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Double free boundary problem for defaultable corporate bond with credit rating migration risks and their asymptotic behaviors

Yuchao Dong111Yuchao Dong acknowledges partial support from National Natural Science Foundation of China (No. 12071333 & No. 12101458) [email protected] Jin Liang222Jin Liang and Claude-Michel Brauner acknowledge partial support from National Natural Science Foundation of China (No. 12071349) [email protected] Claude-Michel Brauner333Jin Liang and Claude-Michel Brauner acknowledge partial support from National Natural Science Foundation of China (No. 12071349) [email protected]
Abstract

In this work, a pricing model for a defaultable corporate bond with credit rating migration risk is established. The model turns out to be a free boundary problem with two free boundaries. The latter are the level sets of the solution but of different kinds. One is from the discontinuous second order term, the other from the obstacle. Existence, uniqueness, and regularity of the solution are obtained. We also prove that two free boundaries are CC^{\infty}. The asymptotic behavior of the solution is also considered: we show that it converges to a traveling wave solution when time goes to infinity. Moreover, numerical results are presented.

keywords:
Traveling wave; Free boundary problem; PDE with discontinuous leading order coefficient; Asymptotic behavior; Credit rating migration risk model
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myclipboard_main \affiliation[1]organization=School of Mathematical Sciences, Tongji University, city=Shanghai 200092, country=China \affiliation[2]organization=Institut de Mathématiques de Bordeaux, Université de Bordeaux, city=33405 Talence, country=France

1 Introduction

Due to the globalization and complexity of financial markets, the credit risks become more and more important and an unstable factor impacting the market, which might cause a crucial crisis. For example, in the 2008 financial tsunami and the 2010 European debt crisis, credit rating migration risk played a key role. The first step to managing the risks is modeling and measuring them. Thus, it has attracted more and more attention both in academics and in industry to understand these risks, especially default risk and credit rating migration risk.

Most credit risk research falls into two kinds of framework, namely structure model and intensity one. The intensity model assumes that the risk is due to some exogenous factors, which are usually modeled by Markov chains, see [36]. In this way, the default and/or migration times are determined by an exogenous transition intensity; see Jarrow, Lando, and Turnbull [22, 21], Duffie and Singleton [12], to mention a few. In the implementation, intensity transition matrices are usually obtained from historical statistical data. However, it is well-known that companies’ current financial status plays a crucial role in default and credit rating migrations. For example, the main reason which caused the 2010 European debt crisis was that the sovereign debts of several European countries reached an unsustainable level due to their poor economical situation. The crisis happened in these countries because of the downgrading of their credit ratings and the subsequent chain reactions. Therefore, Markov chain model alone cannot fully capture the credit risks.

To include the endogenous factor, the structural model comes into consideration for credit risk modeling, which could be traced back to Merton [35] in 1974. In such a kind of models, the reason for credit rating migration and default is related to the firm’s asset value and its obligation. For example, in Merton’s model, it is assumed that the company’s asset value follows a geometric Brownian motion and a default would happen if the asset value drops below the debt at maturity. Thus, the corporate bond, representing the company’s obligation, is a contingent claim of the asset value. Later, Black and Cox [2] extended Merton’s model to the so-called first passage-time model, where the default would happen whenever the asset value reached a given boundary; see also [26, 33, 27, 6, 39] for related works. Dai et al. [10] considered an optimal control problem in the case where a bank’s asset is opaque.

Using the structural model, Liang and Zeng [32] studied the pricing problem of the corporate bond with credit rating migration risk, where a predetermined migration threshold is given to divide asset value into high and low rating regions, in which the asset value follows different stochastic processes. Hu, Liang, and Wu [19] further developed this model, where the migration boundary is a free boundary governed by a ratio of the firm’s asset value and debt. Some theoretical results and traveling wave properties are also obtained in [30]. Li, Zhang, and Hu [28] studied the numerical method for solving related variational inequality. Later, Fu, Chen, and Liang [17] provided more mathematical analysis and detailed description of the free migration boundary. More extension of this model is considered in [31, 43, 40, 41]. Recently, Chen and Liang [9] also considered the case where upgrade and downgrade boundaries are different. The readers can also refer to the survey paper [8] for a summary.

However, the reason behind the credit rating migration is the default possibility; hence, it is natural to consider a model involving both the credit rating migration and default risks. In [41], as the first step, a predetermined default boundary of asset level is considered. In this paper, we will let the default boundary also depend on the ratio between the stock price and bond value. Therefore, the model will contain two free boundaries. Both of these boundaries are the level sets of the solution but of different types. One is from discontinuous leading second order term as in previous credit rating migration works (for example, see [30]); the other is from a more traditional free boundary problem, i.e. obstacle problem. Using PDE techniques, existence, uniqueness, regularity, and asymptotic behavior of the solution are obtained, which from a theoretical perspective insure the rationality of the model. Numerical results support our theoretical approach. The stability of traveling wave equation will be studied in our future work [3] using the techniques of [1, 4, 5].

This paper is organized as follows. In Section 2, the model is established and the pricing problem is reduced to a system of two parabolic PDEs with two free boundaries. In Section 3, for the sake of both uniform estimates and asymptotic behavior, we consider a traveling wave solution to the original problem. In Section 4, we use a penalization method and simultaneously a regularization of the coefficient of the 2nd order term to approximate the free boundary problem by a smooth Cauchy problem depending on a small parameter ε>0\varepsilon>0. A series of lemmas are proved in order to establish estimates which are independent of ε\varepsilon. The key point is that the two approximating free boundaries can be separated by a positive distance independent of ε\varepsilon. In Section 5, the main results are stated, including the existence, uniqueness, and regularity of the solution. In particular, we prove that two free boundaries are CC^{\infty}. The asymptotic behavior of the solution as time tends to infinity is examined in Section 6. Finally, a numerical method and some computational results are presented in Section 7.

2 The Model

2.1 Assumptions

Let (Ω,,P)(\Omega,\mathcal{F},P) be a complete probability space. We assume that the firm issues a corporate bond, which is a contingent claim of its value. The stock price of the firm admits different dynamics for different credit ratings.

Assumption 2.1 (the firm asset with credit rating migration).

Let StS_{t} denote the firm’s value in the risk neutral world. It satisfies

dSt={rStdt+σHStdWt, in high rating region,rStdt+σLStdWt, in low rating region,dS_{t}=\left\{\begin{array}[]{l}rS_{t}dt+\sigma_{H}S_{t}dW_{t},\quad\mbox{ in high rating region,}\\ rS_{t}dt+\sigma_{L}S_{t}dW_{t},\quad\mbox{ in low rating region,}\end{array}\right.

where rr is the risk free interest rate, which is positive constant, and

σH<σL\sigma_{H}<\sigma_{L} (2.1)

represent volatilities of the firm under the high and low credit grades respectively. They are also assumed to be positive constants. WtW_{t} is the Brownian motion which generates the filtration {t}\{{\mathcal{F}}_{t}\}.

It is reasonable to assume (2.1), namely that the volatility in high rating region is lower than the one in the low rating region. The firm issues only one zero coupon corporate bond with face value FF. Let Φt\Phi_{t} denote the discount value of the bond at time tt. Therefore, at the maturity time TT, an investor can get ΦT=min{ST,F}\Phi_{T}=\min\{S_{T},F\}. For simplicity, we assume in the following sections F=1F=1. The rating criterion is based on the ratio between the stock price and liability.

Assumption 2.2 (the credit rating migration time).

High and low rating regions are determined by the proportion between the debt and asset value. The credit rating migration time τ1\tau_{1} and τ2\tau_{2} are the first moments when the firm’s grade is downgraded and upgraded respectively as follows:

τ1=inf{t>0|Φ0/S0<γeδT,Φt/Stγeδ(Tt)},\displaystyle\tau_{1}=\inf\{t>0|\Phi_{0}/S_{0}<\gamma e^{-\delta T},\Phi_{t}/S_{t}\geqslant\gamma e^{-\delta(T-t)}\},
τ2=inf{t>0|Φ0/S0>γeδT,Φt/Stγeδ(Tt)},\displaystyle\tau_{2}=\inf\{t>0|\Phi_{0}/S_{0}>\gamma e^{-\delta T},\Phi_{t}/S_{t}\leqslant\gamma e^{-\delta(T-t)}\},

where Φt=Φt(St,t)\Phi_{t}=\Phi_{t}(S_{t},t) is a contingent claim with respect to StS_{t} and

0<γ<10<\gamma<1 (2.2)

is a positive constant representing the threshold proportion of the debt and value of the firm’s rating. Also

δ>0,\delta>0,

is the so-called credit discount rate. In this paper, we also make the assumption that

12σH2<δ<12σL2.\frac{1}{2}\sigma^{2}_{H}<\delta<\frac{1}{2}\sigma^{2}_{L}. (2.3)

Further, we assume that the bond will default when the stock price is too low, compared with the debt.

Assumption 2.3 (the defaultable corporate bond).

The default time is also determined by the proportion of the debt and asset value. Here, we assume that the default happens whenever

Steδ(Tt)Φt.S_{t}e^{-\delta(T-t)}\leqslant\Phi_{t}.

The default time is defined as

τ=inf{t>0|Φ0>eδTS0,Φteδ(Tt)St}.\tau=\inf\{t>0|\Phi_{0}>e^{-\delta T}S_{0},\Phi_{t}\geqslant e^{-\delta(T-t)}S_{t}\}.

At the default time, the contract is closed and the investor obtains the cash eδ(Tt)Ste^{-\delta(T-t)}S_{t}.

Remark 2.4.

Condition (2.3) is also assumed in [30] to ensure the existence of the travelling wave equation. In finance, if δ\delta is too small or too large, it is possible that the company will always be low rating or high rating. To see this, assume that the stock price is

St=ert120tσ2(u)𝑑u+0tσ(u)𝑑Wu,S_{t}=e^{rt-\frac{1}{2}\int_{0}^{t}\sigma^{2}(u)du+\int_{0}^{t}\sigma(u)dW_{u}},

where σ(s)\sigma(s) is the volatility of the stock taking values in {σH,σL}\{\sigma_{H},\sigma_{L}\} depending on whether the stock is low rating or high rating. The present value of the bond is er(Tt)e^{-r(T-t)}. Then, the company’s discounted debt-to-asset ratio is

eδter(Tt)St=erTe0t(12σ2(u)δ)𝑑u0tσ(u)𝑑Wu.e^{-\delta t}\frac{e^{-r(T-t)}}{S_{t}}=e^{-rT}e^{\int_{0}^{t}(\frac{1}{2}\sigma^{2}(u)-\delta)du-\int_{0}^{t}\sigma(u)dW_{u}}.

If δ<12σH2\delta<\frac{1}{2}\sigma_{H}^{2}, the right hand side will go to \infty as tt\rightarrow\infty with probability 11. This implies that the company will always be low rating in the end. On the other hand, if δ>12σL2\delta>\frac{1}{2}\sigma^{2}_{L}, the right hand side will go to 0 and, hence, the company will always be high rating.

2.2 The Cash Flow

If the bond does not default, once the credit rating migrates before the maturity TT, a virtual substitute termination happens, i.e., the bond is virtually terminated and substituted by a new one with a new credit rating. There is a virtual cash flow of the bond. We denote by ΦH(S,t)\Phi_{H}(S,t) and ΦL(S,t)\Phi_{L}(S,t) the values of the bond in high and low grades respectively, which are functions of SS and tt. Then, they are conditional expectations of the following

ΦH(S,t)=\displaystyle\Phi_{H}(S,t)= E[er(Tt)min(ST,F)𝟏{T<τ1τ}\displaystyle E\Big{[}e^{-r(T-t)}\min(S_{T},F)\cdot{\bf 1}_{\{T<\tau_{1}\wedge\tau\}}
+Steδ(Tτ)er(τt)𝟏{τ<Tτ1}\displaystyle+S_{t}e^{-\delta(T-\tau)}e^{-r(\tau-t)}{\bf 1}_{\{\tau<T\wedge\tau_{1}\}}
+er(τ1t)ΦL(Sτ1,τ1)𝟏{τ1<Tτ}|St=S>1γeδ(Tt)ΦH(S,t)],\displaystyle+e^{-r(\tau_{1}-t)}\Phi_{L}(S_{\tau_{1}},\tau_{1})\cdot{\bf 1}_{\{\tau_{1}<T\wedge\tau\}}\Big{|}S_{t}=S>\frac{1}{\gamma e^{-\delta(T-t)}}\Phi_{H}(S,t)\Big{]}, (2.4)
ΦL(S,t)=\displaystyle\Phi_{L}(S,t)= E[er(Tt)min(ST,F)𝟏{T<τ2τ}\displaystyle E[e^{-r(T-t)}\min(S_{T},F)\cdot{\bf 1}_{\{T<\tau_{2}\wedge\tau\}}
+Steδ(Tτ)er(τt)𝟏{τ<Tτ2}\displaystyle+S_{t}e^{-\delta(T-\tau)}e^{-r(\tau-t)}{\bf 1}_{\{\tau<T\wedge\tau_{2}\}}
+er(τ2t)ΦH(Sτ2,τ2)𝟏{τ2<Tτ}|1eδ(Tt)ΦL(S,t)<St=S<1γeδ(Tt)ΦL(S,t)],\displaystyle+e^{-r(\tau_{2}-t)}\Phi_{H}(S_{\tau_{2}},\tau_{2})\cdot{\bf 1}_{\{\tau_{2}<T\wedge\tau\}}\Big{|}\frac{1}{e^{-\delta(T-t)}}\Phi_{L}(S,t)<S_{t}=S<\frac{1}{\gamma e^{-\delta(T-t)}}\Phi_{L}(S,t)\Big{]}, (2.5)

where 𝟏{event}={1, if “event” happens,0, otherwise. {\bf 1}_{\{event\}}=\left\{\begin{array}[]{ll}1,&\mbox{ if ``event" happens},\\ 0,&\mbox{ otherwise. }\end{array}\right.

2.3 The PDE problem

In the life time of the bond, by Feynman-Kac formula (see, e.g. [11]), it is not difficult to derive that the letting values ΦH\Phi_{H} and ΦL\Phi_{L} satisfy the following system of partial differential equations in their respective life regions:

ΦHt+12σH2S22ΦHS2+rSΦHSrΦH=0,\displaystyle\frac{\partial\Phi_{H}}{\partial t}+\frac{1}{2}{\sigma}^{2}_{H}S^{2}\frac{\partial^{2}\Phi_{H}}{\partial S^{2}}+rS\frac{\partial\Phi_{H}}{\partial S}-r\Phi_{H}=0,
S>1γeδ(Tt)ΦH,t>0,\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad S>\frac{1}{\gamma e^{-\delta(T-t)}}\Phi_{H},\;t>0, (2.6)
ΦLt+12σL2S22ΦLS2+rSΦLSrΦL=0,\displaystyle\frac{\partial\Phi_{L}}{\partial t}+\frac{1}{2}{\sigma}^{2}_{L}S^{2}\frac{\partial^{2}\Phi_{L}}{\partial S^{2}}+rS\frac{\partial\Phi_{L}}{\partial S}-r\Phi_{L}=0,
1eδ(Tt)ΦL<S<1γeδ(Tt)ΦL,t>0.\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\frac{1}{e^{-\delta(T-t)}}\Phi_{L}<S<\frac{1}{\gamma e^{-\delta(T-t)}}\Phi_{L},\;t>0. (2.7)

If the bond life last to maturity, ΦH\Phi_{H} and ΦH\Phi_{H} satisfy the terminal conditions:

ΦH(S,T)=ΦL(S,T)=min{S,F}.\Phi_{H}(S,T)=\Phi_{L}(S,T)=\min\{S,F\}.

Define the function Φ\Phi as

Φ(S,t)={ΦH(S,t), in the high rating region;ΦL(S,t), in the low rating region;Seδ(Tt), in the default region.\begin{split}\Phi(S,t)=\left\{\begin{split}&\Phi_{H}(S,t),\text{ in the high rating region;}\\ &\Phi_{L}(S,t),\text{ in the low rating region;}\\ &Se^{-\delta(T-t)},\text{ in the default region.}\\ \end{split}\right.\end{split}

Then, it satisfies the following variational form

min{Φt+12σ2(Φ,S,t)S22ΦS2+rSΦSrΦ,Φ(S,t)+Seδ(Tt)}=0,\min\Big{\{}\frac{\partial\Phi}{\partial t}+\frac{1}{2}\sigma^{2}(\Phi,S,t)S^{2}\frac{\partial^{2}\Phi}{\partial S^{2}}+rS\frac{\partial\Phi}{\partial S}-r\Phi,\,-\Phi(S,t)+Se^{-\delta(T-t)}\Big{\}}=0,

with

σ(Φ,S,t)=σH𝟏{Φ<γSeδ(Tt)}+σL𝟏{ΦγSeδ(Tt)}.\sigma(\Phi,S,t)=\sigma_{H}{\bf 1}_{\{\Phi<\gamma Se^{-\delta(T-t)}\}}+\sigma_{L}{\bf 1}_{\{\Phi\geqslant\gamma Se^{-\delta(T-t)}\}}.

First, we make some transformation. Let ϕ(x,t)=ertΦ(ex,Tt)\phi(x,t)=e^{rt}\Phi(e^{x},T-t). Then, ϕ\phi satisfies

min{ϕt+12σ2(ertϕ,ex,t)2ϕx2+(r12σ2)ϕx,ϕ(s,t)+ex+(rδ)t}=0.\min\Big{\{}-\frac{\partial\phi}{\partial t}+\frac{1}{2}\sigma^{2}(e^{-rt}\phi,e^{x},t)\frac{\partial^{2}\phi}{\partial x^{2}}+(r-\frac{1}{2}\sigma^{2})\frac{\partial\phi}{\partial x},\,-\phi(s,t)+e^{x+(r-\delta)t}\Big{\}}=0.

As already indicated in [30], it is more convenient to work in the moving coordinate frame

ξ=x+ct,c=rδ,u(ξ,t)=ϕ(x,t).\xi=x+ct,\;c=r-\delta,\;u(\xi,t)=\phi(x,t).

Then, the equation reads

min{ut+12σ2(u)2uξ2+(δ12σ2)uξ,u+eξ}=0.\min\left\{-\frac{\partial u}{\partial t}+\frac{1}{2}\sigma^{2}(u)\frac{\partial^{2}u}{\partial\xi^{2}}+(\delta-\frac{1}{2}\sigma^{2})\frac{\partial u}{\partial\xi},\,-u+e^{\xi}\right\}=0. (2.8)

Let us introduce the weight eξe^{-\xi} and make the further transformation v=eξuv=e^{-\xi}u; we define

:=t+12σ2(v)(2ξ2+ξ)+δ(ξ+1).{\mathcal{L}}:=-\frac{\partial}{\partial t}+\frac{1}{2}\sigma^{2}(v)\Big{(}\frac{\partial^{2}}{\partial\xi^{2}}+\frac{\partial}{\partial\xi}\Big{)}+\delta\Big{(}\frac{\partial}{\partial\xi}+1\Big{)}.

Thus, vv satisfies the following problem:

min{v, 1v}=0,v(ξ,0)=min{1,eξ},\displaystyle\min\left\{{\mathcal{L}}v,\,1-v\right\}=0,\quad v(\xi,0)=\min\{1,e^{-\xi}\}, (2.9)

with

σ(v)=σH𝟏{v<γ}+σL𝟏{vγ}.\sigma(v)=\sigma_{H}{\bf 1}_{\{v<\gamma\}}+\sigma_{L}{\bf 1}_{\{v\geqslant\gamma\}}.

Let us finally define the free boundaries which will play a crucial role throughout the paper, respectively the default boundary

κ^(t):=inf{ξ|v(ξ,t)<1},\hat{\kappa}(t):=\inf\{\xi\,|\,v(\xi,t)<1\},

and the transit boundary

η^(t):=inf{ξ|v(ξ,t)<γ}.\hat{\eta}(t):=\inf\{\xi\,|\,v(\xi,t)<\gamma\}.

Our goal is not only to solve (2.9) but also to study the properties of these boundaries. If the solution is smooth enough, system (2.9) can be rewritten as the free boundary problem

{vt+12σL2(2vξ2+vξ)+δ(vξ+v)=0,κ^(t)<ξ<η^(t);vt+12σH2(2vξ2+vξ)+δ(vξ+v)=0,ξ>η^(t);v(κ^(t)+)=1,vξ(κ^(t)+)=0;v(η^(t)+)=v(η^(t))=γ,vξ(η^(t)+)=vξ(η^(t)).\left\{\begin{split}&-\frac{\partial v}{\partial t}+\frac{1}{2}\sigma_{L}^{2}\Big{(}\frac{\partial^{2}v}{\partial\xi^{2}}+\frac{\partial v}{\partial\xi}\Big{)}+\delta\Big{(}\frac{\partial v}{\partial\xi}+v\Big{)}=0,\quad\hat{\kappa}(t)<\xi<\hat{\eta}(t);\\ &-\frac{\partial v}{\partial t}+\frac{1}{2}\sigma_{H}^{2}\Big{(}\frac{\partial^{2}v}{\partial\xi^{2}}+\frac{\partial v}{\partial\xi}\Big{)}+\delta\Big{(}\frac{\partial v}{\partial\xi}+v\Big{)}=0,\quad\xi>\hat{\eta}(t);\\ &v(\hat{\kappa}(t)+)=1,\quad\frac{\partial v}{\partial\xi}(\hat{\kappa}(t)+)=0;\\ &v(\hat{\eta}(t)+)=v(\hat{\eta}(t)-)=\gamma,\quad\frac{\partial v}{\partial\xi}(\hat{\eta}(t)+)=\frac{\partial v}{\partial\xi}(\hat{\eta}(t)-).\end{split}\right. (2.10)

For convenience, we set

cL=2δσL2,cH=2δσH2.c_{L}=\frac{2\delta}{\sigma^{2}_{L}},\qquad c_{H}=\frac{2\delta}{\sigma^{2}_{H}}.

It follows from (2.3) that cL<1c_{L}<1 and cH>1c_{H}>1.

3 Traveling Wave Solution

In this section, we will consider the steady state of (2.9), i.e. the traveling wave solution for the original problem. In addition to giving the asymptotic behavior of (2.9), the traveling wave equation is also useful for constructing sub-solutions. The traveling wave solution KK satisfies

min{12σ2(K)(dKdξ2+dKdξ)+δ(dKdξ+K), 1K}=0.\min\left\{\frac{1}{2}\sigma^{2}(K)\Big{(}\frac{dK}{d\xi^{2}}+\frac{dK}{d\xi}\Big{)}+\delta\Big{(}\frac{dK}{d\xi}+K\Big{)},\,1-K\right\}=0. (3.1)

Denoting the two free boundaries respectively by κ\kappa^{*} and η\eta^{*}, and assuming that the solution is sufficiently smooth, we may reformulate Equation (3.1) as the following free boundary problem

{d2Kdξ2+dKdξ+cH(dKdξ+K)=0,ξ>η,d2Kdξ2+dKdξ+cL(dKdξ+K)=0,κ<ξ<η,K(κ+)=1,Kξ(κ)=0,K(η+)=K(η)=γ,dKdξ(η+)=dKdξ(η),K(ξ)=1,for ξ<κ, and limξ+eξK(ξ)=1,\left\{\begin{split}&\frac{d^{2}K}{d\xi^{2}}+\frac{dK}{d\xi}+c_{H}\Big{(}\frac{dK}{d\xi}+K\Big{)}=0,\quad\xi>\eta^{*},\\ &\frac{d^{2}K}{d\xi^{2}}+\frac{dK}{d\xi}+c_{L}\Big{(}\frac{dK}{d\xi}+K\Big{)}=0,\quad\kappa^{*}<\xi<\eta^{*},\\ &K(\kappa^{*}+)=1,\quad\frac{{\partial}K}{{\partial}\xi}(\kappa^{*})=0,\\ &K(\eta^{*}+)=K(\eta^{*}-)=\gamma,\quad\frac{dK}{d\xi}({\eta^{*}+})=\frac{dK}{d\xi}({\eta^{*}-}),\\ &K(\xi)=1,\text{for $\xi<\kappa^{*}$, and }\lim_{\xi\rightarrow+\infty}e^{\xi}K(\xi)=1,\end{split}\right. (3.2)

Note that we also add a growth condition at ++\infty due to the financial nature of our problem.

Theorem 3.1.

System (3.2) has a unique solution (K,η,κ)(K,\eta^{*},\kappa^{*}) such that KK belongs to C1([κ,+))C^{1}([\kappa^{*},+\infty)) and the respective restrictions of KK to [κ,η][\kappa^{*},\eta^{*}] and [η,+][\eta^{*},+\infty] are CC^{\infty}.

Proof.

It is elementary to solve the second order system in (3.2):

K(ξ)={eξ+BecHξ,ξ>η,Ceξ+DecLξ,κ<ξ<η.K(\xi)=\left\{\begin{split}&e^{-\xi}+Be^{-c_{H}\xi},\;\xi>\eta^{*},\\ &Ce^{-\xi}+De^{-c_{L}\xi},\;\kappa^{*}<\xi<\eta^{*}.\end{split}\right. (3.3)

From the boundary conditions at κ\kappa^{*}, it comes

Ceκ+DecLκ=1, and CeκcLDecLκ=0.Ce^{-\kappa^{*}}+De^{-c_{L}\kappa^{*}}=1,\text{ and }-Ce^{-\kappa^{*}}-c_{L}De^{-c_{L}\kappa^{*}}=0.

This implies that C=cL1cLeκC=-\frac{c_{L}}{1-c_{L}}e^{\kappa^{*}} and D=11cLecLκD=\frac{1}{1-c_{L}}e^{c_{L}\kappa^{*}}. Then, from K(η)=γK(\eta^{*}-)=\gamma, we have that

cL1cLeκη+11cLecL(ηκ)=γ.-\frac{c_{L}}{1-c_{L}}e^{\kappa^{*}-\eta^{*}}+\frac{1}{1-c_{L}}e^{-c_{L}(\eta^{*}-\kappa^{*})}=\gamma. (3.4)

Define the mapping

Ψ(x):xcL1cLex+11cLecLx,\displaystyle\Psi(x):x\mapsto-\frac{c_{L}}{1-c_{L}}e^{-x}+\frac{1}{1-c_{L}}e^{-c_{L}x}, (3.5)

hence Ψ(x)=cL1cL(execLx)\Psi^{\prime}(x)=\frac{c_{L}}{1-c_{L}}(e^{-x}-e^{-c_{L}x}). Since cL<1c_{L}<1, we have that the mapping Ψ\Psi is decreasing on [0,)[0,\infty). Since Ψ(0)=1\Psi(0)=1 and limx+Ψ(x)=0\lim_{x\rightarrow+\infty}\Psi(x)=0, the transcendental equation (3.4) admits a unique positive solution

ηκ=Ψ1(γ),\eta^{*}-\kappa^{*}=\Psi^{-1}(\gamma), (3.6)

The interface condition [dKdξ]η=0\big{[}\frac{dK}{d\xi}\big{]}_{\eta^{*}}=0 yields that

eη+cHBecHη=cL1cLe(ηκ)+cL1cLecL(ηκ)=γecL(ηκ),e^{-\eta^{*}}+c_{H}Be^{-c_{H}\eta^{*}}=-\frac{c_{L}}{1-c_{L}}e^{-(\eta^{*}-\kappa^{*})}+\frac{c_{L}}{1-c_{L}}e^{-c_{L}(\eta^{*}-\kappa^{*})}=\gamma-e^{-c_{L}(\eta^{*}-\kappa^{*})},

where the last equality is due to (3.6). Combining with the condition γ=K(η+)=eη+BecHη\gamma=K(\eta^{*}+)=e^{-\eta^{*}}+Be^{-c_{H}\eta^{*}}, we have that

B=1cH1ecL(ηκ)+cHη and (cH1)eη=(cH1)γ+ecL(ηκ).B=-\frac{1}{c_{H}-1}e^{-c_{L}(\eta^{*}-\kappa^{*})+c_{H}\eta^{*}}\text{ and }(c_{H}-1)e^{-\eta^{*}}=(c_{H}-1)\gamma+e^{-c_{L}(\eta^{*}-\kappa^{*})}.

This implies that

η=log(γ+1cH1ecLΨ1(γ)).\eta^{*}=-\log\left(\gamma+\frac{1}{c_{H}-1}e^{-c_{L}\Psi^{-1}(\gamma)}\right). (3.7)

Thus, κ\kappa^{*}, B,CB,C and DD are determined. Summarizing, it comes

K(ξ)={eξ+(γeη)ecH(ξη),ξ>η,cL1cLe(ξκ)+11cLecL(ξκ),κ<ξ<η.\displaystyle K(\xi)=\left\{\begin{array}[]{ll}e^{-\xi}+(\gamma-e^{-\eta^{*}})e^{-c_{H}(\xi-\eta^{*})},&\xi>\eta^{*},\\[5.69054pt] -\frac{c_{L}}{1-c_{L}}e^{-(\xi-\kappa^{*})}+\frac{1}{1-c_{L}}e^{-c_{L}(\xi-\kappa^{*})},&\kappa^{*}<\xi<\eta^{*}.\end{array}\right. (3.10)

Some properties of KK are needed in the sections below. We list them in the following proposition.

Proposition 3.2.

(i) for ξ>κ\xi>\kappa^{*}, dKdξ<0\frac{dK}{d\xi}<0, K+dKdξ>0K+\frac{dK}{d\xi}>0, and d2Kdξ2+dKdξ<0\frac{d^{2}K}{d\xi^{2}}+\frac{dK}{d\xi}<0 if ξη\xi\neq\eta^{*};
(ii) γ<K(ξ)<1\gamma<K(\xi)<1 if ξ(κ,η)\xi\in(\kappa^{*},\eta^{*}) and K(ξ)<γ<1K(\xi)<\gamma<1 if ξ>η\xi>\eta^{*};
(iii) for ξκ\xi\geqslant\kappa^{*}, K(ξ)min{1,eξ}K(\xi)\leqslant\min\{1,e^{-\xi}\};
(iv) η\eta^{*} is a decreasing function of γ\gamma. Moreover, limγ0η=+\lim_{\gamma\rightarrow 0}\eta^{*}=+\infty and limγ1η=logcHcH1\lim_{\gamma\rightarrow 1}\eta^{*}=-\log\frac{c_{H}}{c_{H}-1}.

Proof.

(i) It is straightforward to compute

dKdξ={eξcH(γeη)ecH(ξη),ξ>η,cL1cLe(ξκ)cL1cLecL(ξκ),κ<ξ<η.\frac{dK}{d\xi}=\left\{\begin{array}[]{ll}-e^{-\xi}-c_{H}(\gamma-e^{-\eta^{*}})e^{-c_{H}(\xi-\eta^{*})},&\xi>\eta^{*},\\[2.84526pt] \frac{c_{L}}{1-c_{L}}e^{-(\xi-\kappa^{*})}-\frac{c_{L}}{1-c_{L}}e^{-c_{L}(\xi-\kappa^{*})},&\kappa^{*}<\xi<\eta^{*}.\end{array}\right.

Since cL<1c_{L}<1, it holds that dKdξ<0\frac{dK}{d\xi}<0 for κ<ξ<η\kappa^{*}<\xi<\eta^{*}. For ξ>η\xi>\eta^{*}, we rewrite

dKdξ=eηe(ξη)cH(γeη)ecH(ξη).\frac{dK}{d\xi}=-e^{-\eta^{*}}e^{-(\xi-\eta^{*})}-c_{H}(\gamma-e^{-\eta^{*}})e^{-c_{H}(\xi-\eta^{*})}.

With the notation from Theorem 3.1, we have that

cHBecHη=cH(γeη)c_{H}Be^{-c_{H}\eta^{*}}=c_{H}(\gamma-e^{-\eta^{*}})

and

eη+cHBecHη=cL1cLe(ηκ)+cL1cLecL(ηκ)>0.e^{-\eta^{*}}+c_{H}Be^{-c_{H}\eta^{*}}=-\frac{c_{L}}{1-c_{L}}e^{-(\eta^{*}-\kappa^{*})}+\frac{c_{L}}{1-c_{L}}e^{-c_{L}(\eta^{*}-\kappa^{*})}>0.

Since cH>1c_{H}>1, it holds that dKdξ<0\frac{dK}{d\xi}<0 for ξ>η\xi>\eta^{*}. Next, it comes

K+dKdξ={(1cH)(γeη)ecH(ξη),ξ>η,ecL(ξκ),κ<ξ<η,K+\frac{dK}{d\xi}=\left\{\begin{array}[]{ll}(1-c_{H})(\gamma-e^{-\eta^{*}})e^{-c_{H}(\xi-\eta^{*})},&\xi>\eta^{*},\\[5.69054pt] e^{-c_{L}(\xi-\kappa^{*})},&\kappa^{*}<\xi<\eta^{*},\end{array}\right.

and

dKdξ+d2Kdξ2={(cH2cH)(γeη)ecH(ξη),ξ>η,cLecL(ξκ),κ<ξ<η.,\frac{dK}{d\xi}+\frac{d^{2}K}{d\xi^{2}}=\left\{\begin{array}[]{ll}(c_{H}^{2}-c_{H})(\gamma-e^{-\eta^{*}})e^{-c_{H}(\xi-\eta^{*})},&\xi>\eta^{*},\\[5.69054pt] -c_{L}e^{-c_{L}(\xi-\kappa^{*})},&\kappa^{*}<\xi<\eta^{*}.\end{array},\right.

Noting that eη=γ+1cH1ecLΨ1(γ)>γe^{-\eta^{*}}=\gamma+\frac{1}{c_{H}-1}e^{-c_{L}\Psi^{-1}(\gamma)}>\gamma, cL<1c_{L}<1 and cH>1c_{H}>1, we achieve the desired results.

(ii) It follows immediately from (i).

(iii) We know from (ii) that K(ξ)1K(\xi)\leq 1. On the one hand, thanks to (3.7), γeη<0\gamma-e^{-\eta^{*}}<0 hence K(ξ)<eξK(\xi)<e^{-\xi} if ξ>η\xi>\eta^{*} (see (3.10)). On the other hand, note that K+dKdξ>0K+\frac{dK}{d\xi}>0 implies that ξeξK(ξ)\xi\mapsto e^{\xi}K(\xi) is increasing, which indicates that K(ξ)<eξK(\xi)<e^{-\xi} for κ<ξ<η\kappa^{*}<\xi<\eta^{*}.

(iv) Since Ψ1\Psi^{-1} is decreasing with respect to γ\gamma and cH>1c_{H}>1, it follows from (3.7) that η\eta^{*} is decreasing with respect to γ\gamma. It also holds that limγ0Ψ1(γ)=+\lim_{\gamma\rightarrow 0}\Psi^{-1}(\gamma)=+\infty and limγ1Ψ1(γ)=0\lim_{\gamma\rightarrow 1}\Psi^{-1}(\gamma)=0, hence the result. ∎

4 Penalized and Regularized Cauchy Problem

Problem (2.9) has singularities: at v=γv=\gamma due to the indicator function in the definition of σ\sigma; at v=1v=1 as in a usual obstacle problem; and at t=0t=0 because of the lack of regularity of the initial condition. To address these issues, we introduce HεH_{\varepsilon}, βε\beta_{\varepsilon} and ψε\psi_{\varepsilon} which depend upon a small positive parameter ε\varepsilon. These smooth functions are chosen as the following. Let H(s)H(s) be the Heaviside function, i.e., H(s)=0H(s)=0 for s<0s<0 and H(s)=1H(s)=1 for s>0s>0. Then, σ(v)\sigma(v) in (2.9) reads

σ(v)=σH+(σLσH)H(vγ).{\sigma}(v)={\sigma}_{H}+({\sigma}_{L}-{\sigma}_{H})H(v-\gamma).

First, we approximate HH by a CC^{\infty} function HεH_{\varepsilon} such that

Hε(s)=0 for s<ε,Hε(s)=1 for s>0, 0Hε(s)2/ε for <s<.H_{\varepsilon}(s)=0\,\mbox{ for }s<-\varepsilon,\,H_{\varepsilon}(s)=1\mbox{ for }s>0,\,0\leqslant H_{\varepsilon}^{\prime}(s)\leqslant 2/\varepsilon\ \mbox{ for }-\infty<s<\infty.

Second, let βε(y)\beta_{\varepsilon}(y) be a smooth penalty function satisfying the following condition:

βε(y)C(),βε(y)0,βε(y)=0 if yε;\beta_{\varepsilon}(y)\in C^{\infty}(\mathbb{R}),\,\beta_{\varepsilon}(y)\geqslant 0,\,\beta_{\varepsilon}(y)=0\text{ if $y\leqslant-\varepsilon$;}
βε(0)=C02δ;βε(y)0;βε′′(y)0;\beta_{\varepsilon}(0)=C_{0}\geqslant 2\delta;\;\beta^{\prime}_{\varepsilon}(y)\geqslant 0;\;\beta^{\prime\prime}_{\varepsilon}(y)\geqslant 0;
limε0βε(y)=0 if y<0; and limε0βε(y)=+ if y>0.\lim_{\varepsilon\rightarrow 0}\beta_{\varepsilon}(y)=0\text{ if $y<0$; and }\lim_{\varepsilon\rightarrow 0}\beta_{\varepsilon}(y)=+\infty\text{ if $y>0$.}

Let εβ>0\varepsilon_{\beta}>0 be the unique solution of βε(εβ2)=δ\beta_{\varepsilon}(-\frac{\varepsilon_{\beta}}{2})=\delta. It is easy to see that εβ0\varepsilon_{\beta}\rightarrow 0 when ε0\varepsilon\rightarrow 0. Finally, let us define ψε(y):=1+εβψ(y1εβ)\psi_{\varepsilon}(y):=1+\varepsilon_{\beta}\psi(\frac{y-1}{\varepsilon_{\beta}}), where ψC\psi\in C^{\infty}, ψ(y)=0\psi(y)=0 for y1/2y\geqslant 1/2; ψ(y)=y\psi(y)=y for y<1/2y<-1/2 and ψ(y)y, 0ψ(y)1,ψ′′(y)0\psi(y)\leqslant y,\,0\leqslant\psi^{\prime}(y)\leqslant 1,\,\psi^{\prime\prime}(y)\leqslant 0 for 1/2y1/2-1/2\leqslant y\leqslant 1/2.

From the construction of ψε\psi_{\varepsilon}, we have the following lemma.

Lemma 4.1.

(i) For y0y\geqslant 0, 0yψε(y)(1+εβ)0\leqslant y\psi^{\prime}_{\varepsilon}(y)\leqslant(1+\varepsilon_{\beta});   (ii) 0ψε(y)yψε(y)10\leqslant\psi_{\varepsilon}(y)-y\psi^{\prime}_{\varepsilon}(y)\leqslant 1.

Proof.

(i) It is easy to see that yψε(y)=yψ(y1εβ)y\psi^{\prime}_{\varepsilon}(y)=y\psi^{\prime}(\frac{y-1}{\varepsilon_{\beta}}), hence positive for y0y\geqslant 0. Note that ψ(y1εβ)=0\psi^{\prime}(\frac{y-1}{\varepsilon_{\beta}})=0 for y1+εβ2y\geqslant 1+\frac{\varepsilon_{\beta}}{2} and ψ(y1εβ)1\psi^{\prime}(\frac{y-1}{\varepsilon_{\beta}})\leqslant 1 for y1+εβ2y\leqslant 1+\frac{\varepsilon_{\beta}}{2}. Then, we shall have the second inequality.
(ii) Differentiating ψε(y)yψε(y)\psi_{\varepsilon}(y)-y\psi^{\prime}_{\varepsilon}(y), we have that

(ψε(y)yψε(y))=yεβψ′′(y1εβ).(\psi_{\varepsilon}(y)-y\psi^{\prime}_{\varepsilon}(y))^{\prime}=-\frac{y}{\varepsilon_{\beta}}\psi^{\prime\prime}(\frac{y-1}{\varepsilon_{\beta}}). (4.1)

This implies that the minimum is achieved at y=0y=0. Thus,

ψε(y)yψε(y)ψε(0)=0.\psi_{\varepsilon}(y)-y\psi^{\prime}_{\varepsilon}(y)\geqslant\psi_{\varepsilon}(0)=0.

It is easy to verify that ψε(y)yψε(y)=1\psi_{\varepsilon}(y)-y\psi^{\prime}_{\varepsilon}(y)=1 for y<1εβ2y<1-\frac{\varepsilon_{\beta}}{2} or y>1+εβ2y>1+\frac{\varepsilon_{\beta}}{2}. From (4.1), we see that ψε(y)yψε(y)1\psi_{\varepsilon}(y)-y\psi^{\prime}_{\varepsilon}(y)\leqslant 1 for any yy. ∎

Now, for ε\varepsilon small, we consider the following approximated Cauchy problem:

ε[vε]=vεt+12σε2(vε)(2vεξ2+vεξ)+δ(vεξ+vε)=βε(vε1),\begin{split}{{\mathcal{L}}_{\varepsilon}}[v_{\varepsilon}]=-\frac{{\partial}v_{\varepsilon}}{{\partial}t}+\frac{1}{2}{\sigma}^{2}_{\varepsilon}(v_{\varepsilon})\Big{(}\frac{{\partial}^{2}v_{\varepsilon}}{{\partial}\xi^{2}}+\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\Big{)}+\delta\Big{(}\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}+v_{\varepsilon}\Big{)}=\beta_{\varepsilon}(v_{\varepsilon}-1),\end{split} (4.2)

where (ξ,t)ΩT=×(0,T](\xi,t)\in\Omega_{T}=\mathbb{R}\times(0,T], T>0T>0, and

σε(vε)=σH+(σLσH)Hε(vεγ),{\sigma}_{\varepsilon}(v_{\varepsilon})={\sigma}_{H}+({\sigma}_{L}-{\sigma}_{H})H_{\varepsilon}(v_{\varepsilon}-\gamma), (4.3)

together with the initial condition

vε(ξ,0)=ψε(eξ),ξ.v_{\varepsilon}(\xi,0)=\psi_{\varepsilon}(e^{-\xi}),\quad\xi\in\mathbb{R}. (4.4)

Hence, from the definition of ψε\psi_{\varepsilon} in the previous, we have that vε(ξ,0)=1v_{\varepsilon}(\xi,0)=1 for ξlog(1+εβ2)\xi\leqslant-\log(1+\frac{\varepsilon_{\beta}}{2}); vε(ξ,0)=eξv_{\varepsilon}(\xi,0)=e^{-\xi} for ξlog(1εβ2)\xi\geqslant-\log(1-\frac{\varepsilon_{\beta}}{2}). We have the following existence result:

Theorem 4.2.

For ε>0\varepsilon>0 fixed, problem (4.2)-(4.4) has a unique bounded classical solution vεv_{\varepsilon}. Moreover, vεC(×[0,T])v_{\varepsilon}\in C^{\infty}(\mathbb{R}\times[0,T]).

Proof.

First, we turn Equation (4.2) into a quasilinear equation whose principal part is in divergence form:

vεtξa(ξ,vε,vεξ)+A(ξ,vε,vεξ)=0,\frac{{\partial}v_{\varepsilon}}{{\partial}t}-\frac{\partial}{{\partial}\xi}a\big{(}\xi,v_{\varepsilon},\frac{\partial v_{\varepsilon}}{{\partial}\xi}\big{)}+A\big{(}\xi,v_{\varepsilon},\frac{\partial v_{\varepsilon}}{{\partial}\xi}\big{)}=0, (4.5)

with

a(ξ,v,p)=12σε2(v)p,A(ξ,v,p)=βε(v1)δv(12σε2(v)+δ)p+σεσε(v)p2.a(\xi,v,p)=\frac{1}{2}\sigma^{2}_{\varepsilon}(v)p,\quad A(\xi,v,p)=\beta_{\varepsilon}(v-1)-\delta v-\big{(}\frac{1}{2}\sigma^{2}_{\varepsilon}(v)+\delta\big{)}p+\sigma_{\varepsilon}\sigma^{\prime}_{\varepsilon}(v)p^{2}.

One can check that aa and AA satisfy the assumptions of [25, Chapter V, Theorem 8.1]. Thus, there exists a unique bounded solution vεC2+α,1+α2(×[0,T])v_{\varepsilon}\in C^{2+\alpha,1+\frac{\alpha}{2}}(\mathbb{R}\times[0,T]) for any 0<α<10<\alpha<1.444For usual parabolic Hölder spaces, see, e.g., [25, Chapter 1, Section 1],[34, Section 5.1]). Then, σε(vε)\sigma_{\varepsilon}(v_{\varepsilon}) and βε(vε)\beta_{\varepsilon}(v_{\varepsilon}) belong to the same function class. Further Hölder regularity follows from classical results for linear problems (see [25, Chapter IV, Theorem 5.1], [34, Theorem 5.1.10]), which yields that vεC4+α,2+α2(×[0,T])v_{\varepsilon}\in C^{4+\alpha,2+\frac{\alpha}{2}}(\mathbb{R}\times[0,T]). The result follows by bootstrapping. ∎

Remark 4.3.

From the definition of HεH_{\varepsilon} and βε\beta_{\varepsilon}, it is easy to see that σε(vε)=σL\sigma_{\varepsilon}(v_{\varepsilon})=\sigma_{L} when vε>γv_{\varepsilon}>\gamma and βε(vε)=0\beta_{\varepsilon}(v_{\varepsilon})=0 when vε<1εv_{\varepsilon}<1-\varepsilon. Thus, when ε\varepsilon is small enough, at least one of these two equations holds.

4.1 Estimates on the approximating solution

We now proceed to derive necessary estimates on vεv_{\varepsilon} independent of ε\varepsilon, via the the maximum principle for parabolic equations in unbounded domains (see, e.g., [16, Chapter 2], [37, Chapter 7]). These properties will be inherited by the limit vv when taking ε0\varepsilon\to 0 and, thus, are crucial for the analysis of the bond value and free boundaries.

Lemma 4.4.

For ε\varepsilon sufficiently small, it holds in ×[0,T]\mathbb{R}\times[0,T]:

0vεmin(1,eξ).0\leqslant v_{\varepsilon}\leqslant\min(1,e^{-\xi}).
Proof.

Recall that we have introduced a smooth cut-off function ψ\psi in the beginning of this section. Define a function hh as h(y):=εψ(y12ε)+12h(y):=\varepsilon\psi(\frac{y-\frac{1}{2}}{\varepsilon})+\frac{1}{2}. Then, we see that h(y)=12h(y)=\frac{1}{2} for y12(1+ε)y\geqslant\frac{1}{2}(1+\varepsilon); h(y)=yh(y)=y for y12(1ε)y\leqslant\frac{1}{2}(1-\varepsilon) and 0h(y)1,h′′(y)00\leqslant h^{\prime}(y)\leqslant 1,\,h^{\prime\prime}(y)\leqslant 0 for yy\in\mathbb{R}. Thus, it holds that h(y)0h(y)\geqslant 0 if and only if y0y\geqslant 0. Furthermore, one can directly check that |yh(y)h(y)||\frac{yh^{\prime}(y)}{h(y)}| is bounded. Let w=h(vε)w=h(v_{\varepsilon}) and we have that

ε[w]=h(vε)βε(vε1)+12σε2(vε)h′′(vε)(vεξ)2+δ(wh(vε)vε),\mathcal{L}_{\varepsilon}[w]=h^{\prime}(v_{\varepsilon})\beta_{\varepsilon}(v_{\varepsilon}-1)+\frac{1}{2}\sigma^{2}_{\varepsilon}(v_{\varepsilon})h^{\prime\prime}(v_{\varepsilon})(\frac{\partial v_{\varepsilon}}{\partial\xi})^{2}+\delta(w-h^{\prime}(v_{\varepsilon})v_{\varepsilon}),

which can be rewritten as

ε[w]δ(1vεh(vε)h(vε))w=h(vε)βε(vε1)+12σε2(vε)h′′(vε)(vεξ)2.\mathcal{L}_{\varepsilon}[w]-\delta(1-\frac{v_{\varepsilon}h^{\prime}(v_{\varepsilon})}{h(v_{\varepsilon})})w=h^{\prime}(v_{\varepsilon})\beta_{\varepsilon}(v_{\varepsilon}-1)+\frac{1}{2}\sigma^{2}_{\varepsilon}(v_{\varepsilon})h^{\prime\prime}(v_{\varepsilon})(\frac{\partial v_{\varepsilon}}{\partial\xi})^{2}.

Since βε(vε1)=0\beta_{\varepsilon}(v_{\varepsilon}-1)=0 when vε<1εv_{\varepsilon}<1-\varepsilon and h(vε)=0h^{\prime}(v_{\varepsilon})=0 when vε>12(1+ε)v_{\varepsilon}>\frac{1}{2}(1+\varepsilon), we see that h(vε)βε(vε1)=0h^{\prime}(v_{\varepsilon})\beta_{\varepsilon}(v_{\varepsilon}-1)=0 if ε\varepsilon is sufficiently small. Noting that h′′0h^{\prime\prime}\leqslant 0, it holds that ε[w]δ(1vεh(vε)h(vε))w0\mathcal{L}_{\varepsilon}[w]-\delta(1-\frac{v_{\varepsilon}h^{\prime}(v_{\varepsilon})}{h(v_{\varepsilon})})w\leqslant 0. As the coefficient of zeroth order term is bounded, one can apply maximum principle to get that w0w\geqslant 0, which is equivalent to vε0v_{\varepsilon}\geqslant 0.

Next, set w=vε1w=v_{\varepsilon}-1. Then, ww verifies

ε[w]=βε(w)δ=βε(w)βε(0)ww+βε(0)δ.\begin{split}\mathcal{L}_{\varepsilon}[w]&=\beta_{\varepsilon}(w)-\delta=\frac{\beta_{\varepsilon}(w)-\beta_{\varepsilon}(0)}{w}w+\beta_{\varepsilon}(0)-\delta.\end{split}

From the definition of βε\beta_{\varepsilon}, it holds that βε(0)=C02δ\beta_{\varepsilon}(0)=C_{0}\geqslant 2\delta. Hence, this leads to w0w\leqslant 0 according again to the maximum principle.

Finally, Let w=vεeξw=v_{\varepsilon}-e^{-\xi}. Then, it holds that

ε[w]=βε(vε1)=βε(vε1)βε(eξ1)ww+βε(eξ1).\mathcal{L}_{\varepsilon}[w]=\beta_{\varepsilon}(v_{\varepsilon}-1)=\frac{\beta_{\varepsilon}(v_{\varepsilon}-1)-\beta_{\varepsilon}(e^{-\xi}-1)}{w}w+\beta_{\varepsilon}(e^{-\xi}-1).

Noting that w(ξ,0)0w(\xi,0)\leqslant 0 and βε(eξ1)0\beta_{\varepsilon}(e^{-\xi}-1)\geqslant 0, we deduce that w0w\leqslant 0 according to the maximum principle.

Lemma 4.5.

It holds in ΩT\Omega_{T}:

(1+εβ)eδtvεξ<0.-(1+\varepsilon_{\beta})e^{\delta t}\leqslant\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}<0.
Proof.

Differentiating (4.2), it comes

ε[vεξ]=σε(vε)σε(vε)vεξ(2vεξ2+vεξ)+βε(vε1)vεξ.{\mathcal{L}}_{\varepsilon}\Big{[}\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\Big{]}=-{\sigma}_{\varepsilon}(v_{\varepsilon})\,{\sigma}^{\prime}_{\varepsilon}(v_{\varepsilon})\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\,(\frac{{\partial}^{2}v_{\varepsilon}}{{\partial}\xi^{2}}+\frac{{\partial}v_{\varepsilon}}{{\partial}\xi})+\beta^{\prime}_{\varepsilon}(v_{\varepsilon}-1)\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}.

At t=0,t=0, vεξ=eξψε(eξ)\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}={\color[rgb]{0,0,0}-e^{-\xi}\psi^{\prime}_{\varepsilon}(e^{-\xi})}, which lies between (1+εβ)-(1+\varepsilon_{\beta}) and 0 from the proof of Lemma 4.1. By the maximum principle, one can deduce that (1+εβ)eδtvεξ0-(1+\varepsilon_{\beta})e^{\delta t}\leqslant\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\leqslant 0. Furthermore, the strict inequality in ΩT\Omega_{T} holds due to strong maximum principle. ∎

Lemma 4.6.

It holds in ΩT\Omega_{T}:

1vεξ+vε>0.1\geqslant\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}+v_{\varepsilon}>0.
Proof.

By Lemma 4.4 and 4.5, we have the first inequality of the lemma. Then, set w=vεξ+vεw=\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}+v_{\varepsilon}, w(ξ,0)=eξψε(eξ)+ψε(eξ)w(\xi,0)=-e^{-\xi}\psi^{\prime}_{\varepsilon}(e^{-\xi})+\psi_{\varepsilon}(e^{-\xi}). It follows from Lemma 4.1 that w0w\geqslant 0 at t=0t=0. Also, ww verifies

ε[w]+σε(vε)σε(vε)vεξwξ=βε(vε1)(wvε)+βε(vε1).\displaystyle{\mathcal{L}}_{\varepsilon}[w]+{\sigma}_{\varepsilon}(v_{\varepsilon}){\sigma}_{\varepsilon}^{\prime}(v_{\varepsilon})\,\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\frac{{\partial}w}{{\partial}\xi}=\beta_{\varepsilon}^{\prime}(v_{\varepsilon}-1)(w-v_{\varepsilon})+\beta_{\varepsilon}(v_{\varepsilon}-1). (4.6)

Using Taylor expansion of βε(ε)\beta_{\varepsilon}(-\varepsilon) at yy, one has that

0=βε(ε)=βε(y)βε(y)(y+ε)+12βε′′(θ)(y+ε)2.0=\beta_{\varepsilon}(-\varepsilon)=\beta_{\varepsilon}(y)-\beta^{\prime}_{\varepsilon}(y)(y+\varepsilon)+\frac{1}{2}\beta_{\varepsilon}^{\prime\prime}(\theta)(y+\varepsilon)^{2}.

That is,

βε(y)(y+ε)βε(y)0.\beta_{\varepsilon}(y)-(y+\varepsilon)\beta_{\varepsilon}^{\prime}(y)\leqslant 0.

Replacing yy by vε(ξ)1v_{\varepsilon}(\xi)-1 in the above formula, we have,

βε(vε1)(vε1+ε)βε(vε1)0.\beta_{\varepsilon}(v_{\varepsilon}-1)-(v_{\varepsilon}-1+\varepsilon)\beta_{\varepsilon}^{\prime}(v_{\varepsilon}-1)\leqslant 0.

Thus, (4.6) reads

ε[w]σε(vε)σε(vε)vεξwξ+βε(vε1)w\displaystyle-{\mathcal{L}}_{\varepsilon}[w]-{\sigma}_{\varepsilon}(v_{\varepsilon}){\sigma}_{\varepsilon}^{\prime}(v_{\varepsilon})\,\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\frac{{\partial}w}{{\partial}\xi}+\beta_{\varepsilon}^{\prime}(v_{\varepsilon}-1)w
=βε(vε1)vεβε(vε1)vεβε(vε1)(vε1+ε)βε(vε1)=(1ε)βε(vε1)0.\displaystyle=\beta_{\varepsilon}^{\prime}(v_{\varepsilon}-1)v_{\varepsilon}-\beta_{\varepsilon}(v_{\varepsilon}-1)\geqslant v_{\varepsilon}\beta_{\varepsilon}^{\prime}(v_{\varepsilon}-1)-(v_{\varepsilon}-1+\varepsilon)\beta_{\varepsilon}^{\prime}(v_{\varepsilon}-1)=(1-\varepsilon)\beta_{\varepsilon}^{\prime}(v_{\varepsilon}-1)\geqslant 0.

By the strong maximum principle, w>0w>0 in ΩT\Omega_{T}. ∎

Lemma 4.7.

For sufficiently small ε\varepsilon, it holds in ×[0,T]\mathbb{R}\times[0,T]:

2vεξ2+vεξ0.\frac{{\partial}^{2}v_{\varepsilon}}{{\partial}\xi^{2}}+\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\leqslant 0.
Proof.

At t=0,t=0, 2vεξ2+vεξ=eξψε′′(eξ)\frac{{\partial}^{2}v_{\varepsilon}}{{\partial}\xi^{2}}+\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}=e^{-\xi}\psi^{\prime\prime}_{\varepsilon}(e^{-\xi}) is non-positive. Now, consider the function w=vεtδ(vεξ+vε)+βε(vε1)w=\frac{{\partial}v_{\varepsilon}}{{\partial}t}-\delta\big{(}\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}+v_{\varepsilon}\big{)}+\beta_{\varepsilon}(v_{\varepsilon}-1). From the definition of HεH_{\varepsilon} and βε\beta_{\varepsilon}, it is easy to see that σε(vε)=σL\sigma_{\varepsilon}(v_{\varepsilon})=\sigma_{L} when vε>γv_{\varepsilon}>\gamma and βε(vε1)=0{\color[rgb]{0,0,1}\beta_{\varepsilon}(v_{\varepsilon}-1)}=0 when vε<1εv_{\varepsilon}<1-\varepsilon. Thus, we divide the space into two parts {vε<12ε}\{v_{\varepsilon}<1-2\varepsilon\} and {vε12ε}\{v_{\varepsilon}\geqslant 1-2\varepsilon\}
Case 1: vε<12εv_{\varepsilon}<1-2\varepsilon. In this case, we see that βε0\beta_{\varepsilon}\equiv 0. Then, it holds that

ε[w]+σε(vε)σε(vε)(2vεξ2+vεξ)(vεtδvεξ)\displaystyle{\mathcal{L}}_{\varepsilon}[w]+{\sigma}_{\varepsilon}(v_{\varepsilon}){\sigma}_{\varepsilon}^{\prime}(v_{\varepsilon})\Big{(}\frac{{\partial}^{2}v_{\varepsilon}}{{\partial}\xi^{2}}+\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\Big{)}\Big{(}\frac{{\partial}v_{\varepsilon}}{{\partial}t}-\delta\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\Big{)}
=ε[w]+2σε(vε)σε(vε)(w+δvε)w=0.\displaystyle={\mathcal{L}}_{\varepsilon}[w]+\frac{2{\sigma}_{\varepsilon}^{\prime}(v_{\varepsilon})}{{\sigma}_{\varepsilon}(v_{\varepsilon})}(w+\delta v_{\varepsilon})w=0. (4.7)

Case 2: vε12εv_{\varepsilon}\geqslant 1-2\varepsilon. For sufficiently small ε\varepsilon, we have that σεσL\sigma_{\varepsilon}\equiv\sigma_{L}. Then, it holds that

ε[vεt]=βεvεt,ε[vεξ]=βεvεξ,\mathcal{L}_{\varepsilon}[\frac{{\partial}v_{\varepsilon}}{{\partial}t}]=\beta_{\varepsilon}^{\prime}\frac{{\partial}v_{\varepsilon}}{{\partial}t},\mathcal{L}_{\varepsilon}[\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}]=\beta_{\varepsilon}^{\prime}\frac{{\partial}v_{\varepsilon}}{{\partial}\xi},

and

ε[βε]=βεβε+12σε2βε′′(vεξ)2+δβεδβεvε,\mathcal{L}_{\varepsilon}[\beta_{\varepsilon}]=\beta_{\varepsilon}^{\prime}\beta_{\varepsilon}+\frac{1}{2}\sigma^{2}_{\varepsilon}\beta_{\varepsilon}^{\prime\prime}\left(\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\right)^{2}+\delta\beta_{\varepsilon}-\delta\beta^{\prime}_{\varepsilon}v_{\varepsilon},

where we denote βε=β(vε1)\beta_{\varepsilon}=\beta(v_{\varepsilon}-1), βε=β(vε1)\beta^{\prime}_{\varepsilon}=\beta^{\prime}(v_{\varepsilon}-1) and βε′′=β′′(vε1)\beta^{\prime\prime}_{\varepsilon}=\beta^{\prime\prime}(v_{\varepsilon}-1) for simplicity of the notations. Combining above equations, we have that

ε[w]βεw=12σε2βε′′(vεξ)20,\mathcal{L}_{\varepsilon}[w]-\beta_{\varepsilon}^{\prime}w=\frac{1}{2}\sigma^{2}_{\varepsilon}\beta_{\varepsilon}^{\prime\prime}\left(\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\right)^{2}\geqslant 0,

where the last inequality is due to the fact that βε\beta_{\varepsilon} is convex.

Combining these two cases, it holds that

ε[w]+(2σε(vε)σε(vε)(w+δvε)1{v<12ε}βε1{v12ε})w0.\mathcal{L}_{\varepsilon}[w]+\Big{(}\frac{2{\sigma}_{\varepsilon}^{\prime}(v_{\varepsilon})}{{\sigma}_{\varepsilon}(v_{\varepsilon})}(w+\delta v_{\varepsilon})1_{\{v<1-2\varepsilon\}}-\beta_{\varepsilon}^{\prime}1_{\{v\geqslant 1-2\varepsilon\}}\Big{)}w\geqslant 0.

Then, by the maximum principle, w0w\leqslant 0.

Lemma 4.8.

It holds in ×(0,T]\mathbb{R}\times(0,T]:

vεt<0.\frac{{\partial}v_{\varepsilon}}{{\partial}t}<0.
Proof.

Set w=vεtw=\frac{{\partial}v_{\varepsilon}}{{\partial}t}. Then, we see that

ε[w]=σε(ve)σε(ve)(2vεξ2+vεξ)w+βε(vε1)w.\mathcal{L}_{\varepsilon}[w]=-\sigma_{\varepsilon}(v_{e})\sigma^{\prime}_{\varepsilon}(v_{e})(\frac{{\partial}^{2}v_{\varepsilon}}{{\partial}\xi^{2}}+\frac{{\partial}v_{\varepsilon}}{{\partial}\xi})w+\beta^{\prime}_{\varepsilon}(v_{\varepsilon}-1)w. (4.8)

Because vε(ξ,0)=ψε(eξ)v_{\varepsilon}(\xi,0)=\psi_{\varepsilon}(e^{-\xi}), we have that

w(ξ,0)=12σε2(vε)e2ξψε′′(eξ)+δ(ψε(eξ)eξψε(eξ))βε(ψε(eξ)1).\begin{split}w(\xi,0)=\frac{1}{2}\sigma^{2}_{\varepsilon}(v_{\varepsilon})e^{-2\xi}\psi^{\prime\prime}_{\varepsilon}(e^{-\xi})+\delta\big{(}\psi_{\varepsilon}(e^{-\xi})-e^{-\xi}\psi^{\prime}_{\varepsilon}(e^{-\xi})\big{)}-\beta_{\varepsilon}(\psi_{\varepsilon}(e^{-\xi})-1).\end{split} (4.9)

Since ψε′′()0\psi^{\prime\prime}_{\varepsilon}(\cdot)\leqslant 0, the first term is negative. Then, it is easy to check that when eξ<1εβ2e^{-\xi}<1-\frac{\varepsilon_{\beta}}{2}, the second term is zero. Hence, w(ξ,0)0w(\xi,0)\leqslant 0 in this case. Now, it remains only to check the case eξ1εβ2e^{-\xi}\geqslant 1-\frac{\varepsilon_{\beta}}{2}. From Lemma 4.1 and monotonicity of βε\beta_{\varepsilon}, we have that

δ(ψε(eξ)eξψε(eξ))βε(ψε(eξ)1)δβε(εβ2).\delta(\psi_{\varepsilon}(e^{-\xi})-e^{-\xi}\psi^{\prime}_{\varepsilon}(e^{-\xi}))-\beta_{\varepsilon}(\psi_{\varepsilon}(e^{-\xi})-1)\leqslant\delta-\beta_{\varepsilon}(-\frac{\varepsilon_{\beta}}{2}).

According to our choice of εβ\varepsilon_{\beta}, we see that the above term is non-positive. Thus, we proved that w(ξ,0)0w(\xi,0)\leqslant 0, which yields the desired result thanks to the strong maximum principle. ∎

Lemma 4.9.

There are positive constants c1,C2c_{1},C_{2} and C3C_{3}, independent of ε\varepsilon, such that it holds in ×(0,T]\mathbb{R}\times(0,T]

vεtC3C2texp(c1ξ2t).\frac{{\partial}v_{\varepsilon}}{{\partial}t}\geqslant-C_{3}-\frac{C_{2}}{\sqrt{t}}\exp\Big{(}-c_{1}\frac{\xi^{2}}{t}\Big{)}.
Proof.

Since vε(0,0)=1>γv_{\varepsilon}(0,0)=1>\gamma, and by Hölder continuity of the solution (see Theorem 4.2), there exists a ρ>0\rho>0, independent of ε\varepsilon, such that

vε(x,t)>(1+γ)/2 in Bρ,v_{\varepsilon}(x,t)>(1+\gamma)/2\;\mbox{ in }B_{\rho},

where

Bρ={(ξ,t),|ξ|ρ, 0tρ2}.B_{\rho}=\left\{(\xi,t),\,|\xi|\leqslant\rho,\;0\leqslant t\leqslant\rho^{2}\right\}.

Thus, for ε\varepsilon small enough such that ε<(1γ)/2\varepsilon<(1-\gamma)/2, σεσL{\sigma}_{\varepsilon}\equiv{\sigma}_{L} in BρB_{\rho}. We observe that, in BρB_{\rho}, the problem is reminiscent of a vanilla American option, which has a lower estimate (see, e.g., [18])

vεtC2C2texp(c1ξ2t) in Bρ.\frac{{\partial}v_{\varepsilon}}{{\partial}t}\geqslant-C_{2}-\frac{C_{2}}{\sqrt{t}}\exp\Big{(}-c_{1}\frac{\xi^{2}}{t}\Big{)}\mbox{ in }B_{\rho}. (4.10)

Let us refer to Lemma 4.8 for the notation w=vεtw=\frac{{\partial}v_{\varepsilon}}{{\partial}t}. From (4.9), it is easy to verify that w(ξ,0)w(\xi,0) is uniformly bounded from below on |ξ|ρ|\xi|\geqslant\rho. Combining with (4.10), there exists C3>0C_{3}>0 such that w(x,t)C3w(x,t)\geqslant-C_{3} on {|ξ|ρ,t=0}{|ξ|=ρ,0tρ2}{|ξ|ρ,t=ρ2}\{|\xi|\geqslant\rho,t=0\}\cup\{|\xi|=\rho,0\leqslant t\leqslant\rho^{2}\}\cup\{|\xi|\leqslant\rho,t=\rho^{2}\}. The Maximum Principle (see Lemma 4.8) yields that w(ξ,t)C3w(\xi,t)\geqslant-C_{3} in ΩTBρ\Omega_{T}\setminus B_{\rho}. Together with (4.10), we get the desired result. ∎

As an immediate corollary, we have

Lemma 4.10.

There are positive constants C4,C5C_{4},C_{5} and C6C_{6}, independent of ε\varepsilon, such that it holds in ×(0,T]\mathbb{R}\times(0,T]

C4C5texp(c1ξ2t)2vεξ2C6.-C_{4}-\frac{C_{5}}{\sqrt{t}}\exp\Big{(}-c_{1}\frac{\xi^{2}}{t}\Big{)}\leqslant\frac{{\partial}^{2}v_{\varepsilon}}{{\partial}\xi^{2}}\leqslant C_{6}.

4.2 The approximating transit boundary

Let us denote by ηε(t)\eta_{\varepsilon}(t) the approximating transit boundary, which is implicitely defined by the equation

vε(ηε(t),t)=γ.v_{\varepsilon}(\eta_{\varepsilon}(t),t)=\gamma. (4.11)

We will construct the curve tηε(t)t\mapsto\eta_{\varepsilon}(t) via the Implicit Function Theorem. To begin with, we give a lower bound for vεv_{\varepsilon}. Fom Lemma 4.4, it holds that vε(ξ,t)γεv_{\varepsilon}({\xi},t)\leqslant\gamma-\varepsilon when ξlog1γε\xi\geqslant\log\frac{1}{\gamma-\varepsilon}. This implies that σε=σH\sigma_{\varepsilon}=\sigma_{H} when ξlog1γε\xi\geqslant\log\frac{1}{\gamma-\varepsilon} and ηε(t)log1γ\eta_{\varepsilon}(t)\leqslant\log\frac{1}{\gamma}. Then, we give a lower bound for vεv_{\varepsilon}.

Lemma 4.11.

Let (K~,η~,κ~)(\tilde{K},{\color[rgb]{0,0,0}\tilde{\eta}^{*}},\tilde{\kappa}^{*}) be the solution of (3.2) as constructed in Theorem 3.1 with γ\gamma replaced by γ~\tilde{\gamma}. Choose γ~\tilde{\gamma} properly such that η~=log2γ\tilde{\eta}^{*}=\log\frac{2}{{\gamma}}. Then, we have that vεK~(εεβ)eδtv_{\varepsilon}\geqslant\tilde{K}-(\varepsilon\vee\varepsilon_{\beta})e^{\delta t} when ε<γ2\varepsilon<\frac{\gamma}{2}.

Proof.

From Proposition 3.2 (iv), γ~\tilde{\gamma} is well defined. We can rewrite that

12σ22(d2K~dξ2+dK~dξ)+δ(dK~dξ+K~)=δ1{ξκ~},\frac{1}{2}\sigma_{2}^{2}\Big{(}\frac{d^{2}\tilde{K}}{d\xi^{2}}+\frac{d\tilde{K}}{d\xi}\Big{)}+\delta\Big{(}\frac{d\tilde{K}}{d\xi}+\tilde{K}\Big{)}=\delta 1_{\{\xi\leqslant\tilde{\kappa}^{*}\}},

with σ2:=σH1{ξη~}+σL1{ξ<η~}\sigma_{2}:=\sigma_{H}1_{\{\xi\geqslant\tilde{\eta}^{*}\}}+\sigma_{L}1_{\{\xi<\tilde{\eta}^{*}\}}. Let w=vε(K~(εεβ)eδt)w=v_{\varepsilon}-(\tilde{K}-(\varepsilon\vee\varepsilon_{\beta})e^{\delta t}). Then, it holds that

ε[w]=βε(vε1)δ1{ξκ~}+12(σ22σε2)(d2K~dξ2+dK~dξ).\begin{split}{\mathcal{L}}_{\varepsilon}[w]=\beta_{\varepsilon}({v_{\varepsilon}-1})-\delta 1_{\{\xi\leqslant\tilde{\kappa}^{*}\}}+\frac{1}{2}(\sigma_{2}^{2}-\sigma_{\varepsilon}^{2})\Big{(}\frac{d^{2}\tilde{K}}{d\xi^{2}}+\frac{d\tilde{K}}{d\xi}\Big{)}.\end{split}

Since we choose η~=log2γ\tilde{\eta}^{*}=\log\frac{2}{{\gamma}}, it holds that σε2σ22\sigma_{\varepsilon}^{2}\leqslant\sigma^{2}_{2}. Combining with the fact that d2K~dξ2+dK~dξ0\frac{d^{2}\tilde{K}}{d\xi^{2}}+\frac{d\tilde{K}}{d\xi}\leqslant 0, we see that the last term on the right hand side is non-positive. Since K~min{1,eξ}\tilde{K}\leqslant\min\{1,e^{-\xi}\} (see Proposition 3.2 (iii)), βε(K~(εεβ)eδt1)βε(ε)=0\beta_{\varepsilon}(\tilde{K}-(\varepsilon\vee\varepsilon_{\beta})e^{\delta t}-1)\leqslant\beta_{\varepsilon}(-\varepsilon)=0. Thus,

ε[w]βε(vε1)βε(K~εεβ1)ww0.\mathcal{L}_{\varepsilon}[w]-\frac{\beta_{\varepsilon}(v_{\varepsilon}-1)-\beta_{\varepsilon}(\tilde{K}-\varepsilon\vee\varepsilon_{\beta}-1)}{w}w\leqslant 0.

At t=0t=0, vε(ξ,0)=ψε(eξ)v_{\varepsilon}(\xi,0)=\psi_{\varepsilon}(e^{-\xi}). It is easy to see that ψε(eξ)=eξ\psi_{\varepsilon}(e^{-\xi})=e^{-\xi} for eξ1εβ2e^{-\xi}\leqslant 1-\frac{\varepsilon_{\beta}}{2} and ψε(eξ)1εβ2\psi_{\varepsilon}(e^{-\xi})\geqslant 1-\frac{\varepsilon_{\beta}}{2} for eξ1εβ2e^{-\xi}\geqslant 1-\frac{\varepsilon_{\beta}}{2}. For both cases, we have vε(ξ,0)K~(ξ)εεβv_{\varepsilon}(\xi,0)\geqslant\tilde{K}(\xi)-\varepsilon\vee\varepsilon_{\beta}. The desired result follows from the maximum principle. ∎

Theorem 4.12.

For fixed ε>0\varepsilon>0, there exists an decreasing smooth function ηε(t)\eta_{\varepsilon}(t) such that

ηε(0)=log(1γ),κ~<ηε(t)log1γ,\eta_{\varepsilon}(0)=\log(\frac{1}{\gamma}),\quad\tilde{\kappa}^{*}<\eta_{\varepsilon}(t)\leqslant\log\frac{1}{\gamma}, (4.12)

and (4.11) holds for all t[0,T]t\in[0,T].

Proof.

To begin with, we compute

vε(logγ,0)=ψε(elogγ)=ψε(γ)=1+εβψ(γ1εβ).v_{\varepsilon}(-\log\gamma,0)=\psi_{\varepsilon}(e^{\log\gamma})=\psi_{\varepsilon}(\gamma)=1+\varepsilon_{\beta}\psi\big{(}\frac{\gamma-1}{\varepsilon_{\beta}}\big{)}.

Because γ1<0\gamma-1<0, it is clear that γ1εβ<12\frac{\gamma-1}{\varepsilon_{\beta}}<-\frac{1}{2} if εβ\varepsilon_{\beta} small enough, hence ψ(γ1εβ)=γ1εβ\psi\big{(}\frac{\gamma-1}{\varepsilon_{\beta}}\big{)}=\frac{\gamma-1}{\varepsilon_{\beta}} and ψε(γ)=γ\psi_{\varepsilon}(\gamma)=\gamma. Therefore, vε(logγ,0)=γv_{\varepsilon}(-\log\gamma,0)=\gamma. We remind that the function ξvε(ξ,0)\xi\mapsto v_{\varepsilon}(\xi,0) is smooth and non-increasing; however, in some neighborhood of logγ-\log\gamma such that γ1εβ<12\frac{\gamma-1}{\varepsilon_{\beta}}<-\frac{1}{2}, the function vε(ξ,0)v_{\varepsilon}(\xi,0) is decreasing which yields that the initial position of ηε\eta_{\varepsilon} is well-defined by (4.12).

Next, we compute

vεξ(logγ,0)=γψε(γ)=γ<0,\frac{\partial v_{\varepsilon}}{\partial\xi}(-\log\gamma,0)=-\gamma\psi_{\varepsilon}^{\prime}(\gamma)=-\gamma<0,

and (see the proof of Lemma 4.8)

vεt(logγ,0)=βε(γ1)=0.\frac{\partial v_{\varepsilon}}{\partial t}(-\log\gamma,0)=-\beta_{\varepsilon}(\gamma-1)=0.

It is now an exercise to apply the Implicit Function Theorem, which shows that there exist δi,τi>0,i=1,2\delta_{i},\tau_{i}>0,i=1,2, and a unique function φεC([τ1,τ2])\varphi_{\varepsilon}\in C^{\infty}([-\tau_{1},\tau_{2}]) such that, if (ξ,t)[logγδ1,logγ+δ2]×[τ1,τ2](\xi,t)\in[-\log\gamma-\delta_{1},-\log\gamma+\delta_{2}]\times[-\tau_{1},\tau_{2}] verifies vε(ξ,t)=γv_{\varepsilon}(\xi,t)=\gamma, then ξ=φε(t)\xi=\varphi_{\varepsilon}(t). Note that φε\varphi_{\varepsilon} is a decreasing function because

φε(t)=vεt(φε(t),t)(vεξ(φε(t),t))1<0.\varphi^{\prime}_{\varepsilon}(t)=-\frac{\partial v_{\varepsilon}}{\partial t}(\varphi_{\varepsilon}(t),t)\,\Big{(}\frac{\partial v_{\varepsilon}}{\partial\xi}(-\varphi_{\varepsilon}(t),t)\Big{)}^{-1}<0.

As by product, taking the restriction of φε\varphi_{\varepsilon} to [0,τ2][0,\tau_{2}], we have constructed a (small) branch of the curve ηε\eta_{\varepsilon}, of class CC^{\infty}, such that (4.11) holds for all t[0,τ2]t\in[0,\tau_{2}], ηε(0)=γ\eta_{\varepsilon}(0)=\gamma. Lemma 4.11 implies that vε1(εεβ)eδtv_{\varepsilon}\geqslant 1-(\varepsilon\vee\varepsilon_{\beta})e^{\delta t} for ξκ~\xi\leqslant\tilde{\kappa}^{*}. Combining with the fact that vεξ<0\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}<0, we see that κ~<ηε(t)log1γ\tilde{\kappa}^{*}<\eta_{\varepsilon}(t)\leqslant\log\frac{1}{\gamma}.

In view of Lemmas 4.5 and 4.8, we may reiterate the Implicit Function Theorem and continue this branch up to a endpoint achieved at time TT. ∎

Lemma 4.13.

For any T>0T>0, there exists a constant CT>0C_{T}>0, independent of ε\varepsilon, such that supt[0,T]|ηε(t)|CT\sup_{t\in[0,T]}|\eta^{\prime}_{\varepsilon}(t)|\leqslant C_{T}.

Proof.

From the Implicit Function Theorem, it holds:

ηε(t)=vεt(ηε(t),t)(vεξ((ηε(t),t))1.\eta^{\prime}_{\varepsilon}(t)=-\frac{\partial v_{\varepsilon}}{\partial t}(\eta_{\varepsilon}(t),t)\,\Big{(}\frac{\partial v_{\varepsilon}}{\partial\xi}((\eta_{\varepsilon}(t),t)\Big{)}^{-1}.

Note that Lemmas 4.8 and 4.9 implies that vεt\frac{{\partial}v_{\varepsilon}}{{\partial}t} is bounded. To prove the desired results, we only need to show that vεξ(ηε(t),t)c\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}(\eta_{\varepsilon}(t),t)\leqslant-c, for some positive cc. In Lemma 4.7, we proved that 2vεξ2+vεξ0\frac{{\partial}^{2}v_{\varepsilon}}{{\partial}\xi^{2}}+\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\leqslant 0, which implies that eξvεξe^{\xi}\frac{{\partial}v_{\varepsilon}}{{\partial}\xi} is non-increasing in ξ\xi. Since vεv_{\varepsilon} is smooth, there exists a point η^ε(t)(κ~,ηε(t))\hat{\eta}_{\varepsilon}(t)\in(\tilde{\kappa}^{*},\eta_{\varepsilon}(t)) such that

vεξ(η^ε(t),t)=vε(κ~,t)vε(ηε(t),t)κ~ηε(t)=vε(κ~,t)vε(ηε(t),t)ηε(t)κ~.\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}(\hat{\eta}_{\varepsilon}(t),t)=\frac{v_{\varepsilon}(\tilde{\kappa}^{*},t)-v_{\varepsilon}(\eta_{\varepsilon}(t),t)}{\tilde{\kappa}^{*}-\eta_{\varepsilon}(t)}=-\frac{v_{\varepsilon}(\tilde{\kappa}^{*},t)-v_{\varepsilon}(\eta_{\varepsilon}(t),t)}{\eta_{\varepsilon}(t)-\tilde{\kappa}^{*}}.

We have shown that vε(κ~,t)K~(κ~)(εεβ)eδt=1(εεβ)eδtv_{\varepsilon}(\tilde{\kappa}^{*},t)\geqslant\tilde{K}(\tilde{\kappa}^{*})-(\varepsilon\vee\varepsilon_{\beta})e^{\delta t}=1-(\varepsilon\vee\varepsilon_{\beta})e^{\delta t} and ηε(t)log1γ\eta_{\varepsilon}(t)\leqslant\log\frac{1}{\gamma}. This yields that

vεξ(η^ε(t))1(εεβ)eδtγlog1γκ~.\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}(\hat{\eta}_{\varepsilon}(t))\leqslant-\frac{1-(\varepsilon\vee\varepsilon_{\beta})e^{\delta t}-\gamma}{\log\frac{1}{\gamma}-\tilde{\kappa}^{*}}.

Since eξvεξe^{\xi}\frac{{\partial}v_{\varepsilon}}{{\partial}\xi} is non-increasing, it holds that

vεξ(ηε(t),t)eη^ε(t)ηε(t)1(εεβ)eδtγlog1γκ~eκ~log1γ1(εεβ)eδtγlog1γκ~.\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}(\eta_{\varepsilon}(t),t)\leqslant-e^{\hat{\eta}_{\varepsilon}(t)-\eta_{\varepsilon}(t)}\frac{1-(\varepsilon\vee\varepsilon_{\beta})e^{\delta t}-\gamma}{\log\frac{1}{\gamma}-\tilde{\kappa}^{*}}\leqslant-e^{\tilde{\kappa}^{*}-\log\frac{1}{\gamma}}\frac{1-(\varepsilon\vee\varepsilon_{\beta})e^{\delta t}-\gamma}{\log\frac{1}{\gamma}-\tilde{\kappa}^{*}}.

This completes the proof. ∎

From Theorem 4.12 and Lemma 4.13, we see that the sequence (ηε)ε>0(\eta_{\varepsilon})_{\varepsilon>0} is bounded in C1([0,T])C^{1}([0,T]), therefore, extracting a subsequence if necessary, it converges uniformly to a function η^(t)\hat{\eta}(t).

Corollary 4.14.

Extracting a subsequence if necessary, the sequence ηε\eta_{\varepsilon} converges uniformly to a limit η^(t)\hat{\eta}(t).

5 Main Results

5.1 Existence and Uniqueness

Lemmas 4.4-4.7 provide estimates on the approximated solution vεv_{\varepsilon}. By taking a limit as ε0\varepsilon\to 0, we are able to derive the existence of a solution to (2.9)-(2.10).

Theorem 5.1.

(i) For any T>0T>0, there exists a sequence ε0\varepsilon\to 0 such that vεvv_{\varepsilon}\to v a.e. in ×[0,T]\mathbb{R}\times[0,T], vεξvξ\frac{\partial v_{\varepsilon}}{\partial\xi}\to\frac{\partial v}{\partial\xi} a.e. in ×[0,T]\mathbb{R}\times[0,T], vεvv_{\varepsilon}\to v in W1,0(×[0,T])W^{1,0}_{\infty}(\mathbb{R}\times[0,T]) weak-* and W2,1((×[0,T])Q¯ρ)W^{2,1}_{\infty}((\mathbb{R}\times[0,T])\setminus\overline{Q}_{\rho}) weak-*, for any ρ>0\rho>0, where Qρ=(ρ,ρ)×(0,ρ2)Q_{\rho}=(-\rho,\rho)\times(0,\rho^{2}). Moreover, extracting a subsequence if necessary, ηε\eta_{\varepsilon} converges uniformly to η^\hat{\eta};555For Ω×[0,T]\Omega\subset{\mathbb{R}}\times[0,T], Wp2,1(Ω)W^{2,1}_{p}(\Omega), 1<p<1<p<\infty, is the space of elements of Lp(Ω)L^{p}(\Omega) whose derivatives are also in Lp(Ω)L^{p}(\Omega), respectively up to second order in ξ\xi and to first order in tt. W2,1(Ω)W^{2,1}_{\infty}(\Omega) is the space of bounded functions whose derivatives are bounded, respectively up to second order in ξ\xi and first order in tt. W1,0(Ω)W^{1,0}_{\infty}(\Omega) denotes the space of bounded functions whose derivative w.r.t. ξ\xi is also bounded.
(ii) vv is a solution of the original free boundary problem (2.9);
(iii) vv satisfies the estimates of Lemmas 4.4-4.7, and the inequality

2vξ2+vξ0a.e. in×[0,T])Q¯ρ,\frac{{\partial}^{2}v}{{\partial}\xi^{2}}+\frac{{\partial}v}{{\partial}\xi}\leqslant 0\;\,\mbox{a.e. in}\;\,\mathbb{R}\times[0,T])\setminus\overline{Q}_{\rho}, (5.1)

as well as the following growth condition: there exists a constant B>0B>0 such that v(ξ,t)=1v(\xi,t)=1 when ξ<B\xi<-B and v(ξ,t)eξv(\xi,t)\leqslant e^{-\xi} when ξ>B\xi>B, 0tT0\leqslant t\leqslant T.

Proof.

Let (εn)n1(\varepsilon_{n})_{n\geqslant 1} be a sequence converging to 0 when n+n\to+\infty and consider the corresponding solutions (vεn)(v_{\varepsilon_{n}}) of (4.2) and (4.4). For simplicity, we denote vεnv_{\varepsilon_{n}} by vnv_{n}. According to Lemmas 4.4-4.10, we first observe that the sequence (vn)(v_{n}) is bounded in the spaces W1,0(×[0,T])W2,1((×[0,T])Q¯ρ)W^{1,0}_{\infty}(\mathbb{R}\times[0,T])\cap W^{2,1}_{\infty}((\mathbb{R}\times[0,T])\setminus\bar{Q}_{\rho}). Second, the sequence (vn)(v_{n}) is bounded in the space Wp,loc2,1(×[0,T])W^{2,1}_{p,\mbox{\tiny{loc}}}(\mathbb{R}\times[0,T]), 1<p<21<p<2.

Next, let (Am)m1(A_{m})_{m\geqslant 1} be a sequence of positive numbers such that Am+A_{m}\to+\infty as m+m\to+\infty. Let us consider the restriction vnmv_{n}^{m} of vnv_{n} to the interval [Am,Am][-A_{m},A_{m}]. At fixed m1m\geqslant 1, the sequence (vnm)(v_{n}^{m}) is bounded in the space Wp2,1([Am,Am])×[0,T])W^{2,1}_{p}([-A_{m},A_{m}])\times[0,T]) for any 1<p<21<p<2. One can extract a subsequence, denoted by (vnjm)(v_{n_{j}}^{m}), which converges a.e. in [Am,Am]×[0,T][-A_{m},A_{m}]\times[0,T] and weakly in Wp2,1([Am,Am]×[0,T])W^{2,1}_{p}([-A_{m},A_{m}]\times[0,T]), 1<p<21<p<2, as j+j\to+\infty. By a standard diagonal extraction procedure, one can eventually extract a subsequence, say (vnk)(v_{n_{k}}), such that vnkv_{n_{k}} and ξvnk\frac{\partial}{\partial\xi}v_{n_{k}} converge respectively to vv and ξv\frac{\partial}{\partial\xi}v almost everywhere in ×[0,T]\mathbb{R}\times[0,T] as k+k\to+\infty. After a new extraction, vnkvv_{n_{k}}\to v in W1,0(×[0,T])W^{1,0}_{\infty}(\mathbb{R}\times[0,T]) weak-* and W2,1((×[0,T])Q¯ρ)W^{2,1}_{\infty}((\mathbb{R}\times[0,T])\setminus\overline{Q}_{\rho}) weak-*.

It is not difficult to see that vv satisfies the properties of Lemmas 4.4-4.7. Set fε=2vεξ2+vεξf_{\varepsilon}=\frac{{\partial}^{2}v_{\varepsilon}}{{\partial}\xi^{2}}+\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}, fε0×[0,T]f_{\varepsilon}\leqslant 0\in\mathbb{R}\times[0,T] (see Lemma 4.7). According to the above results, fn′′f=2vξ2+vξf_{n^{\prime\prime}}\to f=\frac{{\partial}^{2}v}{{\partial}\xi^{2}}+\frac{{\partial}v}{{\partial}\xi} in L((×[0,T])Q¯ρ)L^{\infty}((\mathbb{R}\times[0,T])\setminus\overline{Q}_{\rho}) weak-* which is non-negative in the distribution sense, and, hence, (5.1) holds.

Since the sequence ηn′′\eta_{n^{\prime\prime}} is bounded in C1([0,T])C^{1}([0,T]) (see Lemma 4.13), a subsequence converges to some η~\tilde{\eta} in C0([0,T])C^{0}([0,T]). More specifically, in the proof of Lemma 4.13, we showed that vεξ(ηε(t),t)c\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}(\eta_{\varepsilon}(t),t)\leqslant-c, where the constant cc is independent of ε\varepsilon. With the estimate of 2vεξ2\frac{{\partial}^{2}v_{\varepsilon}}{{\partial}\xi^{2}} in Lemma 4.10, we deduce that, for any t>0t>0, there exists a small constant Υ\Upsilon independent of ε\varepsilon such that, for x<Υx<\Upsilon,

vn′′(ηn′′(t)+x,t)vn′′(ηn′′(t),t)c2x,v_{n^{\prime\prime}}(\eta_{n^{\prime\prime}}(t)+x,t)-v_{n^{\prime\prime}}(\eta_{n^{\prime\prime}}(t),t)\leqslant-\frac{c}{2}x,

and

vn′′(ηn′′(t)x,t)vn′′(ηn′′(t),t)c2x.v_{n^{\prime\prime}}(\eta_{n^{\prime\prime}}(t)-x,t)-v_{n^{\prime\prime}}(\eta_{n^{\prime\prime}}(t),t)\geqslant\frac{c}{2}x.

Taking the limit as n′′n^{\prime\prime}\to\infty and combining with v(ξ,)v(\xi,\cdot) non-increasing in ξ\xi, we see that v<γv<\gamma if ξ>η~(t)\xi>\tilde{\eta}(t) and v>γv>\gamma if ξ<η~(t)\xi<\tilde{\eta}(t). This yields that η~=η^\tilde{\eta}=\hat{\eta} (see Corollary 4.14).

Moreover, the convergence of ηn′′\eta_{n^{\prime\prime}} to η^\hat{\eta} implies the almost everywhere convergence of σn′′(vn′′)\sigma_{n^{\prime\prime}}(v_{n^{\prime\prime}}) to σ(v)\sigma(v). Hence, we have that n′′[vn′′]\mathcal{L}_{n^{\prime\prime}}[v_{n^{\prime\prime}}] converges to [v]\mathcal{L}[v] in L((×[0,T])Q¯ρ)L^{\infty}((\mathbb{R}\times[0,T])\setminus\overline{Q}_{\rho}) weak-*. This implies that [v]0\mathcal{L}[v]\geqslant 0. It is also easy to verify that [v]=0\mathcal{L}[v]=0 whenever v<1v<1. Thus, vv is a solution to (2.9).

Finally, let us check the growth condition as ξ±\xi\to\pm\infty: according to Lemmas 4.4 and 4.11, vεmin(1,eξ)v_{\varepsilon}\leqslant\min(1,e^{-\xi}) and vεK~(εεβ)eδtv_{\varepsilon}\geqslant\tilde{K}-(\varepsilon\vee\varepsilon_{\beta})e^{\delta t}, respectively. At the limit ε0\varepsilon\to 0, it holds almost everywhere v(ξ,t)min(1,eξ)v(\xi,t)\leqslant\min(1,e^{-\xi}) and v(ξ,t)K~v(\xi,t)\geqslant\tilde{K}, <ξ<+,0tT-\infty<\xi<+\infty,0\leqslant t\leqslant T. Therefore, v(ξ,t)=1v(\xi,t)=1 when ξκ~\xi\leqslant\tilde{\kappa}^{*} and v(ξ,t)eξv(\xi,t)\leqslant e^{-\xi} when ξ0\xi\geqslant 0. ∎

Then, the uniqueness of vv given by Theorem 5.1 is a direct consequence of the following theorem:

Theorem 5.2.

Let vi{ρ>0W2,1((×[0,T])Q¯ρ)}W1,0(×[0,T])v_{i}\in\Big{\{}\bigcap_{\rho>0}W^{2,1}_{\infty}((\mathbb{R}\times[0,T])\setminus\bar{Q}_{\rho})\Big{\}}\cap W^{1,0}_{\infty}(\mathbb{R}\times[0,T]) be a solution to (2.9) satisfying

2viξ2+viξ0,\frac{{\partial}^{2}v_{i}}{{\partial}\xi^{2}}+\frac{{\partial}v_{i}}{{\partial}\xi}\leqslant 0,

for i=1,2i=1,2. Suppose that there exists Bi>0B_{i}>0 such that vi=1v_{i}=1 for ξ<Bi\xi<-B_{i} and vieξv_{i}\leqslant e^{-\xi} for ξ>Bi\xi>B_{i}, i=1,2i=1,2. Then, it holds v1=v2v_{1}=v_{2}.

Proof.

Denote by F=12(σ2(v1)σ2(v2))(2v1ξ2+v1ξ)F=\frac{1}{2}\left(\sigma^{2}(v_{1})-\sigma^{2}(v_{2})\right)\left(\frac{{\partial}^{2}v_{1}}{{\partial}\xi^{2}}+\frac{{\partial}v_{1}}{{\partial}\xi}\right) and 2=t+12σ2(v2)(2ξ2+ξ)+δ(ξ+1)\mathcal{L}_{2}=-\frac{\partial}{\partial t}+\frac{1}{2}\sigma^{2}(v_{2})\Big{(}\frac{\partial^{2}}{\partial\xi^{2}}+\frac{\partial}{\partial\xi}\Big{)}+\delta\Big{(}\frac{\partial}{\partial\xi}+1\Big{)}. We rewrite that

min{2[v1]+F,1v1}=0.\min\{\mathcal{L}_{2}[v_{1}]+F,1-v_{1}\}=0.

Let w=e2δt(v1v2)w=e^{-2\delta t}\left(v_{1}-v_{2}\right). We prove that w0w\geqslant 0. Due to the growth condition on v1v_{1} and v2v_{2}, it holds that limξ±w(ξ,t)=0\lim_{\xi\rightarrow\pm\infty}w(\xi,t)=0 for t[0,T]t\in[0,T]. Therefore if this conclusion is not true, ww will achieve a negative minimum at some point (ξ,t)(\xi^{*},t^{*}). By the parabolic version of Bony’s maximum principle, it holds that

lim sup(ξ,t)(ξ,t)ess{wt12σ2(v2)2wξ2(12σ2(v2)+δ)wξ}0\limsup_{(\xi,t)\rightarrow(\xi^{*},t^{*})}\mathop{ess}\left\{\frac{{\partial}w}{{\partial}t}-\frac{1}{2}\sigma^{2}(v_{2})\frac{{\partial}^{2}w}{{\partial}\xi^{2}}-\left(\frac{1}{2}\sigma^{2}(v_{2})+\delta\right)\frac{{\partial}w}{{\partial}\xi}\right\}\leqslant 0

This is equivalent to

lim sup(ξ,t)(ξ,t)ess{2[v1v2]}δ(v1v2)>0.\limsup_{(\xi,t)\rightarrow(\xi^{*},t^{*})}\mathop{ess}\left\{\mathcal{L}_{2}[v_{1}-v_{2}]\right\}\geqslant-\delta(v_{1}-v_{2})>0.

By the continuity of viv_{i}, we derive σ(v1)σ(v2)\sigma(v_{1})\leqslant\sigma(v_{2}) in a small parabolic neighborhood of (ξ,t)(\xi^{*},t^{*}). It follows that

lim sup(ξ,t)(ξ,t)F(ξ,t)0.\limsup_{(\xi,t)\rightarrow(\xi^{*},t^{*})}F(\xi,t)\geqslant 0.

In this neighborhood, we also have that

2[v1]+F=0 and 2[v2]0,a.e..\mathcal{L}_{2}[v_{1}]+F=0\text{ and }\mathcal{L}_{2}[v_{2}]\geqslant 0,\text{a.e..}

Therefore,

lim sup(ξ,t)(ξ,t)ess{2[v1v2]}lim sup(ξ,t)(ξ,t)F(ξ,t)0,\limsup_{(\xi,t)\rightarrow(\xi^{*},t^{*})}\mathop{ess}\left\{\mathcal{L}_{2}[v_{1}-v_{2}]\right\}\leqslant\limsup_{(\xi,t)\rightarrow(\xi^{*},t^{*})}-F(\xi,t)\leqslant 0,

which is a contradiction. Thus, we proved that w0w\geqslant 0. Similarly, the reverse inequality holds, which yields the uniqueness result. ∎

5.2 Properties of the free boundaries

For the original problem (2.9), we already introduced formally the default boundary κ^\hat{\kappa} and the transit boundary η^\hat{\eta}, see System (2.10). The goal of this subsection is to to define the free boundaries rigorously and prove some basic properties.

The default boundary

Let us remind that vεK~(εεβ)eδtv_{\varepsilon}\geqslant\tilde{K}-(\varepsilon\vee\varepsilon_{\beta})e^{\delta t}, see Lemma 4.11. Taking the limit as ε0\varepsilon\to 0, this implies that vK~v\geqslant\tilde{K}. Since K~=1\tilde{K}=1 for ξκ~\xi\leqslant\tilde{\kappa}^{*}, it holds that the set {ξ|v(ξ,t)<1}\{\xi\,|\,v(\xi,t)<1\} is bounded from below. Now, we are in position to define

κ^(t):=inf{ξ|v(ξ,t)<1}.\displaystyle\hat{\kappa}(t):=\inf\{\xi\,|\,v(\xi,t)<1\}. (5.2)

Then, veξv\leqslant e^{-\xi} indicates that κ^(t)\hat{\kappa}(t) is also bounded from above. Thus, we will have the following result.

Theorem 5.3.

For each t(0,T]t\in(0,T], κ^(t)\hat{\kappa}(t) is well-defined, i.e. we have <κ^(t)<-\infty<\hat{\kappa}(t)<\infty. Moreover, v(ξ,t)=1v(\xi,t)=1 for ξκ^(t)\xi\leqslant\hat{\kappa}(t) and v(ξ,t)<1v(\xi,t)<1 whenever ξ>κ^(t)\xi>\hat{\kappa}(t).

The transit boundary

We remind that η^C0([0,T])\hat{\eta}\in C^{0}([0,T]) is the limit of ηε\eta_{\varepsilon} (see Theorem 5.1). Thus, we will have the following theorem.

Theorem 5.4.

The initial positions of the free boundaries are as follows:

η^(0)=log1γ,κ^(0)=0.\hat{\eta}(0)=\log\frac{1}{\gamma},\quad\hat{\kappa}(0)=0.

Furthermore, κ^(t)\hat{\kappa}(t) and η^(t)\hat{\eta}(t) are non-increasing with respect to tt.

Proof.

On the one hand, we know that ηε(0)=logγ\eta_{\varepsilon}(0)=-\log\gamma and ηε(t)\eta_{\varepsilon}(t) decreasing, see Section 4.2. On the other hand, the properties of κ^(t)\hat{\kappa}(t) follow from Theorem 5.3 and the initial value of vv. ∎

In the following, we will prove the smoothness of the free boundaries. Note that the uniform lower bound in Lemma 4.5 implies that there exists a constant cc such that vε(ξ,t)1+γ2v_{\varepsilon}(\xi,t)\leqslant\frac{1+\gamma}{2} whenever ηε(t)ξc\eta_{\varepsilon}(t)-\xi\leqslant c. Then, one can choose a smooth function ζ\zeta such that ζ(t)<ηε(t)\zeta(t)<\eta_{\varepsilon}(t) and ζηεL[0,T][c/4,c/2]\|\zeta-\eta_{\varepsilon}\|_{L^{\infty}[0,T]}\in[{c}/{4},{c}/{2}] for sufficiently small ε\varepsilon. Therefore, ζ\zeta separates the default boundary κ^\hat{\kappa} and the transit boundary η^\hat{\eta}. So, we can discuss them one by one with cut-off functions being applied when necessary.

We first study the default boundary. The proof is essentially the same as that in [42], where the authors proved the smoothness of free boundary in American option problem. Thus, we just give a sketch of the proof for readers’ convenience. We make the change of variable ξ=ζ(t)+x\xi=\zeta(t)+x and set u(x,t)=v(ζ(t)+x,t)u(x,t)=v(\zeta(t)+x,t). For suitable a,ba,b\in\mathbb{R}, we have that ζ(t)+aκ^(t)ζ(t)+b<η^(t)\zeta(t)+a\leqslant\hat{\kappa}(t)\leqslant\zeta(t)+b<\hat{\eta}(t). It holds that

utL(t1,t2;H1(a,b)),2ut2L2(t1,t2;L2(a,b)),\frac{{\partial}u}{{\partial}t}\in L^{\infty}(t_{1},t_{2};H^{1}(a,b)),\frac{{\partial}^{2}u}{{\partial}t^{2}}\in L^{2}(t_{1},t_{2};L^{2}(a,b)),

which implies the continuity of vtv_{t} at ξ=κ^(t)\xi=\hat{\kappa}(t). From the definition of κ^\hat{\kappa}, one can prove that κ^\hat{\kappa} is continuous in (0,T](0,T]. Applying a result from Cannon et al. [7], we will have that κ^C1((0,T])\hat{\kappa}\in C^{1}((0,T]). Then, we may use the theory of parabolic equations to improve the regularity of v(ξ,τ)v(\xi,\tau) by bootstrapping. Repeating the procedure yields the following result.

Theorem 5.5.

κ^C((0,T])\hat{\kappa}\in C^{\infty}((0,T]).

Next, we consider the smoothness of the transit free boundary η^(t)\hat{\eta}(t). For this purpose, we need the following lemma of the parabolic diffraction problem. The proof is essentially similar to that in [29]; hence, we just give a sketch.

Lemma 5.6.

In the domain Q={a<x<b, 0<t<T}Q=\{a<x<b,\ 0<t<T\}, where a<ba<b are some constants, consider the following initial boundary problem

{ut(Kf(ux+u))x+f1(x,t)ux+f2(x,t)u=0,u(a,t)=ga(t),u(b,t)=gb(t),u(x,0)=ϕ(x),ga(0)=ϕ(a),gb(0)=ϕ(b),\left\{\begin{split}&u_{t}-(K_{f}(u_{x}+u))_{x}+f_{1}(x,t)u_{x}+f_{2}(x,t)u=0,\\ &u(a,t)=g_{a}(t),\ u(b,t)=g_{b}(t),\quad u(x,0)=\phi(x),\\ &g_{a}(0)=\phi(a),\quad g_{b}(0)=\phi(b),\end{split}\right. (5.3)

where ga,gbC2[0,T]g_{a},g_{b}\in C^{2}[0,T], Kf(ϕx+ϕ)(x)C1[a,b],K_{f}(\phi_{x}+\phi)(x)\in C^{1}[a,b], fi(x,t)C([a,b]×[0,T]),f_{i}(x,t)\in C([a,b]\times[0,T]), i=1,2,i=1,2, Kf={μ1, if x>f(t),μ2, if x<f(t),K_{f}=\left\{\begin{array}[]{ll}\mu_{1},&\mbox{ if }x>f(t),\\ \mu_{2},&\mbox{ if }x<f(t),\end{array}\right., a<f(t)<ba<f(t)<b, for t[0,T]t\in[0,T], f(t)C0,1(0,T)f(t)\in C^{0,1}(0,T),a<f(t)<ba<f(t)<b, for t[0,T]t\in[0,T], and μ1,μ2\mu_{1},\mu_{2} are positive constants. Then, the problem (5.3) admits a solution, and

u(f(t),t)=u(f(t)+,t),μ2(u+ux)(f(t),t)=μ1(u+ux)(f(t)+,t).u(f(t)-,t)=u(f(t)+,t),\quad\mu_{2}(u+u_{x})(f(t)-,t)=\mu_{1}(u+u_{x})(f(t)+,t).

Moreover, there exists a positive constant CC and 0<α<10<\alpha<1 depend only on the given data such that

Kf(u+ux)Cα(Q)C.\|K_{f}(u+u_{x})\|_{C^{\alpha}(Q)}\leqslant C.
Proof.

Make the transformation y=xf(t)y=x-f(t), v=ueyv=ue^{y}, then problem (5.3) satisfies

{vt(K0(vy))yf(t)vy+f1vy+(f2f1)v=0,v(af,t)=ga(t)eaf,v(bf,t)=gb(t)ebf,v(y,0)=ϕ(y+f)ey.\left\{\begin{split}&v_{t}-(K_{0}(v_{y}))_{y}-f^{\prime}(t)v_{y}+f_{1}v_{y}+(f_{2}-f_{1})v=0,\\ &v(a-f,t)=g_{a}(t)e^{a-f},\ v(b-f,t)=g_{b}(t)e^{b-f},\quad v(y,0)=\phi(y+f)e^{y}.\end{split}\right. (5.4)

where K0=μ1K_{0}=\mu_{1} if y>0y>0,  μ2\mu_{2} if y<0y<0. By well-known estimates for linear parabolic PDEs with discontinuous coefficients whose principal part is in divergence form (see [25, Chapter III, 5]), and the proof of [29, Theorem 1.1], the claim of this lemma follows. ∎

Now, we are in position to prove the smoothness of η^\hat{\eta}.

Theorem 5.7.

η^C((0,T])\hat{\eta}\in C^{\infty}((0,T]).

Proof.

In a neighborhood of η^\hat{\eta}, vv satisfies the system

{vt+12σH2(2vξ2+vξ)+δ(vξ+v)=0,ξ>η^(t),vt+12σL2(2vξ2+vξ)+δ(vξ+v)=0,ξ<η^(t),v(η^(t)+,t)=v(η^(t),t)=γ,vξ(η^(t)+,t)=vξ(η^(t),t).\left\{\begin{split}&-\frac{{\partial}v}{{\partial}t}+\frac{1}{2}{\sigma}^{2}_{H}\Big{(}\frac{{\partial}^{2}v}{{\partial}\xi^{2}}+\frac{{\partial}v}{{\partial}\xi}\Big{)}+\delta\Big{(}\frac{{\partial}v}{{\partial}\xi}+v\Big{)}=0,\quad\xi>\hat{\eta}(t),\\ &-\frac{{\partial}v}{{\partial}t}+\frac{1}{2}{\sigma}^{2}_{L}\Big{(}\frac{{\partial}^{2}v}{{\partial}\xi^{2}}+\frac{{\partial}v}{{\partial}\xi}\Big{)}+\delta\Big{(}\frac{{\partial}v}{{\partial}\xi}+v\Big{)}=0,\quad\xi<\hat{\eta}(t),\\ &v(\hat{\eta}(t)+,t)=v(\hat{\eta}(t)-,t)=\gamma,\quad v_{\xi}(\hat{\eta}(t)+,t)=v_{\xi}(\hat{\eta}(t)-,t).\end{split}\right. (5.5)

Thus, it holds that

η^(t)=vt(η^(t)+,t)vξ(η^(t)+,t)=vt(η^(t),t)vξ(η^(t),t),\hat{\eta}^{\prime}(t)=-\frac{v_{t}(\hat{\eta}(t)+,t)}{v_{\xi}(\hat{\eta}(t)+,t)}=-\frac{v_{t}(\hat{\eta}(t)-,t)}{v_{\xi}(\hat{\eta}(t)-,t)}, (5.6)

which means that

vt(η^(t)+,t)=vt(η^(t),t).v_{t}(\hat{\eta}(t)+,t)=v_{t}(\hat{\eta}(t)-,t). (5.7)

Set w=vξw=v_{\xi}. From (5.5) and (5.7), it turns out that ww verifies the system

{wt+12σH2(2wξ2+wξ)+δ(wξ+w)=0,ξ>η^(t),wt+12σL2(2wξ2+wξ)+δ(wξ+w)=0,ξ<η^(t),w(η^(t)+,t)=w(η^(t),t),σL2(wx+w)(η^(t)+,t)=σH2(wx+w)(η^(t),t).\left\{\begin{split}&-\frac{{\partial}w}{{\partial}t}+\frac{1}{2}{\sigma}^{2}_{H}\Big{(}\frac{{\partial}^{2}w}{{\partial}\xi^{2}}+\frac{{\partial}w}{{\partial}\xi}\Big{)}+\delta\Big{(}\frac{{\partial}w}{{\partial}\xi}+w\Big{)}=0,\quad\xi>\hat{\eta}(t),\\ &-\frac{{\partial}w}{{\partial}t}+\frac{1}{2}{\sigma}^{2}_{L}\Big{(}\frac{{\partial}^{2}w}{{\partial}\xi^{2}}+\frac{{\partial}w}{{\partial}\xi}\Big{)}+\delta\Big{(}\frac{{\partial}w}{{\partial}\xi}+w\Big{)}=0,\quad\xi<\hat{\eta}(t),\\ &w(\hat{\eta}(t)+,t)=w(\hat{\eta}(t)-,t),\quad\sigma^{2}_{L}(w_{x}+w)(\hat{\eta}(t)+,t)=\sigma^{2}_{H}(w_{x}+w)(\hat{\eta}(t)-,t).\end{split}\right. (5.8)

According to the free boundary condition, ww satisfies a typical Verigin problem, see [29, 38]. In particular, the CC^{\infty} regularity of the free boundary was proved in [29]. Therefore, we may obtain the same result for our problem in a similar manner. To see this, note that the free boundary η^\hat{\eta} is Lipschitz continuous and satisfies

η^(t)\displaystyle\hat{\eta}^{\prime}(t) =\displaystyle= 12σH2(wξ(η^(t)+,t)+w(η^(t)+,t))+δ(w(η^(t)+,t)+γ)w(η^(t)+,t)\displaystyle-\frac{\frac{1}{2}{\sigma}^{2}_{H}\Big{(}w_{\xi}(\hat{\eta}(t)+,t)+w(\hat{\eta}(t)+,t)\Big{)}+\delta\Big{(}w(\hat{\eta}(t)+,t)+\gamma\Big{)}}{w(\hat{\eta}(t)+,t)} (5.9)
=\displaystyle= 12σL2(wξ(η^(t),t)+w(η^(t),t))+δ(w(η^(t),t)+γ)w(η^(t),t),\displaystyle-\frac{\frac{1}{2}{\sigma}^{2}_{L}\Big{(}w_{\xi}(\hat{\eta}(t)-,t)+w(\hat{\eta}(t)-,t)\Big{)}+\delta\Big{(}w(\hat{\eta}(t)-,t)+\gamma\Big{)}}{w(\hat{\eta}(t)-,t)},

which is a kind of Stefan condition, see [23, 24, 14] for references. Applying Lemma 5.6 to problem (5.8) (up to some simple transformation), wξ+wCαw_{\xi}+w\in C^{\alpha} up to the free boundary. Furthermore, by Lemma 4.6, ww has a negative upperbound. Then, the right hand side of (5.9) belongs to CαC^{\alpha}. This implies in turn η^C1+α\hat{\eta}\in C^{1+\alpha}. In this way, by an iteration process, one can further improve the regularity of η^\hat{\eta} and shows eventually that it belongs to C.C^{\infty}.

6 Asymptotic Convergence

In this section, we will prove that vv converges to the traveling wave solution as tt goes to ++\infty. Since vt\frac{{\partial}v}{{\partial}t} is non-positive, we see that, for any tt,

00tvt(ξ,s)𝑑s=v(ξ,t)v(ξ,0)K~(ξ)v(ξ,0).0\geqslant\int_{0}^{t}\frac{{\partial}v}{{\partial}t}(\xi,s)ds=v(\xi,t)-v(\xi,0)\geqslant\tilde{K}(\xi)-v(\xi,0).

Note that for ξ<κ~\xi<\tilde{\kappa}^{*}, v(ξ,0)=K~(ξ)=1v(\xi,0)=\tilde{K}(\xi)=1 and K~(ξ),v(ξ,0)eξ\tilde{K}(\xi),v(\xi,0)\leqslant e^{-\xi} which implies the integrability of K~v(,0)\tilde{K}-v(\cdot,0) over \mathbb{R}. Thus, we have that

00tvt(ξ,s)𝑑s𝑑ξ(K~(ξ)v(ξ,0))𝑑ξ.0\geqslant\int_{-\infty}^{\infty}\int_{0}^{t}\frac{{\partial}v}{{\partial}t}(\xi,s)dsd\xi\geqslant\int_{-\infty}^{\infty}(\tilde{K}(\xi)-v(\xi,0))d\xi.

Letting tt tend to infinity, we get that there exists a constant C>0C>0 such that

00vt(ξ,s)𝑑s𝑑ξC.0\geqslant\int_{-\infty}^{\infty}\int_{0}^{\infty}\frac{{\partial}v}{{\partial}t}(\xi,s)dsd\xi\geqslant-C. (6.1)

Now let vn(ξ,t):=v(ξ,t+n)v^{n}(\xi,t):=v(\xi,t+n) and consider vnv^{n} as a sequence of functions defined on ×[0,1]\mathbb{R}\times[0,1]. Lemmas 4.4-4.10 indicate that it is a bounded sequence in W2,1(×[0,1])W^{2,1}_{\infty}(\mathbb{R}\times[0,1]). As in the proof of Theorem 5.1, via a standard diagonal extraction procedure there exists a function K¯\bar{K} and a subsequence njn_{j} such that such that vnjv_{n_{j}} and ξvnj\frac{\partial}{\partial\xi}v_{n_{j}} converge respectively to K¯\bar{K} and ξK¯\frac{\partial}{\partial\xi}\bar{K} almost everywhere in ×[0,1]\mathbb{R}\times[0,1]. After a new extraction if necessary,

vnjtK¯t,2vnjξ22K¯ξ2 in L(×[0,1])weak-,\frac{{\partial}v^{n_{j}}}{{\partial}t}\to\frac{{\partial}\bar{K}}{{\partial}t},\quad\frac{{\partial}^{2}v^{n_{j}}}{{\partial}\xi^{2}}\to\frac{{\partial}^{2}\bar{K}}{{\partial}\xi^{2}}\;\text{ in }L^{\infty}(\mathbb{R}\times[0,1])\;\mbox{weak-$*$},

Since non-positivity is preserved under weak-* convergence and vt0\frac{{\partial}v}{{\partial}t}\leqslant 0, one can deduce that K¯t0\frac{{\partial}\bar{K}}{{\partial}t}\leqslant 0. Since (6.1) implies that 01vnj(ξ,t)𝑑ξ𝑑t=njnj+1v(ξ,t)𝑑ξ𝑑t0\int_{0}^{1}\int_{-\infty}^{\infty}v^{n_{j}}(\xi,t)d\xi dt=\int_{n_{j}}^{n_{j}+1}\int_{-\infty}^{\infty}v(\xi,t)d\xi dt\rightarrow 0 as nj0n_{j}\rightarrow 0, we have that 01K¯t𝑑ξ𝑑t=0\int_{0}^{1}\int_{-\infty}^{\infty}\frac{{\partial}\bar{K}}{{\partial}t}d\xi dt=0. Combining with the non-positivity of K¯t\frac{{\partial}\bar{K}}{{\partial}t}, it follows that K¯t0\frac{{\partial}\bar{K}}{{\partial}t}\equiv 0 which means that K¯\bar{K} is only a function of ξ\xi. Then, the following properties pass from vv to K¯\bar{K},

K~K¯min{1,eξ},dK¯dξ0, and d2K¯dξ2+dK¯dξ0.\tilde{K}\leqslant\bar{K}\leqslant\min\{1,e^{-\xi}\},\;\,\frac{d\bar{K}}{d\xi}\leqslant 0,\,\text{ and }\,\frac{d^{2}\bar{K}}{d\xi^{2}}+\frac{d\bar{K}}{d\xi}\leqslant 0.

Since η^()\hat{\eta}(\cdot) and κ^()\hat{\kappa}(\cdot) are also non-increasing with respect to tt, they also admit limits at \infty, which are denoted as η¯\bar{\eta} and κ¯\bar{\kappa} respectively. Then, one can verify that K¯(η¯)=γ\bar{K}(\bar{\eta})=\gamma and K¯(κ¯)=1\bar{K}(\bar{\kappa})=1. For any interval II such that I¯(κ¯,η¯)\bar{I}\subset(\bar{\kappa},\bar{\eta}), there exists TT such that I¯(κ^(t),η^(t))\bar{I}\subset(\hat{\kappa}(t),\hat{\eta}(t)) for any t>Tt>T. In II, it holds that

vnt+12σL2(2vnξ2+vnξ)+δ(vnξ+vn)=0.-\frac{{\partial}v^{n}}{{\partial}t}+\frac{1}{2}\sigma_{L}^{2}\left(\frac{{\partial}^{2}v^{n}}{{\partial}\xi^{2}}+\frac{{\partial}v^{n}}{{\partial}\xi}\right)+\delta\left(\frac{{\partial}v^{n}}{{\partial}\xi}+v^{n}\right)=0.

Taking subsequence njn_{j}, we derive that

12σL2(d2K¯dξ2+dK¯dξ)+δ(dK¯dξ+K¯)=0, for ξI.\frac{1}{2}\sigma_{L}^{2}\left(\frac{d^{2}\bar{K}}{d\xi^{2}}+\frac{d\bar{K}}{d\xi}\right)+\delta\left(\frac{d\bar{K}}{d\xi}+\bar{K}\right)=0,\text{ for }\xi\in I.

Since II is arbitrary, it holds that

12σL2(d2K¯dξ2+dK¯dξ)+δ(dK¯dξ+K¯)=0, for κ¯<ξ<η¯.\frac{1}{2}\sigma_{L}^{2}\left(\frac{d^{2}\bar{K}}{d\xi^{2}}+\frac{d\bar{K}}{d\xi}\right)+\delta\left(\frac{d\bar{K}}{d\xi}+\bar{K}\right)=0,\text{ for }\bar{\kappa}<\xi<\bar{\eta}.

Similarly, we can also show that

12σH2(d2K¯dξ2+dK¯dξ)+δ(dK¯dξ+K¯)=0, for ξ>η¯.\frac{1}{2}\sigma_{H}^{2}\left(\frac{d^{2}\bar{K}}{d\xi^{2}}+\frac{d\bar{K}}{d\xi}\right)+\delta\left(\frac{d\bar{K}}{d\xi}+\bar{K}\right)=0,\text{ for }\xi>\bar{\eta}.

Note that K~K¯min{1,eξ}\tilde{K}\leqslant\bar{K}\leqslant\min\{1,e^{-\xi}\} implies that limξeξK¯(ξ)=1\lim_{\xi\rightarrow\infty}e^{\xi}\bar{K}(\xi)=1. Combining with the fact that K¯C1+α\bar{K}\in C^{1+\alpha}, we see that it is a solution to (3.2), i.e.

{d2K¯dξ2+dK¯dξ+cH(dK¯dξ+K)=0,ξ>η¯,d2K¯dξ2+dK¯dξ+cL(dK¯dξ+K)=0,κ¯<ξ<η¯,K¯(κ¯)=1,dK¯dξ(κ¯)=0,K¯(η¯)=K¯(η¯)=γ,dK¯dξ(η¯+)=dK¯dξ(η¯),limξeξK¯(ξ)=1.\left\{\begin{split}&\frac{d^{2}\bar{K}}{d\xi^{2}}+\frac{d\bar{K}}{d\xi}+c_{H}(\frac{d\bar{K}}{d\xi}+K)=0,\xi>\bar{\eta},\\ &\frac{d^{2}\bar{K}}{d\xi^{2}}+\frac{d\bar{K}}{d\xi}+c_{L}(\frac{d\bar{K}}{d\xi}+K)=0,\bar{\kappa}<\xi<\bar{\eta},\\ &\bar{K}(\bar{\kappa})=1,\frac{d\bar{K}}{d\xi}(\bar{\kappa})=0,\\ &\bar{K}(\bar{\eta})=\bar{K}(\bar{\eta}^{*}-)=\gamma,\frac{d\bar{K}}{d\xi}(\bar{\eta}+)=\frac{d\bar{K}}{d\xi}(\bar{\eta}-),\\ &\lim_{\xi\rightarrow\infty}e^{\xi}\bar{K}(\xi)=1.\end{split}\right.

Then, interior estimate implies that K¯\bar{K} is smooth in (κ¯,η¯)(\bar{\kappa},\bar{\eta}) and (η¯,)(\bar{\eta},\infty). Now, from the uniqueness of the solution, we derive that K¯=K\bar{K}=K. Since any sub-sequential limit must be same, the full sequence must converge as nn goes to \infty. We have proved the local convergence of vv. But, noting that v(ξ,t)1v(\xi,t)\equiv 1 for ξ<κ~\xi<\tilde{\kappa}^{*} and v(ξ,t)eξv(\xi,t)\leqslant e^{-\xi}, the convergence is also uniform over \mathbb{R}. Finally, we prove the following result.

Theorem 6.1.

As tt goes to ++\infty, v(,t)v(\cdot,t) converges uniformly to KK.

7 Numerical Results

In this section, we will give some numerical results for illustration. As uu represents the value of the bond, we will come back to (2.8) instead of (2.9) which will give us more clear financial meaning.

7.1 Numerical Scheme

As our problem is non-standard, we will introduce the numerical scheme first. To solve the free boundary problem, we use an explicit-implicit finite difference scheme combined with Newton iteration to solve the penalized equation. The first step is to discretize the equation. Let ti=iΔt,i=0,1,,Mt_{i}=i\Delta t,i=0,1,...,M, and ξj=jΔξ,j=0,1,±2,,±N\xi_{j}=j\Delta\xi,j=0,\rm 1,\pm 2,...,\pm N. Ui,jU_{i,j} will be the approximation of the solution uu of (2.8) at mesh point (ti,ξj)(t_{i},\xi_{j}). Consider the approximating penalized equation

{ut+12σε2(u,ξ)(2uξ2uξ)+δuξ=ε1(ueξ)+,ξ[NΔξ,NΔξ],t0;u(ξ,0)=min{1,eξ};u(NΔξ,t)=1,u(NΔξ,t)=0.\left\{\begin{split}&-\frac{{\partial}u}{{\partial}t}+\frac{1}{2}\sigma^{2}_{\varepsilon}(u,\xi)(\frac{{\partial}^{2}u}{{\partial}\xi^{2}}-\frac{{\partial}u}{{\partial}\xi})+\delta\frac{{\partial}u}{{\partial}\xi}=\varepsilon^{-1}(u-e^{\xi})^{+},\quad\xi\in[-N\Delta\xi,N\Delta\xi],t\geqslant 0;\\ &u(\xi,0)=\min\{1,e^{\xi}\};\\ &u(N\Delta\xi,t)=1,u(-N\Delta\xi,t)=0.\end{split}\right.

Here σε(u,ξ)=σH+(σLσH)Hε(uγeξ)\sigma_{\varepsilon}(u,\xi)=\sigma_{H}+(\sigma_{L}-{\sigma}_{H})H_{\varepsilon}(u-\gamma e^{\xi}) with HεH_{\varepsilon} be a proper smooth function. For numerical convenience, we use the penalty function ε1(ueξ)+\varepsilon^{-1}(u-e^{\xi})^{+}. In the numerical experiment, we choose

Hε(z)={0,zε;6ε5z5+15e4z4+10ε3z3+1,ε<z<0;1,z0,H_{\varepsilon}(z)=\left\{\begin{split}&0,z\leqslant-\varepsilon;\\ &6\varepsilon^{-5}z^{5}+15e^{-4}z^{4}+10\varepsilon^{-3}z^{3}+1,-\varepsilon<z<0;\\ &1,z\geqslant 0,\end{split}\right.

as proposed in [28]. Note that the left hand side is a nonlinear operator since coefficients depend on uu. In the numerical implement, we determine these coefficients with function value from previous time step. For illustration, let us perform discretization at (ti,ξj)(t_{i},\xi_{j}). Denote by σi,j:=σε(Ui,j,ξj)\sigma_{i,j}:=\sigma_{\varepsilon}(U_{i,j},\xi_{j}). The first order term is discretized by the upwind scheme, i.e.

(δσi1,j)uξ(ti,ξj){(δσi1,j)Ui,j+1,Ui,jΔξ, if δσi1,j0;(δσi1,j)Ui,j,Ui,j1Δξ, if δσi1,j<0.(\delta-\sigma_{i-1,j})\frac{{\partial}u}{{\partial}\xi}(t_{i},\xi_{j})\approx\left\{\begin{split}&(\delta-\sigma_{i-1,j})\frac{U_{i,j+1},U_{i,j}}{\Delta\xi},\text{ if $\delta-\sigma_{i-1,j}\geqslant 0$;}\\ &(\delta-\sigma_{i-1,j})\frac{U_{i,j},U_{i,j-1}}{\Delta\xi},\text{ if $\delta-\sigma_{i-1,j}<0$.}\end{split}\right.

We use the fully implicit approximation to the temporal term

ut(ti,ξj)Ui,jUi1,jΔt,\frac{{\partial}u}{{\partial}t}(t_{i},\xi_{j})\approx\frac{U_{i,j}-U_{i-1,j}}{\Delta t},

and the usual discretization for the second order term

2uξ2Ui,j+1+Ui,j12Ui,j(Δξ)2.\frac{{\partial}^{2}u}{{\partial}\xi^{2}}\approx\frac{U_{i,j+1}+U_{i,j-1}-2U_{i,j}}{(\Delta\xi)^{2}}.

Thus, given function value Ui1,U_{i-1,\cdot} at previous time step, current value Ui,U_{i,\cdot} is obtained by solving the following equation

[AiUi,]j=ε1(Ui,jeξj)+[A_{i}U_{i,\cdot}]_{j}=\varepsilon^{-1}(U_{i,j}-e^{\xi_{j}})^{+} (7.1)

for j=0,±1,±2,,±Nj=0,\pm 1,\pm 2,...,\pm N. Here the matrix AiA_{i} is determined by Ui1,U_{i-1,\cdot} and is a sparse MM-matrix due to our discretization scheme.

Now, we have to solve the nonlinear equation (7.1). We adopt the method used by [13] to value American options. For illustration, let us recall the classical Newton iteration for finding the root of a convex function ff. Given an initial guess, the point is updated as

zn+1=znf(zn)f(zn)z_{n+1}=z_{n}-\frac{f(z_{n})}{f^{\prime}(z_{n})}

which is equivalent to say that zn+1z_{n+1} solves

f(zn)+f(zn)(zzn)=0.f(z_{n})+f^{\prime}(z_{n})(z-z_{n})=0.

It is easy to see that the left hand side of above equation is an first order approximation of ff at znz_{n}. Similarly, we can solve (7.1) with Newton iteration. Denote Ui,jkU^{k}_{i,j} as the approximation at (ti,ξj)(t_{i},\xi_{j}) for kkth iteration. Then, Ui,kU^{k}_{i,\cdot} solves the linearized equation

[AiUi,k]j=ε1(Ui,jk1eξj)++ε11{Ui,jk1eξj>0}(Ui,jkUi,jk1).[A_{i}U^{k}_{i,\cdot}]_{j}=\varepsilon^{-1}(U^{k-1}_{i,j}-e^{\xi_{j}})^{+}+\varepsilon^{-1}1_{\{U^{k-1}_{i,j}-e^{\xi_{j}}>0\}}(U^{k}_{i,j}-U^{k-1}_{i,j}). (7.2)

When the difference between Ui,kU^{k}_{i,\cdot} and Ui,k1U^{k-1}_{i,\cdot} is small enough, we stop the iteration and set Ui,U_{i,\cdot} equals Ui,kU^{k}_{i,\cdot}. Moreover, the initial guess Ui,0U^{0}_{i,\cdot} is chosen to be Ui1,U_{i-1,\cdot}.

In summary, we have the following iterative algorithm.

  Algorithm 1 Explicit-Implicit Finite-difference Iterative Algorithm

 

N,M,L,Δt,ΔξN,M,L,\Delta t,\Delta\xi, smooth function Hε()H_{\varepsilon}(\cdot) and tolerance toltol
Initialize U0,j=min{1,eξj}U_{0,j}=\min\{1,e^{\xi_{j}}\}
for  i=1,2,,Mi=1,2,...,M do
     Construct the matrix AiA_{i} according to upwind scheme with
σi,j:=σε(Ui,j,ξj)\sigma_{i,j}:=\sigma_{\varepsilon}(U_{i,j},\xi_{j})
     Set Ui,0=Ui1,U^{0}_{i,\cdot}=U_{i-1,\cdot}
     while True do
         Solve
[AiUi,k]j=ε1(Ui,jk1eξj)++ε11{Ui,jk1eξj>0}(Ui,jkUi,jk1).[A_{i}U^{k}_{i,\cdot}]_{j}=\varepsilon^{-1}(U^{k-1}_{i,j}-e^{\xi_{j}})^{+}+\varepsilon^{-1}1_{\{U^{k-1}_{i,j}-e^{\xi_{j}}>0\}}(U^{k}_{i,j}-U^{k-1}_{i,j}).
         If Ui,kUi,k1max{1,Ui,k1}<tol\frac{\|U^{k}_{i,\cdot}-U^{k-1}_{i,\cdot}\|_{\infty}}{\max\{1,\|U^{k-1}_{i,\cdot}\|_{\infty}\}}<tol, Quit
     end while
     Set Ui,=Ui,kU_{i,\cdot}=U^{k}_{i,\cdot}.
end for

 

7.2 Numerical Results

In the numerical experiment, we set the model parameters as δ=0.03,σl=0.3,σh=0.2\delta=0.03,\sigma_{l}=0.3,\sigma_{h}=0.2 and γ=0.6\gamma=0.6. For discretization, we have Δt=0.01,Δξ=0.001\Delta t=0.01,\Delta\xi=0.001 and N=103N=10^{3}. We also choose ε=108\varepsilon=10^{-8}, tol=104tol=10^{-4}. Having numerically solved (3.6), we are able to plot the traveling equation for (2.8), which is eξK(ξ)e^{\xi}K(\xi).

Refer to caption
Figure 1: Typical traveling wave equation

Next, we plot the numerical solution for (2.8) and compared it with the traveling wave equation in Figure 2.

Refer to caption
Figure 2: Solutions of the free boundary problem at time t=0,50,100,150t=0,50,100,150.

It seems that the solution will converge to the traveling wave equation as tt goes to infinity as the theoretical result indicates. To numerically check this, we compute the solution for large time tt and plot the error between the solution and the traveling wave equation. The result is shown in Figure 3. The error is defined as the supreme norm between the traveling wave equation KK and the value function at time tt. We see that the error is monotone decreasing with respect to tt. The final error is about 3.6×1033.6\times 10^{-3} at time t=1500t=1500.

Refer to caption
Figure 3: Differences between the free-boundary problem and traveling wave equation.

Finally, we plot the default and transit boundaries as a function of tt and compare them with those of traveling wave equation. The result is shown in Figure 4. It is clear that the boundaries are decreasing with respect to tt which is consistent with our previous theoretical analysis. We also see the convergence of two boundaries. \Copyr1major2

Refer to caption
Figure 4: Default and transit boundaries as a function of tt

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