# Smooth Solutions of a Class of Iterative Functional Differential Equations

Authors: Houyu Zhao
Journal: Abstract and Applied Analysis
 (2012)

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## Abstract

By Faà di Bruno’s formula, using the fixed-point theorems of Schauder and Banach, we study the existence and uniqueness of smooth solutions of an iterative functional differential equationx′(t)=1/(c0x[0](t)+c1x[1](t)+⋯+cmx[m](t)).


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## Body

## 1. Introduction

There has been a lot of monographs and research articles to discuss the kinds of solutions of functional differential equations since the publication of Jack Hale's paper [1]. Several papers discussed the iterative functional differential equations of the form (1.1)x′(t)=H(x[0](t),x[1](t),…,x[m](t)),
where x[0](t)=t,x[1](t)=x(t),x[k](t)=x(x[k-1](t)),k=2,…,m. More specifically, Eder [2] considered the functional differential equation(1.2)x′(t)=x[2](t)
and proved that every solution either vanishes identically or is strictly monotonic. Furthermore, Fečkan [3] and Wang [4] studied the equation (1.3)x′(t)=f(x[2](t))
with different conditions. Staněk [5] considered the equation (1.4)x′(t)=x(t)+x[2](t)
and obtained every solution either vanishes identically or is strictly monotonic. Si and his coauthors [6, 7] studied the following equations: (1.5)x′(t)=x[m](t),x′(t)=1x[m](t),(1.6)x′(t)=1c0x[0](t)+c1x[1](t)+⋯+cmx[m](t)
and established sufficient conditions for the existence of analytic solutions. Especially in [8, 9], the smooth solutions of the following equations: (1.7)x′(t)=∑j=1majx[j](t)+F(t),x′(t)=∑j=1maj(t)x[j](t)+F(t),
have been studied by the fixed-point theorems of Schauder and Banach.A smooth function is taken to mean one that has a number of continuous derivatives and for which the highest continuous derivative is also Lipschitz. Letx∈Cn if x′,…,x(n) are continuous, Cn(I,I) is the set in which x∈Cn and maps a closed interval I into I. As in [9], we, using the same symbols, denote the norm (1.8)‖x‖n=∑k=0n‖x(k)‖,‖x‖=max⁡t∈I{|x(t)|},
then Cn(I,R) with ∥·∥n is a Banach space, and Cn(I,I) is a subset of Cn(I,R). For given Mi>0(i=1,2,…,n+1), let (1.9)Ω(M1,…,Mn+1;I)={x∈Cn(I,I):|x(i)(t)|≤Mi,i=1,2,…,n;|x(n)(t1)-x(n)(t2)|≤Mn+1|t1-t2|,t,t1,t2∈I}.For convenience, we will make use of the notation(1.10)xij(t)=x(i)(x[j](t)),x*jk(t)=(x[j](t))(k),
where i,j, and k are nonnegative integers. Let I be a closed interval in R. By induction, we may prove that(1.11)x*jk(t)=Pjk(x10(t),…,x1,j-1(t);…;xk0(t),…,xk,j-1(t)),(1.12)βjk=Pjk(x′(ξ),…,x′(ξ);︷jterms…;x(k)(ξ),…,x(k)(ξ)︷jterms),(1.13)Hjk=Pjk(1,…,1︷jterms;M2,…,M2︷jterms;…;Mk,…,Mk︷jterms),
where Pjk is a uniquely defined multivariate polynomial with nonnegative coefficients. The proof can be found in [8].In order to seek a solutionx(t) of (1.6), in Cn(I,I) such that ξ is a fixed point of the function x(t), that is, x(ξ)=ξ, it is natural to seek an interval I of the form [ξ-δ,ξ+δ] with δ>0.Let us define(1.14)X(ξ;ξ0,…,ξn;1,M2,…,Mn+1;I)={x∈Ω(1,M2,…,Mn+1;I):x(ξ)=ξ0=ξ,x(i)(ξ)=ξi,i=1,2,…,n}.


## 2. Smooth Solutions of (1.6)

In this section, we will prove the existence theorem of smooth solutions for (1.6). First of all, we have the inequalities in the following for all x(t),y(t)∈X:(2.1)|x[j](t1)-x[j](t2)|≤|t1-t2|,t1,t2∈I,j=0,1,…,m,(2.2)‖x[j]-x[j]‖≤j‖x-y‖,j=1,…,m,(2.3)‖x-y‖≤δn‖x(n)-y(n)‖,
and the proof can be found in [9].Theorem 2.1.
LetI=[ξ-δ,ξ+δ], where ξ and δ satisfy
(2.4)ξ≥1|c0|-∑i=1m|ci|,0<δ≤ξ-1|c0|-∑i=1m|ci|,
where |c0|>∑i=0m|ci|, then (1.6) has a solution in
(2.5)X(ξ;ξ0,…,ξn;1,M2,…,Mn+1;I),
provided the following conditions hold:
(i)(2.6)ξ1=ξ-1(∑i=0mci)-1,(2.7)ξk=∑(-1)s(k-1)!s!s1!s2!⋯sk-1!(ξ∑i=0mci)-s-1(∑i=0mciβi11!)s1×(∑i=0mciβi22!)s2⋯(∑i=0mciβik-1(k-1)!)sk-1,
where k=2,…,n, and the sum is over all nonnegative integer solutions of the Diophantine equation s1+2s2+⋯+(k-1)sk-1=k-1 and s=s1+s2+⋯+sk-1,
(ii)(2.8)∑(k-1)!s!s1!s2!⋯sk-1!(ξ-δ)-s-1(|c0|-∑i=1m|ci|)-s-1(∑i=0m|ci|Hi11!)s1×(∑i=0m|ci|Hi22!)s2⋯(∑i=0m|ci|Hik-1(k-1)!)sk-1≤Mk,k=2,…,n,
where s1+2s2+⋯+(k-1)sk-1=k-1 and s=s1+s2+⋯+sk-1,
(iii)(2.9)∑(n-1)!s!s1!s2!⋯sn-1!1!s12!s2⋯(n-1)!sn-1×[(s+1)(ξ-δ)-s-2(|c0|-∑i=1m|ci|)-s-2(∑i=0m|ci|Hi1)s1+1(∑i=0m|ci|Hi2)s2×⋯×(∑i=0m|ci|Hin-1)sn-1+s1(ξ-δ)-s-1(|c0|-∑i=1m|ci|)-s-1(∑i=0m|ci|Hi1)s1-1×(∑i=0m|ci|Hi2)s2+1(∑i=0m|ci|Hi3)s3⋯(∑i=0m|ci|Hin-1)sn-1+⋯+sn-1(ξ-δ)-s-1(|c0|-∑i=1m|ci|)-s-1(∑i=0m|ci|Hi1)s1⋯(∑i=0m|ci|Hin-2)sn-2×(∑i=0m|ci|Hin-1)sn-1-1(∑i=0m|ci|Hin)]≤Mn+1,
where s1+2s2+⋯+(n-1)sn-1=n-1 and s=s1+s2+⋯+sn-1,Proof.
Define an operatorT from X into Cn(I,I) by
(2.10)(Tx)(t)=ξ+∫ξt1c0x[0](s)+c1x[1](s)+⋯+cmx[m](s)ds.
We will prove that for any x∈X,Tx∈X,
(2.11)|(Tx)(t)-ξ|=|∫ξt1∑i=0mcix[i](s)ds|≤((ξ-δ)(|c0|-∑i=1m|ci|))-1|t-ξ|≤δ,
where the second inequality is from (2.4) and x(I)⊆I. Thus, (Tx)(I)⊆I.
It is easy to see that(2.12)(Tx)′(t)=1∑i=0mcix[i](t),
and by Faà di Bruno's formula, for k=2,…,n, we have
(2.13)(Tx)(k)(t)=(1∑i=0mcix[i](t))(k-1)=∑(-1)s(k-1)!s!s1!s2!⋯sk-1!1(∑i=0mcix[i](t))s+1((∑i=0mcix[i](t))′1!)s1×((∑i=0mcix[i](t))′′2!)s2⋯((∑i=0mcix[i](t))(k-1)(k-1)!)sk-1=∑(-1)s(k-1)!s!s1!s2!⋯sk-1!1(∑i=0mcix[i](t))s+1(∑i=0mcix*i1(t)1!)s1×(∑i=0mcix*i2(t)2!)s2⋯(∑i=0mcix*ik-1(t)(k-1)!)sk-1,
where the sum is over all nonnegative integer solutions of the Diophantine equation s1+2s2+⋯+(k-1)sk-1=k-1 and s=s1+s2+⋯+sk-1.
Furthermore, note(Tx)(ξ)=ξ, by (2.6) and (2.7),
(2.14)(Tx)′(ξ)=1∑i=0mcix[i](ξ)=1ξ∑i=0mci=ξ1,(Tx)(k)(ξ)=∑(-1)s(k-1)!s!s1!s2!⋯sk-1!1(ξ∑i=0mci)s+1(∑i=0mcix*i1(ξ)1!)s1×(∑i=0mcix*i2(ξ)2!)s2⋯(∑i=0mcix*ik-1(ξ)(k-1)!)sk-1=∑(-1)s(k-1)!s!s1!s2!⋯sk-1!1(ξ∑i=0mci)s+1(∑i=0mciβi11!)s1×(∑i=0mciβi22!)s2⋯(∑i=0mciβik-1(k-1)!)sk-1=ξk,k=2,…,n,
where s1+2s2+⋯+(k-1)sk-1=k-1 and s=s1+s2+⋯+sk-1. Thus, (Tx)(k)(ξ)=ξk for k=0,1,…,n,
(2.15)|(Tx)′(t)|=|1∑i=0mcix[i](t)|≤((ξ-δ)(|c0|-∑i=1m|ci|))-1≤1=M1,
By (2.8), we have
(2.16)|(Tx)(k)(t)|≤∑(k-1)!s!s1!s2!⋯sk-1!1|∑i=0mcix[i](t)|s+1(∑i=0m|ci||x*i1(t)|1!)s1×(∑i=0m|ci||x*i2(t)|2!)s2⋯(∑i=0m|ci||x*ik-1(t)|(k-1)!)sk-1≤∑(k-1)!s!s1!s2!⋯sk-1!(ξ-δ)-s-1(|c0|-∑i=1m|ci|)-s-1(∑i=0m|ci|Hi11!)s1×(∑i=0m|ci|Hi22!)s2⋯(∑i=0m|ci|Hik-1(k-1)!)sk-1≤Mk,k=2,…,n,
where s1+2s2+⋯+(k-1)sk-1=k-1 and s=s1+s2+⋯+sk-1.
Finally,(2.17)|(Tx)(n)(t1)-(Tx)(n)(t2)|≤∑(n-1)!s!s1!s2!⋯sn-1!1!s12!s2⋯(n-1)!sn-1×|(∑i=0mcix[i](t1))-s-1(∑i=0mcix*i1(t1))s1(∑i=0mcix*i2(t1))s2⋯(∑i=0mcix*in-1(t1))sn-1-(∑i=0mcix[i](t2))-s-1(∑i=0mcix*i1(t2))s1(∑i=0mcix*i2(t2))s2⋯(∑i=0mcix*in-1(t2))sn-1|≤∑(n-1)!s!s1!s2!⋯sn-1!1!s12!s2⋯(n-1)!sn-1×[|1(∑i=0mcix[i](t1))s+1-1(∑i=0mcix[i](t2))s+1|(∑i=0m|ci||x*i1(t1)|)s1×⋯(∑i=0m|ci||x*in-1(t1)|)sn-1+|(∑i=0mcix[i](t2))-s-1|×|(∑i=0mcix*i1(t1))s1-(∑i=0mcix*i1(t2))s1|(∑i=0m|ci||x*i2(t1)|)s2⋯×(∑i=0m|ci||x*in-1(t1)|)sn-1+⋯+|(∑i=0mcix[i](t2))-s-1|(∑i=0m|ci||x*i1(t2)|)s1×(∑i=0m|ci||x*i2(t2)|)s2⋯(∑i=0m|ci||x*in-2(t2)|)sn-2×|(∑i=0mcix*in-1(t1))sn-1-(∑i=0mcix*in-1(t2))sn-1|]≤∑(n-1)!s!s1!s2!⋯sn-1!1!s12!s2⋯(n-1)!sn-1×[(s+1)(ξ-δ)-s-2(|c0|-∑i=1m|ci|)-s-2(∑i=0m|ci|Hi1)s1+1(∑i=0m|ci|Hi2)s2×⋯×(∑i=0m|ci|Hin-1)sn-1+s1(ξ-δ)-s-1(|c0|-∑i=1m|ci|)-s-1(∑i=0m|ci|Hi1)s1-1×(∑i=0m|ci|Hi2)s2+1(∑i=0m|ci|Hi3)s3⋯(∑i=0m|ci|Hin-1)sn-1+⋯+sn-1(ξ-δ)-s-1(|c0|-∑i=1m|ci|)-s-1(∑i=0m|ci|Hi1)s1⋯(∑i=0m|ci|Hin-2)sn-2×(∑i=0m|ci|Hin-1)sn-1-1(∑i=0m|ci|Hin)]|t1-t2|,
where s1+2s2+⋯+(k-1)sk-1=k-1 and s=s1+s2+⋯+sk-1. By (2.9), we see that
(2.18)|(Tx)(n)(t1)-(Tx)(n)(t2)|≤Mn+1|t1-t2|.
Now, we can say that T is an operator from X into itself.
Next, we will show thatT is continuous. Let x,y∈X, then
(2.19)‖Tx-Ty‖n=‖Tx-Ty‖+‖(Tx)′-(Ty)′‖+∑k=2n‖(Tx)(k)-(Ty)(k)‖=max⁡t∈I|∫ξt(1∑i=0mcix[i](s)-1∑i=0mciy[i](s))ds|+max⁡t∈I|1∑i=0mcix[i](t)-1∑i=0mciy[i](t)|+∑k=2nmax⁡t∈I|∑(k-1)!s!s1!s2!⋯sk-1!1!s12!s2⋯(k-1)!sk-1×[1(∑i=0mcix[i](t))s+1(∑i=0mcix*i1(t))s1(∑i=0mcix*i2(t))s2⋯×(∑i=0mcix*ik-1(t))sk-1-1(∑i=0mciy[i](t))s+1(∑i=0mciy*i1(t))s1×(∑i=0mciy*i2(t))s2⋯(∑i=0mciy*ik-1(t))sk-1]|≤δ(ξ-δ)-2(|c0|-∑i=1m|ci|)-2(∑i=0m|ci|‖x[i]-y[i]‖)+(ξ-δ)-2(|c0|-∑i=1m|ci|)-2(∑i=0m|ci|‖x[i]-y[i]‖)+∑k=2nmax⁡t∈I∑(k-1)!s!s1!s2!⋯sk-1!1!s12!s2⋯(k-1)!sk-1×[|(∑i=0mcix[i](t))-s-1-(∑i=0mciy[i](t))-s-1|(∑i=0m|ci||x*i1(t)|)s1⋯×(∑i=0m|ci||x*ik-1(t)|)sk-1+|(∑i=0mciy[i](t))-s-1|×|(∑i=0mcix*i1(t))s1-(∑i=0mciy*i1(t))s1|×(∑i=0m|ci||x*i2(t)|)s2⋯(∑i=0m|ci||x*ik-1(t)|)sk-1+⋯+|(∑i=0mciy[i](t))-s-1|(∑i=0m|ci||y*i1(t)|)s1⋯(∑i=0m|ci||y*ik-2(t)|)sk-2×|(∑i=0mcix*ik-1(t))sk-1-(∑i=0mciy*ik-1(t))sk-1|]≤(ξ-δ)-2(|c0|-∑i=1m|ci|)-2(|c0|+∑i=1mi|ci|)‖x-y‖+∑k=2n∑(k-1)!s!s1!s2!⋯sk-1!1!s12!s2⋯(k-1)!sk-1×[(s+1)(ξ-δ)-s-2(|c0|-∑i=1m|ci|)-s-2(∑i=0m|ci|‖x[i]-y[i]‖)×(∑i=0m|ci|Hi1)s1⋯(∑i=0m|ci|Hik-1)sk-1+s1(ξ-δ)-s-1(|c0|-∑i=1m|ci|)-s-1(∑i=0m|ci|Hi1)s1-1×(∑i=0m|ci|Hi2)s2+1⋯(∑i=0m|ci|Hik-1)sk-1(∑i=0m|ci|‖x[i]-y[i]‖)+⋯+sk-1(ξ-δ)-s-1(|c0|-∑i=1m|ci|)-s-1(∑i=0m|ci|Hi1)s1⋯×(∑i=0m|ci|Hik-2)sk-2(∑i=0m|ci|Hik-1)sk-1-1(∑i=0mciHik)×(∑i=0m|ci|‖x[i]-y[i]‖)]+δn+1(ξ-δ)-2(|c0|-∑i=1m|ci|)-2(|c0|+∑i=1mi|ci|)‖x(n)-y(n)‖,
where s1+2s2+⋯+(k-1)sk-1=k-1 and s=s1+s2+⋯+sk-1.
Moreover, we can find some constantsPk such that
(2.20)∑k=2n∑(k-1)!s!s1!s2!⋯sk-1!1!s12!s2⋯(k-1)!sk-1×[(s+1)(ξ-δ)-s-2(|c0|-∑i=1m|ci|)-s-2(∑i=0m|ci|‖x[i]-y[i]‖)××(∑i=0m|ci|Hi1)s1⋯(∑i=0m|ci|Hik-1)sk-1+s1(ξ-δ)-s-1(|c0|-∑i=1m|ci|)-s-1(∑i=0m|ci|Hi1)s1-1×(∑i=0m|ci|Hi2)s2+1⋯(∑i=0m|ci|Hik-1)sk-1(∑i=0m|ci|‖x[i]-y[i]‖)+⋯+sk-1(ξ-δ)-s-1(|c0|-∑i=1m|ci|)-s-1(∑i=0m|ci|Hi1)s1⋯×(∑i=0m|ci|Hik-2)sk-2(∑i=0m|ci|Hik-1)sk-1-1(∑i=0mciHik)×(∑i=0m|ci|‖x[i]-y[i]‖)]≤∑k=1n-1Pk(ξ,δ,ci,Hij)‖x(k)-y(k)‖,
where
(2.21)Pk(ξ,δ,ci,Hij)=P(ξ,δ;c1,…,cm;H11,…,H1k+1;…;Hm1,…,Hmk+1;)
are the positive constants depend on ξ,δ,ci, and Hij,i=1,…,m;j=1,…,k+1. Then
(2.22)‖Tx-Ty‖n≤(ξ-δ)-2(|c0|-∑i=1m|ci|)-2(|c0|+∑i=1mi|ci|)‖x-y‖+∑k=1n-1Pk(ξ,δ,ci,Hij)‖x(k)-y(k)‖+δn+1(ξ-δ)-2(|c0|-∑i=1m|ci|)-2(|c0|+∑i=1mi|ci|)‖x(n)-y(n)‖≤Γ‖x-y‖n.
Here,
(2.23)Γ=max⁡{(ξ-δ)-2(|c0|-∑i=1m|ci|)-2(|c0|+∑i=1mi|ci|);max⁡1≤k≤n-1{Pk(ξ,δ,ci,Hij)};δn+1(ξ-δ)-2(|c0|-∑i=1m|ci|)-2(|c0|+∑i=1mi|ci|)},k=1,…,n-1. So we can say that T is continuous.
It is easy to see thatX is closed and convex. We now show that X is a relatively compact subset of Cn(I,I). For any x=x(t)∈X,
(2.24)‖x‖n≤‖x‖+∑k=1n‖x(k)‖≤|ξ|+δ+1+∑k=2nMk.
Next, for any t1,t2 in I, we have
(2.25)|x(t1)-x(t2)|≤|t1-t2|.
Hence, X is bounded in Cn(I,I) and equicontinuous on I, and by the Arzela-Ascoli theorem, we know X is relatively compact in Cn(I,I), since Cn(I,I) is the subset of Cn(I,R), and we can say that X is relatively compact in Cn(I,R).
From Schauder's fixed-point theorem, we conclude that(2.26)x(t)=ξ+∫ξt1c0x[0](s)+c1x[1](s)+⋯+cmx[m](s)ds,
for some x=x(t) in X. By differentiating both sides of the above equality, we see that x is the desired solution of (1.6). This completes the proof.Theorem 2.2.
LetI=[ξ-δ,ξ+δ], where ξ and δ satisfy (2.4), then (1.6) has a unique solution in
(2.27)X(ξ;ξ0,…,ξn;1,M2,…,Mn+1;I),
provided the conditions (2.6)–(2.9) hold and Γ<1 in (2.23).Proof.
SinceΓ<1, we see that T defined by (2.10) is contraction mapping on the close subset X of Cn(I,I). Thus, the fixed point x in the proof of Theorem 2.1 must be unique. This completes the proof.Remark 2.3.
By Theorem2.1 or Theorem 2.2, the existence and uniqueness of smooth solutions of an iterative functional differential equation of the form (1.6) can be obtained. If n→+∞, we can also find that the solution is C∞-smooth.
Now, we will show that the conditions in Theorem2.1 do not self-contradict. Consider the following equation:
(2.28)x′(t)=1t+(1/2)x(t)+(1/4)x(x(t)),
where c0=1,c1=(1/2),c2=(1/4), and ξ≥4. Moreover, we take 0<δ≤ξ-4. Then, (2.4) is satisfied, and ξ,δ define the interval I=[ξ-δ,ξ+δ]. Now, take ξ0=ξ,
(2.29)ξ1=47ξ-1,ξ2=-1649ξ-2(1+12ξ1+14ξ12),ξ3=4343ξ-3(4+2ξ1+ξ12)2-449ξ-2ξ2(2+ξ1+ξ12),
then (2.6) and (2.7) are satisfied.
Finally, if we take(2.30)M1=1,M2=28(ξ-δ)-2,M3=392(ξ-δ)-3+16(ξ-δ)-2M2
as positive, and
(2.31)M4=8232(ξ-δ)-4+576(ξ-δ)-3M2+8(ξ-δ)-2(6M22+5M3),
then (2.8) and (2.9) are satisfied.
Thus, we have shown that whenξ0,…,ξ3 and M1,…,M4 are defined as above, then there will be a solution for (2.28) in X(ξ;ξ0,…,ξ3;1,…,M4;I).



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