A STOCHASTIC MOVING BOUNDARY VALUE PROBLEM 1 ...
A STOCHASTIC MOVING BOUNDARY VALUE PROBLEM KUNWOO KIM, CARL MUELLER, AND RICHARD B. SOWERS
Abstract. We consider a stochastic perturbation of a moving boundary problem proposed by Ludford and Stewart and studied by Caffarelli and Vazquez. We prove existence and uniqueness.
1. Introduction Moving boundary problems are one of the important areas of partial differential equations. They provide the correct quantitative description of a wide range of physically interesting phenomena where a system has two phases. However, since the boundary between these phases is defined implicitly by the behavior of the rest of the system, they provide deep mathematical challenges in the areas of existence, uniqueness, and regularity. Our goal here is to study the effect of noise on a specific free boundary problem which was introduced by Stewart [Ste85] and subsequently addressed in the mathematics literature (see [CS05, CV95, Vaz96]). Fix a probability triple (Ω, F , P) and assume that B is a Brownian motion on (Ω, F , P). We consider the SPDE du(t, x) = (1)
lim x&β(t)
∂2u (t, x)dt + αu(t, x)dt + u(t, x) ◦ dBt ∂x2
∂u (t, x) = 1 ∂x u(0, x) = u◦ (x)
x > β(t)
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 more natural than not to include this term). We also assume that the initial function u◦ ∈ C(R) satisfies some specific properties: ◦ • u◦ ≡ 0 on R− , u◦ > 0 on (0, ∞), and limx&0 du dx (x) = 1. • u◦ and its first three derivatives exist on (0, ∞) and are square-integrable (on (0, ∞)).
In (1), ◦dBt represents Stratonovich integration, and the last line means that the boundary between u ≡ 0 and u > 0 is 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] and [CLM06]; 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 dBt 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 means that bubbles where u is positive cannot spontaneously nucleate within the region where u = 0. Our major contributions here are to 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 single Brownian motion, several transformations (the transformations of Lemmas 3.3 and 3.5 and (17)) can transform the problem into a random nonlinear PDE (see (18)). 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 3.2 and the iterative bounds of Lemma 4.4. 1
2. Weak Formulation To see what we mean by (1), let’s replace ◦dB by a smooth path b; the Wong-Zakai result (cf. [KS91, Section 5.2D]) implies that SDE with smoothed versions of dB converge to Stratonovich SDE (and that the Stratonovich interpretation is correct when we do so). Let’s also assume that there is only one interface. Namely, consider the PDE ∂2v ∂v + αv(t, x) + v(t, x)b(t) (t, x) = ∂t ∂x2 ∂v lim (t, x) = 1 x&β◦ (t) ∂x v(0, x) = u◦ (x). x∈R
(2)
x > β◦ (t)
{(t, x) ∈ R+ × R | v(t, x) > 0} = {(t, x) ∈ R+ × R | x > β◦ (t)}. 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 Z Z ∞ def v(t, x)ϕ(t, x)dx. v(t, x)ϕ(t, x)dx = Uϕ (t) = x=β◦ (t)
x∈R
Differentiating, we get that Z ∞ ∂v ∂ϕ ˙ Uϕ (t) = (t, x)ϕ(t, x) + v(t, x) (t, x) dx − v(t, β◦ (t))ϕ(t, β◦ (t))β˙ ◦ (t) ∂t ∂t x=β◦ (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 x > β◦ (t) to rewrite ∂v ∂t . Integrating by parts, we have that Z ∞ ∂2v (t, x)ϕ(t, x)dx 2 x=β◦ (t) ∂x Z ∞ ∂ϕ ∂2ϕ ∂v = lim v(t, x) 2 (t, x)dx. − (t, x)ϕ(t, β◦ (t)) + v(t, β◦ (t)) (t, β◦ (t)) + ∂x ∂x ∂x x&β◦ (t) x=β◦ (t) ∂v . Recombining Again we use the fact that v(t, β◦ (t)) = 0, and we can also use the boundary condition on ∂x things we get the standard formula that Z ∂ϕ ∂2ϕ U˙ ϕ (t) = v(t, x) (t, x) + (t, x) + αϕ(t, x) dx ∂t ∂x2 x∈R Z + v(t, x)ϕ(t, x)dx b(t) − ϕ(t, β◦ (t)). x∈R
Replacing b by ◦dB, we should have the following formulation: that for any ϕ ∈ Cc∞ (R+ × R) and any t > 0, Z Z Z t Z ∂2ϕ ∂ϕ (r, x) + u(t, x)ϕ(t, x)dx = u◦ (x)ϕ(t, x)dx + u(r, x) (r, x) + αϕ(r, x) dx dr ∂t ∂x2 x∈R x∈R r=0 x∈R Z t Z Z t + u(r, x)ϕ(r, x)dx ◦ dBr − ϕ(r, β(r))dr. r=0
The Ito formulation of this would be that Z Z Z u(t, x)ϕ(t, x)dx = u◦ (x)ϕ(t, x)dx + x∈R
x∈R
t
r=0
∂ϕ ∂2ϕ u(r, x) (r, x) + (r, x) + α ˆ ϕ(r, x) dx dr ∂t ∂x2 r=0 x∈R Z t Z Z t + u(r, x)ϕ(r, x)dx dBr − ϕ(r, β(r))dr Z
r=0
where α ˆ =α+
x∈R
1 2. 2
x∈R
r=0
Remark 2.1 Thus 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. We let def
Ft = σ{Bs ; 0 ≤ s ≤ t} for all t ≥ 0; then B is a Brownian motion with respect to {Ft }t>0 and stochastic integration against B will be with respect to this filtration. Definition 2.2. A weak solution of (1) is a predictable path {u(t, ·) | 0 ≤ t < τ } in 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
u(τ 0 , x)ϕ(τ 0 , x)dx =
x∈R
Z
Z
τ0
x∈R
∂2ϕ ∂ϕ (r, x) + α ˆ ϕ(r, x) dx dr (r, x) + ∂t ∂x2 Z τ0 u(r, x)ϕ(r, x)dx dBr − ϕ(r, β(r))dr
Z
u◦ (x)ϕ(0, x)dx+
u(r, x) r=0
Z
x∈R τ0
Z
+ r=0
x∈R
r=0
and where {(t, x) ∈ [0, τ ) × R | u(t, x) > 0} = {(t, x) ∈ [0, τ ) × R | x > β(t)}. Our main existence and uniqueness theorems are the following. The arguments leading up to these results will come together in Section 4. Theorem 2.3 (Existence). A solution of (1) exists. Furthermore, u(t, ·) ∈ C 2 [β(t), ∞) for all t ∈ [0, τ ) and 2 ∂ u τ ≤ inf t ≥ 0 : 2 (t−, β(t)) = ∞ . ∂x Proof. Combine Lemmas 4.6 and 4.7.
We also have uniqueness. Theorem 2.4 (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 4.8.
3. Regularity and a Transformation The proof of Theorems 2.3 and 2.4 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 3.2. First, however, let’s make sure that we understand a bit about regularity; this will illuminate the assumptions needed. 3.1. 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. The following representation result will help us in carrying out this analysis. Define 2 1 x def t > 0, x ∈ R exp − p◦ (t, x) = √ 4t 4πt def (3) p± (t, x, y) = {p◦ (t, x − y) ± p◦ (t, x + y)} eαt = {p◦ (t, x − y) ± p◦ (t, −x − y)} eαt t > 0, x, y ∈ R def
ˆ ˆ pˆ± (t, x, y) = {p◦ (t, x − y) ± p◦ (t, x + y)} eαt = {p◦ (t, x − y) ± p◦ (t, −x − y)} eαt ;
t > 0, x, y ∈ R
the second representations of p± and pˆ± stem from the fact that p◦ is even in its second argument. The distinction between p± and pˆ± naturally lies in the distinction between Ito and Stratonovich calculations. 3
We then have that ∂p± (t, x, y) = ∂t ∂ pˆ± (t, x, y) = ∂t lim p± (t, x, ·) =
(4)
t&0
∂ 2 p± (t, x, y) + αp± (t, x, y) t > 0, x, y ∈ R ∂y 2 ∂ 2 pˆ± (t, x, y) + α ˆ pˆ± (t, x, y) t > 0, x, y ∈ R ∂y 2 lim pˆ± (t, x, ·) = δx ; x ∈ R \ {0}
t&0
the relevant distinction between p+ and p− is their behavior at x = 0. This will come up in the arguments of Lemma 3.2. Lemma 3.1. Let u be a weak solution of (1) and assume that β is continuous. If 0 < t < τ and x > β(t), then Z t Z Bt −Bs Bt (5) u(t, x) = − e p± (t − s, x − β(t), β(s) − β(t))ds + e p± (t, x − β(t), y − β(t))u◦ (y)dy. s=0
y∈R
Furthermore, u(t, ·) is C ∞ on (β(t), ∞). Proof. Fix t > 0, x ∈ R, δ > 0 and c ∈ R and define def
ϕt,x,δ,c (s, y) = pˆ± (t + δ − s, x − c, y − c) Z def Ut,x,δ,c (s) = ϕt,x,δ,c (s, y)u(s, y)dy.
s ∈ [0, t], y ∈ R s β(t), and β is continuous, inf {|t − s| + |(x − β(t)) − (β(t) − β(s))|} = min {|t − s| + |(x − β(t)) − (β(t) − β(s))|} > 0
0≤s≤t
0≤s≤t
inf {|t − s| + |(x − β(t)) + (β(t) − β(s))|} = min {|t − s| + |(x − β(t)) + (β(t) − β(s))|} > 0.
0≤s≤t
0≤s≤t
Thus lim sup ϕt,x,δ,β(t) (s, β(s)) − pˆ± (t − s, x − β(t), β(s) − β(t)) = 0 δ&0 0≤s≤t lim sup ϕt,x,δ,β(t) (0, y) − pˆ± (t, x − β(t), y − β(t)) = 0. δ&0 y∈R
This gives us the claimed representation result. We can then differentiate to get the claimed smoothness. Note that (5) is not an explicit formula for u since the right-hand side of (5) depends on u through β. We also note that the proof effectively converts the Ito integral of (6) into a Stratonovich one, implying that pˆ± is converted back into p± . Next, let’s see what happens if we in fact assume that β is continuously differentiable. It turns out that not only does the boundary behavior of (1) hold pointwise, but we can find an evolution equation for β (which depends on u). To get the general idea of this latter fact, let’s return to our deterministic PDE (2). By definition v(t, β◦ (t)) = 0, so differentating (and using an approximation just to the right of β◦ ) we get that ∂v ∂v (t, β◦ (t)) + (t, β◦ (t))β˙ ◦ (t) = 0. ∂t ∂x Using the PDE for v and the boundary conditions (again, a rigorous proof would require pushing the calculation just a bit to the right of β◦ ), we get that in fact 2 ∂ v ∂2v (7) β˙ ◦ (t) = − (t, β (t)) + αv(t, β (t)) = − (t, β◦ (t)). ◦ ◦ ∂x2 ∂x2 For the SPDE (1) we should have the same result (since the noise term vanishes at the boundary). To proceed, let’s rewrite (5) in a slightly more convenient way. If {u(t, ·) | 0 ≤ t < τ } is a weak solution of (1) and 0 < t < τ , set Z t def A1 (t, ε) = exp [Bt − Bt−s + αs] p◦ (s, ε + β(t) − β(t − s))ds ε ∈ R \ {0} s=0 Z def Bt A± p± (t, ε, y − β(t))u◦ (y)dy. ε∈R 2 (t, ε) = e y∈R
Then some simple manipulation (which reflects the second representation of p± in (3) and the fact that p◦ is even in its second argument) shows that (8)
− u(t, β(t) + ε) = −A1 (t, ε) − A1 (t, −ε) + A+ 2 (t, ε) = −A1 (t, ε) + A1 (t, −ε) + A2 (t, ε).
We in fact have Lemma 3.2. Let {u(t, ·) | 0 ≤ t < τ } be a solution of (1). If β is continuously differentiable, then (9)
∂u (t, x) = 1 x&β(t) ∂x lim
∂2u ˙ (t, x) = −β(t) x&β(t) ∂x2
and
lim
for all t ∈ [0, τ ). Proof. From (8), we have that ∂A1 ∂A1 ∂A+ ∂u 2 (t, β(t) + ε) = − (t, ε) + (t, −ε) + (t, ε) ∂x ∂ε ∂ε ∂ε ∂2u ∂ 2 A1 ∂ 2 A1 ∂ 2 A− 2 (t, β(t) + ε) = − (t, ε) + (t, −ε) + (t, ε). 2 2 2 ∂x ∂ε ∂ε ∂ε2 Note that 2 ∂p◦ 1 x x (10) (t, x) = − √ exp − 3/2 ∂x 4t 2 4π t
∂ 2 p◦ 1 (t, x) = √ 2 ∂x 2 4πt3/2
and 5
2 x2 x − 1 exp − . 2t 4t
Thus ∂A1 (t, ε) = ∂ε
Z
t
exp [Bt − Bt−s − αs] s=0
1 =− √ 2 4π
∂p◦ (s, ε + β(t) − β(t − s))ds ∂x
" # 2 (ε + β(t) − β(t − s)) ε + β(t) − β(t − s) exp − exp [Bt − Bt−s + αs] ds 4s s3/2 s=0
Z
t
= −A˜1,1 (t, ε) − A˜1,2 (t, ε) Z t ∂ 2 p◦ ∂ 2 A1 exp [Bt − Bt−s + αs] (t, ε) = (s, ε + β(t) − β(t − s))ds 2 ∂ε ∂x2 s=0 ( ) Z t 2 (ε + β(t) − β(t − s)) 1 exp [Bt − Bt−s + αs] = √ −1 2s 2 4π s=0 " # 2 (ε + β(t) − β(t − s)) 1 × exp − ds 4s s3/2 = A˜2,1 (t, ε) + A˜2,2 (t, ε) + A˜2,3 (t, ε) where 1 A˜1,1 (t, ε) = √ 2 4π 1 A˜1,2 (t, ε) = √ 2 4π 1 A˜2,1 (t, ε) = √ 2 4π 1 A˜2,2 (t, ε) = √ 2 4π 1 A˜2,3 (t, ε) = √ 4 4π
# 2 (ε + β(t) − β(t − s)) exp [Bt − Bt−s + αs] 3/2 exp − ds 4s s s=0 # " Z t 2 (ε + β(t) − β(t − s)) β(t) − β(t − s) ds exp − exp [Bt − Bt−s + αs] 4s s3/2 s=0 " # 2 Z t 2 (ε + β(t) − β(t − s)) 1 ε ds − 1 exp − exp [Bt − Bt−s + αs] 3/2 2s 4s s s=0 " # Z t 2 ε β(t) − β(t − s) (ε + β(t) − β(t − s)) exp [Bt − Bt−s + αs] 3/2 exp − ds s 4s s s=0 " # Z t 2 2 (ε + β(t) − β(t − s)) (β(t) − β(t − s)) exp − ds. exp [Bt − Bt−s + αs] 4s s5/2 s=0 Z
t
"
ε
Since β is by assumption continuously differentiable, def
|β(t) − β(t − δ)| δ 0 0 and ε ∈ R. We also note that ∂p− ∂p◦ (t, x, 0) = −2 (t, x)eαt ∂y ∂x
t > 0, x ∈ R
and ∂ 2 p− (t, x, y) = ∂y 2
∂ 2 p◦ ∂ 2 p◦ (t, x − y) − (t, x + y) eαt ∂x2 ∂x2 t > 0, x ∈ R, y ∈ R 2 ∂ p◦ ∂ 2 p◦ αt = (t, x − y) − (t, −x − y) e ∂x2 ∂x2 where the last representation uses the fact that p◦ is even in its second argument. Thus
∂p◦ (s, ε)eαs ∂x Z 1 ∂ 2 p◦ + (β(t) − β(t − s))2 eαs (1 − r) 2 (s, ε − r(β(t) − β(t − s))) dr ∂x r=0 Z 1 ∂ 2 p◦ − (β(t) − β(t − s))2 eαs (1 − r) 2 (s, −ε − r(β(t) − β(t − s))) dr. ∂x r=0
p− (s, ε, β(t) − β(t − s)) = −2 (β(t) − β(t − s))
Thus A˜2,1 (t, ε) − A˜2,1 (t, −ε) = A˜a2,1 (t, ε) + A˜b2,1 (t, ε) − A˜b2,1 (t, −ε) + A˜c2,1 (t, ε) − A˜c2,1 (t, ε) 7
where A˜a2,1 (t, ε) = −
Z
t
exp [Bt − Bt−s + αs]
s=0 t Z 1
ε2 −1 2s
β(t) − β(t − s) ∂p◦ (s, ε)ds s ∂x 2
ε2 (β(t) − β(t − s)) ∂ 2 p◦ (s, ε − r(β(t) − β(t − s))) dr ds 2s s ∂x2 s=0 r=0 2 2 Z 1 Z ε2 t β(t) − β(t − s) ∂ p◦ = exp [Bt − Bt−s + αs] (s, ε − r(β(t) − β(t − s))) dr ds 4 s=0 r=0 s ∂x2 Z 1 Z 2 (β(t) − β(t − s)) ∂ 2 p◦ 1 t exp [Bt − Bt−s + αs] A˜c2,1 (t, ε) = − (s, ε − r(β(t) − β(t − s))) dr ds 2 s=0 r=0 s ∂x2 2 2 Z 1 Z β(t) − β(t − s) 1 t ∂ p◦ exp [Bt − Bt−s + αs] =− s 2 (s, ε − r(β(t) − β(t − s))) dr ds. 2 s=0 r=0 s ∂x √ We will again use the transformation u = |ε|/ s. We compute that for ε > 0 2 2 Z t ε β(t) − β(t − s) ε ε 1 a ˜ √ exp [Bt − Bt−s + αs] −1 exp − ds A2,1 (t, ε) = − 3/2 2s s 4s s 2 4π s=0 2 Z ∞ u β(t) − β(t − ε2 /u2 ) 2 2 2 2 = − sgn(ε) −1 √ exp Bt − Bt−ε /u + αε /u 2 ε2 /u2 u=ε/ t 2 u 1 √ du. × exp − 4 4π Thus by dominated convergence, 2 Z ∞ 2 u u 1 ˙ √ du = 0. − 1 exp − lim A˜a2,1 (t, ε) = β(t) ε&0 2 4 4π u=0 Similarly, Z Z 1 β(t) − β(t − ε2 /u2 ) 2 ε ∞ b 2 2 ˜ 2 2 A2,1 (t, ε) = exp Bt − Bt−ε /u + αε /u 2 u=|ε|/√t r=0 ε2 /u2 ∂ 2 p◦ 2 2 × ε3 /u3 ε /u , ε − r(β(t) − β(t − ε2 /u2 )) dr du. 2 ∂x From the second equality of (10), we see that there is a K > 0 such that 2 2 ∂ p◦ K x K (12) ∂x2 (s, x) ≤ s3/2 exp − 8s ≤ s3/2 1 A˜b2,1 (t, ε) = 2
Z
exp [Bt − Bt−s + αs]
for all s ∈ (0, t] and x ∈ R. Assume again that ε < 1/(2K). If u ≤ 1, then 3 3 3 3 ∂ 2 p◦ 2 2 ε /u 2 2 ε /u ε /u , ε − r(β(t) − β(t − ε /u )) ≤ K = K. 3/2 ∂x2 (ε2 /u2 ) On the other hand, if u ≥ 1, we have that 3 3 ∂ 2 p◦ 2 2 2 2 ε /u ε /u , ε − r(β(t) − β(t − ε /u )) ∂x2 " 2 # ε3 /u3 ε − r(β(t) − β(t − ε2 /u2 )) ≤K exp − 3/2 4ε2 /u2 (ε2 /u2 ) " 2 # 2 u2 β(t) − β(t − ε2 /u2 ) u ≤ K exp − 1−r ≤ K exp − 4 ε 8 by again using (11). Combining things together, we see that there is a K > 0 such that 2 Z ∞ √ u ˜b exp − du = Kε 2π; A2.1 (t, ε) ≤ Kε 8 u=0 8
thus indeed lim A˜b2.1 (t, ε) = 0.
ε→0
√ We next turn to A˜c2.1 . We here use the last bound of (12). Since s 7→ 1/ s is integrable on (0, 1], we can use dominated convergence to see that 2 2 Z 1 Z β(t) − β(t − s) 1 t ∂ p◦ c ˜ exp [Bt − Bt−s + αs] lim A2,ε (t, ε) = − s 2 (s, r(β(t) − β(t − s))) dr ds, ε→0 2 s=0 r=0 s ∂x this integral being finite. We have again availed ourselves of the fact that p◦ is even in its second argument. Finally, let’s understand the relevant behavior of A± 2 . We have that Z + ∂A2 ∂p+ lim (t, ε) = eBt (t, 0, y − β(t))u◦ (y)dy ε&0 ∂ε y∈R ∂x Z ∂ 2 p− ∂ 2 A− Bt 2 (t, ε) = e (t, 0, y − β(t))u◦ (y)dy. lim 2 ε&0 ∂ε2 y∈R ∂x From the second expression for p± in (3), we have that ∂p+ ∂p◦ ∂p◦ (t, 0, y) = eαt (t, −y) − (t, −y) = 0. ∂x ∂x ∂x We also have that p− (t, 0, y) = 0 for all t > 0, so ∂p− ∂ 2 p− (t, 0, y) = (t, 0, y) − αp− (t, 0, y) = 0. ∂x2 ∂t Thus in fact
∂A+ 2 (t, ε) = 0 and ε&0 ∂ε Combining things together, we indeed get (9). lim
∂ 2 A− 2 (t, ε) = 0. ε&0 ∂ε2 lim
3.2. A Transformation. The characterization of β given in (9) allows us to rewrite the moving boundaryvalue problem in a more convenient way. The calculation which gives us some analytical traction is found in [Lun04] (see also [Fri64, Ch. 8]). Again, let’s return to our deterministic PDE (2). For all t ≥ 0 and x ∈ R, define v˜(t, x) = v(t, x + β◦ (t)) + e−x . Then v(t, x) = v˜(t, x − β◦ (t)) − exp [−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 (13) (t, x) = (t, x + β◦ (t)) − e−x ∂x ∂x ∂ 2 v˜ ∂2v (t, x) = (t, x + β◦ (t)) + e−x . 2 ∂x ∂x2 We can combine these equations and use the PDE for v to rewrite the evolution of v˜ as
(14)
∂˜ v ∂2v (t, x) = (t, x + β◦ (t)) + αv(t, x + β◦ (t)) + v(t, x + β◦ (t))b(t) ∂t ∂x2 ∂v + (t, x + β◦ (t))β˙ ◦ (t) ∂x ∂ 2 v˜ ∂˜ v −x −x −x = (t, x) − e + α v ˜ (t, x) − e + (t, x) + e β˙ ◦ (t) ∂x2 ∂x + v˜(t, x) − e−x b(t).
Note also that
∂˜ v (t, 0) = 1 − 1 = 0. ∂x Furthermore, v˜(t, 0) = 1 for all t > 0, so evaluating (14) at x = 0 (or more accurately, as x & 0), we get that 0=
∂˜ v ∂ 2 v˜ (t, 0) = (t, 0) − 1 + β˙ ◦ (t). ∂t ∂x2 9
Thus in fact ∂ 2 v˜ (t, 0); β˙ ◦ (t) = 1 − ∂x2 alternately by combining (7) and the last line of (13), we have that 2 ∂ v˜ ∂2v (0, x) − 1 . β˙ ◦ (t) = − 2 (t, β◦ (t)) = − ∂x ∂x2 (15)
Inserting the dynamics of β◦ back into (14) we can collect things and get a PDE for v˜; we have that ∂˜ v ∂˜ v ∂ 2 v˜ ∂ 2 v˜ −x −x −x (t, x) − e + α v˜(t, x) − e + (t, 0) (t, x) = (t, x) + e 1− ∂t ∂x2 ∂x ∂x2 + v˜(t, x) − e−x b(t) t > 0, x > 0 ∂˜ v (t, 0) = 0 ∂x
t>0 def
v˜(0, x) = u ˜◦ (x) = u◦ (x) + e−x
x>0
Replacing b by our Brownian motion B and α by α ˆ , we now get the following. Lemma 3.3. Suppose that {u(t, ·) | 0 ≤ t < τ } ⊂ C(R) ∩ L(R) is a weak solution of (1). Suppose also that β is continuously differentiable and {Ft }t≥0 -adapted. Then u ˜(t, x) = u(t, x + β(t)) + e−x satisfies the integral equation Z ∞ u ˜(t, x) = pˆ+ (t, x, y) u◦ (y) + e−y dy y=0 t
∞
∂2u ˜ ∂u ˜ −y (s, y) + e−y 1− (s, 0) − (ˆ α + 1)e dy ds ∂x ∂x2 s=0 y=0 Z t Z ∞ + pˆ+ (t − s, x, y) u ˜(s, y) − e−y dy dBs Z
Z
pˆ+ (t − s, x, y)
+
(16)
s=0
y=0
for all t > 0 and x > 0. Thanks to Lemmas 3.1 and 3.2, the assumption that β is continuously differentiable ensures that the spatial derivatives of u ˜ on the right-hand side of (16) are well-defined. Proof of Lemma 3.3. Fix x > 0 and T > 0. For t ∈ [0, τ ), define Z ∞ def T U (t) = u ˜(t, y)ˆ p+ (T − t, x, y)dy = AT1 (t) + AT2 (t) y=0
where AT1 (t)
Z
∞
Z
∞
u(t, y + β(t))ˆ p+ (T − t, x, y)dy =
= y=0
u(t, y)ˆ p+ (T − t, x, y − β(t))dy y=β(t)
Z u(t, y)ˆ p+ (T − t, x, y − β(t))dy
= y∈R ∞
AT2 (t) =
Z
e−y pˆ+ (T − t, x, y)dy.
y=0
Using Definition 2.2 and (4), we get that Z ∂ 2 pˆ+ ∂ pˆ+ T (T − t, x, y − β(t)) + (T − t, x, y − β(t)) dA1 (t) = u(t, y) − ∂t ∂y 2 y∈R ∂ pˆ+ ˙ +ˆ αpˆ+ (T − t, x, y − β(t)) − (T − t, x, y − β(t))β(t) dydt ∂y Z + u(t, y)ˆ p+ (T − t, x, y − β(t))dy dBt − pˆ+ (T − t, x, 0)dt y∈R 10
(Z
) ∂ pˆ+ ˙ u(t, y) =− (T − t, x, y − β(t))dy β(t)dt ∂y y=β(t) (Z ) ∞ + u(t, y)ˆ p+ (T − t, x, y − β(t))dy dBt − pˆ+ (T − t, x, 0)dt ∞
y=β(t)
(Z
∞
= y=β(t)
(Z
∂u (t, y)ˆ p+ (T − t, x, y − β(t))dy ∂y
) ˙ β(t)dt )
∞
u(t, y)ˆ p+ (T − t, x, y − β(t))dy
+
dBt − pˆ+ (T − t, x, 0)dt
y=β(t)
Z
∞
∂u ˙ = (t, y + β(t))ˆ p+ (T − t, x, y)dy β(t)dt y=0 ∂y Z ∞ u(t, y + β(t))ˆ p+ (T − t, x, y)dy dBt − pˆ+ (T − t, x, 0)dt + y=0 Z ∞ ∂u ˜ ˙ = (t, y) + e−y pˆ+ (T − t, x, y)dy β(t)dt ∂y y=0 Z ∞ −y + u ˜(t, y) − e pˆ+ (T − t, x, y)dy dBt − pˆ+ (T − t, x, 0)dt. y=0
We have here used the fact that u(t, β(t)) = 0. We have also employed a fairly straightforward generalization of the integral equality in Definition 2.2 to predictable integrands; the continuous differentiability and adaptedness of β allow us to apply this. Combining the characterization of β˙ as in Lemma 3.2 and a calculation as in (15), we get that ∂2u ˜ ˙ (t, 0). β(t) =1− ∂x2 Thus Z t Z ∞ ∂u ˜ ∂2u ˜ AT1 (t) = (s, y) + e−y 1− (s, 0) p ˆ (T − s, x, y)dy ds + ∂y ∂x2 s=0 y=0 Z t Z ∞ Z t −y + u ˜(s, y) − e pˆ+ (T − s, x, y)dy dBs − pˆ+ (T − s, x, 0)ds. s=0
y=0
s=0
A straightforward differentiation, on the other hand, shows that Z ∞ ∂ pˆ+ A˙ T2 (t) = − e−y (T − t, x, y)dy ∂t y=0 Z ∞ Z ∞ 2 ˆ+ −y ∂ p =− e (T − t, x, y)dy − e−y α ˆ pˆ+ (T − t, x, y)dy ∂y 2 y=0 y=0 Z ∞ ∂ pˆ+ = (T − t, x, 0) + pˆ+ (T − t, x, 0) − (ˆ α + 1) e−y pˆ+ (T − t, x, y)dy. ∂y y=0 Note that
∂ pˆ+ ∂y (T
− t, x, 0) = 0. Combine things to get that Z t Z ∞ ∂u ˜ ∂2u ˜ U T (t) = U T (0) + (s, y) + e−y 1− (s, 0) p ˆ (T − s, x, y)dy ds + ∂y ∂x2 s=0 y=0 Z t Z ∞ −y − (ˆ α + 1) e pˆ+ (T − s, x, y)dy ds Z
t
Z
s=0 ∞
+ s=0
y=0
u ˜(s, y) − e−y pˆ+ (T − s, x, y)dy dBs .
y=0
Now let T & t to get the claimed result.
11
Of course (16) is equivalent to the SPDE 2 ∂ u ˜ ∂u ˜ ∂2u ˜ −x −x −x (t, x) + α ˆ u ˜(t, x) − e (t, 0) dt d˜ u(t, x) = −e + (t, x) + e 1− ∂x2 ∂x ∂x2 + u ˜(t, x) − e−x dBt t > 0, x > 0 ∂u ˜ (t, 0) = 0 t>0 ∂x u ˜(0, x) = u ˜◦ (x) = u◦ (x) + e−x .
x>0
We can also find a converse to Lemma 3.3. First of all note the following. Lemma 3.4. Suppose that {˜ u(t, ·) | 0 ≤ t < τ } ⊂ C 2 (R+ ) satisfies (16). Then u ˜(t, 0) = 1 for all 0 ≤ t < τ . Furthermore, u ˜(t, x) > 0 for all t ≥ 0 and x ≥ 0. Proof. Let’s first smooth things out. Fix δ > 0 and define Z ∞ Z ∞ def u ˜δ (t, x) = pˆ+ (δ, x, y)˜ u(t, y)dy = pˆ+ (t + δ, x, y) u◦ (y) + e−y dy y=0
y=0 t
∞
∂u ˜ ∂2u ˜ −y (s, y) + e−y 1− (s, 0) − ( α ˆ + 1)e dy ds ∂x ∂x2 s=0 y=0 Z t Z ∞ + pˆ+ (t + δ − s, x, y) u ˜(s, y) − e−y dy dBs ; Z
Z
pˆ+ (t + δ − s, x, y)
+
s=0
y=0
we have of course used the fact that pˆ+ is a semigroup of integral kernels. For each x > 0, some straightforward computations show that 2 ∂ u ˜δ def d˜ uδ (t, x) = (t, x) + α ˆ u ˜ (t, x) dt δ ∂x2 Z ∞ ∂u ˜ ∂2u ˜ −y + pˆ+ (δ, x, y) (t, y) + e−y 1− (t, 0) − (ˆ α + 1)e dy dt ∂x ∂x2 y=0 Z ∞ −y + pˆ+ (δ, x, y) u ˜(t, y) − e dy dBt . y=0
We now let δ & 0 and use the assumed continuity of u ˜. We also fix ε > 0 and evaluate the result at x = ε. We get that Z t Z t ∂2u ˜ (s, ε)ds + α ˆ u ˜(s, ε)ds u ˜(t, ε) = u ˜◦ (ε) + 2 s=0 s=0 ∂x Z t Z t ∂u ˜ ∂2u ˜ + (s, ε) + e−ε 1− (s, 0) ds − (ˆ α + 1) e−ε ds 2 ∂x ∂x s=0 s=0 Z t + u ˜(s, ε) − e−ε dBs s=0 Z t 2 ∂ u ˜ ∂2u ˜ ∂u ˜ −ε =u ˜◦ (ε) + (s, ε) − (s, 0) (s, ε) + e ds ∂x2 ∂x2 ∂x s=0 Z t Z t ∂u ˜ + (s, ε)ds + u ˜(s, ε) − e−ε dBs . s=0 ∂x s=0 Letting ε & 0, we see that Z
1
(˜ u(s, 0) − 1) dBs
u ˜(t, 0) = 1 + s=0
or alternately Z
1
u ˜(t, 0) − 1 =
(˜ u(s, 0) − 1) dBs s=0
which indeed implies that u ˜(t, 0) = 1 for all t ∈ [0, τ ). 12
To see the positivity, we define def u∗ (t, x) = u ˜(t, x) − e−x e−Bt .
(17)
t ≥ 0, x ≥ 0
Some straightforward calculations show that u∗ satisfies the random PDE ∗ 2 ∗ ∂u ∂ u ∂u∗ ∂ 2 u∗ ∗ Bt (t, x) + αu (t, x) − (t, 0)e (t, x) = (t, x) ∂x ∂x2 ∂x2 ∂x (18) u∗ (t, 0) = e−Bt t>0 u∗ (0, x) = u◦ (x).
t > 0, x > 0
x>0
Note that e−Bt > 0 for all t > 0 and u◦ (x) > 0 for all x > 0. Standard calculations for the heat equation then ensure that indeed u∗ (t, x) > 0 for all t > 0 and x∗ > 0. We then have Lemma 3.5. Suppose that {˜ u(t, ·) | 0 ≤ t < τ } ⊂ C 2 (R+ ) ∩ L1 (R+ ) satisfies (16). Set Z t ∂2u ˜ (s, 0) ds 0≤t 0 and ΨL ∈ C ∞ (R; [0, 1]) such that ΨL (x) = 1 if |x| ≤ L and ΨL (x) = 0 if |x| ≥ L + 1. Set Z ∞ L u ˜1 (t, x) = pˆ+ (t, x, y)˜ u◦ (y)dy y=0
for all t > 0 and x ∈ R and recursively define Z ∞ L (21) u ˜n+1 (t, x) = pˆ+ (t, x, y)˜ u◦ (y)dy +
t
y=0 ∞
∂2u ˜L ∂u ˜L n n (s, y) + e−y 1− (s, 0) ΨL k˜ uL n (s, ·)kH 2 ∂x ∂x s=0 y=0 Z t Z ∞ L −(ˆ α + 1)e−y dy ds + pˆ+ (t − s, x, y) u ˜n (s, y) − e−y dy dBs . t > 0, x > 0
Z
Z
pˆ+ (t − s, x, y)
s=0
y=0
{˜ uL n (t, ·);
For each n ∈ N, t ≥ 0} is a well-defined, adapted, and continuous path in Heven . To study (21), we will use the Neumann heat semigroup. For ϕ ∈ C0∞ (R+ ), t > 0, and x > 0, define Z ∞ def (Tt ϕ)(x) = p+ (t, x, y)ϕ(y)dy. y=0
Lemma 4.2. For each t > 0, Tt has a unique extension from C0∞ (R+ ) to H such that Tt H ⊂ Heven and such that kTt f kH ≤ kf kH for all f ∈ H. Secondly, there is a KA > 0 such that kTt f˙kH ≤
KA kf kH t3/4
for all f ∈ Heven ∩ C 4 (R+ ). 16
Again, we delay the proof until Subsection 4.1. Another convenience will be to rewrite the ds part of (21). Define ¨ ˜ L (ψ) def Ψ = (1 − ψ(0))Ψ L (kψkH ) for all ψ ∈ H. Then L L ∂2u ˜L ∂u ˜n ∂u ˜n n −x L −x ˜ L (˜ (t, 0) ΨL k˜ un (t, ·)kH = (t, x) + e 1− (t, x) + e Ψ uL N (t, ·)) ∂x ∂x2 ∂x for all n ∈ N. For ψ and η in H, let’s also define def
¨ ˜ L )(ψ; η) = −¨ ˙ L (kψkH ) (DΨ η (0)ΨL (kψkH ) + (1 − ψ(0)) Ψ
hψ, ηiH . kψkH
˜ L )(ψ, η) is the Gˆ ˜ L at ψ in the direction of Lemma 4.3. For each ψ and η in H, (DΨ ateaux derivative of Ψ η. Furthermore, there is a KB > 0 such that ˜ (DΨL )(ψ, η) ≤ KB χ[0,L+1] (kψkH )kηkH for all ψ and η in H. Proof. The claim is straightforward.
def
For each n ∈ N, we now define w ˜nL (t, x) = u ˜L ˜L n (t, x) for all x ≥ 0 and t ≥ 0. Clearly n+1 (t, x) − u L 2 sup0≤t≤T E kw ˜1 kH < ∞ for all T > 0. We then write that L w ˜n+1 (t, x) =
4 X
(n)
Aj (t, x)
j=1
where Z
(n)
1
Z
t
Z
∞
pˆ+ (t − s, x, y)
A1 (t, x) = λ=0 1
Z
λ=0 1
Z
s=0 t
y=0
∂w ˜nL ˜L u ˜L ˜nL (s, ·) ds dλ (s, y)dy Ψ n (s, ·) + λw ∂x
∂w ˜nL ˜L u = (s, ·) (x)Ψ ˜L ˜nL (s, ·) ds dλ Tt−s n (s, ·) + λw ∂x λ=0 s=0 L Z 1 Z t Z ∞ ∂w ˜n ∂u ˜n (n) A2 (t, x) = (s, y) + λ L (s, y) dy pˆ+ (t − s, x, y) ∂x ∂x λ=0 s=0 y=0 L L L ˜ × D ΨL u ˜n (s, ·) + λw ˜n (s, ·), w ˜n (s, ·) ds dλ L Z 1 Z t n ∂w ˜L ∂u ˜n ˜L u = (s, ·) + λ (s, ·) (x)DΨ ˜L ˜nL (s, ·), w ˜nL (s, ·) ds dλ Tt−s n (s, ·) + λw ∂x ∂x λ=0 s=0 Z 1 Z t Z ∞ (n) ˜L u pˆ+ (t − s, x, y)e−y dyDΨ ˜L ˜nL (s, ·), w ˜nL (s, ·) ds dλ A3 (t, x) = n (s, ·) + λw Z
Z
s=0 t
y=0
˜L u (Tt−s E)(x)DΨ ˜L ˜nL (s, ·), w ˜nL (s, ·) ds dλ n (s, ·) + λw λ=0 s=0 Z t Z ∞ (n) L A4 (t, x) = pˆ+ (t − s, x, y)w ˜n (s, y)dy dBs =
s=0 t
Z =
y=0
Tt−s w ˜nL (s, ·) (x)dBs
s=0 def
where for convenience we have set E(x) = e−x for all x ≥ 0. Note that the u ˜L ˜nL ’s are all in Heven . n ’s and w An easy calculation gives us that Z t h i Z t L (n) E kA4 (t, ·)k2H = E kTt−s w ˜nL (s, ·)k2H ds ≤ E kw ˜n (s, ·)k2H ds. s=0
s=0 17
We similarly have (using Jensen’s inequality) that Z Z t h i L (n) 2 2 2 2 2 2 kTt−s EkH E kw ˜n (s, ·)kH ds ≤ tKB kEkH E kA3 (t, ·)kH ≤ tKB To bound
and
(n) A2 ,
−3/4
we use the fact that t Z t s=0
for all t > 0. Thus " Z h i (n) 2 E kA1 (t, ·)k2H ≤ KB E
L E kw ˜n (s, ·)k2H ds.
s=0
s=0 (n) A1
t
is locally integrable. More precisely,
1 ds = 4t1/4 (t − s)3/4
2 #
L
∂ w ˜
Tt−s n (s, ·) ds
∂x s=0 H h
2 i 2
L Z t E w
L Z t ˜ (s, ·)
n w ˜ (s, ·) H n 2 2 1/4 2 2 H ≤ 4K K t ≤ KA KB E ds ds. A B s=0 (t − s)3/4 (t − s)3/4 s=0 t
Finally, we have that h i (n) E kA2 (t, ·)k2H " Z 2 #
L L
L
L t
∂ u ˜ ∂ w ˜ n n L 2
Tt−s ˜n (s, ·)kH ds ˜n (s, ·) + λw ˜n (s, ·) H kw χ[0,L+1] u ≤ KB E (s, ·) + λ (s, ·)
∂x ∂x s=0 H " Z 2 # Z t t L+1 E[kw ˜nL (s, ·)k2H ] 2 2 2 2 2 1/4 L ≤ KA KB E ≤ 4K K (L + 1) t k w ˜ (s, ·)k ds ds. H A B n 3/4 (t − s)3/4 s=0 (t − s) s=0 L P∞ def L Lemma 4.4. For each T > 0, we have that n=1 sup0≤t≤T E k˜ un+1 − u ˜L n kH < ∞. Thus P-a.s., u (t, ·) = L limn→∞ uL n (t, ·) exists as a limit in C([0, T ]; H) and u satisfies the integral equation Z ∞ u ˜L (t, x) = pˆ+ (t, x, y)˜ u◦ (y)dy y=0
Z
t
Z
∞
∂u ˜L (s, y) + e−y ∂x
∂2u ˜L 1− (s, 0) ΨL k˜ uL (s, ·)kH 2 ∂x
pˆ+ (t − s, x, y) −(ˆ α + 1)e−y dy ds Z t Z ∞ L + pˆ+ (t − s, x, y) u ˜ (s, y) − e−y dydBs . +
(22)
s=0
y=0
s=0
y=0
t > 0, x > 0
Proof. See also [Wal86, Lemma 3.3]. Fixing T > 0 we collect the above calculations to see that there is a KT > 0 such that Z t E[kw ˜nL (s, ·)k2H ] L E[kw ˜n+1 (t, ·)k2H ] ≤ KT ds (t − s)3/4 s=0 for all t ∈ [0, T ]. Iterating this, we get that n−2 Y E[kw ˜nL (t, ·)k2H ] ≤ KTn−1 t(n−1)/4 B(1 + j/4, 1/4) sup E[kw ˜1L k2H ] 0≤t≤T j=1
where B is the standard Beta function and thus that 1/2 q q n−2 Y (n−1)/2 (n−1)/8 E[kw ˜nL (t, ·)k2H ] ≤ KT t B(1 + j/4, 1/4) sup E[kw ˜1L k2H ]. 0≤t≤T j=1
To show that the terms on the right are summable, we use the ratio test. It suffices to show that 1/2 n−2 1/2 (23) lim KT t1/8 B 1 + , 1/4 = 0. n→∞ 4 18
We calculate that Z
1 n/4
s
B(1 + n/4, 1/4) =
(1 − s)
−3/4
1 1− n
Z ds =
s=0
n/4
s
−3/4
(1 − s)
Z
≤n
sn/4 (1 − s)−3/4 ds
1 s=1− n
s=0 3/4
1
ds +
Z
1 1− n
n/4
s
Z
(1 − s)−3/4 ds
1 s=1− n
s=0
4n3/4 ≤ +4 n+4
1
ds +
1/4 1 . n
This implies (23). The rest of the proof follows by standard calculations.
We can finally show uniqueness. Lemma 4.5. The solution of (22) is unique. def
Proof. Let u1 and u2 be two solutions. Define w ˜ = u1 − u2 . By calculations as above we get that Z t E[kw(t, ˜ ·)k2H ] ≤ KT (t − s)−3/4 E[kw(s, ˜ ·)k2H ]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 E[kw(t, ˜ ·)kH ] ≤ KT (t − s) (s − r)−3/4 E[kw(r, ˜ ·)k2H ]dr ds s=0 r=0 Z t 2 = KT B(1/4, 1/4) (t − r)−2/4 E[kw(r, ˜ ·)k2H ]dr r=0 Z t Z r ≤ KT3 B(1/4, 1/4) (t − r)−2/4 (r − s)−3/4 E[kw(s, ˜ ·)k2H ]ds dr r=0 s=0 Z t = KT3 B(1/4, 1/4)B(1/2, 1/4) (t − s)−1/4 E[kw(s, ˜ ·)k2H ]ds s=0 Z t Z s ≤ KT4 B(1/4, 1/4)B(1/2, 1/4) (t − s)−1/4 (s − r)−3/4 E[kw(r, ˜ ·)k2H ]dr ds s=0 r=0 Z t = KT4 B(1/4, 1/4)B(1/2, 1/4)B(3/4, 1/4) E[kw(r, ˜ ·)k2H ]dr r=0
We can now use Gronwall’s inequality.
Let’s now see what happens as L % ∞. Define the random times def
τL = inf{t ≥ 0 : k˜ uL (t, ·)kH ≥ L}
L>0
def
τ = lim (τL ∧ L). L→∞
Let’s also define def
u ˜(t, x) = lim u ˜L (t ∧ τL , x). L→∞
t ≥ 0, x ≥ 0
Lemma 4.6. We have that lim k˜ u(t, ·)kH = ∞.
t%τ
Define u as in (19)–(20). Then {u(t, ·) | 0 ≤ t < τ } is a weak solution of (1). 0
Proof. Fixing L0 > L we have from the uniqueness claim of Lemma 4.5 that u ˜L (t, ·) = u ˜L (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 have that u ˜(t, ·) = limL→∞ u ˜L (t, ·) for 0 ≤ t < τ . From this and Lemma 3.5, we conclude that {u(t, ·) | 0 ≤ t < τ } as defined by (19)–(20) indeed is a weak solution of (1). The characterization of k˜ u(t, ·)kH at τ − is obvious. In fact, we have a more explicit characterization of τ . 19
Lemma 4.7. We have that
2 ∂ u ˜ lim 2 (t, 0) = ∞. t%τ ∂x
Proof. For each L > 0, define 2 ∂ u ˜ def τL0 = inf t ∈ [0, τ ] | 2 (t−, 0) ≥ L . ∂x
(inf ∅ = τ )
By standard SPDE calculations like we used in Lemma 4.4, we know that (22) has a solution on [0, τL0 ]. Thus in fact τ > τL0 and hence 2 ∂ u ˜ (τL0 , 0) = L. ∂x2 Consequently 2 ∂ u ˜ 0 lim (τL , 0) = ∞. L→∞ ∂x2 Since τL0 ≤ τ , we of course also have that limL→∞ τL0 ≤ τ . On the other hand, ku(t, ·)kH may become large 2 for many reasons other than ∂∂xu˜2 (τL0 , 0) becoming large, so necessarily τ ≤ limL→∞ τL0 . Putting things together, we get that limL→∞ τL = τ . The claimed result now follows. To finish things off, we prove uniqueness. Lemma 4.8 (Uniqueness). If {˜ u(t, ·) | 0 ≤ t < τ } ⊂ H and {˜ u0 (t, ·) | 0 ≤ t < τ 0 } ⊂ H are two solutions of 0 0 (16), then u ˜(t, ·) = u ˜ (t, ·) for 0 ≤ t < min{τ, τ }. Proof. For each L > 0, define 2 2 0 ∂ u ∂ u ˜ ˜ def σL = inf t ∈ [0, τ ∧ τ 0 ) : 2 (t, 0) ≥ L or 2 (t, 0) ≥ L . ∂x ∂x
inf ∅ = τ ∧ τ 0
Then τ ∧ τ 0 ≤ limL→∞ σL . We can use standard uniqueness theory to conclude that u ˜ and u ˜0 coincide on [0, σL ], and we then let L % ∞. 4.1. Proofs. We here give the delayed proofs. We start with the structural claims about H. Proof of Lemma 4.1. The fact that H ⊂ C 2 is well-known; [Eva98]. Fix ϕ ∈ C0∞ (R+ ), x ∈ (0, ∞), and i ∈ {0, 1, 2}. We then have that Z x+1 i Z x+1 i ∂iϕ ∂ϕ ∂ϕ ∂iϕ (x) = (s)ds − (s) − (x) ds i ∂xi ∂xi ∂xi s=x ∂x s=x Z x+1 i Z x+1 Z s Z x+1 i Z x+1 ∂ϕ ∂ϕ ∂ i+1 ϕ ∂ i+1 ϕ = (s)ds − (r)dr ds = (s)ds − (x + 1 − r) i+1 (r)dr i i+1 i ∂x s=x ∂x s=x r=x ∂x s=x ∂x r=x Thus s i sZ x+1 i Z x+1 i+1 ∂ ϕ 2 ∂ ϕ 2 ∂ ϕ ∂xi (s) ds + ∂xi+1 (r) dr ≤ 2kϕkH . ∂xi (x) ≤ s=x
Of course we also have that
r=x
p p (i) ϕ (x) − ϕ(i) (y) ≤ kϕkH |x − y|
so the stated limits at x = 0 exist.
We next study {Tt }t>0 . Proof of Lemma 4.2. The proof relies upon a combination of fairly standard calculations. To begin, fix ϕ ∈ C0∞ (R+ ) and define Z def u(t, x) = p◦ (t, x − y)ϕ(|y|)dy y∈R
Z
∞
Z
0
p◦ (t, x − y)ϕ(y)dy +
= y=0
Z
∞
p◦ (t, x − y)ϕ(−y)dy = y=−∞
{p◦ (t, x − y) + p◦ (t, x + y)} ϕ(y)dy. y=0
20
Thus u(t, x) = (Tt ϕ)(x) for x > 0, and since p◦ is even in its second argument, Z Z u(t, −x) = p◦ (t, −x + y)ϕ(|y|)dy = p◦ (t, −x − y)ϕ(|y|)dy = u(t, x) y∈R
y∈R ∂nu ∂xn (t, 0)
= 0 for all odd n ∈ N; thus Tt ϕ ∈ Heven . so in fact u(t, ·) is even. Thus we indeed have that A standard calculation shows that Tt is a contraction on H. Indeed, for each nonnegative integer n, 2 2 n n+1 Z Z Z ∂ u ∂ ∂ n+2 u ∂nu u d dx ≤ 0 dx = 2 (t, x) (t, x) (t, x)dx = −2 (t, x) n+2 n+1 dt x∈R ∂xn ∂xn x∈R ∂x x∈R ∂x and thus n 2 2 2 n n Z Z ∂ u ∂ u ∂ u dx = 1 dx ≤ 1 dx (t, x) (t, x) (0, x) ∂xn n n 2 x∈R ∂x 2 x∈R ∂x x=0 Z ∞ 2 (n) = ϕ (x) dx.
Z (24)
∞
x=0
Summing these inequalities up for n ∈ {0, 1, 2, 3}, we see that kTt ϕk2H ≤ kϕk2H 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,even (R+ ) and define Z ∞ Z (1) v(t, x) = {p◦ (t, x − y) + p◦ (t, x + y)} ϕ (y)dy = p◦ (t, x − y)ϕ(1) (|y|)dy; y=0
y∈R (1)
(1)
note for future reference that since ϕ (0) = 0, y 7→ ϕ (|y|) is continuous at y = 0. Differentiating, we get that Z Z ∂v ∂p◦ (1) (25) (t, x) = (t, x − y)ϕ (|y|)dy = p◦ (t, x − y)ϕ(2) (|y|) sgn(y)dy. ∂x y∈R ∂x y∈R Thus in particular Z
∞
Z
2
∞
|v(t, x)| dx ≤ x=0
x=0 ∞
(1) 2 ϕ (x) dx
2 Z ∞ (2) 2 ∂v (t, x) dx ≤ (x) ϕ dx. x=0 x=0 ∂x
Z
Differentiating (25) again, we get that Z ∂2v ∂p◦ (t, x) = (t, x − y)ϕ(2) (|y|) sgn(y)dy 2 ∂x y∈R ∂x Z ∂3v ∂ 2 p◦ (t, x) = (t, x − y)ϕ(2) (|y|) sgn(y)dy 3 2 ∂x ∂x y∈R Z ∂p◦ ∂p◦ (t, x)ϕ(2) (0) + (t, x − y)ϕ(3) (|y|)dy. =2 ∂x ∂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 2 Z ∂ v K (2) √ (t, x) ≤ p (2t, x − y) (|y|) ϕ dy ◦ ∂x2 t y∈R 3 K Z 2K ∂ v (3) ≤ √ p◦ (2t, x) ϕ(2) (0) + √ (t, x) p (2t, x − y) (|y|) ϕ dy. ◦ ∂x3 t t y∈R We can now fairly easily conclude from (24) with n = 0 that s sZ 2 Z ∞ 2 ∞ ∂ v K dx ≤ √ ϕ(2) (x) 2 dx. (t, x) ∂x2 t x=0 x=0 21
We also note that sZ
s
∞
p2◦ (2t, x)dx
≤
x=0
1 √ 2πt
∞
Z
x=0
2 1 x 1 √ exp − dx ≤ . 1/4 2t (2πt) 2πt
Thus s
sZ 3 2 ∞ ∂ v K K (2) ϕ(3) (x) 2 dx. √ dx ≤ (t, x) |ϕ (0)| + ∂x3 1/4 3/4 (2π) t t x=0 x=0
Z
∞
Combine things together to get the last claim.
5. Numerical Simulation In this section, we will see from numerical simulations where the moving boundary is. 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 3.5. That is, ( def
(26)
u(t, x) =
u ˜(t, x − β(t)) − exp [− (x − β(t))] x ≥ β(t), 0 ≤ t < τ 0 x < β(t), 0 ≤ t < τ ,
where β(t) is defined as Z (27)
t
β(t) = s=0
∂2u ˜ 1− (s, 0) ds ∂x2
0≤t 0, x > 0
∂u ˜ (t, 0) = 0 t>0 ∂x u ˜(0, x) = u ˜◦ (x) = u◦ (x) + e−x .
x>0
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). Here we first discretize space by using the explicit finite difference scheme, then we can obtain SDE’s. Now we use the Euler-Maruyama Method to find numerical solutions of SDE’s (see [Gai96, Hig02]). 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 ( u◦ (x) =
x+x2 1+x4
0
if x ≥ 0 else
and α = .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. 22
0.9
0.8
0.7
0.6
t
0.5
0.4
0.3
0.2
0.1
0 −4
−3
−2
−1
0 x
1
2
3
4
Figure 1. Weak solution u(t, x)
References [BDP02]
Viorel Barbu and Giuseppe Da Prato. The two phase stochastic Stefan problem. Probab. Theory Related Fields, 124(4):544–560, 2002. [BDPR09] Viorel Barbu, Giuseppe Da Prato, and Michael R¨ ockner. Existence of strong solutions for stochastic porous media equation under general monotonicity conditions. Ann. Probab., 37(2):428–452, 2009. [CLM06] L. A. Caffarelli, K.-A. Lee, and A. Mellet. Homogenization and flame propagation in periodic excitable media: the asymptotic speed of propagation. Comm. Pure Appl. Math., 59(4):501–525, 2006. [CS05] Luis Caffarelli and Sandro Salsa. A geometric approach to free boundary problems, volume 68 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, 2005. [CV95] Luis A. Caffarelli and Juan L. V´ azquez. A free-boundary problem for the heat equation arising in flame propagation. Trans. Amer. Math. Soc., 347(2):411–441, 1995. [DPR04a] Giuseppe Da Prato and Michael R¨ ockner. Invariant measures for a stochastic porous medium equation. In Stochastic analysis and related topics in Kyoto, volume 41 of Adv. Stud. Pure Math., pages 13–29. Math. Soc. Japan, Tokyo, 2004. [DPR04b] Giuseppe Da Prato and Michael R¨ ockner. Weak solutions to stochastic porous media equations. J. Evol. Equ., 4(2):249–271, 2004. [DPRRW06] G. Da Prato, Michael R¨ ockner, B. L. Rozovskii, and Feng-Yu Wang. Strong solutions of stochastic generalized porous media equations: existence, uniqueness, and ergodicity. Comm. Partial Differential Equations, 31(1-3):277– 291, 2006. [Eva98] L. C. Evans. Partial Differential Equations. American Mathematical Society, Providence, RI, 1998. [Fri64] Avner Friedman. Partial differential equations of parabolic type. Prentice-Hall Inc., Englewood Cliffs, N.J., 1964. [Gai96] J. G. Gaines. Numerical experiments with S(P)DE’s. In Stochastic partial differential equations, volume 216 of London Math. Soc. Lecture Note Ser., pages 55–71, Cambridge University Press, Cambridge, 1995. [Hig02] Desmond J. Higham. An algorithmic introduction to numerical simulation of stochastic differential equations. SIAM Rev., 43(3):525–546, 2001. [Kim06] Jong Uhn Kim. On the stochastic porous medium equation. J. Differential Equations, 220(1):163–194, 2006. 23
[KS91] [Lun04] [Ste85] [Vaz96]
[Wal86]
Ioannis Karatzas and Steven E. Shreve. Brownian motion and stochastic calculus. Springer-Verlag, New York, second edition, 1991. Alessandra Lunardi. An introduction to parabolic moving boundary problems. In Functional analytic methods for evolution equations, volume 1855 of Lecture Notes in Math., pages 371–399. Springer, Berlin, 2004. D. S. Stewart. Transition to detonation in a model problem. J. Mech. Theor. Appl., 4(1):103–137, 1985. J. L. Vazquez. The free boundary problem for the heat equation with fixed gradient condition. In Free boundary problems, theory and applications (Zakopane, 1995), volume 363 of Pitman Res. Notes Math. Ser., pages 277–302. Longman, Harlow, 1996. ´ John B. Walsh. An introduction to stochastic partial differential equations. In Ecole d’´ et´ e de probabilit´ es de Saint-Flour, XIV—1984, volume 1180 of Lecture Notes in Math., pages 265–439. Springer, Berlin, 1986.
Department of Mathematics, University of Illinois at Urbana–Champaign, Urbana, IL 61801 E-mail address: [email protected] Department of Mathematics, University of Rochester, Rochester, NY 14627 Department of Mathematics, University of Illinois at Urbana–Champaign, Urbana, IL 61801 E-mail address: [email protected]
24
Abstract. We consider a stochastic perturbation of a moving boundary problem proposed by Ludford and Stewart and studied by Caffarelli and Vazquez. We prove existence and uniqueness.
1. Introduction Moving boundary problems are one of the important areas of partial differential equations. They provide the correct quantitative description of a wide range of physically interesting phenomena where a system has two phases. However, since the boundary between these phases is defined implicitly by the behavior of the rest of the system, they provide deep mathematical challenges in the areas of existence, uniqueness, and regularity. Our goal here is to study the effect of noise on a specific free boundary problem which was introduced by Stewart [Ste85] and subsequently addressed in the mathematics literature (see [CS05, CV95, Vaz96]). Fix a probability triple (Ω, F , P) and assume that B is a Brownian motion on (Ω, F , P). We consider the SPDE du(t, x) = (1)
lim x&β(t)
∂2u (t, x)dt + αu(t, x)dt + u(t, x) ◦ dBt ∂x2
∂u (t, x) = 1 ∂x u(0, x) = u◦ (x)
x > β(t)
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 more natural than not to include this term). We also assume that the initial function u◦ ∈ C(R) satisfies some specific properties: ◦ • u◦ ≡ 0 on R− , u◦ > 0 on (0, ∞), and limx&0 du dx (x) = 1. • u◦ and its first three derivatives exist on (0, ∞) and are square-integrable (on (0, ∞)).
In (1), ◦dBt represents Stratonovich integration, and the last line means that the boundary between u ≡ 0 and u > 0 is 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] and [CLM06]; 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 dBt 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 means that bubbles where u is positive cannot spontaneously nucleate within the region where u = 0. Our major contributions here are to 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 single Brownian motion, several transformations (the transformations of Lemmas 3.3 and 3.5 and (17)) can transform the problem into a random nonlinear PDE (see (18)). 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 3.2 and the iterative bounds of Lemma 4.4. 1
2. Weak Formulation To see what we mean by (1), let’s replace ◦dB by a smooth path b; the Wong-Zakai result (cf. [KS91, Section 5.2D]) implies that SDE with smoothed versions of dB converge to Stratonovich SDE (and that the Stratonovich interpretation is correct when we do so). Let’s also assume that there is only one interface. Namely, consider the PDE ∂2v ∂v + αv(t, x) + v(t, x)b(t) (t, x) = ∂t ∂x2 ∂v lim (t, x) = 1 x&β◦ (t) ∂x v(0, x) = u◦ (x). x∈R
(2)
x > β◦ (t)
{(t, x) ∈ R+ × R | v(t, x) > 0} = {(t, x) ∈ R+ × R | x > β◦ (t)}. 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 Z Z ∞ def v(t, x)ϕ(t, x)dx. v(t, x)ϕ(t, x)dx = Uϕ (t) = x=β◦ (t)
x∈R
Differentiating, we get that Z ∞ ∂v ∂ϕ ˙ Uϕ (t) = (t, x)ϕ(t, x) + v(t, x) (t, x) dx − v(t, β◦ (t))ϕ(t, β◦ (t))β˙ ◦ (t) ∂t ∂t x=β◦ (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 x > β◦ (t) to rewrite ∂v ∂t . Integrating by parts, we have that Z ∞ ∂2v (t, x)ϕ(t, x)dx 2 x=β◦ (t) ∂x Z ∞ ∂ϕ ∂2ϕ ∂v = lim v(t, x) 2 (t, x)dx. − (t, x)ϕ(t, β◦ (t)) + v(t, β◦ (t)) (t, β◦ (t)) + ∂x ∂x ∂x x&β◦ (t) x=β◦ (t) ∂v . Recombining Again we use the fact that v(t, β◦ (t)) = 0, and we can also use the boundary condition on ∂x things we get the standard formula that Z ∂ϕ ∂2ϕ U˙ ϕ (t) = v(t, x) (t, x) + (t, x) + αϕ(t, x) dx ∂t ∂x2 x∈R Z + v(t, x)ϕ(t, x)dx b(t) − ϕ(t, β◦ (t)). x∈R
Replacing b by ◦dB, we should have the following formulation: that for any ϕ ∈ Cc∞ (R+ × R) and any t > 0, Z Z Z t Z ∂2ϕ ∂ϕ (r, x) + u(t, x)ϕ(t, x)dx = u◦ (x)ϕ(t, x)dx + u(r, x) (r, x) + αϕ(r, x) dx dr ∂t ∂x2 x∈R x∈R r=0 x∈R Z t Z Z t + u(r, x)ϕ(r, x)dx ◦ dBr − ϕ(r, β(r))dr. r=0
The Ito formulation of this would be that Z Z Z u(t, x)ϕ(t, x)dx = u◦ (x)ϕ(t, x)dx + x∈R
x∈R
t
r=0
∂ϕ ∂2ϕ u(r, x) (r, x) + (r, x) + α ˆ ϕ(r, x) dx dr ∂t ∂x2 r=0 x∈R Z t Z Z t + u(r, x)ϕ(r, x)dx dBr − ϕ(r, β(r))dr Z
r=0
where α ˆ =α+
x∈R
1 2. 2
x∈R
r=0
Remark 2.1 Thus 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. We let def
Ft = σ{Bs ; 0 ≤ s ≤ t} for all t ≥ 0; then B is a Brownian motion with respect to {Ft }t>0 and stochastic integration against B will be with respect to this filtration. Definition 2.2. A weak solution of (1) is a predictable path {u(t, ·) | 0 ≤ t < τ } in 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
u(τ 0 , x)ϕ(τ 0 , x)dx =
x∈R
Z
Z
τ0
x∈R
∂2ϕ ∂ϕ (r, x) + α ˆ ϕ(r, x) dx dr (r, x) + ∂t ∂x2 Z τ0 u(r, x)ϕ(r, x)dx dBr − ϕ(r, β(r))dr
Z
u◦ (x)ϕ(0, x)dx+
u(r, x) r=0
Z
x∈R τ0
Z
+ r=0
x∈R
r=0
and where {(t, x) ∈ [0, τ ) × R | u(t, x) > 0} = {(t, x) ∈ [0, τ ) × R | x > β(t)}. Our main existence and uniqueness theorems are the following. The arguments leading up to these results will come together in Section 4. Theorem 2.3 (Existence). A solution of (1) exists. Furthermore, u(t, ·) ∈ C 2 [β(t), ∞) for all t ∈ [0, τ ) and 2 ∂ u τ ≤ inf t ≥ 0 : 2 (t−, β(t)) = ∞ . ∂x Proof. Combine Lemmas 4.6 and 4.7.
We also have uniqueness. Theorem 2.4 (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 4.8.
3. Regularity and a Transformation The proof of Theorems 2.3 and 2.4 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 3.2. First, however, let’s make sure that we understand a bit about regularity; this will illuminate the assumptions needed. 3.1. 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. The following representation result will help us in carrying out this analysis. Define 2 1 x def t > 0, x ∈ R exp − p◦ (t, x) = √ 4t 4πt def (3) p± (t, x, y) = {p◦ (t, x − y) ± p◦ (t, x + y)} eαt = {p◦ (t, x − y) ± p◦ (t, −x − y)} eαt t > 0, x, y ∈ R def
ˆ ˆ pˆ± (t, x, y) = {p◦ (t, x − y) ± p◦ (t, x + y)} eαt = {p◦ (t, x − y) ± p◦ (t, −x − y)} eαt ;
t > 0, x, y ∈ R
the second representations of p± and pˆ± stem from the fact that p◦ is even in its second argument. The distinction between p± and pˆ± naturally lies in the distinction between Ito and Stratonovich calculations. 3
We then have that ∂p± (t, x, y) = ∂t ∂ pˆ± (t, x, y) = ∂t lim p± (t, x, ·) =
(4)
t&0
∂ 2 p± (t, x, y) + αp± (t, x, y) t > 0, x, y ∈ R ∂y 2 ∂ 2 pˆ± (t, x, y) + α ˆ pˆ± (t, x, y) t > 0, x, y ∈ R ∂y 2 lim pˆ± (t, x, ·) = δx ; x ∈ R \ {0}
t&0
the relevant distinction between p+ and p− is their behavior at x = 0. This will come up in the arguments of Lemma 3.2. Lemma 3.1. Let u be a weak solution of (1) and assume that β is continuous. If 0 < t < τ and x > β(t), then Z t Z Bt −Bs Bt (5) u(t, x) = − e p± (t − s, x − β(t), β(s) − β(t))ds + e p± (t, x − β(t), y − β(t))u◦ (y)dy. s=0
y∈R
Furthermore, u(t, ·) is C ∞ on (β(t), ∞). Proof. Fix t > 0, x ∈ R, δ > 0 and c ∈ R and define def
ϕt,x,δ,c (s, y) = pˆ± (t + δ − s, x − c, y − c) Z def Ut,x,δ,c (s) = ϕt,x,δ,c (s, y)u(s, y)dy.
s ∈ [0, t], y ∈ R s β(t), and β is continuous, inf {|t − s| + |(x − β(t)) − (β(t) − β(s))|} = min {|t − s| + |(x − β(t)) − (β(t) − β(s))|} > 0
0≤s≤t
0≤s≤t
inf {|t − s| + |(x − β(t)) + (β(t) − β(s))|} = min {|t − s| + |(x − β(t)) + (β(t) − β(s))|} > 0.
0≤s≤t
0≤s≤t
Thus lim sup ϕt,x,δ,β(t) (s, β(s)) − pˆ± (t − s, x − β(t), β(s) − β(t)) = 0 δ&0 0≤s≤t lim sup ϕt,x,δ,β(t) (0, y) − pˆ± (t, x − β(t), y − β(t)) = 0. δ&0 y∈R
This gives us the claimed representation result. We can then differentiate to get the claimed smoothness. Note that (5) is not an explicit formula for u since the right-hand side of (5) depends on u through β. We also note that the proof effectively converts the Ito integral of (6) into a Stratonovich one, implying that pˆ± is converted back into p± . Next, let’s see what happens if we in fact assume that β is continuously differentiable. It turns out that not only does the boundary behavior of (1) hold pointwise, but we can find an evolution equation for β (which depends on u). To get the general idea of this latter fact, let’s return to our deterministic PDE (2). By definition v(t, β◦ (t)) = 0, so differentating (and using an approximation just to the right of β◦ ) we get that ∂v ∂v (t, β◦ (t)) + (t, β◦ (t))β˙ ◦ (t) = 0. ∂t ∂x Using the PDE for v and the boundary conditions (again, a rigorous proof would require pushing the calculation just a bit to the right of β◦ ), we get that in fact 2 ∂ v ∂2v (7) β˙ ◦ (t) = − (t, β (t)) + αv(t, β (t)) = − (t, β◦ (t)). ◦ ◦ ∂x2 ∂x2 For the SPDE (1) we should have the same result (since the noise term vanishes at the boundary). To proceed, let’s rewrite (5) in a slightly more convenient way. If {u(t, ·) | 0 ≤ t < τ } is a weak solution of (1) and 0 < t < τ , set Z t def A1 (t, ε) = exp [Bt − Bt−s + αs] p◦ (s, ε + β(t) − β(t − s))ds ε ∈ R \ {0} s=0 Z def Bt A± p± (t, ε, y − β(t))u◦ (y)dy. ε∈R 2 (t, ε) = e y∈R
Then some simple manipulation (which reflects the second representation of p± in (3) and the fact that p◦ is even in its second argument) shows that (8)
− u(t, β(t) + ε) = −A1 (t, ε) − A1 (t, −ε) + A+ 2 (t, ε) = −A1 (t, ε) + A1 (t, −ε) + A2 (t, ε).
We in fact have Lemma 3.2. Let {u(t, ·) | 0 ≤ t < τ } be a solution of (1). If β is continuously differentiable, then (9)
∂u (t, x) = 1 x&β(t) ∂x lim
∂2u ˙ (t, x) = −β(t) x&β(t) ∂x2
and
lim
for all t ∈ [0, τ ). Proof. From (8), we have that ∂A1 ∂A1 ∂A+ ∂u 2 (t, β(t) + ε) = − (t, ε) + (t, −ε) + (t, ε) ∂x ∂ε ∂ε ∂ε ∂2u ∂ 2 A1 ∂ 2 A1 ∂ 2 A− 2 (t, β(t) + ε) = − (t, ε) + (t, −ε) + (t, ε). 2 2 2 ∂x ∂ε ∂ε ∂ε2 Note that 2 ∂p◦ 1 x x (10) (t, x) = − √ exp − 3/2 ∂x 4t 2 4π t
∂ 2 p◦ 1 (t, x) = √ 2 ∂x 2 4πt3/2
and 5
2 x2 x − 1 exp − . 2t 4t
Thus ∂A1 (t, ε) = ∂ε
Z
t
exp [Bt − Bt−s − αs] s=0
1 =− √ 2 4π
∂p◦ (s, ε + β(t) − β(t − s))ds ∂x
" # 2 (ε + β(t) − β(t − s)) ε + β(t) − β(t − s) exp − exp [Bt − Bt−s + αs] ds 4s s3/2 s=0
Z
t
= −A˜1,1 (t, ε) − A˜1,2 (t, ε) Z t ∂ 2 p◦ ∂ 2 A1 exp [Bt − Bt−s + αs] (t, ε) = (s, ε + β(t) − β(t − s))ds 2 ∂ε ∂x2 s=0 ( ) Z t 2 (ε + β(t) − β(t − s)) 1 exp [Bt − Bt−s + αs] = √ −1 2s 2 4π s=0 " # 2 (ε + β(t) − β(t − s)) 1 × exp − ds 4s s3/2 = A˜2,1 (t, ε) + A˜2,2 (t, ε) + A˜2,3 (t, ε) where 1 A˜1,1 (t, ε) = √ 2 4π 1 A˜1,2 (t, ε) = √ 2 4π 1 A˜2,1 (t, ε) = √ 2 4π 1 A˜2,2 (t, ε) = √ 2 4π 1 A˜2,3 (t, ε) = √ 4 4π
# 2 (ε + β(t) − β(t − s)) exp [Bt − Bt−s + αs] 3/2 exp − ds 4s s s=0 # " Z t 2 (ε + β(t) − β(t − s)) β(t) − β(t − s) ds exp − exp [Bt − Bt−s + αs] 4s s3/2 s=0 " # 2 Z t 2 (ε + β(t) − β(t − s)) 1 ε ds − 1 exp − exp [Bt − Bt−s + αs] 3/2 2s 4s s s=0 " # Z t 2 ε β(t) − β(t − s) (ε + β(t) − β(t − s)) exp [Bt − Bt−s + αs] 3/2 exp − ds s 4s s s=0 " # Z t 2 2 (ε + β(t) − β(t − s)) (β(t) − β(t − s)) exp − ds. exp [Bt − Bt−s + αs] 4s s5/2 s=0 Z
t
"
ε
Since β is by assumption continuously differentiable, def
|β(t) − β(t − δ)| δ 0 0 and ε ∈ R. We also note that ∂p− ∂p◦ (t, x, 0) = −2 (t, x)eαt ∂y ∂x
t > 0, x ∈ R
and ∂ 2 p− (t, x, y) = ∂y 2
∂ 2 p◦ ∂ 2 p◦ (t, x − y) − (t, x + y) eαt ∂x2 ∂x2 t > 0, x ∈ R, y ∈ R 2 ∂ p◦ ∂ 2 p◦ αt = (t, x − y) − (t, −x − y) e ∂x2 ∂x2 where the last representation uses the fact that p◦ is even in its second argument. Thus
∂p◦ (s, ε)eαs ∂x Z 1 ∂ 2 p◦ + (β(t) − β(t − s))2 eαs (1 − r) 2 (s, ε − r(β(t) − β(t − s))) dr ∂x r=0 Z 1 ∂ 2 p◦ − (β(t) − β(t − s))2 eαs (1 − r) 2 (s, −ε − r(β(t) − β(t − s))) dr. ∂x r=0
p− (s, ε, β(t) − β(t − s)) = −2 (β(t) − β(t − s))
Thus A˜2,1 (t, ε) − A˜2,1 (t, −ε) = A˜a2,1 (t, ε) + A˜b2,1 (t, ε) − A˜b2,1 (t, −ε) + A˜c2,1 (t, ε) − A˜c2,1 (t, ε) 7
where A˜a2,1 (t, ε) = −
Z
t
exp [Bt − Bt−s + αs]
s=0 t Z 1
ε2 −1 2s
β(t) − β(t − s) ∂p◦ (s, ε)ds s ∂x 2
ε2 (β(t) − β(t − s)) ∂ 2 p◦ (s, ε − r(β(t) − β(t − s))) dr ds 2s s ∂x2 s=0 r=0 2 2 Z 1 Z ε2 t β(t) − β(t − s) ∂ p◦ = exp [Bt − Bt−s + αs] (s, ε − r(β(t) − β(t − s))) dr ds 4 s=0 r=0 s ∂x2 Z 1 Z 2 (β(t) − β(t − s)) ∂ 2 p◦ 1 t exp [Bt − Bt−s + αs] A˜c2,1 (t, ε) = − (s, ε − r(β(t) − β(t − s))) dr ds 2 s=0 r=0 s ∂x2 2 2 Z 1 Z β(t) − β(t − s) 1 t ∂ p◦ exp [Bt − Bt−s + αs] =− s 2 (s, ε − r(β(t) − β(t − s))) dr ds. 2 s=0 r=0 s ∂x √ We will again use the transformation u = |ε|/ s. We compute that for ε > 0 2 2 Z t ε β(t) − β(t − s) ε ε 1 a ˜ √ exp [Bt − Bt−s + αs] −1 exp − ds A2,1 (t, ε) = − 3/2 2s s 4s s 2 4π s=0 2 Z ∞ u β(t) − β(t − ε2 /u2 ) 2 2 2 2 = − sgn(ε) −1 √ exp Bt − Bt−ε /u + αε /u 2 ε2 /u2 u=ε/ t 2 u 1 √ du. × exp − 4 4π Thus by dominated convergence, 2 Z ∞ 2 u u 1 ˙ √ du = 0. − 1 exp − lim A˜a2,1 (t, ε) = β(t) ε&0 2 4 4π u=0 Similarly, Z Z 1 β(t) − β(t − ε2 /u2 ) 2 ε ∞ b 2 2 ˜ 2 2 A2,1 (t, ε) = exp Bt − Bt−ε /u + αε /u 2 u=|ε|/√t r=0 ε2 /u2 ∂ 2 p◦ 2 2 × ε3 /u3 ε /u , ε − r(β(t) − β(t − ε2 /u2 )) dr du. 2 ∂x From the second equality of (10), we see that there is a K > 0 such that 2 2 ∂ p◦ K x K (12) ∂x2 (s, x) ≤ s3/2 exp − 8s ≤ s3/2 1 A˜b2,1 (t, ε) = 2
Z
exp [Bt − Bt−s + αs]
for all s ∈ (0, t] and x ∈ R. Assume again that ε < 1/(2K). If u ≤ 1, then 3 3 3 3 ∂ 2 p◦ 2 2 ε /u 2 2 ε /u ε /u , ε − r(β(t) − β(t − ε /u )) ≤ K = K. 3/2 ∂x2 (ε2 /u2 ) On the other hand, if u ≥ 1, we have that 3 3 ∂ 2 p◦ 2 2 2 2 ε /u ε /u , ε − r(β(t) − β(t − ε /u )) ∂x2 " 2 # ε3 /u3 ε − r(β(t) − β(t − ε2 /u2 )) ≤K exp − 3/2 4ε2 /u2 (ε2 /u2 ) " 2 # 2 u2 β(t) − β(t − ε2 /u2 ) u ≤ K exp − 1−r ≤ K exp − 4 ε 8 by again using (11). Combining things together, we see that there is a K > 0 such that 2 Z ∞ √ u ˜b exp − du = Kε 2π; A2.1 (t, ε) ≤ Kε 8 u=0 8
thus indeed lim A˜b2.1 (t, ε) = 0.
ε→0
√ We next turn to A˜c2.1 . We here use the last bound of (12). Since s 7→ 1/ s is integrable on (0, 1], we can use dominated convergence to see that 2 2 Z 1 Z β(t) − β(t − s) 1 t ∂ p◦ c ˜ exp [Bt − Bt−s + αs] lim A2,ε (t, ε) = − s 2 (s, r(β(t) − β(t − s))) dr ds, ε→0 2 s=0 r=0 s ∂x this integral being finite. We have again availed ourselves of the fact that p◦ is even in its second argument. Finally, let’s understand the relevant behavior of A± 2 . We have that Z + ∂A2 ∂p+ lim (t, ε) = eBt (t, 0, y − β(t))u◦ (y)dy ε&0 ∂ε y∈R ∂x Z ∂ 2 p− ∂ 2 A− Bt 2 (t, ε) = e (t, 0, y − β(t))u◦ (y)dy. lim 2 ε&0 ∂ε2 y∈R ∂x From the second expression for p± in (3), we have that ∂p+ ∂p◦ ∂p◦ (t, 0, y) = eαt (t, −y) − (t, −y) = 0. ∂x ∂x ∂x We also have that p− (t, 0, y) = 0 for all t > 0, so ∂p− ∂ 2 p− (t, 0, y) = (t, 0, y) − αp− (t, 0, y) = 0. ∂x2 ∂t Thus in fact
∂A+ 2 (t, ε) = 0 and ε&0 ∂ε Combining things together, we indeed get (9). lim
∂ 2 A− 2 (t, ε) = 0. ε&0 ∂ε2 lim
3.2. A Transformation. The characterization of β given in (9) allows us to rewrite the moving boundaryvalue problem in a more convenient way. The calculation which gives us some analytical traction is found in [Lun04] (see also [Fri64, Ch. 8]). Again, let’s return to our deterministic PDE (2). For all t ≥ 0 and x ∈ R, define v˜(t, x) = v(t, x + β◦ (t)) + e−x . Then v(t, x) = v˜(t, x − β◦ (t)) − exp [−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 (13) (t, x) = (t, x + β◦ (t)) − e−x ∂x ∂x ∂ 2 v˜ ∂2v (t, x) = (t, x + β◦ (t)) + e−x . 2 ∂x ∂x2 We can combine these equations and use the PDE for v to rewrite the evolution of v˜ as
(14)
∂˜ v ∂2v (t, x) = (t, x + β◦ (t)) + αv(t, x + β◦ (t)) + v(t, x + β◦ (t))b(t) ∂t ∂x2 ∂v + (t, x + β◦ (t))β˙ ◦ (t) ∂x ∂ 2 v˜ ∂˜ v −x −x −x = (t, x) − e + α v ˜ (t, x) − e + (t, x) + e β˙ ◦ (t) ∂x2 ∂x + v˜(t, x) − e−x b(t).
Note also that
∂˜ v (t, 0) = 1 − 1 = 0. ∂x Furthermore, v˜(t, 0) = 1 for all t > 0, so evaluating (14) at x = 0 (or more accurately, as x & 0), we get that 0=
∂˜ v ∂ 2 v˜ (t, 0) = (t, 0) − 1 + β˙ ◦ (t). ∂t ∂x2 9
Thus in fact ∂ 2 v˜ (t, 0); β˙ ◦ (t) = 1 − ∂x2 alternately by combining (7) and the last line of (13), we have that 2 ∂ v˜ ∂2v (0, x) − 1 . β˙ ◦ (t) = − 2 (t, β◦ (t)) = − ∂x ∂x2 (15)
Inserting the dynamics of β◦ back into (14) we can collect things and get a PDE for v˜; we have that ∂˜ v ∂˜ v ∂ 2 v˜ ∂ 2 v˜ −x −x −x (t, x) − e + α v˜(t, x) − e + (t, 0) (t, x) = (t, x) + e 1− ∂t ∂x2 ∂x ∂x2 + v˜(t, x) − e−x b(t) t > 0, x > 0 ∂˜ v (t, 0) = 0 ∂x
t>0 def
v˜(0, x) = u ˜◦ (x) = u◦ (x) + e−x
x>0
Replacing b by our Brownian motion B and α by α ˆ , we now get the following. Lemma 3.3. Suppose that {u(t, ·) | 0 ≤ t < τ } ⊂ C(R) ∩ L(R) is a weak solution of (1). Suppose also that β is continuously differentiable and {Ft }t≥0 -adapted. Then u ˜(t, x) = u(t, x + β(t)) + e−x satisfies the integral equation Z ∞ u ˜(t, x) = pˆ+ (t, x, y) u◦ (y) + e−y dy y=0 t
∞
∂2u ˜ ∂u ˜ −y (s, y) + e−y 1− (s, 0) − (ˆ α + 1)e dy ds ∂x ∂x2 s=0 y=0 Z t Z ∞ + pˆ+ (t − s, x, y) u ˜(s, y) − e−y dy dBs Z
Z
pˆ+ (t − s, x, y)
+
(16)
s=0
y=0
for all t > 0 and x > 0. Thanks to Lemmas 3.1 and 3.2, the assumption that β is continuously differentiable ensures that the spatial derivatives of u ˜ on the right-hand side of (16) are well-defined. Proof of Lemma 3.3. Fix x > 0 and T > 0. For t ∈ [0, τ ), define Z ∞ def T U (t) = u ˜(t, y)ˆ p+ (T − t, x, y)dy = AT1 (t) + AT2 (t) y=0
where AT1 (t)
Z
∞
Z
∞
u(t, y + β(t))ˆ p+ (T − t, x, y)dy =
= y=0
u(t, y)ˆ p+ (T − t, x, y − β(t))dy y=β(t)
Z u(t, y)ˆ p+ (T − t, x, y − β(t))dy
= y∈R ∞
AT2 (t) =
Z
e−y pˆ+ (T − t, x, y)dy.
y=0
Using Definition 2.2 and (4), we get that Z ∂ 2 pˆ+ ∂ pˆ+ T (T − t, x, y − β(t)) + (T − t, x, y − β(t)) dA1 (t) = u(t, y) − ∂t ∂y 2 y∈R ∂ pˆ+ ˙ +ˆ αpˆ+ (T − t, x, y − β(t)) − (T − t, x, y − β(t))β(t) dydt ∂y Z + u(t, y)ˆ p+ (T − t, x, y − β(t))dy dBt − pˆ+ (T − t, x, 0)dt y∈R 10
(Z
) ∂ pˆ+ ˙ u(t, y) =− (T − t, x, y − β(t))dy β(t)dt ∂y y=β(t) (Z ) ∞ + u(t, y)ˆ p+ (T − t, x, y − β(t))dy dBt − pˆ+ (T − t, x, 0)dt ∞
y=β(t)
(Z
∞
= y=β(t)
(Z
∂u (t, y)ˆ p+ (T − t, x, y − β(t))dy ∂y
) ˙ β(t)dt )
∞
u(t, y)ˆ p+ (T − t, x, y − β(t))dy
+
dBt − pˆ+ (T − t, x, 0)dt
y=β(t)
Z
∞
∂u ˙ = (t, y + β(t))ˆ p+ (T − t, x, y)dy β(t)dt y=0 ∂y Z ∞ u(t, y + β(t))ˆ p+ (T − t, x, y)dy dBt − pˆ+ (T − t, x, 0)dt + y=0 Z ∞ ∂u ˜ ˙ = (t, y) + e−y pˆ+ (T − t, x, y)dy β(t)dt ∂y y=0 Z ∞ −y + u ˜(t, y) − e pˆ+ (T − t, x, y)dy dBt − pˆ+ (T − t, x, 0)dt. y=0
We have here used the fact that u(t, β(t)) = 0. We have also employed a fairly straightforward generalization of the integral equality in Definition 2.2 to predictable integrands; the continuous differentiability and adaptedness of β allow us to apply this. Combining the characterization of β˙ as in Lemma 3.2 and a calculation as in (15), we get that ∂2u ˜ ˙ (t, 0). β(t) =1− ∂x2 Thus Z t Z ∞ ∂u ˜ ∂2u ˜ AT1 (t) = (s, y) + e−y 1− (s, 0) p ˆ (T − s, x, y)dy ds + ∂y ∂x2 s=0 y=0 Z t Z ∞ Z t −y + u ˜(s, y) − e pˆ+ (T − s, x, y)dy dBs − pˆ+ (T − s, x, 0)ds. s=0
y=0
s=0
A straightforward differentiation, on the other hand, shows that Z ∞ ∂ pˆ+ A˙ T2 (t) = − e−y (T − t, x, y)dy ∂t y=0 Z ∞ Z ∞ 2 ˆ+ −y ∂ p =− e (T − t, x, y)dy − e−y α ˆ pˆ+ (T − t, x, y)dy ∂y 2 y=0 y=0 Z ∞ ∂ pˆ+ = (T − t, x, 0) + pˆ+ (T − t, x, 0) − (ˆ α + 1) e−y pˆ+ (T − t, x, y)dy. ∂y y=0 Note that
∂ pˆ+ ∂y (T
− t, x, 0) = 0. Combine things to get that Z t Z ∞ ∂u ˜ ∂2u ˜ U T (t) = U T (0) + (s, y) + e−y 1− (s, 0) p ˆ (T − s, x, y)dy ds + ∂y ∂x2 s=0 y=0 Z t Z ∞ −y − (ˆ α + 1) e pˆ+ (T − s, x, y)dy ds Z
t
Z
s=0 ∞
+ s=0
y=0
u ˜(s, y) − e−y pˆ+ (T − s, x, y)dy dBs .
y=0
Now let T & t to get the claimed result.
11
Of course (16) is equivalent to the SPDE 2 ∂ u ˜ ∂u ˜ ∂2u ˜ −x −x −x (t, x) + α ˆ u ˜(t, x) − e (t, 0) dt d˜ u(t, x) = −e + (t, x) + e 1− ∂x2 ∂x ∂x2 + u ˜(t, x) − e−x dBt t > 0, x > 0 ∂u ˜ (t, 0) = 0 t>0 ∂x u ˜(0, x) = u ˜◦ (x) = u◦ (x) + e−x .
x>0
We can also find a converse to Lemma 3.3. First of all note the following. Lemma 3.4. Suppose that {˜ u(t, ·) | 0 ≤ t < τ } ⊂ C 2 (R+ ) satisfies (16). Then u ˜(t, 0) = 1 for all 0 ≤ t < τ . Furthermore, u ˜(t, x) > 0 for all t ≥ 0 and x ≥ 0. Proof. Let’s first smooth things out. Fix δ > 0 and define Z ∞ Z ∞ def u ˜δ (t, x) = pˆ+ (δ, x, y)˜ u(t, y)dy = pˆ+ (t + δ, x, y) u◦ (y) + e−y dy y=0
y=0 t
∞
∂u ˜ ∂2u ˜ −y (s, y) + e−y 1− (s, 0) − ( α ˆ + 1)e dy ds ∂x ∂x2 s=0 y=0 Z t Z ∞ + pˆ+ (t + δ − s, x, y) u ˜(s, y) − e−y dy dBs ; Z
Z
pˆ+ (t + δ − s, x, y)
+
s=0
y=0
we have of course used the fact that pˆ+ is a semigroup of integral kernels. For each x > 0, some straightforward computations show that 2 ∂ u ˜δ def d˜ uδ (t, x) = (t, x) + α ˆ u ˜ (t, x) dt δ ∂x2 Z ∞ ∂u ˜ ∂2u ˜ −y + pˆ+ (δ, x, y) (t, y) + e−y 1− (t, 0) − (ˆ α + 1)e dy dt ∂x ∂x2 y=0 Z ∞ −y + pˆ+ (δ, x, y) u ˜(t, y) − e dy dBt . y=0
We now let δ & 0 and use the assumed continuity of u ˜. We also fix ε > 0 and evaluate the result at x = ε. We get that Z t Z t ∂2u ˜ (s, ε)ds + α ˆ u ˜(s, ε)ds u ˜(t, ε) = u ˜◦ (ε) + 2 s=0 s=0 ∂x Z t Z t ∂u ˜ ∂2u ˜ + (s, ε) + e−ε 1− (s, 0) ds − (ˆ α + 1) e−ε ds 2 ∂x ∂x s=0 s=0 Z t + u ˜(s, ε) − e−ε dBs s=0 Z t 2 ∂ u ˜ ∂2u ˜ ∂u ˜ −ε =u ˜◦ (ε) + (s, ε) − (s, 0) (s, ε) + e ds ∂x2 ∂x2 ∂x s=0 Z t Z t ∂u ˜ + (s, ε)ds + u ˜(s, ε) − e−ε dBs . s=0 ∂x s=0 Letting ε & 0, we see that Z
1
(˜ u(s, 0) − 1) dBs
u ˜(t, 0) = 1 + s=0
or alternately Z
1
u ˜(t, 0) − 1 =
(˜ u(s, 0) − 1) dBs s=0
which indeed implies that u ˜(t, 0) = 1 for all t ∈ [0, τ ). 12
To see the positivity, we define def u∗ (t, x) = u ˜(t, x) − e−x e−Bt .
(17)
t ≥ 0, x ≥ 0
Some straightforward calculations show that u∗ satisfies the random PDE ∗ 2 ∗ ∂u ∂ u ∂u∗ ∂ 2 u∗ ∗ Bt (t, x) + αu (t, x) − (t, 0)e (t, x) = (t, x) ∂x ∂x2 ∂x2 ∂x (18) u∗ (t, 0) = e−Bt t>0 u∗ (0, x) = u◦ (x).
t > 0, x > 0
x>0
Note that e−Bt > 0 for all t > 0 and u◦ (x) > 0 for all x > 0. Standard calculations for the heat equation then ensure that indeed u∗ (t, x) > 0 for all t > 0 and x∗ > 0. We then have Lemma 3.5. Suppose that {˜ u(t, ·) | 0 ≤ t < τ } ⊂ C 2 (R+ ) ∩ L1 (R+ ) satisfies (16). Set Z t ∂2u ˜ (s, 0) ds 0≤t 0 and ΨL ∈ C ∞ (R; [0, 1]) such that ΨL (x) = 1 if |x| ≤ L and ΨL (x) = 0 if |x| ≥ L + 1. Set Z ∞ L u ˜1 (t, x) = pˆ+ (t, x, y)˜ u◦ (y)dy y=0
for all t > 0 and x ∈ R and recursively define Z ∞ L (21) u ˜n+1 (t, x) = pˆ+ (t, x, y)˜ u◦ (y)dy +
t
y=0 ∞
∂2u ˜L ∂u ˜L n n (s, y) + e−y 1− (s, 0) ΨL k˜ uL n (s, ·)kH 2 ∂x ∂x s=0 y=0 Z t Z ∞ L −(ˆ α + 1)e−y dy ds + pˆ+ (t − s, x, y) u ˜n (s, y) − e−y dy dBs . t > 0, x > 0
Z
Z
pˆ+ (t − s, x, y)
s=0
y=0
{˜ uL n (t, ·);
For each n ∈ N, t ≥ 0} is a well-defined, adapted, and continuous path in Heven . To study (21), we will use the Neumann heat semigroup. For ϕ ∈ C0∞ (R+ ), t > 0, and x > 0, define Z ∞ def (Tt ϕ)(x) = p+ (t, x, y)ϕ(y)dy. y=0
Lemma 4.2. For each t > 0, Tt has a unique extension from C0∞ (R+ ) to H such that Tt H ⊂ Heven and such that kTt f kH ≤ kf kH for all f ∈ H. Secondly, there is a KA > 0 such that kTt f˙kH ≤
KA kf kH t3/4
for all f ∈ Heven ∩ C 4 (R+ ). 16
Again, we delay the proof until Subsection 4.1. Another convenience will be to rewrite the ds part of (21). Define ¨ ˜ L (ψ) def Ψ = (1 − ψ(0))Ψ L (kψkH ) for all ψ ∈ H. Then L L ∂2u ˜L ∂u ˜n ∂u ˜n n −x L −x ˜ L (˜ (t, 0) ΨL k˜ un (t, ·)kH = (t, x) + e 1− (t, x) + e Ψ uL N (t, ·)) ∂x ∂x2 ∂x for all n ∈ N. For ψ and η in H, let’s also define def
¨ ˜ L )(ψ; η) = −¨ ˙ L (kψkH ) (DΨ η (0)ΨL (kψkH ) + (1 − ψ(0)) Ψ
hψ, ηiH . kψkH
˜ L )(ψ, η) is the Gˆ ˜ L at ψ in the direction of Lemma 4.3. For each ψ and η in H, (DΨ ateaux derivative of Ψ η. Furthermore, there is a KB > 0 such that ˜ (DΨL )(ψ, η) ≤ KB χ[0,L+1] (kψkH )kηkH for all ψ and η in H. Proof. The claim is straightforward.
def
For each n ∈ N, we now define w ˜nL (t, x) = u ˜L ˜L n (t, x) for all x ≥ 0 and t ≥ 0. Clearly n+1 (t, x) − u L 2 sup0≤t≤T E kw ˜1 kH < ∞ for all T > 0. We then write that L w ˜n+1 (t, x) =
4 X
(n)
Aj (t, x)
j=1
where Z
(n)
1
Z
t
Z
∞
pˆ+ (t − s, x, y)
A1 (t, x) = λ=0 1
Z
λ=0 1
Z
s=0 t
y=0
∂w ˜nL ˜L u ˜L ˜nL (s, ·) ds dλ (s, y)dy Ψ n (s, ·) + λw ∂x
∂w ˜nL ˜L u = (s, ·) (x)Ψ ˜L ˜nL (s, ·) ds dλ Tt−s n (s, ·) + λw ∂x λ=0 s=0 L Z 1 Z t Z ∞ ∂w ˜n ∂u ˜n (n) A2 (t, x) = (s, y) + λ L (s, y) dy pˆ+ (t − s, x, y) ∂x ∂x λ=0 s=0 y=0 L L L ˜ × D ΨL u ˜n (s, ·) + λw ˜n (s, ·), w ˜n (s, ·) ds dλ L Z 1 Z t n ∂w ˜L ∂u ˜n ˜L u = (s, ·) + λ (s, ·) (x)DΨ ˜L ˜nL (s, ·), w ˜nL (s, ·) ds dλ Tt−s n (s, ·) + λw ∂x ∂x λ=0 s=0 Z 1 Z t Z ∞ (n) ˜L u pˆ+ (t − s, x, y)e−y dyDΨ ˜L ˜nL (s, ·), w ˜nL (s, ·) ds dλ A3 (t, x) = n (s, ·) + λw Z
Z
s=0 t
y=0
˜L u (Tt−s E)(x)DΨ ˜L ˜nL (s, ·), w ˜nL (s, ·) ds dλ n (s, ·) + λw λ=0 s=0 Z t Z ∞ (n) L A4 (t, x) = pˆ+ (t − s, x, y)w ˜n (s, y)dy dBs =
s=0 t
Z =
y=0
Tt−s w ˜nL (s, ·) (x)dBs
s=0 def
where for convenience we have set E(x) = e−x for all x ≥ 0. Note that the u ˜L ˜nL ’s are all in Heven . n ’s and w An easy calculation gives us that Z t h i Z t L (n) E kA4 (t, ·)k2H = E kTt−s w ˜nL (s, ·)k2H ds ≤ E kw ˜n (s, ·)k2H ds. s=0
s=0 17
We similarly have (using Jensen’s inequality) that Z Z t h i L (n) 2 2 2 2 2 2 kTt−s EkH E kw ˜n (s, ·)kH ds ≤ tKB kEkH E kA3 (t, ·)kH ≤ tKB To bound
and
(n) A2 ,
−3/4
we use the fact that t Z t s=0
for all t > 0. Thus " Z h i (n) 2 E kA1 (t, ·)k2H ≤ KB E
L E kw ˜n (s, ·)k2H ds.
s=0
s=0 (n) A1
t
is locally integrable. More precisely,
1 ds = 4t1/4 (t − s)3/4
2 #
L
∂ w ˜
Tt−s n (s, ·) ds
∂x s=0 H h
2 i 2
L Z t E w
L Z t ˜ (s, ·)
n w ˜ (s, ·) H n 2 2 1/4 2 2 H ≤ 4K K t ≤ KA KB E ds ds. A B s=0 (t − s)3/4 (t − s)3/4 s=0 t
Finally, we have that h i (n) E kA2 (t, ·)k2H " Z 2 #
L L
L
L t
∂ u ˜ ∂ w ˜ n n L 2
Tt−s ˜n (s, ·)kH ds ˜n (s, ·) + λw ˜n (s, ·) H kw χ[0,L+1] u ≤ KB E (s, ·) + λ (s, ·)
∂x ∂x s=0 H " Z 2 # Z t t L+1 E[kw ˜nL (s, ·)k2H ] 2 2 2 2 2 1/4 L ≤ KA KB E ≤ 4K K (L + 1) t k w ˜ (s, ·)k ds ds. H A B n 3/4 (t − s)3/4 s=0 (t − s) s=0 L P∞ def L Lemma 4.4. For each T > 0, we have that n=1 sup0≤t≤T E k˜ un+1 − u ˜L n kH < ∞. Thus P-a.s., u (t, ·) = L limn→∞ uL n (t, ·) exists as a limit in C([0, T ]; H) and u satisfies the integral equation Z ∞ u ˜L (t, x) = pˆ+ (t, x, y)˜ u◦ (y)dy y=0
Z
t
Z
∞
∂u ˜L (s, y) + e−y ∂x
∂2u ˜L 1− (s, 0) ΨL k˜ uL (s, ·)kH 2 ∂x
pˆ+ (t − s, x, y) −(ˆ α + 1)e−y dy ds Z t Z ∞ L + pˆ+ (t − s, x, y) u ˜ (s, y) − e−y dydBs . +
(22)
s=0
y=0
s=0
y=0
t > 0, x > 0
Proof. See also [Wal86, Lemma 3.3]. Fixing T > 0 we collect the above calculations to see that there is a KT > 0 such that Z t E[kw ˜nL (s, ·)k2H ] L E[kw ˜n+1 (t, ·)k2H ] ≤ KT ds (t − s)3/4 s=0 for all t ∈ [0, T ]. Iterating this, we get that n−2 Y E[kw ˜nL (t, ·)k2H ] ≤ KTn−1 t(n−1)/4 B(1 + j/4, 1/4) sup E[kw ˜1L k2H ] 0≤t≤T j=1
where B is the standard Beta function and thus that 1/2 q q n−2 Y (n−1)/2 (n−1)/8 E[kw ˜nL (t, ·)k2H ] ≤ KT t B(1 + j/4, 1/4) sup E[kw ˜1L k2H ]. 0≤t≤T j=1
To show that the terms on the right are summable, we use the ratio test. It suffices to show that 1/2 n−2 1/2 (23) lim KT t1/8 B 1 + , 1/4 = 0. n→∞ 4 18
We calculate that Z
1 n/4
s
B(1 + n/4, 1/4) =
(1 − s)
−3/4
1 1− n
Z ds =
s=0
n/4
s
−3/4
(1 − s)
Z
≤n
sn/4 (1 − s)−3/4 ds
1 s=1− n
s=0 3/4
1
ds +
Z
1 1− n
n/4
s
Z
(1 − s)−3/4 ds
1 s=1− n
s=0
4n3/4 ≤ +4 n+4
1
ds +
1/4 1 . n
This implies (23). The rest of the proof follows by standard calculations.
We can finally show uniqueness. Lemma 4.5. The solution of (22) is unique. def
Proof. Let u1 and u2 be two solutions. Define w ˜ = u1 − u2 . By calculations as above we get that Z t E[kw(t, ˜ ·)k2H ] ≤ KT (t − s)−3/4 E[kw(s, ˜ ·)k2H ]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 E[kw(t, ˜ ·)kH ] ≤ KT (t − s) (s − r)−3/4 E[kw(r, ˜ ·)k2H ]dr ds s=0 r=0 Z t 2 = KT B(1/4, 1/4) (t − r)−2/4 E[kw(r, ˜ ·)k2H ]dr r=0 Z t Z r ≤ KT3 B(1/4, 1/4) (t − r)−2/4 (r − s)−3/4 E[kw(s, ˜ ·)k2H ]ds dr r=0 s=0 Z t = KT3 B(1/4, 1/4)B(1/2, 1/4) (t − s)−1/4 E[kw(s, ˜ ·)k2H ]ds s=0 Z t Z s ≤ KT4 B(1/4, 1/4)B(1/2, 1/4) (t − s)−1/4 (s − r)−3/4 E[kw(r, ˜ ·)k2H ]dr ds s=0 r=0 Z t = KT4 B(1/4, 1/4)B(1/2, 1/4)B(3/4, 1/4) E[kw(r, ˜ ·)k2H ]dr r=0
We can now use Gronwall’s inequality.
Let’s now see what happens as L % ∞. Define the random times def
τL = inf{t ≥ 0 : k˜ uL (t, ·)kH ≥ L}
L>0
def
τ = lim (τL ∧ L). L→∞
Let’s also define def
u ˜(t, x) = lim u ˜L (t ∧ τL , x). L→∞
t ≥ 0, x ≥ 0
Lemma 4.6. We have that lim k˜ u(t, ·)kH = ∞.
t%τ
Define u as in (19)–(20). Then {u(t, ·) | 0 ≤ t < τ } is a weak solution of (1). 0
Proof. Fixing L0 > L we have from the uniqueness claim of Lemma 4.5 that u ˜L (t, ·) = u ˜L (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 have that u ˜(t, ·) = limL→∞ u ˜L (t, ·) for 0 ≤ t < τ . From this and Lemma 3.5, we conclude that {u(t, ·) | 0 ≤ t < τ } as defined by (19)–(20) indeed is a weak solution of (1). The characterization of k˜ u(t, ·)kH at τ − is obvious. In fact, we have a more explicit characterization of τ . 19
Lemma 4.7. We have that
2 ∂ u ˜ lim 2 (t, 0) = ∞. t%τ ∂x
Proof. For each L > 0, define 2 ∂ u ˜ def τL0 = inf t ∈ [0, τ ] | 2 (t−, 0) ≥ L . ∂x
(inf ∅ = τ )
By standard SPDE calculations like we used in Lemma 4.4, we know that (22) has a solution on [0, τL0 ]. Thus in fact τ > τL0 and hence 2 ∂ u ˜ (τL0 , 0) = L. ∂x2 Consequently 2 ∂ u ˜ 0 lim (τL , 0) = ∞. L→∞ ∂x2 Since τL0 ≤ τ , we of course also have that limL→∞ τL0 ≤ τ . On the other hand, ku(t, ·)kH may become large 2 for many reasons other than ∂∂xu˜2 (τL0 , 0) becoming large, so necessarily τ ≤ limL→∞ τL0 . Putting things together, we get that limL→∞ τL = τ . The claimed result now follows. To finish things off, we prove uniqueness. Lemma 4.8 (Uniqueness). If {˜ u(t, ·) | 0 ≤ t < τ } ⊂ H and {˜ u0 (t, ·) | 0 ≤ t < τ 0 } ⊂ H are two solutions of 0 0 (16), then u ˜(t, ·) = u ˜ (t, ·) for 0 ≤ t < min{τ, τ }. Proof. For each L > 0, define 2 2 0 ∂ u ∂ u ˜ ˜ def σL = inf t ∈ [0, τ ∧ τ 0 ) : 2 (t, 0) ≥ L or 2 (t, 0) ≥ L . ∂x ∂x
inf ∅ = τ ∧ τ 0
Then τ ∧ τ 0 ≤ limL→∞ σL . We can use standard uniqueness theory to conclude that u ˜ and u ˜0 coincide on [0, σL ], and we then let L % ∞. 4.1. Proofs. We here give the delayed proofs. We start with the structural claims about H. Proof of Lemma 4.1. The fact that H ⊂ C 2 is well-known; [Eva98]. Fix ϕ ∈ C0∞ (R+ ), x ∈ (0, ∞), and i ∈ {0, 1, 2}. We then have that Z x+1 i Z x+1 i ∂iϕ ∂ϕ ∂ϕ ∂iϕ (x) = (s)ds − (s) − (x) ds i ∂xi ∂xi ∂xi s=x ∂x s=x Z x+1 i Z x+1 Z s Z x+1 i Z x+1 ∂ϕ ∂ϕ ∂ i+1 ϕ ∂ i+1 ϕ = (s)ds − (r)dr ds = (s)ds − (x + 1 − r) i+1 (r)dr i i+1 i ∂x s=x ∂x s=x r=x ∂x s=x ∂x r=x Thus s i sZ x+1 i Z x+1 i+1 ∂ ϕ 2 ∂ ϕ 2 ∂ ϕ ∂xi (s) ds + ∂xi+1 (r) dr ≤ 2kϕkH . ∂xi (x) ≤ s=x
Of course we also have that
r=x
p p (i) ϕ (x) − ϕ(i) (y) ≤ kϕkH |x − y|
so the stated limits at x = 0 exist.
We next study {Tt }t>0 . Proof of Lemma 4.2. The proof relies upon a combination of fairly standard calculations. To begin, fix ϕ ∈ C0∞ (R+ ) and define Z def u(t, x) = p◦ (t, x − y)ϕ(|y|)dy y∈R
Z
∞
Z
0
p◦ (t, x − y)ϕ(y)dy +
= y=0
Z
∞
p◦ (t, x − y)ϕ(−y)dy = y=−∞
{p◦ (t, x − y) + p◦ (t, x + y)} ϕ(y)dy. y=0
20
Thus u(t, x) = (Tt ϕ)(x) for x > 0, and since p◦ is even in its second argument, Z Z u(t, −x) = p◦ (t, −x + y)ϕ(|y|)dy = p◦ (t, −x − y)ϕ(|y|)dy = u(t, x) y∈R
y∈R ∂nu ∂xn (t, 0)
= 0 for all odd n ∈ N; thus Tt ϕ ∈ Heven . so in fact u(t, ·) is even. Thus we indeed have that A standard calculation shows that Tt is a contraction on H. Indeed, for each nonnegative integer n, 2 2 n n+1 Z Z Z ∂ u ∂ ∂ n+2 u ∂nu u d dx ≤ 0 dx = 2 (t, x) (t, x) (t, x)dx = −2 (t, x) n+2 n+1 dt x∈R ∂xn ∂xn x∈R ∂x x∈R ∂x and thus n 2 2 2 n n Z Z ∂ u ∂ u ∂ u dx = 1 dx ≤ 1 dx (t, x) (t, x) (0, x) ∂xn n n 2 x∈R ∂x 2 x∈R ∂x x=0 Z ∞ 2 (n) = ϕ (x) dx.
Z (24)
∞
x=0
Summing these inequalities up for n ∈ {0, 1, 2, 3}, we see that kTt ϕk2H ≤ kϕk2H 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,even (R+ ) and define Z ∞ Z (1) v(t, x) = {p◦ (t, x − y) + p◦ (t, x + y)} ϕ (y)dy = p◦ (t, x − y)ϕ(1) (|y|)dy; y=0
y∈R (1)
(1)
note for future reference that since ϕ (0) = 0, y 7→ ϕ (|y|) is continuous at y = 0. Differentiating, we get that Z Z ∂v ∂p◦ (1) (25) (t, x) = (t, x − y)ϕ (|y|)dy = p◦ (t, x − y)ϕ(2) (|y|) sgn(y)dy. ∂x y∈R ∂x y∈R Thus in particular Z
∞
Z
2
∞
|v(t, x)| dx ≤ x=0
x=0 ∞
(1) 2 ϕ (x) dx
2 Z ∞ (2) 2 ∂v (t, x) dx ≤ (x) ϕ dx. x=0 x=0 ∂x
Z
Differentiating (25) again, we get that Z ∂2v ∂p◦ (t, x) = (t, x − y)ϕ(2) (|y|) sgn(y)dy 2 ∂x y∈R ∂x Z ∂3v ∂ 2 p◦ (t, x) = (t, x − y)ϕ(2) (|y|) sgn(y)dy 3 2 ∂x ∂x y∈R Z ∂p◦ ∂p◦ (t, x)ϕ(2) (0) + (t, x − y)ϕ(3) (|y|)dy. =2 ∂x ∂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 2 Z ∂ v K (2) √ (t, x) ≤ p (2t, x − y) (|y|) ϕ dy ◦ ∂x2 t y∈R 3 K Z 2K ∂ v (3) ≤ √ p◦ (2t, x) ϕ(2) (0) + √ (t, x) p (2t, x − y) (|y|) ϕ dy. ◦ ∂x3 t t y∈R We can now fairly easily conclude from (24) with n = 0 that s sZ 2 Z ∞ 2 ∞ ∂ v K dx ≤ √ ϕ(2) (x) 2 dx. (t, x) ∂x2 t x=0 x=0 21
We also note that sZ
s
∞
p2◦ (2t, x)dx
≤
x=0
1 √ 2πt
∞
Z
x=0
2 1 x 1 √ exp − dx ≤ . 1/4 2t (2πt) 2πt
Thus s
sZ 3 2 ∞ ∂ v K K (2) ϕ(3) (x) 2 dx. √ dx ≤ (t, x) |ϕ (0)| + ∂x3 1/4 3/4 (2π) t t x=0 x=0
Z
∞
Combine things together to get the last claim.
5. Numerical Simulation In this section, we will see from numerical simulations where the moving boundary is. 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 3.5. That is, ( def
(26)
u(t, x) =
u ˜(t, x − β(t)) − exp [− (x − β(t))] x ≥ β(t), 0 ≤ t < τ 0 x < β(t), 0 ≤ t < τ ,
where β(t) is defined as Z (27)
t
β(t) = s=0
∂2u ˜ 1− (s, 0) ds ∂x2
0≤t 0, x > 0
∂u ˜ (t, 0) = 0 t>0 ∂x u ˜(0, x) = u ˜◦ (x) = u◦ (x) + e−x .
x>0
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). Here we first discretize space by using the explicit finite difference scheme, then we can obtain SDE’s. Now we use the Euler-Maruyama Method to find numerical solutions of SDE’s (see [Gai96, Hig02]). 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 ( u◦ (x) =
x+x2 1+x4
0
if x ≥ 0 else
and α = .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. 22
0.9
0.8
0.7
0.6
t
0.5
0.4
0.3
0.2
0.1
0 −4
−3
−2
−1
0 x
1
2
3
4
Figure 1. Weak solution u(t, x)
References [BDP02]
Viorel Barbu and Giuseppe Da Prato. The two phase stochastic Stefan problem. Probab. Theory Related Fields, 124(4):544–560, 2002. [BDPR09] Viorel Barbu, Giuseppe Da Prato, and Michael R¨ ockner. Existence of strong solutions for stochastic porous media equation under general monotonicity conditions. Ann. Probab., 37(2):428–452, 2009. [CLM06] L. A. Caffarelli, K.-A. Lee, and A. Mellet. Homogenization and flame propagation in periodic excitable media: the asymptotic speed of propagation. Comm. Pure Appl. Math., 59(4):501–525, 2006. [CS05] Luis Caffarelli and Sandro Salsa. A geometric approach to free boundary problems, volume 68 of Graduate Studies in Mathematics. American Mathematical Society, Providence, RI, 2005. [CV95] Luis A. Caffarelli and Juan L. V´ azquez. A free-boundary problem for the heat equation arising in flame propagation. Trans. Amer. Math. Soc., 347(2):411–441, 1995. [DPR04a] Giuseppe Da Prato and Michael R¨ ockner. Invariant measures for a stochastic porous medium equation. In Stochastic analysis and related topics in Kyoto, volume 41 of Adv. Stud. Pure Math., pages 13–29. Math. Soc. Japan, Tokyo, 2004. [DPR04b] Giuseppe Da Prato and Michael R¨ ockner. Weak solutions to stochastic porous media equations. J. Evol. Equ., 4(2):249–271, 2004. [DPRRW06] G. Da Prato, Michael R¨ ockner, B. L. Rozovskii, and Feng-Yu Wang. Strong solutions of stochastic generalized porous media equations: existence, uniqueness, and ergodicity. Comm. Partial Differential Equations, 31(1-3):277– 291, 2006. [Eva98] L. C. Evans. Partial Differential Equations. American Mathematical Society, Providence, RI, 1998. [Fri64] Avner Friedman. Partial differential equations of parabolic type. Prentice-Hall Inc., Englewood Cliffs, N.J., 1964. [Gai96] J. G. Gaines. Numerical experiments with S(P)DE’s. In Stochastic partial differential equations, volume 216 of London Math. Soc. Lecture Note Ser., pages 55–71, Cambridge University Press, Cambridge, 1995. [Hig02] Desmond J. Higham. An algorithmic introduction to numerical simulation of stochastic differential equations. SIAM Rev., 43(3):525–546, 2001. [Kim06] Jong Uhn Kim. On the stochastic porous medium equation. J. Differential Equations, 220(1):163–194, 2006. 23
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Department of Mathematics, University of Illinois at Urbana–Champaign, Urbana, IL 61801 E-mail address: [email protected] Department of Mathematics, University of Rochester, Rochester, NY 14627 Department of Mathematics, University of Illinois at Urbana–Champaign, Urbana, IL 61801 E-mail address: [email protected]
24
