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Complete Solution For The Time Fractional Diffusion Problem With Mixed Boundary Conditions by Operational Method

A. Aghili
A. Aghili
   | 17 abr 2020
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Publicado en línea: 17 abr 2020
Páginas: 9 - 20
Recibido: 18 dic 2019
Aceptado: 31 ene 2020
DOI: https://doi.org/10.2478/amns.2020.2.00002
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© 2020 A. Aghili, published by Sciendo
This work is licensed under the Creative Commons Attribution 4.0 International License.
Introduction

Fractional partial differential equations provide an excellent model for the description of memory and hereditary properties of various processes and materials. This is the main advantage of fractional partial differential equations in comparison with classical integer - order models, in which such effects are in fact neglected. It is well - known that the mixed boundary value problems occur in the theory of elasticity in connection with punching and crack problems. The main objective of present study is to justify, in a clear fasion, the interesting possibility that fractional methods represent for modelling the dynamics certain phenomena which ordinary models cannot. The solution of the mixed boundary value problems requires considerable mathematical skills. Most mixed boundary value problems are solved using integral transform method or separation of variables [7,13,14]. Transform method are usually led to the problem of solving dual or triple Fourier or Bessel integral equations. For a discussion of such equation see [4,5]. The main goal of this study is to give an updated treatment of this subject. An alternative method of solving mixed boundary value problem involves Green’s function. Conformal mapping is a mathematical technique to solve certain types of mixed boundary value problems [4]. It should be emphasized that the focus of this paper is only on integral transform method for solving fractional partial differential equations. However, some papers have recently presented numerical techniques for this class of problems. In [9], the authors used a q- homotopy analysis transform method to find the solution for fractional Drinfeld - Sokolov - Wilson equation, where fractional derivative defined with Atangana - Baleanu (AB) operator. In [16], P. Veeresha used a numerical technique called q - homotopy analysis transform method to solve a non - linear Fisher’s equation of fractional order. Finally, we list a number of research articles where the background and many applications of numerical methods of solution could be found (see [8,9,12,15,16,17]) and focus mostly on the solution of non - linear equations.

Definitions and Notations

In the last three decades or so, fractional derivatives and notably fractional calculus have played a very important role in the various fields such as chemistry, biology, engineering, economics and signal processing. At this point, it should be pointed out that several definitions have been proposed of a fractional derivative, among those the Riemann - Liouville and Caputo fractional derivatives are the most popular. The differential equations defined in terms of Riemann - Liouville derivatives require fractional initial conditions, whereas the differential equations defined in terms of Caputo derivatives require regular boundary conditions. For this reason, the Caputo fractional derivatives are popular among engineers and scientists.

Definition 1.1

The left Caputo fractional derivative of order α (0 < α < 1) of ϕ(t) is as follows Dac,αϕ(t)=1Γ(1−α)∫at1(t−ξ)αϕ′(ξ)dξ, \matrix{ {D_a^{c,\alpha }\phi (t) = {1 \over {\Gamma (1 - \alpha )}}\int_a^t {1 \over {{{(t - \xi )}^\alpha }}}\phi '(\xi )d\xi ,}}

Definition 1.2

The Laplace transform of the function f (t) is as follows [6] ℒ{f(t)}=∫0∞e−stf(t)dt:=F(s). {\cal L}\{ f(t)\} = \int_0^\infty {e^{ - st}}f(t)dt: = F(s).

If ℒ {f(t) } = F(s), then ℒ−1{F(s)} is given by f(t)=12πi∫c−i∞c+i∞estF(s)ds, f(t) = {1 \over {2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{st}}F(s)ds, where F(s) is analytic in the region Re(s) > c. The parameter s is generally complex, but for the present time it is more convenient to consider it as real. The existence of the Laplace transform will depend upon the parameter s and the function f (t). The above complex integration along vertical line Res = c is known as Bromwich’s integral [6,10]. The real merit of the Laplace transform is revealed by its effect on fractional derivatives. Here we derive a relation between the Laplace transform of the Caputo fractional derivative of a function and the Laplace transform of the function itself [11]. ℒ[D0,tc,αf(t);s]=sαℒ[f(t);s]−sα−1f(0+).0<α<1. {\cal L}[D_{0,t}^{c,\alpha }f(t);s] = {s^\alpha }{\cal L}[f(t);s] - {s^{\alpha - 1}}f({0^ + }).0 < \alpha < 1.

Definition 1.3

A two - parameter Mittag - Leffler function is defined by the series expansion [11] Eα,β(ξ)=∑k=0+∞ξkΓ(αk+1). {E_{\alpha ,\beta }}(\xi ) = \sum\limits_{k = 0}^{ + \infty } {{{\xi ^k}} \over {\Gamma (\alpha k + 1)}}.

Lemma 1.1

The following integral identities hold true [11] 1.∫0+∞e−sttαk+β−1Eα,β(k)(±λtα)dt=k!sα−β(sα∓λ)k+1. 1.\int_0^{ + \infty } {e^{ - st}}{t^{\alpha k + \beta - 1}}E_{\alpha ,\beta }^{(k)}( \pm \lambda {t^\alpha })dt = {{k!{s^{\alpha - \beta }}} \over {{{({s^\alpha } \mp \lambda )}^{k + 1}}}}. 2.∫0+∞e−sttβ−1Eα,β(±λtα)dt=sα−β(sα∓λ). 2.\int_0^{ + \infty } {e^{ - st}}{t^{\beta - 1}}{E_{\alpha ,\beta }}( \pm \lambda {t^\alpha })dt = {{{s^{\alpha - \beta }}} \over {({s^\alpha } \mp \lambda )}}.

Note 3.∫0+∞e−sttαEα,α−1(±λtα)dt=1s(sα∓λ). 3.\int_0^{ + \infty } {e^{ - st}}{t^\alpha }{E_{\alpha ,\alpha - 1}}( \pm \lambda {t^\alpha })dt = {1 \over {s({s^\alpha } \mp \lambda )}}.

Note: With Eα,β(k)(ξ)=dkdξkEα,β(ξ) E_{\alpha ,\beta }^{(k)}(\xi ) = {{{d^k}} \over {d{\xi ^k}}}{E_{\alpha ,\beta }}(\xi ) .

Many problems of physical interest lead to Laplace transform whose inverses are not readily expressed in terms of tabulated functions. Therefore, it is highly desirable to have methods for inversion. In this section an algorithm to invert the Laplace transform is presented [1,2,3]
Remark

In the next Lemmas, we need the following integral representation for the modified Bessel’s function of the second kind 1. K0(aξ)=∫0+∞e−aξcoshzdz, 1.\quad {K_0}(a\xi ) = \int_0^{ + \infty } {e^{ - a\xi \ coshz}}dz,

Lemma 1.2

By using an appropriate integral representation for the modified Bessel’s functions of the second kind of order zero, eλsK0(λ s), show that ℒ−1[eλsK0(λs)]=1t(t+2λ). {{\cal L}^{ - 1}}[{e^{\lambda s}}{K_0}(\lambda s)] = {1 \over {\sqrt {t(t + 2\lambda )} }}.

Proof

In view of the definition1.1 taking the inverse Laplace transform of the given eλsK0(λ s), we obtain f(t)=12πi∫c−i∞c+i∞est(eλsK0(λs))ds, f(t) = {1 \over {2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{st}}({e^{\lambda s}}{K_0}(\lambda s))ds, at this stage, using the following integral representation for K0(λ s). eλsK0(λs)=∫0∞eλs−(λs)coshzdz. {e^{\lambda s}}{K_0}(\lambda s) = \int_0^\infty {e^{\lambda s - (\lambda s) \ coshz}}dz.

By setting relation (8) in (7), we obtain f(t)=12πi∫c−i∞c+i∞est(∫0∞eλs−(λs)coshzdz)ds, f(t) = {1 \over {2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{st}}(\int_0^\infty {e^{\lambda s - (\lambda s) coshz}}dz)ds, let us change the order of integration in relation (9), we arrive at f(t)=∫0∞(12πi∫c−i∞c+i∞es(t−(λcoshz−λ))ds)dz, f(t) = \int_0^\infty ({1 \over {2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{s(t - (\lambda coshz - \lambda ))}}ds)dz, the value of the inner integral is δ (t − (λ coshz − λ)), therefore f(t)=∫0∞δ(t−(λcoshz−λ))dz, f(t) = \int_0^\infty \delta (t - (\lambda \ coshz - \lambda ))dz, after making a change of variable t − (λ coshz − λ) = ψ, and considerable algebra, we obtain f(t)=∫−∞0δ(ψ)λ(t+λ−ψλ)2−1)dψ=1t(t+2λ). f(t) = \int_{ - \infty }^0 {{\delta (\psi )} \over {\lambda \sqrt {{{({{t + \lambda - \psi } \over \lambda })}^2} - 1)} }}d\psi = {1 \over {\sqrt {t(t + 2\lambda )} }}.

In view of the relation (1.3), we obtain the following integral relation ∫0∞e−stt(t+2λ)dt=eλsK0(λs). \int_0^\infty {{{e^{ - st}}} \over {\sqrt {t(t + 2\lambda )} }}dt = {e^{\lambda s}}{K_0}(\lambda s).

Thus, we get the following integral representation for the modified Bessel’s function of the second kind of order zero, K0(λs)=∫0+∞e−(λ+t)st(t+2λ). {K_0}(\lambda s) = \int_0^{ + \infty } {{{e^{ - (\lambda + t)s}}} \over {\sqrt {t(t + 2\lambda )} }}.

Definition 1.4

The Hankel transform of order ν of a function f (t) is given by ℋν[f(t);ρ]=∫0+∞f(t)tJν(ρt)dt=F(ρ). {{\cal H}_\nu }[f(t);\rho ] = \int_0^{ + \infty } f(t){{tJ}_\nu }(\rho t)dt = F(\rho ).

In order for a transformation to be useful in solving boundary value problems, it must have an inverse. The inverse Hankel transform of a function F(ρ) is given by [6] ℋν−1[F(ρ);t]=∫0+∞F(ρ)ρJν(tρ)dρ=f(t). {\cal H}_\nu ^{ - 1}[F(\rho );t] = \int_0^{ + \infty } F(\rho )\rho {J_\nu }(t\rho )d\rho = f(t).

Lemma 1.3

We have the following relation ℋν[1;ρ]=∫0+∞tJν(ρt)dt=νρ2. {{\cal H}_\nu }[1;\rho ] = \int_0^{ + \infty } {{tJ}_\nu }(\rho t)dt = {\nu \over {{\rho ^2}}}.

Proof

Let us start with the following Laplace transform relation ℒ[Jν(λt)]=∫0+∞e−stJν(λt)dt=(s2+λ2−s)νλνs2+λ2, {\cal L}[{J_\nu }(\lambda t)] = \int_0^{ + \infty } {e^{ - st}}{J_\nu }(\lambda t)dt = {{{{(\sqrt {{s^2} + {\lambda ^2}} - s)}^\nu }} \over {{\lambda ^\nu }\sqrt {{s^2} + {\lambda ^2}} }}, taking derivative with respect to the parameter s from both sides of the above relation, we obtain ∫0+∞−te−stJν(λt)dt=−λν[−12(s2+λ2)−32(s2+λ2−s)ν]+........+λν[(s2+λ2)−12(s2+λ2−s)ν−1(ss2+λ2−1)], \matrix{ {\int_0^{ + \infty } - t{e^{ - st}}{J_\nu }(\lambda t)dt = - {\lambda ^\nu }[{{ - 1} \over 2}{{({s^2} + {\lambda ^2})}^{{{ - 3} \over 2}}}{{(\sqrt {{s^2} + {\lambda ^2}} - s)}^\nu }] + ......} \cr {.. + {\lambda ^\nu }[{{({s^2} + {\lambda ^2})}^{{{ - 1} \over 2}}}{{(\sqrt {{s^2} + {\lambda ^2}} - s)}^{\nu - 1}}({s \over {\sqrt {{s^2} + {\lambda ^2}} }} - 1)],}} in the above relation, if we set s = 0 and λ = ρ after simplifying, we obtain ∫0+∞tJν(ρt)dt=νρ2. \int_0^{ + \infty } {{tJ}_\nu }(\rho t)dt = {\nu \over {{\rho ^2}}}.

Note. In special case ν=n+12 \nu = n + {1 \over 2} , we have ∫0+∞tJn+12(ρt)dt=n+12ρ2. \int_0^{ + \infty } {{tJ}_{n + {1 \over 2}}}(\rho t)dt = {{n + {1 \over 2}} \over {{\rho ^2}}}.

Corollary 1.1

The following integral identities hold true 1.  ℋν[δ(t−λ);ρ]=∫0+∞tJν(ρt)δ(t−λ)dt=λJν(λρ). 1.\quad \quad {{\cal H}_\nu }[\delta (t - \lambda );\rho ] = \int_0^{ + \infty } {{tJ}_\nu }(\rho t)\delta (t - \lambda )dt = \lambda {J_\nu }(\lambda \rho ). 2.  ℋν[tν−1e−at;ρ]=∫0+∞tνe−atJν(ρt)dt=(2ρ)νΓ(ν+0.5)π(ρ2+a2)ν+0.5 2.\quad \quad {{\cal H}_\nu }[{t^{\nu - 1}}{e^{ - at}};\rho ] = \int_0^{ + \infty } {t^\nu }{e^{ - at}}{J_\nu }(\rho t)dt = {{{{(2\rho )}^\nu }\Gamma (\nu + 0.5)} \over {\sqrt \pi {{({\rho ^2} + {a^2})}^{\nu + 0.5}}}}

Corollary 1.2

Parseval’s relation for Hankel transform

If F(ρ) and G(ρ) are Hankel transforms of f (t) and g(t), respectively, then we have the following relation ∫0+∞tf(t)g(t)dt=∫0+∞ρF(ρ)G(ρ)dρ. \int_0^{ + \infty } tf(t)g(t)dt = \int_0^{ + \infty } \rho F(\rho )G(\rho )d\rho .

Like the Laplace transform, the Hankel transforms used in a variety of applications. Perhaps the most common usage of the Hankel transform is in the solution of boundary value problems. However, there are other situations for which the properties of the Hankel transform are also very useful, such as in the evaluation of certain integrals.
Example 1.1

By using Parseval’s relation show that ∫0+∞ρ2ν+1dρ[(ρ2+a2)(ρ2+b2)]ν+0.5=πΓ(2ν)4ν(a+b)2νΓ2(ν+0.5). \int_0^{ + \infty } {{{\rho ^{2\nu + 1}}d\rho } \over {{{[({\rho ^2} + {a^2})({\rho ^2} + {b^2})]}^{\nu + 0.5}}}} = {{\pi \Gamma (2\nu )} \over {{4^\nu }{{(a + b)}^{2\nu }}{\Gamma ^2}(\nu + 0.5)}}.

Solution

Let us take f (t) = tν−1e−at and g(t) = tν−1e−bt, with Hankel transforms F(ρ)=(2ρ)νΓ(ν+0.5)π(ρ2+a2)ν+0.5,G(ρ)=(2ρ)νΓ(ν+0.5)π(ρ2+b2)ν+0.5. \matrix{ {F(\rho ) = {{{{(2\rho )}^\nu }\Gamma (\nu + 0.5)} \over {\sqrt \pi {{({\rho ^2} + {a^2})}^{\nu + 0.5}}}},} & {G(\rho ) = {{{{(2\rho )}^\nu }\Gamma (\nu + 0.5)} \over {\sqrt \pi {{({\rho ^2} + {b^2})}^{\nu + 0.5}}}}.}} respectively.

At this point, using Parseval’s relation leads to ∫0+∞e−(a+b)tt2ν−1dt=Γ(2ν)(a+b)2ν=∫0+∞ρ(2ρ)2νΓ2(ν+0.5)π[(ρ2+a2)(ρ2+b2)]ν+0.5dρ. \int_0^{ + \infty } {e^{ - (a + b)t}}{t^{2\nu - 1}}dt = {{\Gamma (2\nu )} \over {{{(a + b)}^{2\nu }}}} = \int_0^{ + \infty } {{\rho {{(2\rho )}^{2\nu }}{\Gamma ^2}(\nu + 0.5)} \over {\pi {{[({\rho ^2} + {a^2})({\rho ^2} + {b^2})]}^{\nu + 0.5}}}}d\rho .

After simplifying, we arrive at ∫0+∞ρ2ν+1dρ[(ρ2+a2)(ρ2+b2)]ν+0.5=πΓ(2ν)4ν(a+b)2νΓ2(ν+0.5). \int_0^{ + \infty } {{{\rho ^{2\nu + 1}}d\rho } \over {{{[({\rho ^2} + {a^2})({\rho ^2} + {b^2})]}^{\nu + 0.5}}}} = {{\pi \Gamma (2\nu )} \over {{4^\nu }{{(a + b)}^{2\nu }}{\Gamma ^2}(\nu + 0.5)}}.

Lemma 1.4

The following integral relations hold true. 1. ∫0+∞xν+1Jν(ρx)dx(x2+a2)λ+1=aν−λρλ2λΓ(λ+1)Kν−λ(aρ). 1.\,\int_0^{ + \infty } {x^{\nu + 1}}{J_\nu }(\rho x){{dx} \over {{{({x^2} + {a^2})}^{\lambda + 1}}}} = {{{a^{\nu - \lambda }}{\rho ^\lambda }} \over {{2^\lambda }\Gamma (\lambda + 1)}}{K_{\nu - \lambda }}(a\rho ). 2. ∫0+∞ρλ+1Jν(xρ)Kν−λ(aρ)=2λΓ(λ+1)xνaν−λ(x2+a2)λ+1. 2.\,\int_0^{ + \infty } {\rho ^{\lambda + 1}}{J_\nu }(x\rho ){K_{\nu - \lambda }}(a\rho ) = {{{2^\lambda }\Gamma (\lambda + 1){x^\nu }} \over {{a^{\nu - \lambda }}{{({x^2} + {a^2})}^{\lambda + 1}}}}. 3. ∫0+∞ρYν(xρ)Kν(aρ)dρ=2π(a−νxνcosπν−aνx−ν2(x2+a2)sinπν. 3.\,\int_0^{ + \infty } \rho {Y_\nu }(x\rho ){K_\nu }(a\rho )d\rho = {{2\pi ({a^{ - \nu }}{x^\nu }\cos \pi \nu - {a^\nu }{x^{ - \nu }}} \over {2({x^2} + {a^2})\sin \pi \nu }}.

Proof

Let us start with the following elementary integral identity 1(x2+a2)λ+1=1Γ(λ+1)∫0+∞e−(x2+a2)uuλdu. {1 \over {{{({x^2} + {a^2})}^{\lambda + 1}}}} = {1 \over {\Gamma (\lambda + 1)}}\int_0^{ + \infty } {e^{ - ({x^2} + {a^2})u}}{u^\lambda }du.

By substituting this integral on the left hand side of the first integral and interchanging the order of integration, we obtain ∫0+∞xν+1Jν(ρx)[1Γ(λ+1)∫0+∞e−(x2+a2)uuλdu]dx=....=12Γ(λ+1)∫0+∞e−a2uuλ(∫0+∞xν+1e−ux2Jν(ρx)dx)du. \matrix{ {\int_0^{ + \infty } {x^{\nu + 1}}{J_\nu }(\rho x)[{1 \over {\Gamma (\lambda + 1)}}\int_0^{ + \infty } {e^{ - ({x^2} + {a^2})u}}{u^\lambda }du]dx = ....} \cr { = {1 \over {2\Gamma (\lambda + 1)}}\int_0^{ + \infty } {e^{ - {a^2}u}}{u^\lambda }(\int_0^{ + \infty } {x^{\nu + 1}}{e^{ - u{x^2}}}{J_\nu }(\rho x)dx)du.}}

At this point, let us evaluate the inner integral by making a change of variable x2 = ϕ and ρ=2β \rho = 2\sqrt \beta , after evaluation of the integral, we arrive at ∫0+∞xν+1Jν(ρx)[1Γ(λ+1)∫0+∞e−(x2+a2)uuλdu]dx=....βν2Γ(λ+1)∫0+∞e−a2uuλ(e−βuuν+1)du=ρν2νΓ(λ+1)∫0+∞e−(a2u+ρ24u)du2uν−λ+1. \matrix{ {\int_0^{ + \infty } {x^{\nu + 1}}{J_\nu }(\rho x)[{1 \over {\Gamma (\lambda + 1)}}\int_0^{ + \infty } {e^{ - ({x^2} + {a^2})u}}{u^\lambda }du]dx = ....} \cr {{{{\beta ^{{\nu \over 2}}}} \over {\Gamma (\lambda + 1)}}\int_0^{ + \infty } {e^{ - {a^2}u}}{u^\lambda }({{{e^{ - {\beta \over u}}}} \over {{u^{\nu + 1}}}})du = {{{\rho ^\nu }} \over {{2^\nu }\Gamma (\lambda + 1)}}\int_0^{ + \infty } {e^{ - ({a^2}u + {{{\rho ^2}} \over {4u}})}}{{du} \over {2{u^{\nu - \lambda + 1}}}}.}}

Making a change of variable a2u = w, after simplifying, we obtain ∫0+∞xν+1Jν(ρx)dx(x2+a2)λ+1=ρλa2(ν−λ)2λΓ(λ+1)[(aρ2)ν−λ∫0+∞e−w−a2ρ24wdw2wν−λ+1]. \int_0^{ + \infty } {x^{\nu + 1}}{J_\nu }(\rho x){{dx} \over {{{({x^2} + {a^2})}^{\lambda + 1}}}} = {{{\rho ^\lambda }{a^{2(\nu - \lambda )}}} \over {{2^\lambda }\Gamma (\lambda + 1)}}[{({{a\rho } \over 2})^{\nu - \lambda }}\int_0^{ + \infty } {e^{ - w - {{{a^2}{\rho ^2}} \over {4w}}}}{{dw} \over {2{w^{\nu - \lambda + 1}}}}].

But, the value of the integral in the braces is Kν−λ (aρ), therefore we get the following ∫0+∞xν+1Jν(ρx)dx(x2+a2)λ+1=ρλaν−λ2λΓ(λ+1)Kν−λ(aρ). \int_0^{ + \infty } {x^{\nu + 1}}{J_\nu }(\rho x){{dx} \over {{{({x^2} + {a^2})}^{\lambda + 1}}}} = {{{\rho ^\lambda }{a^{\nu - \lambda }}} \over {{2^\lambda }\Gamma (\lambda + 1)}}{K_{\nu - \lambda }}(a\rho ).

At this stage, the above relation can be written as Hankel transform of a function as below ℋν[xν(x2+a2)λ+1;ρ]=aν−λ2λΓ(λ+1)ρλKν−λ(aρ). {{\cal H}_\nu }[{{{x^\nu }} \over {{{({x^2} + {a^2})}^{\lambda + 1}}}};\rho ] = {{{a^{\nu - \lambda }}} \over {{2^\lambda }\Gamma (\lambda + 1)}}{\rho ^\lambda }{K_{\nu - \lambda }}(a\rho ).

Thus, by taking the inverse Hankel transform, we obtain ∫0+∞ρλ+1Jν(xρ)Kν−λ(aρ)dρ=2λΓ(λ+1)aν−λxν(x2+a2)λ+1. \int_0^{ + \infty } {\rho ^{\lambda + 1}}{J_\nu }(x\rho ){K_{\nu - \lambda }}(a\rho )d\rho = {{{2^\lambda }\Gamma (\lambda + 1)} \over {{a^{\nu - \lambda }}}}{{{x^\nu }} \over {{{({x^2} + {a^2})}^{\lambda + 1}}}}.

Let us choose λ = 0, after simplifying we get ∫0+∞ρJν(xρ)Kν(aρ)dρ=2xνaν(x2+a2). \int_0^{ + \infty } \rho {J_\nu }(x\rho ){K_\nu }(a\rho )d\rho = {{2{x^\nu }} \over {{a^\nu }({x^2} + {a^2})}}.

Let us consider the following relations, Kν(ξ)=K−ν(ξ),  Yν(ξ)=π2Jν(ξ)cos(πν)−J−ν(ξ)sin(πν). {K_\nu }(\xi ) = {K_{ - \nu }}(\xi ),\quad \quad {Y_\nu }(\xi ) = {\pi \over 2}{{{J_\nu }(\xi )\cos (\pi \nu ) - {J_{ - \nu }}(\xi )} \over {\sin (\pi \nu )}}.

Combination of (38) and (39) leads to the following integral relation ∫0+∞ρYν(xρ)Kν(aρ)dρ=2π(a−νxνcosπν−aνx−ν)2(x2+a2)sinπν. \int_0^{ + \infty } \rho {Y_\nu }(x\rho ){K_\nu }(a\rho )d\rho = {{2\pi ({a^{ - \nu }}{x^\nu }\cos \pi \nu - {a^\nu }{x^{ - \nu }})} \over {2({x^2} + {a^2})\sin \pi \nu }}.

Note: In the above relation, Yν(ξ) stands for the Newmann function of order ν.

Theorem 1.1

Let us assume that ℒ [f (t)] = F(s), then the following relation holds true exp[−exp(−t2)]f(exp−t2)=ℒ−1[∫0+∞ξsJ2s(2ξ)F(ξ)dξ]. \exp [ - \exp ( - {t \over 2})]f(\exp {{ - t} \over 2}) = {{\cal L}^{ - 1}}[\int_0^{ + \infty } {\xi ^s}{J_{2s}}(2\sqrt \xi )F(\xi )d\xi ].

Proof

The right hand side of the above relation can be written as follows ℒ−1[∫0+∞ξsJ2s(2ξ)F(ξ)dξ]=12πi∫c−i∞c+i∞est[∫0+∞ξsJ2s(2ξ)F(ξ)dξ]ds. {{\cal L}^{ - 1}}[\int_0^{ + \infty } {\xi ^s}{J_{2s}}(2\sqrt \xi )F(\xi )d\xi ] = {1 \over {2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{st}}[\int_0^{ + \infty } {\xi ^s}{J_{2s}}(2\sqrt \xi )F(\xi )d\xi ]ds.

In the above double integral, we set ℒ[f(η)]=F(ξ)=∫0+∞e−ξηf(η)dη {\cal L}[f(\eta )] = F(\xi ) = \int_0^{ + \infty } {e^{ - \xi \eta }}f(\eta )d\eta . to obtain ℒ−1[∫0+∞ξsJ2s(2ξ)F(ξ)dξ]=12πi∫c−i∞c+i∞est[∫0+∞ξsJ2s(2ξ)∫0+∞e−ξηf(η)dη]dξds, {{\cal L}^{ - 1}}[\int_0^{ + \infty } {\xi ^s}{J_{2s}}(2\sqrt \xi )F(\xi )d\xi ] = {1 \over {2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{st}}[\int_0^{ + \infty } {\xi ^s}{J_{2s}}(2\sqrt \xi )\int_0^{ + \infty } {e^{ - \xi \eta }}f(\eta )d\eta ]d\xi ds, changing the order of integration leads to ℒ−1[∫0+∞ξsJ2s(2ξ)F(ξ)dξ]=12πi∫c−i∞c+i∞est[∫0+∞f(η)[∫0+∞ξsJ2s(2ξ)e−ξηdξ]dη]ds, {{\cal L}^{ - 1}}[\int_0^{ + \infty } {\xi ^s}{J_{2s}}(2\sqrt \xi )F(\xi )d\xi ] = {1 \over {2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{st}}[\int_0^{ + \infty } f(\eta )[\int_0^{ + \infty } {\xi ^s}{J_{2s}}(2\sqrt \xi ){e^{ - \xi \eta }}d\xi ]d\eta ]ds, after evaluating the inner integral, we get ℒ−1[∫0+∞ξsJ2s(2ξ)F(ξ)dξ]=12πi∫c−i∞c+i∞est[∫0+∞e−1ηη2s+1f(η)dη]ds, {{\cal L}^{ - 1}}[\int_0^{ + \infty } {\xi ^s}{J_{2s}}(2\sqrt \xi )F(\xi )d\xi ] = {1 \over {2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{st}}[\int_0^{ + \infty } {{{e^{ - {1 \over \eta }}}} \over {{\eta ^{2s + 1}}}}f(\eta )d\eta ]ds, at this stage, we change the order of integration and simplifying, we arrive at ℒ−1[∫0+∞ξsJ2s(2ξ)F(ξ)dξ]=∫0+∞e−1ηf(η)[12πi∫c−i∞c+i∞est−(2s+1)lnηds]dη, {{\cal L}^{ - 1}}[\int_0^{ + \infty } {\xi ^s}{J_{2s}}(2\sqrt \xi )F(\xi )d\xi ] = \int_0^{ + \infty } {e^{ - {1 \over \eta }}}f(\eta )[{1 \over {2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{st - (2s + 1)\ln \eta }}ds]d\eta , after simplifying the inner integral, we obtain ℒ−1[∫0+∞ξsJ2s(2ξ)F(ξ)dξ]=∫0+∞η−1e−1ηf(η)[12πi∫c−i∞c+i∞es(t−2lnη)ds]dη, {{\cal L}^{ - 1}}[\int_0^{ + \infty } {\xi ^s}{J_{2s}}(2\sqrt \xi )F(\xi )d\xi ] = \int_0^{ + \infty } {\eta ^{ - 1}}{e^{ - {1 \over \eta }}}f(\eta )[{1 \over {2\pi i}}\int_{c - i\infty }^{c + i\infty } {e^{s(t - 2\ln \eta )}}ds]d\eta , the value of the complex integral is δ (t − 2lnη), therefore ℒ−1[∫0+∞ξsJ2s(2ξ)F(ξ)dξ]=∫0+∞η−1e−1ηf(η)δ(t−2lnη)dη, {{\cal L}^{ - 1}}[\int_0^{ + \infty } {\xi ^s}{J_{2s}}(2\sqrt \xi )F(\xi )d\xi ] = \int_0^{ + \infty } {\eta ^{ - 1}}{e^{ - {1 \over \eta }}}f(\eta )\delta (t - 2\ln \eta )d\eta , making a change of variable ϕ = t − 2lnη, then we get η=et−ϕ2 \eta = {e^{{{t - \phi } \over 2}}} and dη=−12et−ϕ2 d\eta = - {1 \over 2}{e^{{{t - \phi } \over 2}}} , after simplifying we have the following ℒ−1[∫0+∞ξsJ2s(2ξ)F(ξ)dξ]=∫−∞+∞e−eϕ−t2f(eϕ−t2)δ(ϕ)dϕ=e−e−t2f(e−t2). {{\cal L}^{ - 1}}[\int_0^{ + \infty } {\xi ^s}{J_{2s}}(2\sqrt \xi )F(\xi )d\xi ] = \int_{ - \infty }^{ + \infty } {e^{ - {e^{{{\phi - t} \over 2}}}}}f({e^{{{\phi - t} \over 2}}})\delta (\phi )d\phi = {e^{ - {e^{{{ - t} \over 2}}}}}f({e^{{{ - t} \over 2}}}).

Lemman 1.5

Let us assume that ℒ [f (t);s] = F(s), then the following identities hold true 1.  ℒ[f(t3);s]=13π∫0+∞sξK13[2(s3ξ3)]F(ξ)dξ, 1.\quad \quad {\cal L}[f({t^3});s] = {1 \over {3\pi }}\int_0^{ + \infty } \sqrt {{s \over \xi }} {K_{{1 \over 3}}}[2({s \over {3\root 3 \of \xi }})]F(\xi )d\xi , 2.  ℒ[1tf(1t);s]=∫0+∞J0(2su)F(u)du. 2.\quad \quad {\cal L}[{1 \over t}f({1 \over t});s] = \int_0^{ + \infty } {J_0}(2\sqrt {su} )F(u)du. 3.  ℒ[1tf(1t3);s]=∫0+∞J0(2su)[13π∫0+∞sξK13[2(s3ξ3)]F(ξ)dξ]du. 3.\quad \quad {\cal L}[{1 \over t}f({1 \over {{t^3}}});s] = \int_0^{ + \infty } {J_0}(2\sqrt {su} )[{1 \over {3\pi }}\int_0^{ + \infty } \sqrt {{s \over \xi }} {K_{{1 \over 3}}}[2({s \over {3\root 3 \of \xi }})]F(\xi )d\xi ]du.

Proof

See [6].
Example 1.2

The following integral identity holds true λ763e−sλ3=∫0+∞J0(2su)[13π∫0+∞sξK13(2s3ξ3)e−λξdξ]du. {{{\lambda ^{{7 \over 6}}}} \over 3}{e^{ - {s \over {\root 3 \of \lambda }}}} = \int_0^{ + \infty } {J_0}(2\sqrt {su} )[{1 \over {3\pi }}\int_0^{ + \infty } \sqrt {{s \over \xi }} {K_{{1 \over 3}}}({{2s} \over {3\root 3 \of \xi }}){e^{ - \lambda \xi }}d\xi ]du.

Solution

Let us take f (t) = δ (t − λ), then F(s) = e−λs, so we may evaluate ℒ[1tδ(1t3−λ)] {\cal L}[{1 \over t}\delta ({1 \over {{t^3}}} - \lambda )] in two different ways as below, first, directly by the definition of the Laplace transform, we get ℒ[1tδ(1t3−λ)]=∫0+∞e−st1tδ(1t3−λ)dt=λ763e−sλ3. {\cal L}[{1 \over t}\delta ({1 \over {{t^3}}} - \lambda )] = \int_0^{ + \infty } {e^{ - st}}{1 \over t}\delta ({1 \over {{t^3}}} - \lambda )dt = {{{\lambda ^{{7 \over 6}}}} \over 3}{e^{ - {s \over {\root 3 \of \lambda }}}}.

On the other hand, by using part three of the Lemma 1.5, we arrive at λ763e−sλ3=∫0+∞J0(2su)[13π∫0+∞sξK13(2s3ξ3)e−λξdξ]du. {{{\lambda ^{{7 \over 6}}}} \over 3}{e^{ - {s \over {\root 3 \of \lambda }}}} = \int_0^{ + \infty } {J_0}(2\sqrt {su} )[{1 \over {3\pi }}\int_0^{ + \infty } \sqrt {{s \over \xi }} {K_{{1 \over 3}}}({{2s} \over {3\root 3 \of \xi }}){e^{ - \lambda \xi }}d\xi ]du.

Main Results

In the past three decades, considerable research efforts have been expended to study anomalous diffusion using the time fractional equation. Anomalous diffusion transport appears to be a universal experimental phenomenon. A number of works have been published dealing with anomalous transport in fractals and disordered media, glass - forming liquids and colloidal structures. Let us consider the following two-dimensional heat conduction problem that arises during the manufacture of p-n junctions. To the best of the author’s knowledge this kind of fractional mixed boundary value problem is not considered in the literature.

Problem 2.1

Let us consider the following time fractional diffusion problem with mixed boundary conditions ∂c,αu(x,y,t)∂tα=a2Δu(x,y,t),0<α<1. \matrix{ {{{{\partial ^{c,\alpha }}u(x,y,t)} \over {\partial {t^\alpha }}} = {a^2}\Delta u(x,y,t),} & {0 < \alpha < 1.}} where −∞ < x < ∞, 0 < y,t < +∞ and subject to the mixed conditions uy(x,0,t)=0, {u_y}(x,0,t) = 0, u(x,y,0)=0, u(x,y,0) = 0, u(x,π,t)=U0tαΓ(α+1). u(x,\pi ,t) = {{{U_0}{t^\alpha }} \over {\Gamma (\alpha + 1)}}. lim|x|−>+∞u(x,y,t)=limy−>+∞u(x,y,t)=0.   {lim_{|x| - > + \infty }}u(x,y,t) = {lim_{y - > + \infty }}u(x,y,t) = 0.

Note: In this study, it is assumed that the time fractional derivatives have been defined in the sense of the Caputo fractional derivatives.

Solution

In order to solve the above mixed boundary value problem, we reformulating it in cylindrical coordinates, to obtain ∂c,αu(r,ϕ,t)∂tα=a2(urr+1rur+1r2uϕϕ),0<α<1. \matrix{ {{{{\partial ^{c,\alpha }}u(r,\phi ,t)} \over {\partial {t^\alpha }}} = {a^2}({u_{rr}} + {1 \over r}{u_r} + {1 \over {{r^2}}}{u_{\phi \phi }}),} & {0 < \alpha < 1.}} where 0 < ϕ < π, 0 < r,t < +∞ and subject to the mixed conditions uϕ(r,0,t)=0, {u_\phi }(r,0,t) = 0, u(r,π,0)=0, u(r,\pi ,0) = 0, u(r,π,t)=U0tαΓ(α+1). u(r,\pi ,t) = {{{U_0}{t^\alpha }} \over {\Gamma (\alpha + 1)}}. limr−>0u(r,ϕ,t)=limr−>+∞u(r,ϕ,t)=0. {lim_{r - > 0}}u(r,\phi ,t) = {lim_{r - > + \infty }}u(r,\phi ,t) = 0.

The above mixed boundary value problem can be solved via the Laplace transform.

Let us define U(r,ϕ,s)=∫0+∞e−stu(r,ϕ,t)dt. U(r,\phi ,s) = \int_0^{ + \infty } {e^{ - st}}u(r,\phi ,t)dt.

Then the transformed equation becomes sαU(r,ϕ,s)=a2(Urr+1rUr+1r2Uϕϕ),0<α<1. \matrix{ {{s^\alpha }U(r,\phi ,s) = {a^2}({U_{rr}} + {1 \over r}{U_r} + {1 \over {{r^2}}}{U_{\phi \phi }}),} & {0 < \alpha < 1.}} where 0 < ϕ < π, 0 < r,t < +∞ and subject to the mixed conditions Uϕ(r,0,s)=0, {U_\phi }(r,0,s) = 0, U(r,π,0)=0, U(r,\pi ,0) = 0, U(r,π,s)=U0sα+1. U(r,\pi ,s) = {{{U_0}} \over {{s^{\alpha + 1}}}}. limr−>0U(r,ϕ,s)=limr−>+∞U(r,ϕ,s)=0. {lim_{r - > 0}}U(r,\phi ,s) = {lim_{r - > + \infty }}U(r,\phi ,s) = 0.

At this stage, let us choose U(r,ϕ,s)=U0sα+1+V(r,ϕ,s). U(r,\phi ,s) = {{{U_0}} \over {{s^{\alpha + 1}}}} + V(r,\phi ,s).

Then we have the following relations sαa2V(r,ϕ,s)=Vrr+1rVr+1r2Vϕϕ−U0a2s,0<α<1. \matrix{ {{{{s^\alpha }} \over {{a^2}}}V(r,\phi ,s) = {V_{rr}} + {1 \over r}{V_r} + {1 \over {{r^2}}}{V_{\phi \phi }} - {{{U_0}} \over {{a^2}s}},} & {0 < \alpha < 1.}} where 0 < ϕ < π, 0 < r,t < +∞ and subject to the mixed conditions Vϕ(r,0,s)=0, {V_\phi }(r,0,s) = 0, V(r,π,0)=0, V(r,\pi ,0) = 0, limr−>0V(r,ϕ,s)=limr−>+∞V(r,ϕ,s)=0. {lim_{r - > 0}}V(r,\phi ,s) = {lim_{r - > + \infty }}V(r,\phi ,s) = 0.

Now, we express the Fourier series solution to the above equation as follows V(r,ϕ,s)=∑n=0+∞Vn(r,s)cos(n+12)ϕ. V(r,\phi ,s) = \sum\limits_{n = 0}^{ + \infty } {V_n}(r,s)cos(n + {1 \over 2})\phi .

Note that the Equation (77) satisfies the boundary conditions (74),(75) and each Fourier coefficient Vn(r,ϕ,s) is governed by the following non - homogenous second order Bessel’s differential equation. Vn′′+1rVn′−((n+12)2r2+sαa2)Vn=2(−1)nU0πa2(n+12)s {V''_{n}} + {1 \over r}{V'_{n}} - ({{{{(n + {1 \over 2})}^2}} \over {{r^2}}} + {{{s^\alpha }} \over {{a^2}}}){V_n} = {{2{{( - 1)}^n}{U_0}} \over {\pi {a^2}(n + {1 \over 2})s}}

Equation (78) is known as non - homogeneous modified Bessel equation of order ( n+12 n + {1 \over 2} ) with complete solution as below Vn(r,s)=c1Kn+12(rasα)+c2In+12(rasα)+Ψc(r,s), {V_n}(r,s) = {c_1}{K_{n + {1 \over 2}}}({r \over a}\sqrt {{s^\alpha }} ) + {c_2}{I_{n + {1 \over 2}}}({r \over a}\sqrt {{s^\alpha }} ) + {\Psi _c}(r,s), where, Ψc(r,s) is the complementary solution and not known. Since the modified Bessel’s functions Iν(r), Kν(r) are unbounded at origin and infinity respectively, so that in view of the boundary conditions (60), we should have c1 = c2 = 0, therefore, we get the following formal solution Vn(r,s)=Ψc(r,s). {V_n}(r,s) = {\Psi _c}(r,s).

One of the principal uses of the Hankel transform is in the solution of boundary value problems involving cylindrical coordinates. At this stage, we apply the Hankel transform of order ( n+12 n + {1 \over 2} ) to the variable r in (78), this action leads to ℋn+12[Vn′′+1rVn′−(n+12)2r2Vn]−ℋn+12[sαa2Vn]=ℋn+12[2(−1)nU0πa2(n+12)s], {{\cal H}_{n + {1 \over 2}}}[{V''_n} + {1 \over r}{V'_n} - {{{{(n + {1 \over 2})}^2}} \over {{r^2}}}{V_n}] - {{\cal H}_{n + {1 \over 2}}}[{{{s^\alpha }} \over {{a^2}}}{V_n}] = {{\cal H}_{n + {1 \over 2}}}[{{2{{( - 1)}^n}{U_0}} \over {\pi {a^2}(n + {1 \over 2})s}}], after using some properties of the Hankel transform and in view of the Lemma 1.3, we arrive at −ρ2A(ρ,s)−sαa2A(ρ,s)=2U0(−1)nπa2(n+12)s(n+12ρ2), - {\rho ^2}A(\rho ,s) - {{{s^\alpha }} \over {{a^2}}}A(\rho ,s) = {{2{U_0}{{( - 1)}^n}} \over {\pi {a^2}(n + {1 \over 2})s}}({{n + {1 \over 2}} \over {{\rho ^2}}}), from which we deduce that A(ρ,s)=−2(−1)nU0πρ2s(sα+a2ρ2). A(\rho ,s) = - {{2{{( - 1)}^n}{U_0}} \over {\pi {\rho ^2}s({s^\alpha } + {a^2}{\rho ^2})}}.

Inverting this result by means of the Hankel inversion formula, we have Vn(r,s)=Ψc(r,s)=−∫0∞2(−1)nU0πs(sα+a2ρ2)Jn+12(rρ)dρρ, {V_n}(r,s) = {\Psi _c}(r,s) = - \int_0^\infty {{2{{( - 1)}^n}{U_0}} \over {\pi s({s^\alpha } + {a^2}{\rho ^2})}}{J_{n + {1 \over 2}}}(r\rho ){{d\rho } \over \rho }, and V(r,ϕ,s)=−∑n=0+∞[∫0∞2(−1)nU0πs(sα+a2ρ2)Jn+12(rρ)dρρ]cos(n+12)ϕ. V(r,\phi ,s) = - \sum\limits_{n = 0}^{ + \infty } [\int_0^\infty {{2{{( - 1)}^n}{U_0}} \over {\pi s({s^\alpha } + {a^2}{\rho ^2})}}{J_{n + {1 \over 2}}}(r\rho ){{d\rho } \over \rho }]cos(n + {1 \over 2})\phi .

Finally, we get the solution to transformed Equation (67) as follows U(r,ϕ,s)=U0sα+1−2U0π∑n=0+∞(−1)n[∫0∞1s(sα+a2ρ2)ρJn+12(rρ)dρρ]cos(n+12)ϕ, U(r,\phi ,s) = {{{U_0}} \over {{s^{\alpha + 1}}}} - {{2{U_0}} \over \pi }\sum\limits_{n = 0}^{ + \infty } {( - 1)^n}[\int_0^\infty {1 \over {s({s^\alpha } + {a^2}{\rho ^2})\rho }}{J_{n + {1 \over 2}}}(r\rho ){{d\rho } \over \rho }]cos(n + {1 \over 2})\phi , and thus by taking the inverse Laplace transform, we obtain u(r,ϕ,t)U0=tαΓ(α+1)−2π∑n=0+∞(−1)n[∫0∞ℒ−1[1s(sα+a2ρ2)]Jn+12(rρ)dρρ]cos(n+12)ϕ, {{u(r,\phi ,t)} \over {{U_0}}} = {{{t^\alpha }} \over {\Gamma (\alpha + 1)}} - {2 \over \pi }\sum\limits_{n = 0}^{ + \infty } {( - 1)^n}[\int_0^\infty {{\cal L}^{ - 1}}[{1 \over {s({s^\alpha } + {a^2}{\rho ^2})}}]{J_{n + {1 \over 2}}}(r\rho ){{d\rho } \over \rho }]cos(n + {1 \over 2})\phi , at this point, by virtue of the part three of the Lemma 1.1, and the inversion of Equation (87) yields u(r,ϕ,t)U0=tαΓ(α+1)−2tαπ∑n=0+∞(−1)n[∫0∞Eα,α−1(±λtα)Jn+12(rρ)dρρ]cos(n+12)ϕ, {{u(r,\phi ,t)} \over {{U_0}}} = {{{t^\alpha }} \over {\Gamma (\alpha + 1)}} - {{2{t^\alpha }} \over \pi }\sum\limits_{n = 0}^{ + \infty } {( - 1)^n}[\int_0^\infty {E_{\alpha ,\alpha - 1}}( \pm \lambda {t^\alpha }){J_{n + {1 \over 2}}}(r\rho ){{d\rho } \over \rho }]cos(n + {1 \over 2})\phi ,

Note: It is easy to check that u(r,ϕ,0)=uϕ(r,0,t)=0,u(r,π,t)=U0tαΓ(α+1) u(r,\phi ,0) = {u_\phi }(r,0,t) = 0,u(r,\pi ,t) = {{{U_0}{t^\alpha }} \over {\Gamma (\alpha + 1)}}

Conclusions

In this work, the author presents analytical techniques to solve time fractional diffusion problem with mixed bounadry conditions. We consider a generalization of the fractional heat conduction problem in two dimensions that arises during analysis of the impurity atom distribution near the diffusion mask for a planar p - n junction. The article is intended for scientists and researchers of different disciplines of engineering and science dealing with the solutions of fractional mixed boundary value problems. The results reveal that the integral transforms method is very convenient and effective. It is hoped that this study will lead to further investigations in the field and more elegant solutions would be found.
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