ANALYSIS AND NUMERICAL SOLUTION OF AN INVERSE FIRST PASSAGE PROBLEM FROM RISK MANAGEMENT. by Lan Cheng MA, University of Pittsburgh, 2004

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1 ANALYSIS AND NUMERICAL SOLUTION OF AN INVERSE FIRST PASSAGE PROBLEM FROM RISK MANAGEMENT by Lan Cheng MA, University of Pittsburgh, 24 Submitted to the Graduate Faculty of the Department of Mathematics in partial fulfillment of the requirements for the degree of Doctor of Philosophy University of Pittsburgh 25

2 UNIVERSITY OF PITTSBURGH MATHEMATICS DEPARTMENT This dissertation was presented by Lan Cheng It was defended on October 2th 25 and approved by John Chadam, Mathematics Department Xinfu Chen, Mathematics Department David Saunders, Mathematics Department Ivan Yotov, Mathematics Department Soiliou Namoro, Economics Department Dissertation Advisors: John Chadam, Mathematics Department, Xinfu Chen, Mathematics Department ii

3 ANALYSIS AND NUMERICAL SOLUTION OF AN INVERSE FIRST PASSAGE PROBLEM FROM RISK MANAGEMENT Lan Cheng, PhD University of Pittsburgh, 25 We study the following inverse first passage time problem. Given a diffusion process X t and a probability distribution q(t) on [, ), does there exist a boundary b(t) such that q(t) = P[τ t], where τ is the first hitting time of X t to the time dependent level b(t). We formulate the inverse first passage time problem into a free boundary problem for a parabolic partial differential operator and prove there exists a unique viscosity solution to the associated Partial Differential Equation by using the classical penalization technique. In order to compute the free boundary with a given default probability distribution, we investigate the small time behavior of the boundary b(t), presenting both upper and lower bounds first. Then we derive some integral equations characterizing the boundary. Finally we apply Newton-iteration on one of them to compute the boundary. Also we compare our numerical scheme with some other existing ones. iii

4 ACKNOWLEDGEMENTS I wish to thank to my advisors, Professor J. Chadam and X. Chen for their continuous help and contributions of their great ideas for solving the problem. Their encouragements during the inevitable ups and downs of my research work proved to be priceless. I wish to thank to the members of the Dissertation Committee for their comments and suggestions. A special thank D. Saunders for his many hours spent with discussions on the topic and for his careful reading of the manuscript. I wish to thank to R. Stamicar for bringing the thesis topic to my attention. I wish to thank to A. Kreinin for his comments when he participanted in the Math Finance Session of the AMS conference. Also I am indebt to the Department of Mathematics, for its support and hospitality. iv

5 TABLE OF CONTENTS 1. INTRODUCTION VISCOSITY SOLUTIONS AND UNIQUENESS Preliminaries Viscosity solutions Uniqueness EXISTENCE OF A VISCOSITY SOLUTION The Regularization Continuity Estimates and Existence The Differential Equation and the Free Boundary Problem ESTIMATION OF THE FREE BOUNDARY Upper Bounds Lower Bounds Estimation of the Free Boundary INTEGRAL EQUATIONS Derivation of the equations Solutions to the integral equations within finite time Estimation of free boundary by integral equations A NUMERICAL SCHEME Proposed Numerical Scheme Integral Equation Avellaneda-Zhu s Scheme An Alternative Explicit Scheme v

6 6.5 Conditional default probability NUMERICAL SIMULATION Linear boundary Our scheme vs Avellanda and Zhu s Our scheme VS Iscoe and Kreinin s Our scheme vs Peskir, Sacerdaote and Zucca s Example BIBLIOGRAPHY vi

7 1. INTRODUCTION In this thesis we study the following free boundary problem: find a boundary x = b(t) (t > ) and an unknown function w = w(x, t) (x R, t ) such that w t (x, t) = 1 2 (σ2 w x ) x µw x for x > b(t), t >, w(x, t) = p(t) for x b(t), t >, w x (b(t), t) = for x = b(t), t >, w(x, ) = 1 (,) (x) for x R, t =, (1..1) where p(t) is a given survival probability function with the following properties: 1 = p() = lim t p(t), p(t 1 ) p(t 2 ) < t 1 t 2. (1..2) This problem arises from the consideration of the first passage times of diffusion processes to curved boundaries. More specifically, we let X t be the solution of the following stochastic differential equation: dx t = µ(x t, t)dt + σ(x t, t)db t X =, (1..3) where B t is a standard Brownian motion on a filtered probability space satisfying the usual conditions, µ : R R + R and σ : R R + R are smooth bounded functions, σ(x, t) > ε > for all x R, t. For a given function b : R + R we define the first passage time of the diffusion process X t to the curved boundary b(t) to be: τ = inf{t > X t b(t)}. (1..4) Two important problems concerning the first passage time of a diffusion process to a curved boundary are the following: 1

8 1. The first passage problem: Given a boundary function b(t), find the survival probability p(t) that X t does not cross b before or at t. p(t) := P{τ > t}. (1..5) 2. The inverse first passage problem: Given a survival probability function p(t), find a boundary function b(t), such that (1..5) holds. The first passage problem is a classical problem in probability, and is the subject of a rather large literature. It is also fundamental in many applications of diffusion processes to engineering, physics, biology and economics. For a survey of techniques for approximating and computing first passage times to curved boundaries, and a discussion of their applications in the biological sciences, we refer to [14]. For some applications in economics closely related to those that motivated this study (for example, credit protection) we refer to [1]. The work of Peskir [12] and [13] on the first passage problem is of particular relevance for the inverse problem discussed in this paper. In [12], he derived a sequence of integral equations 1 where t n/2 H n ( b(t) t ) + ( ) b(t) b(s) (t s) n/2 H n ṗ(s)ds =, n = 1,, 1,. (1..6) t s H 1 (x) = 1 2π e x2 /2, H n (x) = x H n 1 (z)dz, n =, 1, 2,. In [13], under the assumption that b(t) is C 1 on (, ), decreasing, and concave, he derived the equality 1 ṗ(+) = lim t 2 b(t) b 2 (t) e 2t 2π t3/2 provided that the second or third limit exists. = lim t ḃ(t) 2πt e b2 (t) 2t, The inverse first passage problem is much harder than the direct problem and there are only a few studies about it. These are principally concerned with the numerical calculation 1 In this reference, the derivations are carried out for the case σ 1 and µ, i.e. when X t is a Brownian motion. As mentioned in the reference, the techniques directly extend to other diffusion processes. 2

9 of the boundary b(t) for a given p(t). There is no publication proving the well-posedness (existence and uniqueness) of the boundary given the survival probability. Our interest in the inverse first passage problem originates from Merton s structural model [11] for credit risk management. Consider a company whose asset value is a stochastic process and its debt value is a time depending function. Denote them at time t by A t and D t respectively. Assume the following: 1. The company s initial debt is no larger than it s initial asset value, i.e., D A 2. The company is in default at a time t > if A t D t. 3. A t follows a geometric Brownian motion. It is convenient to use the default index X t and the boundary function b(t) defined by X t := log A t A, b(t) := log D t A. Then X t is a diffusion process satisfying (1..3). In this context, the inverse first passage time problem is the problem of finding the default boundary b(t) given the survival probability function p(t). In deed the default probability q(t) := 1 p(t) of the company can be estimated from the spreads of the bond issued by the company. If the company wants to get some protection from default, then it is very important to know the threshold of the debt value being in default. A free boundary problem for a parabolic partial differential operator is associated with the inverse first-passage problem. In order to formulate the problem in a PDE setting, we introduce a new function w(x, t) being the joint probability that the company does not default before or at t and its default index X t is bigger than x, i.e., w(x, t) := P{X t > x, τ > t}. (1..7) Then the density function of X t when τ > t can be computed by u(x, t) = d dx P{X(t) x, τ > t} = (p(t) w(x, t)) x. (1..8) From (1..3) and the Kolmogorov forward equation, we see that (assuming sufficient regularity) w(x, t) (x R, t ) satisfies (1..1). From this we see the following: 3

10 The first passage problem is to solve (1..1) for p, with given b. The inverse first passage problem is to solve (1..1) for b, with given p. The first passage problem can be solved as follows. From the Kolmogorov forward equation, we obtain the following closed system for u(x, t) u t (x, t) = 1 2 (σ2 u) xx (µu) x for x > b(t), t >, u(b(t), t) = for x b(t), t >, (1..9) u(x, ) = δ(x) for x >, t =, where δ is a Dirac measure concentrated at. Given sufficiently regular b, this system has a unique solution. Then p and ṗ can be computed from the formulas p(t) = b(t) u(x, t)dx t, (1..1) ṗ(t) = 1 2 (σ2 u) x=b(t) x t. (1..11) It is only possible to compute the solution in a closed form in a few special cases. However, there is a large literature on numerical and analytic approximations of the solution. Avellaneda and Zhu [7] were the first to use (1..9) and (1..11) to study the inverse firstpassage problem. They performed a change of variables from X t to Y t = X t b(t), whose financial meaning is the risk-neutral distance-to-default process (RNDD) for the company. Denote by f(y, t) = u(y + b(t), t), the probability density function of Y t when τ > t. Then (1..9) and (1..11) are equivalent to: f t = ḃ(t)f y (µf) y (σ2 f y ) y for y >, t >, f(, t) = for y =, t >, f(y, ) = δ (y b()) for y >, t =, 1 2 σ2 f y (, t) + ṗ(t) = for y =, t >. (1..12) Zucca, Sacerdote and Peskir [15] applied secant method to one of the integral equation (1..6), derived by Peskir [12], with n =. Also they proposed a Monte Carlo algorithm in the same paper, based on a piecewise linear approximation of the boundary. 4

11 In [1], Iscoe and Kreinin reduced the inverse first-passage problem to a sequential estimation of conditional distributions. They applied a Monte Carlo approach to it in a discrete time setting. All the numerical schemes mentioned above will be discussed later in more details and we will do the comparison of all the schemes. In the thesis, we are particularly interested in the following fundamental questions: (1) Given a probability function p(t) satisfying (1..2), does there exist a boundary function b(t)? (2) If there exists a boundary function, how many are there? (3) If there exists a boundary function, how can we compute it numerically? Namely, we are concerned about the well-posedness (existence and uniqueness) and numerical solution of the free boundary problem (1..1). We point out that solutions to (1..1) are not smooth, so that a notion of weak solution has to be used. Instead of using the classical weak solution defined in the distributional sense (see Evans [2]), we use viscosity solutions, introduced by Crandall and Lions [8] in In the thesis, we shall prove the following theorem. Theorem 1. Problem (1..1) is a well-posed problem, i.e., for any given p(t) satisfying (1..2), there exists a unique (weak) solution. The thesis is organized as follows. In Chapter. 2, we provide a definition of the viscosity solution to (1..1) and show there is at most one such solution. In Chapter. 3, we establish the existence of a viscosity solution. In Chapter. 4, we study the asymptotic behavior of the boundary as t by providing explicit upper and lower bounds. When lim sup t 1 p(t) tṗ(t) <, we prove that lim t b(t) 2t log(1 p(t)) = 1. In Chapter. 5, we derive the integral equations for b when σ 1 and µ under the assumption that p is continuous and non-increasing. In Chapter. 6, we proposed our numerical algorithem and introduced the one by Zucca, Sacerdote and Peskir [15], Avellaneda and Zhu [7] and Iscoe and Kreinin [1]. In Chapter. 7, we presented the numerical results, default boundary b, of both the schemes published and ours with the different probability functions. Also we compared the computing speed and the acuracy of all the schemes. 5

12 2. VISCOSITY SOLUTIONS AND UNIQUENESS 2.1 PRELIMINARIES By noticing that w(x, t) < p(t) for all x > b(t) when τ > t, we can state the inverse first passage problem as follows. Find an unknown function w = w(x, t) such that, Lw = when w(, t) < p(t), w(x, t) p(t) for any (x, t) (R (, )), (2.1.1) w(x, ) = 1 (,) (x) for (x, t) (R [, )), where Lw := w t 1 2 (σ2 w x ) x + µw x. Define the free boundary as: b w (t) := inf {x w(x, t) < p(t)}. Noticing that Lw = when w < p and Lw = ṗ when w = p, we can write (2.1.1) as a variational inequality: max{lw, w p} = in R (, ), w(, ) = 1 (,) ( ) on R {}. (2.1.2) For a given p, we define p (t) = lim inf s t p(s), p (t) = lim sup p(s) t. s t Since cumulative probability distribution functions (hence 1 p) are increasing and right continuous, p should be decreasing and right continuous. 6

13 Lemma For any given b(t), p(t) := P{τ > t} is decreasing and right continuous. In particular, p = p. Proof. Denote by A(t) the set of paths whose default time is bigger than t, i.e., A(t) : = {ω τ(ω) > t} = {ω inf{s > X s (ω) b(s)} > t}. Then p(t) = P(A(t)). First we claim that A(t 1 ) A(t 2 ) < t 2 < t 1. (2.1.3) Indeed for any ω A(t 1 ), since inf{s > X s (ω) b(s)} > t 1 > t 2, then ω A(t 2 ). So that (2.1.3) holds and it is followed that p is a decreasing function. Next we prove that p(t) = p (t) for any t >. Let t n := t + 1 (n N), with the above argument, {A(t n n)} is a non-increasing set, i.e., A(t 1 ) A(t 2 ) A(t n ). Then by the property of probability, we have p(t) = P(A(t)) = P( lim n A(t n )) = lim n P (A(t n )) = lim n p(t n ) = lim s t p(t) = p (t). Hence p = p and it follows that p is right continuous function since it is also decreasing. Furthermore, Blumenthal s zero-one law (see, for example [5]) implies that we must have either p() = (in which case the problem is trivial) or p() = 1. Therefore, in the remainder of the thesis, we shall only consider lower semicontinuous p, i.e., p = p for which p() = 1. 7

14 2.2 VISCOSITY SOLUTIONS For a function w defined on R [, ), we define w and w by w (x, t) := lim sup y x, s t w(y, s), w (x, t) := lim inf w(y, s), y x, s t (x, t) R [, ), (x, t) R [, ). A function w is called upper-semi-continuous (USC) if w = w, and lower-semicontinuous (LSC) if w = w. In the sequel, the parabolic open ball B δ (x, t) is defined as: B δ (x, t) := (x δ, x + δ) (t δ 2, t) δ >, (x, t) R [, ). For any cylindrical set of the form D := Ω (s, t) where s < t and Ω R, the parabolic boundary is defined to be: p D := Ω (s, t) Ω {s} Definition 1 (Viscosity Sub, Super, and Solutions). 1. A function w defined on R (, ) is called a (viscosity) subsolution if w = min{p, w } in R (, ), and Lϕ(x, t) whenever ϕ is smooth and w ϕ attains at (x, t) a local maximum on B δ (x, t), where x R and t > δ 2 >. 2. A function w defined on R (, ) is called a (viscosity) supersolution if w = w in R (, ), and max{w(x, t) p(t), Lϕ(x, t)} whenever ϕ is smooth and w ϕ attains at (x, t) a local minimum on B δ (x, t), where x R and t > δ 2 >. 3. A function w defined on R [, ) is called a (viscosity) solution if w is both a subsolution and a supersolution in R (, ), and for all x R, w(x, ) = lim inf y x,t w(y, t) = 1 (,), lim sup w(y, t) = 1 (,]. (2.2.1) y x,t 8

15 Remark Here we use the default that a viscosity solution is LSC, i.e., w = w (Given any point (x, t) R [, ), if x > b(t), then w(x, t) is continuous, hence is LSC. If x b(t), then τ > t implies that X t > x, i.e., P(X t > x, τ > t) = p(t). As p = p, w = w ). Also, the (probabilistically obvious) condition w imposed for super-solutions is to ensure the boundedness of the super-solution, as is usually required. This condition could be relaxed to the assumption that w e A(1+ x 2) for some A >. To prove the uniqueness of the solution to (2.1.2), we first establish a few properties of viscosity solutions. Lemma Let w be a viscosity solution and define Q := {(x, t) R (, ) w(x, t) < p(t)}, Π := Q c = R (, ) \ Q. Then 1. Q is open and w is a smooth solution to Lw = in Q; 2. Π = {(x, t) R (, ) w(x, t) = p(t)} = Π Π 1 Π 2 where Π : = {(x, t) R (, ) w (x, t) = w (x, t) = p(t)}, Π 1 : = {(x, t) R (, ) p (t) > w (x, t) > w (x, t) = p(t)}, Π 2 : = {(x, t) R (, ) p (t) = w (x, t) > w (x, t) = p(t)}. In particular, if p is continuous, then w is continuous in R (, ). Proof. 1. First we show that Q is open and w is continuous in Q. For each (x, t) Q with t >, w(x, t) < p(t). As a supersolution, w(x, t) = w (x, t). As a subsolution, w(x, t) = min{p(t), w (x, t)} < p(t), which implies that w(x, t) = min{p(t), w (x, t)} = w (x, t). Hence w = w = w at (x, t). That is w is continuous at (x, t) and w(x, t) < p(t). Since p is right continuous and decreasing, there exists δ 1 > such that w < p in (x δ 1, x + δ 1 ) (t, t + δ1). 2 As lim sup y x,s t w(y, s) = w (x, t) = w(x, t) < p(t), there exists δ 2 > such that w(y, s) < p(t) < p(s) (y, s) (x δ 2, x + δ 2 ) (t δ 2, t). Then for any (x, t) Q and t >, there exists an open set D δ (x, t) := (x δ, x + δ) (t δ 2, t + δ 2 ) in Q, where δ = min{δ 1, δ 2 } >. Hence, Q is open and w is continuous in Q. 9

16 2. Next we prove Lw = in Q. Let (x, t ) Q with t >. Then w is continuous at (x, t ) and w(x, t ) < p(t ). With the previous argument, there exists D Q such that w is coutinuous and w < p in D. Denote by w the solution to L w = for (x, t) D, w = w for (x, t) p D. Note that w is smooth in D since the boundary and initial condition are continuous. Let (2.2.2) ϕ ε = w ε t + δ 2 t, ε ψε = w + t + δ 2 t ε >. Then ϕ ε and ψ ε are smooth in D (by interior regularity for PDE [2]). Note that w ϕ ε can attain its minimum on D only at the parabolic boundary. To the contrary, suppose this is not true. Since w ϕ ε as t t + δ 2, we assume that the minimum is attained at a interior point of D, say (x, t ). As a supersolution, max{w(x, t ) p(t ), Lϕ ε (x, t )}. Since w p < in D, Lϕ ε (x, t ). However Lϕ ε (x, t ) = L w(x, t ε ) L t + δ 2 t t=t = ε (t + δ 2 t ) <. 2 This is a contradiction. So that ( ) min(w ϕ ε ε ) = min w w + >. D pd t + δ 2 t Thus ϕ ε < w in D. Also note that w ψ ε can attain its maximum on D only at the parabolic boundary. To the contrary, suppose this is not true. Since w ψ ε as t t + δ 2, we assume that the maximum is attained at a interior point of D, say (x, t ). As a subsolution, Lψ ε (x, t ). However Lψ ε (x, t ) = L w(x, t ε ) + L t + δ 2 t t=t = ε (t + δ 2 t ) >. 2 This is a contradiction. So that ( ) max(w ψ ε ε ) = max w w <. D pd t + δ 2 t 1

17 Thus w < ψ ε in D. So that we have w ε t + δ 2 t < w < w + ε t + δ 2 t. Sending ε we obtain w = w in D, which implies that w is a continuous solution to Lw = in Q. 3. Lastly we prove the second assertion of the lemma. Since w p, Π := Q c = {(x, t) R (, ) w(x, t) = p(t)}. For any t >, as a subsolution, p = w = min{p, w } w. As a supersolution w = w. Also w p implies that w p. Thus w = w = p w p, in Π. There are only three possibilities for w : (i) w = p, (ii) w (p, p ) and (iii) w = p > p. Thus the second assertion holds. 4. In particular, if p is continuous, i.e., p = p = p, then Π = Π. That is w = w = w in Π. Hence w is continuous in Π. It follows that w is continuous at R (, ) \ {(, )} The following Lemma characterizes the discontinuities of a solution. Lemma Suppose w is a viscosity solution. Then for each t >, the following hold: 1. w(, t) = w (, t) is continuous in R; 2. for each x R, w (x, t) = min{p(t), w (x, t)} = lim w(y, s), (2.2.3) y x,s t w (x, t) = lim y x,s t w(y, s) p (t); (2.2.4) 3. if w (x, t) < p (t), then for some δ >, w = w in B δ (x, t) and w is a smooth solution to Lw = in B δ (x, t). 11

18 Proof. 1. First we prove the first assertion. For each t >, since w is a supersolution, w(, t) = w (, t). If (x, t) Q, then w(x, t) < p(t) and w is continuous at (x, t) by Lemma (2.2.1). If (x, t) Π, then w(x, t) = p(t). So that we have Since w p, w (x, t) = lim inf w(y, s) lim inf w(y, t). y x,s t y x lim sup y x w(y, t) lim sup p(t) = p(t). y x By using the fact of w is a supersolution and the above inequalities, lim sup w(y, t) p(t) = w(x, t) = w (x, t) lim inf w(y, t). y x y x So that w(, t) = w (, t) is continuous in R. The first assertion follows. 2. Next we prove (2.2.3). For each x R, the first equality is immediate since w is both a subsolution and a supersolution. We prove the second inequality follows by considering separately the cases (x, t) Q and (x, t) Π as in the previous step. If (x, t) Q, then w is continuous at (x, t). So that w (x, t) = min{p(t), w (x, t)} = lim y x,s t w(y, s). If (x, t) Π, then w (x, t) = w(x, t) = p(t) where the first inequality holds since w p so that lim w(y, s) lim inf w(y, s) = w (x, t), y x,s t y x,s t Thus (2.2.3) holds. p(t) = p (t) = lim s t p(t) = lim p(t) y x,s t lim w(y, s). y x,s t 3. Now we prove (2.2.4) when w (x, t) < p (t) and the third assertion. By the upper semicontinuity of w, there exist some positive constants δ and η such that w(, ) < p (t) η in B δ (x, t). (2.2.5) 12

19 Then we claim that w = w in B δ (x, t) p B δ (x, t). (2.2.6) To the contrary, suppose this is not true, i.e., there exists at least one pair of (y, s) B δ (x, t) p B δ (x, t) such that w(y, s) < w (y, s). As a subsolution, w(y, s) = min{p (s), w (y, s)} < w (y, s). Hence p (s) = w(y, s) < p (t) η p (s) η, where the second inequality follows from (2.2.5) and the last inequality holds since p is nonincreasing and right continuous, which implies that p (t) p(s) = p (s) for all s < t. This is a contradiction. So that (2.2.6) holds and w (y, s) = w(y, s) < p (t) η < p (t) (y, s) B δ (x, t). Send y x, s t, we obtain (2.2.4). Next we prove that w is a smooth solution to Lw = in B δ (x, t). As supersolution w = w and by (2.2.6) w = w on p B δ (x, t), hence w is continuous on p B δ (x, t). Also, for any (y, s) p B δ (x, t), w(y, s) < p (t) η p(s) η <, which implies that w(y, s) < p(s). Denote by w the solution to L w = for (x, t) B δ (x, t), w = w for (x, t) p B δ (x, t). Following the same proof as previous lemma, we can show that w w in B δ (x, t). Hence w = w is a smooth solution to Lw = in B δ (x, t). The third assertion and (2.2.4) for the case w (x, t) < p (t) thus follow. 13

20 4. Finally we verify (2.2.4) for the case w (x, t) = p (t). For each small δ >, we compare w in B δ (x, t) with solutions w and w to L w = in B δ, w = w on p B δ, and Lw = in B δ, w = min{w, p (t)} on p B δ, respectively. Note that on p B δ, w w = w and w = min{w, p (t)} min{w, p} = w since p (t) p(s) for any s < t. Simple comparison gives w w w in B δ. By maximum principle, max Bδ w = max pb δ w = max Bδ w max t δ 2 s<t p p (t δ 2 ), min Bδ w = min pb δ w p (t). Then, lim sup w(y, s) lim inf w(y, s) max { w w} p (t δ 2 ) p (t). y x,s t y x,s t B δ (x,t) Send δ, we conclude that lim y x,s t w(y, s) exists. Now need to show that lim w(y, s) = y x,s t w (x, t). (2.2.7) We show it in the following two cases: (i) Suppose w (x, t) = w (x, t). That is w is continuous at (x, t) so that (2.2.7) follows. (ii) Suppose w ( x, t) < w (x, t). By (2.2.3), lim y x,s t w(y, s) = w (x, t). Then we must have w (x, t) lim sup w(y, s) = lim w(y, s). y x,s t y x,s t This complete the proof of (2.2.4). 14

21 2.3 UNIQUENESS Theorem 2 (Uniqueness). There is at most one viscosity solution to (2.1.2). Proof. Suppose w 1 and w 2 are two viscocity solutions to (2.1.2). For each η >, we claim that w 1 (x, t) w 2 (x η, t) (x, t) R [, ). (2.3.1) To the contrary, suppose this is not true, i.e., there exists at least one pair of (x, t ) R [, ) such that w 1 (x, t ) > w 2 (x η, t ). Then for all sufficiently small positive ε, w 1 (x, t ) > w 2 (x η, t ) + ε 4 x 2 + εe t. (2.3.2) Hence fix such a positive ε such that { } 1 ε min 1, ( σ σσ x µ ) (2.3.3) and let g ε (x, t) := w 1 (x, t) w 2 (x η, t) ε 4 x 2 εe t. Then g ε (x, t ) >, (2.3.4) g ε (x, t) p(t) 1 (x, t) R [, ). (2.3.5) So that g ε attains a supremum in R [, ), denoted by M ε := and together with (2.3.4) and (2.3.5), sup g ε (x, t), (x, t) R [, ) < M ε 1. Let {(x n, t n )} n=1 be a sequence in R [, ) such that the supremum M ε is attained along the sequence. By taking a subsequence if necessary, there exist the limits (ˆx, ˆt) := lim n (x n, t n ), α := lim n w 1 (x n, t n ), β := lim n w 2 (x n η, t n ). 15

22 Consequently M ε = lim n g ε (x n, t n ) = α β ε 4ˆx 2 εeˆt. (2.3.6) Since w 1 (, ) 1 and w 2 ( η, ) 1, α 1, β 1. Also α β = lim w 1 (x n, t n ) lim sup w 1 (y, s) = w n 1(ˆx, ˆt); y ˆx,s ˆt = lim w 2 (x n, t n ) lim inf w 2 (y η, s) = w 2 (ˆx η, ˆt). n y ˆx,s ˆt As M ε >, and In a summary, α = M ε + β + ε 4ˆx 2 + εeˆt > β ˆx 2 = α β εeˆt M ε ε 4 α/ε 4 1/ε 4. α w 1(ˆx, ˆt), β w 2 (ˆx η, ˆt), β < α, ˆx < 1/ε 2. Now we show that this is impossible, by excluding the following three possibilities: (i) ˆt = ; (ii) ˆt >, β < p(ˆt); (iii) ˆt >, β p(ˆt). Case (i): Suppose ˆt =. If ˆx η, then ˆx >, which implies that α w 1(ˆx, ) = 1 (, ] (ˆx) =. So that β < α w 1(ˆx, ) =. This is a contradiction. If ˆx < η, then ˆx η <, which implies that β w 2 (ˆx η, ˆt) = 1 (, ) (ˆx η) = 1. So that 1 β < α 1. This is a contradiction too. Thus case (i) is impossible. 16

23 Case (ii) Suppose ˆt > and β < p(ˆt), then w 2 (ˆx η, ˆt) = w 2 (ˆx η, ˆt) β < p(ˆt). Hence (ˆx η, ˆt) Q, where Q := {(x, t) R [, ) w 2 (x, t) < p(t)} is an open set. By Lemma (2.2.1), there exist D := (ˆx δ, ˆx + δ) (ˆt δ 2, ˆt + δ 2 ) Q, where δ >, such that w 2 ( η, ) < p( ) in D and w 2 is a smooth solution to Lw 2 ( η, ) = in D. Let ϕ(x, t) = w 2 (x η, t) + ε 4 x 2 + εe t + (x ˆx) 4 /δ 4 + (t ˆt) 2 /δ 4. Then ϕ is smooth in D and max{w1 ϕ} = sup{w 1 ϕ} D D { = sup gε (x, t) (x ˆx) 4 /δ 4 (t ˆt) 2 /δ 4} D α β ε 4ˆx 2 εeˆt w 1(ˆx, ˆt) w 2 (ˆx η, ˆt) ε 4ˆx 2 εeˆt = w 1(ˆx, ˆt) ϕ(ˆx, ˆt). That is, w 1 ϕ attain at (ˆx, ˆt) a local maximum on D. As w 1 is a subsolution, Lϕ(ˆx, ˆt). However Lϕ(ˆx, ˆt) = Lw 2 (ˆx η, ˆt) + εeˆt ε 4 σ 2 2ε 4ˆx(2σσ x µ) = εeˆt ε 2 (ε 2 σ 2 ) 2ε 2 (ε 2ˆx)(2σσ x µ) ε ε 2 ( σ σσ x µ ) > by the fact that ˆx < 1/ε 2 and (2.3.3). This is a contradiction. Thus case (ii) is impossible. Case (iii): Suppose ˆt > and β p(ˆt). Since p (s) p(ˆt) for any s > ˆt, w 1 (x, s) p(s) p (s) p (ˆt) β x R, s > ˆt. So that sup w 1 (x, s) β s > ˆt. (2.3.7) x R 17

24 Now we claim t n < ˆt for all sufficiently large n, i.e., there exists N N + such that t n < ˆt for each n N. To the contrary, suppose for each N N +, there exists n > N such that t n ˆt. Then by (2.3.7), w 1 (x n, t n ) sup w 1 (x, t n ) β. x R As α > β, there exists ε >, which is independent of n, such that α > β + ε w 1 (x n, t n ) + ε. This is a contradiction to α = lim n w 1 (x n, t n ). Then by taking the subsequence if necessarily and by (2.2.4), we have α β = lim w 1 (x n, t n ) = lim w 1 (y, s) = w n 1(ˆx, ˆt) p (ˆt), y ˆx,s ˆt = lim w 2 (x n, t n ) = lim w 2 (y η, s) = w n 2(ˆx η, ˆt) p (ˆt). y ˆx,s ˆt Also w 2(ˆx η, ˆt) = β < α p (ˆt). By Lemma (3), for some δ >, w 2 = w 2 in B δ (ˆx + η, ˆt) and w 2 is a smooth solution to Lw 2 = in B δ (ˆx + η, ˆt). Let φ(x, t) := w 2(x η, t) + ε 4 x 2 + εe t + (x ˆx) 4 /δ 4 + (t ˆt) 2 /δ 4. Then, max {w1 φ} = sup {w 1 φ} B δ (ˆx,ˆt) (x,t) B δ (ˆx,ˆt) = sup {w 1 (x, t) w 2 (x η, t) ε 4 x 2 εe t (x ˆx) 4 /δ 4 (t ˆt) 2 /δ 4 } (x,t) B δ (ˆx,ˆt) α β ε 4ˆx 2 εeˆt = w 1(ˆx, ˆt) φ(ˆx, ˆt). 18

25 That is w1 φ obtains at (ˆx, ˆt) its local maximum in B δ (ˆx, ˆt). Since w 1 is a subsolution, Lφ(ˆx, ˆt). However Lφ(ˆx, ˆt) = Lw2(ˆx η, ˆt) + εeˆt ε 4 σ 2 2(σσ x 2µ)ε 4ˆx ε ε 2 σ 2 2ε 2 σσ x µ > by the fact that ˆx < 1/ε 2 and (2.3.3). This is a contradiction. Thus case (iii) is impossible. The exclusion of cases (i), (ii) and (iii) implies that (2.3.1) holds for each η >. Sending η and using Lemma (1), i.e., w(, t) is continuous in R, we conclude that w 1 w 2 on Ω. Exchanging the roles of w 1 and w 2, we also have w 2 w 1, so that w 1 w 2. As a product, (2.3.1) and the uniqueness give the following. Corollary The unique solution w, if it exists, is non-increasing in x, i.e., w(x, t) w(x η, t) for all η > and (x, t) R [, ). 19

26 3. EXISTENCE OF A VISCOSITY SOLUTION In this chapter we prove the existence of viscocity solution to (2.1.2) by establishing one. Following the classical penalization technique (see for example Friedman [4]) for variational inequalities, we define a ε-regularization of the problem carefully so that the solution is monotonic in ε and therefore the existence of a limit as ε is automatically guaranteed. For the purpose of showing that the limit is a viscosity solution, we study the regularization and then prove some regularity properties of the solution to the penalized problem, and therefor establish compactness. 3.1 THE REGULARIZATION Following the classical penalization technique for variational inequalities, we consider a semilinear parabolic equation: ( ) Lw ε = β ε 1 (w ε p ε ) in R (, ), w ε (, ) = W ε ( ) on R {}. (3.1.1) where p ε and W ε are the smooth approximations of p and w(, ) = 1 (,) respectively, and β( ) is a smooth function being identically zero in (, ] and strictly increasing and convex in [, ). For definiteness, we take β(s) := max{, s 3 } s R. The particular p ε and W ε are chosen so that the solution w ε is strictly increasing in ε. 2

27 Lemma For any given decreasing function p(t) such that p() = 1 and p(t) for any t >, there exists p ε such that 1. p ε C 1 ([, )), and 1 ε d dt pε (t), and consequently ṗ ε 1 ε. 2. for each t, and consequently d dε pε (t) 2 ε, 1/3 lim ε pε (t) = p(t) = p (t). (3.1.2) Proof. 1. Suppose that p C 1 ([, )). Define p ε as following p ε (t) := (1 z 2 )p(t + ε + εz) dz 3ε 2/3 ε >, t. Then p ε C 1 ([, )). As p is decreasing, i.e., ṗ, so that Also since p 1, d dt pε (t) = 3 4 d dt pε (t) = = 3(1 z2 ) 4ε = 3 4ε 3 4ε 3 4ε (1 z 2 )ṗ(t + ε + εz)dz <. (1 z 2 )ṗ(t + ε + εz)dz p(t + ε + εz) z=1 z= zp(t + ε + εz)dz 4ε 1 2zp(t + ε + εz)dz 2zp(t + ε + εz)dz 2zdz = 3 4ε 1 ε. 21

28 For each t, d dε pε (t) = (1 z 2 )(1 + z)ṗ(t + ε + εz)dz 2ε 1/3 2ε 1/3 <. Hence p ε (t) is decreasing in terms of ε and it is bounded from below by 3ε 2/3. It implies that lim ε p ε (t) exists and can be obtained by lim ε pε (t) = lim p εn (t), where ε n = 1 n n. For each n >, t, when z [ 1, 1], < (1 z 2 )p(t + ε n + ε n z) (1 z 2 )p (t) where (1 z 2 )p (t) is integrable in [ 1, 1]. Using Lebesgue Convergence theorem, we get lim ε pε (t) = lim = 3 4 = n (1 z 2 )p(t + ε n + ε n z)dz (1 z 2 ) lim p(t + ε n + ε n z)dz n (1 z 2 )p (t)dz = p (t). This completes the proof for the case p C 1 ([, )). 2. Suppose p is not a smooth function, then one can choose a sequence of functions {p n } n=1 such that 1. for each t, lim n p n (t) = p(t); 2. for each n >, p n C 1 ([, )); 3. {p n } is uniformly bounded, i.e., there exists M > such that for each n >, p n M. 22

29 Let p ε n(t) := Then for each n >, t, when z [ 1, 1], 1 (1 z 2 )p n (t + ε + εz) dz 3ε 2/3 ε >, t. (1 z 2 )p n (t + ε + εz) (1 z 2 )M. By Lebesgue Convergence theorem lim n p ε n(t) exists, denoted by p ε (t) := lim n p ε n(t) ε >, t. Remark When t =, (3.1.2) yields: lim ε p ε () = p() = 1 and p ε () is a monotone function of ε. We denote by ε > the unique constant such that p ε () =, and in the sequel assume ε (, ε ). Lemma There exists an approximation W ε for w(, ) = 1 (,) such that 1. for each ε >, W ε (x) = x, W ε (x) = p ε () x ε; 2. W ε C 1 (R) and d dx W ε (x) ; 3. for each x R, d dε W ε (x), and consequently, lim W ε = 1 (,). ε 23

30 Proof. We fix a function W ( ) C 1 (R) defined on R that satisfies: W (x) = x, W (x) = 1 x 1, Ẇ x ( 1, ). Set W ε (x) := p ε () W (x/ε) x R. Now we verify that W ε is the approximation we need. 1. For each ε >. If x ε, then x/ε 1. Hence W ε (x) = p ε (). If x, then x/ε. Hence W ε (x) =. The first assertion follows. 2. Since W C 1 (R) and Ẇ, W ε C 1 (R) and The second assertion follows. 3. For each x R, d dx W ε (x) = 1 ε pε ()Ẇ (x/ε). d dε W ε (x) = d dε pε ()W (x/ε) x ε 2 pε ()Ẇ (x/ε). As W is nonnegative and d dε pε () <, d dε pε ()W (x/ε). If 1 < x/ε < then x < and Ẇ (x/ε). Consequently x ε 2 pε ()Ẇ (x/ε). If x/ε and x/ε 1, then Ẇ (x/ε) =. Thus d dε W ε (x). Since W ε is decreasing in terms of ε and bounded, lim ε W ε (x) exists. If x, then otherwise x <, then lim W ε (x) = lim p ε ()W (x/ε) = lim = ; ε ε ε lim W ε (x) = lim p ε () lim W (x/ε) = 1. ε ε ε The last equality holds since x/ε 1 when ε. Hence the third assertion follows. 24

31 Before proving the existence of a solution to problem (3.1.1), we introduce the following functions. 1. Consider a first order linear initial value problem, Lw ε = in R (, ), w ε (, ) = W ε ( ) on R {}. (3.1.3) Since W ε ( ) is a smooth function, the problem admits a unique solution, denoted by w ε (x, t) and it can be expressed as w(x, ε t) = R K(x, t; y, )w(y, ε ) dy = p ε () K(x, t; y, )W (y/ε) dy where K(x, t; y, s) is the fundamental solution associated with the linear operator L. particular, when L = t 1 2 xx, i.e., µ and σ 1, In K(x, t; y, s) = Γ(x y, t s), Γ(x, t) = 1 2πt e x2 /2t. 2. Consider a first order ODE, to: Since β ( ρ ε (t) p ε (t) solution, denoted by ρ ε. ε d dt ρε (t) = β ρ ε () = p ε (). ( ρ ε (t) p ε (t) ε ) in (, ), (3.1.4) ) is a smooth function with respect to ρ ε, (3.1.4) admits a unique smooth Lemma The solution ρ ε to (3.1.4) satisfies the following: Consequently p ε (t) ρ ε (t) p ε (t) + ε ṗ ε 1/3, ρ ε (t), t. ρ ε (t) p ε (t) ε 2/3 and lim ε ρ ε (t) = p(t). 25

32 Proof. Set ϕ(x) := x1 (, ) (x). Let {ϕ n } be a sequence of twice continuously differentiable functions such that lim n ϕ n (x) = ϕ(x) and ϕ n 1. Then lim n ϕ n (x) = 1 (, ) (x) almost everywhere. 1. First we claim that p ε (t) ρ ε (t). Set α(t) := p ε (t) ρ ε (t). Then for α() = and for each t >, α(s) is bounded as s t since α(s) p ε (s) + ρ ε (s) p ε () + ρ ε () = 2p ε (). For each n > and t >, ϕ n (α(t)) ϕ n (α()) = = = = ϕ n (α(s)) α(s)ds ϕ n (α(s)) (ṗ ε (s) ρ ε (s)) ds ( ( ρ ϕ n (α(s)) ṗ ε ε (s) p ε (s) (s) + β ε ( ) ρ ε (s) p ε (s) ϕ n (α(s))β ds ε 1 ε 3 ϕ n(α(s))(α(s) ) 3 ds. )) ds Note that as ϕ n 1 and α(s) is bounded, ϕ n (α(s))(α(s) ) 3 is bounded. Hence using the dominated convergence theorem, we obtain α(t) + = ϕ(α(t)) ϕ(α()) = lim ϕ n (α(t)) ϕ n (α()) n 1 lim n ε ϕ n(α(s))(α(s) ) 3 ds 3 = 1 1 ε 3 α(s)> (α(s))(α(s) ) 3 ds =. Then α + (t) =, which implies that α(t). Hence p ε ρ ε. 2. Now we claim that ρ ε p ε + ε ṗ ε 1/3. 26

33 Set γ(t) := α(t) ε ṗ ε 1/3. Then γ() = ε ṗ ε 1/3 bounded as s t. For each n > and t >, < and for each t >, γ(s) is ϕ n (γ(t)) ϕ n (γ()) = = = ϕ n (γ(s)) γ(s)ds ϕ n (γ(s)) ( ρ ε (s) ṗ ε (s)) ds ( ( ) ) α(t) ϕ n (γ(s)) β ṗ(s) ds ε ( α(t) ϕ n (γ(s)) ε ) ) 3 (ε ṗ ε 1/3 3 + ds ε = = 1 ε ϕ n(γ(s)) ( γ(s)) ( ( α(s)) 2 ) εα(s) ṗ ε 1/3 3 + ε 2 ṗ ε 2/3 2 ds 1 ε ϕ n(γ(s)) ( γ(s)) + ( ( α(s)) 2 ) εα(s) ṗ ε 1/3 3 + ε 2 ṗ ε 2/3 2 ds C(t) ϕ n (γ(s)) ( γ(s)) + ds ε 3 C(t) ε 3 ϕ n (γ(s))γ(s) ds, where C(t) is the constant depending on t. Since γ(s) is bounded, we can use the dominated convergence theorem and obtain γ(t) + = ϕ(γ(t)) ϕ(γ()) = lim ϕ n (γ(t)) ϕ n (γ()) n C(t) 1 ε 3 1 γ(s)> (γ(s)) γ(s) ds =. So that γ(t) + = and it implies that γ(t). Hence ρ ε p ε + ε ṗ ε 1/3. Now we are ready to prove the existence of a solution to problem (3.1.1). Theorem 3. For each ε >, problem (3.1.1) admits a unique smooth (C 2,1 ) solution in R [, ). The solution is continuously differentiable in ε and satisfies, for all ε > and (x, t) R (, ), w ε (x, t) + ρ ε (t) ρ ε () w ε (x, t) min{ρ ε (t), w ε (x, t)}, (3.1.5) 27

34 w ε x(x, t) <, d dε wε (x, t) <. Consequently, the following limit exists w(x, t) := lim ε w ε (x, t) (x, t) R [, ). Proof. Let w ε := min{ρ ε, w ε } and w ε := w ε (x, t) + ρ ε (t) ρ ε (). First we claim that w ε is a supersolution, w ε is a subsolution to (3.1.1) and w ε w ε. Since Lw ε + β( wε pε ε ) = β( wε pε ), w ε ε is a supersolution. Also we have w ε max{w ε (, )} = max x R W ε (x) = max x R pε ()W (x/ε) = p ε () = ρ ε (). (3.1.6) Since Lρ ε + β( ρε p ε ) =, ρ ε is another supersolution. Hence, w ε is a supersolution. Since ε β, the direct computation, together with (3.1.6), gives ( w Lw ε ε p ε ) + β ε ( w = Lw(x, ε t) + Lρ ε ε p ε ) (t) + β [ ε ( ρ ε p ε ) ( ρ ε p ε + w ε ρ ε () ) ] = β β. ε ε Hence w ε is a subsolution, and w ε w ε by (3.1.6). Now we prove that (3.1.1) admits a unique smooth solution in R [, ) satisfying (3.1.5) with the standard method of subsolutions and supersolution [2]. Fix λ 3ε 5/3. Let w ε 1 = w ε, and then given w ε k (k = 1, 2, ) inductively define wε k+1 C2,1 (R (, )) to be the unique solution of the linear initial-value problem to First we claim that ( ) Lwk+1 ε + λwε k+1 = β w ε k p ε + λwk ε in R (, ), ε w ε k+1 (, ) = W ε ( ) on R {}. (3.1.7) w ε = w ε 1 w ε 2 w ε k in R [, ). 28

35 Let v := w ε 2 w ε 1, then Lv ε + λv ε in R (, ), v ε (, ) = on R {}. By the maximum principle max v ε max v ε =, R (, ) R {} so that w ε 1 w ε 2. Now assume inductively w ε k 1 wε k and let v := wε k+1 wε k. Then ( ( ) ( w Lv ε + λv ε ε = β k p ε w ε β k 1 p ε )) + λ(wk ε w ε ε ε k 1) ( = λ 1 ) ε β(ξ) (wk ε wk 1) ε where w ε k pε ε ξ wε k 1 pε. ε Since 3x β(x) 2 x >, = x, and w ε k pε ε ξ wε k 1 pε ε wε p ε ε ρε p ε, ε we obtain that ( ) ) ρ β(ξ) ε p ε 2 (ε ṗ ε 1/ ε 2/3, ε ε where the second inequality follows from lemma (3.1.3) and the third inequality follows from the first assertion of lemma (3.1.1). Thus Lv ε + λv ε in R (, ) v ε (, ) = on R {} 29

36 By the maximum principle max v ε max v ε =, R (, ) R {} so that w ε k wε k+1. Secondly we claim that w ε k w ε in R [, ), for k = 1, 2,. (3.1.8) It is clear that w ε 1 = w ε w ε. Assume now for induction w ε k w ε in R [, ). Let v ε := w ε wk+1 ε, then ( ( ) ( )) w Lv ε + λv ε ε p ε w ε β β k p ε + λ(w ε w ε ε ε k) ( = λ β(ξ) ) (w ε w ε ε k), with the similar argument as above. By the maximum principle so that w ε k+1 wε. Thus (3.1.8) holds. Now we have max v ε max v ε =, R (, ) R {} w ε = w ε 1 w ε 2 w ε k w ε k+1 w in R [, ). Therefore w ε (x, t) := lim k w ε k(x, t) exists and it is bounded. Let Γ(x, t; y, s) be the fundamental solution associated with the operator L + λ. Then the solution to (3.1.7) for each k > can be expressed as: ( ( w ε ) ) wk(x, ε t) = Γ(x, t; y, s) β k 1 (y, s) p(s) λw k 1 (y, s) dyds ε + Γ(x, t; y, )W ε (y)dy. 3

37 As wk ε are bounded, by the Dominated Convergent Theorem, we obtain that w ε (x, t) = + ( Γ(x, t; y, s) β Γ(x, t; y, )W ε (y)dy. ( ) w ε (y, s) p(s) ε ) λw ε (y, s) dyds Then w ε is continuous. Let Ω T := R [, T ]. Then w ε is locally Hölder continuous and uniformly continuous with respect to t. Let w ε be the solution to L w ε + λ w ε = β ( ) w ε p ε ε + λw ε in R (, ), w ε (, ) = W ε ( ) on R {}. Then w ε C 2,1 (Ω T ) and w ε (x, t) = + ( Γ(x, t; y, s) β ( ) w ε (y, s) p(s) Γ(x, t; y, )W ε (y)dy = w ε (x, t). ε ) λw ε (y, s) dyds Hence w ε solves Lw ε = β ( ) w ε p ε in ΩT. Let T, we obtain that w ε (x, t) is the smooth ε solution to (3.1.1). Also since w ε is bounded by w ε and w ε, the uniqueness follows. We see that (3.1.1) admits a unique smooth solution in R [, ) satisfying (3.1.5). We remark that w ε is smooth with respect to ε with the standard arguments. 2. Differentiating the system (3.1.1) with respect to ε we obtain d dε wε (x, ) = d dε W ε (x) x R, L d dε wε + 1 ε β ( w ε p ε ) d ε dε wε = 1 β( w ε p ε ){w ε p ε + ε d } ε 2 ε dε pε, since β, w ε p ε ρ ε p ε ε ṗ ε 1/3 ε 2/3, and d dε pε 2ε 1/3. Then, by the maximum principle, d dε wε < in R (, ). The monotonicity and boundedness of w ε in ε and imply that w = lim ε w ε exists. 31

38 In a similar manner, differentiating the system (3.1.1) with respect to x and let u ε := w ε x, we obtain Au ε + 1 β ( ) w ε p ε u ε =, ε ε u ε (x, ) = d dx wε (x, ) = d dx W ε (x) x R. where Au = Lu σσ x u x +(µ x σσ xx +(σ x ) 2 )u. Since 1 β ( ) w ε p ε >, w ε ε ε x(x, t) = u ε (x, t) > in R (, ). Also note that since w ε is monotonic in ε and bounded, the limit w := lim ε w ε exists and is the solution to Lw = in R (, ), w (, ) = 1 (,). 3.2 CONTINUITY ESTIMATES AND EXISTENCE. In this section, we prove that the limit w := lim ε w ε is the viscosity solution to our variational inequality. In order to do so, we first need to derive some supplementary estimates on the continuity of w. Lemma For each T >, there exists a constant C = C(T ) that depends only on σ and µ such that for all ε (, ε ), < s < t T, and x, y R, Cpε () t w ε x(x, t) w ε x(x, t), (3.2.1) w ε (x, t) w ε (y, s) Cpε () { min{ x y + 2 } t s + ρ ε (s) ρ ε (t). (3.2.2) s, 1} Consequently, the limit w = lim ε w ε satisfies for all < s < t T and x, y R, C t w x (x, t) w x (x, t), (3.2.3) 32

39 w(x, t) w(y, s) C { min{ x y + 2 } t s + p(s) p(t), (3.2.4) s, 1} w(x, t) w(y, s) C { min{ x y + 2 } t s. (3.2.5) s, 1} We remark that when σ 1 and µ, C = C(T ) = (2π) 1/2 for all T. Proof. Differentiating the systems (3.1.1) and (3.1.3) with respect to x, and using the notation from the previous theorem, we find Aw ε x = ε 1 β(ε 1 (w ε p ε )w ε x = Aw ε x in R (, ), w ε x(, ) = W ε x( ) = w ε x(, ) on R {}. since w ε x. Therefore by the maximum principle (the zeroth order term in A is bounded above) w ε x w ε x. Next we estimate the lower bound of w ε x. Differentiating the system (3.1.3) with respect to x, we obtain Aw ε x = in R (, ), w ε x(, ) = W ε x( ) on R {}. This is a linear problem, the solution can be expressed as wx(x, ε t) = R K(x, t; y, )W ε y (y) dy, where K is the fundamental solution associated with the linear operator A. By Friedman [3]), K(x, t; y, s) C e λ(x y) 2 t s 4(t s), where λ is some positive constant. When s =, t K(x, t; y, ) Ce λ(x y)2 4t C. Hence t K(x, t; y, ) has a least upper bounded and we denote it as C = C(T ) = { } sup t K(x, t; y, ). x,y R,<t<T 33

40 As w x and W ε y, then for any < t T, wx(x, ε t) sup { K(x, t; y, )} Wy ε (y) dy C Wy ε (y) dy = Cpε (). x,y R R t R t The estimates for w ε x and w ε x (3.2.1) thus follow. Sending ε, we obtain (3.2.3). Now we estimate the continuity in the time variable. By Theorem 3, w ε x < and w ε (x, t) ρ ε (t) ρ ε () since w ε. We conclude that lim x w ε exists. Similarly, the limit lim x w ε exists, and is nonnegative since w ε. Now since w ε (x, t) min{ρ ε (t), w ε (x, t)} ρ ε (t) t, we can compute R wx(x, ε t) dx = R w ε x(x, t)dx ρ ε (t) lim x w ε (x, t) ρ ε (t) lim x ( w ε (x, t) + ρ ε (t) ρ ε () ) = ρ ε () lim x w ε (x, t) ρ ε () = p ε (). Also note that β(ε 1 (w ε p ε )) β(ε 1 (ρ ε p ε )) = ρ ε (t) t, x R, since β( ) is increasing and w ε ρ ε. For < s < t T denote wx ε s,t = sup wx. ε R [s,t] Since w ε x(x, s) Cpε () s, { } Cp wx ε s,t ε () sup w x (, ξ) sup Cpε (). s ξ t s ξ t ξ s 34

41 Then for each δ >, x+δ {w ε (y, t) w ε x+δ (y, s)} dy = wv(y, ε v) dv dy x δ x δ s x+δ ( = 1 s x δ 2 (σ2 wy) ε y µwy ε β ( ε 1 (w ε p ε ) ) ) dy dv ( 1 ( σ 2 w ε s 2 y) ) x+δ t x+δ dv + t x+δ µ wydy ε dv + β ( ε 1 (ρ ε p ε ) ) dy dv x δ s x δ s x δ (t s) ( ) t x+δ σ 2 wx ε s,t + p ε () µ + ρ ε (v)dy dv s x δ (t s) ( σ 2 w ε x + p ε () µ ) + 2δ(ρ ε (s) ρ ε (t)). Finally, note that for any s, wε (x, s) 1 2δ x+δ x δ w ε (y, s) dy = 1 2δ 1 2δ x+δ x δ x+δ x δ δ 2 wε x(, s). (w ε (x, s) w ε (y, s)) dy y x w ε x(, s) dy Now we are ready to estimate the continuity in the time variable. For any < s < t T, By taking δ = w ε (x, t) w ε (x, s) wε (x, t) 1 x+δ w ε (y, t)dy 2δ + 1 x δ 2δ + 1 x+δ w ε (y, s)dy w ε (x, s)dy 2δ x δ ( wx ε s,t δ + (t ) s) σ2 2δ σ 2 (t s) 2, we then obtain x+δ x δ + (t s) µ p ε () 2δ (w ε (y, t) w ε (y, s)) dy + ρ ε (s) ρ ε (t). w ε (x, t) w ε (x, s) ( σ2 (t s) 2 w ε 2 x s,t + µ ) p ε () + ρ ε (s) ρ ε (t) σ 2 2Cp ε () min{ s, 1} + ρε (s) ρ ε (t). 35

42 As wx(x, ε s) Cpε () s, we have w ε (x, s) w ε (y, s) x y w ε x(, s) Cpε () s x y. Then w ε (x, t) w ε (y, s) w ε (x, t) w ε (x, s) + w ε (x, s) w ε (y, s) Cpε () min{ { } x y + 2 t s + ρ ε (s) ρ ε (t). s, 1} This proves (3.2.2), and (3.2.4) then follows by sending ε. Finally, observe that in estimating the upper bound of w ε (x, s) w ε (x, t) if we keep the term involving the integral of β, then w ε (x, t) w ε (y, s) w ε (x, t) w ε (x, s) + w ε (x, s) w ε (y, s) Cp ε t () min{ { } x y + 2 t s + s s, 1} { } x y + 2 t s. Cp ε () min{ s, 1} We obtain (3.2.5) by sending ε. This completes the proof. We are ready to show the following: x+δ β ( w ε p ε x δ ε 2δ ) dy dv Theorem 4. Assume p( ) defined on [, ) is nonnegative, decreasing and lower semicontinuous, with p() = 1. There exists a unique viscosity solution to 2.1.2, and it can be obtained as the limit w := lim ε w ε. Proof. 1. First we verify that w satisfies the initial condition (2.2.1). Since lim (x) ε = lim p ε ()W (x/ε) = ε if x 1 if x < = 1 (,) w(x, ) = lim ε w ε (x, ) = lim ε W ε (x) = 1 (,). 36

43 Similarly For any t >, from (3.1.5) w (x, ) = lim ε w ε (x, ) = lim ε W ε (x) = 1 (,). ρ ε (t) ρ ε () w ε (, t) w ε (, t). Sending ε, we get p(t) p() w(, t) w (, t), where Then w (x, t) = lim w ε ε (x, t) = K(x, t; y, ) dy. w(, t) w (, t) = sup w(, t) w (, t) p() p(t). x R From (3.1.5), w ε has upper and lower bounds. Sending ε, we obtain that p(t) 1 w (x, t) + p(t) 1 w(x, t) p(t) < 1, since w (x, t). It implies that lim sup w(y, t), y x,t lim inf w(y, t) 1 y x,t Next we claim that lim sup w(y, t) = 1 (,], and lim inf w(y, t) = 1 (,), (3.2.6) y x,t y x,t by considering the following three cases: (i) x <, (ii) x >, (iii) x =. 37

44 Hence Case (i): Suppose x <. For any sequence x n x, and t n, lim w(x, ) w(x n, t n ) n ( ) lim w(x, ) w (x, ) + w (x, ) w (x n, t n ) + w (x n, t n ) w(x n, t n ) n ( 1 w (x n, t n ) + p() p(t n )) = lim 1 K(x n, t n ; y, )dy =. n lim w(y, t) = w(x, ) = 1 x <. y x,t Case (ii): Suppose x >. For any sequence x n x and t n, Hence lim w(x, ) w(x n, t n ) n lim ( w(x, ) w (x, ) + w (x, ) w (x n, t n ) + w (x n, t n ) w(x n, t n ) ) n = lim n ( w (x n, t n ) + p() p(t n )) = lim K(x n, t n ; y, )dy =. n lim w(y, t) = w(x, ) = x >. y x,t Case (iii): Suppose x =. Let t n = 1, x n n = n α, where < α < 1. Note that 2 lim w(x, ) w(x n, t n ) n lim ( w(x, ) w (x, ) + w (x, ) w (x n, t n ) + w (x n, t n ) w(x n, t n ) ) n = lim n ( w (x n, t n ) + p() p(t n )) = lim K(x n, t n ; y, )dy. n By the property of K [3], we have lim n n = lim K(x n, t n ; y, )dy lim n xn σ 2 tn C 2π e x2 /2 dx lim C 2πtn σ 1 σ 2 n1/2 α n (xn y) 2 e 2σ 2 tn 2 dy C 2π e x2 /2 dx =. 38

45 Then Now y n = n α, where < α < 1. Note that 2 lim n lim w(x n, t n ) =. (3.2.7) n K(y n, t n ; y, )dy = lim K(y n, t n ; y, )dy n lim n yn n lim n = lim C 2πtn σ σ 2 tn 1 σ 2 n1/2 α (yn+y) 2 e 2σ 2 tn 2 C 2π e x2 /2 dx Then ( lim K(y n, t n ; y, )dy = lim K(y n, t n ; y, )dy n n Hence = 1 lim n dy C 2π e x2 /2 dx =. K(y n, t n ; y, )dy = 1. ) K(y n, t n ; y, )dy From (3.2.7) and (3.2.8), we can conclude that lim w(y n, t n ) = 1. (3.2.8) n lim sup w(y, t) = 1, y,t lim inf w(y, t) = y,t Thus (3.2.6) holds and (2.2.1) follows. 2. We verify that w is a viscosity solution in R (, ) by considering two cases for each (x, t) R (, ): (i) p(t) w(x, t) >, and (ii) p(t) w(x, t). Case (i): Suppose p(t) w(x, t) >. Let D δ := (x δ, x + δ) (t δ 2, t + δ 2 ) δ >. 39

46 Then for each (y, s) D δ, ρ ε (s) ρ ε (t) ρ ε (s) p ε (s) + p ε (s) p ε (t) + p ε (t) ρ ε (t) p ε (s) p ε (t) + 2ε 2/3, since ρ ε p ε ε 2/3. As p ε ( ) is decreasing 2 (p ε (t) p ε (s)) 2 (p ε (t) p ε (t + δ 2 )) s > t, p ε (t) p ε (s) + p ε (t) p ε (s) = 2 (p ε (t) p ε (t + δ 2 )) s t. Using (3.2.2), we can compute w ε (y, s) p ε (s) w ε (y, s) w ε (x, t) + w ε (x, t) p ε (s) ( )Cδ min{ t δ 2, 1} + ρε (s) ρ ε (t) + w ε (x, t) p ε (s) ( )Cδ min{ t δ 2, 1} + wε (x, t) p ε (t) +p ε (t) p ε (s) + p ε (t) p ε (s) + 2ε 2/3 ( )Cδ min{ t δ 2, 1} + wε (x, t) p ε (t) + 2(p ε (t) p ε (t + δ 2 )) + 2ε 2/3. Then { } lim sup max w ε p ε ε D δ lim sup ε { ( )Cδ min{ t δ 2, 1} + wε (x, t) p ε (t) +2(p ε (t) p ε (t + δ 2 )) + 2ε 2/3 } ( )Cδ min{ t δ 2, 1} + w(x, t) p(t) + 2(p(t) p(t + δ2 )). Then if we take δ small enough, by the assumption of w < p and p is decreasing, { } lim sup max w ε p ε <. ε D δ Thus, for all sufficiently small positive ε, w ε p ε < in D δ. Consequently, Lw ε = β( wε p ε ) = (x, t) ε D δ. 4

47 Observe that in estimating the boundary of w ε (x, s) w ε (x, t) if we keep the term involving the integral of β, then w ε (x, t) w ε (y, s) = x+δ Cp ε () min{ { } β ( ) w ε p ε s x δ ε dydv x y + 2 t s + s, 1} 2δ Cp ε () min{ { } x y + 2 t s s, 1} It implies that w ε is equicontinuous in D δ. point-wise bounded. Indeed Also from (3.1.5), we can derive that w ε is w ε w ε (x, t) ρ ε () 1, and w ε (x, t) w ε (x, t) + ρ ε (t) ρ ε () p ε (t) p ε (), where p ε (t) p ε () is bounded for small ε >. Hence w ε contains a uniformly convergent subsequence w εn. So that its limit w is differentiable with respect to x and t. Similarly we can show that wx ε is differentiable with respect to x. Hence the limit w is then a smooth solution to Lw = in D δ. Case (ii): Suppose w(x, t) p(t). However, w p in R [, ) since w ε ρ ε and lim ε ρ ε (t) = p(t) in R [, ). Hence, we must have w(x, t) = p(t) = min{p(t), w (x, t)}, where the second inequality holds since p w w. From (3.2.5) So that w (x, t) = w(x, t) = p(t). solution hold. w(x, t) w (x, t) = lim sup(w(x, t) w(y, s)) y x,s t ( C { lim sup x y + 2 t s} ) =. y x,s t s Thus the semi-continuity requirements for a viscosity In this case, we clearly have max{w(x, t) p(t), Lϕ(x, t)} for any smooth ϕ. So that w is a supersolution. It remains to verify the differential inequality for subsolutions. To 41