We characterise $u$ to be a solution in terms of

$u_{k}(\cdot)\coloneqq u|_{(kh,(k+1)h)}(kh+\cdot),\quad
k\in\\{0,\ldots,r-1\\}.$

Using the characterisation of $H^{1}(\Omega)$-functions in Remark 13.1 and
keeping Setting 5.2 in mind, we infer that for $u$ to be a solution it is
equivalent that

$-\alpha_{k}u_{k}^{\prime\prime}=-\beta_{k}u_{k}\quad(k\in\\{0,\ldots,r-1\\}).$
(4)

together with the boundary conditions

$\displaystyle u_{0}(0)=0,u_{n}(h)=0,\quad
u_{k}(h)=u_{k+1}(0)\quad(k\in\\{0,\ldots,r-1\\})$
$\displaystyle\alpha_{k}u^{\prime}_{k}(h)=-\alpha_{k+1}u^{\prime}_{k+1}(0)\quad(k\in\\{0,\ldots,r-1\\})$

With the help of the transition matrix $A^{(k)}$ with respect to
$(h,\beta_{k}/\alpha_{k})$, we obtain the following equations replacing the
ODEs in (4)

$A^{(k)}\begin{pmatrix}u_{k}(0)\\\
u^{\prime}_{k}(0)\end{pmatrix}=\begin{pmatrix}u_{k}(h)\\\
u^{\prime}_{k}(h)\end{pmatrix}.$ (5)

Thus, taking the unknowns $x_{k,j}\coloneqq u_{k}(jh)$, $y_{k,j}\coloneqq
u^{\prime}_{k}(jh)$, $k\in\\{0,\ldots,r\\}$ and $j\in\\{0,1\\}$, we obtain the
following set of equations:

$\mathcal{A}\mathbf{x}=0,$ (6)

where $\mathbf{x}=(x^{(0)},x^{(1)})$ with

$\displaystyle x^{(0)}$
$\displaystyle=(x_{0,0},y_{0,0},x_{1,0},y_{1,0},\cdots,x_{r-1,0},y_{r-1,0},x_{r,0},y_{r,0})^{\top}$
$\displaystyle x^{(1)}$
$\displaystyle=(x_{0,1},y_{0,1},x_{1,1},y_{1,1},\cdots,x_{r-1,1},y_{r-1,1},x_{r,1},y_{r,1})^{\top}$

providing shorthands for the left-hand and right-hand boundary values
respectively. The matrix

$\mathcal{A}=(\mathcal{A}_{jk})_{j,k\in\\{0,1\\}}=\begin{pmatrix}\mathcal{A}_{00}&\mathcal{A}_{01}\\\
\mathcal{A}_{10}&\mathcal{A}_{11}\end{pmatrix}$

consists of

$\displaystyle\mathcal{A}_{00}$
$\displaystyle=-1_{2(r+1)}\in\mathbb{K}^{2(r+1)\times 2(r+1)}$
$\displaystyle\mathcal{A}_{01}$
$\displaystyle=\operatorname{diag}(A^{(0)},\ldots,A^{(r)})\in\mathbb{K}^{2(r+1)\times
2(r+1)}$

with $1_{2(r+1)}=\operatorname{diag}(1,\ldots,1)\in\mathbb{K}^{2(r+1)\times
2(r+1)}$. The equation

$\begin{pmatrix}\mathcal{A}_{00}&\mathcal{A}_{01}\end{pmatrix}\begin{pmatrix}x^{(0)}\\\
x^{(1)}\end{pmatrix}=0$

takes account of the equations in (5). The other equations,
$\begin{pmatrix}\mathcal{A}_{10}&\mathcal{A}_{11}\end{pmatrix}\begin{pmatrix}x^{(0)}\\\
x^{(1)}\end{pmatrix}=0$, are the boundary conditions. The formulas are

$\displaystyle\mathcal{A}_{10}$
$\displaystyle=\begin{pmatrix}\mathbf{o}_{2r}^{\top}&|&|\\\
\operatorname{diag}(1,\alpha_{0},1,\alpha_{1},\cdots,1,\alpha_{r-1})&\mathbf{e}^{2(r+1)}_{2(r+1)}&\mathbf{o}_{2(r+1)}\\\
\mathbf{o}_{2r}^{\top}&|&|\end{pmatrix}\in\mathbb{K}^{2(r+1)\times 2(r+1)}$
$\displaystyle\mathcal{A}_{11}$
$\displaystyle=\begin{pmatrix}|&|&\mathbf{o}_{2r}^{\top}\\\
\mathbf{e}^{1}_{2(r+1)}&\mathbf{o}_{2(r+1)}&\operatorname{diag}(-1,\alpha_{1},-1,\alpha_{2},\cdots,-1,\alpha_{r})\\\
|&|&\mathbf{o}_{2r}^{\top}\end{pmatrix}\in\mathbb{K}^{2(r+1)\times 2(r+1)}.$

where the lower indices of $\mathbf{o}$ and $\mathbf{e}$ refer to the
dimension of the respective zero vector and canonical basis vector. The upper
index of $\mathbf{e}$ signifies the position of the $1$.

Any solution $u$ of $(D_{\alpha}+\beta)u=0$ yields a solution of (6) by
evaluating $u$ and $u^{\prime}$ at the boundary. On the other hand, all
solutions $\mathbf{x}$ of (6) yield a solution $u$ of $(D_{\alpha}+\beta)u=0$.
Hence, $D_{\alpha}+\beta$ is one-to-one if and only if
$d\coloneqq\det\mathcal{A}\neq 0$. Thus, the remainder of this proof is
devoted to show that $\det\mathcal{A}\neq 0$ if and only if (iii) holds.
Therefore, we need to compute $d$. For this we expand this determinant with
respect to the last row of $\mathcal{A}$, to obtain

$d=-\det\mathcal{A}^{\prime}$

with $\mathcal{A}^{\prime}$ given by

$\begin{pmatrix}\begin{pmatrix}-1_{2r}&0\\\ 0&\begin{pmatrix}0\\\
-1\end{pmatrix}\end{pmatrix}&\vline&\operatorname{diag}(A^{(0)},\ldots,A^{(r)})\\\
\hline\cr\begin{pmatrix}\mathbf{o}_{2r}^{\top}&|\\\
\operatorname{diag}(1,\alpha_{0},1,\alpha_{1},\cdots,1,\alpha_{r-1})&\mathbf{o}_{2r+1}\end{pmatrix}&\vline&\begin{pmatrix}|&|&\mathbf{o}_{2r}^{\top}\\\
\mathbf{e}^{1}_{2r+1}&\mathbf{o}_{2r+1}&\operatorname{diag}(-1,\alpha_{1},-1,\alpha_{2},\cdots,-1,\alpha_{r})\end{pmatrix}\end{pmatrix}$

Next, we expand with respect to the row involving the $1$ of
$\mathbf{e}^{1}_{2r+1}$ to obtain

$d=\det\mathcal{A}^{\prime\prime}$

with $\mathcal{A}^{\prime\prime}$ reading as follows

$\begin{pmatrix}\begin{pmatrix}-1_{2r}&0\\\ 0&\begin{pmatrix}0\\\
-1\end{pmatrix}\end{pmatrix}&\vline&\begin{pmatrix}\begin{pmatrix}a_{01}^{(0)}\\\
a_{11}^{(0)}\end{pmatrix}&0\\\
0&\operatorname{diag}(A^{(1)},\ldots,A^{(r)})\end{pmatrix}\\\
\hline\cr\begin{pmatrix}\operatorname{diag}(1,\alpha_{0},1,\alpha_{1},\cdots,1,\alpha_{r-1})&\mathbf{o}_{2r}\end{pmatrix}&\vline&\begin{pmatrix}\mathbf{o}_{2r}&\operatorname{diag}(-1,\alpha_{1},-1,\alpha_{2},\cdots,-1,\alpha_{r})\end{pmatrix}\end{pmatrix}$

Next, we expand with respect to the last column of the upper left block
operator matrix und obtain

$d=\det\mathcal{A}^{\prime\prime\prime}$

with $\mathcal{A}^{\prime\prime\prime}$ given by

$\begin{pmatrix}\begin{pmatrix}-1_{2r}\\\
\mathbf{o}_{2r}^{\top}\end{pmatrix}&\vline&\begin{pmatrix}\begin{pmatrix}a_{01}^{(0)}\\\
a_{11}^{(0)}\end{pmatrix}&0&\\\
0&\operatorname{diag}(A^{(1)},\ldots,A^{(r-1)})&0\\\
0&0&\begin{pmatrix}a_{00}^{(r)}&a_{01}^{(r)}\end{pmatrix}\end{pmatrix}\\\
\hline\cr\begin{pmatrix}\operatorname{diag}(1,\alpha_{0},1,\alpha_{1},\cdots,1,\alpha_{r-1})\end{pmatrix}&\vline&\begin{pmatrix}\mathbf{o}_{2r}&\operatorname{diag}(-1,\alpha_{1},-1,\alpha_{2},\cdots,-1,\alpha_{r})\end{pmatrix}\end{pmatrix}$

We now apply the determinant formula provided in Remark 13.1 using the
following block structure

$\displaystyle M=\begin{pmatrix}A&\vline&B\\\ \hline\cr
C&\vline&D\end{pmatrix}=$
$\displaystyle\begin{pmatrix}-1_{2r}&\vline&\begin{pmatrix}\begin{pmatrix}a_{01}^{(0)}\\\
a_{11}^{(0)}\end{pmatrix}&0&|&|\\\
0&\operatorname{diag}(A^{(1)},\ldots,A^{(r-1)})&\mathbf{o}_{2r}&\mathbf{o}_{2r}\end{pmatrix}\\\
\hline\cr\begin{pmatrix}\mathbf{o}_{2r}^{\top}\\\
\operatorname{diag}(1,\alpha_{0},1,\alpha_{1},\cdots,1,\alpha_{r-1})\end{pmatrix}&\vline&\begin{pmatrix}0&\mathbf{o}_{2(r-1)}^{\top}&\begin{pmatrix}a_{00}^{(r)}&a_{01}^{(r)}\end{pmatrix}\\\
\mathbf{o}_{2r}&\operatorname{diag}(-1,\alpha_{1},-1,\alpha_{2},\cdots,-1,\alpha_{r})\end{pmatrix}\end{pmatrix}$

Then

$d=\det\mathcal{B}=\det(D-CA^{-1}B)$

with $\mathcal{B}=D-CA^{-1}B=D+CB$ given by

$\displaystyle\mathcal{B}$
$\displaystyle=\begin{pmatrix}0&\mathbf{o}_{2(r-1)}^{\top}&a_{00}^{(r)}&a_{01}^{(r)}\\\
\mathbf{o}_{2(r-1)}&\operatorname{diag}(-1,\alpha_{1},\ldots,-1,\alpha_{r-1})&&\\\
0&\mathbf{o}_{2(r-1)}^{\top}&-1&0\\\
0&\mathbf{o}_{2(r-1)}^{\top}&0&\alpha_{r}\\\ \end{pmatrix}$
$\displaystyle\quad+\begin{pmatrix}\mathbf{o}_{2r}^{\top}\\\
\operatorname{diag}(1,\alpha_{0},\ldots,1,\alpha_{r-1})\end{pmatrix}\begin{pmatrix}a_{01}^{(0)}&\mathbf{o}_{2(r-1)}^{\top}&|&|\\\
a_{11}^{(0)}&\mathbf{o}_{2(r-1)}^{\top}&\mathbf{o}_{2r}&\mathbf{o}_{2r}\\\
\mathbf{o}_{2(r-1)}&\operatorname{diag}(A^{(1)},\ldots,A^{(r-1)})&|&|\end{pmatrix}$
$\displaystyle=\begin{pmatrix}0&\mathbf{o}_{2(r-1)}^{\top}&a_{00}^{(r)}&a_{01}^{(r)}\\\
\mathbf{o}_{2(r-1)}&\operatorname{diag}(-1,\alpha_{1},\ldots,-1,\alpha_{r-1})&&\\\
0&\mathbf{o}_{2(r-1)}^{\top}&-1&0\\\
0&\mathbf{o}_{2(r-1)}^{\top}&0&\alpha_{r}\\\ \end{pmatrix}$
$\displaystyle\quad+\begin{pmatrix}\mathbf{o}_{2r+1}^{\top}\\\
a_{01}^{(0)}&\mathbf{o}_{2r}^{\top}\\\
\alpha_{0}a_{11}^{(0)}&\mathbf{o}_{2r}^{\top}\\\
0&\operatorname{diag}(A_{\alpha_{1}},\ldots,A_{\alpha_{r-1}})&0&0\end{pmatrix}$
$\displaystyle=\begin{pmatrix}0&\mathbf{o}_{2(r-1)}^{\top}&&&\begin{pmatrix}a_{00}^{(r)}&a_{01}^{(r)}\end{pmatrix}\\\
\begin{pmatrix}a_{01}^{(0)}\\\
\alpha_{0}a_{11}^{(0)}\end{pmatrix}&\begin{pmatrix}-1&0\\\
0&\alpha_{1}\end{pmatrix}&&&\begin{pmatrix}0&0\\\ 0&0\end{pmatrix}\\\
\begin{pmatrix}0\\\ 0\end{pmatrix}&A_{\alpha_{1}}&\begin{pmatrix}-1&0\\\
0&\alpha_{2}\end{pmatrix}&&\begin{pmatrix}0&0\\\ 0&0\end{pmatrix}\\\
\vdots&0&\ddots&\ddots&&\\\ \begin{pmatrix}0\\\
0\end{pmatrix}&0&\cdots&A_{\alpha_{r-1}}&\begin{pmatrix}-1&0\\\
0&\alpha_{r}\end{pmatrix}\end{pmatrix},$

where $A_{\alpha_{j}}\coloneqq\begin{pmatrix}1&0\\\
0&\alpha_{j}\end{pmatrix}A^{(j)}$. Elementary row and column operations
together with the above Remark yield that $d=(\det\mathcal{C})d^{\prime}$ with

$\mathcal{C}=\begin{pmatrix}\begin{pmatrix}-1&\\\
&\alpha_{1}\end{pmatrix}&&&\\\ A_{\alpha_{1}}&\begin{pmatrix}-1&\\\
&\alpha_{2}\end{pmatrix}&&\\\ &\ddots&\ddots&\\\
&&A_{\alpha_{r-1}}&\begin{pmatrix}-1&\\\ &\alpha_{r}\end{pmatrix}\\\
\end{pmatrix}$

and

$d^{\prime}=\begin{pmatrix}o_{2(r-1)}^{\top}&\begin{pmatrix}a_{00}^{(r)}&a_{01}^{(r)}\end{pmatrix}\end{pmatrix}\mathcal{C}^{-1}\begin{pmatrix}\begin{pmatrix}a_{01}^{(0)}&\alpha_{0}a_{11}^{(0)}\end{pmatrix}&o_{2(r-1)}^{\top}\end{pmatrix}^{\top}.$

As $\det\mathcal{C}=(-1)^{r}\alpha_{1}\cdots\alpha_{r}$ is easily computed
(and is always non-zero by the assumptions on $\alpha$); the computation if
$d^{\prime}$ is more involved. Denoting the lower left corner $2\times 2$
block of $\mathcal{C}^{-1}$ by $\mathcal{D}$, we obtain

$d^{\prime}=\begin{pmatrix}a_{00}^{(r)}&a_{01}^{(r)}\end{pmatrix}\mathcal{D}\begin{pmatrix}a_{01}^{(0)}\\\
\alpha_{0}a_{11}^{(0)}\end{pmatrix}.$

Thus, the assertion follows from Proposition 13.3 below. ∎

Before we state and prove Proposition 13.3, we consider an elementary lemma
next, which we state without the easy proof.

######  Lemma 13.2.

Let $A\in\mathbb{K}^{N\times N}$, $D\in\mathbb{K}^{M\times M}$ be invertible
matrices and $C\in\mathbb{K}^{M\times N}$. Then the matrix
$\begin{pmatrix}A&0\\\ C&D\end{pmatrix}$ is invertible and

$\begin{pmatrix}A&0\\\ C&D\end{pmatrix}^{-1}=\begin{pmatrix}A^{-1}&0\\\
-D^{-1}CA^{-1}&D^{-1}\end{pmatrix}$

######  Proposition 13.3.

Let $A_{1},\ldots,A_{n}\in\mathbb{K}^{\ell\times\ell}$ invertible and
$B_{1},\ldots,B_{r-1}\in\mathbb{K}^{\ell\times\ell}$. Then the matrix

$\mathcal{C}\coloneqq\begin{pmatrix}A_{1}&&&\\\ B_{1}&A_{2}&&\\\
&\ddots&\ddots&\\\ &&B_{r-1}&A_{r}\\\ \end{pmatrix}$

is invertible and the lower left $\ell\times\ell$ block of $\mathcal{C}^{-1}$
reads:

$(-1)^{r}A_{r}^{-1}B_{r-1}A_{r-1}^{-1}\cdots B_{1}A_{1}^{-1}.$

###### Proof.

The invertibility of $\mathcal{C}$ follows by induction from Lemma 13.2. It
remains to show the formula. We prove this claim by induction on
$r\in\mathbb{N}$. For $r=1$ there is nothing to show. Assume the result to be
true for $r-1$. By induction hypothesis and Lemma 13.2, the last block row of
$\mathcal{C}^{-1}$ reads in block operator matrix form

$\begin{pmatrix}D&(-1)^{r-1}A_{r}^{-1}B_{r-1}A_{r-1}^{-1}\cdots
B_{2}A_{2}^{-1}&\cdots&A_{r}^{-1}\end{pmatrix}$

The product of this and the first block operator column of $\mathcal{C}$ needs
to equate to $0$, i.e.,

$DA_{1}+(-1)^{r-1}A_{r}^{-1}B_{r-1}A_{r-1}^{-1}\cdots B_{2}A_{2}^{-1}B_{1}=0.$

Hence,

$D=(-1)^{r}A_{r}^{-1}B_{r-1}A_{r-1}^{-1}\cdots B_{2}A_{2}^{-1}B_{1}A_{1}^{-1}$

as required. ∎

###  13.2 Results in Section 6

The technique used for the proof of Theorem 6.5 is akin to parts of the proof
of [43, Theorem 2.13].

###### Proof of Theorem 6.5.

We denote by $-\Delta_{d-1}$ the Dirichlet–Laplace operator on
$L_{2}(\widehat{\Omega})$, which is a strictly positive selfadjoint operator
with compact resolvent; it is well-known that the corresponding eigenvalues
are discrete and form a discrete sequence, see [27, Section 6.5.1]. The
operator $D_{{\rm red},\alpha}-\alpha\Delta_{d-1}$ is essentially self-
adjoint. Indeed, one computes the adjoint using
$(1-\varepsilon\Delta_{d-1})^{-1}\to 1$ as $\varepsilon\to 0$ as follows.
Beforehand, note that $(1-\varepsilon\Delta_{d-1})^{-1}$ commutes with
$(D_{{\rm red},\alpha}-\alpha\Delta_{d-1})$ in the sense that

$(1-\varepsilon\Delta_{d-1})^{-1}(D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})\subseteq(D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})(1-\varepsilon\Delta_{d-1})^{-1};$

this follows from the fact that both $\alpha$ and $D_{{\rm red},\alpha}$ are
independent of the $(x_{2},\ldots,x_{d})$-variables and that
$(1-\varepsilon\Delta_{d-1})^{-1}$ is a (multiple of a) resolvent of
$\Delta_{d-1}$ and, thus, commutes with $\Delta_{d-1}$.

Next, let $\varphi\in\operatorname{dom}((D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})^{*})$; for $\varepsilon>0$ define
$\varphi_{\varepsilon}\coloneqq(1-\varepsilon\Delta_{d-1})^{-1}\varphi$. Then
for all $u\in\operatorname{dom}(-D_{{\rm red},\alpha}-\alpha\Delta_{d-1})$, we
have

$\displaystyle\langle\varphi_{\varepsilon},(D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})u\rangle$
$\displaystyle=\langle(1-\varepsilon\Delta_{d-1})^{-1}\varphi,(D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})u\rangle$
$\displaystyle=\langle\varphi,(1-\varepsilon\Delta_{d-1})^{-1}(D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})u\rangle$
$\displaystyle=\langle\varphi,(D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})(1-\varepsilon\Delta_{d-1})^{-1}u\rangle$
$\displaystyle=\langle(D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})^{*}\varphi,(1-\varepsilon\Delta_{d-1})^{-1}u\rangle$
$\displaystyle=\langle(1-\varepsilon\Delta_{d-1})^{-1}(D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})^{*}\varphi,u\rangle.$

This shows that $\varphi_{\varepsilon}\in\operatorname{dom}((D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})^{*})$ and

$(D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})^{*}\varphi_{\varepsilon}=(1-\varepsilon\Delta_{d-1})^{-1}(D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})^{*}\varphi.$

The strong operator topology convergence of
$(1-\varepsilon\Delta_{d-1})^{-1}\to 1$ implies that
$\varphi_{\varepsilon}\to\varphi$ as $\varepsilon\to 0$ in the graph norm of
$(D_{{\rm red},\alpha}-\alpha\Delta_{d-1})^{*}$. Next, from the above
computation, we deduce using $\alpha\Delta_{d-1}=\Delta_{d-1}\alpha$

$\displaystyle\langle\varphi_{\varepsilon},D_{{\rm red},\alpha}u\rangle$
$\displaystyle=\langle\varphi_{\varepsilon},\alpha\Delta_{d-1}u\rangle+\langle(1-\varepsilon\Delta_{d-1})^{-1}(D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})^{*}\varphi,u\rangle$
$\displaystyle=\langle\alpha\Delta_{d-1}\varphi_{\varepsilon},u\rangle+\langle(1-\varepsilon\Delta_{d-1})^{-1}(D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})^{*}\varphi,u\rangle.$

Hence, since $\operatorname{dom}((D_{{\rm red},\alpha}-\alpha\Delta_{d-1}))$
is dense in the graph space of $D_{{\rm red},\alpha}$ (use again approximation
and commutativity properties of $(1-\varepsilon\Delta_{d-1})^{-1}$), we get

$\varphi_{\varepsilon}\in\operatorname{dom}((D_{{\rm
red},\alpha})^{*})=\operatorname{dom}(D_{{\rm red},\alpha})$

and

$D_{{\rm
red},\alpha}\varphi_{\varepsilon}=\alpha\Delta_{d-1}\varphi_{\varepsilon}+(1-\varepsilon\Delta_{d-1})^{-1}(D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})^{*}\varphi.$

Thus,

$(D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})^{*}\varphi_{\varepsilon}=(1-\varepsilon\Delta_{d-1})^{-1}(D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})^{*}\varphi=D_{{\rm
red},\alpha}\varphi_{\varepsilon}-\alpha\Delta_{d-1}\varphi_{\varepsilon},$

which implies that $D_{{\rm red},\alpha}-\alpha\Delta_{d-1}$ is essentially
self-adjoint. Next, it is elementary that

$D_{{\rm red},\alpha}-\alpha\Delta_{d-1}\subseteq-\Delta_{\alpha}.$

Hence,

$\displaystyle D_{{\rm
red},\alpha}-\alpha\Delta_{d-1}\subseteq\overline{D_{{\rm
red},\alpha}-\alpha\Delta_{d-1}}\subseteq-\Delta_{\alpha}$
$\displaystyle-\Delta_{\alpha}^{*}\subseteq(D_{{\rm
red},\alpha}-\alpha\Delta_{d-1})^{*}=\overline{D_{{\rm
red},\alpha}-\alpha\Delta_{d-1}}.$

Consequently,

$\overline{D_{{\rm red},\alpha}-\alpha\Delta_{d-1}}=-\Delta_{\alpha},$

and the statement follows. ∎

###  13.3 Results in Section 7

###### Proof of Theorem 7.1.

(i)$\Leftrightarrow$(ii) This is a standard fact for self-adjoint operators.
For convenience of the reader, we present an argument invoking von Neumann’s
spectral theorem. By Theorem 3.1 using that $\alpha$ is real,
$D_{\alpha}+\alpha\lambda_{k}$ is a self-adjoint operator. Moreover,
$D_{\alpha}$ has compact resolvent. Hence, by von Neumann’s the spectral
theorem, there is a sequence $V\colon\mathbb{N}\to\mathbb{R}$ and a unitary
operator $U\colon L_{2}(0,1)\to\ell_{2}(\mathbb{N})$ such that
$D_{\alpha}+\alpha\lambda_{k}=U^{*}M_{V}U$, where $M_{V}$ is the operator of
multiplying by $V$ on $\ell_{2}(\mathbb{N})$ with maximal domain. Thus,
$D_{\alpha}+\alpha\lambda_{k}$ is invertible if and only if $M_{V}$ is.
Moreover, in this case, $\|M_{V}^{-1}\|\leqslant C$ is equivalent to
$\|\frac{1}{|V|}\|_{\infty}\leqslant C$. Thus, for all $j\in\mathbb{N}$,
$\frac{1}{|V(j)|}\leqslant C$, thus, for all $\delta\in(-1/C,1/C)$

$|V(j)-\delta|\geqslant|V(j)|-|\delta|\geqslant\frac{1}{|C|}-|\delta|>0,$

implying that
$(-1/C,1/C)\subseteq\rho(M_{V})=\rho(D_{\alpha}+\alpha\lambda_{k})$. If, now,
$(-1/C,1/C)\subseteq\rho(D_{\alpha}+\alpha\lambda_{k})$, it follows that
$\emptyset=(-1/C,1/C)\cap\sigma(D_{\alpha}+\alpha\lambda_{k})=\\{V(j);j\in\mathbb{N}\\}\cap(-1/C,1/C)$.
Thus, $\|(D_{\alpha}+\alpha\lambda_{k})^{-1}\|=\|1/V\|_{\infty}\leqslant C$.

(ii)$\Leftrightarrow$(iii) (ii) holds if, and only if, for all
$\delta\in[-1,1]$, the operator

$D_{\alpha}+\alpha\lambda_{k}+\tfrac{1}{C}\delta=D_{\alpha}+\alpha(\lambda_{k}+\tfrac{1}{\alpha
C}\delta)$

is continuously invertible. Thus, the equivalence asserted follows directly
from Theorem 5.4. ∎

###### Proof of Proposition 7.3.

(a) The formula is a straightforward consequence of the formula for
$p_{\alpha,\beta}$ taking into account that

$A^{(j)}=\begin{pmatrix}\cosh(m_{j}h)&\frac{1}{m_{j}}\sinh(m_{j}h)\\\
m_{j}\sinh(m_{j}h)&\cosh(m_{j}h)\end{pmatrix}.$

for all $j\in\\{0,\ldots,r\\}$ and factoring out $\cosh(m_{j}h)$.

(b) We treat the case $x=0$ first. By induction on $r$, one obtains for
$\mu\geqslant\mu_{0}$

$\begin{pmatrix}1&\alpha_{r}^{-1}\mu^{-1}\end{pmatrix}\begin{pmatrix}1&\alpha_{r-1}^{-1}\mu^{-1}\\\
\alpha_{r-1}\mu&1\end{pmatrix}\cdots\begin{pmatrix}1&\alpha_{1}^{-1}\mu^{-1}\\\
\alpha_{1}\mu&1\end{pmatrix}=\big{(}\prod_{j=1}^{r-1}(1+\alpha_{j+1}^{-1}\alpha_{j})\big{)}\begin{pmatrix}1&\alpha_{1}^{-1}\mu^{-1}\end{pmatrix}.$

Indeed, the base case $r=1$ being trivial, the inductive step is proved by
observing

$\begin{pmatrix}1&\alpha_{r}^{-1}\mu^{-1}\end{pmatrix}\begin{pmatrix}1&\alpha_{r-1}^{-1}\mu^{-1}\\\
\alpha_{r-1}\mu&1\end{pmatrix}=(1+\alpha_{r}^{-1}\alpha_{r-1})\begin{pmatrix}1&\alpha_{r-1}^{-1}\mu^{-1}\end{pmatrix}.$

Thus, we deduce

$\displaystyle\mu w(\mu,0)$
$\displaystyle=\mu\begin{pmatrix}1&\alpha_{r}^{-1}\mu^{-1}\end{pmatrix}\begin{pmatrix}1&\alpha_{r-1}^{-1}\mu^{-1}\\\
\alpha_{r-1}\mu&1\end{pmatrix}\cdots\begin{pmatrix}1&\alpha_{1}^{-1}\mu^{-1}\\\
\alpha_{1}\mu&1\end{pmatrix}\begin{pmatrix}\mu^{-1}\\\
\alpha_{0}\end{pmatrix}$
$\displaystyle=\mu\big{(}\prod_{j=1}^{r-1}(1+\alpha_{j+1}^{-1}\alpha_{j})\big{)}\begin{pmatrix}1&\mu^{-1}\alpha_{1}^{-1}\end{pmatrix}\begin{pmatrix}\mu^{-1}\\\
\alpha_{0}\end{pmatrix}$
$\displaystyle=\prod_{j=0}^{r-1}(1+\alpha_{j+1}^{-1}\alpha_{j}).$

Next, for $x\in[-\mu_{0}/2,\mu_{0}/2]^{r+1}$, $\mu w(\mu,x)$ is a rational
function in $\mu$ with leading order coefficients of both the enumerator and
denominator polynomials coinciding with the respective ones from $\mu
w(\mu,0)$. Since $x$ comes from a compact, thus, bounded subset of
$\mathbb{R}^{r+1}$, the convergence is uniform in $x$.

(c) For the claim, it suffices to show the assertion concerning the limit;
$q(\mu,\cdot)\in\mathcal{O}(\tfrac{1}{\mu})$ is then a consequence of (b). For
all $s>0$

$\tanh(s)-1=\frac{1}{e^{s}\cosh(s)}.$

Thus, for all $j\in\\{0,\ldots,r\\}$

$t_{j}-1=\frac{1}{e^{m_{j}h}\cosh(m_{j}h)}.$

Let $(\mu,x)\in M_{\mu_{0}}$ and $(m_{0},\ldots,m_{r})\coloneqq S(\mu,x)$. For
$j\in\\{0,\ldots,r\\}$ let $s_{j}\in[0,1]$ and consider

$\displaystyle f_{r}(s_{r})$
$\displaystyle\coloneqq\begin{pmatrix}1&\alpha_{r}^{-1}m_{r}^{-1}s_{r}\end{pmatrix}$
$\displaystyle f_{0}(s_{0})$
$\displaystyle\coloneqq\begin{pmatrix}m_{0}^{-1}s_{0}\\\
\alpha_{0}\end{pmatrix}$ $\displaystyle f_{k}(s_{k})$
$\displaystyle\coloneqq\begin{pmatrix}1&\alpha_{k}^{-1}m_{k}^{-1}s_{k}\\\
\alpha_{k}m_{k}s_{k}&1\end{pmatrix},$

where $k\in\\{1,\ldots,r-1\\}$. Then,

$\displaystyle q(\mu,x)-w(\mu,x)$
$\displaystyle=f_{r}(t_{r})f_{r-1}(t_{r-1})\cdots
f_{0}(t_{0})-f_{r}(1)f_{r-1}(1)\cdots f_{0}(1)$
$\displaystyle=\sum_{j=0}^{r}f_{r}(t_{r})f_{r-1}(t_{r-1})\cdots
f_{j+1}(t_{j+1})(f_{j}(t_{j})-f_{j}(1))f_{j-1}(1)\cdots f_{0}(1).$

Since for all $j\in\\{0,\ldots,r\\}$

$\displaystyle\|f_{j}(t_{j})-f_{j}(1)\|$ $\displaystyle\leqslant
c_{j}\mu|t_{j}-1|$ $\displaystyle\|f_{j}(t_{j})\|$ $\displaystyle\leqslant
c_{j}\mu$

for some $c_{j}>0$ (depending on $\mu_{0}$, $\alpha$) independently of $\mu$
and $x$, we obtain with $c\coloneqq\max_{j}c_{j}$

$|\mu(q(\mu,x)-w(\mu,x))|\leqslant(r+1)c^{r+1}\mu^{n+2}\max_{j}|t_{j}-1|.$

Since, however, for all $j\in\\{0,\ldots,r\\}$, $|t_{j}-1|\to 0$ exponentially
fast as $\mu\to\infty$ by our preliminary observation, we get the desired
assertion.

(d) Since $C_{c}(\mathbb{R}^{r+1})\cap C(M_{\mu_{0}})$ consists of uniformly
continuous functions, we obtain that $\overline{C_{c}(\mathbb{R}^{r+1})\cap
C(M_{\mu_{0}})}^{\|\cdot\|_{\infty}}=C_{0}(\mathbb{R}^{r+1})\cap
C(M_{\mu_{0}})=C_{0}(M_{\mu_{0}})$ is uniformly continuous. By (b) and (c), we
obtain uniform continuity of

$M_{\mu_{0}}\ni(\mu,x)\mapsto\mu
q(\mu,x)-\prod_{j=0}^{r-1}(1+\alpha_{j+1}^{-1}\alpha_{j})\in
C_{0}(M_{\mu_{0}}),$

which implies the assertion. ∎

The proof of Theorem 7.5 requires a preparation.

######  Lemma 13.4.

(a) Let $t>0$, $\alpha_{0},\ldots,\alpha_{r}\in\mathbb{R}\setminus\\{0\\}$.
Then for all $\mu>0$,

$p_{\mu\alpha}(t)=p_{\alpha}(t).$

(b) Let $\mu>0$ and $t=\tanh(\mu h)$. Then the following conditions are
equivalent

  1. (i)

$\tilde{q}(\mu,\ldots,\mu)\neq 0$;

  2. (ii)

$p_{\alpha}(t)\neq 0$.

###### Proof.

(a) For all $\mu>0$, we compute

$\begin{pmatrix}\ast&\mu^{-1}\ast\\\
\mu\ast&\ast\end{pmatrix}\begin{pmatrix}\ast&\mu^{-1}\ast\\\
\mu\ast&\ast\end{pmatrix}=\begin{pmatrix}\ast&\mu^{-1}\ast\\\
\mu\ast&\ast\end{pmatrix},$

where the $\ast$-symbols symbolise possibly different arbitrary numbers
independent of $\mu$.

Then, using the preliminary observation, we deduce by induction on $r$, that

$\displaystyle p_{\mu\alpha}(t)$
$\displaystyle=\begin{pmatrix}1&\mu^{-1}\alpha_{r}^{-1}t\end{pmatrix}\begin{pmatrix}\ast&\mu^{-1}\ast\\\
\mu\ast&\ast\end{pmatrix}\begin{pmatrix}t\\\ \mu\alpha_{0}\end{pmatrix}$
$\displaystyle=\begin{pmatrix}\ast&\mu^{-1}\ast\end{pmatrix}\begin{pmatrix}t\\\
\mu\alpha_{0}\end{pmatrix}=\ast.$

(b) We compute, using (a),

$\mu\tilde{q}(\mu,\ldots,\mu)=p_{\mu\alpha}(t)=p_{\alpha}(t),$

which yields the assertion. ∎

###### Proof of Theorem 7.5.

(i)$\Rightarrow$(ii) Put $d=0$ in (3), the equation defining the
$\tilde{q}$-criterion.

(ii)$\Leftrightarrow$(iii), (iv)$\Leftrightarrow$(v) This is Lemma 13.4 (b).

(iv)$\Rightarrow$(i) Since $\chi(\alpha)\neq 0$, by Theorem 7.4, we find
$0<\delta_{1}\leqslant 1$ and $k_{0}\in\mathbb{N}$ such that for all
$k\geqslant k_{0}$ and $x\in[-\delta_{1},\delta_{1}]^{r+1}$, we have
$q(\mu_{k},x)\neq 0$. Consider the Taylor series
$\sum_{\ell=0}^{\infty}c_{\ell}\xi^{\ell}=\sqrt{1+\xi}$ for $|\xi|\leqslant
1$. It is known that the series converges absolutely for $|\xi|\leqslant c$
for all $c<1$. In particular,

$\kappa\coloneqq\sup_{\xi\in[-1/2,1/2]}\sum_{\ell=1}^{\infty}c_{\ell}\xi^{\ell-1}<\infty.$

We find $k_{1}\geqslant k_{0}$ such that for all $k\geqslant k_{1}$ we have
$1/\lambda_{k}\leqslant\min\\{1/2,\frac{\kappa}{\delta_{1}}\\}$. Thus, for
$k\geqslant k_{1}$ and $-1\leqslant s\leqslant 1$ we compute

$\displaystyle|\sqrt{\lambda_{k}+s}-\mu_{k}|$
$\displaystyle=|\sqrt{\lambda_{k}}\sqrt{1+\frac{s}{{\lambda_{k}}}}-\sqrt{\lambda_{k}}|=\sqrt{\lambda_{k}}\big{|}\sum_{\ell=1}^{\infty}c_{\ell}\left(\frac{s}{{\lambda_{k}}}\right)^{\ell}\big{|}\leqslant\frac{1}{\sqrt{\lambda_{k}}}\kappa\leqslant\delta_{1}.$

Since $q(\mu_{k},0)\neq 0$ for all $k\in\\{0,\ldots,k_{1}\\}$, by continuity
of $q$ and $\sqrt{\cdot}$, we find $0<\delta_{0}\leqslant\delta_{1}$ so that
$\alpha_{0},\ldots,\alpha_{r}$ and $(\lambda_{k})_{k}$ satisfy the
$\tilde{q}$-criterion. ∎

###  13.4 Results in Section 12

###### Proof of Lemma 12.2.

Let $q=\operatorname{grad}u,r=\operatorname{grad}v\in
g_{0}(\Omega)=\operatorname{grad}[H_{0}^{1}((0,1)^{d})]$. We compute

$\displaystyle q=(\iota_{0}^{*}\Gamma\iota_{0})^{-1}r$
$\displaystyle\iff\iota_{0}^{*}\Gamma\iota_{0}\iota_{0}^{*}q=\iota_{0}^{*}r$
$\displaystyle\iff\iota_{0}^{*}\Gamma\iota_{0}\iota_{0}^{*}\operatorname{grad}u=\iota_{0}^{*}\operatorname{grad}v$
$\displaystyle\iff\operatorname{div}\iota_{0}\iota_{0}^{*}\Gamma\iota_{0}\iota_{0}^{*}\operatorname{grad}u=\operatorname{div}\iota_{0}\iota_{0}^{*}\operatorname{grad}v$
$\displaystyle\iff\operatorname{div}\Gamma\operatorname{grad}u=\operatorname{div}\operatorname{grad}v$
$\displaystyle\iff
u=(\operatorname{div}\Gamma\operatorname{grad})^{-1}\operatorname{div}\operatorname{grad}v$
$\displaystyle\iff\iota_{0}^{*}q=\iota_{0}^{*}\operatorname{grad}u=\iota_{0}^{*}\operatorname{grad}(\operatorname{div}\Gamma\operatorname{grad})^{-1}\operatorname{div}\iota_{0}\iota_{0}^{*}r.$

Thus,
$(\iota_{0}^{*}\Gamma\iota_{0})^{-1}=\iota_{0}^{*}\operatorname{grad}(\operatorname{div}\Gamma\operatorname{grad})^{-1}\operatorname{div}\iota_{0}$.
Next, we derive a more explicit description of this expression. For this, we
use the polar decomposition for unbounded operators, see, e.g., [40,
Proposition B.8.6] for

$\operatorname{grad}\colon H_{0}^{1}(\Omega)\subseteq L_{2}(\Omega)\to
L_{2}(\Omega)^{d}.$

Thus, we find a unitary operator
$U\colon\operatorname{ran}(|\operatorname{grad}|)\to\operatorname{ran}(\operatorname{grad})$
such that

$\iota_{0}^{*}\operatorname{grad}=U|\operatorname{grad}|;$

as a consequence
$\operatorname{div}\iota_{0}=-(\iota_{0}^{*}\operatorname{grad})^{*}=|\operatorname{grad}|U^{*}$.
Hence, we need to evaluate

$|\operatorname{grad}|(\operatorname{div}\Gamma\operatorname{grad})^{-1}|\operatorname{grad}|.$

As the latter operator is bounded, it suffices to compute its action on the
dense subset $\operatorname{dom}(|\operatorname{grad}|^{2})$. In order to
derive an expression for this operator, we employ the spectral theorem for the
Dirichlet–Laplace, $-|\operatorname{grad}|^{2}=\Delta$ on $L_{2}((0,1)^{d})$
and the perturbed Dirichlet–Laplace
$\Delta^{(\gamma)}\coloneqq\operatorname{div}\Gamma\operatorname{grad}$ on
$L_{2}((0,1)^{d})$. Since $\Gamma$ is a diagonal matrix, it follows that the
orthonormal basis of eigenfunctions

$(e_{k})_{k\in\mathbb{N}^{d}}=((x_{1},\ldots,x_{d})\mapsto\frac{1}{2^{d/2}}\sin(k_{1}\pi
x_{1})\cdots\sin(k_{d}\pi x_{d}))_{k\in\mathbb{N}^{d}}$

for $-\Delta$ is also one for $-\Delta^{(\gamma)}$. The difference, however,
being the eigenvalues, which in the former case are
$\lambda_{k}=\pi^{2}\sum_{m=1}^{d}k_{m}^{2}$ and in the latter
$\lambda_{k}^{(\gamma)}=\pi^{2}(\gamma k_{1}^{2}+\sum_{m=2}^{d}k_{m}^{2})$ for
all $k\in\mathbb{N}^{d}_{>0}$ (in order for $\lambda_{k}^{(\gamma)}>0$, one
requires $\gamma>0$.) Thus,

$|\operatorname{grad}|(\operatorname{div}\Gamma\operatorname{grad})^{-1}|\operatorname{grad}|$

is unitarily equivalent to the diagonal operator on
$L(\ell_{2}(\mathbb{N}^{d}_{>0}))$ of multiplying entry-wise with

$\mathbb{N}_{>0}^{d}\ni
k\mapsto\frac{\sqrt{\sum_{m=1}^{d}k_{m}^{2}}\sqrt{\sum_{m=1}^{d}k_{m}^{2}}}{\gamma
k_{1}^{2}+\sum_{m=2}^{d}k_{m}^{2}}=\frac{\sum_{m=1}^{d}k_{m}^{2}}{\gamma
k_{1}^{2}+\sum_{m=2}^{d}k_{m}^{2}}\qed$

##  14 Conclusion

We presented a new method for homogenisation. This method is predominantly
tailored for self-adjoint and sign-changing coefficients with $0$ contained in
the (essential) spectrum of the corresponding divergence form operator. The
application of this new method required a detailed analysis for when
divergence form as well as certain Sturm–Liouville operators with sign-
changing coefficients lead to continuously invertible operators. The resulting
condition on zeros of certain polynomials was used to identify the inner
spectrum of the considered conductivities. For classical homogenisation
problems, it was shown that these inner spectra behave rather tame comparing
the pre-asymptotics to the homogenised coefficients. In the non-standard
setting, which requires the new homogenisation method, the holomorphic
$G$-limits are entirely unexpected as the homogenisation of a standard
divergence form equation with local but sign-changing coefficients can be
shown to lead in certain cases to a limit equation which is both nonlocal and
– in contrast to the second order equation one started out with – of 4th
order. This is particularly marked by a discontinuous behaviour of the inner
spectra.

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Marcus Waurick
Institute for Applied Analysis
Faculty of Mathematics and Computer Science
TU Bergakademie Freiberg
Freiberg, Germany
Email: marcus.wauhugo<EMAIL_ADDRESS>