ANALYSIS OF THE OPTIMAL EXERCISE BOUNDARY OF AMERICAN ...

arXiv:0712.3323v4 [math.PR] 19 Feb 2008

A NALYSIS OF THE OPTIMAL EXERCISE BOUNDARY OF A MERICAN OPTIONS FOR JUMP DIFFUSIONS ∗ Erhan Bayraktar †

Hao Xing ‡

Abstract In this paper we show that the optimal exercise boundary/ free boundary of the American put option pricing problem for jump diffusions is continuously differentiable (except at the maturity). We also discuss the higher order regularity of the free boundary. Key Words. American put option, jump diffusions, smoothness of the early exercise boundary, integro-differential equations, parabolic differential equations.

1 Introduction Let (Ω, F , P) be a complete probability space hosting a Wiener process W = {Wt ; t ≥ 0} and a Poisson random measure N on R+ × R+ with mean measure λθ(dy)dt (in which θ is a probability measure on R+ ) independent of the Wiener process. Let F = {Ft }t∈[0,T ] be the (augmented) natural filtration of W and N. For a given x, we will consider a Markov process S x = {Stx ; t ≥ 0} of the form Z x x x x (y − 1)N(dt, dy), S0 = x, (1.1) dSt = µSt dt + σSt dWt + St− R+

as the stock price process. We will take µ = r − q + λ − λξ, in which Z yθ(dy) < ∞, ξ=

(1.2)

R+

(a standing assumption). We impose this condition on µ so that the discounted stock prices are martingales. The constant r ≥ 0 is the interest rate, q ≥ 0 is the dividend and the constant ∗ This research is partially supported by NSF Research Grant, DMS-0604491. We would like to thank Baojun Bian and Sijue Wu for helpful discussions. † Department of Mathematics, University of Michigan, 530 Church Street, Ann Arbor, MI 48109. ‡ e-mail: {erhan, haoxing}@umich.edu

1

σ > 0 is the volatility. We should note that at the time of a jump the stock price moves from x x St− to St− Y in which Y is a positive random variable whose distribution is given by θ. When Y < 1 the stock price jumps down, when Y > 1 the stock price jumps up. In the classical Merton jump diffusion model Y = eZ , where Z is a Gaussian random variable. In this framework, we will study the American put option pricing problem. The value function of the American put option is defined by V (x, t) :=

sup E{e−rτ (K − Sτx )+ },

(1.3)

τ ∈S0,T −t

in which S0,T −t is the set of stopping times (of the filtration generated by W and N) taking values in [0, T − t]. The value function V solves a free boundary problem (see Theorem 2.1). The main goal of this paper is to analyze the regularity of the boundary. We will show that the boundary is C 1 except at the maturity T , and C ∞ with an appropriate regularity assumption on the jump distribution θ. For notational simplicity we will first change variables and transform the value function V into u and its free boundary s into b (see (2.5)) and state our results in terms of u and b. The regularity of the free boundary even for the American option pricing problem for geometric Brownian motion is difficult to establish (see the discussion on page 172 of Peskir (2005)) and has only recently been fully analyzed by Chen and Chadam (2007). In the jump diffusion case, Pham (1995) showed under a certain assumption on the parameters (see (3.1) in Section 3 of our paper) that the free boundary is continuous. A breakthrough came when Yang et al. (2006) proved the continuity of free boundary without the condition (3.1). However, in their section 5, under almost the same assumption (see (4.1) in Section 4 of our paper), they analyzed the smoothness properties of the free boundary. When condition (4.1) is violated, the free boundary of the American option for jump diffusions exhibits a unique behavior near the maturity: The boundary exhibits a jump at the maturity (see Theorem 5.3 in Yang et al. (2006), equation (3.2) and Corollary 3.1 in our paper). This behavior of the free boundary was also observed by Levendorskiˇi (2004). The purpose of our paper is to to analyze the smoothness of the free boundary curve when (4.1) is not satisfied. We will see that the exercise boundary is smooth everywhere except the maturity. Yang et al. (2006) use the assumption in (4.1) to establish the H¨older continuity of the free boundary. In Theorem 3.1 we manage to drop this condition. Moreover, we generalize the result in Cannon et al. (1974) to parabolic integro-differential equations in Lemma 4.1. Using this result, we upgrade the regularity of the free boundary from H¨older continuity to continuous differentiability in Section 4. In Section 5, we analyze the higher order regularity of the free boundary making use of a technique Schaeffer (1976) used for the free boundary of a one dimensional Stefan problem on a bounded domain. Bayraktar (2007), Pham (1995) and Yang et al. (2006) show the regularity of the value function V (x, t) using different methodologies. Pham (1995) uses the theory of viscosity solutions (which are essentially developed for studying non-linear partial differential equations), 2

Figure 1: Our results and the relationships among them. A → B means that statement A is used in the proof of statement B.

Section 2: Properties of the value functions Lemma 2.1

Lemma 2.2

Theorem 2.1

Remark 2.4

Proposition 2.1

Section 4: Free boundaries are continuously differentiable Lemma 4.1

Lemma 2.3

Lemma 4.2

Theorem 4.1

Proposition 2.2

Lemma 2.1 Lemma 3.2

Lemma 3.3

Theorem 3.1

Lemma 3.1

Corollary 3.1

Section 3: Free boundaries are H¨older continuous

Lemma 2.2

Lemma 5.2

Corollary 5.1

Theorem 5.1

Lemma 5.1

Remark 5.1

Corollary 5.2

Section 5: Higher order regularity of the free boundaries

Yang et al. (2006) uses the penalty method of converting the free boundary problem into a fixed boundary problem. On the other hand Bayraktar (2007) uses the approximation of V (x, t) by value functions of appropriate free boundary problems/optimal stopping problems for geometric Brownian motion. (Let us denote n-th element of the approximating sequence by Vn (x, t). The sequence (Vn )n≥0 converges to V uniformly and exponentially fast.) In this paper we also show that the free boundaries corresponding to Vn (x, t) have similar regularity with the free boundary of V (x, t). The regularity proofs for these two types of free boundaries are carried out in parallel, not in tandem. Oftentimes the regularity proofs concerning the free boundary corresponding to Vn (x, t) will almost be the same as their counterparts for the free boundary of V (x, t), in which case the proofs will be omitted. The rest of the paper is organized as follows: In Section 2, we will collect several useful properties of the function u and the sequence of functions approximating it, which will be crucial in establishing our main results in the next three sections about the regularity of their free boundaries. In Section 3, we will show that the the free boundaries are H¨older continuous. In Section 4, we will prove the continuous differentiability of the free boundaries. Finally, in Section 5, we will upgrade the regularity of the boundary curves and show that they are infinitely differentiable under an appropriate regularity assumption on the jump distribution. Proofs of some auxiliary results are presented in the Appendix. Our main results are Theorems 3.1, 4.1 and 5.1. In Figure 1 we show the logical flow of the paper, i.e. we show how several results proved in the paper are related to each other. 3

2 Properties of the value functions Theorem 2.1. The value function for the American put option, V (S, t) is the unique classical solution of the following free boundary problem: Z ∂V 1 2 2 ∂2V ∂V + σ S + µS − (r + λ)V + λ V (Sy, t)θ(dy) = 0, S > s(t), (2.1) ∂t 2 ∂S 2 ∂S R+ V (s(t), t) = K − s(t), t ∈ [0, T ), (2.2) ∂ V (s(t), t) = −1, (2.3) ∂S V (S, T ) = (K − S)+ , S ≥ s(T ), (2.4) R in which µ , r − q − λ R+ yθ(dy) + λ.

Proof. Lemma 3.5 of Bayraktar (2007) shows that (2.1)-(2.4) has a unique classical solution. On the other hand, Proposition 3.1, Lemma 3.1 and Proposition 3.2 in Pham (1995) shows that V is a solution of the free boundary problem (2.1)-(2.4), from which the result follows.

Remark 2.1. One should note that Pham (1995) needs a condition for the uniqueness of the solution and Bayraktar (2007) needs an extra condition for the verification theorem to work. However combining the results of these two papers we can avoid the need to make these extra assumptions. In the following, let us change the variables and state (2.1)-(2.4) in a more convenient form:      2t 2t , z = log(y), x = log(S), u(x, t) = V S, T − 2 , b(t) = log s T − 2 σ σ (2.5) σ2 and T0 = 2 T . Let Y be a random variable whose distribution is θ We will denote by ν the distribution of log(Y ). In terms of the new variables and the distribution ν, the free boundary problem (2.1) - (2.4) reduces to   Z ∂u ∂ 2 u 2λ 2µ ∂u 2(r + λ) Lu , u(x + z, t)ν(dz) = 0, − 2− −1 + u− 2 ∂t ∂x σ2 ∂x σ2 σ R x > b(t), (2.6) u(b(t), t) = K − eb(t) , t ∈ (0, T0 ], (2.7) ∂ u(b(t), t) = −eb(t) , (2.8) ∂x u(x, 0) = (K − ex )+ , x ≥ b(0). (2.9) From Theorem 2.1, it is clear the (2.6) - (2.9) has a unique classical solution (u(x, t), b(t)) for x ≥ b(t), t ∈ (0, T0 ]. 4

Remark 2.2. The integral term in (2.6) can also be considered as a driving term, then the integro-differential equation (2.6) can be viewed as the following parabolic differential equation with a driving term, even though the driving term contains the unknown function:   2µ ∂u 2(r + λ) ∂u ∂ 2 u − 2− −1 + u = f (x, t), (2.10) 2 ∂t ∂x σ ∂x σ2 R in which f (x, t) = σ2λ2 R u(x + z, t)ν(dz). This point of view will be useful in the proof of some results in later sections. Bayraktar (2007) constructs a monotone increasing sequence {un }n≥0 that converges to the unique solution u(x, t) of the parabolic integro-differential equations (2.6) - (2.9), uniformly. In this sequence, u0 (x, t) = (K − ex )+ , and each un (x, t) (n ≥ 1) is the unique solution of the following parabolic differential equation:   ∂un ∂ 2 un 2µ ∂un 2(r + λ) LD u n , − − −1 + un = fn (x, t), x > bn (t),(2.11) 2 2 ∂t ∂x σ ∂x σ2 un (bn (t), t) = K − ebn (t) , t ∈ (0, T0 ], ∂ un (bn (t), t) = −ebn (t) , ∂x un (x, 0) = (K − ex )+ , x ≥ bn (0),

in which

2λ fn (x, t) , 2 σ

Z

un−1 (x + z, t)ν(dz),

(2.12)

(2.13) (2.14)

(2.15)

R



and the free boundary bn (t) , log sn T − σ2t2 is defined in terms of sn (·), which is the approximating free boundaries in Bayraktar (2007). Let us define the continuation regions C and Cn and the stopping regions D and Dn of (2.6)(2.9) and (2.11)-(2.14), respectively as follows C , {(x, t)| b(t) < x < +∞, 0 < t ≤ T0 }, D , {(x, t)| − ∞ < x ≤ b(t), 0 < t ≤ T0 }, Cn , {(x, t)| bn (t) < x < +∞, 0 < t ≤ T0 }, Dn , {(x, t)| − ∞ < x ≤ bn (t), 0 < t ≤ T0 }, for all n ≥ 1. Since the monotone increasing sequence {un }n≥0 converges to u uniformly, the approximating free boundary {bn }n≥1 is a monotone decreasing sequence. As a result, we have ∪n≥1 Cn = C and ∩n≥1 Dn = D. In this section, we will study the properties of u and un in both the continuation and the stopping regions. From the equations for u and un , we can see that both themselves and their first derivatives with respect to x are continuous across the free boundaries b(t) and bn (t) respectively. The following theorem shows that their first derivative with respect to t are also continuously across the free boundaries. 5

Proposition 2.1. For any t ∈ (0, T0 ], ∂ u(x, t) = 0. x↓b(t) ∂t lim

(2.16)

Furthermore, if we assume that ∂t fn (x, t) is bounded in R × [ǫ, T0 ],

for any ǫ > 0,

(C)

then ∂ un (x, t) = 0, x↓bn (t) ∂t lim

n ≥ 1.

(2.17)

Proof. See Appendix. Remark 2.3. Later, we will show that that for each n ≥ 1, fn (x, t) satisfies the condition (C) (see Remark 2.4). In the following, we will study ∂t u and ∂t un in the continuation regions C and Cn . It was proven in Bayraktar (2007) that both u(·, t) and un (·, t) are nondecreasing functions. However, one can show both u(·, t) and un (·, t) are actually strictly increasing functions in the continuation regions. Proposition 2.2. Let u(x, t) and un (x, t), n ≥ 1, be solutions of free boundary problems (2.6) - (2.9) and (2.11) - (2.14) respectively, then ∂u (x, t) > 0, (x, t) ∈ C, ∂t ∂un (x, t) > 0, (x, t) ∈ Cn . ∂t

(2.18) (2.19)

Proof. The inequality (2.18) is proved in Proposition 4.1 in Yang et al. (2006) using the Maximum Principle for the integro-differential equations, which can be found in Theorem 2.7 in Chapter 2 of Garroni and Menaldi (1992). However, both of these inequalities can be simultaneously proved using the ordinary Maximum Principle for parabolic differential equations. We n know that w = ∂u and wn = ∂u satisfy the following equations in C and Cn respectively, ∂t ∂t Z 2λ LD w = w(x + z, t)ν(dz), (2.20) σ2 R Z ∂un−1 2λ (x + z, t)ν(dz). (2.21) LD wn = 2 σ R ∂t We will prove the inequality (2.18) in the following, and the proof of (2.19) is similar, using ∂t un−1 ≥ 0. 6

Since w = ∂t u ≥ 0 in R × (0, T0 ), (2.20) implies that LD w ≥ 0. If there is a point (x0 , t0 ) ∈ C such that w(x0 , t0 ) = 0 (i.e. w achieves its non-positive minimum at (x0 , t0 ) ), it follows from the strong Maximum Principle that w(x, t) = 0 in C ∩ R × {0 < t ≤ t0 }. Together with the fact that w(x, t) = 0 in D, we have that w(x, t) = 0 in R × {0 < t ≤ t0 }. As a result, from Z t0 u(x0 , t0 ) − u(x0 , 0) = w(x0 , s)ds = 0, 0

x0 +

we obtain u(x0 , t0 ) = (K − e ) . This contradicts with the definition of the free boundary b(t), because b(t0 ) = max{x ∈ R : u(x, t0 ) = (K − ex )+ } and x0 > b(t0 ).

In order to investigate the regularity of free boundaries in later sections, we need some regularity results for u and un . It has been proven in Bayraktar (2007) that V (S, ·) is uniformly Lipschitz in R+ and V (·, t) is uniformly semi-H¨older continuous in [0, T ]. The following lemma shows the same properties also holds for u(x, t) and un (x, t), the functions that we obtained after the change of variables in (2.5). (The Lipschitz continuity with respect to x is not a priori clear and one needs to check whether ∂x u(x, t) is bounded.) Lemma 2.1. Let u(x, t) be the solution of equation (2.6) - (2.9), then we have |u(x, t) − u(y, t)| ≤ C|x − y|,

1 2

|u(x, t) − u(x, s)| ≤ D|t − s| ,

x, y ∈ R, t ∈ (0, T0 ],

x ∈ R, 0 ≤ t, s ≤ T0 ,

(2.22) (2.23)

where C and D are positive constants that do not depend on x and t. The same statement also holds for un (x, t), n ≥ 1. Proof. The inequality (2.23) is clear, because we have   12 1 2t 2 2s x x ˜ |u(x, t) − u(x, s)| = V (e , T − 2 ) − V (e , T − 2 ) ≤ D |t − s| 2 . 2 σ σ σ

In order to prove (2.22), it suffices to check that ∂x u(x, t) is uniformly bounded in the domain R×(0, T0 ]. Choose a constant X > log K +1, we will first prove ∂x u(x, t) is uniformly bounded in [X, +∞) × (0, T0 ]. Let us consider a cut-off function η(x) ∈ C ∞ (R), such that η(x) = 0 when x ≤ X − 1 and η(x) = 1 when x ≥ X. Using (2.10) we see that v(x, t) = η(x)u(x, t) satisfies LD v = η(x)f (x, t) + f˜(x, t), v(x, 0) = η(x)(K − ex )+ , where 2λ f (x, t) = 2 σ

Z

u(x + z, t)ν(dz), R

    2µ ′′ ′ ∂u ˜ − f (x, t) = − η u + 2η − 1 η ′ u. (2.24) ∂x σ2 7

It is worth noticing that the term η ′ ∂x u in the expression for f˜ vanishes outside a compact domain. Since we also have that u(x, t) ≤ K, both f (x, t) and f˜(x, t) are bounded in R×(0, T0 ]. Let G(x, t; y, s) be the Green function corresponding to the differential operator LD . We can represent v(x, t) in terms of G as v(x, t) =

Z

y +

R

dy G(x, t; y, 0)η(y)(K −e ) +

Z

t

ds

0

Z

R

  dy G(x, t; y, s) f (y, s)η(y) + f˜(y, s) .

(2.25) The first term on the right-hand-side of (2.25) will vanish by the choice of η(y). On the other hand, Green function G(x, t; y, s) satisfies   |x − y|2 −1 , |∂x G(x, t; y, s)| ≤ c(t − s) exp −c t−s for some positive constant c, (see Theorem 16.3 in page 413 of Ladyzhenskaja et al. (1968)). R 2 1 Since R dy exp(−c (x−y) ) ≤ d (t − s)− 2 for some other positive constant d, we have that t−s Z

t

ds 0

Z

R

dy |∂x G(x, t; y, s)| ≤

Z

t

0

1

1

c t2 , ds c˜(t − s)− 2 = 2˜

˜ the Dominated Convergence Theorem implies Using this estimate and the boundness of f and f, that Z t Z ∂x v(x, t) = ds dy ∂x G(x, t; y, s)(f (y, s)η(y) + f˜(y, s)), 0

R

which is uniformly bounded. On the other hand, ∂x v = η ′ u + η∂x u. By our choice of η(x), we have that ∂x u(x, t) is uniformly bounded on [X, +∞) × (0, T0 ].

Moreover, in the stopping region D, we have ∂x u(x, t) = −ex . This implies that 0 > ∂x u(x, t) ≥ −eb(t) ≥ −K. On the other hand, since it is continuous ∂x u is also bounded in the compact closed domain {(x, t)|b(t) ≤ x ≤ X, 0 ≤ t ≤ T0 }. As a result we have that ∂x u(x, t) is uniformly bounded in R × (0, T0 ]. In the rest of this section, we will investigate the boundness of ∂t u(x, t) and its behavior when x → +∞. These two properties will be useful to show that condition (C) holds as well as other several results in Sections 4 and 5 (see e.g. (4.2), proof of Lemma 4.1 and Remark 5.1). Let us first recall the definition of the H¨older spaces on page 7 of Ladyzhenskaja et al. (1968). Definition 2.1. Let Ω be
a domain in R, QT = Ω × (0, T ). For any positive nonintegral real α,α/2 QT is the Banach space of functions v(x, t) that are continuous in QT , number α, H together with continuous derivatives of the form ∂tr ∂xs v for 2r + s < α, and have a finite norm X (α/2) ||v||(α) = |v|(α) + ||∂tr ∂xs v||(0) , x + |v|t 2r+s≤[α]

8

in which ||v||(0) = maxQT |v|, X |v|(α) = < ∂tr ∂xs v >(α−[α]) , x x 2r+s=[α]

< v >βx =

sup (x, t), (x′ , t) ∈ QT |x − x′ | ≤ ρ0

< v >βt =

sup (x, t), (x, t′ ) ∈ QT |t − t′ | ≤ ρ0

(α/2)

|v|t

=

X

( α−2r−s ) 2

< ∂tr ∂xs v >t

;

α−2 log K. We will first prove that ∂t u(x, t) is uniformly bounded in the domain [X0 , +∞) × [0, T0 ]. Let k(x, t) ∈ C0∞ (R × [0, T0 ]) be such that ∂x k(x, t)|x=X0 = ∂x u(x, t)|x=X0 ,

t ∈ [0, T0 ],

and that k(x, 0) = 0, x ∈ R. These two conditions on k are consistent since ∂x u(x, 0)|x=X0 = 0. The function v(x, t) , u(x, t) − k(x, t) satisfies ∂x v(x, t)|x=X0 = 0,

(2.26)

and LD v(x, t) = f (x, t) + g(x, t), 9

x > b(t), t ∈ (0, T0 ],

(2.27)

in which g(x, t) = −LD k(x, t) and f is given by (2.24). Let us define the even extension of v(x, t) with respect to line x = X0 as vˆ(x, t) ,



v(x, t) x ≥ X0 , v(2X0 − x, t) x < X0 .

ˆ t) and gˆ(x). From (2.26), we have vˆ(x, t) ∈ C 2,1 (R × (0, T0 ]) and that We similarly define f(x, it satisfies the equation LD vˆ = fˆ(x, t) + gˆ(x, t), vˆ(x, 0) = 0, x ∈ R.

(x, t) ∈ R × (0, T0 ],

Here the initial condition follows from (2.9) and the choice of X0 and k(x, t). It follows from (2.22) and (2.23) that f (x, t) is uniformly Lipschitz in x and semi-H¨older coninuous in t. So for any x1 < x2 , if we have either x2 ≤ X0 or X0 ≤ x1 , then 2λ fˆ(x1 , t) − fˆ(x2 , t) ≤ 2 C(x2 − x1 ), σ for the same constant C as in (2.22). On the other hand, if x1 < X0 < x2 , then ˆ 1 , t) − fˆ(x2 , t)| ≤ |f(x ˆ 1 , t) − fˆ(X0 , t)| + |f(X ˆ 0 , t) − fˆ(x2 , t)| |f(x 2λ 2λ 2λ C(X − x ) + C(x − X ) = C(x2 − x1 ). ≤ 0 1 2 0 σ2 σ2 σ2 As a result of the last two equations we observe that fˆ(x, t) is uniformly Lipschitz in its first variable. It is also clear that fˆ(x, t) is semi-H¨older continuous in its second variable. Thus, it follows from Definition 2.1 that α fˆ(x, t) ∈ H α, 2 (R × [0, T0 ]),

for some 0 < α < 1.

On the other hand, gˆ(x, t) ∈ H α,α/2 (R × [0, T0 ]), because k(x, t) ∈ C0∞ (R × [0, T0 ]). Then the regularity property of parabolic differential equation (see Theorem 5.1 in page 320 of Ladyzhenskaja et al. (1968)) implies that α

vˆ(x, t) ∈ H 2+α,1+ 2 (R × [0, T0 ]). In particular, u(x, t) ∈ H 2+α,1+α/2 ([X0 , +∞) × [0, T0 ]). As a result, in [X0 , +∞) × [0, T0 ], ∂t u(x, t) is uniformly bounded by the H¨older norm of u(x, t) . Now, the result follows from Proposition 2.1 and continuity of ∂t u(x, t) inside domain {(x, t)|b(t) ≤ x ≤ X0 , ǫ ≤ t ≤ T0 } for any ǫ > 0. The proof of the statement for ∂t un is similar. 10

Remark 2.4. (i) As a result of Lemma 2.2 and the fact that ∂t f1 (x, t) = 0, condition (C) in Proposition 2.1 is always satisfied. (ii) In the statement of Lemma 2.2, t = 0 cannot be included,
2µ i.e.Rlimt→0 ∂t u(x, t) is not uniformly bounded in x ∈ R: ∂t u = ∂x2 u + σ2 − 1 ∂x u − 2(r+λ) u+ σ2 2λ 2 u(x, t) is not bounded as a result of non-smoothness of u(x + z, t)ν(dz), and lim ∂ t→0 x σ2 R the initial value at x = log K. In the following, we will use the previous lemma to analyze the behavior of ∂t u(x, t) as x → +∞. Lemma 2.3. lim ∂t u(x, t) = 0,

x→+∞

t ∈ (0, T0 ];

lim ∂t un (x, t) = 0,

x→+∞

t ∈ [0, T0 ], n ≥ 1.

Proof. Let X0 > log K be the same as in the proof of Lemma 2.2, choose a cut-off function η(x) ∈ C ∞ (R), such that η(x) = 1 when x ≥ 2X0 and η(x) = 0 when x ≤ X0 . Then the function w(x, t) = η(x)∂t u(x, t) satisfies the following Cauchy problem ˜ t), LD w = η(x)h(x, t) + h(x,

(x, t) ∈ R × [t0 , T0 ],

where 2λ h(x, t) = 2 σ

Z

∂t u(x + z, t)ν(dz), R

˜ t) = − (2η ′ ∂x ∂t u + η ′′ ∂t u) − h(x,



 2µ − 1 η ′ ∂t u, σ2

and we choose w(x, t0 ), for some t0 ∈ (0, T0 ), as the initial condition. It follows from Theorem 3.1 in page 346 of Garroni and Menaldi (1992) that w is a classical solution of this Cauchy problem. Let G(x, t; y, s) be the Green function corresponding to the differential operator LD . The solution w(x, t) of the Cauchy problem can be represented as Z Z t Z ˜ s)), (2.28) w(x, t) = dy G(x, t; y, t0 )w(y, t0)+ ds dy G(x, t; y, s)(η(y)h(y, s)+ h(y, R

t0

R

for all (x, t) ∈ R × (t0 , T0 ]. Since the Green function satisfies   c(x − y)2 − 21 |G(x, t; y, s)| ≤ C(t − s) exp − , t−s

(y, s) ∈ R × [0, t).

the first term in (2.28) is bounded, as long as w(y, t0) is uniformly bounded. ˜ to w is given by, The contribution of η ′ ∂x ∂t u (in the expression for h) Z Z ∂2 ∂ ∂ ′ − dyG(x, t; y, s)η (y) u(y, s) = dy [G(x, t; y, s)η ′(y)] u(y, s). ∂y∂s ∂y ∂s R R 11

Now it follows from Lemma 2.2 that both w(x, t0 ) and h(x, t) are uniformly bounded for x ∈ R, t ∈ [t0 , T0 ]. We also have that η ′ and η ′′ vanish outside [X0 , 2X0 ]. Since limx→+∞ G(x, t; y, s) = 0 and it can easily be shown that limx→+∞ ∂y G(s, t; y, s) = 0, the Dominated Convergence Theorem implies that lim w(x, t) = 0, t ∈ (t0 , T0 ]. x→+∞

for arbitrary t0 . As a result we have that limx→+∞ ∂t u(x, t) = 0, t ∈ (0, T0 ]. The proof of the statement for ∂t un (x, t) can be similarly performed using Remark 2.4. In the following three sections, we will use the properties of the value function we have shown in this section to investigate the regularity of free boundaries b(t) and bn (t), n ≥ 1.

3 Free boundaries are H¨older continuous The continuity of the free boundaries for differential equations with or without integral terms have been studied intensively, see e.g. Friedman (1975), Pham (1995), Yang et al. (2006). For the American option in jump diffusions, Pham (1995) showed the continuity of the free boundary under the technical condition Z +∞ r >q+λ (y − 1)θ(dy). (3.1) 1

In Yang et al. (2006), this condition is removed in the proof of continuity. Moreover, in their Theorem 5.3, they showed that ( R log K, r ≥ q + λ R+ (ez − 1)ν(dz) (I) R b(0+) , lim+ b(t) = min{log K, log S0 } = , log S0 , r < q + λ R+ (ez − 1)ν(dz) (II) t→0 (3.2) in which S0 is the unique solution of the following integral equation Z   f (S) , qS − rK + λ (K − Sez )+ − (K − Sez ) ν(dz) = 0. (3.3) R

Note that f (S) is strictly increasing.

For the approximating free boundaries bn (t), n ≥ 1 the same result still holds. Corollary 3.1. For the approximating sequence bn (t), we also have that ( R log K, r ≥ q + λ R+ (ez − 1)ν(dz) (I) R bn (0+) , lim+ bn (t) = min{log K, log S0 } = . log S0 , r < q + λ R+ (ez − 1)ν(dz) (II) t→0 (3.4) 12

In order to prove this corollary, we need the following lemma. Lemma 3.1. Let g(x) = (K − ex )+ . When bn (t) ≤ x < bn (0+) , limt→0+ bn (t) and t ∈ (0, T0 ], we have Z 2λ un−1(x + z, t)ν(dz) > 0. (3.5) LD g(x) − 2 σ R Proof. Our proof is motivated by that of Theorem 3.2 in Friedman and Shen (2002). Let un (x, t) be the solution to the equation (2.11) - (2.14). We will first assume that the inequality (3.5) is not satisfied and derive a contradiction. Let us consider the domain B = {(x, s) ∈ R × R+ : t − ǫ < s < t, bn (t) < x < bn (t − ǫ)} for small enough ǫ > 0 such that Z 2λ un−1 (x + z, t) − LD g(x) ≥ 0, (x, t) ∈ B. LD (un − g) = 2 σ R On the other hand, for (x, t) ∈ B, we have that un (x, t) > g(x) when x > bn (t) and that un (bn (t), t) = g(bn (t)). Now Hopf’s Lemma (see Theorem 2.8 in page 78 of Garroni and Menaldi (1992)) implies that ∂x (un − g)(bn (t), t) > 0 which contradicts (2.13). Proof of Corollary 3.1. For x < log K, we have that   Z   2(r − q) 2r 2λ x x x x+z + x+z Lg(x) = e + − 1 e + (K − e ) − (K − e ) − (K − e ) ν(dz) σ2 σ2 σ2 R   Z   2 x (K − ex+z )+ − (K − ex+z ) ν(dz) = − 2 qe − rK + λ σ R 2 (3.6) = − 2 f (ex ). σ Noticing that un−1 (x + z, t) ≥ g(x + z) ≥ 0 we can write Z 2λ 2 LD g(x) − 2 un−1(x + z, t) ≤ Lg(x) = − 2 f (ex ). σ R σ

(3.7)

Lemma 3.1 implies that f (ex ) < 0 when bn (t) ≤ x < bn (0+). Thus ex < S0 because the function f is strictly increasing. As a result, we have bn (t) < min{log K, log S0 } for any t > 0, which implies bn (0+) ≤ min{log K, log S0 }. (3.8) Now, the corollary results from combining (3.2) and (3.8), since {bn }n≥1 is an monotone decreasing sequence. In the following, the function f (S) in (3.3) and the Maximum Principle will play a crucial role in showing that both b(t) and bn (t) are H¨older continuous. First, using Proposition 2.1, Proposition 2.2 and Hopf’s Lemma (see Theorem 2.8 in page 78 of Garroni and Menaldi (1992)), we can derive the following two auxiliary tools. 13

Lemma 3.2. Let b(t) and bn (t) be the free boundaries in problems (2.6) - (2.9) and (2.11) (2.14) respectively, then b(t) and bn (t) (n ≥ 1) are strictly decreasing for t ∈ (0, T0 ]. Proof. The proof for b(t) is given in Theorem 5.4 in Yang et al. (2006), and the proof for bn (t) is similar. Lemma 3.3. Let b(t) be the free boundary in Lemma 3.2. For any ǫ > 0, if there exists δ > 0 such that for any t1 and t2 satisfying ǫ ≤ t1 < t2 ≤ T0 and t2 − t1 ≤ δ u(b(t1 ), t) − u(b(t1 ), t1 ) ≤ Cǫ (t2 − t1 )α ,

t1 ≤ t ≤ t2 ,

(3.9)

in which 0 < α ≤ 1 and Cǫ is a constant that does not depend on t1 and t2 , then there exists δ ′ ∈ (0, δ] such that α

b(t1 ) − b(t2 ) ≤ Cǫ′ (t2 − t1 ) 2 ,

0 ≤ t2 − t1 ≤ δ ′ ,

(3.10)

in which Cǫ′ is another positive constant that is independent of t1 and t2 . The same result holds for bn (t), n ≥ 1. Proof. We will first prove the statement for b(t). For any t1 and t2 such that ǫ ≤ t1 < t2 ≤ T0 and t2 − t1 ≤ δ, let us consider the domain D , {(x, t) : b(t) < x < b(t1 ), t1 < t < t2 }. (In what follows, we will choose t1 and t2 close to each other, i.e. we will find an appropriate δ ′ such that t2 − t1 ≤ δ ′ .) Let D be the closure of the domain D. In the following, we will show that the function hp i+ 2 α χ(x) = Cǫ (t2 − t1 ) 2 + β(x − b(t1 )) ,

b(t2 ) ≤ x ≤ b(t1 )

(3.11)

satisfies χ(x) ≥ (u − g)(x, t) on the domain D for suitably chosen positive constant β. √

α

It is clear that χ(x) = 0, when x ≤ b(t1 ) − βCǫ (t2 − t1 ) 2 , ξ. We also have χ(b(t1 )) = Cǫ (t2 − t1 )α ≥ u(b(t1 ), t) − g(b(t1 )) for t1 ≤ t ≤ t2 because of assumption (3.9). On the other hand, χ(b(t)) ≥ 0 = u(b(t), t) − g(b(t)). Therefore on the parabolic boundary of the domain D, we have that χ ≥ u − g. We will show that this holds for all (x, t) ∈ D. To this end, we will compare Lχ with L(u − g) using the Maximum Principle. Note that ξ is carefully chosen so that it has a continuous first derivative and a bounded second derivative. These properties of ξ makes the application of the Maximum Principle for weak solutions (see e.g. Corollary 7.4 in Lieberman (1996)) possible. 14

First, for (x, t) ∈ D let us estimate the integral term: Z Z np o2 2λ 2λ α 2 + β(x + z − b(t )) χ(x + z)ν(dz) = C (t − t ) ν(dz) ǫ 2 1 1 σ2 R σ 2 z≥ξ−x Z np o2 α 2λ 2 + βz C (t − t ) ν(dz) ≤ ǫ 2 1 σ 2 z≥ξ−x Z   4λ α 2 2 ≤ C (t − t ) + β z ν(dz) ǫ 2 1 σ 2 z≥ξ−x  4λ  α 2 C (t − t ) + β M . (3.12) ≤ ǫ 2 1 σ2

for a sufficiently large positive constant M. To obtain R the first inequality, we used x ≤ b(t1 ) for (x, t) ∈ D. The third inequality follows, because R ez ν(dz) < +∞. With the estimate (3.12), we can calculate Lχ inside the domain D.     Z 1 2(r + λ) 2µ 2λ 2 Lχ(x) = −2β − − 1 2βχ 2 + χ 1{x>ξ} − 2 χ(x + z, t)ν(dz) σ2 σ2 σ R # "  ( σ2µ2 − 1)2 σ 2 4λ  + 2 β 2 1{x>ξ} − 2 Cǫ (t2 − t1 )α + β 2 M ≥ − 2(r + λ) σ ≥ −Eβ 2 − F (t2 − t1 )α ,

in which E ,

(2µ/σ2 −1)2 σ2 2(r+λ)

+2+

4λ M σ2

(3.13)

and F ,

4λ C σ2 ǫ

are positive constants.

Recall that for any ε > 0, b(ǫ) < min{log K, log S0 } and that the strictly increasing function f defined in (3.3) satisfies f (y) < 0 for y < S0 . Using these observations and (3.6) it can be seen that for any x ≤ b(ǫ) we have Lg(x) = −

2 2 f (ex ) ≥ − 2 f (eb(ǫ) ) > 0. 2 σ σ

Now choosing c = −2/σ 2 f (eb(ǫ) ) > 0 and δ ′ = min{ that Lχ(x)(x) ≥ −c ≥ L(u − g)(x, t),


c 1/α 2F

, δ} and β ≤

(x, t) ∈ D.

(3.14) p

c , 2E

we have

Considering Ψ = χ − u + g, we have LΨ ≥ 0 in D and Ψ ≥ 0 on the parabolic boundary of D. It follows from the Maximum Principle for weak solutions that Ψ ≥ 0 in D, i.e., χ(x) ≥ (u − g)(x, t),

(x, t) ∈ D.

(3.15)

Observe that (u − g)(x, t) = 0 if x ≤ ξ. For any (x, t) ∈ D, since (u − g)(x, t) > 0, we can see that x > ξ. This gives us √ α Cǫ (t2 − t1 ) 2 , 0 < t2 − t1 ≤ δ ′ . (3.16) inf b(t) ≥ b(t1 ) − t1 ≤t≤t2 β 15

We have shown the free boundary b(t) is continuous and strictly decreasing in Lemma 3.2. √ Along with this fact, the inequality (3.16) gives us (3.10) with Cǫ′ = Cǫ /β. The proof for the statement about bn (t) (n ≥ 1) is similar. We will replace (3.14) by   Z 2λ min LD g(x, t) − 2 un−1(x + z, t)ν(dz) > 0, (3.17) (x,t)∈D σ R which follows as a result of Lemma 3.1, and apply the Maximum Principle for the parabolic differential operator LD . Now we are ready to state the main result of this section. Theorem 3.1. Let b(t) be the free boundary in problem (2.6) - (2.9), then for any ǫ > 0 if ǫ ≤ t1 < t2 ≤ T0 , and (t2 − t1 ) is sufficiently small, then 5

b(t1 ) − b(t2 ) ≤ Cǫ (t2 − t1 ) 8 ,

,

(3.18)

in which Cǫ is a positive constant that does not depend on t1 and t2 . The statement also holds for the free boundaries bn (t), n ≥ 1. Proof. We will only prove the H¨older continuity of b(t), the proof for the statement on bn (t) is performed similarly. The proof will follow by applying Lemma 3.3 twice. The first application will show that b(t) is H¨older continuous with exponent 12 . Applying Lemma 3.3 for the second time we will upgrade the H¨older exponent to 58 . As a result of Propositions 2.1 and 2.2 for any ǫ > 0, t1 and t2 satisfying ǫ ≤ t1 < t2 ≤ T0 we have that ∂u (b(t1 ), s)(t − t1 ) ≤ C1 (t2 − t1 ), t1 ≤s≤t ∂t

u(b(t1 ), t) − u(b(t1 ), t1 ) ≤ max

(3.19)

where C1 is a positive constant. Now as a result of Lemma 3.3, we know that there exists a sufficiently small constant δ1 ∈ (0, T0 − ǫ] such that 1

b(t1 ) − b(t2 ) ≤ C1′ (t2 − t1 ) 2 ,

0 ≤ t2 − t1 ≤ δ1 ,

(3.20)

in which C1′ is a positive constant that does not depend on t1 , t2 and δ1 . Recall the estimate (A-7) in Appendix A.1: for any a < b < log K and t ∈ [t1 , t2 ], we have ∂u 1 (x, t) − ∂u (x, t) ≤ C˜ |x − x| 2 , x, x ∈ (a, b), (3.21) ∂t ∂t

in which C˜ is a positive constant that does not depend on t. Taking x = b(t1 ) and x = b(t) in (3.21) and using Proposition 2.1, we obtain 0≤

1 ∂u ˜ 1 ) − b(t2 )| 21 , (b(t1 ), t) ≤ C˜ |b(t1 ) − b(t)| 2 ≤ C|b(t ∂t

16

t1 ≤ t ≤ t2 ,

(3.22)

where the second inequality follows from Lemma 3.2. Combining (3.20) and (3.22), we get 0≤

1 ∂u (b(t1 ), t) ≤ C2 (t2 − t1 ) 4 , ∂t

t1 ≤ t ≤ t2 , 0 ≤ t2 − t1 ≤ δ1 .

(3.23)

As a result u(b(t1 ), t) − u(b(t1 ), t1 ) ≤ max

t1 ≤s≤t2

5 ∂u (b(t1 ), s)(t2 − t1 ) ≤ C2 (t2 − t1 ) 4 . ∂t

(3.24)

Applying Lemma 3.3, we know that there exists δ2 ∈ (0, δ1 ] such that 5

b(t1 ) − b(t2 ) ≤ Cǫ (t2 − t1 ) 8 ,

0 ≤ t2 − t1 ≤ δ2 ,

(3.25)

where Cǫ is a positive constant that does not depend on t1 , t2 and δ2 .

4 Free boundaries are continuously differentiable In this section, we will investigate the continuous differentiability of free boundaries. In Theorem 5.6 in Yang et al. (2006), the authors have shown b(t) ∈ C 1 ((0, T0 ]), with the extra condition Z +∞

r ≥q+λ

(y − 1)θ(dy).

1

(4.1)

We will prove the continuous differentiability of the free boundary without imposing this extra condition. We will also show that bn (t) ∈ C 1 ((0, T0 ]) for all n ≥ 1. Remark 4.1. If condition (4.1) is not satisfied, we can see from (3.2) that there is a gap between limt→0+ b(t) and b(0) = log K. Therefore it is impossible to have b(t) to be even continuous at t = 0. But we shall see that it is continuously differentiable for all t ∈ (0, T0 ]. Let us consider functions ∂t u(x, t) and ∂t un (x, t). Recall that u(x, t) is the solution of (2.6) - (2.9). The derivative w = ∂ut (x, t) satisfies the following partial differential equation LD w = h(x, t), w(b(t), t) = 0,

x > b(t), t ∈ (0, T0 ], lim w(x, t) = 0, t ∈ (0, T0 ],

x→+∞

w(x, 0) = lim ∂t u(x, t), t→0

in which

(4.2)

2λ h(x, t) = 2 σ

Z

R

x ≥ b(0),

∂ u(x + z, t)ν(dz). ∂t 17

(4.3)

When x < b(t), we also have w(x, t) = 0. Given u(x, t) and b(t), (4.2) is a parabolic differential equation for w(x, t). Similarly, from the equation (2.11), we have the following system for wn LD wn = hn (x, t), wn (bn (t), t) = 0,

x > bn (t), t ∈ (0, T0 ], lim wn (x, t) = 0, t ∈ (0, T0 ],

x→+∞

wn (x, 0) = lim ∂t un (x, t), t→0

in which

2λ hn (x, t) = 2 σ

(4.4)

x ≥ bn (0),

∂ un−1 (x + z, t)ν(dz). ∂t

Z

R

(4.5)

In these equations, the boundary conditions for w(x, t) and wn (x, t) along b(t) or bn (t) and at infinity follow from Proposition 2.1 and Lemma 2.3. In order to show the differentiability of the free boundaries, we need to study the behaviors ∂2 ∂2 of ∂x∂t u and ∂x 2 u at the free boundary (by first making sure that the cross derivatives exist in the classical sense), which is carried out in the following two lemmas. 2

2

∂ ∂ Lemma 4.1. (i) As a function of t, ∂x∂t u(b(t)+, t) , limx↓b(t) ∂x∂t u(x, t) is continuous on ∂2 (0, T0 ]. The same statement also holds for ∂x∂t un (bn (t)+, t), n ≥ 1. 2

∂ (ii) Moreover, the function ∂x∂t u(x, t) is continuous for x > b(t), t ∈ (0, T0 ]. The same holds 2 ∂ for ∂x∂t un (x, t), n ≥ 1.

This lemma is a slight generalization of the result in Cannon et al. (1974) to parabolic integro-differential equations. Considering the integral term h in (4.2) as the driving term, this lemma follows from using the same technique presented in Section 1 of Chapter 8 in Friedman (1964). We will postpone this proof to the Appendix and proceed to a result which will be a crucial tool in proving the main result of this section. Lemma 4.2. For t ∈ (0, T0 ] and n ≥ 1 ∂2 ∂2 u(b(t)+, t) , lim u(x, t) > −eb(t) , x↓b(t) ∂x2 ∂x2

∂2 ∂2 u (b (t)+, t) , lim un (x, t) > −ebn (t) . n n x↓bn (t) ∂x2 ∂x2

Proof. We will only prove the first equation. The proof of the second equation can be similarly performed. For any ǫ > 0, let us define the domain B , {(x, t) : b(t) ≤ x ≤ b(ǫ), ǫ ≤ t ≤ T0 }. In B, we have   Z 2(r + λ) x 2λ 2 2µ x x x −1 e + e − 2 ex+z ν(dz) = 2 qex . (4.6) Le = −e − 2 2 σ σ σ R σ This gives us the identity L (u + ex ) =

2 x qe , σ2 18

(x, t) ∈ B.

(4.7)

Taking another derivative with respect to x on both hand side of (4.7), we have   ∂u 2 x L + e = 2 qex ≥ 0, (x, t) ∈ B, ∂x σ

(4.8)

using the bounded convergence theorem (which is used to exchange the derivative and the integral). Note that the term ∂x3 u(x, t) on the left hand side of (4.8) exists and is continuous: L∂x u = 0 and ∂x ∂t u is continuous as a result of Lemma 4.1. Hence all derivatives in the left hand side of (4.8) exist in classical sense. On the other hand, we have (∂x u + ex )(b(t), t) = 0 and (∂x u + ex )(x, t) = S(∂S V + 1) > 0 when (x, t) ∈ B (Recall that V is a convex function dominating the pay-off (K − S)+ and V (0, t) = K.) Applying the Hopf Lemma to (4.8) (see Theorem 2.8 in page 78 of Garroni and Menaldi (1992)), we obtain that ∂2 u(b(t)+, t) + eb(t) > 0, ǫ ≤ t ≤ T0 . 2 ∂x Then the lemma follows, since the choice of ǫ is arbitrary.

(4.9)

We are now ready to state and prove the main theorem of this section. Theorem 4.1. Let b(t) and bn (t) (n ≥ 1) be the free boundaries in the free boundary problems (2.6) - (2.9) and (2.11) - (2.14) respectively. Then b(t) ∈ C 1 ((0, T0 ]) and bn (t) ∈ C 1 ((0, T0 ]). Proof. We will only prove the result for b(t), the proof for bn (t) is the same. First, we will show ∂2 b(t0 ) . Lemma 4.2 b(t) is differentiable at t0 ∈ (0, T0 ]. Let us define ρ = ∂x 2 u(b(t0 )+, t0 ) + e implies that ρ > 0. For sufficiently small ǫ > 0, it follows from (2.8) that   1 ∂ ∂ b(t0 ) b(t0 −ǫ) = 0. u(b(t0 ), t0 ) − u(b(t0 − ǫ), t0 − ǫ) + e −e ǫ ∂x ∂x Applying the Mean Value Theorem yileds  2  ∂ b(t0 ) − b(t0 − ǫ) ∂2 b(t0 )+y u(b(t ) + y, t ) + e = − u(b(t0 − ǫ), t0 − τ ), 0 0 ∂x2 ǫ ∂x∂t

(4.10)

for some y ∈ [0, b(t0 − ǫ) − b(t0 )] and τ ∈ [0, ǫ]. Letting ǫ → 0 in (4.10) and using Lemma 4.1 (ii), we obtain ∂2 u(b(t0 )+, t0 ) b(t0 ) − b(t0 − ǫ) ∂x∂t lim = − ∂2 , (4.11) b(t0 ) ǫ→0 ǫ u(b(t )+, t ) + e 2 0 0 ∂x which implies that b(t) is differentiable since ρ > 0. Moreover, from (2.10) and Proposition 2.1, we have   2(r + λ) 2(µ − r − λ) ∂2 u(b(t)+, t) = K+ − 1 eb(t) − f (b(t), t), ∂x2 σ2 σ2 19

which is clearly a continuous function of t on t(0, T0 ], since b(t) is a continuous function. Along with Lemma 4.1 (i), we can see from (4.11) that b(t) ∈ C 1 ((0, T0 ]).

5 Higher order regularity of the free boundaries In the previous section, we have proved that free boundaries b(t) and bn (t), n ≥ 1, are continuously differentiable. In this section, we will upgrade their regularity. First, let us derive two identities on b′ (t) and b′n (t). Let w = ∂t u and wn = ∂t un , which satisfy equations (4.2) and (4.4) with their boundary and initial conditions respectively. Since b(t) is differentiable, taking derivative with respect to t on both sides of (2.8), we have ∂2 ∂2 ′ u(b(t)+, t)b (t) + u(b(t)+, t) = −eb(t) b′ (t). ∂x2 ∂x∂t

(5.1)

The term ∂x2 u(b(t)+, t) can be represented as ∂2 u(b(t)+, t) = ∂x2



 2(µ − r − λ) 2(r + λ) − 1 eb(t) + K − f (b(t), t). 2 σ σ2

(5.2)

Plugging (5.2) back into (5.1), we obtain b′ (t) = − 2(µ−r−λ) σ2

∂ w(b(t)+, t) ∂x eb(t) + 2(r+λ) K− σ2

f (b(t), t)

,

t ∈ (0, T0 ].

(5.3)

Similarly, for all n ≥ 1 we have that b′n (t) = − 2(µ−r−λ) σ2

∂ w (b (t)+, t) ∂x n n , ebn (t) + 2(r+λ) K − f (b (t), t) 2 n n σ

t ∈ (0, T0 ].

(5.4)

We can see from equations (4.2), (4.4) that w(x, t) and wn (x, t) are solutions of formal Stefan problems in the unbounded continuation regions C and Cn , n ≥ 1, respectively. Schaeffer (1976) gave a proof of the infinite differentiability of the free boundary of a one dimensional x Stefan problem in a bounded domain. By introducing the new variable ξ = b(t) , he reduced the problem into a fixed boundary problem on a bounded domain. However, if we apply the same change of variables we will have unbounded coefficients in the corresponding fixed boundary problem. Instead, we will define ξ , x − b(t),

v(ξ, t) , w(x, t),

in which b(t) is the free boundary in (2.6) - (2.9). The function v(ξ, t) satisfies the following 20

fixed boundary equation,   Z ∂v ∂ 2 v ∂ 2λ 2µ ∂v 2(r + λ) ′ − 2− + b (t) − 1 + v= 2 u(ξ + b(t) + z, t)ν(dz), 2 2 ∂t ∂ξ σ ∂ξ σ σ R ∂t (ξ, t) ∈ (0, +∞) × (0, T0 ], (5.5) v(0, t) = 0, t ∈ (0, T0 ], (5.6) v(ξ, 0) = w(ξ + b(t), 0), ξ ≥ 0. (5.7) Moreover, we have the following identity ′

b (t) =

∂ v(0, t) ∂ξ − 2(µ−r−λ) eb(t) + 2(r+λ) K σ2 σ2

− f (b(t), t)

,

t ∈ (0, T0 ].

(5.8)

Meanwhile, vn (ξ, t) satisfies a similar equation with b(t) replaced by bn (t) and u replaced by un−1 in (5.5). Remark 5.1. Since b(t) ∈ C 1 (0, T0 ], so for any ǫ > 0, b′ (t) is continuous and bounded in [ǫ, T0 ]. R On the other hand, since ∂t u is bounded by Lemma 2.2, so σ2λ2 R ∂t u(ξ + b(t) + z, t)ν(dz) is also bounded when (ξ, t) ∈ [0, +∞) × [ǫ, T0 ]. as a result, it follows from Theorem 2.6 in page 19 of Ladyzhenskaja et al. (1968) that the parabolic differential equation (5.5) has at most one bounded solution. It is clear that v(ξ, t) = ∂t u(x, t) is a bounded solution, so it is the unique bounded solution of (5.5). The following result for parabolic differential equations will be an essential tool in the proof of the main result in this section. Lemma 5.1. Let us assume w(ξ, t) ∈ H 2α,α ([0, +∞) × [δ, T ]) (for some α and δ > 0) satisfies the following equation Z ∂w ∂ 2 w ∂w − 2 +ℓ + cw = d φ(ξ + z, t)ν(dz), (ξ, t) ∈ ((0, +∞) × (δ, T ))(5.9) ∂t ∂ξ ∂ξ R w(0, t) = g(t), t ∈ [δ, T ]. (5.10) R We assume that R φ(ξ + z, t)ν(dz) ∈ H 2α,α ([0, +∞) × [δ, T ]) and that coefficients ℓ,c also belong to H 2α,α ([0, +∞) × [δ, T ]), d is a constant and g(t) ∈ H 1+α ([δ, T ]). Then w(ξ, t) ∈ H 2+2α,1+α ([0, +∞) × [δ ′ , T ], for any δ ′ > δ. Proof. Consider a cut-off function η(t) ∈ C0∞ ((0, T0 ]), such that η(t) = 0 when t ∈ (0, δ] and η(t) = 1 for t ∈ [δ ′ , T ]. The function w(ξ, ˜ t) = η(t)w(ξ, t) satisfies Z ∂ w˜ ∂η ∂ w˜ ∂ 2 w˜ − 2 +ℓ + cw˜ = d η(t)φ(ξ + z, t)ν(dz) + w(ξ, t), (ξ, t) ∈ (0, +∞) × (δ, T ], ∂t ∂ξ ∂ξ ∂t R w(0, ˜ t) = η(t)g(t), t ∈ [δ, T ], w(ξ, ˜ δ) = 0, ξ ≥ 0. 21

From our assumptions we have that Z ∂η d η(t)φ(ξ + z, t)ν(dz) + w(ξ, t) ∈ H 2α,α ([0, +∞) × [δ, T ]), ∂t R η(t)g(t) ∈ H 1+α ([δ, T ]). Moreover, the coefficients of the above differential equation are all inside space H 2α,α ([0, +∞)× [δ, T ]). It follows from regularity estimation for parabolic differential equation (see Theorem 5.2 in page 320 of Ladyzhenskaja et al. (1968)) that w(ξ, ˜ t) ∈ H 2+2α,1+α ([0, +∞) × [δ, T ]), which implies w(ξ, t) ∈ H 2+2α,1+α ([0, +∞) × [δ ′ , T ]) by the choice of η(t). Remark 5.2. We will apply the previous lemma to w(x, t) = ∂t u(x, t). Because the initial condition for u(x, t) is not smooth, limt→0 ∂t u(x, t) is not smooth and we can not apply Theorem 5.2 in page 320 of Ladyzhenskaja et al. (1968) to upgrade the regularity of w directly. This is the reason we work with w˜ in the proof of the previous lemma. In order to apply Lemma 5.1 to (5.5) - (5.8), we need H¨older continuous coefficients and value functions. Let us first show that the coefficients in equation (5.5) are H¨older continuous. Lemma 5.2. Let b(t) be the free boundary in (2.6) - (2.9). Then b(t) ∈ H 1+α ([δ, T0 ]) with 0 < α < 21 for any δ > 0. The same statement also holds for bn (t), n ≥ 1. Proof. For any δ > 0, since b(t) ∈ C 1 (0, T0 ] by Theorem 4.1, the coefficients in equation (5.5) are bounded and continuous in [δ, T0 ]. On the other hand,R because ∂t u(x, t) is bounded ∂ u(ξ + b(t) + z, t)ν(dz) is in R × [δ, T0 ] by Lemma 2.2, the function h(ξ + b(t), t) = σ2λ2 R ∂t also bounded when (ξ, t) ∈ [0, +∞) × [δ, T0 ]. It follows from Theorem 9.1 in page 341 of Ladyzhenskaja et al. (1968) that equation (5.5) has a unique solution v(ξ, t) ∈ Wq2,1 ([0, M] × [δ, T0 ]) for any q > 1 and M > 0. By the Sobolev Embedding Theorem (see, for example, Theorem 2.1 in page 61 of Ladyzhenskaja et al. (1968)), for q > 3, we have v(ξ, t) ∈ H β,β/2 ([0, M] × [δ, T0 ]) with β = 2 − 3q (1 < β < 2). as a result, we have β−1 ∂ β−1 1 v(0, t) ∈ H 2 ([δ, T0 ]), with 0 < < . (5.11) ∂ξ 2 2 Let us analyze the terms in the denominator on the right hand side of (5.8). We have that b(t) ∈ C 1 ([δ, T0 ]) and that Z 2λ f (b(t), t) = 2 u(b(t) + z, t)ν(dz) ∈ C 1 ([δ, T0 ]), σ R since u(x, t) is the classical solution of the free boundary problem (2.6) - (2.9). Moreover, this denominator is also bounded away from 0, because ∂2 2(µ − r − λ) b(t) 2(r + λ) e + K − f (b(t), t) = u(b(t), t) + eb(t) > 0, σ2 σ2 ∂x2 22

t ∈ [δ, T0 ],

where the last inequality follows from Lemma 4.2. With (5.8) and (5.11), it is clear that b′ (t) ∈ H

β−1 2

([δ, T0 ]).

The proof of the regularity of bn (t) is similar. We use the facts that ∂t un−1 (x, t) is bounded in R × [ǫ, T0 ] (Lemma 2.2) and un−1 is the classical solution of (2.11) - (2.14). As a corollary of Lemmas 5.1 and 5.2, we can improve the regularity of the functions u(x, t). Corollary 5.1. Let u(x, t) and b(t) be the classical solution of the free boundary problem (2.6) - (2.9). Then u(ξ + b(t), t) ∈ H 2+2α,1+α ([0, +∞) × [δ ′ , T0 ]) for any δ ′ > 0, with α ∈ (0, 1/2). The same statement also holds for un (x, t), n ≥ 1. Proof. Let ξ = x − b(t), κ(ξ, t) = u(x, t) and φ(ξ + z, t) = u(ξ + b(t) + z, t). Then κ(ξ, t) satisfies a differential equation of the form (5.9) and (5.10) in Lemma 5.1 with g(t) = K − eb(t) (in fact κ satisfies (5.5) when ∂ut in the driving term is replaced by u). Moreover, by Lemma 5.2, the coefficients in this equation (5.9) are inside space H α ([δ, T0 ]) for any δ > 0, and g(t) ∈ H 1+α ([δ, T0 ]). On the other hand, since u(x, t) is uniformly Lipschitz in x ∈ R and uniformly semiH¨older continuous in t ∈R [0, T0 ] (see Lemma 2.1 ), and b(t) is continuously differentiable, it is not hard to see that R u(ξ + b(t) + z, t)ν(dz) ∈ H 2α,α ([0, +∞) × [δ, T0 ]). Moreover, u(ξ + b(t), t) ∈ H 2α,α ([0, +∞) × [δ, T0 ]) again because of Lemma 2.1. Now, the first part of this lemma follows directly from Lemma 5.1. Using the fact that un−1 (x, t) is uniformly Lipschitz in x and uniformly semi-H¨older continuous in t, the proof of the regularity of un can be similarly done. Armed with Lemmas 5.1, 5.2 and Corollary 5.1, we can state and prove the main theorem of this section. Theorem 5.1. Let b(t) be the free boundary in (2.6) - (2.9).R Assume that ν has a density, u i.e. ν(dz) = ρ(z)dz. Let α ∈ (0, 1/2). If ρ(z) satisfies −∞ ρ(z)dz ∈ H 2α (R− ), then 3 b(t) ∈ H 2 +α ([ǫ, T0 ]). On the other hand, if ρ(z) ∈ H ℓ−1+2α (R− ) for ℓ ≥ 1, then b(t) ∈ 3 ℓ H 2 + 2 +α ([ǫ, T0 ]), for any ǫ > 0. Proof. The proof consists of four steps. Step 1. From Lemma 5.2 and Corollary 5.1, we have that b(t) ∈ H 1+α ([δ, T0 ]) and that u(ξ + b(t), t) ∈ H 2+2α,1+α ([0, +∞) × [δ ′ , T0 ]) for any δ ′ > δ > 0 with α ∈ (0, 1/2), which implies that ∂t u(ξ + b(t), t) ∈ H 2α,α ([0, +∞) × [δ ′ , T0 ]) (see Definition 2.1). 23

Step 2. Assume that there is a positive nonintegral real number β with 2β ≤ 2α + ℓ, such that b(t) ∈ H 1+β ([δ, T0 ]), ∂ u(ξ + b(t), t) ∈ H 2β,β ([0, +∞) × [δ ′ , T0 ]), ∂t u(ξ + b(t), t) ∈ H 2+2β,1+β ([0, +∞) × [δ ′ , T0 ]),

(5.12) (5.13) (5.14)

for δ ′ > δ > 0. We will upgrade the regularity exponent from β to 1/2 + β, in steps 2 and 3. Let us analyze ∂t u(ξ + b(t), t). For any integers r, s ≥ 0, 2r + s < 2β, since ∂t u(ξ + b(t) + z, t) = 0 when z ≤ −ξ, we have Z Z +∞ ∂ ∂ ∂s ∂r ∂s ∂r u(ξ + b(t) + z, t)ν(dz) = s r u(ξ + b(t) + z, t)ρ(z)dz s r ∂ξ ∂t R ∂t ∂ξ ∂t −ξ ∂t s−1 X ∂i ∂r ∂ ds−1−i = 1{s≥1} u(ξ + b(t) + z, t) ρ(−ξ) s−1−i ∂ξ i ∂tr ∂t z↓−ξ dξ i=0 Z +∞ s r ∂ ∂ ∂ u(ξ + b(t) + z, t)ρ(z)dz, (5.15) + ∂ξ s ∂tr ∂t −ξ for any ξ ≥ 0. When t is fixed, in the following, we will show Z ∂s ∂r ∂ u(ξ + b(t) + z, t)ν(dz) ∈ H 2β−[2β] ([0, +∞)), s r ∂ξ ∂t R ∂t

for 2r + s = [2β].

(5.16)

For any ξ1 > ξ2 ≥ 0 such that ξ1 − ξ2 ≤ ρ0 , we have s r Z s r Z ∂ ∂ ∂ ∂ ∂ ∂ u(ξ + b(t) + z, t)ν(dz) − u(ξ + b(t) + z, t)ν(dz) 1 2 ∂ξ s ∂tr ∂ξ s ∂tr R ∂t R ∂t   s−1 i ds−1−i X ∂ ∂r ∂ ≤ 1{s≥1} u(ξ + b(t) + z, t) i r s−1−i ρ(−ξ1 ) − ρ(−ξ2 ) ∂ξ ∂t ∂t dξ z↓−ξ i=0  Z +∞ s r  ∂ ∂ ∂ ρ(z)dz u(ξ + b(t) + z, t) − u(ξ + b(t) + z, t) + 1 2 ∂ξ s ∂tr ∂t −ξ2 Z −ξ2 s r ∂ ∂ ∂ ρ(z)dz. + (5.17) u(ξ + b(t) + z, t) 1 ∂ξ s ∂tr ∂t −ξ1

Let us analyze the right hand side of (5.17) term by term. Since s − 1 < 2β − 1 ≤ 2α + ℓ − 1, we have ρ(z) ∈ H 2β−1 (R− ), which implies  s−1 i ds−1−i  X ∂ ∂r ∂ u(ξ + b(t) + z, t) 1{s≥1} s−1−i ρ(−ξ1 ) − ρ(−ξ2 ) i r dξ ∂ξ ∂t ∂t z↓−ξ i=0 ≤ C||∂t u||(2β) |ξ1 − ξ2 |2β−[2β] ,

(5.18)

24

in which C is a positive constant and || · ||(2β) is the H¨older norm (see Definition 2.1). On the other hand, it follows from (5.13) that  Z +∞ s r  ∂ ∂ ∂ ∂ξ s ∂tr ∂t u(ξ1 + b(t) + z, t) − u(ξ2 + b(t) + z, t) ρ(z)dz −ξ2 Z +∞ (2β) 2β−[2β] ≤ ||∂t u|| |ξ1 − ξ2 | ρ(z)dz ≤ ||∂t u||(2β) |ξ1 − ξ2 |2β−[2β] . (5.19) −ξ2

Ru Moreover, because ρ(z) ∈ H ℓ−1+2α (R− ) for ℓ ≥ 1 or −∞ ρ(z)dz ∈ H 2α,α (R− ), we have Ru Ru ρ(z)dz ∈ H ℓ+2α (R− ) for ℓ ≥ 0. In particular, using 2β ≤ 2α+ℓ, we can see −∞ ρ(z)dz ∈ −∞ H 2β−[2β] (R− ). As a result, Z −ξ2 s r ∂ ∂ ∂ ρ(z)dz u(ξ + b(t) + z, t) 1 ∂ξ s ∂tr ∂t −ξ1 Z −ξ2  Z −ξ1 (2β) ˜ t u||(2β) |ξ1 − ξ2 |2β−[2β] , (5.20) ρ(z)dz − ρ(z)dz ≤ C||∂ ≤ ||∂t u|| −∞

−∞

where C˜ is also a positive constant. Plugging the estimates (5.18) - (5.20) into (5.17), we observe that (5.16) holds. When ξ is fixed, using (5.15), it directly follows from (5.13) that Z 2r+s ∂s ∂r ∂ u(ξ + b(t) + z, t)ν(dz) ∈ H β− 2 ([δ ′ , T0 ]), for 2β − 2 < 2r + s < 2β. (5.21) s r ∂ξ ∂t R ∂t Now, (5.16) and (5.21) imply that Z ∂ u(ξ + b(t) + z, t)ν(dz) ∈ H 2β,β ([0, +∞) × [δ ′ , T0 ]). R ∂t

(5.22)

Let v(ξ, t) be a bounded solution of the fixed boundary problem (5.5) - (5.8). The uniqueness in Remark 5.1 implies that ∂ v(ξ, t) = u(ξ + b(t), t). (5.23) ∂t As a result, the assumption (5.13) implies that v(ξ, t) ∈ H 2β,β ([0, +∞) × [δ ′ , T0 ]).

(5.24)

We will apply Lemma 5.1 to (5.5) - (5.8) with φ(ξ + z, t) = ∂t u(ξ + b(t) + z, t), ℓ =
and d = 2λ . Thanks to (5.12), the coefficient l belongs to − σ2µ2 + b′ (t) − 1 , c = 2(r+λ) σ2 σ2 β H ([δ, T0 ]). The other coefficients already happen to reside there since they are constants. Along with (5.22) and (5.24), Lemma 5.1 yields v(ξ, t) ∈ H 2+2β,1+β ([0, +∞) × [δ ′′ , T0 ]) 25

for any δ ′′ > δ ′ > δ,

(5.25)

which implies that

and

1 ∂ v(0, t) ∈ H 2 +β ([δ ′′ , T0 ]), ∂ξ

(5.26)

∂ u(ξ + b(t), t) ∈ H 2+2β,1+β ([0, +∞) × [δ ′′ , T0 ]), ∂t

(5.27)

by (5.23). Using (5.8) and (5.26), we will improve the regularity of b(t) in the following. From (2.24) we have Z 2λ u(b(t) + z, t)ν(dz) f (b(t), t) = σ2 R Z Z 2λ +∞ 2λ 0 = u(b(t) + z, t)ν(dz) + 2 (K − eb(t)+z )ν(dz). (5.28) σ2 0 σ −∞ Along with (5.12) and (5.14), we can see from (5.28) that f (b(t), t) ∈ H 1+β ([δ ′′ , T0 ]).

(5.29) 1

Together with (5.12), (5.26) and (5.29), we can see from the identity (5.8) that b′ (t) ∈ H 2 +β ([δ ′′ , T0 ]) for any δ ′′ > δ ′ . It in turn implies that 3

b(t) ∈ H 2 +β ([δ ′′ , T0 ]).

(5.30)

Step 3. Let us investigate u(ξ + b(t), t). For any r, s ≥ 0, 2r + s < 2 + 2β, we have Z ∂s ∂r u(ξ + b(t) + z, t)ν(dz) ∂ξ s ∂tr R Z +∞ Z −ξ ∂s ∂r ∂s ∂r u(ξ + b(t) + z, t)ρ(z)dz + s r u(ξ + b(t) + z, t)ρ(z)dz = s r ∂ξ ∂t −ξ ∂ξ ∂t −∞ # " s−1 i r X ∂ ∂ ds−1−i ∂i ∂r − u(ξ + b(t) + z, t) u(ξ + b(t) + z, t) ρ(−ξ) = 1{s≥1} s−1−i ∂ξ i ∂tr ∂ξ i ∂tr z↓−ξ z↑−ξ dξ i=0 Z +∞ s r Z −ξ s r ∂ ∂ ∂ ∂ + u(ξ + b(t) + z, t)ρ(z)dz + u(ξ + b(t) + z, t)ρ(z)dz. s r ∂ξ s ∂tr −ξ −∞ ∂ξ ∂t It is worth noticing that ∂ξi ∂tr u(ξ + b(t) + z, t)|z↓−ξ 6= ∂ξi ∂tr u(ξ + b(t) + z, t)|z↑−ξ for some i and r. Following the same arguments that lead up to (5.22), we can show Z u(ξ + b(t) + z, t)ν(dz) ∈ H 2+2β,1+β ([0, +∞) × [δ ′ , T0 ]), (5.31) R

given 1 + 2β ≤ 2α + ℓ − 1. 26

Now, we can apply Lemma 5.1 to the differential equation u(ξ + b(t), t) satisfies, taking (5.14) and (5.30) into account. This results in 3

u(ξ + b(t), t) ∈ H 3+2β, 2 +β ([0, +∞) × [δ ′′′ , T0 ]),

(5.32)

for any δ ′′′ > δ ′′ . As a result, we have improved the regularities from (5.12), (5.13) and (5.14) to (5.30), (5.27) and (5.32), respectively. Step 4. For any ǫ > 0, we apply Steps 2 and 3 inductively starting from β = α in Step 1. Let n be the number of time we apply Steps 2 and 3. Let δ1′ = δ ′ , in which δ ′ > 0 is as in Step 1. Running Step 2 and 3 once, we obtain two constants δ1′′ and δ1′′′ such that (5.30), (5.32) hold ′′′ with β = α. In the n-th time, n ≥ 2, we choose δn′ = δn−1 and δn′′′ > δn′′ > δn′ , such that δn′′′ < ǫ for any n so that [ǫ, T0 ] ⊂ [δn′′′ , T0 ]. n

The application of Step 2 for the n-th time will give us that b(t) ∈ H 1+α+ 2 (R− ). Applying Step 2 for ℓ + 1 and Step 3 for ℓ times the result follows. Remark 5.3. (i) The previous proof has also shown the higher order regularity of u(x, t), i.e. ℓ u(ξ + b(t), t) ∈ H 2+2α+ℓ,1+α+ 2 ([0, +∞) × [ǫ, T0 ]), for any ǫ > 0, under the assumptions of Theorem 5.1, . (ii) Note that b(t) ∈ C 1 ((0, T ]) without any assumption on the density ρ(z). If ρ(z) ∈ 3 H 2m−1+2α (R− ) for some m ≥ 1, then b(t) ∈ H 2 +m+α ([ǫ, T0 ]). From Definition 2.1 and the arbitrary choice of ǫ, we have that b(t) ∈ C m+1 ((0, T0 ]) under this assumption. As a corollary of Theorem 5.1, we have the following sufficient condition for the infinitely differentiability of b(t). Corollary 5.2. Let b(t) be the free boundary in (2.6) - (2.9). Assume that ν has a density, dℓ i.e. ν(dz) = ρ(z)dz. If ρ(z) ∈ C ∞ (R− ) with dz ℓ ρ(z) bounded for each ℓ ≥ 1, then b(t) ∈ C ∞ ((0, T0 ]). Proof. For any m ≥ 1 with ρ(z) ∈ C 2m+1 (R− ) and derivatives of ρ(z) up to order 2m + 1 are bounded, it follows from Definition 2.1 that ρ(z) ∈ H 2m−1+2α (R− ). As a result of Remark 5.3 (ii), we have b(t) ∈ C m+1 ((0, T0 ]). Remark 5.4. There are two well-known examples of jump diffusion models in the literature, Kou’s model and Merton’s model (see Cont and Tankov (1998), p.111), in which the density ρ(z) is double exponential and normal, respectively. For both of these densities, it is easy to see that the conditions for Corollary 5.2 are satisfied. Therefore, the free boundaries in both models are infinitely differentiable. 27

A Appendix A.1

Proof of Proposition 2.1

The proof of (2.16) is given in Theorem 5.1 in Yang et al. (2006), which summarize Lemmas 2.8 and 2.11 in the same paper and use the following result, which is a special case of Lemma 4.1 in page 239 of Friedman (1976): Lemma A-1. For any a < b < log K, 0 < t1 < t2 < T0 , if both u(x, t) and ∂t u(x, t) belong to L2 ((t1 , t2 ); L2 (a, b)), then u(t) belongs to C((t1 , t2 ); L2 (a, b)). In this lemma, L2 ((t1 , t2 ); L2 (a, b)) is the class of L2 maps which map t ∈ (t1 , t2 ) to the Hilbert space L2 (a, b). On the other hand C((t1 , t2 ); L2 (a, b)) is the class of continuous maps which map t ∈ (t1 , t2 ) to L2 (a, b). The proof of (2.17) is similar to that of (2.16): First, we will study the penalty problem associated to the free boundary problem (2.11) - (2.14). Then, we will list some key estimates for the solution of the penalty problem. And finally using Lemma A-1 we will conclude. We will give a sketch of this proof below. Let us consider the following penalty problem LD uǫn + βǫ (uǫn − gǫ ) = fnǫ (x, t), uǫn (x, 0) = gǫ (x), x ∈ R,

x ∈ R, 0 < t < T0 ,

(A-1)

in which 0 < ǫ < 1, gǫ (x) ∈ C ∞ (R) such that gǫ (x) = (K −ex )+ when x satisfies |K −ex | ≥ ǫ. We define fnǫ (x, t) = ζǫ ∗ fn (x, t), where ζǫ is the standard mollifier in x and t (see Evans (1998) Appendix C4 in page 629). As a result, we have fnǫ (x, t) ∈ C ∞ (R×(0, T0)). Moreover, because fn (x, t) is continuous, fnǫ (x, t) uniformly converge to fn (x, t) on any compact domain as ǫ → 0. If fn (x, t) satisfies condition (C), it is easy to see that ∂t fnǫ (x, t) are uniformly bounded for any ǫ > 0. The penalty functions βǫ (x) is a sequence of infinitely differentiable, negative, increasing and concave functions such that βǫ (0) = −Cε ≤ −(r + λ)K − rǫ. The limit of the sequence is ( 0, lim βǫ (x) = ǫ→0 −∞,

x > 0, x < 0.

It is well known that the penalty problem has a classical solution (see page 1009 of Friedman and Kinderlehrer (1975)). Moreover, a proof similar to that of the proof of Theorem 2.1 of Yang et al. (2006) shows that uǫn (x, t) ∈ C ∞ (R × (0, T0 )) ∩ L∞ (R × (0, T0 )). On the other hand, uǫn (x, t) satisfy the following estimates for any a < b < log K, 0 < t1 < 28

t2 < T0 , 2 ∂uǫn (x, t)dx ≤ C, t ∈ [t1 , t2 ], ∂t a Z t2 Z b  2 ǫ 2 ∂ un dxdt ≤ C, ∂x∂t t1 a Z t2 Z b  2 ǫ 2 Z b  2 ǫ 2 ∂ un ∂ un (x, t)dx ≤ C, dxdt + ∂t2 ∂x∂t t1 a a Z b

(A-2) (A-3) t ∈ [t1 , t2 ],

(A-4)

in which C is a constant independent of ǫ. These estimates use similar techniques to the ones used in the proofs of Lemmas 2.8, 2.10 and 2.11 in Yang et al. (2006), given fn (x, t) satisfies the condition (C). (Similar estimates can also be found in Friedman and Kinderlehrer (1975)). We will give the proof for the inequality (A-4) below. The other inequalities can be similarly obtained. Proof of inequality (A-4). Let us consider wn (x, t) = ∂t uǫn (x, t). Since uǫn (x, t) ∈ C ∞ (R × (0, T0 )), it follows from (A-1) that wn (x, t) satisfies LD wn + βǫ′ (uǫn − gǫ )wn =

∂ ǫ f (x, t). ∂t n

(A-5)

Let η(x, t) ∈ C0∞ (R × (0, T0 )), such that η(x, t) = 1 for (x, t) ∈ [a, b] × [t1 , t2 ], and η(x, t) = 0 outside a small neighborhood of [a, b] × [t1 , t2 ]. Multiplying both sides of (A-5) by η 2 ∂t wn and integrating over the domain Ωt = R × (0, t) in which t1 ≤ t ≤ t2 , we obtain 

2

2

 Z Z wn ∂wn 2µ 2 ∂wn ∂wn η 0 = η η dxds − − 1 dxds dxds − ∂x2 ∂t σ2 ∂x ∂t Ωt Ωt Ωt Z Z ∂wn ∂wn 2(r + λ) 2 η wn η 2 βǫ′ (uǫn − gǫ )wn dxds + dxds + 2 σ ∂t ∂t Ωt Ωt Z Z ∂wn ∂ ǫ − η2 f (x, s)dxds ∂t ∂t n Ωt Z Z

2

∂wn ∂t Z Z

Z Z

2∂

, I1 + I2 + I3 + I4 + I5 + I6 ,

where Ij is the j-th term on the left. In the following, we will estimate each Ij separately. In deriving these estimates we will make use of the inequality 1 2 9 A + AB + B 2 ≥ 0, 6 6

(A-6)

for any A, B ∈ R. In the following estimations, C will represent different constants independent 29

of ǫ. I2 = = = ≥

≥

Z Z Z Z 2 ∂η ∂wn ∂wn wn ∂wn 2 ∂wn ∂ wn η 2η dxds = dxds + dxds η − 2 ∂x ∂t ∂x ∂x∂t ∂x ∂x ∂t Ωt Ωt Ωt  2 Z Z Z Z 1 ∂η ∂wn ∂wn ∂wn 2 ∂ η η dxds dxds + 2 2 ∂t ∂x Ωt Ωt ∂x ∂x ∂t  2  2 Z Z Z Z Z ∂wn ∂η ∂wn ∂η ∂wn ∂wn 1 2 η η η dxds (x, t)dx − dxds + 2 2 R ∂x ∂x Ωt ∂t Ωt ∂x ∂x ∂t Z Z  2 ǫ 2  2 Z ∂η 1 ∂wn ∂ un 2 η dxds (x, t)dx − η 2 R ∂x ∂t L∞ ∂x∂t Ωt  2 Z Z  2  2 ǫ 2 Z Z ∂wn ∂η ∂ un 1 2 η −6 dxds − dxds ∂x ∂x∂t 6 ∂t Ωt Ωt   2 2 Z Z Z ∂wn ∂wn 1 1 2 2 η η (x, t)dx − C − dxds. 2 R ∂x 6 ∂t Ωt Z Z

2∂

2

The first four equalities follow from integration by part. The first inequality follows from the ∂η ∂wn n and B = 2 ∂x . The last inequality follows from estimation inequality (A-6) with A = η ∂w ∂t ∂x (A-3). For Ii (i=3, 4, 5), a similar procedure yields  2 Z Z ∂wn 1 2 η I3 ≥ −C − dxds, 6 ∂t Ωt  2 Z Z 1 ∂wn 2 I4 ≥ −C − η dxds, 6 ∂t Ωt  2 Z Z ∂wn 1 2 η dxds. I5 ≥ −C − 6 ∂t Ωt For I6 , we have ∂wn ∂ ǫ f dxds ∂t ∂t n Ωt   2 2 Z Z Z Z 9 ∂ ǫ ∂wn 1 2 2 ≥ − η η dxds f dxds − 6 ∂t n 6 ∂t Ωt Ωt 2  Z Z ∂wn 1 2 dxds. η ≥ −C − 6 ∂t Ωt

I6 = −

Z Z

η2

The first inequality can be obtained using (A-6), whereas to obtain the last inequality, we use the fact that ∂t fnǫ (x, t) is uniformly bounded. Combining all these estimates for Ij , we obtain   2 2 Z Z Z ∂wn ∂wn 1 1 2 2 η η dxds + (x, t)dx ≤ C. 6 ∂t 2 R ∂x Ωt 30



This completes the proof of (A-4).

Using a similar proof to that of Lemma 2.2 of Yang et al. (2006), we can show that uǫn (x, t) is uniformly bounded. Thus there is a subsequence that {uǫnk } converges weakly to un in L2 ((a, b); L2 (t1 , t2 )) for any a < b < log K, 0 < t1 < t < t2 < T0 (see Appendix D in Evans (1998) for an account of the concept of weak convergence). On the other hand, it fol2 uǫ ǫ n n and ∂∂x∂t are uniformly bounded in L2 (a, b), lows from the estimates in (A-2) - (A-4) that ∂u ∂t 2 ǫ 2 ǫ ∂ un and ∂∂tu2n are uniformly bounded in L2 ((a, b); L2 (t1 , t2 )). Therefore there exists a further ∂x∂t subsequence satisfying ǫk

∂un j ∂un ⇀ , ∂t ∂t

ǫk

∂ 2 un j ∂ 2 un ⇀ , ∂x∂t ∂x∂t

ǫk

∂ 2 un j ∂ 2 un ⇀ , ∂t2 ∂t2

where derivatives of un are defined in weak sense (see Appendix D in Evans (1998)). Here, the ǫk convergences are weak convergences. Since ||u|| ≤ lim inf j ||un j || (see Appendix D in Evans (1998) ) (A-2) - (A-4) imply that ∂ 2 un ∈ L2 ((t1 , t2 ); L2 (a, b)). 2 ∂t

∂un ∈ L∞ ((t1 , t2 ); L2 (a, b)), ∂t

Then it follows from Lemma A-1 that the derivative ∂t un exists and is inside the space C((t1 , t2 ); L2 (a, b)). On the other hand, for fixed t ∈ [t1 , t2 ], it also follows from (A-2) and (A-4) and the Sobolev Embedding Theorem (see, for example, Theorem 4 in page 266 of Evans (1998)) that ∂un ∂u n ≤ C|x − x¯|1/2 , x, x¯ ∈ (a, b), (A-7) (x, t) − (¯ x , t) ∂t ∂t in which C is a positive constant that does not depend on t. We already know that ∂t un (·, t) is a continuous map with respect to t, therefore (A-7) implies that ∂un ∈ C((a, b) × (t1 , t2 )). ∂t As a result, we have

∂un ∂un (x, t0 ) = lim− (b(t0 ), t) = 0, x↓b(t0 ) ∂t t→t0 ∂t lim

(A-8)

because (bn (t0 ), t) is inside the stopping region for t < t0 , since bn (t) is decreasing.

A.2

Proof of Lemma 4.1

We will only prove the statement for u, the statement for un can be proved similarly using Lemma 2.2. We will first establish a one to one correspondence between solutions of (4.2) and solutions of a nonlinear integral equation of Volterra type. 31

Lemma A-2. (i) Let G(x, t; y, s) be the Green function associated to the differential operator LD and let us consider the following nonlinear integral equation of Volterra type, Z t 2 X 3 v(t) = − ds v(s)∂x G(b(t), t; b(s), s) + Ni (t), t0 ≤ t ≤ T0 , (A-9) 2 t0 i=1 R +∞ Rt R +∞ where N1 (t) = b(t) dy ∂x G(b(t), t; y, t0 )w(y, t0) and N2 (t) = t0 ds b(s) dy ∂x G(b(t), t; y, s)h(y, s). There exists a unique solution v to (A-9). The function v(t) is continuous. (ii) Let w(x, t) be a classical solution of the equation (4.2) on [t0 , T0 ] with the initial condition w(x, t0 ) = ∂t u(x, t0 ), such that t → ∂x w(b(t)+, t) is continuous. Then there is a one to one correspondence between w(x, t) and v(t). Moreover ∂x w(b(t)+, t) = v(t), t0 ≤ t ≤ T0 . The initial value of equation (4.2) may not be smooth. This is the reason we take w(x, t0 ) = ∂t u(x, t0 ), 0 < t0 < T0 , as the initial condition of (4.2) and consider the differential equation on t ∈ [t0 , T0 ].

Remark A-1. The correspondence in Lemma A-2 is well known for the Stefan problem on heat equation with Lipschitz continuous free boundary (see Section 1 Chapter 8 of Friedman (1964)). Along Friedman’s line of proof, we will extend the correspondence to our parabolic differential equation with H¨older continuous free boundary. Proof of Lemma A-2. Proof of (i). First, because ∂x G(b(t), t; b(s), s) is continuous for s ∈ (0, t), it follows from the classical result on Volterra equations (see Rust (1934)) that the integral equation (A-9) has a unique solution v(t) and it is continuous with respect to t ∈ [t0 , T0 ], as long as Ni (t), i = 1, 2, are continuous with respect to t. It is not hard to show these functions are indeed continuous, using the continuity of b(t) and the following estimates on the Green function G and its derivatives: |∂xℓ G(x, t; y, s)|

− 1+ℓ 2

≤ C(t − s)

|x − y|2 exp −c t−s 



,

  |x − y|2 , exp −c t−s   2+α |x′′ − y|2 |∂x G(x, t; y, s) − ∂x˜ G(˜ x, t; y, s)| ≤ C|x − x ˜|α (t − s)− 2 exp −c , t−s α

|∂x G(x, t; y, s) − ∂x G(x, t˜; y, s)| ≤ C(t − t˜) 2 (t˜ − s)−

2+α 2

where ℓ = 0, 1, s < t˜ < t, |x′′ − y| = |x − y| ∧ |˜ x − y|, 0 < α < 1, C and c are positive constants. These estimates are from Theorem 16.3 in page 413 of Ladyzhenskaja et al. (1968). Proof of (ii) Let us assume that w(x, t) is a classical solution of (4.2). As a result, the following Green’s identity is satisfied   ∂ ∂ ∂ ∂ G(x, t; y, s) w(y, s) − w(y, s) G(x, t; y, s) − (G(x, t; y, s)w(y, s)) ∂y ∂y ∂y ∂s   2µ ∂ + −1 (G(x, t; y, s)w(y, s)) = −G(x, t; y, s)h(y, s), 2 σ ∂y (A-10) 32

where t0 ≤ s < t ≤ T0 , x > b(t) and y > b(s). Integrating both hand side of (A-10) over the domain b(s) < y < +∞, t0 < s < t − ǫ, we obtain Z

t−ǫ

ds lim ∂y w(y, s)G(x, t; y, s) − y→+∞

t0

− −

t−ǫ

Z

t−ǫ

Z

ds ∂y w(b(s)+, s)G(x, t; b(s), s)

t0

ds lim w(y, s)∂y G(x, t; y, s) + y→+∞

t Z 0+∞

t−ǫ

Z

ds w(b(s), s)∂y G(x, t; b(s), s)

t0

dy [G(x, t; y, t − ǫ)w(y, t − ǫ) − G(x, t; y, t0 )w(y, t0)]   Z t−ǫ   2µ + −1 ds lim w(y, s)G(x, t; y, s) − w(b(s), s)G(x, t; b(s), s) y→+∞ σ2 t0 Z t−ǫ Z +∞ =− ds dy G(s, t; y, s)h(y, s). (A-11) b(t−ǫ)

t0

b(s)

In the fifth term on the left of (A-11), we used w(x, t) = 0 when x < b(t). Using the boundary and initial conditions for w(x, t) and the facts that limy→+∞ G(x, t; y, s) = 0 and limy→+∞ ∂y G(x, t; y, s) = 0, letting ǫ → 0, we can write w(x, t) = − +

Z

t

t0 Z t t0

ds ∂x w(b(s)+, s)G(x, t; b(s), s) +

Z

+∞

dy G(x, t; y, t0)w(y, t0)

b(t)

ds

Z

+∞

dy G(x, t; y, s)h(y, s)

b(s)

, −M0 (x, t) + M1 (x, t) + M2 (x, t).

(A-12)

Before differentiating both sides of (A-12) with respect to x, let us recall the jump identity: if ρ(t), t0 ≤ t ≤ T0 , is a continuous function and b(t) is the H¨older continuous with H¨older exponent α > 21 , then for every t0 ≤ t ≤ T0 , ∂ lim x↓b(t) ∂x

Z

t t0

1 ds ρ(s)G(x, t; b(s), s) = ρ(t) + 2

Z

t

t0

ds ρ(s) ∂x G(x, t; b(s), s)|x=b(t) .

(A-13)

This identity can be proved in the similar way as in Lemma 1 in Chapter 8 of Friedman (1964). As commented in the paragraph after Lemma 4.5 in Friedman (1975), the proof of Lemma 1 can go through when we replace Lipschitz free boundary with H¨older continuous free boundary with H¨older exponent α > 21 . Now we will take the derivative of (A-12) with respect to x to obtain 2

X ∂ ∂ w(x, t) = Mi (x, t) ∂x ∂x i=0 33

(A-14)

and let x ↓ b(t). Since ∂x w(b(s)+, s), t0 ≤ s < t, is continuous and b(t) is H¨older continuous with exponent α > 12 (see Theorem 3.1), taking ρ(s) = ∂x w(b(s)+, s) in (A-13) we obtain Z t ∂ ∂ ds ∂x w(b(s)+, s)G(x, t; b(s), s) M0 (t) = lim lim x↓b(t) ∂x t0 x↓b(t) ∂x Z t 1 = ∂x w(b(t)+, t) + ds ∂x w(b(s)+, s)∂x G(b(t), t; b(s), s). (A-15) 2 t0 On the other hand, by Lemma 2.2, w(y, t) and h(y, s) are bounded in R × [t0 , T0 ]. Using the Dominated Convergence Theorem we get Z +∞ ∂ M1 (x, t) = dy ∂x G(b(t), t; y, t0)w(y, t0) , N1 (t), (A-16) lim x↓b(t) ∂x b(t) Z t Z +∞ ∂ M2 (x, t) = ds lim dy ∂x G(b(t), t; y, s)h(y, s) , N2 (t), (A-17) x↓b(t) ∂x t0 b(s) It follows from (A-14) - (A-17) that ∂x w(b(t)+, t) satisfies (A-9). Let us prove the converse. For any solution v(t) of the integral equation (A-9), we can define w(x, t) as follows Z +∞ Z t ds v(s)G(x, t; b(s), s) + w(x, t) := − dy G(x, t; y, t0 )w(y, t0) +

Z

b(t)

t0 t

t0

ds

Z

+∞

dy G(x, t; y, s)h(y, s),

b(s)

t0 ≤ t ≤ T0 , x ≥ b(t),

(A-18)

and w(x, t0 ) := ∂t u(x, t0 ). We will show in the following that w(x, t) is a classical solution of (4.2) and that t → ∂x w(b(t)+, t) is continuous. Now we will show that w(x, t) defined in (A-18) is a classical solution of (4.2) on [t0 , T0 ] with initial condition ∂t u(x, t0 ). By definition w(x, t0 ) = ∂t u(x, t0 ). On the other hand we have that limx→+∞ w(x, t) = 0, which follows from the facts that limx→+∞ G(x, t; y, t0 ) = 0 and v(s), w(y, t0) and h(y, s) are all bounded. Furthermore, using the properties of the Green function and the definition of w (see A-18), we also have that LD w(x, t) = h(x, t) for x > b(t), t ∈ [t0 , T0 ]. Observe that ∂t w , ∂x w and ∂x2 w all exist and are all continuous in this domain. In the following we will show that ∂x w(b(t)+, t) = v(t), which implies the continuity of ∂x w(b(t)+, t). We differentiate w(x, t) with respect to x and let x ↓ b(t). Since v(t) is continuous and b(t) is H¨older continuous with exponent α > 21 , we can apply the jump identity (A-13) with ρ(s) = v(s). Following the steps that lead to (A-9) in the first part of the proof, we obtain Z t 2 X 1 ∂x w(b(t)+, t) = − v(t) − ds v(s)∂x G(b(t), t; b(s), s) + Ni (t). (A-19) 2 t0 i=1 34

Comparing (A-19) to (A-9), we see that ∂x w(b(t)+, t) = v(t), t0 ≤ t ≤ T0 . Then it remains to show that w(b(t), t) = 0, t0 ≤ t ≤ T0 . To this end, since we have already shown LD w = h, w satisfies the Green’s identity given by (A-10). Integrating the identity (A-10) and using (A-18) and the fact that limx→+∞ w(x, t) = 0 we can write     Z t 2µ − 1 G(x, t; b(s), s) = 0, x > b(t), t0 ≤ t ≤ T0 . ds w(b(s), s) ∂y G(x, t; b(s), s) − σ2 t0 (A-20) Let x > b(t). Integrating both sides of (A-20) on [x, +∞) and using the fact that ∂x G = −∂y G, we obtain  Z +∞   Z +∞  Z t 2µ 0 = ds w(b(s), s) − du ∂x G(u, t; b(s), s) − −1 du G(u, t; b(s), s) σ2 t0 x x  Z +∞    Z t 2µ −1 du G(u, t; b(s), s) . = ds w(b(s), s) G(x, t; b(s), s) − σ2 x t0 Taking the derivative with respect to x, letting x ↓ b(t) and using the jump identity (A-13) with ρ(s) = w(b(s), s), we arrive at     Z t 2µ 1 w(b(t), t) = ds w(b(s), s) ∂y G(b(t), t; b(s), s) − − 1 G(b(t), t; b(s), s) . 2 σ2 t0 (A-21) Since b(t) is H¨older continuous with exponent α > 1/2, we have |∂y G(b(t), t; b(s), s)| ≤

C 3

(t − s) 2 −α

.

Therefore both ∂y G(b(t), t; b(s), s) and G(b(t), t; b(s), s) are integrable. Consequently, it follows from (A-20), (A-21) and the Dominated Convergence Theorem that w(b(t), t) = 0, t0 ≤ t ≤ T0 .  Proof of Lemma 4.1. Proof of (i). Let v(t) be the unique continuous solution of the Volterra equation (A-9). Define w(x, t) as in (A-18). The Lemma A-2 shows that w(x, t) is a classical solution to equation (4.2). Let us define Z t u˜(x, t) = u(x, t0 ) + w(x, s)ds, x ≥ b(t), t0 ≤ t ≤ T0 . t0

It is easy to check that u˜(x, t) is a classical solution of the equation (2.6) - (2.9) with initial condition u(x, t0 ). Since (2.6) - (2.9) has a unique solution, we conclude that u(x, t) = u˜(x, t), x ≥ b(t) and t0 ≤ t ≤ T0 . Lemma A-2 also implies that ∂x ∂t u(b(t)+, t) = ∂x w(b(t)+, t) = v(t), 35

t0 ≤ t ≤ T0 ,

which implies that ∂x ∂t u(b(t)+, t), t0 ≤ t ≤ T0 , is continuous. The statement follows since t0 > 0 is arbitrary. ProofR of (ii). Let (x, t) be such that x > b(t). Choosing t0 < t such that b(t0 ) < x, we can see t that t0 ds∂x G(x, t; b(s), s) < +∞. As a result, we have ∂ M0 (x, t) = − ∂x

Z

t

ds ∂x w(b(s)+, s)∂x G(x, t; b(s), s).

t0

We have shown in part (i) that ∂x w(b(s)+, s) is continuous with respect to s. It is easy to show ∂x M0 (x, t) is continuous around a sufficiently small neighborhood of (x, t). One can also show that the functions ∂x Mi (x, t), i ∈ {1, 2} are also continuous by similar means. Thus, it is clear from (A-14) that ∂x ∂t u(x, t) is continuous in this small neighborhood around (x, t). Therefore, the part (ii) of Lemma 4.1 follows, because of the arbitrary choice of x and t.

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