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Research article

Flag-transitive 2-designs with block size 5 and alternating groups

  • Received: 15 November 2024 Revised: 24 April 2025 Accepted: 28 April 2025 Published: 06 May 2025
  • MSC : 05B05, 05B25, 05E18, 20B25

  • This paper contributes to the classification of flag-transitive 2-designs with block size 5. In a recent paper, the flag-transitive automorphism groups of such designs are reduced to point-primitive groups of affine type and almost simple type, and a classification is given of such automorphism groups with sporadic socle. In the present paper, we classify such designs admitting a flag-transitive automorphism group whose socle is an alternating group. We prove that there are precisely six such designs and determine the corresponding automorphism groups.

    Citation: Jiaxin Shen, Yuqing Xia. Flag-transitive 2-designs with block size 5 and alternating groups[J]. AIMS Mathematics, 2025, 10(5): 10308-10323. doi: 10.3934/math.2025469

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  • This paper contributes to the classification of flag-transitive 2-designs with block size 5. In a recent paper, the flag-transitive automorphism groups of such designs are reduced to point-primitive groups of affine type and almost simple type, and a classification is given of such automorphism groups with sporadic socle. In the present paper, we classify such designs admitting a flag-transitive automorphism group whose socle is an alternating group. We prove that there are precisely six such designs and determine the corresponding automorphism groups.



    The globalization of financial markets has been developing rapidly, which requires more on credit risk management. Credit risk, which is attracting more and more attention from people in academics and practices, refers to not only default risk but also credit rating migration risk. The credit rating migration risk is playing a more and more significant role in financial markets and risk management, especially after the outbreak and spread of 2008 financial crisis. In particular, the credit rating migration risk makes difference to the corporate bond pricing.

    There are two traditional models for default risk, involving the structural models and reduced form models. In the reduced form models, the default event is depicted and captured by introducing an exogenous variant. The default time is modeled by a stochastic default intensity in this approach, see [7,14,17] and so forth. The assumption on the structural models is that if the below bound of some insolvency threshold is met by corporate value, default occurs. In the model proposed by Merton [28], a default event may only occur at the maturity. Subsequently, Black and Cox [2] extended Merton's model to a first-passage-time model, where default may occur at any time before the debt maturity, see also [1,3,18,25] and so forth.

    With regard to literatures working on credit rating migration, a commonly adopted approach is the Markov chain, which is captured by transition intensity matrix coming from general statistic data, see [5,8,9] and so forth. The framework of reduced form then can be directly developed for dynamic credit rating migration process, see [6,15,30] and so forth. However, the Markov chain ignores the role played by the corporate value when modelling credit rating migration. In fact, the corporate value is an important factor in the credit rating migration and should be taken into consideration. Accordingly, from the corporate perspective, Liang et al. [20] started to model and analyze credit rating migration risk by structural model based on Merton's model. They set a predetermined migration threshold to divide the corporate value into high and low rating regions, where the corporate value follows different stochastic processes. However in practice, the threshold dividing credit ratings is usually not predetermined. To solve this problem, Hu et al. [11] improved the model proposed by Liang et al. [20]. They determined the migration boundary by the dynamic proportion between corporate debt and corporate value, which results in a free boundary problem. Subsequently, Liang et al. [21] incorporated a risk discount factor, which measures the sensibility of credit rating migration to the proportion, into the model and showed that an asymptotic traveling wave solution exists in the free boundary problem. Problem on credit rating migration in switching macro regions can also be referred to Wu and Liang [36], while credit contingent interest rate swap with credit rating migration can be seen in Liang and Zou [23].

    In particular, the aforementioned works [11,20,21,23,36] only take two credit ratings into consideration in credit rating migration problem. The credit region is divided into the high rating region and the low rating region, which results in only one free boundary in the corresponding free boundary problem. However, in practice, we should notice that there are usually more than two credit ratings used when accessing the corporate credit levels. The Standard & Poor's, an international rating agency, downgraded the long-term sovereign credit rating of Greece from A- to BBB+ on the evening of December 16, 2009. This verifies the fact and inspires us to consider multiple credit ratings in migration problems. Wu and Liang [34] provided some numeric results for multiple credit rating migration problem. Wang et al. [32] presented some theoretical results by showing that the asymptotic traveling wave solution obtained in Liang et al. [21] persists in the free boundary problem with multiple free boundaries. By considering stochastic interest rate in reality [19,24,29], Yin et al. [37] improved the model by replacing the constant interest rate with a stochastic version. This improved model covers the previous works where only two credit ratings are involved with an interest-dependent volatility [22] or multiple credit ratings migration with constant interest rate [32,34]. Then Huang et al. [12] continued to study the bond pricing model with multiple credit rating migration and stochastic interest rate. They contributed to establishing the asymptotic traveling wave solution in the time-heterogeneous free boundary problem with multiple free boundaries.

    The aforementioned models for credit rating migration in the structure framework are based on the Merton's model, i.e., default may only occur at the maturity. However, in practice, default may occur at any time up to maturity [4,26,31]. Wu et al. [35] relaxed the default restriction in the credit rating migration model by setting a predetermined threshold capturing the first-passage time when the default occurs. Once the corporate value falls below the threshold at any time, the default occurs. This results in a free boundary problem subject to a new boundary condition. Again we have to notice that the credit region in their credit rating migration model pricing a defaultable corporate bond is still divided into the high and low rating regions and meanwhile, a constant interest rate is also considered. Hence, motivated by these existing works on the effect of credit rating migration risk when valuating a corporate bond, in this paper, we devote to studying a pricing model for a defaultable corporate bond with both multiple credit rating migration risk and stochastic interest rate. Our work extends the existing works [12,37], where the multiplicity of credit rating and stochasticity of interest rate are involved in their models, by inserting the default risk. Meanwhile, we improve the results of Wu et al. [35], who considered default risk in pricing the corporate bond with only one credit rating migration boundary and constant interest rate, to fit the effects of multiple credit ratings and stochastic interest rate.

    The difficulties in analysis are generated from the joint effects and mutual restrictions among the multiplicity of credit ratings, stochastic volatility of interest rate and presence of default boundary. Besides that the free boundary problem turns into a initial-boundary problem, in particular, it is perplexed by not only a time-dependent and discontinuous coefficient but also a time-dependent process arisen in the problem. In addition to an irreducible barrier boundary, these indeed cause some troubles in deriving necessary estimates and then proving the existence and uniqueness of solution to the free boundary problem. Another contribution is the asymptotic behavior of solution to the free boundary problem. We prove that the solution converges to some steady status, which is the spatially homogeneous solution of an auxiliary free boundary problem, whose coefficients are the long time limits of the time-dependent coefficients in the original free boundary problem. This convergence, which shows us the developing tendency of solution, is established by two steps. In the first step, it is shown that the solution of the original free boundary problem converges to the solution of the auxiliary free boundary problem with time tending to infinity, while in the second step, it is shown that the solution of the auxiliary free boundary problem converges to the steady status. Moreover, the steady status can be solved explicitly. Thus, we present an explicit formula to valuate the defaultable corporate bond with multiple credit rating migration risk and stochastic volatility of interest rate.

    The paper is organized as follows. In Section 2, the pricing model is constructed. In Section 3, the model is reduced into a free boundary problem with initial condition and boundary conditions involving default boundary and migrating boundaries. In Section 4, an approximated free boundary problem is analyzed and some preliminary lemmas for uniform estimates are collected. In Section 5, through the approximated free boundary problem, the existence and uniqueness of solution to the free boundary problem are obtained. In Section 6, it is proved that the solution is convergent to a steady status by a Lyapunov argument. Then we conclude the paper by presenting an explicit pricing formula for the defaultable corporate bond in Section 7.

    In this section, we set up the baseline pricing model for a defaultable corporate bond subject to multiple credit rating migration risk and stochastic interest rate. Some necessary assumptions are put forward as follows.

    Let (Ω,F,P) be a complete probability space. Suppose that the corporation issues a defaultable bond, which is a contingent claim of its value on the space (Ω,F,P).

    Let S denote the corporate value in the risk neutral world. It satisfies the Black-Scholes model

    dS=r(t)Sdt+σ(t)SdWt,

    where r is the time-varying interest rate and σ is the heterogeneous volatility with respect to credit ratings. Wt is the Brownian motion generating the filtration {Ft}. The credit is divided into n ratings. In different credit ratings, the corporation shows different volatilities of its value. We denote the volatility in the i'th rating by σ(t)=σi, i=1,2,,n, and in addition, they satisfy

    0<σ1<σ2<<σn1<σn<,

    which means that in the highest credit rating, the corporation shows the smallest volatility σ1 and in the lowest credit rating, it shows the largest volatility σn. The stochastic interest rate r is supposed to satisfy the Vasicek model [27,33]

    dr=a(t)(θ(t)r)dt+σr(t)dWrt,

    which is widely popular in financial application, where the parameters a, θ, σr are supposed to be positive constants in this paper. σr is the volatility of the interest rate. θ is considered as the central location or the long-term value. a determines the speed of adjustment.

    We suppose that the corporation issues only one defaultable bond with face value F. The effect of corporate value on the bond value is focused on and the discount value of bond is considered. Denote by ϕt the discount value of bond at time t. The corporation exhibits two risks, the default risk and credit rating migration risk. The corporation can default before maturity time T. The default time τd is the first moment when the corporate value falls below the threshold K, namely that

    τd=inf{t>0|S0>K,StK},

    where K<FD(t,T), where 0<D(t,T)<1 is the discount function. Once the corporation defaults, the investors will get what is left. Hence, ϕt(K)=K and at the maturity time T, the investors can get ϕT=min{ST,F}. On the other hand, the credit regions are determined by the leverage γ(t)=ϕt/St. Denote the thresholds of leverage γ(t) by γi, i=1,2,,n1, and they satisfy

    0<γ1<γ2<<γn2<γn1<1.

    The credit rating migration times are the first moments when the corporate credit rating is upgraded or downgraded. They are defined as follows:

    τ1=inf{t>0|ϕ0/S0<γ1,ϕt/Stγ1},
    τn=inf{t>0|ϕ0/S0>γn1,ϕt/Stγn1},
    τi,i+1=inf{t>0|γi1<ϕ0/S0<γi,ϕt/Stγi},i=2,3,,n1,
    τi,i1=inf{t>0|γi1<ϕ0/S0<γi,ϕt/Stγi1},i=2,3,,n1.

    τ1 is the first moment that the corporation degrades from the highest credit rating. τn is the first moment that the corporation upgrades from the lowest credit rating. τi,i+1 is the first moment that the corporation jumps up into the i+1'th credit rating from the i'th credit rating, while τi,i1 is the first moment that the corporation jumps down into the i1'th credit rating from the i'th credit rating.

    Once the credit rating migrates before the maturity T, a virtual substitute termination happens, namely that the bond is virtually terminated and substituted by a new one with a new credit rating [35]. Thus, there is a virtual cash flow of the bond. Denote the bond values in different credit ratings by ϕi(t,S), i=1,2,,n. Then they are the conditional expectations as follows

    ϕ1(t,S)=Et,S[h1(t,T)|ϕ1(t,S)<γ1S],

    where

    h1(t,T)=eTtr(s)dsmin{ST,F}χ(min{τ1,τd}T)+eτ1tr(s)dsϕ2(τ1,Sτ1)χ(t<τ1<min{τd,T})+eτdtr(s)dsKχ(t<τd<min{τ1,T}),

    and for i=2,3,,n1,

    ϕi(t,S)=Et,S[hi(t,T)|γi1S<ϕi(t,S)<γiS],

    where

    hi(t,T)=eTtr(s)dsmin{ST,F}χ(min{τi,i+1,τi,i1,τd}T)+eτi,i+1tr(s)dsϕi+1(τi,i+1,Sτi,i+1)χ(t<τi,i+1<min{τi,i1,τd,T})+eτi,i1tr(s)dsϕi1(τi,i1,Sτi,i1)χ(t<τi,i1<min{τi,i+1,τd,T})+eτdtr(s)dsKχ(t<τd<min{τi,i+1,τi,i1,T})

    and

    ϕn(t,S)=Et,S[hn(t,T)|ϕn(t,S)>γn1S],

    where

    hn(t,T)=eTtr(s)dsmin{ST,F}χ(min{τn,τd}T)+eτntr(s)dsϕn1(τn,Sτn)χ(t<τn<min{τd,T})+eτdtr(s)dsKχ(t<τd<min{τn,T}),

    where χ is the indicative function, satisfying χ=1 if the event happens and otherwise, χ=0.

    Suppose that the correlation between the interest rate and the corporate value is given by dWrtdWt=ρt, 1ρ1. By the Feynman-Kac formula, we can derive that ϕi, i=1,2,,n, are functions of time t, interest rate r and value S. They satisfy the following PDE in their regions

    ϕ1t+σ21S222ϕ1S2+σrσ1ρS2ϕ1Sr+rSϕ1S+2ϕ1r2+a(θr)ϕ1rrϕ1=0,ϕ1<γ1S, (2.1)

    and for i=2,3,,n1,

    ϕit+σ2iS222ϕiS2+σrσiρS2ϕiSr+rSϕiS+2ϕir2+a(θr)ϕirrϕi=0,γi1S<ϕi<γiS, (2.2)

    and

    ϕnt+σ2nS222ϕnS2+σrσnρS2ϕnSr+rSϕnS+2ϕnr2+a(θr)ϕnrrϕn=0,ϕn>γn1S, (2.3)

    with terminal conditions

    ϕi(T,r,S)=min{S,F},i=1,2,,n, (2.4)

    and boundary condition

    ϕn(t,r,K)=K. (2.5)

    The bond value is continuous when it passes a rating threshold, i.e., ϕi=ϕi+1 on the rating migration boundaries, where i=1,2,,n1. Also, if we construct a risk free portfolio π by longing a bond and shorting Δ amount asset value S, i.e., π=ϕΔS and such that dπ=rπ, this portfolio is also continuous when it passes the rating migration boundaries, namely that πi=πi+1 or Δi=Δi+1 on the rating migration boundaries, where i=1,2,,n1. By Black-Scholes theory [16], it is equivalent to

    ϕiS=ϕi+1Sontheratingmigrationboundary,i=1,2,,n1. (2.6)

    Denote by P(t,r) the value of a guaranteed zero-coupon bond with face value 1 at the maturity t=T, where the interest rate follows the Vasicek model. By the Feynman-Kac formula, P(t,r) satisfies the following PDE

    Pt+σ2r22Pr2+a(θr)PrrP=0,r>0,0<t<T,

    with terminal condition P(T,r)=1, whose explicit solution is solved as P(t,r)=eA(Tt) [13], where

    A(Tt)=1a2(B2(Tt)(Tt))(a2θσ2r2)σ2r4aB2(Tt)rB(Tt)

    and

    B(Tt)=1a(1ea(Tt)).

    Take transformations

    y=SP(t,r),ψi(t,y)=ϕi(t,r,S)P(t,r),i=1,2,,n.

    Then ψi, i=1,2,,n satisfy

    ψ1t+ˆσ21y222ψ1y2=0,ψ1<γ1y, (2.7)

    and for i=2,3,,n1,

    ψit+ˆσ2iy222ψiy2=0,γi1y<ψi<γiy, (2.8)

    and

    ψnt+ˆσ2ny222ψny2=0,ψn>γn1y, (2.9)

    where

    ˆσ2i=σ2i+2ρσiσrB(Tt)+σ2rB2(Tt).

    The terminal conditions are given as

    ψi(T,y)=min{y,F},i=1,2,,n, (2.10)

    and the boundary conditions are

    ψn(t,KeA(Tt))=KeA(Tt) (2.11)

    and

    ψi=ψi+1,ψiy=ψi+1y,ontheratingmigrationboundary (2.12)

    for i=1,2,,n1.

    We introduce the standard transformation of variable x=logy, remaining Tt as t, and define

    φ(t,x)=exψi(Tt,ex)inthei'thratingregion.

    Using (2.12), we drive the following equation from (2.7)-(2.9) as

    φtˆσ222φx2ˆσ22φx=0,logKA(t)<x<,t>0, (3.1)

    where ˆσ=ˆσ1 as φ<γ1, ˆσ=ˆσi as γi1<φ<γi for i=1,2,,n1, ˆσ=ˆσn as φ>γn1,

    ˆσ2i(t)=σ2i+2ρσiσrB(t)+σ2rB2(t),i=1,2,,n. (3.2)

    Meanwhile denote by

    ˉˆσ2i=σ2i+2ρσiσra+σ2ra2,i=1,2,,n, (3.3)

    the limits of ˆσ2i(t) as time tends to infinity, i=1,2,,n. Without loss of generality, suppose that F=1 and there holds K<1. Then (3.1) is supplemented with the initial condition

    φ(0,x)=min{1,ex}. (3.4)

    and boundary condition

    φ(t,s(t))=1, (3.5)

    where s(t)=logKA(t). Take u(t,x)=φ(t,x+s(t)). Then u satisfies

    utˆσ222ux2(ˆσ22+˙s(t))ux=0,0<x<,t>0, (3.6)

    where

    ˙s(t)=(σ2r2a2θ)(2B(t)eat1)+σ2r2aB(t)eat+reat

    and ˆσ=ˆσ1 as u<γ1, ˆσ=ˆσi as γi1<u<γi for i=1,2,,n1, ˆσ=ˆσn as u>γn1, with the initial condition

    u(0,x)=min{1,e(x+s(0))}=min{1,e(x+logK)} (3.7)

    and boundary condition

    u(t,0)=1. (3.8)

    The domain will be divided into n rating regions Qi, i=1,2,,n. We will prove that the domain can be separated by n1 free boundaries x=λi(t), i=1,2,,n1. These boundaries are a prior unknown since they should be solved by equations

    u(t,λi(t))=γi,i=1,2,,n1, (3.9)

    where u is also a priori unknown. Since we have assumed that (3.1) is valid cross the free boundaries, (2.12) implies that for i=1,2,,n1,

    u(t,λi(t))=u(t,λi(t)+)=γi,ux(t,λi(t))=ux(t,λi(t)+). (3.10)

    In the work [35] and [12], where the former model is subject to constant interest rate and the later one is subject to stochastic interest rate but without default boundary, the process ˙s(t) in (3.6) is replaced by a constant. The presence of ˙s(t) indeed leads to some technical differences in deriving the estimates in the following argument. We can rewrite the formula of ˙s(t) as

    ˙s(t)=(β+r)eatβe2ata2(βσ2r2a2),

    where

    β=σ2ra32θa+σ2r2a2.

    It is not difficult to analyze and derive that one of the following conditions holds, then there holds ˙s(t)0:

    02a2βσ2r; (3.11)

    or

    βr,a(β+r)2+σ2rβ2a2β2; (3.12)

    or

    r<β<0. (3.13)

    Let H(ξ) be the Heaviside function, namely that H(ξ)=0 for ξ<0 and H(ξ)=1 for ξ0. Then we can rewrite the volatility ˆσ in (3.6) as

    ˆσ=ˆσ1+n1i=1(ˆσi+1ˆσi)H(uγi).

    We approximate H(ξ) by a C function Hϵ(ξ) satisfying

    Hϵ(ξ)=0forξ<ϵ,Hϵ(ξ)=1forξ>0,Hϵ(ξ)0for<ξ<.

    Consider the approximated free boundary problem

    Lϵ[uϵ]uϵtˆσ2ϵ22uϵx2(ˆσ2ϵ2+˙s(t))uϵx=0,0<x<,t>0, (4.1)

    with initial condition

    uϵ(0,x)=min{1,e(x+logK)} (4.2)

    and

    uϵ(t,0)=1, (4.3)

    where

    ˆσϵ=ˆσ1+n1i=1(ˆσi+1ˆσi)Hϵ(uϵγi).

    Problem (4.1)-(4.3) admits a unique classical solution uϵ. Now we proceed to derive some estimates for uϵ.

    Some uniform estimates are presented in this section, which are sufficient to obtain the existence and uniqueness of solution to problem (3.6)-(3.10).

    Lemma 4.1. Let uϵ be the solution of problem (4.1)-(4.3). Suppose that one of the conditions (3.11)-(3.13) holds. Then there holds

    0uϵmin{1,e(x+logK)},0<x<,t>0.

    Proof. It is easy to verify that 0 is the lower solution of uϵ and meanwhile, e(x+logK) and 1 are upper solutions. The result is a direct application of comparison principle.

    Lemma 4.2. Let uϵ be the solution of problem (4.1)-(4.3). Then there exists a constant C>0, independent of ϵ, such that

    Cuϵx0,0<x<,t>0.

    Proof. It is easy to see that ˆσ2ϵ can be written as

    ˆσ2ϵ=ˆσ21+n1i=1(ˆσ2i+1ˆσ2i)Hϵ(uϵγi).

    Differentiating (4.1) with respect to x gives

    Lϵ1[uϵx]Lϵ[uϵx]12n1i=1(ˆσ2i+1ˆσ2i)Hϵ(uϵγi)(2uϵx2+uϵx)uϵx=0.

    It is known that uϵx(0,x)=0 for 0<x<logK and uϵx(0,x)=e(x+logK)0 for x>logK. Since

    uϵ(t,x)uϵ(t,0)x0,

    then letting x0, it holds that uϵx(t,0)0. Thus it follows by maximum principle that there holds uϵx0.

    On the other hand, since ˙s(t) is uniformly bounded, take an appropriate value C>0, such that

    Lϵ[eCx]=(ˆσ2ϵ2C2+(ˆσ2ϵ2+˙s(t))C)eCx<0.

    Clearly, eCx|x=0=1 and eCxmin{1,e(x+logK)} for C sufficiently large. Then there hold uϵ(t,x)eCx and

    uϵ(t,x)uϵ(t,0)xeCx1x.

    Letting x0, we have uϵx(t,0)C. Clearly, there holds

    Lϵ1[C]=C22n1i=1(ˆσ2i+1ˆσ2i)Hϵ(uϵγi)0,

    as ˆσ2i+1>ˆσ2i for i=1,2,,n1. It follows by the comparison principle that there holds uϵxC.

    Lemma 4.3. Let uϵ be the solution of problem (4.1)-(4.3). Suppose that one of the conditions (3.11)-(3.13) holds. Then there exist constants C1, C2, C3 and C4, independent of ϵ, such that

    C3C2texp(C1t|x+logK|2)uϵtC4,0<x<,t>0.

    Proof. Differentiating (4.1) with respect to t gives

    Lϵ[uϵt]12ˆσ2ϵtuϵt+˙s(t)2ˆσ2ϵtuϵx¨s(t)uϵx=0,

    where

    ¨s(t)=2aβe2ata(β+r)eat.

    Since

    ˆσ2ϵt=h1(t)+h2(t)uϵt,

    where

    h1(t)=ˆσ21t+n1i=1(ˆσ2i+1tˆσ2it)Hϵ(uϵγi),

    and

    h2(t)=n1i=1(ˆσ2i+1ˆσ2i)Hϵ(uϵγi),

    we can write

    Lϵ[uϵt]=h2(t)2(uϵt)2+12(h1(t)˙s(t)h2(t)uϵx)uϵt+(¨s(t)˙s(t)h1(t)2)uϵx. (4.4)

    According to the formulas of ˆσ2i, i=1,2,,n, we have h1(t)ˆσ2nt. On the other hand, there exists ˜λϵi(t) such that Hϵ(uϵ(t,˜λϵi(t))γi) attains its maximum and

    h2(t)˜h2(t)max1in1(ˆσ2i+1(t)ˆσ2i(t))Hϵ(uϵ(t,˜λϵi(t))γi).

    Denote by y(t) the solution of the following ODE

    y(t)=˜h2(t)y2(t)+12(ˆσ2nt+C˙s(t)˜h2(t))y(t)+C(|¨s(t)|+˙s(t)2ˆσ2nt),y(0)=y0, (4.5)

    where the constant C is given as the one in Lemma 4.2. At x=logK, 2uϵx2(0,x) produces a Dirac measure of density 1. Thus uϵt(0,x)0 in the distribution sense. In addition, since the second order compatibility condition is satisfied at (0,0), we have uϵt is continuous at (0,0). Meanwhile, uϵt(0,t)=0. By further approximating the initial data with smooth function if necessary, there holds by the comparison principle that

    uϵt(t,x)y(t),0<x<,t>0,

    if we set y0=0. The ODE (4.5) can be solved formally as

    y(t)=q(t)exp(t0p(s)ds),

    where

    p(t)=˜h2(t)y(t)+12(ˆσ2nt+C˙s(t)˜h2(t)),

    and

    q(t)=Ct0(|¨s(r)|+˙s(r)2ˆσ2nr)exp(r0p(τ)dτ)dr.

    Since ˆσ2nt0 as t and Hϵ converges to the Dirac measure as ϵ0, this implies that exp(t0p(s)ds) is uniformly bounded with respect to t. With regard to q(t), in addition to ¨s(t)0 as t, it is known that q(t) is also uniformly bounded. Hence, we conclude that y(t) is uniformly bounded with respect to t.

    On the other hand, since uϵ(0,0)=1>γn1, and by Hölder continuity of solution, there exists a ρ>0, independent of ϵ, such that

    uϵ(t,x)>1+γn12

    for |x|ρ, 0tρ2. Thus for sufficiently small ϵ<12(1γn1), ˆσϵ=ˆσn for |x+logK|ρ, 0tρ2. It follows from the standard parabolic estimates [10] that

    uϵtC2C2texp(C1t|x+logK|2)

    for |x+logK|<ρ2, 0<tρ24. Note that (4.4) can be rewritten as

    Lϵ[uϵt]=h2(t)2(uϵt˙s(t)uϵx)uϵt+h1(t)2uϵt+(¨s(t)˙s(t)h1(t)2)uϵx.

    As ˙s(t) is uniformly bounded, we take a sufficiently large constant C3 such that C3supt0|˙s(t)|C, where constant C is the one given in Lemma 4.2. Then if uϵt<C3, there holds

    Lϵ[uϵt]12ˆσ2ntuϵt(|¨s(t)|+˙s(t)2ˆσ2nt)C.

    Denote by z(t) the solution of the following ODE

    z(t)=12ˆσ2ntz(t)(|¨s(t)|+˙s(t)2ˆσ2nt)C,z(0)=z0,

    which can be solved as

    z(t)=b(t)exp(12t0ˆσ2nsds),

    where

    b(t)=z0Ct0(|¨s(r)|+˙s(r)2ˆσ2nr)exp(12r0ˆσ2nτdτ)dr.

    As ˆσ2nt and ¨s(t) tends to 0 as t, z(t) is uniformly bounded. Moreover, z(t) is also decreasing if z00. Take the initial data |z0| and constant C3 sufficiently large such that

    C3supt0|z(t)|C2+C2texp(C1t|x+logK|2)

    on the boundary

    {|x+logK|=ρ2,0<t<ρ24}{|x+logK|<ρ2,t=ρ24}.

    We claim that the region

    {uϵt<C3}\{|x+logK|<ρ2,0<tρ24}

    is an empty set. If not, on the parabolic boundary of this region, we clearly have uϵtC3, which implies by the comparison principle that

    uϵtz(t)C3

    in this region. This is a contradiction.

    Remark 4.4. In the work [35], it was proved that uϵt0, which is different from the result shown in Lemma 4.3. However, although we get a similar result to Lemma 5.4 in [12], the proof is very different and more technical. This is due to the joint effect of stochastic interest rate and default boundary.

    Corollary 4.5. Let uϵ be the solution of problem (4.1)-(4.3). Suppose that one of the conditions (3.11)-(3.13) holds. Then there exist constants C1, C2, C3 and C4, independents of ϵ, such that

    C3C2texp(C1t|x+logK|2)2uϵx2C4,0<x<,t>t0.

    Denote by λϵi, i=1,2,,n1 the approximated free boundaries, which are the solutions of equations

    uϵ(t,λϵi(t))=γi,i=1,2,,n1. (4.6)

    Then we have the following estimates for the approximated free boundaries.

    Lemma 4.6. Let λϵi, i=1,2,,n1, be the approximated free boundaries defined in (4.6). Suppose that one of the conditions (3.11)-(3.13) holds. Then there exist constants C1, C2, independent of ϵ, such that

    C1λϵn1(t)λϵn2(t)λϵ2(t)λϵ1(t)C2.

    Proof. Since

    uϵ(t,λϵi(t))=γi<γi+1=uϵ(t,λϵi+1(t)),

    which implies that λϵi(t)λϵi+1(t) by Lemma 4.2. From Lemma 4.1, we have

    uϵ(t,x)e(x+logK),

    which implies that

    uϵ(t,x)<γ1forx>logγ1K.

    This means that region {x>logγ1K} is in the highest rating region and hence

    λϵ1(t)C2logγ1K.

    Denote by m=supt0˙s(t) and

    v(x)=1+γn12exp((1+2mσ21)x).

    Then v(0)=12(1+γn1)<1=uϵ(t,0). We can see that v(x)v(0)<1 and

    v(x)ex+logK=1+γn12exp(2mσ21x+logK)<1,

    which implies that

    v(x)<min{1,e(x+logK)}=uϵ(0,x).

    In addition, we have

    Lϵ[v]=1+γn12exp((1+2mσ21)x)(1+2mσ21)(˙s(t)mˆσ2ϵσ21)0.

    By the comparison principle, we have v(x)uϵ(t,x), which implies that

    uϵ(t,x)1+γn12exp((1+2mσ21)x)>γn1

    for

    x<C1σ21σ21+2mlog1+γn12γn1.

    This means that region {x<C1} is in the lowest rating region and hence λϵn1(t)C1.

    Lemma 4.7. Let λϵi, i=1,2,,n1, be the approximated free boundaries defined in (4.6). Suppose that one of the conditions (3.11)-(3.13) holds. Then there exists constant C independent of ϵ, such that

    CdλϵidtC,0<t<T,i=1,2,,n1.

    Proof. Clearly, there holds

    dλϵidt=uϵt(t,λϵi(t))/uϵx(t,λϵi(t)),i=1,2,,n1.

    Since λϵi(0)=logγilogK, i=1,2,,n1, by Lemma 4.3, there is a constant ρ>0 independent of ϵ such that

    λϵi(t)+logKρfor0tρ2,i=1,2,,n1.

    It follows from Lemma 4.3 that

    C0uϵt(t,λϵi(t))C0,i=1,2,,n1,

    where C0 is a constant independent of ϵ. To finish the proof, it is sufficient to prove that

    uϵx(t,λϵi(t))C

    for some positive constant C independent of ϵ. As shown in Lemma 4.2, we have

    Lϵ1[uϵx]=Lϵ[uϵx]+12n1i=1(ˆσ2i+1ˆσ2i)Hϵ(uϵγi)(2uϵx2+uϵx)uϵx=0.

    In addition, there also holds uϵx(0,x)=0 for 0<x<logK and uϵx(0,x)=e(x+logK) for x>logK, and uϵx(t,0)0 for t>0. By Lemmas 4.3 and 4.6, there exists constant R>0 independent of ϵ, such that

    2Rλϵi(t)R1for0<tT,i=1,2,,n1,

    and

    λϵi(t)+logKρfor0tρ2,i=1,2,,n1.

    Consider the region

    Ω={ρ2logK<x<R,0<t<ρ2}{1RxR,ρ2tT}. (4.7)

    The parabolic boundary of this region Ω consists of five line segments. On the initial line segment {t=0,ρ2logKxR}, there holds that uϵx(0,x)=e(x+logK). The remaining four parabolic boundaries {0tT,x=R}{0tρ2,x=ρ2logK}{t=ρ2,1Rxρ2logK}{ρ2tT,x=1R} are completely and uniformly within the highest or lowest rating region (independent of ϵ). Thus by compactness and the strong maximum principle, on these four boundaries, it holds that uϵxˉC>0 for some ˉC independent of ϵ. It follows that

    uϵxmin{1,ˉC}ConΩ, (4.8)

    which completes the proof of the lemma.

    Lemmas 4.1-4.3 and Corollary 4.5 provide uniform estimates for approximated solution uϵ. By taking a limit ϵ0 (along a subsequence if necessary), we derive the existence of solution to problem (3.6)-(3.10). Lemmas 4.6-4.7 show that there are uniform estimates in C1([0,T]) for the approximated free boundaries λϵi, i=1,2,,n1. Therefore, the limits of λϵi as ϵ0 exist, which are denoted by λi, i=1,2,,n1. These λi, i=1,2,,n1, are the free boundaries of the original problem.

    Theorem 5.1. The free boundary problem (3.6)-(3.10) admits a solution (u,λi,i=1,2,,n1) with

    uW1,2([0,T]×(0,)¯Qt0)W0,1([0,T]×(0,))

    for any t0>0, where

    Qt0=(0,t20)×(t0logK,t0logK)

    and λiW1([0,T]), i=1,2,,n1.

    By the classical parabolic theory, it is also clear that the solution is in ni=1C(Ωi), where

    Ω1={(t,x):x>λ1(t),0<tT},

    and for i=2,3,,n1

    Ωi={(t,x):λi(t)<x<λi1(t),0<tT},

    and

    Ωn={(t,x):x<λn1(t),0<tT}.

    Now we prove the uniqueness of solution to the problem (3.6)-(3.10).

    Theorem 5.2. The solution (u,λi,i=1,2,,n1) of the problem (3.6)-(3.10) with

    uW1,2([0,T]×(0,)¯Qt0)W0,1([0,T]×(0,))

    and λiC([0,T]) is unique.

    Proof. Suppose that (u,λi,i=1,2,,n1) and (˜u,˜λi,i=1,2,,n1) are two solutions of the problem (3.6)-(3.10). Then u(t,λi(t))=˜u(t,˜λi(t))=γi, i=1,2,,n1 and

    u(t,λi(t))˜u(t,λi(t))=˜u(t,˜λi(t))˜u(t,λi(t)),i=1,2,,n1.

    Besides, at t=0, λi(0)=˜λi(0)=logKlogγi, i=1,2,,n1. As shown in Lemma 4.7, there exists a constant C>0, such that ux<C and ˜ux<C on the region Ω defined by (4.7). Then by the implicit function theorem, there exists ρ>0, such that when 0<t<ρ,

    |λi(t)˜λi(t)|Cmax0<x<|u(t,x)˜u(t,x)|,i=1,2,,n1, (5.1)

    where C is a positive constant, whose value may change line on line but makes no difference. Let w=u˜u and denote by ˆσ and ˜ˆσ the corresponding coefficients, then w satisfies

    1ˆσ2wt122wx2(˙s(t)ˆσ2+12)wx=(1˜ˆσ21ˆσ2)(˜ut˙s(t)˜ux). (5.2)

    u, ˜u and their derivatives decay exponentially fast to 0 as x. Multiplying (5.2) by w on both sides and integrating x from 0 to gives

    0(wˆσ2wtw22wx2(˙s(t)ˆσ2+12)wwx)dx=0(1˜ˆσ21ˆσ2)(˜ut˙s(t)˜ux)wdx. (5.3)

    Since

    1˜ˆσ21ˆσ20forxn1i=1[λi(t)˜λi(t),λi(t)˜λi(t)],

    and ˜ut and ˜ux are uniformly bounded outside the region ¯Qt0, we conclude that they are bounded for xn1i=1[λi(t)˜λi(t),λi(t)˜λi(t)]. Since w decays exponentially to 0 as x, for any t>0, there exists x0< such that

    max0<x<w2(t,x)=w2(t,x0).

    Take

    ¯w=1ϵx0+ϵx0w(t,x)dx=w(t,x),

    for some x(x0,x0+ϵ). Then there holds

    max0<x<|w(t,x)|22|w(t,x0)¯w|2+2|¯w|2=2(xx0wxdx)2+2ϵ2(x0+ϵx0wdx)22ϵx0+ϵx0(wx)2dx+2ϵx0+ϵx0w2dx2ϵ0(wx)2dx+2ϵ0w2dx. (5.4)

    It follows that

    0(1˜ˆσ21ˆσ2)(˜ut˙s(t)˜ux)wdxCmax0<x<|w(t,x)|n1i=1λi(t)˜λi(t)λi(t)˜λi(t)|1˜ˆσ21ˆσ2|dxCmax0<x<|w(t,x)|n1i=1|λi(t)˜λi(t)|Cmax0<x<|w(t,x)|2(by(5.1))Cϵ0w2dx+ϵ0(wx)2dx. (5.5)

    We now proceed to estimate the left side of (5.3). First, we have

    +0wˆσ2wtdx=λn10wˆσ2nwtdx+n2i=1λiλi+1wˆσ2i+1wtdx+λ1wˆσ21wtdx=g1(t)+g2(t)+n1i=1(1ˆσ2i1ˆσ2i+1)λi(t)w2(t,λi(t))2g1(t)+n1i=1(1ˆσ2i1ˆσ2i+1)λi(t)w2(t,λi(t))2, (5.6)

    where

    g1(t)=λn10w22ˆσ2ndx+n2i=1λiλi+1w22ˆσ2i+1dx+λ1w22ˆσ21dx,

    and

    g2(t)=λn10w22ˆσ4nˆσ2ntdx+n2i=1λiλi+1w22ˆσ4i+1ˆσ2i+1tdx+λ1w22ˆσ41ˆσ21tdx.

    Second, we have

    0w22wx2dx=120(wx)2dx (5.7)

    and

    0(˙s(t)ˆσ2+12)wwxdxϵ0(wx)2dx+Cϵ0w2dx, (5.8)

    as ˙s(t) is uniformly bounded. Combining the above inequalities (5.5)-(5.8), taking into account (5.3), we drive

    g1(t)ϵ0(wx)2dx+Cϵ0w2dx+n1i=1(1ˆσ2i+11ˆσ2i)λi(t)w2(t,λi(t))2ϵ0(wx)2dx+Cϵ0w2dx+Cmax0x|w(t,x)|2Cϵ0(wx)2dx+Cϵ0w2dx.

    It is easy to see that there exists a constant C0>0 such that

    C00w2dxg1(t)Cϵt00(wx)2dxds+Cϵt00w2dxds.

    Then for sufficiently small ϵ, by applying the Gronwall's inequality, we conclude that w0. This proves the uniqueness for 0tρ. A close examination of the proof indicates that the uniqueness result can be extended to any time interval, where ux is strictly negative, which is already verified in (4.8).

    Denote by ψ the solution of the following static problem

    ˉˆσ222ψx2+(ˉˆσ22+θσ2r2a2)ψx=0, (6.1)

    with boundary conditions

    ψ(0)=1,limxψ(x)=0, (6.2)
    ψ(λi)=γi,ψx(λi+)=ψx(λi),i=1,2,,n1, (6.3)

    where ˉˆσ=ˉˆσ1 as ψ<γ1, ˉˆσ=ˉˆσi as γi1<ψ<γi for i=1,2,,n1, ˉˆσ=ˉˆσn as ψ>γn1, and ˉˆσi, i=1,2,,n, are given in (3.3). We suppose that in the i'th rating region, ψ admits the following form

    ψ(x)=pi+qiexp(kix),i=1,2,,n, (6.4)

    where pi, qi and ki are undetermined constants. Substituting (6.4) into (6.1) in the corresponding rating region gives

    ki=σ2ra2ˉˆσ2i2θˉˆσ2i1,i=1,2,,n.

    As it is supposed that one of the conditions (3.11)-(3.13) holds, then ˙s(t)0, which implies that ki<0, i=1,2,,n. Substituting (6.4) into the boundary condition (6.3) gives

    pi+qiexp(kiλi)=γi, (6.5)
    pi+1+qi+1exp(ki+1λi)=γi, (6.6)

    and

    qikiexp(kiλi)=qi+1ki+1exp(ki+1λi) (6.7)

    for i=1,2,,n1. Also, substituting (6.4) into the boundary condition (6.2) gives

    p1=0,pn+qn=1. (6.8)

    It is easy to see that coefficient system (6.5)-(6.8) can be equivalently rewritten as

    logqi+kiλi=log(γipi), (6.9)
    logqi+1+ki+1λi=log(γipi+1), (6.10)
    logqi+logki+kiλi=logqi+1+logki+1+ki+1λi (6.11)

    for i=1,2,,n1, and

    log(γ1p1)=logγ1,logqn=log(1pn). (6.12)

    For i=1,2,,n2, from the equations

    pi+1+qi+1exp(ki+1λi)=γi,pi+1+qi+1exp(ki+1λi+1)=γi+1,

    we can derive their relationship as

    \begin{equation*} \log(\gamma_{i}-p_{i+1}) = \log(\gamma_{i+1}-p_{i+1})+k_{i+1}\lambda_{i}^{*}-k_{i+1}\lambda_{i+1}^{*}. \end{equation*}

    Thus, denote by x_{i} = \log q_{i} for i = 1, 2, \cdots, n , y_{i} = \log(\gamma_{i}-p_{i}) for i = 1, 2, \cdots, n-1 and y_{n} = \log(1-p_{n}) , z_{i} = \log(\gamma_{i}-p_{i+1}) , i = 1, 2, \cdots, n-1 . Then (6.9)-(6.12) can be rewritten as

    \begin{equation} x_{i}-y_{i}+k_{i}\lambda_{i}^{*} = 0,\; i = 1,2,\cdots,n-1, \end{equation} (6.13)
    \begin{equation} x_{i+1}-z_{i}+k_{i+1}\lambda_{i}^{*} = 0,\; i = 1,2,\cdots,n-1, \end{equation} (6.14)
    \begin{equation} x_{i}-x_{i+1}+(k_{i}-k_{i+1})\lambda_{i}^{*}+\log k_{i}-\log k_{i+1} = 0,\; i = 1,2,\cdots,n-1, \end{equation} (6.15)
    \begin{equation} z_{i}-y_{i+1}-k_{i+1}\lambda_{i}^{*}+k_{i+1}\lambda_{i+1}^{*} = 0,\; i = 1,2,\cdots,n-1, \end{equation} (6.16)
    \begin{equation} \lambda_{n}^{*} = 0,\; y_{1} = \log\gamma_{1},\; x_{n} = y_{n}, \end{equation} (6.17)

    where a virtual parameter \lambda_{n}^{*} is added. (6.13)-(6.17) is a linearized system from (6.5)-(6.8) and can be solved according to fundamental linear algebra theory. Thus, we obtain the explicit solution of the static problem (6.1)-(6.3).

    The coefficients in the model of [35] are time-homogeneous and moreover, the solution is decreasing in time. However, both of these are not the cases in our model. We cannot take advantage of the decreasing property of solution to obtain the convergence. In this paper, we are motivated by the idea of [12] and obtain the convergence by two steps. The first step is to show that the solution of problem (3.6)-(3.10) converges to the solution of some auxiliary problem defined below, while in the second step, we show that the solution of auxiliary problem converges to the solution of static problem (6.1)-(6.3). Now define an auxiliary free boundary problem as follows

    \begin{equation} \frac{\partial \bar{u}}{\partial t}-\frac{\bar{\hat{\sigma}}^{2}}{2}\frac{\partial^{2}\bar{u}}{\partial x^{2}}-\bigg(\frac{\bar{\hat{\sigma}}^{2}}{2}+\theta-\frac{\sigma_{r}^{2}}{2a^{2}}\bigg)\frac{\partial \bar{u}}{\partial x} = 0,\; 0 \lt x \lt \infty,\; t \gt 0, \end{equation} (6.18)

    with initial condition \bar{u}_{0} and boundary condition \bar{u}(t, 0) = 1 ,

    \begin{equation} \bar{u}(t,\bar{\lambda}_{i}(t)-) = \bar{u}(t,\bar{\lambda}_{i}(t)+) = \gamma_{i},\; \frac{\partial \bar{u}}{\partial x}(t,\bar{\lambda}_{i}(t)-) = \frac{\partial \bar{u}}{\partial x}(t,\bar{\lambda}_{i}(t)+),\; i = 1,2,\cdots,n-1. \end{equation} (6.19)

    All the results derived above involving the existence, uniqueness and some properties of solution presented in Lemmas 4.1-4.6 hold for solution of problem (6.18)-(6.19). We have to notice that as the presence of default boundary, although we follow the idea of [12], the technical proofs are different, especially in the step, i.e., the convergence from the original solution to the auxiliary solution.

    Since \hat{\sigma}_{i}(t)\rightarrow\bar{\hat{\sigma}}_{i} , i = 1, 2, \cdots, n , and \dot{s}(t)\rightarrow\theta-\frac{\sigma_{r}^{2}}{2a^{2}} as t\rightarrow\infty , then for any \epsilon > 0 , there exists a T > 0 such that for t\geq T ,

    \begin{equation} \frac{1}{\bar{\hat{\sigma}}_{i}^{2}}-\epsilon\leq\frac{1}{\hat{\sigma}_{i}^{2}(t)}\leq\frac{1}{\bar{\hat{\sigma}}_{i}^{2}}+\epsilon,\; \theta-\frac{\sigma^{2}_{r}}{2a^{2}}-\epsilon\leq\dot{s}(t)\leq\theta-\frac{\sigma^{2}_{r}}{2a^{2}}+\epsilon. \end{equation} (6.20)

    Let \bar{u}_{0}(x) = u(T, x) and denote by u_{T}(t, x) = u(t+T, x) for t\geq0 . We have u_{T}(t, \lambda_{i}(t)) = \bar{u}(t, \bar{\lambda}_{i}(t)) = \gamma_{i} , i = 1, 2, \cdots, n-1 , and

    \begin{equation*} u_{T}(t,\lambda_{i}^{T}(t))-\bar{u}(t,\lambda_{i}^{T}(t)) = \bar{u}(t,\bar{\lambda}_{i}(t))-\bar{u}(t,\lambda_{i}^{T}(t)),\; i = 1,2,\cdots,n-1, \end{equation*}

    where \lambda^{T}_{i}(t) = \lambda_{i}(t+T) , i = 1, 2, \cdots, n-1 . Similarly to the proof of Theorem 5.2, by the implicit function theorem, there exists a \rho > 0 , such that when 0 < t < \rho ,

    \begin{equation} |\bar{\lambda}_{i}(t)-\lambda_{i}^{T}(t)|\leq C\max\limits_{0\leq x\leq\infty}|u_{T}(t,x)-\bar{u}(t,x)|,\; i = 1,2,\cdots,n-1, \end{equation} (6.21)

    where C is a positive constant, whose value may change line on line but makes no difference. Let w = u_{T}-\bar{u} . Then w satisfies

    \begin{equation} \frac{1}{\hat{\sigma}_{T}^{2}}\frac{\partial w}{\partial t} = \frac{1}{2}\frac{\partial^{2}w}{\partial x^{2}}+\bigg(\frac{1}{2}+\frac{\dot{s}_{T}(t)}{\hat{\sigma}_{T}^{2}}\bigg)\frac{\partial w}{\partial x}+h_{1}+h_{2}, \end{equation} (6.22)

    where \dot{s}_{T}(t) = \dot{s}(t+T) ,

    \begin{equation*} h_{1} = \bigg(\frac{1}{\bar{\hat{\sigma}}^{2}}-\frac{1}{\hat{\sigma}^{2}_{T}}\bigg)\bigg(\frac{\partial\bar{u}}{\partial t}-\bigg(\theta-\frac{\sigma_{r}^{2}}{2a^{2}}\bigg)\frac{\partial\bar{u}}{\partial x}\bigg), \end{equation*}

    and

    \begin{equation*} h_{2} = \bigg(\dot{s}_{T}(t)-\theta+\frac{\sigma_{r}^{2}}{2a^{2}}\bigg)\frac{\partial\bar{u}}{\partial x}. \end{equation*}

    As u , \bar{u} and their derivatives decay exponentially fast to 0 as x\rightarrow\infty , multiplying (6.22) by w on both sides and integrating x from 0 to \infty gives

    \begin{equation} \int_{0}^{\infty}\frac{w}{\hat{\sigma}_{T}^{2}}\frac{\partial w}{\partial t}dx = \int_{0}^{\infty}\frac{w}{2}\frac{\partial^{2}w}{\partial x^{2}}dx+\int_{0}^{\infty}\bigg(\frac{1}{2}+\frac{\dot{s}_{T}(t)}{\hat{\sigma}_{T}^{2}}\bigg)\frac{\partial w}{\partial x}wdx+\int_{0}^{\infty}h_{1}wdx+\int_{0}^{\infty}h_{2}wdx. \end{equation} (6.23)

    By Lemmas 4.2 and 4.3, we know that \frac{\partial\bar{u}}{\partial t} and \frac{\partial\bar{u}}{\partial x} are uniformly bounded as the initial data is set to be \bar{u}_{0}(x) = u(T, x) for a sufficiently large T . It follows that

    \begin{equation*} \begin{split} \int_{0}^{\infty}h_{1}wdx\leq&\max\limits_{0\leq x\leq\infty}|w(t,x)|\sum\limits_{i = 1}^{n-1}\int_{\lambda_{i}^{T}\wedge\bar{\lambda}_{i}}^{\lambda_{i}^{T}\vee\bar{\lambda}_{i}}\bigg|\frac{1}{\bar{\hat{\sigma}}^{2}}-\frac{1}{\hat{\sigma}_{T}^{2}}\bigg|dx+h_{3}\\ \leq&\max\limits_{0\leq x\leq\infty}|w(t,x)|\sum\limits_{i = 1}^{n-1}|\lambda_{i}^{T}(t)-\bar{\lambda}_{i}(t)|+h_{3}\; (\text{by}\; (6.21))\\ \leq&\max\limits_{0\leq x\leq\infty}|w(t,x)|^{2}+h_{3}, \end{split} \end{equation*}

    where

    \begin{equation*} \begin{split} h_{3} = \bigg(\int_{0}^{\lambda_{n-1}^{T}\wedge\bar{\lambda}_{n-1}}+\sum\limits_{i = 1}^{n-2}\int_{\lambda_{i+1}^{T}\vee\bar{\lambda}_{i+1}}^{\lambda_{i}^{T}\wedge\bar{\lambda}_{i}}+\int_{\lambda_{1}^{T}\vee\bar{\lambda}_{1}}^{\infty}\bigg)\bigg|\frac{1}{\bar{\hat{\sigma}}^{2}}-\frac{1}{\hat{\sigma}_{T}^{2}}\bigg|wdx\leq\epsilon\int_{0}^{\infty}wdx\leq C\epsilon, \end{split} \end{equation*}

    hold by (6.20) and the exponential decay of w . Using (5.4), there holds

    \begin{equation} \int_{0}^{\infty}h_{1}wdx\leq\frac{C}{\epsilon}\int_{0}^{\infty}w^{2}dx+\epsilon\int_{0}^{\infty}\bigg(\frac{\partial w}{\partial x}\bigg)^{2}dx+\epsilon C. \end{equation} (6.24)

    On the other hand, there holds

    \begin{equation} \int_{0}^{\infty}h_{2}wdx\leq\epsilon\int_{0}^{\infty}\bigg|\frac{\partial\bar{u}}{\partial x}\bigg|wdx\leq C\epsilon\int_{0}^{\infty}wdx\leq C\epsilon. \end{equation} (6.25)

    We now proceed to estimate the left side of (6.23) by

    \begin{equation} \begin{split} \int_{0}^{+\infty}\frac{w}{\hat{\sigma}_{T}^{2}}\frac{\partial w}{\partial t}dx = &\int_{0}^{\lambda^{T}_{n-1}}\frac{w}{(\hat{\sigma}_{T})_{n}^{2}}\frac{\partial w}{\partial t}dx+\sum\limits_{i = 1}^{n-2}\int^{\lambda^{T}_{i}}_{\lambda^{T}_{i+1}}\frac{w}{(\hat{\sigma}_{T})_{i+1}^{2}}\frac{\partial w}{\partial t}dx+\int_{\lambda^{T}_{1}}^{\infty}\frac{w}{(\hat{\sigma}_{T})_{1}^{2}}\frac{\partial w}{\partial t}dx\\ = &g'_{1}(t)+g_{2}(t)+\sum\limits_{i = 1}^{n-1}\bigg(\frac{1}{(\hat{\sigma}_{T})_{i}^{2}}-\frac{1}{(\hat{\sigma}_{T})_{i+1}^{2}}\bigg)\frac{1}{2}\frac{d\lambda^{T}_{i}}{dt}w^{2}(t,\lambda^{T}_{i}(t))\\ \geq&g'_{1}(t)+\sum\limits_{i = 1}^{n-1}\bigg(\frac{1}{(\hat{\sigma}_{T})_{i}^{2}}-\frac{1}{(\hat{\sigma}_{T})_{i+1}^{2}}\bigg)\frac{1}{2}\frac{d\lambda^{T}_{i}}{dt}w^{2}(t,\lambda^{T}_{i}(t)), \end{split} \end{equation} (6.26)

    where

    \begin{equation*} g_{1}(t) = \int_{0}^{\lambda^{T}_{n-1}}\frac{w^{2}}{2(\hat{\sigma}_{T})_{n}^{2}}dx+\sum\limits_{i = 1}^{n-2}\int^{\lambda^{T}_{i}}_{\lambda^{T}_{i+1}}\frac{w^{2}}{2(\hat{\sigma}_{T})_{i+1}^{2}}dx+\int_{\lambda^{T}_{1}}^{\infty}\frac{w^{2}}{2(\hat{\sigma}_{T})_{1}^{2}}dx, \end{equation*}

    and

    \begin{equation*} g_{2}(t) = \int_{0}^{\lambda^{T}_{n-1}}\frac{w^{2}}{2(\hat{\sigma}_{T})_{n}^{4}}\frac{\partial(\hat{\sigma}_{T})_{n}^{2}}{\partial t}dx+\sum\limits_{i = 1}^{n-2}\int_{\lambda^{T}_{i+1}}^{\lambda^{T}_{i}}\frac{w^{2}}{2(\hat{\sigma}_{T})_{i+1}^{4}}\frac{\partial(\hat{\sigma}_{T})_{i+1}^{2}}{\partial t}dx+\int_{\lambda^{T}_{1}}^{\infty}\frac{w^{2}}{2(\hat{\sigma}_{T})_{1}^{4}}\frac{\partial(\hat{\sigma}_{T})_{1}^{2}}{\partial t}dx. \end{equation*}

    The remaining terms in (6.23) can be estimated by

    \begin{equation} \int_{0}^{\infty}\frac{w}{2}\frac{\partial^{2}w}{\partial x^{2}}dx = -\frac{1}{2}\int_{0}^{\infty}\bigg(\frac{\partial w}{\partial x}\bigg)^{2}dx \end{equation} (6.27)

    and

    \begin{equation} \int_{0}^{\infty}\bigg(\frac{\dot{s}_{T}}{\hat{\sigma}_{T}^{2}}+\frac{1}{2}\bigg)w\frac{\partial w}{\partial x}dx\leq\epsilon\int_{0}^{\infty}\bigg(\frac{\partial w}{\partial x}\bigg)^{2}dx+\frac{C}{\epsilon}\int_{0}^{\infty}w^{2}dx. \end{equation} (6.28)

    Combining the inequalities (6.24)-(6.28), we have

    \begin{equation*} \begin{split} g'_{1}(t)\leq&\epsilon\int_{0}^{\infty}\bigg(\frac{\partial w}{\partial x}\bigg)^{2}dx+\frac{C}{\epsilon}\int_{0}^{\infty}w^{2}dx+\sum\limits_{i = 1}^{n-1}\bigg(\frac{1}{(\hat{\sigma}_{T})_{i+1}^{2}}-\frac{1}{(\hat{\sigma}_{T})_{i}^{2}}\bigg)\frac{\partial\lambda^{T}_{i}}{\partial t}\frac{w^{2}(t,\lambda^{T}_{i}(t))}{2}+C\epsilon\\ \leq&\epsilon\int_{0}^{\infty}\bigg(\frac{\partial w}{\partial x}\bigg)^{2}dx+\frac{C}{\epsilon}\int_{0}^{\infty}w^{2}dx+C\max\limits_{0\leq x\leq\infty}|w(t,x)|^{2}+C\epsilon\\ \leq&C\epsilon\int_{0}^{\infty}\bigg(\frac{\partial w}{\partial x}\bigg)^{2}dx+\frac{C}{\epsilon}\int_{0}^{\infty}w^{2}dx+C\epsilon. \end{split} \end{equation*}

    It follows that

    \begin{equation*} C_{0}\int_{0}^{\infty}w^{2}dx\leq g_{1}(t)\leq C\epsilon\int_{0}^{t}\int_{0}^{\infty}\bigg(\frac{\partial w}{\partial x}\bigg)^{2}dxds+\frac{C}{\epsilon}\int_{0}^{t}\int_{0}^{\infty}w^{2}dxds+C\epsilon t. \end{equation*}

    Then taking a sufficiently small \epsilon , which means a sufficiently large T , by applying the Gronwall's inequality, we conclude that w = 0 , namely that u_{T}(t, x) = \bar{u}(t, x) for 0\leq t\leq\rho . The result can be extended to any interval of t .

    The convergence from the solution of auxiliary problem (6.18)-(6.19) to the solution of problem (6.1)-(6.3) is proved by a Lyapunov argument, which is similar to the procedure in [12,21,32], but with some necessary modifications to fit the model in this paper. For instance, we have extend the solution with domain [0, \infty) to the whole real line. The first step is to present the formal construction of a Lyapunov function, ignoring the integrability of any arisen integral. The second step is to verify the integrability of those integrals arisen in the formal construction. The third step is to complete the proof of convergence.

    Denote by \bar{U} the extension of \bar{u} , who is the solution of auxiliary problem (6.18)-(6.19), from x\in[0, \infty) to the real line \mathbb{R} , namely that

    \begin{equation*} \bar{U}(t,x) = 1\; \text{for}\; x \lt 0,\; \bar{U}(t,x) = \bar{u}(t,x)\; \text{for}\; x\geq0. \end{equation*}

    Following [12], let V(x, u, q) be a undetermined function and set

    \begin{equation*} E[\bar{U}](t) = \int_{-\infty}^{\infty}V(x,\bar{U}(t,x),\bar{U}_{x}(t,x))dx. \end{equation*}

    Formally, assuming the integrability, we also have

    \begin{equation*} \begin{split} \frac{d}{dt}E[\bar{U}] = &\int_{-\infty}^{\infty}(V_{u}\bar{U}_{t}+V_{q}\bar{U}_{xt})dx\\ = &\int_{-\infty}^{\infty}\bar{U}_{t}(V_{u}-V_{qx}-V_{qu}\bar{U}_{x}-V_{qq}\bar{U}_{xx})dx\\ = &\int_{-\infty}^{\infty}\bar{U}_{t}(V_{u}-V_{qx}-V_{qu}\bar{U}_{x}-V_{qq}\bigg(\frac{2}{\bar{\hat{\sigma}}^{2}}\bar{U}_{t}-\frac{2}{\bar{\hat{\sigma}}^{2}}\bigg(\frac{\bar{\hat{\sigma}}^{2}}{2}+\delta\bigg)\bar{U}_{x}\bigg)\bigg)dx\\ = &-\int_{-\infty}^{\infty}\frac{2}{\bar{\hat{\sigma}}^{2}}V_{qq}\bar{U}_{t}^{2}dx+\int_{-\infty}^{\infty}\bar{U}_{t}(V_{u}-V_{qx}-V_{qu}\bar{U}_{x}+V_{qq}\bigg(1+\frac{2\delta}{\bar{\hat{\sigma}}^{2}}\bigg)\bar{U}_{x}\bigg)dx\\ = &-\int_{-\infty}^{\infty}\frac{2}{\bar{\hat{\sigma}}^{2}}V_{qq}\bar{U}_{t}^{2}dx, \end{split} \end{equation*}

    where \delta = \theta-\frac{\sigma_{r}^{2}}{2a^{2}} , provided taking V satisfying

    \begin{equation} V_{u}-V_{qx}-qV_{qu}+qV_{qq}\bigg(1+\frac{2\delta}{\bar{\hat{\sigma}}^{2}(u)}\bigg) = 0. \end{equation} (6.29)

    Denote by \rho = V_{qq} . Suppose that V(x, u, 0) = V_{q}(x, u, 0) = 0 as in [12]. Then we have

    \begin{equation*} \int_{0}^{q}(q-m)\rho(x,u,m)dm = \int_{0}^{q}(q-m)dV_{q}(x,u,m) = \int_{0}^{q}V_{q}(x,u,m)dm = V(x,u,q). \end{equation*}

    It follows that

    \begin{equation*} V_{u}(x,u,q) = \int_{0}^{q}(q-m)\rho_{u}(x,u,m)dm, \end{equation*}
    \begin{equation*} V_{q}(x,u,q) = \int_{0}^{q}\rho(x,u,m)dm, \end{equation*}
    \begin{equation*} V_{qx}(x,u,q) = \int_{0}^{q}\rho_{x}(x,u,m)dm, \end{equation*}
    \begin{equation*} V_{qu}(x,u,q) = \int_{0}^{q}\rho_{u}(x,u,m)dm, \end{equation*}

    and

    \begin{equation*} qV_{qq} = q\rho = \int_{0}^{q}\frac{d}{dm}(\rho(x,u,m)m)dm = \int_{0}^{q}(\rho(x,u,m)+\rho_{q}(x,u,m))dm. \end{equation*}

    Then (6.29) can be written as

    \begin{equation*} \begin{split} &\int_{0}^{q}(q\rho_{u}(x,u,m)-m\rho_{u}(x,u,m)-\rho_{x}(x,u,m)-q\rho_{u}(x,u,m))dm\\ &+\int_{0}^{q}\bigg(1+\frac{2\delta}{\bar{\hat{\sigma}}^{2}(u)}\bigg)(\rho(x,u,m)+\rho_{q}(x,u,m)m)dm = 0. \end{split} \end{equation*}

    To ensure (6.29) holds, it should be

    \begin{equation} m\rho_{u}(x,u,m)+\rho_{x}(x,u,m)-\bigg(1+\frac{2\delta}{\bar{\hat{\sigma}}^{2}(u)}\bigg)(\rho(x,u,m)+\rho_{q}(x,u,m)m) = 0. \end{equation} (6.30)

    Formally, denote by v the solution of the following equation

    \begin{equation} -v_{xx}-\bigg(1+\frac{2\delta}{\bar{\hat{\sigma}}^{2}(v)}\bigg)v_{x} = 0, \end{equation} (6.31)

    with boundary conditions

    \begin{equation*} v(x_{0}) = u_{0},\; v_{x}(x_{0}) = q_{0}. \end{equation*}

    Then by (6.30) and (6.31), there holds

    \begin{equation*} \frac{d}{dx}\rho = \rho_{x}+\rho_{u}v_{x}+\rho_{q}v_{xx} = \rho_{x}+\rho_{u}v_{x}-\rho_{q}\bigg(1+\frac{2\delta}{\bar{\hat{\sigma}}^{2}(v)}\bigg)v_{x} = \bigg(1+\frac{2\delta}{\bar{\hat{\sigma}}^{2}(v)}\bigg)\rho, \end{equation*}

    which can be solved as

    \begin{equation*} \rho(x_{0},u_{0},q_{0}) = \exp\bigg(\int_{0}^{x_{0}}\bigg(1+\frac{2\delta}{\bar{\hat{\sigma}}^{2}(v(z))}\bigg)dz\bigg). \end{equation*}

    Replacing x_{0} by x , u_{0} by u and q_{0} by q , we have

    \begin{equation} \rho(x,u,q) = \exp\bigg(\int_{0}^{x}\bigg(1+\frac{2\delta}{\bar{\hat{\sigma}}^{2}(v(z))}\bigg)dz\bigg). \end{equation} (6.32)

    Integrating the Lyapunov function and assuming E(t)\geq0 , we have

    \begin{equation*} \int_{t_{0}}^{t}\int_{-\infty}^{\infty}\frac{2}{\bar{\hat{\sigma}}^{2}}\rho(x,\bar{U}(s,x),\bar{U}_{x}(s,x))\bar{U}_{s}^{2}(s,x)dxds = E(t_{0})-E(t)\leq E(t_{0}). \end{equation*}

    Following the formal construction, we proceed to the second step, namely that verify the integrability of those integrals arisen in the formal construction of the Lyapunov function. As indicated by Liang et al. [21], there are two problems, where the first one is that \rho grows exponentially as x\rightarrow\pm\infty , while the second one is that the coefficient in (6.31) is discontinuous and the theory of ODE cannot be applied directly. These two problems will be overcame through the approximated solution of problem (4.1)-(4.3) with all the uniform estimates. We begin this step by defining

    \begin{equation*} E_{R}[\bar{U}_{\epsilon}](t) = \int_{-R}^{R}V_{\epsilon}\bigg(x,\bar{U}_{\epsilon}(t,x),\frac{\partial\bar{U}_{\epsilon}}{\partial x}(t,x)\bigg)dx, \end{equation*}

    for R > 0 , where \bar{U}_{\epsilon} is the approximation of \bar{U} with the coefficient of the approximated problem given as

    \begin{equation*} \bar{\hat{\sigma}}_{\epsilon} = \bar{\hat{\sigma}}_{1}+\sum\limits_{i = 1}^{n-1}(\bar{\hat{\sigma}}_{i+1}-\bar{\hat{\sigma}}_{i})H_{\epsilon}(\bar{U}_{\epsilon}-\gamma_{i}), \end{equation*}

    and V_{\epsilon} satisfies

    \begin{equation*} V_{\epsilon}(x,u,q) = \int_{0}^{q}(q-m)\rho_{\epsilon}(x,u,m)dm,\; V_{\epsilon}(x,u,0) = \frac{\partial V_{\epsilon}}{\partial q}(x,u,0) = 0, \end{equation*}

    and \rho_{\epsilon} is defined by (6.32) with \bar{\hat{\sigma}}^{2} replaced by \bar{\hat{\sigma}}^{2}_{\epsilon} . Thus, (6.31) can be solved on the real line x\in\mathbb{R} and V_{\epsilon} is well defined. Meanwhile, it also satisfies

    \begin{equation*} \frac{\partial V_{\epsilon}}{\partial u}-\frac{\partial^{2}V_{\epsilon}}{\partial q\partial x}-q\frac{\partial^{2}V_{\epsilon}}{\partial q\partial u}+q\frac{\partial^{2}V_{\epsilon}}{\partial q^{2}}\bigg(1+\frac{2\delta}{\bar{\hat{\sigma}}_{\epsilon}^{2}(u)}\bigg) = 0. \end{equation*}

    Then we have

    \begin{equation*} \frac{d}{dt}E_{R}[\bar{U}_{\epsilon}] = \int_{-R}^{R}\bigg(\frac{\partial V_{\epsilon}}{\partial u}\frac{\partial\bar{U}_{\epsilon}}{\partial t}+\frac{\partial V_{\epsilon}}{\partial q}\frac{\partial^{2}\bar{U}_{\epsilon}}{\partial x\partial t}\bigg)dx = \frac{\partial V_{\epsilon}}{\partial q}\frac{\partial\bar{U}_{\epsilon}}{\partial t}\bigg|_{-R}^{R}-\int_{-R}^{R}\frac{2\rho_{\epsilon}}{\bar{\hat{\sigma}}_{\epsilon}^{2}}\bigg(\frac{\partial\bar{U}_{\epsilon}}{\partial t}\bigg)^{2}dx. \end{equation*}

    Lemma 6.1. Let \bar{u}_{\epsilon} be the approximated solution of problem (6.18)-(6.19). Then for any K_{1} > 0 , there exist constants C_{0} , K_{2} > 0 such that for x > C_{2} , there holds

    \begin{equation*} \bigg|\frac{\partial\bar{u}_{\epsilon}}{\partial x}\bigg|+\bigg|\frac{\partial\bar{u}_{\epsilon}}{\partial t}\bigg|\leq C_{0}e^{K_{2}t-K_{1}x}, \end{equation*}

    where C_{2} is the constant given in Lemma 4.5.

    Proof. We know that for x > C_{2}

    \begin{equation*} \mathscr{L}^{\epsilon}_{\sigma_{1}}[\bar{u}_{\epsilon}] = \mathscr{L}^{\epsilon}_{\sigma_{1}}\bigg[\frac{\partial\bar{u}_{\epsilon}}{\partial t}\bigg] = \mathscr{L}^{\epsilon}_{\sigma_{1}}\bigg[\frac{\partial\bar{u}_{\epsilon}}{\partial x}\bigg], \end{equation*}

    where the operator

    \begin{equation*} \mathscr{L}^{\epsilon}_{\sigma_{1}}[\; \cdot\; ] = \frac{\partial}{\partial t}-\frac{\bar{\hat{\sigma}}_{1}^{2}}{2}\frac{\partial^{2}}{\partial x^{2}}-\bigg(\frac{\bar{\hat{\sigma}}_{1}^{2}}{2}+\delta\bigg)\frac{\partial}{\partial x}. \end{equation*}

    By Lemmas 4.2 and 4.3, it has been shown that

    \begin{equation*} \sup\limits_{0 \lt t \lt \infty}\bigg(\bigg|\frac{\partial\bar{u}_{\epsilon}}{\partial t}(t,C_{2})\bigg|+\bigg|\frac{\partial\bar{u}_{\epsilon}}{\partial x}(t,C_{2})\bigg|\bigg)\leq C, \end{equation*}

    for some C > 0 . On the other hand, there holds

    \begin{equation*} \frac{\partial\bar{u}_{\epsilon}}{\partial t}(0,x) = -\delta e^{-(x+\log K)},\; \frac{\partial\bar{u}_{\epsilon}}{\partial x}(0,x) = -e^{-(x+\log K)} \end{equation*}

    for x > C_{2} . Then for any K_{1} > 0 , there exist constants C_{0} , K_{2} > 0 such that

    \begin{equation*} \mathscr{L}^{\epsilon}_{\sigma_{1}}[C_{0}e^{K_{2}t-K_{1}x}]\geq0,\; x \gt C_{2},\; t \gt 0, \end{equation*}

    and C_{0}\geq C , C_{0}e^{-K_{1}x}\geq(\delta\vee1)e^{-(x+\log K)} , which implies that

    \begin{equation*} -C_{0}e^{K_{2}t-K_{1}x}\leq\frac{\partial\bar{u}_{\epsilon}}{\partial x}\leq C_{0}e^{K_{2}t-K_{1}x},\; -C_{0}e^{K_{2}t-K_{1}x}\leq\frac{\partial\bar{u}_{\epsilon}}{\partial t}\leq C_{0}e^{K_{2}t-K_{1}x} \end{equation*}

    for x > C_{2} .

    From the formula of \rho_{\epsilon}

    \begin{equation*} \rho_{\epsilon}(x,u,q) = \exp\bigg(\int_{0}^{x}\bigg(1+\frac{2\delta}{\bar{\hat{\sigma}}_{\epsilon}^{2}(v(z))}\bigg)dz\bigg), \end{equation*}

    we have

    \begin{equation} \exp\bigg(\bigg(1+\frac{2\delta}{\bar{\hat{\sigma}}_{n}^{2}}\bigg)x\bigg)\leq\rho_{\epsilon}(x,u,q)\leq\exp\bigg(\bigg(1+\frac{2\delta}{\bar{\hat{\sigma}}_{1}^{2}}\bigg)x\bigg), \end{equation} (6.33)

    which also clearly implies that

    \begin{equation*} \exp\bigg(\bigg(1+\frac{2\delta}{\bar{\hat{\sigma}}_{n}^{2}}\bigg)x\bigg)q\leq\frac{\partial V_{\epsilon}}{\partial q}(x,u,q)\leq\exp\bigg(\bigg(1+\frac{2\delta}{\bar{\hat{\sigma}}_{1}^{2}}\bigg)x\bigg)q \end{equation*}

    and

    \begin{equation*} \exp\bigg(\bigg(1+\frac{2\delta}{\bar{\hat{\sigma}}_{n}^{2}}\bigg)x\bigg)q^{2}\leq V_{\epsilon}(x,u,q)\leq\exp\bigg(\bigg(1+\frac{2\delta}{\bar{\hat{\sigma}}_{1}^{2}}\bigg)x\bigg)q^{2} \end{equation*}

    for x > 0 . Thus by Lemma 6.1, we can choose K_{1} > 1+\frac{2\delta}{\bar{\hat{\sigma}}_{1}^{2}} such that

    \begin{equation*} \lim\limits_{R\rightarrow\infty}\frac{\partial V_{\epsilon}}{\partial q}(t,R)\frac{\partial\bar{U}_{\epsilon}}{\partial t}(t,R) = 0, \end{equation*}

    and

    \begin{equation*} \lim\limits_{R\rightarrow\infty}\int_{-R}^{R}\frac{2\rho_{\epsilon}}{\bar{\hat{\sigma}}_{\epsilon}^{2}}\bigg(\frac{\partial\bar{U}_{\epsilon}}{\partial t}\bigg)^{2}dx = \int_{-\infty}^{\infty}\frac{2\rho_{\epsilon}}{\bar{\hat{\sigma}}_{\epsilon}^{2}}\bigg(\frac{\partial\bar{U}_{\epsilon}}{\partial t}\bigg)^{2}dx. \end{equation*}

    Then there holds

    \begin{equation*} \int_{t_{0}}^{t}\int_{-\infty}^{\infty}\frac{2\rho_{\epsilon}}{\bar{\hat{\sigma}}_{\epsilon}^{2}}\bigg(\frac{\partial\bar{U}_{\epsilon}}{\partial s}\bigg)^{2}dxds\leq E_{\infty}[\bar{U}_{\epsilon}](t_{0})\leq C, \end{equation*}

    where the constant C is independent of \epsilon , which implies that

    \begin{equation} \int_{t_{0}}^{\infty}\int_{0}^{\infty}\bar{u}_{t}^{2}(s,x)dxds\leq C, \end{equation} (6.34)

    according to (6.28) and the setting of \bar{U} .

    Now we arrive at the third step, completing the proof of convergence from \bar{u} to \psi . Denote by \bar{u}_{m}(t, x) = \bar{u}(t+m, x) and consider \bar{u}_{m} as a sequence of functions on [0, 1]\times(0, \infty) . Since \bar{u}_{m} is a bounded sequence in W^{1, 2}_{\infty}([0, 1]\times(0, \infty)) , we derive by the embedding theorem that there exists a subsequence m_{j} of m and a function \tilde{\psi} such that as m_{j}\rightarrow\infty , there holds

    \begin{equation} \bar{u}_{m_{j}}\rightarrow\tilde{\psi}\; \text{in}\; C^{\frac{(1+\alpha)}{2},1+\alpha}([0,1]\times(0,R)),\; 0 \lt \alpha \lt 1, \end{equation} (6.35)

    for any R > 1 . Furthermore, by taking a further subsequence if necessary, there holds that

    \begin{equation*} \frac{\partial\bar{u}_{m_{j}}}{\partial t}\overset{w^{*}}{\longrightarrow}\tilde{\psi}_{t},\; \frac{\partial^{2}\bar{u}_{m_{j}}}{\partial x^{2}}\overset{w^{*}}{\longrightarrow}\tilde{\psi}_{xx}\; \text{in}\; L^{\infty}([0,1]\times(0,\infty)), \end{equation*}

    and thus

    \begin{equation*} \parallel\tilde{\psi}_{t}\parallel_{L^{\infty}}\leq\liminf\limits_{m\rightarrow\infty}\parallel\frac{\partial\bar{u}_{m_{j}}}{\partial t}\parallel_{L^{\infty}}\leq C,\; \parallel\tilde{\psi}_{xx}\parallel_{L^{\infty}}\leq\liminf\limits_{m\rightarrow\infty}\parallel\frac{\partial^{2}\bar{u}_{m_{j}}}{\partial x^{2}}\parallel_{L^{\infty}}\leq C, \end{equation*}

    for some constant C > 0 . As (6.34) suggests that

    \begin{equation*} \int_{0}^{1}\int_{0}^{\infty}\bigg(\frac{\partial\bar{u}_{m}}{\partial t}\bigg)^{2}dxds = \int_{m}^{m+1}\int_{0}^{\infty}\bar{u}_{t}^{2}dxds\rightarrow0\; \text{as}\; m = m_{j}\rightarrow\infty. \end{equation*}

    we have

    \begin{equation*} \int_{0}^{1}\int_{0}^{\infty}\tilde{\psi}^{2}_{t}dxdt = 0, \end{equation*}

    which implies that \tilde{\psi}_{t}\equiv0 and \tilde{\psi} is independent of t and only depends on x . Now we proceed to prove that \tilde{\psi} satisfies (6.1). Take a test function f\in C_{c}^{\infty}(0, \infty) , then there holds

    \begin{equation} \int_{0}^{\infty}\frac{\partial\bar{u}_{m}}{\partial t}fdx = \int_{0}^{\infty}\frac{1}{2}\bar{\hat{\sigma}}^{2}(\bar{u}_{m})\bigg(\frac{\partial^{2}\bar{u}_{m}}{\partial x^{2}}+\frac{\partial\bar{u}_{m}}{\partial x}\bigg)fdx+\int_{0}^{\infty}\delta\frac{\partial\bar{u}_{m}}{\partial x}fdx. \end{equation} (6.36)

    Clearly, the second term on the right side converges to the corresponding integral of \tilde{\psi} as m = m_{j}\rightarrow\infty . With regard to the first term on the right side of (6.36), there holds

    \begin{equation} \begin{split} &\int_{0}^{\infty}\frac{1}{2}\bar{\hat{\sigma}}^{2}(\bar{u}_{m})\bigg(\frac{\partial^{2}\bar{u}_{m}}{\partial x^{2}}+\frac{\partial\bar{u}_{m}}{\partial x}\bigg)fdx\\ = &\int_{0}^{\infty}\frac{1}{2}\bar{\hat{\sigma}}^{2}(\tilde{\psi})\bigg(\frac{\partial^{2}\bar{u}_{m}}{\partial x^{2}}+\frac{\partial\bar{u}_{m}}{\partial x}\bigg)fdx+\int_{0}^{\infty}\frac{1}{2}(\bar{\hat{\sigma}}^{2}(\bar{u}_{m})-\bar{\hat{\sigma}}^{2}(\tilde{\psi}))\bigg(\frac{\partial^{2}\bar{u}_{m}}{\partial x^{2}}+\frac{\partial\bar{u}_{m}}{\partial x}\bigg)fdx. \end{split} \end{equation} (6.37)

    By the weak-star convergence, the first term on the right side of (6.32) converges to the corresponding integral of \tilde{\psi} . The second term on the right side of (6.37) is bounded by C\int_{0}^{\infty}|\bar{\hat{\sigma}}^{2}(\bar{u}_{m})-\bar{\hat{\sigma}}^{2}(\tilde{\psi})|fdx , which converges to 0 by the dominated convergence theorem. By the convergence (6.35), we have

    \begin{equation*} \int_{0}^{1}\int_{0}^{\infty}\frac{\partial\bar{u}_{m}}{\partial t}fdxdt\rightarrow0\; \text{as}\; m\rightarrow\infty. \end{equation*}

    Thus integrating (6.36) with respect to t over [0, 1] and letting m\rightarrow\infty , there holds

    \begin{equation*} \int_{0}^{\infty}\frac{1}{2}\bar{\hat{\sigma}}^{2}(\tilde{\psi})(\tilde{\psi}_{xx}+\tilde{\psi}_{x})fdx+\int_{0}^{\infty}\delta\tilde{\psi}_{x}fdx = 0. \end{equation*}

    It follows that \tilde{\psi} satisfies (6.1). The convergence (6.35) also suggests that \tilde{\psi} satisfies (6.2). Now suppose that

    \begin{equation} \liminf\limits_{m_{j}\rightarrow\infty}\inf\limits_{0\leq t\leq1}\bar{\lambda}_{i}(t+m_{j}) = \bar{\lambda}_{i}^{min}\leq\bar{\lambda}_{i}^{max} = \limsup\limits_{m_{j}\rightarrow\infty}\sup\limits_{0\leq t\leq1}\bar{\lambda}_{i}(t+m_{j}),\; i = 1,2,\cdots,n-1. \end{equation} (6.38)

    We choose t_{i, j}^{min} , t_{i, j}^{max}\in[0, 1] such that

    \begin{equation*} \inf\limits_{0\leq t\leq1}\bar{\lambda}_{i}(t+m_{j}) = \bar{\lambda}_{i}(t_{i,j}^{min}+m_{j}),\; \sup\limits_{0\leq t\leq1}\bar{\lambda}_{i}(t+m_{j}) = \bar{\lambda}_{i}(t_{i,j}^{max}+m_{j}). \end{equation*}

    Taking the subsequences along which the liminf and limsup in (6.38) are attained, together with the boundary conditions

    \begin{equation*} \bar{u}_{m}(t,\bar{\lambda}_{i}(t+n)) = \gamma_{i},\; i = 1,2,\cdots,n-1, \end{equation*}

    and (6.35), it is deduced that

    \begin{equation*} \tilde{\psi}(\bar{\lambda}_{i}^{min}) = \tilde{\psi}(\bar{\lambda}_{i}^{max}) = \gamma_{i},\; i = 1,2,\cdots,n-1. \end{equation*}

    However, by the uniqueness of solution to static problem (6.1)-(6.3), there should hold

    \begin{equation*} \bar{\lambda}_{i}^{min} = \bar{\lambda}_{i}^{max} = \lambda_{i}^{*},\; i = 1,2,\cdots,n-1, \end{equation*}

    and \psi\equiv\tilde{\psi} . In addition, the uniqueness implies that all subsequences limit should be uniform and thus the full sequence must converge as m\rightarrow\infty .

    In this paper, we study a free boundary problem for pricing a defaultable corporate bond with multiple credit rating migration risk and stochastic interest rate. By using PDE techniques, the existence, uniqueness some regularities of solution are obtained to support the rationality of the model to pricing a defaultable corporate bond. In [35], it is shown that the solution and rating boundaries of the free boundary problem pricing a defaultable corporate bond with constant interest rate are all decreasing with respect to time, which do not hold any more in our model with stochastic interest rate, as the coefficients of model are all time heterogeneous. Furthermore, we present the asymptotic behavior of solution to this pricing model. Asymptotic behaviors of solution to free boundary problems pricing corporate bonds with credit rating migration risk have been analyzed in [12,21,32]. The asymptotic solution of the model with only one migration boundary can be solved explicitly [21], while in the works [12,32], where the models are subject to multiple migration boundaries, it is not the case. However, interestingly, in this paper, although our model is also subject to multiple migration boundaries, it is proved that the asymptotic solution can be solved explicitly. We conclude that if the maturity T is sufficiently large, we can valuate the defaultable corporate bond with multiple credit rating migration risk and stochastic interest rate by an explicit pricing formula as follows:

    \begin{equation*} \phi(t) = S(t)\psi(\log S(t)-\log K), \end{equation*}

    where S(t) is the corporate value, K is default threshold and \psi is the steady status given in Section 6, whose explicit form can be obtained by solving the linear algebraic equation set.

    This work was supported by the National Natural Science Foundation of China (No. 11701115).

    All authors declare no conflict of interest.



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