# The Generalized Pomeron Functional Equation

Authors: Yong-Guo Shi
Journal: Discrete Dynamics in Nature and Society
 (2019)

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## Abstract

This paper investigates the linear functional equation with constant coefficientsφt=κφλt+ft, where both κ>0 and 1>λ>0 are constants, f is a given continuous function on ℝ, and φ:ℝ⟶ℝ is unknown. We present all continuous solutions of this functional equation. We show that (i) if κ>1, then the equation has infinite many continuous solutions, which depends on arbitrary functions; (ii) if 0<κ<1, then the equation has a unique continuous solution; and (iii) if κ=1, then the equation has a continuous solution depending on a single parameter φ0 under a suitable condition on f.


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## Body

## 1. Introduction

Recently, Mickens [1] considered a linear functional equation(1)φt=NφtNm−N−1,where N≥1 is a positive integer, m is a positive real number, and φ is an unknown function with the domain ℝ. When m=2, this equation is called pomeron functional equation. The functional equation comes from some phenomena in physics (cf. [2, 3, 4]). Mickens gives an exact solution of (1) in [1].(2)φt=1−ct1/m,t∈0,∞,1−c−t1/m,t∈−∞,0,where c is any positive constant. Mickens mistakenly thought solution (2) as the general solution. We recall that the general solution of functional equations, which in general depends on arbitrary functions, are quite different from the one of differential equations, which in general depends on arbitrary constants.This paper considers the generalized pomeron functional equation:(3)φt=κφλt+ft,where κ>0 and 1>λ>0, f is a given continuous function on ℝ, and φ:ℝ⟶ℝ is unknown. Note that equation (3) is a linear, inhomogeneous, and functional equation. Bessenyei in [5] and Shi and Gong in [6] considered cyclic cases for some linear functional equations. Brydak in [7] showed the stability of the linear equation φfx=gxφx+Fx. Czerwik in [8] studied the continuous dependence on given functions for solutions. We use the recurrence method to present the general solution of equation (3). Compared with Kuczma’s monograph ([9], Chapter II), our method follows more directly and easily.


## 2. The General Solution to Equation (3)

Note that, ifκ≠1 and φ satisfies (3), then φ0=f0/1−κ. Furthermore, if κ=1 and (3) has a solution, then f0=0.According to the range ofκ, we consider the following three cases to investigate continuous solutions of equation (3).


### 2.1. The Caseκ>1

For this case, equation (3) has infinitely many continuous solutions, which depend on arbitrary functions.Theorem 1.
Choose arbitrarilyt0∈0,+∞ and s0∈−∞,0. Let φ0 and ψ0 be any continuous function defined in λt0,t0 and s0,λs0, respectively, such that(4)φ0t0=κφ0λt0+ft0,ψ0s0=κψ0λs0+fs0.
Defineφ+ on 0,+∞ by(5)φ+t=f01−κ,t=0,φ0t,t∈λt0,t0,φkt,t∈λkt0,λk−1t0,k=±1,±2,…,where(6)φkt=κ−1φk−1tλ+κ−1ftλ,fort∈λkt0,λk−1t0,k=±1,±2,….
Defineφ− on −∞,0 by(7)φ−t=f01−κ,t=0,ψ0t,t∈s0,λs0,ψkt,t∈λk−1s0,λks0,k=±1,±2,…,where(8)ψkt=κ−1ψk−1tλ+κ−1ftλ,fort∈λk−1s0,λks0,k=±1,±2,….
Then,(9)φt=φ+t,t∈0,+∞,φ−t,t∈−∞,0,is the general continuous solution on ℝ of equation (3).Proof.
One can check thatφ defined above satisfies equation (3).
Firstly, we prove thatφ+ is continuous in 0,+∞. By the continuity of φ0 in λt0,t0, we have limt⟶λt0+0φ+t=φ+λt0. And(10)limt⟶λt0−0φ+t=limt⟶t0−0φ1λt=limt⟶t0−0κ−1φ0t+κ−1ft=φ0λt0=φ+λt0.
Thus,φ+ is continuous for t∈λt0,t0. Now 0,+∞=∪λk+1t0,λkt0. Assume that φ+ is continuous for t∈λkt0,λk−1t0, k>0. We have for t∈λk+1t0,λkt0,(11)φk+1t=κ−1φktλ+κ−1ftλ.
Thus,φ+ is continuous for t∈λk+1t0,λkt0. Similarly, we can show that φ+ is continuous for t∈λk+1t0,λkt0 for k<0.
Secondly, we shall prove that(12)limt⟶0+φ+t=f01−κ.
Lett∈0,λt0. Then, t=λnx for some n∈ℕ and x∈λt0,t0. Hence, we obtain(13)φ+t=φ+λnx=φnλnx=κφn−1λn−1x+fλn−1x=κ2φn−2λn−2x+κfλn−2x+fλn−1x=⋮=κnφ0x+κn−1fx+⋯+κfλn−2x+fλn−1x.
Letε>0. Since f is continuous, there exists N∈ℕ such that, for every n∈ℕ with n>N, we have fλnx−f0<ε. Therefore, we obtain(14)∑i=1nκi−1fλn−ix−f01−κ=∑i=1nκi−1fλn−ix−f0∑i=1∞κi−1≤∑i=1n−Nκi−1fλn−ix−f0+∑i=n−N+1nκi−1fλn−ix+f0∑i=n−N+1∞κi−1≤ε∑i=1n−Nκi−1+∑i=n−N+1nκi−1fλn−ix+f0∑i=n−N+1∞κi−1≤ε1−κ+f0+supt≤xft∑i=n−N+1∞κi−1=ε1−κ+κn−N1−κf0+supt≤xft.
Together with (13) and (14), we have (15)limt⟶0+φ+t−f01−κ=limn⟶+∞φ+λnx−f01−κ=0.
Therefore,φ+ is continuous in 0,+∞. Similarly, one can prove that φ− is continuous in −∞,0.
On the other hand, every continuous solution can be obtained in this manner.


### 2.2. The Case0<κ<1

The corresponding homogeneous equation of (3) is given by(16)φt=κφλt.According to the superposition principle for the linear equation, the general solution of (3) is given by the additions of the general solution to the corresponding homogeneous equation (16) and a particular solution of the original inhomogeneous equation (3), cf. [9].Lemma 1.
Supposeκ,λ∈0,1. Then, φ=0 is the only continuous solution of equation (16).Proof.
We use the proof by contradiction.
Assumeφ0≠0. Then, φt≠0 in ℝ. If fact, if there exists a t0 such that φt0=0, then by φt=κφλt,(17)φ0=limn⟶∞φλnt0=limn⟶∞κ−nφt0=0,which contradicts the assumption. Thus, it follows from φt=κφλt that(18)limn⟶∞κ−n=limn⟶∞φλntφt=φ0φt.
This is a contradiction. Therefore,φ0=0.
Assume that there exists an intervalJ=a,b⊂ℝ\0 such that(19)φt>c>0,fort∈J.
Given anε>0, we can find a δ>0 such that(20)φt<εc,fort∈0,δ.
Thus, there exists anN>0 such that λnb<δ for n≥N, which implies the inequality λnt<δ for t∈J and n≥N. Hence, we have(21)κ−n=φλntφt<ε.
This is a contradiction. Therefore,φt≡0 on ℝ.Theorem 2.
Supposeκ,λ∈0,1. Then, equation (3) has a unique continuous solution:(22)φt=∑n=0∞κnfλnt.Proof.
We can obtain the uniqueness from Lemma1. It suffices to prove that (22) is a continuous solution of equation (3).
Choose an arbitraryc∈0,∞. Put t∈−c,c. One can see that λnt∈−c,c for n≥0. Let(23)M=sup0,cft.
We have(24)κnfλnt≤κnM,forn≥0,t∈−c,c.
Consequently, the series∑n=1∞κnfλnt in (22) converges uniformly in −c,c for every c∈0,∞. Therefore, φt in (22) is a continuous function in ℝ. Further,(25)κφλt+ft=∑n=1∞κnfλnt+ft=∑n=0∞κnfλnt=φt.
Thus,φt is the only continuous solution of equation (3).


### 2.3. The Caseκ=1

Ifκ=1, then equation (3) becomes the following form:(26)φt=φλt+ft.Lemma 2.
Supposeφt is a continuous solution of equation (26). Then, (27)φt=η+∑j=0∞fλjt,whereη=φ0.Proof.
Letφ be a continuous solution of equation (26). Then,(28)φt=φλt+ft=φλ2t+fλt+ft=⋯=φλnt+∑j=0n−1fλjt=φ0+∑j=0∞fλjt.
Substitutingt=0 into equation (26), we have φ0=φ0+f0. Consequently, f0=0. □
Note that the series∑j=0∞fλjt in (27) may not converge. Some additional assumption on f should be made to guarantee the existence of continuous solutions of equation (26).
We present one sufficient condition onf in the following theorem.Theorem 3.
Suppose that there exists positive constantsδ, σ, and L such that (29)ft≤Ltσ,fort∈−δ,δ.
Then, equation (26) has a continuous solution.Proof.
Since(30)∑j=0∞fλjt≤∑j=0∞Lλjtσ=Ltσ1−λσ,the series (27) converges uniformly in the interval −c,c for any fixed c∈0,∞. It follows from Lemma 2 that φt in (27) is a continuous solution.


## 2.1. The Caseκ>1

For this case, equation (3) has infinitely many continuous solutions, which depend on arbitrary functions.Theorem 1.
Choose arbitrarilyt0∈0,+∞ and s0∈−∞,0. Let φ0 and ψ0 be any continuous function defined in λt0,t0 and s0,λs0, respectively, such that(4)φ0t0=κφ0λt0+ft0,ψ0s0=κψ0λs0+fs0.
Defineφ+ on 0,+∞ by(5)φ+t=f01−κ,t=0,φ0t,t∈λt0,t0,φkt,t∈λkt0,λk−1t0,k=±1,±2,…,where(6)φkt=κ−1φk−1tλ+κ−1ftλ,fort∈λkt0,λk−1t0,k=±1,±2,….
Defineφ− on −∞,0 by(7)φ−t=f01−κ,t=0,ψ0t,t∈s0,λs0,ψkt,t∈λk−1s0,λks0,k=±1,±2,…,where(8)ψkt=κ−1ψk−1tλ+κ−1ftλ,fort∈λk−1s0,λks0,k=±1,±2,….
Then,(9)φt=φ+t,t∈0,+∞,φ−t,t∈−∞,0,is the general continuous solution on ℝ of equation (3).Proof.
One can check thatφ defined above satisfies equation (3).
Firstly, we prove thatφ+ is continuous in 0,+∞. By the continuity of φ0 in λt0,t0, we have limt⟶λt0+0φ+t=φ+λt0. And(10)limt⟶λt0−0φ+t=limt⟶t0−0φ1λt=limt⟶t0−0κ−1φ0t+κ−1ft=φ0λt0=φ+λt0.
Thus,φ+ is continuous for t∈λt0,t0. Now 0,+∞=∪λk+1t0,λkt0. Assume that φ+ is continuous for t∈λkt0,λk−1t0, k>0. We have for t∈λk+1t0,λkt0,(11)φk+1t=κ−1φktλ+κ−1ftλ.
Thus,φ+ is continuous for t∈λk+1t0,λkt0. Similarly, we can show that φ+ is continuous for t∈λk+1t0,λkt0 for k<0.
Secondly, we shall prove that(12)limt⟶0+φ+t=f01−κ.
Lett∈0,λt0. Then, t=λnx for some n∈ℕ and x∈λt0,t0. Hence, we obtain(13)φ+t=φ+λnx=φnλnx=κφn−1λn−1x+fλn−1x=κ2φn−2λn−2x+κfλn−2x+fλn−1x=⋮=κnφ0x+κn−1fx+⋯+κfλn−2x+fλn−1x.
Letε>0. Since f is continuous, there exists N∈ℕ such that, for every n∈ℕ with n>N, we have fλnx−f0<ε. Therefore, we obtain(14)∑i=1nκi−1fλn−ix−f01−κ=∑i=1nκi−1fλn−ix−f0∑i=1∞κi−1≤∑i=1n−Nκi−1fλn−ix−f0+∑i=n−N+1nκi−1fλn−ix+f0∑i=n−N+1∞κi−1≤ε∑i=1n−Nκi−1+∑i=n−N+1nκi−1fλn−ix+f0∑i=n−N+1∞κi−1≤ε1−κ+f0+supt≤xft∑i=n−N+1∞κi−1=ε1−κ+κn−N1−κf0+supt≤xft.
Together with (13) and (14), we have (15)limt⟶0+φ+t−f01−κ=limn⟶+∞φ+λnx−f01−κ=0.
Therefore,φ+ is continuous in 0,+∞. Similarly, one can prove that φ− is continuous in −∞,0.
On the other hand, every continuous solution can be obtained in this manner.


## 2.2. The Case0<κ<1

The corresponding homogeneous equation of (3) is given by(16)φt=κφλt.According to the superposition principle for the linear equation, the general solution of (3) is given by the additions of the general solution to the corresponding homogeneous equation (16) and a particular solution of the original inhomogeneous equation (3), cf. [9].Lemma 1.
Supposeκ,λ∈0,1. Then, φ=0 is the only continuous solution of equation (16).Proof.
We use the proof by contradiction.
Assumeφ0≠0. Then, φt≠0 in ℝ. If fact, if there exists a t0 such that φt0=0, then by φt=κφλt,(17)φ0=limn⟶∞φλnt0=limn⟶∞κ−nφt0=0,which contradicts the assumption. Thus, it follows from φt=κφλt that(18)limn⟶∞κ−n=limn⟶∞φλntφt=φ0φt.
This is a contradiction. Therefore,φ0=0.
Assume that there exists an intervalJ=a,b⊂ℝ\0 such that(19)φt>c>0,fort∈J.
Given anε>0, we can find a δ>0 such that(20)φt<εc,fort∈0,δ.
Thus, there exists anN>0 such that λnb<δ for n≥N, which implies the inequality λnt<δ for t∈J and n≥N. Hence, we have(21)κ−n=φλntφt<ε.
This is a contradiction. Therefore,φt≡0 on ℝ.Theorem 2.
Supposeκ,λ∈0,1. Then, equation (3) has a unique continuous solution:(22)φt=∑n=0∞κnfλnt.Proof.
We can obtain the uniqueness from Lemma1. It suffices to prove that (22) is a continuous solution of equation (3).
Choose an arbitraryc∈0,∞. Put t∈−c,c. One can see that λnt∈−c,c for n≥0. Let(23)M=sup0,cft.
We have(24)κnfλnt≤κnM,forn≥0,t∈−c,c.
Consequently, the series∑n=1∞κnfλnt in (22) converges uniformly in −c,c for every c∈0,∞. Therefore, φt in (22) is a continuous function in ℝ. Further,(25)κφλt+ft=∑n=1∞κnfλnt+ft=∑n=0∞κnfλnt=φt.
Thus,φt is the only continuous solution of equation (3).


## 2.3. The Caseκ=1

Ifκ=1, then equation (3) becomes the following form:(26)φt=φλt+ft.Lemma 2.
Supposeφt is a continuous solution of equation (26). Then, (27)φt=η+∑j=0∞fλjt,whereη=φ0.Proof.
Letφ be a continuous solution of equation (26). Then,(28)φt=φλt+ft=φλ2t+fλt+ft=⋯=φλnt+∑j=0n−1fλjt=φ0+∑j=0∞fλjt.
Substitutingt=0 into equation (26), we have φ0=φ0+f0. Consequently, f0=0. □
Note that the series∑j=0∞fλjt in (27) may not converge. Some additional assumption on f should be made to guarantee the existence of continuous solutions of equation (26).
We present one sufficient condition onf in the following theorem.Theorem 3.
Suppose that there exists positive constantsδ, σ, and L such that (29)ft≤Ltσ,fort∈−δ,δ.
Then, equation (26) has a continuous solution.Proof.
Since(30)∑j=0∞fλjt≤∑j=0∞Lλjtσ=Ltσ1−λσ,the series (27) converges uniformly in the interval −c,c for any fixed c∈0,∞. It follows from Lemma 2 that φt in (27) is a continuous solution.


## 3. Conclusion

This paper investigates one kind of linear functional equation with constant coefficientsφt=κφλt+ft, where κ>0 and λ∈0,1, f is a given continuous function on ℝ, and φ:ℝ⟶ℝ is unknown. If κ>1, then the equation has infinite many continuous solutions, which depend on arbitrary functions. If 0<κ<1, then the equation has the unique continuous solution:(31)φt=∑n=0∞κnfλnt.Ifκ=1 and the equation has a continuous solution, then f0=0 and(32)φt=η+∑j=0∞fλjt,whereη=φ0.For the last case, we only give one sufficient condition for the existence of continuous solutions. The problem for the necessary and sufficient condition still remains open.



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