
We investigate periodicity of functions related to the Gompertz difference equation. In particular, we derive difference equations that must be satisfied to guarantee periodicity of the solution.
Citation: Tom Cuchta, Nick Wintz. Periodic functions related to the Gompertz difference equation[J]. Mathematical Biosciences and Engineering, 2022, 19(9): 8774-8785. doi: 10.3934/mbe.2022407
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We investigate periodicity of functions related to the Gompertz difference equation. In particular, we derive difference equations that must be satisfied to guarantee periodicity of the solution.
We study the Gompertz difference equation
yΔ(t)=(⊖r)(t)y(t)(K(t)+a+∫t0yΔ(τ)y(τ)Δτ),y(0)=y0, | (1.1) |
as well as periodic functions that arise from it. This is to say when ω∈{1,2,…}, f:N0→R is ω-periodic if f(t+ω)=f(t) for all t∈N0. Here, (⊖r)(t)=−r(t)1+r(t) is the time scales analogue of the growth rate while K(t) is the analogue of the carrying capacity at time t from the traditional continuous Gompertz model. Throughout, we will use notation inspired from time scales calculus for the time scale T=N0, including σ(t)=t+1, yΔ(t)=y(σ(t))−y(t), and the integration symbol representing summation, i.e. ∫baf(t)Δt=b−1∑k=af(k). See the monograph [1] for the usual introduction to dynamic equations on time scales and see the recent texts on first and second order boundary value problems on time scales [2] and its companion book on third, fourth, and higher-order boundary value problems on time scales [3] for more recent books.
In [4], the model (1.1) as well as a second model without the ⊖ was introduced, solved, and bounds of its solutions were established. Three discrete fractional analogues of (1.1) were explored in [5] by changing the difference to a fractional difference and exploring defining the logarithm with a fractional integral. These three models were compared to another existing fractional Gompertz difference equation [6], which was built around using the classical logarithm instead of a time scales logarithm. The solution of (1.1) can be normalized to create a probability distribution which was studied in [7] where bounds on the expected value were derived and a connection between the classical continuous Gompertz distribution with the q-geometric distribution of the second kind was established. An alternative approach to Gompertz equations on time scales appears in [8] which uses the ⊙ operation to define a Gompertz dynamic equation.
Gompertz models have been used to study a number of applications in both discrete and continuous settings. This includes studying the growth rate of tumors [9,10], modeling growth of prey in predator-prey dynamics [11], as well as study the change in cost in adopting new technologies [12,13], effect of seasonality for Gompertz models using time series [14], and the spread of COVID-19 [15,16].
Before introducing our main results, some preliminary definitions and results are in order. Equation (1.1) has the unique solution y(t)=y0ep(t,0), where
p(t)=(⊖r)(t)(ae⊖r(t,0)−∫t0(⊖r)(s)e⊖r(t,σ(s))K(s)Δs−K(t)). | (2.1) |
Here, ef:N0×N0→R, called the discrete exponential, is the unique solution of the initial value problem yΔ=fy,y(0)=1. We often make use of the so-called "simple useful formula, "
ep(σ(t),0)=(1+p(t))ep(t,0). | (2.2) |
when rewriting exponentials.
Time scales integration by parts is given by
∫baf(τ)gΔ(τ)Δτ=f(t)g(t)|ba−∫bafΔ(τ)g(σ(τ))Δτ. | (2.3) |
A function f:N0→C is said to be of exponential order α [17,Definition 4.1] if there is an α∈R with 1+α>0 and a M>0 such that |f(t)|≤Meα(t,s) for all t∈N0. In particular, [17,Lemma 4.4] shows that if f is of exponential order α and |z+1|>1+α, then limt→∞f(t)e⊖z(t,0)=0.
The time scales Laplace transform is given by [1,Section 3.10]
L{f}(z)=∫∞0f(τ)e⊖z(σ(τ),0)Δτ. |
which for T=N0 is a scaled and shifted Z-transform. It's known [18,Theorem 3.2] that if w is a regressive constant and T=N0, then
L{feσw(⋅,s)}(z)=L{f}(z⊖w), | (2.4) |
and if X(t)=∫t0x(τ)Δτ, then [18,Theorem 6.4]
L{X}(z)=1zL{x}(z). | (2.5) |
A well-known identity for the T=N0 delta derivative operator is
f(k+ω)=ω∑j=0(ωj)fΔj(k). | (2.6) |
The Laplace transform for differences of f is given by
L{fΔj}(z)=zjL{x}(z)−j−1∑ℓ=0zℓfΔj−ℓ−1(0). | (2.7) |
The Laplace transform of the shifted argument is useful for the sequel.
Lemma 1. If f is of exponential order α, then
L{f(⋅+ω)}(z)=(z+1)ωL{f}(z)−ω∑j=0(ωj)j−1∑ℓ=0zℓfΔj−ℓ−1(0). | (2.8) |
Proof. Applying the Laplace transform to (2.6) and using (2.7), we have
L{f(⋅+ω)}(z)=ω∑j=0(ωj)L{fΔj}(z)=ω∑j=0(ωj)[zjL{f}(z)−j−1∑ℓ=0zℓfΔj−ℓ−1(0)]=L{f}(z)(ω∑j=0(ωj)zj)−ω∑j=0(ωj)j−1∑ℓ=0zℓfΔj−ℓ−1(0). |
An application of the binomial theorem to the first summation completes the proof.
Now we calculate the discrete Laplace transform of a certain time-dependent delta integral.
Lemma 2. If f is of exponential order α and X(t)=∫t+ωtf(τ)Δτ, then for all z∈C with |1+z|>1+α,
L{X}(z)=1zω−1∑k=0f(k)+(z+1)ω−1zL{f}(z)−1zω∑j=0(ωj)j−1∑ℓ=0zℓfΔj−ℓ−1(0). | (2.9) |
Proof. First use (2.2) to see
e⊖z(k+1,0)=(1+(⊖z))e⊖z(k,0)=11+ze⊖z(k,0). |
Calculate, where k is understood to be the variable,
[k+ω−1∑ℓ=kf(ℓ)]Δ=k+ω∑ℓ=k+1f(ℓ)−k+ω−1∑ℓ=kf(ℓ)=f(k+ω)−f(k). |
Now since X(t)=t+ω−1∑ℓ=tf(ℓ), (2.2) reveals
L{X}(z)=∫∞0(τ+ω−1∑ℓ=τf(ℓ))e⊖z(σ(τ),0)Δτ=11+z∫∞0(τ+ω−1∑ℓ=τf(ℓ))e⊖z(τ,0)Δτ. |
Since (⊖z) is constant and eΔ⊖z=(⊖z)e⊖z, we observe
L{X}(z)=11+z1(⊖z)∫∞0(τ+ω−1∑ℓ=τf(ℓ))eΔ⊖z(τ,0)Δτ=−1z∫∞0(τ+ω−1∑ℓ=τf(ℓ))eΔ⊖z(τ,0)Δτ. |
Apply (2.3) to obtain
L{X}(z)=−1z(τ+ω−1∑ℓ=τf(ℓ))e⊖z(τ,0)|τ=∞τ=0+1z∫∞0(f(τ+ω)−f(τ))e⊖z(σ(τ),0)Δτ. |
Thus we have
L{X}(z)=1zω−1∑k=0f(k)+1zL{f(⋅+ω)}(z)−1zL{f}(z). |
Applying (2.8) to the middle term of the right-hand side completes the proof.
First we establish which functions r yield e⊖r to be ω-periodic.
Lemma 3. The discrete exponential is periodic, meaning
e⊖r(t+ω,0)=e⊖r(t,0) | (3.1) |
if and only if
r(t+ω−1)=−1+t+ω−2∏k=t11+r(k). | (3.2) |
Proof. First calculate
1+(⊖r)(t)=1−r(t)1+r(t)=11+r(t). |
Now, (3.1) becomes
t+ω−1∏k=011+r(k)=t−1∏k=011+r(k), |
hence 1=t+ω−1∏k=t11+r(k), which is equivalent to r(t+ω−1)=−1+t+ω−2∏k=t11+r(k). Since all steps are reversible, the proof is complete. Since the ω-periodicity of e⊖r is equivalent to r satisfying the difference equation (3.2), solving it is of importance.
Lemma 4. If r(0),…,r(ω−2) are known, then the unique solution of (3.2) is ω-periodic.
Proof. Use (3.2) with t=0 to generate the ωth value
r(ω−1)=−1+1(1+r(0))(1+r(1))…(1+r(ω−2)). |
We claim that the function r is ω-periodic.
From (3.2), we obtain
r(t+ω)=−1+t+ω−1∏k=t+111+r(k)=−1+1(1+r(t+1))(1+r(t+2))…(1+r(t+ω−1)). |
But also by (3.2),
1+r(t+ω−1)=t+ω−2∏k=t11+r(k)=1(1+r(t))(1+r(t+1))…(1+r(t+ω−2)). |
Therefore,
![]() |
completing the proof.
By (2.1), when p is ω-periodic, p(t+ω)=p(t) expands to
(⊖r)(t+ω)(ae⊖r(t+ω,0)−∫t+ω0(⊖r)(s)e⊖r(t+ω,σ(s))K(s)Δs−K(t+ω))=(⊖r)(t)(ae⊖r(t,0)−∫t0(⊖r)(s)e⊖r(t,σ(s))K(s)Δs−K(t)). | (3.3) |
Theorem 5. If r solves (3.2), then p(t+ω)=p(t) if and only if
K(t+ω)=K(t)−∫t+ωt(⊖r)(s)e⊖r(t+ω,σ(s))K(s)Δs. |
Proof. By Lemma 3, e⊖r(⋅,0) is ω-periodic. By Lemma 4, r is ω-periodic, hence (⊖r) is also ω-periodic. Using (3.3), we divide by (⊖r)(t+ω) and subtract ae⊖r(t+ω,0) to obtain
−∫t+ω0(⊖r)(s)e⊖r(t+ω,σ(s))K(s)Δs−K(t+ω)=−∫t0(⊖r)(s)e⊖r(t,σ(s))K(s)Δs−K(t). |
Hence
0=K(t+ω)−K(t)+∫t0(⊖r)(s)[e⊖r(t+ω,σ(s))−e⊖r(t,σ(s))]K(s)Δs+∫t+ωt(⊖r)(s)e⊖r(t+ω,σ(s))Δs. | (3.4) |
By the semigroup property and the periodicity of e⊖r(⋅,0),
e⊖r(t+ω,σ(s))−e⊖r(t,σ(s))=[e⊖r(t+ω,0)−e⊖r(t,0)]e⊖r(0,σ(s))=0, | (3.5) |
and applying (3.5) to (3.4) completes the proof.
Theorem 6. If r(t)=r is constant and K is ω-periodic, then p(t+ω)=p(t) if and only if
K(t+ω−1)=ar(1+r)t[1−1(1+r)ω]+11+r∫t0K(s)(1+r)t−s−1Δs−11+r∫t+ω−10K(s)(1+r)t+ω−s−1Δs. |
Proof. From (3.3), since r is constant, so is (⊖r), hence both (⊖r)(t+ω) and (⊖r)(t) can be divided off. Similarly, since K(t+ω)=K(t), those terms also vanish in (3.3). What remains is
ae⊖r(t+ω,0)−(⊖r)∫t+ω0e⊖r(t+ω,σ(s))K(s)Δs=ae⊖r(t,0)−(⊖r)∫t0e⊖r(t,σ(s))K(s)Δs |
Thus,
a(1+r)t+ω+r1+r∫t+ω0K(s)(1+r)t+ω−s−1Δs=a(1+r)t+r1+r∫t0K(s)(1+r)t−σ(s)Δs |
Now
a(1+r)t+ω+r1+r∫t+ω−10K(s)(1+r)t+ω−s−1Δs+rK(t+ω−1)=a(1+r)t+r1+r∫t0K(s)(1+r)t−s−1Δs, |
and solving for K(t+ω−1) completes the proof.
Define α(t,s):=(⊖r)(s)e⊖r(t,σ(s))K(s) and
β(t):=1(⊖r)(t+ω−1)[1−1ep(t+ω−1,t)]+ae⊖r(t+ω−1,0). |
Theorem 7. If r:N0→R, then the function t↦ep(t,0) is ω-periodic if and only if
K(t+ω−1)=β(t)−∫t+ω−10α(t+ω−1,s)Δs. | (4.1) |
Proof. If ep is ω-periodic, then using the semigroup property of ep, we obtain
p(t+ω−1)=−1+1ep(t+ω−1,t). |
By (2.1), this becomes
(⊖r)(t+ω−1)[ae⊖r(t+ω−1,0)−∫t+ω−10α(t+ω−1,s)Δs−K(t+ω−1)]=−1+1ep(t+ω−1,t), | (4.2) |
which we rearrange to obtain (4.1). All steps are reversible so the converse is also true, completing the proof.
We provide a numerical example of Theorem 7 in Figure 1. It is difficult in general to solve (4.1) in closed form, but if r is a constant function, then it may be solved with Laplace transform techniques.
Theorem 8. If r∈Rc(N0,R) and K is of exponential order α, then for all |z+1|>1+α, the Laplace transform of (4.1) is
L{K}(z)=1(z+1)ω−1−r((z⊕r)+1)ω−1(1+r)ω−1(z⊕r)×[L{β}(z)+r(1+r)ω−1(z⊕r)ω−2∑k=0er(σ(k),0)K(k)+ω−1∑j=0(ω−1j)j−1∑ℓ=0zℓ[KΔk−ℓ−1(0)−r(1+r)ω−1(z⊕r)[er(σ(⋅),0)K(⋅)]Δk−ℓ−1(0)]]. |
Proof. By the semigroup and reciprocal properties for the discrete exponential, (4.1) becomes
K(t+ω−1)=β(t)−(⊖r)e⊖r(t+ω−1,0)∫t+ω−10er(σ(s),0)K(s)Δs. |
By (2.8), we know that
L{K(⋅+ω−1)}(z)=(z+1)ω−1L{K}(z)−ω−1∑j=0(ω−1j)j−1∑ℓ=0zℓKΔj−ℓ−1(0) |
Using (2.2), compute
e⊖r(t+ω−1,0)=e⊖r(σω−1(t),0)=(1+(⊖r))ω−2e⊖r(t+1,0)=e⊖r(t+1,0)(1+r)ω−2. | (4.3) |
Let
g(t)=(⊖r)e⊖r(t+ω−1,0)∫t0er(σ(s),0)K(s)Δs=−r1+re⊖r(t+ω−1,0)∫t0er(σ(s),0)K(s)Δs=−r(1+r)ω−1e⊖r(t+1,0)∫t0er(σ(s),0)K(s)Δs |
Using (2.4), (2.5), and (4.3), we compute
L{g}(z)=−r(1+r)ω−1L{∫⋅0er(σ(s),0)K(s)Δs}(z⊕r)=−r(1+r)ω−1(z⊕r)L{er(σ(⋅),0)K(⋅)}(z⊕r)=−r(1+r)ω−1(z⊕r)L{K}(z). |
Now let h(t)=∫t+ω−1t(⊖r)(s)e⊖r(t+ω−1,σ(s))K(s)Δs. Using (2.9),
L{h}(z)=−r1+rL{∫⋅+ω−1⋅e⊖r(⋅+ω−1,σ(s))K(s)Δs}(z)=−r(1+r)ω−1L{e⊖r(σ(⋅),0)∫⋅+ω−1⋅er(σ(s),0)K(s)Δs}(z)=−r(1+r)ω−1L{∫⋅+ω−1⋅er(σ(s),0)K(s)Δs}(z⊕r)=−r(1+r)ω−1[1z⊕rω−2∑k=0er(σ(k),0)K(k)+((z⊕r)+1)ω−1−1z⊕rL{er(σ(⋅),0)K(⋅)}(z⊕r)=−1z⊕rω−1∑j=0(ω−1j)j−1∑ℓ=0zℓ[er(σ(⋅),0)K(⋅)]Δj−ℓ−1(0)]=−r(1+r)ω−1(z⊕r)[ω−2∑k=0er(k+1,0)K(k)+((z⊕r)+1)ω−1−1)L{er(σ(⋅),0)K(⋅)}(z⊕r)=−ω−1∑j=0(ω−1j)j−1∑ℓ=0zℓ[er(σ(⋅),0)K(⋅)]Δj−ℓ−1(0)]. |
One further step applying (2.8) on the second term yields
L{h}(z)=−r(1+r)ω−1(z⊕r)[ω−2∑k=0er(σ(k),0)K(k)+(((z⊕r)+1)ω−1−1)L{K}(z)=−ω−1∑j=0(ω−1j)j−1∑ℓ=0zℓ[er(σ(⋅),0)K(⋅)]Δj−ℓ−1(0)]. |
Therefore we have shown that the Laplace transform of (4.1) is
(z+1)ω−1L{K}(z)−ω−1∑j=0(ω−1j)j−1∑ℓ=0zℓΔj−ℓ−1K(0)=L{β}(z)+r(1+r)ω−1(z⊕r)[ω−2∑k=0er(k+1,0)K(k)+((z⊕r)+1)ω−1L{K}(z)=−ω−1∑j=0(ω−1j)j−1∑ℓ=0zℓ[er(σ(⋅),0)K(⋅)]Δj−ℓ−1(0)]. |
Solving for L{K}(z) completes the proof.
Now we consider the reverse case of Theorem 7 where K is given and r must be solved for.
Theorem 9. If K:N0→R is known, then the function t↦ep(t,0) is ω-periodic if and only if
r(t+ω−1)=⊖(−1+1ep(t+ω−1,t)ae⊖r(t+ω−1,0)−∫t+ω−10α(t+ω−1,s)Δs−K(t+ω−1)). |
Proof. By solving (4.2) for (⊖r)(t+ω−1), we obtain
(⊖r)(t+ω−1)=−1+1ep(t+ω−1,t)ae⊖r(t+ω−1,0)−∫t+ω−10α(t+ω−1,s)Δs−K(t+ω−1), |
and so taking ⊖ of both sides completes the proof, since all steps are algebraically reversible.
We provide a numerical example of Theorem 9in Figure 2.
We have explored periodicity of functions related to the Gompertz difference equation (1.1). In Theorem 5, we found a difference equation that K must satisfy in order for p to be ω-periodic whenever r is itself ω-periodic. Theorem 6 does the same thing, but when r is constant. In Theorem 7, we considered ω-periodicity of solutions of (1.1) and arrived at difference equations that K must solve in order to guarantee it. In Theorem 8 we solved that difference equation in the special case of a constant r using Laplace transform techniques. Finally, in Theorem 9, we instead found a difference equation that r must solve if K is known.
Future work in this area includes the extension of the results to ω-periodic functions on more general time scales as studied in [19,20]. Throughout, we have showcased the basic framework for these results on a more general time scale to aid in such a generalization. The connections between Volterra integral equations and generalizations of (1.1) are of interest, as well as interpreting the function K as a periodic control for population models.
We thank the referees for their valuable contribution in improving this paper. This research was made possible by NASA West Virginia Space Grant Consortium, Training Grant #80NSSC20M0055.
The authors declare that there is no conflict of interest.
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