Conformal Maps, Biharmonic Maps, and the Warped Product

Abstract

In this paper we study some properties of conformal maps between equidimensional manifolds, we construct new example of non-harmonic biharmonic maps and we characterize the biharmonicity of some maps on the warped product manifolds.

1. Introduction.

Let $\varphi :(M^m,g)\to (N^n,h)$ be a smooth map between Riemannian manifolds. Then $\varphi$ is said to be harmonic if it is a critical point of the energy functional :

$$E\left(\varphi \right)=\frac{1}{2}{\int }_K{\left|d\varphi \right|}^{2}d{v}_g$$

(1)

for any compact subset $K\subset M$. Equivalently, ϕ is harmonic if it satisfies the associated Euler-Lagrange equations :

$$\mathsf{\tau }\left(\varphi \right)=T{r}_g\nabla d\varphi =0,$$

(2)

and $\mathsf{\tau }\left(\varphi \right)$ is called the tension field of ϕ. One can refer to [1,2,3,4] for background on harmonic maps. In the context of harmonic maps, the stress-energy tensor was studied in details by Baird and Eells in [5]. The stress-energy tensor for a map $\varphi :(M^m,g)⟶(N^n,h)$ defined by

$$S\left(\varphi \right)=e\left(\varphi \right)g-{\varphi }^{*}h$$

and the relation between $S\left(\varphi \right)$ and $\mathsf{\tau }\left(\varphi \right)$ is given by

$$divS\left(\varphi \right)=-h\left(\mathsf{\tau }\right(\varphi ),d\varphi ).$$

The map ϕ is said to be biharmonic if it is a critical point of the bi-energy functional :

$$E_2\left(\varphi \right)=\frac{1}{2}{\int }_M{\left|\mathsf{\tau }\left(\varphi \right)\right|}^{2}d{v}_g$$

(3)

Equivalently, ϕ is biharmonic if it satisfies the associated Euler-Lagrange equations :

$${\mathsf{\tau }}_2\left(\varphi \right)=-T{r}_g{\left({\nabla }^{\varphi }\right)}^2\mathsf{\tau }\left(\varphi \right)-T{r}_g{R}^N(\mathsf{\tau }\left(\varphi \right),d\varphi )d\varphi =0,$$

(4)

where ${\nabla }^{\varphi }$ is the connection in the pull-back bundle ${\varphi }^{-1}\left(TN\right)$ and, if ${\left(e_i\right)}_{1\le i\le m}$ is a local orthonormal frame field on M, then

$$T{r}_g{\left({\nabla }^{\varphi }\right)}^2\mathsf{\tau }\left(\varphi \right)=\left({\nabla }_{e_i}^{\varphi }{\nabla }_{e_i}^{\varphi }-{\nabla }_{{\nabla }_{e_i}{e}_i}^{\varphi }\right)\mathsf{\tau }\left(\varphi \right),$$

where we sum over repeated indices. We will call the operator ${\mathsf{\tau }}_2\left(\varphi \right)$, the bi-tension field of the map ϕ. In analogy with harmonic maps, Jiang In [6] has constructed for a map ϕ the stress bi-energy tensor defined by

$$S_2\left(\varphi \right)=\left(\frac{-1}{2}{\left|\mathsf{\tau }\left(\varphi \right)\right|}^2+divh\left(\mathsf{\tau }\left(\varphi \right),d\varphi \right)\right)g-2symh\left(\nabla \mathsf{\tau }\left(\varphi \right),d\varphi \right),$$

where

$$symh\left(\nabla \mathsf{\tau }\left(\varphi \right),d\varphi \right)\left(X,Y\right)=\frac{1}{2}\left\{h\left({\nabla }_X\mathsf{\tau }\left(\varphi \right),d\varphi \left(Y\right)\right)+h\left({\nabla }_Y\mathsf{\tau }\left(\varphi \right),d\varphi \left(X\right)\right)\right\},$$

for any $X,Y\in \Gamma \left(TM\right).$ The stress bi-energy tensor was also studied in [7] and those results could be useful when we study conformal maps. The stress bi-energy tensor of ϕ satisfies the following relationship

$$div{S}_2\left(\varphi \right)=h\left({\mathsf{\tau }}_2\left(\varphi \right),d\varphi \right).$$

Clearly any harmonic map is biharmonic, therefore it is interesting to construct non-harmonic biharmonic maps. In [8] the authors found new examples of biharmonic maps by conformally deforming the domain metric of harmonic ones. While in [9] the author analyzed the behavior of the biharmonic equation under the conformal change the domain metric, she obtained metrics $\tilde{g}=e^{2\gamma }$ such that the idendity map $Id:\left(M,g\right)⟶\left(M,\tilde{g}\right)$ is biharmonic non-harmonic. Moreover, in [10] the author gave some extensions of the result in [9] together with some further constructions of biharmonic maps. The author in [11] deform conformally the codomain metric in order to render a semi-conformal harmonic map biharmonic. In [12] the authors studied the case where $\varphi :\left(M^n,g\right)⟶\left(N^n,h\right)$ is a conformal mapping between equidimensional manifolds where they show that a conformal mapping ϕ is biharmonic if and only if the gradient of its dilation satisfies a second order elliptic partial differential equation. We can refer the reader to [13], for a survey of biharmonic maps. In the first section of this paper, we present some properties for a conformal mapping $\varphi :\left(M^n,g\right)⟶\left(N^n,h\right)$, we prove that the stress bi-energy tensor depend only on the dilation (Theorem 1) and we calculate the bitension field of ϕ (Theorem 2). In the last section we study the biharmonicity of some maps on the warped product (Theorem 4 and 5), with this setting we obtain new examples of biharmonic non-harmonic maps.

2. Some properties for conformal maps.

We study conformal maps between equidimensional manifolds of the same dimension $n\ge 3$. Note that by a result in [12], any such map can have no critical points and so is a local conformal diffeomorphism. Recall that a mapping $\varphi :(M^n,g)\to (N^n,h)$ is called conformal if there exist a $C^{\infty }$ function $\lambda :M\to {\mathbb{R}}_{+}^{*}$ such that for any $X,Y\in \Gamma \left(TM\right)$ :

$$h(d\varphi \left(X\right),d\varphi \left(Y\right))={\lambda }^{2}g(X,Y).$$

The function λ is called the dilation for the map ϕ. The tension field and the stress energy tensor for a conformal map are given by (see [14]):

Proposition 1. 

Let $\varphi :(M^n,g)\to (N^n,h)$ be a conformal map of dilation λ, we have

$$\left(i\right)divS\left(\varphi \right)=(n-2){\lambda }^{2}dln\lambda ,$$

(5)

$$\left(ii\right)divh(\mathsf{\tau }\left(\varphi \right),d\varphi )=(2-n)\left(2{\lambda }^2{\left|gradln\lambda \right|}^2+{\lambda }^2\Delta ln\lambda \right).$$

(6)

$$\left(iii\right)\mathsf{\tau }\left(\varphi \right)=(2-n)d\varphi (gradln\lambda ).$$

(7)

$$\left(iv\right){\left|\mathsf{\tau }\left(\varphi \right)\right|}^2={(2-n)}^2{\lambda }^2{\left|gradln\lambda \right|}^2.$$

(8)

Note that the conformal map $\varphi :(M^n,g)\to (N^n,h)$ of dilation λ is harmonic if and only if $n=2$ or the dilation λ is constant.

In the first, wa calculate the stress bi-energy tensor for a conformal map ϕ when we prove that $S_2\left(\varphi \right)$ depend only the dilation.

Theorem 1. 

Let $\varphi :(M^n,g)\to (N^n,h)$ be a conformal map with dilation λ, then we have

$$S_2\left(\varphi \right)=\left(2-n\right){\lambda }^2\left\{\left(\frac{n-2}{2}{\left|gradln\lambda \right|}^2+\Delta ln\lambda \right)g-2\nabla dln\lambda \right\},$$

(9)

and the trace of $S_2\left(\varphi \right)$ is given by

$$Tr{S}_2\left(\varphi \right)=-{\left(2-n\right)}^2{\lambda }^2\left\{\frac{n}{2}{\left|gradln\lambda \right|}^2+\Delta ln\lambda \right\}.$$

(10)

To prove Theorem 1, we need the following Lemma:

Lemma 1. 

Let $\varphi :(M^n,g)\to (N^n,h)$ be a conformal map with dilation λ, then for any function $f\in C^{\infty }\left(M\right)$ and for any $X,Y\in \Gamma \left(TM\right)$, we have

$$\begin{array}{cc}\hfill h\left({\nabla }_{X}d\varphi \left(gradf\right),d\varphi \left(Y\right)\right)& ={\lambda }^2\left(X\left(ln\lambda \right)Y\left(f\right)-Y\left(ln\lambda \right)X\left(f\right)\right)\hfill \\ & +{\lambda }^2\nabla df\left(X,Y\right)+{\lambda }^{2}dln\lambda \left(gradf\right)g\left(X,Y\right).\hfill \end{array}$$

(11)

Proof of Lemma 1. Let $f\in C^{\infty }\left(M\right)$, for any $X,Y\in \Gamma \left(TM\right)$, we have

$$\begin{array}{cc}\hfill h\left({\nabla }_{X}d\varphi \left(gradf\right),d\varphi \left(Y\right)\right)& =X\left({\lambda }^{2}g\left(gradf,Y\right)\right)-h\left(d\varphi \left(gradf\right),{\nabla }_{X}d\varphi \left(Y\right)\right)\hfill \\ & =X\left({\lambda }^2\right)g\left(gradf,Y\right)+{\lambda }^{2}g\left({\nabla }_{X}gradf,Y\right)+{\lambda }^{2}g\left(gradf,{\nabla }_{X}Y\right)\hfill \\ & -h\left(d\varphi \left(gradf\right),\nabla d\varphi \left(X,Y\right)\right)-h\left(d\varphi \left(gradf\right),d\varphi \left({\nabla }_{X}Y\right)\right)\hfill \\ & =X\left({\lambda }^2\right)g\left(gradf,Y\right)+{\lambda }^{2}g\left({\nabla }_{X}gradf,Y\right)+{\lambda }^{2}g\left(gradf,{\nabla }_{X}Y\right)\hfill \\ & -h\left(d\varphi \left(gradf\right),\nabla d\varphi \left(X,Y\right)\right)-{\lambda }^{2}g\left(gradf,{\nabla }_{X}Y\right).\hfill \end{array}$$

Note that

$$g\left({\nabla }_{X}gradf,Y\right)=\nabla df\left(X,Y\right),$$

then we obtain

$$h\left({\nabla }_{X}d\varphi \left(gradf\right),d\varphi \left(Y\right)\right)=2{\lambda }^{2}X\left(ln\lambda \right)Y\left(f\right)+{\lambda }^2\nabla df\left(X,Y\right)-h\left(d\varphi \left(gradf\right),\nabla d\varphi \left(X,Y\right)\right).$$

By similary, we have

$$h\left({\nabla }_{Y}d\varphi \left(gradf\right),d\varphi \left(X\right)\right)=2{\lambda }^{2}Y\left(ln\lambda \right)X\left(f\right)+{\lambda }^2\nabla df\left(X,Y\right)-h\left(d\varphi \left(gradf\right),\nabla d\varphi \left(X,Y\right)\right).$$

Then, we deduce that

$$\begin{array}{cc}\hfill h\left({\nabla }_{X}d\varphi \left(gradf\right),d\varphi \left(Y\right)\right)& =h\left(d\varphi \left(X\right),{\nabla }_{Y}d\varphi \left(gradf\right)\right)\hfill \\ & +2{\lambda }^2\left(X\left(ln\lambda \right)Y\left(f\right)-Y\left(ln\lambda \right)X\left(f\right)\right).\hfill \end{array}$$

(12)

For the term $h\left(d\varphi \left(X\right),{\nabla }_{Y}d\varphi \left(gradf\right)\right)$, we have

$$\begin{array}{cc}\hfill h\left({\nabla }_{Y}d\varphi \left(gradf\right),d\varphi \left(X\right)\right)& =h\left(\nabla d\varphi \left(gradf,Y\right),d\varphi \left(X\right)\right)+{\lambda }^{2}g\left({\nabla }_{Y}gradf,X\right)\hfill \\ & =h\left({\nabla }_{gradf}d\varphi \left(Y\right),d\varphi \left(X\right)\right)-{\lambda }^{2}g\left({\nabla }_{gradf}Y,X\right)\hfill \\ & +{\lambda }^{2}g\left({\nabla }_{Y}gradf,X\right)\hfill \\ & =gradf\left({\lambda }^{2}g\left(X,Y\right)\right)-h\left({\nabla }_{gradf}d\varphi \left(X\right),d\varphi \left(Y\right)\right)\hfill \\ & -{\lambda }^{2}g\left({\nabla }_{gradf}Y,X\right)+{\lambda }^{2}g\left({\nabla }_{Y}gradf,X\right)\hfill \\ & =2{\lambda }^{2}dln\lambda \left(gradf\right)g\left(X,Y\right)-h\left(\nabla d\varphi \left(X,gradf\right),d\varphi \left(Y\right)\right)\hfill \\ & +{\lambda }^{2}g\left({\nabla }_{Y}gradf,X\right).\hfill \end{array}$$

We deduce that

$$\begin{array}{cc}\hfill h\left({\nabla }_{Y}d\varphi \left(gradf\right),d\varphi \left(X\right)\right)& =-h\left({\nabla }_{X}d\varphi \left(gradf\right),d\varphi \left(Y\right)\right)+2{\lambda }^2\nabla df\left(X,Y\right)\hfill \\ & +2{\lambda }^{2}dln\lambda \left(gradf\right)g\left(X,Y\right).\hfill \end{array}$$

(13)

Finally, if we replace (13) in (12), we obtain

$$\begin{array}{cc}\hfill h\left({\nabla }_{X}d\varphi \left(gradf\right),d\varphi \left(Y\right)\right)& ={\lambda }^2\left(X\left(ln\lambda \right)Y\left(f\right)-Y\left(ln\lambda \right)X\left(f\right)\right)\hfill \\ & +{\lambda }^2\nabla df\left(X,Y\right)+{\lambda }^{2}dln\lambda \left(gradf\right)g\left(X,Y\right).\hfill \end{array}$$

This completes the proof of Lemma 1.

Remark 1. 

Let $\varphi :(M^n,g)\to (N^n,h)$ be a conformal map with dilation λ, then if we consider $f=ln\lambda$, the equation (11) gives

$$h\left({\nabla }_{X}d\varphi \left(gradln\lambda \right),d\varphi \left(Y\right)\right)={\lambda }^2\left(\nabla dln\lambda \left(X,Y\right)+{\left|gradln\lambda \right|}^{2}g\left(X,Y\right)\right).$$

Proof of Theorem 1. By definition, the stress bi-energy tensor is given by :

$$S_2\left(\varphi \right)=\left(-\frac{1}{2}{\left|\mathsf{\tau }\left(\varphi \right)\right|}^2+divh\left(\mathsf{\tau }\left(\varphi \right),d\varphi \right)\right)g-2symh\left(\nabla \mathsf{\tau }\left(\varphi \right),d\varphi \right).$$

(14)

Using the equations (2) et (4) for the Proposition 1, we have

$$-\frac{1}{2}{\left|\mathsf{\tau }\left(\varphi \right)\right|}^2+divh\left(\mathsf{\tau }\left(\varphi \right),d\varphi \right)=(2-n){\lambda }^2\left(\frac{n+2}{2}{\left|gradln\lambda \right|}^2+\Delta ln\lambda \right).$$

(15)

Calculate now $symh\left(\nabla \mathsf{\tau }\left(\varphi \right),d\varphi \right)$, we have by definition for any $X,Y\in \Gamma \left(TM\right)$

$$\begin{array}{cc}\hfill symh\left(\nabla \mathsf{\tau }\left(\varphi \right),d\varphi \right)\left(X,Y\right)& =\frac{1}{2}\left(h\left({\nabla }_X\mathsf{\tau }\left(\varphi \right),d\varphi \left(Y\right)\right)+h\left({\nabla }_Y\mathsf{\tau }\left(\varphi \right),d\varphi \left(X\right)\right)\right)\hfill \\ & =\frac{2-n}{2}\left(h\left({\nabla }_{X}d\varphi \left(gradln\lambda \right),d\varphi \left(Y\right)\right)+h\left({\nabla }_Y\left(gradln\lambda \right),d\varphi \left(X\right)\right)\right).\hfill \end{array}$$

By Lemma 1, we have

$$h\left({\nabla }_{X}d\varphi \left(gradln\lambda \right),d\varphi \left(Y\right)\right)={\lambda }^2\left(\nabla dln\lambda \left(X,Y\right)+{\left|gradln\lambda \right|}^{2}g\left(X,Y\right)\right)$$

and

$$h\left({\nabla }_{Y}d\varphi \left(gradln\lambda \right),d\varphi \left(X\right)\right)={\lambda }^2\left(\nabla dln\lambda \left(X,Y\right)+{\left|gradln\lambda \right|}^{2}g\left(X,Y\right)\right),$$

then

$$symh\left(\nabla \mathsf{\tau }\left(\varphi \right),d\varphi \right)\left(X,Y\right)=\left(2-n\right){\lambda }^2\left(\nabla dln\lambda \left(X,Y\right)+{\left|gradln\lambda \right|}^{2}g\left(X,Y\right)\right).$$

(16)

If we substitute (15) and (16) in (14), we conclude that

$$S_2\left(\varphi \right)=\left(2-n\right){\lambda }^2\left\{\left(\frac{n-2}{2}{\left|gradln\lambda \right|}^2+\Delta ln\lambda \right)g-2\nabla dln\lambda \right\}$$

Calculate now the trace of stress bi-energy tensor. Let ${\left(e_i\right)}_{1\le i\le n}$ be an orthonormal frame on M, we have

$$\begin{array}{cc}\hfill T{r}_g{S}_2\left(\varphi \right)& =S_2\left(\varphi \right)(e_i,e_i)\hfill \\ & =\left(2-n\right){\lambda }^2\left(\frac{n-2}{2}{\left|gradln\lambda \right|}^2+\Delta ln\lambda \right)g\left(e_i,e_i\right)\hfill \\ & -2\left(2-n\right){\lambda }^2\nabla dln\lambda \left(e_i,e_i\right)\hfill \\ & =\left(2-n\right)n{\lambda }^2\left(\frac{n-2}{2}{\left|gradln\lambda \right|}^2+\Delta ln\lambda \right)\hfill \\ & -2\left(2-n\right){\lambda }^2\left(\Delta ln\lambda \right)\hfill \\ & =\left(2-n\right){\lambda }^2\left\{\frac{n\left(n-2\right)}{2}{\left|gradln\lambda \right|}^2+\left(n-2\right)\Delta ln\lambda \right\}.\hfill \end{array}$$

Then

$$Tr{S}_2\left(\varphi \right)=-{\left(2-n\right)}^2{\lambda }^2\left\{\frac{n}{2}{\left|gradln\lambda \right|}^2+\Delta ln\lambda \right\}.$$

By calculating the Laplacian of the function ${\lambda }^{\frac{n}{2}}$ and by using

$$\Delta {\lambda }^{\frac{n}{2}}=\frac{n}{2}{\lambda }^{\frac{n}{2}}\left(\frac{n}{2}{\left|gradln\lambda \right|}^2+\Delta ln\lambda \right),$$

we obtain immediately the following corollary

Corollary 1. 

Let $\varphi :(M^n,g)\to (N^n,h)$, $(n\ne 2)$ to be a conformal map of dilation λ, then the trace of $S_2\left(\varphi \right)$ is zero if and only if the function ${\lambda }^{\frac{n}{2}}$ is harmonic.

The bi-tension field of the conformal map is given by

Theorem 2. 

Let $\varphi :(M^n,g)\to (N^n,h)$, ($n\ge 3$) to be a conformal map of dilation λ, then bi-tension field of ϕ is given by :

$${\mathsf{\tau }}_2\left(\varphi \right)=\left(n-2\right)d\varphi \left(H\right)$$

where

$$\begin{array}{cc}\hfill H& =grad\Delta ln\lambda -\frac{\left(n-6\right)}{2}grad\left({\left|gradln\lambda \right|}^2\right)+2Ricc{i}^M\left(gradln\lambda \right)\hfill \\ & -\left(2\left(\Delta ln\lambda \right)+\left(n-2\right){\left|gradln\lambda \right|}^2\right)gradln\lambda .\hfill \end{array}$$

(17)

Remark 2. 

A. Balmus in [9] studied the case where $\varphi =I{d}_M$, she obtained the biharmonicity of the identity map from $\left(M,g\right)$ onto $\left(M,{\lambda }^{2}g\right)$, this case was also studied in [15].

To prove the Theorem 2, we need two Lemmas. In the first Lemma, we give a simple formula of the term $T{r}_g{\left({\nabla }^{\varphi }\right)}^{2}d\varphi \left(grad\gamma \right)$ for a conformal map $\varphi :(M^n,g)\to (N^n,h)$ ($n\ge 3$) of dilation λ and for any function $\gamma \in C^{\infty }\left(M\right)$.

Lemma 2. 

Let $\varphi :(M^n,g)\to (N^n,h)$ ($n\ge 3$) to be a conformal map of dilation λ, then for any function $\gamma \in C^{\infty }\left(M\right)$, we have

$$\begin{array}{cc}\hfill T{r}_g{\left({\nabla }^{\varphi }\right)}^{2}d\varphi \left(grad\gamma \right)& =d\varphi \left(grad\Delta \gamma \right)+4d\varphi \left({\nabla }_{gradln\lambda }grad\gamma \right)+d\varphi \left(Ricc{i}^M\left(grad\gamma \right)\right)\hfill \\ & +\left(\Delta ln\lambda \right)d\varphi \left(grad\gamma \right)-2\left(\Delta \gamma \right)d\varphi \left(gradln\lambda \right)\hfill \\ & -\left(n-2\right)dln\lambda \left(grad\gamma \right)d\varphi \left(gradln\lambda \right).\hfill \end{array}$$

(18)

Proof of Lemma 2. Let $\gamma \in C^{\infty }\left(M\right)$, by definition, we have

$$T{r}_g{\left({\nabla }^{\varphi }\right)}^{2}d\varphi \left(grad\gamma \right)={\nabla }_{e_i}^{\varphi }{\nabla }_{e_i}^{\varphi }d\varphi \left(grad\gamma \right)-{\nabla }_{{\nabla }_{e_i}{e}_i}^{\varphi }d\varphi \left(grad\gamma \right).$$

(19)

(Here henceforth we sum over repeated indices.) Let us start with the calculation of the term ${\nabla }_{e_i}^{\varphi }{\nabla }_{e_i}^{\varphi }d\varphi \left(grad\gamma \right),$ we have

$${\nabla }_{e_i}^{\varphi }d\varphi \left(grad\gamma \right)=\nabla d\varphi \left(e_i,grad\gamma \right)+d\varphi \left({\nabla }_{e_i}grad\gamma \right).$$

It is known that (see [16])

$$\nabla d\varphi \left(e_i,grad\gamma \right)=e_i\left(ln\lambda \right)d\varphi \left(grad\gamma \right)+dln\lambda \left(grad\gamma \right)d\varphi \left(e_i\right)-e_i\left(\gamma \right)d\varphi \left(gradln\lambda \right),$$

then

$$\begin{array}{cc}\hfill {\nabla }_{e_i}^{\varphi }d\varphi \left(grad\gamma \right)& =e_i\left(ln\lambda \right)d\varphi \left(grad\gamma \right)+dln\lambda \left(grad\gamma \right)d\varphi \left(e_i\right)\hfill \\ & -e_i\left(\gamma \right)d\varphi \left(gradln\lambda \right)+d\varphi \left({\nabla }_{e_i}grad\gamma \right).\hfill \end{array}$$

(20)

It follows that

$$\begin{array}{cc}\hfill {\nabla }_{e_i}^{\varphi }{\nabla }_{e_i}^{\varphi }d\varphi \left(grad\gamma \right)& ={\nabla }_{e_i}^{\varphi }\left\{e_i\left(ln\lambda \right)d\varphi \left(grad\gamma \right)\right\}+{\nabla }_{e_i}^{\varphi }\left\{dln\lambda \left(grad\gamma \right)d\varphi \left(e_i\right)\right\}\hfill \\ & -{\nabla }_{e_i}^{\varphi }\left\{e_i\left(\gamma \right)d\varphi \left(gradln\lambda \right)\right\}+{\nabla }_{e_i}^{\varphi }d\varphi \left({\nabla }_{e_i}grad\gamma \right).\hfill \end{array}$$

(21)

We will study term by term the right-hand of this expression. For the first term ${\nabla }_{e_i}^{\varphi }\left\{e_i\left(ln\lambda \right)d\varphi \left(grad\gamma \right)\right\}$, we have

$${\nabla }_{e_i}^{\varphi }\left\{e_i\left(ln\lambda \right)d\varphi \left(grad\gamma \right)\right\}=e_i\left(ln\lambda \right){\nabla }_{e_i}^{\varphi }d\varphi \left(grad\gamma \right)+e_i\left(e_i\left(ln\lambda \right)\right)d\varphi \left(grad\gamma \right).$$

By using the Equation (20), we deduce that

$$\begin{array}{cc}\hfill {\nabla }_{e_i}^{\varphi }\left\{e_i\left(ln\lambda \right)d\varphi \left(grad\gamma \right)\right\}& =e_i\left(ln\lambda \right)e_i\left(ln\lambda \right)d\varphi \left(grad\gamma \right)+e_i\left(ln\lambda \right)dln\lambda \left(grad\gamma \right)d\varphi \left(e_i\right)\hfill \\ & -e_i\left(ln\lambda \right)e_i\left(\gamma \right)d\varphi \left(gradln\lambda \right)+e_i\left(ln\lambda \right)d\varphi \left({\nabla }_{e_i}grad\gamma \right)\hfill \\ & +e_i\left(e_i\left(ln\lambda \right)\right)d\varphi \left(grad\gamma \right),\hfill \end{array}$$

then, we obtain

$$\begin{array}{cc}\hfill {\nabla }_{e_i}^{\varphi }\left\{e_i\left(ln\lambda \right)d\varphi \left(grad\gamma \right)\right\}& ={\left|gradln\lambda \right|}^{2}d\varphi \left(grad\gamma \right)+d\varphi \left({\nabla }_{gradln\lambda }grad\gamma \right)\hfill \\ & +e_i\left(e_i\left(ln\lambda \right)\right)d\varphi \left(grad\gamma \right).\hfill \end{array}$$

(22)

For the second term ${\nabla }_{e_i}^{\varphi }\left\{dln\lambda \left(grad\gamma \right)d\varphi \left(e_i\right)\right\}$, a similar calculation gives

$$\begin{array}{cc}\hfill {\nabla }_{e_i}^{\varphi }\left\{dln\lambda \left(grad\gamma \right)d\varphi \left(e_i\right)\right\}& =dln\lambda \left(grad\gamma \right){\nabla }_{e_i}^{\varphi }d\varphi \left(e_i\right)+e_i\left\{g\left(gradln\lambda ,grad\gamma \right)\right\}d\varphi \left(e_i\right)\hfill \\ & =dln\lambda \left(grad\gamma \right){\nabla }_{e_i}^{\varphi }d\varphi \left(e_i\right)+g\left({\nabla }_{e_i}gradln\lambda ,grad\gamma \right)d\varphi \left(e_i\right)\hfill \\ & +g\left(gradln\lambda ,{\nabla }_{e_i}grad\gamma \right)d\varphi \left(e_i\right)\hfill \\ & =dln\lambda \left(grad\gamma \right){\nabla }_{e_i}^{\varphi }d\varphi \left(e_i\right)+g\left({\nabla }_{grad\gamma }gradln\lambda ,e_i\right)d\varphi \left(e_i\right)\hfill \\ & +g\left({\nabla }_{gradln\lambda }grad\gamma ,e_i\right)d\varphi \left(e_i\right),\hfill \end{array}$$

it follows that

$$\begin{array}{cc}\hfill {\nabla }_{e_i}^{\varphi }\left\{dln\lambda \left(grad\gamma \right)d\varphi \left(e_i\right)\right\}& =dln\lambda \left(grad\gamma \right){\nabla }_{e_i}^{\varphi }d\varphi \left(e_i\right)+d\varphi \left({\nabla }_{grad\gamma }gradln\lambda \right)\hfill \\ & +d\varphi \left({\nabla }_{gradln\lambda }grad\gamma \right).\hfill \end{array}$$

(23)

For the third term ${\nabla }_{e_i}^{\varphi }\left\{e_i\left(\gamma \right)d\varphi \left(gradln\lambda \right)\right\}$, by using the same calculation method and the equation (20), we have

$$\begin{array}{cc}\hfill {\nabla }_{e_i}^{\varphi }\left\{e_i\left(\gamma \right)d\varphi \left(gradln\lambda \right)\right\}& =e_i\left(\gamma \right){\nabla }_{e_i}^{\varphi }d\varphi \left(gradln\lambda \right)+e_i\left(e_i\left(\gamma \right)\right)d\varphi \left(gradln\lambda \right)\hfill \\ & =e_i\left(\gamma \right)e_i\left(ln\lambda \right)d\varphi \left(gradln\lambda \right)+e_i\left(\gamma \right)dln\lambda \left(gradln\lambda \right)d\varphi \left(e_i\right)\hfill \\ & -e_i\left(\gamma \right)e_i\left(ln\lambda \right)d\varphi \left(gradln\lambda \right)+e_i\left(\gamma \right)d\varphi \left({\nabla }_{e_i}gradln\lambda \right)\hfill \\ & +e_i\left(e_i\left(\gamma \right)\right)d\varphi \left(gradln\lambda \right),\hfill \end{array}$$

which gives us

$$\begin{array}{cc}\hfill {\nabla }_{e_i}^{\varphi }\left\{e_i\left(\gamma \right)d\varphi \left(gradln\lambda \right)\right\}& ={\left|gradln\lambda \right|}^{2}d\varphi \left(grad\gamma \right)+d\varphi \left({\nabla }_{grad\gamma }gradln\lambda \right)\hfill \\ & +e_i\left(e_i\left(\gamma \right)\right)d\varphi \left(gradln\lambda \right).\hfill \end{array}$$

(24)

Now let us look at the last term ${\nabla }_{e_i}^{\varphi }d\varphi \left({\nabla }_{e_i}grad\gamma \right)$, a simple calculation gives

$$\begin{array}{cc}\hfill {\nabla }_{e_i}^{\varphi }d\varphi \left({\nabla }_{e_i}grad\gamma \right)& =e_i\left(ln\lambda \right)d\varphi \left({\nabla }_{e_i}grad\gamma \right)+dln\lambda \left({\nabla }_{e_i}grad\gamma \right)d\varphi \left(e_i\right)\hfill \\ & -g\left(e_i,{\nabla }_{e_i}grad\gamma \right)d\varphi \left(gradln\lambda \right)+d\varphi \left({\nabla }_{e_i}{\nabla }_{e_i}grad\gamma \right)\hfill \\ & =2d\varphi \left({\nabla }_{gradln\lambda }grad\gamma \right)-\left(\Delta \gamma \right)d\varphi \left(gradln\lambda \right)\hfill \\ & +d\varphi \left({\nabla }_{e_i}{\nabla }_{e_i}grad\gamma \right),\hfill \end{array}$$

then

$$\begin{array}{cc}\hfill {\nabla }_{e_i}^{\varphi }d\varphi \left({\nabla }_{e_i}grad\gamma \right)& =d\varphi \left({\nabla }_{e_i}{\nabla }_{e_i}grad\gamma \right)+2d\varphi \left({\nabla }_{gradln\lambda }grad\gamma \right)\hfill \\ & -\left(\Delta \gamma \right)d\varphi \left(gradln\lambda \right).\hfill \end{array}$$

(25)

If we replace (22), (23), (24) and (25) in (21), we obtain

$$\begin{array}{cc}\hfill {\nabla }_{e_i}^{\varphi }{\nabla }_{e_i}^{\varphi }d\varphi \left(grad\gamma \right)& =4d\varphi \left({\nabla }_{gradln\lambda }grad\gamma \right)+e_i\left(e_i\left(ln\lambda \right)\right)d\varphi \left(grad\gamma \right)\hfill \\ & +dln\lambda \left(grad\gamma \right){\nabla }_{e_i}^{\varphi }d\varphi \left(e_i\right)-e_i\left(e_i\left(\gamma \right)\right)d\varphi \left(gradln\lambda \right)\hfill \\ & +d\varphi \left({\nabla }_{e_i}{\nabla }_{e_i}grad\gamma \right)-\left(\Delta \gamma \right)d\varphi \left(gradln\lambda \right).\hfill \end{array}$$

(26)

To complete the proof, it remains to investigate the term ${\nabla }_{{\nabla }_{e_i}{e}_i}^{\varphi }d\varphi \left(grad\gamma \right)$, we have

$${\nabla }_{{\nabla }_{e_i}{e}_i}^{\varphi }d\varphi \left(grad\gamma \right)=\nabla d\varphi \left({\nabla }_{e_i}{e}_i,grad\gamma \right)+d\varphi \left({\nabla }_{{\nabla }_{e_i}{e}_i}grad\gamma \right),$$

Therefore, by using the equation (20), we obtain

$$\begin{array}{cc}\hfill {\nabla }_{{\nabla }_{e_i}{e}_i}^{\varphi }d\varphi \left(grad\gamma \right)& ={\nabla }_{e_i}{e}_i\left(ln\lambda \right)d\varphi \left(grad\gamma \right)+dln\lambda \left(grad\gamma \right)d\varphi \left({\nabla }_{e_i}{e}_i\right)\hfill \\ & -{\nabla }_{e_i}{e}_i\left(\gamma \right)d\varphi \left(gradln\lambda \right)+d\varphi \left({\nabla }_{{\nabla }_{e_i}{e}_i}grad\gamma \right).\hfill \end{array}$$

(27)

By substituting (26) and (27) in (19), we deduce

$$\begin{array}{cc}\hfill T{r}_g{\left({\nabla }^{\varphi }\right)}^{2}d\varphi \left(grad\gamma \right)& ={\nabla }_{e_i}^{\varphi }{\nabla }_{e_i}^{\varphi }d\varphi \left(grad\gamma \right)-{\nabla }_{{\nabla }_{e_i}{e}_i}^{\varphi }d\varphi \left(grad\gamma \right)\hfill \\ & =d\varphi \left(T{r}_g{\nabla }^{2}grad\gamma \right)+4d\varphi \left({\nabla }_{gradln\lambda }grad\gamma \right)\hfill \\ & +\left(\Delta ln\lambda \right)d\varphi \left(grad\gamma \right)+dln\lambda \left(grad\gamma \right)\mathsf{\tau }\left(\varphi \right)\hfill \\ & -2\left(\Delta \gamma \right)d\varphi \left(gradln\lambda \right).\hfill \end{array}$$

Finally, using the fact that (see [11])

$$T{r}_g{\nabla }^{2}grad\gamma =grad\Delta \gamma +Ricc{i}^M\left(grad\gamma \right)$$

and

$$\mathsf{\tau }\left(\varphi \right)=\left(2-n\right)d\varphi \left(gradln\lambda \right),$$

we conclude that

$$\begin{array}{cc}\hfill T{r}_g{\left({\nabla }^{\varphi }\right)}^{2}d\varphi \left(grad\gamma \right)& =d\varphi \left(grad\Delta \gamma \right)+4d\varphi \left({\nabla }_{gradln\lambda }grad\gamma \right)+d\varphi \left(Ricc{i}^M\left(grad\gamma \right)\right)\hfill \\ & +\left(\Delta ln\lambda \right)d\varphi \left(grad\gamma \right)-2\left(\Delta \gamma \right)d\varphi \left(gradln\lambda \right)\hfill \\ & -\left(n-2\right)dln\lambda \left(grad\gamma \right)d\varphi \left(gradln\lambda \right).\hfill \end{array}$$

This completes the proof of Lemma 2. Now, in the second Lemma, we will calculate $T{r}_g{R}^N\left(d\varphi \left(grad\gamma \right),d\varphi \right)d\varphi$ for a conformal maps $\varphi :(M^n,g)\to (N^n,h)$ ($n\ge 3$) of dilation λ and for any function $\gamma \in C^{\infty }\left(M\right)$

Lemma 3. 

Let $\varphi :(M^n,g)\to (N^n,h)$ ($n\ge 3$) to be a conformal map of dilation λ, then for any function $\gamma \in C^{\infty }\left(M\right)$, we have

$$\begin{array}{cc}\hfill T{r}_g{R}^N\left(d\varphi \left(grad\gamma \right),d\varphi \right)d\varphi & =d\varphi \left(Ricc{i}^M\left(grad\gamma \right)\right)-\left(n-2\right)d\varphi \left({\nabla }_{grad\gamma }gradln\lambda \right)\hfill \\ & -\left(\Delta ln\lambda +\left(n-2\right){\left|gradln\lambda \right|}^2\right)d\varphi \left(grad\gamma \right)\hfill \\ & +\left(n-2\right)dln\lambda \left(grad\gamma \right)d\varphi \left(gradln\lambda \right)\hfill \end{array}$$

(28)

Proof of Lemma 3. Let $\gamma \in C^{\infty }\left(M\right)$, by definition we have

$$T{r}_g{R}^N\left(d\varphi \left(grad\gamma \right),d\varphi \right)d\varphi =R^N\left(d\varphi \left(grad\gamma \right),d\varphi \left(e_i\right)\right)d\varphi \left(e_i\right)$$

(29)

but we know that (see [16])

$$\begin{array}{cc}\hfill Ri{c}^N\left(d\varphi \left(X\right),d\varphi \left(Y\right)\right)& =Ri{c}^M\left(X,Y\right)+\left(n-2\right)X\left(ln\lambda \right)Y\left(ln\lambda \right)\hfill \\ & -\left(n-2\right){\left|gradln\lambda \right|}^{2}g\left(X,Y\right)\hfill \\ & -\left(n-2\right)\nabla dln\lambda \left(X,Y\right)-\left(\Delta ln\lambda \right)g\left(X,Y\right).\hfill \end{array}$$

Then

$$\begin{array}{cc}\hfill Ri{c}^N\left(d\varphi \left(grad\gamma \right),d\varphi \left(e_i\right)\right)& =Ri{c}^M\left(grad\gamma ,e_i\right)+\left(n-2\right)grad\gamma \left(ln\lambda \right)e_i\left(ln\lambda \right)\hfill \\ & -\left(n-2\right){\left|gradln\lambda \right|}^{2}g\left(grad\gamma ,e_i\right)\hfill \\ & -\left(n-2\right)\nabla dln\lambda \left(grad\gamma ,e_i\right)-\left(\Delta ln\lambda \right)g\left(grad\gamma ,e_i\right)\hfill \end{array}$$

it follows that

$$\begin{array}{cc}\hfill Ri{c}^N\left(d\varphi \left(grad\gamma \right),d\varphi \left(e_i\right)\right)& =Ri{c}^M\left(grad\gamma ,e_i\right)+\left(m-2\right)dln\lambda \left(grad\gamma \right)e_i\left(ln\lambda \right)\hfill \\ & -\left(n-2\right){\left|gradln\lambda \right|}^2{e}_i\left(\gamma \right)-\left(n-2\right)\nabla dln\lambda \left(grad\gamma ,e_i\right)\hfill \\ & -\left(\Delta ln\lambda \right)e_i\left(\gamma \right).\hfill \end{array}$$

(30)

If we replace (30) in (29), we deduce that

$$\begin{array}{cc}\hfill T{r}_g{R}^N\left(d\varphi \left(grad\gamma \right),d\varphi \right)d\varphi & =R^N\left(d\varphi \left(grad\gamma \right),d\varphi \left(e_i\right)\right)d\varphi \left(e_i\right)\hfill \\ & =d\varphi \left(Ricc{i}^M\left(grad\gamma \right)\right)+\left(n-2\right)dln\lambda \left(grad\gamma \right)d\varphi \left(gradln\lambda \right)\hfill \\ & -\left(n-2\right){\left|gradln\lambda \right|}^{2}d\varphi \left(grad\gamma \right)-\left(n-2\right)\nabla dln\lambda \left(grad\gamma ,e_i\right)d\varphi \left(e_i\right)\hfill \\ & -\left(\Delta ln\lambda \right)d\varphi \left(grad\gamma \right).\hfill \end{array}$$

To complete the proof, we will simplify the term $\nabla dln\lambda \left(grad\gamma ,e_i\right)d\varphi \left(e_i\right)$, we obtain

$$\begin{array}{cc}\hfill \nabla dln\lambda \left(grad\gamma ,e_i\right)d\varphi \left(e_i\right)& =\left\{e_i\left(g\left(gradln\lambda ,grad\gamma \right)\right)-dln\lambda \left({\nabla }_{e_i}grad\gamma \right)\right\}d\varphi \left(e_i\right)\hfill \\ & =g\left({\nabla }_{e_i}gradln\lambda ,grad\gamma \right)d\varphi \left(e_i\right)\hfill \\ & =g\left({\nabla }_{grad\gamma }gradln\lambda ,e_i\right)d\varphi \left(e_i\right)\hfill \\ & =d\varphi \left({\nabla }_{grad\gamma }gradln\lambda \right),\hfill \end{array}$$

which finally gives

$$\begin{array}{cc}\hfill T{r}_g{R}^N\left(d\varphi \left(grad\gamma \right),d\varphi \right)d\varphi & =d\varphi \left(Ricc{i}^M\left(grad\gamma \right)\right)-\left(n-2\right)d\varphi \left({\nabla }_{grad\gamma }gradln\lambda \right)\hfill \\ & -\left(\Delta ln\lambda +\left(n-2\right){\left|gradln\lambda \right|}^2\right)d\varphi \left(grad\gamma \right)\hfill \\ & +\left(n-2\right)dln\lambda \left(grad\gamma \right)d\varphi \left(gradln\lambda \right).\hfill \end{array}$$

This completes the proof of Lemma 3. We are now able to prove Theorem 2.

Proof of Theorem 2. By definition, the bi-tension field is given by

$${\mathsf{\tau }}_2\left(\varphi \right)=-T{r}_g{\left({\nabla }^{\varphi }\right)}^2\mathsf{\tau }\left(\varphi \right)-T{r}_g{R}^N\left(\mathsf{\tau }\left(\varphi \right),d\varphi \right)d\varphi .$$

The tension field of the conformal map ϕ is given by

$$\mathsf{\tau }\left(\varphi \right)=\left(2-n\right)d\varphi \left(gradln\lambda \right),$$

it follows that

$${\mathsf{\tau }}_2\left(\varphi \right)=\left(n-2\right)\left(T{r}_g{\left({\nabla }^{\varphi }\right)}^{2}d\varphi \left(gradln\lambda \right)+T{r}_g{R}^N\left(d\varphi \left(gradln\lambda \right),d\varphi \right)d\varphi \right).$$

(31)

By Lemma 2, we have

$$\begin{array}{cc}\hfill T{r}_g{\left({\nabla }^{\varphi }\right)}^{2}d\varphi \left(gradln\lambda \right)& =d\varphi \left(grad\Delta ln\lambda \right)+2d\varphi \left(grad\left({\left|gradln\lambda \right|}^2\right)\right)\hfill \\ & -\left(\Delta ln\lambda \right)d\varphi \left(gradln\lambda \right)+d\varphi \left(Ricc{i}^M\left(gradln\lambda \right)\right)\hfill \\ & -\left(n-2\right){\left|gradln\lambda \right|}^{2}d\varphi \left(gradln\lambda \right).\hfill \end{array}$$

(32)

By using lemma 3 and the fact that ${\nabla }_{gradln\lambda }gradln\lambda =\frac{1}{2}grad\left({\left|gradln\lambda \right|}^2\right)$

$$\begin{array}{cc}\hfill T{r}_g{R}^N\left(d\varphi \left(gradln\lambda \right),d\varphi \right)d\varphi & =d\varphi \left(Ricc{i}^M\left(gradln\lambda \right)\right)-\left(\Delta ln\lambda \right)d\varphi \left(gradln\lambda \right)\hfill \\ & -\frac{\left(n-2\right)}{2}d\varphi \left(grad\left({\left|gradln\lambda \right|}^2\right)\right).\hfill \end{array}$$

(33)

If we replace (32) and (33) in (31), we deduce that

$$\begin{array}{cc}\hfill {\mathsf{\tau }}_2\left(\varphi \right)& =\left(n-2\right)d\varphi \left(grad\Delta ln\lambda \right)-\frac{\left(n-2\right)\left(n-6\right)}{2}d\varphi \left(grad\left({\left|gradln\lambda \right|}^2\right)\right)\hfill \\ & -\left(n-2\right)\left(2\left(\Delta ln\lambda \right)+\left(n-2\right){\left|gradln\lambda \right|}^2\right)d\varphi \left(gradln\lambda \right)\hfill \\ & +2\left(n-2\right)d\varphi \left(Ricc{i}^M\left(gradln\lambda \right)\right).\hfill \end{array}$$

Then the bi-tension field of ϕ is given by :

$${\mathsf{\tau }}_2\left(\varphi \right)=\left(n-2\right)d\varphi \left(H\right)$$

where

$$\begin{array}{cc}\hfill H& =grad\Delta ln\lambda -\frac{\left(n-6\right)}{2}grad\left({\left|gradln\lambda \right|}^2\right)+2Ricc{i}^M\left(gradln\lambda \right)\hfill \\ & -\left(2\left(\Delta ln\lambda \right)+\left(n-2\right){\left|gradln\lambda \right|}^2\right)gradln\lambda .\hfill \end{array}$$

The proof of Theorem 2 is complete. By application of Theorem 2, we get the following result (see [15]).

Theorem 3. 

([12]) Let $\varphi :(M^n,g)\to (N^n,h)$ ($n\ge 3$) to be a conformal map of dilation λ, then ϕ is biharmonic if and only if the dilation λ satisfies

$$\begin{array}{cc}& grad\left(\Delta ln\lambda \right)-\left(2\left(\Delta ln\lambda \right)+(n-2){\left|gradln\lambda \right|}^2\right)gradln\lambda \hfill \\ & +\frac{6-n}{2}grad\left({\left|gradln\lambda \right|}^2\right)+2Ricc{i}^M(gradln\lambda )=0.\hfill \end{array}$$

In particular, we prove that the biharmonicity of the conformal map $\varphi :({\mathbb{R}}^n,g)\to (N^n,h)$ ($n\ge 3$) where the dilation λ is radial $\left(ln\lambda =\alpha \left(r\right),r=\left|x\right|\text{ and}\alpha \in C^{\infty }\left(\mathbb{R},\mathbb{R}\right)\right)$ is equivalent to an ordinary differential equation of the second order. More precisely, we have

Corollary 2. 

Let $\varphi :({\mathbb{R}}^n,g)\to (N^n,h)$ ($n\ge 3$) to be a conformal map of dilation λ when we suppose that $ln\lambda$ is radial $\left(ln\lambda =\alpha \left(r\right),r=\left|x\right|\text{ and}\alpha \in C^{\infty }\left(\mathbb{R},\mathbb{R}\right)\right)$. Then ϕ is biharmonic if and only if $\beta ={\alpha }^{\prime }$ satisfies the following ordinary differential equation :

$${\beta }^{\prime \prime }-\left(n-4\right)\beta {\beta }^{\prime }+\frac{n-1}{r}{\beta }^{\prime }-\frac{n-1}{r^2}\beta -\frac{2\left(n-1\right)}{r}{\beta }^2-\left(n-2\right){\beta }^3=0.$$

(34)

Proof of Corollary 2 Let $\varphi :({\mathbb{R}}^n,g)\to (N^n,h)$ ($n\ge 3$) to be a conformal map of dilation λ such that $ln\lambda =\alpha \left(r\right)$. By Theorem 3, ϕ is biharmonic if and only if the dilation λ satisfies

$$\begin{array}{cc}& grad\left(\Delta ln\lambda \right)-\left(2\left(\Delta ln\lambda \right)+(n-2){\left|gradln\lambda \right|}^2\right)gradln\lambda \hfill \\ & +\frac{6-n}{2}grad\left({\left|gradln\lambda \right|}^2\right)+2Ricc{i}^M(gradln\lambda )=0.\hfill \end{array}$$

A direct calculation gives

$$gradln\lambda ={\alpha }^{\prime }\frac{\partial }{\partial r},$$

$${\left|gradln\lambda \right|}^2={\left({\alpha }^{\prime }\right)}^2,$$

$$grad\left({\left|gradln\lambda \right|}^2\right)=2{\alpha }^{\prime }{\alpha }^{\prime \prime }\frac{\partial }{\partial r},$$

$$\Delta ln\lambda ={\alpha }^{\prime \prime }+\frac{n-1}{r}{\alpha }^{\prime }$$

and

$$grad\left(\Delta ln\lambda \right)=\left({\alpha }^{\prime \prime \prime }+\frac{n-1}{r}{\alpha }^{\prime \prime }-\frac{n-1}{r^2}{\alpha }^{\prime }\right)\frac{\partial }{\partial r}.$$

Therefore ϕ is biharmonic if and only if the function α satisfies the following differential equation

$${\alpha }^{\prime \prime \prime }-\left(n-4\right){\alpha }^{\prime }{\alpha }^{\prime \prime }+\frac{n-1}{r}{\alpha }^{\prime \prime }-\frac{n-1}{r^2}{\alpha }^{\prime }-\frac{2\left(n-1\right)}{r}{\left({\alpha }^{\prime }\right)}^2-\left(n-2\right){\left({\alpha }^{\prime }\right)}^3=0.$$

If we denote $\beta ={\alpha }^{\prime }$, the biharmonicity of ϕ is equivalent to the differential equation

$${\beta }^{\prime \prime }-\left(n-4\right)\beta {\beta }^{\prime }+\frac{n-1}{r}{\beta }^{\prime }-\frac{n-1}{r^2}\beta -\frac{2\left(n-1\right)}{r}{\beta }^2-\left(n-2\right){\beta }^3=0.$$

As a consequence of the Corollary 2, We will present some remarks which we give a particular solutions of the Equation (34) that allows us to construct a biharmonic non-harmonic maps.

Remark 3. 

. Looking for particular solutions of type $\beta =\frac{a}{r}$ ($a\in {\mathbb{R}}^{*}$). By (34), we deduce that $\varphi :({\mathbb{R}}^n,g)\to (N^n,h)$ ($n\ge 3$) is biharmonic if and only if a is a solution of the algebraic equation

$$\left(n-2\right)a^2+\left(n+2\right)a+2n-2=0.$$

This equation has real solutions if and only if $n\in \left\{3,4\right\}$.

• If $n=3$, we find $a=\frac{-5+\sqrt{17}}{2}$ or $a=\frac{-5-\sqrt{17}}{2}$, so $\lambda =C{r}^{-\left(\frac{5-\sqrt{17}}{2}\right)}$ or $\lambda =C{r}^{-\left(\frac{5+\sqrt{17}}{2}\right)}$ ($C\in {\mathbb{R}}_{+}^{*}$). It follows that any conformal map $\varphi :({\mathbb{R}}^3,g)\to (N^3,h)$ of dilation $\lambda =C{r}^{-\left(\frac{5-\sqrt{17}}{2}\right)}$ or $\lambda =C{r}^{-\left(\frac{5+\sqrt{17}}{2}\right)}$ is biharmonic non-harmonic.
• If $n=4$, we find $a=-1$ or $a=-2$, so $\lambda =\frac{C}{r^2}$ or $\lambda =\frac{C}{r}$ ($C\in {\mathbb{R}}_{+}^{*}$). Then, in this case any conformal map $\varphi :({\mathbb{R}}^4,g)\to (N^4,h)$ of dilation $\lambda =\frac{C}{r^2}$ or $\lambda =\frac{C}{r}$ is biharmonic non-harmonic. For example, the inversion $\varphi :\left({\mathbb{R}}^n\backslash \left\{0\right\},g_{{\mathbb{R}}^n}\right)⟶\left({\mathbb{R}}^n\backslash \left\{0\right\},g_{{\mathbb{R}}^n}\right)$ definded by $\varphi \left(x\right)=\frac{x}{{\left|x\right|}^2}$ is a conformal map with dilation $\lambda =\frac{1}{r^2}$. By (34), the inversion is biharmonic non-harmonic if and only if $n=4$.

Remark 4. 

. Looking for particular solutions of type $\beta =\frac{ar}{1+r^2}$ ($a\in {\mathbb{R}}^{*}$). By (34), $\varphi :({\mathbb{R}}^n,g)\to (N^n,h)$ ($n\ge 3$) is biharmonic if and only we have

$$\left(n-2\right)a^2+\left(n+2\right)a+2n-2=0$$

and

$$3\left(n-2\right)a+2n+4=0.$$

These two equations gives $a=-2$ and $n=4$, it follows that the dilation is equal to $\lambda =\frac{C}{r^2+1}$ ($C\in {\mathbb{R}}_{+}^{*}$). Then, all conformal maps $\varphi :({\mathbb{R}}^4,g)\to (N^4,h)$ of dilation $\lambda =\frac{C}{r^2+1}$ are biharmonic non-harmonic. For example, the inverse of the stereographic projection of the sphere $\varphi :{\mathbb{R}}^n⟶S^n$ definded by $\varphi \left(x\right)=\frac{1}{{\left|x\right|}^2+1}\left({\left|x\right|}^2-1,2x\right)$ is a conformal map with dilation $\lambda =\frac{2}{r^2+1}$. By (34), the inverse of the stereographic projection is biharmonic non-harmonic if and only if $n=4$.

The last part of this paper is devoted to the study of biharmonic maps between warped product manifolds, these maps were also studied in [17]. We will give some results of the biharmonicity in other particular cases.

3. Biharmonic maps and the warped product

Let $\left(M^m,g\right)$ and $\left(N^n,h\right)$ two Riemannian manifolds and let $f\in C^{\infty }\left(M\right)$ be a positive function. The warped product $M{\times }_{f}N$ is the product manifolds $M\times N$ endowed with the Riemannian metric $G_f$ defined, for $X,Y\in \Gamma \left(T\left(M\times N\right)\right)$, by

$$G_f\left(X,Y\right)=g\left(d\pi \left(X\right),d\pi \left(Y\right)\right)+{\left(f\circ \pi \right)}^{2}h\left(d\eta \left(X\right),d\eta \left(Y\right)\right),$$

where $\pi :M\times N⟶M$ and $\eta :M\times N⟶N$ are respectively the first and the second projection. The function f is called the warping function of the warped product. Let $X,Y\in \Gamma \left(T\left(M\times N\right)\right)$, $X=\left(X_1,X_2\right)$, $Y=\left(Y_1,Y_2\right).$ Denote by ∇ the Levi-Civita connection on the Riemannian product $M\times N$ . The Levi-Civita connection $\tilde{\nabla }$ of the warped product $M{\times }_{f}N$ is given by

$${\tilde{\nabla }}_{X}Y={\nabla }_{X}Y+X_1\left(lnf\right)\left(0,Y_2\right)+Y_1\left(lnf\right)\left(0,X_2\right)-f^{2}h\left(X_2,Y_2\right)\left(gradlnf,0\right).$$

(35)

In the first, we consider a smooth map $\varphi :\left(M^m,g\right)⟶\left(P^p,k\right)$ and we defined the map $\tilde{\varphi }:\left(M^m{\times }_f{N}^n,G_f\right)⟶\left(P^p,k\right)$ by $\tilde{\varphi }\left(x,y\right)=\varphi \left(x\right)$. We will study the biharmonicity of $\tilde{\varphi }$. By calculating the tension field of $\tilde{\varphi }$, we obtain the following result :

Proposition 2. 

Let $\varphi :\left(M^m,g\right)⟶\left(P^p,k\right)$ be a smooth map. The tension field of the map $\tilde{\varphi }:\left(M^m{\times }_f{N}^n,G_f\right)⟶\left(P^p,k\right)$ defined by $\tilde{\varphi }\left(x,y\right)=\varphi \left(x\right)$ is given by

$$\mathsf{\tau }\left(\tilde{\varphi }\right)=\mathsf{\tau }\left(\varphi \right)+nd\varphi \left(gradlnf\right)$$

(36)

Proof of Proposition 2. Let us choose ${\left\{e_i\right\}}_{1\le i\le m}$ to be an orthonormal frame on M and ${\left\{f_j\right\}}_{1\le j\le n}$ to be an orthonormal frame on N. An orthonormal frame on $M{\times }_{f}N$ is given by $\left\{\left(e_i,0\right),\frac{1}{f}\left(0,f_j\right)\right\}$. Note that in this case we have $d\tilde{\varphi }\left(X,Y\right)=\left(d\varphi \left(X\right),0\right)$ for any $X\in \Gamma \left(TM\right)$ and $Y\in \Gamma \left(TN\right)$. By definition to the tension field, we have

$$\begin{array}{cc}\hfill \mathsf{\tau }\left(\tilde{\varphi }\right)& =T{r}_{G_f}\nabla d\tilde{\varphi }\hfill \\ & ={\nabla }_{\left(e_i,0\right)}^{\tilde{\varphi }}d\tilde{\varphi }\left(e_i,0\right)+\frac{1}{f^2}{\nabla }_{\left(0,f_j\right)}^{\tilde{\varphi }}d\tilde{\varphi }\left(0,f_j\right)\hfill \\ & -d\tilde{\varphi }\left({\tilde{\nabla }}_{\left(e_i,0\right)}\left(e_i,0\right)\right)-\frac{1}{f^2}d\tilde{\varphi }\left({\tilde{\nabla }}_{\left(0,f_j\right)}\left(0,f_j\right)\right).\hfill \end{array}$$

A simple calculation gives

$${\nabla }_{\left(e_i,0\right)}^{\tilde{\varphi }}d\tilde{\varphi }\left(e_i,0\right)={\nabla }_{e_i}^{\varphi }d\varphi \left(e_i\right)$$

and

$${\nabla }_{\left(0,f_j\right)}^{\tilde{\varphi }}d\tilde{\varphi }\left(0,f_j\right)=0,$$

By using the equation (35), we deduce that

$${\tilde{\nabla }}_{\left(e_i,0\right)}\left(e_i,0\right)=\left({\nabla }_{e_i}{e}_i,0\right)$$

and

$${\tilde{\nabla }}_{\left(0,f_j\right)}\left(0,f_j\right)=\left(0,{\nabla }_{f_j}{f}_j\right)-n{f}^2\left(gradlnf,0\right).$$

It follows that

$$\mathsf{\tau }\left(\tilde{\varphi }\right)={\nabla }_{e_i}^{\varphi }d\varphi \left(e_i\right)-d\varphi \left({\nabla }_{e_i}^M{e}_i\right)+nd\varphi \left(gradlnf\right),$$

then, we obtain

$$\mathsf{\tau }\left(\tilde{\varphi }\right)=\mathsf{\tau }\left(\varphi \right)+nd\varphi \left(gradlnf\right).$$

Remark 5. 

If $\varphi :\left(M^m,g\right)⟶\left(P^m,k\right)$ ($m\ge 3$) is a conformal map with dilation λ, the tension field of $\tilde{\varphi }$ is given by

$$\mathsf{\tau }\left(\tilde{\varphi }\right)=\left(2-m\right)d\varphi \left(gradln\lambda \right)+nd\varphi \left(gradlnf\right)=d\varphi \left(gradln\left({\lambda }^{2-m}{f}^n\right)\right).$$

Then $\tilde{\varphi }$ is harmonic if and only if the function ${\lambda }^{2-m}{f}^n$ is constant.

We will now calculate the bitension field of the map $\tilde{\varphi }:\left(M^m{\times }_f{N}^n,G_f\right)⟶\left(P^p,k\right)$.

Theorem 4. 

Let $\varphi :\left(M^m,g\right)⟶\left(P^p,k\right)$ be a smooth map. The bitension field of the map $\tilde{\varphi }:\left(M^m{\times }_f{N}^n,G_f\right)⟶\left(P^p,k\right)$ defined by $\tilde{\varphi }\left(x,y\right)=\varphi \left(x\right)$ is given by

$$\begin{array}{cc}\hfill {\mathsf{\tau }}_2\left(\tilde{\varphi }\right)& ={\mathsf{\tau }}_2\left(\varphi \right)-n\left(T{r}_g{\nabla }^{2}d\varphi \left(gradlnf\right)+T{r}_g{R}^p\left(d\varphi \left(gradlnf\right),d\varphi \right)d\varphi \right)\hfill \\ & -n{\nabla }_{gradlnf}\mathsf{\tau }\left(\varphi \right)-n^2{\nabla }_{gradlnf}d\varphi \left(gradlnf\right).\hfill \end{array}$$

(37)

Proof of Theorem 4. By definition of the bi-tension field, we have

$${\mathsf{\tau }}_2\left(\tilde{\varphi }\right)=-T{r}_{G_f}{\left({\nabla }^{\tilde{\varphi }}\right)}^2\mathsf{\tau }\left(\tilde{\varphi }\right)-T{r}_{G_f}{R}^P\left(\mathsf{\tau }\left(\tilde{\varphi }\right),d\tilde{\varphi }\right)d\tilde{\varphi }$$

(38)

For the first term $T{r}_{G_f}{\left({\nabla }^{\tilde{\varphi }}\right)}^2\mathsf{\tau }\left(\tilde{\varphi }\right)$, we have

$$\begin{array}{cc}\hfill T{r}_{G_f}{\left({\nabla }^{\tilde{\varphi }}\right)}^2\mathsf{\tau }\left(\tilde{\varphi }\right)& ={\nabla }_{\left(e_i,0\right)}^{\tilde{\varphi }}{\nabla }_{\left(e_i,0\right)}^{\tilde{\varphi }}\mathsf{\tau }\left(\tilde{\varphi }\right)+\frac{1}{f^2}{\nabla }_{\left(0,f_j\right)}^{\tilde{\varphi }}{\nabla }_{\left(0,f_j\right)}^{\tilde{\varphi }}\mathsf{\tau }\left(\tilde{\varphi }\right)\hfill \\ & -{\nabla }_{{\tilde{\nabla }}_{\left(e_i,0\right)}\left(e_i,0\right)}^{\tilde{\varphi }}\mathsf{\tau }\left(\tilde{\varphi }\right)-\frac{1}{f^2}{\nabla }_{{\tilde{\nabla }}_{\left(0,f_j\right)}\left(0,f_j\right)}^{\tilde{\varphi }}\mathsf{\tau }\left(\tilde{\varphi }\right).\hfill \end{array}$$

We will study term by term the right-hand of this expression. A simple calculation gives

$$\begin{array}{cc}\hfill {\nabla }_{\left(e_i,0\right)}^{\tilde{\varphi }}{\nabla }_{\left(e_i,0\right)}^{\tilde{\varphi }}\mathsf{\tau }\left(\tilde{\varphi }\right)& ={\nabla }_{\left(e_i,0\right)}^{\tilde{\varphi }}{\nabla }_{\left(e_i,0\right)}^{\tilde{\varphi }}\mathsf{\tau }\left(\varphi \right)+n{\nabla }_{\left(e_i,0\right)}^{\tilde{\varphi }}{\nabla }_{\left(e_i,0\right)}^{\tilde{\varphi }}d\varphi \left(gradlnf\right)\hfill \\ & ={\nabla }_{e_i}^{\varphi }{\nabla }_{e_i}^{\varphi }\mathsf{\tau }\left(\varphi \right)+n{\nabla }_{e_i}^{\varphi }{\nabla }_{e_i}^{\varphi }d\varphi \left(gradlnf\right)\hfill \end{array}$$

and

$${\nabla }_{\left(0,f_j\right)}^{\tilde{\varphi }}{\nabla }_{\left(0,f_j\right)}^{\tilde{\varphi }}\mathsf{\tau }\left(\tilde{\varphi }\right)=0.$$

By using the equation (35), we obtain

$${\nabla }_{{\tilde{\nabla }}_{\left(e_i,0\right)}\left(e_i,0\right)}^{\tilde{\varphi }}\mathsf{\tau }\left(\tilde{\varphi }\right)={\nabla }_{{\nabla }_{e_i}^M{e}_i}^{\varphi }\mathsf{\tau }\left(\varphi \right)+n{\nabla }_{{\nabla }_{e_i}^M{e}_i}^{\varphi }d\varphi \left(gradlnf\right),$$

and

$${\nabla }_{{\tilde{\nabla }}_{\left(0,f_j\right)}\left(0,f_j\right)}^{\tilde{\varphi }}\mathsf{\tau }\left(\tilde{\varphi }\right)=-n{f}^2{\nabla }_{gradlnf}^{\varphi }\mathsf{\tau }\left(\varphi \right)-n^2{f}^2{\nabla }_{gradlnf}^{\varphi }d\varphi \left(gradlnf\right).$$

Then, we deduce that

$$\begin{array}{cc}\hfill T{r}_{G_f}{\left({\nabla }^{\tilde{\varphi }}\right)}^2\mathsf{\tau }\left(\tilde{\varphi }\right)& =T{r}_g{\left({\nabla }^{\varphi }\right)}^2\mathsf{\tau }\left(\varphi \right)+nT{r}_g{\left({\nabla }^{\varphi }\right)}^{2}d\varphi \left(gradlnf\right)\hfill \\ & +n{\nabla }_{gradlnf}^{\varphi }\mathsf{\tau }\left(\varphi \right)+n^2{\nabla }_{gradlnf}^{\varphi }d\varphi \left(gradlnf\right).\hfill \end{array}$$

(39)

To complete the proof, we will simplify the term $T{r}_{G_f}{R}^P\left(\mathsf{\tau }\left(\tilde{\varphi }\right),d\tilde{\varphi }\right)d\tilde{\varphi }$, we have

$$\begin{array}{cc}\hfill T{r}_{G_f}{R}^P\left(\mathsf{\tau }\left(\tilde{\varphi }\right),d\tilde{\varphi }\right)d\tilde{\varphi }& =R^P\left(\mathsf{\tau }\left(\tilde{\varphi }\right),d\tilde{\varphi }\left(e_i,0\right)\right)d\tilde{\varphi }\left(e_i,0\right)\hfill \\ & +\frac{1}{f^2}{R}^P\left(\mathsf{\tau }\left(\tilde{\varphi }\right),d\tilde{\varphi }\left(0,f_j\right)\right)d\tilde{\varphi }\left(0,f_j\right)\hfill \\ & =R^P\left(\mathsf{\tau }\left(\tilde{\varphi }\right),d\tilde{\varphi }\left(e_i,0\right)\right)d\tilde{\varphi }\left(e_i,0\right)\hfill \\ & =R^P\left(\mathsf{\tau }\left(\varphi \right),d\varphi \left(e_i\right)\right)d\varphi \left(e_i\right)\hfill \\ & +n{R}^P\left(d\varphi \left(gradlnf\right),d\varphi \left(e_i\right)\right)d\varphi \left(e_i\right).\hfill \end{array}$$

It follows that

$$T{r}_{G_f}{R}^P\left(\mathsf{\tau }\left(\tilde{\varphi }\right),d\tilde{\varphi }\right)d\tilde{\varphi }=T{r}_g{R}^P\left(\mathsf{\tau }\left(\varphi \right),d\varphi \right)d\varphi +nT{r}_g{R}^P\left(d\varphi \left(gradlnf\right),d\varphi \right)d\varphi .$$

(40)

If we replace (39) and (40) in (38), we obtain

$$\begin{array}{cc}\hfill {\mathsf{\tau }}_2\left(\tilde{\varphi }\right)& ={\mathsf{\tau }}_2\left(\varphi \right)-n\left(T{r}_g{\nabla }^{2}d\varphi \left(gradlnf\right)+T{r}_g{R}^p\left(d\varphi \left(gradlnf\right),d\varphi \right)d\varphi \right)\hfill \\ & -n{\nabla }_{gradlnf}\mathsf{\tau }\left(\varphi \right)-n^2{\nabla }_{gradlnf}d\varphi \left(gradlnf\right).\hfill \end{array}$$

The proof of Theorem 4 is complete.

Remark 6. 

Theorem 4 is a particular result of generalized warped product manifolds (see [18]).

As a consequence, if ϕ is harmonic, we have

Corollary 3. 

Let $\varphi :\left(M^m,g\right)⟶\left(P^p,k\right)$ a harmonic map. The map $\tilde{\varphi }:\left(M^m{\times }_f{N}^n,G_f\right)⟶\left(P^p,k\right)$ defined by $\tilde{\varphi }\left(x,y\right)=\varphi \left(x\right)$ is biharmonic if and only if

$$T{r}_g{\nabla }^{2}d\varphi \left(gradlnf\right)+T{r}_g{R}^P\left(d\varphi \left(gradlnf\right),d\varphi \right)d\varphi +n{\nabla }_{gradlnf}d\varphi \left(gradlnf\right)=0.$$

In particular if $\varphi =I{d}_M$, the first projection $P_1:\left(M^m{\times }_f{N}^n,G_f\right)⟶\left(M^m,g\right)$ defined by $P_1\left(x,y\right)=x$ is biharmonic if and only if (see [17])

$$grad\Delta lnf+\frac{n}{2}grad\left({\left|gradlnf\right|}^2\right)+2Ricc{i}^M\left(gradlnf\right)=0.$$

In the following we shall present an example of biharmonic non-harmonic maps.

Example 1. 

Let $\tilde{\phi }:{\mathbb{R}}^m\backslash \left\{0\right\}{\times }_f{N}^n⟶{\mathbb{R}}^m\backslash \left\{0\right\}$ defined by $\tilde{\phi }\left(x,y\right)=\frac{x}{{\left|x\right|}^2}$ when we suppose that $lnf$ is radial $\left(lnf=\alpha \left(r\right)\right)$. Then by Theorem 4, we deduce that the map $\tilde{\phi }:{\mathbb{R}}^m\backslash \left\{0\right\}{\times }_f{N}^n⟶{\mathbb{R}}^m\backslash \left\{0\right\}$ is biharmonic if and only if the function α satisfies the following differential equation

$$n{\alpha }^{\prime \prime \prime }+\frac{n\left(m-5\right)}{r}{\alpha }^{\prime \prime }-\frac{3n\left(3m-7\right)}{r^2}{\alpha }^{\prime }+n^2{\alpha }^{\prime }{\alpha }^{\prime \prime }-\frac{2n^2}{r}{\left({\alpha }^{\prime }\right)}^2-\frac{8\left(m-2\right)\left(m-4\right)}{r^3}=0.$$

Let $\beta ={\alpha }^{\prime }$, this equation becomes

$$n{\beta }^{\prime \prime }+\frac{n\left(m-5\right)}{r}{\beta }^{\prime }-\frac{3n\left(3m-7\right)}{r^2}\beta +n^2\beta {\beta }^{\prime }-\frac{2n^2}{r}{\beta }^2-\frac{8\left(m-2\right)\left(m-4\right)}{r^3}=0.$$

Looking for particular solutions of type $\beta =\frac{a}{r}$ ($a\in {\mathbb{R}}^{*}$), then $\tilde{\phi }:{\mathbb{R}}^m\backslash \left\{0\right\}{\times }_f{N}^n⟶{\mathbb{R}}^m\backslash \left\{0\right\}$ is biharmonic if and only if

$$3n^2{a}^2+2n\left(5m-14\right)a+8\left(m-2\right)\left(m-4\right)=0.$$

This equation has two solutions $a=\frac{4-2m}{n}$ and $a=\frac{4\left(4-m\right)}{3n}$.

1.

For $a=\frac{4-2m}{n}$, we obtain $f\left(r\right)=C{r}^{\frac{4-2m}{n}}$ and in this case $\tilde{\phi }:{\mathbb{R}}^m\backslash \left\{0\right\}{\times }_f{N}^n⟶{\mathbb{R}}^m\backslash \left\{0\right\}$ is harmonic so biharmonic.

2.

For $a=\frac{4\left(4-m\right)}{3n}$, we obtain $f\left(r\right)=C{r}^{\frac{4\left(4-m\right)}{3n}}$ and in this case $\tilde{\phi }:{\mathbb{R}}^m\backslash \left\{0\right\}{\times }_f{N}^n⟶{\mathbb{R}}^m\backslash \left\{0\right\}$ is biharmonic non-harmonic.

Now, we consider a smooth map $\psi :\left(N^n,g\right)⟶\left(P^p,k\right)$ and we define the map $\tilde{\psi }:\left(M^m{\times }_f{N}^n,G_f\right)⟶\left(P^p,k\right)$ by $\tilde{\psi }\left(x,y\right)=\psi \left(y\right)$. We will study the biharmonicity of $\tilde{\psi }$, we obtain the following result :

Theorem 5. 

Let $\psi :\left(N^n,h\right)\to \left(P^p,k\right)$ be a smooth map, we define $\tilde{\psi }:\left(M^m{\times }_{f^2}{N}^n,G_{f^2}\right)\to \left(P^p,k\right)$ by $\tilde{\psi }\left(x,y\right)=\psi \left(y\right).$ The tension field and the bi-tension field of $\tilde{\psi }$ are given by

$$\mathsf{\tau }\left(\tilde{\psi }\right)=\frac{1}{f^2\circ \pi }\mathsf{\tau }\left(\psi \right)$$

(41)

and

$${\mathsf{\tau }}_2\left(\tilde{\psi }\right)=\frac{1}{f^4\circ \pi }{\mathsf{\tau }}_2\left(\psi \right)-\frac{2}{f^2\circ \pi }\left(\left(\Delta lnf+\left(n-2\right){\left|gradlnf\right|}^2\right)\circ \pi \right)\mathsf{\tau }\left(\psi \right).$$

(42)

Proof of Theorem 5. In the first, we calculate the tension field of of $\tilde{\psi }$. By definition to the tension field, we have

$$\begin{array}{cc}\hfill \mathsf{\tau }\left(\tilde{\psi }\right)& =T{r}_{G_f}\nabla d\tilde{\psi }\hfill \\ & ={\nabla }_{\left(e_i,0\right)}^{\tilde{\psi }}d\tilde{\psi }\left(e_i,0\right)+\frac{1}{f^2\circ \pi }{\nabla }_{\left(0,f_j\right)}^{\tilde{\psi }}d\tilde{\psi }\left(0,f_j\right)\hfill \\ & -d\tilde{\psi }\left({\tilde{\nabla }}_{\left(e_i,0\right)}\left(e_i,0\right)\right)-\frac{1}{f^2\circ \pi }d\tilde{\psi }\left({\tilde{\nabla }}_{\left(0,f_j\right)}\left(0,f_j\right)\right).\hfill \end{array}$$

By using the equation (35), we obtain

$$\mathsf{\tau }\left(\tilde{\psi }\right)=\frac{1}{f^2\circ \pi }{\nabla }_{f_j}^{\psi }d\psi \left(f_j\right)-\frac{1}{f^2\circ \pi }d\psi \left({\nabla }_{f_j}{f}_j\right)=\frac{1}{f^2\circ \pi }\mathsf{\tau }\left(\psi \right),$$

then

$$\mathsf{\tau }\left(\tilde{\psi }\right)=\frac{1}{f^2\circ \pi }\mathsf{\tau }\left(\psi \right).$$

By this expression, we deduce that $\tilde{\psi }$ is harmonic if and only if ψ is harmonic. Now, we will calculate the bi-tension field of $\tilde{\psi }$. By definition, we have

$${\mathsf{\tau }}_2\left(\tilde{\psi }\right)=-T{r}_{G_f}{\left({\nabla }^{\tilde{\psi }}\right)}^2\mathsf{\tau }\left(\tilde{\psi }\right)-T{r}_{G_f}{R}^P\left(\mathsf{\tau }\left(\tilde{\psi }\right),d\tilde{\psi }\right)d\tilde{\psi }.$$

(43)

For the first term $T{r}_{G_f}{\left({\nabla }^{\tilde{\psi }}\right)}^2\mathsf{\tau }\left(\tilde{\psi }\right)$, we have

$$\begin{array}{cc}\hfill T{r}_{G_f}{\left({\nabla }^{\tilde{\psi }}\right)}^2\mathsf{\tau }\left(\tilde{\psi }\right)& ={\nabla }_{\left(e_i,0\right)}^{\tilde{\psi }}{\nabla }_{\left(e_i,0\right)}^{\tilde{\psi }}\mathsf{\tau }\left(\tilde{\psi }\right)+\frac{1}{f^2\circ \pi }{\nabla }_{\left(0,f_j\right)}^{\tilde{\psi }}{\nabla }_{\left(0,f_j\right)}^{\tilde{\psi }}\mathsf{\tau }\left(\tilde{\psi }\right)\hfill \\ & -{\nabla }_{{\tilde{\nabla }}_{\left(e_i,0\right)}\left(e_i,0\right)}^{\tilde{\psi }}\mathsf{\tau }\left(\tilde{\psi }\right)-\frac{1}{f^2\circ \pi }{\nabla }_{{\tilde{\nabla }}_{\left(0,f_j\right)}\left(0,f_j\right)}^{\tilde{\psi }}\mathsf{\tau }\left(\tilde{\psi }\right).\hfill \end{array}$$

A long calculation gives

$${\nabla }_{\left(e_i,0\right)}^{\tilde{\psi }}{\nabla }_{\left(e_i,0\right)}^{\tilde{\psi }}\mathsf{\tau }\left(\tilde{\psi }\right)=\frac{2}{f^2\circ \pi }\left(\left(2{\left|gradlnf\right|}^2-e_i\left(e_i\left(lnf\right)\right)\right)\circ \pi \right)\mathsf{\tau }\left(\psi \right)$$

and

$$\frac{1}{f^2\circ \pi }{\nabla }_{\left(0,f_j\right)}^{\tilde{\psi }}{\nabla }_{\left(0,f_j\right)}^{\tilde{\psi }}\mathsf{\tau }\left(\tilde{\psi }\right)=\frac{1}{f^4\circ \pi }{\nabla }_{f_j}^{\psi }{\nabla }_{f_j}^{\psi }\mathsf{\tau }\left(\psi \right).$$

Finally, by (35), we obtain

$${\nabla }_{{\tilde{\nabla }}_{\left(e_i,0\right)}\left(e_i,0\right)}^{\tilde{\psi }}\mathsf{\tau }\left(\tilde{\psi }\right)=\frac{2}{f^2\circ \pi }\left({\nabla }_{e_i}{e}_i\left(\left(lnf\right)\right)\circ \pi \right)\mathsf{\tau }\left(\psi \right)$$

and

$$\frac{1}{f^2\circ \pi }{\nabla }_{{\tilde{\nabla }}_{\left(0,f_j\right)}\left(0,f_j\right)}^{\tilde{\psi }}\mathsf{\tau }\left(\tilde{\psi }\right)=\frac{1}{f^4\circ \pi }{\nabla }_{{\nabla }_{f_j}{f}_j}^{\psi }\mathsf{\tau }\left(\psi \right)+\frac{2n}{f^2\circ \pi }\left(\left({\left|gradlnf\right|}^2\right)\circ \pi \right)\mathsf{\tau }\left(\psi \right).$$

Which gives us

$$T{r}_{G_f}{\left({\nabla }^{\tilde{\psi }}\right)}^2\mathsf{\tau }\left(\tilde{\psi }\right)=\frac{1}{f^4\circ \pi }T{r}_h{\nabla }^2\mathsf{\tau }\left(\psi \right)-\frac{2}{f^2\circ \pi }\left(\left(\Delta lnf+\left(n-2\right){\left|gradlnf\right|}^2\right)\circ \pi \right)\mathsf{\tau }\left(\psi \right)$$

(44)

Finally for the first term $T{r}_{G_f}{R}^P\left(\mathsf{\tau }\left(\tilde{\psi }\right),d\tilde{\psi }\right)d\tilde{\psi }$, it is easy to verify that

$$T{r}_{G_f}{R}^P\left(\mathsf{\tau }\left(\tilde{\psi }\right),d\tilde{\psi }\right)d\tilde{\psi }=\frac{1}{f^4\circ \pi }T{r}_h{R}^P\left(\mathsf{\tau }\left(\psi \right),d\psi \right)d\psi .$$

(45)

If we substitute (44) and (45) in (43), we obtain

$${\mathsf{\tau }}_2\left(\tilde{\psi }\right)=\frac{1}{f^4\circ \pi }{\mathsf{\tau }}_2\left(\psi \right)-\frac{2}{f^2\circ \pi }\left(\left(\Delta lnf+\left(n-2\right){\left|gradlnf\right|}^2\right)\circ \pi \right)\mathsf{\tau }\left(\psi \right).$$

This completes the proof of Theorem 5. An immediate consequence of Theorem 5 is given by the following corollary :

Corollary 4. 

Let $\psi :\left(N^n,h\right)⟶\left(P^p,k\right)$ a biharmonic non-harmonic map. The map $\tilde{\varphi }:\left(M^m{\times }_f{N}^n,G_{f^2}\right)⟶\left(P^p,k\right)$ defined by $\tilde{\psi }\left(x,y\right)=\psi \left(y\right)$ is biharmonic if and only if the function $f^{n-2}$ is harmonic.