线性代数网课代修|ENGO361代写|ENGO361英文辅导|ENGO361

如果你也在线性代数linearalgebra这个学科遇到相关的难题,请随时添加vx号联系我们的代写客服。我们会为你提供专业的服务。 linearalgebra™长期致力于留学生网课服务,涵盖各个网络学科课程:金融学Finance,经济学Economics,数学Mathematics,会计Accounting,文学Literature,艺术Arts等等。除了网课全程托管外,linearalgebra™也可接受单独网课任务。无论遇到了什么网课困难,都能帮你完美解决! Appendix A Generic results for Steklov eigenfunctionsLet \(M\) be a compact smooth \(n\)-dimensional manifold with smooth boundary \(tial M\), and let \(G\) denote the set of smooth Riemannian metrics on \(M\). In this section we will prove the following theorem.**Theorem**.: _Let \(M\) be a compact smooth \(n\)-dimensional manifold with smooth boundary \(tial M\). Then the subset of \(G\) of smooth metrics on \(M\) for which all Steklov eigenfunctions are Morse functions in \(M\) is residual (hence dense) in \(G\)._We prove Theorem A.1 following the ideas of [12] (see also [13]).We shall need the following transversality theorem (see [12, Transversality ]):**Theorem**.: _Let \(Q,B,X,Y,Y^{\prime}\) be separable Banach manifolds, \(Y^{\prime}\subset Y\), \(X,Y\) finite dimensional. Let \(\pi:Q\to B\) be a \(C^{k}\) Fredholm map of index \(0\). Then, if \(F:Q\times X\to Y\) is a \(C^{k}\) map for \(k>\max\{1,\dim X+\dim Y^{\prime}-\dim Y\}\) and \(F\) is transverse to \(Y^{\prime}\), the set \(\{b\in B:F_{b}:=F_{\pi^{-1}b}\) is transverse to \(Y^{\prime}\}\) is residual in \(B\)._Let \(H^{k}(tial M)\) denote the usual Sobolev spaces of functions in \(L^{2}(tial M)\) with weak derivatives up to order \(k\in\mathbb{N}\) in \(L^{2}(tial M)\). We choose \(k\) sufficiently large so that functions in \(H^{k}(tial M)\) have some regularity (for example, \(C^{1}\)). Following [12], we consider the map\[\varphi:H^{k}(tial M)\times\mathbb{R}\times G\to H^{k-1}(tial M)\]defined by\[\varphi(f,\sigma,g)=(\mathscr{D}-\sigma)f,\]where \(\mathscr{D}\) is the Dirichlet-to-Neumann map, namely \(\mathscr{D}f=tial_{\nu}(Hf)\), where \(Hf\) is the _harmonic extension_ of \(f\), namely, it solves\[\begin{cases}\Delta(Hf)=0&\text{in }M, Hf=f&\text{on }tial M.\end{cases}\]The Laplacian is taken with respect to the metric \(g\).Since we are considering smooth metrics, from [12, Lemma 2.1] we get that \(\varphi_{b}:=\varphi(\cdot,\cdot,b)\) is a Fredholm map of index \(0\). Moreover, it is clear that \(\varphi(f,\sigma,g)=0\) if and only if \(f\) is an eigenfunction of \(\mathscr{D}\) with eigenvalue \(\sigma\) for the metric \(g\) (see also [12, Lemma 2.2]).Consider now the Banach manifold \(Q=\varphi^{-1}(0)\), namely the subset of \(H^{k}(tial M)\times\mathbb{R}\times G\) of all the eigenfunctions, and consider the following map:\[\beta:Q\times M\to TM\]defined by\[\beta(f,\sigma,g,x)=\nabla Hf(x).\]Note that \(f\) is an eigenfunction of \(\mathscr{D}\) with eigenvalue \(\sigma\) for the metric \(g\) if and only if \(Hf\) is a Steklov eigenfunction on \(M\) for the same metric and with the same eigenvalue. From the smoothness assumptions on the metrics of \(G\), we have that \(\beta\) is a smooth map.We will denote the differentiation of \(\varphi\) with respect to the first, second and third parameters respectively by \(D_{1},D_{2},D_{3}\). It is convenient now to describe the tangent space of \(Q\):\[T_{(f,\sigma,g)}Q=\{(v,s,h)\in H^{k}(tial M)\times\mathbb{R}\times T_{g}G:( \mathscr{D}-\sigma)v+sf+(D_{3}\varphi|_{(f,\sigma,g)})(h)=0\}. \tag{19}\]The following two propositions proved in [12, 14] will be crucial in the sequel:**Proposition**.: _Let \((f,\sigma,g)\in Q\) with \(\sigma\neq 0\), and consider the map \(D_{3}\varphi|_{(f,\sigma,g)}:T_{g}G\to H^{k-1}(tial M)\). Then_\[(\operatorname{Im}D_{3}\varphi|_{(f,\sigma,g)})^{\perp}\subseteq\{\psi\in H^{ k-1}(tial M):\operatorname{supp}(\psi)\subseteq f^{-1}(0)\},\]_or, equivalently,_\[\{\psi\in H^{k-1}(tial M):\operatorname{supp}(\psi)\cap f^{-1}(0)=\emptyset \}\subseteq\operatorname{Im}D_{3}\varphi|_{(f,\sigma,g)}.\]_In particular, for any \(\psi\in H^{k-1}(tial M)\) with \(\operatorname{supp}(\psi)\cap f^{-1}(0)=\emptyset\), there exists a smooth function \(\omega\) defined on \(M\) such that_\[D_{3}\varphi|_{(f,\sigma,g)}(\omega g)=\psi.\]Using the previous proposition we get the following:**Proposition**.: \(\operatorname{Im}D\varphi|_{(f,\sigma,g)}=H^{k-1}(tial M)\) _for any \((f,\sigma,g)\in Q\) with \(\sigma\neq 0\). In particular \(0\) is a regular value of \(\varphi\)._We are now in position to prove Theorem A.1.Proof of Theorem A.1.: We wish to apply Theorem A.2 with \(Q=\varphi^{-1}(0)\), \(B=G\), \(X=M\), \(Y=TM\), \(F=\beta\). Note that \(\beta\) is smooth, and the projection map \(\pi:Q\to G\) (restriction to \(Q\) of the projection \(\pi:H^{k}(tial M)\times\mathbb{R}\times G\to G\)) has Fredholm index \(0\) (see [12, Lemma 2.5]). Hence we have to prove that \(\beta\) is transverse to \(Y^{\prime}=0\) the zero section of \(TM\) (i.e., the zero vector field on \(M\)).To do so, let \(\beta_{x}:=\beta(\cdot,\cdot,\cdot,x):Q\to T_{x}M\). We need to prove that \(D\beta_{x}:T_{(f,\sigma,g)}\to T_{x}M\) is surjective when \(\beta_{x}(f,\sigma,g)=0\) (and \(\sigma\neq 0\)). Writing \(D\beta_{x}\) explicitly, since \(H\) is linear, we have\[D\beta_{x}|_{(f,\sigma,g)}(v,s,h)=\nabla Hv(x).\]We recall that \(f\) is an eigenfunction of \(\mathscr{D}\) with eigenvalue \(\sigma\) for the metric \(g\). We also assume that \(\sigma\neq 0\) (that is, we consider non-constant eigenfunctions). Assume now that there exists \(0\neq V\in T_{x}M\) such that \(\langle\nabla Hv(x),V\rangle=0\) for all \(v\) such that \((v,s,h)\in T_{(f,\sigma,g)}Q\). Let us consider the following two situations:* \((f,0,0)\in T_{(f,\sigma,g)}Q\) (in fact, it satisfies (19)). Hence \(D\beta_{x}|_{(f,\sigma,g)}(f,0,0)=\nabla Hf(x)\) and by assumption \(\langle\nabla Hf(x),V\rangle=0\). * From Proposition A.3 it follows that for all \(\psi\in H^{k-1}(tial M)\) with \(\operatorname{supp}(\psi)\cap f^{-1}(0)=\emptyset\), there exists a conformal variation of the metric \(\omega g\) such that \(D_{3}\varphi|_{f,\sigma,g}(\omega g)=\psi\) (\(\omega\) is a smooth function on \(M\)). Let \(\psi\in\operatorname{Ker}\left(\mathscr{D}-\sigma\right)^{\perp}\) be such that \(\operatorname{supp}(\psi)\cap f^{-1}(0)=\emptyset\), and let \(\omega\) the corresponding conformal factor from Porposition A.3. We also have that \(\psi=\sum_{j:\sigma_{j}\neq\sigma}a_{j}f_{j}\), where \(\{f_{j}\}\) is a \(L^{2}(tial M)\) orthonormal basis of \(H^{1/2}(tial M)\) of eigenfunctions of \(\mathscr{D}\). Let us define\[v:=-\sum_{\sigma_{j}\neq\sigma}\frac{a_{j}f_{j}}{\sigma_{j}-\sigma}.\]Then \((v,0,\omega g)\in T_{f,\sigma,g}Q\), in fact \((\mathscr{D}-\sigma)v+D_{3}\phi|_{f,\sigma,g}(\omega g)=-\psi+\psi=0\), so that the condition in (19) is satisfied. Hence\[\langle\nabla v(x),V\rangle=\sum_{\sigma_{j}\neq\sigma}\frac{a_{j}}{\sigma_{j} -\sigma}\langle\nabla Hf_{j}(x),V\rangle=0.\]We recall that \(a_{j}=\int_{tial M}\psi f_{j}\). Since \(\psi\in\operatorname{Ker}\left(\mathscr{D}-\sigma\right)^{\perp}\) with \(\operatorname{supp}(\psi)\cap f^{-1}(0)=\emptyset\) is dense in \(\operatorname{Ker}\left(\mathscr{D}-\sigma\right)^{\perp}\) with respect to the \(L^{2}(tial M)\) norm, we get that \(\langle\nabla Hf_{j}(x),V\rangle=0\) for all \(j\) with \(\sigma_{j}\neq\sigma\).In conclusion, assuming that \(D\beta_{x}\) is not surjective, we get that there exists \(0\neq V\in T_{x}M\) such that\[\langle\nabla H(x),\nabla V\rangle=0\]for all harmonic functions \(H\) in \(M\)图片描述

发表回复

您的电子邮箱地址不会被公开。 必填项已用*标注