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International Journal of Differential Equations
Volume 2011, Article ID 268465, 10 pages
Research Article

Boundary Value Problems with Integral Gluing Conditions for Fractional-Order Mixed-Type Equation

1Kazakh National Pedagogical University named after Abai, 050010 Almaty, Kazakhstan
2University of Santiago de Compostela, 15782 Santiago de Compostela, Spain

Received 18 May 2011; Accepted 11 October 2011

Academic Editor: Shaher M. Momani

Copyright © 2011 A. S. Berdyshev et al. This is an open access article distributed under the Creative Commons Attribution License, which permits unrestricted use, distribution, and reproduction in any medium, provided the original work is properly cited.


Analogs of the Tricomi and the Gellerstedt problems with integral gluing conditions for mixed parabolic-hyperbolic equation with parameter have been considered. The considered mixed-type equation consists of fractional diffusion and telegraph equation. The Tricomi problem is equivalently reduced to the second-kind Volterra integral equation, which is uniquely solvable. The uniqueness of the Gellerstedt problem is proven by energy integrals' method and the existence by reducing it to the ordinary differential equations. The method of Green functions and properties of integral-differential operators have been used.

1. Introduction

Mathematical model of the movement of gas in a channel surrounded by a porous environment was described by parabolic-hyperbolic equation. This was done in the fundamental work of Gel’fand [1]. Modeling of heat transfer processes in composite environment with finite and infinite velocities leads to boundary value problems (BVPs) for parabolic-hyperbolic equations [2]. Omitting the huge amount of works devoted to studying these kinds of equations, we refer the readers to [3, 4].

We would like to note works [510], devoted to the studying of BVPs for parabolic-hyperbolic equations, involving fractional derivatives. In turn, applications of Fractional-order differential equations can be found in the monographs [1115]. We also note some recent papers [1618], related to the fractional diffusion and diffusion-wave equations.

BVP for parabolic-hyperbolic equations with integral gluing condition for the first time was investigated by Kapustin and Moiseev [19] and was generalized for this kind of equation, but with parameters, in the work [20]. Another motivation of the usage of integral gluing conditions comes from the appearance of them in heat exchange processes [21].

The consideration of equations with parameters was interesting because of the possibility of studying some multidimensional analogues of the main BVP via reducing them by Fourier transformation to the BVP for equations with parameters. On the other hand, consideration of equations with parameters will give possibility to study some spectral properties of BVPs for this kind of equations such as the existence of nontrivial solutions for corresponding homogeneous problem at some values of parameters [22].

2. Analog of the Tricomi Problem

Consider an equation𝑢𝑥𝑥𝐷𝛼𝐻(𝑥)+2𝐻(𝑥)0𝑦𝑢=𝜆𝑢(2.1) in the domain Ω=Ω1𝐴𝐴0Ω2. Here Ω1={(𝑥,𝑦)0<𝑥<1,0<𝑦<1}, Ω2 is characteristic triangle with endpoints 𝐴(0,0), 𝐴0(0,1), 𝐶(1/2,1/2), 𝐻(𝑥) is Heaviside function, 𝐷𝛼𝑎𝑡1𝑓(𝑡)=𝑑Γ(𝑛𝛼)𝑑𝑡𝑛𝑡𝑎(𝑡𝑠)𝛼+𝑛1𝑓(𝑠)𝑑𝑠(2.2) is the 𝛼th Riemann-Liouville fractional-order derivative of a function 𝑓 given on interval [𝑎,𝑏], where 𝑛=[𝛼]+1 and [𝛼] is the integer part of 𝛼, and Γ() is the Euler gamma function defined byΓ(𝛼)=0𝑡𝛼1𝑒𝑡𝑑𝑡,𝛼>0.(2.3) For 𝜆>0 and 0<𝛼1 given, we formulate the following problem called the analog of the Tricomi problem.

Problem AT
To find a solution of (2.1), which belongs to the class of functions 𝑊1=𝑢𝐷𝛼10𝑦𝑢𝐶Ω1,𝑢𝑥𝑥,𝐷𝛼0𝑦Ω𝑢𝐶1,𝑢𝑥0±,𝑦𝐻(0;1),𝑢𝐶Ω2𝐶2Ω2,(2.4) satisfying the initial condition lim𝑦0𝑦1𝛼𝑢(𝑥,𝑦)=𝜔(𝑥),0𝑥1(2.5) together with the boundary conditions 𝑢(𝑦/2,𝑦/2)=𝜓1𝑢(𝑦),0𝑦1,(1,𝑦)=𝜓2(𝑦),0𝑦1,(2.6) and the gluing conditions 𝑢(01,𝑦)=Γ(1𝛼)𝑦0𝑢0+,𝑡(𝑦𝑡)𝛼𝑑𝑡,0<𝑦1,𝑦0𝑢𝑥(0,𝑡)𝐽01𝜆(𝑦𝑡)𝑑𝑡=Γ(1𝛼)𝑦0𝑢𝑥0+,𝑡(𝑦𝑡)𝛼𝑑𝑡,0<𝑦<1.(2.7) Here 𝜔(𝑥), 𝜓𝑖(𝑦) (𝑖=1,2) are given functions such as limy0𝑦1𝛼𝜓1(𝑦)=𝜔(0).
Solution of the Cauchy problem for (2.1) in Ω2 defined as 1𝑢(𝑥,𝑦)=2𝜏(𝑦+𝑥)+𝜏(𝑦𝑥)+𝑦+𝑥𝑦𝑥𝜈(𝑡)𝐽0𝜆(𝑦𝑡)2𝑥2𝑑𝑡+𝜆𝑥𝑦+𝑥𝑦𝑥𝜏𝐽(𝑡)1𝜆(𝑦𝑡)2𝑥2𝜆(𝑦𝑡)2𝑥2,𝑑𝑡(2.8) where 𝐽𝑘[] is the first-kind Bessel function of the order 𝑘, 𝜏(𝑦)=𝑢(0,𝑦), 𝜈(𝑦)=𝑢𝑥(0,𝑦).
We calculate 𝑢(𝑦/2,𝑦/2) in order to use condition (2.5): =1𝑢(𝑦/2,𝑦/2)2𝜏(0)+𝜏(𝑦)𝑦0𝜈(𝑡)𝐽0𝑦𝜆𝑡(𝑡𝑦)𝑑𝑡+𝜆2𝑦0𝜏𝐽(𝑡)1𝜆𝑡(𝑡𝑦).𝜆𝑡(𝑡𝑦)𝑑𝑡(2.9) Considering the condition (2.5) and the following integral operator [23] 𝐵𝑛,𝜆𝑚𝑥[]𝑓(𝑥)=𝑓(𝑥)+𝑥𝑚𝑓(𝑡)𝑥𝑚𝑡𝑚1𝑛𝜕𝐽𝜕𝑥0𝜆(𝑡𝑚)(𝑡𝑥)𝑑𝑡,𝑚,𝑛=0,1,(2.10) equality (2.9) can be written as follows 𝜓11(𝑦)=2𝜓1(0)+𝐵0,𝜆0𝑦[𝜏](𝑦)𝑦0𝐵1,𝜆0𝑡[𝜈](𝑡)𝑑𝑡.(2.11) Now we use an integral operator 𝐴𝑛,𝜆𝑚𝑥[]𝑓(𝑥)=𝑓(𝑥)𝑥𝑚𝑓(𝑡)𝑡𝑚𝑥𝑚𝑛𝜕𝐽𝜕𝑡0𝜆(𝑥𝑚)(𝑥𝑡)𝑑𝑡,𝑚,𝑛=0,1,(2.12) which is mutually inverse with the operator (2.10). Applying the operator (2.12) to both sides of (2.11), we obtain 𝐴0,𝜆0𝑦𝜓1=1(𝑦)2𝜓1(0)+𝐴0,𝜆0𝑦𝐵0,𝜆0𝑦[𝜏](𝑦)𝐴0,𝜆0𝑦𝑦0𝐵1,𝜆0𝑡[𝜈](𝑡)𝑑𝑡.(2.13) Considering the following properties of operators (2.10) and (2.12) 𝐴0,𝜆0𝑦𝐵0,𝜆0𝑦[]𝑓(𝑦)=𝑓(𝑦),𝐴0,𝜆0𝑦𝑦0𝐵1,𝜆0𝑡[]=𝑓(𝑡)𝑑𝑡𝑦0𝑓(𝑡)𝐽0𝜆(𝑦𝑡)𝑑𝑡,(2.14) we derive 2𝐴0,𝜆0𝑦𝜓1(𝑦)=𝜓1(0)+𝜏(𝑦)𝑦0𝜈(𝑡)𝐽0𝜆(𝑦𝑡)𝑑𝑡.(2.15) Taking gluing conditions (2.7) into account, we have 𝐷𝛼10𝑦𝜈+(𝑦)=𝐷𝛼10𝑦𝜏+(𝑦)2𝐴0,𝜆0𝑦𝜓1(𝑦)+𝜓1(0).(2.16)
Applying operator 𝐷1𝛼0𝑦 to both sides of (2.16) and considering the following composition rule [11]: 𝐷𝛼𝑎𝑡𝐷𝛽𝑎𝑡𝑓(𝑡)=𝐷𝛼+𝛽𝑎𝑡𝑓(𝑡),𝛽0,(2.17) we get 𝜏+(𝑦)=𝜈+(𝑦)+𝜓1(𝑦),0<𝑦<1,(2.18) where 𝜓1(𝑦)=𝐷1𝛼0𝑦{2𝐴0,𝜆0𝑦[𝜓1(𝑦)]𝜓1(0)}.
Let us consider the following auxiliary problem: 𝑢𝑥𝑥𝐷𝛼0𝑦𝑢𝑢𝜆𝑢=0,𝑥(0,𝑦)=𝜈+(𝑦),𝑢(1,𝑦)=𝜓2(𝑦),lim𝑦0𝑦1𝛼𝑢(𝑥,𝑦)=𝜔(𝑥).(2.19) Solution of this problem can be defined as [24] 𝑢(𝑥,𝑦)=10𝜔(𝜉)𝐺(𝑥,𝑦,𝜉,0)𝑑𝜉𝑦0𝜈++(𝜂)𝐺(𝑥,𝑦,0,𝜂)𝑑𝜂𝑦0𝜓2(𝜂)𝐺𝜉(𝑥,𝑦,1,𝜂)𝑑𝜂𝜆10𝑦0𝑢(𝜉,𝜂)𝐺(𝑥,𝑦,𝜉,𝜂)𝑑𝜉𝑑𝜂,(2.20) where 𝐺(𝑥,𝑦,𝜉,𝜂)=(𝑦𝜂)𝛽12𝑛=𝑒1,𝛽1,𝛽||||𝑥𝜉+2𝑛(𝑦𝜂)𝛽+𝑒1,𝛽1,𝛽||||𝑥+𝜉+2𝑛(𝑦𝜂)𝛽(2.21) is the Green function of the problem (2.19), 𝑒1,𝛽1,𝛽(𝑧)=Φ(𝛽,𝛽,𝑧)=𝑛=0𝑧𝑛𝑛!Γ(𝛽𝑛+𝛽)(2.22) is the function of Wright [25], 𝛽=𝛼/2.
Considering (2.20) as an integral equation regarding the function 𝑢(𝑥,𝑦), we write solution via resolvent of the kernel 𝜆𝐺(𝑥,𝑦,𝜉,𝜂): 𝑢(𝑥,𝑦)=𝑃(𝑥,𝑦)𝑦0𝜈+(𝜂)𝐾1(𝑥,𝑦,𝜂)𝑑𝜂,(2.23) where 𝑃(𝑥,𝑦)=10𝜔(𝜉)𝐺(𝑥,𝑦,𝜉,0)𝑑𝜉+𝑦01010+𝜔(𝜉)𝐺(𝑠,𝑡,𝜉,0)𝑅(𝑥,𝑦,𝜉,0)𝑑𝜉𝑑𝑠𝑑𝑡𝑦0𝜓2𝐺(𝜂)(𝑥,𝑦,1,𝜂)+𝑦𝜂10𝐺𝐾(𝑠,𝑡,1,𝜂)𝑅(𝑥,𝑦,1,𝜂)𝑑𝑠𝑑𝑡𝑑𝜂,1(𝑥,𝑦,𝜂)=𝐺(𝑥,𝑦,0,𝜂)+𝑦𝜂10𝐺(𝑠,𝑡,0,𝜂)𝑅(𝑥,𝑦,0,𝜂)𝑑𝑠𝑑𝑡,(2.24)𝑅(𝑥,𝑦,𝜉,𝜂) is a resolvent of the kernel 𝜆𝐺(𝑥,𝑦,𝜉,𝜂).
From (2.23), tending 𝑥 to 0+, we obtain 𝑢0+,𝑦=𝜏+0(𝑦)=𝑃+,𝑦𝑦0𝜈(𝜂)𝐾10+,𝑦,𝜂𝑑𝜂.(2.25) Considering functional relation (2.18), from (2.25) we get 𝜈+(𝑦)+𝑦0𝜈+(𝜂)𝐾1(𝑦,𝜂)𝑑𝜂=𝜓1(𝑦)𝑃(0,𝑦).(2.26) Equality (2.26) is the second-kind Volterra-type integral equation regarding the function 𝜈+(𝑦). Since kernel 𝐾1(𝑦,𝜂) has weak singularity and functions on the right-hand side are continuous, we can conclude that (2.26) is uniquely solvable [26], and solution can be represented as 𝜈+(𝑦)=Ψ(𝑦)+𝑦0Ψ(𝜂)𝐾2(𝑦,𝜂)𝑑𝜂,(2.27) where Ψ(𝑦)=𝜓1(𝑦)𝑃(0,𝑦), 𝐾2(𝑦,𝜂) is the resolvent of the kernel 𝐾1(𝑦,𝜂).
Once we have obtained 𝜈+(𝑦), considering (2.18) or (2.25) we find function 𝜏+(𝑦). Then using gluing conditions (2.7) we find functions 𝜏(y), 𝜈(y). Finally, we can define solution of the considered problem by the formula (2.23) in the domain Ω1, by formula (2.8) in the domain Ω2.

Hence, we prove the following theorem.

Theorem 2.1. If 𝜔(𝑥)𝐶2[]0,1,𝜓𝑖(𝑦)𝐶1[]0,1𝐶2(0,1)(𝑖=1,2),(2.28) then there exists unique solution of the Problem AT and is defined by formulas (2.23) and (2.8) in the domains Ω1, Ω2, respectively.

3. Analog of the Gellerstedt Problem

We would like to note some related works. Regarding the consideration of Gellerstedt problem for parabolic-hyperbolic equations with constant coefficients we refer the readers to [3] and for loaded parabolic-hyperbolic equations work by Khubiev [27], and also for Lavrent’ev-Bitsadze equation [28].

Consider an equation𝑢0=𝑥𝑥𝐷𝛼0𝑦𝑢𝜆𝑢,Φ0𝑢𝑥𝑥𝑢𝑦𝑦+𝜆𝑢,Φ𝑖,(𝑖=1,2)(3.1) in the domain Φ=(2𝑘=0Φ𝑘)𝐼0, where Φ0 is a domain, bounded by segments 𝐴𝐴0, 𝐵𝐵0, 𝐴0𝐵0 of straight lines 𝑥=0, 𝑥=1, 𝑦=1, respectively; Φ1 is a domain, bounded by the segment AE of the axe 𝑥 and by characteristics of (3.1) 𝐴𝐶1𝑥+𝑦=0, 𝐸𝐶1𝑥𝑦=𝑟; Φ2 is a domain, bounded by the segment 𝐸𝐵 of the axe 𝑥 and by characteristics of (3.1) 𝐸𝐶2𝑥𝑦=𝑟, 𝐵𝐶2𝑥𝑦=1; 𝐼0 is an interval 0<𝑥<1, 𝐼1 is an interval 0<𝑥<𝑟, and 𝐼2 is an interval 𝑟<𝑥<1.

Problem AG
To find a solution of (3.1) from the class of functions 𝑊2=𝑢𝐷𝛼10𝑦𝑢𝐶Φ0,𝑢𝑥𝑥,𝐷𝛼0𝑦Φ𝑢𝐶0,𝑢𝑦𝑥,0±𝐼𝐻0,𝑢𝐶Φ𝑖𝐶2Φ𝑖,(𝑖=1,2)(3.2) satisfying boundary conditions 𝑢(0,𝑦)=𝜑1(𝑦),𝑢(1,𝑦)=𝜑2(𝑦),0𝑦1,(3.3)𝑢𝐴𝐶1𝑥=𝑢2𝑥,2=𝜑3(𝑥),0𝑥𝑟,(3.4)𝑢𝐸𝐶2=𝑢(𝑥+𝑟)2,(𝑟𝑥)2=𝜑4(𝑥),𝑟𝑥1,(3.5) together with gluing conditions lim𝑦0+𝑦1𝛼𝑢(𝑥,𝑦)=lim𝑦0+𝑢(𝑥,𝑦),𝑥𝐼0,(3.6)lim𝑦0+𝑦1𝛼𝑦1𝛼𝑢(𝑥,𝑦)𝑦=𝑥0lim𝑦0+𝑢𝑦(𝑡,𝑦)𝐽0𝛼(𝑥𝑡)𝑑𝑡,𝑥𝐼0{r}.(3.7) Here 𝜑𝑗()(𝑗=1,4) are given functions such as lim𝑦0+𝑦1𝛼𝜑1(𝑦)=𝜑3(0), limy0+𝑦1𝛼𝑢(𝑟,𝑦)=𝜑4(𝑟).

Theorem 3.1. If the following conditions 𝜆0,𝜑𝑖(𝑦)𝐶1[]0,1𝐶2(0,1),𝜑𝑗(𝑥)𝐶1𝐼𝑖𝐶2𝐼𝑖(𝑖=1,2;𝑗=3,4)(3.8) are fulfilled, then the Problem AG has a unique solution.

Proof. Introduce the following designations: lim𝑦0+𝑦1𝛼𝑢(𝑥,𝑦)=𝜏+(𝑥),lim𝑦0𝑢(𝑥,𝑦)=𝜏(𝑥),𝑥𝐼0,lim𝑦0+𝑦1𝛼𝑦1𝛼𝑢(𝑥,𝑦)𝑦=𝜈+(𝑥),lim𝑦0𝑢𝑦(𝑥,𝑦)=𝜈(𝑥),𝑥𝐼0.(3.9)
Solution of the Cauchy problem for (3.1) in the domain Φ𝑖 (𝑖=1,2) in case, when 𝜆0 has a form 1𝑢(𝑥,𝑦)=2𝜏(𝑥+𝑦)+𝜏(𝑥𝑦)+𝑥+𝑦𝑥𝑦𝜈(𝑡)𝐽0𝜆(𝑥𝑡)2𝑦2𝑑𝑡+𝜆𝑦𝑥+𝑦𝑥𝑦𝜏𝐽(𝑡)1𝜆(𝑥𝑡)2𝑦2𝜆(𝑥𝑡)2𝑦2.𝑑𝑡(3.10) Using boundary conditions (3.4), (3.5), and gluing conditions (3.6), (3.7), from (3.10) we obtain 𝜈+(𝑥)=𝜏+(𝑥)+𝜑3(𝑥),𝑥𝐼1,(3.11)𝜈+(𝑥)=𝜏+(𝑥)+𝜑4(𝑥),𝑥𝐼2,(3.12) where 𝜑3(𝑥)=𝜑3(0)𝐴0,𝜆0𝑥2𝜑3(𝑥),𝜑4(𝑥)=𝜑4(𝑟)𝐴0,𝜆𝑟𝑥2𝜑4(𝑥).(3.13) According to [10], tending 𝑦 to +0, from (3.1) we get 𝜈+1(𝑦)=Γ𝜏(1+𝛼)+(𝑥)𝜆𝜏+(𝑥).(3.14) In order to prove the uniqueness of the solution for the Problem AG, we need estimate the following integral: 𝕀=10𝜏+(𝑥)𝜈+(𝑥)𝑑𝑥.(3.15) Considering homogeneous case of the condition (3.3) and taking designation (3.9) into account, after some evaluations we derive 𝕀=10𝜏+(𝑥)2𝜏+𝜆+(𝑥)2𝑑𝑥.(3.16) If 𝜆0, then 𝕀0. On the other hand, if we consider homogeneous cases of (3.11) and (3.12), one can easily be sure that 𝕀0. Hence, we get that 𝕀0. Based on (3.16) we can conclude that 𝜏+(𝑥)=0 for all 𝑥𝐼0. Due to the solution of the first boundary problem [24] we can conclude that 𝑢(𝑥,𝑦)0 in Φ0. Further, according to the gluing conditions and the solution of Cauchy problem, we have 𝑢(𝑥,𝑦)0 in Φ.
Considering functional relations (3.11)–(3.14) and conditions (3.3)–(3.5), we get the following problems: 𝜏+(𝑥)(𝜆+Γ(1+𝛼))𝜏+(𝑥)=𝜑3𝜏(𝑥)Γ(1+𝛼),+(0)=𝜑3(0),𝜏+(𝑟)=𝜑4(𝑟),𝑥𝐼1,𝜏(3.17)+(𝑥)(𝜆+Γ(1+𝛼))𝜏+(𝑥)=𝜑4𝜏(𝑥)Γ(1+𝛼),+(𝑟)=𝜑4(𝑟),𝜏+(1)=lim𝑦+0𝑦1𝛼𝜑2(𝑦),𝑥𝐼1.(3.18) The problems (3.17) and (3.18) are model problems and can be solved directly. After the finding function 𝜏+(𝑥) for all 𝑥𝐼0, functions 𝜈+(𝑥) and 𝜏(𝑥), 𝜈(𝑥) can be defined by formulas (3.14) and (3.6), (3.7), respectively. Finally, solution of the Problem AG can be recovered by formulas (3.10) and (2.23) in the domains Φ𝑖 (𝑖=1,2) and Φ0, respectively, but only with some changes in (2.23), precisely, Green function 𝐺(𝑥,𝑦,𝜉,𝜂) should be replaced by 𝐺(𝑥,𝑦,𝜉,𝜂)=(𝑦𝜂)𝛽12𝑛=𝑒1,𝛽1,𝛽||||𝑥𝜉+2𝑛(𝑦𝜂)𝛽𝑒1,𝛽1,𝛽||||𝑥+𝜉+2𝑛(𝑦𝜂)𝛽,(3.19) which is the Green function of the first boundary problem for the (3.1) in Φ0 [24].
Theorem 3.1 is proved.


This paper is partially supported by the foundation ERASMUSMUNDUS (Project number 155778-EM-1-2009-1-BEERASMUNMUNDUS-ECW-LO9).


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