The application of Kantorovich-Ritz method to the flow of an incompressible potential fluid between two solid walls

ictp
(original), html, JCR/ISI, paper, PDF, works

Abstract

Authors

Doina Bradeanu
Institutul de Calcul

Keywords

?

Paper coordinates

Brădeanu, The application of Kantorovich-Ritz method to the flow of an incompressible potential fluid between two solid walls, Rev. Anal. Numér. Théor. Approx. 25 (1996), no. 1-2, 23–31.

PDF

https://ictp.acad.ro/jnaat/journal/article/view/1996-vol25-nos1-2-art3/Bradeanu

About this paper

Journal

Revue d’analyse Numerique et de Theorie de l’approximation

Publisher Name

Romanian Academy

DOI

https://ictp.acad.ro/jnaat/journal/article/view/1996-vol25-nos1-2-art3

Print ISSN

2457-6794

Online ISSN

ISSN-E 2501-059X

google scholar link

[1] N. M Beliaev and A. A. Riadno, Methods of Theory of Heat Conduction, 1, Moscow, Vyshaia Skola,1982 (in Russian).
[2] P. Brădeanu, Mecanica fluidelor, Ed. Tehnică, Bucureşti, 1973.
[3] L E. Elsgolts, Differential Equations and the Calculus of Variations, Mir Publishers, Moscorw, English transl , 1977.
[4] L. V. Kantorovich and V.l. Krylov, Approximate Method of Higher Analzsis, Fizmatgiz, 1962 (in Russian).
[5] S G. Mikhlin, Variational Methods in Mathematical Physics, Nauka, Moscow, 1970 (in Russian).
[6] S. G. Mikhlin, Ecuaţii cu derivate parţiale, Ed. Ştiinţifică şi Enciclopedică, bucureşti, 1973.

Paper (preprint) in HTML form

jnaat,+Journal+manager,+1996-Bradeanu

THE APPLICATION OF KANTOROVICH-RITZ METHOD TO THE FLOW OF AN INCOMPRESSIBLE POTENTIAL FLUID BETWEEN TWO SOLID WALLS

DOINA BRÄDEANU(Cluj-Napoca)

1. THE FORMULATION OF THE PROBLEM

a) Differential problem. Consider the plane O x y O x y OxyO x yOxy and the bounded domain Ω ¯ = { ( x , y ) ∈ R 2 ∣ − x 0 ≤ x ≤ x 0 , 0 ≤ y ≤ y w ( x ) } Ω ¯ = ( x , y ) ∈ R 2 ∣ − x 0 ≤ x ≤ x 0 , 0 ≤ y ≤ y w ( x ) bar(Omega)={(x,y)inR^(2)∣-x_(0) <= x <= x_(0),0 <= y <= y_(w)(x)}\bar{\Omega}=\left\{(x, y) \in \mathbf{R}^{2} \mid-x_{0} \leq x \leq x_{0}, 0 \leq y \leq y_{w}(x)\right\}Ω¯={(x,y)∈R2∣−x0≤x≤x0,0≤y≤yw(x)}, where y = y w ( x ) y = y w ( x ) y=y_(w)(x)y=y_{w}(x)y=yw(x) is the given equation of the C B ― C B ¯ bar(CB)\overline{C B}CB― boundary, fig 1 . It is assumed that D A ― D A ¯ bar(DA)\overline{D A}DA― and C B ― C B ¯ bar(CB)\overline{C B}CB― boundaries are solid walls and that the ideal incompressible fluid between them irrotationally flows in the positive x x xxx direction with the velocity V → ( x , y ) = ( V x , V y ) V → ( x , y ) = V x , V y vec(V)(x,y)=(V_(x),V_(y))\vec{V}(x, y)=\left(V_{x}, V_{y}\right)V→(x,y)=(Vx,Vy). On D C ― D C ¯ bar(DC)\overline{D C}DC― and A B ― A B ¯ bar(AB)\overline{A B}AB― boundaries the velocity vector is constant and parallel to O x O x OxO xOx axis.
Fig. 1
With these assumptions, the stream function ψ ψ psi\psiψ satisfies Laplace's equation on Ω , Δ ψ = 0 Ω , Δ ψ = 0 Omega,Delta psi=0\Omega, \Delta \psi=0Ω,Δψ=0.
The boundary conditions on the solid walls are ψ ( x , 0 ) = ψ 0 = 0 ψ ( x , 0 ) = ψ 0 = 0 psi(x,0)=psi_(0)=0\psi(x, 0)=\psi_{0}=0ψ(x,0)=ψ0=0 and ψ ( x , y w ( x ) ) == Ψ w ( x ) = q ψ x , y w ( x ) == Ψ w ( x ) = q psi(x,y_(w)(x))==Psi_(w)(x)=q\psi\left(x, y_{w}(x)\right)= =\Psi_{w}(x)=qψ(x,yw(x))==Ψw(x)=q, where q q qqq is a positive constant. It is known from the definition of the stream function that l A ( − x 0 , y ) = ( ∂ ψ / ∂ y ) x = − x 0 = l A − x 0 , y = ( ∂ ψ / ∂ y ) x = − x 0 = l_(A)(-x_(0),y)=(del psi//del y)_(x=-x_(0))=l_{A}\left(-x_{0}, y\right)=(\partial \psi / \partial y)_{x=-x_{0}}=lA(−x0,y)=(∂ψ/∂y)x=−x0= const ≡ k 0 − ≡ k 0 − -=k_(0)^(-)\equiv k_{0}^{-}≡k0−and using the
relation ψ ( − x 0 , 0 ) = 0 ψ − x 0 , 0 = 0 psi(-x_(0),0)=0\psi\left(-x_{0}, 0\right)=0ψ(−x0,0)=0, we obtain that ψ ( − x 0 , y ) = k 0 − y ψ − x 0 , y = k 0 − y psi(-x_(0),y)=k_(0)^(-)y\psi\left(-x_{0}, y\right)=k_{0}^{-} yψ(−x0,y)=k0−y and similarly, ψ ( x 0 , y ) = k 0 + y ψ x 0 , y = k 0 + y psi(x_(0),y)=k_(0)^(+)y\psi\left(x_{0}, y\right)=k_{0}^{+} yψ(x0,y)=k0+y. The flow rate in the x x xxx direction being constant, we have [2]
∫ 0 y w ( − y 0 ) ∂ ψ ∂ y ( − x 0 , y ) d y = ∫ 0 y w ( x 0 ) ∂ ψ ∂ y ( x 0 , y ) d y = q ∫ 0 y w − y 0   ∂ ψ ∂ y − x 0 , y d y = ∫ 0 y w   x 0 ∂ ψ ∂ y x 0 , y d y = q int_(0)^(y_(w)(-y_(0)))(del psi)/(del y)(-x_(0),y)dy=int_(0)^(y_(w))(x_(0))(del psi)/(del y)(x_(0),y)dy=q\int_{0}^{y_{w}\left(-y_{0}\right)} \frac{\partial \psi}{\partial y}\left(-x_{0}, y\right) \mathrm{d} y=\int_{0}^{y_{w}}\left(x_{0}\right) \frac{\partial \psi}{\partial y}\left(x_{0}, y\right) \mathrm{d} y=q∫0yw(−y0)∂ψ∂y(−x0,y)dy=∫0yw(x0)∂ψ∂y(x0,y)dy=q
Hence ψ ψ psi\psiψ satisfies the boundary conditions
(1) ψ ( ± x 0 , y ) = q y y w ( ± x 0 ) , 0 ≤ y ≤ y w (1) ψ ± x 0 , y = q y y w ± x 0 , 0 ≤ y ≤ y w {:(1)psi(+-x_(0),y)=(qy)/(y_(w)(+-x_(0)))","0 <= y <= y_(w):}\begin{equation*} \psi\left( \pm x_{0}, y\right)=\frac{q y}{y_{w}\left( \pm x_{0}\right)}, 0 \leq y \leq y_{w} \tag{1} \end{equation*}(1)ψ(±x0,y)=qyyw(±x0),0≤y≤yw
Then the transformation
(2) ψ ( x , y ) = q y y w ( x ) − v ( x , y ) (2) ψ ( x , y ) = q y y w ( x ) − v ( x , y ) {:(2)psi(x","y)=(qy)/(y_(w)(x))-v(x","y):}\begin{equation*} \psi(x, y)=\frac{q y}{y_{w}(x)}-v(x, y) \tag{2} \end{equation*}(2)ψ(x,y)=qyyw(x)−v(x,y)
is made in equation Δ ψ = 0 Δ ψ = 0 Delta psi=0\Delta \psi=0Δψ=0. The problem reduces to solving the following boundary value problem with homogeneous Dirichlet boundary conditions :
(3) Δ v ( x , y ) = q y d 2 d x 2 1 y w ( x ) , ( x , y ) ∈ Ω (3) Δ v ( x , y ) = q y d 2 d x 2 1 y w ( x ) , ( x , y ) ∈ Ω {:(3)Delta v(x","y)=qy(d^(2))/((d)x^(2))(1)/(y_(w)(x))","(x","y)in Omega:}\begin{equation*} \Delta v(x, y)=q y \frac{\mathrm{~d}^{2}}{\mathrm{~d} x^{2}} \frac{1}{y_{w}(x)},(x, y) \in \Omega \tag{3} \end{equation*}(3)Δv(x,y)=qy d2 dx21yw(x),(x,y)∈Ω
(4) v ( x , 0 ) = 0 , v ( x , y w ( x ) ) = 0 , v ( − x 0 , y ) = 0 , v ( x 0 , y ) = 0 (4) v ( x , 0 ) = 0 , v x , y w ( x ) = 0 , v − x 0 , y = 0 , v x 0 , y = 0 {:(4)v(x","0)=0","quad v(x,y_(w)(x))=0","quad v(-x_(0),y)=0","quad v(x_(0),y)=0:}\begin{equation*} v(x, 0)=0, \quad v\left(x, y_{w}(x)\right)=0, \quad v\left(-x_{0}, y\right)=0, \quad v\left(x_{0}, y\right)=0 \tag{4} \end{equation*}(4)v(x,0)=0,v(x,yw(x))=0,v(−x0,y)=0,v(x0,y)=0
b) Equivalent variational problem. The boundary value problem can now be written in the form of operatorial equation, i.e.
(5) A 0 v = f 0 ( f 0 = − q y ′ d 2 d x 2 1 y w ( x ) , A 0 = − Δ = − ∂ 2 ∂ x 2 − ∂ 2 ∂ y 2 ) (5) A 0 v = f 0 f 0 = − q y ′ d 2 d x 2 1 y w ( x ) , A 0 = − Δ = − ∂ 2 ∂ x 2 − ∂ 2 ∂ y 2 {:[(5)A_(0)v=f_(0)],[(f_(0)=-qy^(')(d^(2))/((d)x^(2))(1)/(y_(w)(x)),quadA_(0)=-Delta=-(del^(2))/(delx^(2))-(del^(2))/(dely^(2)))]:}\begin{gather*} A_{0} v=f_{0} \tag{5}\\ \left(f_{0}=-q y^{\prime} \frac{\mathrm{d}^{2}}{\mathrm{~d} x^{2}} \frac{1}{y_{w}(x)}, \quad A_{0}=-\Delta=-\frac{\partial^{2}}{\partial x^{2}}-\frac{\partial^{2}}{\partial y^{2}}\right) \end{gather*}(5)A0v=f0(f0=−qy′d2 dx21yw(x),A0=−Δ=−∂2∂x2−∂2∂y2)
where we consider that the function f 0 f 0 f_(0)f_{0}f0 belongs to the Hilbert space H = L 2 ( Ω ) H = L 2 ( Ω ) H=L_(2)(Omega)H=L_{2}(\Omega)H=L2(Ω) and the operator A 0 : D ( A 0 ) ⊂ H → H A 0 : D A 0 ⊂ H → H A_(0):D(A_(0))sub H rarr HA_{0}: D\left(A_{0}\right) \subset H \rightarrow HA0:D(A0)⊂H→H has the definition domain
D ( A 0 ) = { v ∈ H | v ∈ C 2 ( Ω ) ∩ C ( Ω ¯ ) , A 0 v ∈ H , v = 0 on ∂ Ω } , D ( A 0 ) ― H = H D A 0 = v ∈ H v ∈ C 2 ( Ω ) ∩ C ( Ω ¯ ) , A 0 v ∈ H , v = 0  on  ∂ Ω , D A 0 ¯ H = H D(A_(0))={v in H|_({:v inC^(2)(Omega)nn C(( bar(Omega))),quadA_(0)v in H,quad v=0quad" on "quad del Omega},quad bar(D(A_(0)))^(H)=H):}D\left(A_{0}\right)=\left\{\left.v \in H\right|_{\left.v \in C^{2}(\Omega) \cap C(\bar{\Omega}), \quad A_{0} v \in H, \quad v=0 \quad \text { on } \quad \partial \Omega\right\}, \quad{\overline{D\left(A_{0}\right)}}^{H}=H}\right.D(A0)={v∈H|v∈C2(Ω)∩C(Ω¯),A0v∈H,v=0 on ∂Ω},D(A0)―H=H
Remark 1. The linear operator A 0 A 0 A_(0)A_{0}A0 is symmetric and positive definite on D ( A 0 ) D A 0 D(A_(0))D\left(A_{0}\right)D(A0), [5]:
( A 0 u , v ) = ∫ Ω ∇ u ⋅ ∇ v d Ω = ( u , A 0 v ) , ( A 0 v , v ) ≥ 0 , ( A 0 v , v ) = 0 ⇔ v = 0 ∃ α 2 > 0 such that ( A 0 v , v ) ≥ α 2 ( v , v ) , α 2 = 1 C F ( Ω ) A 0 u , v = ∫ Ω   ∇ u ⋅ ∇ v d Ω = u , A 0 v , A 0 v , v ≥ 0 , A 0 v , v = 0 ⇔ v = 0 ∃ α 2 > 0  such that  A 0 v , v ≥ α 2 ( v , v ) , α 2 = 1 C F ( Ω ) {:[(A_(0)u,v)=int_(Omega)grad u*grad vdOmega=(u,A_(0)v)","(A_(0)v,v) >= 0","(A_(0)v,v)=0<=>v=0],[EEalpha^(2) > 0" such that "(A_(0)v,v) >= alpha^(2)(v","v)","quadalpha^(2)=(1)/(C_(F)(Omega))]:}\begin{gathered} \left(A_{0} u, v\right)=\int_{\Omega} \nabla u \cdot \nabla v \mathrm{~d} \Omega=\left(u, A_{0} v\right),\left(A_{0} v, v\right) \geq 0,\left(A_{0} v, v\right)=0 \Leftrightarrow v=0 \\ \exists \alpha^{2}>0 \text { such that }\left(A_{0} v, v\right) \geq \alpha^{2}(v, v), \quad \alpha^{2}=\frac{1}{C_{F}(\Omega)} \end{gathered}(A0u,v)=∫Ω∇u⋅∇v dΩ=(u,A0v),(A0v,v)≥0,(A0v,v)=0⇔v=0∃α2>0 such that (A0v,v)≥α2(v,v),α2=1CF(Ω)
where C F ( Ω ) C F ( Ω ) C_(F)(Omega)C_{F}(\Omega)CF(Ω) is the constant in the Friedrichs inequality and (., ) denotes the scalar product in L 2 ( Ω ) L 2 ( Ω ) L_(2)(Omega)L_{2}(\Omega)L2(Ω).
The energetic space H A 0 H A 0 H_(A_(0))H_{A_{0}}HA0 of the operator A 0 A 0 A_(0)A_{0}A0 can be identified with the Sobolev space H 0 1 ( Ω ) H 0 1 ( Ω ) H_(0)^(1)(Omega)H_{0}^{1}(\Omega)H01(Ω) endowed with energetic norm and energetic scalar product, respectively [6]:
H A 0 = H 0 1 ( Ω ) = { v ∈ H 1 ∣ v = 0 on ∂ Ω ( in the trace sense ) } ‖ v ‖ A 0 2 = ∫ Ω | ∇ v | 2 d Ω ; ( u , v ) A 0 = ∫ Ω ∇ u ⋅ ∇ v d Ω H A 0 = H 0 1 ( Ω ) = v ∈ H 1 ∣ v = 0  on  ∂ Ω (  in the trace sense  ) ‖ v ‖ A 0 2 = ∫ Ω   | ∇ v | 2 d Ω ; ( u , v ) A 0 = ∫ Ω   ∇ u ⋅ ∇ v d Ω {:[H_(A_(0))=H_(0)^(1)(Omega)={v inH^(1)∣v=0" on "del Omega(" in the trace sense ")}],[||v||_(A_(0))^(2)=int_(Omega)|grad v|^(2)dOmega;(u","v)_(A_(0))=int_(Omega)grad u*grad vdOmega]:}\begin{gathered} H_{A_{0}}=H_{0}^{1}(\Omega)=\left\{v \in H^{1} \mid v=0 \text { on } \partial \Omega(\text { in the trace sense })\right\} \\ \|v\|_{A_{0}}^{2}=\int_{\Omega}|\nabla v|^{2} \mathrm{~d} \Omega ;(u, v)_{A_{0}}=\int_{\Omega} \nabla u \cdot \nabla v \mathrm{~d} \Omega \end{gathered}HA0=H01(Ω)={v∈H1∣v=0 on ∂Ω( in the trace sense )}‖v‖A02=∫Ω|∇v|2 dΩ;(u,v)A0=∫Ω∇u⋅∇v dΩ
  • According to the properties from Remark 1 and using the theorem of the minimum of the energy functional F 0 ( V ) F 0 ( V ) F_(0)(V)F_{0}(V)F0(V) on H A 0 H A 0 H_(A_(0))H_{A_{0}}HA0, the operatorial equation (5) is equivalent to the following variational problem [6] :
(6) F 0 ( v ) → minimum on D ( F 0 ) = H A 0 (6) F 0 ( v ) →  minimum   on  D F 0 = H A 0 {:(6)F_(0)(v)rarr" minimum "quad" on "quad D(F_(0))=H_(A_(0)):}\begin{equation*} F_{0}(v) \rightarrow \text { minimum } \quad \text { on } \quad D\left(F_{0}\right)=H_{A_{0}} \tag{6} \end{equation*}(6)F0(v)→ minimum  on D(F0)=HA0
where
(7) F 0 ( v ) = ‖ v ‖ A 0 2 − 2 ( f 0 , v ) = ∫ Ω ( | ∇ v | 2 − 2 f 0 v ) d Ω (7) F 0 ( v ) = ‖ v ‖ A 0 2 − 2 f 0 , v = ∫ Ω   | ∇ v | 2 − 2 f 0 v d Ω {:(7)F_(0)(v)=||v||_(A_(0))^(2)-2(f_(0),v)=int_(Omega)(|grad v|^(2)-2f_(0)v)dOmega:}\begin{equation*} F_{0}(v)=\|v\|_{A_{0}}^{2}-2\left(f_{0}, v\right)=\int_{\Omega}\left(|\nabla v|^{2}-2 f_{0} v\right) \mathrm{d} \Omega \tag{7} \end{equation*}(7)F0(v)=‖v‖A02−2(f0,v)=∫Ω(|∇v|2−2f0v)dΩ
Remark 2. The solution v ~ ∈ H A 0 v ~ ∈ H A 0 widetilde(v)inH_(A_(0))\widetilde{v} \in H_{A_{0}}v~∈HA0 of problem (6), which exists and is unique, is the generalized solution of problem (3)-(4) and has the form
v ~ = ∑ k = 1 ∞ ( f 0 , w k ) w k v ~ = ∑ k = 1 ∞   f 0 , w k w k widetilde(v)=sum_(k=1)^(oo)(f_(0),w_(k))w_(k)\widetilde{v}=\sum_{k=1}^{\infty}\left(f_{0}, w_{k}\right) w_{k}v~=∑k=1∞(f0,wk)wk
if { w k } k = 1 ∞ w k k = 1 ∞ {w_(k)}_(k=1)^(oo)\left\{w_{k}\right\}_{k=1}^{\infty}{wk}k=1∞ is an orthonormal and complete system of functions in the energetic space H A 0 H A 0 H_(A_(0))H_{A_{0}}HA0.

2. KANTOROVICH METHOD

This method is based on obtaining an approximate solution V N ′ ∈ H A n V N ′ ∈ H A n V_(N^('))inH_(A_(n))V_{N^{\prime}} \in H_{A_{n}}VN′∈HAn of the form [4], [3] :
(8) v N ′ ( x , y ) = ∑ k = 1 N a k ( x ) ϕ k ( x , y ) (8) v N ′ ( x , y ) = ∑ k = 1 N   a k ( x ) ϕ k ( x , y ) {:(8)v_(N^('))(x","y)=sum_(k=1)^(N)a_(k)(x)phi_(k)(x","y):}\begin{equation*} v_{N^{\prime}}(x, y)=\sum_{k=1}^{N} a_{k}(x) \phi_{k}(x, y) \tag{8} \end{equation*}(8)vN′(x,y)=∑k=1Nak(x)ϕk(x,y)
which is of N N NNN-order approximation for v ~ v ~ widetilde(v)\widetilde{v}v~. Here a k , k = 1 , 2 , … , N a k , k = 1 , 2 , … , N a_(k),k=1,2,dots,Na_{k}, k=1,2, \ldots, Nak,k=1,2,…,N, are unknown functions on [ − x 0 , x 1 ] − x 0 , x 1 [-x_(0),x_(1)]\left[-x_{0}, x_{1}\right][−x0,x1] and ϕ k , k = 1 , 2 , … , N ϕ k , k = 1 , 2 , … , N phi_(k),k=1,2,dots,N\phi_{k}, k=1,2, \ldots, Nϕk,k=1,2,…,N, are known functions on Ω Ω Omega\OmegaΩ belonging to H A 0 H A 0 H_(A_(0))H_{A_{0}}HA0, or to D ( A 0 ) D A 0 D(A_(0))D\left(A_{0}\right)D(A0) if the considered problem can be formulated in D ( A 0 ) D A 0 D(A_(0))D\left(A_{0}\right)D(A0). These functions have to be chosen so that conditions (4) for v N v N v_(N)v_{N}vN be valid.

According to the Kantorovich procedure:

a) The trial functions ϕ k , k = 1 , 2 , … , N ϕ k , k = 1 , 2 , … , N phi_(k),k=1,2,dots,N\phi_{k}, k=1,2, \ldots, Nϕk,k=1,2,…,N, are chosen from a complete (nonorthonormal) system of functions { ϕ k } k = 1 ∞ ϕ k k = 1 ∞ {phi_(k)}_(k=1)^(oo)\left\{\phi_{k}\right\}_{k=1}^{\infty}{ϕk}k=1∞ and satisfy the conditions ϕ k ( x , y ) = 0 ϕ k ( x , y ) = 0 phi_(k)(x,y)=0\phi_{k}(x, y)=0ϕk(x,y)=0 on ∂ Ω ∂ Ω del Omega\partial \Omega∂Ω, excluding the lines x = ± x 0 x = ± x 0 x=+-x_(0)x= \pm x_{0}x=±x0. Then the trial functions can be chosen in the form
(9) ϕ k ( x , y ) = [ y − y w ( x ) ] y k , k = 1 , 2 , … , N (9) ϕ k ( x , y ) = y − y w ( x ) y k , k = 1 , 2 , … , N {:(9)phi_(k)(x","y)=[y-y_(w)(x)]y^(k)","k=1","2","dots","N:}\begin{equation*} \phi_{k}(x, y)=\left[y-y_{w}(x)\right] y^{k}, k=1,2, \ldots, N \tag{9} \end{equation*}(9)ϕk(x,y)=[y−yw(x)]yk,k=1,2,…,N
b) The functions a k a k a_(k)a_{k}ak satisfy the following system of ordinary differential equations, [1] :
(10) ∫ 0 y w ( x ) ( A 0 v N − f 0 ) ϕ i ( x , y ) d y = 0 , i = 1 , 2 , … , N (10) ∫ 0 y w ( x )   A 0 v N − f 0 ϕ i ( x , y ) d y = 0 , i = 1 , 2 , … , N {:(10)int_(0)^(y_(w)(x))(A_(0)v_(N)-f_(0))phi_(i)(x","y)dy=0","quad i=1","2","dots","N:}\begin{equation*} \int_{0}^{y_{w}(x)}\left(A_{0} v_{N}-f_{0}\right) \phi_{i}(x, y) \mathrm{d} y=0, \quad i=1,2, \ldots, N \tag{10} \end{equation*}(10)∫0yw(x)(A0vN−f0)ϕi(x,y)dy=0,i=1,2,…,N
and the conditions
(11) a k ( − x 0 ) = a k ( x 0 ) = 0 , k = 1 , 2 , … , N (11) a k − x 0 = a k x 0 = 0 , k = 1 , 2 , … , N {:(11)a_(k)(-x_(0))=a_(k)(x_(0))=0","quad k=1","2","dots","N:}\begin{equation*} a_{k}\left(-x_{0}\right)=a_{k}\left(x_{0}\right)=0, \quad k=1,2, \ldots, N \tag{11} \end{equation*}(11)ak(−x0)=ak(x0)=0,k=1,2,…,N
obtained from the stationary condition of the functional F 0 F 0 F_(0)F_{0}F0 on V N V N V_(N)V_{N}VN written as
F 0 ( v N ) = ∫ − x 0 x 0 { ∫ 0 y w ( x ) [ ( ∂ v N ∂ x ) 2 + ( ∂ v N ∂ y ) 2 − 2 f 0 v N ] d y } d x ≡ ≡ ∫ − x 0 x 0 G ( x , a 1 ( x ) , … , a N ( x ) , a 1 ′ ( x ) , … , a N ′ ( x ) ) d x F 0 v N = ∫ − x 0 x 0   ∫ 0 y w ( x )   ∂ v N ∂ x 2 + ∂ v N ∂ y 2 − 2 f 0 v N d y d x ≡ ≡ ∫ − x 0 x 0   G x , a 1 ( x ) , … , a N ( x ) , a 1 ′ ( x ) , … , a N ′ ( x ) d x {:[F_(0)(v_(N))=int_(-x_(0))^(x_(0)){int_(0)^(y_(w)(x))[((delv_(N))/(del x))^(2)+((delv_(N))/(del y))^(2)-2f_(0)v_(N)]dy}dx-=],[-=int_(-x_(0))^(x_(0))G(x,a_(1)(x),dots,a_(N)(x),a_(1)^(')(x),dots,a_(N)^(')(x))dx]:}\begin{aligned} F_{0}\left(v_{N}\right) & =\int_{-x_{0}}^{x_{0}}\left\{\int_{0}^{y_{w}(x)}\left[\left(\frac{\partial v_{N}}{\partial x}\right)^{2}+\left(\frac{\partial v_{N}}{\partial y}\right)^{2}-2 f_{0} v_{N}\right] \mathrm{d} y\right\} \mathrm{d} x \equiv \\ & \equiv \int_{-x_{0}}^{x_{0}} G\left(x, a_{1}(x), \ldots, a_{N}(x), a_{1}^{\prime}(x), \ldots, a_{N}^{\prime}(x)\right) \mathrm{d} x \end{aligned}F0(vN)=∫−x0x0{∫0yw(x)[(∂vN∂x)2+(∂vN∂y)2−2f0vN]dy}dx≡≡∫−x0x0G(x,a1(x),…,aN(x),a1′(x),…,aN′(x))dx
i.e. from the Euler-Lagrange equations
d d x ∂ G ∂ a k ′ − ∂ G ∂ a k = 0 with a k ( − x 0 ) = a k ( x 0 ) = 0 , k = 1 , N ― d d x ∂ G ∂ a k ′ − ∂ G ∂ a k = 0  with  a k − x 0 = a k x 0 = 0 , k = 1 , N ¯ (d)/((d)x)(del G)/(dela_(k)^('))-(del G)/(dela_(k))=0quad" with "quada_(k)(-x_(0))=a_(k)(x_(0))=0,quad k= bar(1,N)\frac{\mathrm{d}}{\mathrm{~d} x} \frac{\partial G}{\partial a_{k}^{\prime}}-\frac{\partial G}{\partial a_{k}}=0 \quad \text { with } \quad a_{k}\left(-x_{0}\right)=a_{k}\left(x_{0}\right)=0, \quad k=\overline{1, N}d dx∂G∂ak′−∂G∂ak=0 with ak(−x0)=ak(x0)=0,k=1,N―
By substituting A 0 , V N , f 0 A 0 , V N , f 0 A_(0),V_(N),f_(0)A_{0}, V_{N}, f_{0}A0,VN,f0 and ϕ k ϕ k phi_(k)\phi_{k}ϕk with their expressions and performing some calculations, the differential problem (10)-(11) becomes
(12) ∑ k = 1 N [ α i k ( x ) d 2 a k d x 2 + 2 β i k ( x ) d a k d x + γ i k ( x ) a k ( x ) ] = r i ( x ) , i = 1 , N ― (13) a k ( − x 0 ) = a k ( x 0 ) = 0 , k = 1 , N ― (12) ∑ k = 1 N   α i k ( x ) d 2 a k d x 2 + 2 β i k ( x ) d a k d x + γ i k ( x ) a k ( x ) = r i ( x ) , i = 1 , N ¯ (13) a k − x 0 = a k x 0 = 0 , k = 1 , N ¯ {:[(12)sum_(k=1)^(N)[alpha_(ik)(x)(d^(2)a_(k))/((d)x^(2))+2beta_(ik)(x)(da_(k))/((d)x)+gamma_(ik)(x)a_(k)(x)]=r_(i)(x)","quad i= bar(1,N)],[(13)a_(k)(-x_(0))=a_(k)(x_(0))=0","quad k= bar(1,N)]:}\begin{gather*} \sum_{k=1}^{N}\left[\alpha_{i k}(x) \frac{\mathrm{d}^{2} a_{k}}{\mathrm{~d} x^{2}}+2 \beta_{i k}(x) \frac{\mathrm{d} a_{k}}{\mathrm{~d} x}+\gamma_{i k}(x) a_{k}(x)\right]=r_{i}(x), \quad i=\overline{1, N} \tag{12}\\ a_{k}\left(-x_{0}\right)=a_{k}\left(x_{0}\right)=0, \quad k=\overline{1, N} \tag{13} \end{gather*}(12)∑k=1N[αik(x)d2ak dx2+2βik(x)dak dx+γik(x)ak(x)]=ri(x),i=1,N―(13)ak(−x0)=ak(x0)=0,k=1,N―
where
α i k ( x ) = ∫ 0 y w ( x ) ϕ i ( x , y ) ϕ k ( x , y ) d y = 2 y w i + k + 3 ( x ) ( i + k + 1 ) ( i + k + 2 ) ( i + k + 3 ) β i k ( x ) = ∫ 0 y w ( x ) ϕ i ( x , y ) ∂ ϕ k ∂ x d y = y w i + k + 2 ( x ) y w ′ ( x ) ( i + k + 1 ) ( i + k + 2 ) α i k ( x ) = ∫ 0 y w ( x )   ϕ i ( x , y ) ϕ k ( x , y ) d y = 2 y w i + k + 3 ( x ) ( i + k + 1 ) ( i + k + 2 ) ( i + k + 3 ) β i k ( x ) = ∫ 0 y w ( x )   ϕ i ( x , y ) ∂ ϕ k ∂ x d y = y w i + k + 2 ( x ) y w ′ ( x ) ( i + k + 1 ) ( i + k + 2 ) {:[alpha_(ik)(x)=int_(0)^(y_(w)(x))phi_(i)(x","y)phi_(k)(x","y)dy=(2y_(w)^(i+k+3)(x))/((i+k+1)(i+k+2)(i+k+3))],[beta_(ik)(x)=int_(0)^(y_(w)(x))phi_(i)(x","y)(delphi_(k))/(del x)dy=(y_(w)^(i+k+2)(x)y_(w)^(')(x))/((i+k+1)(i+k+2))]:}\begin{gathered} \alpha_{i k}(x)=\int_{0}^{y_{w}(x)} \phi_{i}(x, y) \phi_{k}(x, y) \mathrm{d} y=\frac{2 y_{w}^{i+k+3}(x)}{(i+k+1)(i+k+2)(i+k+3)} \\ \beta_{i k}(x)=\int_{0}^{y_{w}(x)} \phi_{i}(x, y) \frac{\partial \phi_{k}}{\partial x} \mathrm{~d} y=\frac{y_{w}^{i+k+2}(x) y_{w}^{\prime}(x)}{(i+k+1)(i+k+2)} \end{gathered}αik(x)=∫0yw(x)ϕi(x,y)ϕk(x,y)dy=2ywi+k+3(x)(i+k+1)(i+k+2)(i+k+3)βik(x)=∫0yw(x)ϕi(x,y)∂ϕk∂x dy=ywi+k+2(x)yw′(x)(i+k+1)(i+k+2)
γ i k ( x ) = ∫ 0 y w ( x ) ϕ i Δ ϕ k d y = [ y w ( x ) y w ′ ′ ( x ) i + k + 2 − 2 i k ( i + k − 1 ) ( i + k ) ] y w i + k + 1 ( x ) i + k + 1 r i ( x ) = − ∫ 0 y x ( x ) f 0 ( x , y ) ϕ i ( x , y ) d y = q [ y w ( x ) y w ′ ′ ( x ) − 2 y w ′ 2 ( x ) ] y w i ( x ) ( i + 2 ) ( i + 3 ) γ i k ( x ) = ∫ 0 y w ( x )   ϕ i Δ ϕ k d y = y w ( x ) y w ′ ′ ( x ) i + k + 2 − 2 i k ( i + k − 1 ) ( i + k ) y w i + k + 1 ( x ) i + k + 1 r i ( x ) = − ∫ 0 y x ( x )   f 0 ( x , y ) ϕ i ( x , y ) d y = q y w ( x ) y w ′ ′ ( x ) − 2 y w ′ 2 ( x ) y w i ( x ) ( i + 2 ) ( i + 3 ) {:[gamma_(ik)(x)=int_(0)^(y_(w)(x))phi_(i)Deltaphi_(k)dy=[(y_(w)(x)y_(w)^('')(x))/(i+k+2)-(2ik)/((i+k-1)(i+k))](y_(w)^(i+k+1)(x))/(i+k+1)],[r_(i)(x)=-int_(0)^(y_(x)(x))f_(0)(x","y)phi_(i)(x","y)dy=q([y_(w)(x)y_(w)^('')(x)-2y_(w)^('2)(x)]y_(w)^(i)(x))/((i+2)(i+3))]:}\begin{aligned} & \gamma_{i k}(x)=\int_{0}^{y_{w}(x)} \phi_{i} \Delta \phi_{k} \mathrm{~d} y=\left[\frac{y_{w}(x) y_{w}^{\prime \prime}(x)}{i+k+2}-\frac{2 i k}{(i+k-1)(i+k)}\right] \frac{y_{w}^{i+k+1}(x)}{i+k+1} \\ & r_{i}(x)=-\int_{0}^{y_{x}(x)} f_{0}(x, y) \phi_{i}(x, y) \mathrm{d} y=q \frac{\left[y_{w}(x) y_{w}^{\prime \prime}(x)-2 y_{w}^{\prime 2}(x)\right] y_{w}^{i}(x)}{(i+2)(i+3)} \end{aligned}γik(x)=∫0yw(x)ϕiΔϕk dy=[yw(x)yw′′(x)i+k+2−2ik(i+k−1)(i+k)]ywi+k+1(x)i+k+1ri(x)=−∫0yx(x)f0(x,y)ϕi(x,y)dy=q[yw(x)yw′′(x)−2yw′2(x)]ywi(x)(i+2)(i+3)

3. THE CALCULATION OF THE FIRST-ORDER KANTOROVICH APPROXIMATION

a) The boundary value problem. Consider that the upper solid surface is a flat plate and the equation of the line C B ― C B ¯ bar(CB)\overline{C B}CB― is
y w ( x ) = α 0 x + β 0 , x ∈ [ − x 0 , x 0 ] y w ( x ) = α 0 x + β 0 , x ∈ − x 0 , x 0 y_(w)(x)=alpha_(0)x+beta_(0),quad x in[-x_(0),x_(0)]y_{w}(x)=\alpha_{0} x+\beta_{0}, \quad x \in\left[-x_{0}, x_{0}\right]yw(x)=α0x+β0,x∈[−x0,x0]
In this case the boundary value problem (12)-(13) becomes
− ∑ k = 1 N [ ( i + k ) ! ( i + k + 3 ) ! d x ( y w i + k + 3 ( x ) d a k d x ) − i k ( i + k − 2 ) ! ( i + k + 1 ) ! y w i + k + 1 ( x ) a k ( x ) ] = = q y w ′ 2 ( x ) y w i ( x ) ( i + 2 ) ( i + 3 ) , i = 1 , N ― ; a k ( − x 0 ) = a k ( x 0 ) = 0 , k = 1 , N ― − ∑ k = 1 N   ( i + k ) ! ( i + k + 3 ) ! d x y w i + k + 3 ( x ) d a k d x − i k ( i + k − 2 ) ! ( i + k + 1 ) ! y w i + k + 1 ( x ) a k ( x ) = = q y w ′ 2 ( x ) y w i ( x ) ( i + 2 ) ( i + 3 ) , i = 1 , N ¯ ; a k − x 0 = a k x 0 = 0 , k = 1 , N ¯ {:[-sum_(k=1)^(N)[((i+k)!)/((i+k+3)!dx)(y_(w)^(i+k+3)(x)(da_(k))/((d)x))-(ik(i+k-2)!)/((i+k+1)!)y_(w)^(i+k+1)(x)a_(k)(x)]=],[=q(y_(w)^('2)(x)y_(w)^(i)(x))/((i+2)(i+3))","i= bar(1,N);quada_(k)(-x_(0))=a_(k)(x_(0))=0","quad k= bar(1,N)]:}\begin{gathered} -\sum_{k=1}^{N}\left[\frac{(i+k)!}{(i+k+3)!\mathrm{d} x}\left(y_{w}^{i+k+3}(x) \frac{\mathrm{d} a_{k}}{\mathrm{~d} x}\right)-\frac{i k(i+k-2)!}{(i+k+1)!} y_{w}^{i+k+1}(x) a_{k}(x)\right]= \\ =q \frac{y_{w}^{\prime 2}(x) y_{w}^{i}(x)}{(i+2)(i+3)}, i=\overline{1, N} ; \quad a_{k}\left(-x_{0}\right)=a_{k}\left(x_{0}\right)=0, \quad k=\overline{1, N} \end{gathered}−∑k=1N[(i+k)!(i+k+3)!dx(ywi+k+3(x)dak dx)−ik(i+k−2)!(i+k+1)!ywi+k+1(x)ak(x)]==qyw′2(x)ywi(x)(i+2)(i+3),i=1,N―;ak(−x0)=ak(x0)=0,k=1,N―
If we choose N = 1 N = 1 N=1N=1N=1 (i.e. i = k = 1 i = k = 1 i=k=1i=k=1i=k=1 ) and we write x = x 0 t , y w ( x ) = w ( t ) = α 0 x 0 t + + β 0 , a 1 ( x ) = u ( t ) , t ∈ ( − 1 , 1 ) = I x = x 0 t , y w ( x ) = w ( t ) = α 0 x 0 t + + β 0 , a 1 ( x ) = u ( t ) , t ∈ ( − 1 , 1 ) = I x=x_(0)t,y_(w)(x)=w(t)=alpha_(0)x_(0)t++beta_(0),a_(1)(x)=u(t),t in(-1,1)=Ix=x_{0} t, y_{w}(x)=w(t)=\alpha_{0} x_{0} t+ +\beta_{0}, a_{1}(x)=u(t), t \in(-1,1)=Ix=x0t,yw(x)=w(t)=α0x0t++β0,a1(x)=u(t),t∈(−1,1)=I, we obtain the following ordinary differential equation and the boundary conditions:
(14) A u ≡ − d d t ( w 5 ( t ) d u d t ) + 10 x 0 2 w 3 ( t ) u ( t ) = 5 q w ( t ) w ′ 2 ( t ) , t ∈ ( − 1 , 1 ) (15) u ( − 1 ) = 0 , u ( 1 ) = 0 (14) A u ≡ − d d t w 5 ( t ) d u d t + 10 x 0 2 w 3 ( t ) u ( t ) = 5 q w ( t ) w ′ 2 ( t ) , t ∈ ( − 1 , 1 ) (15) u ( − 1 ) = 0 , u ( 1 ) = 0 {:[(14)Au-=-(d)/((d)t)(w^(5)(t)(du)/((d)t))+10x_(0)^(2)w^(3)(t)u(t)=5qw(t)w^('2)(t)","quad t in(-1","1)],[(15)u(-1)=0","u(1)=0]:}\begin{gather*} A u \equiv-\frac{\mathrm{d}}{\mathrm{~d} t}\left(w^{5}(t) \frac{\mathrm{d} u}{\mathrm{~d} t}\right)+10 x_{0}^{2} w^{3}(t) u(t)=5 q w(t) w^{\prime 2}(t), \quad t \in(-1,1) \tag{14}\\ u(-1)=0, u(1)=0 \tag{15} \end{gather*}(14)Au≡−d dt(w5(t)du dt)+10x02w3(t)u(t)=5qw(t)w′2(t),t∈(−1,1)(15)u(−1)=0,u(1)=0
This problem can be written in the operatorial form
(16) A u = f , (16) A u = f , {:(16)Au=f",":}\begin{equation*} A u=f, \tag{16} \end{equation*}(16)Au=f,
with
A : D ( A ) ⊂ L 2 ( I ) → L 2 ( I ) D ( A ) = { u ∈ L 2 ( I ) ∣ u ∈ C 2 ( I ) ∩ C ( I ¯ ) , A u ∈ L 2 ( I ) , u ( − 1 ) = u ( 1 ) = 0 } f ( t ) = 5 q w ( t ) w ′ 2 ( t ) A : D ( A ) ⊂ L 2 ( I ) → L 2 ( I ) D ( A ) = u ∈ L 2 ( I ) ∣ u ∈ C 2 ( I ) ∩ C ( I ¯ ) , A u ∈ L 2 ( I ) , u ( − 1 ) = u ( 1 ) = 0 f ( t ) = 5 q w ( t ) w ′ 2 ( t ) {:[A:D(A)subL_(2)(I)rarrL_(2)(I)],[D(A)={u inL_(2)(I)∣u inC^(2)(I)nn C(( bar(I))),Au inL_(2)(I),u(-1)=u(1)=0}],[f(t)=5qw(t)w^('2)(t)]:}\begin{gathered} A: D(A) \subset L_{2}(I) \rightarrow L_{2}(I) \\ D(A)=\left\{u \in L_{2}(I) \mid u \in C^{2}(I) \cap C(\bar{I}), A u \in L_{2}(I), u(-1)=u(1)=0\right\} \\ f(t)=5 q w(t) w^{\prime 2}(t) \end{gathered}A:D(A)⊂L2(I)→L2(I)D(A)={u∈L2(I)∣u∈C2(I)∩C(I¯),Au∈L2(I),u(−1)=u(1)=0}f(t)=5qw(t)w′2(t)
Taking into account that the operator A A AAA is linear, symmetric and positive definite on D ( A ) D ( A ) D(A)D(A)D(A) and its energetic space H A = H 0 1 ( I ) H A = H 0 1 ( I ) H_(A)=H_(0)^(1)(I)H_{A}=H_{0}^{1}(I)HA=H01(I) is endowed with the energetic norm ‖ ⋅ ‖ A ‖ ⋅ ‖ A ||*||_(A)\|\cdot\|_{A}‖⋅‖A (and the energetic scalar product ( . , ) A ) , [ 6 ] ( . , ) A , [ 6 ] {:(.,)_(A)),[6]\left.(.,)_{A}\right),[6](.,)A),[6],
‖ u ‖ A 2 = ( u , u ) A = ∫ − 1 1 ( w 5 u ′ 2 + 10 x 0 2 w 3 u 2 ) d t ‖ u ‖ A 2 = ( u , u ) A = ∫ − 1 1   w 5 u ′ 2 + 10 x 0 2 w 3 u 2 d t ||u||_(A)^(2)=(u,u)_(A)=int_(-1)^(1)(w^(5)u^('2)+10x_(0)^(2)w^(3)u^(2))dt\|u\|_{A}^{2}=(u, u)_{A}=\int_{-1}^{1}\left(w^{5} u^{\prime 2}+10 x_{0}^{2} w^{3} u^{2}\right) \mathrm{d} t‖u‖A2=(u,u)A=∫−11(w5u′2+10x02w3u2)dt
it follows that the operatorial equation (i.c. the boundary value problem (14)-(15)) is equivalent to the variational problem
(17) F ( u ) = ‖ u ‖ A 2 − 10 q ∫ − 1 1 w ( t ) w ′ 2 ( t ) u ( t ) d t → minimum on H A (17) F ( u ) = ‖ u ‖ A 2 − 10 q ∫ − 1 1   w ( t ) w ′ 2 ( t ) u ( t ) d t →  minimum on  H A {:(17)F(u)=||u||_(A)^(2)-10 qint_(-1)^(1)w(t)w^('2)(t)u(t)dt rarr" minimum on "H_(A):}\begin{equation*} F(u)=\|u\|_{A}^{2}-10 q \int_{-1}^{1} w(t) w^{\prime 2}(t) u(t) \mathrm{d} t \rightarrow \text { minimum on } H_{A} \tag{17} \end{equation*}(17)F(u)=‖u‖A2−10q∫−11w(t)w′2(t)u(t)dt→ minimum on HA
b) Ritz algorithm. With given trial functions φ k ( 1 ) , k = 1 , n ― [ { φ k } k = 1 ∞ φ k ( 1 ) , k = 1 , n ¯ φ k k = 1 ∞ varphi_(k)(1),k= bar(1,n)[{varphi_(k)}_(k=1)^(oo):}\varphi_{k}(1), k=\overline{1, n}\left[\left\{\varphi_{k}\right\}_{k=1}^{\infty}\right.φk(1),k=1,n―[{φk}k=1∞ is a complete base in H A H A H_(A)H_{A}HA ], we construct the subspace H A ( n ) = span { φ 1 , … , φ n } H A ( n ) = span φ 1 , … , φ n H_(A)^((n))=span{varphi_(1),dots,varphi_(n)}H_{A}^{(n)}=\operatorname{span}\left\{\varphi_{1}, \ldots, \varphi_{n}\right\}HA(n)=span{φ1,…,φn} and we choose for the solution u ~ u ~ tilde(u)\tilde{u}u~ of (17) the Ritz approximate solution u n ∈ H A ( n ) u n ∈ H A ( n ) u_(n)inH_(A)^((n))u_{n} \in H_{A}^{(n)}un∈HA(n), [5] in the form
(18) u n ( t ) = ∑ k = 1 n c k φ k ( t ) , c k ∈ R 1 (18) u n ( t ) = ∑ k = 1 n   c k φ k ( t ) , c k ∈ R 1 {:(18)u_(n)(t)=sum_(k=1)^(n)c_(k)varphi_(k)(t)","quadc_(k)inR^(1):}\begin{equation*} u_{n}(t)=\sum_{k=1}^{n} c_{k} \varphi_{k}(t), \quad c_{k} \in \mathbf{R}^{1} \tag{18} \end{equation*}(18)un(t)=∑k=1nckφk(t),ck∈R1
The unknown coefficients c k , k = 1 , n ― c k , k = 1 , n ¯ c_(k),k= bar(1,n)c_{k}, k=\overline{1, n}ck,k=1,n― are obtained using the Ritz procedure by solving the linear system of algebraic equations (Ritz system)
(19) ∑ k = 1 n K j k c k = b j , j = 1 , n ― (19) ∑ k = 1 n   K j k c k = b j , j = 1 , n ¯ {:(19)sum_(k=1)^(n)K_(jk)c_(k)=b_(j)","j= bar(1,n):}\begin{equation*} \sum_{k=1}^{n} K_{j k} c_{k}=b_{j}, j=\overline{1, n} \tag{19} \end{equation*}(19)∑k=1nKjkck=bj,j=1,n―
where
K j k = ( φ j , φ k ) A , b j = 5 q w ′ 2 ( w , φ j ) L 2 ( I ) ( ( A u n − f , φ j ) L 2 ( I ) = 0 = ∑ k = 1 n c k ( A φ k , φ j ) L 2 ( I ) − ( f , φ j ) L 2 ( I ) ) K j k = φ j , φ k A , b j = 5 q w ′ 2 w , φ j L 2 ( I ) A u n − f , φ j L 2 ( I ) = 0 = ∑ k = 1 n   c k A φ k , φ j L 2 ( I ) − f , φ j L 2 ( I ) {:[K_(jk)=(varphi_(j),varphi_(k))_(A)","b_(j)=5qw^('2)(w,varphi_(j))_(L_(2)(I))],[((Au_(n)-f,varphi_(j))_(L_(2)(I))=0=sum_(k=1)^(n)c_(k)(Avarphi_(k),varphi_(j))_(L_(2)(I))-(f,varphi_(j))_(L_(2)(I)))]:}\begin{gathered} K_{j k}=\left(\varphi_{j}, \varphi_{k}\right)_{A}, b_{j}=5 q w^{\prime 2}\left(w, \varphi_{j}\right)_{L_{2}(I)} \\ \left(\left(A u_{n}-f, \varphi_{j}\right)_{L_{2}(I)}=0=\sum_{k=1}^{n} c_{k}\left(A \varphi_{k}, \varphi_{j}\right)_{L_{2}(I)}-\left(f, \varphi_{j}\right)_{L_{2}(I)}\right) \end{gathered}Kjk=(φj,φk)A,bj=5qw′2(w,φj)L2(I)((Aun−f,φj)L2(I)=0=∑k=1nck(Aφk,φj)L2(I)−(f,φj)L2(I))
Remark 3. Since H = L 2 ( I ) H = L 2 ( I ) H=L_(2)(I)H=L_{2}(I)H=L2(I) is a separable space and H A H A H_(A)H_{A}HA is also separable, there will also exist complete systems of functions { φ k } k = 1 ∞ φ k k = 1 ∞ {varphi_(k)}_(k=1)^(oo)\left\{\varphi_{k}\right\}_{k=1}^{\infty}{φk}k=1∞ in H A H A H_(A)H_{A}HA.
  • The trial functions are chosen to be of the for:
φ k ( t ) = 1 k π sin ⁡ k π t , t ∈ [ − 1 , 1 ] , k = 1 , n ― φ k ( t ) = 1 k π sin ⁡ k π t , t ∈ [ − 1 , 1 ] , k = 1 , n ¯ varphi_(k)(t)=(1)/(k pi)sin k pi t,quad t in[-1,1],quad k= bar(1,n)\varphi_{k}(t)=\frac{1}{k \pi} \sin k \pi t, \quad t \in[-1,1], \quad k=\overline{1, n}φk(t)=1kπsin⁡kπt,t∈[−1,1],k=1,n―
where n n nnn is a given natural number ; φ k ∈ H A φ k ∈ H A varphi_(k)inH_(A)\varphi_{k} \in H_{A}φk∈HA.
The coefficients of the Ritz system (19) are
2 K j k = ∫ − 1 1 [ w 5 ( t ) + 10 x 0 2 j k π 2 w 3 ( t ) ] cos ⁡ ( j − k ) π t d t + + ∫ − 1 1 [ w 5 ( t ) − 10 x 0 j k π 2 w 3 ( t ) ] cos ⁡ ( j + k ) π t d t , j , k = 1 , n ― 2 K j k = ∫ − 1 1   w 5 ( t ) + 10 x 0 2 j k π 2 w 3 ( t ) cos ⁡ ( j − k ) π t d t + + ∫ − 1 1   w 5 ( t ) − 10 x 0 j k π 2 w 3 ( t ) cos ⁡ ( j + k ) π t d t , j , k = 1 , n ¯ {:[2K_(jk)=int_(-1)^(1)[w^(5)(t)+(10x_(0)^(2))/(jkpi^(2))w^(3)(t)]cos(j-k)pi tdt+],[+int_(-1)^(1)[w^(5)(t)-(10x_(0))/(jkpi^(2))w^(3)(t)]cos(j+k)pi tdt","j","k= bar(1,n)]:}\begin{aligned} & 2 K_{j k}=\int_{-1}^{1}\left[w^{5}(t)+\frac{10 x_{0}^{2}}{j k \pi^{2}} w^{3}(t)\right] \cos (j-k) \pi t \mathrm{~d} t+ \\ & +\int_{-1}^{1}\left[w^{5}(t)-\frac{10 x_{0}}{j k \pi^{2}} w^{3}(t)\right] \cos (j+k) \pi t \mathrm{~d} t, j, k=\overline{1, n} \end{aligned}2Kjk=∫−11[w5(t)+10x02jkπ2w3(t)]cos⁡(j−k)πt dt++∫−11[w5(t)−10x0jkπ2w3(t)]cos⁡(j+k)πt dt,j,k=1,n―
In order to calculate these coefficients and the right hand side of (19), the following exact formulas are successively obtained
I m ( 5 ) = ∫ − 1 1 w 5 ( t ) cos ⁡ m π t d t = = ( − 1 ) m m 2 π 2 { ( w 5 ) ′ ( 1 ) − ( w 5 ) ′ ( − 1 ) − 1 m 2 π 2 [ ( w 5 ) ′ ′ ′ ( 1 ) − ( w 5 ) ′ ′ ′ ( − 1 ) ] } = = ( − 1 ) m m 2 ( α 0 x 0 π ) 2 40 β 0 [ β 0 2 + α 0 2 x 0 2 − 6 m 2 ( α 0 x 0 π ) 2 ] , m = 1 , 2 n ― I m ( 3 ) = ∫ − 1 1 w 3 ( t ) cos ⁡ m π t d t = ( − 1 ) m m 2 π 2 [ ( w 3 ) ′ ( 1 ) − ( w 3 ) ′ ( − 1 ) ] = = ( − 1 ) m m 2 12 β 0 ( α 0 x 0 π ) 2 , m = 1 , 2 n ― ( I 0 ( s ) = ∫ − 1 1 w s ( t ) d t ; s = 3 ; s = 5 ) b j = 5 q w ′ 2 j π ∫ − 1 1 w ( t ) sin ⁡ j π t d t = ( − 1 ) j + 1 j 2 10 q π ( α 0 x 0 π ) 3 , j = 1 , n ― I m ( 5 ) = ∫ − 1 1   w 5 ( t ) cos ⁡ m π t d t = = ( − 1 ) m m 2 π 2 w 5 ′ ( 1 ) − w 5 ′ ( − 1 ) − 1 m 2 π 2 w 5 ′ ′ ′ ( 1 ) − w 5 ′ ′ ′ ( − 1 ) = = ( − 1 ) m m 2 α 0 x 0 π 2 40 β 0 β 0 2 + α 0 2 x 0 2 − 6 m 2 α 0 x 0 π 2 , m = 1 , 2 n ¯ I m ( 3 ) = ∫ − 1 1   w 3 ( t ) cos ⁡ m π t d t = ( − 1 ) m m 2 π 2 w 3 ′ ( 1 ) − w 3 ′ ( − 1 ) = = ( − 1 ) m m 2 12 β 0 α 0 x 0 π 2 , m = 1 , 2 n ¯ I 0 ( s ) = ∫ − 1 1   w s ( t ) d t ; s = 3 ; s = 5 b j = 5 q w ′ 2 j π ∫ − 1 1   w ( t ) sin ⁡ j π t d t = ( − 1 ) j + 1 j 2 10 q π α 0 x 0 π 3 , j = 1 , n ¯ {:[I_(m)^((5))=int_(-1)^(1)w^(5)(t)cos m pi tdt=],[=((-1)^(m))/(m^(2)pi^(2)){(w^(5))^(')(1)-(w^(5))^(')(-1)-(1)/(m^(2)pi^(2))[(w^(5))^(''')(1)-(w^(5))^(''')(-1)]}=],[=((-1)^(m))/(m^(2))((alpha_(0)x_(0))/(pi))^(2)40beta_(0)[beta_(0)^(2)+alpha_(0)^(2)x_(0)^(2)-(6)/(m^(2))((alpha_(0)x_(0))/(pi))^(2)]","quad m= bar(1,2n)],[I_(m)^((3))=int_(-1)^(1)w^(3)(t)cos m pi tdt=((-1)^(m))/(m^(2)pi^(2))[(w^(3))^(')(1)-(w^(3))^(')(-1)]=],[=((-1)^(m))/(m^(2))12beta_(0)((alpha_(0)x_(0))/(pi))^(2)","quad m= bar(1,2n)],[(I_(0)^((s))=int_(-1)^(1)w^(s)(t)dt;quad s=3;s=5)],[b_(j)=(5qw^('2))/(j pi)int_(-1)^(1)w(t)sin j pi tdt=((-1)^(j+1))/(j^(2))10 q pi((alpha_(0)x_(0))/(pi))^(3)","quad j= bar(1,n)]:}\begin{gathered} I_{m}^{(5)}=\int_{-1}^{1} w^{5}(t) \cos m \pi t \mathrm{~d} t= \\ =\frac{(-1)^{m}}{m^{2} \pi^{2}}\left\{\left(w^{5}\right)^{\prime}(1)-\left(w^{5}\right)^{\prime}(-1)-\frac{1}{m^{2} \pi^{2}}\left[\left(w^{5}\right)^{\prime \prime \prime}(1)-\left(w^{5}\right)^{\prime \prime \prime}(-1)\right]\right\}= \\ =\frac{(-1)^{m}}{m^{2}}\left(\frac{\alpha_{0} x_{0}}{\pi}\right)^{2} 40 \beta_{0}\left[\beta_{0}^{2}+\alpha_{0}^{2} x_{0}^{2}-\frac{6}{m^{2}}\left(\frac{\alpha_{0} x_{0}}{\pi}\right)^{2}\right], \quad m=\overline{1,2 n} \\ I_{m}^{(3)}=\int_{-1}^{1} w^{3}(t) \cos m \pi t \mathrm{~d} t=\frac{(-1)^{m}}{m^{2} \pi^{2}}\left[\left(w^{3}\right)^{\prime}(1)-\left(w^{3}\right)^{\prime}(-1)\right]= \\ =\frac{(-1)^{m}}{m^{2}} 12 \beta_{0}\left(\frac{\alpha_{0} x_{0}}{\pi}\right)^{2}, \quad m=\overline{1,2 n} \\ \left(I_{0}^{(s)}=\int_{-1}^{1} w^{s}(t) \mathrm{d} t ; \quad s=3 ; s=5\right) \\ b_{j}=\frac{5 q w^{\prime 2}}{j \pi} \int_{-1}^{1} w(t) \sin j \pi t \mathrm{~d} t=\frac{(-1)^{j+1}}{j^{2}} 10 q \pi\left(\frac{\alpha_{0} x_{0}}{\pi}\right)^{3}, \quad j=\overline{1, n} \end{gathered}Im(5)=∫−11w5(t)cos⁡mπt dt==(−1)mm2π2{(w5)′(1)−(w5)′(−1)−1m2π2[(w5)′′′(1)−(w5)′′′(−1)]}==(−1)mm2(α0x0π)240β0[β02+α02x02−6m2(α0x0π)2],m=1,2n―Im(3)=∫−11w3(t)cos⁡mπt dt=(−1)mm2π2[(w3)′(1)−(w3)′(−1)]==(−1)mm212β0(α0x0π)2,m=1,2n―(I0(s)=∫−11ws(t)dt;s=3;s=5)bj=5qw′2jπ∫−11w(t)sin⁡jπt dt=(−1)j+1j210qπ(α0x0π)3,j=1,n―
c) Numerical application. By assuming that x 0 = 1 x 0 = 1 x_(0)=1x_{0}=1x0=1, (i.e. x = 1 x = 1 x=1x=1x=1 ), α 0 = − 1 4 α 0 = − 1 4 alpha_(0)=-(1)/(4)\alpha_{0}=-\frac{1}{4}α0=−14, β 0 = 3 4 β 0 = 3 4 beta_(0)=(3)/(4)\beta_{0}=\frac{3}{4}β0=34, (i.e. y w ( x ) = − 1 4 ( x − 3 ) y w ( x ) = − 1 4 ( x − 3 ) y_(w)(x)=-(1)/(4)(x-3)y_{w}(x)=-\frac{1}{4}(x-3)yw(x)=−14(x−3) ), n = 3 , q = 1 n = 3 , q = 1 n=3,q=1n=3, q=1n=3,q=1 and taking into account the above expressions, we have
K 11 = 1 2 [ I 0 ( 5 ) + I 2 ( 5 ) + 10 π 2 ( I 0 ( 3 ) − I 2 ( 3 ) ) ] = 0.810466 K 22 = 1 2 [ I 0 ( 5 ) − I 4 ( 5 ) + 5 2 π 2 ( I 0 ( 3 ) − I 4 ( 3 ) ) ] = 0.450106 K 33 = 1 2 [ I 0 ( 5 ) − I 6 ( 5 ) + 10 9 π 2 ( I 0 ( 3 ) − I 6 ( 3 ) ) ] = 0.381254 K 12 = 1 2 [ I 1 ( 5 ) + I 3 ( 5 ) + 5 π 2 ( I 1 ( 3 ) − I 3 ( 3 ) ) ] = − 0.069901 K 13 = 1 2 [ I 2 ( 5 ) + I 4 ( 5 ) + 10 3 π 2 ( I 2 ( 3 ) − I 4 ( 3 ) ) ] = 0.020117 K 23 = 1 2 [ I 1 ( 5 ) + I 5 ( 5 ) + 5 3 π 2 ( I 1 ( 3 ) − I 5 ( 3 ) ) ] = − 0.062747 K 11 = 1 2 I 0 ( 5 ) + I 2 ( 5 ) + 10 π 2 I 0 ( 3 ) − I 2 ( 3 ) = 0.810466 K 22 = 1 2 I 0 ( 5 ) − I 4 ( 5 ) + 5 2 π 2 I 0 ( 3 ) − I 4 ( 3 ) = 0.450106 K 33 = 1 2 I 0 ( 5 ) − I 6 ( 5 ) + 10 9 π 2 I 0 ( 3 ) − I 6 ( 3 ) = 0.381254 K 12 = 1 2 I 1 ( 5 ) + I 3 ( 5 ) + 5 π 2 I 1 ( 3 ) − I 3 ( 3 ) = − 0.069901 K 13 = 1 2 I 2 ( 5 ) + I 4 ( 5 ) + 10 3 π 2 I 2 ( 3 ) − I 4 ( 3 ) = 0.020117 K 23 = 1 2 I 1 ( 5 ) + I 5 ( 5 ) + 5 3 π 2 I 1 ( 3 ) − I 5 ( 3 ) = − 0.062747 {:[K_(11)=(1)/(2)[I_(0)^((5))+I_(2)^((5))+(10)/(pi^(2))(I_(0)^((3))-I_(2)^((3)))]=0.810466],[K_(22)=(1)/(2)[I_(0)^((5))-I_(4)^((5))+(5)/(2pi^(2))(I_(0)^((3))-I_(4)^((3)))]=0.450106],[K_(33)=(1)/(2)[I_(0)^((5))-I_(6)^((5))+(10)/(9pi^(2))(I_(0)^((3))-I_(6)^((3)))]=0.381254],[K_(12)=(1)/(2)[I_(1)^((5))+I_(3)^((5))+(5)/(pi^(2))(I_(1)^((3))-I_(3)^((3)))]=-0.069901],[K_(13)=(1)/(2)[I_(2)^((5))+I_(4)^((5))+(10)/(3pi^(2))(I_(2)^((3))-I_(4)^((3)))]=0.020117],[K_(23)=(1)/(2)[I_(1)^((5))+I_(5)^((5))+(5)/(3pi^(2))(I_(1)^((3))-I_(5)^((3)))]=-0.062747]:}\begin{aligned} & K_{11}=\frac{1}{2}\left[I_{0}^{(5)}+I_{2}^{(5)}+\frac{10}{\pi^{2}}\left(I_{0}^{(3)}-I_{2}^{(3)}\right)\right]=0.810466 \\ & K_{22}=\frac{1}{2}\left[I_{0}^{(5)}-I_{4}^{(5)}+\frac{5}{2 \pi^{2}}\left(I_{0}^{(3)}-I_{4}^{(3)}\right)\right]=0.450106 \\ & K_{33}=\frac{1}{2}\left[I_{0}^{(5)}-I_{6}^{(5)}+\frac{10}{9 \pi^{2}}\left(I_{0}^{(3)}-I_{6}^{(3)}\right)\right]=0.381254 \\ & K_{12}=\frac{1}{2}\left[I_{1}^{(5)}+I_{3}^{(5)}+\frac{5}{\pi^{2}}\left(I_{1}^{(3)}-I_{3}^{(3)}\right)\right]=-0.069901 \\ & K_{13}=\frac{1}{2}\left[I_{2}^{(5)}+I_{4}^{(5)}+\frac{10}{3 \pi^{2}}\left(I_{2}^{(3)}-I_{4}^{(3)}\right)\right]=0.020117 \\ & K_{23}=\frac{1}{2}\left[I_{1}^{(5)}+I_{5}^{(5)}+\frac{5}{3 \pi^{2}}\left(I_{1}^{(3)}-I_{5}^{(3)}\right)\right]=-0.062747 \end{aligned}K11=12[I0(5)+I2(5)+10π2(I0(3)−I2(3))]=0.810466K22=12[I0(5)−I4(5)+52π2(I0(3)−I4(3))]=0.450106K33=12[I0(5)−I6(5)+109π2(I0(3)−I6(3))]=0.381254K12=12[I1(5)+I3(5)+5π2(I1(3)−I3(3))]=−0.069901K13=12[I2(5)+I4(5)+103π2(I2(3)−I4(3))]=0.020117K23=12[I1(5)+I5(5)+53π2(I1(3)−I5(3))]=−0.062747
b 1 = − 0.015831 ; b 2 = 0.003957 ; b 3 = − 0.001759 b 1 = − 0.015831 ; b 2 = 0.003957 ; b 3 = − 0.001759 b_(1)=-0.015831;b_(2)=0.003957;b_(3)=-0.001759b_{1}=-0.015831 ; b_{2}=0.003957 ; b_{3}=-0.001759b1=−0.015831;b2=0.003957;b3=−0.001759
if we consider only six decimal digits.
The solution of the Ritz system (19) for n = 3 n = 3 n=3\mathrm{n}=3n=3 is
c 1 = 0.018195 ; c 2 = 0.005465 ; c 3 = − 0.012711 c 1 = 0.018195 ; c 2 = 0.005465 ; c 3 = − 0.012711 c_(1)=0.018195;c_(2)=0.005465;c_(3)=-0.012711c_{1}=0.018195 ; c_{2}=0.005465 ; c_{3}=-0.012711c1=0.018195;c2=0.005465;c3=−0.012711
and then the third order Ritz approximate solution u 3 ( x ) ( ≦ a 1 ( x ) ) u 3 ( x ) ≦ a 1 ( x ) u_(3)(x)( <= a_(1)(x))u_{3}(x)\left(\leqq a_{1}(x)\right)u3(x)(≦a1(x)) has the form
(20) a 1 ( x ) = 1 π ( c 1 sin ⁡ π x + 1 2 c 2 sin ⁡ 2 π x + 1 3 c 3 sin ⁡ 3 π x ) (20) a 1 ( x ) = 1 π c 1 sin ⁡ π x + 1 2 c 2 sin ⁡ 2 π x + 1 3 c 3 sin ⁡ 3 π x {:(20)a_(1)(x)=(1)/(pi)(c_(1)sin pi x+(1)/(2)c_(2)sin 2pi x+(1)/(3)c_(3)sin 3pi x):}\begin{equation*} a_{1}(x)=\frac{1}{\pi}\left(c_{1} \sin \pi x+\frac{1}{2} c_{2} \sin 2 \pi x+\frac{1}{3} c_{3} \sin 3 \pi x\right) \tag{20} \end{equation*}(20)a1(x)=1π(c1sin⁡πx+12c2sin⁡2πx+13c3sin⁡3πx)
The approximation of the stream function given by Kantorovich-Ritz method is
(21) ψ a ( x , y ) = { 1 y w ( x ) − a 1 ( x ) [ y − y w ( x ) ] } y (21) ψ a ( x , y ) = 1 y w ( x ) − a 1 ( x ) y − y w ( x ) y {:(21)psi_(a)(x","y)={(1)/(y_(w)(x))-a_(1)(x)[y-y_(w)(x)]}y:}\begin{equation*} \psi_{a}(x, y)=\left\{\frac{1}{y_{w}(x)}-a_{1}(x)\left[y-y_{w}(x)\right]\right\} y \tag{21} \end{equation*}(21)ψa(x,y)={1yw(x)−a1(x)[y−yw(x)]}y
d) Test. Error. The explicit approximate equations y = f ( x , k ) y = f ( x , k ) y=f(x,k)y=f(x, k)y=f(x,k) of the stream lines (obtained from ψ a ( x , y ) = k , k = ψ a ( x , y ) = k , k = psi_(a)(x,y)=k,k=\psi_{a}(x, y)=k, k=ψa(x,y)=k,k= const., k ∈ [ 0 , 1 ] k ∈ [ 0 , 1 ] k in[0,1]k \in[0,1]k∈[0,1] ) in the Kantorovich-Ritz approximation are found by solving
(22) a 1 ( x ) y 2 − [ 1 y w ( x ) + a 1 ( x ) y w ( x ) ] y + k = 0 (22) a 1 ( x ) y 2 − 1 y w ( x ) + a 1 ( x ) y w ( x ) y + k = 0 {:(22)a_(1)(x)y^(2)-[(1)/(y_(w)(x))+a_(1)(x)y_(w)(x)]y+k=0:}\begin{equation*} a_{1}(x) y^{2}-\left[\frac{1}{y_{w}(x)}+a_{1}(x) y_{w}(x)\right] y+k=0 \tag{22} \end{equation*}(22)a1(x)y2−[1yw(x)+a1(x)yw(x)]y+k=0
Table 1.
x x xxx a 1 ( x ) a 1 ( x ) a_(1)(x)a_{1}(x)a1(x) e ( x , 1 4 ) e x , 1 4 e(x,(1)/(4))e\left(x, \frac{1}{4}\right)e(x,14) e ( x , 1 2 ) e x , 1 2 e(x,(1)/(2))e\left(x, \frac{1}{2}\right)e(x,12) e ( x , 3 4 ) e x , 3 4 e(x,(3)/(4))e\left(x, \frac{3}{4}\right)e(x,34)
-0.75 0.00535 0.00083 0.00110 0.00083
-0.50 0.00576 0.00072 0.00096 0.00072
-0.25 0.00361 0.00037 0.00048 0.00037
0.25 -0.00361 -0.00021 -0.00029 -0.00022
0.50 -0.00576 -0.00026 -0.00035 -0.00026
0.75 -0.00535 -0.00017 -0.00024 -0.00017
x a_(1)(x) e(x,(1)/(4)) e(x,(1)/(2)) e(x,(3)/(4)) -0.75 0.00535 0.00083 0.00110 0.00083 -0.50 0.00576 0.00072 0.00096 0.00072 -0.25 0.00361 0.00037 0.00048 0.00037 0.25 -0.00361 -0.00021 -0.00029 -0.00022 0.50 -0.00576 -0.00026 -0.00035 -0.00026 0.75 -0.00535 -0.00017 -0.00024 -0.00017| $x$ | $a_{1}(x)$ | $e\left(x, \frac{1}{4}\right)$ | $e\left(x, \frac{1}{2}\right)$ | $e\left(x, \frac{3}{4}\right)$ | | ---: | :---: | :---: | :---: | :---: | | -0.75 | 0.00535 | 0.00083 | 0.00110 | 0.00083 | | -0.50 | 0.00576 | 0.00072 | 0.00096 | 0.00072 | | -0.25 | 0.00361 | 0.00037 | 0.00048 | 0.00037 | | 0.25 | -0.00361 | -0.00021 | -0.00029 | -0.00022 | | 0.50 | -0.00576 | -0.00026 | -0.00035 | -0.00026 | | 0.75 | -0.00535 | -0.00017 | -0.00024 | -0.00017 |
taking into account that 0 ≤ y ≤ y w ( x ) 0 ≤ y ≤ y w ( x ) 0 <= y <= y_(w)(x)0 \leq y \leq y_{w}(x)0≤y≤yw(x).
On the other hand, the exact equations of the stream lines are straight lines y = g ( x , k ) y = g ( x , k ) y=g(x,k)y=g(x, k)y=g(x,k), where g ( x , k ) = k 4 ( 3 − x ) g ( x , k ) = k 4 ( 3 − x ) g(x,k)=(k)/(4)(3-x)g(x, k)=\frac{k}{4}(3-x)g(x,k)=k4(3−x).
The approximate stream function (21) obtained by Kantorovich-Ritz method is tested in Table 1 which contains the values a 1 ( x ) a 1 ( x ) a_(1)(x)a_{1}(x)a1(x) from (20) and also the errors e ( x , k ) = f ( x , k ) − g ( x , k ) e ( x , k ) = f ( x , k ) − g ( x , k ) e(x,k)=f(x,k)-g(x,k)e(x, k)=f(x, k)-g(x, k)e(x,k)=f(x,k)−g(x,k) with respect to a stream line (by considering k = 1 4 , k = 1 2 , k = 3 4 k = 1 4 , k = 1 2 , k = 3 4 k=(1)/(4),k=(1)/(2),k=(3)/(4)k=\frac{1}{4}, k=\frac{1}{2}, k=\frac{3}{4}k=14,k=12,k=34 ) for several values of x ∈ [ − 1 , 1 ] x ∈ [ − 1 , 1 ] x in[-1,1]x \in[-1,1]x∈[−1,1]. One can observe that these errors are small and consequently, the exact stream function in the case of two plane walls is well approximated by Ψ a Ψ a Psi_(a)\Psi_{a}Ψa given in (21).

4. CONCLUSIONS

It was shown that the Kantorovich-Ritz method can be successfully applied for the incompressible potential fluid flow between two walls. In order to calculate the approximation of the stream function we obtained :
1 ∘ 1 ∘ 1^(@)1^{\circ}1∘. A mathematical model (3)-(4) for the stream function ψ ψ psi\psiψ of a steady state two-dimensional flow in a bounded domain;
2 ∘ 2 ∘ 2^(@)2^{\circ}2∘. Two-points boundary value problem (12)-(13) for a system of N N NNN ordinary differential equations with respect to the variable coefficients a k ( x ) , k = 1 , N ― a k ( x ) , k = 1 , N ¯ a_(k)(x),k= bar(1,N)a_{k}(x), k=\overline{1, N}ak(x),k=1,N― of the Kantorovich method (for the N N NNN-order approximation) considering the trial functions of the form (9);
3 ∘ 3 ∘ 3^(@)3^{\circ}3∘. For the first order Kantorovich approximation on a domain with two plane solid walls, the Ritz variational method is applied and the Ritz algebraic system is constructed by choosing trigonometric trial functions (in the n n nnn-order Ritz approximation);
4 ∘ 4 ∘ 4^(@)4^{\circ}4∘. In the numerical application we effectively found the Ritz solution (20), the approximate stream function (21) and the approximate equation of the stream lines (22), in order to test the exactity of the Kantorovich-Ritz approximate method (Table 1).

REFERENCES

  1. N.M Beliaev and A.A. Riadno, Methods of Theory of Heat Conduction, 1, Moscow, Vyshaia Skola, 1982 (in Russian).
  2. P. Brădeauu, Mecanica fluidelor, Ed. Tehnică, Bucureşti, 1973.
  3. L.E. Elsgolts, Differential Equations and the Calculus of Variations, Mir Publishers, Moscow, English transl., 1977.
  4. L.V. Kantorovich and V.I. Krylov, Approximate Methods of Higher Analysis, Fizmatgiz, 1962 (in Russian).
  5. S.G. Mikhlin, Variational Methods in Mathematical Physics, Nauka, Moscow, 1970 (in Russian).
  6. S. G. Mikhlin, Ecuatii cu derivate parțiale, Ed. Științifică și Enciclopedică, Bucuresti,1973.
1996

Related Posts