# Exact Solutions for Nonclassical Stefan Problems

Authors: Adriana C. Briozzo; Domingo A. Tarzia
Journal: International Journal of Differential Equations
 (2010)

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## Abstract

We consider one-phase nonclassical unidimensional Stefan problems for a source functionF
which depends on the heat flux, or the temperature on the fixed face x=0. In the first case, we assume a temperature boundary condition, and in the second case we assume a heat flux boundary condition or a convective boundary condition at the fixed face. Exact solutions of a similarity type are obtained in all cases.


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## Body

## 1. Introduction

The one-phase Stefan problem for a semi-infinite material is a free boundary problem for the classical heat equation which requires the determination of the temperature distributionu of the liquid phase (melting problem) or the solid phase (solidification problem) and the evolution of the free boundary x=s(t). Phase change problems appear frequently in industrial processes and other problems of technological interest [1–4].Nonclassical heat conduction problem for a semi-infinite material was studied in [5–11]. A problem of this type is the following:(1.1)(i)ut-uxx=-F(W(t),t),x>0,t>0,(ii)u(0,t)=f(t),t>0,(iii)u(x,0)=h(x),x>0,
where functions f=f(t) andh=h(x) are continuous real functions, and F is a given function of two variables. A particular and interesting case is the following:(1.2)F(W(t),t)=λ0tW(t)(λ0>0),
where W=W(t) represents the heat flux on the boundary x=0, that is W(t)=ux(0,t).Problems of the types (1.1) and (1.2) can be thought of by modelling of a system of temperature regulation in isotropic mediums [10, 11], with a nonuniform source term which provides a cooling or heating effect depending upon the properties of F related to the course of the heat flux (or the temperature in other cases) at the boundary x=0[10].In the particular case of a bounded domain, a class of problems, when the heat source is uniform and belongs to a given multivalued function fromℝ into itself, was studied in [8] regarding existence, uniqueness, and asymptotic behavior. Moreover, in [5] conditions are given on the nonlinearity of the source term Fso as to accelerate the convergence of the solution to the steady-state solution. Other references on the subject are in [7, 12, 13].Nonclassical free boundary problems of the Stefan type were recently studied in [14–16] from a theoretical point of view by using an equivalent formulation through a system of second kind Volterra integral equations [17–19]. A large bibliography on free boundary problems for the heat equation was given in [20].In this paper, firstly we consider a free boundary problem which consists in determining the temperatureu=u(x,t) and the free boundary x=s(t) such that the following conditions are satisfied:(1.3)ρcut-kuxx=-γF(W(t),t),0<x<s(t),t>0,(1.4)u(0,t)=f>0,t>0,(1.5)u(s(t),t)=0,t>0,(1.6)kux(s(t),t)=-ρls.(t),t>0,(1.7)s(0)=0,
where the thermal coefficients k,ρ,c,l,γ>0, the boundary temperature f>0, and the control function F depend on the evolution of the heat flux at the boundary x=0as follows:(1.8)W(t)=ux(0,t),F(W(t),t)=F(ux(0,t),t)=λ0tux(0,t),
where λ0>0 is a given constant.The existence and the uniqueness of the solution of a general free boundary problem of the type (1.3)–(1.8) was given recently in [14, 15]. Moreover, we consider other two free boundary problems which consist in determining the temperature u=u(x,t) and the free boundary x=s(t) such that (1.3), (1.5), (1.6), and (1.7) are satisfied, and in these cases the control function F depends on the evolution of the temperature at the boundary x=0as follows:(1.9)W(t)=u(0,t),F(W(t),t)=F(u(0,t),t)=λ0tu(0,t),λ0>0.
In this case, a heat flux boundary condition(1.10)kux(0,t)=-q0t>0,t>0
or a convective boundary condition(1.11)kux(0,t)=q0t(u(0,t)-f)>0,t>0
can be considered at the fixed face x=0in order to obtain the corresponding explicit solutions.The plan of this paper is the following. In Section2, we show an explicit solution of a similarity type for the nonclassical one-phase Stefan problem (1.3)–(1.7) for a control function F given by (1.8).In Sections3 and 4, we obtain sufficient conditions on data in order to have a similarity type solution to the problems (1.3), (1.5), (1.6), and (1.7), where the control function F is given by (1.9) (instead of (1.8)) and we take into account the heat flux condition (1.10) or the convective condition (1.11) at the fixed face x=0, respectively.The restrictions on data we have obtained for these two free boundary problems with a heat flux boundary condition (1.10) or a convective boundary condition (1.11) at the fixed face x=0 can be interpreted in the same way as we have obtained in the classical Stefan problem with the same boundary conditions in [21, 22] in order to have an instantaneous phase-change problem (see, e.g., sufficient condition λ0<ρc/2γ in Theorems 3.2 and 4.1).


## 2. Explicit Solution to a One-Phase Stefan Problem for a Nonclassical Heat Equation with Control Function of the TypeF(ux(0,t),t)=(λ0/t)ux(0,t) and a Temperature Condition at the Fixed Boundary

We consider the following free boundary problem for a semi-infinite material given by the following conditions:(2.1)ρcut-kuxx=-γF(ux(0,t),t),0<x<s(t),t>0,u(0,t)=f>0,t>0,u(s(t),t)=0,t>0,kux(s(t),t)=-ρls.(t),t>0,s(0)=0,
where the thermal coefficients k,ρ,c,l,γare positive and the control function F, which depends on the evolution of the heat flux at the extremum x=0, is given by (1.8).In order to obtain an explicit solution of a similarity type, we define(2.2)Φ(η)=u(x,t),η=x2at,
where a2=k/ρc is the diffusion coefficient of the phase change material. The problem (2.1) and (1.8) become(2.3)Φ′′(η)+2ηΦ′(η)=2λΦ′(0),0<η<η0,(2.4)Φ(0)=f,(2.5)Φ(η0)=0,(2.6)Φ′(η0)=-2lcη0,
where the dimensionless parameter λis defined by(2.7)λ=γλ0ρca>0,
and the free boundary s(t) must be of the type(2.8)s(t)=2aη0t,
where η0 is an unknown parameter to be determined later. The general solution of the differential equation (2.3) is given by(2.9)Φ(η)=C2+C1[π2erf(η)+2λ∫0ηf1(z)dz],
where C1 andC2are arbitrary constants, and(2.10)erf(x)=2π∫0xexp(-z2)dz,f1(x)=exp(-x2)∫0xexp(r2)dr
are the error function and the Dawson's integral (see [23, page 298] and [24, page 43]), respectively.After some elementary computations, from (2.3), (2.4), and (2.5) we obtain(2.11)Φ(η)=f[1-E(η,λ)E(η0,λ)],0<η<η0,
where(2.12)E(x,λ)=erf(x)+4λπ∫0xf1(r)dr.Taking into account condition (2.6), the unknown parameter η0=η0(λ,Ste)must be the solution of the following equation:(2.13)Steπ[exp(-x2)+2λf1(x)]=x[erf(x)+4λπ∫0xf1(z)dz],x>0,
where Ste=fc/l>0 is the Stefan's number. Equation (2.13) is equivalent to the following one:(2.14)W1(x)=2λW2(x),x>0,
where the real functions W1and W2 are defined by(2.15)W1(x)=Steexp(-x2)-πxerf(x),(2.16)W2(x)=2x∫0xf1(r)dr-Stef1(x).Remark 2.1.
Ifλ=0 (i.e.,λ0=0), then the problem (2.1) and (1.8) represented the classical Lamé-Clapeyron problem [25]. In this case, there exists a unique solution η00of (2.17) (equivalent to (2.13))given by
(2.17)F0(x)=Steπ,x>0,
where
(2.18)F0(x)=xerf(x)exp(x2),
and the explicit solution is given by [2, 23]:
(2.19)u(x,t)=f[1-erf(η)erf(η00)],0<η=x2at<η00,s(t)=2aη00t.In order to solve (2.14), we will study firstly the behavior of function f1. We obtain some preliminary properties.Lemma 2.2.
The Dawson's integral satisfies the following properties:(i)
f1(0)=0,(ii)
f1(+∞)=0,(iii)
(2.20)f1′(x)=1-2xf1(x)={>0if0<x<x1,=0ifx=x1,<0ifx>x1,
where x1≃0.924,f1(x1)≃0.541,(iv)
(2.21)f1′′(x)=-2[1+f1(x)(1-2x2)]={<0if0<x<x2,=0ifx=x2,>0ifx>x2,
where x2≃1.502,f1(x2)≃0.428,(v)
limx→+∞2xf1(x)=1.Proof.
The properties (i)–(iv) have been proved in [23, page 298] (see also [24, pages 42–45])
(v) By the L'Hopital Theorem, we have(2.22)limx→+∞2xf1(x)=limx→+∞2x∫0xexp(r2)drexp(x2)=limx→+∞∫0xexp(r2)dr+xexp(x2)xexp(x2)=limx→+∞(1+∫0xexp(r2)drxexp(x2))=limx→+∞(1+f1(x)x)=1,
then (v) holds.Next, we define the following auxiliary functions:(2.23)φ1(x)=∫0xf1(r)dr,φ2(x)=xφ1(x)=x∫0xf1(r)dr,φ3(x)=xf1(x),φ4(x)=x(2xf1(x)-1)=-xf1′(x),φ5(x)=f1(x)-xf1′(x),φ6(x)=Ste-2(1+Ste)xf1(x).We have the following results.Lemma 2.3.
(a)
Functionφ1satisfies the following properties:
(i)
φ1(0)=0,(ii)
φ1′(x)=f1(x),(iii)
φ1′(0+)=0,(iv)
φ1(+∞)=+∞,(v)
(2.24)φ1′′(x)=f1′(x)=1-2xf1(x)={>0if0<x<x1,=0ifx=x1,<0ifx>x1,(vi)
limx→+∞(φ1(x)/log(x))=1/2,(vii)
limx→+∞φ1(x)f1′(x)=0.(b)
Functionφ4satisfies the following properties:
(i)
φ4(0+)=0-,(ii)
φ4′(x)=-1+4xf1(x)-2x2(2xf1(x)-1),(iii)
φ4(+∞)=0+,(iv)
φ4′(0+)=-1,(v)
φ4′(+∞)=0+,(vi)
φ4(x)=0⇔x=x1(the maximum point off1),(vii)
φ4′(x1)=1.(c)
Functionφ3satisfies the following properties:
(i)
φ3(0+)=0,(ii)
φ3(+∞)=1/2,(iii)
φ3′(x)=f1(x)+x(1-2xf1(x)),(iv)
φ3′(0+)=0,(v)
φ3′(+∞)=0,(vi)
φ3(x1)=x1f1(x1)≃0.4999,(vii)
φ3(x2)=x2f1(x2)≃0.64.(d)
Functionφ2satisfies the following properties:
(i)
φ2(0+)=0,(ii)
φ2(+∞)=+∞,(iii)
φ2′(x)=φ1(x)+xf1(x)>0,for all x>0,(iv)
φ2′(0+)=0,(v)
φ2′(+∞)=+∞,(vi)
φ2′′(x)=2f1(x)-x(2xf1(x)-1),(vii)
φ2′′(+∞)=0,(viii)
φ2′′(0+)=0.(e)
Functionφ5satisfies the following properties:
(i)
φ5(0+)=0,(ii)
φ5(+∞)=0+,(iii)
(2.25)φ5′(x)=-xf1′′(x)={>0if0<x<x2,=0ifx=x2,<0ifx>x2,(iv)
φ5(x)>0,for all x>0.(f)
Functionφ6satisfies the following properties:
(i)
φ6(0+)=Ste>0,(ii)
φ6(+∞)=-1,(iii)
φ6′(x)=-2(1+Ste)φ3′(x),(iv)
φ6′(0+)=0,(v)
φ6′(+∞)=0,(vi)
φ6(x1)=x1f1(x1)≃0.4999,(vii)
φ6(x2)=x2f1(x2)≃0.64.Proof.
(a) Taking into account properties off1, we have
(2.26)φ1′(x)=f1(x)>0,∀x>0,φ1′(0)=f1(0)=0,
and (v) holds. If we consider Lemma 2.2(v), we get φ1(+∞)=+∞ and we have
(2.27)limx→+∞φ1(x)log(x)=limx→+∞xf1(x)=12,
then (iv) and (vi) hold.
To prove (vii), we consider(2.28)φ1(x)f1′(x)=(∫0xf1(r)dr)f1′(x)=f1(c)xf1′(x),
where c=c(x)∈(0,x). Then limx→+∞φ1(x)f1′(x)=0 because limx→+∞xf1′(x)=0 and f1is a bounded function.
(b) From the definition ofφ4, we obtain (i) and (ii). To prove (iii), we have(2.29)φ4(+∞)=limx→+∞x(2xf1(x)-1)=limx→+∞2xf1(x)-11/x=limx→+∞2[f1(x)+x(1-2xf1(x))]1/x2=2limx→+∞[x2f1(x)+x2x(1-2xf1(x))],
then
(2.30)limx→+∞x(2xf1(x)-1)=2limx→+∞[x2x(2xf1(x)-1)-x2f1(x)].
If we suppose that
(2.31)limx→+∞x(2xf1(x)-1)=L>0,
we get
(2.32)L=2limx→+∞[x2x(2xf1(x)-1)-x2f1(x)]=+∞,
which is a contradiction.If we suppose that(2.33)limx→+∞x(2xf1(x)-1)=+∞,
then
(2.34)φ4′(+∞)=limx→+∞-1+4xf1(x)-2x2(2xf1(x)-1)=-∞,
which is also a contradiction. Therefore, limx→+∞x(2xf1(x)-1)=0and (iii) hold.
Taking into account (ii), we haveφ4′(x)=-1+4xf1(x)-2x2(2xf1(x)-1), thenφ4′(0)=-1 and if we consider (iii) we haveφ4′(+∞)=0+. From properties of f1, we have
(2.35)φ4(x)=0⇔2xf1(x)-1=0⇔f1′(x)=0⇔x=x1,
and (vi) holds. Taking into account f1′(x)=1-2xf1(x)=0, we get φ4′(x1)=1.
(c) From Lemmas2.2 and 2.3(b) we get (i)–(vii).
(d) We haveφ2(x)=xφ1(x)=x∫0xf1(r)dr, then from (a) and (b)(iii) we get (i)–(vi).
(e) As we haveφ5(x)=f1(x)-xf1′(x)=f1(x)+φ4(x), then by using the properties of f1and (b) we obtain the properties of φ5.
(f) We haveφ6(x)=Ste-2(1+Ste)xf1(x)=Ste-2(1+Ste)φ3(x), and from the properties of φ3,we obtain (i)–(v).Corollary 2.4.
One has(i)
limx→+∞x2[2xf1(x)-1]=1/2,(ii)
limx→+∞x[x2(2xf1(x)-1)-xf1(x)]=0.Now, we are in conditions to enunciate properties of functionsW1and W2 in order to study after (2.14).Lemma 2.5.
The functionsW1(x)and W2(x), defined by (2.15) and (2.16), respectively, satisfy the following properties.(a)
Properties of functionW1:
(i)
W1(0)=Ste,(ii)
W1(+∞)=-∞,(iii)
limx→+∞(W1(x)/x)=-π,(iv)
limx→+∞(W1(x)+πx)=0,(v)
W1′(x)<0,for all x>0,(vi)
W1(η00)=0, where η00 is the unique solution of (2.17),(vii)
(2.36)W1′′(x)={<0if0<x<x0,=0ifx=x0,<0ifx>x0,
where
(2.37)x0=3+2Ste4(1+Ste),(viii)
W1′′(0+)=-2(3+2Ste)<0.(b)
Properties of functionW2:
(i)
W2(0)=0,(ii)
W2(+∞)=+∞,(iii)
there exists a uniquex4>0such that W2(x4)=0,(iv)
W2′(x)=2∫0xf1(r)dr+2xf1(x)(1+Ste)-Ste,(v)
there exists a uniquex3>0such thatW2′(x3)=0and W2(x3)<0,(vi)
W2′(0+)=-Ste<0,(vii)
W2′(+∞)=+∞,(viii)
W2′′(x)=2(1+Ste)x+2f1(x)[2+Ste-2(1+Ste)x2],(ix)
W2′′(0+)=0,(x)
W2(η00)<0.Proof.
(a) Taking into account the definition of the functionW1,we get (i) and (ii).
(iii) We have(2.38)limx→+∞W1(x)x=limx→+∞[Steexp(-x2)x-πerf(x)]=-π.
(iv) We have(2.39)limx→+∞(W1(x)+πx)=limx→+∞(Steexp(-x2)-πxerf(x)+πx)=limx→+∞(Steexp(-x2)+πxerfc(x))=limx→+∞(Steexp(-x2)+Q(x)exp(-x2))=limx→+∞exp(-x2)(Ste+Q(x))=0,
where Q is the function defined by
(2.40)Q(x)=πxexp(x2)erfc(x),erfc(x)=1-erf(x),
which satisfies the following properties:
(2.41)Q(0)=0,Q(+∞)=1,Q′(x)>0,∀x>0.
(v) We have(2.42)W1′(x)=-πerf(x)-2xexp(-x2)[Ste+1]<0,∀x>0.
(vi) Taking into account (i), (iii), and (v), we get that there exists a unique zero ofW1 which is given byη00,the unique solution of (2.17).
(vii)We have
(2.43)W1′′(x)=-2exp(-x2)[3+2Ste-4(1+Ste)x2],
then
(2.44)W1′′(x)=0⇔4(1+Ste)x2=3+2Ste⇔x=x0=3+2Ste4(1+Ste).
Sincesign(W1′′(x))=sign(4(1+Ste)x2-3-2Ste),then we obtain (vii).
(b) Taking into account Lemmas2.2 and 2.3, we have (i) and (ii).
We can write(2.45)W2′(x)=2∫0xf1(r)dr+2xf1(x)(1+Ste)-Ste=2φ1(x)-φ6(x),
then W2′(0+)=-Ste,W2′(+∞)=+∞and W2′′(x)=2φ1′(x)-φ6′(x) satisfies W2′′(0+)=0.Then (iv), (vi), (vii), (viii), and (ix) hold.
We have(2.46)W2(x)=0⇔2φ2(x)=Stef1(x),
then taking into account the properties of φ2andf1, we get that there exists a unique x4>0such that
(2.47)W2(x)=0,x>0.
Moreover, we have
(2.48)W2(x)={=0ifx=0,<0if0<x<x4,=0ifx=x4,>0ifx>x4.
In the same way, we have
(2.49)W2′(x)=0⇔2φ1(x)=φ6(x).
Then, if we consider the properties of the functions φ1andφ2, we have that there exists a unique x3such that W2′(x3)=0. Moreover, W2(x3)=-2x32f1(x3)-Steφ5(x3)<0 and then (v) holds.
To prove (x), we take into account that
(2.50)W2(x)=2x∫0xf1(r)dr-Stef1(x)=πxerf(x)F(x)-πx∫0xerf(r)exp(r2)dr-Steexp(-x2)F(x)=πexp(-x2)[F0(x)-Steπ]F(x)-πx∫0xerf(r)exp(r2)dr,
where F(x)=∫0xexp(r2)dr andF0 was defined in (2.18). Then by using (2.17), we have
(2.51)W2(η00)=-πη00∫0η00erf(r)exp(r2)dr<0.Lemma 2.6.
For eachλ>0, there exists a unique solution η0 of (2.14). This solution η0=η0(λ) satisfies the following properties:
(2.52)(i)η0(0+)=η00,(ii)η0(+∞)=x4,(iii)η0=η0(λ)isanincreasingfunctiononλ,
where η00 and x4 are the unique solution of (2.17) and (2.47), respectively.Proof.
Taking into account Lemma2.5, we get that there exists a unique solution η0 of (2.14). Let 0<λ1<λ2 be given, taking into account properties of function W2, we obtain that the real functions Z1and Z2 defined by
(2.53)Z1(x)=2λ1W2(x),Z2(x)=2λ2W2(x)
satisfy the following properties:
(2.54)Z2(x)<Z1(x)if0<x<x4,Z2(x)=Z1(x)ifx=x4,Z2(x)>Z1(x)ifx>x4.
Then η0(λ1)<η0(λ2), where η0(λi)is the solution of equation Zi(x)=W1(x),i=1,2.Therefore, η0=η0(λ)is an increasing function on λ. Moreover, we obtain η00<η0(λ)<x4because W2(η00)<0.Then, we have proved the following result.Theorem 2.7.
For eachλ>0, the free boundary problem (2.1), where F is defined by (1.8), has a unique similarity solution of the type
(2.55)u(x,t,λ)=f[1-E(n,λ)E(η0(λ),λ)],0<η=x2at<η0(λ),s(t,λ)=2aη0(λ)t,
where
(2.56)E(η,λ)=erf(η)+4λπ∫0ηf1(r)dr
and η0=η0(λ) is the unique solution of (2.14) with η00<η0(λ)<x4.


## 3. Explicit Solution to a One-Phase Stefan Problem for a Nonclassical Heat Equation with Control Function of the TypeF(u(0,t),t)=(λ0/t)u(0,t) and a Heat Flux Condition at the Fixed Face

In this section, the free boundary problem consists in determining the temperatureu=u(x,t) and the free boundary x=s(t) with a control function F which depends on the evolution of the temperature at the extremum x=0given by the following conditions:(3.1)ρcut-kuxx=-γF(u(0,t),t),0<x<s(t),t>0,kux(0,t)=-q0t>0,t>0,u(s(t),t)=0,t>0,kux(s(t),t)=-ρls.(t),t>0,s(0)=0,
where the coefficient q0>0 characterizes the heat flux on the x=0 [21] and the control function Fis given by (1.9).In order to obtain an explicit solution of a similarity type, we define the same transformation given by (2.2). The problem (3.1) and (1.9) are equivalent to the following one:(3.2)Φ′′(η)+2ηΦ′(η)=ΛΦ(0),0<η<μ0,(3.3)Φ′(0)=-q0*,(3.4)Φ(μ0)=0,(3.5)Φ′(μ0)=-2lcμ0,
where the dimensionless parameters Λand q0* are defined by(3.6)Λ=4γλ0ρc>0,q0*=2aq0k,(3.7)s(t)=2aμ0t
is the free boundary, where μ0 is an unknown parameter to be determined.From (3.2), (3.3), and (3.4), we obtain the similarity solution(3.8)Φ(η)=q0*π2G(μ0,Λ)[erf(μ0)G(η,Λ)-erf(η)G(μ0,Λ)],0<η<μ0,
where(3.9)G(x,Λ)=1+Λ∫0xf1(r)dr=1+Λφ1(x),
and f1 is the Dawson's integral and φ1 is given by (2.23).By condition (3.5), the unknown parameter μ0=μ0(Λ,l,c,q0*)must be solution of the following equation:(3.10)Λerf(x)f1(x)=2πG(x,Λ)[exp(-x2)-2lcq0*x],x>0,
which is equivalent to the following one:(3.11)H2(x)=H3(x),x>0,
where the real functions H2and H3 are defined by(3.12)H2(x)=Λerf(x)f1(x),(3.13)H3(x)=2πG(x,Λ)H1(x),(3.14)H1(x)=[exp(-x2)-2lcq0*x].Remark 3.1.
IfΛ=0(i.e., λ0=0), we have the solution
(3.15)Φ(η)=q0*π2[erf(μ00)-erf(η)],0<η<μ00,
where μ00is the unique solution of the following equation:
(3.16)exp(-x2)=2lcq0*x.In order to solve (3.11), we consider properties of Dawson's integral, error function, and some auxiliary functions, and then we obtain the following result.Theorem 3.2.
For eachλ0<ρc/2γ, the free boundary problem (3.1), where F is defined by (1.9), has a unique similarity solution of the type
(3.17)u(x,t,λ0)=q0aπkG(μ0(λ0),4γλ0/ρc)[erf(x2at)G(μ0(λ0),4γλ0ρc)-erf(μ0(λ0))G(x2at,4γλ0ρc)],0<x2at<μ0(λ0),t>0,s(t,λ0)=2aμ0(λ0)t,t>0,
where μ0=μ0(λ0) is the unique solution of (3.11), 0<μ0(λ0)<μ00.Proof.
We follow a similar method developed in Theorem2.7.


## 4. Explicit Solution to a One-Phase Stefan Problem for a Nonclassical Heat Equation with Control Function of the TypeF(u(0,t),t)=(λ0/t)u(0,t) and a Convective Condition at the Fixed Face

In this section, we consider a similar problem to the one given in Section3 for a convective boundary condition [22, 26] on the fixed face given by(4.1)ρcut-kuxx=-γF(u(0,t),t),0<x<s(t),t>0,kux(0,t)=h0t(u(0,t)-f)>0,t>0,u(s(t),t)=0,t>0,kux(s(t),t)=-ρls.(t),t>0,s(0)=0,
where Fis defined by (1.9) and h0 characterizes the heat transfer coefficients [22, 26]. To solve this problem, we consider again a similarity type solution given by (2.2). Then, the problem (4.1) and (1.9) are equivalent to the following one:(4.2)Φ′′(η)+2ηΦ′(η)=ΛΦ(0),0<η<μ0,(4.3)Φ′(0)=h0*(Φ(0)-f),h0*=2ah0k,(4.4)Φ(μ0)=0,(4.5)Φ′(μ0)=-2lcμ0,
where the dimensionless parameter Λis defined by (3.6) and(4.6)s(t)=2aμ0t
is the free boundary, where μ0 is an unknown parameter to be determined. We obtain the solution(4.7)Φ(η)=h0*fπ2[erf(μ0)G(η,Λ)-erf(η)G(μ0,Λ)]G(μ0,Λ)+(h0*π/2)erf(μ0),0<η<μ0,
where G(x,Λ) is given by (3.9).Taking into account the condition (4.5), the unknown parameter μ0=μ0(Λ,l,c,h0*)must be the solution of the following equation:(4.8)Λerf(x)f1(x)+2Steerf(x)x=2πG(x,Λ)[exp(-x2)-2h0*Stex],x>0,
which is equivalent to(4.9)H2*(x)=H3*(x),x>0,
where(4.10)H2*(x)=H2(x)+2Steerf(x)x,x>0,H3*(x)=2πG(x,Λ)[exp(-x2)-2h0*Stex],x>0,
and the function H2 is defined by (3.12).Similarly to the previous cases, we can enunciate the following result.Theorem 4.1.
(a) For each Λ<2 (λ0<ρc/2γ), the free boundary problem (4.1), where F is defined by (1.9), has a unique similarity solution given by
(4.11)u(x,t,λ0)=-h0afπk[erf(x/2at)G(μ0(λ0),4γλ0/ρc)(h0afπ/k)erf(μ0(λ0))+G(μ0(λ0),4γλ0/ρc)-erf(μ0(λ0))G(x/2at,4γλ0/ρc)(h0afπ/k)erf(μ0(λ0))+G(μ0(λ0),4γλ0/ρc)],0<x2at<μ0(λ0),t>0,s(t,λ0)=2aμ0(λ0)t,t>0,
where μ0=μ0(λ0) is the unique solution of (4.9).
(b) Let M(x)=Λf1(x)and N(x)=2xG(x,Λ) be, there exists a unique solution x*>0 of the equation M(x)=N(x).
For eachΛ>2(λ0>ρc/2γ) such that M(α(Λ))-N(α(Λ))<2/h0*Ste, where 0<α(Λ)<x* satisfies M′(α(Λ))-N′(α(Λ))=0,there exists a unique similarity solution to the free boundary problem (3.1), where F is defined by (1.9). The solution is given by (4.11).



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*Source: 868059-2010-09-14.xml*