The solution of a boundary value problem for the biharmonic equation

ictp
Uncategorized

Abstract

Authors

C. Kalik
Institutul de Calcul

Keywords

?

Paper coordinates

C. Kalik, La solution d’un problème aux limites pour l’équation biharmonique. [Soluția unei probleme cu valori la limită pentru ecuația biarmonică](Romanian) Acad. R. P. Romîne. Fil. Cluj. Stud. Cerc. Mat. 9 1958 135–148.

PDF

https://ictp.acad.ro/scm/journal/article/view/29/29

About this paper

Journal

Studii si Cercetari Matematice

Publisher Name

Academy of the Republic of S.R.

DOI

https://ictp.acad.ro/scm/journal/article/view/29

Print ISSN

1220-269X

Online ISSN

google scholar link

??

Paper (preprint) in HTML form

scm,+1958_201-4_20Kalik
Original text
Rate this translation
Your feedback will be used to help improve Google Translate

SOLVING A LIMIT PROBLEM FOR THE BIHARMONY EQUATION

OFCAROL KALIK

Paper presented at the paper meeting of the Computing Institute of the RPR Academy, Cluj Branch, in November 1958

  1. Starting from 1948, a series of works by G. Fichera, as well as by other mathematicians, appeared, in which the completeness of sets of functions or vectors, chosen in such a way that with their help we can construct the solution of certain limit problems (see for example the work [1] or [2]). Following the path of these works, we will deal with the solution of a limit problem related to the biharmonic equation.
Whether Ω Ω Omega\OmegaΩan arbitrary domain, bounded by the curve Γ Γ Gamma\GammaΓWe will assume that Γ Γ Gamma\GammaΓcan be divided into a finite number of portions, such that each of them can be represented in local coordinates using a function y = φ ( x ) y = φ ( x ) y=varphi(x)y=\varphi(x)y=φ(x), where φ ( x ) φ ( x ) varphi(x)\varphi(x)φ(x)is continuous and its derivative satisfies the Lipschitz condition. We also assume positive radius of curvature ρ 0 ρ 0 rho_(0)\rho_{0}ρ0his/her Γ Γ Gamma\GammaΓWe study the following limit problem: to determine the function you ( x , y ) you ( x , y ) u(x,y)u(x, y)you(x,y)in the field Ω Ω Omega\OmegaΩin such a way that it satisfies the conditions
(1) Δ 2 you ≡ ∂ 4 you ∂ x 4 + 2 ∂ 4 you ∂ x 2 ∂ y 2 + ∂ 4 you ∂ y 4 = 0 in Ω (2) you = f 1 ( S ) on Γ (2') − Δ you + 1 − σ ρ 0 ∂ you ∂ n = f 2 ( S ) on Γ (1) Δ 2 you ≡ ∂ 4 you ∂ x 4 + 2 ∂ 4 you ∂ x 2 ∂ y 2 + ∂ 4 you ∂ y 4 = 0  in  Ω (2) you = f 1 ( S )  on  Γ (2') − Δ you + 1 − σ ρ 0 ∂ you ∂ n = f 2 ( S )  on  Γ {:[(1)Delta^(2)u-=(del^(4)u)/(delx^(4))+2(del^(4)u)/(delx^(2)dely^(2))+(del^(4)u)/(dely^(4))=0" in "Omega],[(2)u=f_(1)(s)" on "Gamma],[(2')-Delta u+(1-sigma)/(rho_(0))(del u)/(del nu)=f_(2)(s)" on "Gamma]:}\begin{align*} \Delta^{2} u \equiv \frac{\partial^{4} u}{\partial x^{4}}+2 \frac{\partial^{4} u}{\partial x^{2} \partial y^{2}}+\frac{\partial^{4} u}{\partial y^{4}} & =0 \text { în } \Omega \tag{1}\\ u & =f_{1}(s) \text { pe } \Gamma \tag{2}\\ -\Delta u+\frac{1-\sigma}{\rho_{0}} \frac{\partial u}{\partial \nu} & =f_{2}(s) \text { pe } \Gamma \tag{2'} \end{align*}(1)Δ2you≡∂4you∂x4+2∂4you∂x2∂y2+∂4you∂y4=0 in Ω(2)you=f1(S) on Γ(2')−Δyou+1−σρ0∂you∂n=f2(S) on Γ
where f 1 ( s ) f 1 ( s ) f_(1)(s)f_{1}(s)f1(S)and f 2 ( s ) f 2 ( s ) f_(2)(s)f_{2}(s)f2(S)are given square integrable functions on Γ , 0 < σ < 1 Γ , 0 < σ < 1 Gamma,0 < sigma < 1\Gamma, 0<\sigma<1Γ,0<σ<1being Poisson's constant, and ν ν nu\nunis the inner normal to Γ Γ Gamma\GammaΓ.
We will briefly summarize the idea followed in solving this problem. Let us note with { v i } v i {v_(i)}\left\{v_{i}\right\}{Vand}a series of biharmonic functions. Based on Green's formula we have
∫ Γ ( ∂ Δ u ∂ ν v i − Δ u ∂ v i ∂ ν + ∂ u ∂ ν Δ v i − u ∂ Δ v i ∂ ν ) d σ = 0 , ( i = 1 , 2 , … ) ∫ Γ   ∂ Δ u ∂ ν v i − Δ u ∂ v i ∂ ν + ∂ u ∂ ν Δ v i − u ∂ Δ v i ∂ ν d σ = 0 , ( i = 1 , 2 , … ) int_(Gamma)((del Delta u)/(del nu)v_(i)-Delta u(delv_(i))/(del nu)+(del u)/(del nu)Deltav_(i)-u(del Deltav_(i))/(del nu))d sigma=0,quad(i=1,2,dots)\int_{\Gamma}\left(\frac{\partial \Delta u}{\partial \nu} v_{i}-\Delta u \frac{\partial v_{i}}{\partial \nu}+\frac{\partial u}{\partial \nu} \Delta v_{i}-u \frac{\partial \Delta v_{i}}{\partial \nu}\right) d \sigma=0, \quad(i=1,2, \ldots)∫Γ(∂Δyou∂nVand−Δyou∂Vand∂n+∂you∂nΔVand−you∂ΔVand∂n)dσ=0,(and=1,2,…)
from where, taking into account (2) and (2'), we obtain the system of integral equations relative to the unknown vector [ ∂ u ∂ ν , − ∂ Δ u ∂ ν ] ∂ u ∂ ν , − ∂ Δ u ∂ ν [(del u)/(del nu),-(delDelta_(u))/(del nu)]\left[\frac{\partial u}{\partial \nu},-\frac{\partial \Delta_{u}}{\partial \nu}\right][∂you∂n,−∂Δyou∂n]:
∫ Γ ˙ [ ∂ u ∂ ν ( − Δ v i + 1 − σ ρ ∂ v i ∂ ν ) − ∂ Δ u ∂ ν v i ] d σ = (3) = ∫ Γ ˙ ( f 2 ∂ v i ∂ ν − f 1 ∂ Δ v i ∂ ν ) d σ = C i , ( i = 1 , 2 , … ) ∫ Γ ˙   ∂ u ∂ ν − Δ v i + 1 − σ ρ ∂ v i ∂ ν − ∂ Δ u ∂ ν v i d σ = (3) = ∫ Γ ˙   f 2 ∂ v i ∂ ν − f 1 ∂ Δ v i ∂ ν d σ = C i , ( i = 1 , 2 , … ) {:[int_(Gamma^(˙))[(del u)/(del nu)(-Deltav_(i)+(1-sigma)/(rho)(delv_(i))/(del nu))-(del Delta u)/(del nu)v_(i)]d sigma=],[(3)=int_(Gamma^(˙))(f_(2)(delv_(i))/(del nu)-f_(1)(del Deltav_(i))/(del nu))d sigma=C_(i)","(i=1","2","dots)]:}\begin{align*} & \int_{\dot{\Gamma}}\left[\frac{\partial u}{\partial \nu}\left(-\Delta v_{i}+\frac{1-\sigma}{\rho} \frac{\partial v_{i}}{\partial \nu}\right)-\frac{\partial \Delta u}{\partial \nu} v_{i}\right] d \sigma= \\ = & \int_{\dot{\Gamma}}\left(f_{2} \frac{\partial v_{i}}{\partial \nu}-f_{1} \frac{\partial \Delta v_{i}}{\partial \nu}\right) d \sigma=C_{i},(i=1,2, \ldots) \tag{3} \end{align*}∫Γ˙[∂you∂n(−ΔVand+1−σρ∂Vand∂n)−∂Δyou∂nVand]dσ=(3)=∫Γ˙(f2∂Vand∂n−f1∂ΔVand∂n)dσ=Cand,(and=1,2,…)
This system of integral equations is called the Riesz-Fischer system. Constructing the series of biharmonic functions in this way { v i } v i {v_(i)}\left\{v_{i}\right\}{Vand}as the string of vectors
{ v i } = { − Δ v i + 1 − σ ρ 0 ∂ v i ∂ ν , v i } , ( i = 1 , 2 , … ) v i = − Δ v i + 1 − σ ρ 0 ∂ v i ∂ ν , v i , ( i = 1 , 2 , … ) {v_(i)}={-Deltav_(i)+(1-sigma)/(rho_(0))(delv_(i))/(del nu),v_(i)},(i=1,2,dots)\left\{v_{i}\right\}=\left\{-\Delta v_{i}+\frac{1-\sigma}{\rho_{0}} \frac{\partial v_{i}}{\partial \nu}, v_{i}\right\},(i=1,2, \ldots){Vand}={−ΔVand+1−σρ0∂Vand∂n,Vand},(and=1,2,…)
to be orthogonal and complete on Γ Γ Gamma\GammaΓin Hilbert's sense, from system (3) we obtain
∂ u ∂ ν = ∑ i = 1 ∞ C i ( − Δ v i + 1 − σ ρ 0 ∂ v i ∂ ν ) s i s i − ∂ Δ u ∂ ν = ∑ i = 1 ∞ C i v i 1 ) ∂ u ∂ ν = ∑ i = 1 ∞   C i − Δ v i + 1 − σ ρ 0 ∂ v i ∂ ν s i s i − ∂ Δ u ∂ ν = ∑ i = 1 ∞   C i v i 1 ) (del u)/(del nu)=sum_(i=1)^(oo)C_(i)(-Deltav_(i)+(1-sigma)/(rho_(0))(delv_(i))/(del nu))si^(si)-(del Delta u)/(del nu)=sum_(i=1)^(oo)C_(i)v_(i)^(1))\frac{\partial u}{\partial \nu}=\sum_{i=1}^{\infty} C_{i}\left(-\Delta v_{i}+\frac{1-\sigma}{\rho_{0}} \frac{\partial v_{i}}{\partial \nu}\right) \stackrel{s i}{s i}-\frac{\partial \Delta u}{\partial \nu}=\sum_{i=1}^{\infty} C_{i} v_{i}^{1)}∂you∂n=∑and=1∞Cand(−ΔVand+1−σρ0∂Vand∂n)SandSand−∂Δyou∂n=∑and=1∞CandVand1)
These series converge to the mean square on Γ Γ Gamma\GammaΓ. Considering also the boundary conditions (2) and (2'), we know Γ Γ Gamma\GammaΓthe four functions u , ∂ u ∂ ν , Δ u u , ∂ u ∂ ν , Δ u u,(del u)/(del nu),Delta uu, \frac{\partial u}{\partial \nu}, \Delta uyou,∂you∂n,Δyouand ∂ Δ u ∂ ν ∂ Δ u ∂ ν (delDelta_(u))/(del nu)\frac{\partial \Delta_{u}}{\partial \nu}∂Δyou∂nwith the help of which we can write the solution to the formulated limit problem, considering the well-known formula:
u ( P ) = 1 ε π ∫ Γ ( u ∂ Δ U ∂ ν − ∂ u ∂ ν Δ U + Δ u ∂ U ∂ ν − ∂ Δ u ∂ ν U ) d σ u ( P ) = 1 ε π ∫ Γ   u ∂ Δ U ∂ ν − ∂ u ∂ ν Δ U + Δ u ∂ U ∂ ν − ∂ Δ u ∂ ν U d σ u(P)=(1)/(epsi pi)int_(Gamma)(u(del Delta U)/(del nu)-(del u)/(del nu)Delta U+Delta u(del U)/(del nu)-(del Delta u)/(del nu)U)d sigmau(P)=\frac{1}{\varepsilon \pi} \int_{\Gamma}\left(u \frac{\partial \Delta U}{\partial \nu}-\frac{\partial u}{\partial \nu} \Delta U+\Delta u \frac{\partial U}{\partial \nu}-\frac{\partial \Delta u}{\partial \nu} U\right) d \sigmayou(P)=1επ∫Γ(you∂ΔU∂n−∂you∂nΔU+Δyou∂U∂n−∂Δyou∂nU)dσ
where U = r ln ⁡ r U = r ln ⁡ r U=r ln rU=r \ln rU=RIn⁡Ris the fundamental solution of the biharmonic equation.
We mention that this method, the idea of ​​which was developed by M. Picone, is of particular interest from a practical point of view, having advantages over most methods of approximate solution of boundary problems. Namely, to calculate approximate solutions we have to calculate the value of the integrals on the boundary Γ Γ Gamma\GammaΓand not on Ω Ω Omega\OmegaΩIn addition, the functions v i v i v_(i)v_{i}Vand, which intervene in the calculations, are harmonic and biharmonic polynomials, respectively their combinations, which also makes the calculation easier.
In point 5 of the paper we will construct the string { v i } v i {v_(i)}\left\{v_{i}\right\}{Vand}in such a way that they satisfy the above conditions.
2. Keeping the notations, as well as the conditions on it Γ Γ Gamma\GammaΓ, from the previous point, we will formulate two theorems that are immediate consequences of the results of G. Fichera [1].
Theorem 1. If φ 1 ( Q ) ∈ L ( Γ ) s s i φ 2 ( Q ) ∈ L ( Γ ) 2 φ 1 ( Q ) ∈ L ( Γ ) s s i φ 2 ( Q ) ∈ L ( Γ ) 2 varphi_(1)(Q)in L(Gamma)ssivarphi_(2)(Q)in L(Gamma)^(2)\varphi_{1}(Q) \in L(\Gamma) s s i \varphi_{2}(Q) \in L(\Gamma)^{2}φ1(Q)∈IT(Γ)SSandφ2(Q)∈IT(Γ)2, then the function
v ( P ) = ∫ Γ φ 1 ( Q ) U ( P , Q ) d σ + ∫ Γ φ 2 ( Q ) ln ⁡ r ( P , Q ) d σ v ( P ) = ∫ Γ   φ 1 ( Q ) U ( P , Q ) d σ + ∫ Γ   φ 2 ( Q ) ln ⁡ r ( P , Q ) d σ v(P)=int_(Gamma)varphi_(1)(Q)U(P,Q)d sigma+int_(Gamma)varphi_(2)(Q)ln r(P,Q)d sigmav(P)=\int_{\Gamma} \varphi_{1}(Q) U(P, Q) d \sigma+\int_{\Gamma} \varphi_{2}(Q) \ln r(P, Q) d \sigmaV(P)=∫Γφ1(Q)U(P,Q)dσ+∫Γφ2(Q)In⁡R(P,Q)dσ
there is for almost everyone P ∈ Γ P ∈ Γ P in GammaP \in \GammaP∈Γand is summable on Γ Γ Gamma\GammaΓAlmost
for everyone M ∈ Γ M ∈ Γ M in GammaM \in \GammaM∈ΓHAVE
lim v ( P ) = ∫ Γ φ 1 ( Q ) U ( M , Q ) d σ + ∫ Γ φ 2 ( Q ) ln ⁡ r ( M , Q ) d σ lim v ( P ) = ∫ Γ   φ 1 ( Q ) U ( M , Q ) d σ + ∫ Γ   φ 2 ( Q ) ln ⁡ r ( M , Q ) d σ lim v(P)=int_(Gamma)varphi_(1)(Q)U(M,Q)d sigma+int_(Gamma)varphi_(2)(Q)ln r(M,Q)d sigma\lim v(P)=\int_{\Gamma} \varphi_{1}(Q) U(M, Q) d \sigma+\int_{\Gamma} \varphi_{2}(Q) \ln r(M, Q) d \sigmalimV(P)=∫Γφ1(Q)U(M,Q)dσ+∫Γφ2(Q)In⁡R(M,Q)dσ
and
lim ∂ v ( P ) ∂ v M = ± π φ 2 ( M ) + ∫ Γ φ 1 ( Q ) ∂ U ( M , Q ) ∂ v M d σ + ∫ Γ φ 2 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ v M d σ 3 ) lim ∂ v ( P ) ∂ v M = ± π φ 2 ( M ) + ∫ Γ   φ 1 ( Q ) ∂ U ( M , Q ) ∂ v M d σ + ∫ Γ   φ 2 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ v M d σ 3 {: lim(del v(P))/(delv_(M))=+-pivarphi_(2)(M)+int_(Gamma)varphi_(1)(Q)(del U(M,Q))/(delv_(M))d sigma+int_(Gamma)varphi_(2)(Q)(del ln r(M,Q))/(delv_(M))dsigma^(3))\left.\lim \frac{\partial v(P)}{\partial v_{M}}= \pm \pi \varphi_{2}(M)+\int_{\Gamma} \varphi_{1}(Q) \frac{\partial U(M, Q)}{\partial v_{M}} d \sigma+\int_{\Gamma} \varphi_{2}(Q) \frac{\partial \ln r(M, Q)}{\partial v_{M}} d \sigma^{3}\right)lim∂V(P)∂VM=±πφ2(M)+∫Γφ1(Q)∂U(M,Q)∂VMdσ+∫Γφ2(Q)∂In⁡R(M,Q)∂VMdσ3)
when P → M P → M P rarr MP \rightarrow MP→Mnormally at Γ Γ Gamma\GammaΓat the point M M MMM, and the limits are summable functions on Γ Γ Gamma\GammaΓ.
Theorem 2. If φ 1 ( Q ) ∈ L ( Γ ) φ 1 ( Q ) ∈ L ( Γ ) varphi_(1)(Q)in L(Gamma)\varphi_{1}(Q) \in L(\Gamma)φ1(Q)∈IT(Γ)and φ 2 ( Q ) ∈ L ( Γ ) φ 2 ( Q ) ∈ L ( Γ ) varphi_(2)(Q)in L(Gamma)\varphi_{2}(Q) \in L(\Gamma)φ2(Q)∈IT(Γ), then the function
w ( P ) = ∫ Γ φ 1 ( Q ) ∂ U ( P , Q ) ∂ ν Q d σ + ∫ Γ φ 2 ( Q ) ∂ ln ⁡ r ( P , Q ) ∂ ν Q d σ w ( P ) = ∫ Γ   φ 1 ( Q ) ∂ U ( P , Q ) ∂ ν Q d σ + ∫ Γ   φ 2 ( Q ) ∂ ln ⁡ r ( P , Q ) ∂ ν Q d σ w(P)=int_(Gamma)varphi_(1)(Q)(del U(P,Q))/(delnu_(Q))d sigma+int_(Gamma)varphi_(2)(Q)(del ln r(P,Q))/(delnu_(Q))d sigmaw(P)=\int_{\Gamma} \varphi_{1}(Q) \frac{\partial U(P, Q)}{\partial \nu_{Q}} d \sigma+\int_{\Gamma} \varphi_{2}(Q) \frac{\partial \ln r(P, Q)}{\partial \nu_{Q}} d \sigmaw(P)=∫Γφ1(Q)∂U(P,Q)∂nQdσ+∫Γφ2(Q)∂In⁡R(P,Q)∂nQdσ
there is for almost everyone P ∈ Γ P ∈ Γ P in GammaP \in \GammaP∈Γand is summable on Γ Γ Gamma\GammaΓAlmost
for everyone M ∈ Γ M ∈ Γ M in GammaM \in \GammaM∈ΓHAVE
lim w ( P ) = ∓ π φ 2 ( M ) + ∫ Γ ˙ φ 1 ( Q ) ∂ U ( M , Q ) ∂ ν Q d σ + ∫ Γ ˙ φ 2 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ ν Q d σ 4 ) lim w ( P ) = ∓ π φ 2 ( M ) + ∫ Γ ˙   φ 1 ( Q ) ∂ U ( M , Q ) ∂ ν Q d σ + ∫ Γ ˙   φ 2 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ ν Q d σ 4 {: lim w(P)=∓pivarphi_(2)(M)+int_(Gamma^(˙))varphi_(1)(Q)(del U(M,Q))/(delnu_(Q))d sigma+int_(Gamma^(˙))varphi_(2)(Q)(del ln r(M,Q))/(delnu_(Q))dsigma^(4))\left.\lim w(P)=\mp \pi \varphi_{2}(M)+\int_{\dot{\Gamma}} \varphi_{1}(Q) \frac{\partial U(M, Q)}{\partial \nu_{Q}} d \sigma+\int_{\dot{\Gamma}} \varphi_{2}(Q) \frac{\partial \ln r(M, Q)}{\partial \nu_{Q}} d \sigma^{4}\right)limw(P)=∓πφ2(M)+∫Γ˙φ1(Q)∂U(M,Q)∂nQdσ+∫Γ˙φ2(Q)∂In⁡R(M,Q)∂nQdσ4)
when P → M P → M P rarr MP \rightarrow MP→Mnormally at Γ Γ Gamma\GammaΓat the point M M MMMand the limit function is summable on Γ Γ Gamma\GammaΓWe note
that the limits in these two theorems exist if the point M M MMMis the Lebesgue point of the functions φ 1 ( Q ) φ 1 ( Q ) varphi_(1)(Q)\varphi_{1}(Q)φ1(Q)and φ 2 ( Q ) φ 2 ( Q ) varphi_(2)(Q)\varphi_{2}(Q)φ2(Q).
3. We introduce the set of functions { u } { u } {u}\{u\}{you}defined in the field Ω Ω Omega\OmegaΩas follows: - function u u uuyoubelongs to the crowd { u } { u } {u}\{u\}{you}, if there are two functions φ 1 ( Q ) ∈ L 2 ( Γ ) φ 1 ( Q ) ∈ L 2 ( Γ ) varphi_(1)(Q)inL_(2)(Gamma)\varphi_{1}(Q) \in L_{2}(\Gamma)φ1(Q)∈IT2(Γ)and φ 2 ( Q ) ∈ L 2 ( Γ ) φ 2 ( Q ) ∈ L 2 ( Γ ) varphi_(2)(Q)inL_(2)(Gamma)\varphi_{2}(Q) \in L_{2}(\Gamma)φ2(Q)∈IT2(Γ)in such a way that
u ( P ) = ∫ Γ φ 1 ( Q ) U ( P , Q ) d σ + ∫ Γ φ 2 ( Q ) ln ⁡ r ( P , Q ) d σ u ( P ) = ∫ Γ   φ 1 ( Q ) U ( P , Q ) d σ + ∫ Γ   φ 2 ( Q ) ln ⁡ r ( P , Q ) d σ u(P)=int_(Gamma)varphi_(1)(Q)U(P,Q)d sigma+int_(Gamma)varphi_(2)(Q)ln r(P,Q)d sigmau(P)=\int_{\Gamma} \varphi_{1}(Q) U(P, Q) d \sigma+\int_{\Gamma} \varphi_{2}(Q) \ln r(P, Q) d \sigmayou(P)=∫Γφ1(Q)U(P,Q)dσ+∫Γφ2(Q)In⁡R(P,Q)dσ
The crowd is characterized by
Theorem 3. If u ∈ { u } u ∈ { u } u in{u}u \in\{u\}you∈{you}, then
a) almost for every M ∈ Γ M ∈ Γ M in GammaM \in \GammaM∈Γequalities occur
lim u ( P ) = μ 1 ( M ) lim ∂ u ( P ) ∂ ν M = δ 1 ( M ) lim Δ ϑ ( P ) = μ 2 ( M ) lim ∂ Δ ϑ ( P ) ∂ ν M = δ 2 ( M ) lim u ( P ) = μ 1 ( M ) lim ∂ u ( P ) ∂ ν M = δ 1 ( M ) lim Δ ϑ ( P ) = μ 2 ( M ) lim ∂ Δ ϑ ( P ) ∂ ν M = δ 2 ( M ) {:[ lim u(P)=mu_(1)(M)],[ lim(del u(P))/(delnu_(M))=delta_(1)(M)],[ limDelta_(vartheta)(P)=mu_(2)(M)],[ lim(delDelta_(vartheta)(P))/(delnu_(M))=delta_(2)(M)]:}\begin{aligned} & \lim u(P)=\mu_{1}(M) \\ & \lim \frac{\partial u(P)}{\partial \nu_{M}}=\delta_{1}(M) \\ & \lim \Delta_{\vartheta}(P)=\mu_{2}(M) \\ & \lim \frac{\partial \Delta_{\vartheta}(P)}{\partial \nu_{M}}=\delta_{2}(M) \end{aligned}limyou(P)=μ1(M)lim∂you(P)∂nM=δ1(M)limΔθ(P)=μ2(M)lim∂Δθ(P)∂nM=δ2(M)
b) for each P ∈ Ω P ∈ Ω P in OmegaP \in \OmegaP∈ΩHAVE
8 π u ( P ) = ∫ Γ [ μ 1 ( Q ) ∂ Δ U ( P , Q ) ∂ ν Q − δ 1 ( Q ) Δ U ( P , Q ) + μ ˙ 2 ( Q ) ∂ U ( P , Q ) ∂ ν Q − (4) − δ 2 ( Q ) U ( P , Q ) ] d σ 8 π u ( P ) = ∫ Γ   μ 1 ( Q ) ∂ Δ U ( P , Q ) ∂ ν Q − δ 1 ( Q ) Δ U ( P , Q ) + μ ˙ 2 ( Q ) ∂ U ( P , Q ) ∂ ν Q − (4) − δ 2 ( Q ) U ( P , Q ) d σ {:[8pi u(P)=int_(Gamma)[mu_(1)(Q)(del Delta U(P,Q))/(delnu_(Q))-delta_(1)(Q)Delta U(P,Q)+mu^(˙)_(2)(Q)(del U(P,Q))/(delnu_(Q))-:}],[(4){:-delta_(2)(Q)U(P,Q)]d sigma]:}\begin{gather*} 8 \pi u(P)=\int_{\Gamma}\left[\mu_{1}(Q) \frac{\partial \Delta U(P, Q)}{\partial \nu_{Q}}-\delta_{1}(Q) \Delta U(P, Q)+\dot{\mu}_{2}(Q) \frac{\partial U(P, Q)}{\partial \nu_{Q}}-\right. \\ \left.-\delta_{2}(Q) U(P, Q)\right] d \sigma \tag{4} \end{gather*}8πyou(P)=∫Γ[μ1(Q)∂ΔU(P,Q)∂nQ−δ1(Q)ΔU(P,Q)+μ˙2(Q)∂U(P,Q)∂nQ−(4)−δ2(Q)U(P,Q)]dσ
c) for each P ′ ∈ C Ω ( C Ω = P ′ ∈ C Ω ( C Ω = P^(')in C Omega(C Omega=P^{\prime} \in C \Omega(C \Omega=P′∈CΩ(CΩ=the complementary set of Ω Ω Omega\OmegaΩrelative to the whole plane) we have
(5) 0 = ∫ Γ [ μ 1 ( Q ) ∂ Δ U ( P ′ , Q ) ∂ ν Q − δ 1 ( Q ) Δ U ( P ′ , Q ) + μ 2 ( Q ) ∂ U ( P ′ , Q ) ∂ ν Q − − δ 2 ( Q ) U ( P ′ , Q ) ] d σ (5) 0 = ∫ Γ   μ 1 ( Q ) ∂ Δ U P ′ , Q ∂ ν Q − δ 1 ( Q ) Δ U P ′ , Q + μ 2 ( Q ) ∂ U P ′ , Q ∂ ν Q − − δ 2 ( Q ) U P ′ , Q d σ {:[(5)0=int_(Gamma)[mu_(1)(Q)(del Delta U(P^('),Q))/(delnu_(Q))-delta_(1)(Q)Delta U(P^('),Q)+mu_(2)(Q)(del U(P^('),Q))/(delnu_(Q))-:}],[{:-delta_(2)(Q)U(P^('),Q)]d sigma]:}\begin{gather*} 0=\int_{\Gamma}\left[\mu_{1}(Q) \frac{\partial \Delta U\left(P^{\prime}, Q\right)}{\partial \nu_{Q}}-\delta_{1}(Q) \Delta U\left(P^{\prime}, Q\right)+\mu_{2}(Q) \frac{\partial U\left(P^{\prime}, Q\right)}{\partial \nu_{Q}}-\right. \tag{5}\\ \left.-\delta_{2}(Q) U\left(P^{\prime}, Q\right)\right] d \sigma \end{gather*}(5)0=∫Γ[μ1(Q)∂ΔU(P′,Q)∂nQ−δ1(Q)ΔU(P′,Q)+μ2(Q)∂U(P′,Q)∂nQ−−δ2(Q)U(P′,Q)]dσ
Conversely, if μ 1 , δ 1 , μ 2 , δ 2 ∈ L 2 ( Γ ) μ 1 , δ 1 , μ 2 , δ 2 ∈ L 2 ( Γ ) mu_(1),delta_(1),mu_(2),delta_(2)inL_(2)(Gamma)\mu_{1}, \delta_{1}, \mu_{2}, \delta_{2} \in L_{2}(\Gamma)μ1,δ1,μ2,δ2∈IT2(Γ)and if these functions satisfy the condition c c ccc, then the function u ( P ) u ( P ) u(P)u(P)you(P)from formula (4) belongs to the set { u } { u } {u}\{u\}{you}.
Demonstration. Let u ∈ { u } u ∈ { u } u in{u}u \in\{u\}you∈{you}Based on Theorems 1 and 2 we have
μ 1 ( M ) = ∫ Γ φ 1 ( Q ) U ( M , Q ) d σ + ∫ Γ φ 2 ( Q ) ln ⁡ r ( M , Q ) d σ δ 1 ( M ) = π φ 2 ( M ) + ∫ Γ φ 1 ( Q ) ∂ U ( M , Q ) ∂ ν M d σ + ∫ Γ φ 2 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ ν Q d σ μ 2 ( M ) = 4 ∫ Γ φ 1 ( Q ) ln ⁡ r ( M , Q ) d σ + 3 ∫ Γ φ 1 ( Q ) d σ δ 2 ( M ) = 4 π φ 1 ( M ) + 4 ∫ Γ φ 1 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ ν Q d σ μ 1 ( M ) = ∫ Γ   φ 1 ( Q ) U ( M , Q ) d σ + ∫ Γ   φ 2 ( Q ) ln ⁡ r ( M , Q ) d σ δ 1 ( M ) = π φ 2 ( M ) + ∫ Γ   φ 1 ( Q ) ∂ U ( M , Q ) ∂ ν M d σ + ∫ Γ   φ 2 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ ν Q d σ μ 2 ( M ) = 4 ∫ Γ   φ 1 ( Q ) ln ⁡ r ( M , Q ) d σ + 3 ∫ Γ   φ 1 ( Q ) d σ δ 2 ( M ) = 4 π φ 1 ( M ) + 4 ∫ Γ   φ 1 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ ν Q d σ {:[mu_(1)(M)=int_(Gamma)varphi_(1)(Q)U(M","Q)d sigma+int_(Gamma)varphi_(2)(Q)ln r(M","Q)d sigma],[delta_(1)(M)=pivarphi_(2)(M)+int_(Gamma)varphi_(1)(Q)(del U(M,Q))/(delnu_(M))d sigma+int_(Gamma)varphi_(2)(Q)(del ln r(M,Q))/(delnu_(Q))d sigma],[mu_(2)(M)=4int_(Gamma)varphi_(1)(Q)ln r(M","Q)d sigma+3int_(Gamma)varphi_(1)(Q)d sigma],[delta_(2)(M)=4pivarphi_(1)(M)+4int_(Gamma)varphi_(1)(Q)(del ln r(M,Q))/(delnu_(Q))d sigma]:}\begin{aligned} & \mu_{1}(M)=\int_{\Gamma} \varphi_{1}(Q) U(M, Q) d \sigma+\int_{\Gamma} \varphi_{2}(Q) \ln r(M, Q) d \sigma \\ & \delta_{1}(M)=\pi \varphi_{2}(M)+\int_{\Gamma} \varphi_{1}(Q) \frac{\partial U(M, Q)}{\partial \nu_{M}} d \sigma+\int_{\Gamma} \varphi_{2}(Q) \frac{\partial \ln r(M, Q)}{\partial \nu_{Q}} d \sigma \\ & \mu_{2}(M)=4 \int_{\Gamma} \varphi_{1}(Q) \ln r(M, Q) d \sigma+3 \int_{\Gamma} \varphi_{1}(Q) d \sigma \\ & \delta_{2}(M)=4 \pi \varphi_{1}(M)+4 \int_{\Gamma} \varphi_{1}(Q) \frac{\partial \ln r(M, Q)}{\partial \nu_{Q}} d \sigma \end{aligned}μ1(M)=∫Γφ1(Q)U(M,Q)dσ+∫Γφ2(Q)In⁡R(M,Q)dσδ1(M)=πφ2(M)+∫Γφ1(Q)∂U(M,Q)∂nMdσ+∫Γφ2(Q)∂In⁡R(M,Q)∂nQdσμ2(M)=4∫Γφ1(Q)In⁡R(M,Q)dσ+3∫Γφ1(Q)dσδ2(M)=4πφ1(M)+4∫Γφ1(Q)∂In⁡R(M,Q)∂nQdσ
It is observed that these functions are square integrable. Substituting these expressions into the second part of equality (4), which we denote by I I IIand, we obtain
I = ∫ Γ ~ φ 1 ( Q ) d σ ∫ Γ [ U ( M , Q ) ∂ Δ U ( P , M ) ∂ v M − ∂ U ( M , Q ) ∂ v M Δ U ( P , M ) + + 4 ln ⁡ r ( M , Q ) ∂ U ( P , M ) ∂ v M − 4 ∂ ln ⁡ r ( M , Q ) ∂ v M U ( P , M ) ] d σ + + ∫ Γ ˙ φ 2 ( Q ) d σ ∫ Γ [ ln ⁡ r ( M , Q ) ∂ Δ U ( P , M ) ∂ v M − ∂ ln ⁡ r ( M , Q ) ∂ v M Δ U ( P , M ) ] d σ + + 3 ∫ Γ ∂ ∂ U ( P , M ) ∂ v M d σ ∫ Γ φ 1 ( Q ) d σ − 4 π ∫ Γ φ 1 ( Q ) U ( P , Q ) d σ − π ∫ Γ ˙ φ 2 ( Q ) Δ U ( P , Q ) d σ I = ∫ Γ ~   φ 1 ( Q ) d σ ∫ Γ   U ( M , Q ) ∂ Δ U ( P , M ) ∂ v M − ∂ U ( M , Q ) ∂ v M Δ U ( P , M ) + + 4 ln ⁡ r ( M , Q ) ∂ U ( P , M ) ∂ v M − 4 ∂ ln ⁡ r ( M , Q ) ∂ v M U ( P , M ) d σ + + ∫ Γ ˙   φ 2 ( Q ) d σ ∫ Γ   ln ⁡ r ( M , Q ) ∂ Δ U ( P , M ) ∂ v M − ∂ ln ⁡ r ( M , Q ) ∂ v M Δ U ( P , M ) d σ + + 3 ∫ Γ ∂   ∂ U ( P , M ) ∂ v M d σ ∫ Γ   φ 1 ( Q ) d σ − 4 π ∫ Γ   φ 1 ( Q ) U ( P , Q ) d σ − π ∫ Γ ˙   φ 2 ( Q ) Δ U ( P , Q ) d σ {:[I=int_( tilde(Gamma))varphi_(1)(Q)d sigmaint_(Gamma)[U(M,Q)(del Delta U(P,M))/(delv_(M))-(del U(M,Q))/(delv_(M))Delta U(P,M)+:}],[{:+4ln r(M,Q)(del U(P,M))/(delv_(M))-4(del ln r(M,Q))/(delv_(M))U(P,M)]d sigma+],[+int_(Gamma^(˙))varphi_(2)(Q)d sigmaint_(Gamma)[ln r(M,Q)(del Delta U(P,M))/(delv_(M))-(del ln r(M,Q))/(delv_(M))Delta U(P,M)]d sigma+],[+3int_(Gamma)^(del)(del U(P,M))/(delv_(M))d sigmaint_(Gamma)varphi_(1)(Q)d sigma-4piint_(Gamma)varphi_(1)(Q)U(P","Q)d sigma-piint_(Gamma^(˙))varphi_(2)(Q)Delta U(P","Q)d sigma]:}\begin{gathered} I=\int_{\tilde{\Gamma}} \varphi_{1}(Q) d \sigma \int_{\Gamma}\left[U(M, Q) \frac{\partial \Delta U(P, M)}{\partial v_{M}}-\frac{\partial U(M, Q)}{\partial v_{M}} \Delta U(P, M)+\right. \\ \left.+4 \ln r(M, Q) \frac{\partial U(P, M)}{\partial v_{M}}-4 \frac{\partial \ln r(M, Q)}{\partial v_{M}} U(P, M)\right] d \sigma+ \\ +\int_{\dot{\Gamma}} \varphi_{2}(Q) d \sigma \int_{\Gamma}\left[\ln r(M, Q) \frac{\partial \Delta U(P, M)}{\partial v_{M}}-\frac{\partial \ln r(M, Q)}{\partial v_{M}} \Delta U(P, M)\right] d \sigma+ \\ +3 \int_{\Gamma}^{\partial} \frac{\partial U(P, M)}{\partial v_{M}} d \sigma \int_{\Gamma} \varphi_{1}(Q) d \sigma-4 \pi \int_{\Gamma} \varphi_{1}(Q) U(P, Q) d \sigma-\pi \int_{\dot{\Gamma}} \varphi_{2}(Q) \Delta U(P, Q) d \sigma \end{gathered}and=∫Γ~φ1(Q)dσ∫Γ[U(M,Q)∂ΔU(P,M)∂VM−∂U(M,Q)∂VMΔU(P,M)++4In⁡R(M,Q)∂U(P,M)∂VM−4∂In⁡R(M,Q)∂VMU(P,M)]dσ++∫Γ˙φ2(Q)dσ∫Γ[In⁡R(M,Q)∂ΔU(P,M)∂VM−∂In⁡R(M,Q)∂VMΔU(P,M)]dσ++3∫Γ∂∂U(P,M)∂VMdσ∫Γφ1(Q)dσ−4π∫Γφ1(Q)U(P,Q)dσ−π∫Γ˙φ2(Q)ΔU(P,Q)dσ
On the other hand, either Q ′ Q ′ Q^(')Q^{\prime}Q′a point on the external normal to Γ Γ Gamma\GammaΓat the point Q Q QQQ. U ( P , Q ′ ) U P , Q ′ U(P,Q^('))U\left(P, Q^{\prime}\right)U(P,Q′)is a regular biharmonic function when P P PPPbelongs to Ω Ω Omega\OmegaΩ, so
8 π U ( P , Q ′ ) = ∫ Γ ˙ [ U ( M , Q ′ ) ∂ Δ U ( P , M ) ∂ v M − ∂ U ( M , Q ′ ) ∂ v M Δ U ( P , M ) + + Δ U ( M , Q ′ ) ∂ U ( P , M ) ∂ v M − ∂ Δ U ( M , Q ′ ) ∂ v M U ( P , M ) ] d σ 8 π U P , Q ′ = ∫ Γ ˙   U M , Q ′ ∂ Δ U ( P , M ) ∂ v M − ∂ U M , Q ′ ∂ v M Δ U ( P , M ) + + Δ U M , Q ′ ∂ U ( P , M ) ∂ v M − ∂ Δ U M , Q ′ ∂ v M U ( P , M ) d σ {:[8pi U(P,Q^('))=int_(Gamma^(˙))[U(M,Q^('))(del Delta U(P,M))/(delv_(M))-(del U(M,Q^(')))/(delv_(M))Delta U(P,M)+:}],[{: quad+Delta U(M,Q^('))(del U(P,M))/(delv_(M))-(del Delta U(M,Q^(')))/(delv_(M))U(P,M)]d sigma]:}\begin{aligned} & 8 \pi U\left(P, Q^{\prime}\right)=\int_{\dot{\Gamma}}\left[U\left(M, Q^{\prime}\right) \frac{\partial \Delta U(P, M)}{\partial v_{M}}-\frac{\partial U\left(M, Q^{\prime}\right)}{\partial v_{M}} \Delta U(P, M)+\right. \\ & \left.\quad+\Delta U\left(M, Q^{\prime}\right) \frac{\partial U(P, M)}{\partial v_{M}}-\frac{\partial \Delta U\left(M, Q^{\prime}\right)}{\partial v_{M}} U(P, M)\right] d \sigma \end{aligned}8πU(P,Q′)=∫Γ˙[U(M,Q′)∂ΔU(P,M)∂VM−∂U(M,Q′)∂VMΔU(P,M)++ΔU(M,Q′)∂U(P,M)∂VM−∂ΔU(M,Q′)∂VMU(P,M)]dσ
Crossing the line when Q ′ → Q Q ′ → Q Q^(')rarr QQ^{\prime} \rightarrow QQ′→Q, we obtain
∫ Γ [ U ( M , Q ) ∂ Δ U ( P , M ) ∂ ν M − ∂ U ( M , Q ′ ) ∂ ν M Δ U ( P , M ) + 4 ln ⁡ r ( M , Q ) ∂ U ( P , M ) ∂ ν M − − 4 ∂ ln ⁡ r ( M , Q ) ∂ ν M U ( P , M ) ] d σ = 12 π U ( P , Q ) − 3 ∫ Γ ∂ U ( P , M ) ∂ ν M d σ ∫ Γ   U ( M , Q ) ∂ Δ U ( P , M ) ∂ ν M − ∂ U M , Q ′ ∂ ν M Δ U ( P , M ) + 4 ln ⁡ r ( M , Q ) ∂ U ( P , M ) ∂ ν M − − 4 ∂ ln ⁡ r ( M , Q ) ∂ ν M U ( P , M ) d σ = 12 π U ( P , Q ) − 3 ∫ Γ   ∂ U ( P , M ) ∂ ν M d σ {:[int_(Gamma)[U(M,Q)(del Delta U(P,M))/(delnu_(M))-(del U(M,Q^(')))/(delnu_(M))Delta U(P,M)+4ln r(M,Q)(del U(P,M))/(delnu_(M))-:}],[{:-4(del ln r(M,Q))/(delnu_(M))U(P,M)]d sigma=12 pi U(P","Q)-3int_(Gamma)(del U(P,M))/(delnu_(M))d sigma]:}\begin{gathered} \int_{\Gamma}\left[U(M, Q) \frac{\partial \Delta U(P, M)}{\partial \nu_{M}}-\frac{\partial U\left(M, Q^{\prime}\right)}{\partial \nu_{M}} \Delta U(P, M)+4 \ln r(M, Q) \frac{\partial U(P, M)}{\partial \nu_{M}}-\right. \\ \left.-4 \frac{\partial \ln r(M, Q)}{\partial \nu_{M}} U(P, M)\right] d \sigma=12 \pi U(P, Q)-3 \int_{\Gamma} \frac{\partial U(P, M)}{\partial \nu_{M}} d \sigma \end{gathered}∫Γ[U(M,Q)∂ΔU(P,M)∂nM−∂U(M,Q′)∂nMΔU(P,M)+4In⁡R(M,Q)∂U(P,M)∂nM−−4∂In⁡R(M,Q)∂nMU(P,M)]dσ=12πU(P,Q)−3∫Γ∂U(P,M)∂nMdσ
Analogously we have
∫ Γ [ ln ⁡ r ( M , Q ) ∂ Δ U ( P , M ) ∂ ν M − ∂ ln ⁡ r ( M , Q ) ∂ ν M Δ U ( P , M ) ] d σ = 8 π ln ⁡ r ( P , Q ) + π Δ U ( P , Q ) . ∫ Γ   ln ⁡ r ( M , Q ) ∂ Δ U ( P , M ) ∂ ν M − ∂ ln ⁡ r ( M , Q ) ∂ ν M Δ U ( P , M ) d σ = 8 π ln ⁡ r ( P , Q ) + π Δ U ( P , Q ) . int_(Gamma)[ln r(M,Q)(del Delta U(P,M))/(delnu_(M))-(del ln r(M,Q))/(delnu_(M))Delta U(P,M)]d sigma=8pi ln r(P,Q)+pi Delta U(P,Q).\int_{\Gamma}\left[\ln r(M, Q) \frac{\partial \Delta U(P, M)}{\partial \nu_{M}}-\frac{\partial \ln r(M, Q)}{\partial \nu_{M}} \Delta U(P, M)\right] d \sigma=8 \pi \ln r(P, Q)+\pi \Delta U(P, Q) .∫Γ[In⁡R(M,Q)∂ΔU(P,M)∂nM−∂In⁡R(M,Q)∂nMΔU(P,M)]dσ=8πIn⁡R(P,Q)+πΔU(P,Q).
We insert these results into the expression of I I IIand, from where
I = 8 π u ( P ) I = 8 π u ( P ) I=8pi u(P)I=8 \pi u(P)and=8πyou(P)
which means that condition (4) is satisfied. The same applies to equality (5); the point Q ′ Q ′ Q^(')Q^{\prime}Q′, which intervenes here in the auxiliary calculations, must be on the inner normal to Γ Γ Gamma\GammaΓat the point Q Q QQQ.
Let us proceed to the proof of the converse statement. Let μ 1 , δ 1 , μ 2 s ̧ i δ ∈ L 2 ( Γ ) μ 1 , δ 1 , μ 2 s ̧ i δ ∈ L 2 ( Γ ) mu_(1),delta_(1),mu_(2)şi delta inL_(2)(Gamma)\mu_{1}, \delta_{1}, \mu_{2} s ̧ i \delta \in L_{2}(\Gamma)μ1,δ1,μ2Şandδ∈IT2(Γ)We consider the system of Fredholm type integral equations:
(6) 1 4 π δ 2 ( M ) = φ 1 ( M ) + λ π ∫ Γ φ 1 ( Q ) δ ln ⁡ r ( M , Q ) ∂ ν M d σ 1 π δ 1 ( M ) = φ 2 ( M ) + λ π ∫ Γ φ 1 ( Q ) ∂ U ( M , Q ) ∂ ν M d σ + λ π ∫ Γ ρ φ 2 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ ν M d σ } (6) 1 4 π δ 2 ( M ) = φ 1 ( M ) + λ π ∫ Γ   φ 1 ( Q ) δ ln ⁡ r ( M , Q ) ∂ ν M d σ 1 π δ 1 ( M ) = φ 2 ( M ) + λ π ∫ Γ   φ 1 ( Q ) ∂ U ( M , Q ) ∂ ν M d σ + λ π ∫ Γ ρ   φ 2 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ ν M d σ {:(6){:[(1)/(4pi)delta_(2)(M)=varphi_(1)(M)+(lambda )/(pi)int_(Gamma)varphi_(1)(Q)(delta ln r(M,Q))/(delnu_(M))d sigma],[(1)/(pi)delta_(1)(M)=varphi_(2)(M)+(lambda )/(pi)int_(Gamma)varphi_(1)(Q)(del U(M,Q))/(delnu_(M))d sigma+(lambda )/(pi)int_(Gamma)^(rho)varphi_(2)(Q)(del ln r(M,Q))/(delnu_(M))d sigma]}:}\left.\begin{array}{l} \frac{1}{4 \pi} \delta_{2}(M)=\varphi_{1}(M)+\frac{\lambda}{\pi} \int_{\Gamma} \varphi_{1}(Q) \frac{\delta \ln r(M, Q)}{\partial \nu_{M}} d \sigma \tag{6}\\ \frac{1}{\pi} \delta_{1}(M)=\varphi_{2}(M)+\frac{\lambda}{\pi} \int_{\Gamma} \varphi_{1}(Q) \frac{\partial U(M, Q)}{\partial \nu_{M}} d \sigma+\frac{\lambda}{\pi} \int_{\Gamma}^{\rho} \varphi_{2}(Q) \frac{\partial \ln r(M, Q)}{\partial \nu_{M}} d \sigma \end{array}\right\}(6)14πδ2(M)=φ1(M)+λπ∫Γφ1(Q)δIn⁡R(M,Q)∂nMdσ1πδ1(M)=φ2(M)+λπ∫Γφ1(Q)∂U(M,Q)∂nMdσ+λπ∫Γρφ2(Q)∂In⁡R(M,Q)∂nMdσ}
We will show that this system has a solution [ φ 1 , φ 2 ] φ 1 , φ 2 [varphi_(1),varphi_(2)]\left[\varphi_{1}, \varphi_{2}\right][φ1,φ2], and with its help the function u ( P ) u ( P ) u(P)u(P)you(P)can be represented by formula (3).
Parallel to (6) let us consider the homogeneous system
( ′ ) φ 1 ( M ) = − λ π ∫ Γ φ 1 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ v M d σ φ 2 ( M ) = − λ π ∫ Γ φ 1 ( Q ) ∂ U ( M , Q ) ∂ v M d σ − λ π ∫ Γ φ 2 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ v M d σ } ( ′ ) φ 1 ( M ) = − λ π ∫ Γ   φ 1 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ v M d σ φ 2 ( M ) = − λ π ∫ Γ   φ 1 ( Q ) ∂ U ( M , Q ) ∂ v M d σ − λ π ∫ Γ   φ 2 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ v M d σ {:('")"{:[varphi_(1)(M)=-(lambda )/(pi)int_(Gamma)varphi_(1)(Q)(del ln r(M,Q))/(delv_(M))d sigma],[varphi_(2)(M)=-(lambda )/(pi)int_(Gamma)varphi_(1)(Q)(del U(M,Q))/(delv_(M))d sigma-(lambda )/(pi)int_(Gamma)varphi_(2)(Q)(del ln r(M,Q))/(delv_(M))d sigma]}:}\left.\begin{array}{l} \varphi_{1}(M)=-\frac{\lambda}{\pi} \int_{\Gamma} \varphi_{1}(Q) \frac{\partial \ln r(M, Q)}{\partial v_{M}} d \sigma \tag{$\prime$}\\ \varphi_{2}(M)=-\frac{\lambda}{\pi} \int_{\Gamma} \varphi_{1}(Q) \frac{\partial U(M, Q)}{\partial v_{M}} d \sigma-\frac{\lambda}{\pi} \int_{\Gamma} \varphi_{2}(Q) \frac{\partial \ln r(M, Q)}{\partial v_{M}} d \sigma \end{array}\right\}(′)φ1(M)=−λπ∫Γφ1(Q)∂In⁡R(M,Q)∂VMdσφ2(M)=−λπ∫Γφ1(Q)∂U(M,Q)∂VMdσ−λπ∫Γφ2(Q)∂In⁡R(M,Q)∂VMdσ}
For λ = 1 λ = 1 lambda=1\lambda=1λ=1the system (6') has as eigenvector only [ 0 , φ 0 ] , φ 0 ( M ) 0 , φ 0 , φ 0 ( M ) [0,varphi_(0)],varphi_(0)(M)\left[0, \varphi_{0}\right], \varphi_{0}(M)[0,φ0],φ0(M)being the only eigenfunction of the first equation in system (6') for λ = 1 λ = 1 lambda=1\lambda=1λ=1We can convince ourselves of this last statement in the following way: we observe that
lim ∂ ln ⁡ r ( M , Q ) ∂ v M = − curbura în M 2 lim ∂ ln ⁡ r ( M , Q ) ∂ v M = −  curbura în  M 2 lim(del ln r(M,Q))/(delv_(M))=-(" curbura în "M)/(2)\lim \frac{\partial \ln r(M, Q)}{\partial v_{M}}=-\frac{\text { curbura în } M}{2}lim∂In⁡R(M,Q)∂VM=− curvature in M2
when Q → M Q → M Q rarr MQ \rightarrow MQ→M, which means that δ ln ⁡ r ( M , Q ) δ ν M δ ln ⁡ r ( M , Q ) δ ν M (delta ln r(M,Q))/(deltanu_(M))\frac{\delta \ln r(M, Q)}{\delta \nu_{M}}δIn⁡R(M,Q)δnMis continuous on Γ Γ Gamma\GammaΓ, so any eigenfunction of the equation satisfies the Lipschitz condition [3].
Let us now assume that the first equation of the system (6') has two linearly independent eigenfunctions g 1 ( M ) g 1 ( M ) g_(1)(M)g_{1}(M)g1(M)and g 2 ( M ) g 2 ( M ) g_(2)(M)g_{2}(M)g2(M). Either
V ( P ) = ∫ Γ ˙ [ C 1 ⋅ g 1 ( Q ) + C 2 g 2 ( Q ) ] ln ⁡ r ( P , Q ) d σ V ( P ) = ∫ Γ ˙   C 1 ⋅ g 1 ( Q ) + C 2 g 2 ( Q ) ln ⁡ r ( P , Q ) d σ V(P)=int_(Gamma^(˙))[C_(1)*g_(1)(Q)+C_(2)g_(2)(Q)]ln r(P,Q)d sigmaV(P)=\int_{\dot{\Gamma}}\left[C_{1} \cdot g_{1}(Q)+C_{2} g_{2}(Q)\right] \ln r(P, Q) d \sigmaV(P)=∫Γ˙[C1⋅g1(Q)+C2g2(Q)]In⁡R(P,Q)dσ
Given that lim ∂ V ( P ) ∂ ν M = 0 lim ∂ V ( P ) ∂ ν M = 0 lim(del V(P))/(delnu_(M))=0\lim \frac{\partial V(P)}{\partial \nu_{M}}=0lim∂V(P)∂nM=0, we have V ( P ) ≡ C V ( P ) ≡ C V(P)-=CV(P) \equiv CV(P)≡Con Ω + Γ Ω + Γ Omega+Gamma\Omega+\GammaΩ+ΓWe choose the constants C 1 C 1 C_(1)C_{1}C1and C 2 C 2 C_(2)C_{2}C2in such a way that we have V ( P ) ≡ 0 V ( P ) ≡ 0 V(P)-=0V(P) \equiv 0V(P)≡0But
∬ C ( grad V ) 2 d τ = ∫ Γ V ∂ V ∂ ν d σ = 0 ∬ C   ( grad V ) 2 d τ = ∫ Γ   V ∂ V ∂ ν d σ = 0 ∬_(C)(grad V)^(2)d tau=int_(Gamma)V(del V)/(del nu)d sigma=0\iint_{C}(\operatorname{grad} V)^{2} d \tau=\int_{\Gamma} V \frac{\partial V}{\partial \nu} d \sigma=0∬C(degreeV)2dτ=∫ΓV∂V∂ndσ=0
so V ( P ) ≡ 0 V ( P ) ≡ 0 V(P)-=0V(P) \equiv 0V(P)≡0in the whole plan. And finally, let it be P ∈ Ω P ∈ Ω P in OmegaP \in \OmegaP∈Ωand P ′ ∈ C Ω P ′ ∈ C Ω P^(')in C OmegaP^{\prime} \in C \OmegaP′∈CΩthen
lim [ ∂ V ( F ) ∂ v M − ∂ V ( P ′ ) ∂ v M ] = 2 π [ C 1 g 1 ( M ) + C 2 g 2 ( M ) ] = 0 , lim ∂ V ( F ) ∂ v M − ∂ V P ′ ∂ v M = 2 π C 1 g 1 ( M ) + C 2 g 2 ( M ) = 0 , lim[(del V(F))/(delv_(M))-(del V(P^(')))/(delv_(M))]=2pi[C_(1)g_(1)(M)+C_(2)g_(2)(M)]=0,\lim \left[\frac{\partial V(F)}{\partial v_{M}}-\frac{\partial V\left(P^{\prime}\right)}{\partial v_{M}}\right]=2 \pi\left[C_{1} g_{1}(M)+C_{2} g_{2}(M)\right]=0,lim[∂V(F)∂VM−∂V(P′)∂VM]=2π[C1g1(M)+C2g2(M)]=0,
which is inconsistent with the hypothesis that functions g 1 ( M ) g 1 ( M ) g_(1)(M)g_{1}(M)g1(M)and g 2 ( M ) g 2 ( M ) g_(2)(M)g_{2}(M)g2(M)are linearly independent. We now proceed to verify the first statement. Let [ ψ 1 , ψ 2 ] ψ 1 , ψ 2 [psi_(1),psi_(2)]\left[\psi_{1}, \psi_{2}\right][ψ1,ψ2]an arbitrary eigenvector of the system (6') when λ = 1 λ = 1 lambda=1\lambda=1λ=1It is easy to see that ψ 1 ≡ 0 ψ 1 ≡ 0 psi_(1)-=0\psi_{1} \equiv 0ψ1≡0and ψ 2 = c φ 0 ψ 2 = c φ 0 psi_(2)=cvarphi_(0)\psi_{2}=c \varphi_{0}ψ2=cφ0Indeed, based on the above ψ 1 = c φ 0 ψ 1 = c φ 0 psi_(1)=cvarphi_(0)\psi_{1}=c \varphi_{0}ψ1=cφ0and then
φ 2 ( M ) = − c π ∫ Γ φ 0 ( Q ) ∂ U ( M , Q ) ∂ v M d σ − 1 π ∫ Γ φ 2 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ v M d σ φ 2 ( M ) = − c π ∫ Γ   φ 0 ( Q ) ∂ U ( M , Q ) ∂ v M d σ − 1 π ∫ Γ   φ 2 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ v M d σ varphi_(2)(M)=-(c)/( pi)int_(Gamma)varphi_(0)(Q)(del U(M,Q))/(delv_(M))d sigma-(1)/(pi)int_(Gamma)varphi_(2)(Q)(del ln r(M,Q))/(delv_(M))d sigma\varphi_{2}(M)=-\frac{c}{\pi} \int_{\Gamma} \varphi_{0}(Q) \frac{\partial U(M, Q)}{\partial v_{M}} d \sigma-\frac{1}{\pi} \int_{\Gamma} \varphi_{2}(Q) \frac{\partial \ln r(M, Q)}{\partial v_{M}} d \sigmaφ2(M)=−cπ∫Γφ0(Q)∂U(M,Q)∂VMdσ−1π∫Γφ2(Q)∂In⁡R(M,Q)∂VMdσ
This equation has a solution only if the free term is orthogonal to the unique eigenfunction of the conjugate homogeneous equation, which in this case is the constant. So the equality must hold
c. K π ∫ Γ d σ ∫ Γ φ 0 ( Q ) ∂ U ( M , Q ) ∂ ν M d σ = 0  c.  K π ∫ Γ   d σ ∫ Γ   φ 0 ( Q ) ∂ U ( M , Q ) ∂ ν M d σ = 0 (" c. "K)/(pi)int_(Gamma)d sigmaint_(Gamma)varphi_(0)(Q)(del U(M,Q))/(delnu_(M))d sigma=0\frac{\text { c. } K}{\pi} \int_{\Gamma} d \sigma \int_{\Gamma} \varphi_{0}(Q) \frac{\partial U(M, Q)}{\partial \nu_{M}} d \sigma=0 c. Kπ∫Γdσ∫Γφ0(Q)∂U(M,Q)∂nMdσ=0
or
− c . K π ∬ C . d τ ∫ Γ φ 0 ( Q ) Δ U ( P , Q ) d σ = 0 − c . K π ∬ C .   d τ ∫ Γ   φ 0 ( Q ) Δ U ( P , Q ) d σ = 0 -(c.K)/(pi)∬_(C.)d tauint_(Gamma)varphi_(0)(Q)Delta U(P,Q)d sigma=0-\frac{c . K}{\pi} \iint_{C .} d \tau \int_{\Gamma} \varphi_{0}(Q) \Delta U(P, Q) d \sigma=0−c.Kπ∬C.dτ∫Γφ0(Q)ΔU(P,Q)dσ=0
From here it follows c = 0 c = 0 c=0c=0c=0, since ∫ Γ φ 0 ( Q ) Δ U ( P , Q ) d σ ≡ ∫ Γ   φ 0 ( Q ) Δ U ( P , Q ) d σ ≡ int_(Gamma)varphi_(0)(Q)Delta U(P,Q)d sigma-=\int_{\Gamma} \varphi_{0}(Q) \Delta U(P, Q) d \sigma \equiv∫Γφ0(Q)ΔU(P,Q)dσ≡constant ≠ 0 ≠ 0 !=0\neq 0≠0As a result ψ 1 ≡ 0 ψ 1 ≡ 0 psi_(1)-=0\psi_{1} \equiv 0ψ1≡0and ψ 2 = c φ 0 ψ 2 = c φ 0 psi_(2)=cvarphi_(0)\psi_{2}=c \varphi_{0}ψ2=cφ0Based on Fredholm's theorems we can state that the conjugate system of (6') also has a single eigenvector, when λ = 1 λ = 1 lambda=1\lambda=1λ=1It is immediately noticeable that [ 1 , 0 ] [ 1 , 0 ] [1,0][1,0][1,0]is this vector. So the necessary and sufficient condition for system (6) to have a solution is that
∫ Γ δ 2 ( Q ) d σ = 0 ∫ Γ   δ 2 ( Q ) d σ = 0 int_(Gamma)delta_(2)(Q)d sigma=0\int_{\Gamma} \delta_{2}(Q) d \sigma=0∫Γδ2(Q)dσ=0
and the general solution is
φ 1 ( M ) = φ 1 ∗ ( M ) ; φ 2 ( M ) = φ 2 ∗ ( M ) + c φ v ( M ) φ 1 ( M ) = φ 1 ∗ ( M ) ; φ 2 ( M ) = φ 2 ∗ ( M ) + c φ v ( M ) varphi_(1)(M)=varphi_(1)^(**)(M);varphi_(2)(M)=varphi_(2)^(**)(M)+cvarphi_(v)(M)\varphi_{1}(M)=\varphi_{1}^{*}(M) ; \varphi_{2}(M)=\varphi_{2}^{*}(M)+c \varphi_{v}(M)φ1(M)=φ1∗(M);φ2(M)=φ2∗(M)+cφV(M)
where [ φ 1 ∗ , φ 2 ∗ ] φ 1 ∗ , φ 2 ∗ [varphi_(1)^(**),varphi_(2)^(**)]\left[\varphi_{1}^{*}, \varphi_{2}^{*}\right][φ1∗,φ2∗]represents a particular solution of the system (6). We will show that
u ( P ) = ∫ Γ φ 1 ( Q ) U ( P , Q ) d σ + ∫ Γ φ 2 ( Q ) ln ⁡ r ( P , Q ) d σ u ( P ) = ∫ Γ   φ 1 ( Q ) U ( P , Q ) d σ + ∫ Γ   φ 2 ( Q ) ln ⁡ r ( P , Q ) d σ u(P)=int_(Gamma)varphi_(1)(Q)U(P,Q)d sigma+int_(Gamma)varphi_(2)(Q)ln r(P,Q)d sigmau(P)=\int_{\Gamma} \varphi_{1}(Q) U(P, Q) d \sigma+\int_{\Gamma} \varphi_{2}(Q) \ln r(P, Q) d \sigmayou(P)=∫Γφ1(Q)U(P,Q)dσ+∫Γφ2(Q)In⁡R(P,Q)dσ
Indeed, either P 0 P 0 P_(0)P_{0}P0an arbitrary point in the domain Ω Ω Omega\OmegaΩ. We define the constant c c cccfrom the condition
u ( P 0 ) = ∫ Γ φ 1 ( Q ) U ( P 0 , Q ) d σ + ∫ Γ φ 2 ( Q ) ln ⁡ r ( P 0 , Q ) d σ u P 0 = ∫ Γ   φ 1 ( Q ) U P 0 , Q d σ + ∫ Γ   φ 2 ( Q ) ln ⁡ r P 0 , Q d σ u(P_(0))=int_(Gamma)varphi_(1)(Q)U(P_(0),Q)d sigma+int_(Gamma)varphi_(2)(Q)ln r(P_(0),Q)d sigmau\left(P_{0}\right)=\int_{\Gamma} \varphi_{1}(Q) U\left(P_{0}, Q\right) d \sigma+\int_{\Gamma} \varphi_{2}(Q) \ln r\left(P_{0}, Q\right) d \sigmayou(P0)=∫Γφ1(Q)U(P0,Q)dσ+∫Γφ2(Q)In⁡R(P0,Q)dσ
note
v ( P ) = u ( P ) − ∫ Γ φ 1 ( Q ) U ( P , Q ) d σ − ∫ Γ φ 2 ( Q ) ln ⁡ r ( P , Q ) d σ v ( P ) = u ( P ) − ∫ Γ   φ 1 ( Q ) U ( P , Q ) d σ − ∫ Γ   φ 2 ( Q ) ln ⁡ r ( P , Q ) d σ v(P)=u(P)-int_(Gamma)varphi_(1)(Q)U(P,Q)d sigma-int_(Gamma)varphi_(2)(Q)ln r(P,Q)d sigmav(P)=u(P)-\int_{\Gamma} \varphi_{1}(Q) U(P, Q) d \sigma-\int_{\Gamma} \varphi_{2}(Q) \ln r(P, Q) d \sigmaV(P)=you(P)−∫Γφ1(Q)U(P,Q)dσ−∫Γφ2(Q)In⁡R(P,Q)dσ
It is obvious that v ( P ) ∈ { u } v ( P ) ∈ { u } v(P)in{u}v(P) \in\{u\}V(P)∈{you}and lim ∂ v ( P ) ∂ v M = lim ∂ Δ v ( P ) ∂ v M = 0 . lim ∂ v ( P ) ∂ v M = lim ∂ Δ v ( P ) ∂ v M = 0 . lim(del v(P))/(delv_(M))=lim(del Delta v(P))/(delv_(M))=0.quad\lim \frac{\partial v(P)}{\partial v_{M}}=\lim \frac{\partial \Delta v(P)}{\partial v_{M}}=0 . \quadlim∂V(P)∂VM=lim∂ΔV(P)∂VM=0.We denote lim v ( P ) == v 1 ( M ) v ( P ) == v 1 ( M ) v(P)==v_(1)(M)v(P)= =v_{1}(M)V(P)==V1(M)and lim Δ v 2 ( P ) = v 2 ( M ) lim Δ v 2 ( P ) = v 2 ( M ) lim Deltav_(2)(P)=v_(2)(M)\lim \Delta v_{2}(P)=v_{2}(M)limΔV2(P)=V2(M)Using equality (5) we have
∫ Γ [ ν 1 ( Q ) ∂ Δ U ( P ′ , Q ) ∂ ν Q + ν 2 ( Q ) ∂ U ( P ′ , Q ) ∂ ν Q ] d σ = 0 ∫ Γ   ν 1 ( Q ) ∂ Δ U P ′ , Q ∂ ν Q + ν 2 ( Q ) ∂ U P ′ , Q ∂ ν Q d σ = 0 int_(Gamma)[nu_(1)(Q)(del Delta U(P^('),Q))/(delnu_(Q))+nu_(2)(Q)(del U(P^('),Q))/(delnu_(Q))]d sigma=0\int_{\Gamma}\left[\nu_{1}(Q) \frac{\partial \Delta U\left(P^{\prime}, Q\right)}{\partial \nu_{Q}}+\nu_{2}(Q) \frac{\partial U\left(P^{\prime}, Q\right)}{\partial \nu_{Q}}\right] d \sigma=0∫Γ[n1(Q)∂ΔU(P′,Q)∂nQ+n2(Q)∂U(P′,Q)∂nQ]dσ=0
But applying the operator Δ Δ Delta\DeltaΔfrom this equality we obtain
∫ Γ ν 2 ( Q ) ∂ ln ⁡ r ( P ′ , Q ) ∂ ν Q d σ = 0 ∫ Γ   ν 2 ( Q ) ∂ ln ⁡ r P ′ , Q ∂ ν Q d σ = 0 int_(Gamma)nu_(2)(Q)(del ln r(P^('),Q))/(delnu_(Q))d sigma=0\int_{\Gamma} \nu_{2}(Q) \frac{\partial \ln r\left(P^{\prime}, Q\right)}{\partial \nu_{Q}} d \sigma=0∫Γn2(Q)∂In⁡R(P′,Q)∂nQdσ=0
Crossing the line when P ′ → M P ′ → M P^(')rarr MP^{\prime} \rightarrow MP′→Mon the normal at the point M M MMMwe have
respectively
4 π ν 1 ( M ) = − 4 ∫ Γ ν 1 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ ν Q d σ − ∫ Γ ν 2 ( Q ) ∂ U ( M , Q ) ∂ ν Q d σ 4 π ν 1 ( M ) = − 4 ∫ Γ   ν 1 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ ν Q d σ − ∫ Γ   ν 2 ( Q ) ∂ U ( M , Q ) ∂ ν Q d σ 4pinu_(1)(M)=-4int_(Gamma)nu_(1)(Q)(del ln r(M,Q))/(delnu_(Q))d sigma-int_(Gamma)nu_(2)(Q)(del U(M,Q))/(delnu_(Q))d sigma4 \pi \nu_{1}(M)=-4 \int_{\Gamma} \nu_{1}(Q) \frac{\partial \ln r(M, Q)}{\partial \nu_{Q}} d \sigma-\int_{\Gamma} \nu_{2}(Q) \frac{\partial U(M, Q)}{\partial \nu_{Q}} d \sigma4πn1(M)=−4∫Γn1(Q)∂In⁡R(M,Q)∂nQdσ−∫Γn2(Q)∂U(M,Q)∂nQdσ
π ν 2 ( M ) = − ∫ Γ ν 2 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ ν Q d σ π ν 2 ( M ) = − ∫ Γ   ν 2 ( Q ) ∂ ln ⁡ r ( M , Q ) ∂ ν Q d σ pinu_(2)(M)=-int_(Gamma)nu_(2)(Q)(del ln r(M,Q))/(delnu_(Q))d sigma\pi \nu_{2}(M)=-\int_{\Gamma} \nu_{2}(Q) \frac{\partial \ln r(M, Q)}{\partial \nu_{Q}} d \sigmaπn2(M)=−∫Γn2(Q)∂In⁡R(M,Q)∂nQdσ
Apart from a constant factor, this system coincides with the conjugate system of (6). So the only solution is [ c , 0 ] [ c , 0 ] [c,0][c, 0][c,0]Applying formula (4) we find
v ( P ) = c 8 π ∫ Γ ∂ Δ U ( P , Q ) ∂ v Q d σ = c 2 π ∫ Γ ∂ ln ⁡ r ( P , Q ) ∂ v Q d σ = c , ( P ∈ Ω ) 5 ) v ( P ) = c 8 π ∫ Γ   ∂ Δ U ( P , Q ) ∂ v Q d σ = c 2 π ∫ Γ   ∂ ln ⁡ r ( P , Q ) ∂ v Q d σ = c , ( P ∈ Ω ) 5 {:v(P)=(c)/(8pi)int_(Gamma)(del Delta U(P,Q))/(delv_(Q))d sigma=(c)/(2pi)int_(Gamma)(del ln r(P,Q))/(delv_(Q))d sigma=c,(P in Omega)^(5))\left.v(P)=\frac{c}{8 \pi} \int_{\Gamma} \frac{\partial \Delta U(P, Q)}{\partial v_{Q}} d \sigma=\frac{c}{2 \pi} \int_{\Gamma} \frac{\partial \ln r(P, Q)}{\partial v_{Q}} d \sigma=c,(P \in \Omega)^{5}\right)V(P)=c8π∫Γ∂ΔU(P,Q)∂VQdσ=c2π∫Γ∂In⁡R(P,Q)∂VQdσ=c,(P∈Ω)5)
But v ( P 0 ) = 0 v P 0 = 0 v(P_(0))=0v\left(P_{0}\right)=0V(P0)=0so v ( P ) ≡ 0 v ( P ) ≡ 0 v(P)-=0v(P) \equiv 0V(P)≡0, which proves our statement.
Lemma 1. If u ∈ { u } u ∈ { u } u in{u}u \in\{u\}you∈{you}then
∫ Γ δ 2 ( Q ) d σ = 0 ∫ Γ   δ 2 ( Q ) d σ = 0 int_(Gamma)delta_(2)(Q)d sigma=0\int_{\Gamma} \delta_{2}(Q) d \sigma=0∫Γδ2(Q)dσ=0
Demonstration. We apply the operator Δ Δ Delta\DeltaΔequality (5):
∫ Γ [ μ 2 ( Q ) ∂ Δ U ( P ′ , Q ) ∂ ν Q − δ 2 ( Q ) Δ U ( P ′ , Q ) ] d σ = 0 ∫ Γ   μ 2 ( Q ) ∂ Δ U P ′ , Q ∂ ν Q − δ 2 ( Q ) Δ U P ′ , Q d σ = 0 int_(Gamma)[mu_(2)(Q)(del Delta U(P^('),Q))/(delnu_(Q))-delta_(2)(Q)Delta U(P^('),Q)]d sigma=0\int_{\Gamma}\left[\mu_{2}(Q) \frac{\partial \Delta U\left(P^{\prime}, Q\right)}{\partial \nu_{Q}}-\delta_{2}(Q) \Delta U\left(P^{\prime}, Q\right)\right] d \sigma=0∫Γ[μ2(Q)∂ΔU(P′,Q)∂nQ−δ2(Q)ΔU(P′,Q)]dσ=0
We integrate over an arbitrary circle C R C R C_(R)C_{R}CRradius R R RRR, which contains the domain Ω Ω Omega\OmegaΩinside it
∫ Γ μ 2 ( Q ) d σ ∫ C R ∂ Δ U ( P ′ , Q ) ∂ ν Q d σ − ∫ Γ ~ δ 2 ( Q ) d σ ∫ C R Δ U ( P ′ , Q ) d σ = 0 Dar ∫ C R Δ U ( P ′ , Q ) d σ = ∫ C R [ 4 ln ⁡ r ( P ′ , Q ) + 3 ] d σ ≡ constantă pe Γ , findcă ∫ Γ   μ 2 ( Q ) d σ ∫ C R   ∂ Δ U P ′ , Q ∂ ν Q d σ − ∫ Γ ~   δ 2 ( Q ) d σ ∫ C R   Δ U P ′ , Q d σ = 0 Dar ∫ C R   Δ U P ′ , Q d σ = ∫ C R   4 ln ⁡ r P ′ , Q + 3 d σ ≡  constantă pe  Γ ,  findcă  {:[int_(Gamma)mu_(2)(Q)d sigmaint_(C_(R))(del Delta U(P^('),Q))/(delnu_(Q))d sigma-int_( tilde(Gamma))delta_(2)(Q)d sigmaint_(C_(R))Delta U(P^('),Q)d sigma=0],[Darint_(C_(R))Delta U(P^('),Q)d sigma=int_(C_(R))[4ln r(P^('),Q)+3]d sigma-=" constantă pe "Gamma","" findcă "]:}\begin{gathered} \int_{\Gamma} \mu_{2}(Q) d \sigma \int_{C_{R}} \frac{\partial \Delta U\left(P^{\prime}, Q\right)}{\partial \nu_{Q}} d \sigma-\int_{\tilde{\Gamma}} \delta_{2}(Q) d \sigma \int_{C_{R}} \Delta U\left(P^{\prime}, Q\right) d \sigma=0 \\ \operatorname{Dar} \int_{C_{R}} \Delta U\left(P^{\prime}, Q\right) d \sigma=\int_{C_{R}}\left[4 \ln r\left(P^{\prime}, Q\right)+3\right] d \sigma \equiv \text { constantă pe } \Gamma, \text { findcă } \end{gathered}∫Γμ2(Q)dσ∫CR∂ΔU(P′,Q)∂nQdσ−∫Γ~δ2(Q)dσ∫CRΔU(P′,Q)dσ=0But∫CRΔU(P′,Q)dσ=∫CR[4In⁡R(P′,Q)+3]dσ≡ constant on Γ, because 
∫ C R ln ⁡ r ( P ′ , Q ) d σ ≡ c ∫ C R   ln ⁡ r P ′ , Q d σ ≡ c int_(C_(R))ln r(P^('),Q)d sigma-=c\int_{C_{R}} \ln r\left(P^{\prime}, Q\right) d \sigma \equiv c∫CRIn⁡R(P′,Q)dσ≡cinside the circle. From here it follows ∫ C R ∂ Δ U ( P ′ , Q ) ∂ ν Q d σ = 0 ∫ C R   ∂ Δ U P ′ , Q ∂ ν Q d σ = 0 int_(C_(R))(del Delta U(P^('),Q))/(delnu_(Q))d sigma=0\int_{C_{R}} \frac{\partial \Delta U\left(P^{\prime}, Q\right)}{\partial \nu_{Q}} d \sigma=0∫CR∂ΔU(P′,Q)∂nQdσ=0therefore ∫ Γ δ 2 ( Q ) d σ = 0 ∫ Γ   δ 2 ( Q ) d σ = 0 int_(Gamma)delta_(2)(Q)d sigma=0\int_{\Gamma} \delta_{2}(Q) d \sigma=0∫Γδ2(Q)dσ=0.
4. We will prove the uniqueness of the solution to the boundary value problem proposed in point 2, relative to the elements of the set { u } { u } {u}\{u\}{you}.
We introduce the functional
(7) F ( u ) = ∫ Γ ˙ ( μ 1 δ 2 − μ 2 δ 1 ) d σ + ( 1 − σ ) ∫ Γ δ 1 2 d σ (7) F ( u ) = ∫ Γ ˙   μ 1 δ 2 − μ 2 δ 1 d σ + ( 1 − σ ) ∫ Γ   δ 1 2 d σ {:(7)F(u)=int_(Gamma^(˙))(mu_(1)delta_(2)-mu_(2)delta_(1))d sigma+(1-sigma)int_(Gamma)delta_(1)^(2)d sigma:}\begin{equation*} F(u)=\int_{\dot{\Gamma}}\left(\mu_{1} \delta_{2}-\mu_{2} \delta_{1}\right) d \sigma+(1-\sigma) \int_{\Gamma} \delta_{1}^{2} d \sigma \tag{7} \end{equation*}(7)F(you)=∫Γ˙(μ1δ2−μ2δ1)dσ+(1−σ)∫Γδ12dσ
defined on the elements of the set { u } . F ( u ) { u } . F ( u ) {u}.F(u)\{u\} . F(u){you}.F(you)is a positive functional, which is obvious if we first observe that
∫ Γ ( μ 1 δ 2 − μ 2 δ 1 ) d σ = ∬ Ω ( Δ u ) 2 d σ ∫ Γ   μ 1 δ 2 − μ 2 δ 1 d σ = ∬ Ω   ( Δ u ) 2 d σ int_(Gamma)(mu_(1)delta_(2)-mu_(2)delta_(1))d sigma=∬_(Omega)(Delta u)^(2)d sigma\int_{\Gamma}\left(\mu_{1} \delta_{2}-\mu_{2} \delta_{1}\right) d \sigma=\iint_{\Omega}(\Delta u)^{2} d \sigma∫Γ(μ1δ2−μ2δ1)dσ=∬Ω(Δyou)2dσ
But this last equality can be verified by replacing the expressions of the functions in the first member μ 1 , δ 1 , μ 2 , δ 2 μ 1 , δ 1 , μ 2 , δ 2 mu_(1),delta_(1),mu_(2),delta_(2)\mu_{1}, \delta_{1}, \mu_{2}, \delta_{2}μ1,δ1,μ2,δ2and performing simple calculations following the path of proving Theorem 3.
Theorem 4. If u ∈ { u } s s i u ∈ { u } s s i u in{u}ssiu \in\{u\} s s iyou∈{you}SSand
u = 0 ре Γ − Δ u + 1 − σ ρ 0 ∂ u ∂ ν = 0 ре Γ u = 0  ре  Γ − Δ u + 1 − σ ρ 0 ∂ u ∂ ν = 0  ре  Γ {:[u=0" ре "Gamma],[-Delta u+(1-sigma)/(rho_(0))(del u)/(del nu)=0" ре "Gamma]:}\begin{aligned} u & =0 \text { ре } \Gamma \\ -\Delta u+\frac{1-\sigma}{\rho_{0}} \frac{\partial u}{\partial \nu} & =0 \text { ре } \Gamma \end{aligned}you=0 re Γ−Δyou+1−σρ0∂you∂n=0 re Γ
then u ≡ 0 u ≡ 0 u-=0u \equiv 0you≡0in Ω Ω Omega\OmegaΩ.
Demonstration. Obviously F ( u ) = 0 F ( u ) = 0 F(u)=0F(u)=0F(you)=0, from where it follows
∬ Ω ( Δ u ) 2 d τ = 0 şi ∫ Γ ˙ ( ∂ u ∂ ν ) 2 d σ = 0 ∬ Ω   ( Δ u ) 2 d τ = 0  şi  ∫ Γ ˙   ∂ u ∂ ν 2 d σ = 0 ∬_(Omega)(Delta u)^(2)d tau=0" şi "int_(Gamma^(˙))((del u)/(del nu))^(2)d sigma=0\iint_{\Omega}(\Delta u)^{2} d \tau=0 \text { şi } \int_{\dot{\Gamma}}\left(\frac{\partial u}{\partial \nu}\right)^{2} d \sigma=0∬Ω(Δyou)2dτ=0 and ∫Γ˙(∂you∂n)2dσ=0
so Δ u = 0 Δ u = 0 Delta u=0\Delta u=0Δyou=0in Ω Ω Omega\OmegaΩand ∂ u ∂ v = 0 ∂ u ∂ v = 0 (del u)/(del v)=0\frac{\partial u}{\partial v}=0∂you∂V=0on Γ Γ Gamma\GammaΓThis means that u u uuyouis the solution to Neumann's problem with the condition ∂ u ∂ ν = 0 ∂ u ∂ ν = 0 (del u)/(del nu)=0\frac{\partial u}{\partial \nu}=0∂you∂n=0on the border; so u ≡ c u ≡ c u-=cu \equiv cyou≡cin Ω Ω Omega\OmegaΩ, and because the condition of the theorem requires that u = 0 u = 0 u=0u=0you=0on Γ Γ Gamma\GammaΓ, follows u ≡ 0 u ≡ 0 u-=0u \equiv 0you≡0.
5. Either P ( ρ , φ ) P ( ρ , φ ) P(rho,varphi)P(\rho, \varphi)P(ρ,φ)and Q ( ρ ′ , φ ′ ) Q ρ ′ , φ ′ Q(rho^('),varphi^('))Q\left(\rho^{\prime}, \varphi^{\prime}\right)Q(ρ′,φ′)two points in the plane. The formula is well known
ln ⁡ r ( P , Q ) = ln ⁡ ρ ′ − ∑ n = 1 ∞ 1 n κ n ρ ′ n cos ⁡ n ( φ − φ ′ ) ln ⁡ r ( P , Q ) = ln ⁡ ρ ′ − ∑ n = 1 ∞   1 n κ n ρ ′ n cos ⁡ n φ − φ ′ ln r(P,Q)=ln rho^(')-sum_(n=1)^(oo)(1)/(n)(kappa^(n))/(rho^('n))cos n(varphi-varphi^('))\ln r(P, Q)=\ln \rho^{\prime}-\sum_{n=1}^{\infty} \frac{1}{n} \frac{\kappa^{n}}{\rho^{\prime n}} \cos n\left(\varphi-\varphi^{\prime}\right)In⁡R(P,Q)=In⁡ρ′−∑n=1∞1nκnρ′nCart⁡n(φ−φ′)
With its help we get
r 2 ln ⁡ r = ln ⁡ ρ ′ ( ρ 2 + ρ ′ 2 − 2 ρ ρ ′ cos ⁡ γ ) + ρ 2 − ρ ρ ′ cos ⁡ γ − ∑ n = 1 ∞ 1 n ( n + 1 ) ρ n + 2 ρ ′ n cos ⁡ n γ + + ∑ n = 2 ∞ 1 n ( n − 1 ) ρ n ρ ′ n − 2 cos ⁡ n γ r 2 ln ⁡ r = ln ⁡ ρ ′ ρ 2 + ρ ′ 2 − 2 ρ ρ ′ cos ⁡ γ + ρ 2 − ρ ρ ′ cos ⁡ γ − ∑ n = 1 ∞   1 n ( n + 1 ) ρ n + 2 ρ ′ n cos ⁡ n γ + + ∑ n = 2 ∞   1 n ( n − 1 ) ρ n ρ ′ n − 2 cos ⁡ n γ {:[r^(2)ln r=ln rho^(')(rho^(2)+rho^('2)-2rhorho^(')cos gamma)+rho^(2)-rhorho^(')cos gamma-sum_(n=1)^(oo)(1)/(n(n+1))(rho^(n+2))/(rho^('n))cos n gamma+],[+sum_(n=2)^(oo)(1)/(n(n-1))(rho^(n))/(rho^('n-2))cos n gamma]:}\begin{gathered} r^{2} \ln r=\ln \rho^{\prime}\left(\rho^{2}+\rho^{\prime 2}-2 \rho \rho^{\prime} \cos \gamma\right)+\rho^{2}-\rho \rho^{\prime} \cos \gamma-\sum_{n=1}^{\infty} \frac{1}{n(n+1)} \frac{\rho^{n+2}}{\rho^{\prime n}} \cos n \gamma+ \\ +\sum_{n=2}^{\infty} \frac{1}{n(n-1)} \frac{\rho^{n}}{\rho^{\prime n-2}} \cos n \gamma \end{gathered}R2In⁡R=In⁡ρ′(ρ2+ρ′2−2ρρ′Cart⁡γ)+ρ2−ρρ′Cart⁡γ−∑n=1∞1n(n+1)ρn+2ρ′nCart⁡nγ++∑n=2∞1n(n−1)ρnρ′n−2Cart⁡nγ
where γ = φ − φ ′ γ = φ − φ ′ gamma=varphi-varphi^(')\gamma=\varphi-\varphi^{\prime}γ=φ−φ′If we introduce the following notations
α 1 = − ρ { cos ⁡ φ sin ⁡ φ ; α n = 1 n ( n − 1 ) ρ n { cos ⁡ n φ sin ⁡ n φ ( n = 2 , 3 , … ) , β 0 = ρ ∘ ; β n = − 1 n ( n + 1 ) ρ n + 2 { cos ⁡ n φ sin ⁡ n φ ( n = 1 , 2 , … ) , γ n = ρ ′ − n { cos ⁡ n φ ′ sin ⁡ n φ ′ ( n = 0 , 1 , … ) , δ n = ρ ′ n − 2 { cos ⁡ n φ ′ sin ⁡ n φ ′ ( n = 1 , 2 , … ) α 1 = − ρ cos ⁡ φ sin ⁡ φ ; α n = 1 n ( n − 1 ) ρ n cos ⁡ n φ sin ⁡ n φ ( n = 2 , 3 , … ) , β 0 = ρ ∘ ; β n = − 1 n ( n + 1 ) ρ n + 2 cos ⁡ n φ sin ⁡ n φ ( n = 1 , 2 , … ) , γ n = ρ ′ − n cos ⁡ n φ ′ sin ⁡ n φ ′ ( n = 0 , 1 , … ) , δ n = ρ ′ n − 2 cos ⁡ n φ ′ sin ⁡ n φ ′ ( n = 1 , 2 , … ) {:[alpha_(1)=-rho{[cos varphi],[sin varphi];quadalpha_(n)=(1)/(n(n-1))rho^(n){[cos n varphi],[sin n varphi]quad(n=2,3,dots),:}],[beta_(0)=rho^(@);quadbeta_(n)=-(1)/(n(n+1))rho^(n+2){[cos n varphi],[sin n varphi]quad(n=1,2,dots),:}],[gamma_(n)=rho^('-n){[cos nvarphi^(')],[sin nvarphi^(')]quad(n=0,1,dots),:}],[delta_(n)=rho^('n-2){[cos nvarphi^(')],[sin nvarphi^(')]quad(n=1,2,dots):}]:}\begin{gathered} \alpha_{1}=-\rho\left\{\begin{array}{l} \cos \varphi \\ \sin \varphi \end{array} ; \quad \alpha_{n}=\frac{1}{n(n-1)} \rho^{n}\left\{\begin{array}{l} \cos n \varphi \\ \sin n \varphi \end{array} \quad(n=2,3, \ldots),\right.\right. \\ \beta_{0}=\rho^{\circ} ; \quad \beta_{n}=-\frac{1}{n(n+1)} \rho^{n+2}\left\{\begin{array}{l} \cos n \varphi \\ \sin n \varphi \end{array} \quad(n=1,2, \ldots),\right. \\ \gamma_{n}=\rho^{\prime-n}\left\{\begin{array}{l} \cos n \varphi^{\prime} \\ \sin n \varphi^{\prime} \end{array} \quad(n=0,1, \ldots),\right. \\ \delta_{n}=\rho^{\prime n-2}\left\{\begin{array}{l} \cos n \varphi^{\prime} \\ \sin n \varphi^{\prime} \end{array} \quad(n=1,2, \ldots)\right. \end{gathered}α1=−ρ{Cart⁡φmy⁡φ;αn=1n(n−1)ρn{Cart⁡nφmy⁡nφ(n=2,3,…),β0=ρ∘;βn=−1n(n+1)ρn+2{Cart⁡nφmy⁡nφ(n=1,2,…),γn=ρ′−n{Cart⁡nφ′my⁡nφ′(n=0,1,…),δn=ρ′n−2{Cart⁡nφ′my⁡nφ′(n=1,2,…)
With their help we can write
again
r 2 ln ⁡ r = β 0 ln ⁡ ρ ′ + ρ ′ − ln ⁡ ρ + 2 α 1 δ 1 ln ⁡ ρ ′ + ∑ n = 0 ∞ β n γ n + ∑ n = 1 ∞ α n δ n r 2 ln ⁡ r = β 0 ln ⁡ ρ ′ + ρ ′ − ln ⁡ ρ + 2 α 1 δ 1 ln ⁡ ρ ′ + ∑ n = 0 ∞   β n γ n + ∑ n = 1 ∞   α n δ n r^(2)ln r=beta_(0)ln rho^(')+rho^(')-ln rho+2alpha_(1)delta_(1)ln rho^(')+sum_(n=0)^(oo)beta_(n)gamma_(n)+sum_(n=1)^(oo)alpha_(n)delta_(n)r^{2} \ln r=\beta_{0} \ln \rho^{\prime}+\rho^{\prime}-\ln \rho+2 \alpha_{1} \delta_{1} \ln \rho^{\prime}+\sum_{n=0}^{\infty} \beta_{n} \gamma_{n}+\sum_{n=1}^{\infty} \alpha_{n} \delta_{n}R2In⁡R=β0In⁡ρ′+ρ′−In⁡ρ+2α1δ1In⁡ρ′+∑n=0∞βnγn+∑n=1∞αnδn
To make the following calculations easier, we introduce shorthand notations
P ( u ) = μ 1 ( Q ) − δ 1 ( Q ) + μ 2 ( Q ) − δ 2 ( Q ) ; G ( v ) = ∂ Δ v ∂ v Q + Δ v + ∂ v ∂ v + v P ( u ) = μ 1 ( Q ) − δ 1 ( Q ) + μ 2 ( Q ) − δ 2 ( Q ) ; G ( v ) = ∂ Δ v ∂ v Q + Δ v + ∂ v ∂ v + v P(u)=mu_(1)(Q)-delta_(1)(Q)+mu_(2)(Q)-delta_(2)(Q);G(v)=(del Delta v)/(delv_(Q))+Delta v+(del v)/(del v)+vP(u)=\mu_{1}(Q)-\delta_{1}(Q)+\mu_{2}(Q)-\delta_{2}(Q) ; G(v)=\frac{\partial \Delta v}{\partial v_{Q}}+\Delta v+\frac{\partial v}{\partial v}+vP(you)=μ1(Q)−δ1(Q)+μ2(Q)−δ2(Q);G(V)=∂ΔV∂VQ+ΔV+∂V∂V+V
P ( u ) . G ( v ) = μ 1 ∂ Δ v ∂ v Q − δ 1 Δ v + μ 2 ∂ v ∂ v Q + δ 2 v P ( u ) . G ( v ) = μ 1 ∂ Δ v ∂ v Q − δ 1 Δ v + μ 2 ∂ v ∂ v Q + δ 2 v P(u).G(v)=mu_(1)(del Delta v)/(delv_(Q))-delta_(1)Delta v+mu_(2)(del v)/(delv_(Q))+delta_(2)vP(u) . G(v)=\mu_{1} \frac{\partial \Delta v}{\partial v_{Q}}-\delta_{1} \Delta v+\mu_{2} \frac{\partial v}{\partial v_{Q}}+\delta_{2} vP(you).G(V)=μ1∂ΔV∂VQ−δ1ΔV+μ2∂V∂VQ+δ2V
Substituting into formula (5) U = r ∙ ln ⁡ r U = r ∙ ln ⁡ r U=r^(∙)ln rU=r^{\bullet} \ln rU=R∙In⁡Rthrough its string above and using the abbreviated notations, equality (5) can be written
ln ⁡ ρ ′ ∫ Γ ˙ P ( u ) ⋅ G ( β 0 ) d σ + 2 δ 1 ln ⁡ ρ ′ ∫ Γ ˙ P ( u ) G ( α 1 ) d σ + ∑ n = o ∞ γ n ∫ Γ ˙ P ( u ) ⋅ G ( β n ) d σ + (8) + ∑ n = 1 ∞ δ n ∫ Γ P ( u ) G ( α n ) d σ = 0 ln ⁡ ρ ′ ∫ Γ ˙   P ( u ) ⋅ G β 0 d σ + 2 δ 1 ln ⁡ ρ ′ ∫ Γ ˙   P ( u ) G α 1 d σ + ∑ n = o ∞   γ n ∫ Γ ˙   P ( u ) ⋅ G β n d σ + (8) + ∑ n = 1 ∞   δ n ∫ Γ   P ( u ) G α n d σ = 0 {:[ln rho^(')int_(Gamma^(˙))P(u)*G(beta_(0))d sigma+2delta_(1)ln rho^(')int_(Gamma^(˙))P(u)G(alpha_(1))d sigma+sum_(n=o)^(oo)gamma_(n)int_(Gamma^(˙))P(u)*G(beta_(n))d sigma+],[(8)+sum_(n=1)^(oo)delta_(n)int_(Gamma)P(u)G(alpha_(n))d sigma=0]:}\begin{align*} \ln \rho^{\prime} \int_{\dot{\Gamma}} P(u) \cdot G\left(\beta_{0}\right) d \sigma & +2 \delta_{1} \ln \rho^{\prime} \int_{\dot{\Gamma}} P(u) G\left(\alpha_{1}\right) d \sigma+\sum_{n=o}^{\infty} \gamma_{n} \int_{\dot{\Gamma}} P(u) \cdot G\left(\beta_{n}\right) d \sigma+ \\ & +\sum_{n=1}^{\infty} \delta_{n} \int_{\Gamma} P(u) G\left(\alpha_{n}\right) d \sigma=0 \tag{8} \end{align*}In⁡ρ′∫Γ˙P(you)⋅G(β0)dσ+2δ1In⁡ρ′∫Γ˙P(you)G(α1)dσ+∑n=a∞γn∫Γ˙P(you)⋅G(βn)dσ+(8)+∑n=1∞δn∫ΓP(you)G(αn)dσ=0
which is valid for every point external to a circle containing within it the Ω Ω Omega\OmegaΩ. Either C R C R C_(R)C_{R}CRsuch a fixed circle. Given that on C R C R C_(R)C_{R}CRfunction string 1 , sin ⁡ φ ′ , cos ⁡ φ ′ , sin ⁡ 2 φ ′ , cos ⁡ 2 φ ′ , … 1 , sin ⁡ φ ′ , cos ⁡ φ ′ , sin ⁡ 2 φ ′ , cos ⁡ 2 φ ′ , … 1,sin varphi^('),cos varphi^('),sin 2varphi^('),cos 2varphi^('),dots1, \sin \varphi^{\prime}, \cos \varphi^{\prime}, \sin 2 \varphi^{\prime}, \cos 2 \varphi^{\prime}, \ldots1,my⁡φ′,Cart⁡φ′,my⁡2φ′,Cart⁡2φ′,…is complete, from equality (8) it follows
∫ Γ ˙ P ( u ) G ( β 0 ) d σ + ln ⁡ R ∫ Γ ˙ P ( u ) G ( β 0 ) d σ = 0 2 R ln ⁡ R ∫ Γ ˙ P ( u ) G ( α 1 ) d σ + ∫ Γ ˙ P ( u ) G ( β 1 ) d σ + R ∫ Γ ˙ P ( u ) G ( α 1 ) d σ = 0 ∫ Γ P ( u ) G ( β n ) d σ + R 2 ∫ Γ ˙ P ( u ) G ( α n ) d σ = 0 ( n = 2 , 3 , … ) ∫ Γ ˙   P ( u ) G β 0 d σ + ln ⁡ R ∫ Γ ˙   P ( u ) G β 0 d σ = 0 2 R ln ⁡ R ∫ Γ ˙   P ( u ) G α 1 d σ + ∫ Γ ˙   P ( u ) G β 1 d σ + R ∫ Γ ˙   P ( u ) G α 1 d σ = 0 ∫ Γ   P ( u ) G β n d σ + R 2 ∫ Γ ˙   P ( u ) G α n d σ = 0 ( n = 2 , 3 , … ) {:[int_(Gamma^(˙))P(u)G(beta_(0))d sigma+ln Rint_(Gamma^(˙))P(u)G(beta_(0))d sigma=0],[2R ln Rint_(Gamma^(˙))P(u)G(alpha_(1))d sigma+int_(Gamma^(˙))P(u)G(beta_(1))d sigma+Rint_(Gamma^(˙))P(u)G(alpha_(1))d sigma=0],[int_(Gamma)P(u)G(beta_(n))d sigma+R^(2)int_(Gamma^(˙))P(u)G(alpha_(n))d sigma=0quad(n=2","3","dots)]:}\begin{gathered} \int_{\dot{\Gamma}} P(u) G\left(\beta_{0}\right) d \sigma+\ln R \int_{\dot{\Gamma}} P(u) G\left(\beta_{0}\right) d \sigma=0 \\ 2 R \ln R \int_{\dot{\Gamma}} P(u) G\left(\alpha_{1}\right) d \sigma+\int_{\dot{\Gamma}} P(u) G\left(\beta_{1}\right) d \sigma+R \int_{\dot{\Gamma}} P(u) G\left(\alpha_{1}\right) d \sigma=0 \\ \int_{\Gamma} P(u) G\left(\beta_{n}\right) d \sigma+R^{2} \int_{\dot{\Gamma}} P(u) G\left(\alpha_{n}\right) d \sigma=0 \quad(n=2,3, \ldots) \end{gathered}∫Γ˙P(you)G(β0)dσ+In⁡R∫Γ˙P(you)G(β0)dσ=02RIn⁡R∫Γ˙P(you)G(α1)dσ+∫Γ˙P(you)G(β1)dσ+R∫Γ˙P(you)G(α1)dσ=0∫ΓP(you)G(βn)dσ+R2∫Γ˙P(you)G(αn)dσ=0(n=2,3,…)
But this system is also valid for any R 1 > R R 1 > R R_(1) > RR_{1}>RR1>R, from which it follows
∫ Γ ˙ P ( u ) G ( α n ) d σ = 0 , ( n = 1 , 2 , … ) ∫ Γ P ( u ) G ( β n ) d σ = 0 , ( n = 0 , 1 , … ) ∫ Γ ˙   P ( u ) G α n d σ = 0 , ( n = 1 , 2 , … ) ∫ Γ   P ( u ) G β n d σ = 0 , ( n = 0 , 1 , … ) {:[int_(Gamma^(˙))P(u)G(alpha_(n))d sigma=0","quad(n=1","2","dots)],[int_(Gamma)P(u)G(beta_(n))d sigma=0","quad(n=0","1","dots)]:}\begin{aligned} & \int_{\dot{\Gamma}} P(u) G\left(\alpha_{n}\right) d \sigma=0, \quad(n=1,2, \ldots) \\ & \int_{\Gamma} P(u) G\left(\beta_{n}\right) d \sigma=0, \quad(n=0,1, \ldots) \end{aligned}∫Γ˙P(you)G(αn)dσ=0,(n=1,2,…)∫ΓP(you)G(βn)dσ=0,(n=0,1,…)
or noting with { v n } v n {v_(n)}\left\{v_{n}\right\}{Vn}RANGE { α n } ∪ { f n } α n ∪ f n {alpha_(n)}uu{f_(n)}\left\{\alpha_{n}\right\} \cup\left\{f_{n}\right\}{αn}∪{fn}:
(9) ∫ Γ P ( u ) G ( v n ) d σ = 0 , ( n = 1 , 2 , … ) (9) ∫ Γ   P ( u ) G v n d σ = 0 , ( n = 1 , 2 , … ) {:(9)int_(Gamma)P(u)G(v_(n))d sigma=0","quad(n=1","2","dots):}\begin{equation*} \int_{\Gamma} P(u) G\left(v_{n}\right) d \sigma=0, \quad(n=1,2, \ldots) \tag{9} \end{equation*}(9)∫ΓP(you)G(Vn)dσ=0,(n=1,2,…)
From the above it follows that system (9) represents the necessary condition for u ∈ { u } u ∈ { u } u in{u}u \in\{u\}you∈{you}. Relation (9) also represents the sufficient condition. Indeed, either μ 1 , δ 1 , μ 2 , δ 2 ∈ L 2 ( Γ ) μ 1 , δ 1 , μ 2 , δ 2 ∈ L 2 ( Γ ) mu_(1),delta_(1),mu_(2),delta_(2)inL_(2)(Gamma)\mu_{1}, \delta_{1}, \mu_{2}, \delta_{2} \in L_{2}(\Gamma)μ1,δ1,μ2,δ2∈IT2(Γ); function
w ( P ′ ) = ∫ Γ ˙ [ μ 1 ( Q ) ∂ Δ U ( P ′ , Q ) ∂ ν Q − δ 1 ( Q ) Δ U ( P ′ , Q ) + μ 2 ( Q ) ∂ U ( P ′ , Q ) ∂ ν Q − − δ 2 ( Q ) U ( P ′ , Q ) ] d σ w P ′ = ∫ Γ ˙   μ 1 ( Q ) ∂ Δ U P ′ , Q ∂ ν Q − δ 1 ( Q ) Δ U P ′ , Q + μ 2 ( Q ) ∂ U P ′ , Q ∂ ν Q − − δ 2 ( Q ) U P ′ , Q d σ {:[w(P^('))=int_(Gamma^(˙))[mu_(1)(Q)(del Delta U(P^('),Q))/(delnu_(Q)):}-delta_(1)(Q)DeltaU(P^('),Q)+mu_(2)(Q)(del U(P^('),Q))/(delnu_(Q))-],[{:-delta_(2)(Q)U(P^('),Q)]d sigma]:}\begin{aligned} w\left(P^{\prime}\right)=\int_{\dot{\Gamma}}\left[\mu_{1}(Q) \frac{\partial \Delta U\left(P^{\prime}, Q\right)}{\partial \nu_{Q}}\right. & -\delta_{1}(Q) \Delta \mathrm{U}\left(P^{\prime}, Q\right)+\mu_{2}(Q) \frac{\partial U\left(P^{\prime}, Q\right)}{\partial \nu_{Q}}- \\ & \left.-\delta_{2}(Q) U\left(P^{\prime}, Q\right)\right] d \sigma \end{aligned}w(P′)=∫Γ˙[μ1(Q)∂ΔU(P′,Q)∂nQ−δ1(Q)ΔU(P′,Q)+μ2(Q)∂U(P′,Q)∂nQ−−δ2(Q)U(P′,Q)]dσ
is biharmonic in C Ω C Ω C OmegaC \OmegaCΩ. But from (9) it follows w ( P ′ ) ≡ 0 w P ′ ≡ 0 w(P^('))-=0w\left(P^{\prime}\right) \equiv 0w(P′)≡0outside the circle C R C R C_(R)C_{R}CRwhich means that w ( P ′ ) ≡ 0 w P ′ ≡ 0 w(P^('))-=0w\left(P^{\prime}\right) \equiv 0w(P′)≡0and in C Ω C Ω C OmegaC \OmegaCΩ.
The final result is formulated in the following theorem:
Theorem 5. The vector sequence { ψ i } = | v i − Δ v i + 1 − σ ρ ∂ v i ∂ ν | ψ i = v i − Δ v i + 1 − σ ρ ∂ v i ∂ ν {psi_(i)}=|v_(i)-Deltav_(i)+(1-sigma)/(rho)(delv_(i))/(del nu)|\left\{\psi_{i}\right\}=\left|v_{i}-\Delta v_{i}+\frac{1-\sigma}{\rho} \frac{\partial v_{i}}{\partial \nu}\right|{ψand}=|Vand−ΔVand+1−σρ∂Vand∂n|is complete in Hilbert's sense on Γ Γ Gamma\GammaΓ.
Demonstration. Let φ φ varphi\varphiφan orthogonal vector on all elements of the string { ψ i } ψ i {psi_(i)}\left\{\psi_{i}\right\}{ψand}. Vector components φ φ varphi\varphiφwe will note them as follows: φ = [ − δ 2 , δ 1 ] φ = − δ 2 , δ 1 varphi=[-delta_(2),delta_(1)]\varphi=\left[-\delta_{2}, \delta_{1}\right]φ=[−δ2,δ1]The orthogonality condition is
∫ Γ [ − δ 2 v i − δ 1 Δ v i + I − σ ρ 0 δ 1 ∂ v i ∂ ν ] d σ = 0 , ( i = 1 , 2 , … ) ∫ Γ   − δ 2 v i − δ 1 Δ v i + I − σ ρ 0 δ 1 ∂ v i ∂ ν d σ = 0 , ( i = 1 , 2 , … ) int_(Gamma)[-delta_(2)v_(i)-delta_(1)Deltav_(i)+(I-sigma)/(rho_(0))delta_(1)(delv_(i))/(del nu)]d sigma=0,quad(i=1,2,dots)\int_{\Gamma}\left[-\delta_{2} v_{i}-\delta_{1} \Delta v_{i}+\frac{I-\sigma}{\rho_{0}} \delta_{1} \frac{\partial v_{i}}{\partial \nu}\right] d \sigma=0, \quad(i=1,2, \ldots)∫Γ[−δ2Vand−δ1ΔVand+and−σρ0δ1∂Vand∂n]dσ=0,(and=1,2,…)
If we consider μ 1 ≡ 0 μ 1 ≡ 0 mu_(1)-=0\mu_{1} \equiv 0μ1≡0and μ 2 = 1 − σ ρ 0 δ 1 μ 2 = 1 − σ ρ 0 δ 1 mu_(2)=(1-sigma)/(rho_(0))delta_(1)\mu_{2}=\frac{1-\sigma}{\rho_{0}} \delta_{1}μ2=1−σρ0δ1, then the previous equalities coincide with (9), so
u ( P ) = 1 8 π ∫ T ˙ [ 1 − σ ρ 0 δ 1 ( Q ) ∂ U ( P , Q ) ∂ v Q − δ 1 ( Q ) Δ U ( P , Q ) − δ 2 U ( P , Q ) ] d σ ∈ { u } u ( P ) = 1 8 π ∫ T ˙   1 − σ ρ 0 δ 1 ( Q ) ∂ U ( P , Q ) ∂ v Q − δ 1 ( Q ) Δ U ( P , Q ) − δ 2 U ( P , Q ) d σ ∈ { u } u(P)=(1)/(8pi)int_(T^(˙))[(1-sigma)/(rho_(0))delta_(1)(Q)(del U(P,Q))/(delv_(Q))-delta_(1)(Q)Delta U(P,Q)-delta_(2)U(P,Q)]d sigma in{u}u(P)=\frac{1}{8 \pi} \int_{\dot{T}}\left[\frac{1-\sigma}{\rho_{0}} \delta_{1}(Q) \frac{\partial U(P, Q)}{\partial v_{Q}}-\delta_{1}(Q) \Delta U(P, Q)-\delta_{2} U(P, Q)\right] d \sigma \in\{u\}you(P)=18π∫T˙[1−σρ0δ1(Q)∂U(P,Q)∂VQ−δ1(Q)ΔU(P,Q)−δ2U(P,Q)]dσ∈{you}
we
lim u ( P ) = 0 , lim Δ u ( P ) = 1 − σ ρ 0 δ 1 ( M ) , lim ∂ u ( P ) ∂ v M = δ 1 ( M ) lim u ( P ) = 0 , lim Δ u ( P ) = 1 − σ ρ 0 δ 1 ( M ) , lim ∂ u ( P ) ∂ v M = δ 1 ( M ) lim u(P)=0,quad lim Delta u(P)=(1-sigma)/(rho_(0))delta_(1)(M),quad lim(del u(P))/(delv_(M))=delta_(1)(M)\lim u(P)=0, \quad \lim \Delta u(P)=\frac{1-\sigma}{\rho_{0}} \delta_{1}(M), \quad \lim \frac{\partial u(P)}{\partial v_{M}}=\delta_{1}(M)limyou(P)=0,limΔyou(P)=1−σρ0δ1(M),lim∂you(P)∂VM=δ1(M)
and lim ∂ Δ u ( P ) ∂ v M = δ 2 ( M ) lim ∂ Δ u ( P ) ∂ v M = δ 2 ( M ) lim(del Delta u(P))/(delv_(M))=delta_(2)(M)\lim \frac{\partial \Delta u(P)}{\partial v_{M}}=\delta_{2}(M)lim∂Δyou(P)∂VM=δ2(M), all considered when P → M ∈ Γ P → M ∈ Γ P rarr M in GammaP \rightarrow M \in \GammaP→M∈Γon the normal.
Noting that
and
lim [ − Δ u + 1 − σ ρ 0 ∂ u ∂ v M ] = − 1 − σ f 0 δ 1 + 1 − σ ρ 0 δ 1 = 0 lim − Δ u + 1 − σ ρ 0 ∂ u ∂ v M = − 1 − σ f 0 δ 1 + 1 − σ ρ 0 δ 1 = 0 lim[-Delta u+(1-sigma)/(rho_(0))(del u)/(delv_(M))]=-(1-sigma)/(f_(0))delta_(1)+(1-sigma)/(rho_(0))delta_(1)=0\lim \left[-\Delta u+\frac{1-\sigma}{\rho_{0}} \frac{\partial u}{\partial v_{M}}\right]=-\frac{1-\sigma}{f_{0}} \delta_{1}+\frac{1-\sigma}{\rho_{0}} \delta_{1}=0lim[−Δyou+1−σρ0∂you∂VM]=−1−σf0δ1+1−σρ0δ1=0
lim u = 0 lim u = 0 lim u=0\lim u=0limyou=0
based on the uniqueness theorem it follows u ≡ 0 u ≡ 0 u-=0u \equiv 0you≡0, so ∂ u ∂ v M = δ 1 = 0 ∂ u ∂ v M = δ 1 = 0 (del u)/(delv_(M))=delta_(1)=0\frac{\partial u}{\partial v_{M}}=\delta_{1}=0∂you∂VM=δ1=0and ∂ Δ u ∂ ν = δ 2 = 0 ∂ Δ u ∂ ν = δ 2 = 0 (del Delta u)/(del nu)=delta_(2)=0\frac{\partial \Delta u}{\partial \nu}=\delta_{2}=0∂Δyou∂n=δ2=0on Γ Γ Gamma\GammaΓwhat does it mean φ ≡ 0 φ ≡ 0 varphi-=0\varphi \equiv 0φ≡0.
RPR Academy Computing Institute, Cluj Branch

SOLUTION OF ONE BOUNDARY PROBLEM FOR BIHARMONIC EQUATION

SHORT CONTENT

Let's denote by Ω Ω Omega\OmegaΩплсскую область, ограницанную красный кривой Г. We assume that the radius of curvature ρ 0 ρ 0 rho_(0)\rho_{0}ρ0crooked Γ Γ Gamma\GammaΓit is a positive function and that's it Γ Γ Gamma\GammaΓit is sufficiently smooth.
In this work, it is proved that the sequence of vector functions
{ v i , − Δ v i + 1 − σ ρ 0 ∂ v i ∂ v } v i , − Δ v i + 1 − σ ρ 0 ∂ v i ∂ v {v_(i),-Deltav_(i)+(1-sigma)/(rho_(0))(delv_(i))/(del v)}\left\{v_{i},-\Delta v_{i}+\frac{1-\sigma}{\rho_{0}} \frac{\partial v_{i}}{\partial v}\right\}{Vand,−ΔVand+1−σρ0∂Vand∂V}
it is full in the middle of the curve Здесь
{ v n } = { α n } U { β n } v n = α n U β n {v_(n)}={alpha_(n)}U{beta_(n)}\left\{v_{n}\right\}=\left\{\alpha_{n}\right\} U\left\{\beta_{n}\right\}{Vn}={αn}U{βn}
and
α 1 = − ρ { cos ⁡ φ sin ⁡ φ ; α n = 1 n ( n − 1 ) ρ n { cos ⁡ n φ sin ⁡ n φ ( n = 2 , 3 , … ) ; γ n = ρ ′ − n { cos ⁡ n φ ′ sin ⁡ n φ ′ ( n = 0 , 1 , … ) α 1 = − ρ cos ⁡ φ sin ⁡ φ ; α n = 1 n ( n − 1 ) ρ n cos ⁡ n φ sin ⁡ n φ ( n = 2 , 3 , … ) ; γ n = ρ ′ − n cos ⁡ n φ ′ sin ⁡ n φ ′ ( n = 0 , 1 , … ) alpha_(1)=-rho{[cos varphi],[sin varphi];alpha_(n)=(1)/(n(n-1))rho^(n){[cos n varphi],[sin n varphi](n=2,3,dots);gamma_(n)=rho^(')-n{[cos nvarphi^(')],[sin nvarphi^(')](n=0,1,dots):}\alpha_{1}=-\rho\left\{\begin{array}{l}\cos \varphi \\ \sin \varphi\end{array} ; \alpha_{n}=\frac{1}{n(n-1)} \rho^{n}\left\{\begin{array}{l}\cos n \varphi \\ \sin n \varphi\end{array}(n=2,3, \ldots) ; \gamma_{n}=\rho^{\prime}-n\left\{\begin{array}{l}\cos n \varphi^{\prime} \\ \sin n \varphi^{\prime}\end{array}(n=0,1, \ldots)\right.\right.\right.α1=−ρ{Cart⁡φmy⁡φ;αn=1n(n−1)ρn{Cart⁡nφmy⁡nφ(n=2,3,…);γn=ρ′−n{Cart⁡nφ′my⁡nφ′(n=0,1,…)
β 0 = ρ 2 ; β n = − 1 n ( n + 1 ) ρ n + 2 { cos ⁡ n φ sin ⁡ n φ ( n = 1 , 2 , … ) ; δ n = ρ ′ n − 2 { cos ⁡ n φ ′ sin ⁡ n φ ′ ( n = 1 , 2 , … ) β 0 = ρ 2 ; β n = − 1 n ( n + 1 ) ρ n + 2 cos ⁡ n φ sin ⁡ n φ ( n = 1 , 2 , … ) ; δ n = ρ ′ n − 2 cos ⁡ n φ ′ sin ⁡ n φ ′ ( n = 1 , 2 , … ) beta_(0)=rho^(2);beta_(n)=-(1)/(n(n+1))rho^(n+2){[cos n varphi],[sin n varphi](n=1,2,dots);delta_(n)=rho^('n-2){[cos nvarphi^(')],[sin nvarphi^(')](n=1,2,dots):}\beta_{0}=\rho^{2} ; \beta_{n}=-\frac{1}{n(n+1)} \rho^{n+2}\left\{\begin{array}{l}\cos n \varphi \\ \sin n \varphi\end{array}(n=1,2, \ldots) ; \delta_{n}=\rho^{\prime n-2}\left\{\begin{array}{l}\cos n \varphi^{\prime} \\ \sin n \varphi^{\prime}\end{array}(n=1,2, \ldots)\right.\right.β0=ρ2;βn=−1n(n+1)ρn+2{Cart⁡nφmy⁡nφ(n=1,2,…);δn=ρ′n−2{Cart⁡nφ′my⁡nφ′(n=1,2,…)where
​ ρ ρ rho\rhoρand φ φ varphi\varphiφthey indicate polar coordinates, whose center is placed within the area Ω , σ Ω , σ Omega,sigma\Omega, \sigmaΩ,σ- Poisson's constant value and ν denotes the internal normal. Some ideas from works [1] and [2] were used to prove the above statement.
Full sequence of vectors
{ v i , − Δ v i + 1 − σ ρ 0 ∂ v i ∂ v } v i , − Δ v i + 1 − σ ρ 0 ∂ v i ∂ v {v_(i),-Deltav_(i)+(1-sigma)/(rho_(0))(delv_(i))/(del v)}\left\{v_{i},-\Delta v_{i}+\frac{1-\sigma}{\rho_{0}} \frac{\partial v_{i}}{\partial v}\right\}{Vand,−ΔVand+1−σρ0∂Vand∂V}
serves to solve the following task: in the regions Ω Ω Omega\OmegaΩI am looking for such a function u ( x , y ) u ( x , y ) u(x,y)u(x, y)you(x,y)for which
Δ 2 u ≡ ∂ 4 u ∂ x 4 + 2 ∂ 4 u ∂ x 2 ∂ y 2 + ∂ 4 u ∂ y 4 = 0 в Ω u = f 1 Δ 2 u ≡ ∂ 4 u ∂ x 4 + 2 ∂ 4 u ∂ x 2 ∂ y 2 + ∂ 4 u ∂ y 4 = 0  в  Ω u = f 1 {:[Delta^(2)u-=(del^(4)u)/(delx^(4))+2(del^(4)u)/(delx^(2)dely^(2))+(del^(4)u)/(dely^(4))=0" в "Omega],[u=f_(1)]:}\begin{gathered} \Delta^{2} u \equiv \frac{\partial^{4} u}{\partial x^{4}}+2 \frac{\partial^{4} u}{\partial x^{2} \partial y^{2}}+\frac{\partial^{4} u}{\partial y^{4}}=0 \text { в } \Omega \\ u=f_{1} \end{gathered}Δ2you≡∂4you∂x4+2∂4you∂x2∂y2+∂4you∂y4=0 in Ωyou=f1
and
− Δ u + 1 − σ ρ 0 ∂ u ∂ v = f 2 − Δ u + 1 − σ ρ 0 ∂ u ∂ v = f 2 -Delta u+(1-sigma)/(rho_(0))(del u)/(del v)=f_(2)-\Delta u+\frac{1-\sigma}{\rho_{0}} \frac{\partial u}{\partial v}=f_{2}−Δyou+1−σρ0∂you∂V=f2
on Γ Γ Gamma\GammaΓwhere f 1 f 1 f_(1)f_{1}f1and f 2 f 2 f_(2)f_{2}f2- given quadratically summable functions on Γ Γ Gamma\GammaΓ.

- LA SOLUTION D'UN PROBLÈME AUX LIMITES POUR L'ÉQUATION BIHARMONIQUE

SUMMARY

I am Ω Ω Omega\OmegaΩa domaine borné par la courbe Γ Γ Gamma\GammaΓ. We suppose that the curve Γ Γ Gamma\GammaΓallows the radius of curvature ρ 0 ρ 0 rho_(0)\rho_{0}ρ0positiv et, en outre, that this curve is sufficiently smooth.
Dans ce travail, on demune que la succession des vecteurs
{ v i , − Δ v i + 1 − σ ρ 0 ∂ v i ∂ v } v i , − Δ v i + 1 − σ ρ 0 ∂ v i ∂ v {v_(i),-Deltav_(i)+(1-sigma)/(rho_(0))(delv_(i))/(del v)}\left\{v_{i},-\Delta v_{i}+\frac{1-\sigma}{\rho_{0}} \frac{\partial v_{i}}{\partial v}\right\}{Vand,−ΔVand+1−σρ0∂Vand∂V}
est complète au sens by Hilbert sur Γ Γ Gamma\GammaΓ, where
{ v n } = { α n } ∪ { β n } et α 1 = − ρ { cos ⁡ φ sin ⁡ φ ; α n = 1 n ( n − 1 ) ρ n { cos ⁡ n φ sin ⁡ n φ ( n = 2 , 3 , … ) ; γ n = ρ ′ − n { cos ⁡ n φ ′ sin ⁡ n φ ′ ( n = 0 , 1 , … ) , β β 0 = ρ 2 ; ρ n = − 1 n ( n + 1 ) ρ n + 2 { cos ⁡ n φ sin ⁡ n φ ( n = 1 , 2 , … ) ; δ n = ρ ′ n − 2 { cos ⁡ n φ ′ sin ⁡ n φ ′ ( n = 1 , 2 , … ) , v n = α n ∪ β n  et  α 1 = − ρ cos ⁡ φ sin ⁡ φ ; α n = 1 n ( n − 1 ) ρ n cos ⁡ n φ sin ⁡ n φ ( n = 2 , 3 , … ) ; γ n = ρ ′ − n cos ⁡ n φ ′ sin ⁡ n φ ′ ( n = 0 , 1 , … ) , β β 0 = ρ 2 ; ρ n = − 1 n ( n + 1 ) ρ n + 2 cos ⁡ n φ sin ⁡ n φ ( n = 1 , 2 , … ) ; δ n = ρ ′ n − 2 cos ⁡ n φ ′ sin ⁡ n φ ′ ( n = 1 , 2 , … ) , {:[{v_(n)}={alpha_(n)}uu{beta_(n)}" et "],[alpha_(1)=-rho{[cos varphi],[sin varphi];alpha_(n)=(1)/(n(n-1))rho^(n){[cos n varphi],[sin n varphi](n=2,3,dots);gamma_(n)=rho^(')-n{[cos nvarphi^(')],[sin nvarphi^(')](n=0,1,dots),:}]:}beta_(beta_(0)=rho^(2);quadrho_(n)=-(1)/(n(n+1))rho^(n+2){[cos n varphi],[sin n varphi](n=1,2,dots);delta_(n)=rho^('n-2){[cos nvarphi^(')],[sin nvarphi^(')](n=1,2,dots),:})\begin{gathered} \left\{v_{n}\right\}=\left\{\alpha_{n}\right\} \cup\left\{\beta_{n}\right\} \text { et } \\ \alpha_{1}=-\rho\left\{\begin{array}{l} \cos \varphi \\ \sin \varphi \end{array} ; \alpha_{n}=\frac{1}{n(n-1)} \rho^{n}\left\{\begin{array}{l} \cos n \varphi \\ \sin n \varphi \end{array}(n=2,3, \ldots) ; \gamma_{n}=\rho^{\prime}-n\left\{\begin{array}{l} \cos n \varphi^{\prime} \\ \sin n \varphi^{\prime} \end{array}(n=0,1, \ldots),\right.\right.\right. \end{gathered} \beta_{\beta_{0}=\rho^{2} ; \quad \rho_{n}=-\frac{1}{n(n+1)} \rho^{n+2}\left\{\begin{array}{l} \cos n \varphi \\ \sin n \varphi \end{array}(n=1,2, \ldots) ; \delta_{n}=\rho^{\prime n-2}\left\{\begin{array}{l} \cos n \varphi^{\prime} \\ \sin n \varphi^{\prime} \end{array}(n=1,2, \ldots),\right.\right.}{Vn}={αn}∪{βn} and α1=−ρ{Cart⁡φmy⁡φ;αn=1n(n−1)ρn{Cart⁡nφmy⁡nφ(n=2,3,…);γn=ρ′−n{Cart⁡nφ′my⁡nφ′(n=0,1,…),ββ0=ρ2;ρn=−1n(n+1)ρn+2{Cart⁡nφmy⁡nφ(n=1,2,…);δn=ρ′n−2{Cart⁡nφ′my⁡nφ′(n=1,2,…),
egg ρ ρ rho\rhoρand φ φ varphi\varphiφare the polar coordinates whose center is located inside the domain Ω ; σ Ω ; σ Omega;sigma\Omega ; \sigmaΩ;σis at constant Poisson values; ν ν nu\nundesignates la normale intérieure par rapport à Γ Γ Gamma\GammaΓ. To demonstrate this assertion, the author is based on some ideas of travaux [1] et [2].
La succession complète des vecteurs { v i , − Δ v i + 1 − σ ρ 0 ∂ v i ∂ v } v i , − Δ v i + 1 − σ ρ 0 ∂ v i ∂ v {v_(i),-Deltav_(i)+(1-sigma)/(rho_(0))(delv_(i))/(del v)}\left\{v_{i},-\Delta v_{i}+\frac{1-\sigma}{\rho_{0}} \frac{\partial v_{i}}{\partial v}\right\}{Vand,−ΔVand+1−σρ0∂Vand∂V}serve to solve the following problem: dans le domaine Ω Ω Omega\OmegaΩwe are looking for the function u ( x , y ) u ( x , y ) u(x,y)u(x, y)you(x,y)for which
Δ 2 u = ∂ 4 u ∂ x 4 + 2 ∂ 4 u ∂ x 2 ∂ y 2 + ∂ 4 u ∂ y 4 = 0 dans Ω , u = f 1 et − Δ u + 1 − σ ρ 0 ∂ u ∂ ν = f 2 sur Γ , Δ 2 u = ∂ 4 u ∂ x 4 + 2 ∂ 4 u ∂ x 2 ∂ y 2 + ∂ 4 u ∂ y 4 = 0  dans  Ω , u = f 1  et  − Δ u + 1 − σ ρ 0 ∂ u ∂ ν = f 2  sur  Γ , {:[Delta^(2)u=(del^(4)u)/(delx^(4))+2(del^(4)u)/(delx^(2)dely^(2))+(del^(4)u)/(dely^(4))=0," dans "Omega","],[u=f_(1)" et "-Delta u+(1-sigma)/(rho_(0))(del u)/(del nu)=f_(2)," sur "Gamma","]:}\begin{array}{ll} \Delta^{2} u=\frac{\partial^{4} u}{\partial x^{4}}+2 \frac{\partial^{4} u}{\partial x^{2} \partial y^{2}}+\frac{\partial^{4} u}{\partial y^{4}}=0 & \text { dans } \Omega, \\ u=f_{1} \text { et }-\Delta u+\frac{1-\sigma}{\rho_{0}} \frac{\partial u}{\partial \nu}=f_{2} & \text { sur } \Gamma, \end{array}Δ2you=∂4you∂x4+2∂4you∂x2∂y2+∂4you∂y4=0 dance Ω,you=f1 and −Δyou+1−σρ0∂you∂n=f2 gray Γ,
egg f 1 f 1 f_(1)f_{1}f1and f 2 f 2 f_(2)f_{2}f2sont deux fonctions données sur Γ Γ Gamma\GammaΓ, summable in squares.

BIBLIOGRAPHY

  1. G. Fichera, Theorems of completeness on the frontier of a domain for many systems of functions. Ann. Mat. pure e appl., 27, 1-28 (1948).
  2. Theorems of completeness connected to the integration of the equation Δ 4 u = f Δ 4 u = f Delta_(4)u=f\Delta_{4} u=fΔ4you=f. Newspapers Mat. Battaglini, 77, 184-199 (1947-1948).
  3. N. M. GUNTER, Potential theory and its application κ κ kappa\kappaκmain tasks of mathematical physics. Moscow, 58-59 (1953).
1958

Related Posts