# The Solution of Initial Boundary Value Problem with Time and Space-Fractional Diffusion Equation via a Novel Inner Product

Authors: Hulya Kodal Sevindir; Ali Demir
Journal: Advances in Mathematical Physics
 (2018)

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## Abstract

The main goal of this study is to find the solution of initial boundary value problem for the one-dimensional time and space-fractional diffusion equation which is a very intriguing topic for many researchers. With the aim of newly defined inner product, which is the main contribution of this study, the analytic solution of the boundary value problem is obtained. The time and space-fractional derivatives are defined in the Caputo sense which is more suitable than Riemann-Liouville sense. We apply the separation of variables method to reduce the problem to two separate fractional ODEs. The generalized solution is constructed/formed in the form of a Fourier series with respect to the eigenfunctions of a certain eigenvalue problem. In order to obtain the coefficients of the Fourier series for the solution, we define a new inner product which is the key point of study.


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## Body

## 1. Introduction

Since PDE of fractional order contributes to modeling for the wide range of processes and systems, including past memories, in various scientific research areas, it has become very intriguing topic for many scientists. Main reason of this trend is that using fractional derivative is global in nature whereas the integer derivative is local in nature. This property makes fractional DEs the best possible choice in modeling physical problems involving past memory and/or delay and attracts growing number of researchers.The Caputo derivatives are more useful than Riemann-Liouville derivatives since the analysis of the mathematical models involving Caputo derivatives gives closer results to the analysis of ones including integer derivatives. There are various studies in literature supporting this conclusion [1–8]. By making use of Mittag–Leffler function, characteristic equations of fractional ODEs are solved and solutions of them are constructed efficiently. In this sense, the Mittag–Leffler function takes the role of the exponential function which is used in the determination of solutions for ODEs with integer derivatives.


## 2. Preliminary Results

In this section, some basic definitions and known results regarding Caputo fractional derivatives are given.Definition 1.
The Caputo fractional derivative ofut of order q where n-1<q<n is given by equation (1)cDqut=1Γn-q∫t0tt-sn-q-1unsds,t∈t0,t0+Twhere unt=dnu/dtn. If q is an integer, then the above definition of fractional derivative coincides with the integer derivative.Definition 2.
If0<q<1, the Caputo fractional derivative of order q can be defined by equation(2)cDqut=1Γ1-q∫t0tt-s-qu′sdswhere 0<q<1.
To determine the solution of eigenvalue problem in the following section, we use the two-parameter Mittag–Leffler function defined as(3)Eα,βλt-t0α=∑k=0∞λt-t0αkΓαk+β,α,β>0where λ is a constant. In particular, for t0=0,α=β=q we obtain (4)Eq,qλtq=∑k=0∞λtqkΓqk+q,q>0.When q=1, Mittag–Leffler function corresponds to usual exponential function, i.e., E1,1λt=eλt. For further reading see [9, 10].
Let us define the following functions which play an important role in the representation of solution for the problem under consideration:(5)sinq⁡μtq=Eq,1iμtq-Eq,1-iμtq2i=∑k=0∞-1kμtq2k+1Γ2k+1q+1and(6)cosq⁡μtq=Eq,1iμtq+Eq,1-iμtq2=∑k=0∞-1kμtq2kΓ2kq+1.Note that for q=1 these functions are usual trigonometric functions sinμt and cosμt.


## 3. Novel Inner Product

In this section, we define a new inner product to determine the solution which satisfies the initial boundary value problem. The missing of such an inner product makes the space-time fractional initial and boundary value problems open problems. This inner product promotes these kinds of problems to be studied theoretically. Moreover, it provides the best understanding possible to develop the theoretical study of space-time fractional initial boundary value problems. By this inner product, the maximum principle for time-space-fractional PDE will be discussed and investigated for a wide range of problems. In the definition of this inner product, we use the functions which satisfy the initial and boundary value problems with integer orders which make the calculations much more easier.Theorem 3.
LetV be a vector space, made up of all linear combinations of sinβ⁡μ(x/b-a)β and cosβ⁡μ(x/b-a)β for fixed β where 0<β≤1,μ∈R on the interval I=a,b, i.e., V=span{sinβ⁡μ(x/b-a)β,cosβ⁡μ(x/b-a)β}. Let T:V→span{⁡sin⁡μx/b-a,cos⁡(μx/b-a)} be a linear transformation which is one-to-one and onto. Hence it has its inverse transformation T-1. The mapping <·,·>:V×V→R is defined as (7)ux;β,vx;β=T-1∫Tux;β.Tvx;βdxx=abwhere Tux;β=u(x;1) and Tvx;β=v(x;1). In other words,(8)ux;β,vx;β=∫ux;1.vx;1dx=wx,1.At this point we transform the function wx,1 to the corresponding function wx,β. Replacing β back and substituting upper and lower values give us the result of the new inner product.(9)ux;β,vx;β=wx,βx=ax=b=wb,β-wa,β.Proof.
Let us show that this mapping satisfies the conditions of an inner product:(1)
<ux;β,ux;β>≥0<ux;β,ux;β>=∫u2x;1dx=wx,1<ux;β,ux;β>=wx,βx=ax=b=wb,β-w(a,β)≥0sincewb,1-w(a,1)≥0 and -1≤sinqx≤1,-1≤cosqx≤1,∀q∈0,1.Moreover<ux;β,ux;β>=0⇔ux;β=0.(2)
<ux;β,vx;β>=∫ux;1vx;1dx=∫vx;1ux;1dx=wx,1<ux;β,vx;β>=wx,βx=ax=b=wb,β-wa,β=<vx;β,ux;β>.(3)
<ux;β+vx;β,z(x;β)>=∫ux;1+vx;1z(x;1)dx=∫ux;1zx;1dx+∫vx;1zx;1dx<ux;β+vx;β,z(x;β)>=w1x,βx=ab+w2x,βx=ab=w1b,β-w1a,β+w2b,β-w2a,β=<ux;β,zx;β>+<vx;β,z(x;β)>.(4)
<αux;β,vx;β>=∫αux;1vx;1dx=α∫ux;1vx;1dx=αwx,1<αux;β,vx;β>=αwx,βx=ab=α(wb,β-wa,β)=α<ux;β,vx;β>.Example 4.
Let us illustrate this inner product by an example in which we take the inner product of Mittag–Leffler functionsEβλ1(βxβ) and Eβλ2(βxβ⁡):(10)Eβλ1βxβ,Eβλ2βxβ=∫eλ11xeλ21xdx=∫eλ11+λ21xdx=eλ11+λ21xλ11+λ21Eβλ1βxβ,Eβλ2βxβ=Eβλ1β+λ2βxβλ1β+λ2βx=ab=Eβλ1β+λ2βbβ-Eβλ1β+λ2βaβλ1β+λ2βNote that this inner product extends orthogonality property between the functions ux;1 and vx;1 to the functions ux;β and vx;β.


## 4. Main Results

In this section, we deal with the following initial boundary value problem involving time and space-fractional PDE:(11)Dtαux,t=aDx2β,0<α<1,1<2β<2,a>0(12)ux,0=uox,0≤x≤l(13)u0,t=0,ul,t=00≤t≤T.The generalized solution of this problem can be obtained in an analytical form by using the separation of variables method. Let a particular solution of (11)-(13) be in the following form:(14)ux,t;α,β=Xx;βTt;α,βwhere 0≤x≤l, 0≤t≤T.Note that the functionX depends on order of fractional derivative with respect to x and function T depends on orders of fractional derivatives with respect to x and t. Substituting (14) into (11) and applying separation of the variables lead to the equation (15)DαTt;α,βaTt;α,β=D2βXx;βXx;β=-λβwhere λ depends on order of the fractional derivative with respect to x. The equations in (15) reduce the problem to two separate ODEs including fractional derivatives with respect to both time and space. The equation on the right with boundary conditions (13) gives the following fractional differential equation:(16)D2βXx;β+λβXx;β=0(17)X0;β=0,Xl;β=0In order to determine the solution of the eigenvalue problem ((16)-(17)) we use the following Mittag–Leffler function:(18)Xx;β=Eβ,1rxβwhich leads to the following characteristic equation for the eigenvalue problem ((16)-(17)):(19)r2+λβ=0.Case 1.
Ifλ<0, we have two real and distinct solutions r=±-λ(β), and the general solution of the eigenvalue problem ((16)-(17)) is represented by (20)Xx;β=c1Eβ,1-λxβ+c2Eβ,1--λxβX0;β=0⟹c1Eβ,10+c2Eβ,10=0⟹c2=-c1which leads to the following solution:(21)Xx;β=c1Eβ,1-λxβ-Eβ,1--λxβ.From the second boundary condition we have (22)Xl;β=0⟹Xl;β=c1Eβ,1-λlβ-Eβ,1--λlβ=0which implies that(23)Eβ,1-λlβ=Eβ,1--λlβ.Thus(24)-λlβ=0⟹λ=0Xx;β=c1Eβ,10-Eβ,10=0which means that there is no solution for the case λ<0.Case 2.
Ifλ=0, the characteristic equation becomes (25)r2=0⟹r1,2=0.Hence the general solution of the eigenvalue problem ((16)-(17)) is represented by(26)Xx;β=c1+c2xββFrom the first boundary condition we have(27)X0;β=0⟹c1=0which leads to the following solution: (28)Xx;β=c2xββ.From the last boundary condition we have(29)Xl;β=0⟹c2xββ=0⟹c2=0which leads to Xx;β=0 which means that there is no solution for λ=0.Case 3.
Ifλ>0, we obtain two complex roots r=∓iλ. Then the general solution of the eigenvalue problem ((16)-(17)) is represented by (30)Xx;β=c1sinβ⁡λxβ+c2cosβ⁡λxβ.Let us check if the boundary conditions are satisfied by the solution. From the first boundary condition we have (31)X0;β=0⟹c1sinβ0+c2cosβ0=0⟹c2=0which gives the following solution:(32)Xx;β=c1sinβ⁡λxβFrom the second boundary condition we have (33)Xl;β=0⟹Xl=c1sinβ⁡λlβ=0which implies that(34)sinβ⁡λlβ=0.Let wnβ=λn(β)lβ. Hence the eigenvalues can be represented in terms of wnβ as follows:(35)λnβ=wn2βl2β,0<w1β<w2β<⋯Thus the solution of the eigenvalue problem is represented in the following form:(36)Xnx;β=cnsinβ⁡wnβxlβ.The equation on the left of (15) for each eigenvalue λn gives the following fractional differential equation:(37)DαTnt;α,β+awn2βl2βTnt;α,β=0By using the similar calculations the solution of (37) is determined in the following form:(38)Tnt;α,β=cnEα-awn2βl2βtα.For each eigenvalue λn we obtain the following function:(39)unx,t;α,β=Xnx;βTnt;α,β=cnsinβ⁡wnβxlβEα-awn2βl2βtαand hence we have the following sum: (40)ux,t;α,β=∑n=1∞unx,t;α,β=∑n=1∞cnsinβ⁡wnβxlβEα-awn2βl2βtαwhich satisfy both the fractional equation (11) and boundary condition (13).
In order to determine the solution which satisfies the initial condition (12) we use the inner product defined in (7).
In (38), replacing t by 0 and using the initial condition (12) we have(41)ux,0;α,β=u0x;α,β=∑n=1∞unx,0;α,β=∑n=1∞cnsinβ⁡wnβxlβ.Via the inner product (7) we will obtain the coefficients cn for n=1,2,3,… as follows:(42)cn=u0x;α,β,sinβ⁡wnβxlβ,n=1,2,3,…


## 5. Illustrative Example

Before solving fractional heat-like problem we first solve the following problem forα=β=1 in order to compare the solutions of these two problems:(43)Dtux,t=14Dx2ux,t,0≤x≤2,t>0ux,0=-sin⁡πx,0≤x≤2u0,t=0,u2,t=00≤t≤TThe solution of this problem is obtained as follows:(44)ux,t=-sin⁡πxexp⁡-π24t.Let us consider the following fractional heat-like problem:(45)Dtαux,t=14Dx2βux,t,0<α<1,1<2β<2,(46)ux,0=-sin⁡πx,0≤x≤2(47)u0,t=0,u2,t=00≤t≤TApplying separation of the variables to (45) leads to the equation (48)4DαTt;α,βTt;α,β=D2βXx;βXx;β=-λβwhere λ depends on β. Hence we get two ODEs having fractional derivatives with respect to time and space separately. The equation on the right of (48) with boundary conditions (47) gives the fractional differential equation (49)D2βXx;β+λXx;β=0(50)X0;β=0,X2;β=0.Using the Mittag–Leffler function Xx;β=Eβ,1(rxβ) we obtain the following characteristic equation r2+μ2(β)=0 where we take λ=μ2. Hence the solution becomes(51)Xx;β=c1sinβ⁡μxβ+c2cosβ⁡μxβ.Using boundary conditions (52)X0;β=0⟹c1sinβ0+c2cosβ0=0⟹c2=0which gives the following solution:(53)Xx;β=c1sinβ⁡μxβ.From the second boundary condition we have (54)X2;β=0⟹X2;β=c1sinβ⁡μ2β=0which implies that(55)sinβ⁡μ2β=0.Let wnβ=μn2β, i.e., sinβ⁡wn=0. Hence the eigenvalues can be represented in terms of wnβ as follows:(56)λnβ=wn2β22β,0<w1β<w2β<⋯Thus the solution of the eigenvalue problem is represented in the following form:(57)Xnx;β=cnsinβ⁡wnβx2β.The equation on the left of (48) for each eigenvalue λn(β) gives the following fractional differential equation:(58)DαTnt;α,β+wn2β22β+2Tnt;α,β=0.By using the similar calculations the solution of (58) is determined in the following form:(59)Tnt;α,β=Eα-wnβ2β+12tα.For each eigenvalue λn we obtain the following function:(60)unx,t;α,β=Xnx;βTnt;α,β=cnsinβ⁡wnβx2βEα-wnβ2β+12tαand thus we have the following sum: (61)ux,t;α,β=∑n=1∞unx,t;α,β=∑n=1∞cnsinβ⁡wnβx2βEα-wnβ2β+12tαwhich satisfy both (45) and boundary condition (47). In order to determine the solution which satisfies the initial condition (46), (62)ux,0;α,β=∑n=1∞unx,0;α,β=∑n=1∞cnsinβ⁡wnβx2β=-sinπx.Via the inner product we obtain the coefficients cn for n=1,2,3,… as follows:(63)cn=-sinπx,sinβ⁡wnβx2β,n=1,2,3,…(64)cn=-∫sin⁡πxsin⁡nπx2dx,n≠2=1n-2πsin⁡n-2πx2-1n+2πsin⁡n+2πx2.Replacing β back and substituting upper and lower limits give us the result of the new inner product.(65)cn=1wn-2sinβ⁡wn-2βx2β-1wn+2sinβ⁡wn+2βx2βx=0x=2=1wn-2sinβ⁡wn-2β-1wn+2sinβ⁡wn+2βThus cn=0,n≠2.Forn=2 we get (66)c2=-sin⁡πx,sinβ⁡w2βx2β,n=1,2,3,…=-∫sin2⁡πxdx=-12x+sin⁡2πx4π=-12xβ+⁡sinβ⁡w4βx/2βw4βx=0x=2=-2β-1.Thus(67)ux,t;α,β=∑n=1∞unx,t;α,β=-2β-1sinβ⁡w2βx2βEα-w2β2β+12tα.Notice that as β and α go to 1, solution (67) goes to the solution (44).


## 6. Conclusion

In this paper, we establish a new inner product which plays the remarkable role in the determination of the solution for the one-dimensional time and space-fractional diffusion problem which is an open problem. Moreover this inner product can be applied in many related problems and brings new insights into the understanding of time and space-fractional problems.



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