在本节中,我们利用积分方程方法建立了直接散射问题解的一些先验估计。消除事故现场\(u^{i}\),很容易发现散射场\(w{1}:=u_{1} -u个^{i} \)在里面\(\欧米茄_{+}\)和发射场\(w{2}:=u{2}\)在里面\(\欧米茄_{-}\)满足以下边值问题:
$$\开始{对齐}&\三角形w{1}+k{1}^{2} w个_{1} =0\quad\text{in}\Omega_{+},\end{aligned}$$
(2.1)
$$\开始{对齐}&\三角形w{2}+k{2}^{2} w个_{2} =0\quad\text{in}\Omega_{-},\end{aligned}$$
(2.2)
$$\开始{对齐}(&w)_{1} -w个_{2} =f_{1},\qquad\frac{\partial w_{1{}{\partical\nu}-\lambda\frac{\ partial w_{2}}{\partial\nu}=f_}2}\quad\text{on}\Gamma,\end{aligned}$$
(2.3)
$$开始{对齐}&w_{1}(x)=\sum_{n\in{\mathbb{Z}}^{2}}w_{n}^{+}\exp\bigl(i\alpha_{n{}\cdot\widetilde{x}+i\beta^{+{_{n} x_{3} \biger),\quad x_{3}>A_{1},\end{aligned}$$
(2.4)
$$开始{对齐}&w_{2}(x)=\sum_{n\in{\mathbb{Z}}^{2}}w_{n}^{-}\exp\bigl(i\alpha_{n{}\cdot\widetilde{x} -i\β^{-}_{n} x个_{3} \biger),\quad x_{3}<A_{2}\end{aligned}$$
(2.5)
一般情况下\(f_{1},f_{2}\在L中^{p}_{\alpha}(\Gamma)\)具有\(1<p \leq 2).\(这里,L^{p}_{\alpha}(\Gamma)(第1页)表示Γ上标量函数的Sobolev空间,假设为α-关于变量的拟周期性x̃在通常的Sobolev空间中配备了标准配置\(L^{p}(\Gamma)\).
在继续之前,我们首先介绍本文其余部分中使用的基本符号。为了简单起见,我们使用\(\Omega _{\pm}\)和Γ再次表示限制在一个周期内的相同集合\(0<x{1},x{2}<2\pi\).对于每个\(h>0),表示为\(欧米茄{+}(h):=欧米茄中的x:x{3}<A{1}+h\}),\(欧米茄{-}(h):=\{x\in\Omega{-}:x_{3}>A_{2} -小时\}\),\(伽马{+}(h):=\{x\in\Omega{+}:x{3}=A{1}+h\})、和\(伽马{-}(h):=\{x\in\Omega{-}:x{3}=A_{2} -小时\}\)分别是。那么,让我们\(H)^{1}_{\alpha}(\Omega _{\pm}(h))和\(升^{p}_{\alpha}(\Omega_{\pm}(h))(第1页)表示上标量函数的Sobolev空间\(\Omega _{\pm}(h)\)假设为α-关于变量的准周期x̃在通常的Sobolev空间中配备了规范\(H^{1}(\Omega_{pm}(H))和\(L^{p}(\Omega_{pm}(h))分别是。让\(H^{1/2}_{\alpha}(γ_{\pm}(H))\)表示的跟踪空间\(H)^{1}_{\alpha}(\Omega _{\pm}(h))、和\(H^{-1/2}{\alpha}(\Gamma{\pm}(H))是的双重空间\(H^{1/2}_{\alpha}(γ_{\pm}(H))\).
我们引入自由空间α-准周期格林函数
$$开始{对齐}G{1}(x,y;k{1})=\frac{i}{8\pi^{2}}\sum{n\in{mathbb{Z}}^{2{}\frac}{1}{\beta^{+}{n}}\exp\bigl(i\alpha_{n}\cdot(\widetilde{x}-\widetilde{y})+i\beta_{n}^{+}\vert x_{3} -年_{3} \vert\bigr),\quad x \neq y \end{对齐}$$
(2.6)
和α-准周期层势算符\(S_{1}\),\(K_{1}\),\(K'_{1}\)、和\(T_{1}\)由定义
$$\开始{aligned}和S_{1}\xi(x)=\int_{\Gamma}G{1}(x,y;k_{1})\xi(y)\,ds(y),\quad x\in\Gamma,\end{aligned}$$
(2.7)
$$\begin{aligned}&K_{1}\xi(x)=\int_{\Gamma}\frac{\partial}{\partitle\nu(y)}G_{1{(x,y;K_{1neneneep)\xi$$
(2.8)
$$\begin{aligned}&K'_{1}\xi(x)=\frac{\partial}{\partical\nu(x)}\int_{\Gamma}G{1}(x,y;K_{1{)\ xi(y)\,ds(y),\quad x\in\Gamma,\end{alinged}$$
(2.9)
$$\begin{aligned}&T_{1}\xi(x)=-\frac{\partial}{\parial\nu。\结束{对齐}$$
(2.10)
注意到\(G_1}(x,y;k_1})-\Phi(x,y;k_1{})\)是平滑的,它是从[8]操作员\(S_{1}:H^{-\压裂{1}{2}}{{\alpha}(\Gamma)\rightarrow H^{\frac{1}}{2\alpha}(\Gamma)\),\(K{1}:H^{\frac{1}{2}}{{\alpha}(\Gamma)\rightarrow H^{\frac{1}}{2{}}{\alfa}(\ Gamma,\(K'{j}:H^{-\frac{1}{2}}{{\alpha}(\Gamma)\rightarrow H^{-\frac}{1}}{\alfa}(\ Gamma、和\(T_{1}:H^{\压裂{1}{2}}{{\alpha}(\Gamma)\rightarrow H^{-\frac{1}}{\alfa}(\伽马)\)都是有界的,其中\(\Phi(x,y;k{1})=\分形{1}{4\pi}\分形{e^{ik_{1}|x-y|}}{|x-y|}\)是亥姆霍兹方程的基本解\(\三角形\ Phi+k_{1}^{2}\ Phi=-\三角形_{y}\)在自由空间\({\mathbb{R}}^{3}\).
定理2.1
对于 \L中的(f_{1},f_{2}^{p}_{\alpha}(\Gamma)\) 具有 \(1<p\leq 2\),有一个独特的解决方案 \L中的((w_{1},w_{2})^{p}_{\alpha}(\Omega_{+}(h))\乘以L^{p}_{\alpha}(\Omega_{-}(h)) 传输问题(2.1)——(2.5)满足估计
$$\开始{aligned}\Vert w_{1}\Vert_{L^{p}_{\alpha}(\Omega_{+}(h))}+\Vert w_{2}\Vert_{L^{p}_{\alpha}(\Omega_{-}(h))}\leq C\bigl(\Vert f_{1}\Vert_{L^{p}_{\alpha}(\Gamma)}+\Vert f_{2}\Vert_{L^{p}_{\alpha}(\Gamma)}\biger),\end{对齐}$$
(2.11)
哪里 \(C>0\) 是独立于 \(f{1},f{2}\),并且取决于 \(G_{j}(\cdot,y;k_{j{),\Omega_{+}(h)\) 具有 \(j=1,2) 和算子的有界性 \(S_{j},K_{j},K'_{j{,j=1,2\),和 \(T_{2} -T型_{1}\) 在里面 \(升^{p}_{\alpha}(\Gamma)\).
此外,如果 \L中的(f_{1},f_{2}^{p}_{\alpha}(\Gamma)\) 具有 \(压裂{4}{3}<p\leq 2),我们有
$$\开始{aligned}\Vert w_{1}\Vert_{L^{2}_{\alpha}(\Omega_{+}(h))}+\Vert w_{2}\Vert_{L^{2}_{\alpha}(\Omega_{-}(h))}\leq C\bigl(\Vert f_{1}\Vert _{L^{p}_{\alpha}(\Gamma)}+\Vert f_{2}\Vert_{L^{p}_{\alpha}(\Gamma)}\bigr)\end{对齐}$$
(2.12)
具有正常数 \(C>0\),独立于 \(f{1},f{2}\),并且取决于 \(G_{j}(\cdot,y;k_{j{),\Omega_{+}(h)\) 具有 \(j=1,2) 和算子的有界性 \(S_{j},K_{j},K'_{j{,j=1,2\) 和 \(T_{2} -T型_{1}\) 在里面 \(升^{p}_{\alpha}(\Gamma)\).
证明
我们寻求问题的解决方案(2.1)–(2.5)以单层和双层组合电势的形式
$$\begin{aligned}&w_{1}(x)=\int_{\Gamma}G_1}(x,y;k_{1{)\varphi_{1neneneep(y)\,ds(y$$
(2.13)
$$\开始{对齐}&w{2}(x)=\int_{\Gamma}G{2](x,y;k{2})\varphi_{1}(y)\,ds(y)+\int_\\Gamma}\frac{\partialG{2}(x,y;k{2])}{\partical\nu(y)}\varphi_2}(y)\,bs(y),\end{aligned}$$
(2.14)
哪里\(G_2}(x,y;k_2})\)定义为(2.6)用波数\(k{1}\)替换为\(k{2}\).
借助层电位的跳跃关系(参见[26]对于中的案例\(L^{p}\)范数),我们得到传输问题(2.1)–(2.5)可以简化为积分方程组
$$\begin{aligned}\begin{pmatrix}\varphi{2}\\varphi}{1}\end{pmmatrix}+L\ begin}\varphi{2}\\varphi{1}\ end{pmatrix}=\ begin{pmatricx}\frac{2}{1+\lambda}f_{1}\\-\frac{2}}{1+/lambda{f_{2}\end{pmatriax}\quad\text{in}L^{p}_{\alpha}(\Gamma)\乘以L^{p}_{\alpha}(\Gamma),\end{aligned}$$
(2.15)
其中操作员L(左)由提供
$$\开始{aligned}L:=\开始{pmatrix}\frac{2}{1+\lambda}(\lambda K_{1}-K_{2})和压裂{2}{1+\lambda}(S_{1} -S型_{2} )\\frac{2\lambda}{1+\lambda}(T_{2} -T型_{1} )&\压裂{2}{1+\lambda}(\lambda K'_{2} -克'_{1})\结束{pmatrix}。\结束{对齐}$$
很容易看出(2.15)由于运算符的紧凑性,属于Fredholm类型\(S_{j},K_{j},K'_{j{,j=1,2\)、和\(T_{2} -T型_{1}\)在里面\(升^{p}_{\alpha}(\Gamma)\)这与散射问题的唯一性(1.2)–(1.6),意味着(2.15)有独特的解决方案\L中的((\varphi_{2},\varphi_1})^{T}^{p}_{\alpha}(\Gamma)\乘以L^{p}_{\alpha}(\Gamma)\)根据估计
$$\开始{aligned}\Vert\varphi_{2}\Vert_{L^{p}_{\alpha}(\Gamma)}+\Vert\varphi_{1}\Vert_{L^{p}_{\alpha}(\Gamma)}\leq C\bigl(\Vert f_{1}\Vert_{L^{p}_{\alpha}(\Gamma)}+\Vert f_{2}\Vert_{L^{p}_{\alpha}(\Gamma)}\bigr)。\结束{对齐}$$
(2.16)
我们接下来要证明\(升^{p}_{\alpha},1<p\leq 2)传输问题解的估计(2.1)–(2.5). 事实上,可以检查
$$\begin{aligned}和\biggl\Vert\int_{\Gamma}\Omega_{+}(h)G_{1}(\cdot,y;k_{1{)\varphi_{1neneneep(y)\,ds(y)\biggr\Vert_{L^{p}_{\alpha}(\Omega_{+}(h))}\\&\quad=\sup_{g\in L^{q}_{\alpha},\Vert g\Vert_{L^{q}_{\alpha}(\Omega_{+}(h))}=1}\biggl\vert\int_{\Omega _{+{(h^{q}_{\alpha},\Vert g\Vert_{L^{q}_{\alpha}(\Omega_{+}(h))}=1}\biggl\vert\int_{\Gamma}\int_{\欧米茄_{+{(h},\垂直G\垂直_{L^{q}_{\alpha}(\Omega_{+}(h))}=1}\sup_{y\in\Gamma}\bigl\VertG_{1}(\ cdot,y;k_{1{)\bigr\Vert_{L^{p}_{\alpha}(\Omega_{+}(h))}\Vert g\Vert_{L^{q}_{\alpha}(\Omega_{+}(h))}\Vert\varphi_{1}\Vert_{L^{p}_{\alpha}(\Gamma)}\\&\quad=\vert\Gamma\vert^{\frac{1}{q}}\sup_{y\in\Gamma}\bigl\VertG{1}^{p}_{\alpha}(\Omega_{+}(h))}\Vert\varphi_{1}\Vert_{L^{p}_{\alpha}(\Gamma)}\leq C\Vert\varphi_{1}\Vert_{L^{p}_{\alpha}(\Gamma)}\end{aligned}$$
(2.17)
和
$$\begin{aligned}&\biggl\Vert\int_{\Gamma}\frac{\partial G_{1}(\cdot,y;k_{1{)}{\partical\nu(y)}\varphi_{2}(y)\,ds(y)\biggr\Vert_{L^{p}_{\alpha}(\Omega_{+}(h))}\\&\quad=\sup_{g\in L^{q}_{\alpha},\Vert g\Vert_{L^{q}_{\alpha}(\Omega_{+}(h))}=1}\biggl\vert\int_{\Omega _{+{(h^{q}_{\alpha},\Vert g\Vert_{L^{q}_{\alpha}(\Omega_{+}(h))}=1}\biggl\vert\int_{\Gamma}\frac{\partial}{\parial\nu(y)}\int_{\ Omega_{+}(h)}G_1}(x,y;k_1})G(x)\,dx\varphi_{2}(y^{q}_{\alpha},\垂直g \垂直_{L^{q}_{\alpha}(\Omega_{+}(h))}=1}\biggl\Vert\frac{\partial}{\parial\nu(y)}\int_{\Omega _{+{(h^{q}_{\alpha}(\Gamma)}\Vert\varphi_{2}\Vert_{L^{p}_{\alpha}(\Gamma)}\\&\quad\leq\sup_{g\in L^{q}_{\alpha},\Vert g\Vert_{L^{q}_{\alpha}(\Omega_{+}(h))}=1}C\Vert g\Vert _{L^{q}_{\alpha}(\Omega_{+}(h))}\cdot\Vert\varphi_{2}\Vert_{L^{p}_{\alpha}(\Gamma)}=C\Vert\varphi_{2}\Vert_{L^{p}_{\alpha}(\Gamma)}\end{aligned}$$
(2.18)
具有\(\frac{1}{p}+\frac{1}{q}=1\)这里,我们使用了体积势算子有界于\(升^{q}_{\alpha}(\Omega_{+}(h))进入之内\(W^{2,q}_{\alpha}(\Omega_{+}(h))具有\(2)(请参见[15,定理9.9]),边界跟踪算子有界于\(W^{1,q}_{\alpha}(\Omega_{+}(h))进入之内\(升^{q}_{\alpha}(\Gamma)\)具有\(2)(请参见[1,定理5.36])。值得注意的是(2.17)–(2.18)仍然适用\(G_{1}(x,\cdot;k_{1{)\)替换为\(G_2}(x,\cdot;k_2})和\(\欧米茄{+}(h)\)替换为\(\欧米茄{-}(h)\)分别是。现在所需的估计(2.11)以下为(2.13)–(2.14)和(2.16)–(2.18). 此外,如果\L中的(f_{1},f_{2}^{p}_{\alpha}(\Gamma)\)具有\(压裂{4}{3}<p\leq 2),通过与(2.17)–(2.18),可以得出所需的结果(2.13). 这就完成了定理的证明。□
推论2.2
对于 \(y_{0}\in\Gamma\),定义顺序 \(y{j}:=y_{0}-\frac{1}{j}\nu(y_{0})\in\Omega_{+}\),\(j\在{\mathbb{N}}\中).让 \((u{1j},u{2j})\) 是散射问题的解(1.2)——(1.6)与入射点源 \(u^{i}=G{1}(x,y{j};k{1}).然后,对于任何 \({\mathbb{R}}\中的h\),我们有
$$\开始{对齐}\Vert u_{1j}\Vert_{L^{2}_{\alpha}(\Omega_{+}(h))}+\Vertu_{2j}\Vert_{L^{2}_{\alpha}(\Omega_{-}(h))}\leqC\end{aligned}$$
(2.19)
统一用于 \(j\在{\mathbb{N}}_{+}\中),哪里 \(C>0\) 是一个常数,取决于 \(G_{j}(\cdot,y;k_{j}),\Omega_{+}(h)\) 具有 \(j=1,2).
证明
很明显\(((u{1j}^{s},u{2j})\)满足问题(2.1)–(2.5)使用边界数据
$$\开始{对齐}f_{1}(j):=-G_{1{(x,y_{j};k_{1neneneep),\qquad f_{2}(j):=-\frac{\partial G_{1neneneei(x,y_{j{;k_1})}{\parial\nu}\quad j\在{\mathbb{N}中。\结束{对齐}$$
很容易看出这一点\L中的(f_{1}(j),f_{2}(j)^{p}_{\alpha}(\Gamma)\)是一致有界的\(j\在{\mathbb{N}}\中)具有\(压裂{4}{3}<p)然后是所需的结果(2.19)遵循定理2.1这证明了推论。□
定理2.3
让 \((u_{1j},u_{2j})\) 是散射问题的解(1.2)——(1.6)对应于入射点源 \(u^{i}=G{1}(x,y{j};k{1}) 具有 \(y{j}\) 在推论中定义2.2.然后,对于任何 \({\mathbb{R}}\中的h\),它认为
$$\开始{对齐}\垂直u_{2j}\垂直_{H^{1}_{\alpha}(\Omega_{-}(h)\set-nuss{\overline{B}})}\leqC\end{aligned}$$
(2.20)
统一用于 \(j\在{\mathbb{N}}_{+}\中).在这里,\(C>0\) 是一个常数,取决于 \(G_{j}(\cdot,y;k_{j{),\Omega_{+}(h)\) 具有 \(j=1,2\) 和一致有界性 \(S_{\Gamma\集合减去{B}}(j)\) 和 \(K_{\Gamma\set-nuse-{B}}(j)\) 在相应的Hilbert空间中,B 一个球能让你满意吗 \(B\supset B_{\delta}\),和 \(B_{\delta}\) 一个小球的中心在 \(y_{0}\) 具有半径 \(增量>0).
证明
定义\(\波浪号{y}(y)_{j} :=y_{0}+\frac{1}{j}\nu(y_{0})\in\Omega_{-}\),让\(w{1}(j):=u{1j}^{s} -G_{1} (x,\颚化符{y}_{j} ;k{1})\)和\(w{2}(j):=u{2j}\),因此\((w{1}(j),w{2}(j))满足问题(2.1)–(2.5)使用边界数据
$$\开始{对齐}&f_{1}(j):=-G_{1{(x,y_{j};k_{1neneneep)-G_{1'(x,\波浪线{y}(y)_{j} ;k{1}),\\&f{2}(j):=-\frac{\partialG{1}(x,y{j};k{1{)}{\paratil\nu}-\frac{\particlG{1}(x,tilde{y}(y)_{j} ;k{1})}{\部分\nu}。\结束{对齐}$$
显然,\L中的(f_{1}(j)^{p}_{\alpha}(\Gamma)\)一致有界\(j\在{\mathbb{N}}\中),其中\(1<p<2)此外,从中可以看出[9,引理4.2]\(f _{2}(j)在C(伽马)中一致有界\(j\在{\mathbb{N}}\中).所以\L中的(f_{2}(j)^{p}_{\alpha}(\Gamma)\)是一致有界的\(j\在{\mathbb{N}}\中),其中\(1<p<2)然后,通过(2.16)在定理中2.1,我们可以得出解决方案\((\varphi_{1},\varphi_2})^{T}\)第页,共页(2.15)满足
$$\开始{aligned}\Vert\varphi_{1}\Vert_{L^{p}_{\alpha}(\Gamma)}+\Vert\varphi_{2}\Vert_{L^{p}_{\alpha}(\Gamma)}\leq C\bigl(\Vert f_{1j}\Vert_{L^{p}_{\alpha}(\Gamma)}+\Vert f_{2j}\Vert_{L^{p}_{\alpha}(\Gamma)}\biger),\quad 1<p<2。\结束{对齐}$$
(2.21)
我们接下来证明操作员\(S_{1j}:L)^{p}_{\alpha}(\Gamma)\rightarrow L^{2}_{\alpha}(\Gamma\set-nuse-{B})一致有界\(j\在{\mathbb{N}}\中),其中\(1<p<2)事实上,通过直接计算,我们可以推断出
$$\begin{aligned}&\biggl\Vert\int_{\Gamma}G_{1}(\cdot,y;k_{1{)\varphi_{1neneneep(y)\,ds(y)\biggr\Vert_{L^{2}(\ Gamma\setminus{B})}\\&\quad=\sup_{\psi\ in L^{2}_{\alpha},\Vert\psi\Vert_{L^{2}_{\alpha}(\Gamma\setminus{B})}=1}\biggl\vert\int_{\Gamma\setminus{B}}\int_{\ Gamma}G_{1}(x,y;k_{1})\varphi_{1(y)\,ds(y)\psi(x)\,dx\biggr\vert\&\quad=\sup_{psi\ in L^{2}_{\alpha},\Vert\psi\Vert_{L^{2}_{\alpha}(\Gamma\set-muse-{B})}=1}\biggl\vert\int_{\Gamma}\int_{\Gamma\set-muss-{B}}G_{1}(x,y;k_1})\psi(x)\,dx\varphi_1}(y)\,ds(y)\biggr\vert\&\quad\leq\vert\Gamma\vert\vert^{frac_1}{q}}\sup_{psi\ in L^{2}_{\alpha},\Vert\psi\Vert_{L^{2}_{\alpha}(\Gamma\set-muse-{B})}=1}\sup_{y\in\Gamma\set-muse-}}\bigl\VertG_{1}(\ cdot,y;k_{1{)\bigr\Vert_{L^{2}_{\alpha}(\Gamma\set-nuse-{B})}\Vert\psi\Vert_{L^{2}_{\alpha}(\Gamma\set-nuse{B})}\Vert\varphi_{1}\Vert_{L^{p}_{\alpha}(\Gamma)}\\&\quad=\vert\Gamma\vert^{\frac{1}{q}}\sup_{y\in\Gamma\set-muse-{B}}\bigl\VertG{1}(\ cdot,y;k_{1})\bigr\vert_{L^{2}_{\alpha}(\Gamma\set-muse-{B})}\Vert\varphi_{1}\Vert_{L_{alpha}^{p}(\ Gamma)}\leq C\Vert\valphi_{1\Vert_{L^{p}_{\alpha}(\Gamma)}。\结束{对齐}$$
(2.22)
在这里,我们使用了以下事实\(G_{1}(\ cdot,y;k_{1{)\)边界平滑\(\Gamma\设置减号{B}\)在第一个不等式中。那么我们有了\(S_{1j}:L^{p}_{\alpha}(\Gamma)\rightarrow L^{2}_{\alpha}(\Gamma\set-nuse-{B})一致有界\(j\在{\mathbb{N}}_{+}\中)此外,通过使用与证明(2.22),可以看出操作员\(S_{ij}\),\(K_{ij}\),\(K'_{ij}\)、和\(T_{ij}\)都是一致有界的\(升^{p}_{\alpha}(\Gamma)\)进入之内\(升^{2}_{\alpha}(\Gamma\set-nuse-{B})对于\(j\在{\mathbb{N}}_{+}\中),\(i=1,2)。还要注意\L中的(f_{1}(j),f_{2}(j)^{2}_{\alpha}(\ Gamma\setminus{B})\)一致有界\(j \ in{\mathbb{N}}_{+}\)这一点,结合方程式(2.15),提供了唯一的解决方案\((\varphi_{1},\varphi_2})^{T}\)第页,共页(2.15)满足这一点\L中的((\varphi_{1},\varphi_2})^{T}^{2}_{\alpha}(\Gamma\set-nuse-{B})\乘以L^{2}_{\alpha}(\Gamma\set-nuse-{B})。从中可以看出(2.14)解决方案\(u{2j}\)变速器故障(2.1)–(2.5)可以重写为
$$开始{对齐}u_{2j}(x)={}&\int_{\Gamma\set-buse-{B}}G_{2}部分G_2}(x,y;k_2})}{\varphi{2}(y)\,ds(y)。\结束{对齐}$$
(2.23)
定义
$$S_{\Gamma\set-muse-{B}}(j)\varphi_{1}:=\int_{\Gamma\set-mose-{B{}}G_{2}(x,y;k_{2{)\varfi_{1{(y)\,ds(y)$$
很容易看出\(S_{\Gamma\集合减号{B}}(j):H^{-\frac{1}{2}}{\alpha}一致有界\(j\在{\mathbb{N}}\中)这与以下事实相结合:\(L中的\varphi_{1}^{2}_{\alpha}(\Gamma\set-nuse-{B})意味着\(q_{1j}(x):=S_{\Gamma\setminus{B}}(j)\varphi_{1}\)满足以下Dirichlet问题:
$$\开始{aligned}\textstyle\begin{cases}\三角形w+k^{2}_{2} w个=0&\text{in}\Omega_{-}\集合减号{B},\\w=q_{1j}\在H^{frac{1}{2}}{{alpha}(tilde{Gamma})和\text{on}\tilde{\Gamma},\\w(x)=\sum_{n\in{mathbb{Z}}^{2}w_{n}^{-}\exp(i\alpha_{n{}\cdot\wid埃蒂尔德{x} -i\β^{-}_{n} x个_{3} )&x{3}<A{2},\end{cases}\displaystyle\end{aligned}$$
(2.24)
哪里\(\tilde{\Gamma}=(\Gamma\set-nuse-{B})\cup(\partialB\cap\Omega_{-})\)然后是Dirichlet问题的适定性(2.24)对于任何\({\mathbb{R}}\中的h\),\(H^{1}中的q_{1j}(\Omega_{-}(H)\setminuse\overline{B}})统一用于\(j\在{\mathbb{N}}_{+}\中).
我们现在定义
$$q_{2j}(x):=\int_{\Gamma\cap B}G_{2}(x,y;k_2})\varphi_{1}(y)\,ds(y)$$
自该地区\(\Omega_{-}\set-nuse-{B}\)与…有正距离\(y_{0}\),发现\(q_{2j}(x)在H^{1}中(\Omega_{-}(H)\set-nuse{\overline{B}})统一用于\(j\在{\mathbb{N}}_{+}\中)。我们进一步定义
$$K{\Gamma\set-muse-{B}}(j)\varphi_{2}:=\int_{\Gamma\set-mose-{B{}}\frac{\partialG{2}(x,y;K_{2{)}{\parial\nu(y)}\varphi_2}(y)\,ds(y)$$
显然,\(K_{\Gamma\集合减号{B}}(j):H^{-\frac{1}{2}}{\alpha}一致有界\(j\在{\mathbb{N}}_{+}\中)然后,根据事实\L中的(\varphi_{2}^{2}_{\alpha}(\Gamma\set-nuse-{B}),我们得到\(q_{3j}(x):=K_{\Gamma\set-muse-{B}}(j)\varphi_{2}\)满足Dirichlet问题(2.24),使用边界数据\(w=q{1j}\)替换为\(w=q{3j}\)关于Γ̃。然后使用与证明中类似的参数\H中的(q_{1j}^{1}_{\alpha}(\Omega_{-}(h)\set-nuss{\overline{B}})收益\H中的(q_{3j}^{1}_{\alpha}(\Omega_{-}(h)\set-nuss{\overline{B}})统一用于\(j\在{\mathbb{N}}_{+}\中)。我们还定义
$$q_{4j}(x):=\int_{\Gamma\cap B}\frac{\partial G_{2}(x,y;k_{2{)}{\parial\nu(y)}\varphi_{2neneneep(y)\,ds(y)$$
的一致有界性\H中的(q_{4j}^{1}_{\alpha}(\Omega_{-}(h)\setminus{\overline{B})\)对于\(j\在{\mathbb{N}}_{+}\中)可以从区域之间的正距离得出结论\((\Omega_{-}(h)\set-nuse-{上划线{B}})\)和\(y_{0}\)最后,得到了预期的结果(2.20)来自以下讨论(2.24). 定理的证明就这样完成了。□