Asymptotic behavior of optimal exercise strategy for a small number of executive stock options

J. Math. Anal. Appl. 472 (2019) 1253–1276

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Journal of Mathematical Analysis and Applications www.elsevier.com/locate/jmaa

Asymptotic behavior of optimal exercise strategy for a small number of executive stock options ✩ Cong Qin a,∗ , Xinfu Chen b , Xin Lai c , Wanghui Yu a a b c

Center for Financial Engineering, Soochow University, Suzhou, Jiangsu 215006, PR China Department of Mathematics, University of Pittsburgh, PA 15260, USA College of Science, Civil Aviation University of China, Tianjin 300300, PR China

a r t i c l e

i n f o

a b s t r a c t

Article history: Received 30 July 2017 Available online 3 December 2018 Submitted by S.A. Fulling

We investigate the asymptotic behavior, as the number of options approaches zero, of the optimal exercise strategy of continuous-exercise perpetual executive stock options. Here the optimal strategy is described in terms of the free boundary of a degenerate variational inequality. To our knowledge, our method based on the convexity of the variational inequality at the initial time is new. Also, the unique formal asymptotic expansions of the free boundary and the solution of the variational inequality are derived. © 2018 Elsevier Inc. All rights reserved.

Keywords: Executive stock options Optimal exercise strategy Free boundary Asymptotic expansion

1. Introduction In this paper, we continue to study the variational inequality   min A[ϕ], Bϕ = 0 in R × (0, ∞),

ϕ(·, 0) = 0,

(1.1)

where ϕ = ϕ(z, a) is unknown, A[ϕ] := Raϕa − ϕzz − νϕz + ϕ2z ,

Bϕ := ϕa − g + ,

subscripts represent partial derivatives, R > 0 and ν are constants, g + := max{g, 0}, and g = ez − 1. Setting φ(z, t) := ϕ(z, et ), the problem can be written as  Rφt = max φzz + νφz − φ2z , Ret g + (z)}

in R2 ,

φ(·, −∞) = 0.

✩ Qin thanks support from the Start-up fund at Soochow University (grant number Q422000118). Chen thanks the support from NSF grant DMS-1008905. * Corresponding author. E-mail address: [email protected] (C. Qin).

https://doi.org/10.1016/j.jmaa.2018.11.075 0022-247X/© 2018 Elsevier Inc. All rights reserved.

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This is a fully non-linear degenerate parabolic equation with initial value given at t = −∞. It is shown in [13] that for the physical relevant solution, ϕ = ∞ if ν  R. When ν < R − 1, Song and Yu [15] show the existence of a strong solution. When ν < R, in [8,9] we prove the existence of a unique classical solution and the existence of a continuous and strictly decreasing function s(·) defined on [0, ∞) such that ⎧ ⎪ ⎨ A[ϕ] = 0 < Bϕ A[ϕ] > 0 = Bϕ ⎪ ⎩ A[ϕ] = 0 = Bϕ

in

N := {(z, a) | a > 0, z < s(a)},

in

T := {(z, a) | a > 0, z > s(a)},

on

Γ := {(z, a) | a > 0, z = s(a)}.

(1.2)

We call Γ the free boundary. In its original context, N is called no-trading region and T the trading region, and optimal strategies depend only on the free boundary. Particularly, when ν < R − 1, the C ∞ regularity of the free boundary has been shown in [12] by converting (1.1) to a Stefan problem. Here we consider the asymptotic behavior of the free boundary s(a) near the origin a = 0, i.e., when the number of ESOs is very small. Problem (1.1) originates from the study of perpetual Executive Stock Options (ESOs) which are American call options granted by a firm with long maturity to an executive as a form of benefit in addition to salary. For option holder, the problem is to find optimal excise strategy. An exercise strategy can be described mathematically by an adaptive stochastic process {A(t)}t0 where A(t) is right-continuous, decreasing, non-negative, and A(0− ) = A; here A(t) is the number of ESO’s holding at time t (after trading), and A is the initial number of shares of ESOs. There are a number of studies for ESOs; see Lambert, Larcker and Verechia [10], Carpenter [1], Hall and Murphy [4], Ingersoll [6], Rogers and Scheinkman [14], Jain and Subramanian [7], Grasselli and Henderson [3], Henderson and Hobson [5], Leung and Sircar [11], Song and Yu [15,16]. The paper [2] by Carpenter, Stanton, and Wallace contains an extensive list of references. Based on the models proposed in [14,15] and our analysis presented in [13], in this paper we continue the study of the value function

V (x, s, A) := lim

T →∞

sup {A(t)}∈M(A)

T 

  + E U x − e−rt S(t) − K dA(t) . 0−

Here, x, s, K, A are initial cash, stock price, strike price, and ESO holding respectively, r  0 is a constant discount rate, U is a given utility (concave increasing) function, and M(A) is the set of all admissible strategies. Assume that the stock price is modeled by a geometric Brownian motion with constant expected return α and constant volatility σ > 0 and that U (x) = −e−γx where γ is positive constant risk aversion. Using dimensionless quantities z = log

s , K

a = γKA,

R=

2r , σ2

ν=

2 σ2  , α − σ2 2

one can show (cf. [13]) that V (x, s, A) = e−γx u(z, a) where u solves   min Raua − uzz − νuz , ua + g + u = 0 in R × (0, ∞),

u(·, 0) = −1.

(1.3)

The dimensionless certainty equivalent is a function ϕ = ϕ(z, a) such that   V = U x + γ −1 ϕ . Then ϕ = − ln(−u) and is a viscosity solution of (1.1). The optimal exercise strategy is to hold the following amount of option at time t:

C. Qin et al. / J. Math. Anal. Appl. 472 (2019) 1253–1276

A

optimal

1255

  erρ a∗ ln S(ρ) K (t) = min A, min , 0ρt γK 

where a = a∗ (z) is the inverse function of z = s(a), with natural extension a∗ = ∞ for z  0 = s(∞) and a∗ = 0 for z > s(0+ ); see [13]. Since options are not exercised when stock price is below strike price, (1.1) can be simplified as  min{Raϕa − ϕzz − νϕz + ϕ2z , ϕa − g = 0 in R × (0, ∞),

ϕ(·, 0) = 0.

(1.4)

In this paper, we develop a new method, which depends on the convexity of the equation (1.1) at the initial time a = 0, to study the asymptotic behavior of the free boundary as a  0. Our main result is the following: Theorem 1. Let R > 0 and (ϕ, s) be the unique solution of (1.1), (1.2). There exist positive constants δ and M that depend only on ν and R such that for every a ∈ (0, δ), the following holds: λ (1) If ν < R − 1, then 0  ln λ−1 − s(a)  M a, where λ > 1 is the positive root of x2 + νx = R; (2) If ν = R − 1, then 1

−M a 8  s(a) − where λ is the positive root of x2 − (1 −

R 2 )x

1 λ 1 ln  Ma6 , 2 a(2 − λ)

= R. Consequently, lim ae2s(a) = a0

(3) If R − 1 < ν < R, then 1

−M a 4  s(a) − ln

λ ; 2−λ

1 1+ν−R  Ma3 ; a

Consequently, lim aes(a) = 1 + ν − R. a0

We shall also derive the limiting profile of ϕ(z, a) as a  0; see Theorems 3, 4, 5 in Sections 2, 3, and 4 respectively. Remark 1.1. When A is small, we have the following approximation of the optimal strategy: (1) when α ∈ (r, r + σ 2 /2),  2(α − r)erρ  A, ; 0ρt σ 2 γS(ρ)

Aoptimal (t) ≈ min

(2) when α = r, denoting by λ the positive root of x2 + (R/2 − 1)x = R, Aoptimal (t) ≈ min

0ρt

(3) when α < r, Aoptimal (t) = 0 whenever S(t) >

 A,

Kλ λ−1 ,

 λKerρ ; (2 − λ)γ S 2 (ρ)

where λ is the positive roots of x2 + νx = R.

Note that when α  r + σ 2 /2, ϕ(z, a) = ∞ for every a > 0, z ∈ R; see [13]. Next, we shall give the formal asymptotic expansion of the solution (ϕ, s) with respect to a. To be more precise, we introduce the following definition:

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Definition 1. Given functions f , {fi }∞ i=0 and a differential operator D. Then (1) We call

∞ i=0

ai fi (x) an asymptotic expansion of f with respect to a if n      ai fi (x) = O(1)an+1 f (x, a) −

∀ n  0,

i=1

and denote by f (x, a) ∼

∞ 

an fn (x) = f0 (x) + af1 (x) + a2 f2 (x) + · · · ;

n=0

(2) We call

∞ i=0

ai fi (x) a formal asymptotic expansion of f with respect to a if Df = 0 and D

n 

 ai fi (x) = O(1)an+1

∀ n  0,

i=0

and denote by f (x, a) 

∞ 

an fn (x) = f0 (x) + af1 (x) + a2 f2 (x) + · · · .

n=0

Due to the asymptotic behavior of the free boundary, we first split the derivation into three scenarios, i.e. ν < R − 1, ν = R − 1 and R − 1 < ν < R. Then, by the key transformation with the corresponding case (see Theorem 2 below), we convert the original free boundary problem for (ϕ, s) to the fix boundary problem for (Φ, ξ). Finally, we obtain the unique formal asymptotic expansion of (Φ, ξ) with respect to a. Thus we shall prove the following: Theorem 2. Let R > 0 and (ϕ, s) be the unique solution of (1.1), (1.2). (1) If ν < R − 1, set ψ = ϕa and ϕ(z, a) = Φ(z − s(a), a),

ψ(z, a) = Ψ(z − s(a), a).

Then there exists a unique formal asymptotic expansion of (Φ, Ψ, s) as a  0: ∞ 

Φ(x, a) 

an+1 Φn (x),

Ψ(x, a) 

n=0

∞ 

an Ψn (x),

s(a) 

n=0

∞ 

an sn ,

n=0

with the leading order Φ0 (x) = Ψ0 (x) =

eλx λ−1

∀ x  0,

s0 = ln

λ ; λ−1

(2) If ν = R − 1, set   ϕ(z, a) = a ez − Φ(z − s(a), a) ,

s(a) = ξ(a) −

1 ln a. 2

Then there exists a unique formal asymptotic expansion of (Φ, ξ) as a  0:

∀ x  0,

C. Qin et al. / J. Math. Anal. Appl. 472 (2019) 1253–1276

Φ(x, a) 

∞ 

n

ξ(a) 

a 2 Φn (x),

n=0

∞ 

n

1257

∀ x ∈ R,

a 2 ξn ,

n=0

with the leading order

Φ0 (x) =

⎧ ⎨1 − ⎩

λ −2x 2(2+λ) e

− 12 e2x+2ξ0 +

x0

if

4 λx 4−λ2 e

if

 ,

ξ0 = ln

x<0

λ ; 2−λ

(3) If R − 1 < ν < R, set ϕ(z, a) = Φ(z − s(a), a),

s(a) = ξ(a) − ln a.

Then there exists a unique formal asymptotic expansion of (Φ, ξ) as a  0: Φ(x, a) 

∞ 

ξ(a) 

an Φn (x),

n=0

∞ 

∀ x ∈ R,

an ξn ,

n=0

with the leading order 

x Φ0 (x) =

w(y)dy,

w(x) =

−∞

λ(1−λ) e−λx −(1−λ) x

(1 − λ)e

x<0 x0

ξ0 = ln(1 − λ).

,

Therefore, the rest of the paper is organized as follows. We divide the proof of Theorem 1 into three parts. In Sections 2–4, we give the proof for the case ν < R − 1, ν = R − 1, and R − 1  ν < R respectively. Then, Sections 5–7 devote to the derivation of the unique formal asymptotic expansion of (Φ, ξ) respectively. 2. The case ν < R − 1 (i.e. α < r) Suppose that (ϕ, s) is the unique classical solution of (1.1), (1.2). Set ψ(z, a) = ϕ(z, a)/a

∀ z ∈ R, a > 0.

From (1.4) we obtain, for (z, a) ∈ R × [0, ∞), min{Raψa + Rψ − ψzz − νψz + aψz2 , aψa + ψ − g} = 0. 2.1. The limit profile Sending a  0, we obtain formally ψ(·, a) → ψ0 (·) which solves H[ψ0 ] := min{Rψ0 − ψ0 − νψ0 , ψ0 − g}

in R

A straightforward verification shows the following. Lemma 2.1. Assume that R > 0 and ν < R − 1. Then (2.1) admits a unique solution given by  ψ0 (z) =

ez − 1

λ if z  b := ln λ−1 ,

(eb − 1)eλ(z−b)

if z < b,

(2.1)

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where λ is the positive root of equation x2 + νx = R, given by λ=

−ν +

 (2 − ν)2 + 4(R − ν − 1) > 1. 2

√

−ν + ν 2 + 4R = 2

2,∞ Remark 2.1. We say ψ0 is a solution of (2.1) if ψ0 ∈ Wloc and (2.1) is satisfied in a.e. sense.

For later application, here we provide a lower bound of ψ0 − g. When z ∈ [0, b],

b ψ0 (z) − g(z) =

[g  (x) − ψ0 (x)]dx = g  (b)

z

 g  (b)e−

b e

1+λ 2 (x−b)

2 sinh

(λ − 1)(b − x) dx 2

z 1+λ 2 b

b (λ − 1)(b − x)dx =

λ − 1 1−λ b λ−1 e 2 (b − z)2 > (b − z)2 ; 2 4

z

here we use (λ − 1)b = (λ − 1) ln(1 +

1 λ−1 )

< 1, so e

1−λ 2 b

 e−1/2 > 1/2.

2.2. Bounds of the solution In this subsection we give upper and lower bounds of the solution which are critical for our study of the asymptotic behavior of the free boundary s. Theorem 3. Let ψ0 , λ, and b be given by Lemma 2.1. Then ϕ = ag in [b, ∞) × [0, ∞) and aψ0 (z) −

(b − z)λa2 2λ2 a3 λ2 a4 − −  ϕ(z, a)  aψ0 (z) (λ − 1)2 3R(λ − 1)3 4R(λ − 1)4

∀ z  b, a  0.

Proof. First of all, since ϕa  g and ϕ(·, 0) = 0, we have ϕ  ag in R × [0, ∞). Next, with φ = aψ0 , we have, for every (z, a) ∈ R × [0, ∞), min{A[φ], Bφ} = min{a[Rψ0 − ψ0 − νψ0 ] + a2 ψ0 2 , ψ0 − g}  min{a[Rψ0 − ψ0 − νψ0 ], ψ0 − g} = 0. By a comparison principle, ϕ(z, a)  φ = aψ0 (z) for all z ∈ R and a  0. Consequently, ϕ(z, a) = ag(z) in [b, ∞) × [0, ∞). For lower bound, let λ2 be the positive root of x2 + νx = 2R. Then one can show that λ < λ2 < 2λ; see [12]. Define, for z  b and some large constants B and C to be chosen later, w0 (z) :=

1 2 g (b){e2λ(z−b) − eλ2 (z−b) }, 2

φ(z, a) := aψ0 (z) + a2 w0 (z) − Ba3 − Ca4 .

Note that 2Rw0 − w0 − νw0 + ψ0 2 = 0 > w0 and |w0 |  λg 2 (b) in (−∞, b). Hence, in (−∞, b] × [0, ∞), A[φ] = Raφa − φzz − νφz + φ2z = a{Rψ0 − ψ0 − νψ0 } + a2 {2Rw0 − w0 − νw0 + ψ0 2 } + a3 {2ψ0 w0 + aw0 2 − 3RB − 4RCa]} = a3 {2ψ0 w0 + aw0 2 − 3RB − 4RCa}  a3 {2g 3 (b)λ2 + ag 4 (b)λ2 − 3BR − 4RCa}  0  A[ϕ],

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provided that we take B = 2λ2 g 3 (b)/(3R) and C = g 4 (b)λ2 /(4R). Note that when z = b, we have w0 (b) = 0 so φ(b, z)  aψ0 (b) = ag(b) = ϕ(b, a). Hence, by comparison, we have φ  ϕ on (−∞, b] × [0, ∞). Upon noting that g(b) = 1/(λ − 1) and w0 (z) > −λg(b)2 (b − z), we obtain the asserted lower bound of ϕ. This completes the proof. 2 2.3. Location of the free boundary Since ϕa = g when z > s(a), we have, for fixed a > 0, ϕ(s(a), 2a) = ϕ(s(a), a) + ag(s(a)). Hence, by the lower bound in Theorem 3, [2a]ψ0 (s(a)) −

λ(b − s(a))[2a]2 λ2 [2a]3 [2a]4 − − 2 3 (λ − 1) 3R(λ − 1) 4R(λ − 1)4

 ϕ(s(a), 2a) = ϕ(s(a), a) + ag(s(a))  aψ0 (s(a)) + ag(s(a)). It then follows that 8λ2 a2 4λ2 a3 (λ − 1) 4λ(b − s(a))a (s(a) − b)2 . + +  ψ0 (s(a)) − g(s(a))  2 3 4 (λ − 1) 3R(λ − 1) R(λ − 1) 4 This implies that

b − s(a) −

4(λ − 1)a2  4a2 λ2  8λa 2 8(λ − 1)2 + .  16 + (λ − 1)3 (λ − 1)6 3R R

Taking the square root we then obtain the following: Lemma 2.2. Suppose R > 0 and ν < R − 1. Let (ϕ, s) be the unique classical solution of (1.1), (1.2). Then there exists a positive constant M depending only on R and ν such that 0  b − s(a)  M a

∀ a > 0.

(2.2)

We remark that since s(a) ∈ (0, b], the above estimate is automatically true when a > b/M . 3. The case ν = R − 1 (i.e. r = α) In this section we turn to the case ν = R − 1. Suppose (ϕ, s) is the unique classical solution of (1.1), (1.2). We make the transformation ξ = z + ln

√

a,

  √ ϕ(z, a) = a ez − ψ(z + ln a, a)

for z ∈ R, a > 0.

(3.1)

Then from (1.4) we have, for (ξ, a) ∈ R × [0, ∞),   √ R 1 G[ψ] := max Raψa − ψξξ + 1 − ψξ + Rψ − e2ξ + 2 aeξ ψξ − aψξ2 , aψa + ψξ + ψ − 1 = 0. (3.2) 2 2 3.1. The limit profile Sending a  0, we obtain formally ψ(·, a) → ψ0 (·) solving H[ψ0 ] := max



 R  1 − ψ0 + 1 − ψ0 + Rψ0 − e2ξ , ψ0 + ψ0 − 1 = 0 in R. 2 2

Then a straightforward verification shows the following.

(3.3)

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Lemma 3.1. Assume that R > 0. Then problem (3.3) admits a unique classical solution given by

ψ0 (ξ) =

 1 − Be−2(ξ−ξ0 ) − 12 e2ξ

+ Ceλ(ξ−ξ0 )

if

ξ  ξ0 ,

if

ξ < ξ0 ,

where λ is the positive root of equation x2 − (1 − R/2)x = R and  ξ0 = ln

λ , 2−λ

B=

λ , 2(2 + λ)

C=

4 . 4 − λ2

Now we investigate the function ψ0 . Lemma 3.2. Assume that R > 0. Let ψ0 , λ, and ξ0 be as in Lemma 3.1. Then (1) 1 < λ < 2, ψ0 ∈ C 2 (R) and 0 < ψ0 < 1; (2) 0 < e−ξ ψ0 < m1 and 0 < eξ ψ0 < m2 , where 4λ m1 = 4 − λ2



2−λ , λ

m2 =

4λ 4 − λ2



λ ; 2−λ

(3) F1 [ψ0 ] := 12 ψ0 + ψ0 − 1 ∈ C 1 (R) and F1 [ψ0 ] = 0

in [ξ0 , ∞),

F1 [ψ0 ]  −λ(ξ − ξ0 )2 e2(ξ−ξ0 )

in (−∞, ξ0 );

(4) F2 [ψ0 ] := −ψ0 + (1 − R/2)ψ0 + Rψ0 − e2ξ ∈ C(R) and F2 [ψ0 ] = 0

in (−∞, ξ0 ],

F2 [ψ0 ]  −4λ(ξ − ξ0 )

in (ξ0 , ∞).

Proof. (1) A straightforward calculation shows that 1 < λ < 2, ψ0 ∈ C 2 (R) and 0 < ψ0 < 1. (2) In [ξ0 , ∞), we have, using λ ∈ (1, 2), λ −2(ξ−ξ0 ) e > 0, 2+λ  λ λeξ0 −(ξ−ξ0 ) λeξ0 λ ξ  e =  m2 ,  e ψ0 = 2+λ 2+λ 2+λ 2−λ  λ λe−ξ0 −3(ξ−ξ0 ) λe−ξ0 2−λ −ξ  e =  m1 .  e ψ0 = 2+λ 2+λ 2+λ λ ψ0 = 2Be−2(ξ−ξ0 ) =

In (−∞, ξ0 ], we have, λ 2(ξ−ξ0 ) 4λ λ(ξ−ξ0 ) e + e 2−λ 4 − λ2

λ λ(ξ−ξ0 ) λ λ(ξ−ξ0 ) λ 2(ξ−ξ0 ) e e = + − e2(ξ−ξ0 )  > 0, e 2+λ 2−λ 2+λ

ψ0 = −e2ξ + λCeλ(ξ−ξ0 ) = −

eξ ψ0  e−ξ ψ0 

4λeξ0 (λ+1)(ξ−ξ0 ) 4λeξ0 e  = m2 , 2 4−λ 4 − λ2 4λe−ξ0 (λ−1)(ξ−ξ0 ) 4λe−ξ0 e  = m1 . 4 − λ2 4 − λ2

Therefore 0 < e−ξ ψ0  m1 and 0 < eξ ψ0  m2 .

(3.4)

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(3) It is clear that F1 [ψ0 ] = 0 in [ξ0 , ∞). In (−∞, ξ0 ],

1 1 − e2ξ + λCeλ(ξ−ξ0 ) − e2ξ + Ceλ(ξ−ξ0 ) − 1 2 2 2 λ(ξ−ξ0 ) λ 2(ξ−ξ0 ) = e e −1− 2−λ 2−λ

F1 [ψ0 ] =

2λ =− 2−λ −

ξ

0 −ξ

−λx

[e

−e

−2x

0

2λe−2(ξ0 −ξ) 2−λ

2λ ]dx = − 2−λ

ξ

0 −ξ

e−2x [e(2−λ)x − 1]dx

0

ξ

0 −ξ

(2 − λ)x dx = −2λe2(ξ−ξ0 ) (ξ − ξ0 )2 . 0

(4) It is clear that F2 [ψ0 ] = 0 in (−∞, ξ0 ]. In [ξ0 , ∞), let x = ξ −ξ0  0. Then F2 [ψ0 ] = Be−2x [4 +2(1 − R2 ) −R] −e2ξ0 e2x +R. Since λ2 −(1 −R/2)λ = R, we have R = 2λ(λ − 1)/(2 − λ). Hence, F2 [ψ0 ] =

λ (3 − 2λ)e−2x − e2x + 2(λ − 1) =: f (x). 2−λ

Note that f (0) = 0 and for x > 0

λ (6 − 4λ)e−2x + 2e2x 2−λ   λ = (8 − 4λ)e−2x + 2 e2x − e−2x 2−λ

λ  (8 − 4λ)(1 − 2x) + 8x 2−λ

λ = 4(2 − λ) + 8(λ − 1)x  4λ. 2−λ

−f  (x) =

Therefore F2 [ψ0 ] = f (x) =

x 0

f  (ρ)dρ  −4λx = −4λ(ξ − ξ0 ) in (ξ0 , ∞).

2

3.2. Bounds of the solution Theorem 4. Assume that R = 1 + ν > 0. Let m1 , m2 be defined in (3.4) and ψ, ψ0 be the solution of (3.2) and (3.3) respectively. Then ψ  ψ0

in R × [0, 4/m21 ],

ψ  ψ0 −

4m2 √ a 3R

in R × [0, ∞).

Consequently, for the unique classical solution of (1.1), 0

√ √ 4m2 a ϕ(z, a) z − e − ψ0 (z + ln a)  a 3R

4  ∀ z ∈ R, a ∈ 0, 2 . m1

Proof. By definition, 

√

G[ψ0 ] = max F2 [ψ0 ] + 2 ae

ξ

ψ0

√  a −ξ   e ψ0 , F1 [ψ0 ] . 1− 2



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Note that ψ0 > 0 and √ √ a −ξ  a e ψ0  1 − m1  0 1− 2 2

if 0  a 

4 . m21

Hence G[ψ0 ]  0 in R × [0, 4/m21 ]. By comparison principle, we have ψ(ξ, a)  ψ0 (ξ) for ξ ∈ R, a ∈ [0, 4/m21 ]. On the other hand, let D = 4m2 /(3R), then √   √ √ √ 3D a + F1 [ψ0 ] G[ψ0 − D a] = max F2 [ψ0 ] − 2m2 a + 2 aeξ ψ0 − aψ0 2 , − 2  max{F2 [ψ0 ], F1 [ψ0 ]} = 0. Then applying the comparison principle again, we have ψ  ψ0 −

4m2 √ a 3R

in R × [0, ∞). 2

3.3. Elliptic differential inequality Suppose (ϕ, s) is the unique solution of (1.1), (1.2). Set ξ∗ (a) := s(a) +

1 ln a ∀ a > 0. 2

(3.5)

Then in (−∞, ξ∗ (a)), we have aψa + 12 ψξ + ψ − 1  0 and √ R 0 = Raψa − ψξξ + 1 − ψξ + Rψ − e2ξ + 2 aeξ ψξ − aψξ2 2 1

√ R  −R ψξ + ψ − 1 − ψξξ + 1 − ψξ + Rψ − e2ξ + 2 aeξ ψξ − aψξ2 = J [ψ], 2 2 where J is an elliptic operator defined by √ J [ζ] = −RF1 [ζ] + F2 [ζ] + 2 aeξ ζ  − aζ  2 √ = −ζ  + [1 − R]ζ  + R + 2 aeξ ζ  − aζ  2 .

(3.6)

In (ξ∗ (a), ∞), we have aψa + 12 ψξ + ψ − 1 = 0 and √ R )ψξ + Rψ − e2ξ + 2 aeξ ψξ − aψξ2 2 1

√ R = −R ψξ + ψ − 1 − ψξξ + (1 − )ψξ + Rψ − e2ξ + 2 aeξ ψξ − aψξ2 = J [ψ]. 2 2

0  Raψa − ψξξ + (1 −

Hence we have the following lemma. Lemma 3.3. Assume that R = 1 + ν > 0 and let (ϕ, s) be solution of (1.1), (1.2). Let J be defined by (3.6), ψ be the solution of (3.2) and ξ∗ be defined as (3.5). Then for each a > 0, J [ψ(·, a)]  0 in (ξ∗ (a), ∞),

J [ψ(·, a)]  0 in (−∞, ξ∗ (a)).

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Fig. 1. The geometric illustration of the method.

3.4. Location of free boundary Lemma 3.4. Suppose R = 1 + ν > 0 and let (ϕ, s) be the solution of (1.1), (1.2). Let λ be the positive root λ of x2 − (1 − R/2)x2 = R, ξ0 = 12 ln 2−λ and ξ∗ be defined as (3.5). Then there exist δ > 0 and M > 0 such that 1

1

−M a 8  ξ∗ (a) − ξ0  M a 6 Consequently, lim ae2s(a) = e2ξ0 = a0

∀ a ∈ (0, δ).

(3.7)

λ . 2−λ

Proof. (i) Suppose that the second inequality in (3.7) is not true for the following M and δ:  m2 16  27K  13  1 1 M = max 1, , , 4 4λ

δ=

1 , M6

2 where K1 = 2m2 +2D(1+|1−R|+2eξ0 +1 ) and D = 4m 3R . Then there exists a ∈ (0, δ) such that ξ∗ (a) > ξ0 +3d 1 1 1 where d = 3 M a 6 ∈ (0, 3 ). Then by Lemma 3.3, we have J [ψ(·, a)]  0 in I = [ξ0 + d, ξ0 + 3d], since ξ∗ (a) > ξ0 + 3d. We now construct a subsolution on I, see Fig. 1 for illustration. √ ¯ where ψ1 (x) := (x − ξ0 − 2d)2 /d2 . Since ψ  ψ0 − D√a Define ψ(ξ) := ψ0 (ξ) − D aψ1 (ξ) for ξ ∈ I, √ and ψ1 = 1 on ∂I = {ξ0 + d, ξ0 + 3d}, we have ψ = ψ0 − D a < ψ(·, a) on ∂I1 . In addition, in I1 using F1 [ψ0 ] = 0 and F2 [ψ0 ] < −4λd (by Lemma 3.2) we obtain

√ J [ψ] = −ψ  + (1 − R)ψ  + R − e2ξ + 2 aeξ ψ  − aψ  2  √  √ = −RF1 [ψ0 ] + F2 [ψ0 ] − aψ  2 + a 2eξ ψ0 + Dψ1 − D(1 − R + 2 aeξ )ψ1 √  2D 2  −4λd + a 2m2 + 2 + D(|1 − R| + 2eξ0 +1 ) d d √ √   3 K1 a 4M λ aK1  −4λd + − 2 − 1 < 0  J [ψ(·, a)]. d2 d 27K1

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Hence, by comparison principle, ψ < ψ(·, a) in I1 . However ψ = ψ0  ψ(·, a) at ξ = ξ0 + 2d which is a contradiction. Hence the second inequality in (3.7) is true. (ii) For the first inequality in (3.7), we use  m2 18  81eK  14  2 1 M = max 1, , , 4 Rλ

δ=

1 , M8

where K2 = 2D(1 + |1 − R| + 2eξ0 ) + (m2 + 2D)2 . If the first inequality in (3.7) is not true, then there exists 1 a ∈ (0, δ) such that ξ∗ (a) < ξ0 − 3d, where d = 13 M a 8 . √ Define ψ(ξ) := ψ0 (ξ) − D aψ2 (ξ) in I2 = [ξ0 − 3d, ξ0 − d], where ψ2 (ξ) := (ξ − ξ0 + d)(ξ0 − 3d − ξ)/d2 . Note that ψ = ψ0  ψ(·, a) on ∂I2 . Also, by Lemma 3.3, J [ψ(·, a)]  0 in I2 . In I2 using F2 [ψ0 ] = 0 and F1 [ψ0 ] < −λd2 e−1 (by Lemma 3.2), one can verify that J [ψ] > 0 on I2 . Thus the comparison principle √ implies ψ > ψ(·, a) in I2 . However, ψ = ψ0 − D a  ψ(·, a) at ξ = ξ0 − 2d which is a contradiction. This completes the proof. 2 4. The case ν ∈ (R − 1, R) (i.e. α ∈ (r, r + σ 2 /2)) In this section we deal with the case R − 1 < ν < R. Let (ϕ, s) be the unique classical solution of (1.2). We use the transformation ξ := z + ln a,

for z ∈ R, a > 0.

ϕ(z, a) = ψ(z + ln a, a)

(4.1)

Then from (1.4) we have, for (ξ, a) ∈ R × (0, ∞),   G[ψ] := min Raψa − ψξξ + (R − ν)ψξ + ψξ2 , aψa + ψξ − eξ + a = 0.

(4.2)

4.1. The limit profile Formally, sending a  0, we obtain ψ(·, a) → ψ0 (·) which solves H[ψ0 ] := min{−ψ0 + [R − ν]ψ0 + ψ0 2 , ψ0 − eξ }

in R.

(4.3)

Then a straightforward verification shows the following lemma. Lemma 4.1. Assume that λ := R − ν ∈ (0, 1). The problem (4.3) admits a unique solution given by

ξ ψ0 (ξ) =

v(z)dz, −∞

v(ξ) =

 eξ

if ξ  ξ0 := ln(1 − λ),

λ(1−λ) e−λ(ξ−ξ0 ) −(1−λ)

if ξ < ξ0 .

Now we study ψ0 . Lemma 4.2. Let ψ0 and ξ0 be given by Lemma 4.1. Then the following holds (1) ψ0 ∈ C 2 (R); (2) F1 [ψ0 ] := ψ0 − eξ ∈ C 1 (R) and F1 [ψ0 ] = 0

in [ξ0 , ∞),

F1 [ψ0 ] 

λ(1 − λ)2 (ξ − ξ0 )2 e2(ξ−ξ0 ) 2

in (−∞, ξ0 );

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(3) F2 [ψ0 ] := −ψ0 + λψ0 + ψ0 2 ∈ C(R) and F2 [ψ0 ] = 0

F2 [ψ0 ]  (1 − λ)2 (ξ − ξ0 )

in (−∞, ξ0 ),

in [ξ0 , ∞).

Proof. (1) Straightforward calculation shows that ψ0 ∈ C 2 (R). (2) From Lemma 4.1, it is clear that F1 [ψ0 ] = 0 in [ξ0 , ∞). In (−∞, ξ0 ), denoting x = ξ − ξ0 < 0 for simplicity, we have F1 [ψ0 ] = ψ0 − eξ =

λ(1 − λ) − eξ e−λ(ξ−ξ0 ) − (1 − λ)

λ(1 − λ) − (1 − λ)eξ−ξ0 − (1 − λ)   1−λ λ − ex(1−λ) + (1 − λ)ex = −λx e − (1 − λ)

x   1−λ = −λx (1 − λ) eρ − eρ(1−λ) dρ e − (1 − λ) =

e−λ(ξ−ξ0 )

0

=

(1 − λ)2 e−λx − (1 − λ)

 (1 − λ) e e 2 λx

0 e

2−λ 2 ρ



 λ λ e− 2 ρ − e 2 ρ dρ

x

2−λ 2 x

0 (−λρ)dρ x

λ λ(1 − λ)2 2 (1+ λ )x λ = (1 − λ)2 eλx e(1− 2 )x x2 = x e 2 2 2 λ(1 − λ)2 2 2x x e ,  2

where we use the fact that ex − e−x  2x for x  0. (3) From Lemma 4.1, it is clear that F2 [ψ0 ] = 0 in (−∞, ξ0 ]. Also in [ξ0 , ∞) we have   F2 [ψ0 ] = −ψ0 + λψ0 + ψ0 2 = eξ − 1 + λ + eξ     = eξ eξ − eξ0 = eξ+ξ0 eξ−ξ0 − 1  (1 − λ)2 (ξ − ξ0 ). This completes the proof. 2 4.2. Bounds of the solution Theorem 5. Suppose R > 0 and R − 1 < ν < R. Let ψ be the solution of (4.2) and ψ0 be as in lemma (4.1). Then ψ0 (ξ) − a  ψ(ξ, a)  ψ0 (ξ)

∀ ξ ∈ R, a  0.

Consequently, the solution of (1.1) satisfies −a  ϕ(z, a) − ψ0 (z + ln a)  0

∀ z ∈ R, a  0.

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Proof. Notice that G[ψ0 ] = min{−ψ0 + λψ0 + ψ0 2 , a + ψ0 − eξ }  H[ψ0 ] = 0, G[ψ0 − a] = min{−Ra − ψ0 + λψ0 + ψ0 2 , ψ0 − eξ }  H[ψ0 ] = 0. The assertion of the lemma then follows by comparison principle. 2 4.3. Elliptic differential inequality Suppose (ϕ, s) is the solution of (1.1), (1.2). Set ∀ a > 0.

ξ∗ (a) := s(a) + ln a

(4.4)

Then in (−∞, ξ∗ (a)), we have aψa + ψξ − eξ + a  0 and 0 = Raψa − ψξξ + λψξ + ψξ2  −R(ψξ − eξ + a) − ψξξ + λψξ + ψξ2 = J [ψ(·, a)], where J is an elliptic operator defined by J [ζ] := −RF1 [ζ] − aR + F2 [ζ] = R(eξ − a) − ζ  − νζ  + ζ  2 .

(4.5)

In (ξ∗ (a), ∞), we have aψa + ψξ − eξ + a = 0 and 0  Raψa − ψξξ + λψξ + ψξ2 = −R(ψξ − eξ + a) − ψξξ + λψξ + ψξ2 = J [ψ(·, a)]. Hence we have the following lemma. Lemma 4.3. Let J be defined by (4.5), ψ be the solution of (4.2) and ξ∗ be defined by (4.4). Then J [ψ(·, a)]  0

in (−∞, ξ∗ (a)),

J [ψ(·, a)]  0

in (ξ∗ (a), ∞)

∀ a > 0.

4.4. Location of free boundary Lemma 4.4. Suppose R > 0 and R − 1 < ν < R. Let λ = R − μ and ξ0 = ln(1 − λ), (ϕ, s) be the unique solution of (1.2), and ξ∗ be defined as (4.4). Then there exist δ > 0 and M > 0 such that 1

1

−M a 4  ξ∗ (a) − ξ0  M a 3

∀ a ∈ (0, δ).

(4.6)

Consequently,  1 − λ   lim s(a) − ln  = 0, a0 a

lim aes(a) = 1 − λ.

a0

Proof. Suppose the second inequality in (4.6) is not true with the following  27(R + 2 + 2|ν| + 4eξ0 +1 ) 13  M = max 1, , (1 − λ)2

δ=

1 . M3

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1

Then there exists a ∈ (0, δ) such that ξ∗ (a) > ξ0 + 3d, where d = 13 M a 3 ∈ (0, 1/3]. By Lemma 4.3, J [ψ(·, a)]  0 in I := [ξ0 + d, ξ0 + 3d]. We now construct a supersolution. Define ψ(ξ) := ψ0 (ξ) − aψ1 (ξ) for ξ ∈ I := [ξ0 + d, ξ0 + 3d], where ψ1 (ξ) := (ξ − ξ0 − d)(ξ0 + 3d − ξ)/d2 . Note that ψ = ψ0  ψ(·, a) on ∂I. Also, in I, using F1 [ψ0 ] = 0 and F2 [ψ0 ]  (1 − λ)2 d (by Lemma 4.2), we have 2

J [ψ] = −R(ψ ξ − eξ + a) − ψ ξξ + λψ ξ + ψ ξ   = −RF1 [ψ0 ] + F2 [ψ0 ] + a − R − ψ1 + (2ψ0 − ν)ψ1 + aψ1 2   2 2  (1 − λ)2 d − a R + 2 + |ν| + 2eξ0 +1 d d  ξ0 +1  2 R + 2 + 2|ν| + 4e (1 − λ) 3 > d − a > 0  J [ψ(·, a)]. d2 R + 2 + 2|ν| + 4eξ0 +1 Hence, comparison principle implies ψ > ψ(·, a) in I1 . However, ψ = ψ0 − a  ψ at ξ = ξ0 + 2d, a contradiction. This contradiction implies that the second inequality in (4.6) is true. Similarly, to prove the first inequality in (4.6), we set  81(R + 6 + 2|ν| + 4eξ0 ) 14  M = max 1, , Rλ(1 − λ)2 e2

δ=

1 . M4 1

If the first inequality in (4.6) is not true, there exists a ∈ (0, δ) such that ξ∗ (a) < ξ0 − 3d, where d = 13 M a 4 . By Lemma 4.3, J [ψ(·, a)]  0 in I := [ξ0 + d, ξ0 + 3d]. We now construct a subsolution. 2 Define ψ(ξ) := ψ0 (ξ) − aψ2 (ξ) in I2 := [ξ0 − 3d, ξ0 − d], where ψ2 := (ξ−ξ0d+2d) . Note that ψ = ψ0 − a  2 ψ(·, a) on ∂I2 . Also, J [ψ] < 0  J [ψ(·, a)] in I2 . Thus the comparison principle implies ψ < ψ(·, a) in I2 . However, ψ = ψ0  ψ(·, a) at ξ = ξ0 − 2d which is a contradiction. This completes the proof of the lemma. 2 Proof of Theorem 1. It is clear that Theorem 1 follows by Lemmas 2.2, 3.4, and 4.4.

2

5. Formal asymptotic expansion in the case ν < R − 1 Suppose ν < R − 1 (i.e. α < r) and (ϕ, s) is the solution of (1.1), (1.2). Set ψ = ϕa , then ψ satisfies, see [9], ⎧ ⎪ ⎪ ⎨Raψa + Rψ − ψzz − νψz + 2ϕz ψz = 0, ∀ z < s(a), a > 0, ∀ z  s(a), a > 0, ψ(z, a) = g(z), ψz (z, a) = g  (z) ⎪ ⎪ ⎩ϕ = ψ, ϕ(·, 0) = 0, ψ(−∞, ·) = 0.

(5.1)

a

5.1. Straighten the free boundary Set x = z − s(a),

ϕ(z, a) = Φ(z − s(a), a),

ψ(z, a) = Ψ(z − s(a), a)

Then from (5.1), (Ψ, s) satisfies the following problem.

∀ z ∈ R, a ∈ [0, ∞).

(5.2)

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⎧ ⎪ ⎪ ⎨LΨ := RaΨa − Ψxx − νΨx + RΨ = f in (−∞, 0) × (0, ∞), Ψx (0, a) = Ψ(0, a) + 1, Ψ(−∞, a) = 0 ∀ a > 0, ⎪ ⎪ ⎩s(a) = ln Ψ (0, a) ∀ a > 0, x

(5.3)

where f = Ras Ψx − 2Φx Ψx . And Φ satisfies Ψ(x, a) = Φa (x, a) − s (a)Φx (x, a),

in (−∞, 0) × (0, ∞),

Φ(·, 0) = 0,

∀a > 0

(5.4)

5.2. Equations We seek the following formal asymptotic expansion of (Φ, Ψ, s) as a  0: Φ(x, a) 

∞ 

an+1 Φn (x) = aΦ0 (x) + a2 Φ1 (x) + · · · ,

(5.5)

an Ψn (x) = Ψ0 (x) + aΨ1 (x) + · · · ,

(5.6)

an sn = s0 + as1 + a2 s2 + · · · .

(5.7)

n=0

Ψ(x, a) 

∞  n=0

s(a) 

∞  n=0

In this subsection, we shall derive the equations for (Φn , Ψn , sn ) of n  0 by the following three steps. 1. From (5.3), we have 0 = LΨ − f 

∞ 

an (Fn − fn ),

n=0

where, for n  0, Fn = −Ψn − νΨn + (n + 1)RΨn fn = ksk Ψn−k − 2

n−1 

Φk Ψn−1−k .

k=0

Therefore, for n  0, we have Ln Ψn := −Ψn − νΨn + (n + 1)RΨn = R

n 

ksk Ψn−k − 2

k=0

n−1 

Φk Ψn−1−k .

(5.8)

k=0

Combined with the boundary condition Ψx (0, a) = Ψ(0, a) + 1, we have Ψ0 (0) − Ψ0 (0) = 1,

Ψn (0) − Ψn (0) = 0

∀ n  1.

(5.9)

Also since Ψ(−∞) = 0, then Ψn (−∞) = 0

∀ n  0.

(5.10)

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2. From the boundary condition s(a) = ln Ψx (0, a), we have s0 = ln Ψ0 (0),

sn =

Ψn (0) + hn−1 Ψ0 (0)

∀ n  1,

(5.11)

where hn−1 is a function of Ψ0 (0), · · · , Ψn−1 (0). 3. From (5.4) we have that 0 = Φa − s Φx − Ψ ∞ n      an (n + 1)Φn − ksk Φn−k − Ψn . n=0

k=0

Therefore, the equation for Φn is (n + 1)Φn =

n 

ksk Φn−k + Ψn

for n  0.

(5.12)

k=0

5.3. The leading order Using the scheme described in the previous subsection, we get that the leading order expansion (Φ0 , Ψ0 , s0 ) satisfies the following problem ⎧   ⎪ in (−∞, 0), ⎪ ⎨L0 Ψ0 = −Ψ0 − νΨ0 + RΨ0 = 0 Ψ0 (0) − Ψ0 (0) = 1, Ψ0 (−∞) = 0, ⎪ ⎪ ⎩s = ln Ψ (0), Φ = Ψ . 0 0 0 0

(5.13)

Lemma 5.1. Suppose ν < R − 1. Then problem (5.13) admits a unique solution (Ψ0 , Φ0 , s0 ) which is given by Ψ0 (x) = Φ0 (x) =

eλx λ−1

∀ x  0,

s0 = ln

λ . λ−1

Proof. First, the general solution to this second order homogeneous ode of Ψ0 is given by Ψ0 (x) = C1 eλx + C2 eλ− x , where λ and λ− are the positive and negative roots of the equation λ2 + νλ = R respectively. Since Ψ0 (−∞) = 0, we have C2 = 0. Also note that λ = 1, thus C1 can be uniquely determined by the boundary condition Ψ0 (0) − Ψ0 (0) = 1. This completes the proof. 2 5.4. High order For the high order expansion (Φn , Ψn , sn ) (n  1), we have ⎧ ⎪ in (−∞, 0), Ln Ψn = nRsn Ψ0 + fn−1 ⎪ ⎪ ⎪ ⎪ ⎨Ψ (0) − Ψn (0) = 0, Ψn (−∞) = 0, n 

Ψ (0) ⎪ sn = Ψn (0) + hn−1 , ⎪ ⎪ 0

⎪ ⎪ ⎩Φn = 1 n ksk Φ n−k + Ψn . k=0 n+1

Here fn−1 is a function of s0 , · · · , sn−1 , Ψ0 , · · · , Ψn−1 , Φ0 , · · · , Φn−1 .

(5.14)

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To solve this system, we consider the function ˆ n (x) = Φ (x) + Φ 0

λ eλ n x , 1 − λn

where λn is the positive root of equation x2 + νx = (n + 1)R. ˆ n = nRΦ , Φ0 = Ψ0 , Φ ˆ  (0) − Φ ˆ n (−∞) = 0. Set Ψ ˜ n = Ψn − sn Φ ˆ n , then we ˆ n (0) = 0 and Φ Note that Ln Φ n 0 have  ˜ n = fn−1 in (−∞, 0), Ln Ψ (5.15) ˜  (0) − Ψ ˜ n (0) = 0, Ψ ˜ n (−∞) = 0. Ψ n

Lemma 5.2. For n  1, problem (5.15) admits a unique solution. Proof. Note that fn−1 = O(1)eλx for n  1 and λ > 1. Then one can check that ±M eλx are supersubsolution provided M > 0 is sufficient large. Thus the existence can be derived from the general theory of super-subsolution. For the uniqueness, it follows by the fact that the second order homogeneous ode with zero boundary conditions admits a unique solution 0. Then the lemma follows by an induction argument. 2 ˜ n is solved, sn can be derived by Once the unique solution Ψ sn =

˜  (0) + Ψ (0)hn−1 Ψ n 0 , ˆ  (0) Ψ (0) − Φ n

0

since ˆ  (0) = Ψ (0) − Ψ (0) − cn λn = Ψ0 (0) − Φ n 0 0

λ > 0. λn − 1

Therefore, we have the following lemma. Lemma 5.3. Assume that ν < R − 1. Then there exists a unique solution (Φn , Ψn , sn ) of the high order equation (5.14). To sum up, we obtain the following theorem: Lemma 5.4. Assume that ν < R − 1, (ϕ, ψ, s) is the unique solution of (1.4), (5.1) and (1.2), and (Φ, Ψ) is defined as (5.2). Then there exists a unique formal asymptotic expansion (5.5), (5.6) and (5.7) of (Φ, Ψ, s) as a  0. Proof. The existence and uniqueness of the formal asymptotic expansion follows by Lemma 5.1, 5.3 and the definition. 2 6. Formal asymptotic expansion in the case ν = R − 1 In this section, we assume that ν = R − 1, and (ϕ, s) is the solution of (1.1), (1.2). 6.1. Transformation We make the following transformation:   ϕ(z, a) = a ez − Φ(z − s(a), a) ,

z = x + s(a),

s(a) = ξ(a) −

1 ln a. 2

(6.1)

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Then (1.4) can be written as F1 [Φ, ξ] = 0 for x < 0, a > 0,

F2 [Φ, ξ] = 0 for x > 0, a > 0,

where F1 [Φ, ξ] := RaΦa − Φxx −

R 2

√ − 1 Φx + RΦ − Raξ  Φx − aΦ2x − e2ξ+2x + 2 aeξ+x Φx ,

1 F2 [Φ, ξ] := aΦa + Φx + Φ − aξ  Φx − 1. 2 6.2. Equations Assume Φ(x, a) and ξ(a) have the following formal expansions as a  0: Φ(x, a) 

∞ 

n

1

a 2 Φn (x) = Φ0 (x) + a 2 Φ1 (x) + · · · ,

(6.2)

n=0

ξ(a) 

∞ 

n

1

a 2 ξn = ξ0 + a 2 ξ1 + · · · .

(6.3)

n=0

Then in (−∞, 0), 0 = F1 [Φ, ξ] 

∞ n=0

n

a 2 Fn , where

R − 1)Φ0 + RΦ0 − e2ξ0 +2x , 2 n+2 R n − Fn = RΦn − Φn − ( − 1)Φn − Rξn Φ0 − 2ξn e2ξ0 +2x + fn−1 2 2 2 F0 = −Φ0 − (

∀ n  1.

− Here fn−1 is the function of the lower order items, i.e., − − fn−1 = fn−1 (Φ1 , Φ2 , · · · , Φn−1 , ξ1 , ξ2 , · · · , ξn−1 ).

In (0, ∞), 0 = F2 [Φ, ξ] 

∞ n=0

n a 2 F˜n , where

1 F˜0 = Φ0 + Φ0 − 1 2 n +2 1 n + Φn + Φn − ξn Φ0 + fn−1 F˜n = 2 2 2 + Here fn−1 is the function of the lower order items. Hence the leading order equation is ⎧ R   2ξ0 +2x ⎪ =0 ⎪ ⎨−Φ0 − ( 2 − 1)Φ0 + RΦ0 − e 1  2 Φ0 + Φ0 − 1 = 0 ⎪ ⎪ ⎩Φ (−∞) = 0, Φ ∈ C 2 (R). 0

∀ n  1.

in (−∞, 0), in (0, ∞),

(6.4)

0

Also the high order equation for n  1 is ⎧ − n+2 R n    2ξ0 +2x ⎪ + fn−1 =0 ⎪ 2 RΦn − Φn − ( 2 − 1)Φn − 2 Rξn Φ0 − 2ξn e ⎪ ⎪ ⎨ + n+2 1  n  2 Φn + 2 Φn − 2 ξn Φ0 + fn−1 = 0 ⎪ ⎪ ⎪ ⎪ ⎩ Φn (−∞) = 0, Φn ∈ C 2 (R).

in (−∞, 0), in (0, ∞),

(6.5)

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Remark 6.1. (1) The condition Φn (−∞) = 0 for n  0 comes from ϕ(−∞) = 0. (2) The C 2 regularity of Φn for n  0 is critical for searching the unique solution. 6.3. Leading order It is easy to see that problem (6.4) admits a unique solution given by

Φ0 (x) =

⎧ ⎨1 − ⎩

λ −2x 2(2+λ) e

− 12 e2x+2ξ0 +

if x  0

4 λx 4−λ2 e

 ,

ξ0 = ln

if x < 0

λ , 2−λ

where λ is the positive root of equation x2 + ( R2 − 1)x = R. 6.4. An auxiliary problem In this subsection, we consider the following problem which is important for searching high order expansion. Lemma 6.1. Assume p, q > 0, r > 0 are constants, and f0  0, f1 (x) = O(1)eλx for x < 0, with 0 < λ <  p+

p2 +4q , 2

and f2 ∈ C((0, ∞)). Then the following problem

⎧   ⎪ ⎪ ⎨−ζ + (p + f0 )ζ + qζ = f1 

ζ + rζ = f2 ⎪ ⎪ ⎩ζ(−∞) = 0,

in (−∞, 0), in (0, ∞),

ζ ∈ C (R), 1

admits a unique solution. Proof. Uniqueness. Suppose there are two solutions ζ1 , ζ2 . Let w = ζ1 − ζ2 . Then w satisfies the following problem ⎧   ⎪ ⎪ ⎨−w + (p + f0 )w + qw = 0 in (−∞, 0), w + rw = 0 ⎪ ⎪ ⎩w(−∞) = 0,

in (0, ∞),

w ∈ C (R). 1

If w = 0, then w(0) = 0. Otherwise, w(x) = 0 for x  0 since the initial value problem admits a unique solution. On the other hand, the maximum principle implies w = 0 in (−∞, 0). Therefore w = 0, which is a contradiction. Without loss of generality, assume w(0) > 0. Then, w (0−) > 0 follows by the Hopf lemma. Also note that w (0+) = −rw(0+) < 0. Hence this contradicts with w ∈ C 1 (R). Therefore, w = 0. Existence. We construct the solution in the following form ˆ + φ(x), ζ(x) = cζ(x) where c is a constant to be determined later, ζˆ and φ are defined in the following ways. Let ζˆ be the solution of the problem: ⎧ ˆ ˆ ˆ ⎪ ⎪ ⎨−ζ + (p + f0 )ζ + q ζ = 0 in (−∞, 0), ˆ ˆ = 1, ζ(−∞) = 0, ζ(0) ⎪ ⎪ ⎩ζ(x) ˆ = e−rx for x > 0.

C. Qin et al. / J. Math. Anal. Appl. 472 (2019) 1253–1276 

p+

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p2 +4q

Note that f0  0 and 0 < λ < , then one can check that ±M eλx are super-subsolutions of the 2 boundary value problem in (−∞, 0) provided that M is large enough. Thus the existence follows by the general super-subsolution theory. Also, the uniqueness comes from the maximum principle. In addition, ˆ JζK(0) = 0, Jζˆ K(0) = −r − ζˆ (0−) < 0 because of the Hopf lemma, where Jf K(x) := f (x+) − f (x−) is the jump of a function f at the point x. On the other hand, define φ as the solution of the following problem ⎧   ⎪ ⎪ ⎨−φ + (p + f0 )φ + qφ = f1 

φ + rφ = f2 ⎪ ⎪ ⎩φ(−∞) = 0,

in (−∞, 0) in (0, ∞),

φ(0) = 0.

Note that in (0, ∞) the  initial value problem admits a unique solution. In (−∞, 0), since f0  0, f1 (x) = p+

p2 +4q

O(1)eλx and 0 < λ < , then the same argument as before shows that the boundary value problem 2 admits a unique solution. ˆ Therefore there exist unique ζˆ and φ. Finally, set c = −Jφ K(0)/Jζˆ K(0), we have ζ(x) = cζ(x) + φ(x) ∈ 1 C (R). This completes the proof. 2 6.5. High order For n  1, set ˆ n + ξn Φ , Φn = Φ 0 ˆ n (x) satisfies the following problem. where ξn is a constant to be determined later, and Φ ⎧ − n+2 ˆ R ˆ  ˆ ⎪ ⎪ 2 RΦn − Φn − ( 2 − 1)Φn + fn−1 = 0 in (−∞, 0), ⎪ ⎪ ⎨ + n+2 ˆ 1 ˆ in (0, ∞), 2 Φn + 2 Φn + fn−1 = 0 ⎪ ⎪ ⎪ ⎪ˆ ⎩ ˆ n ∈ C 1 (R). Φn (−∞) = 0, Φ − Note that fn−1 = O(1)eλx for x < 0 (by induction argument), then Lemma 6.1 implies there exists a ˆ ˆ  K(0) = 0, JΦ K(0) = 0 and JΦ K(0) = 4λ > 0. By taking unique solution Φn . Also, note that JΦ n 0 0

ξn = −

ˆ  K(0) JΦ n ˆ  K(0), = −4λJΦ n JΦ 0 K(0)

we have Φn is the unique solution of (6.5). To sum up, we have the following theorem: Lemma 6.2. Assume that ν = R − 1 and (ϕ, s) is the unique solution of (1.1), (1.2). Let (Φ, ξ) be defined as (6.1). Then there exists a unique formal asymptotic expansion (6.2), (6.3) of (Φ, ξ) as a  0. 7. Formal asymptotic expansion in the case R − 1 < ν < R In this section we try to achieve the formal asymptotic expansion of solution (ϕ, s) when 0 < λ := R−ν < 1.

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7.1. Transformation We make the following transformations at the beginning. Set ϕ(z, a) = Φ(z − s(a), a),

z = x + s(a),

s(a) = ξ(a) − ln a.

(7.1)

Then (1.4) can be written as F1 [Φ, ξ] = 0 for x < 0, a > 0,

F2 [Φ, ξ] = 0 for x > 0, a > 0

(7.2)

where F1 [Φ, ξ] = RaΦa − Φxx + λΦx + Φ2x − Raξ  Φx , F2 [Φ, ξ] = aΦa + Φx − aξ  Φx − ex+ξ(a) + a. 7.2. Equations Assume Φ(a, x) and ξ(a) have the following formal expansion: Φ(a, x) 

∞ 

an Φn (x) = Φ0 (x) + aΦ1 (x) + · · · ,

(7.3)

an ξn = ξ0 + aξ1 + · · · .

(7.4)

n=0

ξ(a) 

∞  n=0

Substituting (7.3) and (7.4) into (7.2), we obtain equations of (Φn , ξn ) for n  0. 1. When x < 0, F1 [Φ, ξ] = RaΦa − Φxx + λΦx + Φ2x − Raξ  Φx = 0 

∞ 

an Fn ,

n=0

then F0 = −Φ0 + λΦ0 + Φ0 2 = 0, F1 = RΦ1 − Φ1 + λΦ1 + 2Φ0 Φ1 − Rξ1 Φ0 = 0, − Fn = RnΦn − Φn + λΦn + 2Φ0 Φn − Rnξn Φ0 + fn−1 =0

n  1,

− where fn−1 is a function of lower order items, i.e., − − fn−1 = fn−1 (Φ1 , Φ2 , · · · , Φn−1 , ξ1 , ξ2 , · · · , ξn−1 ).

2. When x > 0, F2 [Φ, ξ] = aΦa + Φx − aξ  Φx − ex+ξ(a) + a = 0 

∞  n=0

Note that

an F˜n .

C. Qin et al. / J. Math. Anal. Appl. 472 (2019) 1253–1276

ex+ξ(a) = ex+ξ0 e

∞

n=1

an ξn

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∞    = ex+ξ0 1 + an (ξn + δn−1 ) n=1

where δn−1 = δn−1 (ξ1 , ξ2 , · · · , ξn−1 ). Then F˜0 = Φ0 − ex+ξ0 = 0, F˜1 = Φ1 + Φ1 − ξ1 Φ0 − ex+ξ0 ξ1 + 1 = 0, + = 0 n  1, F˜n = nΦn + Φn − nξn Φ0 − ex+ξ0 ξn + fn−1 + here as before, fn−1 is a function of lower order items.

Therefore, the leading order equation is ⎧   2 ⎪ ⎪ ⎨−Φ0 + λΦ0 + Φ0 = 0 Φ ⎪ 0

−e =0 ⎪ ⎩Φ (−∞) = 0, 0 x+ξ0

in

(−∞, 0),

in

(0, ∞),

(7.5)

Φ0 ∈ C (R). 2

And the high order equation for n  1 is ⎧ −      ⎪ ⎪ ⎨RnΦn − Φn + λΦn + 2Φ0 Φn − Rnξn Φ0 + fn−1 = 0 in + =0 in nΦn + Φn − (n + 1)ξn Φ0 + fn−1 ⎪ ⎪ ⎩Φ (−∞) = 0, Φ ∈ C 2 (R), n

(−∞, 0), (0, ∞),

(7.6)

n

− + where fn−1 and fn−1 are functions of lower order items.

7.3. Leading order It is easy to see that the leading order equation (7.5) admits a unique solution given by 

x Φ0 (x) =

w(y)dy, −∞

w(x) =

λ(1−λ) e−λx −(1−λ) x

(1 − λ)e

x<0 x0

,

ξ0 = ln(1 − λ).

  2 Remark 7.1. When x  0, Φ0 = Φ0 = Φ0 = O(1)eλx . And JΦ 0 K(0) = 1 − λ − [λ + 2Φ0 (0)]Φ0 (0) = −(1 − λ) .

7.4. High order For n  1, set ˆ n (x) + ξn Φ (x), Φn = Φ 0 ˆ n satisfies where ξn is a constant to be determined later, and Φ ⎧ −  ˆ ˆ  ˆ ˆ ⎪ ⎪ ⎨RnΦn − Φn + λΦn + 2Φ0 Φn + fn−1 = 0 in (−∞, 0), ˆ + f+ = 0 ˆn + Φ in (0, ∞), nΦ n n−1 ⎪ ⎪ ⎩Φ 1 ˆ n (−∞) = 0, ˆ n ∈ C (R). Φ

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C. Qin et al. / J. Math. Anal. Appl. 472 (2019) 1253–1276

− Note that in (−∞, 0), Φ0 > 0, fn−1 = O(1)eλx and ˆ unique solution Φn . Now by taking

ξn = −

√ λ+ λ2 +4nR 2

> λ, then Lemma 6.1 implies there exists a

ˆ  K(0) ˆ  K(0) JΦ JΦ n n = , JΦ (1 − λ)2 0 K(0)

we have Φn is the unique solution of (7.6). To sum up, we have the following theorem: Lemma 7.1. Let 0 < λ := R − ν < 1, (ϕ, s) be the unique solution of (1.4), (1.2). Assume that (Φ, ξ) is defined as (7.1). Then there exists a unique formal asymptotic expansion (7.3) and (7.4) of (Φ, ξ) as a  0. Proof of Theorem 2. It is clear that Theorem 2 follows by Lemma 5.4, 6.2 and 7.1. 2 References [1] J. Carpenter, The exercise and valuation of executive stock options, J. Financ. Econ. 48 (1998) 127–158. [2] J.N. Carpenter, R. Stanton, N. Wallace, Optimal exercise of executive stock options and implications for firm cost, J. Financ. Econ. 98 (2010) 315–337. [3] M. Grasselli, V. Henderson, Risk aversion and block exercise of executive stock options, J. Econom. Dynam. Control 33 (2009) 109–127. [4] B.J. Hall, K.J. Murphy, Stock option for undiversified executives, J. Account. Econ. 33 (2002) 3–42. [5] V. Henderson, D. Hobson, Perpetual American options in incomplete markets: the infinitely divisible case, Quant. Finance 8 (2008) 461–469. [6] J. Ingersoll, The subjective and objective evaluation of incentive stock options, J. Bus. 79 (2006) 453–487. [7] A. Jain, A. Subramanian, The intertemporal exercise and valuation of employee options, Account. Rev. 79 (2004) 705–743. [8] X. Lai, X. Chen, C. Qin, W. Yu, A variational inequality for optimal exercise perpetual executive stock options, European J. Appl. Math. 29 (2018) 55–77. [9] X. Lai, X. Chen, M. Wang, C. Qin, W. Yu, Mathematical analysis of a perpetual executive Stock option, European J. Appl. Math. 26 (2015) 193–213. [10] R. Lambert, D. Larcker, R. Verrecchia, Portfolio considerations in valuing executive compensation, J. Acc. Res. 29 (1991) 129–149. [11] T. Leung, R. Sircar, Accounting for risk aversion, vesting, job termination risk and multiple exercises in valuation of employee stock options, Math. Finance 19 (2009) 99–128. [12] C. Qin, X. Chen, X. Lai, W. Yu, Regularity of free boundary arising from optimal continuous-exercise perpetual executive stock options, Interfaces Free Bound. 17 (2015) 69–92. [13] C. Qin, X. Chen, X. Lai, W. Yu, A continuous-exercise model for American call options with hedging constraints, working paper, available at SSRN: https://doi.org/10.2139/ssrn.2757541. [14] L.C.G. Rogers, J. Scheinkman, Optimal exercise of executive stock options, Finance Stoch. 11 (2007) 357–372. [15] L. Song, W. Yu, A parabolic variational inequality related to the perpetual American executive stock options, Nonlinear Anal. 74 (2011) 6583–6600. [16] L. Song, W. Yu, A free boundary problem coming from the perpetual American call options with utility, European J. Appl. Math. 24 (2013) 231–271.