#  $f$-biharmonic Riemannian submersions from 3-manifolds

Ze-Ping Wang∗ and Li-Hua Qin  School of Mathematical Sciences,
Guizhou Normal University,
Guiyang 550025,
People’s Republic of China
<EMAIL_ADDRESS>(Wang)

<EMAIL_ADDRESS>(Qin)

(Date: 2/1/2024)

#  $f$-Biharmonic submanifolds in space forms and $f$-biharmonic Riemannian
submersions from 3-manifolds

Ze-Ping Wang∗ and Li-Hua Qin  School of Mathematical Sciences,
Guizhou Normal University,
Guiyang 550025,
People’s Republic of China
<EMAIL_ADDRESS>(Wang)

<EMAIL_ADDRESS>(Qin)

(Date: 2/1/2024)

###### Key words and phrases:

Biharmonic maps, $f$-biharmonic maps, Riemannian submersions, $f$-biharmonic
curves, $f$-biharmonic submanifolds.

###### 1991 Mathematics Subject Classification:

58E20, 53C12

*Supported by the Scientific and Technological Project in Guizhou Province ( Grant no. Qiankehe Platform Talents [2018]5769-04) and by the NSFC (No. 11861022).

## Abstract

> $f$-Biharmonic maps are generalizations of harmonic maps and biharmonic
> maps. In this paper, we obtain some descriptions of $f$-biharmonic curves in
> a space form. We also obtain a complete classification of proper
> $f$-biharmonic isometric immersions of a developable surface in
> $\mbox{${\mathbb{R}}$}^{3}$ by proving that a proper $f$-biharmonic
> developable surface exists only in the case where the surface is a cylinder.
> Based on this, we show that a proper biharmonic conformal immersion of a
> developable surface into $\mbox{${\mathbb{R}}$}^{3}$ exists only in the case
> when the surface is a cylinder. Riemannian submersions can be viewed as the
> dual notion of isometric immersions (i.e., submanifolds). We also study
> $f$-biharmonicity of Riemannian submersions from 3-space forms by using the
> integrability data. Examples are given of proper $f$-biharmonic Riemannian
> submersions and $f$-biharmonic surfaces and curves.

##  1\. Introduction and Preliminaries

All manifolds, maps, tensor fields studied in this paper are assumed to be
smooth unless there is an otherwise statement.

A biharmonic map is a map $\varphi:(M,g)\longrightarrow(N,h)$ between
Riemannian manifolds that is a critical point of the bienergy

$E^{2}\left(\varphi,\Omega\right)=\frac{1}{2}{\int}_{\Omega}\left|\tau(\varphi)\right|^{2}v_{g}$

for every compact subset $\Omega$ of $M$, where $\tau(\varphi)={\rm
Trace}_{g}\nabla{\rm d}\varphi$ is the tension field of $\varphi$ vanishing of
which means the map is harmonic. By computing the first variation of the
functional (see [6]) we see that $\varphi$ is biharmonic if and only if its
bitension field vanishes identically, i.e.,

(1) $\tau_{2}(\varphi):={\rm
Trace}_{g}(\nabla^{\varphi}\nabla^{\varphi}-\nabla^{\varphi}_{\nabla^{M}})\tau(\varphi)-{\rm
Trace}_{g}R^{N}({\rm d}\varphi,\tau(\varphi)){\rm d}\varphi=0,$

where $R^{N}$ is the curvature operator of $(N,h)$ defined by

$R^{N}(X,Y)Z=[\nabla^{N}_{X},\nabla^{N}_{Y}]Z-\nabla^{N}_{[X,Y]}Z.$

$f$-biharmonic maps and their equations: $f$-biharmonic maps are critical
points of the $f$-bienergy functional for maps $\varphi:(M,g)\to(N,h)$ between
Rie- mannian manifolds:

$E_{2,f}(\varphi)=\int_{\Omega}f|\tau(\varphi)|^{2}v_{g},$

where $\Omega$ is a compact domain of $M$ and $f:M\to(0,+\infty)$. The Euler-
Lagrange equation gives the $f$-biharmonic map equation ( see e.g., [9, 11])

(2)
$\tau_{2,f}(\varphi)=-J^{\varphi}(f\tau(\varphi))=f\tau_{2}(\varphi)+(\Delta
f)\tau(\varphi)+2\nabla^{\varphi}_{{\rm grad}f}\tau(\varphi)=0,$

where $\tau(\varphi)$ and $\tau_{2}(\varphi)$ are the tension and bitension
fields of $\varphi$ respectively, $J^{\varphi}$ is the Jacobi operator of the
map defined by $J^{\varphi}(X)=-\\{{\rm
Trace}_{g}(\nabla^{\varphi}\nabla^{\varphi}-\nabla^{\varphi}_{\nabla^{M}})X-{\rm
Trace}_{g}R^{N}({\rm d}\varphi,X){\rm d}\varphi\\}$. Clearly, both harmonic
maps and biharmonic maps are $f$-biharmonic. We also have the following
inclusion relations

$\\{Harmonic\;\;maps\\}\;\subset\;\\{Biharmonic\;\;maps\\}$$\;\subset\;\\{f$-$biharmonic\;\;maps\\}.$

A submanifold in a Riemannian manifold is called an $f$-biharmonic submanifold
if the isometric immersion defining the submanifold is an $f$-biharmonic map.
Similarly, a Riemannian submersion is called an $f$-biharmonic Riemannian
submersion if the Riemannian submersion is an $f$-biharmonic map. We call an
$f$-biharmonic map (respectively, submaniold, Riemannian submersion ) which is
not biharmonic a proper $f$-biharmonic map (respectively, submanifold,
Riemannian submersion ).

$f$-Biharmonic map was first introduced in [9]. Later, $f$-biharmonic
submanifolds were studied in [11] where the author obtained the $f$-biharmonic
curve equation in a space form and also gave a complete classification of
$f$-biharmonic curves in $\mbox{${\mathbb{R}}$}^{3}$. The paper [15] shows
that neither of a circular cone and a part of the standard sphere $S^{2}$ in
$\mbox{${\mathbb{R}}$}^{3}$ is $f$-biharmonic for any $f$, and a constant mean
curvature surface in $\mbox{${\mathbb{R}}$}^{3}$ is $f$-biharmonic if and only
if it is a part of a plane or a circular cylinder. For some recent progress on
$f$-biharmonic submanifolds, we refer the readers to [11, 12, 14] and the
references therein. We will focus to classify (or construct) proper
$f$-biharmonic submanifolds in a space form.

As the dual notion of biharmonic isometric immersions (i.e., biharmonic
submanifolds), biharmonic Riemannian submersions were first studied by using
the integrability data of an adapted frame of the Riemannian submersion in
[15] where the authors showed that a biharmonic Riemannian submersion from a
3-space form into a surface has to be harmonic and also constructed a family
of proper biharmonic Riemannian submersions from $\mbox{${\mathbb{R}}$}^{3}$
provided with a warped product metric. Following [15], we study $f$-biharmonic
Riemannian submersions from 3-manifolds by using the integrability data.
However, it is extremely difficult to find a proper $f$-biharmonic Riemannian
submersion since the function $f$ is a solution of $PDE$. We will focus to
construct examples of proper $f$-biharmonic Riemannian submersions.

In this paper, we first derive $f$-biharmonic curve equation in a general
Riemannian manifold. We then obtain some characterizations of $f$-biharmonic
curves in a space form by giving the explicit functions $f$ and the explicit
curvatures of the curves. For $f$-biharmonic surfaces in
$\mbox{${\mathbb{R}}$}^{3}$, we obtain a complete classification of proper
$f$-biharmonic isometric immersions of a developable surface in
$\mbox{${\mathbb{R}}$}^{3}$ by proving that a proper $f$-biharmonic
developable surface exists only in the case when the surface is a cylinder. An
interesting result is that a cylinder whose directrix takes a proper
$\bar{f}$-biharmonic curve in a 2-space form $N^{2}(K)$ of constant Gauss
curvature $K$ has to be a proper $f$-biharmonic cylinder for
$f=\frac{\psi(v,K)}{c_{1}}\bar{f}$, where $\psi(v,K)$ given by (31) and a
constant $c_{1}>0$. We then construct many examples of proper $f$-biharmonic
cylinders in $\mbox{${\mathbb{R}}$}^{3}$. Based on this, we show that a proper
biharmonic conformal immersion of a developable surface into
$\mbox{${\mathbb{R}}$}^{3}$ exists only in the case where the surface is a
cylinder. Finally, we derive $f$-biharmonic equation for Riemannian
submersions from 3-manifolds by using the integrability data. We then use it
to study $f$-biharmonic Riemannian submersions from a 3-space form. Examples
are given of proper $f$-biharmonic Riemannian submersions.

##  2\.  $f$-Biharmonic submanifolds in space forms

In this section, we characterize $f$-biharmonic curves in a space form by
using the explicit functions $f$ and the explicit curvatures of the curves.
Many examples of proper $f$-biharmonic curves in $\mbox{${\mathbb{R}}$}^{n}$
are obtained. We also give complete classifications of $f$-biharmonic
isometric immersions and biharmonic conformal immersions of a developable
surface into $\mbox{${\mathbb{R}}$}^{3}$. An interesting result is that a
cylinder whose directrix takes a proper $\bar{f}$-biharmonic curve in a
2-space form $N^{2}(K)$ of constant Gauss curvature $K$ has to be a proper
$f$-biharmonic cylinder for $f=\frac{\psi(v,K)}{c_{1}}\bar{f}$, where
$\psi(v,K)$ given by (31) and a constant $c_{1}>0$. Based on this, one
constructs infinitely many examples of proper $f$-biharmonic cylinders and
biharmonic conformal immersions of cylinders into $\mbox{${\mathbb{R}}$}^{3}$

###  2.1. $f$-Biharmonic curves in space forms

A Frenet frame $\\{F_{i}\\}_{i=1,2,\ldots,n}$ associated to an arclength
parametrized curve $\gamma:(a,b)\to(N^{n},h)$ (see e.g., [8]) is an
orthonormal frame which can be described by

$\begin{cases}F_{1}=d\gamma(\frac{\partial}{\partial s})=\gamma^{\prime},\\\
\nabla_{F_{1}}F_{1}=k_{1}F_{2},\\\
\nabla_{F_{1}}F_{i}=-k_{i-1}F_{i-1}+k_{i}F_{i+1},\forall\;i=2,\;3,\;\ldots,\;n-1,\\\
\nabla_{F_{1}}F_{n}=-k_{n-1}F_{n-1},\\\ \end{cases}$

where the functions $\\{k_{1},k_{2},\ldots,k_{n-1}\\}$ are called the
curvatures of $\gamma$. It is well known (see [11]) that the curve is an
$f$-biharmonic curve with a function $f:(a,b)\to(0,+\infty)$ if and only if

(3)
$f(\nabla_{\gamma^{\prime}}^{N}\nabla_{\gamma^{\prime}}^{N}\nabla_{\gamma^{\prime}}^{N}\gamma^{\prime}-R^{N}(\gamma^{\prime},\nabla_{\gamma^{\prime}}^{N}\gamma^{\prime})\gamma^{\prime})+2f^{\prime}\nabla_{\gamma^{\prime}}^{N}\nabla_{\gamma^{\prime}}^{N}\gamma^{\prime}+f^{\prime\prime}\nabla_{\gamma^{\prime}}^{N}\gamma^{\prime}=0.$

With respect to the Frenet frame, the equation (3) can be written as

######  Lemma 2.1.

Let $\gamma:(a,b)\to(N^{n},h)(n\geq 2)$ be a curve parametrized by arc length
into a Riemannian manifold. Then $\gamma$ is $f$-biharmonic if and only if :

(4)
$\left\\{\begin{array}[]{rl}-3\kappa_{1}\kappa^{\prime}_{1}-2\kappa_{1}^{2}f^{\prime}/f=0,\\\
\kappa_{1}^{\prime\prime}-\kappa_{1}\kappa_{2}^{2}-\kappa_{1}^{3}+\kappa_{1}R^{N}(F_{1},F_{2},F_{1},F_{2})+\kappa^{\prime}_{1}f^{\prime\prime}/f+2\kappa^{\prime}_{1}f^{\prime}/f=0,\\\
2\kappa_{1}\kappa_{2}+\kappa_{1}\kappa^{\prime}_{2}+\kappa_{1}R^{N}(F_{1},F_{2},F_{1},F_{3})+2\kappa_{1}\kappa_{2}f^{\prime}/f=0,\\\
\kappa_{1}\kappa_{2}\kappa_{3}+\kappa_{1}R^{N}(F_{1},F_{2},F_{1},F_{4})=0,\\\
k_{1}R(F_{1},F_{2},F_{1},F_{j})=0,\;\;\;\;\;\;j=5,\ldots,n.\end{array}\right.$

###### Proof.

A straightforward computation gives

(5)
$\begin{array}[]{lll}\tau(\gamma)=\nabla_{\gamma^{\prime}}^{N}{\gamma^{\prime}}=\kappa_{1}F_{2},\\\
\nabla_{\gamma^{\prime}}^{N}\nabla_{\gamma^{\prime}}^{N}{\gamma^{\prime}}=-\kappa_{1}^{2}F_{1}+\kappa^{\prime}_{1}F_{2}+\kappa_{1}\kappa_{2}F_{3},\\\
\tau_{2}(\gamma)=\nabla_{\gamma^{\prime}}^{N}\nabla_{\gamma^{\prime}}^{N}\nabla_{\gamma^{\prime}}^{N}{\gamma^{\prime}}-R^{N}(\gamma^{\prime},\nabla_{\gamma^{\prime}}^{N}{\gamma^{\prime}})\gamma^{\prime}\\\
=-3\kappa_{1}\kappa^{\prime}_{1}F_{1}+(\kappa^{\prime\prime}_{1}-\kappa_{1}\kappa_{2}^{2}-\kappa_{1}^{3}+\kappa_{1}R^{N}(F_{1},F_{2},F_{1},F_{2}))F_{2}\\\
+(2\kappa^{\prime}_{1}\kappa_{2}+\kappa_{1}\kappa^{\prime}_{2}+\kappa_{1}R^{N}(F_{1},F_{2},F_{1},F_{3}))F_{3}\\\
+(\kappa_{1}\kappa_{2}\kappa_{3}+\kappa_{1}R^{N}(F_{1},F_{2},F_{1},F_{4}))F_{4}+\sum\limits_{j=5}^{n}\kappa_{1}R^{N}(F_{1},F_{2},F_{1},F_{j})F_{j}.\\\
\end{array}$

Substituting (5) into $f$-biharmonic curve equation (3) and comparing the
coefficients of both sides we obtain (4). From which the lemma follows. ∎

Applying Lemma 2.1, we have

######  Proposition 2.2.

(see [11]) An arclength parametrized curve $\gamma:(a,b)\to N^{n}(C)$ in an
$n$-dimensional space form is an $f$-biharmonic curve if and only if one of
the following cases happens:
(i) $\kappa_{2}=0,f=c_{1}\kappa_{1}^{-3/2}$ and the curvature $\kappa_{1}$
solves the following $ODE$

(6) $3\kappa^{\prime
2}_{1}-2\kappa_{1}\kappa^{\prime\prime}_{1}=4\kappa_{1}^{2}(\kappa_{1}^{2}-C).$

(ii) $\kappa_{2}\neq
0,\;\kappa_{3}=0,\;\kappa_{2}/\kappa_{1}=c_{3},\;f=c_{1}\kappa_{1}^{-3/2}$ and
the curvature $\kappa_{1}$ solves the following $ODE$

(7) $3\kappa^{\prime
2}_{1}-2\kappa_{1}\kappa^{\prime\prime}_{1}=4\kappa_{1}^{2}[(1+c_{3}^{2})\kappa_{1}^{2}-C].$

It is critical to solve the $ODEs$ (6) and (7), to describe $f$-biharmonic
curves in space forms. So we need the following proposition.

######  Proposition 2.3.

For constants $A$ and $C$, solving the following $ODE$

(8) $3y^{\prime 2}-2yy^{\prime\prime}=4y^{2}(Ay^{2}-C),$

we obtain all nonconstant solutions as

$y=\begin{cases}\frac{1}{C_{1}e^{2\sqrt{-C}s}+C_{2}e^{-2\sqrt{-C}s}\pm\sqrt{4C_{1}C_{2}+\frac{A}{C}}},\;\;\;\;\;\;\;\;{\rm
for}\;C<0,\\\
\frac{4C_{1}}{16A+C_{1}^{2}(s+C_{2})^{2}},\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm
for}\;C=0,\\\
\frac{1}{C_{1}\cos(2\sqrt{C}s)+C_{2}\sin(2\sqrt{C}s)\pm\sqrt{C_{1}^{2}+C_{2}^{2}+\frac{A}{C}}},\;{\rm
for}\;C>0,\end{cases}$

where $C_{1}$, $C_{2}$ are constants and $C_{1}^{2}+C_{2}^{2}\neq 0$.

###### Proof.

First of all, one sees that $y=0$ is a solution of (8). For $AC>0$, it is easy
to check that (8) has constant solutions $y=\pm\sqrt{\frac{C}{A}}$.

From now on, we only need to consider that $y(s)$ is a nonconstant solution of
(8). Putting $y=u^{-1}$ and substituting this into (8), we have

(9) $u^{\prime 2}-2uu^{\prime\prime}=4(Cu^{2}-A).$

Setting $p=u^{\prime}=\frac{du}{ds}$, then we have
$p\frac{dp}{du}=u^{\prime\prime}(s)$. Denoting by $\frac{dp}{du}=p^{\prime}$,
then (9) turns into

(10) $(p^{2})^{\prime}-\frac{1}{u}p^{2}+4Cu-4A/u=0,$

which is solved by $p^{2}=-4Cu^{2}+Bu-4A.$ This, together with $u^{\prime}=p$,
implies that

(11) $u^{\prime 2}=-4Cu^{2}+Bu-4A,$

where $B$ is a constant.
We take the derivative of both sides of (11) with respect to $s$ and simplify
the resulting equation to obtain

(12) $u^{\prime\prime}=-4Cu+\frac{1}{2}B.$

For the case $C>0$, we solve (12) to obtain the general solution as

(13)
$\begin{array}[]{lll}u=C_{1}\cos(2\sqrt{C}s)+C_{2}\sin(2\sqrt{C}s)+\frac{B}{8C},\end{array}$

where $C_{1},C_{2}$ are constants. Substituting (13) into (11) and simplifying
the resulting equation we have

(14) $\begin{array}[]{lll}B^{2}=64AC+64C^{2}(C_{1}^{2}+C_{2}^{2}),\end{array}$

which implies that

$\begin{array}[]{lll}u=C_{1}\cos(2\sqrt{C}s)+C_{2}\sin(2\sqrt{C}s)\pm\sqrt{\frac{A}{C}+(C_{1}^{2}+C_{2}^{2})},\\\
({\rm
and\;hence})\;y=\frac{1}{C_{1}\cos(2\sqrt{C}s)+C_{2}\sin(2\sqrt{C}s)\pm\sqrt{\frac{A}{C}+C_{1}^{2}+C_{2}^{2}}},\end{array}$

where constant $8C\sqrt{\frac{A}{C}+C_{1}^{2}+C_{2}^{2}}=\pm B$.

In a similar way, we solve (8) for $C=0$ and $C<0$ respectively, to obtain
$y=\frac{4C_{1}}{16A+C_{1}^{2}(s+C_{2})^{2}}$ and
$y=\frac{1}{C_{1}e^{2\sqrt{-C}s}+C_{2}e^{-2\sqrt{-C}s}\pm\sqrt{4C_{1}C_{2}+\frac{A}{C}}}$
respectively, where $C_{1}$ and $C_{2}$ are constants.

Summarizing all results above we obtain the proposition. ∎

######  Remark 1.

Hereafter, $c_{1}>0$, $C$, $c_{3}$, $C_{3}>0$, $C_{4}$, $C_{5}$, $C_{1}>0$,
$C_{2}>0$ and $4C_{1}C_{2}+\frac{1+c_{3}^{2}}{C}>0$ are assumed to be constant
unless it is otherwise stated. It is convenient to introduce the following new
function

(15)
$\chi(s,c_{3},C)=\begin{cases}\frac{1}{C_{1}e^{2\sqrt{-C}s}+C_{2}e^{-2\sqrt{-C}s}+\sqrt{4C_{1}C_{2}+\frac{1+c_{3}^{2}}{C}}},\;\;\;\;\;\;\;\;{\rm
for}\;C<0,\\\
\frac{4C_{3}}{16(1+c_{3}^{2})+C_{3}^{2}(s+C_{4})^{2}},\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm
for}\;C=0,\\\
\frac{1}{C_{4}\cos(2\sqrt{C}s)+C_{5}\sin(2\sqrt{C}s)+\sqrt{\frac{1+c_{3}^{2}}{C}+(C_{4}^{2}+C_{5}^{2})}},\;{\rm
for}\;C>0.\end{cases}$

As an application of Proposition 2.3, we now give a characterization of a
proper $f$-biharmonic curve in $N^{n}(C)$.

######  Theorem 2.4.

An arclength parametrized curve $\gamma:(a,b)\to N^{n}(C)$ in an
$n$-dimensional space form is a proper $f$-biharmonic curve if and only if one
of the following cases happens:
$(i)$: $\gamma$ is a curve with $\kappa_{2}=0$, $\kappa_{1}=\chi(s,0,C)$ and
$f=c_{1}\kappa_{1}^{-3/2}$, or
$(ii)$: $\gamma$ is a curve with
$\kappa_{3}=0,\;\kappa_{2}/\kappa_{1}=c_{3}\neq 0$,
$\kappa_{1}=\chi(s,c_{3},C)$ and $f=c_{1}\kappa_{1}^{-3/2}$.

###### Proof.

Applying Proposition 2.2 and Proposition 2.3 with $A=1$ or $A=1+c_{3}^{2}$ to
(8), we immediately obtain the theorem. ∎

######  Remark 2.

From Theorem 2.4, we can give the explicit function $f$ for any proper
$f$-biharmonic curve in a space form, in a sense, our result recovers Theorem
4.2 and 4.4 in [11].

By Theorem 2.4, we see that a proper $f$-biharmonic curve in
$\mbox{${\mathbb{R}}$}^{3}$ is either a planar curve or a general helix.

######  Corollary 2.5.

( [11]) An arclength parametrized curve
$\gamma:(a,b)\to\mbox{${\mathbb{R}}$}^{3}$ is a proper $f$-biharmonic curve if
and only if one of the following cases happens:
$(i)$ $\gamma$ is a planar curve with $\kappa_{2}=0$,
$\kappa_{1}(s)=\frac{4C_{1}}{16+C_{1}^{2}(s+C_{2})^{2}}$, and
$f=c_{1}\kappa_{1}^{-3/2}$, or
$(ii)$ $\gamma$ is a general helix with
$\kappa_{1}(s)=\frac{4C_{1}}{16(1+c_{3}^{2})+C_{1}^{2}(s+2C_{2})^{2}}$,
$\kappa_{2}/\kappa_{1}(s)=c_{3}\neq 0$, and $f=c_{1}\kappa_{1}^{-3/2}$.

######  Remark 3.

A proper $f$-biharmonic map $\varphi:(M,g)\to(N,h)$ followed by a totally
geodesic embedding $\psi:(N,h)\to(Q,k)$ is proper $f$-biharmonic map
$\psi\circ\varphi:(M,g)\to(Q,k)$.

In fact, using the fact in [10] (Page 371), we can easily check that
$\tau(\psi\circ\varphi)=d\psi(\tau(\varphi))$,
$\tau^{2}(\psi\circ\varphi)=d\psi\left(\tau^{2}(\varphi)\right)$ and
$\nabla^{\psi\circ\varphi}_{{\rm
grad}f}\tau(\psi\circ\varphi)=d\psi\left(\nabla^{\varphi}_{{\rm
grad}f}\tau(\varphi)\right)$. It follows that
$\tau_{2,f}(\psi\circ\varphi)=d\psi\left(\tau_{2,f}(\varphi)\right)$ and hence
$\tau_{2,f}(\psi\circ\varphi)=0$ iff $\tau_{2,f}(\varphi)=0$.

We know that there is no proper biharmonic curve in
$\mbox{${\mathbb{R}}$}^{n}$ (see e.g., [14]). For some results on biharmonic
curves, we refer the readers to [1, 2, 3, 4, 7, 5, 14] and the references
therein. At the end of this section, we will try to construct some explicit
examples of proper $f$-biharmonic curves in a space form.

######  Proposition 2.6.

(1) If $\rho^{\prime}(s)\neq\pm 1$ solves the $ODE$

(16) $\begin{array}[]{lll}\rho^{\prime\prime}+(1-\rho^{\prime
2})\tan\rho+\frac{\sqrt{1-\rho^{\prime
2}}}{C_{4}\cos(2s)+C_{5}\sin(2s)+\sqrt{1+C_{4}^{2}+C_{5}^{2}}}=0,\end{array}$

then the curve $\gamma:(a,b)\to(S^{2},d\rho^{2}+\cos^{2}\rho d\phi^{2})$ with
$\gamma(s)=(\rho(s),\;\int\frac{\sqrt{1-\rho^{\prime 2}}}{\cos\rho}ds)$
followed by a totally geodesic embedding $\psi:S^{2}\to S^{n}$ is a proper
$f$-biharmonic curve $\tilde{\gamma}=\psi\circ\gamma:(a,b)\to S^{n}$ for
$f=c_{1}\left(C_{4}\cos(2s)+C_{5}\sin(2s)+\sqrt{1+C_{4}^{2}+C_{5}^{2}}\right)^{3/2}$.

(2) If $u^{\prime}(s)\neq\pm 1$ solves the $ODE$

(17) $\begin{array}[]{lll}u^{\prime\prime}(s)-(1-u^{\prime
2})+\frac{\sqrt{1-u^{\prime
2}}}{C_{1}e^{2s}+C_{2}e^{-2s}+\sqrt{4C_{1}C_{2}-1}}=0,\end{array}$

then the curve $\gamma:(a,b)\to(H^{2},du^{2}+e^{2u}dv^{2})$ with
$\gamma(s)=(u(s),\int\frac{\sqrt{1-u^{\prime 2}}}{e^{u}}ds)$ followed by a
totally geodesic embedding $\psi:H^{2}\to H^{n}$ is a proper $f$-biharmonic
curve $\tilde{\gamma}=\psi\circ\gamma:(a,b)\to H^{n}$ for
$f=c_{1}\left(C_{1}e^{2s}+C_{2}e^{-2s}+\sqrt{4C_{1}C_{2}-1}\right)^{3/2}$.

(3) If $x^{\prime}(s)\neq\pm 1$ solves the $ODE$

(18) $\begin{array}[]{lll}x^{\prime\prime}(s)+\frac{4C_{3}\sqrt{1-x^{\prime
2}}}{16+C_{3}^{2}(s+C_{4})^{2}}=0,\end{array}$

then the planar curve
$\gamma:(a,b)\to(\mbox{${\mathbb{R}}$}^{2},dx^{2}+dy^{2})$ with
$\gamma(s)=(x(s),\int\sqrt{1-x^{\prime 2}}ds)$ followed by a totally geodesic
embedding $\psi:\mbox{${\mathbb{R}}$}^{2}\to\mbox{${\mathbb{R}}$}^{n}$ is a
proper $f$-biharmonic curve
$\tilde{\gamma}=\psi\circ\gamma:(a,b)\to\mbox{${\mathbb{R}}$}^{n}$ for
$f=c_{1}\left(\frac{4C_{3}}{16+C_{3}^{2}(s+C_{4})^{2}}\right)^{-3/2}$.

###### Proof.

Let $\gamma:(a,b)\to(S^{2},d\rho^{2}+\cos^{2}\rho d\phi^{2})$,
$\gamma(s)=(\rho(s),\;\phi(s))$ be a proper $f$-biharmonic curve parametrized
by arc length. It follows from the Statement (i) of Theorem 2.4 that the curve
$\gamma$ has the geodesic curvature as

$\begin{array}[]{lll}\kappa_{1}(s)=\chi(s,0,1)=\frac{1}{C_{4}\cos(2s)+C_{5}\sin(2s)+\sqrt{1+C_{4}^{2}+C_{5}^{2}}}.\end{array}$

Suppose that $\theta$ is the angle between the curve $\gamma$ and
$\rho$-curves. Therefore, applying the Liouville’s formula for the geodesic
curvature of curves on the surface $S^{2}$ in $\mbox{${\mathbb{R}}$}^{3}$
yields

(19)
$\begin{array}[]{lll}\theta^{\prime}(s)-\sin\theta\tan\rho(s)=\kappa_{1}(s).\end{array}$

Since $\rho^{\prime}=\cos\theta$ and $\phi^{\prime}\cos\rho=\sin\theta$, then
(19) implies that

$\begin{array}[]{lll}\rho^{\prime\prime}+(1-\rho^{\prime
2})\tan\rho+\kappa_{1}\sqrt{1-\rho^{\prime 2}}=0,\end{array}$

and hence $\phi(s)=\int\frac{\sqrt{1-\rho^{\prime 2}}}{\cos\rho}ds$. Note that
$\kappa_{1}(s)\neq 0$, i.e., $\sin\theta\neq 0$ and hence $\cos\theta\neq\pm
1$, then $\rho^{\prime}\neq\pm 1$. Hence, using Statement (i) of Theorem 2.4
and Remark 3, we obtain the statement (1).

Similar to the statement (1), we can get the statement (2) and (3).

Summarizing all the above results, we obtain the proposition. ∎

We can apply Statement (3) of Proposition 2.6 to construct infinitely many
examples of proper $f$-biharmonic curves in $\mbox{${\mathbb{R}}$}^{n}$.

######  Example 1.

A family of planar curves
$\tilde{\gamma}(s)=(\frac{4\ln|\sqrt{16+C_{3}^{2}s^{2}}+C_{3}s|}{C_{3}}-\frac{4\ln
C_{3}}{C_{3}},\;\frac{\sqrt{16+C_{3}^{2}s^{2}}}{C_{3}},\\\ 0,\ldots,\;0)$ in
$\mbox{${\mathbb{R}}$}^{n}$ are proper $f$-biharmonic curves for
$f=c_{1}(\frac{4C_{3}}{16+C_{3}^{2}s^{2}})^{-3/2}$, where $c_{1}$ and $C_{3}$
are positive constants. Moreover, the curvature
$\kappa_{1}=\frac{4C_{3}}{16+C_{3}^{2}s^{2}}$ and the torsion $\kappa_{2}=0$.
The example recovers the family of proper $f$-biharmonic curves found in [11].

In fact, we solve (2) to obtain the general solutions as

(20)
$\begin{array}[]{lll}x=-\frac{\cos(4C_{3}a_{2})\sqrt{16+C_{3}^{2}(s^{2}+2C_{4}s+C_{4}^{2})}}{C_{3}}-\frac{4\sin(4C_{3}a_{2})\ln|C_{3}s+C_{3}C_{4}+\sqrt{16+C_{3}^{2}(s^{2}+2C_{4}s+C_{4}^{2})}|}{C_{3}}+a_{1},\end{array}$

where $a_{1}$ and $a_{2}$ are constants. By taking $C_{3}>0$,
$4C_{3}a_{2}=-\pi/2$, $a_{1}=-\frac{4\ln C_{3}}{C_{3}}$ and $C_{4}=0$, one
concludes that
$x(s)=\frac{4\ln|\sqrt{16+C_{3}^{2}s^{2}}+C_{3}s|}{C_{3}}-\frac{4\ln
C_{3}}{C_{3}}$ and hence $y(s)=\int\sqrt{1-x^{\prime
2}}ds=\frac{\sqrt{16+C_{3}^{2}s^{2}}}{C_{3}}$. Clearly, the curvature
$\kappa_{1}=\frac{4C_{3}}{\sqrt{16+C_{3}^{2}s^{2}}}$ and the torsion
$\kappa_{2}=0$ in this case. Therefore, we apply Statement (3) of Proposition
2.6 to obtain the example.

We will give a family of proper $f$-biharmonic general helixes in
$\mbox{${\mathbb{R}}$}^{3}$.

######  Proposition 2.7.

If $x^{\prime}(s)\neq\pm\sin\omega$ solves the $ODE$

(21)
$\begin{array}[]{lll}x^{\prime\prime}(s)+\frac{\sqrt{\sin^{2}\omega-x^{\prime
2}}}{\sin\omega}\frac{4C_{3}}{16(1+c_{3}^{2})+C_{3}^{2}(s+C_{4})^{2}}=0,\end{array}$

where $\sin\omega\cos\omega\neq 0$ and $c_{3}=\frac{\cos\omega}{\sin\omega}$
are constants, then a general helix $\gamma:(a,b)\to\mbox{${\mathbb{R}}$}^{3}$
with $\gamma(s)=(x(s),\int\sqrt{\sin^{2}\omega-x^{\prime 2}}ds,s\cos\omega)$
followed by a totally geodesic embedding
$\psi:\mbox{${\mathbb{R}}$}^{3}\to\mbox{${\mathbb{R}}$}^{n}$ is a proper
$f$-biharmonic curve
$\tilde{\gamma}=\psi\circ\gamma:(a,b)\to\mbox{${\mathbb{R}}$}^{n}$ for
$f=c_{1}(\frac{4C_{3}}{16(1+c_{3}^{2})+C_{3}^{2}(s+C_{4})^{2}})^{-3/2}$.

###### Proof.

Let a general helix $\gamma:(a,b)\to\mbox{${\mathbb{R}}$}^{3}$ with
$\gamma(s)=(x(s),y(s),s\cos\omega)$ be a proper $f$-biharmonic curve
parametrized by arclength, where constant $\cos\omega\neq 0$. By Corollary
2.5, the helix has the curvature
$\kappa_{1}(s)=\frac{4C_{1}}{16(1+c_{3}^{2})+C_{1}^{2}(s+2C_{2})^{2}}\neq 0$
and the torsion $\kappa_{2}=c_{3}\kappa_{1}(s)\neq 0$. Since
$\gamma^{\prime}(s)=(x^{\prime}(s),y^{\prime}(s),\cos\omega)$ and
$|\gamma^{\prime}(s)|^{2}=x^{\prime 2}(s)+y^{\prime 2}+\cos^{2}\omega=1$, we
may assume that
$x^{\prime}(s)=\sin\omega\cos\theta(s),\;y^{\prime}=\sin\omega\sin\theta$.
Therefore, we have

(22)
$\begin{array}[]{lll}\gamma^{\prime}=(\sin\omega\cos\theta,\sin\omega\sin\theta,\cos\omega),\;\gamma^{\prime\prime}=\sin\omega\theta^{\prime}(-\sin\theta,\cos\theta,0),\\\
\gamma^{\prime\prime\prime}=\sin\omega\theta^{\prime\prime}(-\sin\theta,\cos\theta,0)-\sin\omega\theta^{\prime
2}(\cos\theta,\sin\theta,0).\end{array}$

It follows that

(23)
$\begin{array}[]{lll}\kappa_{1}(s)=\sin\omega\theta^{\prime},\;\kappa_{2}=\frac{(\gamma^{\prime},\gamma^{\prime\prime},\gamma^{\prime\prime\prime})}{\kappa_{1}^{2}}=\cos\omega\theta^{\prime},\;{\rm
and\;hence}\;c_{3}=\frac{\cos\omega}{\sin\omega}.\end{array}$

Combining these, together with $\kappa_{1}(s)\neq 0$ and $\kappa_{2}\neq 0$,
we have $\sin\omega\cos\omega\neq 0$ and $x^{\prime}\neq\pm\sin\omega$. A
direct computation gives
$x^{\prime\prime}(s)=-\sin\omega\theta^{\prime}\sin\theta=-\theta^{\prime}y^{\prime}=-\theta^{\prime}\sqrt{\sin^{2}\omega-x^{\prime
2}}$. Substituting this into the 1st equation of (23) we obtain (21). Clearly,
$y=\int\sqrt{\sin^{2}\omega-x^{\prime 2}}ds$. From these and using Remark 3,
the proposition follows. ∎

We will look for some special solutions of (21).

######  Example 2.

A family of helixes
$\gamma(s)=(\frac{4\ln|\sqrt{64+2C_{3}^{2}s^{2}}+\sqrt{2}C_{3}s|}{C_{3}}-\frac{4\ln
C_{3}}{C_{3}},\;\frac{\sqrt{64+2C_{3}^{2}s^{2}}}{C_{3}},\frac{\sqrt{2}}{2}s)$
in $\mbox{${\mathbb{R}}$}^{3}$ with the curvature
$\kappa_{1}=\frac{4C_{3}}{32+C_{3}^{2}s^{2}}$ and the torsion
$\kappa_{2}=\frac{4C_{3}}{32+C_{3}^{2}s^{2}}$ are proper $f$-biharmonic curves
for $f=c_{1}(\frac{4C_{3}}{32+C_{3}^{2}s^{2}})^{-3/2}$, where $c_{1}$ and
$C_{3}$ are positive constants. Moreover, a family of curves
$\bar{\gamma}(s)=(\frac{4\ln|\sqrt{64+2C_{3}^{2}s^{2}}+\sqrt{2}C_{3}s|}{C_{3}}-\frac{4\ln
C_{3}}{C_{3}},\;\frac{\sqrt{64+2C_{3}^{2}s^{2}}}{C_{3}},\\\
\frac{\sqrt{2}}{2}s,\;0,\ldots,0)$ in $\mbox{${\mathbb{R}}$}^{n}$ are also
proper $f$-biharmonic curves for
$f=c_{1}(\frac{4C_{3}}{32+C_{3}^{2}s^{2}})^{-3/2}$.

In fact, by taking $\sin\omega=\cos\omega=\frac{\sqrt{2}}{2}$, $c_{3}=1$ and
$C_{4}=0$, (21) becomes

(24)
$\begin{array}[]{lll}x^{\prime\prime}(s)+\frac{\sqrt{\frac{1}{2}-x^{\prime
2}}}{\frac{\sqrt{2}}{2}}\frac{4C_{3}}{32+C_{3}^{2}s^{2}}=0,\end{array}$

which is solved by

$\begin{array}[]{lll}x=-\frac{\cos(4\sqrt{2}C_{3}a_{2})\sqrt{64+2C_{3}^{2}s^{2}}}{2C_{3}}-\frac{4\sin(4\sqrt{2}C_{3}a_{2})\ln|\sqrt{2}C_{3}s+\sqrt{64+2C_{3}^{2}s^{2}}|}{C_{3}}+a_{1},\end{array}$

where $a_{1}$ and $a_{2}$ are constants. Note that $y(s)=\int\sqrt{1-x^{\prime
2}}ds$. Furthermore, taking $C_{3}>0$, $4\sqrt{2}C_{3}a_{2}=-\pi/2$ and
$a_{1}=-\frac{4\ln C_{3}}{C_{3}}$, one finds that
$x(s)=\frac{4\ln|\sqrt{64+2C_{3}^{2}s^{2}}+\sqrt{2}C_{3}s|}{C_{3}}-\frac{4\ln
C_{3}}{C_{3}}$ and hence $y(s)=\int\sqrt{1-x^{\prime
2}}ds=\frac{\sqrt{64+2C_{3}^{2}s^{2}}}{C_{3}}$. Clearly, the curvature
$\kappa_{1}=\frac{4C_{3}}{\sqrt{32+C_{3}^{2}s^{2}}}$ and the torsion
$\kappa_{2}=\frac{4C_{3}}{\sqrt{32+C_{3}^{2}s^{2}}}$ in this case. Thus, we
apply Proposition 2.7 to get the example.

###  2.2.  $f$-Biharmonic isometric immersions and biharmonic conformal
immersions of a developable surface into $\mbox{${\mathbb{R}}$}^{3}$

The equation for $f$-biharmonic hypersurfaces in a space form can be stated as

######  Lemma 2.8.

$($ see e.g.,[11, 13]$)$ A hypersurface $\varphi:M^{m}\to N^{m+1}(C)$ in a
space forms of constant sectional curvature $C$ with mean curvature vector
field $\eta=H\xi$ is $f$-biharmonic if and only if the function $H$ statisfies
the following equation

(25) $\begin{cases}\Delta(fH)-(fH)[|A|^{2}-mC]=0,\\\ A\,({\rm
grad}\,(fH))+\frac{m}{2}(fH){\rm grad}\,H=0,\end{cases}$

where ${\rm Ric}^{N}:T_{q}N\to T_{q}N$ denotes the Ricci operator of the
ambient space defined by $\langle{\rm Ric}^{N}\,(Z),W\rangle={\rm
Ric}^{N}(Z,W)$ and $A$ is the shape operator of the hypersurface with respect
to the unit normal vector field $\xi$.

We are ready to give a characterization of proper $f$-biharmonic cylinders in
$\mbox{${\mathbb{R}}$}^{3}$.

######  Theorem 2.9.

Let $X:(M^{2},g)\to(\mbox{${\mathbb{R}}$}^{3},h={\rm d}x^{2}+{\rm d}y^{2}+{\rm
d}z^{2})$, $X(s,v)=a(s)+vb$ be a cylinder with the mean curvature function
$H$, where $b$ is a unit constant vector in $\mbox{${\mathbb{R}}$}^{3}$ and
$a(s)$ is an immersed regular curve in $\mbox{${\mathbb{R}}$}^{3}$
parametrized by arc length with the geodesic curvature $\kappa_{1}(s)$
satisfying $h(a^{\prime},b)=0$. Then, $X$ is proper $f$-biharmonic if and only
if one of the following cases happens:
$(1)$ $\kappa_{1}$ is a nonzero constant, and the directrix $a(s)$ of the
cylinder is (a part of) a circle with radius $\frac{1}{\sqrt{|\kappa_{1}|}}$.
Moreover, the mean curvature $H=\frac{\kappa_{1}}{2}$ and the function
$f=d_{1}e^{\kappa_{1}v}+d_{2}e^{-\kappa_{1}v}$, where $d_{1}$ and $d_{2}$ are
constants.
or,
$(2)$ $\kappa_{1}(s)$ is nonconstant, and the directrix $a(s)$ of the cylinder
has the geodesic curvature $\kappa_{1}(s)=\kappa_{N}(s,K)$ given by (32) and
the geodesic torsion $\kappa_{2}=0$, where $K$ is a constant. Furthermore, the
mean curvature $H=\frac{\kappa_{N}(s,K)}{2}$ and the function
$f=\psi(v,K)\kappa_{N}^{-3/2}(s,K)$, where $\psi(v,K)$ given by (31).

###### Proof.

A straightforward computation gives

$\begin{array}[]{lll}X_{s}=a^{\prime}(s),\;X_{v}=b,\;X_{ss}=a^{\prime\prime}(s),\;X_{sv}=0,\;X_{vv}=0,\\\
N=\frac{X_{s}\times X_{v}}{|X_{s}\times X_{v}|}=a^{\prime}\times
b,\;\;|a^{\prime\prime}(s)|=\kappa(s),\end{array}$

where $\kappa$ is the curvature of the curve $a(s)$. By a further computation,
the first fundamental form $I$ and the second fundamental form $II$ of the
cylinder given by

$\begin{array}[]{lll}I=g={\rm d}s^{2}+{\rm d}v^{2},\;\;II=\kappa_{N}(s){\rm
d}s^{2},\end{array}$

where $\kappa_{N}$ is the normal curvature of the curve $a(s)$. It follows
that $\\{e_{1}=\frac{\partial}{\partial s}$, $e_{2}=\frac{\partial}{\partial
v},\;\xi=N\\}$ forms an orthonormal frame adapted to the surface with $\xi$
being normal vector field,
$A(e_{1})=II(e_{1},e_{1})e_{1}+II(e_{1},e_{2})e_{2}=\kappa_{N}e_{1}$,
$A(e_{2})=II(e_{2},e_{1})e_{1}+II(e_{2},e_{2})e_{2}=0$ and
$H=\frac{\kappa_{N}}{2}$. Substituting these into the $f$-biharmonic equation
(25) with $m=2$ and $C=0$ we have

(26) $\begin{cases}\kappa_{N}^{2}\frac{\partial\ln f}{\partial
s}=-\frac{3}{2}\kappa_{N}^{\prime}(s)\kappa_{N},\\\
\kappa_{N}^{\prime\prime}(s)-\kappa_{N}^{3}+\kappa_{N}[\frac{\partial^{2}\ln
f}{\partial s^{2}}+\frac{\partial^{2}\ln f}{\partial v^{2}}+(\frac{\partial\ln
f}{\partial s})^{2}+(\frac{\partial\ln f}{\partial v})^{2}]+2\frac{\partial\ln
f}{\partial s}\kappa_{N}^{\prime}(s)=0.\end{cases}$

If $\kappa_{N}=0$, i.e., $H=0$, then the surface is harmonic and hence
biharmonic, not proper $f$-biharmonic. From now on, we assume that
$\kappa_{N}\neq 0$. If $\kappa_{N}$ is a nonzero constant, then (26) turns
into

(27) $\begin{array}[]{lll}\frac{\partial\ln f}{\partial
s}=0,\;\;-\kappa_{N}^{2}+\frac{\partial^{2}\ln f}{\partial
v^{2}}+(\frac{\partial\ln f}{\partial v})^{2}=0,\end{array}$

which is solved by $f=d_{1}e^{v\kappa_{N}}+d_{2}e^{-v\kappa_{N}}$, where
$d_{1}$ and $d_{2}$ are constants. Note that the curve $a(s)$ can be viewed as
(a part of) a circle of radius $\frac{1}{\sqrt{|\kappa_{N}|}}$.
We now assume that $\kappa$ is nonconstant and apply the first equation of
(26) to have

(28) $\ln f=-\frac{3}{2}\ln|\kappa_{N}(s)|+\ln\psi(v),\;{\rm
and\;hence}\;f=\psi(v)\kappa_{N}^{-3/2}(s),$

where $\psi(v)$ is a positive function. Substituting (28) into the second
equation of (26) and simplifying the resulting equation we obtain

(29) $-2\kappa_{N}\kappa_{N}^{\prime\prime}(s)+3\kappa_{N}^{\prime
2}(s)-4\kappa_{N}^{4}+4\kappa_{N}^{2}(s)\frac{\psi^{\prime\prime}(v)}{\psi(v)}=0.$

The equation (29) implies that for any $s,v$, we have

(30) $\begin{array}[]{lll}3\kappa_{N}^{\prime
2}(s)-2\kappa_{N}\kappa_{N}^{\prime\prime}(s)=4\kappa_{N}^{2}(\kappa_{N}^{2}-K)\;\;{\rm
and}\;\;\frac{\psi^{\prime\prime}(v)}{\psi(v)}=K,\end{array}$

where $K$ is a constant. The 2nd equation of (30) is equivalent to
$\psi^{\prime\prime}(v)=K\psi(v)$ solved by

(31)
$\psi(v)=\psi(v,K)=\begin{cases}d_{5}\sin(\sqrt{-K}\;v)+d_{6}\cos(\sqrt{-K}\;v),\;\;\;{\rm
for}\;K<0,\\\ d_{3}v+d_{4},\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm
for}\;K=0,\\\ d_{1}e^{\sqrt{K}\;v}+d_{2}e^{-\sqrt{K}\;v},\;\;\;\;\;\;\;\;{\rm
for}\;K>0,\end{cases}$

where $d_{i}$, $i=1,2,\ldots,6,$ are constants.
An interesting thing is that the 1st equation of (30) happens to be the type
of $ODE$ (8) in Proposition 2.3 when $A=1$ and $C=K$, which is solved by

(32)
$\kappa_{N}(s)=\kappa_{N}(s,K)=\chi(s,0,K)=\begin{cases}\frac{1}{C_{1}e^{2\sqrt{-K}\;s}+C_{2}e^{-2\sqrt{-K}\;s}+\sqrt{4C_{1}C_{2}+\frac{1}{K}}},\;\;\;\;\;\;\;\;{\rm
for}\;K<0,\\\
\frac{4C_{3}}{16+C_{3}^{2}(s+C_{4})^{2}},\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;{\rm
for}\;K=0,\\\
\frac{1}{C_{4}\cos(2\sqrt{K}\;s)+C_{5}\sin(2\sqrt{K}\;s)+\sqrt{\frac{1}{K}+(C_{4}^{2}+C_{5}^{2})}},\;{\rm
for}\;K>0,\end{cases}$

where $C_{i}$ are constants, $i=1,2,\ldots,6$.

It is not difficult to check that
$\nabla_{e_{1}}e_{1}=\kappa_{N}e_{3},\;\nabla_{e_{1}}e_{3}=-\kappa_{N}e_{1}$
and $\nabla_{e_{1}}(-e_{2})=0$, and hence $\\{e_{1},\;e_{3}=N,\;-e_{2}\\}$
forms the Frenet frame along the curve $a(s)$. This implies that the curve
$a(s)$ has the geodesic curvature $\kappa_{1}=\kappa_{N}$ and the geodesic
torsion $\kappa_{2}=0$.

Summarizing all the above results, we complete the proof of the theorem. ∎

######  Remark 4.

Theorem 2.9 recovers the classification result of proper $f$-biharmonc
cylinders with constant mean curvature in [11].

######  Corollary 2.10.

Let $X:(M^{2},g)\to\mbox{${\mathbb{R}}$}^{3}$ be a cylinder with
$X(s,v)=a(s)+vb$ and nonconstant mean curvature function $H$, where $b$ is a
unit constant vector in $\mbox{${\mathbb{R}}$}^{3}$ and $a(s)$ is an immersed
regular curve in $\mbox{${\mathbb{R}}$}^{3}$ parametrized by arc length with
the geodesic curvature $\kappa_{1}(s)$ satisfying $h(a^{\prime},b)=0$. Then,
$X$ is proper $f$-biharmonic if and only if the directrix $a(s)$ is a proper
$\bar{f}$-biharmonic curve in a 2-space form
$N^{2}(K)\subset\mbox{${\mathbb{R}}$}^{3}$ with constant Gauss curvature $K$
for $\bar{f}=c_{1}\kappa_{1}^{-3/2}$, where $c_{1}>0$ is a constant and
$\kappa_{1}(s)=\kappa_{N}(s,K)$ given by (32). Moreover, the function
$f=\frac{\psi(v,K)}{c_{1}}\bar{f}=\psi(v,K)\kappa_{N}^{-3/2}(s,K)$, where
$\psi(v,K)$ given by (31).

###### Proof.

From the proof of Theorem 2.9 and Statement (i) of Proposition 2.2, together
with the assumption that $H$ is nonconstant, one sees that the directrix
$a(s)$ on the surface has the geodesic torsion $\kappa_{2}=0$ and nonconstant
geodesic curvature $\kappa_{1}=\kappa_{N}=2H$ solving the 1st equation of (30)
which is a proper $\bar{f}$-biharmonic curve equation in a space form of
constant sectional curvature $K$ for $\bar{f}=c_{1}\kappa_{1}^{-3/2}$, where
constant $c_{1}>0$. This implies that the directrix $a(s)$ can be viewed as a
proper $\bar{f}$-biharmonic curve in a 2-space form $N^{2}(K)$ of constant
Gauss curvature $K$. Note that $f=\frac{\psi(v,K)}{c_{1}}\bar{f}$, where
$\psi(v,K)$ given by (31). ∎

######  Proposition 2.11.

There is neither a proper $f$-biharmonic cone nor tangent surface in
$\mbox{${\mathbb{R}}$}^{3}$.

###### Proof.

Let $b(s)$ be a curve on a unit sphere parametrized by arc length (i.e., a
spherical curve). Consider a cone
$X:(M^{2},g)\to(\mbox{${\mathbb{R}}$}^{3},h={\rm d}x^{2}+{\rm d}y^{2}+{\rm
d}z^{2})$ into $\mbox{${\mathbb{R}}$}^{3}$ with $X(s,v)=a+vb(s)$ , where $a$
is a constant vector. A simple computation, we obtain the first and the second
fundamental forms $I$ and $II$ of the cone as

$\begin{array}[]{lll}I=g=v^{2}{\rm d}s^{2}+{\rm d}v^{2},\;\;II=vw(s){\rm
d}s^{2},\end{array}$

where $w(s)=b^{\prime\prime}\cdot(b^{\prime}\times b)$. One can check that
$\\{e_{1}=\frac{1}{v}\frac{\partial}{\partial s}$,
$e_{2}=\frac{\partial}{\partial v},\;\xi=b^{\prime}\times b\\}$ constitutes an
orthonormal frame adapted to the surface with $\xi$ being normal to the
surface. By a straightforward computation, we have
$A(e_{1})=II(e_{1},e_{1})e_{1}+II(e_{1},e_{2})e_{2}=\frac{w}{v}e_{1}$,
$A(e_{2})=II(e_{2},e_{1})e_{1}+II(e_{2},e_{2})e_{2}=0$ and
$H=\frac{w(s)}{2v}$. Substituting these into the second equation of
$(\ref{fb})$ and simplifying the resulting equation yields

(33) $H[3e_{1}(H)+2He_{1}(\ln f)]=0,\;He_{2}(H)=H\frac{\partial}{\partial
v}\left(\frac{w(s)}{2v}\right)=0.$

The 2nd equation of (33) implies that $-\frac{H^{2}}{2v^{2}}=0$ and hence
$H=0$, i.e., the cone has to be minimal. It follows that any $f$-biharmonic
cone in $\mbox{${\mathbb{R}}$}^{3}$ has to be harmonic and hence biharmonic,
but not proper $f$-biharmonic.

For a tangent surface in $\mbox{${\mathbb{R}}$}^{3}$, let $a(s)$ be a curve
parametrized by arclength and let $X:(M^{2},g)\to\mbox{${\mathbb{R}}$}^{3}$ be
a tangent surface with $X(s,v)=a(s)+va^{\prime}(s)$. By simple computation,
the first and the second fundamental forms $I$ and $II$ of the tangent surface
given by respectively

(34) $\begin{array}[]{lll}I=g=(1+v^{2}\kappa^{2}){\rm d}s^{2}+2{\rm d}s{\rm
d}v+{\rm d}v^{2},\;\;II=-v\kappa\tau{\rm d}s^{2},\end{array}$

where $\kappa=\kappa(s)$ and $\tau=\tau(s)$ are the curvature and torsion of
$a(s)$ respectively.

One can easily check that
$\\{\varepsilon_{1}=\frac{1}{\sqrt{1+v^{2}\kappa^{2}}}\frac{\partial}{\partial
s},\;\varepsilon_{2}=\frac{1}{v\kappa\sqrt{(1+v^{2}\kappa^{2}}}\frac{\partial}{\partial
s}-\frac{\sqrt{1+v^{2}\kappa^{2}}}{v\kappa}\frac{\partial}{\partial v},\\\
\xi=\frac{a^{\prime\prime}\times a^{\prime}}{\kappa}\\}$ forms an orthonormal
frame adapted to the surface with the normal vector field $\xi$. Suppose that
$A(\varepsilon_{1})=\kappa_{1}e_{1}+\kappa_{12}\varepsilon_{2}$ and
$A(\varepsilon_{2})=\kappa_{21}\varepsilon_{1}+\kappa_{2}\varepsilon_{2}$. A
direct computation gives
$\kappa_{1}=II(\varepsilon_{1},\varepsilon_{1})=-\frac{v\kappa\tau}{1+v^{2}\kappa^{2}}$,
$\kappa_{2}=II(\varepsilon_{2},\varepsilon_{2})=-\frac{\tau}{v\kappa(1+v^{2}\kappa^{2})}$
and $\kappa_{12}=\kappa_{21}=II(e_{1},e_{2})=-\frac{\tau}{1+v^{2}\kappa^{2}}$.
It follows that the mean curvature
$H=\frac{\kappa_{1}+\kappa_{2}}{2}=-\frac{\tau}{2v\kappa}$ and
$\kappa_{1}\kappa_{2}=\frac{\tau^{2}}{(1+v^{2}\kappa^{2})^{2}}$. Note that
$H=0$ is equivalent to $\tau=0$, then the surface is not proper $f$-biharmonic
in this case.

Hereafter, assume that $H\neq 0$. Let
$\\{e_{1}=\cos\theta\varepsilon_{1}+\sin\theta\varepsilon_{2},\;e_{2}=-\sin\theta\varepsilon_{1}+\cos\theta\varepsilon_{2},\;\xi\\}$
be another orthonormal frame adapted to the surface with the normal vector
field $\xi$ such that $A(e_{1})=\lambda_{1}e_{1}$ and
$A(e_{2})=\lambda_{2}e_{2}$, where $\theta$ is the angle between $e_{1}$ and
$\varepsilon_{1}$. Since the tangent surface is flat, it follows from the
Gauss equation of the tangent surface that $\lambda_{1}\lambda_{2}=0$. Without
a loss of generality, we may suppose that $\lambda_{2}=0$ and hence
$\lambda_{1}=2H\neq 0$. We can easily conclude that
$\kappa_{1}=2H\cos^{2}\theta,\;\kappa_{2}=2H\sin^{2}\theta,\;\kappa_{12}=\tau=2H\sin\theta\cos\theta,\;\cos^{2}\theta=\frac{v^{2}\kappa^{2}}{1+v^{2}\kappa^{2}}$
and $\sin^{2}\theta=\frac{1}{1+v^{2}\kappa^{2}}.$ If the tangent surface is
$f$-biharmonic, by (25) with $m=2$ and $C=0$, we have

(35)
$\begin{array}[]{lll}\Delta(fH)-4fH^{3}=0,\;2He_{1}(fH)+fHe_{1}(H)=0,\;fHe_{2}(H)=0,\end{array}$

which implies that $e_{2}(H)=0$. Therefore, we get
$[-\sin\theta\varepsilon_{1}+\cos\theta\varepsilon_{2}](H)=0$, i.e.,
$\sin^{2}\theta(\varepsilon_{1}(H))^{2}-\cos^{2}\theta(\varepsilon_{2}(H))^{2}=0$.
A further computation gives

(36)
$\begin{array}[]{lll}0=\sin^{2}\theta[\varepsilon_{1}(-\frac{\tau}{2v\kappa})]^{2}-\cos^{2}\theta[\varepsilon_{2}(-\frac{\tau}{2v\kappa})]^{2}\\\
=\frac{1}{1+v^{2}\kappa^{2}}\left(\frac{1}{\sqrt{1+v^{2}\kappa^{2}(s)}}\frac{\partial}{\partial
s}(\frac{\tau}{2v\kappa})\right)^{2}-\frac{v^{2}\kappa^{2}}{1+v^{2}\kappa^{2}}\left(\frac{1}{v\kappa\sqrt{(1+v^{2}\kappa^{2}}}\frac{\partial}{\partial
s}(\frac{\tau}{2v\kappa})-\frac{\sqrt{1+v^{2}\kappa^{2}}}{v\kappa}\frac{\partial}{\partial
v}(\frac{\tau}{2v\kappa})\right)^{2}\\\
=-\frac{(\frac{\tau}{\kappa})^{2}}{4v^{4}}-\frac{[(\frac{\tau}{\kappa})^{2}]^{\prime}(s)}{4v^{3}(1+v^{2}\kappa^{2})}.\end{array}$

It is easy to check that (36) implies that

(37)
$\begin{array}[]{lll}\tau^{2}(s)\;v^{2}+[(\frac{\tau}{\kappa})^{2}]^{\prime}(s)\;v+(\frac{\tau}{\kappa})^{2}(s)=0,\end{array}$

which, together with any $s,\;v$, implies that $\tau=0$. This contradicts to
the assumption that $\tau\neq 0$. Combining these, any $f$-biharmonic tangent
surface in $\mbox{${\mathbb{R}}$}^{3}$ is harmonic, but not proper
$f$-biharmonic.

Summarizing all the above results, the proposition follows. ∎

Applying Theorem 2.9 and Proposition 2.11, we have

######  Corollary 2.12.

A proper $f$-biharmonic developable surface in $\mbox{${\mathbb{R}}$}^{3}$
exists only in the case when the surface is either a circle cylinder or a
cylinder containing a directrix with nonconstant geodesic curvature
$\kappa_{1}(s)=\kappa_{N}(s,K)$ given by (32) and the geodesic torsion
$\kappa_{2}(s)=0$.

We will construct a family of cylinders whose directrices are proper
$\bar{f}$-biharmonic curves in $\mbox{${\mathbb{R}}$}^{2}$ for
$\bar{f}=c_{1}\kappa_{N}^{-3/2}(s,0)$, which are proper $f$-biharmonic
cylinders for $f=\frac{\psi(v,0)}{c_{1}}\bar{f}$, where $\psi(v,0)$ and
$\kappa_{N}^{-3/2}(s,0)$ given by (31) and (32) respectively.

######  Example 3.

A family of cylinders $X:(M^{2},g)\to\mbox{${\mathbb{R}}$}^{3}$,
$X(s,v)=(\frac{4\ln|\sqrt{16+C_{3}^{2}s^{2}}+C_{3}s|}{C_{3}}-\frac{4\ln
C_{3}}{C_{3}},\;\frac{\sqrt{16+C_{3}^{2}s^{2}}}{C_{3}},\;v)$ are proper
$f$-biharmonic for
$f=\frac{c_{1}(d_{3}v+d_{4})(16+C_{3}^{2}s^{2})^{3/2}}{(4C_{3})^{3/2}}$, where
$d_{3}$, $d_{4}$, $c_{1}$ and $C_{3}$ are positive constants. This family of
cylinders followed by a totally geodesic embedding
$\phi:\mbox{${\mathbb{R}}$}^{3}\to\mbox{${\mathbb{R}}$}^{n}$ are proper
$f$-biharmonic submanifolds $\phi\circ
X:(M^{2},g)\to\mbox{${\mathbb{R}}$}^{n}$.

In fact, it is easy to check that $X(s,v)=a(s)+vb$, where
$a(s)=(\frac{4\ln|\sqrt{16+C_{3}^{2}s^{2}}+C_{3}s|}{C_{3}}-\frac{4\ln
C_{3}}{C_{3}},\;\frac{\sqrt{16+C_{3}^{2}s^{2}}}{C_{3}},\;0)$ are a family of
curves in $\mbox{${\mathbb{R}}$}^{2}\subset\mbox{${\mathbb{R}}$}^{3}$ and
$b=(0,0,1)$ is a unit constant vector in $\mbox{${\mathbb{R}}$}^{3}$. Clearly,
$|a^{\prime}|^{2}=1$ and $h(a^{\prime},b)=0$. Therefore, by Example 1, any
curve $a(s)$ is a proper $\bar{f}$-biharmonic curve in
$\mbox{${\mathbb{R}}$}^{2}\subset\mbox{${\mathbb{R}}$}^{3}$ for
$\bar{f}=c_{1}(\frac{4C_{3}}{16+C_{3}^{2}s^{2}})^{-3/2}$, where $c_{1}>0$ is a
constant. Then, applying Corollary 2.10, we obtain the example.

We know from [11] that a map $\varphi:(M^{2},g)\to(N^{n},h)$ is an
$f$-biharmonic map if and only if the conformal immersion
$\varphi:(M^{2},f^{-1}g)\to(N^{n},h)$ is a biharmonic map. Therefore, applying
Theorem 2.9 and Proposition 2.11, we have the following corollaries as follows

######  Corollary 2.13.

Under the same assumptions as in Theorem 2.9, then we have
$(1)$ If $\kappa_{1}$ is a constant and
$f=d_{1}e^{\kappa_{1}v}+d_{2}e^{-\kappa_{1}v}>0$, where $d_{1}$ and $d_{2}$
are constants, then the conformal immersion
$X:(M^{2},f^{-1}g)\to\mbox{${\mathbb{R}}$}^{3}$ with $X(s,v)=a(s)+vb$ is
proper biharmonic, or
$(2)$ If $\kappa_{1}(s)=\kappa_{N}(s,K)$ and $\psi(v,K)$ given by (32) and
(31) respectively, and $f=\psi(v,K)\kappa_{N}^{-3/2}(s,K)$, where $K$ is a
constant, then the conformal immersion
$X:(M^{2},f^{-1}g)\to\mbox{${\mathbb{R}}$}^{3}$ with $X(s,v)=a(s)+vb$ is
proper biharmonic.

######  Corollary 2.14.

We have
(1) There can not exist a proper biharmonic conformal immersion of a cone or a
tangent surface into $\mbox{${\mathbb{R}}$}^{3}$.
(2) A conformal immersion of a developable surface into
$\mbox{${\mathbb{R}}$}^{3}$ is proper biharmonic if and only if the surface is
either a circle cylinder or a cylinder containing a directrix with nonconstant
geodesic curvature $\kappa_{1}(s)=\kappa_{N}(s,K)$ given by (32) and the
torsion $\kappa_{2}(s)=0$.

######  Remark 5.

Corollary 2.14 recovers the result in [12] stated that there is no a proper
biharmonic conformal immersion of a circular cone into
$\mbox{${\mathbb{R}}$}^{3}$.

Adopting the same notations as in Example 3 and applying Example 3 and
Corollary 2.13, we have

######  Example 4.

The conformal immersions of a family of cylinders
$X:(M^{2},f^{-1}g)\to\mbox{${\mathbb{R}}$}^{3}$,
$X(s,v)=(\frac{4\ln|\sqrt{16+C_{3}^{2}s^{2}}+C_{3}s|}{C_{3}}-\frac{4\ln
C_{3}}{C_{3}},\;\frac{\sqrt{16+C_{3}^{2}s^{2}}}{C_{3}},\;v)$ are proper
biharmonic with
$f=\frac{c_{1}(d_{3}v+d_{4})(16+C_{3}^{2}s^{2})^{3/2}}{(4C_{3})^{3/2}}$.

##  3\.  $f$-biharmonic Riemannian submersions from 3-manifolds

In this section, we will study $f$-biharmonic Riemannian submersions from
3-manifolds by using the integrability data of an adapted frame of the
Riemannian submersion. We also construct a family of proper $f$-biharmonic
Riemannian submersions from Sol space.

Recall that an orthonormal frame $\\{e_{1},e_{2},e_{3}\\}$ is adapted to a
Riemannian submersion $\pi:(M^{3},g)\to(N^{2},h)$ with $e_{3}$ being vertical,
then we have (see [15])

(38)
$\begin{array}[]{lll}[e_{1},e_{3}]=\kappa_{1}e_{3},\;[e_{2},e_{3}]=\kappa_{2}e_{3},\;[e_{1},e_{2}]=f_{1}e_{1}+f_{2}e_{2}-2\sigma
e_{3},\\\
\nabla_{e_{1}}e_{1}=-f_{1}e_{2},\;\;\nabla_{e_{1}}e_{2}=f_{1}e_{1}-\sigma
e_{3},\;\;\nabla_{e_{1}}e_{3}=\sigma e_{2},\\\
\nabla_{e_{2}}e_{1}=-f_{2}e_{2}+\sigma
e_{3},\;\;\nabla_{e_{2}}e_{2}=f_{2}e_{1},\;\;\nabla_{e_{2}}e_{3}=-\sigma
e_{1},\\\ \nabla_{e_{3}}e_{1}=-\kappa_{1}e_{3}+\sigma
e_{2},\nabla_{e_{3}}e_{2}=-\sigma
e_{1}-\kappa_{2}e_{3},\nabla_{e_{3}}e_{3}=\kappa_{1}e_{1}+\kappa_{2}e_{2},\end{array}$

where $f_{1},\;f_{2},\;\kappa_{1},\;\kappa_{2}\;{\rm and}\;\sigma\in
C^{\infty}(M)$ called the integrability data of the adapted frame of $\pi$.
The bitension field of $\pi$ given by

(39)
$\begin{array}[]{lll}\tau_{2}(\pi)=[-\Delta^{M}\kappa_{1}-f_{1}e_{1}(\kappa_{2})-e_{1}(\kappa_{2}f_{1})-f_{2}e_{2}(\kappa_{2})-e_{2}(\kappa_{2}f_{2})\\\
+\kappa_{1}\kappa_{2}f_{1}+\kappa_{2}^{2}f_{2}+\kappa_{1}\\{-K^{N}+f_{1}^{2}+f_{2}^{2}\\}]\varepsilon_{1}+(-\Delta^{M}\kappa_{2}+f_{1}e_{1}(\kappa_{1})\\\
+e_{1}(\kappa_{1}f_{1})+f_{2}e_{2}(\kappa_{1})+e_{2}(\kappa_{1}f_{2})-\kappa_{1}\kappa_{2}f_{2}-\kappa_{1}^{2}f_{1}+\kappa_{2}\\{-K^{N}+f_{1}^{2}+f_{2}^{2}\\})\varepsilon_{2},\end{array}$

where $d\pi(e_{i})=\varepsilon_{i},\;i=1,2$.

Using an adapted frame and the associated integrability data, the
$f$-biharmonicity of a Riemannian submersion can be described as follows.

######  Proposition 3.1.

Let $\pi:(M^{3},g)\to(N^{2},h)$ be a Riemannian submersion with the adapted
frame $\\{e_{1},\;e_{2},\;e_{3}\\}$ and the integrability data
$\\{f_{1},f_{2},\kappa_{1},\;\kappa_{2},\;\sigma\\}$. Then, the Riemannian
submersion $\pi$ is $f$-biharmonic if and only if

(40)
$\begin{cases}-\Delta^{M}(f\kappa_{1})-2\sum\limits_{i=1}^{2}f_{i}e_{i}(f\kappa_{2})-f\kappa_{2}\sum\limits_{i=1}^{2}\left(e_{i}(f_{i})-\kappa_{i}f_{i}\right)+f\kappa_{1}\left(-K^{N}+\sum\limits_{i=1}^{2}f_{i}^{2}\right)=0,\\\
-\Delta^{M}(f\kappa_{2})+2\sum\limits_{i=1}^{2}f_{i}e_{i}(f\kappa_{1})+f\kappa_{1}\sum\limits_{i=1}^{2}(e_{i}(f_{i})-\kappa_{i}f_{i})+f\kappa_{2}\left(-K^{N}+\sum\limits_{i=1}^{2}f_{i}^{2}\right)=0,\\\
\end{cases}$

where
$K^{N}=R^{N}_{1212}\circ\pi=e_{1}(f_{2})-e_{2}(f_{1})-f_{1}^{2}-f_{2}^{2}$ is
the Gauss curvature of Riemannian manifold $(N^{2},h)$.

###### Proof.

A direct computation using (38) gives

(41)
$\begin{array}[]{lll}\tau(\pi)=\nabla^{\pi}_{e_{i}}d\pi(e_{i})-d\pi(\nabla^{M}_{e_{i}}e_{i})=-d\pi(\nabla^{M}_{e_{3}}e_{3})=-\kappa_{1}\varepsilon_{1}-\kappa_{2}\varepsilon_{2},\\\
\Delta^{M}f=\sum\limits_{i=1}^{3}e_{i}e_{i}(f)+f_{1}e_{2}(f)-f_{2}e_{1}(f)-\kappa_{1}e_{1}(f)-\kappa_{2}e_{2}(f),\\\
2\nabla^{\pi}_{{\rm grad}f}\tau(\pi)=-2\\{\langle{\rm grad}f,{\rm
grad}\kappa_{1}\rangle+\kappa_{2}f_{1}e_{1}(f)+\kappa_{2}f_{2}e_{2}(f)\\}\varepsilon_{1}\\\
-2\\{\langle{\rm grad}f,{\rm
grad}\kappa_{2}\rangle-\kappa_{1}f_{1}e_{1}(f)-\kappa_{1}f_{2}e_{2}(f)\\}\varepsilon_{2}.\end{array}$

Substituting (41) and (39) into $f$-biharmonic map equation (2) and a
straightforward computation gives

$\begin{array}[]{lll}0=\tau_{2,f}(\pi)=-J^{\pi}(f\tau(\pi))\\\
=f\tau^{2}(\pi)+(\Delta^{M}f)\tau(\pi)+2\nabla^{\pi}_{{\rm grad}f}\tau(\pi)\\\
=[-\Delta^{M}(f\kappa_{1})-f_{1}e_{1}(f\kappa_{2})-e_{1}(f\kappa_{2}f_{1})-f_{2}e_{2}(f\kappa_{2})-e_{2}(f\kappa_{2}f_{2})\\\
+f\kappa_{2}(\kappa_{1}f_{1}+\kappa_{2}f_{2})+f\kappa_{1}\\{-K^{N}+f_{1}^{2}+f_{2}^{2}\\}]\varepsilon_{1},\\\
+[-\Delta^{M}(f\kappa_{2})+f_{1}e_{1}(f\kappa_{1})+e_{1}(f\kappa_{1}f_{1})+f_{2}e_{2}(f\kappa_{1})+e_{2}(f\kappa_{1}f_{2})\\\
-f\kappa_{1}(\kappa_{1}f_{1}+\kappa_{2}f_{2})+f\kappa_{2}\\{-K^{N}+f_{1}^{2}+f_{2}^{2}\\}]\varepsilon_{2}.\end{array}$

From which the proposition follows. ∎

When the integrability data $\kappa_{2}=0$ we have the following corollary

######  Corollary 3.2.

Let $\pi:(M^{3},g)\to(N^{2},h)$ be a Riemannian submersion with the adapted
frame $\\{e_{1},\;e_{2},\;e_{3}\\}$ and the integrability data
$\\{f_{1},f_{2},\kappa_{1},\;\kappa_{2},\;\sigma\\}$ with $\kappa_{2}=0$.
Then, the Riemannian submersion $\pi$ is $f$-biharmonic if and only if

(42)
$\begin{cases}-\Delta^{M}(f\kappa_{1})+f\kappa_{1}\\{-K^{N}+f_{1}^{2}+f_{2}^{2}\\}=0,\\\
2f_{1}e_{1}(f\kappa_{1})+2f_{2}e_{2}(f\kappa_{1})+f\kappa_{1}\\{e_{1}(f_{1})++e_{2}(f_{2})\\}-f\kappa_{1}^{2}f_{1}=0.\end{cases}$

######  Example 5.

For any positive constants $A,B$ and the function
$f=Ae^{\sqrt{2}z}+Be^{-\sqrt{2}z}$ globally defined on Sol space, the
Riemannian submersion given by

$\begin{array}[]{lll}\pi:(\mbox{${\mathbb{R}}$}^{3},g_{Sol}=e^{2z}{\rm
d}x^{2}+e^{-2z}{\rm d}y^{2}+{\rm
d}z^{2})\to(\mbox{${\mathbb{R}}$}^{2},e^{-2z}{\rm d}y^{2}+{\rm
d}z^{2}),\;\pi(x,y,z)=(y,z)\end{array}$

is a proper $f$-biharmonic map.

In fact, it is easy to check that the orthonormal frame
$\\{e_{1}=\frac{\partial}{\partial z},\;e_{2}=e^{z}\frac{\partial}{\partial
y},\;e_{3}=e^{-z}\frac{\partial}{\partial x}\\}$ on Sol space is adapted to
the Riemannian submersion $\pi$ with $e_{3}$ being vertical. A straightforward
computation gives

(43)
$\begin{array}[]{lll}[e_{1},e_{3}]=-e_{3},\;[e_{2},e_{3}]=0,\;[e_{1},e_{2}]=e_{2},\;f_{1}=\kappa_{2}=\sigma=0,\;-f_{2}=\kappa_{1}=-1,\\\
({\rm
and\;hence})\;K^{N}=e_{1}(f_{2})-e_{2}(f_{1})-f_{1}^{2}-f_{2}^{2}=-1.\end{array}$

Substituting (43) into (42) yields

(44) $\begin{array}[]{lll}f_{zz}+e^{-2z}f_{xx}-2f=0,\;f_{y}=0.\end{array}$

We find $f=Ae^{\sqrt{2}z}+Be^{-\sqrt{2}z}$ to be a special solution of (44),
where $A$ and $B$ are positive constants. The bitension field
$\tau_{2}(\pi)=-2\frac{\partial}{\partial z}\neq 0$ implies that $\pi$ is not
biharmonic. Thus, we obtain the example.

##  4\. $f$-biharmonic Riemannian submersions from $3$-space forms

A biharmonic Riemannian submersion from a 3-space form into a surface has to
be harmonic (cf.[15]). We want to know whether there is a proper
$f$-biharmonic Riemannian submersion from a 3-space form. Let
$\pi:(M^{3}(c),g)\to(N^{2}(c),h)$ be a Riemannian submersion from a 3-space
form of constant sectional curvature $c$ into a surface with constant Gauss
curvature $c$. It follows from Lemma 3.2 in [15] that we can choose an
orthonormal frame $\\{e_{1},e_{2},e_{3}\\}$ adapted to the Riemannian
submersion with the integrability data
$\\{f_{1},f_{2},\kappa_{1},\kappa_{2},\sigma\\}$ and $\kappa_{2}=0$ such that

(45)
$\begin{array}[]{ccc}e_{1}(\sigma)-2\kappa_{1}\sigma=0,\;e_{1}(\kappa_{1})+\sigma^{2}-\kappa_{1}^{2}=c,\;e_{3}(\sigma)=\kappa_{1}f_{1}=0,\;K^{N}-3\sigma^{2}=c,\\\
e_{3}(\kappa_{1})=e_{2}(\sigma)=e_{2}(\kappa_{1})=0,\;\sigma^{2}-\kappa_{1}f_{2}=c,\;K^{N}=e_{1}(f_{2})-e_{2}(f_{1})-f_{1}^{2}-f_{2}^{2}.\end{array}$

######  Remark 6.

Note that $\kappa_{1}=\kappa_{2}=0$, $\pi$ is harmonic and hence biharmonic.
We now suppose that $\kappa_{1}\neq 0$, $\kappa_{2}=0$ and the Gauss curvature
of the base space $K^{N}=c$. A simple computation using (45) gives

(46) $\begin{array}[]{ccc}\kappa_{1}\neq
0,\;\sigma=f_{1}=e_{2}(\kappa_{1})=e_{3}(\kappa_{1})=0,\\\
\kappa_{1}f_{2}=-c,\;e_{1}(\kappa_{1})=\kappa_{1}^{2}+c,\;K^{N}=e_{1}(f_{2})-f_{2}^{2}=c.\end{array}$

Here, a Riemannian submersion $\pi:(M^{3}(c),g)\to(N^{2}(c),h)$ may be the map
as $\pi_{U}:(M^{3}(c)\supseteq U,g)\to(V\subseteq N^{2}(c),h)$ from a subset
$M^{3}(c)\supseteq U$ to a subset $V\subseteq N^{3}(c)$. In spite of this, we
denote the map $\pi_{U}:(M^{3}(c)\supseteq U,g)\to(V\subseteq N^{2}(c),h)$ as
$\pi:(M^{3}(c),g)\to(N^{2}(c),h)$ later in the rest of this section by abuse
of notations.

######  Theorem 4.1.

Let $\pi:(M^{3}(c),g)\to(N^{2}(c),h)$ be a Riemannian submersion from a
3-space form of constant sectional curvature $c$ into a surface of constant
Gauss curvature $c$ with an adapted frame $\\{e_{1},\;e_{2},\;e_{3}\\}$ and
the integrability data $f_{1},f_{2},\kappa_{1},\;\sigma$ and $\kappa_{2}=0$.
Then, $\pi$ is proper $f$-biharmonic if and only if

(47)
$\begin{cases}-\Delta^{M}(f\kappa_{1})+f\kappa_{1}(-c+c^{2}/\kappa_{1}^{2})=0,\\\
ce_{2}(f)=0,\\\
-\Delta^{M}(\kappa_{1})+\kappa_{1}(-c+c^{2}/\kappa_{1}^{2})\neq 0.\end{cases}$

###### Proof.

Since the Gauss curvature of the base space $K^{N}=c$, $\kappa_{2}=0$ and the
assumption that $\pi$ is proper $f$-biharmonic, then we have $\kappa_{1}\neq
0$. From these and Remark 6, we conclude that $f_{2}=c/\kappa_{1}$ and hence
$e_{2}(f_{2})=0$. Substituting this and (46) into (42) we have
$-\Delta^{M}(f\kappa_{1})+f\kappa_{1}(-c+c^{2}/\kappa_{1}^{2})=0$ and
$ce_{2}(f)=0$. It is not difficult to check that the bitension field
$\tau_{2}(\pi)=[-\Delta^{M}(\kappa_{1})+\kappa_{1}(-c+c^{2}/\kappa_{1}^{2})]\varepsilon_{1}$.
Combining these, then $\pi$ is proper $f$-biharmonic if and only if (47)
holds. From which the theorem follows. ∎

Applying Theorem 4.1 with $c=0$ we get

######  Proposition 4.2.

For the function $f=\rho>0$, then a Riemannian submersion
$\pi:(\mbox{${\mathbb{R}}$}^{3}\backslash\\{0\\},{\rm d}\rho^{2}+{\rm
d}z^{2}+\rho^{2}{\rm
d}\theta^{2})\to(\mbox{${\mathbb{R}}$}^{2}\backslash\\{0\\},{\rm
d}\rho^{2}+{\rm d}z^{2})$ with $\pi(\rho,z,\theta)=(\rho,z)$ is a proper
$f$-biharmonic map.

###### Proof.

It is not difficult to check that the orthonormal frame
$\\{e_{1}=\frac{\partial}{\partial\rho},\;e_{2}=\frac{\partial}{\partial
z},\;e_{3}=\frac{1}{\rho}\frac{\partial}{\partial\theta}\\}$ on
$(\mbox{${\mathbb{R}}$}^{3}\backslash\\{0\\},{\rm d}\rho^{2}+{\rm
d}z^{2}+\rho^{2}{\rm d}\theta^{2})$ is adapted to the Riemannian submersion
$\pi$ with ${\rm d}\pi(e_{3})=0$ and ${\rm
d}\pi(e_{i})=\varepsilon_{i},i=1,2$, where
$\\{\varepsilon_{1}=\frac{\partial}{\partial\rho},\;\varepsilon_{2}=\frac{\partial}{\partial
z}\\}$ forms an orthonormal frame on the base space
$(\mbox{${\mathbb{R}}$}^{2}\backslash\\{0\\},{\rm d}\rho^{2}+{\rm d}z^{2})$. A
straightforward computation gives

(48)
$\begin{array}[]{lll}[e_{1},e_{3}]=-\frac{1}{\rho}e_{3},[e_{2},e_{3}]=0,[e_{1},e_{2}]=0,\;f_{1}=f_{2}=\sigma=\kappa_{2}=0,\;\kappa_{1}=-\frac{1}{\rho}\neq
0.\end{array}$

Using (38) and (48) with $c=0$, (47) reduces to

(49) $\begin{array}[]{ccc}\Delta\left(\frac{f}{\rho}\right)=0.\end{array}$

Looking for a special solution of the form $f=f(\rho)$, then (49) becomes
$\left(\frac{f}{\rho}\right)^{\prime\prime}+\frac{1}{\rho}\left(\frac{f}{\rho}\right)^{\prime}=0,$
which is solved by $\frac{f}{\rho}=C_{1}+C_{2}\ln\rho,$ where $C_{1},C_{2}$
are constants. Taking $C_{1}=1$ and $C_{2}=0$, we obtain $f=\rho$. Thus, we
obtain the proposition. ∎

Applying Theorem 4.1 with $c=-1$ we have

######  Proposition 4.3.

For any positive constants $C_{1}$, $C_{2}$ and the function
$f=C_{1}e^{(1+\sqrt{3})\rho}+C_{2}e^{(1-\sqrt{3})\rho}$, then the Riemannian
submersion $\pi:(H^{3},{\rm d}\rho^{2}+e^{-2\rho}{\rm d}z^{2}+e^{-2\rho}{\rm
d}\theta^{2})\to(H^{2},{\rm d}\rho^{2}+e^{-2\rho}{\rm d}z^{2})$ with
$\pi(\rho,z,\theta)=(\rho,z)$ is a proper $f$-biharmonic map.

###### Proof.

We can easily check that the orthonormal frame
$\\{e_{1}=\frac{\partial}{\partial\rho},\;e_{2}=e^{\rho}\frac{\partial}{\partial
z},\;e_{3}=e^{\rho}\frac{\partial}{\partial\theta}\\}$ on $(H^{3},{\rm
d}\rho^{2}+e^{-2\rho}{\rm d}z^{2}+e^{-2\rho}{\rm d}\theta^{2})$ is adapted to
the Riemannian submersion $\pi$ with ${\rm d}\pi(e_{3})=0$. By a simple
computation yields

(50)
$[e_{1},e_{3}]=e_{3},[e_{2},e_{3}]=0,[e_{1},e_{2}]=e_{2},\;f_{1}=\sigma=\kappa_{2}=0,\;\kappa_{1}=f_{2}=1.$

By (38) and (50), together with $c=-1$, (47) reduces to

(51) $\begin{array}[]{ccc}-\Delta^{M}f+2f=0,\;e_{2}(f)=0.\end{array}$

We can easily check that
$f=C_{1}e^{(1+\sqrt{3})\rho}+C_{2}e^{(1-\sqrt{3})\rho}$ satisfies (51), where
$C_{1}$ and $C_{2}$ are positive constants. Thus, we get the proposition. ∎

As an application of Theorem 4.1 with $c=1$, we have

######  Proposition 4.4.

For a positive function $f=f(\rho)$ defined on an open interval
$I\subset(0,\frac{\pi}{2})$ solves the following $ODE$

(52) $f^{\prime\prime}+\frac{2\cos 2\rho-4}{\sin
2\rho}f^{\prime}+\frac{1+2\sin^{2}\rho}{\sin^{2}\rho}f=0,$

then the Riemannian submersion $\pi:(S^{3}\supset I\times S^{1}\times
S^{1},{\rm d}\rho^{2}+\cos^{2}\rho{\rm d}z^{2}+\sin^{2}\rho{\rm
d}\theta^{2})\to(S^{2}\supset I\times S^{1},{\rm d}\rho^{2}+\cos^{2}\rho{\rm
d}z^{2})$ with $\pi(\rho,z,\theta)=(\rho,z)$ is a proper $f$-biharmonic map.

###### Proof.

One can easily check that the orthonormal frame
$\\{e_{1}=\frac{\partial}{\partial\rho},\;e_{2}=\frac{1}{\cos\rho}\frac{\partial}{\partial
z},\;e_{3}=\frac{1}{\sin\rho}\frac{\partial}{\partial\theta}\\}$ on $(I\times
S^{1}\times S^{1},{\rm d}\rho^{2}+\cos^{2}\rho{\rm d}z^{2}+\sin^{2}\rho{\rm
d}\theta^{2})$ is adapted to the Riemannian submersion $\pi$ with ${\rm
d}\pi(e_{i})=\varepsilon_{i},i=1,2$ and vertical $e_{3}$, where
$\\{\varepsilon_{1}=\frac{\partial}{\partial\rho},\;\varepsilon_{2}=\frac{1}{\cos\rho}\frac{\partial}{\partial
z}\\}$ forms an orthonormal frame on the base space $(I\times S^{1},{\rm
d}\rho^{2}+\cos^{2}\rho{\rm d}z^{2})$. A direct computation gives

(53) $\begin{array}[]{lll}[e_{1},e_{3}]=-\cot\rho
e_{3},[e_{2},e_{3}]=0,[e_{1},e_{2}]=\tan\rho e_{2},\\\
f_{1}=\sigma=\kappa_{2}=0,\;\kappa_{1}=-\cot\rho,f_{2}=\tan\rho.\end{array}$

It is easy to check that
$-\Delta^{M}(\kappa_{1})+\kappa_{1}(-1+1/\kappa_{1}^{2})=\frac{1+\sin^{2}\rho\cos
2\rho}{\sin^{3}\rho\cos\rho}\neq 0$. Substituting (53) and $c=1$ into (47),
then we have

(54)
$\begin{array}[]{ccc}-\Delta^{M}(-f\cot\rho)-f\cot\rho(-1+\tan^{2}\rho)=0\;{\rm
and}\;e_{2}(f)=0.\end{array}$

If $f=f(\rho)$ depends on only the variable $\rho$, then substituting this,
(38) and (53) into (54) and simplifying the resulting equation we obtain (52).
By the theory of $ODE$, we conclude that there is a local solution for (52).
Moreover, if $y$ is a solution of (52) on $I\subset(0,\frac{\pi}{2})$, one
finds $-y$ to be a solution of (52) on $I$. Hence, there is a positive
function solution $f=f(\rho)$ on $I$ for (52). Thus, we obtain the
proposition. ∎

Acknowledgments: Both authors would like to thank Prof. Dr. Ye-Lin Ou for his
guidance, help, and encouragement through many invaluable discussions,
suggestions, and stimulating questions during the preparation of this work. A
part of the work was done when Ze-Ping Wang was a visiting scholar at Yunnan
University in Fall 2023. He would like to express his gratitude to Professor
Han-Chun Yang for his invitation and to Yunnan University for the hospitality.

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