4.A. Calculation of negative eigenvalues in Robin problem


## Appendix 4.A. Calculation of negative eigenvalues in Robin problem

### Variational approach

Analyzing Example 4.2.6 and Example 4.2.7. We claim that

Theorem 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. Consider quadratic forms $$Q (u)= \int_0^l |u'|^2\,dx + \alpha |u(0)|^2 + \beta |u(l)|^2 \label{eq-4.A.1}$$ and $$P (u)= \int _0^l |u|^2\,dx . \label{eq-4.A.2}$$ 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 $$Q_\lambda (u)= Q(u)- \lambda P(u) \label{eq-4.A.3}$$ 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 $\alpha=\beta$

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).