Periodic Functions On Non-Linear Temporal Models

By Alexia Puig

 

Over the course of this semester, I have been reading and working with two of Dr. Devitos papers, A Non-Linear Model For Time and Time Scapes. These papers define time as a partially ordered set of instants. To define a partially ordered set, let I be an infinite set and let ≤ be a relation on the elements of I.[1] Then ≤ is a partial ordering of I, and the pair (I,≤) is a partially ordered set if:

a)      for every i I we have i≤i

b)      if i and j are in I and we have both i≤j and j≤i, then i = j

c)      if i, j and k are in I and we have i≤j and j≤k, then i≤k

There is a duration function that assigns a non-negative real number for each pair of comparable instants.[2] This function is called duration or dur, for short and has the following properties:

1) dur(x,y) = p, where p is the time between x and y (two instants)

2) dur(x,y) = dur(y,x), and if x=y, then dur(x,y) = 0

3) if y is between x and z, meaning x≤y≤z or z≤y≤x, then dur(x,z) = dur(x,y) + dur(y,z).

I have been looking at functions defined on time tracks and studying their mathematical properties. A time track, T, is a non-empty subset of the infinite set of instants and has the following properties[3]:

a) Any two instants on T can be compared.

b) If x, y, and z are on T and y is between x and z, then dur(x,z) = dur(x,y)

+ dur(y,z).

c) Given y on T, and a positive, real number p, there are exactly 2 instants

x and z on T such that dur(x,y) = p and dur(y,z) = p.

I am trying to extend the idea of periodicity and integrability to functions defined on time tracks. Since instants cannot be added, this requires some new ideas. I was successful in extending periodicity using the translation function. This is defined as follows:

Definition 1: There is a translation function, tp, for any fixed number p, which maps T to T as follows:[4]

1)      t0(x) = x for all x T

2)      if p>0, tp(x) = y where y is the unique instant on T such that x<y and dur(x,y) = p

3)      if p<0, tp(x) = z where z is the unique instant on T such that z<x and dur(z,x) = |p|

Then, a real number p is a period of the function f if f[tp(x)] = f(x) for all x. The set of all such numbers is denoted by P(f). We shall say that f is a periodic function if P(f) contains a non-zero number.

Lemma 1: If p and q are contained in the set P(f), then so are -p, p+q and pq

Proof : Suppose that f[t-p(x)] = f(y) where y is the instant in the past such that dur(y,x) = p. Then tp(y) = x, so f[t-p(x)] = f(y) and f[tp(y)] = f(x). But by definition f[tp(y)] = f(y). Therefore f[t-p(x)] = f(x) for all x, which concludes that -p is also a period.

Now to prove that p+q and pq are also included in the set of periods for f, when p and q are both periods:

f[tp(x)] = f(x) and f[tq(x)] = f(x) for all x

f[tp+q(x)] = f[tp[tq (x)]] = f[tq(x)] = f(x) for all x

\p+q P(f)

f[tp-q(x)] = f[tp[t-q (x)]] = f[t-q(x)] = f(x) for all x. Since the first proof shows that -p is a period contained in the set when p is one, so -q is a period when q is contained in the set.

\p-q P(f)

Corollary 1: If p is contained in the set P(f) then so is np, where n is an integer. When p is the smallest, positive member of P(f), these two sets coincide.

To prove that np is also a period contained in the set:

Since this is clearly true for n=1, I will need to prove it for n>1.

Suppose that (n-1)p P(f) and consider np. Since both 1p and (n-1)p are in P(f),

then 1p + np = np is in P(f) by Lemma 1

Now to prove that the sets nq P(f) and q P(f) coincide:

Def.: f[tp(x)] = f(x) for all x and lets say that p is the smallest positive number for which this is true.

Suppose that f[tq(x)] = f(x) for all x and q>0

Then p<q, so divide q = np + r, where r is the remainder (0r<p)

f[tq(x)] = f[tnp+r(x)] = f[tnp [tr(x)]]

Since p is a period so is np as shown above.

Then f[tnp [tr(x)]] = f[tr(x)] for all x, so r is a period.

But since 0r<p, and p is the smallest positive period, r must be 0.

If q<0, then -q P(f) and since -q is positive, -q = np for some integer n. But then

q = (-n)p , so r can be proven to be 0 for q<0 also.

 

I have also been interpreting the integrals of these functions and started by studying the Riemann-Stieltjes Integral, its properties and applications. The Riemann-Stieltjes Integral is defined as follows for two functions: Let g(x) and f(x) be real functions of x defined on the interval [a,b], where a x b, (a=x0<x1<x2<.<xn=b). The limit of: , as max|xj xj+1| 0, where xj-1≤xj*≤xj, is denoted by: , and is called the integral of f with respect to g.[5] There is a Mean Value Theorem for the Riemann-Stieltjes Integral: = f(x)[g(b) g(a)] where ax b.[6] This will be useful below. The advantage here is that f(x) and g(x) are numbers while x is an instant.

Let g be a function such that dur(x,y) = |g(x)-g(y)|. Such a function is defined in Dr. Devitos paper as: D(s(p1), s(p2)) = |p1-p2|.[7] Instead of s, I shall use g

 

Lemma 2: Let f be a continuous and periodic function on the real line, and let a>0 be any element contained in P(f), then for any real x and y, = .

Proof: = = +

Now =

First note that g[tα(u)] = g(u) because α is a constant and g(x) g(y) = dur(x,y), so g[tα(x)] g[tα(y)] = dur(x,y)

Then=

So =

Therefore=

Lemma 3: If a continuous function has arbitrarily small periods, it is a constant

Proof: Consider the integral:

Suppose a and b are instants where a=x0<x1<x2<.<xn=b.

Then the integral can be represented by the sum

where xj-1 xj* xj

The limit of this sum as max[g(xj) g(xj-1)] goes to 0 is defined to be which is the Riemann-Stieltjes Integral.

Then using the Mean Value Theorem: m[g(b)-g(a)] M[g(b)-g(a)] where

m stands for minimum and M stands for maximum.

So

Now a = dur(tα(x),x) = g[tα(x)]-g(x) by the definition of g

Then 1/[ g(x+αn) g(x)] = f(x)/an [ g(x+αn) g(x)] =an f(x)/an = f(x)

where x xx+αn

Since an P(f), then for any x: 1/an = f(x)

Then for any x and y: 1/an = 1/an so f(x) = f(y)

Therefore, f is a constant.

Corollary 2: Any non-constant, continuous, periodic function has a smallest positive period

Proof: P(f[tα(x)]) = P(f(x)) for all x. In Lemma 1 it was proved that the set P+(f) = {a P(f)|a>0} is non-empty, where P+(f) is the set of positive periods for of P(f). Let a0 be the greatest lower bound for this set. If a0 is 0, then a sequence in P+(f) must converge to it, meaning that f would have arbitrarily small periods. But as proved above, if f has arbitrarily small periods, f must be a constant and this cannot be true since f is non-constant. Therefore a0 must be positive.

To prove that a0 P+(f), I shall argue by contradiction. If a0 is not in this set, then there must be some {an } P+(f) such that an =a0. But since f is continuous, = for any fixed x.

\a0 P+(f)

 

 

 

 

References

1. Devito, Carl L. A Non-Linear Model For Time. Astrophysics and Space Science

244 (1996): 357-369.

 

2. Devito, Carl L. Time Scapes. Chaos, Solitons & Fractals Vol. 9 No. 7 (1998):

1105-1114.

 

3. Olmstead, John M. H. Advanced Calculus. New York: Appleton-Century-Crofts, Inc.

1961.

 

4. Widder, David V. Advanced Calculus. Englewood Cliffs, N.J.: Prentice-Hall, Inc.

1947.

 

5. Widder, David Vernon. The Laplace Transform. London: Humphrey Milford Oxford

University Press. 1946.

 

 



[1] Devito, Carl L. Time Scapes.

[2] Devito, Carl L. A Non-Linear Model For Time.

[3] Devito, Carl L. A Non-Linear Model For Time.

[4] Devito, Carl L. A Non-Linear Model For Time.

[5] Widder, David Vernon. The Laplace Transform.

[6] Olmstead, John M. H. Advanced Calculus.

[7] Devito, Carl L. Time Scapes.