Региональный №1(65)2015 гуманит indd Тіркеу нөмірі 204-ж
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- Бұл бет үшін навигация:
- Региональный вестник Востока
- Statement of the problem and finite-differential equations
- Шығыстың аймақтық хабаршысы
- Uniqueness condition of solutions.
F.S. AMENOVA S. Amanzholov East-Kazakhstan State University, Ust-Kamenogorsk, Kazakhstan THE METHOD TO SOLvE iNCOMPRESSiBLE FLUiD EQUATiONS iN ( )
Ψ , vARiABLES in this paper, we consider the question of convergence of solving two-dimensional op- erator-difference problem for incompressible fluid in variables «stream function-vortisity» to solving differential problem. For error of solution estimation of convergence rate was got. iteration algorithm for numerical solution of operator-differenсe equations are examined. Es- timation of convergence rate for iteration algorithm is got. ТеХНИКА, ТеХНОлОГИЯ ЖӘНе ФИЗИКАлыҚ-МАТеМАТИКАлыҚ ҒылыМдАР 11 Региональный вестник Востока
Выпускается ежеквартально Keywords: difference problem, iteration algorithms, estimation of convergence rate. ( ) Ω Ψ , АйНыМАлылАРыНдА СИПАТТАлҒАН СыҒылМАйТыН СҰйыҚТыҚТАРдың ТеңдеУлеРіН ШеШУ ӘдіСі Мақалада «ағын және құйын» айнымалыларында сипатталған сығылмайтын сұйықтықтардың екіөлшемді операторлы-айырымдық есебі шешімінің дифференциалдық есеп шешіміне жинақталу сұрағы қарастырылған. Шешімнің қателігі үшін жинақталу жылдамдығының бағасы алынған. Операторлы-айырымдық есепті сандық шешу үшін итерациялық алгоритм қарастырылған. Қарастырылған итерациялық алгоритмдер үшін жинақталу жылдамдықтары анықталған. түйін сөздер: айырымдық есеп, итерациялық алгоритм, жинақтылық жылдамдығының бағасы. MеТОд РеШеНИЯ УРАВНеНИй НеСЖИМАеМОй ЖИдКОСТИ В ПеРеМеННыХ ( )
Ψ , В данной работе рассматривается вопрос о сходимости решения двумерной операторно-разностной задачи для несжимаемой жидкости в переменных «функция тока, вихрь скорости» к решению дифференциальной задачи. для погрешности реше- ния получена оценка скорости сходимости. Рассмотрен итерационный алгоритм для численного решения операторно-разностных уравнений. Получена оценка скорости сходимости итерационного алгоритма.
сходимости. Sufficient number of scientific publications devoted to issues of numerical so- lution of two-dimensional boundary problems for incompressible fluid equations in “stream function, vorticiy” variables. Descriptions of the most famous computing technologies used at conducting computational experiments for studying different flows of incompressible fluid can be found in monographs [1-3]. it is known, that major difficulties arising during numerical solution of Navier-Stokes equations for incompressible fluid related to implementation of boundary conditions for vorticity. Generally, in practice for finding values of vorticity at the boundary formulas which are approximating adhesion and speed components non-flowing conditions in physical formulation of considered problems [1, 5]. The most famous among them are Tom and Woods formulas having first and second order of accuracy respectively to determine vorticiy at the boundary [1, 5]. Sufficient number of papers devoted to theoretical and practical issues of usage of Tom’s formula at calculations of incompressible fluid flow and most recent of them [6-10]. Statement of the problem and finite-differential equations: in square domain D={0 ≤ x, y ≤ 1} let us study the following system of Navier-Stokes steady-state equa- tions in variables stream function, velocity curl for incompressible fluid [1]: F.S. AMENOvA. 1 (65) 2015. P. 10-20
iSSN 1683-1667 12 Тоқсанына бір рет шығарылады
Шығыстың аймақтық хабаршысы
, ) , ( - y x f x y y x + Ω ∆ = Ψ Ω Ψ Ω ν ∂ ∂ ∂ ∂
(1)
) y , (
, D x ∈ Ω = Ψ ∆ (2)
with following boundary conditions
, 0 = Ψ = Ψ
n ∂ ∂ ∂
(3) where 0 > ν is viscosity coefficient, n is outer normal to domain boundary, ∆ is two-
dimensional Laplace operator, ( )
y x f , is some given function, Ψ is stream function, Ω is velocity curl. For approximation of equations (1), (2) in finite-difference domain ( ) { } 1 , 1 , , , 2 1 − ∈ = N m k h m h k D h , where 1
and
2 h are grid steps in x and
y di-
rections, respectively, we examine the following scheme on symmetrical pattern
, ) (
L h h + Ω ∆ = Ψ Ω ν
(4)
, Ω = Ψ ∆
(5)
where h L is difference operator, which complies with respective approximation of convectional summands of equation (1). The difficulty to examine equation of incompressible fluid in variables ) ,
Ψ
is caused by the absence of boundary conditions for velocity curl while statement of differential problem. Tom’s work [5] was the first to offer a formula of first order of accuracy to define the value of the curl on the wall and it is still used up to now. [6-10]. in our investigation the boundary conditions for velocity curl are taken in the form of Tom’s formulas
1
1 , 2 , 2 , 1 0 , 1 0 − = Ψ − = Ω Ψ = Ω
m h h m N x m N m x m ,
1 , 1 , 2 , 2 , 2 , 2 0 − = Ψ − = Ω Ψ = Ω N k h h N k y N k o k y k
. (6) ENGiNEERiNG, TECHNOLOGy, PHySiCAL AND MATHEMATiCAL SCiENCES 13 Региональный вестник Востока
Выпускается ежеквартально in what follows, it is assumed that the solution of differential problem (1)-(3) has enough smoothness and for difference operator h L the following conditions are satisfied
, ) , ) ( ( 0 v u w ñ v u w L h h h ∆ ∆ ≤
( ) , v , , 0 ) , ) ( ( h h h D u u u w L Ω ∈ ∀ =
(7) where
0 0 > c is a uniformly bounded constant. We observe that for difference problem (4)-(6) the following estimation is cor- rect
f h ≤ Ψ ∆ ν . Uniqueness condition of solutions. Let us show that at condition 1 2 0 < ν
c the solution of the problem (4)-(6) will be unique. Assume that we have two solutions ( ) 1 1 ,Ω Ψ and
( ) 2 2 ,Ω Ψ . Then for differences 2 1 2 1 , Ω − Ω = Ψ − Ψ = Φ Z we have differential problem: , )
) ( 2 1 Z Z L L h h h ∆ = Ψ + Φ Ω ν , Z h = Φ ∆ with boundary conditions , 0
0 = Φ = Φ = Φ = Φ N k k m N m
y kN k y k h Z h Z , 2 0 , 2 0 2 , 2 Φ − = Φ = . N k y N k k y k h Z h Z , 2 0 , 2 0 2 , 2 Φ − = Φ = We have ( ) 2 2 0 2 2 , ) ( Ψ ∆ Φ ∆ ≤ Φ Ψ ≤ Φ ∆ h h h h c Z L ν , ( ) 0 2 2 0 ≤ Φ ∆ Ψ ∆ − h h c ν . Hence, if 0 2 0 > Ψ ∆ −
c ν or 1 2 0 < ν
c , F.S. AMENOvA. 1 (65) 2015. P. 10-20 iSSN 1683-1667 14 Тоқсанына бір рет шығарылады
Шығыстың аймақтық хабаршысы then it should be 0 =
∆ h , i.e. the solution is unique. About convergence of difference problem. We have the following equations
,
( ) ( h h h h h R Z Z L L + ∆ = Ψ + Φ Ω ν (8)
h h Q Z + = Φ ∆ , (9) with boundary conditions ( )
( ) h D y x y x ∂ ∈ = Φ , , 0 , , , 2 , 2 , 1 0 0 , 1 0 m N m N x m N m m x m r h Z r h Z + Φ − = + Φ =
, 2 , 2 , 2 0 0 , 2 0 N k N k y N k k k y k r h Z r h Z + Φ − = + Φ =
(10) where
h h Q R , are residuals of difference equations (4) and (5), respectively, which are defined in the following way ( ) ( )
h h h h h h h D y x y x f L R ∈ + Ω ∆ − Ψ Ω = , , , ) ( ν , h h h h Q Ω − Ψ ∆ = . Currently we will multiply relation (8) by Φ and summarize on the units of grid h D . By doing so we will obtain the following main energy identical equation ( ) (
) ( )
( ) Φ Ψ = Ω + Φ ∆ , , , Z L R Z h h h ν . Taking into account relations (9), (10) and by using summation by part formulas, we have
( ) ( ) ( ) ( ) = Φ + Φ − Φ + Φ − Φ + Φ ∆ ∑ ∑ − = − = 1 1 1 1 1 , 0 , 0 2 , 0 , 0 , , N m N k h N k y N k k y k m N x m N m x m h R h Z Z h Z Z Z ν ν ν
( ) ( ) , ,Φ Ψ = Z L h ENGiNEERiNG, TECHNOLOGy, PHySiCAL AND MATHEMATiCAL SCiENCES 15 Региональный вестник Востока
Выпускается ежеквартально ( )
Φ + Φ − − Φ + Φ + Φ ∆ − Φ ∆ ∑ − = 2 1 1 , , 1 0 , 0 0 , 1 2 2 ,
r h r h Q N m m N x m N m N x m x m m x h h h ν ν ( ) ( ) ( ) , , , 2 2 1 1 1 , , 2 0 , 0 0 , 2 Φ Ψ = Φ + Φ + Φ − − Φ + Φ + ∑ − = Z L R h r h r h h h N k N k y N k N k y k y k k y ν
After simple transformation we will obtain the energy identical equation ∑ ∑ − = − = = Φ + Φ + Φ + Φ + Φ ∆ 1 1 1 2 , 2 0 , 2 1 1 2 2 , 2 0 , 1 2 2 2 N k N k y k y N m m N x m x h h h h h ν ν ν ( ) ( ) ∑ ∑ − = − = + Φ + Φ + Φ + Φ + Φ ∆ 1 1 1 , 0 0 , 1 1 2 2 , 0 0 , ,
k N k N k y k k y N m m N m N x m m x h h h r r h r r Q ν ν ν
( ) ( ) ( ) Φ Ψ + Φ + , , Z L R h h . (11) Summands of energy identical equation (11), by using inequality (7), ε - in- equality we estimate in the following way ( ) , 4 1 , 2 1 2 1
h h h h h Q Q Q ε ε + Φ ∆ ≤ Φ ∆ ⋅ ≤ Φ ∆ ( ) , 4 1 , 2 2 2 2
h h h h R R R ε ε + Φ ∆ ≤ Φ ∆ ⋅ ≤ Φ ( ) ( ) = Φ ∆ ⋅ Ψ ∆ + Φ ∆ ≤ Φ ∆ ⋅ Ψ ∆ ⋅ ≤ Φ Ψ
h h h h h h Q c Z c Z L 0 0 , ) ( ( ) ( ) , 4 1 1 2 3 2 3 0 2 0 + Φ ∆ + Ψ ∆ ≤ Φ ∆ ⋅ + Φ ∆ Ψ ∆ = h h h h h h h Q c Q c ε ε . 4 1 1 1 2 2 0 1 4 1 2 0 , 4 1 1 1 2 0 0 , 1 1 1 2 0 0 , ∑ ∑ ∑ − = − = − = + Φ ≤ Φ = Φ N m m m x N m m m x N m m m x h r h h h h r h h r ε ε ν ν ν By means of inequalities obtained above the identical equation is estimated F.S. AMENOvA. 1 (65) 2015. P. 10-20
iSSN 1683-1667 16 Тоқсанына бір рет шығарылады
Шығыстың аймақтық хабаршысы (11):
( ) ( ) + Ψ ∆ + + ≤ Φ ∆ Ψ ∆ + − − − 2 3 0 2 2 2 1 2 3 0 2 1 4 4 1 4 1
h h h h h Q c R Q c ε ε ε ν ε ε ε ν ν
. 4 4 1 1 1 2 2 0 5 2 1 1 2 2 2 0 4 1 ∑ ∑ − = − = + + + + N k N k k N m m N m h r r h h r r h ε ν ε ν
if we assume that ( ) , 0 1 0 3 0 2 1 > ≥ Ψ ∆ + − − − δ ε ε ε ν ν h c where
0 δ is some constant, then , 3 2 0 h M h ⋅ ≤ Φ ∆ δ where М is a uniformly bounded constant, thus . 2
0 h c h ⋅ ≤ Φ ∆ Further, we consider the questions of numerical solution for difference problem (4)-(6). Direct use of Tom’s boundary conditions for numerical implementation of dif- ference equations (4), (5) causes the necessity to make the relaxation of boundary con- ditions. if we omit this procedure while observing stability conditions then divergence of iteration schemes will be discovered. The most advanced iteration methods are used for difference equations (4) and (5), obtained after the introduction of velocity curl complementary function with homogeneous boundary conditions on the border. Fol- lowing this, we will write the following boundary conditions and equations (4), (5)
( ) , , 0 0 0 0 y x f A h h y x x y + Ψ + Ω ∆ = Ψ Ω − Ψ Ω ν
(12)
( ) ( ) , , , , h h D y x y x ∈ Ω = Ψ ∆ (13)
( )
( ) ( )
, , , 0 , , h D y x y x y x ∂ ∈ = Ω = Ψ
(14) ENGiNEERiNG, TECHNOLOGy, PHySiCAL AND MATHEMATiCAL SCiENCES 17 Региональный вестник Востока
Выпускается ежеквартально where (
( ) − = − = Ψ + − = − = Ψ + = Ψ − − , 1 , 1 , 1 , 1 , 2 , 1 , 1 , 1 , 1 , 2 1 , 1 , 4 2 1 , 1 , 4 1
m N k h N m N k h A km N m m km N k k m k h δ δ ν δ δ ν
= ≠ = . ,1 , , 0 ,
k m k m k δ
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