text:
\section{Introduction}
In this paper we provide a full analysis of two  operators under certain Neumann boundary conditions. More precisely, we consider
\begin{equation}\label{operatori A_i}
	A_iu:=\begin{cases}
		\frac{\partial^2}{\partial x^2}\Bigl (a\frac{\partial^2u}{\partial x^2}\Bigr ) &\text{if }\,\, i=1,\\
		a\frac{\partial^4u}{\partial x^4} &\text{if }\,\, i=2,
	\end{cases}
\end{equation}
where the function $a:[0,1]\to\mathbb{R}_+$ degenerates at $x_0\in [0,1]$.

We are interested in studying this kind of degenerate operators, which represent the subject of numerous papers and books, together with suitable Neumann boundary conditions for two reasons.

First of all, many problems coming from Physics, Biology and Economics are governed by degenerate operators
that often have to do with Neumann boundary conditions.

Secondly, there are real problems connected to freely supported beams (with only fixed end points and without any other condition) and beams supported at only one end, that lead to natural boundary conditions. It is relevant to observe that end point conditions are modular, i.e., it is possible to use different end point conditions at each end of the beam. In particular, the main applications are characterized by the choice of different conditions:
\begin{itemize}
	\item clamped: specifies the position and the derivative;
	\item freely supported: specifies the position (in this case the natural boundary condition is that the second derivative is zero at the end point);
	\item no condition: neither position nor end point are specified (the natural boundary conditions fix the second and third derivatives at the end point to be zero).
\end{itemize}

Let us present some interesting contributions about the boundary conditions for problems associated to the beam analysis.

Starting from a static fourth order Euler-Bernoulli beam equation, we recall \cite{bo yang} where the authors study the fourth order ordinary differential equation
\begin{equation}\label{+}
	u''''(x)=g(x)f(u(x)),\quad\,\,\,\,\,\,\,\,\,\,\,0\le x\le 1,
\end{equation}
which is often referred to as the beam equation. In the formulation of the previous equation, the authors assume that $f:[0,+\infty)\to [0,+\infty)$ and $g:[0,1]\to [0,+\infty)$ are continuous functions and $g$ is such that $\int_0^1 g(t)dt>0$. From the physical point of view, \eqref{+} describes the deflection or deformation of an elastic beam under a certain force, together with the following  boundary conditions
\begin{equation}\label{bound cond}
	u(0)=u'(0)=u''(1)=u'''(1)=0,
\end{equation}
which arise from the study of elasticity.
 These conditions have definite physical meanings; in fact, they mean that the
beam is embedded at the end $x=0$ and free at the end $x=1$.
Furthermore, the boundary conditions (\ref{bound cond}) represent a special case of the so called right focal boundary conditions, for which an extensive research has been done (we refer to  \cite{argai} for a systematic survey of this field; for some recent results on focal boundary value problems, see, e.g., \cite{argai regan} and the references therein).
Equation \eqref{+} is also analyzed in \cite{bo yang2}, where it is coupled with the boundary conditions
\begin{equation*}
	u(0)=u'(0)=u'(1)=u'''(1)=0,
\end{equation*}
which mean that the beam is embedded at $x=0$ and fastened with a sliding clamp at  $x=1$.
In \cite{GUPTA} a boundary value problem under the following boundary conditions is considered:
\begin{equation*}
	u'(0)=u'''(0)=u'''(\pi)=u'(\pi)=0.
\end{equation*}
It means that the beam is fastened with sliding clamps at both ends $x=0$ and $x=\pi$. Similar conditions are considered in \cite{yang zhang} in order to study the existence of non trivial solutions to the semilinear fourth order problem
\begin{equation*}
	\begin{cases}
		u''''(x)-2u''(x)+u(x)=f(x,u(x)),& 0<x<1,\\
		u'(0)=u'(1)=u'''(0)=u'''(1)=0,
	\end{cases}
\end{equation*}
being $f\in\mathcal{C}([0,1]\times\mathbb{R};\mathbb{R})$.
Imposing some ideal conditions, the deflection of a beam rigidly
fastened at left and simply fastened at right led \cite{argai0} to a fourth order non linear differential equation together with two point boundary conditions.
Then \cite{elgindi} takes in consideration different boundary conditions, corresponding to various ways in which the ends of a beam may be supported, and examines a class of  fourth order non linear boundary value problems which govern the equilibrium states of a beam-column. In this case the source of the non linearity comes from a non linear lateral constraint (foundation) and the equilibrium equation is formulated as a fourth order non linear differential equation along with one of the following six sets of boundary conditions:
\begin{equation*}
	\begin{split}
		u(0)=u''(0)=u(1)=u''(1)=0,\\
		u(0)=u''(0)=u(1)=u'(1)=0,\\
		u(0)=u''(0)=u'(1)=u'''(1)=0,\\
		u(0)=u'(0)=u(1)=u'(1)=0,\\
		u(0)=u'(0)=u''(1)=u'''(1)=0,\\
		u(0)=u'(0)=u'(1)=u'''(1)=0.
	\end{split}
\end{equation*}
These conditions represent, respectively, the following situations: both ends are simply-supported; one end is simply-supported and the other is fixed; one end is simply-supported and the other is sliding clamped; both ends are fixed; one end is fixed and the other is free and one end is fixed and the other is sliding clamped. See also \cite{kosmatov}, in which the beam equation is studied together with the boundary conditions
\[
	u(0)=u'(0)=u'(1)=u(1)=0,
\]
which mean that the beam is embedded at both ends $x=0$ and $x=1$.

For other works connected to boundary value problems of the beam equation we refer, for example, to \cite{bai}, \cite{davis},
\cite{graef}, \cite{gupta},
\cite{liu},
\cite{ma2}, \cite{yang}, \cite{yao2} and the references therein.

However,  there are some equations of great interest due to their applications in many engineering fields, such as mechanical engineering and civil engineering, that are not written as a static beam equation, but as a partial differential equation governed by the operators introduced in \eqref{operatori A_i}. In this context, the coefficient $a(x)$ may be thought as the flexural rigidity (for $i=1$) or as the density of the beam (for $i=2$).
Indeed, the problem of the transversely vibrating beam, for which the most important factor is represented by the bending effect, is formulated in terms of the partial differential equation of motion, of the boundary and initial conditions, which give rise to an initial boundary value problem.
More precisely, if $u(t,x)$ denotes the transverse displacement at time $t$ and position $x$ from one
end of the beam taken as the origin, $a(x)$ the flexural rigidity,
and $m(x) > 0$ the lineal mass, the transverse motion of an unloaded thin beam is represented by the following Euler-Bernoulli beam equation
\[
m(x)u_{tt} +(a(x)u_{xx})_{xx} =0,
\]
where $t>0$ and $x \in (0,L)$.
On the other hand, if we consider an external forcing function $f$ and if $E$ is the elastic modulus (or Young's modulus), $I$ is the area of inertia and $\lambda$ is the mass per unit length, then
 the problem of the deflection of the beam can be described by the following fourth order partial differential equation:
	\begin{equation}\label{01}
		\lambda \frac{\partial^2u}{\partial t^2}+\frac{\partial^2}{\partial x^2}\Biggl (EI\frac{\partial^2u}{\partial x^2}\Biggr )=f(u).
	\end{equation}
	The applied load function $f$, which is  sufficiently smooth, can depend on $u$, as in this case, or  on time and position.
If $E, I, \lambda$ are constant, after a rescaling, the previous equation becomes
\begin{equation*}
	\frac{\partial^2u}{\partial t^2}+ \frac{\partial^4u}{\partial x^4}=f(u).
\end{equation*}
Starting from \eqref{01}, a lot of papers are devoted to the study of models coming from the previous one. In particular, in \cite{russell1} and \cite{russell2} Russell notes that  viscous damping models such as
	\begin{equation*}
		\rho\frac{\partial^2u}{\partial t^2}+2\frac{\partial u}{\partial t}+	\frac{\partial^2}{\partial x^2}\Biggl (EI	\frac{\partial^2u}{\partial x^2} \Biggr )=0,
	\end{equation*}
	which produce uniform damping rates, are inadequate if experimentally observed damping
properties are to be incorporated in the model. At the end of the
nineteenth century, Kelvin and Voigt note  that damping rates tend to increase with frequency. Incorporated into
the Euler-Bernoulli beam model their approach yields an equation of the form
	\begin{equation*}
		\rho\frac{\partial^2u}{\partial t^2}+2\rho\frac{\partial^3}{\partial t\partial x^2}\Biggl (EI	\frac{\partial^2u}{\partial x^2} \Biggr )+	\frac{\partial^2}{\partial x^2}\Biggl (EI	\frac{\partial^2u}{\partial x^2} \Biggr )=0.
	\end{equation*}
An interesting variant of a Kelvin-Voigt viscoelastic damped Euler-Bernoulli beam is represented by the following equation
\begin{equation}\label{wp}
	\frac{\partial^2u}{\partial t^2}+\frac{\partial^2}{\partial x^2}\Biggl (a\frac{\partial^3u}{\partial t\partial x^2}\Biggr )+\frac{\partial^2}{\partial x^2}\Biggl (a	\frac{\partial^2u}{\partial x^2}\Biggr )=0,
\end{equation}
for which we appreciate a deep connection with parabolic equations. Indeed, considering $f$ such that
\begin{equation*}
	\frac{\partial f}{\partial t}+\frac{\partial^2}{\partial x^2}\Biggl (a	\frac{\partial^2u}{\partial x^2}\Biggr )=0,
\end{equation*}
equation \eqref{wp} can be rewritten as
\begin{equation}\label{problem1}
	\frac{\partial u}{\partial t}+\frac{\partial^2}{\partial x^2}\Biggl (a	\frac{\partial^2u}{\partial x^2}\Biggr )=f.
\end{equation}
A problem similar to \eqref{problem1} can also be found in the study of thin films and phase field models, see \cite{elliott}. Indeed, in this case the general form is
\[
u_t= \nabla\cdot(f(u)\nabla w),
\]
where
\[
w= -\gamma \Delta u + \varphi(u),
\]
$f$ is non negative, $u>0$ and both $f$ and $\varphi$ are smooth. If we take $\gamma =1$ and after some computations, one can rewrite the previous problem as \eqref{problem1} (for more details see \cite{bertozzi}).

In this respect \cite{CF} and \cite{3} are the first papers that deal with
degenerate fourth order operators. In particular, in \cite{CF} we consider the operator $A_2$ defined in \eqref{operatori A_i}, while in   \cite{3} the operator $A_1$ is studied, both
with Dirichlet boundary conditions. We underline that while in \cite{CF} the degeneracy point belongs to $[0,1]$, in \cite{3} only the interior degenerate case is considered.
Moreover, in \cite{CF} the properties of $A_2$ are used to obtain the well posedness of the associated parabolic problem.

Motivated by the above problems, in this paper we study  the operators $A_i$, $i=1,2$, introduced in \eqref{operatori A_i} with Neumann boundary conditions. We admit two types of degeneracy for $a$, namely weak and strong degeneracy according to the following definitions:
\begin{Definition}
	The function $a$ is weakly degenerate if there exists a point $x_0\in [0,1]$ such that  $a(x_0)=0$, $a>0$ on $[0,1]\setminus \{x_0\}$, $a\in\mathcal{C}[0,1] \cap \mathcal{C}^1([0,1] \setminus \{x_0\})$ and $\frac{1}{a}\in L^1(0,1)$.
\end{Definition}
For example, as $a$, we can consider $a(x)=|x-x_0|^{\alpha}$, $0<\alpha<1$.
\begin{Definition}\label{def strong function}
	The function $a$ is strongly degenerate if there exists a point $x_0\in [0,1]$ such that  $a(x_0)=0$, $a>0$ on $[0,1]\setminus \{x_0\}$, $a\in \mathcal{C}^1([0,1] \setminus \{x_0\}) \cap W^{1, \infty}(0,1)$ and $\frac{1}{a}\notin L^1(0,1)$.

\end{Definition}
For example, as $a$, we can consider $a(x)=|x-x_0|^{\alpha}$, $\alpha\ge 1$.

Regarding the boundary terms considered in this paper, we underline that they are different according to the degeneracy of $a$ and to the {\it form} of the operator. Consider, for example, the operator  in divergence form $A_1$. If the degeneracy point $x_0$ belongs to $(0,1)$, we can consider the following boundary conditions
\begin{equation}\label{condition1}
\begin{cases}
u''(0)=u'''(0)=0,\\
u''(1)=u'''(1)=0.
\end{cases}
\end{equation}
Indeed, since $u$ is $\mathcal C^3$ in a neighbourhood of $x=0$ and $x=1$, the previous terms are well defined. If $x_0=0$, then at $x=1$ we can consider the same conditions as in \eqref{condition1}, but at $0$ we have to require other conditions since we do not know that $u \in \mathcal C^3$  at $x=0$. Actually, the natural boundary conditions at $x=0$ are
\[
(au'')(0)=(au'')'(0)=0,
\]
which make sense since we assume that $au'' \in H^2(0,1)$.
Analogously, if $x_0=1$. On the other hand, take the operator in non divergence form $A_2$; then the boundary conditions are given by \eqref{condition1}  whether the function $a$ is weakly or strongly degenerate and $x_0 \in (0,1)$. Actually \eqref{condition1} are the natural boundary conditions even if $x_0=0$ or $x_0=1$ and the function $a$ is weakly degenerate; on the other hand, in the strongly degenerate case, we require
\[
\begin{cases}
u''(0)=u'''(0)=u''(1)=u(1)=0, & \text{if} \; x_0=1,\\
u''(1)=u'''(1)=u''(0)=u(0)=0, & \text{if} \; x_0=0,
\end{cases}
\]
(see Sections \ref{Sec2} and \ref{Sec3} for more details).

The paper is constituted by two main sections in which we study  the operators in divergence and non divergence form with suitable boundary conditions. We then apply the properties of the operators $A_i$, $i=1,2$, to study the well posedness of the following parabolic problems
\[
	\begin{cases}
		\frac{\partial u}{\partial t}(t,x)+A_iu(t,x)=h(t,x), &(t,x)\in (0,T) \times (0,1),\\

		u(0,x)=u_0(x),&x\in(0,1),
	\end{cases}
\]
where  $T>0$, $u_0$ belongs to a suitable Hilbert space $H$ and $h\in L^2(0,T;H)$.  In particular, $H$ denotes the Hilbert space $L^2(0,1)$, in the divergence form (i.e., $i=1$), and a suitable weighted Hilbert space, in the non divergence one (i.e., $i=2$).

A final comment on the notations: by $C$ we shall denote universal positive constants, which are allowed to vary from line to line; we denote with $u'$ the derivative of a function $u$ depending only on
one variable x, which we assume to vary in $[0, 1]$, while $u_t$, $u_{x}$ are the usual partial derivatives $\frac{\partial u}{\partial t}$, $\frac{\partial u}{\partial x}$, respectively, of a function $u=u(t,x)$.

It is worth noting that in the present work we deal with real function spaces, but the assertions can be easily extended to the complex case.

\section{The operator in divergence form}\label{Sec2}
In this section we consider the operator $A_1u:= (au_{xx})_{xx}$
with suitable boundary conditions and we prove its generation property.

To this aim we discuss two situations: the weakly degenerate case and the strongly degenerate one.

\subsection{Weakly degenerate setting}\label{Section 2}
Throughout this subsection we assume that the function $a$ is weakly degenerate.

In order to deal with the generation property of  $A_1$ we have to consider a suitable domain that will be defined later. To this aim,
inspired by \cite{ACF}, \cite{FGGR} or \cite{fm2013}, we introduce the following (weighted) Hilbert spaces:
\[\begin{aligned}
	H^i_a(0,1):=\{u\in H^{i-1}(0,1):& \; u^{(i-1)} \text{ is absolutely continuous in [0,1]},\\
	& \sqrt{a}u^{(i)}\in L^2(0,1)\},
\end{aligned}\]
endowed with the norms
\begin{equation}\label{normadiv}
	\|u\|^2_{H^i_a(0,1)}:= \sum_{j=0}^{i-1}\|u^{(j)}\|^2_{L^2(0,1)}+ \|\sqrt{a}u^{(i)}\|^2_{L^2(0,1)} \quad \forall \; u \in H^i_a(0,1),
\end{equation}
$i=1,2$. Here  $H^0(0,1):= L^2(0,1)$ and $u^{(0)}=u$. Observe that $H^1_a(0,1)$ is exactly the space considered for the first time in \cite{CMP}.
Furthermore, in \cite[Proposition 2.1]{CF-articolo Wentzell} it is proved that for every $u\in H^i_a(0,1)$ the norm $\|u\|^2_{H^i_{a}(0,1)}$ is equivalent to the following one
\begin{equation}\label{norma tre stanghette}
|||u|||^2:= \|u\|^2_{L^2(0,1)} +  \|\sqrt{a}u^{(i)}\|^2_{L^2(0,1)},
\end{equation}
$i=1,2$.
Then, introducing
\begin{equation*}
	{\mathcal Z}(0,1):=\{u\in H^2_a(0,1): au''\in H^2(0,1)\},
\end{equation*}
define the operator $A_1$ by
\begin{equation}\label{ope1}
	\begin{split}
		A_1u:&=(au'')'',\\
		\forall\; u\in D(A_1):=\{u\in {\mathcal Z}(0,1):\, u''(0)&=u''(1)=0, u'''(0)=u'''(1)=0\},
	\end{split}
\end{equation}
if $x_0\in (0,1)$.
If $x_0\in \{0,1\}$, we consider natural boundary conditions which arise by computation (see Theorem \ref{teorema generazione (div)}). Indeed, the definition of the domain of the operator is slightly different.
More precisely, if $x_0=0$, we consider  as $D(A_1)$ the set
\begin{equation*}
	\begin{split}
		D(A_1):=\{u\in {\mathcal Z}(0,1): u''(1)=u'''(1)=0, (au'')(0)=(au'')'(0)=0\},
	\end{split}
\end{equation*}
while, if $x_0=1$, the domain becomes
\begin{equation*}
	\begin{split}
		D(A_1):=\{u\in {\mathcal Z}(0,1): u''(0)=u'''(0)=0, (au'')(1)=(au'')'(1)=0\}.
	\end{split}
\end{equation*}
Note that in \cite{ACF} a degenerate operator of a second order in divergence form with a Neumann boundary condition of the type $(au_x)(t,0)=0$ and $u(t,1)=0$ is considered  if $x_0=0$; we refer to \cite{BFM} if $x_0 \in (0,1)$.

To prove that $(A_1, D(A_1))$ generates a strongly continuous semigroup the following formula of integration by parts is essential:
\begin{Lemma}\label{lemma gauss-green}
	For all $(u,v)\in {\mathcal Z}(0,1) \times H^2_a(0,1)$ one has
	\begin{equation}\label{gf0}
		\int_{0}^{1}(au'')''v\,dx=[(au'')'v]^{x=1}_{x=0}-[au''v']^{x=1}_{x=0}+\int_{0}^{1}au''v''dx.
	\end{equation}
\end{Lemma}
\begin{proof}
	Take $u \in {\mathcal Z}(0,1)$ and $v \in H^2_a(0,1)$. Clearly, $au'', (au'')' \in H^1(0,1)$ and $v, v'$ are absolutely continuous;
	hence
	\[
	\begin{aligned}
		\int_0^1 (au'')'' v\,dx&=  [(au'')'v]_{x=0}^{x=1}-\int_0^1 (au'')'v'dx\\
		&=  [(au'')'v]_{x=0}^{x=1}-[au''v']_{x=0}^{x=1}+\int_0^1 au''v''dx.
	\end{aligned}
	\]
	Observe that $au''v''  \in L^1(0,1)$ since, by assumption,  $\sqrt{a}u''$ and $\sqrt{a}v''$ belong to $L^2(0,1)$.
\end{proof}

We underline that in order to prove Lemma \ref{lemma gauss-green}, we do not use the assumption $\frac{1}{a} \in L^1(0,1)$. It is sufficient to require $a\in \mathcal C[0,1]$.

\begin{Remark}\label{remark 1}
	Note that:
	\begin{enumerate}
		\item[1)] if $u\in D(A_1)$ and $v\in H^2_a(0,1)$, then the boundary terms $[(au'')'v]^{x=1}_{x=0}$ and $[au''v']^{x=1}_{x=0}$ are equal to $0$ and (\ref{gf0}) becomes
		\begin{equation}\label{gf0 senza bordo}
			\int_{0}^{1}(au'')''v\,dx=\int_{0}^{1}au''v''dx;
		\end{equation}
		\item[2)] the boundary conditions prescribed in the domain $D(A_1)$ play a crucial role to prove the generation property of the operator. Indeed, the proof of the next theorem is based on  (\ref{gf0 senza bordo}).
	\end{enumerate}
\end{Remark}

The next result holds.
\begin{Theorem}\label{teorema generazione (div)}
	The operator $A_1:D(A_1)\to L^2(0,1)$ is self-adjoint and non negative on $L^2(0,1)$ with dense domain. Hence $-A_1$ generates a contractive analytic semigroup of angle $\frac{\pi}{2}$ on $L^2(0,1)$.
\end{Theorem}
\begin{proof}
	In order to show that $A_1$ is non negative and self-adjoint it is sufficient to prove that $A_1$ is symmetric, non negative and $(I+A_1)(D(A_1))=L^2(0,1)$. Indeed, if $A_1$ is non negative and $I+A_1$ is surjective on $D(A_1)$, then $A_1$ is maximal monotone and in this case $A_1$ is symmetric if and only if $A_1$ is self-adjoint. Moreover, by  \cite[Corollary 3.20]{en}, we have that
	$-A_1$ is densely defined and generates a contraction semigroup.

	\underline{$A_1$ is symmetric}: by (\ref{gf0 senza bordo}), for any $u,v\in D(A_1)$, one has
	\begin{equation*}
		\left\langle A_1u, v \right\rangle_{L^2(0,1)}=\int_{0}^{1} (a u'')''v\,dx=\int_{0}^{1}au''v''dx=\left\langle u, A_1v \right\rangle_{L^2(0,1)}.
	\end{equation*}
	\indent
	\underline{$A_1$ is non negative}: again by (\ref{gf0 senza bordo}), for any $u\in D(A_1)$, one has
	\begin{equation*}
		\left\langle A_1u, u \right\rangle_{L^2(0,1)}=\int_{0}^{1}(au'')''u\,dx=\int_{0}^{1} a(u'')^2dx\ge 0.
	\end{equation*}
	\indent
	\underline{$I+A_1$ is surjective}: observe that $H^2_{a}(0, 1)$ equipped with the inner product
	\begin{equation*}
		\left\langle u, v \right\rangle_{H^2_{a}(0, 1)} := \int_{0}^{1} \biggl (u v+au''v''\biggr ) dx,\quad \,\,\,\,\,\,\,\forall\; u, v \in H^2_{a}(0, 1),
	\end{equation*}
	is a Hilbert space. Moreover
	\begin{equation*}
		H^2_{a}(0, 1) \hookrightarrow L^2(0, 1) \hookrightarrow \left(H^2_{a}(0, 1)\right)^*,
	\end{equation*}
	where $\left(H^2_{a}(0, 1)\right)^*$ is the dual space of $H^2_{a}(0, 1)$ with respect to $L^2(0, 1)$. Now, for $f \in L^2(0, 1)$, define the functional $F\in \left(H^2_{a}(0, 1)\right)^*$ given by
	\begin{equation*}
		F(v):= \int_{0}^{1}f v\,dx,\quad \,\,\,\,\,\,\,\forall \;v\in H^2_{a}(0, 1).
	\end{equation*}
	Consequently, by the Lax-Milgram Theorem, there exists a unique $u\in H^2_{a}(0, 1)$ such that
	\begin{equation}\label{Consequence Riesz's Theorem}
		\left\langle u, v \right\rangle_{H^2_{a}(0, 1)}=\int_{0}^{1} f v\,dx \Longleftrightarrow \int_{0}^{1} \biggl (u v+au''v''\biggr ) dx=\int_{0}^{1} f v\,dx,
	\end{equation}
for all $v\in H^2_{a}(0, 1)$, being $\left\langle \cdot, \cdot \right\rangle_{H^2_{a}(0, 1)}$ the scalar product associated to $|||\cdot |||$.
	In particular, since $\mathcal{C}^{\infty}_c(0, 1) \subset H^2_{a}(0, 1)$, (\ref{Consequence Riesz's Theorem}) becomes
	\begin{equation*}
		\int_{0}^{1} au''v''dx = \int_{0}^{1} (f-u)v\,dx,\quad \,\,\,\,\,\,\,\forall	\;v\;\in \mathcal{C}^{\infty}_c(0, 1).
	\end{equation*}
	Thus, the distributional second derivative of $au''$ is equal to $f-u$ a.e. in $(0,1)$ and it is a function in $L^2(0,1)$, that is $au''\in H^2(0,1)$ \cite[Lemma 2.1]{CF} and $u+A_1u=f$.  This implies that $au''$ and $(au'')'$ are continuous functions in $[0,1]$ and, in particular, $u\in {\mathcal Z}(0,1)$. Moreover, it is possible to prove that
	\[
	\begin{cases}
	u''(0)=u''(1)=u'''(0)=u'''(1)=0, & x_0 \in (0,1),\\
	u''(1)=u'''(1)=0, (au'')(0)=(au'')'(0)=0, & x_0=0,\\
	u''(0)=u'''(0)=0, (au'')(1)=(au'')'(1)=0, & x_0=1.
	\end{cases}
	\] In fact, by  (\ref{gf0}) and (\ref{Consequence Riesz's Theorem}), one has that for all $u\in {\mathcal Z}(0,1)$ and $v\in H^2_a(0,1)$ the following relation holds:
	\begin{equation*}
		\int_0^1(au'')''v\,dx-[(au'')'v]^{x=1}_{x=0}+[au''v']^{x=1}_{x=0}=\int_{0}^{1} au''v''dx = \int_{0}^{1} (f-u)v\,dx.
	\end{equation*}
	Thus, $[(au'')'v]_{x=0}^{x=1}-[au''v']_{x=0}^{x=1}=0$ for all $v\in H^2_{a}(0, 1)$. Hence, one can conclude that
	\begin{equation}\label{co}
		(au'')'(1)=(au'')'(0)=0
		\quad \text{  and  } \quad
		(au'')(1)=(au'')(0)=0.
	\end{equation}

	Now, just to fix the idea, assume $x_0\in (0,1)$ and set $m:=au''\in H^2(0,1)$. Since $a (x)\neq 0$ for all $x \neq x_0$, we have
	\begin{equation*}
		m(x)=a(x)u''(x) \Longleftrightarrow \frac{m(x)}{a(x)}=u''(x), \quad \forall \; x \in [0,1] \setminus \{x_0\}.
	\end{equation*}
	Hence $u''(0)= \frac{m(0)}{a(0)} =0$ and $u''(1)= \frac{m(1)}{a(1)} =0$. Moreover,
	\begin{equation*}
		(au'')'(x)=a'(x)u''(x)+a(x)u'''(x), \quad \forall \; x \in [0,1] \setminus \{x_0\},
	\end{equation*}
	and, in particular,
	\begin{equation*}
		(au'')'(1)=0 \Longleftrightarrow a'(1)u''(1)+a(1)u'''(1)=0.
	\end{equation*}
	Analogously,
	$
		a'(0)u''(0)+a(0)u'''(0)=0.
$
	However, we know that $u''(1)=0=u''(0)$ and $a(1)\neq 0 \neq a(0)$, thus
	$
		u'''(1)=u'''(0)=0.
$
	Thus, we can conclude that if $x_0 \in (0,1)$, then
	\begin{equation}\label{su}
		u\in D(A_1)\,\,\,\,\text{ and }\,\,\,\,(I+A_1)(u)=f.
	\end{equation}
	On the other hand, if $x_0=0$, proceeding as before, we have $u''(1)=0=u'''(1)$. This fact, together with \eqref{co}, implies \eqref{su}.
	The same result follows if $x_0=1$.

	In every case, as an immediate consequence of the Stone-von Neumann Spectral Theorem and functional calculus associated with the spectral theorem, one has that the operator $(A_1,D(A_1))$ generates a cosine family and an analytic semigroup of angle $\displaystyle\frac{\pi}{2}$ on $L^2(0, 1)$.
\end{proof}

As a consequence of the previous generation property, one has the following well posedness result which completes the discussion of the problem in the weakly degenerate case giving information on the regularity of the solution itself.

First of all, recall the following definition:
\begin{Definition}
	If $u_0 \in L^2(0,1)$ and $h\in L^2(0,T;L^2(0,1))$, a function $u$ is said to be a weak solution of
	\begin{equation}\label{problema1}
	\begin{cases}
		u_t(t,x)+(au_{xx})_{xx}(t,x)=h(t,x), &(t,x)\in (0,T) \times (0,1),\\

		u(0,x)=u_0(x),&x\in(0,1),
	\end{cases}
\end{equation}
	 with the following boundary conditions
	\[
	(BC)\quad	\begin{cases}
		u_{xx}(t,0)=u_{xxx}(t,0)=u_{xx}(t,1)=u_{xxx}(t,1)=0, &x_0 \in (0,1),\\
		u_{xx}(t,1)=u_{xxx}(t,1)=(au_{xx})(t,0)= (au_{xx})_x(t,0)=0, &x_0=0,\\
		u_{xx}(t,0)=u_{xxx}(t,0)=(au_{xx})(t,1)= (au_{xx})_x(t,1)=0, &x_0=1,
	\end{cases}
	\]
	if
	\begin{equation*}
		u\in \mathcal{C}\Bigl([0,T];L^2(0, 1)\Bigr )\cap L^2\Bigl (0,T;H^2_{a}(0, 1) \Bigr )
	\end{equation*}
	and
	\begin{equation*}
		\begin{split}
			&	\int_0^1u(T,x)\varphi(T,x)\,dx-\int_0^1u_0(x)\varphi(0,x)\,dx-\int_0^T\int_0^1u(t,x)\varphi_t(t,x)\,dx\,dt =\\
			&- \int_0^T\int_0^1a(x)u_{xx}(t,x)\varphi_{xx}(t,x)\,dx\,dt+\int_0^T\int_0^1h(t,x)\varphi(t,x)\,dx\,dt
		\end{split}
	\end{equation*}
	for all $\varphi\in H^1(0,T;L^2(0,1))\cap L^2(0,T;H^2_{a}(0, 1))$.
\end{Definition}
Using the semigroup technique one has the next result.
\begin{Theorem}\label{teo exist and regul caso div}
	For all $h \in L^2(0,T;L^2(0,1))$ and $u_0\in L^2(0,1)$, there exists a unique solution
	\begin{equation*}
		u\in \mathcal{C}\Bigl([0,T];L^2(0, 1)\Bigr )\cap L^2\Bigl (0,T;H^2_{a}(0, 1) \Bigr )
	\end{equation*}
	of \eqref{problema1} satisfying $(BC)$ such that
	\[
		\sup_{t\in [0,T]}\Vert u(t)\Vert^2_{L^2(0, 1)}+\int_0^T\Vert u(t)\Vert^2_{H^2_{a}(0, 1)}dt\le C_T\Bigl (\Vert u_0\Vert^2_{L^2(0, 1)}+\Vert h\Vert^2_{L^2(0,T;L^2(0,1))} \Bigr )
	\]
	for some positive constant $C_T$. Moreover, if $h\in W^{1,1}(0,T;L^2(0,1))$ and $u_0\in H^2_a(0,1)$, then $
		u\in \mathcal{C}^1\Bigl([0,T];L^2(0, 1)\Bigr )\cap \mathcal{C}\Bigl ([0,T];D(A_1) \Bigr).
	$
\end{Theorem}
\subsection{Strongly degenerate setting}
In this subsection we assume that the function $a$ is strongly degenerate. Inspired again by \cite{ACF}, \cite{FGGR} and \cite{fm2013}, we introduce the weighted spaces
\begin{equation}\label{space}
	\begin{aligned}
		H^i_a(0,1):=\{u\in H^{i-1}(0,1):& \; u^{(i-1)} \text{ is locally absolutely continuous in}\\
		& [0,1]\setminus\{x_0\} \text { and } \sqrt{a}u^{(i)}\in L^2(0,1)\},
\end{aligned}\end{equation}
 $i=1,2$, equipped with the norms \eqref{normadiv}. We consider the operator $A_1$ defined in \eqref{ope1} with the same domain as in the weakly degenerate case.
Here ${\mathcal Z}(0,1)$ is as in Section \ref{Section 2}, where in this case the space $H^2_a(0,1)$ is the one defined in \eqref{space}. Thus, since $u\in {\mathcal Z}(0,1)$, $u'$ is locally absolutely continuous in $[0,1]\setminus \{x_0\}$ and not absolutely continuous in $[0,1]$ as for the weakly degenerate case, thus equality (\ref{gf0}) is not true a priori. An idea to prove the Gauss-Green formula given in Lemma \ref{lemma gauss-green} is to characterize $\mathcal Z(0,1)$ and hence $D(A_1)$.
To do this, define
\begin{equation*}
	\begin{split}
		X:=\{u\in H^1(0,1):\,
		&\text{$u'$ is locally absolutely continuous in $[0,1]\setminus\{x_0\}$},\\
		&au, au' \in H^1(0,1), au''\in H^2(0,1), \sqrt{a}u''\in L^2(0,1),\\
		&(au^{(k)})(x_0)=0,\, \text{for all $k=0,1,2$}

		\}.
	\end{split}
\end{equation*}
Using the definition of $X$ one can easily have the next property.
\begin{Lemma}\label{lemma2.2} For all $u \in X$ we have that
	\begin{enumerate}
		\item $|a(x)u(x)| \le \|(au)'\|_{L^2(0,1)}\sqrt{|x-x_0|}$,
		\item $|a(x)u'(x)| \le \|(au')'\|_{L^2(0,1)}\sqrt{|x-x_0|}$,
		\item $|a(x)u''(x)| \le \|(au'')'\|_{L^2(0,1)}\sqrt{|x-x_0|}$
	\end{enumerate}
	for all $x \in [0,1]$.
\end{Lemma}
\begin{proof} We will prove only the last point, the proof of the other points being similar. Let $u \in X$. Since $(au'')(x_0)=0$, then
	\[
	|(au'')(x)|= \left|\int_{x_0}^x (au'')'(s)ds\right| \le
	\|(au'')'\|_{L^2(0,1)}\sqrt{|x-x_0|},
	\]
	for all $ x \in [0,1]$.
\end{proof}

Using the assumptions on $a$, in particular the fact that $\ds\frac{1}{a} \not\in L^1(0,1)$, one can prove the next characterization.
\begin{Proposition}\label{Proposition 2.1}
	The spaces $X$ and $\mathcal Z(0,1)$ coincide.
\end{Proposition}
\begin{proof}
	Obviously, $X\subseteq \mathcal Z (0,1)$.
	Now, we prove $\mathcal Z (0,1)\subset X$. To this aim, let $u\in \mathcal Z(0,1)$; we only need to prove that
	\begin{equation*}
		au\in H^1(0,1),
		\quad
		au'\in H^1(0,1)
	\end{equation*}
	and $(au^{(k)})(x_0)=0,$ for all $k=0,1,2$.

	It is easy to prove that $(au)(x_0)=(au'')(x_0)=0.$  Indeed, since $u\in H^1(0,1)$, we have that $u \in \mathcal C[0,1]$, thus $au \in  \mathcal C[0,1]$ and there exists
	\[
	\lim_{x \rightarrow x_0} (au)(x)=(au)(x_0)=0.
	\]

Now, we prove	 that $(au'')(x_0)=0$. By assumption, we know that $au''\in H^2(0,1)$, hence $au''$  is continuous in $[0,1]$. In particular, there exists $\lim_{x\rightarrow x_0} (au'')(x)= (au'')(x_0)=:N \in \R$. If $N\neq 0$, there exists $C>0$ such that
$
	|(au'')^2(x)|\ge C
$
	for all $x$ in a  neighbourhood of $x_0$, $x \neq x_0$. In particular,
	\[
	|(a(u'')^2)(x)|\ge \frac{C}{a(x)}
	\]
	for all $x$ in a  neighbourhood of $x_0$, $x \neq x_0$. Using the assumption on $a$ one has that $\sqrt{a} u'' \not\in L^2(0,1)$; hence $N=0$.

	Now, we prove that $au \in H^1(0,1)$.
	We already know that $au\in L^2(0,1)$. Furthermore, since $a\in W^{1,\infty}(0,1)$,
	$
		(au)'=a'u+au'\in L^2(0,1).
	$
Moreover,  we easily get that $(au)'$ is the distributional derivative of $au$, and so $au \in H^1(0,1)$. Indeed, assuming for simplicity $x_0 \in (0,1)$, taking $\varphi \in \mathcal C_c^\infty(0,1)$ and $h>0$, one has
	\[
	\begin{aligned}
		\int_0^1(au)'(x)\varphi(x)dx&= \int_0^{x_0-h}(au)'(x)\varphi(x)dx + \int_{x_0-h}^{x_0+h}(au)'(x)\varphi(x)dx\\
		&+ \int_{x_0+h}^1(au)'(x)\varphi(x)dx.
	\end{aligned}\]
	Obviously, by the absolute continuity of the Lebesgue integral, it follows that
	\[
	\lim_{h \rightarrow 0}\int_{x_0-h}^{x_0+h}(au)'(x)\varphi(x)dx=0;
	\]
	moreover
	\[
	\begin{aligned}
		&\lim_{h \rightarrow 0}\int_0^{x_0-h}(au)'(x)\varphi(x)dx =\lim_{h \rightarrow 0}\left(( au\varphi)(x_0-h) - \int_0^{x_0-h}(au)(x)\varphi '(x)dx\right) \\
		&= - \int_0^{x_0}(au)(x)\varphi '(x)dx
	\end{aligned}
	\]
	and
	\[
	\begin{aligned}
		&
		\lim_{h \rightarrow 0}\int_{x_0+h}^1(au)'(x)\varphi(x)dx = \lim_{h \rightarrow 0}\left(- (au\varphi)(x_0+h) - \int_{x_0+h}^1(au)(x)\varphi '(x)dx\right)
		\\
		&=  - \int_{x_0}^1(au)(x)\varphi '(x)dx.
	\end{aligned}
	\]
	Hence, for all  $\varphi \in \mathcal C_c^\infty(0,1)$ it holds
	\[
	\int_0^1(au)'(x)\varphi(x)dx = -\int_0^1(au)(x)\varphi '(x)dx.
	\]
Nothing change if $x_0=0$ or $x_0=1$;	thus $(au)'$ is the distributional derivative of $au$ and so $au \in H^1(0,1)$.

Now, we prove that $\exists \; \lim_{x \rightarrow 0} (au')(x)=(au')(x_0)=0$.
	As a first step, consider $x_0 \in (0,1)$ and take $x<x_0$. Since $u'$ is locally absolutely continuous in $[0,1] \setminus \{x_0\}$, we have that $u'(0)$ and $u'(1)$ are well defined. Moreover,  $au' \in L^2(0,1)$ and
	\[
	(au')' =a'u' + au'' \in L^2(0,1),
	\]
	since $u \in \mathcal Z(0,1)$.
	 Thus
	\begin{equation*}
		(au')(x)=\int_0^x (au')'(t)dt+(au')(0)
	\end{equation*}
	and there exists $\lim_{x \to x_0^-}(au')(x)=(au')(x_0^-)=\int_{0}^{x_0}(au')'(t)dt+(au')(0) =:L \in \R$. Analogously, for $x>x_0$: starting from
	\begin{equation*}
		(au')(x)=-\int_x^1 (au')'(t)dt+(au')(1),
	\end{equation*}
	one can prove that $\exists\;\lim_{x \to x_0^+}(au')(x)=(au')(x_0^+)=-\int_{x_0}^{1}(au')'(t)dt+(au')(1)=:M \in \R$. Proceeding as before, we can prove that $L=M=0$.
	Indeed, if we assume $L, M\neq 0$, then there exists $C>0$ such that
	\begin{equation*}
		|(u')^2(x)|\ge  \frac{C}{a(x)},
	\end{equation*}
	for all $x$ in a  left and in a right  neighbourhood of $x_0$. Using the assumption on $a$, one has that $u' \notin L^2(0,1)$; hence $L=M=0$.
	Thus, there exists $\lim_{x \rightarrow x_0 } (au')(x)= (au')(x_0)=0$.

	If $x_0=0$ or $x_0=1$, one can proceed as in the case $x_0 \in (0,1)$, taking $x>0$ or $x<1$, respectively.

	Proceeding as before, one can prove that  $(au')'$ is the distributional derivative of $au'$  so that $au' \in H^1(0,1)$ and the thesis follows.
\end{proof}
Clearly, the following characterizations hold:
\[
D(A_1):=\{u\in X:\, u''(0)=u''(1)=u'''(0)=u'''(1)=0
\}, \quad \text{if} \;\; x_0 \in (0,1),
\]
\[
	\begin{split}
		D(A_1):=\{u\in X:\; u''(1)=u'''(1)=0, (au'')(0)=(au'')'(0)=0\},\quad \text{if} \;\; x_0 =0,
	\end{split}
\]
\[
	\begin{split}
		D(A_1):=\{u\in X: u''(0)=u'''(0)=0, (au'')(1)=(au'')'(1)=0\},\quad \text{if} \;\; x_0 =1.
	\end{split}
\]

Using Proposition \ref{Proposition 2.1} and Lemma \ref{lemma2.2}, we can prove the formula of integration by parts (\ref{gf0}) also in the strongly degenerate case.
\begin{Lemma}\label{gf0 strong case}
	For all $(u,v)\in {\mathcal Z}(0,1) \times H^2_a(0,1)$ the Gauss-Green formula \eqref{gf0} still holds.
\end{Lemma}
\begin{proof}
	Take $u \in \mathcal Z(0,1)$ and $v \in H^2_a(0,1)$. By definitions of $\mathcal Z(0,1)$ and $H^2_a(0,1)$, one has that $au''\in H^2(0,1)$ and $v \in H^1(0,1)$. Hence, one can integrate by parts, obtaining
	\begin{equation}\label{sg1}
		\int_{0}^{1}(au'')''v\,dx=[(au'')'v]^{x=1}_{x=0}-\int_{0}^{1}(au'')'v'dx.
	\end{equation}
	Now, we consider the term $\int_{0}^{1}(au'')'v'dx$ and take $x_0 \in (0,1)$.  For $\delta >0$, we have
	\begin{equation}\label{integrationS}
		\begin{aligned} \int_0^1 (au'')' v' dx &= (au''v')(x_0- \delta) - (au''v')(0)
			-\int_0^{x_0-\delta}au''v''dx\\& + \int_{x_0-
				\delta}^{x_0+\delta}(au'')'v' dx + (au''v')(1) - (au''v')(x_0+ \delta) \\
				&- \int_{x_0+
				\delta}^1 au''v'' dx.
		\end{aligned}
	\end{equation}
	Moreover, by the  H\"older inequality
	$a u''v''$ and $(au'')' v'$ belong to $L^1(0, 1)$ (observe that $au''v''=\sqrt{a}u'' \sqrt{a}v'' \in L^1(0,1)$). Hence, by the absolute continuity of the Lebesgue integral, one has
	\begin{equation}\label{delta1S}
		\lim_{\delta \rightarrow 0} \int_0^{x_0- \delta}au''v''  dx =
		\int_0^{x_0}au''v''  dx, \quad \lim_{\delta \rightarrow 0} \int_{x_0+
			\delta}^1au''v''  dx = \int_{x_0}^1au''v'' dx
	\end{equation}
	and
	\begin{equation}\label{delta2S}
		\lim_{\delta \rightarrow 0} \int_{x_0- \delta}^{x_0+ \delta}(au'')'v'
		dx =0.
	\end{equation}
	In order to obtain the
	desired result it is sufficient to prove that
	\begin{equation}\label{delta3S}
		\lim_{\delta \rightarrow 0} (au''v')(x_0- \delta) =\lim_{\delta
			\rightarrow 0} (au''v')(x_0+ \delta).
	\end{equation}

	To this aim, we can choose $\delta$ in \eqref{integrationS} such that $v'$ is absolutely continuous in $\mathcal K:=[0, x_0-\delta]\cup [x_0+ \delta, 1]$. Moreover, using the fact that also $au''\in \mathcal C[0,1]$, one has
	\[
	\begin{aligned}
		(au''v')(x_0-\delta)&= \int_0^{x_0-\delta}(au''v')'(s)ds +(au''v')(0)\\
		&=\int_0^{x_0-\delta}[(au'')'v'](s)ds +\int_0^{x_0-\delta}[au''v''](s)ds +(au''v')(0).
	\end{aligned}
	\]
	Proceeding as before, one has that there exists
	$L_1\in \R$ such that
	\[
	\begin{aligned}
		\lim_{\delta \rightarrow 0}(au''v')(x_0-\delta)
		&=\lim_{\delta \rightarrow 0}\left( \int_0^{x_0-\delta}[(au'')'v'](s)ds +\int_0^{x_0-\delta}[au''v''](s)ds \right)\\
		&+(au''v')(0)\\
		& =\int_0^{x_0}[(au'')'v'](s)ds + \int_0^{x_0}(au''v'')(s)ds +(au''v')(0) =:L_1.
	\end{aligned}
	\]
	Analogously, one has
	\[
	\begin{aligned}
		\lim_{\delta \rightarrow 0}(au''v')(x_0+\delta)&=(au''v')(1)-\int_{x_0}^1[(au'')'v'](s)ds-\int_{x_0}^1(au''v'')(s)ds =:L_2.
	\end{aligned}
	\]
	If $L_1 \neq 0$, then there exists $C>0$ such that
$
	|(au''v')(x)| \ge C
$
	for all $x$ in a left  neighbourhood of $x_0$, $x \neq x_0$. Thus, by
	Lemma \ref{lemma2.2},
	\[
	|v'(x)| \ge \frac{C}{|(au'')(x)|}\ge \frac{C_1}{\sqrt{x_0-x}}
	\]
	for all $x$ in a left  neighbourhood of $x_0$, $x \neq x_0$, and for a
	suitable positive constant $C_1$. This implies that $v' \not \in
	L^2(0,1)$. Hence $L_1=0$. Analogously, one can prove that $L_2=0$.
	Thus \eqref{delta3S} holds and
	the thesis follows by \eqref{sg1}-\eqref{delta3S}.

	The case $x_0=0$ or $x_0=1$ can be proved in a similar way.
\end{proof}
Also in the strongly degenerate case, to prove Theorem \ref{teorema generazione (div)}    it is important the equivalence between the two norms $\|\cdot\|_{H^2_a(0,1)}$ and $|||\cdot|||$ (defined in \eqref{norma tre stanghette}). This is done in \cite[Proposition 2.2]{CF-articolo Wentzell} under the additional Hypothesis:
\begin{Assumptions}\label{Ipoaggiuntiva articolo wentzell}
	Assume that there exists $K \in [1,2)$ such that the function
	$
	x \mapsto \frac{|x-x_0|^K}{a}
	$
	is
	\begin{enumerate}
		\item non increasing on the left of $x_0$ and non decreasing on the right of $x_0$, if $x_0 \in (0,1),$
		\item non decreasing on the right of $0$, if $x_0=0,$
		\item non increasing on the left of $1$, if $x_0=1$.
	\end{enumerate}
\end{Assumptions}

Thus, the analogues of Theorems \ref{teorema generazione (div)} and \ref{teo exist and regul caso div} hold under Hypothesis \ref{Ipoaggiuntiva articolo wentzell} if $a$ is strongly degenerate.

\section{The operator in non divergence form}\label{Sec3}
In this section we consider the operator $A_2u:= au_{xxxx}$
with suitable boundary conditions and we prove its generation property.

In this case the spaces that we consider are the same in the weakly or in the strongly degenerate case. In particular, we take into account the following spaces introduced in \cite{CaFrRo}, if the degeneracy occurs on the boundary of the space domain, or in \cite{6} (see also \cite{memoirs}), if the degeneracy point is in the interior:
\begin{equation*}
	L^2_{\frac{1}{a}}(0, 1):=\biggl \{u\in L^2(0, 1):\int_{0}^{1}\frac{u^2}{a}\,dx<+\infty \biggr \}
\end{equation*}
and
\begin{equation*}
	H^i_{\frac{1}{a}}(0, 1):= L^2_{\frac{1}{a}}(0, 1)\cap H^i(0, 1),
\end{equation*}
with the respective norms defined by
\begin{equation*}
	\norm{u}^2_{L^2_{\frac{1}{a}}(0, 1)}:= \int_{0}^{1}\frac{u^2}{a}\,dx, \quad\,\,\,\,\,\,\forall \;u \in L^2_{\frac{1}{a}}(0, 1),
\end{equation*}
and
\begin{equation*}
	\norm{u}_{H^i_{\frac{1}{a}}(0, 1)}^2:=\norm{u}^2_{L^2_{\frac{1}{a}}(0, 1)} + \sum_{j=1}^{i}\| u^{(j)}\|^2_{L^2(0, 1)}, \quad\,\,\,\, \forall \;u \in H^i_{\frac{1}{a}}(0, 1),
\end{equation*}
$i=1,2$.
Observe that for all $u \in H^i_{\frac{1}{a}}(0, 1)$, one can prove that $\|u\|^2_{H^i_{\frac{1}{a}}(0,1)}$ is equivalent to the following one
\[
\|u\|_i^2:= \|u\|^2_{L^2_{\frac{1}{a}}(0,1)} +  \|u^{(i)}\|^2_{L^2(0,1)}, \quad \forall \; u \in H^i_{\frac{1}{a}}(0, 1)
\]
(see, e.g., \cite[Chapter VIII]{brezis} or \cite{CF-articolo Wentzell}).  We underline also that if $u \in H^2_{\frac{1}{a}}(0, 1)$, then $(au)(x_0)=(au')(x_0)=0$ since $a(x_0)=0$.

Also in this section we distinguish between the weakly degenerate case and the strongly degenerate one.

\subsection{Weakly degenerate case}
Assume that the function $a$ is weakly degenerate at a point $x_0 \in [0,1]$ and define the space
\begin{equation}\label{spazioW}
	\mathcal{W}(0,1):=\Bigl \{u\in H^2_{\frac{1}{a}}(0, 1):au''''\in L^2_{\frac{1}{a}}(0, 1)\Bigr \}.
\end{equation}
Clearly, if $u \in \mathcal W(0,1)$, then $u'''' \in L^1(0,1)$; indeed
\[
\int_0^1 u''''(x)dx = \int_0^1 \frac{\sqrt{a}u''''(x)}{\sqrt{a}}dx \le \left\|\frac{1}{a}\right\|_{L^1(0,1)} \|\sqrt{a}u''''\|_{L^2(0,1)}.
\] Thus, we can apply \cite[Lemma 2.1]{CF} obtaining that $u \in W^{4,1}(0,1)$  and, as a consequence, $u \in \mathcal C^3[0,1]$.

In particular,  being $a(x_0)=0$, $(au'')(x_0)=(au''')(x_0)=0$.

Moreover, as in \cite[Proposition 3.1]{CF} for the Dirichlet case, one can prove
that the spaces $H^i_{\frac{1}{a}}(0, 1)$
and $H^i(0, 1)$, $i=1,2$, coincide algebraically and the two norms are equivalent.

Hence, as in Lemma \ref{lemma gauss-green}, one has immediately the following integration by parts.
\begin{Lemma}\label{lemma problema di prova}
	For all $(u,v)\in \mathcal{W}(0,1)\times H^2_{\frac{1}{a}}(0, 1)$ one has
	\begin{equation}\label{GF1}
		\int_{0}^{1}u''''v\,dx=[u'''v]^{x=1}_{x=0}-[u''v']^{x=1}_{x=0}+\int_{0}^{1}u''v''dx.
	\end{equation}
\end{Lemma}
Using the previous spaces, we define the operator $A_2$ by
\[
A_2u:=au''''
\]
for all $u \in
D(A_2)$, where
\begin{equation*}
	\begin{aligned}
		D(A_2):&=\{u\in \mathcal{W}(0,1):\, u''(0)=u''(1)= u'''(0)=u'''(1)=0 \}\\
		&=\left\{ u\in H^2_{\frac{1}{a}}(0,1): u''(0)=u''(1)= u'''(0)=u'''(1)=0\right. \\
		 &\left. \qquad \qquad \qquad \quad
		 \text{ and }au''''\in L^2_{\frac{1}{a}}(0, 1) \right\}.
	\end{aligned}
\end{equation*}
In this case the definition of $D(A_2)$ and, in particular, the natural Neumann boundary conditions
\begin{equation*}
	u''(0)=u''(1)=u'''(0)=u'''(1)=0
\end{equation*}
are independent of the degeneracy point $x_0\in [0,1]$ (recall that $u \in \mathcal C^3[0,1]$, thus all the previous terms are well defined).

\begin{Remark}
	As observed in Remark \ref{remark 1}, we underline that, if $u\in D(A_2)$ and $v\in H^2_{\frac{1}{a}}(0, 1)$, then the boundary terms in \eqref{GF1}, i.e., $[u'''v]^{x=1}_{x=0}$ and $[u''v']^{x=1}_{x=0}$, are equal to $0$ and (\ref{GF1}) becomes
	\begin{equation}\label{rem2}
		\int_{0}^{1}u''''v\,dx=\int_{0}^{1}u''v''dx.
	\end{equation}
\end{Remark}
As in the previous section, this formula will represent the key ingredient in the generation property established in the next theorem.
\begin{Theorem}\label{Theorem 3.1}
	The operator $(A_2,D(A_2))$ is self-adjoint and non negative on $L^2_{\frac{1}{a}}(0, 1)$ with dense domain. Hence $-A_2$ generates a contractive analytic semigroup of angle $\displaystyle\frac{\pi}{2}$ on $L^2_{\frac{1}{a}}(0, 1)$.
\end{Theorem}
\begin{proof}
	The proof is similar to the one of Theorem \ref{teorema generazione (div)}, so we sketch it.

	\underline{$A_2$ is symmetric}: using \eqref{rem2}, for any $u,v\in D(A_2)$, one has
	\begin{equation*}
		\left\langle v, A_2u \right\rangle_{L^2_{\frac{1}{a}}(0, 1)}=\int_{0}^{1}\frac{v a u''''}{a}\,dx=\int_{0}^{1}v''''u\,dx=\left\langle A_2v, u \right\rangle_{L^2_{\frac{1}{a}}(0, 1)}.
	\end{equation*}
	\indent
	\underline{$A_2$ is non negative}: for any $u\in D(A_2)$, by \eqref{rem2},
	\begin{equation*}
		\left\langle A_2u, u \right\rangle_{L^2_{\frac{1}{a}}(0, 1)}=\int_{0}^{1}\frac{a u'''' u}{a}\,dx=\int_{0}^{1}(u'')^2dx\ge 0.
	\end{equation*}
	\indent
	\underline{$I+A_2$ is surjective}:  observe that $H^2_{\frac{1}{a}}(0, 1)$ equipped with the inner product
	\begin{equation*}
		\left\langle u, v \right\rangle_{H^2_{\frac{1}{a}}(0, 1)} := \int_{0}^{1} \biggl (\frac{u v}{a}+u''v''\biggr ) dx\quad \,\,\,\,\,\,\,\forall\; u, v \in H^2_{\frac{1}{a}}(0, 1),
	\end{equation*}
	is a Hilbert space. Moreover
	\begin{equation*}
		H^2_{\frac{1}{a}}(0, 1) \hookrightarrow L^2_{\frac{1}{a}}(0, 1) \hookrightarrow \left(H^2_{\frac{1}{a}}(0, 1)\right)^*,
	\end{equation*}
	where $\left(H^2_{\frac{1}{a}}(0, 1)\right)^*$ is the dual space of $H^2_{\frac{1}{a}}(0, 1)$ with respect to $L^2_{\frac{1}{a}}(0, 1)$.

	Now, for $f \in L^2_{\frac{1}{a}}(0, 1)$, define the functional $F\in \left(H^2_{\frac{1}{a}}(0, 1)\right)^*$ given by

	\begin{equation*}
		F(v):= \int_{0}^{1}\frac{f v}{a}\,dx\quad \,\,\,\,\,\,\,\forall \;v\in H^2_{\frac{1}{a}}(0, 1).
	\end{equation*}

	Consequently, by the Lax-Milgram Theorem, there exists a unique $u\in H^2_{\frac{1}{a}}(0, 1)$ such that
	\begin{equation}\label{Consequence Riesz}
		\left\langle u, v \right\rangle_{H^2_{\frac{1}{a}}(0, 1)}=\int_{0}^{1} \frac{f v}{a}\,dx \Longleftrightarrow \int_{0}^{1} \biggl (\frac{u v}{a}+u''v''\biggr ) dx=\int_{0}^{1} \frac{f v}{a}\,dx,
	\end{equation}
for all $v\in H^2_{\frac{1}{a}}(0, 1)$.	In particular, since $\mathcal{C}^{\infty}_c(0, 1) \subset H^2_{\frac{1}{a}}(0, 1)$, (\ref{Consequence Riesz}) becomes
	\begin{equation*}
		\int_{0}^{1} u''v''dx = \int_{0}^{1} \frac{f-u}{a}\,v\,dx\quad \,\,\,\,\,\,\,\forall	\;v\;\in \mathcal{C}^{\infty}_c(0, 1).
	\end{equation*}

	Thus, the distributional second derivative of $u''$ is equal to $\ds\frac{f-u}{a}$ a.e. in $(0,1)$ and, proceeding as in \cite{CF}, one has
	$au''''\in L^2_{\frac{1}{a}}(0, 1)$; hence $u \in \mathcal W(0,1)$. Coming back to (\ref{Consequence Riesz}) and using the Gauss-Green Identity, it results that
	\begin{equation*}
		[u'''v]^{x=1}_{x=0}=[u''v']^{x=1}_{x=0}=0\quad\,\,\,\,\,\,\,\,\,\,\forall\; v \in H^2_{\frac{1}{a}}(0, 1)
	\end{equation*}
	and one can conclude that $u''(0)=u''(1)=u'''(0)=u'''(1)=0$.

	As a consequence, $u \in D(A_2)$ and
	$
		au''''=f -u\quad \text{a.e. in } (0,1).
$
	Hence $(I+A_2)(u)=f$. The rest of the proof is as in Theorem \ref{teorema generazione (div)}.
\end{proof}

Hence, one has the following well posedness result, for which we premise the following definition:
\begin{Definition}
	If $u_0 \in L^2_{\frac{1}{a}}(0, 1)$ and $h\in L^2(0,T;L^2_{\frac{1}{a}}(0, 1))$, a function $u$ is said to be a weak solution of
	\begin{equation}\label{problema2}
	\begin{cases}
		u_t(t,x)+au_{xxxx}(t,x)=h(t,x), &(t,x)\in (0,T) \times (0,1),\\
		u_{xx}(t,0)=u_{xx}(t,1)=0,&t\in (0,T),\\
		u_{xxx}(t,0)=u_{xxx}(t,1)=0,&t\in (0,T),\\
		u(0,x)=u_0(x),&x\in(0,1),
	\end{cases}
\end{equation}
	if
	\begin{equation*}
		u\in \mathcal{C}\Bigl([0,T];L^2_{\frac{1}{a}}(0, 1)\Bigr )\cap L^2\Bigl (0,T;H^2_{\frac{1}{a}}(0, 1) \Bigr )
	\end{equation*}
	and
	\begin{equation*}
		\begin{split}
			&	\int_0^1\frac{u(T,x)\varphi(T,x)}{a(x)}\,dx-\int_0^1\frac{u_0(x)\varphi(0,x)}{a(x)}\,dx-\int_0^T\int_0^1\frac{u(t,x)\varphi_t(t,x)}{a(x)}\,dx\,dt =\\
			&- \int_0^T\int_0^1u_{xx}(t,x)\varphi_{xx}(t,x)\,dx\,dt+\int_0^T\int_0^1h(t,x)\frac{\varphi(t,x)}{a(x)}\,dx\,dt
		\end{split}
	\end{equation*}
	for all $\varphi\in H^1(0,T;L^2_{\frac{1}{a}}(0, 1))\cap L^2(0,T;H^2_{\frac{1}{a}}(0, 1))$.
\end{Definition}

Therefore, using the semigroup theory, one can prove the following well posedness theorem.
\begin{Theorem}\label{Theorem 3.2}
	For all $h\in L^2(0,T;L^2_{\frac{1}{a}}(0,1))$ and $u_0\in L^2_{\frac{1}{a}}(0, 1)$, there exists a unique solution
	\begin{equation*}
		u\in \mathcal{C}\Bigl([0,T];L^2_{\frac{1}{a}}(0, 1)\Bigr )\cap L^2\Bigl (0,T;H^2_{\frac{1}{a}}(0, 1) \Bigr )
	\end{equation*}
	of (\ref{problema2}) such that
	\[
		\sup_{t\in [0,T]}\Vert u(t)\Vert^2_{L^2_{\frac{1}{a}}(0, 1)}+\int_0^T\Vert u(t)\Vert^2_{H^2_{\frac{1}{a}}(0, 1)}dt\le C_T\left (\Vert u_0\Vert^2_{L^2_{\frac{1}{a}}(0, 1)}+\Vert h\Vert^2_{L^2(0,T;L^2_{\frac{1}{a}}(0,1))} \right)
	\]
	for some positive constant $C_T$. In addition, if $h\in W^{1,1}(0,T;L^2_{\frac{1}{a}}(0,1))$ and $u_0\in H^2_{\frac{1}{a}}(0,1)$, then
	$
		u\in \mathcal{C}^1\Bigl([0,T];L^2_{\frac{1}{a}}(0, 1)\Bigr )\cap \mathcal{C}\Bigl ([0,T];D(A_2) \Bigr ).
$
\end{Theorem}

\subsection{Strongly degenerate case}
In this subsection we will assume that the function $a$ is strongly degenerate. As in the previous subsections, we need to prove a formula similar to \eqref{GF1}. To this aim we need an additional assumption on the function $a$, which is not surprising because already used in other papers and it is satisfied by the prototype function we have in mind.

\begin{Assumptions}\label{hp 3.3}
	Assume that
	there exist $K \in [1,2)$ and $C>0$ such that
	\[
	\frac{1}{a(x)}\le \frac{C}{|x-x_0|^K}
	\]
	for all $x \in [0,1]\setminus\{x_0\}$.
\end{Assumptions}
Observe that the previous hypothesis is obviously satisfied by $a(x):=|x-x_0|^K$, where $K\in [1,2)$, and it is more general then the one made in the divergence case, Hypothesis \ref{Ipoaggiuntiva articolo wentzell}.

\vspace{0.5cm}
Now, we introduce
\begin{equation*}
	X:=\biggl \{u\in H^2_{\frac{1}{a}}(0, 1): u(x_0)=0 \biggr \}.
\end{equation*}

Proceeding as in \cite{CaFrRo} and \cite{FGGR}, one can prove the following result.
\begin{Proposition} \label{Prop 3.2}
	If Hypothesis \ref{hp 3.3} is satisfied, then
	\begin{equation*}
		H^2_{\frac{1}{a}}(0, 1)=X.
	\end{equation*}
\end{Proposition}
We underline that this characterization holds if $x_0 \in [0,1]$.

Now, consider the space $\mathcal W(0,1)$ defined in \eqref{spazioW}.
Clearly, if Hypothesis \ref{hp 3.3} is satisfied, we can rewrite $\mathcal W(0,1)$ as \[
\mathcal{W}(0,1)=\Bigl \{u\in H^2_{\frac{1}{a}}(0, 1):u(x_0)=0 \text{ and } au''''\in L^2_{\frac{1}{a}}(0, 1) \Bigr \}.
\]

The next proposition holds.
\begin{Proposition}\label{regolarità appendice}Assume that the function $a$ is strongly degenerate and Hypothesis \ref{hp 3.3} holds.

	 $\star$ If $x_0\in (0,1)$, then for all $y \in \mathcal W(0,1)$, $y'' \in W^{1,1}(0, x_0)$ and $y'' \in W^{1,1}(x_0,1)$.

$\star$ If $x_0=0$ or $x_0=1$, then for all $y \in \mathcal W(0,1)$, $y'' \in W^{1,1}(0,1)$.
\end{Proposition}
\begin{proof} Assume $x_0=0$, being the other cases similar. Fix $y \in \mathcal W(0,1)$ and take $\delta >0$. Clearly
	\begin{equation}\label{y secondo in delta*}
			y''(\delta)=y''(1)-\int_\delta^1y'''(x)dx,
			\end{equation}
			thus
			\begin{equation}\label{y secondo in delta}
		\begin{aligned}
			y''(\delta)&=y''(1)-\int_\delta^1\Biggl (y'''(1)-\int_x^1y''''(s)ds \Biggr )dx\\
			&	=y''(1)-y'''(1)+\delta y'''(1)+\int_\delta^1y''''(s)(s-\delta)ds.
		\end{aligned}
	\end{equation}
Trivially $\displaystyle\lim_{\delta\to 0}\delta y'''(1)= 0$; moreover
\[
	\int_\delta^1y''''(s)(s-\delta)ds=\int_\delta^1sy''''(s)ds-\delta \int_\delta^1y''''(s)ds.
\]
Since $sy''''(s)\in L^1(0,1)$, $\displaystyle\lim_{\delta\to 0}\int_\delta^1 sy''''(s)ds = \int_0^1 sy''''(s)ds$, by the absolute continuity of the integral,
and\begin{equation*}
	\begin{aligned}
		0&<\delta \int_\delta^1|y''''(s)|ds\le \delta \Biggl (\int_\delta^1\frac{1}{a(s)}ds\Biggr )^{\frac{1}{2}}\Vert\sqrt{a}y''''\Vert_{L^2(0,1)}\\
		&\le\delta^{1- \frac{K}{2}} \Biggl (\int_\delta^1\frac{s^K}{a(s)}ds \Biggr )^{\frac{1}{2}}\Vert\sqrt{a}y''''\Vert_{L^2(0,1)}\le C\delta^{1- \frac{K}{2}}(1-\delta)^{\frac{1}{2}}\Vert\sqrt{a}y''''\Vert_{L^2(0,1)},	\end{aligned}
\end{equation*} for a positive constant $C$.
If we pass to the limit as $\delta\to 0$ in (\ref{y secondo in delta}), we conclude that
\begin{equation*}
	\exists \lim_{\delta\to 0}y''(\delta)= y''(1)- y'''(1)+\int_0^1sy''''(s)ds \in\mathbb{R}.
\end{equation*}
By continuity, it is possible to define
$
	y''(0):=\lim_{\delta\to 0}y''(\delta).
$
In particular, by \eqref{y secondo in delta*},
\begin{equation*}
	\int_0^1y'''(x)dx=y''(1)-y''(0)
\end{equation*}
and the thesis follows.
\end{proof}

As a consequence, one can prove the next Gauss-Green formula.
\begin{Lemma}\label{green} Assume Hypothesis \ref{hp 3.3}.

	 $\star$ If $x_0\in (0,1)$, then for all $(u,v)\in \mathcal W(0,1)\times H^2_{\frac{1}{a}}(0, 1)$, equality \[\int_0^1 u''''vdx= [u'''v]^{x=1}_{x=0}-[u''v']^{x=1}_{x=0}+ [u''v']^{x_0^+}_{x_0^-} +\int_0^1 u''v'' dx\]
holds. Here $u''(x_0^+)= \lim_{\delta\rightarrow 0^+} u''(x_0+\delta)$, $u''(x_0^-)= \lim_{\delta\rightarrow 0^+} u''(x_0-\delta)$ and $v'(x_0^+)= v'(x_0^-)= v'(x_0)$.

$\star$ If $x_0=0$, then for all $(u,v)\in  \mathcal W(0,1)\times H^2_{\frac{1}{a}}(0, 1)$
		\begin{equation*}
		\int_{0}^{1}u''''v\,dx=u'''(1)v(1)-[u''v']^{x=1}_{x=0}+\int_{0}^{1}u''v''dx.
	\end{equation*}

$\star$	If $x_0=1$, then for all $(u,v)\in  \mathcal W(0,1)\times H^2_{\frac{1}{a}}(0, 1)$
\begin{equation*}
	\int_{0}^{1}u''''v\,dx=-u'''(0)v(0)-[u''v']^{x=1}_{x=0}+\int_{0}^{1}u''v''dx.
\end{equation*}

\end{Lemma}

\begin{proof}

Just to fix the idea, we will prove the thesis if  $x_0 \in (0,1)$, being the other proofs similar to this case. Take $\delta >0$ and $\mathcal K:= [0, x_0-\delta ] \cup [x_0+ \delta, 1]$.

	By definition of $\mathcal W(0,1)$, $u''''\in L^2(\mathcal K)$, thus, by \cite[Lemma 2.1]{CF}, $u \in H^4(\mathcal K)$. Hence, we can integrate by parts, obtaining

	\begin{equation*}
		\begin{split}
			&\int_{0}^{1}u''''v\,dx
			=\int_{0}^{x_0-\delta}u''''v\,dx+\int_{x_0-\delta}^{x_0+\delta}u''''v\,dx+\int_{x_0+\delta}^{1}u''''v\,dx \\

			&=\int_{0}^{x_0-\delta}u''v''dx+\int_{x_0-\delta}^{x_0+\delta}u''''v\,dx+\int_{x_0+\delta}^{1}u''v''dx\\
			&-  (u'''v)(0)+(u''v')(0)+  (u'''v)(1)-(u''v')(1)\\
			& + (u'''v)(x_0-\delta)-  (u'''v)(x_0+\delta)- (u''v')(x_0-\delta)+ (u''v')(x_0+\delta).
		\end{split}
	\end{equation*}
	As in \cite[Proposition 2.1]{CF}, one can prove
	\begin{equation*}
		\lim_{\delta\to 0}\int_{0}^{x_0-\delta}u''v''dx=\int_{0}^{x_0}u''v''dx,\,\,\,\,\,\lim_{\delta\to 0}\int_{x_0+\delta}^{1}u''v''dx=\int_{x_0}^{1}u''v''dx
	\end{equation*}
	and
	$
		\lim_{\delta\to 0}\int_{x_0-\delta}^{x_0+\delta}u''''v\,dx=0.
$
By Proposition \ref{regolarità appendice}, $\lim_{\delta \rightarrow0} (u''v')(x_0-\delta)= u''(x_0^-)v'(x_0)$ and $\lim_{\delta \rightarrow0} (u''v')(x_0+\delta)= u''(x_0^+)v'(x_0)$. \\
Moreover,
$\lim_{\delta \rightarrow0} (u'''v)(x_0-\delta)$ $= \lim_{\delta \rightarrow0} (u'''v)(x_0+\delta)=0$. Indeed
\[
\begin{aligned}
|(u'''v)(x_0+\delta)|&\le |v(x_0+\delta) u'''(1)| + \left|v(x_0+ \delta)\int_{x_0+\delta}^1 u''''(s)ds\right|\\
& \le  |v(x_0+\delta) u'''(1)| + C|v(x_0+ \delta)|	\left(\int_{x_0+\delta}^1 (s-x_0)^{-k}ds\right)^{\frac{1}{2}}\\
&\le \begin{cases}
|v(x_0+\delta) u'''(1)| + C \frac{\delta^{\frac{3-K}{2}}}{\sqrt{K-1}} \frac{|v(x_0+ \delta)|}{\delta}, & K \neq 1\\
|v(x_0+\delta) u'''(1)| + C (-\delta^2\log \delta)^{\frac{1}{2}}\frac{|v(x_0+ \delta)|}{\delta},& K=1.
\end{cases}
\end{aligned}
\]
In any case $ \lim_{\delta \rightarrow0} (u'''v)(x_0+\delta)=0$, since $v(x_0)=0$. Analogously, $ \lim_{\delta \rightarrow0} (u'''v)(x_0-\delta)=0$.
\end{proof}

Now, define the operator $A_2$ by
\[
A_2u:=au''''
\]
for all $u \in D(A_2)$, where
\[
\begin{aligned}
	D(A_2):&=\{ u\in \mathcal W(0,1): u''(0)=u''(1)=u'''(0)=u'''(1)=0\},
\end{aligned}
\]
if $x_0 \in (0,1)$,
\[
\begin{aligned}
	D(A_2):&=\{ u\in \mathcal W(0,1):u''(0)=0=u''(1)=u'''(1) \},
\end{aligned}
\]
if $x_0=0$ and
\[
\begin{aligned}
	D(A_2):&=\{ u\in \mathcal W(0,1):u''(1)=0=u''(0)=u'''(0) \},
\end{aligned}
\]
if $x_0=1$.

Observe that if $x_0=0$ and $u \in \mathcal W(0,1)$, then $u''(0)$ is well defined thanks to Proposition \ref{regolarità appendice}. In this case, the other boundary terms in $D(A_2)$ are well defined since  far  from the degeneracy point the function $u \in \mathcal W(0,1)$ is $\mathcal C^3$. Analogously if $x_0=1$ or $x_0 \in (0,1)$.

Thanks to Proposition \ref{Prop 3.2}, the following characterizations of $D(A_2)$ hold under Hypothesis \ref{hp 3.3}:
\[
\begin{aligned}
	D(A_2):&=\{ u\in H^2_{\frac{1}{a}}(0, 1):u(x_0)=0= u''(0)=u''(1)=u'''(0)=u'''(1) \\
	& \qquad  \qquad\qquad\quad \;\; \text{ and } au''''\in L^2_{\frac{1}{a}}(0, 1) \},
\end{aligned}
\]
if $x_0 \in (0,1)$,
\[
\begin{aligned}
	D(A_2):&=\{ u\in H^2_{\frac{1}{a}}(0, 1):u(0)=u''(0)=0=u''(1)=u'''(1) \\
	& \qquad  \qquad\qquad\quad \;\; \text{ and } au''''\in L^2_{\frac{1}{a}}(0, 1) \},
\end{aligned}
\]
if $x_0=0$ and
\[
\begin{aligned}
	D(A_2):&=\{ u\in H^2_{\frac{1}{a}}(0, 1):u(1)=u''(1)=0=u''(0)=u'''(0) \\
	& \qquad  \qquad\qquad\quad \;\; \text{ and } au''''\in L^2_{\frac{1}{a}}(0, 1) \},
\end{aligned}
\]
if $x_0=1$.

Lemma \ref{green} is crucial to prove  the analogues of Theorem \ref{Theorem 3.1} and Theorem \ref{Theorem 3.2} in the strongly degenerate case at least if $x_0=0$ or $x_0=1$; if $x_0 \in (0,1)$ then we refer to \cite[Theorem 2.2]{memoirs}. Indeed we can consider the function $\mathcal L: L^2_{\frac{1}{a}}(0,1) \rightarrow L^2_{\frac{1}{a}}(0,1)$ defined as $\mathcal L(f)= u\in H^2_{\frac{1}{a}}(0,1)$, where $u$ is the unique solution of $\ds \int_0^1 \frac{uv}{a}dx + \int_0^1 u''v'' dx = \ds \int_0^1 \frac{fv}{a}dx$ for all $v \in H^2_{\frac{1}{a}}(0,1)$.

redpajama_set_name: RedPajamaArXiv


text:
Canada Calling
SOCIAL CREDIT PRINCIPLES
Reprinted in THE SOCIAL CREDITER, January 2, 1954.
FOR POLITICAL AND ECONOMIC REALISM: This journal expresses and supports the policy of the Social Credit Secretariat which was founded in 1933 by Clifford Hugh Douglas.
The Social Credit Secretariat is a non-party, non-class organisation neither connected with nor supporting any political party, Social Credit or otherwise.
An address delivered at Swanwick, November, 1924 .
By Major C. H. DOUGLAS
(Revised,* January, 1954)
The financial system, in its control over production, stands to the works or factory system of the world, considered as an economic unit, in the same relation as the planning department of a modern factory does to that factory.
The distribution side of the financial system exercises a function not dissimilar to that of the progress department of a factory.
No discussion of the financial system can serve any useful purpose which does not recognise:–
(a) That a works system must have a definite objective.
(b) That when that objective has been decided upon it is a technical matter to fit methods of human psychology and physical facts, so that the objective will be most easily obtained.
In regard to (a) the policy of the world economic system amounts to a philosophy of life. There are really only three alternative policies in respect to a world economic organisation:–
The first is that it is the end in itself for which man exists.
The second is that while not an end in itself, it is the most powerful means of constraining the individual to do things he does not want to do; e.g., it is a system of Government. This implies a fixed ideal of what the world ought to be.
And the third is that the economic activity is simply a functional activity of men and women in the world; that the end of man, while unknown, is something towards which most rapid progress is made by the free expansion of individuality, and that, therefore, economic organisation is most efficient when it most easily and rapidly supplies economic wants without encroaching on other functional activities.
You cannot spend too much time in making these issues clear to your minds, because until they are clear you are not in a position to offer an opinion on any economic proposal whatever.
In regard to (b) certain factors require to be taken into consideration.
(1) That money has no reality in itself. That in itself it is either gold, silver, copper, paper, cowrie shells, or broken tea cups. The thing which makes it money, no matter of what it is made, is purely psychological, and consequently there is no limit to the amount of money except a psychological limit.
(2) That economic production is simply a conversion of one thing into another, and is primarily a matter of energy. It seems highly probable that both energy and production are only limited by our knowledge of how to apply them.
(3) That in the present world unrest two entirely separate factors are confused. The cry for the democratisation of industry obtains at least 90 per cent of its force from the desire for the democratisation of the proceeds of industry, which, is, of course, a totally different thing. This confusion is assisted by the objective fact that the chief controllers of industry get rich out of their control.
I do not, myself, believe in the democratic control of industry any more than I should believe in the democratic control of a cricket team, while actually playing, and I believe that the idea that the average individual demands a share in the administrative control of industry is a pure myth.
The present world financial system is a Government based on the theory that men should be made to work, and this theory is considerably intermixed with the even stronger contention that the end of man is work. I want you to realise that this is a statement of fact, not a theory. More than 95 per cent of the purchasing-power actually expended in consumption is wages and salaries.
It will therefore be seen that there are two standpoints from which to examine its mechanism. The first considered as a method of achieving its political end of universal work, and the second as a means of achieving some other political end–for instance, the third alternative already mentioned.
Considered as a means of making people work (an aim which is common both to the Capitalist and Socialist Party Politics) the existing financial system, as a system, is probably nearly perfect.
Its banking system, methods of taxation and accountancy counter every development of applied science, organisation, and machinery, so that the individual, instead of obtaining the benefit of these advances in the form of a higher civilisation and greater leisure, is merely enabled to do more work. Every other factor in the situation is ultimately sacrificed to this end of providing him with work, and at this moment the world in general, and Europe in particular, is undoubtedly settling down to a policy of intensive production for export, which must quite inevitably result in a world cataclysm, urged thereto by what is known as the Unemployment Problem.
To blame the present financial system for failing to provide employment is most unfair; if left alone it will continue to provide employment in the face of all scientific progress, even at the cost of a universal world war, in which not only all possible production would be destroyed, but such remnants of the world's population as are left will probably be reduced to the meagre production of the Middle Ages.
Considered as a mechanism for distributing goods, however, the existing financial system is radically defective. In the first place, it does not provide enough purchasing-power to buy the goods which are produced.
I do not wish to enter at any great length into the analysis of why this is so, because it is always a matter of some heated controversy. I have, however, no hesitation whatever in asserting not only that it is so, but that the fact that it is so is the central fact of the existing economic system, and that unless it is dealt with no other reforms are of any use whatever.
And the second feature of equal importance is that considerably less than the available number of individuals, working with modern tools and processes, can produce everything that the total population of the world, as individuals, can use and consume, and that this situation is progressive, that is to say, that year by year a smaller number of individuals can usefully be employed in economic production.
To summarise the matter, the principles which must govern any reform of the financial system, which will at one and the same time avoid catastrophe, and re-orientate world economic policy along the lines of the third alternative, are three in number:–
1. That the cash credits of the population of any country shall at any moment be collectively equal to the collective cash prices for consumable goods for sale in that country, and such cash credits shall be cancelled on the purchase of goods for consumption.
2. That the credits required to finance production shall be supplied, not from savings, but be new credits relating to new production.
3. That the distribution of cash credits to individuals shall be progressively less dependent upon employment. That is to say, that the dividend shall progressively displace the wage and salary.
I may conclude by a few remarks on the position of the banks, in respect of this situation. It is becoming fairly well understood that the banks have the control of the issue of purchasing-power to a very large extent in their hands. The complaint which is levelled at the banks is generally that they pay too large a dividend. Now curiously enough, in my opinion, almost the only thing which is not open to destructive criticism about the banks is their dividend. Their dividend goes to shareholders and is purchasing-power, but their enormous concealed profits, a small portion of which goes in immensely redundant bank premises, etc., do not provide purchasing-power for anyone, and merely aggrandise banks as banks.
But the essential point in the position of the banks, which is so hard to explain, and which is grasped by so few people, is that their true assets are not represented by anything actual at all, but are represented by the difference between a society functioning under centralised and restricted credit and a free society unfettered by financial restrictions.
To bring that perhaps vague generalisation into a more concrete form, the true assets of the banks collectively consist of the difference between the total amount of legal tender, or Government money, which exists, and the total amount of bank credit money, not only which does exist, but which might exist, and which is kept out of existence by the fiat of the banking executive.
*THE SOCIAL CREDITER, Saturday, January 2, 1954.
The Editor, The Social Crediter,
The important brief statement of the principles of Social Credit, which takes its origin from the address delivered by Douglas at Swanwick in November, 1924, was printed in The New Age in its issue for November 20 of that year, has since been several times reprinted and appears in Warning Democracy by C. H. Douglas as Chapter IV of that work (pages 37 to 43 of the third Edition).
A correspondent abroad has on several occasions questioned the accuracy of the newspaper's report of the first paragraph only and has suggested the withdrawal of the printed texts in existence, so far as practicable, and the substitution of an accurate text.
His suggestion was reported in detail to Major Douglas in 1935 and again, to my knowledge, in September, 1945. A written statement was elicited from Major Douglas on the first occasion which is conclusive concerning the authoritative text, but neither then nor later was it ascertained whether the author considered public correction necessary or desirable. Nevertheless, there can be no question but that the authentic text is superior to the published text in completeness of expression. Distortion of meaning is not involved, excepting in regard to an omitted paragraph, the wording of which Major Douglas (1935) reported from memory. Here too, a competent Social Crediter will recognise the emendation proposed by Major Douglas as a paraphrase of what he said on other occasions, and thus not of the order of a modification in any sense of his views.
The corrupt passage of which complaint is made is as follows:–
The financial system is the works or factory system of the world, considered as an economic unit, just as the planning department of a modern factory is of that factory.
Having corrected this passage, (vide complete text infra) Major Douglas went on:–
As far as my memory serves me, there is a further complete paragraph left out, somewhat as follows:– "The distribution side of the financial system exercises a function not dissimilar to that of the Progress department of a factory."
It is not for me to speculate concerning Major Douglas's reasons for not insisting on immediate correction of the text of a brief but important summary of his views, which has been very widely distributed. Even non-technical readers will appreciate that both the statements cited above are in fact similes, and, as such, have only such force as a figure of speech may possess. However this may be, I have authorised the reprinting of the leaflet, "Social Credit Principles," with the wording as corrected by Douglas, with an intimation that the text is a revised text, the date of the revision (January, 1954) being stated with a following reference to this letter of explanation of the circumstances. I have taken this course because I think it would be unwise to distract the attention of any new reader from uncomplicated study of the text as Douglas originally intended it to appear.
I append the revised text.
Tudor Jones,
December 23, 1953 Chairman, Social Credit Secretariat.

redpajama_set_name: RedPajamaCommonCrawl


text:
require "godmin/medium/version"
require "godmin/medium/engine"
require "godmin/medium/helper"

module Godmin
  module Medium
    # Your code goes here...
  end
end

redpajama_set_name: RedPajamaGithub


text:
Q: MVC 5 EF 6 - How to Search a Table Using Related Table Object Criteria I need to display a resulting table of PO Requests which includes the PO Items (this part I figured out) and provide search functionality for all items displayed (this part I got mostly figured out).
Using the code below, which is a slightly stripped down version, I am able to display the data I need and I am able to search for nearly all I need, as long as it is data from the PORequests model (this is the "parent model"). When I try to search using the "child model," POItems, the actual objects (ID, Description, etc.) are not accessible. This was built using DatabaseFirst model and EF6.x DbContextGenerator.
//POItemData ViewModel
using FinanceSearch.Models;
namespace FinanceSearch.ViewModels
{
    public class POItemData
    {
        public PagedList.IPagedList<PORequest> PORequests { get; set; }
        public PagedList.IPagedList<POItem> POItems { get; set; }
    }
}

//POItems Model
namespace FinanceSearch.Models
{
    using System;
    using System.Collections.Generic;

    public partial class POItem
    {
        public int ID { get; set; }
        public int Amount { get; set; }
        public string Description { get; set; }

        public virtual PORequest PORequest { get; set; }
    }
}

//PORequest Model
namespace FinanceSearch.Models
{
    using System;
    using System.Collections.Generic;
    using System.ComponentModel.DataAnnotations;

    public partial class PORequest
    {
        [System.Diagnostics.CodeAnalysis.SuppressMessage("Microsoft.Usage", "CA2214:DoNotCallOverridableMethodsInConstructors")]
        public PORequest()
        {
            this.POAttachments = new HashSet<POAttachment>();
            this.POItems = new HashSet<POItem>();
        }

        //other relevant stuff

        public virtual ICollection<POItem> POItems { get; set; }

    }
}

//Controller
using System;
using System.Collections.Generic;
using System.Data;
using System.Data.Entity;
using System.Linq;
using System.Net;
using System.Web;
using System.Web.Mvc;
using FinanceSearch.Models;
using PagedList;
using System.Web.UI.WebControls;
using System.Data.Entity.SqlServer;
using FinanceSearch.ViewModels;

namespace FinanceSearch.Controllers
{
    public class PORequestsController : Controller
    {
        private Finance db = new Finance();

        public ActionResult Index(int? id, int? page, string sortOrder, string currentFilter, string poNumber,
           string AppropNumber, string ContractNumber, string ItemDescription)
        {

            ViewBag.CurrentSort = sortOrder;
            ViewBag.POSortParm = String.IsNullOrEmpty(sortOrder) ? "PONumber_desc" : "";
            ViewBag.AppropNumberSortParm = sortOrder == "AppropNumber" ? "AppropNumber_desc" : "AppropNumber";
            ViewBag.ContractNumberSortParm = sortOrder == "ContractNumber" ? "ContractNumber_desc" : "ContractNumber";

            if (poNumber != null)
            {
                page = 1;
            }
            else
            {
                ponumber = currentFilter;
            }

            //int pageSize = 3;
            int pageNumber = (page ?? 1);

            var viewModel = new POItemData();

            viewModel.PORequests = db.PORequests
               .Include(i => i.POItems)
               .OrderBy(i => i.ID).ToPagedList(pageNumber, 5);

            if (!String.IsNullOrEmpty(poNumber))
            {
                viewModel.PORequests = viewModel.PORequests.Where(s => s.PONumber.Contains(ponumber)).ToPagedList(pageNumber, 5);
            }
            if (!string.IsNullOrEmpty(AppropNumber))
            {
                viewModel.PORequests = viewModel.PORequests.Where(x => x.AppropNumber.Contains(AppropNumber)).ToPagedList(pageNumber, 5);
            }
            if (!string.IsNullOrEmpty(ContractNumber))
            {
                viewModel.PORequests = viewModel.PORequests.Where(x => x.ContractNumber.Contains(ContractNumber)).ToPagedList(pageNumber, 5);
            }
            if (!string.IsNullOrEmpty(ItemDescription))
            {

            }

In the ItemDescription if statement, something like the following would be nice:
if (!string.IsNullOrEmpty(ItemDescription))
{
viewModel.PORequests = viewModel.PORequests.Where(x => x.POItems.Description.Contains(ItemDescription)).ToPagedList(pageNumber, 5);
}

The x.POItems.Description part doesn't exist in this sense. Apparently related to it being a list. Continuing remainder of code...
            switch (sortOrder)
            {
                case "PONumber_desc":
                    viewModel.PORequests = viewModel.PORequests.OrderByDescending(s => s.PONumber).ToPagedList(pageNumber, 5);
                    break;
                case "AppropNumber":
                    viewModel.PORequests = viewModel.PORequests.OrderBy(s => s.AppropNumber).ToPagedList(pageNumber, 5);
                    break;
                case "AppropNumber_desc":
                    viewModel.PORequests = viewModel.PORequests.OrderByDescending(s => s.AppropNumber).ToPagedList(pageNumber, 5);
                    break;
                case "ContractNumber":
                    viewModel.PORequests = viewModel.PORequests.OrderBy(s => s.ContractNumber).ToPagedList(pageNumber, 5);
                    break;
                case "ContractNumber_desc":
                    viewModel.PORequests = viewModel.PORequests.OrderByDescending(s => s.ContractNumber).ToPagedList(pageNumber, 5);
                    break;
                default:
                    viewModel.PORequests = viewModel.PORequests.OrderBy(s => s.PONumber).ToPagedList(pageNumber, 5);
                    break;
            }

            return View(viewModel);
        }

In the code above, I included my paging and sorting stuff as well, because in the tutorials and answers I have come across, it appears to make a difference in how it all turns out, and is required for this project.
Below is the Index/View I am using to display the results:
//View    (Index)
@model FinanceSearch.ViewModels.POItemData
@using PagedList.Mvc;
<link href="~/Content/PagedList.css" rel="stylesheet" type="text/css" />

@{
    ViewBag.Title = "PORequests";
}

Finance

@*

    @Html.ActionLink("Create New", "Create")

*@

    <button class="btn btn-default" data-toggle="collapse" href="#MainFilter" @*data-target="#MainFilter"*@ aria-multiselectable="true" aria-expanded="true">Filter By...</button>

@*
*@
@using (Html.BeginForm("Index", "PORequests", FormMethod.Get))
{

                    Item:

                    @Html.TextBox("ItemDescription", ViewBag.CurrentFilter as string) //this is where the user would search items using a string

                    PO #:

                    @Html.TextBox("PONumber", ViewBag.CurrentFilter as string)

            <button class="btn btn-primary btn-sm" data-parent="#MainFilter" data-toggle="collapse" href="#SubFilterPOInfo" @*data-target="#SubFilterPOInfo"*@>Filter by PO Info...</button>

                        AppropNumber:

                        @Html.TextBox("AppropNumber", ViewBag.CurrentFilter as string)

                        Contract Number:

                        @Html.TextBox("ContractNumber", ViewBag.CurrentFilter as string)

}

                @Html.ActionLink("PONumber", "Index", new { sortOrder = ViewBag.POSortParm, currentFilter = ViewBag.CurrentFilter })

                @Html.ActionLink("AppropNumber", "Index", new { sortOrder = ViewBag.AppropNumberSortParm, currentFilter = ViewBag.CurrentFilter })

                @Html.ActionLink("ContractNumber", "Index", new { sortOrder = ViewBag.ContractNumberSortParm, currentFilter = ViewBag.CurrentFilter })

        @foreach (var item in Model.PORequests)
            {

                    @*@Html.ActionLink("Edit", "Edit", new { id = item.ID }) |*@
                    @Html.ActionLink("Details", "Details", new { id = item.ID }) |
                    @*@Html.ActionLink("Delete", "Delete", new { id = item.ID })*@

                    //I am able to list the related items in the PORequests table
                    @foreach (var poitem in item.POItems)
                    {
                        @poitem.Description

                    }

                    @Html.DisplayFor(modelItem => item.PONumber)

                    @Html.DisplayFor(modelItem => item.AppropNumber)

                    @Html.DisplayFor(modelItem => item.ContractNumber)

        }

//paging and sorting stuff
    Page @(Model.PORequests.PageCount < Model.PORequests.PageNumber ? 0 : Model.PORequests.PageNumber) of @Model.PORequests.PageCount
    @Html.PagedListPager(Model.PORequests, page => Url.Action("Index", new { page, sortOrder = ViewBag.CurrentSort, currentFilter = ViewBag.CurrentFilter }))

To summarize:
I don't seem to comprehend how to use the variable ItemDescription to search within the POItems list, and get the PORequest table, which includes the POItem table as a related object, to display the results. My other searches all function as expected.
I have searched, on this site in particular, but any of the similar questions involve the view end of it just displaying each item in the list, which I am able to do, or creating/editing functionality, or, perhaps, they are beyond my understanding of MVC enough where they didn't appear to be a solution.
Here are a few such examples:
MVC Code First: One-to-many relationship between own model and SimpleMembership user
EntityFramework 6. How to get related objects?
https://stackoverflow.com/questions/26372533/add-the-list-of-object-to-parent-object-using-mvc-5-and-ef-6
Any help would be greatly appreciated.

A: Drill down to the items in the search expression:
if (!string.IsNullOrEmpty(ItemDescription))
{
    viewModel.PORequests = viewModel.PORequests.Where(x => x.POItems
        .Any(i => i.Description.Contains(ItemDescription)))
        .ToPagedList(pageNumber, 5);
}

redpajama_set_name: RedPajamaStackExchange


text:
Fingercitron (Citrus medica var. sarcodactylis) är en varietet av suckatcitron (C. medica) i familjen vinruteväxter. Varieteten förekommer inte som vild utan har uppstått i kultur.

Citrusfrukter
Vinruteväxter
Fi Växtindex

redpajama_set_name: RedPajamaWikipedia


The watch that a man wears on his wrist says a lot about him. A great watch is not just a symbol of status, but a representation of the wearer's taste. However, many people argue that wrist watches are slowly becoming obsolete. After all, when you can easily consult your phone to know the time, why bother wearing a watch around the wrist? We at Zymak beg to differ. A good-looking, slick watch can be a tasteful embellishment to complete your overall suave look. It will make you stand out from the crowd and you can wear a different watch every day to match your mood. Here are a few more good reasons.
Why wear a Watch in Modern Times?
When going for a meeting, an interview or any formal event, a watch is the perfect accessory to finish off a suit and tie. For formal events, one should stick to wearing a watch with a metal band, rather than plastic, rexin or any other material.
According to Mark Horstman, co-founder of Manager Tools, a good watch is a must when going for an interview, especially because managing time is of utmost importance if you want to create a good first impression. Checking your watch in that setting is a subtle gesture, whereas pulling out your phone to see the time is impolite and creates a poor impression.
And the last reason, but certainly not the least is that a watch is a classic style that can never go out of fashion.
The watches we have at Zymak come in a variety of shapes and sizes. You can choose from contemporary designs to more traditional ones. In terms of dials, there are watches with different shapes, makes and colors, for example round, oval, rectangular, or square. The straps come in a variety of forms too, like leather, metal, or fabric. Watch prices in Bangladesh vary according to the quality of the strap as well as the internal mechanics of the watch.
Are you looking for a slick watch for yourself, or to gift a family member or significant other? No matter what type or price of watch you are looking for, we have a watch for you.
If you are looking for a watch that is stylish and functional, but at the same time, easy on your pocket, you can consider purchasing a Curren watch.
A lot of people in Bangladesh may not be aware of the Curren brand, though it is quite a popular brand worldwide. The Curren Brand is of Chinese origin and was founded in 2001. If you look it up on the internet, you will find millions of positive reviews about this brand. A lot of people avoid buying Chinese products as they believe the quality of the products to be low. However, the quality of the Current brand of watches is far better than other Chinese watches and competitor companies in the low-price watch domain.
In fact, many of its fans and loyal users say that its quality can be compared to the likes of brands RADO, Rolex, Fast Track, Armani, Gucci, Titan, Casio and other popular brands. At Zymak electrnonics we import these watches, so you can rest assured of the authenticity of these watches, knowing that they have been sourced directly from China. You will find a wide selection of Curren watches at Zymak electronics and we are offering them at the best curren watch price in Bangladesh.
These watches also boast of high-quality tempered glass and a high-quality stainless steel body. So, they have a longer life and don't get damaged easily. For those who would like to use branded watches but have a low budget, Curren Brand Watches can be the best choice for them. These watches are stylish, high quality, low price, waterproof and have fashionable belts. Here is a sum up of their best features.
Why should you buy Curren watches?
The belt and dial are both made from real stainless steel, so it is longer lasting and the color does not change. This also makes it resistant to scratches and corrosion.
The face of the watch is covered with tempered glass, so it neither scratches, nor cracks easily. You can ask one of our representatives at the store to demonstrate this for you.
The watch is completely water-resistant. Although many watch companies claim their watches to be water-resistant, few of them actually follow through and fulfill that promise. However, Current watches are guaranteed 100% water-resistant. So, they are durable, long lasting and are ideal to be taken along on long trips.
At Zymak electronics, you will get original Curren watches and you can choose to order online or come down to our shop located in Dhanmondi, Dhaka Bangladesh. If you would like to purchase from our selection of Curren watches online, please feel free to visit our catalog and select the from our wide array of Curren watches of different, designs and colors. We work with the best courier services to ensure a timely delivery and reliable service. We also deliver all over Bangladesh and you can receive your product right at your doorstep without a hassle.
However, if you wish to wear it and see for yourself which one looks good on your wrist, you can visit our shop. Our friendly staff will be delighted to help you pick your ideal watch.
We will also try to find you the best watch according to your budget. Curren watch prices in Bangladesh vary from tk. 250 to tk. 1200. These are, by far, the best current watch prices in bd, we guarantee that. Click here now to get the best watch prices in Bangladesh!


text:
Pacific Discovery School to Open in Ring Mountain Day School Space – Open House Set for Sunday, March 4
Ring Mountain Day School, a private K-8 school, has leased the space from the Mill Valley School District since 2004 but is closing at the end of the 2017-18 school year due to dwindling enrollment.
Scenes from the future Pacific Discovery School, which opens in the Ring Mountain Day School space at 70 Lomita Drive in fall 2018. Courtesy images.
Big changes are on the way for the longtime Mill Valley School District space at 70 Lomita Drive, the home of the private Ring Mountain Day School for the past 14 years in the building adjacent to the campus of Edna Maguire Elementary School.
Pacific Discovery School, a new, private K-8 school founded and headed by Wendy Xa, a former vice president at Goldman Sachs and a mother of two who founded the Mandarin immersion Presidio Knolls School in San Francisco in 2008 as a preschool before expanding it to a K-8 program, is taking over the space at the end of the 2017-18 school year.
The school's administrative team is feverishly planning outreach to the Mill Valley community and beyond, starting with an open house set for this Sunday, March 4, from 1-3pm to promote a curriculum that embraces small class sizes, customized learning plans, on-site learning specialists, outdoor education and collaborative learning.
Pacific Discovery is taking over Ring Mountain's lease with the district, an agreement that runs through 2024, according to Sarah Flowers, Ring Mountain's head of school for the past five years.
"After struggling with enrollment and demographics for a number of years, we couldn't model out the growth we needed in order to be sustainable," Flowers says. "Our board decided that we would have to wind down. We've been helping our families and our teachers find new schools."
With the success of Presidio Knolls behind it, Xa's Pacific Discovery School hopes to step into that void with "a focus on embracing and empowering the individual child to feel confident in who he is versus what others think he should be," says Christina Forté, the school's community outreach director. "That's really the foundation on which this school has been built. Creating, self-aware, happy, empathetic and confident kids…that's going to be the school's measure of success."
Pacific Discovery officials are working with some Ring Mountain families and teachers to stay on with the new school, according to Forté. Flowers says Ring Mountain's current enrollment is 55 students, down from 80-90 in recent years. She says enrollment dipped when its preschool, which served a feeder for K-8 enrollment, closed a few years ago.
"My sense is that small independent schools are all feeling the forces of demographic changes and the shifts of where families live and where they are spending their money," Flowers says. "And the public schools here in Marin are just so good. We're one of the small independent schools that have to close – but we are by no means the last."
Forté first learned about Pacific Discovery School through her friend Kimberly Hawks, the school's director of strategy operations and communications. Their kids go to public school together now.
"I think this school is really special and I decided to join the team to help spread the word," she says. "Wendy is an incredibly impressive person, and she's pulled together a great team of educators and curriculum specialists."
Forté says Xa's idea for Pacific Discovery School came from her desire to customize the learning environment for each child.
"She fleshed out this idea of looking at the child from a very individualistic perspective, focusing on the academic and cognitive ability of course, but also giving equal measure to the social and emotional growth of a child," Forté says. "Customized learning plans for each child make Pacific Discovery a unique offering and one that I feel many parents will find attractive."
Xa felt that Ring Mountain's space lent itself well to the outdoor education component, Forte says, particularly its proximity to places like Muir Woods and Mount Tam. "Learning differences will also be embraced – there will be real support," Forté adds, noting that Pacific Discovery will have a learning specialist on site.
Forté says the school is recruiting students from Novato to San Francisco and plans to run a bus from San Francisco to accommodate those students.
She notes that "there were many things that were happening at Ring Mountain that were pretty special, and we'll be retaining some of those things." That includes the Harkness table, a teaching method by which students sit at a table to discuss ideas in an encouraging, open-minded environment with minimal teacher intervention.
"It's an amazing to watch these kids engage in a richer and more robust discussion in which everyone feels heard and there's a confidence level in every kid," she says.
Xa and her team plan to renovate parts of the Ring Mountain building, including the front facade, Forté says. They'll take over the building in June.
The 411: Pacific Discovery School holds an Open House on Sunday, March 4, 1-3pm. 70 Lomita Drive. REGISTER & MORE INFO. Tours also happen every Tuesday at 9:30am. Sign up for a tour here.
Want to know what's happening around town? Click here to subscribe to the Enjoy Mill Valley Blog by Email!


Of All To Benefit, Says The New Frontier, Glory Sparkles As Crystal Clear!!!
To Metaphysical Ascension Uv Consciousness U Speak Uv, Maybe because Metallic Soil Has Positive Charged Ions, or A Possible Connection founded in Mettalics for a Different Reason Relating to Ascension, a Possible Referenc to Telemtry8 Intertwined with Ascendi g Consciousness'. Either Way, th Message Uv UPLIFTMENT +UNITY AS A COMPONENT UV TH NEW FRONTIER SPOKE YES.


text:
GE to open drive train repair innovation lab
26 February 2014 by Patrick Smith
UNITED STATES: GE is to open a laboratory aimed at improving its processes for the repair of turbine drivetrains.
The lab will work on a number of GE turbine models
The Innovation Lab at GE's existing power generation repair centre in Albany, New York state, will develop innovations for repairs to gearboxes, generators and rotors.
In order to achieve this, the centre will be equipped with rapid prototyping tools such as 3D printers, robotic welding and advanced machining tools. These will be used with the aim of developing new technologies to apply to field service repairs.
It will work on the repair of various gearbox models within GE's range, the company said.
The manufacturer added that iit believes the facility will lower operations and maintenance costs for its customers.
GE claims DC system boosts offshore output by 15%
GE to use new 7.5MW test centre
GE to enhance E.on turbines


text:
Arts History
Van Gogh's fascination with Japan
Japanese art flooded Western Europe when in 1854, America forced Japan to open its borders to trade.
Some of the prints of Japanese woodcuts made it all the way to Vincent Van Gogh in Paris. He grew obsessed with ukyio-e, or "pictures of the world," joyful elements he copied into his own art.
'Seeing with Japanese eye'
Van Gogh amassed a collection of Japanese wood prints in his Paris studio. It was there he started emulating the bright and exotic images of Japanese art, an influence he called Japonaiserie.
"My studio's quite tolerable, mainly because I've pinned a set of Japanese prints on the walls that I find very diverting. You know, those little female figures in gardens or on the shore, horsemen, flowers, gnarled thorn branches.
According to the exhibition of Van Gogh & Japan at the Van Gogh museum in Amsterdam, the artist adopted the 'bold, flat areas of color, bold contour lines, and prominent diagonals.' He even cropped subjects at the edges of pictures and used the Japanese unique play on foreground/background spatial effects.
Van Gogh's Japanese obsession permeated his work. "All my work is based to some extent on Japanese art," he told his brother Theo.
Courtesan (after Eisen) by Vincent van Gogh (1887)
Find out more about Van Gogh's love affair with Japan at the Exhibition Van Gogh & Japan.
Tags art, history, japan, van gogh
← Questioning the potency of graphic design → Craving wilderness/Out to Live


text:
Q: Drawing graphics with an method in another class I want to draw an circle from a method in another Class. But it says invalid parameters on line 41 and 42 in my class. When im done whith this project it should be an analog clock. and this is my first Project im unsing the drawing Event.Im undersatnding a shit and dont know how to fix my Problem. Im new in Forms. I allready tried to draw the circle in the main and it works fine but however it didnt work putting into a class. Im sorry for my classnames they might seem strange because im german.  Thanks for your help.
Thats my class:
    class Ziffernblatt
    {
        Size RectSize;

        Point RectPoint = new Point(5, 10);

        Rectangle Myrect;
        Rectangle MyCircle;

        Pen MyPen = new Pen(Color.Black, 1);
        Pen Invpen = new Pen(Color.White, 1);

        Graphics gObject;

        public Ziffernblatt(Graphics NgObject)
        {
            gObject = NgObject;

        }

        public void Draw(int PosX, int PosY)
        {
            RectSize.Width = PosX / 2;
            RectSize.Height = PosY / 2;

            RectPoint.X = PosX / 2 - RectSize.Width / 2;
            RectPoint.Y = PosY / 2 - RectSize.Height / 2;

            Myrect = new Rectangle(RectPoint, RectSize);
            MyCircle = new Rectangle(RectPoint, RectSize);

           gObject.DrawRectangle(Pens.Red, Myrect);
           gObject.DrawEllipse(Pens.Black, MyCircle);

        }

and this is my main:
 public partial class Form1 : Form
    {

        Ziffernblatt[] Ziffern = new Ziffernblatt[1];
        Graphics gObject;

        public Form1()
        {

            InitializeComponent();
        }

        private void Form1_Load(object sender, EventArgs e)
        {
        }

        private void Form1_Paint(object sender, PaintEventArgs e)
        {
            gObject = e.Graphics;
            Ziffern[0] = new Ziffernblatt(gObject);
        }

        private void Form1_SizeChanged(object sender, EventArgs e)
        {
            Ziffern[0].Draw(this.ClientSize.Width, this.ClientSize.Height);
            Invalidate();
        }
    }

A: You received this error because of you encapsulated the Graphics object to you class. Just call your draw method from the paint event.
The best practice when drawing is performed from the paint method of the Form1. Take to account that the Graphics object will be changed for each paint event.
Your code might look like this:
public partial class Form1 : Form
{
    Ziffernblatt[] Ziffern = new Ziffernblatt[1];

    public Form1()
    {
        InitializeComponent();
        Ziffern[0] = new Ziffernblatt();
    }

    private void Form1_Load(object sender, EventArgs e)
    {
    }

    private void Form1_Paint(object sender, PaintEventArgs e)
    {
        Ziffern[0].Draw(e.Graphics, this.ClientSize.Width, this.ClientSize.Height);
    }

    private void Form1_SizeChanged(object sender, EventArgs e)
    {
        //Ziffern[0].Draw(this.ClientSize.Width, this.ClientSize.Height);
        Invalidate();
    }
}

public class Ziffernblatt
{
    Size RectSize;

    Point RectPoint = new Point(5, 10);

    Rectangle Myrect;
    Rectangle MyCircle;

    Pen MyPen = new Pen(Color.Black, 1);
    Pen Invpen = new Pen(Color.White, 1);

    public Ziffernblatt() //Graphics NgObject)
    {
        //gObject = NgObject;
    }

    public void Draw(Graphics g, int PosX, int PosY)
    {
        RectSize.Width = PosX / 2;
        RectSize.Height = PosY / 2;

        RectPoint.X = PosX / 2 - RectSize.Width / 2;
        RectPoint.Y = PosY / 2 - RectSize.Height / 2;

        Myrect = new Rectangle(RectPoint, RectSize);
        MyCircle = new Rectangle(RectPoint, RectSize);

        g.DrawRectangle(Pens.Red, Myrect);
        g.DrawEllipse(Pens.Black, MyCircle);
    }
}

For additional information about the paint event see Control.Paint
Event


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