Research article

Local Hölder continuity of inverse variation-inequality problem constructed by non-Newtonian polytropic operators in finance

  • Received: 21 August 2023 Revised: 10 October 2023 Accepted: 17 October 2023 Published: 23 October 2023
  • MSC : 35K99, 97M30

  • This paper aims to explore the inverse variation-inequality problems of a specific type of degenerate parabolic operators in a non-divergence form. These problems have significant implications in financial derivative pricing. The study focuses on analyzing the Hölder continuity of weak solutions by employing cut-off factors.

    Citation: Jia Li, Zhipeng Tong. Local Hölder continuity of inverse variation-inequality problem constructed by non-Newtonian polytropic operators in finance[J]. AIMS Mathematics, 2023, 8(12): 28753-28765. doi: 10.3934/math.20231472

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  • This paper aims to explore the inverse variation-inequality problems of a specific type of degenerate parabolic operators in a non-divergence form. These problems have significant implications in financial derivative pricing. The study focuses on analyzing the Hölder continuity of weak solutions by employing cut-off factors.



    In recent years, the study of the following variational inequality has attracted the interest of scholars:

    {min{Lu,uu0}0,(x,t)ΩT,u(0,x)=u0(x),xΩ,u(t,x)=uν=0,(x,t)Ω×(0,T), (1)

    where Lu is a linear parabolic operator or a degenerate parabolic operator. Due to the satisfaction of the condition Lu0 in ΩT, researchers have found it convenient to use the comparison principle to obtain upper bounds for the solutions. This approach has been combined with limit methods [1,2], the Leray-Schauder fixed point theorem [3,4], or semi-discrete methods [5,6] to prove the existence of solutions. Additionally, some scholars have started from weak solutions and obtained integral inequalities for the difference between two weak solutions, analyzing the stability and uniqueness of weak solutions with respect to initial values [7,8,9]. The authors from [10,11,12] have demonstrated the explosive nature of weak solutions under certain special conditions through energy estimates. The authors have obtained Caccioppoli inequalities that match the variational inequality by analyzing integral inequalities of weak solutions in locally cylindrical regions, and subsequently studied the Schauder estimates for weak solutions [13,14].

    In recent years, research on the pricing of financial derivative products with embedded early exercise provisions has found that inverse variation-inequalities, such as the one shown below, are more suitable for

    {min{Lu,uu0}0,(x,t)ΩT,u(0,x)=u0(x),xΩ,u(t,x)=uν=0,(x,t)Ω×(0,T). (2)

    For example, researches from [15,16] analyzed the pricing problem of American options under the Black-Scholes model, and the value was reduced to the free boundary problem of the variation inequality (2). Therefore, the parabolic operator Lu (denoted as LBSu) satisfies

    LBS=tu12σ2xxu+rxuru, (3)

    where σ represents the volatility of the underlying stock of the option, and r is the risk-free interest rate in the financial market. Based on the aforementioned financial background considerations, the author of this study investigates the inverse variation-inequality problems with the degenerate parabolic operator in non-divergence form

    Lu=tuuσΔpuγuσ1|u|p,  p>2,σ>γ>0. (4)

    Additionally, we impose the condition that the initial value u0 satisfies u0W1,p0(Ω). For a recent study on inverse variational inequalities in a different context, please refer to [17]. In that study, an inverse quasi-variational inequality is solved using a dynamical system.

    In this paper, we provide the weak solution to the variational inequality (1) and prove the existence of weak solutions. We also prove that the weak solution satisfies an energy inequality in a local cylindrical region, and based on this, we establish the Holder continuity and Harnack inequality of the weak solution. The study of such conclusions is usually focused on degenerate parabolic equation initial-boundary value problems, and research on variational inequalities is still rare.

    Due to the fact that the inverse variational inequality (2) implies Lu0inΩT, it no longer allows us to determine the upper bound of the solution u and establish the existence of weak solutions through the comparison principle, as in the traditional variational inequality (1). First, this difficulty is overcome by analyzing the energy upper bound of (uM0)+. We can select a suitable M0 as an upper bound for u, which is also an innovative aspect of this paper. Second, this paper also explores the Harnack inequality and Hölder continuity of weak solutions by analyzing the weak solutions of the reverse variational inequality (2) and combining it with the integral inequality of (uk)±. This analysis is further enhanced by the use of the cut-off factor and the selection of an appropriate k, which adds another innovative aspect to this study.

    This section is dedicated to addressing our specific problem: We begin by providing a clear definition of a nonnegative weak solution to Eq (1). To begin, it can be inferred from Eq (1) that

    uu00inΩT. (5)

    In fact, by utilizing inequality (1) once again, we have Lu0 in ΩT. Furthermore, since u(t,x)=u0=0 in Ω×(0,T), when, using the comparison principle, we can conclude that (5) still holds.

    Next, we analyze the upper bound of u. Let us choose a constant M0>0 as a parameter. Multiplying both sides of Lu0 by (uM0)+ and integrating over Ω yields (note that (uM0)+0),

    Ωtu(uM0)+uσΔpu(uM0)+γuσ1|u|p(uM0)+dx0. (6)

    On one hand, when uM0 occurs, t(uM0)+=tu, thereby resulting in

    Ωtu(uM0)+dx=Ωt(uM0)(uM0)+dx=12tΩ(uM0)2+dx. (7)

    On the other hand, when u<M0 occurs, (uM0)+=0 and t(uM0)+=0, leading to

    Ωtu(uM0)+dx=0. (8)

    Combining (6)–(8), we obtain

    12tΩ(uM0)2+dxΩuσΔpu(uM0)++γuσ1|u|p(uM0)+dx0. (9)

    Note that

    ΩuσΔpu(uM0)++γuσ1|u|p(uM0)+dx=0,

    when u<M0, while

    ΩuσΔpu(uM0)++γuσ1|u|p(uM0)+dx=Ωuσ|(uM0)+|p+(σγ)uσ1|u|p(uM0)+dx,

    when u is greater than or equal to M0. Therefore, (9) implies the significance of

    12tΩ(uM0)2+dx+Ωuσ|(uM0)+|p+(σγ)uσ1|u|p(uM0)+dx0. (10)

    Due to uu00 and σγ0,

    Ωuσ|(uM0)+|p+(σγ)uσ1|u|p(uM0)+dx

    is nonnegative. Combining (10), we have

    Ω(uM0)2+dxΩ(u0M0)2+dx. (11)

    Furthermore, due to u0W1,p0, when M0 is sufficiently large,

    Ω(u0M0)2+dx=0.

    In this case,

    uM0inΩT. (12)

    By combining (5) and (12), we can demonstrate that the inverse variational inequality (2) satisfies

    0uM0inΩT. (13)

    Therefore, in [12], before providing a weak solution to the inverse variational inequality (2), we first present a set of maximal monotone maps

    G(λ)={ξ|ξ=0,λ>0;ξ0,λ=0}. (14)

    If ξG(uu0), it is easy to see that when u>u0, ξ=0; and in this case Lu=0. When u=u0, ξ0, and in this case we also have Lu0. This inspires us to use Lu=ξ to construct a weak solution for the variational inequality (2).

    Definition 2.1. A pair (u,ξ) is considered a generalized solution of the inverse variation-inequality (2) if it satisfies the following conditions:

    (a) uL(0,T,H1(Ω)),tuL(0,T,L2(Ω)).

    (b) ξGforany(x,t)ΩT.

    (c) For fixed ν=σ1p+1 and for every test-function φC1(ˉΩT), there exists an equality

    ΩTtuφ+1νp1uν|uν|p2uνφ+σγνp|uν|pφdxdt=ΩTξφdxdt.

    By utilizing (13) and (14), combined with a standard energy method from [2,12], we can establish the existence of a weak solution for the inverse variational inequality (2).

    Theorem 2.1. Assuming that u0W1,p0(Ω) in ΩT, the inverse variational inequality (2) has a solution (u,ξ) within the class defined in Definition 2.1.

    The final part of this section is dedicated to introducing some notation and presenting several previously established results, which will be used in the subsequent proof of the Hölder continuity. The detailed proof can be found in [17].

    Lemma 2.1. Assume that {Yn},n=1,2,3, is a nonnegative sequence satisfying

    Yn+1CbnY1+αn,   C,b>1,α>0.

    If Y0C1/αb1/α2, then Yn0,n.

    Lemma 2.2. Assuming that p2, there exists a positive constant C such that

    ΩT|u|pdxdtC|{u>0}|p/(N+p)||u||pLp(ΩT),

    where C depends only on N and p.

    Along this section, we assume that u is a nonnegative weak solution to Eq (1) with p2. Our objective is to establish an integral inequality, which will be used to determine the Hölder continuity of the weak solution on the domain

    Q=Q(ρ,θ)=Bρ(x0)×(t0θ,t0),

    where ρ and θ are positive undetermined constants. Of course, ρ and θ should be sufficiently small to ensure QΩT. Let us define

    μ+=esssupQ(2R,Rp)u,   μ=essinfQ(2R,Rp)u,   ω=oscQ(2R,Rp)u=μ+μ,
    Rn=12R+12n+1R,   Qn=Q(Rn,dRpn), d(0,1],  

    and introduce the symbol

    kn=μ+12s+1ω+12s+n+1ω,  k+n=μ+12s+1ω12s+n+1ω,

    where s is a nonnegative undetermined constant. We obtain the Hölder estimate for the weak solution of inequality (2) by using the upper bound estimate of esssupQ(12R,d(12R)p)u that includes ω. In order to estimate essinfQ(12R,d(12R)p)u, we construct kn) and simultaneously construct kn)+ to estimate esssupQ(12R,d(12R)p)u. In order to prove the Hölder continuity, s must satisfy the condition s>1.

    Lemma 3.1. Assuming p2 and ν=σ1p+1, one can infer

    (ukn)ν+1(2sω)pνν1(ukn)pν. (15)

    Proof. According to the definition of kn, it is easy to obtain

    (ukn)μ+kn=12s+1ω+12s+n+1ω12sω.

    Since (ukn)pν reaches its maximum, when u takes the value μ+,

    (2sω)pνν1(ukn)pν(2sω)pνν1(ω2s)pν=(ω2s)ν+1 (16)

    holds. At this point, (ukn)ν+1 satisfies

    (ukn)ν+1=(12s+1ω+12s+n+1ω)ν+1=(ω2s)ν+1. (17)

    By combining Eqs (16) and (17), the result is proven to hold.

    In order to achieve the desired outcome, a test function w=ϕp×(uk)ν± is selected, resulting in

    ΩT1νp1uν|uν|p2uν[ϕp×(uk)μ±]+σγνp|uν|p×ϕp×(uk)ν±dxdt=ΩTξφdxdtΩTtu×ϕp×(uk)ν±dxdt. (18)

    Considering that ΩTtu×ϕp×(uk)ν±dxdt is not suitable for integration calculations, the following transformation is performed:

    Ωt(ϕp×(uk)ν+1±)dx=(ν+1)Ωϕp×(uk)ν±utdxdt+pΩϕp1×tϕ×(uk)ν+1±dx. (19)

    In Ωuν|uν|p2uν[ϕp×(uk)ν±]dx, a differential transformation is applied to [ϕp×(uk)ν±], resulting in

    Ωuν|uν|p2uν[ϕp×(uk)ν±]dx=Ωuν×|(uk)ν±|p×ϕpdx+Ω|uν|p2uν×(uk)ν±×uν×ϕpdx. (20)

    Further utilizing the Hölder and Young inequalities, we can obtain the expression

    |Ω|uν|p2uν×(uk)ν±×uνϕpdx|p1pΩuν×|(uk)±|p×ϕpdx+1pΩ(uk)pν±×uν×|ϕ|pdx. (21)

    Consequently, by combining Eqs (18)–(21), we can obtain the equation

    esssupt(t0θ,t0)Ω(ϕp×(uk)ν+1±)dx+1νp1pt0t0θΩuν|(uk)ν±|p×ϕpdxdt+σγνpΩT|uν|p×ϕp×(uk)ν±dxdtpΩϕp1×|tϕ|×(uk)ν+1±dx+Ω(ϕp(x,t0θ)×(u(x,t0θ)k)ν+1±)dx+1pνp1t0t0θΩ|(uk)pν±×uν|ϕ|pdxdt. (22)

    Theorem 3.1 Let u be a weak solution of the inverse variational inequality (2) with p2, then it follows that

    esssupt(t0θ,t0)Ω(ϕp×(uk)ν+1±)dx+1νp1pt0t0θΩuν|(uk)ν±|p×ϕpdxdt+σγνpΩT|uν|p×ϕp×(uk)ν±dxdtpΩϕp1×|tϕ|×(uk)ν+1±dx+1pνp1t0t0θΩ|(uk)pν±×uν|ϕ|pdxdt. (23)

    In (23), we utilize the condition ϕ(x,t0θ)=0, which readily yields

    Ω(ϕp(x,t0θ)×(u(x,t0θ)k)ν+1±)dx=0.

    Additionally, it is worth noting that by selecting suitable ϕ and (uk)± in (22), we can obtain local estimates for the weak solution u, thereby establishing the Harnack inequality and Hölder continuity.

    This section is devoted to analyzing the Hölder continuity of weak solutions of the inverse variational inequality (2). We first examine the local lower bound estimate of weak solutions u of the inverse variational inequality (2), and define a cut-off function ϕn(x,t) on Qn as described in

    ϕn(x,t)={0,(x,t)Qn,1,(x,t)Qn+1. (24)

    Additionally, we assume that ϕn(x,t) satisfies the condition

    |ϕn(x,t)|2nRn, |tϕn(x,t)|2pnRp. (25)

    In (23), (uk)± is set as (uk), while k is chosen as kn, resulting in

    esssupt(t0dR/2,t0)Ω(ϕp×(ukn)ν+1)dx+1pνp1t0t0dR/2Ωuν|(ukn)ν|p×ϕpdxdt+σγνpΩT|uν|p×ϕp×(ukn)νdxdtpΩϕp1×|tϕ|×(ukn)ν+1dx+1pνp1t0t0dR/2Ω|(ukn)pν×uν|ϕ|pdxdt. (26)

    Due to the presence of σγ>0 and ν=σ1p+1>1,

    σγνpΩT|uν|p×ϕp×(ukn)νdxdt0.

    After removing them, we have

    esssupt(t0dR/2,t0)Ω(ϕp×(ukn)ν+1)dx+1pνp1t0t0dR/2Ωuν|(ukn)ν|p×ϕpdxdtp2pnRp(t0t0dR/2Bnϕp1×(ukn)ν+1dxdt+dp2νp1t0t0dR/2Bn|(ukn)pν|dxdt). (27)

    Further analysis of t0t0dR/2Bnϕp1×(ukn)ν+1dxdt+dp2νp1t0t0dR/2Bn|(ukn)pν|dxdt is conducted by applying Lemma 3.1,

    t0t0dR/2Bnϕp1×(ukn)ν+1dxdt+dp2νp1t0t0dR/2Bn|(ukn)pν|dxdt(ω2s)ν+1[p2+dp2νp1(ω2s)pνν1]t0t0dR/2Bn(ukn)ν+1dxdt.

    By substituting the aforementioned results into Eq (27), we can obtain

    esssupt(t0dR/2,t0)Ω(ϕp×(ukn)ν+1)dx+1pνp1t0t0dR/2Ωuν|(ukn)ν|p×ϕpdxdtp2pnRp(ω2s)ν+1[p2+1p2νp1(ω2s)pνν1]t0t0dR/2Bnχ(ukn)>0dxdt. (28)

    For the purpose of facilitating the discussion, let us define An={xBn|ukn}. Consequently, it can be derived from Eq (28) that

    ||(ukn)ϕn||pLp(Qn)p2pnRp(ω2s)ν+1[p2+1p2νp1(ω2s)pνν1]t0t0dR/2|An|dt. (29)

    Applying Lemma 2.2 to ||(ukn)ϕn||pLp(Qn), we obtain

    ||(ukn)||pLp(Qn)||(ukn)ϕn||pLp(Qn)(t0t0dR/2|An|dt)pN+p. (30)

    Lemma 4.1. If u is a weak solution of the inverse variational inequality (2) with p>2, then

    ||(ukn)||pLp(Qn+1)12p(n+2)(ω2s)pt0t0dR/2|An+1|dt.

    Proof. Due to

    kn=μ+12s+1ω+12s+n+1ω,

    it follows that

    ||(ukn)||pLp(Qn+1)=Qn+1(ukn)pdxdtl=n+1Qn+1(klkn)pdxdtQn+1(kn+1kn)pdxdt,

    thereby

    ||(ukn)||pLp(Qn+1)|knkn+1|pt0t0dR/2|An+1|dt. (31)

    Furthermore, due to

    |knkn+1|p=12s+n+1ω12s+n+2ω=12s+n+2ω,

    it follows that

    |knkn+1|p12p(n+2)(ω2s)p,

    thereby

    |knkn+1|pt0t0dR/2|An+1|dt12p(n+2)(ω2s)pt0t0dR/2|An+1|dt. (32)

    By combining Eqs (31) and (32), Lemma 4.1 is proven.

    Continuing the analysis of the lower bound for weak solutions by combining (30) and Lemma 4.1 and substituting the obtained result into (29), it can be easily deduced that

    12p(n+2)(ω2s)pt0t0θ|An+1|dtp2pnRp(ω2s)ν+1[p2+1p2(ω2s)pνν1]t0t0dR/2|An|dt. (33)

    Consequently, simplifying (33) yields

    t0t0θ|An+1|dtp4pRp(ω2s)νp+1[p2+1p2(ω2s)pνν1]4pnt0t0dR/2|An|dt.

    This from Lemma 2.2 implies that t0t0dR/2|An|dt0asn, if

    t0t0θ|{xB12R|uμ+12s+1ω}|dtp2pN+pN+ppRpN+pp(ω2s)(1+ν)p2N+p4p3(N+p)2p2N+p. (34)

    It is worth noting that σ1 when selecting a sufficiently large S such that (34) always holds, thus leading to the following result.

    Theorem 4.1. If σ1, selecting a sufficiently large s>1, it holds that

    uμ+ω2s+1a.e.(x,t)Q(12R,d(12R)p). (35)

    Furthermore, if

    t0t0θ|{xB12R|uμ+12s+1ω}|dtp2pN+pN+ppRpN+pp(ω2s)(1+ν)p2N+p4p3(N+p)2p2N+p,

    then, (35) still holds.

    Next, we analyze the upper bound of the weak solution. By applying Lemma 4.1, it is easy to obtain

    (uk+n)ν+1+(2sω)pνν1(uk+n)pν+. (36)

    Consequently, in Eq (23) we set (uk)± as (uk+n)+ and eliminate the nonpositive term σγνpΩT|uν|p×ϕp×(uk+n)ν+dxdt, resulting in

    esssupt(t0dR/2,t0)Ω(ϕp×(uk+n)ν+1+)dx+1pνp1t0t0dR/2Ωuν|(uk+n)ν+|p×ϕpdxdtp2pnRp(t0t0dR/2Bnϕp1×(uk+n)ν+1+dxdt+dp2νp1t0t0dR/2Bn|(uk+n)pν+|dxdt). (37)

    Note that

    k+n=μ+12s+1ω12s+n+1ω.

    By applying Lemma 3.1, we can obtain

    (uk+n)ν+1+(2sω)pνν1(uk+n)pν+, (38)

    which implies that

    esssupt(t0dR/2,t0)Ω(ϕp×(uk+n)ν+1+)dx+1pνp1t0t0dR/2Ωuν|(uk+n)ν+|p×ϕpdxdtp2pnRp(ω2s)ν+1[p2+1p2νp1(ω2s)pνν1]t0t0dR/2Bnχ(uk+n)+>0dxdt. (39)

    For the sake of convenience in the discussion, we will continue to use the symbol An={xBn|uk+n}, hence

    ||(uk+n)+ϕn||pLp(Qn)p2pnRp(ω2s)ν+1[p2+1p2νp1(ω2s)pνν1]t0t0dR/2|An|dt. (40)

    By utilizing Lemma 2.2, we can obtain the following estimation

    ||(uk+n)+||pLp(Qn)||(uk+n)+ϕn||pLp(Qn)(t0t0dR/2|An|dt)pN+p. (41)

    Following the same approach as in Lemma 4.1, we can deduce that

    ||(uk+n)+||pLp(Qn+1)12p(n+2)(ω2s)pt0t0dR/2|An+1|dt. (42)

    Combining (41) and (42) and substituting the result into (40), we can simplify and deduce that

    t0t0dR/2|An+1|dtC(p,ν)4pnRp(t0t0dR/2|An|dt)1+pN+p. (43)

    Clearly, the equation above and Lemma 2.2 imply that t0t0dR/2|An|dt0asn, if

    t0t0θ|{xB12R|uμ+12s+1ω}|dtp2pN+pN+ppRpN+pp(ω2s)(1+ν)p2N+p4p3(N+p)2p2N+p. (44)

    Thus we have

    uμ+ω2s+1a.e.(x,t)Q(12R,d(12R)p). (45)

    Due to the fact that

    oscQ(12R,d(12R)p)u=esssupQ(12R,d(12R)p)uessinfQ(12R,d(12R)p)u,

    combining Eqs (35) and (45), we obtain

    oscQ(12R,d(12R)p)u(112s)ωa.e.(x,t)Q(12R,d(12R)p). (46)

    Theorem 4.2. (Hölder continuity) For any (x,t)Q(12R,d(12R)p), if σ>1, there exists a nonnegative constant C such that

    oscQ(12R,d(12R)p)uCω.

    Furthermore, if (34) and (44) hold, the above inequality still holds.

    In fact, by choosing

    C=(112s)

    in (46), the conclusion of Theorem 3.3 is evident. Furthermore, by selecting

    C=(2s+1)/(2s1)2,

    we have the following result.

    Theorem 4.3. (Harnack's inequality) Assuming σ1, there exists a nonnegative constant C such that

    esssupQ(12R,d(12R)p)uCessinfQ(12R,d(12R)p)u.

    The Harnack inequality implies the following Hölder modulus estimate, as indicated by the literature in [17].

    Theorem 4.4. (Hölder's modulus estimate) Let uL(0,T;W1,p0(Ω)) be a weak solution of the inverse variation-inequality (2) and σ1. Then, there exists a constant C and β(0,1), for any ΩΩ, such that [u]β,12β;ΩTC.

    This paper aimed to explore a specific type of inverse variation inequality problem

    {min{Lu,uu0}0,(x,t)ΩT,u(0,x)=u0(x),xΩ,u(t,x)=uν=0,(x,t)Ω×(0,T),

    which was formulated using degenerate parabolic operators in non-divergence form

    Lu=tuuσΔpuγuσ1|u|p, p>2,σ>γ>0.

    First, by incorporating (uM0)+ into Lu0 in ΩT, we obtained the following integral inequality:

    12tΩ(uM0)2+dx+Ωuσ|(uM0)+|p+(σγ)uσ1|u|p(uM0)+dx0.

    Subsequently, we derived the upper and lower bounds for the inverse variation inequality problem (2) and utilized them to construct a weak solution for the inverse variation-inequality problem (2).

    Next, in the weak solution, the test function w=ϕp×(uk)ν± was chosen and an integral inequality was obtained using the Hölder and Young inequalities, as shown in Theorem 3.1. Finally, the incorporation of a cut-off factor in Theorem 3.1 yields the Hölder continuity of the weak solution to problem (2), the Harnack inequality and the Hölder modulus estimate.

    There are still some areas in this paper that can be improved. The current study only considered the case where σ>γ, and the existence of weak solutions cannot be proven if σ<γ. Additionally, in the proof process of the Hölder continuity in Section 4, the condition σ>γ was also used to ensure that σγνpΩT|uν|p×ϕp×(uk)ν+dxdt was nonnegative. In Lemma 2.2, the parameter p was restricted to be greater than two. In future research, we will attempt to analyze the impact of these restrictive conditions on the results.

    The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

    The author sincerely thanks the editors and anonymous reviewers for their insightful comments and constructive suggestions, which greatly improved the quality of the paper.

    The authors declare no conflicts of interest.



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