# Applying the Reproducing Kernel Method to Fractional Differential Equations with Periodic Conditions in Hilbert Space

Authors: Hoda Saky; Saeid Abbasbandy; Elyas Shivanian
Journal: Journal of Mathematics
 (2022)

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## Abstract

In this article, the reproducing kernel method is presented for the fractional differential equations with periodic conditions in the Hilbert space. This method gives an approximate solution to the problem. The approximate and exact solutions are displayed in the form of series in the reproduction kernel space. In addition, we provide an error analysis for this technique. The presented method is tested by some examples to show its precision.


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## Body

## 1. Introduction

Fractional differential equations are needed to model and analyze large volumes of problems. FDEs are applied in large number of fields such as fluid mechanics, biology, chemistry, and diffusion [1–7]. Some methods for solving these equations are Laplace transforms [8], Fourier transform [9], Adomian decomposition method [10], finite difference method [11], variational iteration method [12, 13], collocation method [14], and other methods [15–19].Many papers have worked on FDEs with periodic conditions, some of which are listed below. Belmekki et al. have discussed the existence and uniqueness of the solution in [20]. Wei et al. have reviewed the minimal and maximal solutions for periodic problems in [21]. In [22], authors have given monotone iterative techniques for existing solutions. In [23], Javidi and Saedshoar Heris have used the method fractional backward differentiation formulas for(1)λnDtαnyt+λn−1Dtαn−1yt+⋯+λ1Dtα1yt+λyt−τ=ft,with periodic condition y0=yT.In this work, we use reproducing kernel Hilbert space (RKHS) method to solve multiterm FDEs in the form as follows:(2)μnD0+ηnvt+μn−1D0+ηn−1vt+⋯+μ0D0+η0vt=gt,vt,t∈0,T,with periodic condition as follows:(3)v0=vT,where 0≤η0<η1<⋯<ηn<1,T>0, μj∈ℝj=0,1,…,n,μn≠0 in Caputo sense. In [24], the existence and uniqueness of the solution have been proven to this problem by using green’s function.The reproducing kernel method was first used in research on boundary value problems in the early twentieth century. In 1907, Zarmba was the first to introduce the kernel of certain functions and to express their reproducing properties. Since 1980, with the efforts of Cui, the reproducing kernel functions of Hilbert space have been introduced in the form of very simple polynomials. They were able to use methods based on the reproducing kernel space [25–27]. Many researchers use the RKHS method to find approximate solutions to various problems [28–30], and also some new applications of reproducing kernel methods and neural networks in machine learning are found in [31–33]. Very recently, RKHS is applied on fractional differential equations [34–36]. In this paper, the reproducing kernel method is presented for the fractional differential equations with periodic conditions in the Hilbert space. The approximate solution obtained from this method is uniformly convergent to the exact solution.This paper is arranged as follows. Section2 provides some definitions. Analysis of the RKHS method is proposed in Section 3. The convergence of the approximate solution to the exact solution is given in Section 4. Examples are given in Section 5.


## 2. Basic Definitions

We describe some of the symbols and basic definitions used in this article. LetCI,ℝ represent the Banach space of all continuous functions of I=0,T into ℝ, and CmI,ℝ shows the real valued functions on I where the mth order derivative is continuous.Definition 1.
The fractional integral ofg∈CI,ℝ of order η>0 is(4)I0+ηgt=1Γη∫0tgst−s1−ηds,0<t<T.Definition 2.
The Caputo fractional derivative ofg∈CmI,ℝ of order η>0 is(5)D0+ηgt=Im−ηDmgt,m−1<η<m,m∈ℕ,gmt,η=m.Definition 3 (see [37]).
SupposeH is a function Hilbert space, including all real or complex value functions defined on a abstract space X, with the inner product .,.H. For each fixed y∈X, if there exist a function Ry.∈H which satisfies(6)f,RyH=fy,then Ry. is called the reproducing kernel of H and the Hilbert space H is called the reproducing kernel space.Remark 1.
The real value function space(7)W220,T=v|v′is absolutely continuous,v″∈L20,T,v0=vT,is a function Hilbert space with(8)v,zW22=∑i=01vi0zi0+∫0Tv2tz2tdt,(9)vW22=v,vW221/2,where L20,T denotes the set of square Lebesgue integrable functions on 0,T.Remark 2.
The reproducing kernel functionRy. in W220,T can be written as(10)Ryt=∑i=14biyti−1,t≤y,∑i=14ciyti−1,t>y.
It is easy to prove thatRy. is obtained as follows.
From (8), we have(11)v,Ryw22=∑i=01vi0∂iRy0∂ti+∫0Tv2t∂2Ryt∂t2dt.
Using several integration by part of∫0Tv2t∂2Ryt/∂t2dt, we obtain that(12)v,Ryw22=∑i=01vi0∂iRy0∂ti−−1i∂3−iRy0∂t3−i+∑i=01−11−iviT∂3−iRyT∂t3−i+∫0Tv2t∂2Ryt∂t2dt.
IfRy.∈W220,T, then Ry0=RyT; also if v.∈W220,T, then v0=v1. Therefore,(13)v,Ryw22=∑i=01vi0∂iRy0∂ti−−1i∂3−iRy0∂t3−i+∑i=01−11−iviT∂3−iRyT∂t3−i+∫0Tv2t∂2Ryt∂t2dt+b1v0−v1.
Therefore,Ry. satisfies the following generalized differential equation:(14)∂4Ryt∂t4=δt−y,∂2RyT∂t2=0,Ry0+∂3Ry0∂t3+b1=0,∂Ry0∂t−∂2Ry0∂t2=0,∂3RyT∂t3+b1=0,where δ denotes the Dirac delta function. While y≠t, Ryt is the solution of the constant differential equation:(15)∂4Ryt∂t4=0,with the boundary conditions(16)∂2RyT∂t2=0,Ry0+∂3Ry0∂t3+b1=0,∂Ry0∂t−∂2Ry0∂t2=0,∂3RyT∂t3+b1=0.
The characteristic equation for (15) is λ4=0. Therefore, the general solution can be written as (10), where coefficients biy and ciy,i=1,2,3,4, are obtained by solving the following equations:(17)∂mRyt+0∂tm=∂mRyt−0∂tm,m=0,1,2,∂3Ryt+0∂t3−∂3Ryt−0∂t3=1,∂2RyT∂t2=0,Ry0+∂3Ry0∂t3+b1=0,∂Ry0∂t−∂2Ry0∂t2=0,∂3RyT∂t3+b1=0.
Therefore, the reproducing kernel functionRy. in W220,T is obtained as follows:(18)Ryt=112T23+T36T2−6t3T2+12T3−2t3T3+6t3Ty+12tT3y,+6t2T3y+3t3Ty2−18tT2y2−9t2T2y2−t3y3+6tTy3+3t2Ty3,y≤t,112T23+T36T2+12T3+6t3Ty−18t2T2y+12tT3y+3t3Ty2,−9t2T2y2+6tT3y2−t3y3+6tTy3+3t2Ty3−6T2y3−2T3y3,y>t.Remark 3.
The real value function space(19)W210,T=v|vis absolutely continuous,v′∈L20,T,is a function Hilbert space with inner product(20)v,zW21=v0z0+∫0Tv′tz′tdt.
It can be proved thatW210,T is a reproducing kernel Hilbert space and(21)Ryt=t+1,t≤y,y+1,t>y.


## 3. Solution Procedure (2) by RKHS Method

Here, we will construct a linear differential operator and an orthogonal system inW210,1. After that, the RKHS method for obtaining solution (2) with condition (3) is presented.First, by introducing linear operatorL:W220,T⟶W210,T as(22)Lvt=∑j=0nμjD0+ηjvt,then problem (2) will be converted into the following form:(23)Lvt=gt,vt,0≤t≤T,v0=vT.Theorem 1.
The operatorL is a bounded linear operator.Proof.
It can be easily shown thatL is a linear operator. So, we only prove the boundary of L. From (20), we have(24)LvW212=Lv,LvW21=Lv02+∫01Lv′t2dt.
By reproducing property ofRy., we have(25)vt=vy,RtyW22,Lvt=vy,LRtyW22,Lv′t=vy,LRty′W22.
By Schwarz inequality, we get(26)Lvt=vy,LRtyW22≤LRtW22vW22=M1vW22,Lv′t=uy,LRty′W22≤LRt′W22vW22=M2vW22,where M1,M2>0 are positive constants and the proof is completed.
Thus,Lv02≤M12vW222,Lv′t2≤M22 and ∫01Lv′t2dt≤M22vW222.That is,(27)LvW212=Lv02+∫01Lu′t2dx≤M12+M22vW22,where M=M12+M22>0. We will construct a complete system of Ψi.i=1∞ of W220,T by setting Φit=Ttit and Ψit=L⋆Φit, where tii=1∞ is dense on [0, T] and L⋆:W210,T⟶W220,T is conjugate operator of L.Lemma 1 (see [37]).
Iftii=1∞ is dense on [0, T], then Ψi.i=1∞ is a complete system of W220,T and Ψit=LRyt|y=ti.By using the Gram–Schmidt process ofΨi.i=1∞ is obtained the orthonormal basis Ψ¯i.i=1∞ of space W220,T, which satisfies(28)Ψ¯it=∑k=1iρjkΨkt.The coefficientsρjk are positive and given by(29)ρ11=1Ψ1,ρii=1Ψi2−∑k=1i−1dik2,ρij=−∑k=ji−1dikρkjΨi2−∑k=1i−1dik2,where dik=Ψi,Ψk¯W22.(30)vt=∑i=1∞AiΨ¯it,where Ai=∑k=1iρikgtk,vk−1tk are unknown and we will obtain Ai by using Bi. So, suppose v0t=0 and vnt is given by(31)vnt=∑i=1nBiΨ¯it,where Bi of Ψ¯it is given by(32)B1=ρ11gt1,v0t1,v1t=B1Ψ¯1t,B2=∑k=12ρ2kgtk,vk−1tk,v2t=∑i=12BiΨ¯it,⋮Bn=∑k=1nρnkgtk,vk−1tk.Theorem 2 (see [37]).
Lettii=1∞ be dense set in 0,T and the exact solution v. of (23) in space W220,T be unique, then(33)vt=∑i=1∞∑k=1iρikgtk,vtkΨ¯it,and for this problem, approximate solution nth order as follows:(34)vnt=∑i=1n∑k=1iρikgtk,vtkΨ¯it.Theorem 3.
Ifv∈W220,T, then there exists constant C>0 such that(35)vit≤CvW22,i=0,1.Proof. For each t∈0,T, we obtain(36)vt=vξ,RtξW22≤RtξW22vξW22≤C1vξW22,and we also have(37)v′t=vξ,∂Rtξ∂tW22≤∂Rtξ∂tW22vξW22≤C2vξW22,where C1 and C2 are positive constants and C=maxC1,C2.Corollary 1.
The approximate solutionsvn. and vn′. uniformly converge to the exact solutions v. and v′., respectively.Proof.
From Theorem3, for each t∈0,T, we obtain(38)vnit−vit=vnξ−vξ,∂iRtξ∂tW22,≤∂iRtξ∂tW22vnξ−vξW22,≤Civnξ−vξW22,where Ci,i=0,1 are positive constants. Then, if vnξ⟶.W22vξ as n⟶∞, the approximate solutions vnt and vn′t converge uniformly to vt and v′t, respectively.Remark 4.
We apply the following two cases to solve equations (2) and (3) by using the RKHS method.Case 1.
Let (2) be linear and (30) and (31) denote the exact and approximate solutions, respectively.Case 2.
Let (2) be nonlinear; in this case, the solution of (2) is as follows:We can guarantee thatvnt in equation (37) satisfies condition (3).


## 4. Convergence Analysis

In this section, we will show that approximate solutionvn. of equation (37) is convergent to the exact solution v. of equation (2). First, we express the following lemma.Lemma 2.
Ifvnt⟶.W22vt,tn⟶y,n⟶∞ and gt,y is continuous function with respect to t∈0,T and y∈−∞,∞, then gtn,vn−1tn⟶gy,vyasn⟶∞.Proof.
Observe that(39)vn−1tn−vy=vn−1tn−vn−1y+vn−1y−vy≤vn−1tn−vn−1y+vn−1y−vy.
Reproducing property ofRyξ yields that(40)vn−1tn−vn−1y=vn−1ξ,Rtnξ−RyξW22≤vn−1W22Rtnξ−RyξW22.
From the symmetry ofRyξ, result is Rtnξ−Ryξ⟶0. Therefore, vn−1tn−vn−1y⟶0 as tn⟶y,n⟶∞. From Corollary 1, it holds that vn−1y−vy⟶0asn⟶∞. Then, vn−1tn⟶.W22vyasn⟶∞. Because g. is continuous functions, then gtn,vn−1tn⟶gy,vyasn⟶∞.Lemma 3.
Forvnt in equation (37), we have(41)Lvntj=Lvtj=gtj,vj−1tj.Proof. Suppose j≤n, therefore,(42)Lvntj=∑i=1nBiLΨ¯itj=∑i=1nBiLsΨ¯it,ΦjtW21=∑i=1nBiΨ¯it,L∗ΦjtW22=∑i=1nBiΨ¯it,ΨjtW22.
By using orthogonality ofΨ¯iti=1∞, we obtain(43)∑l=1jρjlLvntl=∑i=1nBiΨ¯it,∑l=1jρjlΨltW22,=∑i=1nBiΨ¯it,Ψ¯jtW22,=Bj=∑l=1jρjlgtl,vl−1tl.
Ifj=1, then(44)Lvnt1=gt1,v0t1.
Besides ifj=2, then(45)ρ21Lvnt1+ρ22Lvnt2=ρ21gt1,v0t1+ρ22gt2,v1t2,that is, Lvnt2=gt2,v1t2. By the same manner, it yields that(46)Lvntj=gtj,vj−1tj.
Hence,vt=∑i=1∞Biψ¯it is obtained by taking the limit of equation (37). Therefore, vnt=Pnvt,Pn is an orthogonal projector of W220,1 to span Ψ1,Ψ2,…,Ψn. Then,(47)Lvntj=Lvnt,ΦjtW21=vnt,L∗ΦjtW22=Pnvt,ΨjtW22=vt,PnΨjtW22=vt,ΨjtW22=vt,L∗ΦjtW22=Lvt,ΦjtW21=Lvtj.Theorem 4.
LetvnW22 be bounded and tii=1∞ is dense on 0,T, then n-term approximate solutions vnt in equation (37) converge to(48)vt=∑i=1∞∑k=1iρikgtk,vk−1Ψ¯it.Proof.
Firstly, we prove that the convergence ofvnn=1∞ in equation (37) is convergent in the sense of .W22. From equation (37), it is inferred that vn+1t=vnt+Bn+1Ψ¯n+1t. Since Ψ¯ii=1∞ is orthogonal, hence,(49)vn+1W222=vnW222+Bn+12=⋯=∑i=1n+1Bi2,where Bi=∑k=1iρikgtk,vk−1tk. It holds that vnW222≤vn+1W222.
BecausevnW222 is bounded, vnW22 is convergent as n⟶∞. Therefore, there exists constant c such that ∑i=1∞Bi2=c. The implies that Bi∈l2=B|∑i=1∞Bi2<∞.
Letm>n, then from the orthogonality of vn+1t−vnt, it follows that(50)vm−vnW222=vm−vm−1+vm−1−⋯+vn+1−vnW222,=vm−um−1W222+⋯+vn+1−vnW222,because vm−vm−1W222=Bm2. Consequently,(51)vm−vnW222=∑i=n+1mBi2⟶0asn,m⟶∞.
Hence,W220,T is complete, and then vnt⟶.W22vt, as n⟶∞.
Now, we prove thatvt is the solution of equation (23). Because tii=1∞ is dense on 0,T, for each t∈0,T, there exists a subsequence tnj such that tnj⟶t as j⟶∞.
Sincetjj=1∞ is dense on 0,T, thus for all t∈0,T, there exists a subsequence tnj such that tnj⟶t, as i⟶∞. By Lemma 3, it follows that Lvntnj=gtnj,vnj−1tk. Hence, let j⟶∞, from Lemma 2 and the continuity of g, we have Lvt=gt,vt.Theorem 5.
Assumern=vt−vntW222, where vnt is derived from the RKHS method. Therefore, rn is decreasing in .W22.Proof.
Note that(52)rn2=vt−untW222=∑j=n+1∞vt,Ψ¯jtW22Ψ¯t2=∑j=n+1∞vt,Ψ¯jtW222,rn−12=vt−vn−1tW222=∑j=n∞vt,Ψ¯jtW22Ψ¯jtW222=∑j=n∞vt,Ψ¯jtW222,therefore, rntW22≤rn−1tW22.


## 5. Numerical Tests

We provide three examples to explain the content given, and we realize the validity and accuracy of the RKHS method.Example 1.
In this example, we consider FDE with periodic condition:(53)D0+0.4vt+D0+0.3vt+v2tevt=Γ4Γ3.6t2.6−2Γ3Γ2.6t1.6+Γ2Γ1.6t0.6+Γ4Γ3.7t2.7−2Γ3Γ2.7t1.7+Γ2Γ1.7t0.7+t3−2t2+t2et3−2t2+t,v0=v1,for t∈0,1. The exact solution is vt=t3−2t2+t. We choose 22 points in 0,1, and by using the proposed method, the approximate solution v22 is obtained. Absolute error values are reported in Table 1 for T=1 and ti=i/n,i=1,2,…,n. The graphs of the absolute error and the numerical solution are plotted in Figure 1. Here,(54)L=D0+0.4vt+D0+0.3vt,gt,vt=Γ4Γ3.6t2.6−2Γ3Γ2.6t1.6+Γ2Γ1.6t0.6+Γ4Γ3.7t2.7−2Γ3Γ2.7t1.7+Γ2Γ1.7t0.7+t3−2t2+t2et3−2t2+t−v2tevt,and(55)Ψit=LRyt|y=ti=D0+0.4Rtit+D0+0.3Rtit,finally vt and vnt are obtained from (36) and (37), respectively.Table 1
The absolute error for Example1 with n=22.
tivtv22tAbsolute error0.10.0810.08088250.0002238370.20.1280.1278620.0001723470.30.1470.146840.0001576140.40.1440.1438280.0001502390.50.1250.124830.0001433590.60.0960.09584220.0001348680.70.0630.06285960.0001245060.80.0320.03187940.0001126940.90.0090.008902750.000099706410−1.55353×10−151.55353×10−15Figure 1
Graphs of numerical solution and absolute error withn=22 for Example 1.Example 2.
We consider FDE with periodic condition:(56)D0+0.6vt+D0+0.5vt+D0+0.2vt+sinhvt1−t=Γ3Γ2.4t1.4−Γ2Γ1.4t0.4+Γ3Γ2.5t1.5−Γ2Γ1.5t0.5+Γ3Γ2.8t1.8−Γ2Γ1.8t0.8+sinht2−t1−t,v0=v1,for t∈0,1. The exact solution is vt=t2−t. We choose 20 points in 0,1, and by using the proposed method, we obtained the approximate solution v20. Absolute error values are reported in Table 2 for T=1 and ti=i/n,i=1,2,…,n. The graphs of the absolute error and the numerical solution are plotted in Figure 2.Table 2
The absolute error for Example2 with n=20.
tivtv20tAbsolute error0.1−0.09−0.0922470.002246960.2−0.16−0.1611320.001132370.3−0.21−0.2108520.0008520890.4−0.24−0.2407140.0007137470.5−0.25−0.2505210.000621490.6−0.24−0.2405490.0005491030.7−0.21−0.2104850.0004854260.8−0.16−0.1604240.0004235690.9−0.09−0.09035080.00035081410−7.51086×10−167.51086×10−16Figure 2
Graphs of numerical solution and absolute error withn=20 for Example 2.Example 3.
We consider FDE with periodic condition(57)D0+η1vt+D0+η0vt=Γ3Γ3−η1t2−η1−2Γ2Γ2−η1t1−η1+Γ3Γ3−η0t2−η0−2Γ2Γ2−η0t1−η0,v0=v2,for t∈0,2. The exact solution is vt=t2−2t. Using this method, taking ti=i/n,i=1,2,…,n, for n=10 and T=2, the numerical results are given in Table 3 for η0=0.2andη1=0.5. Three graphs of the approximate solution for η1andη0 are drawn in Figure 3.Table 3
Numerical solution for Example3 with n=10.
tivtv10tη0=0.4,η1=0.5η0=0.5,η1=0.6Absolute error0.2−0.36−0.351811−0.350017−0.345990.008189450.4−0.64−0.637683−0.636476−0.6338140.002317020.6−0.84−0.837922−0.836963−0.8349290.002078290.8−0.96−0.958537−0.95775−0.9560960.00146331−1−0.998869−0.998235−0.996920.001131051.2−0.96−0.959148−0.958648−0.9576240.0008521811.4−0.84−0.839392−0.839020.838270.000608331.6−0.64−0.639624−0.639379−0.6389010.0003763931.8−0.36−0.359955−0.359858−0.3596760.0000452721203.3158×10−141.37181×10−143.4689×10−143.3158×10−14Figure 3
Graph of numerical solution withn=10 for Example 3.


## 6. Conclusion

In this paper, we have proposed the RKHS method to solve fractional differential equations with periodic conditions. This method is a powerful technique for finding approximate solutions. The approximation error and convergence analysis are obtained in the RKHS. We illustrate the efficiency of the method with a few examples.



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