$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\erf}{\operatorname{erf}}$ $\newcommand{\dag}{\dagger}$ $\newcommand{\const}{\mathrm{const}}$ $\newcommand{\arcsinh}{\operatorname{arcsinh}}$ ##[Appendix 4.A. Calculation of negative eigenvalues in Robin problem](id:sect-4.A) ------------------------------------ > 1. [Variational approach](#sect-4.A.1) > 2. [Case $\alpha=\beta$](#sect-4.A.2) ###[Variational approach](id:sect-4.A.1) Analyzing [Example 4.2.6](./S4.2.html#example-4.2.6) and [Example 4.2.7](./S4.2.html#example-4.2.7). We claim that **[Theorem 1.](id:thm-4.A.1)** Eigenvalues are monotone functions of $\alpha$, $\beta$. To prove it we need without proof to accept variational description of eigenvalues of self-adjoint operators bounded from below (very general theory) which in this case reads as: **[Theorem 2.](id:thm-4.A.2)** Consider quadratic forms \begin{equation} Q (u)= \int\_0^l |u'|^2\,dx + \alpha |u(0)|^2 + \beta |u(l)|^2 \label{eq-4.A.1} \end{equation} and \begin{equation} P (u)= \int \_0^l |u|^2\,dx . \label{eq-4.A.2} \end{equation} Then there are at least $N$ eigenvalues which are less than $\lambda$ if and only iff there is a subspace $\mathsf{K}$ of dimension $N$ on which quadratic form \begin{equation} Q\_\lambda (u)= Q(u)- \lambda P(u) \label{eq-4.A.3} \end{equation} is *negative definite* (i.e. $Q\_\lambda (u)\<0$ for all $u\in \mathsf{K}$, $u\ne 0$). Note that $Q(u)$ is montone non-decreasing function of $\alpha,\beta$. Therefore $N(\lambda)$ (the exact number of e.v. which are less than $\lambda$) is montone non-increasing function of $\alpha,\beta$ and therefore $\lambda\_N$ is montone non-decreasing function of $\alpha,\beta$. ###[Case $\alpha=\beta$](id:sect-4.A.2) The easiest way to deal with it would be to note that the hyperbola $\alpha+\beta+\alpha\beta l=0$ has two branches and divides plane into 3 regions and due to continuity of eigenvalue in each of them the number of negative eigenvalues is the same. Consider $\beta=\alpha$, it transects all three regions. Shift coordinate $x$ to the center of interval, which becomes $[-L,L]$, $L=l/2$. Now problem becomes \begin{align} &X''+\lambda X=0,\label{eq-4.A.4}\\\\ &X'(-L)=\alpha X(-L),\label{eq-4.A.5}\\\\ &X'(L)=-\alpha X(L)\label{eq-4.A.6} \end{align} and therefore if $X$ is an eigenfunction, then $Y(x):=X(-x)$ is an eigenfunction with the same eigenvalue. Therefore we can consider separately eigenfunctions which are even functions, and which are odd functions--and those are described respectively by \begin{align} &X''+\lambda X=0,\label{eq-4.A.7}\\\\ &X'(0)=0,\label{eq-4.A.8}\\\\ &X'(L)=-\alpha X(L)\label{eq-4.A.9} \end{align} and \begin{align} &X''+\lambda X=0,\label{eq-4.A.10}\\\\ &X(0)=0,\label{eq-4.A.11}\\\\ &X'(L)=-\alpha X(L).\label{eq-4.A.12} \end{align} Since we are looking at $\lambda=-\gamma^2$ ($\gamma\>0$, we look at $X=\cosh (x \gamma)$ and $X=\sinh (X\gamma)$ respectively (see conditions (\ref{eq-4.A.8}), (\ref{eq-4.A.11})) and then conditions (\ref{eq-4.A.9}), (\ref{eq-4.A.12}) tell us that \begin{align} &\alpha L = -(L\gamma)\tanh (L\gamma),\label{eq-4.A.13}\\\\ &\alpha L= - (L\gamma) \coth (L\gamma)\label{eq-4.A.14} \end{align} respectively. Both functions $z\tanh(z)$ and $z\coth(z)$ are monotone increasing for $z\>0$ with minima at $z=0$ equal $0$ and $1$ respectively. Thus equation (\ref{eq-4.A.13}) has a single solution $\gamma$ iff $\alpha\<0$ and (\ref{eq-4.A.14}) has a single solution $\gamma$ iff $\alpha L \< -1$. Therefore as $\alpha l\<0$ there is one negative eigenvalue with an even eigenfunction and as $2\alpha l+(\alpha l)^2\<0$ comes another negative eigenvalue with an odd eigenfunction. Sure, one can apply a variational arguments above but analysis above has its own merit (mainly learning). _____________ [$\Leftarrow$](./S4.2.html)  [$\Uparrow$](../contents.html)  [$\Rightarrow$](./S4.3.html)