Generation of analytic semigroups with generalized Wentzell boundary condition

Abstract

In this paper we study second order ordinary and partial differential equations with generalized Wentzell boundary condition. We prove, by a perturbation method, that certain second order ordinary differential operators generate an analytic semigroup in $W^{1,p}([0,1])$ and the same result has been extended for the degenerate operator $Au(x):=x(1-x)u''(x)$. Finally, we prove that certain linear partial differential operators of the second order generate analytic semigroups in the space of continuous functions.

1 Introduction

The study of ordinary and partial differential equations with Wentzell boundary conditions was stimulated by the theory of probability in the pioneering paper of Wentzell [17]. He considered a second order elliptic differential operator $A$ in a domain $\Omega$ in ${\mathbb {R}}^n$, with sufficiently smooth boundary $\partial \Omega$, and he looked for the most general supplementary conditions which restrict the given operator $A$ to be the infinitesimal generator of a semigroup corresponding to a Markov process in the domain. In [17] it is proved that, when the domain $\Omega$ has some special shape, the boundary conditions could involve the operator $A$. Generalized Wentzell boundary conditions contain, as special cases, Dirichlet, Neumann and Robin boundary conditions. In a paper by Feller [9] we find a diffusion process involving the operator

$$\begin{aligned} Au(x):=x(1-x)u''(x),\ \ \ x\in (0,1), \end{aligned}$$

(1.1)

with the boundary conditions

$$\begin{aligned} \lim _{x\rightarrow 0^+,\ x\rightarrow 1^-} Au(x)=0. \end{aligned}$$

(1.2)

In [3] it is proved that $A$ with the domain

$$\begin{aligned} \mathcal{D}(A)=\{ u\in C[0,1]\cap C^1(0,1)\ :\ \lim _{x\rightarrow 0^+,\ x\rightarrow 1^-} Au(x)=0\} \end{aligned}$$

(1.3)

is the generator of a $C_0$-contraction semigroup on $C[0,1]$. For this reason in this paper we will consider the operator $x(1-x)u''(x)$. Several motivations to study elliptic operators with Wentzell boundary conditions can also be found in [11, 13–15].

In [7] generation results in suitable $L^p$ spaces were proved. The present paper can be considered an extension of that work to the case $W^{1,p}$. We recall what has been proved in [7], introducing the problem

$$\begin{aligned} \left\{ \begin{array}{clcr} &{}\lambda u -\nabla \cdot (a\nabla u) =f,\ \ \mathrm{in}\ \ \ \Omega ,\\ &{} \nabla \cdot (a\nabla u) +\beta \frac{\partial u}{\partial n} +\gamma u=0, \ \ \mathrm{on} \ \ \partial \Omega , \end{array}\right. \end{aligned}$$

(1.4)

where $\Omega$ is a bounded open set with smooth boundary $\partial \Omega$, $\displaystyle {\frac{\partial u}{\partial n}}$ is the outward normal derivative, $a(x)$ is a strictly positive $C^2$ function for all $x\in \overline{\Omega }$ and $\beta (x) >0$, $\gamma (x) \ge 0$, for all $x\in \partial \Omega$.

We start by pointing out the surprising fact that the presence of the term $\beta \frac{\partial u}{\partial n}$ in the boundary conditions led in [7] to introduce suitable $L^p$ weighted spaces to prove a generation result. More precisely the authors in [7] define $L^p(\overline{\Omega },d\mu )$, where

$$\begin{aligned} d\mu =dx\Big |_{\Omega } \oplus \frac{a }{\beta } \ \ dS\Big |_{\Gamma }\,; \end{aligned}$$

here $dx$ denotes the Lebesgue measure on $\Omega$, while $(a / \beta )\ dS$ denotes the measure with weight $a/ \beta$ on $\Gamma$: it is assumed that $a>0$ in $\Omega \cup \Gamma$, where $\emptyset \ne \Gamma \subset \partial \Omega$ and $a\in C^2(\Omega \cup \Gamma ) \cap C\left( \overline{\Omega }\right)$.

Let $U=(u,v)\in L^p(\overline{\Omega },d\mu )$, where $u:\Omega \rightarrow {\mathbb {C}}$ and $v:\partial \Omega \rightarrow {\mathbb {C}}$ are measurable functions such that

$$\begin{aligned} \int _\Omega |u(x)|^p\, dx+\int _{\partial \Omega }|v(x)|^p\, \frac{a}{ \beta }\ dS \end{aligned}$$

is finite. With the norm

$$\begin{aligned} \Vert U\Vert _{L^p(\overline{\Omega },d\mu )}:=\left[ \int _\Omega |u(x)|^p\, dx +\int _{\partial \Omega }|v(x)|^p\, \frac{a}{ \beta }\ dS\right] ^{1/p}, \end{aligned}$$

$L^p(\overline{\Omega },d\mu )$ becomes a Banach space. Moreover, the space $L^p(\overline{\Omega },d\mu )$ can be identified with

$$\begin{aligned} L^p(\Omega ,dx)\times L^p\Big (\Gamma ,\displaystyle {\frac{a}{\beta }\ dS}\Big ). \end{aligned}$$

This space can be defined in an equivalent way: consider $u\in C\left( \overline{\Omega }\right)$, set $U=(u|_{\Omega }, u|_{\partial \Omega })$ and define $X_p(\overline{\Omega })$ to be the completion of $C(\overline{\Omega })$ with the norm $\Vert U\Vert _{L^p(\overline{\Omega },d\mu )}$. It can be shown that

$$\begin{aligned} X_p\left( \overline{\Omega }\right) =L^p(\overline{\Omega },d\mu ). \end{aligned}$$

In the spaces $C\left( \overline{\Omega }\right)$ the trace of a function has a clear meaning. We recall, for the sake of completeness, the main generation theorem in the space $X_p\left( \overline{\Omega }\right)$ proved in [7] (see Theorem 3.1).

Theorem 1.1

Let $\Omega$ be an open bounded set in ${\mathbb {R}}^n$ with boundary of class $C^2$. Let $a\in C^1(\overline{\Omega })$ with $a>0$ in $\overline{\Omega }$, $Au:=\nabla \cdot (a\nabla u)$, $\Gamma :=\{x\in \partial \Omega : a(x)>0\}\ne \emptyset$. If $\beta$ and $\gamma$ are non negative functions in $C^1(\partial \Omega )$ with $\beta >0$, then $\overline{G}$, the closure of the operator

$$\begin{aligned} G=\left[ \begin{array}{rr} A &{} 0\\ -\beta \frac{\partial }{\partial n}&{}-\gamma \end{array} \right] \end{aligned}$$

(1.5)

with domain

$$\begin{aligned} D_p(G)&:= \Big \{ u\in C^2(\Omega \cup \Gamma )\cap C(\overline{\Omega }):\ Au\in L^p(\Omega , dx),\ \ Au\\&\;\;+\,\beta \frac{\partial u}{\partial n}+\gamma u=0,\ \ on\ \ \Gamma \ \Big \} \end{aligned}$$

generates a $(C_0)$ contraction semigroup on $X_p$, for $p\in [1,\infty )$. The semigroup is analytic if $p\in (1,\infty )$.

This result was significally extended in [6] and [4].

We are able to generalize the above result for ordinary differential operators in the space $W^{1,p}([0,1])$ with Wentzell boundary conditions also for a degenerate case.

The plan of the paper is the following.

• In Sect. 2 we give a generation result in $W^{1,p}([0,1])$ with Wentzell boundary conditions.

• In Sect. 3 we establish a generation result in $C^1([0,1])$.

• In Sect. 4 we prove a generation result for a degenerate operator in $W^{1,p}([0,1])$ with Wentzell boundary conditions.

• In Sect. 5 we consider a linear partial differential operator of the second order and prove a generation result in $C(\overline{\Omega })$ with Wentzell boundary conditions.

We set $I = [0,1]$ and, in the following, we shall always suppose that $a, b, c\in C^{\infty }(I)$ and, for every $x\in I$, $a(x) \ge a_0 > 0$.

Set

$$\begin{aligned} \left( A^{\sharp }u\right) (x) : = a(x) u'' + b(x) u' + c(x) u, \qquad x\in [0,1], \end{aligned}$$

We make the following hypothesis:

• (H1) for every $c_0, c_1 \in {\mathbf {C}}$, there exists a unique solution $u$ of the boundary value problem

  $$\begin{aligned} A^{\sharp } u = 0, \quad u(0) = c_0, \quad u(1) = c_1. \end{aligned}$$

  Then, if $B_0, B_1$ are boundary operators of order not greater than one, the mapping $(A^{\sharp }, B_0, B_1)$ is an isomorphism from $W^{3,p}([0,1])$ onto $W^{1,p}([0,1]) \times {\mathbb {C}} \times {\mathbb {C}}$, by e.g. [16], Theorem 5.5.2.

2 Generation results in $W^{1,p}([0,1])$ with Wentzell boundary conditions

Theorem 2.1

Define, for $p>1$, the operator

$$\begin{aligned} \left\{ \begin{array}{clcr} &{}\mathcal{D}(A)=\{ u\in W^{1,p}(I)\ :\ au''+bu'+cu\in W^{1,p}_0(I)\ \},\\ &{} Au:=au''+bu'+cu,\ \ \ u\in \mathcal{D}(A),\\ \end{array} \right. \end{aligned}$$

(2.1)

and suppose that $A$ satisfies assumption $(H1)$. Then $(A, \mathcal{D}(A))$ generates an analytic semigroup in $W^{1,p}(I)$.

Proof

If $\lambda \in {\mathbb {C}}\setminus \{0\}$ and $f\in W^{1,p}(I)$, consider the boundary problem

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} \lambda u(x)-Au(x)=f(x),\qquad x\in I,\\ &{} Au(j)=0,\ \ j=0,1,\\ \end{array} \right. \end{aligned}$$

(2.2)

and the auxiliary problem

$$\begin{aligned} \left\{ \begin{array}{clcr} &{}AG(x)=0,\ \qquad x\in I,\\ &{} G(j)=f(j),\ \qquad j=0,1. \\ \end{array} \right. \end{aligned}$$

(2.3)

By hypothesis $(H1)$, problem (2.3) has a unique solution $G\in W^{3,p}(I)$. Then we can rewrite Problem (2.2) as

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} \lambda \Big ( u(x)- \frac{G(x)}{\lambda }\Big ) -A\Big ( u(x)- \frac{G(x)}{\lambda }\Big )=f(x)-G(x),\ \qquad x\in I,\\ &{} u(j)- \frac{G(j)}{\lambda }=0,\ \ \ A\Big ( u- \frac{G}{\lambda }\Big )(j)=0,\ \qquad j=0,1.\\ \end{array} \right. \end{aligned}$$

(2.4)

Since $f\in W^{1,p}(I)$ we have $f-G\in W^{1,p}_0(I)$, and the operator

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} \mathcal{D}(A_0)=\{ u\in W_0^{1,p}(I)\ :\ au''+bu'+cu\in W_0^{1,p}(I)\ \},\\ &{} A_0u:=au''+bu'+cu,\ \ \ u\in \mathcal{D}(A_0),\\ \end{array} \right. \end{aligned}$$

(2.5)

generates an analytic semigroup in $W_0^{1,p}(I)$, by [1]. So we have proved that Problem (2.4) has a unique solution $v:=u-G/\lambda \in \mathcal{D}\left( A_0\right)$ and the estimate

$$\begin{aligned} \Vert v\Vert _{W_0^{1,p}(I)} \le \frac{c_p}{|\lambda |} \Vert f-G\Vert _{W_0^{1,p}(I)} \end{aligned}$$

(2.6)

holds, where $c_p$ is a positive real constant independent of $\lambda$. Moreover, for Problem (2.3) we have

$$\begin{aligned} \Vert G\Vert _{W^{1,p}(I)} \le K_p\ \Vert f\Vert _{W^{1,p}(I)} \end{aligned}$$

(2.7)

where $K_p$ is a positive real constant. Then we get the generation estimate

$$\begin{aligned} \left\| u\right\| _{W^{1,p}(I)}&\le \left\| u- \frac{G}{\lambda }\right\| _{W^{1,p}(I)} + \left\| \frac{G}{\lambda }\right\| _{W^{1,p}(I)} \le \frac{c_p}{|\lambda |}\ \Vert f-G\Vert _{W^{1,p}(I)} \nonumber \\&+\,\frac{K_p}{|\lambda |} \left\| f\right\| _{W^{1,p}(I)}. \end{aligned}$$

(2.8)

$\square$

Theorem 2.2

Define, for $p>1$, the operator

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} \mathcal{D}(A_0)=\{ u\in W^{1,p}(I)\ :\ au''\in W^{1,p}(I),\ \ au''+bu'+cu\in W_0^{1,p}(I) \ \}, \\ &{} A_0u:=au'',\ \ \ u\in \mathcal{D}(A_0).\\ \end{array} \right. \end{aligned}$$

(2.9)

Then $\left( A_0, \mathcal{D}(A_0)\right)$ generates an analytic semigroup in $W^{1,p}(I)$.

Proof

The proof is based on a perturbation method and on Theorem 2.1. Consider the operator $B$ defined on $W^{1,p}(I)$ as follows:

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} \mathcal{D}(B)=\mathcal{D}(A_0),\\ &{} Bu:=-bu'-cu,\ \ \ \forall u\in \mathcal{D}(B).\\ \end{array} \right. \end{aligned}$$

(2.10)

Consider the chain of inequalities, for $\varepsilon >0$:

$$\begin{aligned} \Vert Bu\Vert _{L^p(I)}&\le K_0\Vert u'\Vert _{L^p(I)}+M_0\Vert u\Vert _{L^p(I)} \le \displaystyle {\frac{K_0\varepsilon }{a_0} }\Vert au''\Vert _{L^p(I)}+M_{\varepsilon } \Vert u\Vert _{L^p(I)} \nonumber \\&+M_0\Vert u\Vert _{L^p(I)}\nonumber \\&\le \displaystyle {\frac{K_0\varepsilon }{a_0}}\Vert au''+bu'+cu\Vert _{L^p(I)} + \displaystyle {\frac{K_0\varepsilon }{a_0}}\Vert bu'+cu\Vert _{L^p(I)}+M_{\varepsilon } \Vert u\Vert _{L^p(I)}\nonumber \\&+K_0\Vert u\Vert _{L^p(I)}\nonumber \\&= \displaystyle {\frac{K_0\varepsilon }{a_0}}\Vert Au\Vert _{L^p(I)} +\displaystyle {\frac{K_0\varepsilon }{a_0}}\Vert Bu\Vert _{L^p(I)}+\left( M_{\varepsilon } +K_0\right) \Vert u\Vert _{L^p(I)}, \end{aligned}$$

(2.11)

where, here and in the sequel, $K_i$, $M_i$, $M_{\varepsilon }, M_{\varepsilon }'$ and $M_{\varepsilon }''$ denote positive constants. Furthermore,

$$\begin{aligned} \Vert (Bu)'\Vert _{L^p(I)}&\le K_0\Vert u''\Vert _{L^p(I)}+\Vert (b'+c)u'\Vert _{L^p(I)}+\Vert c'u\Vert _{L^p(I)}\nonumber \\&\le K_0\ \varepsilon \ \Vert u'''\Vert _{L^p(I)}+M_{\varepsilon } \Vert u'\Vert _{L^p(I)} +M_1\Vert u'\Vert _{L^p(I)} +M_2\Vert u\Vert _{L^p(I)}\nonumber \\&= K_0\ \varepsilon \ \Vert u'''\Vert _{L^p(I)}+M_{\varepsilon } \Vert u\Vert _{W^{1,p}(I)}\nonumber \\&\le \displaystyle {\frac{K_0 }{a_0}} \ \varepsilon \ \Vert (au'')'-a'u''\Vert _{L^p(I)}+M'_{\varepsilon } \Vert u\Vert _{W^{1,p}(I)}\nonumber \\&\le \displaystyle {\frac{K_0 }{a_0}} \ \varepsilon \ \Vert (au''+bu'+cu)'- (bu'+cu)'\Vert _{L^p(I)} +\frac{K_1 \varepsilon }{a_0}\Vert u''\Vert _{L^p(I)}\nonumber \\&+\, M'_{\varepsilon } \Vert u\Vert _{W^{1,p}(I)}\nonumber \\&\le \displaystyle {\frac{K_0 }{a_0}}\ \varepsilon \ \left( \Vert Au\Vert _{W^{1,p}(I)}+ \Vert (Bu)'\Vert _{L^p(I)}\right) \nonumber \\&+\,\frac{K_1 }{a_0^2}\ \varepsilon \left( \Vert Au\Vert _{L^p(I)} + \Vert Bu\Vert _{L^p(I)} \right) +M'_{\varepsilon } \Vert u\Vert _{W^{1,p}(I)}. \end{aligned}$$

(2.12)

Then, from (2.11)–(2.12), we get the estimates

$$\begin{aligned} \left( 1-\frac{K_0\varepsilon }{a_0}\right) \Vert (Bu)'\Vert _{L^p(I)}&\le M_1\varepsilon \Vert Au\Vert _{W^{1,p}(I)}+M'_{\varepsilon } \Vert u\Vert _{W^{1,p}(I)},\end{aligned}$$

(2.13)

$$\begin{aligned} \left( 1-\frac{K_0\varepsilon }{a_0}\right) \Vert Bu\Vert _{L^p(I)}&\le \displaystyle {\frac{K_0\varepsilon }{a_0}}\Vert Au\Vert _{L^p(I)} +M''_{\varepsilon } \Vert u\Vert _{L^p(I)}. \end{aligned}$$

(2.14)

By taking $\varepsilon$ sufficiently small we get the desired estimate:

$$\begin{aligned} \Vert Bu\Vert _{W^{1,p}(I)} \le M \varepsilon \Vert Au\Vert _{W^{1,p}(I)} +M'''_{\varepsilon } \Vert u\Vert _{W^{1,p}(I)}. \end{aligned}$$

(2.15)

Since $A$ generates an analytic semigroup in $W^{1,p}(I)$, thanks to Theorem 2.1, and $B$ is $A$-bounded with $A$-bound equal to zero, we conclude that $A_0=A+B$ generates an analytic semigroup as well. $\square$

Remark

The above result should be compared with Warma [18].

More generally, we have the following result.

Theorem 2.3

Let $A_1$ be the operator in $W^{1,p}(I)$, with $p>1$,

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} \mathcal{D}\left( A_1\right) =\{ u\in W^{1,p}(I)\ :\ au''+b_1u'+c_1u\in W^{1,p}_0(I)\ \},\\ &{} A_1 u:=au''+bu'+cu,\ \ \ \forall u\in \mathcal{D}\left( A_1\right) ,\\ \end{array} \right. \end{aligned}$$

(2.16)

where $b_1, c_1 \in C^{\infty }(I)$. Then $\left( A_1, D\left( A_1\right) \right)$ generates an analytic semigroup in $W^{1,p}([0,1])$.

Proof

Define the operator $C$ in $W^{1,p}([0,1])$ by

$$\begin{aligned} {\mathcal {D}}\left( C\right) = {\mathcal {D}}\left( A\right) , \qquad Cu = \left( b_1 - b\right) u' + \left( c_1 - c\right) u. \end{aligned}$$

(2.17)

Then $A_1 u = Au + Cu$, where $A$ is the operator defined in Theorem 2.1.

It is well known (see, e.g., [5], Example III.2.2) that the operator $B$, defined by

$$\begin{aligned} {\mathcal {D}}\left( B\right) = W^{1,p}(I), \qquad Bu = u', \end{aligned}$$

is ${\mathcal {A}}$-bounded, with ${\mathcal {A}}$-bound equal to $0$, where

$$\begin{aligned} {\mathcal {D}}\left( {\mathcal {A}}\right) = W^{2,p}(I), \qquad {\mathcal {A}}u = u''. \end{aligned}$$

Therefore, for every $\varepsilon \in \mathbf{R}^+$, we have

$$\begin{aligned} \Vert Cu\Vert _{L^p(I)}&\le K_1\left( \Vert u'\Vert _{L^p(I)} + \Vert u\Vert _{L^p(I)} \right) \\&\le K_1 \left( \varepsilon \Vert u''\Vert _{L^p(I)} + M_{\varepsilon } \Vert u\Vert _{L^p(I)} + \Vert u\Vert _{L^p(I)} \right) \\&\le K_1a_0^{-1}\ \varepsilon \ \Vert au''\Vert _{L^p(I)} + K_2 M_{\varepsilon } \Vert u\Vert _{L^p(I)}\\&\le K_1a_0^{-1}\ \varepsilon \ \Vert au'' + bu' + cu\Vert _{L^p(I)} + K_1a_0^{-1}\ \varepsilon \ \Vert bu'\Vert _{L^p(I)}\\&\quad + K_1a_0^{-1}\ \varepsilon \ \Vert cu\Vert _{L^p(I)} + K_2 M_{\varepsilon } \Vert u\Vert _{L^p(I)}\\&\le K_1a_0^{-1}\ \varepsilon \ \Vert Au\Vert _{L^p(I)} + M'_{\varepsilon } \Vert u\Vert _{W^{1,p}(I)} \le K_1a_0^{-1}\ \varepsilon \Vert Au\Vert _{W^{1,p}(I)}\\&\quad + M'_{\varepsilon } \Vert u\Vert _{W^{1,p}(I)}. \end{aligned}$$

Furthermore,

$$\begin{aligned} \Vert (Cu)'\Vert _{L^p(I)}&\le \Vert Cu\Vert _{W^{1,p}(I)} \le K_3 \left( \Vert u'\Vert _{W^{1,p}(I)} + \Vert u\Vert _{W^{1,p}(I)}\right) \\&\le K_3 \varepsilon \Vert u''\Vert _{W^{1,p}(I)} + M''_\varepsilon \Vert u\Vert _{W^{1,p}(I)} \\&\le K_3 a_0^{-1} \varepsilon \left( \Vert au'' + bu' + cu\Vert _{W^{1,p}(I)} + \Vert bu' + cu\Vert _{W^{1,p}(I)}\right) \\&\quad +\,M''_\varepsilon \Vert u\Vert _{W^{1,p}(I)} \\&\le K_3 a_0^{-1} \varepsilon \Vert Au\Vert _{W^{1,p}(I)} + K_4 a_0^{-1} \varepsilon \Vert u'\Vert _{W^{1,p}(I)} \\&\quad + \left( K_5 a_0^{-1} \varepsilon + M''_\varepsilon \right) \Vert u\Vert _{W^{1,p}(I)}. \end{aligned}$$

Hence,

$$\begin{aligned}&\left( K_3 - K_4 a_0^{-1} \varepsilon \right) \Vert u'\Vert _{W^{1,p}(I)} + \left( K_3 - K_5 a_0^{-1} \varepsilon \right) \Vert u\Vert _{W^{1,p}(I)} \\&\quad \le K_3 a_0^{-1} \varepsilon \Vert Au\Vert _{W^{1,p}(I)} + M'_\varepsilon \Vert u\Vert _{W^{1,p}(I)}. \end{aligned}$$

Taking $\varepsilon$ sufficiently small, we get

$$\begin{aligned} \Vert u\Vert _{W^{1,p}(I)}&\le K_6 \varepsilon \Vert Au\Vert _{W^{1,p}(I)} + M'''_\varepsilon \Vert u\Vert _{W^{1,p}(I)} ,\\ \Vert u'\Vert _{W^{1,p}(I)}&\le K_6 \varepsilon \Vert Au\Vert _{W^{1,p}(I)} + M'''_\varepsilon \Vert u\Vert _{W^{1,p}(I)} \, \end{aligned}$$

whence

$$\begin{aligned} \Vert (Cu)'\Vert _{L^p(I)} \le K_7 \varepsilon \Vert Au\Vert _{W^{1,p}(I)} + M'''_\varepsilon \Vert u\Vert _{W^{1,p}(I)}. \end{aligned}$$

Then,

$$\begin{aligned} \Vert Cu\Vert _{W^{1,p}(I)} \le \varepsilon \Vert Au\Vert _{W^{1,p}(I)} + K_\varepsilon \Vert u\Vert _{W^{1,p}(I)}. \end{aligned}$$

This proves that $C$ is $A$-bounded, with $A$-bound equal to $0$ and hence $A_1 = A + C$ generates an analytic semigroup in $W^{1,p}(I)$. $\square$

3 A generation result in $C^1(I)$

In the monograph [12] it is proved the generation theorem of an analytic semigroup in $C^1_0\left( \overline{\Omega }\right)$, where $\Omega$ is a bounded open subset of $R^N$ with smooth boundary. Here

$$\begin{aligned} C^1_0\left( \overline{\Omega }\right) = \left\{ u\in C^1\left( \overline{\Omega }\right) ; u_{/\partial \Omega } = 0 \right\} . \end{aligned}$$

Let ${\mathcal {L}}(\cdot , D)$ denote a linear second order partial differential operator in $\overline{\Omega }$ with smooth coefficients, whose leading part satifies the uniform ellipticity condition

$$\begin{aligned} \forall x\in \overline{\Omega }, \forall z\in {\mathbb {C}}^n, \quad \sum _{i,j=1}^N a_{ij}(x) z_i \overline{z_j} \ge \alpha _0 |z|^2, \end{aligned}$$

and set

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} \mathcal{D}\left( A_N\right) =\left\{ u\in {\mathcal {D}}(A_{00})\ : A_Nu \in C^1_0\left( \overline{\Omega }\right) \ \right\} ,\\ &{} A_N u:= {\mathcal {L}}(\cdot , D)u.\\ \end{array} \right. \end{aligned}$$

(3.1)

Here

$$\begin{aligned} {\mathcal {D}}(A_{00}) = \left\{ u\in \bigcap _{p=1}^{\infty } W^{2,p}_{loc} (\Omega ) : u,\ {\mathcal {L}}(\cdot , D)u \in C\left( \overline{\Omega }\right) \right\} . \end{aligned}$$

Notice that, if $N=1$, then ${\mathcal {D}}(A_{00}) = \left\{ u\in C^2\left( \overline{\Omega }\right) : u_{/\partial \Omega } = 0 \right\}$.

It is known (see [12], Theorem 3.1.25), that $A_N$ is sectorial in $C^1_0\left( \overline{\Omega }\right)$.

We confine to $N=1$ and define the operator $E$ by

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} \mathcal{D}\left( E\right) =\left\{ u \in C^1_0(I)\ : \alpha u'' \in C^1_0(I) \right\} ,\\ &{} Eu:= \alpha u'', \end{array} \right. \end{aligned}$$

(3.2)

where $\alpha \in C^1(I)$, with $\alpha (x) \ge \alpha _0 > 0$ in $I$.

Theorem 3.1

Under the previous assumptions, the operator $E$ is sectorial in $C^1(I)$.

Proof

Recall that the operator $A_1$ defined above is sectorial in $C^1_0(I)$. Then, if $f\in C^1(I)$, for any $\lambda \in {\mathbb {C}}$, with $\mathrm{Re}\ \lambda$ large enough, there exists a unique $u\in \mathcal{D}\left( A_1\right)$, such that

$$\begin{aligned}&\lambda u(x) - \alpha u''(x) = f(x) - (1 - x) f(0) - xf(1), \nonumber \\&\Vert u\Vert _{C^1_0(I)} \le C|\lambda |^{-1} \Vert f(x) - (1-x) f(0) - xf(1) \Vert _{C^1_0(I)}. \end{aligned}$$

(3.3)

Observe that Eq. (3.3) reads

$$\begin{aligned}&\lambda \left( u(x) + \lambda ^{-1}( (1-x) f(0) + x f(1)) \right) \\&\quad - \alpha (x) \left( u(x) + \lambda ^{-1} ((1-x) f(0) + x f(1)) \right) '' = f(x). \end{aligned}$$

Set $v(x) = u(x) + \lambda ^{-1} ((1-x) f(0) + x f(1))$; then $v\in C^2(I)$ and $\alpha v'' (=\alpha u'') \in C^1_0(I)$. Moreover,

$$\begin{aligned} \Vert v\Vert _{C^1(I)}&\le \Vert u\Vert _{C^1(I)} + 2 |\lambda |^{-1} \Vert f\Vert _{C(I)} \\&\le K |\lambda |^{-1} \Vert f\Vert _{C(I)} + 2 |\lambda |^{-1} \Vert f\Vert _{C(I)} = K_1 |\lambda |^{-1} \Vert f\Vert _{C(I)}. \end{aligned}$$

On the other hand, if $\lambda v_1 - Ev_1 = f = \lambda v_2 - Ev_2$, then $\lambda \left( v_1 - v_2\right) - E \left( v_1 - v_2\right) = 0$, and hence $\alpha \left( v_1 - v_2\right) ''_{/ \partial [0,1]} = 0$, so that $\left( v_1 - v_2\right) (0) = \left( v_1 - v_2\right) (1) = 0$.

Then, $v_1 - v_2 \in \mathcal{D}\left( A_1\right)$ and

$$\begin{aligned} \lambda \left( v_1 - v_2\right) - A_1 \left( v_1 - v_2\right) = 0. \end{aligned}$$

It follows that $v_1 = v_2$ and uniqueness of solutions follows. This implies that $E$ is sectorial in $C^1([0,1])$, as required. $\square$

4 A generation result for a degenerate operator in $W^{1,p}([0,1])$ with Wentzell boundary condition

In this section we show that a second order degenerate operator in $I$ generates an analytic semigroup in $W^{1,p}(I)$, where $1 < p <\infty$. This result must be compared to Theorem 7.9, in the monograph [8]. Here we give an alternative proof of an important case of large interest.

Recall that the norms

$$\begin{aligned} \left( \int _0^1 |u(x)|^p\, dx\right) ^{1/p}+\left( \int _0^1 |u'(x)|^p\, dx\right) ^{1/p} \ \ \ \ \mathrm{and}\ \ \ \ \left( \int _0^1 |u'(x)|^p\, dx\right) ^{1/p} \end{aligned}$$

are equivalent in $W_0^{1,p}(I)$, and define the operator:

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} \mathcal{D}\left( A_{(p)}\right) =\{ u\in W_0^{1,p}(I); \ x(1-x)u''\in W_0^{1,p}(I)\ \},\\ &{}A_{(p)}u=x(1-x)u'',\ \ \ u\in \mathcal{D}\left( A_{(p)}\right) . \end{array} \right. \end{aligned}$$

(4.1)

We prove the following generation result.

Theorem 4.1

The operator $\left( A_{(p)}, \mathcal{D}\left( A_{(p)}\right) \right)$ defined in (4.1) generates an analytic semigroup in $W_0^{1,p}(I)$, for $1<p<\infty$.

Proof

Consider the resolvent equation for $\lambda \in {\mathbb {C}}$, such that $\mathrm{Re} \lambda > 0$, and $f \in W_0^{1,p}(I)$,

$$\begin{aligned} \lambda u(x)-A_{(p)}u(x)=f(x),\ \ \end{aligned}$$

so that

$$\begin{aligned} \displaystyle { \frac{\lambda u(x)}{x(1-x)} }-u''(x)=\displaystyle { \frac{f(x) }{x(1-x)}}. \end{aligned}$$

Note that $f\in W_0^{1,p}(I)$ implies

$$\begin{aligned} \int _0^1 \Big |\displaystyle { \frac{f(x) }{x(1-x)}} \Big |^p\, dx\le c \Vert f\Vert ^p_{ W_0^{1,p}(I)}, \end{aligned}$$

by the Hardy inequality. On the other hand, if $x(1-x)u''=g\in W_0^{1,p}(I)$, then

$$\begin{aligned} |g(x)|\le x^{1/p'}\Vert g\Vert _{L^p(I)},\ \ \ \ |g(x)|\le (1-x)^{1/p'}\Vert g\Vert _{L^p(I)} \end{aligned}$$

so that

$$\begin{aligned} u'(x)-u'(y)=\int _y^x u''(t)\, dt=\int _y^x\displaystyle { \frac{g(t) }{t(1-t)}}\, dt \end{aligned}$$

and in a neighborhood of zero for $y<x$ we have

$$\begin{aligned} |u'(x)-u'(y)|\le \int _y^x\displaystyle { \frac{|g(t)| }{t(1-t)}}\, dt \le c \int _y^x t^{-(1-1/p')}\, dt\le c (x-y)^{1-1/p}, \end{aligned}$$

where $c$, here and in the sequel, denotes a real positive constant. Analogously one has the same bound near $1$. Hence, from

$$\begin{aligned} \lambda \int _0^1 \displaystyle { \frac{|u(x)|^p }{x(1-x)}} \, dx -\int _0^1 u''(x) \,\overline{u(x)}\, |u(x)|^{p-2}\, dx= \int _0^1\displaystyle { \frac{f(x)\, \overline{u(x)} }{x(1-x)}}\, |u(x)|^{p-2}\, dx \end{aligned}$$

by integration by parts, one obtains, using the boundary conditions, if $p \ge 2$:

$$\begin{aligned}&\lambda \int _0^1 \displaystyle { \frac{|u(x)|^p }{x(1-x)}} \, dx +\int _0^1 |u'(x)|^2 \, |u(x)|^{p-2}\, dx\\&\quad +\,(p-2)\int _0^1 u'(x)\overline{u(x)} (\mathrm{Re}( u'(x)\overline{u(x)}) )\, |u(x)|^{p-4}\, dx\\&\quad \le \int _0^1\displaystyle { \frac{f(x)\, \overline{u(x)} }{x(1-x)}}\, |u(x)|^{p-2}\, dx. \end{aligned}$$

We consider the case $p\ge 2$ to show the strategy to obtain suitable estimates for our problem. Attention must be devoted to the case $1<p<2$ that follows with a few modifications of the previous case. Taking real and imaginary parts and integrating by parts yields

$$\begin{aligned}&\left( \mathrm{Re}\ \lambda \right) \int _0^1 \displaystyle { \frac{|u(x)|^p }{x(1-x)}} \, dx +\int _0^1|u'(x)|^2 |u(x)|^{p-2}\, dx\nonumber \\&\quad +\,(p-2)\int _0^1 \left( \mathrm{Re} ( u'(x)\overline{u(x)}) \right) ^2\, |u(x)|^{p-4}\, dx\nonumber \\&\quad = \mathrm{Re}\; \int _0^1 \frac{f(x)\, \overline{u(x)} }{x(1-x)}\, |u(x)|^{p-2}\, dx \end{aligned}$$

(4.2)

and

$$\begin{aligned} \left( \mathrm{Im} \ \lambda \right) \int _0^1 \displaystyle { \frac{|u(x)|^p }{x(1-x)}} \, dx = \mathrm{Im} \ \ \int _0^1\displaystyle { \frac{f(x)\, \overline{u}(x) }{x(1-x)}}\, |u(x)|^{p-2}\, dx. \end{aligned}$$

(4.3)

Then we get, for any $\varepsilon >0$,

$$\begin{aligned}&|\mathrm{Im}\ \lambda | \int _0^1 \displaystyle { \frac{|u(x)|^p }{x(1-x)}}\, dx \!\le \! (p\!-\!2)\ \int _0^1 \left| \mathrm{Im}( u'(x)\overline{u(x)} ) \right| \, \left| \mathrm{Re}( u'(x)\overline{u(x)}) \right| \, |u(x)|^{p-4}\, dx\\&\qquad +\left| \int _0^1\displaystyle { \frac{f(x)\, \overline{u(x)} }{x(1-x)}}\, |u(x)|^{p-2}\, dx\right| \\&\quad \le (p-2)\left( \varepsilon \int _0^1 \left| \mathrm{Im}\left( u'(x)\overline{u(x)}\right) \right| ^2\, |u(x)|^{p-4}\, dx\right. \\&\qquad \left. +\,\frac{1}{\varepsilon } \int _0^1 \left| \mathrm{Re}\left( u'(x)\overline{u(x)}\right) \right| ^2\, |u(x)|^{p-4}\, dx\right) \\&\qquad +\left| \int _0^1\displaystyle { \frac{f(x)\, \overline{u(x)} }{x(1-x)}}\, |u(x)|^{p-2}\, dx\right| . \end{aligned}$$

Let $\eta >0$ be an arbitrary real number. Multiplying the last inequality by $\eta$ and adding it to Eq. (4.2), one gets

$$\begin{aligned}&\left( \mathrm{Re}\lambda +\eta |\mathrm{Im} \lambda |\right) \displaystyle { \int _0^1 \frac{|u(x)|^p }{x(1-x)} \, dx}\\&\quad + \left( p-1- (p-2)\frac{\eta }{\varepsilon }\right) \displaystyle {\int _0^1 \left( \mathrm{Re}\left( u'(x)\overline{u(x)}\right) \right) ^2\, |u(x)|^{p-4}\, dx}\\&\quad +\,(1-(p-2)\varepsilon \eta )\displaystyle { \int _0^1\left( \mathrm{Im} \left( u'(x)\overline{u(x)}\right) \right) ^2|u(x)|^{p-4}\, dx}\\&\quad \le \left| \mathrm{Re}\displaystyle { \int _0^1\displaystyle { \frac{f(x)\, \overline{u(x)} }{x(1-x)}}\, |u(x)|^{p-2}\, dx}\right| \\&\quad +\,\eta \left| \mathrm{Im} \int _0^1\displaystyle { \frac{f(x)\, \overline{u(x)} }{x(1-x)}}\, |u(x)|^{p-2}\, dx\right| . \end{aligned}$$

Since $\eta$ and $\varepsilon$ are arbitrary positive constants, we can choose them so that

$$\begin{aligned} \eta < \varepsilon , \quad \varepsilon ^2 (p-2) < 1\,; \end{aligned}$$

this assures that $p-1- (p-2)\eta /\varepsilon > 0$ and that $1-(p-2)\varepsilon \eta > 0$; then in the sector $\Sigma :=\{ \lambda \in {\mathbb {C}}\ :\ \mathrm{Re}\lambda +\eta |\mathrm{Im}\lambda |\ge \delta _0>0\ \}$, we have the estimates

$$\begin{aligned} \int _0^1 \left( \mathrm{Re}\left( u'(x)\overline{u(x)}\right) \right) ^2\, |u(x)|^{p-4}\, dx&\le c\Vert f\Vert _{W^{1,p}_0(I)}\Vert u\Vert ^{p-1}_{L^p(I)},\\ \int _0^1\left( \mathrm{Im} \left( u'(x)\overline{u(x)}\right) \right) ^2|u(x)|^{p-4}\, dx&\le c\Vert f\Vert _{W^{1,p}_0(I)}\Vert u\Vert ^{p-1}_{L^p(I)}. \end{aligned}$$

Furthermore,

$$\begin{aligned} \mathrm{Re}\lambda \; \int _0^1 \displaystyle { \frac{|u(x)|^p }{x(1-x)}} \, dx&\le c\Vert f\Vert _{W^{1,p}_0(I)}\Vert u\Vert ^{p-1}_{L^p(I)}, \\ |\mathrm{Im }\lambda |\; \int _0^1 \displaystyle { \frac{|u(x)|^p }{x(1-x)}} \, dx&\le c\Vert f\Vert _{W^{1,p}_0(I)}\Vert u\Vert ^{p-1}_{L^p(I)}. \end{aligned}$$

Therefore

$$\begin{aligned} \left( \mathrm{Re}\lambda + |\mathrm{Im} \lambda |\right) \Vert u\Vert ^p_{L^p(I)} \le \left( \mathrm{Re}\lambda + |\mathrm{Im} \lambda |\right) \int _0^1 \displaystyle { \frac{|u(x)|^p }{x(1-x)}} \, dx \le c\Vert f\Vert _{W^{1,p}_0(I)}\Vert u\Vert ^{p-1}_{L^p(I)}, \end{aligned}$$

which gives

$$\begin{aligned} \left( \mathrm{Re}\lambda + |\mathrm{Im} \lambda |\right) \Vert u\Vert _{L^p(I)} \le c\Vert f\Vert _{W^{1,p}_0(I)}. \end{aligned}$$

We need now to estimate $\Vert u'\Vert _{L^p(I)}$. Consider

$$\begin{aligned} \lambda u'(x)-[x(1-x)u''(x)]'=f'(x) \end{aligned}$$

and multiply it by $\overline{u'(x)}|u'(x)|^{p-2}$, so that

$$\begin{aligned}&\lambda \int _0^1|u'(x)|^p\, dx-\big [ x(1-x)u''(x)\overline{u'(x)}|u'(x)|^{p-2}\big ]_0^1\\&\quad +\int _0^1 x(1-x)|u''(x)|^2\, |u'(x)|^{p-2}\, dx\\&\quad +\,(p-2)\int _0^1 x(1-x)\overline{u'(x)}\, u''(x)\, \mathrm{Re}\left( \overline{u'(x)}\, u''(x)\right) \,|u'(x)|^{p-4}\, dx\\&\quad =\int _0^1 f(x)\, \overline{u'(x)}\, |u'(x)|^{p-2}\, dx. \end{aligned}$$

As $x(1-x)u''(x)$ vanishes on the boundary and $u'(x)$ has a finite limit as $x\rightarrow 0$ or $x\rightarrow 1$, we obtain

$$\begin{aligned} \left[ x(1-x)u''(x)\overline{u}'(x)|u'(x)|^{p-2}\right] _0^1=0. \end{aligned}$$

By repeating for $u'$ the arguments above for $u$ we get the estimate, if $\lambda \in \Sigma$,

$$\begin{aligned} |\lambda |\Vert u'\Vert _{L^p(I)} \le c \Vert f'\Vert _{L^p(I)}\,; \end{aligned}$$

so we have the bound, if $\lambda \in \Sigma$,

$$\begin{aligned} |\lambda |\Vert u\Vert _{W^{1,p}_0(I)} \le c \Vert f'\Vert _{W^{1,p}_0(I)}. \end{aligned}$$

Take now $f\in W^{1,p}_0(I)$, so that $f'\in L^p(I)$ and consider the problem

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} \lambda v(x)-(x(1-x)v'(x))'=f'(x),\ \ \ x\in I,\\ &{} x(1-x)v'(x)\in W^{1,p}_0(I), \end{array} \right. \end{aligned}$$

(4.4)

for $\mathrm{Re} \lambda >0$. It is known, from the paper [2], that it has a unique solution such that

$$\begin{aligned} |\lambda |\Vert v\Vert _{L^p(I)} \le c \Vert f'\Vert _{L^p(I)}. \end{aligned}$$

Integrating over $(0,x)$ we obtain

$$\begin{aligned} \lambda \int _0^x v(t)\, dt-x(1-x)v'(x)=f(x). \end{aligned}$$

Set

$$\begin{aligned} u(x)=\int _0^x v(t)\, dt\,; \end{aligned}$$

then

$$\begin{aligned} \lambda u(x)-x(1-x)u''(x)=f(x), \quad 0\le x\le 1, \end{aligned}$$

so $u$ is the desired solution to our problem. Therefore $A_{(p)}$ generates an analytic semigroup in $W_0^{1,p}(I)$. $\square$

We now define the operator $W$, for $1< p < \infty$,

$$\begin{aligned} \left\{ \begin{array}{clcr} &{}\mathcal{D}(W)=\{ u\in W^{1,p}(I)\ :\ x(1-x)u''\in W^{1,p}_0(I)\ \},\\ &{} Wu:=x(1-x)u'',\ \ \forall \ u\in \mathcal{D}(W). \end{array} \right. \end{aligned}$$

(4.5)

Theorem 4.2

The operator $(W, \mathcal{D}(W))$ generates an analytic semigroup in $W^{1,p}(I)$.

Proof

We only need to observe, see Theorem 2.1, that

$$\begin{aligned} x(1-x)u''(x)=0, \ \ \ x\in I, \end{aligned}$$

and

$$\begin{aligned} u(0)=u_0,\ \ u(1)=u_1 \end{aligned}$$

has a unique solution in $C^\infty ([0,1])$. $\square$

5 A generation result in $C(\overline{\Omega })$ with Wentzell boundary conditions

In this case we have a more general result. Let $\Omega$ be a bounded domain in ${\mathbb {R}}^n$ with smooth boundary $\Gamma$. Let us assume the conditions

• $(K1)$ $a_{jk}, b_k, d$ are real $C^\infty (\overline{\Omega })$ functions, with $a_{jk}=a_{kj}$, $k,j=1,...,n$, and $d(x)<0$;

• (K2) $\sum _{j,k=1}^n a_{jk}(x)\xi _j\xi _k\ge c\sum _{j=1}^k\xi _j^2$, $\forall x\in \overline{\Omega }$ and $(\xi _1,...,\xi _n)\in {\mathbb {R}}^n$; here $c>0$.

Let us define the elliptic operator :

$$\begin{aligned} Lu(x)= \sum _{j,k=1}^n \frac{\partial }{\partial x_j}\left( a_{jk}(x)\frac{\partial u(x)}{\partial x_k}\right) +\sum _{k=1}^n b_{k}(x) \frac{\partial u(x)}{\partial x_k} +d(x)u(x) \end{aligned}$$

(5.1)

with domain

$$\begin{aligned} \mathcal{D}(L)=\left\{ u\in \bigcap _{p>1}W^{2,p}_{\mathrm{loc}}(\Omega )\ :\ \ u,\ Lu\in C(\overline{\Omega }) \ \right\} . \end{aligned}$$

(5.2)

We recall the following well known theorem (see [10] Corollary 9.18).

Theorem 5.1

Let $L$ be the operator defined in (5.1)–(5.2), satisfying assumptions $(K1)$ and $(K2)$. Then, for any $\phi \in C(\Gamma )$, the boundary value problem

$$\begin{aligned} \left\{ \begin{array}{clcr} &{}LG(x)=0, \ \ x\in \Omega ,\\ &{}G(x)=\phi (x),\ \ \ x\in \Gamma ,\\ \end{array} \right. \end{aligned}$$

(5.3)

has a unique solution $G\in \mathcal{D}(L)$ and there exists a positive constant $M$ such that

$$\begin{aligned} \Vert G\Vert _{C(\overline{\Omega })}\le M\Vert \phi \Vert _{C(\Gamma )}. \end{aligned}$$

(5.4)

In this section we will use the functional space

$$\begin{aligned} C_0(\overline{\Omega })=\{ u\in C(\overline{\Omega })\ :\ \ u|_{\Gamma }=0\ \} \end{aligned}$$

(5.5)

with its natural norm. Thanks to Corollary 3.1.21 (ii) in Lunardi [12] we observe that operator

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} \mathcal{D}\left( L_0\right) =\left\{ u \in \bigcap \nolimits _{p>1} W^{2,p}_{\mathrm{loc}} (\Omega )\ : \ u, Lu\in C(\overline{\Omega }),\ \ u|_{\Gamma }=0\right\} ,\\ &{} L_0u=Lu,\ \ \ u\in \mathcal{D}\left( L_0\right) , \end{array} \right. \end{aligned}$$

(5.6)

is sectorial in $C(\overline{\Omega })$. This implies that the operator $L_1$

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} \mathcal{D}\left( L_1\right) = \left\{ u\in \bigcap \nolimits _{p>1} W^{2,p}_{\mathrm{loc}}(\Omega )\ :\ u, Lu\in C_0(\overline{\Omega })\ \right\} \\ &{} L_1u=Lu,\ \ \ u\in \mathcal{D}\left( L_1\right) \end{array} \right. \end{aligned}$$

(5.7)

generates an analytic semigroup in $C_0(\overline{\Omega })$. We are now in position to prove the following generation result.

Theorem 5.2

Let $L$ be the operator defined in (5.1), satisfying assumptions $(K1)$ and $(K2)$. Then the operator $L_W$ defined by

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} \mathcal{D}\left( L_W\right) =\left\{ u\in \bigcap \nolimits _{p>1} W^{2,p}_{\mathrm{loc}}(\Omega )\ :\ u\in C(\overline{\Omega }),\ \ Lu\in C_0(\overline{\Omega })\ \right\} ,\\ &{} L_Wu=Lu,\ \ \ u\in \mathcal{D}\left( L_W\right) , \end{array} \right. \end{aligned}$$

(5.8)

generates an analytic semigroup in $C(\overline{\Omega })$.

Proof

Consider the resolvent equation, for $f\in C(\overline{\Omega })$

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} \lambda u(x)- Lu(x)=f(x),\ \ x\in \overline{\Omega },\\ &{} Lu(x)|_{\Gamma }=0, \end{array} \right. \end{aligned}$$

(5.9)

and let $G$ be the solution of the problem

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} LG(x)=0,\ \ x\in \overline{\Omega }, \\ &{} G|_{\Gamma }=f|_{\Gamma }. \end{array} \right. \end{aligned}$$

(5.10)

Problem (5.9), taking into account (5.10), can be transformed into

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} \lambda \Big ( u -\frac{G}{\lambda }\Big ) - L\Big ( u -\frac{G}{\lambda }\Big )=f-G,\ \ x\in \overline{\Omega },\\ &{}\Big ( u -\frac{G}{\lambda }\Big )\Big |_{\Gamma }=0,\ \ \ \ L\Big ( u -\frac{G}{\lambda }\Big )\Big |_{\Gamma }=0. \end{array} \right. \end{aligned}$$

(5.11)

Note that $f-G\in C_0 (\overline{\Omega })$ so that there is a unique $u-\frac{G}{\lambda }\in C_0(\overline{\Omega })$, in view of the previous considerations, that satisfies:

$$\begin{aligned} \left\| u-\frac{G}{\lambda } \right\| _{C_0(\overline{\Omega })}\le \frac{c}{|\lambda |}\, \Vert f-G\Vert _{C_0(\overline{\Omega })}. \end{aligned}$$

(5.12)

On the other hand, if $v:=u-G/\lambda$ satisfies $\lambda v-A_Wv=f-G$, then $u=v+G/\lambda$ belongs to $\mathcal{D}\left( A_W\right)$ and satisfies Problem (5.9). Moreover,

$$\begin{aligned} \Vert u\Vert _{C(\overline{\Omega }) }&\le \left\| u-\frac{G}{\lambda }\right\| _{C(\overline{\Omega })}+\left\| \frac{G}{\lambda }\right\| _{C(\overline{\Omega })} \le \displaystyle {\frac{c}{|\lambda |}} \left( \Vert f-G\Vert _{C(\overline{\Omega })}+\Vert f\Vert _{C(\Gamma )}\right) \end{aligned}$$

(5.13)

$$\begin{aligned}&\le \displaystyle {\frac{c_1}{|\lambda |}} \Big ( \Vert f\Vert _{C(\overline{\Omega })}+2\Vert f\Vert _{C(\Gamma )}\Big ) \le \displaystyle {\frac{c_2}{|\lambda |}} \Vert f\Vert _{C(\overline{\Omega })}, \end{aligned}$$

(5.14)

where $c_i$ denote positive constants, so that $L_W$ generates an analytic semigroup in $C(\overline{\Omega })$. $\square$

We now apply a perturbation argument to obtain the following generation result. Let $\tilde{L}$ be the operator in $C(\overline{\Omega })$ defined by

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} \mathcal{D}\left( \tilde{L}\right) =\left\{ u\in \bigcap _{p>1} W^{2,p}_{\mathrm{loc}}(\Omega )\ : \ \ \ \sum _{j,k=1}^n \frac{\partial }{\partial x_j}\left( a_{jk}(\cdot )\frac{\partial u}{\partial x_k}\right) \right. \\ &{} +\sum _{k=1}^n b_{k}(\cdot ) \frac{\partial u}{\partial x_k} +d(\cdot )u\in C_0\left( \overline{\Omega }) \right\} , \\ &{} \tilde{L}u=\sum _{j,k=1}^n \frac{\partial }{\partial x_j} \left( a_{jk}(x)\frac{\partial u(x)}{\partial x_k}\right) ,\ \ \ u\in \mathcal{D}\left( \tilde{L}\right) . \end{array} \right. \end{aligned}$$

(5.15)

By a perturbation method we get the following generation result.

Theorem 5.3

The operator $\tilde{L}$, defined in (5.15), generates an analytic semigroup in $C(\overline{\Omega })$.

Proof

From Theorem 5.2 we know that $A_W$, defined in (5.8) generates an analytic semigroup in $C(\overline{\Omega })$. Let us introduce the operator

$$\begin{aligned} \left\{ \begin{array}{clcr} &{} \mathcal{D}(\tilde{C})=\mathcal{D}(\tilde{L})=\left\{ u\in \bigcap _{p>1} W^{2,p}_{\mathrm{loc}}(\Omega )\ : \ \ \ \sum _{j,k=1}^n \frac{\partial }{\partial x_j}\left( a_{jk}(\cdot )\frac{\partial u}{\partial x_k}\right) \right. \\ &{}\left. +\sum _{k=1}^n b_{k}(\cdot ) \frac{\partial u}{\partial x_k} +d(\cdot )u\in C_0(\overline{\Omega })\ \right\} ,\\ &{} \tilde{C} u=-\sum _{k=1}^n b_{k}(\cdot ) \frac{\partial u}{\partial x_k} -d(\cdot )u ,\ \ \ u\in \mathcal{D}(\tilde{C}). \end{array} \right. \end{aligned}$$

(5.16)

Then

$$\begin{aligned} \mathcal{D}(\tilde{C})\hookrightarrow W^{2,p}(\Omega ),\ \ \mathrm{for\ any}\ p>1, \end{aligned}$$

and by the Rellich’s imbedding theorem we have

$$\begin{aligned} W^{2,p}(\Omega )\hookrightarrow ^c C^1(\overline{\Omega })\hookrightarrow C(\overline{\Omega }) ,\ \ \mathrm{for\ any}\ p>n. \end{aligned}$$

Applying the Ehrling Lemma (see, e.g. [19], Theorem 7A16) , we have that for every $\varepsilon >0$ there exists $c(\varepsilon )>0$ such that

$$\begin{aligned} \Vert u\Vert _{ C^1(\overline{\Omega })}\le \varepsilon \Vert u\Vert _{ \mathcal{D}(\tilde{C})}+M_{\varepsilon }\Vert u\Vert _{ C(\overline{\Omega })} \end{aligned}$$

in other words $\Vert Cu\Vert _{ C(\overline{\Omega })}$ is estimated by

$$\begin{aligned} \Vert Cu\Vert _{ C(\overline{\Omega })}\le \varepsilon \Vert A_Wu\Vert _{C(\overline{\Omega }) }++M_{\varepsilon }\Vert u\Vert _{ C(\overline{\Omega })}. \end{aligned}$$

Since $\tilde{C}$ is $A_W$-bounded with $A_W$-bound equal to zero, the statement follows. $\square$