text:
\section{Introduction}

     A spray $S$ on an $n$-dimensional  manifold $M$ is a vector field on the tangent bundle $TM$ and arises   from  a system of second-order   ordinary
differential equations and   conversely, a spray can be associated with a system of second-order    ordinary differential equations.
The inverse problem for sprays (or, the so-called Finsler metrizability problem) is the problem of deciding,  for a given spray $S$, whether or not there exists  a Lagrangian or a Finsler function such that $S$ is its  geodesic spray.
The inverse problem for sprays  is   a special case of the inverse problem of the calculus of variations for arbitrary systems of second-order ordinary differential equations.
Therefore,   a   Finsler function $F$ has  a unique geodesic spray but for a given spray, to decide if it is coming from a Finsler function, one has to solve the inverse problem.

\medskip

 In a  Berwald manifold  $(M,F)$ and for any piecewise smooth curve $c(t)$ connecting two points $p,q\in M$, the Berwald parallel translation $P_c$ is linear isometry  between $(T_pM,F_p)$  and $(T_qM,F_q)$ and this is equivalent to that the geodesic spray of $F$ is quadratic.  But, in a Landsberg manifold the   parallel translation $P_c$ along $c$ preserves the induced Riemannian metrics on the slit tangent spaces, that is,  $P_c : (T_pM \backslash \{0\},{g}_p) \longrightarrow (T_qM \backslash \{0\}, {g}_q)$ is an isometry.
   This is equivalent to that the horizontal covariant derivative of the metric tensor of $F$ with respect to the Berwald connection vanishes.
It is clear  that every Berwald manifold  is  Landsbergian. Whether there exists a regular Landsberg manifold,   which is not  Berwaldian,  is a long-standing question in Finsler geometry, which is still open.  To the best of our knowledge, it is not known a concrete  example of a regular non-Berwaldian Landsberg manifold.  In \cite{Asanov}, G. S. Asanov  introduced a class of  examples, arising from   general relativity, of non-Berwaldian Landsberg manifolds, of dimension at least $3$.     In \cite{Shen_example}, Z. Shen  generalized Asanov's class to  the  $(\alpha,\beta)$-metrics of Landsberg type. The difficulty  of finding a regular non-Berwaldian Landsberg space leads   Bao \cite{Bao} to describe them as the unicorns of Finsler geometry. For more details and more examples of non-regular non-Berwaldian Landsberg metrics, we refer to, \cite{ Elgendi-LBp,Elgendi-solutions,Elgendi-ST_condition}.
In dimension two, Zhou  \cite{Zhou} provides non-regular examples of non-Berwaldian Landsberg spherically symmetric surfaces. But it was shown in   \cite{Elgendi-Youssef} that  these surfaces   are  Berwaldian. At the end of this paper we discuss this class in some details.

\medskip

Let $\Omega $ be a convex domain in $\mathbb{R}^n$. A Finsler metric $F$ on $\Omega$  is said to be spherically symmetric if the orthogonal group acts as isometries of $F$, that is, $(\Omega,F)$ is  invariant under all rotations in $\mathbb{R}^n$. In other words, $F$ is spherically symmetric if it is orthogonally invariant in the sense that $F(Ax,Ay)=F(x,y)$, where $A\in O(n)$, for more details see \cite{Guo-Mo}. Spherically symmetric metrics were   introduced by Rutz    \cite{Rutz} where she  generalized the classic Birkhoff theorem in general relativity to the Finslerian case. Recently, spherically symmetric Finsler metrics have been paid  a lot of attention, for example, we refer to \cite{Zhou_Mo,Zhou}.

\medskip

Let $|\cdot|$ and $\langle \cdot , \cdot \rangle$ be the standard Euclidean norm and the inner product on $\mathbb{R}^n$.  A Finsler metric $F$ on $\mathbb{B}^n(r_0)\subset\mathbb{R}^n$   is  spherically symmetric if and only if there exists  a positive $C^\infty$ function $\phi $ such that
$F(x,y)= u \phi(r,s)$
where  $r=|x|$, $u=|y|$ and $s=\frac{\langle x, y \rangle}{|y|}$ and $(x,y)\in T\mathbb{B}^n(r_0)\backslash \{0\}$. The geodesic spray $S$ of $F$ is given by
 $$ G^i=uPy^i+u^2 Qx^i.$$

Now,  since a spherically symmetric metric $F$  is written in the form  $F=u\phi$ then we can talk about $\phi$ rather than $F$. For a given function    $\phi$ we have the geodesic spray and hence the functions $P$ and $Q$. But if     $P$ and $Q$ are given, then we have to solve the inverse problem to decide if there exists a function $\phi$ such that the geodesic spray is given by $P$ and $Q$.

\medskip

In this paper, we solve the inverse problem for the spherically symmetric metrics and find compatibility conditions on  $P$, $Q$ and $\phi$, that is for   given $P$ and $Q$ then the compatibility conditions  provide the function $\phi$ where the geodesic spray is given by $P$ and $Q$ (see Proposition \ref{Comp_C_C_2}).

Depending on the compatibility conditions, we classify all  Landsberg spherically symmetric Finsler metrics; that is, we prove that  a Landsberg spherically symmetric Finsler metric of dimension $n\geq3$ is either Riemannian or the geodesic  spray is given by the functions
$$ P=c_1 s+\frac{c_2}{r^2}\sqrt{r^2-s^2}  , \quad Q=\frac{1}{2}c_0 s^2-\frac{c_2 s}{r^4}\sqrt{r^2-s^2}+c_3,$$
where $c_0$, $c_1$, $c_2$, $c_3$ are arbitrary functions of $r$ (see Theorem \ref{Theorem_A}).   We find out  new non-regular Landsberg spherically symmetric metrics which are not Berwaldian. These metrics are given by $F=u\phi(r,s)$, where $\phi$ is given by
\small{\begin{equation*}
\begin{split}
\phi(r,s)=&(5s^2-r^2)^{\frac{1}{3}}(2s^2-r^2)^{\frac{1}{6}}\exp\Bigg{(} \frac{1}{3}\operatorname{arctanh} \Bigg{(} \frac{\sqrt{5}}{10 }\frac{(\sqrt{5}s+5r)}{\sqrt{r^2-s^2}}\Bigg{)}-\frac{1}{6}\operatorname{arctanh}\Bigg{(}  \frac{\sqrt{2}}{2  }\frac{(\sqrt{2}s+2r)}{\sqrt{r^2-s^2}}\Bigg{)}\\
&
+ \frac{1}{3}\operatorname{arctanh} \Bigg{(} \frac{\sqrt{5}}{2  }\frac{(\sqrt{5}s-5r)}{\sqrt{r^2-s^2}}\Bigg{)}-\frac{1}{6}\operatorname{arctanh}\Bigg{(}  \frac{\sqrt{2}}{  2}\frac{(\sqrt{2}s-2r)}{\sqrt{r^2-s^2}}\Bigg{)}\Bigg{)}
,
\end{split}
\end{equation*}}
\small{\begin{equation*}
\phi(r,s)=\frac{1}{r} (r^4-5r^2s^2+5s^4)^{\frac{1}{4}} \Big{(} \frac{A}{B}\Big{)}^{\frac{5-\sqrt{5}}{20}}\Big{(} \frac{C}{D}\Big{)}^{\frac{5+\sqrt{5}}{20}}\exp\Bigg{(} -\frac{\sqrt{5}}{10}\operatorname{arctanh} \Bigg{(}  \frac{\sqrt{5}(r^2-2s^2)}{r^2}\Bigg{)}\Bigg{)},
\end{equation*}}
where
\begin{align*}
A:&=\left(2\sqrt{r^2-s^2}-  2r+\left(\sqrt{5}   -1\right)s\right)\sqrt{r^2-s^2}-\left(\sqrt{5} -1 \right)rs,\\
B:&= \left(2\sqrt{r^2-s^2}-2r-\left(\sqrt{5}   -1\right)s\right)\sqrt{r^2-s^2}+\left(\sqrt{5} -1  \right)rs,\\
C:&=\left(2\sqrt{r^2-s^2}-2r-\left(\sqrt{5}   +1\right)s\right)\sqrt{r^2-s^2}+\left(\sqrt{5} +1  \right)rs,\\
D:&=\left(2\sqrt{r^2-s^2}-2r+\left(\sqrt{5}   +1\right)s\right)\sqrt{r^2-s^2}-\left(\sqrt{5} +1  \right)rs.\\
\end{align*}
For more details about the above two metrics, see Examples \ref{Ex:1} and \ref{Ex:2}.
  We show that    all (regular or non-regular) Berwald spherically symmetric metrics of dimension $n\geq 3$ are Riemannian  (see Theorem \ref{Th_Riemannian}). Also,  we prove that all regular Landsberg spherically symmetric metrics of dimension $n\geq 3$ are Riemannian  (see Theorem  \ref{Reg_Lands_Riem_D_3}).

 \medskip

Finally, in Section 6,  we focus our attention on  the two-dimensional case. We characterize all Berwaldian and Landsbergian spherically symmetric surfaces. We prove that a spherically symmetric Finsler surface is Berwald if and only if
$$P=b_1 s+ \frac{b_2}{ \sqrt{r^2-s^2}} +\frac{b_3 (r^2-2s^2)}{\sqrt{r^2-s^2}} ,$$
$$ Q= b_0 s^2+\frac{1}{2} b_1+ \frac{b_2 s (r^2-2 s^2)}{r^4 \sqrt{r^2-s^2}} -\frac{b_3 s (3r^2-2s^2)}{r^2\sqrt{r^2-s^2}}-\frac{a  }{r^2} s \sqrt{r^2-s^2},$$
where $a$, $b_0$, $b_1$, $b_2$, $b_3$ are arbitrary functions of $r$ and to be chosen such  that the compatibility conditions are satisfied. We discuss the Landsberg surfaces. Moreover, we give a note on a class of Landsberg surfaces mentioned in \cite{Zhou} which,   in fact, is Berwaldian.

\section{Preliminaries}

Let $M$ be an $n$-dimensional manifold and $(TM,\pi,M)$ be its tangent
bundle. We denote by $\mathcal T\hspace{-1pt}M:=TM\setminus \{0\}$   the set of nonzero tangent
vectors.   Also, $(x^i) $ denote  the local coordinates on the base manifold
$M$ and   $(x^i, y^i)$ are  the induced coordinates on $TM$.
 The vector $1$-form $J$ on $TM$ defined by $J = \frac{\partial}{\partial y^i} \otimes dx^i$ is called the
natural almost-tangent structure of $T M$. The vertical vector field
${\mathcal C}=y^i\frac{\partial}{\partial y^i}$ on $TM$ is called the canonical or the
Liouville vector field.

A vector field $S\in \mathfrak{X}({\mathcal T} M)$ is called a spray if $JS = {\mathcal C}$ and
$[{\mathcal C}, S] = S$. Locally, a spray can be expressed as follows
\begin{equation*}
  \label{eq:spray}
  S = y^i \frac{\partial}{\partial x^i} - 2G^i\frac{\partial}{\partial y^i},
\end{equation*}
where the spray coefficients $G^i=G^i(x,y)$ are $2$-homogeneous
functions in  $y$.

A nonlinear connection on $TM$ is defined by an $n$-dimensional distribution $H : z \in \T M \rightarrow H_z(\T M) $ which  is supplementary to the vertical distribution. This means that for all $z \in \T M$, we have
$$T_z(\T M) = H_z(\T M) \oplus V_z(\T M).$$

Every spray $S $ induces a canonical nonlinear connection through the corresponding horizontal and vertical projectors,
\begin{equation*}
  \label{projectors}
    h=\frac{1}{2}  (Id + [J,S]), \,\,\,\,            v=\frac{1}{2}(Id - [J,S])
\end{equation*}
 With respect to the induced nonlinear connection,  we have two associated  projectors $h$ and $v$ which are  expressed as follows
$$h=\frac{\delta}{\delta x^i}\otimes dx^i, \quad\quad v=\frac{\partial}{\partial y^i}\otimes \delta y^i,$$
$$\frac{\delta}{\delta x^i}=\frac{\partial}{\partial x^i}-G^j_i(x,y)\frac{\partial}{\partial y^j},\quad \delta y^i=dy^i+G^i_j(x,y)dx^j, \quad G^j_i(x,y)=\frac{\partial G^j}{\partial y^i}.$$

If $X\in  \mathfrak{X}(M)$, $ i_{X}$ and $\mathcal{L}_X$ denote the interior product by $X$ and the Lie derivative  in the direction of $X$, respectively.   Every vector $\ell$-form $L$ defines two graded derivations $i_L$ and $d_L$ of the exterior algebra of $M$ such that
 $$i_Lf=0, \quad i_Ldf=df\circ L, \quad d_L:=[i_L,d]=i_L\circ d-(-1)^{\ell-1}di_L,$$
 where $f\in C^\infty(M)$ and $df$ is the exterior derivative of $f$.
  If $X\in \mathfrak{X}(M)$ is a vector field, then $i_{X}$ is simply the
interior product by $X$ and $d_X=\mathcal{L}_X$ the Lie derivative with
respect to $X$.

\medskip

 A Finsler function  on $M$ is defined as follows:

\begin{definition}
A Finsler manifold  of dimension $n$ is a pair $(M,F)$, where $M$ is a  differentiable manifold of dimension $n$ and $F$ is a map, called Finsler function or Finsler metric,   $$F: TM \longrightarrow \mathbb R ,\vspace{-0.1cm}$$  such that{\em:}
 \begin{description}
    \item[(a)] $F$ is smooth and strictly positive on ${\mathcal T} M$ and $F(x,y)=0$ if and only if $y=0$,
    \item[(b)]$F$ is positively homogeneous of degree $1$ in the direction argument $y${\em:}
    $d_{\mathcal{C}} F=\mathcal{L}_{\mathcal C} F=F$,
    \item[(c)] The metric tensor $g_{ij}= \dot{\partial}_i \dot{\partial}_j E$ has maximal rank on ${\mathcal T} M$, where $E:=\frac{1}{2}F^2$ is the energy function.
 \end{description}
 In this case $(M,F)$ is called a regular Finsler space. If $F$ is not smooth or even not defined in some directions, then we call the Finsler function non-regular. In other words, by a non-regular Finsler metric $F$ we mean that $F$ does not satisfy the  condition (a) or (c) on certain directions  of $TM$.
 \end{definition}

 Since the $2$-form $dd_JE$ is non-degenerate,  the Euler-Lagrange equation
\begin{equation*}
  i_Sdd_JE=-dE
\end{equation*}
uniquely determines a spray $S$ on $TM$.  This spray is called the
\emph{geodesic spray} of the Finsler function $F$.

\begin{definition}
  A spray $S$ on a manifold $M$ is called \emph{Finsler metrizable} if there
  exists a Finsler function $F$ such that the geodesic spray of the Finsler
  manifold $(M,F)$ is $S$.
  \end{definition}

  It is known that a spray $S$ is Finsler metrizable if and only if there exists a non-degenerate  solution $F$    for the system
  \begin{equation}
  \label{metrizable_system}
  d_hF=0, \quad d_{\mathcal C} F=F,
  \end{equation}
  where $h$ is the horizontal projector associated to $S$.

  So,  to solve the inverse problem for  a given spray $S$,  we have to find  a non-degenerate solution $F$ for the system \eqref{metrizable_system}.

\section{Spherically symmetric metrics}

Let $|\cdot|$ and $\langle \cdot , \cdot \rangle$ be the standard Euclidean norm and inner product on $\mathbb{R}^n$.  A Finsler metric $F$ on $\mathbb{B}^n(r_0)\subset\mathbb{R}^n$   is  spherically symmetric if and only if there exists a function $\phi:[0,r_0)\times\mathbb{R}^n\to \mathbb{R}$ such that
$$F(x,y)=|y|\phi\left(|x|,\frac{\langle x, y \rangle}{|y|}\right)$$
where $(x,y)\in T\mathbb{B}^n(r_0)\backslash \{0\}$. Or simply we write $F=u\ \phi(r,s)$ where $r=|x|$, $u=|y|$ and $s=\frac{\langle x, y \rangle}{|y|}$.

 \medskip

The spherically symmetric metrics are a special case of the general $(\alpha,\beta)$-metrics. The general $(\alpha,\beta)$-metrics are introduced and studied by C. Yu and H. Zhu \cite{Yu-Zhu}. Therefore, the spherically symmetric metrics $F=u\phi(r,s)$ on $\mathbb{B}^n(r_0)$ are regular if and only if $\phi$ is positive and $C^\infty$ function and
\begin{equation}
\label{Regular_condition}
\phi-s\phi_s>0,\quad \phi-s\phi_s+(r^2-s^2)\phi_{ss}>0
\end{equation}
when $n\geq 3$ or $\phi-s\phi_s+(r^2-s^2)\phi_{ss}>0$  when $n=2$ for all $|s|\leq r<r_0$.
Throughout, the subscript $s$ (resp. $r$) refers to the derivative with respect to $s$ (rep. $r$).
The spherically symmetric Finsler metrics are studied in many papers for example, we refer   to, \cite{Guo-Mo,Zhou_Mo,Zhou}.

\begin{remark}
It should be noted that throughout, the results and formulae work and are valid for both regular and non-regular spherically symmetric metrics unless it is specified. So we consider the most general case,  that is, the non-regular case and when the situation needs, we mention clearly that the metric is regular.
\end{remark}

The components $g_{ij}$ of the metric tensor of the spherically symmetric metric $F=u \phi(r,s)$ are given by
\begin{align}
\label{Eq:g^ij}
g_{ij}=&\sigma_0\ \delta_{ij}+\sigma_1\ \delta_{ik}\delta_{hj}x^kx^h+\frac{\sigma_2}{u} (\delta_{ik}\delta_{hj}+\delta_{jk}\delta_{hi})x^hy^k+\frac{\sigma_3}{u^2}\delta_{ik}\delta_{hj}y^ky^h,
\end{align}
where  $$\sigma_0=\phi(\phi-s\phi_s),\quad \sigma_1= \phi_s^2+\phi\phi_{ss},\quad \sigma_2= (\phi-s\phi_s)\phi_s-s\phi\phi_{ss},  \quad \sigma_3= s^2\phi\phi_{ss}-s(\phi-s\phi_s)\phi_s.$$
 The above formula can be found in \cite{Zhou} and to keep the indices consistent in  both sides of a tensorial  equation,   we  write  for example,   $\delta_{ik}y^k$ when there is a  $y^i$ and $i$ is a lower index.

\medskip

The components $g^{ij}$ of the inverse metric tensor are given by \cite{Zhou} as follows
\begin{align}
\label{Eq:g^ij}
g^{ij}=&\rho_0\delta^{ij}+ \frac{\rho_1}{u^2}y^iy^j+ \frac{\rho_2}{u} (x^iy^j+x^jy^i)+\rho_3x^ix^j,
\end{align}
where
$$
 \rho_0=\frac{1}{\phi(\phi-s\phi_s)}, \quad \rho_1=\frac{(s\phi+(r^2-s^2)\phi_s)(\phi\phi_s-s\phi_s^2-s\phi_{ss})}{\phi^3(\phi-s\phi_s)(\phi-s\phi_s+(r^2-s^2)\phi_{ss})},$$
 $$\rho_2=-\frac{\phi\phi_s-s\phi_s^2-s\phi_{ss}}{\phi^2(\phi-s\phi_s)(\phi-s\phi_s+(r^2-s^2)\phi_{ss})},\quad \rho_3=-\frac{\phi_{ss}}{\phi(\phi-s\phi_s)(\phi-s\phi_s+(r^2-s^2)\phi_{ss})}.$$
  Moreover, the geodesic spray coefficients  $G^i$ of $F$ are  given by
\begin{equation}\label{G}
  G^i=uPy^i+u^2 Qx^i,
\end{equation}
where the functions $P$ and $Q$ have the following formulae
\begin{equation}\label{P,Q}
 Q:=\frac{1}{2 r}\frac{ -\phi_r+s\phi_{rs}+r\phi_{ss}}{\phi-s\phi_s+(r^2-s^2)\phi_{ss}}, \quad P:=-\frac{Q}{\phi}(s\phi +(r^2-s^2)\phi_s)+\frac{1}{2r\phi}(s\phi_r+r\phi_s).
\end{equation}

For the geodesic spray of the  spherically symmetric metric $F$, we have the associated nonlinear connection whose  components $G^i_j$ are defined by  $G^i_j:=\frac{\partial G^i}{\partial y^j}$. Using the following identities:
\begin{equation}
\label{Eq:derivatives}
    \begin{split}
        \frac{\partial r}{\partial x^j}&= \frac{1}{r}\delta_{jk}x^k, \quad  \frac{\partial u}{\partial y^j}= \frac{1}{u}\delta_{jk}y^k,  \quad  \frac{\partial u}{\partial x^j}= 0,         \quad  \frac{\partial r}{\partial y^j}= 0,   \\
         \frac{\partial s}{\partial x^j}&= \frac{1}{u}\delta_{jk}y^k, \quad  \frac{\partial s}{\partial y^j}= \frac{1}{u}\delta_{jk}x^k-\frac{s}{u^2}\delta_{jk}y^k ,   \\
    \end{split}
\end{equation}
the components $G^i_j$ can be calculated as follows
\begin{equation}
\label{Eq:G^i_j}
G^i_j=u P \delta^i_j+P_s \delta_{kj}x^ky^i+  \frac{1}{u}(P-sP_s) \ \delta_{kj}y^ky^i+u Q_s\delta_{jk}x^ix^k+(2Q-sQ_s)\delta_{jk}x^iy^k.
\end{equation}
The components $G^i_{jk\ell}$ of the Berwald curvature are defined by
$$G^i_{jk\ell}=\frac{\partial^2 }{\partial y^\ell \partial y^k} G^i_j.$$
For a spherically symmetric Finsler metric $F=u\ \phi(r,s)$,  we use the identities \eqref{Eq:derivatives} and \eqref{Eq:G^i_j} to find the components $G^i_{jk\ell}$ as follows
\begin{equation*}
    \begin{split}
        G^i_{jk\ell}= & \frac{P_{ss}}{u}(\delta^i_j \delta_{hk}\delta_{t\ell}+\delta^i_\ell \delta_{hk}\delta_{tj}+\delta^i_k \delta_{hj}\delta_{t\ell})x^hx^t+\frac{1}{u}(P-sP_s)(\delta^i_j \delta_{k\ell}+\delta^i_k \delta_{j\ell}+\delta^i_\ell \delta_{jk})\\
        &-\frac{sP_{ss}}{u^2}(\delta^i_j(\delta_{hk}\delta_{t\ell}+\delta_{h\ell}\delta_{tk})+\delta^i_k(\delta_{hj}\delta_{t\ell}+\delta_{h\ell}\delta_{tj})+\delta^i_\ell(\delta_{hk}\delta_{tj}+\delta_{hj}\delta_{tk}))x^hy^t\\
        &-\frac{sP_{ss}}{u^2}(\delta_{jk}\delta_{h\ell}+\delta_{j\ell}\delta_{kh}+\delta_{k\ell}\delta_{jh})x^h y^i
        +\frac{1}{u}(Q_s-sQ_{ss})(\delta_{jk}\delta_{h\ell}+\delta_{j\ell}\delta_{kh}+\delta_{k\ell}\delta_{jh})x^i x^h\\
        &+\frac{1}{u^3}(s^2P_{ss}+sP_s-P)((\delta^i_j\delta_{kh}\delta_{t\ell}+\delta^i_k\delta_{jh}\delta_{t\ell}+\delta^i_\ell\delta_{kh}\delta_{tj})y^hy^t+(\delta_{jk}\delta_{\ell h}+\delta_{j\ell}\delta_{k h}+\delta_{k\ell}\delta_{j h})y^hy^i)\\
        &+\frac{1}{u^5}(3P-s^3P_{sss}-6s^2P_{ss}-3sP_s)\delta_{jh}\delta_{km}\delta_{\ell t} y^hy^my^ty^i+\frac{P_{sss}}{u^2}\delta_{jh}\delta_{k m}\delta_{\ell t} x^hx^mx^ty^i\\
        &+\frac{1}{u^4}(s^2P_{sss}+3sP_{ss})(\delta_{jm}\delta_{kh }\delta_{\ell t} +\delta_{jm}\delta_{kt }\delta_{\ell h}+\delta_{\ell m}\delta_{kh }\delta_{j t}  )x^ty^my^hy^i\\
        &-\frac{1}{u^3}(P_{ss}+sP_{sss})(\delta_{km}\delta_{jh }\delta_{\ell t} +\delta_{jm}\delta_{kh }\delta_{\ell t}+\delta_{jm}\delta_{\ell h }\delta_{k t})x^tx^my^hy^i+\frac{Q_{sss}}{u}\delta_{km}\delta_{jh }\delta_{\ell t}x^hx^mx^tx^i\\
        &+\frac{1}{u^3}(s^2Q_{sss}+sQ_{ss}-Q_s)(\delta_{km}\delta_{jh }\delta_{\ell t}+\delta_{\ell m}\delta_{kh }\delta_{j t}+\delta_{\ell h}\delta_{km }\delta_{j t})x^hx^iy^my^t
        -\frac{sQ_{sss}}{u^2}(\delta_{\ell m}\delta_{jh }\delta_{k t}\\
        &+\delta_{\ell t}\delta_{kh }\delta_{j m}+\delta_{\ell m}\delta_{kh }\delta_{j t})x^hx^ix^my^t+\frac{1}{u^4}  (3sQ_s-3s^2Q_{ss}-s^3Q_{sss})\delta_{\ell m}\delta_{kh }\delta_{j t}y^ty^my^hx^i\\
        &+\frac{1}{u^2}(s^2Q_{ss}-sQ_s)(\delta_{k\ell }\delta_{hj }+\delta_{j\ell }\delta_{hk }+\delta_{kj }\delta_{h\ell })y^hx^i.
    \end{split}
\end{equation*}
The above formula of the Berwald curvature can be found, for example,  in \cite{Zhou_Mo}.  By straightforward calculations, the components $E_{ij}:=G^h_{ijh}$ of the mean Berwald curvature can be calculated as follows:
\begin{equation}
\label{Mean_curv.}
 \begin{split}
E_{ij} =&\frac{1}{u}((n+1)(P-sP_s)+(r^2-s^2)(Q_s-sQ_{ss}))\delta_{ij}+ \frac{1}{u^3}((n+1)(s^2P_{ss}+sP_s-P)\\
&+r^2(s^2Q_{sss}+sQ_{ss}-Q_s)+3s^2Q_s-3s^3Q_{ss}-s^4Q_{sss})\delta_{ih}\delta_{jk} y^h y^k\\
&+\frac{1}{u}((n+1)P_{ss}+2(Q_s-sQ_{ss})+(r^2-s^2)Q_{sss})\delta_{ih}\delta_{jk}x^hx^k\\
&-\frac{s}{u^2}((n+1)P_{ss}+2(Q_s-sQ_{ss})+(r^2-s^2)Q_{sss})(\delta_{ih}\delta_{jk}+\delta_{ik}\delta_{jh}) x^h y^k .
\end{split}
\end{equation}
One can rewrite  $E_{ij}$ as follows
\begin{equation}
\label{E_H}
E_{ij}=\frac{H}{u}\delta_{ij}-\frac{1}{u^3}(sH_s+H) \delta_{hi}\delta_{kj}y^h y^k+\frac{H_s}{su^2}(s(\delta_{hi}\delta_{kj}+\delta_{ki}\delta_{hj})x^hy^k-u \delta_{hi}\delta_{kj}x^hx^k),
\end{equation}
where
 $$H:=(n+1)(P-sP_s)+(r^2-s^2)(Q_s-sQ_{ss}).$$
 Now,   we have the following proposition.
\begin{proposition}
If the mean Berwald curvature vanishes, then either $n=2$ or $H=0$.
\end{proposition}
 \begin{proof}
 Assume that $E_{ij}=0$, then contracting \eqref{E_H} by $\delta^{ij}$, and using the facts that $\delta_{ij}y^iy^j=u^2$, $\delta_{ij}x^ix^j=r^2$ and $\delta_{ij}x^iy^j=\langle x,y\rangle=su$, we have
 $$ s(n-1)H-(r^2-s^2)H_s=0.$$
 Contracting \eqref{E_H} by $x^{i}x^{j}$ implies
 $$ sH-(r^2-s^2)H_s=0.$$
 Subtracting the above two equations,    yields
 $$s(n-2)H=0.$$
 Which holds for all $s$, then we obtain that $n=2$ or $H=0$ and hence the proof is completed.
  \end{proof}

  To avoid confusions, we have to clarify a point   mentioned in \cite[Proposition 2.1]{Zhou} by providing the following remark.
  \begin{remark}
  It is clear that if $H=0$, then the mean Berwald curvature vanishes, that is $E_{ij}=0$. By making use of the above proposition, when $n=2$, the condition $E_{ij}=0$ does not imply  $H=0$ contrary to what mentioned in  \cite{Zhou} and  discussed in \cite{Elgendi-Youssef}.    For the higher dimensions   $n\geq3$,  $E_{ij}=0$ if and only if $H=0$.
  \end{remark}

  The scalar trace $E$ of the mean Berwlad curvature is $E=g^{ij}E_{ij}$.  For a spherically symmetric Finsler metric,   the scalar trace of the mean Berwald curvature is given as follows.
  \begin{proposition}\label{Scalar_E}
  The scalar trace $E$ of the mean Berwald curvature of a spherically symmetric Finsler metric is given by
  $$E=\frac{1}{su}\left(s((n-1)\rho_0+\rho_3(r^2-s^2))H-(r^2-s^2)(\rho_0+\rho_3(r^2-s^2))H_s\right).$$
  \end{proposition}
  \begin{proof}
 By using \eqref{Eq:g^ij} and the fact that $y^iE_{ij}=0$,  the scalar trace of the mean Berwald curvature is calculated as follows
   \begin{equation*}
    \begin{split}
       E=&g^{ij}E_{ij}\\
       =&\frac{\rho_0}{s u}\left((n-1)sH -(r^2-s^2)H_s)+\frac{\rho_3}{s u}(s(r^2-s^2)H-(r^2-s^2)^2H_s\right)\\
       =&\frac{1}{su}\left(s((n-1)\rho_0+\rho_3(r^2-s^2))H-(r^2-s^2)(\rho_0+\rho_3(r^2-s^2))H_s\right).
    \end{split}
\end{equation*}
  \end{proof}

The components $L_{jk\ell}$ of the Landsberg curvature are defined by
$$L_{jk\ell}=-\frac{1}{2} FG^h_{jk\ell} \frac{\partial F}{\partial y^h}.$$
For a spherically symmetric Finsler metric $F=u\ \phi(r,s)$,   the components $L_{jk\ell}$ are given by
\begin{equation}
\label{L_ijk}
    \begin{split}
       L_{jk\ell}=& -\frac{\phi}{2}\Big{(}L_1\delta_{jh}\delta_{km}\delta_{\ell t} x^hx^mx^t+\frac{3sL_2-s^3L_1}{u^3}\delta_{jh}\delta_{km}\delta_{\ell t} y^hy^my^t+L_2(\delta_{k\ell}\delta_{jh}+\delta_{j\ell}\delta_{kh}\\
       &+\delta_{k j}\delta_{\ell h})x^h-\frac{sL_2}{u}(\delta_{k\ell}\delta_{hj}+\delta_{kh}\delta_{\ell j}+\delta_{kj}\delta_{h\ell})y^h  -   \frac{sL_1}{u}(\delta_{m k}\delta_{t\ell}\delta_{hj} +\delta_{kh}\delta_{m j}\delta_{t\ell}\\
       & +\delta_{mj}\delta_{h\ell}\delta_{k t})x^mx^ty^h + \frac{s^2L_1-L_2}{u^2}(\delta_{m k}\delta_{t\ell}\delta_{hj}+\delta_{kh}\delta_{m j}\delta_{t\ell}+\delta_{mj}\delta_{h\ell}\delta_{k t})y^my^tx^h \Big{)}
    \end{split}
\end{equation}
where
$$L_1=  3\phi_sP_{ss}+\phi P_{sss}+(s\phi+(r^2-s^2)\phi_s)Q_{sss},   $$
        $$ L_2=  -s\phi P_{ss}+\phi_s(P-sP_{s})+(s\phi+(r^2-s^2)\phi_s)(Q_{s}-sQ_{ss}).$$

    It is well known that a Finsler metric $F$ is said to be Berwaldain if and only if its Berwald curvature vanishes, that is, $G^h_{ijk}=0$ and this is equivalent to the fact that the geodesic spray of $F$ is quadratic in $y$. Also, $F$ is called Landsbrgain if and only if its Landsberg  curvature vanishes, that is, $L_{ijk}=0$. Paying more attention to the spherically symmetric Finsler metrics of  Landsberg type, we have the following proposition.
\begin{proposition}\label{Landsberg_cond}
If a spherically symmetric Finsler metric $F=u\phi$ is Landsbergian then  either $n=2$ or  $L_1=L_2=0$.
\end{proposition}
 \begin{proof}
 Let $F$ be a Landsberg spherically symmetric Finsler metric, then $L_{jk\ell}=0$.  Contracting \eqref{L_ijk} by $x^j\delta^{k\ell}$ and using the facts that $\delta_{ij}y^iy^j=u^2$, $\delta_{ij}x^ix^j=r^2$ and $\delta_{ij}x^iy^j=\langle x,y\rangle=su$, we have
$$(r^2-s^2) L_1+(n+1)L_2=0.$$
Similarly, contracting \eqref{L_ijk} by $x^jx^kx^\ell$, we get
$$(r^2-s^2) L_1+3L_2=0.$$
Subtracting the above two equations implies
$$(n-2)L_2=0.$$
Therefore,  $n=2$ or  $L_2=0$ and this completes the proof.
  \end{proof}
  By the above proposition together with the fact that if  $L_1=L_2=0$ then $L_{jk\ell}=0$, one can see that   a Landsbergian spherically symmetric Finsler metric of dimension $n\geq 3$ is characterized by the following two conditions
\begin{equation}
\label{Zhou_L_1_2}
    \begin{split}
        L_1= & 3\phi_sP_{ss}+\phi P_{sss}+(s\phi+(r^2-s^2)\phi_s)Q_{sss}=0,   \\
         L_2=& -s\phi P_{ss}+\phi_s(P-sP_{s})+(s\phi+(r^2-s^2)\phi_s)(Q_{s}-sQ_{ss})=0.
    \end{split}
\end{equation}

  Again, to avoid confusions in the class of    the spherically symmetric Finsler metrics of Landsberg type,  we have  the following remark.
  \begin{remark}
  It is clear that if $L_1=L_2=0$ then $L_{jk\ell}=0$ and hence the metric $F$ is Landsbergian. By the above proposition, when $n=2$,  $L_{jk\ell}=0$  does not yield $L_1=L_2=0$ as be discussed in Section 6. For the higher dimensions $n\geq 3$,   $L_{jk\ell}=0$ if and only if $L_1=L_2=0$.
  \end{remark}

\section{The compatibility conditions}

For a non-regular spherically symmetric metric $F=u\phi(r,s)$,  the case  $\phi-s\phi_s=0$ is excluded. If    $\phi-s\phi_s=0$, then  $\phi=f(r) s$ and hence $F=f(r) us=f(r)\langle x,y\rangle$ which is linear in $y$ for an arbitrary function $f(r)$. If the Finsler function is linear in $y$, then the  metric tensor is degenerate everywhere and this is impossible. Moreover, if $\phi_s=0$, then $F=f(r) u$  which is the Euclidean metric or a conformal transformation of the Euclidean metric and this case also is excluded.
\begin{proposition}\label{Q_not_defined}
 Let $F=u\phi(r,s)$ be a spherically symmetric metric, then the geodesic spray of $F$ is not defined  if and only if $\phi-s\phi_s+(r^2-s^2)\phi_{ss}=0$ or equivalently $\phi$ is given by
 $$\phi(r,s)=f_1(r) s+f_2(r) \sqrt{r^2-s^2},$$
 where $f_1(r)$ and $f_2(r)$ are arbitrary functions of $r$.
 \end{proposition}
 \begin{proof}
 By \eqref{P,Q}, it is clear that the function $Q$ is not defined and so the geodesic spray if and only if
  $\phi-s\phi_s+(r^2-s^2)\phi_{ss}=0$. Since the above formula holds for all  $r$  and $s$, then we can write
  $$\frac{\phi_{ss}}{\phi-s\phi_s}=-\frac{1}{r^2-s^2}.$$
  Since $(\phi-s\phi_s)_s=-s\phi_{ss}$,    integrating the above equation with respect to $s$ gives
  $$\phi-s\phi_s=\frac{f_1(r)}{\sqrt{r^2-s^2}}.$$
  Which has the solution
  $$\phi(r,s)=f_1(r) s+f_2(r) \sqrt{r^2-s^2}.$$
 \end{proof}

For a given  spherically symmetric Finsler metric $F=u \phi$, then the geodesic spray \eqref{G} is given by    the functions $P$ and $Q$. But if the functions $P$ and $Q$ are given, then to find the Finsler metric whose    geodesic spray is determined  by $P$ and $Q$,  we have to solve the inverse problem or the metrizability problem. So, we investigate the following two compatibility conditions.
\begin{proposition}
Let $P(r,s)$ and $Q(r,s)$ be given, then the Finsler function $F=u\phi(r,s)$ whose geodesic spray given by $P$ and $Q$ is determined by the function $\phi$ provided that $\phi$ satisfies the following   compatibility conditions:
\begin{equation}
\label{Comp_C_C_2}
    \begin{split}
       C_1:= &(1+sP-(r^2-s^2)(2Q-sQ_s))\phi_s+(s P_s-2P-s(2Q-sQ_s))\phi =0,   \\
        C_2:=& \frac{1}{r}\phi_r-(P+Q_s(r^2-s^2))\phi_s-(P_s+sQ_s) \phi =0.
    \end{split}
\end{equation}
\end{proposition}
\begin{proof}Let $P(r,s)$ and $Q(r,s)$ be two given functions. To find the Finsler function $F=u\phi(r,s)$ whose geodesic spray determined by $P$ and $Q$,  we have to solve the system  $d_hF=0$  \eqref{metrizable_system}.  Locally,  the system  $d_hF=0$  gives rise to
\begin{equation}
\label{Eq:sys_compt}
\frac{\partial (u \phi)}{\partial x^j}-G^i_j\frac{\partial (u \phi)}{\partial y^i}=0.
\end{equation}
We have the following properties
$$\frac{\partial  \phi }{\partial x^i}=\frac{\phi_r}{r}\delta_{ik}x^k+\frac{\phi_s}{u}\delta_{ik}y^k, \quad \frac{\partial  \phi }{\partial y^i}= \frac{\phi_s}{u}\delta_{ik}x^k-\frac{s \phi_s}{u^2}\delta_{ik}y^k . $$
Moreover, since $u\phi$ is homogeneous of degree one in $y$, then $y^j\frac{\partial  (u\phi) }{\partial y^j}=u\phi$. Also, we have $x^j\frac{\partial  (u\phi) }{\partial y^j}=s\phi+(r^2-s^2)\phi_s$. Now,  substituting the above formulae into the system \eqref{Eq:sys_compt}, we have
\begin{align*}
\frac{\partial (u \phi)}{\partial x^j}-G^i_j\frac{\partial (u \phi)}{\partial y^i} =& \frac{u\phi_r}{r}\delta_{jk}x^k+\phi_s\delta_{jk}y^k
 -\Big{(} u P \delta^i_j+P_s \delta_{kj}x^ky^i+  \frac{1}{u}(P-sP_s) \ \delta_{kj}y^ky^i\\
 &+u Q_s\delta_{jk}x^ix^k+(2Q-sQ_s)\delta_{jk}x^iy^k \Big{)}\frac{\partial (u \phi)}{\partial y^i}\\
 =&u\left(\frac{1}{r}\phi_r-P\phi_s-\phi P_s-Q_s(s\phi+(r^2-s^2)\phi_s))\right)\delta_{jk}x^k\\
 &+\left( \phi_s-2P\phi+s\phi P_s+sP\phi_s-(2Q-sQ_s)(s\phi+(r^2-s^2)\phi_s) \right)\delta_{jk}y^k\\
 =&0.
\end{align*}
Now, if there exist two functions $\mu$ and $\eta$  such that
$$\mu \delta_{jk} x^k+\eta \delta_{jk} y^k=0.$$
Then by contracting the above equation by $x^j$, resp. by $y^j$ we get the following two equations
$$r^2 \mu +su \eta=0, \quad s\mu+u\eta=0.$$
Then, we have  $(r^2-s^2) \mu=0$ which holds for all $r$ and $s$, that is, $\mu=0$ and hence $\eta=0$. Using this property, we get the conditions $C_1$ and $C_2$.

Conversely, assume that the conditions $C_1$ and $C_2$ are satisfied. Rewriting $C_1$ and $C_2$ as follows
\begin{equation}
\label{Rewrite_C_1}
\phi_s+s(P_s \phi+P \phi_s)-2P \phi-(s\phi+(r^2-s^2)\phi_s)(2Q-sQ_s)=0,
\end{equation}
\begin{equation}
\label{Rewrite_C_2}
\frac{1}{r}\phi_r-(P\phi_s+P_s\phi)-(s\phi+(r^2-s^2)\phi_s)Q_s =0.
\end{equation}
Adding \eqref{Rewrite_C_1} to the multiplication of \eqref{Rewrite_C_2} by $s$, we have
$$\phi_s+\frac{s}{r}\phi_r-2P\phi -2Q(s\phi+(r^2-s^2)\phi_s)=0.$$
Which implies
$$P=\frac{Q}{\phi}(s\phi+(r^2-s^2)\phi_s)+\frac{1}{2r\phi}(r\phi_s+s\phi_r).$$
By the above formula of $P$, we get
$$P\phi_s+\phi P_s=\frac{1}{2r}(\phi_r+s\phi_{rs}+r\phi_{ss}+2rQ(s\phi_s-\phi-(r^2-s^2)\phi_{ss}))-Q_s(s\phi+(r^2-s^2)\phi_s).$$
Now,  substituting  $P\phi_s+\phi P_s$ into \eqref{Rewrite_C_2}, we obtain that
$$Q=\frac{1}{2r} \frac{-\phi_r+s\phi_{rs}+r\phi_{ss}}{\phi-s\phi_s+(r^2-s^2)\phi_{ss}}.$$
This completes the proof.
\end{proof}

\section{ Landsberg spherically symmetric metrics }

In this section we focus our attention  to the spherically symmetric metrics of Landsberg type. We start by the following result.
\begin{theorem}\label{Theorem_A}
A Landsberg spherically symmetric Finsler metric of dimension $n\geq 3$ is either Riemannian or the geodesic  spray is determined  by the functions
 \begin{equation}
\label{Zhou_P&Q}
P=c_1 s+\frac{c_2}{r^2}\sqrt{r^2-s^2}  , \quad Q=\frac{1}{2}c_0 s^2-\frac{c_2 s}{r^4}\sqrt{r^2-s^2}+c_3,
\end{equation}
where $c_0$, $c_1$, $c_2$, $c_3$ are arbitrary functions of $r$.
\end{theorem}
\begin{proof}
Suppose  that $F=u\phi(r,s)$ be a spherically symmetric metric of dimension $n\geq 3$, then  by Proposition \ref{Landsberg_cond}, the function $\phi$ and  the functions $P$ and $Q$ of the geodesic spray satisfy the Landsberg conditions  \eqref{Zhou_L_1_2}, that is,  we have $L_1=L_2=0$. The condition $L_2=0$ implies that
$$Q_s-sQ_{ss}=-\frac{-s\phi P_{ss}+\phi_s(P-sP_{s})}{s\phi+(r^2-s^2)\phi_s}.$$
It should be noted  that if $s\phi+(r^2-s^2)\phi_s=0$, then by taking the derivative with respect to $s$ implies $\phi-s\phi_s+(r^2-s^2)\phi_{ss}=0$ which is impossible by Proposition \ref{Q_not_defined}. Now, making use of the property that $(Q_s-sQ_{ss})_s=-sQ_{sss}$, then differentiating the above equation with respect to $s$, we get  $Q_{sss}$. Substituting  $Q_{sss}$ into the Landsberg condition $L_1=0$ and straightforward simplifications, we have
$$((r^2-s^2)P_{ss}+(P-sP_s))(s\phi_s^2+s\phi\phi_{ss}-\phi\phi_{s})=0.$$
If $s\phi_s^2+s\phi\phi_{ss}-\phi\phi_{s}=0$, then we can write
$$\frac{-s\phi_{ss}}{\phi -s\phi_s} =-\frac{\phi_s}{\phi}.$$
By integration with respect to $s$, we have
$$\phi -s\phi_s=\frac{f_1(r)}{\phi}.$$
Using the substitution $\phi=s \psi$, then we can rewrite the above equation as follows
$$\psi\psi_s=-\frac{f_1(r)}{s^3}.$$
Which implies
$$\phi=\sqrt{f_1(r)+f_2(r)s^2}.$$
That is, the Finsler function  given by $F=u\phi=\sqrt{f_1(r) u^2+f_2(r)\langle x, y\rangle^2}$  is Riemannian.

If $(r^2-s^2)P_{ss}+(P-sP_s)=0$, then by assuming that $P-sP_s\neq 0$ we can write
$$\frac{-sP_{ss}}{P-sP_s}=\frac{s}{r^2-s^2}.$$
Since $(P-sP_s)_s=-sP_{ss}$, then integrating the above equation with respect to $s$ gives rise to
$$P-sP_s=\frac{ c_1(r)}{\sqrt{r^2-s^2}}.$$
Which has the solution
$$P=c_2(r)s+\frac{c_1(r)}{r^2}\sqrt{r^2-s^2}.$$
Taking the fact that $s\phi+(r^2-s^2)\phi_s\neq 0$, then substituting by the function $P$ into the Landsberg condition $L_2=0$, implies that
$$Q_s-sQ_{ss}=-\frac{c_1(r)}{(r^2-s^2)^{3/2}}.$$
Using the substitution $s\psi=Q_s$, then we can write
$$s^2\psi_{s}=\frac{c_2(r)}{(r^2-s^2)^{3/2}}.$$
Which yields the solution
$$Q_s=c_0(r) s-\frac{c_2(r) (r^2-2s^2)}{r^4\sqrt{r^2-s^2}}.$$
Which has the solution
$$Q=\frac{1}{2}c_0 (r) s^2-\frac{c_2 (r)s}{r^4}\sqrt{r^2-s^2}+c_3(r).$$

In the case where   $P-sP_s=0$ then we have $(r^2-s^2)P_{ss}=0$ for all $r$ and $s$ and hence we obtain that $P_{ss}=0$ and also  the Landsberg condition $L_2=0$, implies that $Q_s-sQ_{ss}=0$ and hence $Q_{sss}=0$. Therefore,  $P=f_1(r)s $ and $Q=f_2(r)s^2+f_3(r)$ which are a special case of the given formulae for $P$ and $Q$. Hence the proof is completed.
 \end{proof}
Based on the above proof, we provide the following remark.
\begin{remark}
It should be noted that  if $P-sP_s=0$ and $Q_s-sQ_{ss}=0$, then $P_{ss}=0$ and $Q_{sss}=0$. That is the Berwald curvature vanishes and hence the metric is Berwaldian. Moreover, if $n\geq 3$ and the Berwald curvature vanishes, then we get $P-sP_s=0$ and $Q_s-sQ_{ss}=0$.  That is, as it was mentioned in \cite{Guo-Mo,Zhou_Mo}, the Berwaldain spherically symmetric Finsler metrics are characterized by $P-sP_s=0$ and $Q_s-sQ_{ss}=0$ provided that $n\geq 3$.

\end{remark}

\begin{theorem}\label{main_theorem}
There exist non-Berwaldian Landsberg spherically symmetric Finsler metrics of dimension $n\geq 3$.
\end{theorem}

\begin{proof}
Let $F=u\phi$ be  a non-Riemannian Landsberg spherically symmetric Finsler metric  of dimension $n\geq 3$, then by Theorem \ref{Theorem_A} the functions  $P$ and $Q$ are given by
\begin{equation*}
P=c_1 s+\frac{c_2}{r^2}\sqrt{r^2-s^2}  , \quad Q=\frac{1}{2}c_0 s^2-\frac{c_2 s}{r^4}\sqrt{r^2-s^2}+c_3.
\end{equation*}
Now,    the functions $P$, $Q$ and $\phi$  must satisfy the compatipility conditons \eqref{Comp_C_C_2}.  Since $\phi$ is non-zero then dividing the conditions $C_1=0$ and $C_2=0$ in  \eqref{Comp_C_C_2} by $\phi$ and substituting by $P$ and $Q$,  straightforward calculations imply  the following two equations:
$$\frac{\phi_s}{\phi}(r^2+(c_1+2c_3)r^2s^2-2c_3r^4+2c_2s\sqrt{r^2-s^2})-(c_1+2c_3)sr^2 -2c_2 \sqrt{r^2-s^2}=0,$$
$$\frac{\phi_s}{\phi}( -c_0r^4s(r^2-s^2)-c_1sr^4-2c_2s^2\sqrt{r^2-s^2})+r^3\frac{\phi_r}{\phi}-c_0s^2r^4-c_1r^4+2c_2s\sqrt{r^2-s^2}=0.$$
Solving the above equations algebraically for $\frac{\phi_s}{\phi}$ and $\frac{\phi_r}{\phi}$, we obtain that
\begin{equation}
\label{Zhou_phi}
    \begin{split}
        \frac{\phi_s}{\phi}= & \frac{(c_1+2c_3)r^2s+2c_2\sqrt{r^2-s^2}}{r^2+(c_1+2c_3)r^2s^2-2c_3r^4+2c_2s\sqrt{r^2-s^2}},   \\
         \frac{\phi_r}{\phi}=&  \frac{(2c_0c_2r^4+4(c_1+c_3)c_2r^2-2c_2)s\sqrt{r^2-s^2}}{r(r^2+(c_1+2c_3)r^2s^2-2c_3r^4+2c_2s\sqrt{r^2-s^2})}   \\
         &+  \frac{c_0c_1r^6s^2+(c_0+4c_1c_3+2c_1^2)r^4s^2-2c_1c_3r^6+c_1r^4}{r(r^2+(c_1+2c_3)r^2s^2-2c_3r^4+2c_2s\sqrt{r^2-s^2})} .
    \end{split}
\end{equation}
It should be noted that the above two formulas are obtained   in \cite[Eq. (4.9)]{Zhou_Mo} by a long and  completely different way.

 Now, $\phi$ must satisfy    the  condition $\phi_{rs}=\phi_{sr}$ so we have to find what conditions on the functions  $c_0$, $c_1$, $c_2$ and $c_3$ so that     $\phi_{rs}=\phi_{sr}$. For this purpose we have
$$\left(\frac{\phi_s}{\phi}\right)_r-\left(\frac{\phi_r}{\phi}\right)_s=\frac{\phi( \phi_{sr}-\phi_{rs})}{\phi^2}.$$
 That is, $\phi_{rs}=\phi_{sr}$ if and only if $\left(\frac{\phi_s}{\phi}\right)_r=\left(\frac{\phi_r}{\phi}\right)_s$. Applying the condition $\left(\frac{\phi_s}{\phi}\right)_r=\left(\frac{\phi_r}{\phi}\right)_s$ on \eqref{Zhou_phi}, we get the following
  \begin{align*}
 & \frac{r^2}{\sqrt{r^2-s^2}(r^2+(c_1+2c_3)r^2s^2-2c_3r^4+2c_2s\sqrt{r^2-s^2})^2}
 \Big{(} s\sqrt{r^2-s^2}(  c_1'r^2+2c_3'r^2+8c_3^2r^3\\
 &-2c_0r^3-2c_1^2r^3+4c_0c_1c_3r^7+4c_0c_3r^5+4c_1^2c_3r^5+8c_1c_3^2r^5-2c_0c_1r^5+2c_1c_3'r^4\\
 &-2c_1'c_3r^4)+( 4c_0c_2c_3r^7+4c_1c_2c_3r^5-8c_2c_3^2s^2r^3+2c_0c_2s^2r^3+2c_1c_2s^2r-4c_2c_3's^2r^2\\
 &-4c_2c_3s^2r+4c_2'c_3s^2r^2 +2c_2'r^2-2c_2's^2-4c_0c_2c_3s^2r^5-4c_1c_2c_3s^2r^3+8c_2c_3^2r^5-2c_0c_2r^5\\
 &-2c_1c_2r^3+4c_2c_3'r^3-4c_2'c_3r^4+4c_2c_3r^3 )\Big{)}=0,
\end{align*}
where  $c_1'$ (resp. $c_2'$, $c_3'$ ) is the  derivative  of $c_1$ (resp. $c_2$, $c_3$) with respect to $r$.
Straightforward simplifications, we have
\begin{align}
  \label{Eq:Lands_com_1}
 s \sqrt{r^2-s^2} A(r)+ (r^2-s^2) B(r)=0,
\end{align}
where
 \begin{align*}
  A(r):=& r^2(4c_0c_1c_3r^5+ 2(2c_0c_3+2c_1^2c_3+4c_1c_3^2-c_0c_1)r^3+2(c_1c_3'-c_1'c_3)r^2\\
  &+2(4c_3^2 -c_0-c_1^2)r+2c_3'+c_1')\\
  B(r):=&4c_0c_2c_3r^5+2c_2(2c_1c_3+4c_3^2-c_0)r^3+4(c_2c_3'-c_2'c_3)r^2+2c_2(2c_3-c_1)r+2c_2'.
\end{align*}
 Since the   equation \eqref{Eq:Lands_com_1} is satisfied for all  $s$, then we must have
$$A(r)=0, \quad  B(r)=0.$$
Without loss of generality, we can assume that $c_1$ and $c_3$ are arbitrary functions in $r$ and solve $A(r)=0$ algebraically for $c_0$ we have
\begin{equation}
\label{Eq:c_0}
c_0=-\frac{4c_1c_3(2c_3+c_1)r^3-2(c_1'c_3-c_1c_3')r^2+2(4c_3^2-c_1^2)r+c_1'+2c_3'}{2r(c_1r^2+1)(2c_3r^2-1)}.
\end{equation}
By substituting  $c_0$ into $B(r)=0$ and solving it for $c_2$ we obtain that
\begin{equation}
\label{Eq:c_2}
c_2=c \sqrt{(c_1r^2+1)(2c_3r^2-1)},
\end{equation}
where $c$ is an arbitrary real constant. So if $c_1=-\frac{1}{r^2}$ or $c_3=\frac{1}{2r^2}$, then $c_2=0$ and this implies that the spray coefficients are quadratic and hence the space is Berwaldian. Moreover,  according to Proposition \ref{Regular_condition}, we have to exclude the case where  $\phi$ satisfies  the equation
  $$ \phi-s\phi_s+(r^2-s^2)\phi_{ss}=0.$$
 Moreover, dividing the above equation by $\phi$, we have
\begin{equation}
\label{Q_defined}
1- \frac{s\phi_s}{\phi}+(r^2-s^2)\frac{\phi_{ss}}{\phi}=0.
\end{equation}
 By using the property that $\left( \frac{\phi_s}{\phi} \right)_s=\frac{\phi_{ss}}{\phi}-\left( \frac{\phi_s}{\phi} \right)^2 $ and substituting from \eqref{Zhou_phi} into \eqref{Q_defined}, we get that
\begin{equation}
\label{Regularity_2}
-\frac{r^4(c_1r^2+1)(2c_3r^2-1)}{(r^2+(c_1+2c_3)r^2s^2-2c_3r^4+2c_2s\sqrt{r^2-s^2})^2}=0.
\end{equation}
That is, the choices $c_1=-\frac{1}{r^2}$ or $c_3=\frac{1}{2r^2}$ must be excluded since it implies the non-existence of the function $Q$. Also, the case where $c_1+2c_3=0$ implies that $c_2=c\sqrt{-(2c_3r^2-1)^2}$ which is a contradiction because $c_2$ is a real-valued  function of $r$.  That is, we have the conditions
  \begin{equation}
\label{Eq:c_1_c3}
c_1\neq -\frac{1}{r^2}, \quad c_3\neq \frac{1}{2r^2}, \quad c_1+2c_3\neq 0.
\end{equation}
Taking the equations \eqref{Eq:c_0}, \eqref{Eq:c_2} and \eqref{Eq:c_1_c3} into account, one can find explicit examples of  non-Berwaldian Landsberg metrics of dimension $n\geq 3$. This can be seen by the Examples \ref{Ex:1} and \ref{Ex:2}.
\end{proof}

\begin{example}\label{Ex:1}
To find an explicit example of a non-Berwaldian Landsberg spherically symmetric metric, consider the functions $c_1$ and $c_3$ are given by
$$c_1=\frac{1}{r^2}, \quad c_3=\frac{1}{r^2}.$$
Then the functions $c_0$  and $c_2$ are calculated by the help of \eqref{Eq:c_0} and \eqref{Eq:c_2} as follows
$$c_0=-\frac{ 3}{ r^4}, \quad c_2=\sqrt{2} c=\frac{1}{2},$$
where we choose the constant $c=\frac{1}{2\sqrt{2}}$.
By substituting the functions $c_0$, $c_1$, $c_2$ and $c_3$ into \eqref{Zhou_phi}, we have
\begin{equation*}
            \frac{\phi_s}{\phi}=  \frac{ 3s+\sqrt{r^2-s^2}}{-r^2+3s^2+ s\sqrt{r^2-s^2}},  \quad \,\,
         \frac{\phi_r}{\phi}= -\frac{ r}{-r^2+3s^2+ s\sqrt{r^2-s^2}} .
    \end{equation*}
Solving the above equations for $\phi$, we obtain that
\small{\begin{equation}\label{phi_1}
\begin{split}
&\phi(r,s)=(5s^2-r^2)^{\frac{1}{3}}(2s^2-r^2)^{\frac{1}{6}}\exp\Bigg{(} \frac{1}{3}\operatorname{arctanh} \Bigg{(} \frac{\sqrt{5}}{10 }\frac{(\sqrt{5}s+5r)}{\sqrt{r^2-s^2}}\Bigg{)}\\
&-\frac{1}{6}\operatorname{arctanh}\Bigg{(}  \frac{\sqrt{2}}{2  }\frac{(\sqrt{2}s+2r)}{\sqrt{r^2-s^2}}\Bigg{)}
+ \frac{1}{3}\operatorname{arctanh} \Bigg{(} \frac{\sqrt{5}}{2  }\frac{(\sqrt{5}s-5r)}{\sqrt{r^2-s^2}}\Bigg{)}-\frac{1}{6}\operatorname{arctanh}\Bigg{(}  \frac{\sqrt{2}}{  2}\frac{(\sqrt{2}s-2r)}{\sqrt{r^2-s^2}}\Bigg{)}\Bigg{)}
.
\end{split}
\end{equation}}
Therefore, the metric $F=u \phi(r,s)$ is a non-Berwladian Landsberg metric with the geodesic spray obtained by the the functions $P$ and $Q$ \eqref{P,Q} and  $\phi$ is given by \eqref{phi_1}.
\end{example}
Another example can be obtained as the following.
\begin{example}\label{Ex:2}
Let the functions $c_1$ and $c_3$ be given by
$$c_1=0, \quad c_3=\frac{1}{r^2}.$$
Then the functions $c_0$  and $c_2$ are calculated the help of \eqref{Eq:c_0} and \eqref{Eq:c_2} as follows
$$c_0=-\frac{ 2}{ r^4}, \quad c_2=  c=\frac{1}{2},$$
where we choose the constant $c=\frac{1}{2 }$.
By substituting the functions $c_0$, $c_1$, $c_2$ and $c_3$ into \eqref{Zhou_phi}, we have
\begin{equation}
\label{Exmp_2}
            \frac{\phi_s}{\phi}=  \frac{ 2s+\sqrt{r^2-s^2}}{-r^2+2s^2+ s\sqrt{r^2-s^2}},  \quad \,\,
         \frac{\phi_r}{\phi}= \frac{ -2s^2-s\sqrt{r^2-s^2}}{r(-r^2+2s^2+ s\sqrt{r^2-s^2})} .
    \end{equation}
Solving the first  equation  for $\phi$ by integrating with respect to $s$, we have
\small{$$\phi(r,s)=a(r) (r^4-5r^2s^2+5s^4)^{\frac{1}{4}} \Big{(} \frac{A}{B}\Big{)}^{\frac{5-\sqrt{5}}{20}}\Big{(} \frac{C}{D}\Big{)}^{\frac{5+\sqrt{5}}{20}}\exp\Bigg{(} -\frac{\sqrt{5}}{10}\operatorname{arctanh} \Bigg{(}  \frac{\sqrt{5}(r^2-2s^2)}{r^2}\Bigg{)}\Bigg{)},$$}
where
\begin{align*}
A:&=\left(2\sqrt{r^2-s^2}-  2r+\left(\sqrt{5}   -1\right)s\right)\sqrt{r^2-s^2}-\left(\sqrt{5} -1 \right)rs,\\
B:&= \left(2\sqrt{r^2-s^2}-2r-\left(\sqrt{5}   -1\right)s\right)\sqrt{r^2-s^2}+\left(\sqrt{5} -1  \right)rs,\\
C:&=\left(2\sqrt{r^2-s^2}-2r-\left(\sqrt{5}   +1\right)s\right)\sqrt{r^2-s^2}+\left(\sqrt{5} +1  \right)rs,\\
D:&=\left(2\sqrt{r^2-s^2}-2r+\left(\sqrt{5}   +1\right)s\right)\sqrt{r^2-s^2}-\left(\sqrt{5} +1  \right)rs.\\
\end{align*}
The function $a(r)$ that satisfies the second equation of \eqref{Exmp_2} is $a(r)=\frac{1}{r}$. That is, the function $\phi$ is given by
 \small{\begin{equation}\label{phi_2}
\phi(r,s)=\frac{1}{r} (r^4-5r^2s^2+5s^4)^{\frac{1}{4}} \Big{(} \frac{A}{B}\Big{)}^{\frac{5-\sqrt{5}}{20}}\Big{(} \frac{C}{D}\Big{)}^{\frac{5+\sqrt{5}}{20}}\exp\Bigg{(} -\frac{\sqrt{5}}{10}\operatorname{arctanh} \Bigg{(}  \frac{\sqrt{5}(r^2-2s^2)}{r^2}\Bigg{)}\Bigg{)}.
\end{equation}}
Therefore, the metric $F=u \phi(r,s)$ is a non-Berwladian Landsberg metric with the geodesic spray obtained by the   functions $P$ and $Q$  and  $\phi$ is given by \eqref{phi_2}.
\end{example}

We end this section by the following two theorems.
\begin{theorem}\label{Th_Riemannian}
All   Berwaldian spherically symmetric metrics of dimension $n\geq3$ are Riemannian.
\end{theorem}
\begin{proof}
Let $F$ be a Berwald spherically symmetric of  dimension $n\geq3$.  Since every Berwald metric is Landsbergian, then the geodesic spray  of $F$ is given by the functions $P$ and $Q$ \eqref{Zhou_P&Q};
$$P=c_1 s+\frac{c_2}{r^2}\sqrt{r^2-s^2}  , \quad Q=\frac{1}{2}c_0 s^2-\frac{c_2 s}{r^4}\sqrt{r^2-s^2}+c_3.$$
 Moreover, since $F$ is Berwaldian then the mean curvature $E_{ij}$ \eqref{Mean_curv.} vanishes. Substituting by
 $$P-sP_s=\frac{c_2}{\sqrt{r^2-s^2}}, \quad Q_s-sQ_{ss}=-\frac{c_2}{(r^2-s^2)^{3/2}},$$
  $$ P_{ss}=-\frac{c_2}{(r^2-s^2)^{3/2}}, \quad  Q_{sss}=\frac{3c_2}{(r^2-s^2)^{5/2}}$$
   into the equation $E_{ij}=0$ implies
$$\frac{n c_2}{u\sqrt{r^2-s^2}}\left(  \delta_{ij}-\frac{r^2}{u^2(r^2-s^2)} \delta_{ih} \delta_{jk}y^iy^j -\frac{1}{r^2-s^2} \delta_{ih} \delta_{jk}x^ix^j +\frac{s}{u(r^2-s^2)}(\delta_{ih} \delta_{jk}+\delta_{ik} \delta_{jh})x^hy^k\right)=0.$$
Contracting the above equation by $\delta^{ij}$ and using the facts that $\delta_{ij}y^iy^j=u^2$, $\delta_{ij}x^ix^j=r^2$ and $\delta_{ij}x^iy^j=\langle x,y\rangle$, we have
$$ n (n-2)c_2 \sqrt{r^2-s^2}=0. $$
From which together with the facts that $n\geq3$ and the equation holds for all $r$ and $s$, we must have $c_2=0$. That is, we have
$$P=c_1 s  , \quad Q=\frac{1}{2}c_0 s^2+c_3.$$
Now, the equation \eqref{Zhou_phi} becomes
\begin{equation}
\label{Zhou_phi_2}
    \begin{split}
        \frac{\phi_s}{\phi}= & \frac{(c_1+2c_3)s }{1+(c_1+2c_3)s^2-2c_3r^2 },   \\
         \frac{\phi_r}{\phi}=&   \frac{c_0c_1r^6s^2+(c_0+4c_1c_3+2c_1^2)r^4s^2-2c_1c_3r^6+c_1r^4}{r(r^2+(c_1+2c_3)r^2s^2-2c_3r^4)} ,
    \end{split}
\end{equation}
where $c_0$ is given by \eqref{Eq:c_0}. Integrating $ \frac{\phi_s}{\phi}$ with respect to $s$ implies
$$\phi=a(r) \sqrt{(c_1+2c_3) s^2-2c_3 r^2+1},$$
where $a(r)$ is to be chosen such that both formulae of  \eqref{Zhou_phi_2} are satisfied, that is, calculating $\frac{\phi_r}{\phi}$ and equaling it with the second formula of \eqref{Zhou_phi_2} we obtain the formula of $a(r)$ as follows
$$a(r)=\exp{\int \frac{r(2c_1c_3 r^2-c_3'r-c_1-2c_3)}{2c_3 r^2 -1} dr}.$$
 Consequently, the metric $F=u\phi= a(r) u  \sqrt{(c_1+2c_3) s^2-2c_3 r^2+1} $ is Riemannian.
\end{proof}

\begin{theorem}\label{Reg_Lands_Riem_D_3}
All regular Landsberg spherically symmetric metrics of dimension $n\geq 3$ are Riemannian.
\end{theorem}
\begin{proof}
Let $F=u\phi$ be a regular Landsberg spherically symmetric metric of dimension $n\geq 3$. Now, by the regularity condition \eqref{Regular_condition} together with the fact that $\phi$ is positive, then  we can write
$$1-s\frac{\phi_s}{\phi}+(r^2-s^2)\frac{\phi_{ss}}{\phi}>0.$$
Moreover, by \eqref{Regularity_2} we get
$$1-s\frac{\phi_s}{\phi}+(r^2-s^2)\frac{\phi_{ss}}{\phi}=-\frac{r^4(c_1r^2+1)(2c_3r^2-1)}{(r^2+(c_1+2c_3)r^2s^2-2c_3r^4+2c_2s\sqrt{r^2-s^2})^2}.$$
By \eqref{Eq:c_2}, we must have  $(c_1r^2+1)(2c_3r^2-1)>0$ and hence we have
$$\phi-s\phi_s+(r^2-s^2)\phi_{ss}<0$$
which is a contradiction and hence the metric is Berwaldian. Thus the result follows by Theorem \ref{Th_Riemannian}.
\end{proof}

  \section{The two-dimensional case}

  This section is devoted to the two-dimensional spherically symmetric Finsler metrics. We characterize all spherically symmetric surfaces of Berwald  or Landsberg  types and then we determine  all      Berwald surfaces. We end this section by a note on a class of surfaces mentioned in \cite{Zhou}.

  We start by the following proposition.

 \begin{proposition}\label{2D_E_E_ij=0}
 For a spherically symmetric surface the mean Berwald curvature $E_{ij}$ and the scalar trace $E$ of the mean Berwald curvature  vanish if and only if $sH-(r^2-s^2)H_s=0$.
 \end{proposition}
 \begin{proof}
 Assume that $E_{ij}$ vanishes then  contracting \eqref{E_H} by $\delta^{ij}$ together with the fact that $n=2$ imply
 $$ sH-(r^2-s^2)H_s=0.$$
 Also, let the scalar $E$ vanish then by    Proposition \ref{Scalar_E}, we have
$$(\rho_0+\rho_3(r^2-s^2))(sH-(r^2-s^2)H_s)=0.$$
By substituting by  the formulae of $\rho_0$ and $\rho_3$ given in \eqref{Eq:g^ij}, we obtain that
$$\frac{sH-(r^2-s^2)H_s}{\phi(\phi-s\phi_s+(r^2-s^2) \phi_{ss})}=0.$$
Therefore,      $sH-(r^2-s^2)H_s=0$.

Conversely, assume that $ sH-(r^2-s^2)H_s=0$, then by Proposition \ref{Scalar_E} it is clear that $E=0$. Now, taking the fact that $r^2-s^2=\frac{(x_1y_2-x_2y_1)^2}{u^2}$ into account, we can conclude that the components $E_{11}$, $E_{12}$ and $E_{22}$ vanish. For example, we have
$$E_{11}=\frac{y_2^2}{su^3}(s u^2 H-(x_1y_2-x_2y_1)^2 H_s)=\frac{y_2^2}{su }(s  H-(r^2-s^2) H_s)=0.$$
Similarly, we can calculate $E_{12}$ and $E_{22}$. Consequently, the proof is completed.
 \end{proof}

 \begin{proposition}\label{2D_L_ijk=0}
A spherically symmetric surface is Landsbergian  if and only if $ (r^2-s^2)L_1+3L_2=0$.
 \end{proposition}
 \begin{proof}
 Assume that the surface is Landsbergian, then     $L_{ijk}=0$. Contracting \eqref{L_ijk} by $x^j\delta^{k\ell}$, we have
  $$(r^2-s^2) L_1+3L_2=0.$$

Conversely, assume that $(r^2-s^2) L_1+3L_2=0$,  one can see that the components $L_{ijk}=0$. For example, we have
$$L_{111}=\frac{\phi}{2u^3}(-x_1^3u^3+3 x_1^2 y_1 s u^2-3x_1 y_1^2 s^2 u+s^3y_1^3)L_1+3(y_1 su^2-x_1 u^3-sy_1^3+x_1y_1^2 u)L_2.$$
Using the formulae of $u$, $r$ and $s$ when $n=2$, one can see that
$$L_{111}=\frac{y_2^3\phi}{2u^4}(x_1y_2 -x_2 y_1)((r^2-s^2)L_1+3 L_2)=0.$$
Similarly, we  can calculate $L_{112}$, $L_{122}$ and $L_{222}$.  As required.
 \end{proof}

  Rewriting  the Landsberg condition for spherically symmetric surfaces and  straightforward calculations imply that
 \begin{equation}
\label{Main_Lands_surface_cond}
\begin{split}
 (r^2-s^2)L_1+3L_2 =& \lambda_1  \phi_s +\lambda_2 \phi, \\
 \end{split}
\end{equation} where
$$\lambda_1:= \frac{1}{s}(sH-(r^2-s^2)H_s), \quad \lambda_2:=(r^2-s^2)K_s-3sK,\quad K:=P_{ss}-Q_s+sQ_{ss}.$$
  Now, we announce our   first result in this section.
  \begin{theorem}
  \label{First_surface_result}
A spherically symmetric Finsler surface is Berwaldian if and only if
$$P=b_1 s+ \frac{b_2}{ \sqrt{r^2-s^2}} +\frac{b_3 (r^2-2s^2)}{\sqrt{r^2-s^2}} ,$$
$$ Q= b_0 s^2+\frac{1}{2} b_1+ \frac{b_2 s (r^2-2 s^2)}{r^4 \sqrt{r^2-s^2}} -\frac{b_3 s (3r^2-2s^2)}{r^2\sqrt{r^2-s^2}}-\frac{a  }{r^2} s \sqrt{r^2-s^2},$$
where $a$, $b_0$, $b_1$, $b_2$, $b_3$ are arbitrary functions of $r$ and to be chosen such  that the  compatibility conditions are satisfied.
\end{theorem}

\begin{proof}
Let $F$ be a Berwald spherically symmetric Finsler surface, then the mean   curvature $E_{ij}$ and the scalar trace $E$   vanish. Then, by Proposition \ref{2D_E_E_ij=0}, we have  $sH-(r^2-s^2)H_s=0$ which has the solution
\begin{equation}
\label{Eq:Berwald_H_1}
H=\frac{a_0(r)}{\sqrt{r^2-s^2}}.
\end{equation}
 Moreover, since every Berwald metric is Landsbergian, then $F$ is Landsbergian. Now,   by Proposition   \ref{2D_L_ijk=0}, we have
$(r^2-s^2)L_1+3L_2=0$ and since $sH-(r^2-s^2)H_s=0$, then   \eqref{Main_Lands_surface_cond} implies
   $$\lambda_2=(r^2-s^2)K_s-3sK=0.$$
  That is, we have
  $$\frac{K_s}{K}=\frac{P_{sss}+sQ_{sss}}{P_{ss}-Q_s+sQ_{ss}}=\frac{3s}{r^2-s^2}.$$
  Then,  we have the solution
  \begin{equation}
\label{Eq:Berwald_H_2}
K=P_{ss}-Q_s+sQ_{ss}=\frac{b_0(r)}{(r^2-s^2)^{3/2}}.
\end{equation}
One can see that $P_{ss}-Q_s+sQ_{ss}=P_{ss}+(sQ_s)_s-2Q_{s}$, then  integrating the above equation with respect to $s$, we have
 \begin{equation}
\label{Eq:Berwald_H_3}
P_s+sQ_s-2Q=\frac{b_0(r) s}{r^2\sqrt{r^2-s^2}}+b(r).
\end{equation}
Now, by using the definition of $H$ together with  the equations \eqref{Eq:Berwald_H_1} and \eqref{Eq:Berwald_H_2}, we can write
$$(r^2-s^2)P_{ss}-3s P_s+3P=\frac{a_1(r)}{\sqrt{r^2-s^2}}.$$
The   solution of the above equation can be written in the form
\begin{equation}
\label{P_berwald_surface}
P=b_1 s+ \frac{b_2}{ \sqrt{r^2-s^2}} +\frac{b_3 (r^2-2s^2)}{\sqrt{r^2-s^2}} ,
\end{equation}
where $b_1$, $b_2$, $b_3$ are arbitrary functions of $r$. To find the function $Q$, we rewrite   \eqref{Eq:Berwald_H_3} as follows
$$s^3\left( \frac{Q}{s^2}\right)_s= -P_s+\frac{b_0(r) s}{r^2\sqrt{r^2-s^2}}.$$
Hence, by integration  with respect to $s$ we can write the function $Q$ as follows
\begin{equation}
\label{Q_berwald_surface}
Q=b_0 s^2+\frac{1}{2} b_1+ \frac{b_2 s (r^2-2 s^2)}{r^4 \sqrt{r^2-s^2}} -\frac{b_3 s (3r^2-2s^2)}{r^2\sqrt{r^2-s^2}}-\frac{a  }{r^2} s \sqrt{r^2-s^2},
\end{equation}
where $b_0$, $a$ are arbitrary functions of $r$.

 Conversely, if $P$ and $Q$ are given by the formulae \eqref{P_berwald_surface} and \eqref{Q_berwald_surface}, then straightforward calculations imply that the Berwald curvature vanishes. Or instead, one can see that  $(r^2-s^2)L_1+3L_2=0$ and $sH-(r^2-s^2)H_s=0$ which    means  that the surface is Landsberian and weekly Berwald ($E_{ij}=0$). But it is known that a Landsberian weekly Berwald surface is Berwaldian and hence the proof is completed.
\end{proof}

     It should be noted that, in \cite{Zhou},   L. Zhou   provided a class of Landsberg spherically symmetric surfaces and he claimed that this class is  not Berwaldian. In fact, this class is Berwaldian as it is shown in  \cite{Elgendi-Youssef}. Also, this can be seen easily since the corresponding  functions $P$ and $Q$   are a special case of  Theorem \ref{First_surface_result}. Moreover, we are going to discuss  another point of Zhou's class related to the compatibility conditions.

 In what follows, to make the comparison easier and  to avoid confusions, we use the terminology and notations of \cite{Zhou}. In the class of metrics  mentioned in  \cite[Theorem 3.3]{Zhou} the  functions $\phi$,  $P$ and $Q$ are not compatible, i.e., the compatibility conditions are not satisfied. We will provide an example showing this non-compatibility soon. The functions $P$ and $Q$ of the class provided in   \cite[Theorem 3.3]{Zhou} are given by
  \begin{equation}
  \label{P_Q_2D}
  P=-\frac{s}{r^2}+\frac{c}{r^2}\sqrt{r^2-s^2}, \quad Q=c_0-\frac{4r^4c_0^2+2r^3c_0'+c^2}{2r^4(2r^2c_0-1)}s^2-\frac{s}{r^4}\sqrt{r^2-s^2},
  \end{equation}
  where $c_0$  is function of  $r$ and   $c\neq 3$ is a constant, $c_0'$ denotes the derivative of $c_0$ with respect to $r$. By the formula of $Q$, it is clear that the case where $c_0=\frac{1}{2r^2}$ is excluded.

  Plugging the functions $P$ and $Q$ given by \eqref{P_Q_2D} into the compatibility conditions \eqref{Comp_C_C_2},  then  straightforward calculations by hand or by using Maple program,  we have the following two equations:
$$\frac{\phi_s}{\phi}(r^2-s^2)( (2c_0r^2-1)\sqrt{r^2-s^2} -(c+1)s)+  (2c_0r^2-1)s\sqrt{r^2-s^2}+2cr^2-(c+1)s^2=0,$$
\begin{equation*}
    \begin{split}
         &\frac{\phi_r}{\phi} (2c_0r^2-1)r^2\sqrt{r^2-s^2}+\frac{\phi_s}{\phi} \Big{(} \sqrt{r^2-s^2}( (r^2-s^2)(4c_0^2sr^4+sc^2)+(2c_0r^2-1)sr^2) \\
         &+  3s^2r^2+2c_0cs^2r^4+2c_0r^6+cr^4-2c_0cr^6-cr^2s^2-6c_0s^2r^4+4c_0s^4r^2-r^4-2s^4\Big{)} \\
         &+\sqrt{r^2-s^2}(4c_0^2s^2r^4+2c_0r^4+c^2s^2-r^2)+2c_0csr^4-sr^2+2c_0sr^4-csr^2-4c_0s^3r^2+2s^3=0.
    \end{split}
\end{equation*}

By solving the above equations  for $\frac{\phi_s}{\phi}$ and $\frac{\phi_r}{\phi}$, we have

 \small{ \begin{equation}
\label{Zhou)_2D_phi}
    \begin{split}
        \frac{\phi_s}{\phi}= &- \frac{(c+1)s^2-(2r^2c_0-1)s\sqrt{r^2-s^2}-2r^2c}{(r^2-s^2)((2r^2c_0-1)\sqrt{r^2-s^2}-(c+1)s)},   \\
         \frac{\phi_r}{\phi}=& - \frac{\sqrt{r^2-s^2}(2c(2c_0r^2-1)(1-c)(r^2-s^2)-r^2+2s^2+4c_0r^4+8c_0^2r^4s^2-4c_0^2r^6-8c_0r^2s^2)}{r(r^2-s^2)(2r^2c_0-1)((c+1)s-(2r^2c_0-1)\sqrt{r^2-s^2})}   \\
         &+ \frac{5cr^2s-2c^3r^2s-4cc_0'r^5s+12cc_0r^2s^3-14cc_0r^4s+4cc_0'r^3s^3+r^2s-4cs^3-2c_0r^4s}{r(r^2-s^2)(2r^2c_0-1)((c+1)s-(2r^2c_0-1)\sqrt{r^2-s^2})}\\
        &-\frac{4c_0r^2s^3+2c^3s^3-2s^3}{r(r^2-s^2)(2r^2c_0-1)((c+1)s-(2r^2c_0-1)\sqrt{r^2-s^2})} .
    \end{split}
\end{equation}}

     Now,  consider  the following example.

\begin{example}  This example is constructed from \cite[Theorem 3.3]{Zhou} and shows a non-compatibility between the functions $P$, $Q$ and $\phi$. Let $c=-1$ and $c_0=\frac{1}{r^2}$, then, by  substitution in the  formula of $ a(r)$, we have
  $$a(r)=\exp\left( \int -\frac{2c_0r^2-1+2c^2-2c}{r(2c_0r^2-1)}dr\right)=\exp\left( \int -\frac{5}{r}dr\right)=\frac{1}{r^5}.$$
  Now, we calculate   $F$  as follows
\begin{align*}
  F & =u\exp \left(\int  \frac{(c+1) s^{2}-\left(2 r^{2} c_{0}-1\right) s \sqrt{r^{2}-s^{2}}-2 r^{2} c}{\left(r^{2}-s^{2}\right)\left(\left(2 c_{0} r^{2}-1\right) \sqrt{r^{2}-s^{2}}-(c+1) s\right)} d s\right) a(r) \\
    & = \frac{u}{r^5}\exp \left(\int \frac{ -  s \sqrt{r^{2}-s^{2}}+2 r^{2}  }{\left(r^{2}-s^{2}\right)\left(\sqrt{r^{2}-s^{2}}\right)} d s\right)\\
    &=\frac{u}{r^5}  \sqrt{r^{2}-s^{2}}\exp \left(  \frac{2 s  }{ \sqrt{r^{2}-s^{2}} } \right).
\end{align*}
Using \eqref{P,Q} and the above formula of  $\phi(r,s)$,  we obtain that
\begin{equation}
\label{P_Q_from_phi}
P= -\frac{s}{r^2}-\frac{3}{4 r^2}\,\sqrt {{r}^{2}-{s}^{2}}, \quad Q=\frac{7}{8 r^2}-\frac{3 s^2}{r^4}-\frac{3s}{4 r^4}\,\sqrt {{r}^{2}-{s}^{2}}
\end{equation}
 On the other hand,  by \eqref{P_Q_2D}, the choice $c=-1$ and $c_0=\frac{1}{r^2}$ leads to
  \begin{equation}
  \label{P_Q_from_c's}
P= -\frac{s}{r^2}-\frac{1}{r^2}\,\sqrt {{r}^{2}-{s}^{2}}, \quad Q=\frac{1}{r^2}-\frac{s^2}{2r^4}-\frac{s}{r^4}\,\sqrt {{r}^{2}-{s}^{2}}
.
\end{equation}
  It is clear that the functions $P$ and $Q$ in the equations \eqref{P_Q_from_phi} and \eqref{P_Q_from_c's} are different. This is because the functions $P$,  $Q$ and $\phi$ do not satisfy the compatibility conditions, one can see that the compatibility conditions are $C_1=0$ and $C_2=\frac{1}{r^2} \phi$.
\end{example}

The problem of the non-compatibility in the previous example comes from the factor $a(r)$. In \cite{Zhou}, it is mentioned that
the function $a(r)$ is given by  $\phi(r,0)=a(r)$ and it is not clear   how   this formula is obtained, in the above example, one can see that $\phi(r,0)=\frac{1}{r^4}\neq a(r)$.
\medskip

In the previous example, by substitution by $c=-1$ and $c_0=\frac{1}{r^2}$ into  \eqref{Zhou)_2D_phi}, we have
$$\frac{\phi_s}{\phi}=-\frac{r^2s-s^3-2r^2\sqrt{r^2-s^2} }{(r^2-s^2)^2}, \quad \frac{\phi_r}{\phi}=-\frac{5r^4-11r^2s^2+6s^4-2r^2s\sqrt{r^2-s^2} }{r(r^2-s^2)^2}.$$
  Solving the above equations, we get that
  $$\phi=\frac{1}{r^6}\sqrt{r^{2}-s^{2}}\exp \left(  \frac{2 s  }{ \sqrt{r^{2}-s^{2}} } \right).$$
  That is, $a(r)=\frac{1}{r^6}$. Moreover,  one can see that   the above formula of $\phi$ together with \eqref{P,Q} yield the same formulae \eqref{P_Q_from_c's} of $P$ and $Q$.

redpajama_set_name: RedPajamaArXiv


text:
Sheridan Smith reunited with Gavin & Stacey co-star in first look image from their big new BBC1 drama
David Hollingsworth Mon, October 29 11:17am
Sheridan Smith plays a supermarket worker whose life is turned upside down when her mum suffers a stroke
The BBC has released a first look image of Sheridan Smith with her old Gavin & Stacey co-star Alison Steadman in their major new drama, Care.
Sheridan Smith stars as Jenny, a mother of two-raising her girls after the departure of her husband Dave.
Alison Steadman plays Mary, Jenny's widowed mother, who is a rock in Jenny's life, looking after the children so Jenny can work in her full-time job at the supermarket.
First look of Alison Steadman and Sheridan Smith in BBC1's Care
But their worlds come crashing down when Mary suffers a devastating stroke and develops dementia.
Jenny is left torn between her own life and caring for her mother. Could another way be possible? Little Boy Blue actress Sinead Keenan also features in Care as Jenny's sister Claire.
The 90-minute drama has been written by Jimmy McGovern, whose long CV includes Broken, The Accused and The Street, and Gillian Juckes, whose real life experiences formed the inspiration for the story.
Jimmy previously said: "A great cast, a wonderful director and a superb co-writer! I'm really looking forward to making this."
Gillian previously said: "It has been a great experience working with Jimmy and a wonderful opportunity to write about a subject very close to my heart."
Alison Steadman previously worked with Sheridan Smith on Gavin & Stacey
Cilla actress Sheridan of course famously worked with Alison on Gavin & Stacey. Sheridan played Smithy's sister Rudi, while Alison was Gavin's mum Pam in the hugely successful comedy series, which ended in 2010.
And Sheridan's a busy woman! She is playing a cleaner in upcoming ITV drama Clean Break and next year she'll star in C4's Adult Material as an ageing porn star. Sheridan has described Adult Material as a "taboo-busting" drama.
Care will be shown on BBC1 later this year.
Sheridan Smith image: PA; Alison picture: PA Wire/PA Images.
THIS familiar face returns to Neighbours with unfinished business today

redpajama_set_name: RedPajamaCommonCrawl


text:
use bazel_protos;
use bytes;
use digest::FixedOutput;
use hashing;
use protobuf::Message;
use sha2::{self, Digest};

pub struct TestData {
  string: String,
}

impl TestData {
  pub fn empty() -> TestData {
    TestData::new("")
  }

  pub fn roland() -> TestData {
    TestData::new("European Burmese")
  }

  pub fn catnip() -> TestData {
    TestData::new("catnip")
  }

  pub fn fourty_chars() -> TestData {
    TestData::new(
      "0123456789012345678901234567890123456789\
       0123456789012345678901234567890123456789",
    )
  }

  pub fn new(s: &str) -> TestData {
    TestData {
      string: s.to_owned(),
    }
  }

  pub fn bytes(&self) -> bytes::Bytes {
    bytes::Bytes::from(self.string.as_str())
  }

  pub fn fingerprint(&self) -> hashing::Fingerprint {
    hash(&self.bytes())
  }

  pub fn digest(&self) -> hashing::Digest {
    hashing::Digest(self.fingerprint(), self.string.len())
  }

  pub fn string(&self) -> String {
    self.string.clone()
  }

  pub fn len(&self) -> usize {
    self.string.len()
  }
}

pub struct TestDirectory {
  directory: bazel_protos::remote_execution::Directory,
}

impl TestDirectory {
  pub fn empty() -> TestDirectory {
    TestDirectory {
      directory: bazel_protos::remote_execution::Directory::new(),
    }
  }

  // Directory structure:
  //
  // /roland
  pub fn containing_roland() -> TestDirectory {
    let mut directory = bazel_protos::remote_execution::Directory::new();
    directory.mut_files().push({
      let mut file = bazel_protos::remote_execution::FileNode::new();
      file.set_name("roland".to_owned());
      file.set_digest((&TestData::roland().digest()).into());
      file.set_is_executable(false);
      file
    });
    TestDirectory { directory }
  }

  // Directory structure:
  //
  // /cats/roland
  pub fn nested() -> TestDirectory {
    let mut directory = bazel_protos::remote_execution::Directory::new();
    directory.mut_directories().push({
      let mut subdir = bazel_protos::remote_execution::DirectoryNode::new();
      subdir.set_name("cats".to_string());
      subdir.set_digest((&TestDirectory::containing_roland().digest()).into());
      subdir
    });
    TestDirectory { directory }
  }

  // Directory structure:
  //
  // /dnalor
  pub fn containing_dnalor() -> TestDirectory {
    let mut directory = bazel_protos::remote_execution::Directory::new();
    directory.mut_files().push({
      let mut file = bazel_protos::remote_execution::FileNode::new();
      file.set_name("dnalor".to_owned());
      file.set_digest((&TestData::roland().digest()).into());
      file.set_is_executable(false);
      file
    });
    TestDirectory { directory }
  }

  // Directory structure:
  //
  // /cats/roland
  // /treats
  pub fn recursive() -> TestDirectory {
    let mut directory = bazel_protos::remote_execution::Directory::new();
    directory.mut_directories().push({
      let mut subdir = bazel_protos::remote_execution::DirectoryNode::new();
      subdir.set_name("cats".to_string());
      subdir.set_digest((&TestDirectory::containing_roland().digest()).into());
      subdir
    });
    directory.mut_files().push({
      let mut file = bazel_protos::remote_execution::FileNode::new();
      file.set_name("treats".to_string());
      file.set_digest((&TestData::catnip().digest()).into());
      file.set_is_executable(false);
      file
    });
    TestDirectory { directory }
  }

  // Directory structure:
  //
  // /feed (executable)
  // /food
  pub fn with_mixed_executable_files() -> TestDirectory {
    let mut directory = bazel_protos::remote_execution::Directory::new();
    directory.mut_files().push({
      let mut file = bazel_protos::remote_execution::FileNode::new();
      file.set_name("feed".to_string());
      file.set_digest((&TestData::catnip().digest()).into());
      file.set_is_executable(true);
      file
    });
    directory.mut_files().push({
      let mut file = bazel_protos::remote_execution::FileNode::new();
      file.set_name("food".to_string());
      file.set_digest((&TestData::catnip().digest()).into());
      file.set_is_executable(false);
      file
    });
    TestDirectory { directory }
  }

  pub fn directory(&self) -> bazel_protos::remote_execution::Directory {
    self.directory.clone()
  }

  pub fn bytes(&self) -> bytes::Bytes {
    bytes::Bytes::from(
      self
        .directory
        .write_to_bytes()
        .expect("Error serializing proto"),
    )
  }

  pub fn fingerprint(&self) -> hashing::Fingerprint {
    hash(&self.bytes())
  }

  pub fn digest(&self) -> hashing::Digest {
    hashing::Digest(self.fingerprint(), self.bytes().len())
  }
}

fn hash(bytes: &bytes::Bytes) -> hashing::Fingerprint {
  let mut hasher = sha2::Sha256::default();
  hasher.input(bytes);
  hashing::Fingerprint::from_bytes_unsafe(hasher.fixed_result().as_slice())
}

redpajama_set_name: RedPajamaGithub


Fairchild's top quality mission is to deliver superior semiconductor items and client worth via innovation, service and manufacturing excellence. If you require to do run flat, the latest low profile tires, or extremely large wheels you will need side help technology which is a lot much more pricey than a simple tire changer. I recommend acquiring a spray can full of it from an automotive hardware store, but just paint thinner and a shop cloth will do. Spray or dab some paint thinner on to the cloth and wipe down the complete tank. On an automotive relay, the control circuit is the one you or the pc activates. Mechanics can turn out to be Automotive Service Excellence certified, which implies they require to pass 1 of the 40 exams provided and have around two years of function experience. I am sorry, the best person to ask that query would be a salvage yard, they have books to cross reference parts. Bear in mind, this is only for dust and bugs that might have gotten stuck to the paint although drying.
The Smoking Tire, hosted by Matt Farah, is the premier location for automotive video reviews and adventures. Their knowledge of personal computer technology will be like that of automotive racers, fighter jet pilots or astronauts. By May 2004, the RM500 million manufacturing facility in Gurun- known as Naza Automotive Manufacturing (NAM) – commenced operations by rolling out a Naza Ria MPV. Either sort you choose will give your paint a thin protection layer and a small added shine. The S75 Pro's Xtend sensor technology enables this device to supply trusted results each time. So you want to know how to airbrush graphics or artwork on your car! You have identified the right spot. I am going to take you via the actions to prepare a car paint surface for artwork.
Repairing electrical circuits: Even if you don't know a lot about electricity, your automobile repair manual will assist you do easy voltage drop tests, troubleshoot relays, find and verify fuses, tell you how to check lamps, connections and other easy components that usually cause trouble.
Find out how to produce your personal distinctive stamps, molds, carving, texture mats, tear-away textures, etched copper or brass sheets, 3D dimensional paint sheets, appliqué with silver clay paper, and several other texturing methods, supplies, tools and gear.
Absolutely nothing a little paint and shiny white vinyl with cool alligator print on it could not fix. As you can see below, there is a progression of the certification procedure to distinguish what level of skill your auto repair technician is at. Just as in the previous, this year's AUTOMOTIVE HUNGARY once again featured the conventional competitors of engineering students, Techtogether Automotive Hungary. Product and operation tests and inspections at different stages of the value chain are made to steer clear of these item recalls by making certain finish-user security and security and compliance with the automotive market requirements. If the auto parts shop tells you that they never make your part any longer, attempt the hardware shop.


Article 95 is repealed by Law No. 1 of 1963.
The President of the Senate shall be appointed by the King. The Senate shall elect two Vice-Presidents. The result of the election shall be submitted to the King for approval. The appointment of the President and the election of the Vice-Presidentsshall be for a period of two years and the President may be re-appointed and the two Vice-Presidents may be elected.
Not be a member of the Royal Family. In addition to the condition prescribed by the Electoral Law.
The meetings of either of the two Chambers shall not be valid unless the majority of the members are present at the opening of the meeting. Neither of the two Chambers may take a decision unless the majority of its members are present at thetime of the decision.
Except in cases where a special majority is required, decisions in each of the Chambers shall be adopted by a majority of the members present. If the vote is equally divided, the proposal in question shall be considered to have been rejected.
A bill which has been rejected by either Chamber may not be reintroduced at the same session.
If a seat becomes vacant in either of the Chambers, it shall be filled within three months by election or appointment in conformity with the provisions of this Constitution. The period of three months shall commence on the date on which theChamber informs the Government of the vacancy. The term of office of a new Senator shall be limited to the term of office of his predecessor. The terms of office of a new member of House of Representatives shall expire upon the termination of the period of office of the Chamber.
No one may present a request to Parliament except in writing. Each Chamber may transmit the petitions addressed to it to the Ministers. The inisters shall be bound to give the Chamber necessary explanations regarding such petitions whenever theChamber so requires.
The meetings of Congress shall be valid only when the absolute majority of the members of each of the Chambers composing the Congress are present.


text:
package org.kavaproject.kavatouch.corefoundation;

import org.kavaproject.kavatouch.internal.CFunction;
import org.kavaproject.kavatouch.internal.Header;
import org.kavaproject.kavatouch.internal.OpaqueType;
import org.kavaproject.kavatouch.util.NotImplementedException;
import org.kavaproject.kavatouch.util.OutArg;

import java.io.OutputStream;
import java.nio.ByteBuffer;

@Header("CFString")
@OpaqueType("CFStringRef")
public class CoreString implements CorePropertyList, CoreType {
    private final String mAdaptee;

    private CoreString(String adaptee) {
        this.mAdaptee = adaptee;
    }

    public static CoreString valueOf(String adaptee) {
        return new CoreString(adaptee);
    }

    @CFunction("CFStringGetBytes")
    public static int getBytes(String string, CoreRange range, CoreStringBuiltInEncodings encoding, int lossByte,
                               boolean isExternalRepresentation, byte[] buffer, OutArg<Integer> usedBufLen) {
        string.getBytes(range.location, range.location + range.length, buffer, 0); //TODO Charset
        return Math.min(range.length, string.length());
    }

    @CFunction("CFStringCreateWithBytes")
    public static CoreString create(byte[] bytes, int numBytes, CoreStringBuiltInEncodings encoding,
                                    boolean isExternalRepresentation) {
        return new CoreString(new String(bytes, 0, numBytes));
    }

    @Override
    public String toString() {
        return mAdaptee;
    }

    @Override
    public CorePropertyList deepCopy(CorePropertyListMutabilityOptions mutabilityOption) {
        throw new NotImplementedException(); //TODO
    }

    @Override
    public ByteBuffer createData(CorePropertyListFormat format, int options) throws RuntimeException {
        throw new NotImplementedException(); //TODO
    }

    @Override
    public long write(OutputStream stream, CorePropertyListFormat format, int options) throws RuntimeException {
        throw new NotImplementedException(); //TODO
    }

    @Override
    public ByteBuffer createXMLData(CorePropertyList propertyList) {
        throw new NotImplementedException(); //TODO
    }

    @Override
    public long writeToStream(OutputStream stream, CorePropertyListFormat format, OutArg<String> errorString) {
        throw new NotImplementedException(); //TODO
    }

    @Override
    public boolean isValid(CorePropertyListFormat format) {
        throw new NotImplementedException(); //TODO
    }
}


text:
Q: My jQuery doesn't change the fontcolor with the .css() function

This is my code, placed at the end of my HTML document, before the end of . I'm trying to make my h3 tag change its color to a light blue when I hover over it. This isn't working in my HTML:

sunil singh

Can anybody tell me what's wrong? I would really appreciate it.

A: There is a syntax error in your code, missing :. You are confusing the method('prop', 'value') with method({ prop: 'value' }) syntax. Also, you're using an O ("Oh") instead of a 0 (zero). The "Oh" is not a hex value.
  $('h3').hover(function() {
      // `this` here refers to the hovered element
      $(this).css({"color": "#00FFFF"});
  });

A: There is a syntax error in the code. You have a comma instead of a colon in the object literal.
There is an error in your color code. You have two O characters instead of two zero digits.
This:
$('h3').css({"color","#OOFFFF"});

should be:
$('h3').css({"color":"#00FFFF"});

Side note: The color #00FFFF is not light blue, it's cyan.
Another side note: You can use CSS for hover effects, then the effect also goes away when you no longer hover over the element. Example (goes in the head element):

A: Try this:
$(document).ready(function () {
      $('h3').hover(function() {
          $(this).css("color", "red");
      });
  });

Added a comma or , instead of a colon or : in css("color", "red")
I made it red as an example, change it for your needs :)
Note: It is better to use a class in your function unless you want all h3 to change color on hovering

redpajama_set_name: RedPajamaStackExchange


text:
Q: EditText all possible number in random output in Android I'm new to Android. I'm developing an app which gives all the possible combination number based on the user input and separated it by comma.

For example the user input is: 1234
The output should be list all the possible number combination from "1234" like
"1234,1243,1324,1342,1423,1432,2134,etc.."

And there's also another EditText that limit the output by setting it to 1 or 2 or 3 or 4(max).
For example the user input is: 2
The output should be list all the possible number combination from "1234" by "2" digits
"12,13,14,21,23,24,31,32,34,41,42,43"

This is my MainActivity.java
package com.example.androidtest1;

import androidx.appcompat.app.AppCompatActivity;

import android.os.Bundle;
import android.view.View;
import android.widget.Button;
import android.widget.CheckBox;
import android.widget.EditText;

import java.util.Random;

public class MainActivity extends AppCompatActivity {
    EditText et_input1, et_output1;
    Button b_generate, b_clear;
    CheckBox checkBox1;
    Random r;
    int min,output;

    @Override
    protected void onCreate(Bundle savedInstanceState) {
        super.onCreate(savedInstanceState);
        setContentView(R.layout.activity_main);

        EditText mEdit = (EditText) findViewById(R.id.et_output1);
        mEdit.setEnabled(false);

        r = new Random();

        et_input1 = (EditText) findViewById(R.id.et_input1);
        et_output1 = (EditText) findViewById(R.id.et_output1);
        b_generate = (Button) findViewById(R.id.b_generate);
        b_clear = (Button) findViewById(R.id.b_clear);
        checkBox1 = (CheckBox) findViewById(R.id.checkBox1);

        b_generate.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                if (et_input1.length()==0){
                  et_input1.setError("Insert numbers");

            }else if(checkBox1.isChecked()){
                    min = Integer.parseInt(et_input1.getText().toString());
                    output = (min);
                    et_output1.setText("" + output + "*");
                    et_input1.setError(null);
                }
                else{
                    et_input1.setError("Check the checkbox");
                }

            }
        });

        b_clear.setOnClickListener(new View.OnClickListener() {
            @Override
            public void onClick(View v) {
                et_output1.getText().clear();
                et_input1.getText().clear();
                return;
            }
        });
           }
    }


text:
Collapse

Accordions

Launching

                            Anim pariatur cliche reprehenderit, enim eiusmod high life accusamus terry richardson ad squid. 3 wolf moon officia aute, non cupidatat skateboard dolor brunch. Food truck quinoa nesciunt.

Deskop Apliaction

                            Anim pariatur cliche reprehenderit, enim eiusmod high life accusamus terry richardson ad squid. 3 wolf moon officia aute, non cupidatat skateboard dolor brunch. Food truck quinoa nesciunt.

Storages

                            Anim pariatur cliche reprehenderit, enim eiusmod high life accusamus terry richardson ad squid. 3 wolf moon officia aute, non cupidatat skateboard dolor brunch. Food truck quinoa nesciunt.

Protection

                            Anim pariatur cliche reprehenderit, enim eiusmod high life accusamus terry richardson ad squid. 3 wolf moon officia aute, non cupidatat skateboard dolor brunch. Food truck quinoa nesciunt.

...

                            ...

...

                            ...


Commercial property refurbishment projects include converting period barns into high specification, contemporary office space and converting warehouses into retail space.
A recent project for the Copas Partnership in Remenham, on Henley on Thames river side, comprised the conversion of an 18th century barn into luxury office space. The works included major structural rearrangement, whilst retaining the original character. Retained components included beams (oak trusses), red terracotta clay peg tiles dating from the late 1800s and old stock bricks. We extensively repointed, using traditional lime mortar and installed new internal doors in oak. To bring the offices up to modern day standards, we installed data cabling, fibre optics, heating and cooling systems and extensive insulation to achieve optimum EPC ratings.
A modern barn conversion combines the best of both worlds: a beautiful looking building with all of the amenities of contemporary life.
The Italian Continental Stores in Maidenhead is a long established family business. When the company outgrew its premises, they asked Farr & Roberts to create a brand new shop, warehouse and set of offices, all within one building. We built a modern, high specification office suite and kept the feel of the original, popular shop by utilising a similar floor to ceiling aisle layout. The building has a striking visual appeal, as a green through colour render was used on the exterior.


text:
asbury-1965
About asbury-1965
June 5 TCM Film Noir Discussions for #NoirSummer for All 13 Films
asbury-1965 replied to Dr. Rich Edwards's topic in Summer of Darkness: Investigating Film Noir
I have taken a previous movie course in college and remember M as a favorite. I thought it decades before its time in its subject matter. The story of a child murderer would not be done in the U.S. or at least not in this way. Johnny Cash was once asked about the most famous line in his song, "Folsom Prison Blues" and how he came up with lyric "I shot a man in Reno just to watch him Die". He said it was the worst thing he could think of for a person to do. Similarly, Fritz Lange could think of nothing worse than a child murderer and his "Man in Black" is first represented in a childs maca
film noir foundation
film restoration
UCLA film & TV Archive
Daily Dose of Darkness #3: Under a Full Moon (The opening scene of The Letter)
asbury-1965 replied to BeachGaBulldog's topic in Summer of Darkness: Investigating Film Noir
I do like the calm beginning, the camera panning slowing past the sleeping plantation worker and then by the light of the moon, we see the woman kill the man and she calls it an accident. I wonder where we are headed. Knowing who the killer is but not why reminds me of Columbo where we, the audience, always knew who done it but it was up to Columbo to figure out why by himself without our help. I have never seen the movie before so I have no clue what the authorities will have to say or in what manner the truth will be revealed and if the truth will only be known by the audience because the
The Letter (1940)
Daily Dose of Darkness #2 The Arrival of a Train (The Opening Scene of La Bete Humaine)
asbury-1965 replied to Screenfiend's topic in Summer of Darkness: Investigating Film Noir
I also thought that was a signal to take a drink from a flask.
NoirSummer
I think this is the longerst train scene without words I have ever seen. I was fascinated to watch all the tasks the two men had perform to go over the terrain and there was a false anxiety that something would happen along the way which was caused by the sights and sounds - the whislte, the dark tunnels, the darkness of the day, the train smoke, the chugging of the wheels, the bridge, the oncoming train and then you arrive at the station. The journey of life can be comparable. You never know what can happen at any moment and then you reach a station and you think you have arrived but what
listening to a pod cast interview with our instructor. http://www.hifilmfest.com/2015/05/27/138-into-the-darkness-preparing-for-tcms-investigating-film-noir-class/


text:
Django application, powers [okthess.gr](http://okthess.gr/).

## Setup

Install requirements:
```
pip3 install -r requirements.txt
```

Then, migrate your database:
```
python3 manage.py migrate
```

Finally, run the Django server:
```
python3 manage.py runserver
```

The Django project is `okthess`. There is one Django app, `main`, which includes
all business logic.

This project uses PostgreSQL.

## Create new i18n strings

Add new strings (with `trans` and `blocktrans` on templates).
Then run this to parse the new strings:
```
python3 manage.py makemessages -i venv -l el
```

Translate them in [`django.po`](/locale/el/LC_MESSAGES/django.po).

Once finished, run this to compile them into the binary file named [`django.mo`](/locale/el/LC_MESSAGES/django.mo).
```
python3 manage.py compilemessages
```

## Infrastructure

This project is deployed to AWS Elastic Beanstalk using [Terraform](https://www.terraform.io/) (v0.10.2).
The infrastructure resources are described at the [`infra`](/infra) directory.
Those resource assume the default VPC exists (including subnets, route tables, ACLs, internet gateway, etc).

```sh
cd infra/
terraform init  # initialize terraform project
terraform validate -var-file terraform.tfvars  # validate tf files
terraform fmt  # format tf files in the default, not very useful, way

# plan against production secret variables
terraform plan -var-file production.tfvars -state=terraform.tfstate -refresh=true -out=terraform.tfplan

# apply the infra now and make the current resources described in the tf files reality
terraform apply -var-file production.tfvars -backup=terraform.tfstate.backup -state=terraform.tfstate
```


text:
Fabrication of a covalently attached self-assembly multilayer film based on CdTe nanoparticles
Hao Zhang, Bai Yang, Ruibing Wang, Gang Zhang, Xueliang Hou, Lixin Wu
A precursor film has been fabricated from TGA (thiolglycolic acid)-stabilized CdTe nanoparticles and NDR (nitro-containing diazoresin) using electrostatic interactions and the standard layer-by-layer assembly method; covalent bonds are formed under ultraviolet irradiation. XPS provided evidence for the presence of CdTe nanoparticles within the polymer ultrathin films. UV-visible spectroscopy and FTIR spectroscopy provide evidence for the formation of a covalent linkage. Moreover, the UV-visible spectroscopy and AFM measurement support the improvement of the stability of the hybrid organic/inorganic film toward polar solvents when the linkages between the nanoparticles and polymer changed from ionic bonds to covalent bonds.
https://doi.org/10.1006/jcis.2001.8136
10.1006/jcis.2001.8136
Fingerprint Dive into the research topics of 'Fabrication of a covalently attached self-assembly multilayer film based on CdTe nanoparticles'. Together they form a unique fingerprint.
Multilayer films Chemical Compounds
Self assembly Chemical Compounds
Spectroscopy Chemical Compounds
Covalent bonds Chemical Compounds
Nanoparticles Chemical Compounds
diazoresin Chemical Compounds
Fabrication Chemical Compounds
Ultrathin films Chemical Compounds
Zhang, H., Yang, B., Wang, R., Zhang, G., Hou, X., & Wu, L. (2002). Fabrication of a covalently attached self-assembly multilayer film based on CdTe nanoparticles. Journal of Colloid And Interface Science, 247(2), 361-365. https://doi.org/10.1006/jcis.2001.8136
Zhang, Hao ; Yang, Bai ; Wang, Ruibing ; Zhang, Gang ; Hou, Xueliang ; Wu, Lixin. / Fabrication of a covalently attached self-assembly multilayer film based on CdTe nanoparticles. In: Journal of Colloid And Interface Science. 2002 ; Vol. 247, No. 2. pp. 361-365.
@article{52f366e141204487a8e95bc232c7aa6b,
title = "Fabrication of a covalently attached self-assembly multilayer film based on CdTe nanoparticles",
abstract = "A precursor film has been fabricated from TGA (thiolglycolic acid)-stabilized CdTe nanoparticles and NDR (nitro-containing diazoresin) using electrostatic interactions and the standard layer-by-layer assembly method; covalent bonds are formed under ultraviolet irradiation. XPS provided evidence for the presence of CdTe nanoparticles within the polymer ultrathin films. UV-visible spectroscopy and FTIR spectroscopy provide evidence for the formation of a covalent linkage. Moreover, the UV-visible spectroscopy and AFM measurement support the improvement of the stability of the hybrid organic/inorganic film toward polar solvents when the linkages between the nanoparticles and polymer changed from ionic bonds to covalent bonds.",
author = "Hao Zhang and Bai Yang and Ruibing Wang and Gang Zhang and Xueliang Hou and Lixin Wu",
doi = "10.1006/jcis.2001.8136",
journal = "Journal of Colloid and Interface Science",
Zhang, H, Yang, B, Wang, R, Zhang, G, Hou, X & Wu, L 2002, 'Fabrication of a covalently attached self-assembly multilayer film based on CdTe nanoparticles', Journal of Colloid And Interface Science, vol. 247, no. 2, pp. 361-365. https://doi.org/10.1006/jcis.2001.8136
Fabrication of a covalently attached self-assembly multilayer film based on CdTe nanoparticles. / Zhang, Hao; Yang, Bai; Wang, Ruibing; Zhang, Gang; Hou, Xueliang; Wu, Lixin.
In: Journal of Colloid And Interface Science, Vol. 247, No. 2, 2002, p. 361-365.
T1 - Fabrication of a covalently attached self-assembly multilayer film based on CdTe nanoparticles
AU - Zhang, Hao
AU - Yang, Bai
AU - Wang, Ruibing
AU - Zhang, Gang
AU - Hou, Xueliang
AU - Wu, Lixin
N2 - A precursor film has been fabricated from TGA (thiolglycolic acid)-stabilized CdTe nanoparticles and NDR (nitro-containing diazoresin) using electrostatic interactions and the standard layer-by-layer assembly method; covalent bonds are formed under ultraviolet irradiation. XPS provided evidence for the presence of CdTe nanoparticles within the polymer ultrathin films. UV-visible spectroscopy and FTIR spectroscopy provide evidence for the formation of a covalent linkage. Moreover, the UV-visible spectroscopy and AFM measurement support the improvement of the stability of the hybrid organic/inorganic film toward polar solvents when the linkages between the nanoparticles and polymer changed from ionic bonds to covalent bonds.
AB - A precursor film has been fabricated from TGA (thiolglycolic acid)-stabilized CdTe nanoparticles and NDR (nitro-containing diazoresin) using electrostatic interactions and the standard layer-by-layer assembly method; covalent bonds are formed under ultraviolet irradiation. XPS provided evidence for the presence of CdTe nanoparticles within the polymer ultrathin films. UV-visible spectroscopy and FTIR spectroscopy provide evidence for the formation of a covalent linkage. Moreover, the UV-visible spectroscopy and AFM measurement support the improvement of the stability of the hybrid organic/inorganic film toward polar solvents when the linkages between the nanoparticles and polymer changed from ionic bonds to covalent bonds.
U2 - 10.1006/jcis.2001.8136
DO - 10.1006/jcis.2001.8136
JO - Journal of Colloid and Interface Science
JF - Journal of Colloid and Interface Science
Zhang H, Yang B, Wang R, Zhang G, Hou X, Wu L. Fabrication of a covalently attached self-assembly multilayer film based on CdTe nanoparticles. Journal of Colloid And Interface Science. 2002;247(2):361-365. https://doi.org/10.1006/jcis.2001.8136


>> according to the Gitlab interface.
> warning, would the patch be bigger or larger?
modifying the Makefile.in to accommodate this.
> variables list at the end of the file, rather than in the first line.
> and that first line can get a bit crowded.
a-f-l-v-p-l or some other function to put it at the bottom?