Tachiana [10, 11] introduced the conception of contravariant almost analytic vector field in a certain almost Hermitian manifold based on the fact that on a compact Kähler manifold, the inner product of a contravariant vector field and a covariant vector field is constant. Later, this conception was generalized to the general almost complex manifolds by several mathematicians [8, 15].Precisely, in an almost complex manifold $ M $ with almost complex structure $ J $ a contravariant almost analytic vector field $ v $ is defined by $ \mathcal{L}_vJ=0, $ where $ \mathcal{L}_v $ denotes the Lie derivative along $ v $. Further, Sawaki and Tamakatsu [9]in 1967 defined and studied an extended contravariant almost analytic vector $ v $ in an almost complex manifold, namely, in a local orthonormal frame it satisfies $ \mathcal{L}_vJ^i_j+λ J^r_jN^i_{rl}v^l=0, $ here λ is $ C^\infty $ scalar function and $ N^i_{rl} $ is Nijenhuis tensor. In addition, Tamakatsu gave a decomposition of this kind of vector fields [12].

As an odd dimensional analogue of contravariant almost analytic vector fields in an almost complex manifold, Sato [7]defined a called contravariant $ C $ -analytic vector $ v $ in a Sasakian manifold with $ (1,1) $ -tensor $ \phi $ by $ (\mathcal{L}_v\phi)\phi=0. $ Following this definition, Eum and Kim [5] provided a definition of contravariant $ C^* $ -analytic vector field $ v $ in a cosymplectic manifold defined by

$(\mathcal{L}_v\phi)\phi=0 \quad\hbox{and}\quad \mathcal{L}_v\xi=0.$

They obtained an unique decomposition of this vector field in compact cosymplectic $ \eta $ -Einstein manifolds. This result is the analogue of a contravariant analytic vector in a compact Kähler space [14] and Sasakian manifold [7], respectively.

Besides, in [4] Deshmukh defined a called $ \phi $ -analytic vector field and studied the case where the induced structure vector $ \xi $ in real hypersurfaces of complex projective space $ CP^{\frac{n+1}{2}} $ is $ \phi $ -analytic.

For an almost contact manifold with almost contact structure $ (\phi,\xi,\eta) $, we may consider a vector field $ v $ such that it leaves $ \phi $ invariant, i.e., $ \mathcal{L}_v\phi=0 $. By a simple calculation we have $ \mathcal{L}_v\xi=\sigma\xi $, where $ \sigma $ is a smooth function. Particularly, Ghosh and Sharma proved that $ \sigma $ is constant in a contact metric manifold [6 Lemma 1].

Motivated by the above background, in the present paper we define a vector field called $ \phi^* $ -analytic vector field, which leaves the $ (1,1) $ -tensor $ \phi $ invariant and satisfies $ \sigma=0 $ in an almost contact metric manifold (see Def.2.1). In Section 2, we will give some basic conceptions and properties, and the main results and proofs are showed in Section 3.

Let $ M^{2n+1} $ be a $ (2n+1) $ -dimensional Riemannian manifold. An almost contact structure on $ M $ is a triple $ (\phi,\xi,\eta) $, where $ \phi $ is a $ (1,1) $ -tensor field, $ \xi $ a unit vector field, $ \eta $ a one-form dual to $ \xi $ satisfying

$\label{eq2.1}\phi^2=-I+\eta\otimes\xi,\quad\eta\circ\phi=0,\quad\phi\circ\xi=0.$

(2.1)

A smooth manifold with such a structure is called an almost contact manifold. It is well-known that there exists a Riemannian metric $ g $ such that

$\label{eq2.2}g(\phi X,\phi Y)=g(X,Y)-\eta(X)\eta(Y)$

(2.2)

for any $ X,Y\in\mathfrak{X}(M) $. It is easy to get from (2.1) and (2.2) that

$\label{eq2.3}g(\phi X,Y)=-g(X,\phi Y),\quad g(X,\xi)=\eta(X).$

(2.3)

Due to (2.3), we can decompose the tangent bundle of an almost contact manifold as the orthonormal sum of the codimension $ 1 $ bundle $ \mathcal{D}={\rm ker}\eta $ and the 1-dimensional foliation defined by $ \xi $. An almost contact structure $ (\phi,\xi,\eta) $ is said to be normal if the Nijenhuis torsion

$\label{eq2.4*} N_\phi(X,Y)=\phi^2[X,Y]+[\phi X,\phi Y]-\phi[\phi X,Y]-\phi[X,\phi Y]+2d\eta(X,Y)\xi$

(2.4)

vanishes for any vector fields $ X,Y $ on $ M $.

A contact metric manifold is an almost contact manifold $ (M,\phi,\xi,\eta,g) $ such that the metric $ g $ satisfies

$\label{eq2.4} d\eta(X,Y)=g(X,\phi Y),$

(2.5)

$ \forall X,Y\in\mathfrak{X}(M) $.Moreover, it is well-known that there exists a $ (1,1) $ -tensor field $ h $ on a contact manifold defined by $ h=\frac{1}{2}\mathcal{L}_\xi\phi $ and it satisfies

$h\xi = 0,\;\;\phi h = - h\phi ,$

(2.6)

${\nabla _X}\xi = - \phi X - AX,$

(2.7)

where $ A=\phi h $.

From (2.5), we have $ \xi\lrcorner d\eta=0 $, which means that $ \xi $ is a geodesic vector field (see [2, Lemma 6.3.3]). If the structure vector field $ \xi $ is Killing, then $ M $ is said to be $ K $ -contact. It is easy to see that $ M $ is $ K $ -contact if and only if $ h=0 $.

Also, if the Nijenhuis tensor of contact manifold $ N_\phi=0 $ then it is said to be Sasakian, and the following formula is also well-known for a Sasakian manifold (see [1])

$\label{eq2.7} (\nabla_X\phi)Y=g(X,Y)\xi-\eta(Y)X.$

(2.8)

Moreover, we know that a Sasakian manifold is automatically $ K $ -contact and in general the converse is not true, but a three dimensional $ K $ -contact manifold is always Sasakian.

Definition 2.1  A vector field $ v $ in an almost contact manifold $ (M,\phi,\xi,\eta,g) $ is called $ \phi^* $ -analytic vector field if it satisfies

$\label{eq:def}\mathcal{L}_v\xi=0\quad\hbox{and}\quad\mathcal{L}_v\phi=0.$

(2.9)

Two equations of (:def) are respectively equivalent to $ [v,\xi]=0 $ and $ [v,\phi X]=\phi[v,X], $ $ \forall X\in\mathfrak{X}(M), $ where $ [\cdot,\cdot] $ denotes by the Lie bracket.

Next, we will give an example of $ \phi^* $ -analytic vector field in almost contact manifold.

Example 2.2  Consider the $ (2n+1) $ -dimensional Euclidean space $ \mathbb{R}^{2n+1} $ equipped with the Cartesian coordinates $ (x_1,\cdots,x_n,y_1,\cdots,y_n,z) $. Define the almost contact structure $ (\phi,\xi,\eta,g) $ by

$\begin{align*} &\phi(\frac{\partial}{\partial x_i}) =-\frac{\partial}{\partial y_i},\quad \phi(\frac{\partial}{\partial y_i})=\frac{\partial}{\partial x_i}+y_i\frac{\partial}{\partial z},\quad\phi(\frac{\partial}{\partial z})=0,\\ &\xi =2\frac{\partial}{\partial z},\quad \eta=\frac{1}{2}(dz-\sum_{i=1}^ny_idx_i),\quad g=\eta\otimes\eta+\frac{1}{4}\sum_{i=1}^n\big((dx_i)^2+(dy_i)^2\big).\end{align*}$

The $ \phi^* $ -analytic vector field $ v $ can be written as $ v=\sum\limits_{i=1}^n(V^i\frac{\partial}{\partial x_i}+\overline{V}^i\frac{\partial}{\partial y_i})+V^z\frac{\partial}{\partial z}. $ By (:def), we derive that $ V^i $ and $ \overline{V}^i $ do not depend on $ z $ and that the following PDEs hold

$\label{3.8}\begin{aligned} &\frac{\partial V^i}{\partial x_j} =\frac{\partial \overline{V}^i}{\partial y_j},\quad\frac{\partial V^i}{\partial y_j} =-\frac{\partial \overline{V}^i}{\partial x_j},\quad y_i\frac{\partial V^i}{\partial y_j} & =\frac{\partial V^z}{\partial y_j},\\ &\overline{V}^j= y_j\frac{\partial V^z}{\partial z}-y_i\frac{\partial \overline{V}^i}{\partial y_j},\quad\frac{\partial V^z}{\partial z} =0.\end{aligned}$

(2.10)

It is clear that there exists a non-zero solution $ V^i=c,\overline{V}^i=0,V^z=H(x_1,\cdots,x_n), $ where $ H $ is a smooth function on $ \mathbb{R}^{2n+1} $ and $ c $ is a non-zero constant.Hence we see that $ c\sum\limits_{i=1}^n\frac{\partial}{\partial x_i}+H(x_1,\cdots,x_n)\frac{\partial}{\partial z} $ is a $ \phi^\ast $ -analytic vector field in $ (\mathbb{R}^{2n+1},\phi,\xi,\eta,g) $.

For any an almost contact manifold, we have the following:

Proposition 2.3  For a $ \phi^* $ -analytic vector field $ v $ in almost contact manifold $ (M,\phi,\xi,\eta,g) $, the following identity holds: $ g(\nabla_Xv,\xi)+g(\nabla_\xi v,X)=0,\forall X\in\mathfrak{X}(M). $

Proof  From the definition of the $ \phi^* $ -analytic vector field we know $ [\xi,v]=0 $, i.e., $ \nabla_v\xi=\nabla_\xi v. $ Using (2.1), a straightforward computation yields

$[\phi^2X,v]=\nabla_{\phi^2X}v-\nabla_v(\phi^2X)=-\nabla_Xv+\nabla_vX-\eta(\nabla_vX)\xi-g(\nabla_\xi v,X)\xi.$

On the other hand,

$\phi[\phi X,v]=\phi^2(\nabla_Xv-\nabla_vX)=-\nabla_Xv+\nabla_vX+\eta(\nabla_Xv)\xi-\eta(\nabla_vX)\xi.$

Then by replacing $ X $ by $ \phi X $ in (:def) and comparing the previous two equations, we complete the proof.

Write $ f=g(\xi,v) $, which is a smooth function, then the following corollaries are easy to obtain from Proposition 2.3.

Corollary 2.4  If $ (M,\phi,\xi,\eta,g) $ is an almost contact manifold with $ \phi^* $ -analytic vector field $ v $, then $ v(f)=0 $. In particular, if $ v\in\mathcal{D} $ the integral curves of $ \xi $ are geodesics.

Proof  The first assertion is obvious by making use of Proposition 2.3 with $ X=v $. If $ v\in\mathcal{D} $, then from Proposition 2.3 with $ X=\xi $, we have $ g(v,\nabla_\xi\xi)=0 $, i.e., $ \nabla_\xi\xi\in {\rm Span}\{\xi\} $, so $ \nabla_\xi\xi=0 $.

Corollary 2.5  Let $ v $ be a $ \phi^* $ -analytic vector in almost contact manifold $ (M,\phi,\xi,\eta,g) $. If $ \eta $ is closed, then $ f $ is constant.

Proof  Applying Proposition 2.3, for any vector field $ X $, we have

$d\eta (X,v) = X(\eta (v)) - v(\eta (X)) - \eta ([X,v]) = g(v,{\nabla _X}\xi ) - g(X,{\nabla _\xi }v) = X(f).$

Corollary 2.6  Let $ (M,\phi,\eta,\xi,g) $ be a contact metric manifold with $ \phi^* $ -analytic field $ v $. Then $ \nabla_v\xi-\phi\nabla_{\phi v}\xi=-2\phi v $ and $ \xi(f)=0. $

Proof  Using (2.7) for $ X=v $ and $ X=\phi v $, respectively, we have

$\nabla_v\xi=-\phi v-Av,\nabla_{\phi v}\xi= v-f\xi-A\phi v.$

Therefore it completes the proof the first assertion in view of(2.6) and the above two equations.

On other hand, in terms of Proposition 2.3 with $ X=\xi $, we have $ g(\nabla_\xi v,\xi)=0 $, that is, $ \xi(f)=0 $.

In this section, we shall suppose that $ (M,\phi,\xi,\eta,g) $ is always a contact metric manifold.

Theorem 3.1  A $ \phi^* $ -analytic vector field $ v $ in a contact metric manifold $ M $ is Killing.

Proof  In view of Proposition 2.3, for any $ X\in\mathfrak{X}(M) $, we have

$(v\lrcorner d\eta)(X)=d\eta(v,X)=-X(f),$

which means that $ v\lrcorner d\eta=-df $. Thus $ \mathcal{L}_vd\eta=d(v\lrcorner d\eta)=0. $ Since for a contact metric manifold the associated metric $ g $ may write as $ d\eta\circ(\phi\otimes\mathbb{I})+\eta\otimes\eta $ by(2.4), where $ \mathbb{I} $ is the identical map on $ TM $, we get $ \mathcal{L}_v g=d\eta\circ(\mathcal{L}_v\phi\otimes\mathbb{I}). $ Therefore we complete the proof from (:def).

Remark 3.2  In fact, a Killing vector field is not necessary $ \phi^* $ -analytic, however, in particular, if the structure vector field $ \xi $ is Killing then the converse of Theorem 3.1 is valid (see [2, Proposition 6.6.12]).

From [13, Theorem 3.4], we have

Corollary 3.3  The $ \phi^* $ -analytic vector field $ v $ in contact metric manifold satisfies $ \Delta v+Qv=0, $ here $ \Delta $ and $ Q $ denote by the Laplace operator and Ricci operator, respectively.

Next we will prove the following conclusion.

Theorem 3.4  Let $ (M,\phi,\eta,\xi,g) $ be a contact metric manifold with $ \phi^* $ -analytic field $ v $. Then $ \phi v $ must be not a $ \phi^* $ -analytic vector field unless zero vector.

To prove this theorem we need the following two lemmas.

Lemma 3.5  Under the assumption of Theorem 3.4, if $ \phi v $ is also a $ \phi^* $ -analytic vector field, then $ \nabla_\xi v=-\phi v $.

Proof  Since $ v\lrcorner d\eta=-df $, using (2.5) we have $ \phi v=Df $, where $ D $ denotes by the gradient operator.Thus if $ \phi v $ is $ \phi^* $ -analytic, $ [\phi X,Df]=\phi[X,Df] $ for any $ X\in\mathfrak{X}(M) $, which implies $ [\xi,Df]=0 $. It reduces that $ hv=0 $ from the definition of $ h $. Hence we complete the proof in terms of (2.7).

Lemma 3.6  Under the assumption of Theorem 3.4, if $ \phi v $ is also a $ \phi^* $ -analytic vector field then

$\label{eq3.9*} (\nabla_X\phi)v=g(v,X)\xi-fX.$

Proof  For any contact metric manifold the following identity (see [2, Lemma 7.3.2]) holds

$2g((\nabla_X\phi)Y,Z) =g(N_\phi(Y,Z),\phi X)+2d\eta(\phi Y,X)\eta(Z)-2d\eta(\phi Z,X)\eta(Y)$

for any $ X,Y,Z\in\mathfrak{X}(M). $

Since $ v $ and $ \phi v $ are $ \phi^* $ -analytic vector fields, we obtain

$\begin{align*} 2g((\nabla_X\phi)v,Y)& =g(N_\phi(v,Y),\phi X)+2d\eta(\phi v,X)\eta(Y)-2d\eta(\phi Y,X)\eta(v) \\ & =g((\mathcal{L}_{\phi v}\phi)Y+\phi(\mathcal{L}_v\phi)Y,\phi X)-2g(\phi^2v,X)\eta(Y)+2g(\phi^2Y,X)\eta(v)\\ &=-2g(\phi^2v,X)\eta(Y)+2g(\phi^2Y,X)\eta(v).\end{align*}$

Hence

$(\nabla_X\phi)v=-g(\phi^2v,X)\xi+\eta(v)\phi^2X=g(v,X)\xi-fX.$

Proof of Theorem 3.4  We assume that $ \phi v $ is also a $ \phi^* $ -analytic vector field, then $ \phi v $ is also Killing because of Theorem 3.1, namely, for any $ X,Y\in\mathfrak{X}(M) $, we have

$g(\nabla_X(\phi v),Y)+g(\nabla_Y(\phi v),X)=0.$

Making use of Lemma 3.6, we obtain

$g(v,X)\eta(Y)+g(v,Y)\eta(X)-2fg(X,Y)+g(\phi\nabla_Xv,Y)-g(\nabla_Yv,\phi X)=0.$

Hence using Theorem 3.1 gives

$\label{eq3.9} g(v,X)\xi+\eta(X)v-2fX+\phi\nabla_Xv+\nabla_{\phi X}v=0.$

(3.1)

By replacing $ \phi X $ by $ X $ in (3.1) and using Lemma 3.5, we get

$\label{eq3.10} g(v,\phi X)\xi-2f\phi X+\phi\nabla_{\phi X}v-\nabla_Xv-\eta(X)\phi v=0.$

(3.2)

Operating $ \phi $ onto (3.2) and using Proposition 2.3, we have

$2fX-4f\eta(X)\xi-\nabla_{\phi X}v+g(v,X)\xi-\phi\nabla_Xv+\eta(X)v=0.$

Thus by comparing with (3.1) it yields that $ 2f\eta(X)\xi-g(v,X)\xi-\eta(X)v=0 $ for any $ X\in\mathfrak{X}(M) $, which implies $ \phi^2v=0 $ by taking $ X=\xi $ in the above formula.So $ \phi v $ must be identically zero. We complete the proof.

For the case where $ M $ is a normal contact metric manifold, i.e., a Sasakian manifold, we have

Theorem 3.7  Let $ v $ be a $ \phi^* $ -analytic vector field in a Sasakian manifold. Then $ f=g(v,\xi) $ is constant and $ v $ is collinear to $ \xi $ with constant length.

Proof  In the proof of Theorem 3.1, we have known $ v\lrcorner d\eta=-df $. Since $ M $ is Sasakian, $ N_{\phi}=0 $. Thus it follows from (2.4) and (2.9) that $ N_{\phi}(X,v)=2d\eta(X,v)\xi=0 $ for every field $ X $, that means that $ df=0 $.

On the other hand, from (2.5) and Proposition 2.3, we know that $ \phi v=Df $. Hence $ v=-\phi Df+f\xi=f\xi $. We complete the proof of theorem.

At last we consider that $ M $ is a three dimensional contact metric manifold. Let $ U $ be the open subset where the tensor $ h\neq0 $ and $ U' $ be the open subset such that $ h $ is identically zero. Thus $ U\cup U' $ is open dense in $ M $. Assume that $ M $ is non-$ K $ -contact, then $ U $ is non-empty and there exits a local orthonormal frame field $ \mathcal{E}=\{e_1,e_2=\phi e_1,\xi\} $ such that $ he_1=\mu e_1 $ and $ he_2=-\mu e_2 $, where $ \mu $ is a positive non-vanishing smooth function of $ M $. With respect to the frame field, we have

Lemma 3.8  [3] Let $ (M^3,\phi,\eta,\xi,g) $ be a contact metric manifold. Then with respect to $ \mathcal{E} $ the Levi-Civita connection $ \nabla $ is given by

$\begin{align*} &\nabla_{e_1}e_1=be_2,\; \nabla_{e_1}e_2=-be_1+(1+\mu)\xi,\;\nabla_{e_1}\xi=-(1+\mu)e_2,\\ &\nabla_{e_2}e_1=-ce_2+(\mu-1) \xi,\;\nabla_{e_2}e_2=ce_1,\; \nabla_{e_2}\xi=(1-\mu)e_1,\\ &\nabla_\xi e_1=-ae_2,\; \nabla_\xi e_2=ae_1,\;\nabla_\xi\xi=0,\end{align*}$

where $ a,b $ and $ c $ are smooth functions.

Theorem 3.9  There does not exist a non-zero $ \phi^* $ -analytic vector field in a three dimensional contact metric manifold.

Proof  First, for $ \mathcal{E} $, $ e_3=\xi $ is globally defined, thus we can define the global frame field, still denoted by $ \{e_1,e_2,e_3=\xi\} $, by lifting to the universal covering space $ \widetilde{M}^3 $ if necessary.Since $ v=-\phi Df+f\xi $, with respect to the orthonormal frame field $ \mathcal{E} $, $ v $ can be also written as $ v=e_2(f)e_1-e_1(f)e_2+f\xi $. In view of Lemma 3.8 and Corollary2.6, we compute

$\label{eq3.11}\begin{aligned} \nabla_\xi v & =\xi(e_2(f))e_1+e_2(f)\nabla_\xi e_1-\xi(e_1(f))e_2+e_1(f)\nabla_\xi e_2 \\ & =\xi(e_2(f))e_1-ae_2(f)e_2-\xi(e_1(f))e_2+ae_1(f)e_1\\ &=\Big(\xi(e_2(f))+ae_1(f)\Big)e_1-\Big(ae_2(f)+\xi(e_1(f))\Big)e_2.\end{aligned}$

(3.3)

Similarly, we have

$\label{eq3.12}\begin{aligned} \nabla_{e_1}v =\Big[e_1(e_2(f))&+be_1(f)\Big]e_1+\Big[be_2(f)-e_1(e_1(f))-f(\mu+1) \Big]e_2-e_1(f)\mu\xi\end{aligned}$

(3.4)

and

$\label{eq3.13}\begin{aligned} \nabla_{e_2}v =\Big[e_2(e_2(f))-ce_1(f)+f(1-\mu)\Big]e_1-\Big[e_2(f)c+ e_2(e_1(f))\Big]e_2+e_2(f)\mu\xi.\end{aligned}$

(3.5)

By Theorem 3.1, we know

$g(\nabla_\xi v,e_1) +g(\nabla_{e_1}v,\xi)=0,\quad g(\nabla_\xi v,e_2) +g(\nabla_{e_2}v,\xi)=0.$

Thus making use of (3.3), (3.4) and(3.5), we obtain

$\label{eq3.15} \left\{ \begin{array}{ll} \xi(e_2(f))=(\mu-a)e_1(f),\\\xi(e_1(f))=(\mu-a)e_2(f). \end{array} \right.$

(3.6)

On the other hand, we notice that for a $ \phi^* $ -analytic vector field, $ \nabla_\xi v=\nabla_v\xi $, i.e.,

$\begin{align*} \nabla_\xi v=e_2(f)\nabla_{e_1}\xi-e_1(f)\nabla_{e_2}\xi =-e_1(f)(1-\mu)e_1-e_2(f)(1+\mu)e_2.\end{align*}$

By comparing with (3.3), we arrive at

$\label{eq3.16*}\left\{ \begin{array}{ll} \xi(e_2(f))=-(a+1-\mu)e_1(f),\\ \xi(e_1(f))=(1+\mu-a)e_2(f). \end{array}\right.$

(3.7)

It follows from (3.6) and (3.7) that $ e_1(f)=e_2(f)=0. $ Thus $ f $ is constant and $ v=f\xi $.

By Theorem 3.1, $ v $ is Killing, thus we find that $ \xi $ is also Killing, namely, $ M $ is $ K $ -contact, so $ h=0 $. It yields a contradiction, thus we complete the proof.

Next we apply Example 2.2 to check both Theorem 3.4 and Theorem3.9.

Example 3.10  We consider $ \mathbb{R}^{2n+1} $ equipped with contact structure $ (\phi,\xi,\eta,g) $ as in Example 2.2. Let $ v=\sum\limits_{i=1}^n(V^i\frac{\partial}{\partial x_i}+\overline{V}^i\frac{\partial}{\partial x_i})+V^z\frac{\partial}{\partial z} $ be a $ \phi^* $ -analytic vector field, then

$\begin{align*}\phi v&=\sum_{i=1}^n(-V^i\frac{\partial}{\partial y_i}+\overline{V}^i\frac{\partial}{\partial x_i})+\sum_{i=1}^ny_i\frac{\partial}{\partial z}.\end{align*}$

If $ \phi v $ is also $ \phi^* $ -analytic, we get

$\begin{align*} y_i\frac{\partial \overline{V}^i}{\partial y_j}=\frac{\partial}{\partial y_j}\sum_{i=1}^n y_i=1\quad\hbox{for all}\quad j.\end{align*}$

But from the fourth equation of (2.10) we find $ \overline{V}^j=1 $ for all $ j $. It comes to a contradiction. Thus $ \phi v $ can not be $ \phi^* $ -analytic. It is consistent with the result of Theorem 3.4.

For the case of three dimension, if $ v $ is a $ \phi^* $ -analytic vector field, then we shall prove that $ v $ is a zero vector field.We know that the $ \phi^* $ -analytic vector field can be written as $ v=-\phi Df+f\xi $, thus a straightforward computation yields

$\label{3.18}\left\{ \begin{array}{ll} V=-\frac{\partial f}{\partial y},& \\ \overline{V}=\frac{\partial f}{\partial x},& \\ -y\frac{\partial f}{\partial y}=\frac{3}{2}f. & \end{array}\right. $

(3.8)

By virtue of the fourth equation of (2.10) and the third equation of \eqref (3.8), we have

$\frac{\partial f}{\partial x}=-y\frac{\partial ^2f}{\partial x\partial y}=\frac{3}{2}\frac{\partial f}{\partial x},$

i.e., $ \overline{V}=\frac{\partial f}{\partial x}=0. $ Moreover, differentiating the third equation of (3.8) with respect to $ y $ gives

$-\frac{\partial f}{\partial y}-y\frac{\partial^2 f}{\partial y^2}=\frac{3}{2}\frac{\partial f}{\partial y}.$

That means that $ V=-\frac{\partial f}{\partial y}=0 $ since the first equation of (3.8) and the second equation of (2.10) imply $ \frac{\partial^2 f}{\partial y^2}=-\frac{\partial V}{\partial y}=\frac{\partial\overline{V}}{\partial x}=0. $ Further, we have $ f=0 $ by using the third equation of (3.8) again, i.e., $ v=0 $, which concides with the conclusion of Theorem 3.9.