Solutions of similarity-type for a nonlinear non-classical Stefan problem with temperature-dependent thermal conductivity and a Robin boundary condition are obtained. The analysis of several particular cases are given when the thermal conductivity L(f) and specific heat N(f) are linear in temperature such that L(f)=α+δf with N(f)=β+γf. Existence of a similarity type solution also obtained for the general problem by proving the lower and upper bounds of the solution.
Citation: Lazhar Bougoffa, Ammar Khanfer. Solutions of a non-classical Stefan problem with nonlinear thermal coefficients and a Robin boundary condition[J]. AIMS Mathematics, 2021, 6(6): 6569-6579. doi: 10.3934/math.2021387
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Solutions of similarity-type for a nonlinear non-classical Stefan problem with temperature-dependent thermal conductivity and a Robin boundary condition are obtained. The analysis of several particular cases are given when the thermal conductivity L(f) and specific heat N(f) are linear in temperature such that L(f)=α+δf with N(f)=β+γf. Existence of a similarity type solution also obtained for the general problem by proving the lower and upper bounds of the solution.
In [1], the authors introduced the following Stefan problem which is governed by a non-classical and nonlinear heat equation with heat source F, thermal coefficients which depend on the temperature and a convective boundary condition at fixed face x=0 (see also [2,3,4]):
{ρ(T)c(T)∂T∂t=∂∂x(k(T)∂T∂x)−F(Z(t),t), 0<x<s(t), t>0,k(T(0,t))∂T(0,t)∂x=h√t(T(0,t)−T∗), h>0,T(s(t),t)=Tm,k(T(s(t),t))∂T(s(t),t)∂x=−ρ0ls′(t),s(0)=0, | (1.1) |
where ρ(T),c(T) and k(T) are the density of the material, its specific heat, and its thermal conductivity, respectively; Tm is the phase-change temperature, ρ0>0 its the constant density of mass at the melting temperature; l>0 is the latent heat of fusion by unity of mass and s(t) is the position of phase change location. Also, F depends on the evolution of the heat flux at the boundary x=0 and assumed that
F(Z(t),t)=F(Z(t),t)=F(∂T(0,t)∂x,t)=λ√t∂T(0,t)∂x, λ>0. | (1.2) |
From the second equation in Pr.(1.1), we see that Tx(0,t)<0. This can be physically justified since F in this particular study represents a heat sink due to "melting ice/water phase change".
To obtain a similarity solution, we introduce the new independent variable
ξ=x2√α0t | (1.3) |
and the dimension-free dependent variable, defined by
f(ξ)=T(x,t)−T∗Tm−T∗. | (1.4) |
In terms of these new variables, the problem takes the following nonlinear boundary value problem [1]
{[L(f(ξ))f′(ξ)]′+2ξN(f(ξ))f′(ξ)=Af′(0), 0<ξ<ξ0,L(f(0))f′(0)=pf(0),f(ξ0)=1,f′(ξ0)=Mξ0, | (1.5) |
where the nonlinear terms L(f(ξ)) and N(f(ξ)) are given by
L(f(ξ))=k((Tm−T∗)f(ξ)+T∗)k0, N(f(ξ))=ρc((Tm−T∗)f(ξ)+T∗)ρ0c0 | (1.6) |
and
A=2λρ0c0k0, p=2√α0h0k0>0, M=2k0kTmSte | (1.7) |
with Ste=c(T∗−Tm)l>0 (Stefan number) and the system parameters k0, ρ0, c0 and α0=k0ρ0c0 are the reference thermal conductivity, density of mass, specific heat and thermal diffusivity, respectively.
Recently, the authors [1] proved the existence and uniqueness of the similarity solution of Pr.(1.1) under the restrictive Lipschitz conditions on the parameters and using techniques of functional analysis.
In this paper, explicit solutions of Pr.(1.1) are obtained when the thermal conductivity and specific heat are linear in temperature. Also, existence solutions of similarity-type are proved for this non-classical Stefan problem with nonlinear thermal conductivity and specific heat by using a technique based on lower and upper bounds of the solution. Our approach is simpler than the approach used in [1] and is more accessible to readers since it doesn't assume Lipschitz conditions on the parameters and doesn't involve tools of functional analysis.
We consider the case where thermal conductivity and specific heat are linear in temperature [2,3,4]. This case has been largely discussed in the literature since it provides a good approximation of the actual values for some material such as water (see for example [5,6,7,8,9]). We also assume that T is continuous differentiable, which is a natural assumption.
Let
L(f(ξ))=kT∗k0+k(Tm−T∗)k0f(ξ), N(f(ξ))=ρcT∗ρ0c0+ρc(Tm−T∗)ρ0c0f(ξ). | (2.1) |
Writing the nonlinear second-order ODE of Pr.(1.5) in a simple form
[(α+δf(ξ))f′(ξ)]′+2ξ(β+γf(ξ))f′(ξ)=Af′(0), 0<ξ<ξ0, | (2.2) |
where α=kT∗k0, β=ρcT∗ρ0c0, δ=k(Tm−T∗)k0 and γ=ρc(Tm−T∗)ρ0c0.
Multiplying both sides of Eq (2.2) by δγ, we obtain
δ[(αγ+δγf(ξ))f′(ξ)]′+2γξ(βδ+δγf(ξ))f′(ξ)=Af′(0)δγ, 0<ξ<ξ0, | (2.3) |
or
[(αγ+δγf(ξ))f′(ξ)]′+2γδξ(βδ+δγf(ξ))f′(ξ)=Af′(0)γ, 0<ξ<ξ0. | (2.4) |
Since αγ=βδ=kT∗k0ρc(Tm−T∗)ρ0c0. Then Eq (2.4) can be written as
[(a+bf(ξ))f′(ξ)]′+2γδξ(a+bf(ξ))f′(ξ)=Af′(0)γ, 0<ξ<ξ0, | (2.5) |
where a=αγ and b=δγ.
By the change of variable
z(ξ)=(a+bf(ξ))f′(ξ). | (2.6) |
Eq (2.5) becomes
z′(ξ)+2γδξz(ξ)=Af′(0)γ. | (2.7) |
The solution of this linear first-order equation is given by
z(ξ)=Af′(0)γφ(ξ)exp(−γδξ2)+z(0)exp(−γδξ2), | (2.8) |
where φ(ξ) is the imaginary error function, that is φ(ξ)=∫ξ0exp(γδt2)dt and z(0) can be determined by using z(0)=(a+bf(0))f′(0)=γ(α+δf(0))f′(0) and the first initial condition of Pr.(1.5) from which we can readily obtain
z(0)=γpf(0). | (2.9) |
Hence
z(ξ)=Af′(0)γφ(ξ)exp(−γδξ2)+γpf(0)exp(−γδξ2). | (2.10) |
From Eq (2.6), we have
(a+bf(ξ))2=2b∫ξ0z(t)dt+(a+bf(0))2. | (2.11) |
Substituting z(ξ) into this equation, we obtain
a+bf(ξ)=√2bAγf′(0)∫ξ0φ(t)exp(−γδt2)dt+2bγpf(0)√δγerf(√γδξ)+(a+bf(0))2. | (2.12) |
Employing now the two boundary conditions of Pr.(1.5): f(ξ0)=1 and f′(ξ0)=Mξ0 to obtain
z(ξ0)=(a+bf(ξ0))f′(ξ0). | (2.13) |
Thus, the condition on ξ0 is given by
(a+b)Mξ0=[Aγf′(0)φ(ξ0)+γpf(0)]exp(−γδξ20). | (2.14) |
Thus
Theorem 2.1. The exact solution of Pr. (1.5) is given by
f(ξ;ξ0)=1b√2bAγf′(0)∫ξ0φ(t)exp(−γδt2)dt+2bγpf(0)√δγerf(√γδξ)+(a+bf(0))2−ab | (2.15) |
subject to the condition (2.14).
Remark 2.2. From f(ξ)=T(x,t)−T∗Tm−T∗, it can be readily seen that 0<f(0)<1. Furthermore, for arbitrary ϵ>0, it can also be shown using classical chain rule on this relation with ξ=x2√α0t that
f′(ϵ)=2√α0tTm−T∗Tx(2√α0tϵ,t)>0. | (2.16) |
This gives 0≤f′(0)=2√α0tTm−T∗Tx(0,t). Using the second equation in Pr.(1.1) we obtain 0≤f′(0)<2√α0hk(T(0,t)). Therefore, from (2.14) we conclude that ξ0>0.
Consequently,
Theorem 2.3. The exact solution of the original Pr.(1.1) is given by
T(x,t)=(Tm−T∗)f(ξ;ξ0)+T∗, ξ=x2√α0t | (2.17) |
subject to the condition (2.14) and the free boundary is given by s(t)=2ξ0√α0t.
Another important case is L(f(ξ))=1+δf(ξ) and N(f(ξ))=1 [1]. For this case Pr.(1.5) becomes
{[(1+δf(ξ))f′(ξ)]′+2ξf′(ξ)=Af′(0), 0<ξ<ξ0,(1+δf(0))f′(0)=pf(0),f(ξ0)=1,f′(ξ0)=2ξ0(1+δ)Ste. | (2.18) |
A result on the existence and uniqueness of solution to the nonlinear boundary value problem Pr.(2.18) when A=0 was proved in [8], where the solution was treated as a Generalized Modified Error.
Writing the nonlinear ODE of Pr.(2.18) as
((1+δf(ξ))(1+δf(ξ))′)′+2ξ(1+δf(ξ))′=Af′(0)δ, 0<ξ<ξ0. | (2.19) |
The transformation u(ξ)=(1+δf(ξ))2 leads to
u″+2ξu′√u=2Af′(0)δ, 0<ξ<ξ0 | (2.20) |
subject to the initial-boundary conditions
{u′(0)=2p(√u(0)−1),u(ξ0)=(1+δ)2,u′(ξ0)=4ξ0δSte. | (2.21) |
In the special case A=0, Eq (2.20) is transformed to a homogeneous equation
u″+2ξu′√u=0, 0<ξ<ξ0. | (2.22) |
Consider the following transformation [9]
v=ξu′u, w=−2ξ2√u. | (2.23) |
Hence
dwdξ=dwdvdvdξ=dwdv(u′u+xu″u−x(u′u)2) and dwdξ=−4ξ√u+ξ2u′u1√u. | (2.24) |
The substitution of these into (2.22) gives
(w−v+1)dwdv=(2v−12)w. | (2.25) |
The substitution χ=w−v+1 brings also Eq (2.25) to the Abel equation of the second kind
χχv=Φ(v)χ+Ψ(v), | (2.26) |
where
Φ(v)=(2v−32) and Ψ(v)=(32−v2−2v), | (2.27) |
which is in general very difficult to handle. Since the explicit solution cannot be found, then we estimate the solution by finding upper and lower bounds of u.
If we assume that u′(ξ)≥0 then
u(0)≤u(ξ)≤u(ξ0), ξ∈[0,ξ0]. | (2.28) |
So that
2ξu′(ξ)√u(ξ0)≤2ξu′(ξ)√u(ξ)≤2ξu′(ξ)√u(0). | (2.29) |
From (2.22), we have
2ξu′(ξ)√u(ξ0)≤−u″(ξ)≤2ξu′(ξ)√u(0). | (2.30) |
Thus
2ξ√u(0)≤u″(ξ)u′(ξ)≤−2ξ√u(ξ0). | (2.31) |
Integrating both sides of (2.31) from ξ to ξ0, we obtain
u′(ξ0)eξ20√u(ξ0)e−ξ2√u(ξ0)≤u′(ξ)≤u′(ξ0)eξ20√u(0)e−ξ2√u(0). | (2.32) |
Integrating again (2.32) from ξ to ξ0, we obtain
u(ξ0)−u′(ξ0)eξ20√u(0)∫ξ0ξe−η2√u(0)dη≤u(ξ)≤u(ξ0)−u′(ξ0)eξ20√u(ξ0)∫ξ0ξe−η2√u(ξ0)dη, | (2.33) |
where u(ξ0)=(1+δ)2 and u′(ξ0)=4ξ0δSte.
The constants u(0) and u′(0) can be estimated from the mixed condition u′(0)=2p(√u(0)−1) and (2.32) (or (2.33)) to find
4ξ0δSteeξ201+δ≤u′(0)≤4ξ0δSteeξ20√u(0). | (2.34) |
The case u′(ξ)≤0 follows in a similar fashion.
Based on this, and in view of the condition u′(0)=2p(√u(0)−1), we have
Lemma 2.4. 1). If A=0 and u′(ξ)≥0, ξ∈[0,ξ0], then there is at least one solution u(ξ) of Pr.(2.20) and (2.21) such that
(1+δ)2−4ξ0δSteeξ20√u(0)∫ξ0ξe−η2√u(0)dη≤u(ξ)≤(1+δ)2−4ξ0δSteeξ201+δ∫ξ0ξe−η21+δdη | (2.35) |
subject to the following conditions
u′(0)≥4ξ0δSteeξ201+δ and u′(0)e−ξ20u′(0)2p+1≤4ξ0δSte. | (2.36) |
2). If A=0 and u′(ξ)≤0, ξ∈[0,ξ0], then there is at least one solution u(ξ) of Pr.(2.20) and (2.21) such that
(1+δ)2−4ξ0δSteeξ201+δ∫ξ0ξe−η21+δdη≤u(ξ)≤(1+δ)2−4ξ0δSteeξ20√u(0)∫ξ0ξe−η2√u(0)dη | (2.37) |
subject to the following conditions
u′(0)≤4ξ0δSteeξ201+δ and u′(0)e−ξ20u′(0)2p+1≥4ξ0δSte. | (2.38) |
Let the thermal conductivity k(θ) given by k(θ)=ρ0c0(a+bθ)2 (see [10]) for a list of references that considered this case.
Thus
L(f(ξ))=k(f(ξ))k0=ρ0c0k0(a+bf(ξ))2, N(f(ξ))=k∗, | (3.1) |
where a,b,k∗ are positive constants. Hence the nonlinear ODE of Pr.(1.5) can be written as
(ρ0c0k0(a+bf(ξ))2f′(ξ))′+2k∗ξf′(ξ)=Af′(0), 0<ξ<ξ0 | (3.2) |
or in the equivalent form
k1((a+bf(ξ))′(a+bf(ξ))2)′+2k2ξ(a+bf(ξ))′=Af′(0), 0<ξ<ξ0, | (3.3) |
where k1=ρ0c0k0b and k2=k∗b.
The change of variable z(ξ)=a+bf(ξ) leads to an equation of the form
k1(z′(ξ)z2(ξ))′+2k2ξz′(ξ)=Af′(0), 0<ξ<ξ0. | (3.4) |
In turn, Eq (3.4) can be reduced, by the introduction of the new independent variable u(ξ)=1z(ξ), where u′(ξ)=−z′(ξ)z2(ξ) and z′(ξ)=−u′(ξ)u2(ξ) to the following form
k1u″(ξ)+2k2ξu′(ξ)u2(ξ)=−Af′(0), 0<ξ<ξ0. | (3.5) |
The case A=0, brings Eq (3.5) to the simpler form
u″+2k2k1ξu′u2=0, 0<ξ<ξ0. | (3.6) |
Consider now the following transformation
v=ξu′u, w=−2ξ2u2. | (3.7) |
A simple computation leads to
dwdξ=dwdvdvdξ=dwdv(u′u+xu″u−x(u′u)2) and dwdξ=−4ξu2+4ξ2u′u3. | (3.8) |
The substitution of these into (3.6) gives
(k1(v−1)−k2w)dwdv+2k1w(1v−1)=0. | (3.9) |
The following transformation χ=w−v+1 reduces Eq (3.9) into the known Abel equation of the second kind (2.26), where Φ(v)=k1−2(1v−1) and Ψ(v)=−2k21(v−1)2v, which is in general very difficult to handle. So that, we will find the lower and upper bounds of the solution u(x).
Proceeding as before, we obtain
Lemma 3.1. 1). If u′(ξ)≥0, ξ∈[0,ξ0] then there is at least one solution u(ξ)) of Eq (3.6) subject to (2.21) such that
(1+δ)2−4ξ0δSteek2ξ20k1u2(0)∫ξ0ξe−k2η2k1u2(0)dη≤u(ξ)≤(1+δ)2−4ξ0δSteek2ξ20k1(1+δ)4∫ξ0ξe−k2η2k1(1+δ)4dη | (3.10) |
subject to the following conditions
u′(0)≥4ξ0δSteek2ξ20k1(1+δ)4 and u′(0)e−k2ξ20k1(u′(0)2p+1)4≤4ξ0δSte. | (3.11) |
2). If u′(ξ)≤0, ξ∈[0,ξ0] then there is at least one solution u(ξ)) of Eq (3.6) subject to (2.21) such that
(1+δ)2−4ξ0δSteek2ξ20k1(1+δ)4∫ξ0ξe−k2η2k1(1+δ)4dη≤u(ξ)≤(1+δ)2−4ξ0δSteek2ξ20k1u2(0)∫ξ0ξe−k2η2k1u2(0)dη | (3.12) |
subject to the following conditions
u′(0)≤4ξ0δSteek2ξ20k1(1+δ)4 and u′(0)e−k2ξ20k1(u′(0)2p+1)4≥4ξ0δSte. | (3.13) |
Rewrite the general nonlinear equation
[L(f(ξ))f′(ξ)]′+2ξN(f(ξ))f′(ξ)=Af′(0), 0<ξ<ξ0 | (4.1) |
in the form
[L(f(ξ))f′(ξ)]′+2ξN(f(ξ))L(f(ξ))[L(f(ξ))f′(ξ)]=Af′(0), 0<ξ<ξ0. | (4.2) |
Let F(ξ)=L(f(ξ))f′(ξ). Thus
F′(ξ)+2ξN(f(ξ))L(f(ξ))F(ξ)=Af′(0), 0<ξ<ξ0. | (4.3) |
Multiplying both sides of Eq (4.3) by exp(2∫ξ0ηN(f(η))L(f(η))dη), we obtain
[F(ξ)exp(2∫ξ0N(f(η))L(f(η))dη)]′=Af′(0)exp(2∫ξ0ηN(f(η))L(f(η))dη). | (4.4) |
Thus
F(ξ)exp(2∫ξ0ηN(f(η))L(f(η))dη)=Af′(0)∫ξ0exp(2∫μ0ηN(f(η))L(f(η))dη)dμ+C, | (4.5) |
where C is a constant of integration and can be found from the condition F(0)=L(f(0))f′(0)=pf(0), that is C=pf(0).
Hence f(ξ) can be immediately expressed by
f(ξ)=f(0)+pf(0)∫ξ0exp(−2∫μ0ηN(f(η))L(f(η))dη)L(f(μ))dμ+Af′(0)∫ξ0exp(−2∫μ0ηN(f(η))L(f(η))dη)L(f(μ))∫μ0exp(2∫θ0θN(f(θ))L(f(θ))dθ)dμ. | (4.6) |
If we assume that
0<Nm≤N(f(η))≤NM and 0<Lm≤L(f(η))≤LM. | (4.7) |
Then a simple computation leads to
1LM∫ξ0exp(−NMLmη2)dη≤∫ξ0exp(−2∫μ0ηN(f(η))L(f(η))dη)L(f(μ))dμ≤1Lm∫ξ0exp(−NmLMη2)dη | (4.8) |
and
∫μ0exp(NmLMθ2)dθ≤∫ξ0exp(2∫μ0θN(f(θ))L(f(θ))dθ)dθ≤∫μ0exp(NMLmθ2)dθ, | (4.9) |
where ∫ξ0exp(−NmLMη2)dη=√πLM2√Nmerf(√NmLm√LMξ). Based on this and in view of the boundary conditions f(ξ0)=1 and f′(ξ0)=Mξ0, we have
Theorem 4.1. If the conditions (4.7) are satisfied, then there exists at least one solution f(ξ) of Pr.(1.5) such that
f1(ξ)≤f(ξ)≤f2(ξ), 0<ξ<ξ0, | (4.10) |
where
f2(ξ)=f(0)+pf(0)Lm√πLM2√Nmerf(√Nm√LMξ)+Af′(0)Lm∫ξ0exp(−NmLMη2)∫η0exp(NMLmθ2)dθdη | (4.11) |
and
f1(ξ)=f(0)+pf(0)LM√πLm2√NMerf(√NM√Lmξ)+Af′(0)LM∫ξ0exp(−NMLmη2)∫η0exp(NmLMθ2)dθdη | (4.12) |
subject to the following conditions
f1(ξ0)≤f(ξ0)=1≤f2(ξ0) | (4.13) |
and
f′1(ξ0)≤f′(ξ0)=Mξ0≤f′2(ξ0). | (4.14) |
The authors would like to acknowledge the support of Prince Sultan University, Saudi Arabia for paying the Article Processing Charges (APC) of this publication.
The authors declare no conflict of interest.
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