On a boundary problem that occurs in a steam boiler project

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C. Kalik
Institutul de Calcul

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C. Kalik, Sur un problème aux limites qui intervient dans un projet d’une chaudière à vapeur. (French) Mathematica (Cluj) 1 (24) 1959 27–34.

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Mathematica Cluj

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Published by the Romanian Academy  Publishing House

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1222-9016

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2601-744X

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1959-Kalik-On-a-Boundary-Problem
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ON A LIMITATION PROBLEM THAT ARISES IN A STEAM BOILER PROJECT

by

C. KALIK

in Cluj
  1. In his work [1], L. NÉMETI addresses a problem that arises when designing a tubular steam boiler with a forced passage. The author proposes a method for calculating the thermal stress in the tube walls. However, calculating the thermal stress requires knowing the temperature, which necessitates solving a boundary value problem. The aim of this work is to provide the solution to this boundary value problem.
First, we introduce the symbols used below and formulate the boundary value problem. Let r , φ , z r , φ , z r,varphi,zr,φ,zcylindrical coordinates in three-dimensional space. The boiler tube is determined by the following inequalities:
r 0 ⩽ r ⩽ r 0 + s ; 0 ⩽ φ ⩽ 2 π ; − L ⩽ x ⩽ + L r 0 ⩽ r ⩽ r 0 + s ; 0 ⩽ φ ⩽ 2 π ; − L ⩽ x ⩽ + L r_(0) <= r <= r_(0)+s;quad0 <= varphi <= 2pi;quad-L <= x <= +Lr0⩽r⩽r0+s;0⩽φ⩽2π;−L⩽x⩽+L
Or r 0 r 0 r_(0)r0is the inner radius, s s ssthe thickness and 2 L 2 L 2L2Lthe length of the tube. The thermal field can be considered to be the same in each section of the tube with the plane φ = φ = varphi=φ=constant. The phenomenon becomes static at a certain point, therefore the function u u uuwhich gives us the temperature values ​​must satisfy the following partial differential equation:
(1.1) HAS ( u ) = 1 r ∂ ∂ r ( r ∂ u ∂ r ) + ∂ 2 u ∂ z 2 = 0 (1.1) HAS ( u ) = 1 r ∂ ∂ r ( r ∂ u ∂ r ) + ∂ 2 u ∂ z 2 = 0 {:(1.1)A(u)=(1)/(r)(del)/(del r)(r(del u)/(del r))+(del^(2)u)/(delz^(2))=0:}(1.1)HAS(u)=1r∂∂r(r∂u∂r)+∂2u∂z2=0
at each point on the tube wall.
The boundary conditions are determined by the following data: outside the tube, a constant regime is maintained such that the heat flux is constant Q Q QQLet's assume that heat does not pass through the extremities z = ± L z = ± L z=+-Lz=±Lof the tube, which means that here the
flow is equal to zero. The water in the tube reaches the level z = 0 z = 0 z=0z=0, while above this level there is steam. Let us also assume that at the inner surface of the tube the heat flux is proportional to the temperature. Based on these data, we obtain the following boundary conditions: ∂ u ∂ v = Q k ∂ u ∂ v = Q k (del u)/(del v)=(Q)/(k)∂u∂v=Qkon the outer surface of the tube, ∂ u ∂ v = 0 ∂ u ∂ v = 0 (del u)/(del v)=0∂u∂v=0at the ends of the tube, and ∂ u ∂ v = h k u ∂ u ∂ v = h k u (del u)/(del v)=(h)/(k)u∂u∂v=hkuon the inner surface of the tube. Here v v vvis the normal external surface of the tube, k k kkis the coefficient of thermal conductivity and h h hhis the heat transfer coefficient. It should be noted that h h hhis equal to the constant h 1 h 1 h_(1)h1for water and at constant h 2 h 2 h_(2)h2for steam. We will write the boundary conditions we just formulated above into a single formula:
(1.2) ∂ u ∂ v − γ u = ψ (1.2) ∂ u ∂ v − γ u = ψ {:(1.2)(del u)/(del v)-gamma u=psi:}(1.2)∂u∂v−γu=ψ
Or γ ⩾ 0 γ ⩾ 0 gamma >= 0γ⩾0, while the measurement of the points for which γ γ gammaγis strictly positive is more g: a ade than zero. Due to the symmetry of the thermal field with respect to φ φ varphiφThe boundary value problem (1.1)-(1.2) is in fact just a problem in the plane. Below we denote by Ω Ω OmegaΩthe planar domain given by the inequalities:
r 0 < r < r 0 + s ; − L < z < + L r 0 < r < r 0 + s ; − L < z < + L r_(0) < r < r_(0)+s;quad-L < z < +Lr0<r<r0+s;−L<z<+L
and by I I IIthe boundary of the domain Ω Ω OmegaΩIt
should be noted that the boundary value problem (1.1)-(1.2) has been studied in several articles [2], [3], [4]. The solution to the boundary value problem is sought by pursuing two approaches: applying the Fourier method and using the theory of functions of complex variables. However, the results obtained have not met the requirements of the technique. This is because the first method yielded an infinite system of linear equations that still needs to be studied. As for the second method, it only provides a solution when the tube thickness is very small, which does not correspond to the technical conditions.
2. We now proceed to the study of the boundary value problem (1.1)-(1.2) using the variational method.
Let's introduce Hilbert space W 2 ( 1 ) ( Ω ) W 2 ( 1 ) ( Ω ) W_(2)^((1))(Omega)W2(1)(Ω)defined as follows: any function v ( r , z ) v ( r , z ) v(r,z)v(r,z)belongs to space W 2 ( 1 ) ( Ω ) W 2 ( 1 ) ( Ω ) W_(2)^((1))(Omega)W2(1)(Ω)if it has all first-order partial derivatives generalized and square-summable (see [5] or [6]). Let us define the norm in this space using the equality
(2.1) ‖ v ‖ W 2 ( 1 ) 2 = ∬ 0 [ ( ∂ v ∂ r ) 2 + ( ∂ v ∂ z ) ] r d Ω + ∫ Γ γ v 2 r d σ (2.1) ‖ v ‖ W 2 ( 1 ) 2 = ∬ 0   [ ( ∂ v ∂ r ) 2 + ( ∂ v ∂ z ) ] r d Ω + ∫ Γ   γ v 2 r d σ {:(2.1)||v||_(W_(2)^((1)))^(2)=∬_(0)[((del v)/(del r))^(2)+((del v)/(del z))]rd Omega+int_(Gamma)gammav^(2)rd sigma:}(2.1)‖v‖W2(1)2=∬0[(∂v∂r)2+(∂v∂z)]rdΩ+∫Γγv2rdσ
Or d Ω d Ω of OmegadΩis the element of the surface and d σ d σ d sigmadσis the element of the arc. The space W 2 ( 1 ) ( Ω ) W 2 ( 1 ) ( Ω ) W_(2)^((1))(Omega)W2(1)(Ω)is complete and separable [5]. We will study the functional
(2.2) F ( v ) = ∬ Ω [ ( ∂ v ∂ r ) 2 + ( ∂ v ∂ z ) 2 ] r d Ω + ∫ Γ γ v 2 r d σ − 2 ∫ Γ ψ ⋅ v ⋅ r d σ (2.2) F ( v ) = ∬ Ω   [ ( ∂ v ∂ r ) 2 + ( ∂ v ∂ z ) 2 ] r d Ω + ∫ Γ   γ v 2 r d σ − 2 ∫ Γ   ψ ⋅ v ⋅ r d σ {:(2.2)F(v)=∬_(Omega)[((del v)/(del r))^(2)+((del v)/(del z))^(2)]rd Omega+int_(Gamma)gammav^(2)rd sigma-2int_(Gamma)psi*v*rd sigma:}(2.2)F(v)=∬Ω[(∂v∂r)2+(∂v∂z)2]rdΩ+∫Γγv2rdσ−2∫Γψ⋅v⋅rdσ
defined on the elements of space W 2 ( 1 ) ( Ω ) W 2 ( 1 ) ( Ω ) W_(2)^((1))(Omega)W2(1)(Ω)The existence of line integrals is guaranteed by SL Sobolev's immersion theorems
Lemma (2.1). The functional (2.2) is bounded internally on the set W 2 ( 1 ) ( Ω ) W 2 ( 1 ) ( Ω ) W_(2)^((1))(Omega)W2(1)(Ω).
Indeed, according to the Cauchy-Buniakovsky inequalities and SL Sobolev's immersion theorems, we can write
∫ Γ ψ ⋅ v ⋅ r d σ ⩽ ‖ ψ ‖ L 2 ( Γ ) ⋅ ‖ v ‖ L 2 ( Γ ) ⩽ ‖ ψ ‖ L 2 ( Γ ) ⋅ ‖ v ‖ W 2 ( 1 ) K 1 = K ⋅ ‖ v ‖ W 2 ( 1 ) ∫ Γ   ψ ⋅ v ⋅ r d σ ⩽ ‖ ψ ‖ L 2 ( Γ ) ⋅ ‖ v ‖ L 2 ( Γ ) ⩽ ‖ ψ ‖ L 2 ( Γ ) ⋅ ‖ v ‖ W 2 ( 1 ) K 1 = K ⋅ ‖ v ‖ W 2 ( 1 ) int_(Gamma)psi*v*rd sigma <= ||psi||_(L_(2)(Gamma))*||v||_(L_(2)(Gamma)) <= ||psi||_(L_(2)(Gamma))*||v||_(W_(2)^((1)))K_(1)=K*||v||_(W_(2)^((1)))∫Γψ⋅v⋅rdσ⩽‖ψ‖L2(Γ)⋅‖v‖L2(Γ)⩽‖ψ‖L2(Γ)⋅‖v‖W2(1)K1=K⋅‖v‖W2(1)
for each v ∈ W 2 ( 1 ) ( Ω ) v ∈ W 2 ( 1 ) ( Ω ) v inW_(2)^((1))(Omega)v∈W2(1)(Ω)The constant K > 0 K > 0 K > 0K>0does not depend on the function v ( r , z ) v ( r , z ) v(r,z)v(r,z)and the standards of space L 2 ( Γ ) L 2 ( Γ ) L_(2)(Gamma)L2(Γ)are calculated using the weighting function r r rrUsing this inequality, we obtain
F ( v ) = ‖ v ‖ W 2 ( 1 ) 2 − 2 ∫ Γ ψ ⋅ v ⋅ r d σ ⩾ ‖ v ‖ W 2 ( 1 ) 2 − 2 K ‖ v ‖ W 2 ( 1 ) = = ( ‖ v ‖ W 2 ( 1 ) − K ) 2 − K 2 ⩾ − K 2 F ( v ) = ‖ v ‖ W 2 ( 1 ) 2 − 2 ∫ Γ   ψ ⋅ v ⋅ r d σ ⩾ ‖ v ‖ W 2 ( 1 ) 2 − 2 K ‖ v ‖ W 2 ( 1 ) = = ( ‖ v ‖ W 2 ( 1 ) − K ) 2 − K 2 ⩾ − K 2 {:[F(v)=||v||_(W_(2)^((1)))^(2)-2int_(Gamma)psi*v*rd sigma >= ||v||_(W_(2)^((1)))^(2)-2K||v||_(W_(2)^((1)))=],[=(||v||_(W_(2)^((1)))-K)^(2)-K^(2) >= -K^(2)]:}F(v)=‖v‖W2(1)2−2∫Γψ⋅v⋅rdσ⩾‖v‖W2(1)2−2K‖v‖W2(1)==(‖v‖W2(1)−K)2−K2⩾−K2
which means that lemma (2.1) is proven.
Let inf F ( v ) = − d F ( v ) = − d F(v)=-dF(v)=−dBy applying the ideas used by S.L. Sobolev ν ∈ W 2 ′ 1 ( Ω ) ν ∈ W 2 ′ 1 ( Ω ) nu inW_(2)^('1)(Omega)ν∈W2′1(Ω)
For the solution to Neumann's problem [5], I will prove the following lemma:
Lemma (2.2). In Hilbert space W 2 ( 1 ) ( Ω ) W 2 ( 1 ) ( Ω ) W_(2)^((1))(Omega)W2(1)(Ω)there is a function u ( r , z ) u ( r , z ) u(r,z)u(r,z)for which
inf v ∈ W 2 ( 1 ) ( Ω ) F ( v ) = F ( u ) = − d inf v ∈ W 2 ( 1 ) ( Ω )   F ( v ) = F ( u ) = − d in f_(v inW_(2)^((1))(Omega))F(v)=F(u)=-dinfv∈W2(1)(Ω)F(v)=F(u)=−d
and for which we have
(2.3) ∬ Ω [ ∂ u ∂ r ∂ v ∂ r + ∂ u ∂ z ∂ v ∂ z ] r d Ω + ∫ Γ γ u v r d σ − ∫ Γ ψ v r d σ = 0 (2.3) ∬ Ω   [ ∂ u ∂ r ∂ v ∂ r + ∂ u ∂ z ∂ v ∂ z ] r d Ω + ∫ Γ   γ u v r d σ − ∫ Γ   ψ v r d σ = 0 {:(2.3)∬_(Omega)[(del u)/(del r)(del v)/(del r)+(del u)/(del z)(del v)/(del z)]rd Omega+int_(Gamma)gamma uvrd sigma-int_(Gamma)psi vrd sigma=0:}(2.3)∬Ω[∂u∂r∂v∂r+∂u∂z∂v∂z]rdΩ+∫Γγuvrdσ−∫Γψvrdσ=0
regardless of v ∈ W 2 ( 1 ) ( Ω ) v ∈ W 2 ( 1 ) ( Ω ) v inW_(2)^((1))(Omega)v∈W2(1)(Ω)
Let us prove lemma (2.2). Let { u n } { u n } {A)}{un}a sequence of functions of space W 2 ( 1 ) ( Ω ) W 2 ( 1 ) ( Ω ) W_(2)^((1))(Omega)W2(1)(Ω)for which
(2.4) limit n → ∞ F ( u n ) = − d (2.4) limit n → ∞   F ( u n ) = − d {:(2.4)lim_(n rarr oo)F{:(u_(n)):}=-d:}(2.4)limitn→∞F(un)=−d
We can immediately verify that
1 2 F ( u n ) + 1 2 F ( u m ) − F ˙ ( u n + u m 2 ) = 1 2 ‖ u n ‖ W 2 ( 1 ) 2 + 1 2 ‖ u m ‖ W 2 ( 1 ) 2 − − ‖ u ˙ n + u m 2 ‖ W 2 ( 1 ) 2 = ‖ u ˙ n − u m 2 ‖ W 2 ( 1 ) 2 1 2 F ( u n ) + 1 2 F ( u m ) − F ˙ ( u n + u m 2 ) = 1 2 ‖ u n ‖ W 2 ( 1 ) 2 + 1 2 ‖ u m ‖ W 2 ( 1 ) 2 − − ‖ u ˙ n + u m 2 ‖ W 2 ( 1 ) 2 = ‖ u ˙ n − u m 2 ‖ W 2 ( 1 ) 2 {:[(1)/(2)F{:(u_(n)):}+(1)/(2)F{:(u_(m)):}-F^(˙)((u_(n)+u_(m))/(2))=(1)/(2)||u_(n)||_(W_(2)^((1)))^(2)+(1)/(2)||u_(m)||_(W_(2)^((1)))^(2)-],[-||(u^(˙)_(n)+u_(m))/(2)||_(W_(2)^((1)))^(2)=||(u^(˙)_(n)-u_(m))/(2)||_(W_(2)^((1)))^(2)]:}12F(un)+12F(um)−F˙(un+um2)=12‖un‖W2(1)2+12‖um‖W2(1)2−−‖u˙n+um2‖W2(1)2=‖u˙n−um2‖W2(1)2
Therefore, it follows from (2.4) that ‖ u n − u m ‖ W 2 ( 1 ) → 0 ‖ u n − u m ‖ W 2 ( 1 ) → 0 ||u_(n)-u_(m)||_(W_(2)^((1)))rarr0‖un−um‖W2(1)→0But space W 2 ( 1 ) ( Ω ) W 2 ( 1 ) ( Ω ) W_(2)^((1))(Omega)W2(1)(Ω)being complete, it follows that the following { u n } { u n } {u_(n)}{un}tends towards a function of space. Let us denote this function by u ( r , z ) u ( r , z ) u(r,z)u(r,z)In light of the above, it is evident that
inf v ∈ W 2 ( 1 ) ( Ω ) F ( v ) = F ( u ) = − d inf v ∈ W 2 ( 1 ) ( Ω )   F ( v ) = F ( u ) = − d i n f_(v inW_(2)^((1))(Omega))F(v)=F(u)=-dinfv∈W2(1)(Ω)F(v)=F(u)=−d
Let's return to equation (2.3). We have
F ( u + λ v ) = F ( u ) + 2 λ { ∬ 0 [ ∂ u ∂ r ∂ v ∂ r + ∂ u ∂ z ∂ v ∂ z ] r d Ω + ∫ Γ γ u v r d σ − ∫ Γ ψ v r d σ } + + λ 2 ‖ v ‖ W 2 ( 1 ) 2 F ( u + λ v ) = F ( u ) + 2 λ { ∬ 0   [ ∂ u ∂ r ∂ v ∂ r + ∂ u ∂ z ∂ v ∂ z ] r d Ω + ∫ Γ   γ u v r d σ − ∫ Γ   ψ v r d σ } + + λ 2 ‖ v ‖ W 2 ( 1 ) 2 {:[F(u+lambda v)=F(u)+2lambda{:{∬_(0)[(del u)/(del r)(del v)/(del r)+(del u)/(del z)(del v)/(del z)]rd Omega+int_(Gamma)gamma uvrd sigma-int_(Gamma)psi vrd sigma}:}+],[+lambda^(2)||v||_(W_(2)^((1)))^(2)]:}F(u+λv)=F(u)+2λ{∬0[∂u∂r∂v∂r+∂u∂z∂v∂z]rdΩ+∫Γγuvrdσ−∫Γψvrdσ}++λ2‖v‖W2(1)2
for any real value of the parameter λ λ lambdaλand for any function v v vvspace W 2 ( 1 ) ( Ω ) W 2 ( 1 ) ( Ω ) W_(2)^((1))(Omega)W2(1)(Ω)By virtue of the fact that this expression must reach its minimum for λ = 0 λ = 0 lambda=0λ=0We obtain the necessary relation.
Theorem (2.1) The function u ( r , z ) u ( r , z ) u(r,z)u(r,z)which achieves the minimum of the functional (2.2) admits partial derivatives of almost any order at nearly every interior point of the domain Ω Ω OmegaΩAt the same time, it satisfies equation (1.1) almost everywhere in Ω Ω OmegaΩ.
Demonstration. Let Ω ′ Ω ′ Omega^(')Ω′an arbitrary domain, internal to the domain Ω , δ Ω , δ Omega,deltaΩ,δthe distance between Ω ′ Ω ′ Omega^(')Ω′And Γ , Q Γ , Q Gamma,QΓ,Qany point of Ω ′ Ω ′ Omega^(')Ω′Let's choose a function g ( ρ ) g ( ρ ) g(rho)g(ρ)which meets the following conditions
g ( ρ ) = { 1 pour ρ ⩽ 1 2 0 pour ρ ⩾ 1 g ( ρ ) = { 1 pour ρ ⩽ 1 2 0 pour ρ ⩾ 1 g(rho)={{:[1"pour"rho <= (1)/(2)],[0"pour"rho >= 1]:}g(ρ)={1Forρ⩽120Forρ⩾1
and which is infinitely differentiable. ρ ρ rhoρdenotes the distance between the points P P PPAnd Q , P Q , P Q,PQ,Pbeing an arbitrary point of the plane r , z r , z r,zr,zWe refer to L ( P , Q ) L ( P , Q ) L(P,Q)L(P,Q)the fundamental solution (in the sense of EE Levi [8]) of equation (1.1).
Let's introduce the function ξ ( p ) ξ ( p ) xi(p)ξ(p)determined as follows
(2.5) ξ ( p ) = [ g ( ϱ h 1 ) − g ( ϱ h 2 ) ] L ( P , Q ) (2.5) ξ ( p ) = [ g ( ϱ h 1 ) − g ( ϱ h 2 ) ] L ( P , Q ) {:(2.5)xi(p)=[g((ϱ)/(h_(1)))-g((ϱ)/(h_(2)))]L(P","Q):}(2.5)ξ(p)=[g(ϱh1)−g(ϱh2)]L(P,Q)
Or h 1 h 1 h_(1)h1And h 2 h 2 h_(2)h2are two constants that satisfy the condition 0 < h 1 < h 2 < δ 0 < h 1 < h 2 < δ 0 < h_(1) < h_(2) < delta0<h1<h2<δWe will demonstrate that the set of functions { ω h ( ρ ) } { ω h ( ρ ) } {omega_(h)(rho)}{ωh(ρ)}Or
(2.6) ω h ( ρ ) = h 2 A [ g ( ϱ h ) L ( P , Q ) ] (2.6) ω h ( ρ ) = h 2 A [ g ( ϱ h ) L ( P , Q ) ] {:(2.6)omega_(h)(rho)=h^(2)A[g((ϱ)/(h))L(P","Q)]:}(2.6)ωh(ρ)=h2HAS[g(ϱh)L(P,Q)]
h h hhvarying between δ δ deltaδand 0, forms a set of regular kernels according to Sobolev's theorem [7]. That is to say, it must be shown that { ω h ( ρ ) } { ω h ( ρ ) } {omega_(h)(rho)}{ωh(ρ)}meets the following conditions:
  1. { ω h ( ρ ) } { ω h ( ρ ) } {omega_(h)(rho)}{ωh(ρ)}are uniformly limited with respect to h h hhand to Q ∈ Ω ′ Q ∈ Ω ′ Q inOmega^(')Q∈Ω′,
  2. each function ω h ( ρ ) ω h ( ρ ) omega_(h)(rho)ωh(ρ)is summable with respect to each coordinate of the points P P PPAnd Q Q QQ,
  3. there are positive numbers γ γ gammaγAnd ε ε epsiεsuch that for each point Q ∈ Ω ′ Q ∈ Ω ′ Q inOmega^(')Q∈Ω′we have
∬ ϱ < h ω h ( ρ ) d Ω p > γ h 2 ∬ ϱ < h   ω h ( ρ ) d Ω p > γ h 2 ∬_(ϱ < h)omega_(h)(rho)dOmega_(p) > gammah^(2)∬ϱ<hωh(ρ)dΩp>γh2
if h < ε h < ε h < epsih<εand
4. outside the circle having the center Q Q QQand the radius h h hhthe function ω h ( ρ ) ω h ( ρ ) omega_(h)(rho)ωh(ρ)is considered identically zero.
Let's check condition 1. For ρ ⩽ h 2 ρ ⩽ h 2 rho <= (h)/(2)ρ⩽h2We have ω h ( ρ ) = 0 ω h ( ρ ) = 0 omega_(h)(rho)=0ωh(ρ)=0by virtue of the definition of the fundamental solution L ( P , Q ) L ( P , Q ) L(P,Q)L(P,Q)Therefore, condition 1 must be checked for ρ > h 2 ρ > h 2 rho > (h)/(2)ρ>h2It is easy to notice that for ρ > h 2 ρ > h 2 rho > (h)/(2)ρ>h2the absolute values ​​of the first-order partial derivatives of the function g ( ϱ h ) g ( ϱ h ) g((ϱ)/(h))g(ϱh)are bounded above by C 1 h C 1 h (C_(1))/(h)C1h, while the absolute values ​​of the second-order partial derivatives are bounded by C 2 h C 2 h (C_(2))/(h)C2h. C 1 C 1 C_(1)C1And C 2 C 2 C_(2)C2being positive constants. The function L ( P , Q ) L ( P , Q ) L(P,Q)L(P,Q)can be written in the form L ( P , Q ) = 1 2 π ln ⁡ 1 ϱ + W ( P , Q ) L ( P , Q ) = 1 2 π ln ⁡ 1 ϱ + W ( P , Q ) L(P,Q)=(1)/(2pi)ln((1)/(ϱ))+W(P,Q)L(P,Q)=12πln⁡1ϱ+W(P,Q)Or W ( P , Q ) W ( P , Q ) W(P,Q)W(P,Q)and all its partial derivatives possess a singularity smaller than the function 1 2 π ln ⁡ 1 ϱ 1 2 π ln ⁡ 1 ϱ (1)/(2pi)ln((1)/(ϱ))12πln⁡1ϱand its derivatives, when ρ → 0 ρ → 0 rho rarr0ρ→0It follows that the absolute values ​​of the first-order partial derivatives, as well as those of the second order, are bounded above by C 3 h C 3 h (C_(3))/(h)C3hrespectively by C 4 h C 4 h (C_(4))/(h)C4h, when ρ > h 2 . C 3 ρ > h 2 . C 3 rho > (h)/(2).C_(3)ρ>h2.C3And C 4 C 4 C_(4)C4are also positive constants. Having in mind the form of equation (1.1), it follows from the above.
| ω h ( ρ ) | < C | ω h ( ρ ) | < C |omega_(h)(rho)| < C|ωh(ρ)|<C
whatever h h hhAnd Q Q QQ. C C CCis a positive constant. Conditions 2 and 4 are easy to verify since the function L ( P , Q ) L ( P , Q ) L(P,Q)L(P,Q)is infinitely differentiable with respect to the coordinates of the points P P PPAnd Q Q QQ; and the function g ( ϱ h ) = 0 g ( ϱ h ) = 0 g((ϱ)/(h))=0g(ϱh)=0When ρ ⩾ h ρ ⩾ h rho >= hρ⩾hLet's calculate the integral of condition 3.
But
∬ ϱ < h ω h ( ρ ) d Ω P p = h 2 ∬ ϱ < h A [ g ( ϱ h ) L ( P , Q ) ] d Ω P ∬ ϱ < h A [ g ( ϱ h ) L ( P , Q ) ] d Ω P = ∬ h 2 < ϱ < h A [ g ( ϱ h ) L ( P , Q ) ] d Ω P = ∫ C h 2 [ ∂ L ∂ v + + L r cos ⁡ ( v , r ) ] d σ P = − ∫ C h 2 ( ∂ L ∂ ϱ + L r ) d σ P → 1 ∬ ϱ < h   ω h ( ρ ) d Ω P p = h 2 ∬ ϱ < h   A [ g ( ϱ h ) L ( P , Q ) ] d Ω P ∬ ϱ < h   A [ g ( ϱ h ) L ( P , Q ) ] d Ω P = ∬ h 2 < ϱ < h   A [ g ( ϱ h ) L ( P , Q ) ] d Ω P = ∫ C h 2   [ ∂ L ∂ v + + L r cos ⁡ ( v , r ) ] d σ P = − ∫ C h 2   ( ∂ L ∂ ϱ + L r ) d σ P → 1 {:[∬_(ϱ < h)omega_(h)(rho)dOmega_(P)p=h^(2)∬_(ϱ < h)A[g((ϱ)/(h))L(P","Q)]dOmega_(P)],[∬_(ϱ < h)A[g((ϱ)/(h))L(P","Q)]dOmega_(P)=∬_((h)/(2) < ϱ < h)A[g((ϱ)/(h))L(P","Q)]dOmega_(P)=int_(C_((h)/(2)))[(del L)/(del v)+],[+(L)/(r)cos(v","r)]dsigma_(P)=-int_(C_((h)/(2)))((del L)/(delϱ)+(L)/(r))dsigma_(P)rarr1]:}∬ϱ<hωh(ρ)dΩPp=h2∬ϱ<hHAS[g(ϱh)L(P,Q)]dΩP∬ϱ<hHAS[g(ϱh)L(P,Q)]dΩP=∬h2<ϱ<hHAS[g(ϱh)L(P,Q)]dΩP=∫Ch2[∂L∂v++Lrcos⁡(v,r)]dσP=−∫Ch2(∂L∂ϱ+Lr)dσP→1
when the radius of the circle C h 2 C h 2 C_((h)/(2))Ch2having the center in Q Q QQtends towards zero uniformly for each Q ∈ Ω ′ Q ∈ Ω ′ Q inOmega^(')Q∈Ω′. Either γ γ gammaγan arbitrary number in the interval ( 0 , 1 ) ( 0 , 1 ) (0,1)(0,1)We deduce from the above that we can determine a number ε > 0 ε > 0 epsi > 0ε>0so that we have
∬ ϱ < h ω h ( ρ ) d Ω P > γ h 2 ∬ ϱ < h   ω h ( ρ ) d Ω P > γ h 2 ∬_(ϱ < h)omega_(h)(rho)dOmega_(P) > gammah^(2)∬ϱ<hωh(ρ)dΩP>γh2
if h < ε h < ε h < epsih<εThe
whole { ω h ( ρ ) } { ω h ( ρ ) } {omega_(h)(rho)}{ωh(ρ)}being a regular set of kernels, as proved above, it follows that the functions
(2.7) u h ( Q ) = ∬ o < h ω h ( ρ ) u ( P ) d Ω P ∬ o < h ω h ( ρ ) d Ω P (2.7) u h ( Q ) = ∬ o < h   ω h ( ρ ) u ( P ) d Ω P ∬ o < h   ω h ( ρ ) d Ω P {:(2.7)u_(h)(Q)=(∬_(o < h)omega_(h)(rho)u(P)dOmega_(P))/(∬_(o < h)omega_(h)(rho)dOmega_(P)):}(2.7)uh(Q)=∬o<hωh(ρ)u(P)dΩP∬o<hωh(ρ)dΩP
almost in every Q ∈ Ω ′ Q ∈ Ω ′ Q inOmega^(')Q∈Ω′tend towards u ( Q ) u ( Q ) u(Q)u(Q)When h → 0 h → 0 h rarr0h→0Now
we can move on to the proof of theorem (2.1). It is obvious that the function ξ ( P ) ξ ( P ) xi(P)ξ(P)belongs to space W 2 ( 1 ) ( Ω ) W 2 ( 1 ) ( Ω ) W_(2)^((1))(Omega)W2(1)(Ω)It is also easy to see that the same function ξ ( P ) ξ ( P ) xi(P)ξ(P)is zero in a band adjacent to the boundary Γ Γ GammaΓTherefore, from (2.3) we deduce
(2.8) ∫ Ω ∫ [ ∂ u ∂ r ∂ ξ ∂ r + ∂ u ∂ z ∂ ξ ∂ z ] r d Ω = 0 (2.8) ∫ Ω   ∫ [ ∂ u ∂ r ∂ ξ ∂ r + ∂ u ∂ z ∂ ξ ∂ z ] r d Ω = 0 {:(2.8)int_(Omega)int[(del u)/(del r)(del xi)/(del r)+(del u)/(del z)(del xi)/(del z)]rd Omega=0:}(2.8)∫Ω∫[∂u∂r∂ξ∂r+∂u∂z∂ξ∂z]rdΩ=0
where by r r rrAnd z z zzthe coordinates of the point are designated P P PPApplying Green's formula to (2.8), we obtain
(2.9) ∬ Ω u A [ ξ ( P ) ] r d Ω = 0 (2.9) ∬ Ω   u A [ ξ ( P ) ] r d Ω = 0 {:(2.9)∬_(Omega)uA[xi(P)]rd Omega=0:}(2.9)∬ΩuHAS[ξ(P)]rdΩ=0
hence, according to (2.6) it follows
(2.10) 1 h 1 2 ∬ ω h 1 ( ρ ) u ( P ) r d Ω P = 1 h 2 2 ∬ ω h 2 ( ρ ) u ( P ) r d Ω P (2.10) 1 h 1 2 ∬ ω h 1 ( ρ ) u ( P ) r d Ω P = 1 h 2 2 ∬ ω h 2 ( ρ ) u ( P ) r d Ω P {:(2.10)(1)/(h_(1)^(2))∬omega_(h_(1))(rho)u(P)rdOmega_(P)=(1)/(h_(2)^(2))∬omega_(h_(2))(rho)u(P)rdOmega_(P):}(2.10)1h12∬ωh1(ρ)u(P)rdΩP=1h22∬ωh2(ρ)u(P)rdΩP
We can immediately see that the two sides of the last equality converge at each point Q ∈ Ω ′ Q ∈ Ω ′ Q inOmega^(')Q∈Ω′where the function (2.7) also converges. But according to equality (2.10) we write
u ( Q ) = 1 h 2 ∬ Q ω h ( ρ ) u ( P ) r d Ω P u ( Q ) = 1 h 2 ∬ Q   ω h ( ρ ) u ( P ) r d Ω P u(Q)=(1)/(h^(2))∬_(Q)omega_(h)(rho)u(P)rdOmega_(P)u(Q)=1h2∬Qωh(ρ)u(P)rdΩP
which shows that the function u ( Q ) u ( Q ) u(Q)u(Q)is infinitely differentiable, the second member of the last equality also being infinitely differentiable.
We will demonstrate that the function u u uusatisfies equation (1.1) almost everywhere in Ω Ω OmegaΩ. Either v ∈ W 2 ( 1 ) ( Ω ) v ∈ W 2 ( 1 ) ( Ω ) v inW_(2)^((1))(Omega)v∈W2(1)(Ω)an arbitrary function equal to zero in a strip neighboring the boundary Γ Γ GammaΓWe deduce from (2.3)
∫ Ω ∫ [ ∂ u ∂ r ∂ v ∂ r + ∂ u ∂ z ∂ v ∂ z ] r d Ω = 0 ∫ Ω   ∫ [ ∂ u ∂ r ∂ v ∂ r + ∂ u ∂ z ∂ v ∂ z ] r d Ω = 0 int_(Omega)int[(del u)/(del r)(del v)/(del r)+(del u)/(del z)(del v)/(del z)]rd Omega=0∫Ω∫[∂u∂r∂v∂r+∂u∂z∂v∂z]rdΩ=0
Applying Green's formula, we obtain
∬ Q v A ( u ) r d Ω = 0 ∬ Q   v A ( u ) r d Ω = 0 ∬_(Q)vA(u)rd Omega=0∬QvHAS(u)rdΩ=0
It follows that A ( u ) = 0 A ( u ) = 0 A(u)=0HAS(u)=0is satisfied almost everywhere in Ω Ω OmegaΩTherefore, theorem (2.1) is proven.
We now have to consider in which direction the boundary condition (1.2) is satisfied and to examine whether the solution of the boundary problem (1.1)-(1.2) is unique.
As expected, condition (1.2) is satisfied in a weak sense. We will denote by { Ω m } { Ω m } {Omega_(m)}{Ωm}a series of domains that tend towards Ω Ω OmegaΩ, When m → ∞ m → ∞ m rarr oom→∞Suppose that the following { Ω m } { Ω m } {Omega_(m)}{Ωm}is monotonous, that is to say Ω m ⊆ Ω m + 1 ⊂ Ω Ω m ⊆ Ω m + 1 ⊂ Ω Omega_(m)subeOmega_(m+1)sub OmegaΩm⊆Ωm+1⊂Ω. Either Γ m Γ m Gamma_(m)Γmthe border of Ω m Ω m Omega_(m)Ωm, which we assume continues on the segment. For a v ∈ W 2 ( 1 ) ( Ω ) v ∈ W 2 ( 1 ) ( Ω ) v inW_(2)^((1))(Omega)v∈W2(1)(Ω)arbitrary we obviously have
∬ Q [ ∂ u ∂ r i v ∂ r + ∂ u ∂ z ∂ v ∂ z ] r d Ω = lim m → ∞ ∫ Q m ∫ m [ ∂ u ∂ r ∂ v ∂ r + ∂ u ∂ z ∂ v ∂ z ] r d Ω = lim m → ∞ ∫ Γ m v ∂ u ∂ v r d σ ∬ Q   [ ∂ u ∂ r i v ∂ r + ∂ u ∂ z ∂ v ∂ z ] r d Ω = lim m → ∞   ∫ Q m   ∫ m   [ ∂ u ∂ r ∂ v ∂ r + ∂ u ∂ z ∂ v ∂ z ] r d Ω = lim m → ∞   ∫ Γ m   v ∂ u ∂ v r d σ ∬_(Q)[(del u)/(del r)(iv)/(del r)+(del u)/(del z)(del v)/(del z)]rd Omega=lim_(m rarr oo)int_(Q_(m))int_(m)[(del u)/(del r)(del v)/(del r)+(del u)/(del z)(del v)/(del z)]rd Omega=lim_(m rarr oo)int_(Gamma_(m))v(del u)/(del v)rd sigma∬Q[∂u∂riv∂r+∂u∂z∂v∂z]rdΩ=limitm→∞∫Qm∫m[∂u∂r∂v∂r+∂u∂z∂v∂z]rdΩ=limitm→∞∫Γmv∂u∂vrdσ
From (2.3) we obtain the following formula which shows in which direction condition (1.2) is satisfied. Namely:
(2.11) lim m → ∞ ∫ Γ m v ∂ n ∂ v r d σ = − ∫ Γ γ u v r d σ + ∫ Γ ψ v r d σ (2.11) lim m → ∞   ∫ Γ m   v ∂ n ∂ v r d σ = − ∫ Γ   γ u v r d σ + ∫ Γ   ψ v r d σ {:(2.11)lim_(m rarr oo)int_(Gamma_(m))v(del n)/(del v)rd sigma=-int_(Gamma)gamma uvrd sigma+int_(Gamma)psi vrd sigma:}(2.11)limitm→∞∫Γmv∂n∂vrdσ=−∫Γγuvrdσ+∫Γψvrdσ
As for the uniqueness of the solution to the boundary value problem, it can be proven in the following way. Suppose there are two solutions u 1 u 1 u_(1)u1And u 2 u 2 u_(2)u2of the problem to the boundaries that are part of the space W 2 ( 1 ) ( Ω ) W 2 ( 1 ) ( Ω ) W_(2)^((1))(Omega)W2(1)(Ω)The function u = u 1 − u 2 u = u 1 − u 2 u=u_(1)-u_(2)u=u1−u2satisfies the condition
lim m → ∞ ∫ Γ m v ∂ u ∂ v r d σ = − ∫ Γ γ u v r d σ lim m → ∞   ∫ Γ m   v ∂ u ∂ v r d σ = − ∫ Γ   γ u v r d σ lim_(m rarr oo)int_(Gamma_(m))v(del u)/(del v)rd sigma=-int_(Gamma)gamma uvrd sigmalimitm→∞∫Γmv∂u∂vrdσ=−∫Γγuvrdσ
By replacing v v vvby u u uuwe have
lim m → ∞ ∫ Γ m ∫ m [ ( ∂ u ∂ r ) 2 + ( ∂ u ∂ z ) 2 ] r d Ω + ∫ Γ γ u 2 r d σ = 0 lim m → ∞   ∫ Γ m   ∫ m   [ ( ∂ u ∂ r ) 2 + ( ∂ u ∂ z ) 2 ] r d Ω + ∫ Γ   γ u 2 r d σ = 0 lim_(m rarr oo)int_(Gamma m)int_(m)[((del u)/(del r))^(2)+((del u)/(del z))^(2)]rd Omega+int_(Gamma)gammau^(2)rd sigma=0limitm→∞∫Γm∫m[(∂u∂r)2+(∂u∂z)2]rdΩ+∫Γγu2rdσ=0
But bearing in mind the condition imposed on γ γ gammaγIn p. 1, it follows from the preceding equality that u ≡ 0 u ≡ 0 u-=0u≡0in Ω Ω OmegaΩ3.
For the approximate calculation of the solution to the boundary value problem (1.1)-(1.2), we will use the generalized Ritz method. We will seek the approximation of the order n n nnof the solution u ( r , z ) u ( r , z ) u(r,z)u(r,z)in the following form:
(3.1) u n ( r , z ) = ∑ p , q = 0 n a p q r p z q (3.1) u n ( r , z ) = ∑ p , q = 0 n   a p q r p z q {:(3.1)u_(n)(r","z)=sum_(p,q=0)^(n)a_(pq)r^(p)z^(q):}(3.1)un(r,z)=∑p,q=0nhaspqrpzq
Let's determine the coefficients a p q a p q a_(pq)haspqso that the function F ( u n ) = F ( a p q ) ( p , q = 0 , … , n ) F ( u n ) = F ( a p q ) ( p , q = 0 , … , n ) F(u_(n))=F(a_(pq))(p,q=0,dots,n)F(un)=F(haspq)(p,q=0,…,n)takes the minimum. We therefore obtain the approximations u n ( r , z ) u n ( r , z ) u_(n)(r,z)un(r,z)by solving the linear system of equations
(3.2) ∂ F ∂ a p q = 0 ( p , q = 0 , … , n ) (3.2) ∂ F ∂ a p q = 0 ( p , q = 0 , … , n ) {:(3.2)(del F)/(dela_(pq))=0quad(p","q=0","dots","n):}(3.2)∂F∂haspq=0(p,q=0,…,n)
It is obvious that the approximate solutions u n u n u_(n)untend towards the exact solution of the boundary value problem in the metric of space W 2 ( 1 ) ( Ω ) W 2 ( 1 ) ( Ω ) W_(2)^((1))(Omega)W2(1)(Ω)and according to SL Sobolev's immersion theorems, we can affirm that the functions u n u n u_(n)untend towards u u uualso in the middle square.

BIBLIOGRAPHY

  1. Németi L., Tensiuni termice în tuburi cu pereți subtiri în cazul unui cîmp termic simetric faţă de axă. Studii și Cercetări ştiințifice, Cluj. t. IV (1953) 64-72.
  2. Călugăreanu G., Asupra unei probleme de propagarea căldurii. Studii şi cercetări științifice, Cluj t. IV (1953) 10-17.
  3. Ionescu DV şi Né meti L., Integrarea uni ecuatii cu derivate partiale care ivtervine in problema calculului tensiunilor termice în tuburile fierbătoare ale cazanelor cu trecere forfată şi ale cazanelor cu radiație. Studii şi Cercetări Ştiințifice Cluj, t. IV (1953) 73-78.
  4. Călugăreanu G. şi Rado F., Asupra uni problem de propagare a căldurii. Bulletin Ştiințific Secțiunea de Ştiințe Matematice şi Fizice t. VI (1954), 17-30.
  5. Соболев С. Л., Unprecedented functional analysis in mathematics физике. Leningrad 1950.
  6. Соболев С. Le. Matt. Инст. им. Стеклова, т. IX. (1935, 39-105.
  7. Schwartz L., Theory of distributions I. Paris 1950.
  8. Levi EE, Sulle equazioni lineari totally ellitiche alle terivate parziali. Returns. of the Circoio Matematico di Palermo. t. 24 (1907) 275-317.
    Received March 5, 1958

ON SIMPLIFYING EXACT PROGRAMS FOR DISCRETE AUTOMATIC MECHANISMS

Abstract

by

Gr. C. MOIST.

in Bucharest

I

An exact program is a set formed by three arrays:
(III) F ( K 0 , Z 0 ) = Z 0 Φ ( K 0 , Z 0 ) = W 0 F ( K 1 , Z 0 ) = Z 1 Φ ( K 1 , Z 0 ) = W 1 F ( K 2 , Z 1 ) = Z 2 Φ ( K 2 , Z 1 ) = W 2 ⋯ ⋯ ⋯ ⋯ ⋅ ⋯ ⋯ F ( K i , Z i − 1 ) = Z i (II) ⋯ Φ ( K i , Z i − 1 ) = W i F ( K n , Z n − 1 ) = Z n ⋯ F ( K 0 , Z n ) = Z 0 Φ ( K n , Z n − 1 ) = W n Z h = Z j ⋯ (III) F ( K 0 , Z 0 ) = Z 0 Φ ( K 0 , Z 0 ) = W 0 F ( K 1 , Z 0 ) = Z 1 Φ ( K 1 , Z 0 ) = W 1 F ( K 2 , Z 1 ) = Z 2 Φ ( K 2 , Z 1 ) = W 2 ⋯ ⋯ ⋯ ⋯ ⋅ ⋯ ⋯ F ( K i , Z i − 1 ) = Z i (II) ⋯ Φ ( K i , Z i − 1 ) = W i F ( K n , Z n − 1 ) = Z n ⋯ F ( K 0 , Z n ) = Z 0 Φ ( K n , Z n − 1 ) = W n Z h = Z j ⋯ {:(III){:[F{:(K_(0)","Z_(0)):}=Z_(0),Phi{:(K_(0)","Z_(0)):}=W_(0)],[F{:(K_(1)","Z_(0)):}=Z_(1),Phi{:(K_(1)","Z_(0)):}=W_(1)],[F{:(K_(2)","Z_(1)):}=Z_(2),Phi{:(K_(2)","Z_(1)):}=W_(2)],[cdots cdots cdots,cdots*cdots cdots],[F{:(K_(i)","Z_(i-1)):}=Z_(i),(II)],[cdots,Phi{:(K_(i)","Z_(i-1)):}=W_(i)],[F{:(K_(n)","Z_(n-1)):}=Z_(n),cdots],[F{:(K_(0)","Z_(n)):}=Z_(0),Phi{:(K_(n)","Z_(n-1)):}=W_(n)],[,],[,],[,],[,],[,],[,],[,Z_(h)=Z_(j)],[,cdots],[,],[,]:}:}(III)F(K0,Z0)=Z0Φ(K0,Z0)=W0F(K1,Z0)=Z1Φ(K1,Z0)=W1F(K2,Z1)=Z2Φ(K2,Z1)=W2⋯⋯⋯⋯⋅⋯⋯F(Ki,Zi−1)=Zi(II)⋯Φ(Ki,Zi−1)=WiF(Kn,Zn−1)=Zn⋯F(K0,Zn)=Z0Φ(Kn,Zn−1)=WnZh=Zj⋯
K 0 , … , K n , W 0 , … , W n + 1 K 0 , … , K n , W 0 , … , W n + 1 K_(0),dots,K_(n),W_(0),dots,W_(n+1)K0,…,Kn,W0,…,Wn+1are given elements; Z 0 , … , Z n Z 0 , … , Z n Z_(0),dots,Z_(n)Z0,…,Znare variables in an unknown domain D ; F ( K , Z ) D ; F ( K , Z ) D;F(K,Z)D;F(K,Z)And Φ ( K , Z ) Φ ( K , Z ) Phi(K,Z)Φ(K,Z)are unknown functions having the arguments K ∈ ( K 0 , … , K n ) , Z ∈ D , F ( K , Z ) ∈ D K ∈ ( K 0 , … , K n ) , Z ∈ D , F ( K , Z ) ∈ D K in(K_(0),dots,K_(n)),Z in D,F(K,Z)in DK∈(K0,…,Kn),Z∈D,F(K,Z)∈D, Φ ( K , Z ) ∈ ( W 0 , … , W n + 1 ) Φ ( K , Z ) ∈ ( W 0 , … , W n + 1 ) Phi(K,Z)in(W_(0),dots,W_(n+1))Φ(K,Z)∈(W0,…,Wn+1).
As an example, we will consider the following exact program: D 1 D 1 D_(1)D1, which is encountered when one wants to build a signaling device at a level crossing
1959

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