In Finsler geometry, there are several important classes of Finsler metrics. The Berwald metrics were first investigated by L. Berwald. By definition, a Finsler metric $F$ is a Berwald metric if the spray coefficients $G^i = G^i(x, y)$ are quadratic in $y\in T_xM$ at every point $x$, i.e.,

$\begin{eqnarray*} G^i=\frac{1}{2}\Gamma^i_{jk}(x)y^jy^k. \end{eqnarray*}$

Riemannian metrics are special Berwald metrics. In fact, Berwald metrics are "almost Riemannian" in the sense that every Berwald metric is affinely equivalent to a Riemannian metric, i.e., the geodesics of any Berwald metric are the geodesics of some Riemannian metric. Weak Berwald spaces were first investigated by Bácsó and Yoshikawa in 2002 [2]. The class of weak Berwald metrics is more generalized than Berwald metrics in [6]. Hence it becomes an important and natural problem to study weak Berwald $(\alpha,\beta)$-metrics. Cui obtained the necessary and sufficient conditions for two important kinds of $(\alpha,\beta)$-metrics in the forms of $F=\alpha+\varepsilon\beta+k\frac{\beta^2}{\alpha}$ and $F=\frac{{{\alpha }^{2}}}{\alpha -\beta }$ to be weak Berwald metrics in [7]. Xiang and Cheng characterized a special class of weak Berwald $(\alpha,\beta)$-metrics in the form of $F= {(\alpha +\beta)}^{m+1}/{\alpha^{m}}$ in [10]. Further, Cheng and Lu studied two kinds of weak Berwald metrics of scalar flag curvature in [4].

The purpose of this paper is to study a special class of weak Berwald $(\alpha,\beta)$-metrics in the form of $F=\alpha+\varepsilon\beta+\beta\ {\rm arctan}(\frac{\beta}{\alpha})$. We have the following:

Theorem 1.1 Let $F=\alpha+\varepsilon\beta+\beta\ \arctan(\frac{\beta}{\alpha})$ be an arctangent Finsler metric on an $n$-dimensional manifold $M$($n\geq3$), where $\varepsilon$ is a constant. Then the following are equivalent:

(a) $F$ has isotropic $S$-curvature, i.e., $\operatorname{S} = (n +1)cF$;

(b) $F$ has isotropic mean Berwald curvature, i.e., $\operatorname{E} =\frac{n+1}{2}cF^{-1}h$;

(c) $\beta$ is a Killing 1-form of constant length with respect to $\alpha$, i.e., $r_{00} = s_0 = 0$;

(d) $F$ has vanished $S$-curvature, i.e., $\operatorname{S} = 0$;

(e) $F$ is a weak Berwald metric, i.e., $\operatorname{E} = 0$,

where $c=c(x)$ is a scalar function on $M$.

By [2], an arctangent Finsler metric $F=\alpha+\varepsilon\beta+\beta\ \arctan(\frac{\beta}{\alpha})$ is of scalar flag curvature with vanishing $S$-curvature if and only if its flag curvature $\operatorname{K} = 0$ and it is a Berwald metric. In this case, $F$ is a locally Minkowski metric. Thus $F$ is a weak Berwald metric with scalar flag curvature, its local structure can be completely determined.

Corollary 1.2  Let $F=\alpha+\varepsilon\beta+\beta\ \arctan(\frac{\beta}{\alpha})$ be an arctangent Finsler metric on an $n$-dimensional manifold $M$ ($n\geq3$), where $\varepsilon$ is a constant. Then $F$ is a weak Berwald metric with scalar flag curvature $\operatorname{K} = \operatorname{K} (x, y)$ if and only if it is a Berwald metric and $\operatorname{K} = 0$. In this case, $F$ must be locally Minkowskian.

In Finsler geometry, $(\alpha ,\beta )$-metrics form a very important and rich class of Finsler metrics. An $(\alpha ,\beta )$-metric is expressed as the following form

$F=\alpha \phi (s),\ s=\frac{\beta }{\alpha},$

where $\alpha$ is a Riemannian metric and $\beta$ is a 1-form. $\phi(s)$ is a positive $C^\infty $ function on an open interval $(-{{b}_{0}},{{b}_{0}})$ and satisfying

$\phi (s)-s{{\phi }^{\prime}}(s)+({{b}^{2}}-{{s}^{2}}){{\phi }^{\prime\prime}}(s)>0,\ \left| s \right|\le b< {{b}_{0}},$

where $b:={{\left\| \beta \right\|}_{\alpha }}$. It is known that $F=\alpha \phi (s)$ is a Finsler metric if and only if ${{\left\| \beta \right\|}_{\alpha }}< {{b}_{0}}$ for any $x\in M$ in [6]. In this paper, we consider a special $(\alpha, \beta)$-metric in the following form:

$\begin{eqnarray} F=\alpha+\varepsilon\beta+\beta\arctan(\frac{\beta}{\alpha}), \end{eqnarray}$

(2.1)

where $\varepsilon$ is an arbitrary constant. We call this metric an arctangent Finsler metric. Let $b_{0} > 0 $ be the largest number such that

$\begin{eqnarray} \frac{1-s^2+2b^2}{(1+s^2)^2}>0,\ \ |s|\leq b <b_{0}, \end{eqnarray}$

(2.2)

so that $F=\alpha+\varepsilon\beta+\beta\ \arctan(\frac{\beta}{\alpha})$ is a Finsler metric if and only if $\beta$ satisfies that $ \| \beta\|_{\alpha} <b_{0} $ for any $x\in M$. Let $ \nabla \beta = b_{i|j}dx^i\otimes dx^j $ denote covariant derivative of $\beta$ with respect to $\alpha$.

Denote

$\begin{eqnarray*}&& {{s}_{ij}}:=\frac{1}{2}({{b}_{i|j}}-{{b}_{j|i}}),\ {{r}_{ij}}:=\frac{1}{2}({{b}_{i|j}}+{{b}_{j|i}}),\ {{s}_{l0}}:={{s}_{li}}{{y}^{i}},\ {{s}_{0}}:={{b}^{l}}{{s}_{l0}},\\ && {{r}_{00}}:={{r}_{ij}}{{y}^{i}}{{y}^{j}},\ {{r}_{i}}:={{r}_{ij}}{{b}^{j}},\ {{r}_{0}}:={{r}_{j}}{{y}^{j}}.\end{eqnarray*}$

Let ${{G}^{i}}(x,y)$ and $G_{\alpha }^{i}(x,y)$ denote the spray coefficients of $F$ and $\alpha$, respectively. We have the following formula for the spray coefficients ${{G}^{i}}(x,y)$ of $F$,

$\begin{eqnarray} {{G}^{i}}=G_{\alpha }^{i}+\alpha Q{{s}^{i}}_{0}+\Theta \{-2\alpha Q{{s}_{0}}+{{r}_{00}}\}\frac{{{y}^{i}}}{\alpha }+\Psi \{-2\alpha Q{{s}_{0}}+{{r}_{00}}\}{{b}^{i}}, \end{eqnarray}$

(2.3)

where

$\begin{eqnarray} \nonumber Q&=&\varepsilon+\varepsilon s^2+\arctan(s)+ s^2\arctan(s)+s,\\ \nonumber \Theta&=&\frac{\varepsilon-s-\varepsilon s^{2}-\arctan(s)\ s^2+\arctan(s)} {2(1+2b^2-s^{2})(1+\varepsilon s+s\arctan(s))},\\ \Psi&=&\frac{1}{1+2b^2-s^{2}}. \end{eqnarray}$

(2.4)

As is well known, the Berwald tensor of a Finsler metric $F$ with the spray coefficients $G^i $ is defined by ${\bf B_y}:=B_{jkl}^i(x,y)dx^j\otimes dx^k\otimes dx^l \otimes\partial_i$, where

$\begin{eqnarray} B_{jkl}^i:=\frac{\partial^3G^i}{\partial y^j\partial y^k\partial y^l}. \end{eqnarray}$

(2.5)

Furthermore, the mean Berwald tensor $\operatorname{E_y} := E_{ij}(x,y)dx^i \otimes dx^j $ is defined by

$\begin{eqnarray} E_{ij}:=\frac{1}{2}B_{mij}^{m}=\frac{1}{2}\frac{\partial^2}{\partial y^i \partial y^j}(\frac{\partial G^m}{\partial y^m}). \end{eqnarray}$

(2.6)

A Finsler metric is called a Berwald metric if the Berwald curvature $\operatorname{B} = 0$. A Finsler metric is called a weak Berwald metric if the mean Berwald curvature $\operatorname{E} = 0$.

The $S$-curvature $S = S(x, y)$ is one of the most important non-Riemannian quantities. For a Finsler metric $F = F(x, y)$ on an $n$-dimensional manifold $M$, the Busemann-Hausdorff volume form $dV_F = \sigma_Fdx^1\wedge\cdots\wedge dx^n$ is given by

$\begin{eqnarray*} \sigma_F(x):=\frac{{\rm Vol}(B^n(1))}{Vol\{(y^i)\in R^n\mid F(x,y)<1\}}. \end{eqnarray*}$

Here Vol denotes the Euclidean volume in $R^n$. The well-known $S$-curvature is given by

$\begin{eqnarray*} S(x,y)=\frac{\partial G^m}{\partial y^m}-y^m\frac{\partial (ln \sigma_F)}{\partial x^m}. \end{eqnarray*}$

Cheng and Shen obtained a formula for the $ S$-curvature of an $(\alpha,\beta)$-metric on an $n$-dimensional manifold $M$ as follows

Lemma 2.1  [5] The $S$-curvature of an $(\alpha,\beta)$-metric is given by

$\begin{eqnarray} \operatorname{S} =\lambda(r_0+s_0)+2(\Psi+QC)s_0+2\Psi r_0-\alpha^{-1}C r_{00}, \end{eqnarray}$

(2.7)

where $ \lambda:=-\frac{f^{\prime}(b)}{bf(b)}$ is a scalar function on $M$ and $C:=-(b^2-s^2)\Psi^{\prime}-(n+1)\Theta$.

The proof contains the following steps:

Step 1 (a) $\Rightarrow$ (b) In fact, (a) $\Rightarrow$ (b) is obvious true.

Step 2 (b) $\Rightarrow$ (a) Assume that (b) holds, which is equivalent to

$\begin{eqnarray} \operatorname{S} =(n+1)\{c F+\eta\}, \end{eqnarray}$

(3.1)

where $\eta$ is a 1-form on $M$. So (a) is equivalent to (b) if and only if $\eta= 0$. Plugging (2.4) and (2.7) into (3.1), we obtain

$\begin{eqnarray} \nonumber && (J_6 \alpha^6+J_5 \alpha^5+J_4 \alpha^4+J_3 \alpha^3+J_2 \alpha^2+J_1 \alpha+J_0)\arctan^2(\frac{\beta}{\alpha})\\ \nonumber && +(K_6\alpha^6+K_5 \alpha^5+K_4 \alpha^4+K_3 \alpha^3+K_2 \alpha^2+K_1\alpha+K_0)\arctan(\frac{\beta}{\alpha})\\ \nonumber &&+M_7 \alpha^7+M_6\alpha^6+M_5 \alpha^5+M_4 \alpha^4+M_3 \alpha^3 \\ &&+M_2\alpha^2+M_1 \alpha+M_0=0, \end{eqnarray}$

(3.2)

where

$\begin{eqnarray*} J_6&=&2\nu s_0(1+2b^2),\ J_5=2\nu c \beta^2 (1+2b^2)^2,\\ J_4&=&2 s_0 \beta^2(4b^2-\nu),\ J_3=-4\nu c \beta^4 (1+2b^2),\\ J_2&=&-2 s_0 \beta^4(-2b^2+5+2nb^2+n),\ J_1=2c\nu \beta^6,\\ J_0&=&2 s_0 \beta^6(n-3),\\ K_6&=&4(1+2b^2)(2\nu c b^2\beta+\nu c \beta +s_0 \varepsilon + \varepsilon n s_0),\\ K_5&=&-8 r_0 b^2 \beta +8\nu\eta b^2\beta-\nu r_{00}+8\lambda r_{0}b^2\beta+4\nu c \varepsilon \beta^2+2\lambda s_{0}\beta\\ && +8\nu\eta b^2 \beta+8\lambda r_{0} b^4 \beta+8\lambda s_{0} b^2 \beta+2\nu\eta\beta-2 \nu b^2r_{00}-4r_{0}\beta\\ && -4s_{0}\beta+8\lambda b^4s_{0}\beta +16\nu c \varepsilon b^2\beta^2+2\lambda r_{0}\beta+16\nu c \varepsilon b^4\beta^2,\\ K_4&=&-4\beta^2(4\nu c \beta b^2+2\nu c \beta + \varepsilon\nu s_{0} -4 b^2 \varepsilon s_{0}),\\ K_3&=&-2\beta^2(4\nu c \varepsilon \beta^2+8\nu c \varepsilon \beta^2 b^2+2\lambda r_{0}\beta+4\nu\eta b^2 \beta+2\lambda s_{0}\beta-2 r_{0} \beta\\ && + 4s_{0}\beta+2\beta n s_{0}+4 n \beta b^2 s_{0}+4\lambda s_{0}b^2\beta+2\nu\eta \beta-4\beta b^2s_{0}+4\lambda r_{0}b^2\beta\\ && -\nu r_{00} -n b^2 r_{00}+b^2r_{00}),\\ K_2&=&-4\beta^4(-\nu c \beta+5 \varepsilon s_{0} + \varepsilon n s_{0}+2\varepsilon n b^2 s_{0}-2b^2\varepsilon s_{0} ),\\ K_1&=&\beta^4(4\nu c \varepsilon \beta^2+4\beta n s_{0}+2 \lambda \beta s_{0} + 2 \lambda \beta r_{0}-12 s_{0} \beta +2 \nu \eta \beta\\ && +3 r_{00}-nr_{00}), \\ K_0&=&4(n-3) \varepsilon \beta^6s_{0},\ M_7=2\nu c (1+2b^2)^2,\\ M_6&=&2(1+2b^2)(2\nu\eta b^2+4\nu c \varepsilon \beta b^2+2\lambda b^2 r_{0}+ 2\lambda b^2 s_{0} -2r_{0} + \lambda r_{0}\\ &&+\varepsilon^2 \nu s_{0} -2 s_{0}+2\nu c \varepsilon \beta +\nu\eta + \lambda s_{0} ),\end{eqnarray*}$

$\begin{eqnarray*} M_5&=& 2\lambda \varepsilon \beta r_{0}+8\lambda b^2\varepsilon\beta s_{0}+2\lambda\varepsilon\beta s_{0}-8\varepsilon\beta b^2 r_{0} +8\lambda\varepsilon\beta b^4 r_{0}+2\nu c \varepsilon^2\beta^2\\ && +8\lambda\varepsilon\beta b^4 s_{0}+ 8 \lambda\varepsilon\beta b^2 r_{0}-\varepsilon \nu r_{00}+8\nu\eta\varepsilon b^2 \beta-8\nu c b^2 \beta^2\\ && +2\nu\eta\varepsilon\beta-4\varepsilon\beta r_{0}+8\nu c \varepsilon^2 b^4 \beta^2+8\nu\eta\varepsilon b^4\beta-4\nu c \beta^2 \\ && + 8 \nu c \varepsilon^2 b^2 \beta^2-2\varepsilon n b^2 r_{00}-4\varepsilon\beta s_{0}-2b^2\varepsilon r_{00},\\ M_4&=&-\beta(8\nu\varepsilon c \beta^2+16 \nu\varepsilon c b^2\beta^2+4 n b^2\beta s_{0} +4\lambda\beta r_{0}+ 2\beta\varepsilon^2 s_{0}\\ && -2 \beta s_{0}-4 b^2 \beta s_{0} +4\nu\eta\beta-4r_{0}\beta+8\nu\eta b^2\beta+2n\varepsilon^2\beta s_{0} -8\varepsilon^2 b^2\beta s_{0}\\ && +2 n \beta s_{0}+8 \lambda b^2\beta s_{0}+4\lambda \beta s_{0} +8\lambda b^2 \beta r_{0} +2 b^2 r_{00} -2nb^2r_{00}-\nu r_{00}),\\ M_3&=&-2\beta^2(2\nu\varepsilon^2 c \beta^2+4 \nu\varepsilon^2 c b^2\beta^2-\nu c \beta^2+4 \lambda\varepsilon b^2 \beta s_{0} + 4\varepsilon\beta s_{0} -2 \varepsilon\beta r_{0} \\ && +2\lambda\varepsilon\beta r_{0}-4\varepsilon b^2\beta s_{0} + 4\nu\eta\varepsilon b^2\beta+2\lambda \varepsilon \beta s_{0} + 2\varepsilon n \beta s_{0} + 2\nu\eta\varepsilon\beta\\ && +4\varepsilon n b^2 \beta s_{0} + 4 \lambda \varepsilon b^2 \beta r_{0}-\varepsilon \nu r_{00}-\varepsilon n b^2 r_{00}+\varepsilon b^2 r_{00}),\\ M_2&=&-\beta^3(-4\nu c \varepsilon \beta^2 + 4 \beta n \varepsilon^2 b^2 s_{0} + 10 \varepsilon^2 \beta s_{0} -2\lambda\beta r_{0} -4 \varepsilon^2 b^2\beta s_{0}\\ && -2n\beta s_{0} -2\nu\eta\beta+6\beta s_{0}-2\lambda\beta s_{0}+2n \varepsilon^2\beta s_{0}-3 r_{00}+n r_{00}),\\ M_1&=&\varepsilon\beta^4(2\nu c \varepsilon \beta^2+2\lambda\beta r_{0}+2\lambda \beta s_{0} -12\beta s_{0} +4 n \beta s_{0}+2\nu\eta\beta \\ &&\ \ +3 r_{00} -n r_{00}),\\ M_0&=&2\varepsilon^2(n-3)\beta^6 s_{0},\ \nu=n+1. \end{eqnarray*}$

Replacing $y^i$ in (3.2) by $-y^i$, we get the following

$\begin{eqnarray} \nonumber && (-J_6 \alpha^6+J_5 \alpha^5-J_4 \alpha^4+J_3 \alpha^3-J_2 \alpha^2+J_1 \alpha-J_0)\arctan^2(\frac{\beta}{\alpha})\\ \nonumber && +(K_6 \alpha^6-K_5 \alpha^5+K_4 \alpha^4-K_3 \alpha^3+K_2 \alpha^2-K_1 \alpha+K_0)\arctan(\frac{\beta}{\alpha})\\ \nonumber &&+M_7 \alpha^7-M_6 \alpha^6+M_5 \alpha^5-M_4 \alpha^4+M_3 \alpha^3\\ &&-M_2 \alpha^2+M_1 \alpha-M_0=0. \end{eqnarray}$

(3.3)

(3.2) $+$ (3.3) yields

$\begin{eqnarray} \nonumber && (J_5 \alpha^5+J_3\alpha^3+J_1 \alpha)\arctan^2(\frac{\beta}{\alpha}) +M_7 \alpha^7+M_5 \alpha^5+M_3 \alpha^3+M_1 \alpha\\ &&+(K_6 \alpha^6+K_4 \alpha^4+K_2 \alpha^2+K_0)\arctan(\frac{\beta}{\alpha})=0. \end{eqnarray}$

(3.4)

(3.2) $-$ (3.3) yields

$\begin{eqnarray} \nonumber && (J_6 \alpha^6+J_4 \alpha^4+J_2 \alpha^2+J_0)\arctan^2(\frac{\beta}{\alpha})+(K_5 \alpha^5+K_3 \alpha^3+K_1\alpha)\arctan(\frac{\beta}{\alpha})\\ && =-M_6\alpha^6-M_4 \alpha^4-M_2\alpha^2-M_0. \end{eqnarray}$

(3.5)

Using Taylor expansion of $\arctan(\frac{\beta}{\alpha})$, we can find that the right side of (3.5) is an integral expression in $y$ and the left side of (3.5) is a fraction expression in $y$, so that we get

$\begin{eqnarray} && (J_6 \alpha^6+J_4 \alpha^4+J_2 \alpha^2+J_0)\arctan(\frac{\beta}{\alpha})+K_5 \alpha^5+K_3 \alpha^3+K_1\alpha=0, \end{eqnarray}$

(3.6)

$\begin{eqnarray} && M_6\alpha^6+M_4 \alpha^4+M_2\alpha^2+M_0=0. \end{eqnarray}$

(3.7)

Similarly, from (3.6), we get the following

$\begin{eqnarray} &&J_6 \alpha^6+J_4 \alpha^4+J_2 \alpha^2+J_0=0, \end{eqnarray}$

(3.8)

$\begin{eqnarray} && K_5 \alpha^4+K_3\alpha^2+K_1=0. \end{eqnarray}$

(3.9)

For the same reason, by (3.4), we have

$\begin{eqnarray} && J_5 \alpha^4+J_3\alpha^2+J_1=0, \end{eqnarray}$

(3.10)

$\begin{eqnarray} && K_6 \alpha^6+K_4 \alpha^4+K_2 \alpha^2+K_0=0, \end{eqnarray}$

(3.11)

$\begin{eqnarray} && M_7 \alpha^6+M_5 \alpha^4+M_3 \alpha^2+M_1 =0. \end{eqnarray}$

(3.12)

(3.10) tells us that $J_1=2(n+1) c \beta^6 $ has the factor $\alpha^2$. Because $\beta^6$ and $\alpha^2$ are relatively prime polynomials of $(y^i)$, we immediately obtain $c=0$.

Now we split the proof into four cases:

(i) $\varepsilon\neq 0$ and $n=3$;

(ii) $\varepsilon\neq 0$ and $n>3$;

(iii) $\varepsilon=0$ and $n>3$;

(iv) $\varepsilon=0$ and $n=3$.

Case i $\varepsilon\neq 0$ and $n=3$.

In this case, $K_0=4(n-3) \varepsilon \beta^6s_{0}=0$. Hence, (3.11) implies that $K_2=-16\varepsilon\beta^4s_0(b^2+2)$ has the factor $\alpha^2$. This implies $s_{0}=0$. By use of $s_{0}=0$ and $c=0$, we have $M_7=M_0=0$.

By (3.7), we obtain the following

$\begin{eqnarray} \nonumber && 2(1+2b^2)(8\eta b^2 +2 \lambda b^2 r_0+\lambda r_0 +4\eta-2 r_0)\alpha^4-\beta(16\eta \beta + 32 \eta b^2 \beta+8 \lambda r_0 b^2 \beta+4\lambda r_0 \beta \\ && -4 r_0 \beta-4 b^2 r_{00}-4r_{00})\alpha^2 +\beta^3(8\eta\beta+2\lambda r_0 \beta)=0. \end{eqnarray}$

(3.13)

By (3.12), we obtain the following

$\begin{eqnarray} \nonumber &&2\varepsilon(1+2b^2)(2\lambda r_0 b^2 \beta + 8\eta b^2 \beta+\lambda r_0\beta+4 \eta \beta-2 r_0 \beta -2 r_{00})\alpha^4-4\varepsilon\beta^2(2 \lambda r_0 b^2\beta\\ &&-r_0\beta+\lambda r_0 \beta+8 \eta b^2 \beta+4 \eta \beta-b^2 r_{00} -2 r_{00})\alpha^2+\varepsilon\beta^4(2\lambda r_0 \beta+8 \eta\beta)=0. \end{eqnarray}$

(3.14)

(3.14)$-$(3.13)$\times \varepsilon \beta$ gives

$\begin{eqnarray} 4\varepsilon [(1+2b^2)\alpha^2-\beta^2]r_{00}=0. \end{eqnarray}$

(3.15)

Because $F$ is non-Riemannian, $(1+2b^2)\alpha^2-\beta^2\neq0$, thus we get

$\begin{eqnarray} r_{00}=0,\ \ r_{0}=0. \end{eqnarray}$

(3.16)

Plugging (3.16) into (2.7) yields $\operatorname{S} =0$. In this case $\eta = 0$.

Case ii $\varepsilon\neq 0$ and $n>3$.

From (3.7), we can see that $M_0=2\varepsilon^2(n-3)\beta^6 s_{0}$ has the factor $\alpha^2$. Since $\varepsilon\neq 0$ and $n>3$, we have $s_{0}=0$. By use of (3.7) and (3.12) and using the same skills in case (i), we obtain

$\begin{eqnarray} r_{00}=0,\ \ r_{0}=0,\ \ \eta = 0,\ \ \operatorname{S} =0. \end{eqnarray}$

(3.17)

Case iii $\varepsilon= 0$ and $n>3$.

By (3.8), we can see that $J_0=2 s_0 \beta^6(n-3)$ has the factor $\alpha^2$. Obviously, we can get $s_{0}=0$.

From (3.7), we get the following

$\begin{eqnarray} && 2(1+2b^2)(2(n+1)\eta b^2+2\lambda b^2 r_0 +\lambda r_0+(n+1)\eta-2 r_0)\alpha^4\\ \nonumber &&-\beta( 4(n+1)\eta\beta+8(n+1)\eta b^2\beta+8\lambda r_0 b^2 \beta+4\lambda r_0 \beta-4 r_0\beta-2n b^2 r_{00}\\ \nonumber && -(n+1) r_{00}+2 b^2 r_{00}) \alpha^2+\beta^3(2(n+1)\eta\beta+2 \lambda r_0\beta-n r_{00}+3 r_{00})=0. \end{eqnarray}$

(3.18)

From (3.9), we have

$\begin{eqnarray} \nonumber && (1+2b^2)(4(n+1)\eta b^2\beta+4 \lambda r_0 b^2 \beta-(n+1) r_{00}-4 r_0 \beta+ 2 (n+1) \eta \beta+2\lambda r_0 \beta)\alpha^4\\ \nonumber &&-2\beta^2(4(n+1)\eta b^2\beta+2\lambda r_0 \beta+2(n+1)\eta\beta+4\lambda r_0 b^2 \beta-2 r_0 \beta-(n+1)r_{00}\\ &&-n b^2 r_{00}+b^2 r_{00})\alpha^2+\beta^4(2\lambda r_0 \beta + 2(n+1)\eta\beta-n r_{00}+3r_{00})=0. \end{eqnarray}$

(3.19)

(3.19)$-$(3.18)$\times \beta$ yields

$\begin{eqnarray} (n+1)[(1+2b^2)\alpha^2-\beta^2]r_{00}=0. \end{eqnarray}$

(3.20)

This implies $r_{00}=0$. For the same reason, we have

$\begin{eqnarray} r_{0}=0,\ \ \eta = 0,\ \ \operatorname{S} =0. \end{eqnarray}$

(3.21)

Case iv $\varepsilon=0$ and $n=3$.

In this case, $J_0=2 s_0 \beta^6(n-3)=0$. (3.8) becomes

$\begin{eqnarray} [(1+2b^2)\alpha^4+(b^2-1)\beta^2\alpha^2-(b^2+2)\beta^4]s_0=0. \end{eqnarray}$

(3.22)

We assert that $s_{0}=0$. Or else, (3.22) tells us that $\beta^4(b^2+2)$ has the factor $\alpha^2$. This implies $\beta=0$, but it is impossible by the assumptions. By using the same methods as case iii, we get that (3.21) holds.

Anyway, we obtain $r_{00}=0,\ s_{0}=0,\ \eta = 0,\ \operatorname{S} =0$. Which implies that $F$ is of isotropic $S$-curvature with $c = 0$.

Step 3 (b) $\Rightarrow$ (c) The proof has been contained in Step 2.

Step 4 (c) $\Rightarrow$(d) When $r_{00}=0$ and $s_{0}=0$, by (2.7), we have $\operatorname{S} =0$.

Step 5  (d) $\Rightarrow$ (e) $\operatorname{S} =0$ implies that $F$ is of isotropic $S$-curvature with $c = 0$. Thus, we obtain $\operatorname{E} =0$ by the equivalence of (a) and (b).

Step 6 (e) $\Rightarrow$ (a) $\operatorname{E} =0$ is equivalent to that F is of isotropic mean Berwald curvature with $c = 0$, that is, (b) holds with $c = 0$. By the equivalence of (a) and (b), we know that $F$ has isotropic $S$-curvature with $c = 0$. This completes the proof. Theorem 1.1 is proved completely.