Approximations to Convolutions of Exponential Functions



Convolutions of Exponential Functions

Background Material and Recent Results

Frank Massey

1. Introduction

In the following prime and * will denote differentiation and convolution with respect to t, i.e. f((t) = (f/(t and f(t) * g(t) = . Let An(t) = An(t;λ1,...,λn) = λ1e-λ1t * … * λne-λnt be the convolution of exponential functions λje-λjt. More precisely, A1(t;λ1) = λ1e-λ1t and An(t;λ1,...,λn) = An-1(t;λ1,...,λn-1)  * λne-λnt for n > 1. An(t) appears in the formula for the amount of an isotope in a radioactive decay chain, see [3]. An(t) is also the density functions of the hypoexponential (or generalized Erlang) probability distribution, see Ross [2]. We are interested in approximating the function An(t) by simpler functions. Some results of Jeffrey Prentis and myself in this direction are in the manuscript [1]. The purpose of this survey is to establish some of the basic properties of these functions and to give some other results that we have not yet put into a formal manuscript. Since An(t) is the solution to a differential equation of the form Error! Bookmark not defined., we begin by establishing some properties of solutions to these equations.

2. Solutions to differential equations of the form

Let x(t) = x(t,() be the solution to the initial value problem

(2.1) x'(t) = ( ( f(t) – x(t) ) x(0) = xo

Assume ( > 0 and f(t) is continuous for all t. x(t) can be expressed as

(2.2) x(t) = xoe-λt + F(t)

where

(2.3) F(t)  =  F(t,()  =   = f(t) * (e-λt

A function is g(t) is unimodal if it has a unique local maximum. We shall show if f(t) is unimodal then so is x(t) and we shall see how the location of the maximum of x(t,() varies with (; see Proposition 2.2. The first proposition describes how x(t,() varies with t and (. It involves x2(t) = x2(t,() = xoe-λt + x(t) * (e-λt which is the solution to

(2.4) x(t) = ( ( x(t) – x2(t) ) x2(0) = xo

In what follows we shall frequently suppress the variable ( in the functions x(t,(), x2(t,() and xλ(t,() and just write x(t), x2(t) and xλ(t).

Proposition 2.1. Let x and x2 be defined by (2.1) and (2.4) and x( = . Then

(a) x'(t) is positive, zero or negative depending on whether x(t) is less than, equal to or greater than f(t).

(b) If f(t) is non-decreasing and x(r) < f(s) for some r and s satisfying r < s, then x(t)  f(s) for some r and s satisfying r < s, then x(t) > f(s).

(d) For each t > 0 one has x(t,() ( f(t) as λ ( (.

(e) x((t)  =  (f(t) - x(t)) * e-λt  =  .

(f) x((t) is positive, zero or negative depending on whether x2(t) is less than, equal to or greater than x(t).

Proof. (a) This follows immediately from (2.1).

(b) It is not hard to show that x(s) = x(r) + . One has  ( [ f(s) - x(r)] < f(s) - x(r) since < 1. The result then follows.

(c) The proof is the same as the proof of (b), but with the inequalities reversed.

(d) We use formula (2.2). One has e-λtxo ( 0 as λ ( ( for each t > 0. So we need to show F(t,() ( f(t) as λ ( ( for each t > 0. (e-λt forms an approximate identity, i.e. (e-λt = (((t) = ((((t) where ((t) = e-t and = 1. One has

+ + .)

The second integral is small if ( is small. For fixed ( the first integral is close to f(t) and the third integral is small for ( large.

(e) If one differentiates (2.1) with respect to ( one obtains

x(t) = ( ( – x((t) )

Since x((0) = 0 one obtains

x((t) = (f(t) - x(t)) * e-λt = =

(f) This follows from (e). //

Proposition 1.2. Let x and x2 be defined by (2.1) and (2.4) and x( = . Suppose

(i) f(t) is non-decreasing for 0 ( t ( T

(ii) f(t) is decreasing for T ( t < (,

(iii) f(t) ( 0 as t ( (,

(iv) xo ≥ 0,

(v) xo < f(t) for 0 < t ( T, and

(vi) if T = 0, then xo < f(0).

Then there is τ = τ(λ) and τ2 = τ2(λ) such that T < τ < τ2 and

(a) x(t) < f(t) and x(t) is strictly increasing for 0 < t  f(t) and x(t) is strictly decreasing for τ  0. We must find δ > 0 such that τ(λ) - (  f(T+(). Since for fixed λ the maximum of x(t, λ) as a function of t occurs at t = τ(λ), one has x(τ(λ), λ) > f(T+(). Since x(τ(λ), λ) = f(τ(λ)), it follows that f(τ(λ)) > f(T+(). Since f(t) is decreasing for t ( T, one has τ(λ) < T + (. This proves (f). //

3. Properties of the functions An(t).

Note that An(t;λ1,...,λn) is a symmetric function of λ1,...,λn since convolution is commutative and associative. We begin with some formulas for An(t;λ1,...,λn).

Proposition 3.1.

(a) A2(t) = = = if λ1 ≠ λ2.

(b) An(t;λ1,...,λn) = c1e-λ1t +(+ cne-λnt with cj = .

(c) An(t; λ,...,λ) =  . In particular, A2(t; λ, λ) =  λ2te-λt.

(d) A(t) = if λ1 ≠ λ2.

(e) A(t; λ,...,λ) = . In particular, A(t, λ, λ) = λ2(1 - λt)e-λt.

Proof. (a) A2(t) = λ1λ2 = λ1λ2 e-λ1t = ,λ1 - λ2) = = . (b) is proved by induction on n. The case n = 2 is just part (a). If we convolute the formula in (b) with λn+1e-λn+1t and use part (a) we get the corresponding formula for An+1(t;λ1,...,λn+1) except for the formula for cn+1. However, the formula for cn+1 follows from the symmetry of An+1(t;λ1,...,λn+1) in λ1,...,λn+1. (c) is also proved by induction on n. The case n = 1 is just the definition of A1(t;λ). Assume the formula is true for n and convolute this formula with λe-λt . We get An+1(t; λ,...,λ) =   \I(0,t, e-λ(t-s) sn-1e-λs ds) =   =   which is the desired formula for An+1(t; λ,...,λ). (d) follows from (a), while (e) follows from (c). //

From now on we assume λ1,...,λn are strictly positive. Let τ = τn(λ1,...,λn) be the value of t for which An(t) assumes its maximum with respect to t. It is not hard to see that τ is well defined; see Propositions 3.2 and 3.3 below. In some cases we will also assume λ1 ≤ λ2 ≤... ≤ λn so that  ≤ ... ≤  ≤ .

The next proposition contains some elementary properties of τ2. It uses the function

g(z) =

which is defined if z ≠ 1 and z is not equal to a non-positive real number.

Proposition 3.2. Let τ = τ2(λ1,λ2).

(a) = - ln(z),(z - 1)2)

(b) g(z) can be extended to be analytic near z = 1 with g(1) = 1 and = - for z = 1.

(c) g(z) is a strictly decreasing function of z for 0 < z < ∞.

(d) If λ1 ≠ λ2, then τ = = where λ* is between λ1 and λ2. In particular, if λ1 < λ2 then .

(e) τ2(λ, λ) = . More generally, τn(λ,...,λ) = .

(f) τ = g( )

(g) τ(λ1,λ2) is a strictly decreasing function of λ2 for 0 < λ2 < ∞.

(h) If λ1 ≠ λ2, then = - ln( ),(λ2 - λ1)2)

(i) If λ1 = λ2 = λ, then =

(j) If one fixes λ1 and lets λ2 vary near λ1 then τ(λ1,λ2) ≈ [ 1 - ]. In other words, a 1% increase in λ2 results in approximately a ½% decrease in τ.

Proof. (a) follows from the formula for g(z). (b) It is well known that ) for | z – 1 | < 1. Therefore g(z) = ) for 0 < | z – 1 | < 1. (b) follows from this.

(c) We shall show that < 0 for 0 < z < ∞. By (b) this is true if z = 1. By part (a) we need to show 1 - - ln(z) < 0 for 0 < z < ∞ and z ≠ 1. If we let x = 1/z, then we need to show 1 - x + ln(x) < 0 for 0 < x < ∞ and x ≠ 1. This is equivalent to ln(x)  1 there exists τn = τn(t, λ1,...,λn) such that

(a) τn-1  1.

(b) An(t) < An-1(t) and An(t) is increasing in t for 0 < t  An-1(t) and An(t) is decreasing in t for τn  0 one has An(t, λ1,...,λn) ( An-1(t, λ1,...,λn-1) as λn ( (.

(f) τn is a continuously differentiable function of λ1,...,λn. If n ( 2 then   1 the function x(t) = An(t) is the solution to (1) with f(t) = An-1(t) and x(0) = 0. Therefore, this Theorem follows inductively from Propositions 2.1 and 2.2. Note that in part (f) we may assume j = n since An(t, λ1,...,λn) is a symmetric function of λ1,...,λn. //

4. Various estimates of An(t).

We are interested in estimates of An(t) from above and below. We begin with the following simple consequence of Proposition 3.1.

Proposition 4.1.

A2(t) ≤ = if λ1 < λ2.

Proof. This follows from Proposition 3.1(a). //

This proposition estimates A2(t) by from above. From Proposition 3.1(a) this estimate will be good to within 0.1% if In particular, if λ2 ≥ 8λ1 and t ≥ then this estimate will be good to within 0.1%. However, if , then the estimate of A2(t) by is not good for t in range of . Instead we might consider approximating A2(t) by for t ≥ . However, I don't think we have A2(t) ≤ λte-λ1t for t ≥ . The reason for this is as follows. If we use the approximation e-(λ2-λ1)t ≈ 1 - (λ2-λ1)t + (λ2-λ1)2t2/2 in the formula , then we get - )] which is larger than λte-λ1t if λ2 is larger than λ1 and t < . The following proposition relates to this.

Proposition 4.2. If λ1 ≤ λ2, then

A2(t) ≤ λte-λ1t for t ≥

Proof. By Proposition 3.1(a) one has A2(t) = where f(λ) = e-λt and By the generalized mean value theorem for derivatives, this is equal to where fλ(λ) = = -te-λt, gλ(λ) = = - and λ1 ≤ λ* ≤ λ2. Therefore A2(t) = . One has (λ∗t)2e-λ*t ≤ (λ1t)2e-λ1t since 2 ≤ λ1t ≤ λ∗t and xe-x is decreasing with x for x ≥ 2. The proposition then follows from this. //

Assume λ1 < λ2. Then the previous two propositions give estimates of A2(t) from above by both and λte-λ1t. However, the second estimate is only good for t ≥ , so let's restrict t to this range. Which is the better estimate for t in this range? In other words, when is < λte-λ1t? It is not hard to see that this is true for t > , where x = . So will be smaller than λte-λ1t for larger values of t. One has x > 1. As x goes from 1 to ∞, the quantity goes from ∞ to 1 and when x = 2 the quantity = 2. So, if λ2 > 2λ1, then will be smaller than λte-λ1t for all t > . On the other hand, if λ1 < λ2 < 2λ1, then will be smaller than λte-λ1t for t > , while λte-λ1t will be smaller than for < t < .

The following proposition generalizes Proposition 3.2 to An(t).

Propostion 4.3. Assume 0 < λ1 ≤ λ2,...,λn. Then

(a) ≤ 0 for t ≥ .

(b) An(t, λ1,...,λn) ≤ tn-1 e-λ1t,(n-1)!) for t ≥ .

Proof. By Proposition 2.2(f) one has ≤ 0 for t ≥ τn+1(λ1,...,λn,λj). By Proposition 2.3(f) one has τn+1(λ1,...,λn,λj) ≤ τn+1(λ1,...,λ1) and by Proposition 3.2(d) one has . Part (a) follows from these facts. To prove part (b), note that part (a) implies An(t, λ1,...,λn) ≤ An(t, λ1,...,λ1) for t ≥ . Part (b) follows from this and the fact that An(t; λ1,...,λ1) =   tn-1 e-λ1t,(n-1)!) which is proved in Proposition 3.1(c). //

5. Another Large Time Approximation.

Assume 0  ................
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