$\displaystyle=0\quad\mbox{in}\quad\Omega\times\mathbb{R}_{>0},\qquad
z(\cdot,0)=z^{0}\quad\mbox{in}\quad\Omega,\qquad\frac{\partial
z}{\partial\mathbf{n}}=0\quad\mbox{on}\quad\partial\Omega\times\mathbb{R}_{>0}.$

The solution is given by the heat semigroup
$(S_{\Delta}(t))_{t\in\mathbb{R}_{\geq 0}}\subset\mathcal{L}(L^{2}(\Omega))$,

$\displaystyle z(x,t)=(S_{\Delta}(t)z^{0})(x),\qquad x\in\Omega.$ (A.1)

Moreover, the heat semigroup defined by (A.1) can be extended to a contraction
semigroup $(S_{\Delta,p}(t))_{t\in\mathbb{R}_{\geq 0}}$ on $L^{p}(\Omega)$ for
each $1\leq p\leq\infty$, which is strongly continuous for $1\leq p<\infty$.
For $1<p<\infty$, the semigroup is analytic and $S_{\Delta,p}(t)$ is compact
for $t>0$. Consequently, the generator $A_{p}:\mathcal{D}(A_{p})\subset
L^{p}(\Omega)\to L^{p}(\Omega)$ of $(S_{\Delta,p}(t))_{t\in\mathbb{R}_{\geq
0}}$ has compact resolvent for $1<p<\infty$.

###### Proof.

Contractivity of the semigroup is shown in [13, Theorem 1.3.9]. By [13,
Theorems 1.4.1, 1.4.2], $(S_{\Delta}(t))_{t\in\mathbb{R}_{\geq 0}}$ is a
strongly continuous semigroup for each $1\leq p<\infty$, which is even
analytic for $p>1$. Compactness of the semigroup operators follows from [13,
Theorem 1.6.3]. Finally, compactness of the resolvent $R(\lambda,A_{p})$ for
each $\lambda\in\rho(A_{p})$ is a consequence of [14, Ch. II, Theorem 4.29]. ∎

For $A_{2}$, the domain $\mathcal{D}(A_{2})$ is given in [13] by the following
characterization of a weak solution, i.e.,

$\displaystyle w\in\mathcal{D}(A_{2})$
$\displaystyle\quad\Longleftrightarrow\quad w\in W^{1,2}(\Omega)\;\text{and
there exists a function}\;f\in L^{2}(\Omega)\;\text{such that}$
$\displaystyle\qquad\qquad\quad(\nabla
w,\nabla\varphi)_{L^{2}(\Omega)}=(f,\varphi)_{L^{2}(\Omega)}\qquad\forall\;\varphi\in
W^{1,2}(\Omega).$

As usual in the weak sense, we identify $A_{2}w=f$ and obtain the graph norm
estimate

$\displaystyle\|w\|^{2}_{W^{1,2}(\Omega)}=\|w\|^{2}_{L^{2}(\Omega)}+(f,w)_{L^{2}(\Omega)}\leq
C\left(\|w\|^{2}_{L^{2}(\Omega)}+\|A_{2}w\|^{2}_{L^{2}(\Omega)}\right)\quad\forall\;w\in\mathcal{D}(A_{2}).$

From this characterization it is clear that the set

$H_{N}^{2}(\Omega):=\\{w\in
W^{2,2}(\Omega)\mid\partial_{\mathbf{n}}w=0\;\text{a.e. in}\;\Omega\\}$

is included in $\mathcal{D}(A_{2})$. In general, elliptic regularity is
restricted for low-regular boundaries such as $\partial\Omega\in C^{0,1}$ [16,
Section 4.4.3]. By the characterization of contraction semigroups via maximal
dissipative generators in [5, Theorem 3.4.5], we obtain
$H_{N}^{2}(\Omega)\subsetneq\mathcal{D}(A_{2})$ for certain Lipschitz
boundaries. As shown in [52, Theorem 2.15], there holds
$\mathcal{D}(A_{2})=H_{N}^{2}(\Omega)$ in case of $\partial\Omega\in C^{1,1}$
since [16, Theorem 2.4.1.3] is used.

Let us next characterize the domain $\mathcal{D}(A_{p})$ of the generator
$A_{p}$ of the heat semigroup on $L^{p}(\Omega)$ for $p>2$. We mention that,
for Lipschitz domains, we might have elements in $\mathcal{D}(A_{p})$ which
are not in $W^{1,p}(\Omega)$, although these functions are Hölder continuous
[33, 50].

######  Lemma A.2.

Let $2\leq p<\infty$ and the semigroup
$(S_{\Delta,p}(t))_{t\in\mathbb{R}_{\geq 0}}$ be defined as in Proposition A.1
with generator $A_{p}:\mathcal{D}(A_{p})\subset L^{p}(\Omega)\to
L^{p}(\Omega)$. Then, $A_{p}$ is the part of the operator $A_{2}$ in
$L^{p}(\Omega)$, i.e.,

$\displaystyle\mathcal{D}(A_{p})=\\{w\in\mathcal{D}(A_{2})\cap
L^{p}(\Omega)\mid A_{2}w\in L^{p}(\Omega)\\}.$ (A.2)

Moreover, for $p>n/2$, the embedding $\mathcal{D}(A_{p})\hookrightarrow
C^{0,\gamma}(\overline{\Omega})$ is continuous for some $\gamma>0$ with

$\displaystyle\|w\|_{C^{0,\gamma}(\overline{\Omega})}\leq
C\left(\|w\|_{L^{p}(\Omega)}+\|A_{p}w\|_{L^{p}(\Omega)}\right)\qquad\forall\;w\in\mathcal{D}(A_{p}).$
(A.3)

###### Proof.

Since the semigroups on $L^{p}(\Omega)$ are consistent due to [13, Theorem
1.4.1], we obtain that the heat semigroup on $L^{p}(\Omega)$ is the
restriction of the semigroup $(S_{\Delta}(t))_{t\in\mathbb{R}_{\geq 0}}$
defined on $L^{2}(\Omega)$ onto $L^{p}(\Omega)$ for $p\geq 2$. Hence, [35, Ch.
4, Theorem 5.5] shows equality (A.2).
The domain $\mathcal{D}(A_{2})$ is implicitly defined by an elliptic problem.
Hence, we focus on the problem $A_{2}w=f$ for $f\in L^{p}(\Omega)$ in order to
prove an embedding for $\mathcal{D}(A_{p})$ endowed with the graph norm. We
apply [33, Proposition 3.6] for $p>n/2$ to obtain an embedding into the Hölder
space $C^{0,\gamma}(\overline{\Omega})$ for a certain $\gamma>0$ which depends
on $\Omega$. Estimate (A.3) follows from the continuous embedding
$L^{p}(\Omega)\hookrightarrow L^{2}(\Omega)$ for $p\geq 2$. ∎

Next we consider a restriction of the Laplace operator to the space
$L^{\infty}(\Omega)$, i.e., we define

$\displaystyle A_{\infty}:\mathcal{D}(A_{\infty})\subset L^{\infty}(\Omega)\to
L^{\infty}(\Omega)$

as the part of $A_{p}$ defined in Lemma A.2 in $L^{\infty}(\Omega)$. From

$\mathcal{D}(A_{\infty})=\\{u\in\mathcal{D}(A_{p})\cap L^{\infty}(\Omega)\mid
A_{p}u\in L^{\infty}(\Omega)\\}$

we infer that this operator is again closed as the generator $A_{p}$ of the
heat semigroup is closed on $L^{p}(\Omega)$ [5, Section 3.10, p. 184].
However, $A_{\infty}$ is not densely defined due to the inclusion
$\mathcal{D}(A_{p})\subset C(\overline{\Omega})$ from Lemma A.2. This means
that $(A_{\infty},\mathcal{D}(A_{\infty}))$ cannot be a generator of a
strongly continuous semigroup by [14, Ch. II, Theorem 1.4]. Note that
continuity of the heat semigroup restricted to $L^{\infty}(\Omega)$ only fails
in $t=0$ and also analyticity is preserved if we exclude $t=0$. In fact,
following [29, Corollary 3.1.24], there exists a corresponding resolvent
estimate for $A_{\infty}$ as for sectorial operators. Unfortunately, this
estimate is shown under higher regularity assumptions on the boundary. But
using heat kernel estimates from [34], we derive a corresponding resolvent
estimate next.

######  Lemma A.3.

The operator $(A_{\infty},\mathcal{D}(A_{\infty}))$ is closed on
$L^{\infty}(\Omega)$. Moreover, there exist an angle
$\frac{\pi}{2}<\omega<\pi$, a closed sector
$\Sigma_{\omega}:=\\{\lambda\in\mathbb{C}\mid|\mathrm{arg}\,\lambda|\leq\omega\\}$
and a constant $M>0$ such that
$\Sigma_{\omega}\setminus\\{0\\}\subset\rho(A_{\infty})$ holds with the
resolvent estimate

$\displaystyle\|R(\lambda,A_{\infty})\|\leq\frac{M}{|\lambda|}\qquad\forall\;\lambda\in\Sigma_{\omega}\setminus\\{0\\}.$

###### Proof.

It is well-known that there exists such a resolvent estimate for the spaces
$L^{p}(\Omega)$ for $p<\infty$ [52, Theorem 2.12]. Nevertheless, constants in
these estimates deteriorate as $p\to\infty$. Hence, we choose another strategy
which is used for continuous functions in [43, Section 5.5]. Therefore, let us
start with the analytic heat semigroup $(S_{\Delta}(t))_{t\in\mathbb{R}_{\geq
0}}$ generated by the Laplace operator defined on $L^{2}(\Omega)$, see
Proposition A.1. [13, Theorem 2.4.4] shows that
$(S_{\Delta}(t))_{t\in\mathbb{R}_{\geq 0}}$ is hypercontractive and maps
$L^{2}(\Omega)$ to $L^{\infty}(\Omega)$ for each $t>0$. Thus for each $t>0$,
$x\in\Omega$, the map $z\mapsto S_{\Delta}(t)z(x)\in L^{2}(\Omega)^{\ast}$ can
be represented by a function $p(t,x,\cdot)\in L^{2}(\Omega)$, the so called
heat kernel, such that

$(S_{\Delta}(t)z)(x)=\int_{\Omega}p(t,x,y)z(y)\;\mathrm{d}y\qquad\forall\;z\in
L^{2}(\Omega),x\in\Omega.$

We use an estimation of the heat kernel from [13, Theorem 3.2.9], more
precisely, there exist constants $C_{1}$, $C_{2}>0$ such that

$0\leq p(t,x,y)\leq
C_{1}m(t)^{-n/2}\exp\left(\frac{-|x-y|^{2}}{C_{2}t}\right)\qquad\forall\;x,y\in\Omega,t\in\mathbb{R}_{>0},$

where $m(t)=\min\\{1,t\\}$. By [34, Theorem 7.2], this estimate can be
extended to some sector $\Sigma_{\theta_{1}}\setminus\\{0\\}\subset\mathbb{C}$
for $0<\theta_{1}<\frac{\pi}{2}$, where
$\Sigma_{\theta_{1}}=\\{\lambda\in\mathbb{C}\mid|\mathrm{arg}\,\lambda|\leq\theta_{1}\\}$.
Indeed, we have $\cos(\theta_{1})\leq\cos(\theta)\leq 1$ for each argument
$\theta=\mathrm{arg}\,z$ of $z\in\Sigma_{\theta_{1}}$ with $\mathrm{Re}\,z>0$.
Hence, according to [34, Theorem 7.2], for each $0<\theta_{1}<\frac{\pi}{2}$
there exist constants $C_{1},C_{2}>0$ such that

$\displaystyle 0\leq p(z,x,y)\leq
C_{1}|z|^{-n/2}\exp\left(\frac{-|x-y|^{2}}{C_{2}|z|}\right)\quad\forall\;x,y\in\Omega,z\in\Sigma_{\theta_{1}}\setminus\\{0\\}.$
(A.4)

Such an estimate for the heat kernel can be used to obtain resolvent
estimates, since

$\displaystyle
R(\lambda,A_{2})=\int_{0}^{\infty}\mathrm{e}^{-\lambda\tau}S_{\Delta}(\tau)\;\mathrm{d}\tau\quad\forall\;\lambda\in\mathbb{C},\mathrm{Re}\,\lambda>0$
(A.5)

holds for the generator $A_{2}$ of the heat semigroup
$(S_{\Delta}(t))_{t\in\mathbb{R}_{\geq 0}}$ from Proposition A.1 [14, Ch. II,
Theorem 1.10]. Such estimates are also derived in [43, Theorem 5.7] under
higher regularity assumptions. Note that the spectrum of the Laplacian $A_{p}$
considered on $L^{p}(\Omega)$ is independent of $2\leq p<\infty$ by [3,
Example 1.2], and there holds
$R(\lambda,A_{2})_{|L^{p}(\Omega)}=R(\lambda,A_{p})$ for each
$\lambda\in\rho(A_{2})=\rho(A_{p})$. Furthermore, as in the proof of
Proposition 2.8, one can show $\rho(A_{p})=\rho(A_{\infty})$ and
$R(\lambda,A_{\infty})=R(\lambda,A_{p})_{|L^{\infty}(\Omega)}$ for
$p>n^{\ast}$. In this way, we use the Laplace transform (A.5) of the heat
semigroup to deduce estimates of $R(\lambda,A_{p})$ for each $2\leq
p\leq\infty$.
Recall from [14, Ch. II, Theorem 3.5] that the contraction property of the
heat semigroup $(S_{\Delta}(t))_{t\in\mathbb{R}_{\geq 0}}$ in $L^{p}(\Omega)$
with $p<\infty$ implies

$\displaystyle\|R(\lambda,A_{2})\|_{\mathcal{L}(L^{p}(\Omega))}\leq\frac{1}{\mathrm{Re}\,\lambda}\quad\forall\;\lambda\in\mathbb{C},\mathrm{Re}\,\lambda>0.$
(A.6)

Taking the limit $p\to\infty$ in this resolvent estimate (A.6), we obtain the
same estimate for the resolvent restricted to $L^{\infty}(\Omega)$, compare
also [52, Ch. 2, Section 2.2]. Hence, as in the proof (b) $\Rightarrow$ (c) of
[35, Ch. 2, Theorem 5.2], it remains to verify a corresponding estimate for
the imaginary part of $\lambda\in\mathbb{C}$ with $\mathrm{Re}\,\lambda>0.$
Such an estimate can be deduced as in the proof (a) $\Rightarrow$ (b) of [35,
Ch. 2, Theorem 5.2] by shifting the path of integration in (A.5) to a ray in
the complex plane. Recall that $A_{p}$ is sectorial in the sense of [14, Ch.
II, Definition 4.1] since the heat semigroup is an analytic contraction
semigroup [14, Ch. II, Theorem 4.6]. Uniform boundedness of the semigroup with
respect to $L^{\infty}(\Omega)$ in some closed sector in $\mathbb{C}$ can be
deduced from [34, Theorem 7.4], using heat kernel estimate (A.4). Following
the proof of [35, Ch. 2, Theorem 5.2], we find a certain angle
$0<\delta<\pi/2$ and for each small $\varepsilon>0$ there is a constant
$M_{\varepsilon}>0$ such that

$\displaystyle\|R(\lambda,A_{2})\|_{\mathcal{L}(L^{\infty}(\Omega))}\leq\frac{M_{\varepsilon}}{|\lambda|}\quad\forall\;\lambda\in\Sigma_{\pi/2+\delta-\varepsilon}\setminus\\{0\\}.$
(A.7)

This is analog to [43, pp. 207-210] and [43, p. 216] for continuous functions
and $\theta_{0}=\frac{\pi}{2}-\delta+\varepsilon$. Finally, recall that
$\rho(A_{2})=\rho(A_{\infty})$ and
$R(\lambda,A_{\infty})=R(\lambda,A_{2})_{|L^{\infty}(\Omega)}$ for
$\lambda\in\Sigma_{\pi/2+\delta-\varepsilon}\setminus\\{0\\}$. This shows that
estimate (A.7) holds for $R(\lambda,A_{\infty})$. ∎

Finally, the Laplace operator can be restricted to the space of uniformly
continuous functions, see [4, 6] and [29, Corollary 3.1.24].

######  Lemma A.4.

Let $A_{2}$ be the generator of the analytic heat semigroup
$(S_{\Delta}(t))_{t\in\mathbb{R}_{\geq 0}}$ defined in Proposition A.1 on
$L^{2}(\Omega)$ with domain $\mathcal{D}(A_{2})\subset W^{1,2}(\Omega)$. Then
the part of the Laplacian $A_{2}$ in $C(\overline{\Omega})$, denoted by
$A_{c}:\mathcal{D}(A_{c})\subset C(\overline{\Omega})\to
C(\overline{\Omega})$, generates a strongly continuous semigroup
$(S_{\Delta,c}(t))_{t\in\mathbb{R}_{\geq 0}}$ of contractions which is even
analytic on $C(\overline{\Omega})$. The domain of $A_{c}$ is defined by the
restriction

$\displaystyle\mathcal{D}(A_{c})$ $\displaystyle=\\{w\in\mathcal{D}(A_{2})\cap
C(\overline{\Omega})\mid A_{2}w\in C(\overline{\Omega})\\}.$ (A.8)

The operator $A_{c}$ is also the part of the operator $A_{p}$ defined in Lemma
A.2 for $p\geq 2$. Moreover, the semigroup
$(S_{\Delta,c}(t))_{t\in\mathbb{R}_{\geq 0}}$ is compact, i.e.,
$S_{\Delta,c}(t)$ is compact for each $t\in\mathbb{R}_{>0}$. Consequently, the
operator $A_{c}$ has compact resolvent on $C(\overline{\Omega})$.

###### Proof.

References and proofs can be found in [6, Proposition 3, Theorem 4] and [4,
Ch. I, Section 2.6]. That $A_{c}$ is a restriction of the Laplace operator
$A_{p}$ defined on $L^{p}(\Omega)$ is a consequence of elliptic regularity and
the fact that $C(\overline{\Omega})\subset L^{p}(\Omega)$ for all $p\geq 2$,
compare Lemma A.2 and Lemma A.3. Finally, compactness of the resolvent
$R(\lambda,A_{c})$ for each $\lambda\in\rho(A_{c})$ is a consequence of [14,
Ch. II, Theorem 4.29]. ∎

##  Appendix B Spectral characterization of the reaction-diffusion-ODE
operator

We already characterized the spectrum of the operator $\mathbf{L}$ in
Proposition 2.3 by splitting the spectrum into the two components

$\displaystyle\sigma(\mathbf{L})=\sigma(\mathbf{A}_{\ast})\mathop{\dot{\cup}}\Sigma$

where $\Sigma$ is a set of eigenvalues and $\sigma(\mathbf{A}_{\ast})$ is the
spectrum of the corresponding multiplication operator $\mathbf{A}_{\ast}$ on
$L^{\infty}(\Omega)^{m}$. In the case where the resolvent set
$\rho(\mathbf{A}_{\ast})$ is connected, then [24, Theorem 10.1.3 (i)]
additionally yields that $\Sigma$ is a discrete set of eigenvalues. However,
when $\rho(\mathbf{A}_{\ast})$ is not a connected set, $\Sigma$ might include
a non-discrete set of the point spectrum of $\mathbf{L}$ [18, Example 4.3]. We
aim to further characterize the set $\Sigma$ of eigenvalues of $\mathbf{L}$
and to show that the spectral bound of $\mathbf{L}$ is determined by the
spectral bound $s(\mathbf{A}_{\ast})$ and the spectral bound of a discrete set
of eigenvalues within the unbounded connected component $\Lambda_{\infty}$ of
$\rho(\mathbf{A}_{\ast})$.

######  Proposition B.1.

Let $m,k\in\mathbb{N}$ and $\mathbf{D}^{v}\in\mathbb{R}_{>0}^{k\times k}$. Let
$\mathbf{L}$ be the partly diffusive operator defined in (2.5) on
$L^{\infty}(\Omega)^{m}\times L^{p}(\Omega)^{k}$ for bounded coefficient
matrices
$\mathbf{A}_{\ast},\mathbf{B}_{\ast},\mathbf{C}_{\ast},\mathbf{D}_{\ast}$ and
a finite $n^{\ast}<p<\infty$. Then

$\Sigma:=\sigma(\mathbf{L})\cap\rho(\mathbf{A}_{\ast})=\sigma_{d}(\mathbf{L})\;\dot{\cup}\;\Sigma_{0}$

where $\sigma_{d}(\mathbf{L})$ is the discrete spectrum of $\mathbf{L}$ and
$\Sigma_{0}$ is a union of bounded (open) connected components of
$\rho(\mathbf{A}_{\ast})$ consisting of non-isolated eigenvalues of
$\mathbf{L}$. In case that $\rho(\mathbf{A}_{\ast})$ is a connected set in
$\mathbb{C}$, then $\Sigma_{0}=\emptyset$.

###### Proof.

By calculations in Corollary 2.4 we infer that, for
$\lambda\in\rho(\mathbf{A}_{\ast})$, invertibility of $\lambda I-\mathbf{L}$
is equivalent to invertibility of the elliptic operator

$\displaystyle\lambda
I-\mathbf{D}^{v}\Delta-\mathbf{D}_{\ast}-\mathbf{C}_{\ast}(\lambda
I-\mathbf{A}_{\ast})^{-1}\mathbf{B}_{\ast}.$

To distinguish discrete parts within the spectral set
$\Sigma=\sigma(\mathbf{L})\cap\rho(\mathbf{A}_{\ast})$ we apply an analytic
Fredholm theorem. For this reason we use that invertibility of this elliptic
operator is equivalent to invertibility of $I-\mathbf{C}(\lambda)$ where

$\displaystyle\mathbf{C}(\lambda)=R(\lambda_{0},L_{1})\left((\lambda_{0}-\lambda)I+\mathbf{C}_{\ast}(\lambda
I-\mathbf{A}_{\ast})^{-1}\mathbf{B}_{\ast}\right)$ (B.1)

for some $\lambda_{0}\in\rho(L_{1})$,
$L_{1}:=\mathbf{D}^{v}\Delta+\mathbf{D}_{\ast}$. From [13, Theorem 1.6.3] and
[14, Ch. II, Theorem 4.29], we obtain compactness of the resolvent
$R(\lambda_{0},L_{1})$ for $L_{1}$ defined on $L^{p}(\Omega)^{k}$. Hence,
$\mathbf{C}(\lambda)$ is compact for each $\lambda\in\rho(\mathbf{A}_{\ast})$
and $\mathbf{C}$ is analytic since the resolvent mapping of
$\mathbf{A}_{\ast}$ is. Note that for
$\lambda\in\rho(\mathbf{A}_{\ast})\cap\rho(L_{1})$ the operator
$\mathbf{C}(\lambda)$ can be chosen similar to [18, Theorem 4.2], i.e.,

$\mathbf{C}(\lambda)=R(\lambda,L_{1})\mathbf{C}_{\ast}(\lambda
I-\mathbf{A}_{\ast})^{-1}\mathbf{B}_{\ast}.$

In the case where $\rho(\mathbf{A}_{\ast})$ is connected, we have seen in
Proposition 2.3 that $\Sigma_{0}=\emptyset$. In the case where the open set
$\rho(\mathbf{A}_{\ast})$ is not connected, there exists an at most countable
number of open connected components $\Lambda_{i}$, $i\in I$, which are
pairwise disjoint and form a partition of $\rho(\mathbf{A}_{\ast})$. The
components are chosen to be maximal, i.e. for all $i,j\in I$ with $i\neq j$
the set $\Lambda_{i}\cup\Lambda_{j}$ is not connected. Moreover, there exists
exactly one unbounded connected component denoted by $\Lambda_{\infty}$ since
$\sigma(\mathbf{A}_{\ast})$ is bounded. The remaining $\Lambda_{i}$, $i\in
I\setminus\\{\infty\\}$, are bounded components. Now, we are able to apply the
analytic Fredholm theorem on each of the components $\Lambda_{i}$. Either
there holds $1\in\sigma(\mathbf{C}(\lambda))$ for all $\lambda\in\Lambda_{i}$
or there exists $\lambda\in\Lambda_{i}$ such that
$1\in\rho(\mathbf{C}(\lambda))$.
In the first case, we recall from Proposition 2.3 that
$\Lambda_{i}\subset\Sigma$ is a set of eigenvalues of $\mathbf{L}$. As
$\Lambda_{i}$ is an open subset of $\mathbb{C}$,
$\Lambda_{i}\subset\sigma_{e6}(\mathcal{L})$ is a set of non-isolated
eigenvalues, hence $\Lambda_{i}\cap\sigma_{d}(\mathbf{L})=\emptyset$. We
define the open set $\Sigma_{0}$ as the disjoint union of all such
$\Lambda_{i}$.
In the second case where $I-\mathbf{C}(\lambda)$ is invertible for some
$\lambda\in\Lambda_{i}$, the Gohberg-Shmul’yan theorem [24, Theorem 2.5.13]
applies and shows that $\Lambda_{i}\cap\Sigma$ is a discrete set of
eigenvalues of $\mathbf{L}$. Due to the separation of the connected components
by the closed set $\sigma(\mathbf{A}_{\ast})$, eigenvalues
$\lambda\in\Lambda_{i}\cap\Sigma$ are also isolated values of
$\sigma(\mathbf{L})$. Since $\lambda I-\mathbf{L}$ is a Fredholm operator of
index zero, we find $\Lambda_{i}\cap\Sigma\subset\sigma_{d}(\mathbf{L})$ for
each such component $\Lambda_{i}$. We recall from Proposition 2.3 that
$\sigma(\mathbf{A}_{\ast})\cap\sigma_{d}(\mathbf{L})=\emptyset$ and, hence, we
characterized all parts of $\sigma_{d}(\mathbf{L})$.
Note that in any case, the unbounded connected component $\Lambda_{\infty}$
satisfies
$\Lambda_{\infty}\cap\sigma_{p}(\mathbf{L})\subset\sigma_{d}(\mathbf{L})$. In
fact, since the multiplication operator $\mathbf{A}_{\ast}$ is bounded, the
operator $I-\mathbf{C}(\lambda)$ resp. $\lambda I-\mathbf{L}$ is invertible
for some large $\lambda\in\rho(\mathbf{L})\cap\rho(\mathbf{A}_{\ast})$ by [14,
Ch. II, Theorem 1.10]. ∎

When $\rho(\mathbf{A}_{\ast})$ is connected, then $\rho(\mathbf{A}_{\ast})$
coincides with the only unbounded connected component $\Lambda_{\infty}$. Then
it is well-known that the spectral bound of the reaction-diffusion-ODE
operator $\mathbf{L}$ is determined by the spectral bound of
$\mathbf{A}_{\ast}$ and the spectral bound of the discrete part
$\Sigma=\Lambda_{\infty}\cap\sigma_{p}(\mathbf{L})$ [32]. In general, the
spectrum $\sigma(\mathbf{L})$ might consist of further discrete or non-
discrete bounded components of $\rho(\mathbf{A}_{\ast})$. However, their
spectral bound can be controlled in terms of the spectral bound of the
surrounding spectrum $\sigma(\mathbf{A}_{\ast})$.

######  Corollary B.2.

Let the assumptions of Proposition B.1 hold true and denote by
$\Lambda_{\infty}$ the unbounded open connected component of
$\rho(\mathbf{A}_{\ast})$. Further, let us denote by

$s_{\infty}:=\sup\\{\mathrm{Re}\,\lambda\mid\lambda\in\Lambda_{\infty}\cap\sigma_{p}(\mathbf{L})\\}$

the spectral bound of the discrete set of eigenvalues within
$\Lambda_{\infty}$. Then the spectral bound of $\mathbf{L}$ is determined by

$\displaystyle s(\mathbf{L})=\max\\{s(\mathbf{A}_{\ast}),s_{\infty}\\}.$ (B.2)

###### Proof.

The spectral bound of $\mathbf{L}$ is given by
$s(\mathbf{L})=\sup\\{\mathrm{Re}\,\lambda\mid\lambda\in\sigma(\mathbf{L})\\}.$
Let us decompose $\rho(\mathbf{A}_{\ast})$ again into open connected
components $\Lambda\subset\mathbb{C}$, where exactly one of these is
unbounded, denoted by $\Lambda_{\infty}$. We define

$s_{b}=\sup\\{\mathrm{Re}\,\lambda\mid\lambda\in\Lambda\cap\sigma_{p}(\mathbf{L})\;\text{for
some bounded connected
component}\;\Lambda\;\text{of}\;\rho(\mathbf{A}_{\ast})\\}.$

By Proposition 2.3 and Proposition B.1 it is clear that
$s(\mathbf{L})=\max\\{s(\mathbf{A}_{\ast}),s_{\infty},s_{b}\\}.$ This proves
$\max\\{s(\mathbf{A}_{\ast}),s_{\infty}\\}\leq s(\mathbf{L})$. For the reverse
inequality we show $s_{b}\leq s(\mathbf{A}_{\ast})$. Therefore, let us take a
bounded open connected component $\Lambda$ of the resolvent set
$\rho(\mathbf{A}_{\ast})$ which is completely surrounded by
$\sigma(\mathbf{A}_{\ast})$. Then
$s(\Lambda):=\sup\\{\mathrm{Re}\,\lambda\mid\lambda\in\Lambda\\}$ cannot be
attained within the open set $\Lambda$, however, the supremum is attained on
the boundary $\partial\Lambda$ which is a subset of
$\sigma(\mathbf{A}_{\ast})$. Thus, we have $s_{b}\leq s(\mathbf{A}_{\ast})$
which proves the claim. ∎

Let us note that the discrete sets $\sigma_{d}(\mathbf{L})$ or
$\Lambda_{\infty}\cap\sigma_{p}(\mathbf{L})$ are not necessarily closed.
However, all accumulation points in $\mathbb{C}$ lie on the boundary of the
connected components in which the discrete sets are located. Since this
boundary is a subset of $\partial\rho(\mathbf{A}_{\ast})$, we infer that these
accumulation points of $\sigma_{d}(\mathbf{L})$ are included in
$\sigma(\mathbf{A}_{\ast})$. Compare also [25, Section 2.1.2] in the case
$\Sigma_{0}=\emptyset$.

##  Appendix C Existence of mild solutions

In the case of continuous functions, existence and uniqueness of mild
solutions for the corresponding equation (2.1) is well established [35, Ch. 6,
Theorem 1.4]. In this section, we present a proper notion and an existence and
uniqueness result of a mild solution to system (2.1) with discontinuous
components, i.e.,

$\displaystyle\frac{\partial\xi}{\partial
t}=\mathbf{L}\xi+\mathbf{N}(\xi),\qquad\xi(0)=\xi^{0}\in
L^{\infty}(\Omega)^{m+k}.$

This is in accordance with the theory developed in [37]. The analytic
semigroup $(\mathbf{T}(t))_{t\in\mathbb{R}_{\geq 0}}$ from Lemma 2.1 can be
restricted to $L^{\infty}(\Omega)^{m+k}$ by Lemma 2.11. However, strong
continuity of the semigroup is lost in the limit $p\to\infty$ [37, Part I,
Lemma 2]. Nevertheless, due to regularizing effects of the heat semigroup, a
solution to equation (2.1) can be obtained by a Picard iteration in
$L^{\infty}(\Omega)^{m+k}$ [37, Part II, Theorem 1]. The result in [37] is
based on an integral representation

$\displaystyle\xi(\cdot,t)=\mathbf{S}(t)\xi^{0}+\int_{0}^{t}\mathbf{S}(t-\tau)\big{(}\mathbf{N}(\xi(\cdot,\tau))+\mathbf{J}\xi(\cdot,\tau)\big{)}\;\mathrm{d}\tau,$
(C.1)

with the analytic contraction semigroup $(\mathbf{S}(t))_{t\in\mathbb{R}_{\geq
0}}$ generated by $\mathbf{D}\Delta$ on $L^{\infty}(\Omega)^{m}\times
L^{p}(\Omega)^{k}$, see (2.4). In order to relate stability properties of the
steady state to its linearization $\mathbf{L}$, it is necessary to obtain an
integral representation with the analytic semigroup
$(\mathbf{T}(t))_{t\in\mathbb{R}_{\geq 0}}$ generated by $\mathbf{L}$, i.e.,

$\displaystyle\xi(\cdot,t)=\mathbf{T}(t)\xi^{0}+\int_{0}^{t}\mathbf{T}(t-\tau)\mathbf{N}(\xi(\cdot,\tau))\;\mathrm{d}\tau.$
(C.2)

To show existence and uniqueness of a mild solution, we use estimate (2.8) for
the nonlinear term $\mathbf{N}$ as well as growth estimate (2.15) for the
semigroup $(\mathbf{T}(t))_{t\in\mathbb{R}_{\geq 0}}$ on
$L^{\infty}(\Omega)^{m+k}$. A solution of equation (2.1) is understood analog
to the mild sense of Rothe, compare [37, Part II, Definition 2].

######  Definition C.1.

Let $0<T\leq\infty$. A mild solution of problem (2.1) on the interval $[0,T)$
for initial datum $\xi^{0}\in L^{\infty}(\Omega)^{m+k}$ is a measurable
function $\xi:\Omega\times[0,T)\to\mathbb{R}^{m+k}$ satisfying for all
$t\in(0,T)$

  * (i)

$\xi(\cdot,t)\in L^{\infty}(\Omega)^{m+k}$ and
$\sup_{s\in(0,t)}\|\xi(\cdot,s)\|_{\infty}<\infty$;

  * (ii)

the integral representation

$\xi(\cdot,t)=\mathbf{T}(t)\xi^{0}+\int_{0}^{t}\mathbf{T}(t-\tau)\mathbf{N}(\xi(\cdot,\tau))\;\mathrm{d}\tau,$

where the integral is an absolutely converging Bochner integral in
$L^{\infty}(\Omega)^{m+k}$.

The local-in-time mild solution fulfills $\xi\in L^{\infty}(\Omega_{T})^{m+k}$
for each finite $T<T_{\max}$, where $[0,T_{\max})$ is the maximal time
interval of existence derived in the next Proposition C.2. This is achieved
via a Picard iteration in the Banach space
$(L^{\infty}(\Omega)^{m+k},\|\cdot\|_{\infty})$, similar to [37, Part II,
Theorem 1]. Boundedness of the solution and norm-continuity of the analytic
semigroup $(\mathbf{T}(t))_{t\in\mathbb{R}_{\geq 0}}$ for $t>0$ yields a mild
solution $\xi\in C([0,T_{\max});L^{\infty}(\Omega)^{m}\times
L^{p}(\Omega)^{k})$. The regularity of the semigroup for $p>n^{\ast}$ even
implies $\xi\in C((0,T_{\max});L^{\infty}(\Omega)^{m}\times
C(\overline{\Omega})^{k})$, as a consequence of Lemma 2.11. We refer to Remark
3.2 for more comments on the regularity of solutions depending on the
regularity of the initial condition.

######  Proposition C.2.

Let the nonlinear operator $\mathbf{N}$ defined in (2.1) satisfy the local
Lipschitz condition 1 and let the semigroup
$(\mathbf{T}(t))_{t\in\mathbb{R}_{\geq 0}}$ defined in Lemma 2.1 satisfy the
growth estimate $\|\mathbf{T}(t)\|_{\infty}\leq M\mathrm{e}^{wt}$ for some
$M\geq 1,w\in\mathbb{R}$. Then the following assertions hold.

  * (i)

For all $\xi^{0}\in L^{\infty}(\Omega)^{m+k}$ there exists a $0<T\leq\infty$,
such that the initial value problem (2.1) has a unique mild solution on the
time interval $[0,T)$.

  * (ii)

If we consider $T=T(\xi^{0})$ as a function of $\xi^{0}$, then it satisfies

$0<\inf\\{T(\xi^{0})\mid\xi^{0}\in
L^{\infty}(\Omega)^{m+k},\|\xi^{0}\|_{\infty}\leq U_{0}\\}$

for all $U_{0}\in[0,\infty)$.

  * (iii)

The existence interval can be chosen to be maximal, i.e., $\mathrm{(i)}$ holds
for $T\leq T_{\max}$, $T_{\max}\leq\infty$, and not for $T>T_{\max}$. If
$T_{\max}<\infty$, the solution $\xi$ satisfies

$\|\xi(\cdot,t)\|_{\infty}\to\infty\qquad(t\nearrow T_{\max}).$

###### Proof.

Let $\xi^{0}\in L^{\infty}(\Omega)^{m+k}$ satisfy $\|\xi^{0}\|_{\infty}\leq
U_{0}$ for some $U_{0}\geq 0$. Choose an arbitrary real number $U>MU_{0}$ and
consider the bounded set
$B=\overline{\Omega}\times\overline{B_{U}(\mathbf{0})}\subset\mathbb{R}^{n}\times\mathbb{R}^{m+k}$
on which the nonlinearity is bounded and Lipschitz continuous with some
constant $L(B)>0$ by Assumption 1. Herein, we denote $B_{U}(\mathbf{0})$ the
open ball with radius $U$ around $\mathbf{0}\in\mathbb{R}^{m+k}$. Choose $T>0$
such that

$MU_{0}\mathrm{e}^{wT}+\mathrm{e}^{(L(B)M+w)T}-1\leq U,$ (C.3)

where this particular choice is due to the following proof. As the case $w\leq
0$ is already considered in [37, Part II, Theorem 1], we assume $w\geq 0$ in
the following.
We use a Picard iteration to obtain the mild solution $\xi$. Define
$(\xi_{i})_{i\in\mathbb{N}}\subset L^{\infty}(\Omega_{T})^{m+k}$ recursively
by

$\displaystyle\xi_{1}(\cdot,t)=\mathbf{T}(t)\xi^{0},\qquad\xi_{i+1}(\cdot,t)=\mathbf{T}(t)\xi^{0}+\int_{0}^{t}\mathbf{T}(t-s)\mathbf{N}(\xi_{i}(\cdot,s))\;\mathrm{d}s$
(C.4)

for $i\in\mathbb{N},t\in[0,T]$. Recall that we use in fact the restricted
semigroup $(\mathbf{T}_{\infty}(t))_{t\in\mathbb{R}_{\geq 0}}$ on
$L^{\infty}(\Omega)^{m+k}$ from Lemma 2.11. In order to show well-definedness
of the approximating sequence $(\xi_{i})_{i\in\mathbb{N}}$, define the
differences

$\eta_{i}:[0,T]\to\mathbb{R}_{\geq 0},\quad
t\mapsto\|(\xi_{i+1}-\xi_{i})(\cdot,t)\|_{\infty}$

for each $i\in\mathbb{N}$. Our aim is to show the following inequalities

$\displaystyle\begin{split}\|\xi_{i}(\cdot,t)\|_{\infty}&\leq U,\\\
\eta_{i}(t)&\leq\alpha\int_{0}^{t}\mathrm{e}^{w(t-s)}\eta_{i-1}(s)\;\mathrm{d}s\qquad\text{where}\quad\eta_{0}:=1,\\\
\sum_{j=1}^{i}\eta_{j}(t)&\leq\mathrm{e}^{(\alpha+w)t}-1,\\\
\eta_{i}(t)&\leq\alpha\int_{0}^{t}\frac{(\alpha
s)^{i-1}}{(i-1)!}\mathrm{e}^{ws}\;\mathrm{d}s\end{split}$ (C.5)

for all $t\in[0,T],i\in\mathbb{N}$ and $\alpha:=ML(B)$ via induction. Norm-
continuity of the semigroup $(\mathbf{T}_{\infty}(t))_{t\in\mathbb{R}_{>0}}$
implies (Bochner) measurability of $\xi_{1}$ by [5, Corollary 1.1.2]. The
growth estimate (2.15) of the semigroup
$(\mathbf{T}_{\infty}(t))_{t\in\mathbb{R}_{\geq 0}}$ yields

$\|\xi_{1}(\cdot,t)\|_{\infty}=\|\mathbf{T}(t)\xi^{0}\|_{\infty}\leq
M\mathrm{e}^{wt}U_{0}<U\qquad\text{for}\quad t\in[0,T],$

where we used the choice of $T$ in (C.3). Continuity of $\mathbf{N}$ implies
that $\mathbf{N}\circ\xi_{1}$ is also (Bochner) measurable. Estimates for the
nonlinearity from Assumption 1 lead us to integrability, i.e.,
$s\mapsto\mathbf{N}(\xi_{1}(\cdot,s))\in
L^{1}([0,T];L^{\infty}(\Omega)^{m+k})$. By [5, Proposition 1.3.4] and
continuity of the semigroup from Lemma 2.11, the convolution in definition of
$\xi_{2}$ is well-defined as a Bochner integral in $L^{\infty}(\Omega)^{m+k}$
or $L^{\infty}(\Omega)^{m}\times C(\overline{\Omega})^{k}$ with the estimate

$\displaystyle\eta_{1}(t)$
$\displaystyle\leq\int_{0}^{t}M\mathrm{e}^{w(t-s)}\|\mathbf{N}(\xi_{1}(\cdot,s))\|_{\infty}\;\mathrm{d}s\leq
L(B)M\int_{0}^{t}\mathrm{e}^{w(t-s)}\;\mathrm{d}s=\alpha\int_{0}^{t}\mathrm{e}^{ws}\;\mathrm{d}s$
$\displaystyle\leq\alpha\int_{0}^{t}\mathrm{e}^{(\alpha+w)s}\;\mathrm{d}s\leq\mathrm{e}^{(\alpha+w)t}-1$

since $\alpha=ML(B)\geq 0$ and $w\geq 0$. This proves the base case for
estimates (C.5). If the estimates hold true for a particular $i\in\mathbb{N}$,
this implies

$\displaystyle\eta_{i+1}(t)$
$\displaystyle\leq\int_{0}^{t}M\mathrm{e}^{w(t-s)}\|\mathbf{N}(\xi_{i+1}(\cdot,s))-\mathbf{N}(\xi_{i}(\cdot,s))\|_{\infty}\;\mathrm{d}s$
$\displaystyle\leq\alpha\int_{0}^{t}\mathrm{e}^{w(t-s)}\eta_{i}(s)\;\mathrm{d}s\leq\alpha^{2}\int_{0}^{t}\mathrm{e}^{w(t-s)}\int_{0}^{s}\frac{(\alpha
r)^{i-1}}{(i-1)!}\mathrm{e}^{wr}\;\mathrm{d}r\;\mathrm{d}s$
$\displaystyle=\frac{\alpha^{2}}{w}\mathrm{e}^{wt}\int_{0}^{t}\frac{(\alpha
r)^{i-1}}{(i-1)!}\;\mathrm{d}r-\frac{\alpha^{2}}{w}\int_{0}^{t}\frac{(\alpha
r)^{i-1}}{(i-1)!}\mathrm{e}^{wr}\;\mathrm{d}r=\alpha\int_{0}^{t}\frac{(\alpha
s)^{i}}{i!}\mathrm{e}^{ws}\;\mathrm{d}s,$

where we used partial integration twice. We demonstrated only the case $w>0$,
$w=0$ is even simpler. By definition of $\xi_{i+1}$ in (C.4) and $T$ in (C.3),
we obtain

$\displaystyle\|\xi_{i+1}(\cdot,t)\|_{\infty}$ $\displaystyle\leq
M\mathrm{e}^{wt}U_{0}+ML(B)\int_{0}^{t}\mathrm{e}^{w(t-s)}\;\mathrm{d}s$
$\displaystyle\leq
M\mathrm{e}^{wT}U_{0}+ML(B)\int_{0}^{t}\mathrm{e}^{(ML(B)+w)s}\;\mathrm{d}s$
$\displaystyle\leq M\mathrm{e}^{wT}U_{0}+\mathrm{e}^{(L(B)M+w)T}-1\leq U$

for all $t\in[0,T]$, since $\alpha=ML(B)\geq 0$ and $w\geq 0$. Finally, it is
left to prove the estimate for the sum of $\eta_{j}$ in (C.5). The induction
step follows from

$\displaystyle\sum_{j=1}^{i+1}\eta_{j}(t)\leq\alpha\int_{0}^{t}\sum_{j=1}^{i+1}\frac{(\alpha
s)^{j-1}}{(j-1)!}\mathrm{e}^{ws}\;\mathrm{d}s\leq\alpha\int_{0}^{t}\mathrm{e}^{(\alpha+w)s}\;\mathrm{d}s\leq\mathrm{e}^{(\alpha+w)t}-1.$

For fixed $t\in[0,T]$, these results imply convergence of

$\xi_{i+1}(\cdot,t)=\sum_{j=1}^{i}(\xi_{j+1}-\xi_{j})(\cdot,t)+\xi_{1}(\cdot,t)\to\xi(\cdot,t):=\sum_{j\in\mathbb{N}}(\xi_{j+1}-\xi_{j})(\cdot,t)+\xi_{1}(\cdot,t)$

as $i\to\infty$. The latter series converges absolutely in
$L^{\infty}(\Omega)^{m+k}$, i.e.,

$\sum_{j\in\mathbb{N}}\|(\xi_{j+1}-\xi_{j})(\cdot,t)\|_{\infty}=\sum_{j\in\mathbb{N}}\eta_{j}(t)\leq\mathrm{e}^{(\alpha+w)t}-1,$

by convergence of the partial sums in (C.5). We even have convergence
$\xi_{i}\to\xi$ in the Banach space $L^{\infty}(\Omega_{T})^{m+k}$ and
$L^{\infty}([0,T];L^{\infty}(\Omega)^{m+k})$. This is a consequence of the
estimates

$\displaystyle\sup_{t\in[0,T]}\|(\xi_{i}-\xi)(\cdot,t)\|_{\infty}$
$\displaystyle\leq\sup_{t\in[0,T]}\sum_{j=i}^{\infty}\eta_{j}(t)\leq\sup_{t\in[0,T]}\sum_{j=i}^{\infty}\alpha\int_{0}^{t}\frac{(\alpha
s)^{j-1}}{(j-1)!}\mathrm{e}^{ws}\;\mathrm{d}s$
$\displaystyle\leq\alpha\int_{0}^{T}\left(\mathrm{e}^{\alpha
s}-\sum_{j=0}^{i-2}\frac{(\alpha
s)^{j}}{(j)!}\right)\mathrm{e}^{ws}\;\mathrm{d}s\to 0\quad(i\to\infty).$

where the exponential series converges uniformly on the bounded interval
$[0,T]$. Using the above estimate of $\xi_{1}$ and inequalities (C.5), we
obtain an estimate for $\xi$,

$\sup_{t\in[0,T]}\|\xi(\cdot,t)\|_{\infty}\leq\sup_{t\in[0,T]}\left(\mathrm{e}^{(\alpha+w)t}-1+M\mathrm{e}^{wt}U_{0}\right)\leq
U.$

Due to uniform convergence, the limit $\xi$ is (Bochner) measurable and a mild
solution of the integral equation

$\xi(\cdot,t)=\mathbf{T}(t)\xi^{0}+\int_{0}^{t}\mathbf{T}(t-s)\mathbf{N}(\xi(\cdot,s))\;\mathrm{d}s\qquad\forall\;t\in[0,T].$

Uniqueness of mild solutions follows from Gronwall’s inequality and local
Lipschitz continuity from Assumption 1. For the proof of (ii) and (iii), we
refer to [37, p. 114]. ∎

Equivalence of the implicit integral formulations (C.1) and (C.2) can be shown
using [47, Equation (4.100)] for $X=L^{p}(\Omega)^{m+k}$. Indeed, by [27,
Section 5.2.2], $\mathbf{D}\Delta$ and $\mathbf{D}\Delta+\mathbf{J}$ generate
analytic semigroups on $L^{p}(\Omega)^{m+k}$ and it can be verified that its
restrictions to the subspace $L^{\infty}(\Omega)^{m}\times L^{p}(\Omega)^{k}$
coincide with $(\mathbf{S}(t))_{t\in\mathbb{R}_{\geq 0}}$ defined in (2.4) and
$(\mathbf{T}(t))_{t\in\mathbb{R}_{\geq 0}}$, respectively [35, Ch. 4, Theorem
5.5]. Continuity of $\mathbf{N}$ from Lemma 2.2 implies that

$h:=\mathbf{N}\circ\xi\in C([0,T];L^{p}(\Omega)^{m+k})$

for a solution $\xi$ of problem (2.1) and [47, Equation (4.100)] applies to
the bounded perturbation $\mathbf{J}$ on $L^{p}(\Omega)^{m+k}$.

##  Appendix D Multiplication operator on uniformly continuous functions

Each matrix-valued function $\mathbf{A}\in C(\overline{\Omega})^{m\times m}$
induces a corresponding multiplication operator

$\mathbf{M}_{\mathbf{A}}:C(\overline{\Omega})^{m}\to
C(\overline{\Omega})^{m},\qquad\mathbf{z}\mapsto\mathbf{A}\mathbf{z}$

where $(\mathbf{A}\mathbf{z})(x):=\mathbf{A}(x)\mathbf{z}(x)$ for each
$\mathbf{z}\in C(\overline{\Omega})^{m}$. Since
$\|\mathbf{M}_{\mathbf{A}}\|\leq\|\mathbf{A}\|_{\infty}$, this is a bounded,
linear operator. Let us simply write $\mathbf{A}$ instead of
$\mathbf{M}_{\mathbf{A}}$ in the following. The knowledge of the spectrum of a
multiplication operator $\mathbf{A}$ allows us to characterize the spectrum of
the partly diffusive operator $\mathbf{L}^{c}$ in Proposition 2.10.

######  Lemma D.1.

Let $\mathbf{A}\in C(\overline{\Omega})^{m\times m}$ for $m\in\mathbb{N},$ and
let $\mathbf{A}$ denote its corresponding bounded multiplication operator on
$C(\overline{\Omega})^{m}$. Then the spectrum $\sigma(\mathbf{A})$ is
essential in the sense that each $\lambda\in\sigma(\mathbf{A})$ is
characterized by the property

$\displaystyle\forall\,\varepsilon>0\;\exists\,\Omega_{\varepsilon}\subset\Omega\;\text{open
set}\;\exists\,\mathbf{v}_{\varepsilon}\in\mathbb{R}^{m}\setminus\\{\mathbf{0}\\}:\|(\lambda
I-\mathbf{A})\mathbf{v}_{\varepsilon}\|_{C(\overline{\Omega_{\varepsilon}})^{m}}\leq\varepsilon|\mathbf{v}_{\varepsilon}|.$
(D.1)

This means
$\sigma(\mathbf{A})=\\{\lambda\in\mathbb{C}\mid\text{condition}\;\eqref{essspectrumC2}\;\text{is
satisfied}\\}=:A_{\mathrm{ess}}$ and, furthermore, $\sigma(\mathbf{A})$
consists of approximate eigenvalues only.

###### Proof.

First, let us verify $A_{\mathrm{ess}}\subset\sigma_{ap}(\mathbf{A})$. We
assume $\lambda\in A_{\mathrm{ess}}$ and consider the sequence
$(\xi_{\varepsilon})_{\varepsilon>0}\subset C(\overline{\Omega})^{m}$ defined
by
$\xi_{\varepsilon}=\eta_{\varepsilon}\mathbf{v}_{\varepsilon}/|\mathbf{v}_{\varepsilon}|$
for a $\eta_{\varepsilon}\in C_{c}^{\infty}(\Omega_{\varepsilon})$ with
$0\leq\eta_{\varepsilon}\leq 1$ and $\eta_{\varepsilon}\equiv 1$ on some small
ball $\overline{B_{\delta}(y)}\subset\Omega_{\varepsilon}$ [8, Lemma 9.3].
This construction implies $\|\xi_{\varepsilon}\|_{C(\overline{\Omega})^{m}}=1$
and $\|(\lambda
I-\mathbf{A})\xi_{\varepsilon}\|_{C(\overline{\Omega})^{m}}=\|(\lambda
I-\mathbf{A})\xi_{\varepsilon}\|_{C(\overline{\Omega_{\varepsilon}})^{m}}\leq\varepsilon.$
Thus, $\lambda\in A_{\mathrm{ess}}$ is also an approximate eigenvalue of
$\mathbf{A}$.
In order to show $\sigma(\mathbf{A})\subset A_{\mathrm{ess}}$, we verify the
converse inclusion $\mathbb{C}\setminus
A_{\mathrm{ess}}\subset\rho(\mathbf{A})$. For $\lambda\notin
A_{\mathrm{ess}}$, we find a constant $\varepsilon>0$ such that for all open
sets $\Omega_{0}\subset\Omega$ and vectors
$\mathbf{v}\in\mathbb{R}^{m}\setminus\\{\mathbf{0}\\}$ there holds

$\|(\lambda
I-\mathbf{A})(\cdot)\mathbf{v}\|_{C(\overline{\Omega_{0}})^{m}}\geq\varepsilon|\mathbf{v}|.$

By contradiction, this implies the pointwise estimate

$|(\lambda
I-\mathbf{A})(x)\mathbf{v}|\geq\varepsilon|\mathbf{v}|\qquad\forall\;x\in\overline{\Omega},\mathbf{v}\in\mathbb{R}^{m}.$

Consequently, the matrices $(\lambda I-\mathbf{A}(x))^{-1}$ imply a bounded
multiplication operator on $C(\overline{\Omega})^{m}$ which turns out to be
the inverse of $\lambda I-\mathbf{A}$, hence $\lambda\in\rho(\mathbf{A})$. ∎

######  Proposition D.2.

Let $\mathbf{A}\in C(\overline{\Omega})^{m\times m}$ for $m\in\mathbb{N}$.
Then the spectrum of the bounded multiplication operator $\mathbf{A}$ on
$C(\overline{\Omega})^{m}$ is given by

$\displaystyle\sigma(\mathbf{A})=\overline{\bigcup_{x\in\Omega}\sigma(\mathbf{A}(x))}.$
(D.2)

###### Proof.

In order to verify equality (D.2), we use a continuity argument. Recall that
$\sigma(\mathbf{A})$ is a closed set in $\mathbb{C}$ by [5, Corollary B.3]. In
view of the closedness of $A_{\mathrm{ess}}=\sigma(\mathbf{A})$, it suffices
to show

$\bigcup_{x\in\Omega}\sigma(\mathbf{A}(x))\subset A_{\mathrm{ess}}$

to obtain one inclusion in (D.2). For this reason consider
$\lambda\in\sigma(\mathbf{A}(x_{0}))$ for some $x_{0}\in\Omega$. Then there
exists a normalized eigenvector $\mathbf{v}_{0}\in\mathbb{R}^{m}$ with
$|\mathbf{v}_{0}|=1$ such that $(\lambda
I-\mathbf{A}(x_{0}))\mathbf{v}_{0}=\mathbf{0}$. By continuity, we obtain a
small neighborhood $\Omega_{\varepsilon}\subset\Omega$ of $x_{0}$ such that

$\sup_{x\in\Omega_{\varepsilon}}|(\lambda
I-\mathbf{A}(x))\mathbf{v}_{0}|\leq\varepsilon=\varepsilon|\mathbf{v}_{0}|,$

hence $\lambda\in A_{\mathrm{ess}}$.
For the reverse inclusion in (D.2), we show that
$\lambda\notin\overline{\bigcup_{x\in\Omega}\sigma(\mathbf{A}(x))}$ implies
$\lambda\notin A_{\mathrm{ess}}$. Indeed, if $\lambda$ is not in the closed
set $\overline{\bigcup_{x\in\Omega}\sigma(\mathbf{A}(x))}$ and $x\in\Omega$,
every matrix $\lambda I-\mathbf{A}(x)$ is invertible in $\mathbb{R}^{m\times
m}$, i.e., we find a constant $C_{x}>0$ such that

$|(\lambda I-\mathbf{A}(x))\mathbf{v}|\geq
C_{x}|\mathbf{v}|\qquad\forall\;\mathbf{v}\in\mathbb{R}^{m}.$

Let us verify that there exists a lower bound $C_{x}\geq C>0$ which is
independent of $x\in\Omega$. We assume to the contrary that there exists a
sequence $(x_{\ell})_{\ell\in\mathbb{N}}\subset\Omega$ and a sequence of
normalized eigenvectors
$(\mathbf{v}_{\ell})_{\ell\in\mathbb{N}}\subset\mathbb{R}^{m}$ with
$|\mathbf{v}_{\ell}|=1$ satisfying $|(\lambda
I-\mathbf{A}(x_{\ell}))\mathbf{v}_{\ell}|\to 0$ as $\ell\to\infty$. However,
this yields a contradiction, since $\lambda$ is arbitrarily close to the
spectrum of the matrices $\mathbf{A}(x_{\ell})$ as $\ell\to\infty$. Uniformity
of the constant $C>0$ implies that $\lambda I-\mathbf{A}$ is bounded from
below on $C(\overline{\Omega})^{m}$ and, thus, $\lambda\notin
A_{\mathrm{ess}}$ by definition (D.1). ∎

In order to characterize the spectrum of the part of the operator $\mathbf{L}$
on $C(\overline{\Omega})^{m+k}$ in Proposition 2.10, we use again that the
spectrum of the multiplication operator $\mathbf{A}_{\ast}$ is essential in
the sense of Wolf. This is a consequence of the following refined
characterization of the spectrum of the multiplication operator $\mathbf{A}$.

######  Proposition D.3.

Let $\mathbf{A}\in C(\overline{\Omega})^{m\times m}$ for $m\in\mathbb{N},$ and
let $\mathbf{A}$ denote its corresponding bounded multiplication operator on
$C(\overline{\Omega})^{m}$. Then the point spectrum of $\mathbf{A}$ is given
by

$\sigma_{p}(\mathbf{A})=\\{\lambda\in\mathbb{C}\mid\Gamma_{\lambda}\;\text{has
non-empty interior}\;\Gamma_{\lambda}^{\circ}\\},$ (D.3)

where the compact set $\Gamma_{\lambda}$ is defined by
$\Gamma_{\lambda}=\\{x\in\overline{\Omega}\mid\det(\lambda
I-\mathbf{A}(x))=0\\}.$ Moreover, the spectrum $\sigma(\mathbf{A})$ is
essential in the sense of Wolf, i.e.,

$\sigma(\mathbf{A})=\sigma_{\mathrm{ess}}(\mathbf{A}):=\\{\lambda\in\mathbb{C}\mid\lambda
I-\mathbf{A}\;\text{is not a Fredholm operator}\\}.$

###### Proof.

Note that $\Gamma_{\lambda}$ is compact by continuity of the determinant and
entries of the matrix $\mathbf{A}(x)$. First, let us verify characterization
(D.3) for $\sigma_{p}(\mathbf{A})$. On the one hand, let
$\lambda\in\sigma_{p}(\mathbf{A})$ with a corresponding eigenfunction
$\mathbf{f}\in C(\overline{\Omega})^{m}\setminus\\{0\\}$. Then
$\mathbf{f}(x)\not=\mathbf{0}$ for some $x\in\overline{\Omega}$. By continuity
of $\mathbf{f}$ and (path-)connectedness of $\Omega\subset\mathbb{R}^{n}$, we
may choose a non-empty open set $\tilde{\Omega}\subset\Omega$ for which
$\mathbf{f}(x)\not=\mathbf{0}$ holds for all $x\in\tilde{\Omega}$. Then, for
each $x\in\tilde{\Omega}$, the vector
$\mathbf{f}(x)\in\mathbb{C}^{m}\setminus\\{\mathbf{0}\\}$ is an eigenvector of
the matrix $\lambda I-\mathbf{A}(x)$. Hence,
$\tilde{\Omega}\subset\Gamma_{\lambda}$ and $\Gamma_{\lambda}$ has non-empty
interior with
$\tilde{\Omega}=\tilde{\Omega}^{\circ}\subset\Gamma_{\lambda}^{\circ}$.
On the other hand, if $\lambda\in\mathbb{C}$ satisfies
$\Gamma_{\lambda}^{\circ}\not=\emptyset$, we can construct eigenvectors
$\mathbf{v}(x)$ corresponding to $\mathbf{A}(x)$ and the constant eigenvalue
$\lambda$ such that the function $\mathbf{v}$ is continuous on some closed
ball $\overline{B_{\delta}(y)}\subset\Gamma_{\lambda}^{\circ}$. This follows
from Gaussian elimination using elementary row operations which consists of
multiplying, adding or swapping continuous entries. Once we obtained such
eigenvectors, we use the idea of proof of [18, Theorem 2.5]. For arbitrary
$\phi\in C(\overline{B_{\delta}(y)})$ with $\phi\not\equiv 0$ on the open ball
$B_{\delta/2}(y)$, we define the function

$\displaystyle\mathbf{f}:\overline{\Omega}\to\mathbb{R}^{m}\setminus\\{\mathbf{0}\\},\quad\mathbf{f}(x)=\begin{cases}\mathbf{v}(x)\phi(x)\eta(x)&\text{for}\quad
x\in B_{\delta}(y),\\\ \mathbf{0}&\text{else}\end{cases}$

for a cut-off function $\eta\in C_{c}^{\infty}(B_{\delta}(y))$ with
$\eta_{|B_{\delta/2}(y)}\equiv 1$. Then, $\mathbf{f}\in
C(\overline{\Omega})^{m}\setminus\\{\mathbf{0}\\}$ satisfies $(\lambda
I-\mathbf{A})\mathbf{f}=\mathbf{0}$, i.e., $\lambda\in\sigma_{p}(\mathbf{A})$.

It remains to show that $\lambda I-\mathbf{A}$ is not Fredholm for all
$\lambda\in\sigma(\mathbf{A})$.
If $\lambda\in\sigma_{p}(\mathbf{A})$, we infer
$\sigma_{p}(\mathbf{A})\subset\sigma_{\mathrm{ess}}(\mathbf{A})$ from an
infinite-dimensional kernel of $\lambda I-\mathbf{A}$, analog to [18,
Proposition 3.2]. Indeed, from above calculations we see that this kernel
contains a subspace which is isomorphic to $C(\overline{B_{\delta/2}(y)})$
and, hence, is infinite-dimensional.
If $\lambda\in\sigma(\mathbf{A})\setminus\sigma_{p}(\mathbf{A})$, the
injective operator $\lambda I-\mathbf{A}$ cannot be bounded from below since
$\lambda\in\sigma_{ap}(\mathbf{A})$ by Lemma D.1. Thus, $\lambda I-\mathbf{A}$
cannot have closed range by [8, Theorem 2.19, Remark 18] and
$\lambda\in\sigma_{\mathrm{ess}}(\mathbf{A})$. ∎

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