RESEARCH PAPER
FRACTIONAL HEAT EQUATION AND THE SECOND LAW OF THERMODYNAMICS Luis V´ azquez 1 , Juan J. Trujillo 2 , M. Pilar Velasco
1
Abstract Dedicated to Prof. Stefan Samko on the occasion of his 70th Anniversary In the framework of second law of thermodynamics, we analyze a set of fractional generalized heat equations. The second law ensures that the heat flows from hot to cold regions, and this condition is analyzed in the context of the Fractional Calculus. MSC 2010: 35Q80, 35R11, 26A33 Key Words and Phrases: second law of thermodynamics, fractional heat equation, fractional derivatives and integrals 1. Introduction The general one dimensional space-fractional diffusion equation ∂β T ∂T =k β, 1 0. (1.13) ∂x Thus this property will restrict the possible fractional operators that can be considered in the generalized Fourier law and the initial and boundary conditions of the considered problem. When ρ = 1 the condition (1.13) is trivial. But for 0 < ρ < 1 it will be analyzed in the next section for two examples. Dxρ T ·
2. Fractional heat equation As it is not easy to prove the condition (1.13) in general, we will study specific cases of the fractional heat equation, that is, we will use a specific fractional derivative and particular initial conditions. Let be the equation ∂t T = k∂x1+ρ T,
0 < ρ ≤ 1.
(2.1)
The solutions of the Cauchy problem associated to this equation are discussed by several authors, with different fractional derivative operators, see [2], [4]. 2.1. With Liouville space-fractional derivative Considering in this case x ∈ R, we take the Liouville derivative operator that has the following expression, see for instance [12], [2]: d n ∞ 1 f (ξ)dξ L α − , x ∈ R, (2.2) ( Dx,∞ f )(x) = Γ(n − α) dx (ξ − x)α−n+1 x for α ∈ C, (α) > 0 and n = [(α)] + 1 (n ∈ N), where [(α)] means the integral part of (α). In this paper, we consider 0 < ρ ≤ 1 and then ∞ d2 1 f (ξ)dξ L 1+ρ , ( Dx,∞ f )(x) = 2 Γ(2 − α) dx x (ξ − x)ρ
x ∈ R.
(2.3)
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L. V´ azquez, J.J. Trujillo, M.P. Velasco So the generalized Cauchy problem in one dimension is: 1+ρ T )(x, t), 0 < ρ ≤ 1, t > 0, x ∈ R, k ∈ R+ (2.4) Dt T (x, t) = k(L Dx,∞
x ∈ R, t > 0.
T (x, 0+) = g(x), lim T (x, t) = 0, x→±∞
(2.5) (2.6)
This problem is solvable and its solution is obtained applying the Laplace transform Lt with respect to t > 0 ∞ u(x, t)e−st dt, x ∈ R, s > 0, (2.7) (Lt u)(x, s) = 0
and the Fourier transform Fx with respect to x ∈ R ∞ u(x, t)eixσ dx, σ ∈ R, (Fx u)(σ, t) =
t > 0,
(2.8)
−∞
and using the following known formula for the Fourier transform of the fractional derivative [2]: β T ))(σ, t) = (iσ)β (Fx T )(σ, t) (Fx (Dx,∞
where
(β > 0),
(2.9)
βπi
(σ ∈ R, β > 0). (2.10) (iσ)β := |σ|β e 2 sign(σ) This is a multi-valued function but the principal value will be taken. So we obtain Fx g(σ) , (2.11) (Fx Lt T )(σ, s) = s − k(iσ)1+ρ and applying the inverse Laplace and Fourier transform γ+i∞ 1 −1 est u(x, s)ds, x ∈ R, s > 0 (2.12) (Lt u)(x, t) = 2πi γ−i∞ (Fx−1 u)(x, t) =
1 2π
(γ = (s) > convergence abscissa) ∞
e−iσx u(σ, t)dσ,
−∞
σ ∈ R,
t > 0,
the solution of this problem is (see [2] pp. 385) γ+i∞ ∞ 1 (Fx g)(σ) −iσx 1 st e ds e dσ T (x, t) = 2πi γ−i∞ 2π −∞ s − k(iσ)1+ρ
(2.13)
(γ ∈ R). (2.14)
That is equivalent to 1 T (x, t) = 2π
∞
−∞
1+ρ t
ek(iσ)
(Fx g)(σ)e−iσx dσ,
(2.15)
provided that the integrals in the right-hand sides of (2.14) and (2.15) exist.
FRACTIONAL HEAT EQUATION
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Also in the case that g is a function in the space S¯ = {ϕ ∈ C ∞ (R) : lim ϕ(m) (x) = 0, m = 0, 1, 2...}, |x|→∞
(2.16)
then we can represent the exponential function as a power series, such that the uniform convergence allows to introduce the integral in the series, and finally the solution has the following form: T (x, t) =
∞ ktj j=0
j!
(1+ρ)j (L Dx,∞ g)(x)
(2.17)
with the condition that this series converges for all x ∈ R and t > 0, a nd considering that this type of functions g ∈ S¯ are continuous and the continuous functions verifies the indices law of the Liouville derivative. Let us check if this solution verifies the condition (1.13). We need to compute the derivatives of the solution and these derivatives can be introduced into the series of the solution (2.17) because we have supposed this series converges: ρ T )(x, t) = (L Dx,∞
∞ ktj
j!
j=0
∂T (x, t) = ∂x
∞ ktj j=0
j!
(1+ρ)j+ρ (L Dx,∞ g)(x),
(1+ρ)j+1 (L Dx,∞ g)(x).
(2.18)
(2.19)
2.1.1. Particular solution for negative exponential initial condition: Attending to this last expression, it is not possible to obtain restrictions of ρ in order to verify the condition (1.13) for all initial condition g. For this reason, we will study the following particular case: T (x, 0+) = e−λx ,
λ > 0, x ∈ R.
(2.20)
¯ it has the following Although this function does not belong to the space S, property: L α Dx,∞ e−λx = λα e−λx , (2.21) and then we can obtain the solution to the problem (2.4-2.5-2.6) easily, as: 1+ρ t−λx
T (x, t) = ekλ
.
(2.22)
Consequently we have 1+ρ t−λx
ρ T )(x, t) = λρ ekλ (L Dx,∞ ∂T 1+ρ (x, t) = −λekλ t−λx , ∂x
,
(2.23) (2.24)
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L. V´ azquez, J.J. Trujillo, M.P. Velasco
∂T 1+ρ (x, t) = −λ1+ρ e2(kλ t−λx) < 0. (2.25) ∂x It is clear that in this case the condition (1.13) is not verified for all value of ρ. ρ T )(x, t) · (L Dx,∞
2.1.2. Particular solution for potential initial condition: If we take the initial condition T (x, 0+) = xγ−1 ,
0 < γ < 1, x ∈ R,
choosing γ such that the following property is verified: Γ(1 + α − γ) γ−α−1 L α x Dx,∞ xγ−1 = , (α − [(α)] + γ) < 1, Γ(1 − γ)
(2.26)
(2.27)
for all value α = (1+ρ)j, α = (1+ρ)j+ρ and α = (1+ρ)j+1, j = 0, 1, 2, 3, ... ¯ the solution (2.17) has the form As this function belongs to the space S, ∞ ktj Γ(1 + (1 + ρ)j − γ) γ−(1+ρ)j−1 x . T (x, t) = j! Γ(1 − γ)
(2.28)
j=0
Then, L
(
ρ Dx,∞ T )(x, t)
∞ ktj Γ(1 + (1 + ρ)j + ρ − γ) γ−(1+ρ)j−ρ−1 x = , (2.29) j! Γ(1 − γ) j=0
∞
ktj Γ(2 + (1 + ρ)j − γ) ∂T (x, t) = xγ−(1+ρ)j−2 . (2.30) ∂x j! Γ(1 − γ) j=0
Now, the condition (1.13) is reduced to j 2 ∞ ∂T kt L ρ (x, t) = cj x2γ−2−j(ρ+1) ( Dx,∞ T )(x, t) · ∂x j!
(2.31)
j=1
which is always positive for ρ = 1, but it is not possible to assure its positivity for 0 < ρ < 1 in x ∈ R because negative values of x with real potentials appear in the result. 3. Conclusions In this paper we give a motivation to introduce a space fractional derivative with parameter β in the heat equation. Such fractional generalization allows to model the anomalous properties of the medium, getting a generalization of the Fourier law. Then we study the second law of thermodynamics in this new fractional model and we introduce two examples where we can see that the use of the mentioned characterization is not trivial: we can find particular cases that do not verify the second law of thermodynamics and cases where it is not possible to assure that whether this law is verified.
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Acknowledgements The authors thank the partial support by the Projects AYA2009-14212C05-05/ESP MTM2010-16499 and Predoctoral Fellowship FPU AP200700864 from the MICINN of Spain. The second author’s work (J.J. Trujillo) is also supported by Project D ID 02-25-2009 (Integral Transform Methods, Special Functions and Applications) with National Science Fund - Ministry of Education, Youth and Science of Bulgaria.
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[12] S.G. Samko, A.A. Kilbas, O.I. Marichev, Fractional Integrals and Derivatives: Theory and Applications. Gordon and Breach Science Publishers, Switzerland (1993). [13] J. Tenreiro Machado, V. Kiryakova, F. Mainardi, Recent history of fractional calculus. Commun. Nonlinear Sci. Numer. Simulat. 16 (2011), 1140–1153; doi:10.1016/j.cnsns.2010.05.027 1
Universidad Complutense de Madrid Departamento de Matem´ atica Aplicada 28040 Madrid, SPAIN e-mails:
[email protected];
[email protected] 2
Universidad de La Laguna Departamento de An´ alisis Matem´ atico 38271 La Laguna, Tenerife, SPAIN e-mail:
[email protected]
Received: January 8, 2011
Please cite to this paper as published in: Fract. Calc. Appl. Anal., Vol. 14, No 3 (2011), pp. 334–342; DOI: 10.2478/s13540-011-0021-9
