$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\supp}{\operatorname{supp}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$ #[Chapter 13. Eigenvalues and eigenfunctions](id:chapter-13) ##[13.1. Variational theory](id:sect-13.1) ____________________________ > 1. [Introduction](#sect-13.1.1) > 1. [Main variational principles](#sect-13.1.1) > 2. [Dependence on parameters](#sect-13.1.2) > 3. [Dependence on domain](#sect-13.1.3) ###[Introduction](id:sect-13.1.1) Consider quadratic forms \begin{gather} Q\_0(u)=\\|u\\|^2=\iiint\_\Omega |u|^\,dx; \label{eq-13.1.1}\\\\ Q(u)= \iiint\_\Omega \\|\nabla u\\|^2\,dx + \iint\_\Sigma \alpha|u|^2\,d\sigma \label{eq-13.1.2} \end{gather} where $\Omega $ is a bounded domain with a smooth boundary $\Sigma=\partial \Omega$. Let us consider a variational problem for $Q(u)$ under restriction $\\|u\\|=1$ and \begin{equation} u\bigr|\_{\Sigma^-}=0 \label{eq-13.1.3} \end{equation} where $\Sigma^-\subset \Sigma$. Constructing functional $Q\_\lambda (u)= Q(u)-\lambda Q(u)$ and taking it variation we arrive to Euler-Lagrange equation \begin{equation} -\Delta u=\lambda u \label{eq-13.1.4} \end{equation} with the boundary conditions \begin{equation} u\bigr|\_{\Sigma^-}=0,\qquad (\partial\_{\boldsymbol{\nu}}u-\alpha u)\bigr|\_{\Sigma^+}=0\qquad \Sigma^+=\Sigma\setminus \Sigma^- \label{eq-13.1.5} \end{equation} where $\boldsymbol{\nu}$ is a unit inner normal. So we arrived to the eigenvalue problem. We need the following theorems from the Real Analysis: **[Theorem 1.](id:thm-13.1.1)** Let $\mathsf{H}$ be a Hilbert space. Let $\\|v\_n\\|\le M$; then there exists a subsequence $v\_{n\_k}$ which *converges weakly* in $\mathsf{H}$: $(v\_{n\_k}-v, \phi)\to 0$ for all $\phi\in \mathsf{H}$. **[Theorem 2.](id:thm-13.1.2)** Let $\Omega$ be a domain of a finite volume with a boundary $\Sigma$. Then any sequence $v\_n$ such that $\\|\nabla v\_n\\||\le M$ and $v\_n|\_\Sigma=0$, contains a subsequence $v\_{n\_k}$ which converges in $L^2(\Omega)$: $\\|v\_{n\_k}-v\\|\to 0$ for some $v\in L^2(\Omega)$, $\nabla v\in L^2(\Omega)$. **[Theorem 3.](id:thm-13.1.3)** Let $\Omega$ be a bounded domain with a smooth boundary. Then a. $v\in L^2(\Omega)$, $\nabla v\in L^2(\Omega)$ imply that $v\in L^2(\Sigma)$; b. any sequence $v\_n$ such that $\\|v\_n\\|+\\|\nabla v\_n\\||\le M$ contains a subsequence $v\_{n\_k}$ which converges in $L^2(\Omega)$ and also converges in $L^2(\Sigma)$: $\\|v\_{n\_k}-v\\|\_{L^2(\Sigma)}\to 0$ to some $v\in L^2(\Omega)\cap L^2(\Sigma)$, $\nabla v\in L^2(\Omega)$; c. For any $\epsilon>0$ there exists $C\_\epsilon$ such that \begin{equation} \\|v\\|\_{L^2(\Sigma)}\le \epsilon \\|\nabla v\\|+ C\_\epsilon \\|v\\|. \label{eq-13.1.6} \end{equation} **[Remark 1.](id:remark-13.1.1)** Actually one need only assume that $\Sigma^+$ is bounded and smooth. ###[Main variational principles](id:sect-13.1.2) **[Theorem 4.](id:thm-13.1.4)** Let $\Omega$ be a bounded domain with the smooth boundary. Then a. there exists a sequence of eigenvalues $\lambda_1\le \lambda\_2\le \lambda\_3\le \ldots $, $\lambda\_k\to \infty$ and a sequence of corresponding eigenfunctions $u\_1,u\_2,\ldots $ which forms a basis in $L^2(\Omega)$; b. Variational problem of minimizing $Q(u)$ under constrains $\\|u\\|=1$ and (\ref{eq-13.1.3}) has solution $u\_1$ and this minimum is $\lambda\_1$. c. Variational problem of minimizing $Q(u)$ under constrains $\\|u\\|=1$, (\ref{eq-13.1.3}) and $(u,u\_1)=(u,u\_2)=\ldots=(u,u\_{n-1})=0$ has solution $u\_n$ and this minimum is $\lambda\_n$. *Proof.* (i) Let us prove first that $u\_1$ exists. Let us consider minimizing sequence $v\_m$ for $Q(u)$ under constrain $\\|v\_m\\|=1$. Observe that due to [Theorem 3](#thm-13.1.3)(c) \begin{equation} \\|\nabla v\\|_{L^2(\Sigma)}^2 \le (1-\epsilon )Q(v)+ C\_\epsilon \\|v\\|^2. \label{eq-13.1.7} \end{equation} Then we have $\\| v\_m\\|+\\|\nabla v\_m\\|\le M$ and in virtue of [Theorem 1](#thm-13.1.1) and either [Theorem 2](#thm-13.1.2) or [Theorem 3](#thm-13.1.3) there is a subsequence $v\_{m\_k}$ converging to some $v$ both in $L^2(\Omega)$ and weakly in $\mathsf{K}=\\{v: \\|v\\|\_{\mathsf{K}}<\infty,\, u|\_{\Sigma^-}=0\\}$ with $\\|v\\|\_{\mathsf{K}}=\bigl(C\_0\\|v\\|^2 +Q(v)\bigr)^{\frac{1}{2}}$. Then $\\|v\\|=1$ and $Q(v)$ is minimal. We skip the proof. (ii) Similarly we prove existence $\lambda\_n,u\_n$ by induction. Now we claim that $\lambda\_n\to \infty$. Indeed if it is not the case then $\\|u\_n\\|=1$ and $\\|\nabla u\_n\\|\le M$ for some $M$; then in virtue of either [Theorem 2](#thm-13.1.2) or [Theorem 3](#thm-13.1.3) there exists a subsequence $u\_{n\_k}$ converging in $L^2(\Omega)$ which is impossible since $u\_{n\_k}$ are mutually orthogonal and $\\| u\_{n\_k}-u\_{n\_l}\\|^2=2$ as $k\ne l$. (iii) Further, observe that $u\_n$ are orthogonal in $\mathsf{K}$. Let us prove that system $u\_n$ is complete in $\mathsf{K}=$. If it is not the case then there exists $u\in \mathsf{K}$ which is orthogonal to all $u\_n$. But then $(u\_n, u)\_{\mathsf{K}}= (-\Delta u\_n, u) + C\_0 (u\_n,u)= (\lambda\_n+C_0) (u\_n,u)$ and therefore $u$ is orthogonal to all $u\_n$ in $L^2(\Omega)$ as well. But then since $\lambda\_k\to \infty$ and $Q(u)< \infty$ we conclude that $u$ must appear as a minimizing element in the sequence which is the contradiction. (iv) Finally, $u\_n$ is complete in $L^2(\Omega)$ as well. It follows from completeness in $\mathsf{K}$ and the fact that $\mathsf{K}$ is dense in $L^2(\Omega)$. The latter proof is elementary but we skip it as well. **[Remark 2.](id:remark-13.1.2)** If $\Sigma=\Sigma^-$ we do not need to assume that $\Sigma$ is smooth or even $\Omega$ is bounded; it is sufficient to assume that it has a finite volume (but even this is not necessary!) **[Corollary 1.](id:corollary-13.1.1)** Consider $n$–dimensional subspace $L\subset \mathsf{H}=\\{u\in C^1(u), u|\_{\Sigma^-}=0\\}$. Let $\mu\_n (L)= \max\_{u\in L:\, \\|u\\|=1} Q(u)$. Then \begin{equation} \lambda\_n =\min\_{L\subset \mathsf{H}: \, \dim(L)=n} \mu\_n (L)= \min\_{L\subset \mathsf{H}: \, \dim(L)=n} \ \max\_{u\in L:\, \\|u\\|=1} Q(u) \label{eq-13.1.8} \end{equation} *Proof.* If $L$ is a span of $u\_1,\ldots,u\_n$ then $\lambda\_n(L)=\lambda\_n$ so the right-hand expression is not greater than the left=hand one. On the other hand, if $L$ is not a span of $u\_1,\ldots,u\_n$ then there exists $u\in L,\, u\ne 0$ which is orthogonal to $u\_1,\ldots,u\_n$ and therefore $Q(u)\ge \lambda\_{n+1}\\|u\\|^2$. ###[Dependence on parameters](id:sect-13.1.3) **[Theorem 5.](id:thm-13.1.5)** a. Eigenvalues $\lambda\_n=\lambda\_n(\alpha)$ does not decrease as $\alpha$ increases; b. Eigenvalues $\lambda\_n=\lambda\_n(\Sigma^-)$ does not decrease as $\Sigma^-$ increases; c. As $\alpha\to +\infty$ these eigenvalues $\lambda\_n=\lambda\_n(\alpha)$ tend to $\lambda\_{D,n}$ which are eigenvalues of Dirichlet problem (i.e. when $\Sigma^-=\Sigma$). *Proof.* a. As $\alpha $ increases then $Q(u)$ also increases and then the right-hand expression in (\ref{eq-13.1.6}) cannot decrease. b. As we increase $\Sigma^-$ we make space $\mathsf{H}$ only smaller and thus make smaller the choice of $n$-dimensional subspaces $L$ so the right-hand expression in (\ref{eq-13.1.6}) cannot decrease. c. Therefore $\lambda\_n \le \lambda\_{D,n}$. Then as $\alpha\to +0$ expressions $\alpha \iint\_{\Sigma}|u\_n(\alpha)|^2\,d\sigma$ are bounded by $\lambda\_{D,n}$ and therefore $u|\_\Sigma\to +0$. Assume that $\lambda\_n(\alpha)\le \mu\_n$ for all $\alpha$. Then by means of Real Analysis one can prove that there is a sequence $\alpha\_k\to +\infty$ such that $u\_n(\alpha\_k)\to v\_n$ and $\\|\nabla v\_n\\|^2 \le \mu\_n$, $v\_n|\_\Sigma=0$ and $\\|v\_n\\|=1$. But then $\mu\_n \ge \lambda\_n$. Therefore $\lambda\_n=\mu\_n$ and $\lambda\_n(\alpha)\to \\lambda\_{D,n}$. ###[Dependence on domain](id:sect-13.1.4) Consider now *only* Dirichlet boundary problem. **[Theorem 6.](id:thm-13.1.6)** Eigenvalues $\lambda\_n=\lambda\_{D,n}(\Omega)$ does not increase as $\Omega$ increases. *Proof.* If $\Sigma^-=\Sigma$ we can identify $\mathsf{H}=\mathsf{H}(\Omega)$ with $\\{u:\, \nabla u\in L^2(\mathbb{R}^d), \ u=0 \ \text{on } \ \mathbb{R}^d\setminus\bar{\Omega}\\}$ where $\bar{\Omega}$ is a closure of $\Omega$. Then \begin{equation\*} Q(u)=\int |\nabla u|^2\, dx,\qquad Q_0(u)=\int | u|^2\, dx \end{equation*} with integrals taken over $\mathbb{R}^d$. Therefore as $\Omega$ increases then $\mathsf{H}(\Omega)$ increases and the choice of $L$ increases, so the right-hand expression in (\ref{eq-13.1.6}) cannot increase. **[Remark 3.](id:remark-13.1.3)** This statement holds only for Dirichlet eigenvalues. **[Theorem 7.](id:thm-13.1.7)** Let vary $\Omega$ by moving by $h$ into direction of $\mathbf{\nu}$ (where $h$ is a small function on $\Sigma$. Then for a simple eignevalue $\lambda\_n$ \begin{equation} \delta \lambda\_{D,n} = \frac{1}{\\|u\_n \\|^2}\int \_\Sigma h|\partial\_{\boldsymbol{\nu}} u\_n|^2 \,d\sigma . \label{eq-13.1.9} \end{equation} *Proof.* Consider $u\_n+\delta u\_n$ matching to a new domain. Since it must be $0$ on the new boundary, modulo smaller terms we conclude that it is $u\_n+\delta u\_n =-h \partial\_{\boldsymbol{\nu}} u\_n $ on the old boundary $\Sigma$ and since $u\_n=0$ there we conclude that \begin{equation} \delta u\_n = -h \partial\_{\boldsymbol{\nu}}u\_n \qquad \text{on }\ \Sigma. \label{eq-13.1.10} \end{equation} On the other hand $(\Delta +\lambda\_n +\delta \lambda\_n)(u\_n+\delta u\_n)=0$ and then modulo smaller terms we conclude that \begin{equation} (\Delta +\lambda\_n) \delta u\_n = -\delta \lambda\_n u\_n. \label{eq-13.1.11} \end{equation} Let us multiply by $u\_n$ and integrate over $\Omega$; the left-hand expression becomes \begin{multline} \int \_\Omega (\Delta +\lambda\_n) \delta u\_n \cdot u\_n \,dx =\\\\ \int \_\Omega \delta u\_n \cdot (\Delta +\lambda\_n) u\_n \,dx -\int \_\Sigma \partial\_{\boldsymbol{\nu}} \delta u\_n \cdot u\_n \,d\sigma +\int \_\Sigma \delta u\_n \cdot \partial\_{\boldsymbol{\nu}} u\_n \,d\sigma= \\\\ -\int \_\Sigma h|\partial\_{\boldsymbol{\nu}} u\_n|^2 \,d\sigma \label{eq-13.1.12} \end{multline} where we used that $ (\Delta +\lambda\_n) u\_n=0$, $u\_n|\_\Sigma=0$ and (\ref{eq-13.1.10}). Meanwhile the right-hand expression becomes $-\delta \lambda\_n \\|u\_n\\|^2$. Since it i must be equal to the right-hand expression of (\ref{eq-13.1.12}) we arrive to (\ref{eq-13.1.9}). --------------- [$\Leftarrow$](../Chapter12/S12.4.html)  [$\Uparrow$](../contents.html)  [$\Rightarrow$](./S13.2.html)