Determination of two unknown thermal coefficients through an inverse one-phase fractional Stefan problem

# Determination of two unknown thermal coefficients through an inverse one-phase fractional Stefan problem

## Abstract

We consider a semi-infinite one-dimensional phase-change material with two unknown constant thermal coefficients among the latent heat per unit mass, the specific heat, the mass density and the thermal conductivity. Aiming at the determination of them, we consider an inverse one-phase Stefan problem with an over-specified condition at the fixed boundary and a known evolution for the moving boundary. We assume that the phase-change process presents latent-heat memory effects by considering a fractional time derivative of order () in the Caputo sense and a sharp front model for the interface. According to the choice of the unknown thermal coefficients, six inverse fractional Stefan problems arise. For each of them, we determine necessary and sufficient conditions on data to obtain the existence and uniqueness of a solution of similarity type. Moreover, we present explicit expressions for the temperature and the unknown thermal coefficients. Finally, we show that the results for the classical statement of this problem, associated with , are obtained through the fractional model when .

Keywords: Inverse Stefan problems, Time fractional Caputo derivative, Unknown thermal coefficients, Explicit solutions, Wright and Mainardi Functions, Over-specified boundary condition.

Mathematics Subject Classification 2010: 26A33, 35C05, 35R11, 35R35, 80A22.

## 1 Introduction

Determination of thermal coefficients for phase-change materials through inverse Stefan problems has been widely studied during the last decades [5, 6, 14, 27, 28, 29]. Especially, phase-change processes involving solidification or melting have been extensively studied because of their scientific and technological applications [1, 2, 4, 7, 8, 10, 18, 26, 31]. A rewiew of a long bibliography on moving and free boundary value problems for the heat equation can be consulted in [30]. Recently, a new sort of Stefan problems including time-fracional derivatives have begun to be studied [13, 15, 33, 34, 35, 23, 24, 25]. Some references in fractional derivatives can be found at [16, 19, 22, 9, 17, 20]. In [23, 24, 25] there are considered free boundary value problems which are obtained by replacing the time derivative in a one-phase Stefan problem by a fractional derivative of order () in the Caputo sense [3], and explicit solutions of similarity type are given for the resultant fractional Stefan problems. A physical interpretation of the problems considered in [23, 24, 25] is given in [35]. In that article, authors derive fractional Stefan problems for phase-change processes by substituting the local expression of the heat flux given by the Fourier law for a new non-local definition. They consider a heat flux given as a weighted sum of local fluxes back in time, which they express in terms of the Riemann-Liouville integral of order () of the local flux given by the Fourier law. They also explain how this change implies that the new model takes into consideration latent-heat memory effects in the evolution of the phase-change process and give to the parameter the physical meaning of being the strength of memory retention. This fractional model reduces to the classical Stefan problem when . The same occurs with the solutions of similarity type given in [23, 24, 25] in the sense that they converge to the similarity solutions of the classical Stefan problems with which they are related to, when . To the authors knowledge, the first use of inverse fractional Stefan problems for the determination of thermal coefficients has been done recently in [32]. In that article the author studies the determination of one unknown thermal coefficient for a semi-infinite material through a fractional one-phase Stefan problem with an over-specified condition at the fixed boundary. Necessary and sufficient conditions on data to obtain the existence and uniqueness of solutions of similarity type are established, and explicit expressions for the the temperature of the material, the free boundary and the unknown thermal coefficient are given. Moreover, it has shown that results through the fractional model reduce to the results previously obtained in [27] for the determination of one unknown thermal coefficient using a classical inverse Stefan problem. Encouraged by [28, 32], we consider here the problem of determining two unknown thermal coefficients through an inverse fractional one-phase Stefan problem for which it is known the evolution of the free boundary. In order to have dimensional coherence in the time fractional heat equation as well as in the fractional Stefan condition, we have included two extra parameters in the model, which are such that:

 μα→1when α→1− (1a) να→1when α→1−. (1b)

In particular, and can be considered equal to 1 with the corresponding physical dimension (see below).

More precisely, we consider the following inverse problem for a one-phase melting process:

 DαT(x,t)=μαλ2Txx(x,t) 00 (2a) T(s(t),t)=Tm t>0 (2b) −kTx(s(t),t)=ναρlDαs(t) t>0 (2c) T(0,t)=T0 t>0 (2d) kTx(0,t)=−q0tα/2 t>0 (2e)

where the unknowns are the temperature [C] of the liquid phase and two thermal coefficients among:

According to [11, 12], the coefficient [] in equation (2a) is a sort of fractional diffusion coefficient, [] being the thermal diffusivity given by:

 λ2=kρc(λ>0).

We assume that the remaining coefficients:

involved in Problem (2), are all known (e.g. through a phase-change experiment). Aiming at the simultaneous determination of two unknown thermal coefficients, we consider that the time evolution of the sharp interface is also known. More precisely, we follow [23, 24, 25] in assuming that it is given by:

 s(t)=σtα/2,t>0, (3)

with []. The operator in (2a) and (2c) represents the fractional time derivative of order in the Caputo sense, which is defined by [3]:

 Dαf(t)=⎧⎪⎨⎪⎩1Γ(1−α)∫t0f′(τ)(t−τ)αdτ%if0<α<1f′(t) if α=1,t>0 (4)

for any , where is the Gamma function defined by:

 Γ(x)=∫+∞0sx−1exp(−s)ds,x>0.

## 2 Solutions of similarity type

We begin with a definition of being a solution to the inverse fractional Stefan problem (2), which is based in the definition established in [23, 24] for the direct case.

###### Definition 2.1.

The triplet given by the temperature and two unknown thermal coefficients among , , and is a solution to the inverse fractional one-phase Stefan problem (2) if:

1. is defined in .

2. , where and .

3. is a continuous function over , where , except maybe in the point , in which it must be satisfied the following condition:

 0≤liminf(x,t)→(0,0)T(x,t)≤limsup(x,t)→(0,0)T(x,t)<∞.
4. For all , exists .

5. The unknown thermal coefficients are positive real numbers.

6. The temperature and the two unknown thermal coefficients verify (2).

###### Remark 1.

We note that the function given by (3) is consistent with the definition of being a solution to a (direct) fractional one-phase Stefan problem given in [23, 24], since it is a positive function belonging to .

Encouraged by [23, 24, 25, 32], in this section we are looking for a solution of similarity type to Problem (2). That is, we look for a temperature function such that:

 T(x,t)=A+B(1−W(−x√μαλtα/2,−α2,1)),00, (5)

where and are real numbers that must be determined, and is the Wright function given by [36, 37]:

 W(z,a,b)=+∞∑k=0zkk!Γ(ak+b),z∈C,a>−1,b∈C. (6)

According to [23, 24], we know that the function given by (5) fulfill conditons 1 to 4 in Definition 2.1 and that it satisfies the fractional diffusion equation (2a). Noting that , it follows from condition (2d) that:

 A=T0. (7)

From this and condition (2b), we have that:

 B=Tm−T01−W(−σ√μαλ,−α2,1). (8)

We observe that because is a strictly decreasing function in [23] and, as we have already noted, .

By taking into consideration that the Wright function satisfies [36, 37]:

 ddzW(z,a,b)=W(z,a,a+b),z∈C,a>−1,b∈C,

and the fractional Caputo derivative of a power function with positive exponent is given by [22]:

 Dαtp=Γ(p+1)Γ(p−α+1)tp−α,t>0,p>0,

it follows from the fractional Stefan condition (2c) that:

 Missing or unrecognized delimiter for \left (9)

where is the Mainardi function, which is defined by [21]:

 Mν(z)=W(−z,−ν,1−ν),z∈C,0<ν<1 (10)

and satisfies for all [23].

Finally, when we consider given by (8), the heat flux boundary condition (2e) implies that it must be satisfied the following equality:

 √μαq0λk[1−W(−σ√μαλ,−α2,1)]=T0−TmΓ(−α2+1). (11)

We have thus proved the following result:

###### Theorem 2.1.

If the moving boundary is defined by (3), then the function given by:

 Missing or unrecognized delimiter for \left (12)

is a solution to Problem (2) with two unknown thermal coefficients among , , and , if and only if these are a solution to the following system of equations:

 ξ[1−W(−ξ,−α2,1)]Mα2(ξ)=c(T0−Tm)Γ(−α2+1)μαναlΓ(α2+1) (13a) 1−W(−ξ,−α2,1)=√kρc(T0−Tm)√μαq0Γ(−α2+1) (13b)

where the dimensionless parameter is defined by:

 ξ=σ√μαλ. (14)

## 3 Existence and uniqueness of solutions of similarity type. Formulae for the two unknown thermal coefficients.

In this section we will look for necessary and sufficient conditions on data to have existence and uniqueness of solution to Problem (2) for each possible choice of the two unknown thermal coefficients, as well as explicit formulae for them. Thanks to Theorem 2.1, it can be done through analysing and solving the system of equations (13) for each pair of unknown thermal coefficients. With the aim of organizing the main results of this section, we will write:

For each , we introduce the real functions , and defined in by:

 Fα(x)=fα(x)x (15a) Gα(x)=xfα(x) (15b) Hα(x)=xfα(x)Mα2(x) (15c)

where

 fα(x)=1−W(−x,−α2,1). (16)

The following result will be useful all throughout this section:

###### Lemma 3.1.

For any , the real functions , and defined in (15) satisfy the following conditions:

 Fα(0+) =1Γ(−α2+1), Fα(+∞) =0, F′α(x) <0∀x>0 (17a) Gα(0+) =0, Gα(+∞) =+∞, G′α(x) >0∀x>0 (17b) Hα(0+) =0, Hα(+∞) =+∞, H′α(x) >0∀x>0. (17c)
###### Proof.

The proof of (17a) was done in [32]. The demonstrations of (17b) and (17c) follow from elementary computations and the following facts:

1. Since , is a positive and strictly increasing function in [23].

2. Since , is a positive and strictly decreasing function in [23].

3. and [9].

###### Theorem 3.1 (Case 1: Determination of l and c).

If the moving boundary is given by (3), then the Problem (2) admits the solution , and given by (12) and:

 c=1ρk⎡⎢ ⎢⎣q0√μαΓ(−α2+1)[1−W(−ξ,−α2,1)]T0−Tm⎤⎥ ⎥⎦2 (18a) l=q20Γ3(−α2+1)[1−W(−ξ,−α2,1)]Mα2(ξ)ναρk(T0−Tm)Γ(α2+1)ξ (18b)

respectively, being the only one solution to the equation:

 Fα(x)=k(T0−Tm)σq0Γ(−α2+1),x>0, (19)

if and only if the following inequality holds:

 k(T0−Tm)σq0<1. (20)
###### Proof.

Isolating from equation (13b), we have that is given by (18a). Now, by combining this with equation (13a), it can be obtained that is given by (18b). It must be noted that the parameter involved in both (18a) and (18b) depends on . Nevertheless, it can be determined without making any reference to as follows. By replacing (18a) in the definition of given in (14), we have that must be a solution to equation (19). It follows from (17a) that the equation (19) admits a solution if and only if its RHS is between and . To complete the proof only remains to observe that this is equivalent to say that inequality (20) must hold and that, when this happens, equation (19) has an only one positive solution. ∎

###### Theorem 3.2 (Case 2: Determination of c and k).

If the moving boundary is given by (3), then the Problem (2) admits the solution , and given by (12) and:

 c=μαναlΓ(α2+1)ξ[1−W(−ξ,−α2,1)](T0−Tm)Γ(−α2+1)Mα2(ξ) (21a) k=q20Γ3(−α2+1)[1−W(−ξ,−α2,1)]Mα2(ξ)ναρl(T0−Tm)Γ(α2+1)ξ (21b)

respectively, being the only one solution to the equation:

 Mα2(x)=νασρlΓ(α2+1)q0Γ2(−α2+1),x>0, (22)

if and only if the following inequality holds:

 νασρlΓ(α2+1)q0Γ(−α2+1)<1. (23)
###### Proof.

By following the same steps as in the demonstration of the Theorem 3.1, it can be shown that and must be given by (21), where the parameter should be a solution to equation (22). Since the Mainardi function is a strictly decreasing function from to in [23], we have that the equation (22) admits a solution if and only if its RHS is between and . This is equivalent to say that inequality (23) must holds. Moreover, when data satisfy (23), equation (22) has an only one positive solution. ∎

###### Theorem 3.3 (Case 3: Determination of l and k).

If the moving boundary is given by (3), then the Problem (2) admits the solution , and given by (12) and:

 k=1ρc⎡⎢ ⎢⎣q0√μαΓ(−α2+1)[1−W(−ξ,−α2,1)]T0−Tm⎤⎥ ⎥⎦2 (24a) l=c(T0−Tm)Γ(−α2+1)Mα2(ξ)μαναΓ(α2+1)ξ[1−W(−ξ,−α2,1)] (24b)

respectively, being the only one solution to the equation:

 Gα(x)=σρc(T0−Tm)μαq0Γ(−α2+1),x>0. (25)
###### Proof.

By proceeding analogously to the proofs of the previous Theorems 3.1 and 3.2, we have that and should be given by (24), being a solution to equation (25). Since the RHS of the equation (25) is a positive number, it follows from (17b) that the equation (25) admits an only one solution for any set of data. ∎

###### Theorem 3.4 (Case 4: Determination of c and ρ).

If the moving boundary is given by (3), then the Problem (2) admits the solution , and given by (12), (21a) and:

 ρ=q20Γ3(−α2+1)[1−W(−ξ,−α2,1)]Mα2(ξ)ναkl(T0−Tm)Γ(α2+1)ξ (26a)

respectively, being the only one solution to the equation (19), if and only if inequality (20) holds.

###### Proof.

It is similar to the demonstration of Theorem 3.1. ∎

###### Theorem 3.5 (Case 5: Determination of l and ρ).

If the moving boundary is given by (3), then the Problem (2) admits the solution , and given by (12), (24b) and:

 Missing or unrecognized delimiter for \right (27a)

respectively, being the only one solution to the equation (19), if and only if inequality (20) holds.

###### Proof.

It is similar to the demonstration of Theorem 3.1. ∎

###### Theorem 3.6 (Case 6: Determination of ρ and k).

If the moving boundary is given by (3), then the Problem (2) admits the solution , and given by (12) and:

 Missing or unrecognized delimiter for \right (28a) k=σq0Γ(−α2+1)[1−W(−ξ,−α2,1)](T0−Tm)ξ (28b)

respectively, being the only one solution to the equation:

 Hα(x)=c(T0−Tm)Γ(−α2+1)μαναlΓ(α2+1),x>0. (29)
###### Proof.

By following the same ideas as in the proof of the Theorem 3.1, we have that and must be given by (28), where should be a solution to equation (29). By noting that the RHS of this equation is a positive number, it follows from (17c) that the equation (29) admits an only one positive solution for any set of data. ∎

Table 1 summarizes the formulae obtained for the two unknown thermal coefficients and the condition that data must verify to obtain them, for each one of the six possible choices of the two unknown thermal coefficients among , , and in Problem (2) when the moving boundary is defined by (3).

## 4 Convergence to the classic case when α→1−

When , the time fractional derivative of order in the Caputo sense of a function coincides with its classical time derivative. Then, if we allow to be equal to 1 in Problem (2) and we consider that case, we obtain that Problem (2) reduces to the classical inverse one-phase Stefan problem studied in [28]. This problem, which we will refer to as Problem (2), is given by the classical diffusion equation:

 Tt(x,t)=λ2Txx(x,t),00, (\ref{1}${}^{\star}$)

the classical Stefan condition:

 −kTx(s(t),t)=ρl˙s(t),t>0 (\ref{3}${}^{\star}$)

and conditions (2b), (2d) and (2e). Of course, to obtain ((\ref{1}${}^{\star}$)) and ((\ref{3}${}^{\star}$)) we have also considered and in (2a) and (2c), respectively. According to the physical interpretation given in [35], as increases its value to 1, the ability of memory retention of the system diminishes to the limit case of no memory retention corresponding to . In this context, classical Stefan problems are representing phase-change processes whose temporal evolution can be described in terms of local in time properties (absence of memory).

The determination of two unknown thermal coefficients through a classical inverse one-phase Stefan problem was done in [28]. In that article, necessary and sufficient conditions on data to obtain existence and uniqueness of solution to Problem (2) are given, together with formulae for the unknown thermal coefficients. In several articles [23, 24, 25, 32] it has been proved the convergence when of the solution to a fractional Stefan problem with to the solution to the associated classical problem obtained by considering . Encouraged by those works, we are interested in this section in proving the convergence when of the results obtained in Section 3 to the ones given in [28].

In order to emphasize the dependence on of the formulae given in Theorems 3.1 to 3.6, we will mention it here explicitly. For example, if we are analysing the convergence of the solution to Problem (2) given in Theorem 3.1, we will refer to it as , and . We will also write to represent the coefficient defined by (14).

The next result will be useful in the following:

###### Lemma 4.1.

For each , the Wright and Mainardi functions are such that:

 Missing or unrecognized delimiter for \left when α →1− (30a) Mα/2(x)→1√πexp(−x24), when α →1−, (30b)

where is the error function, which is defined by:

 erf(x)=2√π∫x0exp(s2)ds,x>0.

See [23]. ∎

###### Theorem 4.1 (Convergence related to Case 1).

If inequality (20) holds, then the solution , , to Problem (2) given in Theorem 3.1 converges to the solution obtained in [28], which is given by:

 T(x,t)=T0+T0−Tmerf(σ⋆λ)erf(x2λ√t),00 (31a) c=kρ(ξ⋆σ⋆)2 (31b) l=q0exp(−ξ⋆2)ρσ⋆ (31c)

where is defined by:

 σ⋆=σ2 (32)

and is the only one solution to the equation:

 erf(x)x=k(T0−Tm)q0σ⋆√π,x>0. (33)
###### Proof.

Taking the limit when into both sides of the equation (19) and using (30b), we obtain the following equation:

 erf(x/2)x=k(T0−Tm)q0σ√π,x>0. (34)

On one hand, we have that the LHS of the equation (34) defines a strictly decreasing function from to in . On the other hand, since inequality (20) holds, the RHS of equation (20) is between and . Therefore, it follows that equation (34) has an only one positive solution. By introducing the parameter defined by (32), we can rewrite equation (34) as follows:

 erf(x/2)x/2=k(T0−Tm)q0σ⋆√π,x>0,

and see that the solution to the equation (19) is such that:

 ξ(α)→2ξ⋆,when α→1− (35)

being the only one solution to the equation (33). From (1), (30) and (35), it follows from elementary computations that:

 c(α)→c,when α→1− (36a) l(α)→l,when α→1− (36b)

where and are given by (31b) and (31c), respectively. Finally, it follows from (1a), (30), (35) and (36) that:

 T(x,t,α)→T(x,t),when α→1−, (37)

for each pair with and , where is given by (31a). We then have proved that the solution to Problem (2) given in Theorem 3.1 converges to the solution to Problem (2) given in [28] when . ∎

###### Remark 2.

We note that inequality (20) can be written as:

 k(T0−Tm)2σ⋆q0<1, (38)

being the parameter defined by (32), which is the condition established in [28] to ensure the existence and uniqueness of the solution to Problem (2) given by (31).

###### Theorem 4.2 (Convergence related to Case 2).

If inequality (23) holds for each , then the solution , , to Problem (2) given in Theorem 3.2 converges to the solution obtained in [28], which is given by (31a) and:

 c=q0√πξ⋆erf(ξ⋆)ρσ⋆(T0−Tm) (39a) k=σ⋆q0√πerf(ξ⋆)(T0−Tm)ξ⋆ (39b)

where is given by (32) and is the only one solution to the equation:

 exp(x2)=q0ρlσ⋆,x>0. (40)
###### Proof.

If we take the limit when side by side of equation (22) and we have into consideration (1b) and (30b), the following equation is obtained:

 exp(x24)=2q0σρl,x>0. (41)

Since inequality (23) holds for all , we have that the following inequality also holds:

 2q0σρl<1. (42)

Therefore, equation (41) admits only one positive solution. We note that equation (41) can be rewritten as:

 exp((x2)2)=q0σ⋆ρl,x>0,

where is given by (32), from which we can see that the solution to equation (22) is such that:

 ξ(α)→2ξ⋆,when α→1−, (43)

being the only one solution to equation (40). It follows now from (1) (30), (43) and elementary computations that:

 c(α)→c (44a) k(α)→k (44b)

where and are given by (21), respectively. Finally, we have from (1a), (30), (43) and (44) that satisfies (37). ∎

###### Remark 3.

By introducing the parameter defined by (32), we have that the inequality (42) can be rewritten as:

 q0ρlσ⋆>1, (45)

which is the condition established in [28] to ensure the existence and uniqueness of the solution to Problem (2) given by (31a) and (39).

###### Theorem 4.3 (Convergence related to Case 3).

If inequality (20) holds, then the solution , , to Problem (2) given in Theorem 3.3 converges to the solution obtained in [28], which is given by (31a) and:

 l=q0exp(−ξ⋆2)ρσ⋆ (46a) k=ρc(σ⋆ξ⋆)2 (46b)

where is given by (32) and is the only one solution to the equation:

 xerf(x)=ρcσ⋆(T0−Tm)q0√π,x>0. (47)
###### Proof.

Taking the limit when into both sides of the equation (25) and having into consideration (1a) and (30a), we obtain the following equation:

 xerf(x2)=σρc(T0−Tm)q0√π,x>0. (48)

Since the LHS of equation (48) defines a strictly increasing function from to in and the RHS of equation (48) is a positive number, it follows that equation (48) has an only one positive solution. By introducing the parameter defined by (32), the equation (48) can be rewritten as:

 x2erf(x2)=σ⋆ρc(T0−Tm)q0√π,x>0,

from which we can see that the solution to the equation (25) is such that:

 ξ(α)→2ξ⋆,when α→1− (49)

being the only one solution to equation (47). The rest of the proof runs as before. ∎

The following results can be proved in the same manner than the previous theorems in this section. Then, we have prefered to omit their proofs.

###### Theorem 4.4 (Convergence related to Case 4).

If inequality (20) holds, then the solution , , to Problem (2) given in Theorem 3.4 converges to the solution obtained in [28], which is given by (31a) and:

 c=klξ⋆2exp(ξ⋆2)q0√π (50a) ρ=q0exp(−ξ⋆2)lσ⋆ (50b)

where is given by (32) and is the only one solution to the equation (33).

###### Theorem 4.5 (Convergence related to Case 5).

The solution , , to Problem (2) given in Theorem 3.5 converges to the solution obtained in [28], which is given by (31a) and:

 l=q0cσ⋆exp(−