A STOCHASTIC STEFAN PROBLEM 1. Introduction Moving Boundary

A STOCHASTIC STEFAN PROBLEM

KUNWOO KIM, ZHI ZHENG, AND RICHARD B. SOWERS

Abstract. We consider a stochastic perturbation of the Stefan problem. The noise is Brownian in time and smoothly correlated in space. We prove existence and uniqueness and characterize the domain of existence.

1. Introduction Moving boundary problems are one of the important areas of partial differential equations. They describe a wide range of physically interesting phenomena where a system has two phases. However, since the boundary between these phases is defined implicitly, they provide deep mathematical challenges. Our goal here is to study the effect of noise on the Stefan problem, which is one of the canonical moving boundary value problems. Fix a probability triple (Ω, F , P) and assume that ζ is a random field which is Brownian in time but which is correlated in space (we will rigorously define ζ in Section 2). We consider the stochastic partial differential equation (SPDE) ∂u ∂2u (t, x) = (t, x) + αu(t, x) + u(t, x)dζ (x) x > β(t) ∂t ∂x2 t ∂u lim (t, x) = −%β˙(t) (1) x&β(t) ∂x u(0, x) = u◦(x) x ∈ R {(t, x) ∈ R+ × R | u(t, x) > 0} = {(t, x) ∈ R+ × R | x > β(t)}. The constant α ∈ R is fixed (we shall later see why it is natural to include this term). We also assume that the initial condition u◦ ∈ C(R) satisfies some specific properties: du◦ • u◦ ≡ 0 on R−, u◦ > 0 on (0, ∞), and limx&0 dx (x) exists. • u◦ and its first three derivatives exist on (0, ∞) and are square-integrable (on (0, ∞)). The last requirement in (1) means that the boundary between u ≡ 0 and u > 0 is exactly the graph of β. In fact, it is not yet clear that (1) makes sense. Differential equations are pointwise statements. Stochastic differential equations are in fact shorthand representations of corresponding integral equations; pointwise statements typically don’t make sense. It will take some work to restate the pointwise stochastic statement in the first line of (1) as a statement about stochastic integrals. There has been fairly little written on the effect of noise on moving boundary problems (see [BDP02], [CLM06], and [KMS]; see also the work on the stochastic porous medium equation in [BDPR09, DPR04a, DPR04b, DPRRW06, Kim06]). We note here that the multiplicative term u in front of the dζt places this work slightly outside of the purview of the theory of infinite-dimensional evolution equations with Gaussian perturbations. The multiplicative term is in fact a natural nonlinearity. It implies that there is only one interface; bubbles where u > 0 cannot spontaneously nucleate where u = 0, and conversely bubbles where u = 0 cannot spontaneously nucleate where u > 0 (see Lemma 5.8). Our work here follows on [KMS], where we investigated the structure of a moving boundary problem driven by a single Brownian motion. The moving boundary problem of interest there a model of detonation. We here focus on the Stefan problem, and consider noise which is spatially correlated. We formulate several techniques which can (hopefully) be applied to a number of stochastic moving boundary value problems. In our particular case, where the randomness comes from a noise which is correlated in space and Brownian

Date: March 16, 2011. This work was supported by NSF grant DMS0705260. R.S. would like to thanks the Departments of Mathematics and Statistics of Stanford University for their hospitality in the Spring of 2010 during a sabbatical stay. The authors would like to thank the anonymous referee for his meticulous reading of the original version of this manuscript. 1 in time, several transformations (the transformations of Lemmas 4.3 and 4.4 and (25)) can transform the problem into a random nonlinear PDE (see (26)). All of these transformations are not in general available when the noise is more complicated, but most of the techniques we develop here should be. Secondly, the irregularity of the Brownian driving force requires some detailed analysis, no matter what perspective one takes; namely in the analysis of Lemma 4.2 and the iterative bounds of Lemma 5.5.

2. The Noise Let’s start by constructing the noise process. Fix η ∈ C∞(R) ∩ L2(R) such that η(n) ∈ C∞(R) ∩ L2(R) for all n ∈ {1, 2, 3} and such that Z (2) η2(y)dy = 1. y∈R Let W be a Brownian sheet. For t ≥ 0 and x ∈ R, deﬁne t def Z Z ζt(x) = η(x − y)W (ds, dy) s=0 y∈R Then ζ is a zero-mean Gaussian ﬁeld with covariance structure given by Z E[ζt(x)ζs(y)] = (t ∧ s) η(x − z)η(y − z)dz z∈R for all s and t in R+ and x and y in R. Thus for each x ∈ R, t 7→ ζt(x) is a Brownian motion, and for each 2 t > 0, x 7→ ζt(x) is in C with derivative given by Z t Z (n) (n) ζt (x) = η (x − y)W (ds, dy) s=0 y∈R for all t ≥ 0 and x ∈ R for n ∈ {0, 1, 2}.

3. Weak Formulation ˙ To see what we mean by (1), let’s replace ζt(x) by a smooth function b : R+×R → R; the Wong-Zakai result (cf. [KS91, Section 5.2D]) implies that this is a reasonable approximation of an SPDE with Stratonovich integration against the noise; we can then convert this into the desired SPDE with Ito integration. Namely, consider the PDE ∂v ∂2v (t, x) = + α v(t, x) + v(t, x)b(t, x) x > β (t) ∂t ∂x2 s ◦ ∂v ˙ (3) lim (t, x) = −%β◦(t) x&β◦(t) ∂x v(0, x) = u◦(x). x ∈ R {(t, x) ∈ R+ × R | v(t, x) > 0} = {(t, x) ∈ R+ × R | x > β◦(t)}.

def 1 where αs = α − 2 (we will see that this corresponds to the Stratonovich analogue of (1)). This will be our starting point. ∞ Let’s see what a weak formulation looks like (see [Fri64, Ch. 8]). Fix ϕ ∈ Cc (R+ × R). Assume that β◦ is differentiable. Define ∞ def Z Z Uϕ(t) = v(t, x)ϕ(t, x)dx = v(t, x)ϕ(t, x)dx. x∈R x=β◦(t) Differentiating, we get that Z ∞ ∂v ∂ϕ ˙ U˙ ϕ(t) = (t, x)ϕ(t, x) + v(t, x) (t, x) dx − v(t, β◦(t))ϕ(t, β◦(t))β◦(t) x=β◦(t) ∂t ∂t

and we can use the fact that v(t, β◦(t)) = 0 to delete the last term. We can also use the PDE for v for ∂v x > β◦(t) to rewrite ∂t . Integrating by parts, we have that

2 Z ∞ ∂2v 2 (t, x)ϕ(t, x)dx x=β◦(t) ∂x ∂v ∂ϕ Z ∞ ∂2ϕ = lim − (t, x)ϕ(t, β◦(t)) + v(t, β◦(t)) (t, β◦(t)) + v(t, x) (t, x)dx. x&β (t) 2 ◦ ∂x ∂x x=β◦(t) ∂x ∂v Again we use the fact that v(t, β◦(t)) = 0, and we can also use the boundary condition on ∂x . Recombining things we get the standard formula that Z ∂ϕ ∂2ϕ U˙ (t) = v(t, x) (t, x) + (t, x) + α ϕ(t, x) dx ϕ ∂t ∂x2 s x∈R Z ˙ + v(t, x)ϕ(t, x)b(t, x)dx + %ϕ(t, β◦(t))β◦(t). x∈R ∞ Replacing b by our noise, we should have the following formulation: that for any ϕ ∈ Cc (R+ × R) and any t > 0, Z Z u(t, x)ϕ(t, x)dx = u◦(x)ϕ(0, x)dx x∈R x∈R Z t Z ∂ϕ ∂2ϕ + u(r, x) (r, x) + (r, x) + α ϕ(r, x) dx dr ∂r ∂x2 s r=0 x∈R Z Z t Z t ˙ + u(r, x)ϕ(r, x) ◦ dζr(x)dx + % ϕ(r, β◦(r))β◦(r)dr x∈R r=0 r=0 The Ito formulation of this would be that Z Z u(t, x)ϕ(t, x)dx = u◦(x)ϕ(0, x)dx x∈R x∈R Z t Z ∂ϕ ∂2ϕ + u(r, x) (r, x) + (r, x) + αϕ(r, x) dx dr ∂r ∂x2 r=0 x∈R Z Z t Z t ˙ + u(r, x)ϕ(r, x)dζr(x)dx + % ϕ(r, β◦(r))β◦(r)dr. x∈R r=0 r=0 Remark 3.1 The structure of the SPDE (1) is invariant under Ito and Stratonovich formulations; this is the motivation for including α in (1) We can now formally define a weak solution of (1). In this definition, we allow for blowup. Define def Ft = σ{W (s, y): s ≤ t, y ∈ R} for all t ≥ 0. Definition 3.2. A weak solution of (1) is a nonnegative predictable path {u(t, ·) | 0 ≤ t < τ} ⊂ C(R)∩L1(R), ∞ where τ is a predictable stopping time with respect to {Ft}t>0, such that for any ϕ ∈ Cc (R+ × R) and any finite stopping time τ 0 < τ, Z Z 0 0 u(τ , x)ϕ(τ , x)dx = u◦(x)ϕ(0, x)dx x∈R x∈R 0 Z τ Z ∂ϕ ∂2ϕ (4) + u(r, x) (r, x) + (r, x) + αϕ(r, x) dx dr ∂r ∂x2 r=0 x∈R Z Z τ 0 Z τ 0 ˙ + u(r, x)ϕ(r, x)dζr(x)dx + % ϕ(r, β(r))β(r)dr x∈R r=0 r=0 and where (5) {(t, x) ∈ [0, τ) × R | u(t, x) > 0} = {(t, x) ∈ [0, τ) × R | x > β(t)} where β is a semimartingale. Our main existence and uniqueness theorems are the following. The arguments leading up to these results will come together in Section 5. 3 Theorem 3.3 (Existence). There exists a predictable path {u(t, ·) | 0 ≤ t < τ} ⊂ C(R) ∩ L1(R) which satisfies (4), and u(t, ·) ∈ C1[β(t), ∞) for all t ∈ [0, τ) and ∂u τ ≤ inf t ≥ 0 : (t−, β(t)) = ∞ . ∂x Furthermore, if u(t, ·) ∈ C2[β(t), ∞) for all t ∈ [0, τ), then it satisfies (5).

Proof. Combine Lemmas 5.9 and 5.10. We also have uniqueness.

Theorem 3.4 (Uniqueness). Suppose that {u1(t, ·); 0 ≤ t < τ1} and {u2(t, ·); 0 ≤ t < τ2} are two solutions of (1). Assume that for i ∈ {1, 2}, the map x 7→ ui(t, x + βi(t)) has three generalized square-integrable derivatives on (0, ∞). Then u1(t, ·) = u2(t, ·) for 0 ≤ t < min{τ1, τ2}. Proof. The proof follows from Lemma 5.11. 4. Regularity and a Transformation The proof of Theorems 3.3 and 3.4 will hinge upon a transformation of (1) into a nonlinear integral equation on a fixed (as opposed to an implicitly defined) domain; we will address this in Subsection 4.2. First, however, let’s make sure that we understand a bit about regularity; this will illuminate the assumptions needed. 4.1. Regularity. While regularity of moving boundary-value problems is an incredibly challenging area (see [CS05]), we can make some headway. Namely, if we assume enough regularity for the boundary, we can get better control of the sense in which the boundary behavior holds. We start by rewriting (4) using heat kernels. Define 2 def 1 x p◦(t, x) = √ exp − t > 0, x ∈ R 4πt 4t def αt αt p±(t, x, y) = {p◦(t, x − y) ± p◦(t, x + y)} e = {p◦(t, x − y) ± p◦(t, −x − y)} e t > 0, x, y ∈ R

the second representation of p± stems from the fact that p◦ is even in its second argument. We then have that 2 ∂p± ∂ p± (t, x, y) = (t, x, y) + αp±(t, x, y) t > 0, x, y ∈ R (6) ∂t ∂y2 lim p±(t, x, ·) = δx ± δ−x; x ∈ R \{0} t&0 the relevant distinction between p+ and p− is their behavior at x = 0, namely, ∂2n−1p ∂2np + (t, 0, y) = − (t, 0, y) = 0 ∂x2n−1 ∂x2n def for all n ∈ N = {1, 2 ... }, all t > 0 and all y ∈ R. This will come up in the arguments of Lemma 4.2 and Lemma 4.3. Let’s next understand integration against ζ. Let {β(t) | t ≥ 0} be an R-valued predictable and continuous function. Let {f(t) | t ≥ 0} be a second R-valued predictable and continuous function, which is also bounded. We define Z t def X j f(s)dζs(β(s)) = lim f ζ(j+1)/N (β(j/N)) − ζj/N (β(j/N)) N→∞ N s=0 0≤j≤btNc Z t Z = f(s)η(β(s) − y)W (ds, dy). s=0 y∈R this being a limit in L2. Thanks to (2), Z t Z t Z dζs(β(s)) = η(β(s) − y)W (ds, dy) s=0 s=0 y∈R 4 x is a Brownian motion. Therefore we can define a Brownian motion Bt for each fixed x as Z t x def (7) Bt = dζs(x + β(s)). s=0

Note also that dζt(β(t)) is not the total derivative of ζt(β(t)); i.e., Z t ζt(β(t)) − ζ0(β(0)) 6= dζs(β(s)). s=0

To understand the total derivative, we must also include the spatial variation of ζt : Z Z t Z ˙ d [ζt(β(t))] = η(β(t) − y)W (dy, dt) + η˙(β(t) − y)β(t)W (dy, ds)dt. y∈R s=0 y∈R This implies that Z t Z t ˙ ˙ (8) ζt(β(t)) − ζ0(β(0)) = dζs(β(s)) + ζs(β(s))β(s)ds. s=0 s=0

Lemma 4.1. Suppose that {u(t, ·) | 0 ≤ t < τ} ⊂ C(R) ∩ C1([β(t), ∞)) is a weak solution of (1) with u(t, ∞) = 0 for 0 ≤ t < τ. Suppose also that β is continuously differentiable and {Ft}t≥0-adapted. Then u(t, x + β(t)) satisfies the integral equation Z ∞ u(t, x + β(t)) = p+(t, x, y)u◦(y)dy y=0 Z t Z ∞ ∂u + p (t − s, x, y) (s, y + β(s))β˙(s)dy ds + ∂x (9) s=0 y=0 Z ∞ Z t + p+(t − s, x, y)u(s, y + β(s))dζs(y + β(s))dy y=0 s=0 Z t ˙ + % p+(t − s, x, 0)β(s)ds s=0 for all t < τ and x > 0. Proof. Fix x > 0 and T > 0. For t ∈ [0, τ ∧ T ), define Z ∞ T def U (t) = u(t, y + β(t))p+(T − t, x, y)dy. y=0 Using Definition 3.2, we have Z ∞ Z ∞ T U (t) = u(t, y + β(t))p+(T − t, x, y)dy = u(t, y)p+(T − t, x, y − β(t))dy y=0 y=β(t) Z = u(t, y)p+(T − t, x, y − β(t))dy y∈R T T T = U (0) + A1 (t) + A2 (t) where Z t Z ∂p ∂2p AT (t) = u(r, y) − + (T − r, x, y − β(r)) + + (T − r, x, y − β(r)) 1 ∂t ∂y2 r=0 y∈R ∂p +αp (T − r, x, y − β(r)) − + (T − r, x, y − β(r))β˙(r) dydr + ∂y Z t ˙ + % p+(T − r, x, β(r) − β(r))β(r)dr, r=0 Z Z t T A2 (t) = u(r, y)p+(T − r, x, y − β(r))dζr(y)dy y∈R r=0 5 Then Z ∂p ∂2p dAT (t) = u(t, y) − + (T − t, x, y − β(t)) + + (T − t, x, y − β(t)) 1 ∂t ∂y2 y∈R ∂p +αp (T − t, x, y − β(t)) − + (T − t, x, y − β(t))β˙(t) dydt + ∂y ˙ + %p+(T − t, x, 0)β(t)dt (Z ∞ ) ∂p+ ˙ ˙ = − u(t, y) (T − t, x, y − β(t))dy β(t)dt + %p+(T − t, x, 0)β(t)dt y=β(t) ∂y (Z ∞ ) ∂u ˙ ˙ = (t, y)p+(T − t, x, y − β(t))dy β(t)dt + %p+(T − t, x, 0)β(t)dt y=β(t) ∂y Z ∞ ∂u ˙ ˙ = (t, y + β(t))p+(T − t, x, y)dy β(t)dt + %p+(T − t, x, 0)β(t)dt y=0 ∂y

We have here used the fact that u(t, β(t)) = 0. Thus Z t Z ∞ Z t T ∂u ˙ ˙ A1 (t) = p+(T − s, x, y) (s, y + β(s))β(s)dy ds + % p+(T − s, x, 0)β(s)ds. s=0 y=0 ∂y s=0 T Now we consider A2 (t). Using the stochastic Fubini theorem [Wal86, Theorem 2.6], we obtain Z Z t T A2 (t) = u(r, y)p+(T − r, x, y − β(r))dζr(y)dy y∈R r=0 Z Z t Z = u(r, y)p+(T − r, x, y − β(r)) η(y − z)W (dr, dz)dy y∈R r=0 z∈R Z t Z Z = u(r, y)p+(T − r, x, y − β(r))η(y − z)dy W (dr, dz) r=0 z∈R y∈R Z t Z (Z ∞ ) = u(r, y)p+(T − r, x, y − β(r))η(y − z)dy W (dr, dz) r=0 z∈R y=β(r) Z t Z Z ∞ = u(r, y + β(r))p+(T − r, x, y)η(y + β(r) − z)dy W (dr, dz) r=0 z∈R y=0 Z ∞ Z t = u(r, y + β(r))p+(T − r, x, y)dζr(y + β(r))dy. y=0 r=0 Combine things to get that Z t Z ∞ T T ∂u ˙ U (t) = U (0) + p+(T − s, x, y) (s, y + β(s))β(s)dy ds s=0 y=0 ∂y Z ∞ Z t + p+(T − s, x, y)u(s, y + β(s))dζs(y + β(s))dy y=0 s=0 Z t ˙ + % p+(T − s, x, 0)β(s)ds s=0 Now let T & t to get the claimed result. Note that (9) is not an explicit formula for u since the right-hand side of (9) depends on u through β. Also, ∂u 2 in order that u(t, ∞) = 0, 0 ≤ t < τ, it suffices that u(t, ·), ∂x (t, ·) ⊆ L ((β(t), ∞)), which is guaranteed in 2 2 R ∞ 0 2 R ∞ 2 R ∞ 0 2 the Picard Iteration part. This can be seen by |f (∞)| ≤ |f (M)| + M 2|ff | ≤ |f (M)| + M f + M (f ) 2 for M > 0, and lim infM→∞ f (M) = 0. 6 The value of Lemma 4.1 is that it implies that if β is continuously differentiable, then the Stefan boundary condition of (1) holds pointwise. Lemma 4.2. Let {u(t, ·) | 0 ≤ t < τ} be a solution of (1) with u(t, ∞) = 0 for 0 ≤ t < τ. and let u be continuously differentiable in x for x > β(t). Assume also that   Z ∞ ∂nu 2 (10)  sup (t, x) dx < ∞. E  n  0≤t<τ x=β(t) ∂x n∈{0,1} If β is continuously differentiable, then ∂u (11) lim (t, x) = −%β˙(t) x&β(t) ∂x for all t ∈ [0, τ). Proof. To see this, let’s rewrite (9) in a slightly more convenient way. If {u(t, ·) | 0 ≤ t < τ} is a weak solution of (1) and 0 < t < τ, set ∞ def Z A1(t, ε) = p+(t, ε, y)u◦(y)dy y=0 Z t Z ∞ def α(t−s) ∂u ˙ A2(t, ε) = e p◦(t − s, y + ε) (s, y + β(s))β(s)dy ds s=0 y=0 ∂x Z t Z Z ∞ def α(t−s) A3(t, ε) = e p◦(t − s, y + ε)u(s, y + β(s))η(y + β(s) − z)dy W (ds dz) s=0 z∈R y=0 Z t def ˙ A4(t, ε) = % p+(t − s, ε, 0)β(s)ds. 0 Note that Z t Z ∞ ∂u ˙ p+(t − s, ε, y) (s, y + β(s))β(s)dy ds = A2(t, ε) + A2(t, −ε) s=0 y=0 ∂x Z ∞ Z t p+(t − s, ε, y)u(s, y + β(s))dζs(y + β(s))dy = A3(t, ε) + A3(t, −ε) y=0 s=0 Thus u(t, β(t) + ε) = A1(t, ε) + A2(t, ε) + A2(t, −ε) + A3(t, ε) + A3(t, −ε) + A4(t, ε) and hence ∂u ∂A ∂A ∂A ∂A ∂A ∂A (t, β(t) + ε) = 1 (t, ε) + 2 (t, ε)+ 2 (t, −ε) + 3 (t, ε)+ 3 (t, −ε) + 4 (t, ε). ∂x ∂ε ∂ε ∂ε ∂ε ∂ε ∂ε

∂p+ Since ∂x (t, 0, y) = 0, we have ∂A lim 1 (t, ε) = 0. ε&0 ∂ε Next note that ∂p 1 x x2 ◦ (t, x) = − √ exp − . ∂x 2 4π t3/2 4t Thus ∂A Z t Z ∞ ∂p ∂u 2 (t, ε) = eα(t−s) ◦ (t − s, y + ε) (s, y + β(s))β˙(s)dy ds ∂ε s=0 y=0 ∂x ∂x " # 1 Z t Z ∞ y + ε (y + ε)2 ∂u = − √ eα(t−s) exp − (s, y + β(s))β˙(s)dy ds 3/2 2 4π s=0 y=0 (t − s) 4(t − s) ∂x 1 Z t eα(t−s) Z ∞ y2 ∂u √ √ √ ˙ = − √ y exp − (s, y t − s − ε + β(s))β(s)dy ds 2 4π s=0 t − s y=ε/ t−s 4 ∂x 7 Dominated convergence then implies that

∂A 1 Z t eα(t−s) Z ∞ y2 ∂u √ lim 2 (t, ε) = − √ √ y exp − (s, y t − s + β(s))β˙(s)dy ds ε→0 ∂ε 2 4π s=0 t − s y=0 4 ∂x

Turning to A3, we can integrate by parts and using the fact that u(t, β(t)) = 0, we get that

∂A Z t Z Z ∞ ∂p 3 (t, ε) = eα(t−s) ◦ (t − s, y + ε)u(s, y + β(s))η(y + β(s) − z)dy W (ds dz) ∂ε ∂x s=0 z∈R y=0 Z t Z Z ∞ ∂u = − eα(t−s) p (t − s, y + ε) (s, y + β(s))η(y + β(s) − z) ◦ ∂x s=0 z∈R y=0 +u(s, y + β(s))η ˙(y + β(s) − z)} dy W (ds dz).

Since Z T Z ∞ 2 sup p◦(t, y + ε)dy dt < ∞ ε∈(0,1) t=0 y=0 for all T > 0, we can use dominated convergence and (10) to get that

Z t Z Z ∞ ∂A3 α(t−s) ∂u lim (t, ε) = − e p◦(t − s, y) (s, y + β(s))η(y + β(s) − z) ε→0 ∂ε ∂x s=0 z∈R y=0 +u(s, y + β(s))η ˙(y + β(s) − z)} dy W (ds dz).

αt Finally we consider A4. Since p+(t, x, 0) = 2e p◦(t, x), we get that

∂p eα(t−s) x x2 + (t, x) = − √ exp − . ∂x 4π t3/2 4t Therefore ∂A Z t ε ε2 4 (t, ε) = −% eα(t−s) √ exp − β˙(s)ds 3/2 ∂x s=0 4π(t − s) 4(t − s) Z ∞ 2 2% αε2/u2 u ˙ 2 2 = −√ √ e exp − β(t − ε /u )du 4π u=ε/ t 4 Since β is continuously diﬀerentiable, dominated convergence ensures that

∂A 2%β˙(t) Z ∞ u2 lim 4 (t, ε) = − √ exp − du = −%β˙(t). ε&0 ∂x 4π u=0 4

Collecting things together, we get the claim.

4.2. A Transformation. The characterization of β given in (11) allows us to rewrite the moving boundary- value problem in a more convenient way. The calculation which gives us some analytical traction is found in [Lun04] (see also [Fri64, Ch. 8]). Let’s again return to our deterministic PDE (3). For all t ≥ 0 and x ∈ R, deﬁnev ˜(t, x) = v(t, x + β◦(t)); then v(t, x) =v ˜(t, x − β◦(t)). Assuming that β◦ is diﬀerentiable, we have that for x > 0 and t > 0, ∂v˜ ∂v ∂v (t, x) = (t, x + β (t)) + (t, x + β (t))β˙ (t) ∂t ∂t ◦ ∂x ◦ ◦ ∂v˜ ∂v (t, x) = (t, x + β (t)) ∂x ∂x ◦ ∂2v˜ ∂2v (t, x) = (t, x + β (t)). ∂x2 ∂x2 ◦ 8 We can combine these equations and use the PDE for v to rewrite the evolution ofv ˜ as ∂v˜ ∂2v (t, x) = (t, x + β (t)) + α v(t, x + β (t)) + v(t, x + β (t))b(t, x + β (t)) ∂t ∂x2 ◦ s ◦ ◦ ◦ ∂v (12) + (t, x + β (t))β˙ (t) ∂x ◦ ◦ ∂2v˜ ∂v˜ = (t, x) + α v˜(t, x) + (t, x)β˙ (t) +v ˜(t, x)b(t, x + β (t)). ∂x2 s ∂x ◦ ◦ ˙ ∂v Inserting the boundary condition that −%β◦(t) = limx&β◦(t) ∂x (t, x) back into (12), we have that ∂v˜ ∂2v˜ 1 ∂v˜ ∂v˜ (t, x) = (t, x) + α v˜(t, x) − (t, 0) (t, x) +v ˜(t, x)b(t, x + β (t)) t > 0, x > 0 ∂t ∂x2 s % ∂x ∂x ◦ v˜(t, 0) = 0 t > 0

v˜(0, x) = u◦(x) x > 0 1 ∂v˜ β˙ (t) = − (t, 0) t > 0 ◦ % ∂x def Replacing b by ζ and αs by α, we should be able to write down a nonlinear SPDE foru ˜(t, x) = u(t, x+β(t)). We now get the following.

Lemma 4.3. Suppose that {u(t, ·) | 0 ≤ t < τ} ⊂ C(R) ∩ L1(R) is a solution of (1) with u(t, ∞) = 0 such 1 ∂u 1 that u(t, ·) ∈ C ([β(t), ∞))) and ∂x (t, ·) ∈ L ([β(t), ∞)) for each 0 ≤ t < τ and (10) holds. Suppose also that β is continuously differentiable and {Ft}t≥0-adapted. Then u˜(t, x) = u(t, x + β(t)) satisfies the integral equation Z ∞ u˜(t, x) = p−(t, x, y)u◦(y)dy y=0 1 Z t Z ∞ ∂u˜ ∂u˜ (13) − p−(t − s, x, y) (s, 0) (s, y)dy ds % s=0 y=0 ∂x ∂x Z ∞ Z t + p−(t − s, x, y)˜u(s, y) dζs(y + β(s))dy y=0 s=0 for all t ∈ [0, τ) and x > 0 where 1 Z t ∂u˜ β(t) = − (s, 0)ds % s=0 ∂x for all t ∈ [0, τ). Proof. The proof is very similar to that of Lemma 4.1. Fix x > 0 and T > 0. For t ∈ [0, τ ∧ T ), define Z ∞ T def T T U (t) = u˜(t, y)p−(T − t, x, y)dy = A1 (t) + A2 (t). y=0 Using Definition 3.2, we have Z ∞ Z ∞ T U (t) = u(t, y + β(t))p−(T − t, x, y)dy = u(t, y)p−(T − t, x, y − β(t))dy y=0 y=β(t) Z = u(t, y)p−(T − t, x, y − β(t))dy y∈R T T T = U (0) + A1 (t) + A2 (t) where Z t Z ∂p ∂2p AT (t) = u(r, y) − − (T − r, x, y − β(r)) + − (T − r, x, y − β(r)) 1 ∂t ∂y2 r=0 y∈R ∂p +αp (T − r, x, y − β(r)) − − (T − r, x, y − β(r))β˙(r) dydr − ∂y 9 Z t ˙ + % p−(T − r, x, β(r) − β(r))β(r)dr, r=0 Z Z t T A2 (t) = u(r, y)p−(T − r, x, y − β(r))dζr(y)dy y∈R r=0

Using (6) and the fact that p−(T − t, x, 0) = 0, we get that (Z ∞ ) T ∂p− ˙ dA1 (t) = − u(t, y) (T − t, x, y − β(t))dx β(t)dt y=β(t) ∂y (Z ∞ ) ∂u ˙ = (t, y)p−(T − t, x, y − β(t))dx β(t)dt y=β(t) ∂y Z ∞ ∂u ˙ = (t, y + β(t))p−(T − t, x, y)dx β(t)dt y=0 ∂y Z ∞ ∂u˜ ˙ = (t, y)p−(T − t, x, y)dx β(t)dt. y=0 ∂y

We have here used the fact that u(t, β(t)) = 0. Combining the characterization of β˙ as in Lemma 4.2, we get that 1 ∂u˜ β˙(t) = − (t, 0). % ∂x Thus Z t Z ∞ T 1 ∂u˜ ∂u˜ A1 (t) = − p−(T − s, x, y) (s, 0) (s, y)dy ds. % s=0 y=0 ∂x ∂y T Now we consider A2 (t). Again using the stochastic Fubini theorem, we obtain Z Z t T A2 (t) = u(r, y)p−(T − r, x, y − β(r))dζr(y)dy y∈R r=0 Z Z t Z = u(r, y)p−(T − r, x, y − β(r)) η(y − z)W (dr, dz)dy y∈R r=0 z∈R Z t Z Z = u(r, y)p−(T − r, x, y − β(r))η(y − z)dy W (dr, dz) r=0 z∈R y∈R Z t Z (Z ∞ ) = u(r, y)p−(T − r, x, y − β(r))η(y − z)dy W (dr, dz) r=0 z∈R y=β(r) Z t Z Z ∞ = u(r, y + β(r))p−(T − r, x, y)η(y + β(r) − z)dy W (dr, dz) r=0 z∈R y=0 Z ∞ Z t = u(r, y + β(r))p−(T − r, x, y)dζr(y + β(r))dy y=0 r=0 Z ∞ Z t = u˜(r, y)p−(T − r, x, y)dζr(y + β(r))dy. y=0 r=0 Combine things to get that Z t Z ∞ T T 1 ∂u˜ ∂u˜ U (t) = U (0) − (s, 0) (s, y)p+(T − s, x, y)dy ds % s=0 y=0 ∂x ∂y Z ∞ Z t + u˜(s, y)p+(T − s, x, y)dζr(y + β(s))dy. y=0 s=0 Now let t % T to get the claimed result. 10 We can also obtain a SPDE foru ˜. By (6) and (13), we have that for 0 < t < τ and x > 0 Z ∞ 2 ∂ p− du˜(t, x) = 2 (t, x, y) + αp−(t, x, y) u◦(y)dy dt y=0 ∂x Z t Z ∞ 2 1 ∂ p− ∂u˜ ∂u˜ − 2 (t − s, x, y) + αp−(t − s, x, y) (s, 0) (s, y)dy ds dt % s=0 y=0 ∂x ∂x ∂x Z ∞ Z t 2 ∂ p− + 2 (t − s, x, y) + αp−(t − s, x, y) u˜(s, y) dζs(y + β(s))dy dt y=0 s=0 ∂x 1 ∂u˜ ∂u˜ − (t, 0) (t, x)dt +u ˜(t, x)dζ (x + β(t)). % ∂x ∂x t

Since p−(t, 0, y) = 0, we have the following SPDE ∂2u˜ 1 ∂u˜ ∂u˜ du˜(t, x) = (t, x) + αu˜(t, x) − (t, 0) (t, x) dt ∂x2 % ∂x ∂x

+u ˜(t, x)dζt(x + β(t)) 0 < t < τ, x > 0 (14) u˜(t, 0) = 0 0 < t < τ

u˜(0, x) = u◦(x) x > 0 1 ∂u˜ β˙(t) = − (t, 0) 0 < t < τ % ∂x We can also ﬁnd a converse to Lemma 4.3.

1 1 Lemma 4.4. Suppose that {u˜(t, ·) | 0 ≤ t < τ} ⊂ C (R+) ∩ L (R+) with u˜(t, ∞) = 0 for 0 ≤ t < τ satisﬁes (13) and is positive. Set 1 Z t ∂u˜ (15) β(t) = − (s, 0) ds 0 ≤ t < τ % s=0 ∂x and deﬁne ( def u˜(t, x − β(t)) x ≥ β(t), 0 ≤ t < τ (16) u(t, x) = 0 x < β(t), 0 ≤ t < τ Then {u(t, ·) | 0 ≤ t < τ} is a weak solution of (1).

∞ Proof. Fix ϕ ∈ Cc (R+ × R) and define for 0 ≤ t < τ, Z Z ∞ Z ∞ A(t) def= ϕ(t, x)u(t, x)dx = ϕ(t, x)˜u(t, x − β(t))dx = ϕ(t, x + β(t))˜u(t, x)dx x∈R x=β(t) x=0 To see the evolution of A, we fix δ > 0 and define ∞ def Z u˜δ(t, x) = p−(δ, x, y)˜u(t, y)dy x ≥ 0 y=0 ( def u˜δ(t, x − β(t)) x ≥ β(t), 0 ≤ t < τ uδ(t, x) = 0 x < β(t), 0 ≤ t < τ Then define Z ∞ δ def δ δ δ A (t) = ϕ(t, x + β(t))˜uδ(t, x)dx = A1(t) + A2(t) + A3(t) x=0 where Z t Z ∞ Z ∞ δ A1(t) = ϕ(t, x + β(t))p−(t + δ − s, x, y)ξ(s, y)dy dx ds s=0 x=0 y=0 Z ∞ Z ∞ Z t δ A2(t) = ϕ(t, x + β(t))p−(t + δ − s, x, y)˜u(s, y)dζs(y + β(s)) dy dx x=0 y=0 s=0 11 Z ∞ Z ∞ δ A3(t) = ϕ(t, x + β(t))p−(t + δ, x, y)˜u◦(y)dy dx x=0 y=0 where 1 ∂u˜ ∂u˜ ξ(t, x) = − (t, 0) (t, x) % ∂x ∂x We also note that we can rewrite the evolution of β as 1 ∂u β˙(t) = − (t, β(t)). t ∈ [0, τ) % ∂x Thus Z t Z ∞ Z ∞ δ ∂ϕ ∂ϕ ˙ dA1(t) = (t, x + β(t))p−(t + δ − s, x, y) + (t, x + β(t))p−(t + δ − s, x, y)β(t) s=0 x=0 y=0 ∂t ∂x ∂p +ϕ(t, x + β(t)) − (t + δ − s, x, y) ξ(s, y)dy dx ds dt ∂t Z ∞ Z ∞ + ϕ(t, x + β(t))p−(δ, x, y)ξ(t, y)dy dx dt x=0 y=0 Z t Z ∞ Z ∞ ∂ϕ = (t, x + β(t))p−(t + δ − s, x, y) s=0 x=0 y=0 ∂t ∂2p +ϕ(t, x + β(t)) − (t + δ − s, x, y) + αϕ(t, x + β(t))p (t + δ − s, x, y) ξ(s, y)dy dx ds ∂x2 − Z t Z ∞ Z ∞ ∂ϕ ˙ + (t, x + β(t))p−(t + δ − s, x, y)ξ(s, y)dy dx dsβ(t) dt s=0 x=0 y=0 ∂x Z ∞ Z ∞ + ϕ(t, x + β(t))p−(δ, x, y)ξ(t, y)dy dx dt. x=0 y=0 Similar calculations show that Z ∞ Z ∞ Z t δ ∂ϕ dA2(t) = (t, x + β(t))p−(t + δ − s, x, y) x=0 y=0 s=0 ∂t ∂2p +ϕ(t, x + β(t)) − (t + δ − s, x, y) + αϕ(t, x + β(t))p (t + δ − s, x, y) u˜(s, y)dζ (y + β(s)) dy dx ∂x2 − s Z ∞ Z ∞ Z t ∂ϕ ˙ + (t, x + β(t))p−(t + δ − s, x, y)˜u(s, y)dζs(y + β(s)) dy dx β(t) dt x=0 y=0 s=0 ∂x Z ∞ Z ∞ + ϕ(t, x + β(t))p−(δ, x, y)˜u(t, y)dζt(y + β(t)) dy dx x=0 y=0 and finally Z ∞ Z ∞ δ ∂ϕ dA3(t) = (t, x + β(t))p−(t + δ, x, y) x=0 y=0 ∂t ∂2p +ϕ(t, x + β(t)) − (t + δ, x, y) + αϕ(t, x + β(t))p (t + δ, x, y) u˜ (y)dy dx ∂x2 − ◦ Z ∞ Z ∞ ∂ϕ ˙ + (t, x + β(t))p−(t + δ, x, y)û◦(y)dy dx β(t) dt. x=0 y=0 ∂x

Adding these expressions together and using the deﬁnition ofu ˜δ and (13), we get that Aδ(t) − Aδ(0) ( Z t Z ∞ ∂ϕ Z ∞ Z s Z ∞ = + αϕ (s, x + β(s)) p−(s + δ, x, y)˜u0(y)dy + p−(s + δ − r, x, y)ξ(r, y)dydr s=0 x=0 ∂t y=0 r=0 y=0 12 Z ∞ Z s ) + p−(s + δ − r, x, y)˜u(r, y)dζr(y + β(r))dy dxds y=0 r=0 Z t Z ∞ ( Z ∞ 2 Z s Z ∞ 2 ∂ p− ∂ p− + ϕ(s, x + β(s)) 2 (s + δ, x, y)˜u0(y)dy + 2 (s + δ − r, x, y)ξ(r, y)dydr s=0 x=0 y=0 ∂x r=0 y=0 ∂x Z ∞ Z s 2 ) ∂ p− + 2 (s + δ − r, x, y)˜u(r, y)dζr(y + β(r))dy dxds y=0 r=0 ∂x ( Z t Z ∞ ∂ϕ Z ∞ Z s Z ∞ + (s, x + β(s)) p−(s + δ, x, y)˜u0(y)dy + p−(s + δ − r, x, y)ξ(r, y)dydr s=0 x=0 ∂x y=0 r=0 y=0 Z ∞ Z s ) ˙ + p−(s + δ − r, x, y)˜u(r, y)dζr(y + β(r))dy dxβ(s)ds y=0 r=0 Z t Z ∞ Z ∞ + ϕ(s, x + β(s))p−(δ, x, y)ξ(s, y)dy dx ds s=0 x=0 y=0 Z ∞ Z ∞ Z t + ϕ(s, x + β(s))p−(δ, x, y)˜u(s, y)dζs(y + β(s)) dy dx x=0 y=0 s=0 Z t Z ∞ ∂ϕ = + αϕ (s, x + β(s))˜uδ(s, x)dx s=0 x=0 ∂t Z ∞ 2 Z ∞ ∂ u˜δ ∂ϕ ˙ + ϕ(s, x + β(s)) 2 (s, x)dx + (t, x + β(t))˜uδ(s, x)dxβ(s) ds x=0 ∂x x=0 ∂x Z t Z ∞ Z ∞ + ϕ(s, x + β(s))p−(δ, x, y)ξ(s, y)dy dx ds s=0 x=0 y=0 Z ∞ Z ∞ Z t + ϕ(s, x + β(s))p−(δ, x, y)˜u(s, y)dζs(y + β(s)) dy dx x=0 y=0 s=0 Z t Z ∞ ∂ϕ = + αϕ (s, x + β(s))uδ(s, x + β(s))dx s=0 x=0 ∂t Z ∞ 2 Z ∞ ∂ uδ ∂ϕ ˙ + ϕ(s, x + β(s)) 2 (s, x + β(s))dx + (s, x + β(s))uδ(s, x + β(s))dxβ(s) ds x=0 ∂x x=0 ∂x Z t Z ∞ Z ∞ + ϕ(s, x + β(s))p−(δ, x, y)ξ(s, y)dx dy ds s=0 x=0 y=0 Z ∞ Z ∞ Z t + ϕ(s, x + β(s))p−(δ, x, y)˜u(s, y)dζs(y + β(s)) dy dx x=0 y=0 s=0 Z t Z ∞ ∂ϕ ∂2ϕ = + 2 + αϕ (s, x + β(s))uδ(s, x + β(s))dx s=0 x=0 ∂t ∂x Z ∞ ∂u − ϕ(s, x + β(s)) δ (s, x + β(s))dxβ˙(s) x=0 ∂x ∂u ∂ϕ −ϕ(s, β(s)) δ (s, β(s)) + (s, β(s))u (s, β(s)) − ϕ(s, β(s))u (s, β(s))β˙(s)dx ds ∂x ∂x δ δ Z t Z ∞ Z ∞ + ϕ(s, x + β(s))p−(δ, x, y)ξ(s, y)dy dx ds s=0 x=0 y=0 Z ∞ Z ∞ Z t + ϕ(s, x + β(s))p−(δ, x, y)˜u(s, y)dζs(y + β(s)) dy dx. x=0 y=0 s=0

13 By definition of p−, we conclude that uδ(s, β(s)) = 0. In addition, we also have that ∂u˜ Z ∞ ∂p lim δ (t, x) = lim − (δ, x, y)˜u(t, y)dy δ→0 ∂x δ→0 y=0 ∂x Z ∞ ∂p = lim − + (δ, x, y)˜u(t, y)dy δ→0 y=0 ∂y Z ∞ ∂u˜ = lim p+(δ, x, y) (t, y)dy δ→0 y=0 ∂y ∂u˜ = (t, x). ∂x Upon letting δ & 0 and rearranging things, we indeed get a weak solution of (1). 5. A Picard Iteration Our main task now is to show that we can indeed solve (13). The main complication is that (13) is fully 1 ∂u˜ nonlinear due to the presence of the % ∂x (t, 0) term in the drift and the shift by β in the evaluation of the integral against ζ. If we turn off the noise, we can do this via semigroup theory as in [Lun04]. The noise, however, complicates things, as we need to respect the rules of Ito integration and (unless we want to use more advanced theories of stochastic integrals) integrate against predictable functions. Our approach will be to set up a functional framework in which we can use Picard-type iterations to show ∞ existence and uniqueness. As usual, C0 (R+) is the collection of infinitely smooth functions on [0, ∞) which asymptotically vanishes at infinity. Define next

∞ def n ∞ (n) o C0,odd(R+) = ϕ ∈ C0 (R+) | ϕ (0) = 0 for all even n ∈ N ; ∞ ∞ in other words, C0,odd(R+) are those elements of C0 (R+) which can be extended to an odd element of ∞ ∞ C (R) (namely, consider the map y 7→ sgn(y)ϕ(|y|). For all ϕ ∈ C0 (R+), deﬁne v u 2 Z def uX (i) 2 kϕkH = t ϕ (x) dx. i=0 x∈(0,∞) ∞ ∞ Let H be the closure of C0 (R+) with respect to k · kH and let Hodd be the closure of C0,odd(R+) with respect to k · kH . We also deﬁne sZ def 2 kϕkL = |ϕ(x)| dx x∈(0,∞)

for all square-integrable functions on R+. Of course H and Hodd are Hilbert spaces (H is more commonly 2 written as H ; i.e., it is the collection of functions on R+ which possess two weak square-integrable derivatives). The important aspect of H is the following fairly standard result. Lemma 5.1. We have that H ⊂ C1. More precisely, for any ϕ ∈ H, we have that

(i) sup ϕ (x) ≤ 2kϕkH . x∈R+ i∈{0,1}

(i) def (i) Finally, for i ∈ {0, 1}, ϕ (0) = limx&0 ϕ (x) is well-defined. The proof is in Subsection 5.1. We next need some truncation functions to control the effects of various nonlinearities. In particular, ∂u˜ we need to prevent ∂x (t, 0) from becoming too big; Picard iterations in general allow only linear growth of ∞ various coefficients. Fix L > 0, which we will use as a truncation parameter. Let ΨL ∈ C (R; [0, 1]) be monotone decreasing such that ΨL(x) = 1 if |x| ≤ L and ΨL(x) = 0 if |x| ≥ L + 1 (and thus |ΨL| ≤ 1). In other words, ΨL is a cutoff function with support of width L + 1. Define Z ∞ L u˜1 (t, x) = p−(t, x, y)u◦(y)dy y=0 14 for all t > 0 and x ∈ R and recursively define Z t L L def 1 ∂uñ βn (t) = − (s, 0)ds t > 0 % s=0 ∂x Z ∞ L uñ+1(t, x) = p−(t, x, y)u◦(y)dy y=0 (17) Z t Z ∞ L L 1 ∂uñ ∂uñ L − p−(t − s, x, y) (s, 0) (s, y)ΨL kuñ (s, ·)kH dy ds % s=0 y=0 ∂x ∂x Z t Z ∞ L L L + p−(t − s, x, y)˜un (s, y)ΨL kuñ (s, ·)kH dζs(y + βn (s))dy. t > 0, x > 0 s=0 y=0 L For each n ∈ N, {uñ (t, ·); t ≥ 0} is a well-defined, adapted, and continuous path in Hodd. ∞ To study (17), we will use the Dirichlet heat semigroup. For ϕ ∈ C0 (R+), t > 0, and x > 0, define ∞ def Z (Ttϕ)(x) = p−(t, x, y)ϕ(y)dy. y=0 ∞ Lemma 5.2. For each t > 0, Tt has a unique extension from C0 (R+) to H such that TtH ⊂ Hodd and αt such that kTtfkH ≤ e kfkH for all f ∈ H. Secondly, there is a KA > 0 such that K kT f˙k ≤ A kfk t H t3/4 H 3 for all f ∈ Hodd ∩ C (R+). Again, we delay the proof until Subsection 5.1. Another convenience will be to rewrite the ds part of (17). Set 1 Ψ˜ a (ψ) def= − ψ˙(0)Ψ (kψk ) and Ψ˜ b (ψ) def= Ψ (kψk ) L % L H L L H for all ψ ∈ H. Then ∂u˜L 1 ∂u˜L ∂u˜L − n (t, x) n (t, 0)Ψ ku˜L(t, ·)k = n (t, x)Ψ˜ a (˜uL (t, ·)) ∂x % ∂x L n H ∂x L N L ˜ b L ΨL kuñ (t, ·)kH = ΨL(˜uN (t, ·)) for all n ∈ N. For ψ and η in H, let’s also define ˜ a def 1 1 ˙ ˙ hψ, ηiH (DΨL)(ψ, η) = − η˙(0)ΨL(kψkH ) − ψ(0)ΨL(kψkH ) % % kψkH ˜ b def ˙ hψ, ηiH (DΨL)(ψ, η) = ΨL(kψkH ) . kψkH ˜ a ˜ a Lemma 5.3. For each ψ and η in H, (DΨL)(ψ, η) is the Gâteaux derivative of ΨL at ψ in the direction of ˜ b ˜ b η and similarly (DΨL)(ψ, η) is the Gâteaux derivative of ΨL at ψ in the direction of η. Furthermore, there is a KB > 0 such that

˜ a (DΨL)(ψ, η) ≤ KBχ[0,L+1](kψkH )kηkH

˜ b (DΨL)(ψ, η) ≤ KBχ[0,L+1](kψkH )kηkH for all ψ and η in H and L > 0. Proof. The claim is straightforward. L def L L For each n ∈ N, we now definew ñ (t, x) =u ñ+1(t, x) − uñ (t, x) for all x ≥ 0 and t ≥ 0. Clearly L 2 sup0≤t≤T E kw˜1 kH < ∞ for all T > 0. We then write that 5 L X (n) wñ+1(t, x) = Aj (t, x) j=1 15 where Z 1 Z t Z ∞ L (n) ∂wñ ˜ a L L A1 (t, x) = p−(t − s, x, y) (s, y)dy ΨL uñ (s, ·) + λwñ (s, ·) ds dλ λ=0 s=0 y=0 ∂x Z 1 Z t L ∂wñ ˜ a L L = Tt−s (s, ·) (x)ΨL uñ (s, ·) + λwñ (s, ·) ds dλ λ=0 s=0 ∂x Z 1 Z t Z ∞ L L (n) ∂uñ ∂wñ A2 (t, x) = p−(t − s, x, y) (s, y) + λ (s, y) dy λ=0 s=0 y=0 ∂x ∂x ˜ a L L L × DΨL uñ (s, ·) + λwñ (s, ·), wñ (s, ·) ds dλ Z 1 Z t L L ∂uñ ∂wñ ˜ a L L L = Tt−s (s, ·) + λ (s, ·) (x)DΨL uñ (s, ·) + λwñ (s, ·), wñ (s, ·) ds dλ λ=0 s=0 ∂x ∂x Z 1 Z t Z ∞ (n) L A3 (t, x) = p−(t − s, x, y)w ñ (s, y) λ=0 s=0 y=0 ˜ b L L L L L × ΨL uñ (s, ·) + λwñ (s, ·) dζs y + βn (s) + λ(βn+1(s) − βn (s)) dy dλ Z 1 Z t Z ∞ (n) L L A4 (t, x) = p−(t − s, x, y) uñ (s, y) + λwñ (s, y) λ=0 s=0 y=0 ˜ b L L L L L L × DΨL uñ (s, ·) + λwñ (s, ·), wñ (s, ·) dζs y + βn (s) + λ(βn+1(s) − βn (s)) dy dλ Z 1 Z ∞ Z t Z (n) L L ˜ b L L A5 (t, x) = p−(t − s, x, y) uñ (s, y) + λwñ (s, y) ΨL uñ (s, ·) + λwñ (s, ·) λ=0 y=0 s=0 z∈R L L L L L × βn+1(s) − βn (s) η˙ y − z + βn (s) + λ(βn+1(s) − βn (s)) W (ds, dz)dy dλ.

˙ ∂ζt L L where ζt(x) = ∂x (x). Note that theu ˜n ’s andw ˜n ’s are all in Hodd. (n) (n) −3/4 To bound A1 and A2 , we use the fact that t is locally integrable. More precisely, Z t 1 ds = 4t1/4 3/4 s=0 (t − s) for all t > 0. Thus

" Z t L 2# h (n) 2 i ∂w˜n E kA1 (t, ·)kH ≤ E Tt−s (s, ·) ds s=0 ∂x H  2 h 2 i t L t w˜L(s, ·) Z w˜ (s, ·) Z E n H ≤ K2 n H ds ≤ 4K2 t1/4 ds. AE  3/4  A 3/4 s=0 (t − s) s=0 (t − s) Similarly, we have that

h (n) 2 i E kA2 (t, ·)kH " Z t L L 2# 2 ∂uñ ∂wñ L L L ≤ KBE Tt−s (s, ·) + λ (s, ·) χ[0,L+1] uñ (s, ·) + λwñ (s, ·) H kwñ (s, ·)kH ds s=0 ∂x ∂x H " 2# Z t L + 1 Z t [kw˜L(s, ·)k2 ] ≤ K2 K2 kw˜L(s, ·)k ds ≤ 4K2 K2 (L + 1)2t1/4 E n H ds. A BE 3/4 n H A B 3/4 s=0 (t − s) s=0 (t − s)

To bound A(n), A(n), and A(n), we first rewrite them. For z ∈ , define η (y) def= η(y − z) for all y ∈ . 3 4 5 R z R Then Z t Z Z 1 Z ∞ (n) L ˜ b L n A3 (t, x) = p−(t − s, x, y)w ñ (s, y)ΨL uñ (s, ·) + λw˜L(s, ·) s=0 z∈R λ=0 y=0 L L L ×η y − z + βn (s) + λ(βn+1(s) − βn (s)) dy dλ W (ds, dz) 16 Z t Z Z 1 L = Tt−s wñ (s, ·)η L L L (x) z−βn (s)−λ(βn+1(s)−βn (s)) s=0 z∈R λ=0 ˜ b L L o ×ΨL uñ (s, ·) + λwñ (s, ·) dλ W (ds, dz) Z t Z Z 1 Z ∞ (n) L L ˜ b L L L A4 (t, x) = p−(t − s, x, y) uñ (s, y) + λwñ (s, y) DΨL uñ (s, ·) + λwñ (s, ·), wñ (s, ·) s=0 z∈R λ=0 y=0 L L L ×η y − z + βn (s) + λ(βn+1(s) − βn (s)) dy dλ W (ds, dz) Z t Z Z 1 L L = Tt−s uñ (s, y) + λwñ (s, y) η L L L (x) z−βn (s)−λ(βn+1(s)−βn (s)) s=0 z∈R λ=0 ˜ b L L L o ×DΨL uñ (s, ·) + λwñ (s, ·), wñ (s, ·) dλ W (ds, dz) Z t Z Z 1 Z ∞ (n) L L ˜ b L L A5 (t, x) = p−(t − s, x, y) uñ (s, y) + λwñ (s, y) ΨL uñ (s, ·) + λwñ (s, ·) s=0 z∈R λ=0 y=0 L L L L L × βn+1(s) − βn (s) η˙ y − z + βn (s) + λ(βn+1(s) − βn (s)) dy dλ W (ds, dz) Z t Z Z 1 L L = Tt−s uñ (s, ·) + λwñ (s, ·) η˙ L L L (x) z−βn (s)−λ(βn+1(s)−βn (s)) s=0 z∈R λ=0 ˜ b L L o L L ×ΨL uñ (s, ·) + λwñ (s, ·) dλ βn+1(s) − βn (s) W (ds, dz) We will use the following bound on the interaction between η and the H-norm. Lemma 5.4. There is a K > 0 such that Z kfη(k)k2 dy ≤ Kkfk2 y H H y∈R for all f ∈ H and k ∈ {0, 1}.

Proof. The structure of η ensures that there is anη ˆ ∈ L2(R) such that |η(n)(x)| ≤ ηˆ(x) for all n ∈ {0, 1, 2, 3} and x ∈ R. Thus for all x ∈ R, k ∈ {0, 1}, and n ∈ {0, 1, 2},

n (k) (n) X n (j) (k+n−j) (fη ) (x) = f (x)η (x − y) y j j=0 n 2 X n X ≤ |f (j)(x)|ηˆ(x − y) ≤ 2 |f (j)(x)|ηˆ(x − y) j j=0 j=0 Thus v s u 2 Z X Z Z kfη(k)k2 dy ≤ 6u |f (j)(x)|2ηˆ2(x − y)dx dy y H t y∈R j=0 y∈R x∈R v u 2 uX Z 2 (j) 2 ≤ 6kηˆkL (R)t |f (x)| dx. j=0 x∈R The claim follows. def (n) Fix T > 0 and set K1 = exp(2|α|T ). Let’s first bound A3 . For k ∈ {0, 1, 2},   ∂kT w˜L(s, ·)η k (n) Z t Z Z 1 t−s n z−βn(s)−λ(βL (s)−βL(s)) ∂ A3  L n+1 n k (t, x) = k (x) ∂x s=0 z∈  λ=0 ∂x R  ˜ b L L o ×ΨL uñ (s, ·) + λwñ (s, ·) dλ W (ds, dz); 17 thus for 0 ≤ t ≤ T  2  Z ∞ ∂kA(n) 3 (t, x) dx E  k  x=0 ∂x   ∂kT w˜L(s, ·)η Z t Z Z ∞ Z 1 t−s n z−βL(s)−λ(βL (s)−βL(s))  n n+1 n = E  k (x) s=0 z∈ x=0  λ=0 ∂x R  2 ˜ b L L o ×ΨL un (s, ·) + λwñ (s, ·) dλ dx dz ds  2  ∂kT w˜L(s, ·)η Z 1 Z t Z Z ∞ t−s n z−βL(s)−λ(βL (s)−βL(s))  n n+1 n  ≤ E  (x) dx dz ds dλ  ∂xk  λ=0 s=0 z∈R  x=0 

Z 1 Z t Z " 2 # L ≤ E Tt−s wñ (s, ·)η L L L dz ds dλ z−βn (s)−λ(βn+1(s)−βn (s)) λ=0 s=0 z∈R H Z 1 Z t Z " 2 # L ≤ K1 E wñ (s, ·)η L L L dz ds dλ z−βn (s)−λ(βn+1(s)−βn (s)) λ=0 s=0 z∈R H Z 1 Z t h L 2 i ≤ K1K E wñ (s, ·) H ds dλ λ=0 s=0 Z t h L 2 i ≤ K1K E wñ (s, ·) H ds s=0 (n) The bound on A4 is similar.   ∂kT u˜L(s, ·) + λw˜L(s, ·) η k (n) Z t Z Z 1 t−s n n z−βL(s)−λ(βL (s)−βL(s)) ∂ A4  n n+1 n k (t, x) = k (x) ∂x s=0 z∈  λ=0 ∂x R  ˜ b L L L o ×DΨL uñ (s, ·) + λwñ (s, ·), wñ (s, ·) dλ W (ds, dz). Consequently for 0 ≤ t ≤ T  2  Z ∞ ∂kA(n) 4 (t, x) dx E  k  x=0 ∂x   ∂kT u˜L(s, ·) + λw˜L(s, ·) η Z t Z Z ∞ Z 1 t−s n n z−βL(s)−λ(βL (s)−βL(s))  n n+1 n = E  k (x) s=0 z∈ x=0  λ=0 ∂x R  2 ˜ b L L L o ×DΨL uñ (s, ·) + λwñ (s, ·), wñ (s, ·) dλ dx dz ds  2 ∂kT u˜L(s, ·) + λw˜L(s, ·) η Z 1 Z t Z Z ∞ t−s n n z−βL(s)−λ(βL (s)−βL(s))  n n+1 n ≤ E  (x) dx  ∂xk λ=0 s=0 z∈R  x=0

2 ˜ b L L L × DΨL uñ (s, ·) + λwñ (s, ·), wñ (s, ·) dz ds dλ

18 Z 1 Z t Z " 2 L L ≤ E Tt−s uñ (s, ·) + λwñ (s, ·) η L L L z−βn (s)−λ(βn+1(s)−βn (s)) λ=0 s=0 z∈R H 2 ˜ b L L L × DΨL uñ (s, ·) + λwñ (s, ·), wñ (s, ·) dz ds dλ

Z 1 Z t Z " 2 2 L L ≤ K1KB E uñ (s, y) + λwñ (s, y) η L L L z−βn (s)−λ(βn+1(s)−βn (s)) λ=0 s=0 z∈R H L L L 2 i ×χ[0,L+1] uñ (s, ·) + λwñ (s, ·) H wñ (s, ·) H dz ds dλ Z 1 Z t 2 2 h L 2 i ≤ K1KKB(L + 1) E wñ (s, ·) H ds dλ λ=0 s=0 Z t 2 2 h L 2 i ≤ K1KKB(L + 1) E wñ (s, ·) H ds s=0 (n) L L To bound A5 , we first bound βn+1 − βn . We have that Z t L L L L 1 ∂uñ+1 ∂uñ βn+1(t) − βn (t) ≤ (s, 0) − (s, 0) ds % s=0 ∂x ∂x s Z t Z t 2 L 2 L 2 ≤ kwñ (s, ·)kH ds ≤ t kwñ (s, ·)kH ds % s=0 % s=0 For k ∈ {0, 1, 2}, we then have that   ∂kT u˜L(s, ·) + λwñ(s, ·) η˙ k (n) Z t Z Z 1 t−s n L z−βL(s)−λ(βL (s)−βL(s)) ∂ A5  n n+1 n k (t, x) = k (x) ∂x s=0 z∈  λ=0 ∂x R  ˜ b L L o L L ×ΨL uñ (s, ·) + λwñ (s, ·) dλ βn+1(s) − βn (s) W (ds, dz). Hence for 0 ≤ t ≤ T  2  Z ∞ ∂kA(n) 5 (t, x) dx E  k  x=0 ∂x    ∂kT u˜L(s, ·) + λwñ(s, ·) η˙ Z t Z Z ∞ Z 1 t−s n L z−βL(s)−λ(βL (s)−βL(s))   n n+1 n = E  k (x) s=0 z∈  x=0  λ=0 ∂x R  2 ˜ b L L o L L 2 ×ΨL uñ (s, ·) + λwñ (s, ·) dλ dx βn+1(s) − βn (s) ds dz  2 ∂kT u˜L(s, ·) + λw˜L(s, ·) η˙ Z 1 Z t Z Z ∞ t−s n n z−βL(s)−λ(βL (s)−βL(s))  n n+1 n ≤ E  (x) dx  ∂xk λ=0 s=0 z∈R  x=0

2 ˜ b L L L L 2 × ΨL u˜n (s, ·) + λw˜n (s, ·) βn+1(s) − βn (s) ds dz dλ

Z 1 Z t Z " 2 L L ≤ E Tt−s (˜un (s, ·) + λwñ (s, ·))˙η L L L z−βn (s)−λ(βn+1(s)−βn (s)) λ=0 s=0 z∈R H 2 ˜ b L L L L 2 × ΨL uñ (s, ·) + λwñ (s, ·) βn+1(s) − βn (s) ds dz dλ

19 Z 1 Z t Z " 2 L L ≤ K1 E (˜un (s, ·) + λwñ (s, ·))˙η L L L z−βn (s)−λ(βn+1(s)−βn (s)) λ=0 s=0 z∈R H 2 ˜ b L L L L 2 × ΨL uñ (s, ·) + λwñ (s, ·) βn+1(s) − βn (s) ds dz dλ Z 1 Z t 2 h L L 2i ≤ K1K(L + 1) E βn+1(s) − βn (s) ds dλ λ=0 s=0 2 Z t 4K1K(L + 1) t L 2 ≤ 2 E kwñ (s, ·)kH ds. % s=0

P∞ L L L def Lemma 5.5. For each T > 0, we have that n=1 sup0≤t≤T E kuñ+1 − uñ kH < ∞. Thus P-a.s., u˜ (t, ·) = L L limn→∞ uñ (t, ·) exists as a limit in C([0,T ]; H) and u satisfies the integral equation Z ∞ L u˜ (t, x) = p−(t, x, y)˜u◦(y)dy y=0 Z t Z ∞ L L 1 ∂u˜ ∂u˜ L (18) − p−(t − s, x, y) (s, 0) (s, y)ΨL ku˜ (s, ·)kH }dy ds % s=0 y=0 ∂x ∂x Z ∞ Z t L L + p−(t − s, x, y)˜u (s, y)ΨL ku˜ (s, ·)kH dζs(y + β(s))dy. t > 0, x > 0 y=0 s=0 where t L def 1 Z ∂u β(t) = − (s, 0)ds. ρ s=0 ∂x Proof. See also [Wal86, Lemma 3.3]. Fixing T > 0 we collect the above calculations to see that there is a KT,L > 0 such that Z t [kw˜L(s, ·)k2 ] [kw˜L (t, ·)k2 ] ≤ K E n H ds E n+1 H T,L 3/4 s=0 (t − s) for all t ∈ [0,T ]. Iterating this, we get that

n−2  L 2 n−1 (n−1)/4 Y  L 2 E[kwñ (t, ·)kH ] ≤ KT,L t B(1 + j/4, 1/4) sup E[kw˜1 kH ] j=0  0≤t≤T where B is the standard Beta function and thus that  1/2 q n−2  q L 2 (n−1)/2 (n−1)/8 Y L 2 E[kwñ (t, ·)kH ] ≤ KT,L t B(1 + j/4, 1/4) sup E[kw˜1 kH ]. j=0  0≤t≤T To show that the terms on the right are summable, we use the ratio test. It suffices to show that n − 2 1/2 (19) lim K1/2 t1/8 B 1 + , 1/4 = 0. n→∞ T,L 4 We calculate that Z 1 Z 1/2 Z 1 B(1 + n/4, 1/4) = sn/4(1 − s)−3/4ds = sn/4(1 − s)−3/4ds + sn/4(1 − s)−3/4ds s=0 s=0 s=1/2 1 Z 1− n Z 1 ≤ n3/4 sn/4ds + (1 − s)−3/4ds 1 s=0 s=1− n 4n3/4 1 1/4 ≤ + 4 . n + 4 n This implies (19). The rest of the proof follows by Jensen’s inequality and standard calculations. We can finally show uniqueness. 20 Lemma 5.6. The solution of (18) is unique.

def Proof. Let u1 and u2 be two solutions. Deﬁnew ˜ = u1 − u2. By calculations as above we get that Z t 2 −3/4 2 E[kw˜(t, ·)kH ] ≤ KT,L (t − s) E[kw˜(s, ·)kH ]ds. s=0 We can iterate this inequality several times to get (cf. [Wal86, Theorem 3.2]) Z t Z s 2 2 −3/4 −3/4 2 E[kw˜(t, ·)kH ] ≤ KT,L (t − s) (s − r) E[kw˜(r, ·)kH ]dr ds s=0 r=0 Z t 2 −2/4 2 = KT,LB(1/4, 1/4) (t − r) E[kw˜(r, ·)kH ]dr r=0 Z t Z r 3 −2/4 −3/4 2 ≤ KT,LB(1/4, 1/4) (t − r) (r − s) E[kw˜(s, ·)kH ]ds dr r=0 s=0 Z t 3 −1/4 2 = KT,LB(1/4, 1/4)B(1/2, 1/4) (t − s) E[kw˜(s, ·)kH ]ds s=0 Z t Z s 4 −1/4 −3/4 2 ≤ KT,LB(1/4, 1/4)B(1/2, 1/4) (t − s) (s − r) E[kw˜(r, ·)kH ]dr ds s=0 r=0 Z t 4 2 = KT,LB(1/4, 1/4)B(1/2, 1/4)B(3/4, 1/4) E[kw˜(r, ·)kH ]dr r=0 We can now use Gronwall’s inequality. We can also show non-negativity. Deﬁne the random time def L τL = inf t ≥ 0 : ku˜ (t, ·)kH ≥ L for L > 0.

L Lemma 5.7. The solution u˜ (t, x) of (18) is nonnegative for 0 ≤ t ≤ τL and x ≥ 0.

L def L −αt L Proof. Letu ˜α(t, x) =u ˜ (t, x)e . Since we have that ΨL(ku˜ (t, ·)kH ) = 1 for 0 ≤ t ≤ τL, from (18) we obtain that for 0 ≤ t ≤ τL ∂2u˜L 1 ∂u˜L ∂u˜L du˜L(t, x) = α (t, x) − eαt α (t, 0) α (t, x) dt α ∂x2 % ∂x ∂x L x +u ˜α(t, x)dBt L u˜α(t, 0) = 0 L u˜α(0, x) = u◦(x) > 0, x where Bt is defined as in (7). We will then follow the approach used in [Cho08] to show non-negativity of L L ∞ u˜α which implies non-negativity ofu ˜ . To start, fix a nonnegative and nonincreasing η ∈ C (R) such that η(u) = 2 if u ≤ −1 and η(u) = 0 if u ≥ 0. Define u r def Z Z ϕ(u) = η(s)ds dr u ∈ R r=0 s=0 def 2 u L Finally define ϕε(u) = ε ϕ ε for all u ∈ R. Fixing x ∈ R+ and applying Ito’s formula to {ϕε(˜uα(t, x)) : 0 ≤ t < τL}, we have that

Z t∧τL 2 L L L ∂ u˜α ϕε(˜uα(t ∧ τL, x)) − ϕε(u◦(x)) = ϕ˙ ε(˜uα(s, x)) 2 (s, x)ds s=0 ∂x Z t∧τL L L 1 ∂u˜α αs L ∂u˜α (20) − (s, 0)e ϕ˙ ε(˜uα(s, x)) (s, x) ds % s=0 ∂x ∂x Z t∧τL Z t∧τL 1 L L 2 L L x + ϕ¨ε(˜uα(s, x)) u˜α(s, x) ds + ϕ˙ ε(˜uα(s, x))˜uα(s, x)dBs . 2 s=0 s=0 21 Here ϕε(u◦(x)) = 0 since u◦ ≥ 0. Then (20) implies that Z t∧τL ∂2u˜L L −t∧τL L α −s ϕε(˜uα(t ∧ τL, x))e − ϕε(u◦(x)) = ϕ˙ ε(˜uα(s, x)) 2 (s, x)e ds s=0 ∂x 1 Z t∧τL ∂u˜L ∂u˜L − α (s, 0) ϕ˙ (˜uL(s, x)) α (s, x) e(α−1)sds % ∂x ε α ∂x (21) s=0 Z t∧τL Z t∧τL 1 L L 2 −s L L −s x + ϕ¨ε(˜uα(s, x)) u˜α(s, x) e ds + ϕ˙ ε(˜uα(s, x))˜uα(s, x)e dBs 2 s=0 s=0 Z t∧τL L −s − ϕε(˜uα(s, x))e ds. s=0 ∞ Next fix a nonincreasing $ ∈ C (R+) such that $(x) = 1 for x ≤ 1 and $(x) = 0 for x ≥ 2. For each def N ∈ N, define $N (x) = $(x/N). Let’s now do several things. Let’s multiply (21) with $N . Let’s then integrate in space, and finally take expectations. We get that

Z ∞ Z t∧τL L −t∧τL N,ε −s E ϕε(˜uα(t ∧ τL, x))$N (x)e dx = E A1 (s)e ds (22) x=0 s=0 Z t∧τL L Z t∧τL 1 ∂u˜α αs N,ε −s 1 N,ε −s − E (s, 0)e A2 (s)e ds + E A3 (s)e ds % s=0 ∂x 2 s=0 where Z ∞ 2 L N,ε def L ∂ u˜α A1 (s) = ϕ˙ ε(˜uα(s, x)) 2 (s, x)$N (x)dx x=0 ∂x Z ∞ L N,ε def L ∂u˜α A2 (s) = ϕ˙ ε(˜uα(s, x)) (s, x)$N (x)dx x=0 ∂x Z ∞ N,ε def n L L 2 L o A3 (s) = ϕ¨ε(˜uα(s, x)) u˜α(s, x) − 2ϕε(˜uα(s, x)) $N (x)dx. x=0

We need some bounds on ϕε. Deﬁne k · kC as the sup norm over R. First note thatϕ ¨(u) − 2χR− = η(u) − 2χR− . This implies that for all u ∈ R,

|ϕ¨(u) − 2χR− | ≤ kη − 2kC χ[−1,0](u), |ϕ˙(u) − 2uχR− | ≤ kη − 2kC (23) 2 |ϕ(u) − u χR− | ≤ kη − 2kC |u|; the ﬁrst bound is direct, and the second two follow by integration. Note also that since η is bounded, 1 (24) |ϕ¨(u)| ≤ kηk , |ϕ˙(u)| ≤ kηk |u|, and |ϕ(u)| ≤ kηk u2. C C 2 C Let’s understand the behavior of the various terms of (22) as N → ∞ and then ε → 0. From the last 2 bound of (23), we have that limε→0 ϕε(x) = x χR− (x). Thanks to the last bound of (24), we can use dominated convergence and thus conclude that Z ∞ L −t∧τL lim lim E ϕε(˜uα(t ∧ τL, x))$N (x)e dx ε→0 N→∞ x=0 ∞ Z 2 L L −t∧τL = E u˜α(t ∧ τL, x) χR− (˜uα(t ∧ τL, x))e dx . x=0 N,ε We next consider A1 (s). Integrating by parts and using the boundary conditions at x = 0, we have that Z ∞ L L 2 Z ∞ L L N,ε u˜α(s, x) ∂u˜α 1 u˜α(s, x) ∂u˜α x A1 (s) = − η (s, x) $N (x)dx − εϕ˙ (s, x)$ ˙ dx. x=0 ε ∂x N x=0 ε ∂x N The ﬁrst term is nonpositive since η and $ are nonnegative. We can also see that

Z ∞ L L Z ∞ L 1 u˜α(s, x) ∂u˜α x 2 L ∂u˜α εϕ˙ (s, x)$ ˙ dx ≤ k$˙ kC kηkC u˜α(s, x) (s, x) dx N x=0 ε ∂x N N x=0 ∂x 22 Z ∞ L 2 1 L 2 ∂u˜α ≤ k$˙ kC kηkC u˜α(s, x) + (s, x) dx. N x=0 ∂x Thus Z t∧τL N,ε −s lim lim E A1 (s)e ds ≤ 0. ε→0 N→∞ s=0 Thirdly, another integration by parts gives us that Z ∞ N,ε 1 L x A2 (s) = − ϕε(˜uα(s, x))$ ˙ dx. N x=0 N and thus Z ∞ N,ε 1 L 2 |A2 (s)| ≤ k$˙ kC kηkC u˜α(s, x) dx. 2N x=0 Thus Z t∧τL L ∂u˜α αs N,ε −s lim lim E (s, 0)e A2 (s)e ds = 0. ε→0 N→∞ s=0 ∂x N,ε Let’s ﬁnally bound A3 . Note that for every u ∈ R, 2 2 2 lim ϕ¨ε(u)u − 2ϕε(u) = 2u χ (u) − 2u χ (u) = 0. ε→0 R− R− In light of the ﬁrst and last bounds of (24), we can use dominated convergence to see that

Z t∧τL N,ε −s lim lim E A3 (s)e ds = 0. ε→0 N→∞ s=0 Combining things together, we finally get that ∞ Z 2 L L −t∧τL E u˜α(t ∧ τL, x) χR− (˜uα(t ∧ τL, x))e dx ≤ 0. x=0 This implies the claimed result. Let’s now see what happens as L % ∞. Define the random time def τ = lim (τL ∧ L). L→∞ Let’s also define def L u˜(t, x) = lim u˜ (t ∧ τL, x). t ≥ 0, x ≥ 0 L→∞ Lemma 5.8 (Positivity). u˜(t, x) > 0 for all t ≥ 0 and x > 0. Proof. We first define a transformation:

∗ def (25)u ˜ (t, x) =u ˜(t, x) exp[−ζt(x + β(t))]. t ≥ 0, x ≥ 0

Since ζt(x + β(t)) is an Ito process (recall (8)), by applying Ito’s formula and plugging in (14) we get ∗ du˜ (t, x) = d{u˜(t, x) exp[−ζt(x + β(t))]} 1 = exp[−ζ (x + β(t))] du˜(t, x) +u ˜(t, x) −dζ (x + β(t)) − ζ˙ (x + β(t))β˙(t)dt + dt t t t 2

− u˜(t, x) exp[−ζt(x + β(t))]dt ∂2u˜ h i ∂u˜ = exp[−ζ (x + β(t))] (t, x) + α − ζ˙ (x + β(t))β˙(t) u˜(t, x) + β˙(t) (t, x) dt. t ∂x2 s t ∂x

Since ζt is smooth in space, we also have the following equalities: ∂u˜ ∂u˜∗ (t, x) = exp[ζ (x + β(t))] · (t, x) + ζ˙ (x + β(t))˜u∗(t, x) ; ∂x t ∂x t ∂2u˜ (t, x) = exp[ζ (x + β(t))] ∂x2 t 23 ∂2u˜∗ ∂u˜∗ h i × (t, x) + 2ζ˙ (x + β(t)) (t, x) +u ˜∗(t, x) ζ˙ (x + β(t))2 + ζ¨ (x + β(t)) . ∂x2 t ∂x t t Putting things together, we have thatu ˜∗(t, x) satisfies the random PDE ∂u˜∗ ∂2u˜∗ h i ∂u˜∗ (t, x) = (t, x) + 2ζ˙ (x + β(t)) + β˙(t) (t, x) ∂t ∂x2 t ∂x h ˙ 2 ¨ i ∗ (26) + ζt(x + β(t)) + ζt(x + β(t)) + αs u˜ (t, x), t > 0, x > 0 u˜∗(t, 0) = 0, t > 0 ∗ u˜ (0, x) = u◦(x). x > 0 ∗ Now suppose there exists t0 > 0, x0 > 0 such thatu ˜(t0, x0) =u ˜ (t0, x0) = 0. In light of Lemma 5.7, since ∗ ∗ u˜ (t, x) ≥ 0, t ≥ 0, x ≥ 0, we have (t0, x0) is a minimum point ofu ˜ (t, x). By applying Strong Parabolic ∗ def Maximum Principle (c.f. Theorem 2, page 309, [McOwen96]) upon −u˜ (t, x) on Ut0,x0 = (0, t0+1)×[0, x0+1] ∗ with M = 0, we obtain thatu ˜ (t, x) ≡ 0, (t, x) ∈ Ut0,x0 , which contradicts the fact that u◦(x) > 0, x > 0 and ∗ the continuity ofu ˜ (t, x) at t = 0. Therefore we have thatu ˜(t, x) > 0, t ≥ 0, x > 0. Lemma 5.9. We have that lim ku˜(t, ·)kH = ∞. t%τ Define u as in (15)–(16). Then {u(t, ·) | 0 ≤ t < τ} is a weak solution of (1). 0 L0 L Proof. Fixing L > L we have from the uniqueness claim of Lemma 5.6 thatu ˜ (t, ·) =u ˜ (t, ·) for 0 ≤ t ≤ τL. 0 Thus τL0 ≥ τL for all L > L, and so τ = limL→∞ τL = limL→∞(τL ∧ L) and τ is predictable. We also L have thatu ˜(t, ·) = limL→∞ u˜ (t, ·) for 0 ≤ t < τ. From this, Lemma 5.8, and Lemma 4.4, we conclude that {u(t, ·) | 0 ≤ t < τ} as defined by (15)–(16) indeed is a weak solution of (1). The characterization of ku˜(t, ·)kH at τ− is obvious. In fact, we have a more explicit characterization of τ. Lemma 5.10. We have that

∂u˜ lim (t−, 0) = ∞. t%τ ∂x Proof. For each L > 0, deﬁne 0 def ∂u˜ τ = inf t ∈ [0, τ) | (t, 0) ≥ L . (inf ∅ = τ) L ∂x Thus in fact τ > τ 0 and hence L ∂u˜ 0 (τ , 0) = L. ∂x L Consequently

∂u˜ 0 lim (τL, 0) = ∞. L→∞ ∂x 0 0 Since τL < τ, we of course also have that limL→∞ τL ≤ τ. On the other hand, ku˜(t, ·)kH may become ∂u˜ 0 0 large for many reasons other than ∂x (τL, 0) becoming large, so necessarily τ ≤ limL→∞ τL. Putting things 0 together, we get that limL→∞ τL = τ. The claimed result now follows. To finish things off, we prove uniqueness. Lemma 5.11 (Uniqueness). If {u˜(t, ·) | 0 ≤ t < τ} ⊂ H and {u˜0(t, ·) | 0 ≤ t < τ 0} ⊂ H are two solutions of (13), then u(t, ·) = u0(t, ·) for 0 ≤ t < min{τ, τ 0}. Proof. For each L > 0, define 0 def 0 ∂u˜ ∂u˜ 0 σL = inf t ∈ [0, τ ∧ τ ): (t, 0) ≥ L or (t, 0) ≥ L . inf ∅ = τ ∧ τ ∂x ∂x 0 0 Then τ ∧ τ = limL→∞ σL. We can use standard uniqueness theory to conclude thatu ˜ andu ˜ coincide on [0, σL], and we then let L % ∞. 24 5.1. Proofs. We here give the delayed proofs. We start with the structural claims about H.

1 ∞ Proof of Lemma 5.1. The fact that H ⊂ C is well-known; [Eva98]. Fix ϕ ∈ C0 (R+), x ∈ (0, ∞), and i ∈ {0, 1}. We then have that ∂iϕ Z x+1 ∂iϕ Z x+1 ∂iϕ ∂iϕ i (x) = i (s)ds − i (s) − i (x) ds ∂x s=x ∂x s=x ∂x ∂x Z x+1 ∂iϕ Z x+1 Z s ∂i+1ϕ Z x+1 ∂iϕ Z x+1 ∂i+1ϕ = i (s)ds − i+1 (r)dr ds = i (s)ds − (x + 1 − r) i+1 (r)dr s=x ∂x s=x r=x ∂x s=x ∂x r=x ∂x Thus s s i Z x+1 i 2 Z x+1 i+1 2 ∂ ϕ ∂ ϕ ∂ ϕ i (x) ≤ i (s) ds + i+1 (r) dr ≤ 2kϕkH . ∂x s=x ∂x r=x ∂x Of course we also have that Z y (i) (i) (i+1) p ϕ (x) − ϕ (y) ≤ ϕ (r)dr ≤ kϕkH |x − y| r=x so the stated limits at x = 0 exist.

We next study {Tt}t>0. Proof of Lemma 5.2. The proof relies upon a combination of fairly standard calculations. ∞ To begin, ﬁx ϕ ∈ C0 (R+) and deﬁne ∞ 0 def Z Z Z u(t, x) = p◦(t, x − y) sgn(y)ϕ(|y|)dy = p◦(t, x − y)ϕ(y)dy − p◦(t, x − y)ϕ(−y)dy y∈R y=0 y=−∞ Z ∞ = {p◦(t, x − y) − p◦(t, x + y)} ϕ(y)dy. y=0

Thus u(t, x) = (Ttϕ)(x) for x > 0, and since p◦ is even in its second argument, Z Z u(t, −x) = p◦(t, −x − y) sgn(y)ϕ(|y|)dy = p◦(t, x + y) sgn(y)ϕ(|y|)dy y∈R y∈R Z = − p◦(t, x − y) sgn(y)ϕ(|y|)dy = −u(t, x) y∈R ∂nu so in fact u(t, ·) is odd. Thus we indeed have that ∂xn (t, 0) = 0 for all even n ∈ N; thus Ttϕ ∈ Hodd. A standard calculation shows that Tt is a contraction on H. Indeed, for each nonnegative integer n, Z n 2 Z n+2 n Z n+1 2 d ∂ u ∂ u ∂ u ∂ u (t, x) dx = 2 (t, x) (t, x)dx = −2 (t, x) dx ≤ 0 dt ∂xn ∂xn+2 ∂xn ∂xn+1 x∈R x∈R x∈R and thus Z ∞ n 2 Z n 2 Z n 2 ∂ u 1 ∂ u 1 ∂ u (t, x) dx = (t, x) dx ≤ (0, x) dx ∂xn 2 ∂xn 2 ∂xn (27) x=0 x∈R x∈R Z ∞ 2 (n) = ϕ (x) dx. x=0 2 2 ∞ Summing these inequalities up for n ∈ {0, 1, 2, 3}, we see that kTtϕkH ≤ kϕkH for all ϕ ∈ C0 (R+). This ∞ implies that Tt is a contraction on C0 (R+) and has the claimed extension. ∞ To proceed, fix ϕ ∈ C0,odd(R+). Note that thus y 7→ ϕ(|y|) is continuous. Define Z ∞ Z (1) (1) v(t, x) = {p(t, x − y) − p◦(t, x + y)} ϕ (y)dy = p◦(t, x − y) sgn(y)ϕ (|y|)dy y=0 y∈R We can now fairly easily conclude from (27) with n = 0 that Z ∞ Z ∞ 2 2 (1) v (t, x)dx ≤ ϕ (x) dx. x=0 x=0 25 Differentiating and integrating by parts as needed, we get that ∂v Z ∂p (t, x) = ◦ (t, x − y) sgn(y)ϕ(1)(|y|)dy ∂x ∂x y∈R Z (1) (2) = −2p◦(t, x)ϕ (0) − p◦(t, x − y)ϕ (|y|)dy y∈R ∂2v ∂p Z ∂p (t, x) = −2 ◦ (t, x)ϕ(1)(0) − ◦ (t, x − y)ϕ(2)(|y|)dy ∂x2 ∂x ∂x y∈R ∂p Z = −2 ◦ (t, x)ϕ(1)(0) − p (t, x − y) sgn(y)ϕ(3)(|y|)dy. ∂x ◦ y∈R We now note that there is a K > 0 such that

∂p◦ K (t, x) ≤ √ p◦(2t, x) ∂x t for all t > 0 and x ∈ R. Thus Z ∂v (1) (2) (t, x) ≤ 2p◦(t, x) ϕ (0) + p◦(t, x − y) ϕ (|y|) dy ∂x y∈R 2 Z ∂ v 2K (1) (3) (t, x) ≤ √ p◦(2t, x) ϕ (0) + p◦(t, x − y) ϕ (|y|) dy. ∂x2 t y∈R Note now that s s Z ∞ 1 Z ∞ 1 x2 1 p2(t, x)dx = √ √ exp − dx ≤ . ◦ 1/4 x=0 4πt x=0 πt t (4πt) Combining this and (27) with n = 0, we get that s Z 2 s Z ∞ ∂v 2 (1) (t, x) dx ≤ ϕ (0) + 2 |ϕ(2)|2dy ∂x (4πt)1/4 x∈R y=0 s Z 2 2 s Z ∂ v 2K (1) (t, x) dx ≤ √ ϕ (0) + 2 |ϕ(3)(y)|2dy ∂x2 1/4 x∈R t(8πt) y∈R Combine things together to get the last claim.

6. Numerical Simulation In this section, we will see from numerical simulations where the boundary is and how it is moving. In general, it is difficult to simulate the SPDE (1) directly since we need to find a solution of a stochastic heat equation and at the same time we need to trace the position of the moving boundary. Here we can avoid this difficulty since we have the explicit formula for the solution u in Lemma 4.4. That is, ( u˜(t, x − β(t)) x ≥ β(t), 0 ≤ t < τ u(t, x) def= 0 x < β(t), 0 ≤ t < τ, where (˜u, β) is a solution of the SPDE ∂2u˜ 1 ∂u˜ ∂u˜ du˜(t, x) = (t, x) + αu˜(t, x) − (t, 0) (t, x) dt ∂x2 % ∂x ∂x

+u ˜(t, x)dζt(x + β(t)) t > 0, x > 0 u˜(t, 0) = 0 t > 0 (28) u˜(0, x) = u◦(x) x > 0 1 ∂u˜ β˙(t) = − (t, 0) t > 0 % ∂x β(0) = 0. 26 Therefore we first need to solve the SPDE (28) numerically in order to obtain the moving boundary β(t) and then the weak solution u(t, x). We first discretize space by using the explicit finite difference scheme. Here we can also approximate η by simple functions which converge to η in L2(R)(see [Wal86]). As a result, we can have an approximation of ζt(x). Note that ζt(x) is a Brownian motion for each fixed x, however, it is spatially correlated. Now we use the Euler-Maruyama type method to discretize time(see [Gai96, Hig02]). Then we can get a numerical solution of (28). Since there is a stability issue for parabolic PDE, we note that ∆t/(∆x)2 < 1/2, where ∆t is a time step and ∆x is a space step. Figure 1 is a simulation with initial condition ( x+x2 1+( x )4 if x ≥ 0 u◦(x) = 2 0 else and α = .5, % = .5. We can clearly see that there are two phases separated by the black line, which is the moving boundary, and how u is changing on the colored region where u > 0. Furthermore, we can see that the boundary is moving left. This just follows from the positivity of the solution u(t, x) for x ≥ β(t) and the Stefan boundary condition of (1).

Figure 1. Weak solution u(t, x)

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Department of Mathematics, University of Illinois at Urbana–Champaign, Urbana, IL 61801 E-mail address: [email protected]