#  Minimal period problems for brake orbits of nonlinear autonomous reversible
semipositive Hamiltonian systems

Duanzhi Zhang

School of Mathematical Sciences and LPMC, Nankai University
Tianjin 300071, People’s Republic of China Partially supported by National
Science Foundation of China (10801078, 11171314), LPMC of Nankai University.
E-mail<EMAIL_ADDRESS>
###### Abstract

In this paper, for any positive integer $n$, we study the Maslov-type index
theory of $i_{L_{0}}$, $i_{L_{1}}$ and $i_{\sqrt{-1}}^{L_{0}}$ with
$L_{0}=\\{0\\}\times{\bf R}^{n}\subset{\bf R}^{2n}$ and $L_{1}={\bf
R}^{n}\times\\{0\\}\subset{\bf R}^{2n}$. As applications we study the minimal
period problems for brake orbits of nonlinear autonomous reversible
Hamiltonian systems. For first order nonlinear autonomous reversible
Hamiltonian systems in ${\bf R}^{2n}$, which are semipositive, and
superquadratic at zero and infinity we prove that for any $T>0$, the
considered Hamiltonian systems possesses a nonconstant $T$ periodic brake
orbit $X_{T}$ with minimal period no less than $\frac{T}{2n+2}$. Furthermore
if $\int_{0}^{T}H^{\prime\prime}_{22}(x_{T}(t))dt$ is positive definite, then
the minimal period of $x_{T}$ belongs to $\\{T,\;\frac{T}{2}\\}$. Moreover, if
the Hamiltonian system is even, we prove that for any $T>0$, the considered
even semipositive Hamiltonian systems possesses a nonconstant symmetric brake
orbit with minimal period belonging to $\\{T,\;\frac{T}{3}\\}$.

MSC(2000): 58E05; 70H05; 34C25
Key words: symmetric, brake orbit, semipositive and reversible, Maslov-type
index, minimal period, Hamiltonian systems.

##  1 Introduction and main results

In this paper, let $J=\left(\begin{array}[]{cc}0&-I_{n}\\\
I_{n}&0\end{array}\right)$ and $N=\left(\begin{array}[]{cc}-I_{n}&0\\\
0&I_{n}\end{array}\right)$, where $I_{n}$ is the identity in ${\bf R}^{n}$ and
$n\in{\bf N}$. We suppose the following condition

(H1) $H\in C^{2}({\bf R}^{2n},{\bf R})$ and satisfies the following reversible
condition

$\displaystyle H(Nx)=H(x),\qquad\forall x\in{\bf R}^{2n}.$

We consider the following problem:

$\displaystyle\dot{x}=JH^{\prime}(x),\qquad x\in{\bf R}^{2n},$ (1.1)
$\displaystyle x(-t)=Nx(t),\;\;x(T+t)=x(t),\qquad\forall t\in{\bf R}.$ (1.2)

A solution $(T,x)$ of (1.1)-(1.2) is a special periodic solution of the
Hamiltonian system (1.1). We call it a brake orbit and $T$ the period of $x$.
Moreover, if $x({\bf R})=-x({\bf R})$, we call it a symmetric brake orbit. It
is easy to check that if $\tau$ is the minimal period of $x$, there must holds
$x(t+\frac{\tau}{2})=-x(t)$ for all $t\in{\bf R}$.

Since 1948, when H. Seifert in [47] proposed his famous conjecture of the
existence of $n$ geometrically different brake orbits in the potential well in
${\bf R}^{n}$ under certain conditions, many people began to study this
conjecture and related problems. Let ${}^{\\#}\tilde{\mathcal{O}}({\Omega})$
and ${}^{\\#}\tilde{\mathcal{J}}_{b}({\Sigma})$ the number of geometrically
distinct brake obits in ${\Omega}$ for the second order case and on ${\Sigma}$
for the first order case respectively. S. Bolotin proved first in [7](also see
[8]) of 1978 the existence of brake orbits in general setting. K. Hayashi in
[27], H. Gluck and W. Ziller in [25], and V. Benci in [5] in 1983-1984 proved
${}^{\\#}\tilde{\mathcal{O}}({\Omega})\geq 1$ if $V$ is $C^{1}$,
$\bar{{\Omega}}=\\{V\leq h\\}$ is compact, and $V^{\prime}(q)\neq 0$ for all
$q\in\partial{{\Omega}}$. In 1987, P. Rabinowitz in [45] proved that if $H$ is
$C^{1}$ and satisfies the reversible conditon, ${\Sigma}\equiv H^{-1}(h)$ is
star-shaped, and $x\cdot H^{\prime}(x)\neq 0$ for all $x\in{\Sigma}$, then
${}^{\\#}\tilde{\mathcal{J}}_{b}({\Sigma})\geq 1$. In 1987, V. Benci and F.
Giannoni gave a different proof of the existence of one brake orbit in [6].

In 1989, A. Szulkin in [49] proved that ${}^{\\#}\tilde{{\cal
J}_{b}}(H^{-1}(h))\geq n$, if $H$ satisfies conditions in [43] of Rabinowitz
and the energy hypersurface $H^{-1}(h)$ is $\sqrt{2}$-pinched. E. van Groesen
in [26] of 1985 and A. Ambrosetti, V. Benci, Y. Long in [1] of 1993 also
proved ${}^{\\#}\tilde{\mathcal{O}}({\Omega})\geq n$ under different pinching
conditions.

In [42] of 2006, Long , Zhu and the author of this paper proved that there
exist at least $2$ geometrically distinct brake orbits on any central
symmetric strictly convex hypersuface ${\Sigma}$ in ${\bf R}^{2n}$ for $n\geq
2$. Recently, in [35], Liu and the author of this paper proved that there
exist at least $[n/2]+1$ geometrically distinct brake orbits on any central
symmetric strictly convex hypersuface ${\Sigma}$ in ${\bf R}^{2n}$ for $n\geq
2$, if all brake orbits on ${\Sigma}$ are nondegenerate then there are at
least $n$ geometrically distinct brake orbits on ${\Sigma}$. For more details
one can refer to [42], [35] and the reference there in.

In his pioneering paper [43] of 1978, P. Rabinowitz proved the following
famous result via the variational method. Suppose $H$ satisfies the following
conditions:

($\rm{H}1^{\prime}$) $H\in C^{1}({\bf R}^{2n},{\bf R})$.

(H2) There exist constants $\mu>2$ and $r_{0}>0$ such that

$0<\mu H(x)\leq H^{\prime}(x)\cdot x,\quad\forall|x|\leq r_{0}.$

(H3) $H(x)=o(|x|^{2})$ at $x=0$.

(H4) $H(x)\geq 0$ for all $x\in{\bf R}^{2n}$.

Then for any $T>0$, the system (1.1) possesses a non-constant $T$-periodic
solution. Because a $T/k$ periodic function is also a $T$-periodic function,
in [43] Rabinowitz proposed a conjecture that under conditions (H1′) and
(H2)-(H4), there is a non-constant solution possessing any prescribed minimal
period. Since 1978, this conjecture has been deeply studied by many
mathematicians. A significant progress was made by Ekeland and Hofer in their
celebrated paper [16] of 1985, where they proved Rabinowitz’s conjecture for
the strictly convex Hamiltonian system. For Hamiltonian systems with convex or
weak convex assumptions, we refer to [2]-[3], [12]-[13], [15]-[17], [41],
[20]-[23], and references therein for more details. For the case without
convex condition we refer to [37]-[39] and Chapter 13 of [41] and references
therein. A interesting result is for the semipositive first order Hamiltonian
system, in [18] G. Fei, S.-T. Kim, and T. Wang proved the existence of a T
periodic solution of system (1.1) with minimal period no less than $T/2n$ for
any given $T>0$.

Note that in the second order Hamiltonian systems there are many results on
the minimal problem of brake orbits such us [37]-[39] and [50]. For the even
first order Hamiltonian system, in [51], the author of this paper studied the
minimal period problem of semipositive even Hamiltonian system and gave a
positive answer to Rabinowitz’s conjecture in that case. In [19], G. Fei,
S.-T. Kim, and T. Wang proved the same result for second order Hamiltonian
systems.

So it is natural to consider the minimal period problem of brake orbits in
reversible first order nonlinear Hamiltonian systems. In [32], Liu have
considered the strictly convex reversible Hamiltonian systems case and proved
the existence of nonconstant brake orbit of (1.1) with minimal period
belonging to $\\{T,T/2\\}$ for any given $T>0$.

Since [51], we also hope to obtain some interesting results in the even
Hamiltonian system for the minimal period problem of brake orbits.

It can be found in many papers mentioned above that the Maslov-type index
theory and its iteration theory play a important role in the study of minimal
period problems in Hamiltonian systems. In this paper we study some
monotonicity properties of Maslov-type index and apply it to prove our main
results.

In this paper we denote by $\mathcal{L}({\bf R}^{2n})$ and
$\mathcal{L}_{s}({\bf R}^{2n})$ the set of all real $2n\times 2n$ matrices and
symmetric matrices respectively. And we denote by $y_{1}\cdot y_{2}$ the usual
inner product for all $y_{1},\;y_{2}\in{\bf R}^{k}$ with $k$ being any
positive integer. Also we denote by ${\bf N}$ and ${\bf Z}$ the set of
positive integers and integers respectively.

Let ${\rm Sp}(2n)=\\{M\in\mathcal{L}({\bf R}^{2n})|M^{T}JM=J\\}$ be the
$2n\times 2n$ real symplectic group. For any $\tau>0$, Set
$\mathcal{P}_{\tau}=\\{{\gamma}\in C([0,\tau],{\rm
Sp}(2n))|{\gamma}(0)=I_{2n}\\}$ and $S_{\tau}={\bf R}/(\tau{\bf Z})$.

For any ${\gamma}\in\mathcal{P}_{\tau}$ and ${\omega}\in{\bf U}$, where ${\bf
U}$ is the unit circle of the complex plane ${\bf C}$, the Maslov-type index
$(i_{\omega}({\gamma}),\nu_{\omega}({\gamma}))\in{\bf Z}\times\\{0,1,...2n\\}$
was defined by Long in [40]. We have a brief review in Appendix of Section 6.

For convenience to introduce our results, we define the following (B1)
condition, since the Hamiltonian systems considered here are reversible, this
condition is natural.

(B1) Condition. For any $\tau>0$ and $B\in C([0,\tau],\mathcal{L}_{s}({\bf
R}^{2n})$ with the $n\times n$ matrix square block form
$B(t)=\left(\begin{array}[]{cc}B_{11}(t)&B_{12}(t)\\\
B_{21}(t)&B_{22}(t)\end{array}\right)$ satisfying
$B_{12}(0)=B_{21}(0)=0=B_{12}(\tau)=B_{21}(\tau)$, We will call $B$ satisfies
the condition (B1).

Throughout this paper, we denote by

$\displaystyle L_{0}=\\{0\\}\times{\bf R}^{n}\subset{\bf R}^{2n},\quad
L_{1}={\bf R}^{n}\times\\{0\\}\subset{\bf R}^{2n}.$ (1.3)

The definitions of Maslov-type indices
$(i_{\sqrt{-1}}^{L_{0}}({\gamma}),\nu_{\sqrt{-1}}^{L_{0}}({\gamma}))$ and
$(i_{L_{j}}({\gamma}),\nu_{L_{j}}({\gamma}))\in{\bf Z}\times\\{0,1,...,n\\}$
for $j=0,1$ and ${\gamma}\in\mathcal{P}_{\tau}(2n)$ with $\tau>0$ can be found
in [42] and Section 2 below. Also for $B\in C([0,\tau],\mathcal{L}_{s}({\bf
R}^{2n})$ satisfies condition (B1), the definitions of
$(i_{\sqrt{-1}}^{L_{0}}(B),\nu_{\sqrt{-1}}^{L_{0}}(B))$ and
$(i_{L_{j}}(B),\nu_{L_{j}}(B))\in{\bf Z}\times\\{0,1,...,n\\}$ for $j=0,1$ and
${\gamma}\in\mathcal{P}_{\tau}(2n)$ can be found in Section 2 and references
therein.

For any $B\in C([0,\tau],\mathcal{L}_{s}({\bf R}^{2n}))$, denote by
${\gamma}_{B}$ the fundamental solution of the following problem:

$\displaystyle\dot{{\gamma}}_{B}(t)$ $\displaystyle=$ $\displaystyle
JB(t){\gamma}_{B}(t),$ (1.4) $\displaystyle{\gamma}_{B}(0)$ $\displaystyle=$
$\displaystyle I_{2n}.$ (1.5)

Then ${\gamma}_{B}\in\mathcal{P}_{\tau}$. We call ${\gamma}_{B}$ the
symplectic path associated to $B$.

Definition 1.1. If $H\in C^{2}({\bf R}^{2n},{\bf R})$ is a reversible
function, for any $x_{\tau}$ be a $\tau$-periodic brake orbit solution of
(1.1), let $B(t)=H^{\prime\prime}(x(t))$, we define
${\gamma}_{x_{\tau}}={\gamma}_{B}|_{[0,\frac{\tau}{2}]}$ and call it the
symplectic path associated to $x_{\tau}$. We define

$i_{L_{0}}(x_{\tau})=i_{L_{0}}({\gamma}_{x_{\tau}}),\qquad\nu_{L_{0}}(x_{\tau})=i_{L_{0}}({\gamma}_{x_{\tau}}).$
(1.6)

Moreover, if $H$ is even and $x_{\tau}$ is a $\tau$-periodic symmetric brake
orbit solution of (1.1), let $B(t)=H^{\prime\prime}(x(t))$, we define
${\gamma}_{x_{\tau}}={\gamma}_{B}|_{[0,\frac{\tau}{4}]}$ and call it the
symplectic path associated to $x_{\tau}$. We define

$i_{\sqrt{-1}}^{L_{0}}(x_{\tau})=i_{\sqrt{-1}}^{L_{0}}({\gamma}_{x_{\tau}}),\qquad\nu_{\sqrt{-1}}^{L_{0}}(x_{\tau})=i_{\sqrt{-1}}^{L_{0}}({\gamma}_{x_{\tau}}).$
(1.7)

Definition 1.2. For any $\tau$-period and $k\in{\bf N}\equiv\\{1,2,...\\}$, we
define the $k$ times iteration $x^{k}$ of $x$ by

$x^{k}(t)=x(t-j\tau),\quad j\tau\leq t\leq(j+1)\tau,\quad 0\leq j\leq k.$
(1.8)

As in [35], for any ${\gamma}\in\mathcal{P}_{\tau}$ and $k\in{\bf
N}\equiv\\{1,2,...\\}$, in this paper the $k$-time iteration ${\gamma}^{k}$ of
${\gamma}\in\mathcal{P}_{\tau}(2n)$ in brake orbit boundary sense is defined
by $\tilde{{\gamma}}|_{[0,k\tau]}$ with

$\displaystyle\tilde{{\gamma}}(t)=\left\\{\begin{array}[]{l}{\gamma}(t-2j\tau)(N{\gamma}(\tau)^{-1}N{\gamma}(\tau))^{j},\;t\in[2j\tau,(2j+1)\tau],j=0,1,2,...\\\
N{\gamma}(2j\tau+2\tau-t)N(N{\gamma}(\tau)^{-1}N{\gamma}(\tau))^{j+1}\;t\in[(2j+1)\tau,(2j+2)\tau],j=0,1,2,...\end{array}\right.$
(1.11)

The followings are our main results of this paper.

Theorem 1.1. Suppose that $H$ satisfies conditions (H1)-(H4) and

(H5) $H^{\prime\prime}(x)$ is semipositive definite for all $x\in{\bf
R}^{2n}$.

Then for any $T>0$, the system (1.1)-(1.2) possesses a nonconstant $T$
periodic brake orbit solution $x_{T}$ with minimal period no less that
$\frac{T}{2n+2}$. Moreover, for $x=(x_{1},x_{2})$ with $x_{1},x_{2}\in{\bf
R}^{n}$, denote by $H^{\prime\prime}_{22}(x)$ the second order differential of
$H$ with respect to $x_{2}$, if

$\int_{0}^{\frac{T}{2}}H^{\prime\prime}_{22}(x_{T}(t))\,dt>0,$ (1.12)

then the minimal period of $x_{T}$ belongs to $\\{T,\frac{T}{2}\\}$.

Remark 1.1. (Theorem 1.1 of [32]) Suppose that $H$ satisfies conditions
(H1)-(H4) and if $x_{T}$ satisfies

(H5′) $\int_{0}^{\frac{T}{2}}H^{\prime\prime}(X_{T}(t))\,dt>0$.

Then the minimal period of $x_{T}$ belongs to $\\{T,\frac{T}{2}\\}$.

In the case $n=1$, the result can be better, i.e., the following

Theorem 1.2. For $n=1$, suppose that $H$ satisfies conditions (H1)-(H4).

Then for any $T>0$, the system (1.1)-(1.2) possesses a nonconstant $T$
periodic brake orbit solution with minimal period belong to
$\\{T,\frac{T}{2}\\}$.

Consider the minimal period problem for $H(x)=\frac{1}{2}B_{0}x\cdot
x+\hat{H}(x)$, where $B_{0}\in\mathcal{L}_{s}({\bf R}^{2n})$. This is
motivated by [18], [22], and [43], where in [18] $B_{0}$ was considered to be
semipositive, in [22] and [43] $B_{0}$ was considered to be positive.

We have the following general result.

Theorem 1.3. Let $2n\times 2n$ be real semipositive matrix $B_{0}={\rm
diag}(B_{11},B_{22})$ with $B_{11}$ and $B_{22}$ being $n\times n$ matrix.
Assume $H(x)=\frac{1}{2}B_{0}x\cdot x+\hat{H}(x)$ for all $x\in{\bf R}^{2n}$,
and $\hat{H}$ satisfies conditions (H1)-(H5).

Then for any $T>0$, (1.1) possesses a nonconstant $T$-periodic brake orbit
$x_{T}$ with minimal period no less than
$\frac{T}{2i_{L_{0}}(B_{0})+2\nu_{L_{0}}(B_{0})+2n+2}$, where we see $B_{0}$
as an element in $C([0,T/2],\mathcal{L}_{s}({\bf R}^{2n}))$ satisfies
condition (B1).

Remark 1.2. In section 3, we will show
$i_{L_{0}}(B_{0})+\nu_{L_{0}}(B_{0})\geq 0$.

As a direct consequence of Theorem 1.3, we have the following Corollary 1.1.

Corollary 1.1. For $T>0$ such that $i_{L_{0}}(B_{0})+\nu_{L_{0}}(B_{0})=0$,
where we see $B_{0}$ as an element in $C([0,T/2],\mathcal{L}_{s}({\bf
R}^{2n})$ satisfies condition (B1), under the same assumptions of Theorem 1.2,
the system (1.1) possesses a nonconstant $T$-periodic brake orbit with minimal
period no less that $\frac{T}{2n+2}$.

We can also prove the following Corollary 1.2 of Theorem 1.3.

Corollary 1.2. If $B_{0}\neq 0$, then for $0<T<\frac{\pi}{||B_{0}||}$ with
$||B_{0}||$ being the operator norm of $B_{0}$, under the same condition of
Theorem 1.2, possesses a nonconstant $T$-periodic brake orbit $x_{T}$ with
minimal period no less than $\frac{T}{2n+2}$. Moreover , if

$\displaystyle\int_{0}^{\frac{T}{2}}H^{\prime\prime}_{22}(x_{T}(t))\,dt>0,$

then the minimal period of $x_{T}$ belongs to $\\{T,\frac{T}{2}\\}$.

Theorem 1.4. Suppose that $H$ satisfies conditions (H1)-(H5) and

(H6) $H(-x)=H(x)$ for all $x\in{\bf R}^{2n}$.

Then for any $T>0$, the system (1.1)-(1.2) possesses a nonconstant symmetric
brake orbit with minimal period belonging to $\\{T,T/3\\}$.

Theorem 1.5. Let $2n\times 2n$ be real semipositive matrix $B_{0}={\rm
diag}(B_{11},B_{22})$ with $B_{11}$ and $B_{22}$ being $n\times n$ matrix,
assume $H(x)=\frac{1}{2}B_{0}x\cdot x+\hat{H}(x)$ for all $x\in{\bf R}^{2n}$,
and $\hat{H}$ satisfies conditions (H1)-(H6). Then for any $T>0$, the system
(1.1)-(1.2) possesses a nonconstant symmetric brake orbit $x_{T}$ with minimal
period no less than
$\frac{T}{4(i_{\sqrt{-1}}^{L_{0}}(B_{0})+\nu_{\sqrt{-1}}^{L_{0}}(B_{0}))+7}$.
Moreover, if $i_{\sqrt{-1}}^{L_{0}}(B_{0})+\nu_{\sqrt{-1}}^{L_{0}}(B_{0})$ is
even, then the minimal period of $x_{T}$ is no less than
$\frac{T}{4(i_{\sqrt{-1}}^{L_{0}}(B_{0})+\nu_{\sqrt{-1}}^{L_{0}}(B_{0}))+3}$
where we see $B_{0}$ as an element in $C([0,T/4],\mathcal{L}_{s}({\bf
R}^{2n})$ satisfies condition (B1).

Remark 1.3. In section 3, we will show that $i_{\sqrt{-1}}^{L_{0}}(B_{0})\geq
0$, hence $i_{\sqrt{-1}}^{L_{0}}(B_{0})+\nu_{\sqrt{-1}}^{L_{0}}(B_{0})\geq 0$.

As a direct consequence of Theorem 1.5, we have the following Corollary 1.3.

Corollary 1.3. For $T>0$ such that
$i_{\sqrt{-1}}^{L_{0}}(B_{0})+\nu_{\sqrt{-1}}^{L_{0}}(B_{0})=0$, under the
same assumptions of Theorem 1.4, the system (1.1) possesses a nonconstant
symmetric brake orbit with minimal period belonging to $\\{T,T/3\\}$.

We can also prove the following Corollary 1.4 of Theorem 1.5.

Corollary 1.4. If $B_{0}\neq 0$, then for $0<T<\frac{\pi}{||B_{0}||}$ with
$||B_{0}||$ being the operator norm of $B_{0}$, under the same condition of
Theorem 1.5, the system (1.1) possesses a nonconstant symmetric brake orbit
with minimal period belonging to $\\{T,T/3\\}$.

This paper is organized as follows. In section 2, we study the Maslov-type
index theory of $i_{L_{0}}$, $i_{L_{1}}$ and $i_{\sqrt{-1}}^{L_{0}}$. We
compute the difference between $i_{L_{0}}({\gamma})$ and
$i_{L_{1}}({\gamma})$. In Section 3, we study the relation between the Maslov-
type index $(i_{\sqrt{-1}}^{L_{0}}(B),\nu_{\sqrt{-1}}^{L_{0}}(B))$ for $B\in
C([0,\tau],\mathcal{L}_{s}({\bf R}^{2n})$ satisfies condition (B1) and the
Morse indices of the corresponding Galerkin approximation. As applications we
get some monotonicity properties of $i_{L_{0}}(B)$, $i_{L_{1}}(B)$ and
$i_{\sqrt{-1}}^{L_{0}}(B)$ and we prove Theorem 3.2 which is very important in
the proof of Theorems 1.4-1.5. In Section 4, based on the preparations in
Sections 2 and 3 we prove Theorems 1.1-1.3 and Corollary 1.2. In Section 5, we
prove Theorems 1.4-1.5 and corollary 1.4. In Section 6, we give a briefly
review of $(i_{\omega},\nu_{\omega})$ index theory with ${\omega}\in{\bf U}$
for symplectic paths starting with identity as appendix.

##  2 Maslov-type index theory associated with Lagrangian subspaces

###  2.1 A brief review of index function $(i_{L_{j}},\nu_{L_{j}})$ with
$j=0,1$ and $(i_{\sqrt{-1}}^{L_{0}},\nu_{\sqrt{-1}}^{L_{0}})$

Let

$F={\bf R}^{2n}\oplus{\bf R}^{2n}$ (2.1)

possess the standard inner product. We define the symplectic structure of $F$
by

$\\{v,w\\}=(\mathcal{J}v,w),\;\forall v,w\in F,\;{\rm
where}\;\mathcal{J}=(-J)\oplus J=\left(\begin{array}[]{cc}-J&0\\\
0&J\end{array}\right).\;$ (2.2)

We denote by ${\rm Lag}(F)$ the set of Lagrangian subspaces of $F$, and equip
it with the topology as a subspace of the Grassmannian of all $2n$-dimensional
subspaces of $F$.

It is easy to check that, for any $M\in{\rm Sp}(2n)$ its graph

${\rm Gr}(M)\equiv\left\\{\left(\begin{array}[]{c}x\\\
Mx\end{array}\right)|x\in{\bf R}^{2n}\right\\}$

is a Lagrangian subspace of $F$.

Let

$\displaystyle V_{1}=\\{0\\}\times{\bf R}^{n}\times\\{0\\}\times{\bf
R}^{n}\subset{\bf R}^{4n},\quad V_{2}={\bf R}^{n}\times\\{0\\}\times{\bf
R}^{n}\times\\{0\\}\subset{\bf R}^{4n}.$ (2.3)

By Proposition 6.1 of [35] and Lemma 2.8 and Definition 2.5 of [42], we give
the following definition.

Definition 2.1. For any continuous path ${\gamma}\in\mathcal{P}_{\tau}(2n)$,
we define the following Maslov-type indices:

$\displaystyle i_{L_{0}}({\gamma})=\mu^{CLM}_{F}(V_{1},{\rm
Gr}({\gamma}),[0,\tau])-n,$ (2.4) $\displaystyle
i_{L_{1}}({\gamma})=\mu^{CLM}_{F}(V_{2},{\rm Gr}({\gamma}),[0,\tau])-n,$ (2.5)
$\displaystyle\nu_{L_{j}}({\gamma})=\dim({\gamma}(\tau)L_{j}\cap L_{j}),\qquad
j=0,1,$ (2.6)

where we denote by $i^{CLM}_{F}(V,W,[a,b])$ the Maslov index for Lagrangian
subspace path pair $(V,W)$ in $F$ on $[a,b]$ defined by Cappell, Lee, and
Miller in [11].

For $\omega=e^{\sqrt{-1}\theta}$ with $\theta\in{\bf R}$, we define a Hilbert
space $E^{\omega}=E^{\omega}_{L_{0}}$ consisting of those $x(t)$ in
$L^{2}([0,\tau],{\bf C}^{2n})$ such that $e^{-\theta tJ}x(t)$ has Fourier
expending

$e^{-\frac{\theta t}{\tau}J}x(t)=\sum_{j\in{\bf Z}}e^{\frac{j\pi
t}{\tau}J}\left(\begin{array}[]{cc}0\\\ a_{j}\end{array}\right),\;a_{j}\in{\bf
C}^{n}$

with

$\|x\|^{2}:=\sum_{j\in{\bf Z}}\tau(1+|j|)|a_{j}|^{2}<\infty.$

For $\omega=e^{\sqrt{-1}\theta}$, $\theta\in(0,\pi)$, we define two self-
adjoint operators $A^{\omega},B^{\omega}\in\mathcal{L}(E^{\omega})$ by

$\displaystyle(A^{\omega}x,y)=\int^{1}_{0}\langle-J\dot{x}(t),y(t)\rangle
dt,\;\;(B^{\omega}x,y)=\int^{1}_{0}\langle B(t)x(t),y(t)\rangle dt$

on $E^{\omega}$. Then $B^{\omega}$ is also compact.

Definition 2.2. We define the index function

$\displaystyle
i_{\omega}^{L_{0}}(B)=I(A^{\omega},\;\;A^{\omega}-B^{\omega})\equiv-{\rm
sf}\\{A^{\omega}-sB^{\omega},0\leq s\leq 1\\},$
$\displaystyle\nu_{\omega}^{L_{0}}(B)=m^{0}(A^{\omega}-B^{\omega}),\;\forall\,\omega=e^{\sqrt{-1}\theta},\;\;\theta\in(0,\pi),$

where the definition of ${\rm sf}$ of spectral flow for the path of bounded
self-adjoint linear operators one can refer to [53] and references their in.

By (3.21) of [35], we have

$i_{L_{0}}(B)\leq i^{L_{0}}_{{\omega}}(B)\leq i_{L_{0}}(B)+n.$ (2.7)

Lemma 2.1. For ${\omega}=e^{\sqrt{-1}\theta}$ with $\theta\in(0,\pi)$, let
$V_{\omega}=L_{0}\times(e^{\theta J}L_{0})\subset{\bf R}^{4n}\equiv F$. There
holds

$i_{\omega}^{L_{0}}(B)=\mu_{F}^{CLM}(V_{\omega},{\rm
Gr}({\gamma}_{B}),[0,\tau]).$ (2.8)

Proof. Without loss of generality we can suppose the $C^{1}$ path ${\rm
Gr}({\gamma}_{B})$ of Lagrangian subspaces intersects $V_{\omega}$ regularly
(otherwise we can perturb it slightly with fixed endow points such that they
intersects regularly and the index dose not change by the homotopy invariant
property $\mu_{F}^{CLM}$ ), where the definition of intersection form can be
found in [46]. We denote by $\mu^{BF}$ the maslov index defined by Booss and
Furutani in [9].

By the spectral flow formula of Theorem 5.1 in [9] or Theorem 1.5 of [10] (cf.
also proof of Proposition 2.3 of [52]), we have

$\displaystyle{\rm sf}\\{A^{\omega}sB^{\omega},0\leq s\leq 1\\}$ (2.9)
$\displaystyle=$ $\displaystyle\mu^{BF}({\rm
Gr}({\gamma}_{B}),V_{\omega},[0,\tau])$ $\displaystyle=$
$\displaystyle\mu^{BF}((I\oplus e^{-\sqrt{-1}\theta J}){\rm
Gr}({\gamma}_{B}),(I\oplus e^{-\sqrt{-1}\theta J})V_{\omega},[0,\tau])$
$\displaystyle=$ $\displaystyle\mu^{BF}((I\oplus e^{-\sqrt{-1}\theta J}){\rm
Gr}({\gamma}_{B}),V_{1},[0,\tau])$ $\displaystyle=$
$\displaystyle-m^{-}(-{\Gamma}((I\oplus e^{-\sqrt{-1}\theta J}){\rm
Gr}({\gamma}_{B}),V_{1},0))+\sum_{0<t<\tau}{\rm sign}(-{\Gamma}((I\oplus
e^{-\sqrt{-1}\theta J}){\rm Gr}({\gamma}_{B}),V_{1},t))$
$\displaystyle+m^{+}(-{\Gamma}((I\oplus e^{-\sqrt{-1}\theta J}){\rm
Gr}({\gamma}_{B}),V_{1},\tau))$ $\displaystyle=$
$\displaystyle-m^{+}({\Gamma}((I\oplus e^{-\sqrt{-1}\theta J}){\rm
Gr}({\gamma}_{B}),V_{1},0))-\sum_{0<t<\tau}{\rm sign}({\Gamma}((I\oplus
e^{-\sqrt{-1}\theta J}){\rm Gr}({\gamma}_{B}),V_{1},t))$
$\displaystyle+m^{-}({\Gamma}((I\oplus e^{-\sqrt{-1}\theta J}){\rm
Gr}({\gamma}_{B}),V_{1},\tau))$ $\displaystyle=$
$\displaystyle-\mu_{F}^{CLM}(V_{1},(I\oplus e^{-\sqrt{-1}\theta J}){\rm
Gr}({\gamma}_{B}),[0,\tau])$ $\displaystyle=$
$\displaystyle-\mu_{F}^{CLM}((I\oplus e^{\sqrt{-1}\theta J})V_{1},{\rm
Gr}({\gamma}_{B}),[0,\tau])$ $\displaystyle=$
$\displaystyle-\mu_{F}^{CLM}(V_{\omega},{\rm Gr}({\gamma}_{B}),[0,\tau]),$

where in the fourth equality we have used Theorem 2.1 in [9] and the property
of index $\mu^{RS}$ for symplectic paths defined in [46](cf also (2.6)-(2.8)
of [52]), in the sixth equality we have used Lemma 2.6 of [42], in the second
and seventh equalities we used the symplectic invariance property of index
$\mu^{BF}$ and $\mu_{F}^{CLM}$ respectively.

Definition 2.3. Let $B\in C([0,\tau],\mathcal{L}_{s}({\bf R}^{2n})$ and
${\gamma}_{B}$ be the symplectic path associated to $B$. We define

$\displaystyle i_{{\omega}}^{L_{0}}({\gamma}_{B})=i_{{\omega}}^{L_{0}}(B),$
(2.10)
$\displaystyle\nu_{{\omega}}^{L_{0}}({\gamma}_{B})=\nu_{{\omega}}^{L_{0}}(B).$
(2.11)

By Lemma 2.1, in general we give the following definition.

Definition 2.4. For any ${\gamma}\in\mathcal{P}_{\tau}(2n)$ and
${\omega}=e^{\sqrt{-1}\theta}$ with $\theta\in(0,\pi)$, we define

$\displaystyle i_{\omega}^{L_{0}}({\gamma})=\mu_{F}^{CLM}(V_{\omega},{\rm
Gr}({\gamma}_{B}),[0,\tau]),$
$\displaystyle\nu_{\omega}^{L_{0}}({\gamma})=\dim\left({\gamma}(\tau)L_{0}\cap
e^{\sqrt{-1}\theta J}L_{0}\right).$ (2.12)

For any ${\gamma}\in\mathcal{P}_{\tau}(2n)$, we define a new symplectic path
$\tilde{{\gamma}}\in\mathcal{P}_{\tau}(2n)$ by

$\tilde{{\gamma}}(t)=\left\\{\begin{array}[]{l}I_{2n},\quad
t\in[0,\frac{\tau}{3}],\\\ {\gamma}(3t-\tau),\quad
t\in[\frac{\tau}{3},\frac{2\tau}{3}],\\\ {\gamma}(\tau),\quad
t\in[\frac{2\tau}{3},\tau].\end{array}\right.$ (2.13)

So we can perturb $\tilde{{\gamma}}$ slightly to a $C^{1}$ path
$\hat{{\gamma}}$ such that $\hat{{\gamma}}$ is homotopic to $\tilde{{\gamma}}$
with fixed end points and $\hat{{\gamma}}(t)=I_{2n}$ for
$t\in[0,\frac{\tau}{6}]$ and $\hat{{\gamma}}(t)={\gamma}(\tau)$ for
$t\in[\frac{5\tau}{6},\tau]$. Set
$\hat{B}(t)=-J\dot{\hat{{\gamma}}}(t)(\hat{{\gamma}}(t))^{-1}$. So we have

$\hat{B}(0)=\hat{B}(\tau)=0.$ (2.14)

Then this $\hat{B}\in C([0,\tau],\mathcal{L}_{s}({\bf R}^{2n})$ and satisfies
condition (B1). Also we have $\hat{{\gamma}}$ is is homotopic to ${\gamma}$
with fixed end points. So we have

$\displaystyle
i_{1}(\hat{{\gamma}}^{k})=i_{1}({\gamma}^{k})=i_{1}({\gamma}_{\hat{B}}^{k}),\qquad\forall
k\in{\bf N},$ (2.15)
$\displaystyle\nu_{1}(\hat{{\gamma}}^{k})=\nu_{1}({\gamma}^{k})=\nu_{1}({\gamma}_{\hat{B}}^{k}),\qquad\forall
k\in{\bf N}$ (2.16)

and

$\displaystyle
i_{L_{0}}(\hat{{\gamma}}^{k})=i_{L_{0}}({\gamma}^{k})=i_{L_{0}}({\gamma}_{\hat{B}}^{k}),\qquad\forall
k\in{\bf N},$ (2.17)
$\displaystyle\nu_{L_{0}}(\hat{{\gamma}}^{k})=\nu_{L_{0}}({\gamma}^{k})=\nu_{L_{0}}({\gamma}_{\hat{B}}^{k}),\qquad\forall
k\in{\bf N}.$ (2.18)

Also by the property of index $\mu_{F}^{CLM}$ and Definition 2.4 have

$\displaystyle
i_{\sqrt{-1}}^{L_{0}}({\gamma}^{k})=i_{\sqrt{-1}}^{L_{0}}(\hat{{\gamma}}^{k})=i_{\sqrt{-1}}^{L_{0}}({\gamma}_{\hat{B}}^{k}),\qquad\forall
k\in{\bf N},$
$\displaystyle\nu_{\sqrt{-1}}^{L_{0}}({\gamma}^{k})=\nu_{\sqrt{-1}}^{L_{0}}(\hat{{\gamma}}^{k})=\nu_{\sqrt{-1}}^{L_{0}}({\gamma}_{\hat{B}}^{k}),\qquad\forall
k\in{\bf N}.$

Hence, in [35] the authors essentially proved the following Bott-type
iteration formula.

Theorem 2.1. (Theorem 4.1 of [35]) Let ${\gamma}\in\mathcal{P}_{\tau}(2n)$ and
$\omega_{k}=e^{\pi\sqrt{-1}/k}$. For odd $k$ we have

$\displaystyle
i_{L_{0}}(\gamma^{k})=i_{L_{0}}(\gamma^{1})+\sum_{i=1}^{(k-1)/2}i_{\omega_{k}^{2i}}(\gamma^{2}),$
$\displaystyle\nu_{L_{0}}(\gamma^{k})=\nu_{L_{0}}(\gamma^{1})+\sum_{i=1}^{(k-1)/2}\nu_{\omega_{k}^{2i}}(\gamma^{2}),$

and for even $k$, we have

$\displaystyle
i_{L_{0}}(\gamma^{k})=i_{L_{0}}(\gamma^{1})+i^{L_{0}}_{\sqrt{-1}}(\gamma^{1})+\sum_{i=1}^{k/2-1}i_{\omega_{k}^{2i}}(\gamma^{2}),\;$
$\displaystyle\nu_{L_{0}}(\gamma^{k})=\nu_{L_{0}}(\gamma^{1})+\nu^{L_{0}}_{\sqrt{-1}}(\gamma^{1})+\sum_{i=1}^{k/2-1}\nu_{\omega_{k}^{2i}}(\gamma^{2}).$

Obviously we also have

$i_{L_{0}}({\gamma})\leq i^{L_{0}}_{\sqrt{-1}}({\gamma})\leq
i_{L_{0}}({\gamma})+n.$ (2.19)

###  2.2 The Bott-type iteration formula for
$(i_{\sqrt{-1}}^{L_{0}},\nu_{\sqrt{-1}}^{L_{0}})$

In order to study the minimal period problem for Even reversible Hamiltonian
systems, we need the iteration formula of the Maslov-type index of
$(i_{\sqrt{-1}}^{L_{0}},\nu_{\sqrt{-1}}^{L_{0}})$ for symplectic paths
starting with identity. We use Theorem 2.1 to obtain it.

Precisely we have the following Theorem.

Theorem 2.2. Let ${\gamma}\in\mathcal{P}_{\tau}(2n)$ and
$\omega_{k}=e^{\pi\sqrt{-1}/k}$. For odd $k$ we have

$\displaystyle
i_{\sqrt{-1}}^{L_{0}}(\gamma^{k})=i_{\sqrt{-1}}^{L_{0}}(\gamma^{1})+\sum_{i=1}^{(k-1)/2}i_{\omega_{k}^{2i-1}}(\gamma^{2}),$
(2.20)
$\displaystyle\nu_{\sqrt{-1}}^{L_{0}}(\gamma^{k})=\nu_{\sqrt{-1}}^{L_{0}}(\gamma^{1})+\sum_{i=1}^{(k-1)/2}\nu_{\omega_{k}^{2i-1}}(\gamma^{2}),$
(2.21)

and for even $k$, we have

$\displaystyle
i_{\sqrt{-1}}^{L_{0}}(\gamma^{k})=\sum_{i=1}^{k/2}i_{\omega_{k}^{2i-1}}(\gamma^{2}),\;$
(2.22)
$\displaystyle\nu_{\sqrt{-1}}^{L_{0}}(\gamma^{k})=\sum_{i=1}^{k/2}\nu_{\omega_{k}^{2i-1}}(\gamma^{2}).$
(2.23)

Proof. For odd $k$, since ${\gamma}^{2k}=({\gamma}^{k})^{2}$, by Theorem 2.1
we have

$\displaystyle
i_{L_{0}}({\gamma}^{2k})=i_{L_{0}}({\gamma}^{k})+i_{\sqrt{-1}}^{L_{0}}({\gamma}^{k}),$
(2.24)
$\displaystyle\nu_{L_{0}}({\gamma}^{2k})=\nu_{L_{0}}({\gamma}^{k})+\nu_{\sqrt{-1}}^{L_{0}}({\gamma}^{k}).$
(2.25)

Also by Theorem 2.1 we have

$\displaystyle
i_{L_{0}}(\gamma^{k})=i_{L_{0}}(\gamma^{1})+\sum_{i=1}^{(k-1)/2}i_{\omega_{k}^{2i}}(\gamma^{2}),$
(2.26)
$\displaystyle\nu_{L_{0}}(\gamma^{k})=\nu_{L_{0}}(\gamma^{1})+\sum_{i=1}^{(k-1)/2}\nu_{\omega_{k}^{2i}}(\gamma^{2}),$
(2.27) $\displaystyle
i_{L_{0}}(\gamma^{2k})=i_{L_{0}}(\gamma^{1})+i_{\sqrt{-1}}^{L_{0}}({\gamma})+\sum_{i=1}^{k-1}i_{\omega_{2k}^{2i}}(\gamma^{2}),$
(2.28)
$\displaystyle\nu_{L_{0}}(\gamma^{2k})=\nu_{L_{0}}(\gamma^{1})+\nu_{\sqrt{-1}}^{L_{0}}({\gamma})+\sum_{i=1}^{k-1}\nu_{\omega_{2k}^{2i}}(\gamma^{2}).$
(2.29)

Since ${\omega}_{k}={\omega}_{2k}^{2}$, by (2.24), (2.28) minus (2.26) yields
(2.20). By (2.25), (2.29) minus (2.27) yields (2.21).

For even k, by similar argument we obtain (2.22) and (2.23). The proof of
Theorem 2.2 is complete.

###  2.3 The difference of $i_{L_{0}}({\gamma})$ and $i_{L_{1}}({\gamma})$.

The precise difference of $i_{L_{0}}({\gamma})$ and $i_{L_{1}}({\gamma})$ for
${\gamma}\in\mathcal{P}_{\tau}$ with $\tau>0$ is very important in the proof
of the main results of this paper. In this subsection we use the H$\ddot{{\rm
o}}$rmander index (cf. [14]) to compute it. Note that in [42], in fact we have
already proved that $|i_{L_{0}}({\gamma})-i_{L_{1}}({\gamma})|\leq n$.

For any $P\in{\rm Sp}(2n)$ and $\varepsilon\in{\bf R}$, we set

$\displaystyle
M_{\varepsilon}(P)=P^{T}\left(\begin{array}[]{cc}\sin{2{\varepsilon}}I_{n}&-\cos{2{\varepsilon}I_{n}}\\\
-\cos{2{\varepsilon}}I_{n}&-\sin
2{\varepsilon}I_{n}\end{array}\right)P+\left(\begin{array}[]{cc}\sin{2{\varepsilon}}I_{n}&\cos{2{\varepsilon}}I_{n}\\\
\cos{2{\varepsilon}}I_{n}&-\sin 2{\varepsilon}I_{n}\end{array}\right).$ (2.34)

Then we have the following theorem.

Theorem 2.3. For ${\gamma}\in\mathcal{P}_{\tau}$ with $\tau>0$, we have

$i_{L_{0}}({\gamma})-i_{L_{1}}({\gamma})=\frac{1}{2}{\rm
sgn}M_{\varepsilon}({\gamma}(\tau)),$ (2.35)

where ${\rm sgn}M_{\varepsilon}({\gamma}(\tau))$ is the signature of the
symmetric matrix $M_{\varepsilon}({\gamma}(\tau))$ and ${\varepsilon}>0$ is
sufficiently small.

we also have,

$(i_{L_{0}}({\gamma})+\nu_{L_{0}}({\gamma}))-(i_{L_{1}}({\gamma})+\nu_{L_{1}}({\gamma}))=\frac{1}{2}{\rm
sign}M_{\varepsilon}({\gamma}(\tau)),$ (2.36)

where ${\varepsilon}<0$ and $|{\varepsilon}|$ is sufficiently small.

Proof. By the first geometrical definition of the Maslov-type index in Section
4 of [11], there exists an ${\varepsilon}>0$ small enough such that

$V_{1}\cap e^{-{\varepsilon}\mathcal{J}}{\rm Gr}({\gamma}(0))=\\{0\\},\qquad
V_{2}\cap e^{-{\varepsilon}\mathcal{J}}{\rm Gr}({\gamma}(\tau))=\\{0\\}.$
(2.37)

By definition 2.1, we have

$\displaystyle
i_{L_{0}}({\gamma})=\mu^{CLM}_{F}(V_{1},e^{-{\varepsilon}\mathcal{J}}{\rm
Gr}({\gamma}),[0,\tau])-n,$ (2.38) $\displaystyle
i_{L_{1}}({\gamma})=\mu^{CLM}_{F}(V_{2},e^{-{\varepsilon}\mathcal{J}}{\rm
Gr}({\gamma}),[0,\tau])-n.$ (2.39)

Define ${\gamma}_{1}(t)=e^{-{\varepsilon}\mathcal{J}}{\rm Gr}({\gamma}(t))$
and ${\gamma}_{2}(t)=e^{-{\varepsilon}\mathcal{J}}{\rm Gr}({\gamma}(\tau-t))$
for $t\in[0,\tau]$. Then ${\gamma}_{1}$ and ${\gamma}_{2}$ are two paths of
Lagrangian subspaces of the symplectic space $(F,\mathcal{J})$ defined in
(2.1) and (2.2). ${\gamma}_{1}$ connects $e^{-{\varepsilon}\mathcal{J}}{\rm
Gr}({\gamma}(0))$ and $e^{-{\varepsilon}\mathcal{J}}{\rm Gr}({\gamma}(\tau))$
and is transversal to $V_{1}$ and $V_{2}$. ${\gamma}_{2}$ connects
$e^{-{\varepsilon}\mathcal{J}}{\rm Gr}({\gamma}(\tau))$ and
$e^{-{\varepsilon}\mathcal{J}}{\rm Gr}({\gamma}(0))$ and is transversal to
$V_{1}$ and $V_{2}$. Denote by ${\gamma}$ the catenation of the paths
${\gamma}_{1}$ and ${\gamma}_{2}$. By Definition 3.4.2 of the
$H\ddot{o}rmande\;index$ $s(M_{1},M_{2};L_{1},L_{2})$ on p. 66 of [14] and
(2.38)-(2.39), we have

$\displaystyle s(V_{1},V_{2};e^{-{\varepsilon}\mathcal{J}}{\rm
Gr}({\gamma}(0)),e^{-{\varepsilon}\mathcal{J}}{\rm Gr}({\gamma}(\tau)))$
(2.40) $\displaystyle=$ $\displaystyle\langle{\gamma},{\alpha}\rangle$
$\displaystyle=$
$\displaystyle\mu^{CLM}_{F}(V_{1},{\gamma}_{1})+\mu^{CLM}_{F}(V_{2},{\gamma}_{2})$
$\displaystyle=$
$\displaystyle\mu^{CLM}_{F}(V_{1},e^{-{\varepsilon}\mathcal{J}}{\rm
Gr}({\gamma}))-\mu^{CLM}_{F}(V_{2},e^{-{\varepsilon}\mathcal{J}}{\rm
Gr}({\gamma}))$ (2.41) $\displaystyle=$ $\displaystyle
i_{L_{0}}({\gamma})-i_{L_{1}}({\gamma}),$ (2.42)

where ${\alpha}$ is the Maslov-Arnold index defined in Theorem 3.4.9 on p. 64
of [14]. Since ${\gamma}_{1}$ and ${\gamma}_{2}$ are transversal to $V_{1}$
and $V_{2}$ (2.40) holds, (2.41) holds from the definition of ${\gamma}_{1}$
and ${\gamma}_{2}$.

In the proof of Theorem 3.3 of [42], we have proved that for ${\varepsilon}>0$
small enough, there holds

${\rm sgn}(V_{1},e^{-{\varepsilon}\mathcal{J}}{\rm Gr}(I_{2n});V_{2})=0,$
(2.43)

where ${\rm sgn}(W_{1},W_{3};W_{2})$ for 3 Lagrangian spaces with $W_{3}$
transverses to $W_{1}$ and $W_{2}$ is introduced in Definition 3.2.3 on p. 67
of [14]. Note that by Claim 1 below, we can prove (2.43) at once.

Claim 1. For ${\varepsilon}>0$, small enough, there holds

${\rm sign}(V_{1},e^{-{\varepsilon}\mathcal{J}}{\rm
Gr}({\gamma}(\tau));V_{2})={\rm sgn}(M_{\varepsilon}({\gamma}(\tau))).$ (2.44)

Proof of Claim 1. In fact,

$e^{-\mathcal{J}}{\rm
Gr}({\gamma}(\tau))=\left\\{\left(\begin{array}[]{cc}e^{{\varepsilon}J}&0\\\
0&e^{-{\varepsilon}J}{\gamma}(\tau)\end{array}\right)\left(\begin{array}[]{c}p\\\
q\\\ p\\\ q\end{array}\right)=\left(\begin{array}[]{c}cp-sq\\\ sp+cq\\\
(c,s){\gamma}(\tau)(p,q)^{T}\\\
(-s,c){\gamma}(\tau)(p,q)^{T}\end{array}\right);\quad p,q\in{\bf
R}^{n}\right\\},$ (2.45)

where we denote by $c=\cos{\varepsilon}I_{n}$ and $s=\sin{\varepsilon}I_{n}$.
Hence the transformation $A:V_{1}\mapsto e^{-\mathcal{J}}{\rm Gr}(I_{2n})$
satisfies

$\displaystyle A(0,-sp-cq,0,-(-s,c){\gamma}(\tau)(p,q)^{T})$
$\displaystyle=(cp-
sq,sp+cq,(c,s){\gamma}(\tau)(p,q)^{T},(-s,c){\gamma}(\tau)(p,q)^{T}),\quad\forall
p,q\in{\bf R}^{n},$ (2.46)

where $A$ is introduced in Definition 3.4.3 of ${\rm sign}(M_{1},M_{2};L)$ on
p. 67 of [14]. For the convenience of our computation, we rewrite (2.46) as
follows.

$\displaystyle A\left(-\left(\begin{array}[]{cc}0&0\\\
s&c\end{array}\right)\left(\begin{array}[]{c}p\\\
q\end{array}\right),-\left(\begin{array}[]{cc}0&0\\\
-s&c\end{array}\right){\gamma}(\tau)\left(\begin{array}[]{c}p\\\
q\end{array}\right)\right)$ (2.55)
$\displaystyle=\left(\left(\begin{array}[]{cc}c&-s\\\
s&c\end{array}\right)\left(\begin{array}[]{c}p\\\
q\end{array}\right),\left(\begin{array}[]{cc}c&s\\\
-s&c\end{array}\right){\gamma}(\tau)\left(\begin{array}[]{c}p\\\
q\end{array}\right)\right).$ (2.64)

Then for $p_{1},p_{2},q_{1},q_{2}\in{\bf R}^{n}$, the symmetric bilinear form
$Q(V_{2}):(x,y)\mapsto\mathcal{J}(Ax,y)$ on $V_{1}$ defined in Definition
3.4.3 on p. 67 of [14] satisfies:

$\displaystyle Q(V_{2})\left(\left(-\left(\begin{array}[]{cc}0&0\\\
s&c\end{array}\right)\left(\begin{array}[]{c}p\\\
q\end{array}\right),-\left(\begin{array}[]{cc}0&0\\\
-s&c\end{array}\right){\gamma}(\tau)\left(\begin{array}[]{c}p\\\
q\end{array}\right)\right)\right)$ (2.73) $\displaystyle=$
$\displaystyle\left\langle((-J)\oplus J)\left[\left(\begin{array}[]{cc}c&-s\\\
s&c\end{array}\right)\left(\begin{array}[]{c}p\\\
q\end{array}\right),\left(\begin{array}[]{cc}c&s\\\
-s&c\end{array}\right){\gamma}(\tau)\left(\begin{array}[]{c}p\\\
q\end{array}\right)\right],\right.\;$ (2.91)
$\displaystyle\left[-\left(\begin{array}[]{cc}0&0\\\
s&c\end{array}\right)\left.\left(\begin{array}[]{c}p\\\
q\end{array}\right),-\left(\begin{array}[]{cc}0&0\\\
-s&c\end{array}\right){\gamma}(\tau)\left(\begin{array}[]{c}p\\\
q\end{array}\right)\right]\right\rangle.$ $\displaystyle=$
$\displaystyle\left\langle\left[\left(\begin{array}[]{cc}0&s\\\
0&c\end{array}\right)J\left(\begin{array}[]{cc}c&-s\\\
s&c\end{array}\right)-{\gamma}(\tau)^{T}\left(\begin{array}[]{cc}0&-s\\\
0&c\end{array}\right)J\left(\begin{array}[]{cc}c&s\\\
-s&c\end{array}\right){\gamma}(\tau)\right]\left(\begin{array}[]{c}p\\\
q\end{array}\right),\;\left(\begin{array}[]{c}p\\\
q\end{array}\right)\right\rangle.$ (2.104) $\displaystyle=$
$\displaystyle\left\langle\left[\left(\begin{array}[]{cc}sc&-s^{2}\\\
c^{2}&-sc\end{array}\right)+{\gamma}(\tau)^{T}\left(\begin{array}[]{cc}sc&s^{2}\\\
-c^{2}&-sc\end{array}\right){\gamma}(\tau)\right]\left(\begin{array}[]{c}p\\\
q\end{array}\right),\;\left(\begin{array}[]{c}p\\\
q\end{array}\right)\right\rangle.$ (2.113)

Let
$\tilde{M}_{\varepsilon}({\gamma}(\tau))=\left(\begin{array}[]{cc}sc&-s^{2}\\\
c^{2}&-sc\end{array}\right)+{\gamma}(\tau)^{T}\left(\begin{array}[]{cc}sc&s^{2}\\\
-c^{2}&-sc\end{array}\right){\gamma}(\tau)$. Then by definition of the
symmetric bilinear form $Q(V_{2})$, $\tilde{M}_{\varepsilon}({\gamma}(\tau)$
is an invertible symmetric $2n\times 2n$ matrix. We define

$M_{\varepsilon}({\gamma}(\tau))=2\tilde{M}_{\varepsilon}({\gamma}(\tau))=\tilde{M}_{\varepsilon}({\gamma}(\tau))+\tilde{M}_{\varepsilon}^{T}({\gamma}(\tau)).$
(2.114)

Then we have

$\displaystyle
M_{\varepsilon}({\gamma}(\tau))={\gamma}(\tau)^{T}\left(\begin{array}[]{cc}\sin{2{\varepsilon}}I_{n}&-\cos{2{\varepsilon}I_{n}}\\\
-\cos{2{\varepsilon}}I_{n}&-\sin
2{\varepsilon}I_{n}\end{array}\right){\gamma}(\tau)+\left(\begin{array}[]{cc}\sin{2{\varepsilon}}I_{n}&\cos{2{\varepsilon}}I_{n}\\\
\cos{2{\varepsilon}}I_{n}&-\sin 2{\varepsilon}I_{n}\end{array}\right).$
(2.119)

It is clear that

${\rm sgn}Q(V_{2})={\rm sgn}\tilde{M}_{\varepsilon}({\gamma}(\tau))={\rm
sgn}M_{\varepsilon}({\gamma}(\tau)).$ (2.120)

By the definition of ${\rm sgn}(V_{1},e^{-{\varepsilon}\mathcal{J}}{\rm
Gr}({\gamma}(\tau));V_{2})$, we have

${\rm sgn}(V_{1},e^{-{\varepsilon}\mathcal{J}}{\rm
Gr}({\gamma}(\tau));V_{2})={\rm sgn}Q(V_{2}).$ (2.121)

Then (2.44) holds from (2.120) and (2.121), and the proof of Claim 1 is
complete.

Thus by (2.42), (2.43) and Claim 1, we have

$\displaystyle i_{L_{0}}({\gamma})-i_{L_{1}}({\gamma})$ $\displaystyle=$
$\displaystyle s(V_{1},V_{2};e^{-{\varepsilon}\mathcal{J}}{\rm
Gr}({\gamma}(0)),e^{-{\varepsilon}\mathcal{J}}{\rm Gr}({\gamma}(\tau)))$
$\displaystyle=$ $\displaystyle\frac{1}{2}{\rm
sgn}(V_{1},e^{-{\varepsilon}\mathcal{J}}{\rm
Gr}({\gamma}(\tau));V_{2})-\frac{1}{2}{\rm
sgn}(V_{1},e^{-{\varepsilon}\mathcal{J}}{\rm Gr}({\gamma}(0));V_{2})$
$\displaystyle=$ $\displaystyle\frac{1}{2}{\rm
sgn}(V_{1},e^{-{\varepsilon}\mathcal{J}}{\rm
Gr}({\gamma}(\tau));V_{2})-\frac{1}{2}{\rm
sgn}(V_{1},e^{-{\varepsilon}\mathcal{J}}{\rm Gr}(I_{2n});V_{2})$
$\displaystyle=$ $\displaystyle\frac{1}{2}{\rm
sgn}(V_{1},e^{-{\varepsilon}\mathcal{J}}{\rm Gr}({\gamma}(\tau));V_{2})$
$\displaystyle=$ $\displaystyle\frac{1}{2}{\rm
sgn}M_{\varepsilon}({\gamma}(\tau)).$

Here in the second equality, we have used Theorem 3.4.12 of on p. 68 of [14].
Thus (2.35) holds.

Choose ${\varepsilon}<0$ such that $|{\varepsilon}|$ is sufficiently small, by
the discussion of $\mu^{CLM}_{F}$ index we have

$\displaystyle
i_{L_{0}}({\gamma})=\mu^{CLM}_{F}(V_{1},e^{-{\varepsilon}\mathcal{J}}{\rm
Gr}({\gamma}),[0,\tau])-\nu_{L_{0}}({\gamma}),$ (2.122) $\displaystyle
i_{L_{1}}({\gamma})=\mu^{CLM}_{F}(V_{2},e^{-{\varepsilon}\mathcal{J}}{\rm
Gr}({\gamma}),[0,\tau])-\nu_{L_{1}}({\gamma}).$ (2.123)

Then by the same proof as above, we have

$i_{L_{0}}({\gamma})+\nu_{L_{0}}({\gamma})-i_{L_{1}}({\gamma})-\nu_{L_{1}}({\gamma})=\frac{1}{2}{\rm
sgn}M_{\varepsilon}({\gamma}(\tau)),$ (2.124)

where ${\varepsilon}<0$ is small enough. Hence (2.36) holds. The proof of
Theorem 2.3 is complete.

We have the following consequence.

Corollary 2.1. (Theorem 2.3 of [35]) For ${\gamma}\in\mathcal{P}_{\tau}(2n)$
with $\tau>0$, there hold

$\displaystyle|i_{L_{0}}({\gamma}))-i_{L_{1}}({\gamma}))|\leq
n,\quad|i_{L_{0}}({\gamma})+\nu_{L_{0}}({\gamma})-i_{L_{1}}({\gamma})-\nu_{L_{1}}({\gamma})|\leq
n.$ (2.125)

Moreover if ${\gamma}(1)$ is a orthogonal matrix then there holds

$i_{L_{0}}({\gamma})=i_{L_{1}}({\gamma}).$ (2.126)

Proof. (2.125) holds directly from Theorem 2.3, so we only need to prove
(2.126). Since ${\gamma}(\tau)$ is an orthogonal and symplectic matrix, we
have

${\gamma}^{T}(\tau)J{\gamma}(\tau)=J,\quad{\gamma}^{T}(\tau){\gamma}(\tau)=I_{2n}.$
(2.127)

So we have

${\gamma}(\tau)J=J{\gamma}(\tau),\quad{\gamma}(\tau)^{T}J=J{\gamma}(\tau)^{T}.$
(2.128)

It is easy to check that for any ${\varepsilon}\in{\bf R}$, there holds

$J\left(\begin{array}[]{cc}\sin{2{\varepsilon}}I_{n}&\pm\cos{2{\varepsilon}I_{n}}\\\
\pm\cos{2{\varepsilon}}I_{n}&-\sin
2{\varepsilon}I_{n}\end{array}\right)J=\left(\begin{array}[]{cc}\sin{2{\varepsilon}}I_{n}&\pm\cos{2{\varepsilon}I_{n}}\\\
\pm\cos{2{\varepsilon}}I_{n}&-\sin 2{\varepsilon}I_{n}\end{array}\right).$
(2.129)

Hence by (2.128) and (2.129), we have

$\displaystyle JM_{\varepsilon}({\gamma}(\tau))J$ $\displaystyle=$
$\displaystyle
J\left[{\gamma}(\tau)^{T}\left(\begin{array}[]{cc}\sin{2{\varepsilon}}I_{n}&-\cos{2{\varepsilon}I_{n}}\\\
-\cos{2{\varepsilon}}I_{n}&-\sin
2{\varepsilon}I_{n}\end{array}\right){\gamma}(\tau)+\left(\begin{array}[]{cc}\sin{2{\varepsilon}}I_{n}&\cos{2{\varepsilon}}I_{n}\\\
\cos{2{\varepsilon}}I_{n}&-\sin 2{\varepsilon}I_{n}\end{array}\right)\right]J$
(2.134) $\displaystyle=$ $\displaystyle
J{\gamma}(\tau)^{T}\left(\begin{array}[]{cc}\sin{2{\varepsilon}}I_{n}&-\cos{2{\varepsilon}I_{n}}\\\
-\cos{2{\varepsilon}}I_{n}&-\sin
2{\varepsilon}I_{n}\end{array}\right){\gamma}(\tau)J+J\left(\begin{array}[]{cc}\sin{2{\varepsilon}}I_{n}&\cos{2{\varepsilon}}I_{n}\\\
\cos{2{\varepsilon}}I_{n}&-\sin 2{\varepsilon}I_{n}\end{array}\right)J$
(2.139) $\displaystyle=$
$\displaystyle{\gamma}(\tau)^{T}J\left(\begin{array}[]{cc}\sin{2{\varepsilon}}I_{n}&-\cos{2{\varepsilon}I_{n}}\\\
-\cos{2{\varepsilon}}I_{n}&-\sin
2{\varepsilon}I_{n}\end{array}\right)J{\gamma}(\tau)+J\left(\begin{array}[]{cc}\sin{2{\varepsilon}}I_{n}&\cos{2{\varepsilon}}I_{n}\\\
\cos{2{\varepsilon}}I_{n}&-\sin 2{\varepsilon}I_{n}\end{array}\right)J$
(2.144) $\displaystyle=$
$\displaystyle{\gamma}(\tau)^{T}\left(\begin{array}[]{cc}\sin{2{\varepsilon}}I_{n}&-\cos{2{\varepsilon}I_{n}}\\\
-\cos{2{\varepsilon}}I_{n}&-\sin
2{\varepsilon}I_{n}\end{array}\right){\gamma}(\tau)+\left(\begin{array}[]{cc}\sin{2{\varepsilon}}I_{n}&\cos{2{\varepsilon}}I_{n}\\\
\cos{2{\varepsilon}}I_{n}&-\sin 2{\varepsilon}I_{n}\end{array}\right)$ (2.149)
$\displaystyle=$ $\displaystyle M_{\varepsilon}({\gamma}(\tau)).$ (2.150)

So we have

$M_{\varepsilon}({\gamma}(\tau))J=-JM_{\varepsilon}({\gamma}(\tau)).$ (2.151)

Thus for any $x\in{\bf R}^{2n}$ and ${\lambda}\in{\bf R}$ satisfying

$M_{\varepsilon}({\gamma}(\tau))x={\lambda}x.$ (2.152)

By (2.151) we have

$M_{\varepsilon}({\gamma}(\tau))(Jx)=-JM_{\varepsilon}({\gamma}(\tau))x=-{\lambda}(Jx).$
(2.153)

Since for ${\varepsilon}>0$ small enough $M_{\varepsilon}({\gamma}(\tau))$ is
an invertible symmetric matrix, by (2.153) we have

$m^{+}(M_{\varepsilon}({\gamma}(\tau)))=m^{-}(M_{\varepsilon}({\gamma}(\tau)))=n$
(2.154)

which yields

${\rm
sgn}M_{\varepsilon}({\gamma}(\tau))=m^{+}(M_{\varepsilon}({\gamma}(\tau)))-m^{-}(M_{\varepsilon}({\gamma}(\tau)))=0.$
(2.155)

Then (2.126) holds from Theorem 2.3.

Lemma 2.2. For a symplectic path $P:[0,\tau]\to{\rm Sp}(2n)$ with $\tau>0$, if
for $j=0,1$ there holds $\nu_{L_{j}}(P(t))=constant$ for all $t\in[0,\tau]$,
then for ${\varepsilon}>0$ small enough we have

${\rm sgn}M_{\varepsilon}(P(0))={\rm sgn}M_{\varepsilon}(P(\tau)).$ (2.156)

Proof. Since ${\rm Sp}(2n)$ is path connected, we can choose a path
${\gamma}\in\mathcal{P}_{\tau}$ with ${\gamma}(\tau)=P(0)$. By Proposition
2.11 of [42] and the definition of $\mu_{j}$ for $j=1,2$ in [42], we have

$\mu_{F}^{CLM}(V_{j},{\rm Gr}(P),[0,\tau])=0,\qquad j=0,1.$ (2.157)

So by the Path Additivity and Reparametrization Invariance properties of
$\mu_{F}^{CLM}$ in [11], we have

$\displaystyle i_{L_{j}}(P*{\gamma})$ $\displaystyle=$
$\displaystyle\mu_{F}^{CLM}(V_{j},{\rm Gr}(P*{\gamma}),[0,\tau])-n$ (2.158)
$\displaystyle=$ $\displaystyle\mu_{F}^{CLM}(V_{j},{\rm
Gr}({\gamma}),[0,\tau])+\mu_{F}^{CLM}(V_{j},{\rm Gr}(P),[0,\tau])-n$
$\displaystyle=$ $\displaystyle\mu_{F}^{CLM}(V_{j},{\rm
Gr}({\gamma}),[0,\tau])-n$ $\displaystyle=$ $\displaystyle
i_{L_{j}}({\gamma}),$

where the definition of joint path $\eta*\xi$ is given by (6.1) in Section 6
below. Then by Theorem 2.3 we have

$\displaystyle i_{L_{0}}({\gamma})-i_{L_{1}}({\gamma})=\frac{1}{2}{\rm
sgn}(M_{\varepsilon}(P(0))),$ (2.159) $\displaystyle
i_{L_{0}}(P*{\gamma})-i_{L_{1}}(P*{\gamma})=\frac{1}{2}{\rm
sgn}(M_{\varepsilon}(P(\tau))).$ (2.160)

Then (2.156) holds from (2.158)-(2.160). The proof of Lemma 2.2 is complete.

Remark 2.1. It is easy to check that for $n_{j}\times n_{j}$ symplectic matrix
$P_{j}$ with $j=1,2$ and $n_{j}\in{\bf N}$, we have

$\displaystyle M_{\varepsilon}(P_{1}\diamond
P_{2})=M_{\varepsilon}(P_{1})\diamond M_{\varepsilon}(P_{2}),$
$\displaystyle{\rm sgn}M_{\varepsilon}(P_{1}\diamond P_{2})={\rm
sgn}M_{\varepsilon}(P_{1})+{\rm sgn}M_{\varepsilon}(P_{2}).$

By direct computation according to Theorem 2.3 and Corollary 2.1, for
${\gamma}\in\mathcal{P}_{\tau}(2)$, $b>0$, and ${\varepsilon}>0$ small enough
we have

$\displaystyle{\rm sgn}M_{\varepsilon}(R(\theta))=0,\quad{\rm
for}\;\theta\in{\bf R},$ (2.161) $\displaystyle{\rm
sgn}M_{\varepsilon}(P)=0,\quad{\rm if}\;P=\pm\left(\begin{array}[]{cc}1&b\\\
0&1\end{array}\right)\;{\rm or}\;\pm\left(\begin{array}[]{cc}1&0\\\
-b&1\end{array}\right),$ (2.166) $\displaystyle{\rm
sgn}M_{\varepsilon}(P)=2,\quad{\rm if}\;P=\pm\left(\begin{array}[]{cc}1&-b\\\
0&1\end{array}\right),$ (2.169) $\displaystyle{\rm
sgn}M_{\varepsilon}(P)=-2,\quad{\rm if}\;P=\pm\left(\begin{array}[]{cc}1&0\\\
b&1\end{array}\right).$ (2.172)

Also we give a example as follows to finish this section

${\rm sgn}M_{\varepsilon}(P)=2,\quad{\rm
if}\;P=\pm\left(\begin{array}[]{cc}2&-1\\\ -1&1\end{array}\right).$ (2.173)

##  3 Relation between $i_{L_{0}}$, $i_{L_{1}}$, $i_{\sqrt{-1}}^{L_{0}}$ and
the corresponding Morse indices, and their monotonicity properties.

In [31], Liu studied the relation between the $L$-index of solutions of
Hamiltonian systems with $L$-boundary conditions and the Morse index of the
corresponding functional defined via the Galerkin approximation method on the
finite dimensional truncated space at its corresponding critical points. In
order to prove the main results of this paper, in this section we use the
results of [31] to study some monotonicity properties of $i_{L_{0}}$ and
$i_{L_{1}}$. We also study the index $i_{\sqrt{-1}}^{L_{0}}(B)$ with $B$ being
a continuous symmetric matrices path satisfying condition (B1) defined in
Section 1 and the Morse index of the corresponding functional defined via the
Galerkin approximation method. Then as applications we study some monotonicity
properties of $i_{\sqrt{-1}}^{L_{0}}(B)$ which will be important in the proof
of Theorems 1.4-1.5 in Section 5 below.

For any $\tau>0$ and $B\in C([0,\tau/4],\mathcal{L}_{s}({\bf R}^{2n}))$ (in
order to apply the results in this section conveniently Section 5, we always
assume $B\in C([0,\tau/4],\mathcal{L}_{s}({\bf R}^{2n})$) satisfying condition
(B1). We extend $B$ to $[0,\frac{\tau}{2}]$ by

$B(\frac{\tau}{4}+t)=NB(\frac{\tau}{4}-t)N,\;\forall t\in[0,\frac{\tau}{4}].$
(3.1)

Then since $B(\frac{\tau}{2})=B(0)$, we can extend it
$\frac{\tau}{2}$-periodically to ${\bf R}$, so we can see $B$ as an element in
$C(S_{\tau/2},\mathcal{L}_{s}({\bf R}^{2n}))$.

Let $E_{\tau}=\\{x\in W^{1/2,2}(S_{\tau},{\bf
R}^{2n})|\,x(-t)=Nx(t)\;a.e.\;t\in{\bf R}\\}$ with the usual norm and inner
product denoted by $||\cdot||$ and $\langle\cdot\rangle$ respectively.

By the Sobolev embedding theorem, for any $s\in[1,+\infty)$, there is a
constant $C_{s}>0$ such that

$||z||_{L^{s}}\leq C_{s}||z||,\quad\forall z\in E_{2\tau}.$ (3.2)

Note that $B$ can also be seen as an element in
$C(S_{\tau},\mathcal{L}_{s}({\bf R}^{2n}))$. We define two selfadjoint
operators $A_{\tau}$ and $B_{\tau}$ on $E_{\tau}$ by the following bilinear
forms

$\displaystyle\langle A_{\tau}x,y\rangle=\int_{0}^{\tau}-J\dot{x}\cdot
y\,dt,\qquad\langle B_{\tau}x,y\rangle=\int_{0}^{\tau}B(t)x\cdot y\,dt.$ (3.3)

Then $A_{\tau}$ is a bounded operator on $E_{\tau}$ and dim $\ker A_{\tau}=n$,
the Fredholm index of $A_{\tau}$ is zero, and $B_{\tau}$ is a compact operator
on $E_{\tau}$.

Set

$E_{\tau}(j)=\left\\{z\in E_{\tau}\left|z(t)={\rm exp}(\frac{2j\pi
t}{\tau}J)a+{\rm exp}(-\frac{2j\pi t}{\tau}J)b,\;\forall t\in{\bf R};\;\forall
a,\,b\in L_{0}\right.\right\\}.$

and

$E_{\tau,m}=E_{\tau}(0)+E_{\tau}(1)+\cdots+E_{\tau}(m).$

Let ${\Gamma}_{\tau}=\\{P_{\tau,m}:m=0,1,2,...\\}$ be the usual Galerkin
approximation scheme w.r.t. $A_{\tau}$, just as in [31], i.e.,
${\Gamma}_{\tau}$ is a sequence of orthogonal projections satisfies:

(1) $E_{\tau,0}=P_{\tau,0}E_{\tau}=\ker
A_{\tau},\;E_{\tau,m}=P_{\tau,m}E_{\tau}$ is finite dimension for $m\geq 0$;

(2) $P_{\tau,m}\to x$ as $m\to\infty$ for any $x\in E_{\tau}$;

(3) $P_{\tau,m}A_{\tau}=A_{\tau}P_{\tau,m}$, $\forall m\geq 0$.

For $d>0$, we denote by $M^{+}_{d}(\cdot)$, $M^{-}_{d}(\cdot)$ and
$M^{0}_{d}(\cdot)$ the eigenspace corresponding to the eigenvalue ${\lambda}$
belong to $[d,+\infty)$, $(-\infty,-d]$ and $(-d,d)$ respectively, and
$M^{+}(\cdot)$, $M^{-}(\cdot)$ and $M^{0}(\cdot)$ the positive, negative and
null subspace of of the selfadjoint operator defining it respectively. For any
bounded selfadjoint linear operator on $E$, We denote
$L^{\\#}=(L|_{ImL})^{-1}$, and we also denote by
$P_{\tau,m}LP_{\tau,m}=(P_{\tau,m}LP_{\tau,m})|_{E_{\tau,m}}:E_{\tau,m}\to
E_{\tau,m}$.

Similarly we define two subspaces of $E_{\tau}$ by $\hat{E}=\\{x\in
E|x(t+\frac{\tau}{2})=-x(t),a.e.\,t\in{\bf R}\\}$ and $\tilde{E}=\\{x\in
E|x(t+\frac{\tau}{2})=x(t),a.e.\,t\in{\bf R}\\}$ be the symmetric ones and
$\frac{\tau}{2}$-periodic ones of $E_{\tau}$ respectively.

We define two selfadjoint operators $\hat{A}$ and $\hat{B}$ on $\hat{E}$ by
the following bilinear forms

$\displaystyle\langle\hat{A}x,y\rangle=\int_{0}^{\tau}-J\dot{x}\cdot
y\,dt,\qquad\langle\hat{B}x,y\rangle=\int_{0}^{\tau}B(t)x(t)\cdot y(t)\,dt.$
(3.4)

Then $\hat{A}$ is a bounded Fredholm operator on $\hat{E}$ and dim
$\ker\hat{A}=0$, the Fredholm index of $\hat{A}$ is zero. $\hat{B}$ is a
compact operator on $\hat{E}$.

For any positive integer $m$, we define

$\hat{E}_{m}={\Sigma}_{j=1}^{m}E_{\tau}(2j-1).$

For $m\geq 1$, let $\hat{P}_{m}$ be the orthogonal projection from $\hat{E}$
to $\hat{E}_{m}$. Then $\\{\hat{P}_{m}\\}$ is a Galerkin approximation scheme
w.r.t. $\hat{A}$.

Theorem 3.1. For any $B(t)\in C([0,\frac{\tau}{4}],\mathcal{L}_{s}({\bf
R}^{2n}))$ satisfying condition (B1) and
$0<d\leq\frac{1}{4}||(A_{\tau}-B_{\tau})^{\\#}||^{-1}$, there exists $m^{*}>0$
such that for $m\geq m^{*}$ there hold

$\displaystyle\dim M_{d}^{+}(\hat{P}_{m}(\hat{A}-\hat{B})\hat{P}_{m})$
$\displaystyle=$ $\displaystyle mn-
i_{\sqrt{-1}}^{L_{0}}(B)-\nu_{\sqrt{-1}}^{L_{0}}(B),$ (3.5) $\displaystyle\dim
M_{d}^{-}(\hat{P}_{m}(\hat{A}-\hat{B})\hat{P}_{m})$ $\displaystyle=$
$\displaystyle mn+i_{\sqrt{-1}}^{L_{0}}(B),$ (3.6) $\displaystyle\dim
M_{d}^{0}(\hat{P}_{m}(\hat{A}-\hat{B})\hat{P}_{m})$ $\displaystyle=$
$\displaystyle\nu_{\sqrt{-1}}^{L_{0}}(B).$ (3.7)

Proof. The method of the proof here is similar as that of Theorem 2.1 in [51].

For any positive integer $m$, we define

$\tilde{E}_{m}=\sum_{j=0}^{m}E_{\tau}(2j).$

For $m\geq 1$, let $\tilde{P}_{m}$ be the orthogonal projection from
$\tilde{E}$ to $\tilde{E}_{m}$. Then $\\{\tilde{P}_{m}\\}$ is a Galerkin
approximation scheme w.r.t. $\tilde{A}$.

For any $y\in\hat{E}_{m}$ and $z\in\tilde{E}_{m}$, it is easy to check that

$\displaystyle\langle(P_{\tau,m}(A_{\tau}-B_{\tau})P_{\tau,m}y,z)\rangle=0.$
(3.8)

So we have the following $P_{\tau,m}(A_{\tau}-B_{\tau})P_{\tau,m}$ orthogonal
decomposition

$E_{\tau,2m}=\hat{E}_{m}\oplus\tilde{E}_{m}.$ (3.9)

Similarly, we have the following $A_{\tau}-B_{\tau}$ orthogonal decomposition

$E_{\tau}=\hat{E}\oplus\tilde{E}.$ (3.10)

Hence, under above decomposition we have

$(A_{\tau}-B_{\tau})=(\hat{A}-\hat{B})\oplus(\tilde{A}-\tilde{B}).$ (3.11)

Thus

$\displaystyle||(A_{\tau}-B_{\tau})^{\\#}||^{-1}\leq||(\hat{A}-\hat{B})^{\\#}||^{-1}$
(3.12)
$\displaystyle||(A_{\tau}-B_{\tau})^{\\#}||^{-1}\leq||(\tilde{A}-\tilde{B})^{\\#}||^{-1}$
(3.13)

By the definitions of $M_{d}^{*}(\cdot)$ for
$P_{\tau,2m}(A_{\tau}-B_{\tau})P_{\tau,2m}$,
$\hat{P}_{m}(\hat{A}-\hat{B})\hat{P}_{m}$, and
$\tilde{P}_{m}(\tilde{A}-\tilde{B})\tilde{P}_{m}$ with $*=+,-,0$. So for
$*\in\\{+,-,0\\}$ we have

$\dim M_{d}^{*}(P_{\tau,2m}(A_{\tau}-B_{\tau})P_{\tau,2m})=\dim
M_{d}^{*}(\hat{P}_{m}(\hat{A}-\hat{B})\hat{P}_{m})+\dim
M_{d}^{*}(\tilde{P}_{m}(\tilde{A}-\tilde{B})\tilde{P}_{m}).$ (3.14)

Note that, the space $E_{\tau}$ and the operators $A_{\tau}$, $B_{\tau}$ and
$P_{\tau,m}$ are also defined in the same way. So by the definition we see
that $\tilde{E}$ is the $\tau$-periodic extending of $E_{\tau}$ from
$S_{\tau}$ to $S_{2\tau}$, and $\tilde{E}_{m}$ is the $\tau$-periodic
extending of $E_{\tau,2m}$ from $S_{\tau}$ to $S_{2\tau}$ too.

Thus we have

$||(A_{\tau}-B_{\tau})^{\\#}||^{-1}=||(\tilde{A}-\tilde{B})^{\\#}||^{-1}.$
(3.15)

By (3.13) and (3.15) we have

$||(A_{2\tau}-B_{2\tau})^{\\#}||^{-1}\leq||(A_{\tau}-B_{\tau})^{\\#}||^{-1}.$
(3.16)

For $*\in\\{+,-,0\\}$ we have

$\dim
M_{d}^{*}(P_{\tau,m}(A_{\tau}-B_{\tau})P_{\tau,m})=M_{d}^{*}(\tilde{P}_{m}(\tilde{A}-\tilde{B})\tilde{P}_{m}).$
(3.17)

Then for $0<d\leq\frac{1}{4}||(A_{\tau}-B_{\tau})^{\\#}||^{-1}$, by Theorem
2.1 in [31] there exists $m_{1}>0$ such that for $m\geq m_{1}$ we have

$\displaystyle\dim M_{d}^{+}(P_{\tau,2m}(A_{\tau}-B_{\tau})P_{\tau,2m})=2mn-
i_{L_{0}}({\gamma}_{B}^{2})-\nu_{L_{0}}({\gamma}_{B}^{2}),$ (3.18)
$\displaystyle\dim
M_{d}^{-}(P_{\tau,2m}(A_{\tau}-B_{\tau})P_{\tau,2m})=2mn+n+i_{L_{0}}({\gamma}_{B}^{2}),$
(3.19) $\displaystyle\dim
M_{d}^{0}(P_{\tau,2m}(A_{\tau}-B_{\tau})P_{\tau,2m})=\nu_{L_{0}}({\gamma}_{B}^{2}).$
(3.20)

By (3.16), we have $0<d\leq\frac{1}{4}||(A_{\tau}-B_{\tau})^{\\#}||^{-1}$. By
Theorem 2.1 in [31] again there exists $m_{2}>0$, such that for $m\geq m_{2}$
we have

$\displaystyle\dim M_{d}^{+}(P_{\tau,m}(A_{\tau}-B_{\tau})P_{\tau,m})=mn-
i_{L_{0}}({\gamma}_{B})-\nu_{L_{0}}({\gamma}_{B})),$ (3.21) $\displaystyle\dim
M_{d}^{-}(P_{\tau,m}(A_{\tau}-B_{\tau})P_{\tau,m})=mn+n+i_{L_{0}}({\gamma}_{B})),$
(3.22) $\displaystyle\dim
M_{d}^{0}(P_{\tau,m}(A_{\tau}-B_{\tau})P_{\tau,m})=\nu_{L_{0}}({\gamma}_{B})).$
(3.23)

Let $m^{*}=\max\\{m_{1},m_{2}\\}$. Then for $m\geq m^{*}$, all of
(3.18)-(3.23) hold.

So by (3.14), (3.17), and (3.18)-(3.23) we have

$\displaystyle\dim M_{d}^{+}(\hat{P}_{m}(\hat{A}-\hat{B})\hat{P}_{m})$
$\displaystyle=$ $\displaystyle
mn-(i_{L_{0}}({\gamma}_{B}^{2})-i_{L_{0}}({\gamma}_{B}))-(\nu_{L_{0}}({\gamma}_{B}^{2})-\nu_{L_{0}}({\gamma}_{B})),$
(3.24) $\displaystyle\dim M_{d}^{-}(\hat{P}_{m}(\hat{A}-\hat{B})\hat{P}_{m})$
$\displaystyle=$ $\displaystyle
mn+i_{L_{0}}({\gamma}_{B}^{2})-i_{L_{0}}({\gamma}_{B}),$ (3.25)
$\displaystyle\dim M_{d}^{0}(\hat{P}_{m}(\hat{A}-\hat{B})\hat{P}_{m})$
$\displaystyle=$
$\displaystyle\nu_{L_{0}}({\gamma}_{B}^{2})-\nu_{L_{0}}({\gamma}_{B}).$ (3.26)

Thus (3.5)-(3.7) hold from (3.24)-(3.26), Definition 2.3, and Theorem 2.2. The
proof of Theorem 3.1 is complete.

Remark 3.1. Let any $B\in C([0,\frac{\tau}{4}],\mathcal{L}_{s}({\bf R}^{2n}))$
be a constant matrix path satisfying condition (B1). By Theorem 5.1 of [42],
for $d=0$ the same conclusions of Theorem 2.1 of [31] still holds . Hence for
$d=0$ the same conclusions of Theorem 3.1 still hold, i.e., there exists
$m^{*}>0$ such that for $m\geq m^{*}$ there hold

$\displaystyle\dim M^{+}(\hat{P}_{m}(\hat{A}-\hat{B})\hat{P}_{m})$
$\displaystyle=$ $\displaystyle mn-
i_{\sqrt{-1}}^{L_{0}}(B)-\nu_{\sqrt{-1}}^{L_{0}}(B),$ $\displaystyle\dim
M^{-}(\hat{P}_{m}(\hat{A}-\hat{B})\hat{P}_{m})$ $\displaystyle=$
$\displaystyle mn+i_{\sqrt{-1}}^{L_{0}}(B),$ $\displaystyle\dim
M^{0}(\hat{P}_{m}(\hat{A}-\hat{B})\hat{P}_{m})$ $\displaystyle=$
$\displaystyle\nu_{\sqrt{-1}}^{L_{0}}(B).$

In the following, we study some monotonicity of the the Maslov-type
$i_{\sqrt{-1}}^{L_{0}}$ index. In this paper, for any two symmetric matrices
$B_{1}$ and $B_{2}$, we say $B_{1}>B_{2}$ if $B_{1}-B_{2}$ is positive
definite and we say $B_{1}\geq B_{2}$ if $B_{1}-B_{2}$ is semipositive.
Similarly for two symmetric matrix paths $B_{1}$, $B_{2}\in
C([0,\tau],\mathcal{L}_{s}(R^{2n}))$, we say $B_{1}>B_{2}$ if
$B_{1}(t)-B_{2}(t)$ is positive definite for all $t\in[0,\tau]$ and we say
$B_{1}\geq B_{2}$ if $B_{1}(t)-B_{2}(t)$ is semipositive definite for all
$t\in[0,\tau]$.

Lemma 3.1. For any $\tau>0$ and $B_{1},\;B_{2}\in
C([0,\frac{\tau}{4}],\mathcal{L}_{s}({\bf R}^{2n}))$ satisfying condition
(B1). If $B_{1}\geq B_{2}$, then there hold

$i_{\sqrt{-1}}^{L_{0}}(B_{1})\geq i_{\sqrt{-1}}^{L_{0}}(B_{2})$ (3.27)

and

$\displaystyle i_{\sqrt{-1}}^{L_{0}}(B_{1})+\nu_{\sqrt{-1}}^{L_{0}}(B_{1})\geq
i_{\sqrt{-1}}^{L_{0}}(B_{2})+\nu_{\sqrt{-1}}^{L_{0}}(B_{2}).$ (3.28)

Moreover, if

$\int_{0}^{\frac{\tau}{4}}(B_{1}(t)-B_{2}(t))dt>0,$ (3.29)

then there holds

$\displaystyle i_{\sqrt{-1}}^{L_{0}}(B_{1})\geq
i_{\sqrt{-1}}^{L_{0}}(B_{2})+\nu_{\sqrt{-1}}^{L_{0}}(B_{2}).$ (3.30)

Proof. Let the space $\hat{E}$ and the orthogonal projection operator
$\hat{P}_{m}$ be the ones defined in Section 2. Correspondingly we define the
compact operators $\hat{B}_{1}$ and $\hat{B}_{2}$. By Theorem 3.1, for $d>0$
small enough, there exists $m^{*}>0$ such that

$\displaystyle\dim M_{d}^{+}(\hat{P}_{m}(\hat{A}-\hat{B}_{1})\hat{P}_{m})$
$\displaystyle=$ $\displaystyle mn-
i_{\sqrt{-1}}^{L_{0}}(B_{1})-\nu_{\sqrt{-1}}^{L_{0}}(B_{1}),$ (3.31)
$\displaystyle\dim M_{d}^{-}(\hat{P}_{m}(\hat{A}-\hat{B}_{1})\hat{P}_{m})$
$\displaystyle=$ $\displaystyle mn+i_{\sqrt{-1}}^{L_{0}}(B_{1}),$ (3.32)
$\displaystyle\dim M_{d}^{0}(\hat{P}_{m}(\hat{A}-\hat{B}_{1})\hat{P}_{m})$
$\displaystyle=$ $\displaystyle\nu_{\sqrt{-1}}^{L_{0}}(B_{1}).$ (3.33)

and

$\displaystyle\dim M_{d}^{+}(\hat{P}_{m}(\hat{A}-\hat{B_{2}})\hat{P}_{m})$
$\displaystyle=$ $\displaystyle mn-
i_{\sqrt{-1}}^{L_{0}}(B_{2})-\nu_{\sqrt{-1}}^{L_{0}}(B_{2}),$ (3.34)
$\displaystyle\dim M_{d}^{-}(\hat{P}_{m}(\hat{A}-\hat{B}_{2})\hat{P}_{m})$
$\displaystyle=$ $\displaystyle mn+i_{\sqrt{-1}}^{L_{0}}(B_{2}),$ (3.35)
$\displaystyle\dim M_{d}^{0}(\hat{P}_{m}(\hat{A}-\hat{B}_{2})\hat{P}_{m})$
$\displaystyle=$ $\displaystyle\nu_{\sqrt{-1}}^{L_{0}}(B_{2}).$ (3.36)

If $B_{1}\geq B_{2}$, we have
$\hat{P}_{m}(\hat{A}-\hat{B}_{1})\hat{P}_{m}\leq\hat{P}_{m}(\hat{A}-\hat{B}_{2})\hat{P}_{m}$,
So

$\dim M_{d}^{-}(\hat{P}_{m}(\hat{A}-\hat{B}_{1})\hat{P}_{m})\geq\dim
M_{d}^{-}(\hat{P}_{m}(\hat{A}-\hat{B}_{2})\hat{P}_{m}).$ (3.37)

Then by (3.32) and (3.35), (3.27) holds. Also we have

$\dim M_{d}^{+}(\hat{P}_{m}(\hat{A}-\hat{B}_{1})\hat{P}_{m})\leq\dim
M_{d}^{+}(\hat{P}_{m}(\hat{A}-\hat{B}_{2})\hat{P}_{m}).$ (3.38)

Then by (3.31) and (3.34), (3.28) holds.

If $\int_{0}^{\frac{\tau}{4}}(B_{1}(t)-B_{2}(t))dt>0$, then

$\hat{P}_{m}(\hat{A}-\hat{B}_{1})\hat{P}_{m}<\hat{P}_{m}(\hat{A}-\hat{B}_{2})\hat{P}_{m}.$
(3.39)

So we have

$\dim M_{d}^{-}(\hat{P}_{m}(\hat{A}-\hat{B}_{1})\hat{P}_{m})\geq\dim
M_{d}^{-}(\hat{P}_{m}(\hat{A}-\hat{B}_{2})\hat{P}_{m})+M_{d}^{0}(\hat{P}_{m}(\hat{A}-\hat{B}_{2})\hat{P}_{m}).$
(3.40)

Then by (3.32), (3.35) and (3.36), (3.30) holds and the proof of Lemma 3.1 is
complete.

Corollary 3.1. For any $\tau>0$ and $B\in
C([0,\frac{\tau}{4}],\mathcal{L}_{s}({\bf R}^{2n}))$ satisfying condition (B1)
and $B\geq 0$, there holds

$i_{\sqrt{-1}}^{L_{0}}(B)\geq 0.$ (3.41)

proof. By Lemma 3.1, we have

$i_{\sqrt{-1}}^{L_{0}}(B)\geq i_{\sqrt{-1}}^{L_{0}}(0).$ (3.42)

Then the conclusion holds from the fact that

$i_{\sqrt{-1}}^{L_{0}}(0)=i_{\sqrt{-1}}^{L_{0}}({\gamma}_{0})=0,$ (3.43)

Where ${\gamma}_{0}$ is the identity symplectic path.

By Theorem 2.1 of [31] and the Remark below Theorem 2.1 in [31] and the
similar proof of Lemma 3.1 we have the following lemma.

Lemma 3.2. If $\tau>0$ and $B_{1},\;B_{2}\in
C([0,\frac{\tau}{4}],\mathcal{L}_{s}({\bf R}^{2n}))$ satisfying condition (B1)
and $B_{1}\geq B_{2}$, then for $j=0,1$ there hold

$i_{L_{j}}(B_{1})\geq i_{L_{j}}(B_{2})$ (3.44)

and

$\displaystyle i_{L_{j}}(B_{1})+\nu_{L_{j}}(B_{1})\geq
i_{L_{j}}(B_{2})+\nu_{L_{j}}(B_{2}).$ (3.45)

Moreover, if $\int_{0}^{\frac{\tau}{4}}(B_{1}(t)-B_{2}(t))dt>0$, then there
holds

$\displaystyle i_{L_{j}}(B_{1})\geq i_{L_{j}}(B_{2})+\nu_{L_{j}}(B_{2}).$
(3.46)

Since $i_{L_{j}}(0)=-n$ and $\nu_{L_{j}}(0)=n$ for $j=0,1$, a direct
consequence of Lemma 3.2 is the following

Corollary 3.2. If $\tau>0$ and $B\in C([0,\frac{\tau}{2}],\mathcal{L}_{s}({\bf
R}^{2n}))$ satisfying condition (B1) and $B\geq 0$, then for $j=0,1$ there
hold

$i_{L_{j}}(B)+\nu_{L_{j}}(B)\geq 0,\qquad i_{L_{j}}(B)\geq-n.$ (3.47)

Moreover if $\int_{0}^{\frac{\tau}{2}}B(t)dt>0$, there holds

$i_{L_{j}}(B)\geq 0.$ (3.48)

Moreover we can give a stronger version of Corollary 3.2, i.e., the following
Lemma 3.3.

Lemma 3.3. Let $\tau>0$ and $B\in C([0,\frac{\tau}{2}],\mathcal{L}_{s}({\bf
R}^{2n}))$ with the $n\times n$ matrix square block form
$B(t)=\left(\begin{array}[]{cc}B_{11}(t)&B_{12}(t)\\\
B_{21}(t)&B_{22}(t)\end{array}\right)$ satisfying condition (B1) and $B\geq
0$.

If $\int_{0}^{\frac{\tau}{2}}B_{22}(t)dt>0$, there holds

$i_{L_{0}}(B)\geq 0.$ (3.49)

If $\int_{0}^{\frac{\tau}{2}}B_{11}(t)dt>0$, there holds

$i_{L_{1}}(B)\geq 0.$ (3.50)

Proof. Without loss of generality, assume ${\lambda}>0$ such that

$\int_{0}^{\frac{\tau}{2}}B_{22}(t)\geq{\lambda}I_{n}.$ (3.51)

Also we can extend $B$ to $[0,\tau]$ by

$B(\frac{\tau}{2}+t)=NB(\frac{\tau}{2}-t)N,\;\forall t\in[0,\frac{\tau}{2}].$
(3.52)

Then since $B(\tau)=B(0)$, we can extend it $\tau$-periodically to ${\bf R}$,
so we can see $B$ as an element in $C(S_{\tau},\mathcal{L}_{s}({\bf
R}^{2n}))$. Then we have

$\int_{0}^{\tau}B_{22}(t)\geq 2{\lambda}I_{n}.$ (3.53)

For any $m\in{\bf N}$, we define two subspaces of $E$ as follows

$E^{-}_{\tau,m}=\left\\{z\in E_{\tau}\left|z(t)=\sum_{j=1}^{m}{\rm
exp}(-\frac{2j\pi t}{\tau}J)b_{j},\;\forall t\in{\bf R};\;\forall b_{j}\in
L_{0}\right.\right\\},$ $E_{\tau}(0)=\left\\{z\in E_{\tau}\left|z(t)\equiv
b,\;b\in L_{0}\right.\right\\}.$

Then for any $z=\alpha x+\beta y\in E_{\tau}(0)\oplus E^{-}_{\tau,m}$ with
$\alpha^{2}+\beta^{2}=1$ and $||x||=||y||=1$, we have

$\displaystyle\langle(A_{\tau}-B_{\tau})z,z\rangle$ $\displaystyle=$
$\displaystyle\langle(A_{\tau}-B_{\tau})(\alpha x+\beta y),\alpha x+\beta
y\rangle$ (3.54) $\displaystyle=$ $\displaystyle-\beta^{2}\langle
A_{\tau}y,y\rangle-\langle B_{\tau}(\alpha x+\beta y),\alpha x+\beta y\rangle$
$\displaystyle\leq$ $\displaystyle-||A_{\tau}^{\\#}||^{-1}\beta^{2}-\langle
B_{\tau}(\alpha x+\beta y),\alpha x+\beta y\rangle.$

Since $B\geq 0$, note that $x(t)\equiv b=(0,b_{1})\in L_{0}$ for all $t\in
S_{\tau}$ with $\tau|b_{1}|^{2}=1$, we have

$\displaystyle\langle B_{\tau}(\alpha x+\beta y),\alpha x+\beta y\rangle$
(3.55) $\displaystyle=$ $\displaystyle\int_{0}^{\tau}(\alpha^{2}Bx\cdot
x+\beta^{2}By\cdot y+2\alpha\beta Bx\cdot y)\,dt$ $\displaystyle\geq$
$\displaystyle\alpha^{2}\int_{0}^{\tau}Bx\cdot
x\,dt+\beta^{2}\int_{0}^{\tau}By\cdot
y\,dt-2|\alpha||\beta|(\int_{0}^{\tau}Bx\cdot
x\,dt)^{1/2}(\int_{0}^{\tau}By\cdot y\,dt)^{1/2}$ $\displaystyle\geq$
$\displaystyle\alpha^{2}\int_{0}^{\tau}Bx\cdot
x\,dt+\beta^{2}\int_{0}^{\tau}By\cdot
y\,dt-\frac{1}{1+{\varepsilon}}\alpha^{2}\int_{0}^{\tau}Bx\cdot
x\,dt-(1+{\varepsilon})\beta^{2}\int_{0}^{\tau}By\cdot y\,dt$ $\displaystyle=$
$\displaystyle\frac{{\varepsilon}\alpha^{2}}{1+{\varepsilon}}\int_{0}^{\tau}Bx\cdot
x\,dt-{\varepsilon}\beta^{2}\int_{0}^{\tau}By\cdot y\,dt$ $\displaystyle=$
$\displaystyle\frac{{\varepsilon}\alpha^{2}}{1+{\varepsilon}}\left(\int_{0}^{\tau}B(t)dt\right)b\cdot
b-{\varepsilon}\beta^{2}\int_{0}^{\tau}By\cdot y\,dt$ $\displaystyle=$
$\displaystyle\frac{{\varepsilon}\alpha^{2}}{1+{\varepsilon}}\left(\int_{0}^{\tau}B_{22}(t)dt\right)b_{1}\cdot
b_{1}-{\varepsilon}\beta^{2}\int_{0}^{\tau}By\cdot y\,dt$ $\displaystyle\geq$
$\displaystyle\frac{{\varepsilon}\alpha^{2}}{1+{\varepsilon}}2{\lambda}|b_{1}|^{2}-{\varepsilon}\beta^{2}||B_{\tau}||\,||y||^{2}$
$\displaystyle=$
$\displaystyle\frac{2{\varepsilon}{\lambda}\alpha^{2}}{(1+{\varepsilon})\tau}-{\varepsilon}\beta^{2}||B_{\tau}||$

for any ${\varepsilon}>0$.

Let
${\varepsilon}=\min\\{1,\frac{||A_{\tau}^{\\#}||^{-1}||B_{\tau}||^{-1}}{2}\\}$.
By (3.54) and (3.55), we have

$\displaystyle\langle(A_{\tau}-B_{\tau})z,z\rangle$ $\displaystyle\leq$
$\displaystyle-||A_{\tau}^{\\#}||^{-1}\beta^{2}-\frac{2{\varepsilon}{\lambda}\alpha^{2}}{(1+{\varepsilon})\tau}+{\varepsilon}\beta^{2}||B_{\tau}||$
(3.56) $\displaystyle\leq$
$\displaystyle-\frac{||A_{\tau}^{\\#}||^{-1}\beta^{2}}{2}-\frac{{\varepsilon}{\lambda}\alpha^{2}}{\tau}$
$\displaystyle\leq$ $\displaystyle-d_{0}(\alpha^{2}+\beta^{2})$
$\displaystyle=$ $\displaystyle-d_{0},$

where
$d_{0}=\min\\{\frac{||A_{\tau}^{\\#}||^{-1}}{2},\,\frac{{\varepsilon}{\lambda}}{\tau}\\}=\min\\{\frac{||A_{\tau}^{\\#}||^{-1}}{2},\,\frac{{\lambda}}{\tau},\,\frac{{\lambda}||A_{\tau}^{\\#}||^{-1}||B_{\tau}||^{-1}}{2\tau}\\}$.
Note that $d_{0}$ is independent of $m$, so for
$0<d\leq\min\\{d_{0},\frac{||(A_{\tau}-B_{\tau})^{\\#}||^{-1}}{4}\\}$, by
Theorem 2.1 of [31] there exists $m^{*}>0$ such that, for $m\geq m^{*}$, we
have

$\dim M^{-}_{d}(P_{\tau,m}(A_{\tau}-B_{\tau})P_{\tau,m})=mn+n+i_{L_{0}}(B).$
(3.57)

By (3.56) we have

$\displaystyle\dim
M^{-}_{d}(P_{\tau,m}(A_{\tau}-B_{\tau})P_{\tau,m})\geq\dim(E_{\tau}(0)\oplus
E^{-}_{\tau,m})=mn+n.$ (3.58)

Then by (3.57) and (3.58) we have $i_{L_{0}}(B)\geq 0$.

For $\int_{0}^{\frac{\tau}{2}}B_{11}(t)dt>0$, by similar proof we have
$i_{L_{1}}(B)\geq 0$. The proof of Lemma 3.3 is complete.

Now we give the following Theorem 3.2 which will play a important role in the
proof of our main results in Section 5. This results implies that the
corresponding Maslov-type index of a periodic symmetric solution of a first
order even semipositive Hamilton increases with the increasing of the
iteration time of the solution.

Theorem 3.2. If $\tau>0$ and $B\in C([0,\frac{\tau}{4}],\mathcal{L}_{s}({\bf
R}^{2n}))$ satisfying condition (B1) and $B\geq 0$, then for any two positive
integers $p>q$ there holds

$i_{\sqrt{-1}}^{L_{0}}({\gamma}_{B}^{p})\geq
i_{\sqrt{-1}}^{L_{0}}({\gamma}_{B}^{q}).$ (3.59)

Proof. Extend ${\gamma}_{B}(t)$ to $[0,\frac{p\tau}{4}]$ as
${\gamma}_{B}^{p}$, we still denote it by ${\gamma}_{B}$. By definition of
$i_{\sqrt{-1}}^{L_{o}}$ and the Path additivity and Symplectic invariance
property of $\mu_{F}^{CLM}$ in [11], we have

$\displaystyle
i_{\sqrt{-1}}^{L_{0}}({\gamma}_{B}^{p})-i_{\sqrt{-1}}^{L_{0}}({\gamma}_{B}^{q})$
(3.60) $\displaystyle=$ $\displaystyle\mu_{F}^{CLM}(L_{0}\times JL_{0},{\rm
Gr}({\gamma}_{B}),[0,\frac{p\tau}{4}])-\mu_{F}^{CLM}(L_{0}\times JL_{0},{\rm
Gr}({\gamma}_{B}),[0,\frac{q\tau}{4}])$ $\displaystyle=$
$\displaystyle\mu_{F}^{CLM}(L_{0}\times JL_{0},{\rm
Gr}({\gamma}_{B}),[\frac{q\tau}{4},\frac{p\tau}{4}])$ $\displaystyle=$
$\displaystyle\mu_{F}^{CLM}(L_{0}\times L_{0},{\rm
Gr}(-J{\gamma}_{B}),[\frac{q\tau}{4},\frac{p\tau}{4}]).$

By the first geometrical definition of the index $\mu_{F}^{CLM}$ in section 4
of [11], there is a ${\varepsilon}>0$ small enough such that

$\displaystyle(e^{-{\varepsilon}\mathcal{J}}{\rm
Gr}(-J{\gamma}_{B}(\frac{p\tau}{4}))\cap(L_{0}\times
L_{0})=\\{0\\}=(e^{-{\varepsilon}J}{\rm
Gr}({\gamma}_{B}(\frac{q\tau}{4}))\cap(L_{0}\times L_{0})$ (3.61)

and

$\displaystyle\mu_{F}^{CLM}(L_{0}\times L_{0},{\rm
Gr}(-J{\gamma}_{B}),[\frac{q\tau}{4},\frac{p\tau}{4}])$ (3.62)
$\displaystyle=$ $\displaystyle\mu_{F}^{CLM}(L_{0}\times
L_{0},e^{-{\varepsilon}\mathcal{J}}{\rm
Gr}(-J{\gamma}_{B}),[\frac{q\tau}{4},\frac{p\tau}{4}])$ $\displaystyle=$
$\displaystyle\mu_{F}^{CLM}(L_{0}\times L_{0},{\rm
Gr}(-e^{-{\varepsilon}J}J{\gamma}_{B}e^{-{\varepsilon}J}),[\frac{q\tau}{4},\frac{p\tau}{4}]),$

where in the second equality we have used Symplectic invariance property of
$\mu_{F}^{CLM}$ index in [11]. Choose a $C^{1}$ path
${\gamma}\in\mathcal{P}_{\frac{p\tau}{4}}$ such that
${\gamma}(t)=-e^{-{\varepsilon}J}J{\gamma}_{B}e^{-{\varepsilon}J}$ for all
$t\in[\frac{q\tau}{,}\frac{p\tau}{4}]$. Denote by
$D(t)=-J\dot{{\gamma}}(t){\gamma}(t)^{-1}$ for $t\in[0,\frac{p\tau}{4}]$. For
$t\in[\frac{q\tau}{,}\frac{p\tau}{4}]$, by direct computation we have

$\displaystyle
D(t)=-J\frac{d}{dt}(-e^{-{\varepsilon}J}J{\gamma}e^{-{\varepsilon}J})(-e^{-{\varepsilon}J}J{\gamma}e^{-{\varepsilon}J})^{-1}=-Je^{-{\varepsilon}J}B(t)e^{{\varepsilon}J}J.$
(3.63)

Since $B\geq 0$ we have $D(t)\geq 0$ for $t\in[q\tau,p\tau]$ and $D\in
C([0,\frac{p\tau}{4}],\mathcal{L}_{s}({\bf R}^{2n}))$. For $s\geq 0$, we
define $D_{s}(t)=D(t)+sI_{2n}$ and symplectic path ${\gamma}_{s}(t)$ by

$\displaystyle\frac{d}{dt}{\gamma}_{s}(t)=JD_{s}(t){\gamma}_{s}(t),\quad
t\in[0,\frac{p\tau}{4}]$ $\displaystyle{\gamma}_{s}(0)=I_{2n}.$

It is clear that

${\gamma}_{0}={\gamma}.$ (3.64)

By the same argument of step2 of the proof of Theorem 5.1 in [42], we have

$\displaystyle-J\frac{d}{ds}{\gamma}_{s}(t)({\gamma}_{s}(t))^{-1}>0,\quad{\rm
for}\,t=\frac{p\tau}{4},\frac{q\tau}{4}.$ (3.65)

By (3.61) and definition of ${\gamma}_{s}$ we have

$\nu_{L_{0}}({\gamma}_{0}(\frac{p\tau}{4}))=0=\nu_{L_{0}}({\gamma}_{0}(\frac{q\tau}{4})).$
(3.66)

So by (3.65), there is a ${\sigma}>0$ small enough such that

$\nu_{L_{0}}({\gamma}_{s}(\frac{p\tau}{4}))=0=\nu_{L_{0}}({\gamma}_{s}(\frac{q\tau}{4})),\quad\forall
s\in[0,{\sigma}].$ (3.67)

So we have

$\displaystyle\mu_{F}^{CLM}(L_{0}\times L_{0},{\rm
Gr}({\gamma}_{s}(\frac{p\tau}{4})),s\in[0,{\sigma}])=0,$
$\displaystyle\mu_{F}^{CLM}(L_{0}\times L_{0},{\rm
Gr}({\gamma}_{s}(\frac{q\tau}{4})),s\in[0,{\sigma}])=0.$ (3.68)

By the Homotopy invariance with respect to end points and Path additivity
properties of $\mu_{F}^{CLM}$ index in [11], we have

$\displaystyle\mu_{F}^{CLM}(L_{0}\times L_{0},{\rm
Gr}({\gamma}_{s}(\frac{p\tau}{4})),s\in[0,{\sigma}])+\mu_{F}^{CLM}(L_{0}\times
L_{0}a,{\rm Gr}({\gamma}_{\sigma}(t)),t\in[\frac{q\tau}{4},\frac{p\tau}{4}])$
(3.69) $\displaystyle=$ $\displaystyle\mu_{F}^{CLM}(L_{0}\times L_{0},{\rm
Gr}({\gamma}_{0}(t)),t\in[\frac{q\tau}{4},\frac{p\tau}{4}])+\mu_{F}^{CLM}(L_{0}\times
L_{0},{\rm Gr}({\gamma}_{s}(\frac{p\tau}{4})),s\in[0,{\sigma}]).$

So by (3.60), (3.62), (3.64),(3.68) and (3.69), we have

$i_{\sqrt{-1}}^{L_{0}}({\gamma}_{B}^{p})-i_{\sqrt{-1}}^{L_{0}}({\gamma}_{B}^{q})=\mu_{F}^{CLM}(L_{0}\times
L_{0},{\rm Gr}({\gamma}_{\sigma}(t)),t\in[\frac{q\tau}{4},\frac{p\tau}{4}]).$
(3.70)

Since $D(t)\geq 0$ for $t\in[\frac{q\tau}{4},\frac{p\tau}{4}]$, we have

$D_{\sigma}(t)>0,\quad\forall t\in[\frac{q\tau}{4},\frac{p\tau}{4}].$ (3.71)

So by the proof of Lemma 3.1 of [42] and Lemma 2.6 of [42], we have

$\mu_{F}^{CLM}(L_{0}\times L_{0},{\rm
Gr}({\gamma}_{\sigma}(t)),t\in[\frac{q\tau}{4},\frac{p\tau}{4}])=\sum_{t\in[\frac{q\tau}{4},\frac{p\tau}{4})}\nu_{L_{0}}({\gamma}_{\sigma}(t))\geq
0.$ (3.72)

Thus by (3.70) and (3.72), (3.59) holds. The proof of Theorem 3.1 is complete.

By similar proof of Theorem 3.2 we have the following Theorem 3.3.

Theorem 3.3. If $\tau>0$ and $B\in C([0,\frac{\tau}{4}],\mathcal{L}_{s}({\bf
R}^{2n}))$ satisfying condition (B1) and $B\geq 0$, then for $j=0,1$ and any
two positive integers $p\geq q$ there holds

$i_{L_{j}}({\gamma}_{B}^{p})\geq i_{L_{j}}({\gamma}_{B}^{q}).$ (3.73)

##  4 Proof of Theorems 1.1-1.3 and Corollary 1.2

In this section we study the minimal period problem for brake orbits of the
reversible Hamiltonian system (1.1) and complete the proof of Theorems 1.1-1.3
and Corollary 1.2.

For $T>0$, we set $E=W^{1/2,2}(S_{T},{\bf R}^{2n})$ with the usual norm and
inner product denoted by $||\cdot||$ and $\langle\cdot\rangle$ respectively,
and two subspaces of $E$ by $E_{T}=\\{x\in W^{1/2,2}(S_{\tau},{\bf
R}^{2n})|\,x(-t)=Nx(t)\;a.e.\;t\in{\bf R}\\}$ and $\check{E}_{T}=\\{x\in
W^{1/2,2}(S_{\tau},{\bf R}^{2n})|\,x(-t)=-Nx(t)\;a.e.\;t\in{\bf R}\\}$. Then
we have

$E=E_{T}\oplus\check{E}_{T}.$ (4.1)

As in Section 3, we define two selfadjoint operators $A_{T}$ on $E_{T}$ by the
same way as (3.3). We also define two selfadjoint operators $\check{A}_{T}$ on
$\check{E}_{T}$ by the following bilinear form:

$\displaystyle\langle\check{A}_{{\bf T}}x,y\rangle=\int_{0}^{T}-J\dot{x}\cdot
y\,dt.$ (4.2)

Then $A_{T}$ is a bounded operator on $E_{T}$ and dim $\ker A_{T}=n$, the
Fredholm index of $A_{T}$ is zero, and $\check{A}_{T}$ is a bounded operator
on $\check{E}_{T}$ and dim $\ker\check{A}_{T}=n$, the Fredholm index of
$\check{A}_{T}$ is zero.

Set

$E_{T}(j)=\left\\{z\in E_{T}\left|z(t)={\rm exp}(\frac{2j\pi t}{T}J)a+{\rm
exp}(-\frac{2j\pi t}{T}J)b,\;\forall t\in{\bf R};\;\forall a,\,b\in
L_{0}\right.\right\\},$ $E_{T,m}=E_{T}(0)+E_{T}(1)+\cdots+E_{T}(m)$

and

$\check{E}_{T}(j)=\left\\{z\in E_{T}\left|z(t)={\rm exp}(\frac{2j\pi
t}{T}J)a+{\rm exp}(-\frac{2j\pi t}{T}J)b,\;\forall t\in{\bf R};\;\forall
a,\,b\in L_{1}\right.\right\\},$
$\check{E}_{T,m}=\check{E}_{T}(0)+\check{E}_{T}(1)+\cdots+\check{E}_{T}(m).$

Let $P_{T,m}$ be the orthogonal projection from $E_{T}$ to $E_{T,m}$ and
$\check{P}_{T,m}$ be the orthogonal projection from $\check{E}_{T}$ to
$\check{E}_{T,m}$ for $m=0,1,2,...$, then
${\Gamma}_{T}=\\{P_{T,m}:m=0,1,2,...\\}$ and
$\check{{\Gamma}}_{T}=\\{\check{P}_{T,m}:m=0,1,2,...\\}$ are the usual
Galerkin approximation schemes w.r.t. $A_{T}$ and $\check{A}_{T}$
respectively.

For $z\in{E_{T}}$, we define

$f(z)=\frac{1}{2}\langle A_{T}z,z\rangle-\int_{0}^{T}H(z)dt.$ (4.3)

It is well known that $f\in C^{2}(E_{T},{\bf R})$ whenever,

$H\in C^{2}({\bf R}^{2n})\qquad{\rm and\qquad}|H^{\prime\prime}(x)|\leq
a_{1}|x|^{s}+a_{2}$ (4.4)

for some $s\in(1,+\infty)$ and all $x\in{\bf R}^{2n}$.

By similar argument of Lemma 4.1 of [51], looking for $T$-periodic brake orbit
solutions of (1.1) is equivalent to look for critical points of $f$.

In order to get the information about the Maslov-type indices, we need the
following theorem which was proved in [24, 28, 48].

Theorem 4.1. Let $W$ be a real Hilbert space with orthogonal decomposition
$E=X\oplus Y$, where $\dim X<+\infty$. Suppose $f\in C^{2}(W,{\bf R})$
satisfies (PS) condition and the following conditions:

(i) There exist $\rho,\;\delta>0$ such that $f(w)\geq\delta$ for any $w\in W$;

(ii) There exist $e\in\partial B_{1}(0)\cap Y$ and $r_{0}>\rho>0$ such that
for any $w\in\partial Q$, $f(w)<\delta$ where $Q=(B_{r_{0}}(0)\cap
X)\oplus\\{re:0\leq r\leq r_{0}\\}$, $B_{r}(0)=\\{w\in W:||w||\leq r\\}$.

Then (1) $f$ possesses a critical value $c\geq\delta$, which is given by

$c=\inf_{h\in{\Gamma}}\max_{w\in Q}f(h(w)),$

where ${\Gamma}=\\{h\in C(Q,E):h=id\;{\rm on}\;\partial Q\\}$;

(2) There exists $w_{0}\in\mathcal{K}_{c}\equiv\\{w\in
E:\,f^{\prime}(w)=0,\,f(w)=c\\}$ such that the Morse index $m^{-}(w_{0})$ of
$f$ at $w_{0}$ satisfies

$m^{-}(w_{0})\leq\dim X+1.$

Proof of Theorem 1.3. For any given $T>0$, we prove the existence of
$T$-periodic brake solution of (1.1) whose minimal period satisfies the
inequalities in the conclusion of Theorem 1.2. We divide the proof into five
steps.

Step 1. We truncate the function $\hat{H}$ suitably and evenly such that it
satisfies the growth condition (4.4). Hence corresponding new reversible
function $H$ satisfies condition (4.4).

We follow the method in Rabinowitz’s pioneering work [43] (cf. also [18], [44]
and [51]). Let $K>0$ and $\chi\in C^{\infty}({\bf R},{\bf R})$ such that
$\chi\equiv 1$ if $y\leq K$, $\chi\equiv 0$ if $y\geq K$ and
$\chi^{\prime}(y)<0$ if $y\in(K,K+1)$, Where $K$ will be determined later. Set

$\hat{H}_{K}(z)=\chi(|z|)\hat{H}(z)+(1-\chi(|z|))R_{K}|z|^{4}$ (4.5)

and

$H_{K}(z)=\frac{1}{2}B_{0}x\cdot x+\hat{H}_{K}(z),$ (4.6)

where the constant $R_{K}$ satisfies

$R_{K}\geq\max_{K\leq|z|\leq K+1}\frac{H(z)}{|z|^{4}}.$ (4.7)

Then $H_{K}\in C^{2}({\bf R}^{2n},{\bf R})$. Since $\hat{H}$ satisfies (H3),
$\forall\varepsilon>0$, there is a $\delta_{1}>0$ such that
$\hat{H}_{K}(z)\leq\varepsilon|z|^{2}$ for $|z|\leq\delta_{1}$. It is easy to
see that $H_{K}(z)|z|^{4}$ is uniformly bounded as $|z|\to+\infty$, there is
an $M_{1}=M_{1}(\varepsilon,K)$ such that $\hat{H}_{K}(z)\leq M_{1}|z|^{4}$
for $|z|\geq\delta_{1}$. So

$\hat{H}_{K}(z)\leq\varepsilon|z|^{2}+M_{1}|z|^{4},\quad\forall z\in{\bf
R}^{2n}.$ (4.8)

Set

$\displaystyle
f_{K}(z)=\frac{1}{2}\langle{A_{T}}z,z\rangle-\int_{0}^{T}H_{K}(z)dt,\qquad\forall
z\in\hat{E}.$

Then $f_{K}\in C^{2}(E_{T},{\bf R})$ and

$\displaystyle
f_{K}(z)=\frac{1}{2}\langle({A_{T}}-{B_{0}}_{T})z,z\rangle-\int_{0}^{T}\hat{H}_{K}(z)dt,\qquad\forall
z\in\hat{E},$

where ${B_{0}}_{T}$ is the selfadjoint linear compact operator on ${E_{T}}$
defined by

$\displaystyle\langle{B_{0}}_{T}z,z\rangle=\int_{0}^{T}B_{0}z(t)\cdot
z(t)\,dt.$

Step 2. For $m>0$, let $f_{Km}=f|E_{T,m}$. We show $f_{Km}$ satisfies the
hypotheses of Theorem 4.1.

We set

$\displaystyle X_{m}=M^{-}(P_{T,m}({A_{T}}-{B_{0}}_{T})P_{T,m})\oplus
M^{0}(P_{T,m}({A_{T}}-{B_{0}}_{T})P_{T,m}),$ $\displaystyle
Y_{m}=M^{+}(P_{T,m}({A_{T}}-{B_{0}}_{T})P_{T,m}).$

For $z\in Y_{m}$, by (4.8), (3.2), and the fact that
$P_{T,j}{B_{0}}_{T}=P_{T,j}{B_{0}}_{T}$ for $j>0$, we have

$\displaystyle f_{Km}(z)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\langle({A_{T}}-{B_{0}}_{T})z,z\rangle-\int_{0}^{T}\hat{H}_{K}(z)dt$
(4.9) $\displaystyle\geq$
$\displaystyle\frac{1}{2}||({A_{T}}-{B_{0}}_{T})^{\\#}||^{-1}||z||^{2}-(\varepsilon||z||_{L^{2}}^{2}+M_{1}||z||_{L^{4}}^{4})$
$\displaystyle\geq$
$\displaystyle\frac{1}{2}||({A_{T}}-{B_{0}}_{T})^{\\#}||^{-1}||z||^{2}-(\varepsilon
C_{2}^{2}+M_{1}C_{4}^{4}||z||^{2})||z||^{2},$

where $C_{2}$ and $C_{4}$ are constants for $s=2,\,4$ for the Sobolev
embedding of inequality (3.2), and they are independent of $m$ and $K$.

So if choose $\varepsilon>0$ small enough such that $\varepsilon
C_{2}^{2}<\frac{1}{4}||(A_{T}-{B_{0}}_{T})^{\\#}||^{-1}$, then there exists
$\rho=\rho(K)>0$ small enough and $\delta=\delta(K)>0$, which are independent
of $m$, such that

$f_{m}(z)\geq\delta,\qquad\forall z\in\partial B_{\rho}(0)\cap Y_{m}.$ (4.10)

Let $e\in B_{1}(0)\cap Y_{m}$ and set

$\displaystyle Q_{m}=\\{re:0\leq r\leq r_{1}\\}\oplus(B_{r_{1}}(0)\cap
X_{m}),$

where $r_{1}$ will be determined later. Let $z=z_{-}+z_{0}\in B_{r_{1}}(0)\cap
X_{m}$, we have

$\displaystyle f_{Km}(z+re)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\langle({A_{T}}-{B_{0}}_{T})z_{,}z_{\rangle}+\frac{1}{2}r^{2}\langle({A_{T}}-{B_{0}}_{T})e,e\rangle-\int_{0}^{T}\hat{H}_{K}(z+re)dt$
(4.11) $\displaystyle\leq$
$\displaystyle\frac{1}{2}||{A_{T}}-{B_{0}}_{T}||r^{2}-\frac{1}{2}||(A_{T}-{B_{0}}_{T})^{\\#}||^{-1}||z_{-}||^{2}-\int_{0}^{T}\hat{H}_{K}(z+re)dt.$

Since $\hat{H}$ satisfies (H2) we have

$\displaystyle\hat{H}_{K}(x)\geq a_{1}|x|^{\alpha}-a_{2},\qquad\forall
x\in{\bf R}^{2n},$

where $\alpha=\min\\{\mu,4\\}$, $a_{1}>0$, $a_{2}$ are two constants
independent of $K$ and $m$. Then there holds

$\int_{0}^{T}\hat{H}_{K}(z+re)dt\geq
a_{1}\int_{0}^{T}|z+re|^{\alpha}-Ta_{2}\geq
a_{3}(||z_{0}||_{L^{\alpha}}^{\alpha}+r^{\alpha})-a_{4},$ (4.12)

where $a_{3}$ and $a_{4}$ are constants independent of $K$ and $m$. By (4.11)
and (4.12) we have

$\displaystyle
f_{Km}(z+re)\leq\frac{1}{2}||{A_{T}}-\hat{B_{0}}||r^{2}-\frac{1}{2}||(A-B_{0})^{\\#}||^{-1}||z_{-}||^{2}-a_{3}(||z_{0}||_{L^{\alpha}}^{\alpha}+r^{\alpha})+a_{4}.$

Since $\alpha>2$ there exists a constant $r_{1}>\rho>0$, which are independent
of $K$ and $m$, such that

$f_{Km}\leq 0,\qquad\forall z\in\partial Q_{m}.$ (4.13)

Then by Theorem 4.1, $f_{Km}$ has a critical value $c_{Km}$, which is given by

$c_{Km}=\inf_{g\in{\Gamma}_{m}}\max_{z\in Q_{m}}f_{Km}(g(z)),$ (4.14)

where ${\Gamma}_{m}=\\{g\in C(Q_{m},\hat{E}_{m}|g=id;{\rm on}\;\partial
Q_{m}\\}$. Moreover there is a critical point $x_{Km}$ of $f_{Km}$ which
satisfies

$\displaystyle m^{-}(x_{Km})\leq\dim X_{m}+1.$ (4.15)

Step 3. We prove that there exists a $T$-periodic brake orbit solution $x_{T}$
of (1.1) which satisfies $i_{L_{0}}(x_{T})\leq
i_{L_{0}}(B_{0})+\nu_{L_{0}}(B_{0})+1$.

Note that $id\in{\Gamma}_{m}$, by (4.11) and condition (H4), we have

$\displaystyle c_{Km}\leq\sup_{z\in
Q_{m}}f_{Km}(z)\leq\frac{1}{2}||{A_{T}}-{B_{0}}_{T}||r_{1}^{2}.$

Then $\\{c_{Km}\\}$ possesses a convergent subsequence, we still denote it by
$\\{c_{Km}\\}$ for convenience. So there is a $c_{K}\in[\delta,]$ such that
$c_{Km}\to c_{K}$.

By the same arguments as in section 6 of [44] we have $f_{K}$ satisfies
$(PS)_{c}^{*}$ condition for $c\in{\bf R}$, i.e., any sequence ${z_{m}}$ such
that $z_{m}\in E_{T,m}$, $f_{Km}^{\prime}(z_{m})\to 0$ and $f_{Km}(z_{m})\to
c$ possesses a convergent subsequence in $E_{T}$. Hence in the sense of
subsequence we have

$\displaystyle x_{Km}\to x_{K},\qquad f_{K}(x_{K})=c_{K},\qquad
f^{\prime}_{K}(x_{K})=0.$ (4.16)

By similar argument in [44], $x_{K}$ is a classical nonconstant symmetric
$T$-periodic solution of

$\displaystyle\dot{x}=JH_{K}^{\prime}(x),\quad x\in{\bf R}^{2n}.$

Set $B_{K}(t)=H^{\prime\prime}_{K}(x_{K}(t))$, Then $B_{K}\in
C([0,T/2],\mathcal{L}_{s}({\bf R}^{2n}))$ and satisfies condition (B1). Let
${B_{K}}_{T}$ be the operator defined by the same way of the definition of
${B_{0}}_{T}$. It is easy to show that

$\displaystyle||f^{\prime\prime}(z)-({A_{T}}-{B_{K}}_{T})||\to 0\qquad{\rm
as}\;\;||z-x_{K}||\to 0.$

So for $0<d\leq\frac{1}{4}||(A_{T}-B_{K_{T}})^{\\#}||^{-1}$, there exists
$r_{2}>0$ such that

$\displaystyle||f_{Km}^{\prime\prime}(z)-P_{T,m}({A_{T}}-{B_{K}}_{T})P_{T,m}||\leq||f^{\prime\prime}(z)-({A_{T}}-{B_{K}}_{T})||\leq\frac{1}{2}d,\;\forall
z\in\\{z\in E_{T}:||z-x_{K}||\leq r_{2}\\}.$

Then for $z\in\\{z\in E_{T}:||z-x_{K}||\leq r_{2}\\}\cap E_{T,m}$, $\forall
u\in M^{-}_{d}(P_{T,m}({A_{T}}-{B_{K}}_{T})P_{T,m})\setminus\\{0\\}$, we have

$\displaystyle\langle f_{Km}^{\prime\prime}(z)u,u\rangle$ $\displaystyle\leq$
$\displaystyle\langle
P_{T,m}({A_{T}}-{B_{K}}_{T})P_{T,m}u,u\rangle+\|f_{Km}^{\prime\prime}(z)-P_{T,m}({A_{T}}-{B_{K}}_{T})P_{T,m}\|\|u\|^{2}$
$\displaystyle\leq$ $\displaystyle-\frac{1}{2}d\|u\|^{2}.$

So we have

$m^{-}(f_{Km}^{\prime\prime}(z))\geq\dim
M^{-}_{d}(P_{T,m}({A_{T}}-{B_{K}}_{T})P_{T,m}).$ (4.17)

By Theorem 2.1 of [31] and Remark 3.1, there is $m^{*}>0$ such that for $m\geq
m^{*}$ we have

$\displaystyle\dim X_{m}=mn+n+i_{L_{0}}(B_{0})+\nu_{L_{0}}(B_{0}),$ (4.18)
$\displaystyle\dim
M^{-}_{d}(P_{T,m}({A_{T}}-{B_{K}}_{T})P_{T,m})=mn+n+i_{L_{0}}(B_{K}).$ (4.19)

Then by (4.15), (4.16), and (4.17)-(4.19), we have

$\displaystyle i_{L_{0}}(B_{K})\leq i_{L_{0}}(B_{0})+\nu_{L_{0}}(B_{0})+1.$

By the similar argument as in the section 6 of [44], there is a constant
$M_{2}$ independent of $K$ such that $||x_{K}||_{\infty}\leq M_{2}$. Choose
$K>M_{2}$. Then $x_{K}$ is a non-constant symmetric $T$-periodic solution of
the problem (1.1). From now on in the proof of Theorem 1.3, we write $B=B_{K}$
and $x_{T}=x_{K}$. Then $x_{T}$ is a non-constant symmetric $T$-periodic
solution of the problem (1.1), and $B$ satisfies

$\displaystyle i_{L_{0}}(x_{T})=i_{L_{0}}(B)\leq
i_{L_{0}}(B_{0})+\nu_{L_{0}}(B_{0})+1.$ (4.20)

Since $x_{T}$ obtained in Step 3 is a nonconstant and symmetric $T$-period
solution, its minimal period $\tau=\frac{T}{k}$ for some $k\in{\bf N}$.

We denote by $x_{\tau}=x_{T}|_{[0,\tau]}$, then it is a brake orbit solution
of (1.1) with the minimal $\tau$ and $X_{T}=x_{\tau}^{k}$ being the $k$ times
iteration of $x_{\tau}$. As in Section 1, let ${\gamma}_{x_{T}}$ and
${\gamma}_{x_{\tau}}$ be the symplectic path associated to $(\tau,x)$ and
$(T,x_{T})$ respectively. Then ${\gamma}_{x_{\tau}}\in
C([0,\frac{\tau}{2}],{\rm Sp}(2n))$ and ${\gamma}_{x_{T}}\in
C([0,\frac{T}{2}],{\rm Sp}(2n))$. Also we have
${\gamma}_{x_{T}}={\gamma}_{x_{\tau}}^{k}$.

Step 4. We prove that

$\displaystyle
i_{L_{1}}({\gamma}_{x_{\tau}})+\nu_{L_{1}}({\gamma}_{x_{\tau}})\geq 1.$

We follow the way of the proof of Theorem 1.2 of [18]. By the same way as
$\check{E}_{T}$ and $\check{A}_{T}$ we can define the space $\check{E}_{\tau}$
and the operator $\check{A}_{\tau}$ on it. Also we can define the orthogonal
projection $\check{P}_{\tau},m$ and the subspaces $\check{E}_{\tau,m}$ for
$m=0,1,2,...$. Let $\check{B}_{\tau}$ be the selfadjoint linear compact
operator on ${\check{E}_{T}}$ defined by:

$\displaystyle\langle\check{B}_{\tau}z,z\rangle=\int_{0}^{\tau}B(t)z(t)\cdot
z(t)\,dt,\quad\forall z\in\check{E}_{\tau}.$

For $z\in\check{E}_{\tau}$, set

$\displaystyle
f_{\tau}(z)=\frac{1}{2}\langle(\check{A}_{\tau}-\check{B}_{\tau})z,z\rangle=\frac{1}{2}\langle\check{A}_{\tau}z,z\rangle-\frac{1}{2}\int_{0}^{\tau}H^{\prime\prime}(x_{\tau}(t))z\cdot
z\,dt$

and

$\displaystyle f_{\tau m}(w)=f_{\tau}(w),\qquad\forall
w\in\check{E}_{\tau,m}.$

Let

$X=\\{z\in L_{1}|\,B_{0}z=0\;{\rm
and}\;\hat{H}^{\prime\prime}(x_{\tau}(t))z=0,\;\forall t\in{\bf R}\\}$

and $Y$ be the orthogonal complement of $X$ in $L_{1}$, i.e., $L_{1}=X\oplus
Y$. Since
$H^{\prime\prime}(x_{\tau}(t))=B_{0}+\hat{H}^{\prime\prime}(x_{\tau}(t))$, by
(H4) it is easy to see that there exists ${\lambda}_{0}>0$ such that

$\displaystyle\int_{0}^{\tau}H^{\prime\prime}(x_{\tau}(t))z_{0}\cdot
z_{0}\,dt\geq{\lambda}_{0}||z_{0}||,\qquad\forall z_{0}\in Y.$

Thus for any $z=z_{-}+z_{0}\in\check{P}_{\tau,m}M^{-}(\check{A}_{\tau})\oplus
Y$ with $||z||=1$, we have

$\displaystyle f_{\tau m}(z)$ $\displaystyle=$
$\displaystyle\frac{1}{2}\langle(\check{A}_{\tau}-\check{B}_{\tau})z,z\rangle=\frac{1}{2}\langle\check{A}_{\tau}z_{-},z_{-}\rangle-\frac{1}{2}\int_{0}^{\tau}H^{\prime\prime}(x_{\tau}(t))z\cdot
z\,dt$ $\displaystyle\leq$
$\displaystyle-\frac{1}{2}||\check{A}_{\tau}^{\\#}||^{-1}||z_{-}||^{2}-\frac{1}{2}\int_{0}^{\tau}H^{\prime\prime}(x_{\tau}(t))z_{0}\cdot
z_{0}\,dt-\int_{0}^{\tau}H^{\prime\prime}(x_{\tau}(t))z_{-}\cdot z_{0}\,dt$
$\displaystyle\leq$
$\displaystyle-\frac{1}{2}||\check{A}_{\tau}^{\\#}||^{-1}||z_{-}||^{2}-\frac{{\lambda}_{0}}{2}||z_{0}||^{2}+\max_{t\in[0,\tau]}||H^{\prime\prime}(x_{\tau}(t))||\,||z_{-}||\,||z_{0}||.$
(4.22)

Since

$\displaystyle||z_{-}||\,||z_{0}||\leq\frac{{\varepsilon}}{4}||z_{-}||^{2}+\frac{1}{{\varepsilon}}||z_{0}||^{2},\quad\forall{\varepsilon}>0.$

By choosing ${\varepsilon}$ suitably one can see that there exists $0<c_{0}<1$
with $|1-c_{0}|$ small enough such that if $||z_{0}||\leq c_{0}$,

$f_{\tau m}(z)\leq-\frac{{\lambda}_{0}}{4}c_{0}^{2}.$ (4.23)

When $||z_{0}||\leq c_{0}$, we have $||z_{-}||^{2}\geq 1-c_{0}^{2}$. By (4)
and (H4)

$\displaystyle f_{\tau
m}(z)\leq-\frac{1}{2}||\check{A}_{\tau}^{\\#}||^{-1}||z_{-}||^{2}\leq-\frac{1}{2}||\check{A}_{\tau}^{\\#}||^{-1}(1-c_{0}^{2}).$

Hence we always have

$f_{\tau m}(z)\leq-c||z||^{2},\qquad\forall
z\in\check{P}_{\tau,m}M^{-}(\check{A}_{\tau})\oplus Y,$ (4.24)

where
$c=\max\\{\frac{{\lambda}_{0}}{4}c_{0}^{2},\frac{1}{2}||\check{A}_{\tau}^{\\#}||^{-1}(1-c_{0}^{2})\\}$
is independent of $m$. Let

$d=\min\\{\frac{1}{4}||(\check{A}_{\tau}-\check{B}_{\tau})^{\\#}||^{-1},\frac{c}{2}\\}.$

By (4.24) and Theorem 2.1 of [31] and Remark 3.1 and the definition of
$i_{L_{1}}({\gamma}(x_{\tau}))$, for $m$ large enough, we have

$\displaystyle mn+n+i_{L_{1}}({\gamma}(x_{\tau}))$ $\displaystyle=$
$\displaystyle\dim
M^{-}_{d}(\check{P}_{\tau,m}(\check{A}_{\tau}-\check{B}_{\tau})\check{P}_{\tau,m})$
(4.25) $\displaystyle\geq$
$\displaystyle\dim(\check{P}_{\tau,m}M^{-}(\check{A}_{\tau})\oplus Y)$
$\displaystyle=$ $\displaystyle mn+n-\dim X,$

which implies that

$i_{L^{1}}({\gamma}(x_{\tau}))\geq-\dim X.$ (4.26)

Since $x_{\tau}$ is a nonconstant brake solution of (1.1), by the definition
of $X$ we have

$\nu_{L^{1}}({\gamma}(x_{\tau}))\geq\dim X+1.$ (4.27)

Hence by (4.26) and (4.27) we have

$i_{L^{1}}({\gamma}(x_{\tau}))+\nu_{L^{1}}({\gamma}(x_{\tau}))\geq 1.$ (4.28)

Step 5. Finish the proof of Theorem 1.3.

By Theorem 2.1 and Theorem 6.2 below (also Theorem 2.6 of [32]) we have

$\displaystyle i_{L_{0}}({\gamma}_{x_{\tau}}^{k})\geq
i_{L_{0}}({\gamma}_{{x_{\tau}}})+\frac{k-1}{2}(i_{1}({\gamma}^{2})+\nu_{1}({\gamma}^{2})-n),\quad{\rm
if}\,k\in 2{\bf N}-1,$ (4.29) $\displaystyle
i_{L_{0}}({\gamma}_{x_{\tau}}^{k})\geq
i_{L_{0}}({\gamma}_{{x_{\tau}}})+i_{\sqrt{-1}}^{L_{0}}({\gamma}_{{x_{\tau}}})+(\frac{k}{2}-1)(i_{1}({\gamma}^{2})+\nu_{1}({\gamma}^{2})-n),\quad{\rm
if}\,k\in 2{\bf N}.$ (4.30)

Since $B_{0}$ is semipositive and $\hat{H}$ satisfies (H4), by Corollary 3.2,
we have

$i_{L_{0}}({\gamma}_{x_{\tau}})+\nu_{L_{0}}({\gamma}_{x_{\tau}})\geq 0.$
(4.31)

By Proposition C of [42] and the definitions of $i_{L_{0}}$ and $i_{L_{1}}$ we
have

$\displaystyle i_{1}({\gamma}^{2})=i_{L_{0}}({\gamma})+i_{L_{1}}({\gamma})+n,$
$\displaystyle\nu_{1}({\gamma}^{2})=\nu_{L_{0}}({\gamma})+\nu_{L_{1}}({\gamma}).$

So by (4.28) and (4.31) we have

$i_{1}({\gamma}^{2})+\nu_{1}({\gamma}^{2})-n\geq 1.$ (4.32)

So by (4.29), (4.30) and (4.32) we have

$\displaystyle i_{L_{0}}({\gamma}_{x_{\tau}}^{k})\geq
i_{L_{0}}({\gamma}_{{x_{\tau}}})+\frac{k-1}{2},\quad{\rm if}\,k\in 2{\bf
N}-1,$ (4.33) $\displaystyle i_{L_{0}}({\gamma}_{x_{\tau}}^{k})\geq
i_{L_{0}}({\gamma}_{{x_{\tau}}})+\frac{k-1}{2},\quad{\rm if}\,k\in 2{\bf N}.$
(4.34)

By (4.20) and the definition of ${\gamma}_{x_{\tau}}$ we have

$i_{L_{0}}({\gamma}_{{x_{\tau}}})^{k})\leq
i_{L_{0}}(B_{0})+\nu_{L_{0}}(B_{0})+1.$ (4.35)

By Corollary 3.2, we have

$i_{L_{0}}({\gamma}_{{x_{\tau}}})\geq-n.$ (4.36)

So by (4.33)-(4.36) we have

$k\leq 2(i_{L_{0}}(B_{0})+\nu_{L_{0}}(B_{0}))+2n+4.$ (4.37)

Claim 2. $k$ can not be $2(i_{L_{0}}(B_{0})+\nu_{L_{0}}(B_{0}))+2n+3$ and
$2(i_{L_{0}}(B_{0})+\nu_{L_{0}}(B_{0}))+2n+4$.

Hence by Claim 2, $k\leq 2(i_{L_{0}}(B_{0})+\nu_{L_{0}}(B_{0}))+2n+2$, and
Theorem 1.3 holds.

Proof of Claim 2. We first show that $k$ can not be
$2(i_{L_{0}}(B_{0})+\nu_{L_{0}}(B_{0}))+2n+3$. Otherwise, we have

$k=2(i_{L_{0}}(B_{0})+\nu_{L_{0}}(B_{0}))+2n+3.$ (4.38)

The equality in (4.29) holds, then by (4.32), in this case there must hold
that

$\displaystyle i_{1}({\gamma}^{2})+\nu_{1}({\gamma}^{2})-n=1$ (4.39)

and

$\displaystyle i_{L_{0}}({\gamma}_{{x_{\tau}}})=-n.$ (4.40)

By Corollary 3.2 again we have that

$\nu_{L_{0}}({\gamma}_{{x_{\tau}}})=n.$ (4.41)

Also by (4.39) we have

$\displaystyle
i_{L^{1}}({\gamma}(x_{\tau})+\nu_{L^{1}}({\gamma}(x_{\tau}))=1.$

Denote by $\nu_{L^{1}}({\gamma}(x_{\tau}))=r$. Then we have

$\displaystyle i_{L^{1}}({\gamma}(x_{\tau}))=1-r,$ (4.42)
$\displaystyle\nu_{L^{1}}({\gamma}(x_{\tau}))\geq 1.$ (4.43)

By (4.40) and (4.42) we have

$i_{L_{0}}({\gamma}_{{x_{\tau}}})-i_{L_{1}}({\gamma}_{{x_{\tau}}})=r-n-1.$
(4.44)

So we can write
${\gamma}_{{x_{\tau}}}(\frac{\tau}{2})=\left(\begin{array}[]{cc}A&0\\\
C&D\end{array}\right)$ with $A,C,D$ to be $n\times n$ real matrices. Hence by
(4.2) of [35] we have

$\displaystyle{\gamma}_{{x_{\tau}}}^{2}(\tau)=N{\gamma}_{{x_{\tau}}}(\frac{\tau}{2})^{-1}N{\gamma}_{{x_{\tau}}}(\frac{\tau}{2})=\left(\begin{array}[]{cc}D^{T}A&0\\\
C^{T}A&A^{T}D\end{array}\right).$ (4.47)

Since ${\gamma}_{x_{\tau}}(\frac{\tau}{2})$ is a symplectic matrix we have

$\displaystyle A^{T}D=D^{T}A=I_{n},\quad C^{T}A=A^{T}C.$

So we have

$\displaystyle{\gamma}_{{x_{\tau}}}^{2}(\tau)=\left(\begin{array}[]{cc}I_{n}&0\\\
C^{T}A&I_{n}\end{array}\right).$ (4.50)

Note that here $C^{T}A$ is a symmetric matrix and $A$ is invertible. So by
(4.43) there exists a orthogonal matrix $Q$ such that

$Q(C^{T}A)Q^{T}={\rm
diag}(0,0,...,0,{\lambda}_{1},{\lambda}_{2},...,{\lambda}_{p},{\lambda}_{p+1},...,{\lambda}_{n-p-r})$
(4.51)

with ${\lambda}_{j}>0$ for $j=1,2,...,p$ and ${\lambda}_{j}<0$, for
$j=p+1,p+2,...,n-p-r$, where $1\leq p\leq n-r$. Then it is easy to check that
$(I_{2})^{\diamond^{r}}\diamond N_{1}(1,-1)^{\diamond^{p}}\diamond
N_{1}(1,1)^{\diamond^{(n-p-r)}}\in{\Omega}^{0}({\gamma}_{{x_{\tau}}})$ with
${\Omega}^{0}({\gamma}_{x\tau})$ to be defined in Section 6 below. Then by
Theorem 6.2 below or Theorem 2.6 of [32], when the equality in (4.29) holds,
there must hold $p=n-r$. Hence we have

$\displaystyle Q(C^{T}A)Q^{T}={\rm
diag}(0,0,...,0,{\lambda}_{1},{\lambda}_{2},...,{\lambda}_{n-r}),$ (4.52)
$\displaystyle{\lambda}_{j}>0,\qquad{\rm for}\;j=1,2,...,n-r.$ (4.53)

Case 1. If ${\rm det}A>0$, then there exists a invertible matrix path
$\rho(s)$ for $s\in[0,\frac{\tau}{2}]$ connecting it and $I_{n}$ such that
$\rho(0)=I_{n}$ and $\rho(1)=A$.

We define a symplectic path $\phi_{1}$ by

$\displaystyle\phi_{1}(s)=\left(\begin{array}[]{cc}\rho(s)^{-1}&0\\\
0&\rho(s)^{T}\end{array}\right)\left(\begin{array}[]{cc}A&0\\\
C&D\end{array}\right),\quad\forall s\in[0,\frac{\tau}{2}].$ (4.58)

Then $\nu_{L_{j}}(\phi_{1}(s)=constant$ for $j=0,1$ and
$s\in[0,\frac{\tau}{2}]$. So by Definition 2.5 and Lemma 2.8 and Proposition
2.11 of [42], for $j=1,2$ we have

$\mu_{F}^{CLM}(V_{j},{\rm Gr}(\phi_{1}),[0,\frac{\tau}{2}])=0.$ (4.59)

Also we have $\phi_{1}(0)=\left(\begin{array}[]{cc}A&0\\\
C&D\end{array}\right)$ and $\phi_{1}(0)=\left(\begin{array}[]{cc}I_{n}&0\\\
A^{T}C&I_{n}\end{array}\right)$.

Note that we can always choose the orthogonal matrix $Q$ in (4.52) such that
${\rm det}Q=1$ (otherwise we replace it by ${\rm diag}(-1,1,...,1)Q$). Then
there exists a invertible matrix path $\rho_{2}(s)$ for
$s\in[0,\frac{\tau}{2}]$ connecting it and $I_{n}$ such that
$\rho_{2}(0)=I_{n}$ and $\rho_{2}(\frac{\tau}{2})=Q$. We define a symplectic
path $\phi_{2}$ by

$\displaystyle\phi_{2}(s)=\left(\begin{array}[]{cc}I_{n}&0\\\
\rho_{2}(s)A^{T}C\rho_{2}(s)^{T}&I_{n}\end{array}\right),\quad\forall
s\in[0,\frac{\tau}{2}].$ (4.62)

Then $\nu_{L_{j}}(\phi_{2}(s)=constant$ and for $j=0,1$ and
$s\in[0,\frac{\tau}{2}]$. So by Definition 2.5 and Lemma 2.8 and Proposition
2.11 of [42] again, for $j=1,2$ we have

$\mu_{F}^{CLM}(V_{j},{\rm Gr}(\phi_{2}),[0,\frac{\tau}{2}])=0.$ (4.63)

Also we have

$\displaystyle\phi_{2}(0)=\left(\begin{array}[]{cc}I_{n}&0\\\
A^{T}C&I_{n}\end{array}\right)$ (4.66)
$\displaystyle\phi_{2}(\frac{\tau}{2})=\left(\begin{array}[]{cc}I_{n}&0\\\
QA^{T}CQ^{T}&I_{n}\end{array}\right)=(I_{2})^{\diamond^{r}}\diamond
N_{1}(1,{\lambda}_{1})\diamond\cdots\diamond N_{1}(1,{\lambda}_{n-r}).$ (4.69)

By the Reparametrization invariance and Path additivity of the Maslov index
$\mu_{F}^{CLM}$ in [11] and (4.59) and (4.63), for $j=1,2$ we have

$\displaystyle\mu_{F}^{CLM}(V_{j},{\rm
Gr}({\gamma}_{x_{\tau}}),[0,\frac{\tau}{2}])=\mu_{F}^{CLM}(V_{j},{\rm
Gr}(\phi_{2}*(\phi_{1}*{\gamma}_{{x_{\tau}}})),[0,\frac{\tau}{2}]),$

where the joint path $\phi_{2}*(\phi_{1}*{\gamma}_{{x_{\tau}}})$ is defined by
(6.1). So by definition for $j=0,1$ we have

$i_{L_{j}}({\gamma}_{{x_{\tau}}})=i_{L_{j}}(\phi_{2}*(\phi_{1}*{\gamma}_{{x_{\tau}}})).$
(4.70)

Then by Theorem 2.3 and (4.69) we have

$\displaystyle
i_{L_{0}}({\gamma}_{{x_{\tau}}})-i_{L_{1}}({\gamma}_{{x_{\tau}}})=\frac{1}{2}{\rm
sgn}M_{\varepsilon}((I_{2})^{\diamond^{r}}\diamond
N_{1}(1,{\lambda}_{1})^{T}\diamond\cdots\diamond
N_{1}(1,{\lambda}_{n-r})^{T}).$ (4.71)

By Remark 2.1 and the computations (2.161)-(2.172) at the end of Section 2,
for ${\varepsilon}>0$ small enough we have

${\rm sgn}M_{\varepsilon}((I_{2})^{\diamond^{r}}\diamond
N_{1}(1,{\lambda}_{1})^{T}\diamond\cdots\diamond
N_{1}(1,{\lambda}_{n-r})^{T})=2(r-n).$ (4.72)

So we have

$i_{L_{0}}({\gamma}_{{x_{\tau}}})-i_{L_{1}}({\gamma}_{{x_{\tau}}})=r-n,$
(4.73)

which contradicts to (4.44).

Case 2. If ${\rm det}A<0$, then there exists a invertible matrix path
$\rho(s)$ for $s\in[0,\frac{\tau}{2}]$ such that $\rho(0)={\rm
diag}(-1,1,1,...,1)$ and $\rho(1)=A$. by similar arguments we can show that

$\displaystyle
i_{L_{0}}({\gamma}_{{x_{\tau}}})-i_{L_{1}}({\gamma}_{{x_{\tau}}})=\frac{1}{2}{\rm
sgn}M_{\varepsilon}((-I_{2})\diamond(I_{2})^{\diamond^{(}r-1)}\diamond
N_{1}(1,{\lambda}_{1})\diamond\cdots\diamond N_{1}(1,{\lambda}_{n-r}))=r-n,$
(4.74)

which still contradicts to (4.44).

Hence we have proved that $k$ can not be
$2(i_{L_{0}}(B_{0})+\nu_{L_{0}}(B_{0}))+2n+3$. By the same argument we can
prove that $k$ can not be $2(i_{L_{0}}(B_{0})+\nu_{L_{0}}(B_{0}))+2n+4$. Thus
Claim 2 is proved and the proof of Theorem 1.3 is complete.

Proof of Theorem 1.1. Note that this is the case $B_{0}=0$ of Theorem 1.3.
Then by Theorem 1.3 and the fact that $i_{L_{0}}(0)=-n$ and
$\nu_{L_{0}}(0)=n$, the minimal period of $x_{T}$ is no less than
$\frac{T}{2n+2}$. In the following we prove that if (1.12) holds then the
minimal period of $x_{T}$ belongs to $\\{T,\frac{T}{2}\\}$.

Let $x_{T}$ is the $k$-time iteration of $x_{\tau}$ with $\tau$ being the
minimal period of $x_{\tau}$ and $\tau=\frac{T}{k}$. Then by the proof of
Theorem 1.3 with $B_{0}=0$ we have (4.28), (4.29) and (4.30) hold. Since
(1.12) holds, by Lemma 3.3 we have

$i_{L_{0}}({\gamma}_{x\tau})\geq 0.$ (4.75)

So by (4.29) if $k$ is odd, we have

$1\geq 0+\frac{k-1}{2}.$ (4.76)

Hence $k\leq 3$. Now we prove that $k$ can not be $3$, other wise we have

$\displaystyle i_{L_{0}}({\gamma}_{x_{\tau}})=0,$ (4.77)
$\displaystyle\nu_{L_{0}}({\gamma}_{x_{\tau}})=0,$ (4.78) $\displaystyle
i_{L_{1}}({\gamma}_{x_{\tau}})+\nu_{L_{1}}({\gamma}_{x_{\tau}})=1.$ (4.79)

And by Theorem 2.1 and Theorem 6.2 we have

$\displaystyle 1\geq
i_{L_{0}}({\gamma}_{x_{\tau}}^{3})=i_{L_{0}}({\gamma}_{x_{\tau}})+i_{e^{2\pi/3}}({\gamma}_{x_{\tau}}^{2})\geq(i_{1}({\gamma}_{x_{\tau}}^{2})-\nu_{1}({\gamma}_{x_{\tau}}^{2})-n)\geq
1.$ (4.80)

Then all the equalities of (4.80) hold. By Lemma 6.2 and 2 of Theorem 6.2
again, there exist $p\geq 0$, $q\geq 0$ with $p+q\leq n$ and
$0<\theta_{1}\leq\theta_{2}\leq...\leq\theta_{n-(p+q)}\leq 2\pi/3$ such that

$(I_{2})^{\diamond^{p}}\diamond N_{1}(1,-1)^{\diamond^{q}}\diamond
R(\theta_{1})\diamond R(\theta_{2})\diamond...\diamond
R(\theta_{n-p-q})\in{\Omega}^{0}(({\gamma}^{2}_{x_{\tau}})(\tau)),$ (4.81)

where ${\Omega}^{0}(M)$ for a symplectic matrix $M$ is defined in Section 6\.
By (4.81) we have

$-1\notin{\sigma}(({\gamma}^{2}_{x_{\tau}})(\tau)).$ (4.82)

Now we denote by
${\gamma}_{x_{\tau}}(\frac{\tau}{2})=\left(\begin{array}[]{cc}A&B\\\
C&D\end{array}\right)$ with $A,B,C,D$ are all $n\times n$ matrices.

Claim 1. Both $D$ and $A$ are invertible.

We first prove $D$ is invertible. Otherwise, there exists a $n\times n$
invertible matrix $P$ such that $P^{-1}DP=\left(\begin{array}[]{cc}0&0\\\
0&R\end{array}\right)$ and $R$ is a $(n-r)\times(n-r)$ matrix with $r\geq 1$.
So we have

$\displaystyle\left(\begin{array}[]{cc}P^{T}&0\\\
0&P^{-1}\end{array}\right)\left(\begin{array}[]{cc}A&B\\\
C&D\end{array}\right)\left(\begin{array}[]{cc}(P^{-1})^{T}&0\\\
0&P\end{array}\right):=\left(\begin{array}[]{cc}\tilde{A}&\tilde{B}\\\
\tilde{C}&\tilde{D}\end{array}\right)$ (4.91)

with $\tilde{D}=\left(\begin{array}[]{cc}0&0\\\ 0&R\end{array}\right)$. Since
$\left(\begin{array}[]{cc}\tilde{A}&\tilde{B}\\\
\tilde{C}&\tilde{D}\end{array}\right)$ is a symplectic matrix, we have

$\tilde{A}^{T}D-\tilde{C}^{T}\tilde{B}=I_{n}.$ (4.92)

Since $\tilde{D}=\left(\begin{array}[]{cc}0&0\\\ 0&R\end{array}\right)$,
$\tilde{B}^{T}\tilde{D}$ and $\tilde{A}^{T}\tilde{D}$ both have form

$\displaystyle\tilde{B}^{T}\tilde{D}=\left(\begin{array}[]{cc}0&*\\\
0&*\end{array}\right),\;\;\tilde{A}^{T}\tilde{D}=\left(\begin{array}[]{cc}0&*\\\
0&*\end{array}\right).$ (4.97)

So by (4.92) and (4.97) we have

$\tilde{A}^{T}\tilde{D}+\tilde{C}^{T}\tilde{B}=2\tilde{A}^{T}\tilde{D}-I_{n}=\left(\begin{array}[]{cc}-I_{r}&*\\\
0&*\end{array}\right).$ (4.98)

By direct computation and (4.97) and (4.98) we have

$\displaystyle N\left(\begin{array}[]{cc}\tilde{A}&\tilde{B}\\\
\tilde{C}&\tilde{D}\end{array}\right)^{-1}N\left(\begin{array}[]{cc}\tilde{A}&\tilde{B}\\\
\tilde{C}&\tilde{D}\end{array}\right)$ (4.103) $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}P^{T}&0\\\
0&P^{-1}\end{array}\right)N\left(\begin{array}[]{cc}A&B\\\
C&D\end{array}\right)^{-1}N\left(\begin{array}[]{cc}A&B\\\
C&D\end{array}\right)\left(\begin{array}[]{cc}(P^{-1})^{T}&0\\\
0&P\end{array}\right)$ (4.112) $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}\tilde{D}^{T}\tilde{A}+\tilde{B}^{T}\tilde{C}&2\tilde{B}^{T}\tilde{D}\\\
2\tilde{A}^{T}\tilde{C}&\tilde{A}^{T}\tilde{D}+\tilde{C}^{T}\tilde{B}\end{array}\right)$
(4.115) $\displaystyle=$ $\displaystyle\left(\begin{array}[]{cccc}*&*&0&*\\\
*&*&0&*\\\ *&*&-I_{r}&*\\\ *&*&0&*\end{array}\right)$ (4.120)

Since by (4.2) o f[35] we have

$\displaystyle{\gamma}^{2}_{x_{\tau}}(\tau)=N\left(\begin{array}[]{cc}A&B\\\
C&D\end{array}\right)^{-1}N\left(\begin{array}[]{cc}A&B\\\
C&D\end{array}\right),$ (4.125)

by (4.112) and (4.120) we have

$\displaystyle-1\in{\sigma}({\gamma}^{2}_{x_{\tau}}(\tau)),$

which contradicts to (4.82). Thus we have proved that $D$ is invertible.
Similarly we can prove $A$ is invertible, and Claim 1 is proved.

Claim 2. There exists a invertible $n\times n$ real matrix $Q$ with ${\rm
det}Q>0$ such that

$Q^{-1}(B^{T}C)Q={\rm
diag}({0,0,...,0,{\lambda}_{1},{\lambda}_{2},...{\lambda}_{n-r}})$ (4.126)

with $r=\nu_{L_{1}}({\gamma}_{x_{\tau}})$ and ${\lambda}_{i}\in(-1,0)$ for
$i=1,2,...,n-r$.

In fact

$\displaystyle{\gamma}^{2}_{x_{\tau}}(\tau)$ $\displaystyle=$ $\displaystyle
N\left(\begin{array}[]{cc}A&B\\\
C&D\end{array}\right)^{-1}N\left(\begin{array}[]{cc}A&B\\\
C&D\end{array}\right)$ (4.131) $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}D^{T}&B^{T}\\\
C^{T}&A^{T}\end{array}\right)\left(\begin{array}[]{cc}A&B\\\
C&D\end{array}\right)$ (4.136) $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}I+2B^{T}C&2B^{T}D\\\
2A^{T}C&I+2C^{T}B\end{array}\right).$ (4.139)

Since $B$ and $D$ are both invertible, for any ${\omega}\in{\bf C}$, we have

$\displaystyle\left(\begin{array}[]{cc}I_{n}&0\\\
-\frac{1}{2}(I_{n}+2C^{T}B-{\omega}I_{n})D^{-1}(B^{T})^{-1}&I_{n}\end{array}\right)\left(\begin{array}[]{cc}I+2B^{T}C-{\omega}I_{n}&2B^{T}D\\\
2A^{T}C&I+2C^{T}B-{\omega}I_{n}\end{array}\right)$ (4.144) $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}I+2B^{T}C-{\omega}I_{n}&2B^{T}D\\\
-\frac{1}{2}(I_{n}+2C^{T}B-{\omega}I_{n})D^{-1}(B^{T})^{-1}(I+2B^{T}C-{\omega}I_{n})+2A^{T}C&0\end{array}\right).$
(4.147)

So we have

$\displaystyle{\rm det}({\gamma}^{2}_{x_{\tau}}(\tau)-{\omega}I_{2n})$
$\displaystyle=$ $\displaystyle{\rm det}(B^{T}D){\rm
det}((I_{n}+2C^{T}B-{\omega}I_{n})D^{-1}(B^{T})^{-1}(I+2B^{T}C-{\omega}I_{n})-4A^{T}C)$
(4.148) $\displaystyle=$ $\displaystyle{\rm
det}(D[(I_{n}+2C^{T}B-{\omega}I_{n})D^{-1}(B^{T})^{-1}(I+2B^{T}C-{\omega}I_{n})-4A^{T}C]B^{T})$
$\displaystyle=$ $\displaystyle{\rm
det}(D[I_{n}+2C^{T}B-{\omega}I_{n})D^{-1}(B^{T})^{-1}(I+2B^{T}C-{\omega}I_{n}]B^{T}-4DA^{T}CB^{T})$
$\displaystyle=$ $\displaystyle{\rm
det}((I+2B^{T}C-{\omega}I_{n})^{2}-4(1+CB^{T})CB^{T})$ $\displaystyle=$
$\displaystyle{\rm det}({\omega}^{2}I_{n}-2{\omega}(I+2CB^{T})+I).$

By (4.81) we have

${\sigma}({\gamma}^{2}_{x_{\tau}}(\tau))\subset{\bf U}.$ (4.149)

So for ${\omega}\in{\bf U}$ by (4.148) we have

${\rm
det}({\gamma}^{2}_{x_{\tau}}(\tau)-{\omega}I_{2n})=(-4)^{n}{\omega}^{n}{\rm
det}(CB^{T}-\frac{1}{2}({\rm Re}\,{\omega}-1)).$ (4.150)

Hence by (4.81) again we have ${\sigma}(CB^{T})\subset(-1,0]$, moreover there
exists a invertible $n\times n$ matrix $S$ such that

$S^{-1}CB^{T}S={\rm
diag}(0,0,...,0,{\lambda}_{1},{\lambda}_{2},...,{\lambda}_{n-r}).$ (4.151)

with $r=\nu_{L_{1}}({\gamma}_{x_{\tau}})$ and ${\lambda}_{i}\in(-1,0)$ for
$i=1,2,...,n-r$. Since $S^{-1}CB^{T}S=(B^{T}S)^{-1}B^{T}C(B^{T}S)$, let
$Q=B^{T}S$, if ${\rm det}Q<0$ we replace it by $B^{T}S{\rm
diag}(-1,1,1,...,1)$, Claim 2 is proved.

Continue the proof of Theorem 1.1.

If ${\rm det}B>0$, there is a continuous symplectic matrix path joint
$\left(\begin{array}[]{cc}B^{-1}&0\\\ 0&B^{T}\end{array}\right)$ and $I_{2n}$.
Since

$\displaystyle\left(\begin{array}[]{cc}B^{-1}&0\\\
0&B^{T}\end{array}\right)\left(\begin{array}[]{cc}A&B\\\
C&D\end{array}\right)=\left(\begin{array}[]{cc}B^{-1}A&I_{n}\\\
B^{T}C&B^{T}D\end{array}\right).$ (4.158)

By Lemma 2.2, for ${\varepsilon}>0$ small enough, we have

${\rm sgn}M_{\varepsilon}\left(\left(\begin{array}[]{cc}A&B\\\
C&D\end{array}\right)\right)={\rm
sgn}M_{\varepsilon}\left(\left(\begin{array}[]{cc}B^{-1}A&I_{n}\\\
B^{T}C&B^{T}D\end{array}\right)\right).$ (4.159)

If ${\rm det}B<0$, there is a continuous symplectic path joint
$\left(\begin{array}[]{cc}B^{-1}&0\\\ 0&B^{T}\end{array}\right)$ and
$(-I_{2})\diamond I_{2(n-1)}$. By direct computation we have

$\displaystyle{\rm sgn}M_{\varepsilon}\left(\left(\begin{array}[]{cc}A&B\\\
C&D\end{array}\right)\right)={\rm
sgn}M_{\varepsilon}\left(\left((-I_{2})\diamond
I_{2(n-1)}\right)\left(\begin{array}[]{cc}A&B\\\
C&D\end{array}\right)\right).$ (4.164)

So by Lemma 2.2 again we have (4.159) holds. So whenever ${\rm det}(B)>0$ or
not, (4.159) always holds.

Denote by $\left(\begin{array}[]{cc}P^{T}&0\\\
0&P^{-1}\end{array}\right)\left(\begin{array}[]{cc}B^{-1}A&I_{n}\\\
B^{T}C&B^{T}D\end{array}\right)\left(\begin{array}[]{cc}P&0\\\
0&(P^{-1})^{T}\end{array}\right)=\left(\begin{array}[]{cc}\tilde{A}&I_{n}\\\
\tilde{C}&\tilde{D}\end{array}\right)$. By Claim 2, we have

$\tilde{C}={\rm
diag}(0,0,..,0,{\lambda}_{1},{\lambda}_{2},...,{\lambda}_{n-r}).$ (4.165)

Since $\left(\begin{array}[]{cc}\tilde{A}&I_{n}\\\
\tilde{C}&\tilde{D}\end{array}\right)$ is a symplectic matrix, we have
$\tilde{A}$ and $\tilde{D}$ are both symmetric and have the follow forms:

$\displaystyle\tilde{A}=\left(\begin{array}[]{cc}A_{11}&0\\\
0&A_{22}\end{array}\right),\quad\tilde{D}=\left(\begin{array}[]{cc}D_{11}&0\\\
0&D_{22}\end{array}\right),$ (4.170)

where $A_{11}$ and $D_{11}$ are $r\times r$ invertible matrices, $A_{22}$ and
$D_{22}$ are $(n-r)\times(n-r)$ invertible matrices. So we have

$\left(\begin{array}[]{cc}\tilde{A}&I_{n}\\\
\tilde{C}&\tilde{D}\end{array}\right)=\left(\begin{array}[]{cc}A_{11}&I_{r}\\\
0&D_{11}\end{array}\right)\diamond\left(\begin{array}[]{cc}A_{22}&I_{n-r}\\\
{\Lambda}&D_{22}\end{array}\right),$ (4.171)

where ${\Lambda}={\rm diag}({\lambda}_{1},{\lambda}_{2},...,{\lambda}_{n-r})$.

Since $N\left(\begin{array}[]{cc}A_{11}&I_{r}\\\
0&D_{11}\end{array}\right)^{-1}N\left(\begin{array}[]{cc}A_{11}&I_{r}\\\
0&D_{11}\end{array}\right)=\left(\begin{array}[]{cc}I_{r}&2D_{11}\\\
0&I_{r}\end{array}\right)$, by (4.81) $D_{11}$ is negative definite. So we can
joint it to $-I_{r}$ by a invertible symmetric matrix path. Then by Lemma 2.2,
Remark 2.1, and computations below Remark 2.1 in Section 2, we have

$\displaystyle{\rm
sgn}M_{\varepsilon}\left(\left(\begin{array}[]{cc}A_{11}&I_{r}\\\
0&D_{11}\end{array}\right)\right)$ $\displaystyle=$ $\displaystyle{\rm
sgn}M_{\varepsilon}\left(\left(\begin{array}[]{cc}-I_{r}&I_{r}\\\
0&-I_{r}\end{array}\right)\right)$ $\displaystyle=$ $\displaystyle r\,{\rm
sgn}M_{\varepsilon}(N_{1}(-1,1))$ $\displaystyle=$ $\displaystyle 2r.$ (4.177)

Since $M_{\varepsilon}\left(\left(\begin{array}[]{cc}A_{22}&I_{n-r}\\\
{\Lambda}&D_{22}\end{array}\right)\right)$ is invertible for
${\varepsilon}=0$, for ${\varepsilon}>0$ small enough, we have

$\displaystyle{\rm
sgn}M_{\varepsilon}\left(\left(\begin{array}[]{cc}A_{22}&I_{n-r}\\\
{\Lambda}&D_{22}\end{array}\right)\right)={\rm
sgn}M_{0}\left(\left(\begin{array}[]{cc}A_{22}&I_{n-r}\\\
{\Lambda}&D_{22}\end{array}\right)\right)$ (4.182) $\displaystyle=$
$\displaystyle{\rm sgn}\left\\{\left(\begin{array}[]{cc}A_{22}&{\Lambda}\\\
I_{n-r}&D_{22}\end{array}\right)\left(\begin{array}[]{cc}0&-I_{n-r}\\\
-I_{n-r}&0\end{array}\right)\left(\begin{array}[]{cc}A_{22}&I_{n-r}\\\
{\Lambda}&D_{22}\end{array}\right)+\left(\begin{array}[]{cc}0&I_{n-r}\\\
I_{n-r}&0\end{array}\right)\right\\}$ (4.191) $\displaystyle=$
$\displaystyle{\rm
sgn}\left\\{2\left(\begin{array}[]{cc}-A_{22}{\Lambda}&-{\Lambda}\\\
-{\Lambda}&-D_{22}\end{array}\right)\right\\}$ (4.194) $\displaystyle=$
$\displaystyle{\rm sgn}\left(\begin{array}[]{cc}-A_{22}{\Lambda}&-{\Lambda}\\\
-{\Lambda}&-D_{22}\end{array}\right).$ (4.197)

Since $\left(\begin{array}[]{cc}A_{22}&I_{n-r}\\\
{\Lambda}&D_{22}\end{array}\right)$ is a symplectic matrix, we have

$\displaystyle A_{22}D_{22}-{\Lambda}=I_{n-r},$ $\displaystyle
A_{22}{\Lambda}={\Lambda}A_{22}.$ (4.198)

Hence

$\displaystyle
A_{22}^{-1}{\Lambda}-D_{22}=A_{22}^{-1}({\Lambda}-A_{22}D_{22})=-A_{22}^{-1}.$

So we have

$\displaystyle\left(\begin{array}[]{cc}I_{n-r}&0\\\
-A_{22}^{-1}&I_{n-r}\end{array}\right)\left(\begin{array}[]{cc}-A_{22}{\Lambda}&-{\Lambda}\\\
-{\Lambda}&-D_{22}\end{array}\right)\left(\begin{array}[]{cc}I_{n-r}&-A_{22}^{-1}\\\
0&I_{n-r}\end{array}\right)$ (4.205) $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}-A_{22}{\Lambda}&0\\\
0&A_{22}^{-1}{\Lambda}-D_{22}\end{array}\right)$ (4.208) $\displaystyle=$
$\displaystyle\left(\begin{array}[]{cc}-A_{22}{\Lambda}&0\\\
0&-A_{22}^{-1}\end{array}\right).$ (4.211)

By (4.198), there exist invertible matrix $R$ such that

$\displaystyle R^{-1}A_{22}R={\rm
diag}(\alpha_{1},\alpha_{2},...,\alpha_{n-r}),\quad\alpha_{i}\in{\bf
R}\setminus\\{0\\},\;i=1,2,...,n-r,$ (4.212) $\displaystyle
R^{-1}{\Lambda}R={\rm
diag}({\lambda}_{i_{1}},{\lambda}_{i_{2}},...,{\lambda}_{i_{n-r}}),\;\;\\{i_{1},i_{2},...,i_{n-r}\\}=\\{1,2,...,n-r\\}.$
(4.213)

So we have

$\displaystyle R^{-1}(-A_{22}{\Lambda})R={\rm
diag}(-{\lambda}_{i_{1}}\alpha_{1},-{\lambda}_{i_{2}}\alpha_{2},...,-{\lambda}_{i_{n-r}}\alpha_{n-r}),$
(4.214) $\displaystyle R^{-1}(-A_{22}^{-1})R={\rm
diag}(-\frac{1}{\alpha_{1}},-\frac{1}{\alpha_{2}},...,\frac{1}{\alpha_{n-r}}).$
(4.215)

Since ${\lambda}_{i}\in(-1,0)$ for $i=1,2,...,n-r$, by (4.212)-(4.215) we have

${\rm sgn}(-A_{22}{\Lambda})+{\rm sgn}(-A_{22}^{-1})=0.$ (4.216)

Hence by (4.197), (4.211) and (4.216) we have

$\displaystyle{\rm
sgn}M_{\varepsilon}\left(\left(\begin{array}[]{cc}A_{22}&I_{n-r}\\\
{\Lambda}&D_{22}\end{array}\right)\right)={\rm sgn}(-A_{22}{\Lambda})+{\rm
sgn}(-A_{22}^{-1})=0.$ (4.219)

Since ${\rm det}Q>0$ we can joint it to $I_{n}$ by a invertible matrix path.
Hence by Lemma 2.2 and Remark 2.1, (4.171), (4.177) and (4.219), we have

$\displaystyle{\rm
sgn}M_{\varepsilon}\left(\left(\begin{array}[]{cc}B^{-1}A&I_{n}\\\
B^{T}C&B^{T}D\end{array}\right)\right)$ $\displaystyle=$ $\displaystyle{\rm
sgn}M_{\varepsilon}\left(\left(\begin{array}[]{cc}A_{11}&I_{r}\\\
0&D_{11}\end{array}\right)\right)+{\rm
sgn}M_{\varepsilon}\left(\left(\begin{array}[]{cc}A_{22}&I_{n-r}\\\
{\Lambda}&D_{22}\end{array}\right)\right)$ $\displaystyle=$ $\displaystyle
2r+0$ $\displaystyle=$ $\displaystyle 2r.$ (4.227)

Then by Theorem 2.3, (4.159) and (4.227) we have

$i_{L_{0}}({\gamma}_{x_{\tau}})-i_{L_{1}}({\gamma}_{x_{\tau}})=r.$ (4.228)

However by (4.77), (4.79) and $\nu_{L_{1}}({\gamma}_{x_{\tau}})=r$ we have

$\displaystyle
i_{L_{0}}({\gamma}_{x_{\tau}})-i_{L_{1}}({\gamma}_{x_{\tau}})=r-1,$

which contradicts to (4.228).

Thus we have prove that $k$ can not be $3$. So if $k$ is odd, it must be $1$.
By the same proof we have if $k$ is even, it must be $2$. Then
$\tau\in\\{T,\frac{T}{2}\\}$. The proof of Theorem 1.1 is complete.

Proof of Corollary 1.2. Since $0<T<\frac{\pi}{||B_{0}||}$, there is
$\varepsilon>0$ small enough such that

$\displaystyle 0\leq
B_{0}\leq||B_{0}||I_{2n}<(\frac{\pi}{T}-\varepsilon)I_{2n}.$

It is easy to see that

$\displaystyle{\gamma}_{(\frac{\pi}{T}-\varepsilon)I_{2n}}(t)={\rm
exp}(({\frac{\pi}{T}-\varepsilon})tJ)\quad\forall t\in[0,\frac{T}{2}].$

So we have

$\displaystyle\nu_{L_{0}}({\gamma}_{(\frac{\pi}{T}-\varepsilon)I_{2n}})=0,$
$\displaystyle i_{L_{0}}((\frac{\pi}{T}-\varepsilon)I_{2n})=0.$

Then by (5.40) and Lemma 3.1 and Corollary 3.1 we have

$\displaystyle 0\leq i_{-1}(B_{0})+\nu_{-1}(B_{0})\leq
i_{-1}((\frac{\pi}{T}-\varepsilon)I_{2n})=0.$

So we have

$\displaystyle i_{-1}(B_{0})+\nu_{-1}(B_{0})=0.$

Hence by the same proof of Theorem 1.1, the conclusions of Corollary 1.2
holds.

Remark 4.1. Under the same conditions of Theorem 1.3, if
$\int_{0}^{\frac{T}{2}}H^{\prime\prime}_{22}(x_{T}(t))\,dt>0$, by the same
proof of Theorem 1.1, we have

$\displaystyle\tau\geq\frac{T}{2(i_{L_{0}}(B_{0})+\nu_{L_{0}}(B_{0}))+2}.$

Moreover, if $0<T<\frac{\pi}{||B_{0}||}$ or
$i_{L_{0}}(B_{0})+\nu_{L_{0}}(B_{0})=0$, we have $\tau\in\\{T,\frac{T}{2}\\}$.

Proof of Theorem 1.2. This is the case $n=1$ and $B_{0}=0$ of Theorem 1.3, by
the proof Theorem 1.3, for any $T>0$ we obtain an T-periodic brake solution
$x_{T}$ satisfies

$\displaystyle i_{L_{0}}({\gamma}_{x_{T}})\leq 1.$ (4.229)

If it’s minimal period is $\tau=T/k$ for some $k\in{\bf N}$, we denote
$x_{\tau}=x_{T}|_{[0,\tau]}$. Then by the proof of Theorem 1.3 we have

$i_{1}({\gamma}_{x\tau}^{2})+\nu_{1}({\gamma}_{{x_{\tau}}}^{2})\geq 2.$
(4.230)

In the following we prove Theorem 1.2 in 2 steps.

Step 1. For $k=2p+1$ for some $p\geq 0$, we prove that $p=0$.

Firstly by the proof of Theorem 1.3 we have

$1\geq i_{L_{0}}({\gamma}_{{x_{\tau}}}^{2p+1})\geq
p(i_{1}({\gamma}_{{x_{\tau}}}^{2})+\nu_{1}({\gamma}_{{x_{\tau}}}^{2})-1)+i_{L_{0}}({\gamma}).$
(4.231)

We divide the argument into three cases.

Case 1. $i_{1}({\gamma}_{x\tau}^{2})+\nu_{1}({\gamma}_{{x_{\tau}}}^{2})=2$. If
$\nu_{1}({\gamma}_{{x_{\tau}}}^{2})=1$, then
$i_{1}({\gamma}_{{x_{\tau}}}^{2})=1\in 2{\bf Z}+1$. By Lemma 6.3, we have
$N_{1}(1,1)\in{\Omega}^{0}({\gamma}_{{x_{\tau}}}^{2}(\tau))$. Since

$1=i_{1}({\gamma}_{{x_{\tau}}}^{2})=i_{L_{0}}({\gamma}_{{x_{\tau}}})+i_{L_{1}}({\gamma}_{x\tau})+1.$
(4.232)

By Corollary 2.1 we have

$|i_{L_{0}}({\gamma}_{{x_{\tau}}})-i_{L_{1}}({\gamma}_{{x_{\tau}}})|\leq 1.$
(4.233)

Then by (4.232) and (4.233) we have

$i_{L_{0}}({\gamma}_{{x_{\tau}}})=i_{L_{1}}({\gamma}_{{x_{\tau}}})=0.$ (4.234)

So by Theorem 2.1, Lemma 6.2, and (6.21), we have

$\displaystyle i_{L_{0}}({\gamma}_{{x_{\tau}}}^{3})$ $\displaystyle=$
$\displaystyle
i_{L_{0}}({\gamma}_{{x_{\tau}}})+i_{e^{2\pi\sqrt{-1}/3}}({\gamma}_{x\tau}^{2})$
(4.235) $\displaystyle=$ $\displaystyle
i_{L_{0}}({\gamma}_{{x_{\tau}}})+i_{1}({\gamma}_{{x_{\tau}}}^{2})+S_{N_{1}(1,1)}(1)$
$\displaystyle=$ $\displaystyle 0+1+1$ $\displaystyle=$ $\displaystyle 2>1\geq
i_{L_{0}}({\gamma}_{x\tau}^{2p+1}).$

Then by Theorem 3.3 we have

$\displaystyle 2p+1<3.$

Hence $p=0$.

If $\nu_{1}({\gamma}_{{x_{\tau}}}^{2})=2$, then
$i_{1}({\gamma}_{{x_{\tau}}}^{2})=0$. But now
${\gamma}_{{x_{\tau}}}^{2}(\tau)=I_{2}$, by Lemma 6.3
$i_{1}({\gamma}_{{x_{\tau}}}^{2})\in 2{\bf Z}+1$, which yields a
contradiction. So this case can not happen. So in Case 1, we have proved
$p=0$.

Case 2. $i_{1}({\gamma}_{x\tau}^{2})+\nu_{1}({\gamma}_{{x_{\tau}}}^{2})=3$.

If $\nu_{1}({\gamma}_{{x_{\tau}}}^{2})=1$, then

$\displaystyle i_{1}({\gamma}_{x\tau}^{2})=2\in 2{\bf Z}.$ (4.236)

By Lemma 6.3 we have
$N_{1}(1,-1)\in{\Omega}^{0}(({\gamma}_{{x_{\tau}}}^{2})(\tau))$. So if $p\geq
1$, by Theorem 3.3, Theorem 2.1, Lemma 6.2 and (6.21), we have we have

$\displaystyle 1\geq i_{L_{0}}({\gamma}_{{x_{\tau}}}^{2p+1})$
$\displaystyle\geq$ $\displaystyle i_{L_{0}}({\gamma}_{{x_{\tau}}}^{3})$
(4.237) $\displaystyle=$ $\displaystyle
i_{L_{0}}({\gamma}_{{x_{\tau}}})+i_{e^{2\pi\sqrt{-1}/3}}({\gamma}_{{x_{\tau}}}^{2})$
$\displaystyle=$ $\displaystyle
i_{L_{0}}({\gamma}_{{x_{\tau}}})+i_{1}({\gamma}_{x\tau}^{2})+S_{N_{1}(1,-1)}(1)$
$\displaystyle\geq$ $\displaystyle-1+2+0$ $\displaystyle=$ $\displaystyle 1.$

So there must hold

$\displaystyle i_{L_{0}}({\gamma}_{x\tau})=-1.$

Then by Corollary 2.1 we have

$\displaystyle i_{L_{1}}({\gamma}_{{x_{\tau}}})\leq 0.$

So we have

$\displaystyle
i_{1}({\gamma}_{{x_{\tau}}}^{2})=i_{L_{0}}({\gamma}_{{x_{\tau}}}+i_{L_{1}}({\gamma}_{{x_{\tau}}})+1\leq-1+0+1=0,$

which contradicts (4.236). Thus we have $p=0$.

If $\nu_{1}({\gamma}_{{x_{\tau}}}^{2})=2$, then

$i_{1}({\gamma}_{{x_{\tau}}}^{2})=1,\quad{\gamma}_{x_{\tau}}^{2}(\tau)=I_{2}.$
(4.238)

If $p\geq 1$, by Theorem 3.3, Theorem 2.1, Corollary 3.2, Lemma 6.2 and
(6.21), we have we have

$\displaystyle 1\geq i_{L_{0}}({\gamma}_{{x_{\tau}}}^{2p+1})$
$\displaystyle\geq$ $\displaystyle i_{L_{0}}({\gamma}_{{x_{\tau}}}^{2+1})$
$\displaystyle=$ $\displaystyle
i_{L_{0}}({\gamma}_{{x_{\tau}}})+i_{e^{2\pi\sqrt{-1}/3}}({\gamma}_{{x_{\tau}}}^{2})$
$\displaystyle=$ $\displaystyle
i_{L_{0}}({\gamma}_{{x_{\tau}}})+i_{1}({\gamma}_{{x_{\tau}}}^{2})+S_{I_{2}}(1)$
$\displaystyle\geq$ $\displaystyle-1+1+1$ $\displaystyle=$ $\displaystyle 1.$

So there must hold

$\displaystyle i_{L_{0}}({\gamma}_{x\tau})=-1.$

Then by Corollary 2.1 we have

$\displaystyle i_{L_{1}}({\gamma}_{{x_{\tau}}})\leq 0.$

So we have

$\displaystyle
i_{1}({\gamma}_{x\tau}^{2})=i_{L_{0}}({\gamma}_{{x_{\tau}}}+i_{L_{1}}({\gamma}_{{x_{\tau}}})+1\leq-1+0+1=0,$

which contradicts (4.238). Thus we have $p=0$.

Case 3. $i_{1}({\gamma}_{{x_{\tau}}}^{2})+\nu_{1}({\gamma}_{x_{\tau}}^{2})\geq
4$.

In this case
$i_{1}({\gamma}_{{x_{\tau}}}^{2})+\nu_{1}({\gamma}_{{x_{\tau}}}^{2})-1\geq 3$.
By Corollary 3.2 we have $i_{L_{0}}\geq-1$. So by (4.231) we have

$p\leq 2/3,$ (4.239)

which yields $p=0$. So we finish Step 1.

Step 2. For $k=2p+2$ for some $p\geq 0$, we prove that $p=0$.

In fact, apply Bott-type iteration formula of Theorem 2.1 to the the case of
the iteration time equals to 4 and note that by Corollary 3.1
$i_{\sqrt{-1}}({\gamma}_{{x_{\tau}}})\geq 0$. Then by the same argument of
Step 1, we can prove that $p=0$.

Thus by Steps 1 and 2, Theorem 1.2 is proved.

A natural question is that can we prove the minimal period is $T$ in this way?
We have the following remark.

Remark 4.2. Only use the Maslov-type index theory to estimate the iteration
time of the $T$-periodic brake solution $x_{T}$ obtained by the first 4 steps
in the proof of Theorem 1.3 with $B_{0}=0$, we can not hope to prove $T$ is
the minimal period of $x_{T}$. Even $H^{\prime\prime}(z)>0$ for all $z\in{\bf
R}^{2n}\setminus\\{0\\}$. For $n=1$ and $T=4\pi$, we can not exclude the
following case:

$\displaystyle x_{T}(t)=\left(\begin{array}[]{c}\sin t\\\ \cos
t\end{array}\right),$ (4.242) $\displaystyle H^{\prime}(x_{T}(t))=x_{T}(t),$
$\displaystyle H^{\prime\prime}(x_{T}(t))\equiv I_{2n}.$

It is easy to check that ${\gamma}_{x_{T}}(t)=R(t)$ for $t\in[0,2\pi]$. Hence
by Lemma 5.1 of [30] or the proof of Lemma 3.1 of [42] we have

$\displaystyle
i_{L_{0}}({\gamma}_{x_{T}})=\sum_{0<s<2\pi}\nu_{L_{0}}({\gamma}_{x_{T}})(s)=1.$

In this case the minimal period of $x_{T}$ is $\frac{T}{2}$. Similarly for
$n>1$ we can construct examples to support this remark.

##  5 Proof of Theorems 1.4-1.5 and Corollary 1.4

In this section we study the minimal period problem for symmeytric brake orbit
solutions of the even reversible Hamiltonian system (1.1) and complete the
proof of Theorems 1.4-1.5 and Corollary 1.4.

For $T>0$, let $E_{T}=\\{x\in W^{1/2,2}(S_{\tau},{\bf
R}^{2n})|\,x(-t)=Nx(t)\;a.e.\;t\in{\bf R}\\}$ with the usual $W^{1/2,2}$ norm
and inner product. Correspondingly $\hat{E}$ and $\tilde{E}$ are defined to be
the symmetric ones and the $\frac{T}{2}$-periodic ones in $E_{T}$
respectively. Also $\\{P_{T,m}\\}$ and $\\{\hat{P}_{m}\\}$ are the Galerkin
approximation scheme w.r.t. $A_{T}$ and $\hat{A}$ respectively, where
$\\{P_{T,m}\\}$, $\\{\hat{P}_{m}\\}$, $A_{T}$, and $\hat{A}$ are defined by
the same way as in Section 2, we only need to replace $\tau$ by $T$.

For $z\in E_{T}$, we define

$f(z)=\frac{1}{2}\langle A_{T}z,z\rangle-\int_{0}^{T}H(z)dt.$ (5.1)

For $z\in\hat{E}$, we define

$\hat{f}(z)=\frac{1}{2}\langle\hat{A}z,z\rangle-\int_{0}^{T}H(z)dt.$ (5.2)

We have the following lemma.

Lemma 5.1. Let $z\in\hat{E}$. If $\hat{f}^{\prime}(z)=0$, then
$f^{\prime}(z)=0$.

Proof. Let $z\in\hat{E}$ and $\hat{f}^{\prime}(z)=0$. So for any $y\in\hat{E}$
we have

$\langle\hat{f}^{\prime}(z),y\rangle=\int_{0}^{T}J\dot{z}(t)\cdot
y(t)\,dt-\int_{0}^{T}H^{\prime}(z(t))\cdot y(t)\,dt=0,\quad\forall
y\in\hat{E}.$ (5.3)

Since $H$ is even and $z\in\hat{E}$, we have

$H^{\prime}(z(t+\frac{T}{2}))=H^{\prime}(-z(t))=-H^{\prime}(z(t)).$ (5.4)

So $H^{\prime}(z)\in\hat{E}$ and

$\langle f^{\prime}(z),y\rangle=\int_{0}^{T}J\dot{z}(t)\cdot
y(t)\,dt-\int_{0}^{T}H^{\prime}(z(t))\cdot y(t)\,dt=0,\quad\forall
y\in\tilde{E}.$ (5.5)

By (5.4) and (5.5), we have

$\langle f^{\prime}(z),y\rangle=\int_{0}^{T}J\dot{z}(t)\cdot
y(t)\,dt-\int_{0}^{T}H^{\prime}(z(t))\cdot y(t)\,dt=0,\quad\forall y\in
E_{T}.$ (5.6)

Hence $f^{\prime}(z)=0$

By Lemma 5.1 and arguments in the proof of Theorem 1.3 in Section 4, to look
for the $T$-period symmetric solutions of (1.1) is equivalent to look for
critical points of $\hat{f}$.

Proof of Theorem 1.5. For any given $T>0$, we prove the existence of
$T$-periodic symmetric brake orbit solution of (1.1) whose minimal period
satisfies the inequalities in the conclusion of Theorem 1.5. Since the proof
of existence of $T$-periodic symmetric brake orbit solution $x_{T}$ of (1.1)
is similar to that of the proof of Theorem 1.3, we will only give the sketch.
We divide the proof into several steps.

Step 1. Similarly as Step 1 in the proof of Theorem 1.3, for any $K>0$ we can
truncate the function $\hat{H}$ suitably and evenly to $\hat{H}_{K}$ such that
it satisfies the growth condition (4.4). Correspondingly we obtain a new even
and reversible function $H_{K}$ satisfies condition (4.4).

Set

$\hat{f}_{K}(z)=\frac{1}{2}\langle\hat{A}z,z\rangle-\int_{0}^{T}H_{K}(z)dt,\qquad\forall
z\in\hat{E}.$ (5.7)

Then $\hat{f}_{K}\in C^{2}(\hat{E},{\bf R})$ and

$\hat{f}_{K}(z)=\frac{1}{2}\langle(\hat{A}-\hat{B}_{0})z,z\rangle-\int_{0}^{T}\hat{H}_{K}(z)dt,\qquad\forall
z\in\hat{E},$ (5.8)

where $\hat{B}_{0}$ is the selfadjoint linear compact operator on $\hat{E}$
defined by

$\langle\hat{B}_{0}z,z\rangle=\int_{0}^{T}B_{0}z(t)\cdot z(t)\,dt.$ (5.9)

Step 2. For $m>0$, let $\hat{f}_{Km}=\hat{f}|\hat{E}_{m}$, where
$\hat{E}_{m}=\hat{P}_{m}\hat{E}$. Set

$\displaystyle X_{m}=M^{-}(\hat{P}_{m}(\hat{A}-\hat{B}_{0})\hat{P}_{m})\oplus
M^{0}(\hat{P}_{m}(\hat{A}-\hat{B}_{0})\hat{P}_{m}),$ $\displaystyle
Y_{m}=M^{+}(\hat{P}_{m}(\hat{A}-\hat{B}_{0})\hat{P}_{m}).$

By the same argument of Step 2 in the proof of Theorem 1.3, we can show that
$\hat{f}_{K}m$ satisfies the hypotheses of Theorem 4.1. Moreover, we obtain a
critical point $x_{Km}$ of $\hat{f}_{Km}$ with critical value $C_{K}m$ which
satisfies

$\displaystyle m^{-}(x_{Km})\leq\dim X_{m}+1.$ (5.10)

and

$\delta\leq C_{Km}\leq\frac{1}{2}||\hat{A}-\hat{B}_{0}||r_{1}^{2},$ (5.11)

where $\delta$ is a positive number depending on $K$ and $r_{1}>0$ is
independent of $K$ and $m$.

Step 3. We prove that there exists a symmetric $T$-periodic brake orbit
solution $x_{T}$ of (1.1) which satisfies

$i_{\sqrt{-1}}^{L_{0}}({\gamma}_{x_{T}})\leq
i_{\sqrt{-1}}^{L_{0}}(B_{0})+\nu_{\sqrt{-1}}^{L_{0}}(B_{0})+1.$ (5.12)

From the proof of Theorem 1.3 we have $f_{K}$ satisfies $(PS)_{c}^{*}$
condition for $c\in{\bf R}$, by the same proof of Lemma 5.1, we have
$\hat{f}_{K}$ satisfies $(PS)_{c}^{*}$ condition for $c\in{\bf R}$, i.e., any
sequence ${z_{m}}$ such that $z_{m}\in\hat{E}_{m}$,
$\hat{f}_{Km}^{\prime}(z_{m})\to 0$ and $\hat{f}_{Km}(z_{m})\to c$ possesses a
convergent subsequence in $\hat{E}$. Hence in the sense of subsequence we have

$x_{Km}\to
x_{K},\qquad\hat{f}_{K}(x_{K})=c_{K},\qquad\hat{f}^{\prime}_{K}(x_{K})=0.$
(5.13)

By similar argument as in [44], $x_{K}$ is a classical nonconstant symmetric
$T$-periodic solution of

$\dot{x}=JH_{K}^{\prime}(x),\quad x\in{\bf R}^{2n}.$ (5.14)

Set $B_{K}(t)=H^{\prime\prime}_{K}(x_{K}(t))$, Then $B_{K}\in
C(S_{T/2},\mathcal{L}_{s}({\bf R}^{2n}))$. Let $\hat{B}_{K}$ be the operator
defined by the same way of the definition of $\hat{B}_{0}$. It is easy to show
that

$||\hat{f}^{\prime\prime}(z)-(\hat{A}-\hat{B}_{K})||\to 0\qquad{\rm
as}\;\;||z-x_{K}||\to 0.$ (5.15)

So for $0<d<\frac{1}{4}||(A_{T}-B_{K_{T}})^{\\#}||^{-1}$, there exists
$r_{2}>0$ such that

$||\hat{f}_{Km}^{\prime\prime}(z)-\hat{P}_{m}(\hat{A}-\hat{B}_{K})\hat{P}_{m}||\leq||\hat{f}^{\prime\prime}(z)-(\hat{A}-\hat{B}_{K})||<\frac{1}{2}d,\qquad\forall
z\in\\{z\in\hat{E}:||z-x_{K}||\leq r_{2}\\}.$ (5.16)

Then for $z\in\\{z\in\hat{E}:||z-x_{K}||\leq r_{2}\\}\cap\hat{E}_{m}$,
$\forall u\in
M^{-}_{d}(\hat{P}_{m}(\hat{A}-\hat{B}_{T})\hat{P}_{m})\setminus\\{0\\}$, we
have

$\displaystyle\langle\hat{f}_{Km}^{\prime\prime}(z)u,u\rangle$
$\displaystyle\leq$
$\displaystyle\langle\hat{P}_{m}(\hat{A}-\hat{B}_{K})\hat{P}_{m}u,u\rangle+\|\hat{f}_{Km}^{\prime\prime}(z)-\hat{P}_{m}(\hat{A}-\hat{B}_{K})\hat{P}_{m}\|\|u\|^{2}$
$\displaystyle\leq$ $\displaystyle-\frac{1}{2}d\|u\|^{2}.$

So we have

$m^{-}(\hat{f}_{Km}^{\prime\prime}(z))\geq\dim
M^{-}_{d}(\hat{P}_{m}(\hat{A}-\hat{B}_{K})\hat{P}_{m}).$ (5.17)

By Theorem 3.1, Remark 3.1, there is $m^{*}>0$ such that for $m\geq m^{*}$ we
have

$\displaystyle\dim
X_{m}=mn+i_{\sqrt{-1}}^{L_{0}}(B_{0})+\nu_{\sqrt{-1}}^{L_{0}}(B_{0}),$ (5.18)
$\displaystyle\dim
M^{-}_{d}(\hat{P}_{m}(\hat{A}-\hat{B}_{K})\hat{P}_{m})=mn+i_{\sqrt{-1}}^{L_{0}}(B_{K}).$
(5.19)

Then by (5.10), (5.13), and (5.17)-(5.19), we have

$i_{\sqrt{-1}}^{L_{0}}(B_{K})\leq
i_{\sqrt{-1}}^{L_{0}}(B_{0})+\nu_{\sqrt{-1}}^{L_{0}}(B_{0})+1.$ (5.20)

By the similar argument as in the section 6 of [44], there is a constant
$M_{3}$ independent of $K$ such that $||x_{K}||_{\infty}\leq M_{3}$. Choose
$K>M_{3}$. Then $x_{K}$ is a non-constant symmetric $T$-periodic brake orbit
solution of the problem (1.1). From now on in the proof of Theorem 1.2, we
write $B=B_{K}$ and $x_{T}=x_{K}$. Then $x_{T}$ is a non-constant symmetric
$T$-periodic solution of the problem (1.1), and $B$ satisfies

$i_{\sqrt{-1}}^{L_{0}}({\gamma}_{x_{T}})=_{\sqrt{-1}}^{L_{0}}(B)\leq
i_{\sqrt{-1}}^{L_{0}}(B_{0})+\nu_{\sqrt{-1}}^{L_{0}}(B_{0})+1.$ (5.21)

Step 4. Finish the proof of Theorem 1.5.

Since $x_{T}$ obtained in Step 3 is a nonconstant and symmetric $T$-period
brake orbit solution, its minimal period $\tau=\frac{T}{4r+s}$ for some
nonnegative integer $r$ and $s=1$ or $s=3$. We now estimate $r$.

We denote by $x_{\tau}=x_{T}|_{[0,\tau]}$, then it is a symmetric period
solution of (1.1) with the minimal $\tau$ and $X_{T}=x_{\tau}^{4r+s}$ being
the $4r+s$ times iteration of $x_{\tau}$. As in Section 1, let
${\gamma}_{x_{T}}$ and ${\gamma}_{x_{\tau}}$ the symplectic path associated to
$(\tau,x)$ and $(T,x_{T})$ respectively. Then ${\gamma}_{x_{\tau}}\in
C([0,\frac{\tau}{4}],{\rm Sp}(2n))$ and ${\gamma}_{x_{T}}\in
C([0,\frac{T}{4}],{\rm Sp}(2n))$. Also we have
${\gamma}_{x_{T}}={\gamma}_{x_{\tau}}^{4r+s}$, which is the $4r+s$ times
iteration of ${\gamma}_{x_{\tau}}$.

By (5.21) we have

$i_{\sqrt{-1}}^{L_{0}}({\gamma}_{x_{\tau}}^{4r+s})\leq
i_{\sqrt{-1}}^{L_{0}}(B_{0})+\nu_{\sqrt{-1}}^{L_{0}}(B_{0})+1.$ (5.22)

Since $x_{\tau}$ is also a nonconstant symmetric periodic solution of (1.1).
It is clear that

$\displaystyle\nu_{-1}(x_{\tau}^{2})$ $\displaystyle\geq$ $\displaystyle 1.$
(5.23)

Since $\hat{H}$ satisfies condition (H5) and $B_{0}$ is semipositive, by
Corollary 3.1 of [51] (also by Theorem 6.2) we have

$i_{-1}({\gamma}_{x_{\tau}}^{2})\geq 0.$ (5.24)

By Corollary 3.2 of [51] (cf. aslo [29]), we have

$i_{1}({\gamma}_{x_{\tau}}^{2})+\nu_{1}({\gamma}_{x_{\tau}}^{2})\geq n.$
(5.25)

It is easy to see that

${\gamma}_{x_{\tau}}^{4}(\frac{\tau}{2}+t)={\gamma}_{x_{\tau}}^{2}(t)\,{\gamma}_{x_{\tau}}^{2}(\frac{\tau}{2}),\qquad\forall
t\in[0,\frac{\tau}{2}].$ (5.26)

So by Theorem 6.1 of Bott-type iteration formula we have

$\displaystyle
i_{1}({\gamma}_{x_{\tau}}^{4})+\nu_{1}({\gamma}_{x_{\tau}}^{4})$
$\displaystyle=$ $\displaystyle