Existence of an exact solution for a one-phase Stefan problem with nonlinear thermal coefficients from Tirskii’s method

Nonlinear Analysis 67 (2007) 1989–1998 www.elsevier.com/locate/na

Existence of an exact solution for a one-phase Stefan problem with nonlinear thermal coefficients from Tirskii’s method Adriana C. Briozzo, Mar´ıa Fernanda Natale, Domingo A. Tarzia ∗,1 Depto de Matem´atica, F.C.E., Universidad Austral, Paraguay 1950, S2000FZF Rosario, Argentina Received 5 April 2006; accepted 25 July 2006

Abstract The mathematical analysis of a one-phase Lam´e–Clapeyron–Stefan problem with nonlinear thermal coefficients following [G.A. Tirskii, Two exact solutions of Stefan’s nonlinear problem, Sov. Phys. Dokl. 4 (1959) 288–292] is obtained. Two related cases are considered; one of them has a temperature condition on the fixed face x = 0 and the other one has a flux condition of the type √ −q0 / t (q0 > 0). We obtain in both cases sufficient conditions for data in order to have the existence of an explicit solution of a similarity type which is given by using a double fixed point. c 2007 Elsevier Ltd. All rights reserved.

MSC: 35R35; 80A22; 35C05; 45G10 Keywords: Stefan problem; Free boundary problem; Nonlinear thermal coefficients; Explicit solution; Nonlinear integral equation; Melting

1. Introduction The Lam´e–Clapeyron–Stefan problem is nonlinear even in its simplest form due to the free boundary conditions. If the thermal coefficients of the material are temperature dependent, we have a double nonlinear free boundary problem. The present study provides the existence of an exact solution of the similarity type to a one-phase melting problem. We consider the following free boundary problem for a semi-infinite material [1,2]: ρ(T )c(T )Tt = (k(T )Tx )x , T (0, t) = Tb T (s(t), t) = Tm

0 < x < s(t)

(1) (2) (3)

k (T (s(t), t)) Tx (s(t), t) = −ρ0 l s 0 (t)

(4)

s(0) = 0

(5)

∗ Corresponding author. Tel.: +54 341 5223000; fax: +54 341 5223001.

E-mail addresses: [email protected] (A.C. Briozzo), [email protected] (M.F. Natale), [email protected] (D.A. Tarzia). 1 CONICET, Argentina. c 2007 Elsevier Ltd. All rights reserved. 0362-546X/$ - see front matter doi:10.1016/j.na.2006.07.047

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where T = T (x, t) is the temperature of the liquid phase; ρ(T ), c(T ) and k(T ) are the body’s density, its specific heat, and its thermal conductivity, respectively; Tm is the phase change temperature, Tb > Tm is the temperature on the fixed face x = 0; ρ0 > 0 is the constant density of mass at the melting temperature; l > 0 is the latent heat of fusion by unit of mass and s(t) is the position of the phase change location. This problem was first considered in [3] where the integral equation (19) was obtained but no mathematical result is given. The plan of the paper is the following. In Section 2 we prove the existence of at least one explicit solution of a similarity type for the problem (1)–(5) by using a double fixed point for the integral equation (19) and the transcendental equation (21) under a certain hypothesis for data. In Section 3 we consider an analogous problem (1) and (3)–(5) and the temperature condition (2) will be replaced by the following flux condition: √ k(T (0, t))Tx (0, t) = −q0 / t (6) √ at the fixed face x = 0 where q0 is a positive constant. Here −q0 / t denotes the prescribed flux on the boundary x = 0 which is of the type imposed in [4]. Furthermore, this kind of heat flux on the fixed boundary has also been considered in several applied problems, e.g. [5–7]. We prove the existence of at least one explicit solution of a similar type for all thermal conditions. Different methods in order to prove the existence of a solution for the one-phase Stefan problem were considered: integral equation [8–12]; retarding the argument [13]; by the limit of a sequence of approximating solutions [14,15]. 2. The one-phase Stefan problem with nonlinear thermal coefficients and temperature boundary condition on the fixed face If we define the following transformation: θ (x, t) = (T (x, t) − Tm )/ (Tb − Tm )

(7)

then the problem (1)–(5) becomes N (θ )θt = α0 (L(θ )θx )x ,

0 < x < s(t)

(8)

θ (0, t) = 1 θ (s(t), t) = 0

(9) (10)

k(Tm )θx (s(t), t) = −ρ0 l s 0 (t)/ (Tb − Tm )

(11)

s(0) = 0

(12)

where N (T ) = ρ(T )c(T )/ (ρ0 c0 ) , L(T ) = k(T )/k0 and k0 , ρ0 , c0 and α0 = k0 / (ρ0 c0 ) are the reference thermal conductivity, density of mass, specific heat and thermal diffusivity respectively. Now we assume a similarity solution of the type √
(13) θ (x, t) = f (η), η = x/ 2 α0 t . Taking into account that problem (8)–(12) is a classical Stefan-like problem with nonlinear thermal coefficient, the free boundary condition (10) implies that the free boundary s(t) must be of the type √ s(t) = 2η0 α0 t (14) where η0 is a positive parameter to be determined later. Therefore, the conditions (8)–(11) become the following:  0 L( f ) f 0 (η) + 2ηN ( f ) f 0 (η) = 0, 0 < η < η0

(15)

f (0) = 1 f (η0 ) = 0

(16) (17)

f 0 (η0 ) = −2η0 α0 ρ0l/ [k(Tm )(Tb − Tm )] .

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The problem (15)–(17) is equivalent to the following nonlinear integral equation of Volterra type: f (η) = 1 − Φ [η, L( f ), N ( f )] /Φ [η0 , L( f ), N ( f )]

(19)

where Φ is given by Z η √ E(t, f )/L( f )(t) dt Φ [η, L( f ), N ( f )] := (2/ π ) 0   Z x s N ( f (s))/L( f (s)) ds . E(x, f ) := exp −2

(20)

0

The condition (18) becomes  √  E(η0 , f ) Φ [η0 , L( f ), N ( f )] = η0l π [c0 (Tb − Tm )]

(21)

and then the following theorem holds. Theorem 1. The√solution of the free boundary problem (1)–(5) is given by (14) and T (x, t) = Tm + (Tb − Tm ) f (η),
with η = x/ 2 α0 t where the function f = f (η) and the coefficient η0 > 0 must satisfy the nonlinear integral equation (19) and the condition (21) respectively.  Firstly, in order to prove the existence of the solution to the system (19) and (21) we will obtain some preliminary results. Then we will prove that the integral equation (19) has a unique solution for any given η0 > 0 by using a fixed point theorem. Secondly, in order to solve the problem (1)–(5) we will consider Eq. (21). For convenience of notation, we will define Φ [η, f ] ≡ Φ [η, L( f ), N ( f )]. We suppose that there exists Nm , N M , L m , L M positive constants such as L m ≤ L(T ) ≤ L M ,

Nm ≤ N (T ) ≤ N M .

(22)

Furthermore, we assume that the dimensionless thermal conductivity and specific heat are Lipschitz functions, i.e., e that verify the following: there exist positive constants e L and N

∞ |L(g) − L(h)| ≤ e L kg − hk , ∀g, h ∈ C 0 R+ R+ (23) 0 ∩L 0

e kg − hk , ∀g, h ∈ C 0 R+ ∩ L ∞ R+ . |N (g) − N (h)| ≤ N (24) 0 0 Then we get: Lemma 2. We have exp(−N M x 2 /L m ) ≤ E(x, f ) ≤ exp(−Nm x 2 /L M ),

∀x > 0.

Lemma 3. For 0 < η < η0 we have p p L m /N M erf( N M /L m η)/L M ≤ Φ [η, f ] p p ≤ L M /Nm erf( Nm /L M η)/L m .

(26)

Proof. Taking into account Lemma 2 we have Z η Z √ √ Φ [η, f ] ≤ (2/( π L m )) E(v, f ) dv ≤ (2/ π L m ) 0

η

  exp −Nm v 2 /L M dv 0 Z η p √ = (2 L M /Nm /( π L m )) exp(−t 2 ) dt 0 p p = ( L M /Nm /L m )erf( Nm /L M η).

Analogously we can obtain the other inequality.

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We consider C 0 [0, η0 ], the space of continuous real functions defined on [0, η0 ], with its norm k f k = maxη∈[0,η0 ] | f (η)|. Lemma 4. Let η0 be a given positive real number. We suppose that the dimensionless thermal conductivity and specific heat verify conditions (22)–(24). Then, for all f, f ∗ ∈ C 0 [0, η0 ] we have

 
E [η, f ] − E η, f ∗ ≤ (η2 /L m ) N e + NM e L/L m f ∗ − f ,

∀η ∈ (0, η0 ) .

(27)

Proof. If we consider the following inequality: |exp(−x) − exp(−y)| ≤ |x − y| ,

∀x, y ≥ 0,

then we get   E [η, f ] − E η, f ∗  Z η   Z η  ∗ ∗ u N ( f (u))/L( f (u)) du − exp −2 u N ( f (u))/L( f (u))du = exp −2 0 0 Z η Z η ∗ ∗ ≤ 2 u N ( f (u))/L( f (u)) du − u N ( f (u))/L( f (u)) du 0 Z 0η N ( f (u))/L( f (u)) − N ( f ∗ (u))/L( f ∗ (u)) u du ≤2 0


e + NM e ≤ (η2 /L m ) N L/L m f ∗ − f .  Lemma 5. Let η0 be a given positive real number. We suppose that (22)–(24) holds. For all f, f ∗ ∈ C 0 [0, η0 ] , 0 < η < η0 we have   Φ [η, f ] − Φ η, f ∗

√ e + NM e (28) ≤ (2η/(L 2m π ))(( N L/L m )η2 /3 + e L) f ∗ − f . Proof. (i) We have   Φ [η, f ] − Φ η, f ∗   Z v Z √
η ≤ 2/ π u N ( f (u))/L( f (u)) du exp −2 0   0Z v  ∗ ∗ − exp −2 u N ( f (u))/L( f (u))u du L( f (v)) dv 0  Z v  Z √
η  ∗ ∗ ∗ + 2/ π 1 L( f (v)) − 1/L( f (v)) exp −2 u N ( f (s))/L( f (s)) du dv 0

0

≡ T1 (η) + T2 (η). It follows from (23) that Z √
η 1/L( f (v)) − 1/L( f ∗ (v)) dv T2 (η) ≤ 2/ π 0

Z

√
η L( f ∗ (v)) − L( f (v)) / L( f (v))L( f ∗ (v)) dv ≤ 2/ π 0

√ ≤ [2e L η/( π L 2m )] f ∗ − f . Taking into account Lemma 4 we have that the term T1 (η) can also be bounded in the following way:

(29)

A.C. Briozzo et al. / Nonlinear Analysis 67 (2007) 1989–1998

√
T1 (η) ≤ 2/ π

Z

1993

η

  E [v, f ] − E v, f ∗ /L m dv 0 Z 




η  2 √ e + NM e ≤ 4/ π L m v /L m N L/L m f ∗ − f dv 0 Z


η 2 √ 2 ∗ e + NM e ≤ (2/( π L m )) f − f N v dv L/L m √

= (2η /(3 3

0

π L 2m ))


e + NM e N L/L m f ∗ − f .

Therefore, we obtain (28) by using (29) and (30).

(30)



Theorem 6. Let η0 be a given positive real number. We suppose that (22)–(24) holds. If η0 satisfies the following inequality:  q ! ! 5/2 Nm η L N erf η e M 0 0 η02 M LM 4 N L M e+ (31) +e L <1 β(η0 ) := √ N q  NM L 3 2 Nm π 4 m η L erf m

Lm

0

then there exists a unique solution f ∈ C 0 [0, η0 ] of the integral equation (19). Proof. Let W : C 0 [0, η0 ] −→ C 0 [0, η0 ] be the operator defined by W ( f )(η) = 1 − Φ [η, L( f )] /Φ [η0 , L( f )] ,

f ∈ C 0 [0, η0 ].

(32)

The solution to the equation (19) is the fixed point of the operator W , that is W ( f (η)) = f (η),

0 < η < η0 .

(33)

Let us have f, f ∗ ∈ C 0 [0, η0 ]; then we obtain



W ( f ) − W ( f ∗ ) = Maxη∈[0,η ] W ( f (η)) − W ( f ∗ (η)) 0    
 
≤ Max Φ η, f ∗ Φ [η0 , f ] − Φ η0 , f ∗ Φ [η, f ] / Φ [η0 , f ] Φ η0 , f ∗ η∈[0,η0 ]     ≤ A Max Φ η, f ∗ Φ [η0 , f ] − Φ η0 , f ∗ Φ [η, f ] η∈[0,η0 ]     ≤ A Max Φ η, f ∗ Φ [η0 , f ] − Φ η0 , f ∗ η∈[0,η0 ]
    + Φ η0 , f ∗ Φ η, f ∗ − Φ [η, f ] where   p  A = N M L 2M / L m erf2 η0 N M /L m > 0.

(34)

Finally, for Lemmas 3 and 5 and taking into account that 0 < η < η0 , we have



W ( f ) − W ( f ∗ ) ≤ β (η0 ) f ∗ − f . Then W is a contraction operator and therefore there exists a unique solution of the integral equation (19) if the condition (31) is satisfied.  Remark 1. The solution f to the integral equation (19), given by Theorem 6, depends on the real number η0 > 0. For convenience in the notation from now on we take f (η) = f η0 (η) = f (η0 , η),

0 < η < η0 , η0 > 0.

Let Ω be the set defined by  Ω = η0 ∈ R+ /β (η0 ) < 1  = η0 ∈ R+ /there exists a solution of Eq. (19) .



(35)

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Lemma 7. If 2L 2M e L/L 3m < 1 there exists a positive number β(η0 ) < 1

(36) η0∗

such that

if 0 < η0 < η0∗ ,

β(η0 ) ≥ 1

if η0 ≥ η0∗ .

L/L 3m , β(+∞) = +∞ and β 0 (η0 ) > 0 ∀η0 > 0. Then Ω = Proof. We have β(0) = 2L 2M e ∗ β(η0 ) = 1. 


0, η0∗ where

To prove the existence of the solution to the Eq. (21), we define the real function H (x) := E(x, f )/Φ [x, f ] ,

x >0

where f is the solution to Eq. (19) given by Theorem 6. Lemma 8. If (22) holds, then function H (x) verifies: (i) H2 (x) ≤ H (x) ≤ H1 (x) where p p p H1 (x) := L M N M exp(−Nm x 2 /L M )/( L m erf(x N M /L m )), p p p H2 (x) := L m Nm exp(−N M x 2 /L m )/( L M erf(x Nm /L M )); (ii) H (0) = +∞, H (+∞) = 0. Proof. By Lemmas 2 and 3 we have (i). Moreover H1 and H2 are decreasing functions which satisfy Hi (0) = +∞, Hi (+∞) = 0 (i = 1, 2); therefore (ii) holds.  Theorem 9. Eq. (21) has at least one solution η0 . Moreover, if x0 is the unique solution to equation √ H1 (x) = xl π / (c0 (Tb − Tm )) , x > 0, and x0 < η0∗ then η0 ∈ Ω with η0 < x0 . Proof. √ Eq. (21) ⇐⇒ H (x) = xl π / (c0 (Tb − Tm )) ,

x >0

and then, by Lemma 8, there exists at least one solution η0 > 0 of Eq. (21). Due to the properties of H1 (x) the equation √ H1 (x) = xl π / (c0 (Tb − Tm )) , x > 0, (37) has a unique solution x0 . Furthermore η0 < x0 and since β is an increasing function, then we have β(x0 ) < β(η0∗ ) = 1, and then we have β(η0 ) < 1, that is η0 ∈ Ω .  Remark 2. The solution x0 to Eq. (37) can be expressed as follows: p p x0 := M −1 (L M N M c0 (Tb − Tm )/( π L m l))

(38)

where p M(x) := x erf( N M /L m x) exp(x 2 Nm /L M ) is an increasing real function. Then we have p p β(x0 ) < 1 ⇐⇒ β(M −1 (L M N M c0 (Tb − Tm )/( π L m l))) < 1. And so we have the following theorem.



A.C. Briozzo et al. / Nonlinear Analysis 67 (2007) 1989–1998

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Theorem 10. (i) If N and L verify the conditions (22)–(24) and (36) and β(x0 ) < 1 where x0 is defined by (38), then there exists at least one solution of the problem (1)–(5) where the free √ boundary s(t) is given by (14) and the temperature is given by T (x, t) = Tm + (Tb − Tm ) f (η), with η = x/2 α0 t where f is the unique solution to the integral equation (19) and η0 is given by Theorem 9. (ii) If N and L verify the conditions (22)–(24) and (36) then there exists at least one solution to the problem (1)–(5) for all latent heats of fusion l > l0 for given other parameters where l0 is given by p p l0 := L M N M c0 (Tb − Tm )/( π L m M(η0∗ ))
where η0∗ > 0 is characterized by the condition β η0∗ = 1.  Remark 3. The existence of a solution to the problem (1)–(5) is given for large latent heat of fusion (∀l > l0 ) if conditions (22)–(24) and (36) for the thermal coefficients are verified. This is equivalent to saying that there exists a solution for all small Stefan numbers Ste, i.e. ∀Ste < Ste0 where Ste = c0 (Tb − Tm )/l;

Ste0 = c0 (Tb − Tm )/l0 .

(39)

3. Solution to the free boundary problem with a heat flux condition on the fixed face In this section we consider the problems (1)–(5), but condition (2) will be replaced by condition (6). If we define the following transformation: θ (x, t) = (T (x, t) − Tm ) /Tm

(T (x, t) = Tm + Tm θ (x, t))

(40)

then the problem to solve becomes N (θ )θt = α0 (L(θ )θx )x ,

0 < x < s(t) √ k (Tm (θ (0, t) + 1)) θx (0, t) = −q0 /(Tm t) θ (s(t), t) = 0 k(Tm )θx (s(t), t) = −ρ0 ls 0 (t)/Tm

(41)

s(0) = 0.

(45)

(42) (43) (44)

Now we assume a similarity solution given by (13). Then the free boundary condition (10) implies that the free boundary s(t) must be of the type (14) where η0 is a positive parameter to be determined later. Therefore, the conditions (41)–(45) become the following:  0 L( f ) f 0 (η) + 2ηN ( f ) f 0 (η) = 0, 0 < η < η0 (46) √ 0 k(Tm ( f (0) + 1)) f (0) = −2 α0 q0 /Tm (47) f (η0 ) = 0

(48)

f (η0 ) = −2η0 α0 ρ0 / (k(Tm )Tm )

(49)

s(0) = 0.

(50)

0

We have that the problem (46)–(48) is equivalent to the following nonlinear integral equation of Volterra type: √ f (η) = lη0 π (Φ [η0 , L( f ), N ( f )] − Φ [η, L( f ), N ( f )]) / (c0 Tm E(η0 , f )) (51) where Φ and E were defined in (20). The condition (47) becomes √ E(η0 , f ) = l ρ0 η0 α0 /q0 .

(52)

Theorem 11. The solution to the free boundary problem (1) and (3)–(6) is given by (14) and T (x, t) = Tm + √ Tm f (η) , η = x/(2 α0 t) where the function f = f (η) and the coefficient η0 > 0 must satisfy the nonlinear integral equation (51) and the condition (52). 

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Theorem 12. Let η0 be a given positive real number. We suppose that (22)–(24) hold. If η0 satisfies the following inequality: ! !  (   4η02l NM e Lη02 η02 Nm 2 2 NM 2 e e γ (η0 ) = exp N+ η 2 exp − η +L Lm 0 LM 0 Lm 3 c0 Tm L 2m s s ! !) NM e Lη02 LM Nm e+ N + η0 erf η0 <1 (53) Nm LM Lm then there exists a unique solution to the integral equation (51). Proof. Let R : C 0 [0, η0 ] −→ C 0 [0, η0 ] be the operator defined by √ R( f )(η) = lη0 π (Φ [η0 , f ] − Φ [η, f ]) / (c0 Tm E(η0 , f )) ,

f ∈ C 0 [0, η0 ], 0 < η < η0 .

(54)

The solution to the equation (51) is the fixed point of the operator R, that is R( f (η)) = f (η),

0 < η < η0 .

(55)

Let us have f, f ∗ ∈ C 0 [0, η0 ]; then we obtain R( f ) − R( f ∗ )
√ ≤ lη0 π / c0 Tm E(η0 , f )E(η0 , f ∗ )   
 · E(η0 , f ) Φ η0 , f ∗ − Φ η, f ∗ + E(η, f ) (Φ [η, f ] − Φ [η0 , f ])
√ ≤ lη0 π (U1 + U2 + U3 ) / c0 Tm E(η0 , f )E(η0 , f ∗ ) where  
U1 = E(η0 , f ) Φ η0 , f ∗ − Φ [η0 , f ] 
 U2 = E(η0 , f ∗ ) Φ η0 , f ∗ − Φ [η0 , f ]     U3 = E(η0 , f )Φ [η0 , f ] − E(η0 , f ∗ )Φ η0 , f ∗ − E(η0 , f )Φ η, f ∗ + E(η0 , f ∗ )Φ [η, f ] . Taking into account Lemmas 2–5 we obtain



U2 ≤ U2∗ f − f ∗ , U1 ≤ U1∗ f − f ∗ ,

U3 ≤ U3∗ f − f ∗

where  . √        e + NM e U1∗ = 2η0 L 2m π exp −Nm η02 /L M L N Lη02 /L m η02 /3 + e  . √        e + NM e U2∗ = 2η0 L 2m π exp −Nm η02 /L M N Lη02 /L m η02 /3 + e L p p √ e + NM e Lη02 /L m ). U3∗ = 2U1 + [4η02 L M /Nm /(L 2m π )]erf( Nm /L M η0 )( N Finally, we have



R( f ) − R( f ∗ ) ≤ γ (η0 ) f ∗ − f . Then, there exists a unique solution to the integral equation (51) if condition (53) is verified (i.e. R is a contraction operator).  Lemma 13. Function γ = γ (η), given by ! !  (   4η2l Nm 2 NM e L η2 e 2 NM 2 e η 2 exp − η N+ +L γ (η) = exp Lm LM Lm 3 c0 Tm L 2m s s ! !) e N L LM Nm M e+ N , η>0 + η erf η Nm LM Lm

(56)

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satisfies the following properties: (i) γ (0) = 0, (ii) γ (+∞) = +∞, (iii) γ is an increasing function and ∃e η > 0 / γ (η) < 1 ∀η ∈ (0, e η). Theorem 14. Eq. (52) has at least one solution η0 . Moreover, if x1 is the unique solution to the equation √  I1 (x) = l ρ0 x α0 q0 , x > 0 and satisfies γ (x1 ) < 1, then γ (η0 ) < 1.
Proof. By Lemma 2 we have I2 (x) ≤ E(x, f ) ≤ I1 (x) where I1 (x) = exp −Nm x 2 /L M and I2 (x) =
exp −N M x 2 /L m . Since I1 (0) = I2 (0) = +∞, I1 (+∞) = I2 (+∞) = 0 and they are decreasing functions, then there exists at least one solution η0 to Eq. (52). √  Moreover, η0 < x1 , where x1 is the unique solution to equation I1 (x) = l ρ0 x α0 q0 , x > 0. Furthermore, if γ (x1 ) < 1 then γ (η0 ) < 1.   √

Remark 4. We note that x1 = M∗−1 q0 lρ0 η0 α0 , where   M∗ (x) := x exp Nm x 2 /L M and then   √

 < 1.  γ (x1 ) < 1 ⇐⇒ γ M∗−1 q0 lρ0 η0 α0

(57)

Theorem 15. √ (i) If N and L satisfy (22)–(24) and (57), then there exists at least one solution of (41)–(45) given by s(t) = 2η0 α0 t and  √
T (x, t) = Tm (1 + f (η)) , η = x 2 α0 t where f and η0 satisfy (51) and (52). (ii) If N and L verify the conditions (22)–(24) and (57), then there exists at least one solution to the problem (41)–(45) for all latent heats of fusion l > l0∗ for given other parameters where l0∗ is given by
 √ η) l0∗ := q0 ρ0 η0 α0 M∗ (e where e η > 0 is characterized by the condition γ (e η) = 1.



Remark 5. The existence of a solution to the problem (41)–(45) is given for large latent heat of fusion (∀l > l0∗ ) if conditions (22)–(24) and (57) for the thermal coefficients are verified. Two examples with an explicit solution are well known. Example 1. In the particular case N = L = 1, the solution of integral equation (19) is given by [16,17]  f (η) = 1 − erf(η) erf(η0 ) , 0 < η < η0 ,

(58)

where η0 > 0 is the unique solution to the equation √ x erf(x) exp(x 2 ) = Ste/ π , x > 0

(59)

where Ste is the Stefan number defined by (39).    Example 2. In [18,19], the case of ρ(T ) = ρ0 , c(T ) = c0and k(T ) = k0 1 + ζ (T − Tm ) (Tb − Tm ) = k0 [1 + ζ θ ] was considered, that is N (T ) = 1 and L (θ) = k(T ) k0 = 1 + ζ θ . In this case, the solution is given by √
T (x, t) = Tb + (Tm − Tb ) Ψδ (η) /Ψδ (η0 ) , 0 < η < η0 , η = x/ 2 α0 t √ s(t) = 2η0 α0 t

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where δ and η0 must satisfy the following equations: ζ = δΨδ (η0 ) (1 + δΨδ (η0 )) Ψδ0 (η0 ) / (η0 Ψδ (η0 )) = 2/Ste where Ψ = Ψδ is the error function defined as the unique solution to the ordinary differential problem  0 (1 + δΨ (η)) Ψ 0 (η) + 2ηΨ 0 (η) = 0
Ψ 0+ = 0, Ψ (+∞) = 1. Note that if ζ = 0 we obtain Example 1. Other examples with nonlinear thermal coefficients and an explicit solution of a similarity type for the corresponding free boundary problem have been obtained in [20–28]. Acknowledgments This paper was partially sponsored by the projects PIPNo 5379 from CONICET-UA, Rosario (Argentina), Project No 53900/4 from Fundaci´on Antorchas (Argentina), ANPCYT PICT #03-11165 from Agencia (Argentina) and “Fondo de ayuda a la investigaci´on” from Universidad Austral (Argentina). References [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12] [13] [14] [15] [16] [17] [18] [19] [20] [21] [22] [23] [24] [25] [26] [27] [28]

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