# Existence and Uniqueness Theorems for a Fractional Differential Equation with Impulsive Effect under Band-Like Integral Boundary Conditions

Authors: Zihan Gao; Tianlin Hu; Huihui Pang
Journal: Advances in Mathematical Physics
 (2020)

---

## Abstract

In this paper, we consider a class of nonlinear Caputo fractional differential equations with impulsive effect under multiple band-like integral boundary conditions. By constructing an available completely continuous operator, we establish some criteria for judging the existence and uniqueness of solutions. Finally, an example is presented to demonstrate our main results.


---

## Body

## 1. Introduction

Researches on fractional differential equations have witnessed an unprecedented boom in recent years on account of the far-reaching application in various subjects, such as physics, biology, nuclear dynamics, chemistry, etc., for more details, see [1–3] and the references therein. Considering the impulse effect in the continuous differential equation can quantify the impact of the instantaneous mutation of the model and provide a theoretical basis for the practical application. Therefore, impulsive differential equation problems also attract great attention from scholars. For the theories of impulsive differential equations, the readers can refer to [4–7]. In addition, there have been some excellent results concerning the existence, uniqueness, and multiplicity of solutions or positive solutions to some nonlinear fractional differential equations with various nonlocal boundary conditions. As for some recent bibliographies, we refer readers to see [8–11] and the reference therein.Yang and Zhang in [12] studied the following impulsive fractional differential equation(1)cD0+αxt=ft,xt,t∈J=0,1,t≠tk,Δxt=ξk=Ikxξk,Δx′t=ξk=I¯kxξk,k=1,2,…,m,x0=hx,x1=gx,wherecD0+α is the Caputo fractional derivative, α∈R,1<α≤2. f:0,1×R→R is a continuous function, Ik,I¯k are continuous functions, gx=maxjxξj/λ+xξj, hx=minjxξj/κ+xξj. By transforming the boundary value problem into an equivalent integral equation and employing some fixed point theorems, existence result is obtained.The research results of fractional differential equations with integral boundary conditions are also quite rich, and the research on those questions remains as a hotpot among many scholars in recent years. We refer readers to see [13–16] and the reference therein.In [13], Song and Bai considered the following boundary value problem of fractional differential equation with Riemann–Stieltjes integral boundary condition(2)D0+αut+λft,ut,ut=0,0<t<1,n−1<α≤n,uk0=0,0≤k≤n−2,u1=∫01usdAs,wheren−1<α≤n,λ>0,D0+α is the Riemann–Liouville fractional derivative, A is a function of bounded variation, ∫01usdAs denotes the Riemann–Stieltjes integral of u with respect to A. By the use of fixed point theorem and the properties of mixed monotone operator theory, the existence and uniqueness of positive solutions for the problem are acquired.Moreover, Zhao and Liang in [14] added impulsive effect to fractional equations with integral boundary conditions and discussed the existence of solutions(3)tkDtαut=ft,u,u′,Dα−1u,t≠tk,ΔDα−1utkt=ξk=Ikutk,k=1,2,…,m,u0=u′0=0,u′1=∫0ηgs,usds,wheretkDtα is the Riemann-iouville fractional derivative of order 2<α≤3,Jk=tk,tk+1,k=1,2,…,m.f∈CJ×R3,R,Ik∈CR,R,0<η<1,g∈CJ×R,R. By applying the contraction mapping principle and the fixed point theorem, some sufficient criteria for the existence of solutions are obtained.Inspired by the works above, we will study the impulsive fractional differential equation with band-like integral boundary conditions(4)cD0+αxt=ft,xt,cD0+βxt,t∈0,1,t≠ξk,(5)−Δxt=ξk=Ikxξk,k=1,2,…,n,x0=x1=∑i=0nαi∫ξiξi+1xtgtdt,wherecD0+α, cD0+β are the Caputo fractional derivatives of order 1<α≤2,0<β<1,J=0,1,Jk=ξk,ξk+1,f∈CJk×R2,  R, for k=0,1,2,…,n, and αi is a nonnegative constant, g∈C0,1,R+ satisfying 0<∑i=0nαi∫ξiξi+1gtdt<1,0=ξ0<ξ1<⋯<ξn<ξn+1=1, xξk+=limh→0+xξk+  h, and xξk−=limh→0−xξk+h represent the right and the left limits of xξk at t=ξk,Δx|t=ξk=xξk+−xξk−.Ikxξk∈CR,  R. By using the Leray–Schauder alternative theorem and the Banach contraction mapping principle, the existence and uniqueness theorems of solutions to problem (4) can be established.We emphasize that the discontinuous points caused by impulse are just the upper and lower limits of the band-like integral values in the boundary conditions of (4). In other words, the value of the unknown function at the endpoint of the interval [0,1] is related to the linear combination of the integral values of the unknown function between the discontinuous points.Another thing worth mentioning is that despite the complicated boundary conditions and the interference of the impulse, we use a piecewise function to represent the operatorF in a concise form based on the form of the Green’s function and accurately estimate the upper bound of its absolute value, which is fully prepared for the establishment of the main theorem.Accordingly, the conclusions we reached are extensive results compared with the reference [4–7, 15–20] and a meaningful supplement to the theory of impulsive fractional differential equations.


## 2. Preliminaries

In this section, we present some definitions, lemmas, and some prerequisite results that will be used to prove our results.Definition 1 [19].
The Riemann–Liouville fractional integral of orderα>0 of a function f:0,  ∞→R is defined as(6)I0+αft=1Γα∫0tt−sα−1fsds,
if the right-hand side is pointwise defined on0,  ∞, where Γα is the Euler gamma function satisfying Γα=∫0∞tα−1e−tdt, for α>0.Definition 2 [16].
The Caputo fractional derivative of orderα>0 for a function f:0,  ∞→R is defined as(7)cD0+αft=1Γn−α∫0tfnst−sn−α−1ds,
wheren=α+1 and α stands for the largest integer that not greater than α.Lemma 1.
Forh∈L10,1, the solution of the fractional differential equation cD0+αut+ht=0,0<t<1 can be expressed as(8)ut=−1Γα∫0tt−sα−1hsds+c1+c2t+⋯+cntn−1,0<t<1,
whereci∈R, for i=1,2,…,n.Lemma 2.
For anyv∈L10,1, the following boundary value problem(9)cD0+αxt=−vt,t∈Jk,(10)−Δxt=ξk=Ikxξk,k=1,2,…,n,x0=x1=∑i=0nαi∫ξiξi+1xtgtdt,
has a unique solution(11)xt=Av1t+Av2t+B1v+B2v+∑k=1nGt,ξkIkxξk+1Δ∑i=0nαi∫ξiξi+1∑k=1nGt,ξkIkxξkdt,
where(12)Av1t=−1Γα−1∫ξkt∫0ss−τα−2vτdτds,fort∈Jk,k=0,1,2,…,n,Av2t=1Γα−1∑k=1n+1Gt,ξk∫ξk−1ξk∫0ss−τα−2vτdτds,fort∈Jk,k=0,1,2,…,n,B1v=−1ΔΓα−1∑i=0nαi∫ξiξi+1∫ξit∫0ss−τα−2vτdτdsdt,B2v=1ΔΓα−1∑i=0nαi∫ξiξi+1∑k=1n+1Gt,ξk∫ξk−1ξk∫0ss−τα−2vτdτdsdt,
where(13)Gt,ξk=t−1,0<ξk<t,k=1,2,…,n,t,t≤ξk<1,k=1,2,…,n,Δ=1−∑ni=0αi∫ξiξi+1gtdt.Proof.
From equation (9), through calculation we have(14)x′t=−1Γα−1∫0tt−sα−2vsds+a,
wherea is an arbitrary real constant.
Fort∈J0, according to (15), we can obtain(15)xt=−1Γα∫0tt−sα−1vsds+b+at,
whereb is an arbitrary real constant.
Fort∈J1, based on (15) and (16), we have(16)xt=−1Γα−1∫ξ1t∫0ss−  τα−2vτdτds−1Γα∫0t1ξ1−  sα−1vsds+at+b−I1xξ1.
Analogously, fort∈Jk,k=2,3…,n, it holds that(17)xt=−1Γα−1∫ξkt∫0ss−τα−2vτdτds−1Γα−1∑k=1n∫ξk−1ξk∫0ss−τα−2vτdτds+at+b−∑ξk<tIkxξk.
Sincex0=x1=b, together with (16) and (18), we receive that(18)a=−1Γα−1∑k=1n+1∫ξk−1ξk∫0ss−τα−2vτdτds+∑k=1nIkxξk.
Substitutinga into (18), and based on the form of Green’s function, we get(19)xt=b−1Γα−1∫ξkt∫0ss−τα−2vτdτds+1Γα−1∑k=1n+1Gt,ξk∫ξkt∫0ss−  τα−2vτdτds+∑k=1nGt,ξkIkxξk,fort∈Jk.
Subject to (20), using boundary conditions of (10), we have(20)b=b∑i=0nαi∫ξiξi+1gtdt−∑i=0nαi∫ξiξi+11Γα−1∫ξit∫0ss−τα−2vτgtdτdsdt+∑i=0nαi∫ξiξi+11Γα−1∑k=1n+1Gt,ξk∫ξkt∫0ss−  τα−2vτgtdτdsdt+∑i=0nαi∫ξiξi+1∑k=1nGt,ξkIkxξkgtdt.
Consequently,(21)b=−1ΔΓα−1∑i=0nαi∫ξiξi+1∫ξit∫0sgts−τα−2vτdτdsdt+1ΔΓα−1∑i=0nαi∫ξiξi+1∑k=1n+1Gt,ξk∫ξkt∫0ss−τα−2gtvτdτdsdt+1Δ∑i=0nαi∫ξiξi+1gt∑k=1nGt,ξkIkxξkdt,
whereΔ=1−∑i=0nαi∫ξiξi+1gtdt. In what follows, we always assume that 0<∑i=0nαi∫ξiξi+1gtdt<1.
From (18)–(20), and (22), it can be received that(22)xt=Av1t+Av2t+B1v+B2v+∑k=1nGt,ξkIkxξk+1Δ∑i=0nαi∫ξiξi+1∑k=1nGt,ξkIkxξkdt,
whereAv1t,Av2t,B1v,B2v are denoted by (12).
DefineX={xt:xt∈CJk,cD0+βxt∈CJ,xξk+,xξk− exit and xξk−=xξk,k=0,1,…,n}. Obviously, X is a Banach space endowed with the norm xX=x+cD0+βx=supt∈0,1xt+supt∈0,1|cD0+βxt|.Lemma 3.
For anyv∈L10,1, the following results are true
(1)
A
v
1
t
≤
v
/
Γ
α
+
1
1
−
ξ
k
α, for t∈Jk,k=0,1,2,…,n
(2)
A
v
2
t
≤
v
/
Γ
α
+
1
, for t∈Jk,k=0,1,2,…,n
(3)
B
1
v
≤
v
/
Δ
Γ
α
+
1
∑
i
=
0
n
α
i
1
/
α
+
1
ξ
i
+
1
α
+
1
−
ξ
i
α
+
1
+
ξ
i
+
1
ξ
i
+
1
α
−
ξ
i
α
(4)
B
2
v
≤
v
/
Δ
Γ
α
+
1
∑
i
=
0
n
α
i
ξ
i
+
1
−
ξ
i
.Proof.
Fort∈Jk,k=0,1,2,…,n, we have(23)Av1t≤1Γα−1∫ξk1∫0ss−τα−2vτdτds≤vΓα+11−ξkα,(24)Av2t≤1Γα−1∑k=1n+1Gt,ξk∫tk−1ξk∫0ss−  τα−2vτdτds≤vΓα+1∑k=1n+1ξkα−ξk−1α=vΓα+1.
According to (24) and (25), we get(25)B1v≤1ΔΓα−1∑i=0nαi∫ξiξi+1∫ξit∫0ss−  τα−2vτdτdsdt≤vΔΓα−1∑i=0nαi∫ξiξi+1tα−ξiαdt=vΔΓα+1∑i=0nαi1α+1ξi+1α+1−ξiα+1−ξiαξi+1+ξiα+1=vΔΓα+1∑i=0nαi1α+1ξi+1α+1−ξiα+1+ξi+1ξi+1α−ξiα
and(26)B2v≤vΔΓα−1∑i=0nαi∫ξiξi+1∑k=1n+1Gt,ξk∫ξk−1ξk∫0ss−  τα−2dτdsdt≤vΔΓα+1∑i=0nαi∫ξiξi+1∑k=1n+1ξkα−ξk−1αdt=vΔΓα+1∑i=0nαiξi+1−ξi.Apparently,|Gt,ξk|≤1, for t∈0,1 and k=1,…,n, so in view of Lemma 3, and combing with (11), we can write(27)xt≤2vΓα+1+vΔΓα+1∑i=0nαiξi+1−ξi+vΔΓα+1∑i=0nαi1α+1ξi+1α+1−ξiα+1+ξi+1ξi+1α−ξiα+∑k=1nIkxξk+1Δ∑i=0nαi∫ξiξi+1∑k=1nIkxξkdt.Lemma 4 (the Leray–Schauder alternative theorem).
LetF:X→X be a completely continuous operator (i.e., a map that restricted to any bounded set in X is compact). Let(28)εF=x∈X:x=λFx,0<λ<1.
Then either the setεF is unbounded, or F has at least one fixed point.
The operatorF:X→X is defined by(29)Fxt=Afx1t+Afx2t+B1fx+B2fx+∑k=1nGt,ξkIkxξk+1Δ∑i=0nαi∫ξiξi+1∑k=1nGt,ξkIkxξkdt,
wherefx=ft,xt,cD0+αxt. Accordingly, we know that(30)Fx′t=−1Γα−1∫0tt−sα−2ft,xt,cD0+βxtds+−1Γα−1∑k=1n+1∫ξk−1ξk∫0ss−τα−2ft,xt,cD0+βxtdτds+∑k=1nIkxξk
and(31)cD0+βFxt=∫0t1Γn−βFnst−sn−β−1ds=∫0tt−s−βΓ1−βF′xsds.Lemma 5.
The operatorF:X→X is completely continuous.Proof.
The operatorF is continuous in view of the continuity of Gt,ξk, ft,xt,cD0+βxt, and Ikx. Let Ω⊂X be bounded. Then there are positive constants T1 and T2 such that (32)ft,xt,cD0+αxt≤T1,Ikxξk≤T2,forx∈Ω.
For convenience, we set(33)T=maxT1,T2,R=1Γα+1Γα+1+n,N=2Γα+1+1ΔΓα+1∑i=0nαiξi+1−ξi+1ΔΓα+1∑i=0nαi1α+1ξi+1α+1−ξiα+1+ξi+1ξi+1α−ξiα+n1+1Δ∑i=0nαiξi+1−ξi.
For anyx∈Ω, we have(34)Fxt≤1Γα−1∫ξkt∫0ss−τα−2fxτdτds+1Γα−1∑k=1n+1Gt,ξk∫ξk−1ξk∫0ss−τα−2fxτdτds+1ΔΓα−1∑i=0nαi∫ξiξi+1∫ξit∫0ss−τα−2fxτdτdsdt+1ΔΓα−1∑i=0nαi∫ξiξi+1∑k=1n+1Gt,ξk∫ξk−1ξk∫0ss−τα−2fxτdτdsdt+1Δ∑i=0nαi∫ξiξi+1∑k=1nGt,ξkIkxξkdt+∑k=1nGt,ξkIkxξk≤T12−ξkαΓα+1+1ΔΓα+1∑i=0nαiξi+1−ξi+1ΔΓα+1∑i=0nαi1α+1ξi+1α+1−ξiα+1+ξi+1ξi+1α−ξiα+nT21Δ∑i=0nαiξi+1−ξi+1≤TN.
Meanwhile, forx∈Ω, we can get(35)Fx′t≤T11Γα+1Γα+1+n≤TR.
Furthermore, forx∈Ω, we have(36)cD0+βFxt=1Γ1−β∫0tFx′st−s−βds≤TRΓ1−β∫0tt−s−βds≤TRΓ1−β11−β=TRΓ2−β.
Hence, the following result can be derived(37)FxX=Fx+cD0+βFx≤TN+RΓ2−β.
Thus, we have shown the operatorF is uniformly bounded.
Next, we will show thatF is equicontinuous. Let t1,t2∈Jk with t1≤t2, then we have(38)Fxt2−Fxt1≤T11Γα+1t2α−t1α+T11Γα+1t2−t1+T11ΔΓα+1t2α−t1α∑i=0nαiξi+1−ξi+T11ΔΓα+1t2−t1∑i=0nαiξi+1−ξi+nT2t2−t11Δ∑i=0nαiξi+1−ξi+T2t2−t1n≤t2−t12T1Γα+1+2T1ΔΓα+1∑i=0nαiξi+1−ξi+nT21Δ∑i=0nαiξi+1−ξi+T2n
and(39)Fx′t2−Fx′t1=1Γα−1∫0t2s−τα−2fxτdτ−1Γα−1∫0t1s−τα−2fxτdτ≤T11Γαt2α−1−t1α−1≤T11Γαt2−t1.
So we have(40)cD0+βFxt2−cD0+βFxt1=∫0t2t2−s−βΓ1−βFx′sds−∫0t1t2−s−βΓ1−βFx′sds=Fx′sΓ2−βt21−β−t11−β≤T1t2−t12ΓαΓ2−βt1β.
Hence, we can get(41)Fxt2−Fxt1X≤t2−t12T1Γα+1+2T1ΔΓα+1∑i=0nαiξi+1−ξi+nT21Δ∑i=0nαiξi+1−ξi+T2n+T1t2−t1ΓαΓ2−βt1β,
which implies thatFxt2−Fxt2X→0 as t2→t1. Therefore, the operator F is equicontinuous, and the operator F is completely continuous.


## 3. Main Results

In the following discussion, we assume that the following hypotheses are valid, whereρi,Lj and Mj are positive constants, for i=1,…,4, j=1,2.(H1)
|ft,x,  y|≤ρ1+ρ2|x|+ρ3|y|(H2)
|Ikxξk|≤ρ4|xξk|, for k=1,…,n(H3)
|ft,x,  y−ft,x1,y1|≤L1|x−x1|+|y−y1|(H4)
|Iku−Ikv|≤L2u−  v, for k=1,…,n(H5)
M1=supt∈0,1|ft,0,0|,M2=Ik0, for k=1,…,n.The first result is based on the Letay–Schauder alternative theorem.Theorem 1.
Assume that (H1) and (H2) hold. In addition it is assumed that(42)ρ0N+RΓ2−β<1,
whereρ0=max{ρ2,ρ3,ρ4}. Then boundary value problems (4) and (5) have at least one solution.Proof.
It will be verified that the setε=x∈Xx=λFx,0≤λ≤1 is bounded. Let x∈ε, then x=λFx. For all t∈0,1, we have(43)xt=λFxt.
According to (H1), (H2) and Lemma 3, for t∈Jk,k=0,1,…,n, we have(44)xt=λFxt≤Fxt.
and(45)Fxt≤Afx1t+Afx2t+B1fx+B2fx+∑k=1nGt,ξkIkxξk+1Δ∑i=0nαi∫ξiξi+1∑k=1nGt,ξkIkxξkdt≤ρ1+ρ2xt+ρ3cD0+βxt2−ξkαΓα+1+1ΔΓα+1∑i=0nαiξi+1−ξi+1ΔΓα+1∑i=0nαi1α+1ξi+1α+1−ξiα+1+ξi+1ξi+1α−ξiα+nρ4xξk1Δ∑i=0nαiξi+1−ξi+1≤ρ1+ρ0xXN.
Analogously, we have(46)Fx′t≤ρ1+ρ2xt+ρ3cD0+βxt1Γα+1Γα+1+ρ4xξkn≤ρ1+ρ0xXR,
accordingly, we can get(47)cD0+βFxt=∫0t1Γ−βF′st−s−βds≤ρ1+ρ0xXR∫0tt−s−βΓ1−βds=ρ1+ρ0xXRΓ2−β.
Hence, we have(48)xX=Fx+cD0+βFx≤ρ1+ρ0xXN+RΓ2−β,
which means thatρ0N+R/Γ2−β<1 and ε is bounded. Therefore, by Lemma 4, the operator F has at least one fixed point. So boundary value problems (4) and (5) have at least one solution.
Next, we will prove the uniqueness of solutions to boundary value problems (4) and (5) via the Banach contraction mapping principle.Theorem 2.
Suppose that (H3)–(H5) are true, in addition that(49)L<N+RΓ2−β−1,
then there is a unique solution for boundary value problem (4) and (5).Proof.
For convenience, we denote(50)L=maxL1,L2,M=maxM1,M2.
We setBθ=x∈X:xX≤θ, for x∈Bθ, on the basis of (H1) and (H3), we have(51)ft,xt,cD0+βxt≤ft,xt,cD0+βxt−ft,0,0+ft,0,0≤L1x+cD0+βx+M1=L1xX+M1.
According to (H4) and (H5), we have(52)Ikxk≤Ikxk−Ik0+Ik0≤L2x+M2≤L2xX+M2.
So forx∈Bθ, we have(53)Fxt≤L1xX+M12−ξkαΓα+1+1ΔΓα+1∑i=0nαiξi+1−ξi+1ΔΓα+1∑i=0nαi1α+1ξi+1α+1−ξiα+1+ξi+1ξi+1α−ξiα+L2xX+M2n1+1Δ∑i=0nαiξi+1−ξi≤LxX+MN.
On the other hand, we get(54)Fx′t≤L1xX+M11Γα+1Γα+1+∑k=1nL2xX+M2≤LxX+M1Γα+1Γα+1+n=Lθ+MR,
further,(55)cD0+βFxt=∫0t1Γn−βF′st−s−βds≤Lθ+MR∫0tt−s−βΓ1−βds=1Γ2−βLθ+MR.
In consequence,(56)FxX=Fx+cD0+βFx≤Lθ+MN+1Γ2−βLθ+MR=Lθ+MN+RΓ2−β.
Therefore, provided(57)θ≥MNΓ2−β+RΓ2−β−LNΓ2−β+R,
we haveFBθ⊂Bθ.
Forx,y∈X,t∈Jk,k=0,1,…,n, we obtain(58)Fxt−Fyt≤1Γα−1∫01∫0ss−τα−2fxτ−fyτdτds+1Γα−1∑k=1n+1Gt,ξk∫ξk−1ξk∫0ss−τα−2fxτ−fyτdτds+1ΔΓα−1∑i=0nαi∫ξiξi+1∫ξkt∫0ss−τα−2fxτ−fyτdτdsdt+1ΔΓα−1∑i=0nαi∫ξiξi+1∑k=1n+1Gt,ξk∫ξk−1ξk∫0ss−τα−2fxτ−fyτdτdsdt+∑k=1nGt,ξkIkxk−Ikyk+1Δ∑i=0nαi∫ξiξi+1∑k=1nGt,ξkIkxk−Ikykdt.
According to (H3)–(H5), we get(59)Fxt−Fyt≤L1x−yX2−ξkαΓα+1+1ΔΓα+1∑i=0nαiξi+1−ξi+1ΔΓα+1∑i=0nαi1α+1ξi+1α+1−ξiα+1+ξi+1ξi+1α−ξiα+L2x−yXn1+1Δ∑i=0nαiξi+1−ξi≤Lx−yXN.
Similarly, it holds that(60)F′xt−F′yt≤Lx−yXR,cD0+βFxt−cD0+βFyt≤Lx−yXRΓ2−β.
Based on the above derivation, we conclude that(61)Fx−FyX≤Lx−yXN+Lx−yXRΓ2−β≤Lx−yXN+RΓ2−β,
So if(62)L<N+RΓ2−β−1,
then boundary value problem (4) and (5) has one and only one solution.Example 1.
Consider the following fractional order boundary value problem(63)cD0+1.5xt=ft,xt,cD0+0.5xt,t∈0,1,t≠ξ1,
with multistripe and band-like boundary conditions(64)−Δxt=ξ1=I1xξ1,x0=x1=∑i=01αi∫ξiξi+1xtgtdt,
where(65)ft,xt,cD0+0.5xt=112t2+1xt+112t2+1cD0+0.5xt+112,gt=t,Ikxξ1=116+116xξ1,ξ0=0,ξ1=12,ξ2=1,α0=14,α1=16.
Clearly,(66)fxt≤112xt+112cD0+0.5xt+112,Ixξ1≤116+116xξ1,ft,x,y−ft,x1,y1≤112x−x1+y−y1,I1u−I1v≤116u−v,M1=supt∈0,1ft,0,0=112,M2=Ik0=116.
It is easy to verify that (H1)–(H5) hold. And by calculation, the following results can be obtained,(67)ρ1=112,ρ2=112,ρ3=112,ρ4=116,ρ0=112,L1=M1=112,L2=M2=116,L=M=112.
Furthermore, we have(68)Δ=0.9063,R=2.8811,N=3.0941,ρ0N+RΓ2−β=0.5288<1.
By Theorem1, boundary value problems (63) and (64) have at least one solution. We also have(69)N+RΓ2−β−1=0.1576>L.
By Theorem2, boundary value problems (63) and (64) have a unique solution.



---

*Source: 6360128-2020-01-28.xml*