Research article

Solutions of a non-classical Stefan problem with nonlinear thermal coefficients and a Robin boundary condition

  • Received: 26 February 2021 Accepted: 06 April 2021 Published: 15 April 2021
  • MSC : 35R35, 80A22

  • 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.



    A mapping f:UV is called additive if f satisfies the Cauchy functional equation

    f(x+y)=f(x)+f(y) (1.1)

    for all x,yU. It is easy to see that the additive function f(x)=ax is a solution of the functional equation (1.1) and every solution of the functional equation (1.1) is said to be an additive mapping. A mapping f:UV is called quadratic if f satisfies the quadratic functional equation

    f(x+y)+f(xy)=2f(x)+2f(y) (1.2)

    for all x,yU. A mapping f:UV is quadratic if and only if there exist a symmetric biadditive mapping B:U2V such that f(x)=B(x,x) and this B is unique, refer (see [1,10]). It is easy to see that the quadratic function f(x)=ax2 is a solution of the functional equation (1.2) and every solution of the functional equation (1.2) is said to be a quadratic mapping.

    Mixed type functional equation is an advanced development in the field of functional equations. A single functional equation has more than one nature is known as mixed type functional equation. Further, in the development of mixed type functional equations, atmost only few functional equations have been obtained by many researchers (see [3,6,9,11,12,16,17,22,24]).

    Let G be a group and H be a metric group with a metric d(.,.). Given ϵ>0 does there exists a δ>0 such that if a function f:GH satisfies d(f(xy),f(x)f(y))<δ for all x,yG, then is there exist a homomorphism a:GH with d(f(x),a(x))<ϵ for all xG? This problem for the stability of functional equations was raised by Ulam [23] and answerd by Hyers [7]. Later, it was developed by Rassias [20], Rassias [18,21] and Gvuruta [5].

    The probabilistic modular space was introduced by Nourouzi [14] in 2007. Later, it was developed by K. Nourouzi [4,15].

    Definition 1.1. Let V be a real vector space. If μ:VΔ fulfills the following conditions

    (ⅰ) μ(v)(0)=0,

    (ⅱ) μ(v)(t)=1 for all t>0, if and only if v=γ (γ is the null vector in V),

    (ⅲ) μ(v)(t)=μ(v)(t),

    (ⅳ) μ(au+bv)(r+t)μ(v)(r)μ(v)(t)

    for all u,vV, a,b,r,tR+, a+b=1, then a pair (V,μ) is called a probabilistic modular space and (V,μ) is b-homogeneous if ρ(av)(t)=μ(v)(t|a|b) for all vV,t>0, aR{0}. Here Δ is g:RR+ the set of all nondecreasing functions with inftRg(t)=0 and suptRg(t)=1. Also, the function min is denoted by .

    Example 1.2. Let V be a real vector space and μ be a modular on X. Then a pair (V,μ) is a probabilistic modular space, where

    μ(v)(t)={tt+ρ(x),t>0,vV0,t0,vV.

    In 2002, Rassias [19] studied the Ulam stability of a mixed-type functional equation

    g(3i=1xi)+3i=1g(xi)=1ij3g(xi+xj).

    Later, Nakmalachalasint [13] generalized the above functional equation and obtained an n-variable mixed-type functional equation of the form

    g(ni=1xi)+(n2)ni=1g(xi)=1ijng(xi+xj)

    for n>2 and investigated its Ulam stability.

    In 2005, Jun and Kim [8] introduced a generalized AQ-functional equation of the form

    g(x+ay)+ag(xy)=g(xay)+ag(x+y)

    for a0,±1.

    In 2013, Zolfaghari et al. [25] investigated the Ulam stability of a mixed type functional equation in probabilistic modular spaces. In the same year, Cho et al. [2] introduced a fixed point method to prove the Ulam stability of AQC-functional equations in β-homogeneous probabilistic modular spaces.

    Motivated from the notion of probabilistic modular spaces and by the mixed type functional equations, we introduce a new mixed type functional equation satisfied by the solution f(x)=x+x2 of the form

    n1i=1,j=i+1(f(2xi+xj))+f(2xn+x1)2[n1i=1,j=i+1(f(xi+xj))+f(xn+x1)]=ni=1f(xi), (1.3)

    for nN and investigate its Ulam stability in probabilistic modular spaces.

    This paper is organized as follows: In Section 1, we provide a necessary introduction of this paper. In Sections 2 and 3, we obtain the general solution of the functional equation (1.3) in even case and in odd case, respectively. In Sections 4 and 5, we investigate the Ulam stability of (1.3) in probabilistic modular space by using fixed point theory for even and odd cases, respectively and the conclusion is given in Section 6.

    Let U and V be real vector spaces. In this section, we obtain the general solution of a mixed type functional equation (1.3) for even case of the form

    n1i=1,j=i+1(f(2xi+xj))+f(2xn+x1)2[n1i=1,j=i+1(f(xi+xj))+f(xn+x1)]=ni=1f(xi) (2.1)

    for nN.

    Theorem 2.1. Let f:UV satisfy the functional equation (2.1). If f is an even mapping, then f is quadratic.

    Proof. Assume that f:UV is even and satisfies the functional equation (2.1). Replacing (x1,x2,,xn) by (0,0,,0) and by (x1,0,,0) in (2.1), we obtain f(0)=0 and

    f(2x1)=4f(x1) (2.2)

    for all x1U, respectively. Again, replacing (x1,x2,x3,,xn) by (x1,x1,0,,0) in (2.1), we have

    f(3x1)=9f(x1) (2.3)

    for all x1U. Now, from (2.2) and (2.3), we get

    f(nx1)=n2f(x1),

    for all x1U. Replacing (x1,x2,x3,x4,,xn) by (x1,x2,x2,0,,0) in (2.1), we obtain

    f(2x1+x2)+f(x1+2x2)=4f(x1+x2)+f(x1)+f(x2) (2.4)

    for all x1,x2U. Replacing (x1,x2,x3,x4,,xn) by (x1,x2,0,0,,0) in (2.1), we get

    f(2x1+x2)+f(x2)=2f(x1+x2)+2f(x1) (2.5)

    for all x1,x2U. Replacing x2 by x2 in (2.5), using the evenness of f and again adding the resultant to (2.5), we get

    f(2x1+x2)+f(2x1x2)+2f(x2)=2f(x1+x2)+2f(x1x2)+4f(x1) (2.6)

    for all x1,x2U. Replacing (x1,x2) by (x1+x2,x1x2) in (2.6), we get

    f(3x1+x2)+f(x1+3x2)=4f(x1+x2)2f(x1x2)+8f(x1)+8f(x2) (2.7)

    for all x1,x2U. Letting (x1,x2) by (x1,x1+x2) in (2.5), we get

    f(3x1+x2)+f(x1+x2)=2f(2x1+x2)+2f(x1) (2.8)

    for all x1,x2U. Replacing x1 by x2 and x2 by x1 in (2.8), we have

    f(x1+3x2)+f(x1+x2)=2f(x1+2x2)+2f(x2) (2.9)

    for all x1,x2U. Now, adding (2.8) and (2.9), we obtain

    f(3x1+x2)+f(x1+3x2)+2f(x1+x2)=2f(2x1+x2)+2f(x1+2x2)+2f(x1)+2f(x2) (2.10)

    for all x1,x2U. Using (2.4), (2.7) and (2.10), we obtain (1.2). Hence the mapping f is quadratic.

    Let U and V be real vector spaces. In this section, we obtain the general solution of a mixed type functional equation (1.3) for even case of the form

    n1i=1,j=i+1(f(2xi+xj))+f(2xn+x1)2[n1i=1,j=i+1(f(xi+xj))+f(xn+x1)]=ni=1f(xi) (3.1)

    for nN.

    Theorem 3.1. Let f:UV satisfy the functional equation (3.1). If f is an odd mapping, then f is additive.

    Proof. Assume that f is odd and satisfies the functional equation (3.1). Replacing (x1,x2,,xn) by (0,0,,0) and (x1,0,,0) in (3.1), we obtain f(0)=0 and

    f(2x1)=2f(x1) (3.2)

    for all x1U, respectively. Again, replacing (x1,x2,x3,,xn) by (x1,x1,0,,0) in (3.1), we have

    f(3x1)=9f(x1) (3.3)

    for all x1U. Now, from (3.2) and (3.3), we get

    f(nx1)=nf(x1)

    for all x1U. Replacing (x1,x2,x3,x4,,xn) by (x1,x2,0,0,,0) in (3.1), we get

    f(2x1+x2)2f(x1+x2)=f(x2) (3.4)

    for all x1,x2U. Replacing x2 by x2 in (3.4), using the oddness of g and again adding the resultant to (3.4), we get

    f(2x1+x2)+f(2x1x2)=2f(x1+x2)+2f(x1x2) (3.5)

    for all x1,x2U. Replacing (x1,x2) by (x1+x2,x1x2) in (3.5), we get

    f(3x1+x2)+f(x1+3x2)=4f(x1)+4f(x2) (3.6)

    for all x1,x2U. Replacing x1 by x2 and x2 by x1 in (3.4), we have

    f(2x1+x2)+f(x1+2x2)=4f(x1+x2)f(x1)f(x2) (3.7)

    for all x1,x2U. Replacing (x1,x2) by (x1,x1+x2) in (3.4), we get

    f(3x1+x2)2f(2x1+x2)=f(x1+x2) (3.8)

    for all x1,x2U. Replacing x1 by x2 and x2 by x1 in (3.8) and adding the resultant equation to (3.8), we obtain

    f(3x1+x2)+f(x1+3x2)2f(2x1+x2)2f(x1+2x2)=2f(x1+x2) (3.9)

    for all x1,x2U. Using (3.6), (3.7) and (3.9), we obtain (1.1). Hence the mapping f is additive.

    In this section, we prove the Ulam stability of the n-variablel mixed type functional equation (1.3) for even case in probabilistic modular spaces (PM-spaces) by using fixed point technique.

    For a mapping f:M(V,μ), consider

    Se(x,y)=n1i=1,j=i+1(f(2xi+xj))+f(2xn+x1)2[n1i=1,j=i+1(f(xi+xj))+f(xn+x1)]ni=1f(xi)

    for nN.

    Theorem 4.1. Let M be a linear space and (V,μ) be a μ-complete b-homogeneous PM-space. Suppose that a mapping f:M(V,μ) satisfies an inequality

    μ(Se(x1,x2,,xn))ρ(x1,x2,,xn)(t) (4.1)

    for all x1,x2,,xnM and a given mapping ρ:M×MΔ such that

    ρ(2ax,0,,0)(22baNt)ρ(x,0,,0)(t) (4.2)

    for all xM and

    ρ(2amx1,2amx2,,2amxn)(22bamt)=1 (4.3)

    for all x1,x2,,xnM and a constant 0<N<12b. Then there exists a unique quadratic mapping T:M(V,μ) satisfying (2.1) and

    μ(T(x)f(x))(t22bNa12(12bN))ρ(x,0,,0)(t) (4.4)

    for all xM.

    Proof. Replacing (x1,x2,,xn) by (x,0,,0) in (4.1), we obtain

    μ(f(2x)22f(x))(t)ρ(x,0,,0)(t) (4.5)

    for all xM. This implies

    μ(f(2x)22f(x))(t)=μ(f(2x)22f(x))(22bt)ρ(x,0,,0)(22bt) (4.6)

    for all xM. Replacing x by 21x in (4.6), we obtain

    μ(f(21x)22f(x))(t)=μ(f(x)22f(21x))(t22b)ρ(21x,0,,0)(22bN1Nt22b)ρ(x,0,,0)(22bN1t). (4.7)

    From (4.6) and (4.7), we obtain

    μ(f(2ax)22af(x))(t)Ψ(x)(t):=ρ(x,0,,0)(22bNa12t) (4.8)

    for all xM.

    Consider P:={f:M(U,μ)|f(0)=0} and define η on P as follows:

    η(f)=inf{l>0:μ(f(x))(lt)Ψ(x)(t),xM}.

    It is simple to prove that η is modular on N and indulges the Δ2-condition with 2b=κ and Fatou property. Also, N is η-complete (see [25]). Consider the mapping Q:PηPη defined by QT(x):=T(2ax)22a for all TPη.

    Let f,jPη and l>0 be an arbitrary constant with η(fj)l. From the definition of η, we get

    μ(f(x)j(x))(lt)Ψ(x)(t)

    for all xM. This implies

    μ(Qf(x)Qj(x))(Nlt)=μ(22af(2ax)22aj(2ax))(Nlt)=μ(f(2ax)j(2ax))(22baNlt)Ψ(2ax)(22baNt)Ψ(x)(t)

    for all xM. Hence η(QfQj)Nη(fj) for all f,jPη, which means that Q is an η-strict contraction. Replacing x by 2ax in (4.8), we have

    μ(f(22ax)22af(2ax))(t)Ψ(2ax)(t) (4.9)

    for all xM and therefore

    μ(22(2a)f(22ax)22af(2ax))(Nt)=μ(22af(22ax)f(2ax))(22baNt)Ψ(2ax)(22baNt)Ψ(x)(t) (4.10)

    for all xE. Now

    μ(f(22ax)22(2a)f(x))(2b(Nt+t))μ(f(22ax)22(2a)f(2ax)22a)(Nt)μ(f(2ax)22af(x))(t)Ψ(x)(t) (4.11)

    for all xM. In (4.11), replacing x by 2ax and 2b(Nt+t) by 22βa2b(N2t+Nt), we obtain

    μ(f(23ax)22(2a)f(2ax))(22ba2b(N2t+Nt))Ψ(2ax)(22bjNt)Ψ(x)(t) (4.12)

    for all xM. Therefore,

    μ(f(23ax)23(2a)f(2ax)22a)(2b(N2t+Nt))Ψ(x)(t) (4.13)

    for all xM. This implies

    μ(f(23ax)23(2a)f(x))(2b(2b(N2t+Nt)+t))μ(f(23ax)23(2a)f(2ax)22a)(2b(N2t+Nt))μ(f(2ax)22af(x))(t)Ψ(x)(t) (4.14)

    for all xM. Generalizing the above inequality, we get

    μ(f(2amx)22(am)f(x))((2bN)m1t+2bm1i=1(2bN)i1t)Ψ(x)(t) (4.15)

    for all xM and a positive integer m. Hence we have

    η(Qmff)(2bN)m1+2bm1i=1(2bN)i12bmi=1(2bN)i12b12bN. (4.16)

    Now, one can easily prove that {Qm(f)} is ηconverges to TPη (see [25]). Thus (4.16) becomes

    η(Tf)2b12bN, (4.17)

    which implies

    μ(T(x)f(x))(2b12bNt)Ψ(x)(t)=ρ(x,0,,0)(2b22bNa12t) (4.18)

    for all xM and hence we have

    μ(T(x)f(x))(t22bNa12(12bN))ρ(x,0,,0)(t) (4.19)

    for all xM and hence the inequality (4.4) holds. One can easily prove the uniqueness of T (see [25]).

    In this section, we prove the Ulam stability of the n-variable mixed type functional equation (1.3) for odd case in probabilistic modular spaces (PM-spaces) by using fixed point technique.

    For a mapping f:M(U,μ), consider

    So(x1,x2,,xn)=n1i=1,j=i+1(f(2xi+xj))+f(2xn+x1)2[n1i=1,j=i+1(f(xi+xj))+f(xn+x1)]+ni=1f(xi)

    for nN.

    Theorem 5.1. Let M be a linear space and (V,μ) be a μ-complete b-homogeneous PM-space. Suppose that a mapping f:M(V,μ) satisfies an inequality

    μ(So(x1,x2,,xn))ρ(x1,x2,,xn)(t) (5.1)

    for all x1,x2,,xnM and a given mapping ρ:M×MΔ such that

    ρ(2ax,0,,0)(2baNt)ρ(x,0,,0)(t) (5.2)

    for all xM and

    ρ(2amx1,2amx2,,2amxn)(2bamt)=1 (5.3)

    for all x1,x2,,xnM and a constant 0<N<12b. Then there exists a unique additive mapping A:M(V,μ) satisfying (3.1) and

    μ(A(x)f(x))(t2bNa12(12bN))ρ(x,0,,0)(t) (5.4)

    for all xM.

    Proof. Replacing (x1,x2,,xn) by (x,0,,0) in (5.1), we obtain

    μ(f(2x)2f(x))(t)ρ(x,0,,0)(t) (5.5)

    for all xM. This implies

    μ(f(2x)2f(x))(t)=μ(f(2x)2f(x))(2bt)ρ(x,0,,0)(2bt) (5.6)

    for all xM. Replacing x by 21x in (5.6), we obtain

    μ(f(21x)21f(x))(t)=μ(f(x)2f(21x))(t2b)ρ(21x,0,,0)(2bN1Nt2b)ρ(x,0,,0)(2bN1t). (5.7)

    From (5.6) and (5.7), we obtain

    μ(f(2ax)2af(x))(t)Ψ(x)(t):=ρ(x,0,,0)(2bNa12t) (5.8)

    for all xM.

    Consider P:={f:M(U,μ)|f(0)=0} and define η on P as follows:

    η(f)=inf{l>0:μ(f(x))(lt)Ψ(x)(t),xM}.

    It is simple to prove that η is modular on N and indulges the Δ2-condition with 2b=κ and Fatou property. Also, N is η-complete (see [25]). Consider a mapping Q:PηPη defined by QA(x):=A(2ax)2a for all APη.

    Let f,jPη and l>0 be an arbitrary constant with η(fj)l. From the definition of η, we get

    μ(f(x)j(x))(lt)Ψ(x)(t)

    for all xM. This implies

    μ(Qf(x)Qj(x))(Nlt)=μ(2af(2ax)2aj(2ax))(Nlt)=μ(f(2ax)j(2ax))(2baNlt)Ψ(2ax)(2baNt)Ψ(x)(t)

    for all xM. Hence η(QfQj)Nη(fj) for all f,jPη, which means that Q is an η-strict contraction. Replacing x by 2ax in (5.8), we get

    μ(f(22ax)2af(2ax))(t)Ψ(2ax)(t) (5.9)

    for all xM and thus

    μ(22af(22ax)2af(2ax))(Nt)=μ(2af(22ax)f(2ax))(2baNt)Ψ(2ax)(2baNt)Ψ(x)(t), (5.10)

    for all xE. Now

    μ(f(22ax)22af(x))(2b(Nt+t))μ(f(22ax)22af(2ax)2a)(Nt)μ(f(2ax)2af(x))(t)Ψ(x)(t) (5.11)

    for all xM. In (5.11), replacing x by 2ax and 2b(Nt+t) by 2ba2b(N2t+Nt), we obtain

    μ(f(23ax)22af(2ax))(2ba2b(N2t+Nt))Ψ(2ax)(2bjNt)Ψ(x)(t) (5.12)

    for all xM. Therefore,

    μ(f(23ax)23af(2ax)2a)(2b(N2t+Nt))Ψ(x)(t) (5.13)

    for all xM. This implies

    μ(f(23ax)23af(x))(2b(2b(N2t+Nt)+t))μ(f(23ax)23af(2ax)2a)(2b(N2t+Nt))μ(f(2ax)2af(x))(t)Ψ(x)(t) (5.14)

    for all xM. Generalizing the above inequality, we have

    μ(f(2amx)2amf(x))((2bN)m1t+2bm1i=1(2bN)i1t)Ψ(x)(t) (5.15)

    for all xM and a positive integer m. Hence we have

    η(Qmff)(2bN)m1+2bm1i=1(2bN)i12bmi=1(2bN)i12b12bN. (5.16)

    Now, one can easily prove that {Qm(f)} is η-convergent to APη (see [25]). Thus (4.16) becomes

    η(Af)2b12bN, (5.17)

    which leads

    μ(A(x)f(x))(2b12bNt)Ψ(x)(t)=ρ(x,0,,0)(2b2bNa12t) (5.18)

    for all xM and hence we have

    μ(A(x)f(x))(t2bNa12(12bN))ρ(x,0,,0)(t) (5.19)

    for all xM and hence the inequality (5.4) holds. One can easily prove the uniqueness of A (see [25]).

    In this paper, we introduced a new n-variable mixed type functional equation satisfied by the solution f(x)=ax+bx2. Mainly, we obtained its general solution and investigated its Ulam stability in PM-spaces by using fixed point theory and we hope that this research work is a further improvement in the field of functional equations.

    This work was supported by Incheon National University Research Grant 2020-2021.

    The authors equally conceived of the study, participated in its design and coordination, drafted the manuscript, participated in the sequence alignment, and read and approved the final manuscript.

    The authors declare that they have no competing interests.



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