# Analytic Solutions of a Second-Order Functional Differential Equation with a State Derivative Dependent Delay

Authors: Jiraphorn Somsuwan; Keaitsuda Maneeruk Nakprasit
Journal: International Journal of Mathematics and Mathematical Sciences
 (2015)

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## Abstract

We investigate an analytic solution of the second-order differential equation with a state derivative dependent delay of the formx″(z)=x(p(z)+bx′(z)). Considering a convergent power series g(z) of an auxiliary equation γ2g″(γz)g′(z)=[g(γ2z)-p(g(γz))]γg′(γz)(g′(z))2+p′′(g(z))(g′(z))3+γg′(γz)g″(z) with the relation p(z)+bx′(z)=g(γg-1(z)), we obtain an analytic solution x(z). Furthermore, we characterize a polynomial solution when p(z) is a polynomial.


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## Body

## 1. Introduction

The functional differential equation,(1)xmz=fz,xm1z-τ1z,xm2z-τ2z,…,xmkz-τkz,where all mi≥0, τi≥0, provides a mathematical model for a physical or biological system in which the rate of change of system is determined not only by its present state, but also by its history (see [1, 2]). In recent years, many authors studied the existence and the uniqueness of an analytic solution of a variety of these equations. In 1984, Eder [3] classified solutions of the functional differential equation x′(z)=x(x(z)) by using the Banach fixed point theorem. Let a and b be nonzero complex constants and p(z) a complex function. The first-order functional differential equation x′(z)=x(az+bx(z)), x′(z)=x(p(z)+bx(z)), and x′(z)=x(p(z)+bx′(z)) has been studied by Si and Cheng [4], Qiu and Liu [5], and Zhang [6], respectively.In 2001 [7], Si and Wang investigated the existence of analytic solution of the second-order functional differential equation:(2)x′′z=xaz+bxz.In 2009, Liu and Li [8] studied the equation(3)x′′z+cx′z=xaz+bxz.Observe that (3) can be reduced to (2) by setting c=0.Next, the equation(4)x′′z=xaz+bx′zhas been studied by Si and Wang [9].In order to obtain analytic solutions of (4), they constructed a corresponding auxiliary equation with parameter γ. The existence of solutions of an auxiliary equation depends on the condition of a parameter γ, such as γ is in the unit circle and γ is a root of unity which satisfies the Diophantine condition.In this paper, we study the existence of analytic solutions of the second-order differential equation with a state derivative dependent delay of the form(5)x′′z=xpz+bx′z.If p(z)=az, then (5) reduces to (4).Note that in this paper, we will study three cases of parameterγ in a corresponding auxiliary equation. One of them is the case that γ is a root of unity satisfying the Brjuno condition.To construct an auxiliary equation, we set(6)yz=pz+bx′z.Then(7)xz=xz0+1b∫z0zys-psds,where z0 is a complex constant. In particular, we have(8)xyz=xz0+1b∫z0yzys-psds.Applying relations (6) and (8) to (5), we obtain(9)1by′z-p′z=xz0+1b∫z0yzys-psds.We construct the corresponding equation by differentiating both sides of (9) with respect to z. This yields(10)y′′z-p′′z=yyz-pyzy′z.


## 2. Analytic Solutions of (10)

Consider the auxiliary equation(11)γ2g′′γzg′z=gγ2z-pgγzγg′γzg′z2+p′′gzg′z3+γg′γzg′′zwith initial value conditions g(0)=0 and g′(0)=η≠0, where γ,η are complex numbers. Observe that if g(z) is an analytic solution of (11), then (10) has an analytic solution of the from y(z)=g(γg-1(z)). Equation (11) can be reduced equivalently to the integro-differential equation(12)γg′γz=g′zγ+∫0zγg′γsgγ2s-γg′γspgγs+p′′gsg′sds,where g(0)=0 and g′(0)=η≠0. To construct analytic solutions of (12), we separate our study on the conditions of the parameter γ as follows:(H1)
0<γ<1;(H2)
γ=e2πiθ, where θ∈R∖Q is a Brjuno number; that is, B(θ)=∑k=0∞log⁡qk+1/qk<∞, where {pk/qk} denotes the sequence of partial fraction of the continued fraction expansion of θ;(H3)
γ=e2πiq/p for some integers p∈N with p≥2 and q∈Z∖{0}, and γ≠e2πil/k for all 1≤k≤p-1 and l∈Z∖{0}.From now on, we letp(z) be an analytic function in a neighborhood of the origin. Then we represent p(z) by a power series ∑n=0∞pnzn.Theorem 1.
Letγ satisfy condition(H1). Then (11) has an analytic solution(13)gz=∑n=1∞cnznin a neighborhood of the origin such that g(0)=0, g′(0)=η, where η is a nonzero complex number.Proof.
Sincep(z) is analytic in a neighborhood of the origin, there exists a constant ρ such that pn≤ρn-1 for n≥1. Substituting (13) into (12) and comparing coefficients of zn(n=1,2,…), we get(14)γc1=γc1,2γ2c2=2γc2-γc12p0+2c12p2and in general for n≥1(15)n+2γn+2-γcn+2=∑k=1n∑i=1kin-k+1γ2k-i+2k+1·cick-i+1cn-k+1-∑k=1n+1kp0γn-k+2ckcn-k+2-∑k=1n∑i=1kin-k+1γk+1k+1·cicn-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1pmcl1⋯clm+∑k=1n+12kp2ckcn-k+2+∑k=1n∑i=1kin-k+1k+1·cicn-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1pm+2m+2m+1cl1⋯clm.The first expression allows us to choose c1=η≠0 and the second expression implies c2=(2p2-γp0)c12/2γ(γ-1). Consequently, the sequence {cn}n=3∞ is successively determined by the last expression in a unique manner. This implies that (11) has a formal power series solution.
Next, we show that the power seriesg(z) converges in a neighborhood of the origin. Since |pn|≤ρn-1 and limn→∞⁡1/γn+1-1=-1 for 0<γ<1, there exists a positive constant M such that(16)cn+2≤M∑k=1n∑i=1kcick-i+1cn-k+1+2∑k=1n+1ckcn-k+2+∑k=1n∑i=1kcicn-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1cl1⋯clm+∑k=1n∑i=1kcicn-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1m+2m+1cl1⋯clm.Let us define a power series ∑n=1∞anzn, where a positive sequence {an}n=1∞ is determined by a1=η, a2=2p2-γp0c12/2γγ-1 and for n≥1(17)an+2=M∑k=1n∑i=1kaiak-i+1an-k+1+2∑k=1n+1akan-k+2+∑k=1n∑i=1kaian-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1al1⋯alm+∑k=1n∑i=1kaian-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1m+2m+1al1⋯alm.It follows that cn≤an for n≥1. That is, ∑n=1∞anzn is a majorant series of ∑n=1∞cnzn. We show that ∑n=1∞anzn is analytic in a neighborhood of the origin. Note that if we let A(z)=∑n=1∞anzn, then(18)Az=∑n=1∞anzn=ηz+a2z2+M∑n=1∞∑k=1n∑i=1kaiak-i+1an-k+1zn+2+2∑n=1∞∑k=1n+1akan-k+2zn+2+∑n=1∞∑k=1n∑i=1kaian-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1al1⋯almzn+2+∑n=1∞∑k=1n∑i=1kaian-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1m+2m+1al1⋯almzn+2=ηz+2p2-γp0c122γγ-1z2+MAz3+2Az2-2η2z2+Az31-Az+2Az33-3Az+Az21-Az3.Consider the equation(19)Tz,A=A-ηz-2p2-γp0c122γγ-1z2-MA3+2A2-2η2z2+A31-A+2A33-3A+A21-A3=0.Since T is continuous in a neighborhood of the origin, T(0,0)=0, and TA′0,0=1≠0, the implicit function theorem implies that there exists a unique function A(z) which is analytic in a neighborhood of the origin with a positive radius. Because A(z) is a majorant series of g(z), g(z) is also analytic in a neighborhood of the origin with a positive radius. This completes the proof.Now, we consider an analytic solutiong(z) of the auxiliary equation (11) in the case of γ satisfies condition (H2). In order to study the existence of g(z) under the Brjuno condition, we will recall the definition of Brjuno number and some basic facts. As stated in [10], for a real number θ, we let [θ] be an integer part of θ and let {θ}=θ-[θ] be a fractional part of θ. Observe that if θ is an irrational number, then it has a unique expression of Gauss’s continued fraction(20)θ=d0+θ0=d0+1d1+θ1=⋯,denoted simply by θ=[d0;d1,…,dn,…], where dj’s and θj’s are calculated by the following algorithm:(a)
d0=[θ], θ0={θ};(b)
dn=1/θn-1, θn=1/θn-1 for all n≥1.Define the sequences {pn}n∈N and {qn}n∈N by the following recursive relation:(21)q-2=1,q-1=0,qn=dnqn-1+qn-2;p-2=0,p-1=1,pn=dnpn-1+pn-2.Note that pn/qn=[d0;d1,…,dn]. For each θ∈R∖Q, we consider an arithmetical function B(θ)=∑n≥0log⁡qn+1/qn. When θ satisfies condition B(θ)<+∞, we call it a Brjuno number. Consider θ=[d0;d1,…,dn,…] in which for each n≥0, dn+1≤cedn, where c is a positive constant. We can show that θ is a Brjuno number but is not a Diophantine number. Therefore, Brjuno condition is weaker than the Diophantine condition. So, condition (H2) contains both Diophantine condition and a part of γ near resonance.Letθ∈R∖Q and {qn}n∈N be the sequence of partial denominators of Gauss’s continued fraction for θ. As in [10], let(22)Ak=n≥0:nθ≤18qk,Ek=max⁡qk,qk+14,ηk=qkEk.Let Ak∗ be the set of integers j≥0 such that either j∈Ak or for some j1 and j2 in Ak, with j2-j1<Ek, one has j1<j<j2 and qk divides j-j1. For any nonnegative integer n, we define(23)lkn=max⁡1+ηknqk-2,mnηk+n1qk-1,where mn=max⁡j:0≤j≤n,j∈Ak∗. Let hk:N→R+ be a function defined by(24)hkn=mn+ηknqk-1ifmn+qk∈Ak∗,lknifmn+qk∉Ak∗.Let gk(n)≔max⁡{hk(n),[n/qk]}, and let k(n) be defined by condition qk(n)≤n≤qk(n)+1. Note that k(n) is nondecreasing.Lemma 2 (Davie’s lemma [11]).
Let(25)Kn=nlog⁡2+∑k=0kngknlog⁡2qk+1.Then(a)
there is a universal constantξ>0 (independent of n and θ) such that(26)Kn≤nBθ+ξ,(b)
K(n1)+K(n2)≤K(n1+n2) for all n1 and n2,(c)
-log⁡γn-1≤K(n)-K(n-1).Theorem 3.
Assume thatγ satisfies condition(H2). Then there exists an analytic solution(27)gz=∑n=1∞cnznof (11) in a neighborhood of the origin such that g(0)=0, g′(0)=η, where η is a nonzero complex number.Proof.
We now imitate the proof of Theorem1 with approximate new bound. The sequence {cn}n=1∞ is defined similar to the proof of Theorem 1. Note that c1=η≠0 and c2=(2p2-γp0)c12/2γ(γ-1). Since γ=1 and p(z) is analytic near the origin, there exists a positive constant N so that for n≥1(28)cn+2≤Nγn+1-1∑k=1n∑i=1kcick-i+1cn-k+1+2∑k=1n+1ckcn-k+2+∑k=1n∑i=1kcicn-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1cl1⋯clm+∑k=1n∑i=1kcicn-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1m+2m+1cl1⋯clm.To construct a governing series of g(z), we let {Bn}n=1∞ be a nonnegative sequence determined by B1=η, B2=2p2-γp0η2/2γγ-1 and for all n≥1(29)Bn+2=N∑k=1n∑i=1kBiBk-i+1Bn-k+1+2∑k=1n+1BkBn-k+2+∑k=1n∑i=1kBiBn-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1Bl1⋯Blm+∑k=1n∑i=1kBiBn-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1m+2m+1Bl1⋯Blm.From this construction, we can demonstrate that a power series B(z)=∑n=1∞Bnzn satisfies the implicit functional equation(30)Rz,Bz=Bz-ηz-a2z2-NBz3+2Bz2-2η2z2+Bz31-Bz+2Bz33-3Bz+Bz21-Bz3=0with R(0,0)=0 and RB′(0,0)=1≠0. This yields the power series B(z) converges in a neighborhood of the origin. Hence, there exists a positive constant T such that Bn≤Tn for n≥1.
LetK be a function defined in Lemma 2. By mathematical induction, we can show that for n∈N∪{0}(31)cn+1≤Bn+1eKn.Lemma 2 yields limn→∞⁡(cn+1)1/n≤TeB(θ)+ξ. This implies that g(z) has a convergence radius at least (TeB(θ)+ξ)-1. The proof is completed.Finally, we consider the case thatγ satisfies condition (H3). In this case, γ is not only on the unit circle, but also a root of unity. Let {Dn}n=1∞ be a sequence defined by D1=η, D2=ΓA with A=2p2-γp0c12/2γ, Γ=max⁡{1/γ-1,1/γ2-1,…,1/γp-1-1}, and(32)Dn+2=ΓN∑k=1n∑i=1kDiDk-i+1Dn-k+1+2∑k=1n+1DkDn-k+2+∑k=1n∑i=1kDiDn-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1Dl1⋯Dlm+∑k=1n∑i=1kDiDn-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1m+2m+1Dl1⋯Dlm,where N is a positive constant defined as in the proof of Theorem 3.Theorem 4.
Assume thatγ satisfies condition(H3). Let g(z)=∑n=1∞cnzn be a power series determined by c1=η≠0, c2=(2p2-γp0)c12/2γ, and(33)n+2γγn+1-1cn+2=Θn,γ,n=1,2,…,where(34)Θn,γ=∑k=1n∑i=1kin-k+1γ2k-i+2k+1·cick-i+1cn-k+1-∑k=1n+1kckp0γn-k+2cn-k+2-∑k=1n∑i=1kin-k+1γk+1k+1·cicn-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1pmcl1⋯clm+∑k=1n+12kp2ckcn-k+2+∑k=1n∑i=1kin-k+1k+1·cicn-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1pm+2m+2m+1cl1⋯clm.
IfΘ(vp-1,γ)=0 for v=1,2,…, then (11) has an analytic solution g(z) in a neighborhood of the origin such that g(0)=0, g′(0)=η, and g(vp+1)(0)=(vp+1)!ηvp+1, where ηvp+1 is an arbitrary constant satisfying ηvp+1≤Dvp+1, where {Dn}n=1∞ is defined as in (32).
Otherwise, ifΘ(vp-1,γ)≠0 for some v=1,2,…, then (11) has no analytic solution in a neighborhood of the origin.Proof.
Observe that ifη=0 then g(z)≡0 is a trivial analytic solution of (11). So, we consider only the case η≠0.
IfΘ(vp-1,γ)≠0 for some positive number v, then (vp+1)γ(γvp-1)cvp+1≠0. But (H3) implies γvp-1=0, which is a contradiction. This concludes that (11) has no analytic solution in a neighborhood of the origin.
Assume thatΘ(vp-1,γ)=0 for v=1,2,…. Then (vp+1)γ(γvp-1)cvp+1=0. So, there are infinitely many choices of cvp+1. Choose cvp+1=ηvp+1 so that ηvp+1≤Dvp+1, where Dvp+1 is defined in (32).
Note thatγn+1-1≤Γ for n≠vp-1, where Γ=max⁡{1/γ-1,1/γ2-1,…,1/γp-1-1}. We can see that(35)cn+2≤ΓN∑k=1n∑i=1kcick-i+1cn-k+1+2∑k=1n+1ckcn-k+2+∑k=1n∑i=1kcicn-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1cl1⋯clm+∑k=1n∑i=1kcicn-k+1∑l1+⋯+lm=k-i+1m=1,2,…,k-i+1m+2m+1cl1⋯clmfor n≠vp-1,v=1,2,….
Likewise, the remaining proof is similar to one of Theorem1. Consider the implicit functional equation(36)Hz,R=R-ηz-2p2-γp0c122γ·z2-NR3+2R2-2η2z2+R31-R+2R33-3R+R21-R3=0.Since H(0,0)=0, HR′(0,0)=1≠0, the implicit function theorem implies that there exists a unique function R(z) which is analytic in a neighborhood of the origin with a positive radius. We can show that the power series R(z)=∑n=1∞Dnzn in which {Dn}n=1∞ is determined by (32) satisfies (36). Moreover, cn≤Dn for n≥1. That is, R(z) is a majorant series of g(z). Then g(z) converges in a neighborhood of the origin. This completes the proof.Theorem 5.
Letg(z)=∑n=1∞cnzn be an analytic solution in a neighborhood of the origin of (11), with g(0)=0, g′(0)=η≠0, which is obtained from Theorem 1, Theorem 3, or Theorem 4. Then (10) has an analytic solution of the form y(z)=g(γg-1(z)) in a neighborhood of the origin.Proof.
Sinceg′(0)=η≠0, g-1(z) is analytic in a neighborhood of g(0)=0.
Lety(z)=g(γg-1(z)). Then(37)y′′z-p′′z=γ2g′′γg-1zg′g-1z2-γg′γg-1zg′′g-1zg′g-1z3-p′′z=γ2g′′γg-1zg′g-1z-γg′γg-1zg′′g-1z-p′′zg′g-1z3g′g-1z3=gγ2g-1z-pgγg-1zγg′γg-1zg′g-1z2g′g-1z3=gγg-1gγg-1z-pgγg-1z·γg′γg-1zg′γg-1z=yyz-pyzy′z.That is, y(z)=g(γg-1(z)) is an analytic solution of (10). The proof is completed.


## 3. Analytic Solutions of (5) via (10)

In this section, we construct an analytic solution of (5) from an analytic solution of (10). Assume that x(z) is an analytic solution of the functional differential equation (5) in a neighborhood of the origin. Since x(z) is analytic in a neighborhood of the origin, x(z) can be represented by Taylor’s series(38)xz=∑n=0∞xn0n!·zn=x0+x′0z+x′′0z22!+⋯.Let p(z)=∑n=0∞pnzn, where p0=α and p1=β and pn=p(n)(0)/n! for n≥2. Since x′(z)=1/b(y(z)-p(z)), we have x′(0)=-α/b. Since x′′(z)=1/b(y′(z)-p′(z)) and y′(0)=γ, it follows that x′′(0)=1/b(γ-β). This implies that γ-β/b=x′′(0)=x(p(0)+bx′(0))  = x(α-b·α/b)=x(0).Sincex′′′(z)=x′(p(z)+bx′(z))·(p′(z)+bx′′(z)), we have x′′′(0)=-αγ/b.By using mathematical induction, we can show that(39)xpz+bx′zm=∑i=1mpimp′z+bx′′z,p′′z+bx′′′z,…,pmz+bxm+1zxipz+bx′z,for m=1,2,3,…, where pim(1≤i≤m) is a polynomial with nonnegative coefficients.Therefore, the explicit form of an analytic solution of our equation is(40)xz=γ-βb+-αb·z+12!·γ-βb·z2+13!·-αγbz3+∑m=2∞Γmm+2!·zm+2,where Γm denotes x(m+2)(0).


## 4. Polynomial Solutions of (5)

In this section, we letp(z) be a polynomial. Then, we investigate the polynomial solution of (5).Theorem 6.
For a polynomialp(z), the equation(41)x′′z=xpz+bx′zhas a nontrivial polynomial solution if and only if p(z)=p0 with p0≠0 or p(z)=p0+p1z with p1≠0.Proof.
Necessity. Assume that x(z)=∑k=0nxkzk is a nontrivial polynomial solution of (41). Let p(z)=∑k=0mpkzk with pm≠0.
Observe thatx′′(z)=0 when n=0. This implies x(z)≡0. From now on, we let n≥1.
We consider3 cases.
Case 1 (m=0). That is, p(z)=p0≠0. Equation (41) becomes(42)2x2+6x3z+⋯+nn-1xnzn-2=x0+x1qz+⋯+xnqzn,where q(z)=(p0+bx1)+2bx2z+⋯+nbxnzn-1.
Ifn=1, then (42) changes to 0=x0+x1(p0+bx1).
Next, we considern≥2.
Comparing coefficient ofzn(n-1) in (42), we have xn=0.
Equation (42) is reduced to(43)2x2+⋯+n-1n-2xn-1zn-3=x0+x1qz+⋯+xn-1qzn-1,where q(z)=(p0+bx1)+2bx2z+⋯+(n-1)bxn-1zn-2.
Comparing coefficient ofz(n-1)(n-2) in (43), we have xn-1=0. Then, repeating the above method, we obtain xn-2=⋯=x2=0 and also have 0=x0+x1(p0+bx1).
By choosing an arbitrary nonzerox1, say η, both situations yield a nontrivial solution x(z)=-η(p0+bη)+ηz.
Case 2 (m=1). That is, p(z)=p0+p1z, where p1≠0.
Equation (41) becomes(44)2x2+6x3z+⋯+nn-1xnzn-2=x0+x1qz+⋯+xnqzn,where q(z)=(p0+bx1)+(p1+2bx2)z+3bx3z2+⋯+nbxnzn-1.
By comparing coefficient of constant term andz in (44), we obtain x(z)≡0 for n=1.
Next, we considern≥2.
Comparing coefficient ofzn(n-1) in (44), we have xn(nbxn)n=0, which implies xn=0. Therefore, (44) is reduced to(45)2x2+6x3z+⋯+n-1n-2xn-1zn-3=x0+x1qz+⋯+xn-1qzn-1,where q(z)=(p0+bx1)+(p1+2bx2)z+3bx3z2+⋯+(n-1)bxn-1zn-2.
Comparing coefficient ofz(n-1)(n-2) in (45), we have xn-1=0. Continuing this process, we obtain 2x2=x0+x1(q(z))+x2(q(z))2, where q(z)=(p0+bx1)+(p1+2bx2)z. By comparing coefficient of z2 together with x2≠0, we obtain a nontrivial solution x(z)=-p1/b-η(p0+bη) +p1/2b(p0+bη)2+ηz-p1/2bz2, where η is an arbitrary constant. Note that if x2=0 then x(z)≡0.
Case 3 (m≥2). We consider 2 subcases.
Subcase 3.1 (m<n-1). Equation (41) becomes(46)2x2+6x3z+⋯+nn-1xnzn-2=x0+x1qz+⋯+xnqzn,where q(z)=(p0+bx1)+⋯+(pm+(m+1)bxm+1)zm+(m+2)bxm+2zm+1+⋯+nbxnzn-1. By using method of undetermined coefficient, we obtain xn=⋯=xm+2=0. Consequently, (46) is reduced to(47)2x2+6x3z+⋯+m+1mxm+1zm-1=x0+x1qz+⋯+xm+1qzm+1,where q(z)=(p0+bx1)+⋯+(pm+(m+1)bxm+1)zm.
Comparing coefficient ofzm(m+1) in (47), we have xm+1(pm+(m+1)bxm+1)m+1=0. If xm+1=0, then xm=⋯=x0=0. That is, x(z)≡0 which is a contradiction. Therefore, pm+(m+1)bxm+1=0. Substituting this relation in (46) and repeating this process, we get xm-k=-pm-(k+1)/(m-k)b for k=-1,…,m-2. Using this fact in (47) and then comparing the coefficient of zi(i=1,…,m-1), we obtain x3=⋯=xm+1=0. This yields pm=0, which is a contradiction.
Subcase 3.2 (m≥n-1). Equation (41) becomes(48)2x2+6x3z+⋯+nn-1xnzn-2=x0+x1qz+⋯+xnqzn,where q(z)=(p0+bx1)+⋯+(pn-1+nbxn)zn-1+pnzn+⋯+pmzm. Comparing coefficient of znm in (48) together with pm≠0, we have xn=0. Then, repeating the above method, we obtain x(z)≡0 which is a contradiction.
Thus, (41) has a nontrivial polynomial solution when p(z)=p0 with p0≠0 or p(z)=p0+p1z with p1≠0.
Sufficiency. It follows from the proof of Cases 1 and 2 in necessity part.



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*Source: 904679-2015-10-05.xml*