(l − 0, t) = Q0.
108 Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 2
Time Frequency Method of Solving One Boundary Value Problem...
Suggesting that in (14)
P̄ (t) = P̄ , Q̄(t) = Q̄, α1 = −1, α2 = α3 = 0, β1 = 1, β2 = β3 = 0,
F 1
δ = F 2
δ = 1, χ1 = P 2(l − 0, t), χ2 = −Q2(l − 0, t),
at the point (l + 0, t) we have (boundary layer [8])
P (l + 0, t) = P (l − 0, t) + P 2(l − 0, t)P̄ = Pl+0 = const,
Q(l + 0, t) = Q(l − 0, t)−Q2(l − 0, t)Q̄ = Ql+0 = const.
(33)
Similarly to the above, it is easy to show that in the domain [l, 2l]× [0, T ] ,
P (x, t) = Pl+0 − 2a2
F2
Ql+0x, Q(x, t) = Ql+0. (34)
According to (34), Q(x, t) is constant on [l, 2l]× [0, T ]. This makes it difficult to
find the CHR λ. Besides, the expression for P (x, t) contains a2(λ). This requires
the necessity to have an additional information on the history of the pressure
both at x = 0 (gas supply) and at x = 2l (gas measurement). Let
P̃ (0, t) = P̃0(0) = P̃0 , P̃ (2l, t) = P̃2l . (35)
Let us write the functional (30) for the pressure P (x, t),
RP (λ) =
T∫
0
[
P (2l, t)− P̃ (2l, t)
]2
dt, (36)
where at the point x = 2l from (34) we have
P (2l, t) = Pl+0 − 2a2
F2
Ql+0 2l. (37)
Taking into account that there is no dependence from t either in (35) or (37), the
functional (36) takes the form
RP (λ) = (P (2l, t)− P̃2l)2T = (Pl+0 − 2a2(λ)
F2
Ql+0 l − P̃2l)2T. (38)
Now, similarly as in (31), we have the gradient of the functional in the form
a2(λ) =
F2
4lQ(l + 0)
(Pl+0 − P̃2l). (39)
Taking into account (39), in (17), for λ, we have
λ = −2D2g
ω0 ω2
+
F2D2
2ω2lQ(l + 0)
(Pl+0 − P̃2l). (40)
Journal of Mathematical Physics, Analysis, Geometry, 2016, vol. 12, No. 2 109
F.A. Aliev, N.A. Aliev, and A.P. Guliev
Now we will show the above on the numerical example. Let for 0 ≤ x < l, l =
1485m, c1 = 331m/c, ρ1 = 0, 717 : 3/m3, F1 = 0.006 , λ1 = 0.01, D1 =
0.0876, g = 9.81,ω1 = 48.6390, and for l ≤ x < 2l,
c = 850, D2 = 0.073, F2 = 0.0042, ω2 = 0.0546, λ2 = 0.23, Q̄ = 100, P̄ = 100.
(41)
Performing the corresponding calculations using the values of the latter parame-
ters, we have:
P0 − 2a1
F1
Q0l = −2.466 · 103,
Pl+0 = Pl−0 + P 2
l−0 · P̄ = 1.3494 · 1015,
Pl−0 = P0 − 2
a1
F1
Q0l = −3.6734 · 106,
Ql+0 = Q0 −Q2
0Q̄ = −2495,
Q0 = 0.001, P̄ = 0.00135, P̃ (2l, t) = P̃2l = 1.34 · 1015.
Then, for the value P̃2l = 1.34 · 1015 from (40), we can find λ∗ in the form
λ∗ = 2.2999 · 10−1 which coincides with λ2 from (41) with an accuracy of 10−3.
Thus, the stated problems can be extended to the more difficult problem (1)–(3).
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|
| id | nasplib_isofts_kiev_ua-123456789-140549 |
| institution | Digital Library of Periodicals of National Academy of Sciences of Ukraine |
| issn | 1812-9471 |
| language | English |
| last_indexed | 2025-12-07T15:22:08Z |
| publishDate | 2016 |
| publisher | Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України |
| record_format | dspace |
| spelling | Aliev, F.A. Aliev, N.A. Guliev, A.P. 2018-07-10T13:53:53Z 2018-07-10T13:53:53Z 2016 Time Frequency Method of Solving One Boundary Value Problem for a Hyperbolic System and Its Application to the Oil Extraction / F.A. Aliev, N.A. Aliev, A.P. Guliev // Журнал математической физики, анализа, геометрии. — 2016. — Т. 12, № 2. — С. 101-112. — Бібліогр.: 27 назв. — англ. 1812-9471 DOI: doi.org/10.15407/mag12.02.101 Mathematics Subject Classification 2000: 65M38, 35L02, 35L40, 58J45, 58J90 https://nasplib.isofts.kiev.ua/handle/123456789/140549 We consider the boundary value problem, where the motion of the object is described by the two-dimensional linear system of partial differential equations of hyperbolic type where a discontinuity is at a point within the interval that defines the phase coordinate x. Using the method of series and Laplace transformation in time t (time-frequency method), an analytical solution is found for the determination of debit Q(2l, t) and pressure P(2l, t), which can be effective in the calculation of the coefficient of hydraulic resistance in the lift at oil extraction by gas lift method where l is the well depth. For the case where the boundary functions are of exponential form, the formulas for P(2l, t) and Q(2l, t) depending only on t are obtained. It is shown that at constant boundary functions, these formulas allow us to determine the coefficient of hydraulic resistance in the lift of gas lift wells, which determines the change in the dynamics of pollution. en Фізико-технічний інститут низьких температур ім. Б.І. Вєркіна НАН України Журнал математической физики, анализа, геометрии Time Frequency Method of Solving One Boundary Value Problem for a Hyperbolic System and Its Application to the Oil Extraction Article published earlier |
| spellingShingle | Time Frequency Method of Solving One Boundary Value Problem for a Hyperbolic System and Its Application to the Oil Extraction Aliev, F.A. Aliev, N.A. Guliev, A.P. |
| title | Time Frequency Method of Solving One Boundary Value Problem for a Hyperbolic System and Its Application to the Oil Extraction |
| title_full | Time Frequency Method of Solving One Boundary Value Problem for a Hyperbolic System and Its Application to the Oil Extraction |
| title_fullStr | Time Frequency Method of Solving One Boundary Value Problem for a Hyperbolic System and Its Application to the Oil Extraction |
| title_full_unstemmed | Time Frequency Method of Solving One Boundary Value Problem for a Hyperbolic System and Its Application to the Oil Extraction |
| title_short | Time Frequency Method of Solving One Boundary Value Problem for a Hyperbolic System and Its Application to the Oil Extraction |
| title_sort | time frequency method of solving one boundary value problem for a hyperbolic system and its application to the oil extraction |
| url | https://nasplib.isofts.kiev.ua/handle/123456789/140549 |
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