Article Search
eISSN 0454-8124
pISSN 1225-6951

### Article

Kyungpook Mathematical Journal 2022; 62(4): 615-640

Published online December 31, 2022

### Time-dependent Double Obstacle Problem Arising from European Option Pricing with Transaction Costs

Jehan Oh* and Namgwang Woo

Department of Mathematics, Kyungpook National University, Daegu 41566, Republic of Korea
e-mail : jehan.oh@knu.ac.kr and wng3717@knu.ac.kr

Received: October 15, 2022; Accepted: November 28, 2022

In this paper, we investigate a time-dependent double obstacle problem associated with the model of European call option pricing with transaction costs. We prove the existence and uniqueness of a Wp,loc2,1 solution to the problem. We then characterize the behavior of the free boundaries in terms of continuity and values of limit points.

Keywords: double obstacle problem, parabolic partial differential equation, time-dependent obstacle, free boundary, option pricing.

This paper concerns a double obstacle problem arising from the model of European call option pricing with transaction costs. Since transaction costs have been generally reduced and will be reduced in many countries (see 10, page 187] and [10, page 535]), we substitute the time-dependent transaction costs λ(t) and μ(t) for the usual λ and μ which are a constant fraction of the purchase price of the stock. To be specific, we consider the case that λ(t) and μ(t) diminish over time and analyze the value Q(y,S,t) satisfying

minyQ+γ(1+λ¯(t))SQer(Tt),yQ+γ(1μ¯(t))SQer(Tt),  tQ+σ22S2SSQ+αSSQ=0,y,S>0,0t<T,Q(y,S,T)=exp{γc(y,S)},

where

c(y,S)=(1+λ0)yS,if y<0,(1μ0)yS,if y0,

and σ>0,α>r0,γ>0 are constants. Also, we define

λ¯(t):=(λ0+1)eq(Tt)1,  μ¯(t):=(μ0+1)eq¯(Tt)1,

where 0λ¯(t),μ¯(t)<1 for all 0tT, λ0:=λ¯(T), μ0:=μ¯(T), and the range of q and q¯ is to be chosen later.

For the double obstacle problem arising from European option pricing with constant transaction costs, we refer the reader to [4].

Since (1.1) is a backward parabolic problem, we transform it to a forward parabolic problem. Letting τ=Tt, we have

maxyQ+γ(1+λ(τ))SQerτ,yQ+γ(1μ(τ))SQerτ,  τQσ22S2SSQαSSQ=0,y,S>0,0<τT,Q(y,S,0)=exp{γc(y,S)},

where λ(τ)=(λ0+1)eqτ1,μ(τ)=(μ0+1)eq¯τ1.

Using the transformation described in Section 2, we obtain the time-dependent double obstacle problem:

τVLzV=0, if 1μ(τ)<V<1+λ(τ);τVLzV0, if V=1+λ(τ);τVLzV0, if V=1μ(τ);V(z,0)=1+λ0, if z<0,1μ0, if z0,

where

LzV:=1γz(LV*)  =σ22z2zzV+σ2+αrzzV+(αr)Vγσ2zVzzV+V.

There are various studies on the double obstacle problems. Yang and Yi [13] studied the double obstacle problem associated with European option pricing with transaction costs. Dai and Yi [3] studied the free boundaries of the parabolic double obstacle problem arising from the optimal investment problem of a Constant Relative Risk Aversion (CRRA) investor who faces proportional transaction costs. Furthermore, Chen et al [2] analyzed the double obstacle problem related to a time-dependent Hamilton-Jacobi-Bellman equation with gradient constraints. However, few attempts have been made to analyze the time-dependent parabolic double obstacle problem employing the variational inequality approach. Besides the papers mentioned above, there is a vast literature related to the problem with transaction costs, see for instance, [1, 6] and the references therein.

The aim of the paper is to characterize the free boundaries of problem (1.3). Indeed, we obtain the existence and uniqueness of a Wp,loc2,1 solution for (1.3) using a penalty method. We also show the limits of the free boundaries and the continuity of the free boundaries, which is motivated by [12] and [13]. The main difficulty in carrying out this construction is that free boundaries are not always monotonic. To overcome this problem, we employ a transformation y=x+k(τ), v¯(y,τ)=v(x,τ), where k(τ) is chosen later. This guarantees that the free boundaries are monotonic. Hence, we can see the continuity of the free boundaries.

The present paper is organized as follows. In Section 2, we prove the existence and uniqueness of a Wp,loc2,1 solution for (1.3). In Section 3, we analyze the behavior of two free boundaries. In Section 4, we establish the equivalence between the double obstacle problem (1.3) and the original problem (1.2).

### 2. The Existence and Uniqueness of Solution to the Problem (1.3)

In this section, we show the existence and uniqueness of a Wp,loc2,1 solution to the problem (1.2). First, we prove that the problem (1.3) implies the problem (1.2). Since a positive value Q can be inferred from the reality of the setup, or from results in the paper, we let U=lnQ. Then U satisfies the following equations:

maxyU+γ(1+λ(τ))Serτ,yU+γ(1μ(τ))Serτ,        τUσ22S2SSUαSSUσ22SSU2=0,                  y,S>0,0<τT,U(y,S,0)=γc(y,S).

Also, letting z=erτyS and V*(z,τ)=U(y,S,τ) shows the equalities:

yU=erτSzV*,  SU=erτyzV*,
τU=τV*+rzzV*,  SSU=e2rτy2zzV*.

Using equalities (2.2) and (2.3), we have

yU(y,S,τ)+γ(1+λ(τ))Serτ=erτSzV*(z,τ)+γ(1+λ(τ)),yU(y,S,τ)+γ(1μ(τ))Serτ=erτSzV*(z,τ)+γ(1μ(τ)),τUσ22S2SSUαSSUσ22 S S U2=τV*LV*,

where

LV*=σ22z2zzV*+(αr)zzV*+σ22(zzV*)2.

Therefore, V*=V*(z,τ) satisfies

maxzV*+γ(1+λ(τ)),zV*+γ(1μ(τ)),τV*LV*=0,                      z,0<τT,V*(z,0)=γ(1+λ0)z,z<0,γ(1μ0)z,z0.

Differentiating with respect to z in (2.4), we get

zLV*=σ22z2zzzV*+αr+σ2zzzV*+(αr)zV*    +σ2zzV*zzzV*+zV*.

If we denote

V(z,τ)=1γzV*(z,τ),

then we have

LzV:=1γz(LV*)  =σ22z2zzV+σ2+αrzzV+(αr)Vγσ2zVzzV+V,

which is equivalent to (1.4). The proof is completed by showing that (2.5) implies (1.2); we do this in Section 4.

Now, note that the operator Lz is degenerate at z=0. Using the Fichera Theorem from [9], the problem (1.2) can be divided into two parts: the problems in the domains {z<0} and {z>0} independently. Furthermore, we see that V(z,τ)=1+λ(τ) is the solution of problem (1.2) in the domain {z<0}. Indeed,

(τLz)[1+λ(τ)]=λ(τ)(αr)(1+λ(τ))+γσ2z(1+λ(τ))2      =(q(αr))(1+λ(τ))+γσ2z(1+λ(τ))20,

provided that 0qαr.

From now on, we only consider the problem (1.2) in the domain {z>0}. Let z=ex and v(x,τ)=V(z,τ). Then v(x,τ) satisfies

τvLxv=0, if 1μ(τ)<v<1+λ(τ),x,0<τT,τvLxv0, if v=1μ(τ),x,0<τT,τvLxv0, if v=1+λ(τ),x,0<τT,v(x,0)=1μ0,x,

where

Lxv=σ22xxv+αr+σ22xv+(αr)vγσ2exvxv+v.

Since the domain is unbounded, we confine our attention to (1.2) in a bounded domain (n,n)×(0,T). In order to do so, set ΩT=×(0,T] and ΩTn=(n,n)×(0,T]. Let us consider the following problem in ΩTn:

τvnLxvn=0, if 1μ(τ)<vn<1+λ(τ) and (x,τ)ΩTn;τvnLxvn0, if vn=1μ(τ) and (x,τ)ΩTn;τvnLxvn0, if vn=1+λ(τ) and (x,τ)ΩTn;xvn(x,τ)=0,x=±n,0τT;vn(x,0)=1μ0,nxn.

Lemma 2.1. For any fixed n, there exists a unique solution vnC(Ω¯Tn)Wp2.1 ΩTn to the problem (2.9), where 1<p<+. Moreover,

vnxvn0  and  τvnμ(τ) a.e. in ΩTn.

Proof. We consider a penalty approximation of the problem (2.9):

τvε,nLxvε,n+βεvε,n(1μ(τ))        βεvε,n+(1+λ(τ))=0 in ΩTn,xvε,n(x,τ)=0,x=±n,0τT,vε,n(x,0)=1μ0,nxn,

where

βε(t)C2(,+);βε(t)0,βε(t)0,βε(t)0,t;βε(0)=C0,C0max{γσ2en(1μ0)2+λ(0),(αr)(1+λ(T))},

and moreover,

limε0+βε(t)=0,t>0,,t<0.

For simplicity, we let βε():=βε(vε,n1+μ(τ)) and βε():=βε(vε,n+1+λ(τ)) when no confusion can arise.

Following standard procedure, we can use the Leray-Schauder fixed point theorem, we get the existence and uniqueness of the solution of (2.11). Next, we show that 1μ(τ)vε,n1+λ(τ). We define the operator T by

Tv:=τvLxv+βεv(1μ(τ))βεv+(1+λ(τ)).

From the definition of C0, we obtain

T[1μ(τ)]=μ(τ)(αr)(1μ(τ))+γσ2ex(1μ(τ))2+βε(0)βε(μ(τ)+λ(τ))=μ(τ)(αr)(1μ(τ))+(γσ2ex(1μ(τ))2C0)μ(0)(αr)(1μ(T))λ(0)0

for sufficiently small ε. Combining the above inequality and the initial and boundary conditions, we get 1μ(τ)vε,n by the comparison principle. Similarly, from the definition of C0, we have

T[1+λ(τ)]=λ(τ)(αr)(1+λ(τ))+γσ2ex(1+λ(τ))2+βε(μ(τ)+λ(τ))βε(0)=λ(τ)+γσ2ex(1+λ(τ))2+(C0(αr)(1+λ(τ)))λ(0)+γσ2en(1+λ0)20

for sufficiently small ε, which proves vε,n1+λ(τ).

Next, we prove vε,nxvε,n0. To prove this, we differentiate Tvε,n=0 with respect to x, and let W:=xvε,n. Then W satisfies

τWσ22xxWαr+σ22xW(αr)W+γσ2exvε,nxW+3vε,nW+W2+βε()W+βε()W  =γσ2exvε,n20, in ΩTn,W(x,τ)=0, on pΩTn.

Using the maximum principle, we get xvε,n0 in ΩTn.

On the other hand, we define the operator T^ by

T^w:=τwσ22xxwαr+σ22xw(αr)w          +γσ2exvε,nxw+3vw+w2+βε()w+βε()w.

Then, we have

T^[W]=γσ2exvε,n2.

To make calculations easier, we will use two properties related to the operator T^: For each w1=w1(x,τ) and w2=w2(x,τ),

• 1. T^[w1+w2]=T^[w1]+T^[w2]+2γσ2exw1w2.

• 2. T^[kw1]=kT^[w1]+(k2k)γσ2exw12 for each k.

Set W^:=W+vε,n. From the equation (2.14), we have

T^[W^vε,n]=γσ2exvε,n2.

Using the property 1 with respect to T^, we obtain

T^[W^]+T^[vε,n]+2γσ2exW^(vε,n)=γσ2exvε,n2.

Using the property 2 with respect to T^, we get

T^[W^]=T^[vε,n]+2γσ2exW^vε,n3γσ2exvε,n2  =τvε,nσ22xxvε,nαr+σ22xvε,n(αr)vε,n  +γσ2exvε,nxvε,n+3vε,n2+vε,n2+βε()vε,n+βε()vε,n  +2γσ2exW^vε,n3γσ2exvε,n2  =(τLx)[vε,n]βε()+βε()+3γσ2exvε,n2+βε()vε,n+βε()vε,n  +2γσ2exW^vε,n3γσ2exvε,n2  =βε()+βε()+βε()vε,n+βε()vε,n+2γσ2exW^vε,n.

This gives

T^[W^]T^[0]=βε()+βε()+βε()vε,n+βε()vε,n.

We claim that T^[W^]T^[0]0. From the definition of βε, we get

βε()βε()=βε(η)(2vε,n+1+λ(τ)+1μ(τ)),

where η is a real number between vε,n+1+λ(τ) and vε,n+1μ(τ). There are only the following three possibilities:

• (i) If vε,n=1+λ(τ)+1μ(τ)2, then βε()+βε()=0. It follows that

T^[W^]T^[0]0.

• (ii) If vε,n>1+λ(τ)+1μ(τ)2, then

βε(vε,n1+μ(τ))βε(η)βε(vε,n+1+λ(τ))

by the monotonicity of βε. Hence,

T^[W^]T^[0]=βε(η)(2vε,n+1+λ(τ)+1μ(τ))+γσ2exvε,n2+βε()vε,n+βε()vε,nβε()(2vε,n+1+λ(τ)+1μ(τ))+γσ2exvε,n2+βε()vε,n+βε()vε,n=βε()(vε,n+1+λ(τ)+1μ(τ))+γσ2exvε,n2+βε()vε,n0.

• (iii) If vε,n<1+λ(τ)+1μ(τ)2, then

βε(vε,n1+μ(τ))βε(η)βε(vε,n+1+λ(τ))

by the monotonicity of βε. Hence,

T^[W^]T^[0]=βε(η)(2vε,n+1+λ(τ)+1μ(τ))+γσ2exvε,n2+βε()vε,n+βε()vε,nβε()(2vε,n+1+λ(τ)+1μ(τ))+γσ2exvε,n2+βε()vε,n+βε()vε,n=βε()(vε,n+1+λ(τ)+1μ(τ))+γσ2exvε,n2+βε()vε,n0.

By (i), (ii) and (iii), the proof of the claim is complete.

Since

W^(x,0)=1μ0,nxn,W^(x,τ)=v(x,τ),x=±n,τ[0,T],

from the comparison principle, we see that W^(x,τ)0 in ΩTn, and xvε,n+vε,n0 is proved.

Next, we claim that τvε,nμ(τ). Let w=τvε,n and w˜=w+μ(τ). Differentiating (2.11) with respect to τ, we obtain

τwσ22xxwαr+σ22xw(αr)w+γσ2exvε,nxw+xvε,n+2vε,nw+βε()(w+μ(τ))βε()(w+λ(τ))=0, in ΩTn,xw(x,τ)=0,x=±n,0τT,w(x,0)=(αr)(1μ0)γσ2(1μ0)2ex+βε(0)0

by the definition of βε. For simplicity of notation, we let T˜ stand for the operator:

T˜w:=τwσ22xxwαr+σ22xw(αr)w  +γσ2exvε,nxw+xvε,n+2vε,nw+βε()w+βε()w.

Then, we have

T˜[w˜μ(τ)]=βε()μ(τ)+βε()λ(τ).

From the linearity of T˜, we get

T˜[w˜]=T˜[μ(τ)]βε()μ(τ)+βε()λ(τ).  =μ(τ)(αr)μ(τ)+γσ2ex(xvε,n+2vε,n)μ(τ)  +βε()μ(τ)+βε()μ(τ)βε()μ(τ)+βε()λ(τ)  =(q¯(αr))μ(τ)+γσ2ex(xvε,n+2vε,n)μ(τ)+βε()μ(τ)+βε()λ(τ).

If we assume that q¯αr or q¯=0, then we see from the inequality τvε,nvε,n that

T˜[w˜]0=T˜[0].

By (2.16), we have

xw˜(x,τ)=0,x=±n,w˜(x,0)=(αr)(1μ0)γσ2(1μ0)2ex+βε(0)+μ(0)0.

Combining (2.17) and (2.18), we get

w˜0 if and only if τvε,nμ(τ).

From C0βε(vε,n1+μ(τ))0 and C0βε(vε,n+1+λ(τ))0, we see that

vε,nWp,loc2,1(ΩTn)c,

where c is independent of ε and n. Using a Cα-estimate, we have

vε,nCα,α/2(Ω¯Tn)c

for some constant c>0 which is independent of ε. Then we deduce that

vε,nvn in Wp,loc2,1(ΩTn) weakly,
vε,nvn in C(Ω¯Tn)

as ε0+, where vn is the solution to the problem (2.9). Moreover, vε,nxvε,n0 and τvε,nμ(τ) become the inequalities (2.10) as ε0+.

Finally, we prove the uniqueness of a solution. Suppose that v1,v2 are two solutions to the problem (2.11) and that the set

N:=(x,τ):v1(x,τ)<v2(x,τ),|x|<n,0<τT

is nonempty. Then if (x,τ)N, we have

v1(x,τ)<1+λ(τ) implies that τv1Lxv10,v2(x.τ)>1μ(τ) implies that τv2Lxv20.

Define v*=v2v1. Then v* satisfies

τv*σ22xxv*αr+σ22xv*(αr)v*    +γσ2exv2xv*+v1+v2+xv1v*0, in N,v*(x,0)=0, on pN{|x|<n},xv*(x,0)=0, on pN{|x|=n},

where pN is the parabolic boundary of the domain N. Using the ABP maximum principle (see [7] and [11]), we get v*0 in N, which contradicts the definition of N.

Theorem 2.2. There exists a unique solution vC(Ω¯T)Wp,loc2,1(ΩTR) to problem (2.9) for all R>0, 1<p<+. Also,

xv0 in ΩT;  τvμ(τ) a.e. in ΩT.

Moreover, for any fixed K, vCα,α/2((,K)×(0,T)¯) for some 0<α<1, with

|v|Cα,α/2((,K)×(0,T)¯)CK,

where CK is a positive constant depending on K.

Proof. Since the solution vn of the problem (2.9) belongs to Wp,loc2,1(ΩTn), we rewrite problem (2.9) as

τvnLxvn=f(x,τ), in ΩTn,vn(x,τ)=0,x=±n,0τT,vn(x,0)=1μ0,nxn,

where

f(x,τ)=χvn=1+λ(τ)(x,τ)λ(τ)(αr)(1+λ(τ))+γσ2ex(1+λ(τ))2+χvn=1μ(τ)(x,τ)μ(τ)(αr)(1μ(τ))+γσ2ex(1μ(τ))2.

Then we see that

|f(x,τ)|c(R) for (x,τ)ΩTR,

where the constant c(R) depends on R, but is independent of n. Therefore, for any fixed R>0, we choose n>R. Then we have the following Wp2,1 uniform estimate in Ω¯TR :

vnWp2,1(ΩTR)CvnL(ΩTR)+(1μ0)+f(x,τ)L(ΩTR)    C(R)

for some constant C(R) which is independent of n. Letting n+, we have a subsequence:

vnvR in Wp2,1(ΩTR) weakly  and vnvR in C( Ω¯TR) as n+.

Define v=vR if x[R,R]. Then it follows that v is well-defined and v is the solution of problem (2.7).

Now, we prove (2.24). Note that

τvLxv=f(x,τ),inΩT,v(x,0)=1μ0,x.

Since f(x,τ) is bounded on (,K)×(0,T), (2.24) follows from the standard Cα theory of parabolic equation. The proof of the uniqueness is the same as that of Lemma 2.1.

### 3. Characterizations of the Free Boundaries

In this section, we mainly consider the problem (1.2). We define

BR=(z,τ):V(z,τ)=1+λ(τ)   (buy region),NR=(z,τ):1μ(τ)<V(z,τ)<1+λ(τ)   (no transaction region),SR=(z,τ):V(z,τ)=1μ(τ)   (sell region).

Note that

(τLz)(1+λ(τ))=λ(τ)(αr)(1+λ(τ))+γσ2z(1+λ(τ))20, in BR,(τLz)(1μ(τ))=μ(τ)(αr)(1μ(τ))+γσ2z(1μ(τ))20, in SR.

Also, we deduce that

BR(z,τ):z(αr)(1+λ(τ))λ(τ)γσ2(1+λ(τ))2,
SR(z,τ):z(αr)(1μ(τ))+μ(τ)γσ2(1μ(τ))2.

We remark that

(αr)(1+λ(τ))λ(τ)γσ2(1+λ(τ))2(αr)(1+λ(τ))+μ(0)γσ2(1+λ(τ))2(αr)(1μ0)+μ(0)γσ2(1μ0)2(αr)(1μ(τ))+μ(τ)γσ2(1μ(τ))2.

On the other hand, from the inequalities (2.23) and (2.24), we see that

τVμ(τ),  zV=exxV0,
|V(z,τ)|Cτα/2[0,T]CK,  0<zK,

where CK is a constant depending on K. From the second inequality in (3.3), we can define the free boundaries:

zb(τ)=supz:V(z,τ)=1+λ(τ),  0<τT,zs(τ)=infz:V(z,τ)=1μ(τ),  0<τT.

From the first inequality in (3.3), we deduce that zs(τ) is increasing, but zb(τ) is not always increasing.

Lemma 3.1. There is a constant Ms>0 such that

0zb(τ)(αr)(1+λ(τ))λ(τ)γσ2(1+λ(τ))2,
(αr)(1μ(τ))+μ(τ)γσ2(1μ(τ))2zs(τ)Ms,

where Ms is independent of T.

Proof. Since V(z,τ)=1+λ(τ) for all z<0, zb(τ)0. From (3.1) and (3.2), we see that the second parts of (3.5) and the first parts of (3.6) hold.

Next, we claim that zs(τ)Ms. First, we introduce the stationary problem of (1.2):

LzW(z)=0,if 1μ(T)<W(z)<1+λ(T),z+,LzW(z)0,if W(z)+1+λ(T),z+,LzW(z)0,if W(z)=1μ(T),z+.

By the Fichera Theorem in [9], we consider the problem without the boundary value at z=0. Using the similar way in Section 2, we can show the existence and uniqueness of the solution to problem (3.7): For each R>0 and 1<p<+,

WC1(+)Wp21R,R.

Furthermore, we have W(z)0. Then we can define

SR*:=z+:W(z)=1μ(T).

Set W*(z,τ):=W(z). If 1μ(τ)<W*<1+λ(τ) for each (z,τ)+×(0,T], then 1μ(T)<W(z)<1+λ(T). From the stationary problem (3.7), we have τW*LzW*=LzW(z)=0.

As a result, we deduce that

τW*LzW*=0,if 1μ(τ)<W*<1+λ(τ),(z,τ)+×(0,T],τW*LzW*0,if W*=1μ(τ),(z,τ)+×(0,T],τW*LzW*0,if W*=1+λ(τ),(z,τ)+×(0,T].

Applying the comparison principle, we get V(z,τ)W*(z,τ)=W(z).

Next, we claim that there exists a constant Ms>0 such that

[Ms,+)×{τ=T}SR*×{τ=T}SR.

If it holds, then [Ms,+)×[0,T]SR by the first parts of the inequality (3.3). It means that zs(τ)Ms, which completes the proof.

To prove the claim, it suffices to show that [Ms,+)SR*. Suppose the claim were not. Then we could find that if 1μ(T)<W(z)<1+λ(T),

ddzσ22z(zW)+ασ22r(zW)γσ22(zW)2=0.

Then we have

z(zW)+2(αr)σ2 1(zW)γ(zW)2=C¯,

where C¯ is an unknown constant. Define

W^:=zW+12γ12(αr)σ2.

Then zW^γW^2=C, where C=C¯14γ212(αr)σ2 2.

If C<0, zW^γW^2=C12, where C12=C. By solving the ODE, we can see that

W^=C1γ21C2z2γC11,W=1zW^+αrγσ212γ.

As z+, W0, which contradicts 1μ(τ)W1+λ(τ).

If c>0, zW^γW^2=C12, where C12=C. By solving the ODE, we get

W^=C1γtanC1γlnz+C2,W=1zW^+αrγσ212γ.

Then we obtain liminfz+W=0, which contradicts 1μ(τ)W1+λ(τ).

If C=0, by solving the ODE, then we obtain

W^=1C2+γlnz,W=1zW^+αrγσ212γ.

As z+, W0, which is a contradiction. Therefore, there exists a constant Ms>0 which the above claim holds.

Lemma 3.2. There exist z0>0 and τ0>0 such that (0,z0)×(0,τ0)NR and that

all partial derivatives of  V(z,τ) are bounded on (0,z0)×(0,τ0).}

Proof. Using V(z,0)=1μ0 and (3.4), for any fixed K>0, we observe that there exists τ0>0 such that

V(z,τ)<1+λ(τ),  (z,τ)(0,K)×(0,τ0).

On the other hand, from (3.2), for 0<z0<(αr)(1μ(τ))+μ(τ)γσ2(1μ(τ))2, we get

V(z,τ)>1μ(τ),  (z,τ)(0,z0).

Combining (3.10) and (3.11), we obtain

τVLzV=0, in 0,z0×0,τ0,Vz0,τC0,τ0,V(z,0)=1μ0,0<z<z0.

Set x=lnz,x0=lnz0, and v(x,τ)=V(z,τ). Then we have

τvσ22xxvαr+σ22xv(αr)v+γσ2exvxv+v=0,vx0,τC0,τ0,v(x,0)=1μ0,x,x0.

Using (2.24) and Schauder theory in [9], we see that

vC2+α,1+α/2,x0×0,τ0¯Cx0,

where Cx0 depends on x0. Employing a bootstrap argument, it gives that all partial derivatives of v(x,τ) are bounded on (,x0)×(0,τ0).

We proceed to show that zV(z,τ)=exxv(x,τ) is bounded. Differentiating with respect to x in (3.13), we get

τxvσ22xxxvαr+σ22xxv(αr)xv  +γσ2ex x v+v2+vxxv+xv=0.

Set W=exxv. Then

τWσ22xxWαr+32σ2xW2α2r+σ2W  =γσ2xv+v2+vxxv+xv,(x,τ),x0×0,τ0,Wx0,τC0,τ0,W(x,0)=0,x,x0.

Since the right-hand side of the equation is bounded, zV=exxv is bounded. In the same manner, we can see that zzV=zW=e2x(xxvxv) is bounded. Furthermore, all partial derivatives of V(z,τ) are bounded on (0,z0)×(0,τ0) by the bootstrap argument.

Theorem 3.3. zs(τ)C[0,T]C(0,T] and is strictly increasing with zs(0)=z*, where

z*=(αr)(1μ0)+μ(0)γσ2(1μ0)2.

Proof. First, we show (3.14). Recalling (3.2), we see that

z*(αr)(1μ0)+μ(0)γσ2(1μ0)2.

We suppose that z*>(αr)(1μ0)+μ(0)γσ2(1μ0)2. Then there exists z2<z* such that

τVLzV=0,(z,τ)(αr)(1μ0)+μ(0)γσ2(1μ0)2,z2×(0,T),V(z,0)=1μ0,(αr)(1μ0)+μ(0)γσ2(1μ0)2zz2.

Therefore, τV(z,0)=(αr)(1μ0)γσ2z(1μ0)2<μ(0), which contradicts the first part of the inequalities (3.3). We complete the proof of (3.14).

Next, we claim that zs(τ) is strictly increasing in (0,T]. Suppose that zs(τ) is not strictly increasing in (0,T]. Then zs(τ1)=zs(τ2)=z0 for some z0+ and 0<τ1<τ2T. And there exists z1<z0 satisfying (z1,τ)NR for all τ[τ1,τ2]. Set D=(z1,z0)×(τ1,τ2). Then we see that

V(z0,τ)=1μ(τ),τ[τ1,τ2],τV(z0,τ)=μ(τ),τ[τ1,τ2].

Since V>1μ(τ) in D, we observe that

τVLzV=0, in D.

We show that the strong maximum principle implies that τVμ(τ) or zτV(z0,τ)<0 for any τ(τ1,τ2). Differentiating (3.16) with respect to τ, we obtain

TW:=τWσ22z2zzW(σ2+αr)zW(αr)W  +γσ2z[2VW+(zV)W+zVzW]=0   in D.

Assume that W=μ(τ2) is a minimum at (z,τ)(z1,z0)×(τ1,τ2) and that W>μ(τ2) is a maximum at (z*,τ*)(z1,z0)×(τ1,τ2). Since TW=0 in D, Wμ(τ2) in D{ττ} or zW(z0,τ)<0 for all τ[τ1,τ2] by the strong maximal principle.

Since zW(z0,τ)=z(1μ(τ))=0 for all τ[τ1,τ2], we see that

Wμ(τ2)   in D{ττ}.

This means that τV obtains its minimum value at interior point of NR{ττ}. Then we have τVμ(τ2) in NR{ττ} by the strong maximum principle. If q¯=0, we already know the fact that zs(τ) is strictly increasing. Hence, we assume that q¯0. Then, on the segment {z0}×(τ1,τ),

τV(z,τ)=μ(τ2)=μ(τ),

which is a contradiction. Therefore, zs(τ) is strictly increasing.

We claim that zs(τ) is continuous. Suppose that zs(τ) is not continuous in (0,T]. Then there exist τ0(0,T], z0(0,+), and small ε0,δ0 such that

zs(τ0ε)z0 and zs(τ0+ε)z0+δ0

for all ε(0,ε0). Let D=(z0,z0+δ)×(τ0,τ0+ε) so that DNR. Since zs(τ) is strictly increasing, we observe that z*<z0, i.e., (αr)(1μ(τ0))+μ(τ0)<γσ2z0(1μ(τ0))2. In D, we see that

(τLz)[V](τLz)[1μ(τ)]=μ(τ)+(αr)(1μ(τ))γσ2z(1μ(τ))2=μ(τ)+(αr)(1μ(τ))γσ2z0(1μ(τ))2+γσ2(z0z)(1μ(τ))2μ(τ)+(αr)(1μ(τ))γσ2z0(1μ(τ0))2+γσ2(z0z)(1μ(τ))2=[μ(τ)+(αr)(1μ(τ))][μ(τ0)+(αr)(1μ(τ0))]+γσ2(z0z)(1μ(τ))2=μ(τ)μ(τ0)+(αr)(μ(τ0)μ(τ))+γσ2(z0z)(1μ(τ))2=(q¯(αr))(μ(τ)μ(τ0))+γσ2(z0z)(1μ(τ))2.

Here, we assume that q¯αr. Now (3.17) becomes

(τLz)[V](τLz)[1μ(τ)]γσ2(z0z)(1μ(τ))2.

τV(z,τ0)<μ(τ),

which is impossible. Therefore, zs(τ) is continuous in (0,T]. Employing a method developed by Friedman in [5], we can see that zs(τ)C(0,T].

Remark 3.4. So far, we have assumed that

• 1. 0qαr,

• 2. q¯=0 or q¯=αr.

We next investigate the limit of the solution and the values of the free boundary zb(τ) near τ=0.

Theorem 3.5. We have limz0+V(z,τ)=V0(τ) with

• 1. If αr<q, then zb(τ)=0 for each τ[0,τ*],

where τ*=1(αr)qln1+λ01μ0, and

V0(τ)=(1μ0)e(αr)τ,0ττ*,1+λ(τ),τ>τ*.

• 2. If αr=q, then zb(τ)=0 for each τ[0,τ0], where τ0>0 is the number in Lemma 3.2, and

V0(τ)=(1μ0)e(αr)τ,  0ττ0.

Proof. Since all partial derivative of V(z,τ) are bounded on (0,z0)×(0,τ0), we see that there exists V0(τ)C[0,T] such that

limz0+V(z,τ)=V0(τ),  limz0+τV(z,τ)=V0(τ).

Applying (3.9) and letting z0+ in (3.12), we deduce that

V0(τ)(αr)V0(τ)=0,0<τ<τ0,V0(0)=1μ0.

Then we obtain

V0(τ)=(1μ0)e(αr)τ,0<τ<τ0.

If αr<q and let V0(τ*)=1+λ(τ*), then we have

(1μ0)e(αr)τ=(1+λ0)eqτ*.

In short, it follows that

τ*=1(αr)qln1+λ01μ0,

which deduces (3.19). If αr=q, then we see

V0(τ)=(1μ0)e(αr)τ,0τT,

which implies (3.20).

Next, we prove that zb(τ)C(0,T]. In order to do so, we use a transformation v¯(y,τ):=v(x,τ), where y=x+k(τ). Fix T1(0,T) and consider the problem (2.7) only on domain ×[0,T1]. Let MΩT1n be a domain defined by

M:=(n,x1(τ))×(0,T1),

which x1(τ) is infx:v(x,τ)=1. Similarly,

Mε:=(n,x1ε(τ))×(0,T1),

which x1ε(τ) is infx:vε,n(x,τ)=1. Set some parts of the boundary of M as follows:

1M:=(x,τ)ΩT1n¯:V(x,τ)=1,  2M:={n}×(0,T1).

Similarly, set

1Mε:=(x,τ)ΩT1 n¯:vε,n(x,τ)=1.

The task is now to find τv¯(y,τ)λ(τ) in M. If it holds, this makes it possible that z¯b(τ) is monotonic, where z¯b(τ) is corresponding to zb(τ).

Lemma 3.6. In M, xvε,nk(τ)τvε,nλ(τ).

Proof. Step 1. Let y=x+k(τ) and v¯(y,τ)=v(x,τ).

Then we calculate

xv(x,τ)=yv¯(y,τ),xxv(x,τ)=yyv¯(y,τ),τv¯(y,τ)=τv(x,τ)k(τ)xv(x,τ).

We claim that τv(x,τ)k(τ)xv(x,τ)λ(τ). For simplicity, let

w1=τvε,n,w2=xvε,n,
w=k(τ)xvε,n,Q=w1w=τvε,nk(τ)xvε,n,

where k(τ)=k1λ(τ)+k2μ(τ) such that k1,k2 are to be chosen later. Differentiating (2.11) with respect to τ, we obtain

τw1σ22xxw1αr+σ22xw1(αr)w1+γσ2exvε,nxw1+xvε,n+2vε,nw1+βε()(w1+μ(τ))βε()(w1+λ(τ))=0, in ΩT1n.

Differentiating (2.11) with respect to x, we have

τw2σ22xxw2αr+σ22xw2(αr)w2+γσ2exvε,nxw2+2vε,nw2+vε,nxvε,n+w2xvε,n+vε,n2+βε()w2+βε()w2=0, in ΩT1n.

Multiplying (3.22) by k(τ), we get

τwσ22xxwαr+σ22xw(αr)w+γσ2exvε,nxw+2vε,nw+k(τ)vε,nxvε,n+(xvε,n)w+k(τ)vε,n2+βε()w+βε()wk(τ)xvε,n=0, in ΩT1n.

Subtracting (3.23) from (3.24), we see that

T[Q]:=τQσ22xxQαr+σ22xQ(αr)Q  +γσ2exvε,nxQ+(xvε,n)Q+2vε,nQ+βε()Q+βε()Q  =γσ2ex[k(τ)vε,nxvε,n+k(τ)vε,n2]  βε()μ(τ)+βε()λ(τ)k(τ)xvε,n, in ΩT1n.

Substituting Q=λ(τ) into (3.24), we have

T[λ(τ)]=λ(τ)(αr)λ(τ)+γσ2ex[λ(τ)xvε,n+2λ(τ)vε,n]      +βε()λ(τ)+βε()λ(τ).

Combining (3.24) with (3.25), we see that

T[Q]T[λ(τ)]=γσ2ex[k(τ)vε,nxvε,n+k(τ)vε,n2λ(τ)xvε,n2λ(τ)vε,n]k(τ)xvε,nλ(τ)+(αr)λ(τ)βε(vε,n1+μ(τ))[μ(τ)+λ(τ)]=:I1+I2+I3+I4,

where

I1=γσ2ex[k(τ)vε,nxvε,n+k(τ)vε,n2λ(τ)xvε,n2λ(τ)vε,n],I2=k(τ)xvε,n0,I3=λ(τ)+(αr)λ(τ)=(q+αr)λ(τ)0,I4=βε(vε,n1+μ(τ))[μ(τ)+λ(τ)].

We claim that

T[Q]T[λ(τ)]0.

Since I2 and I3 are nonnegative and I40 in Mε for sufficiently small ε>0, we have to show that I1 is nonnegative. Now, we see the boundary condition in Mε. Define T^ by

T^[w]:=τwσ22xxwαr+σ22(αr)w  +γσ2exvxw+3vw+w2+v2.

Differentiating (2.7) with respect to x, we obtain

T^[xv]=0,   in M.

Let A=(y,τ):xcyx+c,(x,τ)1M be a subset such that ANR for sufficiently small constant c. From the strong maximum principle, there exists δ1>0 such that

xvδ1,   in  A.

Then, for sufficiently small ε>0,

xvε,nδ12,   in 1Mε.

Also, let B=(y,τ):xcyx+c,(x,τ)2MNR be a subset such that BNR for sufficiently small constant c. From the strong maximum principle, there exists δ2>0 such that

xvδ2,   in B.

Then, for sufficiently small ε>0,

xvε,nδ22,   in 2MNR.

Choose

k(τ):=1δ[4λ(τ)+2μ(τ)],

where δ:=min{δ1,δ2,δ3}>0. Here, δ3>0 is to be selected later. Therefore, we have

τvε,nk(τ)xvε,nμ(τ)k(τ)δ2        =μ(τ)+2λ(τ)+μ(τ)        λ(τ).

Since τvε,nk(τ)xvε,n=λ(τ) on the boundary 2MBR, combining (3.26) and (3.27) yields

τvε,nk(τ)xvε,n0   in 1M2M.

On the other hand, we get

T˜[xv+v]:=T^[xv+v]2γσ2exv[xv+v]0   in M.

From the strong minimum principle, there exists δ3>0 such that

xv+vδ3   in M.

Then,

xvε,n+vε,nδ32   in Mε.

Therefore, in the domain Mε,

I1=γσ2ex[k(τ)vε,n(xvε,n+vε,n)λ(τ)(xvε,n+2vε,n)]γσ2exk(τ)vε,n δ3 22λ(τ)vε,n12γσ2exvε,nk(τ)δ4λ(τ)12γσ2exvε,n[2μ(τ)]0.

Therefore, we complete the proof of claim. Note that

Q(x,τ)=w1k(τ)w2λ(τ), if (x,τ)1M2M,Q(x,0)=(w1k(τ)w2)(x,0)=(αr)(1μ0)[γσ2(1 μ 0 )2ex+βε(0)]kλ(τ)0.

Then

Q(x,0)λ(0)=(αr)(1μ0)+[C0γσ2(1μ0)2exλ(0)]0.

Therefore, we conclude that Q(x,τ)λ(τ) in ΩT1n.

We consider the domain S:=(y,τ):yx1(τ)+k(τ). From Lemma 3.6, we see that τv¯(y,τ)λ(τ) in S.

Theorem 3.7. There holds zb(τ)C(0,T].

Proof. Suppose that theorem is false. From the problem (2.7) and Lemma 3.6, we see that there exists (y1,y2)×(0,τ1)S such that

τv¯Lyv¯=0,(y,τ)(y1,y2)×(0,τ1),v¯(y,τ1)=1+λ(τ1),y[y1,y2],

where

Lyv¯:=σ22yyv¯+(αr+σ22k(τ))yv¯+(αr)v¯γσ2eyk(τ)v¯(yv¯+v¯).

Set w¯:=yv¯. Then w¯ satisfies

τw¯σ22yyw¯(αr+σ22k(τ))yw¯(αr)w¯+γσ2eyk(τ)[3v¯w¯+w¯yv¯+v¯yw¯]  =γσ2eyk(τ) v ¯ 20,(y,τ)(y1,y2)×(0,τ1),w¯(y,0)=0,y[y1,y2].

It follows that w¯ achieves non-negative maximum on τ=τ1. By the maximum principle, yv¯=w¯0 in (y1,y2)×(0,τ1). Then w¯0 in NRy{ττ1}, where

NRy:=(y,τ)S:1μ(τ)<v¯(y,τ)<1+λ(τ).

Therefore, by monotonicity, there exist τ*<τ1 and y*<y1 such that y*=zb¯(τ*), where

zb¯(τ)=supy:v¯(y,τ)=1+λ(τ).

Since w¯0 in NRy{ττ1}, we have v¯(y,τ)=1+λ(τ*) on the line {τ=τ*}, which is a contradiction. Hence, we see that zb¯(τ) is continuous, which deduces that zb(τ)C(0,T].

Theorem 3.8. There holds zb(τ)C(0,T].

Proof. Since k(τ) is smooth, the proof of Theorem 3.8 follows directly from the fact that z¯b(τ)C(0,T], which is clear from [13].

The equivalence of the double obstacle problem and the original problem is discussed in this section. According to (2.6), there should be two functions A(τ) and B(τ) such that

V*(z,τ)=A(τ)γ(1+λ(τ)),if (z,τ)BR,B(τ)γ(1μ(τ)),if (z,τ)SR.

Since τV*LV*=0 on z=zb(τ), recalling (2.4) we obtain

A(τ)=γλ(τ)zb(τ)γ(αr)(1+λ(τ))zb(τ)+σ22γ2(1+λ(τ))2zb2(τ).

Since V*(z,0)=γ(1+λ0)z,A(0)=0. Therefore, by the Fundamental Theorem of Calculus, we have

A(τ)=0τγλ(t)zb(t)γ(αr)(1+λ(t))zb(t)+γ2σ2 2(1+λ(t))2zb(t)2dt.

From the integral of (2.6) with respect to z, we get

V*(z,τ)=A(τ)γ0z V(ξ,τ)dξ.

Lemma 4.1. V*(z,τ),τV*(z,τ),zzV*, z2zzV*C(×[0,T]). Moreover,

V*(z,τ)=A(τ)γ(1+λ(τ))z,if zzb(τ),B(τ)γ(1μ(τ))z,if zzs(τ),

where A(τ) is given by (4.2) and

B(τ)=A(τ)γ0zs(τ)V(ξ,τ)dξ+γ(1μ(τ))zs(τ).}

Proof. First we prove (4.4). If zzb(τ), we obtain from (4.3) that

V*(z,τ)=A(τ)γ0z (1+λ(τ))dξ=A(τ)γ(1+λ(τ))z.

If zzs(τ), we have

V*(z,τ)=A(τ)γ0zs(τ) V(ξ,τ)dξγzs(τ)z (1μ(τ))dξ  =A(τ)γ0zs(τ) V(ξ,τ)dξ+γ(1μ(τ))zs(τ)γ(1μ(τ))z  =B(τ)γ(1μ(τ))z,(by(4.5)).

Now, we show the smoothness of V*(z,τ). Since zb(τ)C[0,T], A(τ)C1[0,T] by (4.2). Furthermore, VL(×[0,T]) and is continuous with respect to τ. As a result, V*(z,τ)C(×[0,T]) by (4.3).

Next, we prove τV*(z,τ)C(×[0,T]). Indeed,

τV*(z,τ)=A(τ)0z τ V(ξ,τ)dξ.

It is clear that τV* is continuous across z=0 by (4.6). On the other hand, (4.6) can be rewritten as

τV*(z,τ)=A(τ)γzb (τ)z τV(ξ,τ)dξγλ(τ)zb(τ).

If zzb(τ), then

τV*(z,τ)=A(τ)γλ(τ)zb(τ)    =γ(αr)(1+λ(τ))zb(τ)+γ2σ22(1+λ(τ))2zb2(τ).

If zb(τ)zzs(τ), then

zb(τ)z τV(ξ,τ)dξ  =zb(τ)z L zV(ξ,τ)dξ  =zb(τ)z ξσ2 2ξ2ξV(ξ,τ)+(αr)ξV(ξ,τ)γσ2 2(ξV(ξ,τ))2 dξ  = σ 2 2 ξ 2 ξ V(ξ,τ)+(αr)ξV(ξ,τ)γ σ 2 2 (ξV(ξ,τ)) 2 ξ=zb(τ)ξ=z.

Combining (4.7) and (4.9), we get

τV*(z,τ)=A(τ)γλ(τ)zb(τ)γ σ22 ξ2 ξ V(ξ,τ)+(αr)ξV(ξ,τ)γ σ22 (ξV(ξ,τ))2 ξ=zb(τ)ξ=z=γ σ 2 2z2zV(z,τ)+(αr)zV(z,τ)γ σ 2 2 (zV(z,τ))2.

Combining (4.8) and (4.10) implies that τV*(z,τ) is continuous across z=zb(τ). Moreover, by (4.10),

limzzs(τ)τV*(z,τ)=γ(αr)(1μ(τ))zs(τ)+γ2σ22(1μ(τ))2zs2(τ).

On the other hand, if zzs(τ),

τV*(z,τ)=B(τ)+γμ(τ)z=A(τ)γ0zs(τ)τV(ξ,τ)dξ+γ(zzs(τ))μ(τ)=A(τ)γσ22ξ2ξV(ξ,τ)+(αr)ξV(ξ,τ)γσ22(ξV(ξ,τ))2ξ=zb(τ)ξ=zs(τ)+γ(zzs(τ))μ(τ)=γ(zzs(τ))μ(τ)+γ2σ22zs2(τ)(1μ(τ))2γ(αr)zs(τ)(1μ(τ)).

If z=zs(τ), then

τV*(z,τ)=γ(αr)(1μ(τ))zs(τ)+γ2σ22(1μ(τ))2zs2(τ).

Therefore, τV*(z,τ) is continuous across z=zs(τ). As a result, we notice that z[zV(z,τ)] is bounded on ×[0,T], and that zzV*(z,τ)=γzV(z,τ) is continuous on ×[0,T]. Furthermore,

z2zzV*(z,τ)=γz2zV(z,τ)=γzz[zV(z,τ)]V(z,τ)

is continuous on ×[0,T]. This completes the proof of Lemma 4.1.

Theorem 4.2. V*(z,τ), defined by (4.4), is the solution of the problem (2.5). In detail,

τV*LV*0, in ×(0,T);
γ(1+λ(τ))zV*γ(1μ(τ)), in ×(0,T);
τV*LV*=0, if γ(1+λ(τ))<zV*<γ(1μ(τ));
V*(z,0)=γ(1+λ0)z, if z<0,γ(1μ0)z, if z0.

Proof. Since zV*=γV and 1μ(τ)V1+λ(τ), we obtain (4.13). In detail,

zV*=γ(1+λ(τ)), if zzb(τ),γ(1+λ(τ))<zV*<γ(1μ(τ)), if zb(τ)<z<zs(τ),zV*=γ(1μ(τ)), if zzs(τ).

Combining A(0)=0 and the initial value of V with (4.3) yields (4.15). Next, we show (4.14). Since Lz(γV)=1γz(LV*), we have

z(τV*LV*)=0 if γ(1+λ(τ))<zV*<γ(1μ(τ)).

Moreover, we get

τV*LV*=0   on z=zb(τ).

Combining (4.16) and (4.17) gives (4.14). Finally, we know that

zτV*LV*=γτVLzV0, if zzb(τ);=0, if zb(τ)<z<zs(τ);0, if zzs(τ),

which proves (4.12).

1. M. Akian, J. L. Menaldi and A. Sulem, On an investment-consumption model with transaction costs, SIAM J. Control Optim., 34(1)(1996), 329-364.
2. X. Chen, Y. Chen and F. Yi, Parabolic variational inequality with parameter and gradient constraints, J. Math. Anal. Appl., 385(2)(2012), 928-946.
3. M. Dai and F. Yi, Finite-horizon optimal investment with transaction costs: a parabolic double obstacle problem, J. Differential Equations, 246(4)(2009), 1445-1469.
4. M. H. A. Davis, V. G. Panas and T. Zariphopoulou, European option pricing with transaction costs, SIAM J. Control Optim., 31(2)(1993), 470-493.
5. A. Friedman, Parabolic variational inequalities in one space dimension and smooth-ness of the free boundary, J. Functional Analysis, 18(1975), 151-176.
6. P. Guasoni, Optimal investment with transaction costs and without semimartingales, Ann. Appl. Probab., 12(4)(2002), 1227-1246.
7. G. M. Lieberman. Second order parabolic differential equations. River Edge, NJ: World Scientific Publishing Co.,Inc.; 1996.
8. OECD: Organisation for Economic Co-Operation and Development. OECD Economic Outlook, Organization for Economic Co-operation and Development (OECD). 81 edition. Paris Cedex, France; 2007.
9. O. A. Oleĭnik and E. V. Radkevič. Second order equations with nonnegative characteristic form, Translated from the Russian by Paul C. Fife. New York-London: Plenum Press; 1973.
10. R. Pollin, D. Baker and M. Schaberg, Securities transaction taxes for US financial markets, Eastern Economic Journal, 29(4)(2003), 527-558.
11. K. Tso, On an Aleksandrov-Bakel'man type maximum principle for second-order parabolic equations, Comm. Partial Differential Equations, 10(5)(1985), 543-553.
12. Z. Yang and F. Yi, Valuation of European installment put option: variational inequality approach, Commun. Contemp. Math., 11(2)(2009), 279-307.
13. F. Yi and Z. Yang, A variational inequality arising from European option pricing with transaction costs, Sci. China Ser. A, 51(5)(2008), 935-954.