α | 1 | 1 | 1 | 0.75 | 0.25 | 0.5 | 0.5 | 0.5 | |
t | 1 | 2 | e | e | e | e | 2 | 1 | |
Q | 1.38 | 1.35 | 2.94 | 3.58 | 4.58 | 4.15 | 1.90 | 0.20 |
The objective of this work is to evaluate the effects of smoking hashish on some blood parameters (Red blood cells, Hemoglobins, Hematocrits, MCV, MCH, MCHC, White blood cells, Urea, Creatinine, Cholesterol, Triglycerides, Liver Aminotransferases) among young people from the Beni Mellal region, Morocco.
Blood samples are collected from 30 male, fasting, voluntary and anonymous individuals in three groups; 10 non-smokers (controls), 10 moderate chronic smokers (5 joints/day (5 j/d) or less for a period not exceeding 3 years) and 10 intense chronic smokers (15 joints/day or more for a period equal to or greater than 5 years).
The results obtained show that with the duration and number of joints consumed, a decrease in the level of HB, RBCs, HT, platelets, urea, cholesterol, triglycerides, SGOT and SGPT while there is an increase in MCH, MCHC, eosinophils and creatinine.
These results could be explained by the impact of the active ingredients in the joints, including THC in cannabis and the components of smoked cigarettes in conjunction with cannabis resin on consumer nutrition.
Citation: Hicham Rahmi, Ben Yamine Mallouki, Fatiha Chigr, Mohamed Najimi. The effects of smoking Haschich on blood parameters in young people from the Beni Mellal region Morocco[J]. AIMS Medical Science, 2021, 8(4): 276-290. doi: 10.3934/medsci.2021023
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The objective of this work is to evaluate the effects of smoking hashish on some blood parameters (Red blood cells, Hemoglobins, Hematocrits, MCV, MCH, MCHC, White blood cells, Urea, Creatinine, Cholesterol, Triglycerides, Liver Aminotransferases) among young people from the Beni Mellal region, Morocco.
Blood samples are collected from 30 male, fasting, voluntary and anonymous individuals in three groups; 10 non-smokers (controls), 10 moderate chronic smokers (5 joints/day (5 j/d) or less for a period not exceeding 3 years) and 10 intense chronic smokers (15 joints/day or more for a period equal to or greater than 5 years).
The results obtained show that with the duration and number of joints consumed, a decrease in the level of HB, RBCs, HT, platelets, urea, cholesterol, triglycerides, SGOT and SGPT while there is an increase in MCH, MCHC, eosinophils and creatinine.
These results could be explained by the impact of the active ingredients in the joints, including THC in cannabis and the components of smoked cigarettes in conjunction with cannabis resin on consumer nutrition.
Mean corpuscular volume;
Mean corpuscular hemoglobin;
Mean corpuscular hemoglobin concentration;
Hemoglobin;
Red blood cells;
Hematocrit;
(serum) Glutamic oxaloacetic transaminase;
Serum glutamic-pyruvic transaminase;
Tetrahydrocannabinol
In the recent years, it has been realized that fractional calculus has an important role in various scientific fields. Fractional differential equations (FDE), which is a consequence of the development of fractional calculus, have attracted the attention of many researchers working in different disciplines ([28]). Scientific literature has witnessed the appearance of several kinds of fractional derivatives, such as the Riemann-Liouville fractional derivative, Caputo fractional derivative, Hadamard fractional derivative, Grünwald-Letnikov fractional derivative and Caputo-Fabrizio etc (for more details, see [11,13,16,21,22,44,48,51,54]). It is worthy mentioning here that almost all researches have been conducted within Riemann-Liouville or Caputo fractional derivatives, which are the most popular fractional differential operators.
J. Hadamard suggested a construction of fractional integro-differentiation which is a fractional power of the type (tddt)α. This construction is well suited to the case of the half-axis and is invariant relative to dilation ([53,p. 330]). The dilation is interpreted in various forms in relation to the field of application. Furthermore, Riemann-Liouville fractional integro-differentiation is formally a fractional power (ddt)α of the differentiation operator (ddt) and is invariant relative to translation if considered on the whole axis. On the other hand, the investigations in terms of Hadamard or Grünwald-Letnikov fractional derivatives are comparably considered seldom.
The boundary value problems defined by FDE have been extensively studied over the last years. Particularly, the study of solutions of fractional differential and integral equations is the key topic of applied mathematics research. Many interesting results have been reported regarding the existence, uniqueness, multiplicity and stability of solutions or positive solutions by means of some fixed point theorems, such as the Krasnosel'skii fixed point theorem, the Schaefer fixed point theorem and the Leggett-Williams fixed point theorem. However, most of the considered problems have been treated in the frame of fractional derivatives of Riemann-Liouville or Caputo types ([12,14,15,41,45]). The qualitative investigations with respect to Hadamard derivative have gained less attention compared to the analysis in terms of Riemann-Liouville and Caputo settings. Recent results on Hadamard FDE can be consulted in ([1,4,5,7,10,17,42,43,52]).
The physical phenomena in fluctuating environments are adequately described using the so called Langevin differential equation (LDE) which was proposed by Langevin himself in [31,1908] to give an elaborated interpretation of Brownian motion. Indeed, LDE is a powerful tool for the study of dynamical properties of many interesting systems in physics, chemistry and engineering ([9,32,57]). The generalized LDE was introduced later by Kubo in [29,1966], where a fractional memory kernel was incorporated into the equation to describe the fractal and memory properties. Since then the investigation of the generalized LDE has become a hot research topic. As a result, various generalizations of LDE have been offered to describe dynamical processes in a fractal medium. One such generalization is the generalized LDE which incorporates the fractal and memory properties with a disruptive memory kernel. This gives rise to study fractional Langevin equation ([36]). As the intensive development of fractional derivative, a natural generalization of the LDE is to replace the ordinary derivative by a fractional derivative to yield fractional Langevin equation (FLE). The FLE was introduced by Mainardi and Pironi in earlier 1990s ([40]). Afterwards, different types of FLE were introduced and studied in [2,3,8,19,30,34,37,38,39,47,50,60,61,62]. In [3], the authors studied a nonlinear LDE involving two fractional orders in different intervals with three-point boundary conditions. The study of FLE in frame of Hadamard derivative has comparably been seldom; see the papers [27,56] in which the authors discussed Sturm-Liouville and Langevin equations via Caputo-Hadamard fractional derivatives and systems of FLE of Riemann-Liouville and Hadamard types, respectively.
In the paper by Kiataramkul et al. [27]: Generalized Sturm–Liouville and Langevin equations via Hadamard fractional derivatives with anti-periodic boundary conditions. In particular, the authors initiate the study of the existence and uniqueness of solutions for the generalized Sturm-Liouville and Langevin fractional differential equations of Caputo-Hadamard type ([21]), with two-point nonlocal anti-periodic boundary conditions, by applying the Banach contraction mapping principle. Moreover, two existence results are established via Leray-Schauder nonlinear alternative and Krasnosleskii's fixed point theorem. In addition, the article by W. Sudsutad et al. [56]: Systems of fractional Langevin equations of Riemann-Liouville and Hadamard types subject to the nonlocal Hadamard and standard Riemann-Liouville with multi-point and multi-term fractional integral boundary conditions, respectively. In particular, the authors also studied the existence and uniqueness results of solutions for coupled and uncoupled systems are obtained by Banach's contraction mapping principle, Leray-Schauder's alternative.
In the present work, we study the existence, uniqueness and stability of solutions for the following FLE with Hadamard fractional derivatives involving local boundary conditions
{Dα1(D2+λ2)x(t)=f(t,x(t),Dα1[x](t)),t∈(1,e),D2x(1)=x(1)=0,x(e)=βx(ξ),ξ∈(1,e], | (1.1) |
where 0<α<1, λ,β>0, such that
sinλ(e−1)≠βsinλ(ξ−1), |
Dα1 denotes the Hadamard fractional derivative of order α, D is the ordinary derivative and
f:[1,e]×C([1,e],R)×C([1,e],R)→C([1,e],R), |
is a continuous function.
Our approach is new and is totally different from the ones obtained in [27,56] in the sense that different fractional derivatives, ordinary and Hadamard fractional order, are accommodated. Different boundary conditions are associated to problem (1.1) such as three point local boundary conditions and associating different fixed point theorems. It is worthwhile to mention that the nonlinear term f in papers [27,56] is independent of fractional derivative of unknown function x(t). But the opposite case is more difficult and complicated. The dependence of the solution on the parameters is discussed, which has not been investigated in [27,56]. It is worth mentioning here that Ulam and generalized Ulam-Hyers-Rassias stability results have not been considered in [27,56]. Furthermore, the presented work illustrates a numerical simulation obtained through a discretization methods for the evaluation of the Hadamard derivative.
Our method differs from that used by [27,56] in our emphasis on the Schaefer fixed point theorem is utilized to investigate existence results for problem (1.1). We also employ the generalization Gronwall inequality techniques to prove the Ulam stability for problem (1.1), and we use important classical and fractional techniques such as: integration by parts in the settings of Hadamard fractional operators, right Hadamard fractional integral, method of variation of parameters, mean value theorem, Dirichlet formula, differentiating an integral, incomplete Gamma function and discretization methods. To the best of the authors' knowledge, there is no work in literature which treats local boundary value problems on mixed type ordinary differential equations involving the Hadamard fractional derivative using the above mentioned techniques.
The rest of the paper is organized as follows: In Section 2, we introduce some notations, definitions and lemmas that are essential in our further analysis. In Section 3, we systemically analyze problem (1.1). An equivalent integral equation is constructed for problem (1.1) and some infra structure are furnished for the use of fixed point theorems. The main results of existence and stability are discussed in Sections 4 and 5, respectively. We prove the main results via the implementation of some fixed point theorems and Ulam's approach. We study the solution's dependence on parameters in Section 6. Indeed, we give an affirmative response to the question on how the solution varies when we change the order of differential operator, the initial values or the nonlinear term f. In Section 7, some illustrative examples along with graphical representations are presented to prove consistency with our theoretical findings.
In this section we introduce notations, lemmas, definitions and preliminary facts which are used throughout this paper. In terms of the familiar Gamma function Γ(t), the incomplete Gamma function γ(α,t) and its complement Γ(α,t) are defined by (see, for details, [16,20])
γ(α,t)=∫t0τα−1e−τdτ,ℜe(t)>0, |arg(t)|<π, |
and
Γ(α,t)=∫∞tτα−1e−τdτ, |
for all complex t. For fixed α, γ(α,t) is an increasing function of t with limt→∞γ(α,t)=Γ(α). The classical Riemann-Liouville fractional integral of order α for suitable function x is defined as
Jαa[x](t):=Jαa+[x(τ)](t)=1Γ(α)∫ta(t−τ)α−1x(τ)dτ, | (2.1) |
for 0<a<t and ℜe(α)>0. The corresponding left-sided Riemann-Liouville fractional derivative of order α is defined by
Dαa[x](t)=1Γ(n−α)(ddt)n∫ta(t−τ)n−α−1x(τ)dτ, | (2.2) |
for α∈[n−1,n). However, the left and right Hadamard fractional integrals of order ℜe(α)>0, for suitable function x, introduced essentially by J. Hadamard fractional integral in [18,1892], are defined by
Jαa+[x](t)=1Γ(α)∫ta(lntτ)α−1x(τ)dττ, | (2.3) |
and
Jαb−[x](t)=1Γ(α)∫bt(lnτt)α−1x(τ)dττ, | (2.4) |
respectively. Definition (2.3) is based on the generalisation of the nth integral
Jna[x](t)=∫tadτ1τ1∫τ1adτ2τ2⋯∫τn−1ax(τn)dτnτn≡1Γ(n)∫ta(lntτ)n−1x(τ)dττ, |
where n=[ℜe(α)]+1 and [ℜe(α)] means the integer part of ℜe(α). Hadamard also proposed [18,53] a definition of the fractional integral as
Jαa[x](t)=tαΓ(α)∫10(1−s)α−1x(ts)ds. | (2.5) |
It should be emphasized that expression (2.5) contains x(ts) in place of x(s). Therefore we can consider the term s>0 as a variable that describes dilation. As a consequence, using the change of variables τ=ts, would results in the definition of the classical Riemann-Liouville fractional integral. It should be noted that in order to describe the change of dilation we can use the operator Υs (see [53,p. 330]) such that (Υsx)(t)=x(exp(ts)) where s>0. It is known that the dilation of Euclidean geometric figures changes in size while the shape is unchanged. The connection
Jαa[x]=Υ−1sJαaΥs[x],(Υsx)(t)=x(exp(ts)), | (2.6) |
allows us to extend various properties of operators Jαa to the case of operators Jαa. It is directly checked that such connections for the operators (2.5) and (2.1) are given by the relations (2.6). The corresponding left-sided Hadamard fractional derivative of order α is defined by
Dαa[x](t)=δn1Γ(n−α)∫ta(lntτ)n−α−1x(τ)dττ, | (2.7) |
where α∈[n−1,n) and δn=(tD)n is the so-called δ-derivative and D≡ddt.
Firstly, from the above definitions, we see the difference between Hadamard derivative and the Riemann–Liouville one. As a clarification, the aforementioned derivatives differ in the sense that the kernel of the integral in the definition of the Hadamard derivative contains a logarithmic function, while the Riemann-Liouville integral contains a power function. On the other hand, the Hadamard derivative is viewed as a generalization of the operator (tD)n, while the Riemann–Liouville derivative is considered as an extension of the classical Euler differential operator (D)n. Secondly, we observe that formally the relationship between Hadamard-type derivatives and Riemann-Liouville derivatives is given by the change of variable t→ln(t), leading to the logarithmic kernel.
Supposedly one can reduce the theorems and results to the corresponding ones of Hadamard-type derivatives by a simple change of variables and functions. It is possible to reduce a formula by such a change of operations but not the precise hypotheses under which a formula is valid. As an illustration, the function x(t)=sint is obviously uniformly continuous, but not ln-uniformly continuous on R+, while the function x(t)=sin(lnt) is ln-uniformly continuous but not uniformly continuous on R+. However, the two notions are equivalent on every bounded interval [a,b] with a>0. Besides, the Hadamard derivative (also integral) starts at the initial time a which is bigger than zero, but the Riemann–Liouville derivative (also integral) often begins at the origin (or any other real number). Under certain precise conditions, an equivalence could be obtained between a problem involving Hadamard derivative to another defined using a Riemann Liouville derivative.
Lemma 2.1. [28] Let ℜe(α)>0, n=[ℜe(α)]+1 and x∈C[a,+∞)∩L1[a,+∞), then the Hadamard fractional differential equation Dαa[x](t)=0, has a solution
x(t)=n∑k=1ck(lnta)α−k. |
Further, the following formulas hold
{JαaDαa[x](t)=x(t)−n∑k=1ck(lnta)α−k,DαaJαa[x](t)=x(t), | (2.8) |
where ck∈R,(k=1,2,…,n) are arbitrary constants.
Lemma 2.2. ([21]) If 0<α<1, then
Dαa[x](t)=1Γ(1−α)∫ta(lntτ)−αδ[x(τ)]dττ+x(a)Γ(1−α)(lnta)−α. |
Theorem 2.3. ([6]) Consider the continuous function x:[a,b]→R belongs to C2[a,+∞) and let ΔT=1nlnba for n≥1. Denote the time and space grid by
tN=aexp(NΔT)=an√(ba)N, | (2.9) |
and xN=x(tN) for N∈{0,1,2,⋯,n}. Then for all N∈{1,2,⋯,n},
Dαa[x](tN)=˜Dαa[x](tN)+O(ΔT), |
where
˜Dαa[x](tN)=x(a)Γ(1−α)(lntNa)−α+ζN∑k=1(ταN−k+1)x(tk)−x(tk−1)exp(kΔT).tk, |
and limΔT→0O(ΔT)=0, here (ταk)=k1−α−(k−1)1−α and
ζ=(ΔT)1−αa[1−exp(−ΔT)]Γ(2−α). |
Lemma 2.4. ([26]) If α,β>0, then the following equality holds
Jαa[τβ](t)=β−αtβΓ(α)γ(α,βlnta), |
where a>0 is the starting point in the interval. In particular, for a=0,
Jα0[τβ](t)=β−αtβ. |
The following discussion is essential for our further investigation.
Remark 2.5. If α,β>0, for t∈[1,e]. Then
i) It is easy to verify that
Jα1[τβ](t)≤β−α(βlnt)ααΓ(α)tβ=(lnt)αΓ(α+1)tβ. |
ii) The function Jα1[sinλ(t−1)] is continuous as a result of the continuity of sin function. Furthermore and according to (2.3), we have
Jα1[sinλ(τ−1)](t)≤Jα1[1](t)=1Γ(1+α)(lnt)α. |
Note that
Jα1[sinλ(τ−1)](1)=limt→1+|Jα1[sinλ(τ−1)](t)|=0. | (2.10) |
iii) From Lemma 2.2, we have
Dα1[sinλ(τ−1)](t)=J1−α1[δsinλ(τ−1)](t)+0Γ(1+α).(lnt)−α=λΓ(1−α)∫t1(lntτ)−ατcosλ(τ−1)dττ≤λJ1−α1[τ](t)≤λγ(1−α,lnt)Γ(1−α)t≤λtΓ(2−α)(lnt)1−α. |
Remark 2.6. If α>0, for t∈[1,e]. Then, using the elementary inequality (lns)α≤sα, we obtain the inequality
0≤ρα(t)=∫t1(lns)αds≤1α+1maxt∈[1,e]{tα+1−1}=eα+1α+1. | (2.11) |
Utilizing the particular case of the Fubini's theorem, one can deduce that
∫t1Jα1[x](s)ds=1Γ(α)∫t1ρα−1(ts)x(s)ds. | (2.12) |
Indeed, interchanging the order of integration with the help of (2.3) and (2.4) and it follows that
∫t1Jα1+[x](s)ds=∫t1s×Jα1[x](s)dss=1Γ(α)∫t1s×(∫s1(lnsτ)α−1x(τ)dττ)dss.=1Γ(α)∫t1x(τ)(∫tτs(lnsτ)α−1dss)dττ=∫t1x(τ)Jαt−[τ](s)ds. |
If we take v=sτ, then
∫t1x(τ)Jαt[τ](s)ds=1Γ(α)∫t1x(τ)(∫tτ1(lnv)α−1dv)dτ=1Γ(α)∫t1x(τ)ρα−1(tτ)dτ. |
Following [48], we bring a formula generalizing the well-known rule of differentiating an integral with respect to its upper limit which serves also as a parameter of the integrand
ddt∫q(t)p(t)G(t,τ)dτ=∫q(t)p(t)∂∂tG(t,τ)dτ+G(t,q(t))dq(t)dt−G(t,p(t))dp(t) dt. | (2.13) |
From (2.7), we have for α∈(0,1) and t∈(a,b) that
Dαa[∫saG(s,τ)dτ](t)=1Γ(1−α)tddt∫ta(lnts)−α[∫saG(s,τ)dτ]dss. |
Interchanging the order of integration and applying Dirichlet formula, we obtain
Dαa[∫saG(s,τ) dτ](t)=1Γ(1−α)tddt∫ta(lnts)−α[∫saG(s,τ)dτ]dss=tddt∫ta(1Γ(1−α)∫tτ(lnts)−αG(s,τ)dss)dτ=tddt∫taJ1−ατ[G(s,τ)](t)dτ=∫tat∂∂tJ1−ατ[G(s,τ)](t)dτ+tlimτ→t−aJ1−ατ[G(s,τ)](t)=∫taDατ[G(s,τ)](t)dτ+tlimτ→t−a J1−ατ[G(s,τ)](t). |
In particular, we get
Dαa[∫saG(s,τ)h(τ)dτ](t)=∫taDατ[G(s,τ)](t)h(τ)dτ+tlimτ→t−a(h(τ)J1−ατ[G(s,τ)](t)). | (2.14) |
To simplify the presentation, we let
fx(t)=f(t,x(t),Dα1[x](t)),g(t−s)=sinλ(t−s). | (2.15) |
In virtue of equation (2.14), we deduce that
Dα1(∫s1g(s−τ)Jα1[fx](τ)dτ)(t)=∫t1Dατ[g(s−τ)](t)Jα1[fx](τ)dτ+tlimτ→t−1(Jα1[fx](τ)J1−ατ[G(s,τ)](t)). |
Applying a suitable shift in the fractional operators with lower terminal τ, we deduce the next property [23,24].
Property 2.1. Let 0<α<1, Dα1[g]∈L1(1,e) and Jα1[fx]∈C(1,e). Then we have
Dα1(∫t1g(t−s)Jα1[fx](s)ds)(t)=∫t1Dα1[g(s−1)](τ)Jα1[fx](t−τ+1)dτ+tJα1[fx](t)limτ→1+(J1−α1[g(s−1)](τ)). |
In the literature, we can read the following Schaefer fixed point theorem.
Lemma 2.7. [21,55] Let E be a Banach space and assume that Ψ:E→E is a completely continuous operator. If the set
Λ={x∈E:x=μΨx:0<μ<1}, |
is bounded, then Ψ has a fixed point in E.
The next result is a generalization of Gronwall inequality due to Pachpatte ([46]).
Lemma 2.8. Let u∈C(I,R+), ˜a(t,s), ˜b(t,s)∈C(D,R+) and ˜a(t,s), ˜b(t,s) are nondecreasing in t for each s∈I, where I=[˜α,˜β], R+=[0,∞),
D={(t,s)∈I×I:˜α≤s≤t≤˜β}, |
and suppose that
u(t)≤k+∫t˜α˜a(t,s)u(s)ds+∫˜β˜α˜b(t,s)u(s)ds, |
for t∈I, where k≥0 is a constant. If
p(t)=∫˜β˜α˜b(t,s)exp(∫s˜α˜a(s,τ)dτ)ds<1, |
for t∈I, then
u(t)≤k1−p(t)exp(∫t˜α˜a(t,s)ds). |
The following hypotheses will be used in the sequel:
H1: There exist a constant Ni>0 (i=1,2) such that
|f(t,x1,˜x1)−f(t,x2,˜x2)|≤N1|x1−x2|+N2|˜x1−˜x2|, |
for each t∈[1,e] and all xi,˜xi∈R.
H2: There exists a constant L>0 such that |f(t,x,˜x)|≤L, for each t∈[1,e] and all x,˜x∈R.
In order to study the nonlinear problem (1.1), we first consider the associated linear problem and obtain its solution:
Dα1(D2+λ2)[x](t)=h(t), |
for 0<α≤1, where h is a continuous function on [1,e].
Lemma 3.1. The general solution of the linear differential equation
(D2+λ2)x(t)=˜x(t), | (3.1) |
for t∈[1,e], is given by
x(t)=1λ∫t1sinλ(t−s)˜x(s)ds+c1cosλt+c2sinλt, |
where c1,c2 are unknown arbitrary constants.
Proof. Assume that x(t) satisfies (3.1), then the method of variation of parameters implies the desired results.
Lemma 3.2. Let 0<α<1, h∈C([1,e],R). Then the unique solution of the linear problem
{Dα1[˜x](t)=h(t),˜x(1)=0, | (3.2) |
for t∈(1,e), is equivalent to the integral equation
˜x(t)=Jα1[h](t)=1Γ(α)∫t1(lntτ)α−1h(τ)dττ. | (3.3) |
Proof. Applying Lemma 2.1, we may reduce (3.2)-a to an equivalent integral equation
˜x(t)=Jα1[h](t)+c0(lnt)α−1, |
where c0∈R. In view of the boundary condition ˜x(1)=0, we have c0=0, thus (3.3) holds.
Lemma 3.3. Let h∈C([1,e],R),α∈(0,1] and 1<ξ<e. Then the fractional problem
{Dα1(D2+λ2)[x](t)=h(t),x(1)=D2[x](1)=0,x(e)=βx(ξ), | (3.4) |
has a unique solution given by
x(t)=1λ∫t1sinλ(t−s)[1Γ(α)∫s1(lnsτ)α−1h(τ)dττ]ds+βΔsinλ(t−1)∫ξ1sinλ(ξ−s)×[1Γ(α)∫s1(lnsτ)α−1h(τ)dττ]ds−1Δsinλ(t−1)∫e1sinλ(e−s)[1Γ(α)∫s1(lnsτ)α−1h(τ)dττ]ds, | (3.5) |
where
Δ=λ(sinλ(e−1)−βsinλ(ξ−1))≠0. | (3.6) |
Proof. Assuming
(D2+λ2)[x](t)=˜x(t) |
and then applying Lemma 3.1 when 0<α<1, we get
x(t)=1λ∫t1sinλ(t−s)˜x(s)ds+c1cosλt+c2sinλt. |
By the boundary condition x(1)=0 and privous equation, we conclude that
c1cosλ=−c2sinλ. | (3.7) |
On the other hand, x(e)=βx(ξ), combining with
x(e)=1λ∫e1sinλ(e−s)˜x(s)ds+c1cosλe+c2sinλe, |
and
x(ξ)=1λ∫ξ1sinλ(ξ−s)˜x(s)ds+c1cosλξ+c2sinλξ, |
yield
c2=cosλΔ(β∫ξ1sinλ(ξ−s)˜x(s)ds−∫e1sinλ(e−s)˜x(s)ds), |
where Δ is given by (3.6). If λ=(2k+1)π2, k=0,1,…, then c2=0, and by (3.7), we get
c1=2(2k+1)π[cos(2k+1)πe2−βcos(2k+1)πξ2]−1×[β∫ξ1sin(2k+1)π2(ξ−s)˜x(s)ds−∫e1sin(2k+1)π2(e−s)˜x(s)ds], |
otherwise, we find
c1=−sinλΔ(β∫ξ1sinλ(ξ−s)˜x(s)ds−∫e1sinλ(e−s)˜x(s)ds). |
The above two expressions of c1 are equivalent for the particular choice of λ. Substituting these values of c1 and c2 in (3.7) and applying Lemma 3.2, we finally obtain (3.5). So, the unique solution of problem (3.4) is given by (3.5). Conversely, let x(t) be given by formula (3.5), operating D2 on both sides and using (2.13), we get
D2x(t)=−λ∫t1sinλ(t−s)[1Γ(α)∫s1(lnsτ)α−1h(τ)dττ]ds+1Γ(α)∫t1(lntτ)α−1h(τ)dττ−βλ2Δsinλ(t−1)∫ξ1sinλ(ξ−s)[1Γ(α)∫s1(lnsτ)α−1h(τ)dττ]ds+λ2Δsinλ(t−1)∫e1sinλ(e−s)[1Γ(α)∫s1(lnsτ)α−1h(τ)dττ]ds. |
Hence
(D2+λ2)[x](t)=1Γ(α)∫t1(lntτ)α−1h(τ)dττ. |
Operating Dα1 on the above relation and using (2.8), we obtain the first equation of (3.4). Further, it is easy to get that all conditions in (3.4) are satisfied. The proof is completed.
By virtue of Lemma 3.3, we get the following result.
Lemma 3.4. Let 0<α<1, λ>0. Then the problem (1.1) is equivalent to the integral equation
x(t)=1λ∫t1g(t−s)Jα1[fx](s)ds+1Δg(t−1)[β∫ξ1g(ξ−s)Jα1[fx](s)ds−∫e1g(e−s)Jα1[fx](s)ds]. | (3.8) |
For convenience, we define the following functions
ϕx(t)=∫t1g(t−s)Jα1[fx](s)ds | (3.9) |
and
Hx(ξ,β)=1Δ(βϕx(ξ)−ϕx(e)). | (3.10) |
Then, the integral equation (3.8) can be written as
x(t)=1λϕx(t)+Hx(ξ,β)g(t−1). | (3.11) |
From the expressions of (3.5) and (3.8), we can see that if all conditions in Lemmas 3.3 and 3.4 are satisfied, then the solution is a continuous solution of the boundary value problem (1.1). Let C=C([1,e],R) be a Banach space of all continuous functions defined on [1,e] endowed with the usual supremum norm. Consider the space defined by
E={x:x∈C,Dα1[x]∈C}, |
equipped with the norm ‖x‖E=‖x‖+‖Dα1[x]‖, then (E,‖.‖E) is a Banach space. On this space, by virtue of Lemma 3.4, we may define the operator Ψ:E→E by
Ψx(t)=1λϕx(t)+Hx(ξ,β)g(t−1), |
where g(t−1), ϕx(t) and Hx(ξ,β) defined by (2.15), (3.9) and (3.10) respectively. Then
Dα1[Ψx](t)=1λDα1[ϕx](t)+Hx(ξ,β)Dα1[g(t−1)]. | (3.12) |
By virtue of Property 2.1 and Eq (2.10) in Remark 2.5, we get the following
Dα1[ϕx](t)=∫t1Dα1[g(τ−1)](s)Jα1[fx](t−s+1)ds. | (3.13) |
The continuity of the functional f would imply the continuity of Ψx and Dα1[Ψx]. Hence the operator Ψ maps the Banach space E into itself. This operator will be used to prove our main results. Next section, we employ fixed point theorems to prove the main results of this paper. In view of Lemma 3.4, we transform problem (1.1) as
x=Ψx,x∈E. | (3.14) |
Observe that problem (1.1) or (3.8) has solutions if the operator Ψ in (3.14) has fixed points. For computational convenience, we set the notations:
0≤ρ(t):=1λρα(t)+1|Δ|(βρα(ξ)+ρα(e))≤Mρ, | (3.15) |
and
0≤σα(t):=∫t1s(lns)1−α(ln(t−s+1))αds≤Mσ, | (3.16) |
where
Mρ:=1λ+1|Δ|(β+1),Mσ:=max{∫t1s2−α(t−s+1)αds:t∈[1,e]}, | (3.17) |
and
Q≥1Γ(1+α)[Mρ+1Γ(2−α)(Mσ+λβρα(ξ)+ρα(e)|Δ|e2−α)]. | (3.18) |
In this section, we establish the existence and uniqueness results via fixed point theorems.
Theorem 4.1. Assume that f:[1,e]×C×C→C is a continuous function that satisfies (H1). If we suppose
N=max{N1,N2},NQ<1, | (4.1) |
where Q is defined in (3.18), then problem (3.14) has a unique solution in E.
Proof. To prove this theorem, we need to prove that the operator Ψ has a fixed point in E. So, we shall prove that Ψ is a contraction mapping on E. For any x,˜x∈E and for each t∈[1,e], we have
|Ψ˜x(t)−Ψx(t)|≤1λ|ϕ˜x(t)−ϕx(t)|+|H˜x(ξ,β)−Hx(ξ,β)||g(t−1)|, | (4.2) |
where x(t) and ˜x(t) are defined in Lemma 3.4. From assumption (H1) and Eqs (3.9) and (4.1), we obtain
|ϕ˜x(t)−ϕx(t)|=|∫t1g(t−s)Jα1[f˜x(τ)−fx(τ)](s)ds|≤supt∈[1,e]|f˜x(t)−fx(t)||∫t1Jα1[1]ds|≤supt∈[1,e]N(|˜x(t)−x(t)|+|Dα1˜x(t)−Dα1x(t)|)Γ(1+α)∫t1(lns)αds≤ρα(t)Γ(1+α)N(M1+M2), | (4.3) |
where ρα(t) is given by (2.11) and
M1=supt∈[1,e]|˜x(t)−x(t)|,M2=supt∈[1,e]|Dα1˜x(t)−Dα1x(t)|. |
Similarly, we can obtain |ϕ˜x(ξ)−ϕx(ξ)| and |ϕ˜x(e)−ϕx(e)|. Then
|H˜x(ξ,β)−Hx(ξ,β)|≤1|Δ|[β|ϕ˜x(ξ)−ϕx(ξ)|+|ϕ˜x(e)−ϕx(e)|]≤βρα(ξ)+ρα(e)|Δ|Γ(1+α)N(M1+M2). | (4.4) |
Linking (4.2), (4.3) and (4.4), for every x, ˜x∈E, we get
|Ψ˜x(t)−Ψx(t)|≤ρ(t)Γ(1+α)N(M1+M2), |
where ρ(t) is given by (3.15). Consequently, it yields that
‖Ψ˜x−Ψx‖≤Q1N(‖˜x−x‖+‖Dα1˜x−Dα1x‖), | (4.5) |
with
Q1≥max{ρ(t)Γ(1+α):t∈[1,e]}. | (4.6) |
On the other hand, we observe that
|Dα1[Ψ˜x](t)−Dα1[Ψx](t)|≤1λ|Dα1[ϕ˜x](t)−Dα1[ϕx](t)|+|H˜x(ξ,β)−Hx(ξ,β)||Dα1[g(t−1)]|. | (4.7) |
By (3.13), we have
|Dα1[ϕ˜x](t)−Dα1[ϕx](t)|=|∫t1[Dα1[g(t−1)](s)][(Jα1[f˜x−fx])(t−s+1)]ds|≤supt∈[1,e]|f˜x(t)−fx(t)|∫t1|Dα1[g(t−1)](s)|Jα1[1](t−s+1)ds. | (4.8) |
Taking into account that
R11(t)=∫t1|Dα1[g(t−1)](s)|Jα1[1](t−s+1)ds≤∫t1λ|γ(1−α,lns)Γ(1−α)s|(ln(t−s+1))αΓ(1+α)ds≤λΓ(2−α)Γ(1+α)∫t1s(lns)1−α(ln(t−s+1))αds≤λΓ(2−α)Γ(1+α)σα(t), | (4.9) |
we have,
|Dα1[ϕ˜x](t)−Dα1[ϕx](t)|≤λσα(t)Γ(2−α)Γ(1+α)N(M1+M2), | (4.10) |
where σα(t) is given by (3.16). Therefore, from (4.7), (4.7) and (4.10), we have
|Dα1[Ψ˜x](t)−Dα1[Ψx](t)|≤N(M1+M2)Γ(1+α)(σα(t)Γ(2−α)+R12(t)), | (4.11) |
where
R12(t)=βρα(ξ)+ρα(e)|Δ|λt|γ(1−α,lnt)|Γ(1−α)≤λ(βρα(ξ)+ρα(e))|Δ|Γ(2−α)(lnt)1−αt. | (4.12) |
This gives
‖Dα1[Ψ˜x]−Dα1[Ψx]‖≤Q2N(‖˜x−x‖+‖Dα1˜x−Dα1x‖), | (4.13) |
with
Q2≥1Γ(1+α)Γ(2−α)max{σα(t)+λβρα(ξ)+ρα(e)|Δ|(lnt)1−αt:t∈[1,e]}. | (4.14) |
By (4.5) and (4.13), we can write
‖Ψ˜x−Ψx‖E≤QN‖˜x−x‖E, | (4.15) |
with Q≥Q1+Q2. Combining (4.1) with (4.15), we conclude that Ψ is contractive on E. As a consequence of Banach fixed point theorem, we deduce that Ψ has a unique fixed point which is a solution of our problem in E.
Corollary 4.2. Let the assumptions of the Theorem 4.1 be fulfilled. If we suppose that (4.1) holds, with Q is defined as
Q=1Γ(1+α)[Mρ+1Γ(2−α)(Mσ+λβ+1|Δ|e)], | (4.16) |
then, problem (3.14) has a unique solution in E.
Let Br⊂E be bounded, i.e., there exists a positive constant r>0 such that ‖x‖E<r for all x∈Br. then Br is a closed ball in the Banach space E, hence it is also a Banach space. The restriction of Ψ on Br is still a contraction by Theorem 4.1. Then, problem (3.14) has a unique solution in Br if Ψ(Br)⊂ Br.
Theorem 4.3. Assume that f:[1,e]×C×C→C is a continuous function that satisfies (H1). If we suppose that (4.1) holds, with Q is defined in (3.18), then problem (3.14) has a unique solution in Br.
Proof. Now we show that Ψ(Br)⊂Br, that is ‖Ψx‖E≤r whenever ‖x‖E≤r. Denoting
Lb=N1supt∈[1,e]|x(t)|+N2supt∈[1,e]|Dα1x(t)|+L0, |
where L0=max{|f(t,0,0|:t∈[1,e]}. Observe that
|fx(t)|=|fx(t)−f0(t)+f0(t)|≤|fx(t)−f0(t)|+|f0(t)|≤Lb. |
So, we have
|ϕx(t)|=|∫t1g(t−s)Jα1[fx−f0+f0](s)ds|≤supt∈[1,e](|fx(t)−f0(t)|+|f0(t)|)∫t1Jα1[1](s)ds≤ρα(t)Γ(1+α)Lb |
and
|Hx(ξ,β)|≤βρα(ξ)+ρα(e)Γ(1+α)|Δ|Lb. | (4.17) |
Then |Ψx(t)|≤ρ(t)Γ(1+α)Lb. Therefore,
‖Ψx‖≤Q1(N(‖x‖+‖Dα1x‖)+L0), | (4.18) |
where Q1 is given by (4.6). On the other hand, we have
|Dα1[Ψx](t)|≤|Dα1[ϕx](t)|λ+β|ϕx(ξ)|+|ϕx(e)||Δ||Dα1(g(t−1))|. | (4.19) |
Thanks to (H1), it yields that
|Dα1[ϕx](t)|≤|∫t1Dα1[g(t−1)](s)Jα1[fx−f0+f0](t−s+1)ds|≤Lb[∫t1[Dα1[g(t−1)](s)]Jα1[1](t−s+1)ds]. |
This gives
|Dα1[ϕx](t)|≤λσα(t)Γ(1+α)Γ(2−α)Lb, | (4.20) |
where σα(t) is given by (3.16). Consequently, by (4.17), (4.19) and (4.20), we have
‖Dα1[Ψx]‖≤Q2(N(‖x‖+‖Dα1x‖)+L0), | (4.21) |
where Q2 is given by (4.14). Using (4.18) and (4.21), we obtain
‖Ψx‖E≤Q(N(‖x‖+‖Dα1x‖)+L0), |
and we find that ‖Ψx‖E≤Q(Nr+L0)≤r, where we choose r≥L0(Q−1−N)−1. Hence, the operator Ψ maps bounded sets into bounded sets in Br, therefore Ψ is a contraction. Thus, the conclusion of the theorem follows by the contraction mapping principle.
Corollary 4.4. Assume that f:[1,e]×C×C→C is a continuous function that satisfies (H1). If we suppose N=max{N1,N2} and
L0(Q−1−N)−1<r, |
where Q is defined in (4.16). Then, problem (3.14) has a unique solution in Br.
Our second result will use the Scheafer fixed point theorem.
Theorem 4.5. The problem (3.14) has at least one solution defined on E, whenever assumption (H2) be hold.
Proof. The proof will be given in several steps.
Step 1: We show that Ψ is continuous. Let us consider a sequence {xn}∈E converging to x. For each t∈[1,e], we have
|Ψxn(t)−Ψx(t)|≤1λ|ϕxn(t)−ϕx(t)|+|Hxn(ξ,β)−Hx(ξ,β)||g(t−1)|, |
where
|ϕxn(t)−ϕx(t)|=|∫t1g(t−s)Jα1[fxn−fx](s)ds|≤ρα(t)Γ(1+α)|fxn(t)−fx(t)|. | (4.22) |
Similarly, we can obtain
|Hxn(ξ,β)−Hx(ξ,β)|≤βρα(ξ)+ρα(e)|Δ|Γ(1+α)|fxn(t)−fx(t)|. | (4.23) |
Thus, from (4.19), (4.22) and (4.23), we have
|Ψxn(t)−Ψx(t)|≤ρ(t)Γ(1+α)|fxn(t)−fx(t)|. | (4.24) |
If (t,x)∈[1,e]×E, xn→x as n→∞ and f is continuous, then (4.24) gives
‖Ψxn−Ψx‖→0, | (4.25) |
as n→∞. On the other hand, from (4.11) we observe that
|Dα1[Ψxn](t)−Dα1[Ψx](t)|≤1Γ(1+α)(σα(t)Γ(2−α)+R12)|fxn(τ)−fx(τ)|, |
where R12 is given by (4.12). Thus
‖Dα1[Ψxn]−Dα1[Ψx]‖→0, | (4.26) |
as n→∞. Since the convergence of a sequence implies its boundedness, therefore, there exists r>0 such that ‖xn‖≤r,‖x‖≤r and hence f is uniformly continuous on the compact set
{(t,x(t),Dα1[x](t)):t∈[1,e],‖x‖≤r1,‖Dα1[x]‖≤r2}. |
By (4.25) and (4.26), we can write ‖[Ψxn]−[Ψx]‖E→0 as n→∞. This shows that Ψ is continuous.
Step 2: Now we show that the operator Ψ:E→E maps bounded sets into bounded sets in E. Let Br⊂E be bounded, i.e., there exists a positive constant r>0 such that ‖x‖E≤r for all x∈Br. Let
L=max{|f(t,x(t),Dα1[x](t))|:t∈[1,e], 0<‖x‖≤r,‖Dα1[x]‖≤r}, |
then, for x∈Br, we have
|ϕx(t)|=|∫t1g(t−s)Jα1[fx](s)ds|≤ρα(t)Γ(1+α)L | (4.27) |
and
|Hx(ξ,β)|≤βρα(ξ)+ρα(e)|Δ|Γ(1+α)L. | (4.28) |
Then from (4.27) and (4.28), we get |Ψx(t)|≤ρ(t)Γ(1+α)L. Therefore,
‖Ψx‖≤Q1L. | (4.29) |
According to Property 2.1, we should have
|Dα1[ϕx](t)|=|∫t1Dα1[g(t−1)](s)Jα1[fx](t−s+1)ds|≤λσα(t)Γ(1+α)Γ(2−α)L. | (4.30) |
Consequently, by (3.12), (4.28) and (4.30), we have
‖Dα1Ψx‖≤Q2L. | (4.31) |
Using (4.29) and (4.31), we obtain ‖Ψx‖E≤QL. Hence, the operator Ψ maps bounded sets into bounded sets in E. Next we show that Ψ maps bounded sets into equicontinuous sets of Br.
Step 3: In this step, we show that Ψ(Br) is equicontinuity. Let t1,t2∈[1,e] such that t1<t2. Then we obtain
|Ψx(t2)−Ψx(t1)|≤1λ|ϕx(t2)−ϕx(t1)|+|Hx(ξ,β)||g(t2−1)−g(t1−1)|. | (4.32) |
We can show that
|ϕx(t2)−ϕx(t1)|=|∫t21g(t2−s)Jα1[fx](s)ds−∫t11g(t1−s)Jα1[fx](s)ds|≤∫t11|g(t2−s)−g(t1−s)||Jα1[fx](s)|ds+∫t2t1|g(t2−s)||Jα1[fx](s)|ds≤L∫t11|g(t2−s)−g(t1−s)|Jα1[1](s)ds+L∫t2t1|g(t2−s)|Jα1[1](s)ds≤L∫t11|λ∫t2t1cosλ(τ−s)dτ|Jα1[1](s)ds+L∫t2t1Jα1[1](s)ds≤LΓ(1+α)[λ|t2−t1|∫t11(lns)αds+∫t2t1(lns)αds]. | (4.33) |
Hence
|ϕx(t2)−ϕx(t1)|≤LΓ(2+α)[λ|t2−t1||tα+11−1|+|tα+12−tα+11|]. | (4.34) |
It is easy to find that
|g(t2−1)−g(t1−1)|=|λ∫t2t1cosλ(τ−1)dτ|≤λ|t2−t1|. |
Therefore by (4.28), (4.32), (4.33) and (4.34) we have
|Ψx(t2)−Ψx(t1)|≤1λ|ϕx(t2)−ϕx(t1)|+|Hx(ξ,β)||g(t2−1)−g(t1−1)|≤LΓ(2+α)[|t2−t1||tα+11−1|+1λ|tα+12−tα+11|]+λLβρα(ξ)+ρα(e)|Δ|Γ(1+α)|t2−t1|. | (4.35) |
We have also,
|Dα1[Ψx](t2)−Dα1[Ψx](t1)|≤1λ|Dα1[ϕx](t2)−Dα1[ϕx](t1)|+|Hx(ξ,β)||Dα1[g(t2−1)]−Dα1[g(t1−1)]|. | (4.36) |
Thus, we obtain
|Dα1[ϕx](t2)−Dα1[ϕx](t1)|≤|∫t21Dα1[g(τ−1)](s)Jα1[fx](t2−s+1)ds−∫t11Dα1[g(τ−1)](s)Jα1[fx](t1−s+1)ds|≤∫t11|Dα1[g(τ−1)](s)||Jα1[fx](t2−s+1)−Jα1[fx](t1−s+1)|ds+∫t2t1|Dα1[g(τ−1)](s)|Jα1[fx](t2−s+1)ds. |
We find that
|Jα1[fx](t2−s+1)−Jα1[fx](t1−s+1)|≤L|Jα1[1](t2−s+1)−Jα1[1](t1−s+1)|=LΓ(α)|∫t2−s+11(lnt2−s+1τ)α−1dττ−∫t1−s+11(lnt1−s+1τ)α−1dττ|=LΓ(α)∫t1−s+11|(lnt2−s+1τ)α−1−(lnt1−s+1τ)α−1|dττ+LΓ(α)∫t2−s+1t1−s+1(lnt2−s+1τ)α−1dττ≤LΓ(1+α)[2(lnt2−s+1t1−s+1)α+(lnt2−s+1)α−(lnt1−s+1)α]. | (4.37) |
Note that
|Jα1[fx](t2−s+1)−Jα1[fx](t1−s+1)|, |
is independent of x. Therefore
|Dα1[ϕx](t2)−Dα1[ϕx](t1)|≤LΓ(1+α)∫t11[2(lnt2−s+1t1−s+1)α+(ln(t2−s+1))α−(ln(t1−s+1))α]ds+λL(σα(t2)−σα(t1))Γ(2−α)Γ(1+α). | (4.38) |
In accordance with (4.35), (4.36), (4.37) and (4.38), we deduce that
‖Ψx(t2)−Ψx(t1)‖+‖Dα1[Ψx](t2)−Dα1[Ψx](t1)‖→0, |
as |t2−t1|→0. Hence the sets of functions {Ψx(t):x∈Br} and
{Dα1[Ψx](t):x∈Br}, |
are bounded in Br and equicontinuous on [1,e]. Thus, by the Arzelá–Ascoli Theorem, the mapping Ψ is completely continuous on E.
Step 4: In the last step, it remains to show that the set defined by
Λ={x∈E:x=μΨx for some 0<μ<1}, |
is bounded. Let x be a solution. Then, for t∈[1,e] and using the computations in proving that Ψ is bounded, we have |x(t)|=|μ(Ψx)(t)|. Let x∈Λ, x=μΨx for some 0<μ<1. Thus, by (4.29), for each t∈[1,e], we have
‖x‖=μ‖Ψx‖≤‖Ψx‖≤Q1L, | (4.39) |
for μ∈(0,1). On the other hand, by (4.31), we have
‖Dα1[x]‖=μ‖Dα1[Ψx]‖≤Q2L. | (4.40) |
It follows from (4.39) and (4.40) that ‖x‖E=‖x‖+‖Dα1[x]‖≤QL<∞, where Q defined by (3.18). This implies that the set Λ is bounded independently of μ∈(0,1). Therefore, Λ is bounded. As a conclusion of Schaefer fixed point theorem, we deduce that Ψ has at least one fixed point, which is a solution of (3.14). The proof is completed.
Remark 4.6. Let ε>0 and choose a number η>0 such that
ε>2LλΓ(2+α)min{κ1(η),κ2(η)}+2λLβρα(ξ)+ρα(e)|Δ|Γ(1+α)η, | (4.41) |
where
κ1(η)=λη(eα+1−1)+e(α+1)ηα,κ2(η)=λη(ηα+1−1)+((2η)α+1−1). |
Assume t1,t2∈[1,e]; t1<t2 such that t2−t1≤η. It obvious that t2>η and there are two possibilities for t1 and η.
Case 1: For η≤t1<t2≤e, by means of mean value theorem of differentiation implies that there exists t∈(t1,t2), such that
tα+12−tα+11=(α+1)(t2−t1)tα≤η(α+1)ttα−1≤e(α+1)ηα, |
whence, we obtain
|ϕx(t2)−ϕx(t1)|≤LΓ(2+α)[λη(eα+1−1)+e(α+1)ηα]=LΓ(2+α)κ1(η). |
Case 2: For 1≤t1<η<t2≤e and so t2<2η. These imply that
|ϕx(t2)−ϕx(t1)|≤LΓ(2+α)[λη(ηα+1−1)+((2η)α+1−1)]=LΓ(2+α)κ2(η). |
Combining Case 1 and 2, we obtain
|ϕx(t2)−ϕx(t1)|≤LΓ(2+α)min{κ1(η),κ2(η)}. |
Now, it is obvious by (4.35) and (4.41) that |Ψx(t2)−Ψx(t1)|≤ε2. A similar argument can be applied to obtain
|Dα1[Ψx](t2)−Dα1[Ψx](t1)|≤ε2. |
Corollary 4.7. Suppose that the conditions of Theorem 4.5 hold, with Q is defined as (4.16). Then, problem (3.14) has at least one solution defined on [1,e].
For the study of Hyers-Ulam-Rassias and generalized Ulam-Hyers-Rassias stabilities of problem (1.1) on a compact interval [1,e], we adopt the following definitions [22,35,49,58,59].
Definition 5.1. Problem (1.1) is Ulam-Hyers-Rassias stable with respect to φ∈C([1,e],R+) if for each ϵ>0 and each solution ˜x of the inequality
|Dα1(D2+λ2)˜x(t)−f(t,˜x(t),Dα1[˜x](t))|≤ϵφ(t), | (5.1) |
for t∈[1,e], there exists a real number cφ>0 and a solution x of problem (1.1) such that
|˜x(t)−x(t)|≤ϵcφφ(t), |
for t∈[1,e]. Particularly, in the case that φ is identity function on [1,e], problem (1.1) is called Ulam-Hyers stable. Moreover, if there exists ψ∈C(R+,R+),ψ(0)=0, such that |˜x(t)−x(t)|≤ψ(ϵ), for t∈[1,e], then problem (1.1) is called generalized Ulam-Hyers stable.
Definition 5.2. Problem (1.1) is generalized Ulam-Hyers-Rassias stable with respect to a function φ∈C([1,e],R+) if for each solution ˜x of the inequality
|Dα1(D2+λ2)˜x(t)−f(t,˜x(t),Dα1[˜x](t))|≤φ(t), | (5.2) |
for t∈[1,e], there exist a real number cφ>0 and a solution x of problem (1.1) such that |˜x(t)−x(t)|≤cφφ(t), for t∈[1,e].
Remark 5.3. A function ˜x∈C is a solution of the inequality (5.1) if and only if there exists a function h∈C (which depends on ˜x) such that for all t∈[1,e],
(i) |h(t)|≤ϵφ(t).
(ii) Dα1(D2+λ2)˜x(t)=f(t,˜x(t),Dα1[˜x](t))+h(t).
Lemma 5.4. Let 0<α<1, if ˜x∈C is a solution of the inequality (5.1)(or (5.2)) then ˜x is a solution of the following integral inequality
|˜x(t)−1λϕ˜x(t)−H˜x(ξ,β)g(t−1)|≤ϵω(t), | (5.3) |
for t∈[1,e], where
˜x(1)=0=D2˜x(1),˜x(e)=β˜x(ξ), | (5.4) |
for ξ∈(1,e] and
ω(t)=1λ∫t1Jα1[φ](s)ds+β|Δ|∫ξ1Jα1[φ](s)ds+1|Δ|∫e1Jα1[φ](s)ds. |
Proof. By using Remark 5.3-ii, we have
Dα1(D2+λ2)˜x(t)=f(t,˜x(t),Dα1[˜x](t))+h(t). |
In accordance with Lemma 3.4, we deduce ˜x(t)=1λϕ˜x,h(t)+H˜x,h(ξ,β)g(t−1), where
ϕ˜x,h(t)=∫t1g(t−s)Jα1[f˜x+h](s)ds |
and H˜x,h(ξ,β)=1Δ(βϕ˜x,h(ξ)−ϕ˜x,h(e)). Hence
˜x(t)=1λϕ˜x(t)+H˜x(ξ,β)g(t−1)+1λ∫t1g(t−s)Jα1[h](s)ds+g(t−1)βΔ∫ξ1g(ξ−s)Jα1[h](s)ds−1Δg(t−1)∫e1g(e−s)Jα1[h](s)ds. |
Accordingly, we easily deduce equation (5.3).
By virtue of Remark 5.3-i, it can be easily seen that
|ϕ˜x,h(t)−ϕx(t)|≤|ϕ˜x(t)−ϕx(t)|+|∫t1g(t−s)Jα1[h](s)ds|≤|ϕ˜x(t)−ϕx(t)|+ϵ∫t1Jα1[φ](s)ds. |
Similar arguments can be applied as in (4.3) to deduce that
|ϕ˜x(t)−ϕx(t)|≤Nρα(t)Γ(1+α)‖˜x−x‖E. |
Now, it is obvious that
|ϕ˜x,h(t)−ϕx(t)|≤Nρα(t)Γ(1+α)‖˜x−x‖E+ϵ∫t1Jα1[φ](s)ds, | (5.5) |
and
|H˜x,h(ξ,β)−Hx(ξ,β)|≤1|Δ|(β|ϕ˜x,h(ξ)−ϕx(ξ)|+|ϕx(e)−ϕ˜x,h(e)|)≤N|Δ|Γ(1+α)(βρα(ξ)+ρα(e))‖˜x−x‖E+ϵ|Δ|(β∫ξ1Jα1[φ](s)ds+∫e1Jα1[φ](s)ds). | (5.6) |
Theorem 5.5. Assume that the conditions of Theorem 4.1 and (5.1), (5.4) hold. Then, problem (1.1) is Ulam-Hyers-Rassias stable with respect to positive constant functions. Particularly, problem (1.1) is Ulam-Hyers stable and generalized Ulam-Hyers stable.
Proof. Using Theorem 4.1, there exists a unique solution x∈C of problem (1.1) that is given by integral equation (3.11). Let ˜x∈C be any solution of the inequality (5.1), then by (5.5) and (5.6), we have
|˜x(t)−x(t)|=|1λϕ˜x,h(t)+H˜x,h(ξ,β)g(t−1)−1λϕx(t)−Hx(ξ,β)g(t−1)|≤1λ|ϕ˜x,h(t)−ϕx(t)|+|H˜x,h(ξ,β)−Hx(ξ,β)||g(t−1)|≤Nρα(t)Γ(1+α)‖˜x−x‖E+ϵ[1λ∫t1Jα1[φ](s)ds+β|Δ|∫ξ1Jα1[φ](s)ds+1|Δ|∫e1Jα1[φ](s)ds], |
where x(t) and ρ(t) are as in (3.8) and (3.15). Thus
‖˜x−x‖E≤ϵΓ(1+α)Γ(1+α)−NMρ[1λ∫e1Jα1[φ](s)ds+β|Δ|∫ξ1Jα1[φ](s)ds+1|Δ|∫e1Jα1[φ](s)ds], |
since by (3.18), we get 0<NMρΓ(1+α)<1. Then, for each t∈[1,e],
|˜x(t)−x(t)|≤ϵ[1λ∫t1Jα1[φ](s)ds+β|Δ|(1+Nρα(t)Γ(1+α)−NMρ)∫ξ1Jα1[φ](s)ds+(1|Δ|+1λNρα(t)Γ(1+α)−NMρ+1|Δ|Nρα(t)Γ(1+α)−NMρ)∫e1Jα1[φ](s)ds]. |
Accordingly, to satisfy the inequality |˜x(t)−x(t)|≤ϵcφφ(t), we have to pose that φ is a constant function on [1,e]. Hence, if φ(t)=c>0, t∈[1,e], then any finite positive constant
cφ≥cρα(e)λΓ(1+α)+cβρα(ξ)|Δ|Γ(1+α)[1+Nρα(e)Γ(1+α)−NMρ]+cρα(e)Γ(1+α)[1|Δ|+1λNρα(t)Γ(1+α)−NMρ+1|Δ|Nρα(e)Γ(1+α)−NMρ], |
will satisfy the problem. Thus, the fractional boundary value problem (1.1) is Ulam-Hyers-Rassias with respect to a constant function. The Ulam-Hyers stability can be obtained by putting φ=1, and hence generalized Ulam-Hyers stable with ψ as identity function.
In the next result, we prove the (generalized) Ulam-Hyers-Rassias stability in terms of a function.
Theorem 5.6. Assume that the conditions of Theorem 4.1 and (5.1) hold. Then, problem (1.1) is Ulam-Hyers-Rassias stable with respect to φ provided that
φ(t)≥β|Δ|Γ(α)∫ξ1ρα−1(ξs)φ(s)ds+1|Δ|Γ(α)∫e1ρα−1(es)φ(s)ds+1λΓ(α)∫t1ρα−1(ts)φ(s)ds, | (5.7) |
and supt∈[1,e]p(t)<1, where
p(t)=∫e1(βN1|Δ|Γ(α)ρα−1(ξs)+βN2|Δ|+N1|Δ|Γ(α)ρα−1(es)+N2|Δ|)×exp(N1λΓ(α)∫sαρα−1(sτ)dτ+N2λ(s−1))ds. |
Proof. Let us denote by x∈C the unique solution of the problem (1.1). Let ˜x∈C be a solution of the inequality (5.1), with
˜x(1)=x(1), ˜x(e)=x(e). | (5.8) |
By modifying the estimate (5.5), we have
|ϕ˜x,h(t)−ϕx(t)|≤N1∫t1Jα1[|˜x−x|](s)ds+N2∫t1Jα1Dα1[|˜x−x|](s)ds+∫t1Jα1[h](s)ds. | (5.9) |
The above inequality implies
|H˜x,h(ξ,β)−Hx(ξ,β)|≤1|Δ|[β|ϕ˜x,h(ξ)−ϕx(ξ)|+|ϕ˜x,h(e)−ϕx(e)|]≤βN1|Δ|∫ξ1Jα1[|˜x−x|](s)ds+βN2|Δ|∫ξ1|˜x(s)−x(s)|ds+N1|Δ|∫e1Jα1[|˜x−x|](s)ds+N2|Δ|∫e1|˜x(s)−x(s)|ds+β|Δ|∫ξ1Jα1[h](s)ds+1|Δ|∫e1Jα1[h](s)ds. | (5.10) |
Taking into account (2.12), (5.9) and (5.10), lead to
|˜x(t)−x(t)|≤1λ|ϕ˜x,h(t)−ϕx(t)|+|H˜x,h(ξ,β)−Hx(ξ,β)||g(t−1)|≤N1λ∫t1Jα1[|˜x−x|](s)ds+βN1|Δ|∫ξ1Jα1[|˜x−x|](s)ds+N1|Δ|∫e1Jα1[|˜x−x|](s)ds+N2λ∫t1|˜x(s)−x(s)|ds+βN2|Δ|∫ξ1|˜x(s)−x(s)|ds+N2|Δ|∫e1|˜x(s)−x(s)|ds+1λ∫t1Jα1[h](s)ds+β|Δ|∫ξ1Jα1[h](s)ds+1|Δ|∫e1Jα1[h](s)ds≤∫t1(N1λΓ(α)ρα−1(ts)+N2λ)|˜x(s)−x(s)|ds+ξ∫1(βN1|Δ|Γ(α)ρα−1(ξs)+βN2|Δ|)|˜x(s)−x(s)|ds+e∫1(N1|Δ|Γ(α)ρα−1(es)+N2|Δ|)|˜x(s)−x(s)|ds+1Γ(α)∫e1(β|Δ|ρα−1(ξs)+1|Δ|ρα−1(es)+1λρα−1(ts))h(s)ds. |
Hence
|˜x(t)−x(t)|≤k+∫t1˜a(t,s)|˜x(s)−x(s)|ds+e∫1˜b(t,s)|˜x(s)−x(s)|ds, |
where
k=1Γ(α)∫e1(β|Δ|ρα−1(ξs)+1|Δ|ρα−1(es)+1λρα−1(es))h(s)ds, |
and
˜a(t,s)=(N1λΓ(α)ρα−1(ts)+N2λ),˜b(t,s)=βN1|Δ|Γ(α)ρα−1(ξs)+βN2|Δ|+N1|Δ|Γ(α)ρα−1(es)+N2|Δ|. |
In virtue of Lemma 2.8, we deduce that
|˜x(t)−x(t)|≤k1−p(t)exp(∫t1a(t,s)ds), |
for t∈[1,e]. Problem (1.1) is Ulam-Hyers-Rassias stable with respect to φ≥1ϵ|h|, φ must satisfy the inequality (5.7). In this case, we get |˜x(t)−x(t)|≤cφϵφ(t), where
cφ=maxt∈[1,e][11−p(t)exp(∫t1(N1λΓ(α)ρα−1(ts)+N2λ)ds)]. |
This completes the proof.
Theorem 5.7. Assume the conditions of Theorem 4.1 and (5.2) hold. Then, problem (1.1) is generalized Ulam-Hyers-Rassias stable with respect to φ provided for any t∈[1,e] that
φ(t)≥1λ∫t1Jα1[φ](s)ds+β|Δ|∫ξ1Jα1[φ](s)ds+1|Δ|∫e1Jα1[φ](s)ds+1Γ(1+α)−NMρ[N‖φ‖ρ(t)(ρα(e)λ+βρα(ξ)|Δ|+ρα(e)|Δ|)]+1(Γ(1+α)−NMρ)[Γ(2−α)(Γ(1+α)−NMρ)−N(Mσ+λ(βρα(ξ)+ρα(e))|Δ|e2−α)]−1×N2‖φ‖(ρα(e)λ+βρα(ξ)|Δ|+ρα(e)|Δ|)(Mσ+λ(βρα(ξ)+ρα(e))|Δ|e2−α)ρ(t)+Nρ(t)[Γ(2−α)(Γ(1+α)−NMρ)−N(Mσ+λ(βρα(ξ)+ρα(e))|Δ|e2−α)]−1×[∫t1s(lns)1−αJα1[φ](t−s+1)ds+λβe2−α|Δ|∫ξ1Jα1[φ](s)ds+λe2−α|Δ|∫e1Jα1[φ](s)ds]. | (5.11) |
Proof. Let us denote by x∈C([1,e],R) the unique solution of the problem (1.1). Let ˉx∈C be a solution of the inequality (5.2), with (5.8). It follows
|ϕˉx(t)−ϕx(t)|=|∫t1g(t−s)Jα1[fˉx−fx](s)ds|≤N1∫t1Jα1[|ˉx−x|](s)ds+N2∫t1Jα1[|Dα1[ˉx]−Dα1[x]|](s)ds≤N1∫t1Jα1[|ˉx−x|](s)ds+N2ρα(t)Γ(1+α)‖Dα1[ˉx]−Dα1[x]‖. |
On the other hand, we have, for each t∈[1,e],
|H˜x(ξ,β)−Hx(ξ,β)|≤1|Δ|[β|ϕ˜x(ξ)−ϕx(ξ)|+|ϕ˜x(e)−ϕx(e)|]≤N2(ρα(e)+βρα(ξ))|Δ|Γ(1+α)‖Dα1[ˉx]−Dα1[x]‖+βN1|Δ|∫ξ1Jα1[|ˉx−x|](s)ds+N1|Δ|∫e1Jα1[|ˉx−x|](s)ds. |
Hence by Lemma 5.4, for each t∈[1,e], we get
|ˉx(t)−x(t)|≤|ˉx(t)−1λϕˉx(t)−Hˉx(ξ,β)g(t−1)|+1λ|ϕˉx(t)−ϕx(t)|+|Hˉx(ξ,β)−Hx(ξ,β)||g(t−1)|≤ω(t)+N1λ∫t1Jα1[|ˉx−x|](s)ds+βN1|Δ|∫ξ1Jα1[|ˉx−x|](s)ds+N1|Δ|∫e1Jα1[|ˉx−x|](s)ds+N2ρα(t)λΓ(1+α)‖Dα1[ˉx]−Dα1[x]‖+N2(ρα(e)+βρα(ξ))|Δ|Γ(1+α)‖Dα1[ˉx]−Dα1[x]‖≤ω(t)+N1ρ(t)Γ(1+α)‖ˉx−x‖+N2ρ(t)Γ(1+α)‖Dα1[ˉx]−Dα1[x]‖≤Ω+N1MρΓ(1+α)‖ˉx−x‖+N2MρΓ(1+α)‖Dα1[ˉx]−Dα1[x]‖, |
where
Ω=‖φ‖ρα(e)λ+β‖φ‖ρα(ξ)|Δ|+‖φ‖ρα(e)|Δ|. |
Then (see (3.17))
‖ˉx−x‖≤ΩΓ(1+α)Γ(1+α)−NMρ+NMρΓ(1+α)−NMρ‖Dα1[ˉx]−Dα1[x]‖. |
Accordingly, we get
|ˉx(t)−x(t)|≤ω(t)+ΩNρ(t)Γ(1+α)−NMρ+Nρ(t)Γ(1+α)−NMρ‖Dα1[ˉx]−Dα1[x]‖. |
We are going now to get an estimate for ‖Dα1[ˉx]−Dα1[x]‖. It is obvious that
|Dα1[ˉx](t)−Dα1[x](t)|≤1λ|Dα1[ϕˉx,h](t)−Dα1[ϕx](t)|+|Hˉx,h(ξ,β)−Hx(ξ,β)||Dα1[g(t−1)]|, |
where
|Dα1[ϕˉx,h](t)−Dα1[ϕx](t)|=|∫t1Dα1[g(t−1)](s)Jα1[fˉx−fx+h](t−s+1)ds|≤N1∫t1|Dα1[g(t−1)](s)|Jα1[|ˉx−x|](t−s+1)ds+N2‖Dα1[ˉx]−Dα1[x]‖∫t1|Dα1[g(t−1)](s)|Jα1[1](t−s+1)ds+∫t1|Dα1[g(t−1)](s)|Jα1[|h|](t−s+1)ds≤λNΩσα(t)Γ(2−α)(Γ(1+α)−NMρ)+λNσα(t)Γ(2−α)(Γ(1+α)−NMρ)‖Dα1[ˉx]−Dα1[x]‖+λΓ(2−α)∫t1s(lns)1−αJα1[|h|](t−s+1)ds. |
Also, we get
|H˜x,h(ξ,β)−Hx(ξ,β)|≤β|Δ|(|ϕ˜x(ξ)−ϕx(ξ)|+∫ξ1Jα1[|h|](s)ds)+1|Δ|(|ϕ˜x(e)−ϕx(e)|+∫e1Jα1[|h|](s)ds)≤ΩN(βρα(ξ)+ρα(e))|Δ|(Γ(1+α)−NMρ)+N(βρα(ξ)+ρα(e))|Δ|(Γ(1+α)−NMρ)‖Dα1[ˉx]−Dα1[x]‖+β|Δ|∫ξ1Jα1[|h|](s)ds+1|Δ|∫e1Jα1[|h|](s)ds. |
Hence, we deduce that
|Dα1[ˉx](t)−Dα1[x](t)|≤1λ|Dα1[ϕˉx,h](t)−Dα1[ϕx](t)|+|Hˉx,h(ξ,β)−Hx(ξ,β)||Dα1[g(t−1)]|≤NΩ(σα(t)+λt(lnt)1−α(βρα(ξ)+ρα(e))|Δ|)Γ(2−α)(Γ(1+α)−NMρ)+N(σα(t)+λt(lnt)1−α(βρα(ξ)+ρα(e))|Δ|)Γ(2−α)(Γ(1+α)−NMρ)‖Dα1[ˉx]−Dα1[x]‖+1Γ(2−α)∫t1s(lns)1−αJα1[|h|](t−s+1)ds+λtΓ(2−α)(lnt)1−αβ|Δ|∫ξ1Jα1[|h|](s)ds+λtΓ(2−α)(lnt)1−α1|Δ|∫e1Jα1[|h|](s)ds≤NΩ(Mσ+λ(βρα(ξ)+ρα(e))|Δ|e2−α)Γ(2−α)(Γ(1+α)−NMρ)+N(Mσ+λ(βρα(ξ)+ρα(e))|Δ|e2−α)Γ(2−α)(Γ(1+α)−NMρ)‖Dα1[ˉx]−Dα1[x]‖+1Γ(2−α)∫t1s(lns)1−αJα1[|h|](t−s+1)ds+λβt|Δ|Γ(2−α)(lnt)1−α∫ξ1Jα1[|h|](s)ds+λt|Δ|Γ(2−α)(lnt)1−α∫e1Jα1[|h|](s)ds. |
Then
‖Dα1[ˉx]−Dα1[x]‖≤NΩ(Mσ+λ(βρα(ξ)+ρα(e))|Δ|e2−α)Γ(2−α)(Γ(1+α)−NMρ)−N(Mσ+λ(βρα(ξ)+ρα(e))|Δ|e2−α)+(Γ(1+α)−NMρ)Γ(2−α)(Γ(1+α)−NMρ)−N(Mσ+λ(βρα(ξ)+ρα(e))|Δ|e2−α)×(∫e1s(lns)1−αJα1[|h|](e−s+1)ds+λβe2−α|Δ|∫ξ1Jα1[|h|](s)ds+λe2−α|Δ|∫e1Jα1[|h|](s)ds). |
If |h|≤φ, we get
|ˉx(t)−x(t)|≤1λ∫t1Jα1[φ](s)ds+β|Δ|∫ξ1Jα1[φ](s)ds+1|Δ|∫e1Jα1[φ](s)ds+ΩNρ(t)Γ(1+α)−NMρ+Nρ(t)Γ(1+α)−NMρ×[NΩ(Mσ+λ(βρα(ξ)+ρα(e))|Δ|e2−α)Γ(2−α)(Γ(1+α)−NMρ)−N(Mσ+λ(βρα(ξ)+ρα(e))|Δ|e2−α)+(Γ(1+α)−NMρ)Γ(2−α)(Γ(1+α)−NMρ)−N(Mσ+λ(βρα(ξ)+ρα(e))|Δ|e2−α)]×[∫t1s(lns)1−αJα1[φ](t−s+1)ds+λβe2−α|Δ|∫ξ1Jα1[φ](s)ds+λe2−α|Δ|∫e1Jα1[φ](s)ds]. |
Hence by the given condition (5.11), the equation (1.1) is generalized Ulam-Hyers-Rassias stable with respect to φ.
For f Lipschitz in the second and the third variables, the solution's dependence on the order of the differential operator, the boundary values and the nonlinear term f are discussed in this section. We show that the solutions of two equations with neighbouring orders will (under suitable conditions on their right hand sides f) lie close to one another.
Theorem 6.1. Suppose that the conditions of Theorem 4.1 hold. Let x(t), xϵ(t) be the solutions, respectively, of problems (1.1) and
Dα−ϵ1(D2+λ2)x(t)=f(t,x(t),Dα1[x](t)), | (6.1) |
for t∈(0,1) and ϵ>0, with the boundary conditions (1.1)-b, where 0<α−ϵ<α<1. Then there exists a constant kϵ>0 such that
‖x−xϵ‖E≤kϵ‖f‖∗, | (6.2) |
where ‖f‖∗=supϵ‖fxϵ‖ and fxϵ(t):=f(t,xϵ(t),Dα1[xϵ](t)).
Proof. By Lemma 3.4 and equation (3.11), we can obtain
xϵ(t)=1λϕxϵ(t)+Hxϵ(ξ,β)g(t−1), |
is the solution of (6.1) with the boundary conditions in (1.1), where
ϕxϵ(t)=∫t1g(t−s)Jα−ϵ1[fxϵ](s)ds |
and Hxϵ(ξ,β)=1Δ(βϕxϵ(ξ)+ϕxϵ(e)). Then
|ϕxϵ(t)−ϕx(t)|=|∫t1g(t−s)Jα−ϵ1[fxϵ](s)ds−∫t1g(t−s)Jα1[fx](s)ds|≤|∫t1g(t−s)Jα1[fxϵ−fx](s)ds|+|∫t1g(t−s)[Jα−ϵ1[fxϵ](s)−Jα1[fxϵ](s)]ds|≤‖fxϵ−fx‖∫t1|g(t−s)|Jα1[1](s)ds+‖fxϵ‖∫t1|g(t−s)||Jα−ϵ1[1](s)−Jα1[1](s)|ds≤N‖x−xϵ‖EΓ(1+α)∫t1(lns)αds+‖fxϵ‖∫t1|(lns)α−ϵΓ(1+α−ϵ)−(lns)αΓ(1+α)|ds. |
This leads to
|ϕxϵ(t)−ϕx(t)|≤NΓ(1+α)ρα(t)‖x−xϵ‖E+ϱϵ(t)‖fxϵ‖, |
with
ϱϵ(t)=∫t1|(lns)α−ϵΓ(1+α−ϵ)−(lns)αΓ(1+α)|ds. |
In a similar manner, we can get
|Hxϵ(ξ,β)−Hx(ξ,β)|≤NΓ(1+α)1|Δ|[βρα(ξ)+ρα(e)]‖x−xϵ‖E+1|Δ|[βϱϵ(ξ)+ϱϵ(e)]‖fxϵ‖. |
Then
|x(t)−xϵ(t)|≤NΓ(1+α)ρ(t)‖x−xϵ‖E+ϱ(t)‖fxϵ‖, | (6.3) |
with ϱ(t)=1λϱϵ(t)+1|Δ|[βϱϵ(ξ)+ϱϵ(e)]. On the other hand,
|Dα1[xϵ](t)−Dα1[x](t)|≤1λ|Dα1[ϕxϵ](t)−Dα1[ϕx](t)|+|Hxϵ(ξ,β)−Hx(ξ,β)||Dα1[g(t−1)]|. |
By (4.8), we have
|Dα1[ϕxϵ](t)−Dα1[ϕx](t)|=|∫t1Dα1[g(t−1)](s)[Jα−ϵ1[fxϵ]−Jα1[fx]](t−s+1)ds|=|∫t1Dα1[g(t−1)](s)×[Jα−ϵ1[fxϵ]−Jα1[fx]+Jα1[fxϵ]−Jα1[fxϵ]](t−s+1)ds|.|Dα1[ϕxϵ](t)−Dα1[ϕx](t)|≤|∫t1Dα1[g(t−1)](s)Jα1[fxϵ−fx](t−s+1)ds|+|∫t1Dα1[g(t−1)](s)(Jα−ϵ1[fxϵ]−Jα1[fxϵ])(t−s+1)ds|≤N‖x−xϵ‖E∫t1|Dα1[g(t−1)](s)|Jα1[1](t−s+1)ds+‖fxϵ‖∫t1|Dα1[g(t−1)](s)||(Jα−ϵ1[1]−Jα1[1])(t−s+1)|ds. |
Then
|Dα1[ϕxϵ](t)−Dα1[ϕx](t)|≤Nσα(t)‖x−xϵ‖E+υϵ(t)‖fxϵ‖, |
with
υϵ(t)=∫t1|Dα1[g(t−1)](s)||Jα−ϵ1[1]−Jα1[1](t−s+1)|ds. | (6.4) |
Then the expression above becomes
|Dα1[xϵ](t)−Dα1[x](t)|≤1λNσα(t)‖x−xϵ‖E+υϵ(t)‖fxϵ‖+NΓ(1+α)1|Δ|[βρα(ξ)+ρα(e)]‖x−xϵ‖E+1|Δ|[βϱϵ(ξ)+ϱϵ(e)]‖fxϵ‖|Dα1[g(t−1)]|≤N[1λσα(t)+1Γ(1+α)1|Δ|[βρα(ξ)+ρα(e)]|Dα1[g(t−1)]|]‖x−xϵ‖E+[1λυϵ(t)+1|Δ|[βϱϵ(ξ)+ϱϵ(e)]|Dα1[g(t−1)]|]‖fxϵ‖. |
Then
|Dα1[xϵ](t)−Dα1[x](t)|≤NC22(t)‖x−xϵ‖E+C33(t)‖fxϵ‖, | (6.5) |
with
C22(t)=[1λσα(t)+1Γ(1+α)1|Δ|[βρα(ξ)+ρα(e)]|Dα1[g(t−1)]|],C33(t)=[1λυϵ(t)+1|Δ|[βϱϵ(ξ)+ϱϵ(e)]|Dα1[g(t−1)]|]. | (6.6) |
Moreover, from (6.3), (6.5), we deduce that
|x(t)−xϵ(t)|+|Dα1[xϵ](t)−Dα1[x](t)|≤N[1Γ(1+α)ρ(t)+C22(t)]‖x−xϵ‖E+[ϱ(t)+C33(t)]‖fxϵ‖. |
Finally, we get the inequality
‖x−xϵ‖E≤supt∈[1,e][ϱ(t)+C33(t)]1−Nsupt∈[1,e][1Γ(1+α)ρ(t)+C22(t)]‖f‖∗, |
which is exactly the required inequality (6.2), where
kϵ=supt∈[1,e][ϱ(t)+C33(t)]1−Nsupt∈[1,e][1Γ(1+α)ρ(t)+C22(t)]. | (6.7) |
Theorem 6.2. Suppose that the conditions of Theorem 4.1 hold. Let x(t),xϵ(t) be the solutions, respectively, of the problems (1.1) and
Dα1(D2+λ2)x(t)=f(t,x(t),Dα1[x](t)+ϵhϵ(t), |
for t∈(1,e) and hϵ∈C, with boundary conditions (1.1)-b, where ϵ<0. Then ‖x−xϵ‖E=O(ϵ).
Proof. In accordance with Lemma 3.4, we have
ϕxϵ(t)=∫t1g(t−s)Jα1[fxϵ+ϵhϵ](s)ds |
and
|ϕxϵ(t)−ϕx(t)|=|∫t1g(t−s)Jα1[fxϵ+ϵhϵ](s)ds−∫t1g(t−s)Jα1[fx](s)ds|≤|∫t1g(t−s)Jα1[fxϵ−fx](s)ds|+ϵ|∫t1g(t−s)Jα1[hϵ](s)ds|≤1Γ(1+α)(‖fxϵ−fx‖+ϵ‖hϵ‖)∫t1(lns)αds≤N‖x−xϵ‖E+ϵ‖hϵ‖Γ(1+α)ρα(t). | (6.8) |
and
|Hx(ξ,β)−Hxϵ(ξ,β)|≤1|Δ|[β|ϕxϵ(ξ)−ϕx(ξ)|+|ϕxϵ(e)−ϕx(e)|]≤1|Δ|N‖x−xϵ‖E+ϵ‖hϵ‖Γ(1+α)[βρα(ξ)+ρα(e)]. | (6.9) |
From (6.8) and (6.9), we derive
|x(t)−xϵ(t)|≤ρ(t)Γ(1+α)(N‖x−xϵ‖E+ϵ‖hϵ‖). |
On the other hand,
|Dα1[ϕxϵ](t)−Dα1[ϕx](t)|=|∫t1Dα1[g(t−1)](s)(Jα1[fxϵ+ϵhϵ]−Jα1[fx])(t−s+1)ds|≤|∫t1Dα1[g(t−1)](s)Jα1[fxϵ−fx](t−s+1)ds|+ϵ|∫t1Dα1g(t−1)(s)Jα1[hϵ](t−s+1)ds|≤(N‖x−xϵ‖E+ϵ‖hϵ‖)∫t1|Dα1[g(t−1)](s)|Jα1[1](t−s+1)ds |
and
|Dα1[ϕxϵ](t)−Dα1[ϕx](t)|≤R11(t)Γ(1+α)(N‖x−xϵ‖E+ϵ‖hϵ‖), |
where R11 is given by (4.9). Hence, we obtain
|Dα1[xϵ](t)−Dα1[x](t)|≤R11(t)λΓ(1+α)(N‖x−xϵ‖E+ϵ‖hϵ‖)+1|Δ|N‖x−xϵ‖E+ϵ‖hϵ‖Γ(1+α)[βρα(ξ)+ρα(e)]|Dα1[g(t−1)]|, |
and
|x(t)−xϵ(t)|+|Dα1[xϵ](t)−Dα1[x](t)|≤NΓ(1+α)(ρ(t)+[βρα(ξ)+ρα(e)]|Dα1[g(t−1)]||Δ|+R11(t)λ)‖x−xϵ‖E+ϵ‖hϵ‖Γ(1+α)(ρ(t)+[βρα(ξ)+ρα(e)]|Dα1[g(t−1)]||Δ|+R11(t)λ). |
Consequently
‖x−xϵ‖E≤ϵNQ1−NQ‖h‖∗, |
where Q is given by (3.18) and ‖h‖∗=sup0<ϵ‖hϵ‖. It is obvious that ‖x−xϵ‖E=O(ϵ).
Let us introduce small perturbation in the boundary conditions of (1.1) such that
x(1)=0=D2x(1),x(e)=βx(ξ)+ϵ, | (6.10) |
for ξ∈(1,e].
Theorem 6.3. Assume the conditions of Theorem 4.1 hold. Let x(t), xϵ(t) be respective solutions, of the problems (1.1) and the boundary conditions (1.1)-a with (6.10). Then
‖x−xϵ‖E=O(ϵ). |
Proof. Similar arguments as in the proof of Lemma 3.4, may lead to the solution of equations (1.1)-a and (6.10) that has the following form
xϵ(t)=1λ∫t1sinλ(t−s)Jα1[fxϵ]ds+βΔsinλ(t−1)∫ξ1sinλ(ξ−s)Jα1[fxϵ]ds−1Δsinλ(t−1)∫e1sinλ(e−s)Jα1[fxϵ]ds+ϵλsinλ(t−1)Δcosλ. |
Therefore
xϵ(t)=1λϕxϵ(t)+Hxϵ(ξ,β)g(t−1)+ϵλsinλ(t−1)Δcosλ, |
and Δcosλ≠0, where
ϕxϵ(t)=∫t1g(t−s)Jα1[fxϵ](s)ds |
and Hxϵ(ξ,β)=1Δ(βϕxϵ(ξ)+ϕxϵ(e)). As before, we find that
|x(t)−xϵ(t)|≤Nρ(t)Γ(1+α)‖x−xϵ‖E+ϵλ|Δ||sinλ(t−1)cosλ|, |
and
|Dα1[xϵ](t)−Dα1[x](t)|≤R11(t)λΓ(1+α)N‖x−xϵ‖E+ϵ|λDα1sinλ(t−1)Δcosλ|+1|Δ|N‖x−xϵ‖EΓ(1+α)[βρα(ξ)+ρα(e)]|Dα1[g(t−1)]|. |
Hence
|x(t)−xϵ(t)|+|Dα1[xϵ](t)−Dα1[x](t)|≤NΓ(1+α)[ρ(t)+[βρα(ξ)+ρα(e)]|Dα1[g(t−1)]||Δ|+R11(t)λ]‖x−xϵ‖E+ϵλ|Δcosλ|[|sinλ(t−1)|+λΓ(2−α)t(lnt)1−α]. |
Consequently
‖x−xϵ‖E≤ϵλ|Δcosλ|(1−NQ)[1+λe2−αΓ(2−α)]. |
It is obvious that ‖x−xϵ‖E=O(ϵ).
In this section, we present some examples to illustrate the validity and applicability of the main results.
Example 7.1. Consider problem (1.1) with
f(t,x,˜x)=1/61+|x|+|˜x|. | (7.1) |
Then f fulfills the Lipschitz condition (H1) such that λ=2, β=2, ξ=32,
N=max{N1,N2}=16. |
In Table 1, we show values of α, t and Q. Thus NQ<1. Hence, by Theorem 4.1, the problem (1.1) with (7.1) has a unique solution on [1,e].
α | 1 | 1 | 1 | 0.75 | 0.25 | 0.5 | 0.5 | 0.5 | |
t | 1 | 2 | e | e | e | e | 2 | 1 | |
Q | 1.38 | 1.35 | 2.94 | 3.58 | 4.58 | 4.15 | 1.90 | 0.20 |
Example 7.2. Consider problem (1.1) with
f(t,x,˜x)=58(sinx+cosx)+˜x | (7.2) |
or
f(t,x,˜x)=14(sinx+cosx)+76˜x. | (7.3) |
Then f fulfills the Lipschitz condition (H1), where λ=4, β=2, ξ=32, N=max{N1,N2}=54 or 76. In Table 2, we show values of α, t, Q, 54Q and 76Q. Thus, the condition (4.1) holds. Again, taking N=max{N1,N2}=54 or 76, we have NQ<1. Note that all the assumptions of the Theorem 4.3 holds. Therefore problem (1.1) has a unique solution on E.
α | 1.000 | 1.000 | 0.75 | 0.50 | 0.25 | |
t | 1.030 | 1.002 | 1.03 | 1.03 | 1.03 | |
Q | 0.860 | 0.780 | 0.57 | 0.44 | 0.39 | |
5Q/4 | 1.008>1 | 0.980 | 0.72 | 0.56 | 0.49 | |
7Q/6 | 0.940<1 | 0.910 | 0.67 | 0.52 | 0.46 |
Example 7.3. Consider problem (1.1) with
f(t,x,˜x)=d1(t)sin[d2(t)(x+˜x)]+d3(t)cos[d4(t)(x+˜x)], | (7.4) |
for di∈C[1,e] with i=1,2,3,4, that fulfils (H1) with
N1=N2=|d1(t)d2(t)|+|d3(t)d4(t)|=1, |
for example d1(t)=d3(t)=14, d2(t)=d4(t)=2. Thus, we can put β=2, ξ=32, N=1, L0=14. In Table 3, one can find some values of α, λ, t, Q and r, where r and Q are as defined in Theorem 4.3.
α | 1.00 | 1.00 | 0.750 | 0.750 | 0.500 | 0.500 | 0.25 | 0.250 | |
λ | 1.35 | 7.76 | 7.760 | 9.600 | 7.760 | 9.600 | 7.80 | 9.600 | |
t | 1.00 | 1.50 | 1.500 | 1.000 | 1.500 | 1.000 | 2.00 | 1.000 | |
Q | 0.89 | 0.78 | 0.250 | 0.120 | 0.270 | 0.150 | 0.79 | 0.180 | |
r≥ | 2.03 | 0.89 | 0.084 | 0.035 | 0.093 | 0.045 | 0.94 | 0.054 |
Hence, by Theorem 4.3, problem (1.1) with (7.4) has a unique solution on Br.
Example 7.4. (Illustrative example on stability) Consider the FLE problem (1.1) with Hadamard fractional derivatives involving nonlocal boundary conditions:
{D381(D2+π2)x(t)=f(t,x(t),D381[x](t)),t∈(1,154),D2x(1)=x(1)=0,x(154)=2x(115), | (7.5) |
with
f(t,x,˜x)=x25(t+1)2(|x|+3)+|x|5(t+1)2+sin˜x4(t+3)2−2, | (7.6) |
where α=38, λ=π, β=2 and ξ=115. It is obvious that
sinπ(154−1)=0.7071≠−1.1756=2sinπ(115−1), |
So
Δ=λ(sinλ(e−1)−βsinλ(ξ−1))=5.9146≠0. |
See the Figure 1. Also, we have
|f(t,x1,˜x1)−f(t,x2,˜x2)|≤16135|x1−x2|+149|˜x1−˜x2|, |
for each t∈[1,154] and all xi,˜xi∈R, here N1=16135 and N2=149. Put φ(t)=t2t2+1 and
N=max{16135,149}=16135. |
By using Eq (2.11), we obtain
ρα(t)≤138+1(154)38+1=4.4770 |
and using MatLab program,
ρα(154)=∫1541(lns)αds=2.4371,ρα(115)=∫1151(lns)αds=0.8463. |
Also, by applying Eq (3.17), we obtain
Mρ=1π+3|5.9146|=0.8255, |
Then we get
cφ≥cρα(e)λΓ(1+α)+cβρα(ξ)|Δ|Γ(1+α)[1+Nρα(e)Γ(1+α)−NMρ]+cρα(e)Γ(1+α)[1|Δ|+1λNρα(t)Γ(1+α)−NMρ+1|Δ|Nρα(e)Γ(1+α)−NMρ]. |
Then the assumptions of Theorem 5.5 are satisfied. Then, problem (1.1) is Ulam-Hyers stable and generalized Ulam-Hyers stable.
Example 7.5. (Illustrative example on solution dependence) Consider the FLE problem (1.1) with Hadamard fractional derivatives involving nonlocal boundary conditions:
{D791(D2+(4π3)2)x(t)=f(t,x(t),D791[x](t)),t∈(1,125),D2x(1)=x(1)=0,x(125)=73x(94), | (7.7) |
and
D79−ϵ1(D2+(4π3)2)x(t)=f(t,x(t),D791[x](t)), | (7.8) |
for t∈(0,1) and ϵ>0 with
f(t,x,˜x)=|x+4|(sin2(πt)+14)5+t+(˜x2+4)(sin2(3πt)+6)(t+2)(|˜x+1|)+|˜x|(cos2(3πt)+2)(t+0.5)2+134, | (7.9) |
where α=79, λ=4π3, β=73, ξ=94 and 0<79−ϵ<α<1. It is obvious that
sinπ(154−1)=0.7071≠−1.1756=2sinπ(115−1), |
So
Δ=λ(sinλ(e−1)−βsinλ(ξ−1))=6.7606≠0. |
See Figure 2. Also, we have
|f(t,x1,˜x1)−f(t,x2,˜x2)|≤4121|x1−x2|+275|˜x1−˜x2|, |
for each t∈[1,125] and all xi,˜xi∈R, here N1=4121 and N2=275. Put φ(t)=t2t2+1,
N=max{4121,275}. |
By using Eq (2.11), we obtain
ρα(t)≤179+1(125)79+1=2.6671 |
and
ρα(125)=∫1251(lns)αds=0.7953,ρα(94)=∫941(lns)αds=0.6639. |
Also, by applying Eq (3.17), we obtain
Mρ=1π+3|6.7606|=0.7620, |
Then for ϵ=0.1, by using MatLab program, we get
ϱϵ(e)=ϱϵ(125)=∫t1|(lns)79−ϵΓ(1+79−ϵ)−(lns)αΓ(1+α)|ds=0.1886,ϱϵ(ξ)=ϱϵ(94)=∫t1|(lns)79−ϵΓ(1+79−ϵ)−(lns)αΓ(1+α)|ds=0.1761 |
and
ϱ(t)=34πϱϵ(t)+1|6.7606|[7×0.17613+ϱϵ(0.1886)]. |
Then the assumptions of Theorem 6.1 are satisfied. In addition to, by applying Eqs (6.6) and (6.7), we can calculate C22(t), C33(t) and kϵ.
α=17 | α=12 | α=79 | ||||||||
n | tN | x(tN) | (a) | (b) | (a) | (b) | (a) | (b) | ||
0 | 1.0000 | 0.0000 | 0.0000 | 0.0000 | 0.0000 | 0.0000 | 0.0000 | 0.0000 | ||
1 | 1.0296 | 0.0292 | −0.6481 | 20.4865 | −2.0421 | 20.3037 | −5.7442 | 20.6579 | ||
2 | 1.0601 | 0.0584 | 0.3210 | 20.6028 | 1.2297 | 20.4569 | 2.3563 | 20.9298 | ||
3 | 1.0915 | 0.0875 | 1.1548 | 20.8158 | 3.1716 | 20.6972 | 5.8459 | 21.1952 | ||
4 | 1.1238 | 0.1167 | 1.9214 | 21.0045 | 4.6113 | 20.9107 | 7.9468 | 21.4028 | ||
5 | 1.1571 | 0.1459 | 2.6447 | 21.0604 | 5.7818 | 21.0022 | 9.4115 | 21.4832 | ||
⋮ | ⋮ | ⋮ | ⋮ | ⋮ | ⋮ | ⋮ | ⋮ | ⋮ | ||
20 | 1.7926 | 0.5836 | 11.5237 | 19.1220 | 14.8965 | 19.3122 | 16.5953 | 19.4690 | ||
21 | 1.8456 | 0.6128 | 12.0495 | 19.0026 | 15.3004 | 19.1885 | 16.8158 | 19.3276 | ||
22 | 1.9003 | 0.642 | 12.5704 | 18.5841 | 15.6925 | 18.7749 | 17.0254 | 18.8997 | ||
23 | 1.9566 | 0.6712 | 13.0867 | 18.0950 | 16.0739 | 18.2893 | 17.2254 | 18.3996 | ||
24 | 2.0145 | 0.7004 | 13.5986 | 17.8788 | 16.4453 | 18.0647 | 17.4166 | 18.1580 | ||
25 | 2.0742 | 0.7296 | 14.1063 | 18.0394 | 16.8076 | 18.2061 | 17.5999 | 18.2808 | ||
26 | 2.1356 | 0.7587 | 14.6102 | 18.2910 | 17.1613 | 18.4376 | 17.7761 | 18.4949 | ||
27 | 2.1988 | 0.7879 | 15.1104 | 18.2508 | 17.5070 | 18.3843 | 17.9457 | 18.4272 | ||
28 | 2.2639 | 0.8171 | 15.6071 | 17.9090 | 17.8453 | 18.0343 | 18.1094 | 18.0647 | ||
29 | 2.3310 | 0.8463 | 16.1005 | 17.6773 | 18.1766 | 17.7921 | 18.2675 | 17.8109 | ||
30 | 2.4000 | 0.8755 | 16.5908 | 17.8398 | 18.5014 | 17.9399 | 18.4206 | 17.9478 |
Now, we describ discretization method and use Theorem 2.3 for this example. Fix n≥1 for N∈{1,⋯,n}, define
tN=aexp(ΔT)=exp(ΔT), |
with
ΔT=1nlnba=1nln125 |
and [a,b]=[1,125].
Also,
(ταk)=k1−α−(k−1)1−α=k1−79−(k−1)1−79=k29−(k−1)29 |
and
ζ=(ΔT)1−αa[1−exp(−ΔT)]Γ(2−α)=(1nln125)29[1−exp(−1nln125)]Γ(29). |
Thus, for f(t,x(t),D791[x](t)), we get
D791[x](tN)=˜D791[x](tN)+O(1nln125), |
where
˜D791[x](tN)=x(a)Γ(1−α)(lntNa)−α+ζN∑k=1(ταN−k+1)x(tk)−x(tk−1)exp(kΔT).tk=x(1)Γ(1−79)(lntN)−79+ζN∑k=1(τ79N−k+1)x(tk)−x(tk−1)exp(knln125).tk=x(1)Γ(29)(lntN)−79+ζN∑k=1(τ79N−k+1)x(tk)−x(tk−1)exp(knln125).tk, |
The Langevin equation has been proposed to describe dynamical processes in a fractal medium in which the fractal and memory properties with a dissipative memory kernel are incorporated. However, it has been realized that the classical Langevin equation failed to describe the complex systems. Thus, the consideration of LDE in frame of fractional derivatives becomes compulsory. As a result of this interest, several results have been revealed and different versions of LDE have been under study. In this paper, we have presented some results dealing with the existence and uniqueness of solutions for boundary value problem of nonlinear Langevin equation involving Hadamard fractional order. As a first step, the boundary value problem is transformed to a fixed point problem by applying the tools of Hadamard fractional calculus. Based on this, the existence results are established by means of the Schaefer's fixed point theorem and Banach contraction principle.
We claim that the results of this paper is new and generalize some earlier results. For instance, by taking α=1 in the results of this paper which can be considered a special case of a simple Jerk Chaotic circuit equation see [33]. The paper presented a discuss on the Ulam-Hyers-Rassias and generalized Ulam-Hyers-Rassias stabilities of the solution of the FLD using the generalization for the Gronwall inequality. We present an example to demonstrate the consistency to the theoretical findings. We also analyze the continuous dependence of solutions all on its right side function, initial value condition and the fractional order for FDE. Using these results, the properties of the solution process can be discussed through numerical simulation. We hope to consider this problem in a future work.
Data sharing not applicable to this article as no data sets were generated or analyzed during the current study.
The authors declare that the study was realized in collaboration with equal responsibility. All authors read and approved the final manuscript.
The second author was supported by Bu-Ali Sina University. J. Alzabut would like to acknowledge the support of Prince Sultan University for paying the Article Processing Charges (APC) of this publication.
The authors declare that they have no competing interests.
1 function [ParamMatrix]= discretization_method3(alpha, a, e, lambda, n, x_t) |
2 [xalpha, yalpha]=size(alpha); |
3 DeltaT=log(e/a)/n; |
4 ParamMatrix(1, 1) = 0; |
5 ParamMatrix(1, 2) = a; |
6 for j=3:2+7*yalpha |
7 ParamMatrix(1, j) = 0; |
8 end; |
9 |
10 column=3; |
11 j=1; |
12 while j < =yalpha |
13 for N=1:n |
14 ParamMatrix(N+1, 1) = N; |
15 tN=a*exp(N*DeltaT); |
16 ParamMatrix(N+1, 2) = tN; |
17 end; |
18 for N=1:n |
19 zeta = round(DeltaT^(1-alpha(j))/(a * (1- exp((-1)*DeltaT))*gamma(2-alpha(j))), 6); |
20 ParamMatrix(N+1, column) = zeta; |
21 ParamMatrix(N+1, column+5) =round(eval(subs(x_t, ParamMatrix(N+1, 2))), 6); |
22 s=0; |
23 k=1; |
24 while k < =N |
25 taukalpha= (N-k+1)^(1-alpha(j)) - (N-k)^(1-alpha(j)); |
26 y2=eval(subs(x_t, ParamMatrix(k+1, 2))); |
27 y1=eval(subs(x_t, ParamMatrix(k, 2))); |
28 s = s + taukalpha*(y2-y1)*ParamMatrix(k+1, 2)/exp(k*DeltaT); |
29 k=k+1; |
30 end; |
31 A=eval(subs(x_t, a))* (log(ParamMatrix(N+1, 2)/a))^((-1)*alpha(j)); |
32 HadamardD_xtN= round(A + zeta*s, 6); |
33 ParamMatrix(N+1, column+1) = HadamardD_xtN; |
34 end; |
35 for N=1:n |
36 s=0; |
37 k=1; |
38 while k < =N |
39 taukalpha= (N-k+1)^(1-alpha(j)) - (N-k)^(1-alpha(j)); |
40 y2=lambda^2* eval(subs(x_t, ParamMatrix(k+1, 2))); |
41 y1=lambda^2* eval(subs(x_t, ParamMatrix(k, 2))); |
42 s = s + taukalpha*(y2-y1)*ParamMatrix(k+1, 2)/exp(k*DeltaT); |
43 k=k+1; |
44 end; |
45 A=eval(subs(x_t, a))* (log(ParamMatrix(N+1, 2)/a))^((-1)*alpha(j)); |
46 HadamardD_xtN = round(A + zeta*s, 6); |
47 ParamMatrix(N+1, column+2) = HadamardD_xtN; |
48 end; |
49 for N=1:n |
50 s=0; |
51 k=1; |
52 while k < =N |
53 taukalpha= (N-k+1)^(1-alpha(j)) - (N-k)^(1-alpha(j)); |
54 y2=eval(subs(diff(x_t, 2), ParamMatrix(k+1, 2))); |
55 y1=eval(subs(diff(x_t, 2), ParamMatrix(k, 2))); |
56 s = s + taukalpha*(y2-y1)*ParamMatrix(k+1, 2)/exp(k*DeltaT); |
57 k=k+1; |
58 end; |
59 A=eval(subs(diff(x_t, 2), a))* (log(ParamMatrix(N+1, 2)/a))^((-1)*alpha(j)); |
60 HadamardD_xtN= round(A + zeta*s, 5); |
61 ParamMatrix(N+1, column+3) = HadamardD_xtN; |
62 ParamMatrix(N+1, column+4) = ParamMatrix(N+1, column+2)+ParamMatrix(N+1, column+3); |
63 ParamMatrix(N+1, column+6) = abs(ParamMatrix(N+1, column+5) +4)*((sin(pi*ParamMatrix(N+1, 2)))^2+14)/(5+ParamMatrix(N+1, 2)) + ((ParamMatrix(N+1, column+1))^2+4)*((sin(3*pi*ParamMatrix(N+1, 2)))^2 +6) / ((ParamMatrix(N+1, 2) +2) * (abs(ParamMatrix(N+1, column+1)+1))) + abs(ParamMatrix(N+1, column+1)) * ((cos(pi*ParamMatrix(N+1, 2)))^2+10)/((ParamMatrix(N+1, 2)+0.5)^2) + 13/4; |
64 end; |
65 j=j+1; |
66 column=column+7; |
67 end; |
68 end |
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1 function [ParamMatrix]= discretization_method3(alpha, a, e, lambda, n, x_t) |
2 [xalpha, yalpha]=size(alpha); |
3 DeltaT=log(e/a)/n; |
4 ParamMatrix(1, 1) = 0; |
5 ParamMatrix(1, 2) = a; |
6 for j=3:2+7*yalpha |
7 ParamMatrix(1, j) = 0; |
8 end; |
9 |
10 column=3; |
11 j=1; |
12 while j < =yalpha |
13 for N=1:n |
14 ParamMatrix(N+1, 1) = N; |
15 tN=a*exp(N*DeltaT); |
16 ParamMatrix(N+1, 2) = tN; |
17 end; |
18 for N=1:n |
19 zeta = round(DeltaT^(1-alpha(j))/(a * (1- exp((-1)*DeltaT))*gamma(2-alpha(j))), 6); |
20 ParamMatrix(N+1, column) = zeta; |
21 ParamMatrix(N+1, column+5) =round(eval(subs(x_t, ParamMatrix(N+1, 2))), 6); |
22 s=0; |
23 k=1; |
24 while k < =N |
25 taukalpha= (N-k+1)^(1-alpha(j)) - (N-k)^(1-alpha(j)); |
26 y2=eval(subs(x_t, ParamMatrix(k+1, 2))); |
27 y1=eval(subs(x_t, ParamMatrix(k, 2))); |
28 s = s + taukalpha*(y2-y1)*ParamMatrix(k+1, 2)/exp(k*DeltaT); |
29 k=k+1; |
30 end; |
31 A=eval(subs(x_t, a))* (log(ParamMatrix(N+1, 2)/a))^((-1)*alpha(j)); |
32 HadamardD_xtN= round(A + zeta*s, 6); |
33 ParamMatrix(N+1, column+1) = HadamardD_xtN; |
34 end; |
35 for N=1:n |
36 s=0; |
37 k=1; |
38 while k < =N |
39 taukalpha= (N-k+1)^(1-alpha(j)) - (N-k)^(1-alpha(j)); |
40 y2=lambda^2* eval(subs(x_t, ParamMatrix(k+1, 2))); |
41 y1=lambda^2* eval(subs(x_t, ParamMatrix(k, 2))); |
42 s = s + taukalpha*(y2-y1)*ParamMatrix(k+1, 2)/exp(k*DeltaT); |
43 k=k+1; |
44 end; |
45 A=eval(subs(x_t, a))* (log(ParamMatrix(N+1, 2)/a))^((-1)*alpha(j)); |
46 HadamardD_xtN = round(A + zeta*s, 6); |
47 ParamMatrix(N+1, column+2) = HadamardD_xtN; |
48 end; |
49 for N=1:n |
50 s=0; |
51 k=1; |
52 while k < =N |
53 taukalpha= (N-k+1)^(1-alpha(j)) - (N-k)^(1-alpha(j)); |
54 y2=eval(subs(diff(x_t, 2), ParamMatrix(k+1, 2))); |
55 y1=eval(subs(diff(x_t, 2), ParamMatrix(k, 2))); |
56 s = s + taukalpha*(y2-y1)*ParamMatrix(k+1, 2)/exp(k*DeltaT); |
57 k=k+1; |
58 end; |
59 A=eval(subs(diff(x_t, 2), a))* (log(ParamMatrix(N+1, 2)/a))^((-1)*alpha(j)); |
60 HadamardD_xtN= round(A + zeta*s, 5); |
61 ParamMatrix(N+1, column+3) = HadamardD_xtN; |
62 ParamMatrix(N+1, column+4) = ParamMatrix(N+1, column+2)+ParamMatrix(N+1, column+3); |
63 ParamMatrix(N+1, column+6) = abs(ParamMatrix(N+1, column+5) +4)*((sin(pi*ParamMatrix(N+1, 2)))^2+14)/(5+ParamMatrix(N+1, 2)) + ((ParamMatrix(N+1, column+1))^2+4)*((sin(3*pi*ParamMatrix(N+1, 2)))^2 +6) / ((ParamMatrix(N+1, 2) +2) * (abs(ParamMatrix(N+1, column+1)+1))) + abs(ParamMatrix(N+1, column+1)) * ((cos(pi*ParamMatrix(N+1, 2)))^2+10)/((ParamMatrix(N+1, 2)+0.5)^2) + 13/4; |
64 end; |
65 j=j+1; |
66 column=column+7; |
67 end; |
68 end |
1 function [ParamMatrix]= discretization_method3(alpha, a, e, lambda, n, x_t) |
2 [xalpha, yalpha]=size(alpha); |
3 DeltaT=log(e/a)/n; |
4 ParamMatrix(1, 1) = 0; |
5 ParamMatrix(1, 2) = a; |
6 for j=3:2+7*yalpha |
7 ParamMatrix(1, j) = 0; |
8 end; |
9 |
10 column=3; |
11 j=1; |
12 while j < =yalpha |
13 for N=1:n |
14 ParamMatrix(N+1, 1) = N; |
15 tN=a*exp(N*DeltaT); |
16 ParamMatrix(N+1, 2) = tN; |
17 end; |
18 for N=1:n |
19 zeta = round(DeltaT^(1-alpha(j))/(a * (1- exp((-1)*DeltaT))*gamma(2-alpha(j))), 6); |
20 ParamMatrix(N+1, column) = zeta; |
21 ParamMatrix(N+1, column+5) =round(eval(subs(x_t, ParamMatrix(N+1, 2))), 6); |
22 s=0; |
23 k=1; |
24 while k < =N |
25 taukalpha= (N-k+1)^(1-alpha(j)) - (N-k)^(1-alpha(j)); |
26 y2=eval(subs(x_t, ParamMatrix(k+1, 2))); |
27 y1=eval(subs(x_t, ParamMatrix(k, 2))); |
28 s = s + taukalpha*(y2-y1)*ParamMatrix(k+1, 2)/exp(k*DeltaT); |
29 k=k+1; |
30 end; |
31 A=eval(subs(x_t, a))* (log(ParamMatrix(N+1, 2)/a))^((-1)*alpha(j)); |
32 HadamardD_xtN= round(A + zeta*s, 6); |
33 ParamMatrix(N+1, column+1) = HadamardD_xtN; |
34 end; |
35 for N=1:n |
36 s=0; |
37 k=1; |
38 while k < =N |
39 taukalpha= (N-k+1)^(1-alpha(j)) - (N-k)^(1-alpha(j)); |
40 y2=lambda^2* eval(subs(x_t, ParamMatrix(k+1, 2))); |
41 y1=lambda^2* eval(subs(x_t, ParamMatrix(k, 2))); |
42 s = s + taukalpha*(y2-y1)*ParamMatrix(k+1, 2)/exp(k*DeltaT); |
43 k=k+1; |
44 end; |
45 A=eval(subs(x_t, a))* (log(ParamMatrix(N+1, 2)/a))^((-1)*alpha(j)); |
46 HadamardD_xtN = round(A + zeta*s, 6); |
47 ParamMatrix(N+1, column+2) = HadamardD_xtN; |
48 end; |
49 for N=1:n |
50 s=0; |
51 k=1; |
52 while k < =N |
53 taukalpha= (N-k+1)^(1-alpha(j)) - (N-k)^(1-alpha(j)); |
54 y2=eval(subs(diff(x_t, 2), ParamMatrix(k+1, 2))); |
55 y1=eval(subs(diff(x_t, 2), ParamMatrix(k, 2))); |
56 s = s + taukalpha*(y2-y1)*ParamMatrix(k+1, 2)/exp(k*DeltaT); |
57 k=k+1; |
58 end; |
59 A=eval(subs(diff(x_t, 2), a))* (log(ParamMatrix(N+1, 2)/a))^((-1)*alpha(j)); |
60 HadamardD_xtN= round(A + zeta*s, 5); |
61 ParamMatrix(N+1, column+3) = HadamardD_xtN; |
62 ParamMatrix(N+1, column+4) = ParamMatrix(N+1, column+2)+ParamMatrix(N+1, column+3); |
63 ParamMatrix(N+1, column+6) = abs(ParamMatrix(N+1, column+5) +4)*((sin(pi*ParamMatrix(N+1, 2)))^2+14)/(5+ParamMatrix(N+1, 2)) + ((ParamMatrix(N+1, column+1))^2+4)*((sin(3*pi*ParamMatrix(N+1, 2)))^2 +6) / ((ParamMatrix(N+1, 2) +2) * (abs(ParamMatrix(N+1, column+1)+1))) + abs(ParamMatrix(N+1, column+1)) * ((cos(pi*ParamMatrix(N+1, 2)))^2+10)/((ParamMatrix(N+1, 2)+0.5)^2) + 13/4; |
64 end; |
65 j=j+1; |
66 column=column+7; |
67 end; |
68 end |