valuation of european installment put options — variational inequality...
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Valuation of European Installment Put Options
— Variational Inequality Approach 1
Zhou Yang and Fahuai Yi
School of Mathematical Sciences, South China Normal University,
Guangzhou 510631, China
Abstract: In this paper we consider a parabolic variational inequalityarising from the valuation of European installment put options. We provethe existence and uniqueness of the solution to the problem. Moreover, weobtain C∞ regularity and the bounds of the free boundary. Eventually weshow its numerical result by the binomial method.Key words: Free boundary, variational inequality, option pricing, Euro-pean installment put options.2000 Mathematics Subject Classification: Primary 35R35.
1. Introduction
In this paper we consider a parabolic variational inequality arising from the pricing
model of European continuous installment put options. P (S, t), the price of an European
installment put options, satisfies⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩
∂tP + σ2
2S2∂SSP + (r − q)S∂SP − rP = L∗,
if P > 0 and (S, t) ∈ (0,+∞) × (0, T ],
∂tP + σ2
2S2∂SSP + (r − q)S∂SP − rP ≤ L∗,
if P = 0 and (S, t) ∈ (0,+∞) × (0, T ],
P (S, T ) = (K − S)+, S ∈ [ 0,+∞),
(1.1)
where σ, r, L∗ and K are positive constants and q is a non-negative constant.
We will give the financial and stochastic background of this problem and its heuristic
derivation of its model in Section 2.
If L∗ = 0 in the problem (1.1), it is the model of an standard European put option,
which has an explicit analytic formula of solution and does not have free boundary at
all (see[10]). But if L∗ > 0, the problem becomes very complicated so that it has a non-
monotonic free boundary, see Theorem 1.3 in this paper.
There are some papers about installment options, such as [5], [6], [9], [12], in which
the authors develop some models and consider numerical analysis. Moreover, we had
1The project is supported by National Natural Science Foundation of China(No.10671075), NationalNatural Science Foundation of Guangdong province(No.5005930), and University Special Research Fundfor Ph.D. Program (20060574002).
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considered the free boundary problems about American installment call options in [14]
and European installment call options in [15]. Though their models are similar, their
properties are different, which lead to the different methods applied to them. American
installment call has two monotonic free boundaries, whereas the corresponding European
call has a non-monotonic free boundary if q > 0, and European installment put is more
even complicated, its free boundary disappearing on finite time. Moreover there is no
symmetry relation between the free boundary of European installment put options and
the one of the corresponding call.
Since (1.1) is a degenerate backward parabolic problem, we transform it into a familiar
forward non-degenerate parabolic variational inequality. So let
V (x, τ) = P (S, t)/K, τ = T − t, x = ln(S/K) = lnS − lnK, L = L∗/K, (1.2)
then we have⎧⎪⎨⎪⎩∂τV − LxV = −L, if V > 0 and (x, τ) ∈ IR× (0, T ],
∂τV − LxV ≥ −L, if V = 0 and (x, τ) ∈ IR× (0, T ],
V (x, 0) = (1 − ex)+, x ∈ IR,
(1.3)
where
LxV =σ2
2∂xxV + (r − q − σ2
2)∂xV − rV. (1.4)
The monotonicity and regularity of the free boundary are important to the free bound-
ary problem. Particularly, the behavior ∂τV ≥ 0 is crucial to investigate the regularity.
Based on this behavior it can be deduced that ∂τV is continuous across the free boundary
and C∞ regularity of the free boundary follows (in one dimensional case see [8] and in
higher dimension see [3]). Without the condition ∂τV ≥ 0, the newest progress is that
∂τV is continuous for almost time (see [1]) and the free boundary possesses C∞ regularity
locally around some points which are energetically characterized (see [4]). These results
manifest that the analysis for regularity of the free boundary is not thoroughly solved.
If we cannot get ∂τV ≥ 0, how can we determine the regularity of the free boundary?
In [15], we apply a transformation y = x− kτ, v(y, τ) = V (x, τ), then ∂τv = ∂τV + k∂xV
and since ∂xV ≥ 0, we can let k be large enough so that ∂τv ≥ 0. It means that we
deform the non-monotonic free boundary into a monotonic one. In the present case, if we
apply a similar transformation y = x+ kτ, v(y, τ) = V (x, τ), then we have
v(y, 0) = V (x, 0) = 1 − ex = 1 − ey, ∀ y < 0.
Furthermore, Theorem 1.3 in this paper implies that
∂τv(y, 0) = (∂τV − k∂xV )(x, 0) = Lx(1 − ex) − L+ kex = (q + k)ey − (r + L), if y < 0.
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It is clear that for any fixed k, ∂τv(y, 0) < 0 if y is small enough. So, we cannot get
∂τv(y, 0) ≥ 0 in the full domain IR× [ 0, T ] by the method in [15].
Luckily, the property of the free boundary is only dependent on that of the solution
V in the local domain near the free boundary. So, it is sufficient to prove ∂τv ≥ 0 near
the free boundary, which is the key idea in this paper.
In this paper, we first give a heuristic derivation of the model in the next section, then
establish the existence and uniqueness of the solution in Section 3 by:
Theorem 1.1: Problem (1.3) admits a unique solution V and V ∈W 2,1p (ΩR
T \Bδ(P0)) ∩C(ΩT ) ∩ L∞(ΩT ) for any fixed R > δ > 0, 1 < p < +∞; where ΩT = IR × (0, T ], P0 =
(0, 0), Bδ(P0) = {(x, τ) : y2 + τ 2 ≤ δ2}. Moreover, ∂xV ∈ C(ΩT ),
0 ≤ V ≤ 1, (1.5)
∂xV ≤ 0. (1.6)
If we denote
CRx = {(x, τ) : V (x, τ) > 0} (continuation region),
DRx = {(x, τ) : V (x, τ) = 0} ( default region),
then (1.6) implies that V is monotonic decreasing with respect to x, and we can define
the free boundary as
sx(τ) = inf{x : V (x, τ) = 0}, τ > 0, between CRx and DRx.
Moreover,
CRx = {x < sx(τ)}, DRx = {x ≥ sx(τ)}.In Section 4, we show some properties of the free boundary:
Theorem 1.2: The free boundary sx(τ) possesses the following properties (see Figure 1){s∗x(τ) ≤ sx(τ) ≤ X, τ ∈ (0, T0),
sx(τ) = −∞, τ ∈ [T0 + ∞),(1.7)
limτ→T−
0
sx(τ) = −∞. (1.8)
Here
T0 =1
rlnL+ r
L, s∗x(τ) = ln
[L
r
(er( T0−τ) − 1
)]−(r−q)τ, X =−1
α2ln
((r + L)(α1 − α2)
Lα1
),
where α1 > 0 and α2 < 0 are the roots of the following algebraic equation
σ2
2λ2 + (r − q − σ2
2)λ− r = 0.
3
Theorem 1.3: The free boundary sx(τ) is not monotonic with sx(0)�= lim
τ→0+sx(τ) = 0
(see Figure 1).
�
�τ
x•O
•τ ∗•T0
CRx DRx
sx(τ)
V = 0V > 0
Figure 1. The free boundary sx(τ)
Theorem 1.4: The free boundary sx(τ) is decreasing with respect to r, L, and increasing
with respect to q.
In Section 5, we construct a transformation to deform the original free boundary into
a new monotonic curve and prove its monotonicity and smoothness by considering the
local property of the new solution. Eventually, we show that:
Theorem 1.5: sx(τ) ∈ C[ 0, T0) ∩ C∞(0, T0).
In the last section, we provide numerical result applying the binomial method.
2. Heuristic derivation of the model
An European put option is a contract which gives the owner the right but not the
obligation to sell an asset at a fixed price K at the expiry date T . Because the holder of
the option possibly makes a profit, but impossibly stands a loss for the option, the holder
must pay some premium for it.
In a conventional European put option contract (see [2]), the holder pays the premium
entirely up-front and acquires the right. In a continuous installment European put op-
tion contract, the holder pays a smaller up-front premium and then a constant stream of
installments at a certain rate per unit time. However, the holder can choose at any time
to give up the option contract and stop making installment payments, i.e., the option
contract is canceled.
There are some papers about installment options. Particularly, there is a variational
inequality model in [5] and some numerical results about the model. In the following, we
deduce a parabolic variational inequality model by their idea.
We consider a standard model for a perfect market, continuous trading, no-arbitrage
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opportunity, a constant interest rate r > 0, and an underlying asset with constant con-
tinuous dividend yield q ≥ 0 and price S following a geometric Brownian motion
dS = rSdt+ σSdB, (2.1)
where B is a Wiener process on a risk neutral probability space, σ > 0 is the constant
volatility.
Let P (S, t) denote the value of an European continuous installment put option and
L∗ be the continuous install rate. Applying Ito’s formula to P (S, t) and combining (2.1),
we obtain the dynamics
dP =
(∂P
∂t+σ2
2S2∂
2P
∂S2+ rS
∂P
∂S
)dt+ σS
∂P
∂SdB. (2.2)
We construct a �-hedging portfolio consisting of one continuous installment option and
−� shares of the underlying asset. The value of this portfolio is
Π = P −�S, (2.3)
then, its dynamics is given by
dΠ = dP −�Sqdt−�dS − L∗dt ≤ rΠdt = r(P −�S)dt. (2.4)
Combining (2.1) and (2.2), we get
0 ≥ dΠ − r(P −�S)dt =
(∂P
∂t+σ2
2S2∂
2P
∂S2+ rS(
∂P
∂S−�) + (r − q)�S − rP − L∗
)dt
+ σS
(∂P
∂S−�
)dB. (2.5)
Set � = ∂SP , and the coefficient of dB vanishes. The portfolio is instantaneously risk-
free, then we see that P (S, t) satisfies
∂tP +σ2
2S2∂SSP + (r − q)S∂SP − rP ≤ L∗. (2.6)
What’s more, in continuation region, P (S, t) satisfies
∂tP +σ2
2S2∂SSP + (r − q)S∂SP − rP = L∗. (2.7)
At expiration date T , if S < K, the holder of the option will sell the asset with price S
at K and he can make a profit K − S. On the other hand, if S ≥ K, he will not exercise
the option and stand any loss. Hence, the value of the option is (K −S)+ at maturity T .
That is
P (S, T ) = (K − S)+. (2.8)
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The owner must keep paying premiums to keep the option alive. And if S is large
enough at some time t , the present value of the expected pay-off may be less than that
of the remaining installments, then the holder would allow the option to lapse and stop
paying installment payments. Hence, the value of the option is 0, when (S, t) lies in
default region.
It is clear that P (S, t) ≥ 0 in continuation region. Otherwise, the owner can choose
to give up the option for more wealth. Hence we have
P (S, t) ≥ 0. (2.9)
From the above deduction, we see that the following equality always holds in the two
regions: default region, continuation region[∂tP +
σ2
2S2∂SSP + (r − q)S∂SP − rC − L∗
]P (S, τ) = 0. (2.10)
Then we see that the value of the European continuous-installment put option P (S, t)
satisfies (2.6), (2.8)-(2.10), that is (1.1). Moreover, from the smooth fit conditions [11],
we know that P, ∂SP are continuous. In this paper, we will consider the W 2,1p, loc solution
and free boundary of the problem (1.1).
3. Existence and uniqueness of W 2,1p, loc solution of the problem
(1.3)
In this section, we utilize the penalty method to prove the existence and uniqueness
of W 2,1p, loc ∩ C ∩ L∞ solution to the problem (1.3).
Since the problem lies in an unbounded domain ΩT , we first consider the approximation
problem in the bounded domain ΩnT = (−n, n) × (0, T ], n ∈ IN\{0},⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩
∂τVn −LxVn = −L, if Vn > 0 and (x, τ) ∈ ΩnT ,
∂τVn −Lxvn ≥ −L, if Vn = 0 and (x, τ) ∈ ΩnT ,
Vn(−n, τ) = 1 − e−n, Vn(n, τ) = 0, τ ∈ [ 0, T ],
Vn(x, 0) = (1 − ex)+, x ∈ [−n, n ].
(3.1)
As the standard penalty method, we utilize the following problems to approximate to
the problem (3.1)⎧⎪⎨⎪⎩∂τVε, n −LxVε, n + βε(Vε, n) = −L, in Ωn
T ,
Vε, n(−n, τ) = 1 − e−n, Vε, n(n, τ) = 0, τ ∈ [ 0, T ],
Vε, n(x, 0) = πε(1 − ex), x ∈ [−n, n ],
(3.2)
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where
ε > 0 is small enough, βε(s) ∈ C ∞(IR),
βε(s) ≤ 0, 0 ≤ β ′ε(s) ≤ 2L/ε, β ′′
ε (s) ≤ 0,
and
βε(s) =
{0, s ≥ 2ε,
2Lεs− 3L− rε, s ≤ 3ε/2.
(3.3)
From its definition, it is clear that
βε(ε) = −L− rε, βε(0) = −3L− rε. (3.4)
Since the initial value (1 − ex)+ is not smooth, we utilize πε(s) to smooth it, here
πε(s) =
{s, s ≥ ε,
0, s ≤ −ε, (3.5)
and πε(s) ∈ C ∞(IR), 0 ≤ π′ε(s) ≤ 1, π′′
ε (s) ≥ 0, limε→0+
πε(s) = s+.
Lemma 3.1: For any fixed n ∈ IN\{0}, the problem (3.1) admits a unique solution Vn
and Vn ∈ C(ΩnT ) ∩W 2,1
p (ΩnT\Bδ(P0)) for any fixed 1 < p < +∞, 0 < δ < n. Moreover, if
n is large enough, we have
0 ≤ Vn ≤ 1 − e−n, (3.6)
∂xVn ≤ 0. (3.7)
Proof: As [7], it is not difficult to get the existence of the W 2,1p solution to the problem
(3.2) by applying Schauder fixed point theorem, and get its uniqueness by the comparison
principle. Here, we omit the details. Next, we prove the existence of the W 2,1p solution to
the problem (3.1).
We first estimate Vε, n. From (3.2) and (3.4), we have⎧⎪⎨⎪⎩∂τ0 −Lx 0 + βε(0) = −3L− rε < −L = ∂τVε, n −LxVε, n + βε(Vε, n), in Ωn
T ,
0 ≤ 1 − e−n = Vε, n(−n, τ), 0 = Vε, n(n, τ), τ ∈ [ 0, T ],
0 ≤ πε(1 − ex) = Vε, n(x, 0), x ∈ [−n, n ].
So, 0 is a subsolution to the problem (3.2). Hence, we have Vε, n ≥ 0.
Moreover, if 2ε < 1 − e−n, we can deduce that⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩∂τ (1 − e−n) − Lx (1 − e−n) + βε(1 − e−n) = r(1 − e−n) > −L
= ∂τVε, n − LxVε, n + βε(Vε, n), in ΩnT ,
1 − e−n = Vε, n(−n, τ), 1 − e−n ≥ 0 = Vε, n(n, τ), τ ∈ [ 0, T ],
1 − e−n ≥ πε(1 − ex) = Vε, n(x, 0), x ∈ [−n, n ].
7
It implies 1 − e−n is a supersolution to the problem (3.2). Hence, if 2ε < 1 − e−n, we
obtain that
0 ≤ Vε, n ≤ 1 − e−n. (3.8)
From the properties of βε, we have that if ε is small enough, then
−3L− 1 ≤ −3L− rε = βε(0) ≤ βε(Vε, n) ≤ βε(1 − e−n) ≤ 0.
By W 2,1p and Cα, α/2 (0 < α < 1) estimates of parabolic problem, we conclude that
|Vε, n|W 2,1p (Ωn
T \Bδ(P0)) ≤ C,
|Vε, n|Cα, α/2(ΩnT ) ≤ C,
where C is independent of ε. It implies that, possibly a subsequence,
Vε, n ⇀ Vn in W 2,1p (Ωn
T\Bδ(P0)) weakly and Vε, n → Vn in C(ΩnT ), as ε → 0+.
We can prove that Vn is the solution of the problem (3.1) by the standard penalty
method in [7]. Let ε→ 0+ in (3.8), then (3.6) is obvious.
In order to obtain (3.7), we need (3.9)
∂xVε, n ≤ 0. (3.9)
Derivative (3.2) with respect to x, and denote W as ∂xVε, n, then W satisfies⎧⎪⎨⎪⎩∂τW − LxW + β ′
ε(Vε, n)W = 0, in ΩnT ,
W (−n, τ) ≤ 0, W (n, τ) ≤ 0, τ ∈ [ 0, T ],
W (x, 0) = −π′ε(1 − ex)ex ≤ 0, x ∈ [−n, n ],
where W (±n, τ) ≤ 0 are deduced from estimate (3.8). Then, the comparison principle
implies (3.9), and (3.7) is the consequence of (3.9).
At last, we prove the uniqueness. Suppose V1 and V2 are two W 2.1p, loc(Ω
nT )
⋂C(Ωn
T )
solutions to the problem (3.1) and denote
N = {(y, τ) : V1(x, τ) < V2(x, τ), −n < x < n, 0 < τ ≤ T}.
Suppose N is not empty, then
V2(x, τ) > 0, ∂τV2(x, τ) − LxV2(x, τ) = −L, ∀ (x, τ) ∈ N .
Denote W = V2 − V1, then W satisfies{∂τW − LxW ≤ 0, (x, τ) ∈ N ,
W (x, 0) = 0, on ∂pN .
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where ∂pN is the parabolic boundary of domain N . Applying the A-B-P maximum
principle (see [13]), we have W ≤ 0 in N , which contradicts the definition of N . �
Proof of Theorem 1.1: Applying
(∂τ −Lx) 0 = 0,
we rewrite the problem (3.1) as⎧⎪⎨⎪⎩∂τVn −LxVn = f(x, τ), in Ωn
T ,
Vn(−n, τ) = 1 − e−n, Vn(n, τ) = 0, τ ∈ [ 0, T ],
Vn(x, 0) = (1 − ex)+, x ∈ [−n, n ].
(3.10)
Vn ∈W 2.1p, loc( Ωn
T ) implies
f(x, τ) ∈ Lploc( Ωn
T ), f(x, τ) = I{Vn>0} (−L) a.e. in ΩnT , (3.11)
where IA denotes the indicator function of the set A.
Hence, for any fixed R > δ > 0, if n > R, combining (3.6), we have the following W 2.1p
uniform interior estimates in domain ΩRT
‖Vn‖W 2.1p (ΩR
T \Bδ(P0) )
≤ C(‖Vn‖L∞(ΩRT ) + ‖(1 − ex)+‖C2([−R,−δ ]) + ‖f(x, τ)‖L∞(ΩR
T ) ) ≤ C, (3.12)
where C depends on R, but is independent of n. Let n→ ∞, then we have that, possibly
a subsequence,
Vn ⇀ VR in W 2.1p (ΩR
T \Bδ(P0)) weakly as n→ +∞.
Moreover, the Sobelev imbedding theorem and Cα, α/2 estimate implies that
Vn → VR in C(ΩRT ), ∂xVn → ∂xVR in C(ΩR
T \Bδ(P0)) as n→ +∞. (3.13)
By the standard diagonal argument, extract a subsequence {V (n)n } such that for any
m ∈ IN\{0}, there holds
V (n)n → Vm in W 2.1
p (ΩmT \B1/m(P0)) weakly; V (n)
n → Vm in C(ΩmT ), as n→ +∞.
It is clear that Vm+1 = Vm in [−m,m ] × [ 0, T ], so, we can define V = Vm if
(x, τ) ∈ [−m,m ] × [ 0, T ], what’s more, it is not difficult to deduce that V is the so-
lution to the problem (1.3). Furthermore, (3.13) implies that ∂xV ∈ C(ΩT ).
(1.5), (1.6) is the consequences of (3.6), (3.7) respectively. The proof of the uniqueness
is same as the proof in Lemma 3.1. �
9
4. Some characterizations of the free boundary
In this section, we establish the bounds of the free boundary, which is the preparation
for proving smoothness of the free boundary in the next section. Moreover, we prove
the free boundary is not monotonic with respect to τ , but monotonic with respect to
parameters r, q and L.
In the first, we estimate the bounds of the free boundary through the following three
lemmas and Theorem 1.2.
Lemma 4.1: If τ ≥ T0 (T0 was defined in 1.2), V (x, τ) ≡ 0 (see Figure 1), it means that
CRx ⊂ {0 < τ < T0}, DRx ⊃ {τ ≥ T0}, sx(τ) = −∞, ∀ τ ≥ T0. (4.1)
Proof: Define
W (x, τ) =
⎧⎨⎩1
r[(L+ r)e−rτ − L], 0 ≤ τ ≤ T0,
0, T0 < τ ≤ T.
We claim that W possesses the following four properties⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
(I) W ∈W 2,1p (IR× (0, T )) ∩ C(IR× [ 0, T ]),
(II) W (x, τ) ≥ 0, (x, τ) ∈ IR× (0, T ],
(III) W (x, 0) = 1 > (1 − ex)+ = V (x, 0), x ∈ IR,
(IV) ∂τW − LxW ≥ −L, a.e. in IR× (0, T ).
In fact, from the definition of T0 in Theorem 1.2, we have
W (x, T0) =1
r
[(L+ r) exp
{−r
(1
rlnL+ r
L
)}− L
]=
1
r
[(L+ r)
L
L+ r− L
]= 0,
then property (I) is obvious. Moreover, if 0 < τ ≤ T0, then we deduce
∂τW (x, τ) = −(L+ r)e−rτ ≤ 0.
Combining W (x, T0) = 0, we have property (II). It is easy to check property (III) from the
definition of W . Next, we manifest property (IV) according to the following two cases.
In one case of 0 < τ < T0,
∂τW − LxW = −(L+ r)e−rτ + (L+ r)e−rτ − L = −L;
in the other case of T0 < τ < T ,
∂τW − LxW = 0 ≥ −L.
10
So, we testify properties (I)−−(IV). In the following, we utilize the properties to prove
W ≥ V .
Otherwise, Ω = {W < V } is nonempty, then we have
V > W ≥ 0, ∂τV −LxV = −L, in Ω.
Property (IV) implies that
∂τ (W − V ) −Lx(W − V ) ≥ 0, in Ω.
In the case that the parabolic boundary of Ω lies in IR × (0, T ], from the continuity
of V, W , we deduce that
W − V = 0,
in the case that the parabolic boundary of Ω lies on {τ = 0}, from property (III), we
deduce that
W − V > 0.
Applying the A-B-P maximum principle (see [13]), we have
W ≥ V, in Ω,
which contradicts the definition of Ω. So, we achieve that W ≥ V .
Combining W (x, τ) = 0, τ ≥ T0, it is clear that
V (x, τ) ≤ 0, ∀ τ ≥ T0.
Since V ≥ 0, we have
V (x, τ) ≡ 0, ∀ τ ≥ T0.
�Lemma 4.2: If x < s∗x(τ), 0 < τ < T0 (s∗x(τ) was defined in 1.2), V (x, τ) > 0, it means
that (see Figure 2)
sx(τ) ≥ s∗x(τ), ∀ 0 < τ < T0. (4.2)
Proof: If 0 ≤ τ < T0, then we have
L
r
(er( T0−τ) − 1
)>L
r(1 − 1) = 0,
which means the definition of s∗x(τ) in Theorem 1.2 is reasonable. Denote
w(x, τ) = (1 − ex+(r−q)τ ) +r + L
r(e−rτ − 1), (x, τ) ∈ IR× [ 0, T0 ].
11
We claim that w possesses the following four properties⎧⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎩
(I) w ∈ C 2,1(IR× [ 0, T0 ]),
(II) w(x, 0) = 1 − ex ≤ V (x, 0), x ∈ IR,
(III) ∂τw −Lxw ≤ −L, (x, τ) ∈ IR× (0, T0 ],
(IV) w(x, τ) > 0 (x, τ) ∈ (−∞, s∗x(τ) ) × [ 0, T0).
�
�τ
x•O
•T0
CRx
DRx
sx(τ)
s∗x(τ)
V = 0V > 0
Figure 2. The lower bound of sx(τ)
In fact, from the definition of w, it is easy to get (I), (II). Now, we check properties
(III) and (IV). Since
∂τw − Lxw
= −(r − q)ex+(r−q)τ − (r + L)e−rτ +
(σ2
2+ r − q − σ2
2
)ex+(r−q)τ + rw
= −(r + L)e−rτ + r − rex+(r−q)τ + (r + L)e−rτ − (r + L)
= −L− rex+(r−q)τ < −L,
we show property (III). Moreover, if 0 ≤ τ < T0, x < s∗x(τ), we have
w(x, τ) > 1 − es∗x(τ)+( r−q)τ +r + L
r
(e−rτ − 1
)= 1 − exp
{ln
[L
r(er( T0−τ) − 1)
]}+r + L
re−rτ −
(1 +
L
r
)= −L
rer( T0−τ) +
L
r+r + L
re−rτ − L
r= −L
r
r + L
Le−rτ +
r + L
re−rτ = 0.
Therefor, we get property (IV).
By properties (I), (II), (III) and the comparison principle, we have
V ≥ w in ΩT0 .
Combining property (IV), the conclusion in Lemma 4.2 is obvious. �
12
Lemma 4.3: If x ≥ X, V (x, τ) = 0(X was defined in 1.2), it means that (see Figure 3)
CRx ⊂ {x ≤ X, 0 < τ ≤ T0}, DRx ⊃ {x > X}, sx(τ) ≤ X. (4.3)
�
�τ
x•X
•x
•τx
•O
•T0
CRx DRx
sx(τ)
V = 0V > 0
Figure 3. The upper bound of sx(τ)
Proof: Denote
W (x, τ) =
⎧⎨⎩L
r(α1 − α2)
[α1e
α2(x−X) − α2eα1(x−X)
] − L
r, x ≤ X,
0, x > X.
We claim that W possesses the following three properties⎧⎪⎪⎨⎪⎪⎩(I) W ∈W 2,1
p (IR× (0, T )) ∩ C(IR× [ 0, T ]),
(II) ∂τW −LxW ≥ −L, a.e. in IR× (0, T ),
(III) W (x, τ) ≥ (1 − ex)+, (x, τ) ∈ IR× [ 0, T ].
In fact, it is easy to check that
W (X, τ) =L
r(α1 − α2)(α1 − α2) − L
r= 0,
∂xW (X−, τ) =L
r(α1 − α2)(α1α2 − α2α1) = 0.
They imply property (I). Since α1, α2 are the characteristic roots to the ODE LxW =
0, we have
∂τW −LxW = r
(−Lr
)= −L, ∀ x < X.
Combining W (x, τ) = 0 if x ≥ X, we get property (II). From the definition of W and
α1 > 0, α2 < 0, we have
∂xW (x, τ) =Lα1α2
r(α1 − α2)
[eα2(x−X) − eα1(x−X)
]< 0, ∀ x < X.
13
On the other hand, if we notice that X > 0, α2 < 0, from the definition of X, it is
not difficult to check that
W (0, τ) ≥ L
r(α1 − α2)α1e
−α2X − L
r=
Lα1
r(α1 − α2)
r + L
L
α1 − α2
α1
− L
r
=r + L
r− L
r= 1.
So, W (X, τ) ≥ 1 if x ≤ 0. Hence, property (III) is obvious.
From properties (I)−−(III), as the method in Lemma 4.1, we can deduce W ≥ V and
the conclusion in the lemma. �
Proof of Theorem 1.2: From Lemma 4.1−− 4.3, it is sufficient to prove (1.8).
If we show that for any x small enough, there exists some τx such that (see Figure 3)
0 < τx < T0 and V (x, τ) ≡ 0, ∀ τ ∈ [ τx, T0 ], (4.4)
then we conclude that sx(τ) ≤ x for any τx ≤ τ ≤ T0 and
lim supτ→T−
0
sx(τ) ≤ x.
Taking x→ −∞, then we have (1.8).
In the following, we prove (4.4). Fix x0 < 0 and define
ϕ(x) =
⎧⎪⎨⎪⎩Le2x0
2 r(α1 − α2)
[α1e
α2(x−x0) − α2eα1(x−x0)
]+
(1 − 2r + L
2re2x0
), x ≤ x0,
1 − e2x0 , x > x0,
where α1, α2 are presented in the definition in Theorem 1.2. We claim that ϕ possesses
the following three properties if x0 is small enough,⎧⎪⎪⎨⎪⎪⎩(I) ϕ ∈W 2
p, loc(IR) ∩ C(IR),
(II) −Lxϕ ≥ r − 2r+L2e2x0 , a.e. in IR,
(III) ϕ(x) > (1 − ex)+, x ∈ IR.
(4.5)
In fact, we can check that
ϕ(x0) =Le2x0
2 r(α1 − α2)(α1 − α2) +
(1 − 2r + L
2re2x0
)= 1 − e2x0 ,
ϕ ′(x0−) =Le2x0
2 r(α1 − α2)(α1α2 − α2α1) = 0.
then property (I) is obvious. Since α1, α2 are the characteristic roots to the ODE Lxϕ = 0,
we have
−Lxϕ = r
(1 − 2r + L
2re2x0
)= r − 2r + L
2e2x0 , ∀ x < x0,
−Lxϕ = r(1 − e2x0) ≥ r − 2r + L
2e2x0 , ∀ x > x0.
14
In consequence, we have property (II). Next, we prove property (III).
In the case that x ≥ x0, recalling x0 < 0, we have
ϕ(x) = 1 − e2x0 ≥ 1 − ex0 ≥ (1 − ex)+.
In the case that 2x0 ≤ x < x0, noting α1 > 0, α2 < 0, we deduce
ϕ ′(x) =Lα1α2e
2x0
2 r(α1 − α2)
(eα2(x−x0) − eα1(x−x0)
)<
Lα1α2e2x0
2 r(α1 − α2)(1 − 1) = 0,
ϕ(x) > ϕ(x0) = 1 − e2x0 > 1 − ex = (1 − ex)+.
In the case that x < 2x0, if x0 is small enough, it is easy to see that
ϕ ′′(x) =Le2x0
2 r(α1 − α2)
(α1α
22e
α2(x−x0) − α21α2e
α1(x−x0))> 0,
ϕ ′(x) < ϕ ′(2x0) =Lα1α2
2 r(α1 − α2)(eα2x0 − eα1x0) e2x0 < −e2x0 < −ex = (1 − ex) ′
x,
ϕ(x) − (1 − ex)+ > ϕ(2x0) − (1 − e2x0)+ > ϕ(x0) − (1 − e2x0) = 0.
In all, we have property (III) if x0 is small enough. To prove (4.4), we need another
function
ψ(τ) =
⎧⎨⎩2(r + L) − (2r + L)e2x0
2 r
(e−rτ − 1
), 0 ≤ τ ≤ τ0,
e2x0 − 1, τ0 < τ ≤ T0,
where τ0 satisfies
1 − e−rτ0 =2r(1 − e2x0)
2(r + L) − (2r + L)e2x0.
Since
0 < 1 − e−rτ0 <2r(1 − e2x0)
2(r + L) − (2r + 2L)e2x0=
r
r + L= 1 − e−rT0 ,
we see that 0 < τ0 < T0, which means the definition of ψ(τ) is reasonable. Moreover,
ψ(τ0) = −2(r + L) − (2r + L)e2x0
2 r
2r(1 − e2x0)
2(r + L) − (2r + L)e2x0= e2x0 − 1,
which implies that
ψ(τ) ∈W 1p ( [ 0, T0 ] ) ∩ C( [ 0, T0 ] ). (4.6)
Denote W (x, τ) = ψ(τ)+ϕ(x), we claim that W possesses the following five properties⎧⎪⎪⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎪⎪⎩
(I) W ∈W 2,1p, loc(IR× [ 0, T0 ]) ∩ C(IR× [ 0, T0 ]),
(II) W (x, 0) = ϕ(x) > (1 − ex)+, x ∈ IR,
(III) W (x, τ) ≥ 0, (x, τ) ∈ IR× [ 0, T0 ],
(IV) ∂τW − LxW ≥ −L, a.e. in IR× (0, T0 ],
(V) W (x0, τ) = 0 τ ∈ [ τ0, T0 ].
(4.7)
15
In fact, we can get property (I) and (II) from (4.5) and (4.6). Moreover, since
ψ ′(τ) =
[−(r + L) +
(r +
L
2
)e2x0
]e−rτ < 0, ϕ ′(x) ≤ 0,
We have
W (x, τ) ≥ ψ(τ0) + ϕ(x0) = (e2x0 − 1) + (1 − e2x0) = 0, ∀ (x, τ) ∈ IR× [ 0, T0 ],
which is property (III). Next, we prove property (IV). From property (II) in (4.5), we
deduce that
∂τW − LxW = ψ′ + rψ −Lxϕ ≥ −(r + L) +
(r +
L
2
)e2x0 + r − 2r + L
2e2x0 = −L,
then we have property (IV). From the definition of ϕ, ψ, W , we see that
W (x0, τ) = ψ(τ) + ϕ(x0) = (e2x0 − 1) + (1 − e2x0) = 0, ∀ τ0 ≤ τ ≤ T0.
So, we have property (V). As the method in the proof of Lemma 4.1, from the
properties in (4.7), we conclude that
W ≥ V in ΩT0 .
Combining property (V) in (4.7) and V ≥ 0, we have (4.4) and the conclusion in the
theorem. �
Next, we show the free boundary is not monotonic by two lemmas and Theorem 1.3.
Lemma 4.4: There exists a τ ∗ > 0 such that V (0, τ) > 0 for any 0 < τ < τ ∗ (see Figure
1), which means that
{x = 0, 0 < τ < τ ∗} ⊂ CRx, sx(τ) > 0 for 0 < τ < τ ∗.
Proof: It is clear that the solution to the problem (1.3), V (x, τ) satisfies{∂τV − LxV ≥ −L, (x, τ) ∈ IR× (0, T ],
V (x, 0) = (1 − ex)+, x ∈ IR.(4.8)
Next, we consider the following Cauchy problem{∂τV1 −LxV1 = −L, (x, τ) ∈ IR× (0, T ],
V1(x, 0) = (1 − ex)+, x ∈ IR.(4.9)
Applying the comparison principle to problems (4.8) and (4.9), we have
V (x, τ) ≥ V1(x, τ).
16
If we can prove that there exists a τ ∗ > 0 such that V1(0, τ) > 0 for any 0 < τ < τ ∗,
then the conclusion in the lemma is obvious. For it, we denote
V2(x, τ) = V1(x, τ) − L
r(e−rτ − 1), (4.10)
then (4.9) implies that V2(x, τ) satisfies{∂τV2 −LxV2 = 0, (x, τ) ∈ IR× (0, T ],
V2(x, 0) = (1 − ex)+, x ∈ IR.(4.11)
This is the Cauchy problem which the price of the standard European put option
satisfies. Its solution has an explicit formula (see [10])
V2(x, τ) = e−rτN(−d2) − ex−qτN(−d1), (4.12)
where
N(−d1) =1√2π
∫ −cd1
−∞e−η2/2dη, d1 =
x+ (r − q + σ2/2)τ
σ√τ
, d2 =x+ (r − q − σ2/2)τ
σ√τ
.
It is easy to check that
∂τV2(0, τ) = −re−rτN(−d2) − e−rτn(d2)r − q − σ2/2
2σ√τ
+ qe−qτN(−d1)
+ e−qτn(d1)r − q + σ2/2
2σ√τ
≥ −r + e−qτn(d1)σ
2√τ,
where
n(d1) =1√2π
e−d21/2, d1 =
(r − q + σ2/2)√τ
σ, d2 =
(r − q − σ2/2)√τ
σ.
Here we had utilized the the equality e−qτn(d1) = e−rτn(d2). Letting τ → 0+, we have
d1 → 0, n(d1) → 1/√
2π, ∂τV2(0, τ) → +∞,
From (4.10), we deduce that
∂τV1(0, τ) → +∞, as τ → 0+.
Combining
V1(0, 0) = 0 and V1 ∈ C(IR× [ 0, T ]),
we see that there really exists a τ ∗ > 0 such that V1(0, τ) > 0 for any 0 < τ < τ ∗. So, the
conclusion in the lemma is correct too. �
Lemma 4.5:
lim supτ→0+
sx(τ) ≤ 0.
17
Proof: It is sufficient to prove there exists a δ > 0 such that (see Figure 4)
sx(τ) ≤ s∗∗x (τ)�= (1 + ln 2)
(σ2τ
2 ln 2
)1/3
+ qτ, ∀ 0 < τ < δ. (4.13)
�
�τ
xs∗∗x (τ)
•O•δ
•T0
CRx
DRxsx(τ)
Figure 4. The asymptotic property of sx(τ) as τ → +∞
Fix t ∈ (0, δ ] and apply a transformation
z = x+
(r − q − k2σ2
2
)τ, V (z, τ) = e(r−k2σ2/2)τV (x, τ), (4.14)
where k satisfies
k3 =2 ln 2
σ2t. (4.15)
We choose δ small enough such that for any t ∈ [ 0, δ ], k in (4.15) is large enough and
satisfies
k > 1,k2σ2
2− r > 0,
L
(1 + k2)σ2( ek − 1 ) ≥ 1. (4.16)
We can check that V satisfies⎧⎪⎪⎨⎪⎪⎩∂τ V − LzV = −Le(r−k2σ2/2)τ , if V > 0 and (z, τ) ∈ IR× (0, t ],
∂τ V − LzV ≥ −Le(r−k2σ2/2)τ , if V = 0 and (z, τ) ∈ IR× (0, t ],
V (z, 0) = (1 − ez)+, z ∈ IR,
(4.17)
where
LzV =σ2
2
(∂zzV + (k2 − 1) ∂zV − k2V
). (4.18)
For proving (4.13), define
W (z, τ) =
⎧⎪⎨⎪⎩L
k2σ2
{1
1 + k2
[e−k2(z−1/k) + k2ez−1/k
]− 1
}, (z, τ) ∈ (−∞, 1/k ] × [ 0, t ],
0, (z, τ) ∈ (1/k,+∞) × [ 0, t ].
18
We claim that W possesses the following three properties⎧⎪⎪⎨⎪⎪⎩(I) W ∈W 2,1
p (IR× (0, t)) ∩ C(IR× [ 0, t ]),
(II) ∂τW −LzW ≥ −Le(r−k2σ2/2)τ , a.e. in IR× (0, t),
(III) W (z, τ) ≥ (1 − ez)+, (z, τ) ∈ IR× [ 0, t ].
In fact, it is clear that
W (1/k, τ) =L
k2σ2
[1
1 + k2(1 + k2) − 1
]= 0,
∂zW ((1/k)−, τ) =L
k2σ2
[1
1 + k2(−k2 + k2)
]= 0,
so, property (I) is obvious. Since −k2 and 1 are the characteristic roots to the ODE
LzW = 0, combining (4.15) and (4.16), we see that if (z, τ) ∈ (−∞, 1/k) × (0, t), then
∂τW − LzW = −k2σ2
2
L
k2σ2= −L
2= −Le− ln 2 ≥ −Le− ln 2/k = −Le−k2σ2t/2
≥ −Le−k2σ2τ/2 ≥ −Le(r−k2σ2/2)τ .
Moreover, in the other case of (z, τ) ∈ (1/k,∞) × (0, t), we still have
∂τW −LzW = 0 ≥ −Le(r−k2σ2/2)τ ,
Therefore, we achieve property (II). From the definition of W and (4.16), we can see that
if (z, τ) ∈ (−∞, 1/k) × [ 0, t ]
∂zW (z, τ) =L
k2σ2
−k2
1 + k2
[e−k2(z−1/k) − ez−1/k
]= − L
(1 + k2)σ2
[e−k2(z−1/k) − ez−1/k
]< 0,
∂zW (0, τ) = − L
(1 + k2)σ2( ek − e−1/k ) ≤ − L
(1 + k2)σ2( ek − 1 ) ≤ −1,
∂zzW (z, τ) =L
(1 + k2)σ2
[k2e−k2(z−1/k) + ez−1/k
]≥ 0.
Combining ∂zW (1/k, τ) = 0, we see that when 0 ≤ τ ≤ t
∂zW (z, τ) < 0, ∀ z < 1/k, ∂zW (z, τ) < −1 < −ez, ∀ z < 0.
Since W (1/k, τ) = 0, we have that if 0 ≤ τ ≤ t
W (z, τ) > 0, ∀ z < 1/k, W (z, τ) > 1 − ez, ∀ z < 0.
Hence, we get property (III).
19
By properties (I) −−(III), as the method in the proof of Lemma 4.1, we have W ≥ V .
Combining
W (z, τ) = 0, ∀ (z, τ) ∈ (1/k,+∞) × [ 0, t ],
we deduce
V (z, τ) = 0, ∀ (z, τ) ∈ (1/k,+∞) × [ 0, t ].
From transformation (4.14), it is clear that
V (x, τ) = 0, ∀ (x, τ) ∈{
(x, τ) : x >1
k−
(r − q − k2σ2
2
)τ, 0 ≤ τ ≤ t
}.
So, by (4.15), for any 0 < t < δ, we have
sx(t) ≤ 1
k−
(r − q − k2σ2
2
)t ≤ 1
k+ qt +
k2σ2
2t =
1 + ln 2
k+ qt
= (1 + ln 2)
(σ2t
2 ln 2
)1/3
+ qt,
which is (4.13). Hence, we have the conclusion in the lemma. �
Proof of Theorem 1.3: Lemma 4.5 implies that
lim supτ→0+
sx(τ) ≤ 0.
From Lemma 4.4, we see that there exists a τ ∗ > 0 such that
sx(τ) > 0, ∀ 0 < τ < τ ∗.
So, the free boundary sx(τ) is not monotonically decreasing with (see Figure 3)
sx(0) = limτ→0+
sx(τ) = 0.
By Lemma 4.1, we have
sx(τ) = −∞, ∀ τ > T0.
So, the free boundary sx(τ) is not monotonically increasing (see Figure 4). Hence, the
free boundary sx(τ) is not monotonic. �
In the following, we consider the relationship between the free boundary and param-
eters r, q and L by Theorem 1.4.
Proof of Theorem 1.4: Suppose V1 is the solution to (1.3), where r is r1, V2 is the
solution to (1.3), where r is r2 and r1 > r2. By (1.5) and (1.6), we deduce
∂xV1 − V1 ≤ 0, (x, τ) ∈ IR× (0, T ],
20
then V1 satisfies⎧⎪⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎪⎩
∂τV1 − σ2
2∂xxV1 − (r2 − q − σ2
2)∂xV1 + r2V1 = −L+ (r1 − r2)(∂xV1 − V1) ≤ −L,
if V1 > 0 and (x, τ) ∈ IR× (0, T ],
∂τV1 − σ2
2∂xxV1 − (r2 − q − σ2
2)∂xV1 + r2V1 ≥ −L+ (r1 − r2)(∂xV1 − V1),
if V1 = 0 and (x, τ) ∈ IR× (0, T ],
V1(x, 0) = (1 − ex)+, x ∈ IR.
By the comparison principle for the solution of variational inequality with respect to
nonhomogeneous term, we have V1 ≤ V2. If we set CR1, CR2 as the continuation regions
of V1, V2 respectively, then we have CR1 ⊂ CR2. So, it is clear that the free boundary
sx(τ) is decreasing with respect to r.
Applying the same method, we can prove the other conclusions in the theorem. �
5. Smoothness of the free boundary
In this section, we will apply transformation (5.8) to deform the free boundary into
a new monotonic curve. Moreover, we consider the new problem only in a local domain
{(y, τ) : A1+kτ ≤ y ≤ A2+kτ, 0 ≤ τ ≤ T1} rather than the full domain IR×[ 0, T ], where
k, A1, A2, T1 are some constants, see Lemma 5.1. The transformation and localization
are the key ideas in the paper.
By Lemma 4.1, we consider the variational inequality only on domain IR × [ 0, T0)
because V ≡ 0 if τ ≥ T0. Fix T1 ∈ (0, T0) and consider the problem (1.3) only on domain
IR× [ 0, T1 ]. For our transformation and localization, we make a preparation
Lemma 5.1: There exist some constants k, A1, A2 such that (see Figure 5)
{(sx(τ), τ) : 0 ≤ τ ≤ T1} ⊂ (A1 + 1, A2 − 1) × [ 0, T1 ], (5.1)
(∂τV − k∂xV ) ≥ 0, a.e. in [A1, A2 ] × [ 0, T1 ]. (5.2)
�
�τ
x•O •A2•A1
•T0
•T1�M2�
�M1�
CRx DRx
sx(τ)
Figure 5. The original free boundary sx(τ)
21
Proof: We choose a A1 < s∗x(T1) − 1 and choose a A2 > X + 1, where s∗x(τ), X were
defined in Theorem 1.2. From Theorem 1.2, (5.1) is obvious.
Next, we prove (5.2). If we denote Mn = (A1 − 1/n,A1 + 1/n)× (0, T1 ], then Lemma
4.2 implies M1 ⊂ CRx and
∂τ (∂xV ) − Lx(∂xV ) = 0, in M1. (5.3)
From (1.5), (1.6) and the strong maximum principle, we see
V > 0, ∂xV < 0, in M1.
Moreover, from the initial value in the problem (1.3), we have
V (x, 0) = 1 − ex ≥ 1 − eA1+1 > 0, ∀x ∈ [A1 − 1, A1 + 1 ],
∂xV (A1, 0) = ∂x(1 − ex)|x=A1 = −eA1 < 0.
Since V, ∂xV ∈ C(M1), there exists a δ > 0 such that
V (x, τ) ≥ δ, in M2, (5.4)
∂xV (A1, τ) ≤ −δ, ∀ τ ∈ [ 0, T1 ]. (5.5)
Since the equation (5.3) holds in domain M2, applying the C2+α,1+α/2 interior estimates
of parabolic problem, we deduced that ∂τV (A1, τ) ∈ C[ 0, T1 ], which means there exists
a positive constant C such that
∂τV (A1, τ) ≥ −C, ∀ τ ∈ [ 0, T1 ].
Combining (5.5), we see that there exists a constant k > 0 such that⎧⎪⎨⎪⎩(∂τV − k∂xV )(A1, τ) ≥ 1, ∀ τ ∈ [ 0, T1 ],
(∂τV − k∂xV )(x, 0) = −L+ (LxV − k∂xV )(x, 0) = −(L+ r) + (k + q)ex
≥ −(L+ r) + (k + q)eA1−1/2 ≥ 0, ∀ x ∈ [A1 − 1/2, A1 + 1/2 ].
(5.6)
Here we had utilized the equation and the initial value in the problem (1.3). For proving
(5.2), we come back to the problem (3.2) and let
W = ∂τV − k∂xV, Wε, n = ∂τVε, n − k∂xVε, n.
Without loss of generalization, we suppose n > max{−A1, A2}+ 1, 2ε < δ. (5.4) and
the definition of βε imply that
βε(Vε, n) = 0 in M2,
{∂τWε, n −LxWε, n = 0, in M
2,
Wε, n(x, 0) = −(L+ r) + (k + q)ex, x ∈ [A1 − 1
2, A1 + 1
2
].
22
Repeating the W 2,1p interior estimates of parabolic problem, we can deduce that there
exists a subsequence, still denoted by itself,
Wε, n(A1, τ) →W (A1, τ) in C[ 0, T1 ].
Then the first inequality in (5.6) implies that
Wε, n(A1, τ) ≥ 0, ∀ τ ∈ [ 0, T1 ].
Combining (3.2), we see that in domain [A1, n ] × [ 0, T1 ], Wε, n(x, τ) satisfies⎧⎪⎨⎪⎩∂τWε, n −LxWε, n + β ′
ε(Vε, n)Wε, n = 0,
Wε, n(A1, τ) ≥ 0, Wε, n(n, τ) ≥ 0, τ ∈ [ 0, T1 ],
Wε, n(x, 0) = −L+ Lxπε(1 − ex) − βε(πε(1 − ex)) + kπ′ε(1 − ex)ex, x ∈ [A1, n ],
where Wε, n(n, τ) ≥ 0 comes from Vε, n(n, τ) = 0 and (3.9). Moreover, if k is large enough,
the definitions of βε and πε imply that
Wε, n(x, 0)
= −L+σ2
2π′′
ε (1 − ex)e2x + (k + q − r)π′ε(1 − ex)ex − rπε(1 − ex) − βε(πε(1 − ex))
≥
⎧⎪⎨⎪⎩−L− rπε(1 − ex) − βε(πε(1 − ex)) ≥ −L− rε− βε(ε) = 0, 1 − ex ≤ ε,
−L+ (k + q − r)ex − r(1 − ex) = −r − L+ (k + q)ex
≥ −r − L+ (k + q)eA1 ≥ 0, x ≥ A1 and 1 − ex > ε.
(5.7)
Applying the maximum principle, we have
Wε, n(x, τ) ≥ 0, ∀ (x, τ) ∈ [A1, n ] × [ 0, T1 ] ⊃ [A1, A2 ] × [ 0, T1 ].
Let ε→ 0+, n→ ∞, then (5.2) is obvious. �
Thanks to Lemma 5.1, we can apply transformation (5.8) to deform the free boundary
into a monotonicity curve and prove the new free boundary is smooth. In the first, we
give the transformation
y = x+ kτ, v(y, τ) = V (x, τ), (5.8)
where k is the constant in Lemma 5.1. Then the problem (1.3) is equivalent to problem⎧⎪⎨⎪⎩∂τv − Lyv = −L, if v > 0 and (y, τ) ∈ IR× [ 0, T1 ],
∂τv − Lyv ≥ −L, if v = 0 and (y, τ) ∈ IR× [ 0, T1 ],
v(y, 0) = (1 − ey)+, y ∈ (−∞,+∞),
(5.9)
where
Lyv =σ2
2∂yyv − (r − q − σ2
2− k)∂yv − rv.
23
Respectively set sy(τ), CRy, DRy as the counterparts of sx(τ), CRx, DRx under
transformation (5.8) and denote
M = {(y, τ) : A1 + kτ ≤ y ≤ A2 + kτ, 0 ≤ τ ≤ T1}.
From Lemma 5.1 , (1.6) and transformation (5.8), it is clear that (see Figure 6)
{(sy(τ), τ) : 0 ≤ τ ≤ T1} ⊂ M, (5.10)
∂τv = (∂τV − k∂xV ) ≥ 0, a.e. in M, (5.11)
∂yv(y, τ) = ∂xV (x, τ) ≤ 0, ∀ (y, τ) ∈ M. (5.12)
�
�τ
y•O
•T1
•A1
•A2
��
��
��
��
�
��
��
��
��
�
CRy DRy
sy(τ)
Figure 6. The deformed free boundary sy(τ)
We consider the problem (5.9) constrained in domain M:⎧⎪⎪⎪⎪⎨⎪⎪⎪⎪⎩∂τv − Lyv = −L, if v > 0 and (y, τ) ∈ M,
∂τv − Lyv ≥ −L, if v = 0 and (y, τ) ∈ M,
v(A1 + kτ, τ) = V (A1, τ), v(A2 + kτ, τ) = 0, τ ∈ [ 0, T1 ],
v(y, 0) = (1 − ey)+, y ∈ [A1, A2 ].
(5.13)
Theorem 5.1: sy(τ) ∈ C[ 0, T1 ] ∩ C∞(0, T1 ] and is strictly increasing with sy(0) = 0
(see Figure 6).
Proof: We divide the proof into 4 steps.
Step 1: From (5.11) and (5.12), we see that sy(τ) is non-decreasing in [ 0, T1 ].
Step 2: Prove sy(τ) is continuous in [ 0, T1 ] and sy(0) = 0.
In the first we prove sy(τ) is continuous in [ 0, T1 ]. Otherwise, there exists a domain
(y1, y2) × (τ0, T1) (A1 < y1 < y2 < A2, 0 ≤ τ0 < T1) such that{∂τv − Lyv = −L, (y, τ) ∈ (y1, y2) × (τ0, T1),
v(y, τ0) = 0, y ∈ (y1, y2).
Then we have ∂τv(y, τ0) = −L < 0 for any y1 < y < y2, which contradicts (5.11).
In the same way we can prove sy(0) ≤ 0. Moreover, since v(y, 0) = (1 − ey)+ > 0 if
24
y < 0, we have sy(0) = 0.
Step 3: Prove sy(τ) is strictly increasing in [ 0, T1 ].
Otherwise there exists a domain (y1, y2) × (τ1, τ2) (A1 < y1 < y2 < A2, 0 < τ1 < τ2 ≤T1) such that{
∂τv − Lyv = −L, (x, τ) ∈ (y1, y2) × (τ1, τ2),
v(y2, τ) = 0, τ ∈ (τ1, τ2).
Then, W = ∂τv satisfies{∂τW − LyW = 0, (x, τ) ∈ (y1, y2) × (τ1, τ2),
W (y2, τ) = 0, τ ∈ (τ1, τ2).
Since W = ∂τv ≥ 0, W achieves its non-positive minimum at y = y2. Applying the
strong maximum principle we have ∂yW (y2, τ) < 0 for any τ ∈ (τ1, τ2). On the other
hand, we can deduce that ∂yv(y2, τ) = 0 for any τ ∈ (τ1, τ2) by ∂yv = ∂xV ∈ C(ΩT1). It
follows that ∂yW (y2, τ) = ∂τyv(y2, τ) = 0 for any τ ∈ (τ1, τ2), thus we get a contradiction.
Hence sy(τ) is strictly increasing in (0, T1 ].
Step 4: Next, we prove sy(τ) ∈ C∞(0, T1 ]. In fact, based on (5.11) and applying the
method developed by Friedman(1975) in [8], it is not difficult to deduce sy(τ) ∈ C0,1(0, T1 ].
Moreover sy(τ) ∈ C∞(0, T1 ] is obvious by bootstrap argument. �
From transformation (5.8), Theorem 1.5 is obvious.
6. Numerical methods and results
Starting from the problem (1.3), we have{min{ ∂τV − σ2
2∂xxV − (r − q − σ2
2)∂xV + rV + L, V } = 0, x ∈ IR, τ ∈ (0, T ],
V (x, 0) = (1 − ex)+, x ∈ IR.(6.1)
Given mesh size �x, �τ > 0, V nj = V (j�x, n�τ) represents the value of numerical
approximation at (j�x, n�τ), then the P.D.E is changed into the following difference
equation⎧⎪⎪⎪⎨⎪⎪⎪⎩min{ V n
j −V n−1j
�τ− σ2
2
V n−1j+1 −2V n−1
j +V n−1j−1
�x2 − (r − q − σ2
2)
V n−1j+1 −V n−1
j−1
2�x+ rV n
j
+L, V nj } = 0,
V 0j = (1 − ej�x)+.
(6.2)
It means⎧⎪⎪⎨⎪⎪⎩V n
j = max{ 11+r�τ
[(1 − σ2�τ�x2 )V n−1
j + σ2+(r−q−σ2/2)�x2�x2 �τV n−1
j+1
+σ2−(r−q−σ2/2)�x2�x2 �τV n−1
j−1 − L�τ ], 0 },V 0
j = max{0, 1 − ej�x}.(6.3)
25
Choosing σ2�τ(�x)−2 = 1, we have⎧⎪⎪⎨⎪⎪⎩V n
j = max{ 11+r�τ
[(12
+ r−q−σ2/22σ
√�τ)V n−1j+1 + (1
2− r−q−σ2/2
2σ
√�τ)V n−1j−1
−L�τ ], 0 },V 0
j = max{0, 1 − ej�x}.(6.4)
Denote u = eσ√�τ , d = u−1, ρ = er�τ , p = (ρe−q�τ − d)(u− d)−1.
Then, it is clear as �τ → 0+, there holds
1
2+r − q − σ2/2
2σ
√�τ = p+ o(
√�τ), 1
1 + r�τ =1
ρ+O(�τ 2).
Neglecting a higher order of√�τ , we obtain{
V nj = max { 1
ρ[ pV n−1
j+1 + (1 − p)V n−1j−1 − L�τ ], 0 },
V 0j = max { 0, 1 − uj}.
(6.5)
Consider the point (x, τ) = (j�x, n�τ), then
V nj = V (x, τ), V n−1
j = V (x, τ−�τ), V n−1j+1 = V (x+�x, τ−�τ), V n−1
j−1 = V (x−�x, τ−�τ).
Then we get Figure 7-9:
Plot of the optimal default boundary sx(τ) as a function of time τ . The parameter values
used in the calculations are q = 0.2, σ = 0.8, L = 0.2, T = 3, n = 600. s1x(τ), s
2x(τ)
are respectively the free boundary of the problem (1.3) when r1 = 0.6, r2 = 0.2. Observe
that s1x(τ) ≤ s2
x(τ) ≤ 1, s1x(0) = s2
x(0) = 0, s1x(τ) and s2
x(τ) are not monotonic. The
numerical result is coincided with that of our proof (see Figure 7).
26
Figure 7. The numerical figure with different r
Plot of the optimal default boundary sx(τ) as a function of time τ . The parameter values
used in the calculations are q = 0.1, σ = 0.8, r = 0.15, T = 2, n = 600. s1x(τ), s
2x(τ) are
respectively the free boundary of the problem (1.3) when L1 = 0.5, L2 = 0.45. Observe
that s1x(τ) ≤ s2
x(τ) ≤ 1, s1x(0) = s2
x(0) = 0, s1x(τ) and s2
x(τ) are not monotonic. The
numerical result is coincided with that of our proof (see Figure 8).
Figure 8. The numerical figure with different L
Plot of the optimal default boundary sx(τ) as a function of time τ . The parameter values
used in the calculations are r = 0.35, σ = 0.8, L = 0.2, T = 3, n = 600. s1x(τ), s
2x(τ)
are respectively the free boundary of the problem (1.3) when q1 = 0.4, q2 = 0.3. Observe
that s2x(τ) ≤ s1
x(τ) ≤ 1, s1x(0) = s2
x(0) = 0, s1x(τ) and s2
x(τ) are not monotonic. The
numerical result is coincided with that of our proof (see Figure 9).
27
Figure 9. The numerical figure with different q
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