Scientific journal

European Journal of Natural History

ISSN 2073-4972

ИФ РИНЦ = 0,301

Let´s consider the using a method of fictitious area for the system of not evolutional type, which will be a model for us in filtering problem of two-phase incompressible fluid taking with capillary forces.

Let D - a certain plain simply connected area with enough smooth interface. In the cylinder D_T = {D_x[0 < t ≤ T]}, with side surface S = {γX[0 < t ≤ T]}, there is being searched the solution of mixed Cauchy problem:

 (1)

In (1) k = k(x) > 0, f_i = f_i(x, t), λ_i = const > 0, and necessary for justification of the method of fictitious areas additional requirements for input data of the problem (1) will be specified in the process of explanation. First of all we mention that if

2R = u₁ - u₂,

P = λ₁u₁ - λ₂u₂ (2)

The initial problem (1) is decomposed into two independent problems:

div(kgrand P) + f₁ + f₂ = 0, P_|s = 0 (3)

and

2R(x, 0) = ψ(x), R_|s = 0 (4)

For all that the time t comes in the problem (3) as a parameter Therefore, justification of the method of fictitious areas at the differential level might be carried out for each of the problems (3), (4). Also, it is necessary to note that if in the initial problem (1) instead of uniform boundary condition of the first type there will be examined the uniform boundary condition of the second type, so instead of P_|s = 0 и R_|s = 0 in (3) and (4) accordingly we will have

 (5)

For the problem (3), (4), if , , true estimates are:

 (6)

In accordance with the method of fictitious areas, let´s add the initial area D with a area D₁ up to the area D₀ = D u D₁, with boundary Γ = ∂D₁ u γ, S⁰ = {ΓX[0 < t ≤ T]}. In the compound area D₀ let´s study additional problems:

   (7)

 (8)

On the break curve S we lay down fitting conditions

 (9)

ε > 0 - series expansion parameter, Q - parameter, possessing the value: Q = 1 or Q = -1,

 (10)

For the problem (3), (4) α < 0, and for the same problems, but with boundary conditions (5) α > 0. Through  the normal derivative has been represented. Further , and the signs of minus and plus mean that counterpart is a limiting value by the tending of x to γ inside or outside of D. The auxiliary problems (6)-(8) have the transparent physical sense. The absolute permeability is small (α > 0) or big (α < 0) depending on the type of boundary condition of the initial problem in fictitious area. As regards the input data in the fictitious area D₁, R is an analog of capillary pressure and that´s why equality R to zero means that in the fictitious area there is not only displacing phase. The fitting condition in (6), (7) means that for the transfer through γ (γ - line of factors´ break) phase pressures and phase rates are continuous.

For the solution of the problem (8)-(10), (7), (9), (10) the true estimates are:

 (11)

 (12)

Where  и , и correspond to the solution of the problems (8)-(10), (7), (9), (10) by Q = 1 and Q = -1. Further the problem (8)-(10) is called the problem I, problem (7), (9), (10) - II.

Now the solution of the problem I we will search in the form of power series on the parameter ε, α = -1.

Let  в D_T,  в .

Where we put  formally in the problem I, so then we will get the system relatively to V_m and W_m:

(13)

By m ≤ 1

Let´s suppose that functions in (13) are met the conditions of , , k = 1, 2, ..., so the following theorem is right.

 (14)

Where R_ε - solution of the problem I.

Theorem proving. From the theory of uniform boundary problems and conditions of matching we have

 (15)

where constants C₁, C₂ depend on areas D, D₁ and factors of initial problem and don´t depend on ε.

Applying the theory of trails in Sobolev spaces

Then from (6) and (15) it follows

 (16)

where C₅ = C₁·C₂·C₃·C₄.

Let , thenseriesВ₁is absolutely converging in . For getting equalities (14) we multiply (13) by ε^m and sum on m, we have:

(17)

Si it follows that R_ε = B₁ in D_T, R_ε = B₂ in  by 0 < ε < ε₀.

From the theorem it follows unique existence of the solution of the additional problems (13), and the estimates are:

..(18)

Here it is  by Q = 1,  by Q = -1, from absolute convergence of the series B₁ and B₂ it follows (11), let´s bring it.

 (19)

Where R - solution (4), , - solution (8) by Q = 1 and Q = -1, correspondently.

Theorem proving. From the theorem 1it follows the following expansion:

 (20)

Here it is , - solution (13) by Q = 1.

Applying the theorem 1 for  it is true:

 (21)

Here it is , - solution (13) by Q = -1, it is easy to see that - solutions (4).

Let , so the function  satisfies the following problem:

So from this we have

 or

Let´s suppose that

 ,

so then the function  is the solution of the following problem:

From this we have

, then

Even it supposes that

, ,

we will get

, .

Continuing this process by m ≥ 2 we have:

, if m - even , if m - uneven (22)

The from (21), using (19), (20), we will get in D_T

 (23)

Using the expansion (22) and estimates (17) we have come to (11):

Estimate (12) is got in much the same way.

Accuracy of received bilateral approximation in this case is limited by the estimate (11), (12). If only to get bilateral estimates R, P, with specified accuracy ε^P, we will use the idea of Richardson extrapolation.

Let´s make extrapolated solutions , being a linear combination , with some weight:

 (24)

Concrete view off coefficients β_m depends on choice of sequence ε > ε₁ > ... > ε_p > 0 and accuracy figure p. The more spread choice is:

 (25)

By which coefficients β_m are in the explicit form

 (26)

And it satisfies the conditions

By this way of the task ε_m, β_m we find that

where .

In much the same way it is

Let p - uneven, then  and, it means,

(27)

With the help of this statement the theorem has been proved.

 (28)

Where p - uneven, and  are defined on the formula (26).

The same estimates are received for the function Р, and power (2) for the functions u₁, u₂.

The received results allow simulating the processes of oil extraction with the use of production and forcing wells for water blockage of formation under test.