Persistence of Degenerate Lower Dimensional Invariant Tori in Reversible Systems with Bruno Non-degeneracy Conditions

Abstract

In this paper we consider a class of degenerate reversible systems with Bruno non-degeneracy conditions, and prove the persistence of a lower dimensional invariant torus, whose frequency vector is only a small dilation of the prescribed one.

Appendix

1.1 Solving the Homological Equations

Recall that $N=N_0+N_l+N_h$, where $N_0=((1+\nu )\omega _0,0,0,0)$, $|\nu |<1/2$ and

$$\begin{aligned} N_l=(My+\tilde{e}v,~0,~e^Ty+bv,~au),~~N_h=\sum _{2\le |\beta |\le 2d+1}N_\beta \bar{w}^\beta . \end{aligned}$$

Moreover, $N=(N^1,0,N^3,N^4)$. Suppose that $a>0,~b>0$ and $\omega _0$ is a Diophantine vector with (1.4) holding. Let $P_L(w)=\sum _{|\beta |\le 2d+1} P_{\beta }(x){\bar{w}}^\beta$ and

$$\begin{aligned} \Vert P_L\Vert _{s,r }\le \varepsilon ,~\Vert P_L\Vert _{s,\varepsilon }\le \varepsilon ^{2d+2}. \end{aligned}$$

Moreover, N,  P are reversible vector-fields with respect to $G: (x,y, u, v) \rightarrow (-x,y,-u,v)$,

$$\begin{aligned} N(\bar{w})=-{\mathcal {D}}G\cdot N(G(\bar{w})),~P(w)=-{\mathcal {D}}G\cdot P(G(w)). \end{aligned}$$

The following two lemmas are used to solve the homological equations (3.13) and (3.14), respectively.

Lemma A.1

Consider the above N and P. There exist

$$\begin{aligned} W(w)=\sum _{|\beta |\le 2d+1} W_{\beta }(x){\bar{w}}^\beta \ \text{ and } \ \tilde{R}(w)=\sum _{2d+2\le |\beta |\le 4d+1} \tilde{R}_{\beta }(x){\bar{w}}^\beta \end{aligned}$$

such that

$$\begin{aligned} -{\mathcal {D}}W\cdot N+{\mathcal {D}}N\cdot W+P_L=[P_L]+\tilde{R} \end{aligned}$$

(A.1)

with $[W]=[\tilde{R}]=0$ and

$$\begin{aligned} \Vert W\Vert _{s-\sigma ,r}\lessdot \eta ,~~\Vert W\Vert _{s-\sigma ,\varepsilon }\lessdot \ \eta \varepsilon ^{2d+1} , \ \ \Vert \tilde{R}\Vert _{s-\sigma , r }\lessdot \eta /\sigma , \end{aligned}$$

where $\eta =\frac{\varepsilon }{\alpha ^{\mu }\sigma ^{\kappa }}\le 1,$ where

$$\begin{aligned} \mu =(d+1)(4d+6),~ \kappa =\mu \tau +(d+1)(3d+5)n+3(2d+1). \end{aligned}$$

Moreover, $W(w)={\mathcal {D}}G\cdot W(G(w)).$

Lemma A.2

There exist $Z=(0,S,0,0)$ and $\hat{N}=(\hat{N}^1,0, \hat{N}^3,\hat{N}^4)$ satisfying

$$\begin{aligned} -{\mathcal {D}}Z\cdot N+{\mathcal {D}}N \cdot Z+[P_L]= \hat{N}+\hat{R} \end{aligned}$$

(A.2)

with

$$\begin{aligned} S=\sum _{|\beta |\le 2d+1}S_{\beta }\bar{w}^\beta ,~\hat{N}=\sum _{|\beta |\le 2d+1}\hat{N}_{\beta }{\bar{w}}^{\beta }, \ \hat{R}=\sum _{2d+2\le |\beta |\le 4d+1}\hat{R}_{\beta }{\bar{w}}^{\beta } \end{aligned}$$

and

$$\begin{aligned} |Z({\bar{w}})|\lessdot \varepsilon ^{2d+2}, \ |\hat{N}({\bar{w}})|\lessdot \varepsilon ^{2d+2}, \ \forall |{\bar{w}}|\le \varepsilon , \end{aligned}$$

and

$$\begin{aligned} |\hat{R}({\bar{w}})|\lessdot \varepsilon , \ \forall |{\bar{w}}|\le r. \end{aligned}$$

Moreover, $Z(u,v,y)=Z(-u,v,y).$

Idea of proofs of Lemmas A.1and  A.2.

Since homological Eqs. (A.1) and (A.2) involve some higher order terms of ${\bar{w}}$, it is more complicated than that in traditional KAM steps. For convenience of readers, we give some important steps of the proofs and omit some tedious computations, which are in the same way as in [38].

First let ${\mathcal {K}}=({\mathcal {K}}^1,{\mathcal {K}}^2,{\mathcal {K}}^3,{\mathcal {K}}^4)$ with ${\mathcal {K}}=H,~P,~\tilde{R},~\hat{N}$ and $\hat{R}$.

For Lemma A.1, let

$$\begin{aligned} LW=-(1+\nu )\partial _{\omega _0}W-\partial _uW\cdot (e^Ty+bv)-a\partial _vW\cdot u. \end{aligned}$$

and

$$\begin{aligned} H(W)=-W_x(My+\tilde{e}v)-{\mathcal {D}}W\cdot N_h+DN_h\cdot W. \end{aligned}$$

Note that the order of $\bar{w}$ in H(W) is at least one higher than the order of $\bar{w}$ in W.

Let $W=(X,Y,U,V)$. Then (A.1) can be written as the following four equations,

$$\begin{aligned} LX&=-(MY+\tilde{e}V)-H^1+[P_L^1]-P_L^1+\tilde{R}^1. \end{aligned}$$

(A.3)

$$\begin{aligned} LY&=-H^2+[P_L^2]-P_L^2+\tilde{R}^2, \end{aligned}$$

(A.4)

$$\begin{aligned} \left\{ \begin{aligned} LU+bV&=-H^3+[P_L^3]-P_L^3+\tilde{R}^3-e^TY,\\ LV+aU&=-H^4+[P_L^4]-P_L^4+\tilde{R}^4. \end{aligned}\right. \end{aligned}$$

(A.5)

It is easy to see that

$$\begin{aligned} H^1=H^1(X,Y,U,V),~H^2=H^2(Y),~H^3=H^3(Y,U,V),~H^4=H^4(Y,U,V). \end{aligned}$$

So (A.4) can be first solved independently, then substitute Y into (A.5), where (U, V) should be solved together, finally substitute (U, V) into (A.3) to solve X.

We solve each of the equations in (A.3)–(A.5) layer by layer according to the order of ${\bar{w}}$ from low to high. We first consider all m-order terms of ${\bar{w}}$: $\{{\bar{w}}^{\beta }, |\beta |=m\}$. When we solve all the m-order terms in $P^j$, $H^j, (j=1,2,3,4)$ can produce some new $(m+1)$-order terms, which are joined together with all the $(m+1)$-order terms in $P^j$ into the next layer. Thus, the Eqs. (A.3) and (A.4) are reduced to an equation for W, (W=X or Y):

$$\begin{aligned} LW=P, \end{aligned}$$

(A.6)

where $P=\sum _{i+j+|l|=m}P_{ijl}(x)u^iv^jy^l$. The equations in (A.5) are reduced to an equation of ${\tilde{W}}=(U, V)$:

$$\begin{aligned} L\tilde{W} +\tilde{B}{\tilde{W}}=\tilde{P}, \end{aligned}$$

(A.7)

where $\tilde{B}{\tilde{W}}=(bV, aU)$ and $\tilde{P}=\sum _{i+j+|l|=m} \tilde{P}_{ijl}(x)u^iv^jy^l.$

When we consider all m-order terms of ${\bar{w}}$, we start with 0-order terms of y and m-order terms of (u, v), then 1-order terms of y and $(m-1)$-order terms of (u, v), and so on. In this process, the order of y is from low to high one by one. Note that $e^TyW_u$ and $e^Ty\tilde{W}_u$ can make some new terms with the order of y increasing by one and the order of (u, v) decreasing by one, these terms can be considered in the next step.

By the above analysis, when the order of y is fixed, we require the following results:

• (t1) Let $P=\sum _{i+j=m}P_{ij}(x)u^iv^j.$ There exists a unique solution $W=\sum _{i+j=m}W_{ij}(x)u^iv^j$ such that

  $$\begin{aligned} -(1+\nu )\partial _{\omega _0}W-\partial _uW\cdot bv-a\partial _vW\cdot u=P. \end{aligned}$$

  Moreover, if $P=-P\circ \tilde{G}$, then $W=W\circ \tilde{G}$ and if $P=P\circ \tilde{G}$, then $W=-W\circ \tilde{G}$, where $\tilde{G}: (x, u, v) \rightarrow (-x, -u, v)$.

• (t2) Let $\tilde{P}=\sum _{i+j=m} \tilde{P}_{ij}(x)u^iv^j$. There exists a unique solution ${\tilde{W}}=\sum _{i+j=m}{\tilde{W}}_{ij}(x)u^iv^j$ such that

  $$\begin{aligned} -(1+\nu )\partial _{\omega _0}\tilde{W}-\partial _u\tilde{W}\cdot bv-a\partial _v\tilde{W}\cdot u +\tilde{B}{\tilde{W}}= \tilde{P}, \end{aligned}$$

  Moreover, let $J=\mathrm diag(-1,1),$ if $\tilde{P}=-J\tilde{P}\circ \tilde{G}$, then $\tilde{W}=J\tilde{W}\circ \tilde{G}.$

The above equations are somewhat standard homological linear equations as in the traditional KAM steps and can be solved easily. Here we omit the details.

Then we can solve (A.6) and (A.7) with the following results.

• (s1) There exists a unique solution $W=\sum _{i+j+|l|=m}W_{ijl}(x)u^iv^jy^l$ such that (A.6) holds. Moreover, if $P=-P\circ G,$ then $W=W\circ G;$ if $P=P\circ G$, then $W=-W\circ G$.

• (s2) There exists a unique solution $\tilde{W}=\sum _{i+j+|l|=m}\tilde{W}_{ijl}(x)u^iv^jy^l$ such that (A.7) holds. Moreover, if $\tilde{P}=-J\tilde{P}\circ G$, then $\tilde{W}=J\tilde{W}\circ G.$

For Lemma A.2, the Eq. (A.2) is much simpler. Write as:

$$\begin{aligned} -\partial _uS{\cdot }\left( e^Ty+bv+N^3_h\right) -\partial _vS{\cdot }\left( au+N^4_h\right) +[P^2_L]=\hat{R}^2 \end{aligned}$$

(A.8)

and

$$\begin{aligned} \partial _yN^jS+[P^j_L]=\hat{N}^j+\hat{R}^j,\quad j=1,3,4. \end{aligned}$$

(A.9)

These equations are independent of x, so there is no problem of small divisors. In the same way as above we require the following results.

• (n1) Let $f(u,v)=\sum _{i+j=m}c_{i}u^iv^j$. There exists a solution $S=\sum _{i+j=m}S_{i}u^iv^j$ such that

  $$\begin{aligned} au\partial _vS+bv\partial _uS=f(u,v). \end{aligned}$$

  Moreover, if $f(u,v)=-f(-u,v)$, then $S(u,v)=S(-u,v)$.

• (n2) Let $g(u,v,y)=\sum _{i+j+|l|=m}g_{ijl}u^iv^jy^l$. There exists $S=\sum _{i+j+|l|=m}S_{ijl}(x)u^iv^jy^l$ such that

  $$\begin{aligned} e^Ty\partial _uS+au\partial _vS+bv\partial _uS=g(u,v,y). \end{aligned}$$

  Moreover, if $g(u,v,y)=-g(-u,v,y)$, then $S(u,v,y)=S(-u,v,y)$.

The above results can be easily proved and we omit the details. After finding S for (A.8), substitute S into (A.9) to obtain $\hat{N}^j$ and $\hat{R}^j$, $(j=1,3,4)$.

1.2 Stability of Minimal Points of Real Analytic Functions

In [36], the authors obtained a result on the stability of minimal points of real analytic functions of one variable, which is summarized in the following Lemma A.3.

At first, let $\varepsilon >0$ be a smallness scalar and $q=\varepsilon ^{1+\delta }$, where $\delta =\frac{1}{2d+1}.$ Consider ${\mathcal {H}}(z)=F(z)+G(z)$, which is real analytic for $|z|\le r$ with $F(z)=\sum _{j=2}^{2d+2}h_jz^j \ \text{ and } \ \ G(z)=\sum _{j= 2d+3}^{\infty }h_jz^j.$ Let $|{\mathcal {H}}(z)|\le \bar{B},~\forall |z|\le r$ and assume that

$$\begin{aligned} \begin{aligned} r_0/2\le r \le&r_0, \ \rho \ge \varepsilon ^{\delta }, \ r_0/2\le r-2\rho ,\ h_{2d+2}\ge h>0,\\&|F(z)|\le B_1, \ |G(z)|\le B_2, ~\forall |z|\le r, \end{aligned} \end{aligned}$$

(A.10)

where $r_0,d,h,\bar{B},B_1,B_2$ are fixed constants. Let $\hat{{\mathcal {H}}}(z)=\sum _{j\ge 1}\hat{h}_jz^j$ be real analytic for $|z| \le r$. If

$$\begin{aligned} |\hat{{\mathcal {H}}}(z)| \lessdot \ \varepsilon , \ \ |z|\le r; \ \ \ |\hat{{\mathcal {H}}}(z)|\le \varepsilon ^{2d+3}, \ |z|\le \varepsilon , \end{aligned}$$

then $\hat{{\mathcal {H}}}$ is called an $\varepsilon$-small perturbation. We say 0 is a stable critical point of ${\mathcal {H}}$ under $\varepsilon$-small perturbation, if for any $\varepsilon$-small perturbation $\hat{{\mathcal {H}}}$, $\tilde{{\mathcal {H}}}={\mathcal {H}}+\hat{{\mathcal {H}}}$ has a critical point $x_0$ with $|x_0|\le c\varepsilon ^{1+\delta }$, where $c>0$ is a constant only depending on $(r_0,d,h,\bar{B})$.

The following Lemma A.3 says that there is a sufficiently large neighbour of 0 such that if ${\mathcal {H}}$ has 0 as a minimum in this neighbour, then 0 is a stable minimum point of ${\mathcal {H}}$, moreover, $\tilde{{\mathcal {H}}}={\mathcal {H}}+\hat{{\mathcal {H}}}$ has a stable small minimum point under much smaller perturbations.

Lemma A.3

[36] Consider the above $({\mathcal {H}}, r, \varepsilon , \rho )$ with (A.10) holding. Moreover,

$$\begin{aligned} \min _{|x|\le \lambda ^{2d+1} q} {\mathcal {H}}(x)={\mathcal {H}}(0)=0. \end{aligned}$$

(A.11)

Then, there exists a sufficiently large $\lambda$ only depending on (d, h) and a sufficiently small $\varepsilon _0$ only depending on $( r_0, d, h, B_1, B_2, \lambda )$, such that if $\hat{{\mathcal {H}}}=\sum _{j\ge 1}\hat{h}_jz^j$ is $\varepsilon$-small with $\varepsilon \le \varepsilon _0$, then 0 is a stable minimum point of ${\mathcal {H}}$ under $\varepsilon$-small perturbation. More precisely, $\tilde{{\mathcal {H}}}={\mathcal {H}}+\hat{{\mathcal {H}}}$ has a minimum point $x_*$ such that $|x_*|\le \lambda ^{2d+\frac{1}{4}}q$. Moreover, let ${\mathcal {H}}_+(x)={\tilde{{\mathcal {H}}}}(x+x_*)-{\tilde{{\mathcal {H}}}}(x_*),$ then ${\mathcal {H}}_+(z)$ is analytic in $|z|\le r-2\rho$ and 0 is a minimum point of ${\mathcal {H}}_+$ with

$$\begin{aligned} \min _{|x|\le \frac{1}{2} \lambda ^{\frac{1}{2}} q} {\mathcal {H}}_+(x)={\mathcal {H}}_+(0)=0. \end{aligned}$$

(A.12)

Write ${\mathcal {H}}_+=F_++G_+$ in the same way as ${\mathcal {H}}(z)=F+G$. Then we have

$$\begin{aligned} |F_{+}(z)-F(z)|\le c \varepsilon ,\ \ |G_+(z)-G(z)|\le c \varepsilon , ~if~ |z|\le r-2\rho , \end{aligned}$$

(A.13)

where the constant c depends on $(r_0,d, h, B_1, B_2,\lambda )$.

Lemma A.3 concerns with one equation of simple variable. Below we consider a system of equations of multiple variables with only one degenerate direction, which arises from our KAM steps.

Now let us first recall the problem on equilibrium point in our KAM iteration. At the initial step, from (1.10), we have the equations:

$$\begin{aligned} f_0= {\mathcal {M}}^0y_{\flat }+ \left( \tilde{e}^0_{\flat }-\omega _{01}^{-1}\omega _{0\flat }\tilde{e}_1^0\right) v,~~ g_0=e_{\flat }^{0T}y_\flat + b^{0}v+v^{2d+1} \end{aligned}$$

where $\det ({\mathcal {M}}^0)\ne 0$ and $b^0=e_\flat ^{0T}{{\mathcal {M}}^{0}}^{-1} (\tilde{e}^0_\flat -\omega _{01}^{-1}\omega _{0\flat }\tilde{e}_1^0).$ Let $H_0=(f_0, g_0): (y_\flat , v) \rightarrow H_0(y_\flat , v)\in {\mathbb {R}}^{n}.$ From $f_0(y_\flat ,v)=0$, we obtain $y_\flat =-{{\mathcal {M}}^0}^{-1} (\tilde{e}^0_{\flat }-\omega _{01}^{-1}\omega _{0\flat }\tilde{e}_1^0)v$. Then insert it to $g_0(y_\flat ,v)$, we have $g_0(y_\flat (v),v)=v^{2d+1}.$ Then 0 is a stable zero of $g_0$.

At any later KAM step, we consider

$$\begin{aligned} H=(f, g)=\sum _{1\le |i|+l\le 2d+1}H_{il}y_\flat ^{i}v^{l}, \end{aligned}$$

(A.14)

where (f, g) are defined in (3.6). From the assumption (3.4), we have

$$\begin{aligned} |H(y_\flat ,v)-H_0(y_\flat ,v)|\le c_0 \varepsilon _0, \ \forall |y_\flat |\le r,~|v|\le r, \end{aligned}$$

(A.15)

where $c_0>0$ is a constant.

If H has a stable zero, we hope to prove that after an $\varepsilon$-small perturbation $\hat{H}$, $H+\hat{H}$ still has a stable zero. The result is given in the following Lemma A.4.

Before stating Lemma A.4, we first give an idea of how to find a stable zero of H.

From (A.15), $f(y_\flat ,v)=0$ uniquely determines an implicit function $y_\flat =y_\flat (v)$. Substitute $y_\flat (v)$ into $g(y_\flat , v)=0$ and let

$$\begin{aligned} {\mathcal {H}}(v)=\int _0^v g\bigl (y_\flat (v), v\bigr )\, dv, \ \ |v|\le r. \end{aligned}$$

(A.16)

To find a stable zero of H is equivalent to find a stable critical point of ${\mathcal {H}}(v)$. If all assumptions in Lemma A.3 hold for ${\mathcal {H}}$, then 0 is a stable critical point of ${\mathcal {H}}$, which implies that 0 is a stable zero of H.

Now, we give the lemma.

Lemma A.4

Consider the above $(H_0,H, r, \varepsilon , \rho )$ with (A.15) holding. Let ${\mathcal {H}}(v)$ be given in (A.16) and suppose that $({\mathcal {H}}, r, \varepsilon ,\rho )$ satisfies (A.10) and (A.11). Let $\tilde{H}=H+\hat{H}$, where ${\hat{H}}(y_\flat ,v)=({\hat{f}}, {\hat{g}})=\sum _{0\le |i|+l\le 2d+1} \hat{H}_{il}y_\flat ^{i}v^{l}$ and $|\hat{H}(y_\flat ,v)|\le \varepsilon ^{2d+2},~\forall |y_\flat |<\varepsilon ,~|v|<\varepsilon$. Then there exists a sufficiently small ${\tilde{\varepsilon }}_0>0$, if $\varepsilon \le {\tilde{\varepsilon }}_0$, then the followings hold true:

(i) $\tilde{H}$ has a zero $(y_{\flat 0},v_0)$ such that $|y_{\flat 0}|\lessdot \varepsilon ^{1+\delta },~|v_0|\lessdot \varepsilon ^{1+\delta }$. Let $H_+=\tilde{H}(y+y_{\flat 0},v+v_0)$, then $H_+(y_\flat ,v)$ has the same form as H in (A.14) and satisfies

$$\begin{aligned} |H_+(y_\flat ,v)-H(y_\flat ,v)|\lessdot \varepsilon ,~(y_\flat ,v)\in D(r-2\rho ). \end{aligned}$$

(ii) Define ${\mathcal {H}}_+$ by $H_+$ in the same way as ${\mathcal {H}}$ by H in (A.16). Then (A.12) and (A.13) hold for ${\mathcal {H}}_+$.

Proof

Here we only give the idea of the proof. The estimates can be easily obtained and we omit the details, for more details we can refer to [38].

Write as $\tilde{H}=(\tilde{f},\tilde{g})$. First by solving the non-degenerate equation $\tilde{f}(y_\flat ,v)=0$, we obtain $y=y_\flat (v)$. Then insert it into $\tilde{g}(y_\flat ,v)$ and let $\tilde{{\mathcal {H}}}(v)=\int _0^v \tilde{g}(y_\flat (v), v)\, dv.$ In the same way as the proof of Lemma A.3 in [36], write as $\tilde{{\mathcal {H}}}={\mathcal {H}}+\hat{{\mathcal {H}}}$, where $\hat{{\mathcal {H}}}$ is an $\varepsilon$-small perturbation. By Lemma A.3, we can obtain the result (i). Note that ${\mathcal {H}}_+$ defined by $H_+$ in the same way as ${\mathcal {H}}$ by H in (A.16) is just the one obtained by Lemma A.3. So again by Lemma A.3, it follows the result (ii). $\square$