# 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_ADDRESS>Jin Liang222Jin Liang and Claude-Michel Brauner acknowledge partial support
from National Natural Science Foundation of China (No. 12071349)
<EMAIL_ADDRESS>Claude-Michel Brauner333Jin Liang and Claude-Michel
Brauner acknowledge partial support from National Natural Science Foundation
of China (No. 12071349<EMAIL_ADDRESS>
###### 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
$C^{\infty}$. 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

myclipboard_main [1]organization=School of Mathematical Sciences, Tongji
University, city=Shanghai 200092, country=China [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 $\varepsilon>0$. A series of lemmas are proved in order to establish
estimates which are independent of $\varepsilon$. The key point is that the
two approximating free boundaries can be separated by a positive distance
independent of $\varepsilon$. 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 $C^{\infty}$. 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 $(\Omega,\mathcal{F},P)$ 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 $S_{t}$ denote the firm’s value in the risk neutral world. It satisfies

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

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

$\sigma_{H}<\sigma_{L}$ (2.1)

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

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 $F$. Let $\Phi_{t}$ denote the
discount value of the bond at time $t$. Therefore, at the maturity time $T$,
an investor can get $\Phi_{T}=\min\\{S_{T},F\\}$. For simplicity, we assume in
the following sections $F=1$. 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 $\tau_{1}$ and $\tau_{2}$
are the first moments when the firm’s grade is downgraded and upgraded
respectively as follows:

$\displaystyle\tau_{1}=\inf\\{t>0|\Phi_{0}/S_{0}<\gamma e^{-\delta
T},\Phi_{t}/S_{t}\geqslant\gamma e^{-\delta(T-t)}\\},$
$\displaystyle\tau_{2}=\inf\\{t>0|\Phi_{0}/S_{0}>\gamma e^{-\delta
T},\Phi_{t}/S_{t}\leqslant\gamma e^{-\delta(T-t)}\\},$

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

$0<\gamma<1$ (2.2)

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

$\delta>0,$

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

$\frac{1}{2}\sigma^{2}_{H}<\delta<\frac{1}{2}\sigma^{2}_{L}.$ (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

$S_{t}e^{-\delta(T-t)}\leqslant\Phi_{t}.$

The default time is defined as

$\tau=\inf\\{t>0|\Phi_{0}>e^{-\delta T}S_{0},\Phi_{t}\geqslant
e^{-\delta(T-t)}S_{t}\\}.$

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

######  Remark 2.4.

Condition (2.3) is also assumed in [30] to ensure the existence of the
travelling wave equation. In finance, if $\delta$ 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

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

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

$e^{-\delta
t}\frac{e^{-r(T-t)}}{S_{t}}=e^{-rT}e^{\int_{0}^{t}(\frac{1}{2}\sigma^{2}(u)-\delta)du-\int_{0}^{t}\sigma(u)dW_{u}}.$

If $\delta<\frac{1}{2}\sigma_{H}^{2}$, the right hand side will go to $\infty$
as $t\rightarrow\infty$ with probability $1$. This implies that the company
will always be low rating in the end. On the other hand, if
$\delta>\frac{1}{2}\sigma^{2}_{L}$, 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 $T$, 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  $\Phi_{H}(S,t)$ and
$\Phi_{L}(S,t)$ the values of the bond in high and low grades respectively,
which are functions of  $S$ and $t$. Then, they are conditional expectations
of the following

$\displaystyle\Phi_{H}(S,t)=$ $\displaystyle
E\Big{[}e^{-r(T-t)}\min(S_{T},F)\cdot{\bf 1}_{\\{T<\tau_{1}\wedge\tau\\}}$
$\displaystyle+S_{t}e^{-\delta(T-\tau)}e^{-r(\tau-t)}{\bf
1}_{\\{\tau<T\wedge\tau_{1}\\}}$
$\displaystyle+e^{-r(\tau_{1}-t)}\Phi_{L}(S_{\tau_{1}},\tau_{1})\cdot{\bf
1}_{\\{\tau_{1}<T\wedge\tau\\}}\Big{|}S_{t}=S>\frac{1}{\gamma
e^{-\delta(T-t)}}\Phi_{H}(S,t)\Big{]},$ (2.4) $\displaystyle\Phi_{L}(S,t)=$
$\displaystyle E[e^{-r(T-t)}\min(S_{T},F)\cdot{\bf
1}_{\\{T<\tau_{2}\wedge\tau\\}}$
$\displaystyle+S_{t}e^{-\delta(T-\tau)}e^{-r(\tau-t)}{\bf
1}_{\\{\tau<T\wedge\tau_{2}\\}}$
$\displaystyle+e^{-r(\tau_{2}-t)}\Phi_{H}(S_{\tau_{2}},\tau_{2})\cdot{\bf
1}_{\\{\tau_{2}<T\wedge\tau\\}}\Big{|}\frac{1}{e^{-\delta(T-t)}}\Phi_{L}(S,t)<S_{t}=S<\frac{1}{\gamma
e^{-\delta(T-t)}}\Phi_{L}(S,t)\Big{]},$ (2.5)

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

###  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 $\Phi_{H}$ and $\Phi_{L}$
satisfy the following system of partial differential equations in their
respective life regions:

$\displaystyle\frac{\partial\Phi_{H}}{\partial
t}+\frac{1}{2}{\sigma}^{2}_{H}S^{2}\frac{\partial^{2}\Phi_{H}}{\partial
S^{2}}+rS\frac{\partial\Phi_{H}}{\partial S}-r\Phi_{H}=0,$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad S>\frac{1}{\gamma
e^{-\delta(T-t)}}\Phi_{H},\;t>0,$ (2.6)
$\displaystyle\frac{\partial\Phi_{L}}{\partial
t}+\frac{1}{2}{\sigma}^{2}_{L}S^{2}\frac{\partial^{2}\Phi_{L}}{\partial
S^{2}}+rS\frac{\partial\Phi_{L}}{\partial S}-r\Phi_{L}=0,$
$\displaystyle\qquad\qquad\qquad\qquad\qquad\qquad\frac{1}{e^{-\delta(T-t)}}\Phi_{L}<S<\frac{1}{\gamma
e^{-\delta(T-t)}}\Phi_{L},\;t>0.$ (2.7)

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

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

Define the function $\Phi$ as

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

Then, it satisfies the following variational form

$\min\Big{\\{}\frac{\partial\Phi}{\partial
t}+\frac{1}{2}\sigma^{2}(\Phi,S,t)S^{2}\frac{\partial^{2}\Phi}{\partial
S^{2}}+rS\frac{\partial\Phi}{\partial
S}-r\Phi,\,-\Phi(S,t)+Se^{-\delta(T-t)}\Big{\\}}=0,$

with

$\sigma(\Phi,S,t)=\sigma_{H}{\bf 1}_{\\{\Phi<\gamma
Se^{-\delta(T-t)}\\}}+\sigma_{L}{\bf 1}_{\\{\Phi\geqslant\gamma
Se^{-\delta(T-t)}\\}}.$

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

$\min\Big{\\{}-\frac{\partial\phi}{\partial
t}+\frac{1}{2}\sigma^{2}(e^{-rt}\phi,e^{x},t)\frac{\partial^{2}\phi}{\partial
x^{2}}+(r-\frac{1}{2}\sigma^{2})\frac{\partial\phi}{\partial
x},\,-\phi(s,t)+e^{x+(r-\delta)t}\Big{\\}}=0.$

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

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

Then, the equation reads

$\min\left\\{-\frac{\partial u}{\partial
t}+\frac{1}{2}\sigma^{2}(u)\frac{\partial^{2}u}{\partial\xi^{2}}+(\delta-\frac{1}{2}\sigma^{2})\frac{\partial
u}{\partial\xi},\,-u+e^{\xi}\right\\}=0.$ (2.8)

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

${\mathcal{L}}:=-\frac{\partial}{\partial
t}+\frac{1}{2}\sigma^{2}(v)\Big{(}\frac{\partial^{2}}{\partial\xi^{2}}+\frac{\partial}{\partial\xi}\Big{)}+\delta\Big{(}\frac{\partial}{\partial\xi}+1\Big{)}.$

Thus, $v$ satisfies the following problem:

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

with

$\sigma(v)=\sigma_{H}{\bf 1}_{\\{v<\gamma\\}}+\sigma_{L}{\bf
1}_{\\{v\geqslant\gamma\\}}.$

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

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

and the transit boundary

$\hat{\eta}(t):=\inf\\{\xi\,|\,v(\xi,t)<\gamma\\}.$

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

$\left\\{\begin{split}&-\frac{\partial v}{\partial
t}+\frac{1}{2}\sigma_{L}^{2}\Big{(}\frac{\partial^{2}v}{\partial\xi^{2}}+\frac{\partial
v}{\partial\xi}\Big{)}+\delta\Big{(}\frac{\partial
v}{\partial\xi}+v\Big{)}=0,\quad\hat{\kappa}(t)<\xi<\hat{\eta}(t);\\\
&-\frac{\partial v}{\partial
t}+\frac{1}{2}\sigma_{H}^{2}\Big{(}\frac{\partial^{2}v}{\partial\xi^{2}}+\frac{\partial
v}{\partial\xi}\Big{)}+\delta\Big{(}\frac{\partial
v}{\partial\xi}+v\Big{)}=0,\quad\xi>\hat{\eta}(t);\\\
&v(\hat{\kappa}(t)+)=1,\quad\frac{\partial
v}{\partial\xi}(\hat{\kappa}(t)+)=0;\\\
&v(\hat{\eta}(t)+)=v(\hat{\eta}(t)-)=\gamma,\quad\frac{\partial
v}{\partial\xi}(\hat{\eta}(t)+)=\frac{\partial
v}{\partial\xi}(\hat{\eta}(t)-).\end{split}\right.$ (2.10)

For convenience, we set

$c_{L}=\frac{2\delta}{\sigma^{2}_{L}},\qquad
c_{H}=\frac{2\delta}{\sigma^{2}_{H}}.$

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

##  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 $K$ satisfies

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

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

$\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 $(K,\eta^{*},\kappa^{*})$ such that $K$
belongs to $C^{1}([\kappa^{*},+\infty))$ and the respective restrictions of
$K$ to $[\kappa^{*},\eta^{*}]$ and $[\eta^{*},+\infty]$ are $C^{\infty}$.

###### Proof.

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

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

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

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

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

$-\frac{c_{L}}{1-c_{L}}e^{\kappa^{*}-\eta^{*}}+\frac{1}{1-c_{L}}e^{-c_{L}(\eta^{*}-\kappa^{*})}=\gamma.$
(3.4)

Define the mapping

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

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

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

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

$e^{-\eta^{*}}+c_{H}Be^{-c_{H}\eta^{*}}=-\frac{c_{L}}{1-c_{L}}e^{-(\eta^{*}-\kappa^{*})}+\frac{c_{L}}{1-c_{L}}e^{-c_{L}(\eta^{*}-\kappa^{*})}=\gamma-e^{-c_{L}(\eta^{*}-\kappa^{*})},$

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

$B=-\frac{1}{c_{H}-1}e^{-c_{L}(\eta^{*}-\kappa^{*})+c_{H}\eta^{*}}\text{ and
}(c_{H}-1)e^{-\eta^{*}}=(c_{H}-1)\gamma+e^{-c_{L}(\eta^{*}-\kappa^{*})}.$

This implies that

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

Thus, $\kappa^{*}$, $B,C$ and $D$ are determined. Summarizing, it comes

$\displaystyle
K(\xi)=\left\\{\begin{array}[]{ll}e^{-\xi}+(\gamma-e^{-\eta^{*}})e^{-c_{H}(\xi-\eta^{*})},&\xi>\eta^{*},\\\\[5.69054pt]
-\frac{c_{L}}{1-c_{L}}e^{-(\xi-\kappa^{*})}+\frac{1}{1-c_{L}}e^{-c_{L}(\xi-\kappa^{*})},&\kappa^{*}<\xi<\eta^{*}.\end{array}\right.$
(3.10)

∎

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

######  Proposition 3.2.

(i) for $\xi>\kappa^{*}$, $\frac{dK}{d\xi}<0$, $K+\frac{dK}{d\xi}>0$, and
$\frac{d^{2}K}{d\xi^{2}}+\frac{dK}{d\xi}<0$ if $\xi\neq\eta^{*}$;
(ii) $\gamma<K(\xi)<1$ if $\xi\in(\kappa^{*},\eta^{*})$ and $K(\xi)<\gamma<1$
if $\xi>\eta^{*}$;
(iii) for $\xi\geqslant\kappa^{*}$, $K(\xi)\leqslant\min\\{1,e^{-\xi}\\}$;
(iv) $\eta^{*}$ is a decreasing function of $\gamma$. Moreover,
$\lim_{\gamma\rightarrow 0}\eta^{*}=+\infty$ and $\lim_{\gamma\rightarrow
1}\eta^{*}=-\log\frac{c_{H}}{c_{H}-1}$.

###### Proof.

(i) It is straightforward to compute

$\frac{dK}{d\xi}=\left\\{\begin{array}[]{ll}-e^{-\xi}-c_{H}(\gamma-e^{-\eta^{*}})e^{-c_{H}(\xi-\eta^{*})},&\xi>\eta^{*},\\\\[2.84526pt]
\frac{c_{L}}{1-c_{L}}e^{-(\xi-\kappa^{*})}-\frac{c_{L}}{1-c_{L}}e^{-c_{L}(\xi-\kappa^{*})},&\kappa^{*}<\xi<\eta^{*}.\end{array}\right.$

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

$\frac{dK}{d\xi}=-e^{-\eta^{*}}e^{-(\xi-\eta^{*})}-c_{H}(\gamma-e^{-\eta^{*}})e^{-c_{H}(\xi-\eta^{*})}.$

With the notation from Theorem 3.1, we have that

$c_{H}Be^{-c_{H}\eta^{*}}=c_{H}(\gamma-e^{-\eta^{*}})$

and

$e^{-\eta^{*}}+c_{H}Be^{-c_{H}\eta^{*}}=-\frac{c_{L}}{1-c_{L}}e^{-(\eta^{*}-\kappa^{*})}+\frac{c_{L}}{1-c_{L}}e^{-c_{L}(\eta^{*}-\kappa^{*})}>0.$

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

$K+\frac{dK}{d\xi}=\left\\{\begin{array}[]{ll}(1-c_{H})(\gamma-e^{-\eta^{*}})e^{-c_{H}(\xi-\eta^{*})},&\xi>\eta^{*},\\\\[5.69054pt]
e^{-c_{L}(\xi-\kappa^{*})},&\kappa^{*}<\xi<\eta^{*},\end{array}\right.$

and

$\frac{dK}{d\xi}+\frac{d^{2}K}{d\xi^{2}}=\left\\{\begin{array}[]{ll}(c_{H}^{2}-c_{H})(\gamma-e^{-\eta^{*}})e^{-c_{H}(\xi-\eta^{*})},&\xi>\eta^{*},\\\\[5.69054pt]
-c_{L}e^{-c_{L}(\xi-\kappa^{*})},&\kappa^{*}<\xi<\eta^{*}.\end{array},\right.$

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

(ii) It follows immediately from (i).

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

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

##  4 Penalized and Regularized Cauchy Problem

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

${\sigma}(v)={\sigma}_{H}+({\sigma}_{L}-{\sigma}_{H})H(v-\gamma).$

First, we approximate $H$ by a $C^{\infty}$ function $H_{\varepsilon}$ such
that

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

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

$\beta_{\varepsilon}(y)\in
C^{\infty}(\mathbb{R}),\,\beta_{\varepsilon}(y)\geqslant
0,\,\beta_{\varepsilon}(y)=0\text{ if $y\leqslant-\varepsilon$;}$
$\beta_{\varepsilon}(0)=C_{0}\geqslant
2\delta;\;\beta^{\prime}_{\varepsilon}(y)\geqslant
0;\;\beta^{\prime\prime}_{\varepsilon}(y)\geqslant 0;$
$\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 $\varepsilon_{\beta}>0$ be the unique solution of
$\beta_{\varepsilon}(-\frac{\varepsilon_{\beta}}{2})=\delta$. It is easy to
see that $\varepsilon_{\beta}\rightarrow 0$ when $\varepsilon\rightarrow 0$.
Finally, let us define
$\psi_{\varepsilon}(y):=1+\varepsilon_{\beta}\psi(\frac{y-1}{\varepsilon_{\beta}})$,
where $\psi\in C^{\infty}$, $\psi(y)=0$ for $y\geqslant 1/2$; $\psi(y)=y$ for
$y<-1/2$ and $\psi(y)\leqslant y,\,0\leqslant\psi^{\prime}(y)\leqslant
1,\,\psi^{\prime\prime}(y)\leqslant 0$ for $-1/2\leqslant y\leqslant 1/2$.

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

######  Lemma 4.1.

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

###### Proof.

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

$(\psi_{\varepsilon}(y)-y\psi^{\prime}_{\varepsilon}(y))^{\prime}=-\frac{y}{\varepsilon_{\beta}}\psi^{\prime\prime}(\frac{y-1}{\varepsilon_{\beta}}).$
(4.1)

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

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

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

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

$\begin{split}{{\mathcal{L}}_{\varepsilon}}[v_{\varepsilon}]=-\frac{{\partial}v_{\varepsilon}}{{\partial}t}+\frac{1}{2}{\sigma}^{2}_{\varepsilon}(v_{\varepsilon})\Big{(}\frac{{\partial}^{2}v_{\varepsilon}}{{\partial}\xi^{2}}+\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\Big{)}+\delta\Big{(}\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}+v_{\varepsilon}\Big{)}=\beta_{\varepsilon}(v_{\varepsilon}-1),\end{split}$
(4.2)

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

${\sigma}_{\varepsilon}(v_{\varepsilon})={\sigma}_{H}+({\sigma}_{L}-{\sigma}_{H})H_{\varepsilon}(v_{\varepsilon}-\gamma),$
(4.3)

together with the initial condition

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

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

######  Theorem 4.2.

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

###### Proof.

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

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

with

$a(\xi,v,p)=\frac{1}{2}\sigma^{2}_{\varepsilon}(v)p,\quad
A(\xi,v,p)=\beta_{\varepsilon}(v-1)-\delta
v-\big{(}\frac{1}{2}\sigma^{2}_{\varepsilon}(v)+\delta\big{)}p+\sigma_{\varepsilon}\sigma^{\prime}_{\varepsilon}(v)p^{2}.$

One can check that $a$ and $A$ satisfy the assumptions of [25, Chapter V,
Theorem 8.1]. Thus, there exists a unique bounded solution $v_{\varepsilon}\in
C^{2+\alpha,1+\frac{\alpha}{2}}(\mathbb{R}\times[0,T])$ for any
$0<\alpha<1$.444For usual parabolic Hölder spaces, see, e.g., [25, Chapter 1,
Section 1],[34, Section 5.1]). Then, $\sigma_{\varepsilon}(v_{\varepsilon})$
and $\beta_{\varepsilon}(v_{\varepsilon})$ 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
$v_{\varepsilon}\in C^{4+\alpha,2+\frac{\alpha}{2}}(\mathbb{R}\times[0,T])$.
The result follows by bootstrapping. ∎

######  Remark 4.3.

From the definition of $H_{\varepsilon}$ and $\beta_{\varepsilon}$, it is easy
to see that $\sigma_{\varepsilon}(v_{\varepsilon})=\sigma_{L}$ when
$v_{\varepsilon}>\gamma$ and $\beta_{\varepsilon}(v_{\varepsilon})=0$ when
$v_{\varepsilon}<1-\varepsilon$. Thus, when $\varepsilon$ 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 $v_{\varepsilon}$ independent
of $\varepsilon$, 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 $v$ when taking $\varepsilon\to 0$
and, thus, are crucial for the analysis of the bond value and free boundaries.

######  Lemma 4.4.

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

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

###### Proof.

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

$\mathcal{L}_{\varepsilon}[w]=h^{\prime}(v_{\varepsilon})\beta_{\varepsilon}(v_{\varepsilon}-1)+\frac{1}{2}\sigma^{2}_{\varepsilon}(v_{\varepsilon})h^{\prime\prime}(v_{\varepsilon})(\frac{\partial
v_{\varepsilon}}{\partial\xi})^{2}+\delta(w-h^{\prime}(v_{\varepsilon})v_{\varepsilon}),$

which can be rewritten as

$\mathcal{L}_{\varepsilon}[w]-\delta(1-\frac{v_{\varepsilon}h^{\prime}(v_{\varepsilon})}{h(v_{\varepsilon})})w=h^{\prime}(v_{\varepsilon})\beta_{\varepsilon}(v_{\varepsilon}-1)+\frac{1}{2}\sigma^{2}_{\varepsilon}(v_{\varepsilon})h^{\prime\prime}(v_{\varepsilon})(\frac{\partial
v_{\varepsilon}}{\partial\xi})^{2}.$

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

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

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

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

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

$\mathcal{L}_{\varepsilon}[w]=\beta_{\varepsilon}(v_{\varepsilon}-1)=\frac{\beta_{\varepsilon}(v_{\varepsilon}-1)-\beta_{\varepsilon}(e^{-\xi}-1)}{w}w+\beta_{\varepsilon}(e^{-\xi}-1).$

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

∎

######  Lemma 4.5.

It holds in $\Omega_{T}$:

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

###### Proof.

Differentiating (4.2), it comes

${\mathcal{L}}_{\varepsilon}\Big{[}\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\Big{]}=-{\sigma}_{\varepsilon}(v_{\varepsilon})\,{\sigma}^{\prime}_{\varepsilon}(v_{\varepsilon})\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\,(\frac{{\partial}^{2}v_{\varepsilon}}{{\partial}\xi^{2}}+\frac{{\partial}v_{\varepsilon}}{{\partial}\xi})+\beta^{\prime}_{\varepsilon}(v_{\varepsilon}-1)\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}.$

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

######  Lemma 4.6.

It holds in $\Omega_{T}$:

$1\geqslant\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}+v_{\varepsilon}>0.$

###### Proof.

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

$\displaystyle{\mathcal{L}}_{\varepsilon}[w]+{\sigma}_{\varepsilon}(v_{\varepsilon}){\sigma}_{\varepsilon}^{\prime}(v_{\varepsilon})\,\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\frac{{\partial}w}{{\partial}\xi}=\beta_{\varepsilon}^{\prime}(v_{\varepsilon}-1)(w-v_{\varepsilon})+\beta_{\varepsilon}(v_{\varepsilon}-1).$
(4.6)

Using Taylor expansion of $\beta_{\varepsilon}(-\varepsilon)$ at $y$, one has
that

$0=\beta_{\varepsilon}(-\varepsilon)=\beta_{\varepsilon}(y)-\beta^{\prime}_{\varepsilon}(y)(y+\varepsilon)+\frac{1}{2}\beta_{\varepsilon}^{\prime\prime}(\theta)(y+\varepsilon)^{2}.$

That is,

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

Replacing $y$ by $v_{\varepsilon}(\xi)-1$ in the above formula, we have,

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

Thus, (4.6) reads

$\displaystyle-{\mathcal{L}}_{\varepsilon}[w]-{\sigma}_{\varepsilon}(v_{\varepsilon}){\sigma}_{\varepsilon}^{\prime}(v_{\varepsilon})\,\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\frac{{\partial}w}{{\partial}\xi}+\beta_{\varepsilon}^{\prime}(v_{\varepsilon}-1)w$
$\displaystyle=\beta_{\varepsilon}^{\prime}(v_{\varepsilon}-1)v_{\varepsilon}-\beta_{\varepsilon}(v_{\varepsilon}-1)\geqslant
v_{\varepsilon}\beta_{\varepsilon}^{\prime}(v_{\varepsilon}-1)-(v_{\varepsilon}-1+\varepsilon)\beta_{\varepsilon}^{\prime}(v_{\varepsilon}-1)=(1-\varepsilon)\beta_{\varepsilon}^{\prime}(v_{\varepsilon}-1)\geqslant
0.$

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

######  Lemma 4.7.

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

$\frac{{\partial}^{2}v_{\varepsilon}}{{\partial}\xi^{2}}+\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\leqslant
0.$

###### Proof.

At $t=0,$
$\frac{{\partial}^{2}v_{\varepsilon}}{{\partial}\xi^{2}}+\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}=e^{-\xi}\psi^{\prime\prime}_{\varepsilon}(e^{-\xi})$
is non-positive. Now, consider the function
$w=\frac{{\partial}v_{\varepsilon}}{{\partial}t}-\delta\big{(}\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}+v_{\varepsilon}\big{)}+\beta_{\varepsilon}(v_{\varepsilon}-1)$.
From the definition of $H_{\varepsilon}$ and $\beta_{\varepsilon}$, it is easy
to see that $\sigma_{\varepsilon}(v_{\varepsilon})=\sigma_{L}$ when
$v_{\varepsilon}>\gamma$ and
${\color[rgb]{0,0,1}\beta_{\varepsilon}(v_{\varepsilon}-1)}=0$ when
$v_{\varepsilon}<1-\varepsilon$. Thus, we divide the space into two parts
$\\{v_{\varepsilon}<1-2\varepsilon\\}$ and $\\{v_{\varepsilon}\geqslant
1-2\varepsilon\\}$
Case 1: $v_{\varepsilon}<1-2\varepsilon$.  In this case, we see that
$\beta_{\varepsilon}\equiv 0$. Then, it holds that

$\displaystyle{\mathcal{L}}_{\varepsilon}[w]+{\sigma}_{\varepsilon}(v_{\varepsilon}){\sigma}_{\varepsilon}^{\prime}(v_{\varepsilon})\Big{(}\frac{{\partial}^{2}v_{\varepsilon}}{{\partial}\xi^{2}}+\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\Big{)}\Big{(}\frac{{\partial}v_{\varepsilon}}{{\partial}t}-\delta\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\Big{)}$
$\displaystyle={\mathcal{L}}_{\varepsilon}[w]+\frac{2{\sigma}_{\varepsilon}^{\prime}(v_{\varepsilon})}{{\sigma}_{\varepsilon}(v_{\varepsilon})}(w+\delta
v_{\varepsilon})w=0.$ (4.7)

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

$\mathcal{L}_{\varepsilon}[\frac{{\partial}v_{\varepsilon}}{{\partial}t}]=\beta_{\varepsilon}^{\prime}\frac{{\partial}v_{\varepsilon}}{{\partial}t},\mathcal{L}_{\varepsilon}[\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}]=\beta_{\varepsilon}^{\prime}\frac{{\partial}v_{\varepsilon}}{{\partial}\xi},$

and

$\mathcal{L}_{\varepsilon}[\beta_{\varepsilon}]=\beta_{\varepsilon}^{\prime}\beta_{\varepsilon}+\frac{1}{2}\sigma^{2}_{\varepsilon}\beta_{\varepsilon}^{\prime\prime}\left(\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\right)^{2}+\delta\beta_{\varepsilon}-\delta\beta^{\prime}_{\varepsilon}v_{\varepsilon},$

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

$\mathcal{L}_{\varepsilon}[w]-\beta_{\varepsilon}^{\prime}w=\frac{1}{2}\sigma^{2}_{\varepsilon}\beta_{\varepsilon}^{\prime\prime}\left(\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}\right)^{2}\geqslant
0,$

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

Combining these two cases, it holds that

$\mathcal{L}_{\varepsilon}[w]+\Big{(}\frac{2{\sigma}_{\varepsilon}^{\prime}(v_{\varepsilon})}{{\sigma}_{\varepsilon}(v_{\varepsilon})}(w+\delta
v_{\varepsilon})1_{\\{v<1-2\varepsilon\\}}-\beta_{\varepsilon}^{\prime}1_{\\{v\geqslant
1-2\varepsilon\\}}\Big{)}w\geqslant 0.$

Then, by the maximum principle, $w\leqslant 0$.

∎

######  Lemma 4.8.

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

$\frac{{\partial}v_{\varepsilon}}{{\partial}t}<0.$

###### Proof.

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

$\mathcal{L}_{\varepsilon}[w]=-\sigma_{\varepsilon}(v_{e})\sigma^{\prime}_{\varepsilon}(v_{e})(\frac{{\partial}^{2}v_{\varepsilon}}{{\partial}\xi^{2}}+\frac{{\partial}v_{\varepsilon}}{{\partial}\xi})w+\beta^{\prime}_{\varepsilon}(v_{\varepsilon}-1)w.$
(4.8)

Because $v_{\varepsilon}(\xi,0)=\psi_{\varepsilon}(e^{-\xi})$, we have that

$\begin{split}w(\xi,0)=\frac{1}{2}\sigma^{2}_{\varepsilon}(v_{\varepsilon})e^{-2\xi}\psi^{\prime\prime}_{\varepsilon}(e^{-\xi})+\delta\big{(}\psi_{\varepsilon}(e^{-\xi})-e^{-\xi}\psi^{\prime}_{\varepsilon}(e^{-\xi})\big{)}-\beta_{\varepsilon}(\psi_{\varepsilon}(e^{-\xi})-1).\end{split}$
(4.9)

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

$\delta(\psi_{\varepsilon}(e^{-\xi})-e^{-\xi}\psi^{\prime}_{\varepsilon}(e^{-\xi}))-\beta_{\varepsilon}(\psi_{\varepsilon}(e^{-\xi})-1)\leqslant\delta-\beta_{\varepsilon}(-\frac{\varepsilon_{\beta}}{2}).$

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

######  Lemma 4.9.

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

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

###### Proof.

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

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

where

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

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

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

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

As an immediate corollary, we have

######  Lemma 4.10.

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

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

###  4.2 The approximating transit boundary

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

$v_{\varepsilon}(\eta_{\varepsilon}(t),t)=\gamma.$ (4.11)

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

######  Lemma 4.11.

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

###### Proof.

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

$\frac{1}{2}\sigma_{2}^{2}\Big{(}\frac{d^{2}\tilde{K}}{d\xi^{2}}+\frac{d\tilde{K}}{d\xi}\Big{)}+\delta\Big{(}\frac{d\tilde{K}}{d\xi}+\tilde{K}\Big{)}=\delta
1_{\\{\xi\leqslant\tilde{\kappa}^{*}\\}},$

with
$\sigma_{2}:=\sigma_{H}1_{\\{\xi\geqslant\tilde{\eta}^{*}\\}}+\sigma_{L}1_{\\{\xi<\tilde{\eta}^{*}\\}}$.
Let
$w=v_{\varepsilon}-(\tilde{K}-(\varepsilon\vee\varepsilon_{\beta})e^{\delta
t})$. Then, it holds that

$\begin{split}{\mathcal{L}}_{\varepsilon}[w]=\beta_{\varepsilon}({v_{\varepsilon}-1})-\delta
1_{\\{\xi\leqslant\tilde{\kappa}^{*}\\}}+\frac{1}{2}(\sigma_{2}^{2}-\sigma_{\varepsilon}^{2})\Big{(}\frac{d^{2}\tilde{K}}{d\xi^{2}}+\frac{d\tilde{K}}{d\xi}\Big{)}.\end{split}$

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

$\mathcal{L}_{\varepsilon}[w]-\frac{\beta_{\varepsilon}(v_{\varepsilon}-1)-\beta_{\varepsilon}(\tilde{K}-\varepsilon\vee\varepsilon_{\beta}-1)}{w}w\leqslant
0.$

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

######  Theorem 4.12.

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

$\eta_{\varepsilon}(0)=\log(\frac{1}{\gamma}),\quad\tilde{\kappa}^{*}<\eta_{\varepsilon}(t)\leqslant\log\frac{1}{\gamma},$
(4.12)

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

###### Proof.

To begin with, we compute

$v_{\varepsilon}(-\log\gamma,0)=\psi_{\varepsilon}(e^{\log\gamma})=\psi_{\varepsilon}(\gamma)=1+\varepsilon_{\beta}\psi\big{(}\frac{\gamma-1}{\varepsilon_{\beta}}\big{)}.$

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

Next, we compute

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

and (see the proof of Lemma 4.8)

$\frac{\partial v_{\varepsilon}}{\partial
t}(-\log\gamma,0)=-\beta_{\varepsilon}(\gamma-1)=0.$

It is now an exercise to apply the Implicit Function Theorem, which shows that
there exist $\delta_{i},\tau_{i}>0,i=1,2$, and a unique function
$\varphi_{\varepsilon}\in C^{\infty}([-\tau_{1},\tau_{2}])$ such that, if
$(\xi,t)\in[-\log\gamma-\delta_{1},-\log\gamma+\delta_{2}]\times[-\tau_{1},\tau_{2}]$
verifies $v_{\varepsilon}(\xi,t)=\gamma$, then $\xi=\varphi_{\varepsilon}(t)$.
Note that $\varphi_{\varepsilon}$ is a decreasing function because

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

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

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 $T$. ∎

######  Lemma 4.13.

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

###### Proof.

From the Implicit Function Theorem, it holds:

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

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

$\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}(\hat{\eta}_{\varepsilon}(t),t)=\frac{v_{\varepsilon}(\tilde{\kappa}^{*},t)-v_{\varepsilon}(\eta_{\varepsilon}(t),t)}{\tilde{\kappa}^{*}-\eta_{\varepsilon}(t)}=-\frac{v_{\varepsilon}(\tilde{\kappa}^{*},t)-v_{\varepsilon}(\eta_{\varepsilon}(t),t)}{\eta_{\varepsilon}(t)-\tilde{\kappa}^{*}}.$

We have shown that
$v_{\varepsilon}(\tilde{\kappa}^{*},t)\geqslant\tilde{K}(\tilde{\kappa}^{*})-(\varepsilon\vee\varepsilon_{\beta})e^{\delta
t}=1-(\varepsilon\vee\varepsilon_{\beta})e^{\delta t}$ and
$\eta_{\varepsilon}(t)\leqslant\log\frac{1}{\gamma}$. This yields that

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

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

$\frac{{\partial}v_{\varepsilon}}{{\partial}\xi}(\eta_{\varepsilon}(t),t)\leqslant-e^{\hat{\eta}_{\varepsilon}(t)-\eta_{\varepsilon}(t)}\frac{1-(\varepsilon\vee\varepsilon_{\beta})e^{\delta
t}-\gamma}{\log\frac{1}{\gamma}-\tilde{\kappa}^{*}}\leqslant-e^{\tilde{\kappa}^{*}-\log\frac{1}{\gamma}}\frac{1-(\varepsilon\vee\varepsilon_{\beta})e^{\delta
t}-\gamma}{\log\frac{1}{\gamma}-\tilde{\kappa}^{*}}.$

This completes the proof. ∎

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

######  Corollary 4.14.

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

##  5 Main Results

###  5.1 Existence and Uniqueness

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

######  Theorem 5.1.

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

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

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

###### Proof.

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

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

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

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

$v_{n^{\prime\prime}}(\eta_{n^{\prime\prime}}(t)+x,t)-v_{n^{\prime\prime}}(\eta_{n^{\prime\prime}}(t),t)\leqslant-\frac{c}{2}x,$

and

$v_{n^{\prime\prime}}(\eta_{n^{\prime\prime}}(t)-x,t)-v_{n^{\prime\prime}}(\eta_{n^{\prime\prime}}(t),t)\geqslant\frac{c}{2}x.$

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

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

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

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

######  Theorem 5.2.

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

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

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

###### Proof.

Denote by
$F=\frac{1}{2}\left(\sigma^{2}(v_{1})-\sigma^{2}(v_{2})\right)\left(\frac{{\partial}^{2}v_{1}}{{\partial}\xi^{2}}+\frac{{\partial}v_{1}}{{\partial}\xi}\right)$
and $\mathcal{L}_{2}=-\frac{\partial}{\partial
t}+\frac{1}{2}\sigma^{2}(v_{2})\Big{(}\frac{\partial^{2}}{\partial\xi^{2}}+\frac{\partial}{\partial\xi}\Big{)}+\delta\Big{(}\frac{\partial}{\partial\xi}+1\Big{)}$.
We rewrite that

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

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

$\limsup_{(\xi,t)\rightarrow(\xi^{*},t^{*})}\mathop{ess}\left\\{\frac{{\partial}w}{{\partial}t}-\frac{1}{2}\sigma^{2}(v_{2})\frac{{\partial}^{2}w}{{\partial}\xi^{2}}-\left(\frac{1}{2}\sigma^{2}(v_{2})+\delta\right)\frac{{\partial}w}{{\partial}\xi}\right\\}\leqslant
0$

This is equivalent to

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

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

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

In this neighborhood, we also have that

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

Therefore,

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

which is a contradiction. Thus, we proved that $w\geqslant 0$. 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 $\hat{\kappa}$ and the transit boundary $\hat{\eta}$, 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
$v_{\varepsilon}\geqslant\tilde{K}-(\varepsilon\vee\varepsilon_{\beta})e^{\delta
t}$, see Lemma 4.11. Taking the limit as $\varepsilon\to 0$, this implies that
$v\geqslant\tilde{K}$. Since $\tilde{K}=1$ for
$\xi\leqslant\tilde{\kappa}^{*}$, it holds that the set
$\\{\xi\,|\,v(\xi,t)<1\\}$ is bounded from below. Now, we are in position to
define

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

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

######  Theorem 5.3.

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

#### The transit boundary

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

######  Theorem 5.4.

The initial positions of the free boundaries are as follows:

$\hat{\eta}(0)=\log\frac{1}{\gamma},\quad\hat{\kappa}(0)=0.$

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

###### Proof.

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

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
$c$ such that $v_{\varepsilon}(\xi,t)\leqslant\frac{1+\gamma}{2}$ whenever
$\eta_{\varepsilon}(t)-\xi\leqslant c$. Then, one can choose a smooth function
$\zeta$ such that $\zeta(t)<\eta_{\varepsilon}(t)$ and
$\|\zeta-\eta_{\varepsilon}\|_{L^{\infty}[0,T]}\in[{c}/{4},{c}/{2}]$ for
sufficiently small $\varepsilon$. Therefore, $\zeta$ separates the default
boundary $\hat{\kappa}$ and the transit boundary $\hat{\eta}$. 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 $\xi=\zeta(t)+x$ and set
$u(x,t)=v(\zeta(t)+x,t)$. For suitable $a,b\in\mathbb{R}$, we have that
$\zeta(t)+a\leqslant\hat{\kappa}(t)\leqslant\zeta(t)+b<\hat{\eta}(t)$. It
holds that

$\frac{{\partial}u}{{\partial}t}\in
L^{\infty}(t_{1},t_{2};H^{1}(a,b)),\frac{{\partial}^{2}u}{{\partial}t^{2}}\in
L^{2}(t_{1},t_{2};L^{2}(a,b)),$

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

######  Theorem 5.5.

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

Next, we consider the smoothness of the transit free boundary $\hat{\eta}(t)$.
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 $Q=\\{a<x<b,\ 0<t<T\\}$, where $a<b$ are some constants,
consider the following initial boundary problem

$\left\\{\begin{split}&u_{t}-(K_{f}(u_{x}+u))_{x}+f_{1}(x,t)u_{x}+f_{2}(x,t)u=0,\\\
&u(a,t)=g_{a}(t),\ u(b,t)=g_{b}(t),\quad u(x,0)=\phi(x),\\\
&g_{a}(0)=\phi(a),\quad g_{b}(0)=\phi(b),\end{split}\right.$ (5.3)

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

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

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

$\|K_{f}(u+u_{x})\|_{C^{\alpha}(Q)}\leqslant C.$

###### Proof.

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

$\left\\{\begin{split}&v_{t}-(K_{0}(v_{y}))_{y}-f^{\prime}(t)v_{y}+f_{1}v_{y}+(f_{2}-f_{1})v=0,\\\
&v(a-f,t)=g_{a}(t)e^{a-f},\ v(b-f,t)=g_{b}(t)e^{b-f},\quad
v(y,0)=\phi(y+f)e^{y}.\end{split}\right.$ (5.4)

where $K_{0}=\mu_{1}$ if $y>0$, $\mu_{2}$ if $y<0$. 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 $\hat{\eta}$.

######  Theorem 5.7.

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

###### Proof.

In a neighborhood of $\hat{\eta}$, $v$ satisfies the system

$\left\\{\begin{split}&-\frac{{\partial}v}{{\partial}t}+\frac{1}{2}{\sigma}^{2}_{H}\Big{(}\frac{{\partial}^{2}v}{{\partial}\xi^{2}}+\frac{{\partial}v}{{\partial}\xi}\Big{)}+\delta\Big{(}\frac{{\partial}v}{{\partial}\xi}+v\Big{)}=0,\quad\xi>\hat{\eta}(t),\\\
&-\frac{{\partial}v}{{\partial}t}+\frac{1}{2}{\sigma}^{2}_{L}\Big{(}\frac{{\partial}^{2}v}{{\partial}\xi^{2}}+\frac{{\partial}v}{{\partial}\xi}\Big{)}+\delta\Big{(}\frac{{\partial}v}{{\partial}\xi}+v\Big{)}=0,\quad\xi<\hat{\eta}(t),\\\
&v(\hat{\eta}(t)+,t)=v(\hat{\eta}(t)-,t)=\gamma,\quad
v_{\xi}(\hat{\eta}(t)+,t)=v_{\xi}(\hat{\eta}(t)-,t).\end{split}\right.$ (5.5)

Thus, it holds that

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

which means that

$v_{t}(\hat{\eta}(t)+,t)=v_{t}(\hat{\eta}(t)-,t).$ (5.7)

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

$\left\\{\begin{split}&-\frac{{\partial}w}{{\partial}t}+\frac{1}{2}{\sigma}^{2}_{H}\Big{(}\frac{{\partial}^{2}w}{{\partial}\xi^{2}}+\frac{{\partial}w}{{\partial}\xi}\Big{)}+\delta\Big{(}\frac{{\partial}w}{{\partial}\xi}+w\Big{)}=0,\quad\xi>\hat{\eta}(t),\\\
&-\frac{{\partial}w}{{\partial}t}+\frac{1}{2}{\sigma}^{2}_{L}\Big{(}\frac{{\partial}^{2}w}{{\partial}\xi^{2}}+\frac{{\partial}w}{{\partial}\xi}\Big{)}+\delta\Big{(}\frac{{\partial}w}{{\partial}\xi}+w\Big{)}=0,\quad\xi<\hat{\eta}(t),\\\
&w(\hat{\eta}(t)+,t)=w(\hat{\eta}(t)-,t),\quad\sigma^{2}_{L}(w_{x}+w)(\hat{\eta}(t)+,t)=\sigma^{2}_{H}(w_{x}+w)(\hat{\eta}(t)-,t).\end{split}\right.$
(5.8)

According to the free boundary condition, $w$ satisfies a typical Verigin
problem, see [29, 38]. In particular, the $C^{\infty}$ 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
$\hat{\eta}$ is Lipschitz continuous and satisfies

$\displaystyle\hat{\eta}^{\prime}(t)$ $\displaystyle=$
$\displaystyle-\frac{\frac{1}{2}{\sigma}^{2}_{H}\Big{(}w_{\xi}(\hat{\eta}(t)+,t)+w(\hat{\eta}(t)+,t)\Big{)}+\delta\Big{(}w(\hat{\eta}(t)+,t)+\gamma\Big{)}}{w(\hat{\eta}(t)+,t)}$
(5.9) $\displaystyle=$
$\displaystyle-\frac{\frac{1}{2}{\sigma}^{2}_{L}\Big{(}w_{\xi}(\hat{\eta}(t)-,t)+w(\hat{\eta}(t)-,t)\Big{)}+\delta\Big{(}w(\hat{\eta}(t)-,t)+\gamma\Big{)}}{w(\hat{\eta}(t)-,t)},$

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), $w_{\xi}+w\in
C^{\alpha}$ up to the free boundary. Furthermore, by Lemma 4.6, $w$ has a
negative upperbound. Then, the right hand side of (5.9) belongs to
$C^{\alpha}$. This implies in turn $\hat{\eta}\in C^{1+\alpha}$. In this way,
by an iteration process, one can further improve the regularity of
$\hat{\eta}$ and shows eventually that it belongs to $C^{\infty}.$ ∎

##  6 Asymptotic Convergence

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

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

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

$0\geqslant\int_{-\infty}^{\infty}\int_{0}^{t}\frac{{\partial}v}{{\partial}t}(\xi,s)dsd\xi\geqslant\int_{-\infty}^{\infty}(\tilde{K}(\xi)-v(\xi,0))d\xi.$

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

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

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

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

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

$\tilde{K}\leqslant\bar{K}\leqslant\min\\{1,e^{-\xi}\\},\;\,\frac{d\bar{K}}{d\xi}\leqslant
0,\,\text{ and }\,\frac{d^{2}\bar{K}}{d\xi^{2}}+\frac{d\bar{K}}{d\xi}\leqslant
0.$

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

$-\frac{{\partial}v^{n}}{{\partial}t}+\frac{1}{2}\sigma_{L}^{2}\left(\frac{{\partial}^{2}v^{n}}{{\partial}\xi^{2}}+\frac{{\partial}v^{n}}{{\partial}\xi}\right)+\delta\left(\frac{{\partial}v^{n}}{{\partial}\xi}+v^{n}\right)=0.$

Taking subsequence $n_{j}$, we derive that

$\frac{1}{2}\sigma_{L}^{2}\left(\frac{d^{2}\bar{K}}{d\xi^{2}}+\frac{d\bar{K}}{d\xi}\right)+\delta\left(\frac{d\bar{K}}{d\xi}+\bar{K}\right)=0,\text{
for }\xi\in I.$

Since $I$ is arbitrary, it holds that

$\frac{1}{2}\sigma_{L}^{2}\left(\frac{d^{2}\bar{K}}{d\xi^{2}}+\frac{d\bar{K}}{d\xi}\right)+\delta\left(\frac{d\bar{K}}{d\xi}+\bar{K}\right)=0,\text{
for }\bar{\kappa}<\xi<\bar{\eta}.$

Similarly, we can also show that

$\frac{1}{2}\sigma_{H}^{2}\left(\frac{d^{2}\bar{K}}{d\xi^{2}}+\frac{d\bar{K}}{d\xi}\right)+\delta\left(\frac{d\bar{K}}{d\xi}+\bar{K}\right)=0,\text{
for }\xi>\bar{\eta}.$

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

$\left\\{\begin{split}&\frac{d^{2}\bar{K}}{d\xi^{2}}+\frac{d\bar{K}}{d\xi}+c_{H}(\frac{d\bar{K}}{d\xi}+K)=0,\xi>\bar{\eta},\\\
&\frac{d^{2}\bar{K}}{d\xi^{2}}+\frac{d\bar{K}}{d\xi}+c_{L}(\frac{d\bar{K}}{d\xi}+K)=0,\bar{\kappa}<\xi<\bar{\eta},\\\
&\bar{K}(\bar{\kappa})=1,\frac{d\bar{K}}{d\xi}(\bar{\kappa})=0,\\\
&\bar{K}(\bar{\eta})=\bar{K}(\bar{\eta}^{*}-)=\gamma,\frac{d\bar{K}}{d\xi}(\bar{\eta}+)=\frac{d\bar{K}}{d\xi}(\bar{\eta}-),\\\
&\lim_{\xi\rightarrow\infty}e^{\xi}\bar{K}(\xi)=1.\end{split}\right.$

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

######  Theorem 6.1.

As $t$ goes to $+\infty$, $v(\cdot,t)$ converges uniformly to $K$.

##  7 Numerical Results

In this section, we will give some numerical results for illustration. As $u$
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 $t_{i}=i\Delta
t,i=0,1,...,M$, and $\xi_{j}=j\Delta\xi,j=0,\rm 1,\pm 2,...,\pm N$. $U_{i,j}$
will be the approximation of the solution $u$ of (2.8) at mesh point
$(t_{i},\xi_{j})$. Consider the approximating penalized equation

$\left\\{\begin{split}&-\frac{{\partial}u}{{\partial}t}+\frac{1}{2}\sigma^{2}_{\varepsilon}(u,\xi)(\frac{{\partial}^{2}u}{{\partial}\xi^{2}}-\frac{{\partial}u}{{\partial}\xi})+\delta\frac{{\partial}u}{{\partial}\xi}=\varepsilon^{-1}(u-e^{\xi})^{+},\quad\xi\in[-N\Delta\xi,N\Delta\xi],t\geqslant
0;\\\ &u(\xi,0)=\min\\{1,e^{\xi}\\};\\\
&u(N\Delta\xi,t)=1,u(-N\Delta\xi,t)=0.\end{split}\right.$

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

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

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

$(\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

$\frac{{\partial}u}{{\partial}t}(t_{i},\xi_{j})\approx\frac{U_{i,j}-U_{i-1,j}}{\Delta
t},$

and the usual discretization for the second order term

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

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

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

for $j=0,\pm 1,\pm 2,...,\pm N$. Here the matrix $A_{i}$ is determined by
$U_{i-1,\cdot}$ and is a sparse $M$-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 $f$.
Given an initial guess, the point is updated as

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

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

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

It is easy to see that the left hand side of above equation is an first order
approximation of $f$ at $z_{n}$. Similarly, we can solve (7.1) with Newton
iteration. Denote $U^{k}_{i,j}$ as the approximation at $(t_{i},\xi_{j})$ for
$k$th iteration. Then, $U^{k}_{i,\cdot}$ solves the linearized equation

$[A_{i}U^{k}_{i,\cdot}]_{j}=\varepsilon^{-1}(U^{k-1}_{i,j}-e^{\xi_{j}})^{+}+\varepsilon^{-1}1_{\\{U^{k-1}_{i,j}-e^{\xi_{j}}>0\\}}(U^{k}_{i,j}-U^{k-1}_{i,j}).$
(7.2)

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

In summary, we have the following iterative algorithm.

Algorithm 1 Explicit-Implicit Finite-difference Iterative Algorithm

$N,M,L,\Delta t,\Delta\xi$, smooth function $H_{\varepsilon}(\cdot)$ and
tolerance $tol$

Initialize $U_{0,j}=\min\\{1,e^{\xi_{j}}\\}$

for $i=1,2,...,M$ do

Construct the matrix $A_{i}$ according to upwind scheme with
$\sigma_{i,j}:=\sigma_{\varepsilon}(U_{i,j},\xi_{j})$

Set $U^{0}_{i,\cdot}=U_{i-1,\cdot}$

while True do

Solve
$[A_{i}U^{k}_{i,\cdot}]_{j}=\varepsilon^{-1}(U^{k-1}_{i,j}-e^{\xi_{j}})^{+}+\varepsilon^{-1}1_{\\{U^{k-1}_{i,j}-e^{\xi_{j}}>0\\}}(U^{k}_{i,j}-U^{k-1}_{i,j}).$

If
$\frac{\|U^{k}_{i,\cdot}-U^{k-1}_{i,\cdot}\|_{\infty}}{\max\\{1,\|U^{k-1}_{i,\cdot}\|_{\infty}\\}}<tol$,
Quit

end while

Set $U_{i,\cdot}=U^{k}_{i,\cdot}$.

end for

###  7.2 Numerical Results

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

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.

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

It seems that the solution will converge to the traveling wave equation as $t$
goes to infinity as the theoretical result indicates. To numerically check
this, we compute the solution for large time $t$ 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
$K$ and the value function at time $t$. We see that the error is monotone
decreasing with respect to $t$. The final error is about $3.6\times 10^{-3}$
at time $t=1500$.

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

Finally, we plot the default and transit boundaries as a function of $t$ 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 $t$
which is consistent with our previous theoretical analysis. We also see the
convergence of two boundaries. r1major2

Figure 4: Default and transit boundaries as a function of $t$

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