Periodic orbits and Sturm–Liouville theory

I thinks there is some problem related to the solution of a 2 order differential equation given by Sturm-Liouville system which is nontrivial.

It is well-know that the power of Sturm-Liouville theory see  wiki, is due to it is some kind of “spectral decomposition” in the solution space.

Two kind of problem is interesting, one is the eigenvalue estimate, both upper bound and lower bound, this already investigated in ESTIMATING THE EIGENVALUES OF STURM-LIOUVILLE. PROBLEMS BY APPROXIMATING THE DIFFERENTIAL EQUATION.

I post two problem here, this is a product due to a random walk along the boundary of topology and the analysis,

Problem 1.

Fix a set A=\{k_1<k_2<...<k_l\}, is there a 2 order ordinary differential equation given by Sturm–Liouville theory  such that the eigenfunction f_{k} is periodic if and only if k\in A?

There is also some weak version of this and a infinity version of this.

Of course we have the following map, from the high order ordinary differential equation to the 1 order differential equation in high dimension. But the key point is that it is not a bijection! The Frobenius condition play a crucial role.

Problem 2.

There is a homotopy in the moduli space of differential equation, and we could define a direct product operator in this space, and we consider the topology defamation of the eigenfunction, could there be some equality, one side of it explain the topology information, the other side explain the spectral (or analysis) information?

There is another interesting problem.


Rotation number


Consider compact 1 dimension dynamic system.

We focus on S_1, it does not mean S_1 is the only compact 1 dimensional system , but it is a typical example.

T: S_1\to S_1.

If T is a homomorphism then T stay the order of S_1 (by continuous and the zero point theorem). That is just mean:


(may be do a reflexion e^{2\pi i\theta}\to e^{-2\pi i\theta}).

In the homomorphism case. We try to define the rotation number to describe the expending rate of the dynamic system.

T: \mathbb S_1\to \mathbb S_1 i.e. T:\mathbb R/\mathbb Z\to \mathbb R/\mathbb Z. Lifting to,

\hat T:\mathbb R\to \mathbb R.

How to realize the lifting?

Step1: Periodic extend T:\mathbb S_1\to \mathbb S_1 to T': \mathbb R\to \mathbb S_1. (Regard \mathbb S_1 as [0,2\pi]).

Step2: Consider the “flow” of T':\mathbb R\to \mathbb S_1. we get \hat T:\mathbb R\to \mathbb R.

The rotation number is defined as:

\rho (T)=\limsup_{n\to \infty}\frac{\hat T^n(x)}{n}.

Following we will shall it is independent of the choice of x and \rho (T)=\lim_{n\to \infty}\frac{\hat T^n(x)}{n} in fact.

It is not difficult to proved the following property:


1.If T' is conjugate (in fact semi-conjugate is enough ). Then 1.If T' is conjugate (in fact semi-conjugate is enough ). Thenrotation number of T' equal to rotation number of T.

2.If T' is conjugate to T. Then rotation number of T' equal to
rotation number of T.



T'(y)=\Psi\circ T\circ \Psi^{-1}(y)=\Psi(\Psi^{-1}(y)+t(\Psi^{-1}(y))))

Example: T: x\to x+\alpha, \alpha\in \mathbb R. It is not difficult to prove the rotation number of T is \alpha.


1)For n\geq 1 we have that \rho(T^n)=\rho(T) (mod 1).

2).If T has a periodic point, i.e. x\in S_1,\exists n\in \mathbb N^*, T^{n}(x)=x. Then $latex\rho (T)$ is rational.

3) T:\mathbb R/\mathbb Z \to \mathbb R/\mathbb Z has no periodic point then \rho(T) is irrational.

4) The limit actually exists and we have:  \rho(T)=\lim_{n\to \infty}\frac{\hat T^n(x)}{n} (mod 1).


pf of 1):

\rho(T^n)=\lim_{k\to \infty}\frac{(\hat T^n)^{k}(x)}{k}

=\lim_{k\to \infty}n\frac{(\hat T)^{nk}(x)}{nk}

=n\rho (T).

Used the property |x-y|<k \leftrightarrow |T^{\omega}(x)-T^{\omega}(y)|<k+1, \forall \omega\in N^*, \forall k\in \mathbb Z^{+}.



pf of 2):

It is not difficult to prove \rho(T) is independent with the choice of x. So choose x to be the periodic point.

Remark: but the inverse of 2) is not true. For example:

x\to x+\frac{1}{2}+\frac{1}{100}sin(4\pi x).

This dynamic system has both periodic points(\{0,\frac{1}{2}\},\{\frac{1}{4},\frac{3}{4}\}) and non-periodic pint (maybe orbits generated by \{\frac{1}{\sqrt{2}}\}).

pf of 3):

If not. Assume \rho(T) is rational number \frac{q}{p}. Take any point x\in \mathbb S_1, then:

\lim_{n\to \infty}\frac{\hat T^n(x)}{n}=\frac{q}{p}.

\Longrightarrow \lim_{n\to \infty}\frac{(\hat T^p)(x)}{n}=q.

\Longrightarrow \lim_{n\to \infty}\frac{(\hat T^p-q)^n(x)}{n}=0.

Now assume \hat T^p-q=\widetilde T.

Then \widetilde x>x. $\forall x\in \mathbb S_1$ (if \widetilde x<x, \forall x\in \mathbb S_1, take reflection x\to -x).

And there do not exists n\in \mathbb N^* such that \widetilde T^nx>x+1. If not, we could prove rotation number is large than \frac{1}{n} lead a contradiction.

So \{\widetilde T^nx\}_{n=1}^{\infty} is a bounded monotonically increasing sequences in \mathbb S_1, it limits point z\in \mathbb S_1 must satisfied \widetilde T^n (z)=z.

pf of 4):

Using the point wise approximation inequality induced from the monotonically and stay ordering property of \mathbb S_1 by T.


Assume \rho(T) is irrational.

1. Let n_1,n_2,m_1,m_2\in \mathbb Z, and x,y\in \mathbb R. If \hat T^{n_1}(x)+m_1<\hat T^{n_2}(x)+m_2, then hat T^{n_1}(y)+m_1<\hat T^{n_2}(y)+m_2.

2. The bijection n\rho (T)+m\to \hat T^n(0)+m between the set \Omega=\{n\rho(T)+m| n,m\in \mathbb Z\} and \Gamma=\{\hat T^{n}(0)+m,n,m\in \mathbb Z\} precise the natural ordering on \mathbb R.


This corollary is not difficult to prove use the established property.

 Denjoy’s theorem


If T: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z is a minimal orientation presenving homomorphism with irrational rotation number \rho then T is topologically conjugate to the standard rotation R_{\rho}: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z.

leave as a ex.

For T: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z, T': \mathbb R/\mathbb Z\to \mathbb R. We define the variation of log|T'|: \mathbb R/\mathbb Z\to \mathbb R by:


sup\{\sum_{i=0}^{n-1}|log|T'|(x_{i+1})-log|T'|(x_i)|: 0=x_0<x_1<...<x_n=1\}

We say that the logarithm of |T'| has bounded variation if this value Var(log|T'|) is finite.

Denjoy’s theorem:

If T: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z is a C^1 orientation preserving homomorphism of the circle with derivative of standard variation and irrational rotation number \rho=\rho(T) then T:\mathbb R/\mathbb Z\to \mathbb R/\mathbb Z is topologically conjugate to the standard rotation :

R_{\rho}:\mathbb R/\mathbb Z\to \mathbb R/\mathbb Z.

Due to the upper proposition we only need show T:\mathbb R/\mathbb Z\to \mathbb R/\mathbb Z is minimal. Proof pf minimal is splitting to following two sub lemmas.


If T has irrational rotation number and there are a constant C>0 and a sequences of integers q_n\to \infty such that the map: T: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z Satisfy : |(T^{q_n})'(x)||(T^{-q_n})'(x)|\geq C Then T: \mathbb R/\mathbb Z\to \mathbb R/\mathbb Z is minimal.




Fix x\in \mathbb R/\mathbb Z and write x_n=T^n(x), for x\in \mathbb Z There exists an increasing sequences q_n\to \infty of natural number such that the intervals (x_0,x_{q_n}),(x_1,x_{q_n+1}),...,(x_i,x_{q_n+i}),...,(x_{q_n},x_{2q_n}) are all disjoint.


Paradox and problem 


\rho(T)>0 because of existence of fix point.

T_{\alpha}=T+\alpha for \alpha< sup_x|T_x-x|.

Is \rho(T_{\alpha})=0 always true for \alpha \in R?

If it is right, then there is a contradiction with argument (*), but for what type of dynamic system T?

T_{\alpha}=T+\alpha satisfied \rho(T_{\alpha})=\rho(T)+\alpha. for all \alpha\in \mathbb R?


If T is not homomorphism but T:x\to x+g(x) induced g(x)=x-f(x), f(x) is striating increasing, Is the limit of \lim_{x\to \infty}\frac{\hat T(x)}{n} always exists? it could not be increase with x.



Baragar-Bourgain-Gamburd-Sarnak conjecture

M is the markov triple (x,y,z):

x^2+y^2+z^2=xyz and (x,y,x)\in \mathbb Z^3  \ \ \ \  (*).

It is easy to see:

R_1: (x,y,z)\to (3yz-x,y,z).

map markov triple to markov triple.

This is also true for R_2,R_3. and the transform R_1,R_2,R_3 and permutation a classical result of markov claim that all solution of  (*) could be generated from (1,1,1). I get a similar result for a similar algebraic equation 1 half years ago when consider a Q version of problem about 1-form given by Xu Bin.

Now  we know the graph with root (1,1,1) and with node generate by transform R_1\cup R_2 \cup R_3 \cup S_3 is connected.

The B-B-G-S conjecture is is the connected property still true for prime p surfficed  large?

Metric entropy 2

I am reading the article “ENTROPY THEORY OF GEODESIC FLOWS”.

Now we focus on the upper semi-continuouty of the metric entropy map. The object we investigate is (X,T,\mu), where \mu is a T-invariant measure.

The insight to make us interested to this kind of problem is a part of variational problem, something about the existence of certain object which combine a certain moduli space to make some quantity attain critical value(maximum or minimum). The most simple example maybe Isoperimetric inequality and Dirichlet principle of Laplace. Any way, to establish such a existence result a classical approach is to proof the upper semi-continuouty and bounded for associate energy of the problem. In our case the semi-continuouty will be some thin about the regularity of the entropy map:

E:M(X,T)\to h_{\mu}.

We define the entropy at infinity:

sup_{(\mu_n)}limsup_{\mu_n\to 0}h_{\mu_n}(T)

Where (u_n)_{n=1}^{\infty} varies in all sequences of measure coverage to 0 in the sense for all A\subset M, A measurable then \lim_{n\to \infty} \mu_{n}(A)=0.

Compact case

we say some thing about the compact case, In this case we have finite partition with smaller and smaller cubes, this could be understand as a sequences of smaller and smaller scales. A example to explain the differences is \mathbb N^{\mathbb N},\sigma, shift map on countable alphabet.

Because of this thing, there is a good sympolotic model, i.e.  h-expension, and it generalization  asymptotically  h-expension equipped on a compact metric space $X$ have been proved to be that the corresponding entropy map is upper semi-continous.

In particular C^{\infty} diffeomorphisms on compact manifold is asymptotically h-expensive.



Natural problem but I do not understand very well:

Why it is natural to assume the measure to be probability measure in the non-compact space?


Non-compact case

(X,d) metric space

T:X\longrightarrow X is a continuous map.

d_n(x,y)=\sup_{0\leq k\leq n-1}d(T^kx,T^ky), then d_{n} is still a metric.

Easy to see \frac{1}{n}h_{\mu}(T^n)=h_{\mu}(T). This identity could be proved by the cretition of entropy by \delta-seperate set and \delta-cover set.


Kapok theorem:

X compact, for every ergodic measure \mu the following formula hold:

h_{\mu}(T)=\lim_{\epsilon \to 0}limsup_{n\to \infty}\frac{1}{n}logN_{\mu}(n,\epsilon,\delta).

Where h_{\mu}(T) is the measure theoretic entropy of \mu.

Riquelme proved the same formula hold for Lipchitz maps on topological manifold.



Let M_e(X,T) defined the moduli space of T-invariant portability measure.

Let M_(X,T) defined the moduli space of ergodic T-invariant probability measure.

Simplified entropy formula:

(X,d,T) satisfied simplified entropy formula if \forall \epsilon >0 surfaced small and \forall \delta\in (0,1), \mu\in M _e(X,T).

h_{\mu}(T)=\limsup_{n\to \infty}\frac{1}{n}log(N_{\mu}(n,\epsilon,\delta)).

Simplified entropy inequality:

If \epsilon>0 suffciently small, \mu \in M_{e}(X,T), \delta\in (0,1).

h_{\mu}(T)\leq \limsup_{n\to \infty}\frac{1}{n}log(N_{\mu}(n,\epsilon,\delta)).

Weak entropy dense:

M_e(X,T) is weak entropy dense in M(X,T). \forall \lambda>0, \forall \mu\in M(X,T), \exists \mu_n\in M_e(X,T), satisfied:

  1. \mu_n\to \mu weakly.
  2. h_{\mu_n}(T)>h_{\mu}(T)-\lambda, \forall \lambda>0.

Metric entropy 1

Some basic thing, include the definition of metric entropy is introduced in my early blog.

Among the other thing, there is something we need to focus on:

1.Definition of metric entropy, and more general, topological entropy.

2.Spanning set and separating set describe of entropy.

3.amernov theorem:


Now we state the result of Margulis and Ruelle:

Let M be a compact riemannian manifold, f:M\to M is a diffeomorphism and \mu is a f-invariant measure.

Entropy is always bounded above by the sum of positive exponents;i.e.,

h_{m}(f)\leq \int_{i}\lambda_i^{+}(x)dimE_i(x)dm(x).

Where dimE_i(x) is the multiplicity of \lambda_i(x) and a^{+}=max(a,0).

Pesin show the inequality is in fact an equality if f\in C^2 and m is equivalent to the Riemannian measure on M. So this is also sometime known as Pesin’s formula.

F.Ledrappier and L.S.Young generate the result of Pesin.

One of their main result is:

f:M\to M is a C^2 diffemoephism, where M is a compact riemanian manifold, f is compatible with the Lesbegue measure on M, and


If and only if on the canonical defined quation manifold $M/W_{\mu}$, i.e. the manifold mod unstable manifold $W_{\mu}$, the induced conditional measure m_{\xi} is absolute continuous.

Remark: according to my understanding, the equality just mean in some sense we have the inverse estimate:

h_{m}(f,\mu)\geq \int_{M}\lambda_idim(V_i)dm.

This result maybe just mean near the fix point of f,i.e. the place charge the topology of the foliation, we have the inverse estimate. Such a inverse estimate will lead a control of the singularity of the push forward measure m_{\xi} on the quation manifold.  So m_{\xi} have good regularity. But this idea is not complete to solve the problem.

Now we begin to get a geometric explain and which will lead a rigorous proof of the inequality:

h_{m}(f)\leq \int_{i}\lambda_i^{+}(x)dimE_i(x)dm(x).

At first we could observe that the long time average \lim_{n\to \infty}\frac{1}{n}log||Df^n|| of Df could be diagonal. Assume after diagonal the eigenvalue is

\lambda_1\leq \lambda_2\leq \lambda_3\leq...\leq \lambda_{n-1}\leq \lambda_n.

This eigenvalue could divide into 3 parts: <0,=0,>0.

This will lead to a direct sum decomposition of the tangent bundle TM:

TM\simeq E_{u}\otimes E_s\otimes E_c.

Where $E_u$ is the part corresponding to the eigenvalue>0, For this part we consider the more refinement decomposition:

E_u=\otimes_{k=1}^rV_k, V_k is the eigenvector space of \lambda_k. The dimension of $V_k$ is $dim V_k$.

On the other hand, we have a equality of metric entropy:

h_{m}(f)=\frac{1}{n}h_{m}(f^n)=\sup_{\alpha\in partition \ set}\frac{1}{n}h_m(f^n,\alpha).

For the later one, \alpha is a measurable partition of M, then \alpha could always be refine to a smaller partition \beta, and we have:

h_{m}(f,\alpha)\leq h_{m}(f,\beta).

Now we arrive the central place of the proof:

every partition could be refine by a partition with boundary of almost all cubes is parallel to the foliation. So  we focus ourselves on the portion \beta and all boundary of cubes in \beta is parallel to the eigenvector.

Under this situation, we need only estimate the numbers of \vee_{i=1}^nT^i\beta. Estimate it is not very difficult. we need only observe the following two thing:


\lim{n\to \infty} exists a.e. in M. So this lead to the definition of foliation almost everywhere, and except a measurable zero set. In fact this set is the set of fix point of M under f.


After a rescaling, every point which is not a fix point of f could be understand as it is far away from fix points. Then the foliation could be understand as  a product space locally. The flow with the direction which the eigenvalue is less than 1 cold not change \vee_{i=1}^nT^i\beta. The direction with eigenvalue equal to 1 is just transition and just change the number of \vee_{i=1}^nT^i\beta with polynomial growth. But the central thing is the direction with eigenvalue large than one and will make \vee_{i=1}^nT^i\beta change with viscosity e^{\lambda_i}. and we product it and get :

h_{m}(f,\mu)\leq \int_{M}\lambda_i dim(V_i)dm.

In fact the proof only need f to be C^1



Covering a non-closed interval by disjoint closed intervals

this note will talk about the Ostrowski representation and approximation by continue fraction.

As well-known,by the Weyl criterion,\{n\alpha\} is uniformly distribution in [0,1] iff \alpha\in R-Q.

i.e. we have:\forall 0\leq a\leq b\leq 1,we have:

\lim_{N\to \infty}|\{1\leq n\leq N|\{n\alpha\}\in [a,b]\}|=(b-a)N+o(N).

but this will not give the effective version.i.e. we do not the the more information about the decay of o(N).

we will give a approach of effective version of \alpha with smooth condition by give another proof of the uniformly distribution (in fact to to decomposition the interval [a,b] in to a finite sums of special intervals).and get the result:

D_N=\int_{M}D_N(\theta)d\mu=\int_Msup_{0<a<b<1}|\sum_{n=1}^{N}\chi_{(a,b)}(\{\theta n \})-N(b-a)|d\mu\sim O(log N)

if the term in the continuous fraction of \alpha have a up bound.this is so called \alpha is smooth.

Interval map

1.period 3 induce chaos

theorem:if a interval map T:I\to I have a period 3 point x,then \forall n\in N^*,there is a period n point for T.


n=1 case. trivial

n>1,n\neq 3 case:

the key point is to consider the structure of monotone interval contain previous one with fix length.

this will easy to lead a proof.


2.a work of J.Milnor and W.Thurston.

N(T^n) defined as the number of monotone interval of the map T^n.

theorem:h(T)=lim_{n\to infty}\frac{1}{n}log N(T^n).


3.monotone Markov map

this structure have two property:

1.piesewise monotone and $C^1$,the derive has control!


there is a relative dynamic system with this a shift map with a relative n\times n matrix A.

this two dynamic system have a lot of relation,the key one is:

the topological entropy of monotone Markov map is just the unique maximum eigenvalue of A.

and some byproduct…


Flat surface 1

Topological point of view:

in topological point of view a flat surface is a topological space M with a (ramified in nontrivial case) map\pi:M\longrightarrow T^2.

and the map satisfied:\pi is not ramified on \pi^{-1}(T^2-\{0\}) is not ramified and defined a covering map.

Geometric-analytic point of view:

we begin with compact connected oriented surface M,and a nonempty finite subset \Sigma=\{A_1,...,A_n\} of M.and

translation surface of type k:

translation structure:

complex structure:




T:X\longrightarrow X
介绍了三个简单例子,包括S_1上的加倍映射,旋转映射以及X_k=\Pi_{n\in Z}\{1,2,...,k\} 上的平移映射。
加倍映射会出现在微分流形中一些函数f的singular point,也就是hess f=0的地方附近的环绕数计算,还有一些scalling变换或者是一些多尺度的问题里。\\
平移映射我不是很懂,第二章中讲的Van der warden定理的证明是一个好例子,动力系统中的回复定理主要是用来刻画这些动力系统内蕴的算术性质的,basic ideal是如下事实:\\

动力系统往往具有transitivity的性质,加倍映射的例子用二进制分解构造,平移映射构造transitivity point的方法与之雷同,旋转映射情形这是初等的。\\

2.U open,TU=U \Longrightarrow U=\emptyset或者U是一个稠密集\\
3.U,V开集,\exists N\in N^*,T^n U \cap V\neq \emptyset \\
4.\{x\in X|\{T^nx\}_{n\in Z}dense\}是一个G_{\delta}集合
证明都是标准的,提两个关键点,第一点是注意到\cup_{n\in Z}T^n U这个集合是$T$不变的,第二点是注意到transitive point可以通过选取一组开集集进行描述从而有集合等式:
\{x\in X|\{T^nx\}_{n\in Z}dense\}=\cap_{n\in N^*}\cap_{k\in N^*}\cup_{m\in Z} T^mB_{\frac{1}{k}}(x_n)
minimality定义是动力系统T:X\rightarrow X所有的点都是transitivity point。\\


1.zorn lemma,2.minimal性质的第二点\\


\exists x\in X,\exists \{n_i\},\lim_{i\to \infty}T^{n_i}x=x
步骤三:说明T^i不变的集合满足zorn lemma,所以有最小元\\

\section{Birkhoff回复定理蕴含Van Der Warden定理}
\subsection{Van Der Warden定理与它的动力系统解释}










猜想:存在一个T^i:X\longrightarrow X ,$T^iT^j=T^jT^i$,并不存在满足某种”一致”minimal的子空间。但是我们知道多重回复定理是对的。\\


拓扑熵的定义可复杂了,拓扑熵是一个描述拓扑动力系统复杂程度的量。顺序是先引入标准定义和基本性质,然后给出一个计算方法,再然后引入spanning set和separeting set,利用这两种集合引入等价的定义方法,再证明amernov定理:h(T^m)=mh(T),最后证明动力系统之间的半共轭会导致熵之间的不等式。
















$f(x),g(x)\in C_{c}^{\infty}(R^n)$
\[f*g(x)=\int_{R^n}f(\xi)g(x-\xi)d\xi \]
\[x=(x_1,x_2,…,x_n)\in R^n\]
\[f*g(x)=\int_{R^n}f(\xi)g(x-\xi)d\xi\sim\sum_{k_1=-\infty}^{\infty}…\sum_{k_n=-\infty}^{\infty}\frac{f(x_1-k_1,x_2-k_2,…,x_n-k_n)}{k_1^2+k_2^2+…+k_n^2+1} \]
\[\sum_{k_1=-\infty}^{\infty}…\sum_{k_n=-\infty}^{\infty}\frac{f(x_1-k_1,x_2-k_2,…,x_n-k_n)}{k_1^2+k_2^2+…+k_n^2+1}=\sum_{\xi\in Z^n}f(x-\xi)g(\xi)=\int_{\xi\in R^n}f(x-\xi)\delta g(\xi) d\xi\]\\
(Young inequality)
$f\in L_1(R^n)$,$g\in L^p(R^n)$:
\[ ||f*g||_{p}\leq ||f||_1||g||_{p} \]

(hardy-litterwood-soblev inequality)
$p,r>1$,$0<\lambda<n$,$\frac{1}{p}+\frac{\lambda}{n}+\frac{1}{r}=2$,$f\in L^p(R^n),h\in L^r(R^n)$.exists a constant C,$C\sim n,\lambda,p$.
\[|\int_{R^n}\int_{R^n} f(x)|x-y|^{\lambda}g(y)dxdy|\leq C(n,\lambda,p)||f||_p||h||_r\]

t \to 0^+情况的技巧


\section{Caldron Zygmund算子的谱}
我们需要刻画Caldron Zygmund算子作用在某个区域上之后产生的谱会携带多少区域的形状的信息。通过在热核中令t\longrightarrow 0这个会化简为简单的情况,再加上凸性。
第二条是热核对t求任意次导以后是Caldero ́ n Zygmund算子,对这个算子卷积上一个具备C^1Boudary正则性的区域上特征函数的的谱我们有没有好的刻画,这其中能不能蕴含这个区域的几何信息。