# The General Solution of Differential Equations with Caputo-Hadamard Fractional Derivatives and Noninstantaneous Impulses

Authors: Xianzhen Zhang; Zuohua Liu; Hui Peng; Xianmin Zhang; Shiyong Yang
Journal: Advances in Mathematical Physics
 (2017)

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## Abstract

Based on some recent works about the general solution of fractional differential equations with instantaneous impulses, a Caputo-Hadamard fractional differential equation with noninstantaneous impulses is studied in this paper. An equivalent integral equation with some undetermined constants is obtained for this fractional order system with noninstantaneous impulses, which means that there is general solution for the impulsive systems. Next, an example is given to illustrate the obtained result.


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## Body

## 1. Introduction

Impulsive differential equations are used in modeling of biology and physics and engineering and so forth to describe abrupt changes of the state at certain instants. However, the classical impulsive models in which instantaneous impulses were mainly considered in most of the existing papers can not describe some processes such as evolution processes in pharmacotherapy. Hence, Hernández and O’Regan [1] and Pierri et al. [2] presented and studied a kind of differential equations with noninstantaneous impulses. Next, the fractional differential equations with noninstantaneous impulses were considered in [3, 4].Recently, Hadamard fractional calculus is getting attention which is an important part of theory of fractional calculus [5]. The works in [6–11] made development in fundamental theorem of Hadamard fractional calculus. A Caputo-type modification of Hadamard fractional derivative which is called Caputo-Hadamard fractional derivative was given in [12], and its fundamental theorems were proved in [13, 14].Furthermore, some works in [15–21] uncover that there is general solution for several fractional differential equations with instantaneous impulses. Therefore, we will try to consider the general solution for differential equations with Caputo-Hadamard fractional derivatives and noninstantaneous impulses:(1)DC-Ha+qut=ft,ut,t∈sk,tk+1,k=0,1,…,N,ut=gkt,ut,t∈tk,sk,k=1,2,…,N,ua=ua∈C,where q∈C, R(q)∈(0,1), and a>0 and DC-Ha+q is the left-side Caputo-Hadamard fractional derivative of order q. f:[a,T]×C→C and gk:(tk,sk]×C→C (here k=1,2,…,N) are some appropriate functions, and gk denote noninstantaneous impulses. a=t0=s0<t1≤s1≤t2≤⋯≤tN≤sN≤tN+1=T.Firstly, we only considerDC-Ha+qu(t)=f(t,u(t)) in each interval (sk,tk+1](k=0,1,…,N) in (1), and then(2)
D
C
-
H
a
+
q
u
t
=
f
t
,
u
t
,
for
t
∈
s
k
,
t
k
+
1
=
D
C
-
H
s
k
+
q
u
t
=
f
t
,
u
t
,
for
t
∈
s
k
,
t
k
+
1
⟺
u
t
=
u
s
k
+
1
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q
∫
s
k
t
l
n
t
τ
q
-
1
f
τ
,
u
τ
d
τ
τ
,
for
t
∈
s
k
,
t
k
+
1
.Substituting (2) into (1), we obtain(3)u~t=ua+1Γq∫atln⁡tτq-1fτ,uτdττ,fort∈a,t1,gkt,ut,fort∈tk,sk,k=1,2,…,N,gksk,usk+1Γq∫sktln⁡tτq-1fτ,uτdττ,fort∈sk,tk+1,k=1,2,…,N.In fact, u~(t) satisfies conditions of fractional derivative and noninstantaneous impulses in system (1). However, we will illustrate that u~(t) is not an equivalent integral equations of system (1).For system (1), we have(4)
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s
t
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m
1
→
for
t
∈
(
t
k
,
s
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∀
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∈
1,2
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…
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k
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=
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u
a
=
u
a
∈
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,
⟺
D
C
-
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a
+
q
u
t
=
f
t
,
u
t
,
t
∈
a
,
T
,
u
a
=
u
a
∈
C
,
u
s
i
n
g
L
e
m
m
a
5
⟺
(5)
u
t
=
u
a
+
1
Γ
q
∫
a
t
l
n
t
τ
q
-
1
g
τ
,
u
τ
d
τ
τ
,
for
t
∈
a
,
T
.On the other hand, letting gk(t,u(t))=ua+1/Γ(q)∫atln⁡t/τq-1f(τ,u(τ))dτ/τ for t∈(tk,sk] and all k∈{1,2,…,N} in (3), we get(6)u~t=ua+1Γq∫atln⁡tτq-1fτ,uτdττ,fort∈a,t1,ua+1Γq∫atln⁡tτq-1fτ,uτdττfort∈tk,sk,k=1,2,…,N,ua+1Γq∫askln⁡skτq-1fτ,uτdττ+∫sktln⁡tτq-1fτ,uτdττ,fort∈sk,tk+1,k=1,2,…,N.If u~(t) is equivalent to system (1), then (5) is equivalent to (6). Therefore, some unfit equations can be obtained as(7)∫atln⁡tτq-1fτ,uτdττ=∫askln⁡skτq-1fτ,uτdττ+∫sktln⁡tτq-1fτ,uτdττ,and here t∈(sk,tk+1], k=1,2,…,N. Therefore, u~(t) is not equivalent with system (1), and u~(t) will be regarded as anapproximate solution of system (1).Next, consideringDC-Ha+qu(t)=f(t,u(t)) for t∈(sk,tk+1] (k=0,1,…,N) in whole interval [a,T], we have(8)
D
C
-
H
a
+
q
u
t
=
f
t
,
u
t
,
for
t
∈
s
k
,
t
k
+
1
⊂
a
,
T
⟺
u
t
=
C
k
+
u
a
+
1
Γ
q
∫
a
t
ln
⁡
t
τ
q
-
1
f
τ
,
u
τ
d
τ
τ
,
for
t
∈
s
k
,
t
k
+
1
⊂
a
,
T
,
h
e
r
e
C
k
a
r
e
s
o
m
e
c
o
n
s
t
a
n
t
s
.Substituting (8) into (1), we obtain C0=0 and(9)Ck=gksk,usk-ua-1Γq∫askln⁡skτq-1fτ,uτdττ,k=1,2,…,N.Hence, substituting (8) and (9) into (1), we get(10)ut=ua+1Γq∫atln⁡tτq-1fτ,uτdττ,fort∈a,t1,gkt,ut,fort∈tk,sk,k=1,2,…,N,gksk,usk+1Γq∫atln⁡tτq-1fτ,uτdττ-∫askln⁡skτq-1fτ,uτdττfort∈sk,tk+1,k=1,2,…,N.Obviously, (10) satisfies the conditions in system (1) and(11)limgkt,ut=ua+1/Γq∫atln⁡t/τq-1fτ,uτdτ/τfort∈tk,sk,∀k∈1,2,…,NEq.10⟺limgkt,ut=ua+1/Γq∫atln⁡t/τq-1fτ,uτdτ/τfort∈tk,sk,∀k∈1,2,…,Nsystem1.Therefore, (10) is a solution of system (1).Remark 1.
Equation (10) is only aparticular solution of system (1) because it does not contain the important part ∫skt(ln⁡t/τ)q-1f(τ,u(τ))dτ/τ of the approximate solution u~(t).Next, some definitions and conclusions are introduced in Section2, the equivalent integral equation will be given for differential equation with Caputo-Hadamard fractional derivatives and noninstantaneous impulses in Section 3, and an example will show that there exists the general solution for this fractional differential equation with noninstantaneous impulses in Section 4.


## 2. Preliminaries

Definition 2 (see [5, p. 110]).
Let0≤a≤b≤∞ be finite or infinite interval of the half-axis R+. The left-sided Hadamard fractional integral of order α∈C(R(α)>0) of function φ(x) is defined by(12)JHa+αφx=1Γα∫axln⁡xsα-1φsdss,a<x<b,where Γ(·) is the Gamma function.The left-sided Caputo-Hadamard fractional derivativeDC-Ha+αφ(x) is presented in [12] by the following.Definition 3 (see [12, p. 4]).
LetR(α)≥0 and n=[R(α)]+1 and φ∈φ:[a,b]→C:δ(n-1)φ(x)∈AC[a,b], 0<a<b<∞. Then DC-Ha+αφ(x) exist everywhere on [a,b] and if α∉N0,(13)DC-Ha+αφx=1Γn-α∫axln⁡xsn-α-1δnφsdss=JHa+n-αδnφx,where differential operator δ=xd/dx and δ0y(x)=y(x).Lemma 4 (see [12, p. 5]).
LetR(α)>0, n=[R(α)]+1, and φ∈C[a,b]. If R(α)≠0 or α∈N, then (14)DC-Ha+αJHa+αφx=φx.Lemma 5 (see [12, p. 6]).
Letφ∈ACδn[a,b] or Cδn[a,b] and α∈C, then(15)JHa+αDC-Ha+αφx=φx-∑k=0n-1δkφak!ln⁡xak.Lemma 6 (see [15, p 4]).
Let0<R(q)<1 and ξ is a constant. A function u(t):[a,T]→C is general solution of the system (16)DC-Ha+qut=ht,ut,t∈a,T,t≠tkk=1,2,…,m,Δut=tk=utk+-utk-=Δkutk-∈C,k=1,2,…,m,ua=ua,ua∈C,if and only if u(t) satisfies the fraction integral equation(17)ut=ua+1Γq∫atlntsq-1hdss,fort∈a,t1,ua+∑i=1kΔiuti-+1Γq∫atln⁡tsq-1hdss+ξ∑i=1kΔiuti-Γq∫atiln⁡tisq-1hdss+∫titln⁡tsq-1hdss-∫atln⁡tsq-1hdssfort∈tk,tk+1,k=1,2,…,m.provided that the integral in (17) exists, and here h=h(s,u(s)).


## 3. Main Result

Theorem 7.
Letξk (here k=1,2,…,N) be some arbitrary constants. System (1) is equivalent to(18)
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t
=
u
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+
1
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ln
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t
τ
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-
1
f
d
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r
t
∈
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1
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k
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k
=
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k
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d
τ
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×
∫
a
s
k
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⁡
s
k
τ
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-
1
f
d
τ
τ
+
∫
s
k
t
l
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∫
a
t
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d
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τ
f
o
r
t
∈
s
k
,
t
k
+
1
,
k
=
1,2
,
…
,
N
,provided that the integral in (18) exists, and here f=f(τ,u(τ)).Proof.
Step  1 (Necessity). We will verify that (18) satisfies all conditions of system (1). For convenience, we divide this section into three steps.
Step  1.1. Equation (18) satisfies the fractional derivative in system (1).
By (18), for t∈(sk,tk+1] (here k=0,1,…,N), we get(19)DC-Ha+qut=DC-Ha+qgksk,usk-1Γq∫askln⁡skτq-1fdττ+1Γq∫atln⁡tτq-1fdττ+ξkΓqgksk,usk-ua-1Γq∫asklnskτq-1fdττ×∫askln⁡skτq-1fdττ+∫sktln⁡tτq-1fdττ-∫atln⁡tτq-1fdττ=ft,utt∈sk,tk+1+ξkΓqgksk,usk-ua-1Γq∫askln⁡skτq-1fdττ×DC-Hsk+q∫sktln⁡tτq-1fdττ-DC-Ha+q∫atln⁡tτq-1fdττt∈sk,tk+1=ft,utt∈sk,tk+1.So, (18) satisfies the fractional derivative in system (1).
Step  1.2. We can verify that (18) satisfies noninstantaneous impulses and initial value in system (1).
Step  1.3. Verify that (18) satisfies a hidden condition of system (1).
Lettinggk(t,u(t))=ua+1/Γ(q)∫atln⁡t/τq-1fdτ/τ for t∈(tk,sk] and all k∈{1,2,…,N} in (18), we obtain(20)
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+
1
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,
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=
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,
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1
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d
τ
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+
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k
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q
g
k
s
k
,
u
s
k
-
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a
-
1
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×
∫
a
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1
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+
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⁡
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1
f
d
τ
τ
-
∫
a
t
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⁡
t
τ
q
-
1
f
d
τ
τ
for
t
∈
s
k
,
t
k
+
1
,
k
=
1,2
,
…
,
N
,
=
u
a
+
1
Γ
q
∫
a
t
ln
⁡
t
τ
q
-
1
f
d
τ
τ
for
t
∈
a
,
t
1
,
u
a
+
1
Γ
q
∫
a
t
ln
⁡
t
τ
q
-
1
f
d
τ
τ
for
t
∈
t
k
,
s
k
,
k
=
1,2
,
…
,
N
,
u
a
+
1
Γ
q
∫
a
t
ln
⁡
t
τ
q
-
1
f
d
τ
τ
for
t
∈
s
k
,
t
k
+
1
,
k
=
1,2
,
…
,
N
.Moreover, it is sure that (20) is the solution of system (4) by Lemma 5. Thus, (18) satisfies all conditions of system (1).
Step  2 (Sufficiency). We will verify that the solution of system (1) satisfies (18). For convenience, we divide this section into three steps.
Step  2.1. Verify that the solution of system (1) satisfies (18) in intervals (a,t1] and (t1,s1].
By Lemma5, the solution of (1) satisfies(21)ut=ua+1Γq∫atlntτq-1fdττfort∈a,t1,and u(t)=g1(t,u(t)) for t∈(t1,s1].
Step 2.2. Verify that the solution of system (1) satisfies (18) in intervals (s1,t2] and (t2,s2].
Fort∈(s1,t2], the approximate solution (by the above discussion about u~(t)) is given as(22)u~t=g1s1,us1+1Γq∫s1tln⁡tτq-1fdττfort∈s1,t2,with error e1(t)=u(t)-u~(t) for t∈(s1,t2]. By the particular solution (10), the exact solution u(t) of system (1) satisfies(23)limg1s1,us1-ua-1/Γq∫as1ln⁡s1/τq-1fdτ/τ→0ut=ua+1Γq∫atln⁡tτq-1fdττfort∈s1,t2.Thus, (24)limg1s1,us1-ua-1/Γq∫as1ln⁡s1/τq-1fdτ/τ→0e1t=limg1s1,us1-ua-1/Γq∫as1ln⁡s1/τq-1fdτ/τ→0ut-u~t=1Γq∫atln⁡tτq-1fdττ-∫as1ln⁡s1τq-1fdττ-∫s1tln⁡tτq-1fdττ.Therefore, suppose(25)e1t=1Γqγg1s1,us1-ua-1Γq∫as1ln⁡s1τq-1fdττ×∫atln⁡tτq-1fdττ-∫as1ln⁡s1τq-1fdττ-∫s1tln⁡tτq-1fdττ,where γ is an undetermined function with γ(0)=1. Thus,(26)ut=u~t+e1t=g1s1,us1-1Γq∫as1ln⁡s1τq-1fdττ+1Γq∫atln⁡tτq-1fdττ+1Γq1-γg1s1,us1-ua-1Γq∫as1ln⁡s1τq-1fdττ×∫as1ln⁡s1τq-1fdττ+∫s1tln⁡tτq-1fdττ-∫atln⁡tτq-1fdττfort∈s1,t2.On the other hand, letting t1→s1, we get(27)limt1→s1DC-Ha+qut=ft,ut,t∈sk,tk+1,k=0,1,ut=g1t,ut,t∈t1,s1,ua=ua∈C,=DC-Ha+qut=ft,ut,t∈sk,tk+1,k=0,1,us1=g1s1,us1,ua=ua∈C,=DC-Ha+qut=ft,ut,t∈sk,tk+1,k=0,1,us1+-us1-=g1s1,us1-ua-1Γq∫as1ln⁡s1τq-1fdττ,ua=ua∈C.By using Lemma 6 to (27), we get 1-γ(z)=ξ1z, ∀z∈C, and here ξ1 is an arbitrary constant. Thus,(28)ut=g1s1,us1-1Γq∫as1ln⁡s1τq-1fdττ+1Γq∫atln⁡tτq-1fdττ+ξ1Γqg1s1,us1-ua-1Γq∫as1ln⁡s1τq-1fdττ×∫as1ln⁡s1τq-1fdττ+∫s1tln⁡tτq-1fdττ-∫atln⁡tτq-1fdττfort∈s1,t2.And ut=g2t,ut for t∈(t2,s2].
Step  2.3. Verify that the solution of system (1) satisfies (18) in intervals (sk,tk+1] and (tk+1,sk+1].
The approximate solution ast∈(sk,tk+1] is given by(29)u~t=gksk,usk+1Γq∫sktln⁡tτq-1fdττfort∈sk,tk+1,k=1,2,…,N,with error ek(t)=u(t)-u~t for t∈(sk,tk+1]. Moreover, by the particular solution (10), the exact solution of system (1) satisfies(30)limgksk,usk-ua-1/Γq∫askln⁡sk/τq-1fdτ/τ→0ut=ua+1Γq∫atln⁡tτq-1fdττfort∈sk,tk+1.Thus,(31)limgksk,usk-ua-1/Γq∫askln⁡sk/τq-1fdτ/τ→0ekt=limgksk,xsk-ua-1/Γq∫askln⁡sk/τq-1fdτ/τ→0ut-u~t=1Γq∫atln⁡tτq-1fdττ-∫askln⁡skτq-1fdττ-∫sktln⁡tτq-1fdττ.Therefore, suppose(32)ekt=ϖgksk,usk-ua-1Γq∫askln⁡skτq-1fdττlimgksk,usk-ua-1/Γq∫askln⁡sk/τq-1fdτ/τ→0ekt=-1Γqϖgksk,usk-ua-1Γq∫askln⁡skτq-1fdττ×∫askln⁡skτq-1fdττ+∫sktln⁡tτq-1fdττ-∫atln⁡tτq-1fdττ,where ϖ is an undetermined function with ϖ(0)=1. Therefore,(33)ut=u~t+ekt=gksk,usk-1Γq∫askln⁡skτq-1fdττ+1Γq∫atln⁡tτq-1fdττ+1Γq1-ϖgksk,usk-ua-1Γq∫askln⁡skτq-1fdττ×∫askln⁡skτq-1fdττ+∫sktln⁡tτq-1fdττ-∫atln⁡tτq-1fdττfort∈sk,tk+1.On the other hand, consider a special case of system (1) as (34)limtk→skDC-Ha+qut=ft,ut,t∈si,ti+1,i=0,1,…,k,ut=ua+1Γq∫atln⁡tτq-1fτ,uτdττ,t∈ti,si,i=1,2,…,k-1,ut=gkt,ut,t∈tk,sk,ua=ua∈C,=DC-Ha+qut=ft,ut,t∈si,ti+1,i=0,1,…,k,ut=ua+1Γq∫atln⁡tτq-1fτ,uτdττ,t∈ti,si,i=1,2,…,k-1,usk=gksk,usk,ua=ua∈C,=DC-Ha+qut=ft,ut,t∈si,ti+1,i=0,1,…,k,ut=ua+1Γq∫atln⁡tτq-1fτ,uτdττ,t∈ti,si,i=1,2,…,k-1,usk+-usk-=gksk,usk-ua-1Γq∫askln⁡skτq-1fτ,uτdττ,ua=ua∈C.Using Lemma 6 for system (34), we obtain 1-ϖ(z)=ξkz, ∀z∈C, and here ξk is an arbitrary constant. Thus(35)ut=gksk,usk-1Γq∫askln⁡skτq-1fdττ+1Γq∫atln⁡tτq-1fdττ+ξkΓqgksk,usk-ua-1Γq∫askln⁡skτq-1fdττ×∫askln⁡skτq-1fdττ+∫sktln⁡tτq-1fdττ-∫atln⁡tτq-1fdττfort∈sk,tk+1.
By “Sufficiency” and “Necessity,” system (1) is equivalent to (18). The proof is completed.


## 4. Examples

In this section, we will give an impulsive fractional system to illustrate that there exists general solution for fractional differential equations with noninstantaneous impulses.Example 1.
Let us consider the following impulsive linear fractional system:(36)DC-H1+1/2ut=ln⁡t,t∈1,π2∪π,2π,ut=sin⁡t,t∈π2,π,u1=1.By Theorem 7, system (36) has general solution(37)
u
t
=
1
+
1
Γ
1
/
2
∫
1
t
ln
⁡
t
τ
1
/
2
-
1
ln
⁡
τ
d
τ
τ
,
for
t
∈
1
,
π
2
,
sin
⁡
t
,
for
t
∈
π
2
,
π
,
-
1
Γ
1
/
2
∫
1
π
ln
⁡
π
τ
1
/
2
-
1
ln
⁡
τ
d
τ
τ
+
1
Γ
1
/
2
∫
1
t
ln
⁡
t
τ
1
/
2
-
1
ln
⁡
τ
d
τ
τ
+
ξ
Γ
1
/
2
-
1
-
1
Γ
1
/
2
∫
1
π
ln
⁡
π
τ
1
/
2
-
1
ln
⁡
τ
d
τ
τ
×
∫
1
π
ln
⁡
π
τ
1
/
2
-
1
ln
⁡
τ
d
τ
τ
+
∫
π
t
ln
⁡
t
τ
1
/
2
-
1
ln
⁡
τ
d
τ
τ
-
∫
1
t
ln
⁡
t
τ
1
/
2
-
1
ln
⁡
τ
d
τ
τ
,
for
t
∈
π
,
2
π
,where ξ is a constant. After some elementary computation, (37) can be rewritten by(38)
u
t
=
1
+
4
3
π
ln
⁡
t
3
/
2
,
for
t
∈
1
,
π
2
,
sin
⁡
t
,
for
t
∈
π
2
,
π
,
-
4
3
π
ln
⁡
π
3
/
2
+
4
3
π
ln
⁡
t
3
/
2
t
>
1
+
2
ξ
3
π
-
1
-
4
3
π
ln
⁡
π
3
/
2
×
2
ln
⁡
π
3
/
2
+
ln
⁡
t
π
1
/
2
2
ln
⁡
t
+
ln
⁡
π
t
>
π
-
2
ln
⁡
t
3
/
2
t
>
1
,
for
t
∈
π
,
2
π
.Next, let us verify that (38) satisfies all conditions of system (36). By Definition 3, we have(39)DC-H1+1/243πln⁡t3/2=1Γ1-1/243π∫1tln⁡ts1-1/2-1δln⁡s3/2dss=1Γ1/243π∫1tln⁡ts1-1/2-132ln⁡s1/2dss=ln⁡tDC-H1+1/2ln⁡tπ1/22ln⁡t+ln⁡πt>π=1Γ1-1/2∫1tln⁡ts1-1/2-1δln⁡sπ1/22ln⁡s+ln⁡πt>πdss=1Γ1/2∫πtln⁡ts1-1/2-1δln⁡sπ1/22ln⁡s+ln⁡πdss=1Γ1/2∫πtln⁡ts1-1/2-12ln⁡sπ1/2+12ln⁡sπ-1/22ln⁡s+ln⁡πdss=1Γ1/2∫πtln⁡ts1-1/2-12ln⁡sπ1/2+12ln⁡sπ-1/22ln⁡sπ+3ln⁡πdss=3Γ1/2∫πtln⁡ts1-1/2-1ln⁡sπ1/2dss+3/2ln⁡πΓ1/2∫πtln⁡ts1-1/2-1ln⁡sπ-1/2dss=32Γ12ln⁡tπ+32Γ12ln⁡π=32Γ12ln⁡tt>π.Therefore, for t∈(1,π/2] and t∈(π,2π] in (38), we have (40)DC-H1+1/2ut=DC-H1+1/21+43πln⁡t3/2=ln⁡t,fort∈1,π2,DC-H1+1/2utt∈π,2π=DC-H1+1/2-43πln⁡π3/2+43πln⁡t3/2t>1+2ξ3π-1-43πln⁡π3/2×2ln⁡π3/2+ln⁡tπ1/22ln⁡t+ln⁡πt>π-2ln⁡t3/2t>1t∈π,2π=DC-H1+1/243πln⁡t3/2t>1+2ξ3π-1-43πln⁡π3/2×ln⁡tπ1/22ln⁡t+ln⁡πt>π-2ln⁡t3/2t>1t∈π,2π=ln⁡tt>1+2ξ3π-1-43πln⁡π3/232Γ12ln⁡tt>π-32Γ12ln⁡tt>1t∈π,2π=ln⁡tt>1t∈π,2π,fort∈π,2π.Thus, (38) satisfies fractional derivative and noninstantaneous impulses in system (36). Therefore, (38) is general solution of system (36).



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