Let $(M^n, g)$ be an $n$-dimensional complete Riemannian manifold. The drifting Laplacian is defined by $\Delta_f=\Delta-\nabla f\nabla, $ where $f$ is a smooth function on $M.$ The $N$-Bakry-Emery Ricci tensor is defined by

$ {\rm Ric}_{f}^{N}={\rm Ric}+\bigtriangledown ^{2}f-\frac{1}{N}df\otimes df $

for $0\leq N<\infty$ and $N=0$ if and only if $f=0, $ where $f$ is some smooth function on $M, $ $\nabla^{2}$ is the Hessian and Ric is the Ricci tensor. The $\infty$-Bakry-Emery Ricci tensor is defined by

$ {\rm Ric}_{f}={\rm Ric}+\bigtriangledown ^{2}f. $

In particular, ${\rm Ric}_{f}=\lambda g$ is called a gradient Ricci soliton which is extensively studied in Ricci flow.

In this paper, we want to study positive solutions of the nonlinear elliptic equation with the drifting Laplacian

$ \Delta_f u+au\log u=0 $

(1.1)

on an $n$-dimensional complete Riemannian manifold $(M^n, g), $ where $a$ is a nonzero constant. When $f=$ constant, the above equation (1.1) reduces to

$ \Delta u+au\log u=0. $

(1.2)

Equation (1.2) is closely related to Ricci soliton [9] and the famous Gross Logarithmic Sobolev inequality [6]. Ma [9] first studied the positive solutions of equation (1.2) and derived a local gradient estimate for the case $a<0$. Then the gradient estimate for the case $a>0$ is obtained in [4] and [15] by studying the related heat equation of (1.2). More progress of this and related equations can be found in [2, 8, 10, 13, 14] and the references therein. Recently, inspired by the method used by Brighton in [1], Huang and Ma [7] derived local gradient estimates of the Li-Yau type for positive solutions of equations (1.2). These estimates are different from those in [4, 9, 15]. Using these estimate, they can easily get some Liouville type theorems. We want to generalize their results to equation (1.1) and we obtain the following results

Theorem 1.1  Let $(M^n, g)$ be an $n$-dimensional complete Riemannian manifold with ${\rm Ric}_{f}^{N}(B_P(2R))\geq-K, $ where $K$ is a nonnegative constant. Assuming that $u$ is a positive solution of the nonlinear elliptic eq. (1.1). Then on $B_p(R), $ we have the following inequalities

(1) If $a>0, $ then

$ |\nabla{u}|\leq M\sqrt{{\frac{(n+N+3)^2} {2(n+N)}}{(a+K)}+{\frac{{{(n+N+2)} {{c}_{1}^{2}}+{(n-1+{\sqrt{{n}{K}}}{R})} {{c}_1}+{{c}_2}}}{R^2}}}; $

(1.3)

(2) If $a<0, $ then

$ |\nabla{u}|\leq M\sqrt{{\frac{(5n+5N+6)^2} {36(n+N)}}{\max{\{0, a+K}\}} +{\frac{1}{{R}^2}}{\bigg[{({\frac{75n+75N+12}{6}})}{{c}_{1}^{2}} +{(n-1+{\sqrt{{n}{K}}}{R})} {{c}_1}+{{c}_2}}\bigg]}, $

(1.4)

where $M={\sup\limits_{x\in{{{B}_{p}}{(2R)}}}}{u(x)}, $ the ${{c}_1}$ and ${{c}_2}$ is a positive constants.

Let $R\rightarrow\infty, $ we have the following gradient estimates on complete noncompact Riemannian manifolds.

Corollary 1.2  Let $(M^n, g)$ be an $n$-dimensional complete noncompact Riemannian manifold with ${{\rm Ric}_{f}^{N}}\geq-K, $ where $K$ is a nonnegative constant. Assuming that $u$ is a positive solution of the nonlinear elliptic eq. (1.1). Then the following inequalities hold

(1) if $a>0, $ then

$ |\nabla{u}|\leq{\frac{(n+N+3)M} {\sqrt{2(n+N)}}}{\sqrt{a+K}}; $

(1.5)

(2) if $a<0, $ then

$ |\nabla{u}|\leq{\frac{(5n+5N+6)M} {6{\sqrt{n+N}}}}\sqrt{{\max{\{0, a+K}\}}}, $

(1.6)

where $M=\sup\limits_{x\in{M}^n}u(x).$

In particular, for $a<0, $ if $a\leq-K, $ then $\max{\{0, a+K\}}=0$. Thus, (1.5) implies $|\nabla{u}|\leq{0}$ whenever $u$ is a bounded positive solution of the nonlinear elliptic (1.1). Hence $u\equiv 1$. Therefore the following Liouville-type result follows.

Corollary 1.3  Let $(M^n, g)$ be an $n$-dimensional complete noncompact Riemannian manifold with ${{\rm Ric}_{f}^{N}}\geq-K, $ where $K$ is a nonnegative constant. Assuming that $u$ is a bounded positive solution of (1.1) with $a<0$. If $a\leq-K, $ then $u\equiv 1$.

In particular, we have the following conclusion

Corollary 1.4   Let $(M^n, g)$ be an $n$-dimensional complete noncompact Riemannian manifold with ${{\rm Ric}_{f}^{N}}\geq0.$ Assuming that $u$ is a bounded positive solution defined of (1.1) with $a<0, $ then $u\equiv 1$.

The above results are obtained under the assumption that ${{\rm Ric}_{f}^{N}}$ is bounded by below. We can also obtain similar results under the assumption that ${{\rm Ric}_{f}}$ is bounded by below.

Theorem 1.5  Let $(M^n, g)$ be an $n$-dimensional complete Riemannian manifold with ${\rm Ric}_{f}(B_P(2R))\geq-(n-1)H, $ and ${|\nabla{f}|}\leq K, $ where $K$ and $H$ is a nonnegative constant. Assuming that $u$ is a positive solution of the nonlinear elliptic eq. (1.1) on $B_p(2R).$ Then on $B_p(R), $ the following inequalities hold

(1) if $a>0, $ then

$ |\nabla{u}|\leq M\sqrt{{\frac{(n+3)^2} {2n}}{\bigg[a+(n-1)H\bigg]} +{\frac{{{(n+2)}{{c}_{1}^{2}}+{[(n-1)(1+{\sqrt{HR}})+KR)]} {{c}_1}+{{c}_2}}}{R^2}}}; $

(1.7)

(2) if $a<0, $ then

$ |\nabla{u}|\leq M\\\sqrt{{\frac{(5n+6)^2} {36n}}{\max{\{0, a+(n-1)H}\}} +{\frac{1}{{R}^2}}{\bigg\{{({\frac{75n+12}{6}})}{{c}_{1}^{2}} +{\bigg[(n-1)(1+\sqrt{HR})+{KR}\bigg]} {{c}_1}+{{c}_2}}\bigg\}}, $

(1.8)

where $M={\sup\limits_{x\in{{{B}_{p}}{(2R)}}}}{u(x)}, $ the ${{c}_1}$ and ${{c}_2}$ is a positive constants.

Corollary 1.6  Let $(M^n, g)$ be an $n$-dimensional complete noncompact Riemannian manifold with ${\rm Ric}_{f}\geq-(n-1)H, $ and ${|\nabla{f}|}\leq K, $ where $K$ and $H$ is a nonnegative constant. Assuming that $u$ is a positive solution of the nonlinear elliptic eq. (1.1) on the following inequalities hold

(1) if $a>0, $ then

$ |\nabla{u}|\leq{\frac{(n+3)M}{\sqrt{2n}}}{\sqrt{a+(n-1)H}}; $

(1.9)

(2) if $a<0, $ then

$ |\nabla{u}|\leq{\frac{(5n+6)M} {6{\sqrt{n}}}}\sqrt{{\max{\{0, a+(n-1)H}\}}}, $

(1.10)

where $M=\sup\limits_{x\in{M}^n}u(x).$

In particular, for $a<0, $ if $a\leq-(n-1)H, $ then $\max{\{0, a+(n-1)H\}}=0$. Thus, (1.9) implies $|\nabla{u}|\leq{0}$ whenever $u$ is a bounded positive solution to (1.1). Hence, that $u\equiv 1$. Therefore, the following Liouville-type result follows

Corollary 1.7  Let $(M^n, g)$ be an $n$-dimensional complete noncompact Riemannian manifold with ${\rm Ric}_{f}\geq-(n-1)H, $ and ${|\nabla{f}|}\leq K, $ where $K$ and $H$ is a nonnegative constant. Assuming that u is a bounded positive solution of (1.1) with $a<0$. If $a\leq-(n-1)H, $ then $u\equiv 1$.

In particular, we have the following conclusion.

Corollary 1.8  Let $(M^n, g)$ be an $n$-dimensional complete noncompact Riemannian manifold with ${{\rm Ric}_{f}}\geq0.$ Assuming that $u$ is a bounded positive solution of (1.1) with $a<0, $ then $u\equiv1$.

Now we are in the position to give the proof of Theorem 1.1. First we recall the following key lemma.

Lemma 2.1 Let $(M^n, g)$ be an $n$-dimensional complete Riemannian manifold with ${\rm Ric}_{f}^{N}(B_P(2R))\geq-K, $ where $K$ is a nonnegative constant. Assuming that $u$ is a positive solution to nonlinear elliptic eq. (1.1) on $B_p(2R).$ Then on $B_p(R), $ the following inequalities hold

(1) If $a>0, $ then

$ {\frac{1}{2}}{\Delta_f {|\nabla{h}|}^2} \geq{\frac{5(n+N)}{18}}{\frac{{|\nabla{h}|}^4}{{h}^2}}- {\frac{(n+N)}{3}}{\frac{\nabla{h}}{h}}{\nabla{({|\nabla{h}|}^2})} -{(a+K)}{{|\nabla{h}|}^2}, $

(2.1)

where $h=u^{\frac{3}{n+N+3}}.$

(2) If $a<0, $ then

$ {\frac{1}{2}}{\Delta_f {|\nabla{h}|}^2} \geq{\frac{37(n+N)}{36}}{\frac{{|\nabla{h}|}^4}{{h}^2}}- {\frac{5(n+N)}{6}}{\frac{\nabla{h}}{h}}{\nabla{({|\nabla{h}|}^2})} -{(a+K)}{{|\nabla{h}|}^2}, $

(2.2)

where $h=u^{\frac{6}{5(n+N)+6}}.$

Proof of Lemma 2.1 Let $h={u}^\epsilon, $ where $\epsilon\neq0$ is a constant to determined. Then we have

$ \log h=\epsilon\log u. $

A simple calculation implies

$ \begin{array}{lll} \Delta_f {h}=\Delta_f {({u}^\epsilon)} &=&{\epsilon}(\epsilon-1){u}^{\epsilon-2}| \nabla{u}|^2+{\epsilon} {u}^{\epsilon-1}\Delta_f {u} \\ &=&{\epsilon}(\epsilon-1){u}^{\epsilon-2}|\nabla{u}|^2-{a} {\epsilon}{u}^\epsilon\log u\\ &=&\frac{\epsilon-1}{\epsilon} {\frac{|\nabla{h}|^2}{h}} -a h \log h.\end{array} $

(2.3)

Therefore we get

$ \begin{array}{lll} \nabla{h}{\nabla}\Delta_f {h} &=&\nabla h{\nabla{\bigg({\frac{\epsilon-1}{\epsilon}} {\frac{{|\nabla{h}|}^2}{h}}-{a}{h}{\log h}\bigg)}} \\ &=&{\frac{\epsilon-1}{\epsilon}}\nabla {h}\bigg(\frac{{\nabla ({|\nabla{h}|}^2)h}-{{|\nabla{h}|}^2}\nabla{h}}{h^2}\bigg) -{a}{\nabla{h}}{\bigg(\nabla{h}{\log h}+h\nabla{\log h}\bigg)}\\ &=&{\frac{\epsilon-1}{\epsilon h}}{\nabla{h}} {\nabla{({|\nabla{h}|}^2)}}-{\frac{\epsilon-1}{\epsilon}} {\frac{{|\nabla{h}|}^4}{{h}^2}}-{a}{h}{\log h} {\frac{{|\nabla{h}|}^2}{h}}-{a}{{|\nabla{h}|}^2}.\end{array} $

(2.4)

Applying (2.3) and (2.4) into the famous Bochner formula to $h$, we have

$ \begin{array}{lll} {\frac{1}{2}}{\Delta_f {|\nabla{h}|}^2} &\geq&{\frac{1}{n+N}}{({\Delta_f {h}})^2}+\nabla{h}{\nabla} {\Delta_f {h}}+{{Ric}_{f}^{N}}({\nabla{h}}, {\nabla{h}})\\ &\geq&{\frac{1}{n+N}}{\bigg(\frac{\epsilon-1}{\epsilon} {\frac{|\nabla{h}|^2}{h}} -a h \log h\bigg)^2}+\nabla{h}{\nabla}\Delta_f {h}-{K}{{|\nabla{h}|}^2}\\ &=&\bigg[\frac{(\epsilon-1)^2}{(n+N){\epsilon^2}}-\frac{\epsilon-1}{\epsilon}\bigg] {\frac{{|\nabla{h}|}^4}{h^2}}-{a{\bigg[{\frac{2(\epsilon-1)} {(n+N)\epsilon}}+1}\bigg]}{h}{\log h}{\frac{{|\nabla{h}|}^2}{h}}\\ &&+{\frac{a^2}{n+N}}{({h}{\log h})^2}+{\frac{\epsilon-1}{\epsilon}} {\frac{\nabla{h}}{h}}{\nabla(|\nabla{h}|^2)}-{(a+K)}{{|\nabla{h}|}^2}.\end{array} $

(2.5)

Now we let

$ {a{\bigg[{\frac{2(\epsilon-1)}{(n+N)\epsilon}}+1}\bigg]}\geq0. $

(2.6)

Then for a fixed point $p$, if there exist a positive constant $ \delta$ such that ${h}{\log h}\leq\delta{\frac{{|\nabla{h}|}^2}{h}}, $ then (2.5) becomes

$ \begin{array}{lll} {\frac{1}{2}}{\Delta_f {|\nabla{h}|}^2} &\geq&\bigg[\frac{(\epsilon-1)^2}{(n+N){\epsilon^2}}-\frac{\epsilon-1}{\epsilon}-{a{\delta} {\bigg({\frac{2(\epsilon-1)}{(n+N)\epsilon}}+1}\bigg)}\bigg]{\frac{{|\nabla{h}|}^4}{h^2}}\\ &+&{\frac{a^2}{n+N}}{({h}{\log h})^2}+{\frac{\epsilon-1}{\epsilon}}{\frac{\nabla{h}}{h}} {\nabla(|\nabla{h}|^2)}-{(a+K)}{{|\nabla{h}|}^2}\\ &\geq&\bigg[\frac{(\epsilon-1)^2}{(n+N){\epsilon^2}}-\frac{\epsilon-1}{\epsilon}-{a{\delta} {\bigg({\frac{2(\epsilon-1)}{(n+N)\epsilon}}+1}\bigg)}\bigg] {\frac{{|\nabla{h}|}^4}{h^2}}\\ &&+{\frac{\epsilon-1}{\epsilon}}{\frac{\nabla{h}}{h}}{\nabla(|\nabla{h}|^2)} -(a+K){|{|\nabla{h}|}^2}.\end{array} $

On the contrary, at the point $p$, if ${h}{\log h}\geq\delta{\frac{{|\nabla{h}|}^2}{h}}, $ then (2.5) becomes

$ \begin{array}{lll} {\frac{1}{2}}{\Delta_f {|\nabla{h}|}^2} &\geq&\bigg[\frac{(\epsilon-1)^2}{(n+N){\epsilon^2}}-\frac{\epsilon-1} {\epsilon}\bigg]{\frac{{|\nabla{h}|}^4}{h^2}}+\bigg[{\frac{{a}^2}{n+N}}-{\frac{a}{\delta}} {{\bigg({\frac{2(\epsilon-1)}{(n+N)\epsilon}}+1}\bigg)}\bigg]{({h}{\log h})^2}\\ &&+{\frac{\epsilon-1}{\epsilon}}{\frac{\nabla{h}}{h}}{\nabla(|\nabla{h}|^2)} -{(a+K)}{{|\nabla{h}|}^2}\\ &\geq&\bigg\{\bigg[\frac{(\epsilon-1)^2}{(n+N){\epsilon^2}}-\frac{\epsilon-1}{\epsilon}\bigg] +{\delta^2}\bigg[{\frac{{a}^2}{n+N}}-{\frac{a}{\delta}}{{\bigg({\frac{2(\epsilon-1)} {(n+N)\epsilon}}+1}\bigg)}\bigg]\bigg\}{\frac{{|\nabla{h}|}^4}{h^2}}\\ &&+{\frac{\epsilon-1}{\epsilon}}{\frac{\nabla{h}}{h}}{\nabla(|\nabla{h}|^2)} -{(a+K)}{{|\nabla{h}|}^2}\\ &\geq&\bigg\{\bigg[\frac{(\epsilon-1)^2}{(n+N){\epsilon^2}}-{\frac{\epsilon-1}{\epsilon}}\bigg] -{a}{\delta}{\bigg[{\frac{2(\epsilon-1)}{(n+N)\epsilon}}+1\bigg]}\bigg\} {\frac{{|\nabla{h}|}^4}{h^2}}\\ &&+{\frac{\epsilon-1}{\epsilon}}{\frac{\nabla{h}}{h}}{\nabla(|\nabla{h}|^2)} -{(a+K)}{{|\nabla{h}|}^2}\end{array} $

(2.7)

as long as

$ {\frac{{a}^2}{n+N}}-{\frac{a}{\delta}} {{\bigg[{\frac{2(\epsilon-1)}{(n+N)\epsilon}}+1}\bigg]}>0. $

(2.8)

In order to obtain the bound of ${|\nabla{h}|}$ by applying the maximum principle to (2.7), it is sufficient to choose the coefficient of $\frac{\mid \nabla h\mid^{4}}{h^2}$ in (2.7) is positive, that is

$ {\bigg[{\frac{{(\epsilon-1)}^2}{{(n+N)}{{\epsilon}^2}} -{\frac{\epsilon-1}{\epsilon}}\bigg]}-{a{\delta}}{\bigg[{\frac{2(\epsilon-1)} {(n+N)\epsilon}}+1}\bigg]}>0. $

(2.9)

Then we divide it into two cases.

Case 1 $a>0.$ In this case, provided $\epsilon\in\bigg({\frac{2}{n+N+2}}, {\frac{6}{{(5-\sqrt{13})}{(n+N)}+6}}\bigg), $ there will exist an $\delta$ satisfying (2.6), (2.8) and (2.9). In particular, we choose

$ \epsilon={\frac{3}{n+N+3}} $

and

$ \delta={\frac{n+N}{2a}}. $

Then (2.7) becomes

$ {\frac{1}{2}}{\Delta_f {|\nabla{h}|}^2} \geq{\frac{5(n+N)}{18}}{\frac{{|\nabla{h}|}^4}{{h}^2}}- {\frac{(n+N)}{3}}{\frac{\nabla{h}}{h}}{\nabla{({|\nabla{h}|}^2})} -{(a+K)}{{|\nabla{h}|}^2}. $

(2.10)

Case 2 $a<0.$ In this case, provided $\epsilon\in\bigg({\frac{6}{{(5+\sqrt{13})}{(n+N)}+6}}, {\frac{2}{n+N+2}}\bigg)$, there will exist an $\delta$ satisfying (2.6), (2.8) and (2.9). In particular, we choose

$ \epsilon={\frac{6}{5(n+N)+6}} $

and

$ \delta={\frac{-3(n+N)}{4a}}. $

Then (2.7) becomes

$ {\frac{1}{2}}{\Delta_f {|\nabla{h}|}^2} \geq{\frac{37(n+N)}{36}}{\frac{{|\nabla{h}|}^4}{{h}^2}}- {\frac{5(n+N)}{6}}{\frac{\nabla{h}}{h}}{\nabla{({|\nabla{h}|}^2})} -{(a+K)}{{|\nabla{h}|}^2}. $

(2.11)

Now we begin to prove Theorem 1.1 which will follow by applying comparison theorems and Bochner formula to an appropriate function $h.$

Proof of Theorem 1.1 We first prove the case of $a>0.$ Let $m$ be a cut-off function such that $m(r)=1$ for $r\leq1, $ $m(r)=0$ for $r\geq2, $ $0\leq{m(r)}\leq1, $ and

$ 0\geq{m^{-\frac{1}{2}}}(r)m^{'}(r)\geq-c_1, ~~m^{''}(r)\geq-c_2 $

for positive constants $c_1$ and $c_2.$ Denote by $\rho(x)=d(x, p)$ the distance between $x$ and $p$ in $(M^n, g).$ Let

$ \phi(x)=m\bigg(\frac{\rho(x)}{R}\bigg). $

Making use of an argument of Calabi [3] (see also Cheng and Yau [5]), we can assume without loss of generality that the function $\phi$ is smooth in $B_p(2R).$ Then we have

$ {\frac{{|\nabla{\phi}|}^2}{\phi}}\leq{\frac{{c_1}^2}{R^2}}. $

(2.12)

It was shown by Qian [11] that

$ \Delta_f(\rho^2)\leq{n\bigg\{1+\sqrt{1+{\frac{4K\rho^2}{n}}}\bigg\}}. $

Hence we have

$ \begin{array}{lll} \Delta_f{\rho} &=&{\frac{1}{2\rho}\bigg(\Delta_f(\rho^2)-2{|\nabla{\rho}|}^2\bigg)}\leq{\frac{n-2}{2\rho}}+{\frac{n}{2\rho}}\bigg(1+\sqrt{1+{\frac{4K\rho^2}{n}}}\bigg)\\ &=&\frac{n-1}{\rho}+\sqrt{nK}. \end{array} $

It follows that

$ \begin{array}{lll} \Delta_f{\phi} &=&{\frac{m^{''}(r)|\nabla{\rho}|^2}{R^2}}+{\frac{m^{'}(r)\Delta_f{\rho}}{R}}\\ &\geq&-{\frac{(n-1+\sqrt{nK}{R}){c_1}+{c_2}}{R^2}}.\\\end{array} $

(2.13)

Define $G=\phi{{|\nabla{h}|}^2}, $ we will use the maximum principle for $G$ on ${B}_{p}(2R).$ Assume $G$ achieves its maximum at the point $x_0\in{{{B}_{p}}{(2R)}}$ and assume $G(x_0)>0$ (otherwise this is obvious). Then at the point $x_0$, it holds that

$ \Delta_f {G}\leq0, \ \ \ \ {\nabla{({|\nabla{h}|}^2})} =-{\frac{{|\nabla{h}|}^2}{\phi}}{\nabla{\phi}}. $

Using (2.1) in Lemma 2.1, we obtain

$ \begin{array}{lll}0 &\geq&\Delta_f {G}\\ &=&\phi\Delta_f {({|\nabla{h}|}^2)}+{|\nabla{h}|}^2{\Delta_f {\phi}} +2\nabla{\phi} {\nabla{({|\nabla{h}|}^2})}\\ &=&\phi\Delta_f {({|\nabla{h}|}^2)}+{\frac{\Delta_f {\phi}}{\phi}}{G} -2{\frac{{|\nabla{\phi}|}^2}{\phi^2}}{\nabla{\phi}}\\ &\geq&2\phi\bigg[{\frac{5(n+N)}{18}}{\frac{{|\nabla{h}|}^4}{{h}^2}}- {\frac{n+N}{3}}{\frac{\nabla{h}}{h}}{\nabla{({|\nabla{h}|}^2})} -{(a+k)}{{|\nabla{h}|}^2}\bigg] +{\frac{\Delta_f {\phi}}{\phi}}{G} -2{\frac{{|\nabla{\phi}|}^2}{\phi^2}}{\nabla{\phi}}\\ &=&{\frac{5(n+N)}{9}}{\frac{G^2}{\phi{h}^2}}+ {\frac{2(n+N)G}{3\phi}}{\nabla{\phi}}{\frac{\nabla{h}}{h}}-{2(a+K)}{G} +{\frac{\Delta_f {\phi}}{\phi}}{G}-2{\frac{{|\nabla{\phi}|}^2}{\phi^2}}{G}, \end{array} $

(2.14)

where the second inequality used (2.10). Multiplying both sides of (2.14) by ${\frac{\phi}{G}}, $ we obtain

$ {\frac{5(n+N)}{9}}{\frac{G}{h}^2}\leq -{\frac{2(n+N)}{3}}{\nabla{\phi}}{\frac{\nabla{h}}{h}}+{2\phi(a+K)}- \Delta_f {\phi}+2{\frac{{|\nabla{\phi}|}^2}{\phi}}. $

(2.15)

Then using the Cauchy inequality, we have

$ \begin{array}{lll} -{\frac{2(n+N)}{3}}{\nabla{\phi}}{\frac{\nabla{h}}{h}} &\leq&{\frac{2(n+N)}{3}}{|\nabla{\phi}|}{\frac{|\nabla{h}|}{h}}\leq{\frac{(n+N)}{3\varepsilon}}{\frac{|\nabla{\phi}|^2}{\phi}} +{\frac{(n+N)\varepsilon}{3{h^2}}}{\phi}|\nabla{h}|^2\\ &=&{\frac{(n+N)}{3\varepsilon}}{\frac{|\nabla{\phi}|^2}{\phi}} +{\frac{(n+N)\varepsilon}{3{h^2}}}{G}, \end{array} $

where $\varepsilon\in({0}, {\frac{5}{3}})$ is a positive constant. Taking the above inequality into (2.15), we have

$ \begin{array}{lll} {\frac{(5-3\varepsilon)(n+N)}{9}}{\frac{G}{h^2}} &\leq&{2\phi(a+K)}-\Delta_f {\phi}+(2+\frac{n+N} {3\varepsilon}){\frac{{|\nabla{\phi}|}^2}{\phi}}\\ &\leq&{2(a+K)}-\Delta_f {\phi}+(2+\frac{n+N}{3\varepsilon}) {\frac{{|\nabla{\phi}|}^2}{\phi}}.\end{array} $

(2.16)

In particular, choosing $\varepsilon={\frac{1}{3}}$ in (2.16) and using (2.12) and (2.13), we have

$ \begin{array}{lll}{\frac{4(n+N)G}{9{h^2}}} &\leq&{2(a+K)}-\Delta_f {\phi}+(n+N+2){\frac{{|\nabla{\phi}|}^2}{\phi}}\\ &\leq&2(a+K)+{\frac{{{(n+N+2)}{{c}_{1}^{2}} +{(n-1+{\sqrt{{n}{K}}}{R})} {{c}_1}+{{c}_2}}}{R^2}}. \end{array} $

So for $x_0\in{{{B}_{p}}{(R)}}, $ we have

$ \begin{array}{lll}{\frac{4(n+N)}{9}}{G(x)} &\leq&{\frac{4(n+N)}{9}}{G(x_0)}\\ &\leq&h^2(x_0)\bigg[\frac{(n+N+3)^2}{2(n+N)}(a+K)+{\frac{{{(n+N+2)}{{c}_{1}^{2}} +{(n-1+{\sqrt{{n}{K}}}{R})} {{c}_1}+{{c}_2}}}{R^2}}\bigg]. \end{array} $

This shows

$ {|\nabla{u}|}^2{(x)}\leq {M}^2{\bigg[{\frac{(n+N+3)^2} {2(n+N)}}{(a+K)}+{\frac{{{(n+N+2)}{{c}_{1}^{2}} +{(n-1+{\sqrt{{n}{K}}}{R})} {{c}_1}+{{c}_2}}}{R^2}}\bigg]} $

and

$ |\nabla{u}|\leq M\sqrt{{\frac{(n+N+3)^2} {2(n+N)}} {(a+K)}+{\frac{{{(n+N+2)}{{c}_{1}^{2}} +{(n-1+{\sqrt{{n}{K}}}{R})} {{c}_1}+{{c}_2}}}{R^2}}}, $

where $M={\sup\limits_{x\in{{{B}_{p}}{(2R)}}}}{u(x)}.$ This yields the desired inequality (1.3) of Theorem 1.1.

Next, we prove the case $a<0.$ Define $\overline{G}=\phi{{|\nabla{h}|}^2}, $ we will use the maximum principle for $\overline{G}$ on ${{B}_{p}}{(2R)}.$ Assume $\overline{G}$ achieves its maximum at the point $\overline{x_0}\in{{B}_{p}}{(2R)}$ and assume $\overline{G}(\overline{x_0})>0$ (otherwise this is obvious). Then at the point $\overline{x_0}, $ it holds that

$ \Delta_f {\overline{G}}\leq0, \ \ \ {\nabla{({|\nabla{h}|}^2})} =-{\frac{{|\nabla{h}|}^2}{\phi}}{\nabla{\phi}}. $

In a similar way as the case $a>0$, we have

$ \begin{array}{lll}0 &\geq&\Delta_f {\overline{G}}\\ &=&\phi\Delta_f {({|\nabla{h}|}^2)}+{\frac{\Delta_f {\phi}}{\phi}}{\overline{G}} -2{\frac{{|\nabla{\phi}|}^2}{\phi^2}}{\overline{G}}\\ &\geq&2\phi\bigg[{\frac{37(n+N)}{36}}{\frac{{|\nabla{h}|}^4}{{h}^2}}- {\frac{5(n+N)}{6}}{\frac{\nabla{h}}{h}}{\nabla{({|\nabla{h}|}^2})}-{(a+K)}{{|\nabla{h}|}^2}\bigg]+{\frac{\Delta_f {\phi}}{\phi}}{\overline{G}} -2{\frac{{|\nabla{\phi}|}^2}{\phi^2}}{\overline{G}}\\ &=&{\frac{37(n+N)}{18}}{\frac{{\overline{G}}^2}{\phi{h}^2}}+ {\frac{5(n+N)\overline{G}}{3\phi}}{\nabla{\phi}}{\frac{\nabla{h}}{h}}-2(a+K)\overline{G}+{\frac{\Delta_f {\phi}}{\phi}}{\overline{G}} -2{\frac{{|\nabla{\phi}|}^2}{\phi^2}}{\overline{G}}, \end{array} $

(2.17)

where the second inequality used (2.11). Multiplying both sides of (2.17) by ${\frac{\phi}{\overline{G}}}$, we obtain

$ {\frac{37(n+N)}{18}}{\frac{\overline{G}^2}{{h}^2}} \leq -\frac{5(n+N)}{3}{\nabla{\phi}}{\frac{\nabla{h}}{h}}+2\phi(a+K) -\Delta_f {\phi}+2{\frac{{|\nabla{\phi}|}^2}{\phi}}. $

(2.18)

Using Cauchy inequality, we can get

$ \begin{array}{lll} -\frac{5(n+N)}{3}{\nabla\phi}{\frac {\nabla h}{h}} \leq\frac{5(n+N)}{3}{|\nabla\phi|}{\frac {|\nabla h|}{h}}\leq{\frac{5(n+N)}{6\varepsilon}}{\frac{|\nabla{\phi}|^2}{\phi}} +{\frac{5(n+N)\varepsilon}{6{h^2}}}{\overline{G}}, \end{array} $

where $\varepsilon\in({0}, {\frac{37}{15}})$ is a positive constant. Taking the above inequality into (2.18) gives

$ \begin{array}{lll} {\frac{(37-15\varepsilon)(n+N)}{18}}{\frac{\overline{G}}{h^2}} &\leq&{2\phi(a+K)}-\Delta_f {\phi}+\bigg(2+\frac{5(n+N)} {6\varepsilon}\bigg){\frac{{|\nabla{\phi}|}^2}{\phi}}\\ &\leq&2\max{\{0, a+K}\}-\Delta_f {\phi}+\bigg(2+\frac{5(n+N)} {6\varepsilon}\bigg){\frac{{|\nabla{\phi}|}^2}{\phi}}.\end{array} $

Hence, choosing $\varepsilon=\frac{1}{15}$ in (2.16) and using (2.12) and (2.13), we obtain

$ {\frac{2(n+N)\overline{G}^2}{h^2}} \leq 2\max{\{0, a+K}\}+{\frac{1}{{R}^2}}{\bigg[{({\frac{75n+75N+12}{6}})}{{c}_{1}^{2}} +{(n-1+{\sqrt{{n}{K}}}{R})} {{c}_1}+{{c}_2}}\bigg]. $

Therefore, it holds on ${{B}_{p}}{(R)}, $

$ |\nabla{u}|\leq {M}\sqrt{{\frac{(5n+5N+6)^2} {36(n+N)}}{\max{\{0, a+K}\}} +{\frac{1}{{R}^2}}{\bigg[{({\frac{75n+75N+12}{6}})} {{c}_{1}^{2}} +{(n-1+{\sqrt{{n}{K}}}{R})} {{c}_1}+{{c}_2}}\bigg]}. $

This concludes the proof of inequality (1.4) of Theorm 1.1.

Now we are in the position to give a brief proof of Theorem 1.5.

Skept of the Proof of Theorem 1.5 Noticing that we have the following Bochner formula to $h$ with ${\rm Ric}_{f}$,

$ {\frac{1}{2}}{\Delta_f {|\nabla{h}|}^2} ={|\nabla^2{h}|}^2+\nabla{h}{\nabla}{\Delta_f {h}}+{Ric}_{f} ({\nabla{h}}, {\nabla{h}}), $

then (2.5) becomes

$ \begin{array}{lll} {\frac{1}{2}}{\Delta_f {|\nabla{h}|}^2} &=&{|\nabla^2{h}|}^2+\nabla{h}{\nabla}{\Delta_f {h}}+{Ric}_{f}({\nabla{h}}, {\nabla{h}})\\ &\geq&{\frac{1}{n}}{\bigg(\frac{\epsilon-1}{\epsilon} {\frac{|\nabla{h}|^2}{h}}-a h \log h\bigg)^2} +\nabla{h}{\nabla}\Delta_f {h}-{(n-1)H}{{|\nabla{h}|}^2}\\ &=&\bigg[\frac{(\epsilon-1)^2}{n{\epsilon^2}}-\frac{\epsilon-1}{\epsilon}\bigg] {\frac{{|\nabla{h}|}^4}{h^2}}-{a{\bigg[{\frac{2(\epsilon-1)}{n\epsilon}}+1}\bigg]} {h}{\log h}{\frac{{|\nabla{h}|}^2}{h}}\\ &&+{\frac{a^2}{n}}{({h}{\log h})^2}+{\frac{\epsilon-1}{\epsilon}}{\frac{\nabla{h}}{h}} {\nabla(|\nabla{h}|^2)}-{\bigg[a+(n-1)H\bigg]}{{|\nabla{h}|}^2}.\end{array} $

Moreover, the comparison theorem holds true in the following form (see Theorem 1.1 in [12]): if ${\rm Ric}_f\geq-K$ and $|\nabla{f}|\leq K, $ we have

$ \Delta_f {\rho}\leq (m-1)\sqrt{H}\coth(\sqrt{H}\rho)+K. $

Hence $\Delta_f {\rho}\leq (m-1)(\frac{1}{\rho}+\sqrt{H})+K.$ So (2.12) and (2.13) also hold true in almost the same forms

$ {\frac{{|\nabla{\phi}|}^2}{\phi}}\leq{\frac{{c_1}^2}{R^2}} $

and

$ -\Delta_f {\phi}\leq{\frac{[(n-1)(1+\sqrt{HR})+{KR}]{c_1}+{c_2}}{R^2}}. $

Noticing the above facts, the proof of Theorem 1.5 is the same to that of Theorem 1.1, so we omit it here.