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. Devito’s 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 Î 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 Î 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 p–q

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 p–q 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  Î 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  Î 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 Î 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  Î P(f) and q  Î P(f) coincide:

Def.: f[tp(x)] = f(x) for all x and let’s 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 (0£r<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 0£r<p, and p is the smallest positive period, r must be 0.

If q<0, then -q Î 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  a£x £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. Devito’s 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£ x£x+αn

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

Then for any x and y:  1/anò = 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 Î 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 Î P+(f), I shall argue by contradiction.  If a0 is not in this set, then there must be some {an } Í P+(f) such that an =a0.  But since f is continuous,  =  for any fixed x.

\a0 Î 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 CalculusNew York: Appleton-Century-Crofts, Inc.

1961.

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

1947.

5.  Widder, David Vernon.  The Laplace TransformLondon: 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.”