2.1 Dynamical approach to flag curvature

To define the concept of flag curvature in a generalized Finsler structure, it is preferable to introduce the dynamical approach developed by P. Foulon [Reference Foulon11], see also [Reference Huang and Mo13].

The vertical endomorphism $\mathcal {V}$ is the unique $(1,1)$ tensor on $\Sigma $ that satisfies the following equations:

$$\begin{align*}\mathcal{V}\left(\left[\xi,v\right]\right)=-v,\quad \mathcal{V}(\xi)=\mathcal{V}\left(v\right)=0, \quad \forall v\in V\Sigma. \end{align*}$$

The horizontal endomorphism $\mathcal {H}$ is the unique $(1,1)$ tensor on $\Sigma $ that satisfies the following equations:

$$ \begin{align*} &\mathcal{H}\left(v\right)=-\left[\xi,v\right]-\frac{1}{2}\mathcal{V}\left[\xi,\left[\xi,v\right]\right],\\ &\mathcal{H}\left(\mathcal{H}(v)\right)=\mathcal{H}(\xi)=0, \quad \forall v\in V\Sigma. \end{align*} $$

The image of $\mathcal {H}$ is the horizontal subbundle of $T\Sigma $ . So $T\Sigma $ can be decomposed into the direct sum of the horizontal subbundle, the vertical subbundle, and the line bundle spanned by $\xi $ . We shall denote by $H\Sigma $ the set of smooth sections of the horizontal subbundle.

The Riemann curvature tensor or Jacobi endomorphism $\mathcal {R}$ is defined on the vertical subbundle as follows:

$$\begin{align*}\mathcal{R}(w)=\mathcal{V}\mathcal{H}\left[\xi,\mathcal{H}(w)\right],\quad \forall w\in V\Sigma. \end{align*}$$

Notice that in [Reference Foulon11], the Jacobi endomorphism $\mathcal {R}$ is also defined for horizontal vectors, and $\mathcal {R}(\xi )=0$ . However, only the vertical component is essential for us to define flag curvature.

The following $(0,2)$ tensor h is a Riemannian metric on the vertical subbundle, known as the angular metric:

$$\begin{align*}h(u, v) = \operatorname{\mathrm{d}}\omega([\xi, u], v) = \operatorname{\mathrm{d}}\omega(u, \mathcal{H}(v)),\quad \forall u, v\in V\Sigma. \end{align*}$$

Finally, the flag curvature K is given by

$$\begin{align*}K(v) = \frac{h(\mathcal{R}(v), v)}{h(v,v)},\quad \forall v\in V\Sigma\setminus\{0\}. \end{align*}$$

Notice that $\mathcal {R}$ is self-adjoint with respect to h, so the flag curvature of a generalized Finsler structure is identically zero if and only if $\mathcal {R}\equiv 0$ .

2.2 Two-dimensional case

In two dimensions, the above scenario is significantly simplified. First, for a generalized Finsler structure on a surface, the indicatrix at each point is simply a strongly convex curve toward the origin. Moreover, the generalized Finsler structure reduces to a classical one if all the indicatrices are simple closed curves. Let us recall a simple criterion for strong convexity in [Reference Bao, Chern and Shen2, Chapter 4].

Lemma 2.4 An immersed curve $\ell : \mathbb {R}\to \mathbb {R}^2$ is strongly convex toward the origin if it never passes through the origin and satisfies the following condition:

$$\begin{align*}\frac{\det(\ell',\ell")}{\det(\ell, \ell')}> 0,\quad \forall t\in\mathbb{R}. \end{align*}$$

Now, the vertical subbundle is of rank one, so there is a globally defined vertical vector field $e_3$ satisfying $h(e_3, e_3) = 1$ . Put $e_2 = \mathcal {H}(e_3)$ ; then $e_2$ is a globally defined horizontal vector field. In this way, $\{e_1=\xi , e_2, e_3\}$ becomes a global frame field on $\Sigma $ , known as the Berwald frame. Notice that $e_3$ is unique up to a minus sign. The dual coframe field $\{\omega _1=\omega , \omega _2, \omega _3\}$ satisfies the following structure equations (see [Reference Bryant6, Reference Bryant, Huang and Mo8]):

(1) $$ \begin{align} \begin{aligned} \operatorname{\mathrm{d}}\omega_1&=-\omega_2\wedge\omega_3,\\ \operatorname{\mathrm{d}}\omega_2&=-\omega_3\wedge\omega_1-I\omega_2\wedge\omega_3,\\ \operatorname{\mathrm{d}}\omega_3&=-K\omega_1\wedge\omega_2-J\omega_2\wedge\omega_3, \end{aligned} \end{align} $$

where I, J, and K are known as the Cartan scalar, the Landsberg curvature, and the Gauss curvature, respectively. In two dimensions, the flag $(P,y)$ can only be $(T_xM, y)$ , so the flag curvature $K(P, y)$ can be written as $K(y)$ , and it is referred to as the Gauss curvature.

2.3 Flat Finsler structures

Since we are mainly interested in the $K=0$ (flat) case, we will now review some relevant results in this context. A basic model of flat Finsler space is $\mathbb {R}^n$ equipped with a Minkowski norm; it is called a Minkowski space. A Finsler manifold is called locally Minkowski if every point has a neighborhood that is isometric to an open subset of a Minkowski space. For a locally Minkowski space, the Cartan scalar is bounded, and the Landsberg curvature vanishes. A classical theorem by Akbar-Zadeh [Reference Bao, Chern and Shen2] states that if the flat Finsler structure is complete and the Cartan scalar is bounded, then it is a locally Minkowski space.

When $n=2$ , the Bryant normal form provides the local model of a flat Finsler structure (see [Reference Bryant6]). It expresses the Berwald coframe using two arbitrary functions of two variables.

(2) $$ \begin{align} \begin{aligned} \omega_1 &= \operatorname{\mathrm{d}} y - x\operatorname{\mathrm{d}} z,\\ \omega_2 &= \hat{P}^{-1}\operatorname{\mathrm{d}} x + (\hat{P} y + \hat{Q})\operatorname{\mathrm{d}} z,\\ \omega_3 &= \hat{P} \operatorname{\mathrm{d}} z, \end{aligned} \end{align} $$

where $(x, y, z)$ is an adapted local coordinate system on $\Sigma $ , $\hat {P}$ and $\hat {Q}$ are arbitrary functions of x and z, and $\hat {P}\neq 0$ . Notice that when $\hat {P}$ and $\hat {Q}$ are only functions of x, the corresponding Finsler structure admits a Killing vector field $\frac {\partial {}}{\partial {z}}$ , since the transformations $(x,y,z)\mapsto (x,y,z+c)$ leave the above coframe field unchanged.

Under the assumption that the Finsler structure admits a Killing vector field, Bryant, Huang, and Mo [Reference Bryant, Huang and Mo8] derived another local normal form.

(3) $$ \begin{align} \begin{aligned} \omega_1 &= \operatorname{\mathrm{d}} \tilde{y} + v(\tilde{x}) \operatorname{\mathrm{d}} \tilde{x} + \tilde{x}\operatorname{\mathrm{d}}\tilde{z},\\ \omega_2 &= -u(\tilde{x})^{-1}\operatorname{\mathrm{d}} \tilde{x} + \tilde{y} u(\tilde{x}) \operatorname{\mathrm{d}}\tilde{z},\\ \omega_3 &= u(\tilde{x}) \operatorname{\mathrm{d}}\tilde{z}, \end{aligned} \end{align} $$

where $(t,a,b)$ is a local coordinate system on $\Sigma $ , u and v are functions of a, and $u\neq 0$ .

The above two normal forms are related as follows. Let $V(a) = \int v(a)\operatorname {\mathrm {d}} a$ ; then the change of coordinates

$$\begin{align*}\tilde{z} = z,\quad \tilde{x} = -x, \quad \tilde{y} = y - V(-x) \end{align*}$$

transforms the Bryant–Huang–Mo normal form into the Bryant normal form, with the constraint that $\hat {P}$ and $\hat {Q}$ depend only on x. Thus, it is fair to say that these two normal forms are equivalent when the Finsler structure admits a Killing field.

3.1 A motivating example

In the proof of Proposition 1.2, we have seen that for each fixed t, the vector field

$$\begin{align*}\hat{Y}=\varphi_{t*}\frac{\partial{}}{\partial{y}} \end{align*}$$

has a unit length on M. Thus, at any point $p\in M$ , the curve

$$\begin{align*}t\mapsto\gamma(t) := \left(\varphi_{t*}\frac{\partial{}}{\partial{y}}\right)_p \end{align*}$$

traces a portion of the indicatrix at point p.

For example, when $P=1$ and $Q=0$ , the vector field X has the following expression:

$$\begin{align*}X = y\frac{\partial{}}{\partial{x}} - x\frac{\partial{}}{\partial{y}}. \end{align*}$$

It is defined on the entire plane $M=\mathbb {R}^2$ and it generates the $S^1$ action $\varphi _t$ on M, where

$$\begin{align*}\varphi_t(x,y) = (x\cos t + y\sin t, -x\sin t + y\cos t),\quad \forall (x,y)\in M. \end{align*}$$

Now, let $Y=\frac {\partial {}}{\partial {y}}$ , then it is easy to see that

$$\begin{align*}\gamma(t) = (\varphi_{t*}Y)_p = \sin t \left.\frac{\partial{}}{\partial{x}}\right|_p + \cos t \left.\frac{\partial{}}{\partial{y}}\right|_p. \end{align*}$$

Thus, at each point $p\in \mathbb {R}^2$ , the curve $\gamma (t) = (\varphi _{t*}Y)|_p$ is a circle of radius one, known as the indicatrix of the Euclidean metric.

Motivated by the above example, we restate Theorem 1.3 as follows.

Theorem 3.1 Let P and Q be two smooth functions defined near $0$ in $\mathbb {R}$ such that ${P(x)>0}$ for all x. Let M be the maximal plane region where the vector field $X = (yP(x)+Q(x))\frac {\partial {}}{\partial {x}}-x\frac {\partial {}}{\partial {y}}$ is defined. Let $\varphi _t$ be the local flow generated by X with a flow domain $\Sigma \subset M\times \mathbb {R}$ . Define a map $\iota : \Sigma \to TM$ as follows:

(6) $$ \begin{align} \iota(p, t) := \left(\varphi_{t*}\frac{\partial{}}{\partial{y}}\right)_p,\quad \forall (p,t)\in\Sigma. \end{align} $$

Then $\iota $ is a generalized Finsler structure on M with vanishing flag curvature, and X is a Killing field. Moreover, if P and Q are defined on $\mathbb {R}$ , P is bounded, and Q grows sublinearly, then M covers the entire $\mathbb {R}^2$ , and the generalized Finsler structure is complete.

The proof of this theorem shall consists the rest of this section and it is divided into three subsections.

3.2 The strong convexity of the indicatrices

To show that $\iota $ is a generalized Finsler structure, we only need to prove that ${\gamma (t) := \iota (p,t)}$ is a strongly convex curve toward the origin in $T_pM$ for each $p\in M$ . Let $Y := \frac {\partial {}}{\partial {y}}$ ; then direct computation shows that

$$\begin{align*}-[X, Y] = P\frac{\partial{}}{\partial{x}},\quad [X,[X,Y]] = (PQ'-QP'))\frac{\partial{}}{\partial{x}} - P\frac{\partial{}}{\partial{y}}. \end{align*}$$

Thus, it is easy to obtain

$$ \begin{align*} \det(Y, -[X,Y]) = \left|\begin{matrix} 0 & P \\ 1 & 0 \end{matrix}\right| = -P,\\ \det(-[X,Y], [X,[X,Y]]) = \left|\begin{matrix} P & PQ'-QP' \\ 0 & -P \end{matrix}\right| = -P^2. \end{align*} $$

Notice that these two determinants are negative at any point.

By successively using [Reference Kobayashi and Nomizu16, Corollary 1.10], we have

$$\begin{align*}\frac{\operatorname{\mathrm{d}}}{\operatorname{\mathrm{d}} t} (\varphi_{t*} Y) = -\varphi_{t*}[X, Y],\quad \frac{\operatorname{\mathrm{d}}^2}{\operatorname{\mathrm{d}} t^2}(\varphi_{t*}Y) = \varphi_{t*}[X,[X,Y]]. \end{align*}$$

Restricting to the point p, we have

$$\begin{align*}\det\left(\gamma(t),\gamma'(t)\right)=det\left((\varphi_{t*}Y)_p,-(\varphi_{t*}[X,Y])_p\right)=\det(\varphi_{t*})\det(Y,-[X,Y]), \end{align*}$$

where the last two determinants are evaluated at the point $\varphi _{-t}(p)$ . Similarly, we have

$$\begin{align*}\det\left(\gamma'(t),\gamma"(t)\right) = \det(\varphi_{t*})\det(-[X,Y], [X,[X,Y]]). \end{align*}$$

Thus

$$\begin{align*}\frac{\det\left(\gamma(t),\gamma'(t)\right)}{\det\left(\gamma'(t),\gamma"(t)\right)} = \frac{\det(Y, [Y,X])}{\det([Y,X], [X,[X,Y]])} = \frac{1}{P}, \end{align*}$$

where the function P is evaluated at the x-coordinate of the point $\varphi _{-t}(p)$ . Since $P>0$ , we have proved that the curve $\gamma (t)$ is strongly convex toward the origin (see Lemma 2.4).

3.3 Computation of flag curvature

Now we compute the flag curvature of the above generalized Finsler structure.

3.3.1 Hilbert form

As above, let $(x,y,t)$ be the natural coordinate system on the flow domain $\Sigma $ . We define another coordinate system $(u,v,s)$ on $\Sigma $ as follows:

$$\begin{align*}(u,v) = \varphi_{-t}(x,y),\quad s=t. \end{align*}$$

By the definition of the flow, we know that

(7) $$ \begin{align} u' = -vP(u)-Q(u),\quad v' = u, \end{align} $$

where ${}'$ denotes derivative with respect to t.

Let $\pi :TM\to M$ be the natural projection, then $\pi \circ \iota (x,y,t)=(x,y)$ . Recall that at each point $p=(x,y)$ , the indicatrix is parameterized by the curve

$$\begin{align*}\gamma(t) = (\varphi_{t*}Y)_p, \end{align*}$$

and it satisfies

$$\begin{align*}\gamma'(t) = -(\varphi_{t*}[X,Y])_p. \end{align*}$$

Now, since the $1$ -form $\operatorname {\mathrm {d}} y$ satisfies

$$\begin{align*}(\operatorname{\mathrm{d}} y)(Y) = 1,\quad (\operatorname{\mathrm{d}} y)[-X,Y] = 0, \end{align*}$$

we find that the $1$ -form $\ell ^* = (\varphi _{-t}^*\operatorname {\mathrm {d}} y)_p \in T_p^*M$ satisfies

$$\begin{align*}\ell^*(\gamma) = 1,\quad \ell^*(\gamma') = 0. \end{align*}$$

As a result, the Hilbert form $\omega $ is given by

(8) $$ \begin{align} \omega = (\pi\circ\iota)^*(\ell^*) = \operatorname{\mathrm{d}} v - v_t \operatorname{\mathrm{d}} t = \operatorname{\mathrm{d}} v - u \operatorname{\mathrm{d}} s. \end{align} $$

3.3.2 Riemann curvature tensor

The geodesic spray $\xi $ is the unique vector field on $\Sigma $ determined by

$$\begin{align*}\omega(\xi) = 1, \quad \operatorname{\mathrm{d}}\omega(\xi,\;\cdot\;) = 0. \end{align*}$$

Since $\omega = \operatorname {\mathrm {d}} v - u \operatorname {\mathrm {d}} s$ , it is easy to see that $\xi = \frac {\partial {}}{\partial {v}}$ .

Within the coordinate system $(u,v,s)$ , the vertical vector field $\frac {\partial {}}{\partial {t}}$ is expressed as

$$\begin{align*}V := \frac{\partial{}}{\partial{t}} = vP(u)\frac{\partial{}}{\partial{u}} - u\frac{\partial{}}{\partial{v}} + \frac{\partial{}}{\partial{s}}. \end{align*}$$

Direct calculation shows that

$$\begin{align*}[\xi,V]=P(u)\frac{\partial{}}{\partial{u}},\quad [\xi,[\xi,V]]=0. \end{align*}$$

Consequently, we have

$$\begin{align*}\mathcal{H}(V)=-[\xi, V] - \frac{1}{2}\mathcal{V}[\xi,[\xi,V]] = -P(u)\frac{\partial{}}{\partial{u}}. \end{align*}$$

Therefore, the Riemann curvature tensor $\mathcal {R}$ is given by

$$\begin{align*}\mathcal{R}(V) = \mathcal{V}\mathcal{H}[\xi,\mathcal{H}(V)] =\mathcal{V}\mathcal{H}\left[\frac{\partial{}}{\partial{v}},-P(u)\frac{\partial{}}{\partial{u}}\right]=0, \end{align*}$$

meaning that the generalized Finsler surface we constructed is indeed a flat one.

3.4 Completeness

Recall that a generalized Finsler structure $(M,\Sigma ,\iota )$ is called complete if the geodesic spray vector field on $\Sigma $ is complete.

Lemma 3.4 Let P and Q be smooth functions on $\mathbb {R}$ . If P is bounded and Q grows sublinearly, i.e., there exist positive real numbers $C_1$ and $C_2$ such that

$$\begin{align*}|P(x)|\leq C_1,\quad |Q(x)|\leq C_2|x|,\quad \forall x\in\mathbb{R}, \end{align*}$$

then the vector field $X=(yP(x)+Q(x))\frac {\partial {}}{\partial {x}}-x\frac {\partial {}}{\partial {y}}$ is complete on $\mathbb {R}^2$ .

Proof To prove that the vector field X is complete on $\mathbb {R}^2$ , we need to show that for any $(x_0,y_0)\in \mathbb {R}^2$ , the solution of the initial value problem

$$\begin{align*}(x,y)'=(yP(x)+Q(x),-x),\quad (x,y)|_{t=0} = (x_0, y_0) \end{align*}$$

is defined for all $t\in \mathbb {R}$ . Let $C_3 = (2C_1^2 + 2C_2^2 + 1)^{1/2}$ , then one can easily verify that

$$\begin{align*}\big((yP(x)+Q(x))^2 + (-x)^2\big)^{1/2} \leq C_3\cdot (x^2+y^2)^{1/2}. \end{align*}$$

By using Theorem 2.17 in [Reference Teschl22, p. 53], the above estimate guarantees that the solution of the above initial value problem is defined for all $t\in \mathbb {R}$ .

Together the above two lemmas, we have proved the last assertion in Theorem 1.3. This finishes the proof of Theorem 1.3.

4.1 A sufficient condition

Let P and Q be smooth functions near $0$ in $\mathbb {R}$ and $P> 0$ . The vector field $X = (yP(x)+Q(x))\frac {\partial {}}{\partial {x}}-x\frac {\partial {}}{\partial {y}}$ has a unique singular point at $(0,-Q(0)/P(0))$ . By suitably translating the coordinates along the y-axis, we may assume the singular point is $(0,0)$ , i.e., $Q(0)=0$ .

Recall that for a vector field X, the singular point $(0,0)$ is called an isochronous center if all the integral curves near $(0,0)$ are closed and have a constant period. The maximal domain enclosing all such integral curves is called the period annulus.

Now we are ready to derive the following proposition.

Proof Let p be a point in the period annulus M. Let $\gamma (t)=\big (\varphi _{t*}Y)_p = \varphi _{t*}Y_{\varphi _{-t}(p)}$ as before. We already know from the proof of Theorem 1.3 that $\gamma (t)$ travels clockwise around the origin (see Remark 3.2 at the end of Section 3.2). Lemma 4.1 ensures that $\gamma (t)$ is smooth and closed. Therefore, we only need to prove now that $\gamma (t)$ is a simple curve in $T_pM$ . This amounts to showing that the curve $\gamma (t)$ , $0\leq t\leq T$ , has a winding number of $-1$ .

First, consider the case $p=(0,0)$ . In this situation, we have $\varphi _{-t}(p)= p$ and ${\gamma (t) = \varphi _{t*}Y_p}$ . By the definition of flow, $\varphi _{t+s} = \varphi _t\circ \varphi _s$ , we have $\varphi _{(t+s)*} = \varphi _{t*}\varphi _{s*}$ . Since the maps $\varphi _{t*}:T_pM\to T_pM$ are linear, we find that $t\mapsto \varphi _{t*}$ is a representation of ${S^1 = \mathbb {R}/T\mathbb {Z}}$ on $T_pM$ . Consequently, as the orbit through $Y_p$ , the curve $\gamma (t)$ must be an ellipse. Moreover, since T is the least period of $\varphi _t$ , it must also be the least period of $\gamma (t)$ . Thus, we have proved that $\gamma (t)$ has a winding number of $-1$ when $p=(0,0)$ .

Next, we consider a general point p in the period annulus. Suppose ${\gamma (t) = h_1(t)\left .\frac {\partial {}}{\partial {x}}\right |_p + h_2(t)\left .\frac {\partial {}}{\partial {y}}\right |_p}$ , then we have the following formula for the winding number (see [Reference Berger and Gostiaux4]):

$$\begin{align*}W = \frac{1}{2\pi}\int_0^T \frac{h_1 h_2' - h_2 h_1'}{h_1^2 + h_2^2} \operatorname{\mathrm{d}} t. \end{align*}$$

Since $h_1$ and $h_2$ in the integrand depend continuously on p, the integral W must also be a continuous function on M; but we know that the winding number must be an integer for closed curves, so $W \equiv -1$ .

4.2 A necessary condition

The following proposition suggests that $(0,0)$ being an isochronous center is also the necessary condition for the above generalized Finsler structure to become a classical Finsler structure, at least in a neighborhood of $(0,0)$ .

Proof It is easy to verify that the conclusion holds if $(0,0)$ is not even a center of X.

Meanwhile, if $(0,0)$ is a non-isochronous center of X, then the function ${T:M\to \mathbb {R}}$ , where $T(p)$ is the least period of the orbit containing p, is bound to be differentiable and non-constant on any open set U and its open subset ${U_0:=U\backslash \left \{(x,y)\in U\,|\,yP(x)+Q(x)=0\right \}}$ . As the directional derivative of T along the vector X is already zero, we must have $\frac {\partial {T}}{\partial {y}}\neq 0$ on $U_0$ .

Let $p = (x,y)\in U_0$ . Consider a curve $c(t)$ in M satisfying $c(0)=p$ and $c'(0)=Y_p$ as well as its image under $\varphi _{T(p)}$ . Expanding near $t=0$ , we have $T(c(t)) = T(p) + t\frac {\partial {T}}{\partial {y}}(p) + o(t)$ . Thus

$$\begin{align*}\begin{aligned} \varphi_{T(p)}\left(c(t)\right)&=\varphi_{T(p)-T(c(t))}(c(t))\\ &=\varphi_{-t\frac{\partial{T}}{\partial{y}}(p)-o(t)}(c(t))\\ &=c(t) - t\frac{\partial{T}}{\partial{y}}(p)\cdot X_p + o(t). \end{aligned} \end{align*}$$

Consequently,

$$\begin{align*}(\varphi_{T(p)*}Y)_p=\varphi_{T(p)*} c'(0) = Y_p - \frac{\partial{T}}{\partial{y}}(p)\cdot X_p. \end{align*}$$

By using a similar argument, we have

$$\begin{align*}(\varphi_{kT(p)*}Y)_p = Y_p - k\frac{\partial{T}}{\partial{y}}(p)X_p,\quad k\in\mathbb{N}. \end{align*}$$

Thus, since $\gamma (0) \neq \gamma (kT(p))$ , the curve $\gamma (t)$ cannot be closed.

Even if the vector field X admits an isochronous center, one can similarly prove that for any open set U outside the isochronous period annulus of X, there exists a point $p\in U$ such that the indicatrix at p is not closed. Thus, the above two propositions imply that, for the generalized Finsler structure on M to be a classical one, X must have an isochronous center and M must be included in the isochronicity period annulus. Moreover, combining this result with the last assertion of Theorem 1.3, we find that the constructed generalized structure is classical and complete if and only if $(0,0)$ is a global isochronous center on $\mathbb {R}^2$ , i.e., the entire $\mathbb {R}^2$ is the period annulus.

4.3 Rotational symmetry

At this point, we have almost finished the proof of Theorem 1.4. It remains to show that X admits an isochronous center if and only if it generates an $SO(2)$ action, but this is indeed a conceptual transition between dynamical systems and differential geometry.

If X admits an isochronous center, then its flow $\varphi _t$ has a constant period T, i.e., $\varphi _{t+T} = \varphi _{t}$ holds for all $t\in \mathbb {R}$ . This means that the action of the flow on M reduces to $\mathbb {R}/T\mathbb {Z} = S^1$ .

Conversely, if X generates an $SO(2)=S^1$ action on M, then every orbit has a constant period T. Together with the fact that X has a unique singular point, we find that X must have an isochronous center. The proof of Theorem 1.4 ends here.

In the following, we shall discuss the difference between rotational symmetry and spherical symmetry. Let us begin with the following definition.

The difference between rotational symmetry and spherical symmetry is better illustrated by the following two-dimensional example: Let us consider the usual $SO(2)$ or $O(2)$ action on $\mathbb {R}^2$ . Let

$$\begin{align*}F(x_1,x_2; y_1,y_2) = |y|\phi(s), \end{align*}$$

where $|y| = \sqrt {(y_1)^2+(y_2)^2}$ is the Euclidean norm, and $s = (x_1y_2 - x_2y_1)/|y|$ . If the one-variable function $\phi $ satisfies some open conditions, then F is a Finsler metric defined in a neighborhood of $(0,0)$ . It is easy to check that the corresponding Finsler structure is rotationally symmetric.

$$\begin{align*}F(Ax, Ay) = F(x,y),\quad \forall A\in SO(2),\; x\in\mathbb{R}^2, y\in T_x\mathbb{R}^2\simeq\mathbb{R}^2, \end{align*}$$

but in general it is not spherically symmetric because the above relation does not hold for any $A\in O(n)$ . Actually, this metric is spherically symmetric only if $\phi $ is an even function. Indeed, a generalized Finsler structure on $\mathbb {R}^2$ is just an assignment of indicatrices (strongly convex curves) to the tangent spaces. So along the (positive) $x_1$ -axis, there is a family of such strongly convex curves. If the Finsler structure is rotationally symmetric, then this family of curves can be freely assigned, and it completely determines the Finsler structure since the indicatrices at any point $(x_1,x_2)$ can be obtained by rotating the corresponding curve at $(|x|, 0)$ . However, to make the Finsler structure spherically symmetric, this family of curves has to be symmetric about the $x_1$ -axis because the reflection on the $x_1$ -axis is an element of $O(2)$ .

However, in dimensions $\geq 3$ , one cannot distinguish between rotational symmetry and spherical symmetry by looking at the expression of the Finsler metric F, because in both cases, the Finsler metric can be expressed as $F = |y|\cdot \phi (|x|,\langle x,y\rangle /|y|)$ in some local coordinate system. For a proof of this fact, one may consult [Reference Huang and Mo12], in which the proof of Proposition 3.1 also works for rotationally symmetric Finsler metrics.

5.1 Isochronicity conditions when $Q=0$

When $Q=0$ , the integral curves of X are given by solutions of the dynamical system

(9) $$ \begin{align} x'=yP(x),\quad y'=-x. \end{align} $$

Let $\kappa = P(0)^{-1/2}$ , and let $b(x)$ be the solution to the following initial value problem:

(10) $$ \begin{align} \kappa + b(x) - xb'(x) = P\Big(\frac{x}{\kappa+b(x)}\Big)\cdot(\kappa+b(x))^3,\quad b(0)=0. \end{align} $$

By the following change of variables:

$$\begin{align*}x=\frac{\tilde{x}}{\kappa+b(\tilde{x})},\quad y=\tilde{y}, \end{align*}$$

we can rewrite the above system (9) as

(11) $$ \begin{align} \tilde{x}'=\frac{\tilde{y}}{\kappa+b(\tilde{x})},\quad \tilde{y}'=\frac{-\tilde{x}}{\kappa+b(\tilde{x})}. \end{align} $$

The isochronicity condition of this system will follow in the next two lemmas.

Proof Let $f(x) = \lambda (x) + \lambda (-x) - 2\lambda (0)$ , then $f(x)$ is even and $f(0) = 0$ . Moreover, the condition implies that

$$\begin{align*}\int_0^{\pi} f(r\cos\theta)\operatorname{\mathrm{d}}\theta = 0. \end{align*}$$

By changing variables $x=r\cos \theta $ , one can rewrite the above equation as

$$\begin{align*}\int_{-r}^{r} f(x) (r^2-x^2)^{-1/2} \operatorname{\mathrm{d}} x = 0. \end{align*}$$

Since f is even, we also have

$$\begin{align*}\int_0^r f(x) (r^2 - x^2)^{-1/2} \operatorname{\mathrm{d}} x = 0. \end{align*}$$

Now, we use induction to show that $\int _0^r f(x) x^{2m}(r^2 - x^2)^{-1/2} \operatorname {\mathrm {d}} x = 0$ holds for any nonpositive integer m. The base case $m=0$ is already established above. Suppose it holds for some $m\geq 0$ , then we have

$$ \begin{align*} &\int_0^r f(x) x^{2m+2}(r^2 - x^2)^{-1/2} \operatorname{\mathrm{d}} x \\ =& - \int_0^r f(x) x^{2m}(r^2 - x^2)(r^2 - x^2)^{-1/2} \operatorname{\mathrm{d}} x \\ =& -\int_0^r f(x) x^{2m}(r^2 - x^2)^{1/2} \operatorname{\mathrm{d}} x \\ =& -\int_0^r \operatorname{\mathrm{d}} x \int_x^r s f(x) x^{2m}(s^2 - x^2)^{-1/2} \operatorname{\mathrm{d}} s\\ =& -\int_0^r \operatorname{\mathrm{d}} s \int_0^s s f(x) x^{2m}(s^2 - x^2)^{-1/2} \operatorname{\mathrm{d}} x = 0. \end{align*} $$

Thus, the equality holds for all nonpositive integers m. From the above deduction, we also have

$$\begin{align*}\int_0^r f(x)x^{2m}(r^2 - x^2)^{1/2} \operatorname{\mathrm{d}} x = 0. \end{align*}$$

Since any continuous even function can be approximated by linear combinations of $1$ , $x^2$ , $x^4$ , $\dots $ , we conclude from the above equation that

$$\begin{align*}\int_0^r f(x)(r^2-x^2)^{1/2} g(x) \operatorname{\mathrm{d}} x = 0 \end{align*}$$

holds for any even function $g(x)$ . Taking $g(x) = f(x)(r^2 - x^2)^{1/2}$ then shows that $f(x) = 0$ on $[0,r]$ .

Lemma 5.3 Suppose $\lambda $ is a smooth function satisfying $\lambda (0)>0$ , then the necessary and sufficient condition for $(0,0)$ to be an isochronous center of the system

$$\begin{align*}x'=-\frac{y}{\lambda(x)}, \quad y'=\frac{x}{\lambda(x)} \end{align*}$$

is that $\lambda (x)-\lambda (0)$ is an odd function.

Proof The integral curves of this system obviously share the same shapes as those of the system $x'=-y$ , $y'=x$ , so $(0,0)$ is a center of this system. By changing to polar coordinates, i.e., $x=r\cos \theta $ , $y=r\sin \theta $ , this system can be written as

$$\begin{align*}r' = 0,\quad \theta' = \frac{1}{\lambda(r\cos\theta)}. \end{align*}$$

Thus, the period of an orbit passing through $(r,0)$ is given by

$$\begin{align*}T(r)=\int_0^{T(r)} \operatorname{\mathrm{d}} t = \int_{0}^{2\pi}\lambda(r\cos\theta)\operatorname{\mathrm{d}}\theta. \end{align*}$$

The point $(0,0)$ is an isochronous center if and only if $T(r)$ is independent of r. By Lemma 5.1, this happens if and only if $\lambda (x)-\lambda (0)$ is an odd function.

Using this lemma, we know that the vector field $X=yP(x)\frac {\partial {}}{\partial {x}}-x\frac {\partial {}}{\partial {y}}$ possesses an isochronous center if and only if the function b is odd. What’s more, after picking a positive constant $\kappa $ and an odd function b, we can easily solve for P from the relations

(12) $$ \begin{align} P(x)=(\kappa+b(\tilde{x}) - \tilde{x}b'(\tilde{x}))(\kappa + b(\tilde{x}))^{-3},\quad x = \tilde{x}/(\kappa + b(\tilde{x})). \end{align} $$

Example 5.4 If we take $\kappa =1$ and $b(x) = -\varepsilon x$ , then one can solve the above equations to get

$$\begin{align*}P(x)=(1+\varepsilon x)^3. \end{align*}$$

In this case, we have the following well-known polynomial Abel system (see [Reference Volokitin and Ivanov23, Theorem 8], see also [Reference Boussaada, Chouikha and Strelcyn5, Theorem 6.2]):

$$\begin{align*}x' = y(1+\varepsilon x)^3,\quad y' = -x, \end{align*}$$

where $\varepsilon \in (-1,1)$ .

Example 5.5 Set $b(x)=\varepsilon x/(1+x^2)$ , then we have

$$\begin{align*}P(x)=\frac{(z^2+1)^2}{(z^2+\varepsilon z+1)^2}, \end{align*}$$

where z is determined by the relation $x=z(z^2+1)/(z^2+\varepsilon z+1)$ . From the above expression, it is easy to see that $P(x)$ is bounded on R when $\varepsilon \in (-2,2)$ . Thus, by Theorems 1.3 and 1.4, the corresponding Finsler structure is defined on the entire $\mathbb {R}^2$ , and it is complete.

5.2 Isochronicity conditions when $Q\neq 0$

In the general case, integral curves of X are given by solutions of

$$\begin{align*}x'=yP(x)+Q(x),\quad y'=-x, \end{align*}$$

where, without loss of generality, we can assume that $P(0)=1$ and $Q(0)=0$ .

This time, using a change of variables

$$\begin{align*}u=-\int_0^x\frac{1}{P(s)}ds,\quad v=y+\frac{Q(x)}{P(x)}, \end{align*}$$

we can turn this system into a Liénard system

(13) $$ \begin{align} u'=-v,\quad v'=A(u)+vB(u), \end{align} $$

where the functions A and B are determined by the relations

(14) $$ \begin{align} A(u)=-x, \quad B(u)=\frac{Q'(x)P(x)-P'(x)Q(x)}{P(x)}. \end{align} $$

The necessary and sufficient conditions for Liénard systems to have isochronous centers were studied by Amel’kin and Rudenok, who, in 2018, proved the following theorem (see [Reference Amel’kin and Rudenok1, Theorem 18] or [Reference Rudenok18, Theorem 6]).

Theorem 5.6 [Reference Amel’kin and Rudenok1, Reference Rudenok18] If A and B are analytic near $0$ , then the Liénard system (13) possesses an isochronous center at $(0,0)$ if and only if there exists a function $\alpha $ and an odd function b, such that

$$ \begin{align*} A(u) &= \alpha'(u)\Big(\alpha(u) + \frac{1}{\alpha^3(u)}\Big(\int_0^{\alpha(u)}zb(z)\operatorname{\mathrm{d}} z\Big)^2\Big),\\ B(u) &= \alpha'(u) b(\alpha(u)), \end{align*} $$

where the function $\alpha $ is invertible near $0$ and its inverse function $\alpha ^{-1}$ satisfies that $\alpha ^{-1}(x)-x$ is even.

To obtain concrete examples, we present the following proposition.

Proposition 5.7 Let A, B, P, and Q be defined as above; then we have the following relations:

(15) $$ \begin{align} \begin{aligned} x &= -A(u), \\ P(x) &= A'(u),\\ Q(x)/P(x) &= -\int_0^u B(s)\operatorname{\mathrm{d}} s. \end{aligned} \end{align} $$

Combining Theorem 5.6 and Proposition 5.7, we have proved Theorem 1.5. Now we shall present several examples.

Example 5.8 Set $\alpha (u)=u$ and $b(z)=z$ in Theorem 5.6, then we get

$$ \begin{align*} A(u) &= u + u^{-3}\big(\int_0^u z^2 \operatorname{\mathrm{d}} z\big)^2 = u + u^3/9,\\ B(u) &= u. \end{align*} $$

From the relations (15), we have

$$\begin{align*}\begin{aligned} P(x)&=1+u^2/3,\\ Q(x)&=-\frac{1}{2}u^2(1+u^2/3), \end{aligned} \end{align*}$$

where $x=-A(u)=-u-u^3/9$ .

Example 5.9 Set $\alpha (u)=u$ and $b(z)=\sin z$ in Theorem 5.6, then we have

$$\begin{align*}\begin{aligned} P(x)&=1+u^{-4}(\sin u-u\cos u)\left(2u^2\sin u-3(\sin u-u\cos u)\right),\\ Q(x)&=-P(x)(1-\cos u), \end{aligned} \end{align*}$$

where u is determined by the relation $x=u+(\sin u-u\cos u)^2u^{-3}$ . One can verify that this relationship is globally invertible. From the above expression, it is seen that $P(x)$ and $Q(x)$ are bounded on $\mathbb {R}$ . Therefore, the corresponding Finsler structure is complete on $\mathbb {R}^2$ .

5.3 The Bryant–Huang–Mo normal form

So far, our discussions are based on the Bryant normal form (2). If we use the Bryant–Huang–Mo normal form (3) instead, similar results can be obtained. Since these two normal forms only differ by a coordinate transformation, the corresponding results are identical in content, but different in expressions. For example, the vector field to be considered has the following form:

(16) $$ \begin{align} X = -y\frac{\partial{}}{\partial{x}}+\big(xu(x)^{-2}+yv(x)\big)\frac{\partial{}}{\partial{y}}, \end{align} $$

where $u(x)$ and $v(x)$ are the arbitrary functions appeared in (3). We may again use its flow $\varphi _t$ to push the vector field $Y = \frac {\partial {}}{\partial {y}}$ to generate a generalized Finsler structure. This generalized Finsler structure reduces to a classical one, if and only if X admits an isochronous center. However, to find the isochronicity condition, we do not need to do any coordinate transformations, because it is already a Liénard system.

$$\begin{align*}x'=-y, \quad y'=xu(x)^{-2} + yv(x). \end{align*}$$

Thus, Theorem 5.6 directly gives the expressions of u and v when the system is isochronous. Both normal forms suggest the following corollary holds.